INTEGRAL EQUATIONS AND BOUNDARY VALUE PROBLEMS · PREFACE This book on ‘‘Linear integral equations and boundary value problems’’ has been specially written as per latest UGC

INTEGRAL EQUATIONSAND

BOUNDARY VALUE PROBLEMS[with Green’s function technique and its applications]

[For M.A./M.Sc. (Mathematics) and M.Sc. (Physics) students of all IndianUniversities/Institutions according to latest U.G.C model curriculum and

various engineering and professional examinations such asGATE, C.S.I.R NET/JRF and SLET etc.]

Dr. M.D. RAISINGHANIA M.Sc., Ph.D.

Formerly, Head of Mathematics DepartmentS.D. (Postgraduate) College

Muzaffarnagar (U.P.)

S. CHAND & COMPANY PVT. LTD.(AN ISO 9001: 2008 COMPANY)

RAM NAGAR, NEW DELHI-110 055

S. CHAND & COMPANY PVT. LTD.(An ISO 9001 : 2008 Company)Head Office: 7361, RAM NAGAR, NEW DELHI - 110 055

Phone: 23672080-81-82, 9899107446, 9911310888Fax: 91-11-23677446

Shop at: schandgroup.com; e-mail: [email protected] :

AHMEDABAD : 1st Floor, Heritage, Near Gujarat Vidhyapeeth, Ashram Road, Ahmedabad - 380 014,Ph: 27541965, 27542369, [email protected]

BENGALURU : No. 6, Ahuja Chambers, 1st Cross, Kumara Krupa Road, Bengaluru - 560 001,Ph: 22268048, 22354008, [email protected]

BHOPAL : Bajaj Tower, Plot No. 2&3, Lala Lajpat Rai Colony, Raisen Road, Bhopal - 462 011,Ph: 4274723, 4209587. [email protected]

CHANDIGARH : S.C.O. 2419-20, First Floor, Sector - 22-C (Near Aroma Hotel), Chandigarh -160 022,Ph: 2725443, 2725446, [email protected]

CHENNAI : No.1, Whites Road, Opposite Express Avenue, Royapettah, Chennai - 600014 Ph. 28410027, 28410058, [email protected]

COIMBATORE : 1790, Trichy Road, LGB Colony, Ramanathapuram, Coimbatore -6410045,Ph: 2323620, 4217136 [email protected] (Marketing Office)

CUTTACK : 1st Floor, Bhartia Tower, Badambadi, Cuttack - 753 009, Ph: 2332580; 2332581,[email protected]

DEHRADUN : 1st Floor, 20, New Road, Near Dwarka Store, Dehradun - 248 001,Ph: 2711101, 2710861, [email protected]

GUWAHATI : Dilip Commercial (Ist floor), M.N. Road, Pan Bazar, Guwahati - 781 001,Ph: 2738811, 2735640 [email protected]

HYDERABAD : Padma Plaza, H.No. 3-4-630, Opp. Ratna College, Narayanaguda, Hyderabad - 500 029,Ph: 27550194, 27550195, [email protected]

JAIPUR : 1st Floor, Nand Plaza, Hawa Sadak, Ajmer Road, Jaipur - 302 006,Ph: 2219175, 2219176, [email protected]

JALANDHAR : Mai Hiran Gate, Jalandhar - 144 008, Ph: 2401630, 5000630, [email protected] : Kachapilly Square, Mullassery Canal Road, Ernakulam, Kochi - 682 011,

Ph: 2378740, 2378207-08, [email protected] : 285/J, Bipin Bihari Ganguli Street, Kolkata - 700 012, Ph: 22367459, 22373914,

[email protected] : Mahabeer Market, 25 Gwynne Road, Aminabad, Lucknow - 226 018, Ph: 4076971, 4026791,

4065646, 4027188, [email protected] : Blackie House, IInd Floor, 103/5, Walchand Hirachand Marg, Opp. G.P.O., Mumbai - 400 001,

Ph: 22690881, 22610885, [email protected] : Karnal Bagh, Near Model Mill Chowk, Nagpur - 440 032, Ph: 2720523, 2777666

[email protected] : 104, Citicentre Ashok, Mahima Palace , Govind Mitra Road, Patna - 800 004, Ph: 2300489,

2302100, [email protected] : 291, Flat No.-16, Ganesh Gayatri Complex, IInd Floor, Somwarpeth, Near Jain Mandir,

Pune - 411 011, Ph: 64017298, [email protected] (Marketing Office)RAIPUR : Kailash Residency, Plot No. 4B, Bottle House Road, Shankar Nagar, Raipur - 492 007,

Ph: 2443142,Mb. : 09981200834, [email protected] (Marketing Office)RANCHI : Flat No. 104, Sri Draupadi Smriti Apartments, (Near of Jaipal Singh Stadium) Neel Ratan Street,

Upper Bazar, Ranchi - 834 001, Ph: 2208761, [email protected] (Marketing Office)SILIGURI : 122, Raja Ram Mohan Roy Road, East Vivekanandapally, P.O., Siliguri, Siliguri-734001,

Dist., Jalpaiguri, (W.B.) Ph. 0353-2520750 (Marketing Office) [email protected]: No. 49-54-15/53/8, Plot No. 7, 1st Floor, Opp. Radhakrishna Towers,

Seethammadhara North Extn., Visakhapatnam - 530 013, Ph-2782609 (M) 09440100555,[email protected] (Marketing Office)

© Copyright ReservedAll rights reserved. No part of this publication may be reproduced or copied in any material form (includingphoto copying or storing it in any medium in form of graphics, electronic or mechanical means and whetheror not transient or incidental to some other use of this publication) without written permission of the copyrightowner. Any breach of this will entail legal action and prosecution without further notice.Jurisdiction : All disputes with respect to this publication shall be subject to the jurisdiction of the Courts,tribunals and forums of New Delhi, India only.

First Edition 2007Subsequent Editions 2009, 2010, 2011, 2012Sixth Revised Edition 2013

ISBN : 81-219-2805-2 Code : 14C 544PRINTED IN INDIA

By Rajendra Ravindra Printers Pvt. Ltd., 7361, Ram Nagar, New Delhi -110 055and published by S. Chand & Company Pvt. Ltd., 7361, Ram Nagar, New Delhi -110 055.

PREFACE TO THE THIRD EDITION

Reference to the latest papers of various universities and GATE have been inserted at properplaces. More additional problems have been inserted in the miscellaneous set of problems givenat the end of the book.

I hope that these changes will make the material more accessible and attractive to the reader.All valuable suggestions for further improvement of the book will be highly appreciated.

M.D. Raisinghania

PREFACE TO THE SIXTH EDITION

Reference to the latest papers of GATE and various universities have been inserted at properplaces. Solutions of some new problems are also given.

Suggestions for further improvement of the book will be gratefully received.M.D. Raisinghania

Disclaimer : While the authors of this book have made every effort to avoid any mistake or omission and have used their skill,expertise and knowledge to the best of their capacity to provide accurate and updated information. The author and S. Chand doesnot give any representation or warranty with respect to the accuracy or completeness of the contents of this publication and areselling this publication on the condition and understanding that they shall not be made liable in any manner whatsoever. S.Chandand the author expressly disclaim all and any liability/responsibility to any person, whether a purchaser or reader of this publicationor not, in respect of anything and everything forming part of the contents of this publication. S. Chand shall not be responsible forany errors, omissions or damages arising out of the use of the information contained in this publication.Further, the appearance of the personal name, location, place and incidence, if any; in the illustrations used herein is purelycoincidental and work of imagination. Thus the same should in no manner be termed as defamatory to any individual.

PREFACE TO THE FOURTH EDITION

New matter and latest questions of various universities have been added at appropriate places.In addition to this, the following new useful topics have been added.

Appendix A: Boundary value problems and Green’s identities.Appendix B: Two and three dimensional Dirac delta functionsAppendix C: Additional topics and problems based on Green’s functionsI hope that these changes will make the material of this book more useful to the reader.Suggestions for further improvement of the book will be gratefully received.

M.D. Raisinghania

PREFACE

This book on ‘‘Linear integral equations and boundary value problems’’ has been speciallywritten as per latest UGC model carriculum for MA/M.Sc. students of all Indian universities/institutions. In addition, this book will prove very useful for students preparing for various engi-neering and professional examinations such as GATE, C.S.I.R. NET/JRF and SLET etc.

The author possesses a very long and rich experience of teaching mathematics and has firsthand experience of the problems and difficulties that students generally face.

The silent features of this book are :* The matter has been presented in a simple and lucid language, so that students them-

selves shall be able to understand the solutions of the problems.* Each chapter opens with necessary definitions and complete proofs of the standard

results and theorems. These in turn are followed by solved examples which have beenclassified in various types and methods. This classification will help the students torevise the subject matter at the time of examination without losing any confidence.

* Care has been taken not to omit important steps so that the students can understandevery thing without the guidance of a teacher. Furthermore, a set of unsolved exercisesis given in each chapter to instill confidence in the students.

In view of these special features, it is sincerely hoped that the book will surely serve itspurpose.

I am grateful to Shri Ravindra Kumar Gupta, Managing Director, Shri Navin Joshi, GeneralManager and Shri R.S. Saxena (Adviser, Publishing) for showing keen interest throughout thepreparation of the book. My sincere thanks are due to Shri Shishir Bhatnagar for bringing the bookin an excellent form.

All valuable suggestions for further improvement of the book will be highly appreciated.

M.D. Raisinghania

COMPLETE SYLLABUS FOR INTEGRAL EQUATIONS AND BOUNDARY VALUE PROBLEMS

AS PER LATEST U.G.C. MODEL CARRICULUMFOR M.A./M.SC MATHEMATICS OF ALL INDIAN UNIVERSITIES/INSTITUTIONS

Definitions of integral equations and their classification. Eigenvalues and eigenfunctions.Fredholm integral equations of second kind with separable kernels. Reduction to a system of alge-braic equations. An approximate method.

Method of successive approximations. Iterative schemes for Fredholm integral equations of thesecond kind. Conditions of uniform convergence and uniqueness of series solution. Resolvent kerneland its results. Application of iterative scheme to Volterra integral equations of the second kind.

Classical Fredholm theory. Fredholm theorems.

Integral transform methods. Fourier transform. Convolution integral. Application to Volterraintegral equations with convolution-type kernels.

Abel’s equations. Inversion formula for singular integral equation with kernel of the typeh(s) – h(t), 0 < a < 1. Cauchy’s principal value of singular integrals. Solution of Cauchy-type integralequation. The Hilber kernel. Solution of the Hilbert-type singular integral equation.

Symmetric kernels. Complex Hilbert space. Orthonormal system of functions. Fundamentalproperties of eigenvalues and eigenfunctions for symmetric kernels Expansion in eigenfunction andbilinear form. Hilbert Schmidt theorem and some immediate consequences. Solutions of integralequations with symmetric kernels.

Definition of a boundary value problem for an ordinary differential equation of the secondorder and its reduction to a Fredholm integral equation of the second kind. Dirac delta function.Green’s function approach to reduce boundary value problems of a self-adjoint differential equationwith homogeneous boundary conditions to integral equation forms. Auxiliary problem satisfied byGreen’s function. Integral equation formulations of boundary value problems with more general andinhomogeneous boundary conditions. Modified Green’s function.

Integral representation for the solution of the Laplace’s and Poisson’s equations. Newtoniansingle-layer and double layer potentials. Interior and exterior Dirichelet and Neumann boundaryvalue problems for Laplace’s equation. Green’s function for Laplace’s equation in a space as well asin a space bounded by a ground vessel. Integral equation formulation of boundary value problems forLaplace’s equation. Poisson’s integral formula. Green’s function for the space bounded by groundedtwo parallel plates or an infinite circular cylinder.

Perturbation techniques and its applications to mixed boundary value problems. Two part andthree part boundary value probelms.

Solutions of electrostatic problems involving a charged circular and annular disc, a sphericalcap, an annular spherical cap in a free space or a bounded space.

REFERENCES:1. R.P. Kanwal, Linear integral equations. Theory and techniques. Academic Press, NewYork,

1971

2. S.G. Mikhlin, Linear integral equations (translated from Russian). Hindustan Book Agency,1960

3. I.N. Sneddon, Mixed boundary value problems in potential theory, North Holland, 1966

4. I. Stakgold. Boundary value probelms of mathematical physics, Vol. I and II, Macmillan 1969.

Dedicated to the memory of

my parents

CONTENTS1. PRELIMINARY CONCEPTS 1.1 - 1.12

1.1 Introduction 1.11.2 Abel’s problem 1.11.3 Integral equation. Definition 1.21.4 Linear and non-linear integral equations 1.21.5 Fredholm integral equation 1.3

(i) Fredholm integral equation of the first kind 1.3(ii) Fredholm integral equation of the second kind 1.3

(iii) Fredholm integral equation of the third kind 1.3(iv) Homogeneous Fredholm integral equation 1.3

1.6 Volterra integral equation 1.3(i) Volterra integral equation of the first kind 1.3

(ii) Volterra integral equation of the third kind 1.3(iii) Volterra integral equation of the second kind 1.4(iv) Homogeneous Volterra integral equation 1.4

1.7 Singular integral equation 1.41.8 Special kinds of kernels 1.4

(i) Symmetric kernel 1.4(ii) Separable or degenerate kernel 1.4

1.9 Integral equation of the convolution type 1.51.10 Iterated kernels or functions 1.51.11 Resolvent kernel or reciprocal kernel 1.51.12 Eigenvalues (or characteristic values or characteristic numbers). Eigenfunctions

(or fundamental functions) 1.61.13 Leibnit’z rule of differentiation under integral sign 1.61.14 An important formula for converting a multiple integral into a single ordinary

integral 1.61.15 Regularity conditions 1.7

Square-integrable function or 2-function 1.7

1.16 The inner or scalar product of two functions 1.81.17 Solution of an integral equation. Definition 1.81.18 Solved example based on Art. 1.17 1.8

2. CONVERSION OF ORDINARY DIFFERENTIAL EQUATIONS INTO INTEGRAL EQUATIONS 2.1 - 2.22

2.1 Introduction 2.12.2 Initial value problem 2.12.3 Method of converting an initial value problem into a Volterra integral equation 2.12.4 Alternative method of converting an initial value problem into a Volterra integral

equation 2.72.5 Boundary value problem 2.142.6 Method of converting a boundary value problem into a Fredholm integral equation 2.14

3. HOMOGENEOUS FREDHOLM INTEGRAL EQUATIONS OF THE SECOND KIND WITH SEPARABLE (OR DEGENERATE) KERNELS 3.1 - 3.24

3.1 Characteristic values (or Characteristic numbers or eigenvalues). Characteristicfunctions (or eigenfunctions) 3.1

(vii)

3.2 Solution of homogeneous Fredholm integral equation of the second kind withseparable (or degenerate) kernels 3.1

3.3 Solved examples based on Art 3.1 and Art 3.2 3.34. FREDHOLM INTEGRAL EQUATIONS OF THE SECOND KIND WITH SEPARABLE

(OR DEGENERATE) KERNELS 4.1 - 4.304.1 Solution of Fredholm integral equations of the second kind with separable (or

degenerate) kernels 4.14.2 Solved examples based on Art. 4.1 4.34.3 Fredholm alternative 4.20

Fredholm theorem 4.21Fredholm alternative theorem 4.25

4.4 Solved examples based on Art. 4.3 4.254.5 An approximate method 4.29

5. METHOD OF SUCCESSIVE APPROXIMATIONS 5.1 - 5.685.1 Introduction 5.15.2 Iterated kernels or functions 5.15.3 Resolvent (or reciprocal) kernel 5.1

5.4 Theorem. To prove that ( , ) ( , )b

m raK x t K x y= ∫ K

m – r (y, t) dy 5.2

5.5 Solution of Fredholm integral equation of the second kind by successivesubstitutions 5.3

5.6 Solution of Volterra integral equation of the second kind by successive substitutions 5.55.7 Solution of Fredholm integral equation of the second kind by successive

approximations. Iterative method (iterative scheme). Neumann series 5.75.8 Some important theorems 5.115.9 Solved examples based on solution of Fredholm integral equation of the second kind

by successive approximations (or iterative method) 5.125.10 Reciprocal functions 5.29

Volterra solution of Fredholm integral equation of the second kind 5.305.11 Solution of Volterra integral equation of the second kind by successive

approximations (or iterative method). Neumann series 5.35

5.12 Theorem. To prove that ( , ; ) ( , ) ( , ) ( , ; )x

tR x t K x t K x z R z t dzλ = + λ λ∫ 5.37

5.13 Solved examples based on solution of Volterra integral equation of the second kindby successive approximation (or iterative method) 5.38

5.14 Solution of Volterra integral equation of the second kind when its kernel is of someparticular forms 5.56

5.15 Solution of Volterra integral equation of the second kind by reducing to differentialequation 5.62

5.16 Volterra integral equation of the first kind 5.635.17 Solution of Volterra integral equation of first kind 5.65

6. CLASSICAL FREDHOLM THEORY 6.1 - 6.396.1 Introduction 6.16.2 Fredholm’s first fundamental theorem 6.16.3 Solved examples based on Fredholm’s first fundamental theorem 6.66.4 Fredholm’s second fundamental theorem 6.326.5 Fredholm’s third fundamental theorem 6.36

(viii)

7. INTEGRAL EQUATIONS WITH SYMMETRIC KERNELS 7.1 - 7.48 7.1 Introduction 7.17.1 (a) Symmetric kernels 7.17.1 (b) Regularity conditions 7.17.1 (c) The inner or scalar product of two functions 7.27.1 (d) Schwarz inequality. Minkowski inequality. 7.27.1 (e) Complex Hilbert space 7.27.1 (f) An orthonormal system of functions 7.37.1 (g) Riesz-Fischer’s theorem 7.47.1 (h) Some useful results 7.57.1 (i) Fourier series of a general character 7.57.1 (j) Some examples of the complete orthogonal and orthonormal systems 7.6

7.1 (k) A complete two-dimensional orthonormal set over the rectangle ,a x b≤ ≤

c t d≤ ≤

7.7

7.2 Some fundamental properties of eigen values and eigenfunctions for symmetrickernels 7.7

7.3 Expansion in eigenfunctions and bilinear form 7.157.4 Hilber-Schmidt theorem 7.177.5 Definite kernels and Mercer’s theorem 7.207.6 Schmidt solution of non-homogeneous Fredholm integral equation of the second

kind with continuous, real and symmetric kernel 7.217.7 Solved example based on Art. 7.6 7.247.8 Solution of the Fredholm integral equation of the first kind with symmetric kernel 7.407.9 Solved example based on Art. 7.8 7.41

7.10 Approximations of a general 2

-kernel (not necessarily symmetric) by a separablekernel 7.44

7.11 Operator method in the theory of integral equations 7.44 8 SINGULAR INTEGRAL EQUATIONS 8.1 - 8.24

8.1 Singular integral equation 8.18.2 The solution of the Abel integral equation 8.18.3 General form of the Abel singular integral equation 8.38.4 Another general form of the Abel singular integral equation 8.5

Weakly singular kernel 8.68.5 Solved examples 8.68.6 Cauchy principal value for integrals 8.9

Cauchy’s general and principal values. Singular integrals 8.9

H

. .o

lder condition 8.10The definition of Cauchy principal value for the contour 8.10

8.7 The Cauchy integrals 8.11Plemelj formulas 8.11Poincare-Bertrand transformation formula 8.11

8.8 Solution of the Cauchy-type singular integral equation 8.138.9 The Hilbert kernel 8.16

Hilbert formula 8.178.10 Solution of the Hilbert type singular integral equation of the second kind 8.188.11 Solution of the Hibert-type singular integral equation of the first kind 8.21

9. INTEGRAL TRANSFORM METHODS 9.1 - 9.259.1 Introduction 9.19.2 Some useful results about Laplace transform 9.1

(ix)

9.3 Some special types of integral equations 9.5(i) Integro-differential equation 9.5(ii) Integral equation of convolution type 9.5

9.4 Application of Laplace transform to determine the solution of Volterra integralequation with convolution-type kernels. Working rule 9.5

9.5 Solved examples based on Art. 9.2 to Art 9.4 9.79.6 Some useful results about Fourier transforms 9.179.7 Application of Fourier transform to determine the solution of integral equations 9.189.8 Hilbert transform 9.199.9 Infinite Hilbert transform 9.21

9.10 Mellin transform 9.239.11 Solution of Fox’s integral equation 9.23

10. SELF ADJOINT OPERATOR, DIRAC DELTA FUNCTION AND SPHERICAL HORMONICS 10.1 - 10.1410.1 Introduction 10.110.2 Adjoint equation of second order linear differential equation 10.110.3 Self adjoint equation 10.110.4 Solved examples based on Art. 10.2 and Art 10.3 10.310.5 Green’s formula 10.410.6 The Dirac delta function 10.510.7 Shifting property of Dirac delta function 10.610.8 Derivatives of Dirac delta function 10.710.9 Relation between Dirac delta function and Heaviside unit function 10.7

10.10 Alternative forms of representing Dirac delta function 10.810.11 Spherical harmonics 10.810.12 Bessel functions 10.1311. APPLICATIONS OF INTEGRAL EQUATIONS AND GREEN’S FUNCTIONS TO ORDINARY DIFFERENTIAL EQUATIONS 11.1 - 11.62

11.1 Introduction 11.111.2 Green’s function 11.111.3 Conversion of a boundary value problem into Fredholm integral equation.

Solution of a boundary value problem 11.411.4 An important special case of result of Art. 11.2 11.511.5 Solved example based on construction of Green’s function (based on Art. 11.2

and Art 11.4) 11.1011.6 Solved examples based on result 1 of Art. 11.3 11.1811.7 Solved examples based on result 2 of Art. 11.3 11.2211.8 Solved examples based on result 3 of Art. 11.3 11.3211.9 Linear integral equations in cause and effect. The influence function 11.37

11.10 Green’s function approach for converting an initial value problem into an integralequation 11.40

11.11(a) Green’s function approach for converting a boundary value problem into anintegral equation. An alternative procedure 11.43

11.11(b) Integral equation formulation for the boundary value problem with more generaland inhomogeneous boundary conditions Working rule 11.45

11.12 Modified Green’s function or Generalized Green’s function 11.4811.13 Working rule for construction of modified Green’s function 11.5111.14 Solved examples based on Art. 11.13 11.52

(x)

12 APPLICATIONS OF INTEGRAL EQUATIONS TO PARTIAL DIFFERENTIAL EQUATIONS 12.1 - 12.3912.1 Introduction 12.112.2 Integral representation formulas for the solutions of the Laplace and Poisson

equations 12.212.3 Solved examples based on Art. 12.2 12.712.4 Green’s function approach 12.9

12.4 A The method of images 12.1412.5 Solved example based on Art. 12.4 and 12.4 A 12.1412.6 The Helmholtz equation 12.1812.7 Solved examples based in Art 12.6 12.19

ADDITIONAL RESULTS ON GREEN’S FUNCTION AND ITS APPLICATIONS12.8 Additional results about Green’s function 12.2312.9 The theory of Green’s function for Laplace’s equation 12.26

12.10 Construction of Green’s function with help of the method of images 12.3112.11 Green’s function for the two dimensional Laplace’s equation 12.3412.12 Construction of the Green’s function with the help of the method of images 12.3613. APPLICATIONS OF INTEGRAL EQUATIONS TO MIXED BOUNDARY VALUE PROBLEMS 13.1 - 13.2413.1 Introduction 13.113.2 Two-part boundary value problems 13.113.3 Three-part boundary value problems 13.813.4 Generalized two-part boundary value problems 13.1413.5 Generalized three-part boundary value problems 13.1713.6 Appendix 13.23

14. INTEGRAL EQUATION PERTURBATION TECHNIQUES 14.1 - 14.1714.1 Introduction 14.114.2 Working rule for solving an integral equation by perturbation techniques 14.114.3 Applications of perturbation techniques to electrostatics 14.314.4 Applications of perturbation techniques to low-Reynolds number hydrodynamics 14.6

14.4 A Steady Stokes flow 14.614.4 B Boundary effects of Stokes flow 14.714.4 C Longitudinal oscillations of solids in Stokes flow 14.814.4 D Steady rotary Stokes flow 14.914.4 E Rotary oscillations in Stokes flow 14.1114.4 F Oseen flow - Translation motion 14.1414.4 G Oseen flow - Rotary motion 14.15

APPENDIX A. Boundary Value problems and Green’s identities A.1 - A.2A.1 Some useful notation A.1A.2 Boundary value problems for Laplace equation A.1

Classification of boundary value problems for Laplace equation A.1A.3 Green’ identities A.2

APPENDIX B. Two and three dimensional Dirac delta functions B.1 - B.2B.1 Introduction B.1B.2 Two-dimensional Dirac delta function B.1B.3 Three-dimensional Dirac delta function B.1B.4 Dirac delta function in general curvilinear coordinates in two-dimensions B.2B.5 Dirac delta function in general curvilinear coordinates in three-dimensions B.2

(xi)

APPENDIX C. Additional topics and problems based on Green’s functions C.1 - C.27C.1 The eigenfunction method for computing Green’s function for the given

Dirichlet boundary value problem C.1C.2 The space form of the wave equation (or Helmholtz equation) C.3C.3 Helmholtz’s theorem C.4C.4 Application of Green’s function in determining the solution of the wave equation C.5C.5 Determination of the Green’s function for the Helmholtz equation for the

half-space

0z ≥

C.6C.6 Solution of one-dimensional wave equation using the Green’s function technique. C.9C.7 Solution of one-dimensional inhomogeneous wave equation using the

Green’s function technique C.11C.8 Solution of one-dimensional heat equation using the Green’s function technique C.18C.9 Solution of one-dimensional inhomogeneous heat equation involving an external

heat source using Green’s function technique C.20C.10 The use of Green’s function in the determination of the solution of the

solution of heat equation (or the diffusion equation) C.22C.11 The use of Green’s function in the determination of the solution of heat equation

(or diffusion equation) for infinite rod C.24APPENDIX D. Additional problems based on modified (or generalised)

Green’s function D.1–D.10D.1 Additional problems based on Art. 11.12, Art. 11.13 and Art. 11.14 of chapter 11 D.1D.2 Extension of the theory of Art. 11.13 of chapter 11 to the case when the

associated self adjoint system has two linearly indendent solutions inplace of exactly non-zero solution. D.6

Miscellaneous problems on the entire book M.1 – M.7

Index I.1 – I. 5

(xii)

GENERAL NOTATIONS[Numbers refer the page on which the explanation first appeared]

AT Transpose of matrix A 4.23B (x, y) Beta function 8.1

( )D λ

Fredholm determinant 6.7

( , ; )D x t λ

Fredholm minor 6.7

or div∇⋅ A A

divergence of vector A 12.1exp a exponential of a, i.e., ea 12.18

E (x, t) fundamental solution or free space solution 12.2

0( , )E x x

Green’s function for Laplace’s equation 10.11

F Fourier transform 9.17F–1 inverse Fourier transform 9.17F

cFourier cosine transform 9.17

Fc–1 inverse Fourier cosine transform 9.18

Fs

Fourier sine transform 9.17F

s–1 inverse Fourier sine transform 9.17

| | f | | norm of the function f 1.8(f, g) inner (or scalar) product of f and g 1.8

u∇

or grad u gradient of scalar point function u 12.1G (x, t) Green’s function 11.1G

M(x, t) modified Green’s function 11.49

H (x – a) Heaviside (step or unit) function 9.2

(1) (2)( ), ( )H z H zα α

Hankel functions or Bessel functions of the third kind 10.14I unit or identity matrix 4.21

( )I zα modified Bessel function 10.14

Jn (x) Bessel function of the first kind 7.6

K Fredholm operator 7.2( )K zα modified Bessel function 10.14

K (x, x) trace of symmetric kernel 7.8K (x, t) kernel of an integral equation 1.2

( , )K x t complex conjugate of K (x, t) 1.4

Kn(x, t) iterated kernel 1.5

ln x

loge

x

3.23

£2– square integrable 1.7l.u.b. least upper bound 7.45

L Laplace transform 9.1L–1 inverse Laplace transform 9.3M Mellin transform 9.23

M–1 inverse Mellin transform 9.23

( )N zα

Neumann function 10.14

( )b

aP f x dx∫ or

*

( )b

af x dx∫ principal value of integral 8.9

Pn (x) Legendre polynomial 7.6

(xiii)

( )mn

P x associated legendre function 10.11

p1, p Green’s vector 12.19

�

Reynold’s number 12.21

r<

max (r, r0) 10.13

r>

min (r, r0) 10.13

( , ; )R x t λ

or

( , ; )x tΓ λ

Resolvent (or reciprocal) kernel 1.5

T1, T Green’s tensor 12.19

q velocity vector 12.19

( , )mn

Y θ φ

spherical harmonics 10.11

( , )mn

Y θ φ

complex conjugate of

( , )mn

Y θ φ

10.11

W (y1, y

2, ..., y

x) Wronskian of y

1, y

2, ..., y

n11.3

( )xΓ

Gamma function 8.2

( )xδ Dirac delta function 10.5

okδ

Kronecker delta 12.8

2∇

Laplacian 12.2

�

approximately 3.10

∀for all 7.1

THE GREEK ALPHABETalpha

α

A nu

ν

Nbeta

β

B xi

ξΞ

gamma

γΓ

omicron o Odelta

δΔ

pi

π∏

epsilon

ε

E rho

ρ

Pzeta

ζ

Z sigma

σΣ

eta

η

H tau

τ

Ttheta

θΘ

upsilon

υϒ

iota

ι

I phi

φΦ

kappa

κ

K chi

χ

Xlambda

λΛ

psi

ψΨ

mu

μ

M omega

ωΩ

(xiv)

CHAPTER 1

Preliminary Concepts1.1 INTRODUCTION.

Many physical problems of science and technology which were solved with the help of theoryof ordinary and partial differential equations can be solved by better methods of theory of integralequations. For example, while searching for the representation formula for the solution of lineardifferential equation in such a manner so as to include boundary conditions or intitial conditionsexplicitly, we arrive at an integral equation. The solution of the integral equation is much easierthan the orginal boundary value or initial value problem. The theory of integral equations is veryuseful tool to deal with problems in applied mathematics, theoretical mechanis, and mathematicalphysics. Several situations of science lead to integral equations, e.g., neutron diffusion problem andradiation transfer problem etc.1.2. ABEL’S PROBLEM.

We propose to give an example of a situationwhich leads to an integral equation. Consider thefollowing problem in mechanis.

Consider a given smooth curve in a vertical planeand suppose a material point start from rest at any pointP under the influence of gravity along the curve. Let Tbe the time taken by the particle from P to the lowestpoint O. Treat O as the origin of coordinates, the x-axisvertically upward, and the y-axis horizontal. Let thecoordinates of P and Q be (x, y) and ( , ) respectively..Let arc OQ = s.

Then the velocity of the particle at Q is given by

2 ( )ds g xdt

so that .2 ( )

Q

P

dstg x

Hence, .2 ( )

P

Q

dsTg x

... (1)

If the shape of the curve is given, then s can be expressed in terms of and hence ds can beexpressed in terms of . So, let

ds = u ( ) d .

from (1), 0

( ) .2 ( )

x u dTg x

... (2)

Able treated the above problem in modified form by finding that curve for which the time Tof descent is a given function of x, say f (x). Thus, we are led to the problem of finding theunknown function u from the equation

1.1

Q ( )

P (x, y)x

y

O

s

Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/

1.2 Preliminary Concepts

0

1( ) ( ) .2 ( )

xf x u d

g x

... (3)

Equation (3) is called Able integral equation.1.3. INTEGRAL EQUATION. DEFINITION. [Meerut 2005, 08, 12]

An integral equation is an equation is which an unknown function appears under one ormore integral signs.

For example, for a x b, a t b, the equations

( , ) ( ) ( )b

aK x t y t dt f x ... (1)

( ) – ( , ) ( ) ( )b

ay x K x t y t dt f x ... (2)

and 2( ) ( , ) [ ( )] ,b

ay x K x t y t dt ...(3)

where the function y(x), is the unknown function while the functions f (x) and K (x, t) are knownfunctions and , a and b are constants, are all integral equations. The above mentioned functionsmay be complex-valued functions of the real variables x and t.1.4. LINEAR AND NON-LINEAR INTEGRAL EQUATIONS. DEFINITIONS.

An integral equation is called linear if only linear operations are performed in it upon theunknown function. An integral equation which is not linear is known as a non-linear integralequation.By writing either

( ) ( , ) ( )b

aL y K x t y t dt or ( ) ( ) ( , ) ( ) ,

b

aL y y x K x t y t dt

we can easily verify that L is a linear integral operator. In fact, for any constants c1 and c2, we haveL {c1 y1 (x) + c2 y2 (x)} = c1 L {y1 (x)} + c2 L {y2 (x)},

which is well known general criterion for a linear operator. In this book, we shall study only linearintegral equations.

For example, the integral equations (1) and (2) of Art. 1.3 are linear integral equations whilethe integral equation (3) is non-linear integral equation.

The most general type of linear integral equation is of the form

( ) ( ) ( ) ( , ) ( ) ,a

g x y x f x K x t y t dt ... (1)

where the upper limit may be either variable x or fixed. The functions f, g and K are knownfunctions while y is to be determined; is a non-zero real or complex, parameter. The functionK (x, t) is known as the kernel of the integral equation.

Remark 1. The constant can be incorporated into the kernel K (x, t) in (1). However, inmany applications represents a significant parameter which may take on various values in adiscussion being considered. For theoretical discussion of integral equations, plays an importantrole.

Remark 2. If ( ) 0,x ! (1) is known as linear integral equation of the third kind. When

( ) 0,x ! (1) reduces to

( ) ( , ) ( ) 0,a

f x K x t y t dt ... (2)


Preliminary Concepts 1.3

which is known as linear integral equation of the first kind. Again, when ( ) 1,x ! (1) reduces to

( ) ( ) ( , ) ( ) ,a

y x f x K x t y t dt ... (3)

which is known as linear integral equation of the second kind.In the present book, we shall study in details equations of the form (2) and (3) only. In next

two articles, we discuss special cases of (2) and (3).1.5. FREDHOLM INTEGRAL EQUATION. DEFINITION. (Kanpur 2010, 2011)

A linear integral equation of the form

( ) ( ) ( ) ( , ) ( ) ,b

ag x y x f x K x t y t dt ... (1)

where a, b are both constants, f (x) g (x) and K (x, t) are known functions while y (x) is unknownfunction and is a non-zero real or complex parameter, is called Fredholm integral equation ofthird kind. The function K (x, t) is known as the kernel of the integral equation.

The following special cases of (1) are of our main interest.(i) Fredholm integral equation of the first kind.A linear integral equation of the form (by setting g (x) = 0 in (1))

( ) ( , ) ( ) 0,b

af x K x t y t dt ... (2)

is known as Fredholm integral equation of the first kind.(ii) Fredholm integral equation of the second kind.A linear integral equation of the form (by setting g (x) = 1 in (1))

( ) ( ) ( , ) ( ) ,b

ay x f x K x t y t dt ... (3)

is known as Fredholm integral equation of the second kind.(iii) Homogeneous Fredholm integral equation of the second kind.A linear integral equation of the form (by setting f (x) = 0 in (3)).

( ) ( , ) ( ) ,b

ay x K x t y t dt ... (4)

is known as the homogeneous Fredholm integral equation of the second kind.1.6. VOLTERRA INTEGRAL EQUATION. DEFINITION.

A linear integral equation of the form

( ) ( ) ( ) ( , ) ( ) ,x


where a, b are both constants, f (x), g (x) and K (x, t) are known functions while y (x) is unknownfunction; is a non-zero real or complex parameter is called Volterra integral equation of thirddkind. The function K (x, t) is known as the kernel of the integral equation.

The following special cases of (1) are of our main interest.(i) Volterra integral equation of the first kind.A linear integral equation of the form (by setting g (x) = 0 in (1))

( ) ( , ) ( ) 0,x




is known as Volterra integral equation of the first kind.(ii) Volterra integral equation of the second kind.A linear integral equation of the form (by setting g (x) = 1)

( ) ( ) ( , ) ( ) ,x


is known as Volterra integral equation of the second kind.(iii) Homogeneous Voterra integral equation of the second kind.A linear integral equation of the form (by setting f (x) = 0 is (3))

( ) ( , ) ( ) ,x


is known as the homogeneous Volterra integral equation of the second kind.1.7. SINGULAR INTEGRAL EQUATION. DEFINITION. [Meerut 2008]

When one or both limits of integration become infinite or when the kernel becomes infinite atone or more points within the range of integration, the integral equation is known as singularintegral equation. For example, the integral equations

| |( ) ( ) ( )x ty x f x e y t dt

and0

1( ) ( ) ,0 1( )

xf x y t dt

x t

are singular integral equations.1.8. SPECIAL KINDS OF KERNELS.

The following special cases of the kernel of an integral equation are of main interest and weshall frequently come across with such kernels throughout the discussion of this book.

(i) Symmetric kernal. Definition.A kernel K (x, t) is symmetric (or complex symmetric or Hermitian) if

K (x, t) = K (t, x)where the bar donates the complex conjugate. A real kernel K (x, t) is symmetric if

K (x, t) = K (t, x).For example, sin (x + t), log (x t), x2t2 + xt + 1 etc. are all symmetric kernels. Again, sin (2x

+ 3t) and x2t3 + 1 are not symmetric kernels.Again i (x – t) is a symmetric kernel, since in this case, if K (x, t) = i (x – t), then k (t, x) =

i(t – x) and so ( , )K t x = – i (t – x) = i (x – t) = K (x, t). On the other hand, i (x + t) is not a

symmetric kernel, since in this case, if K (x, t) = i (x + t), then ( , ) ( )K t x i t x = – i (t + x) =

– K (x, t) and so ( , ) ( , )K x t K x t

(ii) Separable or degenerate kernel. Definition. [Meerut 2000]A kernel K (x, t) is called separable if it can be expressed as the sum of a finite number of

terms, each of which is the product of a function of x only and a function of t only, i.e.,

1( , ) ( ) ( ).

n

i iiK x t g x h t

... (1)

Remark. The functions gi (x) can be regarded as linearly independent, otherwise the numberof terms in relation (1) can be further reduced. Recall that the set of functions gi (x) is said to belinearly independent, if c1 g1 (x) + c2 g2 (x) + ... + cn gn (x) = 0, where c1, c2, ... cn are arbitraryconstants, then c1 = c2 = ..... = cn = 0.



1.9. INTEGRAL EQUATIONS OF THE CONVOLUTION TYPE. DEFINITION.Consider an integral equation in which the kernel K (x, t) is dependent solely on the difference

x – t, i.e., K (x, t) = K (x – t), ... (1)

where K is a certain function of one variable. Then integral equations

( ) ( ) ( ) ( ) ,x


and ( ) ( ) ( ) ( )b


are called integral equations of the convolution type. K (x – t) is called difference kernel.Let y1 (x) and y2 (x) be two continuous functions defined for x 0. Then the convolution or

Faltung of y1 and y2 is denoted and defined by

1 2 1 2 1 20 0

( ) ( ) ( ) ( ) .*x x

y y y x t y t dt y t y x t dt ... (4)

The integrals occuring in (4) are called the convolution integrals.Note that the convolution defined by relation (4) is a particular case of the standard convolution.

1 2 1 2 1 2( ) ( ) ( ) ( ) .*y y y x t y t dt y t y x t dt

... (5)

By setting y1 (t) = y2 (t) = 0, for t < 0 and t > x, the integrals in (4) can be obtained fromthose in (5).1.10. ITERATED KERNELS OR FUNCTIONS. DEFINITION.

(i) Consider Fredholm integral equation of the second kind

( ) ( ) ( , ) ( )b


Then, the iterated kernels Kn (x, t), n =1, 2, 3, ... are defined as follows :

and

1

1

( , ) ( , )

( , ) ( , ) ( , ) , 2, 3, ...b

n na

K x t K x t

K x t K x z K z t dz n

... (2)

(ii) Consider Volterra integral equation of the second kind

( ) ( ) ( , ) ( ) .x


Then, the iterated kernals Kn (x, t), n = 1, 2, 3 ... are defined as follows :

and

1

1

( , ) ( , )

( , ) ( , ) ( , ) , 2, 3,...x

n nt

K x t K x t


... (4)

1.11. RESOLVENT KERNEL OR RECIPROCAL KERNEL. DEFINITION.Suppose solution of integral equations

( ) ( ) ( , ) ( )b




and ( ) ( ) ( , ) ( )x


be respectively

( ) ( ) ( , ; ) ( ) ,b

ay x f x R x t f t dt ... (3)

and ( ) ( ) ( , ; ) ( ) ,x

ay x f x x t f t dt ... (4)

then R (x, t; ) or (x, t; ) is called the resolvent kernel or reciprocal kernel of the givenintegral equation.1.12 EIGENVALUES (OR CHARACTERISTIC VALUES OR CHARACTERISTIC NUM-BERS). EIGENFUNCTIONS (OR CHARACTERISTIC FUNCTIONS OR FUNDAMENTALFUNCTIONS). DEFINITIONS.

Consider the homogeneous Fredholm integral equation

( ) ( , ) ( ) .b


Then (1) has the obvious solution y (x) = 0, which is called the zero or trivial solution of (1).The values of the parameter for which (1) has a non-zero solution ( ) 0y x are called eigenvaluesof (1) or of the kernel (x, t), and every non-zero solution of (1) is called on eigenfunctioncorresponding to the eigen value .

Remark 1. The number = 0 is not an eigenvalue since for = 0 it follows from (1) thaty(x) = 0.

Remark 2. If y (x) is an eigenfunction of (1), then c y (x), where c is an arbitrary constant, isalso an eigenfunction of (1), which corresponds to the same eigenvalue .

Remark 3. A homogeneous Fredholm integral equation of the second kind may, generally,have no eigenvalue and eigenfunction, or it may not have any real eigenvalue or eigenfunction.1.13.LEIBNITZ’S RULE OF DIFFERENTIATION UNDER INTEGRAL SIGN

Let F (x, t) and /F x be continuous functions of both x and t and let the first derivativesof G (x) and H (x) be continuous. Then

( ) ( )

( ) ( )( , ) [ , ( )] [ , ( )]

H x H x

G x G x

d F dH dGF x t dt dt F x H x F x G xdx x dx dx

... (1)

Particular Case : If G and H are absolute constants, then (1) reduces to

( , )H H

G G

d FF x t dt dtdx x

... (2)

1.14.AN IMPORTANT FORMULA FOR CONVERTING A MULTIPLE INTEGRAL INTO ASINGLE ORDINARY INTEGRAL.

1( )( ) ( ) .( 1)!

nx xn

a a

x ty t dt y t dtn

Note that the integral on the L.H.S. is a multiple integral of order n while the integral on the

R.H.S is ordinary integral of order one.



Proof. Let 1( ) ( ) ( ) ,x n

na

I x x t y t dt ... (1)

where n is a positive integer and a is constant.Differentiating (1) with respect to x and using Leibnitz’s rule, we have

2 1 –1 0( 1) ( ) ( ) ( ) ( ). ( – 0) (0)x n n nn

a

dI dx dn x t y t dt x x y x x ydx dx dx

i.e., d In /dx = (n – 1) In–1, n > 1 ... (2)

From (1), 1 ( )x

aI y t dt so that 1 ( )

dI y xdx

... (3)

Now, differentiating (2) with respect to x successively k times, we haved k In /dxk = (n – 1) (n – 2) ... (n – k) In–k, n > k ... (4)

Using (4) for k = n – 1, we haved n–1 In /dxn–1 = (n– 1)! I1 ... (5)

Differentiating (5) w.r.t. ‘x’ and using (3), we obtaind nIn/dxn = (n – 1)! y (x) ... (6)

From (1), (4) and (5), it follows that In (x) and its first n – 1 derivatives all vanish when x = a.Hence using (3) and (6), we obtain

1 1 1( ) ( )x

aI x y t dt

22 1 2 2 1 1 2( ) ( ) ( )

x x t

a a aI x I t dt y t dt dt

Proceeding likewise, we obtain

3 21 1 2 1( ) ( 1)! ... ( ) ...

nx t t tn n n

a a a aI x n y t dt dt dt dt ... (7)

Combining (1) and (7), we obtain

3 2 11 1 2 1

1... ( ) ... ( ) ( )( 1)!

nx t t t x nn n

a a a a ay t dt dt dt dt x t y t dt

n

... (8)

From (8), we obtain

1( )( ) ( )( 1)!

nx xn

a a

x ty t dt y t dtn

1.15. Regularity conditions.

In this book we shall deal with functions which are either continuous, or integrable or square-integrable. We know that if an integral sign is used, the Lebesgue integral is understood.Furthermore, if a function is Riemann-integrable, it is also Lebesgue integrable. However thereexist functions that are Lebesgue-integrable but not Riemann-integrable. Fortunately, we shall notcome across with such functions in this book.

Square-integrable function or -function. Definition.A given function y (x) is said to be square-integrable if

2| ( ) |b

ay x dx ...(i)



The regularity conditions on the kernel K (x, t) as a function of two variables are similar.Thus, K (x, t) is an -function if

(i) for each set of values of x, t in the square a x b, a t b,

2( , )b b

a aK x t dx dt ... (ii)

(ii) for each value of x in a x b,

2( , )b

aK x t dt ... (iii)

(iii) for each value of t in a t b,

2( , )b

aK x t dx ... (iv)

1.16.THE INNER OR SCALAR PRODUCT OF TWO FUNCTIONS.The inner or scalar product (f, g) of two complex -functions f and g of a real variable x,

a x b, is defined as

( , ) ( ) ( ) ,b

af g f x g x dx ... (i)

where the bar denotes the complex conjugate.The given functions f and g are called orthogonal if their inner product is zero, i.e., if

(f, g) = 0, i.e., ( ) ( ) 0b

af x g x dx

The norm of a function f (x) is denoted by || f (x) || and is defined as

1/ 2 1/ 22|| ( ) || ( ) ( ) | ( ) |

b b

a af x f x f x dx f x dx ... (ii)

A function f (x) is called normalized if || f (x) || = 1. From this definition, it follows that a nonnull function (whose norm is not zero) can be normalized by dividing it by its norm.

In our subsequent analyis, we shall require is following two inequalities :Schwarz inequality | (f, g) | || f || || g ||Minkowski inequality || f + g || || f || + || g ||

1.17.SOLUTION OF AN INTEGRAL EQUATION. DEFINITION.Consider the linear integral equations :

( ) ( ) ( ) ( , ) ( )b


and ( ) ( ) ( ) ( , ) ( )x


A solution of the integral equation (1) or (2) is a function y (x), which, when substituted intothe equation, reduces it to an identity (with respect to x).1.18.SOLVED EXAMPLES BASED ON ART 1.17

Ex. 1. Show that the function y(x) = (1 + x2)–3/2 is a solution of the Voterra integral equation

2 20

1( ) ( )1 1

x ty x y t dtx x

[Kanpur 2009; Meerut 2003]



Sol. Given integral equation is 2 20

1( ) ( )1 1

x ty x y t dtx x

... (1)

Also, given y (x) = (1 + x2)–3/2 ... (2)From (2), y (t) = (1 + t2)–3/2 ... (3)

Then, R.H.S. of (1) 2 3/ 22 20

1 (1 ) ,1 1

x t t dtx x

using (3)

23/ 2

2 2 0

1 1 1(1 ) .21 1

xu du

x x

(on putting t2 = u and 2tdt = du)

21/ 2

2 20

1 1 1 (1 ). .2 ( 1/ 2)1 1

xu

x x

2

2 2 1/ 2 2 2 2 1/ 20

1 1 1 1 1 1 11 1 (1 ) 1 1 (1 )

x

x x u x x x

= (1 + x2)–3/2 = y (x), by (2)= L.H.S. of (1)

Hence (2) is a solution of given integral equation (1).Ex. 2. Show that the function y (x) = xex is a solution of the Volterra integral equation.

0( ) sin 2 cos( ) ( )

xy x x x t y t dt [Meerut 2009, 10, 11; Kanpur 2005, 10]

Sol. Given integral equation is0

( ) sin 2 cos( ) ( ) .x

y x x x t y t dt ... (1)

Also, given y (x) = x ex. ... (2)From (1) y (t) = t et. ... (3)Again, we know the following standard results :

2 2sin( ) [ sin ( ) cos ( )]ax

ax ee bx c dx a bx c b bx ca b

... (4)

and 2 2cos( ) [ cos ( ) sin ( )].ax

ax ee bx c dx a bx c b bx ca b

... (5)

Then R.H.S. of (1)

0 0

sin 2 {cos( ) } sin 2 cos ( )x xt tx x t te dt x t e t x dt

0

0

sin 2 {cos ( ) sin ( )} 1. cos ( ) sin ( ) ,2 2

xt txe ex t t x t x t x t x dt

[Integrating by parts and using formula (5)]

0 0

sin cos ( ) sin( )x xx t tx xe e t x dt e t x dt



0 0

sin cos ( ) sin ( ) sin ( ) cos ( )2 2

x xt tx e ex xe t x t x t x t x

[using formulas (4) and (5)]

1 1sin (cos sin ) ( sin cos ) ( ), by(2)2 2 2 2

x xx xe ex xe x x x x xe y x

= L.H.S. of (1).

Hence (2) is a solution of (1).Ex. 3. Show that y (x) = cos 2x is a solution of the integral equation

0( ) cos 3 ( , ) ( )y x x K x t y t dt

wheree sin cos , 0

( , )cos sin , .

x t x tK x t

x t t x

[Garhwal 1998, Kanpur 2005, 08, 09; Meerut 2004, 2008, 2012]

Sol. Given integral equation is 0

( ) cos 3 ( , ) ( ) ,y x x K x t y t dt

... (1)

where sin cos , 0

( , )cos sin , .

x t x tK x t

x t t x

... (2)

Also given, y (x) = cos 2x ... (3)From (3), y (t) = cos 2t ... (4)Then, R.H.S. of (1)

0cos 3 ( , ) ( ) ( , ) ( )

x

xx K x t y t dt K x t y t dt

0cos 3 cos sin cos 2 sin cos cos 2 ,

x

xx x t t dt x t t dt

by (2) and (4)

0cos 3cos cos 2 sin 3sin cos 2 cos

x


0

3 3cos cos (sin 3 sin ) sin (cos3 cos )2 2

x


0

3 1 3 1cos cos cos3 cos sin sin3 sin2 3 2 3

x

xx x t t x t t

3 1 1 3 1cos cos cos3 cos 1 sin sin 3 sin2 3 3 2 3

x x x x x x x 2 21 3cos (cos3 cos sin 3 sin ) (cos sin ) cos

2 2x x x x x x x x

1 3 1 3cos (3 ) cos 2 cos 2 cos 22 2 2 2

x x x x x

= cos 2x = y (x), by(3) = L.H.S.of (1).Hence (3) is a solution of (1).Ex. 7. Show that the function y (x) = sin ( x / 2) is a solution of the Fredholm integral

equation 2 1

0( ) ( , ) ( ) ,

4 2xy x K x t y t dt

where the kernel K (x, t) is of the form

(1/ 2) (2 ), 0( , )

(1/ 2) (2 ), 1.x t x t

K x tt x t x

[Kanpur 2011; Meerut 2005]



Sol. Given integral equation is2 1

0( ) ( , ) ( ) ,

4 2xy x K x t y t dt

... (1)

where (1/ 2) (2 ), 0

( , )(1/ 2) (2 ), 1.

x t x tK x t

t x t x

... (2)

Given y (x) = sin ( x / 2). ... (3)From (3), y (t) = sin ( t / 2). ... (4)Then, L.H.S. of (1)

2 1

0sin ( , ) ( ) ( , ) ( ) ,

2 4

x

x

x K x t y t dt K x t y t dt using (3)

2 1

0

1 1sin (2 ) sin (2 ) sin ,2 4 2 2 2 2

x

x

x t tt x dt x t dt by (2) and (4)

2 2 1

0sin (2 ) sin (2 )sin

2 8 2 8 2

x

x

x t x tx t dt t dt

2

00

(2 ) cos( / 2 cos ( / 2)sin 12 8 / 2 / 2

x xx x t tt dt

12 1cos ( / 2) cos ( / 2)(2 ) ( 1)

8 / 2 / 2xx

x t tt dt

2

20

(2 ) 2 sin( / 2)sin cos2 8 2 ( / 2)

xx x x x t

12

22(2 ) sin( / 2)cos

8 2 ( / 2) x

x x x x 2

2(2 ) 2 4sin cos sin

2 8 2 2x x x x x

2

2 22(2 ) 4 4cos sin

8 2 2x x x x

1sin 1 (2 ) R.H.S. of (1).2 2 2 2 2x x x xx

Hence (3) is a solution of (1).

EXERCISEVerify that the given functions are solutions of the corresponding integral equations.

1.0

( ) 1 ; ( )x x ty x x e y t dt x (Kanpur 2007) 2.

0

1 ( )( ) ;2

x y ty x dt xx t

3. 3 2

0( ) 3; ( ) ( ) .

xy x x x t y t dt (Kanpur 2011)



4.3

0( ) ; ( ) sinh( ) ( ) .

6

xxy x x y x x x t y t dt 5.

0( ) ; ( ) sin 2 cos( ) ( )

xx sy x xe y x e x x t y t dt

6.3 3

2 5/ 22 2 2 20

3 2 3 2( ) /(1 ) ; ( ) ( ) .3(1 ) (1 )

xx x x x ty x x x y x y t dtx x

7. y(x) = ex (cos ex – ex sin ex); 2 2 2

0( ) (1– )cos 1 sin 1 {1– ( – ) } ( )

xx x xy x xe e x t e y t dt 8.

1

0( ) ; ( ) sin ( ) 1.xy x e y x xt y t dt

9. 2

0( ) cos ; ( ) ( ) cos ( ) sin .

xy x x y x x t t y t dt x

10. ( )

0( ) ; ( ) 4 ( ) ( 1)x x t xy x xe y x e y t dt x e

11. 0

2sin( ) 1 ; ( ) cos( ) ( ) 1.(1 / 2)

xy x y x x t y t dt

12.1

1/ 20

1 ( )( ) ; 1( )

y ty x dtx x t

[Kanpur 2006]

13.4( ) sin ,cy x x

(c being an arbitary constant) : 2

0

4 sin( ) sin ( ) 0.ty x x y t dtt

14.1 3/ 2

0( ) ; ( ) ( , ) ( ) (4 7),

15xy x x y x K x t y t dt x x where

(1/ 2) (2 ), 0( , )

(1/ 2) (2 ), 1.x t x t

K x tt x t x

15.1

0( ) (2 2 / 3); ( ) 2 ( ) 2x x t xy x e x y x e y t dt x e [Kanpur 2006, 10]

16.1

0( ) 1; ( ) ( 1) ( )xt xy x y x x e y t dt e x

17. For what value of , the function y (x) = 1+ x is a solution of the integral equation

0( )

x x tx e y t dt ?

Hint : Proceed as in solved Ex. 1 on page 1.8 Ans. 1 .


CHAPTER 2

Conversion of Ordinarydifferential equations intointegral equations2.1. INTRODUCTION

While searching for the representation formula for the solution of an ordinary differentialequation in such a manner so as to include the boundary conditions or initial conditions explicitly,we always arrive at integral equations. Thus, a boundary value or an initial value problem is convertedto an integral equation. Later on in this chapter, the reader will notice that an initial value probelmis always converted into a Volterra integral equation and a boundary value problem is alwaysconverted into a Fredholm integral equation. After converting an initial value or a boundary valueproblem into an integral equation, it can be solved by shorter methods of solving integral equations.2.2. INITIAL VALUE PROBLEM. DEFINITION.

When an ordinary differential equation is to be solved under conditions involving dependentvariable and its derivative at the same value of the independent variable, then the problem underconsideration is said to be an initial value problem.

For example, d2y/dx2 + y = x, y (0) = 2, y (0) = 3 ... (1)and d2y/dx2 + y = x, y (1) = 2, y (1) = 2 ... (2)are both intial value problems. Note that in (1), the same value x = 0 of the independent variable isinvolved whereas in (2), the same value x = 1 of the independent variable is involved.2.3. METHOD OF CONVERTING AN INITIAL VALUE PROBLEM INTO A VOLTERRAINTEGRAL EQUATION.

This method is illustrated with the help of the following solved examples.Ex. 1. Convert the following differential eqaution into integral equation :

0 (0) (0) 0.y y when y y

Sol. Given ( ) ( ) 0,y x y x ... (1)with initial conditions (0) = 0 ... 2(a)

and (0) 0y ... 2(b)

From (1), ( ) ( )y x y x ... (3)Integrating both sides of (3) w.r.t. ‘x’ from 0 to x, we have

0 0

( ) ( )x x

y x dx y x dx or 0 0( ) ( )

xxy x y x dx

or 0

( ) – (0) ( )x

y x y y x dx or 0

( ) – ( ) , using 2( )x

y x y x dx b ...(4)

2.1


2.2 Conversion of Ordinary differential equation into integral equation

Integrating both sides of (4) w.r.t. ‘x’ from 0 to x, we have

2

0 0( ) ( )

x xy x dx y x dx or 2

0 0( ) ( )

xxy x y x dx

or 2

0( ) (0) ( )

xy x y y x dx or 2

0( ) ( ) ,

xy x y t dt using 2 (a)

or 0

( ) ( ) ( )x

y x x t y t dt , using result of Art. 1.14.

which is the desired integral equation.Ex. 2. Convert the following differential equation into an integral equation :

( ), (0) 1, (0) 0y x y f x y y

Sol. Given ( ) ( ) ( )y x x y x f x ... (1)with initail conditions y (0) = 1 ... 2(a)and (0) 0.y ... 2(b)

From (1), ( ) ( ) ( )y x f x x y x ...(3)Integrating both sides of (3) w.r.t. ‘x’ from 0 to x, we have

0 0

( ) [ ( ) ( )]x x

y x dx f x x y x dx or 00

[ ( )] [ ( ) ( )]xxy x f x x y x dx

or 0

( ) (0) [ ( ) ( )]x

y x y f x x y x dx or 0

( ) [ ( ) ( )] ,x

y x f x x y x dx using (2b) ... (4)


2

0 0( ) [ ( ) ( )]

x xy x dx f x xy x dx or 2

00

[ ( )] [ ( ) ( )]xxy x f x xy x dx

or 2

0( ) (0) [ ( ) ( )]

xy x y f x xy x dx or 2

0( ) 1 [ ( ) ( )] ,

xy x f t t y t dt using 2(a)

or0

( ) 1 ( )[ ( ) ( )] ,x

y x x t f t t y t dt using result of Art. 1.14

which is the required integral equation.Ex. 3. Convert the following initial value problem into an integral equation :

2

0 02 ( ) ( ) ( ), ( ) , ( ) .d y dyA x B x y f x y a y y a ydxdx

Sol. Given ( ) ( ) ( ) ( ) ( ) ( )y x A x y x B x y x f x ... (1)with initial conditions : y (a) = y0 ... 2(a)

and 0( )y a y ... 2(b)

From (1), ( ) ( ) ( ) ( ) ( ) ( ).y x f x B x y x A x y x ... (3)Integrating both sides of (3) w.r.t. ‘x’ from a to x, we have

[ ( ) [ ( ) ( ) ( )] ( ) ( )x xx

aa a

y x f x B x y x dx A x y x dx


Conversion of Ordinary differential equations into integral equations 2.3

or ( ) ( ) [ ( ) ( ) ( )] [ ( ) ( )] ( ) ( )x xx

aa a

y x y a f x B x y x dx A x y x A x y x dx

[on integrating by parts the second terms on R.H.S.]

or 0( ) [ ( ) ( ) ( )] ( ) ( ) ( ) ( ) ( ) ( ) ,x x

a ay x y f x B x y x dx A x y x A a y a A x y x dx

by 2 (b)

or 0 0( ) ( ) ( ) ( )y x y A x y x A a y { ( ) ( ) ( ) ( ) ( )}x

af x B x y x A x y x dx , by 2(a) ... (4)

Integrating both sides of (4) w.r.t. ‘x’ from a to x, we have

0 0( ) [ ( )] ( ) ( )x x x

a a ay x dx y y A a dx A x y x dx 2{ ( ) ( ) ( ) ( ) ( )}

x

af x B x y x A x y x dx

or 0 0( ) [ ( )]( ) ( ) ( )xx

a ay x y y A a x a A x y x dx 2{ ( ) ( ) ( ) ( ) ( )}

x

af t B t y t A t y t dt

or 0 0( ) ( ) [ ( )]( ) ( ) ( )x

ay x y a y y A a x a A t y t dt ( ){ ( ) ( ) ( ) ( ) ( )}

x

ax t f t B t y t A t y t dt

[using result of Art. 1.14]

or 0 0 0( ) [ ( )]( ) ( ) ( )x

ay x y y y A a x a x t f t dt [ ( ) ( ){ ( ) ( )}] ( ) ,

x

aA t x t B t A t y t dt

which is Volterra integral equation of the second kind.

Ex. 4. Convert sin xy x y e y x with initial conditions y (0) = 1, (0) 1y to aVolterra integral equation of the second kind. Conversely, derive the original differential equationwith the initial conditions from the integral equation obtained. [Meerut 2002, 04, 07, 11]

Sol. Given ( ) sin ( ) ( )xy x x y x e y x x ... (1)

with initial conditions : y(0) = 1 ... 2(a)

and (0) 1y ...2(b)

From (1), ( ) ( ) sin ( ).xy x x e y x x y x ... (3)


0 0 0 0

( ) ( ) sin ( )x x x xxy x dx x dx e y x dx x y x dx

or 2

0 00 0

[ ( )] ( ) [sin ( )] cos ( )2

x xx x xxy x e y x dx x y x x y x dx [Integrating by parts the third term on R.H.S.]

2

0 0( ) (0) ( ) sin ( ) cos ( )

2

x xxxy x y e y x dx x y x x y x dx

or 2

0( ) 1 sin ( ) ( cos ) ( ) ,

2

x xxy x x y x e x y x dx using 2 (b)



or2

0( ) –1 sin ( ) – ( cos ) ( )

2

x xxy x x y x e x y x dx ... (4)

Integrating both sides of (4) w.r.t. ‘x’ from 0 to x, we have2

2

0 0 0 0( ) 1 sin ( ) ( cos ) ( )

2

x x x x xxy x dx dx x y x dx e x y x dx

or3

20

0 00

[ ( )] sin ( ) ( cos ) ( )6

xx xx txy x x t y t dt e t y t dt

or 3

0 0( ) (0) sin ( ) ( ) ( cos ) ( )

6

x x txy x y x t y t dt x t e t y t dt , using result of Art. 1.14

or3

0( ) –1 – {sin – ( – )( cos )} ( ) , by 2( )

6

x txy x x t x t e t y t dt a

or3

0( ) 1 [sin ( ) ( cos )] ( ) ,

6

x txy x x t x t e t y t dt ... (5)

which is the required Volterra integral equation of the second kind.Second part : Derivation of the given differential equation together with given initial conditions

from integral equation (5) :Differentiating both sides of (5) w.r.t. ‘x’, we get

2

0( ) 1 [sin ( ) ( cos )] ( )

2

x tx dy x t x t e t y t dtdx

or2

0( ) 1 [{sin ( ) ( cos )} ( )]

2

x txy x t x t e t y t dtx

[sin ( ) ( cos )] ( )x dxx x x e x y xdx

0 0[sin 0 ( 0) ( cos0)] (0) dx e ydx

[using Leibnitz’s rule of differentiation under integral sign (refer Art. 1.13)]

or2

0( ) 1 ( cos ) ( ) sin ( ).

2

x txy x e t y t dt x y x ... (6)

Differentiating both sides of (6) with respect to‘x’ weget

0( ) cos ( ) sin ( ) ( cos ) ( )

x tdy x x x y x x y x e t y t dtdx

or0

( ) cos ( ) sin ( ) {( cos ) ( )}x ty x x x y x x y x e t y t dt

x

0 0( cos ) ( ) ( cos0) (0)x dx de x y x e ydx dx

, using Leibnitz’s rule

or ( ) cos ( ) sin ( ) [0 ( cos ) ( ) 0]xy x x x y x x y x e x y x

or ( ) sin ( ) ( ) ,xy x x y x e y x x ... (7)

which is the same as given differential equation (1).



Putting x = 0 on both sides of (5) and (6), we easily obtain

y (0) = 1 and (0) 1.y ... (8)(7) and (8) together give us the given differential equation and initial conditions.

Ex. 5. Convert ( ) 3 ( ) 2 ( ) 4siny x y x y x x with initial conditions y(0) = 1, (0) 2y into a Volterra integral equation of the second kind. Conversely, derive the original differentialequation with initial conditions from the integral equation obtained.

[Meerut 2003, 06; Kanpur 2009]

Sol. Given ( ) 3 ( ) 2 ( ) 4siny x y x y x x ... (1)with initial conditions : y (0) = 1 ... 2(a)and (0) 2y ... 2(b)

From (1), ( ) 4sin 2 ( ) 3 ( )y x x y x y x ... (3)Integrating both sides of (3) w.r.t. ‘x’ from 0 to x, we have

0 0 0 0

( ) 4 sin 2 ( ) 3 ( )x x x x

y x dx x dx y x dx y x dx or 0 0 00

( ) 4 cos 2 ( ) 3 ( )xx x xy x x y x dx y x

or 0

( ) (0) 4( cos 1) 2 ( ) 3[ ( ) (0)]x

y x y x y x dx y x y or

0( ) 2 4 4cos 2 ( ) 3 ( ) 3,

xy x x y x dx y x using 2 (a) and 2 (b)

or 0

( ) 1 4 cos 3 ( ) 2 ( )x

y x x y x y x dx ... (4)


2

0 0 0 0 0( ) 4 cos 3 ( ) 2 ( )

x x x x xy x dx dx x dx y x dx y x dx

or 20 0 0 0

( ) 4 sin 3 ( ) 2 ( )x xx xy x x x y x dx y t dt

or0 0

( ) (0) 4sin 3 ( ) 2 ( ) ( ) ,x x

y x y x x y t dt x t y t dt by result of Art. 1.14

or 0

( ) 1 4sin [3 2( )] ( ) ,x

y x x x x t y t dt using 2 (a) ... (5)

which is the required Volterra integral equation of the second kind.Second Part : Derivation of the given differential equation together with given initial

conditions from integral equation (5).Differentiating both sides of (5) w.r.t. ‘x’, we get

0

( ) 1 4cos [3 2( )] ( )xdy x x x t y t dt

dx

or 0

( ) 1 4 cos [{3 2( )} ( )]x

y x x x t y t dtx

0[3 2( )] ( ) [3 2( 0)] (0)dx dx x y x x ydx dx

[using Leibnitz’s rule of differentiation under integral sign (refer Art. 1.13]



or0

( ) 1 4 cos ( 2) ( ) 3 ( )x

y x x y t dt y x or

0( ) 1 4cos 3 ( ) 2 ( ) .

xy x x y x y t dt ... (6)

Differentiating both sides of (6) w.r.t. ‘x’, we get

0( ) 4sin 3 ( ) 2 ( )

xdy x x y x y t dtdx

or

0

0( ) 4sin 3 ( ) 2 ( ) ( ) (0)x dx dy x x y x y t dt y x y

x dx dx by Leibnitz’s rule

( ) 4sin 3 ( ) 2[0 ( ) 0]y x x y x y x

or ( ) 3 ( ) 2 ( ) 4sin ,y x y x y x x ... (7)which is the same as the given differential equation.

Putting x = 0 in (5), we get y(0) = 1. Further putting x = 0 in (6), we get(0) 1 4 3 (0)y y or (0) 1 4 3 2,y using 2(a)

Thus, y(0) = 1 and (0) 2.y ... (8)(7) and (8) together give us the given differential equation and initial conditions.

Ex. 6. The initial value problem corresponding to the integral equation 0

( ) 1 ( )x

y x y t dt is

(a) – 0, (0) 1y y y (b) 0, (0) 0y y y

(c) – 0, (0) 0y y y (d) 0, (0) 1y y y [GATE 2001]

Sol. Ans (a) Given 0

( ) 1 ( ) .x

y x y t dt ...(1)

Differentiating both sides of (1) with respect to x and using the Leibnitz’s rule of differentiationunder the sign of integral (refer Art. 1.13), we obtain

0

( )( ) 0 ( ) – (0)x y t dx doy x dt y x y

x dx dx

or ( ) ( ),y x y x i.e., – 0y y ...(2)

From (1), 0

0(0) 1 ( ) 1,y y t dt i.e., (0) 1y ...(3)

(2) and (3) show that result (a) is true.

EXERCISE-2A 1. (a) Show that, if y (x) satisfies the differential equation (d2y/dx2) + xy = 1 and the conditions

(0) (0) 0,y y then y also satisfies the Volterra equation 2

0

1( ) ( ) ( ) .2

xy x x t t x y t dt

(b) Prove that the converse of the preceding statement is also true. (Meerut 2011)2. (a) If ( ) ( ),y x F x and y satisfies the initial conditions y(0) = y0 and 0(0) ,y y show

that 0 00

( ) ( ) ( ) .x

y x y xy x t F t dt



(b) Verify that this expression satisfies the prescribed differential equation and intialconditions.

3. Convert ( ) 2 ( ) 3 ( ) 0y x xy x y x with initial conditions y(0) = 1, (0) 0y to aVolterra integral equation of the second kind. Conversely, derive the original differential equation

with initial conditions from the integral equation obtained. Ans.0

( ) 1 ( ) ( )x

y x x t y t dt 4. Reduce the following initial value problem into an integral equation

2

2 0, (0) 1, (0) 1.d y dyx y y ydxdx

Ans. 0

( ) 1 ( )x

y x x t y t dt [Kanpur 2007, 11; Meerut 2009]

5. Show that the solution of the Volterra equation 0

( ) 1 ( ) ( )x

y x t x y t dt satisfies the

differential equation ( ) ( ) 0y x y x and the boundary conditions y (0) = 1, (0) 1y

2.4. ALTERNATIVE METHOD OF CONVERTING AN INITIAL VALUE PROBLEM INTOA VOLTERRA INTEGRAL EQUATION.This method is somewhat simpler than the method outlined in Art. 2.3. However, the method

explained in Art. 2.3 is very useful in problem where we are required to derive the original differentialequation together with initial conditions from the integral equation obtained.

Consider the ordinary linear differential equation of order n :

1 2

1 21 2( ) ( ) ... ( ) ( )n n n

nn n nd y d y d ya x a x a x y xdx dx dx

... (1)

with the initial conditions( 1)

0 1 2 1( ) , ( ) , ( ) , . . .,. . . ( ) ,nny a q y a q y a q y a q ... (2)

where the functions a1 (x), ..., an(x) and (x) are defined and continuous in .a x b In order to reduce the initial value problem (1)–(2) to the Volterra integral equation, we

introduce an unknown function u (x). Thus, we takedny/dxn = u (x) ...(An)

Integrating both sides of equation (An) w.r.t. ‘x’ from a to x, we have

–1

–1 ( )xn x

n aa

d y u x dxdx

or

1( 1)

1 ( ) ( )n xn

n a

d y y a u x dxdx

or1

11 ( ) ,n x

nn a

d y u x dx qdx

using (2) 1...( )nA

or 1

11 ( )n x

nn a

d y u t dt qdx

... (An–1)

Integrating both sides of equation 1( )nA w.r.t. ‘x’ from a to x, we have

22

12 ( )xn x x

nn a aa

d y u x dx q dxdx



or 2

( 2) 212 ( ) ( )

n x xnn an a

d y y a u x dx q xdx

or2

21 22 ( ) ( ) ,

n xn nn a

d y u x dx x a q qdx

using (2) 2...( )nA

or2

21 22 ( ) ( )

n xn nn a

d y u t dt x a q qdx

or2

1 22 ( ) ( ) ( )n x

n nn a

d y x t u t dt x a q qdx

... (An–2 )

[using result of Art. 1.14]Integrating both sides of equation 2( )nA w.r.t. ‘x’ from a to x, we have

33

1 23 ( ) ( )xn x x x

n nn a a aa

d y u x dx q x a dx q dxdx

or3 2

( 3) 31 23

( )( ) ( ) [ ]2

xn xn xn n an a

a

d y x ay a u x dx q q xdx

or3 2

31 2 33

( ) ( )( ) ,2! 1!

n xn n nn a

d y x a x au x dx q q qdx

using (2) 3...( )nA

or3 2

31 2 33

( ) ( )( )2! 1!

n xn n nn a

d y x a x au t dt q q qdx

or3 2 2

1 2 33( ) ( ) ( )( )

2! 2! 1!

n xn n nn a

d y x t x a x au t dt q q qdx

... (An–3)

[using result of Art. 1.14 again]Ans so on. Finally, we arrive at :

2 2 3

1 2( ) ( ) ( )( )( 2)! ( 2)! ( 3)!

n n nxn n

a

dy x t x a x au t dt q qdx n n n

+ ... + q2 (x – a) + q1 ... (A1)

and1 1 2

1 2( ) ( ) ( )( )( 1)! ( 1)! ( 2)!

n n nxn n

a

x t x a x ay u t dt q qn n n

+ ... + q1 (x – a) + q0 ... (A0)

Multiplying (An), (An–1), ...., (A1) and (A0) by 1, a1(x), ..., an–1 (x) and an (x) respectively andadding, we get

1

1 1( ) ... ( ) ( )n n

nn nd y d ya x a x y u xdx dx

–1 1 –2 –1 2( ) { ( – ) } ( ) ...

n n nq a x q x a q a x

–1

0 1 –1( – )( – ) ... ( )

( –1)!

n

n nx aq q x a q a x

n

2

1 2 3( )( ) ( ) ( ) ( )

2!

x

a

x ta x x t a x a x

1( )... ( ) ( )

( 1)!

n

nx t a x u t dtn



or ( ) ( ) ( ) ( , ) ( ) ,x

ax u x x K x t u t dt ... (3)

where we have used (1) and assumed the following :1

1 1 –2 –1 2 0 1 1( )( ) ( ) { ( – ) } ( ) ... ( – ) ... ( )

( 1)!

n

n n n n nx ax q a x q x a q a x q q x a q a x

n

... (4)

and 1

1 2( )( , ) ( ) ( ) ( ) ... ( )( 1)!

n

nx tK x t a x x t a x a xn

... (5)

Again, let ( ) ( ) ( ).x x f x ... (6)

Using (6), (3) reduces to ( ) ( ) ( , ) ( ) ,x

au x f x K x t u t dt ... (7)

which is the required Volterra integral equational of the second kind. Thus, the initial value problem(1) – (2) has been converted into Volterra integral equation of the second kind (7).

SOLVED EXAMPLES BASED ON ART. 2.4Ex. 1. Form an integral equation corresponding to the differential equation 0,y xy y

with the initial conditions : y(0) = 1, (0) 0.y Sol. Given differential equation is d2y/dx2 + x(dy/dx) + y = 0, ... (1)

subject to the initial conditions : y (0) = 1 ... 2(a)

and (0) 0.y ... 2(b)Suppose that d2y/dx2 = u (x) ... (A2)Integrating both sides of (A2) w.r.t. ‘x’ from 0 to x, we have

00( )

x xdy u x dxdx or

0(0) ( )

xdy y u x dxdx

or0

( ) ,xdy u x dx

dx using 2 (b) 1...( )A

or 0

( )xdy u t dt

dx ... (A1)

Integrating both sides of 1( )A w.r.t. ‘x’ from 0 to x, we have

2

0( ) (0) ( )

xy x y u x dx or 2

0( ) 1 ( ) ,

xy x u t dt using 2 (a)

or0

( ) 1 ( ) ( )x

y x x t u t dt , using result of Art. 1.14 ... (A0)

Putting values of d2y/dx2, dy/dx and y given by (A2), (A1) and (A0) respectively in (1), we get

0 0( ) ( ) 1 ( ) ( ) 0

x xu x x u t dt x t u t dt or

0 0( ) 1 ( ) ( ) ( ) 0

x xu x xu t dt x t u t dt

or0

( ) 1 { ( )} ( ) 0x

u x x x t u t dt or 0

( ) 1 (2 ) ( ) ,x

u x x t u t dt ... (3)

which is the required Volterra integral equation of the second kind.



Ex. 2. Form an integral equation corresponding to the differential equation (d2y/dx2)– sin x (dy / dx) + exy = x, with the initial conditions y(0) = 1, (0) 1.y

Sol. Given differential equation is d2y/dx2 – sin x (dy / dx) + exy = x, ... (1)subject to the initial conditions : y(0) = 1 ... 2(a)

and (0) 1y ... 2(b)Suppose that d2y/dx2 = u (x). ... (A2)Integrating (A2) w.r.t. ‘x’ from 0 to x, we get

00( )

x xdy u x dxdx or

0(0) ( )

xdy y u x dxdx

or0

1 ( ) ,xdy u x dx

dx by 2 (b) or

01 ( )

xdy u x dxdx

1...( )A

or0

1 ( ) .xdy u t dt

dx ... (A1)

Integrating 1( )A w.r.t. ‘x’, we get

20 0 0

( ) ( )xx xy x x u x dx or 2

0( ) (0) ( )

xy x y x u t dt

or 2

0( ) 1 ( ) ,

xy x x u t dt using 2 (a)

or0

( ) 1 ( ) ( ) ,x

y x x x t u t dt using result (1) of Art. 1.14 ... (A0)

Putting values of d2y/dx2, dy/dx and y given by (A2), (A1) and (A0) respectively in (1), we get

0 0

( ) sin [ 1 ( ) ] [1 ( ) ( ) ]x xxu x x u t dt e x x t u t dt x

or0 0

( ) sin (1 ) sin ( ) ( ) ( )x xx xu x x x e x x u t dt e x t u t dt

or0

( ) sin (1 ) [sin ( )] ( ) ,xx xu x x x e x x e x t u t dt ... (3)

which is the required Volterra integral equation of the second kind.Ex. 3. Form the integral equation corresponding to the following differential equation with

the given initial conditions : 2 0; (0) 1/ 2, (0) (0) 1.y xy y y y

Sol. Given differential equation is d3y/dx3 – 2xy = 0, ... (1)subject to the initial conditions : y (0) = 1/2, ... 2(a)

(0) 1y ... 2(b)

and (0) 1y ... 2(c)Suppose that d3y/dx3 = u (x) ... (A3 )Integrating (A3) w.r.t. ‘x’ from 0 to x, we get

2

2 00

( )x

xd y u x dxdx

or

2

2 0(0) ( )

xd y y u x dxdx



or2

2 01 ( )

xd y u x dxdx

, by 2 (c) 2...( )A

or2

2 01 ( ) .

xd y u t dtdx

... (A2)

Integrating 2( )A w.r.t. ‘x’ from 0 to x, we get

2

0 00( )

x x xdy dx u x dxdx or 2

0(0) ( )

xdy y x u x dxdx

or 2

01 ( ) ,

xdy x u x dxdx

by 2 (b) 1...( )A

or 2

01 ( )

xdy x u t dtdx

or

01 ( ) ( ) ,

xdy x x t u t dtdx

using result (1) of Art. 1.14 ... (A1)

Integrating 1( )A w.r.t. ‘x’, from 0 to x, we get

3

0 0( ) (0) (1 ) ( )

x xy x y x dx u x dx or 2 3

00

1 1( ) ( ) ,2 2

x xy x x x u t dt by 2 (a)

or2

2

0

1 1 ( )( ) ( ) ,2 2 2!

x x ty x x x u t dt using result of Art. 1.14 ... (A0)

Puting values of d3y/dx3 and y given by (A3) and (A0) respectively in (1), we have

2 2

0

1 1 1( ) 2 ( ) ( ) 02 2 2

xu x x x x x t u t dt or 2 2

0( ) (1 2 ) ( ) ( )

xu x x x x x x t u t dt

or 2 2

0( ) ( 1) ( ) ( ) ,

xu x x x x x t u t dt

which is the required integral equation.Ex. 4. Form an integral equation corresponding to the differential equation

2( ) 1xy x y x x y x e with initial conditions : (0) 1 (0), (0) 0.y y y [Meerut 2009]

Sol. Given differential equation is d3y/dx3 + x (d2y/dx2) + (x2 – x) y = x ex + 1 ... (1)subject to the initial conditions : y (0) = 1, ... 2(a)

(0) 1y ...2 (b)

and (0) 0y ... 2(c)Suppose that d3y/dx3 = u (x). ... (A3)Integrating (A3) w.r.t. ‘x’ from 0 to x, we get

2

2 00

( )x

xd y u x dxdx

or

2

2 0(0) ( )

xd y y u x dxdx

or2

2 0( ) ,

xd y u x dxdx

by 2 (c) 2...( )A



or2

2 0( )

xd y u t dtdx

... (A2)


2

00( )

x xdy u x dxdx or 2

0(0) ( )

xdy y u x dxdx

or 2

01 ( ) ,

xdy u x dxdx

by 2 (b) 1...( )A

or 2

01 ( )

xdy u t dtdx

or 0

1 ( ) ( ) ,xdy x t u t dt

dx using result of Art. 1.14 ... (A1)


3

0 0( ) (0) ( )

x xy x y dx u x dx or 3

0( ) 1 ( ) ,

xy x x u t dt by 2(a)

or2

0

( )( ) 1 ( ) ,2!

x x ty x x u t dt using result of Art 1.14 ... (A0)

Putting values of d3y/dx3, d2y/dx2 and y (x) given by (A3), (A2) and (A0) respectively in (1),

we get2 2

0 0

1( ) ( ) ( – ) 1 ( ) ( ) 12

x x xu x x u t dt x x x x t u t dt x e

or 2 2 2

0 0

1( ) 1 ( ) ( 1) ( ) – ( ) ( ) ( )2

x xxu x x e x x x x u t dt x x x t u t dt

or2 2 2

0

1( ) 1 ( 1) ( ) ( ) ( ) ,2

xxu x x e x x x x x x t u t dt which is the required integral equation.

Ex. 5. Prove that the linear differential equation of second order2

1 22 ( ) ( ) ( ),d y dya x a x y F xdxdx

with initial conditions y(0) = C0 and 1(0)y C can be transformed into non-homogeneous Volterra’ssintegral equation of second kind. [GATE 2006]

Sol. Given differential equation is2

1 22 ( ) ( ) ( ),d y dya x a x y F xdxdx

... (1)

subject to the boundary conditions : y (0) = C0 ... 2 (a)

and 1(0)y C ... 2 (a)Suppose that d2y / dx2 = u (x). ... (A2)Integrating (A2) w.r.t. ‘x’ from 0 to x, we get

00( )

x xdy u x dxdx or

0(0) ( )

xdy y u x dxdx

or 10

( ) ,xdy C u x dx

dx by 2 (b) 1...( )A



or 10

( ) .xdy C u t dt

dx ... (A1)


21

0 0( ) (0) ( )

x xy x y C dx u x dx or 2

0 10

( ) ( ) ,x

y x C C x u t dt by 2 (a)

or 0 10

( ) ( ) ,x

y x C C x x t u dt using result of Art. 1.14 ... (A0)

Putting values of d2 / dx2, dy/dx and y given by (A2), (A1) and (A0) respectively in (1), we get

1 1 2 0 10 0

( ) ( ) [ ( ) ] ( ) [ ( ) ( ) ] ( )x x

u x a x C u t dt a x C C x x t u t dt F x or 1 1 0 1 2( ) ( ) ( ) ( ) ( )u x F x C a x C C x a x 1 2

0 0( ) ( ) ( ) ( ) ( )

x xa x u t dt a x x t u t dt

or 1 1 0 1 2 1 20

( ) ( ) ( ) ( ) ( ) [ ( ) ( ) ( )] ( ) ,x

u x F x C a x C C x a x a x a x x t u t dt which is the required non-homogeneous Volterra’s integral equation of second kind.

EXERCISE-2 (b)Reduce the following initial value problems into Volterra integral equations of the second kind :

1. 0; (0) 0, (0) 1.y y y y Ans. 0

( ) ( ) ( ) , where ( )x

u x x x t u t dt u x y 2. 0; (0) 1y y y Ans.

0( ) 1 ( ) , where ( )

xu x u t dt u x y

3. cos : (0) 0, (0) 1y y x y y

Ans. 0

( ) cos ( ) ( ) , where ( )x

u x x x x t u t dt u x y 4. 5 6 0; (0) 0, (0) 1.y y y y y (Kanpur 2008)

Ans.0

( ) 6 5 (5 6 6 ) ( ) , where ( )x

u x x x t u t dt u x y 5. ( ) 3 ( ) 2 ( ) 4 sin , (0) 1, (0) 2.y x y x y x x y y

Ans.0

( ) 4 ( sin 2) [3 2( )] ( ) , where ( )x

u x x x x t u t dt u x y 6. cos ; (0) 0, (0) 0y y x y y Ans.

0( ) cos ( ) ( ) , where ( )

xu x x x t u t dt u x y

7. 2 3 0, (0) 1, (0) 0.y xy y y y Ans.0

( ) 3 (5 3 ) ( ) , where ( )x

u x x t u t dt u x y 8. 1, (0) (0) 0.y xy y y (Kanpur 2009)

Ans.0

( ) 1 ( ) ( ) , where ( )x

u x x x t u t dt u x y 9. 2(1 ) cos ; (0) 0, (0) 2y x y x y y

Ans. 2 2

0( ) cos 2 (1 ) (1 ) ( ) ( ) , where ( )

xu x x x x x x t u t dt u x y



10. ( ) ( ) ( ), (0) 1, (0) 0.y x y x F x y y (Meerut 2012)

Ans.0

( ) ( ) ( ) ( ) , where ( )x

u x F x x t u t dt u x y 11. Show that a linear differential equation with constant coefficients reduces, under any initial

conditions, to a Volterra integral of the second kind with kernel dependent solely on the difference(x – t) of arguments (integral equation of the closed cycle or equation of the Faltung type, orconvolution type).

12. Establish relation between linear differential equation of order n and Volterra integralequation of the second kind.

13. Reduce linear differential equation of order n into a Volterra integral equation the secondkind. Hence prove that the solution of Volterra integral equation is unique.

14. Show that linear differential equation d ny/dxn + a1(x) (d n–1y/dxn–1) + ... + an (x) y = (x)can be converted into volterra integral equation of the type

1

1 20

( )( ) ( ) ( ) ( ) ... ( ) ( ) ( ).( 1)!

nxn

x tu x a x a x x t a x u t f xn

2.5. BOUNDARY VALUE PROBLEM, DEFINITION.When an ordinary differential equation is to be solved under conditions involving dependent

variable and its derivatives at two different values of independent variable, then the problem underconsideration is said to be a boundary value problem.

For example d2y / dx2 + y = 0, y (a) = y, y (b) = y2is a boundary value problem. Note that here different values x = a and x = b of the independentvariable x are involved.2.6. METHOD OF CONVERTING A BOUNDARY VALUE PROBLEM INTO A

FREDHOLM INTEGRAL EQUATION.We explain the method with help of the following solved examples.Ex. 1. (a) Reduce the following boundary value problem into an integral equation :

2 2/ 0d y dx y with y (0) = 0, y (l) = 0 [Kanpur 2000]

(b) Also recover the B.V.P. from the integral equation. [Meerut 2000]

Sol. Given ( ) ( ) 0y x y x ... (1)with boundary conditions y (0) = 0 ... 2(a)and y(l) = 0. ... 2(b)

From (1), ( ) ( )y x y x ... (3)Integrating both sides of (3) w.r.t. ‘x’ from 0 to x, we have

0 0( ) ( )

x xy x dx y x dx or 0 0

( ) ( )xxy x y x dx

or0

( ) (0) ( ) .x

y x y y x dx ... (4)

Let (0) ,y C a constant ... (5)

Using (5), (4) gives0

( ) ( )x

y x C y x dx ... (6)



Integrating both sides of (6) w.r.t. ‘x’, from 0 to x, we get

2

0 0 0( ) ( )

x x xy x dx C dx y x dx or 2

0 0( ) ( )

xxy x Cx y t dt or

0( ) (0) ( ) ( ) ,

xy x y Cx x t y t dt using result of Art. 1.14

or 0

( ) 0 ( ) ( ) ,x

y x Cx x t y t dt by 2 (a) or 0

( ) ( ) ( ) .x

y x Cx x t y t dt ...(7)

Putting x = l in (7), we get

0( ) ( ) ( )

ly l Cl l t y t dt or

00 ( ) ( ) ,

lCl l t y t dt using 2 (b)

or0

( ) ( )l

C l t y t dtl

... (8)

Using (8), (7) reduces to

0 0( ) ( ) ( ) ( ) ( )

l xy x x l t y t dt x t y t dt

l

... (9)

or0 0

( )( ) ( ) ( ) ( )l xx l ty x y t dt x t y t dt

l

or0 0

( ) ( )( ) ( ) ( ) ( ) ( )x l x

x

x l t x l ty x y t dt y t dt x t y t dtl l

or0

( ) ( )( ) ( ) ( ) ( )x l

x

x l t x l ty x x t y t dt y t dtl l

or0

( ) ( ) ( )( ) ( ) ( )x l

x

x l t l x t x l ty x y t dt y t dtl l

or0

( ) ( )( ) ( ) ( )x l

x

t l x x l ty x y t dt y t dtl l

or0

( ) ( , ) ( ) ,l

y x K x t y t dt ... (10)

where( / ) ( ), if 0

( , )( / ) ( ), if l

t l l x t xK x t

x l l t x t

... (11)

(10) is the required Fredholm integral equation, where K (x, t) is given by (11)(b) Refer part (c) of the next Ex. 2.

Ex. 2. (a) If ( ) ( ) 0,y x y x and y satisfies the end condition y (0) = 0, y (l) = 0, show that

0 0( ) ( ) ( ) ( ) ( ) .

l xxy x l t y t dt x t y t dtl

(b) Show that the result of part (a) can be written as0

( ) ( , ) ( ) ,l

y x K x t y t dt

where( / ) ( ), when

( , )( / ) ( ) when .t l l x t x

K x tx l l t t x



(c)Verify directly that the expression obtained satisfies the prescribed differential equationand end conditions.

Sol. (a) Given ( ) ( ) 0y x y x ... (1)with the boundary conditions y(0) = 0 ... 2 (a)and y(l) = 0 ... 2 (b)

Now proceed as in Ex. 1 upto equation (9), i.e.,

0 0( ) ( ) ( ) ( ) ( ) .

l xxy x l t y t dt x t y t dtl

... (i)

Part (b) Proceed as in Ex. 1. Equations (10) and (11) give the required results.Part (c) We shall now proceed with integral equation (9) of solution of Ex. 1 and obtain the

given differential equation (1) together with given boundary conditions 2 (a) and 2 (b) as follows :Re-writing (9), we have

0 0


l

... (ii)

Putting x = 0 and x = l by turn in (ii), we gety (0) = 0 and y (l) = 0 (iii)

Differentiating both sides of (ii) w.r.t. ‘x’, we get

0 0

( )( ) ( ) ( ) ( ) ,l xd x l t dy x y t dt x t y t dt

dx l dx ... (iv)

Using Leibnitz’s rule (refer results (1) and (2) of Art 1.13), (iv) reduces to

0 0


x l x

0( ) ( ) ( 0) (0)dx dx x y x x ydx dx

or0 0

( )( ) ( ) ( ) .l xl ty x y t dt y t dt

l ... (v)

Differentiating both sides of (v) w.r.t. ‘x’, we get

0 0

( )( ) ( ) ( ) .l xd l t dy x y x dt y t dt

dx l dx ... (vi)

Using Leibrintz’s rule again, (vi) reduces to

0 0

( ) 0( ) ( ) { ( )} ( ) (0)l xl t dx dy x y t dt y t dt y x y

x l x dx dx

[Using results (1) and (2) of Art. 1.3]

or ( ) 0 0 ( ) 0y x y x or ( ) ( ) 0y x y x ... (vii)Equation (vii) together with boundary conditions (iii) show that expression (i) satisfies the

prescribed differential equation and end conditions.Ex. 3. If y(x) is continuous and satisfies

1

0( ) ( , ) ( ) ,y x K x t y t dt where

(1 ) , 0( , )

(1 ) , 1.t x x t

K x tx t t x

then prove that y(x) is also the solution of the boundary value problemd2y/dx2 + y = 0, y (0) = 0, y (1) = 0.

Sol. Given1

0( ) ( , ) ( ) ,y x K x t y t dt ... (1)



where(1 ) , 0

( , )(1 ) , 1.

t x x tK x t

x t t x

... (2)

Re-writing (1), we have1

0( ) ( , ) ( ) ( , ) ( )

x

xy x K x t y t dt K x t y t dt

or1

0( ) (1 ) ( ) (1 ) ( ) , by (2)

x

xy x t x y t dt x t y t dt ... (3)

Putting x = 0 and x = 1 by turn in (3), we gety(0) = 0 and y(1) = 0. ...(4)

Differentiating both sides of (3), w.r.t. ‘x’, we get1

0(1 ) ( ) (1 ) ( )

x

x

dy d dt x y t dt x t y t dtdx dx dx

0

0{ (1 ) ( )} (1 ) ( ) 0xdy dx dt x y t dt x x y x

dx x dx dx

1{ (1 – ) ( )}

xx t y t dt

x

(1)0 – (1– ) ( ) ,d dxx x y xdx dx

using Leibnitz’s rule

or1

0( ) (1 ) ( )

x

x

dy t y t dt t y t dtdx

... (5)

Differentiating both sides of (5) w.r.t. ‘x’, we get2

2 0

0{ ( )} ( ) 0xd y dx dt y t dt x y x

x dx dxdx

1 (1){ (1 ) ( )} 0 (1 ) ( )x

d dxt y t dt x y xx dx dx

[using Leibntiz’s rule again]

or2

2d ydx

= – x y (x) – (1 – x) y (x) or 2

2 0d y ydx

...(6)

(4) and (6) show that if y(x) satisfies (1), then y(x) is also the solution of the boundaryvalue problem

d2y/dx2 + y = 0, y(0) = y(1) = 0Ex. 4. Transform d2y/dx2 + xy = 1, y(0) = y(1) = 0 into an integral equation.Also recover the boundary value problem you obtain. [Meerut 2001, 10, 12]Sol. Given ( ) ( ) 1y x x y x ... (1)

with boundary conditions y(0) = 0 ... 2(a)and y(1) = 1. ... 2(b)

From (1), ( ) 1 ( ).y x x y x ... (3)Integrating both sides of (3) w.r.t. ‘x’, from 0 to x, we get

0 0 0( ) ( )

x x xy x dx dx x y x dx or 0 0

( ) ( )xxy x x x y x dx

or0

( ) (0) ( ) .x

y x y x x y x dx ... (4)

Let (0) .y c ... (5)



Using (5), (4) gives0

( ) ( ) .x

y x c x x y x dx ... (6)

Integrating both sides of (6) w.r.t. ‘x’, from 0 to x, we get

2

0 0 0( ) ( ) ( )

x x xy x dx c x dx x y x dx or 2 2

0 00

1( ) ( )2

x xxy x cx x t y t dt

or 2

0

1( ) (0) ( ) ( ) ,2

xy x y cx x x t t y t dt using result of Art. 1.14.

or 2

0

1( ) ( ) ( ) ,2

xy x cx x x t t y t dt by 2 (a) ... (7)

Putting x = 1 in (7), we have1

0

1(1) (1 ) ( )2

y c t t y t dt or 1

0

11 (1 ) ( ) ,2

c t t y t dt by 2 (b)

or1

0

1 (1 ) ( ) .2

c t t y t dt ...(8)

Using (8), (7) reduces to1 2

0 0

1 1( ) (1 ) ( ) ( ) ( )2 2

xy x x t t y t dt x x t t y t dt

or1

0 0

1( ) (1 ) (1 ) ( ) ( ) ( )2

xy x x x xt t y t dt t x t y t dt

or 1

0 0

1( ) (1 ) (1 ) ( ) (1 ) ( ) ( ) ( )2

x x

xy x x x xt t y t dt xt t y t dt t x t y t dt

or1

0

1( ) (1 ) ( ){ } (1 ) ( )2

x

xy x x x t y t x xt x t dt xt t y t dt

or12

0

1( ) (1 ) (1 ) ( ) (1 ) ( )2

x

xy x x x t x y t dt xt t y t dt

or1

0

1( ) (1 ) ( , ) ( ) ,2

y x x x K x t y t dt ... (9)

where2 (1 ), when( , )

(1 ), whent x t xK x txt t t x

... (10)

To recover the B.V.P. from the integral equation given by (9) and (10) :Re-writing (9), we have

12

0

1( ) ( ) ( , ) ( ) ( , ) ( )2

x

xy x x x K x t y t dt K x t y t dt

or12 2

0

1( ) ( ) (1 ) ( ) (1 ) ( ) ,2

x

xy x x x t x y t dt xt t y t dt ... (11)

Differentiating both sides of (11) with respect to x and using Leibnitz’s rule of differentiationunder the integral sign (see Art. 1.13), we obtain



2 2

0

1 0( ) (1 2 ) ( ) ( ) (1 ) ( ) (0) (0)2

x dx dy x x t y t dt x x y x ydx dx

1 2 2(1)( ) ( ) (1 1) (1) (1 ) ( )x

d dxt t y t dt x y x x y xdx dx

or

12 2

0

1( ) (1 2 ) ( ) ( ) ( )2

x

xy x x t y t dt t t y t dx ... (12)

Differentiating both sides of (12) w.r.t. ‘x’ and using Leibnitz rule as before, we have2

2

0

{ ( )} 0( ) 1 ( ) (0) (0)x t y t dx dy x x y x y

x dx dx

21 2{( ) ( )} 1(1 1) (1) ( ) ( )x

t t y t d dxy x x y xx dx dx

or 2 2( ) 1 ( ) ( ) ( )y x x y x x x y x or 1y x y ... (13)

From (11),0 12

0 0(0) 0 (1 0) ( ) (0) ( ) 0y t y t dt y t dt ... (14)

and1 12

0 1(1) 1 (1 1) ( ) (1 ) ( ) 1y t y t dt t t y t dt ... (15)

Thus, we have recovered the given boundary value problem with help of (13), (14) and (15).Ex. 5. Obtain Fredholm integral equation of second kind corresponding to the boundary

value problem 2 2/ , (0) 0, (1) 1.d dx x Also recover the boundary value problemfrom the integral equation obtained [Kanpur 2011; Meerut 2005]

Sol. Given ( ) ( )x x x ... (1)

with boundary conditions : (0) 0 ... (2)

and (1) 1. ... (3)Integrating both sides of (1) w.r.t. ‘x’ from 0 to x, we get

0 0 0( ) ( )

x x xx dx xdx x dx or

2

0 00

( ) ( )2

xxx xx x dx

or2

0( ) (0) ( )

2

xxx x dx ... (4)

Assume that (0) ,C C being a constant ... (5)

Using (5), (4) yields 2

0( ) ( ) .

2

xxx C x dx ... (6)

Integrating both sides of (6) w.r.t. ‘x’, we obtain2

2

0 0 0 0 0( ) ( )

2

x x x x xxx dx C dx dx x dx or 3 2

0 0 0 0 0( ) [ ] [ /6] ( )

x xx x xx C x x t dt



or 3

0( ) (0) ( / 6) ( ) ( ) ,

xx C x x x t t dt using result of Art. 1.14

or 3

0( ) ( / 6) ( ) ( ) ,

xx C x x x t t dt by (2) ... (7)

Putting x = 1 in (7) and using (3), we obtain1

0

11 (1 ) ( )6

C t t dt or1

0

5 (1 ) ( )6

C t t dt Substituting the above value of C in (7), we obtain

31

0 0

5( ) (1 ) ( ) ( ) ( )6 6

xxx x t t dt x t t dt

or3 1

0 0

5( ) (1 ) ( ) ( ) ( ),6 6

xx xx x t t dt x t t dt

or 3( ) (5 ) / 6x x x 1

0 0(1 ) ( ) (1 ) ( ) ( ) ( )

x x

xx t t dt x t t x t t dt

or 3( ) (5 ) / 6x x x 1

0( ) ( ) ( ) (1 ) ( )

x

xx xt x t t dt x t t dt

or13

0( ) (5 ) / 6 (1 ) ( ) (1 ) ( )

x

xx x x t x t dt x t t dt

or 13

0( ) (5 ) / 6 ( , ) ( ) ,x x x K x t t dt ... (8)

where (1 ), when 0( , )

(1 ), when 1t x t x

K x tx t x t

... (9)

The required integral equation is given by (8) and (9).To recover the B.V.P from the integral equation given by (8) and (9) :Re-writing (8), we have

13

0( ) (5 ) / 6 ( , ) ( ) ( , ) ( )

x

xx x x K x t t dt K x t t dt

or 13

0( ) (5 ) / 6 (1 ) ( ) (1 ) ( ) , using (9)

x

xx x x t x t dt x t t dt ... (11)

Differentiating both sides of (11) w.r.t. ‘x’ and using Leibnitz’s rule, we obtain

2

0

{ (1 ) ( )}( ) (5 3 ) / 6 (1 ) ( ) 0x t x t dxx x dt x x x

x dx

1 { (1 ) ( )} 0 (1 ) ( )x

x t t dxdt x x xx dx

or12

0

1( ) (5 3 ) ( ) ( ) (1 ) ( )6

x

xx x t t dt t t dt ... (12)



Differentiating both sides of (12) w.r.t. ‘x’ and using Leibnitz’s rule, we obtain

0

{( ) ( )}( ) ( ) ( ) 0x t t dxx x dt x x

x dx

1 { (1 ) ( )} 0 (1 ) ( )x

t t dxdt x xx dx

or ( ) ( ) ( ) ( )x x x x x x x or x ... (13)

From (11), (0) 0 and (1) 1 ... (14)Thus, we have recovered the given boundary value problem with help of (13) and (14).

Ex. 6. The integral equation 1

0 0( ) ( ) ( ) (1 ) ( )

xy x x t y t dt x t y t dt is equivalent to :

(a) 0, (0) 0, (1) 0y y y y (b) 0, (0) 0, (0) 0y y y y

(c) 0, (0) 0, (1) 0y y y y (d) 0, (0) 0, (0) 0y y y y [GATE 2003]

Sol. Ans. (a) Given : 1

0 0( ) ( ) ( ) (1 ) ( )

xy x x t y t dt x t y t dt ... (1)

Differentiating both sides of (1) w.r.t. ‘x’ and using Leibnitz’s rule, we have

0

{( ) ( )} 0( ) ( ) ( ) ( 0) (0)x x t y t dx dy x dt x x y x x y

x dx dx

1

0

{ (1 ) ( )}x t y t dtx

or1

0 0( ) ( ) (1 ) ( )

xy x y t dt t y t dt ... (2)

Differentiating both sides of (2) w.r.t. ‘x’ and using Leibnitz’s rule, we obtain1

0 0

( ) 0 {(1 ) ( )}( ) ( ) (0)x y t dx d t y ty x dt y x y dt

dx dx dx x

or ( ) 0 ( ) 0 0y x y x or 0y y ... (3)

From (1), 0 1

0 0(0) ( ) ( ) 0 (1 ) ( ) 0y t y t dt t y t dt ... (4)

and 1 1

0 0(1) (1 ) ( ) (1 ) ( ) 0y t y t dt t y t dt ... (5)

Thus, the given integral equation is equivalent to the boundary value problem given by (3),(4) and (5). Hence alternative (a) is true.

EXERCISE-2C1. Convert the boundary value problem 0, (0) 1, (1) 0,y y y y into an integral equation.

Ans. 1

0( ) 1 ( , ) ( )y x K x t y t dt , where

,( , ) .

,t t x

K x tx t x

2. If y(x) has continuous first and second derivatives and satisfies the boundary value problemd2y/dx2 + y = 0, y(0) = 0, y(1) = 0, then show that y(x) is continuous and satisfies the homogeneouslinear integral equation,



1

0( ) ( , ) ( ) ,y x K x t y t dt where

(1 ) , for 0( , ) .

(1 ) , for 1t x x t

K x tx t t x

(Kanpur 2007)

3. (a) If ( ) ( )y x F x , and y satisfies the end conditions y(0) = 0 and y(1) = 0, show that

1

0 0( ) ( ) ( ) (1 ) ( )

xy x x t F t dt x t F t dt .

(b) Show that the result of part (a) can be written in the form

1

0( ) ( , ) ( ) ,y x K x t F t dt where

( ), when( , )

( 1), whent x t t x

K x tx t t x

.

(c) Verify directly that the expression obtained satisfies the prescribed differential equationand end conditions.

4. Show that the boundary value problem 0, (0) 0, (1) 1,y Ay By y y where A and B are constants, leads to the integral equation

1

0( ) ( , ) ( ) ,y x K x t y t dt where

(1 ) , when( , )

(1 ) , when .Bt x Ax A t x

K x tBx t Ax t x

5. Transform the boundary value problem d2y/dx2 + y = x, y(0) = 0, (1) 0y to a Fredholmintegral equation. [Meerut 2000, 2008]

Ans. 13

0

1( ) ( 3 ) ( , ) ( ) ,6

y x x x K x t y t dt where , 1

( , ), 1x x

K x tt x

6. Show that the boundary value problem d2y/dx2 + y = x, y(0) = y ( ) = 0 can be connectedinto an integral equation

3 2

0

1( ) (1 ) ( , ) ( ) ,6

y x x K x t y t dt

where ( / ) ( ), when

( , )( / 1) ( ), whenx x t t x

K x tx x t t x

7. Reduce the following boundary value problem into an integral equation.

0, (0) 0, (1) (1) 0.y y y y v y

Ans. 1

0( ) ( , ) ( ) ,

1xy x K x t y t dt

v

where

1 (1 ) ,1( , )

[1 (1 )] ,1

v x t t xvK x t

x v t t xv

8. Reduce the boundary value problem 0, (0) ( / 2) 0y y y y to an integral equation.

Ans. / 2

0( ) ( , ) ( ) ,y x K x t y t dt

where

[1 (2 / )] , 0( , )

[1 (2 / )] , / 2.t x x t

K x tx t t x

9. Find the Fredholm integral equation of second kind corresponding to the boundary-valueproblem (d2y/dx2) + y = 0, y(0) = 0, y(l) = 0.

Also recover the boundary-value problem from the integral equation you obtain.


CHAPTER 3

Homogeneous FredholmIntegral Equations of theSecond Kind with Separableor Degenerate Kernels3.1 CHARACTERISTIC VALUES (OR CHARACTERISTIC NUMBERS OR

EIGENVALUES). CHARACTERISTIC FUNCTIONS (OR EIGENFUNCTIONS).Consider a homogeneous Fredholm integral equation of the second kind :

( ) ( , ) ( ) .b


Then (1) has always the obvious solution y (x) = 0, which is known as zero or trivialsolution of (1). The values of the parameter for which (1) has non-zero (or non-trivial) solution

( ) 0/y x are known as the eigenvalues of (1) or of the kernel K (x, t). Further, if ( )x is continuousand ( ) 0/x on the interval (a, b) and

0( ) ( , ) ( ) ,b

ax K x t t dt ... (2)

then ( )x is known as an eigenfunction of (1) corresponding to the eigenvalue 0 .

Remark 1. The number 0 is not an eigenvalue since for 0, (1) yields y (x) = 0,which is a zero solution.

Remark 2. If the kernel K (x, t) is continuous in the rectangle : , ,R a x b a t b and thenumbers a and b are finite, then to every eigenvalue there exist a finite number of linearlyindependent eigenfunctions; the number of such functions is known as the index of the eigenvalue.Different eigenvalues have different indices.

Remark 3. If ( )x is an eigenfunction of (1) corresponding to eigenvalue 0 then ( )C x

is also eigenfunction of (1) corresponding to the same eigenvalue. Here C is an arbitrary constant.Remark 4. A homogeneous Fredholm integral equation may, generally, have no eigenvalues

and eigenfunctions or it may not have any real eigenvalue and eigenfunction.3.2 SOLUTION OF HOMOGENEOUS FREDHOLM INTEGRAL EQUATION OF THE

SECOND KIND WITH SEPARABLE (OR DEGENERATE) KERNEL. [Meerut 2000]Consider a homogeneous Fredholm integral equation of the second kind :

( ) ( , ) ( ) .b


3.1


3.2 Homogeneous Fredholm Integral Equations of the Second Kind with Reparative

Since kernel K (x, t) is separable, we take

1( , ) ( ) ( ).

n

i iiK x t f x g t

... (2)


1( ) ( ) ( ) ( )

nb

i iiay x f x g t y t dt

or 1

( ) ( ) ( ) ( ) .n b

i ii ay x f x g t y t dt

... (3)

[Interchanging the order of summation and integration]

Let ( ) ( ) , where 1, 2,..., .b

i iag t y t dt C i n ... (4)

Using (4), (3) reduces to1

( ) ( ),n

i iiy x C f x

... (5)

where constants Ci (i = 1, 2, ..., n) are to be determined in order to find solution of (1) in the formgiven by (5).

We now proceed to evaluate Ci’s as follows :Multiplying both sides of (5) successively by g1(x), g2 (x), ..., gn (x) and integrating over the

interval (a, b), we have

1 11( ) ( ) ( ) ( ) ,

nb b

i iia ag x y x dx C g x f x dx

... (A1)

2 21( ) ( ) ( ) ( ) ,

nb b

i iia ag x y x dx C g x f x dx

... (A2)

... ... ... ... ... ...

and1

( ) ( ) ( ) ( ) .nb b

n i n iia ag x y x dx C g x f x dx

... (An)

Let ( ) ( ) , where , 1, 2,..., .b

j i j iag x f x dx i j n ...(6)

Using (4) and (6), (A1) reduces to

1 11

n

i iiC C

or

1 1 11 2 12 1[ ... ]n nC C C C

or 11 1 12 2 1(1 ) ... 0.n nC C C

Similarly, we may simplify (A2), ..., (An). Thus, we obtain the following system of homogeneouslinear equations to determine C1, C2, ... Cn.

11 1 12 2 1(1 ) ... 0n nC C C ... (B1)

21 1 22 2 2(1 ) ... 0n nC C C ... (B2)

... ... .... ... .... ...

... ... .... ... .... ...

1 1 2 2... (1 ) 0.n n n n nC C C ... (Bn)


Homogeneous Fredholm Integral Equations of the Second Kind with separable 3.3

The determinant ( )D of this system is

11 12 1

21 22 2

1 2

1 ...

1 ...( ) .

: : ... :... 1

n

n

n n nn

D

... (7)

If ( ) 0,D the system of equations (B1), (B2), ..., (Bn) has only trivial solution C1 = C2 = ...= Cn = 0 and hence from (5) we notice that (1) has only zero or trivial solution y (x) = 0. However,if ( ) 0,D at least one of the Ci’s can be assigned arbitrarily, and the remaining Ci’s can bedetermined accordingly. Hence when ( ) 0,D infinitely many solutions of the integral equation(1) exist.

Those values of for which ( ) 0D are called the eigenvalues, and any non-trivial solutionof (1) is called a corresponding eigenfunction of (1).

The eigenvalues of (1) are given by ( ) 0,D i.e.,

11 12 1

21 22 2

1 2

1 ...

1 ...0.

... ... ... ...... 1

n

n

n n nn

... (8)

So the degree of equation (8) in is .m n It follows that if integral equation (1) hasseparable kernel given by (2), then (1) has at the most n eigenvalues.3.3. SOLVED EXAMPLES BASED ON ART. 3.1 AND 3.2

Ex. 1. Solve the homogeneous Fredholm equation1

0( ) ( )x ty x e e y t dt [Kanpur 2005, 2007, 10, 11]

ORFind the eigenvalues and eigenfunctions of the homogeneous integral equation.

1

0( ) ( )x ty x e e y t dt

Sol. Given 1

0( ) ( )x ty x e e y t dt or

1

0( ) ( ) .x ty x e e y t dt ... (1)

Let1

0( ) .tc e y t dt ... (2)

Then (1) reduces to ( ) .xy x c e ... (3)

From (3), ( ) .ty t c e ... (4)

Using (4), (2) becomes1

0( )t tc e c e dt



or12

2

0

( 1).2 2

te cc c e

or 21 ( 1) 0.2

c e ... (5)

If c = 0 then (4) gives y (x) = 0. We, therefore, assume that for non-zero solution of (1), 0.c Then (5) gives

21 ( / 2) ( 1) 0e or 22 /( 1),e ... (6)which is an eigenvalue of (1).

Putting the value of given by (6) in (3), the corresponding eigenfunction is given by

2( ) 2 /( 1) xy x c e e

Hence, corresponding to eigenvalue 2/ (e2 – 1) there corresponds the eigenfunction ex.Remark. While writing eigenfunction the constant 2c/ (e2 – 1) is taken as unity.

Ex. 2. Show that the homogeneous integral equation 1

0( ) (3 2) ( ) 0y x x t y t dt has

no characteristic numbers and eigenfunctions. (Kanpur 2010, Meerut 2011)

Sol. Given1

0( ) (3 2) ( )y x x t y t dt or

1

0( ) (3 2) ( ) .y x x t y t dt ... (1)

Let 1

0( ) .C t y t dt ... (2)

Then (1) reduces to ( ) (3 2).y x C x ... (3)

From (3), ( ) (3 2).y t C t ... (4)Using (4), (2) becomes

1

0(3 2)C C t t dt or

13 20

C C t t or C = 0.

From (3) ( ) 0,y x which is zero solution of (1). Hence for any , (1) has only zero

solution ( ) 0.y x Therefore, (1) does not possess any characteristic number or eigenfunction.Remark. Note that the kernel K (x, t) = (3x – 2) t of the above example is not symmetric.

Thus we have shown that a kernel which is not symmetric does not necessarily have a characteristicconstant. On the other hand, it will be shown in Art. 7.2 of Chapter 7 that a Fredholm homogeneousintegral equation with symmetric kernel possesses at least one characteristic constant.

Ex. 3. Find the eigenvalues and the corresponding eigenfunctions of the homogeneous

integral equation 1

0( ) sin cos ( ) .y x x t y t dt (Kanpur 2009)

Sol. Given 1

0( ) sin cos ( )y x x t y t dt or

1

0( ) sin cos ( ) .y x x t y t dt ... (1)

Let1

0cos ( ) .C t y t dt ... (2)

Then (1) reduces to ( ) sin .y x C x ... (3)

From (3), ( ) sin .y t C t ... (4)




0cos ( sin )C t c t dt or

1

0sin 2

2CC t dt

or1

0

cos 2 1 1 .2 2 2 2 2C t CC

Hence C = 0 and so from (3), ( ) 0.y x Thus for any , (1) has only zero solution ( ) 0.y x Therefore, (1) does not possess any characteristic number or eigenfunction.

Ex. 4. (a) The values of for which the integral equation 1

0( ) (6 – ) ( )y x x t y t dt has a

non-trivial solution, are given by the roots of the equations :

(a) 2(3 –1) (2 ) – 0 (b) (3 –1) (2 ) 2 0

(c) 2(3 –1) (2 ) – 4 0 (d) 3(3 –1) (2 ) 0 [GATE 2004]

Sol. Ans (c). Given 1 1

0 0( ) 6 ( ) – ( )y x x y t dt t y t dt ...(1)

Let1

10

( )C y t dt ...(2)

and1

20

( )C t y t dt ...(3)

Then (1) yields 1 2( ) 6 –y x x C C ...(4)

From (4), 1 2( ) 6 –y t t C C ..(5)

Using, (5), (2) becomes1

1 1 20

(6 – )C t C C dt or 2 1

1 1 2 0[3 – ]C C t C t or 1 1 23 –C C C

Thus, 1 2(3 –1) – 0C C ...(6)


2 1 20

(6 – )C t t C C dt or 3 2 1

2 1 2 0[2 – (1/ 2) ]C C t C t or 2 1 22 – (1/ 2)C C C

Thus, 1 2– 4 (2 ) 0C C ...(7)Now, we have a system of homogeneous linear equations (6) and (7) to determine C1 and C2.

For non-trivial solution of the given integral equation, the system (6) – (7) must possess non-trivialsolution and so we must have

3 –1 –0

– 4 2

or 2(3 –1)(2 ) – 4 0,

which gives the desired values of for the required non-trivial solution of the given integralequation.



Ex. 4. (b) Find the eigenvalues and the corresponding eigenfunctions of the integral equation

1 2

0( ) (2 4 ) ( ) .y x xt x y t dt [MEERUT 2007]

Sol. Given1 2

0( ) (2 4 ) ( )y x xt x y t dt

or1 12

0 0( ) 2 ( ) 4 ( ) .y x x t y t dt x y t dt ... (1)

Let1

1 0( )C t y t dt ... (2)

and1

2 0( ) .C y t dt ... (3)

Then (1) reduces to 21 2

( ) 2 4 .y x C x C x ...(4)

From (4), 21 2

( ) 2 4 .y t C t C t ...(5)Using (5), (2) becomes

1 21 1 20

(2 4 )C t C t C t dt or1 12 3

1 20 0[1 2 ] 4 0C t dt C t dt

or 1 2(1 2 / 3) 0.C C ... (6)Again, using (5), (3) becomes

1 2

2 1 20(2 4 )C C t C t dt or

1 1 21 20 0

2 [1 4 ] 0C dt C t dt or 1 2 (1 4 /3) 0.C C ... (7)

Thus, we have a system of homogeneous linear equations (6) and (7) for determining C1 andC2. For non-zero solution of this system of equations, we must have

(1 2 / 3)0

(1 4 / 3)

or 22 41 1 0

3 3

or 2 6 9 0 so that 3, 3

Hence the eigenvalues are 1 23, 3.

To determine eigenfunction corresponding to 1 3

Putting 1 3 in (6) and (7), we get

3C1 – 3C2 = 0 ... (8)

and 1 23 3 0C C ... (9)(8) or (9) give C1 = C2. Hence from (4), we have

2 21 1 1

( ) 2 ( 2 ) 6 ( 2 ).y x C x x C x x

Taking – 6C1 = 1, the eigenfunction is (x – 2x2).Hence eigenfunction corresponding to eigenvalue 1 2 3 is x – 2x2.



Ex. 5. (a) Solve the homogeneous Fredholm integral equation of the second kind :

2

0( ) sin( ) ( )y x x t y t dt

Sol. Given2

0( ) sin( ) ( )y x x t y t dt

or 2

0( ) (sin cos cos sin ) ( )y x x t x t y t dt

or 2 2

0 0( ) sin cos ( ) cos sin ( ) .y x x t y t dt x t y t dt

... (1)

Let 2

1 0cos ( )C t y t dt

... (2)

and 2

2 0sin ( ) .C t y t dt

... (3)

Then (1) reduces to 1 2( ) sin cos .y x C x C x ... (4)

From (4), 1 2( ) sin cos .y t C t C t ... (5)


1 1 20cos ( sin cos )C t C t C t dt

or 2 2

1 21 0 0

sin 2 (1 cos 2 )2 2C C

C t dt t dt

or 2 2

1 21

0 0

cos 2 sin 22 2 2 2

C Ct tC t

or 1 20C C or 1 2 0.C C ... (6)


2 1 20sin ( sin cos )C t C t C t dt

or 2 2

1 22 0 0

(1 cos 2 ) sin 22 2C C

C t dt t dt

or 2 2

1 22

0 0

sin 2 cos 22 2 2 2C Ct tC t

or 2 1C C or 1 2 0.C C ... (7)Thus, we have a system of homogeneous linear equations (6) and (7) for determining C1 and

C2. For non-zero solution of this system of equations, we must have

10

1

or 2 21 0 so that 1 .

Hence the eigenvalues are given by 1 1/ and 2 1/ . ... (8)

To determine eigenfunction corresponding to 1 1/ .

Putting 1 1/ in (6) and (7), we get

C1 – C2 = 0 ... (9)and C1 – C2 = 0 ... (10)

Both (9) and (10) give C2 = C1. Hence from (4), we have

1 1( ) (1/ ) sin (1/ ) cosy x C x C x or 1( ) ( / ) (sin cos ).y x C x x



Taking 1( / ) 1,C the required eigenfunction y1 (x) is given by

1( ) sin cos .y x x x ... (11)

To determine eigenfunction corresponding to 2 1/ .

Putting 2 1/ in (6) and (7), we getC1 + C2 = 0 ... (12)

and C1 + C2 = 0. ... (13)Both (12) and (13) give C2 = – C1. Hence from (4), we have

1 1( ) ( 1/ ) sin ( 1/ ) ( ) cosy x C x C x or ( ) ( / ) (sin cos ).y x C x x

Taking 1( / ) 1,C the required eigenfunction y2 (x) is given by

2 ( ) sin cosy x x x ... (14)From (8), (11) and (14), the required eigenvalues and eigenfunctions are given by

1 11/ , ( ) sin cosy x x x and 2 21/ , ( ) sin cos .y x x x

Ex.5. (b) The eigenvalues of the integral equation 2

0( ) sin( ) ( )y x x t y t dt

aree

(a) 1/ 2 , –1/ 2 (b) 1/ , –1/ (c) , – (d) 2 , – 2 [GATE 2005]Sol. Ans. (b). Refer Ex. 5(a)

Ex.5. (c) The integral equation 2

0( ) sin ( ) ( )y x x t y t dt

has

(a) two solutions for any value of (b) unique solution for every value of (c) infinitely many solutions for only one values of (d) infinitely many solutions for two values of [GATE 2003]Sol. (d). Refer Ex. 5(a).Ex. 6. Find the eigenvalues and eigenfunctions of the homogeneous integral equation

2 3

0( ) (cos cos 2 cos3 cos ) ( ) .y x x t x t y t dt

[Meerut 2000, 01, 03, 08, 09, 10; Kanpur 2005, 06]

Sol. Given 2 3

0( ) (cos cos 2 cos3 cos ) ( )y x x t x t t dt

or 2 3

0 0( ) cos cos 2 ( ) cos3 cos ( )y x x t y t dt x t y t dt

... (1)

Let 1 0cos 2 ( )C t y t dt

... (2)

and 32 0

cos ( ) .C t y t dt

... (3)


( ) cos cos3 .y x C x C x ... (4)



From (4), 21 2

( ) cos cos3 .y t C t C t ... (5)

Using (5), (2) becomes 21 1 20

cos 2 ( cos cos3 ) .C t C t C t dt

or 2

1 20 0[1 cos 2 cos ] cos 2 cos3 0.C t t dt C t t dt

... (6)

Now, 2

0cos 2 cost t dt

2

0 0 0

(1 cos 2 ) 1 1cos 2 cos 2 cos 22 2 2

tt dt t dt t dt

00 0

1 sin 2 1 1 cos 4 1 sin 40 .2 2 2 2 4 4 4

t t tdt t ... (7)

Again, 0 0 0

1 1 sin 5cos 2 cos3 [cos5 cos ] sin 0.2 2 5

tt t dt t t dt t ... (8)

Using (7) and (8), (6) reduces to 1 2(1 / 4) 0. 0C C ... (9)

Again, using (5), (3) becomes 3 22 1 20

cos ( cos cos3 )C t C t C t dt

or 5 3

1 20 0cos [ cos cos3 1] 0.C t dt C t t dt

... (10)

Now, 5

0cos 0,t dt

as 5 5cos ( ) cost t ... (11)

and 3

0cos cos3t t dt

0

1 cos3 (cos3 3cos )4

t t t dt

3 3[ cos3 4cos 3cos cos (1/ 4) (cos3 3cos )]t t t t t t

2

0 0

1 3cos 3 cos3 cos4 4

t dt t t dt

0 0

1 1 cos6 3 cos 4 cos4 2 4 2

t t tdt dt

0 0

1 sin 6 3 sin 4 sin .8 6 8 4 8

t tt t

3

0cos cos3 .

8t t dt

... (12)

Using (11) and (12), (10) reduces to

1 20. ( / 8 1) 0C C or1 20. (1 /8) 0C C ... (13)

For non-zero solution of the system of equations (9) and (13), we must have

1 ( / 4) 00

0 1 ( /8)

or 1 1 0

4 8

4 / or 8 / .

Hence the eigenvalues of (1) are 1 4 / and 2 8/ . ... (14)

Determination of eigenfunction corresponding to the eigenvalue 1 4/ .



Putting 1 4 / . in (9) and (13), we have

0. C1 + 0. C2 = 0 ...(15)and 0. C1 + (1/2) × C2 = 0 ... (16)

Solving (15) and (16), C2 = 0 and C1 is arbitrary. Putting these values in (4), we get2 2

1 1( ) cos (4 / ) cos .y x C x C x

Setting 1(4 / ) 1,C the eigenfunction = 2

1( ) cos .y x x

Determination of eigenfunction corresponding to the eigenvalue 2 8/ .

Putting 2 8/ in (9) and (13), we have

– C1 + 0. C2 = 0 ... (17)and 0. C1 + 0. C2 = 0 ... (18)

Solving (17) and (18), C1 = 0 and C2 is arbitrary. Putting these values in (4), we get

2 2( ) cos3 (8 / ) cos3y x C x C x

Setting 2(8 / ) 1,C the eigenfunction 2 ( ) cos3 .y x x

Hence 21( ) cosy x x and 2 ( ) cos3y x x are the required eigenfunctions corresponding to

the eigenvalues 1 4 / and 2 8/ respectively..

Ex. 7. Find the eigenvalues and eigenfunctions of the homogeneous integral equation2

1

1( ) ( ) .y x xt y t dtxt

[Merrut 2006]

Sol. Given2

1

1( ) ( )y x xt y t dtxt

or2 2

1 1

1( ) ( ) ( ) .y x x t g t dt g t dtx t

... (1)

Let2

1 1( )C t y t dt ... (2)

and2

2 1

1 ( ) .C y t dtt

... (3)

Then (1) reduces to 1 2( ) ( / )y x C x C x ... (4)

From (4), 1 2( ) ( / )y t C t C t ... (5)Using (5), (2) becomes

22

1 11

CC t C t dt

t

23

211 2 1 2

1

8 1 (2 1)3 3 3tC C t C C

or1 2(1 7 / 3) 0.C C ... (6)

Using (5), (3) becomes

212 22

12 1 1 2 1 211

1 1(2 1) 11 2

C tC C t dt C t C C Ct t



or 1 2(1 / 2) 0.C C ... (7)For non-zero solution of the system of equations (6) and (7), we must have

1 (7 / 3)0

1 ( / 2)

or 27 11 1 03 2

or 217 11 06 6

or 2 17 6 0 so that 2 1/ 2[17 {(17) 24} ] / 2 (17 265) / 2 Hence the required eigenvalues are

1 2(17 265) / 2 16.6394, (17 265) / 2 0.3606. ... (8)

Here the symbol stands for approximately

Determination of eigenfunction corresponding to the eigenvalue 1 16.6394.

Putting 1 16.6394 in (6) and (7), we get

[1 – (7/3) × (16.6394)] C1 – 16.6394 C2 = 0 ... (9)and – 16.6394 C1 + [1 – (1/2) × (16.6394)] C2 = 0. ... (10)

Both (9) and (10) reduce to 2 12.2732 .C C ... (11)

Using (11) in (4), the eigenfunction y1(x) corresponding to eigenvalue 1 16.6394 is

given by

1 1 1 1 1 1 1( ) ( / ) ( 2.2732 ) [ 2.2732 (1/ )]y x C x x C C x x

or 1( ) [ 2.2732 (1/ )].y x x x [Taking 1C1 = 16.6394 C1 = 1]

Determination of eigenfunction corresponding to the eigenvalue 2 0.3606.

Putting 2 0.3606 in (6) and (7), we get

[1 – (7/3) × (0.3606)] C1 – 0.3606 C2 = 0 ... (12)and – 0.3606 C1 + [1 – (1/2) × (0.3606)] C1 = 0. ... (13)

Both (12) and (13) reduce to 2 10.4399 .C C ... (14)

Using (14) in (4), the eigenfunction y2 (x) corresponding to eigenvalue 2 0.3606 is

given by

2 2 1 2 1( ) ( / ) (0.4399 )y x C x x C or 2 2 1( ) [ 0.4399 (1/ )]y x C x x

or 2 ( ) [ 0.4399 (1/ )].y x x x [Taking 2C1 = 0.4399 C1 = 1]

Thus eigenvalues are 1 16.6394 and 2 0.3606 and the corresponding eigenfunctionsare y1(x) = [x – 2.2732 × (1/x)] and y2 (x) = [x + 0.4399 × (1/x)].

Ex. 8. Show that the homogeneous integral equation1

0( ) ( ) ( )y x t x x t y t

does not have real eigenvalues and eigenfunctions.



Sol. Given1

0( ) ( ) ( )y x t x x t y t dt

or1 1

0 0( ) ( ) ( ) .y x x t y t dt x t y t dt ... (1)

Let1

1 0( )C t y t dt ... (2)

and1

2 0( ) .C t y t dt ... (3)

Then (1) reduces to 1 2( ) .y x C x C x ... (4)

From (4), 1 2( ) .y t C t C t ... (5)


1

1 1 20( )C t C t C t dt or

1 15/ 2 3

1 1 20 0

5 / 2 3t tC C C

or1 2

21 0.5 3

C C

...(6)


2 1 20( )C t C t C t dt

or1 12 5/ 2

2 1 20 0

2 5 / 2t tC C C

or 1 221 0

2 5C C

... (7)

For non-zero solution of the system of equation (6) and (7), we must have

1 (2 / 5) / 3( ) 0

/ 2 1 (2 / 5)D

or22 21 1 0

5 5 6

or

21 0

150

or 2 150 0 so that 150,i

showing that ( ) 0D for any real value of . Hence the system of equations (6) and (7) has

unique solution C1 = C2 = 0 for all real . Hence, from (4), ( ) 0,y x which is zero solution.Hence, the given equation does not have real eigenvalues and eigenfunctions.

Ex. 9. Find the eigenvalues and eigenfunctions of the homogeneous integral equation1 3 2

1( ) (5 4 3 ) ( ) .y x x t x t x t y t dt

(Meerut 2012)

Sol. Given1 3 2

1( ) (5 4 3 ) ( ) .y x x t x t x t y t dt

or1 13 2

–1 1( ) 5 ( ) (4 3 ) ( ) .y x x t y t dt x x t y t dt

... (1)



Let1 3

1 –1( ) ,C t y t dt ... (2)

and 1

2 –1( ) .C t y t dt ... (3)


( ) 5 (4 3 ).y x C x C x x ... (4)

From (4), 21 2

( ) 5 (4 3 ).y t C t C t t ... (5)Using (5), (2) becomes

1 3 2

1 1 2–1[5 (4 3 )]C t C t C t t dt

115 6 5

1 2–1 –1

5 4 35 6 5t t tC C

or 1 1 22 (6 / 5)C C C or 1 2(1 2 ) (6 /5) 0.C C ... (6)Using (5), (3) becomes

1

2 1 2–1[5 (4 3 )]C t C t C t t dt

113 4 3

1 2–1 –1

5 4 3 .3 4 3t t tC C

or 2 1 2(10 / 3) 2C C C or 1 2(10 / 3) (1 2 ) 0.C C ... (7)For non-zero solution of the system of equations (6) and (7), we must have

1 2 (6 / 5)0

(10 / 3) 1 2

or 2 2(1 2 ) 4 0 or 1 4 0.

Hence the only eigenvalue is 1/ 4. To determine the corresponding eigenfunction, weproceed as follows :

Putting 1/ 4 in (6) and (7), we have

1 2(1/ 2) – (3 /10) 0C C ... (8)

and – 1 2(5 / 6) (1/ 2) 0C C ... (9)

Both (8) and (9) lead us to 1 2(3/ 5) .C C ... (10)

Putting 1/ 4 and using (10), (4) gives

22 2

1 3 1( ) 5 (4 3 )4 5 4

y x C x C x x

or2

23( )2

y x C x x

Setting C2 = 1, 2( ) (3/ 2) .y x x x

Hence the required eigenvalue is 1/ 4 and the corresponding eigenfunction is2( ) (3/ 2) .y x x x

Ex. 10. Find the eigenvalues and eigenfunctions of the homogeneous equation

0( ) ( , ) ( ) ,y x K x t y t dt

where

cos sin , 0( , )

cos sin , .x t x t

K x tt t t x



[Note : Whenever homogeneous equation is given in the above special form of K (x, t), wereduce the given integral equation into differential equation together with two boundary conditions.Then, we solve the resulting boundary value problem (*Strum-Liouville problem) to determineeigenvalue and eigenfunctions.]

Sol. Given 0

( ) ( , ) ( ) .y x K x t y t dt

... (1)

wherecos sin , 0

( , )cos sin , .

x t x tK x t

t x t x

... (2)


( ) ( , ) ( ) ( , ) ( )x


or0

( ) ( cos sin ) ( ) ( cos sin ) ( ) , using (2)x

xy x t x y t dt x t y t dt

... (3)


0

( ) ( cos sin ) ( ) ( cos sin ) ( )x

x

d dy x t x y t dt x t y t dtdx dx

or 0

( ) { cos sin ( )} cos sin ( )x dxy x t x y t dt x x y x

x dx cos 0 sin (0) dox y

dx

{ cos sin ( )}x

x t y t dtx

cos sin ( ) cos sin ( )d dxx y x x y xdx dx

[using Leibnitz-Rule of differentiation under the sign of integration see (Art. 1.13)]

or0

( ) ( cos cos ) ( ) cos sin ( )x

y x t x y t dt x x y x ( sin sin ) ( ) cos sin ( )

xx t y t dx x x y x

0

( ) ( cos cos ) ( ) ( sin sin ) ( ) .x


... (4)


0( ) ( cos cos ) ( ) ( sin sin ) ( )

x

x


or 2

0( ) { cos cos ( )} cos ( )

x dxy x t x y t dt x y xx dx

0cos 0 cos (0) dx ydx

{ sin sin ( )}x

x t y t dtx

2sin sin ( ) sin ( )d dxx y x y x

dx dx

2

0( cos sin ) ( ) cos ( )

xt x y t dt x y x 2( cos sin ) ( ) sin ( )

xx t y t dt x y x

0( ) [ ( cos sin ) ( ) ( cos sin ) ( ) ]

x


( ) ( ),y x y x using (3)

( ) ( 1) ( ) 0.y x y x ... (5)* For more details of Strum-Liouville problem, refer author’s part II of Ordinary and partial

differential equations, published by S.Chand & Co., New Delhi.



Putting x in (3), we get ( ) 0.y ... (6)

Again putting x = 0 in (4), we get (0) 0.y ... (7)Now, we shall solve Strum-Liouville problem given by (5), (6) and (7) by the usual procedure

to get eigenvalues and the corresponding eigenfunctions.Three cases arise :

Case I. Let 1 0 so that 1. Then (5) reduces to ( ) 0y x whose solution is

( ) .y x Ax B ... (8)From (8), y (x) = A. ... (9)Putting x in (8) and using (6), we get 0 .A B ... (10)

Putting 0x in (9) and using (7), we get 0 = A ... (11)

Solving (10) and (11), A = 0, B = 0.

Hence (8) gives y (x) = 0, which is not an eigenfunction and so 1 is not an eigenvalue.

Case II. Let 21 , where 0. Then (5) reduces to 2( ) ( ) 0y x y x whose

solution is ( ) .x xy x A e Be ... (12)

From (12), ( ) x xy x A e B e ... (13)

Putting x in (12) and using (6), we get 0 .Ae Be ... (14)

Putting x = 0 in (13) and using (7), we get 0 A B

or 0 = A – B, as 0 ... (15)Solving (14) and (15). A = B = 0.

Hence (8) gives y (x) = 0, which is not an eigenfunction and so 1 is not an eigen value.

Case III. Let 21 , where 0. Then (5) reduces to 2( ) ( ) 0y x y x whose

solution is ( ) cos sin .y x A x B x ... (16)

From (16), ( ) sin cos .y x A x B ... (17)

Putting x in (16) and using (6), we get 0 cos sin .A B ... (18)Again, putting x = 0 in (17) and using (7), we get

0 B or B = 0, as 0 ... (19)

Using (19), (18) gives cos 0.A ... (20)

Now, we must take 0,A otherwise A = 0 and B = 0 will give ( ) 0y x as before andhence we shall not get eigenfunction.

(20) gives cos 0 so that (2 1) ( / 2),n where n is an integer. Then

1/ 2n But 21 so that 21 .

Hence the eigenvalues are given by 2 21 1 ( 1/ 2) .n n ... (21)



Putting B = 0 and ( 1/ 2)n in (16), corresponding eigenfunctions yn (x) are given by

yn (x) = A cos (n + 1/2) x or ( ) cos( 1/ 2) ,ny x n x taking A = 1.

Hence the required eigenfunctions yn (x) with the corresponding eigenvalues n are given

by2( ) cos ( 1/ 2) , 1 ( 1/ 2) ,n ny x n x n where n is an integer..

Ex. 11. Determine the eigenvalues and eigenfunctions of the homogeneous integral equation1

0( ) ( , ) ( ) ,y x K x t y t dt where

( 1), 0 ,( , )

( 1), 1.x t x t

K x tt x t x

Sol. Given1

0( ) ( , ) ( ) ,y x K x t y t dt ... (1)

where( 1), 0 ,

( , )( 1), 1.

x t x tK x t

t x t x

... (2)

Re-writing (1), we have 1

0( ) [ ( , ) ( ) ( , ) ( ) ]

x


or1

0( ) ( 1) ( ) ( 1) ( ) , using (2)

x

xy x t x y t dt x t y t dt ... (3)

Differentiating (3) w.r.t. ‘x’ and using Leibinitz’s rule of differentiating under integral sign(refer Art. 1.13), we have

1

0( ) ( ) ( 1) ( ) 0 ( 1) ( ) 0 ( 1) ( )

x

xy x t y t dt x x y x t y t dt x x y x

or1

0( ) ( ) ( 1) ( ) .

x

xy x t y t dt t y t dt ... (4)

Differentiating (4) w.r.t. ‘x’ and using Leibnitz rule as before, we have( ) 0 ( ) 0 0 0 ( 1) ( )y x x y x x y x

or ( ) ( ) 0.y x y x ... (5)Putting x = 0 and x = 1 by turn in (3), we get

(0) 0y ... (6A)

and (1) 1.y ... (6B)We shall solve (5) subject to boundary conditions (6A) and (6B) to determine the required

eigenvalues and eigenfunctions.Three cases arise :Case I. Let 0. Then (5) reduces to y (x) = 0 whose general solution is

y (x) = Ax + B. ... (7)Putting x = 0 in (7) and using (6A), we get 0 = B. ... (8)Again, putting x = 1 in (7) and using (6B), we get 0 = A + B ... (9)Solving (8) and (9), A = B = 0. Hence (7) gives y (x) = 0, which is not an eigenfunction and

so 0 is not an eigenvalue.

Case II. Let 2 , where 0. Then (5) reduces to 2( ) ( ) 0y x y x whose

general solution is ( ) .x xy x Ae Be ... (10)



Putting x = 0 in (10) and using (6A), we get 0 = A + B. ... (11)

Again, putting x = 1 in (10) and using (6B), we get 0 .Ae Be ... (12)

Solving (11) and (12), A = B = 0. Hence (7) reduces to ( ) 0,y x which is not an eigenfunctionand hence 2 does not give eigenvalues.

Case III. Let 2 , where 0. Then (5) reduces to 2( ) ( ) 0y x y x whose

general solution ( ) cos sin .y x A x B x ... (13)Putting x = 0 in (13) and using (6A), we get 0 = A. ... (14)Again, putting x = 1 in (13) and using (6B), we get0 cos sin .A B or sin 0,B using (14) ... (15)

But 0,B otherwise B = 0 and A = 0 will give ( ) 0y x by (13) and so we shall not get aneigenfunction. Hence (15) gives

sin 0 so that , 1, 2, 3, ...n n

The required eigenvalues are given by2 2 2 , 1, 2, 3, ...n n n

From (13), the corresponding eigenfunctions yn (x) are given by( ) sinny x B n x ( 0, )A n

or ( ) sin ,ny x n x taking B = 1.Thus the required eigenvalues and eigenfunctions are given by

2 2 , ( ) sin , 1, 2, 3, ...n nn y x n x n

Ex. 12. Determine the eigenvalues and eigenfunctions of the homogeneous integral equation1

0( ) ( , ) ( ) ,y x K x t y t dt where ( 1), 0 ,

( , )( 1), 1.

t x x tK x t

x t t x

Sol. Given1

0( ) ( , ) ( ) ,y x K x t y t dt ... (1)

where( 1), 0 ,

( , )( 1), 1.

t x x tK x t

x t t x

... (2)


0( ) [ ( , ) ( ) ( , ) ( ) ]

x


or1

0( ) ( 1) ( ) ( 1) ( ) , using (2)

x

xy x x t y t dt t x y t dt ... (3)

Differentiating (3) w.r.t. ‘x’ and using Leibnitz’s rule of differentiating under integral sign(refer Art. 1.13), we have

1

0( ) ( 1) ( ) ( 1) ( ) 0 ( )

x

xy x t y t dt x x y x t y t dt 0 ( 1) ( )x x y x

or1

0( ) ( 1) ( ) ( ) .

x




Differentiating (4) w.r.t. ‘x’ and using Leibnitz’s rule as before, we have( ) 0 ( 1) ( ) 0 0 0 ( )y x x y x x y x

or ( ) ( ) 0.y x y x ... (5)Putting x = 0 in (3) and (4), we get

1

0(0) ( )y t y t dt ... (6)

and1

0(0) ( ) .y t y t dt ... (7)

Putting x = 1 in (3) and (4), we get1

0(1) ( 1) ( ) .y t y t dt ... (8)

and1

0(1) ( 1) ( ) .y t y t dt ... (9)

From (6) and (7), (0) (0).y y ... (10)

From (8) and (9), (1) (1).y y ... (11)We shall now solve (5) under boundary conditions (10) and (11) by usual method.Three cases arise :Case I. Let 0. Then (5) reduces to ( ) 0y x whose general solution is

( ) .y x Ax B ... (12)From (12), y(x) = A. ... (13)Putting x = 0 in (12) and (13), we get

(0)y B and (0) .y A (10) reduces to B = A. ... (14)Putting x = 1 in (12) and (13), we get

(1)y A B and (1) .y A ... (15) (11) reduces to A + B = A so that B = 0. ... (16)Solving (14) and (16), A = B = 0. Hence (12) gives y (x) = 0, which is not an eigenfunction

and so 0 is not an eigenvalue.

Case II. Let 2 , where 0. Then (5) reduces to 2( ) ( ) 0,y x y x whose

general solution is ( ) x xy x Ae Be ... (17)

From (17) ( ) .x xy x A e B e ... (18)Putting x = 0 in (17) and (18), we get

y (0) = A + B and (0) .y A B

(10) reduces to A B A B

or (1 ) (1 ) 0.A B ... (19)Putting x = 1 in (17) and (18), we get

(1)y Ae Be and (1) .y A e B e



(11) reduces to Ae Be A e B e

or (1 ) (1 ) 0.Ae Be ... (20)For non-trivial solution of (19) and (20), we must have

1 10

(1 ) (1 )e e

or (1 ) (1 ) (1 ) (1 ) 0e e

or (1 ) (1 ) ( ) 0e e or 2(1 ) (1 ) sinh 0. ... (21)

Since 0 by assumption, sinh 0. So (21) reduces to

(1 ) (1 ) 0 so that 1 or 1.

When 1, (19) and (20) reduces to A.0 + 2B = 0 and A.0 + 2Be–1 = 0 ... (22)

Solving (22), B = 0 and A is arbitrary constant. (17) gives y (x) = Aex. ... (23)Next, when 1, (19) and (20) reduces to

2A + B.0 = 0 and 2Ae + B.0 = 0Solving these, A = 0 and B is an arbitrary constant.

(17) gives ( ) .xy x Be ... (24)Setting A = 1 in (23) or B = 1 in (24), the required eigenfunction is ex which correspond to

eigenvalue 2 2 2(1) ( 1) 1.

Case III. Let 2 , where 0. Then (5) reduces to 2( ) ( ) 0,y x y x whose

general solution is ( ) cos sin .y x A x B x ... (25)

From (25), ( ) sin cos .y x A x B x ... (26)Putting x = 0 in (25) and (26), we get

(0)y A and (0) .y B

(10) reduces to .A B ... (27)Putting x = 1 in (25) and (26), we get

(1) cos siny A B and (1) sin cos .y A B

(11) reduces to cos sin sin cosA B A B ... (28)Putting value of A given by (27) in (28), we get

2cos sin sin cosB B B B or 2(1 ) sin 0.B ... (29)

But 0B for otherwise from (27), A = 0 when B = 0 and it gives ( ) 0y x by (25). Thus,we do not get eigenfunction when B = 0.

Again 2(1 ) 0, for otherwise 21 0 would give 2 1 which is not possible as

is real and so 2 cannot be negative.

(29) reduces to sin 0, giving , 1, 2,3, ...n n



2 2 2 , 1, 2,3, ...n n

Putting n and A B in (25), we get

( ) cos sin ( cos sin ).y x B n x B n x B n x n x

Setting 1, we have ( ) cos sin .B y x n x n x

Hence the required eigenvalues 0 , n and the corresponding eigenfunctions y0 (x), yn (x)

are given by 2 20 0

1, ( ) , , ( ) cos sin , 1, 2,3, ...xn ny x e n y x n x n x n

Ex. 13. Determine the eigenvalues and eigenfunctions of the homogeneous integral equation

1

0( ) ( , ) ( ) ,y x K x t y t dt where

sinh , 0( , )

sinh , 1.

t

x

e x x tK x t

e t t x

Sol. Given1

0( ) ( , ) ( ) ,y x K x t y t dt ... (1)

where sinh , 0

( , )sinh , 1.

t

x

e x x tK x t

e t t x

... (2)


0( ) [ ( , ) ( ) ( , ) ( ) ]

x


or1

0( ) sinh ( ) sinh ( ) , using (2)

x x t

xy x e t y t dt e x y t dt ... (3)

Differentiating (3) w.r.t., ‘x’ and using Leibnitz’s rule of differentiating under integral sign(refer Art. 1.13), we have

0( ) [ ( sinh ) ( ) sinh ( ) 0]

x x xy x e t y t dt e x y x 1

[ ( cosh ) ( ) 0 sinh ( )]t x

xe x y t dt e x y x

or1

0( ) sinh ( ) cosh ( ) .

x x t

xy x e t y t dt e x y t dt ... (4)

Differentiating (4) w.r.t. ‘x’ and using Leibnitz rule of differentiating under integral sign asbefore, we have

0( ) ( sinh ) ( ) sinh ( ) 0

x x xy x e t y t dt e x y x 1

[ ( sinh ) ( ) 0 cosh ( ) ]t x

xe x y t dt e x y x

or1

0sinh ( ) sinh ( )

x x t

xy e t y t dt e x y t dt ( ) (sinh cosh )xe y x x x

( ) ( ) ,2 2

x x x xx e e e ey x e y x

by (3)

( ) ( ) ( ) ( )x xy x e y x e y x y x



or ( ) (1 ) ( ) 0.y x y x ... (5)Putting x = 0 in (3), y (0) = 0. ... (6)Putting x = 1 in (3) and (4), we get

1 1

0(1) sinh ( )y e t y t dt ... (7)

and1 1

0(1) sinh ( ) .y e t y t dt ... (8)

Adding (7) and (8), (1) (1) 0.y y ... (9)We shall now solve (5) under boundary conditions (6) and (9). Three cases arise :Case I. Let 0. Then (5) reduces to ( ) 0y x , whose general solution is

y (x) = Ax + B. ... (10)From (10) y (x) = A. ... (11)Putting x = 0 in (10) and using B.C. (6), we get 0 = B .... (12)Putting x = 1 in (10) and (11), we get y (1) = A + B and y (1) = A. ... (13) B.C. (9) becomes A + B + A = 0 ... (14)Solving (12) and (14), A = B = 0. So by (10), y (x) = 0, which is not an eigenfunction.

Case II. Let 21 , where 0. Then (5) reduces to 2( ) ( ) 0,y x y x whose

general solution is ( ) x xy x Ae Be ... (15)

From (15), ( ) x xy x A e B e ... (16)

Putting x = 0 in (15) and using B.C. (6), we get 0 = A + B or B = –A ... (17)Putting x = 1 in (15) and (16), we get

(1)y Ae Be and (1) .y A e B e

B.C. (9) becomes

0Ae Be A e B e or 0Ae Ae A e A e , by (17)

or [ ( )] 0A e e e e or [2 sinh 2 cosh ] 0.A

A = 0 and so from (17), B = 0. With A = B = 0, (15) gives ( ) 0,y x which is not aneigenfunction.

Case III. Let 21 , where 0. Then (5) reduces to 2( ) ( ) 0,y x y x whose

general solution is ( ) cos sin .y x A x B x ... (18)

From (18) ( ) sin cos .y x A x B x ... (19)Putting x = 0 in (18) and using B.C. (6), we get 0 = A. ... (20)Putting x = 1 and A = 0 in (18) and (19), we get

(1) siny B and (1) cos .y B

B.C. (9) becomes sin cos 0B B or (sin cos ) 0.B ... (21)If B = 0, then with A = 0, (18) reduces to y (x) = 0, which is not an eigenfunction. So we take

0B and hence (21) gives



sin cos 0 or tan , ... (22)

which is a trigonometrical equation in . Let ( 1, 2, 3, ...)n n be the positive roots of (22). WithA = 0, (18) reduces to

( ) siny x B x or ( ) sin ,y x x taking B = 1.

Again, 21 . Hence the required eigenvalues n and the corresponding

eigenfunctions yn (x) are gives by 2 21n n and ( ) sin , 1, 2, 3, ...,n ny x n where n arethe positive roots of (22).

Ex. 14. The eigenvalue of the Fredholm integral equation 1 20

( ) ( )y x x t y t dt is

(a) –2 (b) 2 (c) 4 (d) – 4 [GATE 2011]

Solution. Ans. (c). Given 120

( ) ( )y x x t y t dt ... (1)

Let 1

0( )c t y t dt ... (2)

Then, (1) yields y(x) = cx2 so that y(t) = ct2 ... (3)

Using (3), (2) reduces to 1 30

( ) / 4c ct dt c so that c(4 – ) = 0 ... (4)

If c = 0, then (4) gives y(x) = 0. We, therefore, assume that for nonzero solution of (1), 0.c Hence, (4) reduce to 4 – = 0 or = 4, which is the required eigenvalue.

EXERCISE1. Solve the following homogeneous integral equations :

(i)1

0( ) ( ) .y x y t dt (ii)

0

1( ) sin ( ) .2

y x x y t dt

(iii)10

0

1( ) ( ) .50

y x t y t dt (iv)1

2 0

1( ) 2 ( ) .1

x ty x e e y t dte

2. Determine the eigenvalues and the eigenfunctions of the following homogeneous integralequations :

(i)/ 4 2

0( ) sin ( ) .y x x y t dt

(ii)

2

0( ) sin cos ( ) .y x x t y t dt

(iii)2

0( ) sin sin ( ) .y x x t y t dt

(iv)

1 3 2

1( ) (5 4 ) ( ) .y x xt x t y t dt

(v)1 2 2

0( ) (45 9 ) ( ) .y x x ln t t ln x y t dt where ln x = logex

(vi)0

( ) cos ( ) ( ) .y x x t y t dt

(vii)1

1( ) ( cosh sinh ) ( )y x x t t x y t dt



(viii)1 2

1( ) ( cosh sinh ) ( )y x x t t x y t dt

(x)

10

( ) ( )xx e d (Kanpur 2008)

(ix) 1

1( ) ( cosh cosh ) ( ) .y x x t t x y t dt

3. Show that the integral equation 0

( ) (sin sin 2 ) ( )y x x t y t dt

has no eigenvalues.

4. Find the eigenvalues and eigenfunctions of the following homogeneous integral equations1

0( ) ( , ) ( ) ,y x K x t y t dt where

(i) sin cos , 0 ,( , )

sin cos , .x t x t

K x tt x t x

(ii) sin cos , 0 ,( , )

sin cos , / 2.x t x t

K x tt x t x

(iii) ( 1) ( 2) , 0 ,( , )

( 1) ( 2), 1x t x t

K x tt x t x

(iv) sin sin ( 1), ,

( , )sin sin ( 1), .

x t x tK x t

t x t x

(v)sin ( / 4) sin ( / 4), 0

( , )sin ( / 4) sin ( / 4), .

x t x tK x t

t x t x

(vi) | |( , ) , 0 1, 0 1.x tK x t e x t

5. What do you understand by eigenvalues and eigenfunctions of the integral equation

( ) ( , ) ( ) .b

ay x K x t y t dt Prove that the integral equation with degenerate kernel

1( , ) ( ) ( )

n

r rrK x t a x b t

has at most n eigenvalues.

ANSWERS

1. (i) ( ) 0y x (ii) ( ) 0y x (iii) ( ) 0y x (iv) ( ) 0y x .

2. (i) 28 /( 2), ( ) sin .y x x

(ii) Eigenvalues and eigenfunctions do not exist.

(iii) 1/ , ( ) sin .y x x (iv) 21/ 2, ( ) (5 / 2) (10 /3)y x x x

(v) There are no real eigenvalues and real eigenfunctions.

(vi) 1 1 2 22 / , ( ) sin ; 2 / , ( ) cos .y x x y x x

(vii) / 2, ( ) sinh .e y x x

(viii) Eigenvalues and eigenfunctions do not exist.(ix) There are no real eigenvalues and real eigenfunctions.(x) = 1, (x) = ex



4. (i) 2( 1/ 2) 1; ( ) sin( 1/ 2) , 1, 2, 3, ...n nn y x n x n

(ii) 24 –1; ( ) sin 2 , 1, 2, 3, ...n nn y x nx n

(iii) 2(1/ 3) ; ( ) sin cos , 1,2,3,...n n n n n ny x x x n

where n is a root of the equation (1/ ) 2cot .

(iv) 2(1 ) cosec 1; ( ) sin{ ( )}, 1, 2,3, ...n n n ny x x n

where n are roots of the equation tan 2 tan1.

(v) 21 ; ( ) sin cos , 1,2,3,...n n n n n ny x x x n

where n are the roots of the equation 2cot (1/ ).

(vi) 2(1 ) / 2; ( ) sin cos , 1, 2,3,...n n n n ny x x x n

where n are roots of the equation 2cot (1/ ).


CHAPTER 4

Fredholm Integral Equations ofthe Second Kind With Separable(or Degenerate) Kernels4.1 SOLUTION OF FREDHOLM INTEGRAL EQUATION OF THE SECOND KIND

WITH SEPARABLE (OR DEGENERATE) KERNEL. [Meerut 2000, 01, 05]Consider a Fredholm integral equation of the second kind :

( ) ( ) ( , ) ( ) .b


Since kernel K (x, t) is separable, we take

1( , ) ( ) ( ).

n

i iiK x t f x g t

... (2)

where the functions fi (x) are assumed to be linearly independent.


( ) ( ) [ ( ) ( )] ( )nb

i iiay x f x f x g t y t dt

or1

( ) ( ) ( ) ( ) ( ) .n b

i ii ay x f x f x g t y t dt

... (3)

[Interchanging the order of summation and integration]

Let ( ) ( ) , ( 1,2, ..., )b

i iag t y t dt C i n ... (4)


( ) ( ) ( ),n

i iiy x f x C f x

... (5)

where constants Ci (i = 1, 2, 3, ... n) are to be determined in order to find solution of (1) in the formgiven by (5). We now proceed to evalueate Ci’s as follows :

Re-writing (5),1

( ) ( ) ( )n

j jjy x f x C f x

so that1

( ) ( ) ( )n

j jjy t f t C f t

... (6)

Substituting the values of y (x) and y (t) given by (5) and (6) respectively in (3), we have

1 1 1( ) ( ) ( ) ( ) ( ) ( ) ( )

n n nb

i i i i j ji i jaf x C f x f x f x g t f t C f t dt

or 1 1( ) ( ) ( ) ( ) ( ) ( )

n nb b

i i i i j i ji ja aC f x f x g t f t dt C g t f t dt

... (7)

4.1


4.2 Fredholm Integral Equations of the Second Kind With Separable (or Degenerate) Kernels

We now introduce the following useful notations :

( ) ( )b

i iag t f t dt and ( ) ( ) , , 1, 2, ...,

b

i j i jag t f t dt i j n ... (8)

where i and i j are known constants. Then (7) may be simplified as

1 11

( ) ( )n n n

i i i i i j ji ji

C f x f x C

or

1( ) – 0

n

i i i i j jjf x C C

But the functions fi (x) are linearly independent, therefore

1

0, 1,2,3, ...,n

i i i j jjC C i n

or1

, 1,2,3, ...,n

i i j j ijC C i n

... (9)

Taking i = 1 in (9), we have

1 11

n

i j jjC C

or 1 11 1 12 2 1 1( ... )n nC C C C

or 11 1 12 2 1 1(1 ) ... .n nC C C

Similarly, we may simplify (A2), (A3), ... (An). Thus, we obtain the following system of linearequations to determine C1, C2, ..., Cn.

11 1 12 2 1 1(1 ) ... .n nC C C ... (B1)

21 1 22 2 2 2(1 ) ... .n nC C C ... (B2)... ... ... ... .... ...... ... ... ... .... ...

and 1 1 2 2 ... (1 ) .n n nn n na C C C ... (Bn)

The determinant ( )D of the system (B1), (B2), ..., (Bn) of linear equations is given by

11 12 1

21 22 2

1 2

1 . . .

1 . . .( ) ,

: : : : . . . : :. . . 1

n

n

n n nn

D

... (10)

which is a polynomial in of degree at most n. Again, ( )D is not identically zero, since, when

0, ( ) 1.D To discuss the solution of (1), the following situations arise :Situation I. When at least one right member of the system (B1), ..., (Bn) is nonzero.

The following two cases arise under this situation :

(i) If ( ) 0,D then a unique nonzero solution of the system (B1), ..., (Bn) exists and so (1)has a unique nonzero solution given by (5).

(ii) If ( ) 0,D then the equations (B1), ... (Bn) have either no solution or they possessinfinite solutions and hence (1) has either no solution or infinite solutions.


Fredholm Integral Equations of the Second Kind With Separable (or Degenerate) Kernels 4.3

Situation II. When 0f(x) = . Then (8), shows that 0j for j = 1, 2, ..., n. Hence the

equations (B1), ... (Bn) reduce to a system of homogeneous linear equations.The following two cases arise under this situation :(i) If ( ) 0,D then a unique zero solution C1 = C2 = .... = Cn = 0 of the system (B1), ...,

(Bn) exists and so from (5), we see that (1) has only unique zero solution y (x) = 0.(ii) If ( ) 0,D then the system (B1), ... (Bn) possess infinite nonzero solutions and so (1)

has infinite nonzero solutions. Those values of for which ( ) 0D are known as the eigenvalues(or characteristic constants or values) and any nonzero solution of the homogeneous Fredholm

integral equation ( ) ( , ) ( )b

ay x K x t y t dt (with a convenient choice of the arbitrary constant or

constants) is known as a corresponding eigenfunction (or characteristic function) of integral equation.For more discussion of theory and problems based on situation II, refer chapter 2.Situation III. When ( ) 0,f x but

1 2( ) ( ) 0, ( ) ( ) 0, ..., ( ) ( ) 0,b b b

na a ag x f x dx g x f x dx g x f x dx

i.e., f (x) is orthogonal to all the functions g1 (t), g2 (t), ..., gn (t), then (8) shows that

1 20, 0, ..., 0n and hence the equations (B1), ..., (Bn) reduce to a system of homogeneouslinear equations.

The following two cases arise under this situation : (i) If ( ) 0,D then a unique zero solution C1 = C2 = ... Cn = 0 of the system (B1), ..., (Bn)

exists and so from (5), we see that (1) has only unique solution y (x) = f (x).(ii) If ( ) 0,D then the system (B1), ..., (Bn) possess infinite nonzero solutions and so (1)

has infinite nonzero solutions. The solutions corresponding to the eigenvalues of are now expressedas the sum of f (x) and arbitrary multiples of eigenfunctions.4.2. SOLVED EXAMPLES BASES ON ART. 4.1.

Ex. 1. The solution of the integral equation 1 20

( ) ( )g s s s u g u du is given by

(a) ( ) 3 / 4g t t (b) ( ) 4 / 3g t t (c) ( ) 2 / 3g t t (d) ( ) 3 / 2g t t [GATE 1999]

Sol. Ans (b) Given 1 20

( ) ( )g s s s u g u du ...(1)

Let 1 20

( )C u g u du ...(2)

Using (2), (1) yields ( ) (1 )g s s Cs s C ...(3)

From (3), ( ) (1 )g u u C ...(4)

Using (4), (2) yields1 13 4

00(1 ) (1 ) / 4C u C du C u

or (1 ) (1/ 4)C C so that C = 1/3

Hence, (3) g(s) = (1 1/3) 4 /3 and so ( ) 4 /3.s s g t t



Ex. 2(a). Solve : 1

0( ) 2 ( ) .x x ty x e e e y t dt [Meerut 2009]

Sol. Given1

0( ) 2 ( ) .x x ty x e e e y t dt or

1

0( ) 2 ( ) .x x ty x e e e y t dt ... (1)

Let1

0( ) .tC e y t dt ... (2)

Using (2), (1) reduces to ( ) 2 (1 2 ).x x xy x e C e e C ... (3)

From (3), ( ) (1 2 ).ty t e C ... (4)Using (4), (2) becomes

121

00

[ . (1 2 )] (1 2 )2

tt t eC e e C dt C

21(1 2 ) ( 1)

2C e

or 2 21[1 ( 1)] ( 1)2

C e e or2

21 ,

2 [1 ( 1)]eC

e

where 21 .

1e

Putting this value of C in (3), we get2

21( ) 1 2

2{1 ( 1)}x ey x e

e

or 2( ) ,1 ( 1)

xey xe

where 2

1 .1e

which is the required solution of given integral equation.

Ex. 2(b). Solve :0

( ) cos sin ( ) .y x x x y t dt

[Kanpur 2008]

Sol. Given0

( ) cos sin ( ) .y x x x y t dt

or 0

( ) cos sin ( )y x x x y t dt

... (1)

Let 0

( )C y t dt

... (2)

Using (2), (1) becomes ( ) cos sin .y x x C x ... (3)

From (3), ( ) cos sin .y t t C t ... (4)

Using (4), (2) reduces to 0

(cos sin )C t C t dt

0 0sin cost C t

or 0 [ cos cos 0]C C or C = 2 C

or (1– 2 ) 0C so that 0, if 1/ 2C

Hence by (3), the required solution is ( ) cos ,y x x provided 1/ 2.

Ex.3. Solve :/ 2 2

0( ) 2 4 sin ( ) .y x x x y t dt

[Kanpur 2007]

Sol. Given / 2 2

0( ) 2 4 sin ( )y x x x y t dt

or / 22

0( ) 2 4sin ( )y x x x y t dt

... (1)



Let/ 2

0( )C y t dt

... (2)

Using (2), (1) becomes 2( ) 2 4 siny x x C x ... (3)

From (3), 2( ) 2 4 siny t t C t ... (4)Using (4), (2) becomes

/ 2 2

0(2 4 sin )C t C t dt

/ 2/ 220 0

1 cos 242

tt t C dt

or/ 22 2

0

sin 224 2 2

tC C t

or 2

24 2

C C or

2

4( 1)C

Putting this value of C in (3), the required solution of given integral equation is2 2sin( ) 2 .

1xy x x

Ex. 4. Solve :1

0( ) ( ) ( ) .y x f x xt y t dt

Sol. Given1

0( ) ( ) ( ) .y x f x xt y t dt

or1

0( ) ( ) ( )y x f x x t y t dt ... (1)

Let1

0( ) .C t y t dt ... (2)

Then (1) reduces to ( ) ( )y x f x C x ... (3)

From (3), ( ) ( )y t f t C t ... (4)Using (4), (2) reduces to

1

0[ ( ) ]C t f t Ct dt or C

131

00

( )3tt f t dt C

or

1

0( )

3CC tf t dt

or1

01 ( )

3C t f t dt

or 1

0

3 ( ) ,3

C t f t dt where 3.

Putting this value of C in (3), the required solution is1

0

3( ) ( ) ( ) ,3

xy x f x t f t dt

where 3

Ex. 5. Invert the integral equation :2

0( ) ( ) (sin cos ) ( ) .y x f x x t y t dt

OR

Find the solution of the integral equation2


[Kanpur 2006]

Sol. Given2




or2

0( ) ( ) sin cos ( ) .y x f x x t y t dt

... (1)

Let2

0cos ( ) .C t y t dt

... (2)

Then (1) reduces to ( ) ( ) siny x f x C x ... (3)

From (3), ( ) ( ) siny t f t C t ... (4)


0cos [ ( ) (sin )C t f t t dt

or2 2

0 0cos ( ) sin 2

2CC t f t dt t dt

or

22

0 0

cos 2cos ( )2 2C tC t f t dt

or2

0

1 1cos ( )2 2 2CC t f t dt

or2

0cos ( ) .C t f t dt

Putting this value of C in (3), the required solution is2

0( ) ( ) sin ( ) cosy x f x x f t t dt

or2

0( ) ( ) (sin cos ) ( ) .y x f x x t f t dt

Ex. 6. Solve the Fredholm integral equation of the second kind :1 2 2

0( ) ( ) ( ) .y x x xt x t y t dt

Sol. Given1 2 2

0( ) ( ) ( ) .y x x xt x t y t dt

or1 12 2

0 0( ) ( ) ( ) .y x x x t y t dt x t y t dt ... (1)

Let 1 2

1 0( )C t y t dt ... (2)

and1

2 0( )C t y t dt ... (3)

Using (2) and (3), (1) reduces to 21 2

( ) .y x x C x C x ... (4)

From (4), 21 2

( ) .y t t C t C t ... (5)Using (5), (2) reduces to

1 2 21 1 20

( )C t t C t C t dt 14 54

1 2

04 4 5

C t C tt

1 214 4 5

C C

or 1 2(20 5 ) 4 5.C C ... (6)



Next, using (5), (2) reduces to

1 2

2 1 20( )C t t C t C t dt

13 431 2

03 3 4

C t C tt

= 1 213 3 4

C C

or 1 24 (12 – 3 ) 4.C C ... (7)Solving (6) and (7) for C1 and C2, we get

2160

240 120C

and 22

80 .240 120

C

Putting these values of C1 and C2 in (4), the required solution is2

2 2(60 ) 80( )

240 120 240 120x xy x x

or

2

2(240 60 ) 80( )

240 120x xy x

Ex. 7. Solve :1

0( ) 1 (1 ) ( ) .x ty x e y t dt [Kanpur 2007]

Sol. Given1

0( ) 1 (1 ) ( )x ty x e y t dt or

1 1

0 0( ) 1 ( ) ( )x ty x y t dt e e y t dt ... (1)

Let1

1 0( )C y t dt ... (2)

and1

2 0( ) .tC e y t dt ... (3)

Using (2) and (3), (1) reduces to 1 2( ) 1 .xy x C C e ... (4)

From (4), we have 1 2( ) 1 ty t C C e ... (5)


1 1 20(1 )tC C C e dt or

1

1 1 2 0tC t C t C e

or C1 = 1 + C1 + C2 (e – 1) or C2 = – 1 / (e – 1). ... (6)Using (5), (3) reduces to

1

2 1 20(1 )t tC e C C e dt

12

1 20

2

tt t ee C e C

2211 ( 1) ( 1)

2C

e C e e

or 2

11 11 ( 1) ,

1 2 ( 1)ee C e

e e

using (6)

or1

1 ( 1)( 1)1 2

eC ee

or

2

212 3.

2 ( 1)e eC

e

... (7)

Using (6) and (7) in (4), the required solution is2

22 3( ) 1

12( 1)

xe e ey xee

or2

22 1( ) .

12( 1)

xe e ey xee



or2

22 1 2 ( 1)( ) .

2( 1)

xe e e ey xe

Ex. 8. Solve :12 2 2

1( ) (1 ) ( ) ( ) .y x x xt x t y t dt

Sol. Given 12 2 2

1( ) (1 ) ( ) ( )y x x xt x t y t dt

or 1 12 2 2

1 1( ) (1 ) ( ) ( ) .y x x x t y t dt x t y t dt

... (1)

Let1

1 1( )C t y t dt

... (2)

and1 2

2 1( ) .C t y t dt

... (3)

Using (2) and (3), (1) reduces to 2 21 2

( ) (1 ) .y x x C x C x ... (4)

From (4), 2 21 2

( ) (1 ) .y t t C t C t ... (5)Using (5), (2) reduces to

1 2 21 1 21

[(1 ) ]C t t C t C t dt

or1 2

1 1 21[1 (2 ) (1 ) ]C t C t C t dt

or1 1 12 3 4

1 1 21 1 1

(2 ) (1 )2 3 4t t tC C C

or 1 1(2 / 3) (2 )C C so that C1 = 4 ... (6)Using (5), (3) reduces to

1 2 2 22 1 21

[(1 ) ]C t t C t C t dt

1 2 2

1 21[1 (2 ) (1 ) ]t C t C t dt

or1 1 13 4 5

2 1 2–1 1 1

(2 ) (1 )3 4 5t t tC C C

or 2 22 / 3 (1 ) (2 / 5)C C or C2 = 16/9. ... (7)Using (6) and (7), (4) gives the required solution

2 2( ) (1 ) 4 (16 /9)y x x x x or 2( ) 1 6 (25 / 9) .y x x x

Ex.9. Solve :0

( ) cos sin( ) ( ) .y x x x t y t dt

[Merrut 2007]

Sol. Given0

( ) cos sin( ) ( ) .y x x x t y t dt

or

0( ) cos (sin cos cos sin ) ( )y x x x t x t y t dt

or0 0

( ) cos sin cos ( ) cos sin ( ) .y x x x t y t dt x t y t dt

... (1)



Let 1 0cos ( )C t y t dt

... (2)

and 2 0sin ( ) .C t y t dt

... (3)

Using (2) and (3), (1) reduces to 1 2( ) cos sin cos .y x x C x C x ... (4)

From (4), 1 2( ) cos sin cos .y t t C t C t ... (5)Using (5), (2) reduces to

1 1 20cos (cos sin cos )C t t C t C t dt

2

2 10

1[(1 ) cos sin 2 ]2

C t dt C t dt

or 11 2 0 0

1 cos 2(1 ) sin 22 2

CtC C dt t dt

2 1

0 0

1 sin 2 cos 22 2 2 2

C Ct tt

or 1 2(1 ) ( / 2)C C or 1 22 .C C ... (6)Using (5), (3) reduces to

2 1 20sin (cos sin cos )C t t C t C t dt

2 1

0 0

1sin 2 (1 cos 2 )

2 2C C

t dt t dt

or 2 12

0 0

1 cos 2 sin 22 2 2 2

C Ct tC t

or 2 1( ) / 2C C ... (7)Solving (6) and (7) for C1 and C2, we get

2 2 2 2 21 2

(2 ) /(4 ) and ( ) / (4 ).C C

Putting these values of C1 and C2 in (4), the required solution is2 2

2 2 2 22 sin cos( ) cos4 4

x xy x x

or

2 2

2 2 2 22 sin( ) cos 1

4 4xy x x

2 2( ) (4 cos 2 sin ) / (4 )y x x x

Ex. 10. Solve :1

0( ) ( ) ( ) ( ) .y x f x x t y t dt

Sol. Given1

0( ) ( ) ( ) ( ) .y x f x x t y t dt

or1 1

0 0( ) ( ) ( ) ( ) .y x f x x y t dt t y t dt ... (1)

Let1

1 0( )C y t dt ... (2)

and1

2 0( ) .C t y t dt ... (3)

Using (2) and (3), (1) reduces to 1 2( ) ( ) .y x f x x C C ... (4)

From (4), 1 2( ) ( ) .y t f t t C C ... (5)




1

1 1 20[ ( ) ]C f t tC C dt or

121 101 1 20

0

( )2tC f t dt C C t

or1 1 1 2( / 2) ,C f C C ... (6)

where1

1 0( ) .f f t dt ... (7)


1

2 1 20[ ( ) ]C t f t tC C dt or

1 13 21

2 1 200 0

( )3 2t tC t f t dt C C

or 2 2 1 2( / 3) ( / 2)C f C C ... (8)

where1

2 0( ) .f t f t dt ... (9)

Re-writing (6) and (8), we have

1 2 1(2 ) 2 2C C f ... (10)

and 1 2 22 3 (2 ) 6 .C C f ... (11)Solving (10) and (11) for C1 and C2, we get

1 2 1 22 21 2

6 ( 2) 12 4 6 ( 2)and

12 12 12 12

f f f fC C

Putting these values of C1 and C2 in (4), the required solution is

1 2 1 22 2

{6 ( 2) 12 } 4 6 ( 2)( ) ( )

12 12 12 12

x f f f fy x f x

or1 2

2

{6 ( 2) 4 } {6 ( 2) 12 }( ) ( )

12 12

f x f xy x f x

or1

2 0( ) ( ) {6 ( 2) 4 } ( )

12 12y x f x x f t dt

1

0{6( – 2) –12 } ( )x t f t dt

or1

2 0( ) ( ) {6 ( 2) 4 } ( )

12 12y x f x x f t dt

1

0{6( 2) 12 } ( )x t f t dt

or1

2 0( ) ( ) {6( ) ( ) 12 4 } ( )

12 12y x f x x t xt f t dt

or1

20

6( 2) ( ) 12 4( ) ( ) ( ) .12 12

x t xty x f x f t dt

Ex. 11. Solve :

1 2 2

1( ) ( ) ( ) ( ) .y x f x xt x t y t dt

Find its resolvent kernel also.

[Meerut 2009; Kanpur 2006]

Sol. Given1 2 2

1( ) ( ) ( ) ( )y x f x xt x t y t dt



or1 12 2

1 1( ) ( ) ( ) ( ) .y x f x x t y t dt x t y t dt

... (1)

Let1

1 1( )C t y t dt

... (2)

and1 2

2 1( ) .C t y t dt

... (3)

Using (2) and (3), (1) reduces to 21 2

( ) ( ) .y x f x C x C x ... (4)

From (4), 21 2

( ) ( ) .y t f t C t C t ... (5)Using (5), (2) reduces to

1 21 1 21

[ ( ) ]C t f t C t C t dt

1 13 41

1 211 1

( )3 4t tt f t dt C C

or1

11 1

2( )

3C

C t f t

or

1

1 1

21 ( )3

C t f t dt

or1

1 1

3 ( ) .3 2

C t f t dt

... (5)


1 2 2

2 1 21[ ( ) ]C t f t C t C t dt

1 14 51 21 21

1 1

( )4 5t tt f t dt C C

or 1 2 2

2 1

2( )

5C

C t f t dt

or 1 2

2 1

21 ( )5

C t f t dt

or

1 22 1

5 ( ) .5 2

C t f t dt

... (7)

Using (6) and (7) in (4), the required solution is21 1 2

1 1

3 5( ) ( ) ( ) ( )3 2 5 2

x xy x f x t f t dt t f t dt

or2 21

1

3 5( ) ( ) ( ) .3 2 5 2

xt x ty x f x f t dt

... (8)

The required resolvent kernel ( , : )R x t is given by 2 23 5( , : ) .

3 2 5 2xt x tR x t

Ex. 12. Solve the integral equation 2( ) ( cos sin cos sin ) ( ) .y x x t t x x t y t dt x

Sol. Given 2( ) ( cos sin cos sin ) ( )y x x x t t x x t y t dt

or 2( ) cos ( ) sin ( )y x x x t y t dt x t y t dt

cos sin ( ) .x t y t dt

... (1)



Let 1 cos ( ) ,C t y t dt

... (2)

22 ( ) ,C t y t dt

... (3)

and 3 sin ( ) .C t y t dt

... (4)

Using (2), (3) and (4), (1) reduces to

1 2 3( ) sin cos .y x x C x C x C x ... (5)

From (5), 1 2 3( ) sin cos .y t t C t C t C t ... (6)

Using (6), (2) reduces to 1 1 2 3cos ( sin cos )C t t C t C t C t dt

or 21 1 2 3(1 ) cos sin cos cosC C t t dt C t t C t dt

or 21 3 0

0 0 2 cosC C t dt

[ t cos t and sin t cos t are odd functions whereas cos2 t in an even function]

or 1 3 0

(1 cos 2 )22

tC C dt

or 1 30

sin 22

tC C t

1 3 0.C C ... (7)

Using (6), (3) reduces to 22 1 2 3[ sin cos ]C t t C t C t C t dt

or 3 2 22 1 2 3(1 ) sin cosC C t dt C t t dt C t t dt

or 22 3 0

2 cosC C t t dt

23 0 0

2 sin 2 sin ,C t t t t dt integrating by parts

or 2 3 0– 4 sinC C t t dt

3 0 0

4 [( ) ( cos )] ( cos )C t t t dt

or 2 3 04 cosC C t dt

or 02 3 34 4 sinC C C t

or 2 34 0.C C ... (8)

Using (6), (4) reduces to3 1 2 3sin ( sin cos )C t t C t C t C t dt

or 23 1 2 3(1 ) sin sin sin cosC C t t dt C t dt C t t dt

or 23 1 20 0

2 (1 ) sin 2 sin 0C C t t dt C t dt



or 03 1 20 0

1 cos 22 (1 ) ( cos ) ( cos ) 22

tC C t t t dt C dt

or 3 1 0 20

sin 22 (1 ) [sin ]2

tC C t C t

or 3 1 2 1 2 32(1 ) or 2 2C C C C C C ... (9)Solving (7), (8) and (9) for C1, C2 and C3, we have

2 2

2 2 2 2 2 21 2 32 8 2, and .

1 2 1 2 1 2C C C

Putting these values of C1, C2 and C3 in (5), the required solution is2 2 2 2

2 2 2 2 2 22 8 sin 2 cos( )

1 2 1 2 1 2x x xy x x

x

or 2 22( ) ( 4 sin cos ).

1 2y x x x x x

Ex. 13. Show that the integral equation2

0

1( ) ( ) sin ( ) ( )y x f x x t y t dt

possesses no solution for f (x) = x, but that it possesses infinitely many solutions when f (x) = 1.

Sol. Given2

0

1( ) ( ) sin ( ) ( )y x f x x t y t dt

or2

0

1( ) ( ) (sin cos cos sin ) ( )y x f x x t x t y t dt

or2 2

0 0

sin cos( ) ( ) cos ( ) sin ( ) .x xy x f x t y t dt t y t dt

... (1)

Let 2


... (2)

and2

2 0sin ( ) .C t y t dt

... (3)

Using (2) and (3), (1) reduces to 1 2( ) ( ) ( / ) sin ( / ) cosy x f x C x C x ...(4)

We now discuss two particular cases as mentioned in the problem.Case I. Let f(x) = x. Then (4) reduces to

1 2( ) ( / ) sin ( / ) cosy x x C x C x ...(5)

From (5), 1 2( ) ( / ) sin ( / ) cosy t t C t C t ...(6)Using (6), (2) becomes

21 2

1 0

sin coscos

C t C tC t t dt



2 2 2

1 20 0 0

cos sin 2 (1 cos 2 )2 2C C

t t dt t dt t dt

or 2 222 1 2

01 0 0 0

cos 2 sin 2sin sin2 2 2 2

C Ct tC t t t dt t

or 2 201 cos [2 0]

2C

C t

or C1 – C2 = 0 ... (7)

Again using (6), (3) becomes

21 2

2 0

sin cossin

C t C tC t t dt

2 2 21 2

0 0 0sin (1 cos 2 ) sin 2

2 2C C

t t dt t dt t dt

or 2 222 1 2

02 0 0 0

sin 2 cos 2cos ( cos )2 2 2 2C Ct tC t t t dt t

or 2

02 12 sin ( / 2 ) (2 0)C t C or 1 2 2 .C C ... (8)

The system of equations (7) and (8) is inconsistent and so it possesses no solution.Hence C1 and C2 cannot be determined and so (5) shows that the given integral equation

possesses no solution when f (x) = x.Case II. Let f (x) = 1. Then (4) reduces to

1 2( ) 1 ( / ) sin ( / ) cosy x C x C x ...(9)

From (9), 1 2( ) 1 ( / ) sin ( / ) cosy t C t C t ...(10)Using (6), (2) becomes

21 2

1 0

sin coscos 1

C t C tC t dt

2 2 21 2

0 0 0cos sin 2 (1 cos 2 )

2 2C C

t dt t dt t dt

or 2 2

2 1 201

0 0

cos 2 sin 2sin2 2 2 2

C Ct tC t t

or 1 20 0 ( / 2 ) (2 0)C C or C1 = C2 ... (11)Again using (6), (3) becomes

21 2

2 0

sin cossin 1

C t C tC t dt

2 2 21 2

0 0 0sin (1 cos 2 ) sin 2

2 2C C

t dt t dt t dt

or2 2

2 1 22 0

0 0

sin 2 cos 2[ cos ]2 2 2 2

C Ct tC t t

or 2 10 ( / 2 ) (2 0) 0C C or C1= C2. ... (12)From (11) and (12), we see that C1 = C2 = C , (say). Here C is an arbitrary constant. Thus,

the system (11) – (12) has infinite number of solutions C1 = C and C2 = C . Putting these valuesin (9), the required solution of given integral equation is

( ) 1 ( / ) (sin cos )y x C x x or ( ) 1 (sin cos ).y x C x x

where ( / )C C is another arbitrary constant. Since C is an arbitrary constant, we have infinitelymany solutions of (1) when f (x) = 1.



Ex. 14. Solve the following integral equation and discuss all its possible cases1

0( ) ( ) (1 3 ) ( ) .y x f x xt y t dt [Meerut 2002, 04, 06, 08, 10, 11, 12]

Sol. Given1

0( ) ( ) (1 3 ) ( ) .y x f x xt y t dt ... (1)

or1 1

0 0( ) ( ) ( ) 3 ( )y x f x y t dt x t y t dt ... (2)

Let 1

1 0( )C y t dt ... (3)

and1

2 0( ) .C t y t dt ... (4)


1 2( ) ( ) 3 .y x f x C x C ... (5)

From (5), 1 2( ) ( ) 3 .y t f t C tC ... (6)Using (6) and (3) becomes

1

1 1 20[ ( ) 3 ]C f t C t C dt or

121 101 1 20

0

( ) 32tC f t dt C t C

or 1

1 1 20

3( )2

C f x dx C C ... (7)


1

2 1 20[ ( ) 3 ]C t f t C tC dt or

1 12 31

2 1 200 0

( ) 32 3t tC t f t dt C C

or1

2 1 20

1( ) .2

C x f x dx C C ... (8)

Re-writing (7) and (8), we have

1

1 2 0

3(1 ) ( )2

C C f x dx ... (9)

and1

1 2 0

1 (1 ) ( ) .2

C C x f x dx ... (10)

The determinant of coefficients of the system of equations (9) and (10) is given by

2

3112( ) (4 ).41 1

2

D

... (11)

Hence a unique solution of the system (9) and (10) exists if and only if ( ) 0D i.e.,2. When ( ) 0,D C1 and C2 can be determined by solving (9) and (10). By putting the

values of C1 and C2 so obtained in (5), the required unique solution of (1) can be obtained. In



particular, if f (x) = 0 and 2, the only zero solution C1 = C2 = 0 is obtained from (9) and (10)and hence we get trivial solution ( ) 0x for (1). The numbers 2 are the eigenvalues of theproblem.

If 2, equations (9) and (10) become1

1 2 03 ( )C C f x dx ... (12)

and1

1 2 03 ( ) .C C x f x dx ... (13)

While if 2, equations (9) and (10) become

1

1 2 0

1 ( )3

C C f x dx ... (14)

and1

1 2 0( ) .C C x f x dx ... (15)

Equations (12) and (13) are incompatible (i.e., possess no solution) unless the given functionf (x) satisfies the condition

1 1

0 0( ) ( )f x dx x f x dx or

1

0(1 ) ( ) .x f x dx ... (16)

When condition (16) is satisfied, the equations (12) and (13) are redundant (i.e., identicaland hence possess infinitely many solutions.)

Similarly, equations (14) and (15) are incompatible unless1 1

0 0

1 ( ) ( )3

f x dx x f x dx or 1

0(1 3 ) ( ) 0.x f x dx ... (17)

When condition (17) is satisfied, the equations (14) and (15) are again redundant.We now discuss solution of (1). Two cases arise :Case I. When f (x) = 0. Then (1) reduces to homogeneous equation

1

0( ) (1 3 ) ( ) .y x xt y t dt ... (18)

Then if 2, (1) has only trivial solution ( ) 0,y x as mentioned above.

For non-trivial solution of (18), we have 2. Hence the eigenvalues are 2.

To find eigenfunction corresponding to 2, we use (12) and (13) with f (x) = 0. Thesegive C1 = 3C2 and so (5) becomes

y (x) = 2 (3C2 – 3x C2) = 6C2 (1– x) = A (1 – x),where A (= 6C2) is an arbitrary constant.

Thus the function 1– x (or any conveneint non-zero multiple of that function) is theeigenfunction corresponding to the eigenvalue 2.

Next, to find eigenfunction corresponding to 2, we use (14) and (15). With f (x) = 0,these give C1 = C2 and so (5) becomes

1( ) 2 (1 3 ) (1 3 ),y x C x B x

where B = (– 2C1) is an arbitrary constant.Thus the function 1 – 3x (or any convenient non-zero multiple of that function) is the

eigenfunction corresponding to the eigenvalue 2.



Case II. Let ( ) 0.f x Then (1) is non-homogeneous integral equation. Three cases arise.

(i) When 2. (1) possesses a unique solution as explained above.

(ii) When 2.= Equations (12) and (13) show that no solution exists unless f (x) is orthogonalto 1 – x over the relevant interval (0, 1), that is, unless f (x) is orthogonal to the eigenfunctioncorresponding to 2. When f (x) satisfies this restriction, equations (12) and (13) are identicaland these give us

1

1 2 03 ( ) .C C f x dx

Putting this value of C1 in (5), we get

1

2 20( ) ( ) [3 ( ) ] 3y x f x C f x dx x C

or1

20( ) ( ) 2 ( ) 6 (1 ),y x f x f x dx C x as 2

or1

0( ) ( ) 2 ( ) (1 ),y x f x f x dx A x ... (19)

where A = (6C2) is an arbitrary constant.

Thus, if 2 and 1

0(1 ) ( ) 0,x f x dx the given equation (1) possesses infinitely many

solutions given by (19).

(iii) When – 2. Equation (14) and (15) show that no solution exists unless f (x) isorthogonal to 1 – 3x over the relevant interval (0, 1), that is, unless f (x) is orthogonal to theeigenfunction corresponding to 2. When f (x) satisfies this restriction, equations (14) and(15) are identical and these give us

1

1 2 0

1 ( )3

C C f x dx Putting this value of C1 in (5), we get

1

2 20

1( ) ( ) [ ( ) ] 33

y x f x C f x dx xC

or1

20

2( ) ( ) ( ) 2 (1 3 ),3

y x f x f x dx C x as 2

or1

0

2( ) ( ) ( ) (1 3 ),3

y x f x f x dx B x ... (20)

where 2( 2 )B C is an arbitrary constant.

Thus, if 2 and 1

0(1 3 ) ( ) 0,x f x dx the given equation (1) possesses infinitely many

solutions given by (20).



EXERCISE 4 (a)1. Solve the following integral equations :

(i)11 sin

1( ) tan ( ) .xy x x e y t dt

(ii)

/ 2

0( ) sin sin cos ( ) .y x x x t y t dt

(iii)1

0( ) sec tan ( )y x x x y t dt (iv)

1 12 0

1( ) cos ( ) .1

y x t y t dtx

(v)12

0( ) sec ( ) .y x x y t dt (vi)

/ 4

/ 4( ) tan ( ) coty x t y t dt x

[Kanpur 2007]

(vii)1

0( ) cos( ln ) ( ) 1.y x q t y t dt (viii)

1

0

1( ) ln ( ) 1.p

y x y t dtt

(p > – 1).

2. Solve the following integral equations :

(i)1

0

6( ) ( ln ln ) ( ) (1 4 ), where ln log5 ey x x t t x y t dt x t t

(ii)2

0( ) | | sin ( ) .y x t x y t dt x

(iii)1

0( ) (1 ) ( ) .y x x x t y t dt (iv)

0( ) (1 sin sin ) ( ) .y x x x t y t dt

[Meerut 2006, 10, 11; Kanpur 2011] [Kanpur 2008]

(v)1 3

0( ) (4 ) ( ) .y x xt x y t x (vi)

1 2

0( ) – (4 ) ( )y x xt x y t dt x

3. Express the solution of the integral equation 1

0( ) ( ) (1 3 ) ( )y x f x xt y t dt in the form

1

0( ) ( ) ( , ; ) ( ) ,y x f x x t f t dt when 2.

4. (a) Show that the characteristic values of for the equation 2

0( ) sin ( ) ( )y x x t y t dt

are 1 1/ and 2 1/ , with corresponding characteristic functions of the form y1(x) = sin x+ cos x and y2 (x) = sin x – cos x.

(b) Obtain the most general solution of the equation 2

0( ) ( ) sin ( ) ( )y x f x x t y t dt

when f (x) = x and when f (x) = 1, under the assumption that 1/ .

(c) Prove that the equation 2

0

1( ) ( ) sin ( ) ( )y x f x x t y t dt

possesses no solution when f (x) = x, but that it possesses infinitely many solutions when f (x) = 1.Determine all such solutions.

5. Consider the equation 2

0( ) ( ) cos ( ) ( ) .y x f x x t y t dt

(a) Determine the characteristic values of and the characterisitc functions.

(b) Express the solution in the form 2

0( ) ( ) ( , ; ) ( ) ,y x f x x t f t dt

when is not characteristic value.



(c) Obtain the general solution (when it exists) if f (x) = sin x, considering all possiblecases.

6. Solve the equation ( )( ) 1 ( ) ,i w x ty x e y t dt

considering separately all exceptional

cases.

7. Obtain an approximate solution of the integral equation 12

0( ) sin ( ) ( ) ,y x x x t y t dt

by replacing sin (x t) by the first two terms of its power series development3( )sin ( ) ( ) ...

3!xtx t x t

8. Solve the integral equation 2

0( ) ( ) cos ( ) ( )y x f x x t y t dt

and find the condition

that f (x) must satisfy in order that this equation has a solution when is an eigenvalue. Obtain thegeneral solution if f (x) = sin x considering all possible cases.


0( ) cos3 cos( ) ( ) .x x x t t dt Discuss all the cases.

[Meerut 2000]

10. Find the eigenvalues of the equation 2

0( ) ( ) sin( ) ( )u x f x x t u t dt

(Kanpur 20111)

ANSWERS1. (i) y (x) = tanx. (ii) ( ) {2 /(2 )} sin , 2.y x x

(iii) ( ) sec tan { /(1 )} sec1, 1.y x x x

(iv)2

2

1( ) , 1.8 ( 1)1

y xx

(v) 2( ) sec { /(1– )}tan1, 1y x x .

(vi) ( ) cot ( ) / 2.y x x (vii) 2 2( ) (1 ) /(1 ).y x q q

(viii) ( ) 1/{1 ( 1)}.y x p

2. (i)2 2

26 2 ( / 4) ln( ) (1 4 )5 1 (29 / 48)

x xy x x

, where ln logex x

(ii) 3( ) sin .y x x x (iii) 2( ) [10 (6 ) ]12 24

y x x x

(iv)2

21( ) 2 (1– / 2) (1– 2 )sin2(1 ) (1 / 2) 4

y x x x

(v) 3 2( ) [15(4 ) 30 ] / (60 65 4 ).y x x

(vi) 2 2( ) [6 (3 ) 9 ] / (18 18 ).y x x x

3. 24 3( , ; ) [1 ( ) 3(1 ) ], 2.

24x t x t xt

4. (b) 2 2

2 2 2 22 2( ) ; ( ) sin cos ; ( ) 1; ( ) 1.

1 1f x x y x x x x f x y x

(c) ( ) 1; ( ) 1 (cos sin ).f x y x C x x



5. (a) 1 1 2 21/ , ( ) cos ; 1/ , ( ) sin .y x x y x x

(b) 2 2cos( ) cos( ) 1( , ; ) , if = ± .

1x t x tx t

(c) sin( )

1xy x

if 1

1( ) sin cos ,2

y x x A x A arbitrary, if 1 ;

No solution if 1

6.2 sin( ) 1 ,

(1 2 )i xy x e

if 1 , 0;2

2( ) 1 ,1 2

i xy x e

if 1 , 0;

2

No solution if 1/ 2 .

7. 2 3( ) 0.363 0.039 ,y x x x x when stands for approximately..10. Required eigenvalues are 1/ and – (1/).

4.3 FREDHOLM ALTERNATIVEIn Art. 4.1, we have seen that, if the kernel is separable, the problem of solving an integral

equation of the second kind reduces to that of solving an algebraic system of equations. Althoughthe integral equation with separable kernel are not found frequently in practice, yet the resultsderived for such equations are essential to study integral equations of more general type.Furthermore, any reasonably well-behaved kernel can be expressed as an infinite series of degeneratekernels.

When an integral equation cannot be solved in closed form, then we have to use approximatemethods to solve a given integral equation. However any approximate method can be employedwith confidence only if the existence of the solution is known in advance. The Fredholm theoremsproved in this chapter will provide an assurance for the existence of the solution of a given integralequation. The basic theorems of the general theory of integral equations given by Fredholm,correspond to the basic theorems of linear algebraic systems. Fredholm’s classical theory will beexplained in Chapter 6 for general kernels. In the present chapter we shall study integral equationswith separable kernels and make use of the well known results of linear algebra.

Proceed exactly as in Art. 4.1 upto equation (10). *Then proceed as follows. We see that therequired solution of (1) depends on the determinant ( )D given by (10).

Two case arises :Case (i) If ( ) 0,D then the system of equations (B1), (B2) ...., (Bn) has only one solution,

given by Cramer’ rule.

1 1 2 2( ... ) / ( ),i i i ni nC D D D D i = 1, 2, ..., n. ... (11)

where Dki denotes the cofactor of the (k, i) the element of the determinant ( ).D Substituting thevalue of Ci given by (11) in (5), the unique solution of integral equation (1), is given by

1 1 2 21

...( ) ( ) ( )

( )

ni i ni n

ii

D D Dy x f x f x

D

... (12)

Also observe that the corresponding homogeneous integral equation

( ) ( , ) ( )b

ay x k x t y t dt ... (13)

has only the trivial solution y (x) = 0

* In the entire discussion of the present article for equations (1) to (10), please refer Art. 4.1.



Substituting the values of 1 2, , ..., n given by (8) in (12), the unique solution of (1) can bere-written as

1 2 21( ) ( ) ( ) ( ) ... ( ) ( ) ( )

( )

nb

i i i ni n iiay x f x D g t D g t D g t f x f t dt

D

... (14)

Let us consider the following determinant of (n + 1)th order

1 2

1 11 12 1

2 21 22 2

1 2

0 ( ) ( ) ......... ( )

( ) 1 ......... –

( , : ) ( ) 1 .........

.....................................................................( ) ......... 1

n

n

n

n n n nn

f x f x f x

g t

D x t g t

g t

... (15)

Now, by developing the determinant (15) by the elements of the first row and the correspondingminors of the first column, we have

1 2 21( ) ( ) ... ( ) ( ) ( , , )

n

i i i ni n iiD g t D g t D g t f x D x t

Hence (14), can be re-written as

( , : )( ) ( ) ( )( )

b

a

D x ty x f x f t dtD

or ( ) ( ) ( , : ) ( ) ,b


where ( , : ) ( , : ) / ( )R x t D x t D ... (17)

The function ( , : )R x t is known as the resolvent (or reciprocal) kernel of the given integral

equation (1). We note that the only possible singular points of ( , : )R x t in the -plane are theroots of the equation ( ) 0,D i.e., the eigenvalues of the kernel K (x, t).

In view of the above discussion, we have the following basic Fredholm theorem.Fredholm Theorem. The inhomogeneous Fredholm integral equation (1) with a separable

kernel has unique solution, given by (16). The resolvent kernel ( , : )R x t is given by the quotient(17) of two polynomials.

Case (ii) If ( ) 0,D then (1) has no solution in general, because an algebraic system with

( ) 0D can be solved only for some particular values of 1 2, ,..., .n In order to discuss thissituation, we re-write the system of equations (B1), (B2), ..., (Bn) in matrix form as follows :

( ) ,I A C B ... (18)

where

11 12 1 1 1

21 22 2 2 2

1 2

.....

....., , ,

..... .......... ..... ..... ..........

n

n

n nn n nn

C

CA B C

C

... (19)



and I is the unit (or identity) matrix of order n. Since ( ) 0,D for each nontrivial solution of thehomogeneous system

( ) 0I A C ... (20)there corresponds a nontrivial solution (an eigenfunction) of the homogeneous integral equation(13). From linear algebra, we know that if is equal to certain eigenvalue 0 for which the

determinant 0 0( ) | |D I A has the rank ,1 ,p p n then there are r (= n – p) linearlyindependent solutions of the algebraic system (B1), (B2), ..., (Bn). r is known as the index of the

eigenvalue 0 . The same result holds for the homogeneous integral equation (13). Let these r

linearly independent solutions be also denoted by 01 02 0( ), ( ), ..., ( )ry x y x y x and let us assume thatthey have been normalized. Then, to each eigenvalue 0 of index r (= n – p), there corresponds a

solution 0 ( )y x of (3) of the form

0 1( ) ( ),

r

k o kky x a y x

where ak are arbitrary constants.Let m be the multiplicity of the eigenvalue 0 , i.e., ( ) 0D has m equal roots . Then

from linear algebra, we know that by using the elementary transformations on the determinant| |,I A we shall get at most m + 1 identical rows and this maximum is obtained only when A is

symmetric. It follows that the rank p of 0( )D is greater than or equal to n – m and hence

( ),r n p n n m i.e., ,r m

where the equality holds only when i j j i

Thus we have proved the Fredholm theorem, namely, if 0 is a root of multiplicity 1m of the equation ( ) 0,D then the homogeneous equation (13) has r linearly independentsolutions; r is the index of the eigenvalue such that 1 .r m

The number r and m are known as the geometric multiplicity and algebraic multiplicityrespectively. Since ,r m if follows that the algebraic multiplicity of an eigenvalue is always greaterthan or equal to its geometric multiplicity.

We now proceed with the situation when the inhomogeneous Fredholm integral equation (1)has solutions even when ( ) 0.D To this end, we first define the transpose (or adjoint) of theequation (1). The integral equation

( ) ( ) ( , ) ( )b

az x f x K t x z t dt ... (21)

is known as the transpose (or adjoint) of the integral equation (1). Note that the relation between(1) and its transpose (21) is symmetric, since (1) is the transpose of (21).

Since the separable kernel K (x, t) of (1) is given by (2), it follows that the kernel K (t, x) ofthe transposed equation has the expansion

1

( , ) ( ) ( )n

i iiK t x f t g x

... (22)



Proceeding as before in Art. 4.1 and the present article, we shall arrive at the algebraic system

( ) ,TI A C B ... (23)

where AT denotes the transpose of matrix A and where Ci and i are now given by thefollowing relations

( ) ( )b

i iaC f t y t dt and ( ) ( )

b

i iaf t f t dt ... (24)

Note carefully that the determinant ( )D for the system (23) is the same function (10) exceptthat there has been interchange of rows and columns in view of the interchange in the functionsfi and gi. Hence, the eigenvalues of the transposed equation (21) are the same as those of theoriginal equation (1). It follows that the transposed equation (21) also possesses a unique solutionwhenever (1) does.

On the otherhand, from linear algebra, we know that the eigenfunctions of the homogeneoussystem

( ) 0,TI A C

are different from the corresponding eigenfunctions of the system (20). The same result is alsoapplicable to the eigenfunctions of the transposed integral equation. Since the index r of eigenvalue

0 is the same in both these systems, the number of linearly independent eigenfunctions is also r

for the transposed system. Let these r linearly independent solutions be denoted by 01 02 0, , ..., rz z z

and let us also assume that they have been normalized. Then, any solution 0( )z x of the transposedhomogeneous integral equation

( ) ( , ) ( )b

az x K t x z t dt ... (25)

corresponding to the eigenvalue 0 is of the form

0 00 1( ) ( ),

r

i iz x b z x

where bi are arbitrary constants.

We now prove that eigenfunctions ( )y x and ( )z x corresponding to distinct eigenvalues 1

and 2 , respectively, of (13) and its transpose (25) are orthogonal.

Indeed, by definition of eigenfunction, we have

1( ) ( , ) ( )b

ay x K x t y t dt ... (26)

and 2( ) ( , ) ( )b

az x K t x z t dt ... (27)

Multiplying both sides of (26) by 2 ( )z x and then integrating w.r.t. ‘x’ over the interval(a, b), we get

2 1 2( ) ( ) ( ) ( , ) ( )b b b

a a ay x z x dx z x K x t y t dt dx



or 2 1 2( ) ( ) ( ) ( , ) ( )b b b

a a ay x z x dx y t K x t z x dx dt

[on interchanging the order of integration]

or 2 1 2( ) ( ) ( ) ( , ) ( )b b b

a a ay x z x dx y x K t x z t dt dx

... (28)

Again, multiplying both sides of (27) by ( )y x and then integrating w.r.t. ‘x’ over theinterval (a, b), we get

1 1 2( ) ( ) ( ) ( , ) ( )b b b

a a ay x z x dx y x K t x z t dt dx

... (29)

From (28) and (29), we obtain

2 1( ) ( ) ( ) ( )b b

a ay x z x dx y x z x dx or 2 1( – ) ( ) ( ) 0

b

ay x z x dx

Since 2 1, we have ( ) ( ) 0,b

ay x z x dx

showing that the eigenfunction ( )y x of (13) and eigenfunction ( )z x of (25) are orthogonal.

We now proceed to discuss the solution of (1) for the case ( ) 0.D In what follows, we

shall prove that the necessary and sufficient condition for (1) to possess a solution for 0 ,

a root of ( ) 0,D is that f (x) be orthogonal to the r eigenfunctions 0iz of the transposedequation (25).

Proof. The condition is necessary. Suppose that (1) for 0 admits a certain solutiony(x). Then, we have

0( ) ( ) ( , ) ( )b

ay x f x K x t y t dt or 0( ) ( ) ( , ) ( ) .

b

af x y x K x t y t dt

Multiplying both sides of the above equation by 0 ( )iz x and then integrating w.r.t. ‘x’ overthe interval (a, b), we get

0 0 0 0( ) ( ) ( ) ( ) ( ) ( , ) ( )b b b b

i i ia a a af x z x dx y x z x dx z x K x t y t dt dx

0 0 0( ) ( ) ( ) ( , ) ( )b b b

i ia a ay x z x dx y t K x t z x dx dt

[on interchanging the order of integration]

0 0 0 0( ) ( ) ( ) ( ) ( ) ( , ) ( )b b b b

i i ia a a af x z x dx y x z x dx y x K t x z t dt dx

= 0, ... (30)

because 0 and 0 ( )iz x are eigenvalues and corresponding eigenfunctions of the transposed equation.The condition is sufficient. In what follows we shall make use of well known results of

linear algebra Actually, the corresponding condition of orthogonality for linear-algebraic systemshows that the inhomogeneous system (18) reduces to only n – r independent equations. This



implies that the rank of the matrix ( )I A is exactly p (= n – r) and hence the system (B1), (B2),..., (Bn) or (18) is soluble. Substituting this solution in (5), we arrive at the required solution for (1).

Finally, we note that the difference of any two solutions of (1) is a solution of the homogeneousequation (13). Hence, the most general solution of (1) has the form.

y (x) = Y (x) + k1 y01 (x) + k2 y02 (x) + ... + kr yor (x),where Y (x) is a suitable linear combination of the functions f1 (x), f2 (x), ..., fn (x).

We have thus proved the theorem that, if 0 is a root of multiplicity 1m of the equation

( ) 0,D then the inhomogeneous equation (1) has a solution if and only if f (x) is orthogonal toall the eigenfunctions of the transposed equation.

Combining all the results of this article, we have the following theorem.Fredholm Alternative Theorem. Either the integral equation

( ) ( ) ( , ) ( )b

ay x f x K x t y t dt ... (i)

with fixed possesses one and only one solution y (x) for arbitrary - functions f (x) andK (x, t), in particular the solution y = 0 for f = 0; or the homogeneous equation

( ) ( , ) ( )b

ay x K x t y t dt ... (ii)

possesses a finite number r of linearly independent solutions yoi , i = 1, 2, ..., r. In the first case, thetransposed inhomogeneous equation

( ) ( ) ( , ) ( )b

az x f x K t x z t dt ... (iii)

also possesses a unique solution. In the second case, the transposed homogeneous equation

( ) ( , ) ( )b

az x K t x z t dt ... (iv)

also has r linearly independent solutions 0 , 1, 2, ..., ;iz i r the inhomogeneous integral equation

(i) has a solution if and only if the given function f (x) satisfies the r conditions

0( ) ( ) 0, 1, 2, ...,b

iaf x z x dx i r ... (v)

In this case the solution of (i) is determined only upto an additive linear combination 01

.r

i iik y

We now illustrate the above theorem with help of the following solved examples.4.4 SOLVED EXAMPLES BASED ON ART. 4.3.

Ex. 1. Show that the integral equation2

0

1( ) ( ) sin( ) ( )y x f x x t y t dt

possesses no solution for f (x) = x, but that it possesses infinitely many solutions when f (x) = 1.

Sol. Given2

0

1( ) ( ) sin( ) ( )y x f x x t y t dt

... (1)

Comparing (1) with2

0( ) ( ) ( , ) ( ) ,y x f x K x t y t dt

... (2)

we have 1/ and ( , ) sin( ) sin cos cos sinK x t x t x t x t



Let K (x, t) = f1 (x) g1 (t) + f2 (x) g2 (t). So, for the given problem, we havef1 (x) = sin x, f2 (x) = cos x, g1 (t) = cos t, g2 (t) = sin t.

But2

0( ) ( ) , , 1,2i j i jg t f t dt i j

222 2

11 1 10 00

sin( ) ( ) cos sin 02

tg t f t dt t t dt

2 2 2212 1 20 0 0

1 cos 2( ) ( ) cos2

tg t f t dt t dt dt

Similarly, 21 and 22 0.

11 12 2 2

21 22

1 1 –( ) 1 .

1 1D

The eigenvalues are given by ( ) 0D i.e., 2 21 0

the eigenvalues are 1 1/ and 2 1/ .

But here (1) contains 1 1/ . Thus ( ) 0D and hence (1) will not possess a uniquesolution.

It follows that (1) will possess either no solution or infinite number of solutions. We nowproceed to examine these facts.

Let us find the eigenfunctions of the transposed equation (note that the kernel is symmetric)2

0

1( ) sin( ) ( )y x x t y t dt

... (3)

The algebraic system corresponding to (3) (given by 1( ) and 2( ) of Art. 4.1) is

11 1 12 2(1 ) 0C C and – 21 1 22 2(1 ) 0C C

i.e.,1 2 0C C and

1 2 0C C ... (4)

Here 1/ . Hence (4) gives C1 = C2.

Then the corresponding eigenfunction 1( )z x is given by2

1 1 1 2 21( ) ( ) (1/ ) ( ) ( )i ii

z x C f x C f x C f x

... (5)

1 1 1( ) (1/ ) ( sin cos ) (sin cos )z x C x C x C x x , where 1 /C C

When 1/ . Hence (4) given C1 = – C2.

As, before the corresponding eigenfunction 2 ( )z x is given by

2 ( ) (sin cos ),z x C x x when 1 /C C

Discussion of solution of (1) when f (x) = x. Then, we have2 2

10 0( ) ( ) (sin cos ) 0f x z x dx Cx x x dx

and2 2

20 0( ) ( ) (sin cos ) 0.f x z x dx C x x x dx



Hence (1) will possess infinite solution given by

1 1 2 2( ) ( ) ( ) ( )y x f x C z x C z x

i.e., 1 2( ) (sin cos ) (sin cos )y x x C C x x C C x x

i.e., ( ) cos sin ,y x x A x B x

where 1 2( )A C C C C and 1 2( – )B C C C C are arbitrary constants.Discussion of solution of (1) when f (x) = 1. Then, we have

2 2

10 0( ) ( ) (sin cos ) 0,f x z x dx C x x dx

showing that f (x) is not orthogonal to 1( )z x and hence (1) will possess no solution.Remark. For an alternative solution, refer Ex. 13 on page 4.13.

Ex. 2. Solve 1

0( ) ( ) (1 3 ) ( )y x f x xt y t dt .

Sol. Given1 1

0 0( ) ( ) ( ) 3 ( )y x f x y t dt x t y t dt ... (1)

Let1

1 0( )C y t dt ...(2)

and1

2 0( ) .C t y t dt ... (3)


1 2( ) ( ) 3y x f x C x C ... (4)

From (4) 1 2( ) ( ) 3y t f t C t C ... (5)

Let1

10( )f t dt f and

1

20( )t f t dt f ... (6)


1

1 1 20( ) 3C f t C t C dt or

1 1 1 2(3 / 2) ,C f C C by (6)

or 1 2 1(1 ) (3/ 2)C C f ... (7)Using (5), (3) reduces to

1

2 1 20( ) 3C t f t C t C dt or 2 2 1 2(1/ 2) ,C f C C by (6)

or 1 2 2(1/ 2) (1 )C C f ... (8)

Here21 3 / 2 1( ) (4 )

/ 2 1 4D

... (9)

Therefore, the inhomogeneous equation (5) will have a unique solution if and only if 2. The required solution can be obtained by solving (7) and (8) for C1 & C2 and substituting thevalues of 1C and C2 so obtained in (4).



Then the homogeneous equation1

0( ) (1 3 ) ( )y x xt y t dt

has only the trivial solution.Let us now examine the case when is equal to one of the eigenvalues and examine the

eigenfunctions of the transposed homogeneous equation (note that the kernel is symmetric)1

0( ) (1 3 ) ( )y x xt y t dt ... (10)

The algebraic system corresponding to (10) is given by (7) and (8) with f1 = f2 = 0, i.e., by

1 2(1 ) (3/ 2) 0C C and 1 2(1/ 2) (1 ) 0C C ... (11)

Case (i) When 2, then the algebraic system (11) gives C1 = 3C2. Then, the corresponding

eigenfunction 1( )z x is given by (5) with f (x) = 0 and 2.

1 1 2 1( ) 2 6 6 (1 ) (1 ),z x C xC C x C x

where 1( 6 )C C is an arbitrary constant. It follows from the Fredholm alternative theorem that the

integral equation1

0( ) ( ) 2 (1 3 ) ( )y x f x x t y t dt

will possess a solution if f (x) satisfies the condition1

10( ) ( ) 0,f x z x dx i.e.,

1

0(1 ) ( ) 0x f x dx

Case (ii) When 2, then the algebraic system (11) given C1 = C2. Then, the corresponding

eigenfunction 2 ( )z x is given by (5) with f (x) = 0 and 2.

2 1 1 1( ) 2 6 2 (1 3 ),z x C xC C x

where C = (–2C1) is an arbitrary constant. It follows from the Fredholm alternative theorem thatthe integral equation

1

0( ) ( ) 2 (1 3 ) ( )y x f x x t y t dt

will possess a solution if f (x) satisfies the condition1

20( ) ( ) 0,f x z x dx i.e.,

1

0(1 3 ) ( ) 0x f x dx

Remark. For an alternative solution, refer Ex. 14 on page 4.15

EXERCISE- 4 (b)


0( ) ( ) cos( ) ( )y x f x x t y t dt

and find the condition

that f (x) must satisfy in order that this equation has a solution when is an eigenvalue. Obtain thegeneral solution if f (x) = sin x, considering all possible cases.

2. Solve the integral equation ( – )( ) 1 ( ) ,iw x ty x e y t dt

considering separately all the

exceptional cases.



4.5 AN APPROXIMATE METHOD.We propose to describe a useful method for finding approximate solutions of some special

type of integral equations. We shall explain this method with help of the following examples.

Example 1. Solve the integral equation 1

0( ) ( 1) ( )x xty x e x x e y t dt

Solution. Let us approximate the kernel by the sum of the first three terms in its Taylor series :2 3( ) ( )( , ) ( 1) 1 1

11 21 31xt xt xt xtK x t x e x

i.e., 2 3 2 4 3( , ) (1/ 2) (1/ 6) ,K x t x t x t x t which is a separable kernel.Then, the given integral equation takes the form

1 2 3 2 4 3

0

1 1( ) ( )2 6

xy x e x x t x t x t y t dt ...(1)

or3 41 1 12 2 3

0 0 0( ) ( ) ( ) ( )

2 6x x xy x e x x t y t dt t y t dt t y t dt ... (2)

Let1

1 0( )C t y t dt ... (3)

1 22 0

1 ( )2

C t y t dt ... (4)

1 33 0

1 ( )6

C t y t dt ... (5)

Then (2) gives 2 3 41 2 3

( ) xy x e x C x C x C x ... (6)

From (6), 2 3 41 2 3

( ) ty t e t C t C t C t ... (7)Substituting the value of y (t) given by (7) in (3), we get

1 2 3 41 1 2 30

( )tC t e t C t C t C t dt or

1 1 1 1 12 3 4 51 1 2 30 0 0 0 0

tC te dt t dt C t dt C t dt C t dt or

1

1 1 2 30(1/ 3) ( / 4) ( / 5) ( / 6)t tC te e C C C

or 1 2 3(5 / 4) (1/5) (1/ 6) 2 / 3C C C ... (8)Substituting the value of y (t) given by (7) in (4), we get

1 2 2 3 42 1 2 30

1 ( )2

tC t e t C t C t C t dt or

1 1 1 1 12 3 4 5 631 22 0 0 0 0 0

1 12 2 2 2 2

t CC CC t e dt t dt t dt t dt t dt

or12 31 2

2 0

1 1( ) ( ) (2 ) ( ) (2) ( )2 8 10 12 14

t t t CC CC t e t e e

[using chain rule of integration by parts to evaluate the first integral]



or C1/10 + (13/12) × C2 + C3/14 = – (1/2) × (e – 2e + 2e – 2) + 1/8or (1/5) × C1 + (13/6) × C2 + (1/7) × C2 = (9/4) – e ... (9)

Substituting the value of y (t) given by (7) in (5), we get1 3 2 3 4

3 1 2 30

1 ( )6

tC t e t C t C t C t dt

or1 1 1 1 13 4 5 6 731 2

3 0 0 0 0 0

1 16 6 6 6 6

t CC CC t e dt t dt t dt t dt t dt

or13 2

3 0(1/ 6) ( ) ( ) (3 ) ( ) (6 ) ( ) (6) ( ) 1/ 30t t t tC t e t e t e e

– (1/36) × C1 – (1/42) × C2 – (1/48) × C3[using chain rule of integration by parts to evaluate the first integral]

or 3 1 2 2(1/ 6) ( 3 6 6 6) 1/ 30 (1/36) (1/ 42) (1/ 48)C e e e e C C C

or 1 2 2(1/ 36) (1/ 42) (49 / 48) / 3 (29 /36)C C C e

or 1 2 2(1/ 6) (1/ 7) (49 /8) 2 (29 /5)C C C e ... (10)Solving (8), (9) and (10) leads to (after simplifications)C1 = – 0.5010, C2 = – 0.1671, and C3 = – 0.0422 ...(11)With these valuese, (6) gives the required approximate solution of (1) as

2 3 4( ) 0.5010 0.1671 0.0422xy x e x x x x ... (12)Now as usual, we prove that the exact solution of given equation

1

0( ) ( 1) ( )x xty x e x x e y t dt ... (13)

is given by y (x) = 1 ... (14)From (14), y (t) = 1. Then, we have

R. H.S. of (13) 1

0( 1)x xte x x e dt

110

0

( 1)xt

x x xee x x xt e x e xx

= 1 = y (x) = L.H.S. of (13)

Hence y (x) = 1 is the exact solution of (13).Using the approximate solution (12) for x = 0, x = 0.5 and x = 1.0, the values of y (x) arey (0) = 1.0000, y (0.5) = 1.0000 and y (1) = 1.0080 ... (15)

which agrees with exact solution (14) rather closely

Example 2. Consider the integral equation12

0( ) (sin ) ( ) ;y x x xt y t dt

Replacing sin xt by the first two terms of its power-series development, namely,3sin ( ) / 31 ........ ,xt xt xt obtain an approximate solution.

Solution. Proceed as in example 1.


CHAPTER 5

Method of successiveapproximations5.1 INTRODUCTION

We already know that ordinary first-order differential equation are solved by the well-knownPicard method of successive approximations. In this chapter we shall study an iterative schemebased on the same principle for linear integral equations of the second order. Throughout ourdiscussion we shall assume that the functions f(x) and K(x, t) involved in an integral equation are

- functions.5.2 ITERATED KERNELS OR FUNCTIONS. DEFINITION.

(i) Consider Fredholm integral equation of the second kind

( ) ( ) ( , ) ( ) .b


Then, the iterated kernels Kn (x, t), n = 1, 2, 3, ... are defined as follows :K1(x, t) = K (x, t), ... (2a)

1

1

and ( , ) ( , ) ( , ) , 2, 3, ...

or ( , ) ( , ) ( , ) , 2, 3, ...

bn na

bn n

a



... (2b)

(ii) Consider Volterra integral equation of the second kind

( ) ( ) ( , ) ( )x


Then the iterated kernels Kn (x, t), n = 1, 2, 3, ... are defined as follows : K1 (x, t) = K (x, t) ... (3a)

1

1

and ( , ) ( , ) ( , ) , 2, 3,...

or ( , ) ( , ) ( , ) , 2, 3, ...

xn nt

xn n

t



... (3b)

5.3 RESOLVENT KERNEL OR RECIPROCAL KERNEL.(i) Suppose solution of Fredholm integral equation of the second kind

( ) ( ) ( , ) ( )b


5.1


5.2 Method of Successive Approximations

takes the form ( ) ( ) ( , ; ) ( ) ,b

ay x f x R x t f t dt ... (2a)

or ( ) ( ) ( , ; ) ( ) ,b

ay x f x x t f t dt ... (2b)

then R (x, t; ) or (x, t; ) is known as the resolvent kernel of (1).(ii) Suppose solution of Volterra integral equation of the second kind

( ) ( ) ( , ) ( )x


takes the form ( ) ( ) ( , ; ) ( )x

ay x f x R x t f t dt ... (4a)

or ( ) ( ) ( , ; ) ( )x

ay x f x x t f t dt ... (4b)

then R (x, t: ) or (x, t; ) is known as the resolvent kernel of (3)5.4. Theorem. The mth iterated kernel Km (x, t) satisfies the relation

( , ) ( , ) ( , ) ,b

m r m ra

K x t K x y K y t dy where r is any positive integer less than m.

Proof. The mth iterated kernel Km(x, t) is defined asK1 (x, t) = K (x, t), ... (1)

and 1( , ) ( , ) ( , ) , 2, 3, ...b

m ma

K x t K x s K s t ds m ... (2)

Re-writing (2), we have 1 1 1 1( , ) ( , ) ( , ) .b

m ma

K x t K x s K s t ds ... (3)

Replacing m by m – 1 in (2), we have

1 2 2 2 2 2( , ) ( , ) ( , ) ( , ) ( , )b b

m m ma a

K x t K x s K s t ds K x s K s t ds or 1 1 1 2 2 2 2( , ) ( , ) ( , ) .

bm m

aK s t K s s K s t ds ... (4)


1 1 2 2 2 2 1( , ) ( , ){ ( , ) ( , ) }b b

m ma a

K x t K x s K s s K s t ds ds or 1 1 2 2 2 2 1( , ) ( , ) ( , ) ( , )

b bm m

a aK x t K x s K s s K s t ds ds

Proceeding likewise, we obtain

1 1 2 2 3 1 1 1 2 1( , ) ... ( , ) ( , ) ( , ) ... ( , ) ...b b b

m m ma a a

K x t K x s K s s K s s K s t ds ds ds or 1 1 2 1 1 –1 –1 2 1

( –1) orderintegral

( , ) ... ( , ) ( , )... ( , ) ( , )... ... ( , ) ...b b b

m r r r r m ma a am th

K x t K x s K s s K s s K s s K s t ds ds ds

... (5)


Method of Successive Approximations 5.3

Note that R.H.S. of (5) is a multiple integral of order m – 1.Proceeding as above, we may also write.

1 1 2 1 2 11

( 1) orderintegral

( , ) ... ( , ) ( , )... ( , ) ... ...(6)b b b

r r ra a ar th

K x y K x u K u u K u y du du du

and 1 1 2 1 – –1

( – –1) orderintegral

( , ) ... ( , ) ( , )... ( , ) ... .b b b

m r m r m ra a a

m r th

K y t K y K K t d d d ! "# # # # # # # ... (7)

Now, ( , ) ( , )b

r m ra

K x y K y t dy

1 1 2 1 1 2 1... ( , ) ( , )... ( , ) ...b b b b

r ra a a aK x u K u u K u y du du du

1 1 2 1 1 2 1... ( , ) ( , )... ( , ) ... , by (6) and (7)b b b

m r m ra a aK y k K t d d d dy

# # # # # # #

or 1 1 2 1( , ) ( , ) ... ( , ) ( , )... ( , )b b b b

r m r ra a a a

K x y K y t dy K x u K u u K u y 1 1 2 1 1 2 1 1 2 1( , ) ( , )... ( , ) ... ...m r m r rK y K K t d d d dydu du du # # # # # # # ... (8)

[on changing the order of integration]Note that the order of the multiple integral on R.H.S. of (8) is 1 + (r – 1) + (m – r – 1), that

is, m – 1. We have already proved that the order of the multiple integral on R.H.S. of (5) is also

m – 1. Thus, multiple integrals involved in Km (x, t) and ( , ) ( , )b

r m ra

K x y K y t dy are both of the

same order, namely, (m – 1) th.Now, changing the variables of integrations in (8) without changing the limits of integration

according to the following scheme

1 2 1 1 2 1

2 1 11 1 2

... ...

r m r

r mr rr

u u u y

s s ss s ss

# # #

we obtain

1 1 2 –1

( –1) orderintegral

( , ) ( , ) ... ( , ) ( , )... ( , )b b b b

r m r r ra a a a

m th

K x y K y t dy K x s K s s K s s

× K (sr,sr+1) ... K (sm–1,t) dsm–1 ... ds2 ds1. ... (9)

From (5) and (9), we obtain ( , ) ( , ) ( , ) .b

m r m ra

K x t K x y K y t dy 5.5. SOLUTION OF FREDHOLM INTEGRAL EQUATION OF THE SECOND KIND BY

SUCCESSIVE SUBSTITUTIONS. [Garhwal 1996; Meerut 2000, 02]

Theorem. Let ( ) ( ) ( , ) ( )b




be given Fredholm integral equation of the second kind. Suppose that (i) Kernel K(x, t) | 0, is real and continuous in the rectangle R, for which

, .a x b a t b Also let | K (x, t) | M, in R. ...(2)

(ii) f (x) | 0, is real and continuous in the interval I, for which .a x b Also, let | f (x) | N, in I. ...(3) (iii) is a constant such that | | < 1/M (b – a) ...(4)Then (1) has a unique continuous solution in I and this solution is given by the absolutely

and uniformly convergent series

21 1 1( ) ( ) ( , ) ( ) ( , ) ( , ) ( ) ...

b b b

a a ay x f x K x t f t dt K x t K t t f t dt dt ... (5)

Proof. Re-writing (1), we have

1 1 1( ) ( ) ( , ) ( ) .b


Replacing x by t in (6), we get 1 1 1( ) ( ) ( , ) ( ) .b

ay t f t K t t y t dt ... (7)

Substituting the above value of y (t) in (1), we get

1 1 1( ) ( ) ( , ) ( ) ( , ) ( ) .b b

a ay x f x K x t f t K t t y t dt dt

or 21 1 1( ) ( ) ( , ) ( ) ( , ) ( , ) ( ) .

b b b

a a ay x f x K x t f t dt K x t K t t y t dt dt ... (8)

Re-writing (7), we have 2 2 2( ) ( ) ( , ) ( ) .b


Replacing t by t1 in (9), we have 1 1 1 2 2 2( ) ( ) ( , ) ( ) .b


Substituting the above value of y (t1) in (8), we get

( ) ( ) ( , ) ( )b

ay x f x K x t f t dt 2

1 1 1 2 2 2 1( , ) ( , ) ( ) ( , ) ( )b b b

a a aK x t K t t f t K t t y t dt dt dt

or 21 1 1( ) ( ) ( , ) ( ) ( , ) ( , ) ( )

b b b

a a ay x f x K x t f t dt K x t K t t f t dt dt

31 1 2 2 2 1( , ) ( , ) ( , ) ( ) .

b b b

a a aK x t K t t K t t y t dt dt dt ... (11)

Proceeding likewise, we have

21 1 1( ) ( ) ( , ) ( ) ( , ) ( , ) ( )

b b b


1 2 1 1 1 1... ( , ) ( , )... ( , ) ( ) ...b b bn

n n n na a a

K x t K t t K t t f t dt dt dt + Rn + 1(x), ... (12)

where Rn + 1 (x) 11( , ) ( , )...

b bn

a aK x t K t t 1 1( , ) ( ) ... .

bn n n n

aK t t y t dt dt dt ... (13)

Now, let us consider the following infinite series



21 1 1( ) ( , ) ( ) ( , ) ( , ) ( ) ...

b b b

a a af x K x t f t dt K x t K t t f t dt dt ... (14)

In view of the assumptions (i) and (ii), each term of the series (14) is continuous in I. Itfollows that the series (14) is also continuous in I, provided it converges uniformly in I.

Let Un(x) denote the general term of the series (14) i.e., let

1 2 1 1 1 1( ) ( , ) ( , )... ( , ) ( ) ...b b bn

n n n n na a a

U x K x t K t t K t t f t dt dt dt ...(15)

From (15), we have

1 2 1 1 1 1| ( ) | | ( , ) ( , )... ( , ) ( ) ... |b b bn

n n n n na a a

U x K x t K t t K t t f t dt dt dt | ( ) | | | ( ) ,n n n

nU x NM b a using (2) and (3) ...(16)

The series of which this is a general term converges only when| | M (b – a) < 1 or | | < 1/M (b – a)

which holds in view of assumption (iii).It follows that the series (14) converges absolutely and uniformly when condition (4) holds.If (1) has a continuous solutions, clearly it must be expressed by (12). If y (x) is continuous

in I, | y (x) | must have a maximum value Y. Thus,

| ( ) | .y x Y ... (17)Now, from (13), we have

11 1 1 1| ( ) | | ( , ) ( , ).... ( , ) ( ) ... |

b b bnn n n n n

a a aR x K x t K t t K t t y t dt dt dt

1 1 1

1| ( ) | | | ( )n n nnR x YM b a , using (2) and (17)

Since (4) holds, so 1lim ( ) 0.nnR x

It follows that the function y(x) satisfying (12) is the continuous function given by the series(14). This proves what we wished to prove.5.6. SOLUTION OF VOLTERRA INTEGRAL EQUATION OF THE SECOND KIND BY

SUCCESSIVE SUBSTITUTIONS. [Garhwal 1996; Meerut 2000, 01]

Theorem. Let ( ) ( ) ( , ) ( )x


be given Volterra integral equation of the second kind. Suppose that

(i) kernel K (x, t) 0, is real and continuous in the rectangle R, for which , .a x b a t b Also, let | K (x, t) | M, in R. ... (2)

(ii) f (x) 0, real and continuous in the interval I, for which .a x b Also, let | f (x) | N, in I. ... (3)

(iii) is a constant. ... (4)Then (1) has a unique continuous solution in I and this solution is given by the absolutely

and uniformly convergent series



21 1 1( ) ( ) ( , ) ( ) ( , ) ( , ) ( ) ...

x x t

a a ay x f x K x t f t dt K x t K t t f t dt dt ... (5)

Proof. Re-writing (1), we have

1 1 1( ) ( ) ( , ) ( ) .x


Replacing x by t in (6), we get y (t) 1 1 1( ) ( , ) ( ) .t

af t K t t y t dt ... (7)

Substituting the above value of y (t) in (1), we get

1 1 1( ) ( ) ( , ) ( ) ( , ) ( ) .x t

a ay x f x K x t f t K t t y t dt dt

or 21 1 1( ) ( ) ( , ) ( ) ( , ) ( , ) ( ) .

x x t

a a ay x f x K x t f t dt K x t K t t y t dt dt ... (8)

Re-writing (7), we have 2 2 2( ) ( ) ( , ) ( ) .t


Replacing t by t1 in (9), we have1

1 1 1 2 2 2( ) ( ) ( , ) ( ) .t


Substituting the above value of y (t1) in (8), we get

( ) ( ) ( , ) ( )x

ay x f x K x t f t dt

121 1 1 2 2 2 1( , ) ( , )[ ( ) ( , ) ( ) ]

x t t

a a aK x t K t t f t K t t y t dt dt dt

or12

1 1 1( ) ( ) ( , ) ( ) ( , ) ( , ) ( )x x t


131 1 2 2 2 1( , ) ( , ) ( , ) ( ) .

x t t

a a aK x t K t t K t t y t dt dt dt ... (11)

Proceeding likewise, we have

121 1 1( ) ( ) ( , ) ( ) ( , ) ( , ) ( ) ...

x x t


21 2 1 1 1 1 1( , ) ( , )... ( , ) ( ) .... ( ),

nx t tnn n n n n

a a aK x t K t t K t t f t dt dt dt R x

... (12)

where11

1 1 1 1( ) ( , ) ( , ).... ( , ) ( ) ... .nx t tn

n n n n na a a

R x K x t K t t K t t y t dt dt dt

... (13)

Now, let us consider the following infinite series

21 1 1( ) ( , ) ( ) ( , ) ( , ) ( ) ...

x x t

a a af x K x t f t dt K x t K t t f t dt ... (14)

In view of the assumptions (i) and (ii), each term of the series (14) is continuous in I. Itfollows that the series (14) is also continuous in I, provided it converges uniformly in I. Let Un (x)denote the general term of the series (14), i.e., let

1( ) ( , ) ( , )x tm

na a

U x K x t K t t ...... 2

2 1 1 1 1( , ) ( ) ... .nt

n n n na

K t t f t dt dt dt

... (15)

From (15) we have



21 2 1 1 1 1| ( ) | | ( , ) ( , )... ( , ) ( ) ... |

nx t tnn n n n n

a a aU x K x t K t t K t t f t dt dt dt

( )| ( ) | | | ,

!

nn n

nx aU x NM

n

using (2) and (3)

or( )| ( ) | | | ,

!

nn n

nb aU x NM a x b

n

or[ ( )]| ( ) | | |

!

nn

nM b aU x N

n

, a x b ... (16)

Clearly, the series, for which the positive constant | | [ ( )]!

n nN M b an

is the general

expression for the nth term, is convergent for all values of , , , ( ).N M b a So from (16), it followsthat the series (14) converges absolutely and uniformly.

If (1) has a continuous solution, clearly it must be expressed by (12). If y (x) is continuous inI, | y(x) | must have a maximum value Y. Thus,

| ( ) | .y x Y ... (17)Now, from (13), we have

111 1 1 1| ( ) | | ( , ) ( , )... ( , ) ( ) ... |

nx t tnn n n n n

a a aR x K x t K t t K t t y t dt dt dt

1 1

1 1 1 11

( ) ( )| ( ) | | | | | , ( )( 1)! ( 1)!

n nn n n n

nx a b aR x YM YM a x b

n n

Hence 1lim ( ) 0.nnR x

It follows that the function y (x) satisfying (12) is the continuous function given by the series(14). This proves the desired result.5.7. SOLUTION OF FREDHOLM INTEGRAL EQUATION OF THE SECOND KIND BY

SUCCESSIVE APPROXIMATIONS. ITERATIVE METHOD (ITERATIVE SCHEME).NEUMANN’S SERIES. [Meerut 2006]Consider Fredholm integral equation of the second kind

( ) ( ) ( , ) ( ) .b


As a zero-order approximation to the required solution y (x), let us takey0 (x) = f (x) ... (2)

Further, if yn (x) and yn – 1 (x) are the nth order and (n – 1)th-order approximations respectively,then these are connected by

1( ) ( ) ( , ) ( ) .b

n na

y x f x K x t y t dt ... (3)

We know that the iterated kernels (or iterated functions) ( , ), ( 1, 2, 3,...)nK x t n are defined byK1 (x, t) = K (x, t) ... (4A)

and 1( , ) ( , ) ( , ) .b

n na

K x t K x z K z t dz ...(4B)

Putting n = 1 in (3), the first-order approximation y1 (x) is given by



1 0( ) ( ) ( , ) ( ) .b


But from (2), y0(t) = f (t) ... (6)Substituting the above value of y0(t) in (5), we get

1( ) ( ) ( , ) ( ) .b

ay x f x K x t f t dt ... (7)

Putting n = 2 in (3), the second-order approximation y2(x) is given by

2 1( ) ( ) ( , ) ( ) .b

ay x f x K x t y t dt

or 2 1( ) ( ) ( , ) ( ) .b

ay x f x K x z y z dz ... (8)

Replacing x by z in (7), we get 1( ) ( ) ( , ) ( ) .b

ay z f z K z t f t dt ... (9)

Substituting the above value of y1(z) in (8), we get

2 ( ) ( ) ( , )[ ( ) ( , ) ( ) ]b b

a ay x f x K x z f z K z t f t dt dz

or 22 ( ) ( ) ( , ) ( ) ( , )[ ( , ) ( ) ]

b b b

a a ay x f x K x z f z dz K x z K z t f t dt dz ... (10)

or2

2 ( ) ( ) ( , ) ( ) ( ) ( , ) ( , )b b b

a a ay x f x K x t f t dt f t K x z K z t dz dt

[On changing the order of integration in third term on R.H.S of (10)]

or 22 1 2( ) ( ) ( , ) ( ) ( , ) ( ) ,

b b

a ay x f x K x t f t dt K x t f t dt using (4A) and (4B)

or2

2 1( ) ( ) ( , ) ( ) .

bmmm a

y x f x K x t f t dt

... (11)

Proceeding likewise, we easily obtain by Mathematical induction the nth approximate solutionyn(x) of (1) as

1( ) ( ) ( , ) ( ) .

n bmn mm a


... (12)

Proceeding to the limit as n , we obtain the so called Neumann series.

1( ) lim ( ) ( ) ( , ) ( )

bmn mn m a

y x y x f x K x t f t dt

... (13)

We now determine the resolvent kernel (or reciprocal kernal) ( , ; )R x t or ( , ; )x t interms of the iterated kernels Km(x, t). For this purpose, by changing the order of integration andsummation in the so called Neumann series (13), we obtain

1

1( ) ( ) [ ( , )}] ( ) .

b mmma


... (14)

Comparing (14) with ( ) ( ) ( , ; ) ( ) ,b


here 11

( , ; ) ( , ).mmm

R x t K x t

... (16)



Determination of the conditions of convergence of (13).Consider the partial sum (12) and apply the Schwarz inequality [refer Art. 1.16, Chapter 1] to

the general term of this series. This leads us to

2 2 2| ( , ) ( ) | | ( , ) | | ( ) | .b b b

m ma a a

K x t f t dt K x t dt f t dt ... (17)

Let D = norm of 1/ 2

2( ) | ( ) | .b

af t f t dt ... (18)

Further, let 2mC denote the upper bound of the integral 2| ( , ) | ,

bm

aK x t dt

so that 2 2| ( , ) |b

m ma

K x t dt C ... (19)


2 2 2| ( , ) ( ) |b

m ma

K x t f t dt C D ... (20)

Now, applying the Schwarz inequality to relation

1( , ) = ( , ) ( , ) ,b

m ma

K x t K x z K z t dzwe get 2 2 2

1| ( , ) | | ( , ) | | ( , ) | ,b b

m ma a

K x t K x z dz K z t dz

which when integrated with respect to t, gives

2 2 21| ( , ) | ,

bm m

aK x t dt B C ... (21)

where 2 2| ( , ) | .b b

a aB K x t dx dt ... (22)

The inequality (21) gives rise to the recurrence relation2 2 2 2

1 .mmC B C ... (23)

Using (20) and (23), we get 2 2 2 2 21| ( , ) ( ) |

b mm

aK x t f t dt C D B ... (24)

showing that the general term of the partial sum (12) has a magnitude less than the quantity1

1 | | .m mDC B Hence the inifinte series (13) converges faster than the geometic series withcommon ratio | | B. It follows that, if the condition

| | 1B or 1/ 22

1| || ( , ) |

b b

a aK x t dx dt

... (25)

is satisfied, then the series (13) will be uniformly covergent.Uniqueness of solution for a given .If possible, let (1) possess two solutions y1(x) and y2(x). Then we have

1 1( ) ( ) ( , ) ( )b


and 2 2( ) ( ) ( , ) ( )b


Let y1(x) – y2(x) = (x). ... (28)Subtracting (27) from (26), we have



1 2 1 2( ) ( ) ( , )[ ( ) ( )]b

ay x y x K x t y t y t dt

or ( ) ( , ) ( ) ,b

ax K x t t dt using (28) ... (29)

which is homogeneous integral equation. Applying the Schwarz inequality to (29), we have

2 2 2 2| ( ) | | | | ( , ) | | ( ) |b b

a ax K x t dt t dt

... (30)

Integrating with respect to x, (30) gives

2 2 2 2| ( ) | | | | ( , ) | | ( ) |b b b b

a a a ax dx K x t dx dt x dx

or 2 2 2 2| ( ) | | | | ( ) | ,b b

a ax dx B x dx by (22)

or 2 2 2(1 | | | ( ) | 0,b

aB x dx ... (31)

giving ( ) 0x , using (25)or y1(x) – y2(x) = 0 or y1(x) = y2(x),showing that (1) has a unique solution

From the uniqueness of the solution of (1), we now proceed to show that the resolvent kernelR (x, t; ) is also unique. If possible, let equation (1) have, with 0 , two resolvent kernels

1 0( , ; )R x t and 2 0( , ; ).R x t In view of the uniqueness of the solution (1), an arbitrary functionf (x) satisfies the identity

0 1 0 0 2( ) ( , ; ) ( ) ( ) ( , ; ) ( ) .b b

a af x R x t f t dt f x R x t f t dt ... (32)

Setting F (x, t; 0 ) = R1 (x, t; 0 ) – R2 (x, t; 0 ), (32) reduces to

0( , ; ) ( ) 0,b

aF x t f t dt ... (33)

for an arbitrary function f (t). Let us choose f (t) = 0( , ; )F x t with fixed x. Here 0( , ; )F x t

denotes the complex conjugate of F(x, t; 0 ). Then (33) reduces to

20| ( , ; ) | 0

b

aF x t dt

F (x, t; 0 ) = 0 R1 (x, t; 0 ) – R2 (x, t; 0 ) = 0 R1 (x, t; 0 ) = R2 (x, t; 0 )showing that the resolvent kernel in unique.

The above analysis can be summed up in the following basic theorem.Theorem. To each -kernel K(x, t), there corresponds a unique resolvent kernel

R (x, t; ) which is an analytic function of , regular at least inside the circle | | < B–1, andrepresented by the power series

11

( , ; ) ( , ).mmm

R x t K x t



Furthermore, if f (x) is also an -function, then the unique -solution of the Fredholmequation

( ) ( ) ( , ) ( )b


valid in the circle 1| | B is given by the formula

( ) ( ) ( , ; ) ( ) .b

ay x f x R x t f t dt

5.8 SOME IMPORTANT THEOREMSTheorem I. Let R (x, t; ) be the resolvent (or reciprocal) kernel of a Fredholm integral equation.

( ) ( ) ( , ) ( ) ,b


then the resolvent kernel satisfies the integral equation

( , ; ) ( , ) ( , ) ( , ; ) .b

aR x t K x t K x z R z t dz

Proof. We know that R (x, t; ) is given by

11

( , ; ) ( , ),mmm

R x t K x t

... (1)

where iterated kernels (or functions) are given byK1 (x, t) = K (x, t) ... (2A)

and 1( , ) ( , ) ( , ) .b

m ma

K x t K x z K z t dz ... (2B)

Now, from (1), we have

11 2

( , ; ) ( , ) ( , )mmm

R x t K x t K x t

112

( , ) ( , ) ( , )bm

mm aK x t K x z K z t dz

, using (2A) and (2B)

1( , ) ( , ) ( , )

bnnn a

K x t K x z K z t dz

(Setting m – 1 = n)

1( , ) ( , ) ( , )

bmmm a


11

( , ) ( , ) ( , )bm

mm aK x t K x z K z t dz

1

1( , ) [ ( , )] ( , )

b mmma

K x t K z t K x z dz

[on changing the order of summation and integration]

( , ) ( , ; ) ( , ) ,b

aK x t R z t K x z dz using (1)

( , ; ) ( , ) ( , ) ( , ; ) .b

aR x t K x t K x z R z t dz

Theorem II. The series for the resolvent kernel R (x,t; ).

1

1( , ; ) ( , )m

mmR x t K x t

... (1)

is absolutely and uniformly convergent for all values of x and t in the circle 1| | .B



Proof. In addition to the assumptions of Art. 5.7, we shall use the following inequality

2 2| ( , ) | ,b

aK x t dx E E = constant. ... (2)

(2) follows from one of the conditions for the kernel K(x, t) to be an kernel. We have, therecurrence formula

1( , ) ( , ) ( , )b

m ma

K x t K x z K z t dz ... (3)

Applying the Schwarz inequality to (3), we have

2 2 21| ( , ) | | ( , ) | | ( , ) |

b bm m

a aK x t K x z dz K z t dz

... (4)

Using inequality (23) of Art. 5.7, (4) reduces to 11| ( , ) | m

mK x t C EB

so that 1 1 11| ( , ) | ( ),m m m

mK x t C E B showing that the series (1) is dominated by the geometric series with the general term

1 11 ( )m mC E B and hence the result

Theorem III. The resolvent kernel satisfies the integro-differential equation.

( , ; ) / ( , ; ) ( , ; )b

aR x t R x z R z t dz

Proof. We have 1 11 1

( , ; ) ( , ; ) ( , ) ( , )b b m n

m nm na aR x z R z t dz K x z K z t dz

In view of the result of theorem II, both the series on R.H.S. are absolutely and uniformlyconvergent. Hence we can multiply the series under the integral sign and integrate it term by term.

Thus, we obtain 21 1

( , ; ) ( , ; ) ( , )b m n

m nm naR x z R z t dz K x t

... (1)

[Using the result ( , ) ( , ) ( , )]b

m n m na

K x z K z t dz K x tOn setting m + n = p and changing the order of summation on R.H.S. of (1), we obtain

–1

– 2 –21 1 2 1

( , ) ( , )p

m n pm n pm n p n

K x t K x t

–22

( , ; )( –1) ( , )ppp

R x tp K x t

...(2)

From (1) and (2), we get the required result.5.9 SOLVED EXAMPLES BASED ON SOLUTION OF FREDHOLM INTEGRAL EQUA-

TION OF SECON KIND BY SUCCESSIVE APPROXIMATIONS OR ITERATIVEMETHOD. (Refer Art. 5.7)

Type1: Determination of iterated kernels (or functions) for ( ) ( ) ( ) ( )b

ay x = f x + K x,t y t dt.

Ex. 1. Find the iterated kernels (or functions) for the following kernels :(i) ( , ) sin( 2 ), 0 2 ,0 2 .K x t x t x t (ii) K (x, t) = ex cos t; a = 0, b = .

[Meerut 2008](iii) K (x, t) = x + sin t; a = – , b = . (iv) K (x, t) = x – t; a = 0, b = 1.

[Kanpur 2007]



Sol. (i) Iterated kernel Kn (x, t) are given byK1 (x, t) = K (x, t) ... (1)

and2

10

( , ) ( , ) ( , ) , ( 2, 3, ...)n nK x t K x z K z t dz n

... (2)

[ 0 2 , 0 2 0 and 2 ]x t a b From (1), K1 (x, t) = K (x, t) = sin (x – 2t). ... (3)Putting n = 2 in (2), we have

22 1

0( , ) ( , ) ( , )K x t K x z K z t dz

2

0sin( 2 )sin( 2 ) ,x z z t dz

using (3)

2

0

1 [cos( 2 3 ) cos( 2 )]2

x t z x t z dz

2

0

1 1 sin( 2 3 ) sin( 2 )2 3

x t z x t z

= 0, on simplification. K2 (x, t) = 0 ... (4)

Putting n = 3 in (2), we have2

3 20

( , ) ( , ) ( , ) 0.K x t K x z K z t dz

[ K2 (z, t) = 0, by (4)]

Thus, K1 (x, t) = sin (x – 2t) and Kn (x, t) = 0 for n = 2, 3, 4, ...Part (ii) Iterated kernels Kn (x, t) are given by

K1 (x, t) = K (x, t) ... (1)

and 10

( , ) ( , ) ( , ) .n nK x t K x z K z t dz

... (2)

From (1), K1 (x, t) = K (x, t) = ex cos t. ... (3)Putting n = 2 in (2), we have

2 10 0

( , ) ( , ) ( , ) cos cos ,x zK x t K x z K z t dz e z e t dz

using (3)

2 200

cos cos cos (cos sin )1 1

zx z x ee t e z dz e t z z

2 2cos ( cos sin )ax

ax ee bx dx a bx b bxa b

cos { (1/ 2) (1/ 2)}.xe t e ...(4)

12

1( , ) ( 1) cos .2

xeK x t e t

Next, putting n = 3 in (2), we have

3 20

( , ) ( , ) ( , )K x t K x z K z t dz

0

1cos cos2

x zee z e t dz

, using (3) and (4)

0

1 1 1cos cos cos ,2 2 2

x z xe e ee t e z dz e t

as before

2

23

1( , ) ( 1) cos2

xeK x t e t

... (5)



ans so on Noting (3) (4) and (5), we see that the iterated kernels are given by1

1 1( , ) ( 1) cos , 1, 2, 3, ...2

nn x

neK x t e t n

Part (iii) Iterated kernels are given byK1 (x, t) = K (x, t) ... (1)

and 1( , ) ( , ) ( , ) .n nK x t K x z K z t dz

... (2)

From (1), K1 (x, t) = K (x, t) = x + sin t. ... (3)Putting n = 2 in (2), we have

2 1( , ) ( , ) ( , ) ( sin ) ( sin )K x t K x z K z t dz x z z t dz

sin sin sin sinx z dz t z dz x t dz z z dz

2

sin cos sin cos cos2zx t z x t z z z z dz

[Integrating the last integral by method of integration by parts]= 2 x sin t + 2 , on simplification K2 (x, t) = 2 (1 + x sin t). ... (4)


3 2( , ) ( , ) ( , ) ( sin ){2 (1 sin )}K x t K x z K z t dz x z z t dz

2 ( sin sin sin sin )x xz t z z z t dz

2

2 2 sin 2 cos 2 sin cos cos2zx z x t z t z z z dz

2 24 4 sin .t

K3 (x, t) = 4 2 (x + sin t) = 4 2 K1 (x, t), using (3) ... (5)Now, putting n = 4 in (2), we have

4 3( , ) ( , ) ( , ) .K x t K x z K z t dz

2( sin ){4 ( sin )} ,x z z t dx

using (3) and (5)

24 ( sin ) ( sin )x z z t dz

24 2 (1 sin ),x t as before

K4 (x, t) = 4 2 K2(x, t), using (4) ...(6)Proceeding like wise, we have

K5 (x, t) = 4 2 K3 (x, t) = 16 4 K1 (x, t), by (5) ...(7)

and K6 (x, t) = 4 2 K4 (x, t) = 16 4 K2 (x, t), by (6) ... (8)Observing (3), (5), (7) etc. and also (4), (6), (8) etc. we can write the required iterted kernels

Kn (x, t) as follows :



If n = 2m–1, then K2m–1(x, t) = ( 2 )2m–2(x + sin t), m = 1, 2, 3, ... ... (9)If n = 2m, then K2m (x, t) = ( 2 )2m–1(1 + x sin t), m = 1, 2, 3, ... ... (10)Part (iv). Iterated kernels Kn (x, t) are given by

K1 (x, t) = K (x, t) ..(1)

and1

10

( , ) ( , ) ( , ) .n nK x t K x z K z t dz ... (2)

From (1), K1 (x, t) = K (x, t) = x – t. ... (3)Putting n = 2 in (2), we have

1 12 1

0 0( , ) ( , ) ( , ) ( ) ( ) ,K x t K x z K z t dz x z z t dz by (3)

12 31 2

00

[( ) ] ( )2 3z zx t z z xt dz x t xt z

21( , ) .

2 3x tK x t xt

... (4)

Next, putting n = 3 in (2), we have1

3 20

( , ) ( , ) ( , )K x t K x z K z t dz 1

0

1( ) ,2 3

z tx z zt dt using (3) and (4)

1 2

0

1 1 12 2 3 2 2 3x t tz xt z t x dz

12 3

0

1 1 12 2 2 3 2 2 2 3z x t z txt t xz

1 1 1 1 12 2 2 3 3 2 2 3

x t txt t x 12 12 12

x t x t

3 1( , ) (1/12) ( ) (1/12) ( , ),K x t x t K x t ... (5)Now, putting n = 4 in (2), we have

1 14 3

0 0( , ) ( , ) ( , ) ( ) ,

12z tK x t K x z K z t dz x z dz by (3) and (5)

1

0

1 1 1( ) ( ) ,12 12 2 3

x tx z z t dz xt as before

4 21 1 1( , ) ( , )

12 2 3 12x tK x t xt K x t

... (6)

Similarly, we find2

5 3 11 1( , ) ( , ) ( , ),

12 12K x t K x t K x t

by (5) ... (7)

and2

6 4 21 1( , ) ( , ) ( , ),

12 12K x t K x t K x t

by (6) ... (8)

and so on.



Observing (3), (5), (7) etc. and also (4), (6), (8) etc., we can write the required iterated kernelsKn (x, t) as follows:

If n = 2m – 1, then K2m–1 (x, t) 1

1( 1) ( ), 1, 2, 3, ...12

m

m x t m

... (9)

If n = 2m, then K2m (x, t) 1

1( 1) 1 , 1, 2, ...

2 312

m

mx t xt m

... (10)

Type 2. Determination of the resolvent kernel or reciprocal kernel R (x, t : ) or (x, t; ).

If Kn (x, t) be iterated kernels then

1=1

( ) = ( ) ( )m-mm

R x,t; x,t; K x,t .

Ex. 2. Determine the resolvent kernels for the Fredholm integral equation having kernels :(i) K (x, t) = ex + t; a = 0, b = 1. [Kanpur 2007, 08, 11]

(ii) K (x, t) = (1 + x) (1 – t); a = – 1, b = 1. [Kanpur 2006, 10, Meerut 2004, 2012]Sol. (i) Iterated kernels Km (x, t) are given by

K1 (x, t) = K (x, t). ...(1)

and1

10

( , ) ( , ) ( , )m mK x t K x z K z t dz ... (2)

From (1), K1 (x, t) = K (x, t) = ex + t. ... (3)Putting n = 2 in (2), we have

1 12 1

0 0( , ) ( , ) ( , ) ,x z x tK x t K x z K z t dz e e dz using (3)

1 21 2 2

0 0

1 12 2

x t z x t z x t ee e dz e e e , ... (4)

Putting n = 3 in (2), we have21 1

3 10 0

1( , ) ( , ) ( , )2

x z z t eK x t K x z K z t dz e e dz

22 21 2

0

1 1 ,2 2

x t z x te ee e dz e

as before ... (5)

and so on. Observing (3), (4) and (5), we may write12 1( , ) ,

2

mx t

meK x t e

m = 1, 2, 3, ... ... (6)

Now, the required resolvent kernel is given by12

1 1

1 1

1( , ; ) ( , )2

mm m x t

mm m

eR x t K x t e

12

1

( 1)2

mx t

m

ee

... (7)

But1 22 2 2

1

( 1) ( 1) ( 1)1 ....2 2 2

m

m

e e e

which is an infinite geometric series with common ratio 2{ ( 1)}/ 2.e



12

2 21

( 1) 1 2 ,2 1 { ( 1)}/ 2 2 ( 1)

m

m

ee e

provided2

2( 1) 21 or | |

2 1e

e

... (9)


22( , ; ) ,

2 ( 1)

x teR x te

provided 2

2| |1e

... (10)

Part (ii) Iterated Kernels Km (x, t) are given byK1 (x, t) = K (x, t) ... (1)

and1

11

( , ) ( , ) ( , ) .m mK x t K x z K z t dz

... (2)

From (1), K1 (x, t) = K (x, t) = (1 + x) (1 – t). ... (3)Putting n = 2 in (2), we have

1 12 1

1 1( , ) ( , ) ( , ) (1 ) (1 ) (1 ) (1 ) ,K x t K x z K z t dz x z z t dz

by (3)

11 2 3

1 1

1(1 ) (1 ) (1 ) (1 ) (1 )3

x t z dz x t z z

K2 (x, t) = (2/3) × (1 + x) (1 – t). ... (4)Next, putting n = 3 in (3), we have

1 13 21 1

2( , ) ( , ) ( , ) (1 ) (1 ). (1 ) (1 )3

K x t K x z K z t dz x z z t dz

21 2

1

2 2(1 ) (1 ) (1 ) (1 ) (1 ),3 3

x t z dz x t

as before. ... (5)

and so on. Observing (3), (4) and (5), we may write12( , ) (1 ) (1 ).

3

m

mK z t x t

...(6)

Now, the required resolvent kernel is given by1

1 1

1 1

2( , ; ) ( , ) (1 ) (1 ),3

mm m

mm mR x t K x t x t

by (6)

1

1

2(1 ) (1 )3

m

mx t

... (7)

But1 2 3

1

2 2 2 21 .....3 3 3 3

m

m

which is an infinite geometric series with common ratio (2 / 3).

1

1

2 1 3 ,3 1 (2 / 3) 3 2

m

m

... (8)



provided 2 /3 1 or | | 3 / 2 . ... (9)Using (8) and (9), (7) reduces to

3(1 ) (1 )( , ; ) ,

3 2x tR x t

provided 3| |2

... (10)

Type 3 : Solution of Fredholm integral equation with help of the resolvent kernel.

Working Rule: Let ( ) ( ) ( , ) ( )b


be given Fredholm integral equation. Let Km (x, t) be mth iterated kernel and let R (x, t; ) be theresolvent kernel of (1). Then we have

1

1( , ; ) ( , )m

mmR x t K x t

... (2)

Suppose the sum of infinite series (2) exists and so R (x, t, ) can be obtained in the closedform. Then, the required solution of (1) is given by

( ) ( ) ( , ; ) ( ) .b


Ex. 3. Solve 1/ 2

0( ) ( )y x x y t dt [Kanpur 2005, Meerut 2006, 08]

Sol. Given1/ 2

0( ) ( ) .y x x y t dt ... (1)

Comparing (1) with1/ 2

0( ) ( ) ( , ) ( ) ,y x f x K x t y t dt

we have f (x) = x, 1, K (x, t) = 1. ... (2)Let Km (x, t) be the mth iterated kernel. Then, we have

K1 (x, t) = K (x, t) ... (3)

and1/ 2

10

( , ) ( , ) ( , ) .m mK x t K x z K z t dz ... (4)

From (1), K1 (x, t) = K (x, t) = 1, by (2) ... (5)Putting m = 2 in (4), we have

1/ 2 1/ 22 1

0 0( , ) ( , ) ( , )K x t K x z K z t dz dz 1/ 2

01 .2

z ... (6)

Next, putting m = 3 in (4), we have1/ 2

3 20

( , ) ( , ) ( , )K x t K x z K z t dz 1/ 2

0

1 ,2

dz by (5) and (6)

= (1/2)2, ... (7)and so on. Observing (5), (6) and (7), we find

Km (x, t) = (1/2)m–1 ... (8)Now, the resolvent kernel R (x, t; ) is given by

1–1

1 1

1( , ; ) ( , ) ,2

mm

mm mR x t K x t

using (2) and (8) ... (9)

But1 2 3

1

1 1 1 11 .....2 2 2 2

m

m

which is an infinite geometric series with common ratio 1/2.



1

1

1 1 2.2 1 (1/ 2)

m

m

Substituting the above value in (9), we have R (x, t; ) = 2 ... (10)Finally, the required solution of (1) is given by

1/ 2

0( ) ( ) ( , ; ) ( )y x f x R x t f t dt or

1/ 2

0( ) (2 ) ,y x x t dt by (2) and (10)

or1/ 220

( ) 2 / 2 (1/ 4)y x x t x

Ex. 4. Solve 1

0

1 1 1( ) ( ) ,2 2 2

xy x e e y t dt Sol. Given

1

0

1 1 1( ) ( ) ,2 2 2

xy x e e y t dt ... (1)

Comparing (1) with1

0( ) ( ) ( , ) ( ) ,y x f x K x t y t dt

we have 1 1 1( ) , , ( , ) 1.2 2 2

xf x e e K x t ... (2)

Let Km (x, t) by the mth iterated kernel. Then, we haveK1 (x, t) = K (x, t) ... (3)

and1

10

( , ) ( , ) ( , ) .m mK x t K x z K z t dz ... (4)

From (1), K1 (x, t) = K (x, t) = 1, by (2) ... (5)Putting m = 2 in (4) and using (5), we have

1 12 1

0 0( , ) ( , ) ( , )K x t K x z K z t dz dz = 1. ... (6)

Next, putting m = 3 in (4), and using (5) and (6), we have1 1

3 20 0

( , ) ( , ) ( , )K x t K x z K z t dz dz = 1 ... (7)

and so on. Observing (5), (6) and (7), we findKm (x, t) = 1 for m = 1, 2, 3, ... ... (8)

Now, the resolvent kernel R (x, t; ) is given by1

1

1 1

1( , ; ) ( , ) ,2

mm

mm mR x t K x t

using (2) and (8) ... (9)

2 31 1 1 11 ...2 2 2 1 (1/ 2)

R (x, t; ) = 2. ... (10)Finally, the required solution of (1) is given by

1

0( ) ( ) ( , ; ) ( )y x f x R x t f t dt



or1

0

1 1 1 1 1( ) 2 ,2 2 2 2 2

x ty x e e e e dt using (2) and (10)

or1

0

1 1 1( )2 2 2 2

x t ety x e e e t

or 1 1 1 1( ) 1 , . ., ( )2 2 2 2

x x xy x e e e e e i e y x e

Ex. 5. Solve the following integral equations by the method of successive approximations :

(i)1

0

5 1( ) ( ) .6 2xy x xt y t dt [Meerut 2000, 01, 02]

(ii) 1

0( ) ( ) .y x x xt y t dt

(iii) / 2

0

1( ) sin ( ) .4 4xy x x xt y t dt

[Kanpur 2005]

Sol. (i) Given 1

0

5 1( ) ( ) .6 2xy x xt y t dt ... (1)

Comparing (1) with1

0( ) ( ) ( , ) ( )y x f x K x t y t dt

we have5 1( ) , , ( , )6 2xf x K x t xt ... (2)

Let Km (x, t) by the mth iterated kernel. ThenK1 (x, t) = K (x, t) ... (3)

and1

10

( , ) ( , ) ( , ) .m mK x t K x z K z t dz ... (4)

From (1), K1 (x, t) = K (x, t) = xt ... (5)Putting m = 2 in (4), we have

1 1 1 22 1

0 0 0( , ) ( , ) ( , ) ( ) ( ) ,K x t K x z K z t dz xz zt dz xt z dz by (5)

K2 (x, t) = (1/3) × xt ...(6)Next, putting m = 3 in (4), we have

1 13 2

0 0

1( , ) ( , ) ( , ) ( ) ,3

K x t K x z K z t dz x z zt dz using (5) and (6)

21 2

0

1 1 ,3 3

xt z dz xt ... (7)

and so on. Observing (5), (6) and (7), we findKm (x, t) = (1/3)m–1 × xt. ... (8)

Now, the resolvent kernel R (x, t; ) is given by

11

( , ; ) ( , )mmm

R x t K x t

1 1

1

1 1 ,2 3

m m

mxt

by (2) and (8)



1 2

1

1 1 11 ...6 6 6 1 (1/ 6)

m

m

xtxt x t

R (x, t ; ) = (6/5) × xt. ... (9)Finally, the required solution of (1) is given by

1

0( ) ( ) ( , ; ) ( )y x f x R x t f t dt or

1

0

5 1 6 5( ) ,6 2 5 6x xt ty x dt

by (2) and (9)

or

131 2

00

5 1 5( )6 2 6 2 3x x x ty x x t dt

or

5( ) .6 6x xy x x

(ii) Proceed like part (i). Required solution is y (x) = (3x)/(3 – ), where | | < 3.(iii) Proceed like part (i). Required solution is y (x) = sin x.

Ex. 6. Using iterative method, solve1

0( ) ( ) ( ) .x ty x f x e y t dt

Sol. Given1

0( ) ( ) ( ) .x ty x f x e y t dt ... (1)

Comparing (1) with ( ) ( ) ( , ) ( ) ,b


we have f (x) = f (x), = , K (x, t) = ex – t ... (2)Let Km (x, t) be the mth interated kernel. Then

K1 (x, t) = K (x, t) ... (3)

and1

10

( , ) ( , ) ( , ) .m mK x t K x z K z t dz ... (4)

From (1), K1 (x, t) = K (x, t) = ex – t ... (5)Putting m = 2 in (4), and using (5), we have

1 12 1

0 0( , ) ( , ) ( , ) x z z tK x t K x z K z t dz e e dz

1

0.x t x te dz e ... (6)

Next, putting m = 3 in (4), we have1 1

3 20 0

( , ) ( , ) ( , ) ,x z z tK x t K x z K z t dz e e dz by (5) and (6)

– ,x te as before. ... (7)and so on. Observing (5), (6) and (7), we find

Km (x, t) = ex – t for m = 1, 2, 3, ... ... (8)Now, the resolvent kernel R (x, t; ) is given by

1 1 1

1 1 1( , ; ) ( , )m m x t x t m

mm m mR x t K x t e e

2(1 ...)x te

( , ; ) /(1– ),x tR x t e provided | | < 1.... (9)

Finally, the required solution of (1) is given by1

0( ) ( ) ( , ; ) ( )y x f x R x t f t dt or

1

0

1( ) ( ) ( )1

x ty x f x e f t dt



or1

0( ) ( ) ( ) ,

1x ty x f x e f t dt

where | | < 1.

Ex. 7. Solve by the method of successive approximation :1

0

3 1 1 1( ) ( ) .2 2 2 2

x xy x e xe t y t dt [Meerut 2010, G.N.D.U., Amritsar 2004, Kanpur 2005]

Sol. Given1

0

3 1 1 1( ) ( ) .2 2 2 2

x xy x e xe t y t dt ... (1)

Comparing (1) with1

0( ) ( ) ( , ) ( ) ,y x f x K x t y t dt

we have 3 1 1 1( ) , , ( , ) .2 2 2 2

x xf x e xe K x t t ... (2)

Let Km (x, t) by the mth iterated kernel. Then, we haveK1 (x, t) = K (x, t) ... (3)

and1

10

( , ) ( , ) ( , ) .m mK x t K x z K z t dz ... (4)

From (1), K1 (x, t) = K (x, t) = t. ... (5)Putting m = 2 in (4), we have

121 12 1

0 00

1( , ) ( , ) ( , ) .2 2zK x t K x z K z t dz zt dz t t

... (6)

Putting m = 3 in (4), and using (2) and (6), we have

13 2

0( , ) ( , ) ( , )K x t K x z K z t dz

21

0

1 12 2

z t dz t ... (7)

and so on. Observing (5), (6) and (7), we find1( , ) (1/ 2) .m

mK x t t ... (8)

Now, the resolvent kernel R (x, t; ) is given by1 1

11 1

1 1( , ; ) ( , ) . ,2 2

m mm

mm mR x t K x t t

by (2) and (8)

1 2

1

1 1 1 1 41 ... .4 4 4 1 (1/ 4) 3

m

m

tt t t

... (9)

Finally, the required solution of (1) is given by

1

0( ) ( ) ( , ; ) ( )y x f x R x t f t dt

or 1

0

3 1 1 1 4 3 1 1( ) ,2 2 2 2 3 2 2 2

x x t tty x e xe e te dt by (2) and (9)

1 1 12

0 0 0

3 1 1 1 12 2 2 3 3

x x t te xe te dt t e dt t dt



121 11200 0

0

3 1 1 1 122 2 2 3 3 2

x x t t t te xe te dt t e t e dt

1 1

0 0

3 1 1 2 12 2 2 3 3 6

x x t tee xe te dt t e dt 1

0

3 1 2 52 2 3 3 3

x x tee xe t e dt

11

0 0

3 1 2 52 2 3 3 3

x x t tee xe t e e dt

1

0

3 1 2 52 2 3 3 3

x x tee xe e e

3 1 2 5 ( 1)2 2 3 3 3

x x ee xe e e

y (x) = (3/2) × ex – (1/2) × xex – (1/3) × e + 1

Ex. 8. By iterative method, solve 0

( ) 1 sin ( ) ( ) .y x x t y t dt

Sol. Given

0( ) 1 sin ( ) ( ) .y x x t y t dt

... (1)

Comparing (1) with0

( ) ( ) ( , ) ( ) ,y x f x K x t y t dt

we have f (x) = 1, , , K (x, t) = sin (x + t). ... (2)

Let Km (x, t) be the mth iterated kernel. Then, we have K1 (x, t) = K (x, t) ... (3)

and 10

( , ) ( , ) ( , ) .m mK x t K x z K z t dz

... (4)

From (1), K1 (x, t) = K (x, t) = sin (x + t) ... (5)Putting m = 2 in (4), we have

2 10 0

( , ) ( , ) ( , ) sin ( )sin ( )K x t K x z K z t dz x z z t dz

0 0

1 1 1[cos ( ) cos (2 ) kernel cos ( ) sin (2 )2 2 2

x t z x t dz z x t z x t

1 1 1[ cos ( ) sin ( ) sin ( )] cos ( ).2 2 2 2

x t x t x t x t ... (6)

Next, putting m = 3 in (4), we have

3 20

( , ) ( , ) ( , )K x t K x z K z t dz

0sin( ) cos ( ) ,

2x z z t dz

by (5) and (6)

0

[sin(2 ) sin ( )]4

z x t x t dz



0

1 cos (2 ) sin ( )4 2

z x t z x t

1 1cos ( ) cos ( ) sin ( )

4 2 2x t x t x t

= 2

sin( )2

x t

... (7)

Now, putting m = 4 in (4), we have

4 30

( , ) ( , ) ( , )K x t K x z K z t dz

2

0sin ( ).sin ( ) ,

2x z z t dz

by (5) and (7)

2

0sin ( ) sin ( )

2x z z t dz

2

cos ( ),2 2

x t

as before

K4 (x, t) = ( /2)3 × cos (x – t). ... (8)Next, putting m = 5 in (4), we have

3

5 40 0

( , ) ( , ) ( , ) sin ( ).cos ( ) ,2

K x t K x z K z t dz x z z t dz

by (5) and (8)

3 3

0sin ( ) cos ( ) sin ( ),

2 2 2x z z t dz x t

as before

K5 (x, t) = ( /2)4 × sin (x + t), ... (9)and so on. Taking advantage of symmetry and noting that all odd iterated kernels involve sin (x + t)whereas all even iterated kernels involve cos (x – t), we may, write

K6 (x, t) = ( /2)5 × cos (x – t), K7 (x, t) = ( /2)6 × sin (x + t) etc. ... (10)Now, the resolvent kernel R (x, t; ) is given by

11

( , ; ) ( , )mmm

R x t K x t

= K1 (x, t) + K2 (x, t) + 2 K3 (x, t) + 3 K4 (x, t) + ...

= K1 (x, t) + 2 K3 (x, t) + 4 K5 (x, t) + ...

+ [ K2 (x, t) + 2 K4 (x, t) + 4 K6 (x, t)+ ....]

2 4

sin ( ) 1 ...2 2

x t

2 4

cos( ) 1 ...2 2 2

x t

[using (5), (6), (8),(9) and (10) etc]

2 4

sin ( ) cos ( ) 1 ...2 2 2

x t x t

21sin ( ) cos ( ) . ,

2 1 ( / 2)x t x t

provided 2or | |

2

Thus, 2 2( , ; ) 2{2sin( ) cos( – )}/(4 )R x t x t x t ... (11)Finally, the required solution of (1) is given by

0

( ) ( ) ( , ; ) ( )y x f x R x t f t dt



2 2 0

21 {2sin ( ) cos( )} ,4

x t x t dt

by (2) and (11)

02 221 2cos ( ) sin ( )

4x t x t

2 221 { 2 cos ( ) sin ( ) 2cos sin }

4x x x x

2 2

4( ) 1 (2cos sin ),4

y x x

where

2| | .

Ex. 9. Consider1

0( ) 1 (1 3 ) ( ) .y x xt y t dt

Evaluate the resolvent kernel. For what values of the solution does not exist. Obtainsolution of the above integral equation.

Sol. Given1

0( ) 1 (1 3 ) ( ) .y x xt y t dt ... (1)

Comparing (1) with1

0( ) ( ) ( , ) ( )y x f x K x t y t dt

we have f (x) = 1, = 1, K (x, t) = 1 – 3xt. ... (2)Let Km (x, t) be the mth iterated kernel. Then, we have

K1 (x, t) = K (x, t) ... (3)

and1

10

( , ) ( , ) ( , ) .m mK x t K x z K z t dz ... (4)

From (1), K1 (x, t) = K (x, t) = 1 – 3xt. ... (5)Putting m = 2 in (4), we have

1 12 1

0 0( , ) ( , ) ( , ) (1 3 ) (1 3 )K x t K x z K z t dz xz zt dz

121 2 3

00

3{1 3 ( ) 9 } ( ) 32zz x t x t z dz z x t xt z

= 1 – (3/2) × (x + t) + 3xt. ... (6)Next, putting m = 3 in (4), we have

1 13 2

0 0

3( , ) ( , ) ( , ) (1 3 ) 1 ( ) 32

K x t K x z K z t dz xz z t zt dz

1

0

3 1(1 3 ) 1 32 2

xz t z t dz

1 2

0

3 1 3 11 3 92 2 2 2

t z t x xt xz t dt

12 3

0

3 3 1 3 11 32 2 2 2 2

t z z t x xt xz t



3 3 1 3 1 11 3 (1– 3 )2 2 2 2 2 4

t t x xt x t xt

K3 (x, t) = (1/4) × K1 (x, t), using (5) ... (7)Now, putting m = 4 in (4), we have

1 14 3

0 0

1( , ) ( , ) ( , ) (1 3 ). (1 3 ) ,4

K x t K x z K z t dz xz zt dz by (5) and (7)

1

0

1 1 3(1 3 ) (1 3 ) 1 ( ) 3 ,4 4 2

xz zt dz x t xt as before

4 21 3 1( , ) 1 ( ) 3 ( , ).4 2 4

K x t x t xt K x t

by (6) ... (8)

Next, putting m = 5 in (4), we have1 1

5 40 0

1 3( , ) ( , ) ( , ) (1 3 ). 1 ( ) 34 2

K x t K x z K z t dz xz z t zt dz

1

0

1 3 1 1(1 3 ) 1 ( ) 3 (1 3 ),4 2 4 4

xz z t zt dz xt as before

K5 (x, t) = (1/4)2 × (1 – 3xt) = (1/4)2 × K1 (x, t), by (5) ... (9)and so on. On observing (5), (6), (7), (8) and (9), we find that all odd iterated kernels involveK1 (x, t) and all even iterated kernels invalue K2 (x, t). Hence, by symmetry, we may write

K6 (x, t) = (1/4)2 × K2 (x, t), K7 (x, t) = (1/4)3 × K1 (x, t), and so on. ... (10)Now, the resolvent kernel R (x, t; ) is given by

1

1( , ; ) ( , )m

mmR x t K x t

= K1 (x, t) + K2 (x, t) + 2 K3 (x, t) + 3 K4 (x, t) ....

= K1 (x, t) + 2 K3 (x, t) + 4 K5 (x, t) + ... + [K2 (x, t) + 2 K4 (x, t) + 4 K6 (x, t) + ...]2 4

1 1 12( , ) ( , ) ( , ) ...4 4

K x t K x t K x t

2 4

2 2 22( , ) ( , ) ( , ) ...4 4

K x t K x t K x t

[using (7), (8), (9), (10)]

2 22 2 2 21 2( , ) 1 ( / 4) / 4 ... ( , ) 1 ( / 4) / 4 ...K x t K x t

22 21 2{ ( , ) ( , )} 1 ( / 4) / 4 ...K x t K x t

21 2 2

1{ ( , ) ( , )} , provided 41 ( / 4)

K x t K x t

or | | < 2

24 31 3 1 ( ) 3

24xt x t xt

( , ; )R x t 24 3 11 3

2 24x t x x

.... (11)

Finally, the required solution of (1) is given by1

0( ) ( ) ( , ; ) ( )y x f x R x t f t dt

or1

2 0

4 3 1( ) 1 1 32 24

y x x t x x dt , by (2) and (11)



1

22

0

4 3 3 11 12 2 24

x t t x x

24 3 31 1

2 2 24x x x 2

4 3 31 12 44

x

2

2 24 4 6 4 (4 6 )1

44 4x x

24 2 (2 3 )( ) , provided | | 2.

4xy x

Since solution exists only when | | < 2, it follows that (1) will not possess solution if | | > 2.Type 4 : Solution of Fredholm integral equation when the resolvent kernel cannot be obtained

in closed form i.e., the sum of infinite series occuring in the formula of the resolvent kernelcannot be determined. In such integral equation, we use the method of successive approximationsto find solutions upto third order.

Working Rule : Let the given Fredholm integral equation of the second kind be

( ) ( ) ( , ) ( )b


As zero-order approximation, we take y0 (x) = f (x) ... (2)If nth order approximation be yn (x), then

1( ) ( ) ( , ) ( ).b

n nay x f x K x t y x ...(3)

With help of (2) and (3), we easily obtain y1 (x) y2 (x) and y3 (x).Remark. Sometimes the zero-order approximation is mentioned in the problem. In that case,

we modify equation (2) according to data of the problem.

Ex. 10. (a) Solve the following integral equation1

0( ) 1 ( ) ( ) ,y x x t y t dt

by the method of successive approximation to third order.

(b) Solve the integral equation 1

0( ) 1 ( ) ( )y x x t y t dt by the method of successive

approximation upto second order for y0(x) = 1 (Kanpur 2009)

Sol. (a). Given1

0( ) 1 ( ) ( ) ,y x x t y t dt ... (1)

Let y0 (x) denote the zero-order approximation, Then, we may takey0 (x) = 1. ... (2)

If yn (x) denotes the nth order approximation, then we know that1

10( ) 1 ( ) ( ) .n ny x x t y t dt ... (3)

Putting n = 1 in (3),1 1

1 00 0

( ) 1 ( ) ( ) 1 ( ) , by (2)y x x t y t dt x t dt

or 1( )y x1

2

0

1 11 1 .2 2

xt t x ... (4)




1

2 10( ) 1 ( ) ( )y x x t y t dt

1

0

11 ( ) 1 ,2

x t t dt by (4)

1

01 ( ) 1

2x t t dt

1 2

01 1 1 ,

2 2x t x t dt

12 3

0

1 1 12 2 2 3

t tx t x

11 1 12 2 2 3

x x 21 71 .

2 12x x

... (5)

Finally, putting n = 3 in (3), we have

1

3 201 ( ) ( )y x t y t dt

1 2

0

1 71 ( ) 12 12

x t t t dt

21

0

71 ( ) 1 (1 )2 12

x t t dt

2 21 2 2

0

7 71 1 1 (1 )2 12 2 12

x t x x t dt

12 2 2

2 3

0

7 7 11 1 1 (1 )2 12 2 2 12 3

tx t x x t

2 22 27 7 11 1 1 (1 ).

2 12 2 2 12 3x x x

2 3

31 7 13 5( ) 12 12 12 8

y x x x x

(b) Do as in part (a). We are given y0 (x) = 1. The required solution to second order is givenby (5) of part (a).

Ex. 11. Solve the inhomogeneous Fredholm integral equation of the second kind1

0( ) 2 ( ) ( ) ,y x x x t y t dt by the method of successive approximations to the third order by

taking y0 (x) = 1.

Sol. Given1

0( ) 2 ( ) ( ) ,y x x x t y t dt ... (1)

Let y0 (x) denote the zero-order approximation, then given that y0 (x) = 1. ... (2)If yn (x) denotes the nth order approximation, then we know that

11

0( ) 2 ( ) ( ) .n ny x x x t y t dt ... (3)

Putting n = 1 in (3),1

1 00( ) 2 ( 1) ( )y x x x y t dt

1

02 ( 1) ,x x dt using (2)



or

12

10

1( ) 2 2 .2 2ty x x tx x x

... (4)


1

2 10

( ) 2 ( ) ( )y x x x t y t dt 1

0

12 ( ) 2 ,2

x x t t t dt by (4)

1

02 ( ) (2 )

2x x t t dt

1 2

02 2 (2 )

2 2xx t x x t dt

12 3

0

2 2 (2 )2 2 2 3

x t t tx x x

1 12 2 (2 )

2 2 2 3xx x x

22 723 12

x x x

... (5)

Finally, putting n = 3 in (3), we have

13 20( ) 2 ( ) ( )y x x x t y t dt

1 2

0

2 72 ( ) 23 12

x x t t t t dt

21 2

0

2 72 ( ) (2 )3 12

x x t t dt

2 21 2 2 2

0

2 7 2 72 2 (2 )3 12 3 12

x x t x x x t dt

12 2 2 32 2

0

2 7 2 72 2 (2 )3 12 2 3 12 3

t tx xt x x x

2

2 2 22 7 1 2 7 12 2 (2 )3 12 2 3 12 3

x x x x x

or2 32 7 2 13 5( ) 2

3 6 3 12 8y x x x x x

5.10. RECIPROCAL FUNCTIONS.

Let Kn (x, t), n = 1, 2, 3, ... be iterated kernels (or functions) given byK1 (x, t) = K (x, t) ... (1)

and 1( , ) ( , ) ( , ) .b

n naK x t K x z K z t dz ... (2)

Let – k (x, t) = K1 (x, t) + K2 (x, t) + ... + Kn (x, t) + ... ... (3)

Let K (x, t) be real and continuous in a rectangle R, for which , .a x b a t b LetK (x, t) 0 and let M be the maximum value of |K (x, t)| in R, that is, | K (x, t) | M in R. Then, if M(b – a) < 1, we easily find that the infinite series (3) for k (x, t) is absolutely and uniformlyconvergent. Hence, k (x, t) is real and continuous in R.



We know that ( , ) ( , ) ( , ) .b

p q p qa

K x t K x z K z t dz ... (4)

Now, re-writing (3), we have– k (x, t) – K1 (x, t) = K2 (x, t) + K3 (x, t) + ... + Kn (x, t) + ... ... (5)

or – k (x, t) – K (x, t) = K2 (x, t) + K3 (x, t) + ... + Kn (x, t) + ... ... (6)Using (4), (6) may be written as

1 1 1 2( , ) ( , ) ( , ) ( , ) ( , ) ( , ) ...b b

a ak x t K x t K x z K z t dz K x z K z t dz

1 1 2 3( , )[ ( , ) ( , ) ( , ) ...]b

aK x z K z t K z t K z t dz

Thus, – ( , ) – ( , )k x t K x t ( , ) ( , ) ,b

aK x z k z t dz using (1) and (3) ... (7)

Again, using (4), (6) may be written in another form as follows :

1 1 2 1( , ) ( , ) ( , ) ( , ) ( , ) ( , ) ...b b

a ak x t K x t K x z K z t dz K x z K z t dz

1 2 3 1[ ( , ) ( , ) ( , ) ...] ( , )b

aK x z K x z K x z K z t dz

1( , ) ( , )b

ak x z K z t dz , using (1) and (3) ... (8)

From (7) and (8), we have

( , ) ( , ) ( , ) ( , )b

ak x t K x t K x z k z t dz ... (9A)

and ( , ) ( , ) ( , ) ( , ) .b

ak x t K x t k x z K z t dz ... (9B)

Two functions K (x, t) and k (x, t) are known as reciprocal if they are both real and continuousin R and if they satisfy the condition (9A) or (9B).

Theorem I.If K (x, t) is real and continuous in R, there exists a reciprocal functions k (x, t)given by

– k (x, t) = K1 (x, t) + K2 (x, t) + K3 (x, t) + ...where K1 (x, t), K2 (x, t), ... are iterated functions (or kernels), provided that M (b – a) < 1, whereM is the maximum value of | K (x, t) | in rectangle R, for which a x b and a t b.

Proof. Left as an exercise.Theorem II. Volterra’s solution of Fredholm Integral equation of the second kind

Given. ( ) ( ) ( , ) ( ) .b


If (i) K (x, t) is real and continuous in rectangle R, for which a x b and a t b.K (x, t) 0.

(ii) f (x) is real and continuous in I and f (x) | 0.(iii) If a function k (x, t) reciprocal to K (x, t) exists, then the integral equation (1) has a

unique continuous solution in I given by

( ) ( ) ( , ) ( ) .b

ay x f x k x t f t dt ... (2)



Proof. Re-writing (1), we have ( ) ( ) ( , ) ( ) .b


Replaing x by t in (3), we have ( ) ( ) ( , ) ( ) .b

ay t f t K t z y z dz ... (4)

Multiplying both sides of (4) by k (x, t) and then integrating both sides w.r.t. ‘t’ from a to b,we have

( , ) ( ) ( , ) ( ) ( , ){ ( , ) ( ) }b b b b

a a a ak x t y t dt k x t f t dt k x t K t z y z dz dt

or ( , ) ( ) ( , ) ( ) ( ){ ( , ) ( , ) }b b b b

a a a ak x t y t dt k x t f t dt y z k x t K t z dt dz ... (5)

[changing the order of integration]Since k (x, t) and K (x, t) are reciprocal functions, we have by definition

( , ) ( , ) ( , ) ( , ).b

ak x t K t z dt k x z K x z ... (6)


( , ) ( ) ( , ) ( ) ( )[ ( , ) ( , )]b b b

a a ak x t y t k x t f t dt y z k x z K x z dzdt

or ( , ) ( ) ( , ) ( ) ( , ) ( ) ( , ) ( )b b b b

a a a ak x z y z k x t f t dt k x z y z dz K x z y z dzdz

or 0 ( , ) ( ) ( , ) ( )b b

a ak x t f t K x t y t dt ... (7)

From (1) ( , ) ( ) ( ) ( )b

aK x t y t y x f x ... (8)

Using (8), (7) becomes 0 ( , ) ( ) ( ) ( )b

ak x t f t dt y x f x

or ( ) ( ) ( , ) ( ) ,b

ay x f x k x t f t dt

showing that if (1) has a continous solution, then it is given by (2) and it is unique solution.We now show that the expression of y (x) given by (2) is, indeed, a solution of (1). Re-

writing (2), we have

( ) ( ) ( , ) ( )b

af x y x k x t f t dt ... (9)

Clearly, (9) is an integral equation for the determination of f (x). The function reciprocal tok (x, t) is K (x, t). Hence, making use of the method of getting solution (2) from (1), we find that if(9) has a conitnuous solution, it is unique and is given by

( ) ( ) ( , ) ( )b

af x y x K x t y t dt or ( ) ( ) ( , ) ( ) ,

b


which is the equation (1) from which we started. Hence we find that (1) is satisfied by the value ofy (x) given by (2).



ILLUSTRATIVE SOLVED EXAMPLES

Ex. 1. Solve 1

0

1( ) ( ) ( ) .2

x ty x f x e y t dt

Sol. Given1

0

1( ) ( ) ( ) .2

x ty x f x e y t dt ... (1)

Comparing (1) with1

0( ) ( ) ( , ) ( ) ,y x f x K x t y t dt

we have K (x, t) = (1/2) × ex–t. ... (2)Let k (x, t) be the reciprocal function of K (x, t). Then if K1 (x, t), K2 (x, t), ... be iterated

functions, then

1 2 3( , ) ( , ) ( , ) ( , ) ...k x t K x t K x t K x t ... (3)Now, the solution of (1) is given by

1

0( ) ( ) ( , ) ( ) .y x f x k x t f t dt ... (4)

We know that iterated functions are given byK1 (x, t) = K (x, t) ... (5)

and1

10( , ) ( , ) ( , ) , ( 2, 3, ...)n nK x t K x z K z t dz n ... (6)

From (5) and (2), K1 (x, t) = K (x, t) = (1/2) × ex–t. ... (7)Putting n = 2 in (6), we have

12 10( , ) ( , ) ( , )K x t K x z K z t dz

1 1

2 20 0

1 1 1 1. .2 2 2 2

x z z t x t x te e dz e dz e ... (8)


1 13 2 20 0

1 1( , ) ( , ) ( , ) .2 2

x z z tK x t K x z K z t dz e e dz 1

3 30

1 1 ,2 2

x t x te dz e ... (9)

and so on, Observing (7), (8) and (9), we may writeKn (x, t) = (1/2n) × ex–t, n = 1, 2, 3 ... ... (10)

Substituting the above values of K1 (x, t), K2 (x, t), K3 (x, t), ... etc. in (3), we have

12 3

1 1 1( , ) ...2 2 2

x x t x tk x t e e e 12

1 1 1 1 11 ...2 2 2 1 (1/ 2)2

x x te e

– k (x, t) = ex – t or k (x, t) = – ex – t. ... (11)Substituting the above value of k (x, t) in (4), we have

1

0( ) ( ) ( ) ,x ty x f x e f t dt

which is the required solution.Ex. 2. Solve the following integral equations.

(i)1

0( ) ( )y x x y t dt Ans. y (x) = x



(ii) 1

0

5 1( ) ( )6 2xy x xt y t dt Ans. y (x) = x

(iii) 1

0

3 1 1 1( ) ( ) .2 2 2 2

x xy x e xe t y t dt Ans.3 1 1( ) – – 12 2 3

x xy x e xe e

(iv) / 2

0

1( ) sin ( ) .4 4xy x x xt y t dt

Ans. y (x) = sin x

Sol. Solve as in the above solved Ex. 1.

EXERCISE 5A1. Find the iterated kernels Kn (x, t) of the following kernels for specified a and b.

(i) K (x, t) = x – t; a = – 1, b = 1.

(ii) K (x, t) = sin (x – t); a = 0, b = / 2 for n = 2, 3(iii) K (x, t) = (x – t)2; a = –1, b = 1 for n = 2, 3. (Kanpur 2008, 09)(iv) K (x, t) = xet; a = 0, b = 1.(v) K (x, t) = e|x – t|; a = 0, b = 1 for n = 2

(vi) K (x, t) = e| x | + t. a = – 1, b = 1 for n = 2

Ans. (i) 1 1

2 1 24 4 1( , ) ( ), ( , ) 2( 1) , 1, 2, 3, ...3 3 3

m mm

m mK x t x t K x t xt m

(ii) 22 3

1 1 1( , ) sin ( ) cos ( ), ( , ) (4 )sin ( ).2 4 16

K x t x t x t K x t x t

(iii) K2 (x, t) = (2/3) × (x + t)2 + 2x2t2 + (4/3) × xt + (2/5).K3 (x, t) = (56/45) × (x2 + t2) + (8/3) × x2 t2 – (32/9) × xt + (8/15).

(iv) Kn (x, t) = xet, n = 1, 2, 3, ...

(v)

2

2 2

( 1) , 02( , )

( 1) , 1.2

x t x tt x

x t x tx t

e e t x e x tK x t

e e x t e t x

(vi)2

2 2

(1/ 2) ( 1) , 1 0( , )

(1/ 2) ( 1) , 0 1.

t x

t x

e e xK x t

e e x

2. (a) Construct the resolvent kernels for the following kernels for specified a and b.

(i) K (x, t) = sin x cos t; a = 0, b = / 2. [Kanpur 2006] (ii) K (x, t) = x et; a = –1, b = 1.(iii) K (x, t) = x2t2; a = – 1, b = 1. (iv) K (x, t) = xt; a = – 1, b = 1.(v) K (x, t) = sin x cos t + cos 2x sin 2t; a = 0, b = 2 .

(vi) K (x, t) = 1 + (2x – 1) (2t – 1); a = 0, b = 1.(vii) K (x, t) = xt + x2t2; a = – 1, b = 1

Ans. (i) 2sin cos( , ; ) ;| | 2.2

x tR x t

(ii) 1

( , ; ) ;| | .2 2

txe eR x te

(iii) R (x, t; ) = (5x2t2)/(5 – 2 ); | | < 5/2.(iv) R (x, t; ) = (3xt)/(3 – 2 ); | | < 3/2. (v) R (x, t; ) = sin x cos t + cos 2x sin 2t.



(vi) 1 3(2 1) (2 1)( , ; ) ;| | 11 3

x tR x t

(vii)

2 23 5( , ; ) ;| | 3 / 23 2 5 2

xt x tR x t

2. (b) Find the resolvent kernel associated with the following kernels :(i) |x – t|, in the interval (0, 1) (ii) e – | x – t |, in the interval (0, 1)

(iii) cos (x + t), in the interval (0, 2 ).

3. Solve 1

0

5 1 1( ) ( ) ( ) ,6 9 3xy x x t y t dt by the method of successive approximations.

[Ans. y (x) = x.]

4. Consider the integral equation1

0( ) 1 ( ) .y x xt y t dt

(i) Make use of the relation 1| | B to show that the iterative procedure is valid for

| | 3. (ii) Show that the iterative procedure leads formally to the solution

2 3( ) 1 [( / 2) ( / 6) ( /18) ...].y x x

(iii) Use the method of chapter 4 to obtain the exact solutiony (x) = 1 + [3 x/2 (x – )], 3.

5. Explain the method of solving a Fredholm integral equation by the method of successiveapproximation and hence solve the integral equation

1/ 2

0( ) ( ) .y x x y t dt [Ans. y (x) = x + constant]

6. Solve the integral equation by the method of iterated kernel1

0

5 1( ) ( ) .6 2xy x xt y t dt [Ans. y (x) = x.]

7. Find the iterated kernel of the kernel K (x, t) = x + sin t, in the interval ( , ) .[Hint. Here a = – , b = . Refer solved Ex. 1. part (iii) of Art. 5.9]

8. Explain the method of successive substitution for solving Fredholm integral equation ofsecond kind and show that the solution is unique. How does this method differ from the method ofsuccessive approximation ?

9. Define iterated kernels. Prove that the nth iterated kernel Kn (x, t) satisfies the relation

( , ) ( , ) ( , ) , .b

n m n maK x t K x z K z t dz m n

10. Define resolvent kernel and find the resolvent kernel of the kernel K (x, t) = 1 – 3xt in(0, 1).

11. Explain the iterative method (method of successive approximation) of solving the Fredholmintegral equation of the second kind

( ) ( ) ( , ) ( )b


and obtain the sufficient condition for the convergence of the iterative procedure.12. What do you understand by iterated function, reciprocal function and resolvent kernel. If

k (x, t) and K (x, t) and reciprocal functions, then prove that

( , ) ( , ) ( , ) ( , ) .b

aK x t k x t K x z k z t dz




0( ) 1 ( ) ( ) ,y x x t y t dt by the method of successive

approximations and show that the estimate afforded by the relation 1| | B is conservative inthis case.

14. Obtain the radius of convergence of the Neumann series when the function f (x) and thekernel K (x, t) are continuous in the interval (a, b).

5.11.SOLUTION OF VOLTERRA INTEGRAL EQUATION OF THE SECOND KIND BYSUCCESSIVE APPROXIMATIONS. ITERATIVE METHOD. NEUMANN SERIES.

[Meerut 2000, 01, 03, 07]Consider Volterra integral equation of the second kind

( ) ( ) ( , ) ( ) .x


As a zero-order approximation to the required solution y (x), let us takey0 (x) = f (x). ... (2)

Further, if yn (x) and yn – 1 (x) are the nth order and (n – 1)th order approximations respectively,then these are connected by

1( ) ( ) ( , ) ( ) .x

n na

y x f x K x t y t dt ... (3)

We know that the iterated kernels (or iterated functions) Kn (x, t), (n = 1, 2, 3, ...) are defined byK1 (x, t) = K (x, t) ... (4A)

and 1( , ) ( , ) ( , ) .x

n nt

K x t K x z K z t dz ... (4B)

Putting n = 1 in (3), the first-order approximation y1 (x) is given by

1 0( ) ( ) ( , ) ( ) .x


But, from (2), y0 (t) = f (t) ... (6)Substituting the above value of y0 (t) in (5), we get

1( ) ( ) ( , ) ( ) .x

ay x f x K x t f t dt ... (7)

Putting n = 2 in (3), the second-order approximation y2 (x) is given by

2 1( ) ( ) ( , ) ( )x


or 2 1( ) ( ) ( , ) ( ) .x


Replacing x by z in (7), we have

1( ) ( ) ( , ) ( ) .z

ay z f z K z t f t dt ... (9)

Substituting the above value of y1 (z) in (8), we get

2 ( ) ( ) ( , )[ ( ) ( , ) ( ) ]x z

a ay x f x K x z f z K z t f t dt dz



or 2 ( ) ( ) ( , ) ( )x

ay x f x K x z f z dz 2 ( , )[ ( , ) ( ) ] .

x z

z a t aK x z K z t f t dt dz

... (10)

Now, consider the double integral on the R.H.S. of (10). The limits of integration are given byt = a, t = z, z = a, z = x. Clearly the region of integration is the triangle ABC as shown in thefollowing figure. In double integral, clearly strips have been taken parallel to t-axis (see strip RS)

When we wish to change the order of integration in the above mentioned double integral, weshall take strips parallel to z-axis (see strip PQ). Then for the same region (triangle ABC), we seethat the limits of z are z = t to z = x and limits for t are t = a to t = x.

( , )[ ( , ) ( ) ] ( ) [ ( , ) ( , ) ] .x z x x

z a t a t a z tK x z K z t f t dt dz f t K x z K z t dz dt

Using the above equivalent value of the double integral in (10), we obtain

2 ( ) ( ) ( , ) ( )x

ay x f x K x z f z dz 2 ( ) [ ( , ) ( , ) ]

x x

t a z tf t K x z K z t dz dt

or 22 2( ) ( ) ( , ) ( ) ( ) ( , )

x x

a ay x f x K x t f t dt f t K x t dt

2 1

2

For 2 in (4 ), ( , ) ( , ) ( , )

or ( , ) ( , ) ( , ) , by (4 )

x

t

x

t

n B K x t K x z K z t dz

K x t K x z K z t dz A

or 22 1 2( ) ( ) ( , ) ( ) ( , ) ( ) ,

x x

a ay x f x K x t f t dt K x t f t dt using (4A).

or2

2 1( ) ( ) ( , ) ( ) .

xmmm a


... (11)

Proceeding likewise, we easily obtain by Mathematical induction the nth approximate solutionyn (x) of (1) given by

1( ) ( ) ( , ) ( ) .

n xmn mm a


... (12)

Proceeding to the limit as n , we obtain the so called Neumann series

1( ) lim ( ) ( ) ( , ) ( ) .

xmn mn m a

y x y x f x K x t f t dt

... (13)



We now determine the resolvent kernel (or reciprocal kernel) R (x, t; ) or (x, t; ) interms of the iterated kernels Km (x, t). For this purpose, by changing the order of integration andsummation (13), we obtain

11

( ) ( ) [ ( , )] ( ) .x m

mmay x f x K x t f t dt

... (14)

Comparing (14) with ( ) ( ) ( , ; ) ( ) ,x


here 1

1( , ; ) ( , ).m

mmR x t K x t

... (16)

The series (16) converges absolutely and uniformly when K (x, t) is continuous in R.5.12. AN IMPORTANT THEOREM.

Let R (x, t ; ) be the resolvent (or reciprocal) kernel of a Volterra integral equation.

( ) ( ) ( , ) ( ) ,x


then the resolvent kernel satisfies the integral equation

( , ; ) ( , ) ( , ) ( , ; ) .x

tR x t K x t K x z R z t dz (Meerut 2010; Kanpur 2007)

Proof. We know that R (x, t, ) is given by

1

1( , ; ) ( , ),m

mmR x t K x t

... (1)

where iterated kernels (or functions) are given byK1 (x, t) = K (x, t) ... (2A)

and 1( , ) ( , ) ( , )x

m mtK x t K x z K z t dz ... (2B)


11 2

( , ; ) ( , ) ( , )mmm

R x t K x t K x t

11

2( , ) ( , ) ( , ) ,

xmmm t


using (2A) and (2B)

1( , ) ( , ) ( , ) ,

xnnn t


setting m – 1 = n

1( , ) ( , ) ( , ) ,

xmmm t


11

( , ) ( , ) ( , )xm

mm tK x t K x z K z t dz

1

1( , ) [ ( , )] ( , )

x mmmt

K x t K z t K x z dz

[on changing the order of summation and integration]

( , ) ( , ; ) ( , ) ,x

tK x t R z t K x z dz using (1)

( , ; ) ( , ) ( , ) ( , ; ) .x

tR x t K x t K x z R z t dz



5.13. SOLVED EXAMPLES BASED ON SOLUTION OF VOLTERRA INTEGRALEQUATION OF SECOND KIND BY SUCCESSIVE APPROXIMATIONS(OR ITERATIVE METHOD).

Type 1. Determination of resolvent kernel or reciprocal kernel for Volterra integral

equation . ( ) ( ) ( ) ( )x

ay x f x + K x,t y t dt

Ex. 1. Find the resolvent kernel of the Volterra integral equation with the kernel K (x, t) = 1.[Kanpur 2005, 2006]

Sol. Iterated kernels Kn (x, t) are given byK1 (x, t) = K (x, t) ... (1)

and 1( , ) ( , ) ( , ) , 1, 2, 3, ...x

n ntK x t K x z K z t dz n ... (2)

Given K (x, t) = 1. ... (3) From (1) and (3), K1 (x, t) = K (x, t) = 1. ... (4)Putting n = 2 in (2), and using (4), we have

2 1( , ) ( , ) ( , )x x

t tK x t K x z K z t dz dz .x

tz x t ... (5)


3 2( , ) ( , ) ( , ) 1.( ) ,x x

t tK x t K x z K z t dz z t dz by (4) and (5)

2 2( ) ( )2 2!

x

t

z t x t

... (6)

Now, putting n = 4 in (2), we have2

4 3( )( , ) ( , ) ( , ) 1. ,

2!

x x

t t

z tK x t K x z K z t dz dz by (4) and (6)

3 31 ( ) ( )2! 3 3!

x

t

z t x t

... (7)

and so on. Observing (4), (5), (6) and (7) etc, we find by mathematical induction, that1( )( , ) , 1, 2, 3,...

( 1)!

n

nx tK x t nn

... (8)

Now, by the definition of the resolvent kernel, we have

11

( , ; ) ( , )mmm

R x t K x t

2

1 2 3( , ) ( , ) ( , ) ...K x t K x t K x t

2 3( ) [ ( )] [ ( )]1 ...,1! 2! 3!x t x t x t

by (8)

( ) .x te Ex. 2. Find the resolvent kernel of the Volterra integral equation with the kernel

K (x, t) = ex – t.Sol. Iterated kernels Kn (x, t) are given by

K1 (x, t) = K (x, t) ... (1)



and 1( , ) ( , ) ( , ) , 2, 3, ...x

n nt

K x t K x z K z t dz n ... (2)

Given K (x, t) = ex – t. ... (3) From (1) and (3), K1 (x, t) = K (x, t) = ex – t. ... (4)Putting n = 2 in (2), we have

2 1( , ) ( , ) ( , )x

tK x t K x z K z t dz

( ).x xx z z t x t x t

t te e dz e dz e x t ... (5)

Next, putting n = 3 in (2), and using (5) we have

3 2( , ) ( , ) ( , )x

tK x t K x z K z t dz ( – ) ( )

x xx z z t x tt t

e z t e dz e z t dz 2 2( ) ( )

2 2!

xx t x t

t

z t x te e

... (6)

Now, putting n = 4 in (2) and using (6), we have2

24 3

( )( , ) ( , ) ( , ) ( )2! 2!

x tx x xx z z t

t t t

z t eK x t k x z k z t dz e e dz z t dz

3 3( ) ( ) ,2! 3 3!

xx tx t

t

e z t x te

... (7)

and so on. Observing (4), (5), (6) and (7) etc. we find by methematical induction, that1( )( , ) , 1, 2, 3, ...

( 1)!

nx t

nx tK x t e nn

... (8)


11

( , ; ) ( , )mmm

R x t K x t

= K1 (x, t) + K2 (x, t) + 2 K3 (x, t) + ...

2( ) [ ( )] ...,1! 2!

x t x t x tx t x te e e by (8)

2( ) [ ( )]1 ...1! 2!

x t x t x te

( ) ( ) ( )x t x t x t x te e e ( ) (1 ) .x te

Ex. 3. Find the resolvent kernel of the Volterra integral equation with the kernelK (x, t) = (2 + cos x) / (2 + cos t). (Meerut 2010, 11)

Sol. Iterated kernels Kn (x, t) are given byK1 (x, t) = K (x, t) ... (1)

and 1( , ) ( , ) ( , ) , 2, 3, ...x

n nt

K x t K x z K z t dz n ... (2)

Given K (x, t) = (2 + cos x) / (2 + cos t) ... (3)



From (1) and (3), K1 (x, t) = K (x, t) (2 cos ) /(2 cos )x t . ...(4)Putting n = 2 in (2) and using (4), we have

2 1( , ) ( , ) ( , )x


2 cos 2 cos 2 cos.2 cos 2 cos 2 cos

x x

t t

x z xdz dzz t t

2 cos ( ).2 cos

x x tt

... (5)


3 2( , ) ( , ) ( , )x


2 cos 2 cos. ( ) ,2 cos 2 cos

x

t

x z z t dzz t

by (3) and (5)

22 cos 2 cos ( )( )

2 cos 2 cos 2

xx

tt

x x z tz t dzt t

22 cos ( ). .2 cos 2!

x x tt

... (6)

Now, putting n = 4 in (2), we have

4 3( , ) ( , ) ( , )x


22 cos 2 cos ( ). . ,2 cos 2 cos 2!

x

t

x z z t dzz t

by (3) and (6)

3 32 cos 1 ( ) 2 cos ( ). . ,

2 cos 2! 3 2 cos 3!

x

t

x z t x x tt t

... (7)

and so on. Observing (4), (5), (6) and (7) etc. we find by mathematical intuction that

12 cos ( )( , ) . , 1, 2, 3,

2 cos ( 1)!

n

nx x tK x t nt n

... (8)

Now, by the definition of the resolvent kernel, we get

1

1( , ; ) ( , )m

mmR x t K x t

= K1 (x, t) + K2 (x, t) + 2 K3 (x, t) + ...

22 cos 2 cos ( ) 2 cos { ( )}. . ...,2 cos 2 cos 1! 2 cos 2!

x x x t x x tt t t

by (8)

2 32 cos ( ) { ( )} { ( )}1 ...2 cos 1! 2! 3!

x x t x t x tt

( )2 cos .

2 cosx tx e

t

EXERCISE 5BFind the resolvent kernels for Volterra integral equations with the following kernels.

1. 2 2( , ) .x tK x t e Ans. 2 2 ( ) .x t x te e

2.2

21( , ) .1

xK x tt

Ans.

2( )

21 .1

x tx et

3. K (x, t) = ax – t (a > 0). Ans. ax – t e (x – t)

4. cosh( , ) .cosh

xK x tt

Ans. ( )cosh .cosh

x tx et

5. K (x, t) = x – t. Ans. (1/ ) sinh {( ) }, ( 0)x t



Type 2. Solution of Volterra integral equation with help of the resolvent kernel.

Working Rule: Let ( ) ( ) + ( ) ( )x

ay x f x K x,t y t dt ... (1)

be given Volterra integral equation. Let Km (x, t) be the mth iterated kernel and let R (x, t; ) bethe resolvent kernel of (1). Then, we have

-11

( ) = ( ).mmm=

R x,t; K x,t ... (2)

Suppose the sum of infinite series (2) exists and so R (x, t; ) can be obtained in the closedform. Then the required solution of (1) is given by

( ) = ( ) + ( ) ( )x

ay x f x R x,t; f t dt. ... (3)

Ex. 4. With the aid of the resolvent kernel, find the solution of the integral equation2 2 2

0( ) ( ) .

xx x ty x e e y t dt [Meerut 2004, 11]

Sol. Given2 2 2

0( ) ( ) .

xx x ty x e e y t dt ... (1)

Comparting (1) with0

( ) ( ) ( , ) ( ) ,x

y x f x K x t y t dt we have 2 2 2

( ) , 1, ( , ) . x x tf x e K x t e ... (2)Let Km (x, t) be the mth iterated kernel. then, we have

K1 (x, t) = K (x, t) ... (3)

and 1( , ) ( , ) ( , ) .x

m mtK x t K x z K z t dz ... (4)

From (2) and (3), K1 (x, t) = K (x, t) =2 2

.x te ... (5)Putting m = 2 in (4), we have

2 2 2 2

2 1( , ) ( , ) ( , ) ,x x x z z t

t tK x t K x z K z t dz e e dz by (5)

2 2 2 2( ).

xx t x t

te dz e x t ... (6)

Next, putting m = 3 in (4), we have2 2 2 2

3 2( , ) ( , ) ( , ) . ( ) ,x x x z z t

t tK x t K x z K z t dz e e z t dz by (5) and (6)

2 2 2 2 2 22 2( ) ( )( ) .2 2!

xxx t x t x t

tt

z t x te z t dz e e

... (7)


4 3( , ) ( , ) ( , )x


2 22 2 2 2 2 22 3 3( ) ( ) ( ) ,

2! 2! 3 3!

xx tx x z z t x t

tt

z t e z t x te e dz e

... (8)



and so on. Observing (5), (6), (7) and (8) etc., by mathematical induction, we have

2 2 1( )( , ) , 1, 2, 3, ...

( 1)!

mx t

mx tK x t e mm

... (9)


1 21 2 31

( , ; ) ( , ) ( , ) ( , ) ( , ) ...mmm

R x t K x t K x t K x t K x t

2 2 2 2 2 2 2( ) ( ) ...,1! 2!

x t x t x tx t x te e e using (2) and (9)

2 2 2( ) ( )1 ...1! 2!

x t x t x te

2 2x t x te e ... (10)


0( ) ( ) ( , ; ) ( )

xy x f x R x t f t dt

or2 2 2 2

0( ) ,

xx x t x t ty x e e e e dt using (2) and (10)

or2 2 2 2

00( )

x xx x x t x x x ty x e e e dt e e e or 2 2 2 2 2

( ) [ 1] –x x x x x x x xy x e e e e e e or 2( ) .x xy x e

Ex. 5. Solve the following integral equation by successive approximation

0( ) ( ) ( ) .

x x ty x f x e y t dt and find the resolvent kernel.

Sol. Given0

( ) ( ) ( )x x ty x f x e y t dt ... (1)

Comparing (1) with0

( ) ( ) ( , ) ( ) ,x

y x f x K x t y t dt here K (x, t) = ex – t. ... (2)

Proceed as in solved Ex. 2. and show that

( ) (1 )( , ; ) x tR x t e ... (3)Now, the required solution of (1) is given by

0( ) ( ) ( , ; ) ( )


or ( ) (1 )

0( ) ( ) ( ) ,

x x ty x f x e f t dt by (3).

Ex. 6. By means of resolvent kernel, find the solution of

0

2 cos( ) sin ( ) .2 cos

xx xy x e x y t dtt

[Meerut 2009]

Sol. Given 0

2 cos( ) sin ( ) .2 cos

xx xy x e x y t dtt

... (1)



Comparing (1) with0

( ) ( ) ( , ) ( ) ,x

y x f x K x t y t dt we have f (x) = ex sin x, 1, K (x, t) 2 cos

2 cosxt

... (2)

Proceed as in solved Ex. 3. and show that

2 cos( , ; )2 cos

x txR x t et

[Note that here 1 , by (2)] ... (3)

The required solution is given by

0( ) ( ) ( , ; ) ( )


0

2 cossin sin ,2 cos

xx x t txe x e e t dtt

using (2) and (3)

0

sinsin (2 cos )2 cos

xx x te x x e dtt

0sin (2 cos )[log (2 cos )]x x xe x e x t

2 cos( ) sin (2 cos ) log

3x x xy x e x e x

Ex. 7. Solve y (x) = sin x + 2 0

( ) ,x x te y t dt [G.N.D.U. Amritsar 2004,

Meerut 2000, 02, 03, 12; Kanpur 2009]

Sol. Given0

( ) sin 2 ( ) .x x ty x x e y t dt ... (1)

Comparing (1) with0

( ) ( ) ( , ) ( ) ,x

y x f x K x t y t dt we have f (x) = sin x, 2, K (x, t) = ex – t. ... (2)

Proceed as in Ex. 2. and show that R (x, t; ) = e(x – t) (1+ ) = e3 (x – t). [ 2, by (2)]

Now, the required solution of (1) is

0( ) ( ) ( , ; ) ( )


3( ) 3 3

0 0sin 2 sin sin 2 sin

x xx t x tx e t dt x e e t dt

3

3

0

sin 2 ( 3sin cos )10

xtx ex e t t

as 2 2

( sin – cos )sinax

ax e a bx b bxe bx dxa b

or y(x) 3

3sin ( 3 sin cos ) 15

xxex e x x or 31 2 1( ) sin cos

5 5 5xy x e x x

Ex. 8. With helop of the resolvent kernel, find the solution of the integral equation2

220

1( ) 1 ( ) .1

x xy x x y t dtt

[Amritsar 2004, Meerut 2000, 08]



Sol. Given2

220

1( ) 1 ( ) .1

x xy x x y t dtt

... (1)

Comparing (1) with0

( ) ( ) ( , ) ( )x

y x f x K x t y t dt we have f (x) = 1 + x2, 1, K (x, t) = (1 + x2)/(1 + t2). ... (2)

Let Km (x, t) be the mth iterated kernel. Then, we haveK1 (x, t) = K (x, t) ... (3)

and 1( , ) ( , ) ( , ) .x

m mt

K x t K x z K z t dz ... (4)

From (2) and (3), K1 (x, t) = K (x, t) = (1 + x2) / (1 + t2) ... (5)Putting m = 2 in (4), we have

2 2

2 1 2 21 1( , ) ( , ) ( , ) . ,1 1

x x

t t

x zK x t K x z K z t dz dzz t

by (5)

2 2

2 21 1 ( ).1 1

x

t

x xdz x tt t

... (6)

Next, putting m = 3 in (4), we have

2 2

3 2 2 21 1( , ) ( , ) ( , ) . .( ) ,1 1

x x

t t

x zK x t K x z K z t dz z t dzz t

by (5) and (6)

2 2 2 2 2

2 2 21 1 ( ) 1 ( )( ) . .

2 2!1 1 1

xx

tt

x x z t x x tx t dzt t t

... (7)

Now, putting m = 4 in (4), we have2 2 2

4 3 2 21 1 ( )( , ) ( , ) ( , ) . . ,

2!1 1

x x

t t

x z z tK x t K x z K z t dz dzz t

by (5) and (7)

2 2 3 2 32

2 2 21 1 1 1 ( 1) 1 ( )( ) ,

2! 2! 3 3!1 1 1

xx

tt

x x z x x tz t dzt t t

... (8)

and so on. Observing (5) (6), (7) and (8) etc. by mathematical induction, it follows that2 1

21 ( )( , ) , 1, 2, 3, ...

( 1)!1

m

mx x tK x t m

mt

... (9)

Now, by the definition of the resolvent kernel R (x, t; ), we have

11 1

( , ; ) ( , ) ( , ),mm mm m

R x t K x t K x t

[ 1]

= K1 (x, t) + K2 (x, t) + K3 (x, t) + ...2 2 2 2

2 2 21 1 ( ) 1 ( ) ...

1! 2!1 1 1x x x t x x tt t t

2 2 2

2 21 ( ) ( ) 11 ... .

1! 2!1 1x tx x t x t x e

t t

... (10)




0

( ) ( ) ( , ; ) ( )x

y x f x R x t f t dt 2

2 22

11 (1 ) .,1

x x t

t

xx e t dtt

using (2) and (10)

2 2 2 200

1 (1 ) 1 (1 )x xx t x tx x e e dt x e x e

= 1 + x2 + ex (1 + x2) [–e–x + 1] = 1 + x2 – (1 + x2) + ex (1 + x2) y (x) = ex (1 + x2).

Ex. 9. Solve 0

( ) 1 ( ) .x

y x y t dt [Kanpur 2007, 08]

Sol. Given0

( ) 1 ( ) .x

y x y t dt ...(1)

Comparing (1) with0

( ) ( ) ( , ) ( ) ,x

y x f x K x t y t dt we have f (x) = 1, 1, K (x, t) = 1. ... (2)

Proceeding as in Ex. 1, we have( )( , ; ) x t x tR x t e e [ 1, by (2)] ... (3)

Now, the required solution of (1) is given by

0( ) ( ) ( , ; ) ( )


or0

( ) 1 ,x x ty x e dt using (1)

00

1 1 1 [ 1] 1 1 .x xx t x t x x xe e dt e e e e e

y (x) = ex.

Ex. 10. Solve 0

( ) ( ) ( ) .x

y x x t x y t dt

Sol. Given0

( ) ( – ) ( ) .x

y x x t x y t dt ...(1)

Comparing (1) with y(x) = f (x) + 0

( , ) ( ) ,x

K x t y t dtwe have f (x) = x, 1, K (x, t) = t – x. ... (2)

Let ( , )mK x t be the mth iterated kernel. Then

1( , ) ( , )K x t K x t ...(3)

and 1( , ) ( , ) ( , ) , 2,3,...x

m mtK x t K x z K z t dz m ...(4)



From (2) and (3), 1( , ) ( , ) –K x t K x t t x ...(5)Putting m = 2 in (4) and using (5), we get

2 1( , ) ( , ) ( , ) ( – )( – )x x

t tK x t K x z K z t dz z x t z dz

2 2( – ) ( – )( – ) – (–1)2 2

xx

tt

z x z xt z dz

, integrating by parts

32 ( – )1 1( – )

2 2 3

xx

tt

z xz x dz

K2 (x, t) = – (t – x)3 / 3! ... (6)Next, putting m = 3 in (4), we have

3

3 2( )( , ) ( , ) ( , ) ( ) ,

3!

x x

t t

t zK x t K x z K z t dz z x dz

by (5) and (6)

4 431 1 ( ) ( )( ) ( ) ( ) 1.

3! 3! ( 4) ( 4)

xx x

t tt

t z t zz x t z dz z x dz

541 1 ( )( )

4.3! 4.3! ( 5)

xx

tt

t zt z dz

K3 (x, t) = (t – x)5 / 5 !, ... (7)

and so on. Observing (5), (6) and (7) etc. by mathematical induction, we have2 1

1 ( )( , ) ( 1) , 1, 2, 3, ...(2 1)!

mm

mt xK x t m

m

... (8)

Now, by the definition of the resolvent kernel

1

1 1( , ; ) ( , ) ( , )m

m mm mR x t K x t K x t

[ 1, by (2)]

= K1 (x, t) + K2 (x, t) + K3 (x, t) + ....3 5( ) ( ) ( ) .. sin ( )

1! 3! 5!t x t x t x t x

... (9)


0 0

( ) ( ) ( , ; ) ( ) sin ( )x x

y x f x R x t f t dt x t t x dt , by (2) and (9)

0 00{ cos ( )} 1.{ cos ( )} sin ( )

xx xx t t x t x dt x x t x y (x) = sin x.

Ex. 11. Solve 0

( ) 1 ( ) ( ) .x

y x t x y t dt [Meerut 2008, 10]

Sol. Comparing (1) with0

( ) 1 ( ) ( ) ,x

y x t x y t dt ... (1)

here f (x) = 1, 1, K (x, t) = t – x. ... (2)



Proceed as in solved Ex. 10 to show that R (x, t; ) = sin (t – x) ... (3)Now the required solution is given by

0( ) ( ) ( , ; ) ( )

xy x f x R x t f t dt or

0( ) 1 sin ( ) ,

xy x t x dt by (2) and (3)

or 0( ) 1 cos ( ) 1 1 cos .xy x t x x or cosy x

Ex. 12. Solve 0

( ) cos 2 ( ) ( ) .x

y x x x t x y t dt

Sol. Given0

( ) cos 2 ( ) ( ) .x

y x x x t x y t dt ... (1)

Comparing (1) with0

( ) ( ) ( ) ( ) ,x

y x f x t x y t dt here f (x) = cos x – x – 2, 1, K (x, t) = t – x, ... (2)Proceeding as in solved Ex. 10, we have R (x, t; ) = sin (t – x) ... (3)Now, the required solution of (1) is given by

0( ) ( ) ( , ; ) ( )


0cos 2 sin ( ) (cos 2) , using (2)

xx x t x t t dt

0 0 0cos 2 sin ( ) cos sin ( ) 2 sin ( )

x x xx x t x t dt t t x dt t x dt

0

1cos 2 [sin (2 ) – sin ]2

xx x t x x dt 0 0

sin ( ) 2 sin ( )x xt t x dt t x dt

0

1 cos (2 )cos 2 sin2 2

xt xx x t x

0 00cos ( ) 1.{ cos ( )} 2 cos ( )

xx xt t x t x dt t x

[Integrating by parts the 2nd integral]

01 cos coscos 2 sin sin ( ) 2 (1 cos )2 2 2

xx xx x x x x t x x = cos x – x – 2 – (1/2) × x sin x + x + sin (–x) + 2 – 2 cos x y (x) = – cos x – sin x – (x/2) × sin x.

EXERCISE 5CSolve the following integral equations :

1.0

( ) ( ) .xx x ty x e e y t dt Ans. y (x) = e2x.

2.0

( ) .3 3 ( ) .xx x ty x x y t dt Ans. y (x) = 3x (1 – e–x)



3.2 2

0( ) 1 2 ( ) .

x x ty x x e y t dt Ans. 2

( ) 2 .x xy x e x

4.2 2 22

0( ) 2 ( ) .

xx x x ty x e e y x dt Ans.2 2( ) (1 2 ).x xy x e x

5. 2 0

1( ) sin ( ) ( ) .1

xy x x t y t dt

x

Ans. 1 22

1 1( ) tan log (1 )21

y x x x xx

6.2 / 2 ( )

0( ) ( ) .

xx x ty x x e e y t dt Ans.2 / 2( ) ( 1) 1.xy x e x

7. ( )

0( ) sin ( ) ( ) .

xx x ty x e e x t y t dt Ans. y (x) = e–x [1 + (x2 / 2)].

8.0

( ) ( ) .xx x ty x e e y t dt Ans. y (x) = 1.

Type 3. Solution of Volterra integral equation when the sum of the infinite series occuing in theformula for resolvent kernel cannot be obtained in closed form. In such probems we use thefollowing formula for solution, which is also known as Neumann series :

m 1

( ) = ( ) + ( ) ( )x

mm= ay x f x K x,t f t dt, ... (i)

where Km (x, t) is the mth iterated kernel. Here (i) is solution of given Volterra integral equation

( ) = ( ) + ( ) ( )x

ay x f x K x,t y t dt. ... (ii)

Ex. 13. Find the Neumann series for the solution of the integral equation

0( ) 1 ( ) ( ) .

xy x x x t y t dt

Sol. Given0

( ) 1 ( ) ( ) .x

y x x x t y t dt ... (1)

Comparint (1) with0

( ) ( ) ( , ) ( ) .x

y x f x K x t y t dt here f (x) = 1 + x, , K (x, t) = x – t. ... (2)

Let Km (x, 1) be the mth iterated kernel. ThenK1 (x, t) = K (x, t) ... (3)

and 1( , ) ( , ) ( , ) .x


From (2) and (3), K1 (x, t) = K (x, t) = x – t. ... (5)Putting m = 2 in (4) and using (5), we have

2 1( , ) ( , ) ( , ) ( ) ( ) ,x x

t tK x t K x z K z t dz x z z t dz using (5)

2 2( ) ( 1) ( )( ) .2 2

xx

tt

z t z tx z dz

3 3

22

1 1 ( ) ( )( , ) ( )2 2 3 3!

xx

tt

z t x tK x t z t dz

... (6)



Putting m = 3 in (4), we have3

3 2( )( , ) ( , ) ( , ) ( )

3!

x x

t t

z tK x t K x z K z t dz x z dz , using (6)

4 4( ) ( )( ) ( 1)4.3! 4.3!

xx

tt

z t z tx z dz

5 5

41 1 ( ) ( )( ) ,4! 4! 5 5!

xx

tt

z t x tz t dz

... (7)

and so on.Now the Neumann series for the solution of (1) is given by

1 0( ) ( ) ( , ) ( ) , . .,

xmmm

y x f x K x t f t dt i e

21 2

0 0( ) 1 ( , ) (1 ) ( , ) (1 )

x xy x x K x t t dt K x t t dt 3

30

( , ) (1 ) ...,x

K x t t dt [using (2)]

3 52 3

0 0 0

( ) ( )1 ( ) (1 ) (1 ) (1 ) ...3! 5!

x x xx t x tx x t t dt t dt t dt

2 2 2 4

00 0

( ) ( ) ( )1 (1 ) (1 )( 2) ( 2) 3! ( 4)

x xxx t x t x tx t dt t

4 3 6 6

0 00

( ) ( ) ( )(1 ) ...( 4) 5! ( 6) ( 6)

xx xx t x t x tdt t dt

2 3 2 4 5 3 6 7

0 0 0

1 ( ) 1 ( ) 1 ( )1 ...2 2 ( 3) 3! 4 4 ( 5) 5! 6 6 ( 7)

x x xx x t x x t x x tx

2 3 2 4 5 3 6 71 ...

2 6 6 4 20 120 6 42x x x x x xx

or2 3 4 5 6 7

2 3( ) 1 ...2! 3! 4! 5! 6! 7!x x x x x xy x x

... (8)

Remark 1. In particular case, if 1, then (8) reduces to2 3 4

( ) 1 ... .2! 3! 4!

xx x xy x x e ... (9)

Remark 2. In above solved example, the resolvent kernel is given by

1 21 2 31

( , ; ) ( , ) ( , ) ( , ) ( , ) ...mmm


3 52( ) ( )( ) ...,

3! 5!x t x tx t



whose sum cannot be obtained in closed form. Hence solution cannot be obtained by the usual

formula0

( ) ( ) ( , ; ) ( ) .x

y x f x R x t f t dt We, therefore, find solution by using Neumann series. Due to the same reason, in the following

example we have to use Neumann series.

Ex. 14. Solve the Volterra integral equation 0

( ) 1 ( ) .x

y x xt y t dt Sol. Given

0( ) 1 ( ) .

xy x xt y t dt ... (1)

Comparing (1) with0

( ) ( ) ( , ) ( ) ,x

y x f x K x t y t dt here f (x) = 1, 1, K (x, t) = xt. ... (2)

Let Km (x, t) be the mth terated kernel. Then K1 (x, t) = K (x, t) ... (3)

and ( , ) ( , ) ( , ) .x


From (2) and (3), K1 (x, t) = K (x, t) = xt. ... (5)Putting m = 2 in (4), we have

3

2 1( , ) ( , ) ( , ) ( ) ( ) ,3

xx x x

t t tt

zK x t K x z K z t dz xz zt dz xt

by (5)

3 3 4 4( / 3) ( ) (1/ 3) ( ).xt x t x t xt ... (6)Next, putting m = 3 in (4) we have

3 2( , ) ( , ) ( , )x

tK x t K x z K z t dz 3 3( ). ( ) ,

3

x

t

ztxz z t dz by (5) and (6)

6 3 35 2 3( )

3 3 6 3

xx

tt

xt xt z z tz z t dz

6 3 3 6 6

3 6 3 6 3xt x x t t t

7 4 4 7(1/18) ( – 2 )x t x t xt ... (7)Next, putting m = 4 in (4), we have

4 3( , ) ( , ) ( , )x

tK x t K x z K z t dz 7 4 4 71( ) ( 2 )

18

x

txz z t z t zt dz

8 5 4 2 7( 2 )18

x

t

x z t z t z t dz 9 6 4 3 7

18 9 3 3

x

t

x z t z t z t

9 6 4 3 7 10 10 1010 7 4 4 7 101 ( 3 3 ),

18 9 3 3 9 3 3 162x x t x t x t t t t x t x t x t xt

.. (8)

and so on.Now the solution (1) is given by the Neuamnn series

1 0

( ) ( ) ( , ) ( )xm

mmy x f x K x t f t dt

1 0

1 ( , ) ,xm

mmK x t dt

using (2)



1 2 30 0 01 ( , ) ( , ) ( , ) ...

x x xK x t dt K x t dt K x t dt

4 4 7 4 4 7

0 0 0

1 11 ( ) ( 2 )3 18

x x xxt dt x t xt dt x t x t xt dt

10 7 4 4 7 10

0

1 ( 3 3 ) ...162

xx t x t x t xt dt

2 4 2 5 7 2 4 5 8

0 0 0

1 1 212 3 2 5 18 2 5 8

x x xt x t xt x t x t xtx

10 2 7 5 4 8 11

0

1 3 3162 2 5 8 11

xx t x t x t xt

3 6 6 9 9 9 12 12 12 121 1 2 1 3 31 ...

2 3 2 5 18 2 5 8 162 2 5 8 11x x x x x x x x x x

or 3 6 9 12

( ) 1 ...2 2.5 2.5.8 2.5.8.11x x x xy x

Type 4. The method of successive approximations for solving Volterra integral equationof the second kind :

.0( ) = ( ) ( ) ( )

xy x f x + K x,t y t dt ... (1)

Working Rule: Let f (x) be continuous in [0, a] and K (x, t) be continuous for 0 0x a, t x.

We start with some function y0 (x) continuous in [0, a]. Replacing y (t) on R.H.S of (1)by y0 (x), we obtain

1 00

( ) = ( ) + ( ) ( )x

y x f x K x,t y t dt. ... (2)

y1 (x) given by (2) is itself continuous in [0, a]. Proceeding likewise we arrive at a sequenceof functions y0 (x), y1 (x), ..., yn (x), ..., where

–10( ) = ( ) + ( ) ( )

xn ny x f x K x,t y t dt. ... (3)

In view of continuity of f (x) and K (x, t), the sequence {yn (x)} converges, as n toobtain the solution y (x) of given integral equation (1).

Remark. As a particular case, when y0 (x) = f (x), we obtain the so called Neumann series(refer Art. 5.11).

Ex. 15. Using the method of successive approximations, solve the integral equation

0( ) 1 ( ) ,

xy x y t dt taking y0 (x) = 0. [Kanpur 2007]

Sol. Given0

( ) 1 ( ) ,x

y x y t dt ... (1)

and y0 (x) = 0. ... (2)



Comparing (1) with0

( ) ( ) ( , ) ( ) ,x

y x f x K x y y t dt here f (x) = 1, 1, K (x, t) = 1. ... (3)

The nth order approximation is given by

10

( ) ( ) ( , ) ( )x

n ny x f x K x t y t dt or 10

( ) 1 ( ) ,x

n ny x y t dt using (3) ... (4)

Putting n = 1 in (4) and using (5), we have

1 00 0

( ) 1 ( ) 1 (0)x x

y x y t dt dt = 1. ... (5)

Next, putting n = 2 in (4) and using (5), we have

2 10 0

( ) 1 ( ) 1x x

y x y t dt dt 01 [ ] 1 .xt x ... (6)

Now, putting n = 3 in (4) and using (6), we have

2

3 20 0

0

( ) 1 ( ) 1 (1 ) 12

xx x ty x y t dt t dt t

21

2!xx ... (7)

Next, putting n = 4 in (4) and using (7), we have2

4 30 0

( ) 1 ( ) 1 12!

x x ty x y t dt t dt

2 3

12! 3!x xx ... (8)

and so on, Observing (5), (6), (7), (8) etc, we find2 3 1

( ) 1 ...1! 2! 3! ( 1)!

n

nx x x xy x

n

... (9)

Making n , we find the required solution is given by( ) lim ( )nn

y x y x

or2 3

( ) 1 ...ad inf,1! 2! 3!x x xy x or y (x) = ex.


0( ) 1 ( ) ,

xy x x y t dt taking y0 (x) = 1. [Kanpur 2010, 11; Meerut 2011; 2012]

Sol. Given 0

( ) 1 ( ) ,x

y x x y t dt ... (1)

and y0 (x) = 1. ... (2)

Comparing (1) with0

( ) ( ) ( , ) ( ) ,x

y x f x K x t y t dt here f (x) = 1 + x, 1, K (x, t) = 1. ... (3)




10( ) ( ) ( , ) ( )

xn ny x f x K x t y t dt

or 10( ) 1 ( ) ,

xn ny x x y t dt by (3) ... (4)

Putting n = 1 in (4), we have

1 00 0( ) 1 ( ) 1 ,

x xy x x y t dt x dt by (2)

y1(x) 01 [ ] 1 1.xx t x x ... (5)Next, putting n = 2 in (4) and using (5), we have

2 10 0

( ) 1 – ( ) 1 – 1x x

y x x y x dt x dt ...(6)

and so on. Observing (5) and (6), we find that .( ) 1,ny x for n = 1, 2, 3,... .....(7)

Hence, the required solution of (1) is given by ( ) lim ( ) 1nny x y x

Ex.17. Using the method of successive approximations, solve the integral equation

00

( ) – ( – ) ( ) , ( ) 0.x

y x x x t y t dt y x [Kanpur 2006]

Sol. Given0

( ) – ( – ) ( )x

y x x x t y t dt ...(1)

and 0 ( ) 0y x ...(2)

Comparing (1) with0

( ) ( ) ( , ) ( ) ,x

y x f x K x t y t dt here ( ) , –1, and ( , ) – .f x x K x t x t ...(3)

The nth approximation ( )ny x is given by

10

( ) ( ) ( , ) ( )x

n ny x f x K x t y t dt or 10( ) ( ) ( )

xn ny x x x t y t dt by (3) ... (4)

Putting n = 1 in (4) and using (2), we have

1 00 0

( ) ( ) ( ) ( ) (0)x x

y x x x t y t dt x x t dt = x ... (5)

Next putting n = 2 in (4), we have

2 10 0

( ) ( ) ( ) ( ) ,x x

y x x x t y t dt x x t t dt by (5)

or 2 3 3 3 3

20

( ) –2 3 2 3 3!

xxt t x x xy x x x x

... (6)

Now, putting n = 3 in (4), we have3

3 20 0

( ) ( ) ( ) ( ) ,6

x x ty x x x t y t dt x x t t dt

by (6)



3 4 2 4 3 52

00

6 6 2 24 3 30

xx xt t xt xt t tx xt t dt x

or 3 5 3 5

3( )2 24 3 30x x x xy x x

3 5 3 5,

6 120 3! 5!x x x xx x ... (7)

and so on. Observing (5), (6) and (7) we find that

3 5 2 1

1( ) ... ( 1) .3! 5! (2 1)!

nn

nx x xy x x

n

The required solution y (x) of (1) is given by y (x) = limn yn (x), i.e.,

3 5 7 2 11( ) ... ( 1) ...ad inf

3! 5! 7! (2 1)!

nnx x x xy x x

n

or y (x) = sin x.


0( ) 1 ( ) ( ) ,

xy x x t y t dt taking y0 (x) = 1.

Sol. Given0

( ) 1 ( ) ( )x

y x x t y t dt ... (1)

and y0 (x) = 0. .... (2)

Comparing (1) with0

( ) ( ) ( , ) ( )x

y x f x K x t y t dt here f (x) = 1, 1, K (x, t) = x – t. ... (3)


10( ) ( ) ( , ) ( )

xn ny x f x K x t y t dt

or 10( ) 1 ( ) ( ) ,

xn ny x x t y t dt by (3) ... (4)


1 00( ) 1 ( ) ( ) 1,

xy x x t y t dt by (2) ... (5)


2 10 0

( ) 1 ( ) ( ) 1 ( ) ,x x

y x x t y t dt x t dt by (5)

2 22

0

1 12 2

xt xxt x

2 21 1 .

2 2!x x

... (6)

Now, putting n = 3 in (4) and using (6), we have

2

3 20 0

1( ) 1 ( ) ( ) 1 ( ) 12

x xy x x t y t dt x t t dt

3 2 4

2 3

00

1 11 12 2 6 2 8

xx xt t tx xt t t dt xt



or4 2 4 2 4

23 ( ) 1 1

6 2 8 2 24x x x x xy x x

2 41 ,

2! 4!x x

... (7)

and so on. Observing (5), (6) and (7), we have

2 4 2 2

( ) 1 ... .2! 4! (2 2)!

n

nx x xy x

n

Now, the required solution y (x) of (1) is given by ( ) lim ( )nny x y x

or2 4 2 2

( ) 1 ... ...ad inf2! 4! (2 2)!

nx x xy xn

or ( ) coshy x x

Ex. 19. The integral equation 0

( ) ( ) ( )x

y x x x t y t dt is solved by the method of successiveapproximations. Starting with initial approximation y0(x) = x, the second approximation is given by

(a) y2(x) = x + x3/3! + x5/5! (b) y2(x) = x + x3/3!(c) y2(x) = x – x3/3! (d) y2(x) = x – x3/3! + x5/5! [GATE 2005]

Sol. Given0

( ) ( ) ( )x

y x x x t y t dt ... (1)

and y0(x) = x ... (2)

Comparing (1) with 0

( ) ( ) ( , ) ( ) ,x

y x f x K x t y t dt we get

f (x) = x, 1 and K(x, t) = x – t ... (3)The nth order approximation is given by

10( ) ( ) ( , ) ( )

x

n ny x f x K x t y t dt

i.e., 10

( ) ( ) ( ) ,x

n ny x x x t y t dt

using (3) ... (4)


1 00 0( ) ( ) ( ) ( ) ,

x xy x x x t y t dt x x t t dt by (2)

or 2 31 0( ) /2 /3

xy x x xt t or y1(x) = x – x3/6 ... (5)


32 10 0( ) ( ) ( ) ( ) ( /6 ) ,

x xy x x x t y t dx x x t t t dt by (5)

or2 3 4 5

2 3 42 0

0

( ) ( / 6 / 6)2 3 24 30

xx xt t xt ty x x xt t xt t dt x

or 3 3 5 5 3 52( ) ( / 2 / 3 / 24 / 30) / 6 /120y x x x x x x x x x

Thus, y2(x) = x – x3/3! + x5/5!



EXERCISE 5 D1. Using the method of successive approximations, solve the following integral equation with

given value of y0 (x) of zero-order approximation :

(i) 00

( ) 1 ( ) ( ) , ( ) 0.x

y x x t y t dt y x Ans. y (x) = cos x.

(ii) 00( ) 1 ( ) , ( ) 1.

xy x x y t dt y x x Ans. y (x) = 1.

(iii) 200

1( ) ( ) , ( ) 1.2

xy x x x y t dt y x Ans. y (x) = x.

(iv) 20

0

1( ) ( ) , ( ) .2

xy x x x y t dt y x x Ans. y (x) = x.

(v) 2 200

1 1( ) ( ) , ( ) .2 2

xy x x x y t dt y x x x Ans. y (x) = x.

(vi) 00( ) 1 ( ) ( ) , ( ) 1.

xy x x x t t dt y x [Meerut 2007] Ans. y (x) = ex.

(vii) 00

( ) 2 2 ( ) , ( ) 1.x

y x x y t dt y x Ans. y (x) = 2.

(viii) 00( ) 2 2 ( ) , ( ) 2.

xy x x y t dt y x Ans. y (x) = 2.

(ix) 200

( ) 2 2 ( ) , ( ) 2.x

y x x x y t dt y x Ans. y (x) = 2.

(x) 200

( ) 2 2 ( ) , ( ) 2 .x

y x x x y t dt y x x Ans. y (x) = 2.

(xi) 3 200

1( ) 2 ( ) , ( ) .2

xy x x x y t dt y t x Ans. y (x) = x2 – 2x.

2. Find an approximate solution of the integral equation 0

( ) sinh ( ) ,x t xy x x e y t dt by the

method of iteration.3. Prove that the resolvent kernel for a Volterra integral equation of the second kind is an

entire function of for any given (x, t).

5.14 SOLUTION OF VOLTERRA INTEGRAL EQUATION OF THE SECOND KIND

0( ) = ( ) + ( ) ( )

xy x f x K x,t y t dt, ...(1)

when its kernel K (x, t) is of some particular forms.Particular Form I*Suppose that the kernel K (x, t) is a polynomial of degree (n – 1) in t,

which can always be represented in the form2 1

0 1 2 1( ) ( )( , ) ( ) ( ) ( ) ( ) ...

2! ( 1)!

n

nx t x tK x t a x a x x t a x a

n

... (2)

* For proof, refer ‘‘A course in mathematical analysis’’, Volume III, Part II, by EdouardGoursat, Art. 94, page 7.



Then the resolvent kernel R (x, t; ) of (1) is given by

1 ( , ; )( , ; ) ,n

nd g x tR x t

dx

... (3)

where g (x, t; ) is a solution of the differential equation

1 2

0 1 11 2( ) ( ) ... ( ) 0,n n n

nn n nd g d g d ga x a x a x gdx dx dx

... (4)

satisfying the conditions

and

2 2

2 2

1

1

... 0 when ,

1 when

n

n

n

n

dg d g d gg x tdx dx dx

d g x tdx

... (5)

The required solution of (1) is given by

( ) ( ) ( , ; ) ( ) .x


Particular Form II. Suppose that the kernel K (x, t) is a polynomial of degree (n – 1) in t,which can always be represented in the form

K (x, t) = b0 (t) + b1 (t) (t – x) + b2 (t) 2 1

1( ) ( )... ( )

2! ( 1)!

n

nt x t xb t

n

...(2)

Then the resolvent kernel R (x, t; ) of (1) is given by

1 ( , ; )( , ; ) ,n

nd h x tR x t

dt

...(3)

where h (x, t; ) is a solution of the differential equation1 2

0 1 11 2( ) ( ) ... ( ) 0n n n

nn n nd h d h d hb t b t b t hdt dt dt

...(4)

satisfying the conditions

and

2 2

2 2

1

1

... 0 when ,

1 when .

n

n

n

n

dh d h d hh t xdt dt dt

d h t xdt

...(5)

The required solution of (1) is again given by (6).

Ex. 1. Solve 0

( ) 29 6 [5 6 ( )] ( ) .x

y x x x t y t dt [Meerut 2006, 10]

Sol. Given0

( ) 29 6 [5 6 ( )] ( )x

y x x x t y t dt ... (1)

Comparing (1) with0

( ) ( ) ( , ) ( ) ,x

y x f x K x t y t dt here f (x) = 29 + 6x and 1. ... (2)and K (x, t) = 5 – 6 (x – t). ... (3)



Let K (x, t) = a0 (x) + a1 (x) (x – t). ... (4)Comparing (3) and (4), a0 (x) = 5 and a1 (x) = – 6. ... (5)Then the resolvent kernel R (x, t; ) of (1) is given by

2

2( , ; )( , ; ) d g x tR x tdx

or

2 2

2 2( , ;1)( , ;1) ,d g x t d gR x t

dx dx as 1. ... (6)

where g (x, t; 1) satisfies the differential equation

2

0 12 ( ) ( ) 0,d g dga x a x gdxdx

where 1

or 2

2 5 6 0d g dg gdxdx

or (D2 – 5D + 6) g = 0, when /D d dx ... (7)

satisfying the conditions g = 0, when x = t. ... (8A)and dg / dx = 1, when x = t. ... (8B)

Now, the auxiliary equation of (7) is D2 – 5D + 6 = 0 giving D = 3, 2.Hence the general solution of (7) is given by

g = Ae3x + Be2x. ... (9)From (9) dg/dx = 3Ae3x + 2Be2x. ... (10)Putting x = t in (9) and (10) and using (8A) and (8B), we obtain

0 = Ae3t + Be2t ... (11)and 1 = 3Ae3t + 2Be2t. ... (12)

Solving (11) and (12), we have 3tA e and B = – e–2t. ... (13)Subsituting these values in (9), we have

g = g (x, t; 1) = e3(x – t) – e2 (x – t) ... (14)Differentiating both sides of (14) w.r.t. ‘x’, we have

3( ) 2( )/ 3 2x t x tdg dx e e ... (15)

Differentiations both sides of (15) again w.r.t. ‘x’, we have2 2 3( ) 2( )/ 9 4 .x t x td g dx e e ... (16)

Using the above value of 2 2/d g dx in (6), we have

R (x, t; 1) = 9 e3 (x – t) – 4e2 (x – t). ... (17)Hence the required solution of (1) is

0( ) ( ) ( , ; ) ( )

xy x f x R x t f t dt 0

29 6 ( , ;1) (29 6 ) ,x

x R x t t dt by (2)

3( ) 2( )

029 6 [9 4 ] (29 6 ) ,

x x t x tx e e t dt using (17)

3( ) 2( )0

29 6 (29 6 ){ 3 2 }xx t x tx t e e

3( ) 2( )

06{ 3 2 }

x x t x te e dt [on integration by parts]



= 29 + 6x + ( 29 + 6x) (– 3 + 2) – 29 ( – 3e3x + 2e2x) – 6 3( ) 2( )0

xx t x te e

= 29 + 6x – 29x – 6x – 29 (–3e3x + 2e2x) – 6 [ 0 – (e3x – e2x)] y (x) = 93 e3x – 64 e2x.

Ex. 2. Solve 2 2

0( ) 1 2 4 [3 6 ( ) 4( ) ] ( ) .

xy x x x x t x t y t dt

Sol. Given 2 2

0( ) 1 2 4 [3 6 ( ) 4( ) ] ( ) .

xy x x x x t x t y t dt ... (1)

Comparing (1) with0

( ) ( ) ( , ) ( ) ,x

y x f x K x t y t dt here f (x) = 1 – 2x – 4x2 and 1, ... (2)and K (x, t) = 3 + 6 (x – t) – 4 (x – t)2. ... (3)

Let K (x, t) = a0 (x) + a1 (x) (x – t) + a2 (x) 2( )

2!x t

. ... (4)

Comparing (3) and (4), a0 (x) = 3, a1 (x) = 6, a2 (x) = – 8. ... (5)Then the resolvent kernel R (x, t; ) of (1) is given by

3

3( , ; )( , ; ) d g x tR x tdx

or

3 3

3 3( , ;1)( , ;1) ,d g x t d gR x tdx dx

as 1. ... (6)

where g (x, t; 1) satisfies the equation

3 2

0 1 23 2( ) ( ) ( ) 0d g d g dga x a x a x gdxdx dx

or

3 2

3 23 6 8 0d g d g dg gdxdx dx

or (D3 – 3D2 – 6D + 8) g = 0 /D d dx ... (7)satisfying the conditions g = 0, dg / dx = 0 when x = t. ... (8A)and d2g / dx2 = 1 when x = t. ... (8B)

Now, the auxiliary equation of (7) is3 2– 3 – 6 8 0, . .,D D D i e (D – 1) (D + 2) (D – 4) = 0 giving D = 1, –2, 4.

Hence the general solution of (7) is g = Aex + Be–2x + Ce4x. ... (9)

From (9) dg / dx = Aex – 2Be–2x + 4 Ce4x ... (10)and d2g / dx2 = Aex + 4Be–2x + 16 Ce4x. ... (11)

Putting x = t in (9), (10) and (11) and using (8A) and (8B), we have

2 4

2 4

2 4

0 ,

0 2 4 ,

and 1 4 16 .

t t t

t t t

t t t

Ae Be Ce

Ae Be Ce

Ae Be Ce

... (12)

Solving (12) for A, B, C, we have2 4(1/ 9) , (1/18) , (1/18) .t t tA e B e C e ... (13)



Substituting these values in (9), we have

2 2 4 4( , ;1) (1/ 9) (1/18) (1/18) .x t x t x tg g x t e e e ... (14)Differentiating both sides of (14) w.r.t. ‘x’, we get

2 2 4 41 1 1 .9 9 9

x t x t x tdg e e edx

... (15)

Again, differentiating both sides of (15) w.r.t. ‘x’, we get2

2 2 4 42

1 2 8 .9 9 9

x t x t x td g e e edx

... (16)

Finally, differentiating both sides of (16) w.r.t. ‘x’, we get3

2 2 4 43

1 4 32 .9 9 9

x t x t x td g e e edx

... (17)

Using the above value in (6), we have2( ) 4( )( , ; ) (1/9) (4 / 9) (32 /9) .x t x t x tR x t e e e ... (18)

Hence the required solution of (1) is

0( ) ( ) ( , ; ) ( )


or 2 2

0( ) 1 2 4 ( , ;1) (1 2 4 ) ,

xy x x x R x t t t dt by (2)

or 2( ) 1 2 4y x x x 2 2( ) 4( )

0

1 4 32(1 2 4 )9 9 9

x x t x t x tt t e e e dt , by (18)

2 2 2( ) 4( )

0

1 2 81 2 4 (1 2 4 )9 9 9

xx t x t x tx x t t e e e

2( ) 4( )

0

1 2 8( 2 8 ) ,9 9 9

x x t x t x tt e e e dt integrating by parts

2 2 2 41 2 8 1 2 81 2 4 (1 2 4 )

9 9 9 9 9 9x x xx x x x e e e

2( ) 4( )

0

1 2 8(2 8 )9 9 9

x x t x t x tt e e e dt

2 41 2 89 9 9

x x xe e e 2( ) 4( )

0

1 1 2(2 8 )9 9 9

xx t x t x tt e e e

2( ) 4( )

0

1 1 289 9 9

x x t x t x te e e dt

2 41 2 8 1 1 2(2 8 )

9 9 9 9 9 9x x xe e e x

2 4 2( ) 4( )

0

1 1 2 1 1 12 89 9 9 9 18 18

xx x x x t x t x te e e e e e



2 4 2 41 4 4 1 1 1 1 1 18

9 9 9 9 18 18 9 18 18x x x x x xe e e e e e

y (x) = ex.

Ex. 3. Solve 0

( ) cos 2 ( ) ( ) .x

y x x x t x y t dt Sol. Given

0( ) cos 2 ( ) ( ) .

xy x x x t x y t dt ... (1)

Comparing (1) with0

( ) ( ) ( , ) ( ) ,x

y x f x K x t y t dt here f (x) = cos x – x – 2. 1, ... (2)and K (x, t) = (t – x) ... (3)

Let K (x, t) = b0 (t) + b1 (t) (t – x). ... (4)Comparing (3) and (4), b0 (t) = 0, b1 (t) = 1 ... (5)Then the resolvent kernel R (x, t; ) of (1) is given by

2

21 ( , ; )( , ; ) d h x tR x t

dt

or2 2

2 2( , ;1)( , ;1) ,d h x t d hR x tdt dt

as 1 ... (6)

where h or h (x, t; 1) satisfies the differential equation

or2

0 12 ( ) ( ) 0d h dhb t b t hdtdt

or

2

2 (0 ) 0,d h hdt

using (2) and (5)

or (D2 + 1) h = 0, where / ,D d dt ... (7)satisfying the conditions.

h = 0 at t = x, ... (8A)and dh / dt = 1 at t = x. ... (8B)

Now, the auxiliary equation of (7) is D2 + 1 = 0 so that D = 0 .iHence the general solution of (7) is

h = A cos t + B sin t. ... (9)From (9) dh / dt = – A sin t + B cos t. ... (10)Putting t = x in (9) and (10) and using (8A) and (8B), we have

0 = A cos x + B sin x ... (11)and 1 = – A sin x + B cos x. ... (12)

Solving (11) and (12) for A and B, we getA = – sin x and B = cos x. ... (13)

Substituting these values in (9), we geth = – sin x cos t + cos x sin t or h = sin (t – x) ... (14)Differentiating (14) w.r.t. ‘t’, we get dh / dt = cos (t – x) ... (15)Again, differentiating (15) w.r.t. ‘t’, we get d2h / dt2 = – sin (t – x). ... (16)Using this value in (6), we have

R (x, t; 1) = – { – sin (t – x) } = sin (t – x). ... (17)Hence the required solution is given by

0( ) ( ) ( , ; ) ( )




or 0

( ) cos 2 ( , ;1) (cos 2) ,x

y x x x R x t t t dt using (2)

0cos 2 sin ( ) (cos 2) ,

xx x t x t t dt using (17)

0 0cos 2 sin ( ) cos ( 2) sin ( )

x xx x t x t dt t t x dt

0

1cos 2 [sin (2 ) sin ]2

xx x t x x dt 0[( 2){ cos ( )}]xt t x

0( 1){ cos ( ) }

xt x dt [In last term, integrating by parts]

0

1 cos (2 )cos 2 sin2 2

xt xx x t x 0( 2) 2cos [sin ( )]xx x t x

1 cos coscos sin sin ( )2 2 2

x xx x x x

or ( ) – cos – ( / 2) sin – sin .y x x x x x

EXERCISES 5EFind the resolvent kernels of Voltera integral equations with the following kernels (taking

1 ).1. K (x, t) = 2 – (x – t). Ans. ex – t (x – t + 2).

2. K (x, t) = –2 + 3 (x – t). Ans. 3( )(1/ 4) (9 / 4) .x t x te e

3. K (x, t) = 2x Ans. 2 22 .x tx e

4. 4 2 8( )( , ) .2 1 2 1x x tK x tx x

Ans. 2

2( )2 2

4 1 8 42(2 1) 4 1

x tt et t

5.15. SOLUTION OF VOLTERRA INTEGRAL EQUATION OF THE SECOND KIND BYREDUCING TO DIFFERENTIAL EQUATION.The whole procedure will be clear by following example.

Ex. 1. Solve 2 2

0( ) 1 2 4 [3 6 ( ) 4 ( ) ] ( ) .

xy x x x x t x t y t dt [Meerut 2012]

Sol. Given 2 2

0( ) 1 2 4 [3 6 ( ) 4 ( ) ] ( ) .

xy x x x x t x t y t dt ... (1)

Differentiating both sides of (1) w.r.t. x and using Leibnitz’s rule of differentiation under thesign of integration (refer Art. 1.13), we have

0( ) 2 8 {6 8 ( )} ( ) 3 ( ).

xy x x x t y t dt y x ... (2)

Again, differentiating both sides of (2) w.r.t. ‘x’, and using Leibnitz’s rule as before, we get

0( ) 8 8 ( ) 6 ( ) 3 ( ).

xy x y t dt y x y x ... (3)

Finally, differentiating both sides of (3) w.r.t. ‘x’, and using Leibnitz’s rule as before, we get



( ) 8 ( ) 6 ( ) 3 ( ).y x y x y x y x ... (4)Putting x = 0 in (1), we have y (0) = 1. ... (5)Next, putting x = 0 in (2), we have

(0)y = – 2 + 3y (0) = – 2 + 3 = 1, using (5) ... (6)Now, putting x = 0 in (3), we have

y (0) = – 8 + 6y (0) + 3y (0) = – 8 + 6 + 3 = 1,using (5) and (6) ... (7)Re-writing (4), we have

3 2

3 23 6 8 0d y d y dy ydxdx dx

or (D3 – 3D2 – 6D + 8) = 0. ... (8)

We shall solve the reduced differential equation by using the prescribed initial conditions (5),(6) and (7). The auxiliary equation of (8) is

D3 – 3D2 – 6D + 8 = 0 giving D = 1, 4, – 2.Hence the general solution of (8) is

y (x) = Aex + Be4x + Ce–2x. ... (9)

From (9), ( )y x = Aex + 4Be4x – 2Ce–2x. ... (10)

From (10), ( )y x = Aex + 16Be4x + 4Ce–2x. ... (11)Putting x = 0 in (9), (10) and (11) and using (5) (6) and (7), we obtain

1 = A + B + C, ... (12)1 = A + 4B – 2C, ... (13)

and 1 = A + 16B + 4C. ... (14)Solving (12), (13) and (14) for A, B, C, we obtain A = 1, B = 0, C = 0With these values, (9) gives the required solution y (x) = ex.

5.16.VOLTERRA INTEGRAL EQUATION OF THE FIRST KIND.Theorem. Volterra integral equation of the first kind can be converted to a Volterra integral

equation of the second kind.Proof. Let the given Volterra integral equation of the first kind be

0( , ) ( ) ( ),

xK x t y t dt f x ... (1)

where y (x) is the unknown function.Suppose that the Kernel K (x, t) and all its partial derivatives occuring in the problem are

continuous. Then clearly condition f (0) = 0 is necessary for (1) to possess a continuous solution.Moreover, if the kernel K (x, t) possess a continuous derivative ( , ) / ,K x t x the first member of(1) also possess a continuous derivatives; the same must then be true of f (x). Assuming that thiscondition is satisfied, we differentiate both sides of (1) w.r.t. ‘x’, and use Leibnitz’s rule ofdifferentiation under the sign of integration (refer Art. 1.13) to obtain

0

0( , ) ( ) ( , ) ( ) ( ,0) (0) ( )x

xdx dK x t y t dt K x x y x K x y f xdx dx

, where ( , )( , )xK x tK x t

x

or0

( , ) ( ) ( , ) ( ) ( ).x

xK x t y t dt K x x y x f x ... (2)

Conversely, every solution of (2) also satisfies (1), since the two members are zero at x = 0,and their derivatives are identical.

Let K (x, x) 0 at any point of the basic interval [0, a]. Then dividing both sides of (2) by



K (x, x), we get

0

1 ( )( , ) ( ) ( )( , ) ( , )

xx

f xK x t y t dt y xK x x K x x

or 0

( , )( )( ) ( ) .( , ) ( , )

xxK x tf xy x y t dt

K x x K x x

... (3)

which is a Volterra integral equantion of the second kind.On the other hand, if K (x, x) 0, then we do not get (3). In fact, when K (x,x) 0, (2) is again

an integral equation of the first kind, which can be dealt in a similarly fashion. Now, withK (x, x) 0, (2) becomes

0( , ) ( ) ( ).

xxK x t y t dt f x ... (4)

Suppose that the kernel K (x, t) possesses continuous second order partial derivative2 2( , ) / .K x t x Differentiating both sides. of (4) w.r.t. ‘x’ and using again Leibnitz’s rule, we have

0

0( , ) ( ) ( , ) ( ) ( ,0) (0) ( ),x

x xdx dK x t y t dt K x x y x K x y f xdx dx

where 2

2( , )( , )x

K x tk x tx

or0

( , ) ( ) ( , ) ( ) ( ).x

x xK x t y t dt K x x y x f x ... (5)

As before, if ( , ) 0,xK x x then dividing (5) by ( , ),xK x x we have

0

( ) ( , )( ) ( ) .( , ) ( , )

x

x x

f x K x ty x y t dtK x x K x x

... (6)

which is again a Volterra integral equation of the second kind.

On the other hand if ( , ) 0,xK x x (5) reduces to integral equation of the first kind and usethe same procedure again, and so on. In this manner we have a sequence to successive derivatives

of K (x, t) w.r.t. ‘x’, until we arrive at a derivative ( 1) ( , )pxK x t which is not identically zero for

x = t. In order that (1) have a continuous solution, it would be necessary for f (x) to possess continuousderivatives ( ), ( ),f x f x ..., f (p – 1) (x) which are all zero for x = 0. In such a situation, the first(p – 1) equations obtained by differentiating both sides of (1) w.r.t. ‘x’ are satisfied for x = 0. If

( )pxK (x, t) is also continuous, f (p) (x) must also be continuous and, differentiating both sides w.r.t.

x, one more times, we arrive at the integral equation, as before,

( ) ( 1) ( )

0( , ) ( ) ( , ) ( ) ( ).

x p p px xK x t y t dt K x x y x f x ... (7)

If |( 1) ( , ) 0,pxK x t then (7) reduces to as before

( )( )

( 1) ( 1)0

( , )( )( ) ( )( , , ) ( , )

pp x xp p

x x

K x tf xy x y t dtK x x K x x

... (8)

which is again a Volterra integral equation of the second kind. Now, (8) has a unique solution. Infact in going back step by step, we can easily prove that the solution so obtained satisfies all the



intermediate equations as well as (1).5.17.SOLUTION OF VOLTERRA INTEGRAL EQUATION OF THE FIRST KIND.

For this purpose we shall first convert the given volterra integral equation of the first kindinto a Volterra integral equation of the second as explained in Art. 6.16. Then we adopt the usualmethods of solving a Volterra integral equation of the second kind.

ILLUSTRATIVE SOLVED EXAMPLES.Ex. 1. Solved the following Volterra integral equation of the first kind.

0( ) ( ) , (0) 0.

x x tf x e y t dt f

Sol. Given 0

( ) ( )x x te y t dt f x ... (1)

and f (0) = 0. ... (2)Here the Kernel ex – t and its partial derivative 1 /xe x are continuous. Condition (2) is

necessary for (1) to possess a continuous solution. Again, since /x te x is continuous, the samemust be true for f (x).

Differentiating both sides of (1) w.r.t. ‘x’ and using Leibnitz’s rule of differention under thesign of integration, we get

0

0

( ) 0( ) ( ) (0) ( )x tx x x xe dx dy t dt e y x e y f xx dx dx

or 0

( ) ( ) ( ) ,x x ty x f x e y t dt ... (3)

which is a Volterra integral equation of the second kind.

Comparing (3) with0

( ) ( ) ( , ) ( ) ,x

y x F x K x t y t dt we have ( ) ( ), 1, ( , ) x tF x f x K x t e ... (4)

Let Km (x, t) be the mth iterated Kernel. ThenK1 (x, t) = K (x, t) ... (5)

and 1( , ) ( , ) ( , ) , ( 2, 3, ...)x

m mtK x t K x z K z t dz m ... (6)

From (4) and (5), K1 (x, t) = K (x, t) = ex – t. ... (7)Putting m = 2 in (6), we have

2 1( , ) ( , ) ( , ) ,x x x z z t

t tK x t K x z K z t dz e e dz by (7)

[ ]xx t x z x

tt

e dz e z = ex – t (x – t). ... (8)

Next, putting m = 3 in (6) and using (8), we have

3 2( , ) ( , ) ( , )x

tK x t K x z K z t dz . ( ) ( )

x xx z z t x tt t

e e z t dz e z t dz



2 2( ) ( )2 2!

xx t x t

t

z t x te e

... (9)

Now, putting m = 4 in (6) and using (9), we have

4 3( , ) ( , ) ( , )x


22( ) ( )

2! 2!

x tx xx z z t

t t

z t ee e dz z t dz

3 3( ) ( ) ,

2! 3 3!

xx tx t

t

e z t z te

... (10)

and so on. So by mathematical induction, we have

1( )( , ) , 1, 2, 3, ...

( 1)!

mx t

mz tK x t e mm

... (11)

Now, the resolvent kernel R (x, t, ) of (3) is given by

1 1

1 1( , ; ) ( , ) ( 1) ( , )m m

m mm mR x t K x t K x t

[ 1 , by (4)]

= K1 (x, t) – K2 (x, t) + K3 (x, t) – K4 (x, t) + ...

2 3( ) ( )( ) ... ad inf,

2! 3!x t x t x t x tx t x te e x t e e

using (11)

2 3

( )( ) ( ) ( )1 ... ad inf 11! 2! 3!

x t x t x tx t x t x te e e

... (12)

Hence, the required solution is given by

0( ) ( ) ( , ; ) ( )

xy x F x R x t F t dt 0

( ) ( ) ,x

f x f t dt by (4) and (12)

0( ) [ ( )] ( ) [ ( ) (0)] ( ) ( ),xf x f t f x f x f f x f x using (2)

( ) ( ) ( ).y x f x f x

Ex. 2. Solve 0

( ) sin .x x te y t dt x

Sol. Given0

( ) sin .x x te y t dt x ... (1)

Taking f (x) = sin x, we have f (0) = sin 0 = 0. ... (2)We proceed as in solved Ex. 1.Differentiating both sides of (1) w.r.t. ‘x’, we get

0( ) ( ) cos

x x te y t dt y x x or 0

( ) cos ( ) .x x ty x x e y t dt ... (3)

Comparing (3) with0

( ) ( ) ( , ) ( ) .x

y x F x K x t y t dt here F (x) = cos x, 1, K (x, t) = ex – t. ... (4)

As in solved Ex. 1, we have R (x, t; ) = 1.Hence the required solution of (1) is



0( ) ( ) ( , ; ) ( )

xy x F x R x t F t dt

or 0

( ) cos cosx

y x x t dt or 0( ) cos [sin ] cos sin .xy x x t x x

Ex. 3. Solve 0

( ) ( ), (0) 0.x x ta y t dt f x f

Sol. Given0

( ) ( ),x x ta y t dt f x ... (1)

with f (0) = 0. ... (2)Differentiating both sides of (1) w.r.t. ‘x’, we get

0log ( ) ( ) ( )

x x tea a y t dt y x f x

or0

( ) ( ) log ( ) .x x t

ey x f x a a y t dt ... (3)

Comparing (3) with0

( ) ( ) ( , ) ( ) ,x

y x F x K x t y t dt here ( ) ( ), log , ( , ) .x t

eF x f x a K x t a ... (4)

Let Km (x, t) be the mth iterated Kernel. ThenK1 (x, t) = K (x, t) ... (5)

and 1( , ) ( , ) ( , )x

m mt

K x t K x z K z t dz ... (6)

From (4) and (5), K1 (x, t) = K (x, t) = ax – t. ... (7)Putting m = 2 in (6) and using (7), we have

2 1( , ) ( , ) ( , )x x xx z z t x t

t t tK x t K x z K z t dz a a dz a dz [ ] ( ).x t x x t

ta z a x t ... (8)Next, putting m = 3 in (6) and using (8), we have

3 2( , ) ( , ) ( , ) ( )x x x z z t

t tK x t K x z K z t dz a a z t dz

2 2( ) ( )( ) .

2 2!

xxx t x t x t

tt

z t x ta z t dz a a

... (9)


2

4 3( )( , ) ( , ) ( , ) , using (9)

2!

x x x z z t

t t

z tK x t K x z K z t dz a a dz

3 32 ( ) ( )( ) ,

2! 2! 3 3!

xx t x tx x t

tt

a a z t x tz t dz a

... (10)

and so on. By, mathematical induction, we now obtain



1( )( , ) ,( 1)!

mx t

mx tK x t am

(m = 1, 2, 3, ...) ... (11)

Now, the resolvent kernel of (3) is given by

1 21 2 31

( , ; ) ( , ) ( , ) ( , ) ( , ) ...mmm


22 ( )log . ( ) (log ) . ... ad inf,

2!x t x t x t

e ex ta a a x t a a

by (4) and (11)

2 2( ) log [( ) log ]1 ...ad inf

1! 2!

a ax t e ex t x t

a

( )log ( ) .aex tx t x t x ta e a a

log[ ]nem me n

R (x, t; ) = 1. ... (12)Hence the required solution is

0( ) ( ) ( , ; ) ( )

xy x F x R x t F t dt 0

( ) log ( ) ,xa

ef x f t dt using (12)

0( ) log [ ( )] . ( ) log [ ( ) (0)]a x ae ef x f t f x f x f

( ) ( ) ( ) log ,aey x f x f x by (2).

EXERCISE 5FSolve the following Volterra integral equations of the first kind (1 to 5) by first reducing them toVolterra integral equations of the second kind :

1.0

3 ( ) .x x t y t dt x Ans. y (x) = 1 – x loge 3.

2.3

2 2

0(1 ) ( ) .

2

x xx t y t dt Ans. 2( ) .xy x xe

3. 2 2 2

0(2 ) ( ) .

xx t y t dt x Ans. 2 / 2( ) .xy x xe

4.2 / 2

0sin ( ) ( ) 1.

x xx t y t dt e Ans.2 / 2( ) ( 2) 1.xy x e x

5.0

( )x x te y t dt x [Kanpur 2006, 11; Meerut 2006, 07, 08, 111] Ans. ( ) 1 .y x x

6. Change the following Volterra integral equation of first kind into integral equation of second

kind : 0

cos( )u(t) x

x t dt x [Meerut 2008] Ans.0

( ) 1 sin ( ) ( ) .x

u x x t u t dt


CHAPTER 6

Classical Fredholm Theory6.1. INTRODUCTION.

In chapter 5, we obtained the solution of the Fredholm integral equation of the second kind

( ) ( ) ( , ) ( )b


as a uniformly convergent power series in the parameter for | | suitably small. Fredholm derived

the solution of (1) in general form which is valid for all values of the parameter . He gave threeimportant results which are known as Fredholm’s first, second and third fundamental theorems. Inthe present chapter we propose to discuss these theorems.6.2. FREDHOLM’S FIRST FUNDAMENTAL THEOREM. [Meerut 2004]

The non-homogeneous Fredholm integral equation of the second kind

( ) ( ) ( , ) ( ) ,b


where the functions f (x) and y (t) are integrable, has a unique solution

( ) ( ) ( , : ) ( ) ,b


where the resolvent kernel ( , : )R x t is given by

( , : ) ( , : ) / ( )R x t D x t D ... (3)

with ( ) 0,D is a mermorphic function of the complex variable , being the ratio of two entireefunctions defined by the series

111

1

, , ...,( )( , : ) ( , ) ... ..., , ...,!

p ppp

p

x z zD x t K x t K dz dz

t z zp

... (4)

and 1

111

, ...,(– )( ) 1 ... ... ,, ...,!

p ppp

p

z zD K dz dz

z zp

... (5)

both of which converge for all values of . Also, note the following symbol for the determinantformed by the values of the kernel at all points (xi , t i)

1 1 1 2 1

1 22 1 2 2 2

1 2

1 2

( , ) ( , ) ... ( , ), , ...,( , ) ( , ) ... ( , ), , ...,: : ... :

( , ) ( , ) ... ( , )

n

nn

n

n n n n

K x t K x t K x tx x xK x t K x t K x t

Kt t t

K x t K x t K x t

... (6)

which is known as the Fredholm determinant.6.1


6.2 Classical Fredholm Theory

In particular, the solution of the Fredholm homogeneous equation

( ) ( , ) ( )b


is identically zero.Proof. We divide the interval (a, b) into n equal parts, x1 = t1 = a, x2 = t2 = a + h, ..., xn = tn = a + (n – 1) h, ... (8)

where h = (b – a) / n. Thus, we get the approximate formula

1( , ) ( ) ( , ) ( ).

nb

j jjaK x t y t dt h K x x y x

... (9)

Hence (1) reduces to 1

( ) ( ) ( , ) ( ),n

j jjy x f x h K x x y x

... (10)

which must hold for all values of x in the interval (a, b). Using (10) at the n points of division xi, i= 1, 2, ..., n, we arrive at the system of equations

1

( ) ( ) ( , ) ( ), 1,2,..., .n

i i i j jjy x f x h K x x y x i n

... (11)

Let us introduce the following symbols :

( ) , ( ) , ( , ) .i i i i i j ijy x y f x f K x x K ... (12)

Then (11) gives an approximation for (1) in terms of the system of n linear equations

1, 1,2, ...,

n

i ij j ijy h K y f i n

... (13)

which contains n unknown quantities 1 2, ,... .ny y yRe-writing (13), we have

11 1 12 2 1 1

21 1 22 2 2 2

1 1 2 2

(1 ) ...

(1 ) ...

... ... ... ...

... ... ... ..... (1 )

n n

n n

n n nn n n

h K y h K y h K y f

h K y h K y h K y f

h K y h K y h K y f

...(13)

Let

11 12 1

21 22 2

1 2

1 ....

1 ....( )

: : .... :.... 1

n

nn

n n nn

h K h K h K

h K h K h KD

h K h K h K

... (14)

The solutions y1, y2, ..., yn of the system of equations (13) or (13) can be obtained as the

ratios of certain determinants, with the determinant ( )nD as the denominator provided ( ) 0.nD


Classical Fredholm Theory 6.3

We now expand the determinant ( )nD in powers of the quantity ( ).h Clearly, the constantterm is unity. The term containing ( )h in the first power is the sum of all the determinantscontaining only one column , 1,2, ... .vh K n

Taking the contribution from all the columns

v = 1, 2, ..., n, we find that the total contribution is

1.

n

vvvh K

The term containing 2( )h is the sum of all the determinants containing two columns withthat factor. This gives rise to the determinants of the form

2( ) ,pp pq

qp qq

K Kh

K K

where (p, q) is an arbitrary pair of integers taken from the sequence 1, 2, ..., n with p < q.

Next, the term containing the 3( )h is the sum of the determinants of the form

3( ) ,pp pq pr

qp qq qr

rp rq rr

K K K

h K K K

K K K

where (p, q, r) is an arbitrary triplet of integers taken from the sequence 1, 2, ..., n with p < q < r.

Proceeding likewise we may obtain the remaining terms in the expansion of ( ).nD This

leads of the following expansion of the ( ) :nD

2 3

1 , 1 , , 1

( ) ( )( ) 1 ...2 ! 3 !

pp pq prn n npp pq

n vv qp qq qrv p q p q rqp qq

rp rp rr

K K KK Kh hD h K K K KK K

K K K

1 1 1 2 1

2 1 2 2 2

1 1

1 2

, ,..., 1

...

...( )...... ... ... ... ... .

!... ... ... ...

...

n

n

n

n n n n

p p p p p p

p p p p p pn n

p p p

p p p p p p

K K K

K K Kh

n

K K K

... (15)

Using the symbol (6), (15) may be written as2 3

1 , 1 , , 1

, , ,( ) ( )( ) 1 ( , ) ..., , ,2 ! 3!

n n np q p q rn p pp p q p q r

p q p q r

x x x x xh hD h K x x K Kx x x x x

... (16)

Since ( ) / , 0h b a n n h and each term of the sum (16) tends to some single,double, triple integral etc. We thus obtain



2 31 2 31 2

1 2 1 2 31 2 1 2 3

, ,,( ) 1 ( , ) ...

, , ,2 ! 3 !

b b b b b b

a a a a a a

x x xx xD K x x dx K dx dx K dx dx dx

x x x x x

... (17)

(17) is known as the Fredholm’s first series.

Hilbert has shown that the sequence ( ) ( )nD D in the limit. Again Fredholm proved the

convergence of the series (17) for all values of by using the fact that the kernel K (x, t) isbounded and integrable function. Thus, ( )D is an entire function of the complex variable.

If ( , ; )R x t be the resolvent kernel, then we wish to find the solution of (1) in the form

( ) ( ) ( , : ) ( ) ,b


where we expect ( , : )R x t to be the quotient

( , : ) ( , : ) / ( )R x t D x t D ... (19)

where ( , : )D x t is the sum of certain functional series and is yet to be determined. We know that

the resolvent kernel ( , : )R x t satisfies the following relation :

( , : ) ( , ) ( , ) ( , : ) .b

aR x t K x t K x z R z t dz ... (20)

Using (19), (20) may be re-written as

( , : ) ( , : )( , ) ( , )( ) ( )

b

a

D x t D z tK x t K x z dzD D

or ( , : ) ( , ) ( ) ( , ) ( , : ) .b

aD x t K x t D K x z D z t dz ... (21)

From the form of the series (17) for ( ),D it follows that we expect the solution of (21) inthe form of a power series in the parameter :

0 1

( )( , : ) ( , ) ( , ).!

b

ppD x t B x t B x t

p

... (22)

To this end, we re-write (17) as

1

( )( ) 1 ,!

p

ppD C

p

... (23)

where 1 21

1 2

, , ...,... , ... .

, , ...,b b p

p pa ap

x x xC K dx dx

x x x

... (24)

Now, substituting the series for ( , : )D x t and ( )D from (22) and (23) in (21) and comparingthe coefficients of equal powers of , leads us to the following important recursion relations

B0 (x, t) = K (x, t) ... (25)

and 1( , ) ( , ) ( , ) ( , ) .b

p p paB x t C K x t p K x z B z t dz

... (26)



Now, we propose to prove that for each p, (p = 1, 2, 3, ...)

1 21

1 2

, , , ...,( , ) ... ... .

, , , ...,b b p

p pa ap

x z z zB x t K dz dz

t z z z

... (27)

First, observe that for p = 1, (26) taken the form

1 1 0( , ) ( , ) ( , ) ( , )b

aB x t C K x t K x z B z t dz

( , ) ( , ) ( , ) ( , ) ,b b

a aK x t K z z dz K x z K z t dz using (24) and (25)

,b

a

x zK dz

t z

... (28)

showing that (27) holds for p = 1.Now, we shall show that (27) holds for all the integral values. To this end, we expand the

determinant under the integral sign in the relation :

1

1 1 1 1

1 2

1 2

1

( , ) ( , ) ... ( , )

( , ) ( , ) ... ( , ), , , ..., ... ... ... ... ,, , , ..., ... ... ... ...

( , ) ( , ) ... ( , )

...

p

p

p

p

p p p p

K x t K x z K x z

K z t K z z K z zx z z z

Kt z z z

K z t K z z K z z

... (29)

with respect to the elements of the given row, transposing in turn the first column one place to theright, integrating both sides, and using (24); proof of (27) follows by mathematical induction.

Using (22), (25) and (27), we arrive at the so called Fredholm second series :

111

1

, , ...,( )( , : ) ( , ) ... ..., , ...,!

p b b ppp a a

p

x z zD x t K x t K dz dz

t z zp

... (30)

The series (30) converges for all values of .In view of (23) and (30) we observe that both terms of the quotient (19) have been determined

and hence the existence of a solution to (1) is established for a bounded and integrable kernelK (x, t), provided that ( ) 0.D Furthermore since the both terms of the quotient (19) are entirefunctions of the parameter , we conclude that ( , : )R x t must be a mermorphic function of (i.e.,an analytic function whose singularities may only be the poles, which in the present case are zerosof the divisior ( )D ).

In the end we propose to show that the solution in the form obtained by Fredholm is uniqueand is given by (18). Before doing this, we find that the integral equation (20) satisfied by ( , ; )R x t

is valid for all values of for which ( ) 0.D From chapter 5, we already know that (20) holdsfor 1| | ,B where

1/ 22| ( , ) |

b b

a aB K x t dx dt



Since both sides of (20) are thus found to be mermorphic, the result follows. To establish theuniqueness of the solution of (1), we assume that y (x) is a solution of (1) provided that ( ) 0.D

Re-writing (1), ( ) ( ) ( , ) ( ) .b

ay z f z K z t y t dt ... (31)

Multiplying both sides of (31) by ( , ; )R x t and then integrating both sides w.r.t. ‘z’ from ato b, we get

( , ; ) ( ) ( , ; ) ( ) [ ( , ; ) ( , ) ] ( ) .b b b b

a a a aR x z y z dz R x z f z dz R x z K z t dz y t dt ... (32)

Using (20), we have ( , ; ) ( , ) ( , ; ) ( , )b

aR x t K x t R x z K z t dz

or ( , ; ) ( , ) ( , ; ) ( , ).b

aR x z K z t dz R x t K x t ... (33)


( , ; ) ( ) ( , ; ) ( ) [ ( , ; ) ( , )] ( )b b b

a a aR x z y z dz R x z f z dz R x t K x t y t dt

or ( , ; ) ( ) ( , ; ) ( ) ( , ; ) ( ) ( , ) ( )b b b b

a a a aR x t y t dt R x t f t dt R x t y t dt K x t y t dt

or ( , ) ( ) ( , ; ) ( )b b

a aK x t y t dt R x t f t dt ... (34)

From (1), ( ) ( )( , ) ( ) .

b

a

y x f xK x t y t dt

... (35)


( ) ( ) ( , ; ) ( )b

a

y x f x R x t f t dt

or ( ) ( ) ( , ; ) ( ) ,b

ay x f x R x t f t dt

but this form is unique.In particular, the solution of (7) is identically zero.

6.3. SOLVED EXMAPLES BASED ON FREDHOLM’S FIRST FUNDAMENTAL THEOREM.We shall use the following results :

For the Fredholm integral equation ( ) ( ) ( , ) ( ) ,b


the resolvent kernel is given by ( , ; ) ( , ; ) / ( )R x t D x t D ... (2)

where1

( )( , ; ) ( , ) ( , )!

m

mmD x t K x t B x t

m

... (3)

and1

( )( ) 1 .!

m

mmD C

m

... (4)



where

1

1 1 1 1

1 2

1

( , ) ( , ) ... ( , )

( , ) ( , ) ... ( , )

( , ) ... ... ... ... ... ... ,... ... ... ...( , ) ( , ) ... ( , )

n

nb b

n na an

n n n n

K x t K x z K x z

K z t K z z K z z

B x t dz dz dz

K z t K z z K z z

... (5)

and

1 1 1 2 1

2 1 2 2 2

1 2

1 2

( , ) ( , ) ... ( , )

( , ) ( , ) ... ( , )

... ... ... ... ... ... .... ... ... ...( , ) ( , ) ... ( , )

n

nb b

n na an

n n n n

K z z K z z K z z

K z z K z z K z z

C dz dz dz

K z z K z z K z z

...(6)

The function ( , ; )D x t is called the Fredholm minor and ( )D is called the Fredholmdeterminant.

Results (5) and (6) have been written in these forms with help of results (6), (24) and (27) ofArt. 6.2.

Alternative procedure of calculating Bm ( x, t ) and Cm. The following results will beused . 0 1C ...(7)

1 ( , ) , 1b

p paC B s s ds p

... (8)

0 ( , ) ( , ).B x t K x t ... (9)

1( , ) ( , ) ( , ) ( , ) , 1.b

p p paB x t C K x t p K x z B z t dz p

... (10)

After getting ( , ; ),R x t the required solution is given by

( ) ( ) ( , ; ) ( )b


Result (8) follows easily from results (5) and (6)Ex. 1. Using Fredholm determinants, find the resolvent kernel, when

( , ) , 0, 1.tK x t xe a b [Kanpur 2007, Meerut 2001]Sol. First write down all equations (1) to (6) given in Art 6.3.Here K (x, t) = xet. ... (7)

From (5), 1

1

1 111 1 10 0

1 1 1 1 1

( , ) ( , )( , )

( , ) ( , )

zt

zt

K x t K x z xe xeB x t dz dz

K z t K z z z e z e , by (7)

= 0, since the determinant under the integral sign vanishes

1 21 1

2 1 1 1 1 2 1 20 0

2 2 1 2 2

( , ) ( , ) ( , )

( , ) ( , ) ( , ) ( , )

( , ) ( , ) ( , )

K x t K x z K x z

B x t K z t K z z K z z dz dz

K z t K z z K z z



1 2

1 2

1 2

1 1

1 1 1 1 20 0

2 2 2

z zt

z zt

z zt

xe xe xe

z e z e z e dz dz

z e z e z e


Since B1 (x, t) = B2 (x, t) = 0, it follows that ( , ) 0, for 1.nB x t n


1 1 1 11 11 1 1

1 1 1 1 1 1 1 10 0 00 0( , ) ( 1) 1.z z z zC K z z dz z e dxz z e e dz e e e e

1 2

1 2

1 1 1 11 1 1 2 1 12 1 2 1 20 0 0 0

2 1 2 2 2 2

( , ) ( , )

( , ) ( , )

z z

z z

K z z K z z z e z eC dz dz dz dz

K z z K z z z e z e

= 0, since the determinant under the integral sign vanishes.It follows that Cm = 0 for all 2.m

Now, (3) gives 21 2

( , ; ) ( , ) ( , ) ( / 2!) ( , ) ....D x t K x t B x t B x t

= xet, on putting values of K (x, t), B1 (x, t) etc.

and (4) gives 21 2

( ) 1 ( / 2!) ....D C C

1 , on putting values of C1, C2, etc.

Hence (2) yields ( , ; )( , ; ) .( ) 1

tD x t xeR x tD

Alternative method. We shall use results (7), (8), (9) and (10) of Art. 6.3 to compute

( , ; )R x t as follows. First write down these results for complete solution yourself.Here C0 = 1, ... (i)

and B0 (x, t) = K (x, t) = xet. ... (ii)

From (8), 1 1 11

1 0 00 0 0( , ) s s sC B s s ds se ds se e ds

1

0( 1) 1.se e e e

... (iii)

From (10),1

1 1 00( , ) ( , ) ( , ) ( , )B x t C K x t K x z B z t dz

1 1

0 0( ) 1,t z t t t z t txe xe ze dz xe xe ze dz xe xe as above

1 ( , ) 0.B x t ... (iv)

Then, from (8),1

2 10( , ) 0,C B s s ds by (iv) ... (v)

From (10),1

2 2 10( , ) ( , ) 2 ( , ) ( , ) 0,B x t C K x t K x z B z t dz by (iv) and (v)

( , ) 0mB x t for all 1m and 0mC for all 2.m



Now proceed as before to determine ( , ; ).R x t

Important Observation. The reader will find that the above alternative method is a short cut.However, he should find the required quantities strictly in the following order :

0 0 1 1 2 2, ( , ), , ( , ), , ( , )C B x t C B x t C B x t and so on.

Ex.2. (a) Using Fredholm determinant, find the resolvent kernel of the integral equation1

0( ) ( ) ( ) , ( 1)ty x f x xe y t dt and hence solve it.

(b) Explain the method of Fredholm determinant for the solution of Fredholm integralequation and hence solve the integral equation

1––1

( ) ( )x ty x e x e y t dt [Meerut 2001, 03]

Sol. (a) Given1

0( ) ( ) ( ) .ty x f x xe y t dt Here K (x, t) = xet.

Proceed as in solved Ex. 1 and obtain ( , ; ) ( ) /(1 )tR x t xe

Hence the required solution is1

0( ) ( ) ( , ; ) ( )y x f x R x t f t dt

or1

0( ) ( ) ( )

1

txey x f x f t dt or

1

0( ) ( ) ( ) .

1txy x f x e f t dt

(b) For the first part, refer Art. 6.2. Proceed as in Ex. 1 to prove that

( , ; ) ( ) /(1– ), 1.tR x t xe Hence the required solution is given by

1 1– – – ––1 –1

( ) ( , ; ) or ( )1 –

tx t x txey x e R x t e dt y x e e dt

or 1––1( )

1 –x xy x e t

or – 2( ) , 11–

x xy x e if

Ex.3. Find the resolvent kernel and solution of1

0( ) ( ) ( ) ( ) .y x f x x t y t dt

Sol. Given1

0( ) ( ) ( ) ( ) .y x f x x t y t dt ... (1)

Comparing (1) with1

0( ) ( ) ( , ) ( ) ,y x f x K x t y t dt

here K (x, t) = x + t. ... (2)The resolvent kernel ( , ; )R x t is given by ( , ; ) ( , ; ) / ( ),R x t D x t D ... (3)

where1

( )( , ; ) ( , ) ( , )!

m

mmD x t K x t B x t

m

... (4)

1

( )( ) 1 ,!

m

mmD C

m

... (5)

C0 = 1, ...(6)



B0 (x, t) = K (x, t) = x + t, by (2) ...(7)

1

10( , ) , 1p pC B s s ds p

... (8)

and1

10( , ) ( , ) ( , ) ( , ) .p p pB x t C K x t p K x z B z t dz

... (9)

From (8),

121 1

1 00 00

( , ) 2 2 1.2sC B s s ds s ds

... (10)

From (9), 1


1

0( ) ( ) ,x t x z z t dz by (2) and (7)

121 2 3

00

1[ ( , ) ] ( )3 2

zx t z z x t xt dz x t z x t xtz

1 1 1 1( ) ( ) .3 2 2 3

x t x t xt x t xt ... (11)

From (8), 1 1 2

2 10 0

1 1( , ) ( )2 3

C B s s ds s s s ds 12 3

0

12 3 3s s s

Thus, 2 (1/ 2) – (1/3) – (1/ 3) – (1/ 6)C ... (12)

From (9),1

2 2 10( , ) ( , ) 2 ( , ) ( , )B x t C K x t K x z B z t dz

1

0

1 1 1( ) 2 ( ) ( ) ,6 2 3

x t x z z t zt dz by (2), (11) and (12)

1

0

1 1 1 1( ) 2 ( )6 2 2 3

x t x z z t t dz 1 2

0

1 1 1 1 1 1( ) 26 2 2 3 2 2 3

xx t z t z t xt x t dz 13 2

0

1 1 1 1 1( ) 26 3 2 2 2 3 2 2 3

z z x tx t t t xt xz

1 1 1 1 1 1 1( ) 26 3 2 2 2 3 2 2 3

x tx t t t xt x = 0, on simplification. ... (13)

Since B2 (x, t) = 0, it follows from (8) and (9) thatC3 = C4 = C5 = ... = 0 ... (14)

and 3 4 5( , ) ( , ) ( , ) ... 0.B x t B x t B x t ... (15)Using the above values of Cp and Bp (x, t), from (4) and (5), we have

11 1( , ; ) ( , ) ( , ) ( )2 3

D x t K x t B x t x t x t xt



and 2

21 2

1( ) 1 1 .2 ! 12

D C C

Putting these values in (3), we have

2

(1/ 2) ( ) 1/ 3( , ; ) .

1 ( /12)x t x t xt

R x t

The required solution of (1) is given by 1

0( ) ( ) ( , ; ) ( )y x f x R x t f t dt

or 1

20

(1/ 2) ( ) 1/ 3( ) ( ) ( )

1 ( /12)x t x t xt

y x f x f t dt

Ex. 4. Solve1

0( ) 1 (1 3 ) ( ) .y x xt y t dt

Sol. Given1

0( ) 1 (1 3 ) ( )y x xt y t dt ... (1)


0( ) ( ) ( , ) ( ) ,y x f x K x t y t dt

here 1, f (x) = 1 and ( , ) 1 3 .K x t xt ... (2)

The resolvent kernel ( , ; )R x t is given by

( , ; ) ( , ; ) / ( ),R x t D x t D ... (3)

where 1

( )( , ; ) ( , ) ( , ),!

m

mmD x t K x t B x t

m

... (4)

1

( )( ) 1 ,!

m

mmD C

m

... (5)

C0 = 1 ... (6)B0 (x, t) = K (x, t) = 1 – 3xt. ... (7)

1

10( , ) , 1p pC B s s ds p

... (8)

and 1

10( , ) ( , ) ( , ) ( , ) , 1.p p pB x t C K x t p K x z B z t dz p

... (9)

From (7) and (8), 1 1 2

1 00 0( , ) (1 3 ) ,C B s s ds s ds

130

0.s s ... (10)

From (9), 1


1

0(1 3 ) (1 3 ) ,xz zt dz by (2), (7) and (10)

11 2 3 2

0 0

3[9 3 ( ) 1] 3 ( )2

xtz z x t dz xt z z x t z



B1 (x, t) 3 (3/ 2) ( ) 1 .xt x t ... (11)

From (8), 1 1 2

2 10 0( , ) . (3 3 1) ,C B s s ds s s ds by (11)

13 2

0

3 3 11 12 2 2

s s s ... (12)

From (9),1


1

0

1 3(1 3 ) 2 (1 3 ) 3 ( ) 1 ,2 2

xt xz zt z t dz using (11) and (12)

1

0

1 1 3(1 3 ) 2 (1 3 ) 3 12 2 2

xt xz z t t dz

1 2

0

1 1 1 3 3(1 3 ) 2 9 3 12 2 2 2 2

xt x t z z t x xt t dz 1

3 2

0

1 1 3 1 3 3(1 3 ) 2 3 12 2 2 2 2 2

xt x t z z t x xt t z

1 1 3 1 3 3(1 3 ) 2 3 12 2 2 2 2 2

xt x t t x xt t = 0, on simplification. ... (13)

Since B2 (x, t) vanishes, it follows that( , ) 0 for 3pB x t p ... (14)

and 0 for 3.pC p ... (15)

Putting the above values in (4) and (5), we have

1( , ; ) ( , ) ( , )D x t K x t B x t [ here 1 ]

1 3 3 (3/ 2) ( ) 1 2 (3 / 2) ( )xt xt x t x t ... (16)

and 1 2( ) 1 (1/ 2 !)D C C [ 1 ]

( ) 1 (1/ 4) 3/ 4D ... (17)

From (3), ( , ; ) 2 (3/ 2) ( ) 2( , ; ) 4 3( ) .( ) (3 / 4) 3

D x t x tR x t x tD

... (18)

The required solution is given by1

0( ) ( ) ( , ; ) ( )y x f x R x t f t dt

or 1

0

2( ) 1 4 3 ( ) ,3

y x x t dt using (2) and (18)

or

12

0

2 3 2 3 8 6( ) 1 4 3 1 4 3 .3 2 3 2 3

t xy x t xt x



Ex. 5. Determine ( )D and ( , ; )D x t and hence solve the integral equation

10

0( ) ( ) .xy x e xt y t dt [Kanpur 2005, 08]

Sol. Given 10

0( ) ( ) .xy x e xt y t dt

Comparing (1) with10

0( ) ( ) ( , ) ( ) ,y x f x K x t y t dt

we have f (x) = ex and K (x, t) = xt ... (2)The resolvent kernel is given by

( , ; ) ( , ; ) / ( ),R x t D x t D ... (3)

where1

( )( , ; ) ( , ) ( , ),!

m

mmD x t K x t B x t

m

... (4)

1

( )( ) 1 ,!

m

mmD C

m

... (5)

C0 = 1, ... (6) B0 (x, t) = K (x, t) = xt, by (2), ... (7)

10

10( , ) , 1p pC B s s ds p

... (8)

and 10

10( , ) ( , ) ( , ) ( , ) , 1.p p pB x t C K x t p K x z K z t dz p

... (9)

From (7) and (8)10 10 2

1 00 0( , )C B s s ds s ds

103 3

0

10 .3 3s

... (10)

From (9), 10


3 10

0

10 ( ) ( ) ,3

xt x z z t dz by (2), (7) and (10)

103 3

0

10 0.3 3

zxt xt

... (11)

Since B1 (x, t) = 0, from (8) and (9), it follows thatCp = 0 for 2,p ... (12)

and Bp (x, t) = 0 for 2.p ... (13)Putting the above values in (4) and (5), we have

21 2

( , ; ) ( , ) – ( , ) ( / 2!) ( , ) ... ,D x t K x t B x t B x t xt

and 2 31 2

( ) 1 ( / 2!) ... 1 (10 / 3).D C C

From (3), 3( , ; ) .1 (10 / 3)

xtR x t

... (14)



The required solution of (1) is given by 10

0( ) ( ) ( , ; ) ( )y x f x R x t f t dt

or 10

30( ) ,

1 (10 / 3)x txty x e e dt

using (2) and (14)

10

3 01 (10 / 3)x txe t e dt

1010

3 0 0,

1 (10 / 3)x t txe t e e dt integrating by parts

1010 10 103 30

3 310 (10 1)3 10 3 10

x t xx xe e e e e e

103

3( ) (1 9 ).3 10

x xy x e e

Ex.6. Solve10

4( ) sin ( ) .y x x x y t dt

Sol. Given10

4( ) sin ( ) .y x x x y t dt ... (1)

Comparing (1) with10

4( ) ( ) ( ) ,y x f x x y t dt we have

f (x) = sin x and K (x, t) = x. ... (2)The resolvent kernel is given by ( , ; ) ( , ; ) / ( ),R x t D x t D ... (3)

where 1

( )( , ; ) ( , ) ( , ),!

m

mmD x t K x t B x t

m

... (4)

1

( )( ) 1 ,!

m

mmD C

m

... (5)

C0 = 1, ... (6)B0 (x, t) = K (x, t) = x, by (2) ... (7)

10

14( , ) , 1p pC B s s ds p

... (8)

and 10

14( , ) ( , ) ( , ) ( , ) , 1p p pB x t C K x t p K x z B z t dz p

... (9)

From (7) and (8), 10 10

1 04 4( , )C B s s ds s ds

1022 2

4

1 (10 4 ) 42.2 2s

... (10)

From (9),10

1 1 04( , ) ( , ) ( , ) ( , )B x t C K x t K x z K z t dz

10

442 ,x xz dz by (2), (7) and (10)

1024

42 / 2 42 42 0.x x z x x ... (11)



Since B1 (x, t) = 0, (8) and (9) show thatCp = 0 for 2,p ... (12)

and ( , ) 0pB x t for 2.p ... (13)


21 2

( , ; ) ( , ) ( , ) ( / 2!) ( , ) ...D x t K x t B x t B x t x

and 21 2

( ) 1 ( / 2!) ... 1 42 .D C C

From (3), ( , ; ) /(1– 42 ).R x t x ... (14)

The required solution of (1) is given by10

4( ) ( ) ( , ; ) ( )y x f x R x t f t dt

or10

4( ) sin sin ,

1 42xy x x t dt

by (2) and (14)

104sin cos sin [cos 4 cos10]

1 42 1 42x xx t x

2 sin 7 sin 3( ) sin

1 42xy x x

Ex.7. Determine ( )D and ( , ; )D x t and hence solve the integral equation

1

0( ) 2 ( )x x ty x e e e y t dt [Meerut 2002, 06, 08, 10]

Sol. Given1

0( ) 2 ( ) .x x ty x e e e y t dt ... (1)

Comparing (1) with1

0( ) ( ) ( , ) ( ) ,y x f x K x t y t dt

here f (x) = ex and K (x, t) = 2ex et. ... (2)The resolvent kernel is given by ( , ; ) ( , ; ) / ( ),R x t D x t D ... (3)

where1

)( , ; ) ( , ) ( , ),!

m

mmD x t K x t B x t

m

... (4)

1

( )( ) 1 ,!

m

mmD C

m

... (5)

C0 = 1, ... (6)B0 (x, t) = K (x, t) = 2ex et, ... (7)

1

10( , ) , 1p pC B s s ds p

... (8)

and 1


... (9)

From (7) and (8), 1 1

1 00 0( , ) 2 s sC B s s ds e e ds

12 20

2 / 2 1.se e ... (10)



From (9) 1


1 12 2 2

0 02 ( 1) (2 ) (2 ) 2 ( 1) 4x t x z z t x t x t ze e e e e e e dz e e e e e e dz

12 2 2 20

2( 1) 4 / 2 2( 1) 2 ( 1) 0.x t x t z x t x te e e e e e e e e e e e ... (11)

Since B1 (x, t) = 0, (8) and (9) show thatCp = 0 for 2.p ... (12)

and ( , ) 0pB x t for 2.p ... (13)

Putting the above values in (4) and (5), we have2

1 2( , ; ) ( , ) ( , ) ( / 2!) ( , ) ... 2 .x tD x t K x t B x t B x t e e

and 2 21 2

( ) 1 ( / 2!) ... 1 ( 1)D C C e

From (3), 22( , ; ) .

1 ( 1)

x te eR x te

... (14)


0( ) ( ) ( , ; ) ( )y x f x R x t f t dt

or1

20

2( ) . ,1 ( 1)

x tx te ey x e e dt

e

by (2) and (14)

or 121 2

2 200

2 221 ( 1) 1 ( 1)

x x tx t xe e ee e dt e

e e

or2

2 2( 1)( ) .

1 ( 1) 1 ( 1)

x xx e e ey x e

e e

Ex. 8. Solve0

( ) 1 sin ( ) ( ) .y x x t y t dt

Sol. Given

0( ) 1 sin ( ) ( ) .y x x t y t dt

... (1)

Comparing (1) with0

( ) ( ) ( , ) ( ) ,y x f x K x t y t dt

here f (x) = 1 and K (x, t) = sin (x + t). ... (2)

The resolvent kernel is given by ( , ; ) ( , ; ) / ( ),R x t D x t D ... (3)

where1

( )( , ; ) ( , ) ( , ),!

m

mmD x t K x t B x t

m

... (4)

1

( )( ) 1 ,!

m

mmD C

m

... (5)

C0 = 1, ... (6)



B0 (x, t) = K (x, t) = sin (x + t), ... (7)

10( , ) , 1p pC B s s dz p

... (8)

and 10( , ) ( , ) ( , ) ( , ) , 1p p pB x t C K x t p K x z B z t dz p

... (9)

From (7) and (8), 1 00 0 0

cos 2( , ) sin 2 0.2

sC B s s ds s ds ... (10)

From (9), 1 1 00( , ) ( , ) ( , ) ( , )B x t C K x t K x z B z t dz

0sin ( ) sin ( )x z z t dz

by (2), (7) and (10)

0 0

1 1 1cos ( ) cos ( 2 ) cos ( ) sin ( 2 )2 2 2

x t x t z dz z x t x t z

1 1 1cos ( ) sin ( ) sin ( ) cos ( ).2 2 2 2

x t x t x t x t ... (11)

From (8) and (11), 2

02 10 0( , ) .

2 2 2C B s s ds ds s

... (12)

From (9), 2 2 10

( , ) ( , ) 2 ( , ) ( , ) .B x t C K x t K x z B z t dz

2

0sin( ) 2 sin ( ) cos ( ) ,

2 2x t x z z t dz

using (2), (11) and (12)

2

0sin ( ) {sin (2 ) sin ( )} .

2 2x t z x t x t dz

0

cos (2 )sin ( ) sin ( )2 2 2

z x tx t z x t

2 1 1sin ( ) cos ( ) cos( ) sin ( )2 2 2 2

x t x t x t x t B2 (x, t) = 0 ... (13)Since B2 (x, t) = 0, (8) and (9) show that

Cp = 0 for 3p ... (14)

and Bp (x, t) = 0 for 3.p ... (15)Putting the above values in (4) and (5), we have

2 3

1 2 3( , ; ) ( , ) ( , ) ( , ) ( , ) ...2 ! 3 !

D x t K x t B x t B x t B x t

sin ( ) ( / 2) cos ( )x t x t

and 2 3

1 2 3( ) 1 ...2 ! 3 !

D C C C

2 21 .

4



From (3),2 2

sin ( ) ( / 2) cos ( )( , ; )1 ( / 4)

x t x tR x t

... (16)


( ) ( ) ( , ; ) ( )y x f x R x t f t dt

or 2 20

sin ( ) ( / 2) cos ( )( ) 1 ,1 ( / 4)

x t x ty x dt

using (2) and (16)

2 20

41 cos ( ) sin ( )24

x t x t

2 24 1 11 cos sin cos sin

2 24x x x x

or 2 24( ) 1 (2 cos sin ).

4y x x x

Ex. 9. Solve 1 1/ 2

0( ) ( ) ( ) .y x x xt xt y t dt

Sol. Given 1 1/ 2

0( ) ( ) ( ) .y x x xt xt y t dt ... (1)


0( ) ( ) ( , ) ( ) ,y x f x K x t y t dt

here f (x) = x and K (x, t) = xt + (xt)1/2. ... (2)The resolvent kernel ( , ; )R x t is given by ( , ; ) ( , ; ) / ( ),R x t D x t D ... (3)

where1

( )( , ; ) ( , ) ( , ),!

m

mmD x t K x t B x t

m

... (4)

1

( )( ) 1 ,!

m

mmD C

m

... (5)

C0 = 1, ... (6)

B0 (x, t) = K (x, t) = xt + (xt)1/2, ... (7)1

10( , ) , 1p pC B s s ds p

... (8)

and1

10( , ) ( , ) ( , ) ( , ) , 1.p p pB x t C K x t p K x z B z t dz p

... (9)

From (7) and (8), 1 1 2

1 00 0( , ) ( )C B s s ds s s ds

13 20

/ 3 / 2 5 / 6.s s ... (10)

From (9),1


11/ 2 1/ 2 1/ 2

0

5 { ( ) } { ( ) }{ ( ) }6

xt xt xz xz zt zt dz , by (2) and (7)



11/ 2 2 3/2 3/ 2 1/ 2

0

5 { ( ) } [ ( ) ]6

xt xt xtz x t z t x z z xt dz 13 5 / 2 2

1/ 2 1/ 2

0

5 { ( ) } ( ) ( )6 3 (5 / 2) 2

xt z z zxt xt x t t x xt

1/ 2 1/ 25 2 1{ ( ) } ( ) ( )6 3 5 2

xtxt xt x t t x xt

1/ 21 1 2( ) ( ).2 3 5

xt xt x t t x ... (11)

From (8), 1

2 10( , )C B s s ds

1 2

0

1 1 2 ( ) ,2 3 5

s s s s s s ds by (11)

13 2 5 / 21 2 3/ 2

00

1 1 4 4 1 1 8 .2 3 5 6 6 5 (5 / 2) 6 6 25

s s ss s s ds

= 1 / 75. ... (12)

From (9), 1


11/ 2 1/ 2 1/ 2

0

1 1 1 2{ ( ) } 2 { ( ) } ( ) ( )75 2 3 5

xt xt xz xz zt zt z t t z dz

[using (2), (11) and (12)]

11/ 2 2 3/ 2 2 3/ 2

0

1 1 2 2{ ( ) } 2 .75 2 3 5 5

x t x t xtxt xt xt z z z z

3/ 2 1/ 2 1/ 2 3/ 21 1 2 2( ) ( )

2 3 5 5t x z z xt xt z t x z dz

3 5/ 2 3 5 / 2

1/ 21 2 2 4{ ( ) } 275 6 15 15 25

xt z x t z x t z x tzxt xt

15/ 2 1/ 2 2 1/ 2 25/ 2

0

( ) 4( )5 6 25 5

t x z xt z xt t x zz

1/ 21/ 21 2 2 4 ( ){ ( ) } 2

75 6 15 15 25 5 6xt x t x t xt t x xtxt xt

1/ 24 ( ) 025 5

t xxt

... (13)

Since B2 (x, t) = 0, (8) and (9) show thatCp = 0 for 3p ... (14)

and Bp (x, t) = 0 for 3.p ... (15)Putting the above values in (4) and (5), we get

21 2

( , ; ) ( , ) ( , ) ( / 2 !) ( , ) ...D x t K x t B x t B x t



1/ 2 1/ 21 1 2( ) ( ) ( )

2 3 5xt xt xt xt x t t x

and2 2

1 25( ) 1 ... 1 .

2 ! 6 150D C C

from (3), 1/ 2 1/ 2

2

( ) (1/ 2) (1/ 3) ( ) (2 / 5) ( )( , ; )

1 (5 / 6) (1/150)

xt xt xt xt x t t xR x t

... (16)

Hence the required solution of (1) is1

0( ) ( ) ( , ; ) ( )y x f x R x t f t dt

or 1/ 2 1/ 2

1

20

( ) (1/ 2) (1/ 3) ( ) (2 / 5) ( )( )

1 (5 / 6) (1/150)

xt xt xt xt x t t xy x x t dt

[using (2) and (16)]

13 5/ 2 3 5/ 2 5 / 2 3

20

2 2 4 23 5 6 15 25 151 (5 / 6) (1/150)

xt x t xt x t x t x tx

22 2 4 2

3 5 6 15 25 151 (5 / 6) (1/150)x x x x x xx

Thus, 2

2150 (60 75 ) 21( ) ,

125 150x x x xy x

on simplification.

Ex.10. Using Fredholm determinants, find the resolvent kernel of the following kernels :(i) ( , ) 2 , 0 1, 0 1.K x t x t x t

(ii) 2 2( , ) , 0 1, 0 1.K x t x t xt x t (iii) ( , ) 1 3 , 0 1, 0 1.K x t xt x t [Kanpur 2006](iv) ( , ) sin cos , 0 2 , 0 2 .K x t x t x t (v) ( , ) sin sin , 0 2 , 0 2 .K x t x t x t

Sol. (i) Here ( , ) 2K x t x t ...(1)

1 11 1

1 1 10 01 1 1 1 1 1

( , ) ( , ) 2 2( , )

( , ) ( , ) 2 2

K x t K x z x t x zB x t dz dz

K z t K z z z t z z

1 1 2

1 1 1 1 1 1 1 10 0[ (2 ) (2 ) (2 )] [ (2 ) (4 2 2 )]z x t x z z t dz z x t xz xt z z t dz

13 21 2 1 11 1 1 10

0

2[2 (2 4 ) 2 ] ( 2 2 ) 2

3 2

z zz z x t x t xt dz x t xt z

B1 (x, t) = (2/3) – x – t + 2xt. ... (2)

1 2 1 21 1 1 1

2 1 1 1 1 2 1 2 1 1 1 1 2 1 20 0 0 0

2 2 1 2 2 2 2 1 2 2

( , ) ( , ) ( , ) 2 2 2

( , ) ( , ) ( , ) ( , ) 2 2 2

( , ) ( , ) ( , ) 2 2 2

K x t K x z K x z x t x z x z

B x t K z t K z z K z z dz dz z t z z z z dz dz

K z t K z z K z z z t z z z z

= 0 (simplify yourself)



Hence ( , ) 0pB x t for 3.p ... (3)

1 1

1 1 1 1 1 1 10 0( , ) (2 )C K z z dz z z dz

12

10

1 1 .2 2

z ... (4)

1 1 1 1 1 2

2 1 20 02 1 2 2

( , ) ( , )

( , ) ( , )

K z z K z zC dz dz

K z z K z z

1 1 1 1 1 21 20 0

2 1 2 2

2 2 1 ,2 2 3

z z z zdz dz

z z z z

... (5)

[on simplification]

1 1 1 2 1 31 1 1

3 2 1 2 2 2 3 1 2 30 0 0

3 1 3 2 3 3

( , ) ( , ) ( , )

( , ) ( , ) ( , )

( , ) ( , ) ( , )

K z z K z z K z z

C K z z K z z K z z dz dz dz

K z z K z z K z z

1 1 1 2 1 31 1 1

2 1 2 2 2 3 1 2 30 0 0

3 1 3 2 3 3

2 2 2

2 2 2

2 2 2

z z z z z z

z z z z z z dz dz dz

z z z z z z = 0, on simplification ... (6)

Hence Cp = 0 for all 4p

Now,11

( )( , ; ) ( , ) ( , ) ( , ) ( , )!

m

mmD x t K x t B x t K x t B x t

m

2 2 /3 2x t x t xt

and 2 2

1 21

( ) 1( ) 1 1 1! 2 ! 2 6

m

mmD C C C

m

2

2 2 / 3 2( , ; )( , ; ) .( ) 1 / 2 / 6

x t x t xtD x tR x tD

Part (ii) Hint. Student is adviced to complete the solution.

Here 2 2( , ) .K x t x t xt

22 2 21 1 1

1 12 2 2201 1 1 1 1 1

( , )x t xt x z xz

B x t dzz t z t z z z z

1 ,4 3 5

x t xtxt

on simplification

Also, B2 (x, t) = 0. (verify yourself)

( , ) 0pB x t for all 3.p

Again, 1 1 3 3

1 1 1 1 1 1 10 0( , ) ( ) 0.C K z z dz z z dz

2 21 1 1 2 1 2

2 1 22 2 2 20 02 1 2 1 2 2 2 2

0 120

z z z zC dz dz

z z z z z z z z

, on simplifying



Also, C3 = 0 (verify yourself)

Cp = 0 for all 4.p

2 21

1( , ; ) ( , ) ( , ) ...4 3 5

x t x tD x t K x t B x t x t xt xt

and 2

1 2( ) 1 ... 12 ! 240

D C C

2 2

2

1( , ; ) 4 3 5( , ; )

( ) 1 ( /140)

x t xtx t x t x tD x tR x t

D

Part (iii) Hint. Student is adviced to compete the solution.Here K (x, t) = 1 + 3xt ... (1)

1 121 10

1 1

1 3 1 3( , )

1 3 1 3

xt xz

B x t dzz t z

3 ( ) 3 1 ,2

x t xt (on simplification) ...(2)

Also, B2 (x, t) = 0. (verify yourself).

Bp (x, t) = 0 for all 3.p ... (3)

Next,1 1 2

1 1 1 1 1 10 0( , ) (1 3 ) 2. C K z z dz z dz

21 1 1 1 2

2 1 220 02 1 2

1 3 1 3 1 .21 3 1 3

z z zC dz dz

z z z

(simplify yourself)

Also, C3 = 0. (verify yourself)

Cp = 0 for all 4.p

( , ; ) ( , ) ( , ) ... 1 3 [(3 / 2) ( ) 3 1]D x t K x t K x t xt x t xt

and 2 21 2

( ) 1 ( / 2 !) ... 1 2 ( / 4).D C C

2( , ; ) 1 3 [(3/ 2) ( ) 3 1]( , ; )

( ) 1 2 (1/ 4)D x t xt x t xtR x t

D

Part (iv) Here K (x, t) = sin x cos t. ... (1)

2 11 10

1 1 1

( , ) ( , )( , )

( , ) ( , )

K x t K x zB x t dz

K z t K z z2 1

101 1 1

sin cos sin cos,

sin cos sin cos

x t x zdz

z t z z by (1)


Bp (x, t) = 0 for all 2.p

Next, 2 2

1 1 1 1 1 1 10 0( , ) sin cos 0.C K z z dz z z dz

Cp = 0 for all 2.p



Now, 1

( )( , ; ) ( , ) ( , )!

m

mmD x t K x t B x t

m

1( , ) ( , ) ... sin cos K x t B x t x t , using the above values

and1

( )( ) 1!

m

mmD C

m= 11 ... 1, C using the above values

( , ; ) ( , ; ) / ( ) sin cos .R x t D x t D x t Prove (v) Hint. Students are required to complete the solution.Here ( , ) sin sin . K x t x t ... (1)

2 11 10

1

sin sin sin sin( , )

sin sin 0

x t x z

B x t dzz t

(1 2 sin sin ), x t (simplify yourself)Also, B2 (x, t) = 0. (verify yourself)

Bp (x, t) = 0 for all 3.p

Again,2 2

1 1 1 1 1 1 10 0( , ) (sin sin ) 0.

C K z z dz z z dz

2 2 1 1 1 2 22 1 20 0

2 1 2 2

sin sin sin sin4

sin sin sin sin

z z z z

C dz dzz z z z (simplify yourself).

Cp = 0 for all 3.p

11

( )( , ; ) ( , ) ( , ) ( , ) ( , )!

m

mmD x t K x t B x t K x t B x t

m

sin sin (1 2 sin sin ). x t x t

and 2

1 21

( )( ) 1 1 ...! 2 !

m

mmD C C C

m2 21 4 .

2 2( , ; ) sin sin (1 2 sin sin )( , ; )

( ) 1 4

D x t x t x tR x tD

Ex.11. Evaluate the resolvent kernels of the following integrals :(i) ( , ) sin cos ; 0 2 , 0 2 . K x t x t x t

(ii) ( , ) 2 ; 0 1, 0 1 K x t x t x t [Meerut 2011, 2012; Kanpur 2005, 09]

(iii) ( , ) 1; 1 1, 1 1 K x t x t x t

(iv) ( , ) ; 0 1, 0 1 x tK x t e x t

(v) 2( , ) 4 ; 0 1, 0 1 K x t xt x x t

(vi) ( , ) sin( ); 0 2 , 0 2 K x t x t x t

(vii) ( , ) sinh ; 1 1, 1 1 K x t x t x t



(viii) 2 2( , ) ; 0 1, 0 1 K x t x t xt x t

(ix) ( , ) sin sin ; 0 2 , 0 2 K x t x t x t

(x) ( , ) 1 3 ; 0 1, 0 1. K x t xt x t

Sol. (i) Here ( , ) sin cos .K x t x t ... (1)

The resolvent kernel ( , ; )R x t is given by ( , ; ) ( , ; ) / ( ) R x t D x t D ... (2)

where1

( )( , ; ) ( , ) ( , ),!

m

mmD x t K x t B x t

m... (3)

1

( )( ) 1 ,!

m

mmD C

m... (4)

C0 = 1, ... (5) B0 (x, t) = K (x, t) = sin x cos t, ... (6)

2

10( , ) , 1

p pC B s s ds p ... (7)

and 2


... (8)

From (7), 22 2 2

1 00 0 0 0

1 1 cos 2( , ) sin cos sin 2 0.2 2 2

sC B s s ds s s ds s ...(9)

From (8), 2 2

1 1 00 0( , ) ( , ) ( , ) ( , ) (sin cos ) (sin cos )B x t C K x t K x z B z t dz x z z t dz

2

0sin cos sin cos 0.

x t z z dz ... (10)

Since B1 (x, t) = 0, (7) and (8) show that Bp (x, t) = 0 and Cp = 0 for all 2.p ... (11)

Putting the above values in (3) and (4), we have( , ; ) ( , ) sin cos D x t K x t x t and ( ) 1. D

From (2) ( , ; ) sin cos .R x t x t Part (ii) Given K (x, t) = x – 2t. ... (1)

The resolvent kernel is given by ( , ; ) ( , ; ) / ( ) R x t D x t D ... (2)

where1

( )( , ; ) ( , ) ( , ),!

m

mmD x t K x t B x t

m... (3)

1

( )( ) 1 ,!

m

mmD C

m ... (4)

C0 = 1, ... (5) B0 (x, t) = K (x, t) = x – 2t. ... (6)

1

10( , ) , 1

p pC B s s ds p ... (7)



1

10( , ) ( , ) ( , ) ( , )

p p pB x t C K x t p K x z B z t dz ... (8)

From (7), 1 1

1 00 0

1( , ) ( 2 )2

C B s s ds s s ds ... (9)

From (8),1

1 1 00( , ) ( , ) ( , ) ( , ) B x t C K x t K x z B z t dz

1

0

1 ( 2 ) ( 2 ) ( 2 ) ,2

x t x z z t dz using (1), (6) and (9)

13 21 2

00

1 1 2( 2 ) { 2 ( 4 ) 2 } ( – 2 ) ( 4 ) 22 2 3 2

z zx t z z x t xt dz x t x t xtz

1 2 1( 2 ) ( 4 ) 22 3 2

x t x t xt

B1 (x, t) = (2/3) + 2xt – x – t. ... (10)

From (7),131 1 2 2

2 10 00

2 2 2( , ) 2 2 ,3 3 3

sC B s s ds s s ds s s by (3)

C2 = (2/3) + (2/3) – 1 = 1/3. ... (11)

From (8), 1

2 2 10( , ) ( , ) 2 ( , ) ( , ) B x t C K x t K x z B z t dz

1

0

1 2[( – 2 ) 2] ( 2 ) 2 ,3 3

x t x z zt z t dz using (1), (10) and (11)

1 2 2

0

1 2 4( 2 ) 2 2 4 2 23 3 3

xx t xzt xz xt z z t z zt dz 12 2 2 3

2 2

0

1 2 2 4 2( 2 ) 23 3 2 3 3 3

xz xz z z t zx t xtz xtz z t

1 4 4 4( 2 ) 2 2 4 2 0.3 3 3 3

xx t xt x xt t t ... (12)

Since B2 (x, t) = 0, (7) and (8) show that Bp (x, t) = 0 and Cp = 0 for all 3p


1( , ; ) ( , ) ( , ) ... 2 2 / 3 2D x t K x t B x t x t xt x t

and 2 21 2

( ) 1 ( / 2!) ... 1 ( / 2) ( / 6).D C C

from (2), 2

2 2 / 3 2( , ; ) .

1 ( / 2) ( / 6)x t xt x t

R x t

Part (iii) Given K (x, t) = x + t + 1. ... (1)The resolvent kernel is given by ( , ; ) ( , ; ) / ( ) R x t D x t D ... (2)



where1

( )( , ; ) ( , ) ( , ),!

m

mmD x t K x t B x t

m... (3)

1

( )( ) 1 ,!

m

mmD C

m ... (4)

C0 = 1 ... (5) B0 (x, t) = K (x, t) = x + t + 1. ... (6)

1

11( , ) , 1

p pC B s s ds p ... (7)

and 1

11( , ) ( , ) ( , ) ( , ) , 1

p p pB x t C x t p K x z B z t dz p ... (8)

From (6) and (7), 1 1

1 01 1( , ) (1 2 )C B s s ds s ds

121

2.

s s ... (9)

From (8), 1

1 1 01( , ) ( , ) ( , ) ( , )

B x t C x t K x z B z t dz

1

12 ( 1) ( 1) ( 1) ,

x t x z z t dz by (2), (6) and (9)

1 2

12 ( 1) [ ( 2 ) ( 1) ( 1)]

x t z z t x x t dz

13 2

1

2 ( 1) ( 2 ) ( 1) ( 1)3 2z zx t t x x t z

2( 1) 2 / 3 2 ( 1) ( 1)x t x t

2 ( 1) 2 / 3 2 ( 1) 2 1/ 3 .x t xt x t xt ... (10)

From (7) and (10), 1 1 2

2 11 1

1( , ) 23

C B s s ds s ds

13

1

1 82 .3 3 3s s

... (11)

From (8), 1

2 2 11( , ) ( , ) 2 ( , ) ( , )

B x t C K x t K x z B z t dz

1

1

8 1( 1) 2 ( 1) { 2 } ,3 3

x t x z zt dz using (1), (10) and (11)

1 2

1

8 1 1( 1) 4 ( 1)3 3 3

x t z t z xt t x dz

13 2

1

8 1 1( 1) 4 1 ( 1)3 3 2 3 3

z t zx t xt x z

8 2 2( 1) 4 ( 1) 0.3 3 3

tx t x ... (12)




Putting these values in (3) and (4), we get

1( , ; ) ( , ) ( , ) 1 2 1/ 3D x t K x t B x t x t xt

and 2 21 2

( ) 1 ( / 2!) 1 2 (4 / 3)D C C

2

1 2 1/ 3( , ; ) .

1 2 (4 / 3)x t xt

R x t

Part (iv). Given K (x, t) = e x – t. ... (1)The resolvent kernel ( , ; )R x t is given by ( , ; ) ( , ; ) / ( ) R x t D x t D ... (2)

where1

( )( , ; ) ( , ) ( , )!

m

mmD x t K x t B x t

m... (3)

1

( )( ) 1 ,!

m

mmD C

m ... (4)

C0 = 1, ... (5)B0 (x, t) = K (x, t) = ex – t, ... (6)

1

10( , ) , 1

p pC B s s ds p ... (7)

and 1

10( , ) ( , ) ( , ) ( , ) , 1.

p p pB x t C K x t p K x z B z t dz p ... (8)

From (7), 1 1 1

1 00 0 0( , ) 1. s sC B s s ds e ds ds ... (9)

From (8),1


1

0, x t x z z te e e dz using (1), (6) and (9)

1

00. x t x t x t x te e dz e e ... (10)

Since B1 (x, t) = 0, (7) and (8) show thatBp (x, t) = 0 and Cp = 0 for all 2.p ... (11)

Putting these values in (3) and (4), we get

( , ; ) ( , ) x tD x t K x t e and 1( ) 1 1 . D C

From (2), ( , ; ) /(1– ).x tR x t e

Part (v) Given 2( , ) 4 . K x t xt x ... (1)

The resolvent kernel ( , ; )R x t is given by ( , ; ) ( , ; ) / ( ), R x t D x t D ... (2)

where 1

( )( , ; ) ( , ) ( , ),!

m

mmD x t K x t B x t

m... (3)



1

( )( ) 1 ,!

m

mmD C

m ... (4)

C0 = 1, ... (5) B0 (x, t) = K (x, t) = 4 xt – x2, ... (6)

1

10( , ) , 1

p pC B s s ds p ... (7)

and1

10( , ) ( , ) ( , ) ( , ) , 1.


From (6) and (7),1 1 12 2 3

1 0 00 0( , ) (4 ) 1.C B s s ds s s ds s ... (9)

From (8), 1


12 2 2

04 (4 ) (4 ) , xt x xz x zt z dz by (1), (6) and (9)

12 3 2 2 2

04 [ 4 ( 16 ) 4 ] xt x xz z x xt x t z dz

12 4 3 2 20

4 ( / 3) ( 16 ) 2xt x xz z x xt x z

2 2 2 2 21 4 44 ( 16 ) 2 2 .3 3 3

xt x x x xt x t x t x x xt ... (10)

From (7), 1 1 3 2 2

2 10 0

4 4( , ) 2 , using (10)3 3

C B s s ds s s s s ds

14 2 33

0

4 4 1 4 1 4 1 .2 9 2 9 2 9 2 9 9

s s ss ... (11)

From (8), 1

2 2 10( , ) ( , ) 2 ( , ) ( , ) B x t C K x t K x z B z t dz

12 2 2 2

0

1 4 4(4 ) 2 (4 ) 29 3 3

xt x xz x z t z z zt dz

[using (1), (10) and (11)]12 2 2

0

1 4 4(4 ) 2 (4 ) 2 19 3 3

txt x xz x z t z dz

12 3 2 2 2

0

1 4 4 4 4(4 ) 2 4 2 4 1 2 19 3 3 3 3

t txt x x t z z x x t x z dz

13 22 4 2 2

0

1 4 4 4 4(4 ) 2 2 4 1 2 19 3 3 3 3 2 3

z t x txt x x t z x x t z

22 21 4 1 4 4 4(4 ) 2 2 4 1 2 1

9 3 3 3 3 2 2

t x txt x x t x x t = 0.



Since B2 (x, t ) = 0, (7) and (8) show that Bp (x, t) = 0 and Cp = 0 for all 2.p ... (12)

Putting these values in (3) and (4), we have

2 2 21

4 4( , ; ) ( , ) ( , ) 4 23 3

D x t K x t B x t xt x x t x x xt

and 2 21 2

( ) 1 ( / 2!) 1 ( /18).D C C

from (2),

2 2 2

2

4 44 23 3( , ; ) .

1 ( /18)

xt x x t x x xtR x t

Part (vi). Try yourself Ans. 2 2

sin ( ) cos ( )( , ; ) .1

x t x tR x t

Part (vii). Try yourself Ans. 1

1 2sinh 2 ( sinh )( , ; ) .

1 4

x t e x tR x t

e

Part (viii). Try yourself Ans.

2 2

2

14 3 5( , ; ) .

1 ( / 240)

x t xtx t xt xR x t

Part (ix). Try yourself Ans. 2 2sin sin (1 2 sin sin )( , ; ) .

1 2

x t x tR x t

Part (x). Try yourself Ans. 2

3( )1 3 3 12( , ; ) .

1 2 ( / 4)

x txt xtR x t

Ex.12. For the integral equation ( ) ( ) ( , ) ( ) , b


compute ( )D and ( , ; )D x t for the following kernels for the specified limits a and b. (i) K (x, t) = 1, a = 0, b = 1. [Kanpur 2007] (ii) K (x, t) = – 1, a = 0, b = 1.(iii) ( , ) sin , 0, . K x t x a b (iv) K (x, t) = xt, a = 0, b = 10. [Kanpur 2006](v) K (x, t) = t, a = 0, b = 10. (vi) K (x, t) = x, a = 4, b = 10. [Meerut 2009](vii) K (x, t) = 2ex et, a = 0, b = 1. (viii) K (x, t) = g (x), a = a, b = b.(ix) K (x, t) = g (t), a = a, b = b. (x) K (x, t) = x – t, a = 0, b = 1.Sol. (i) Given K (x, t) = 1. ... (1)

We know that1

( )( , ; ) ( , ) ( , ),!

m

mmD x t K x t B x t

m... (2)

1

( )( ) 1 ,!

m

mmD C

m... (3)

C0 = 1, ... (4)B0 (x, t) = K (x, t) = 1, ... (5)

1

10( , ) , 1

p pC B s s ds p ... (6)



and1

10( , ) ( , ) ( , ) ( , ) , 1


From (5) and (6), 1 1

1 00 0( , ) 1. C B s s ds ds ... (8)

From (1), (5), (7) and (8), 1


1

01 0.dz ... (9)


Putting the above values in (2) and (3), we get( , ; ) ( , ) 1 and ( ) 1 .D x t K x t D

Part (ii). Ans : Proceed as in part (i). Ans. ( , ; ) 1, ( ) 1 . D x t DPart (iii) Given K (x, t) = sin x. ... (1)

We know that1

( )( , ; ) ( , ) ( , ),!

m

mmD x t K x t B x t

m ... (2)

1

( )( ) 1!

m

mmD C

m... (3)

C0 = 1 ... (4) B0 (x, t) = K (x, t) = sin x. ... (5)

10

( , ) , 1

p pC B s s ds p ... (6)

and 10( , ) ( , ) ( , ) ( , ) , 1


From (5) and (6), 1

1 0 00 0( , ) sin [ cos ] 2.

C B s s ds s ds s ... (8)

From (7) 1 1 00( , ) ( , ) ( , ) ( , )

B x t C K x t K x z B z t dz

0

2 sin sin sin ,

x x z dz using (1), (5) and (8)

02 sin sin [ cos ] 2 sin 2 sin 0. x x z x x ... (9)

Since B1 (x, t) = 0, (6) and (7) show thatBp (x, t) = 0 and Cp = 0 for all 2.p ... (10)

Putting the above values in (2) and (3), we get( , ; ) ( , ) sin and ( ) 1 2 .D x t K x t x D

Part (iv). Proceed as in part (i) Ans. 3( , ; ) , ( ) 1 (10 / 3) . D x t xt D

Part (v). Given K (x, t) = t. ... (1)

We know that1

( )( , ; ) ( , ) ( , ),!

m

mmD x t K x t B x t

m... (2)



1

( )( ) 1 ,!

m

mmD C

m... (3)

C0 = 1, ... (4) B0 (x, t) = K (x, t) = t, ... (5)

10

–10( , ) , 1p pC B s s ds p ...(6)

and 10

10( , ) ( , ) ( , ) ( , ) , 1.


From (6), 1010 2

1 0 0/ 2 50.C s ds s ... (8)

From (7), 10 10

1 1 00 0( , ) ( , ) ( , ) ( , ) 50 B x t C K x t K x z B z t dz t zt dz , by (1), (5) and (8)

1020

50 / 2 50 50 0.t t z t t ... (9)

Since B1 (x, t) = 0, (6) and (7) show that Bp (x, t) = 0 and Cp = 0 for all 2.p ...(10)

Putting the above values in (2) and (3), we get ( , ; ) ( , ) , ( ) 1 50 . D x t K x t t D

Part (vi). Try yourself Ans. ( , ; ) D x t x and ( ) 1 42 . D

Part (vii). Try yourself Ans. ( , ; ) 2 x tD x t e e and 2( ) 1 ( 1). D e

Part (viii). Try yourself Ans. ( ) 1 ( ) . b

aD g t dt

Part (ix). Try yourself Ans. ( ) 1 ( ) . b

aD g t dt

EXERCISES1. State and prove Fredholm’s first fundamental theorem. [Meerut 2004]2. State and prove first and second series for the non-homogeneous integral equation of

second kind.3. Use Fredholm determinants to find the resolvent kernel.

( , ; ) ( , ; ) / ( ) R x t D x t Dof the kernel K (x, t) = xet under the limits of integration a = 0, b = 1. Hence solve the equation

1

0( ) ( ) . x tx e xe t dt Ans. ( , ; ) ;

1

txeR x t solution is ( )1

x xtx e

4. Solve : 12

0( ) sec ( ) y x x y t dt Ans.

2sec tan11

y x

5. Determine the resolvent kernel and hence solve the following integral equations :

(i)2

0( ) 1 sin ( ), ( ) .

y x x t y t dt Ans. y (x) = 1.



(ii)2

0( ) cos 2 sin cos ( ) .

y x x x t y t dt Ans. y (x) = cos 2x.

(iii)1

0( ) ( ) . x x ty x e e y t dt Ans.

1( ) .2

xy x e

(iv)1

0( ) (2 ) ( ) .

6

xy x x t y t dt Ans. 2

21 (6 2)( ) .6 3 6

x xy x x

(v)1 2

0( ) (4 ) ( ) . y x x xt x y t dt Ans. 2

3 (2 3 6)( ) .18 18

x xy x

6.4. FREDHOLM’S SECOND FUNDAMENTAL THEOREM.If 0 is a zero of multiplicity m of the function ( )D , then the homogeneous integral equation

0( ) ( , ) ( ) b


possesses at least one, and the most m, linearly independent solutions

1 1 1

01 1 1

, ..., , , ...,( ) ,

, ..., , , , ...,i i r

i ri i r

x x x x xy x D

t t t t t

1,2, ..., ; 1 .i r r m ... (2)

not identically zero. Any other solution of this equation is a linear combination of thesesolutions. Here, we have to remember the following definition of the Fredholm minor

1 11 2 1 2

111 2 1 2 1 1

, .., , , ...,, , ..., , , ..., ( ) ... ... ,, , ..., , , ..., , ...., , ,...,!

p b b n pn nn pp a a

n n n p

x x z zx x x x x xD K K dz dz

t t t t t t t t z zp

... (3)where {xi} and {ti}, i = 1, 2, ... n, are two sequences of arbitrary variables. Series (3) convergesfor all values of and hence it is an entire function of .

Proof. We shall first show that every zero of ( )D is a pole of the resolvent kernel ( , ; )R x tgiven by

( , ; ) ( , ; ) / ( ) R x t D x t D , ... (4)

the order of this pole being at most equal to the order of the zero of ( ).DThe Fredholm’s first series is

111

1

, ...,( )( ) 1 ... ..., ...,!

p b b p

pp a ap

x xD K dx dx

x xp... (5)

and the Fredholm’s second series is

111

1

, , ... ,( )( , ; ) ( , ) ... ... ., , ...!

p b b p

pp a ap

x x xD x t K x t K dx dx

t x xp ... (6)

Differentiating both sides of (5) w.r.t. ‘ ’ and interchanging the indices of the variables of

integration, we obtain ( ) – ( , ; )b

aD D x x dx ...(7)



showing that if 0 is a zero of order k of ( ),D then it is a zero of order k – 1 of ( ) D and hence

0 may be zero of order at most k – 1 of the entire function ( , ; ).D x t It follows that 0 must be

the pole of the quotient (4) of order at most k. In particular, if 0 is a simple pole of ( ),D then

0 0( ) 0, ( ) 0, D D and 0 is a simple pole of the resolvent kernel. Again, (7) shows that

( , ; ) 0. D x t Keeping in mind this particular case and the following equation.

( , ; ) ( , ) ( ) ( , ) ( , ; ) , b

aD x t K x t D K x z D z t dz

we find that if ( ) 0 D and ( , ; ) 0, D x t then ( , ; ),D x t as a function of x, is solution of (1)

and so is ( , ; ), D x t being an arbitrary constant.Next, we take the general case when is a zero of an arbitrary multiplicity m, that is,

when ( ) ( )0 0 0 0

( ) 0, ( ) 0, ..., ( ) 0 , ..., ( ) 0,r mD D D D ... (8)

where the superscript r stands for the differential of order r, r = 1, 2, ..., m – 1.Differentiating series (5) n times and comparing it with the series (3), we obtain the following

relation

11

1

,...,( ) ( 1) ... ... ,, ...,

n b b nnn n na a

n

x xd D D dx dxx xd

... (9)

showing that, if 0 is a zero of multiplicity m of the function ( ),D then the following

condition is valid for the Fredholm minor of order m for that value of 0 :

1 20

1 2

, , ...,0.

, ,...,m

mm

x x xD

t t t

... (10)

In this connection note that we may get minors of order lower than m which also do notidentically vanish.

We now determine the relation among the minors that corresponds to the resolvent formula

( , ; ) ( , ) ( , ) ( , : ) . b

aR x t K x t K x z R z t dz ... (11)

Expanding the determinant under the integral sign in (3), we obtain

1 1 1 2 1 1 1 1

2 1 2 2 2 2 1 2

1 2 1

1 1 1 2 1 1 1 1

( , ) ( , ) ... ( , ) ( , ) ... ( , )

( , ) ( , ) ... ( , ) ( , ) ... ( , )

: : : : :( , ) ( , ) ... ( , ) ( , ) ... ( , )

( , ) ( , ) ... ( , ) ( , ) ... ( , )

: : : : :( ,

n p

n p

n n n n n n p

n p

p

K x t K x t K x t K x z K x z



K z t K z t K z t K z z K z z

K z t1 2 1) ( , ) ... ( , ) ( , ) ... ( , )p p n p p pK z t K z t K z z K z z

... (12)



by elements of the first row and integrating p times with respect to z1, z2, ..., zp for 1,p we get

1 11

1 1

, ..., , , ...,... ...

, ..., , , ...,b b n p

pa an p

x x z zK dz dz

t t z z

111

( 1) ( , )n

hhh

K x t

2 11 2

1 1 1 1

, ..., , ..., , , ...,... ...

, ..., , , ... , , ...,b b h n p

pa ah h n p

x x x z zK dz dz dz

t t t t z z

11

( 1)p

h nh

2 1 21 1

1 1 1 1 1

, ..., , , , ..., ,...,... ( , ) ...

, ..., , , , ... , , ...,

b b n h ph pa a

n n h h p

x x z z z zK x z K dz dz

t t t z z z z

... (13)

Observe carefully that the symbols for the determinant K on the right hand side of (13) do notcontain the variable x1 in the upper sequence and the variables th or zh in the lower sequence. Again,by transposing the variable xh in the upper sequence to the first place by means of h + n – 2transpositions, it follows that all the components of the second sum on the right side are equal.Accordingly, (13) can be re-written as

1 11

1 1

, ..., , , ...,... ...

, ..., , , ...,b b n p

pa an p

x x z zK dz dz

t t z z

111

( 1) ( , )n

hhh

K x t

2 11

1 1 1 1

, ..., , , ...,... ...

, ..., , , ..., , , ...,b b n p

pa ah h n p

x x z zK dz dz

t t t t z z

2 1 11 1 1

1 2 1 1

, , ..., , , ...,( , ) ... ... .

, , ..., , , ...,

b b b n ppa a a

n p

z x x z zp K x z K dz dz dz

t t t z z ... (14)

where we have omitted the subscript h from z.Substituting (14) in (13), we see that Fredholm minor satisfies the following integral equation:

21 211 1 11 1 1 11 1 2

, ..., ,, ... , , , ...,( 1) ( , ) ( , )

, ..., , ,, ... , , , ...,

n bnn nhn n n nh ah h nn n

x xx x z z xD K x t D K x z D dz

t t t tt t t t t

... (15)

It may be observed that the expansion by the elements of any other row also leads to asimilar identity, with z placed at the corresponding place.

Expanding the determinant (12) with respect to the first column and proceeding as above, weobtain the integral equation

1 1 11 111 1 11

1 22

, ..., , , ...,, ..., , ...,( 1) ( , ) ( , ) .

, ... , , , ...,, ... ,

n bh h nn nhn h n nh a

n nn

x x x xx x x xD K x t D K z t D dz

t t z t tt t

... (16)

and a similar result can be obtained by expanding by any other columnThe relations (15) and (16) hold for all values of . We now show that (15) can be used to

find the solution of (1) for the special case when 0 is an eigenvalue. Suppose that 0 is

a zero of multiplicity m of ( ).D Then the minor | 0mD and even the minors D1, D2, ..., Dm – 1



may not identically vanish. Suppose Dr is the first minor in the sequence D1, D2, ..., Dm – 1 suchthat | 0.rD Then the number r must be between 1 and m and is the index of the eigenvalue.Further, it follows that Dr – 1 = 0. Then (15) shows that

1 21 0

1 2

, , ...,( )

, , ...,r

rr

x x xy x D

t t t

... (17)

is a solution of (1). Substituting x at different points of the upper sequence in the minor Dr, weobtain r nontrivial solutios yi (x), i = 1, 2, ..., r, of (1), which are usually written as

1 1 10

1

1 1 10

1

, ..., , , ,...,

, ...,( ) , 1, 2, ...,

, ..., , , ,...,

, ..., ,

i i rr

ri

i i i rr

r

x x x x xD

t tx i r

x x x x xD

t t

... (18)

in which the denominator is non-zero.

We now establish that solutions i given by (18) are linearly independent. To this end notethat if we put two of the arguments xi equal in the determinant (12), this is equivalent to putting tworows equal, and hence the determinant vanishes. It follows that, in (18), ( ) 0 k ix for ,i k

whereas ( ) 1. k kx Now, if we have a relation of the form 0, k kkC then putting x = xi, we

obtain 0iC and so the solutions i are linearly independent. This system of solutions i isknown as fundamental system of eigenfunctions of 0 and any linear combination of these functionsgives a solution of (1).

Conversely, we can prove that any solution of (1) is a linear combination of

1 2( ), , ..., ( ). vx x For this purpose we define kernel ( , ; )H x t as follows* :

11 0

1

, , ...,( , ; )

, , ...,r

rr

x x xH x t D

t t t

10

1

, ..., ...,

, ..., ...,r

rr

x xD

t t

. ... (19)

Putting n = r and adding extra arguments x and t in (10), we arrive at

1 11 0 1 0

1 1

, , ..., , ...,( , )

, , ..., , ...,r r

r rr r

x x x x xD K x t D

t t t t t

1 1 101

1 2

, , ..., , ...,( 1) ( , )

, , ..., ,

r h h rhh rh

r

x x x x xK x t D

t t t

10 1 0

1

, , ...,( , ) .

, , ...,

b r

rar

x x xK z t D dz

z t t ... (20)

* The Kernel ( , ; )H x t will correspond to the resolvent kernel ( , ; )R x t of Art. 6.1.



Transposing the variable x from the first place to the place between the variables xh – 1 andxh + 1 in every minor Dr in (20) and then dividing both sides of (20) by the constant

10

1

, ...,0,

, ..., ,

rr

r

x xD

t t we have

0 1( , ; ) ( , ) ( , , ) ( , ) ( , ) ( ).

rb

h hhaH x t K x t H x z K z t dz K x t x ... (21)

Suppose that y (x) is an arbitrary solution of (1). Multiplying both sides of (21) by y (t) andthen integrating both sides w.r.t. ‘t’ from a to b, we obtain

10 0

( )( )( ) ( , ; ) ( ) ( , ; ) ( ),rb b h

hha a

y xy xy t H x t dt y z R x z dz x

... (22)

where we have used (1) in all terms except the first ; we have also assumed that

0 ( , ) ( ) ( ). b

h haK x t y t dt y x

Cancelling the equal terms in (22), we obtain

1( ) ( ) ( )

r

h hhy x y x x ... (23)

This completes the proof of the Fredholm’s second fundamental theorem.6.5 FREDHOLM THIRD FUNDAMENTAL THEOREM.

For an inhomogeneous equation

0( ) ( ) ( , ) ( ) .b

ay x f x K x t y t dt ...(1)

to possess a solution in the case ( ) 0,

D it is necessary and sufficient that the given function

f (x) be orthogonal to all the eigenfunctions ( ), 1, 2, ..., ,iz x i v of the transposed homogenous

equation corresponding to the eigenvalue 0 . The general solution has the form

1 21 0

1 2

11 2

01 2

, , , ...,

, , , ... ,( ) ( ) ( ) ( ),

, , ...,

, , ... ,

rr rb r

h hhar

rr

x x x xD

t t t ty x f x f t dt C x

x x xD

t t t

... (2)

where ( )i x are given by

1 –1 10

1

1 –1 10

1

,..., , , , ... ,

,..............................,( ) , 1,2,...

, ..., , , , ...,

,...............................,

i i rr

ri

i i i rr

r

x x x x xD

t tx i r

x x x x xD

t t

...(2)

Proof. Consider ( ) ( ) ( , ) ( ) . b

ay x f x K x t y t dt ...(1)



We know that the trasponse (or adjoint) of (1) is given by

( ) ( ) ( , ) ( ) .b

az x f x K t x z t dt ... (3)

Then for the transposed equation (3), Fredholm’s first series ( )D and the Fredholm’s secondseries ( , ; )D t x are given by

1

111

, ...,( )( ) 1 ... ...,...,!

p b b p

pp a ap

z zD K dz dz

z zp... (4)

and 111

1

, , ...,( )( , ; ) ( , ) ... ..., ,...,!

p b b p

pp a ap

t z zD t x K t x K dz dz

x z zp... (5)

From, the above fact, it follows that the kernels of (1) and (2) possess the same eigenvalues.Moreover, the resolvent kernel of (2) is given by

( , ; ) ( , ; ) / ( ) R t x D t x D ... (6)

and hence the solution of (2) is

( , ; )( ) ( ) ( ) ,( )

b

a

D t xz x f x f t dtD

... (7)

provided is not an eigenvalue.Again it is obvious that not only the transposed kernel has the eigenvalues as the orginal

kernel of (1) , but also the index r of each of the eigen values is equal. Furthermore the eigenfunctions

of the transposed equation for an eigenvalue 0 are given by

10

1 1 1

10

1 1 1

, .................................,

, ..., , , ...........,( ) ,

, .................................,

, ..., , , ...........,

rr

i i r

ir

ri i i r

x xD

t t t t tz t

x xD

t t t t t

... (8)

where the values (x1, ..., xr) and (t1, ..., tr) are so chosen that the denominator in (8) does not vanish.Substituting r in different places in the lower sequence of this formula, we obtain a linearly

independent system of r eigenfunctions. Again, we know that each i is orthogonal to each iz withdifferent eigenvalues.

Suppose that a solution ( )y x of (1) exists. Then multiplying (1) by each member ( )kz x ofthe above-mentioned system of functions and integrating w.r.t. x from a to b, we get

( ) ( ) ( ) ( ) ( , ) ( ) ( )b b b b

k k ka a a af x z x dx y x z x dx K x t y t z x dx dt

( ) [ ( ) ( , ) ( ) ] 0.b b

k ka ay x dx z x K t x z t dt ... (9)



Since ( )z x

is an eigenfunctions of the transposed equation, we have

( ) ( , ) ( )b

kaz x K t x z t dt

.. (10)

Using (10), (9) reduces to ( ) ( ) 0,b

kaf x z x dx ... (11)

showing that a necessary condition for (1) to have a solution is that the inhomogeneous term f (x)be orthogonal to each solution of the transposed homogeneous equation.

Conversely, we shall prove that the condition (11) of orthogonality is sufficient for the existenceof a solution. In what follows, we shall also obtain an explicit solution in that case. At this stage weneed to define a kernel ( , ; )H x t as follows :

11 0

1

, , ...,( , ; )

, , ...,r

rr

x x xH x t D

t t t

10

1

, ...,,

, ...,r

rr

x xD

t t

... (12)

where we have assumed that 0rD and that r is the index of the eigenvalue 0 .

To prove the required result we shall show that if the orthogonality condition is satisfied,then the function

0 0( ) ( ) ( , ; ) ( ) b

ay x f x H x t f t dt ... (13)

is a solution. To this end, putting the above value of y (x) in (1), we have

0 0 0( ) ( , ; ) ( ) ( , ) [ ( ) ( , ; ) ( ) ] b b b

a a af x H x t dt f x K x t f t H t z f z dz dt

or 0( ) [ ( , ; ) ( , ) ( , ) ( , ; ) ] 0.

b b

a af t dt H x t K x t K x z H z t dz ... (14)

Recall the procedure of getting the equation (21) of Art. 6.4. Proceeding like-wise, we canobtain its ‘‘transpose’’.

0 1

( , ; ) ( , ) ( , ) ( , ; ) ( , ) ( ).rb

h hhaH x t K x t K x z H z t dz K x t z t

... (15)

Substituting this in (14) and making use of the orthogonality condition, we shall arrive at anidentity and thus we prove what we wished to prove.

Now, the difference of any two solutions of (1) is a solution of the homogeneous equation.Therefore, the most general solution of (1) is

0 1( ) ( ) ( , ; ) ( ) ( ).

rb

h hhay x f x H x t f t dt C x

MISLLANEOUS EXERCISE1. Prove that every zero of Fredholm function ( )D is a pole of the resolvent kernel

( , ; ) ( , ; ) / ( ). R x t D x t DProve also that the order of this pole is at most equal to the order of the zero of the

denominator ( ).D [Meerut 2007]Hint : Proof is contained in Fredholm’s second fundamental theorem.

2. State and prove Fredholm’s second fundamental theorem.



3. State and power Fredholm’s third fundamental theorem.4. Use the method of this chapter to find the resolvent kernels for the following integral

equations

(i)1

0( ) ( ) | | ( ) y x f x x t y t dt (ii)

1 | |

0( ) ( ) ( ) x ty x f x e y t dt

(iii)2

0( ) ( ) cos ( ) ( )

y x f x x t y t dt

5. Show by using the method of this chapter that the resolvent kernel for the integral equationwith kernel K (x, t) = 1 – 3xt in the integral (0, 1) is

2( , ; ) [4 / (4 )] [1 (1/ 2) ( ) 3 (1 ) ], 2.R x t x t xt

6. Solve the following homogeneous equations :

(i)0

1( ) sin ( ) ( )2

y x x t y t dt (ii)

1

2 0

1( ) 2 ( )1

x ty x e e y t dt

e

7. State and prove Fredholm first and second series for non-homogeneous Fredholm integralequation of the second kind. [Meerut 2000, 02]Hint : Refer results (17) and (30) of Art. 6.2 for Fredholm first and second seriesrespectively.


CHAPTER 7

Integral equations withsymmetric kernels

7.1. INTRODUCTION7.1 (a) Symmetric kernels.

A kernel is called symmetric if it coincides with its own complex conjugate. Such a kernel ischaracterized by the identity

K (x, t) = K (t, x),where the bar denotes the complex conjugate.

If the kernel is real, then its symmetry is defined by the identity K (x, t) = K (t, x).An integral equation with a symmatric kernel is called a symmetric equation.Remark. We have already discussed eigenvalues and eigenfunctions for integral equations in

chapter 3, 4 and 6. We have established that the eigenvalues of an integral equations are the zerosof certain determinant. In this process we have seen that there are many kernels for which there areno eigenvalues (see solved examples 3 and 4 of Art 3.3, chapter 3).

However in this chapter we shall prove that for a symmetric kernel that is not identicallyzero, at least one eigenvalue will always exist. This is an important characteristic of symmetrickernels.

An eigenvalue is simple if there is only one corresponding eigenfunction, otherwise the eigenvalues are degenerate. The spectrum of the kernel K (x, t) is the set of all its eigenvalues. Thus, asdiscussed above the spectrum of a symmetric kernel is never empty.7.1 (b). Regularity conditions.

In our study we shall mainly deal with functions which are either continuous, or integrableor square-integrable. When an integral sign is used, the Lebesgue integral is understand. Also,note that if a function is Riemann-integrable, it is also Lebesgue-integrable. By a square-

integrable function f (x), we mean that 2| ( ) |b

af x dx ... (1)

A square-integrable function f (x) is called an -function. The regularity conditions on thekernel K (x, t) are identical. It is an -function if the following three conditions are satisfied :

(i) 2| ( , ) | , [ , ], [ , ],b b

a aK x t dx dt x a b t a b ... (2)

(ii) 2| ( , ) | , [ , ],b

aK x t dt x a b ... (3)

(iii) 2| ( , ) | , [ , ].b

aK x t dx t a b ... (4)

Here stands for every (or for all).7.1


7.2 Integral Equation with Symmetric Kernels

7.1. (c) The inner or scalar product of two functions.The inner or scalar product of two complex -functions and of real variable x,

,a x b is denoted by ( , ) and is defined as ( , ) ( ) ( ) .b

ax x dx ... (5)

where the bar denotes the complex conjugate.

Two functions are called orthogonal if their inner product is zero, that is, and are

orthogonal if ( , ) = 0 i.e., ( ) ( ) 0.b

ax x dx ... (6)

The norm of a function (x) is denoted by || ( ) ||x and is given by the relation.

1/ 2 2 1/ 2|| ( ) || [ ( ) ( ) ] [ | ( ) | | .b b

a ax x x dx x dx ... (7)

A function (x) is said to be normalized if || ( ) || 1.x If follows that a nonnull function(whose norm is not zero) can always be normalized by dividing it by its norm.

7.1. (d) Schwarz inequality. If (x) and (x) are -functions, then

| ( , ) | || || || || ... (8)

Minkowski inequality. If (x) and (x) are -functions, then

|| || || || || || . ... (9)

7.1. (e) Complex Hilbert space. (Kanpur 2011)We present review of some important properties of the complex Hilbert space (a, b). The

same discussion will remain applicable to real space as a particular case.A linear space of infinite dimension with inner product (or scalar product) (x, y) which is a

complex number is called a complex Hilbert space if it satisfies the following three axioms :(i) the definiteness axiom (x, x) > 0 for 0,x

(ii) the linearity axiom 1 2 1 2( , ) ( , ) ( , ),x x y x y x y

where and are arbitrary complex numbers.(iii) the axiom of (Hermitian) symmetry (y, x) = ( , ),x y

where the bar denotes the complex conjugate.

Let H be the set of complex-valued functions (x) defined in the interval (a, b) such that

2| ( ) | .b

ax dx

Then H is linear and complex Hilbert space (a, b) (or ). The norm of function generates

the natural metric 1/ 2( , ) || || ( , ) .d ... (10)

A metric space is called complete if every Cauchy sequence of functions in this space is aconvergent sequence. A Hilbert space is an inner-product linear space that is complete in its naturalmetric. The completeness of spaces plays an important role in the theory of linear operatorssuch as the Fredholm operator K, defined as


Integral Equation with Symmetric Kernels 7.3

( , ) ( ) .b

aK K x t t dt ... (11)

The operator adjoint to K is ( , ) ( ) .b

aK K t x t dt ... (12)

The operators (11) and (12) are connected as follows : ( , ) ( , ).K K ... (13)which can be easily proved as follows :

L.H.S. of (13)

= ( , ) ( , ) ( ) ( ) ,b b

a aK K x t t dt x dx

using results (5) and (11)

( , ) ( ) ( )b b

a aK x t x dx t dt

, charging the order of integration

( , ) ( ) ( )b b

a aK t x t dt x dx

[Re-naming the variables x and t as t as x respectively]

( , ) ( ) ( ) ,b b

a aK t x t dt x dx

where the bar denotes the complex conjugate

( , ),K using results (5) and (11)= R.H.S. of (13)

For a symmetric kernel, (13) reduces to ( , ) ( , ),K K ... (14)that is, a symmetric operator is self-adjoint. Further, we find that permutation of factors in a scalarproduct is equivalent to taking the complex conjugte, that is, , ).( , ) (K K Combining thisfact with (14), we find that, for a symmetric kernel, the inner product ( , )K is always real. Theconverse of this is also true.7.1 (f) An orthonormal system of functions.

A finite or an infinite set {k(x)} defined on an interval a n b is said to be an orthogonalset if

( , ) 0i j or ( ) ( ) 0, .b

i jax x dx i j ... (15)

If none of the elements of this set is a zero vector, then it is called a proper orthogonal set.The set { ( )}i x is orthonormal if

0, ,( , ) ( ) ( )

1, .

bi j i ja

i jx x dx

i j

... (16)

Any function (x) for which || (x) || = 1 is said to be normalized.

Given a finite or an infinite (denumerable) independent set of functions 1 2{ , , ..., , ...},k we can construct an orthonormal set 1 2{ , , ..., , ...}k by Gram-Schmidt method as follows.

Let 1 1 1/ || || .

To obtain 2 , we define 2 2 2 1 1( ) ( ) ( , ) .w x x

Then function w2 is orthogonal to 1. Hence 2 can be constructed by setting



2 2 2/ || || .w w Proceding in this manner, we obtain

1

1( ) ( ) ( , ) , where / || || .

k

k k i i k k kiw x x w w

Again, if we are given a set of orthogonal functions, we can convert it into an orthonormalset simply by dividing each function by its norm.

Starting from an arbitrary orthonormal system, it is possible to construct the theory of Fourierseries. Suppose we wish to obtain the best approximation of an arbitrary function ( )x in terms ofa linear combination of an orthonormal set 1 2( , , ..., ).n By the best approximation, we meanthat we choose the coefficients 1 2, , ..., n such as to minimize

2

1 1|| ||, . ., || ||

n n

i i i ii ii e

Now, for any 1 2, , ..., ,n we have

2 2 2 2

1 1 1|| || || || | ( , ) | | ( , ) | .

n n n

i i i i ii i i ... (17)

Clearly, the minimum can be attained by setting ( , )i i ia (say). The numbers ai areknown as Fourier coefficients of the functions ( )x relative to the orthonormal system { }.i Inthat case, the relation (17) may be written as

2 2 21 1

|| || || || | | .n n

i i ii ia

... (18)

Since the quantity on the L.H.S. of (18) is nonnegative, we obtain

2 2

1| | || || ,

n

iia

which, for the infinite set { },i leads to the Bessel inequality

Assuming that we are given an infinite orthonormal system { ( )}i x in space and a

sequence of constants { }i , then the convergence of the series 21| |nn

is clearly a necessary

condition for the existence of an function f (x) whose Fourier coefficients with respects to thesystem i are i . It is to be noted that this condition also turns out to be a sufficiend conditionas proved in the Riesz Fischer theorem given below.7.1 (g) Riesz-Fischer Theorem. [Meerut 2004]

If { ( )}i x is a given orthonormal system of functions in space and { }i is a given

sequence of complex numbers such that the series 2

1| |ii

converges, then there exists a unique

function f (x) for which i are the Fourier coefficients with respect to the orthonormal system

{ ( )}i x and to which the Fourier series converges in the mean, that is,

1|| ( ) ( ) || 0

n

i iif x x

as .n

Proof. Consider the sequence of partial sums



1( ) ( ).

n

n i iiS x x

... (1)

and 2 2 2 21 2| ( ) ( ) | | | | | ... | | .

bn m n n n n m

aS x S x dx ... (2)

Given that the series 21| |ii

converges. Hence, corresponding to each 0, there must

exist such that

|| ( ) ( ) || , ,n m nS x S x n m (arbitrary) ... (3)

Moreover, given that i are the Fourier coefficients of the function f (x) with regard to givenorthonormal system of functions { ( )}.i x Hence we may write

2 2

1 1|| ( ) ( ) || || ( ) || | | 0

n n

i i i ii if x x f x C

as .n ... (5)

Now, by Bessel’s inequality, ( ) ( ) .b

i i iaC f x x dx ... (6)

It follows that the Fourier series 1

( )i iiC x

of the function f (x) with regard to the given

sequence of system of functions { ( )}i x is convergent in the mean to that function, that is,

1

|| ( ) ( ) || 0n

i iif x C x

as .n

7.1 (h) Some useful results.If an orthonormal system of functions i exist in space in such a manner that every other

element of this space can be represented linearly in terms of this system, then it is known as anorthonormal basis. It has been proved that the concepts of an orthonormal basis and a completesystem of orthogonal functions are equivalent. In fact, if any of following conditions are satisfied,the orthonormal set 1 2{ , , ...}n is complete.

(i) For every function in space, ( , )i i i ii ia [ ( , )i ia ]

(ii) For every function in space, 2 21

|| || | ( , ) | ,ii

which is known as Parseval’s identity.(iii) The only function in space for which the Fourier coefficients vanish is the trivial (or

zero) function.(iv) There exists no function in space such that 1 2{ , , , ..., , ...}n is an orthonormal set.

The equivalence of the above four conditions can be easily proved.7.1. (i) Fourier series of a general character.

In practice, we have to deal with Fourier series of a somewhat more general character. Letr (t) be a continuous, real and nonnegative function in the interval (a, b). We shall say that thefunctions ( )t and ( )t are orthogonal with weight r (t) if

( ) ( ) ( ) 0.b

ar t t t dt



The set of function 1 2{ ( ), ( ), ..., ( ), ...}nt t t will be said to be orthonormal with weight r (t)

if its members satisfy the relation0, if

( ) ( ) ( )1, if

bi ja

i jr t t t dt

i j

The Fourier expansions in terms of such functions are dealt with by introducing a new inner

product as follows : ( , ) ( ) ( ) ( )b

ar t t t dt

with the corresponding norm1/ 2

|| || ( ) ( ) ( )b

ra

r t t t dt

The space of functions for which || ||r is a Hilbert space and all the above results of Art.7.3 and 7.4 hold.7.1 (j) Some examples of the complete orthogonal and orthonormal systems.

(i) The system 1/ 2( ) (2 ) ,inxn x e where n takes every integer value from to , is

orthonormal in the interval ( , ). (ii) The functions 1, cos x, cos 2x, cos 3x, .... form an orthogonal system in the interval (0, ).

Again the functions sin x, sin 2x, sin 3x, ... also form an orthogonal system in (0, ).(iii) The Legendre polynomials given by

P0 (x) = 1, 21 ( 1)( ) ,

2 !

n n

n n nd xP x

n dx

n = 1, 2, 3, ...

are orthogonal in the interval (–1, 1). It can be shown that

1

1

0, if( ) ( )

2 /(2 1), ifm nm n

P x P x dxn m n

(iv) The Chebychev polynomials 1 1( ) 2 cos ( cos ),nnT x n x n = 0, 1, 2, 3, ... are orthogonal

with weight 2 1/ 2( ) 1/(1 )r x x in the interval (–1, 1). They can be normalized by multiplying

Tn (x) by the quality 2 1 1/ 2(2 / ) .n

(v) Let Jn (x) denote the Bessel function of the first kind and order n, and let ,i n denote its

positive zeros, i = 1, 2, 3, ..., Let us assume that n > –1. The system of functions ,( ),n i nJ x i = 1,2, 3, ... is orthogonal with weight function r (x) = x in the interval (0, 1). These functions satisfy the

relation1

, , 20 1 ,

0, if( ) ( )

( ) ifn i n n j nn i n

i jx J x J x dx

J i j

The systems (i) – (v) are complete. The systems mentioned above play an important rolein many practical applications.



7.1 (k) A complete two-dimensional orthonormal set over the rectangle .a x b, c t d

Theorem. Let { ( )}i x be a complete orthonormal set over ,a x b and let { ( )}i t be acomplete orthonormal set over .c t d Then, the set 1 1 1 2 2 1( ) ( ), ( ) ( ),..., ( ) ( ),...x t x t x t

is a complete two-dimensional orthonormal set over the rectangle , .a x b c t d

Proof. In order to show that the sequence of two dimensional functions { ( ) ( )}i jx t in an

orthonormal set, one has to integrate over the rectangle ,a x b .c t d The completenesscan be established by showing that every continuous function f (x, t) with finite norm || f (x, t) ||,whose Fourier coefficients with respect to the set { ( ) ( )}i jx x are all zero vanishes identicallyover the rectangle , .a x b c t b

7.2. SOME FUNDAMENTAL PROPERTIES OF EIGENVALUES AND EIGENFUNCTIONSFOR SYMMETRIC KERNELS.

Theorem I. If a kernel is symmetric then all its iterated kernels are also symmetric.Proof. Let kernel K (x, t) be symmetric. Then, by definition

K (x, t) = K (t, x). ... (1)By definition, the iterated kernels Kn (x, t) n = 1, 2, 3, ... are defined as follows :

K1 (x, t) = K (x, t), ... (2)

1( , ) ( , ) ( , ) ,b

n naK x t K x z K z t dz n = 2, 3, ... ... (3)

and 1( , ) ( , ) ( , ) ,b

n naK x t K x z K z t dz n = 2, 3, ... (3)

We shall use mathematical induction to prove the required result.

Now, 2 1( , ) ( , ) ( , ) ,b

aK x t K x z K z t dz by (3)

( , ) ( , ) ,b

aK x z K z t dz by (2) ( , ) ( , ) ,

b

aK z x K t z dz by (1)

1( , ) ( , ) ( , ) ( , ) ,b b

a aK t z K z x dz K t z K z x dz by (2)

2 ( , ),K t x by (3)

Thus, 2 2( , ) ( , ),K x t K t xshowing that K2 (x, t) is symmetric. Hence the required result is true for n = 1 and n = 2.

Let Kn (x, t) be symmetric for n = m. Then by definition, we haveKm (x, t) = mK (t, x). ... (4)

We shall now prove that Kn (x, t) is also symmetric for n = m + 1, i.e., Km + 1 (x, t) = 1mK (t, x). ... (5)

L.H.S. of (5) ( , ) ( , ) ,b

maK x z K z t dz by (3) ( , ) ( , ) ,

bma

K z x K t z dz by (1) and (4)

1( , ) ( , ) ( , ), by (3)b

m maK t z K z x dz K t x = R. H. S. of (5).



Thus iterated Kernel Kn (x, t) is symmetric for n = 1 and n = 2. Moreover, Kn (x, t) is symmetricfor n = m + 1 whenever it is symmetric for n = m. Hence, by the mathematical induction, Kn (x, t) issymmetric for n = 1, 2, 3, ...

Remark. Since ( , ) ( , ),K x x K x x it follows that the trace K(x, x) of a symmetric kernel isalways real. Likewise, we can prove that the traces of all iterated kernels are also real.

Theorem II Hilbert Theorem. Every symmetric kernel with a norm not equal to zero hasat least one eigenvalue. [Meerut 2000, 01, 03, 05, 06]

ORIf the kernel K (x, t) is symmetrical and not identically equal to zero, then it has at least one

eigen value. [Kanpur 2008]There are several proofs of this important theorem, each of them serves as the basis of a

certain method of calculating approximately the eigen value. One such method will be discussedlater on. We shall show that theorem II is a corollory of Art. 7.11 (f) on page 7.47

However, for the present, we shall present the proof the above theorem II for a special caseof real kernel. This is stated and proved as follows.

Theorem II A. If the kernel K (x, t) is real, symmetrical and not identically equal zero,then it has at least one eigenvalue. [Meerut 1998, 2001]

*Proof. From Fredholm’s first fundamental theorem if ( ) 0,D then Fredholm integral

equation, ( ) ( ) ( , ) ( )b


has a unique solution given by ( , ; )( ) ( ) ( )( )

b

a

D x ty x f x f tD

dt ... (2)

To begin with we consider the expansion of D (x, t; )/D ( ) as a power series in .Since (0) 1 0,D it follows that we can determine a constant such that if | | , then

( ) 0, | | .D ... (3)Now, since ( )D is permanently convergent, it can be expanded into a power series which

will be convergent for | | . Thus, we have

20 1 21/ ( ) ..., | | .D d d d ... (4)

Since D (x, t; ) is a permanently convergent power series in , it follows that the product{1/ ( )} ( , ; )D D x t can also be expended into a power series in .

1

1( , ; ) / ( ) ( , ) .n

nnD x t D g x t

... (5)

The series in (5) is uniformly convergent with respect to x and t in rectangle, ,a x b a t b for | | .

Substituting (5) in (2), we have

1

1( ) ( ) ( , ) ( ) , | | .

b nnna

y x f x g x t f t dt

... (6)

Since the series 1

( , ) nnn

g x t

is uniformly convergent, the order of the integration and

summation in (6) can be interchanged. We, therefore, have

*Reader is advised to consult chapter 6 in order to understand the proof that follows.



1

( ) ( ) ( , ) ( ) , | | .bn

nn ay x f x g x t f t dt

... (7)

We known that by the method of successive appronimations (refer Art. 5.7, Chapter 5) thesolution of (1) is given by

1

( ) ( ) ( , ) ( ) ,bn

nn ay x f x K x t f t dt

... (8)

where Kn (x, t) are the iterated kernels given by

1

1

( , ) ( , ),

and ( , ) ( , ) ( , ) .

b

n na

K x t K x t

K x t K x z K z t dz... (9)

Then the coefficients gn (x, t) can be obtained by comparing (7) and (8).

But the solution (8) is valid if | | 1 /{ M (b – a)}, where M is the maximum value of

| K (x, t) | on rectangle : , .R a x b a t b

Let1min ,

( )r

M b a

. ... (10)

Then, clearly, (7) and (8) will be valid simultaneously for | | .r Since (1) has uniquesolution, it follows that (7) and (8) represent the same function. It follows that the coefficients ofcorresponding powers of must be equal.

( , ) ( ) ( , ) ( ) ,b b

n na ag x t f t dt K x t f t dt (x) ... (11)

where the notation (x) is used to mean uniformly as to x.

{ ( , ) ( , )} ( ) 0,b

n na

K x t g x t f t dt (x) ... (12)

Now, Kn (x, t) and gn (x, t) are independent of f (x) and, for a given value of x,Kn (x, t) – gn (x, t) = M (t) ... (13)

is real and continuous function of t. Moreover, since (12) is true for an arbitrary continuous functionf (x), we may take f (t) = M (t). Then (12) reduces to

2{ ( )} 0.b

aM t dt ... (14)

(14) shows that M (t) 0 on [a, b]. ... (15) gn (x, t) = Kn (x, t), (x, t) ... (16)

where the notation (x, t) is used to mean uniformly as to x and t.

Using (16), (5) becomes 11

( , ; ) / ( ) ( , ) , | | .nnn

D x t D K x t r

... (17)

Since (17) is valid for t = x. We, therefore, have

11

( , ; ) / ( ) ( , ) , | |nnn

D x x D K x x r

... (18)

and 1( , ) nnK x x is uniformly convergent on [a, b].



Integrating both sides of (18) w.r.t. ‘x’ from a to b, we have

1

1

1 ( , ; ) [ ( , ) ] .( )

b b nnna a

D x x dx K x x dxD

... (19)

Interchanging the order of integration and summation in (19) on account of the uniform

convergence of –11

( , ) ,nn

nK x x

we have

1

1 ( , ; ) ( , )( )

b bnna an

D x x dx K x x dxD

...(20)

In what follows, we shall use the following symbol.

( , ) ,b

n na

K x x dx U a constant. ... (21)

We know that (refer Eq. (7) of Art 6.4 in Chapter 6) ( , ; ) ( ).b

aD x x dx D ... (22)

Using (21) and (22), (20) becomes

1

( )( )

nnn

D UD

or 1

1

( )( )

nnn

D UD

or 10

( ) ,| |( )

nnn

D U rD

... (23)

If possible, suppose that K (x, t) has no eigenvalue, that is, that ( ) 0D has no root,real or imaginary. Then the quotient ( ) / ( )D D can be directly expended into a power series

0

( ) ,( )

nnn

D CD

... (24)

which is clearly a permanently convergent series.

But (23) is valid for | | .r Hence 10 0,| | ,n n

n nn nC U r

... (25)

so that Cn = Un + 1 ... (26)

and hence 10n

nnU

is permanently convergent and so 10

| | | |nnnU

is permanently convergent

and also 110

| | | |nnnU

is permanently convergent. ... (27)

It follows that the series formed by omitting any number of terms from (27) is also

permanently convergent. Accordingly, 220

| | | | nnn

U

... (28)

is permanently convergent.In the above discussion we have proved that (28) in permanently convergent by assuming

that K (x, t) had no eigen value. It follows that if, for a given kernel K (x, t) we can show that thecorresponding series (28) is not permanently convergent, we will have proved that K (x, t) has atleast one eignvalue.

Now, we know that if K (x, t) is symmetric, then|( , ) 0nK x t on ; , .R a x b a t b ... (29)

Now, 2 2 ( , ) .b

n naU K x x dx ... (30)



Also, we know that ( , ) ( , ) ( , ) .b

p q p qaK x t K x z K z t dz ... (31)

2 ( , ) ( , ) ( , ) ( , ) ,b

n n n n naK x x K x x K x z K z x dz by (31)

( , ) ( , ) ,b

n na

K x z K x z dz [ K (x, t) is symmetric Kn (x, t) is symmetric Kn (z, x) = Kn (x, z)]

2 2[ ( , )] [ ( , )] .b b

n na aK x z dz K x t dt

(30) reduces to 22 [ ( , )] .

b bn na a

U K x t dt dx ... (32)

Again, we have

2 ( 1) ( 1) 1 1( , ) ( , ) ( , ) ( , ) ,b

n n n n na

K x x K x x K x t K t x dt by (31)

= 1 1( , ) ( , )b

n naK x t K x t dt

[ Kn + 1 (t, x) is symmetric Kn + 1 (t, x) = Kn + 1 (x, t)]

from (30), 2 1 1( , ) ( , ) .b b

n n na aU K x t K x t dt dx ... (33)

Now, By Schwarz’s inequality*, we have2

1 1( , ) ( , )b b

n na a

K x t K x t dt dx 2 2

1 1[ ( , )] [ ( , )] .b b b b

n na a a a

K x t dt dx K x t dt dx ...(34)

Using (31) and (33), (34) becomes 22 2 2 2 2 .n n nU U U ... (35)

Since 2 2 20, 0,n nU U dividing both sides of (35) by 2 2 2 ,n nU U we have

2 2 2

2 2 2.n n

n n

U UU U

... (36)

Putting successively n = 2, 3, ..., n is (36), we obtain

2 2 2 6 4

2 2 2 4 2...n n

n n

U U U UU U U U

so that2 2 4

2 2, ( )n

n

U U nU U

... (37)

where the notation (n) is used to mean uniformly as to n

Let 221 1

| | ,nn nn n

U V

... (38)

so that 22 | | n

n nV U ... (39)

*Schwarz’s inequality. If ( , )x y and ( , )x y are real and continuous on rectangle ; , ,R a x b a y b

then 2 2 2b b b b b bdx dy dx dy dx dya a a a a a



We now apply ratio test to the series (38) in order to evaluate the radius of convergence interms of . To, this end, we note that

2 21 2 2 4

2 2| | | |, ( ).,n n

n n

V U U nV U U using (37)

Hence the series (38) is divergent when

24 2

2 4| | 1 or | | .U U

U U

If follows that the series (38) is not a permanently convergent power series in . Then asexplained earlier, we have proved the theorem.

Theorem. III. The eigenvalues of a symmetric kernel are real.[Meerut, 2010, 12; Kanpur 2008]

ORIf K(x, t) is real, symmetric, continuous and identically not equal to zero, then all the

characteristic constants (eigenvalues) are real. [Kanpur 2009, 10; Meerut 2001, 07, 11]Proof. Let and ( )x be an eigenvalue and a corresponding eigenfunction of the kernel

K (x, t). Then, by definition

( ) ( , ) ( )b

ax K x t t dt ... (1)

Multiplying (1) by ( )x and integrating with respect to x from a to b.

( ) ( ) ( , ) ( ) ( )b b b

a a ax x dx K x t t dt x dx

... (2)

By definition of Fredholm operator K, we have

( , ) ( )b

aK K x t t dt ... (3)

Also, norm of 1/ 2

( ) || ( ) || ( ) ( )b

ax x x x dx

... (4)

Using (3), (4) and the definition of inner product (see Art. 7.1 (c)), (2) reduces to2|| ( ) || ( , )x K so that 2|| ( ) || /( , )x K

Since both the numerator and denominator are real, it follows that is also real and thus therequired result is proved.

Theorem IV. The eigenfunctions of a symmetric kernel, corresponding to differenteigenvalues are orthogonal [Meerut 2001, 08, 09, 10]

OR

The fundamental functions (i.e. eigenfunctions) ( )m x and ( )n x of the symmetric kernel

K (x, t) for corresponding eigenvalues m and ( )n m n are orthogonal in the domain (a, b).(Meerut 2011)

Proof. Since ( )m x and ( )n x are eigenfunctions corresponding to eigenvalues

andm n respectively, where .m n Then, by definition, we have



( ) ( , ) ( )b

m m ma

x K x t t dt ... (1)

and ( ) ( , ) ( )b

n n nax K x t t dt ... (2)

Since n is real, (2) may be re-written as ( ) ( , ), ( )b

n n na

x K x t t dt ... (3)

Since K (x, t) is symmetric, we have ( , ) ( , )K x t K t x ... (4)

Using (4), (3) may be re-written as ( ) ( , ) ( )b

n n na

x K t x t dt ... (5)

Interchanging x and t in (5), we have ( ) ( , ) ( )b

n n na

t K x t x dx ... (6)

Multiplying both sides of (1) by ( )n x and then integrating the both sides w.r.t. ‘x’ from a tob, we have

( ) ( ) ( , ) ( ) ( )b b b

m n m m na a a

x x dx K x t t dt x dx

( , ) ( ) ( )b b

m n ma a

K x t x dx t dt

[on changing the order of integration]

( / ) ( ) ( ) ,b

m n m nat t dt by (6)

( ) ( ) ( ) ( )b b

n m n m m na a

x x dx x x dx or ( ) ( ) ( ) 0,

bn m m na

x x dx i.e., ( ) ( , ) 0n m m n ... (7)

Since , ( ) 0n m n m and so (7) reduces to ( , ) 0,m n

showing that the eigenfunctions andm n are orthogonal.Theorem V. The multiplicity of any nonzero eigenvalue is finite for every symmetric kernel

for which 2| ( , ) |b b

a aK x t dx dt is finite.

Proof. Let the functions 1 2( ), ( ), ..., ( )...nx x x be the linearly independenteigenfunctions which correspond to a non-zero eigenvalue . Using the Gram-Schmidt procedure,we can find linear combinations of these functions which form an orthonormal system { ( )}.ku x

Then, the corresponding complex conjugate system { ( )}ku x also forms an orthonormal system.Let

( , ) ( ),i iiK x t a u t where 1( , ) ( ) ( ),

bi i i

aa K x t u t dt u x

... (1)

be the series associated with kernel K (x, t) for a fixed x. Applying Bessel’s inequality to this

series, we get 2 22

1| ( , ) | | ( ) | ,b

iiaK x t dt u x

...(2)



Integrating both sides of (2) with respect to x, we have

22

1| ( , ) |b b

ia aK x t dxdt

or 22| ( , ) | ,

b b

a a

mK x t dx dt ... (3)

where m is the multiplicity of . Since the L.H.S. of (3) is finite, it follows that m is finite.Theorem VI. The sequence of eigenfunctions of a symmetric kernel can be made orthonormal.Proof. Corresponding to a certain eigenvalue, let there be m lineraly independent

eigenfunctions. Using the linearity property of the integral operator, every linear combination ofthese functions is also an eigenfunction. Hence, on applying the well known Gram-Schmidt procedure,we can get equally numerous eigenfunctions which are orthonormal. Again, for different eigenvalues,the corresponding eigenfunctions are orthogonal and can be easily normalized. Combining thesetwo facts, the complete proof of the Theorem VI follows.

Theorem VII. The eigenvalues of a symmetric kernel form a finite or an infinite sequence{ }n with no finite limit point.

Proof. On including each eigenvalue in the sequence a number of times equal to its multiplicity,

we have 221

1 | ( , ) | .b b

n a anK x t dx dt

... (1)

Let {uk (x)} be the orthonormal eigenfunctions corresponding to different (nonzero) eigenvalues.i Then proceeding as in theorem V and using the Bessel inequality, we obtain

22

1 | ( , ) | .b b

i a aiK x t dx dt

...(2)

Hence, if there exists an enumerable infinity of ,i then we must have

21

1 ,n n

...(3)

Therefore lim (1/ ) 0i and is the only limit point of the eigenvalues.Theorem VIII The set of eigenvalues of the second iterated kernel coincide with the set of

squares of the eigenvalues of the given kernel.

Proof. Let be an eigenvalue of kernel K (x, t) corresponding to the eigenfunction ( ).x Then,by definition, we have

or ( ) 0,K I K ... (1)where I is the identity operator.

Operating both sides of (1) with the operator ( ),I K we obtain

2 2( – ) 0I K or 22( ) ( , ) ( ) ,

b

ax K x t t dt ... (2)

showing that 2 is an eigenvalue of the Kernel K2 (x, t).

Conversely, let 2µ be an eigenvalue of the kernel K2 (x, t) with ( )x as the correspondingeigenfunction. Then, we have

2 2( ) 0I K or ( ) ( ) 0.I K I K ... (3)



If is an eigenvalue of kernel K, then the theorem VIII is proved. If not, let us suppose

that ( ) ( )I K x ... (4)

Using (4), (3) reduces to ( ) ( ) 0.I K x ... (5)

Since is not an eigenvalue of K by our assumption, (5) shows that ( ) 0,x or

equivalently ( ) 0.I K Thus, is an eigenvalue of the kernel K and thus the theorem isproved.

Remark I. The result can be extended to the n th iterate. The set of eigenvalues of thekernel Kn (x, t) coincide with the set of n th powers of the eigenvalues of the kernel K (x, t).

Remark 2. While proving above theorem VIII, the symmetry of the kernel K (x, t) has notbeen assumed.

Theorem IX. If 1 is the smallest eigenvalue of the kernel K, then

1

| ( , ) ( ) ( ) |1 ( , )max max| | || || || ||

b b

a aK x t t x dt dxK

... (17)

or, equivalently, 11/ || || max ( , ), || || 1.K ... (18)

Proof. The maximum value is attained when ( )x is an eigenfunction of the symmetric kernel corresponding to the smallest eigenvalue of the kernel K as shown in Art. 7.11 (f) on page7.47.7.3. EXPANSION IN EIGENFUNCTIONS AND BILINEAR FORM

In this article we propose to deal with results concerning the expansion of a symmetric kerneland of functions represented in some sense by the kernel, in terms of its eigenfunctions and theeigenvalues.

Let K (x, t) be a nonnull, symmetric kernel which has a finite or an infinite number ofeigenvalues (always real and nonzero). We order them in the sequence

1 2, , ..., n ...(1)in such a manner that each eigenvalue is repeated as many times as its multiplicity. Also, wedenumerate these eigenvalues in the order that corresponds to their absolute values, that is,

1 2 10 | | | | ...... | | | | ...n n

Let 1 2( ), ( ), ..., ( ), ...nx x x ... (2)be the sequence of eigenfunctions corresponding to the eigenvalues gives by the sequence (1). Wefurther agree to arrange these eigenfunctions in such as manner that they are not repeated and arelinearly independent in each group corresponding to the same eigenvalue. It follows that to eacheigenvalues k in (1) there corresponds just one eigenfunction ( )x in (2). Suppose that theseeigenfunctions have been orthonormalized (refer theorem VI of Art. 7.2).

Suppose that a symmetric kernel has at least one eigenvalue, say 1. Then 1( )x isthe corresponding eigenfunction. It follows that the second ‘‘truncated’’ symmetric kernel

(2) 1 1

1

( ) ( )( , ) ( , ) x tK x t K x t

... (3)

is nonnull and it will also possess at least one eigenvalue 2 (we select the smallest in case thereare more than one eigenvalues) with corresponding normalized eigenfunction 2 ( ).x The function

1 2( ) ( )x x even if 1 2 , since



(2)1( , ) ( )

b

aK x t t dt 1

1 1 11

( )( , ) ( ) ( ) ( )b b

a a

xK x t t dt t t dt

1 1

1 1

( ) ( ) 0.x x

Similarly, the third ‘‘trucated’’ symmetric kernel

2

(3) (2) 2 212

( ) ( ) ( ) ( )( , ) ( , ) ( , )k k

x t x tK x t K x t K x t

... (4)

yields the third eigenvalue 3 and the corresponding normalized eigenfunction 3 ( ).x

Proceeding likewise we finally arrive at the two following possibilities :(i) the above process terminates after n steps, that is, ( 1) ( , ) 0,nK x t and the kernel K (x, t) is

a degenerate kernel, given by1

( ) ( )( , ) ,n

k k

x tK x t

... (5)

(ii) the above process can be continued indefinitely and there are infinite number of eigenvaluesand eigenfunctions.

Remark 1. We have denoted the least eigenvalue and the corrresponding eigenfunction ofK(n) (x, t) as n and ,n which are the n th eigenvalue and the n th eigenfunction in the sequences(1) and (2). This fact can be justified with help of theorem II which is given below.

Remark 2. We shall investigate in the next article whether the bilinear form (5) is valid forthe case when the kernel K (x, t) has infinite eigenvalues and eigenfunctions.

Theorem I. Let the sequence { ( )}k x be all the eigenfunctions of a symmetric kernel,

with { }k as the sequence of corresponding eigenvalues. Then, the series 2

21

| ( ) |nn n

x

converges

and its sum is bounded by 21 ,C which is an uppear bound of the integral 2| ( , ) | .

b

aK x t dt

Proof. The Fourier coefficients an of the function K (x, t) with fixed x, with respect to theorthonormal system { ( )}n x are given by

( )( , ) ( ) .

bn

n na n

xa K x t t dt

... (6)

Using Bessel’s inequatlity, we now obtain2

2 2121

| ( ) | | ( , ) | .bn

n an

x K x t dt C

... (7)

Theorem II. Let the sequence ( )n x be all the eigenfunctions of a symmetric kernel K (x, t),

with { }n as the corresponding eigenvalues. Then, the truncated kernel

( 1)1

( ) ( )( , ) ( , )n

n m mm m

x tK x t K x t

has the eigenvalues 1 2, ( ), ...,n n x to which correspond the eigenfunctions 1 2( ), ( ), ...n nx x The kernel K (n + 1) (x, t) has no other eigenvalues or eigenfunctions.



Proof. (i) We begin with fact that the integral equation

( 1)( ) ( , ) ( ) 0b na

x K x t t dt ... (8)

is equivalent to1

( )( ) ( , ) ( ) ( , ) 0

nbm

mma m

xx K x t t dt dt

... (9)

Writing j and ( ) ( ), 1jx x j n on L.H.S. of (9) and using the orthogonalityconditon, we obtain

( ) ( , ) ( ) 0,b

j j ja

x K x t t dt ... (10)

Hence ( )j x and j for 1j n are eigenfunctions and eigenvalues of the kernel K (n + 1) (x, t).

(ii) Let and ( )x be an eigenvalue and eigenfunction of the kernel K (n + 1) (x, t) so that

1

( )( ) ( ) ( , ) 0.

nm

mm m

xx K x

... (11)

Taking the scalar product of (11) with ( ),j x j n and using the orthonormality of the

( ),j x we have ( , ) ( , ) ( / ) ( , ) 0.j j j jK ... (12)

Now, ( , ) ( , ) (1/ ) ( , ).j j j jK K ... (13)

Using (13), (12) reduces to ( , ) ( / ) {( , ) ( , )} ( , ) 0.j j j j j ... (14)

In view of (14), we find that the last term in the L.H.S. of (11) vanishes and hence (11)

reduces to ( ) ( , ) ( ) 0,b

ax K x t t dt ... (15)

showing that and (x) are eigenvalue and eigenfunction of the kernel K (x, t) and that , .j j n

In fact, we see that is orthogonal to all , ,j j n and ( )x and are always contained in thesequences { ( )}x and { }, 1,k k n respectively..

Remark. Combining the results of the above two theorems I and II, we easily find that, if thesymmetric kernel K has only a finite number of eigenvalues, then it must be saparable. The prooffollows by noting that K (n + 1) (x, t) then has no eigenvalues and therefore it must be null. Hence,

we must have 1

( ) ( )( , )

nm m

m m

x tK x t

In chapter 4, we have already proved that every separable kernel has only a finite number ofeigenvalues. Combining these two results, we have the following theorem.

Theorem III. A necessary and sufficient condition for a symmetric kernel to be separableis that it must possess a finite number of eigenvalues.7.4. HILBERT-SCHMIDT THEOREM. [Kanpur 2008, 09, 10; Meerut 2000, 01, 02, 07]

If f (x) can be written in the form

( ) ( , ) ( )b

af x K x t h t dt ... (1)



where K (x, t) is a symmetric kernel and h (t) is an function, then f (x) can be expandedin an absolutely and uniformly convergent Fourier series with respect to the orthonormal systemof eigenfunctions 1 2( ), ( ), ..., ( ), ...nx x x of the kernel K (x, t) :

1

( ) ( ),n nnf x f x

... (2)

where ( , ).n nf f ... (3)The Fourier coefficients of the function f (x) are related to the Fourier coefficients hn of the

function h (x) by the relations/n n nf h ... (4)

and ( , ),n nh h ... (5)where n are the eigenvalues of the kernel K (x, t).

Proof. Let K (x, t) be a nonnull, symmetric kernel which has a finite or an infinite number ofeigenvalues (always real and non-zero). We order them is the sequence

1 2, , ...., ,...n ...(6)in such a way that each eigenvalue is repeated as many times as its mulltiplicity. We further agreeto denumerate these eigenvalues in the order that corresponds to their absolute values, i.e.,

1 2 10 | | | | ... | | | | ...n n Let 1 2( ), ( ), ..., ( ), ...nx x x ... (7)

be the sequence of eigenfunctions corresponding to the eigenvalues given by the sequence (6) andarranged in such a way that they are no longer repeated and are linearly independent in each groupcorresponding to the same eigenvalue. Thus, to each eigenvalue k in (6) there corresponds just

one eigenfunction ( )k x in (2). Further, suppose that eigenfunctions ( )k x in (2) have beenorthonormalilzed. (refer theorem VI of Art 7.2)

Now, the Fourier coefficients of the function f (x) with respect to the orthonormal system{ ( )}k x are given by

n( , ) ( , ) ( , ) (1/ ) ( , ) /n n n n n n nf f Kh h K h ,by using the self-adjoint property of the operator and the relation .n n nK Hence, the Fourierseries for f (x) is given by

1 1( ) ( ) ( ).n

n n nn n n

hf x f x x

... (8)

We now estimate the remainder term of the series (8) as shown below.2 2

221 1 1

( ) | ( ) |n p n p n pk k

k kk n k n k nk k

x xh h

22

21 1

( ) .n p

kkk n k k

xh

... (9)

Using the relation (7) of Art. 7.3, we see that the above series is bounded. Moreover, since

h (x) is an function, it follows that the series 21 kk

h

is convergent and the partial sum 2

1

n p

kk nh

can be made arbitrarily small. Hence, the series (8) converges absolutely and uniformly.We now proceed to show that the series (8) converges to f (x) in the mean. For this purpose,

we denote its partial sum as 1

( ) ( )n

mn mm m

hx x

... (10)

and estimate the value of || ( ) ( ) || .nf x x



Now, 1

( ) ( ) ( )n

mn mm m

hf x x Kh x

( 1)

1

( , )( ) ,

nmm

mm m

hKh x K h

... (11)

where K (n + 1) is the truncated kernel. From (11), we have2 ( 1) 2 ( 1) ( 1)|| ( ) ( ) || || || ( , )n n n

nf x x K h K h K h

( 1)( 1) ( 1)2( , ) ( , ),nn nh K K h h K h ... (12)

where we have used the self-adjoint property of the kernel K (n + 1) and also the relation( 1)( 1) ( 1)2 .nn nK K K

We know that the set of eigenvalues of the second iterated kernel coincide with the set ofsquares of the eigenvalues of the given kernel (refer theorem VIII of Art. 7.2). Using this property

and theorem II of Art. 7.3., we see that the least eigenvalue of the kernel ( 1)2nK is equal to 2

1.nAgain, using the theroem IX of Art 7.2., we obtain

( 1)2

21

( , )1 max ,( , )

n

n

h K hh h

... (13)

where we have omitted the modulus sign from the scalar product ( 1)2( , ),nh K h because it is a positive

quantity.Combining (12) and (13), we obtain

( 1)2 212|| ( ) ( ) || ( , ) ( , ) / .n

n nf x x h K h h h ... (14)

Since 1 ,n (14) || ( ) ( ) || 0 .nf x x as n ... (15)Now, we have the relation

|| ( ) ( ) || || ( ) ( ) || || ( ) ( ) ||,n nf x x f x x x x ... (16)

where ( )x is the limit of the series with partial sum .n

As shown above, the first terms on the R.H.S. of (16) tends to zero. To show that the secondterm of R.H.S. of (16) also tends to zero, we proceed as follows :

Since the series (8) converges uniformly, we have, for an arbitrarily small and positive ,| ( ) ( ) | ,n x x when n is sufficiently large.

Therefore, 1/ 2|| ( ) ( ) || ( )n x x b a || ( ) ( ) || 0n x x as .n

Thus, we see that both the terms on R.H.S. of (16) tend to zero as .n Therefore, from(16), we have ( ) ( )f x x as required.

Remark. It is to be noted that we assumed neither the convergence of the Fourier series h (x)nor the completeness of the orthonormal system while proving Hilbert-Schmidt theorem.

We now show that the Hilbert-Schmidt theorem easily leads to the bilinear form of the type(5) given in Art. 7.3. By definition, we have

1( , ) ( , ) ( , ) ,b

m maK x t K x z K z t dz m = 2, 3, ..., ... (17)

which is of the form (1) with h (x) = Km (x, t); t fixed. The Fourier coefficient ak (t) of Km (x, t) withrespect to the system of eigenfunctions { ( )}k x of K (x, t) is given by



–( ) ( , ) ( ) ( )b m

k m k k kaa t K x t x dx t

Hence, with help of Hilbert-Schmidt theorem, we find that all the iterated kernels Km (x, t);2,m of a symmetric kernel K (x, t) can be represented by the absolutely and uniformly convergent

series

–

1( , ) ( ) ( )m

m k k kkK x t x t

... (18)

Replacing t by x in (18) and then integrating with respect to x from a to b, we have

–1

( , ) ( ) ( )b bm

m k k kka aK x x dx x x dx

... (19)

But 1/ 2

( ) ( ) || ( ) || 1,b

k k xa

x x dx x ... (20)

because ( )k x is a normalized functionUsing (20), (19) reduces to

1( , ) ,

bmk m mk a

K x x dx A

... (21)

where Am is the trace of the iterated kernel Km.Using the Riesz-Fischer theorem and relation (21) with m = 2, it follows that the series

1

( ) ( )k kk k

x t

... (22)

converges in the mean to a symmetric kernel K (x, t) which, treated as a Fredholm kernel,possesses exactly the sequence of numbers { }k as eigenvalues.

7.5. DEFINITE KERNELS AND MERCER’S THEOREMNonnegative-definite kernel. Definition. A symmetric kernel K (x, t) is said to be

nonnegative-definite if ( , ) 0K for every function .

Positive-definite kernel. Definition. A symmetric kernel K (x, t) is said to be positivedefinite if ( , ) 0K and ( , ) 0K is null.

Nonpositive-definite kernel. Definition. A symmetric kernel K(x, t) is said to benonpositive-definite if ( , ) 0K for every function .

Negative-definite kernel. Defintion. A symmetric kernel K(x, t) is said to be negative-definite if ( , ) 0K and ( , ) 0K is null.

Indefinite kernel. A symmetric kernel that does not fall into any of the above mentionedfour types of kernels, is known as indefinte kernel.

The following theorem is an immediate consequence of the Hilbert-Schmidt theorem.Theorem (i) A nonnull, symmetric kernel K (x, t) is nonnegative if and only if all its

eigenvalues are positive(ii) It is positive definite if and only if the above condition is satisfied and, in addition,

some (and therefore every) full orthonormal system of eigen functions of K (x, t) is complete.



Proof (i) Using the Hilbert-Schmidt theorem, we have

1

( , )( ) ( )n

nn nf x K x

... (1)

Taking the inner product of both sides of (1) with , we have

2

1

| , |( , ) n

n nK

... (2)

If 0n for each n, then (2) ( , ) 0K for all .

In addition, if any 0,n then –1( , ) 0.n n nK

Thus, the part (i) of the theorem is proved.(ii) Let K (x, t) be nonnegative-definite. From (2), we have

( , ) 0 ( , ) 0 nK for all n,

showing that the kernel K (x, t) will be positive-definite if and only if

( , ) 0 for all 0.n

Now using the condition (refer condition (iv), Art. 7.1 (h)) for the completeness of anorthonormal system, part (ii) of the theorem easily follows.

We now state without proof, the following theorem.Mercer’s theorem. If a nonnull, symmetric kernel is quasi-definite (that is, when all but

a finite number of eigenvalues are of one sign) and continuous, then the series 1

1 nn

is convergent

and1

( ) ( )( , ) ,n nn n

x tK x t

the series being uniformly and absolutely convergent.Remark. The result of the above theorem gives the exact conditions for the bilinear form (5)

given in Art. 7.3 to be extended to an infinite series.We have to note that the continuity of the kernel is an important condition for the theorem to

be true.7.6. SCHMIDT’S SOLUTION OF NON-HOMOGENEOUS FREDHOLM INTEGRAL

EQUATION OF THE SECOND KIND [Meerut 2001, 02, 04, 06]

( ) ( ) ( , ) ( ) ,b


where K (x, t) is continuous, real and symmetric and is not an eigenvalue.Statement of Hilbert-Schmidt theorem :Let F (x) be generated from a continuous function y (x) by the operator

( , ) ( ) ,b

aK x t y t dt

where K (x, t) is continuous, real and symmetric, so that

( ) ( , ) ( ) .b

aF x K x t y t dt ... (2)



Then F (x) can be represented over interval (a, b) by a linear combination of the normalizedeigenfunctions of homogeneous integral equation

( ) ( , ) ( ) ,b


having K (x, t) as its kernel.Procedure of solution of (1). Re-writing (1), we have

( ) ( ) ( , ) ( ) .b


Since (4) is of the form (2), it follows by Hilbert-Schmidt theorem

1( ) ( ) ( ), .m mm

y x f x a x a x b

...(4)

where ( )m x (m = 1, 2, 3, ...) are the normalized eigenfunctions of homogeneous integral equation

( ) ( , ) ( )b


Let m (m = 1, 2, 3, ...) be the corresponding eigenvalues of (5)

Let , 1, 2, 3, ...m m ... (6)

Since ( )m x (m = 1, 2, ...) are normalized, we have

0, ,( ) ( )

1, .

bm na

m nx x dx

m n

... (7)

Multiplying both sides of (4) by ( )m x and then integrating w.r.t. ‘x’ from a to b, we get

( ) ( ) ( ) ( )b b

m ma a

y x x dx f x x dx 2

1 1 2( ) ( ) ( ) ( ) ...... ( ) ( ) ...b b b

m m m m ma a aa x x dx a x x dx a x x dx ... (8)

Let ( ) ( )b

m ma

C y x x dx ... (9)

and ( ) ( ) .b

m maf f x x dx ... (10)

Making use of (7), (9) and (10) in (8), we obtain Cm – fm = am ... (11)Now, multiplying both sides of (1) by ( )m x and then integrating w.r.t. ‘x’ from a to b, we have

( ) ( ) ( ) ( ) ( , ) ( ) ( )b b b b

m m ma a a a

y x x dx f x x dx K x t y t dt x dx

or ( ) ( , ) ( )b b

m m ma a

C f y t K x t x dx dt [using (9) and (10) and also

changing the order of integration in double integral on R.H.S]

or ( ) ( , ) ( )b b

m m ma a

C f y t K t x x dx dt ... (12)

[since K (x, t) is symmetric, so K (x, t) = K (t, x)] Since ( )m x is eigenfunction corresponding to the eigenvalue m of (5), by definition, we get



( ) ( , ) ( )b

m m max K x t t dt or ( ) ( , ) ( ) .

bm m ma

x K x z z dz

or ( ) ( , ) ( )b

m m ma

t K t z z dz or ( ) ( , ) ( )b

m m ma

t K t x x dx

or( )

( , ) ( ) .b

mm

a m

tK t x x dx

... (13)


( ) ( )bm

m ma m

y t tC f dt

or ( ) ( )b

m m mam

C f y x x dx

or ( / ) .m m m mC f C using (9) ... (14)

From (11), Cm = am + fm. ... (15)Eliminating Cm from (14) and (15), we get

( )m m m m mm

a f f a f

or 1m mm m

a f

,m mm

a f

...(16)

where m and so am is well-defined.

Substituting the above value of am in (4) , the required solution of (1) is given by

1( ) ( ) ( )m

mm m

fy x f x x

or ( ) ( ) ( ).mmm m

fy x f x x

... (17)

From (10), ( ) ( ) .b

m ma

f f t t dt ... (18)

Using (18), (17) may be re-written as( )

( ) ( ) ( ) ( )b

mmm am

xy x f x f t t dt

or( ) ( )( ) ( ) ( )

b m mma m

x ty x f x f t dt

or ( ) ( ) ( , ; ) ( )b


where the resolvent kernel ( , ; )R x t is given by

( ) ( )( , ; ) m m

m m

x tR x t

... (20)

Three important cases arise :Case I. Unique Solutin. If condition (6) is satisfied, (16) gives well defined value of am for

substituting in (4) . Thus, solution (17) exists uniquely if and only if does not take on aneigenvalue.



Case II. No Solution. Let ,k where k is the k th eigenvalue and also let

0kf ,that is, ( ) ( ) 0,b

ka

f x x dx i.e., eigenfunction ( )k x is not orthogonal to f (x). Then, because of the presence of the term

( )k k

k

f x ... (21)

in (17), we find that no solution exists since the term (21) is undefined.Case III. Infinitely many solutions exist. Let ,k where k is the k th eigenvalue and

also let fk = 0 ,that is, ( ) ( ) 0,b

kaf x x dx

i.e., eigenfunction ( )k x is orthogonal to f (x). Then, (14) reduces to (for m = k)

0 ( / )k kC C or Ck = Ck,which is a trivial identity and hence imposes no restriction on Ck. From (16) it then follows that thecoefficient ak of ( )k x in (17), which formally assumes the form 0/0, is truely arbitrary. Hence, were-write solution (17) as follows :

( ) ( ) ( ) ( ),mk mm m

fy x f x A x x ... (22)

where dash implies that we should neglect m = k in the summation and A is an arbitrary constant.(22) shows that we arrive at infinetely many solutions of (1).7.7. SOLVED EXAMPLES BASED ON ART. 7.6.

Ex. 1. Solve the symmetric integral equation12 2 2

1( ) ( 1) ( ) ( ) ,y x x xt x t y t dt

by using Hilbert-Schmidt theorem. [Kanpur 2010; Meerut 2009, 2012]

Sol. Given12 2 2

1( ) ( 1) ( ) ( ) ,y x x xt x t y t dt

... (1)

Comparing (1) with1 2 2

1( ) ( ) ( ) ( ) ,y x f x xt x t y t dt

... (2)

here f (x) = (x + 1)2 and 1. ... (3)We begin with determining eigenvalues and the corresponding normalized eigenfunctions of

1 2 2

1( ) ( ) ( )y x xt x t y t dt

... (4)

Re-writing (4),1 12 2

1 1( ) ( ) ( ) .y x x t y t dt x t y t dt

...(5)

Let1

11

( )C t y t dt

... (6)

and1 2

2 1( ) .C t y t dt

... (7)



Then (5) reduces to 21 2( ) .y x C x C x ... (8)

From (8), 21 2( ) .y t C t C t ... (9)


1 21 1 2

1( )C t C t C t dt

or

1 13 4

1 1 21 1

3 4t tC C C

or 11

2 03

CC or 1 2

21 0. 0.3

C C

... (10)

Again, using (9), (7) becomes

1 2 22 1 21

( )C t C t C t dt

or

1 14 5

2 1 21 1

4 5t tC C C

or 22

205

CC or 1 2

20. 1 0.5

C C

... (11)

Equations (10) and (11) have a nontrivial solution only if

1 (2 / 3) 0( ) 0

0 1 (2 / 5)D

or {1 (2 / 3)}{1 (2 / 5)} 0 giving 3/ 2 or 5 / 2.

Hence the required eigenvalues are 1 23/ 2 and 5 / 2 ... (12)

Determination of eigenfunction corresponding to ! " #$%&

Putting 1 3/ 2 in (10) and (11), we obtain

C1.0 + 0.C2 = 0 and 1 22 30. 1 0,5 2

C C

Hence C2 = 0 and C1 is arbitrary. Putting these values in (8) and noting that 3/ 2, wehave the required eigenfunction y1 (x) given by

y1 (x) = (3/2) × C1 x.Setting (3/2) × C1 = 1, we may take y1 (x) = x.Now, the corresponding nonmalized eigenfunction 1( )x is given by

1

1 1/ 2 1/ 2 1/ 21 1 132 21 11 1

( )( )/ 3{ ( )}

y x x xxxy x dx x dx

Thus,1/ 2

13 6( ) .2 2(2 / 3)

x xx x

... (13)

Determination of eigenfunction corresponding to % " '$%&

Putting 2 5 / 2 in (10) and (11), we obtain

1 22 51 0. 03 2

C C and 0.C1 + 0.C2 = 0,



Hence C1 = 0 and C2 is arbitrary. Putting these values in (8) and noting that 5 / 2, wehave the required eigenfunction y2 (x) given by

y2 (x) = (5/2) × C2 x2.

Setting (5/2) × C2 = 1, we may take y2 (x) = x2.Now, the corresponding normalized eigenfunction 2 ( )x is given by

222

2 1/ 2 11 4 1/ 222 11

( ) 10( ) .2[ ]{ ( )}

y x xx xx dxy x dx

... (14)

Also, 1 1 2

1 11 1

6( ) ( ) ( 1) ,2

f f x x dx x x dx

by (3) and (13)

14 3 21 2

11

6 6 2 2 6( 2 1) .2 2 4 3 2 3

x x xx x x dx

... (15)

and 1 1 2 2

2 21 1

10( ) ( ) ( 1) ,2

f f x x dx x x dx

by (3) and (14)

15 4 3

1

10 2 8 10.2 5 4 3 15

x x x

... (16)

From (3), 1. Also 1 23/ 2 and 5 / 2. Hence 1 and 2 . Therefore, (I) will possess a unique solution given by (refer case I, Art. 7.6)

2

1( ) ( ) ( )m

mm m

fy x f x x

or

22

1( ) ( 1) ( ),

1m

mm m

fy x x x

using (3)

or2 1 1 2 2

1 2

( ) ( )( ) ( 1)1 1

f x f xy x x

or2

2 (2 6 / 3) ( 6 / 2) (8 10 /15) ( 10 / 2)( ) ( 1)(3/ 2) 1 (5 / 2) 1

x xy x x

or y (x) = (x + 1)2 + 4x + (16/9) × x2 = x2 + 2x + 1 + 4x + (16/9) × x2

or y (x) = (25/9) × x2 + 6x + 1.Ex. 2. Using Hilbert-Schmidt theorem, find the solution of the symmetric integral equation

12 2 2

1

3( ) 1 ( ) ( ) .2

y x x xt x t y t dt

[Kanpur 2009,11; Meerut 2000]

Sol. Given12 2 2

1

3( ) 1 ( ) ( ) .2

y x x xt x t y t dt

... (1)

Comparing (1) with1 2 2

1( ) ( ) ( ) ( ) ,y x f x xt x t y t dt

... (2)

here f (x) = x2 + 1 and 3/ 2. ... (3)



We begin with determining eigenvalues and the corresponding normalized eigenfunctions of

1 2 2

1( ) ( ) ( ) .y x xt x t y t dt

... (4)

Equation (4) is same as equation (4) of solved Ex. 1. So proceed as before and obtain

1 23/ 2 5 / 2, 21 2( ) ( 6) / 2 and ( ) ( 10) / 2.x x x x ... (6)

Also,1

1 11

( ) ( )f f x x dx

14 21 2

11

6 6( 1). 0.2 2 4 2

x x xx dx

... (7)

and 1

2 21( ) ( )f f x x dx

12 5 31 2

11

10 10 8 10( 1) .2 2 5 3 15

x x xx dx

... (8)

Here 1 23/ 2 and .

Since 1 and f1 = 0, hence infinitely many solutions of (1) exist and are given by (refercase III, Art. 7.6)

2

1 1( ) ( ) ( ) ( ),m

mm m

fy x f x A x x

... (9)

where dash in the above sum means that the term with m = 1 must be neglected.

(9) takes the form 21 2

2( ) ( ) ( ) ( )fy x f x A x x

or2

2 6 3 (8 10 /15) 10( ) 12 2 (5 / 2) (3/ 2) 2

x xy x x A

using (3), (5), (6) and (8)

or y (x) = x2 + 1 + Cx + 4x2 or y (x) = 5x2 + Cx + 1,where ( 6) / 2)C A is an arbitrary constant.

Ex. 3. Solve the following symmetric integral equation with the help of Hilbert-Schmidt

theorem : 0

( ) 1 cos ( ) ( ) .y x x t y t dt

[Meerut, 2010, 11]

Sol. Given0

( ) 1 cos ( ) ( ) .y x x t y t dt

... (1)

Comparing (1) with1

0( ) ( ) cos ( ) ( ) ,y x f x x t y t dt ... (2)

here ( ) 1, and .f x ... (3)We begin with determining eigenvalues and the corresponding normalized eigenfunctions of

0( ) cos ( ) ( )y x x t y t dt

... (4)

Re-writing (4),0

( ) (cos cos – sin sin )y x x t x t dt

or

0 0( ) cos cos ( ) sin sin ( ) .y x x t y t dt x t y t dt

... (5)

Let 1 0cos ( )C t y t dt

... (6)

and 2 0sin ( ) .C t y t dt

... (7)



Then (5) reduces to 1 2( ) cos siny x C x C x ... (8)

From (8), 1 2( ) cos sin .y t C t C t ... (9)

Using (9), (6) becomes 1 1 20cos ( cos sin )C t C t C t dt

or 1 21 0 0

(1 cos 2 ) sin 22 2C CC t dt t dt

1 2

00

sin 2 cos 22 2 4C Ctt t

or 1 1( ) / 2C C or 1 2(2 ) 0. 0.C C ... (10)

Next, using (9), (7) becomes 2 1 20

sin ( cos sin )C t C t C t dt

or 1 22 0 0

sin 2 (1 cos 2 )2 2C CC t dt t dt

1 2

00

sin 2cos 22 2 2C C tt t

or 2 2( ) / 2C C or 1 20. (2 ) 0.C C ... (11)Equations (10) and (11) have a nontrivial solution only if

2 0( ) 0

0 2D

or (2 ) (2 ) 0, so that 2 / or 2 / .

Hence the required eigenvalues are 1 22 / and 2 / . ... (12)

Determination of eigenfunction corresponding to ! " %$ &

Putting 1 2 / in (10) and (11), we obtain0.C1 + 0.C2 = 0 and 0.C1 + 4C2 = 0,

Hence C2 = 0 and C1 is arbitrary. Putting these values in (8) and noting that 2 / , we

have the eigenfunction y1 (x) given by 1 1( ) (2 / ) cos .y x C x

Setting 1(2 / ) 1,C we may take y1(x) = cos x.

The corresponding normalized eigenfunction 1( )x is given by

11 1/ 2 1/ 2 1/ 2

2 21

0 0 0

( ) cos cos( )1 cos 2{ ( )} cos

2

y x x xxxy x dx x dx dx

1/ 2

1/ 2

0

cos cos 2 cos ./ 21 sin 2

2 2

x x xxx

... (13)

Determination of eigenfunction corresponding to . 2 = –2/

Putting 2 2 / in (10) and (11), we get 4C1 + 0.C1 = 0 and 0.C1 + 0.C2 = 0,



Hence C1 = 0 and C2 is arbitrary. Putting these values in (8) and noting that 2/ , we

have the eigenfunction y2 (x) given by 2 2( ) ( 2 / ) sin .y x C x

Setting 2(2 / ) 1,C we can take y2 (x) = sin x.

The corresponding normalized eigenfunction 2 ( )x is given by

22 1/ 2 1/ 2 1/ 2

2 220 0 0

( ) sin sin( )1 cos 2{ ( )} sin

2

y x x xxxy x x dx

1/ 2

1/ 2

0

sin sin 2 sin ./ 21 sin 2

2 2

x x xxx

... (14)

Also,1/ 2

1 10 0

2( ) ( ) cos ,f f x x dx x dx by (3) and (13)

1/ 202 / sin 0x ... (15)

and1/ 2

2 20 0

2( ) ( ) sin ,f f x x dx x dx by (3) and (14)

1/ 2 1/ 202 / cos 2 2 / .x ... (16)

Three cases arise :Case I : Let 1 and 2 . Then (1) will possess unique solution given by

2

1( ) ( ) ( ),m

mm m

fy x f x x

by case I, Art. 7.6

or 1 1 2 21 2

( ) 1 ( ) ( ),y x f x f x

by (3)

or 1/ 2 1/ 2

1( ) 2 2( ) 1 0 2 sin(2 / ) (2 / )

xy x x

or 4 sin( ) 1 .2

xy x

... (17)

Case II. Let 2 2 / . Since 2 0,f so (1) possesses no solution. (refer case II, Art. 7.6)

Case III. Let 1 2 / . Since f1 = 0, there exist infinitely many solutions given by (refercase III, Art. 7.6)

2

1 1( ) ( ) ( ) ( ),m

mm m

fy x f x A x x

... (18)

where dash in employed to indicate that the term corresponding to m = 1 in the sum is to beomitted. Accordingly, (18) reduces to

1 2 22

( ) ( ) ( ) ( )y x f x A x f x



or1/ 2 1/ 2 1/ 22 (2 / ) 2 2( ) 1 cos 2 sin

(2 / ) (2 / )y x A x x

or ( ) 1 cos (2 / ) sin ,y x C x x when ( 4 2 / )C is an arbitrary constant.

Ex. 4. Solve the symmetric integral equation ( ) ( ) ( ) ( ) ( ) .b

ay x f x k x k t y t dt

Sol. Given ( ) ( ) ( ) ( ) ( ) .b

ay x f x k x k t y t dt ... (1)

We begin with determining eigenvalues and the corresponding normalized eigenfunctions of

( ) ( ) ( ) ( ) .b

ay x k x k t y t dt ... (2)

Re-writing (2), ( ) ( ) ( ) ( ) .b

ay x k x k t y t dt ... (3)

Let ( ) ( ) .b

aC k t y t dt ... (4)

Then (3) reduces to ( ) ( ).y x C k x ... (5)

From (5), ( ) ( ).y t C k t ... (6)Using (6), (4) becomes

[ ( ). ( )b

aC k t C k t dt or

21– { ( )} 0b

aC k t dt ... (7)

For eigenfunction of (2) , clearly 0.C Hence (7) gives

21 { ( )} 0b

ak t dt or 2

1 1/[ { ( )} ],b

ak x dx ... (8)

which is the only eigenvalue of (2). Putting this value of in (5), the corresponding eigen functiony1 (x) is given by

21( ) ( ) /[ { ( )} ].

b

ay x C k x k x dx

Setting 2/[ { ( )} ] 1,b

aC k x dx we can take y1 (x) = k (x). ... (9)

Hence the corresponding normalized eigenfunction 1( )x is given by

11 1/ 2 1/ 2

2 21

( ) ( )( ) .{ ( )} { ( )}

b b

a a

y x k xxy x dx k x dx

... (10)

Also, 1 1( ) ( )b

af f x x dx

1/ 22( ) ( ) /[ { ( )} ] ,b b

a af x k x k x dx dx

by (10)

Thus, 2 1/ 21 [ ( ) ( ) ] /[ { ( )} ] .

b b

a af f x k x dx k x dx ... (11)



Three cases arise :Case I. Let 1. Then (1) will possess unique solution given by

1

1( ) ( ) ( )m

mm m

fy x f x x

or 1 11

( ) ( ) ( )y x f x f x

or2 1 2 1/ 2 2 1/ 2

0

( ) ( ) ( )( ) ( )[ { ( )} ] [ { ( )} ] [ { ( )} ]

b

ab b b

a a

f x k x dx k xy x f xk x dx k x dx k x dx

[using (8), (10) and (11)]

or 2

( ) ( ) ( )( ) ( ) .

1 { ( )}

b

ab

a

k x f x k x dxy x f x

k x dx

... (12)

Case II. Let 1. Suppose that f (x) be not orthogonal to 1( )x , this is,

1 1( ) ( ) 0.b

af f x x dx

Then (1) possesses no solution.Case III. Let 1. Suppose that f (x) orthogonal to 1( )x , that is,

1 1( ) ( ) 0b

af f x x dx

Then (1) possesses infinitely many solutions given by

1( ) ( ) ( ),y x f x A x A being an arbitrary constant.

or 2 –1/ 2( ) ( ) . ( ).[ { ( )} ] ,b

ay x f x A k x k x dx by (10)

or ( ) ( ) ( ),y x f x Ck x where 2 1/ 2[ { ( )} ]b

aC A k x dx = new arbitrary constant.

Ex. 5. Determine the eigenvalues and the corresponding eigenfunctions of the equation2

0( ) ( ) sin ( ) ( ) ,y x f x x t y t dt

where f (x) = x. Obtain the solution of this equation when is not an eigenvalue.

Sol. Given2

0( ) sin ( ) ( ) .y x x x t y t dt

... (1)

Comparing (1) with2

0( ) ( ) sin ( ) ( ) ,y x f x x t y t dt

... (2)

here f (x) = x and ... (3)We begin with determining eigenvalues and the corresponding normalized eigenfunctions of

2

0( ) sin ( ) ( ) .y x x t y t dt

... (4)



Re-writing (4), 2

0( ) (sin cos cos sin ) ( ) .y x x t x t y t dt

or 2 2

0 0( ) sin cos ( ) cos sin ( )y x x t y t dt x t y t dt

... (5)

Let2


... (6)

and2

20

sin ( ) .C t y t dt

... (7)

Then (5) reduces to 1 2( ) sin cos .y x C x C x ... (8)

From (8), 1 2( ) sin cos .y t C t C t ... (9)

Using (9), (6) becomes 21 1 2

0cos ( sin cos )C t C t C t dt

or2 2

1 21

0 0sin 2 (1 cos 2 )

2 2C CC t dt t dt

221 20

0

sin 2cos 22 2 2C C tt t

or 1 2C C or 1 2 0.C C .. (10)


2 1 20

sin ( sin cos )C t C t C t dt

or2 2

1 22

0 0(1 cos 2 ) sin 2

2 2C CC t dt t dt

or 2 1C C or 1 2 0.C C ...(11)Equations (10) and (11) have a non-trivial solution only if

1( ) 0

1D

or 2 21 0 so that 1/ or 1/ .

Hence the required eigenvalue are 1 1/ and 2 1/ .

Determination of eigenfunction corresponding to . 1 = 1/

Putting 1 1/ in (10) and (11), we getC1 – C2 = 0 and C1 – C2 = 0,

giving C2 = C1. Putting C2 = C1 in (8), the required eigenfunction y1 (x) is given by

1 1( ) ( / ) (sin cos ).y x C x x

Setting 1( / ) 1,C we may take y1 (x) = sin x + cos x.

Hence the corresponding normalized eigenfunction 1( )x is given by

11 1/ 2 1/ 22 22 2

10 0

( ) sin cos( ){ ( )} (sin cos )

y x x xxy x dx x x dx



1/ 2 1/ 22 2

00

sin cos sin cos sin cos2cos 2(1 sin 2 )

2

x x x x x x

xx dx x

...(13)

Determination of eigenfunction corresponding to . 2 = –1/

Putting 2 1/ in (10) and (11), we getC1 + C2 = 0 and – C1 – C2 = 0,

giving C2 = – C1. Putting C2 = – C1 in (8), the required eigenfunction y2 (x) is given by

2 1( ) ( / ) (sin cos ).y x C x x

Setting 1( / ) 1,C we may take y2 (x) = sin x – cos x.

Hence the corresponding normalized eigenfunction 2 ( )x is given by

2

2 2 22 1/ 2 2 1/ 22

0 0

( ) sin cos( )[ { ( )} ] [ (sin cos ) ]

y x x xxy x dx x x dx

1/ 2 1/ 22 2

00

sin cos sin cos sin – cos2cos 2(1 sin 2 )

2

x x x x x x

xx dx x

... (14)

Also, 2

1 10

( ) ( )f f x x dx

2

0

(sin cos ) ,2

x x x dx

by (3) and (13)

22

0 0

1 (sin cos ) (sin cos )2

x x x x x dx

20

1 2 cos sin 2 ,2

x x

and 2

2 20( ) ( )f f x x dx

2

0

(sin cos ) ,2

x x x dx

by (3) and (14)

22

0 0

1 ( cos sin ) ( cos sin )2

x x x x x dx

20

1 2 sin cos 2 .2

x x

Given that 1 and 2 . Hence (1) will possess unique solution given by2

1( ) ( ) ( )m

mm m

fy x f x x

or 2

1( ) ( ),m

mm m

fy x x x

using (3)

or1 1 2 2

1 2

( ) ( )( ) f x f xy x x



or ( 2 ) (sin cos ) ( 2 ) (sin cos )( )

{(1/ ) } 2 {( 1/ ) } 2x x x xy x x

or (sin cos ) (sin cos )( ) –

(1 ) (1 )x x x xy x x

or 1 1 1 1( ) sin cos

1 1 1 1y x x x x

or2 2

2 2 2 22 sin 2 cos( )

1 1x xy x x

Ex. 6. Using Hilbert-Schmidt method, solve 1

0( ) ( , ) ( ) ,y x x K x t y t dt where

( 1), 0 ,( , )

( 1), 1.x t x t

K x tt x t x

[Kanpur 2005; Meerut 2001, 02, 05]

Sol. Given 1

0( ) ( , ) ( ) ,y x x K x t y t dt ... (1)

where ( 1), 0 ,

( , )( 1), 1.

x t x tK x t

t x t x

... (2)


0( ) ( ) ( , ) ( ) .y x f x K x t y t dt

here, f (x) = x. and . ... (3)We begin with determining eigenvalues and the corresponding normalized eigenfunctions of

1

0( ) ( , ) ( )y x K x t y t dt ... (4)


0( ) [ ( , ) ( ) ( , ) ( ) ]

x


or1

0( ) ( 1) ( ) ( 1) ( )

x

xy x t x y t dt x t y t dt , using (2) ... (5)

Differentiating both sides of (5) w.r.t. ‘x’ and using Leibnitz’s rule (see Art. 1.13), we get1

0( ) ( 1) ( ) ( 1) ( )

x

x


or0

( ) { ( 1) ( )} ( 1) ( )x dxy x t x y t dt x x y x

x dx

0.0.( 1) (0) dx ydx

1 1{ ( 1) ( )} (1 1) (1)x

dx t y t dt x yx dx

( 1) ( ) dxx x y x

dx

or1

0( ) ( ) ( 1) ( ) ( 1) ( ) ( 1) ( )

x


or1

0( ) ( ) ( 1) ( ) .

x




Differentiating both sides of (6) w.r.t. ‘x’, we get1

0( ) ( ) ( 1) ( )

x

x

d dy x t y t dt t y t dtdx dx

or0

0( ) { ( )} ( ) .0. (0)x dx dy x t y t dt x y x y

x dx dx

1{ ( 1) ( )}

xt y t dt

x

1(1 1) (1) ( 1) ( ) ,d dxy x y xdx dx

using Leibnitz-rule again

or ( ) ( ) ( 1) ( ).y x x y x x y x

or ( ) ( )y x y x or ( ) ( ) 0.y x y x ... (7)Putting x = 0 in (5), we get y (0) = 0. ... (8)Putting x = 1 in (5) we get y (1) = 0. ... (9)Now, we shall solve Strum-Liouville problem given by differentail equation (7) together with

boundary conditions (8) and (9) by the usual procedure to get eigenvalues and the correspondingeigenfunctions.

Three cases arise :

Case I. Let 0. Then (7) reduces to ( ) 0y x whose general solution is y (x) = Ax + B. ... (10)

Putting x = 0 in (10) and using (8), we get B = 0. ... (11)Putting x = 1 in (10) and using (9), we get 0 = A + B. ... (12)Solving (11) and (12), A = B = 0. Hence (10) gives y (x) = 0, which is not an eigenfunction

and so 0 is not an eigenvalue.

Case II. Let 2 , where 0. Then (7) reduces to 2( ) – ( ) 0,y x y x whose

general solution is –( )x xy x Ae Be

. ...(13)

Putting x = 0 in (13) and using (8), we get 0 = A + B. ...(14)

Putting x = 1 in (13) and using (9), we get –0 Ae Be . ...(15)

Solving (14) and (15), A = B = 0. Hence (13) gives ( ) 0,y x which is not an eigenfunction.

Case III. Let 2µ , where µ 0. Then (7) reduces to 2( ) µ ( ) 0y x y x whose generalsolution is y (x) = A cos µx + B sin µx. ... (16) Putting x = 0 in (16) and using (8), we get 0 = A. ... (17)

Putting x = 1 in (16) and using (9), we get 0 = A cos µ + B sin µ. ... (18)Using (17), (18) gives B sin µ = 0 ... (19)Now, we must take 0,B otherwise A = 0 and B = 0 will give y (x) = 0 as before and hence

we shall not get eigenfunction. Since 0,B (19) reduces to sin µ = 0.

µ ,n when n is any integer. Hence 2 2 2µ n

Hence the required eigenvalue n are given 2 2 , 1, 2, 3, ...n n n ... (21)

Putting A = 0 and µ n in (16), we get ( ) sin .y x B n x



Setting B = 1, required eigenfunctions yn (x) are given by

( ) sin , 1, 2, 3, ...ny x n x n ... (22)

The normalized eigenfunctions ( )n x are given by

1/ 2 1/ 21 12 2

0 0

( ) sin( )

{ ( )} sin

nn

n

y x n xxy x n x dx

1/ 21

0

sin

1 cos 22

n x

n x dx

1/ 21

0

sin sin 2 sin .1/ 21 sin 2

2 2

n x n x n xn xx

n

... (23)

Now, 1

0( ) ( )n nf f x x dx

1

0( ) ( 2 sin ) ,x n x dx by (3) and (23)

1 1

00

cos cos2 n x n xx dxn n

1

0

cos 12 cosn n x dxn n

1

102 2

( 1) 1 ( 1) 22 sin .n n

n xn nn

... (24)

Now, two cases arise.Case (i). Let be not an eigenvalue, that is, . for n 1, 2, 3, ...n Then (1) will possess unique solution given by

1( ) ( ) ( )n

nn n

fy x f x x

or

1

2 21

( 1) 2 2 sin( )n

n

n xy x xn n

or 2 21

2 ( 1) sin( ) .( )

n

n

n xy x xn n

Case (ii) Let 2 2 ,n n n = 1, 2, 3, ... Then since from (24), 0nf for n = 1, 2, 3, ...Hence (1) will possess no solution.

Ex. 7. Solve the symmetric integral equation 1

0( ) ( , ) ( ) ,xy x e K x t y t dt where

sinh sinh ( 1) , 0 ,sinh 1

( , )sinh sinh ( 1) , 1.

sinh 1

x t x tK x t

t x t x

[Kanpur 2005]

Sol. Given1

0( ) ( , ) ( ) ,xy x e K x t y t dt ... (1)



where


( , )sinh sinh ( 1) , 1.

sinh 1

x t x tK x t

t x t x

... (2)

Comparing (1) with1

0( ) ( ) ( , ) ( ) ,y x f x K x t y t dt ... (3)

here f (x) = ex and ... (4)We begin with determination of eigenvalues and the corresponding normalized eigenfunctions

of the homogeneous integral equation

1

0( ) ( , ) ( ) .y x K x t y t dt ...(5)


0( ) ( , ) ( ) ( , ) ( )

x


or1

0

sinh sinh ( 1) ( ) sinh sinh ( 1) ( )( ) , by (2)sinh 1 sinh 1

x

x

t x y t x t y ty x dt dt ... (6)

Differentiating both sides of (6) w.r.t. ‘x’ and using Leibnitz’s-rule (see Art. 1.13), we have

0

sinh cosh ( 1) ( ) sinh sinh ( 1) ( )( )sinh 1 sinh 1

x t x y t x x y xy x dt 1 cosh sinh ( 1) ( ) sinh sinh ( 1) ( )

sinh 1 sinh 1x

x t y t x x y xdt

or1

0

sinh cosh ( 1) ( ) cosh sinh ( 1) ( )( )sinh 1 sinh 1

x

x

t x y t x t y ty x dt dt ...(7)

Differentiating both sides of (7) w.r.t. ‘x’ and using Leibnitz-rule (Art. 1.13), we have

0

sinh sinh ( 1) ( ) sinh cosh ( 1) ( )( )sinh 1 sinh 1

x t x y t x x y xy x dt 1 sinh sinh ( 1) ( ) cosh sinh ( 1) ( )

sinh 1 sinh 1x

x t y t x x y xdt

or ( )( ) ( ) [sinh cosh ( 1) cosh cosh ( 1)]sinh 1y xy x y x x x x x , using (6)

or ( )( ) ( ) sinh{ ( 1)}

sinh 1y xy x y x x x

or ( ) ( )y y x y x or ( ) (1 ) ( ) 0y x y x ... (8)Putting x = 0 in (6), we get y (0) = 0. ... (9)Putting x = 1 in (6), we get y (1) = 0. ... (10)Three cases arise :

Case I. Let 1 0, that is, 1. Then (8) reduces to ( ) 0y x whose general solution isy (x) = Ax + B. ... (11)



Putting x = 0 in (11) and using (9), we get 0 = B. ... (12)Putting x = 1 in (11) and using (10), we get 0 = A + B. ... (13)Solving (12) and (13), A = B = 0. Hence (11) gives y (x) = 0, which is not an eigenfunction and

so 1 is not an eigenvalue.

Case II. Let 21 µ , where µ 0. Then (8) reduces to 2µ ( ) 0y y x whose general

solution is µ µ( ) x xy x A e B e ... (14)

Putting x = 0 in (14) and using (9), we get 0 = A + B ... (15)Putting x = 1 in (14) and using (10), we get 0 = Aeµ + B e–µ. ... (16)Solving (15) and (16), A = B = O. Hence (14) gives y (x) = 0, which is not an eigenfunction.

Case III. Let 21 µ , where µ 0. Then (8) reduces to 2( ) µ ( ) 0y x y x whosegeneral solution is y (x) = A cos µx + B sin µx. ... (17)

Putting x = 0 in (17) and using (19), we get 0 = A. ... (18)Putting x = 1 in (17) and using (10), we get 0 = A cos µ + B sin µ. ... (19)Using (18), (19) gives B sin µ = 0. ... (20)Now, we must take 0,B otherwise A = 0 and B = 0 will give y (x) = 0 as before and hence

we shall not get eigenfunction. Since 0,B (20) reduces to sin µ = 0.

µ ,n where n in any integer.. ... (21)

2 2 21 µ n or 2 2(1 ).n

Hence the required eigenvalues n are given by

2 2(1 ), 1, 2, 3, ...n n n ... (22)

Putting A = 0 and µ n in (17), ( ) sin .y x B n x

Setting B = 1, the required eigenfunctions yn (x) are given by

( ) sin , 1, 2, 3, ...ny x n x n ... (23)


1/ 2 1/ 21 12 2

0 0

( ) sin( ){ ( )} sin

nn

n

y x n xxy x dx n x dx

= 1/ 21

0

sin

1 cos 22

n x

n x dx

1/ 21

0

sin sin 2 sin .(1/ 2)1 sin 2

2 2

n x n x nxn xx

n

... (24)

Now, 1 1

0 0( ) ( ) .( 2 sin ) ,x

n nf f x x dx e n x dx by (4) and (24)

1

2 20

2 (sin cos1

xe n x n n xn

, as 2 2sin ( sin cos )

axax ee bx dx a bx b bx

a b



2 22 [ cos ( )]

1e n n n

n

2 22{1 ( 1) }, 1, 2, 3, ...1

nn e nn

... (25)

Now, two cases arise

Case (i). Let be not an eigenvalue, that is, ,n for n = 1, 2, 3, ...Then (1) will possess unique solution given by

1( ) ( ) ( )n

nn n

fy x f x x

2 2 2 21

2 {1 ( 1) } 2 sin1 1

nx

n

n e n xen n

or 2 2 2 21

{1 ( 1) }sin( ) 2(1 ) (1 )

nx

n

n e n xy x en n

Case (ii). Let 2 21 , 1, 2, 3, ...n n n Then since from (25), 0nf for n = 1, 2, 3, ...,hence (1) will possess no solution.

Ex. 8. Given that the eigenvalues of the integral equation 2

0( ) cos( ) ( )y x x t y t dt

are

1/ and –1/ with respective eigenfunctions cos x and sin .x Then the integral equation2

0( ) sin cos cos( ) ( )y x x x x t y t dt

has

(a) unique solution for 1/ (b) unique solution for –1/

(c) unique solution for (d) no solution for – . [GATE 2004]

Sol. Ans (c). Refer cases I, II and III of Art. 7.6 on pages 7.23 and 7.24. If 1/ or –1/which are eigenvalues, then by cases II and III either there is no solution or there exist infinitelymany solutions. Thus, results (a) and (b) are wrong. If – , which is not an eigenvalue, then bycase I there exists unique solution. So result (d) is wrong. But if , which is not an eigenvalue,then by case I, the there exist a unique solution. Thus result (c) is correct

Ex. 9. Let , 0( , ) .

0, Otherwisex t t x

K x t

Then the integral equation 1

0( ) 1 ( , ) ( ) hasy x K x t y t dt

(a) unique solution for every value of (b) no solution for any value of

(c) a unique solution for finitely many values of only(d) infinitely many solutions for finitely many values of [GATE 2003]Sol. Ans (d). Refer cases I, II and III of Art. 7.6 on pages 7.23 and 7.24 carefully. It follows

that results (a), (b) and (c) are incorrect while result (d) is correct.

EXERCISE1. State and prove Hilbert-Schmidt theorem. [Merrut 2000, 01, 02, 07]2. State and prove Riesz-Fischer theorem. [Merrut 2004]3. State Hilbert-Schmidt theorem. Derive Schmidt’s solution of the integral equation

( ) ( ) ( , ) ( ) ,b

ay x f x K x t y t dt where K (x, t) is symmetric and is not an eigenvalue.



4. Using Hilbert-Schmidt theorem, solve the following symmetric integral equations :

(i)1

1 20

( ) ( 1) ( ) , , .y x x x y t dt 2(6 12) 4( ) .

12 12xy x

Ans.

(ii)1

0( ) (1 3) ( 6 4 3 ( ) ( ) .y x x x t y t dt

Ans. ( ) (1 3) (1 3) 1 3 / 2 .y x x C x where C is an arbitrary constant.

(iii)1

0( ) (1 3) ( 6 4 2) ( ) ( ) .y x x x t y t dt Ans. ( ) (1 3) (1 3) 1 3 / 2 .y x x C x x where C is an arbitrary constant.

(iv)1

0( ) ( ) , ( 1).y x x y t dt Ans. ( ) /{2 (1 )}y x x

(v)1

0

1( ) ( ) .2

y x x y t dt Ans.1( )2

y x x C

5. Determine the eigenvalues and the corresponding eigenfunctions of the equation2

0( ) ( ) cos( ) ( ) ,y x F x x t y t dt

where is not an eigenvalue.

Ans. Eigenvalues are 1 21/ , 1/ and corresponding eigenfunctions are

1 2( ) (cos ) / , ( ) (sin ) / .x x x x x cos cos sin sin( , ; )1 1

x t x tR x t

6. State and prove Hilber-Schmidt theorem and use it for finding the solution of the symmetric

integral equation 1

0( ) cos ( , ) ( ) ,x x K x t t dt where ( 1) , 0

( , )( 1) , 1.x t x t

K x tt x t x

7. Show that solution of the integral equation 1

0( ) ( , ) ( ) ,xy x e K x t y t dt where

(1 ) , 0( , )

(1 ) , 1t x t x

K x tx t x t

is 2 2 2 21

{1 ( 1) } sin( ) 2(1 ) ( )

nx

n

n e n xy x en n

8. Show that the solution of the integral equation

12 2

0( ) sin ( , ) ( ) ,y x m x n K x t y t dt m n where

(1 ) , 0( , )

(1 ) , 1t x t x

K x tx t x t

is 2 2 2( ) sin sin ( sin ) /( ),y x m x C n t n m x n m where C is an arbitrary constant.

7.8. SOLUTION OF THE FREDHOLM INTEGRAL EQUATION OF THE FIRST KINDWITH SYMMETRIC KERNEL. [Meerut 2005]Consider the Fredholm integral equation of the first kind

( ) ( , ) ( ) ,b




where the kernel K (x, t) is a known symmetric kernel and y (t) is unknown function while f (x) is aknown function. Suppose that the sequence of eigenvalues { }n of its kernel K (x, t) and corresponding

eigenfunctions { ( )}n x are known and arranged as in relations (6) and (7) of Art. 7.4.From the relation (4) of Art. 7.4, we have

/n n nf y so that n n ny f ... (2)In view of the Reisz-Fischer theorem (refer Art. 7.1 (g), there exist the following two situations :

(i) If the infinite series 2 21 n nn

f

...(3)

diverges, then (1) has no solution(ii) If the infinite series (3) converges, then there exists a unique function y (x) which is

the solutin of (1). Then the solution of (1) can be determined by taking the limit in mean

1( ) lim ( )

m

n n nm ny x f x

... (4)

7.9. SOLVED EXAMPLES BASED ON ART. 7.8Ex. 1. Solve the symmetric Fredholm integral equation of the first kind

1

0( , ) ( ) ( ),K x t y t f x where

,

(1 ),( , )

(1 ) x t

x t x tK x t

t x

Sol. Given1

0( , ) ( ) ( ),K x t y t dt f x ... (1)

where,

(1 ),( , )

(1 ) x t

x t x tK x t

t x

... (2)

We shall first find the eigenvalues and the corresponding eigenfunction of the kernel K (x, t).To this end we solve the associated homogeneous Fredholm integral equation

1

0( ) ( , ) ( )y x K x t y t dt ... (3)

From (3),1

0( ) ( , ) ( ) ( , ) ( )

x


or 1

0( ) (1 ) ( ) (1 ) ( ) ,

x

xy x t x y t dt x t y t dt by (2) ... (4)

Differentiating both sides of (4) w.r.t. ‘x’ and using Leibnitz’s rule on the R.H.S., we have

1

0( ) ( ) (1 ) ( ) (1 ) ( ) (1 ) ( )

x


or1

0( ) ( ) (1 ) ( ) .

x


Differentiating both sides of (5) w.r.t. ‘x’ and using Leibnitz’s rule on the R.H.S., we have( ) ( ) (1 ) ( )y x x y x x y x or ( ) ( ) 0y x y x

or 2( ) 0,D y where /D d dx ... (6)Putting x = 0 in (4), we get y (0) = 0 ... (7)Putting x = 1 in (4), we get y (1) = 0 ... (8)



Now, we shall solve Strum-Liouville problem given by (6) together with boundary conditions(7) and (8) by the usual procedure to find the required eigenvalues and the corresponding normalizedeigenfunctions.

Three cases arise :

Case I. Let 0. Then (6) reduces to 0y whose solution is y (x) = Ax + B ... (3)

Putting x = 0 and x = 1 by turn in (9) and using (7) and (8), we get B = 0 and A + B = 0 so that A = B = 0.

Hence (9) gives y (x) = 0 which is not an eigenfunction.

Case II. Let 2 , 0. Then (6) reduces to (D2 – µ2) y = 0 whose solution isy (x) = Aeµx + Be–µx ... (10)

Putting x = 0 and x = 1 by turn in (10) and using (7) and (8), we haveA + B = 0 and Aeµ + Be–µ = 0 so that A = B = 0.Hence (10) given y (x) = 0 which is not an eigenfunction.

Case III. Let 2 , 0. Then (6) reduces to (D2 + µ2) y = 0 whose solution is

y (x) = A cos µx + B sin µx. ... (11)Putting x = 0 in (11) and using (7), we get A = 0 ... (12)Putting x = 1 in (11) and using (8), we get B sin µ = 0 ... (13)Now, we must take 0,B otherwise A = B = 0 will give y (x) = 0 and so eigenfunction will

not exist. So for the existence of an eigenfunction, we take 0.B Then (13) reduces to sin µ = 0 so that , 1, 2, 3, ...n n

Hence the required eigenvalues n are given by2 2 , 1, 2, 3, ...n n n ... (14)

Then from (11), setting B = 1, the corresponding eigenfunctions yn (x) are given by( ) sin , 1, 2, 3, ...ny x n x n


1 2 1/ 2

0

( ) sin( ) 2 sin ,|| ( ) || [ sin ]

nn

n

y x n xx n xy x x x dx

on simplifications ... (5)

From result (3) of Art. 7.4, we have1 1

0 0( , ) ( ) ( ) 2 ( ) sinn n nf f f x x dx f x n x dx ... (16)

Then, the given integral equation (1) has a solution of class if and only if the infinite

series 2 2

1 n nnf

i.e., 4 4 2

1 nnn f

converges, where fn is given by (16).

Ex. 2. Solve the Poisson’s integral equation2 2

20

1 ( )( ) , 0 2 , 0 12 1 2 cos ( )

y df

Solution. Comparing the given integral with2

0( ) ( , ) ( ) ,f K y d



we get 2 2 –1( , {(1 ) / 2 } {1 2 cos ( ) } ,K which is clearly a symmetric kernel. Expanding it, we get

1

1 1( , cos { ( )}2

nn

K n

... (1)

We can easily show that (1) leads to2

0( , ) 1K d

so that

2 1/ 2 1/ 2

0(2 ) ( , ) (2 ) .K d

... (2)

0 1 and 1/ 20 ( ) (2 ) ,x

where 0 is an eigenvalue of ( , )K and 0 ( )x is the corresponding normalized eigenfunction.Similarly for n = 1, 2, 3, ..., using the well known formulas

2

0( , ) cos cosnK n d n

and

2

0( , ) sin sin ,nK n d n

we can easily show that1/ 2 1/ 2

2 1 2 2 1 2; ( ) cos ; ( ) sin , 1, 2, 3, ...nn n n nx nx x nx n ... (3)

where n is nth eigenvalue and ( )n x is the corresponding normalized eigenfunction of ( , ).K

We know that the given Poisson’s integral equation has unique solution provided the

series 2 21 n nn

f

converges, where fn can be evaluated with help of result (3) of Art. 7.4, namely

( , ).n nf f

Then we can show that that the given integral equation has an solution if and only if the

infinite series 2 2 2

1( ) / ,n

n nna b

where

2

0

1 ( ) cosna f n d

and

2

0

1 ( ) sinnb f n d

, converges.

Ex. 3. Find a Fourier series solution for the integral equation2

21 1( ) ( ) , 0 1,

2 1 2 cos ( )f x y t dt x

x t

... (1)

Sol. We can show that

2

2 1

1 1 1 12 21 2 cos ( )

nnx t

(cos nx cos nt + sin nx sin n t) ... (2)

and the series is absolutely convergent.We know that the Fourier series for f (x) is given by

01

( ) ( cos sin )2 n nn

af x a nx b nx

... (3)

where 1 1( ) cos , 0 and ( ) sin 0n na f x nx dx n b f x n x dx n

... (4)

Let 01

( ) ( cos sin )2 n nn

cy t c nt d nt

... (5)



Then, we have

01 1

1 , (cos cos sin sin ) ( cos sin2 2

nn nn n

cnx nt nx nt c nt d nt dt

01

( cos sin )2

nn nn

cc nx d nx

an = cn n and bn = dn n

Hence the solution of the given integral equation (1) is given by the series

01

( cos sin )2

nn nn

a a nt b nt

... (6)

provided (6) converges.

7.10.APPROXIMATION OF A GENERAL KERNEL (NOT NECESSARILYSYMMETRIC) BY A SEPARABLE KERNEL.In Art. 4.5, we approximated an analytic kernel x (ext – 1) by a separable kernel. In the

present article, we propose to prove that it is possible to approximate every kernel in the meanby a separable kernel.

In what follows we shall use the result of Art. 7.1 (k), namely, the availability of a two-dimensional complete othonormal set.

Let K (x, t) be an kernel and let { ( )}i x be an arbitrary, complete, orthonormal set over

.a x b Then, the set { ( ) ( )}i jx t is a complete orthonormal set over the square, .a x b a t b Here the bar denotes the complex conjugate. The Fourier expansion of the

kernel in this set is given by

, 1( , ) ( ) ( ),

n

ij i ji jK x t K x t

... (1)

where the Ki j are the Fourier coefficients given by

( , ) ( ) ( )b b

i j i ja a

K K x t x t dx dt ... (2)

Parseval’s identity gives 2 2

, 1| ( , ) | | |

b bi ji ja a

K x t dx dt K

... (3)

Define a separable kernel k (x, t) as follows :

, 1( , ) ( ) ( )

n

i j i ji jk x t K x t

... (4)

Then, we can easily show that

2 2, 1

| ( , ) ( , ) | | |b b

i ji j na aK x t k x t dx dt K

... (5)

Since the series (3) is convergent by hypothesis, the sum in (5) can be made as small as werequire by choosing a sufficiently large n.

This prove what we wished to prove.7.11.THE OPERATOR METHOD IN THE THEORY OF INTEGRAL EQUATIONS.

In the present article we propose to deal with a Fredholm integral equation by using theconcepts of functional analysis. We have already dealt with the properties of a function space inArt. 7.1. (e). Recall the transformation or Fredholm operator K given by



( , ) ( )b

aK K x t t dt ... (1)

Let be a constant. Then, we can show that 1 2 1 2( ) and ( ) ( ),K K K K K

showing that K is a linear operator.The operator K is said to be bounded if || || || ||K M for an kernel K (x, t), an

function and a constant M.

The norm || ||K of K is defined as

|| || . . . (|| || / || ||) or || || . . . || ||,|| || 1,K l u b Ky y K l u b Ky y ... (2)where l.u.b. stands for least upper bound. The two criterions given in (2) are equivalent.

A transformation K is said to be continuous in an space if, whenever { }n is a sequence

in the domain of K with limit , then .nK K A transformation is said to be continuous in theentire domain of K if it is continuous at every point therein. It is to be noted that a lineartransformation is continuous if it is bounded.7.11 (a)To show that the operator K given by (1) is bounded

Proof. We begin with the relation

( ) ( , ) ( )b

ax K K x t t dt

Then, 2 2 2 2| ( ) | | ( , ) ( ) | [ | ( , ) | ] [ | ( ) | ]b b b

a a ax K x t t dt K x t dt t dt

[using Schwarz inequality]

or 2 2 2| ( ) | || || | ( , ) |b

ax K x t dt

2 2 2| ( ) | || || | ( , ) |

b b b

a a ax dx K x t dx dt

2 2 2|| || || || | ( , ) |

b b

a aK x t dx dt

2 1/ 2|| || || || || || { | ( , ) | }

b b

a aK K x t dx dt

2 1/ 2|| || { | ( , ) | } ,b b

a aK K x t dx dt as || || || || ||K K ... (3)

showing that the operator K is bounded.7.11 (b) The concept of complete continuity.

Recall that a set S of elements is said to be compact if a subsequence having a limit can beextracted from any sequence of elements of S. Now, an operator is said to be a completely continuousif it transforms a bounded set into a compact set. Clearly, a completely continuous operator iscontinuous (and therefore bounded), but the converse is not true. Again any bounded operator Kwhose range is finite dimensional is completely continuous because it transforms a bounded set in

[a, b) into a bounded finite-dimensional set which is always compact. Fortunately, number ofthe integral operators, that arise in applications are completely continuous. We shall now establilshthis fact.



7.11 (c) To show that a seperable kernel K (x, t) given by

1( , ) ( ) ( ),

n

i iiK x t f x g t

... (4)

where fi (x) and gi (t) are functions, is completely continuous.Proof. For each function y (x)

1 1( ) ( ) ( ) ( ),

n nbi i i ii ia

K y f x g t y t dt C f x

... (5)

where ( ) ( )b

i iaC g t y t dt ... (6)

(5) shows that the range of K is a finite dimensional subspace of (a, b).

Again1 1

|| || || ( ) || | | || ||n n

i i i ii iKy C f x C f

or1

|| || || || || | ( ) | | ( ) | ,n b

i ii aKy f g t y t dt

by (6) ... (7)

Using the Schwarz inequality (see Art. 7.1 (d)), (7) gives

|| || || ||,Ky M y ... (8)

where1|| || || ||

n

i iiM f g

It follows that K is a bounded operator with finite-dimensional range and hence is completelycontinuous.

7.11. (d). To show that an kernel K (x, t) is completely continuousProof. To this end we shall use the result of an important theorem, namely, if K can be

approximated in norm by a completely continuous operator, then K is compeltely continuous. Byusing this theorem the required result follows because an kernel can always be approximatedby a separable kernel (see Art. 7.1 (a)) and separable kernel is completely continuous as just provedin 7.11 (c).

7.11 (e). To show that the norms of K and of its adjoint K are equalProof. We know that (refer result (13) in Art. 7.1 (e))

( , ) ( , ),K K ... (9)

which is valid for each pair of functions , .

Replacing by K in (9) and then using the Schwerz inequality, we get

( , ) ( , ) || || || ||,K K K K K K

remembering that ( , )K K is a nonnegative real number..

2|| || || || || || || || or || || || || || ||K K K K K

|| || || || || || ||K K || || || ||K K ... (10)

Next replacing by K in (9) and proceeding as before, we obtain

|| || || ||K K ... (11)



Then, (10) and (11) || || || ||,K K as required ... (12)7.11. (f). Theorem. The reciprocal of the modulus of the eigenvalue with the smallest modulus

for a symmetric kernel K is equal to the maximum value of | ( , ) |K with || || 1.

[Refer theorem IX of Art. 7.2.. The result of this theorem will now be proved in this Art.7.11 (f)]

Proof. We can easily show that an upper bound for the reciprocal of the eigenvalue can beeasily obtained because, for the eigenvalue problem ,K we have

2( , ) (1/ ) ( , ) (1/ ) || ||K

2 2(1/ ) || || ( , ) || || || || || || || ||K K K ... (13)

From (3) and (13), 1/ 2

2| 1/ | | ( , ) | ,b b

a aK x t dx dt ... (14)

from which an upper bound of the reciprocal of the modulus of the eigenvalue can be obtained.When the kernel is also symmetric, the following result from the theory of operators can

be used. ‘‘If K is a symmetric and completely continuous operator, at least one of the numbers|| K || or – || K || is the reciprocal of an eigenvalue of K and no other eigenvalue of K has smallerabsolute value’’

Using the definition of || K || and the fact that a symmetric kernel generates completelycontinuous operator, we have proved the required result stated in theorem 7.11 (f).

Corollary. Every symmetric kernel with a norm not equal to zero has at least one eigenvalue.Proof. The result follows from theorem of Art. 7.11 (f) [Merrut 2000, 01, 02, 06]Remark. We have already proved a special case of the above result for real symmetric kernel

is theorem II of Art. 7.2.7.11 (g) Procedure for getting the eigenvalues and eigenfunctions arranged in thesequences (1) and (2) as given in Art. 7.3.

Suppose the first eigenvalue 1 and the corresponding eigenfunction are known. Then, toevaluate 2 and 2 , we shorten the kernel K by subtracting the factor 1 1 1( )/ from it. Then, fromtheorem II of Art. 7.3, we know that the kernel

(2)1 1( ) /K K

satisfies all the requirements of a symmetric kernel. Proceeding as explained earlier, we see thatat least one of the numbers || K2 || or – || K2 || is the reciprocal of 2.

Proceeding likewise, the process can be continued until all the eigenvalues and eigenfunctionsare obtained.

Remarks. The only draw back in the above process is that, to find the nth eigenvalues, wehave to obtain the first (n – 1) eigenvalues.

MISCELLANEOUS EXERCIESE ON CHAPTER 7

1. Compute the iterated kernels for symmetric kernel 11

( , ) sin sin .n

K x t n n x n t

2. Consider the eigenvalue problem1

1( ) (1 | |) ( )y x x t y t dt



Differentiate under the integral sign to obtain the corresponding differential equation andthe boundary conditions. Show that the kernel of this integral equation is positive.

3. Compute the eigenvalues and eigenfunctions of the symmetric kernel K (x, t) = min (x, t) inthe basic interval 0 < x < 1, 0 < t < 1.

4. Show that the kernel K (x, t), 0 < x < 1, 0 < t < 1,

(1 ),( , )

(1 ) ,x t x t

K x tt x x t

has the bilinear form 21

sin sin( , ) 2( )n

n x n tK x tn

Hence deduce that 2 21(1/ ) / 6

nn

5. Use the Gram-Schmidt process to orthogonalize 1, x, x2, x3 in the interval 1 1x andthus compute eigenvalues and eigenfunctions of the symmetric kernel 1 + xt + x2 t2 + x3 t3.

6. Consider the kernel K(x, t) = log [ 1 – cos (x – t)], 0 2 , 0 2 .x t Prove that

(i) it is a symmetric kernel(ii) K (x, t) = – log 2 + 2 log [ 1 – ei (x – t)]

1 1

cos cos sin sinlog 2 2 2n n

nx nt nx ntn n

(iii) its eigenvalues are 0 1/(2 log 2), / 2 ,n n n = 1, 2, ... with eigenfunctions

0 ( ) , ( ) cos sin ,nx C x A nx B nx A, B and C are constants.

7. Which one of the following sets of functions is not orthogonal (with respect to theL2– inner product) over the given interval

(a) {sin : , – }nx n x N (b) {cos : , – }nx n x N

2 1/2 2 1( ) { : , –1 1} ( ) { : , –1 1}.n nc x n x d x n x N N [GATE 2010]

Solution. Ans. (d). Use definition of orthogonality given in Art. 7.1 (c), page 7.2. Here,

for ,n m– –

–

1 1 sin ( – ) sin( )sin sin {cos ( – ) –cos( ) } – 02 2 –

n m x n m xnx mx dx n m x n m x dxn m n m

Similarly, –

cos cos 0nx mx dx

and 12 2 21 2 1/2 2 1/2

–1–1

02 2 2

n mn m xx x dx

n m

But, we have

12 2 31 2 1 2 1–1

–1

2 0.2 2 3 2 2 3

m nm n xx x dx

m n m n

Hence, the sets (A), (B) and (C) are orthogonal while the set (D) is not orthogonal.


CHAPTER 8

Singular Integral Equations8.1. SINGULAR INTEGRAL EQUATION. DEFINITION. [Meerut 2001, 06, 07]

An integral equation in which the range of integration is infinite, or in which the kernel isdiscountinuous, is known as a singular integral equation. Thus, for example, the equations

0 0 0

( )( ) sin ( ) ( ) , ( ) ( ) , and ( )xxt y tf x xt y t dt f x e y t dt f x dt

x t

are all singular integral equations of the first kind.Remark. Singular integular equations possess very unusual properties.

8.2. THE SOLUTION OF THE ABEL INTEGRAL EQUATION, NAMELY,

0

( )( ) , 0 1.( )

x y tf x dtx t

... (1)

in which f (x) is a known function while y (t) is to be determined. [Kanpur 2006, Meerut 2004]

Multiplying both sides of (1) by 11/( )u x and then integrating w.r.t. ‘x’ from 0 to u, we

obtain 1 10 0 0

( ) 1 ( ) .( ) ( ) ( )

u x u t x

x t

f x dx y t dt dxu x u x x t

... (2)

Consider the double integral on right side of (2). Thisintegral is to be first integrated in the t-direction from t = 0 tot = x and then the resulting integral is to be integrated in thex-direction from x = 0 to x = u. The region of integration isthe triangular area OAB. In the integral under consideration,the area OAB is divided in strips parallel to t-axis (for examplestrip PQ). To reverse the order of integration, we have to firstintegrate with respect to x regarding t as constant and thenwith respect to t. This is done by dividing the above mentionedarea OAB in strips parallel to the x-axis (for example, P Q ).Thus, we note that first we must integrate from x t to x u in x-direction and afterwords in thet-direction from 0t to .t u Thus, changing the order of integration on right side of (2), weobtain

1 10 0

( ) ( ) .( ) ( ) ( )

u t u x u

t x t

f x dx dxy t dtu x u x x t

... (3)

Let 1 .( ) ( )

x u

x t

dxIu x x t

... (4)

Put (u – x)/(u – t) = y ... (5A) that is, – ( – ) .x u u t y .. (5B)

8.1


8.2 Singular Integral Equations

From (5B) – ( – ) .dx u t dy ... (6)Using (5A), (5B) and (6) in (4), we have

0

1 11

( )( ) { ( ) }

y

y

u t dyIu t y u u t y t

1

1 10

( )( ) ( ) (1 )

u t dyu t y u t y

1 11 1 (1 ) 1

0 0(1 ) (1 )y y dy y y dy

( ,1 ),B by the definition of the Beta function

( ) (1 )( 1 ) sin

, ... (7)

where () is the usual gamma function.

From (4) and (7), 1 .sin( ) ( )

x u

x t

dxu x x t

... (8)

Substituting (8) in (3), we have 10 0

( ) ( ) .sin( )

u uf x dx y t dtu x

or 10 0

sin ( )( ) .( )

u u f x dxy t dtu x

... (9)

Differentiating both sides of (9) w.r.t. u and using Leibnitz’s rule of differentiation under thesign of integration (refer Art. 1.13), we have

10

sin ( )( )( )

ud f xy u dxdu u x

... (10)

Replacing u by t on both sides of (10), we get

10

sin ( )( ) .( )

td f xy t dxdu t x

... (11)

which is required solution of (1).

Example. Solve the singular integral equation 1/ 20

( ) .( )

x y t dtxx t

[Kanpur 2006, 10; Meerut 2006, 09, 12]

Sol. Given 1/ 20

( ) .( )

x y t dtxx t

... (1)

Comparing (1) with the Abel singular integral equation

0

( )( )( )

x y t dtf xx t

... (2)

here ( ) and 1/ 2.f x x ... (3)We know that the solution of (2) is given by

10

sin ( )( ) .( )

td f xy t dxdt t x

... (4)


Singular Integral Equations 8.3

Putting values of f (x) and from (3) in (4), the required solution is given by

1/ 20

sin ( / 2)( )( )

td xy t dxdt t x

or 1( ) ,dIy t

dt

... (5)

where 1/ 20.

( )

t xI dxt x

... (6)

Put t – x = z2 so that dx = – 2z dz. ... (7)

From (6), 20 0 2( )( 2 ) 2 ( )

t t

t z z dzI t z dzz

03 3/ 22 / 3 2[ (1/ 3) ] (4 / 3) .

ttz z t t t t t

Putting the above value of I in (5), we get

3/ 2 1/ 24 1 4 3 2( ) .3 3 2

I d ty t t tdt

8.3. GENERAL FORM OF THE ABEL SINGULAR INTEGRAL EQUATION.

It is given by ( )( ) ,0 1,[ ( ) ( )]

x

a

y t dtf xh x h t

... (1)

where h (t) is a strictly montonically increasing and differentiable in (a, b) and ( ) 0.h t Determination of solution of (1). To solve (1), we consider an integral I given by

1( ) ( ) .

[ ( ) ( )]

x

a

h u f u duIh x h u

... (2)

From (1),( )( ) .

[ ( ) ( )]

u

a

y t dtf uh u h t

... (3)

Substituting the above value of f (u) in (2), we obtain

1( ) ( ) .

[ ( ) ( )] [ ( ) ( ) ]

u x t u

u a t a

h u y t dtI duh x h u h u h t

... (4)

Clearly, the double integral on the right hand side of(4) is to be first integrated in the t-direction from t a tot u and then the resulting integral is to be integrated inthe u-direction from u a to .u x The region ofintegration is the triangular area ABC. In the integral underconsideration, the area OAB is divided in strips parallel tot-axis (for example, strip PQ). To reverse the order ofintegration, we have to first integrate with respect to uregarding t as constant and then with respect to t. This isdone by dividing the above mentioned area ABC in stripsparallel to u-axis (for example, P Q ). Thus, we note thatfirst we must integrate from u t to u x in u-directionand afterwards in the t-direction from t a to .t x Thus,changing the order of integration on right hand side of (4),we obtain.

u

QC( )a x,

PP Q

t = a

t = u

u = xB(x, x)

O (0, 0)t

u = aA( )a a,



1( )( ) .

[ ( ) ( )] [ ( ) ( )]

t x u x

t a u t

h u duI y t dth x h u h u h t

... (5)

Re-writing (5), we have ( ) ,t x

t aI J y t dt

... (6)

where 1( ) .

[ ( ) ( )] [ ( ) ( )]

u x

u t

h u duJh x h u h u h t

... (7)

We shall now simplify integral J.Put { ( ) – ( )}/{ ( ) – ( )} ,h x h u h x h t y ... (8)

so that h (u) = h (x) – [h (x) – h (t) ] y. ... (9)From (9) ( ) [ ( ) ( ) ] .h u du h x h t dy ... (10)Using (8), (9) and (10) in (7), we have

0

1 11

[ ( ) ( )][ ( ) ( )] [ ( ) { ( ) ( )} ( )]

h x h t dyJh x h t y h x h x h t y h t

1

1 10

[ ( ) ( )][ ( ) ( )] [ ( ) ( )] (1 )

h x h t dyh x h t y h x h t y

1 11 1 (1 ) 1

0 0(1 ) (1 )y y dy y y dy


.sin

... (11)

Substituting the above value of J in (6), we have

( ) ( ) .sin sin

x x

a aI y t dt y t dt

... (12)

Equating the two values of I from (2) and (12), we have

1( ) ( )( )

sin [ ( ) ( )]

x x

a a

h u f u duy t dth x h u

or 1sin ( ) ( )( ) .

[ ( ) ( )]

x x

a a

h u f u duy t dth x h u

... (13)

Differentiating both sides of (13) w.r.t. ‘x’ and using Leibntiz’s rule of differentiation under thesign of integration (refer Art. 1.13), we have

1sin ( ) ( )( ) .

[ ( ) ( )]

x

a

d h u f u duy xdx h x h u

... (14)

Replacing x by t on both sides of (14), we have

1sin ( ) ( )( ) ,

[ ( ) ( )]

t

a

d h u f u duy tdt h t h u

... (15)

which is the required solution of (1).



8.4. ANOTHER GENERAL FORM OF THE ABEL SINGULAR INTEGERAL EQUATION.

It is given by ( )( ) , 0 1[ ( ) ( )]

b

x

y t dtf xh t h x

... (1)

where a < x < b and h (t) is a monotonically increasing function.Determination of solution of (1). To solve (1), we consider an integral I given by

1( ) ( ) .

[ ( ) ( )]

b

x

h u f u duIh t h x

... (2)

From (1),( )( ) .

[ ( ) ( )]

b

u

y t dtf uh t h u

... (3)

Substituting the above values of f (u) in (2), we obtain

1( ) ( ) .

[ ( ) ( )] [ ( ) ( )]

u b t b

u x t u

h u y t dtI duh u h x h t h u

... (4)

Clearly, the double integral on the right hand side of (4)is to be integrated in the t-direction from t u to t b andthen the resulting integral is to be integrated in theu-direction from u x to .u b The region of integration isthe triangular area ABC. In the integral under consideration,the area ABC is divided in strips paralles to t-axis (for example,strip PQ). To reverse the order of integration, we have to firstintegrate with respect to u regarding t as constant and thenwith respect to t. This is done by dividing the above mentionedarea ABC in strips parallel to u-axis (for example, strip P Q ).Thus, we note that first we must integrate from u x to u t inu-direction and afterwards in the t-direction from t x to

.t b Thus, changing the order of integration on right handside of (4), we obtain

1( )( ) .

[ ( ) ( )] [ ( ) ( )]

t b u t

t x u x

h u duI y t dth u h x h t h u

... (5)

Re-writing (5), we have ( ) ,t b

t xI J y t dt

... (6)

where 1( ) .

[ ( ) ( )] [ ( ) ( )]

u t

u x

h u duJh u h x h t h u

... (7)

We shall now simplify J.Put { ( ) ( )}/{ ( ) – ( )}h u h x h t h x y ... (8)

so that h (u) = h (x) + [h (t) – h (x)] y. ... (9)From (9), ( ) { ( ) – ( )}h u du h t h x dy ...(10)

Using (8), (9) and (10) in (7), we have1

1 10

[ ( ) ( )][ ( ) ( )] [ ( ) ( ) { ( ) ( )} ]

h t h x dyJh t h x y h t h x h t h x y

u

P

A ( , )x x

O (0, 0)

t = b

C ( , )b b

t = u

Q

B u = xt

(b, x)

Q

P

u= b



1

1 10

[ ( ) ( )][ ( ) ( )] [ ( ) ( )] (1 )

h t h x dyh t h x y h t h x y

1 11 1 (1 ) 1

0 0(1 ) (1 )y y dy y y dy


) .sin

... (11)

Substituting the above value of J in (6), we have

( ) ( ) .sin sin

t b b

t x xI y t dt y t dt

... (12)

Equating the two values of I from (2) and (12), we have

1( ) ( )( )

sin [ ( ) ( )]

b b

x x

h u f u duy t dth u h x

or 1sin ( ) ( )( ) .

[ ( ) ( )]

b b

x x

h u f u duy t dth u h x

... (13)

Differentiating both sides of (13) w.r.t. ‘x’ and using Leibnitz’s rule of differentiation under thesign of integration (refer Art. 1.13), we have

1sin ( ) ( )( ) .

[ ( ) ( )]

b

x

d h u f u duy xdx h u h x

... (14)

Replacing x by t in (14) and multiplying both sides by (–1), we have

1sin ( ) ( )( ) .

[ ( ) ( )]

b

t

d h u f u duy tdt h u h t

... (15)

Weakly singular kernel. Definition. Consider a Fredholm integral equation with the kernel

of the form ( , )( , ) , 0 1

| |H x tK x tt x

... (1)

where H (x, t) is a bounded function. Then the kernal (1) is known as weakly singular.A weakly singular kernal can be transformed to a kernel which is bounded. This is done by

means of iterated kernels. It has been established that if the singular kernel has the form (1), thenthere always exist a positive number p0, dependent in , such that, for p > p0, the iterated kernelKp (x, t) is bounded.8.5. SOLVED EXAMPLES.

Ex. 1. Solve the integral equation 1/ 2( )( ) , 0 .

(cos cos )

x

a

y t dtf x a x bt x

[Meerut 2003, 10, 11]

Sol. Given that 1/ 2( )( ) ,

(cos cos )

x

a

y t dtf xt x

... (1)

Re-writing (1), we have 1/ 2( )( ) .

[(1 cos ) (1 cos )]

x

a

y t dtf xx t

... (2)



Comparing (2) with general form of Abel integral equation (refer Art. 8.3)

( )( ) ,0 1[ ( ) ( )]

x

a

y t dtf xh x h t

... (3)

where h (t) is a strictly monotonically increasing and differentiable function in (a, b) and ( ) 0h t inthis interval, we have 1/ 2, ( ) 1 cos .h t t

Clearly h (t) is a strictly monotonically increasing function in (0, ).We know that (refer equation (15) Art. 8.3) the solution of (3) is given by

1sin ( ) ( )( ) .

[ ( ) ( )]

t

a


... (4)

Putting 1/ 2, h (t) = 1 – cos t, h (u) = 1 – cos u and ( ) sinh u u du in (4), the requiredsolution of (1) is

1/ 2sin ( / 2) sin ( )( )

[(1 cos ) (1 cos )]

t

a

d u f u duy tdt t u

or 1/ 21 sin ( )( ) , .

(cos cos )

t

a

d u f u duy t a t bdt u t

Ex. 2. Solve 1/ 2( )( ) ,0 .

(cos cos )

b

x

y t dtf x a x bx t

Sol. Given that 1/ 2

( )( ) ,(cos cos )

b

x

y t dtf xx t

0 a x b ... (1)

Re-writing (1), we have 1/ 2( )( ) .

[(1 cos ) (1 cos )]

b

x

y t dtf xt x

... (2)

Comparing (2) with general form of Abel integral equation (refer Art. 8.4)( )( ) ,0 1.

[ ( ) ( )]

b

x

y t dtf xh t h x

... (3)

where h (t) is a monotonically increasing function, we have 1/ 2, ( ) 1 cos .h t t Clearly h (t) is a monotonically increasing function,We known that (refer equation (15) in Art. 8.4) the solution of (3) is given by

1sin ( ) ( )( ) .

[ ( ) ( )]

b

t


... (4)

Putting 1/ 2, ( ) 1 cos , ( ) 1 cosh t t h u u and ( ) sinh u u du in (4), the requiredsolution of (1) is

1/ 2sin ( / 2) sin ( )( )

[1 cos (1 cos )]

b

t

d u f u duy tdt u t

or 1/ 21 sin ( )( ) , .

(cos cos )

b

t

d u f u duy t a t bdt t u

Ex. 3. Solve the integral equation 2 2

( )( ) , 0 1; .( )

x

a

y t dtf x a x bx t

[Kanpur 2011; Meerut 2002, 03]



Sol. Given that 2 2( )( ) , 0 1 ;

( )

x

a

y t dtf x a x bx t

... (1)


( )( ) , 0 1

[ ( ) ( )]

x

a

y t dtf xh x h t

where h (t) is a strictly monotonically increasing and differentiable function in (a, b) and( ) 0h t in this interval, we have

h(t) = t2. ... (3)Clearly h (t) is a strictly monotonically increasing and differentiable function.We know that (refer equation (15) in Art. 8.3) the solution of (2) is given by

1sin ( ) ( )( ) .

[ ( ) ( )]

t

a


... (4)

Putting h (t) = t2, h (u) = u2 and ( ) 2h u u du in (4), the required solution is given by

2 2 1sin 2 ( )( ) ,

( )

t

a


or 2 2 12sin ( )( ) , .

( )

t

a


... (5)

Remark. The result (5) remains valid when 0.a Hence, the solution of the integral equation

2 20

( )( ) , 0 1( )

x y t dtf xx t

... (6)

is 2 2 10

2sin ( )( ) .( )

td u f u duy tdt t u

... (7)

Ex. 4. Solve the integral equation 2 2( )( ) , 0 1; .

( )

b

x

y t dtf x a x bt x

[Meerut 2004, 07]

Sol. Given that 2 2( )( ) , 0 1; .

( )

b

x

y t dtf x a x bt x

... (1)


( )( ) , 0 1

[ ( ) ( )]

b

x

y t dtf xh t h x

... (2)

where h (t) is a monotonically increasing function, we have h (t) = t2. ... (3)

Clearly, h (t) is a strictly monotically increasing function.We known that (refer equation (15) in Art. 8.4) the solution of (2) is given by

1sin ( ) ( )( ) .

[ ( ) ( )]

b

t


... (4)

Putting h (t) = t2, h (u) = u2 and ( ) 2h u u du in (4), the required solution is given by

2 2 1sin 2 ( )( )

( )

b

t

d u f u duy tdt u t



or 2 2 12sin ( )( ) .

( )

b

t

d u f u duy tdt u t

... (5)

Remark. The result (5) remains valid when .b Hence from (1) and (5), it follows that thesolution of the integral equation

2 2( )( ) ,0 1,

( )x

y t dtf xt x

... (6)

is 2 2 12sin ( )( ) .

( )t

d u f u duy tdt u t

... (7)

8.6. CAUCHY PRINCIPAL VALUE FOR INTEGRALS. [Meerut 2001]

According to Riemann, the theory of the definite integrals ( )b

af x dx is based on two

assumptions, namely, (i) the integrad f (x) is bounded and (ii) The range of integration (a, b) isfinite. These are called ordinary integrals.

Cauchy extended the theory of Riemann integration to include the following exceptionalcases :

(i) When f (x) becomes infinite at one of the limits of integration. (ii) When f (x) becomes infinite for one or more values of x between the range of integration.(iii) When one or both the limits of integration are infinite.Such integrals are called improper integrals. Integrals of these forms may have a finite value,

it may be infinite or indeterminate depending upon the function f (x) and the limits a and b.Cauchy’s general and principal values. Singular integrals. In case of an improper integral

( ) ,b

af x dx where f (x) is unbounded at x = c, but is bounded in each of the intervals 1( , )a c

and 2( , )c b where 1 2and are arbitrary small positive numbers. Then, the limit

1

1 22

00

( ) lim ( ) ( ) ,b c b

a a cf x dx f x dx f x dx

... (1)

if it exists, is called the general value of the improper integral. Here it is understood that 1 and

2 tend to zero independently. But it may happen that the limit (1) does not exist when 1 and 2tend to zero independently of each other, but it exists when 1 2 , say. Such a limit is knownas the principal value of the integral and is usually denoted by

( )b

aP f x dx or

*( ) .

b

af x dx

*

0( ) ( ) lim ( ) ( ) .

b b c b

a a a cP f x dx f x dx f x dx f x dx

... (2)

Similarly, the general value of ( )f x dx

is defined by the limit,

1

1 22

1/

0, 1/0

( ) lim ( )f x dx f x dx

... (3)



and the corresponding principal value is given by1/

0 1/( ) lim ( ) .P f x dx f x dx

... (4)

The limits (2) and (4) are also known as singular integrals. Such singular integrals exist whenthe integrand f (x) satisfies the following regularity condition.

H . .o lder Condition. A functions f (x) is said to satisfy the H

..o lder condition if there exist

constants and , 0 1,k such that, for every pair of points x1, x2 lying in the range ,a x b we have

1 2 1 2| ( ) ( ) | | | .f x f x k x x ... (5)

A function satisfying the H..o lder condition is known as H

..o lder continuous.

In particular, when 1, then condition (5) is known as Lipschitz condition.

The H ..o lder condition can be extended to functions of more than one variable. Thus, the

kernel K (x, t) is H ..o lder continuous with respect to both variables if there exist constants

and , 0 1,k such that

1 1 2 2 1 2 1 2| ( , ) ( , ) | [| | | | ],K x t K x t k x x t t ... (6)

where (x1, t1) and (x2, t2) lie within the range of definition.The definition of Cauchy principal value for contour integrals.We known that a contour integral of a complex-valued

function with a pole z0 on the contour does not exist. Howeverit may have the Cauchy principal value which will be definednow. To this end, let C be a closed or open regular curve (referadjoining figure). We enclose the point z0 by a small circle ofradius with centre at z0. Let C denote the part of the contouroutside this circle.

If a complex-valued function f (z) is integrable along ,C however small the positive number, then the limit.

0lim ( ) ,

Cf z dz

... (7)

if it exists, is known as the Cauchy principal value and is denoted as

( )C

P f z dz or *

( ) .C

f z dz ... (8)

In what follows, we shall study the contour integrals of the Cauchy type, that is,( ) .

C

f dz

... (9)

From the theory of functions of complex variable, we known that if f (z) satisfies the H ..older

condition 1 2 1 2| ( ) ( ) | | | ,f z f z k z z ... (10)

where z1, z2 is any pair of points on the curve C, while k and are constants such that 0 1, thenthe integral (9) exists for all points z on the curve C, except perhaps its end points. The function



f1(z) defined by

1( )( )

C

ff z dz

... (11)

is also H..o lder continuous which possesses the similar properties as possessed by the corresponding

real functions.The definition (10) can be extended to complex-valued function of more than one variable as

already done for real-valued functions.

The functions ( )f occuring in the integral (9) is known as the density of the Cauchy integral.

8.7. THE CAUCHY INTEGRALS.

The integral equation 1 ( )( ) ,

2 C

yf z di z

... (1)

where C is a regular curve, is known as a Cauchy-type integral. First of all, we shall discuss (1)when C is a closed contour.

Plemelj formulas. Let ( )y be a H ..older continuous function of a point on a regular closed

contour C and let a point z tend, in an arbitrary manner, from inside or outside the contour C, tothe point t on this contour; then the integral (I) tends to the limit

*1 1 ( )( ) ( ) ,2 2 C

yf t y t di t

... (2)

or*1 1 ( )( ) ( ) ,

2 2 C

yf t y t di t

... (3)

respectively. The formulas (2) and (3) are known as Plemelj formulas. We adopt the standardconvention of counterclockwise traversal of the closed contour C. It follows that the boundary

value ( )f t relates to the values of the Cauchy integral inside the region bounded by C, while the

second boundary value ( )f t relates to the value in the outside region.Poincare-Bertrand transformation formula.

Let y (t) be H..o lder continuous function and let C be a closed countour. Then

* *1

21 1

1 ( ) 1 ( ).4(2 ) C C

d y d y tti

... (4)

Proof. Let*

11 ( )( )

2 C

yy t di t

... (5)

and*

12

( )1( )2 C

yy t di t

... (6)

be two singular integrals, where y1 and y2 are H ..o lder continuous functions.

Now, we shall obtain an iterated integral connecting y2 and y. To this end, consider the Cauchy-

type integrals 1 ( )( )

2 C

yf z di z

... (7)

and1

1( )1( ) .

2 C

yf z di z

...(8)



Using the Plemelj formula (2) and integrals (7) and (8), we arrive at the limiting values

*1 1 ( )( ) ( ) ,

2 2 C

yf t y t di t

... (9)

and *

11 1

( )1 1( ) ( ) .2 2 C

yf t y t di t

... (10)

Comparing (5) and (9), we have

1( ) ( ) (1/ 2) ( ).y t f t y t ... (11)Again, comparing (6) and (10), we have

2 1 1( ) ( ) (1/ 2) ( ).y t f t y t ... (12)

From (11), 1( ) ( ) (1/ 2) ( ).y f y ... (13)

Putting the above value of 1( )y in (8), we get

11 ( ) (1/ 2) ( )( )

2 C

f yf z di z

or 11 ( ) 1 ( )( ) .

2 4C C

f yf z d di z i z

... (14)

Consider the first integral on R.H.S. of (14). Since its density ( )f is the limiting value of f(z), which is regular inside C, using the Cauchy integral formula, we have

1 ( ) ( ).2 C

f d f zi z

... (15)

Comparing the second integral on R.H.S. of (14) with the integral in (7), we see that

1 ( ) 1 ( ).4 2C

y d f zi z

... (16)

Using (15) and (16), (14) becomes 1 1( ) ( ) (1/ 2) ( ) or ( ) (1/ 2) ( ).f z f z f z f z f z ... (17)

From (17), 1 ( ) (1/ 2) ( ).f t f t ... (18)Now, from (12), we obtain

2 1 1( ) ( ) (1/ 2) ( )y t f t y t

or 2 ( ) (1/ 2) ( ) (1/ 2) [ ( ) (1/ 2) ( )],y t f t f t y t by (11) and (18)

or 2 ( ) (1/ 4) ( ).y t y t ... (19)

Now, from (6), we have*

1 12 1

1

( )1( ) .2 C

yy t di t

... (20)

Similarly, from (5), we have*

1 11

1 ( )( ) .2 C

yy di

... (21)



Substituting the value of 1 1( )y from (21) in (20), we have

* *2 1

1 1

1 1 1 ( )( )2 2C C

yy t d di t i

or* *

12

1 1

1 ( ) 1 ( ), using (19)4(2 ) C C

d y d y tti

... (22)

which is the desired Poincare-Bertrand transformation formula.Remark. The reader should note carefully that in the formula (22), it is not allowed to change

the order of integration. Thus, while solving double integral, first integration is performed withrespect to and then the resulting integral is integrated with respect to 1.

8.8. SOLUTION OF THE CAUCHY-TYPE SINGULAR INTEGRAL EQUATION.Two cases arise :Case I. When there is closed contour C.We are to solve the integral equation of the second kind.

* ( )( ) ( ) ,C

b yay t f t di t

... (1)

where a and b are known complex constants, ( )y is a H ..o lder-continuous function, and C is a

regular closed contour.

We introduce an operator L defined as * ( )( ) .

2 C

b yLy ay t di t

... (2)

Re-writing (1), * ( )( ) ( )C

b yay t d f ti t

or Ly = f (t), using the definition of operator L ... (3)

Now, we define an ‘‘adjoint’’ operator

*1

11

( )( ) .C

gbMg ag t di t

... (4)

From (3), we have M [Ly] = Mf

or * ( )( ) ,C

b yM ay t d Mfi t

by (2) ... (5)

Let* ( )( ) ( )C

b yg t ay t di t

... (6)

so that*

1 11

( )( ) ( ) .C

b yg ay di

... (7)

Using (6), (5) becomes Mg = Mf

or **

1 11 1

1 1

( ) ( )( ) ( ) ,C C

g fb bag t d af t di t i t

using the definition of operator M



or* **

1 11 1 1

( ) 1 ( )( ) ( )C C C

b y b b ya ay t d ay d di t i t i

*1

11

( )( )C

fbaf t di t

or* *2 1 1

1

( )( )( )C C

y dab y d aba y ti t i t

* * *2 1 1

121 1 1

( )1 ( ) ( )( ) C C C

d fy bb d af t dt i ti

or* *2 ( ) ( )( )C C

ab y d ab y da y ti t i t

* *2 1

21 1

1 ( )4(2 ) C C

d yb dti

*

11

1

( )( )C

fbaf t di t

or*2 2 1

11

( )1( ) 4 ( ) ( ) ,4 C

fba y t b y t af t di t

using the Poincare-Bertrand formula

or*2 2 ( )( ) ( ) ( ) .C

b fa b y t a f t di t

... (8)

Supposing that 2 2( ) 0,a b (8) gives*

2 2 2 2( )( ) ( ) .

( ) C

a b fy t f t dta b a b i

... (9)

Particular Case. Putting a = 0 in (9), the solution of the Cauchy-type integral equation ofthe first kind

* ( )( )C

b y tf t di t

... (10)

is *1 ( )( ) .C

fy t db i t

... (11)

Deduction. Putting b = 1 in (10) and (11), we see that the solution of*1 ( )( )C

y tf t di t

... (12)

is*1 ( )( ) .C

fy t di t

... (13)

(12) and (13) exhibit the reciprocity of these relations.Case II. When there is unclosed contour. The Riemann-Hilbert problem.Plemelj formulas (2) and (3) of Art. 8.7 still hold for an arc also when we define the plus and

minus directions as follows. To this end, we supplement the arc L with another arc L so as to forma closed contour .L L Then, the interior and exterior of this closed contour stand for the plus andminus directions. Accordingly, for an arc C, we obtain



*1 1 ( )( ) ( )

2 2 C

yf t y t di t

... (14)

and *1 1 ( )( ) ( ) .

2 2 C

yf t y t di t

... (15)

Re-writing (14) and (15), we have y (t) = f + (t) – f – (t) ... (16)

and*1 ( ) ( ) ( ).C

y dt f t f ti t

... (17)

Let a function w(t) be prescribed on an arc L and let it satisfy the H ..o lder condition on L.

Then, we wish to obtain a function W(z) analytic for all points z on L such that it satisfies theboundary (or jump) condition

( ) ( ) ( ), .W t W t w t t L ... (18)

The problem posed in (18) is a special case of the so called Riemann-Hiblert problem,which requires the determination of a function W (z) analytic for all point z not lying on L such

that, for t on L, ( ) ( ) ( ) ( ),W t Z t W t w t ... (19)

where ( )w t and Z (t) are given complex-valued functions.By substituting the formulas (16) and (17) in the integral-equation

* ( )( ) ( ) ,C

b yay t F t di t

... (20)

it follows that the solution of (20) is reduced to solving the Riemann-Hilbert problem (a + b) f + (t) – (a – b) f – (t) = F (t). ... (21)

Let L be a regular unclosed curve. Then the solution of the sigular integral equation (1) is

2 2 1( )( )

– ( ) ( )m ma f t ky ta b t t

*

2 2 ( ) ,( )

m m

C

b t dft ta b i

... (22)

where and are the beginning and end points of the contour C and the number m is given by

1 log ,2

a bmi a b

... (23)

and the quantity k is an arbitrary constant and is suitably closen so that y (t) is bounded at or at .Particular Case. The solution of the first kind (by setting b = 1 also without any loss of

generality),*1 ( )( ) ,C

yf t di t

...(24)

is given by (22) with a = 0, b = 1. Putting a = 0, b = 1 in (23) yields m = 1/2 and so (22) gives therequired solution of (24) as

1/ 2 1/ 2*

1/ 21 ( )( )

[( ) ( )] C

k t fy t dti t tt t

... (25)

Example : Prove that the solution of the integral equation

2 2

2 2 2 1

0

2( ) ( ) ( ) , 0 1ttf t t t y d

... (i)



is 2 1

2 2 1 2 2

0( ) ( ) ( )

1

tt dy t t f ddt

... (ii)

Hint : The required solution can be easily obtained by comparing (i) and (ii) with equations(24) and (25) respectively of Art. 8.8, and by setting 0, 1.

8.9. THE HILBERT KERNEL. DEFINITION. [Kanpur 2011, Meerut 2001, 04, 07, 09]

A kernel of the form ( , ) cot ,2

t xK x t

where x and t are real variables, is known as the Hilbert kernel.Consider the integral equation

* 2

0( ) ( ) ( , ) cot ( ) .

2t xy x f x F x t y t dt

... (2)

where f (x) and F (x, t) are known continuos functions of period 2 .Then the integral equation (2) is equivalent to the Cauchy-type integral equation

* ( , )( ) ( ) ( ) ,C

Gy f y d

... (3)

where and are complex variables and the contour C is the circumference of the unit disc withthe centre at the point z = 0.

Let and denotes the points on the boundary C corresponding to the arguments x and t,respectively. Then, we have

and .i x i te e ... (4)

From (4), i td i e dt or ,d i dt ( )i te

so that (1/ ) .d i dt ... (5)

Now, we havei t

i t i xd i e dt

e e

... (6)

Also, ( ) / 2 ( ) / 2

( ) / 2 ( ) / 21 cot2 2 2 2 2

i t x i t x

i t x i t xt x i i e e i

e e

( ) / 2 ( ) / 2

( ) / 2 ( ) / 2 12

i t x i t x

i t x i t xi e e

e e

( ) / 2

( ) / 2 ( ) / 22

2

i t x

i t x i t xi e

e e

/ 2 / 2

/ 2 / 2 / 2 / 2 .i t ix i t

i t i x i t i x i t i xe e i ei

e e e e e e

... (7)

Using (6) and (7), we have

1 cot2 2 2

i t

i t i xi e dt d t x i dt

e e

or 2 cot2

d t x dt i dt

or 2 cot ,

2d t x ddt

using (5)

or 2cot2

t x d ddt or cot .

2t x ddt

... (8)



With help of (8), we find that (2) reduces to the form (3).The Hilbert kernel is also related to the Poisson kernel in the integral representation formula

for a harmonic function U (r, x) :

22

20

1 1( , ) ( ) ,2 1 2 cos ( )

rU r x u t dtr r t x

... (9)

inside the disc r < 1. The function u (t) = U (1, t) is the prescribed value of the harmonic functionon the circumference C of the disc.

Putting andi x i tz re e ... (10)

in (9), we obtain as before 1( , ) Re ( ) ,

2 C

z dU r x u ti z

... (11)

where Re denotes the real part of the expression that follows.Again, let V (r, x) be the function that is harmonic conjugate to U (r, s). Then, we have

U (r, x) + i V (r, x) 1 ( ) ,

2 C

z du ti z

... (12)

such that V (r, x) = 0 at the centre of the disc, that is, 0( , ) 0rV r x ... (13)

Then, the function V (r, x) is uniquely defined.

When 0,r so that z tends a point of the circumference C from within the disc, we canapply the Plemelj formula (refer formula (4) in Art. 8.7) to the analytic function given by (12) andobtain (keeping (8) in view)

* 2

0

1( ) ( ) cot ,2 2

t xv x u t dt ... (14)

where v (x) = V (1, x), is the limiting value of the harmonic function on C. Thus, (14) gives therelation between the limiting values of the conjuagte harmonic functions U (r, x) and V (r, x) on thecircumference.

We now state and prove an important formula, namely,Hilbert formula. To prove that [Meerut 2001, 07, 09]

* *2 2

2 0 0

1 cot ( ) cot2 24

x ty t dt d

2

0

1( ) ( ) .2

y x y t dt

... (15)

Proof. Consider the following integral equations with Hilbert kernel :

* 2

10

1( ) ( ) cot .2 2

x ty x y t dt ... (16)

and * 2

2 10

1( ) ( ) cot .2 2

x ty x y t dt ... (17)

Let U (r, x), U1 (r, x) and U2 (r, x) be the functions which are harmonic inside the disc r < 1,and whose values on the circumference 1r are equal respectively to y (x), y1 (x) and y2 (x).Then, from (9), it follows that U1 (r, x) is harmonic conjugate to U (r, x) and U2 (r, x) is harmonicconjugate to U1 (r, x). Using the well known Cauchy-Riemann equations of the theory of complexvariables, we have



2 2/ – ( / ), and / ( / )U r U r U x U x

giving U2 (r, x) = – U (r, x) + k. ... (18)where k is an arbitrary constant. To determine k, we use (13) and obtain

2 2

00 0

1 1[ ( , )] (1, ) ( ) ,2 2rk U r x U t dt y t dt

... (19)

where we have used the mean-value property of the harmonic function.

From (18) and (19),2

20

1( , ) ( , ) ( ) .2

U r x U r x y t dt

... (20)

Putting 1r in (20), we have2

2 0

1(1, ) (1, ) ( ) .2

U x U x y t dt

... (21)

But U (1, x) = y (x) and U2 (1, x) = y2 (x). ... (22)

From (21) and (22),2

20

1( ) ( ) ( ) .2

y x y x y t dt

... (23)

From (17), * 2

2 10

1( ) ( ) cot2 2

t xy x y t dt

* 2

2 10

1( ) ( ) cot .2 2

xy x y d ... (24)

Now, from (16), * 2

10

1( ) ( ) cot2 2

t xy x y t dt

* 2

10

1( ) ( ) cot .2 2

ty y t dt ... (25)

Substituting the values of 1( )y given by (25) in (24), we have

* *2 22

0 0

1 1( ) cot ( ) cot2 2 2 2

x ty x y t dt d

or** 2 2

2 2 0 0

1( ) cot ( ) cot2 24

x ty x y t dt d ... (26)

Equating the two values of y2 (x) given by (23) and (26), we have

* *2 2

2 0 0

1 cot ( ) cot2 24

x ty t dt d

2

0

1( ) ( ) ,2

y x y t dt

... (27)

which is the required Hilbert formula.8.10.SOLUTION OF THE HILBERT-TYPE SINGULAR INTEGRAL EQUATION OF THE

SECOND KIND, NAMELY,* 2

0( ) ( ) ( ) cot ,

2 2b t xay x f x y t dt

... (1)

where a and b are complex constants. [Kanpur 2005; Meerut 2002, 04, 06]



We introduce an operator L defined as* 2

0( ) ( ) cot .

2 2b t xLy a y x y t dt

... (2)

Re-writing (1), we have* 2

0( ) ( ) cot ( )

2 2b t xa y x y t dt f x

or Ly = f (x), using the definition of operator L ... (3)

Now, we define an ‘‘adjoint’’ operator* 2

0( ) ( ) cot .

2 2b t xMg a g x g t dt

... (4)

From (3), we have MLy = Mf

or * 2

0( ) ( ) cot ,

2 2b t xM a y x y t dt Mf

using (2) ... (5)

Let * 2

0( ) ( ) ( ) cot .

2 2b t xg x a y x y t dt

... (6)

* 2

0( ) ( ) ( ) cot .

2 2b xg x a y x y d

* 2

0( ) ( ) ( ) cot .

2 2b tg t a y t y d

... (7)

Using (6), (5) reduces to Mg = Mf

or* * 22

0 0( ) ( ) cot ( ) ( ) cot

2 2 2 2b t x b t xa g x g t dt a f x f t dt

... (8)

[using the definition (4) of operator M]

Let* 2

0( ) ( ) ( ) cot .

2 2b t xF x a f x f t dt

... (9)

Using (9), (8) reduces to* 2

0( ) ( ) cot ( ).

2 2b t xa g x g t dt F x

... (10)

Substituting thevalues of g (x) and g (t) given by (6) and (7) in (10), we have* 2

0( ) ( ) cot .

2 2b t xa a y x y t dt

* *2 2

0 0cot ( ) ( ) cot ( )

2 2 2 2b t x b ta y t y d dt F x



or** 2 22

0 0( ) ( ) cot ( ) cot

2 2 2 2ab t x ab t xa y x y t dt y t dt

* *2 22

2 0 0

1 )cot ( ) cot ( )2 24

t x tb y d dt F x

or22 2

0

1( ) ( ) ( ) ( ),2

a y x b y x y d F x using the Hilbert formula of Art. 8.9.

or2 22 2

0( ) ( ) ( ) ( )

2ba b y x y t dt F x

Assuming that 2 2( ) 0, we havea b

2 2

2 2 2 2 0

1( ) ( ) ( ) ,2 ( )

by x F x y t dta b a b

... (11)

which is Fredholm integral equation of the second kind with separable kernel. To solve (11), wenow proceed by usual method*.

Let2

0( ) .C y t dt

... (12)

From (11),2

2 2 2 21( ) ( ) .

2 ( )b Cy x F x

a b a b

... (13)

From (13),2

2 2 2 21( ) ( ) .

2 ( )b Cy t F t

a b a b

... (14)

Substituting the value of y (t) as given by (14) in (12), we have

22

2 2 2 20

( )2 ( )

F t b CC dta b a b

22 202 2 2 20

1 ( )2 ( )

b CF t dt ta b a b

or22

2 2 2 20

1 2( ) or2 ( )

b CC F t dta b a b

2 2

2 2 2 2 0

11 ( )bC F t dta b a b

or2 2

2 0 0

1 1( ) ( )C F t dt f t dtaa

[ from (9), F (t) = a f (t)] ... (15)

Substituting this value of C in (13), we have2 2

2 2 2 2 0

( )( ) ( )2 ( )

F x by x f t dta b a a b

or* 2

2 2 0

1( ) ( ) ( ) cot2 2b t xy x a f x f t dt

a b

2 2

2 2 0( ) ,

2 ( )b f t dt

a a b

using (9)

* We adopt here the usual method outlined in Art. 4.1 and Art. 4.2 of chapter 4.



or * 2

2 2 2 2 0( ) ( ) ( ) cot

22 ( )a b t xy x f x f t dt

a b a b

2 2

2 2 0( ) ,

2 ( )b y t dt

a a b

... (16)

which is the required solution of given integral equation (1).8.11. SOLUTION OF THE HILBERT-TYPE SINGULAR INTEGRAL EQUATION OF THE

FIRST KIND, NAMELY,

* 2

0( ) ( ) cot .

2 2b t xf x y t dt

... (1)

To solve (1), let us reconsider (1) with the constant b ineorporated in y (t). Thus, we shall

first solve * 2

0

1( ) ( ) cot .2 2

t xf x y t dt ... (2)

Re-writing (2), we have * 2

0

1( ) ( ) cot .2 2

xf x y d ... (3)

Replacing x by t in (3), we have * 2

0

1( ) ( ) cot .2 2

tf t y d ... (4)

Multiplying both sides of (4) by 1 cot

2 2t x

and then integrating both sides w.r.t. ‘t’ from

0 to 2 , we have* * *2 2 2

20 0 0

1 1( ) cot cot ( ) cot2 2 2 24

t x t x tf t dt y d dt

or * 2 2

0 0

1 1( ) cot ( ) ( )2 2 2

t xf t dt y x y d ... (5)

[using Hilbert formula of Art. 8.9.]

Let * 2

0

1( ) – ( ) cot .2 2

t xF x f t dt ... (6)


0

1( ) ( ) ( )2

F x y x y d

or 2

0

1( ) ( ) ( ) ,2

y x F x y t dt

... (7)

which is Fredholm integral-equation of the second kind with separable kernel. We now proceed tosolve (7) by the usual method*.

Let2

0

1 ( ) .2

C y t dt

... (8)

* We shall adopt the method outline in Art. 4.1 and 4.1 of chapter 4



Using (8), (7) reduces to y (x) = F (x) + C .... (9)From (9), y (t) = F (t) + C. ... (10)Substituting the value of y (t) given by (10) in (8), we have

2

0

1 [ ( ) ]2

C F t C dt

or

2 2

0 0

1 ( )2 2

CC F t dt dt

or2

0

1 ( ) 22 2

CC F t dt

or

2

0( ) 0F t dt

or

2

0( ) 0F x dx

... (11)

By virtue of relation (7), it follows that (11) holds for all values of the function f (x), Hence, Cmust be an arbitrary constant and so we find that infinite number of solutions of (2) exist and aregiven by (9), that is,

y (x) = C + F (x) or * 2

0

1( ) ( ) cot ,2 2

t xy x C f t dt by (6) ... (12)

By substituting (12) in (2), we find that f (x) given by (12) satisfies (2), if and only if,

2

0( ) 0,f x dt

... (13)

showing that the necessary and sufficient condition for (2) to possess a solution is that condition(13) must hold good.

Deduction. To find solution of the integral equation (1)


0

1( ) [ ( )] cot2 2

t xf x by t dt

or 2

0

1( ) ( ) cot ,2 2

t xf x Y t dt ... (14)

where Y (t) = b y (t) ... (15)Now, proceed as above from (2) upto (12) after replacing y (t) by Y (t) and obtain

* 2

0

1( ) ( ) cot2 2

t xY x C f t dt ... (16)

But from (15), Y (x) = b y (x). ... (17)

Using (17), (16) reduces to* 2

0

1( ) ( ) cot2 2

t xb y x C f t dt

or * 2

0

1( ) ( ) cot ,2 2

C t xy x f t dtb b

or * 2

0

1( ) ( ) cot ,2 2

t xy x C f t dtb

... (18)

where ( / )C C b is another arbitrary constant. (18) gives the desired solution of (1).Alternative method. Use the result of Art. 8.9, where we have connected the Hilbert kernel

with the Cauchy kernel. To this end, write y (e i t) = y (t), etc. and suppose that y (t) and f (t) are



periodic functions with period 2 . Also, replace f (t) by f (t)/i. Then the formulas (12) and (13) ofArt. 8.8 with help of the transformation (7) of Art. 8.9 give rise to the following reciprocal relations

* 2 2

0 0

1 ( ) cot ( ) ( )2 2 2

t x iy t dt y t dt f x ... (19)

and * 2 2

0 0

1 ( ) cot ( ) – ( )2 2 2

t x if t dt f t dt y x ... (20)

Using the above pair of equations, the solution of (2) can be deduced.From the pair (19) – (20), we can prove that for periodic functions f (x) and y (x), if the

condition 2

0( ) 0f t dt

is satisfied, then we must also have

2

0( ) 0.y t dt

Example : Solve the integral equation* 2

1 0

1( cos sin ) ( ) cos2 2n nn

t xa nx b nx y t dt

Hint. Here the function

1( ) ( cos sin )n nn

f x a nx b nx

is a periodic function with period

2 . Also, we have 2

0( ) 0f t dt

Hence, from the reciprocal pair (19) – (20), it follows that* 2

10

1( ) ( cos sin ) cos2 2n nn

t xy x a nt b nt dt

or

1( ) ( sin cos ),n nn

y x a nx b nx

on simplification.

EXERCISE

1. Solve 22 2 1/32

( ) , 2 4.( )

x y t dtx xx t

2. Solve 42

2 2 1/ 3( ) , 2 4.

( )x

y t dtx xt x

3. Solve the integral equation* 2

0

1( ) sin ( ) cot .2 2

t xy x x y t dt [Merrut 2008]

4. If a, b, c, d are real constants, solve the following integral equations :

(i) 21/ 20

( ) .( )

x y t dtax bxx t

(ii) 2 3

1/ 21

( ) , 1 2.(cos cos )

x y t dta bx cx dx xt x

(iii) 22 3

1/ 2( ) , 1 2.

(cos cos )x

y t dta bx cx dx xx t

5. Solve the integral equation2

12 2 1/ 20 2

( ), 0 ,1 ( )( ), .( )

x f x x bt y t dtf x b xx x t

; where b is a constant



6. Solve the integral equation 12 2 1/ 2

2

( ), 0 ,( )( ), .( )x

f x x ay t dtxf x a xt x

7. Prove that the solution of the integral equation2 2

2 1 2 2 1

0

2( ) ( ) ( ) , 0 1,xxf x t x t y t dt

is2 –1 1 2 2 1 2 2

0( ) ( ) ( ) .

1t dy t u t u f u du

dt

8. Prove that the solution of the integral equation2

2 2 1 2 2 12( ) ( ) ( ) , 0 1,x

xf x t x t y t dt

is2 2 1

2 1 2 2( ) – ( ) ( ) .1 t

t dy t u u t f u dudt


*2 2 22 2

0 0( ) ( ) ( ) ( ) ( ) cot

2 2 2b b t xa b y x y t dt af x f t dt

is

* 2

2 2 2 2 0( ) ( ) ( ) cot

22 ( )a b t xy x f x f t dt

a b a b

2 2

2 2 0( ) ,

2 ( )b f t dt

a a b

proved 2 2( ) 0.a b

10. Show that the solution of the integral equation * – ( – )1

0( ) ( )

–

i weF w f w d ww

!

!!

is given by1/ 2 1/ 2* – ( )1

2 0

1 1–( ) – ( )1–

i ww ew f w F dw w

!"!! !#

! ! "


CHAPTER 9

Integral Transform Methods9.1 INTRODUCTION

The integral transform methods are very convenient in solving integral equations of somespecial forms. Suppose that a relationship of the form

( ) ( , ) ( , ) ( )b b

a ay x x z K z t y t dt dz ... (1)

be known to be valid and that this double integral can be evaluated as an iterated integral. Then,

from (1), it follows that if ( ) ( , ) ( )b

aF x K x t y t dt ... (2)

we also have ( ) ( , ) ( ) .b

ay x x t F t dt ... (3)

Thus, if (2) is regarded as an integral equation in y, a solution is given by (3), whereas if (3)is regarded as an integral equation in F a solution is given by (2). It is conventional to refer to oneof the function as the transform of the second function, and to the second function as an inversetransform of the first. Thus, for example, the Fourier integral

1( ) ( ) .

2

i p x i pty x e e y t dt dp

leads to the reciprocal realtions

1( ) ( )2

i xtF x e y t dt ... (4)

and –

1( ) ( ) .2

i x ty x e F t dt

... (5)

The function F (x) is known as the Fourier transform of y (t) and y (x) is called the inverseFourier transform of F (t).9.2 SOME USEFUL RESULTS ABOUT LAPLACE TRANSFORM*.

(1) Laplace transform. Definition. Given a function F (t) defined for all real 0,t theLaplace transform of F (t) is a function of a new variable p given by

0{ ( )} ( ) ( ) ( ) .ptL F t F p f p e F t dt

... (1)

The Laplace transform of F (t) is said to exist if the integral (1) converges for some value ofp, otherwise it does not exist.

9.1

* For more details please refer ‘‘Advanced Differential equations’’ or “Integral transform” by Dr. M.D.Raisinghania, published by S. Chand & Co., New Delhi.


9.2 Integral Transform Methods

(2) Table of Laplace transform of some elementary functions :

S. No. F (t) { ( )} or ( ) or ( )L F t F p f p

1. 1 1/p, p > 0.

2. tn, n > – 1 11) / , 0.nn p p

3. tn n !/pn + 1, p > 0.(n is positive integer)

4. eat 1/(p – a), p > a.5. sin at a/(p2 + a2), p > 0.6. cos at p/(p2 + a2), p > 0.7. sinh at a/(p2 – a2), p > | a |.8. cosh at p/(p2 – a2), p > | a |.9. J0 (at) 2 21/ ( )p a

10. Jn (at) 2 2 2 2[ ( ) ] / ( )n np a p a p a 11. ( )t a e–ap

12. ( )erf t 1/[ 1]p p

(3) Linearity property of Laplace transforms. If c1 and c2 be constants, then

1 1 2 2 1 1 2 2{ ( ) ( )} { ( )} { ( )}.L c F t c F t c L F t C L F t

(4) First translation (or shifting) theorem.

If L {F (t)} = F (p), then { ( ) } ( ).atL e F t F p a

(5) Unit step function or Heaviside’s unit function. Definition. It is denoted and defined as

0, if( )

1, if .t a

H t at a

Note : L {H (t – a) } = (1/p) × e–ap

(6) Second translation (or shifting) theorem.

If { ( )} ( ),L F t F p then { ( ) ( )} ( ).apL F t a H t a e F p OR

If { ( )} ( )L F t F p and ( ),

( )0,F t a t a

G tt a

then { ( )} ( ).apL G t e F p(7) Change of scale property

If { ( )} ( ),L F t F p then { ( )} (1/ ) ( / ).L F at a F p a (8) Laplace transform of derivatives : (i) { ( )} { ( ) } (0) :L F t p L F t F In particular, if F (0) = 0, then { ( )} { ( ) }L F t p L F t

(ii) 2{ ( )} { ( ) } (0) (0),L F t p L F t p F F and so on.


Integral Transform Methods 9.3

(9) Multiplication by positive integral powers of t.

(i) If { ( )} ( ),L F t F p then { ( )} ( )dL t F t F pdp

(ii) If { ( )} ( ),L F t F p then { ( )} ( 1) ( ).n

n nn

dL t F t F pdp

(10) Division by t.

If { ( )} ( ),L F t F p then ( ) ( ) ,p

F tL F p dpt

provided the integral exists.

(11) Initial value theorem :0

lim ( ) lim ( ).t p

F t p F p

Final value theorem :0

lim ( ) lim ( ).t p

F t p F p

(12) Laplace transform of periodic function. Given that F (t) is a periodic function withperiod a, that is, F (t + na) = F (t), for n = 1, 2, 3, ... Then, we have

0

1{ ( ) } ( ) .1

a appaL F t e F t dt

e

(13) Inverse Laplace transform. Definition. Let { ( )} ( ).L F t F p Then F (t) is called an

inverse Laplace transform of ( ),F p and we write 1( ) { ( )}.F t L F p L–1 is known as the inverseLaplace transformation operator.

(14) Table of inverse Laplace transform of some functions

S. No. ( )F p 1 { ( )}L F p

1 1/p 1

2 1/pn + 1, n > – 1 / 1)nt n

3 1/pn + 1 (n is positive integer) tn/n !

4 1/(p – a) eat

5 1/(p2 + a2) (sin at)/a

6 p/(p2 + a2) cos at

7 1/(p2 – a2) (sinh at)/a

8 p/(p2 – a2) cosh at

9 2 21/ ( )p a J0 (at)

10 2 2 2 2[ ( ) ] / ( )n np a p a p a Jn (at)

11 e–ap ( )t a

12 1/[ ( 1)]p p ( )erf t

(14) Linearity property of inverse Laplace transforms.1 1 1

1 1 2 2 1 1 2 2{ ( ) ( )} { ( )} { ( )}.L c F p C F p c L F p c L F p



(15) Heaviside expansion theorem (or formula). Given that ( )F p and ( )G p are polynomials

in p, the degree of ( )F p being less that of ( ),G p and if 1 2( ) ( ) ( )...( ),nG p p p p

where 1 2, , ... n are distinct constants, real or complex, then 11

( )( ) .( ) ( )

rn tr

r r

FF pL eG p G

(16) First translation (or shifting) theorem for inverse Laplace transform.

If 1 { ( )} ( ),L F p F t then 1 { ( )} ( ).atL F p a e F t (17) Second translation (or shifting) theorem for inverse Laplace transform.

If 1 { ( )} ( ),L F p F t then 1 { ( )} ( ) ( )apL e F p F t a H t a

or 1 ( ),{ ( )}

0,ap F t a t a

L e F pt a

Here H (t – a) is Heaviside’s unit function defined in result (5).(18) Change of scale property for inverse Laplace transform.

If 1 { ( )} ( ),L F p F t then 1 { ( )} (1/ ) ( / ).L F ap a F t a (19) Inverse Laplace transform of derivatives

If 1 { ( )} ( ),L F p F t then 1 ( ) ( 1) ( ),n

n nn

dL F p t F tdp

where n = 1, 2, 3, ...

(20) Inverse Laplace transform of integrals.

If 1{ ( )} ( ),L F p F t then1 ( )( ) .

p

F tL F p dpt

(21) Multiplication by p. Let F (0) = 0 and if 1{ ( )} ( ),L F p F t then 1{ ( )} ( ).L p F p F t

Again, if ( 1)(0) (0) ... (0) 0,nF F F then 1{ ( )} ( ).n

nn

dL p F p F tdt

(22) Division by powers of p. If 1{ ( )} ( ),L F p F t then

(i) 1

0

( ) ( )tF pL F t dt

p

(ii) 1

0 0 0

( ) ... ( ) .t t t n

nF pL F t dt

p

(23) Convolution (or Faltung). Definition. The convolution of F (t) and G (t) is denoted and

defined as0

( ) ( )*t

F G F x G t x dx or0

( ) ( ) .*t

F G F t x G x dx (24) Convolution theorem or Convolution property.

If, 1{ ( )} ( )L F p F t and 1{ ( )} ( ),L G p G t

then 1

0{ ( ) ( )} ( ) ( ) *

tL F p G p F x G t x dx F G

OR

1

0{ ( ) ( )} ( ) ( ) .*

tL F p G p F t x G x dx F G



Note : { } ( ) ( ).*L F G F p G p

i.e.0 0

{ ( ) ( ) } { ( ) ( ) } ( ) ( ).t t

L F x G t x dx L F t x G x dx F p G p (25) Complex inversion formula (or integral) for the Laplace transform.

Let { ( )} ( ),L F t F p then 1 { ( )} ( ) , 0,

i i pti

L F p e F p dp t

where the integration is to be performed along a line p in the complex plane where p = x + iy.The real number is chosen so that p lies to the right of all the singularities.

9.3. SOME SPECIALTYPES OF INTEGRAL EQUATIONS(i) Integro-differential equation. Definition. [Meerut 2008]An integral equation in which various derivatives of the unknown function y (t) can also be

present is said to be an integro-differential equation.For example, the following integral equation is an integro-differential equation.

0( ) ( ) ( ) sin ( ) ( ) .

ty t y t f t t x y x dx

(ii) Integral equation of convolution type. Definition.The integral equation

0( ) ( ) ( ) ( ) ,

ty t f t K t x y x dx

in which the kernel K (t – x) is a function of the difference (t – x) only, is known as integral equationof the convolution type. Using the definition of convolution, we may re-write it as

( ) ( ) ( ) ( ).*y t f t K t y t

9.4. APPLICATION OF LAPLACE TRANSFORM TO DETERMINE THE SOLUTIONSOF VOLTERRA INTEGRAL EQUATIONS WITH CONVOLUTION-TYPE KERNELS.WORKING RULE.(i) Consider the Volterra integral equation of the first kind

0( ) ( ) ( ) ,

tF t K t x Y x dx ... (1)

where the kernel K (t – x) depends only on the difference (t – x).Let { } ( ), { } ( ) and { } ( ). L Y Y p L K K p L F F p ... (2)Applying the Laplace transform to both sides of (1), we get

( ) { ( ) ( )}*L F L K t Y t

or ( ) ( ) ( ),F p K p Y p by the convolution theorem

or ( ) ( ) / ( ).Y p F p K p ... (3)Applying the inverse Laplace transform to both sides of (3), we get

1( ) { ( ) / ( )}.Y t L F p K p ... (4)(ii) Consider the Volterra integral equation of the second kind

0( ) ( ) ( ) ( ) .

tY t F t K t x Y x dx ... (5)

Applying the Laplace transform to both sides of (3), we have { } { } { ( ) ( )}*L y L F L K t Y t



or ( ) ( ) ( ) ( ),Y p F p K p Y p using (2) and the convolution theorem

or ( ) [1 ( )] ( )Y p K p F p or ( )( ) .1 ( )

F pY pK p

... (6)

Applying the inverse Laplace transform to both sides of (6), we have

1 ( )( ) .1 ( )

F pY t LK p

... (7)

(iii) Suppose we want the resolvent kernel of (5) in which the kernel K (t – x) depends onlyon the difference (t – x). Before doing so, we first show that, if the original kernel K (t, x) is adifference kernel, then so is the resolvent kernel.

We know that the resolvent kernel R (t, x) is given by (refer Art 5.11, Chapter 5)

1 21( , ) ( , ) ( , ) ( , ) ...mm

R t x K t x K t x K t x

... (8)

[Note that here 1. So we use symbol R (t, x) is place of usual symbol ( , ; ).R t x ]We know that the iterated kernels are given by (refer Art 5.11, Chapter 5)

K1 (t, x) = K (t, x) ... (9)

and 1( , ) ( , ) ( , ) .t

n nxK t x K t z K z x dz ... (10)

Here, by assumption, we have K (t, x) = K (t – x) by (9), K1 (t, x) = K (t, x) = K (t – x). ... (11)Putting n = 2 in (10), we have

2 1( , ) ( , ) ( , ) ( ) ( ) ,t t

x xK t x K t z K z x dz K t z K z x dz using (11)

0

( ) ( ) ,t x

K t x u K u du

putting u = z – x

showing that K2 (t, x) depends only on the difference (t – x). Proceeding likewise we can show thatK3 (t, x), K4 (t, x), ... also depend only on the difference (t – x). From (8), it now follows that theresolvent kernel will also depend only on the difference (t – x) and so we may write

R (t, x) = R (t – x). ... (12)We know that solution of (5) is given by [Reger Art 11.5, Chapter 5]

0( ) ( ) ( , ) ( )

tY t F t R t x F x dx

or0

( ) ( ) ( ) ( ) ,t

Y t F t R t x F x dx by (12) ... (13)

Let { } ( ), { } ( ) and { } ( ).L Y Y p L F F p L R R p ... (14)Applying the Laplace transform to both sides of (13), we have

{ } { } { ( ) ( )}*L Y L F L R t F t

or ( ) ( ) ( ) ( ),Y p F p R p F p using (14) and the convolution theorem ... (15)

Using (6), (15) reduces to( ) ( ) [1 ( )]

1 ( )F p F p R p

K p



or1 ( )( ) 1 .

1 ( ) 1 ( )K pR p

K p K p

... (16)


1 ( )( )1 ( )

K pR t x LK p

... (17)

Substituting the value of R (t – x) given by (17) in (13) we shall get the desired solution of (5).9.5. SOLVED EXAMPLES BASED ON ARTICLES 9.2 TO 9.4.

Ex. 1. Solve the Abel’s equation 1/30

( ) (1 ).( )

t Y x dx t tt x

Sol. Given 1/30

( ) (1 ).( )

t Y x dx t tt x

or 1/3 2( ) .*Y t t t t ... (1)

[using the definition of convolution]Applying the Laplace transform to both sides of (1), we have

L {Y (t)} L {t–1/3} = L {t} + L {t2}, using convolution theorem.

or(1/3) 1 2 3

( 1/ 3 1) 1 2!{ ( )}L Y tp p p

or 2 / 3 2 3

(2 / 3) 1 2{ ( )}L Y tp p p

or 4 / 3 7 / 31 1 2{ ( )} .

(2 / 3)L Y t

p p

... (2)

Taking inverse Laplace transform of both sides of (2), we get

1 14 /3 7 /3

1 1 1( ) 2(2 / 3)

Y t L Lp p

(4 / 3) 1 (7 / 3) 11 2(2 / 3) (4 / 3) (7 / 3)

t t

1/ 3 4/31 2(2 / 3) (1/ 3) (1/ 3) (4 /3) (1/ 3) (1/ 3)

t t

1/3 1/33 3 3 31 1(2 / 3) (1/ 3) 2 / sin( / 3) 2

t t t t

[ ( ) (1 ) / sinn n n (1/ 3) (2 / 3) / sin( / 3)]

1/ 3(3 3 / 4 ) (2 3 )t t

Ex. 2. Solve the Abel’s integral equation : 2

0

( ) 1 .( )

t Y x dx t tt x

Sol. Re-writing the given equation, we have 1/ 2 2( ) 1 .*Y t t t t

Applying the Laplace transform to both sides of (1) and using the convolution theorem, wehave L { Y (t)} L {t–1/2} = L {1} + L {t} + L {t2}

or 1/ 2 2 3(1/ 2) 1 1 2!{ ( )}L Y t

pp p p

or 1/ 2 3/ 2 5 / 2{ ( )} 1/ 1/ 1/ 2 /L Y t p p p ... (2)


1 1 11/ 2 3/ 2 5 /2

1 1 1 1( ) 2Y t L L Lp p p

1/ 2 1/ 2 3/ 21 2(1/ 2) (3/ 2) (5 / 2)t t t



1/ 2 1/ 2 3/ 21 2

(1/ 2) (3/ 2) (1/ 2)t t t

3/ 21/ 2 1/ 21 82

3tt t

Ex. 3. Solve the Volterra integral equation of the first kind : 0

( ) ( ) 16sin 4 .tY x Y t x dx t

(Kanpur 2007, 09; Meerut 2004, 05, 12)Sol. Given integral equation can be re-written as ( ) ( ) 16 sin 4 .*Y t Y t t ... (1)Applying the Laplace transform to (1) and using convolution theorem, we have

L {Y (t) L {Y (t)} = 16 L {sin 4t}

or 22 2

4 64[ { ( )}] 1616 16

L Y tp p

or 2 28{ ( )} .

( 4 )L Y t

p

... (2)

Taking inverse Laplace transform, (2) gives 102 2

1( ) 8 { } 8 (4 ).( 4 )

Y t L J tp

Ex. 4. Solve the Volterra integral equation of the second kind 2

0( ) ( ) sin ( ) .

tY t t Y u t u du

(Kanpur 2010; Meerut 2008, 2011)

Sol. Re-writing the given integral equation, 2( ) ( ) sin .*Y t t Y t t ... (1)Applying the Laplace transform to (1) and using the convolution theorem, we have

L {Y (t)} = L {t2} + L {Y (t) } L {sin t} or 3 2 22! 1{ ( )} { ( )}

1L Y t L Y t

p p

or 2 31 2{ ( )} 1

1L Y t

p p

or2

2 32{ ( )}

1pL Y t

p p

or 3 5{ ( )} 2 / 2 /L Y t p p ... (2)Inverting, (2) reduces to

1 13 5

1 1( ) 2 2Y t L Lp p

or2 4 4

2( ) 2 2 .2! 4! 12t t tY t t

Ex. 5. Solve the integral equation0

( ) 1 ( ) sin ( )t

Y t Y x t x dx and verify your solution.(Meerut 2007)

Sol. The given integral equation can be re-written as ( ) 1 ( ) sin .*y t Y t t ... (1)Applying the Laplace Transform to both sides of (1), we have

L {Y (t)} = L {1} + L {Y (t) * sin t}or L {Y (t)} = (1/p) + L {Y (t)}. L{sin t}, by the convolution theorem

or 21 1{ ( )} { ( )}

1L Y t L Y t

p p

or 21 1(1 ) { ( )}

1L Y t

pP

or2

21{ ( )}

1p L Y t

pp

or 2

3 31 1 1{ ( )} pL Y t

pp p

... (2)

Inverting, (2) reduces to2 2

1 13

1 1( ) 1 1 .2! 2t tY t L L

p p

... (3)



Verification of solution (3) : We wish to show that the solution (3) satisfies the given integral

equation0

( ) 1 ( ) sin ( ) .t

Y t Y x t x dx ... (4)

From (3), Y (x) = 1 + (x2/2)

R.H.S. of (4) 2

01 (1 / 2)sin ( )

tx t x dx 2

0 01 (1 / 2)cos ( ) cos ( )

ttx t x x t x dx

20 0

1 1 ( / 2) cos {[ sin ( )] sin ( ) }ttt t x t x t x dx

2 2 202 ( / 2) cos cos ( ) 2 ( / 2) cos (1 cos ) 1 ( / 2)tt t t x t t t t = Y (t), by (3)

= L.H.S. of (4),showing that (3) is solution of given integral equation (4).

Ex. 6. Solve: 0

( ) sin 2 ( ) cos( ) .t

Y t a t Y x t x dx Sol. Re-writing the given integral equation, ( ) sin 2 ( ) cos .*Y t a t Y t t ... (1)

Taking the Laplace transform of both sides of (1), we getL {Y (t)} = a L {sin t} – 2L {Y (t) * cos t}

or 2 2{ ( )} 2 { ( )} ,1 1

a pL Y t L Y tp p

by the convolution theorem

or 2 221 { ( )}

1 1p aL Y t

p p

or 2{ ( )}( 1)

aL Y tp

... (2)

Inverting, (2) redues to

1 12 2

1 1( ) ,( 1)

tY t a L a e Lp p

by first shifting theorem

or Y (t) = a t e–t.

Ex. 7. Solve the equation 0

( ) 2 cos ( ) ( )ttY t e t x Y x dx by using Laplace transform

(Meerut 2006, 09)

Sol. Re-writing the given integral equation, ( ) 2 ( ) cos .*tY t e Y t t ... (1)

Applying the Laplace transform to (1) and using the convolution theorem, we have

L {Y (t)} = L {e–t} – 2L {Y (t)} L {cos t} or 2

1{ ( )} 2 { ( )}1 1

pL Y t L Y tp p

or 22 11 { ( )}

11p L Y t

pp

or

2

2( 1) 1{ ( )}

11p L Y t

pp

or 2 2

3 31 [( 1) 1] 1{ ( )}

( 1) ( 1)p pL Y tp p

... (2)



Inverting, (2) yields 2

13

[( 1) 1] 1( )( 1)

pY t Lp

21

3( 1) 1t pe L

p

[by first shifting theorem]

21 1 1

3 2 32 2 1 2 2t p pe L e L

pp p p

1 1 12 3

1 1 12 2te L L Lp p p

= e–t [1 – 2t + 2 × (t2 / 2 !] = e–t (1 – 2t + t2) = e–t (1 – t)2

Ex. 8. Solve : 0

( ) 2 cos ( ) ( )t

Y t t t x Y x dx (Kanpur 2005, 08)

Sol. Re-writing given equation, ( ) 2 ( ) cos .*Y t t Y t t .... (1)Taking Laplace transform of (1) and using the convolution theorem, we get

2 21 2{ ( )} { } 2 { ( )} {cos } { ( )}

1pL Y t L t L Y t L t L Y t

p p

or2 22 11 { ( )

1p L Y t

p p

or

2

2 21{ ( ) .

( 1)pL Y t

p p

... (2)

Inverting, (2) gives

1 1 12 2 2 2 2 2

1 1 1 1( ) .( 1) ( 1) ( 1) ( 1)

Y t L L Lp p p p p p

... (3)

Now,1 1

2 21 1

( 1)tL e L

p p

, by first shifting theorem

= et × t ... (4)Using result (22) of Art. 9.2, we have

1

2 00 0

1 1( 1)

t ttu u uL u e du e u e dup p

0

tt ut e e

( 1) ( 1) 1.t t tt e e e t

1

2 0 0 0 0

1 1 [ ( 1) 1]( 1)

t t t tu u uL e u du ue du e du dup p p

= et (t – 1) + 1 – (et – 1) + t = et (t – 1) + t + 2. ... (5)

Using (4) and (5) in (3), Y (t) = t et + et (t – 2) + t + 2 = 2et (t – 1) + 2 + t.Ex. 9. Solve the integro-differenial equation :

0( ) sin ( )cos ,

tY t t Y t x x dx where Y (0) = 0.

(Kanpur 2009)Sol. Re-writing the given equation, we have ( ) sin ( ) cos .*Y t t Y t t ... (1)Also given that Y (0) = 0. ... (2)Applying the Laplace transform to both sides of (1), we get

{ ( )} {sin } { ( ) cos }*L Y t L t L Y t t

or 21{ ( )} (0) { ( )} {cos },

1pL Y t Y L Y t L t

p

by the convolution theorem

or 2 21{ ( )} { ( )} ,

1 1pp L Y t L Y t

p p

using (2)



or 2 21 1(1 ) { ( )}

1 1pL Y t

p p

or

3

2 21{ ( )}

1 1p L Y t

p p

or L {Y (t)} = 1/p3. ... (3)

Inverting, (3) reduces to Y (t) = L–1 {1/p3} = t2/2 ! = t2 / 2.

Ex. 10. Solve 0

( ) ( ) cos , (0) 4.t

Y t t Y t x x dx Y (Kanpur 2005, 08)

Sol. Re-writing the given integral equation, we have ( ) ( ) cos*Y t t Y t t ... (1)Also, given that Y (0) = 4. ... (2)Applying the Laplace transform to (1) and using the convolution theorem, we have

{ ( )} { } { ( )} {cos }L Y t L t L Y t L t or 2 21{ ( )} (0) { ( )}

1pp L Y t Y L Y t

p p

or 2 21 11 { ( )} 4 ,

1p L Y t

p p

using (2)

or3

2 21{ ( )} 4

1p L Y t

p p

or

2

3 21 1{ ( )} 4pL Y t

p p

or2 2

3 5 3 54( 1) 1 4 5 1{ ( )} .p pL Y t

pp p p p

... (3)

Inverting, (3) yields 1 1 3 1 5( ) 4 {1/ } 5 {1/ } {1/ }Y t L p L p L p or Y (t) = 4 + 5 × (t2 / 2!) + (t4 / 4!) = 4 + (5t2 / 2) + (t4 / 24).


( ) .t t xt e Y x dx

Sol. Re-writing the given integral equation, ( ) .* tt Y t e ... (1)Taking the Laplace transform of both sides of (1) and using the convolution theorem, we get

L {t} = L {Y (t)}. L {et}

or 2

1 1{ ( )}1

L Y tpp

or 21{ ( )} pL Y t

p

or 2

1 1{ ( )L Y tp p

... (2)

Inverting, 1 1 2( ) {1/ } {1/ } 1 .Y t L p L p t

Ex. 12. Solve : 00sin ( ) ( ) .

tt J t x Y x dx (Meerut, 2010; Kanpur 2005)

Sol. Re-writing the given integral equation, 0sin ( ) ( ).*t Y t J t ... (1)Taking the Laplace transform of both sides of (1) and using the convolution theorem, we get

L {sin t} = L {Y (t)}. L {J0 (t)} or 2 21 1{ ( )}

1 ( 1)L Y t

p p

or 2{ ( )} 1/ ( 1)L Y t p ... (2)

Inverting (2), 1 20( ) {1/ ( 1)} ( ).Y t L p J t



Ex. 13. Solve the Able integral equation0

( )( ) , 0 1.( )

t Y xF t dxt x

Sol. Re-writing the given Able integral equation, we have ( ) ( ) .*F t Y t t ... (1)Taking the Laplace transform of both sides of (1) and using the convolution theorem, we get

{ ( )} { ( )} { }L F t L Y t L t

or 1(1 )( ) { ( )} ,F p L Y tp

where ( ) { ( )}F p L F t

or1 ( ){ ( )} { ( ) ( )}

(1 ) ( ) (1 )p F p pL Y t p F p

{ ( ) ( )},( / sin )

p F p F p

as ( ) (1 )sin

1sin { }. { ( )}p L t L F t

1sin { ( )},*

p L t F t

by convolution theorem,

1

0

sin( ) ( ) ,

tpL t x F x dx

by definition of convolution

1

0

sin ( ) ( ) .t

p L t x F x dx ... (2)

Let 1

0( ) ( ) ( )

tG t t x F x dx ... (3)

From (3), G (0) = 0. ... (4)Now, { ( )} { ( )} (0) { ( )},L G t p L G t G p L G t using (4)

{ ( )} ( )dp L G t L G tdt

or1

0( ) ( ) ( ) ,

t dp L t x F x dx L G tdx

by (3) ... (5)

Using (5), (2) reduces tosin{ ( )} ( )dL Y t L G t

dt

Inverting, 1

0

sin sin( ) ( ) ( ) ( ) .td dY t G t t x F x dx

dt dt ... (6)

Note : We have already got the above solution in Art. 8.2 of chapter 8 by a different method.Ex. 14. Find the resolvent kernel of the Volterra integral equation and hence its solution

0

( ) ( ) ( ) ( ) .t

Y t F t t x Y x dx (Kanpur 2006, Meerut 2005)

Sol. Re-writing the given integral equation, ( ) ( ) ( ) .*Y t F t Y t t ... (1)Applying the Laplace transform to both sides of (1) and using the convolution theorem, we getL {Y (t)} = L {F (t)} + L {Y (t)}. L {t} or L {Y (t)} = L {F (t)} + L {Y (t)}× (1/p2)

or 211 { ( )} { ( )}L Y t L F tp

or

2

2{ ( )} { ( )}.1

pL Y t L F tp

... (2)



Let R (t – x) be the resolvent kernel of the given integral equation. Then, we know that therequired solution is given by

0( ) ( ) ( ) ( )

tY t F t R t x F t dt ... (3)

or ( ) ( ) ( ) ( ).*Y t F t R t F t ...(3)

Applying the Laplace transform to both sides of (3) and using the convolution theorem,we have

L {Y (t)} = L {F (t)} + L {R (t). L {F (t)}

or2

2 { ( )} [1 { ( )}] { ( )},1

p L F t L R t L F tp

by (2)

or2

21 { ( )}1

pL R tp

or 2

2{ ( )} 11

pL R tp

2

11p

... (4)

Inverting, R (t) = sinh t so that R (t – x) = sinh (t – x), ... (5)giving the required resolvent kernel,

Substituting the above value of R (t – x) in (3), the required solution is

0( ) ( ) sinh ( ) ( ) .

tY t F t t x F t dt

Ex. 15. Determine the resolvent kernel and hence solve the integral equation

0( ) ( ) ( ) .

t t xY t F t e Y x dx Sol. Re-writing the giving integral equation, ( ) ( ) ( ) .* tY t F t Y t e ... (1)Applying the Laplace transform to both sides of (1) and using the convolution theorem, we getL {Y (t)} = L {F (t)} + L {Y (t)}. L {et} or L {Y (t)} = L {F (t)} + L {Y (t)} x {(1/(p – 1)}

or11 { ( )} { ( )}

1L Y t L F t

p

or 1{ ( )} { ( )}.2

pL Y t L F tp

... (2)

Let R (t – x) be the resolvent kernel of the given integral equation. Then, we know that therequired solution is given by

0( ) ( ) ( ) ( )

tY t F t R t x F t dt ... (3)

or ( ) ( ) ( ) ( ).*Y t F t R t F t ...(3)

Applying the Laplace transform to both sides of (3)and using the convolution theorem, we getL {Y (t)} = L {F (t)} + L {R (t)}. L {F (t)}

or1 { ( )} [1 { ( )}] { ( )},2

p L F t L R t L F tp

by (2)

or 11 { ( )}2

pL R tp

or 1 1{ ( )} 1

2 2pL R tp p

Inverting, 1 2( ) {1/( 2)} tR t L p e so that R (t – x) = e2 (t – x) ... (4)giving the required resolvent kernel.



Substituting the above value of R (x – t) in (3), the required solution is

2( )

0( ) ( ) ( )

t t xY t F t e F t dt Ex. 16. Solve the integral equation 2 2

0( ) ( ) ( ) , 0

sf s K s u y u du s

Sol. To re-write the given equation in standard form, we introduce two new variables t and xas follows :

s = t1/2 and u = x1/2 so that ds = (1/2) × t–1/2 dt and du = (1/2) × x–1/2dx ... (1)

and obtain 1/ 2 1/ 2 1/ 2

0( ) { ( ) ( ) (1/ 2) }

tf t K t x y x x dx ... (2)

Let f (t1/2) = F (t) and (1/2) × x–1/2 × y (x1/2) = Y (x) ... (3)Using (3), (2) reduces to standard form

0

( ) ( ) ( ) , 0t

F t K t x Y x dx t where the kernel K (t – x) depends only on the difference (t – x). Re-writting it, we get

( ) ( ) ( ),*F t K t Y t by definition of convolution ... (4)

Let { ( )} ( ), { ( )} ( ) and { ( )} ( )L Y t Y p L K t K p L F t F p Applying the Laplace transform to both sides of (4), we get

{ ( )} { ( ) ( )}*L F t L K t Y t or L { F (t)} = L {K (t)} L {Y (t)}, by the convolution theorem

or( )( ) ( ) ( ) or ( )( )

p F pF p K p Y p Y pp K p

... (5)

Let ( ) 1/ ( )H p p K p ... (6)

and let 1{ ( )} ( )L H p H t ... (7)

Using (6), (5) reduces to ( ) ( ) ( )Y p p H p F p

or0

( ) ( ) ( ) ,t

Y p p L H t x F x dx by result (24) of Art. 9.2

Thus,0

( ) . ( ) ( ) ,tdY p L H t x F x dx

dt by result 8 of Art 9.2

which by inversion yields the solution in terms of t and x

0( ) ( ) ( )

tdY t H t x F x dxdt

... (8)

In order to get the solution in terms of the old variables s and u, we use (1) and (3) and get

1/ 2 2 2 1/ 21/ 2 0

1 1( ) { ( ) ( ) 2 }22

sdy t H s u f x u dus dst

or 2 2

0

( ) 1 ( ) ( ) ,2

sy s d u H s u f u dus s ds

by (3)

or 2 2

0( ) 2 ( ) ( ) ,

sdy s u f u H s u duds

... (9)

which is the required solution of the given integral equation.



Solution of the given equation for particular values of the general kernel K (s2 – u2).

Particular Case I. Let ( ) , 0 1.K t t Then the given integral equation takes the form

2 20

( )( ) , 0 1.( )

s y uf s dus u

... (10)

Here 1( ) { ( )} { } (1 ) /K p L K t L t p

11 1( )

( ) (1 ) (1 )p pH p

p K p p

1

1 11 1 1( ) { ( )}(1 ) (1 ) ( )

tH t L H p Lp

or1

( ) ,( / sin )

tH t

as ( ) (1 )sin

1( ) (sin ) /H t t so that H (s2 – u2) = 2 2 1(sin ) / ( )s u ... (11)Substituting the value of H (s2 – u2) given by (11) in (9), the required solution of (10) is

2 2 10

2sin ( )( )( )

sd u f uy s duds s u

... (12)

which has already been obtained by using different method in Ex. 3 on page 8.7.Particular Case II. Let K (t) = t–1/2 cos ( t1/2), where is a constant. Proceed as usual and

verify that 21/ 2 1/ 2 / 4( ) pK p p e

and 21 1 1/ 2 1/ 2 / 4 1 1/ 2 1/ 2( ) { ( )} { } cosh( )pH t L H p L p e t t Hence we find that solution of the integral equation

2 2 1/ 2

2 2 1/ 20

cos[ ( ) ]( ) ( ) , 0( )

s s uf s y u du ss u

... (13)

is given by2 2 1/ 2

2 2 1/ 20

2 cosh[ ( ) ]( ) ( )( )

sd s uy s u f u duds s u

... (14)

Here (13) and (14) are valid for 0 s

Ex. 17. Solve the integral equation 00

( ) ( ) ( ) ( )t

Y t F t J t x Y x dx Solution. Given 0

0( ) ( ) ( ) ( )

tY t F t J t x Y x dx ... (1)

Here we shall use the method explained in Art. 10.4 (ii) for solving

0( ) ( ) ( ) ( )

tY t F t K t x Y x dx ... (2)

Comparing (2) with (1), we have

0( ) ( )K t J t so that 2 1/ 2( ) { ( )} /(1 ) .K p L K t p ... (3)



We know that solution of (2) is given by

0( ) ( ) ( ) ( )

tY t F t R t x F x dx ... (4)

where R (t – x) is resolvent kernel of (2).(4) may be written as ( ) ( ) ( ) ( )}.*Y t F t R t F t Applying the Laplace transform to both sides, we get

{ ( )} { ( )} { ( ) ( )}*L Y t L F t L R t F t

or ( ) ( ) ( ) ( )Y p F p R p F p ... (5)

Here ( ) { ( )}, ( ) { ( )}, ( ) { ( )} and ( ) { ( )}Y p L Y t F p L F t K p L K t R p L R t

Similarly, (2) yields ( ) ( ) ( ) ( )Y p F p K p Y p

so that ( ) ( ) /(1 ( ))Y p F p K p ... (6)

From (5) and (6), ( ) /(1 ( )) ( ) ( ) ( )F p K p F p R p F p

or11 ( )

1 ( )R p

K p

so that

( )( )1 ( )

K pR pK p

... (7)

From (3) and (7), 2 1/ 2

2 1/ 2 2 1/ 2/(1 )( )

1 /(1 ) (1 )pR p

p p

1 1 2 1/ 2( ) { ( )} [ /{(1 ) }]R t L R p L p

2 1/ 2 12 1/ 2 0

( ){sin (1 ) } ( )(1 )

t J us u duu

2

2 1/2 2 1/22 1/2{[cos (1 ) ]} sin[(1 ) ]

(1 )t t

... (8)

Replacing t by t – x in (8), we get R (t – x) and then solution of (1) can be easily obtainedEx. 18. Solve the inhomogeneous Abel integral equation

0

( )( ) ( ) , 0 1( )

t Y xY t F t dxt x

Hint. Proceed as in Ex. 17 by a similar method. Here ( )K t t and hence

1( ) { ( )} (1 )K p L K t p

and 1 1( ) (1 ) /[1 (1 ) ]R p p p

so that1

1

1

[ (1 ) ]( ) { ( )}[ (1 )]

n

n

tR t L R pt n

Hence the required solution is given by0

( ) ( ) ( ) ( )t

Y t F t R t x F x dx

i.e.,1

10

[ (1 ) ( ) ]( ) ( ) ( )

( ) [ (1 )]

nt n

t xY t F t F x dx

t x n



9.6. SOME USEFUL RESULTS ABOUT FOURIER TRANSFORMS.(1) The (complex) Fourier transform. Definition. Given a function Y (x) defined for all x

in the interval ,x the Fourier transform of Y (x) is a function of a new variable p given by

{ ( )} ( ) ( ) .ipxF Y x Y p e Y x dx

... (1A)

The function Y (x) is then called inverse Fourier transform of F{Y (x)} or ( )Y p and is written

as 1 1( ) { ( ( )} or ( ) { ( )},Y x F F Y x Y x F Y p and is given by

1 1( ) { ( )} ( )2

ipxY x F Y p e Y p dp

... (2A)

Remark 1. Some authors also define (1A) and (2A) in the following manner :

{ ( )} ( ) ( ) .ipxF Y x Y p e Y x dx

... (1B)

and 1 1( ) { ( )} ( ) .2

ipxY x F Y p e Y p dp

... (2B)

Remark 2. Some authors define (1A) and (2A) in the so called symmetric form as follows.1{ ( ) ( ) ( )2

ipxF Y x Y p e Y x dx

... (1C)

and1 1( ) { ( )} ( ) .

2ipxY x F Y p e Y p dp

... (2C)

(2) The (inifinite) Fourier sine transform. Definition. The Fourier sine transform of Y (x),0 x is denoted and defined as follows :

0{ ( )} ( ) ( ) sin .s sF Y x Y p Y x px dx

... (1A)

Then, the corresponding inversion formula is given by

1

0

2( ) { ( )} ( ) sin .s s sY x F Y p Y p px dp

... (2A)

Remark. Some authors define (1A) and (2A) in the so called symmetric form :

1/ 2

0{ ( )} ( ) (2 / ) ( ) sins sF Y x Y p Y x px dx

... (1B)

and 1 1/ 2

0( ) { ( )} (2 / ) ( ) sin .s s sY x F Y p Y p px dp

... (2B)

(3) The (infinite) Fourier cosine transform. Definition. The Fourier cosine transform ofY (x), 0 x is denoted as defined as follows :

0{ ( )} ( ) ( ) cos .c cF Y x Y p Y x px dx

... (1A)

Then, the corresponding inversion formula is given by

1

0

2( ) { ( )} ( ) cos .c cY x F Y p Y p px dp

... (2A)



Remark. Some authors define (1A) and (2A) in the so called symmetric form :

1/ 2

0{ ( )} ( ) (2 / ) ( ) cosc cF Y x Y p Y x px dx

... (1B)

and 1 1/ 2

0( ) { ( )} (2 / ) ( ) cos .c c cY x F Y p Y p px dp

... (2B)

(4) Linearity property of Fourier transforms :(i) F {c1 Y1 (x) + c2 Y2 (x)} = c1 F {Y1 (x)} + c2 F {Y2 (x)}.

(ii) Fs {c1 Y1 (x) + c2 Y2 (x)} = c1 Fs {Y1 (x)} + c2 Fs {Y2 (x)}.(iii) Fc {c1 Y1 (x) + c2 Y1 (x)} = c1 Fc {Y1 (x)} + c2 Fc {Y2 (x)}.

(5) Change of scale property.

(i) If { ( )} ( ),F Y x Y p then { ( )} (1/ ) ( / ).F Y ax a Y p a

(ii) If { ( )} ( ),s sF Y x Y p then { ( )} (1/ ) ( / ).s sF Y ax a Y p a

(iii) If { ( )} ( ),c cF Y x Y p then { ( )} (1/ ) ( / ).c cF Y ax a Y p a

(6) Convolution or Faltung. The convolution of two functions G (x) and H (x), where,x is denoted and defined as

( ) ( )*G H G x H t x dx

or ( ) ( ) .*G H G t x H x dx

(7) Convolution theorem or convolution property. The Fourier transform of the convolutionof G (x) and H (x) is the product of the Fourier transforms of G (x) and H (x), i.e.,

{ } { ( )}. { ( )}.*F G H F G x F H x

(8) Shifting property. If ( )Y p is the complex Fourier transform of Y(x), then complex

Fourier transform of Y (x – a) is ( ).ipae Y p

9.7. APPLICATION OF FOURIER TRANSFORM TO DETERMINE THE SOLU-TIONS OF INTEGRAL EQUATIONS :

The whole procedure will be clear from the following examples.


1 , 0 1( ) cos

0, 1.p p

F x px dxp

by using Fourier transform. (Meerut 2006, 08, 09, 11; Kanpur 2005, 09)

Sol. Let1 , 0 1

( )0, 1.c

p pF p

p

... (1)

Then the given integral equation can be re-written as 0

( ) ( ) cos .cF p F x px dx

... (2)

By definition of Fourier cosine transform, we see that ( )cF p is Fourier cosine transform ofF (x). Hence, using the corresponding inversion formula, we have

0

2( ) ( ) coscF x F p px dp

1

0 1

2 ( ) cos ( ) cosc cF p px dp F p px dp



1

0 1

2 (1 ) cos (0) cos ,p px dp px dp by (1)

1 11

00 0

2 sin sin 2{[(1 ) ] ( 1) } sinpx pxp dp px dpx x x

10 2 2

2 cos 2 2 (1 cos )[ ] ( cos 1) .px xxx x x x

Ex. 2. Solve the integral equation :0

1, 0 1( ) sin 2, 1 2.

0, 2.

pF x xp dx p

p

(Meerut 2005)

Sol. Let1, 0 1

( ) 2, 1 2.0, 2.

s

pF p p

p

... (1)

Then the given integral equation can be re-written as 0

( ) ( ) sin .sF p F x px dx

... (2)

By definition of Fourier sine transform, we see that ( )sF p is Fourier sine transform of F (x).Hence, using the corresponding inversion formula, we have

0

2( ) ( ) sinsF x F p px dp

1 2

0 1 2

2 ( )sin ( ) sin ( ) sins s sF p px dp F p px dp F p px dp

1 2

0 1 2

2 sin (2) sin (0) sin ,px dp px dp px dp using (1)

1 2

0 1

cos cos2 2 cos 1 cos 2 cos2 2px px x x x

x x x x

(2 / ) (1 cos 2cos 2 ).x x x

Ex. 3. Solve : 0

( ) cos .pF x px dx e

[Meerut 2010, 12; Kanpur 2007, 08, 10]

Sol. By definition of the Fourier cosine transform, we see e–p is Fourier cosine transform of F(x) Hence, using the corresponding inversion formula, we have

0 0

2 2( ) ( ) cos cospcF x F p px dp e px dp

[ Here ( ) { ( )}]pce F p F F x

20

2 ( cos sin )1

pe px x pxx

as 2 2

( cos sin )cosax

ax e a bx b bxe bx dxa b

Thus, 2( ) 2 / (1 ).F x x

9.8. HILBERT TRANSFORM.The finite Hilbert transform of a function y() is defined as

*

0

1 sin( ) ( )cos cos

f y d

... (1)



with the inverse*

0 0

1 sin 1( ) ( ) ( )cos cos

y f d y d

... (2)

Deduction of various forms of the Hilbert transform pairs from equations (1) and (2)First alternative form of Hilbert transform pairUsing the well known principle of mathematical induction, we can easily prove that

*

0

cos sincos cos sin

n d n

... (3)

It is to be noted that for n = 0 and n = 1, (3) can be easily verified.Now, (1) and (2) ( ) ( )f f and ( ) ( )y y ... (4)

Let us assume that ( ) ( )F f so that ( ) ( )F F ... (5)

0

1 ( ) ,y d C

where C = constant. ... (6)

and ( ) ( )Y y C so that ( ) ( )Y y C ... (7)From (7), it follows that

0 0 0

1 1 1( ) ( )Y d y d C d

= C – C = 0, using (6) ... (8)

Now,* *

0 0

1 sin 1 sin( ) ( ( ) ) ,cos cos cos cos

Y d y C d

by (7)

* * *

0 0 0

1 sin ( ) sin sin( ) ,cos cos cos cos cos cos

y d C d C df

by (1)

( ),F by above relations

Thus,*

0

1 sin( ) ( )cos cos

F Y d

... (9)

Replacing by ( ) / 2 ( ) / 2 in (9), we get

* *

0 0

1 1( ) cot ( ) cot ( )2 2 2 2

F Y d Y d

... (10)

Replace by in the first integral on R.H.S. of (10) and use the relation (7). Then, oncombining the resulting integral with the second integral on the R.H.S. of (10), we obtain

*1( ) cot ( ) .2 2

F Y d

... (11)

Likewise, now start with (2) (in place of (1) and proceed exactly as before, Then, we shall get* *

0

1 1( ) cot ( ) cot ( )2 2 2 2

Y F d F d

... (12)

Replace by in the second integral on the R.H.S. of (12) and use the relation (5). Thenon combining with the first integral on the R.H.S. of (12), we have

*1( ) cot ( )2 2

Y F d

... (13)

The relations (11) and (13) give us the first alternative form of finite Hilbert transform pair.



Second alternative form of Hilbert transform pairRe-writing (10), we have

* * *

0 0

1 1 1( ) 1 cot ( ) 1 cot ( ) ( )2 2 2 2

F Y d Y d Y d

or * *

0 0

1 1( ) 1 cot ( ) 1 cot ( ) ,2 2 2 2

F Y d Y d

by (8)

Combining the first integral with the second integral by replacing by and using (7), theabove equation reduces to

*1( ) 1 cot ( )2 2

F Y d

... (14)

Starting with (12) and proceeding as before, we get*1( ) 1 cot ( )

2 2Y F d

... (15)

It may noticed that the transform pair (14) – (15) is exactly the reciprocal pair of Hilbert typesingular integral equations already discussed in Art. 8.10 of chapter 8 except for a trivial adjustmentof the symbols and the range of integration. The relations (14) and (15) give us a second alternativeform of finite Hilbert transform pair.

Third alternative form of Hilbert transform pairThis transform pair, which is nonsingular, is deduced from the pair (1) - (2) by introducing

new variables u and v as follows :

cos , cos .u v 1

2 1/ 2( ) (cos )( )

sin (1 )f f up u

u

and1

2 1/ 2( ) (cos )( ) ,

sin (1 )y y uq u

u

Then (1) takes the form*

0

( ) 1 1 ( ) sinsin cos cos sinf y d

or*1

1

1 ( )( ) , 1 1q v dvp u uu v

... (16)

(6) is also known as airfoil equation.Similarly, (2) takes the form

1/ 22*1

2 2 1/ 21

1 1 ( )( ) , 1 11 (1 )

v p v Cq u dv uv uu u

... (17)

where1

1

1 ( ) ,C q v dv

which is an arbitrary constant.

It may be noted that the above pair (16) – (17) is a particular case of the pair of integralequations (24) - (25) of Art. 8.8 of chapter 8. The relations (16) and (17) give us a third alternativeform of finite Hilbert transform pair.9.9. INFINITE HILBERT TRANSFORM. DEFINITION

The infinite Hilbert transform in defined as*1 ( )( ) y tf x dt

t x

and its inverse is defined by*1 ( )( ) f ty x dt

t x



Example 1. Solved the homogeneous integral equation*1

1

( ) 0q v dvu v

Sol. Using the third alternative form of Hilbert transform pair (16) - (17), we see that here p (u)

= 0 and hence the required solution from (17) is given by 2 1/ 2( ) /(1 ) .q u C u

EXERCISE

1. Solve 0

( ) 1 ( ) ( ) .t

Y t t x Y x dx (Meerut 2012) Ans. Y (x) = cos x

2. Solve the integal equation 2 2 1/2( )( )

( )t

Y xF t t dxx t


1

2 2 1/ 2( )( ) 2

( )t

x Y x dxF tx t

is

1

2 2 1/ 21 ( )( )

( )t

d x F x dxY tt dt x t

Find the solution for the following two special cases : (i) F (t) = 2t2/(1 – t2)1/2 (ii) F (t) = t2

4. Solve the Abel integral equation of the second kind 1/ 2 / 41/ 20

( )( )( )

ta t i Y x dxY t t et x

5. Solve the integral equation 10( ) ( ) ( ) ( )

tY t F t J t x Y x dx

6. Find the resolvent kernel of the integral equation 2 2

0( ) ( ) ( ) ( )

tY t F t t x Y x dx

Hint : Use the method of Art. 9.4.7. Use the infinite Hilbert transform pair and solve the integral equation

*

21 ( )

1Y t dtx tx

8. With the help of finite Hilbert transform, solve

22 2

2 ( ) ,l

l

tY tx dxx t

assuming that Y (t) = – Y (–t)

9. With the help of finite Hilbert transform, solve

0 0

( )(log | |) log | |l

lY t dtax b l x x

t x

subject to the conditions : 0 0(0) , ( ) , (0) , ( )l lY Y l Y Y Y l Y

10. Solve the integral equation *1 ( )sin Y tx d t

t x

[Hint : Let us consider the following integral :

* i te dt ix t

(residues of the poles on the t-axis)

* (cos sin ) (cos sin )t i t dt i x i x

x t



Separating the real parts, we have1 cos sint dt x

x t

... (i)

Re-writing the given integral equation,1 ( ) sinY t dt x

x t

... (ii)

Comparing (i) and (ii), the required solution is given by Y (t) = cos t

11. Solve the integral equation *

0 0

1 ( ) ,lr

rr

Y t dta xt x

where ar are given constants.

Hint. Use the substitutions : ( / 2) (1 cos ), ( / 2) (1 cos )x l t l

9.10. *MELLIN TRANSFORM. DEFINITION.

Given a function Y (x), defined for 0 < x < , the Mellin transform of Y (x) is a function ofnew variable p and is denoted and defined by

1

0[ ( ); ] ( ) ( )pM Y x p y p x Y x dx

Clearly, if C is a constant, then M [ C Y (x) ; p] = C M [Y (x) : p]Inverse Mellin transform. If y (p) be the Mellin transform of Y (x), then Y (x) is called the

inverse Mellin transform of y (p) and then we write 1{ ( )} ( ).M y p Y x

Convolution theorem for Mellin transform. Let k(p) and y (p) be the Mellin transforms ofK (x) and Y (x) respectively. Then

0( ) ( ) ( ) (1 ).M K xt Y t dt k p y p

... (1)

Proof L.H.S. of (1)1

0 0( ) ( ) ,px K x t Y t dt dx

by definition of Mellin transform

1

0 0( ) ( )

pu duK u Y t dtt t

[on taking new variable u such that

x t = u so that du = t dx and so x = u/t and dx = (1/t)du]

1 (1 ) 1

0 0( ) ( )p pu K u du t Y t dt

= k (p) y (1 – p), by definition of Mellin transform= R.H.S. of (1)

Hence the result9.11 SOLUTION OF FOX’S INTEGRAL EQUATION, NAMELY

0( ) ( ) ( ) ( ) , 0 .Y x F x K x t Y t dt x

... (1)

containing a special type of kernel K (x t).Let y (p), f (p) and k (p) be Mellin transforms of Y (x), F (x) and K (x) respectively. Again,

from result (1) of Art. 9.10, we have

0( ) ( ) ( ) (1 )M K x t Y t dt k p y p

... (2)

* For details, please refer Author’s Integral transform, published by S.Chand & Co., New Delhi



Taking the Mellin transform of both sides of (1) and using (2), we havey (p) = f (p) + k (p) y (1 – p) ... (3)

Replacing p by 1 – p in (3), we havey (1 – p) = f (1 – p) + k (1 – p) y (p) ... (4)

From (3), y (1 – p) = {y (p) – f (p)}/k (p) ... (5)Substituting the value of y (1 – p) given by (5) in (4), we get

{y (p) – f (p)} / k (p) = f (1 – p) + k (1 – p) y (p)

or1 ( )( ) (1 ) (1 )( ) ( )

f py p k p f pk p k p

so that( ) ( ) (1 )( )1 ( ) (1 )

f p k p f py pk p k p

... (6)

from which we will get the reqired solution of (1) provided we can compute Y (x) from its Mellintransform y (p).

Solved example. Solve the Fox’s integral equation

0( ) ( ) ( ) ( ) , 0Y x F x K x t Y t dt x

... (1)

with the kernel 1/ 2( ) (2 / ) sinK x x (Kanpur 2005) ... (2)

Solution. Let y (p), f (p) and k (p) be Mellin transform of Y (x), F (x) and K (x) respectively.Then, solution of (1) can be obtained from the equation (refer eq. (equation) of Art. 9.11)

( ) ( ) (1 )( )1 ( ) (1 )

f p k p f py pk p k p

... (3)

By definition, 1 1

0 0

2( ) ( ) sin ,p pk p x K x dx x x dx

by (2)

or2( ) ( )sin

2pk p p

... (4)

1

0

*[ [sin ] sin ( )sin ( / 2)]pM x x x dx p p

From (4)2 (1 ) 2(1 ) (1 ) sin (1 )sin

2 2 2p pk p p p

so that2(1 ) (1 ) cos

2pk p p

... (5)

22( ) (1 ) ( ) (1 ) sin cos ,

2 2p pk p k p p p

by (4) and (5)

2

sin ,sin

pp

as ( ) (1 )sin

p pp

Thus, 2( ) (1 )k p k p ... (6)

* Refer Mellin transform in Author’s Integral transform, published by S.Chand & Co. New Delhi




2 2( ) 2( ) ( ) sin (1 )

21 1f p py p p f p

... (7)

Taking inverse Mellin transform of both sides of (7), we have,

12 2

( ) 2( ) ( ) sin (1 )21 1

F x pY x M p f p ... (8)

We can easily verify that

0sin( ) ( ) ( ) sin (1 )

2pM x t F t dt p f p

1

0( )sin (1 ) sin( ) ( )

2pM p f p xt F t dt

... (9)

Using (9), (8) reduces to 2 2 0

( ) 2( ) sin ( ) ( ) , 01 1F xY x x t F t dt x

which is the required solution of the given equation (1).

Miscellaneous problems on Chapter 9

1. Solve the integral equation by using Laplace transform 0

( ) 1 ( ) ( ) .x

x x t t dt (Kanpur 2009)

2. Solve the Abel’s equation 0

( )1

x tdt x

x t

. (Kanpur 2009,10)


CHAPTER 10

Self Adjoint Operator,Dirac Delta Function and

Spherical Harmonics10.1 INTRODUCTION

In this chapter we propose to introduce some important concepts which will be used in thesubsequent chapters. We shall often refer to the definitions and associated results of self adjointoperator, Dirac delta function and spherical harmonics.10.2.ADJOINT EQUATION OF SECOND ORDER LINEAR DIFFERENTIAL EQUATION

Consider the second order homogeneous linear differential equationa0 (x) (d2y/dx2) + a1 (x) (dy/dx) + a2 (x) y = 0, ... (1)

where a0 (x) has a continuous second order derivative, a1 (x) has a continuous first order derivative,a2 (x) is continuous and a0 (x) > 0 on .a x b

Let L be a differential operator defined by2 2

0 1 2( ) ( / ) ( ) ( / ) ( )L a x d dx a x d dx a x ... (2)

Then (1) can be re-written as ( ) 0,L y x a x b ... (3)and the adjoint operator M of L is defined as

2

0 1 22( ) { ( ) ( )} { ( ) ( )} ( ) ( )d dM y x a x y x a x y x a x y xdxdx

... (4)

Also, M y (x) = 0, i.e., 2

0 1 22 { ( ) ( )} { ( ) ( )} ( ) ( )d da x y x a x y x a x y xdxdx

= 0 ... (5)

is known as a adjoint of (1).10.3.SELF ADJOINT EQUATION.

If the adjoint of any linear homogeneous equation is identical with the equation itself, thenthe given equation is known as self adjoint equation.

Theorem I.The necessary and sufficient condition that the second order homogeneous lineardifferential equation a0 (x) (d2y/dx2) + a1 (x) (dy/dx) + a2 (x) y = 0, where a0 (x) is continuouslydifferentiable positive function and a1(x) has a continuous first order derivative on a x b to beself adjoint is that 0 1( ) ( )a x a x on ,a x b where prime denotes differentiation w.r.t. ‘x’.

Proof. By definition, the adjoint equation ofa0 (x) (d2y/dx2) + a1(x) (dy/dx) + a2 (x) y = 0 ... (1)

10.1


10.2 Self Adjoint Operator, Dirac Delta Function and Spherical Harmonics

is2

0 1 22 [ ( ) ] [ ( ) ] ( ) 0d y da x y a x y a x ydxdx

... (2a)

i.e., 20 0 1 0 1 2( ) ( / ) {2 ( ) ( )} ( / ) { ( ) ( ) ( )} 0,a x d y dx a x a x dy dx a x a x a x y ... (2b)

where prime denotes differentiation with respect to x.The condition is necessary. Let (1) be a self adjoint equation. Then (2b) must be identical

with (1) and hence we must have 0 1 12 ( ) ( ) ( )a x a x a x ... (3)

and 0 1 2 2( ) ( ) ( ) ( )a x a x a x a x ... (4)

From (4), 0 1( ) ( )a x a x so that 0 1( ) ( ) ,a x a x C where C is a constant. ... (5)

Substituting the value of 0 ( )a x as given by (5) in (3), we have2 {a1 (x) + C} – a1 (x) = a1 (x) so that C = 0.Hence (5) yields 0 1( ) ( )a x a x Hence the condition is necessary..

The condition is sufficient. Suppose that for (1), we have 0 1( ) ( )a x a x ... (6)

Then, 0 1 1 1 12 ( ) ( ) 2 ( ) ( ) ( ),a x a x a x a x a x by (6) ... (7)

and 0 1 2 1 1 2 2( ) ( ) ( ) ( ) ( ) ( ) ( ),a x a x a x a x a x a x a x by (6) ... (8)Using (7) and (8), (2b) reduces to

a0 (x) (d2y/dx2) + a1 (x) (dy/dx) + a2 (x) y = 0which is identical with (1). Hence the condition is sufficient.

Corollary. If a0 (x) (d2y/dx2) + a1 (x) (dy/dx) + a2 (x)y = 0 is self adjoint, then it can be

re-written as 0 2( ) ( ) 0.d dya x a x ydx dx

Proof. Given that a0 (x) (d2y/dx2) + a1(x) (dy/dx) + a2 (x)y = 0 ... (1)is self adjoint equation. For this, the necessary condition is 0 1( ) ( ).a x a x ... (2)

Substituting the value of a1(x) from (2) in (1), we obtain2

0 0 22( ) ( ) ( ) 0,d y dya x a x a x ydxdx

i.e., 0 2( ) ( ) 0d dya x a x ydx dx

Theorem II. If the coefficients a0 (x) , a1(x), a2(x) in the equation a0(x) (d2y/dx2)+ a1(x) (dy/dx) + a2 (x) y = 0 are continuous on a x b and 0 ( ) 0,a x then it can be trans-

formed into the equivalent self adjoint equation ( ) ( ) 0,d dyp x q x ydx dx

where1

0

( )( ) exp

( )a x

p x dxa x

and2 1

1 0

( ) ( )( ) exp

( ) ( )a x a x

q x dxa x a x

Here exp (a) stands for ea

Proof. Given a0(x) (d2y/dx2) + a1(x) (dy/dx) + a2 (x) y = 0 ... (1)

Multiplying both sides of (1) by 1

0 0

( )1 exp{ },( ) ( )

a x dxa x a x we get

21 1 1 2 1

20 0 0 0 0

( ) ( ) ( ) ( ) ( )exp{ } exp{ } exp{ } 0( ) ( ) ( ) ( ) ( )

a x a x a x a x a xd y dydx dx dy ya x a x a x dx a x a xdx

... (2)


Self Adjoint Operator, Dirac Delta Function and Spherical Harmonics 10.3

Comparing (2) with 2 20 1 2( )( / ) ( )( / ) ( ) 0,A x d y dx A x dy dx A x y we have ... (3)

10

0

( )( ) exp

( )a x

A x dxa x

and1 1

10 0

( ) ( )( ) exp

( ) ( )a x a x

A x dxa x a x

Then, we have 0 1( ) ( ),A x A x

which is the condition for (3) to be a self adjoint equation. Hence (2) is a self adjoint equation andit can be re-written as

1 2 1

0 0 0

( ) ( ) ( )exp exp 0

( ) ( ) ( )a x a x a xdyd dx dx y

dx a x dx a x a x

or ( ) ( ) 0,

dyd p x q x ydx dx

where1

0

( )( ) exp

( )a x

p x dxa x

and2 1

0 0

( ) ( )( ) exp

( ) ( )a x a x

q x dxa x a x

Theorem III. If M is the adjoint operator of operator L, then L is the adjoint of M, i.e.,

the adjoint of the adjoint operator is the operator itself.Proof. Left as an exercise for the reader.

10.4 SOLVED EXAMPLES BASED ON ART. 10.2 AND 10.3Ex. 1. Find the adjoint equation of x2 (d2y/dx2) + (2x3 + 1) (dy/dx) + y = 0.Sol. Given x2 (d2y/dx2) + (2x3 + 1) (dy/dx) + y = 0 ... (1)

Comparing (1) with a0(x) (d2y/dx2) + a1 (x) (dy/dx) + a2(x) y = 0,we have a0(x) = x2, a1(x) = 2x3 + 1 and a2(x) = 1 ... (2)

Now, the required adjoint equation is of the form

2

0 1 22 [ ( ) ] [ ( ) ] ( ) 0d da x y a x y a x ydxdx

or 2

2 32 ( ) {(2 1) } 0d dx y x y y

dxdx

or 2 3( ) 6 (2 1) 0d d dyx y xy x ydx dx dx

or 2 2 32 6 (2 1) 0d dy dyxy x x y x y

dx dx dx

or 2 2 2 2 32 2 ( / ) 2 ( / ) ( / ) 6 (2 1) ( / ) 0y x dy dx x dy dx x d y dx x y x dy dx y or x2 (d2y/dx2) + (4x – 2x3 – 1) (dy/dx) + 3y (1 – 2x2) = 0

Ex. 2. Show that x2 (d2y/dx2) – 2x (dy/dx) + 2y = 0 is not a self adjoint equation. Transform itinto an equivalent self adjoint equation.

Sol. Given x2 (d2y/dx2) – 2x (dy/dx) + 2y = 0 ... (1)Comparing (1) with a0(x) (d2y/dx2) +a1(x) (dy/dx) + a2(x) y = 0, ... (2)

here a0(x) = x2, a1(x) = –2x and a2(x) = 2 ... (3)

Since 0 1( ) 2 ( ),a x x a x it follows that (1) is not a self adjoint equation.We know that (2) can be transformed into an equivalent self adjoint equation by multiplying

its both sides by a factor

1

0 0

( )1 exp( ) ( )

a xdx

a x a x i.e., 2 2

1 2exp ,x dxx x

2

1. ., exp ( 2 log ),i e xx

i.e. 22 log(1/ ) xx e

i.e., (1/x2) × x–2, i.e., 1/x4

Now, multiplying both sides of (1) by 1/x4, we getx–2 (d2y/dx2) – 2x–3 (dy/dx) + 2x–4 y = 0 ... (4)

Comparing (4) with A0(x) (d2y/dx2) + A1(x) (dy/dx) + A2(x) y = 0,



we have A0 (x) = x–2 and A1 (x) = –2 x–3 so that 30 1( ) 2 ( ).A x x A x

Hence (4) is the required self adjoint equation.

EXERCISE

1. Find the adjoint equation of 2 7 8 0x y xy y , where /y dy dx

2. Show that 2(1 ) 2 ( 1) 0x y xy n n y is self adjoint .

3. Transform (tan ) 0y x y y into an equivalent self adjoint equation.4. Show that each of the following equations are self adjoint.

(i) 2

2sin cos 2 0d y dyx x xdxdx

(ii) 2

2 2 31 1 1 0x d y dy y

x dxdx x x

Answers 1. 2 3 3 0x y x y y 3. (cos ) (sin ) (cos ) 0x y x y x y

10.5 GREEN’S FORMULA.We know that if L is the differential operator

2 20 1 2( ) [ ( )( / ) ( ) ( / ) ( )] ( ),L u x a x d dx a x d dx a x u x a < x < b ... (1)

where A (x) is continuously differentiable, positive function, then its adjoint operator M is defined as2

0 1 22( ) { ( ) ( )} { ( ) ( )} ( ) ( ),d dM v x a x v x a x v x a x v x a x bdxdx

... (2)

From (1) and (2), we have

v Lu – u Mv2 2

0 1 2 0 1 22 2[ ] [ ( ) ( ) ]d u du d dv a a a u u a v a v a vdx dxdx dx

2

0 1 0 12 { ( )} ( )d u du d d dv a va u a v u a vdx dx dx dxdx

0 1 0 0 1 1( ) ( )dv a u v a u u a v a v u a v a vdx

0 1 0 0 0 1 1( 2 ) ( )v a u v a u u a v a v a v u a v a v

0 0 1 0( ) ( ) ( )a vu uv a vu uv uv a a 1 0( ) ( )a a uv u v

0 1 0( ) ( )d da v u u v u v a adx dx

Thus, 0 1 0( ) ( ) ( )b b

a a

dv Lu u Mv dx a v u u v u v a a dxdx

or 0 1 0( ) ( ) ( )b b

aav Lu u Mv dx a v u u v u v a a ... (3)

(3) is called Green’s formula for the operator L.An important particular case : Let L be a self adjoint operator defined by

2

2( ) ( ) ( ) ( )d d d dp dL p x q x p x q xdx dx dx dxdx

... (4)

Let u and v be two solutions of L y (x) = 0, Then since M = L for self adjoint operator, we haveLv (x) = Mv (x). Also, here a0(x) = p (x) and 1( ) ( ).a x p x Hence the Green’s formula (3) for self adjointoperator given by (4) takes the following form



( ) [ ( ) ( )]b b

aa

v Lu u Lv dx p x vu v u , 1 0as ( ) ( ) 0a a p x p x ... (5)

Thus, [ ( ) ( )]b b b

aa av Lu dx u L v dx p x vu v u ... (6)

which is known as Green’s formula for the operator L given by (4).10.6 THE DIRAC DELTA FUNCTION.

The Dirac delta function is of general interest, and is frequently encountered in highly abstractmathematics, in theoretical physics, and in the description of concentrated forces in solid and fluidmechanics, and in practical subjects such as electrical engineering. The Dirac delta functioncorresponds to the physical ideal of something concentrated into a point in space or an instant intime; for example Dirac delta function can be used to represent a point mass or a point charge, animpulse given to a dynamical system or a pulse produced by a sudden electrical discharge in somecircuit. The Dirac delta function and its derivative play an important role in the solution of initialand boundary value problems.

Dirac Delta function Definition (Kanpur 2010, 11)Consider the function having the following property :

1/ 2 , | |( )

0, | |x

xx

... (1)

Then,1( ) 1

2x dx dx

... (2)

The limit of ( )x as 0 is denoted by ( ),x that is,

0( ) lim ( )x x

... (3)

Then ( )x is known as the Dirac delta function.With help of (1), (2) and (3), the Dirac delta functions is defined as follows :

, if 0( )

0, if 0x

xx

... (4)

and ( ) 1x dx

.. (5)

The adjoining figure shows the profile of ( ).x Wee

generally call ( )x as Dirac delta function for historical rea-sons while it is not a function in the usual mathematical sense,which requires a function to have a definite value (or values) ateach point of a certain domain. For this reason Dirac has calledthe delta function as ‘‘improper function’’ and has emphasisedthat it may be used in mathematical analyis only when no in-consistency can possibly arise from its application. For appli-cation in initial and boundary value problems, we now pro-ceed to derive the formal properties of the Dirac delta func-tion. It should be clearly noted, however, that these propertiesare purely formal.

From the definition of the Direc delta function, we have

, if( )

0, ifx a

x ax a

... (6)

1/2

– 0

x



and ( ) 1x a dx

... (7)

We also note that1, if ( , )

( )0, if ( , )

b

a

c a bx c dx

c a b

... (8)

10.7.SHIFTING PROPERTY OF DIRAC DELTA FUNCTION.

To show that ( ) ( ) (0),f x x dx f

where f (x) is a continuous and bounded function.

[Himachal Pradesh, 2008]

Proof We have1/ 2 , if | |

( )0, if | |

xx

x

... (1)

Here 1( ) ( ) ( ) ,2

f x x dx f x dx

using (1)

(1/ 2 ) { ( )} ( ), wheref [using mean value theorem]

Clearly, when 0, we have 0. Hence, taking limits of both sides of the aboveequation as 0 and 0, we have

0 0

lim ( ) ( ) lim ( )f x x dx f

or ( ) ( ) (0).f x x dx f

... (2)

With a simple change of variable, (2) transforms to

( ) ( ) ( ),f x x a dx f a

... (3)

showing that the operation of multiplying f (x) by ( )x a and integrating over all x is merelyequivalent to substituting a for x in the original function. Symbolically we may write

( ) ( ) ( ) ( )f x x a f a x a ... (4)While using relation (4), we must clearly understand that (4) has a meaning only in the sense

that its two sides give equivalent results when used as factors in an integrand. As a particular caseof (4) with a = 0 and f (x) = x, we have ( ) 0x x ... (5)

In a similar way we can prove the relations ( ) ( )x x ... (6)and ( ) (1/ ) ( ), 0ax a x a ... (7)

We now prove that ( ) ( ) ( ).x y y a dy x a

... (8)

Proof. We have

0

( ){ ( ) ( ) } ( ) { ( ) ( ) }f x x y y a dy dx y a f x x y dx dy

[Changing the order of integration]

( ) ( ) ,y a f y dy

using shifting property (3)

Thus, ( ){ ( ) ( ) } ( ) ( ) ,f x x y y a dy dx f x x a dx



which formally shows that ( ) ( ) ( )x y y a dy x a

Remark .It is worthwhile to note that the range of integration in relation (3) is not necessareyto be to . It can be over any domain (a, b) surrounding the point at which ( )x is not zero.Clearly limits in these integration need not be mentioned and one may understand that the domainof integration is a subtle one.

10.8.DERIVATIVES OF DIRAC DELTA FUNCTIONWe now propose to examine the interpretation that has to be kept in mind while dealing with

the ‘‘derivatives’’ of ( ).x Suppose that ( )x exists and that both it and ( )x can be treated asordinary functions in the rule for integration by parts. Then, we have

( ) ( ) ( ) ( ) ( ) ( )f x x dx f x x f x x dx

0 (0),f using shifting theorem and definition of ( )x

Thus, ( ) ( ) (0)f x x dx f

... (1)

Repeating this process, we obtain ( ) ( )( ) ( ) ( 1) (0)n n nf x x dx f

... (2)

Also, we have ( ) ( ) ( )f x x a dx f a

... (3)

and ( ) ( )x x x ... (4)10.9.RELATION BETWEEN DIRAC DELTA FUNCTION AND THE HEAVISIDE UNITFUNCTION.

By definition, the Heaviside unit function H (x –a) is given by

0, if( )

1, ifx a

H x ax a

... (1)

Now, ( ) ( )U x a f x dx

( ) ( ) ( ) ( ) ,U x a f x U x a f x dx

integrating by parts

( ) ( ) ( ) ( ) ( )a

af U x a f x dx U x a f x dx

, using definition (1)

( ) 0 ( ) ,a

f f x dx using definition (1) again

( ) [ ( ) ( )] ( ),f f f a f a ... (2)where we have assumed that f (x) is both continuous and bounded.

Again, we have ( ) ( ) ( )f x x a dx f a

... (3)

From (2) and (3), ( ) ( ) ( ) ( )U x a f x dx x a f x dx

from which we formally obtain the required relation ( ) ( )U x a x a ... (4)which with a simple change of vairable transforms to ( ) ( )U x x ... (5)



10.10 ALTERNATIVE FORMS OF REPRESENTING DIRAC DELTA FUNCTION ( ).x

Sometimes it is useful to employ an explicit expression for ( )x as the limit of a sequence

of analytic function, for example, sin( ) limL

x Lxx

... (1)

Then, at x = 0, the limiting value of (sin ) /x L x is equal to /L and its value oscillates withperiod 2 / L when x increases. Its integrand, taken from to is unity and is independent ofthe value of L. Clearly, limit {(sin ) / }

Lx L x

has all the properties of Dirac delta function defined

by the equation (3) of Art. 10.6

An integral representation of the Dirac delta function is 1( )

2iuxx e du

... (2)We now proceed to prove (2). Indeed, we have

sinlim lim 2 limixL ixLL iux

L L LL

e e L xe dui x x

2 ( ),x using (1)

1( )2

iuxx e dux

or 1( ) (cos sin )

2x ux i ux du

... (3)

Separating the real parts, (3) gives1 cos ( )

2ux du x

... (4)

(4) is most commonly used as an explicit expression for ( ).x

Again, ( )x is given by a limit of the functions

, 0 | | 1/( ) ,

0, for all othernn x n

f xx

where n = 1, 2, 3, 4... ... (5)

10.11 SPHERICAL HARMONICSIn terms of spherical coordinates (refer figure),

Laplace’s equation 2 0u takes the form

2 2 2

2 2 2 2 2 2 22 1 cot 1 0,

sinu u u u u

r rr r r r

which can be re-written as2

22 2 2 2 2

1 1 1sin 0.sin sin

u u urr rr r r

... (1)

Let a solution of (1) be of the form( , , ) ( ) ( , ),u r R r ... (2)

where R is function r only and is function of and only. Substituting the value of u as givenby (2) in (1), we obtain

22

2 21 1sin 0

sin sinRr R R

r r

or2

22

1 1 1sinsin sin

RrR r r

... (3)

r

P r( )

z

y

x

O



Now the left-hand side of (3) is a function only of the variable r, while the right-hand side isa function only of the variables and . Since these variables are independent, it follows that lefthand side and right hand side must be a constant which we shall denote by . We thus, obtain

2

21 1 1sin

sin sin

or2

21sin sin

sin

... (4)

or sin ,S ... (5)where S is the differential operator defined by

2

21sin

sinS

... (6)

The same equation (4) also arises when separating Helmholtz’s equation in sphericalcoordinates. It can be proved that S is a formally self-adjoint operator.

We now discuss the boundary conditions to be associated with (4). Suppose it is requiredthat (4) must hold on the entire surface of a sphare (0 , 0 2 ). Then ( , ) should becontinuous and have a continuous gradient on the surface. To this end, we impose the followingboundary conditions:

0 2 0( ,0 ) ( , 2 ), / /

and (0, ) ( , ) finite

... (7)

The first two conditions imply that the planes 0 and 2 are on the same physicalplane. The third condition of finiteness at 0 and is imposed becuase these values of aresingular for the differential equation. It follows that some solutions are infinite at 0 and andhence these solutions must be rejected. As usual, it can be proved that the eigenvalues of (5) arereal and non-negative, and that eigenfunctions corresponding to different values of eigenvalues areorthogonal on the surface of unit sphere.

In many branches of physics and enginerring there is interest in the equation (4) solutions ofwhich are known as spherical harmonics. We now proceed to find the solutions (eigenfunctions) of(4) by using the well known method of separation of variables.

Let a solution of (4) be of the form ( , ) ( ) ( ), ... (8)Inserting this expression into (4), we have

2

21sin sin

sind d dd d d

or2sin sin sind d

d d

... (9)

Now the left hand side of (9) is a function only of the variable , while the right-hand side is a

function only of the variable . Since these two variables are independent, it follows that left-handside and right hand side must separately be a constant which we shall denote by . Thus, we obtain

and 2sin sin sind d

d d

Thus, 0 ... (10)



and2sin sin sind d

d d

... (11)

In view of (7) and (8), we see that the boundary conditions for (10) are(0) (2 ) and (0) (2 ) ... (12)

Hence, in order to get non-trivial solutions of (10), we set2 , 0, 2, 3, ......m m ... (13)

and then the corresponding solution is given by

( ) , 0, 1, 2, .....ime m ... (14)Putting µ = m2 in (11) and re-writting it, we obtain

2sin sin 0, 0

sind d md d

... (15)

In (15), we make the change of variable cosx so that sin .dx d Then, the interval

0 will transform into –1 < x < 1. We, also observe that

1sin

d dd dx

so that 2 2sin sin (1 ) .d d dxd dx dx

Accordingly, (15) takes the form2

22(1 ) , 1 1

1d d mx xdx dx x

... (16)

With m as a given integer, this is an eigenvalue problem in the parameter . For number ofvalues of , it is known that there is no solution which is finite at x = 1, and x = –1 (these valuesof x correspond, respectively, to 0 and ). Only by setting ( 1),n n n = | m |, | m | + 1,| m | + 2, ...., we arrive at a solution which is finite at x = 1 and at x = –1. This solution is the

*associated Legendre function | | (cos )mnP

In particular, if m = 0, the permissible values of n are 0, 1, 2, ..., and the corresponding

eigenfunctions are the ordinary *Legendre polynomials (cos )nP i.e., 0 (cos ).nP

Thus,the only eigenvalues of (4) are( 1), 0, 1, 2,...n n n ... (17)

and to the eigenvalue n (n + 1) correspond 2n + 1 eigenfunctions| |( , ) (cos ), | |m im m

nnY e P m n ... (18)

For example, for 2, we have the three eigenfunctions1 1

1 1 1(cos ), (cos ) and (cos )i ie P e P P

As already discussed, it follows from (4) and the boundary conditions that the set

( , ), | | , 0, 1, 2, ...mnY m n n ... (19)

is an orthogonal set on the surface of the unit sphere. Moreover, it can be proved that (19) iscomplete and hence there can be no other eigenfunctions of (4).

*For Legendre function and associated Legendre function, refer authors ‘‘Advanced Differential Equations’’or ‘Ordinary and Partial Differential equations’, published by S.Chand & Co. New Delhi.



Since the element of area on the surface is sin ,d d we obtain2

0 0sin ( , ) ( , ) 0,m

nd Y Y d

unless andn m ... (20)

where bar denotes the complex conjugate. The normalization integral is defined as2 2

, 0 0sin ( , )m

nm nN d Y d

... (21)

and we can easily verify that ,4 ( | |)!

2 1 ( | |)!m nn mN

n n m

... (22)

If ( , )f is any function regular on the surface of the unit sphere, we obtain

0( , ) ( , ),

nm

nmnn m nf f Y

... (23)

where2

0 0,

1 sin ( , ) ( , )mnmn

m nf d f Y d

N

... (24)

As a particular case, when f is independent of , we obtain expansion involving ordinaryLegendre polynomials

0( ) (cos ),nnn

f f P

where

0

2 1 ( ) (cos ) sin2 nn

nf f P d

... (25)

Spherical expansion of the free-space.* Green’s function for Laplace’s equation.Suppose a unit point source be placed at the point 0 0( )P x (whose spherical polar coordinates

are 0 0 0( , , )r where 00 00, 0 andr ) and P(x) be an arbitrary point in free space. Let( , , )r be spherical polar coordinates of P. Then the free-space Green’ function E (x; x0) forLaplace’s equation is given by

000 00

1( ; ) ( , , ; , , )4 | |

E E r r

x xx x ... (26)

such that E (x; x0) satisfies

00 02 ( ) ( ) ( )1 .sin sin

r rEr S Er r

... (27)

where S is the operator defined by (6).

For 0 ,r r E can be expanded in a series of spherical harmonics

,0( , ),

nm

m n nn m nE E Y

... (28)

where Em,n depends on 0 00, , , .r r Using (25), we find

2, 0 0,

1 sin ( , )mm n n

m nE d E Y d

N

... (29)

To find Em,n, we multiply both sides of (27) by ,(1/ ) sin ( , )mnm nN Y and then integrate

from 0 to and for 0 to 2 . Thus, we obtain

*The remaining part of the present article can be understood after reading chapters 12 & 13.



22, 0 00 0, ,

1 1( , ) ( , )m mm n n n

m n m n

d dr E d Y S E d Ydr dr N N

... (30)

Using (17) and the fact that ( , )mnY is a solution of (5), we have

( , ) ( 1) sin ( , ).m mnnS Y n n Y ... (31)

Integrating by parts, using the fact that S is symmetric and keeping (31) in view, (30) yieldsthe following ordinary differential equation

2, , 0 0 0

,

1( 1) ( , ) ( )mnm n m n

m n

d dr E n n E Y r rdr dr N

... (32)

The functions rn and r–n–1 are linearly independent solutions of the homogeneous equation,first of which is bounded at r = 0 and the other at .r Using the standard arguments for one-dimensional problems, we obtain

0 0 1,

,

( , ),

(2 1)

mn n n

m nm n

YE r r

n N

where 0min( , )r r r and 0min( , ).r r r Therefore, we have

0 010 0 ,

( , ) ( , )( ; )

(2 1)

m mnn n n n

n m n m n

Y YE r r

n N

x x ... (33)

In particular, if the source is placed at a point where 0 0, that is, on the positive z-axis,then (33) can be further simplified as shown below.

For the present case, 0 | |0(0, ) (1),imm m

nnY e P

where | | 0, 0(1)

1, 0m

n

mP

m

and hence 10 0

1( ; ) (cos )4

n nnn

E r r P

x x (source at 0 00, r ) ... (34)

If we now return the source to its arbitrary position in (33), the angle in (34) becomes the

angle between the points 0 0(1, , ) and (1, , ). It can be easily shown that

0 0 0cos cos cos sin sin cos( ) ... (35)and hence (33) can be re-written as

10 0

1( ; ) (cos )4

n nnn

E r r P

x x ... (36)

Comparing (33) and (36), we obtain the so called addition theorem for Legendre functions :

0( ) | | | |0(cos ) (cos )

(cos )( | |)!/( | |)!

im m mn n nn m n

e P PP

n m n m

... (37)

Finally, we discuss a special case of result (34). Suppose the source is placed at r0 = 1,

0 0. Then, we have 2 1/ 2 0

1 (cos ), 1(1 2 cos )

nnn

r P rr r

... (38)



and 12 1/ 2 0

1 (cos ), 1(1 2 cos )

nnn

r P rr r

... (39)

Differentiating both sides of (38) with respect to r, we get2

2 1/ 2 0

1 (2 1) (cos ), 1(1 2 cos )

nnn

r n r P rr r

... (40)

10.12. BESSEL FUNCTIONS*

Let 2 be an arbitrary given complex number and let z be a complex variable. Then Bessel’ss

equation of order is given by2

0d duz z udz dz z

... (1)

By virtue of the above definition, the Bessel’s equations of order and are exactly thesame. Any solution of (1) is known as cylinder function.

The Bessel function of order , ( )J z is defined by the series

2

0

( / 2)( ) ( / 2) ,! ( )!

k

k

zJ z zk k

... (2)

which is a solution of (1). Infinite series in (2) is convergent for all values of z.When is not an integer, the second solution of (1) is given by

2

0

( / 2)( ) ( / 2)! ( )!

k

k

zJ z zk k

...(3)

When is an integer (positive or negative), we have

( ) ( 1) ( )J z J z ... (4)

However when is not an integer, ( )J z and ( )J z are two independent solutions of (1).

It can be shown that the Neumann function cos ( ) ( )

( )sinJ z J z

N z

... (5)

and the Hankel functions (1) ( ) ( ),H J z i N z ... (6)

(2) ( ) ( ),H J z i N z ... (7)are all solutions of (1).

The general solution of (1) is a linear combination of two independent solutions. To this endwe consider the following two cases :

Case I. If is not an integer, then any two of the functions (1) (2), , andJ J N H H areindependent. Accordingly, the general solution (1) can be written as a linear combinations of anytwo of these five functions.

Case II. If is an integer (say ,n where n is any positive integer), then and nnJ J are

not independent due to the relation (4). The function ( )N z defined by (5) by taking the limit

*For more details, refer, chapter dealing with Bessel functions in author’s ‘‘Advanced Differential Equations’’or ‘‘Ordinary and Partial Differentail Equation’’ published by S.Chand & Co., New Delhi.



n is still a solution of (1), and is independent of ( ),nJ Thus any two of the functions(1) (2), , , nn nnJ N H H are independent. Accordingly, the general solution of (1) can be written as a

linear combinations of any two of these four functions.

The modified Bessel functions

Bessel’s modified equation is given by 2

0,d duz z udz dz z

... (8)

which differs from Bessel equation (1) only by the sign of the term zu. Since the formal substitutiont = iz reduces (8) to (1), we expect that the solutions of (8) will be cylinder functions of argument iz.

By convention, the modified Bessel function, 2

0

( / 2)( ) ( / 2)! ( 2 )!

k

k

zI z zk k

... (9)

and the Macdonald function( ) ( )

( )2 sin

I z I zK z

... (10)

are taken as the basic independent solutions of (8). These solutions are real when and z are real andpositive. The functions ( )I z and ( )K z are constant multipler of ( )J z and (1)( )H z respectively..

The behaviour of cylinder functions at zero and at infinity.

If has positive real part or if 0, ( )J z is the only solution of (1) that is bounded in a

neighbourhood of z = 0. Similarly, ( )I z is the only solution of (8) that is bounded at the origin.

Again, as ,z we have

(i)1/ 22 1( ) cos

2 2J z z

z

(ii)1/ 22 1( ) sin

2 2N z z n

z

(iii)1/ 2

(1) 2 exp2 4

H i zz

(iv)

1/2(2) 2 exp

2 4H i z

z

(v)1/ 2

1( )(2 )

zI z ez

(vi)1/ 2

( )2

zK z ez

EXERCISE

1. Define Dirac delta function and show that

(i) ( ) ( ) ( )t a f t dt f a

(ii) ( ) ( ) (0)t f t dt f

(iii) ( ) ( ) ( )t a f t dt f a

(iv) ( ) ( ) (0)t f t dt f

[Himanchal 2008, 09]

2. Prove that ( ) ( )t t [Himanchal 2009]


CHAPTER 11

Applications of Integralequations and Green’s function

to ordinary differential equations11.1 INTRODUCTION.

We have already dealt with conversion of initial and boundary value problems into integralequations in chapter 2. In the present chapter we shall consider the initial and boundary problemsagain in a different context. We shall introduce the concept of Green’s function and utilize it inconverting initial and boundary value problems into integral equations. Sometimes we shall be ableto solve the given initial and boundary value problems completely with help of Green’s function.The reader is advised to study chapter 10 before starting with the present chapter.11.2 GREEN’S FUNCTION. DEFINITION. [Meerut 2002]

Consider a linear homogeneous differential equation of order n :L [y] = 0, ... (1)

where L is the differential operator1

10 1( ) ( ) ... ( ),n n

n n nd dL p x p x p xdx dx

... (2)

where the functions p0 (x), p1 (x), ... , pn (x) are continuous on [a, b], 0 ( ) 0p x on [a, b] and theboundary conditions are

Vk (y) = 0, (k = 1, 2, 3, ..., n) ... (3)

where (1) ( – 1) ( – 1)( ) ( ) ( ) ... ( )n nk k k kV y y a a y a a y a

(1) ( 1) ( 1)( ) ( ) ... ( ),n nk k ky b y b y b ... (4)

where the linear forms V1, ..., Vn in y (a), ( )y a , ........ y (n – 1) (a), y (b), ( )y b , ..., y (n – 1) (b) arelinearly independent.

Suppose that the homogeneous boundary- value problem given by (1) to (4) has only a trivialsolution ( ) 0.y x Then Green’s function of the boundary value problem (1) to (4) is the functionG (x, t) constructed for any point t, a < t < b, and which has the following four properties :

(i) In each of the intervals [a, t) and (t, b] the function G (x, t), considered as a function of x,is a solution of (1), that is,

L [G] = 0. ... (5)(ii) G (x, t) is continuous and has continuous derivatives with respect to x upto order (n – 2)

inclusive for . a x b

11.1


11.2 Applications of integral equations and Green’s function to ordinary differential equation

(iii) (n – 1)th derivative of G (x, t) with respect to x at the point x = t has discontinuity of thefirst kind, *the jump being equal to –1/ p0 (t), that is,

1 1

1 100 0

1 .( )

n n

n nx t x t

G Gp tx x

... (6)

(iv) G (x, t) satisfies the boundary conditions (3), that is,Vk (G) = 0. (k = 1, 2, ..., n) ... (7)

Theorem. If the boundary value problems given by (1) to (4) has only the trivial solution

( ) 0,y x then the operator L has a unique Green’s function G (x, t).

Proof. Suppose that y1 (x), y2 (x), ..., yn (x) be linearly independent solutions of the equation(1). Then, by virtue of property (i) of Green’s function, the unknown Green’s function G (x, t) musthave the following representation on the intervals [a, t) and (t, b] :

1 1 2 2

1 1 2 2

( ) ( ) ... ( ), ,( , )

( ) ( ) ... ( ), ,n n

n n

a y x a y x a y x a x tG x t

b y x b y x b y x t x b

... (8)

where a1, a2, ..., an, b1, b2, ..., bn are some functions of t.Again, by virtue of property (ii), the continuity of the function G (x, t) and of its first n – 2

derivatives with respect to x at the point x = t gives rise to the following relations :[b1 y1 (t) + ... + bn yn (t)] – [a1 y1 (t) + ... + an yn (t)] = 0, ... (A1)

1 1 1 1[ ( ) ... ( )] [ ( ) ... ( )] 0,n n n nb y t b y t a y t a y t ... (A2)

... ... ... ... ... ... .... ... ... ...

... ... ... ... ... ... .... ... ... ...( 2) ( 2) ( 2) ( 2)

1 1 1 1[ ( ) ... ( )] [ ( ) ... ( )] 0.n n n n

n n n nb y t b y t a y t a y t ... (An– 1)

Finally, by virtue of property (iii), we obtain

( 1) ( 1) ( 1) ( 1)

1 1 1 10

1[ ( ) ... ( )] [ ( ) ... ( )] .( )

n n n nn n n nb y t b y t a y t a y t

p t ...(An)

Let Ck (t) = bk (t) – ak (t), where k = 1, 2, ..., n. ... (9)Re-writing (A1), (A2), ..., (An) with help of (9), we obtain a system of linear equations in Ck (t) :

C1 y1 (t) + ... + Cn yn (t) = 0, ... (B1)

C1 y1 (t) + ... + ( )n nC y t = 0, ... (B2)

.............................................................. ..............................................................

( 2) ( 2)1 1

( ) ... ( )n nn nC y t C y t = 0, ... (Bn – 1)

and ( 1) ( 1)1 1

( ) ... ( )n nn nC y t C y t

0

1 .( )p t

... (Bn)

* The reader must note carefully that some authors consider this jump with opposite sign (i.e., 1/p0 (t))and this implies a change of G (x, t) to – G (x, t).


Applications of Integral equations and Green’s function to ordinary differential equation 11.3

The determinant D of the system (B1), ..., (Bn) is given by

1 2

1 2

1 2( 2) ( 2) ( 2)1 2( 1) ( 1) ( 1)1 2

( ) ( ) ... ( )

( ) ( ) ... ( )

... ... ... ... ( , , ...., ),

( ) ( ) ... ( )

( ) ( ) ... ( )

n

n

nn n n

nn n n

n

y t y t y t

y t y t y t

D W y y y

y t y t y t

y t y t y t

... (10)

where W (y1, y2, ..., yn) is the Wronskian of the functions y1, ..., yn. Since y1, y2, ..., yn are taken aslinearly independent, it follows that

1 2( , , ..., ) 0. nD W y y y ... (11)

Since 0D if follows that the system of equations (B1), ..., (Bn) possess a unique solutionfor Ck, k = 1, 2, ..., n.

We now proceed to find the functions ak (t) and bk (t). For this purpose, we use the property(iv) of the Green’s function. We first write Vk (y) in the following form

Vk (y) = Pk (y) + Qk (y), ... (12)

where (1) ( 1) ( 1)( ) ( ) ( ) ... ( ), n nk k k kP y y a y a y a ... (13)

and (1) ( 1) ( 1)( ) ( ) ( ) ... ( ).n nk k k kQ y y b y b y b ... (14)

Then, by virtue of the equation (7) of the property (iv), we haveVk (G) = a1 Pk (y1) + ... + an Pk (yn) + b1 Qk (y1) + ... + bn Qk (yn) = 0, k = 1, 2, ..., n ... (15)From (9), ak = bk – Ck, k = 1, 2, ..., n. ... (16)Using (16), (15) reduces to

(b1 – C1) Pk (y1) + ... + (bn – Cn) Pk (yn) + b1Qk (y1) + ... + bn Qk (yn) = 0, k = 1, 2, ..., n ... (17)or b1 [Pk (y1) + Qk (y1)] + ... + bn [Pk (yn) + Qk (yn)] = C1 Pk (y1) + ... + Cn Pk (yn)or b1 Vk (y1) + ... + bn Vk (yn) = C1 Pk (y1) + ... + Cn Pk (yn), k = 1, 2, ..., n, by (12) ... (18)

Here (18) is a linear system of n equations for determination of n quantities b1, ..., bn. Sincewe have assumed the linear independence of the forms V1, V2, ..., Vn, it follows that the determinantof the system (18) is non-zero, that is,

1 1 1 2 1

2 1 2 2 2

1 2

( ) ( ) ... ( )

( ) ( ) ... ( )0.

...... ...... ... ......( ) ( ) ... ( )

n

n

n n n n

V y V y V y

V y V y V y

V y V y V y

...(19)

By virtue of (19), we see that the system of equaions (18) possess a unique solution in b1 (t),b2 (t), ..., bn (t). Already, we have seen that unique values of C1,(t) ..., Cn (t) exist. Hence (16)shows that the quantities a1 (t), ..., an (t) are defined uniquely.

Thus we have established the existence and uniqueness of Green’s function G (x, t). We havealso indicated a procedure for constructing the Green’s function in the above proof of the theorem.

Remark 1. If the boundary value problem (1) to (4) is self-adjoint, then Green’s function issymmetric, that is, G (x, t) = G (t, x). The converse is also true. For definition and examples of self-adjoint equations, refer chapter 10.



Remark 2. If at one of the extremities of an interval [a, b] the coefficient of the highestderivative vanishes, for example p0 (a) = 0, then the natural boundary condition for boundedness ofthe solution at x a is imposed, and at the other extemity the ordinary boundary condition isspecified. Refer solved example 3 in Art. 11.5 for more details.11.3 CONVERSION OF A BOUNDARY VALUE PROBLEM INTO FREDHOLM

INTEGRAL EQUATION. SOLUTION OF A BOUNDARY VALUE PROBLEMIn what follows, we shall use the following notations :

1

10 1( ) ( ) ... ( ).

n n

n n nd dL p x p x p xdx dx

(1) ( 1) ( 1)( ) ( ) ( ) ... ( )n nk k k kV y y a y a y a (1) ( 1) ( 1)( ) ( ) ... ( ).n n

k k ky b y b y b

Suppose G (x, t) is Green’s function of the boudary value problemL [y] = 0, ... (1)

Vk (y) = 0, k = 1, 2, 3, ..., n, ... (2)involving homogeneous boundary conditions (2) at the end points x = a and x = b of an interval

. a x bResult 1. Consider the boundary value problem

[ ] ( ) 0, L y x ... (3)Vk (y) = 0, k = 1, 2, ..., n, ... (4)

involving the same homogeneous boundary conditions as in (2). Here ( ) x is a direct function of x.*Then solution of the boundary value problem (3)—(4) is given by the formula

( ) ( , ) ( ) . b

ay x G x t t dt ... (5)

Result 2. Consider the boundary-value problem

[ ] ( ) 0, L y x ... (6)Vk (y) = 0, k = 1, 2, ..., n, ... (7)

involving the same homogeneous boundary conditions as in (2).

In this result we assume that ( ) x is not a given direct function of x. However, ( ) xmay also depend upon x indirectly by also involving the unknown function y (x), and so beingexpressible in the form ( ) ( , ( )). x x y x ... (8)

Then the boundary-value problem (6)—(7) can be reduced to the following integral equation

( ) ( , ) ( , ( )) . b

ay x G x t t y t dt ... (9)

Particular case of Result 2 :Let ( ) ( ) ( ) ( ), x r x y x f x where is a parameter. Then, we see that the boundary

value problem [ ] ( ) ( ) ( ) L y r x y x f x ... (10)Vk (y) = 0, k = 1, 2, ..., n ... (11)

reduces to the following integral equation

( ) ( , ) ( ) ( ) ( , ) ( ) , b b

a ay x G x t r t y t dt G x t f t dt ... (12)

* A proof of this result for a particular case will be presented in Art. 11.4.



where G (x, t) is the relevant Green’s function. In (12), G (x, t) r (t) is not symmetric unless thefunction r (t) is a constant. However, if we write

{r (x)}1/2 y (x) = Y (x),under the assumption that r (x) is non-negative over (a, b), as is usually the case in practice, theequation (12) can be re-written in the form

* *1/ 2

( )( ) ( , ) ( ) ( , ) ,{ ( )}

b b

a a

f tY x K x t Y t dt K x t dtr t

where K * (x, t) = G (x, t) {r (x) r (t)}1/2 is a symmetric kernel. We have already seen the importanceof a symmetric kernel in Chapter 7.

Result 3. When the prescribed end conditions are not homogeneous, we shall use amodified method as explained below :

In this case, let G (x, t) denote the Green’s function corresponding to the associatedhomogeneous end conditions. We now search for a function P (x) such that the relation

( ) ( ) ( , ) ( )b

ay x P x G x t t dt ... (13)

is equivalent to the differential equation ( ) ( ) 0, L y x ... (14)together with the prescribed nonhomogeneous end conditions.

Since ( , ) ( ) ( ), b

aL G x t t dt x ... (15)

the requirement that (13) imply (14) leads us toL [P (x)] = 0. ... (16)

Furthermore, since the second term in (13) satisfies the associated homogeneous end conditions,we conclude that function P (x) in (13) must be the solution of (16) which satisfies the prescribednonhomogeneous end conditions. When G (x, t) exists, then P (x) always exists.11.4 AN IMPORTANT SPECIAL CASE OF RESULTS OF ART. 11.2

Consider a linear homogeneous differential equation of order two :

( ) ( ) 0,Ly x x a x b ... (1)where L is the self adjoint differential operator

2

2 ,

d d d dp dL p q p qdx dx dx dxdx

... (2)

together with homogeneous boundary conditions

1 1( ) ( ) 0 y x y a ... (3a)

and 2 2( ) ( ) 0 y b y b ... (3b)

with the usual assumption that at least one of 1 and 1 and one of 2 and 2 are non-zero.

Here ,p p and q are given real valued continuous functions defined on [a, b] such that p (x)is non-zero* on [a, b].

* It p (x) = 0 at an end point, the corresponding appropriate end condition may require merely thaty (x) remains finite at that point. Refer Ex. 3., on page 11.12 of Art 11.5.



Then, by definition, a function G (x, t) is called a Green’s function for L y (x) = 0 if, for a givent.

1

2

( , ),( , )

( , ),

G x t if a x tG x t

G x t if t x b ... (4)

where G1 and G2 are such that(i) The functions G1 and G2 satisfy the equation L G = 0 in their respective intervals of

definition, that is,

1 0, L G when a x t ... (5a)

and 2 0, L G when t x b ... (5b)(ii) G1 satisfies the boundary condition (3a) whereas G2 satisfies the condition (3b).(iii) The function G (x, t) is continuous at x = t, i.e., G1 (t, t) = G2 (t, t) ... (6)(iv) The derivative of G (x, t) has a discontinuity of magnitude –1/p(t) at x = t, that is,

2 1/ / 1/ ( )x t

G x G x p t

... (7)Condition (7) can also be re-written as

0 0

1( )x t x t

G Gx x p t

... (7)

We now use the above definition to construct the Green’s function for boundary value problemgiven by (2), (3a) and (3b) as follows :

Suppose that y = u (x) is a nontrivial solution of the associated equationLy = 0. ... (8)

Furthermore, suppose that y = u (x) satisfies the given homogeneous condition (3a). Similarly,let y = v (x) be a nontrivial solution of (8), which satisfies the given condition (3b). It follows thatthe properties (i) and (ii) are satisfied if we take G1 (x) = C1 u (x) and G2 (x) = C2 v (x), where C1and C2 are constants. Thus, we have

1

2

( ), ,( , )

( ), .

C u x a x tG x t

C v x t x b ... (9)

We shall now determine C1 and C2 by fulfilling the properties (iii) and (iv).By virtue of property (iii), we have

C1 u (t) = C2 v (t) or C2 v (t) – C1 u (t) = 0. ... (10)Again by the virtue of property (iv), we have

C2 v (t) – C1 u (t) = – 1/ p (t). ... (11)The determinant D of the system of equations (10) and (11) is given by

( ) ( )[ ( ), ( )].

( ) ( )v t u t

D W u t v tv t u t

... (12)

Suppose that u (x) and v (x) be two linearly independent solutions of (8). Then, we knownthat the Wronskian W [u (t), v (t)] of linearly independent functions u and v does not vanish.Accordingly, (12) reduces to

[ ( ), ( )] 0, D W u t v t ... (13)showing that (10) and (11) possess a unique solution.

Now, solving (10) and (11) by cross- multiplication we have

2 1 1 ./ ( ) / ( )C C

u p t v p t v u u v

... (14)



We shall now prove the Abel’s formula, namely,u v – u v = A/p (t). ... (15)

By our assumption u and v are solutions of

Ly = 0 or2

2 0, (2)d y dp dyp q y bydx dxdx

or p y + py + qy = 0 or (p y) + q y = 0. ... (16) (p u) + q u = 0 ... (17)

and (p v) + q v = 0 ... (18)Multiplying (17) by v and (18) by u, we get

v (p u) + q u v = 0 ... (19)and u (p v) + q u v = 0. ... (20)

Substracting (19) from (20), we haveu (p v) – v (p u) = 0 or [p (u v – v u)] = 0. ... (21)

Integrating (21), ( ) ,p u v u v A when A is a constant of integration.Thus, u v – u v = A/ p(t),

which proves the Abel’s formula (15).

Using (15), (14) reduces to 2 1 1/ ( ) / ( ) / ( )

C Cu p t v p t A p t

so that C1 = – (1/ A) v (t) and C2 = – (1/ A) u (t).Substituting these values of C1 and C2 in (9), we obtain

(1/ ) ( ) ( ),( , )

(1/ ) ( ) ( ), ,A u x v t a x t

G x tA u t v x t x b

... (22)

where A is a constant, independent of x and t, which is determined by Abel’s formula (15).From (22), it follows that G (x, t) is symmetric, i.e., G (x, t) = G (t, x) ... (23)We now state and prove the main result of this article.

Theorem. Let the Green’s function for ( ) 0L y x be given by (22). Then y (x) is a solution ofboundary value problem given by (1), (3a) and (3b) if and only if

( ) ( , ) ( ) b

ay x G x t t dt ... (24)

Proof. Let the relation (24) be true. Then we shall show that y (x) is a solution of theboundary value problem given by (1), (3a) and (3b).

Re-writing (24), we have ( ) ( , ) ( ) ( , ) ( ) x b

a xy x G x t t dt G x t t dt

or 1( ) ( ) ( ) ( ) ( ) ( ) ( ) ,x b

a xy x v x u t t dt u x v t t dt

A by (22) ... (25)

Differentiating both sides of (25) w.r.t. ‘x’ and using Leibnitz’s rule of differentiation under thesign of integration (refer Art. 1.13), we have

1( ) ( ) ( ) ( ) ( ) ( ) ( )x

ay x v x u t t dt u x v x x

A ( ) ( ) ( ) – ( ) ( ) ( )

b

xu x v t t dt u x v x x



or 1( ) ( ) ( ) ( ) ( ) ( ) ( )

x b

a xy x v x u t t dt u x v t t dt

A ... (26)

Differentiating both sides of (26) with respect of ‘x’ and using Leibnitz’s rule, we get

1( ) ( ) ( ) ( ) ( ) ( ) ( ) x

ay x v x u t t dt v x u x x

A( ) ( ) ( ) ( ) ( ) ( )

b

xu x v t t dt u x v x x

or ( )( ) ( ) ( ) ( ) ( )xy x v x u x u x v xA

1 ( ) ( ) ( ) ( ) ( ) ( )

x b

a xv x u t t dt u x v t t dt

A

or ( ) 1( ) ( ) ( ) ( ) ( ) ( ) ( ) ,( )

x b

a x

xy x v x u t t dt u x v t t dtp x A by (15) ... (27)

Now, L y (x) = p (x) y (x) + p (x) y (x) + q (x) y (x), by definition (2)Substituting the values of y (x), y (x) and y (x) given by (25), (26) and (27) in the above

equation, we get

( )( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )

x b

a x

p xL y x x v x u t t dt u x v t t dtA

( ) ( ) ( ) ( ) ( ) ( ) ( )x b

a x

p x v x u t t dt u x v t t dtA ( ) ( ) ( ) ( ) ( ) ( ) ( )

x b

a x

q x v x u t t dt u x v t t dtA

or 1( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) x

aL y x x p x v x p x v x q x v x u t t dt

A

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) b

xp x u x p x u x q x u x v t t dt

or 1( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )x b

a xL y x x p v q v u t t dt pu qu v t t dt

A

or ( ) ( ), L y x x using (17) and (18)

or ( ) ( ) 0, L y x x ... (28)Further, from the relations (25) and (26), we easily obtain

and( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

b

ab

a

A y a u a v t t dt

A y a u a v t t dt

... (29)

Since u (x) satisfies the boundary condition (3a), it follows from (29) that y (x) also satisfiesthe boundary condition (3a). Similarly we can show that y (x) satisfies the boundary condition (3b).

Thus, we have shown that if G (x, t) is given by (22), then y (x) is a solution of the boundaryvalue problem given by (1), (3a) and (3b).

Conversely, let y (x) satisfy (1) together with boundary conditions (3a) and (3b). Then, weshall prove that y (x) satisfies (24), where G (x, t) is given by (22).

From (1), ( ) ( )L y x x so that ( , ) ( ) ( , ) ( )G x t x G x t Ly x

( , ) ( ) ( , ) ( ) b b

a aG x t x dx G x t L y x dx ( , ) ( ) ( , ) ( )

t b

a tG x t L y x dx G x t L y x dx



or 1 2( , ) ( ) ( , ) ( ) ( , ) ( )

b t b

a a tG x t x dx G x t L y x dx G x t L y x dx , using (4) ... (30)

Now, Green’s formula for self adjoint operator L (refer result (5) of Art. 10.5 of chapter 10 byusing x1 and x2 in place of a and b respectively) is given by

2 211

( ) ( – )x b x

xx av Lu dx u Lv dx p x vu v u ... (31)

Using the Green’s formula (31), we have

1( , ) ( )t

aG x t Ly x dx 1

1 1

( , )( ) ( ) ( , ) ( ) ( )

tt

aa

G x tLy x LG dx p x G x t y x y x

x

11

( , )( ) ( , ) ( ) ( )

x t

G x tp t G t t y t y t

x

... (32)

[on using the properties of G1 (x, t)]Similarly, we have

2 ( , ) ( )b

tG x t Ly x dx 2

2 2

( , )( ) ( ) ( , ) ( ) ( )

bb

tt

G x tLy x LG dx p x G x t y x y x

x

2

2

( , )( ) ( , ) ( ) ( )

x t

G x tp t G t t y t y t

x

... (33)

[on using the properties of G2 (x, t)]Using (32) and (33), we have

R.H.S. of (30) 1 2 2 1( ) ( ) ( , ) ( , ) ( ) ( ) ( / / )x tp t y t G t t G t t p t y t G x G x

( ) ( ) 0 ( ) ( ) { 1/ ( )}, p t y t p t y t p t using (6) and (7)= – y (t)

Hence (30) reduces to

( , ) ( ) ( ), b

aG x t x dx y t i.e., ( ) ( , ) ( )

b

ay x G t x t dt

[on interchanging x and t]

or ( ) ( , ) ( ) , b

ay x G x t t dt using (23)

Thus, we have shown that if y (x) satisfy (1) together with (3a) and (3b), then y (x) mustsatisfy (24).

Remark 1. Relation (24) defines the solution of the boundary value problem given by (1),(3a) and (3b) when is a given direct function of x, whereas (24) constitutes an equivalent integralequation formulation of the problem when involves y. In other words, if involves y, then (24)is not a solution of given boundary value problem. In that case (24) merely is a non-linear integralequation which is equivalent to the given boundary value problem.

Remark 2. Consider inhomogeneous system

1 1 2 2( ) ( ) 0; ( ) ( ) 0, ( ) ( ) 0.L y x x y a y a y b y b ... (i)and the associated completely homogeneous system.

1 1 2 2( ) 0, ( ) ( ) 0, ( ) ( ) 0,L y x y a y a y b y b ... (ii)

where2

20 1 2( ) ( ) d dL a x a a x

dxdxand a < x < b.



If (ii) has only the trivial solution, then (i) has always a unique solution and Green’s functionexists for (ii).

If (ii) has a non-trivial solution, then (i) either has no solution or else has many solutions, butnever just one.

For example, consider the completely homogeneous systemy (x) = 0, 0 < x < 1; y (0) = 0, y (1) = 0. ... (iii)

Solving y (x) = 0, we have y (x) = A and y (x) = Ax + B.Using y (0) = 0 and y (1) = 0, we get A = 0 and so (iii) has non-trivial solution y (x) = B,

where B is an arbitrary constant. Now, consider the related inhomogeneous systemy (x) = 1, 0 < x < 1; y (0) = 0, y (1) = 0 ... (iv)

Solving y (x) = 1, we have y (x) = x + A and y (x) = B + Ax + x2/2Using y (0) = 0, we get A = 0 so that y (x) = xUsing y (1) = 0, y (x) = x 0 = 1, which is absurd. Hence the system (iv) has no solution.Consider the following another inhomogeneous system related to (iii)

( ) cos , 0 1; (0) 0, (1) 0y x x x y y ... (v)

Solving 2( ) cos , ( ) (1/ ) sin , ( ) (1/ ) cosy x x y x x A y x x Ax B

Using y (0) = 0, we get A = 0. So, ( ) (1/ ) sin ,y x x which also satisfies the condition

y (1) = 0. Hence the solution of (v) is given by 2( ) (1/ ) cos , y x B x where B is an arbitraryconstant. Hence the system (v) has the infinite set of solutions.

A modified Green function which is appropriate to the above mentioned exceptional situationswill be discussed in Art. 11.12.11.5 SOLVED EXAMPLES BASED ON CONSTRUCTION OF GREEN’S FUNCTION

(REFER ART. 11.2 AND 11.4)Ex. 1. Find the Green’s function of the boundary value problem. 0, (0) ( ) 0. y y y lSol. Given boundary value problem is y = 0, ... (1)with the boundary conditions : y (0) = 0 ... (2a)

and y (l) = 0. ... (2b)The general solution of (1) is y (x) = Ax + B. ... (3)Putting 0x in (3) and using (2a), we get B = 0 ... (4)

Next, putting x l in (3) and using (2b), we get 0 = Al + B. ... (5)

Solving (4) and (5), we get 0.A B Hence (3) yields only the trivial solution ( ) 0y x forthe given boundary value problem. Therefore, the Green’s function exists and is given by

1 2

1 2

, 0 ,( , )

, .

a x a x tG x t

b x b t x l

... (6)

In addition to the above property (6), the proposed Green’s function must satisfy the followingthree properties :

(i) G (x, t) is continuous at ,x t that is,

1 2 1 2b t b a t a or 1 1 2 2( ) ( ) 0b a t b a ... (7)

(ii) The derivative of G has a discontinuity of magnitude – 1/ p0(t) at the point x = t, wherep0 (x) = coeff of the highest order derivative in (1) = 1.



0 0( / ) ( / ) 1x t x tG x G x

or 1 1 1b a . ... (8)

(iii) G (x, t) must satisfy the boundary conditions (2a) and (2b), that is,

(0, ) 0G t so that 2 0a ... (9)

and ( , ) 0G l t so that 1 2 0.b l b ... (10)

Using (8), (7) becomes – t + b2 – a2 = 0. ... (11)Solving (8), (9), (10) and (11), we have

2 2 10, , / ,a b t b t l 1 1 ( / ).a t l

1 2 1 ( ).

t xa x a x l tl l and 1 2 ( ).

t tb x b x t l xl l

Substituting the above values in (6), the required Green’s function of the given boundaryproblem is given by

( / ) ( ), 0( , )

( / ) ( ), .

x l l t x tG x t

t l l x t x l... (12)

Ex.2. Find the Green’s function for the boundary value problem2 2 2/ 0, (0) (1) 0.d y dx y y y (Kanpur 2008)

Sol. Given boundary value problem is2 0 y y or 2 2( ) 0, where /D y D d dx ... (1)

with the boundary conditions (0) 0,y ... (2a)

and (1) 0.y ... (2b)

The auxiliary equation of (1) is 2 2 0D so that .D i

Hence the general solution of (1) is ( ) cos sin . y x A x B x ... (4)

Putting 0x in (3) and using B.C. (2a), we get 0.A Again, putting x = 1 in (3) and B.C. (2b), we get

0 cos sin .A B ... (5)

Solving (4) and (5), we get 0.A B Hence (3) yields only the trivial solution ( ) 0y x forthe given boundary value problem. Therefore, the Green’s function exists and is given by

1 2

1 2

cos sin , 0 ,( , )

cos sin , 1.

a x a x x tG x t

b x b x t x... (6)

In addition to the above property (6), the proposed Green’s function must also satisfy thefollowing three properties.

(i) G (x, t) is continuous at ,x t that is,

1 2 1 2cos sin cos sin b t b t a t a t

or 1 1 2 2( ) cos ( ) sin 0. b a t b a t ... (7)



(ii) The derivative of G has a discontinuity of magnitude – 1/p0 (t) at the point ,x t where

0 ( )p x coefficient of highest order derivative in (1) = 1.

0 0( / ) ( / ) 1x t x tG x G x

or 1 2 1 2( sin cos ) ( sin cos ) 1 b t b t a t a t

or 1 1 2 2( ) sin ( ) cos (1/ ). b a t b a t ... (8)(iii) G (x, t) must satisfy the boundary conditions (2a) and (2b), that is,

(0, ) 0G t so that 1 0.a ... (9)

and (1, ) 0G t so that 1 2cos sin 0. b b ... (10)

Let 1 1 1b a C and 2 2 2 .b a C ... (11)Then (7) and (8) may written as

1 2cos sin 0 0 C t C t ... (12)

and 1 2sin cos (1/ ) 0. C t C t ... (13)Solving (12) and (13) by cross-multiplication rule, we have

1 22 2

1(1/ ) sin (1/ ) cos cos sin

C Ct t t t

so that 1 2(1/ ) sin (1/ ) cos .C t C t

1 1 (1/ ) sin , b a t by (11) ... (14)

and 2 2 (1/ ) cos , b a t by (11) ... (15)Solving (9), (10), (14), (15), we have

1 1 2 21 sin cos sin cos cos sin ( 1)0, sin , ,

sin sin sint t t ta b t b a

... (16)

Now, 1 2sin ( 1) sincos sin

sint xa x a x

and 1 2sin cos sin sin coscos sin

sin

t x t xb x b x

sin sin ( 1) .sin

t x

Substituting the above values in (6), the required Green’s function is given by

sin ( 1) sin , 0 ,sin

( , )sin sin ( 1) , 1.

sin

t x x tG x t

t x t x

Ex. 3. Construct Green’s function for the differential equation 0xy y for the following

conditions. y (x) is bounded as 0,x and (1) ' (1), 0.y y [Kanpur 2006, Meerut 2005, 10]

Sol. Given boundary value problem is : 20 or 0xy y x y xy



or 2 2 2( / ) ( / ) 0x d y dx x dy dx or 2 2( ) 0, / .x D xD y D d dx ... (1)

with the boundary conditions : y (x) is bounded as 0x ... (2a)

and (1) (1), 0.y y ... (2b)To solve linear homogeneous differential equation (1), we proceed by the usual method.Put x = ez so that log x = z. ... (3)

Then xD = D1 and x2 D2 = D1 (D1 – 1), where 1 / .D d dz ... (4)

Using (4), (1) reduces to [D1 (D1 – 1) + D1] y = 0 or 21

0D y ... (5)

The auxiliary equation of (5) is 21

0D so that D1 = 0, 0. Hence the solution is

y = Az + B or y (x) = A log x + B, by (3) ... (6)Now, from (6), y (x) = A / x. ... (7)From (6) and (7), y (1) = B and y (1) = A.

Putting these values in (2b), we get B A

In view of B.C. (2a), we must take A = 0 in (6). Then A = 0 and 0.B A B

Thus 0.A B Hence (6) yields only the trivial solution ( ) 0.y x Therefore, the Green’ssfunction exists and is given by

1 2

1 2

log , 0 ,( , )

log , 1.

a x a x tG x t

b x b t x ... (8)

In addition to the above property (8), the proposed Green’s function must also satisfy thefollowing three properties :

(i) G (x, t) is continuous at x = t, that is,b1 log t + b2 = a1 log t + a2 or (b1 – a1) log t + (b2 – a2) = 0. ... (9)

(ii) The derivative of G has a discontinuity of magnitude – 1/p0 (t) at the point x = t, wherep0 (x) = coefficient of the highest power of x in the given differential equation = x. Thus, we have

0 0( / ) ( / ) –(1/ )x t x tG x G x t

or (b1/t) – (a1/t) = – (1/t)

or 1 1( ) 1. b a ... (10)

(iii) G (x, t) given by (8) must satisfy the boundary condition (2a) and (2b).

For (2a), G (x, t) must be bounded as 0x i.e., (a1 log x + a2) must be bounded at 0x ,which is possible only if we take a1 = 0, ... (11)

For (2b), we must have (1, ) (1, )G t G t

i.e., 1 2 1 1log 1 ( / )xb b b x

or 2 1.b b ... (12)

Solving (9), (10), (11) and (12), we get

1 1 20, 1,a b b and 2 log . a t

Substituting the above values in (8), the required Green’s function is

log , 0 ,( , )

log , 1.

t x tG x t

x t x



Ex. 4. Construct Green’s function for the homogeneous boundary value problem4 4/ 0, (0) (0) (1) (1) 0.d y dx y y y y [Kanpur 2009; Meerut 2000,03,04]

Sol. Given d4/y/ dx4 = 0 or 4 0, / ,D y D d dx ... (1)

with the boundary conditions y (0) = 0, ... (2a)y (0) = 0, ... (2b)y (1) = 0, ... (2c)

and y (1) = 0. ... (2d)The auxiliary equation of (1) is D4 = 0 so that D = 0, 0, 0, 0.Hence the general solution of (1) is y (x) = A + Bx + Cx2 + Dx3, ... (3)

where A, B, C, D are arbitrary constants.From (3), y (x) = B + 2 Cx + 3Dx2. ... (4)Putting 0x and 1x in (3) and using (2a) and (2c), we have

y (0) = A and y (1) = A + B + C + Dor A = 0 ... (5)and A + B + C + D = 0. ... (6)

Putting x = 0 and 1x in (4) and using (2b) and (2d), we have

(0)y B and (1) 2 3y B C D

or B = 0 ... (7)and B + 2C + 3D = 0 ... (8)

Solving (5), (6), (7) and (8), we get 0.A B C D Hence (3) yields only the trivial

solution ( ) 0y x for the given boundary value problem. Therefore, the Green’s functions existsand is given by

2 31 2 3 4

2 31 2 3 4

, 0 ,( , )

, 1.

a a x a x a x x tG x t

b b x b x b x t x

... (9)


(i) 2 2( , ), / , / G x t G x G x are continuous at .x t Thus, we have

2 3 2 31 2 3 4 1 2 3 4

,b b t b t b t a a t a t a t ... (10)

2 22 3 4 2 3 4

2 3 2 3 ,b b t b t a a t a t ... (11)

and 2 4 2 42 6 2 6 .b b t a a t ... (12)

(ii) The derivative 3 3/ G x of G has a discontinuity of magnitude – 1/ p0 (t) at the point

,x t where 0 ( )p x coefficient of the highest order derivative in (1) = 1. Thus,

3 3 3 30 0

( / ) ( / )x t x tG x G x

= –1

or 6b4 – 6a4 = – 1. ... (13)



(iii) Green’s function must satisfy the boundary conditions (2a), (2b), (2c) and (2d). Thus we have,G (0, t) = 0 so that a1 = 0, ... (14)G (0, t) = 0 so that a2 = 0, .. (15)G (1, t) = 0 so that b1 + b2 + b3 + b4 = 0, ... (16)G (1, t) = 0 so that b2 + 2b3 + 3b4 = 0 ... (17)

Let Ck = bk – ak, k = 1, 2, 3, 4.Then (10), (11), (12) and (13) may be re-written as

C1 + C2 t + C3 t2 + C4t3 = 0, ... (19)

C2 + 2C3t + 3C4t2 = 0, ... (20)

2C3 + 6C4 t = 0 ... (21) 6 C4 = – 1, ... (22)

Solving (19), (20), (21) and (22), we have3 2

1 2 3 4/ 6, / 2, / 2, 1/ 6.C t C t C t C ... (23)

In view of (18), relations (23) take the forms :b1 – a1 = t3 /6, ... (24)b2 – a2 = – t2/2, ... (25)b3 – a3 = t/2, ... (26)

and b4 – a4 = – 1/6. ... (27)Solving (14), (15), (16), (17), (24), (25), (26) and (27), we have

2 3 2 31 2 3 4

3 2 3 2 2 31 2 3 4

0, 0, / 2 / 2; 1/ 6 / 2 / 3

/ 6, / 2, / 2 , / 2 / 3

a a a t t t a t t

b t b t b t t b t t

... (28)

Substituting the above values given by (28) in (9), the required Green’s function is3 2 3

2 2 3

3 2 32 2 3

1 , 02 2 6 2 3

( , )1 , 1.

2 2 6 2 3

t t t tx t x x t

G x tx x x xt x t t x

... (29)

Ex.5. Construct the Green’s function for the boundary value problem d2y/dx2 – y = 0;y (0) = y (0) and ( ) ( ) 0. y l y l

Sol. Given boundary value problem isd2y/dx2 – y = 0 or (D2 – 1) y = 0, where / .D d dx ... (1)

with boundary conditions : y (0) = y (0) ... (2a)

and ( ) ( ) 0 y l y l ... (2b)The general solution of (1) is y (x) = Aex + B e– x ... (3)From (3), y (x) = Aex – Be– x ... (4)From (3) and (4), y (0) = A + B, y (0) = A – B, y (l) = Ael + Be–l and y (l) = Ael – Be–l

(2a) A + B = A – B so that B = 0 ... (5)

Using B = 0, (2b) 0 l lAe Ae giving A = 0.

Thus, A = B = 0 and so (3) yields only the trivial solution y (x) = 0 for the given boundaryvalue problem. Therefore, the Green’s function G (x, t) exists and is given by



1 2

1 2

, 0( , )

,

x x

x x

a e a e x tG x t

b e b e t x l ... (6)

where a1, a2, b1 and b2 are unknown functions of t. In addition to the above property (6), theproposed Green’s function must satisfy the following three properties :

(i) G (x, t) is continuous at x = t, that is,b1e

t + b2 e–t = a1 et + a2 e–t so that (b1 – a1) et + (b2 – a2) e

–t = 0 ... (7)(ii) The derivative of G (x, t) has a discontinuity of magnitude –1/p0 (t) at the point x = t,

where p0 (x) = coefficient of the highest order derivative in (1) = 1, that is,

0 0 0( / ) ( / ) 1/ ( )

x t x tG x G x p t

or (b1 et – b2 e–t) – (a1 et – a2e–t) = – 1 or (b1 – a1) e

t – (b2 – a2) e–t = –1 ... (8)

(iii) The Green’s function must satisfy the prescribed boundary conditions (2a) and (2b).

Now, (2a) 0 0 1 2 1 20 0( ( , )) ( / )

x x x xx x x x

G x t G x a e a e a e a e

Thus, a1 + a2 = a1 – a2 so that a2 = 0 ... (9)

and (2b) 1 2 1 2( ( , )) ( / ) 0 0x x x xx l x l x l x l

G x t G x b e b e b e b e

Thus, 1 2 1 2( ) 0l l l lb e b e b e b e or

1 2(1 ) (1 ) 0 l lb e b e ... (10)

Setting b1 – a1 = C1 and b2 – a2 = C2 , (7) and (8) yieldC1 et + C2 e–t = 0 and C1 et – C2 e–t = –1

Solving for C1 and C2, C1 = –(1/2) × e–t and C2 = (1/2) × et

i.e., b1 – a1 = – (1/2) × e–t and b2 – a2 = (1/2) × et ... (11)Solving (9), (10) and (11), we have

a2 = 0, 21 ,2

tb e 21

1 1 ,2 1

t lb e 2

11 1 12 1 2

t l ta e e

Substituting these values in (6), the required Green’s function G (x, t) is given by

2

2

1 1 1 , 02 1 2

( , )1 1 1 ,2 1 2

x t l x l

x t l x l

e e x tG x t

e e t x l

, where | | 0

EXERCISEIn each of the following boundary value problem examine whether a Green’s function exists and ifit does, construct it.

1. 0; (0) (1), (0) (1).y y y y y

[Ans. Since y (0) = 0 has an infinity of solutions y (x) = C satisfying the given boundarycondition y (0) = y (1) and y (0) = y (1), it follows that Green’s function does not exist for the givenboundary value problem.]

2. y = 0; y (0) = 0, y (1) = y (1) [Ans. No.]3. y + y = 0; y (0) = y (1), y (0) = y (1). [Ans. No.]



4. 0; (0) ( ) 0.y y y y [Ans. No.]5. y = 0; y (0) = y (1) = 0, y (0) + y (1) = 0. [Ans. No.]6. y = 0 ; y (0) = y (1), y (0) = y (1). [Kanpur 2006, 10; Meerut 2008, 12]

Ans. 1 (2 ), 0( , )

(1 ) 1, 1.

t x t x tG x t

x t t x

7. d4y/dx4 = 0, y (0) = y (0) = y (1) = y(1) = 0. Ans. 2

2

(1/ 6) ( 3 ), 0( , )

(1/ 6) ( 3 ), 1.

x x t x tG x t

t t x t x

8. y = 0; y (0) = y (1) = 0; y (0) = y (1).

Ans.(1/ 2) (1 ) ( 2 ), 0

( , )(1/ 2) [ (2 ) (2 ) ], 1.

t x xt t x tG x t

t x x t t t x

9. y = 0; y (0) = y (1) = 0; y (0) = y (1). Ans.( / 2) ( ) (1 ), 0

( , )( / 2) ( ) ( 1), 1.x x t t x t

G x tt t x x t x

10. 2 0 ( 0); (0) (1) 0.y k y k y y Ans.

sinh ( 1) sinh , 0sinh

( , )sinh sinh ( 1) , 1.

sinh

k t kx x tk k

G x tkt k x t xk k

11. 0; (0) (1), (0) (1). y y y y y y Ans.

1 1 1cos cosec , 02 2 2

( , )1 1 1cos cosec , 1.2 2 2

x t x tG x t

t x t x

12. (0) 0; (0) (0), (1) (1).y y Ay y B y Ans.

( 1) [ ( 1) 1] , 0 ,( , )

( 1) [ ( 1) 1] , 1.

Ax B t x tA B AB

G x tAt B x t x

A B AB

13. 2 2 0; ( )x y xy y x is bounded for 0, (1) (1).x y y

Ans.(1/ ) 1, 0 ,

( , )(1/ ) 1, 1.

t x tG x t

x t x

14. 3 4 4 2 3 3 2 2( / ) 6 ( / ) 6 ( / ) 0; ( )x d y dx x d y dx x d y dx y x is bounded as 0x

and (1) (1) 0.y y Ans.

2

2

( 1)log 1 , 0 ,2( , )

( 1)log 1 , 1.2

e

e

x tt t x ttG x t

t xx x t xx

15. 2 0, ( )x y xy y y x is bounded as 0x , y (1) = 0

Ans.2( / 2) {(1/ ) 1}, 0 ,( , )

(1/ 2) {(1/ ) }, 1.x t x tG x t

x x t x

16. (1/ ) 0; (0) xy y x y y is finite, y (1) = 0 Ans. ( / 2) {(1/ ) 1}, 0 ,

( , )( / 2) {(1/ ) }, 1.x t x t

G x tt x x t x



17. 2 2 0, (0)x y xy n y y is finite, y (1) = 0

Ans.2

(1/ 2 ) {( / ) ( ) }, 0( , )

( / 2 ) {( / ) ( ) }, 1.

n n

n

nt x t xt x tG x t

t nt t x xt t x

18. 2 (log 1) 0, (0) ex x y xy y y is finite, y (1) = 0.

Ans.2 2

2

log, 0

(log 1)( , )

log, 1.

(log 1)

e

e

e

e

x tx t

t tG x t

xt x

t t

19. 2(1 ) 0; (0) 0, (1)d dyx y ydx dx

is finite. Ans.

1 1log , 02 1

( , )1 1log , 1.2 1

e

e

x x tx

G x tt t xt

20. 0; (0)xy y y is bounded, y (l) = 0. Ans.log ( / ), 0 ,

( , )log ( / ), 1.

e

e

l t x tG x t

l x t x

11.6 SOLVED EXAMPLES BASED ON RESULT 1 OF ART. 11.3.Ex.1. Using Green’s function, solve the boundary value problem

, (0) ( / 2) 0.y y x y y [Kanpur 2007, Meerut 2006]Sol. Given boundary value problem (y + y) – x = 0 ... (1)

with the boundary conditions : (0) ( / 2) 0 y y ... (2)Consider the associated boundary value problem :

y + y = 0 or 2( 1) 0, /D y D d dx ... (3)

subject to the boundary conditions y (0) = 0 ... (4a)

and ( / 2) 0. y ... (4b)We shall first find the Green’s function of the above mentioned boundary value problem

given by (3), (4a) and (4b).The auxiliary equation of (3) is D2 + 1 = 0 so that D = ± i.Hence the general solution of (3) is y (x) = A cos x + B sin x. ... (5)Putting x = 0 in (5) and using B.C. (4a), we get A = 0. ... (6)

Putting / 2 x in (5) and using B.C. (4b), we get B = 0. ... (7)

From (6) and (7), A = B = 0. Hence (5) yields only the trivial solution ( ) 0.y x Therefore,Green’s function exists for the boundary value problem given by (3), (4a) and (4b) and it is given by

1 2

1 2

cos sin , 0( , )

cos sin , / 2.

a x a x x tG x t

b x b x t x... (8)

In addition to the above property (8), the proposed Green’s functions must also satisfy thefollowing three properties :



(i) G (x, t) is continuous at x = t, that is,b1 cos t + b2 sin t = a1 cos t + a2 sin t

or (b1 – a1) cos t + (b2 – a2) sin t = 0. ... (9)(ii) The derivative of G has a discontinuity of magnitude – 1/p0 (t) at the point x = t, where

p0 (x) = coefficient of the highest order derivative in (3) = 1, that is,

0 0( / ) ( / ) 1x t x tG x G x

or – b1 sin t + b2 cos t – (– a1 sin t + a2 cos t) = – 1or – (b1 – a1) sin t + (b2 – a2) cos t = – 1. ... (10)

(iii) G (x, t) must satisfy the boundary conditions (4a) and (4b), that is,G (0, t) = 0 so that a1 = 0 ... (11)

and ( / 2, ) 0 G t so that b2 = 0. ... (12)Let b1 – a1 = C1 and b2 – a2 = C2. ... (13)Then (9) and (10) may be written as

1 2cos sin 0 0 C t C t ... (14)

and 1 2sin cos 1 0. C t C t ... (15)Solving (14) and (15) by cross-multiplication method, we have

1 22 2

1sin cos cos sin

C Ct t t t

Hence C1 = sin t and C2 = – cos t.

1 1 sin , b a t by (13) ... (16)

and2 2 cos ,b a t by (13) ... (17)

Solving (11), (12), (16) and (17), we havea1 = 0, b2 = 0, b1 = sin t, a2 = cos t.

Substituting the above values in (6), we have

cos sin , 0( , )

sin cos , / 2.

t x x tG x t

t x t x... (18)

Then we known that the solution of the given boundary value problem (1)—(2) is given by/ 2

0( ) ( , ) ( ) ,

y x G x t t dt ... (19)

where ( ) x x so that ( ) . t t [Refer result 1 of Art. 11.3]Hence the required solution is given by

/ 2 / 2

0 0( ) ( , ) ( , ) ( , )

x

xy x G x t t dt t G x t dt t G x t dt

/ 2

0sin cos cos sin ,

x

xt t x dt t t x dt

using (18)

/ 2

0cos sin sin cos ,

x

xx t t dt x t t dt

/ 2/ 20 0

cos [ cos ] ( cos ) sin [ sin ] sinxx

x xx t t t dt x t t t dt



/ 20cos cos sin sin / 2 sin [cos ]x

xx x x t x x x t

cos ( cos sin ) sin {( / 2) sin cos } x x x x x x x x2 2cos sin cos ( / 2) sin sin sin cosx x x x x x x x x

Thus, ( ) ( / 2) sin .y x x x

Ex. 2. Using Green’s function, solve the boundary value problem y – y = x, y (0) = y (1) = 0.[Meerut 2004, 09]

Sol. Given y – y = x, y (0) = y (1) = 0.Consider the associated boundary value problem :

20 or ( 1) 0, /y y D y D d dx ... (1)

with the boundary conditions : y (0) = 0. ... (2a)and y (1) = 0. ... (2b)

The auxiliary equation of (1) is D2 – 1 = 0 so that D = ± 1.Hence the general solution of (1) is y (x) = A cosh x + B sinh x. ... (3)Putting x = 0 in (3) and using B.C. (2a), we get A = 0. ... (4)Again, putting x = 1 in (3) and using B.C. (2b), we get

0 = A cosh 1 + B sinh 1. ... (5)

Solving (4) and (5), we get A = B = 0. Hence (3) yields only the trivial solution ( ) 0y x forthe boundary value problem given by (1), (2a) and (2b). Therefore, Green’s function exists and isgiven by

1 2

1 2

cosh sinh , 0 ,( , )

cosh sinh , 1.

a x a x x tG x t

b x b x t x ... (6)


(i) G (x, t) is continuous at x = t, this is,

1 2 1 2cosh sinh cosh sinh b t b t a t a t

or 1 1 2 2( ) cosh ( ) sinh 0 b a t b a t ... (7)

(ii) The derivative of G has a discontinuity of magnitude –1/p0 (t) at the point x = t, wherep0 (x) = coefficient of the highest order derivative in (1) = 1, that is,

0 0( / ) ( / ) 1x t x tG x G x

or b1 sinh t + b2 cosh t – (a1 sinh t + a2 cosh t) = – 1.or (b1 – a1) sinh t + (b2 – a2) cosh t + 1 = 0 ... (8)

(iii) G (x, t) must satisfy the boundary conditions (2a) and (2b), that is,G (0, t) = 0 so that a1 = 0 ... (9)

and G (1, t) = 0 so that b1 cosh 1 + b2 sin 1 = 0. ... (10)Let b1 – a1 = C1 and b2 – a2 = C2. ... (11)Then (7) and (8) may be written as

C1 cosh t + C2 sinh t + 0 = 0 ... (12)and C1 sinh t + C2 cosh t + 1 = 0. ... (13)



Solving (12) and (13) by cross-multiplication method, we get

1 22 2

1 .sinh cosh cosh sinh

C Ct t t t

Hence C1 = sinh t and C2 = – cosh t. [ cosh2 t – sinh2 t = 1] b1 – a1 = sinh t, by (11) ... (14)

and b2 – a2 = – cosh t, by (11) ... (15)Solving (9), (10), (14) and (15), we have

1 1 2 2sinh cosh 1 sinh cosh 10, sinh , cosh ,

sinh 1 sinh 1t ta b t a t b ... (16)

Now,1 2

sinh cosh 1cosh sinh 0 cosh sinhsinh 1

ta x a x t x

sinh (sinh cosh 1 cosh sinh 1) sinh sinh ( 1)sinh 1 sinh 1

x t t x t

and 1 2sinh cosh 1cosh sinh sinh cosh sinh

sinh 1

tb x b x t x x

sinh (sinh cosh 1 cosh sinh 1) sinh sinh ( 1) .sinh 1 sinh 1

t x x t x

Substituting the above values in (6), the Green’s function of the boundary value problemgiven by (1), (2a) and (2b) is given by


( , )sinh sinh ( 1) , 1.

sinh 1

x t x tG x t

t x t x

... (17)

Hence the solution of the given boundary value problem y – y – x = 0, y (0) = y (1) = 0 is1

0( ) ( , ) ( ) y x G x t t dt or

1

0( ) ( , ) y x G x t t dt

[Refer result 1 of Art. 11.3. Here ( ) so that ( )x x t t ]

or1

0( ) ( , ) ( , )

x

xy x G x t t dt G x t t dt

1

0

sinh sinh ( 1) sinh sinh ( 1) ,sinh 1 sinh 1

x

x

t t x t x tdt dt using (17)

1

0

sinh ( 1) sinhsinh sinh ( 1)sinh 1 sinh 1

x

x

x xt t dt t t dt

0 0

sinh ( 1) [ cosh ] coshsinh 1

xxx t t t dt

11sinh [ cosh ( 1)] cosh ( 1)sinh 1

x x

x t t t dt , integrating by parts

1

0sinh ( 1) sinhcosh [sinh ] 1 cosh ( 1) [sinh ( 1)]

sinh 1 sinh 1

xx

x xx x t x x t



sinh ( 1) sinh( cosh sinh ) [1 cosh ( 1) sinh ( 1)]sinh 1 sinh 1

x xx x x x x x

sinh [sinh ( 1) cosh cosh ( 1) sinh ]sinh 1 sinh 1

x x x x x x sinh sinh ( 1 )

sinh 1 sinh 1

x x x x

sinh sinh sinh 1 sinh( ) sinh ( 1) or ( ) ,sinh 1 sinh 1 sinh 1 sinh 1 sinh 1

x x x x xy x y x x

which is the required solution of the given boundary value problem.

EXERCISE1. Solve the following boundary-value problems using Green’s functions :

(i) 2 cos ; (0) (1), (0) (1).y y x y y y y Ans. (1/ 4 ) (2 1) siny x x

(ii) 2 ; (0) ( / 2) 0.y y x y y Ans. 2 22 cos (2 / 4) sin 2y x x x

(iii) 2 sinh 1; (0) (1) 0.y y y y Ans. y = 2 [sinh x – sinh (x – 1) – sinh 1]

(iv) 2 ; (0) (0), ( ) ( ) 0.xy y e y y y l y l Ans. y = sinh x + ex (l – x)

(v) ; (1) ( ) 0.xy y x y y e Ans. 2 2(1/ 4) [(1 ) log 1]ey e x x

(vi) 4 4/ 1 ; (0) (0) (1) (1) 0.d y dx y y y y Ans. 2 2( / 24) ( 4 6)y x x x

2. Transform the problem 4 4/ ( ) 0, (0) (0) (1) (1) 0d y dx x y y y y to therelation

1

0( ) ( , ) ( ) , y x G x t t dt where

2 2

2 2

(1/ 6) (1 ) (2 3 ), when( , )

(1/ 6) (1 ) (2 3 ), when .

x t xt x t x tG x t

t x xt t x x t

3. Find Green’s function for the boundary value problem 2 2( / ) ( / ) ,d u dx du dx x

(0) (1) 0.u u Hence solve it. [Meerut 2002, 07]

11.7 SOLVED EXAMPLES BASED ON RESULT 2 OF ART. 11.3.Ex.1. Reduce the boundary-value problem , (0) ( / 2) 0 y y x y y to an integral

equation.Sol. Given boundary-value problem is 0, (0) ( / 2) 0.y y x y y ... (1)We shall first find Green’s function of the following associated boudary value problem :

y = 0 or D2 y = 0, /D d dx ... (2)with boundary conditions. y (0) = 0 ... (3)and ( / 2) 0. y ... (4)

The auxiliary equation of (2) is D2 = 0 so that D = 0, 0. So the general solution of (2) isy (x) = Ax + B. ... (5)

Putting x = 0 is (5) and using B.C. (3), we get B = 0. ... (6)Next, putting / 2 x in (5) and using B.C. (4), we get 0 ( / 2) .A B ... (7)

From (6) and (7), A = B = 0. Hence (5) yields only the trivial solution ( ) 0.y x Therefore,Green’s function G (x, t) exists for the associated boundary value problem given by (2), (3) and (4)and is given by



1 2

1 2

, 0( , )

, / 2.

a x a x tG x t

b x b t x ... (8)

In addition to the above property (8), the proposed Green’s function must also satisfy thefollowing properties :

(i) G (x, t) is continuous at x = t, that is,b1t + b2 = a1t + a2 or (b1 – a1) t + b2 – a2 = 0. ... (9)

(ii) The derivative of G has a discontinuity of magnitude – 1 / p0 (t) at the point x = t,where p0 (x) = coefficient of the highest order derivative in (2) = 1, that is,

0 0( / ) ( / ) 1x t x tG x G x

or b1 – a1 = – 1. ... (10)(iii) G (x, t) must satisfy the boundary condition (2) and (4), that is,G (0, t) = 0 so that a2 = 0 ... (11)

and ( / 2, ) 0 G t so that 1 2( / 2) 0. b b ... (12)Using (10), (9) gives – t + b2 – a2 = 0 ... (13)Solving (10), (11), (12) and (13), we have

2 2 1 10, , (2 / ), 1 (2 / ).a b t b t a t

1 2 1 2{1 (2 / )} and {1 (2 / )} .a x a t x b x b x t

Substituting the above values in (8), we have

{1 (2 / )} , 0 ,( , )

{1 (2 / )} , / 2.

t x x tG x t

x t t x ... (14)

Comparing 0 y y x with ( ) 0, y x we have

( ) ( ) x y x x so that ( ) ( ) . t y t t ... (15)Also, we known that, if G (x, t) is Green’s function of the boundary-value problem given by

(2), (3), (4) then the boundary value problem (1) can be reduced to the following integral equation[refer equation (8) in result 2 of Art. 11.3]

/ 2 / 2

0 0( ) ( , ) ( ) ( , ) [ ( ) ]

y x G x t t dt G x t y t t dt , using (15)

or/ 2 / 2

0 0( ) ( , ) ( ) ( , ) .

y x G x t y t dt t G x t dt ... (16)

Now, we have/ 2 / 2

0 0( , ) ( , ) ( , )

x

xt G x t dt t G x t dt t G x t dt

/ 22

0

2 21 1 ,

x

x

x tt dt t x dt using (14)

/ 22 3 2 3/ 22

00

2 2 2 21 13 2 3

x

x

x

x

x t x t t tt dt x t dt x

3 2 2 2 3 3 22 21

3 8 12 2 3 6 24x x x x x xx



Substituting the above value in (16), we obtain the required integral equation3 2/ 2

0( ) ( , ) ( ) .

6 24

x xy x G x t y t dt where G (x, t) is given by (14).

Ex.2. Reduce the following boundary value problem into an integral equation with help ofGreen’s function : 0, y y y (0) = 0, 2(1) (1) 0 y v y

Sol. Given boundary value problem is

0, y y y (0) = 0, 2(1) (1) 0 y v y ... (1)We shall first find the Green’s function of the following associated boundary value problem :

y = 0 or D2y = 0, where /D d dx ... (2)subject boundary condition y (0) = 0 ... (3)and y (1) + v2 y (1) = 0 ... (4)

The general solution of (2) is y (x) = Ax + B. ... (5)From (5), y (x) = A ... (6)Putting x = 0 in (5) and using (3), we have B = 0Hence (5) reduces to y (x) = Ax ... (7)From (6) and (7), (1)y A and y (1) = A

Substituting these values in (1), we get A (1 + v2) = 0 so that A = 0 if 2 1. vThus, A = B = 0. Hence (5) yields only the trivial solution y (x) = 0. Therefore, Green’s

function G (x, t) exist for the associated boundary value given by (2), (3) and (4), and is given by

1 2

1 2

, 0( , )

, 1.

a x a x tG x t

b x b t x... (8)

where a1, a2, b1, b2 are unknown functions of t. In addition to the above property (8), the proposedGreen function must satisfy the following three properties :

(i) G (x, t) is continuous at x = t, that is,b1t + b2 = a1t + a2 or (b1 – a1) t + (b2 – a2) = 0 ... (9)

(ii) The derivative of G (x, t) has a discontinuity of the magnitude – 1/p0 (t) at x = t, wherep0 (x) = coefficient of the highest order derivative in (2) = 1. Thus, we must have

0 0 0( / ) ( / ) 1 / ( )

x t x tG x G x p x

i.e., b1 – a1 = – 1 ... (10)(iii) The Green’s function must satisfy the prescribed boundary conditions (2a) and (2b).Now, (2a) (G (x, t))x = 0 = 0 a2 = 0 ... (11)

and (2b) 1 2 1( / ) [ ( , )] 0

x xG x v G x t

1 2 1 2 11[ ] 0

xxb v b x b

Thus, b1 + v2 (b1 + b2) = 0. ... (12)Solving (9), (10), (11) and (12), we have

a2 = 0, b2 = t, b2 = – (v2t)/ (1 + v2), 1 2 2 2{1 (1 )} / (1 ), 1 a v t v v

Now, 2 21 2 1 2

2 2

1 (1 ) 1 (1 )and

1 1v t v x

a x a x b x b tv v

... (13)




2 2

2 2

[1 (1 )] /(1 ), 0( , )

[1 (1 )] /(1 ), 1

v t x v x tG x t

v x t v t x

... (14)

Comparing 0 y y with ( ) 0, y x we have

( ) ( ) x y x so that ( ) ( ) t y t ... (15)Refer equation (8) in result 2 of Art. 11.3. According to this result, if G (x, t) is Green’s

function of boundary value problem given by (2), (3) and (4), then the given boundary value problemvalue problem (1) can be transformed into the following integral equation

1

0( ) ( , ) ( ) y x G x t t dt i.e.,

1

0( ) ( , ) ( ) , y x t G x t y t dt

where G (x, t) is given by (14).Ex.3. Reduce the boundary-value problem y + y = x, y (0) = 0, y (1) = 0 to a Fredholm

integral equation.Sol. Given boundary value problem is

y + y – x = 0, y (0) = 0, y (1) = 0. ... (1)We shall first find Green’s function of the following associated boundary-value problem

0y or D2y = 0, / .D d dx ... (2)with boundary conditions y (0) = 0 ... (3)and y (1) = 0. ... (4)

The auxilliary equation of (2) is D2 = 0 so that D = 0, 0.Hence the general solution of (2) is y (x) = Ax + B. ... (5)Putting x = 0 is (5) and using B.C. (3), we get B = 0. ... (6)From (5), we have y (x) = A.Putting x = 1 in above relation and using B.C. (4), we get A = 0. ... (7)From (6) and (7), A = B = 0. Hence (5) yields only the trivial solution ( ) 0.y x Therefore,

Green’s function G (x, t) exists for the associated boundary- value problem given by (2), (3) and(4) and is given by

1 2

1 2

, 0( , )

, 1

a x a x tG x t

b x b t x... (8)


(i) G (x, t) is continuous at x = t, that is,b1t + b2 = a1t + a2 or t (b1 – a1) + b2 – a2 = 0. ... (9)

(ii) The derivative of G has a discontinuity of magnitude –1/p0 (t) at the point x = t, wherep0 (x) = coeff. of the highest order derivative in (2) = 1, that is

0 0( / ) ( / ) 1x t x tG x G x

or b1 – a1 = – 1. ... (10)

(iii) G (x, t) must satisfy the boundary conditions (3) and (4), that is,G (0, t) = 0 so that a2 = 0 ... (11)

and G (1, t) = 0 so that b1 = 0. ... (12)From (9) and (10), b2 – a2 = t. ... (13)Solving (10), (11), (12) and (13), we have a2 = 0, b1 = 0, b2 = t, a1 = 1.



Substituting these values in (8), we get , 0( , )

1.

x x tG x t

t t x... (14)

Comparing y + y – x = 0 with ( ) 0, y x we get

( ) ( ) x y x x so that ( ) ( ) . t y t t ... (15)Also, we know that, if G (x, t) is Green’s function of the associated boundary-value problem

(given by (2), (3) and (4)), then the given boundary value problem (1) can be reduced to thefollowing Fredholm integral equation. [Refer equation (8) in result 2 of Art. 11.3]

1 1

0 0( ) ( , ) ( ) ( , ) [ ( ) ] y x G x t t dt G x t y t t dt

or1 1

0 0( ) ( , ) ( ) ( , ) . y x G x t y t dt t G x t dt ... (16)

Now, we have1

0( , ) t G x t dt

1 12

0 0( , ) ( , ) ,

x x

x xt G x t dt t G x t dt t dt xt dt using (14)

13 2 32 3

0

1(1 ) (3 ).3 2 3 2 6

x

x

t t x xx x x x

Substituting the above value in (16), we obtain the required Fredholm integral equation

1 3

0

1( ) ( , ) ( ) (3 ).6

y x G x t y t dt x x Ex.4. Transform the problem 1,y xy y (0) = y (1) = 0 to the integral equation

1

0

1( ) ( , ) ( ) (1 ),2

y x G x t t y t dt x x

where G (x, t) = x (1 – t) when x < t and G (x, t) = t (1 – x) when x > t.Sol. Given boundary-value problem is 1 0, (0) (1) 0.y xy y y ... (1)We shall first find Green’s function of the following associated boundary-value problem

y = 0 or D2y = 0, /D d dx ... (2)with boundary conditions y (0) = 0. ... (3)and y (1) = 0. ... (4)

The auxiliary equation of (2) in D2 = 0 so that D = 0, 0. So general solution of (2) isy (x) = Ax + B. ... (5)

Putting x = 0 in (5) and using B.C. (3), we get 0 = B. ... (6)Putting x = 1 in (5) and using B.C., (4), we get 0 = A + B. ... (7)From (6) and (7), A = B = 0. Hence (5) yields only the trivial solution ( ) 0.y x Therefore,

Green’s function G (x, t) exists for the associated boundary-values problem (given by (2), (3) and(4)) and is given by

1 2

1 2

, 0( , )

, 1.

a x a x tG x t

b x b t x... (8)




(i) G (x, t) is continuous at x = t, that is,b1t + b2 = a1t + a2 or t (b1 – a1) + b2 – a2 = 0. ... (9)

(ii) The detervative of G has a discontinuity of magnitude –1/p0 (t) at the point x = t, wherep0 (x) = coefficient of the highest order derivative in (2) = 1, that is,

0 0( / ) ( / ) 1x t x tG x G x

or b1 – a1 = – 1. ... (10)

(iii) G (x, t) must satisfy the boundary conditions (3) and (4), that is,G (0, t) = 0 so that a2 = 0 ... (11)

and G (1, t) = 0 so that b1 + b2 = 0. ... (12)From (9) and (10), b2 – a2 = t. ... (13)Solving (10), (11), (12), and (13), we have a2 = 0, b2 = t, b1 = – t, a1 = 1 – t.Substituting these values in (8), we have

(1 ), 0( , )

(1 ), 1.

x t x tG x t

t x t x... (14)

Comparing y + xy – 1 = 0 with ( ) 0, y x we get

( ) ( ) 1 x xy x so that ( ) ( ) 1. t t y t ... (15)Also, we know that, if G (x, t) is Green’s function of the associated boundary-value problem

(given by (2), (3) and (4)), then the given boundary value problem (1) can be reduced to thefollowing integral equation. (refer equation (8) in result 2 of Art. 11.3)

1 1

0 0( ) ( , ) ( ) ( , ) [ ( ) 1] , y x G x t t dt G x t ty t dt using (15)

or1 1

0 0( ) ( , ) ( ) ( , ) .y x G x t t y t dt G x t dt ... (16)

Now,1 1 1

0 0 0( , ) ( , ) ( , ) (1 ) (1 ) ,

x x

x xG x t dt G x t dt G x t dt t x dt x t dt by (14)

12 2 2 2

0

1(1 ) (1 ) 12 2 2 2 2

x

x

t t x xx x t x x x

2 3 3 2

2 1 (1 ).2 2 2 2 2 2 2

x x x x x xx x x

Substituting the above value in (16), we obtain the required integral equation1

0

1( ) ( , ) ( ) (1 ),2

y x G x t t y t dt x x where G (x, t) is given by (14)

Ex.5. Reduce the Bessel equation 2 2 2 2( / ) ( / ) ( 1) 0x d y dx x dy dx x y with endconditions y (0) = 0, y (1) = 0, to a Fredholm integral equation. [Kanpur 2005, 08; Meerut 2000, 04, 05]

Sol. Given 2

2 22 0

d y dyx x x y ydxdx

or 2

2 0 d y dy yx xy

dx xdx

or1 0.d dx y xy

dx dx x

... (1)



Comparing (1) with ( ) ( ) ( ) L y r x y x f x

i.e., ( ) ( ) ( ).

d dyp q y r x y x f xdx dx ... (2)

we have p (x) = x, q (x) = – 1/x, r (x) = x, f (x) = 0. ... (3)We shall first find Green’s function for the boundary value problem

Ly = 0 or1 0,

d dx ydx dx x ... (4)

with boundary conditions y (0) = 0 ... (5)and y (1) = 0 ... (6)

We know that the required Green’s function G (x, t) is given by (refer Art. 11.4)(1/ ) ( ) ( ), 0

( , )(1/ ) ( ) ( ), 1.

A u x v t x tG x t

A u t v x t x ... (7)

where A is given by the Abel’s formula / ( ) ( ) ( ) ( ) ( ),A p t u t v t v t u t ... (8)u (t) and v (t) being two independent solution of (4) satisfying boundary conditions (5) and (6)respectively.

Re-writing (4), we have 0

d dy yxdx dx x or

2

2 0 d y dy yx

dx xdx

or 2 2 2( / ) ( / ) 0x d y dx x dy dx y or (x2 D2 + xD – 1) y = 0, /D d dx ... (9)Equation (9) is homogeneous differential equation. To solve (9), we proceed as follows.Let x = ez so that z = log x. ... (10)Also, let 1 / .D d dz Then, we have

1xD D and 2 21 1

( 1). x D D D ... (11)Using (11), (9) reduces to

[D1 (D1 – 1) + D1 – 1] y = 0 or 21

( 1) 0. D y ... (12)

The auxiliary equation of (12) is 21

1 0 D so that D1 = 1, – 1Hence the general solution of (12) is y = C1 ez + C2 e–z = C1 ez + C2 (e

z)–1

or y (x) = C1 x + C2 x–1 using (10) ... (13)As a solution for which B.C. (5) is satisfied, we may take y = u (x), where

u (x) = x. ... (14)and as a solution for which (6) is satisfied, we may take y = v (x), where

v (x) = (1/x) – x. ... (15)The Wronskian W [u (x), v (x)] of u (x) and v (x) is given by

2 2

(1/ )( ) ( ) 1 1[ ( ), ( )] 1 1( ) ( ) 1 (1/ ) 1

x x xu x v xW u x v x x

u x v x xx x

(1/ ) (1/ ) (2 / ) 0,x x x x x

showing that u (x) and v (x) are independent solutions of (4).From (3), (14) and (15), we have p (t) = t, u (t) = t, v (t) = (1/t) – t. ... (16)



Substituting these values in (8), we have

21 1 1 11 1A d dtt t t t

t dt t t dt tt

A/t = – 2/t so that A = – 2. ... (17)

Using (14), (15), (16) and (17), 2( ) ( ) 1 (1 )2 2

u x v t x xt tA t t

... (18)

and 2( ) ( ) 1 (1 ).

2 2

u t v x t tx xA x x ... (19)

Using (18) and (19), (7) becomes2

2

( / 2 ) (1 ), 0( , )

( / 2 ) (1 ), 1.

x t t x tG x t

t x x t x

... (20)

Hence the boundary value problem ( ) ( ) ( ), (0) (1) 0Ly r x y x f x y y reduces to the integral equation (refer equation (12) in Art. 11.3)

1 1

0 0( ) ( , ) ( ) ( ) ( , ) ( ) y x G x t r t y t dt G x t f t dt

or1

0( ) ( , ) ( ) ( ) . y x G x t r t y t dt [ f (x) = 0 f (t) = 0, by (3)]

Ex.6. (a) Show that the Green’s function for the Bessel operator of order n,2

, ( 0)d dy nLy x y ndx dx x

relevant to the end conditions y (0) = y (1) = 0, is of the form

2

2

( / 2 ) (1 ) when( , )

( / 2 ) (1 ) when ,

n n n

n n n

x nt t x tG x t

t nx x x t

(b) Use the result of part (a) to reduce the problem2

2 2 22 ( ) 0, (0) (1) 0

d y dyx x x n y y ydxdx

to an integral equation, when 0.n

Sol. Part (a) We shall find Green’s function for the boundary-value problem

Ly = 0 or2

0,

d d nx ydx dx x

... (1)

with the boundary conditions y (0) = 0 ... (2)and y (1) = 0. ... (3)

We know that the required Green’s functions G (x, t) is given by (refer Art. 11.4)(1/ ) ( ) ( ), 0

( , )(1/ ) ( ) ( ), 1,

A u x v t x tG x t

A u t v x t x

... (4)

where A is given by the Abel’s formula / ( ) ( ) ( ) ( ) ( ), A p t u t v t v t u t ... (5)u (t) and v (t) being two independent solutions of (1) satisfying boundary conditions (5) and (6)respectively.

Comparing (1) with the differential operator ,

d dL p qdx dx

... (6)



we have p (x) = x so that p (t) = t. ... (7)Re-writing (1), we have

20

d dy n yxdx dx x or

2 2

2 0 d y dy n yx

dx xdxor 2 2 2 2( / ) ( / ) 0x d y dx x dy dx n y or 2 2 2( ) 0, /x D xD n y D d dx ... (8)which is homogeneous differential equation. To solve it, we proceed as follows :

Let x = ez so that z = log x. ... (9)Also, let 1 / .D d dz Then, we have

1xD D and 2 21 1

( 1).x D D D ... (10)Using (10), (8) reduces to

[D1 (D1 – 1) + D1 – n2] y = 0 or 2 21

( ) 0. D n y ... (11)

The auxiliary equation of (11) is 2 21

0 D n so that D1 = n, – n.

Hence the general solution of (11) is

1 2 1 2( ) ( ) nz nz z n z ny c e c e c e c e

or 1 2( ) , n ny x c x c x using (9) ... (12)

As a solution for which boundary condition (2) is satisfied, we may take y = u (x), whereu (x) = xn, ... (13)

and as a solution for which boundary condition (3) is satisfied, we may take y = v (x), wherev (x) = (1/xn) – xn. ... (14)

The Wronskian of u (x) and v (x) is given by

1 1 1

( ) ( ) (1/ )[ ( ), ( )]

( ) ( ) ( / )

n n n

n n n

u x v x x x xW u x v x

u x v x nx n x nx

1 2 1 2 1 2 11 0,n n n n n

nn n n nx n x n x nx nx

x x xx

showing that u (x) and v (x) are two independent solutions of (1).From (13) and (14), u (t) = tn and v (t) = (1/tn) – tn. ... (15)

Using (7) and (15), (5) becomes

1 11

1

n n n n

n nA nt nt nt tt t t

or 2 1 2 1 n nA n nnt ntt t t

or A/t = (– 2n)/t so that A = – 2n. ... (16)Using (13), (14), (15) and (16), we have

2( ) ( ) 1 ( / 2 ) (1 )2

nn n n n

nu x v t x t x nt t

A n t... (17)

and 2( ) ( ) 1 ( / 2 ) (1 ).2

nn n n n

nu t v x t x t nx x

A n x... (18)

Using (17) and (18), (4) becomes2

2

( / 2 ) (1 ), ,( , )

( / 2 ) (1 ), .

n n n

n n n

x nt t x tG x t

t nx x x t... (19)

which gives the required Green’s function.



Part (b). Given differential equation is2

2 2 22 ( ) 0

d y dyx x x n ydxdx

or2

0 dy dy nx y xydx dx x

or ( ) 0. Ly xy x ... (20)

Comparing (20) with ( ) ( ) ( ), Ly r x y x f x ... (21)here r (x) = x and f (x) = 0 so that r (t) = t and f (t) = 0 ... (22)

We know that, if G (x, t) is Green’s function of the associated boundary value problem0, (0) 0, (1) 0, Ly y y then the boundary value problem ( ) ( ) ( ) Ly r x y x f x can be reduced

to the integral equation (refer equation (12) in Art. 11.3)1 1

0 0( ) ( , ) ( ) ( , ) ( ) y x G x t r t dt G x t f t dt

or1

0( ) ( , ) ( ) , y x G x t t y t dt [ f (t) = 0, by (22)].

which is the required integral equation.

EXERCISEReduce the following boundary-value problems to integral equations :

1. 2 ; (0) ( / 2) 0.y y x y y Ans.4 3 / 2

0( ) ( , ) ( ) ,

12 90x xy x G x t y t dt

where (1 2 / ) , 0( , )

(1 2 / ) , / 2.t x x t

G x tx t t x

2. ; (0) (1) 0.xy y e y y Ans.1

0( ) 1 ( , ) ( ) , xy x e ex x G x t y t dt

where (1 ) , 0

( , )(1 ) , 1.

t x x tG x t

x t t x

3. ; (0) (0), (1) (1).xy y e y y y y

Ans.1

0( ) ( , ) ( ) , xy x e G x t y t dt where (1 ) , 0 ,

( , )(1 ) , 1.

x t x tG x t

t x t x

4. 0, (0) 0, (1) (1) 0.y y y y ky Ans.1

0( ) ( , ) ( ) ,

1

xy x G x t y t dt

k

where

1 (1 ) , 0 ,1

( , )1 (1 ) , 1.

1

k t x x tk

G x tk x t t x

k

5. 2 1; (0) (1), (0) (1).y y x y y y y Ans. y (x) = (1/6) × (2x3 + 3x2 – 17x – 5)1

0( , ) ( ) , G x t y t dt where

(2 ) 1 , 0 ,( , )

(1 ) 1, 1.

t x t x tG x t

t x t x



6. 2( / 4) cos ( / 2) ; ( 1) (1) , ( 1) (1).y y y x y y y y [Meerut 2005, 2008]

Ans. ( ) sin2

x xy x

1

2 1

2 cos ( , ) ( ) ,2

x G x t y t dt where

1 sin ( ), 1 ,2( , )

1 sin ( ), 1.2

x t x tG x t

t x t x

7. 2 ; (0) (1) 0, (0) (1).y y x y y y y Ans. 2( ) (1/12) ( 1) ( 1)y x x x x 1

0( , ) ( ) , G x t y t dt where

1/ 2 ( ) ( 1), 0 ,( , )

1/ 2 ( ) ( 1), 1.x t x t x t

G x tt t x x t x

8. 4 4/ 1; (0) (0) 0, (1) (1) 0.dy dx y y y y y

Ans.2

2( ) ( 4 6)24

xy x x x

1

0( , ) ( ) , G x t y t dt where

2

2

( 3 ) / 6, 0( , )

( 3 ) / 6, 1.

x x t x tG x t

t t x t x

11.8 SOLVED EXAMPLES BASED ON RESULT 3 OF ART. 11.3. THE CASE OFNONHOMOGENEOUS END CONDITIONS.Ex.1. Reduce the following boundary value problem into an integral equation

y + xy = 1, y (0) = 0, y (l) = 1.Sol. Given boundary value problem is y + xy = 1, y (0) = 0, y (l) = 1. ... (1)Notice that the prescribed end conditions y (0) = 0 and y (l) = 1 are not homogeneous

because the boundary conditions are respectively 0 and 1, which are unequal. We, therefore, proceedas follows.

We first consider the associated boundary value problem with homogenoeous boundarycondition, that is,

y = 0 or 2 0, /D y D d dx ... (2)

with boundary conditions y (0) = 0 ...(3)and y (l) = 0. ... (4)

We now determine Green’s function for the above associated boundary value problem.The general solution of (2) is y (x) = Ax + B. ... (5)Putting x = 0 and x = l by turn in (5) and using B.C. (3) and (4), we have

0 = B and 0 = Al + B ... (6)

From (6), A = B = 0. Hence (5) yields only the trivial solution ( ) 0.y x Therefore, Green’ssfunction G (x, t) exists and is given by

1 2

1 2

, 0 ,( , )

, .

a x a x tG x t

b x b t x l... (7)



(ii) The derivative of G has a discontinuity of magnitude –1/p0 (t) at the point x = t, wherep0 (x) = coefficient of the highest order derivative in (2) = 1, that is,

0 0( / ) ( / )x t x tG x G x

= –1 or b1 – a1 = –1 ... (9)



(iii) G (x, t) must satisfy the boundary conditions (3) and (4), that is,G (0, t) = 0 so that a2 = 0. ... (10)

and G (l, t) = 0 so that b1l + b2 = 0. ... (11)Using (9), (8) gives b2 – a2 = t. ... (12)Solving (9), (10), (11) and (12), we have a2 = 0, b2 = t, b1 = – t / l, a1 = 1 – (t / l).

1 2 1 ( )

t xa x a x l tl l ... (13)

and 1 2 ( ). tx tb x b t l xl l ... (14)


( / ) ( ), 0( , )

( / ) ( ), .x l l t x t

G x tt l l x t x l

... (15)

Comparing (1) with ( ) 0, (0) 0, ( ) 1,y x y y l ... (16)

here ( ) ( ) 1 x xy x so that ( ) ( ) 1. t ty t ... (17)Then, we know that (16) can be reduced to the following integral equation (refer equation (13)

in result 3 of Art. 11.3)

0( ) ( ) ( , ) ( )

ly x P x G x t t dt or

0( ) ( ) ( , ) [ ( ) 1]

ly x P x G x t ty t dt

or0 0

( ) ( ) ( , ) ( ) ( , ) , l l

y x P x G x t t y t dt G x t dt .. (18)

where G (x, t) is given by (15) and P (x) is the solution of the following boundary value problemwith nonhomogeneous end conditions :

P (x) = 0 or D2 P (x) = 0, /D d dx ... (19)with the boundary conditions P (0) = 0 and P (l) = 1. ... (20)

The general solution of (19) is P (x) = A x + B. ... (21)Putting x = 0 and x = l successively in (21) and using (20), we get

0 = B and 1 = lA + B,so that B = 0 and A = 1/l. Putting these values in (21), we have

P (x) = (x / l). ... (22)

Again, 0 0

( , ) ( , ) ( , ) l x l

xG x t dt G x t dt G x t dt

2 2

00

( ) ( )2 2 2

x lx l

xx

t x l x t x tl x dt l t dt ltl l l , using (15)

2 2 2 2 22 2 ( ).

2 2 2 2 2 2 2 2

l x x l x x xl x l x xx l lx lx l xl l l ... (23)

Using (22) and (23) is (18), we get1

0( ) ( ) ( , ) ( ) ,

2

x xy x l x G x t ty t dtl

which is the required integral equation, where G (x, t) is given by (15).

Ex.2. Transform the problem 2 2/ , (0) 1, (1) 0d y dx y x y y to a Fredholm integralequation.



Sol. Given y + y = x, y (0) = 1, y (1) = 0, ... (1)which possesses nonhomogeneous end conditions.

We consider the associated boundary value problem with homogeneous boundaryconditions,

y = 0 or D2 y = 0, / .d d dx ... (2)with boundary conditions y(0) = 0 ...(3)and y (1) = 0. ... (4)

We now determine Green’s function for the above associated boundary-value problem.The general solution of (2) is y (x) = Ax + B. ... (5)From (5), y (x) = A. ... (5)Putting x = 0 in (5) and using B.C. (3), we get 0 = B. ... (6)Next, putting x = 1 in (5) and using B.C. (4), we get 0 = A. ... (6)From (6) and (6), A = B = 0. Hence (5) yields only the trivial solution ( ) 0.y x Therefore,

Green’s functions G (x, t) exists and is given by

1 2

1 2

, 0( , )

, 1.

a x a x tG x t

b x b t x... (7)

In addition to the above property (7), the proposed Green’s function must also satisfy thefollowing properties.


(ii) The derivative of G has a discontinuity of magnitude –1/p0 (t) at the point x = t, wherep0 (x) = coefficient of the highest order derivative in (2) = 1, that is

0 0( / ) ( / ) 1x t x tG dx G dx

or b1 – a1 = – 1. ... (9)(iii) G (x, t) must satisfy the boundary conditions (3) and (4), that is,

G (0, t) = 0 so that a2 = 0 ... (10)and G (1, t) = 0 so that b1 = 0. ... (11)

Using (9), (8) reduces to b2 – a2 = t. ... (12)Solving (9), (10), (11) and (12), we have a2 = 0, b1 = 0, a1 = 1, b2 = t. a1x + a2 = x ... (13)

and b1x + b2 = t. ... (14)

Using (13) and (14), (7) reduces to, 0

( , ), 1.

x x tG x t

t t x... (15)

Comparing (1) with ( ) 0, (0) 1, (1) 0,y x y y ... (16)

here ( ) ( ) x y x x so that ( ) ( ) . t y t t ... (17)Then, we know that (16) can be reduced to the following integral equation

1

0( ) ( ) ( , ) ( ) y x P x G x t t dt

or1

0( ) ( ) ( , ) { ( ) }y x P x G x t y t t dt

or1 1

0 0( ) ( ) ( , ) ( ) ( , ) , y x P x G x t y t dt t G x t dt ... (18)

where G (x, t) is given by (15) and P (x) is the solution of the following boundary value problemwith nonhomogeneous end conditions :



P (x) = 0 or D2P (x) = 0, / .D d dx ... (19)with the boundary conditions P (0) = 1and P (1) = 0. ... (20)

The general solution of (19) is P (x) = Ax + B ... (21)From (21), P (x) = A. ... (22)Putting is x = 0 in (21) and using B.C. P (0) = 1, we get 1 = B. ... (23)Putting x = 1 in (22) and using B.C. P (1) = 0, we get 0 = A. ... (24)Using (23) and (24), (21) gives P (x) = 1. ... (25)

Now,1 1

0 0( , ) ( , ) ( , )

x

xt G x t dt t G x t dt t G x t dt

13 2 312 2

00

(1 ).3 2 3 2

xx

xx

t t x xt dt xt dt x x

... (26)

Using (25) and (26) in (18), we get31 2

0( ) 1 ( , ) ( ) (1 )

3 2

x xy x G x t y t dt x

or13

0

1( ) ( 3 6) ( , ) ( ) ,6

y x x x G x t y t dt

which is the required integral equation.Ex.4. Reduce the following boundary value problem to a Fredholm integral equation :

1, (0) 0, ( ) 0 y xy y y l .

Sol. Given boundary value problem is 1, (0) 0, ( ) 0.y xy y y l ... (1)Since the prescribed end conditions y (0) = 0, y (l) = 1 are not homogeneous, we proceed as

explained in result 3 of Art 11.3.Consider the associated boundary value problem with homogeneous boundary conditions,

namely, y = 0 or D2y = 0, where /D d dx ... (2)with boundary conditions y (0) = 0 ... (3)and y (l) = 0 ... (4)

We now proceed to find the Green’s function of the above associated boundary value problem.The general solution of (2) is y = Ax + B. ... (5)Putting x = 0 and x = l by turn in (5) and using (3) and (4), we have

0 = B and 0 = Al + B ... (6)Solving (6), A = B = 0. Hence (5) yields only the trivial solution y (x) = 0. Therefore, Green’s

function G (x, t) exists and is given by

1 2

1 2

, 0( , )

,

a x a x tG x t

b x b t x l ... (7)



(ii) The derivative of G has a discontinuity of magnitude –1/p0 (t) at the point x = t, wherep0 (x) = coefficient of the highest order in (2) = 1, that is,



0 0( / ) ( / ) 1

x t x tG x G x or b1 – a1 = – 1 ... (9)

(iii) G (x, t) must satisfy the boundary conditions (3) and (4), that isG (0, t) = 0 so that a2 = 0 ... (10)

and G (l, t) = 0 so that b1l + b2 = 0 ... (11)Using (9), (8) gives b2 – a2 = t. ... (12)Solving (9), (10), (11) and (12), we have a2 = 0, b2 = t, b1 = – t / l and a1 = 1 – (t / l)Hence a1x + a2 = (1 – t / l)x = (x / l) (l – t) ... (13)

and b1x + b2 = – (t / l) x + t = (t / l) (l – x) ... (14)Using (13) and (14), (7) reduces to

( / ) ( ), 0( , )

( / ) ( ),x l l t x t

G x tt l l x t x l

... (15)

Comparing (1) with ( ) 0, (0) 0, ( ) 1y x y y l ... (16)

here ( ) ( ) 1 x xy x so that ( ) ( ) 1 t t y t ... (17)Then (16) can be reduced to the following integral equation (refer equation (13) in result 3 of

Art. 11.3)

0( ) ( ) ( , ) ( )

ly x P x G x t t dt or

0( ) ( ) ( , ) { ( ) 1}

ly x P x G x t t y t dt

or0 0

( ) ( ) ( , ) ( ) ( , )l l

y x P x t G x t y t dt G x t dt ... (18)

where P (x) is the solution of the following boundary value problem with non-homogeneous boundaryconditions : P (x) = 0 or D2 P (x) = 0 ... (19)with the boundary conditious P (0) = 0 and P (l) = 1 ... (20)

The general solution of (19) is P (x) = Ax + B. ... (21)Putting x = 0 and x = l successively in (21) and using (20), we have

0 = B and 1 = lA + B giving B = 0 and A = 1/l.Substituting these values in (21), P (x) = (x / l) ... (22)

Now, 0 0 0

( , ) ( , ) ( , ) ( ) ( ) ,l x l x l

x x

t xG x t dt G x t dt G x t dt l x dt l t dtl l

by (15)

2 2 2 2

2 2

02 2 2 2 2

x l

x

l x t x t l x x l xlt x l lxl l l l

2 2 2

( )2 2 2 2 2

x xl x l x xlx l xl ... (23)

Using (22) and (23) in (18), we have1

0( ) ( , ) ( ) ( )

2

x xy x t G x t y t dt l xl

or 2

0( ) ( / 2 ) (2 ) ( , ) ( )

ly x x l l xl t G x t y t dt

which is the required Fredholm integral equation.



11.9 LINEAR INTEGRAL EQUATIONS IN CAUSE AND EFFECT. THE INFLUENCEFUNCTION.

Linear integral equations arise most frequently in physical problems as a result of the possibilityof superimposing the effects due to several causes. To understand the procedure, let x and t bevariables. Suppose each of these variables takes on all values in a certain common interval orregion . For example, x and t may be regarded as each representing position (in space of one,two or three dimensions) and time. Furthermore, let a distribution of causes be active over theregion , and let us suppose that we wish to study the resultant distribution of effects in .

If the effect at x due to a unit cause concentrated at t is denoted by the function G (x, t), thenthe differential effect at x due to a uniform distribution of cause of intensity c (t) over an elementaryregion (t, t + dt) is given by c (t) G (x, t)dt. It follows that the effect e (x) at x, due to a distributionof causes c (t) over the entire region is given by the integral

( ) ( , ) ( ) ,

e x G x t c t dt ... (1)

provided that superposition is valid, that is, if the effect due to the sum of two different causes is(exactly or approximately) the sum of the effects due to each of the causes.

The function G (x, t), representing the effect at x due to a unit concentrated cause at t, is saidto be the influence function of the problem. Clearly, influence function is either identical with orproportioanal to the Green’s function defined in Art. 11.2, when the definition is applicable.

If the distribution of causes is prescribed, and if the influence function is known, (1) may beused to find the effect by direct integration. On the other hand suppose we wish to find a distributionof causes which can produce a known or required effect distribution then for this purpose (1) canbe treated as a Fredholm integral equation of the first kind to find out c. Also the kernel of theintegral equation is the influence function G (x, t) of the problem.

Now, let us consider another type of physical problem. Suppose that the problem underconsideration prescribes neither the cause nor the effect separately, but required that they satisfy acertain linear relation of the form

( ) ( ) ( ), c x f x e x ... (2)where f is a given function or zero and is a constant.

Eliminating the effect e from (1) and (2), we obtain the following Fredholm integral equationof second kind for finding the cause distribution :

( ) ( ) ( , ) ( ) .

c x f x G x t c t dt ... (3)

Similarly, by eliminating the cause c from (1) and(2), we obtain the following Fredholm integral equationof the second kind for finding the effect distribution.

( ) ( , ) ( ) ( , ) ( ) .

e x G x t f t dt G x t e t dt ... (4)

Both cause and effect can be obtained by solvingeither (3) or (4), and using (2).

In order to understand such derivations,consider the following simple example :

Let us examine small deflections of a string fixeda point O (x = 0) and L (x = l), under a loadingdistribution of intensity p (x).

Y

Y

X X

TTt

P

1

Q1

O L



Assume that the string is initially so tightly stretched that nouniformity of the tension, due tosmall deflections, can be neglected. If a unit concentrated load is applied in the y direction at anarbitrary poing t, the string will then be deflected into two linear parts OP and LP intersecting at P(x = t). Assume that (approximately) uniform tension T acts along PO as well as PL. Then,considering equilibrium of string in y-direction, we obtain

1 2sin sin 1. T T ... (5)But for small deflections (and slopes), we have the following results (approximately).

and

1 1

2 2

sin tan,

sin tan

PQOQ tPQLQ l t

... (6)

where is the maximum deflection of the string, at the loaded point t. Substituting the above

approximate values of 1sin and 2sin in (5), we obtain

1

T

t l tso that

1 ( ). t l tTl ... (7)

The equation of linear part OP, joining ( , )P t and O (0, 0) is given by

00

y x

t or . xyt ... (8)

Again, the equation of linear part LP, joints ( , )P t and L (l, 0) is given by

00 ( )

y x l

t l or .

l xyl t ... (9)

Using (8) and (9), the equation of the corresponding deflection curve OPL is given by

( / ), when .[( ) / ( )], when .

x t x ty

l x l t x t ... (10)

Using (7), (10) shows that the influence function G (x, t) (for small deflection) is given by

( / ) ( ), when ,( , )

( / ) ( ), when .

x Tl l t x tG x t

t Tl l x x t ... (11)

Hence, by superposition, the deflection y (x) due to a loading distribution p (x) is given by

0( ) ( , ) ( ) .

ly x G x t p t dt ... (12)

If the deflection is prescribed, (12) is an integral equation of the first kind for finding thenecessary loading distribution.

Again, assume that the string is rotating uniformly about the x-axis, with angular velocity ,and that in addition a continuous distribution of loading g (x) is imposed in the direction radiallyoutward from the axis of revolution. Let the linear mass density of the string be denoted by ( ). xThen the total effective load intensity is given by

2( ) ( ) ( ) ( ). p x x y x g x ... (13)




2

0 0( ) ( , ) ( ) ( ) ( , ) ( ) .

l ly x G x t t y t dt G x t g t dt ... (14)

Re-writing (14), we have

2

0( ) ( , ) ( ) ( ) ( , ) ( ) ( )

x l

xy x G x t t y t dt G x t t y t dt

0( , ) ( ) ( , ) ( )

x l

xG x t g t dt G x t g t dt

or 2

0( ) ( ) ( ) ( ) ( ) ( ) ( )

x l

x

t xy x l x t y t dt l t t y t dtTl Tl

0

( ) ( ) ( ) ( ) ,x l

x

t xl x g t dt l t g t dtTl Tl

using (11) ... (15)

From (15), y (0) = 0 and y (l) = 0 ... (16)Differentiating both sides of (15) w.r.t. ‘x’ and using Leibnitz’s rule, we have

2

0

( )( ) ( ) ( ) ( ) ( ) x t x l xy x t y t dt x y x

Tl Tl( )( ) ( ) ( ) ( )

l

x

l t x l xt y t dt x y xTl Tl

0

( ) ( )( ) ( ) ( ) ( )

x l

x

t x l x l t x l xg t dt g x g t dt g xTl Tl Tl Tl

or2

0( ) ( ) ( ) ( ) ( )

x l

x

t l ty x t y t dt t y t dtTl Tl

0

( ) ( ) .

x l

x

t l tg t dt g t dtTl Tl ... (17)

Differentiating both sides of (17) w.r.t. ‘x’ and using Leibnitz’s rule as before, we have

2( ) ( ) ( ) ( ) ( ) ( ) ( ).x l x x l xy x x y x x y x g x g xTl Tl Tl Tl

or 2( ) ( / ) ( ) ( ) (1/ ) ( ).y x T x y x T g x ... (18)From (16) and (18), we find that the integral equation (14) can be reduced the following

boundary value problem2 2 2( / ) 0, (0) 0, ( ) 0.T d y dx y g y y l ... (19)

Re-writing (19), we have 0,L y ... (20)

where 2 2( / )L T d dx and 2 y g ... (21)We known that the Green’s function of the problem would then be that function which

satisfies T (d2y/dx2) = 0 except at x = t, which vanishes when x = 0 and x = l, which is continuousat x = t, and for which the radical force resultant T (dy/dx) decreases abruptly by unity at x = t.These are exactly the conditions which determines the influence function.

EXERCISE1. Define ‘influence function.’ Prove that the deflection ( ) x due to a loading distribution of

intensity p (x) is given as 0

( ) ( , ) ( ) ,t

x G x t p t dt where ( , )G x t is the influence function.

2. Define influence function and determine the deflection ( ) x of a string due to a loadingdistribution of intensity p (x).

3. Define influence function. [Meerut 2004, 05]



11.10 GREEN’S FUNCTION APPROACH FOR CONVERTINGAN INITIALVALUEPROBLEM INTO AN INTEGRAL EQUATION

Consider the following initial value problem

( )d dyp qy f xdx dx

... (1)

y (a) = 0, y (a) = 0 ... (2)

Let2

2 ,

d d d dp dL p q p qdx dx dx dxdx

... (3)

which is a self-adjoint differential operator. Here the function p (x) is continuously differentiableand positive and q (x) and f (x) are continuous in a given interval (a, b).

The associated homogeneous second-order equation

Ly = 0, i.e., 0

d dyp qydx dx ... (4)

has exactly two linearly independent solutions u (x) and v (x) which are twice differentiable in theinterval a < x < b. Any other solution of (4) is a linear combination of u (x) and v (x), i.e., y (x)= c1u (x) + c2 v (x), where c1 and c2 are constants.

For the self adjoint operator L, the Green’s formula is given by (refer Art. 10.5, Chapter 10)

( ) ( ) ( )b b

aav Lu u L v dx p x vu u v ... (5)

In order to convert the initial value problem (1)—(2) into an integral equation, we considerthe function w(x) given by

( ) ( ) ( ) ( ) ( ) ( ) ( ) . x x

a aw x u x v t f t dt v x u t f t dt ... (6)

Differentiating both sides of (6) w.r.t. ‘x’ we have,

( ) ( ) ( ) ( ) ( ) ( ) ( )x x

a a

dw x u x v t f t dt u x v t f t dtdx

( ) ( ) ( ) ( ) ( ) ( ) x x

a a

dv x u t f t dt v x u t f t dtdx

or0

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) x x

aw x u x v t f t dt u x v x f x v x u t f t dt

– v (x) u (x) f (x), by Leibnitz’s rule (see Art. 1.13)

or ( ) ( ) ( ) ( ) ( ) ( ) ( ) x x

a aw x u x v t f t dt v x u t f t dt ... (7)

From (6) and (7), we have w (a) = w (a) = 0 ... (8)Now, u and v are solutions of (4)

( ) 0 d pu qudx and ( ) 0

d pv qvdx

( ) d pu qudx and ( )

d pv qvdx ... (9)

Using the value of w (x) given by (7), we have

( )d pwdx

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) x x

a a

d p x u x v t f t dt p x v x u t f t dtdx



( ) ( ) ( ) ( ) ( ) x x

a a

d dpu v t f t dt pu v t f t dtdx dx

( ) ( ) ( ) ( ) ( ) x x

a a

d dpv u t f t dt pv u t f t dtdx dx

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) x x

a a

d dpu v t f t dt pu x v x f x pv u t f t dtdx dx

– pv (x) u (x) f (x), by Leibnitz’s rule

( ) ( ) ( ) ( ) ( ) ( )x x

a aqu v t f t dt qv u t f t dt p u v uv f x

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) x x

a a

d pw q u x v t f t dt v x u t f t dt p u v uv f xdx

or ( ) ( ) ( ) ( ) ( ), d pw q x w x p u v uv f xdx using (6) ... (10)

Now,

( ) d p u v u vdx

( ) ( ) ( ) ( )d d dpv u pu v pv u pv u pu v pu vdx dx dx

( ) ( ) 0,qv qu v using (9)

Thus, ( ) 0 d p uv u vdx so that p (uv – uv) = A, ... (11)

where A is a constant. (11) is Abel’s formula (refer page 11.7)

Also, ( , ),u v

uv u v W u vu v

... (12)

where W (u, v) is the Wronskian of u and v. Since u and v are linearly independent solutions of (4),

we have ( , ) 0W u v uv u v ... (13)Using (11), (10) may be re-written as

( )

d dwp qw A f xdx dx or

( / ) ( )

d d w A wp q f xdx dx A ... (14)

where w (a) = w (a) = 0, by (8) ... (15)Comparing (14) with (1), we have y = – w/A so that w = – Ay. Substituting this value of w

in (6), we obtain

( ) ( ) ( ) ( ) ( ) x

aAy u x v t v x u t f t dt or

( ) ( ) ( ) ( )( ) ( )

x

a

u t v x v t u xy x f t dtA

or ( ) ( , ) ( ) x

ay x R x t f t dt ... (16)

where R (x, t) = (1/A) {v (x) u (t) – u (x) v (t)} ... (17)From (17), we find that R (x, t) = – R (t, x) ... (18)One can easily verify that, for a fixed value of t, the function R (x, t) is completely

characterized as the solution of the initial value problem



( ) ( ) ( ),

d dRL R p x q x R x tdx dx ... (19)

[ R ]x = t = 0, (dR/dx)x = t = 1/p (t), ... (20)

where ( ) x t is the Dirac delta function (refer Art. 10.6 in chapter 10)This function describes the influence on the value of y at x due to a concentrated disturbance

at t. It is called the influence function. The function G (x, t) defined by

0,( , )

( , ),x t

G x tR x t x t

... (21)

is known as the causual Green’s function.Remark. When the values of y (a) and y (a) are prescribed to be other than zero, then we

simply add a suitable solution Au (x) + Bv (x) of (4) to the integral equation (16) to get volterraintegral equation of the second kind of the form

( ) ( ) ( ) ( , ) ( ) . x

ay x Au x B v x R x t f t dt ... (22)

The constants A and B are evaluated by using the prescribed initial conditions.Example 1. Convert the initial value problem

y + y = f (x), 0 < x < 1, y (0) = y (0) = 0into an integral equation.

Sol. Given y + y = f (x), 0 < x < 1 ... (1)with initial conditions : y (0) = 0, y (0) = 0 ... (2)

Comparing (1) with ( ),

d dyp qy f xdx dx

here p = 1 = q

The associated homogeneous equation of (1) is

y + y = 0 or (D2 + 1) y = 0, where /D d dx ... (3)Its general solution is y = A cos x + B sin x.Let u = cos x and v = sin x, ... (4)where u and v are linearly independent solution of (3).Now, A = p (uv – uv) = p [cos2 x + sin2 x] = p = 1 R (x, t) = (1/ A) {v (x) u (t) – u (x) v (t)} = sin x cos t – cos x sin t = sin (x – t)Hence the given initial value problem reduces to the integral equation

0

( ) ( , ) ( ) , x

y x R x t f t dt i.e.,0

( ) sin ( ) ( )x

y x x t f t dt ... (5)

Example 2. Convert the initial value problemy + y = f (x), 0 < x < 1, y (0) = 1, y (0) = – 1

into a Volterra integral equation of the second kind.Sol. Refer remark at the end of Art. 11.10.Here the values of y (0) and y (0) are prescribed to be other than zero, hence the given initial

value problem will transform into Volterra integral equation of the second kind of the form

0( ) ( ) ( ) ( , ) ( )

xy x Au x B v x R x t f t dt ... (1)



Proceed as in Ex. 1 and show that u (x) = cos x, v (x) = sin x and R (x, t) = sin (x – t). So

(1) reduces to0

( ) cos sin sin ( ) ( ) x

y x A x B x x t f t dt ... (2)

Putting y = 0 in (2) and using the given condition y (0) = 1, we get A = 1.Now, differentiating both sides of (2) w.r.t. ‘x’ and using Leibnitz’s rule (see Art. 1.13), we

obtain0

( ) sin cos cos( ) ( ) . x

y x A x B x x t f t dt ... (3)

Putting x = 0 in (3) and using the given condition y (0) = – 1, we get B = – 1.Putting A = 1 and B = – 1 in (2), the required Volterra integral equation is given by

0( ) cos sin sin ( ) ( )

xy x x x x t f t dt

11.11(a) GREEN’S FUNCTION APPROACH FOR CONVERTINGA BOUNDARYVALUE PROB-LEM INTO AN INTEGRAL EQUATION. AN ALTERNATIVE PROCEDURE. (COMPAREWITH ART. 11.3)Consider the following boundary value problem

( ),

d dyp qy f x a x bdx dx ... (1)

y (a) = 0, y (b) = 0 ... (2)

Let2

2 ,d d d dp dL p q p qdx dx dx dxdx

... (3)

which is a self-adjoint differential operator. Here the function p (x) is continuously differentiableand positive and q (x) and f (x) are continuous in a given interval (a, b).

The associated homogeneous second order equation

Ly = 0, i.e., 0

d dyp qydx dx ... (4)

has exactly two linearly independent solution u (x) and v (x) which are twice differentiable in theinterval . a x b Any other solution of (4) is a linear combination of u (x) and v (x), i.e., y (x)= C1 u (x) + C2 v (x) where C1 and C2 are constants.

Suppose that the given boundary value problem (1)—(2) possess its general solution as anintegral equation of the form

1 2( ) ( , ) ( ) ( ) ( ), x

ay x R x t f t dt C u x C v x ... (5)

where R (x,t) = (1/A) {v (x) u (t) – u (x) v (t)} ... (6)and A = p (x) {u (x) v (x) – u (x) v (x)} ... (7)

[Refer equations (11) and (17) of Art. 11.10]Putting x = a and x = b in succession in (5) and using the boundary conditions (2), we have

C1 u (a) + C2 v (a) = 0 ... (8)

and 1 2( ) ( ) ( , ) ( )b

aC u b C v b R b t f t dt ... (9)



Let ( ) ( ) ( ) ( ) 0 D u a v b v a u b ... (10)Then the system of equations (8) and (9) give a unique pair of constants C1 and C2 given by

1[ ( ) / ] ( , ) ( )

b

aC v a D R b t f t dt [ ( ) / ] ( , ) ( ) [ ( ) / ] ( , ) ( )

x b

a xv a D R b t f t dt v a D R b t f t dt ...(11)

2 [ ( ) / ] ( , ) ( ) b

aC u a D R b t f t dt [ ( ) / ] ( , ) ( ) [ ( ) / ] ( , ) ( )

x b

a xu a D R b t f t dt u a D R b t f t dt ... (12)

Substituting the values of C1 and C2 given by (11) and (12) in (5), we obtain

( ) [ ( , ) (1/ ) { ( ) ( ) ( ) ( )} ( , )] ( ) x

ay x R x t D v a u x u a v x R b t f t dt

(1/ ) { ( ) ( ) ( ) ( )} ( , ) ( ) .b

xD v a u x u a v x R b t f t dt ... (13)

From (6), R (b, t) = (1/A) {v (b) u (t) – u (b) v (t)} ... (14)Now, (1/D) {v (a) u (x) – u (a) v (x)} R (b, t)

= (1/AD) {v (a) u (x) – u (a) v (x)} {v (b) u (t) – u (b) v (t)}, by (14) ... (15)and R (x, t) + (1/D) {v (a) u (x) – u (a) v (x)} R (b, t) = (1/A) {v (x) u (t) – u (x) v (t)} + (1/AD) {v (a) u (x) – u (a) v (x)} {v (b) u (t) – u (b) v (t)}

[using (6) and (14)] = (1/ AD) [{v (x) u (t) – u (x) v (t)} {u (a) v (b) – v (a) u (b)}

+ {v (a) u (x) – u (a) v (x)} {v (b) u (t) – u (b) v (t)}], using (10) = (1/AD) {u (t) v (a) – v (t) u (a)} {u (x) v (b) – v (x) u (b)} ... (16)

[on simplification]Using (15) and (16), we define the so called Green’s function G (x, t) as follows :

(1/ ) { ( ) ( ) ( ) ( )}{ ( ) ( ) ( ) ( )},( , )

(1/ ) { ( ) ( ) ( ) ( )}{ ( ) ( ) ( ) ( )},

AD u t v a v t u a u x v b v x u b t xG x t

AD u x v a v x u a u t v b v t u b t x... (17)


( ) ( , ) ( ) b

ay x G x t f t dt ... (18)

From (17), we see that G (x, t) is symmetric, i.e., G (x, t) = G (t, x) ... (19)Again, G (x, t) satisfies, for all t, the following auxiliary problem:

( ) ( ) ( )

d dGLG p x q x G x tdx dx

... (20)

[G (x, t)]x = a = 0, [G (x, t)]x = b = 0 ... (21)[G (x, t)]x = t + 0 = [G (x, t)]x = t – 0 ... (22)

0 0( / ) ( / ) {1/ ( )}x t x tG x G x p t

... (23)

where ( ) x t is the Dirac delta function (refer Art. 10.6 of chapter 10). Here [G (x, t)]x = t + 0 standsfor the limit of G (x, t) as x approaches t from the right and there are similar meanings for the restof such expressions occuring in (22) and (23).

The condition (22) states that the Green’s function is continuous at x = t. Again, the condition(23) implies that the derivative of G has a discontinuity of magnitude –1/p (t) at x = t. The condition(22) and (23) are known as the matching conditions.



Remark 1. The relation (23) can be deduced from the relations (19) and (20) as follows.The value of the jump in the derivative of G (x, t) can be obtained by integrating (20) over a

small interval ( , )t x and by remembering that the indefinite integral of ( ) x t is the Heavisidefunction H (x – t), l (See Print header also Art. 10.9 of chapter 10). Thus, we obtain

( , ) ( , )( ) ( ) ( , ) ( ) ( )x

t

G x t G t xp t q x G x t dx p t H x tx x

When x traverses the source point t, then on the right side Heaviside function has a unit

jump discontinuity. Since other terms are continuous functions of x, it follows that /G x has att, a jump discontinuity as given by (23).

Remark 2. From the properties (20)—(23) of the Green’s function G (x, t), it follows that( , ) / G x a t satisfies the following system of equations :

( , ) ( , )( ) ( ) 0,

( , ) / 1/ ( ), ( , ) / 0

d d G x a G x ap x q x a x bdx dx t t

G a a t p a G b a t

... (24)

Similarly, ( , ) / G x b t satisfies the following system of equation

( , ) ( , )( ) ( ) 0, ,

( , ) / 0, ( , ) / 1/ ( )

d d G x b G x bp x q x a x bdx dx t t

G a b t G b b t p b

... (25)

Hence, it can be easily shown that the boundary value problem

( ) ( ) ( ), ( ) , ( )d dyp x q x y f x y a y bdx dx

... (26)

has the solution

( , ) ( , )( ) ( , ) ( ) ( ) ( ) .

b

a

G x a G x by x G x t f t dt p a p bt t

... (27)

11.11 (b) INTEGRAL- EQUATION FORMULATION FOR THE BOUNDARYVALUE PROBLEMWITH MORE GENERAL AND INHOMOGENEOUS BOUNDARY CONDITIONS. WORKINGRULE

Consider the following boundary value problem

1 1 2 2

( ) ( ),

( ) ( ) , ( ) ( )

d dyp x qy f xdx dx

y a v y a y b v y b

... (28)

The above boundary value problem is solved in exactly the same way as the system (1)—(2).We can easily show that the Green’s function for the system (28) can also be derived provided thedeterminant

1 1 2 2 1 1 2 2( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 0 D u a v u a v b v v b v a v a u b v u b

where u (x) and v (x) are the solutions of the homogeneous equation



( ) 0

d dyp x qydx dx ... (29)

Green’s function has the following five properties :

(i) ( ) ( ) ( )d GLG p x q x G x tdx x

... (30)

(ii) 1 1 2 2 0x a x bx a x b

G Gv G v Gx x

... (31)

(iii) 0 0[ ] [ ]x t x tG G

... (32)

(iv) 0 0( / ) ( / ) 1/ ( )x t x tG x G x p t

... (33)

(v) G (x, t) = G (t, x), i.e., G (x, t) is symmetric ... (34)With help of the Green’s function, the boundary value problem (28) has the unique solution.

1 2

( ) ( )( ) ( , ) ( ) ( , ) ( , )b

a

p a p by x G x t f t dt G x a G x b ... (35)

provided 1 and 2 do no vanish. If 1 0, then the factor 1(1/ ) ( , ) G x a is replaced by

1(1/ ) ( ( , ) / ).G x a t Similarly, if 2 0, then 2(1/ ) ( , )G x b is replaced by – 2(1/ ) ( ( , ) / ).G x b t

Since 0,D we cannot have both 1 and 1 or both 2 and 2 equal to zero. When

0, the relation (35) reduces to relation (18).An important deduction. The well known Strum-Liouville problem is the following boundary

value problem involving a parameter

1 1 2 2

( ) ( ) ( ) ( )

( ) ( ) 0, ( ) ( ) 0

d dyp x q x y r x y f xdx dx

y a y a y b y b

... (36)

The values of for which (36) has a nontrivial solution are known as the eigenvalues andthe corresponding solutions are known as the eigenfunctions. If p (a) = p (b), the boundary conditionin (36) are replaced by the periodic boundary conditions ( ) ( ), ( ) ( ). y a y b y a y b

Using the relation (35), the solution of (36) is given by

( ) ( ) ( , ) ( ) ( , ) ( ) , b b

a ay x r t G x t y t dt G x t f t dt ... (37)

which is a Fredholm integral equation of the second kind.We observe that the kernel K (x, t) of (37) is the product r (t) G (x, t). While the condition (v)

of the Green’s function shows that G (x, t) is symmetric, the product r (t) K (x, t) is not symmetric

unless r (t) is a constant. However, if we write 1/ 2( ) ( ) ( ),r x y x Y x under the assumption thatr (x) is nonnegative over (a, b) as is usually the case in practice, equation (37) can be re-written in

the form 1/ 2

( )( ) ( , ) ( ) ( , )( )

b b

a a

f tY x K x t Y t dt K x t dtr t



where 1/ 2( , ) ( ) ( ) ( , ).K x t r x r t G x t

and hence ( , )K x t possesses the same symmetry as G (x, t).Remark. We have seen that the discussion so for regarding the boundary value problem

(1)—(2), (26) or (28) is based on the assumption that

( ) ( ) ( ) ( ) 0. D u a v b v a u bHowever, if D = 0, then the system of homogeneous equation

1 2 1 2( ) ( ) 0, ( ) ( ) 0C u a C v a C u b C v b

will yield nonzero values of C1 and C2 and the function w (x) = C1 u (x) + C2 v (x) satisfies thecompletely homogeneous boundary value problem.

( ) ( ) 0,

d dwp x q x wdx dx w (a) = 0, w (b) = 0 ... (38)

It follows that if y is a solution of (1), (26), or (28), then so is y + cw for any constant c. Thisimplies that these systems will not possess a unique solution. This is not all. There is an additionalconsistency condition which must be satisfied for these systems to possess a solution. To this end,we proceed with system (26), i.e.,

f (x) = (py) + qy ... (39)with boundary condition ( ) y a and ( ) y b ... (40)

Multiplying (39) by w (x) and then integrating w.r.t. ‘x’ from a to b, we obtain

( ) ( ) ( ) ( ) b b

a aw x f x dx w x py qy dx ( ) ( )

b b

a aw x py dx w q y dx

( ) ( ) b bb

a a aw py w py dx w q y dx , integrating by parts

( ) ( ) ( ) ( ) ( ) ( ) ( )b b

a aw b p b y b w a p a y a w p y dx w q y dx ( ) ( ) ,

b bba a a

w p y w p y dx w q y dx using (38)

( ) ( ) ( ) ( ) ( ) ( ) ( ) b

aw b p b y b w a p a y a w p wq ydx

( ) ( ) ( ) ( ), w a p a w b p b using (38)

Thus, ( ) ( ) ( ) ( ) ( ) ( ) b

aw x f x dx w a p a w b p b ... (41)

Hence, in order that (26) may possess a solution, the given function f (x) must satisfy theconsistency condition (41).

As a particular case 0, we see that the system (1)—(2) wil possess a solution if the

consistency condition ( ) ( ) 0b

aw x f x dx ... (42)

is satisfied.We have thus shown that if D = 0, then the boundary value problems either have no solution

or many solutions; but never a unique solution.



Illustration Solved Example Reduce the boundary value problem 1,y xy (0) 0, ( ) 1y y l to Fredholm integral equation.

Sol. Given 1 , y xy ... (1)subject to the boundary conditions : y (0) = 0, y (l) = 1. ... (2)

[We have already dealt with the present problem in Ex. 4 base 11.35. However, we shall againgive an alternative method of dealing with boundary value problem (1)—(2), with help of results(24)—(27) of remark 2 of Art. 11.11.(a)]

For the Green’s function G (x, t) proceed yourself upto result (15) of Ex. 4, base 11.35.

Thus,( / ) ( ), 0

( , )( / ) ( ),

x l l t x tG x t

t l l x t x l ... (3)

Comparing (1) with (26) of Art. 11.11(a), here( ) 1, ( ) 1 , 0, 0, , 1.p x f x xy a b l ...(4)

Using result (27) of Art. 11.11(a), we see that the given boundary value problem transforms to

0( ) (1 ) ( , ) ( , ) /

ly x xy G x t dt G x l t

or0 0

( ) ( , ) ( ) ( , ) ( , ) / l l

y x G x t dt t y t G x t dt G x l t ... (5)

Now, ( , ) / G x l t must satisfy the system of equations (25) of Art. 11.11(a), which forthe present problem reduce to

2

2( , ) (0, ) ( , )0, 0, 1d G x l G l G l lt t tdx

... (6)

whose solution is ( , ) / / G x l t x l ... (7)

Now, 0 0

( , ) ( , ) ( , ) l x l

xG x t dt G x t dt G x t dt

0( ) ( ) ,

x l

x

t xl x dt l t dtl l

by (3)

2 2

02 2

x l

x

l x t x tltl l

2 2 22( )

2 2 2

x l x x l xl lxl l

2 2 2

( )2 2 2 2 2

x xl x l x xlx l xl ... (8)

With help of (7) and (8), (5) reduces to

0( ) ( ) ( ) ( , )

2

lx xy x l x t y t G x t dtl

or 2

0( ) ( / 2 ) (2 ) ( , ) ( ) ,

ly x x l l xl G x t t y t dt

which is the desired Fredholm integral equation.11.12 MODIFIED (OR GENERALIZED) GREEN’S FUNCTION

Consider the inhomogeneous equation ( )Ly x ...(1)where L is the self-adjoint operator defined by

( ) ( ), .

d dL p x q x a x bdx dx ... (1)



We have discussed in remark of Art. 11.11 (b) that if the homogeneous equation L y = 0 withthe prescribed boundary conditions at x = a and x = b has a nontrivial solution w(x), then thecorresponding inhomogeneous equation either has no solution or many solutions depending on theconsistency condition. Accordingly, the Green’s function, as defined in Art. 11.11 (b), does not

exist, because ( ) ( ) 0,b

ax t w x dx a t b

In what follows, we shall discuss a method of constructing the Green’s function for this type ofproblem. Such a function is known as the modified Green’s function and it will be denoted by GM (x,t). We start by selecting a normalized solution of the completely homogeneous system

so that 2 ( ) 1b

aw x dx ... (2)

By a generalized Green’s function, we mean a function GM (x, t) satisfying the followingproperties :

(i) Gm (x, t) satisfies the differential equation( , ) ( ) ( ) ( )

ML G x t x t w x w t ... (3)

This amounts to introducting an addition source density so that the consitency condition is

satisfied, as shown below ( ) ( ) ( ) ( ) 0.b

ax t w x w t w x dx

(ii) GM (x, t) satisfies the prescribed homogeneous boundary conditions.(iii) GM (x, t) is continuous at x = t.(iv) The x derivative of GM (x, t) possesses a jump of magnitude 1 / p(t) as the point x = t is

crossed in the positive x-direction, that is,

0 0( / ) ( / ) 1 / ( )M x t M x tG x G x p t

It follows that the construction of modified Green’s function is entirely similar to that for theordinary Green’s function described in Art. 11.4, but the modified Green’s function is not uniquelydetermined. We can add c w (x) to a Green’s function without violating any of the above fourproperties. It is often convenient to choose a particular modified Green’s function which is symmetricfunction of x and t. To this end, we consider two functions GM (x, t1) and GM (x, t2) which satisfythe equations

1 1 1( , ) ( ) ( ) ( )ML G x t x t w t w x ... (4)

and 2 2 2( , ) ( ) ( ) ( )ML G x t x t w t w x ... (5)

respectively along with prescribed (same for both) homogeneous boundary condition.Multiply (4) by GM (x, t2) and (5) by GM (x, t1), subtract, and integrate with respect to x from

a to b. Thus, we obtain

2 1 1 2{ ( , ) ( , ) ( , ) ( , )}b

M M M MaG x t L G x t G x t L G x t dx

2 1 1 2( , ) ( ) ( , ) ( )b b

M Ma aG x t x t dx G x t x t dx

1 2 2 1( ) ( , ) ( ) ( ) ( , ) ( )b b

M Ma aw t G x t w x dx w t G x t w x dx

Now, using the Green’s formula (refer relation (5) in Art. 10.5) and shifting property of Diracdelta function (refer Art. 10.7) , we arrive at

1 2 2 1 1 2 2 1( , ) ( , ) ( ) ( , ) ( ) ( ) ( , ) ( ) 0b b

M M M Ma aG t t G t t w t G x t w x dx w t G x t w x dx ... (5)

If we impose the condition :(v) GM (x, t) satisfies the property



( , ) ( ) 0,b

MaG x t w x for every t.

then, (5) reduces toGM (t1, t2) – GM (t2, t1) = 0 so that GM (t1, t2) = GM (t2, t1),

showing GM (x, t) will be symmetric. It then follows that the Green’s function is uniquely determined.Remark. The Green’s function can be defined in an equivalent form which does not involve

the Dirac delta function ( ) x t as follows.Suppose that a problem, consisting of the differential equation L y = (py) + qy = 0 and

homogeneous conditions at the ends of an interval (a, b), is satisfied by a normalized non-trivial

solution y = w (x) so that 2( ) 1.b

aw x dx Then the generalized Green’s function is defined as a

function GM (x, t) which, when considered as a function of x for a fixed number t, possesses thefollowing properties :

(i) GM satisfies the differential equation L GM = – w (x) w (t) in the sub-intervals (a, t) and (t, b).(ii) GM satisfies the prescribed end conditions.

(iii) GM is continuous at x = t.(iv) 0 0

( / ) ( / ) 1 / ( )M x t M x t

G x G x p t

(v) GM satisfies the condition ( , ) ( ) 0b

MaG x t w x dx

Method of reducing the inhomogeneous equationL y = (x) ... (6)

with prescribed homogeneous end conditions into an integral equation when the followingconsistency condition is satisfied

( ) ( ) .b

ax w x dx ... (7)

Multiplying (3) by y (x) and (6) by GM (x, t) and then subtracting, we get( , ) ( ) ( ) ( ) ( ) ( ) ( ).

M M MG Ly y L G G x t x x t y x w x w t y x

Now, integrating both sides of the above equation w.r.t. ‘x’ from a to b, we obtain

( , ) ( ) ( ) ( ) ( ) ( ) ( )b b b b

M M Ma a a aG L y y L G dx G x t x dx x t y x dx w x w t y x dx

or ( , ) ( ) ( ) ( )b b

M M Ma aG Ly y LG dx G x t x dx y t c w t ... (8)

where ( ) ( )b

ac w x y x dx an unknown constant ... (9)

and ( ) ( ) ( ),b

ax t y x dx y t by shifting property (refer Art. 10.7, Chapter 10)

Now, the L.H.S. of (8) vanishes because of the Green’s formula (refer result (5) of Art. 10.5,chapter 10) and the boundary conditions. Hence (8) reduces to

( ) ( , ) ( ) ( )b

May t G x t x dx c w t ... (10)

By interchanging x and t in (10), we obtain

( ) ( , ) ( ) ( )b

May x G t x t dt c w x ... (11)



For a symmetric Green’s function, we have GM (t, x) = GM (x, t)Therefore, (11) reduces to

( ) ( , ) ( ) ( )b

May x G x t t dt c w x ... (12a)

Using (9), (12a) can be re-written as

( ) ( , ) ( ) ( ) ( ) ( )b b

Ma ay x G x t t dt w x w x y x dx ... (12b)

Remark. When w (x) = 0, (12) reduces to (18) of Art. 11.11(a).11.13 WORKING RULE FOR CONSTRUCTION OF MODIFIED GREEN’S FUNCTION

Given an inhomogeneous equation with boundary conditions :

1 1 2 2( ); ( ) ( ) 0, ( ) ( ) 0Ly x y a y a y b y b ... (1)

Consider a linear homogeneous equation of order two

Ly = 0, where2

2( ) ( )d dp dL p x q xdx dxdx

... (1)

together with homogeneous boundary conditions

1 1( ) ( ) 0 y a y a ... (2a)

2 2( ) ( ) 0 y b y b ... (2b)

with usual assumption that at least one of 1 and 1 and one of 2 and 2 are non-zero.

Suppose that the homogeneous boundary value problem given by (1), (2a) and (2b) has anon-trivial solution y (x).

Then, || y (x) || = norm of 1/ 2

2( ) ( )

b

ay x y x dx ... (3)

Let w (x) = y (x)/ || y (x) ||so that w (x) is non-trivial normalized solution of the boundary value problem given by (1), (2a) and(2b). Clearly, by definition, we have

|| w (x) || = 1 so that 2{ ( )} 1b

aw x dx ... (4)

Then, by definition GM (x, t) is called the modified Green’s function of the given boundaryvalue problem if it satisfies the differential equation

( ) ( ) ( )M

L G x t w x w t ... (5)

For ,x t (5) reduces to L GM = – w (x) w (t) ... (5)

For a given t, let 1

2

( , ),( , )

( , ),M

G x t if a x tG x t

G x t if t x b

... (6)

where G1 and G2 are such that(i) The functions G1 and G2 satisfy the equation (5) in their respective intervals of definition,

that is 1 ( ) ( ), L G w x w t a x t ... (7a)



2 ( ) ( ), L G w x w t t x b ... (7b)

(ii) G1 satisfies the boundary condition (2a) whereas G2 satisfies the boundary condition (2b).(iii) The function GM (x, t) is continuous at x = t, i.e., G1 (t, t) = G2 (t, t) ... (8)(iv) The derivative of GM (x, t) with respect to x at the point x = t has a discontinuity of

the first kind, the jump being equal to 1/ p (t). Here p (x) is the coeff. of d2y/dx2 in (1).

Thus, 0 0( / ) ( / ) 1/ ( )

M x t M x tG x G x p t

... (9)

(v) In order that GM (x, t) may be symmetric, we must have

( , ) ( ) 0b

MaG x t w x dx ... (10)

Method of reducing the inhomogeneous differential equation (1) with prescribed homogeneousbondary condition into an integral equation.

The required integral equation is given by

( ) ( , ) ( ) ( )b

May x G x t t dt c w x , where c is an arbitrary constant. ... (11a)

(11a) may also be re-written in the form

( ) ( , ) ( ) ( ) ( ) ( )b b

Ma ay x G x t t dt w x w x y x dx ... (11b)

Consistency condition for existence of the desired integral equation (11a) or (11b) is given by :

( ) ( ) 0b

ax w x dx ... (12)

11.14 SOLVED EXAMPLES BASED ON ART. 11.13Ex.1. Find the modified Green’s function for the system

y + f (x) = 0, y (0) = y (l) = 0, 0 < x < land hence transform this boundary value problem into an integral equation.

Sol. Given – ( ), (0) ( ) 0, 0y f x y y l x l ... (1)

Here 2 2( / )d dx is a self adjoint operator

Consider the associated self adjoint system – 0, 0 y x l ... (2)with boundary conditions : y(0) = 0 ... (3a)

y (l) = 0 ... (3b)The general solution of (2) is y (x) = Ax + B ... (4)From (4), y (x) = A ... (5)Putting x = 0 and x = l in (5) and using (3a) and (3b), we get A = 0. Hence the boundary

value problem given by (2), (3a) and (3b) has a non trivial solution y (x) = B, where B is an arbitraryconstant.

Here, 1/ 2 1/ 2

2 2

0 0|| ( ) || norm of ( ) ( )

l ly x y x y x dx B dx B l

Let ( ) ( ) / || ( ) || / 1/ , w x y x y x B B l l ... (6)

so that w (x) is a non-zero normalized solution of the boundary value problem given by (2), (3a)

and (3b). Clearly, 20

( ) 1l

w x dx ... (7)



Then, for x t the required modified Green’s function GM (x, t) must satisfy the equation– d2GM /dx2 = –w (x) w (t) or d2GM /dx2 = 1/l, by (6) ... (8)

The general solution of (8) is of the form GM (x, t) = Ax + B + x2/2l

Hence, we take 2

1 22

1 2

/ 2 , 0( , )

/ 2 ,M

a x a x l x tG x t

b x b x l t x l

... (9)

From (9), 1

1

/ , 0/

/ ,M

a x l x tG x

b x l t x l

... (10)

In addition to the above property (9), the proposed modified Green’s function must satisfythe following properties :

(i) Since GM (x, t) must satisfy the boundary conditions (3a) and (3b), (10) gives

0( / ) 0M xG x

and ( / ) 0M x lG x

a1 = 0 and b1 + 1 = 0 so that a1 = 0 and b1 = – 1 ... (11)(ii) Gm (x, t) is continuous at x = t, that is

a1 t + a2 + t2/2l = b1 t + b2 + t2/2l so that (a1 – b1) t + a2 – b2 = 0 ... (12)(iii) The derivative of GM (x, t) with respect to x at the point x = t has a discontinuity of the

first kind, the jump being 1/ p(t), where p (x) is the coefficient of y in (1), i.e., p (x) = –1. Thus,

0 0/ / 1/ ( )M Mx t x t

G x G x p t

i.e., b1 + t/l – (a1 + t/l) = – 1 or a1 – b1 = 1 ... (13)From (12) and (13), t + a2 – b2 = 0 so that b2 = a2 + t ... (14)Substituting values of a1, b1 and b2 from (11) and (14) in (9), we get

22

22

/ 2 , 0( , )

/ 2 , .M

a x l if x tG x t

a x t x l if t x l

... (15)

(iv) In order that GM (x, t) may be symmetric, we have

0( , ) ( ) 0

l

MG x t w x dx or0

( , ) 0,l

MG x t dx as 1( ) w xl

or0

( , ) ( , ) 0t l

M MtG x t dx G x t dx

or2 2

2 200,

2 2

t l

t

x xa dx a x t dxl l

using (15)

or 3 2 32 20

/ 6 / 2 / 6 0t l

ta x x l a x x t x x l

or 3 2 2 2 2 32 2 2

/ 6 / 2 / 6 ( / 2 / 6 ) 0a t t l a l l t l l a t t t t l

or a2 l = l2/3 – t l + t2/2 or a2 = l/3 – t + t2/2lSubstituting the above value of a2 in (15), the symmetric modified Green’s functionGM (x, t) is given by

2 2

2 2

/ 3 ( ) / 2 , 0( , )

/ 3 ( ) / 2 ,M

l t x t l if x tG x t

l x x t l if t x l

... (16)



which may also be re-written as

2 2 , 0( , )

,3 2M

t if x tl x tG x tx if t x ll

... (17)

The above result (16) could have been obtained by inspecting (15) and making a judiciouschoice of a2.Second part. Transformation of the given boundary value problem into an integral equation.

Refer result (11a) of Art. 11.13. The required integral equation is given by

( ) ( , ) ( )l

MM

cy x G x t f t dtl

or0

( ) ( , ) ( ) ,l

My x c G x t f t dt where ( / )c c l is an arbitrary constant and GM (x, t) is given by (16) or (17).

Ex.2. Find the modified Green’s function for the system y = 0, –1 < x < 1 subject to theboundary conditions y (–1) = y (1) and y (–1) = y (1).

Sol. Given self adjoint system – 0, 1 1 y x ... (1)with boundary conditions : y (–1) = y (1) ... (2a)and y (–1) = y (1) ... (2b)

The general solution of (1) is y (x) = Ax + B ... (3)From (3), y (x) = A ... (4)From (3) and (4), y (– 1) = – A + B, y (1) = A + B, y (–1) = y (1) = A ... (5)From (2a) and (5) , we get –A + B = A + B so that A = 0 ... (6)Then, (2b), (5) and (6) A = A = 0.Hence the given boundary value problem has a non-trivial solution y (x) = B, where B is an

arbitrary constant.

Here || y (x) || = norm of 1/ 2 1/ 21 12 2

1 1( ) ( ) 2

y x y x dx B dx B

Let ( ) ( ) / || ( ) || / 2 1/ 2 w x y x y x B B ... (7)

so that w (x) is a non-zero normalized solution of the given boundary value problem. Clearly,

1 2

1( ) 1.

w x dx

Then for x t the required modified Green’s function GM (x, t) must satisfy the equation2 2/ ( ), ( )Md G dx w x w t or 2 2/ 1/ 2,Md G dx by (7) ... (8)

The general solution of (8) is of the form GM (x, t) = Ax + B + x2/4

Hence, we take2

1 22

1 2

/ 4, 1( , )

/ 4, 1M

a x a x if x tG x t

b x b x if t x

... (9)

From (9),1

1

/ 2, 1/

/ 2, 1M

a x if x tG x

b x if t x

... (10)

In addition to the above property (9), the proposed modified Green’s function GM(x, t) mustsatisfy the following properties :



(i) GM (x, t) is conditnuous at x = t, that is,a1t + a2 + t2/4 = b1t + b2 + t2/4 so that a2 – b2 = t (b1 – a1) ... (11)

(ii) Since GM (x, t) must satisfy (2a), we must have– a1 + a2 + 1/4 = b1 + b2 + 1/4 so that a2 – b2 = a1 + b1, by (9) ... (12)

Again, GM (x, t) must satisfy (2b), hence we haveb1 + 1/2 = a1 – 1/2 or b1 – a1 = – 1, by (10) ... (13)

(iii) The derivative of GM (x, t) with respect to x at the point x = t has a discontinuity of thefirst kind, the jump being 1/p (t), where p (x) is the coefficient of y in (1), i.e., p (x) = –1. Thus,

0 0/ / 1 / ( )M Mx t x t

G x G x p t

i.e., b1 + t/2 – (a1 + t/2) = –1 or b1 – a1 = – 1

which is the same relation as (13). Thus, we see that the jump condition on /MG x is automaticallysatisfied.

From (11) and (13), a2 – b2 = – t so that b2 = a2 + t ... (14)Again, from (13), b1 = a1 – 1 ... (15)Substituting the values of b2 and b1 given by (14) and (15) respectively in (12), we have

a2 – (a2 + t) = a1 + a1 – 1 so that a1 = (1 – t)/2 ... (16)Substituting the values of b2 and b1 given by (14) and (15) respectively in (9), we have

21 2

21 2

/ 4, 1( , )

( 1) / 4, 1M

a x a x if x tG x t

a a t x if t x

... (17)

(iv) In order that GM (x, t) may be symmetric, we have1

1( , ) ( ) 0MG x t w x dx

or

1

1( , ) 0,MG x t dx

as 1( )

2w x

or1

1( , ) ( , ) 0

t


or2 21

1 2 2 110,

4 4

t

t

x xa x a dx a t a x x dx

by (17)

or12 3 2 2 3

1 2 2 11( / 2) /12 ( / 2) / 2 /12 0

t

ta x a x x a x t x a x x x

or a1 (t2/2) + a2 t + t3/12 – (a1/2 – a2 – 1/12) + a2 + t + a1/2 – 1/2 + 1/12

– (a2 t + t2 + a1 (t2/2) – t2 /2 + t3/12) = 0

or 2a2 = t2/2 – t + 1/3 or a2 = t2/4 – t/2 + 1/6 ... (18)Substituting the value of a1 and a2 given by (16) and (18) respectively in (17), the symmetric

Green’s function is given by2 2

2 2

(1 ) / 2 / 4 / 2 1/ 6 / 4, 1( , )

(1 ) / 2 / 4 / 2 1/ 6 / 4, 1M

t x t t x if x tG x t

t x t t t x if t x

11 1( 1) 1

2 2 t ta x x



or2 2

2 2

/ 4 / 4 / 2 ( ) / 2 1/ 6, 1( , )

/ 4 / 4 / 2 ( ) / 2 1/ 6, 1M

t x xt x t if x tG x t

t x xt x t if t x

or2

2

( ) / 4 ( ) / 2 1/ 6, 1( , )

( ) / 4 ( ) / 2 1/ 6, 1M

x t x t if x tG x t

x t x t if t x

which can also be re-written as2( , ) (1/ 4) ( ) (1/ 2) | | 1/ 6.mG x t x t x t

Ex.3. Transform the following boundary value problem into an integral equation

2(1 ) 0, ( 1) (1)d dyx y where y and ydx dx

are both finite

Sol. Given boundary value problem is – 2(1 ) , 1 1d dyx y xdx dx

... (1)

with boundary conditions: y (1) = finite and y (–1) = finite ... (2)

Here the operator 2( / ){(1 ) ( / )}d dx x d dx is a self adjoint operator..Consider the associated self adjoint system :

– 2(1 ) 0, 1 1

d dyx xdx dx

... (3)

and boundary conditions y (–1) = finite ... (4a)and y (1) = finite ... (4b)

Integrating (3), we have (1 – x2) (dy/dx) = ASeparating variables, dy = {A / (1 – x2)} dx.

Integrating it, ( ) log , 1 11

xy x A B xx ... (4)

Since (4) has to satisfy the boundary condition (4a) and (4b), we must have A = 0.Hence the boundary value problem given by (3), (4a) and (4b) has a non-trivial solution y (x)

= B, where B is an arbitrary constant

Here, || y (x) || = norm of 1/ 2 1/ 21 12 2

1 1( ) ( ( )) 2

y x y x dx B dx B

Let ( ) ( ) / || ( ) || / 2 1/ 2w x y x y x B B ... (5)

so that w (x) is a non-zero normalized solution of the given boundary value problem. Clearly,

1 2

1( ) 1.

w x dx

Then, for ,x t the required modified Green’s function GM (x, t) must satisfy the equation

2 . 1(1 ) ( ) ( ) ,2

Md Gd x w x w tdx dx

using (5) ... (6)

Integrating (6), (1 – x2) (d GM / dx) = x/2 + A

or 2 22 (1 ) 1Md G x A

dx x x

or 2 2

1 ( 2 )4 1 1M

x dx Ad G dxx x



Integrating it, 21 1( , ) log (1 ) log4 2 1M

A xG x t x Bx

or GM (x, t) = – (1/4) × {log (1 + x) + log (1 – x)} + (A/2) × {log (1 + x) – log (1 – x)} + Bor GM (x, t) = (A/2 – 1/4) log (1 + x) – (A/2 + 1/4) log (1 – x) + B

Hence, we take

1 1 2

1 1 2

( / 2 1/ 4) log (1 ) ( / 2 1/ 4) log (1 ) 1( , )

( / 2 1/ 4) log (1 ) ( / 2 1/ 4) log (1 ) , 1M

a x a x a if x tG x t

b x b x b if t x

... (7)

In addition to the above property (7), the proposed modified Green’s function GM (x, t) mustsatisfy the following conditions.

(i) By virtue of boundary conditions (4a) and (4b), GM (x, t) must be finite at x = – 1 and x= 1. Accordingly, we must have a1 = 1/2 and b1 = – 1/2. Hence (7) reduces to

2

2

(1/ 2) log (1 ), 1( , )

(1/ 2) log (1 ), 1M

a x if x tG x t

b x if t x

... (8)

(ii) GM (x, t) is continuous at x = t, that is,a2 – (1/2) × log (1– t) = b2 – (1/2) × log (1 + t), using (8)

so that a2 – b2 = (1/2) × log (1 – t) – (1/2) × log (1 + t) ... (9)(iii) The derivative of GM (x, t) with respect to x at the point x = t has a discontinuity of the

first kind, the jump being 1/p (t), where p (x) is the coefficient of y in (3), i.e., p (x) = –(1 – x2).[Note that (3) may be re-written as –(1– x2) y + 2xy = 0]

i.e., 0 0/ / 1/ ( )M Mx t x t

G x G x p t

... (10)

But from (8),1/ 2 (1 ), 1( 1) / 2 (1 ), 1

MG x x tx t xx

Hence, (10) reduces to

21 1 1

2 (1 ) 2 (1 ) 1t t t

or 2 2

1 1 ,1 1t t

showing that the jump condition at x = t is automatically satisfied.(iv) In order that GM (x, t) may be symmetrical, we have

1

1( , ) ( ) 0MG x t w x dx

or

1

1

1( , ) 0, as ( )2MG x t dx w x

or1

1( , ) ( , ) 0

t


or1

2 21

1 1log (1 ) log (1 ) 0,2 2

t

ta x dx b x dx

using (8) ... (11)

Now, 21

1 log (1 )2

ta x dx

21 1

1 1 log(1 )2

t t

a dx x dx

12 1 1

1 ( )log (1 )2 1

tt t xa x x x dxx

, integrating by parts



2 1

1 1 ( )( 1) log (1 ) log 22 2 1

t x dxa t t t

x

2 1

1 1 1( 1) log 1 log 2 12 2 2 1

tta t t dxx

121 1( 1) log (1 ) log 2 log (1 )

2 2 2tta t t x x

21 1 1 1( 1) log (1 ) log 2 log (1 ) log 2

2 2 2 2 2 2

t ta t t t

= a2 (t + 1) – (t/2) × log (1 – t) – (1/2) × log (1 – t) + t/2 + 1/2 ... (12)

and1

21 log (1 )2

tb x dx

1 1

21 1 log (1 )2

t tb dx x dx

11 1

21 log (1 )2 1

tt t

xb x x x dxx , integrating by parts

1

21 1 (1 ) 1(1 ) log 2 log (1 )2 2 1

t

xb t t t dxx

1

21 1 1(1 ) log 2 log (1 ) 12 2 2 1

t

tb t t dxx

121 1(1 ) log 2 log (1 ) log (1 )2 2 2 t

tb t t x x

21 1 1 1(1 ) log 2 log(1 ) log 2 log (1 )2 2 2 2 2 2

t tb t t t

2 (1 ) ( / 2) log(1 ) (1/ 2) log(1 ) / 2 1/ 2 log 2 b t t t t t ... (13)Using (12) and (13), (11) reduces to a2 (t + 1) – (t/2) × log (1 – t) – (1/2) × log (1– t) + t/2 + 1/2

+ b2 (1 – t) + (t/ 2) × log (1 + t) + (1/2) × log (1 + t) – t/2 + 1/2 – log 2 = 0

or 2 21 1( 1) (1 ) log 2 1 log (1 ) log (1 )

2 2t ta t b t t t

or 2 2 2log 2 1 (1/ 2) log (1 ) log (1 ) a b t t ... (14)

Solving (9) and (14), we obtaina2 = log 2 – 1/2 – (1/2) × log (1 + t), b2 = log 2 – 1/2 – (1/2) × log (1 – t)Substituting these values in (8), the symmetric modified Green’s function is given by

log 2 1/ 2 (1/ 2) log ( 1 ) log (1 ) , 1( , )

log 2 1/ 2 (1/ 2) log (1 ) log(1 ) , 1M

x t x tG x t

x t t x

or

log 2 1/ 2 (1/ 2) log (1 ) (1 )}, 1( , )

log 2 1/ 2 (1/ 2) log (1 ) (1 )}, 1M

x t x tG x t

x t t x

... (15)



which may also be re-written as

(1/ 2) log {(1 ) (1 )}, 11( , ) log 2(1/ 2) log {(1 ) (1 )}, 12M

x t x tG x t

x t t x

...(15)

Transformation of given boundary value problem given by (1)—(2) into an integral equation.Comparing (1) with ( ),Ly x here ( ) . x yRefer result (11b) of Art. 11.13. The required integral equation is given by

1 1

1 1( ) ( , ) ( ) ( ) ( ) ( )My x G x t t dt w x w x y x dx

i.e.,1 1

1 1

1( ) ( , ) ( ) ( ) ,2My x G x t y t dt y x dx

as

1( )2

w x

which can also be re-written as1

1( ) ( , ) ( ) ,My x G x t y t dt c

where c is an arbitrary constant given by1

1

1 ( )2

c y x dx

Ex.4. Show that the boundary value problem

2(1 ) 2 ( ), (1), ( 1)x y xy y f x y y finite, transforms into the integral equation1 1 1

1 1 1

1( ) ( , ) ( ) ( , ) ( ) ( ) .2M My x G x t y t dt G x t f t dt y t dt

Also show that1 1

1 1

1 ,

y dt f dt

where GM (x, t) is Green’s function of associated boundary value problem.Sol. Given boundary value problem is

2(1 ) ( ), 1 1, (1) finite and ( 1)d dyx y f x x y ydx dx

finite ... (1)

Here the operator 2( / ){(1 )( / )}d dx x d dx is a self adjoint operator..

Consider the associated self adjoint system is

– 2(1 ) 0, 1 1, (1) finite and ( 1)d dyx x y ydx dx

finite ... (2)

Proceed as in Ex. 3. to show that modified Green’s function of (2) is given by relation (15) of Ex.3. Comparing (1) with ( ), here ( ) ( ) ( )Ly x x y x f x

Refer result (11b) of Art. 11.13. The required integral equation is given by

1 1

1 1( ) ( , ) ( ) ( ) ( ) ( )My x G x t t dt w x w x y x dx

or 1 1

1 1( ) ( , ) ( ) ( ) ( ) ( ) ( )My x G x t y t f t dt w x w t y t dt

or 1 1 1

1 1 1

1( ) ( , ) ( ) ( , ) ( ) ( )2M My x G x t y t dt G x t f t dt y t dt

( ) 1/ 2 ( ) 1/ 2 w x w t



MISCELLANEOUS EXERCISE ON CHAPTER 111. Find the boundary value problem that is equaivalent the integral equation

1

1( ) (1 | |) ( )

y x x t y t dt

2. Transform the boundary value problem 1/ 2( / 2 ) 0, y y x x y y (0) = y (1) = 0 tointegral equation.

3. Convert the initial value problem 2 0, (0) (0) 0y x y y y into a Volterra integral equation.

4. Reduce the differential equation 2( ) 0,y P x y is a known positive constant,

with end condition (0) 0, ( / ) y y y at x = x0 into a Fredholm integral equation.5. Show that the Green’s function for y = 0, y (0) + y (1) = 0, y (0) + y (1) = 0 is

1 , 0( , )

1 , 1

t x tG x t

x t x

Hence solve the B.V.P. ( ), (0) (1) 0, (0) (1) 0 y f x y y y y where

(i) ( ) sin f x x (ii) ( ) , 0 1 xf x e x (iii) f (x) = x

6. Find the modified Green’s function for the system 0, (0) (1), (0) (1). y y y y y y

7. Transform the boundary value problem ( ), (0) ( ) 0, y y f x y y into an integralequation.

8. Reduce the boundary value problem2 2

32 2(4 ) (4 ) 0, (0) 0 (1), (0) 0 (1)

d d yx x y y y y ydx dx

to a Fredholm

integral equation.9. Find Green’s function for the initial value problem m (d2y / dx2) = f (x), with initial conditions

y (0) = y(0) = 0 Ans.0 , 0

( , )( ) / ,

x tG x t

x t m t x

10. Find the solution of the initial value problem y + w2y = f (x), y (0) = y1 and y (0) = y2.

Ans. Green’s function 0, 01( , ) sin ( ) ( )(1/ ) sin ( ),

x tG x t w x t H x t

w w x t t xw

and solution is1 20

1( ) sin ( ) ( ) ( ) cos sinx

y x w x t H x t f t dt y w x y wxw

,

where H (x – t) is the well known Heaviside unit function. (refer chapter (10)).11. Solve the boundary value problem y – y = – 2ex with boundary conditions y (0) = y (0)

and y (l) + y (l) = 0.

Ans. ( ) ( ) sinh wherexy x l x e x ( )

(1/ 2) , 0( , ) ,

(1/ 2) ,

x t

x t

e x tG x t

e t x l

12. Reduce the boundary value problem 2 / 4 cos / 2 , ( 1) (1) y y y x y y andy (–1) = y (1) to an integral equation.



Ans.1

21

2( ) ( , ) ( ) sin cos ,2 2

x x xy x G x t y t dt

where

(1/ ) sin ( ) / 2 , 1( , )

(1/ ) sin ( ) / 2 , 1

x t x tG x t

t x t x

13. Reduce the following boundary value problem to an integral equation2 1 y y x with y (0) = y (1), y (0) = y (1).

Ans. 1 3 2

0

1( ) ( , ) ( ) 2 3 17 5 ,6

y x G x t y t dt x x x

where ( 2) ( 1) , 0

( , )( 1) 1 , 1

t x t x tG x t

t x t x

14. Construct the Green’s function for the boundary value problem( ) 0, (0) (1) 0y x y y and (0) (1) y y

Ans. (1/ 2) ( 1) ( 2 ), 0( , )

(1/ 2) [ (2 ) ( 2) ], 1x t x xt t x t

G x tx x t t t x

15. Find the modified Green’s function for the system L = d2/dx2 with y (0) = – y (1). Thissystem is not self-adjoint.

16. (a) Show that a sufficient condition for the modified Green’s function to be symmetric is

( , ) ( ) 0,b

MaG x t w x dx

where GM (x, t) and w (x) have the same meaning as explained in Art. 11.12. Show that there are problems where this condition is not necessary.(b) Apply the result of part (a) to find the modified Green’s function for the system

L = (d2/dx2) + 1 with (0) (2 ), (0) (2 ).y y y y 17. Find the consistency condition for the following systems,

(i) 2 2( / ) ( ); (0) , (1)d y dx f x y y

(ii) 2 2/ ( ); (0) (2 ) , (0) (2 )d y dx y f x y y y y 18. Obtain the modified Green’s function for the systems

(i) ( ) 0, (0) ( ) 0, 0ky f x y y l x l

(ii) ( ) 0, ( ) ( ); ( ) ( );ky f x y l y l y l y l l x l and hence transform these boundary value problems into respective integral equations.

Ans. (i) 2 2 / , if 0

( , )/ , if3 2M

t k x tl x tG x tx k t x lk kl

; 0

( ) ( , ) ( )l

My x c G x t f t dt

(ii) GM(x, t) = 1/6k + (x–t)2/4kl + | x – t |/2k ; ( ) ( , ) ( )l

Mly x c G x t f t dt

19. Obtain the modified Green’s function for the following boundary value problems:(i) ( ); (0) ( ) 0, 0y ky f x y y x

(ii) ( ), ( ) ( ); ( ) ( ),y f x y y y y x



(iii) ( ), (0) 0, (1) (1), 0 1y f x y y y x

(vi) 2 ( ), (0) ( ) 0; 0y k y f x y y x

(v) 2 ( ), (0) (1) 0; 0 1y y f x y y x 20. Develop the theory of modified Green’s function in the case of a self-adjoint system where

the completely homogeneous system has two linearly independent solutions w1(x) andw2 (x) (hence every solution of the homogeneous equation satisfies the boundary conditions.)

21. Obtain the modified Green’s function for2 ( ); (0) (2 ), (0) (2 ); 0 2y k y f x y y y y x (Kanpur 2011)

Hint. For solution of the above exercise 21 and other problems based on modified (or generalized)Green’s function, refer Appendeix D of this book.


CHAPTER 12

Applications of integral equations to partial differential equations12.1 INTRODUCTION

We have already discussed the applications of integral equations to ordinary differentialequations. But the most important applications of integral equations arise in finding the solutionsof boundary value problems associated with partial differential equations of the second order. Ithas been shown that the boundary value problems for equations of elliptic type can be reduced toFredholm integral equations, while the boundary value problems for equations of parabolic andhyperbolic types can be reduced to Volterra integral equations. However in the present chapter weshall deal with the linear partial differential equations of the elliptic type. In particular, we shallfocus our attention to the Laplace, Poisson and Helmholtz equations because some importantachievements have been arrived from their study.Occurrence of the Laplace, Poisson and Helmholtz equations.

Various physical phenomena are governed by the well known Laplace, Poisson and Helmholtzequations. A few of them, frequently encountered in applications are : steady heat conduction,irrotiational flow of an ideal fluid, distribution of gravitational potential, electrostatics, dielectrics,magnetostatics, steady corrents, surface waves on a fluid etc.

While dealing with elliptic partial differential equations, the following three types of boundaryvalue problems arise :

(i) The Dirichlet problem. In this case we prescribe the value of the solution on the boundary.(ii) The Neumann problem. In this case we prescribe the normal derivative of the solution on

the boundary.(iii) Mixed boundary value problem. In this case we prescribe the Dirichlet condition on some

parts and the Neumann condition on the other parts of the boundary.Types of boundary conditions while dealing with an elliptic equation

(i) Dirichlet condition : When the value of solution is presecribed on the boundary surface.(ii) Neumann condition : When the value of the normal derivative of the solution is prescribed

on the boundary surface.(iii) Mixed boundary condition : When we prescribe the Dirichlet condition on some parts and

Neumann condition on the other parts of the body. Sometimes a linear combination of solution andits normal derivative is prescribed on the boundary surface.Some important mathematical tools.

Suppose that A is a continuously differentiable vector field and u and v have partial derivativesof the second order which are continuous in the bounded region R. Let S denote the boundary of Rand let n denote the unit normal outward to S. Then, we have

(i) Divergence theorem. R S

div dV dS A A n

12.1


12.2 Application of integral equation partial differential equations

(ii) Green’s first identity 2 ( )R R S

du dV grad u grad dV u dSdn

!

! !

(iii) Green’s second identity 2 2( ) ,R S

d uu u dV u dSdn n

!

! ! !

where the Laplacian operator 2 , in cartesian coordinates x, y, z, is defined as

2 2 2 2 2 2 2/ / /x y z

and / n denotes differentiation along the outward-drawn normal on S.

The equation 2 0u is called Laplace’s equation and a function satisfying Laplace’ssequation is called a harmonic function.

Poisson equation is given by 2 4u where is given function of the position.

Some notations to be used in further studyBoldface letter such as x shall stand for the triplet (x1, x2, x3). The quantities Ri and Re will

stand for the regions interior and exterior to S, respectively. In what follows we shall not write dSx

or dS to signify the integration with respect to the variable x or . In place of this, we shallmerely write dS and it shall be clear from the context as to what the variable of integration is.12.2 INTEGRAL REPRESENTATION FORMULAS FOR THE SOLUTIONS OF THELAPLACE AND POISSON EQUATIONS.

We begin with the fundamental solution (or free-space solution) ( ; )E x which satisfies

2 ( )E x ... (1)

and vanishes at infinity. The function ( ; )E x can be interpreted as the electrostatic potential at an

arbitrary field point x due to a unit charge at the source point . We know that ( ; )E x is given by

( ; ) 1/ 4 1/ 4 | |E r x x ... (2)For the two-dimensional case, we have

( ; ) (1/ 2 ) log (1/ ) (1/ 2 ) log (1/ | |),E r x x ... (3)

where 1 2 1 2( , ) and ( , )x x x The fundamental solution is employed to find the solution of the Poisson equation

2 4 .u ... (4)

as follows. Multiply (1) by u(x), (4) by ( , ),E x subtract, integrate over the region Ri, and useGreen’s second identity and the shifting property of the Dirac delta function (refer Art 10.7 ofchapter 10). After relabeling and ,x we finally obtain

1 1 1 1( )

4 4R S S

uu dV u dS dSr n r r n x ... (5)

If the values of , and /u u n involved in formula (5) are known, for example, let

[ ]Su and [ / ]Su n ... (6)then (5) can be re-written as

1 1 1 ( )( ) ( )4 4R S S

Qu P dV Q dS dSr n r r ... (7)


Application of integral equation partial differential equations 12.3

where P is the field point x and Q is a point on S. The formulas (5) and (7) can be used to findvarious important properties of the harmonic function.

With reference to (7), we define some important terms as follows :

The Newtonian or the volume potential. The integral ( / ) ,R

r dV is known the Newtonian

potential. For the Newtonian potential ( / ) ,R

u r dV we have the following properties :

(i) 2 0,u for points P in Re.(ii) For points P within Ri, the integral is improper but it converges and admits two differntials

under the integral sign if the function is sufficiently smooth; the result is 2 4 ( ).u P

The simple or single-layer potential. The integral ( / )S

r dS is known as the single-layer

potential with charge (or source) density . The single-layer potential ( / )S

u r dS has the

following properties :

(i) 2 0,u outside S.

(ii) The integral becomes improper at the surface S but converges uniformly if S is singular.Also, this integral remains continuous as we pass through S.

(iii) Consider the derivative of u taken in the direction of a line normal to the surface S in theoutward direction from S. Then, we have

2cos ( , )2 ( ) ( )

| |SP

u P Q dSn

x n

x ... (8)

and 2cos ( , )2 ( ) ( ) ,

| |SP

u P Q dSn

x n

x ... (9)

where P+ and P– signify that we approach S from Ri and Re respectively, and where both andx areon S. Subtracting (8) from (9), we obtain

1 ,

4 P P

u un n

... (10)

giving the jump of the normal derivative of u across S.

The double-layer potential. The integral ( / ) (1/ )S

n r dS is known as the double layer

potential with dipole of density . The double-layer potential ( / ) (1/ )S

u n r dS has the

following properties.

(i) 2 0u outside S.(ii) The integral becomes improper at the surface but it converges if the surface S is regular.



(iii) While passing through S, the double-layer potential undergoes a discontinuity such that

2cos ( , )[ ] 2 ( ) ( )

| |PS

u P Q dS

x n

x

... (11)

and 2cos( , )[ ] 2 ( ) ( ) ,P S

u P Q dS

x n

x

... (12)

where P+ and P– signify that we approach S from Ri and Re, respectively, and where both andx areon S. Subtracting (12), from (11), we have

(1/ 4 ) [ ] [ ] ,P Pu u

... (13)giving the jump of u across S.

(iv) The normal derivative remains continuous as S is crossed.Interior and exterior Dirichlet problems.

For the solution of a boundary value problem for an elliptic equation, we cannot prescribe uand u n arbitrarily on S. Hence, equation (7) does not allow us to construct a solution of (4)such that u shall itself have arbitrary values of S and also arbitrary values of its normal derivativethere. It follows that there exist the following two types of boundary value problems for ellipticequations.

(i) The Dirichlet problem. In such a problem the value of the solution is prescribed on S.(ii) The Neumann problem. In such a problem the value of the normal derivative of the solution

is prescribed on S.We begin with the Dirichlet problem. Consider first the Dirichlet problem for the region

exterior to the unit sphere in three dimensions. In addition to prescribing boundary values on thesurface of the unit sphere, it will be necessary to impose some sort of boundary conditions atinfinity in order to get a unique solution. To illustrate the need for a boundary condition at infinity,we observe that the functions 1( ) 1u x and 2 ( ) 1/u rx are both hormonic for r > 1 and assumethe same value 1 on the surface of the unit sphere. Hence if we require that the solution vanishes atinfinity, then 2 ( )u x is the required solution. Indeed, it has been proved that, when we solve theDirichlet problem for the exterior of the unit sphere (by expanions in spherical harmonics) suchthat the potential vanishes at infinity, then we arrive at the following nature of the solution

[ ] (1/ )u O r and 2( / ) (1/ )u r O r ... (14)

With help of these considerations and the value of the fundamental solution, it has beenshown that

lim 0,r S

u EE u dSr r

... (15)

on the surface of the sphere of radius r.We now proceed to define and analyze the exterior and interior Dirichlet problems for an

arbitrary surface S.The exterior Dirichlet problem. Definition.

Let Re be the exterior of a bounded region Ri in three dimensions, where the boundary of Riis denoted by S. The exterior Dirichet problem is the boundary value problem

2 20, ; , (1/ ), ( / ) (1/ )e e e e eSu R u f u O r u r O r x ... (16)

where ( )f x is a given continuous function on S.



The interior Dirichlet problem. Definition.Let Ri be a bounded interior region and let S denotes the boundary of Ri. The interior

Dirichlet problem is the boundary value problem.

2 0, ,i iu R x ,i Su f ... (17)where f (x)is a given continuous function on S.

Solution of the interior Dirichlet problem (17). Suppose that such a solution u is thepotential of a double-layer with density (which is as yet unknown)

2( ) cos( , )( )i S

u dSr

x nx ... (18)

In order that ui may satisfy the boundary condition [ ] ( ),i Su f x we use relation (12) andobtain the following Fredholm integral equation of the second kind for ( ) :P

1( ) ( ) ( , ) ( ) ,2 S

P f P K P Q Q dS ... (19)

where the kernel K (P, Q) is given by

2( , ) {cos ( , )}/ 2 | | ,K P Q x n x ... (20)

and ( )P x and ( )Q are both on S.Now, we solve (19) for and substitute this solution in (18) to obtain the desired solution of

the Dirichlet problem given by (17).Proceeding likewise, the Dirichlet problem for an external domain bounded internally by S

can be reduced to the solution of a Fredholm integral equation of the second kind.We now proceed to find an integral-equation formulation of the exterior and interior boundary

value problems (16) and (17) in a composite medium when ( )f x is the same function in both theseproblems.

We know that the free space fundamental solution ( ; )E x satisfies

2 ( ),E x for all and .x ... (21)Multiplying (17) by E, (21) by ui, adding, integrating and using Green’s second identity, we obtain

( ), when0, when

i iii

S e

u Ru EE u dSRn n

... (22)

where n is the outward normal to Ri on S.Similarly, we multiply (16) by E, (21) by ue, add and integrate over the region bounded

internally by S and externally by a sphere Sr and apply Green’s second identity. Noting that thecontribution from Sr vanishes as r in view of the boundary conditions at infinity, we arrive at

0, when( ), when

iee

S e e

Ru EE u dSu Rn n

... (23)

where we have used the fact that the outward normal to Re and S is now in the – n direction.We now add (22) and (23). Since ui and ue are both equal to f on S, their contributions cancel.

Thus, we obtain

( ), if( ; )

( ), ifi ii e

S e e

u Ru uE dSu Rn n

x

... (24)



where .Sx Using the relations (2) and (10) and relabelling x and we obtain

( ), if( )( ), if| |

i i

S e e

u RdS

u R

x xx xx

... (25)

that is, a single-layer potential with unknown charge density . Finally, using the boundaryconditions

[ ] [ ] ,i S e Su u f ... (26)in (25), we arrive at the Fredholm integral equation of the first kind

( )( ) ,

| |Sf dS

x

x ... (27)

with both x and on S. We now solve (27) for the unknown ( ) and substitute this value of

( ) in (25) to calculate ( ) and ( ).i eu ux xInterior and exterior Neumann problems.

In this case we are required to find the solution of Laplace or Poisson equation when thenormal derivative is prescribed on S.The exterior Neumann problem. Definition

Let Re be the exterior of a bounded region Ri in three dimensions, where the boundary of Riis doneted by S. The exterior Neumann problem is the boundary value problem :

2 0, ,e eu R x ( / ) ,e Su n f [ ] 0eu ... (28)

The interior Neumann problem, Definition.Let Ri be a bounded interior region and let S denote the boundary of Ri. The interior Neumann

problem is the boundary value problem

2 0, ,i iu R x ( / ) ,i Su n f ... (29)

where n is the outward normal to Ri and f is a given continuous function on S.We first observe that (29) cannot have a solution for every f. This is clear from the physical

interpretation of (29) as a steady-state heat conduction problem. We have no sources in the regionRi and the heat flow is prescribed on S. These conditions are consistent with the steady state onlyif the total heat flow through S vanishes. Accordingly, a solution of (29) will exist only if

( ) 0, .S

f dS S ... (30)

Alternatively, we now propose to show that (30) is a necessary condition for a solution of(29) to exist.

Indeed, recall that ui is harmonic in Ri, that is, 2 0iu ... (31)

(31) implies that2( ) 0

ii

Ru dV ... (32)

Since 2 ,div grad (32) reduces to 0i

iR

div grad u dV ... (33)

Using the divergence theorem, (33) reduces to

0,iS

grad u dS n i.e., 0iS

u dSdn



or ( ) 0, ,S

f dS S

where we have used the boundary condition ( / ) ,i Su n f which is involved in (29).If the condition (30) is satisfied, it can be shown that (29) has a solution, but that the solution

is no longer unique. Clearly, to any particular solution of (29) we can add an arbitrary constant andthe resulting function still satisfies (29).

Remark. For the exterior Neumann problem (28) in three dimensions no such restriction as(30) is needed. The problem (28) has unique solution.

The exterior and interior Neumann problems can be reduced to integral equations by using amethod employed for the corresponding Dirichlet problem.

Now, we propose to find a solution of the interior Neumann problem (29) in the form of thepotential of a simple layer

( ) ,iS

Qu dSr

... (34)

which is a hormonic function in Ri. Clearly, (34) will be a solution of (29) if the density is sochosen that

( / ) ( ),i Pu n f P P S

... (35)Using the relation (9), we obtain

2cos ( , )( ) 2 ( ) ( ) ,i

SP

uf P P Q dSn r

x n

or 21 cos ( , )( ) ( ) ( ) ,

2 2SP f P Q dS

r

x n... (37)

showing that ( )P is a solution of the Fredholm integral equation of the second kind (37).Exsercise. Show that the solution of the exterior Neumann problem also leads to a similar

integral equation. Also, give the integral equation formulation of the problem (28) and (29) in acomposite medium when f is the same function in both these problems. Proceeding as for thecorresponding Dirichlet problem, obtain a Fredholm integral equation of the first kind. Also, showthat instead of a single layer potential, a double-layer potential is obtained.

Solution. Left as an exercise12.3 SOLVED EXAMPLES BASED ON ART 12.2

Ex. 1. Obtain electrostatic potential due to a thin circular disc.Sol. Take S to be a circular disc of radius a and let V be the potential prescribed on S. With

origin on the centre of the disc and z axis normal to the plane of the disc, we shall use cylindricalpolar coordinates ( , , ).z Then the given disc occupies the region z = 0, 0 a for all . Without

any loss of generality, let the potential V on the disc be ( ) ( ) cos ,nf n where n is an arbitraryinteger, because we can use the Fourier superposition principle. It follows that the charge density

will also be of the form ( ) ( ) cos .n n Then equation (27) of Art. 12.2 reduces to

( ) ( )( ) cos ,

| |n

disc

f n dS

x ... (1)

where 1( , ,0) and ( , ,0).t x Re-writing (1) we have

( )2( ) 1 1

2 2 1/ 20 0 1

( ) cos( ) cos{ 2 cos ( )}

nan t t n d dtf nt t

... (2)



Let 1 so that 1 and 1 .d d Then, we have2

1 12 2 1/ 20 1

cos{ 2 cos ( )}

n dt t

2 2

2 2 1/ 2 2 2 1/ 2cos ( ) cos cos

{ 2 cos } ( 2 cos )n d n n d

t t t t

0 2 2

2 2 1/ 2 2 2 1/ 2 2 2 1/ 20 2

cos cos cos cos cos cos( 2 cos ) ( 2 cos ) ( 2 cos )

n n d n n d n n dt t t t t t

= I1 + I2 + I3, say ... (3)

Now,0 2

1 2 2 1/ 2 2 2 1/ 22

cos cos cos cos( 2 cos ) ( 2 cos )

n n d n n dIt t t t

[Putting 2 so that d d ]

or2

1 2 2 1/ 22

cos cos ,( 2 cos )

n n dIt t

using standard properties of definite integrals.

Thus, I1 = – I3 and hence (3) reduces to

2

1 13 2 3 22 2 1/ 20 1

cos .{ 2 cos( )}

n d I I I It t

Hence, 2 2

1 12 2 1/ 2 2 2 1/ 20 01

cos coscos{ 2 cos( )} ( 2 cos )

n d n dnt t t t

... (4)

Using (4), (2) yields( )2( )

2 2 1/ 20 0

( ) cos( )( 2 cos )

nan t t n d dtft t

... (5)

We now use the following expansion formula :

2 2 1/ 2

0 0( 2 cos ) (2 ) cos ( ) ( ) ( ) ,ok k kk

t t k J u J tu du

... (6)

where ok is the well known Kronecker delta defined by

1, if 00, if 0ok

kk

... (7)

Using the expansion formula (6) in (5) and using the orthogonality of cosine functions, wearrive at the following Fredholm integral equation

( ) ( )00

( ) ( ) ( , ) ,an nf t t K t dt ... (8)

where the kernel 0 ( , )K t is given by

0 0( , ) 2 ( ) ( )n nK t J u J tu du

... (9)

Remark. For an annular disc of radius b and outer radius a, the formula corresponding to (8)

takes the form ( )0( ) ( ) ( , )

an nb

f t t K t dt ... (10)

Ex. 2. Solve the integral equation (27) of Art. 12.2 when S is a unit sphere and sin cosf .

Sol. In terms of spherical coordinates ( , , ),r we know that the elementary area on the

surface of unit sphere is sin .d d Hence the integral equation (27) of Art. 12.2 may be re-written as



2

1 1 1 11

0 0

sin ( , )sin cos

| |d

d

x ... (1)

In what follows we shall use some results of spherical harmonics*

1 10,0 ,

( , ) ( , )1| |

m mnn n

nn m n m n

Y YNN

x ... (2)

where ( , )mnY are the spherical harmonics (refer Art 10.11 in chapter 10) and

2 2,

0 0

4 ( | |)!sin | ( , ) |2 1 ( | |)!

mm n n

n mN d Y dn n m

... (3)

Again, let us assume that

1 1 , 1 10( , ) ( , ),

nm

m n nm m nY

... (4)

and observe that 1 11 1sin cos (1/ 2) ( , ) ( , )Y Y ... (5)

Substituting the values given by (2), (3) and (4) in (1) and also using the orthogonalityproperties of the spherical harmonics, we have

1,1 1, 1 3/8 . ... (6)

and , 0m n for all other values of m and n. ... (7)Using (6) and (7), (4) reduces to

11( , ) (3/ 4 ) (cos ) cos ,P ... (8)

where 11P is an assosiated Legendre function.**

12.4 GREEN’S FUNCTION APPROACHThe Green’s function is an arbitrary function which plays the same role in the integral

formulation of partial differential equations as it plays in the case of ordinary differential equation(refer chapter 11). The Green’s function depends on the form of the differential equation, theboundary coditions and the region.Green’s function for the negative Laplacian.

Let R be an open, bounded and three-dimensional space and let its boundary be denoted byS. The Green’s function for the negative Laplacian in R is the solution ( , )G x of the boundaryvalue problem

2 ( ),G x andx in R; G = 0, on Sx ... (1)

Unless otherwise stated, all differentiations are with respect to the coordinates of x.The boundary value problem (1) has a simple interpretation in electrostatics or in steady heat

conduction. We can view ( , )G x as the temperature at any point x in R due to a unit source

located at , when the boundary temperature is required to vanish.

Alternatively, ( , )G x can be regarded as the electrostatic potential due to a unit charge at when the boundary potential vanishes (which is the case, for instance, if S is a grounded metallic

* For details, refer Art 10.11 in chapter 10**Refer Chapter 9, page 297 part II of ‘‘Advanced Differential Equations’’ by Dr. M.D. Raisinghania, publishedby S. Chand & Co. New Delhi.



shell). In view of this electrostatic interpretation of ( , ),G x it follows that ( , )G x is the sum ofthe potential of the unit source at in free space and of the potential due to the charge induced onS (this charge is present on S because of the requirement that the potential vanishes there). In anynumber of dimensions we can write

( , ) ( , ) ( , )G E x x x ! ... (2)where E is the free-space fundamental solution satisfying

2 ( ),E x for all andx ... (3)and ! is a harmonic function which satisfies the boundary value problem

2 0, ;R x! for onE S x! ... (4)

Thus the problem of finding ( , )G x is reduced to that of finding a harmonic function ! in R,assuming certain special boundary values on S. For ready reference, we have

(1/ 2 ) log (1/ | | in two dimensions1/ 4 | | in three dimensions

E

xx

Note also that the fundamental solution ( ; )E x is the free-space Green’s function.Some properties of Green’s function

Property 1. The Green’s function exists and is unique

Proof. Recall that the Green’s function is given by (2) where ( ; )E x is given by (5). Also

note that ( ; )E x exists and is unique and so we need to show that ! exists and is unique. But (4)is a Dirichlet problem and so the existence and uniqueness of ! follows from the existence anduniqueness of the Dirichlet problem (4) for Laplace’s equation with continuous boundary values.

Property 2 : The Green’s function is symmetric, i.e., ( , ) ( , )G G

Proof. Let ( , )G x and ( )G x, be the Green’s functions for the region R corresponding to

region R corresponding to sources located at and , respectively and S be the boundary of R.Then, by definition of the Green’s function, we have

2 ( , ) ( ),G x x and in ;Rx ( , ) 0,G x x on S ... (6)

and 2 ( , ) ( ),G x x and in ;Rx ( , ) 0,G x x on S ... (7)

Multiplying the differential equation in (6) by ( , )G x , the differential equation in (7) by

( , ),G x subtracting and then integrating over R, we finally obtain

2 2{ ( , ) ( , ) ( , ) ( , )} ( , ) ( ) ( , ) ( )}R R

G G G G dV G G dV x x x x x x x x

Using Green’s second identity (refer Art 12.1) on the L.H.S. and the shifting property ofDirac delta function (refer Art. 10.7) on R.H.S., we obtain

( , ) ( , )( , ) ( , ) ( , ) ( , ),

S

G GG G dS G G

n n

x xx x ... (8)

where n is the outward normal to S at x and dS is the surface element on S at x.Now, from (6) and (7), we see that ( , )G x and ( , )G x vanish when x is on S. Hence (8)

reduces to



0 ( , ) ( , )G G so that ( , ) ( , )G G

showing that ( , )G is a symmetric function of its arguments.

Physicists refer to the symmetric property ( , ) ( , ), ,G G in R as the reciprocity

principle. We now give its physical significance : the potential at due to unit source at is equal

to the potential at due to a unit source at .Property 3. The Green’s function is positiveProof. Draw a small circle R of radius with centre

at the source point as shown in the adjoining figure. LetR R denote the part of R excluding .R In

,R R ( , )G x is harmonic at x. The boundary ofR R consists of S and the spherical surface| | . x On S, ( , )G x vanishes; since ( , )G x ispositively infinite at , ( , )Gx x will be positive on| | x if is chosen sufficiently small.

In view of the maximum principle for harmonic functions* ( , )G x must be strictly positive

in .R R Since R can be made as small as we please, ( , )G x is strictly positive in R.Property 4. In three or more dimensions, we have

0 ( , ) ( , ) ,G E x x x R Proof. Left as an exercise.Solution of the Dirichlet problem. The Green’s function plays its principal role to find the

solution of the Dirichlet problem for the Poisson equation2 ( ) 4 ( ), ; , .u u f S x x x R x ... (9)

Here ( ) x is given function defined in the given region R and f is a given function on theboundary S of R.

Again, we know that the Green’s function ( , )G x for the negative Laplacian in an open, boundedregion R in three dimensional space with boundary S is the solution of the boundary value problem

2 ( ), 0, .G G S x x ... (10)Multiplying the differential equation in (9) by G, the differential equation in (10) by u, subtracting

and then integration over R, we finally obtain

2 2( ) ( , ) 4 ( ) ( , ) ( ) ( )R R R

u G G u dV G dV u dV x x x x x x

Using Green’s second identity (refer Art. 12.1) on L.H.S. and the shifting property of Diracdelta function (refer Art. 10.7 of chapter 10) on R.H.S., we obtain

( , )( ) ( , ) 4 ( ) ( , ) ( ),

S R

G uu G dS G dV un n

xx x x x

where n is the outward normal to S at x.

*The maximum principle for harmonic function. Let R be a bounded, open region; S its boundary and Ris the closed region which is the union of R and S. If u is harmonic in R and continuous in R , the maximumand minimum values of u in R are both attained on S. Moreover, if u is not constant in R, the extremalvalues are attained on S.

R

S

R – R



Since u = f and ( , ) 0G x on S, we obtain

( , )( ) 4 ( ) ( , ) ( )S R

Gf dS G dV un

xx x x

Interchanging x and , the above equation reduces to

( , )( ) 4 ( , ) ( ) ( )

R S

Gu G dV f dSn

xx x

or ( , )( ) 4 ( , ) ( ) ( )

R S

Gu G dV f dSn

xx x

... (11)

[ ( , )G x is symmetric, i.e., ( , ) ( , )G Gx x ]

For the particular-case 0, (11) reduces to

( , )( ) ( )

S

Gu f dSn

xx

... (12)

When f = 1 on S, then the solution u of the Laplace’s equation is clearly u = 1 for the interiorDirichlet problem. Substituting this value of u in (12), we find

( , ) 1,S

G dSn

x for every Rx ... (13)

As an application of result (12), consider the following example.Poisson integral formula

The Green’s function for the Laplace equation, when the surface S is a sphere of radius a willbe obtained by expressing it as source and image point combination. Let ( )P x be any pointwithin the sphere such that ,OP where O is the centre of the sphere. Let ( )P x be the inverse

point of P such that OP Then, by definition of the inverse point we have 2 .a Let ( )Q beany point on S. Let PQ = r and .P Q r

Now, 2a 2OP OP OQ

/ /OP OQ OQ OP ... (i)

In triangles OPQ and ,OP Q QOP QOP and condition (i) is satisfied. Hence, the triangles OPQand OP Q are similar and therefore

/ /r r a so that 1/ /r a r ... (14)Now, for three dimensional case (2) reduces to

( ) 1/ 4 | | ( ; )G x, x x ! ... (15)Hence in view of the relations (1) and (15), the value of the Green’s function is given by

1 1( , )

4aG P Q

r r

... (16)

With help of this Green’s function, we now proceed to solve the interior Dirichlet problemfor the sphere

2 0,u r < a; ( , )u f on r = a ... (17)We now proceed to find the value of /G n which is required in formula (12).



Using cosing formula of trigonometry, from triangles OPQ and ,OP Q we have

2 2 2 2 cos( , )a r ar x ... (18)

and 2 2 2 2 cos ( , ),a r ar x ... (19)From (16), we have

2 21 1

4G r a rn n nr r

2 21 cos ( , ) cos ( , )

4a

r r

x x

2 2 2 2 2 2

3 31 ( ) ,

4 2 2a r a a rar a r

using (18) and (19)

2 2 2 4 2 2 2 2 2

3 3 3 31 ( / ) ( / ) ,

4 2 2 ( / )a r a a r aar a r

2

2and andr a ar aa rr r

2 2 2 2 2 2 2 2

3 3 31

4 2 2 4a r a r aar ar a r

Substituting the above value of / n in (12), we finally obtain

2 2 2

2 2 1/ 2( ) ( , ) sin( )

4 { 2 cos ( , )}a a f d du P

a a

x ... (20)

which is known as Poisson integral formulaSolution of the Neumann problemIn order to extend the above analysis to the Neumann problem

2 0, ,u R x / ( )Su n f x ... (21)

we define the Green’s function ( , )G x by the boundary value problem (compare it with (1) forDirichlet problem)

2 ( , ) ( );G x x / 0,SG n ... (22)where R is an open, bounded and three dimensional space and let its boundary be denoted by S. nis the outward normal to S at x and f is a given function.

Multiplying the differential equation in (21) by G, the differential equation in (22) by u,subtracting and then integrating over R, we finally obtain

2 2( ) ( , ) ( ) ( )R R

u G G u dV u dV x x x x ... (23)

Using Green’s second identity (refer Art. 12.1) on L.H.S. and the shifting property of Diracdelta function (refer Art. 10.7) on R.H.S, (23) reduces to

( ) ( , ) ( ).S

G uu G dS un dn

x x ... (24)

From (21) and (22), / 0SG n and ( / ) ( ). HenceSu n f x (24) yields

( , ) ( ) ( )S

G f dS u x x ... (25)



Interchanging andx , (25) reduces to

( ) ( , ) ( )S

u G f dS x x ... (26)

Since ( , )G x is symmetric, so ( , ) ( , ).G Gx x

Hence, from (26), ( ) ( , ) ( )S

u G f dS x x ... (27)

For existence of a solution of the Neumann problem, the prescribed function f (x)must satisfythe consistency condition (refer result (30) of Art 12.2)

( ) 0,S

f dS S ... (28)

Finally, we propose to discuss the interior and exterior Dirichlet problems for a body S enclosedwithin a surface :S

2 0,iu ,iRx i Su f ... (29)

2 0, ,e eu R x ,e Su f 0e Su ... (30)

The Green’s function ( , )G x satisfies the auxiliary problem (on absorbing the factor 4 in G):2 4 ( ),G x [ ] 0SG ... (34)

Now, we proceed exactly in the same manner as we did in Art. 12.2 while deriving theformula (27) of Art. 12.2. Then, the above the relations (29), (30) and (31) give rise to

( ) ( , ) ( ) ,S

f G dS x x ... (32)

which reduces to (27) of Art. 12.2 for an unbounded medium.12.4.A. THE METHOD OF IMAGES

Before solving the Dirichlet’s problem with help of Green’s function technique, we shalldiscuss the method of images. This method is frequently used in the development of electrostatics.

Working rule for the method of images : Investigate the effect of a certain type of singularityat some point P of a given region R together with the influence of another source type of singularitylocated at a point P outside the region R under consideration. P is chosen as the optical imageof P in the boundary of the given region R.

Note. The method of images can be used only while dealing with very simple geometries.12.5 SOLVED EXAMPLES BASED ON ART. 12.4 AND 12.4 A

Ex. 1. Discuss electrostatic potential problem of a conducting disc bounded by two parallelplanes.

Sol. The given problem is an extension of the problem considered in Ex. 1 of Art. 12.3. In whatfollows we shall use the notations of Ex. 1 of Art. 12.3.

Let the given parallel planes be z = b and z = –c, where b > 0 and c > 0. Then the boundaryvalue problem takes the following form

2 ( , , ) 0 in ,V z D ... (1)( )( , , 0) ( ) cos , 0nV f n a ... (2)

and ( , , ) 0, , ,V z z b z c ... (3)where D is the region between the disc and the parallel planes.

Clearly, the Green’s function ( ; )G x corresponding to the boundary value problem (1) – (3)must satisfy the auxiliary system.



2 ( , ) 4 ( ), 0G G x x on z = b, z = –c. ... (4)

To compute ( , )G x , we shall use the well known method of images (see Art. 12.4 A).

Accordingly, for a positive unit charge at the source point 1 1( , , ),t z the image system consistsof positive unit charge at the points.

1 1[ , , 2 ( ) ),n t n b c z 1, 2, 3, ....n ... (5)and negative unit charge at points

1 1[ , , 2 ( ) 2 ],n t n b c c z 0, 1, 2, 3, ....n ... (6)as shown in the following figure

Hence the value of the Green’s function is given by

1 1

1 1 1 1( , )| | | | | | | |n n nn n n

G

x

x x x x

... (7)

Now, using the identity 1| |0

0

1 ( ) ,| |

u z zJ u e du

x ... (8)

where 2 2 1/ 21{ 2 cos ( )} ,t t ... (9)

(7) reduces to

1| 2 ( ) |0 10

1( , ) ( )| |

u z n b c zG J u e

x

x

1| 2 ( ) |

1

u z n b c ze

1| 2 ( ) 2 |

0

u z n b c c ze

1| 2 ( ) 2 |

1

u z n b c c ze du

... (10)

Summing the geometric series which are involved in (10) and simplifying, we finally get

1 1( ) ( )

00

1 sinh ( ) sinh ( )( , ) ( )| | sinh ( )

c z u n z ue u z b e u z cG J u duu b c

xx

... (11)

We know that the expansion of 0 ( )J u is given by

0 10( ) (2 ) cos ( ) ( ) ( )ok k kk

J u k J u J ut

... (12)

where 0, if 0

Kronecker delta1, if 0ok

kk

... (13)

Using (12), (11) takes the form

( )1 10

( , ) 1/ | | (2 ) cos ( ) ( , , , ),kokk

G k G t z z

x x where ... (14)

( )1( , , , )kG t z z

1 1( ) ( )

0

sinh ( ) sinh ( ) ( ) ( )sinh ( )

u c z u b z

k ke u z b e u z c J u J ut du

u b c

... (15)



Multiplying (1) by G and (4) by V, subtracting, integrating and using Green’s second identity,we obtain as usual

1( , , ) ,4 S S

V GV z G V dSn n

... (16)

where S+ and S– are the upper and lower parts of the disc, respectively. Using the fact that on thesurfaces S+ and S–, the value of the outward normal is 1( / )z and 1/ z respectively and alsousing the boundary conditions (2), (16) yields

2( ) ( )

1 1 10 0( ) cos ( ) ( , , , ,0,0) cos ,

an nf n t t G t n d dt

... (17)

where ( )

11 1

1( ) cos4

n V Vt nz z

... (18)

Now, we assume that 1 . Then, we proceed exactly as done in getting relation (5)from relation (2) in example 1 of Art. 12.3. This will give us the following relation :

1

2( ) ( ) ( ) ( )

0 0 00

cos( ) ( ) ( ) 2 ( , ,0,0)| |

a an n n n

z z

n df t t dt t t G t dt

x ... (19)

where we have substituted the value of G given by (14).Re-writing (19), we have

( ) ( )10

( ) ( ) ( , ) ,an nf t t K t dt ... (20)

where 2 ( )

1 2 2 1/ 20

cos( , ) 2 ( , , 0, 0)( 2 cos )

nn dK t G tt t

... (21)

Remark 1. When b and ,c (21) reduces to the formula (8) of Art. 12.3.Remark 2. For an annular disc of inner radius b and the outer radius a, the formula which

corresponds to (21) is given by

( ) ( )1( ) ( ) ( , )

an n

bf t t K t d ... (22)

which contains the same kernel as in (21).Ex. 2. Discuss a electrostatic potential problem of an axially symmetric conductor placed

symmetrically inside a cylinder of radius b.

Sol. We take cylindrical polar co-ordinates ( , , )z with origin at the centre of the conductorand z axis along its axis of symmetry (which is also the axis of the cylinder). In order to simplifyour calculations, we choose V = 1 on the surface of the conductor. Then, from relation (32), of Art.12.4, we get Fredholm integral equation of the form

1 ( , ) ( ) ,S

G dS x , Sx ... (1)

where G satisfies the system2 4 ( ),G x G = 0 on b ... (2)

Now, the differential equation for ( , )G x involving cylindrical polar coordinates is given by

2 2

12 2 21 1 4 ( ) ( ) ( )G G G t z z

z

... (3)



where 1 . Using the definition of the Green’s function given in Art. 12.4, we know that

1( , ) ( , ) (1/ | |)G G x x x ... (4)

is finite in the limit as .x We now proceed to find the solution of (3) with help of the Fourier series expansions

( )11

,( ; ) (2 ) cos ( , , )kokk

zG k g t z

x .. (5)

where 2( )

0

1 ( ; ) cos2

kg G k d

x .. (6)

Multiplying (3) by (1/ 2 )cos k and integrating with respect to from 0 to 2 , we have

2 2 ( )( ) ( )

1 12 21 2 ( ) ( )

kk kk gg g z z

z

... (7)

Now, we take the Fourier transform of equation (7) by setting

( ) ( )1( ) ;2

k i u z kF g e g dz

( ) ( )1 ( )

2k iu z kg e F g du

... (8)

Then, the system (2) reduces to

1

22 ( ) ( ) 2 2 ( )

12

( )

2( ) ( ) ( ) ( ) ( )

( ) 0,

iu zk k k

k

d dF g F g u k F g edd

F g b

... (9)

The boundary value problem (9) can be easily solved by the method and notations of chapter11 and the solution so obtained is then inverted to give*

1( )( ) 1 ( ) ( ) ( ) / ( ) ( ) ( )iu z zkk k k k k kg e K u I u K ub I ub I u I ut du

... (10)

where Ik and Kk are modified Bessel functions. Finally, from (5) and (10), value of G is given by

1( )

0

1( , ) (2 ) cos ( ) ( )iu z zok k kk

G k e K u I u

x

( ) / ( ) ( ) ( )k k k kK ub I ub I u I ut du ... (11)

When ,b 1/ | |,G x and so (11) reduces to

1( )

0

1 1 (2 ) cos ( ) ( )| |

iu z zok k kk

k e K u I u du

x ... (12)

Combining (11) and (12), we obtain

( )10

( ; ) 1/ | | (2 ) cos ( , , , )kokk

G k G t z z

x x ... (13)

where 1( )( )1

1( , , , ) ( ) ( ) ( ) / ( )iu z zkk k k kG t z z e I u I ut K ub I ub du

1( )

0

1 ( ) ( ) ( ) / ( )iu z zk k k ke I u I ut K ub I ub du

10 ( ) ( ) ( ) ( ) / ( )iu z z

k k k ke I u I ut K ub I ub du

... (14)

* In equations (10), (11) and (12), we have used the following notations : max( , ); min( , )t t



Substituting –u for u in the second integral on the R.H.S. of (14) and noting that

( ) ( 1) ( ),kk kI z I z ( ) ( 1) ( )k

k kK z K z

and hence ( ) / ( ) ( ) / ( ).k k k kK ub I ub K ub I ub Then (14) reduces to

( )1 1

0

( ) ( ) ( )2( , , , ) cos[ ( )]( )

k k k k

k

I u I ut K ubG t z z u z z du

I ub

... (15)

Now, we use the fact that the conductor is axially symmetric.Then, since the Green’s functionis independent of for an axially symmetric body, so it leaves only one term in the series (13) :

Thus, ( )1( , ) 1/ | | ( , , , )oG G t z z x x ... (16)

which is of the form (4). Substituting (16) in (1), we can easily obtain the required integral equation.

Remark. When ,b we obtain as a particular case equation (5) of solved example 1 of

Art. 12.3 for ( ) ( ) 1nf

12.6 THE HELMHOLTZ EQUATIONIn this article we propose to extend the discussion the previous two articles to study the

Helmholtz equation

2( ) 0.u ... (1)

The free-space Green’s function or the fundamental solution ( ; )E x is the solution of thespherically symmetric differential equation

2 ( ),E E x ... (2)and which vanishes at infinity. Such a solution in three dimensions is given by

exp ( | | ) exp ( )( ; ) ,

4 | | 4i irE

r

xxx

where | |r x ... (3)

We now discuss three situations :Case (i) Let be a complex number. Then is selected to be that root of that has a

positive imaginary part so that ( ; )E x vanishes exponentially at infinity..

Case (ii) Let be real and positive, that is, 2 , real, Then the solution

exp( | |) exp ( )( ; ) ,4 | | 4

i iwrEr

xx

x

... (4)

is selected such that 0. This represents an outgoing wave if we adjoin the factor e–iwt.

Case (iii). Let be real and negative. Then as in case (i), is selected to be that rootof that has a positive imaginary part so that ( ; )E x vanishes exponentially at infinity. For the

particular case 2 ,k where k is real and positive, (3) reduces to

( ; ) /(4 )krE e r x ... (5)The solutions which corresponding to (3), (4) and (5) in two dimensions are given by

(1)0( / 4) (| | / ),i H x (1)

0( / 4) ( | |)i H x and 0(1/ 2 ) ( | |),K k x

respectively (1)0H and K0 being well known Hankel and modified Bessel functions, respectively..



The integral representation formula for the solution of the inhomogeneous equation2 2( ) 4 .k u ... (6)

can be obtained with help of (2) and (6) by using Green’s identity and is given by

1 1( )4 4

kr kr kr

R S S

e e e uu dV u dS dSr n r r n

x ... (7)

The interpretation of three integrals occuring in (7) as volume, single-layer, and double-layerpotentials is the same as for the corresponding formulas in Art. 12.2. Again, the properties of thesepotentials are also same as given in Art. 12.2. For example, the formulas that correspond to (8) and(9) of Art. 12.2 are

2 ( ) ( ) ,kr

SP

u eQ dSn n r

... (8)

where ( )kr

S

eu Q dSr

... (9)

Again, the formulas that correspond to (11) and (12) of Art. 12.2 are

2 ( ) ( ) ,kr

P S

eu P Q dSn r

... (10)

where ( )kr

S

eu Q dSn r

... (11)

The remaining notations are same as in Art. 12.2.The integral reperesentation of solutions of the exterior and interior Dirichlel and Neumann

problems can be obtained in on analogous way. This will be illustrated in solved examples ofphysical interest in next Art. 12.7.12.7 SOLVED EXAMPLES BASED ON ART. 12.6

Example 1. Discuss steady Stokes flow in an unbounded medium.Solution. The Stokes flow equations*

2 ,p q 0 q ... (1)

govern the slow, steady flow of incompressible fluids. These have been made dimensionless withhelp of the free-stream velocity ‘u’ and a characteristic length ‘a’ inherent in the given physicalproblem. Here q and p denote the velocity vector and pressure respectively.

Let S be the surface of a given solid moving in the fluid; then the boundary conditions are ( ) , ,S 1q x e x ( ) as 0q x x ... (2)

where e1 is the direction of motion of the given solid B (say) taken to be in the x1 direction.The boundary value problem (1) – (2) can be converted into a Fredholm integral equation of

the first kind by defining the Green’s tensor 1T (or T1ik) and Green’s vector p1 (or p1j), which satisfythe mathematical system

2 ( ), 1 1T p I x ... (3)

*Refer Chapter 14 in ‘‘Fluid Dynamics’’ by Dr. M. D. Raisinghania, published by, S. Chand & Co. NewDelhi



0, 1T as , 1 0T x ... (4)

where ij I the Kronecker delta.It follows by direct verification that the system (3) – (4) has the representation formulas

21 (1/8 ) ( ),grad grad T I 2

1 (1/8 ) ,grad p ... (5)

4 2 2 8 ( ) x ... (6)Clearly, an appropriate solution of the biharmonic equation (6) is given by

| | .r x Hence, we obtain

21 (1/ 8 ) | | | |grad grad T I x x ... (7)

and 21 (1/8 ) | | .grad p x ... (8)

We now proceed to find the integral equation equivalent to the boundary value problemgiven by (1) and (2). To this end, we take the scalar product of (1) by 1T and of (3) by q and use theusual steps of subtracting and integrating. The integral so obtained will contain terms p1T and

1q p which can be re-written to using the following identities.

1 1( ) , q p q p 1 1( ) ,p p T T ... (9)where we have made use of the following results

0 q and 1 0 T ... (10)Proceeding as indicated above, we finally obtain

11 1( )

Sp dS

n n

Tqq x n T q np ... (11)

In view of equation (3) and the divergence theorem, we conclude that, if q is constant on S,then (11) takes the following simple form

1( )S

dS q x f T ... (12)

where ( / )n p f q n ... (13)

Now, making use of the boundary condition 1( ) q x e when Sx as given by (2), in (12),the desired integral equation is given by

1S

d S 1e f T ... (14)

Example 2. By following the method and notations of example 1, prove that the integralrepresentation formula for the velocity vector when the fluid is bounded by a vessel is

( ) p p dSn n

q Tq x n T q n

Substituting the boundary condition 1q e on S gives the Fredholm integral equation

1S

dS e f T

Here T satisfies the boundary value problem2 ( ), 0, on .grad p 0T I x T T

Solution. Left as an exercise.



Example 3. Discuss steady Oseen flowSolution. The Oseen equations are governed by*

21( / ) , 0,x p q q q ... (1)

which have been made dimensionless with the help of the free-stream velocity ‘u’ and a characteristiclength ‘a’ inherent in the given problem. The quantities , v, q and p stand for the Reynoldsnumber, the coefficient of kinematic viscosity, velocity vector and pressure respectively. Also, wehave

/ua v ... (2)Let S be the surface of a given solid moving in the fluid; then the boundary conditions are

1( ) , ,S q x e x ( ) as 0q x x ... (3)where 1e is the direction of motion of the given solid B (say) taken to be in the x1-direction.

The boundary value problem (1) – (3) can be converted into a Fredholm integral equation ofthe first kind by defining the Green’s tensor T and Green’s vector p which satisfy the mathematicalsystem

21( / ) ( )x T p T I x ... (4)

0, T 0T as ,x ... (5)where ij I the Kronecker delta ... (6)

It follows by direct verification that the system (4) – (5) has the representation formulas 2(1/8 ) ( )grad grad T I ... (7)

21(1/ 8 ) ( / )grad x p ... (8)

2 21( / ) 8 ( )x x ... (9)

We now proceed to solve (9) for . Using the following well known formula2 (1/ | |) 4 ( ) x x ... (10)

in (9), we obtain

2 2 21( / ) 2 (1/ | |),x x ... (11)

showing that if satisfies2

1( / ) 2 / | |,x x ... (12)then (9) is satisfied.

On the other hand, we can re-write (9) as

2 21( / ) 8 ( )x x ... (13)

Let 2 ... (14)Also, we have the following identity

1 1 1 1( ) ( )2 2 21( )[ ] [ 2 ( / )]x xe e x ... (15)

Using (14) and (15), (13) may be re-written as

1 1 1 1( ) ( )2 2( ) 8 ( ),x xe e x ... (16)where 2 ... (17)

In view of the nature of the Dirac delta function, note that the factor 1 1( )xe can affect theequation (16) only at 1 1,x where its value is unity. Hence, (16) reduces to

* Refer chapter 14 in ‘‘Fluid Dynamics’’ by Dr. M.D. Raisinghania, published by S.Chand & Co., New Delhi.



21 1

2( , ) exp | | | | ( )| | | |

x

x x

x

... (18)

From (12) and (18), it follows that

1 11

1 1 exp | | | | ( )| | | |

xx

xx

... (19)

Let 1 1| | ( / | |) ( )s x x ... (20)

1 1 1 1

1

( ) ( ) | || | | | | |x xs

x

x x

1 1| | | || | ( )

| | | | | |sx

xx x

... (21)

We know that1 1

sx s x

... (22)

Then, from (19), (20), (21) and (22), we have

| |/ 1 | |ss e s

or| |

0

1 1| |

ts e dtt

... (23)

Then, the Green’s tensor T and Green’s vector p can be determined.In order to get the integral equation equivalent to the boundary value problem (1) – (3), we

proceed exactly in the same way as we did to get equation (11) of example 1 on page 12.19. Then,for the present problem, the integral equation formula is given by

1( ) ( )S

p p R n dSn n

q Tq x T n n q T q ... (24)

where n1 is the x1 component of the outward normal.Using the boundary condition 1q e on S (as given by (3)), the relation (4) and the divergence

theorem, (24) takes the following simple form

1 ,S

dS e T f ... (25)

where ( / )n p f q n ... (26)

EXERCISE1. (a) Discuss the single-layer and double-layer potentials in two-dimensional potential theory

by starting with the formula ( ; ) (1/ 2 ) log(1/ | |)E x x

in place of ( ; ) 1/ 4 | |E x x as employed in Art. 12.2.(b) With help of the results derived in part (a), prove that the solution of the interior

Dirichlet problem in two dimensions can be written as

cos( ) ( ) ,| |C

x dlx

where is the angle between ( )x and ;n and dl denotes the element of the arc length alongthe arc C.

2. Show that the solutions of the integral equations 1 11

( , )cos cos

| |SdS

x



and 1 11 1 1 1

( , ) 10 cos ( , ) | | cos ,| | 2S S

dS X dS

x

where S is the surface of a thin circular disc of unit radius, are2 2 1/ 22 / (1 ) , 2 2 2 1/ 2(2 ) / 3 (1 ) .

3. Let S be the surface of a unit sphere. Then show that the solution of the integral equation

(i) 1 11

( , )cos cos

| |S

zz dS

x is 1

25 (cos )

12P

(ii) 2 1 1 1

1 1 1 1( , )1 3cos (cos ) | | cos cos

2 8 | |S S

zz P dS dS

xx

is 1 11 3(1/ 4 ) (3/ 2) (cos ) (7 /15) (cos )P P

In the above relations, ( , , )z are cylindrical polar coordinates.

Hints : (i) Use the formula 2

1

1| | (cos ),2 3 2 1

nnn

xr x Pn n

x

1 1/ , min ( , ), max ( , )x r r r r r r r r

where 1 1 1cos cos cos sin sin cos ( )

while ( , , )r and 1 1 1( , , )r are the spherical polar coordinates of and ,x respectively..

(ii) On the surface of the unit sphere, sin and cosz 4. Starting with the Cauchy integral formula for an analytic function

1 ( )( ) ,2 C

f tf z dti t z

where C is the circumference | z | = a and z is in the interior of C, and using the formula

1 ( )0 ,2 *C

f t dti t z

2* ,az z

which is a result of Cauchy theorem because the image part z* (of z) is exterior to C, derive thePoisson integral formula in a plane :

2 2 2

2 20

( , )( , )2 2 cos( )

a f a dua a

ADDITIONAL RESULTS ON GREEN’S FUNCTION AND ITS APPLICATIONS

12.8. Additional results about Green’s functionIn order to find analytical solution of the boundary value problem, the Green’s function

method is one of the conveneient techniques. Consider the differential equation L u (x) = f (x), ... (1)

where L is an ordinary linear differential operator, f (x) is a known function, while u (x) is anunknown function. In order to solve (1), one method is to get the operator L–1 in the form of anintegral operator with a kernel G (x, t) such that

1( ) ( ) ( , ) ( )u x L f x G x t f t dt ... (2)



Here the kernal G (x, t) of this linear operator is known as Green’s function for the differentialoperator. It follows that the solution of the non-homogeneous differential equation (1) can be easilyobtained, once the Green’s function for the problem is known.

Applying the differential operator L to both sides of equation (2), we have

1( ) { ( )} { ( , )} ( )f x L L f x L G x t f t dt ... (3)

From the shifting property of Dirac delta function (refer Art. 10.7), we know that

( ) ( ) ( )f x x t f t dt ... (4)

Comparing (3) and (4), we have ( , ) ( ),L G x t x t ... (5)

where ( )x t is a Dirac delta function. The solution of (5) is known as a singularity solution of(1).

In what follows, we present an example for explaining the inversion of a differential operator.Consider the boundary value problem :

d2u/dx2 = f (x), u (0) = 0 and u (1) = 0 ... (6)For this problem, equation (5) reduces to

2 2/ ( )L G d G dx x t ... (7)

We know that (refer Art. 10.10), ( ) ( )x t H x t ... (8)

Then, from (7) and (8), 2 2/ ( )d G dx H x t ... (9)Integrating (9), dG / dx = H (x – t) + c1 (t), ... (10)

where c1(t) is an arbitrary function. Integrating (10), we have

1 2( , ) ( ) ( ) ( ),G x t H x t dx c t x c t where c2 (t) is another arbitrary function.

or G (x, t) = (x – t) H (x – t) + c1 (t) x + c2 (t), ... (11)

( ) ( ) ( ), by Art.10.10H x t dx x t H x t

Now, from (2) and (11), we have

1 20( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )

xu x x t H x t f t dt x c t f t dt c t f t dt

... (12)

Putting x = 0 in (12) and using the boundary condition u (0) = 0, we have

20 0 0 ( ) ( )c t f t dt

so that 2 ( ) 0c t

Putting x = 1 and c2 (t) = 0 in (12) and using the boundary condition u (1) = 0, we have

1

100 (1 ) ( ) ( ) ( )t f t dt c t f t dt

1

(1 ), 0 1( )

0 , otherwiset t

c t

Substituting the values of c1(t) and c2(t) in (12), we have

1

0 0( ) ( ) ( ) ( ) (1 ) ( )

xu x x t H x t f t dt x t f t dt ... (13)

Comparing (13) with the equation (2), we get the kernel of the integral operator, which iscalled Green’s function or source function, given by



G (x, t) = (x – t) H (x – t) – x (1 – t) , 0 1,t ... (14)satisfying the boundary conditions : G (0, t) = G (1, t) = 0 ... (15)

Extension of the above concepts to partial differential equations.

Let us consider ( ) ( ),L u fX X ... (16)

where L is some linear partial differential operator in three independent variables x, y, z and X is a

vector in three dimensional space given by ( , , ).x y zX Let ( , , )x y z X be another vector in

three-dimensional space. Then the Green’s function may be denoted by ( , )G X X which satisfiesthe equation*

( ; ) ( )L G X X X X ... (17)

On expansion, (17) may be re-written as

( , , ; , , ) ( ) ( ) ( )L G x y z x y z x x y y z z ... (18)

Here the expression ( ) X X is the generalisation of the concept of Dirac delta function

in three dimensional space and ( ; )G X X represents the effect at the point X due to a source

function or delta function input applied at .X

Multiplying equation (17) on both sides by ( )f X and integrating over the volume V with

respect to ,X we obtain

( ; ) ( ) ( ) ( ) ( )X X

X XV V

L G f dV f dV f

X X X X X X X

Comparing it with (16), we have

( ) ( ; ) ( )X

XV

u G f dV

X X X X ... (19)

which is the solution of (16). This leads to the simple definition that if a function ( , , ; , , )u x y z z y z

is a fundamental solution of the equation, for example, 2 0,u then u is a solution of the non-homogeneous equation

2 ( , , ; , , )u x y z x y z The above concept can be easily extended to higher dimensions. Thus, the Green’s function

technique can be applied, in principle, to get the solution of any linear non-homogeneous partialdifferential equation. We have given formula (17) for the solution of a non-homoegeneous partialdifferential equation. However, in practice, construction of the Green’s function is not always veryeasy. In order to provide a motivation for constructing Green’s function, we now proceed to presentsome singularity solutions (or fundamental solutions) to the well known operators.

Let us find the fundamental solution for a three-dimensional potential problem that is governedby the differential equation

2 ( )u X or grad div ( ).u X ... (20)where u can be interpreted, for example, as the electrostatic potential. We propose to get a solutiondepending only on the source distance | | .r X Then, for r > 0, u (r) satisfies

*Physical interpretation of equation (17) in heat conduction or electrostatics. ( ; )G X X can be regarded asthe temperature (the electrostatic potential) at any point X in three dimensional space due to a unit source (ora unit charge) at .X



2 22

1 0uu rr rr

Integrating it, we obtain u = (A/r) + B ... (21)Since the potential vanishes at infinity (i.e., 0u as r ), (21) B = 0.Then, (21) reduces to u = A/r ... (22)Integrating (20) over a small sphere R of radius whose surface is denoted by , we get

(div grad ) 1R

u dV

or / 1,ru r dS

using the divergence theorem.

Substituting the value of u from (22) in the above relation, we get2/ 1

rA r dS

or 2( / ) 1A dS

or 2 2( / ) 4 1A

Thus, 1/ 4A Substituting this value of A in (22), the singularity solution or the fundamental solution of

2 0u is given by (1/ 4 )u r ... (23)Next, let us find the fundamental solution for a two-dimensional potential problem that is

governed by (20). As before, we propose to get a solution depending only on the source distance| | .r X Then, for r > 0 and the present two dimensional case, (20) reduces to

2 1 0uu rr r r

... (24)

Integrating it, we get log ,eu A r B ... (25)where A and B are arbitrary constants.

If we write u = 2 q log (1/r) ... (26)with q as constant, then (25) is satisfied except possibly at the origin, where u is not defined. Thissolution has the property that if C is any circle with centre at the origin, the flux of u through thatcircle is 4 .q It therefore corresponds to a uniform line density q along the z-axis which appearsas a point singularity in the two-dimensional theory.

Note 1 : If ( , , )x y zr and ( , , )x y z r be two distinct points in three dimensional space,

then the singularity solution or the fundamental solution of the Lapace equation 2 0u is given by

1/(4 | |)u r r ... (26)

Note 2 : If ( , )x yr and ( , )x y r be two distinct points in two dimensional space, thenthe singularity solution or the fundamental solution of the Laplace equation 2 0u is given by

2 log (1/ | )u q r r | ... (27)

12.9 THE THEORY OF GREEN’S FUNCTION FOR LAPLACE’S EQUATIONLet us reconsider the interior Dirichlet problem (refer Art. 12.2). To start with, we assume

that the values of u and /u n are known at every point of S of a finite region V and that2 0u within V ... (1)



Here, n is the unit vector normal to S drawnoutwards from V and / n denotes differentiation inthat direction.

We now proceed to find the solution ( )u r ofour problem at point P with position vector r. Let Cbe a sphere with centre at P and radius . Let bethe region which is exterior to C and interior to S.Further, let the boundary of be denoted by and

1/ | |,u r r ... (2)

where r is another point Q either in or on theboundary .

If u and u are twice continuously differentiable functions in and have first order derivativeson , then by Green’s theorem in region , we have

2 2( )u uu u dS u u u u dVn n

... (3)

where dS and dV denote elementary surface area on and elementary volume of respectively..

From (1) and (2), it follows that 2 2 0u u within . So, in the region , (3) reduces to

0u uu u dSn n

or 0,C S

u u u uu u dS u u dSn n n n

where C denotes the surface of the circle with centre at P and radius . Substituting the value ofu given by (2) in the above relation, we obtain

1 1 ( )( )

| | | |C

uu dSn n

rrr r r r

1 1 ( )( ) 0,| | | |S

uu dSn n

rrr r r r ... (4)

where the normals n are in the directions shown in the figure 1.From the figure 1, we see that when Q is on C, we have

1 1

| |

r r and 21 1

| |n

r r ... (5)

Also, dS the surface area on 2 sinC d d ... (6)

Further, on C, ( ) ( )u u du r r

or ( ) ( ) ( / ) ( / ) ( / )u u x u x y u y z u z r r

or ( ) ( ) sin cos sin sin cosu u uu ux y z

r r

Thus, ( ) ( ) ( ) onu u O C r r ... (7)

and hence( ) ( ) ( )u u On n

r r

... (8)

VQ O

CPn

n

n

n

r

r

Figure 1



Now, 22

1 1( ( ) ( ) sin

| |C Cu dS u O d d

n

r ) r

r r , using (5), (6) and (7)

( ) sin ( )C

u d d O r2

0 0( ) sin ( )u d d O

r 4 ( ) ( )u O r ... (9)

and 21 ( ) 1 ( ) ( ) sin ( )

| |C C

u udS O d d On n

r rr r ... (10)

[using (5), (6) and (8)]Substituting the results given by (9) and (10) in (4) and letting tend to zero, we obtain

1 1 ( )4 ( ) ( ) 0

| | | |S

uu u dSn n

rr r

r r r r

Thus, 1 1 ( ) 1( ) ( )

4 | | | |S

uu u dSn n

rr rr r r r ... (11)

Therefore the value of u at an interior point of theregion V can be determined in terms of the values of u and

/u n on the boundary S.We now proceed to show that a similar result also

holds in the case of the exterior Dirichlet probelm. In thiscase we take the region (as chosen is case of the interiorDirichlet problem) to be region bounded by S, a sphere withcentre at P and radius and a sphere with centre at theorigin and large radius R as shown in the figure 2. Takingthe directions of the normals to be as indicated in figure 2and proceeding as in the case of the interior Dirichletproblem, we shall arrive at

1 ( ) 14 ( ) ( ) ( )| | | |S

uu O u dSn n

rr rr r r r 2

1 0u u dSR n R

... (12)

Letting 0 and ,R we find that the solution (11) is valid in the case of the exterior

Dirichlet problem provided that Ru and 2 ( / )R u n remain finite as .R

From equation (3), it appears that to get a solution of Dirichlet problem we need to know notonly the value of the function u but also the value of / .u n But this is not so, as can be seen bythe introduction of the concept of a Green’s function defined as follows.

We define a Green’s function ( , ),G r r by the equation

( , ) ( , ) 1/ | |,G H r r r r r r ... (13)

where the function ( , )H r r satisfies the relations2 2 2 2 2 2( / / / ) ( , ) 0,x y z H r r ... (14)

i.e. the function ( , )H r r is harmonic in V

and ( , ) 1/ | | 0H r r r r on S ... (15)Thus, the Green’s function for the Dirichlet problem involving the Laplace operator is a

function ( , )G r r which satisfies the following properties :

V O

S

P

C

R

n

n

n

Figure 2



(i) 2 ( , ) ( )G r r r r in V ... (16)

(ii) ( , ) 0G r r on S ... (17)

(iii) ( , )G r r is symmetric, i.e., ( , ) ( , )G G r r r r ... (18)(iv) G is continuous, but /G n has a discontinuity at the point r, which is given by the equation

0lim 1

C

G dSn

... (19)

Following the method already used in the derivation of (11) and replacing u by ( , ),G r r wecan prove that

1 ( ) ( , )( ) ( , ) ( )

4 S

u Gu G u dSn n

r r rr r r r ... (20)

From (13) and (15), we have ( , ) 0G r r on S. Hence, from (20), it follows that the solutionof the Dirichlet problem is given by the relation

1 ( , )( ) ( ) ,

4 S

Gu u dSn

r rr r ... (21)

showing that the solution of the Dirichlet problem can be reduced to the determination of theGreen’s function ( , ).G r r

Physical interpretation of the Green’s function. Let S be a grounded electrical conductor(boundary potential zero). Then if a unit charge is located at the source point ,r then ( , )G r r is thesum of the potential at the point r due to the charge at the source point r in free space and thepotential due to the charges induced on S. Thus

( , ) ( , ) 1/ | |G H r r r r r r

It follows that the property (i) given by equation (16) implies that 2 0G everywhere exceptat the source point r.

Proof of some important properties of Green’s functionTheorem I. The Green’s function 1 2( , )G r r has the symmetric property, 1 2 2 1( , ) ( , )G Gr r r r

i.e. if P1 and P2 are two points within a finite region bounded by a surface S, then the value at P2of the Green’s function for the point P1 and the surface S is equal to the value at P1 of the Green’sfunction for the point P2 and the surface S.

Proof. The proof of the theorem depends upon the following Lemma which we now state andprove.

Lemma. Consider a sphere with centre at the origin and radius ‘a’. By using the divergence

theorem to the sphere show that 2 (1/ ) 4 ( ),r r where ( )r is a Dirac delta function.

Proof. Applying the divergence theorem to (1/ ),r we get

ˆ(1/ ) (1/ ) ,V S

r dV r dS n ... (23)

where n̂ is an outward drawn normal to the given sphere. If ( , , ),u u r then

1 sinˆ ˆ ˆru u uur r r

e e e

2

2 21 1 1ˆ(1/ ) 4 4r

S S Sr dS dS dS a

r r r a e



Note that 2 (1/ )r has the following three properties

(i) It vanishes if 0r (ii) It is undefined at the origin

(iii) Its integral over any sphere with centre at the origin is 4 .

If follows that 2 1/ 4 ( )V

r dV r ... (23)

Hence the lemma.Proof of the theorem. We define

1( , ) ( , ),

| |G H

r r r r

r r ... (24)

where ( , ),H r r is harmonic so that2 ( , ) 0H r r ... (25)

Then, by the above lemma, we have

2 1 4 ( )

| |

r rr r ... (26)

From (24), (25) and (26), we obtain

2 2 21( , ) ( , ) 4 ( )| |

G H r r r r r r

r r ... (27)

Suppose that P1 and P2 are two points with position vectors 1 2andr r respectively, which lie

in the interior of a finite region V bounded by a surface S. Let r be the position vector of anyvariable point Q lying in V.

Let u = Green’s function for 1 1( , )P G r r

and u Green’s function for 2 2( , )P G r r

Then 1( , ) 0G r r on S ...(28)

and 21 1( , ) 4 ( ),G r r r r by (27) ... (29)

Also, 2( , ) 0G r r on S ... (30)

and 22 2( , ) 4 ( ),G r r r r by (27) ... (31)

If u and u are twice continuously differentiable functions in V and have first order derivativeson S, then by Green’s theorem in region V, we have

2 2( ) ,V S

u uu u u u dV u u dSn n ... (32)

where n is the unit vector normal to dS drawn outwards from V and / n denotes differentiation inthat direction.

Since 1( , )u G r r and 2( , ),u G r r (32) yields

2 21 2 2 1( , ) ( , ) ( , ) ( , )

VG G G G dV r r r r r r r r

2 11 2

( , ) ( , )( , ) ( , )

S

G GG G dS

n n

r r r rr r r r

or 1 2 2 14 ( , ) ( ) ( ) ( ) 0,V

G G dV r r r r r r r r using (28), (29), (30), (31)



or 1 2 2 1( , ) ( ) ( , ) ( )V V

G dV G dV r r r r r r r r

or 1 2 2 1( , ) ( , ),G Gr r r r using the shifting property of Direc delta function (See Art. 10.7).Hence the result.Theorem II. If the Green’s function G is continuous and /G n has discontinuity at r, in

particular, then to show that

0

lim 1C

G dSn

Proof. As shown in figure 1, let C be a sphere with radius and bounded by .C From (16),

it follows that G satisfies2 ( )G r r ... (33)

Integrating both sides of (33) over the sphere C, we obtain

2 1C

G dV ... (34)

Re-writing (34), 20

lim 1C

G dV

or

0lim 1,

C

G dSn

using the divergence theorem

12.10 Construction of the Green’s function with help of the method of images.The following examples will illustrate the method of images given in Art 12.4A on page 12.14.Example 1. Use Green’s function technique to solve the Dirichlet’s problem for a semi-

infinite space. (Kanpur 2011)Solution. The Dirichlet’s problem in the given semi-infinite

space is governed by the following boundary value problem :2 0, 0 , ,u x y z ... (1)

u = f (y, z) on x = 0 ... (2)and 0 as ,u r ... (3)

where 2 2 2 1/ 2( )r x y z By definition, the Green’s function for the given problem

must satisfy the following conditions :( , ) ( , ) 1/ | |G H r r r r r r ... (4)

where 2 2 2 2 2 2( / / / ) ( , ) 0x y z H r r .. (5)

and ( , ) 0G r r on the plane x = 0 ... (6)Suppose that Q with position vector , is the image in the plane x = 0 (i.e., yz-plane) of the

point P with position vector r. If we take ( , ) 1/ | | ,H r r r ... (7)

then it is obvious that equation (5) is satisfied.From the figure-1, we find that PP QP whenever P lies on x = 0. Thus | | | | r r r on

the plane x = 0. Then, from (4) and (7), we see that

1 1 1 1( , ) 0,

| | | | | | | |G

r r

r r r r r r rshowing that equation (6) is also satisfied.

Thus, with help of the method of images, the required Green’s function of the present



problem is given by1 1( , ) ,

| | | |G

r r

r r r ... (8)

where if ( , , ),x y zr then ( , , ).x y z Take ( , , ),x y z r ... (9)The solution of the Dirichlet problem described by (1) – (3) is given by (refer equation (21) of

Art. 12.9)1 ( , )( ) ( ) ,

4 S

Gu u dSn

r rr r ... (10)

where dS is the surface element of the given semi-infinite space.From (8) and (9), we have

1/ 2 1/ 22 2 2 2 2

1 1( , )( ) ( ) ( ) ( ) ( ) ( )

Gx x y y z z x x y y z z

r r

Since 1/ 2 1/ 22 2 2 2 2 2

( , ) 1 1 ,( ) ( ) ( ) ( ) ( ) ( )

Gn x x x y y z z x x y y z z

r r

it follows that on the plane 0x

2 2 2 3/ 2( , ) 2

{ ( ) ( ) }G x

n x y y z z

r r

... (11)

Substituting the value of ( , ) /G n r r as given by (11) and noting that ( ) ( , ),u f y z r (10)takes the form

2 2 2 1/ 2( , )( , , ) ,

2 { ( ) ( ) }x f y z dy dzu x y z

x y y z z

... (12)

giving the solution of the Dirichlet problem given by (1) - (3). If the nature of the function ( , )f y z isexplicitly known, then (12) can be integrated to find the final solution.

Example 2. Obtain the solution of the interior Dirichlet problem for a sphere using theGreen’s function method and hence derive the Poisson integral formula.

Solution. The interior Dirichlet problem for a sphere of radius a is governed by the followingboundary value problem :

2 0, 0 , 0 , 0 2u r a ... (1)

( , , ) ( , )u a f on r = a ... (2)By definition, the Green’s function for the given problem must satisfy the following conditions

: ( , ) ( , ) 1/ | |G H r r r r r r ... (3)

2 2 2 2 2 2( / / / ) ( , ) 0x z z H r r ... (4)

and ( , ) 0,G r r on the surface of the sphere r = a ... (5)

Let ( , , )P r be a point inside the given sphere, where we place a unit charge with positionvector r and let its inverse point with respect to the sphere be Q with position vector . Then

OP r and .OQ

Let r be the position vector of an arbitrary point P inside the sphere such

that .OP r Then | | , | |r OP r OP r r and | | .OQ Since Q is the inverse point of P, we have



OP × OQ = a2 so that OQ = a2 / OP = a2/r. ... (6)

Hence spherical polar coordinates ofP and Q may be taken as ( , , )r and

2( / , , )a r respectively..

If Qbe a variable point on the surfaceof the sphere, then from similar trianglesOQ P and ,OQ Q we have

/ / /PQ QQ r a a so that ( / ) ,PQ r a QQ which is valid for all points on the spherical surface. Hence, the harmonic function is given by

( , )

| || | | |

a a aHrr Q Q r OQ OQ

r rr ... (7)

But 2 2 2( / ) ( / ) ( / ) ( / )OQ OQ r a r r a r

r r r

(7) yields 2 2( , )| ( / ) |

aHr a r

r rr r

... (8)

Then it is obvious that equation (4) is satisfied. Again, from a well-known proposition ofelementary geometry it is known that if P lies on the surface of the sphere, then

( / )P P r a Q P and hence equation (5) is also satisfied. Therefore, the Green’s function for thepresent problem is given by

2 21 ( / )( , )

| | | ( / ) |a rG

a r

r r

r r r r ... (9)

or 1 ( / ) 1 ( / )( , )| | | |

a r a rGR RPP QP

r r ... (10)

where | |PP PP R

, say and | | ,QP QP R

sayBy cosine law in solid geometry, we have

2 2 2 2 ~( ) 2 cosPP R r r rr ... (11)

2 2 2 2 2 2 ~ ,( ) ( / ) 2 ( / ) cosQP R a r r a r r ... (12)

where cos cos cos sin sin cos ( ) ... (13)From equation (10), we have

3

2 2 3 31 ( / ) 1G G R a r R R a R RR R

n r r r r r rR R R R

... (14)

But, we also have / / / .PP QP R R r a Hence (14) yields

3

31G R a r RR R

n r r a rR

... (15)

Now, from equations (11) and (12), we have

Q

Oa

P(r , , )

R

Q ( /r, , )a2 R

P(r, , )



~2 2 2 cosRR r rr

and 2 ~2 2 2 ( / ) cosRR r a rr

Substituting the above values in (14), we obtain2 2

3 2~ ~

1 ( cos ) cosr a

G r aa r an rR a

2 2 2

3 31 r r aa

aR aR

Thus, 2 2

2 2 3/ 2~,

( 2 cos )r a

G r an a r a ar

using (11) ... (16)

The solution of the Dirichet problem described by (1) and (2) is given by (refer eqaution (21)of Art. 12.9)

1 ( , )( ) ( )

4 S

Gu u dSn

r rr r ... (17)

where dS is the surface element of sphere of radius a and centre at origin O.Substituting the value of ( , ) /G n r r as given by (16), noting that ( ) ( , ),u r f and

2 sindS a d d = surface element of sphere of radius a, (17) takes the form

2 2 22

2 2 3/ 20 0 ~

1 ( , ) ( 1) ( ) sin( , , )4 ( 2 cos )

f a r a d du ra r a a r

or2 2 2

2 2 3/ 20 0 ~

( ) ( , ) sin( , , ) ,4 ( 2 cos )

a a r f du r dr a a r

... (18)

where ~ is defined by equation (13)Hence the solution of the interior Dirichlet problem described by (1) and (2) is given by (18).

The integral on the right side of (18) is known as the Poisson integral. The result (18) is known asthe Poisson integral formula. (Compare this with equation (20) of Art. 12.4)

Exercise : Show that the solution of the corresponding exterior Dirichlet problem is given by

2 2 2

2 2 3/20 0 ~

( ) ( , )sin( , , )4 ( 2 cos )

a r a f du r dr a ra

Solution. Left as an exercise.12.11 GREEN’S FUNCTION FOR THE TWO-DIMENSIONAL LAPLACE EQUATION.

The theory of the Green’s function for the two-dimensional Laplace equation may be developedalong the lines similar to those of Art. 12.9. We shall use Green’s identity given below.

Green’s identity of calculus : If P (x, y) and Q (x, y) are functions defined inside and on theboundary C of the closed area D, then by Green’s theorem, we have

( )D C

Q P dS Pdx Qdydx y

... (1)

Putting ( / )P u u y and ( / )Q u u x in (1), and making use of the fact that

( / ) ( / ) /u x dy u y dx u n and 2 2 2 2 2/ /x y for two dimensions,

we get2 ,

D D C

u u u u uu u dS dS u dsx x y y n

... (2)

where /u n denotes the derivative of u in the direction of the outward normal to the boundary



curve C and ds denotes the elementary arc length of the curve C.On interchanging u and u in (2), we have

2

D D C

u u u u uu u dS dS u dsx x y y n ... (3)

Subtracting (3) from (2), we find

2 2( ) ,

D C

u uu u u u dS u u dsn n ... (4)

which is known as Green’s identity.Consider a point P (x, y) inside the region D bounded by the smooth curve C. Assume that

u (x, y) is a harmonic function in D. Draw a circle Cwith centre P and small radius .

Let1log ,

| |u

r r ... (6)

where and ,x y x y r i j r i j ... (7)

Here ( , )x y are coordinates of any point Q in D. From

(6), we note that u is a harmonic function in any region thatdoes not contain the point P (x, y). This function is known asthe fundamental solution of Laplace’s equation in twodimensions.

Let R be the region which is exterior to C and interior to C. Further let the boundary of R bedenoted by . Then, we have

2 0u and 2 0u in R. ... (8)If u and u are twice continuously differentiable functions in R and have first order derivatives

on , then by Green’s identity (4), we have

2 2

R

u uu u ds u u u u dSn n

... (9)

where s is measured in the directions shown in the figure and ds is an elementary arc of .

From (9), 0,u uu u dsn n

using (8)

or 0C C

u u u uu u ds u u dsn n n n

Substituting the value of u given by (6) in the above relation, we have

1 1( , ) log log| | | |C

uu x y dsn n

r r r r

1 1( , ) log log 0,

| | | |C

uu x y dsn n

r r r r ... (10)

where the normals n are in the directions shown in the figure.Proceeding as in the three dimensional case (refer Art. 12.9), we can prove that

1log 2 ( , ) ( )

| |Cu ds u x y O

n

r r ... (11)

C C

n

nP(x, y) Q(x , y )

R



and 1log 2 log ,

| |C

u ds Mn

r r ... (12)

where M is an upper bound of / .u r Substituting the results given by (11) and (12) in (10) and letting tend to zero, we obtain

1 1 ( , ) 1( , ) log ( , ) log

2 | | | |C

u x yu x y u x y dsn n r r r r ... (13)

The equation (13) would seem at first right to indicate that to obtain a solution of Dirichletproblem we need to know not only the value of the function u but also the value of / .u n Thatthis is not in fact so can be demonstrated by introduction of the concept of a Green’s function intwo dimensions defined as follows :

We define a Green’s function ( , ; , )G x y x y by the equation

1( , ; , ) ( , ; , ) log ,| |

G x y x y W x y x y r r ... (14)

where the function ( , ; , )W x y x y satisfies the relations2 2 2 2( / / ) ( , ; , ) 0x y W x y x y ... (15)

and ( , ; , ) 0G x y x y on C. ... (16)

Applying Green’s identity (4) for ( , )u x y and ( , ; , )W x y x y and noting that u and W are

harmonic functions (i.e., 2 20 and 0u W ), we have

C

W uo u W dsn n

or1 ( , )( , ) 0

2 C

W u x yu x y W dsn n

... (17)

Subtracting (17) from (13), we have

1 ( , ) 1 1( , ) log ( , ) log2 | | | |C

u x yu x y W u x y W dsn n

r r r r

or 1 ( , , , )( , ) ( , )2 C

G x y x yu x y u x y dsn

... (18)

1on , log 0, by (14) and (16)| |

C G W r r

Hence the solution of the Dirichlet problem2 0u within D, u = f (x, y) on C

is given by (18) in the form

1 ( , ; , )( , ) ( , ) ,

2 C

G x y x yu x y f x y dsn

... (19)

when n is the outward drawn normal to the boundary curve C.12.12 CONSTRUCTION OF THE GREEN’S FUNCTION WITH HELP OF THE

METHOD OF IMAGES (Refer Art. 12.4A)The following examples will illustrate the method of images given in Art. 12.4A. We shall

deal with two special cases of result (18) of Art. 12.11.



Example 1. (a) Use Green’s function technique tosolve Dirichlet’s problem for a half plane.

Solution. The Dirichlet’s problem in the given semi-infinite half plane is governed by the following boundaryvalue problem :

2 0u for 0,x y ... (1) u = f (y) on x = 0 ... (2)

and 0u as x ... (3)By definition, the Green’s function for the given problem must satisfy the following conditions :

1( , ; , ) ( , ; , ) log ,

| |G x y x y W x y x y

r r ... (4)

where 2 2 2 2( / / ) ( , ; , ) 0x y W x y x y ... (5)and ( , , , ) 0 on CG x y x y ... (6)

Suppose that ( , )Q x y with position vector , is the image in the plane x = 0 (i.e., y-axis) ofthe point P (x, y) with position vector r. Let ( , )P x y be any point in given region with positionvector .r Then, from (4), we have

log(1/ ),G W PP as | | PP r r ... (7)

If we take1 1log log ,

| |W

P Q

r ... (8)

[ | | | | | | ]OQ OP P Q P Q

r then it is obvious that equation (5) is satisfied.

From the figure, we find that PP QP whenever P lies on x = 0. Then, from (7) and (8), weobserve that

log (1/ ) log (1/ ) log(1/ ) log (1/ ) 0,G QP PP PP PP showing that G = 0 on C and hence (6) is also satisfied.

Thus, with help of the method of images, the required Green’s function for the presentproblem is given by

( , ; , ) log (1/ ) log (1/ ) log ( / )G x y x y QP PP QP PP

or2 2

2 21 ( ) ( )( , ; , ) log2 ( ) ( )

x x y yG x y x yx x y y

... (9)

Here, the outward drawn normal to the boundary is in the direction of the x-axis. Hence,

2 2 2 20 0

1 2 22 ( ) ( ) ( ) ( )x x

G G x xn x x x y y x x y y

Thus, 2 2/ (2 ) ( )G n x x y y ... (10)

Substituting the value of /G n as given by (10) and noting that ( , ) ( )u x y f y so that( , ) ( ),u x y f y relation (18) of Art. 12.11 reduces to

2 2 2 21 2 ( )( , ) ( ) ,

2 ( ) ( )x x f y dyu x y f y dy

x y y x y y

giving the solution of the Dirichlet problem given by (1) – (3) for the given half plane. If the natureof the function ( )f y is explicitly known, then the above equation can be integrated to find thefinal solution.



Example 1. (b) Solve 2 u = 0 in the upper half plane defined by 0, ,y x usingGreen’s function method, subject to the condition u = f(x) on y = 0.

Hint. Proceeds as in example 1(a).

Ans. Green’s function 2 2

2 21 ( ) ( )( ) log2 ( ) ( )

ex x y yGx x y y

,r r and solution is given by

2 2( )

( , )( )

f xyu x y dxx x y

Example 2. Obtain the solution of the interior Dirichlet problem for a circle using the

Green’s method.Solution. The interior Dirichlet problem for a circle of radius a this is governed by the

following boundary value problem.2 0,u r a ... (1)

( )u f on r = a ... (2)By definition, the Green’s function for the given

problem must satisfy the following conditions :1( , ; , ) ( , ; , ) log

| |G r r W r r

r r ... (3)

2 0W ... (4)and ( , ; , ) 0,G r r on the circular boundary r = a ... (5)

Let ( , )P r be a point inside the given circle, where we place a unit charge with positionvector r and let its inverse point with respect to the circle be Q with position vector . Then

OP

r and .OQ

Let r be the position vector of an arbitrary point P (r, ) inside the circle

such that .OP

r Then, let

| | ,r OP r | |r OP r and | | .OQ Since Q is the inverse point of P, we have

OP × OQ = a2 so that OQ = a2/OP = a2/r ... (6)

Hence polar coordinates of P and Q may be taken as ( , )r and 2( / , )a r respectively..

If we take log ,r P QWa

... (7)

then it can be verified that 2 0W and so (4) is satisfied.

Since | | ,PP r r from (3) and (7), it follows that the required Green’s function may betaken as

1( , ; , ) log log logr P Q r P QG r ra PP a P P

... (8)

On the circle r = a, we have

log log log log1 0,( / )

P Q P Q aGP P r a P Q a

O

r

a

P(r, )

(r , ) Q( /r, )a2

X

P

2



showing that the condition (5) is also satisfied.Using cosine formula of Trignometry, we have

2 2 2( ) 2 cos ( )PP r r r r

and 2 2 2 2 2( ) ( / ) 2 ( / ) cos ( )P Q r a r r a r Inserting these results into equation (8), we have

2 2 2 2 1/ 2

2 2 1/ 2{ ( / ) 2 ( / ) cos ( )}( , ; , ) log

{ 2 cos ( )}r r a r r a rG r ra r r rr

or 2 2 2 2

2 21 / 2 cos ( )( , ; , ) log2 2 cos ( )

a r r a rrG r rr r rr

... (9)

Now on C, 2 2

2 2( )

{ 2 cos ( ) }r a

G G a rn r a a ar r

.. (10)

Substituting the value of /G n as given by (10) and noting that ( , ) ( )u r f so that( , ) ( ),u r f result (18) of Art 12.11 reduces to

2 2 2

2 20

( )( , ) ,2 2 cos ( )

a r f du ra ar r

... (11)

giving the solution of the Dirichlet problem given by (1) and (2) for a circle. If the nature of thefunction ( )f is explicitly known, then the above equation (11) can be integrated to find the finalsolution.


13.1

CHAPTER 13

Applications of integral equationsto mixed boundary value problems13.1 INTRODUCTION.

In the present chapter we propose to study some of the mixed bounders value problems whichoccur in physical sciences rather frequently. We use the term ‘‘mixed’’ boundary value problem todistinguish this type of problem from the ‘‘uniform’’ problems of Dirichlet and Neumann. Recall thata probelm in potential theory is known as a Dirichlet problem if the potential function whose forminside a region R is to be determined is prescribed at each point of the entire surface S boundingthe region R, and a Neumann probelm if its normal derivative is prescribes at each point of theentire surface S bounding the region R. In potential theory a typical problem of mixed kindwould be one in which the potential function is prescribed over a part of the boundary, and itsnormal derivative is prescribed over the remaining part.

Various mathematical techniques have been employed to solve mixed boundary value problems.In the present chapter, however, we shall discuss an integral-equation method applicable to most ofthe mixed boundary value problems.13.2 TWO-PART BOUNDARY VALUE PROBLEMS

Consider a Fredholm integral equation of the first kind of the form.

00( , ) ( ) ( ), 0

aK t g t dt f a ... (1)

where the function ( )f and the kernel 0 ( , )K t are known and g (t) is to be determined.The solution of various mixed boundary value problems in potential theory, elastostatics,

steady heat conduction, the flow of perfect fluid, and various other problems of equiblirium statesare known to depend upon the solution of (1). The boundaries occuring in such problems are thoseof solids such as circular discs, elliptic discs, spherical caps, and spheroidal caps.WORKING RULE FOR SOLVING (1)

We know that solution of (1), in general, cannot be easily obtained because (1) is a Fredholmintegral equation of the first kind. However, it is possible to reduce the solution of (1) to that of apair of Volterra integral equations of the first kind with simple kernels. This reduction is obtainedfor every kernel 0 ( , )K t which for all g (t) satisfies the relation

20 1 2 2 2 3

0 0( , ) ( ) ( ) ( , ) { ( )} ( , ) ( ) ( ) , 0

a aK t g t dt h K h K t g t h t dt d a

... (2)

where h1, h2, h3, and K2 are known functions. We also suppose that the kernel K2 is such that theVolterra integral equations

20( , ) ( ) ( ), 0K t g t dt f a

... (3)


13.2 Applications of integral equations to mixed boundary value problems

and 2 ( , ) ( ) ( ), 0a

K t g t dt f a

... (4)

possess explicit unique solutions for g in terms of f, for all arbitrary differential functions f.Now, (1) can be easily solved with help of two functions ( )S and ( )C which are defined

as follows :

2 2 3( ) ( ) ( , ) ( ) ( ) , 0a

S h K t g t h t dt a

... (5)

and 1 2 20

( ) ( ) ( , ) ( ) ( ) , 0f h K C h d a

... (6)

Using (2) and (6), (1) may be re-written as

1 2 2 2 2 30

( ) ( , ) ( ) ( ) ( , ) ( ) ( ) .a

h K h h K t g t h t dt d

1 2 20, 0( ) ( , ) ( ) ( ) ah K C h d

... (7)

Replacing by in (5), we have

2 2 3( ) ( ) ( , ) ( ) ( )a

S h K t g t h t dt

... (8)


1 2 2 1 2 20 0( ) ( , ) ( ) ( ) ( ) ( , ) ( ) ( ) , 0h K h S d h K C h d a

Thus, ( ) ( ), 0S C a ... (9)

Hence, (6) reduces to 1 2 20

( ) ( ) ( , ) ( ) ( ) , 0f h K S h d a

... (10)

Noting the assumptions already made about the solution of the Volterra integral equations (3)and (4), we first solve (10) to find the function ( ).S Substitute this value of ( )S in (5). Nextstep will be to invert (5) and obtain the required unknown function g (t).

The following solved examples will illustrate the above entire working rule.Example 1. Solve the integral equation

1 10 0( ) ( ) ( ) , 0 ,

at t J u J ut du dt a

where ( )t is the unknown function and J1 (x) is the Bessel function. (See Art. 10.12)Solution. Comparing the given equation with (1), we have

0 1 10( ) ( ), ( ) , ( , ) ( ) ( )g t t t f K t J u J u t du

... (11)

Now, the kernel 0 ( , )K t satisfies (2) because, for all g (t), we have

00( , ) ( )

aK t g t dt 1 10 0

( ) ( ) ( )a

g t J u J u t du dt

3/2

1/ 2 1/ 22 2 1/ 2 2 2 1/ 20 0 0 0

( ) ( ) ( )2( )( ) ( )

a t J u J u v v dv d du dtug tt t v

2 2 1/ 2 2 2 1/ 20 0 0

2 ( ) ( ) ( )( ) ( )

a tg t v v dv d dtt t v


Applications of integral equations to mixed boundary value problems 13.3

2min ( , )

2 2 2 2 1/ 20 0

2 ( )( ) ( )

a tg t d dtt t

2

2 2 1/ 2 2 2 1/ 20

2 ( ) , 0 ,( ) ( )

a g t dt d at t

... (12)

While arriving at the final form (12), we have employed the well known first Sonine integral(1/ 2)1/ 2

(1/ 2)2 2 1/ 20

( )2 1( )( )

nn

n n

J uuJ u dw ... (13)

and the well known relation 1/ 20

( )( ) ( ) ,( )

vu J u J u v duv

... (14)

where is the well known Dirac delta function. Weehave also used the shifting property of Dirac deltafunction (refer Art. 10.7, chapter 10) and changedthe order of integration as explained in the adjoiningfigure.Comparing (2) and (12), the values of the functionsh1, h2, h3 and K2 are given by

1 2 32 2 1/ 2

2

( ) 2 / , ( ) , ( ) 1/

and ( , ) ( )

h h h

K t t

... (15)

Morever the simple form of the kernel K2 willsurely help in the process of the inversion of integralequations (3) and (4) (refer examples 3 and 4 in Art.8.5, chapter 8). It follows that the present method isapplicable to the given problem.

Now, if we take 2 2 1/ 2( )( )

( )

a t dtSt

... (16)

and 2 2 1/ 20

2 ( ) ,( )

S d

... (17)

then we can easily verify that the given integral equation (1) is identically satisfied.

Inversion of (17) yields 3

2 2 1/ 20( ) 2

( )d t dtSd t

... (18)

Hence (16) reduces to 2 2 1/ 2( )2

( )

a t dtt

... (19)

Inverting (19), we have

2 2 1/ 2 2 2 1/ 24 4( )

( ) ( )

ad u dud u a

... (20)

Remark. The reader should go through Art. 8.5 of Chapter 8 and then use the method toarrive at results (18) and (20).

Example 2. Solve the integral equation

00

( ) ( ) ( , ) , 0a

f g t K t dt a ... (i)



where 0 0( , ) 2 ( ) ( )n nK t J u J u t du

.. (ii)

Solution. Proceeding as in working rule of Art. 13.2, we have, for all g(t)

00

( , ) ( )a

K t g t dt 0 02 ( ) ( ) ( )

an ng t J u J ut du dt

1/ 21/ 2 1/ 2

2 2 1/ 2 2 2 1/ 20 0 0 0

( ) ( ) ( )22 ( )( ) ( ) ( )

na t n nn

J u J uv v dv d du dtug tt t v

2 2 1/ 2 2 2 1/ 20 0 0

4 ( ) ( ) ( )( ) ( )

na t

n ng t v v dv d dtt t v

2min ( , )

2 2 1/ 2 2 2 1/ 20 0

4 ( )( ) ( )

na t

n ng t d dtt t

2

2 2 1/ 2 2 2 1/ 20

4 ( ) , 0 ,( ) ( )

n a

n ng t dt d a

t t

... (iii)

where we have used the following standard results(1/ 2)1/ 2

(1/ 2)2 2 1/ 20

( )2 1( )( )

nn

n n

J uuJ u dw [Sonine Integral] ... (iv)

and 1/ 20

( )( ) ( ) ,( )

vu J u J uv duv

... (v)

where ( )v is the well known Dirac delta function.Since (i) is difficult to solve, we reduce its solution to that of a pair of Volterra integral

equations of the first kind with simple kernels. The desired reduction is obtained for every kernel

0 ( , )K t which for all g (t) satisfies the relation

20 1 2 2 2 3

0 0( , ) ( ) ( ) ( , )[ ( )] ( , ) ( ) ( ) , 0

a aK t g t dt h K h K t g t h t dt d a

... (vi)

where h1, h2, h3 and K2 are known functions. It is further assumed that the kernel K2 is such that theVolterra integral equations

20( , ) ( ) ( ), 0

aK t g t dt f a ... (vii)

and 2 ( , ) ( ) ( ), 0a

K t g t dt f a

... (viii)

possess explicit unique solutions for g in terms of f, for all arbitrary differential functions f.Comparing (iii) and (vi), the values of the functions h1, h2, h3 and K2 are given by

2 2 1/ 21 2 3 2( ) 4 / , ( ) , ( ) , ( , ) ( )n n nh h h K t t ... (ix)

Again the form of K2 given by (ix) ensures the inversion of (vii) and (viii) and so the workingrule of Art. 10.2 can be employed to solve the given problem. Accordingly, in order to solve (i), wedefine two functions ( )S and ( )C as follows :



2 2 3 2 2 1/ 2( )( ) ( ) ( , ) ( ) ( ) , 0

( )

a ann

g t dtS h K t g t h t dt at t

... (x)

1 2 2 2 2 1/ 20 0

4 ( )( ) ( ) ( , ) ( ) ( ) , 0( )

n

nS df h K S h d a

... (xi)

Inverting (xi), we have (refer Chapter 8 for details)1

2 2 1/ 20

1 ( )( ) ,2 ( )

n

nd t f t dtSd t

... (xii)

Next, inverting (x), we find the value of the unknown function g(t) in terms of ( ) :S

1

2 2 1/ 22 ( )( )

( )

n na

t

t d S dg tdt t

... (xiii)

We now substitute the value of ( )S given by (xii) in (xiii) and get the desired value of g(t).

Particular case : Let the disc be kept at a unit potential so that ( ) 1f and n = 0. Then,(xii) and (xiii) reduce to

2 2 1/ 20

1 1( )2 2( )

d t dtSd t

... (xiv)

and 2 2 2 1/ 2 2 2 2 1/ 21 1( ) , 0

( ) ( )

a

t

d dg t t adt t a t

... (xv)

Remark. We now show that the capacity C of the disc can be obtained without finding thevalue of the unknown function g. Indeed, we know that

02 ( )

aC g t dt ... (xvi)

For the particular case, putting n = 0 in (xiii) we have

2 2 1/ 22 ( )( )

( )

a

t

d S dg tdt t

... (xvii)

Substituting the value of g (t) from (xvii) in (xvi), we have

0

4 ( ) 2 / ,a

C S d a using (xiv)

Example 3. Obtain electrostatic potential due to a spherical cap.Solution. In terms of spherical polar coordinates

( , , ),r let the given spherical cap ABC be defined by r =a, 0 , 0 2 as shown in the adjoining figure.Consider the axially symmetric case so that the potentialon the cap can be expressed by a function of only, say

( ).f Thus, we are to solve the boundary value probelmgiven by

2 ( , , ) 0V r in D ... (i)

and ( , , ) ( ), 0 ; 0 2 ,V a f ... (ii)



where D is the rigion exterior to the cap. Then, it can be shown that the integral representationformula for (i) is given by

221

0 0

( )( , , ) sin ,tV r a t d dtR

... (iii)

where ( )t is the charge density at the pont 1( , , )Q a t on the cap and

2 2 1/ 21( 2 cos ) , cos cos cos sin sin cos( )R r a ar t t

Using the boundary condition (ii), (iii) reduces to the Fredholm integral equation of the first kind

200

( ) sin ( ) ( , ) , 0 ,f a t t K t dt

... (iv)

where2

10 2 2 1/ 20( , )

(2 2 cos )dK t

a a y

... (v)

Now, expanding the integrand in (v) in terms of the spherical harmonics ( , )mnY (refer Art.

10.17, chapter 10), we obtain

12 2 1/ 2 0

( | |)! ( , ) ( , )1 1 ,( | |)!(2 2 cos )

m mn nn n

n m n

n m Y Y ta n ma a

... (vi)

where the bar denotes the complex conjugate.From (v) and (vi), we obtain

0 0

2( , ) (cos ) (cos ),n nnK t P P t

a

... (vii)

where Pn is the Legendre polynomial.Using the well known Mehler-Dirichlet integral

1/ 20

2 cos {( 1/ 2) }(cos )(cos cos )n

n dP

and a standard result

0cos{( 1/ 2) } cos{( 1/ 2) } ( / 2) ( ), 0 , 0

nn n v v v

the 0 ( , )K t given by (vii) can be re-written in the integral form given by

0 1/ 2 1/ 20 0

2 ( )( , )(cos cos ) (cos cos )

t v dv dK ta v t

... (viii)

According to working rule of Art. 13.2, the solution of

00

( , ) ( ) ( ), 0a

K t g t dt f ... (ix)

is obtained by reducing its solution to that of a pair of Volterra integral equations of the first kindwith simple kernels. The proposed reduction is obtained for every kernel 0 ( , )K t that for all g (t)satisfies the relation

20 1 2 2 2 3

0 0( , ) ( ) ( ) ( , ) [ ( )] ( , ) ( ) ( ) , 0K t g t dt h K h K t g t h t dt d

... (x)

where h1, h2, h3, and K2 are known functions. It is further assumed that the kernel K2 is such thatthe Volterra integral equations

20( , ) ( ) ( ), 0K t g t dt f

... (xi)



and 2 ( , ) ( ) ( ), 0K t g t dt f

... (xii)

possess explicit unique solutions for g in terms of f, for all arbitrary functions f. Then, (1) is solvedby introducing a function ( )S given by

2 2 3( ) ( ) ( , ) ( ) ( ) , 0S h K t g t h t dt

... (xiii)

and 1 2 20

( ) ( ) ( , ) ( ) ( ) , 0f h K S h d

... (xiv)

Comparing (iv) and (ix), here 2( ) sin ( )g t a t t ... (xv)

For the given problem, the relation that corresponds to (x) is given by

min( , )

0 1/ 2 1/ 20 0 0

2( , ) ( ) ( ) , 0(cos cos ) (cos cos )

t d dtK t g t dt g t Qa t

or 0 1/ 2 1/ 20 0

2 1 ( )( , ) ( ) , 0(cos cos ) (cos cos )

g t dt dK t g t dta t

... (xvi)

Comparing (x) and (xvi), we have

1 2 3( ) 2 / , ( ) ( ) 1,h a h h 1/ 22 ( , ) (cos cos )K t t

Substituting the above values in (xiii) & (xiv) and using (xv), we obtain2

1/ 20

( ) sin( )(cos cos )a t t dtS

t

... (xvii)

and 1/ 20

2 ( )( )(cos cos )

S dfa

... (xviii)

The Volterra integral equations (xvii) and (xviii) can easily be solved as done in solved examples1 and 2 of Art. 8.5, chapter 8. Proceeding as indicated above, the solution of (xviii) yields

1/ 20

( ) sin( )2 (cos cos )a d f dS

d

... (xix)

Likewise, the solution of (xvii) yields

2 1/ 21 ( )sin( )sin (cos cos )t

d S dtdta t t

... (xx)

Substituting the value of ( )S given by (xix) in (xx), we get the required value of theunknown function ( )t of the integral equation (iv).

Remark. The capacity C of the solid can be obtained without first finding the charge density( ).t We know that

2

02 sin ( )C a t t dt

or 2

2 1/ 20

1 ( ) sin2 ,(cos cos )t

d S d dtC adta t

by (xx)

Thus, 1/ 20 0

( ) sin2 2 2 ( ) cos( / 2)(1 cos )S dC S d

... (xxi)

Particular case When the cap is kept at a unit potential, so that, ( ) 1,f then (xix) reduces

to 1/ 20

sin( ) cos2 2(cos cos ) 2a d d aS

d



Substituting the above value of ( )S in (xxi), we obtain

2

0 02 2 cos (1 cos ) ( sin ).

22a a aC d d

EXERCISE

1. Extend the analysis of solved example 2 in Art. 13.2 to the following two exercises (i) The disc is bounded by a grounded cylindrical vessel of radius b such such that / 1.a b The disc and the cylinder have the common axis.

(ii) The disc is placed symmetrically between two grounded parallel plates z b such that

/ 1.a b 2. Instead of the whole disc, consider the case of an annular disc and extend the analyis ofsolved example 2 in Art. 13.2 accordingly. Do the same to the problems in exercises. 1(i) and 1(ii). 3. Solve exercises 1(i) and 1(ii) when the solid is a spherical cap instead of the circular disc.

13.3. THREE-PART BOUNDARYVALUE PROBLEMSA three-part boundary value problem has an integral representation formula of the form

0 ( , ) ( ) ( ),a

bK t g t dt f b a ... (1)

where b and a are two given numbers such as the inner and outer radii of an annular disc or thebounding angles of an annular spherical cap. The function g is unknown while the functions f andthe kernel K0 are known.

Suppose that 1 2( ) ( ) ( )rrr

f a f f

... (2)

where 1 0( ) , 0r

rrf a a

... (3)

and 1

2 ( ) ,rrr

f a b

... (4)

We also define two functions 1( )g and 2 ( )g such that

1 2

0, 0( ) ( ) ( ),

0,

bg g g b a

a

... (5)

In view of the relations (2), (3), (4) and (5), we see easily see that (1) splits into the followingintegral equations :

0 1 10( , ) ( ) ( ), 0K t g t dt f a

... (6)

and 0 2 20( , ) ( ) ( ),K t g t dt f b

... (7)

We now proceed as in Art. 13.2. Suppose that the kernel 0 ( , )K t in such that for all g (t) it

satisfies the following relation for 0 :



211 2 12 2 13

00

0 221 2 22 2 23

0

( ) ( , ) [ ( )] ( , ) ( ) ( )( , ) ( )

( ) ( , ) [ ( )] ( , ) ( ) ( )

h K h K t g t h t dt dK t g t dt

h u h K t g t h t dt d

... (8)

where ijh (i = 1, 2; j = 1, 2, 3) and the kernel K2 are known functions.

We also assume that the kernel K2 is such that the Volterra integral equations

20( , ) ( ) ( ), 0K t g t dt f

... (9)

and 2 ( , ) ( ) ( ), 0K t g t dt f

... (10)

possess unique solutions for g in terms of all arbitrary differentiable functions f.Combining (6) with the first part of the relation (8), we have

211 2 12 2 1 13 10

( ) ( , ) ( ) ( , ) ( ) ( ) ( ), 0h K h K t g t h t dt d f a

... (11)

Again, combining (7) with the second part of the relation (8), we have

221 2 22 2 2 23 2

0( ) ( , ){ ( )} ( , ) ( ) ( ) ( ),h K h K t g t h t dt d f b

... (12)

In order to solve (1), we define six functions S1, S2, T1, T2, C1 and C2 as follows :

112 2 1 13

1

( ), 0( ) ( , ) ( ) ( )

( ),S a

h K t g t h t dtT a

... (13)

222 2 2 23

0 2

( ), 0( ) ( , ) ( ) ( )

( ),T b

h K t g t h t dtS b

... (14)

11 2 1 12 10( ) ( , ) ( ) ( ) ( ), 0 ,h K C h d f a

... (15)

and 21 2 2 22 2( ) ( , ) ( ) ( ) ( ), .h K C h d f b

... (16)

The above mentioned integral equation are similar to (9) and (10), whose solutions are assumedto be known.

With help of (11), (13) and (15), we obtain

11 2 12 1 11 2 1 120 0( ) ( , ) ( ) ( ) ( ) ( , ) ( ) ( ) , 0h K h S d h K C h d a

... (17)

or 1 1( ) ( ), 0S C a ... (18)Again, with help of (12), (4) and (16), we obtain

2 2( ) ( ),S C b ... (19)Using (18) and (19), (15) and (16) may be re-written as

11 2 1 12 10( ) ( , ) ( ) ( ) ( ), 0h K S h d f a

... (20)

and 21 2 2 22 2( ) ( , ) ( ) ( ) ( ),h K S h d f b

.. (21)



The functions S1 and S2 can be obtained in terms of the known functions f1 and f2 fromequations (20) and (21). Hence we are left with determination of only two functions T1 and T2. Tothis end, we use relation (5) and (13) and obtain

12 2 2 13 1( ) ( , ) ( ) ( ) ( ),h K t g t h t dt T a

... (22)

Similarly, from (5) and (14), we obtain

22 2 1 23 20

( ) ( , ) ( ) ( ) ( ), 0h K t g t h t dt T b

... (23)

Next step is to invert (14) for getting the value of g2 (t) in terms of T2 and S2 and substitutethis value of g2 (t) in (22). Thus, we shall arrive at an integral equation containing two unknownfunctions T1 and T2. Similarly, from the relations (13) and (23), we arrive at another integral equationcontaining two unknowns T1 and T2. Both these equations are Fredholm integral equations andhence they can be easily solved by well known iterative method. (refer chapter 5)

Now, we illustrate the above analysis with help of the following solved example.Example. Solve the integral equation

1 10

( ) ( ) ( ) ,a

bt t J u J u t du dt b a

... (24)

[This equation embodies the solution of the torsion of an isotropic and homogeneous elastichalf-space due to a uniformly rotating annular disc with inner radius b and outer radius a]

Solution. Comparing (24) with (1) and (5), we have

1 1 2 2

1 2

( ) ( ), ( ) ( ); ( ) ( )( ) , ( ) , 0 ; ( ) 0,

g t t t g t t t g t t tf f a f b

... (25)

and 0 1 10( , ) ( ) ( ) ,K t J u J u t du

... (26)

where 1 2

0, 0( ) ( ) ( ),

0,

bb a

a

... (27)

Also, the kernel 0 ( , )K t must satisfy the condition (8). Therefore, for all g (t), we have

0 1 10 0 0( , ) ( ) ( ) ( ) ( ) ,K t g t dt g t J u J u t du dt

by (26)

1/ 21/ 2 1/ 2

2 2 1/ 2 2 2 1/ 20 0 0 0

( ) ( ) ( )2( )( ) ( )

t J u J uv v dv d du dtug tt t v

2 2 1/ 2 2 2 1/ 20 0 0

2 ( ) ( ) ( )( ) ( )

tg t v v dv d dtt t v

2min( , )

2 2 1/ 2 2 2 1/ 20 0

2 ( )( ) ( )

tg t d dtt t

2

2 2 1/ 2 2 2 1/ 20

2 ( ) , 0( ) ( )

g t dt dt t

... (28)



and 00( , ) ( )K t g t dt

1 10 0

( ) ( ) ( )g t J u J u t du dt

3/ 2 3/ 2

1/ 2 2 2 1/ 2 2 2 1/ 20 0

( ) ( )2( )( ) ( ) ( )t

J u J uv dv d du dtu tg tv v t

2 2 / 2 2 2 1/ 20

2 ( )( )( ) ( )t

v dv d dtt g tv v t

2 2 2 1/ 2 2 2 1/ 20 max( , )

2 ( )( ) ( )t

d dtt g tt

2 2 2 1/ 2 2 2 1/ 20

2 1 ( ) , 0( ) ( )

t g t dt dt

... (29)

where we have used the following formulas(1/ 2)1/ 2

(1/ 2)2 2 1/ 20

( )2 1( ) ,( )

nn

n n

J uuJ u [First sonine Integral]

1/ 2

0( ) ( ) ( ) /( )u J u J uv du v v

and{ (1/ 2)}1/ 2

(1/ 2)2 2 1/ 2

( )2( )( )

nnn

nJ uuJ u d

In addition to the application of these three formulas, we have changed the order of integration

in the steps leading to formulas (28) and (29) as explained in following figures (i) and (ii)

Comparing (28) and (29) with (8), we have

11 12 13 212 2 1/ 2

22 23 2

( ) 2 / , ( ) , ( ) 1/ , ( ) 2 / ,

( ) 1/ , ( ) , and ( , ) ( )

h h h h

h h K t t

... (30)

Since the kernel K2 is such that the Volterra integral equations (9) and (10) can be easilysolved, it follows that the method prescribed in article 13.3 is applicable for solving the givenproblem.



The system of integral equations that corresponds to the system (13), (14), (20), (21), (22) and(23) for the present problem are given by

112 2 1/ 2

1

( ), 0( )( ),( )

S at dtT at

... (31)

222

2 2 1/ 20 2

( ), 0( )1( ),( )

T bt t dtS bt

... (32)

12 2 1/ 20

( )2 , 0( )

S d a

... (33)

22 2 1/ 2( )2 0,

( )S d b

... (34)

212 2 1/ 2

( ) ( ),( )

t dt T at

.. (35)

21

22 2 1/ 20

( )1 ( ), 0( )t t dt T b

t

... (36)

Using methods employed in examples 3 and 4 of Art. 8.5 (chapter 8) and example 16 of Art. 9.5(chapter 9), the above integral equations (31), (32), (33) and (34) can be inverted to yield

1 11 2 2 1/ 2 2 2 1/ 2

( ) ( )2( )( ) ( )

a

a

S u du T u dudd u u

... (37)

2 22 2

2 2 2 2 1/ 2 2 2 1/ 20

( ) ( )2( )( ) ( )

b

b

u T u du u S u dudd u u

... (38)

3

1 2 2 1/ 20( ) 2 ,

( )d t dtSd t

... (39)

2 ( ) 0S ... (40)Substitute, the values of S1(u) and S2(u) as given by (39) and (40) in (37) and (38) respectively.

Then substituting the resulting values of 2 ( )t and 1 ( )t so obtained in (35) and (36) respectively,,we finally obtain

21 2 2 2 1/ 2 2 2 3/ 20

2( ) ( )( ) ( )

b dt duT u T ut t t u

2 2 2

2 2 11/ 2 2 20

( ) (1/ 2, 1; 5 / 2; / )1 ,( ) (5 / 2) ( )

b u T u F u du au

... (41)

and3 3

12 2 2 1/ 2 2 2 1/ 2 2 2 1/ 2 2 2 3/ 20 0

( )4 2( )( ) ( ) ( ) ( )a

T u du dtt dt tTt a t t u t

2 3 2 2

2 12 2 2 2 2 28 3 5 41, ; ;

2 23 ( ) ( )a aFa a

2 221 2 1

1/ 2 2 2( ) (1/ 2, 1; 5 / 2; / ) , 0

( ) (5 / 2) ( )a

T u F u du bu u

... (42)



In the above relations (41) and (42), 2F1 stands for the well known hypergeometric function.In the process of getting these relations, we have also used the following well known relationscontaining the hypergeometric functions

1/ 2 2

2 12 2 1/ 2 2 2 3/ 2 2 2 2 21 5, 1; ; ,2 2( ) ( ) 2 (5 / 2) ( )

dt uF ut t t u u

1/ 2 3 2 23

2 12 2 1/ 2 2 2 3/ 2 2 20

(1/ 2, 1; 5 / 2; / ),( ) ( ) 2 (5 / 2) ( )

F ut dt ut u t u u

3 1/ 2 3 2 2

2 12 2 1/ 2 2 2 1/ 2 2 2 2 2 2 20

( ) 3 5 41, ; ; ,2 2( ) ( ) 2 (5 / 2) ( ) ( )

t dt a aF at a t a a

Here (41) and (42) are two simultaneous Fredholm Fredholm integral equations of the second

kind. We now proceed to solve them approximately by iteration with help of introduction of theparameter / ,b a such that 1. In fact, it is known that the hypergeometric function 2F1involved in (41) and (42) can be reduced to an elementary function as follows.

22 2

2 1 2 31 5 3, 1; ; 2 ( ) log ,2 2 4

x y y xF xy y x x yy xy x

... (43)

Hence (41) and (42) can be transformed into simple forms given by1

1 2 2 2 20

1 2 1( ) ( ) log , 1uT a T bu duu uu

... (44)

2 2 2 2 2

2 2 12 2 2 2 2 28 3 5 4( ) 1, ; ;

2 23 (1 ) (1 )aT b F

1 2 2 21

1 2( ) log , 0 1u uT au duuu

... (45)

We first solve (45) by noting that2

22 1 1

3 51, ; ; 1 32 2 2 3

n

n

xF xn

... (46)

Using (46), the first iteration for T2 is given by4 4

2 2 62

8 2( ) ( )3 5

aT b O

... (47)

Substituting this value of T2 in (44), we can solve the resulting equation approximately toyield

5 2 24

1 2 3 532 1 2 6( ) 1 ( ) ,

745 7aT a O

... (48)

While arriving at the above approximations (47) and (48), we have included only those termswhich are required to find the torque experienced by the annulus up to 9( ).O

Now, substituting the values of S1, S2, T1 and T2 as given by (39), (40), (47) and (48) in (37)and (38), we have

1/ 25 2 3 21

1 2 2 1/ 2 2 3 24 ( / ) 16 2 3( ) 1 sin 1

7{1 ( / } 45 2a a a

aa a

-1/ 22

22 1a



3/2 1/ 22 5 2 21

5 2 23 15 sin 2 1 9 128

a aa a a

1/ 229

28 1 ( )Oa

... (49)

and1/ 2 1/ 22 2

12 2 2 2

4 3( ) 2 sin 1 2 13

b b bb

3/2 1/ 2 1/ 23 2 2 2 21 5

2 2 2 215 sin 2 1 9 1 8 1 ( )

5b b b b O

bb

... (50)

Substituting the values of 1( ) and 2 ( ) as given by (49) and (50) in the relation

1 2( ) ( ) ( ),

we get the required solution of the integral equation (24).13.4. GENERALIZED TWO-PART BOUNDARYVALUE PROBLEMS

Consider an integral equation of a more general type such as

10

( ) ( , ) ( ), 0a

g t K t dt f a ... (1)

where the kernels 1( , )K t can be perturbed on the kernel 0 ( , )K t of Art. 13.2. Then (1) can alsobe solved by employing the present method. Hence, in order to solve (1), we split 1( , )K t as

1 0( , ) ( , ) ( , ),K t K t G t ... (2)

where we assume that the kernel ( , )G t is in some sense smaller then 0 ( , ).K t From (1) and (2),we obtain

00 0( , ) ( ) ( ) ( , ) ( ) , 0

a aK t g t dt f G t g t dt a ... (3)

The integral equation (3) is a Fredholm integral equation of the first kind and is therefore,in general, difficult to solve. However, it is possible to reduce the solution of (3) to that of a pairof Volterra integral equations of the first kind with simple kernels. This reduction is obtained forevery kernel 0 ( , )K t which for all g (t) satisfies the relation

20 1 2 2 2 3

0 0( , ) ( ) ( ) ( , ) { ( )} ( , ) ( ) ( ) , 0

a aK t g t dt h p K h K t g t h t dt d a

... (4)

where h1, h2, h3 and K2 are known functions. It is further assumed that the K2 is such that theVolterra integral equations

20

( , ) ( ) ( ), 0K t g t dt f a

... (5)

and 2 ( , ) ( ) ( ), 0a

K t g t dt f a

... (6)

possess explicit unique solutions for g in terms of f, for all arbitrary differentiable functions f.From (3) and (4), we have

21 2 2 2 3

0 0( ) ( , ){ ( )} ( , ) ( ) ( ) ( ) ( , ) ( ) , 0

a ah K h K t g t h t dt d f G t g t dt a

... (7)



Now, (1) can be easily solved with help of three functions ( ), ( )S C and ( , )L v which aredefined as follows :

2 2 3( ) ( ) ( , ) ( ) ( ) , 0a

S h K t g t h t dt a

... (8)

1 2 20

( ) ( ) ( , ) ( ) ( ) , 0f h K C h d a

... (9)

1 3 2 2 2 20 0( , ) ( ) ( ) ( , ) ( , ), ( ) ( ) ( , )

tG t h h t K K v t h h v L v dv d

... (10)

We now proceed to put the right hand side of (7) in the form of the left hand sidewith help of (8), (9) and (10) as we did in Art. 13.2. Then, the integral on the right side of (7) takesthe form

1 3 2 2 2 20 0 0 0( , ) ( ) ( ) ( ) ( ) ( , ) ( , ) ( ) ( ) ( , )

a a tG t g t dt g t h h t K K v t h h v L v dv d dt

1 2 2 2 2 30 0

( ) ( , ) ( ) ( , ) ( ) ( , ) ( ) ( ) , 0a a

vh K h L v h v K v t g t h t dt dv d a

... (11)

where it is assumed that various orders of integration can be interchanged.Substituting (8), (9) and (11) in (7), we get

1 2 2 1 2 20 0( ) ( , ) ( ) ( ) ( ) ( , ) ( ) ( )h K h S d h K C h d

1 2 20 0( ) ( , ) ( ) ( , ) ( ) , 0

ah K h L v S v dv d a

... (12)

(12) 0

( ) ( ) ( , ) ( ) , 0a

S C L v S v dv a ... (13)

which is a Fredholm integral equation of the second kind. Solve (13) for ( )S and substitute thevalue of ( )S so obtained in (8) and then invert the resulting equation to get the required functiong (t) as solution of the given integral equation (1).

We now proceed to illustrate the above analysis with the following examples :Example 1. Solve the integral equation

1 10 0

( ) (1/ ){ ( ) ( )} , 0at t u J u J ut du dt a

... (14)

where2 2 1/ 2

2 2 1/ 2

( ) ,

( ) ,

i k u k u

u k u k

... (15)

[The above equation can be used to solve the problem of the torsional oscillations of anisotropic and homogeneous elastic half-space due to a rigid circular disc of radius a which isperforming simple harmonic oscillations]

Solution. Comparting (14) with (1), we have

1 1 10( ) ( ), ( ) , ( , ) (1/ ){ ( ) ( )}g t t t f K t u J u J ut d u

... (16)

We now split 1( , )K t as in (2) with

0 1 10

( , ) ( ) ( ) ,K t J u J ut du

... (17)



and 1 10

( , ) ( / 1) ( ) ( )G t u J u J u t du

3/ 2

1/ 2 1/ 22 2 1/ 2 2 2 1/ 20 0 0

2 ( ) ( / 1) ( ) ( )( ) ( )

t v u u J uv J u du dv dt t

... (18)

Using the procedure of Art. 13.4 and making use of the known results of Art. 13.2, weobtain, as usual, the following results :

2 2 1/ 21 2 3 2( ) 2 / , ( ) , ( ) 1/ , ( , ) ( )h h h K t t ... (19)

1/ 21/ 2 1/ 2

0( , ) ( ) ( / 1) ( ) ( ) ,L v v u u J u v J u du

... (20)

2 2 1/ 2( )( )

( )

a t dtSt

... (21)

2 2 1/ 20

2 ( )( )

C d

... (22)

and0

( ) ( ) ( , ) ( ) , 0a

S C L v S v dv a ... (23)

Now, inverting (22), we obtain (as in Art. 13.2)3

2 2 1/ 20( ) 2 .

( )d t dtC

d t

... (24)

Substituting the above value of ( )C in (23), we have

0( ) 2 ( , ) ( ) , 0

aS L v S v dv a ... (25)

Converting the infinite integral (20) into a finite integral (refer Appendix at the end of thischapter), we have

1/ 2 (1)1/ 2 2 2 21/ 21/ 20

(1)1/ 2 2 2 2 1/ 21/ 2 1/ 20

( ) /( ( ) ( ) ,( , )

( ) /( ) ( ) ( ) ,

k

k

i v u k u H u v J u du vL v

i v u k u J u v H u du v

... (26)

where (1)1/ 2H is a Hankel function of the first kind. This form of the kernel is useful for small values

of k.We now proceed to solve (25) approximately for small values of ak, which is dimensionless

parameter. To this end, we re-write (25) as1

0( ) 2 ( , ) ( ) , 0 1.S a a a L av a S av dv ... (27)

Using the series expansions for the Hankel and Bessel functions, (26) yields the followingapproximate value of the kernel ( , ).a L av a



2 3 4 2 3 5 3 3

6 4 3 2 5 7 5 3 3 5

8

2 3 4 2 3 5 3 3

6 4 2 3 5

( ) / 2 (4 ) / 3 (3 ) /16 8 ( ) / 45

(5 10 ) / 384 4 (3 10 3 ) /1575

( ),( , )

( ) / 2 (4 ) / 3 (3 ) /16 8 ( ) / 45

(5 10 ) / 384 4

a i v v i v v

v v i v v v

O va L av a

v i v v v i v v

v v v

7 5 3 3 5

8

(3 10 3 ) /1575

( ),

i v v v

O v

... (28)

where = ak. By using the usual iteration method to (27), we arrive at an approximate value for( )S a given by

3 5 7 81 3 5 7( ) 2 { ( ) ( ) ( ) ( ) ( )}S a a C C C C O ... (29)

where2 3 4 5 7

61 2

4 19 53 16 143 8051( ) 1 ,4 9 192 225 3840 5880081

i i iC 2 4 5 6 7

3( ) /12 /96 / 45 11 / 2304 47 / 4200C i i 4 6 7 6

5 7( ) / 960 / 3840 /1680 , ( ) / 80640C i C Now, inverting (21), we obtain

1

2 2 1/ 2 2 2 2 1/ 2/

2 ( ) 2 ( )( )( ) { ( / )}

a

a

d S u du d S au dud du u a

... (30)


2 3 22 4

2 2 1/ 2 2 2(4 ) / 1 1 4 13 1( ) 1 1 1

6 6 9 120 60{1 ( / )}a i

a a a

22

21 1

360 a

25 2 2 26

2 2 2 258 2 16 17 53 11 1 1225 45 1680 5040 168081

ia a a

32 7

21 10931

25200 7350i

a

22 28

2 213 11 1 ( )525 630

Oa a

... (31)

Remark When 0, (31) reduces to the result (20) of Art. 13.2.Example 2. Solve the integral equation

101 ( ) ( , ) , 0

at g t K t dt a , where 1 0 0

0( , ) (1/ ) { ( ) ( )}K t u J u J u t du

Hint : To solve the given integral equation, we split the kernel K1 as

1 0( , ) ( , ) ( , )K t K t G t

where 0 0 00( , ) ( ) ( ) ,K t J u J u t du

0 00

( , ) ( / 1) ( ) ( ) .G t u y J u J ut du

Now proceed as in example 1 for getting the solution.

10.5. GENERALIZED THREE-PART BOUNDARYVALUE PROBLEMSConsider an integral equation of a more general type such as

1( , ) ( ) ( ),a

bK t g t dt f b a ... (1)



where the kernels 1( , )K t can be perturbed on the kernel 0 ( , )K t of Art. 13.3. In order to solve (1),we split the kernel 1( , )K t as

1 0( , ) ( , ) ( , )K t K t G t ... (2)where we assume that the kernel ( , )G t is in some sense smaller then 0 ( , ).K t

Suppose that 1 2( ) ( ) ( ),rrr

f a f f

... (3)

where 1 0( ) , 0r

rrf a a

... (4)

and 1

2 ( ) ,rrr

f a b

... (5)

We also define two functions 1( )g and 2 ( )g such that

1 2

0, 0( ) ( ) ( ),

0,

bg g g b a

a

... (6)

Using (2), (3), (4), (5) and (6), the integral equation (1) becomes equivalent to the followingpair of integral equations

0 1 1 10 0( , ) ( ) ( ) ( , ) ( ) , 0K t g t dt f G t g t dt a

... (7)

0 2 2 20 0( , ) ( ) ( ) ( , ) ( ) ,K t g t dt f G t g t dt b

... (8)

We assume that the kernel 0 ( , )K t is such that for all g (t) it satisfies the relation

211 2 12 2 130

00 2

21 2 22 2 230

( ) ( , ){ ( )} ( , ) ( ) ( ) , 0( , ) ( )

( ) ( , ) { ( )} ( , ) ( ) ( ) , 0

h K h K t g t h t dt dK t g t dt

h K h K t g t h t dt d

... (9)

where hij (i = 1, 2,; j = 1, 2, 3), and the kernel K2 are known functions. Moreover, we suppose thatthe kernel K2 is such that the Volterra integral equations

20( , ) ( ) ( ), 0K t g t dt f

... (10)

and 2 ( , ) ( ) ( ), 0K t g t dt f

... (11)

possess unique solutions for g in terms of all arbitrary differentiable functions f.Now we extend the analysis of Art. 10.4 and define two new kernels 1( , )L v and 2( , )L v

such that

11 13 2 2 12 12 10 0

21 23 2 2 22 22 2

( ) ( ) ( , ) ( , ) ( ) ( ) ( , ) ,( , )

( ) ( ) ( , ) ( , ) ( ) ( ) ( , ) ,

t

t

h h t K K v t h h v L v dv dG t

h h t K K t v h h v L v dv d

... (12)

Hence, the integrals on the right side of the relations (7) and (8) can be reduced in thefollowing forms :

10( , ) ( )G t g t dt

1 11 13 2 2 12 12 10 0 0

( ) ( ) ( ) ( , ) ( , ) ( ) ( ) ( , )t

g t h h t K K v t h h v L v dv d dt



11 2 12 1 12 2 1 130 0( ) ( , ) ( ) ( , ) ( ) ( , ) ( ) ( ) , 0

vh K h L v h v K v t g t h t dt dv d a

... (13)

and 20

( , ) ( )G t g t dt

2 21 23 2 2 22 22 20

( ) ( ) ( ) ( , ) ( , ) ( ) ( ) ( , )t

g t h h t K K t v h h v L v dv d dt

21 2 22 2 22 2 2 230 0( ) ( , ) ( ) ( , ) ( ) ( , ) ( ) ( ) ,

vh K h L v h v K t v g t h t dt dv d b

... (14)

where it is assumed that various orders of integration may be interchanged.Using (7), (13) and the first part of (9), we obtain

211 2 12 2 1 13 1 11 2 12

0 0( ) ( , ) { ( )} ( , ) ( ) ( ) ( ) ( ) ( , ) ( )h K h K t g t h t dt d f h K h

1 12 2 1 130( , ) ( ) ( , ) ( ) ( ) , 0

vL v h v K v t g t h t dt dv d a

... (15)

Similarly, (8), (14) and the second part of (9) yields

221 2 22 2 2 23 2 21 2 220

( ) ( , ) { ( )} ( , ) ( ) ( ) ( ) ( ) ( , ) ( )h K h K t g t h t dt d f h K h

2 22 2 2 230 0( , ) ( ) ( , ) ( ) ( ) ,

vL v h v K t v g t h t dt dv d b

... (16)

In order to solve (1), we define six functions S1, S2, T1, T2, C1 and C2 as follows :

112 2 1 13

1

( ), 0( ) ( , ) ( ) ( )

( ),S a

h K t g t h t dtT a

... (17)

222 2 2 23

0 2

( ), 0( ) ( , ) ( ) ( )

( ),T b

h K t g t h t dtS b

... (18)

11 2 1 12 10( ) ( , ) ( ) ( ) ( ), 0h K C h d f a

... (19)

21 2 2 22 2( ) ( , ) ( ) ( ) ( ),h K C h d f b

... (20)

Now, we use (17), (18), (19) and (20) in (15) and (16) and proceed as in Art. 10.3. Then, asbefore, the functions C1 and C2 can be obtained in terms of f1 and f2. Again, it can be shown thatthe four unknown functions S1, S2, T1 and T2 satisfy the following two simultaneous Fredholmintegral equations of the second kind :

1 1 1 1 1 10( ) ( , ) ( ) ( ) ( , ) ( ) , 0

a

aS L v S v dv C L v T v dv a

... (21)

and 2 2 2 2 2 20

( ) ( , ) ( ) ( ) ( , ) ( ) ,b

bS L v S v dv C L v T v dv b

... (22)

From relations (6) and (17), we have

12 2 2 13 1( ) ( , ) ( ) ( ) ( ),h K t g t h t dt T a

... (23)

Similarly, from relations (6) and (18), we have

22 2 1 23 20

( ) ( , ) ( ) ( ) ( ), 0h K t g t h t dt T b

... (24)



When the values of g1 and g2 are substituted in terms of S1, T1, S2 and T2 in (23) and (24), wearrive at two Fredholm integral equations in addition to two Fredholm integral equation (21) and(22) which have been already obtained. Thereby, we have obtained a system of four Fredholmintegral equations for getting four functions S1, S2, T1 and T2. The system is solved by iteration asin the previous articles of this chapter.

We now illustrate the above analyis with the following example.Example. Solve the integral equation

1 10( ) (1/ ){ ( ) ( )} ,

a

bt t u J u J ut du dt b a

... (25)

where2 2 1/ 2

2 2 1/ 2

( ) ,

( ) ,

i k u k u

u k u k

... (26)

[Equation (25) arises in the problem of torsional oscillations of an elastic half-space due to arigid annular disc]

Solution. Comparing (25) with (1), we have

1 1 10( ) ( ), ( ) , ( , ) (1/ ){ ( ) ( )}g t t t f K t u J u J ut du

... (27)

We now split 1( , )K t as in (2) with

0 1 10

( , ) ( ) ( ) ,K t J u J ut du

... (28)

and 1 10( , ) ( / 1) ( ) ( )G t u J up J ut du

3/2

1/ 2 1/ 22 2 1/ 2 2 2 1/ 20 0 0

2 ( ) ( / 1) ( ) ( )( ) ( )

t v u u J uv J u du dv dwt t

... (29)

For the present problem the relations (28) and (29) of Art. 13.3 are valid and it can beproved that ( , )G t is given by

1 10

( , ) ( / 1) ( ) ( )G t u J u J ut du

3/ 2

1/ 2 1/ 22 2 1/ 2 2 2 1/ 20 0 0

1/ 2

3/2 3/22 2 1/ 2 2 2 1/ 2 0

2 ( ) ( / 1) ( ) ( )( ) ( )

2 ( ) ( / 1) ( ) ( )( ) ( )

t

t

v u u J uv J u du dv dt t

t v u u J uv J u du dv dv t

... (30)

which corresponds to equation (12). Hence, the functions hij (i = 1, 2,; j = 1, 2, 3) and the kernel K2are given by

11 12 13 212 2 1/ 2

22 23 2

( ) 2 / , ( ) , ( ) 1/ ( ) 2 / ,

( ) 1/ , ( ) and ( , ) ( )

h h h h

h h K t t

... (31)

Again the kernels 1( , )L v and 2 ( , )L v are given by

1/ 21 1/ 2 1/ 2

0( , ) ( ) ( / 1) ( ) ( )L v v u u J uv J u du

... (32)

and 1/ 22 3/ 2 3/ 20( , ) ( ) [ ( / ) 1] ( ) ( )L v v u u J uv J u du

... (33)



Observe that 1( , )L v is the same as ( , )L v as given by (20) in Art. 13.4Proceeding as before four Fredholm integral equations of the second kind for the unknown

functions S1, S2, T1 and T2 are given by2 2 2

2 2 11 2 1/ 2 2 20

( ) (1/ 2, 1; 5 / 2; / )1( ) ( ) ,( ) (5 / 2)

b u T u F u duT l au

... (34)

2 221 2 1

2 1 1/ 2 2 2( ) (1/ 2, 1; 5 / 2; / )( ) ( ) , 0

( ) (5 / 2) ( )a

T u F u duT l bu u

... (35)

1 1 1 1 10( ) ( , ) ( ) 2 ( , ) ( ) , 0

a

aS L v S v dv L v T v dv a

... (36)

2 2 2 2 20

( ) ( , ) ( ) ( , ) ( ) ,b

bS L v S v dv L v T v dv b

... (37)

where2

11 2 2 1/ 2 2 2 1/ 20

( )2( ) , 0( ) ( )

a

t

S u du dtt dl bdtt u t

... (38)

22

2 2 2 2 1/ 2 2 2 1/ 2( )2 1( ) ,

( ) ( )

t

b

u S u du dtdl adtt t t u

.. (39)

We now proceed to solve equation (34), (35), (36) and (37) approximately by iteration whenthe parameters ka and b/a are small.

In the above equations (34) and (35) 2F1 stands for the hypergeometric function. Indeed, thehypergeometric functions 2F1 occuring under the integral signs in these equations is reducible to anelementary function :

22 2

1 2 31 5 32 ,1; , 2 ( ) log ,2 2 4

x y y xF xy y x x yy xy x

... (40)

Using (40), the equation (34) can be re-written in the form

1

1 2 2 2 2 20

1 2 1( ) ( ) ( ) log , 1uT a l a T bu duu uu

... (41)

where / .b a Likewise, (35) may be re-written as

2 1 1 2 2 21

1 2( ) ( ) ( ) log , 0 1u uT b l b T au duuu

... (42)

Note that the parameters involved in the present problem are,ak ,bk / / .b a ... (43)

and the discussion, which now follows, is based on the assumption that

( )O so that 2( ).O ... (44)Now, re-writing (36), we have

1

1 1 1 1 10 1

( ) ( , ) ( ) 2 ( , ) ( ) , 0 1S a a L av a S av dv a a L av a T av dv p

... (45)

In order to solve (45), we assume that

1 1 1( ) ( ) ( ),S a X a W a ... (46)



where1

1 1 10( ) 2 ( , ) ( ) , 0 1X a a a L av a X av dv ... (47)

and 1

1 1 1 1 11 0

( ) ( , ) ( ) ( , ) ( ) , 0 1W a a L av a T av dv a L av a W av dv

... (48)

As already observed L and L1 are identical and hence the integral equation (47) is same asequation (27) of Art. 13.4. Hence, 1( )X a is given by the expression on the R.H.S. of equation(29) of Art. 13.4.

Proceeding likewise, the integral equation (37) can re-written as1

2 2 2 2 21 0( ) ( , ) ( ) ( , ) ( ) , 1S b b L bv b S bv dv b L bv b T bv dv

... (49)

whose kernel can be reduced to the following form (refer Appendix at the end of this chapter)

(1)1/ 2 2 2 2 1/ 23/2 3/20

2(1)1/ 2 2 2 2 1/ 2

3/2 3/ 20

( ) /( ) ( ) ( ) ,( , )

( ) /( ) ( ) ( ) ,

k

k

i v u k u J u H uv du vL v

i v u k u J uv H u du v

... (50)

Using the expansions for Bessel and Hankel functions, we arrive at the approximate formula

2 2 2 2

2 2 2 2 2

( / 6 ) ( ) ,( , )

( / 6 ) ( ) ,

v O vb L bv b

v O v

... (51)

The functions occuring in the system of equations (34)- (51) must be obtained in the orderX1, l1, T2, S2, l2, T1, W1, S1. We have already obtained X1. We now proceed to compute the otherfunctions of the sequence of functions already mentioned. The required results, obtained by oneiteration, are given by

2 2 2 3 2 24

18 4 2( ) 1 ( ) , 0 1

3 3 9 5a il b O

... (52)

72 1( ) ( ) ( ), 0 1T b l b O ... (53)

2 44

24 1( ) ( ) ,1

45aS b O

... (54)

2 52

2 28 1( ) ( ) , 1

45al a O

... (55)

5 2 3 2 23

1 2 3 532 1 2 6( ) 1 ( ) , 1

4 3 745 7aT a O

... (56)

2 5

1 28( ) { ( )}, 0 1

45aW a O

... (57)

2 5 2 81 1( ) ( ) (8 ) /(45 ) ( ), 0 1S a X a a O .. (58)

In the above approximation, we have included only those terms that are required to compute

the value of the torque experienced by the annular disc to 8( )O



The values of g1 and g2 are obtained by inverting (17) and (18) with T2, S2, T1 and S1 as givenby (53), (54), (56) and (58) respectively. Then, using (6), the value of g can be evaluated. Next, using

(27), we get 1( ) ( ).g

Remark. The method explained in the previous articles must be modified (as indicated below)while dealing with problems connected with spherical caps and annular spherical caps (or sphericalrings). In this connection we have the following results :

Replace by , by , by , bya b and equation (2) of Art. 13.3 by

1 2( ) tan( / 2) ( ) ( )rrr

f a f f

... (i)

where 1 0( ) tan( / 2) , 0r

rrf a

... (ii)

and 1

2 ( ) tan( / 2) , ,rrr

f a

... (iii)

where is the polar angle and , are the bounding angles of the annular cap.13.6 APPENDIX

With help of the complex integration method, show that

01 ( ) ( )uu J uv J u du

2(1)

2 2 1/ 20

2(1)

2 2 1/ 20

( ) ( ) ,( )

( ) ( ) ,( )

k

k

ui H uv J u du vk u

ui J uv H u du vk u

.. (1)

where2 2 1/ 2

2 2 1/ 2

( ) ,

( ) ,

i k u k u

u k u k

and , 0.v ... (2)

Proof. Let v and let the complex plane be .s i

Integrating 2 2 2 1/ 2 (1){ /( ) } ( ) ( )s s k s H sv J s

around the circle C1 in the upper right quadrant passingover the branch point s = k, as shown in the adjoiningfigure, we obtain

1

2(1)

2 2 1/ 2 ( ) ( ) 0,( )C

s s H sv J s dss k

because there are no singularies within this contour. Let

, 0 and .R Then the contributions from thecorresponding arcs tend to zero. Hence, we obtain

2 2

(1) (1)2 2 1/ 2 2 2 1/ 20

( ) ( ) ( ) ( )( ) ( )

k

kH v J d H v J d

i k k

20 (1)2 2 1/ 2 ( ) ( ) 0

( )i i H i v J i d

i k

.. (3)

Similarly, integrating 2 2 2 1/ 2 (2)/( ) ( ) ( )s s k s H s v J s around a contour C2 in the lowerright hand quadrant and passing under the branch point s = k, we obtain



2 2(2) (2)

2 2 1/ 2 2 2 1/ 20( ) ( ) ( ) ( )

( ) ( )

k

kH v J d H v J d

i k k

20 (2)

2 2 ( ) ( ) 0( )

i i H i v J i di

... (4)

We know that (1) (2)( ) ( ) ( ) ( )H i v J i H i v J i ... (5)Using (5) and adding (3) and (4), we obtain, for v

2

2 2 1/ 2 ( ) ( )( )k

J v J dk

2

2 2 1/ 20 0( ) ( ) ( ) ( ) ,

( )

k kJ v J d Y v J d

k

... (6)

where Y is the Bessel functions of the second kind. Using (3) and (6), we have

L.H.S. of (1) 0

1 ( ) ( )uu J uv J u du

2 2

2 2 1/ 2 2 2 1/ 20( ) ( ) ( ) ( )

( ) ( )

k

k

iu uu J uv J u du u J uv J u duk u u k

2

2 2 1/ 20( ) ( ) ( )

( )

k ui J uv iY uv J u duk u

2(1)

2 2 1/ 20( ) ( ) , ,

( )

k ui H uv J u du vu k

which proves the first part of formula (1).Second part of formula (1) can be proved similarly.


CHAPTER 14

Integral EquationPerturbation Techniques

14.1 INTRODUCTIONIn chapter 13, we have discussed method to solve the Fredholm integral equations of the first

kind by converting them to Volterra integral equations and to Fredholm integral equations of thesecond kind. Since we had to deal with probelms involving one variable of integration, the requiredformulation was rather simple. Hence in order to solve the integral equations relating to boundariessuch as a cylinder or a sphere, we require new techniques. In the present chapter, we propose todiscuss three-dimensional problems and present approximate techniques for solving them. Theanalysis for the corresponding plane probelms is simpler once the technique is well understood.14.2 WORKING RULE FOR SOLVING AN INTEGRAL EQUATION BY

PERTURBATION TECHNIQUES.We propose to present approxiamate techniques for solving the Fredholm integral equations

of the first kind

( ) ( , ) ( ) ,S

f P K P Q g Q dS P S ... (1)

with P x and .Q In chapter 13, we have already observed that certain perturbation parameters are naturally

involved in physical problems. Let be such a parameter occuring in (1). Then, we expand all thethree functions K, f and g in power series in as shown below

20 1 2 ....K K K K ... (2)

20 1 2 ....f f f f ... (3)

20 1 2 ....g g g g ... (4)

Substituting the values of K, f and g as given by (2), (3) and (4) in (1) and equating equalpowers of , we arrive at the following set of integral equations

0 0 0 ,S

K g dS f0 1 1 1 0 ,

S SK g dS f K g dS ... (6)

0 2 2 1 1 2 0S S S

K g dS f K g dS K g dS ... (7)

and so on.The following three conditions must be satisfied in order that the present technique be

applicable in solving the given integral equation (1).

14.1


14.2 Integral Equation Perturbation Techniques

(i) K0 (P, Q) is the dominant part of K (P, Q) as in chapter 13.(ii) The integral equation (5) can be solved.

(iii) The functions g0, g1, .... are such that the integrals occuring on the right side of equations(6), (7) etc. can be easily evaluated.

It is worth noting that, in the present technique, we have only to solve (5) because the otherintegral equations (6), (7) etc. in the sequence have the same kernel.

Sometimes, an approximation to order () can be very easily obtained. Let the functiion K1occuring in the expansion (2) be a constant C (say). Then (1) can be re-written as

20( ) ( , ) ( ) ( ),

Sf P f K P Q g Q dS O ... (8)

where ( ) ,S

f C g Q dS ... (9)

is an unknown constant. Then, to order , (8) is similar to (5), whose solution is assumed known.Hence (8) can also be solved. The quantity f can then be evaluated from different considerations.

The occurrance of a constant C can be demonstrated by the kernel

2exp{ | |} exp ( ) 1( , ) ( ),| |i i rK P Q i O

r r

xx

where | |r x and exp (a) = ea. Clearly, here 0 ( , ) 1/K P Q r and C = i.A special case of the above analysis. Let K (P, Q) = K0 (P, Q) + C, where C is a constant.

Assume that solution G (P) of the integral equation

0 ( , ) ( ) 1,S

K P Q G Q dS P S ... (10)

is known. Then, suppose we are required to solve the integral equation

0{ ( , )} ( ) 1,S

C K P Q g Q dS ... (11)

which is obtained by setting K (P, Q) = K0 (P, Q) + C and f (P) = 1 in (1). Re-writing (11), we get

0 ( , ) ( ) 1 ( )S S

K P Q g Q dS C g Q dS ... (12)

Clearly, the value of the integral on the right side of (12) is a constant, although as yetunknown. It follows that the R.H.S of (12) is a constant D (say). Then, (12) reduces to

0 ( , ) ( ) ,S

K P Q g Q dS D ... (13)

where 1 ( )S

D C g Q dS ... (14)

Re-writing (13), we have 0 ( , ){ ( ) / } 1S

K P Q g Q D dS ... (15)

Comparing (15) with (10), where the value of G (Q) is known, we getg (Q)/D = G (Q) or g (Q) = D G (Q) ... (16)

Integrating both sides of (16) over the surface S, we obtain

( ) ( )S S

g Q dS D G Q dS or ( ) 1 ( ) ( ) ,S S S

g Q dS C g Q dS G Q dS using (14)

or 1 ( ) ( ) ( )S S S

C G Q dS g Q dS G Q dS


Integral Equation Perturbation Techniques 14.3

or ( ) ( ) 1 ( )S S S

g Q dS G Q dS C G Q dS ... (17)

Substituting the value of D as given by (14) in (16), we have

( ) 1 ( ) ( )S

g Q C g Q dS G Q ... (18)

Substituting the value of ( )S

g Q dS as given by (17) in (18), we get

( )

( ) 1 ( )1 ( )

S

S

C G Q dSg Q G Q

C G Q dS

or( )( )

1 ( )S

G Qg QC G Q dS

so that ( ) ( ) 1 ( )/ Sg P G P C G Q dS ... (19)

which gives the desired solution g (P) of the integral equation (11).While dealing with the above special case, we took f (P) = 1 and K (P, Q) = K0 (P, Q) + C in

(1), where C is a constant. We now extend the above analysis by taking a separable kernel withfinite terms of the form

0 1( , ) ( , ) ( ) ( )

n

i iiK P Q K P Q Q P

... (22)

where the ( )i Q are known. Then, (1) takes the form

0 1( , ) ( ) ( ) ( ) ( ).

n

i iiSK P Q Q P g Q dS f P

... (23)

Now, suppose that we know the solutions of integral equations

0 0( , ) ( ) ( ),S

K P Q G Q dS f P P S ... (24)

and 0 ( , ) ( ) ( ), , 1, 2, ...,i iS

K P Q G Q dS P P S i n ... (25)

then we propose to show that (23) can also be solved. To this end we re-write (23) as

0 1( , ) ( ) ( ) ( ), ,

n

i iiSK P Q g Q dS f P C P P S

... (26)

where ( ) ( ) .i iS

C Q g Q dS ... (27)

are constants, although as yet unknown. The rest of the steps are exactly similar to those employedin solving (11)

Remark. The techniques of the Art. 14.2 can be applied to solve problems in various disciplinesof mathmatical physics and engineering. In what follows, we shall let ( ; )G x be the generic notationfor the Green’s function as in chapters 11and 12. In view of the expansion (2) for the kernel K, weshall write 0 ( ; )G x for ( ; ).E x

14.3. APPLICATIONS OF PERTUACTION TECHNIQUES TO ELETROSTATICSSuppose that there are conductors with surfaces S1 and S2; S1 is completely contained in S2

and is kept at a unit potential, while the potential on S2 is zero. Let ‘a’ denote a characteristiclength of S1 and b denote the minimum distance between a point of S1 and a point of S2. Then theperturbation parameter can be taken as /a b which is assumed to be much smaller than unity..



In Art. 12.4, we presented an integral representation formula for the electrostatic potential inthe region D between S2 and S1:

1( ) ( , ) ( ) ,

Sf P G P Q Q dS ... (1)

in terms of the Green’s function G (P, Q) and the charge density . Applying the boundary conditionon S1, (1) reduces to

111 ( , ) ( ) ,

SG P Q Q dS P S ... (2)

As explained in working rule of Art. 14.2, we writeG (P, Q) = G0 (P, Q) + G1 (P, Q) ... (3)

where G0 (P, Q) is the free-space Green’s function and G1 (P, Q) is the perturbation term. Using(3), (2) reduces to

1 1

0 1 11 ( , ) ( ) ( , ) ( ) , ,S S

G P Q Q dS G P Q Q dS P S ... (4)

Now, in the absence of the conductor S2, there will be only the first integral on the R.H.S. of(4). It follows, that the second integral on the R.H.S. of (4) must represent the effect of the conductorS2 on the potential of S1. Using the hypothesis of Art. 14.2, we assume that we can solve (4) in theabsence of the second integral in it.

It can be easily seen that it is always possible to introduce a constant A and write

1 2( , ) ( , ),G P Q A G P Q ... (5)

where 2 ( , ) ( ).G P Q O A For example, one possible value of A is the value G1 (P, Q) for anarbitrary pair of points P and Q on S1. Using (5), (4) reduces

1 1 10 21 ( , ) ( ) ( ) ( , ) ( )

S S SG P Q Q dS A Q dS G P Q Q dS ... (6)

Let us introduce a new charge density , defined by

1( ) ( ) 1 ( ) .

SP P A Q dS

... (7)

or1

1 ( ) ( ) ( )S

A Q dS P P

Integration of both sides of the above equation over the surface S1 yields

1 1 11

S S SA dS dS dS

or1 1 1

1S S S

dS A dS dS

1 1 1

1S S S

dS dS A dS ... (8)

Again, re-writing (6) in terms of the density , we have

1 10 21 ( , ) ( ) ( , ) ( ) .

S SG P Q Q dS G P Q Q dS ... (9)

From the above discussion, it follows that the second integral on right side of (9) is 2( )O times

the first one. Accordingly, if we neglect the terms of this order, then is the electrostic chargedensity on S1 when it is raised to a unit potential in free space. Hence we conclude that (8) gives



the capacity C of the condenser formed by S1 and S2 in terms of the free-space capacity C0 of S1,

that is, 1 20 0/ (1 ) ( )C C AC O ... (10)

or 20 0/ 1 ( ).C C AC O ... (11)

Suppose we interpret A as the value of G1 (P, Q) for any pair of points P, Q on S1. Then, thecapacity C given by (11) is exactly the same that would have been obtained had we used theperturbation procedure (2) – (6). The formula (10) is very useful for finding the electrostatic capacitybecause in many problems it is possible to show that, by a suitable choice of A, the formula (10) isvalid for much higher order in .

We now illustrate the above analyis by help of the following example.Example. Consider a sphere of radius a placed with its centre on the axis of an infinite

cylinder of radius b. We have already studied an integral equation formulation of a general axiallysymmetric problem of this type in solved example 2 of Art. 12.5, chapter 12. In terms of cylindericalpolar co-ordinates ( , , ),z we have proved that

( )1 1 1 1 0 1 1 110( , , ; , , ) (2 ) ( , ; , ) cos ( )r

rrG z z G z z r

... (12)

where ( ) 111 0

( ) ( ) ( )2 cos ( ) ,( )

r r r r

r

I u I u K u bG u z z duI ub

... (13)

where ( , , )z and 1 1 1( , , ),z respectively are cylindrical polar coordinates of points P and Q

respectively. Due to axial symmetry, we require only the term (0)G in (12).We now proceed to find the constant A in the relation (6). To this end, we take

cosz a and sin ,a ... (14)where is the angle between Oz and OP, O being the origin. Then (13) reduces to

1(cos cos )(0) 0 0 1 01 0 0

( sin ) ( sin ) ( )2 Re( )

iuaI u a I u a K ubG e du

I ub

... (15)

where the ‘‘Re’’ means that we take the real part of the expression. In addition, we use the formula

cos0 0

( sin )( sin ) (cos ),!

niua

nn

iuaI ua e Pn

... (16)

where Pn are the Legendre polynomials. Using (16), (15) reduces to

(0) 0111 , 00 0

( )( sin )2 ( sin )( , ) Re ( 1) ( ) (cos ) (cos )! ! ( )

mnm m n

n mm n

K ubuauaG P Q i P P dun m I ub

... (17)

011

0 0

( )( sin )2 ( sin )Re ( 1) ( ) (cos ) (cos )! ! ( )

mnm m n m n

n mK vi P P dv

b n m I v

!

10 0

1 Re (cos ) (cos ) ,n nn n n nn n

A P A Pb

... (18)

where v = ub and /a b (the demensionless parameter of the problem). Also, the constants. AAnare given by the formula

00 0

( )2 ( 1) ( )( )

m m n m nm n

K vA A i v dv

I v

... (19)



From (19) we conclude that when (m + n) is odd, Am An is an imaginary quantity.Using (18) and (19), the desired constant A is given by

20 0

0 0

( )2( )

A K vA dvb b I v

... (20)

Substituting the value of A as given by (20) in (10), we obtain

10 0/ 1 (2 / ) (0) ,C C b C I ... (21)

where2

00 0

( )(2 ) (2 1) ,

( )

mv K vI m m dvI v

... (22)

for which numerical tables are available.

EXERCISE1. Instead of spherical coordinates (14) of Art. 14.3, take the oblate-spherical coordinates

,z a 1/ 22 2(1 ) (1 )ae

and solve the electrostatic potential problem for the case of an oblate spheriod placed symmetricallyinside a grounded infinite cylinder of radius b. Show that the capacity of this condenser is

12 46

1 12 ( ) (2) 2( ) (4)1 (0) 0( ),

81 225sin sinae e e I e IC I

e e

where the quantities I (2m) and defined by (22) of Art. 14.3

Hint. Use the expansion 0 ( ) ( ) ( ) ( ),iuzn n nI u e B u P P i

where the Pn are Legendre polynomials.2. Using a method similar to exercise 1, solve the electrostatic problem for a prolate spheroid.3. Solve exercises 1 and 2 when the spheroids are placed between two grounded paralled

plates.14.4 APPLICATIONS OF PERTURBATION TECHNIQUES TO LOW-REYNOLDSNUMBER HYDRODYNAMICS

The flow of an incompressible viscous fluid is governed by the following two types oflinearized equations :

1. Stokes equations2. Oseen equations

In this article we propose to use the perturbation techniques to study these two kinds of flows.14.4A STEADY STOKES FLOW.

We have already studied steady stokes flow in an unbounded motion is solved example 1, ofArt. 12.7. Please read carefully that example before proceeding further. Recall that, for a freespace, the boundary value problem is

2 ,p q 0 q ... (1)

1 1, on ;Sq e 1 0 at ;q ... (2)where this system has been made dimensionless with help of the uniform speed U of the givensolid B and with its characteristic length a. Here ,q p and 1e denote the velocity vector, pressureand the unit vector along the x1-axis respectively.



We have proved that the integral-equation formula for the boundary value problem (1)- (2) interms of Green’s tensor 1T and Green’s vector p1 is given by

11 1 1. , ,

SdS P S e f T ... (3)

where ( / )n p f q n ... (4)2

1 (1/8 ) ( | | grad grad | |) T I x x ... (5)

21 (1/8 ) grad | | p x ... (6)

and 1 ij I the Kronecher delta ... (7)

The corresponding formula for the resistance F (The subscript signifies that an infinitemass of fluid is under consideration) on the given body B can now be obtained by noting that thestress tensor has the value

{ ( ) }p I q q

where ( )q stands for the transpose of .q Using (4), we get

1SdS F f

We now find relation between F and the so called resistance tensor , which is defined to

be such that the force extered on a body with uniform velocity u is u ,Thus, we have

,F F e u ... (9)

where e is the unit vector in the direction of u and | | .F F

From our study of fluid dynamics, the solutions of various boundary value problems forsteady Stokes flow in an unbounded medium are already known. Hence the solution of the integralequation (3) can be found for these problems. It follows that the tensor 1T corresponds to the kernelK0 of Art. 14.2.

In what follows, we propose to show how the correction term may be obtained for morecomplicated cases with help of technique outlined in Art. 14.2.14.4.B. BOUNDARY EFFECTS OF STOKES FLOW

As shown in example 2 of Art. 12.7, the presence of the boundary S2 requires the introductionof a new tensor T and a corresponding vector p. These quantities satisfy the equations

2 ( ), 0, 0T p I x T T on S2 ... (10)

When S2 tends to infinity T and p reduce to 1T and 1p as given above. According to the

present scheme we write 1 2 T T T and 1 2 , p p p where 2T and 2p satisfy the homogeneouspart of the system (10). It follows that the integral equation which is equivalent to the presentproblem is given by

1 11 1 2( )

S Sds d S e f T f T T ... (11)

Let P, Q and the origin be taken on S1. Let be the parameter that gives the ratio of a, thestandard geometric length of the given solid B, to the minimum distance between a point of S1 anda point of S2. Using Taylor’s theorem, we have



0 0 32 2 2 2( ) [ ] ( ),O 0 0x xT T r T T ... (12)

where 02 2 (0,0)T T and the subscript zero on the implies differentiation with respect to the

components of . Taking only the first-order terms of the relation (12) in (11), we obtain

1

01 2 1 ,

Sd S e F T f T ... (13)

where 1S

dS F f

Here, F is the resistance experienced by the given body B in the bounded medium.Since the integral equation (13) has the same kernel as that of (3), it follows (13) can be

regarded as giving the velocity field in an unbounded fluid when the given body B moves with

uniform velocity 1 02e + F T . Using the concept of the resistance tensor , as defined above, we

obtain the force formula F given by0

1 2( ) F e F T ... (14)

Solving (14), we obtain 1 0 11 2.[ ]

F e T ... (15)

Since replacement of F by F produces error of order 2 and hence, to order , (14)reduces to

01 2( ) F e F T ... (16)

Let us define the principal axes of the resistance of the given body B in such a manner thatwhen B moves parallel to one of them in an infinite mass of fluid, the force is in the direction ofmotion. They are the unit eigenvectors of the resistance tensor . We denote them by

1 2 3, andi i i such that

1 21 1 2 2 3 3 3 i i i i i i ... (17)

We now decompose 02T into components with these eigennectors as the basis. Moreover, we

set 1 1.e i Substituting these experessions in (14), we derive

/ 1/(1 )F F F ... (18)

where is independent of the form of S1.

14.4C LONGITUDINAL OSCILLATIONS OF SOLIDS IN STOKES FLOWWe now propose to use the present analysis to compute an approximate value of the velocity

field generated and the resistance experienced by a solid of any shape which is executing slowlongitudinal vibrations in an unbounded viscous fluid. Suppose that the given body B oscillates

about some mean position with velocity 1iwtUe e and q and p have the same time dependence. Then,

we know that the dimensionless Stokes equations for the steady state vibrations are given by2 2 0,p iM q q 0,div q ... (19)

where M2 = a2 w/v is the rotational Reynolds number and v is the coefficient of kinematic viscosity.The integral representation formulas are the same as for the steady Stokes flow (refer Art.

14.4 A), while T and p now satisfy the equations



2 2 ( ),i M p T T I x div T = 0, ... (20)

and 0T as .x Equations (20) are satisfied when T and p are given by2 ,grad grad T I 2 2( )grad i M p ... (21)

2 2 2( ) ( )i M x ... (22)

21 exp[ {(1 ) / 2} | |]

4 | |i M

i M

xx

... (23)

Then, 21 {(1 ) / 6 2} ( ),i M O M T T I ... (24)

where T1 is given by (5).

We now substitute the boundary value 1q e in the integral representation formula for thesystem (19) and note that, in view of (20) and Green’s theorem, we finally obtain

1

2 ,iS R

p d S i M dVn

T n T ... (25)

where Ri is the interior of S1. Then, we get the following Fredholm integral equation of the firstkind for getting f

1

21 1 1{(1 ) / 6 2} ( ),

Si M dS O M P S e F T f ... (26)

Proceeding as before, we have the formula2

1 1{(1 ) / 6 2} ( ) ( )i M O M F e e ... (27)

When the given body B moves parallel to one of its axes of resistance (which is chosen asthe x1-axis in our coordinate system), then (27) reduces to

11 {(1 ) / 6 2}F i M F F e ... (28)

In particular, for a sphere, 6F aU in physical units, where a is the radius of the sphereand is the shear viscosity of the fluid. Then, in physical units, formula (28) yields

216 {1 ( / 2) (1 )} ( )aU M i O M F e ... (29)

14.4D STEADY ROTARY STOKES FLOW.Since the pressure is taken to be constant for the rotation of axially symmetric bodies, the

steady Stokes equations take the simple form:2 , ,div 0 0q q p = const. ... (30)

We choose the z-axis of cylindrical polor coordinates ( , , )z to be the axis of symmetry ofthe given body. Let the streamlines be circles lying in planes perpendicular to Oz. Hence q has anonzero component ( , )z! in the directions only and is independent of . The equation ofcontinuity is thus satisfied automatically, while the equation of motion (30) reduces to

2 2

2 2 21 0,

z

! ! ! !... (31)



which has been made dimensionless with a as the typical velocity. Here, is the uniform angularvelocity of the body and ‘a’ is its characteristic length. The boundary conditions are

1on ;S ! 20 on ,S! ... (32)where S1 is the surface of the rotating body and S2 is the bounding surface.

From (31), we can easily verify that the function

( , , ) ( , ) cosw z z ! ... (33)

is harmonic. Hence it can be represented in terms of a source density 1( ) cosQ spread over S1,

where 1 1 1( , , )z are the coordinates of Q and ( )Q is independent of 1. Therefore, the integralrepresentation formula (32) of Art. 12.4 of chapter 12 can be employed for the harmonic function.Thus, we obtain

11( , , ) ( , ) ( ) cos ,

Sw z G P Q Q dS ... (34)

where P is the arbitrary point in the region between S1 and S2. In view of (32), we have ! on S1and hence we obtain

1(1)

1( , ) ( ) ,C

G P Q Q ds P S ... (35)

where (1/ ) G(1) (P, Q) is the coefficient of 1cos ( ) in the Fourier expansion of G (P, Q) and dsstands for the element of the arc length measured along the curve C. Here C is the bounding curveof S1 lying in the meridian plane.

We have already dealt with the decomposition

1( , ) (1/ | |) ( , )G P Q x G P Q ... (36)

where G1 (P, Q) is finite in the limit as .Q P Likewise, we can decompose the Fourier componentG(1) into the sum

(1) (1)(1)0 1 ,G G G ... (37)

where (1)0G arises from the Fourier expansion of 1/ | |x and (1)

1G arises from the expansion ofG1. Hence, (35) may be re-written as

(1) (1)1 10 1C CG ds G ds ... (38)

Let b denote the minimum distance between a point of S1 and a point of S2. Then we take asmall perturbation parameter / .a b Clearly, the second integral on R.H.S. of (38) is at least oforder of the first integral. For geometric configuration for which

(1)1 21 ( ),G A G ... (39)

where A is a constant and G2 is of order ,A (38) reduces to

(1) 2 21 1 1 20 ,

C C CG ds A ds G ds ... (40)

or(1) 2

1 1 20 ,C C

G ds G ds ... (41)

where 211

CA ds ... (42)



It follows that now we have a situation already dealt in Art. 14.3, that is, represents, with

an error which is at most of order 2 , an appropriate source density for the body rotating in aninfinite mass of fluid.

The tangential stress component on the surface S1 in the direction of increasing is given by

,n

!... (43)

where / n denotes differentiation along the normal drawn outward to S1. Also, from the analysis

of chapter 12, it is known that the source density ( )Q on S1 is related to ! by

4 ( )Qn

!... (44)

Thus, 4 . ... (45)Using this value of the stress component, the value of the frictional torque N can be obtained.

It is given by 2 28C

N d s ... (46)

To find the relation between N and the torque N is an unbounded fluid integrate both sidesof the relation (42) around the meridian section C of the axially symmetric body. The relation is

given by 121 ( / 8 ) ,N N A N

... (47)

with an error of order 2 . By a suitable choice of A, the formula (47) is valid in many cases to amuch higher order in .

Remark. Equation (47) can be illustrated with many useful configurations. For example, thecase of a sphere which is symmetrically placed in an infinite cylindrical shell can be discussed as inthe analysis of Art. 14.3. It can be shown that (47) yields

311 ( /8 ) ,N N N a H ... (48)

where Hk is given by the integral

2 10 1

( )2 ( 1)(2 )! ( )

kk

kK x

H x dxk I x

... (49)

14.4E ROTARY OSCILLATIONS IN STOKES FLOWSetting p = constant in (19), the equations governing the steady-state rotary oscillations (with

circular frequency w) of axially symmetric solids in an incompressible viscous fluid are2 2( ) 0,i M q 0 q ... (50)

Let the z-axis of cylindrical polar coordinates ( , , )z be the axis of symmetry of the givensolid bodies. Assuming that the streamlines are circles lying in planes perpendicular to Oz, then q

has a non-zero component ( , )z! in the direction only and is independent of . Then, theequation of continuity is thus satisfied automatically, and the differential equations (50) reduces to

2 22

2 2 21 0,

dz

! ! ! !! ... (51)



where 2 2i M ... (52)

We now discuss the present analyis for 1. The boundary values are given by

1on ,S ! 20 on ,S! ... (53)where S1 is the surface of the oscillating body and S2 is the bounding surface.

Let cos .w ! ... (54)Then (51) and (53) reduce to the boundary value problem given by

2 2( ) 0,w ... (55)

1cos onw S ... (56a)w = 0 on S2. (56b)

The Green’s function ( ; )G x appropriate to the boundary value problem (55) - (56) is given by

2 2( ) ( ; ) 4 ( ),G x x 2

0.SG ... (57)

Thus, 1exp ( | |)( ; ) ( ; ),

| |G G

xx xx

... (58)

where 1( ; )G x is finite in the limit as . x

Following the analysis of Art. 12.6, the integral representation formula for ( )w x is given by

11 1 1 1( ) ( , ) cos ( ; ) , , ,

Sw z G dS S R x x x ... (59)

where R is the region between S1 and S2, and 1 1( , )z is given by the formula (44). Using theboundary condition (56a), the desired Fredholm integral equation is given by

11 1 1cos ( , ) cos ( ; ) ,

Sz G dS x with x and on S1. ... (60)

Let (1)1 11 ( , ; , )G z z be the coefficient of 1cos ( ) in the Fourier expansion of

1( ; ).G x Then the integration over 1, reduces the above integral to

11 1 1

exp ( | |)cos ( , ) cos| |S

z d S

x

x

(1)

1 1 1 1 11cos ( , ) ( , ; , )C

z G z z d s ... (61)

in the notation of equation (35)We now expand as the perturbation series

nnn

... (62)

in equation (61). By direct expansion of the Green’s function, we can prove that (1) 31 ( ),G O

where is the ratio of the characteristic length of the vibrating body to the distance of its centrefrom the nearest point of S2. Let / (1).q O



Equating like powers on both sides of (61) and rejecting terms that trivially vanish, we obtain

1

0 1 11

( , )cos cos .

| |S

zdS

x ... (63a)

1

1 1 11

( , )0 cos

| |S

zdS

x ... (63b)

1 1

2 1 11 0 1 1 1

( , ) 10 cos ( ) | | cos| | 2S S

zdS z dS

xx

... (63c)

1

3 1 11 1 1

( , ) 10 cos ( , ) | | cos| | 2S

dS z dS

x

x

1

20 1 1 0 1 1 1 1 1

1 ( , ) | | cos cos ( , ) ( , ; )6 S C

z dS z H z z ds x ... (63d)

and so on, where(1) 2 4

1 1 1 11 ( , ; ) ( , ; ) ( )G z z H z z O ... (64)

Relations (63a) – (63d) show that the source densities 0 1 2 3, , , etc can be obtained bysolving potential probelms in free space of the form already dealt in chapter 12.

The velocity field and the frictional torque can also be easily obtained. In fact, using theformula (43), the torque N is given by

2 32S C

N dS dsn n

! !

... (65)

Using the relation (44), (62) and (65), we get

2 2 2 3 40 1 2 38 ( ) ( )

CN ds O ... (66)

Remark. Since the potential problems of the type given in equations (63a) – (63d) can besolved for various configurations such as a sphere, a spheroid, a lens, and a thin circular disc, wecan solve our problems for all these geometric shapes. We now illustrate the above analyis by thefollowing example :

Example. Suppose a thin circular disc is vibrating about its axis in a viscous fluid which iscontained in an infinite circular cylinder. Assume that the axes of the disc and the cylinder coincide.

Solution. By following the steps of solved example 2, page 12.16 of Art. 12.5, the Green’sfunction for an infinite cylinder , 0z b is given by

0 10

exp { | |) 2( ; ) (2 ) cos ( )| | nn

G n

xx

x

2 21 1 2 2 1/ 2

( ) ( ) ( ) cos {( ) ( )}( ) ( )

nn n

n

K ub u duI u p I u u z zI u b u

... (67)

Using (67), (1)1G for the disc 1, 0 2 , 0z is given by

3 3

(1) 51 111 2 2 1/ 2

1

( )( , ) ( )2 ( ) ( )q

K y y dyG OI y y q

... (68)



or 31 1( , ) (1/ 2 ) ( ),H q A q ... (69)

where A (q) denotes the infinite integral in (68).Using the methods of Art. 12.4 and 12.5, the integral equations (63a) to (63b) can be solved

to yield

and

2

0 1 22 2 1/ 2 2 2 1/ 2

3 2 3 2 1/ 2

2 (2 ), 0,(1 ) 3 (1 )

4 2 ( ) , 0 133 (1 )

A qq

... (70)

Substituting the above values of 0 1 2 3, , and in (66), the required value of N (in physicalunits) is given by

2 3 33 4 5

232 4 4 ( )1 ( , )3 5 9 3

iwtA qN a e O

... (71)

14.4F OSCEEN FLOW-TRANSLATIONAL MOTIONThe slow motion past a solid is governed by the dimensionless Oseen equations (refer solved

example 3 of Art. 12.7)2

1( / ) ,x p T q div q = 0 ... (72)

1 1on ;Sq e 2on ,S 0q ... (73)where S1 is the surface of the solid body and S2 is the bounding surface. Then, using the analyis ofsolved 3, page 12.21 of Art. 12.7, the Fredholm integral equation of the first kind which is equivalentto the boundary value problem (72) - (73) is given

1 ,S

dS e T f ... (74)

where the Green’s tensor Tand the Green’s vector p are now defined as2(1/8 ){ }grad grad T I ... (75)

21(1/8 ) ( ( / )}grad x p ... (76)

| |

0

1 1| |

ts e dtt

... (77)

1 1| | ( / | |) ( )s x x ... (78)

( / )n p f q n ... (79)

Now, 2 3 21 1 1 1 .. 1 ...

1! 2! 3! 2! 3!

te t t t t tt t

... (80)

Using (80) and expanding in (77) in terms of the Reynolds number, the relation (75)reduces to

1 ( ),O T T ... (81)



where 1T is given by

21 (1/8 ) ( | | | |)grad grad T I x x ... (82)

The rest of the analyis is similar to the one given in the previous article.14.4G OSEEN FLOW-ROTARY MOTION

With the help of the present technique, we now proceed to find the solutions of the Oseenequations for the steady rotations of axially symmetric solids.

For the rotation of axially symmetric bodies, the pressure is taken to be constant and theOseen equations reduce to the following simple form :

21( / ) ,x 0q q div q = 0 ... (83)

Let the z-axis of cylindrical polar coordinates ( , , )z with (z = x1) be the axis of symmetryof the given bodies. Assuming that the streamlines are circles lying in planes perpendicular to Oz,then q has a nonzero component ( , )V z in the direction only and is independent of . TheTheequation of continuity is thus satisfied automatically, while the boundary value problem becomes

2 2

2 2 21 2 0V V V V Vc

zz

... (84)

where c = Ua / 2v = / 2. Equation (84) has been made dimensionless with U as the typicalvelocity and ‘a’ as the characteristic length of the body.

The boundary conditions on V are given by

1on ;V S V = 0 on S2, ... (85)where S1 is the surface of the rotating solid and S2 is the bounding surface.

Let ( , )c zV e z ! ... (86)

Then, the substitution (86) transforms the boundary value problem (84) - (85) in the form

2 22

2 2 21 0c

z

! ! ! !! ... (87)

1on ;c ze S ! 20 on S! ... (88)

[Note that (87) is the same as (51) of Art 14.4 E with replaced by c. whereas the boundaryconditions (88) are different from those of (53) of Art. 14.4 E]

By repeating the algebraic steps (54) upto (61) of Art. 14.4 E, we finally obtain the Fredholmintegral equation

1

1 1 1( , ) cos exp ( | |)cos

| |c z

S

z ce dS

x

x

(1)1 1 1 1 11cos ( , ) ( , ; , )

Cz G z z ds ... (89)

from which ( , )z can be obtained, where ( , )z is defined by

4 ( )Qn

!... (90)



In order to find solution of (89), we take the expansionn

nnc ... (91)

and also expend cosc ze in power series of c. On comparing the equal powers of c in (89), wefinally get the following integral equations of potential theory

1

0 1 11

( , )cos cos

| |S

zdS

x ... (92a)

1

1 1 11

( , )cos cos

| |S

zz dS

x ... (92b)

1 1

22 1 1

1 0 1 1 1( , )cos 1cos ( , ) | | cos

2 | | 2S S

zz dS z dS

xx

... (92c)

1 1

33 1 1

1 1 1 1 1( , )cos 1cos ( , ) | | cos

6 | | 2S S

zz dS z dS

xx

1

20 1 1 1 0 1 1 1 1 1

1 ( , ) | | cos cos ( , ) ( , ; , )6 S C

z dS z H z z ds x ... (92d)

where 1 1( , ; )H z z is defined by the relation.(1) 2 4

1 1 1 11 ( , ; , ) ( , ; ) ( )G z z c H z z O c ... (93)

Consider the particular case of a thin circular disk 0, 1.z Then the system of equations(92a) – (92d) is the same as the system (63a) – (63d) of Art. 14.4 E. It follows that the solution forsteady rotation problem for the disk in Oseen flow is the same as the corresponding solution for thesteady state vibrations in Stokes flow (refer Art 14.4 E). For example, the value of the torque N forthe problem under consideration can be deduced from the formula (71) of Art 14.4 E. It is gives by

3 2 3 34 5

232 4 4 ( )1 ( , ),

3 5 9 3a c c A qN O c

... (94)

where is the uniform angular velocity of the solid.For solids of other geometric shapes, we have to solve the integral equations (92a) – (92d)

with non-zero left side. We illustrate this fact by the following example.Example. Suppose a sphere of radius a be rotating. For spherical geometry, we shall use

the spherical polar coordinates. For this problem, the value of the Green’s function ( ; )G x is the

same as (67) of Art. 14.4 E with replaced by c. The corresponding values of (1)1G and 1( , )H are

given by

2 3(1) 51 1

11 2 2 1/ 21

sin sin ( )( , ) ( )2 ( ) ( )q

K y y dyG OI y y q

... (95)

and13

sin sin ( )( , )2

A qHq

... (96)

Using the method of chapter 12, the source densities 0 1 2 3, , , and can be obteined from(92a)-(92d). The final results are



10 1(3 / 4 ) (cos ),P 1

1 2(5 /12 ) (cos )P ... (97)

1 12 1 3(1/ 4 ){(3 / 2) (cos ), (7 /15) (cos )}P P ... (98)

1 33 1(3/ 4 ) (cos ) {(1/ 3) (1/ 2 ) ( )},P q A q where 0 . ... (99)

For the present problem torque N is given by

2 3 2 3 40 1 2 38 ( ) ( )

CN c c c ds O c ... (100)

It is same as (66) of Art. 14.4 E with replaced by c

Substituting the values of 0 1 2 3, , and as given by (97) – (98), we obtain (in physicalunits)

2 3 33 4 54 ( )8 1 ( , )

15 3 2c c A qN a O c

... (101)

EXERCISE1. Find the torque experienced by a sphere which is rotating uniformly in Oseen flow and is

bounded by a pair of parallel walls .z c Evaluate also the velocity field.2. Extend the analyis of the steady Oseen flow in this chapter to the case of the steady-state

vibrations of axially symmetric solids in Oseen flow.


APPENDIX A

BOUNDARY VALUE PROBLEMS AND GREEN’S IDENTITIESA.1 Some useful notations

In what follows, we shall use the following notations:(i) If a function f C(n), then it will imply that all derivatives of order n of f are continuous.

Again, if f C(0), then it will imply that f is a continuous function.(ii) Let V denote a bounded closed region in three dimensional space. Again, let the set of all

boundary points of V be denoted by S. Then, the set of all interior points of V together with the set ofboundary points S is denoted by .V Thus, V V S.A.2 Boundary value problems for Laplace equation

If the given problem satisfies Laplace equation in a bounded region V in three dimensions andalso satisfies the prescribed boundary conditions on the boundary S of the region V, then such aproblem is known as a boundary value problem (which is also written as BVP).

Classifications of boundary value problems for Laplace equationWe discuss three main types of boundary value problems for Laplace equation.

(i) Boundary value problem of first kind or the Dirichlet problem (Meerut 2011)If f is a continuous function prescribed on the boundary S of some finite region V, then the

problem of finding a function u (x, y, z) such that 2 0u within V and u = f on S is known as theinterior Dirichlet problem.

Similarly, if f is a continuous function prescribed on the boundary S of a finite simplyconnected region V, then the problem of finding a function u (x, y, z) which satisfies 2 0u out-side V and is such that u = f on S is known as the exterior Dirichlet problem.

(ii) Boundary value problem of second kind or the Neumann problem (Meerut 2011)If f in a continuous function which is defined uniquely at each point of the boundary S of a finite

region V, then the problem of finding a function u (x, y, z) such that 2 0u within V and its normalderivative /u n = f at every point of S is known the interior Neumann problem.

Similarly, if f is a continuous function prescribed at each point of the (smooth) boundary S of abounded simply connected region V, then the problem of finding a function u(x, y, z) satisfying

2 0u outside V and /u n f at every point of S is known as the exterior Neumann problem(iii) Boundary value problem of third kind or the Churchill problemIf f is continuous function prescribed on the boundary S of a finite region V, then the problem of

finding a function u (x, y, z) such that 2 0u within V and /u n hu f at every point of S(Here, h is constant and 0h ) is known as the interior Churchill problem.

Similarly, if f is a continuous function prescribed at each point of S of a finite region V, then theproblem of finding a function u (x, y, z) such that 2 0u outside V and /u u hu f at everypoint of S is known as the exterior churchill problem.A.3 Green’s identities

(i) Green’s first identity Let S be a closed surface in the (x, y, z) space and let V be the closed region bounded by S in

which F is a vector belonging to C(1) in V and continuous on V. Then, by Gauss divergencetheorem, we have

. ,V S

dV = n dS F F

where dV is an element of volume, dS is an element of surface area and n is the outward drawnnormal to the surface S.

A.1


A.2 Appendix A

Let F = f u, where f is a vector function and u in a scalar function of position. Then, (1) reducesto

( )V

u dV f =su dS n f …(2)

By a vector identity, we have ( )u u u f f f …(3)

From (2) and (3), ( )V

u u dV f f =S

u dS n f

or V

u dV f =S Vu dS u dV n f f …(4)

Replacing f by v in (4), we have

V s V

v u dV u vdS u v dV n

or 2

V s Vv u dV u vdS u v dV n … (5)

Since vn is the derivative of v in the direction of n, it is convenient to introduce the notation/v v n n in (5). Then, (5) reduces to

2 ,V S V

dvv u dV u dS u v dVdn

… (6)

which is known as Green’s first identity.A particular case of Green’s first identity. Setting v = u in (6), we have

2 2( )

V S V

uu dV u dS u udVn

…(7)

(ii) Green’s second identity. Interchanging u and v in (6), we get

2

V S V

uu v dV v dS v u dVn

or 2– ,

V S V

uv u dV v dS v u dVn

…(8)

where we have used the fact that .v u u v Subtracting (8) from (6), we get

o = 2 2( )S V

v uu v dS u v v u dVn n

and hence 2 2( )V

u v v u dV ,S

v uu v dSn n … (9)

which is known as second Green’s identity. Clearly, for (9) to be true, both u and v must possesscontinuous second order derivatives.


APPENDIX B

TWO AND THREE DIMENSIONAL DIRAC DELTA FUNCTIONSB.1. Introduction

We have already introduced the concept of one-dimensional Dirac delta function in chapter 10.In what follows, we propose to extend the definition to two and three dimensions.B.2. Two dimensional Dirac delta function. Definition

Let f (x, y) be an arbitrary continuous function over the region S containing the point (, ).Then, two-dimensional Dirac delta function (x – , y – ) is defined as the function satisfying thefollowing condition

( , ) ( , ) ( , )S

x y f x y dx dy f …(1)

Observe that (x – , y – ) is a formal limit of a sequence of ordinary functions, that is,(x – , y – )

0lim ( )r

... (2)where r2 = (x – )2 + (y – )2.

Again, we have ( ) ( ) ( , ) ( , )S

x y f x y dx dy f …(3)

Comparing (1) and (2), ( , ) ( ) ( ),x y x y …(4)showing that a two-dimensional Dirac delta function can be expressed as the product of two one-dimensional Dirac delta functions.B.3. Three-dimension Dirac delta function. Definition

Let f (x, y, z) be an arbitrary continuous function in the region V containing the point (, , ).Then, three dimensional Dirac delta function (x –y –z –)is defined as the function satisfy-ing the following condition

( , , ) ( , , )V

x y z dx dy dz f …(1)Observe that (x – , y – , z – ) is a formal limit of a sequence of ordinary functions, that is, (x – , y – , z – ) =

0lim ( )r

...(2)where r2 = (x – )2 + (y – )2 + (z – )2

Again, we have ( , , ) ( , , ) ( , , )V

x y z f x y z dx dy dz f …(3)

Comparing (1) and (2), (x – , y – , z – ) = (x – ) (y – ) (z – ),showing that a three-dimensional Dirac Delta function can be expressed as the product of three on-dimensional Dirac delta functions.

Example. Show that the three dimensional Dirac delta function (x, y, z; , , ) can be writtenas (x, y, z; ,,) = (x – ) (y – ) (z – ).

Solution. Suppose that the three dimensional Dirac delta function (x, y, z; , , ) is defined by

( , , ) ( , , ; , , )F x y z d d d

F (x, y, z), …(1)

where F (x, y, z) is continuous function. If (x – ), (y – ) and (z – ) are three one-dimensionalDirac delta functions, then we have

( , , ) ( ) ( ) ( ) ( , , )F x y z d d d F x y z

…(2)

B.1


B.2 Appendix B

Comparing (1) and (2), (x, y, z; , , ) = (x – ) (y – ) (z – )B.4. Dirac delta function in general curvilinear coordinates in two dimensions

Consider a transformation from cartesian coordinates x, y to curvilinear coordinates , satis-fying the relations

x = f (, ) and y = g (, ), ... (1)where f and g are single-valued, continuously differentiable functions of their arguments. Underthe transformation (1), let = b1, = b2 correspond to x = a1, y = a2. Again, we have

/ / ( , ) ,/ / ( , )

dx f f d d df g Jdy g g d d d

where J is the Jacobian of transformation. When we transform the coordinates satisfying (1), thenthe equation

1 2( , ) ( ) ( )F x y x a y a dx dy = F (a1, a2) …(2)

takes the form 1 2 1 2( , ) [ ( , ) ] [ ( , ) ] | | ( , )F f g f a g a J d d F a a …(3)

From (3), it follows that the Dirac delta function [ f (, ) – a1] [ f (, ) – a2] | J | assigns toany test function F ( f, g) the value of that test function at the points where f = a1, g = a2, i.e., at thepoints = b1, = b2. Hence, we obtain

[ f (, ) – a1)] [g (, ) – a2)] | J | = (x – a1) ( y – a2) | J | = ( – b1) ( – b2)Thus, (x – a1) (y – a2) = [ ( – b1) ( – b2)] / | J | …(4)An important particular case : If P (x0, y0) is a point in rectangular cartesian coordinates

corresponding to the point P (r0, 0) in polar coordinates, then show that (x – x0) (y – y0) = [ (r – r0) ( – 0)] / rSolution. In polar coordinates, we have x = r cos and y = r sin . Thus, here take = r and

= . Also, here a1 = x0, a2 = y0, b1 = r0 and b2 = 0 Furthermore, here we have x = f (r, ) = r cos and y = g(r, ) = r sin …(1)

Then, J =/ // /

f r fg r g =

cos sinsin cos

rr

= r cos2 + r sin2 = r, using (1)

From Art B.4, we have (x – a1) (y – a2) = 1 2{ ( ) ( )}/ | |b b J …(2)

Substituting the values of a1 (= x0), a2 (= y0), b1 (= r0), b2 (= 0) J = r and noting that here = r and = , (2) reduces to

(x – x0) (y – y0) = { (r – r0) ( – 0)} / rB.5 Dirac delta function in general curvilinear coordinates in three dimensions

Consider a transformation from cartesian coordinates x, y, z to curvilinear coordinates , , satisfying the relations

x = f (, , ), y = g (, , ) and z = h (, , ), …(1)where f, g and h are single-valued, continuously differentiate functions of their arguments. Underthe transformation (1), let = b1, = b2, = b3 correspond to x = a1, y = a2, z = a3. Again, we have

/ / /( , , )/ / /( , , )

/ / /

dx f f f df g hdy g g g d

dz h h h d

,d dd J dd d


Appendix B B.3

where J is the Jacobian of transformation. When we transform the coordinates satisfying (1), thenthe equation

1 2 3( , , ) ( ) ( ) ( )F x y z x a y a z a dx dy dz = F(a1, a2, a3) … (2)

becomes

1 2 3( , , ) ( , , ) [ ( , , ) ] [ ( , , ) ] | |F f g h f a g a h a J d d d = F (a1, a2,a3) … (3)

Now, as discussed in Art B.4, we have [ f (, , ) – a1] [g (, , ) – a2], [h (, , ) – a3] | J | = (x – a1) (y – a2) (z – a3) | J |

= ( – b1) ( – b2) ( – b3)Thus, (x – a1) (y – a2) (z – a3) = [ ( – b1) ( – b2) ( – b3)]/| J | … (4)Some important particular cases.Case I. If P (x0, y0, z0) is a point in rectangular cartesian coordinates corresponding to the

point P (r0, 0, zx) in cylindrical coordinates, then show that (x – x0) (y – y0) (z – z0) = { (r – r0) ( – 0) (z – z0)} / r

Solution. In cylindrical polar coordinates, we have x = r cos , y = r sin and z = zThus, here take = r, = and = z. Also here a1 = x0, a2 = y0, a3 = z0, b1 = r0, b2 = 0

and b3 = z0. Further, here we havex = f (r, , z) = r cos , y = g (r, , z) = r sin and z = h (r, , z) = z

Then, J =

/ / // / // / /

f r f f zg r g g zh r h h z

=cos sin 0sin cos 0

0 0 1

rr

= r cos2 + r sin2 = r, by (1)

From Art. B.5, we have (x – a1) (y – a2) (y – a3) = { ( – b1)( – b2) (– b3)} / | J | …(2)Substituting the values of a1 (= x0), a2 (= y0), a3(= z0), b1 (= r0), b2 (= 0), b3 (= z0), J = r and

noting that here = r, = and = z, (2) reduce to (x – x0) (y – y0) (z – z0) = {(r – r0) ( – 0) (z – z0)}/rCase II. If P (x0, y0, z0) is a point in rectangular cartesian coordinates corresponding to the

point P (r0, 0, 0) in spherical polar coordinates, then show that (x – x0) (y – y0) (z – z0) = { (r – r0) ( – 0) ( – 0)} / r2 sin Solution. In spherical polar coordinates, x = r sin cos , y = r sin sin , z = r cos

Thus, here take = r, = and = . Also, here a1 = x0, a2 = y0, a3 = z0, b1 = r0, b2 = 0 andb3 = 0. Further, have we havex = f (r, , ) = r sin cos , y = g (r, , ) = r sin sin , z = h (r, , ) = r cos … (1)

Then, J =

/ / / sin cos cos cos – sin sin/ / / sin sin cos sin sin cos ,/ / / cos sin 0

f r f f r rg r g g r rh r h h r

by (1)

= cos (r2 sin cos cos2+ r2 sin cos sin2 ) + r sin (r sin2 cos2 + r sin2 sin2 )= cos × r2 sin cos + r2 sin3 = r2 sin (cos2 + sin2 ) = r2 sin

From Art. B.5, (x – a1) (y – a2) (z – a3) = { ( – b1) (– b2) (– b3)}/| J | … (2)Substituting the values of a1 (= x), a2 (= y0), a3 (= z0), b1 (= r0), b2 (= 0), b3 (= 0), J = r and

noting that here = r, = , = , (2) reduce to(x – x0) (y – y0) (z – z0) = { (r – r0) ( – 0) ( –0)} / r2 sin


APPENDIX C

ADDITIONAL TOPICS AND PROBLEMS BASED ON GREEN’S FUNCTION[Note: The reader is advised to study chapter 12 before reading the matter of this Appendix C.

However, references to this chapter have been given at proper places]C.1 The eigenfunction method for computing Green’s function for the given Dirichletboundary value problem

Let D be a region in the xy-plane bounded by a simple closed curve C. Consider the Dirichletboundary value problem described by

2 2 2 2 2/ / , ,u u x u y f x y D …(1)

subject to the boundary condition : u = g, x, y e C …(2)Now, from the definition of Green’s function (Refer Art. 12.8 of chapter 12), the Green’s function

G (x, y; , ) for the problem given by (1) and (2) must satisfy the following relations :

2 ( , ), ,G x y x y D …(3)

and 0, , ,G x y C … (4)where (x – , y – ) is two-dimensional Dirac delta function (refer Art. B. 2 in Appendix B)

Let us consider the eigen value problem associated with the operator 2 in the domain D, thatis,

2 0, ,x y D … (5)

and 0, ,x y C … (6)

Let mn be the *eigenvalues and mn (x, y) be the corresponding eigenfunctions of the eigen-value problem given by (1) and (2) .

Assume that G (x, y; , ) and (x –, y – ) possess the following Fourier series expansions interms of the eigenfunctions mn (x, y):

G (x, y; , ) = ( , ) ( , )mn mnm n

A x y …(7)

and ( , ) ( , ) ( , )mn mnm m

x y B x y ...(8)

where Bmn = 2( , )1 ( , ) ( , )

|| ( , ) |||| ( , ) ||mn

mnD mnmnx y x y dx dy

x yx y

…(9)

and 2 2|| ( , ) || ( , )mn mnD

x y x y dx dy …(10)

Since mn (x, y) is an eigenfunction corresponding to the eigenvalue mn, hence by definitions ofeigenfunction and eigenvalue, (5) yields

2 ( , ) ( , ) 0mn mn mnx y x y … (11)Substituting the values of G(x, y, , ) and (x – , y – ) given by (7) and (8) in (3) and (4), we have

2 ( , ) ( , ) ( , ) ( , )mn mn mn mnm n m n

A x y B x y

* Refer chapter 15 of Part I in author’s ‘‘Advanced Differential equations’’ published by S. Chand and Co., NewDelhi

C.1


C.2 Appendix C

or 2( , ) ( , ) { ( , ) ( , )}/mn mn mn mnm n m n

A x y x y 2|| ( , ) ||mn x y , using (9)

or – ( , ) ( , ) { ( , ) ( , )}mn mn mn mn mnm n m nA x y x y 2|| ( , ) ||mn x y , using (11)

The above result yields ( , ) –{ ( , )mn mnA 2|| ( , ) || }mn mn x y ...(12)

Substituting the above values of Amn (, ) given by (2) in (7), the required Green’s function ofthe given Dirichlet problem is given by

G (x, y; , ) = –{ ( , ) ( , )}mn mnm nx y 2|| ( , ) ||mn mn x y …(13)

An illustrative example. Find the Green’s function for the Dirichlet problem on the rectangle D: 0 x a, 0 y b described by the partial differential equation 2( ) 0u in D and u = 0 onthe boundary C of D.

Solution. Given boundary value problem is given by

2 2 2 2/ / 0, ,u x u y u x y D … (1)

subject to the boundary conditions: u (x, 0) = u (x, a) = 0, 0 x a … (2a)

and u(0, y) = u(b, y) = 0, 0 y b ...(2b)

Now, *we proceed to find the eigenfunctions and thecorresponding eigenvalues of the boundary value prob-lem given by (1), (2a) and (2b).

Suppose that (1) has solutions of the form u(x, y) = X(x) Y(y), ...(3)

where X is a function x alone and Y that of Y aloneSubstituting this value of u in (1) we haveXY + XY + XY = 0 or X/X = – ( + Y/Y) ... (4)Since x and y are independent variables, (4) is true if each side of (4) is a constant (= – , say)

such that X/ X = –, that is, X + X = 0 … (5)

and –( + Y/Y) = –, that is, Y + ( – ) Y = 0 … (6)Using (2 a), (3) yields X (0) Y (y) = 0 and X (a) Y (y) = 0 ...(7)Since Y (y) = 0 leads to trivial solution u = 0, so we take Y (y) / 0. Then, (7) yields X (0) = 0 and X (a) = 0 … (8)We now proceed to find the eigenfunctions and the corresponding eigenvalues of the boundary

value problem given by (5) and (8). Three cases arise :Case I. Let = 0. Then, solution of (5) is X (x) = Ax + B … (9)Using the boundary conditions (8), (10) yields 0 = B and 0 = Aa + B giving A = B = 0.

Hence X(x) = 0 and so u = 0 which is a trivial solution. So we reject = 0.Case II. Let = –v2, where v 0. Then, solution of (5) is X (x) = A evx + Be–vx, A and B being arbitrary constants … (10)Using the boundary conditions (8), (10) yields 0 = A + B and 0 = Aeav + Be–av. Solving these

equations, we get A = B = 0 so that X (0) = 0 and hence u = 0, which is a trivial solution. Hence, wereject = –v2.

* Refer Chapter 15 of part I in author’s “Advanced Differential equations” published by S. Chand and Co, NewDelhi

y

x

x = 0 x = aD

y = b

y = 00


Appendix C C.3

Case III. Let = v2, where v 0. Then, solution of (5) is X(x) = A cos vx + B sin vx, A and B being arbitrary constants …(11)Using the boundary conditions (8), (11) yields 0 = A and 0 = A cos v a + B sin vaHence, we have A = 0 and sin va = 0,

where we have taken B 0, since A = B = 0 yields u = 0 which is a trivial solution, Now,sin va = 0 va = m, m = 1, 2, 3,… vm = m /a, m = 1, 2, 3,... … (12)Corresponding to eigenvalues vm given by (12), The desired real valued eigenfunctions are

given by Xm = sin (m x / a), m = 1, 2, 3, …Next, using (2 b), (3) yields X (x) Y (0) = 0 and X(x) Y(b) = 0 … (13)Since X (x) = 0 leads to trivial solution u = 0, so we take X (x) 0. Then, (13) yields Y (0) = 0 and Y (b) = 0 … (14)As before, now proceed to find the eigenfunctions and the corresponding eigenvalues of the

boundary value problem given by (6) and (14). Then, we get the eigenfunctions Yn(y) = sin(n y /b), n= 1, 2, 3… corresponding to eigenvalues given by 2 2 2

2( / ), 1,2,3m n b n Let mn (x, y) denote the eigenfunctions corresponding to the eigenvalues mn for the given

boundary value problem given by (1), (2a) and (2b). Then, we havemn (x, y) = Xm(x) Yn (y) = sin (m x/a) sin (n y / b), m = 1, 2, 3, …, n = 1, 2, 3, … …(15)

mn = 2 2 2 2 2 2 2 2 2 2/ / /mv n b m a n b = 2 2 2 2 2( / / ),m a n b m = 1, 2, 3, …, n= 1, 2, 3 …(16)Here, the norm of mn (x, y), i.e., | | mn (x, y) | |, is given by

2|| ( , ) ||mn x y = 2 2

0 0sin sin

a b

x y

m x n y dx dya b

=

2 2

0 0sin sin

a bm x n ydx dya b

=0 0

0

1 2 1 2 1 sin(2 / ) sin(2 / )1 cos 1 cos2 2 4 (2 / ) (2 / )

baa b

c

m x n y m x a n y bdx dy x ya b m a n b

Thus, | | mn (x, y) | |2 = (a / 2) × (b / 2) = ab / 4 … (17)Now, from Art. C.l, the Green’s function G (x, y; , ) is given by

G (x, y; , ) = – 2{ ( , ) ( , )}/ || ( , ) ||mn mn mn mnm nx y x y …(18)

Substituting the values of mn (, ), mn (x, y), mn and || mn (x, y) || with help of (15), (16) and (17)in (18), the required Green’s function G (x, y; , ) for the given Dirichlet problem is given by

G (x, y; , ) = 2 2 2 2 21 1

4 sin ( / )sin ( / ) sin ( / ) sin ( / )( / / )m n

ab m x a n y b m a n bm a n b

C.2 The space form of the wave equation or Helmholtz equation

Consider the wave equation in three dimensions2 2 2 2/ ,u t c u or 2 2 2 2 2 2 2 2 2/ ( / / / )u t c u x u y u z … (1)

Suppose that (1) has solutions of the form u (x, y, z, t) = (x, y, z) T (t), … (2)

where (x, y, z) is function of x, y, z only whereas T(t) is function of t alone. Substituting the abovevalue of u in (1), we have

(d2T / dt2) = c2T 2 2 2 2 2 2( / / / )x y z or T = 2 2c T


C.4 Appendix C

Then, 2 2/ / ,T c T where T = d2T / dt2 ...(3)

Now, the L.H.S. of (3) is a function of t alone whereas the R.H.S. of (3) is function of x, y and z.Hence (3) is true only if its each side is a constant (= –, say). Then, (3) yields

2 / giving 2 0, …(4)

which is known as the space form of the wave equation or Helmholtz equation.C.3 Helmholtz’s Theorem

Let (r) be a solution of the space form of the wave equation 2 2 0k whose partialderivatives of the first and second orders are continuous within the volume V and on the closedsurface S bounding V. Again, let r be the position vector of any point P outside the region V and letr be the position vector of a point P lying in the region V. Then,

| | | |1 ( ) ( )4 | | | |

ik ik

s

e e dSn n

r r r rr rr r r r

=( ), if

0, if V

V

r rr

where n is the outward normal to S.Proof Let u be a solution of the Helmholtz

equation

2 2 0u k u …(1)

in the region V and let all the singularities of ulie outside the closed region V boundedby S as shown in the adjoining figure (i). Consider the singularity solution u of (1) given by

u = {e ik |r – r |} | r – r | ...(2)From second Green’s identity (refer Art. A. 3 in Appendix A), we have

2 2( )

V S

u uu u u u dV u u dSn u … (3)

Since P (r) lies outside the region V, | r – r | 0, and hence from (2), it follows that we can find2u for all r V.

Setting u = and u = in (3), we have

2 2( )V

dV = SdS

n n …(4)

Putting u= in (1) gives 2 = –k2. Again, from (2), =u= | | |ik |e r r r r , Substituting thesevalues in (4), we obtain

| | | |

2 2( )| | | |

ik ik

V

e e k dV

r r r r

r r r r=

| | | | ( )( )| | | |

ik ik

S

e e dSn n

r r r r rrr r r r …(5)

Now, | | | | | |

2 2 2( )| | | | | |

ik ik ike e eik k

r r r r r r

r r r r r r… (6)

Using (6), we see that L.H.S. of (5) vanishes. Hence, (5) reduces to

| | | | ( )( ) 0,

| | | |

ik ik

S

e e r dSn n

r r r rr

r r r r … (7)

which is true when the point P (r) lies outside the surface S as shown in figure (i).

n

r

n

r

S

PO

VP

Figure (i)


Appendix C C.5

We now proceed with the other situation when P (r) lies insidethe surface S as shown in adjoining figure (ii). We draw a small spherewith centre at P (r) and radius . Let R denote the region which isexterior to C and interior to S. Proceeding as before, on applyingsecond Green’s identity to the region R and noting that| r – r | 0, we obtain

| | | | ( )( ) 0| | | |

ik ik

R

e e dSn n

r r r r rrr r r r

or| | | | | | | |( ) ( )( ) ( ) 0

| | | | | | | |

ik ik ik ik

C S

e e e edS dSn n n n

r r r r r r r rr rr rr r r r r r r r

… (18)Note that when the point P(r) lies on the surface of the sphere C, then | r – r | = . The direction

of normals n is as shown in figure (ii).Now, on the surface of the sphere C, we have| r – r | = , (r) = (r) + O (), dS = 2 sin d d

( ) ( ) ( ),P

On n

r r | | | |

21, –

| | | |

ik ik ik ik ikike e e e e e

n

r r r r

r r r r

Using (9), the first on the L.H.S. of (8)

= 2 2

20 0

1 ( )( ) ( ) ( ) sinik

ik

P

ik eO e O d dn

rr

=2

0 0

( ){ ( ) ( )}(1 ) ( ) sinik ik

PO ik e e O d d

n

rr

=2

0 0( ) sin ,d d

r on letting 0

= 200( ) cos [ ] 4 ( ) r r

Substituting the above value of the L.H.S. of (8) in complete equation (8), we obtain

| | | | ( )4 ( ) ( ) 0

| | | |

ik ik

S

e e dSn n

r r r r rr rr r r r

| | | |1 ( ) ( ) ( )

4 | | | |

ik ik

S

e e dSn n

r r r rr r rr r r r … (10)

Now, on combining the results (7) and (10), we have

| | | | ( ), if 1 ( ) ( )

0, if 4 | | | |

ik ik

S

Ve e dSVn n

r r r r r rr rrr r r r … (11)

C.4 Application of Green’s function in determining the solution of the waveequation.

Consider the space form of the wave equation, namely,

2 2 0k ...(1)

n

n

nP

SR

r

r PO

Figure (ii)

...(9)


C.6 Appendix C

In what follows, we propose to show how the solution of (1) under certain boundary conditionscan be made to depend on the determination of an appropriate Green’s function G (r, r).

Let G (r, r) be a Green’s function such that

(i) G (r, r) satisfies (1), i.e., 2 2( , ) ( , ) 0G k G r r r r … (2)

(ii) G(r, r) is finite and continuous with respect to either the variables x, y, z or to the variablesx, y, z for the points r, r belonging to a region V which is bounded by a closed surface S except inthe neighbourhood of the point r, where it has a singularity of the same type as{eik | r – r|}/ | r – r | as r r.

Now, proceeding as in the derivation of equation (10) of Art. C.3, we can prove that, if r is theposition vector of a point within V, then

1 ( ) ( , )( ) ( , ) ( ) ,

4 S

GG dSn n

r r rr r r r … (3)

where n is the outward-drawn normal to the surface S.We now proceed to discuss two important particular cases of the above result (3).Particular Case 1: When G (r, r) = G1 (r, r), where G1 (r, r) satisfies the boundary condition

G1 (r, r) = 0 for r S, then (3) takes the form

1( , )1( ) ( )

4 S

G dSn

r rr r … (4)

Using formula (4), the value of at any point r within S can be calculated in terms of the valuesof on the boundary S

Particular case 2 : When G (r, r1 = G2 (r, r), where G2(r, r) satisfies the boundary condition

2 ( , ) / 0G n r r for r S, then (3) takes the form

21 ( )( ) ( , )

4 SG dS

n

rr r r … (5)

Using formula (5), the value of at any point r within S can be calculated in term of the value of/ n on the boundary S.

C.5 Determination of the Green’s function for the Helmholtz equation for thehalf-space z 0 (Meerut 2006)

In the given problem, the boundary is the xy-plane.Let be the position vector of the imageP (x, y, –z) in the xy-plane of the point P (x, y, z) with po-sition vector r. Again, let r be the position vector of anypoint Q (x, y, z) of the half-space z 0. When Q lies onthe boundary, then clearly, QP = QP so that| r – r | = | – r |. Thus, for the present problem, we takeG1 (r, r ) = {eik | r – r |}/ | r – r |

– {e ik | – r |} / | – r| … (1)We now proceed to verify that G1 (r, r) satisfies

the Helmholtz equation 2 2 0.k Thus, we wishto prove that

2 2 2 2 2 2( / / / )x y z G1 (r, r) + k2 G1 (r, r) = 0 … (2)

O

Z

X

r

P(x, y, z)

Q(x, y, z ) r

Y

P(x, y, –z)


Appendix C C.7

Let R1 = | r – r | = {(x – x)2 + (y – y)2 + (z – z)2}1/2 … (3) R2 = | – r | = {(x – x)2 + (y – y)2 + (z + z)2}1/2 … (4)Using (3) and (4), (1) reduce to G1(r, r) = (1 / R1) × eikR1 – (1 / R2) × eikR2 … (5)

From (5), 1 1/ ( ) /R x x x R and 2 2/ ( ) /R x x x R … (6)

Differentiating (5) partially w.r.to x, we have

1 1 2 21 1 1 2 22 2

1 21 2

( , ) 1 1–ikR ikR ikR ikRG R R R Rik ike e e ex R x x R x xR R

r r

= 1 22 2

1 1 2 21 2

1 1 ,ikR ikRik x x ik x xe eR R R RR R

using (6)

Thus, 1 212 3 2 31 1 2 2

( , ) 1 1( ) ( )ikR ikRG ik ike x x e x xx R R R R

r r… (7)

Now, differentiating (7) partially w.r.to x, we have2 1

12

( , )Gx

r r= 1 11 1

2 3 3 41 1 1 1

1 2 3( ) ( )ikR ikRR Rik ikik e x x e x xx xR R R R

+ 1 2 22 3 2 31 1 2 2

1 1 ( )ikR ikR Rik ike ik e x xxR R R R

2 223 4 2 32 2 2 2

2 3 1( )ikR ikRRik ike x x exR R R R

or 2

12

( , )Gx

r r=

12 2

2 3 3 4 2 31 11 1 1 1 1 1

1 ( ) 2 3 ( ) 1 ikRik x x ik x x ikik eR RR R R R R R

22 2

2 3 3 4 2 32 22 2 1 1 2 2

1 ( ) 2 3 ( ) 1 ikRik x x ik x x ikik eR RR R R R R R

… (8)

Similarly, 2

22

( , )Gy

r r= 1

2 2

2 3 3 4 2 31 11 1 1 1 1 1

1 ( ) 2 3 ( ) 1 ikRik y y ik y y ikik eR RR R R R R R

+ 22 2

2 3 3 4 2 32 22 2 2 2 2 2

1 ( ) 2 3 ( ) 1 ikRik y y ik y y ikik eR RR R R R R R

… (9)

Now, from (3) and (4), 1 1/ ( ) /R z z z R and 2 2/ ( ) /R z z z R … (10)Differentiating (5) partially w.r.to z, we have

1 1 2 21

1 1 1 2 22 2

1 21 2

( , ) 1 1ikR ikR ikR ikRG R R R Rik ike e e ez R z z R z zR R

r r

or 1 212 3 2 31 1 2 2

( , ) 1 1( ) ( ),ikR ikRG ik ike z z e z zz R R R R

r r using (10) … (11)


C.8 Appendix C

Now, differentiating (11) partially w.r.t z and simplifying, we get2

12

( , )Gz

r r= 1

2 2

2 3 3 4 2 31 11 1 1 1 1 1

1 ( ) 2 3 ( ) 1 ikRik z z ik z z ikik eR RR R R R R R

22 2

2 3 3 4 2 32 22 2 2 2 2 2

1 ( ) 2 3 ( ) 1 ikRik z z ik z z ikik eR RR R R R R R

… (12)

Now, 2 2 2 2 2 2 2 2 2 21 1 1 1( / / / ) ( , ) ( , ) / ( , ) / ( , ) /x y z G G x G y G z r r r r r r r r

=2 2 2

2 311 1

1 ( ) ( ) ( )ik x x y y z zikRR R

3 41 1

2 3ikR R

12 2 2

2 31 1 1

( ) ( ) ( ) 13 ikRx x y y z z ik eR R R

– 2 2 2

2 322 2

1 ( ) ( ) ( )ik x x y y z zikRR R

3 42 2

2 3– ikR R

22 2 2

2 32 2 2

( ) ( ) ( ) 13 ikRx x y y z z ik eR R R

[Using (8), (9) and (12)]

=2 21 1

2 3 3 4 2 31 11 1 1 1 1 1

1 2 3 13R Rik ik ikikR RR R R R R R

1ikRe

22

2 322 2

1 RikikRR R

22

3 4 2 322 2 2 2

2 3 13Rik ikRR R R R

eikR2, using (3) and (4)

= 1 221 2{(1/ ) (1/ ) }ikR ikRk R e R e = –k2G1 (r, r), using (5)

Thus, 2 2 2 2 2 2( / / / )x y z G1 (r, r) + k2G, (r, r) = 0, i.e., 2 21 1( , ) ( , ) 0,G k G r r r r

showing that G1 (r, r) satisfies the Helmholtz equation.Again, as already discussed, when Q(r) lies on the boundary S of the half-space z 0, we have

|r – r| = | – r | and hence, we have G1(r, r) = 0 for r SThen, from particular case 1 of Art. C.4, we have

(r) = 1( , )1 ( ) ,2 S

G dSn

r rr … (13)

where we have taken 2 in place of 4 in the result of Art C.4. for the present problem.Here bounding surface S is the xy-plane and so here, we have dS= dx dy … (14)

Also, 1( , )Gn

r r

= 1

0

( , )

z

Gz

r r= 1 2

2 3 2 31 1 2 2

1 1ikR ikRik ike z e zR R R R

, using (11)

= 22 1 ,

ikRz e ikRR

as on boundary xy-plane, R1 = R2 = R, say

Thus, 1( , ) 2ikRG e

n z R

r r … (15)


Appendix C C.9

Let f (x, y) be the value of on the boundary xy-plane, Then, using (14) and (15), we obtain

(r) =1 ( , )

2

ikR

x y

ef x y dx dyz R

or (r) = 1 ( , ) ,2

ikR

x y

ef x y dx dyz R

where z > 0 … (16)

Re-wring (16), 1( , , ) ( , )

2

ihR

x y

ex y z f x y dx dyz R

Exercise. Find the Green’s function for the Helmholtz equation for the half plane x 0 andhence solve the Helmholtz equation.

Hint. Proceed as in Art C.5. Then, solution of the Helmholtz equation is given by

(x, y, z) = (r) =– –

1 ( , )2

ikR

y z

ef y z dy dzx R

where = f (y, z) on the boundary plane x = 0C.6. Solution of one-dimensional wave equation using the Green’s functiontechnique

Consider one-dimensional wave equation

2 2 2 2 2/ (1/ ) ( / ), 0 , 0,u x c u t x a t … (1)

where u (x, t ) is the deflection of the string. We now solve (1) with help of *method of separation ofvariables (or product method) under the following boundary and initial conditions :

Boundary conditions : u (o, t) = u (a, t) = 0, t 0 … (2)Initial conditions : u (x, 0) = f (x), 0 x a … (3a)

ut(x, 0) = 0( / ) tu t = g(x), 0 x a … (3b)

Suppose that (1) has solutions of the form u (x, t) = X (x) T (t) … (4)where X (x) is function of x alone and T (t) is function of t alone.

Substituting the value of u given by (4) in (1) we haveXT = (1/c2) × XT or X /X = T /c2T … (5)Since x and t are independent variables, (5) can be true if each side is equal to the same constant,

say . Then, (5) yieldsX – X = 0 … (6)

and T – c2T = 0 … (7)Using (2), (4) yields X (0) T(t) = 0 and X(a) T(t) = 0 … (8)Since T (t) = 0 yields to u = 0, so we suppose that T (t) / 0. Then, (8) gives X (0) = 0 and X (a) = 0 … (9)We now solve (6) under the boundary conditions (9). Three cases arise:Case I. Let = 0, Then, solution of (6) is X(x) = Ax + B, … (10)

where A and B are arbitrary constants. Using the boundary conditions (9), (10) gives 0 = B and0 = Aa + B. These equations give A = B = 0 and so X (x) = 0. This leads to u = 0 which does not satisfythe initial conditions (3a) and (3b). So we reject = 0

* Refer Art. 1.9 and Art 2.10 A in part III of author’s ‘‘Advanced Differential equations’’, publised by S. Chandand Co., New Delhi


C.10 Appendix C

Case II. Let = 2, where 0. Then, solution of (6) is X(x) = Aex + Be–x, A and B being arbitrary constants … (11)Using the boundary conditions (9), (11) gives 0 = A + B and 0 = Aea + Be–a … (12)Solving (12), A = B = 0 so that X(x) = 0. This leads to u = 0, which does not satisfy (3a) and (3b).

So we reject = 2.Case III. Let = – 2, where 0. Then, solution of (1) is given by X(x) = A cos x + B sin x, A and B being arbitrary constants … (13)Using the boundary conditions (9), (3) yields 0 = A and 0 = A cos a + B sin a from which we

have A = 0 and sin a = 0,where we have taken B 0, since otherwise X (x) = 0 and so u = 0 which does not satisfy (3 a) and (3b).

Now, sin a = 0 a = n = n / a, n = 1, 2, 3, … … (14)

Hence non-zero solution Xn (x) of (10) are given by Xn(x) = Bn sin ( / )n x a … (15)

Using (14), we have = –2 = – (n22) / a2. Hence, (7) reduces to T + (n22c2 / a2 ) T = 0, whose general solution is given by

Tn(t) = Cn cos (nct/a) + Dn sin(nct/a), Cn and Dn being arbitrary constants … (16)Thus, un (x, t) = Xn (x) Tn(t) = {En cos (n ct/a) + Fn sin (nct/a)} sin (n x/a),

are solutions of (1) satisfying the boundary condition (2). Here En (= Cn Bn) and Fn (= Bn Dn) are newarbitrary constants. In order to obtain a solution also satisfying the initial conditions (3a) and (3b),we consider more general solution u (x, t) by the method of superposition

u (x, t) =1

( , )nn

u x t

or u (x, t) =

1

cos sin sinn nn

n ct n ct n xE Fa a a

… (17)

Differentiating (17) partially w.r. ‘t’, we have

1

sin cos sinn n

n

n cE n cFu n ct n ct n xt a a a a a

… (18)

Putting t = 0 in (17) and (18) and using initial conditions (3a) and (3b), we get

f (x) =1

sin( / )nn

E n x a

and g (x) =1

( / )sin( / ),nn

n cF a n x a

… (19)

which are Fourier sine series expansions for f (x) and g (x) respectively. Hence, we get

En =0

2 ( ) sina n xf x dx

a a

… (20)

and 0

2 ( ) sinann cF n xg x dx

a a a

so that Fn =2

n c 0( ) sin

a n xg x dxa

… (21)

Thus, the general solution of (1) is given by (17) and En and Fn are given by (20) and (21).

From (20) and (21), 0

2 ( )sina

nnE f d

a a

and 0

2 ( )sina

nnF g d

n c a

...(22)

Substituting the values of En and Fn given by (22) in (17), we obtain

0 01

2 2( , ) ( )sin cos ( )sin sin sina a

n

n n ct n n ct n xu x t f d g da a a n c a a a


Appendix C C.11

Interchanging the operations of summation and integration, the above equation reduces to

0 01 1

2 2 1( , ) sin sin cos ( ) sin sin sin ( )a a

n n

n x n n ct n x n n ctu x t f d g da a a a c n a a a

... (23)

We now define Green’s function ( , ; , )G x t as follows:

1

2 1 ( – )( , ; , ) sin sin sinn

n x n n c tG x tc n a a a

... (24)

It can be easily verified that the above series converges for all values of x, t, and .

From (24), 1

2 ( – )( , ; , ) ( , ; , ) / sin sin cost n

n x n n c tG x t G x t ta a a a

...(25)

Putting = 0 in (24) and (25), we have

1 1

2 1 2( , ; ,0) sin sin sin , ( , ; , 0) sin sin costn n

n x n n ct n x n n ctG x t G x tc n a a a a a a a

Using the above results, the series solution (17) of the given initial boundary value problemcan be re-written in terms of Green’s function as

0 0

( , ) ( , ; , 0) ( ) ( , ; , 0) ( )a a

tu x t G x t f d G x t g d ...(26)

Observe that ( , ; , )G x t as a function of x satisfies the boundary conditions. That is, we have

G (o, t; , ) = 0 and G (a, t; , ) = 0, t 0Further, G (x, t; , ) = 0, 0 x a

G (, t; x, ) = G (x, t; , ) for all x and G (x, ; , t) = – G (x, t; , ) for all t and

The function G (x, t; , ) defined by (24) is known as the Green’s function of the given initialboundary value problem given by (1), (2), (3a) and (3b).C.7. Solution of one-dimensional inhomogeneous wave equation using the Green’sfunction technique

In many physical problems we obtain an inhomogeneous wave equation when there is a forcefunction operating on the medium. Suppose that an external force density (force per unit length ofthe vibrating string) f (x, t) is applied to the string which is independent of the deflection u(x, t) ofthe vibrating string. In such a situation we arrive at an equation containing a source term. Thisequation is said to be an inhomogeneous wave equation.

Consider one-dimensional inhomogeneous wave equation

2 2 2 2 2/ ( / ) ( , ), 0 , 0u t c u x f x t x a t ... (1)

where u(x, t) represents the deflection of the string. We now solve (1) under the following boundaryand initial conditions:

Boundary conditions: u(0, t) = u (a, t), t > 0 ...(2)Initial conditions : u (x, 0) = g(x), 0 x a ...(3a)

and ut(x, 0) = h(x), 0 x a ...(3b)


C.12 Appendix C

We now proceed to express the solution of the above problem as Fourier sine series is x, that is,

1

( , ) ( ) sin( / ),nn

u x t u t n x a

... (4)

where we have taken undetermined functions un(t) of time as multiplying constants. We have se-lected the above form for u(x, t), since this form automatically satisfies both the boundary conditionsgiven by (2). We now proceed to determine the unknown functions un(t) in such a manner so that theresulting value of u(x, t) given by (4) may satisfy (1) and initial conditions (3a) and (3b). To this end,we assume the Fourier sine series expansions of f(x, t), g(x) and h(x) given by

1

( , ) ( ) sin( / ),nnf x t f t n x a

where ...(5a)

0

2( ) ( , ) sin( / )a

nf t f t n a da

...(5b)

1

( ) sin( / ),nng x g n x a

where ...(6a)

0

2 ( ) sin ( / )a

ng g n a da

...(6b)

1

( ) sin( / ),nnh x h n x a

where ...(7a)

0

2 ( ) sin( / )a

nh h n a da

...(7b)

From (4), 2

21 1

2 2 2

2 21 1

( ) sin , ( ) sin ,

( ) cos and – ( ) sin

n nn n

n nn n

u n x u n xu t u tt a at

u n n x u n n xu t u tx a a ax a

... (8)

Substituting the values of f(x, t), 2 2/u x and 2 2/u t given by (5a) and (8) in (1), we get

2 2 2

21 1 1

( ) sin – ( ) sin ( ) sinn n nn n n

n x n c n x n xu t u t f ta a aa

or 2 2 2

21sin ( ) ( ) – ( ) 0n n n

n

n x n cu t u t f ta a

... (9)

Since sin ( / )n x a are linearly independent functions, (9) is satisfied only if, the expression in

parenthesis vanishes ,n N that is,

2 2 2 2( ) ( / ) ( ) – ( ) 0n n nu t n c a u t f t or 2 2 2 2( ) ( / ) ( ) ( ),n n nu t n c a u t f t ...(10)

which is a second order differential equation whose general solution will contain two arbitraryconstants. In order to determine these two arbitrary constants, we require two initial conditions.These conditions can be determined with the help of the initial conditions (3a) and (3b).


Appendix C C.13

Putting t = 0 in (4) and using, the initial condition (3a), we have

1( ) (0) sinn

n

n xg x ua

or

1 1sin (0) sin ,n n

n n

n x n xg ua a

using (6a) ... (11)

Comparing the coefficients of sin( / ),n x a (11) (0)n nu g ...(12)

Differentiating both sides of (4) partially w.r.t. ‘t’, we have

1

( , ) ( , ) ( ) sint nn

u x t n xu x t u tt a

...(13)

Putting t = 0 in (13) and using the initial condition (3b), we have

1( ) (0) sinn

n

n xh x ua

or 1 1

sin (0) sin ,n nn n

n x n xh ua a

using (7a) ...(14)

Comparing the coefficients of sin( / ),n x a (14) (0)n nu h ...(15)

Thus, the original problem of solving partial differential equation (1) with initial conditions (3a)and (3b) has been reduced to that of solving an ordinary differential equation (10) with initial condi-tions (12) and (15).

We now propose to use the superposition principle. Accordingly, we adopt the following proce-dure to solve (10) subject to the initial conditions (12) and (15). We now split the solution un(t) intotwo parts bn(t) and cn(t), i.e., we assume that

( ) ( ) ( ),n n nu t b t c t ... (16)

where bn(t) is the solution of 2 2 2 2( ) ( / ) ( ) ( ),n n nb t n c a b t f t ...(17)

subject to the initial conditions : (0) (0) 0n nb b ...(18)

while ( )nc t is the solution of 2 2 2 2( ) ( / ) ( ) 0,n nc t n c a c t ... (19)

subject to the initial conditions: (0) (0) 0n nc c ... (20)

We now proceed to determine the proposed values of bn(t) and cn(t) one by one:Determination of solution bn(t) of initial value problem given by (17) and (18)In what follows, the results of Art. 9.2 of chapter 9 will be used.Taking Laplace transform of both sides of (17), we obtain

2 2 2 2( ) ( / ) { ( )} { ( )}n n nL b t n c a L b t L f t

or 2 2{ ( )}– (0) – (0) ( / ) { ( )} { ( )}n n n n np L b t pb b n c a L b t L f t , using result 8 of page 9.2

or 2 2{ ( / ) { ( )} { ( )},n np n c a L b t L f t using (18)

or 2 21 sin( / ){ ( )} { ( )} { ( )}

( / )( / )n n nn ct aL b t L f t L L f t

n c ap n c a

[Using result 5 of table given on page 9.3]


C.14 Appendix C

Using the convolution theorem (refer result (24) on page 9.4), the above equation yields

0 0

sin{ ( – ) / } ( – )( ) ( ) sin ( )( / )

t tn n n

n c t a a n c tb t f d f dn c a n c a

...(21)

Determination of solution cn(t) of initial value problem given by (19) and (20)

Let / .D d dt Then, (20) becomes 2 2 2 2 2{ ( / )} ( ) 0,nD n c a c t ... (22)

whose auxiliary equation is 2 2 2 2 2/ 0D n c a giving D = ± i(nc/a).

Hence, the general solution of (22) is given by

( ) cos( / ) sin( / ),nc t A n ct a B n ct a A and B being arbitrary constants ...(23)

From (23), ( ) – ( / ) sin( / ) ( / ) cos( / )nc t An c a n ct a Bn c a n ct a ...(24)

Putting t = 0 in (23) and (24) and using initial conditions given by (20), we get nA g and

( / ).nB h a n c Hence, (23) reduces to

( ) cos( / ) ( / )sin( / )n n nc t g n ct a h a n c n ct a ... (25)

From (4) and (16), 1 1

( , ) ( ) sin ( ) cosn nn n

n x n xu x t b t c ta a

... (26)

Substituting the values of ( )nb t and ( )nc t given by (21) and (25) in (26), we have

01 1

( – )( , ) sin sin ( ) cos sin sint n

n nn n

h aa n c t n x n ct n ct n xu x t f d gn c a a a n c a a

... (27)

Let 1( , )u x t and 2 ( , )u x t denote respectively the first and second terms on R.H.S. of (27). Thus,

1 01

( – )( , ) sin sin ( )t

nn

a n c t n xu x t f dn c a a

... (28)

and 21

( , ) cos sin sinnn

n

h an ct n ct n xu x t ga n c a a

... (29)

Thus, u1(x, t) represents the forced vibrations of the string under the influence of an externalforce whereas u2(x, t) represents the solution of the problem of freely vibrating string with the giveninitial conditions.

Re-writing (5b), we have 0

2( ) ( , ) sina

nnf f d

a a

... (30)

Substituting the value of ( )nf given by (30) in (28), we have

1 0 01

2 ( – )( , ) sin sin sin ( , )t l

n

a n c t n x nu x t f d da n c a a a

... (31)

We now define the Green’s function as

1

2 1 ( – )( , , – ) sin sin sinn

n c t n x nG x tc n a a a

... (32)

Then, (31) reduces to 1 0 0( , ) ( , , – ) ( , )

t au x t G x t f d ... (33)


Appendix C C.15

Physical interpretation of the Green’s function G(x, , t – )

Recall that ( , )f x t represents the external force per unit length of the vibrating string, beingthe mass per unit length of the string. In what follows, suppose that the external force f(x, t) is nonzero only in the small neighbourhood of the point 0 and also during the small time interval around the time point 0 and zero otherwise. Then, the force F() acting on the string in the interval is given by

0

0

/2

– /2( ) ( , )F f d

... (34)

Let I denote the impulse due to the concentrated force F(). Recall that impulse I of this localisedforce may be obtained by an integral of the force with time. Since the force is non zero only over ashort duration of time we have

0 0 0

0 0 0

/2 /2 /2

– /2 – /2 – /2( ) ( , ) ,I F d d f d

using (34) ... (35)

Since u1(x, t) represents the part of the solution of (1) due to the external force, hence thedisplacement u1(x, t) due to impulse I is given by (33).

Recall that we have already assumed that the external force f(x, t) is non zero only in the smallneighbourhood of the point 0 and also during the small time interval around the timepoint 0. Again, note that the function ( , , – )G x t is a well defined function in the neighbourhoodof the point (0, 0). In view of these two facts, G(x, 0, t – 0) can be taken outside the integral for verysmall values of and by applying the mean value theorem. Thus, using mean value theorem,(33) yields approximate displacement given by

0 0

0 0

/2 /21 0 0 0 0– /2 – /2( , ) ( , , – ) ( , ) ( , , – ) ,Iu x t G x t f d d G x t

using (35) ... (36)

Hence, it follows that the Green’s function 0 0( , , – )G x t can be interpreted as the displace-ment at x at time t due to a unit impulse (measured in the units of ) applied at the point 0 at the time0. In this context, note that 0 0( , , – ) 0G x t for 0 ,t because there cannot be any displace-ment even before the external force is applied.

From the above discussion it is obvious that the effect of a disturbed force of density ( , )f x tper unit length of the string can be obtained by integrating the small displacements due to appliedforce over the entire length of the string and over the entire duration of the applied force, that is,

1 0 0( , ) ( , , – ) ( , ) ,

t au x t G x t f d d ... (37)

which is exactly the equation (33).An important particular case. Fundamental solution of the given wave equation.Let f(x, t) the a two-dimensional Dirac delta function, i.e., let f(x, t) = (x – 0) (t – 0)

so that f(, ) = ( – 0) ( – 0) ... (38)Substituting the above value of f(, ) given by (38) in (33), we get

0 0 0 00 0( , ) ( , , – ) ( – ) ( – ) ( , , – ),

t au x t G x t d d G x t [Using result (3) of Art. B.2 of Appendix B]


C.16 Appendix C

showing that the Green’s function can be regarded as a solution of the given equation (1) for the twodimensional Dirac delta source function. In other words, the Green’s function can be treated assolution of the following initial boundary value problem

2 2 2 2 2/ ( / ) ( – ) ( – )u t c u x x t

with boundary conditions u(0, t) = u(a, t) = 0and initial conditions u(x, 0) = ut(x, 0) = 0, for 0 < x < a

In view of the above fact, Green’s function is also known as the fundamental solution of thegiven wave equation for the given boundary conditions. The advantage of finding the fundamentalsolution or the Green’s function for a particular set of boundary conditions lies in the fact thatsolution of the inhomogeneous equation for any source function f(x, t) can be found by simplyevaluating the integral of type involved in (33), and not solve the problem for a different source. Inthis context, we must make sure that the new problem under consideration must have the samegeometry and the same boundary conditions, because the Green’s function is determined for aparticular geometry of the problem and the boundary conditions.

An illustrative solved example. Solve the following inhomogeneous wave equation with a timedependent external force:

2 ( ) cos , 0 , 0tt xxu c u g x kt x a t with homogenous initial and boundary conditions,

namely, (0, ) ( , ) 0, 0; ( , 0) ( , 0) 0tu t u a t t u x u x , 0 .x a Solve for u(x, t) assuming

( / ) .k n c a n N

Solution. Given 2 2 2 2 2/ ( / ) ( ) cos , 0 , 0u t c u x g x kt x a t ... (1)

with boundary conditions: (0, ) ( , ) 0, 0u t u a t t ... (2)

initial conditions: ( , 0) 0, 0u x x a ... (3a)

( , 0) 0, 0tu x x a ... (3b)

Let the required solution be expressed as a Fourier sine series in x, that is,

1

( , ) ( ) sin ,nn

n xu x t u ta

... (4)

where we have taken undetermined functions un(t) of time as multiplying constants. We have se-lected the above form for u(x, t) because this form automatically satisfies both the prescribed bound-ary conditions (2). We now proceed to determine un(t) so that the resulting solution u(x, t) given by(4) may also satisfy (1) and the prescribed initial conditions. To this end we assume the Fourier sineseries expansion of the external force ( , ) ( ( ) cos )f x t g x kt given by

1

( , ) ( ) cos ( ) sinnn

n xf x t g x kt f ta

... (5)

where 0 0

2 2 cos( ) ( ) cos sin ( )sin cosa a

n nn kt nf t g kt d g d g kt

a a a a

... (6)

where 0

2 ( ) sina

nng g d

a a

... (7)


Appendix C C.17

From (4), 2

21 1

2 2 2

2 21 1

( ) sin , ( ) sin ,

( ) cos – ( ) sin

n nn n

n nn n


u n n x u n n xu t and u tx a a ax a

... (8)

Substituting the values of g(x) cos kt, 2 2/u t and 2 2/u x given by (5) and (8) in (1), we get

2 2 2

21 1 1

( ) sin – ( ) sin ( ) sinn n nn n n

n x n c n x n xu t u t f ta a aa

or 2 2 2

21sin ( ) ( ) – cos 0,n n n

n

n x n cu t u t g kta a

using (6) ... (9)

Since sin( / )n x a are linearly independent functions, (9) is satisfied only if, the expression in

parenthesis vanishes n N, that is,

2 2 2 2( ) ( / ) ( ) – cos 0n n nu t n c a u t g kt or 2( ) ( / ) ( ) cosn n nu t n c a u t g kt ... (10)

Let / .D d dt Then, (4) becomes 2 2{ ( / ) } ( ) cosn nD n c a u t g kt ... (11)

The auxiliary equation of (11) is 2 2( / ) 0D n c a giving ( / )D n c a

Hence, C.F. of (11) = 1 2cos( / ) sin( / ),c n ct a c n ct a c1 and c2 being arbitrary constants.

Particular integral (P.I.) or particular solution of (11) is given by

2 21. .

( / )P I

D n c a

2 2

1cos g cos .( / ) –n ng kt ktn c a k

Hence, the general solution of (11) is given by un(t) = C.F. + P.I., i.e.,

2 2cos( ) cos sin

( / ) –n

n n ng ktn ct n ctu t C D

a a n c a k ... (12)

In order to determine c1 and c2, we require two initial conditions. This can be done by usinginitial conditions (3a) and (3b) as follows.

Putting t = 0 in (4) and using the initial condition (3a), we have

1

0 (0) sin( / )nn

u n x a

giving (0) 0nu ... (13)

From (4), 1

( , ) / ( ) sin( / )t nn

u x t u t u t n x a

... (14)

Putting t = 0 in (14) and using the initial condition (3b), we have

1

0 (0) sin( / )nn

u n x a

giving (0) 0nu ... (15)

From (12), 2 2sin

( ) – sin cos –( / ) –

n n nn

C n c D n c g k ktn ct n ctu ta a a a n c a k

... (16)


C.18 Appendix C

Putting t = 0 in (16) and using initial condition (15), we get Dn = 0.Next, putting t = 0 and Dn= 0 in (12) and using initial condition (12), we get

2 20 /{( / ) – }n nC g n c a k so that 2 2– / {( / ) – }n nC g n c a k

Substituting the above values of Cn and Dn in (12), we have

2 2( ) cos – cos ,( / ) –

nn

g n ctu t ktan c a k ... (17)

where gn is the nth Fourier coefficient of g(x) given by (7).C.8. Solution of one-dimensional heat equation using the Green’s function technique.

Consider the heat flow problem in a finite rod of length a described by the partial differentialequation

2 2/ ( / ), 0 , 0u t k u x x a t ... (1)

where /k K is the thermal diffusivity of the material of the rod. Also, K, , are respectively,,thermal conductivity, density and specific heat of the material of the rodwith boundary conditions: u (0, t) = u (a, t) = 0, t > 0 … (2)and initial condition: u (x ,0) = f (x), 0 x a … (3)

We now proceed to solve the above problem with help of *method of separation (or productmethod).

Suppose that (1) has solutions of the form u (x, t) = X (x) T (t), … (4)where X (x) is a function of x alone and T (t) that of t alone. Substituting this value of u in (1) gives

XT = kX T so that X /X = T /kT … (5)Since x and t are independent variables, (5) can be true only if each side is equal to the same

constant, say µ. Hence, (5) leads to the following two ordinary differential equations: X – X = 0 … (6)and = kT … (7)

Using (2), (4) gives X(0) T(t) = 0 and X(a) T (t) = 0 … (8)

Since T(t) = 0 leads to u = 0 which does not satisfy (3). Hence, we take T(t) | 0. Then, by (8) X (0) = 0 and X (a) = 0 … (9)We now solve (6) under the boundary conditions (9). Three cases arise:Case I: Let = 0. Then, solution of (6) is X(x) = Ax + B … (10)Using the boundary conditions (9), (10) yields 0 = B and 0 = Aa + B. Solving these equations, we

get A = B = 0. Hence X (x) = 0 and so u = 0 which does not satisfy (3). So we reject = 0.Case II: Let = 2, where 0. Then, the general solution of (6) is X (x) = Aex + Be–x, A and B being arbitrary constants … (11)Using the boundary conditions (9), (11) yields 0 =A + B and 0 =Ae + Be–a. Solving these equations,

we get A = B = 0. Hence X (x) = 0 and so u = 0 which does not satisfy (3). So reject = 0.Case III: Let = –2, where 0. Then, the solution of (6) is X(x) = Acos x + B sin x, A and B being arbitrary constants … (12)

* Refer Art. 2.3 A in part III in author’s “Advanced Differential Equations”, published by S. Chand and Co.,New Delhi


Appendix C C.19

Using the boundary conditions (9), (12) yields A = 0 and 0 = A cos a + B sin aso that A = 0 and sin a = 0, … (13)which is obtained by assuming that B 0 because if A = B = 0, then X (x) = 0 and so u = 0 which doesnot satisfy (3).

Now, sin a = 0 a = n, n = 1, 2, 3 … = n/a, n = 1, 2, 3 … (14)Hence nonzero solutions Xn(x) of (6) are given by … (15) Xn (x) = Bn sin (n x / a), n = 1, 2, 3, …

Using (14), (7) yields 2 2

2 ,dT n k dtT a

since µ = –2 =

2 2

2na

whose general solution is given by

2 2 2( / ) ,n k a tn nT C e n = 1, 2, 3, … … (16)

From (15) and (16), un (x, t) = Xn(t) Tn (t) = En sin (nx / a) 2 2 2( / ) ,n k a te n = 1, 2, 3, …

are solutions of (1), satisfying the boundary conditions (2). Here En (= BnCn) are new arbitraryconstants.

Using the principle of superposition, the general solution of (1) is given by

2 2 2( / )

1 1

( , ) ( , ) sin( / ) n k a tn n

n n

u x t u x t E n x a e

… (17)

The series represented by (17) can be shown to be uniformly convergent, and it is termwisedifferentiable.

Substituting t = 0 in (17) and using the initial condition (3), we have

f (x) =1

sin( / ),nn

E n x a

which is Fourier sine series. Hence, the constants En are given by

0 0

2 2( ) sin ( )sin , 1, 2, 3,a a

nn x n yE f x dx f y dy n

a a a a

… (18)

Substituting the above value of En in (17), we obtain the following representation of the solutionof (1)

2 2 2– ( / )

01

2( , ) ( ) sin sina n k a t

n

n y n xu x t f y dy ea a a

or u (x, t) =2 2 2( / )

01

2 sin sin ( )a n k a t

n

n x n ye f y dya a a

… (19)

Let us define a function G (x, y, t) as follows :

2 2 2( / )

1

2( , , ) sin sin , 0 , 0 , 0n k a t

n

n x n yG x y t e x a y a ta a a

… (20)

Using (20), the solution of the given boundary value problem is given by

u (x, t) =0

( , , ) ( ) , 0 , 0aG x y t f y dy x a t … (21)


C.20 Appendix C

The function G (x, y, t) defined by (20) is known as the Green’s function for the given heatequation.

It can be easily seen that the Green’s function G (x, y, t) possess the following properties :

(i)2 2

2 2( , , ) ( , , ) ( , , ) , 0 , 0 , 0G x y t G x y t G x y tk k x a y a t

t x y

… (22)

(ii) G (0, y, t) = G (a, y, t) for 0 y a, t > 0 … (23)(iii) G (x, y, t) = G (y, x, t) for 0 x a, 0 y a, t > 0 … (24)Observe that the equation (22) can be derived from equation (20) by term-by-term differentia-

tion. Hence, it follows that the Green’s function G (x, y, t) satisfies the heat equation, SinceG (x, y, t) is symmetric, it also satisfies the boundary conditions.Physical interpretation of the Green’s function G(x, y, t)

The Green’s function G(x, y, t) can be interpreted as the effect on the temperature at x at time t dueto a heat source of unit magnitude (measured in unit of ) applied instantaneously at point y at timet = 0. It follows that the effect of the initial condition at t = 0 may be regarded as equivalent to that ofan instantaneous heat source.C.9. Solution of one-dimensional inhomogeneous heat equation involving an exter-nal heat source using the Green’s function technique

We propose to solve one-dimensional inhomogeneous heat equation

2 2/ ( / ) ( , ), 0 , 0u t k u x f x t x a t ... (1)

with the boundary conditions: (0, ) ( , ) 0, 0u t u a t t ... (2)

and the initial condition: ( , 0) 0, 0u x x a ... (3)

Here /k K is the thermal diffusivity of the material of the rod. Also, K, , are respectively,,thermal conductivity, density and specific heat of the material of the rod. Again, here f(x, t) isproportional to the volumetric heat source.

We now proceed to express the solution of the above problem as Fourier sine series in x, that is,

1

( , ) ( ) sin ,nn

n xu x t u ta

... (4)

where we have taken undetermined functions ( )nu t of time as multiplying constants. We haveselected the above form for ( , ),u x t since this form automatically satisfies both the boundary condi-tions given by (2). We now proceed to determine the unknown functions ( )nu t in such a manner sothat the resulting value of ( , )u x t given by (4) may satisfy (1) and initial condition (3). To this end, weassume the Fourier sine series expansions of f(x, t) given by

1

( , ) ( ) sin( / ),nn

f x t f t n x a

where ... (5)

0

2( ) ( , )sin( / )a

nf t f t n a da

... (6)

From (4), 2 2 2

2 21 1 1

( ) sin , ( ) cos , ( ) – sinn n nn n n

u n x u n n x u n n xu t u t u tt a x a a ax a

Substituting the above values of f(x, t), /u t and 2 2/u x in (1), we get

2 2

21 1 1

( )sin – ( ) sin ( ) sinn n nn n n

n x n n x n xu t k u t f ta a aa


Appendix C C.21

or 2 2 2

1sin ( ) ( / ) ( ) – ( ) 0n n n

n

n x u t kn a u t f ta

... (7)

Since sin ( / )n x a are linearly independent functions, the above equation (7) is satisfied only if,

the expression in parenthesis vanishes n N that is,2 2 2( ) ( / ) ( ) – ( ) 0n n nu t kn a u t f t or 2 2 2( ) / ( / ) ( ) ( ),n n ndu t dt kn a u t f t ... (8)

which is a linear differential equation whose integrating factor (I.F.) is given by2 2 2( / )kn a dt

e i.e.,

2 2 2( / )kn t ae and hence the general solution of (8) is given by2 2 2 2 2 2( / ) ( / )

0( ) ( ) ,

tkn t a kn t an n n nu t e f t e dt c c being an arbitrary constant ... (9)

In order to determine the value of cn, we require one initial condition. This condition can bedetermined with the help of the initial condition (3). Putting t = 0 in (4) and using the initial condition(3), we have

1

0 (0) sinnn

n xua

(0) 0,nu n N ... (10)

Now putting t = 0 in (9) and using (10), we get c = 0. Then, (9) gives2 2 2 2 2 2–( / ) –( / )

0( ) ( )

tkn a kn anu t e f e d or

2 2 2– ( / )( – )0

( ) ( )t kn a t

n nu t e f d ... (11)

Substituting the above value of (0)nu in (4), the required solution is given by

2 2 2–( / )( – )

01

( , ) ( ) sint kn a t

nn

n xu x t e f da

... (12)

From (6), 0

2( ) ( , ) sina

nn yf f y dy

a a

... (13)

Using (13), (12) yields 2 2 2–( / )( – )

0 01

2( , ) ( , ) sin sint akn a t

n

n y n xu x t e f y dy da a a

Interchanging the order of summation and integration, the above equation reduces to

2 2 2–( / )( – )

0 01

2( , ) sin sin ( , )t a kn a t

n

n x n yu x t e f y dy da a a

... (14)

We now define the Green’s function as

2 2 2–( / )( – )

1

2( , , – ) sin sin , 0 , 0 , 0kn a t

n

n x n yG x y t e x a y a ta a a

... (15)

Using the above definition (15), the required solution in terms of Green’s function can bere-written as

0 0

( , ) ( , , – ) ( , )t a

u x t G x y t f y dy d ... (16)

Remarks. Observe that equation (1) of Art. C.8 can be obtained from equation (1) of Art. C.9 bysetting f(x, t) = 0. Again, we observe that equation (3) of Art. C.9 can be obtained from equation of (3)of Art. C.8 by setting f(x) = 0. Furthermore, we observe that the Green’s function defined by (15) inArt. C.9 can be obtained from the Green’s function defined by (20) in Art. C.8 by replacing t by t – .Thus, it follows that the Green’s functions for both the problems, i.e. a source free problem with initial


C.22 Appendix C

condition (See Art. C.8), and that involving an external heat source, i.e., inhomogeneous equationwith zero initial condition (See Art. C.9), are identical. Therefore, it follows that the initial conditioncan be regarded as equivalent to an instantaneous heat source at time = 0.

Physical interpretation of the Green’s function G(x, y, t – )In what follows, suppose that the external source function f(y, ) is non zero only in the small

neighbourhood y of the point y0 and also during the small time interval around the time pointand zero otherwise. It follows that the heat source is switched on only in small neighbourhood yof the point y0 and also during the small time interval around the time We know thatthe function f(y, ) represents the volumetric heat source in units of . In other words,

( , ) ( , ) / ,f y F y where ( , )F y represents the volumetric heat source. Let Q denote the totalheat which is pumped into the system. Then, we have

0 0

0 0

/2 /2

0 0 – /2 – /2( , ) ( , )

t a y y

y yQ d F y dy d f y dy

... (17)

Using (16), the temperature u(x, t) for 0 / 2t is given by

0 0

0 0

/2 /2

– /2 – /2( , ) ( , , – ) ( , )

y y

y yu x t d G x y t f y dy

... (18)

Recall that we have assumed that f(y, ) is non zero only in the small neighbourhood y of thepoint y0 and also during the small time interval around the time point

and zero otherwise.Furthermore, the Green’s function G(x, y, t – ) is a well defined function in a very smallneighbourhood of the point (y0, 0). Hence, using mean value theorem, we can pull out G(x, y0, t – 0)from the integral for very small values of y and Thus, from (18), an approximate temperature isgiven by

0 0

0 0

/2 /20 0 0– /2 – /2

( , ) ( , , – ) ( , ) ( , , – ) ,y y

o y y

Qu x t G x y t d f y dy G x y t

using (17)

This can be interpreted as the influence of an instantaneous heat source at time 0 at point y0.Hence, it follows that the Green’s function G(x, y0, t – 0) represents the temperature at point x at timet due to an instantaneous heat source of unit magnite (meaured in units of , i.e., Q/ = 1) appliedat point y0 at time 0.

Remark. Observe that (1) is a linear partial differential equation. Hence, if the magnitude of theexternal source function is doubled, then the temperature u(x, t) is also doubled everywhere, and soon.C.10. The use of Green’s function in the determination of the solution of heatequation (or the diffusion equation)

Here we propose to find the solution u (r, t) of the heatequation

2/u t k u ...(1)in the volume V, which is bounded by the simple surface S,subject to boundary condition u (r, t) = f (r, t), if r S … (2)and initial condition u (r, 0) = g (r), if r V … (3)where r is the position vector of a point P in volume V and r isthe position vector of another point Q in V as shown in theadjoining figure.

n

S

n

P

Qr

r

O

V


Appendix C C.23

For the given initial boundary value problem given by (1), (2) and (3), we define the Green’sfunction G (r, r, t – t), t > t which satisfies the equation

2( , , ) / ( , , )G t t t k G t t r r r r … (4)

the boundary condition G (r, r, t – t) = 0, if r S … (5)

and the initial condition that lim ( , , ) 0t t

G t t

r rat all points of V except at the point r where G (r, r, t – t) takes the form

2

3/21 | |exp

4 ( )8{ ( )} k t tk t t

r r… (6)

Since G (r, r, t – t) depends only on t, it follows that it can be treated as a function of t – t.Therefore, equation (4) can be re-written in the following equivalent form :

2( , , ) / ( , , ) 0G t t t k G t t r r r r … (7)

From these equations, the physical interpretation of Green’s function G(r, r, t – t) is as follows:G(r, r t – t ) is the temperature at the point P(r) at time t due to an instantaneous point source ofunit strength generated at time t at the point P (r), the solid being initially at zero temperature, and itssurface being maintained at zero temperature.

Observe the time t lies within the interval of t for which equations (1) and (2) are valid. Therefore,(1) and (2) may be re-written in the form

2/ ,u t k u t t … (8) u (r, t) = f (r, t), if r S … (9)

Now, we have ( )uG G uu Gt t t

= 2 2( ),k G u u G by (7) and (8) … (10)

Let be an arbitrary positive constant Now, from (10), we have

2 2

0 0

( ) ( )t t

V V

uG dV dt k G u u G dV dtt

...(11)

Now, interchanging the order of integration on the left hand side of (11) and applying secondGreen’s identity (refer Art. A.3 in Appendix A) on the right hand side of (11), we have

0 0

( )t t

V S

d uG u Gdt dV k G u dS dtdt n n

… (12)

where / n denotes differentiation along the outward-drawn normal to S.From (2) and (5), we have G = 0 on S and u (r, t) = f (r, t) on S.

R.H.S of (12) =0

( , )t

S

Gk f t dS dtn

r ... (13)

Now, L.H.S. of (12) = 0 0[ ]t

t ttV V V tuG dV uG dV uG dV

= u (r, t) ( , , ) ( , ) ( ) , by (3)V Vt t

G t t dV G t g dV

r r r r , r

Using the expression (6) for G ( r, r, t – t), we can easily verify that2

0 3/21 |[ ( , , )] exp 1, as 0

48( )t tV V

G t t dV dVk

r r |r r


C.24 Appendix C

Thus, the L.H.S of (12) = u (r, t) – ( , , ) ( )V

G t g dV r r r … (14)

Now, from (12), (13) and (14), we obtain

0

( , ) ( , , ) ( ) ( , )t

V S

Gu t G t g dV k f t dS dtn

r r r r r

giving 0

( , ) ( ) ( , , ) ( , ) ,t

V S

Gu t g G t dV k f t dS dtn

r r r r r … (15)

which is the solution of the boundary value problem given by (1), (2) and (3)An illustrative example. If the surface z = 0 of the semi-infinite solid z 0 is maintained at

temperature f (x, y, t) for t > 0, and if the initial temperature of the solid is g (x, y, z), determine thedistribution of temperature in the solid.

Hint. Let V denote the half space z 0 and let Sdenote the entire xy-plane. Let P (r) be any point in thesemi-infinite solid z 0 and let Q (r) be another point inthe same solid. Let Q () be the image of Q in the xy-plane. Then, the appropriate Green’s function for thepresent problem can be taken as

G (r, r, t – t) =

2

3/21 | |exp

4 ( )8 ( ) k t tk t t

r r

2| |exp

4 ( )k t t

r

For this Green’s function, we have2 2 2

3/2 5/ 2 3/ 20

( ) ( )exp4 ( )8 ( )z

G G z x x y y zn z k t tk t t

Hence, using result (15) of Art. C. 7, the required solution is given by

u (r ,t) =2 2| | / 4 | | / 4

3 /21 ( )( )

8( )kt kt

Vg e e dV

k

r r rr

2 2 2

3/2 5/20

( , , ) ( ) ( )exp4 ( )8( ) ( )

t

S

z f x y t x x y y z dx dy dtk t tk t t

C.11. The use of Green’s function in the determination or the solution of heatequation (or diffusion equation) for infinite rod

Determine Green’s function for the problem of heat flow in an infinite rod described by partialdifferential equation 2 2/ ( / ), , 0u t k u x x t subject to the initial condition u (x, 0)= f (x), .x Hence, find the solution of the given problem in terms of Green’s function.

Solution. Given 2 2/ ( / ), , 0u t k u x x t … (1)

and the initial condition u (x, 0) = f (x), .x … (2)* Suppose that (1) has solution of the form u (x, t) = X (x) T(t), … (3)

where X (x) is a function of x alone and T (t) that of t alone. Substituting this value of u in (1) yields XT = k X T or X / X = T / kT … (4)

O

Z

X

r

Q(x , y , z )

P(x, y, z)

r

Y

Q (x , y , –z )

* For more details refers chapters 1 and 2 in part III of author’s “Advanced Differential Equation”, publishedby S. Chand and Co, Neal Delhi.


Appendix C C.25

Since x and t are independent variables, (4) can only be true if its each side is equal to the sameconstant, say . Hence, (4) yields the following two ordinary differential equations ;

X – X = 0 … (5)and T = kT … (6)

Solving (1), T = C ekt, C being an arbitrary constant … (7)When > 0, (7) shows that T and hence u (x, t) grows exponentially with time. In view of realistic

physical considerations, it is appropriate to suppose that f (x) 0 as | x | , while| u (x, t) | < M as x. Therefore, in order that u (x, t) remains bounded, must be negative and sowe take = –2, where 0. Thus, (5) takes the form X + 2X = 0 whose general solution is

X = C1 cosx + C2 sin x, C1 and C2 being arbitrary constants … (8)

Now, setting = –2, (7) yields 2–k tT Ce … (9)

Therefore a solution u (x, t, ) of (1) is given by u (x, t, ) = XT = (A cos x + B sin x)e–k2t, by (3) (8) and (9) … (10)

where A (= C1C) and B (= C2C) are new arbitrary constantsSince f (x) is not necessarily a periodic function, we have to use Fourier integral in place of

Fourier series in our given boundary value problem. Again, since A and B are arbitrary, we may treatthem as functions of and write A = A() and B = B(). In the given problem, since no boundarycondition is provided to limit our choice of , hence we must consider all possible values. Using theprinciple of superposition, this summation of the product solutions leads to the relation

u (x, t) =2

0 0( , , ) { ( ) cos ( )sin } k tu x t d A x B x e d

… (11)

which is the general solution of (1).Putting t = 0 in (11) and using the initial condition (2), we have

f (x) =0

{ ( ) cos ( )sin }A x B x d

...(12)

From the **Fourier integral theorem, we have

f (x) =0

1 ( )cos ( )f y x y dy d

or f (x) =

0

1 ( )(cos cos sin sin )f y x y x y dy d

or f (x) =0

1 cos ( )cos sin ( )sinx f y y dy x f y y dy d

… (13)

Let A () =1 ( ) cosf y y dy

and 1( ) ( )sinB f y y dy

… (14)


f (x) =0

{ ( ) cos ( )sin }A x B x d

… (15)

Comparing equations (12) and (15), equation (12) can be re-written as

f (x) =0

1 ( )cos ( )f x x y dy d

… (16)

** Refer Art. 1.3 in part IV B in author’s “Advanced Differential Equations”, published byS. Chandand Co., New Delhi


C.26 Appendix C

Then, from equation (11), we have

u (x, t) =2

0

1 ( ) cos ( ) k tf y x y e dy d

… (17)

Changing the orders of integration on R.H.S of (17), we have

u (x, t) =2

0

1 ( ) cos ( )k tf y e x y d dy

… (18)

From Integral Calculus, 2 2

0cos 2

2z be bz dz e

...(19)

Assume that z = ,kt dz kt d and 2bz = (x – y) … (20)

Then, from (20), 2 ( )b kt x y so that b = (x – y) / 2 kt … (21)


2 2{( ) / 4 }

0cos ( )

2k t x y kte x y kt d e

so that 2 2{( ) / 4 }

1/ 20cos ( )

(4 )k t x y kte x y d e

kt

… (22)

Using (22), (18) yields u (x, t) =2{( ) / 4 }

1/ 21 ( )

(4 )x y ktf y e dy

k t

… (23)

Hence, if f (y) is bounded for all real values of y, then the required solution of the boundary valueproblem is given by (23)

Let G (x – y, kt) =2{( ) / 4 }

1/ 21

(4 )x y kte

k t

… (24)


u (x, t) = ( , ) ( ) ,G x y kt f y dy

… (25)

The function G (x – y, kt) defined by (24) is known as the Green’s function of heat transfer in aninfinite rod.

Exercise1. Use the method of images and prove that the harmonic Green’s function for the

half-space z 0 is G (r, r) = 1/(4r) – 1/(4r), where r2 = (x – x)2 + (y – y)2 + (z – z)2 andr2 = (x – x)2 + (y – y)2 + (z + z)2.

2. Show that the Green’s function G (x, y, t) for the heat flow problem in semi-infinite roddescribed by the partial differential equation 2 2/ ( / ),u t k u x x > 0, t > 0 subject to boundarycondition u (0, t) = 0, t > 0 and initial condition u (x, 0) = f (x), x > 0 is given in the form

G (x, y, t) =2 2{( ) /4 } {( ) /4 } 1/2[ ] / (4 )x y t x y te e t

3. Using Green’s function technique solve 2 0u in the circle | r | < 3 subject to the condi-tion u = f () on the circle | r | = 3.

Ans. Green’s function = G (r, r) =2 2

2 21 81 18 cos( )log2 9{ 2 cos( )}e

r r rrr r rr


Appendix C C.27

and solution is given by u (r, ) =2 2

20

9 ( )2 9 6 cos( )

r f dr r

4. Find the Green’s function of the Dirichlet problem for semi-infinite space y 0,– < x < , –< z < and u = f (x, z) on y = 0 and solve this problem.

Ans. Green’s function = G (r, r) = {(x – x)2 + (y – y)2 + (z – z)2}–1/2– {(x – x)2 + (y + y)2

+ (z – z)2 }–1/2 and solution is given by u(x, y, z) = 2 2 2 3/ 2( , )

2 {( ) ( ) }x z

y f x z dx dzx x y z z

5. Find the Green’s function for the following initial boundary value problem:2 2 2 2 2/ ( / ) ( , ) 0 , 0u t c u x f x t x a t with initial conditions u(x, 0) = ut(x, 0) = 0,

0 < x < a and boundary conditions u(0, t) =ux (a, t) = 0, t > 0.6. Determine the Green’s function for the Robin’s problem on the quarter infinite plane

described by 2u = f (x, y), x > 0, y > 0 subject to the conditions u = g (y) on x = 0 and/ ( )u n h y on y = 0Solution. In xy-plane, let us consider a singularity at

(x, y). Then, we consider its image over one side, again reflectthe image over the image of that side, and so on. This process isstopped after an even number of reflections because then wecome back to the original domain. Let Q (–x, y) be the image ofP (x, y) in the y-axis and let R (–x, –y) and S (x, –y) be theimages of Q and P in the x-axis. Then the Green’s functionG (x, y) is obtained by placing three unit charges at properplaces. We place, a positive charge at (–x, y) and negativecharges at (–x, –y) and (x, –y) Then, we obtain

G (x, y) =2 2 2 2

2 2 2 21 {( ) ( ) }{( ) ( ) }log

4 {( ) ( ) }{( ) ( ) }ex x y y x x y yx x y y x x y y

… (2)

It can easily be verified that (i) 2 0G except at the source point (x, y)(ii) G = 0 on x = 0 (iii) / 0G n on y = 0

y

P(x , y )

S(x – y )

y

R(–x – y )

Q(–x y )

x xO


C.28 Appendix C

Page C.12

From (4), 2

21 1

2 2 2

2 21 1

( ) sin , ( ) sin ,

( ) cos and – ( ) sin

n nn n

n nn n


u n n x u n n xu t u tx a a ax a

... (8)

Page C.17

From (4), 2

21 1

2 2 2

2 21 1

( ) sin , ( ) sin ,

( ) cos – ( ) sin

n nn n

n nn n


u n n x u n n xu t and u tx a a ax a

... (8)

Page C.21We now define the Green’s function as

Page C.22

( , ) ( , ) / ,f y F y where ( , )F y represents the volumetric heat source. Let Q denote the totalheat which is pumped into the system. Then, we have

Page D.1

0 0

( ) ( , ) ( ) ( ) ( ) ( )l l

My x G x t t dt w x w x y x dx

or 0 0

1 1( ) ( , ){ ( ) ( )} ( )l l

my x G x t f t y t dt y x dxl l

[Note that from Ex. 1, page 11.52, ( ) 1/w x l ]

or 0 0

( ) ( , ) ( ) ( , ) ( ) ,l l

M My x G x t f t dt G x t y t dt C

where 0

1 ( )l

C y x dxl

is an arbitrary constant.

Page D.4

0( ) ( , ) ( ) ( )

lMy x G x t x dt C w x or 0

( ) ( , ) ( )2

lM

Cy x G x t f t dt


Appendix C C.29

Page D.6

0( ) ( ) 0w x x dx

or

0sin ( ) 0x f x dx

or

0sin ( ) 0]t f t dt

Page D.7For ,x t (6) reduces to 1 1 2 2– ( ) ( ) – ( ) ( )MLG w x w t w x w t ...(7)

Page D.10

or 2

1 1 0

sin( ) ( / 2) cos ( / 2) sin – cos ( )2

kxy x E kx F kx t kt f t dtk

where 1 2( / 2 / 2 )a E E k and 2( / 2 – / 2 )b F F k are new arbitrary constants


APPENDIX D

ADDITIONAL PROBLEMS BASED ON MODIFIED (OR GENERALIZED)GREEN’S FUNCTION

[Note: The reader is advised to study the entire matter given on pages 11.48–11.62 of chapter 11before reading the matter of this Appendix D. However, references to the above mentioned matter isgiven at proper places]D.1 Additional problems based on Art. 11.12, 11.13 and 11.14 of Chapter 11

Ex.1. Obtain the generalized Green’s function for the boundary value problem y = 0,y= (0) = y(l) = 0, 0 x l.

Sol. Given 2 2– ( / ) 0, (0) ( ) 0, 0d y dx y y l x l ...(1)

Here – (d2/dx2) is a self adjoint operator and hence (1) is a self adjoint system. Refer Ex. 1, page11.52. Start from equation (2) and proceed upto equation (17). Thus, the required generalized Green’sfunction GM(x, t) is given by

2 2 , 0( , ) –

,3 2Ml if x tl x lG x tx if t x ll

Ex.2. Transform the boundary value problem – (d2y/dx2 + y) = f(x), y(0) = y(l) = 0, 0 x linto an integral equation.

Sol. Given – (d2y/dx2) = ( ) , (0) ( ) 0, 0f x y y y l x l ...(1)Here – (d2y/dx2) is a self adjoint operator. Consider the associated self adjoint system

– 0, 0y x l ... (2)

with the boundary conditions : (0) 0y ...(3a)

and ( ) 0y l ... (3b)

Now, proceed exactly as in Ex. 1,, page 11.52 starting with equation (2) upto equation (17). Then,the modified Green’s function GM(x, t) of the given boundary value problem is given by

2 2 , 0( , –

,3 2Ml if x tl x lG x tx if t x ll

Comparing (1) with ( ),Ly x here ( ) ( ) ( ).x f x y x Hence, using result (111b) of Art. 11.13,the required integral equation is given by

0 0

( ) ( , ) ( ) ( ) ( ) ( )l l

My x G x t t dt w x w x y x dx

or 0 0

1 1( ) ( , ){ ( ) ( )} ( )l l

my x G x t f t y t dt y x dxl l

[Note that from Ex. 1, page 11.52, ( ) 1/w x l ]

or 0 0

( ) ( , ) ( ) ( , ) ( ) ,l l

M My x G x t f t dt G x t y t dt C

where 0

1 ( )l

C y x dxl

is an arbitrary constant.

D.1


D.2 Appendix D

Ex.3. Determine the modified Green’s function of the system 2(1– ) ( ), –1 1d dyx f x xdx dx

with boundary conditions that y(x) is finite at x = 1 and x = –1. Hence transform the givenboundary value problem into an integral equation.

Sol. Given 2– (1– ) – ( ),d dyx f xdx dx

–1 1x ... (1)

with boundary conditions: y(1) = finite and f(–1) = finite ...(2)

Here 2–( / ){(1– )( / )}d dx x d dx is a self adjoint operator

Consider the associated self adjoint system 2– (1– ) 0, –1 1d dyx xdx dx

...(3)

with boundary conditions: y(–1) = finite ...(4a)and y(1) = finite ...(4b)

Now, proceed as in Ex. 3. page 11.56 starting with (3) and stop at (15). Thus, the modified Green’sfunction GM(x, t) of the given boundary value problem is given by

(1 / 2) log{(1 – ) / (1 )}, –11( , ) log 2 – –(1/ 2) log{(1 ) / (1– )}, 12M

x t x tG x t

x t t x

Comparing (1) with ( ),Ly x we have ( ) – ( )x f x . Hence, using result (111b) of Art. 11.13,the required integral equation is given by

1 1

–1 –1( ) ( , ) ( ) ( ) ( ) ( )my x G x t t dt w x w x y x dx

or 1 1

–1 –1

1( ) – ( , ) ( ) ( ) ,2my x G x t f t dt y x dx as 1( )

2w x

or 1 1

–1 –1

1( ) – ( , ) ( ) , where = ( )2My x C G x t f t dt C y x dx

Ex.4. Find the generalized Green’s function of the boundary value problem –k(d2y/dx2) = f(x)with (0) ( ) 0, 0 .y y l x l Hence transform the given boundary value problem into anintegral equation.

Sol. Given 2 2– ( / ) ( ), (0) ( ) 0, 0k d y dx f x y y l x l ...(1)

Here 2 2– ( / )k d dx is a self adjoint operator. Consider the associated self adjoint system :

– 0, 0ky x l ...(2)

with boundary conditions: (0) 0y ...(3a)

and ( ) 0y l ...(3b)

The general solution of (2) is ( )y x Ax B ...(4)

From (4), ( )y x A ...(5)

Putting x = 0 and x = l in (5) and using (3a) and (3b), we get A = 0. Hence the boundary valueproblem given by (2), (3a) and (3b) has a non-zero solution y(x) = B, where B is an arbitrary constant


Appendix D D.3

Here, 1/22

0|| ( ) || norm of ( ) [ ( )]

ly x y x y x dx B l

Let ( ) ( )/ || ( ) || / ( ) 1/w x y x y x B B l l ...(6)

Thus, w(x) is a non-zero normalized solution of the boundary value problem given by (2), (3a)

and (3b). Clearly 20[ ( )] 1

lw x dx ... (7)

Then, for ,x t the required generalized Green’s function GM(x, t) must satisfy

2 2– ( / ) – ( ) ( )Mk d G dx w x w t or 2 2/ 1/Md G dx l ... (8)

The general solution of (8) is of the form 2( , ) / 2MG x t Ax B x kl

Hence, we take 2

1 22

1 2

/ 2 , 0( , )

/ 2 ,M

a x a x kl x tG x t

b x b x kl t x l

...(9)

From (9), 1

1

/ , 0/ ,

M a x kl x tGb x kl t x lx

...(10)

In addition to the above property (9), the proposed generalized Green’s function must satisfythe following properties.

(i) Since GM(x, t) must satisfy the boundary conditions (3a) and (3b), (10) gives

0( / ) 0M xG x and ( / ) 0M x lG x

Hence, a1 = 0 and b1 + (1/k) = 0 so that a1 = 0 and b1 = – (1/k) ... (11)(ii) GM(x, t) is continuous at x = t, that is,a1t + a2 + t2/2kl = b1t + b2 + t2/2k or b2 = a2 + (t/k), using (11) ... (12)

(iii) 0 –0( / ) – ( / ) 1/ ( ),M x t M x tG x G x p t where p(x) = – k ...(13)

or b1 + t/kl – (a1 + t/kl) = – (1/k), using (10)or – (1/k) = – (1/k), using (11). Thus, (13) is identically satisfied.

Substituting the values of a1, b1 and b2 given by (11) and (12) in (9), we get

2

22

2

/ 2 , 0( , )

( – ) / / 2 ,M

a x kl x tG x t

a t x k x kl t x l

...(14)

(iv) In order that GM(x, t) may be symmetric, we must have

0( , ) ( ) 0

lMG x t w x dx or

0( , ) 0,

lMG x t dx as 1( )w x

l

or 0

( , ) ( , ) 0t l

M MtG x t dx G x t dx or

2 2

2 20– ,

2 2t l

t

x t x xa dx a dxkl k k kl

using (14)

or 3 2 3

2 20

– 06 2 6

t l

t

x tx x xa x a xkl k k kl

or 2 2 2 2 2 3

2 2 2– – – – 06 2 6 2 6t tl l l t t ta l a l a tkl k k k k k kl


D.4 Appendix D

or 2 2

2 –2 3t tl la lk k k

so that 2 2

21 –

2 3t la tl

kl

Substituting the above value of a2 in (14), we have

2 2 2

2 2 2

1 – , 02 3 2

( , )1 –– ,

2 3 2

M

t l xtl x tkl kl

G x tt l t x xtl t x l

kl k kl

or 2 2

2 2

( ) / 2 – ( / ) ( / 3 ), 0( , )

( ) / 2 – ( / ) ( / 3 ),M

x t kl t k l k x tG x t

x t kl x k l k t x l

... (15)

Using result (11a) of Art. 11.13, the required integral equation is given by

0( ) ( , ) ( ) ( )

lMy x G x t x dt C w x or 0

( ) ( , ) ( )2

lM

Cy x G x t f t dt

or 0

( ) ( , ) ( ) ,l

My x A G x t f t dt A being an arbitrary constant

Ex.5. Transform the boundary value problem ( ),y y f x (0) ( ) 0,y y into an integralequation.

Sol. Given 2 2– ( / ) – ( ), (0) ( ) 0, 0d y dx y f x y y x ...(1)

Here 2 2– ( / 1)d y dx is a self adjoint operator..

Consider the associated self adjoint system – ( ) 0, 0y y x ...(2)

with the boundary conditions: y(0) = 0 ...(3a)and y() = 0 ...(3b)

The general solution of (2) is ( ) cos siny x A x B x ...(4)

Putting x = 0 and x = in (4) and using (3a) and (3b), we get A = 0. Hence, the boundary valueproblem given by (2), (3a) and (3b) has a non-trivial solution ( ) sin ,y x B x B being an arbitraryconstant

Here, 1/21/2 1/2

2 2 20 0 0

1– cos 2|| ( ) || [( ( )] ( sin )2

xy x y x dx B x dx B dx

= 1/21/2

0/ 2 – (sin 2 ) / 4 ( / 2)B x x B

Let 1/2( ) ( )/ || ( ) || (2 / ) sinw x y x y x x ...(5)so that w(x) is a non-zero normalized solution of the problem given by (2), (3a) and (3b). Clearly,

20

[ ( )] 1w x dx

... (6)

Then for ,x t the modified Green’s function of the given boundary value problem must satisfy

2 2– ( / ) – ( ) ( )M Md G dx G w x w t or 2( 1) (2 / ) sin sin ,MD G x t using (6) ...(7)

where / .D d dx The complementary function (= C.F.) of (7) is sin cosC x D x


Appendix D D.5

Again, the particular integral (P.I.) of (7)

2 21 2 2sin 1 2sin cossin sin sin – – sin cos

21 1t t x x xx t x t x

D D

Thus, the general solution of (7) is of the form

( , ) sin cos – ( / ) sin cosMG x t C x D x x t x

Hence, we take ( , ) – sin cosMxG x t t x

1 1

2 2

sin cos , 0sin cos ,

C x D x x tC x D x t x

... (8)

In addition to the above property (8), the proposed modified Green’s function must satisfy thefollowing properties.

(i) Since GM(x, t) must satisfy the boundary conditions (3a) and (3b), (8) gives G(0, t) = 0 and G(t) = 0

Hence, D1 = 0 and 2sin – 0t D so that D1 = 0 and 2 sin .D t So, (8) yields

1

2

sin , 0( , ) – sin cos

sin cos sin ,MC x x txG x t t xC x x t t x

...(9)

(ii) GM(x, t) must be continuous at x = t. Hence, (9) gives

1 2sin sin cos sinC t C t t t so that 2 1 – cosC C t ...(10)

(iii) We must have 0 –0( / ) – ( / ) 1/ ( ),M x t M x tG x G x p t where p(t) = – 1 ...(11)

From (9), 1

2

cos , 01– sin cos sin sincos – sin sin ,

M C x x tG xt x t xC x x t t xx

Hence, (11) reduces to 22 1cos – sin – cos –1C t t C t or 2

1 2( – )cos cosC C t t

Thus, 1 2– cosC C t so that 2 1 – cosC C t ...(12)

Observe that relations (10) and (12) are identical. Using (12), (9) reduces to

1

1

sin , 0( , ) – sin cos

( – cos )sin cos sin ,MC x x txG x t t xC t x x t t x

or 10, 0

( , ) – sin cos sin– sin ( – ),M

x txG x t t x C xx t t x

or 1( , ) – ( / ) sin cos sin – sin ( – ) ( – ),MG x t x t x C x x t H x t ...(13)

where H(x – t) is the usual Heaviside unit function which is defined as

0, if

( – )1, if > t

x tH x t

x

...(14)

In order that GM(x, t) may be symmetric, we must have

0

( , ) ( ) 0MG x t w x dx

... (15)


D.6 Appendix D

The unknown constant C1 occurring in (13) can be determined with help of (5), (13) and (15).

Comparing (1) with ( ),Ly x we have ( ) – ( ).x f x Hence using result (111a) of Art. 11.13, therequired integral equation is given by

0

( ) ( ) ( , ) ( ) ,My x E w x G x t t dt

E being an arbitrary constant

or 1/20

( ) (2 / ) sin – ( , ) ( ) ,My x E x G t x f t dt

using (5) ...(16)

[ GM(x, t) is symmetric GM(x, t) = GM(t, x)]

From (13), 1( , ) – ( / ) sin cos sin – sin( – ) ( – )MG t x t x t C t t x H t x ...(17)

Substituting the above value of GM(t, x) in (16), we have

1/210

( ) (2 / ) sin – {– ( / ) sin cos sin – sin ( – ) ( – )} ( )y x E x t x t C t t x H t x f t dt

or 1/210 0

sin( ) (2 / ) sin cos ( ) – sin ( )xy x E x t t f t dt C t f t dt

0

( ) sin( – ) ( – ) ( ) sin( – ) ( – )x

xf t t x H t x dt f t t x H t x dt

or 1/21( ) (2 / ) sin {(sin ) / } – 0 0 sin ( – ) ( )

xy x E x x F C t x f t dt

... (18)

[ 0cos ( ) constant ,t t f t dt F

say and for the existence of integral equation, the following

consistency condition must be satisfied:

0( ) ( ) 0w x x dx

or

0sin ( ) 0x f x dx

or

0sin ( ) 0]t f t dt

Let 1/2(2 / ) / .E F k Then, (18) may be re-written as

( ) sin sin( – ) ( ) ,x

y x k x t x f t dt k

being an arbitrary constant

D.2 Extension of the theory of Art. 11.13 of chapter 11 to the case when the asso-ciated adjoint system has two linearly independent solutions y1(x) and y2(x) in placeof exactly one non-zero solutionRevised working rule of construction of modified (or generalized) Green’s function

Consider an inhomogeneous equation with boundary conditions:

( ),Ly x 1 1( ) ( ) 0,y a y a 2 2( ) ( ) 0y b y b ...(1)

Here L is a self adjoint system. Consider the following associated self adjoint system:

0,Ly where 2

2( ) ( )d dp dL p x q xdx dxdx

...(2)

with boundary conditions: 1 1( ) ( ) 0y a y a ... (3a)

and 2 2( ) ( ) 0y b y b ...(3b)with the usual assumption that one of 1 and1 and one of 2 and 2 are non-zero.


Appendix D D.7

Suppose that the homogeneous boundary value problem given by (2), (3a) and (3b) possesstwo non-trivial linearly independent solutions y1(x) and y2(x).

Then, 1/22

1 1 1|| ( ) || norm of ( ) ( )b

ay x y x y x dx

and 1/22

2 2 2|| ( ) || norm of ( ) ( )b

ay x y x y x dx

Let 1 1 1( ) ( )/ || ( ) ||w x y x y x and 2 2 2( ) ( )/ || ( ) ||,w x y x y x ...(4)

so that w1(x) and w2(x) are two non-trivial normalized solutions of the boundary value given by (2),(3a) and (3b). Clearly, we have

21[ ( )] 1

b

aw x dx and 2

2[ ( )] 1b

aw x dx ... (5)

Then, by definition of modified Green’s function GM(x, t) of the given boundary value problem,GM(x, t) satisfies the differential equation

1 1 2 2( – ) – ( ) ( ) – ( ) ( )MLG x t w x w t w x w t ...(6)

For ,x t (6) reduces to 1 1 2 2– ( ) ( ) – ( ) ( )MLG w x w t w x w t ...(7)

For a given t, let 1

2

( , ), if( , )

( , ), ifMG x t a x t

G x tG x t t x b

...(8)

where G1 and G2 are such that(i) The functions G1 and G2 satisfy the equations in their respective intervals of definition, that

is, 1 1 1 2 2– ( ) ( ) – ( ) ( ),LG w x w t w x w t a x t ... (9a)

2 1 1 2 2– ( ) ( ) – ( ) ( ),LG w x w t w x w t t x b ... (9b)(ii) G1 satisfies the boundary conditions (3a) whereas G2 satisfier the boundary condition (3b)

(iii) 0 –0( / ) – ( / ) 1/ ( )M n t M n tG x G x p t must be satisfied

(iv) In order that GM(x, t) may be symmetric, we must have

1( , ) ( ) 0b

MaG x t w x and 2( , ) ( ) 0

bMa

G x t w x dx ... (9)

(v) The given boundary value problem can be reduced to an integral equation only if the follow-ing so-called consistency conditions are satisfied:

1( ) ( ) 0b

ax w x dx and 2( ) ( ) 0

b

ax w x dx ... (10)

The required integral equation is given by

1 1 2 2( ) ( , ) ( ) ( ) ( )b

May x G x t t dt C w x C w x ...(11)

or 1 1( ) ( , ) ( ) ( ) ( ). ( )b b

Ma ay x G x t t dt w x w x y x dx 2 2( ) ( ) ( )

b

aw x w x y x dx ...(12)

An illustrative example. Obtain the modified Green’s function of the boundary value problem2 2 2/ ( )d y dx k y f x subject to the boundary conditions (0) (2 )y y and (0) (2 ),y y

0 2 ,x k being a non-zero integral. Hence, convert the given boundary value problem into anintegral equation. [Kanpur 2011]


D.8 Appendix D

Sol. Given 2 2 2– ( / ) – ( ), (0) (2 ), (0) (2 ), 0 2d y dx k y f x y y y y x ...(1)

Here – (d2/d2 + k2) is a self adjoint operatorConsider the associated self adjoint system

2 2 2 2 2– ( / ) 0 . . ( ) 0, where D / , 0 2d y dx k y i e D k y d dx x ...(2)

subject to the boundary condition: (0) (2 )y y ...(3a)

and (0) (2 )y y ...(3b)

We easily verify that 1( ) cosy x kx and 2 ( ) siny x kx are two linearly independent non-zerosolutions of the boundary value problem given by (2), (3a) and (3b). Then, we have

1|| ( ) ||y x = 1/21/2 21/2 1/22 2 22 2

10 0 00

1 cos 2 sin 2[ ( )] cos2 2 4

kx x kxy x dx kx dx dx

Thus, 1|| ( ) ||y x = . Similarly, , 2|| ( ) ||y x

Let 1 1 1( ) ( )/ || ( ) ||w x y x y x and 2 2 2( ) ( )/ || ( ) || .w x y x y x Then, we have

1( ) (cos ) / 2w x kx and 2 ( ) (sin ) / 2w x kx ...(4)

For ,x t ( , )MG x t must satisfy 2 2 21 1 2 2– ( / ) – ( ) ( ) – ( ) ( )M Md G dx k G w x w t w x w t

or 2 2( ) (1/ ) cos cos (1/ ) sin sinMD k G kx kt kx kt ...(5)

C.F. of (5) = cos sin ,a kx b kx a and b being arbitrary constants

and P.I. (5) = 2 21 1 [cos cos sin sin ]kx kt kx kt

D k

1 1cos sin sin (– cos )

2 2x xkt kx kt ktk k

( / 2 ) sin ( – )x k k x t

We take 1 1

1 1

cos sin , 0( , ) sin ( – )

cos sin , 22MA kx B kx x txG x t k x tC kx D kx t xk

...(6)

From (6), 1 1

1 1

– sin cos , 0sin ( – ) cos ( – )– sin cos , 22 2

M A k kx B k kx x tG k x t x k x tC k kx B k kx t xx k

...(7)

The desired modified Green’s function ( , )MG x t must satisfy the following properties:

(i) ( , )MG x t satisfies (3a) (0, ) (2 , )M MG t G t

1 –(1/ ) sinA k kt C so that 1 1– – (1/ ) sinA C k kt ...(8)

( , )MG x t satisfies (3b) 0( / )M xG x = 2( / )M xG x

1 1cosB k kt D k 1 1(1/ ) cosB k kt D 1 1– (1/ ) cosB D k kt ...(9)

(ii) ( , )MG x t is continuous at x = t. Hence, we have

1 1 1 1cos sin cos sinA kt B kt C kt D kt 1 1 1 1( – )cos ( – ) sin 0A C kt B D kt

– (1/ ) sin cos (1/ ) cos sin 0,k kt kt k kt kt using (8) and (9)which is identically satisfied.


Appendix D D.9

(iii) ( , )MG x t satisfies 0 –0( / ) – ( / ) 1/ ( )M x t M x tG x G x p t

1 1 1 1– sin cos – (– sin cos ) –1C k kt D k kt A k kt B k kt

1 1 1 1( – ) sin – ( – )cos – (1/ )A C kt B D kt k

2 2– (1 / ) sin – (1/ ) cos – (1/ )k kt k kt k –(1/k) = –(1/k), using (8) and (9)

which is identically satisfied.Substituting the values of A1 and B1 given by (8) and (9) in (6), we have

1 1

1 1

{ – (1/ ) sin }cos { (1/ ) cos } sin , 0sin ( – )( , )cos sin , 22M

C k kt kx D k kt kx x tx k x tG x tC kx D kx t xk

1 1

1 1

cos sin (1/ ) sin ( – ), 0sin ( – )cos sin , 22

C kx D kx k k x t x tx k x tC kx D kx t xk

1 1(1/ ) sin ( – ), 0sin ( – )cos sin0, 22

k k x t x tx k x tC kx D kxt xk

1 1( / 2 1), 0sin ( – )cos sin

/ 2 , 2x x tk x tC kx D kx

x t xk

Thus, 1 1( , ) cos sin (1/ ) sin ( – ){ / 2 ( – )]MG x t C kx D kx k k x t x H x t ...(10)

where H(x – t) is Heavide unit function defined by

0, if

( – )1, if

x tH x t

x t

...(11)

We know that GM(x, t) satisfies the following conditions.

210

( , ) ( ) 0MG x t w x dx

and 2

20( , ) ( ) 0MG x t w x dx

...(12)

Using (10), (11) and (12), we can determine C1 and D1 occurring in (11).

Comparing (1) with Ly = (x), we have ( ) – ( ).x f x Hence, using the result (11) of Art. D.2, therequired integral equation is given by

21 1 1 2 0

( ) ( ) ( ) ( , ) ( ) ,My x E w x F w x G x t t dt

E1 and F1 being arbitrary constants

or 2

1 1 0( ) ( / 2) cos ( / 2) sin – ( , ) ( ) ,My x E kx F kx G t x f t dt

using (4) ...(13)

[ ( , )MG x t is symmetric ( , ) ( , )]M MG x t G t x

From (10), 1 1( , ) cos sin (1/ ) sin ( – ){ / 2 ( – )}MG t x C kt D kt k k t x t H t x

Substituting the above value of ( , )MG t x in (13), we have

21 1 1 10

( ) ( / 2) cos ( / 2) sin – { cos sin (1/ ) sin ( – ){ / 2 ( – )} ( )y x E kx F kx C kt D kt k k t x t H t x f t dt


D.10 Appendix D

or 2 2

1 1 1 10 0( ) ( / 2) cos ( / 2) sin – cos ( ) – sin ( )y x E kx F kx C kt f t dt D kt f t dt

2 2

0 0

1 1– sin ( – ) ( ) – sin ( – ) ( – ) ( )2

t k x t f t dt k x t H t x f t dtk k

The given boundary value problem can be reduced to the above integral equation only if the

following consistency conditions are satisfied :

2 21 20 0( ) ( ) ( ) ( ) 0,w x x dx w x x dx

i.e.,

2 2

0 0cos ( ) sin ( ) 0kx x dx kx x dx

...(14)

Using (14), the above integral equation reduces to2

1 1 0

1( ) ( / 2) cos ( / 2) sin – (sin cos – cos sin ) ( )2

y x E kx E kx t kx kt kx kt f t dtk

2

0

1– sin ( – ) ( – ) ( ) sin ( – ) ( – ) ( )x

xk x t H t x f t dt k x t H t x f t dt

k

or 2

1 1 0

sin( ) ( / 2) cos ( / 2) sin – cos ( )2

kxy x E kx F kx t kt f t dtk

2 2

0

cos 1sin ( ) – 0 sin ( – ) ( )2 x

kx t kt f t dt k x t f t dtk k

or 1 1 2 2( ) ( / 2) cos ( / 2) sin – ( / 2 ) sin ( / 2 ) cosy x E kx F kx F k kx E k kx

21– sin ( – ) ( ) ,x

k x t f t dtk

... (15)

where 2

20cos ( ) constantt k f t dt F

say and

220

sin ( ) constantt kt f t dt E

, say

Thus, the required integral equation (15) reduces to

or 21( ) cos sin – sin ( – ) ( ) ,x

y x a kx b kx k x t f t dtk

where 1 2( / 2 / 2 )a E E k and 2( / 2 – / 2 )b F F k are new arbitrary constants


MISCELLANEOUS PROBLEMS ONTHE ENTIRE BOOK

1. Solution of the initial value problem (d2y/dx2) + a1(x) (dy/dx) + a2(x) y = F(x), 0 x 1,y (0) = C0, (dy/dx)x = 0 = C1, where a1 (x), a2(x) and F (x) are continuous functions on[0, 1] may be reduced, in general, to a solution of(a) Fredholm integral equation of first kind (b) Volterra integral equation of first kind(c) Fredholm integral equation of second kind (d) Volterra integral equation of second kindHint : Ans. (d). Refer Art. 2.4. [Gate 2006]

2. Using method of degenerate kernels, determine the solution of the integral equation

12 1/ 2 10

( ) (1 ) cos ( ) )x x d [Kanpur 2006]

Solution. Let1 10

cos ( )C d ...(1)

Then the given integral equation reduces to 2 1/ 2( ) (1 )x x C ...(2)

From (2), 2 1/ 2( ) (1 ) C ...(3)

Using (3), (1) yields 1 2 1/ 2 10{(1 ) }cosC C d

or1 11 2 1/ 2 10 0

cos (1 ) (1 cos )C d C d

or11 11 2 1 2 1/ 2

0 0 0( 1/ 2) (cos ) cos (1 )C C d

or12 2 1/ 2 20

/ 8 0 (1 ) /8C C C

or 2 2(1 /8 so that / 8(1 ), 1C C

Hence, from (2), the required solution is 2 1/ 2 2( ) (1 ) / 8(1 )x x

3. Solve 1

0( ) 2 sin log ( )x x x t dt [Kanpur 2006]

Solution. Re-writing, the given-equation is given by

1

0( ) 2 sin log ( )x x x t dt ...(1)


M.2 Miscellaneous problems on the entire book

Let1

0( )C t dt ...(2)

Using (2), (1) yields ( ) 2 sin logx x C x ...(3)

From (3), ( ) 2 sin logt t C t ...(4)

Using (3), (2) yields1

0(2 sin log )C t C t dt

or112

0 0sin log 1 ,C t C t dt C I ...(5)

where 1 11

00 0

1sin log sin log cos logI t dt t t t t dtt

or 1 11

00 0

1cos log cos log ( sin cos )I t dt t t t t dtt

or 1 or 2 1 or 1/ 2I I I I ...(16)

Using (6), (5) yields 1 ( ) / 2 giving 2 / (2 )C C C Hence, from (3), the required solution is given by

( ) 2 2 / (2 ) sin log , 2x x x

4. Which of the following function is a solution of Fredholm type integral equation

0( ) ( ) ?

xf x x x f t dt (a) 2x/3 (b) 3x/3 (c) 3x/4 (d) 4x/3.

[Ans. (b)] [Gate 2006]5. Which of the following function is a solution of Volterra type integral equation

0( ) sin ( ) ( )

xf x x x t f t dt . (a) x + x3/3 (b) x – x3/3 (c) x + x3/6 (d) x – x3/6

[Ans. (c)] [Gate 2006]

6. Show that ( , ; ) ( ) ( , ) ( , ; )x

R x K x K x R x dz

is the solution of a non-

homogenous Volterra integral equation of second kind 0

( ) ( ) ( , ) ( ) ,x

x f x K x d where ( , ; )R x is the resolvent kernel of the equation. [Kanpur 2007]

Hint : Refer Art. 5.12. page 5.37.7. If u(x) has continous first and second order derivatives and satisfies the boundary value

problem d2u/dx2 + u = 0 with (0) ( ) 0,u u l then ( )u x is continuous and satisfies the

homogenous linear integral equation 1

0( ) ( , ) ( ) ,u x G x u d where

( / ) (1 ),0( , )

( / ) (1 ),l x x

G xx l x l

[Kanpur 2007]

Hint : Refer solved Ex. 1 (a), page 2.14.


Miscellaneous problems on the entire book M.3

8. Solved the following integral equations :

(i) 1 2 20

( ) (5 3) ( )xx e x d [Kanpur 2007]

(ii) 2 / 2 1 11/ 2 ( )x x te e t dt

[Meerut 2007]

Ans. (i) 2( ) ( 2) (5 3).xx e e x

9. Find the iterated kernels for the following kernel : ( , ) sin ; , ,K x x a b wherea and b are the limits of the integral; [Kanpur 2007]Hint : Refer part (iii) of Ex. 1, page 5.12.

10. The value of for which the integral equation 1

0( ) ( ) ,x tu x e u t dt has a non-trivial

solution is (a) –2 (b) – 1 (c) 1 (d) 2 [GATE 2007]Hint. Ans. (c). Proceed as in solved Ex. 1, page 3.3.

11. The integral equation 1

0( ) cos sec sinh , 0 1x t t s ds t t has

(a) no solution (b) a unique solution(c) more than one but finitely many solutions (d) infinitely many solutions

[GATE 2008]

12. Suppose 2

0( ) ( ) sin ( ) , [0, 2 ]y x y t x t dt x

has eigenvalues 1/ and 1/

with corresponding eigenfunctions 1( ) sin cosy x x x and 2 ( )y x sin cosx x

respectively. Then the integral equation 2

0

1( ) ( ) ( )y x f x y t

sin ( ) , [0, 2 ]x t dt x

has a solution when f(x) = (a) 1 (b) cos x (c) sin x (d) 1 + sin x + cos x. [GATE 2007]

13. The integral equation 2 31 2 3

0( ) sin ( ) ( ) ,s tx t t s t e x s ds 0 1, 0t , R

has a solution for (a) all non zero values of (b) no value of (c) only countables many + ve values of (d) only countabls many-ve values of

[GATE 2008]14. Define (i) Integral equation (ii) Singular integral equation (iii) Integro-differential equation.

[Meerut 2008]15. Obtain the Fredholm integral equation of second kind corresponding to boundary value

problems d2u/dx2 + u = x, u (0) = 0, u(1) = 1 [Kanpur 2008]

16. Show that ( , ; ) ( , ) ( , ) ( , ; )x

R x K x K x z R z dz

is the solution of a non-

homogeneous Volterra integral equation of second kind 0

( ) ( ) ( , ) ( )x

x f x K x d ,

where ( , , )R x is the resolvent kernel of the equation. [Kanpur 2007]

17. Solve the integral equation equation 1 2 20

( ) (5 3) ( )xx e x [Kanpur 2007]



18. Find the first two iterated kernels of the kernel 2( , ) ( ) ; 1, 1.K x x a b

Hint. Refer Ex. 1 (iii), page 5.33 [Kanpur 2008, 09]


( ) 1 ( ) ( )x

x x d with 0 ( ) 0x [Kanpur 2008]

20. Find the iterated kernels of the function upto third order : ( , ) , 0, 1x tK x t e a b Hint: Do as in Ex. 2(i), page 5.16 upto equation (6). (Kanpur 2009)

Ex. 21. Show that the longitudinal vibrations of a rod fixed at one end and free at the otherrepresent the Fredholm integral equation with Schmidt kernel. [Meerut 2005, 07, 09]

[Sol. The longitudinal vibrations of a rod fixed at one end and free at the other are governedby the following boundary value problem;

( ) ( )y x F x , ... (1)where F(x) is a known continuous function satisfying the boundary conditions :

y (a) = 0 ... (2a)and ( ) 0y b ... (2b)

Integrating both sides of (1) w.r.t. ‘x’ from a to x, we get

( ) ( )xx

a ay x F x dx or ( ) ( ) ( )

x

ay x y a F x dx

or 1 1( ) ( ) where ( )x

ay x F x dx c c y a ... (3)

Integrating both sides of (3) w.r.t. ‘x’ from a to x, we get

2 21 1( ) ( ) [ ] or ( ) ( ) ( ) ( ) ( )

x x x xx xaa a a a a

y x F x dx c x y x y a F t dt c x a or 1( ) ( ) ( ) ( )

x

ay x x t F t dt c x a ... (4)

[on using the boundary condition (2a) and the result of Art. 1.14]

Re-writing (3), we have 1( ) ( )x

ay x F t dt c ... (5)

Putting x = b in (5) and using the boundary condition (2b), we get

10 ( )b

aF t dt c so that 1 ( )

b

ac F t dt ... (6)

Substituting the value of c1 given by (6) in (4), we have

( ) ( ) ( ) ( ) ( )x b

a ay x x t F t dt x a F t dt

or ( ) ( ) ( ) ( ) ( )x b

a ay x x t F t dt a x F t dt

or ( ) ( ) ( ) ( ) ( ) ( ) ( )x x b

a a xy x x t F t dt a x F t dt a x F t dt

or ( ) {( ) ( )} ( ) ( ) ( )x b

a xy x x t a x F t dt a x F t dt

or ( ) ( ) ( ) ( ) ( )x b

a xy x a t F t dt a x F t dt ... (7)

which can be re-written as

( ) ( , ) ( ) ,b

ay x G x t F t dt ... (8)

where,

( , )a x if x t

G x ta t if x t

... (9)

From (8) and (9), it follows that the longitudinal vibrations of a rod fixed at one end and freeat the other represent the Fredholm integral equation (8) with Schmidt kernel.



Ex. 22. The resolvent kernel for the integral equation 2log

( ) ( ) ( )x t xu x F x e u t dt is

(a) cos (x – t) (b) 1 (c) et – x (d) e2(t – x)

[GATE, 2009]Sol. Ans. (d). Proceed as in Ex. 2, Page 5.38.Ex. 23. For a continuous function ( ), 0 1,f t t the integral equation

10

( ) ( ) 3 ( )y t f t t s y s ds has

(a) Unique solution if 10 ( ) 0sf s ds (b) no solution if 1

0 ( ) 0s f s ds

(c) Infinitely many solutions if 10 ( ) 0s f s ds

(d) Infinitely many solutions if 10 ( ) 0s f s ds [GATE 2010]

Sol. Ans. (c) Given y(t)= 10

( ) 3 ( )f t t s y s ds or y(t) =

10

( ) 3 ( )f t t s y s ds ...(1)

Let C = 10

( )s y s ds ...(2)

Using (2), (1) yields y(t) = f(t) + 3 Ct ...(3)From (3), y(s) = f(s) + 3 Cs ...(4)

Using (4), (2) yields C = 11 1 3

0 0 0( ) 3 ( ) 3 / 3s f s Cs ds s f s ds C s

Thus, we have C = 10

( )s f s Cwhich is satisfied by infinitely many values of C if

1

0( ) 0.s f s ds Hence, the given integral

equation (1) has infinitely many solutions if 10

( ) 0.s f s ds Ex. 24. Common data for the following two questions (i) and (ii):

Consider the Fredholm integral equation u (x) = 1

0( )tx xe u t dt

(i). The resolvent kernel R (x, t; ) for this integral equation is(a) (xet) / (1 – ) (b) (xet) / (1 + ) (c) (xet) / (1 + 2) (d) (xet) / (1 – 2) [GATE 2012](ii). The solution of this equation is(a) (x + 1) / (1 – ) (b) x2 / (1 – 2) (c) x / (1 + 2) (d) x / (1 – ) [GATE 2012]

Sol. (i) Ans. (a). Given u (x) = 1

0( )tx xe u t dt …(1)

Comparing (1) with u(x) = f (x) + 1

0( , ) ( ) ,K x t u t dt …(2)

we have u(x) = x, = and K (x, t) = xet …(3)Let Km (x, t) be the mth iterated kernel. Then, we have

K1(x, t) = K (x, t) = xet, using (3)



and Km (x, t) = 1

10

( , ) ( , ) ,mK x z K z t dz where m = 2, 3, 4, …(4)

Putting m = 2 in (5) and using (4), we have

K2 (x, t) = 1

10

( , ) ( , )K x z K z t dz = 1

0

z txe ze dz = 1

0

t zxe ze dz

or K2 (x, t) = 10[( )( ) (1)( )] ,t z zxe z e e on using chain rule of integration by parts

Thus, K2 (x, t) = xet[e – e – (0 – 1)] = xet …(6)Next, putting m = 3 in (5) and using (6), we have

K3(x, t) = 1

20

( , ) ( , )K x z K z t dz = 1

0

z txe ze dz = xet, as before …(7)

and so on. Hence, using Mathematical induction, we have Km (x, t) = xet, for m = 1, 2, 3 … (8)By definition, the required resolvent kernel R (x, t; ) is given by

R(x, t; ) =

1

1

( , )mm

m

K x t =

1

1

,m t

m

x e using (8)

Thus, R (x, t; ) = xet (1 + + 2 + …) = (xet) / (1 – ) …(9)(ii). Ans. (d). As usual, the required solution of (1) is given by

u(x) = f(x) + 1

0( , ; ) ( )R x t f t dt =

1

0,

1

txex t dt since here f (x) = x f (t) = t

or u (x) =

1

01tx

x t e dt =

10[( )( ) (1)( )] ,

1t tx

x t e e as before

or u (x) = x + { (x) / (1 – )} × 1 or u (x) = x / (1 – ).

Ex. 25 The function (x) satisfying the integral equation 0( )

xxe d = x2/2 is

(a) x2/2 (b) x + (x2/2) (c) x – (x2 / 2) (d) 1 + (x2/2)[GATE 2012]

Sol. Ans. (c). Proceed as explained in Art. 5.17 and examples based on it given on page 5.65.

Given 0( )

xxe d = x2/2 …(1)

Differentiating both sides of (1) w.r.t. ‘x’ and using Leibnitz’s rule of differentiation under theintegral sign, we have

0

( )xx ex

() d + ex – x (x) dxdx

– ex–0 (0) 0d

dx or (x) = 0

( ) ,x

xx e d … (2)

which is a Volterra integral equation of the second kind.

Comparing (2) with (x) = f (x) + 0( , ) ( ) ,

xK x d …(3)

we have f (x) = x, = –1 and K (x, ) = ex – …(4)



Let Km (x, ) be the mth iterated kernel. Then, we haveK1 (x, )= K(x, ) = ex – , using (3) …(5)

and Km (x, ) =

1( , ) ( , ) ,x

mK x z K z dz where m = 2, 3, 4, …(6)


K2(x, ) =

1( , ) ( , )xK x z K z dz =

x

x z ze e dz =

x

xe dz

Thus, K2 (x, ) = xxe z = (x – ) ex – …(7)


K3 (x, ) =

2( , ) ( , )xK x z K z dz =

( )

xx z ze e z dz =

( )

xxe z dz

Thus, K3 (x, )= ex –

2( )

2

xz

= ex –

2( )2

x…(8)

Next, putting m = 3 in (6) and using (8), we have

K4(x, ) =

2( , ) ( , )x

K x z K z dz =

2( )2

xx z z z

e e dz =

2( )

2

x xez dz

Thus, K4(x, ) =

3( )

2 3

xxe z = 3( )

,3!

x ze …(9)

and so on, Hence, using Mathematical induction, we have

Km (x, )=

1( ),

( 1)!

mx z

em

where m = 1, 2, 3, … …(10)

Hence, by definition, the resolvent kernel R (x, t; ) of (2) is given by

R (x, ; ) =

1

1

( , )mm

m

K x =

1

1

( 1) ( , ),mm

m

K x since from (3); = –1

= K1 (x, ) – K2 (x, ) + K3(x, ) – K4 (x, ) + … + ad inf

= ex

3( ) ( ) ( )1 inf

1! 2! 1!x x x

ad = ex – e–(x – ) = 1

Hence, as usual, the required function (x) satisfying (1) is given by

(x) = f (x) + 0( , ; ) ( /

xR x f d = 0

xx d = x – (x2/2)


INDEX

I.1

Ableintegral equation, 1.1, 8.1problem, 1.1

adjoint, 4.21operator, 10.1

algebraic multiplicity, 4.22analytic function, 5.10approximate method, 4.29associated Legendre function, 10.10

Besselequation, 10.13function, 7.6, 10.13inequality, 7.4modified function, 10.14

Beta function, 8.2biharmonic equation, 12.20bilinear form, 7.15boundary effects, 14.7boundary value problem (B.V.P.), 2.14branch point, 13.23

Capacity ofcircular disc, 13.5condenser, 14.5conductor, 14.4sphere, 13.7

Cauchygeneral value, 8.9integral, 8.11integral formula, 8.12principal value, 8.9principal value for contour

integrals, 8.10Reimann equation, 8.17sequence, 7.2type singular integral equation, 8.13

cause, 11.37causual Green’s function, 11.42characteristic

functions, 1.6

numbers, 1.6values, 1.6

charge density, 12.7Chebychev polynomial, 7.6Churchill, A.2.compact set, 7.45complete metric space, 7.2completely continuous, 7.45completeness, 7.2complex

Fourier transform, 9.17Hilbert space, 7.2inversion formula, 9.5inversion integral, 9.5symmetric, 1.4

consistency condition, 11.47contour integral, 8.10convergence in mean, 7.4convergent sequence, 7.2convolution, 1.5

integral, 1.5property, 9.4theorem, 9.4

cylinder function, 10.13

Deflection, 11.38degenerate kernel, 1.4density of the Cauchy integral, 8.11difference kernel, 1.5Dirac delta function, 10.5

in general curvilinear coordinates inthree dimensions, B.6.in general curvilinear coordinates in twodimensions, B.4.shifting property of, 10.6three-dimensional, B.1two-dimensional, B.1

Dirichletcondition, 12.1exterior problem, 12.4interior problem, 12.5problem, 12.1

divergence theorem, 12.1


I.2 Integral Equations and Boundary Value Problems

Effect, 11.37eigenfunction, 1.6eigenvalue, 1.6

index of, 3.1multiplicity of, 4.22

electrostatic potential of,annular disc, 12.8axially symmetric conductor, 12.16circular disc, 12.7spherical cap, 13.8

electrostatics, 12.9

Faltung, 1.5final value theorem, 9.3first shifting theorem, 9.2first translation theorem, 9.2Fourier

coefficients, 7.4cosine transform, 9.17integral, 9.1series, 7.5series expansion, 12.17sine transform, 9.17transform, 9.17

Fox’s integral equation, 9.24Fredholm

alternative, 4.20alternative theorem, 4.25classical theory, 6.1determinant, 6.1, 6.7first fundamental theorem, 6.1first series, 6.4minor, 6.7operator, 7.2second fundamental theorem, 6.32second series, 6.5theorem, 4.21third fundamental theorem, 6.36

Fredholm integral equation,homogeneous, 1.3of first kind, 1.3of second kind, 1.3of third kind, 1.3

free space solution, 12.2

fundamentalfunction, 1.6solution, 12.2

Gamma function, 8.2general value of improper integral, 8.9Generalized Green function, 11.48geometric

multiplicity, 4.22series, 5.12

Gram-Schmidt procedure, 7.3Green’s

first identity, 12.2, A.3formula, 10.4second identity, 12.2, A.4tensor, 12.19vector, 12.19

Green’s function, 11.1approach, 12.9causual, 11.42construction of, 11.3for two dimensional Laplace transform,12.34free space, 12.10modified, 11.49

Hankel function, 10.13harmonic

conjugate, 8.17function, 8.17

heat conduction, 12.9Heaviside

expansion formula, 9.4expansion theorem, 9.4unit function, 9.2

Helmholtzequation, 12.18theorem, .C. 7

Hermitian kernel, 1.4

Hilbertformula, 8.17kernel, 8.16Schmidt theorem, 7.17space, 7.2type singular integral equation, 8.18


Integral Equations and Boundary Value Problems I.3

Hilbert transformfinite, 9.19infinite, 9.21

Hilbert type singular integral equation, 8.18Holder

condition, 8.10continuous, 8.10

hypergeometric function, 13.13

Improperfunction, 10.5integral, 8.9

index of eigenvalue, 3.1influence function, 11.37initial value

problem, 2.1theorem, 9.3

inner product, 1.8integrable function, 1.7integral equation, 1.2

Cauchy type, 8.13convolution type, 1.5Fox, 9.24Fredholm, 1.3Hilbert type, 8.18homogeneous, 1.3linear, 1.2non-linear, 1.2of convolution type, 1.5of first kind, 1.3of second kind, 1.3of third kind, 1.2singular, 1.4Volterra, 1.3

integral representation formula, 11.9integral transform methods, 9.1integro-differential equation, 9.5intensity, 11.37inverse Laplace transform, 9.3iterated

function, 1.5kernel, 1.5

iterativemethod, 5.7, 5.35scheme, 5.7, 5.35

Kernel, 1.2complex symmetric, 1.4degenerate, 1.4Hermition, 1.4indefinite, 7.20iterated, 1.5negative definite, 7.20non-negative definite, 7.20non-positive definite, 7.20positive definite, 7.20reciprocal, 1.5resolvent, 1.5separable, 1.4symmetric, 1.4

truncated symmetric, 7.15kinematic viscosity, 12.21kronecker delta, 12.8

2

function, 1.7kernel, 7.1space, 7.2

Laplaceequation, 12.2transform, 9.1

Laplace transform, 9.1of derivatives, 9.2of periodic function, 9.3

Laplacian operator, 12.2Lebesgue

integrable, 1.7integral, 1.7

Legendrefunction, 10.11polynomial, 7.6

Leibnitz’s rule, 1.6linear

independence, 1.4operator, 1.2

linear integral equation offirst kind, 1.3


I.4 Integral Equations and Boundary Value Problems

second kind, 1.3third kind, 1.2

linearity property ofFourier transform, 9.18inverse Laplace transform, 9.3Laplace transform, 9.2

Lipschitz condition, 8.10loading distribution, 11.37low Reynold’s number 14.6

Macdonald function, 10.14matching condition, 11.44maximum principle for harmonicfunction, 12.11Mehler-Dirichlet integral, 13.6Mellin transform, 9.23

inverse of, 9.23convolution theorem for, 9.23

Mercer’s theorem, 7.21mermorphic function, 6.1method of images, 12.14method of successive approximations for

Fredholm integral equation, 5.7Volterra integral equation, 5.35

metricnatural, 7.2space, 7.2

Minkowski inequality, 1.8mixed boundary value problem, 13.1modified

Bessel function, 10.14Green’s function, 11.49

multiplicity, 4.22

Natural metric, 7.2Naumann

condition, 12.1exterior problem, 12.6function, 10.13interior problem, 12.6problem, 12.1series, 5.7, 5.35

Newtonian potential, 12.3non-linear integral equation, 1.2

non-trivial solution, 1.6non-zero solution, 1.6norm, 1.8normalized, 1.8

Operatoradjoint, 10.1bounded, 7.45completely continuous, 7.45Fredholm, 7.2linear, 1.2method, 7.44self-adjoint, 10.1

ordinary differential equation, 2.1orthogonal system/set, 7.3orthogonality of

eigenfunction, 7.12two functions, 1.8

orthonormalset, 7.3system, 7.3functions, 7.3

Oseen flow, 12.21rotary motion in, 14.15translation motion in, 14.14

Parseval’s identity, 7.5partial differential equation,

elliptic, 12.1hyperbolic, 12.1parabolic, 12.1

periodic function, 9.3permanently convergent, 7.10perturbation

parameter, 14.1techniques, 14.1

Picard’s method, 5.1Plemelj formulas, 8.11Poincare-Bertrand transformation formulae,

8.11Poisson’s

equation, 12.2integral formula, 12.12

pole, 13.23


Integral Equations and Boundary Value Problems I.5

potential layerdouble, 12.3single, 12.3volume, 12.3

pressure, 12.19principal axes of resistance, 14.8Reality of eigenvalues, 7.12reciprocal

function, 5.29kernel, 1.5

reciprocity principle, 12.11regular curve, 8.10regularity conditions, 1.7resolvent kernel, 1.5Reynolds number, 12.21Reimann

Hilbert problem, 8.14integrable, 1.7

Reisz-Fischer theorem, 7.4

Scalar product, 1.8Schmidt solution of Fredholm integral-

equation, 7.21Schwarz inequality, 1.8second shifting theorem for

inverse Laplace transform, 9.2Laplace transform, 9.2

second translation theorem forinverse Laplace transform, 9.2Laplace transform, 9.2

self adjoint equation, 10.2separable kernel, 1.4shifting property of Fourier transform, 9.18singular integral, 8.10

equation, 1.4, 8.1equation weakly, 8.6

singularity solution, 12.24solution by successive approximations for

Fredholm integral equation, 5.7Volterra integral equation, 5.35

solution by successive substitutions forFredholm integral equation, 5.3Volterra integral equation 5.5

Solution ofan integral equation, 1.8Volterra integral equation of the firstkind 5.63

sourcedensity, 14.10function, 12.24

space form of the wave equation, C.7.spectrum of kernel, 7.1spherical harmonics, 10.9square integrable functions, 1.7Stokes flow,

boundary effect, 14.7in unbounded medium, 12.19logitudinal oscillations, 14.8rotary motion in, 14.9rotary oscillations in, 14.11steady, 14.6

Strum-Liouville problem, 3.14symmetric kernel, 1.4

Taylor’s theorem, 14.7three part boundary value problem, 13.8

generalized, 13.17trace of a kernel, 7.8transpose, 4.22trivial solution, 1.6truncated symmetric kernel, 7.15two part boundary value problem, 13.1

generalized, 13.14

Unit step function, 9.2unit source, 12.9

Velocity vector, 12.19Volterra integral equation

homogeneous, 1.4of first kind, 1.3of second kind, 1.4of third kind, 1.3

volume potential, 12.3

Weakly singular kernel, 8.6Wronskian, 11.3

Zero solution, 1.6


Documents

INTEGRAL EQUATIONS AND BOUNDARY VALUE PROBLEMS · PREFACE This book on ‘‘Linear integral equations and boundary value problems’’ has been specially written as per latest UGC