An Introduction to Integral Transforms

An Introduction to Integral Transforms

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An Introduction to Integral Transforms

Baidyanath Patra, Ph. D.Ex-Professor, Bengal Engineering Science University

(Recently named as IIEST, Shibpur)

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Dedication

I dedicate this book to my beloved wife, Mrs. Minu Patra, for standing beside me throughout my career and during the writing of this book. She has been my inspi-ration and motivation to expand my knowledge and without her selfless support and encouragement this book would not have been possible.


Preface

Integral transform is one of the powerful tools in Applied Mathematics. This book provides an introduction to this subject for solving boundary value and initial value problems in mathematical physics, Engineering Science and related areas.

During my fifty years teaching and research experiences at the Indian Institute of Engineering Science and Technology, Shibpur, formerly known as Bengal Engineering College and Bengal Engineering & Science University, I have been introduced to the subject of Integral Transforms.

Here an attempt has been made to cover the basic theorems of commonly used various integral transform techniques with their applications.

The transforms that have been covered in this book in detail are Fourier Transfoms, Laplace Transforms, Hilbert and Stieltjes Transforms, Mellin Transforms, Hankel Transforms, Kontorovich-Lebedev Transforms, Legendre Transforms, Mehler-Fock Transforms, Jacobi-Gegenbauer-Laguerre- Hermite Transforms, and the Z Transform.

This textbook targets the graduate and some advanced undergraduate students. My endeavor will be considered fruitful if it serves the need of at least one reader.

Date : January, 2016 B. Patra


Acknowledgement

I would like to take this opportunity to express my gratitude to my respected teacher, Professor S C Dasgupta, D. Sc.. He introduced me to this subject and developed my career in this field of research with special emphasis to application of Integral Transforms.

I am indebted to my many research students who worked under my supervision on this subject. I also acknowledge the authors of all the books that I read and results of which have been used in preparation of this book.

I would also like to thank my sons and their families for their constant encouragement during the writing of this book. Special thanks goes to my grandsons for their support for this work despite all the time it took me away from them.

During the process of writing, I have been greatly inspired by the textbooks of I. N. Sneddon and L. K. Debnath on this subject. Many other books and research papers, all of which are not possible to be mentioned individually, have also been consulted during this process.

I am thankful to the publishers, Levant Books for their unfailing co-operation in bringing this book to light within a reasonable amount of time.


Contents

1 FOURIER TRANSFORM 1

1.1 Introduction. . . . . . . . . . . . . . . . . . . . . . . . . . 1

1.2 Classes of functions . . . . . . . . . . . . . . . . . . . . . . 2

1.3 Fourier Series and Fourier Integral Formula . . . . . . . . 2

1.4 Fourier Transforms . . . . . . . . . . . . . . . . . . . . . . 6

1.4.1 Fourier sine and cosine Transforms. . . . . . . . . . 7

1.5 Linearity property of Fourier Transforms. . . . . . . . . . 8

1.6 Change of Scale property. . . . . . . . . . . . . . . . . . . 9

1.7 The Modulation theorem. . . . . . . . . . . . . . . . . . . 10

1.8 Evaluation of integrals by means of inversion theorems. . 11

1.9 Fourier Transform of some particular functions. . . . . . . 13

1.10 Convolution or Faltung of two integrable functions. . . . . 20

1.11 Convolution or Falting or Faltung Theorem for FT. . . . . 21

1.12 Parseval’s relations for Fourier Transforms. . . . . . . . . 23

1.13 Fourier Transform of the derivative of a function. . . . . . 26

1.14 Fourier Transform of some more useful functions. . . . . . 30

1.15 Fourier Transforms of Rational Functions. . . . . . . . . . 36

1.16 Other important examples concerning derivative of FT. . 37

1.17 The solution of Integral Equations of Convolution Type. . 47

1.18 Fourier Transform of Functions of several variables. . . . . 53

1.19 Application of Fourier Transform to Boundary Value Prob-lems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55

2 FINITE FOURIER TRANSFORM 79

2.1 Introduction. . . . . . . . . . . . . . . . . . . . . . . . . . 79

2.2 Finite Fourier cosine and sine Transforms. . . . . . . . . . 79

2.3 Relation between finite Fourier Transform of the deriva-tives of a function. . . . . . . . . . . . . . . . . . . . . . . 81

2.4 Faltung or convolution theorems for finite Fourier Trans-form. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82

2.5 Multiple Finite Fourier Transform. . . . . . . . . . . . . . 85

2.6 Double Transforms of partial derivatives of functions. . . . 86

2.7 Application of finite Fourier Transforms to boundary valueproblems. . . . . . . . . . . . . . . . . . . . . . . . . . . . 87

3 THE LAPLACE TRANSFORM 102

3.1 Introduction. . . . . . . . . . . . . . . . . . . . . . . . . . 102

3.2 Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . 103

3.3 Sufficient conditions for existence of Laplace Transform. . 103

3.4 Linearity property of Laplace Transform. . . . . . . . . . 104

3.5 Laplace transforms of some elementary functions . . . . . 105

3.6 First shift theorem. . . . . . . . . . . . . . . . . . . . . . . 107

3.7 Second shift theorem. . . . . . . . . . . . . . . . . . . . . 107

3.8 The change of scale property. . . . . . . . . . . . . . . . . 107

3.9 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . 108

3.10 Laplace Transform of derivatives of a function. . . . . . . 110

3.11 Laplace Transform of Integral of a function . . . . . . . . 112

3.12 Laplace Transform of tnf(t) . . . . . . . . . . . . . . . . . 113

3.13 Laplace Transform of f(t)/t . . . . . . . . . . . . . . . . 114

3.14 Laplace Transform of a periodic function. . . . . . . . . . 115

3.15 The initial-value theorem and the final-value theorem ofLaplace Transform. . . . . . . . . . . . . . . . . . . . . . . 116

3.16 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . 117

3.17 Laplace Transform of some special functions. . . . . . . . 121

3.18 The Convolution of two functions. . . . . . . . . . . . . . 131

3.19 Applications . . . . . . . . . . . . . . . . . . . . . . . . . . 132

4 THE INVERSE LAPLACE TRANSFORM ANDAPPLICATION 141

4.1 Introduction. . . . . . . . . . . . . . . . . . . . . . . . . . 141

4.2 Calculation of Laplace inversion of some elementary func-tions. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 143

4.3 Method of expansion into partial fractions of the ratio oftwo polynomials . . . . . . . . . . . . . . . . . . . . . . . 145

4.4 The general evaluation technique of inverse Laplace trans-form. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 153

4.5 Inversion Formula from a different stand point : The Tri-comi’s method. . . . . . . . . . . . . . . . . . . . . . . . . 158

4.6 The Double Laplace Transform . . . . . . . . . . . . . . . 161

4.7 The iterative Laplace transform. . . . . . . . . . . . . . . 166

4.8 The Bilateral Laplace Transform. . . . . . . . . . . . . . . 166

4.9 Application of Laplace Transforms. . . . . . . . . . . . . . 168

5 Hilbert and Stieltjes Transforms 220

5.1 Introduction. . . . . . . . . . . . . . . . . . . . . . . . . . 220

5.2 Definition of Hilbert Transform . . . . . . . . . . . . . . . 220

5.3 Some Important properties of Hilbert Transforms. . . . . 221

5.4 Relation between Hilbert Transform and Fourier Transform.225

5.5 Finite Hilbert Transform. . . . . . . . . . . . . . . . . . . 226

5.6 One-sided Hilbert Transform. . . . . . . . . . . . . . . . . 227

5.7 Asymptotic Expansions of one-sided Hilbert Transform. . 228

5.8 The Stieltjes Transform. . . . . . . . . . . . . . . . . . . . 230

5.9 Some Deductions. . . . . . . . . . . . . . . . . . . . . . . . 231

5.10 The Inverse Stieltjes Transform. . . . . . . . . . . . . . . . 232

5.11 Relation between Hilbert Transform and Stieltjes Trans-form. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 234

6 Hankel Transforms 238

6.1 Introduction. . . . . . . . . . . . . . . . . . . . . . . . . . 238

6.2 The Hankel Transform. . . . . . . . . . . . . . . . . . . . 238

6.3 Elementary properties . . . . . . . . . . . . . . . . . . . . 238

6.4 Inversion formula for Hankel Transform. . . . . . . . . . . 242

6.5 The Parseval Relation for Hankel Transforms. . . . . . . . 244

6.6 Illustrative Examples: . . . . . . . . . . . . . . . . . . . . 245

7 Finite Hankel Transforms 260

7.1 Introduction. . . . . . . . . . . . . . . . . . . . . . . . . . 260

7.2 Expansion of some functions in series involving cylinderfunctions : Fourier-Bessel Series. . . . . . . . . . . . . . . 260

7.3 The Finite Hankel Transform. . . . . . . . . . . . . . . . . 262

7.4 Illustrative Examples. . . . . . . . . . . . . . . . . . . . . 263

7.5 Finite Hankel Transform of order n in 0 � x � 1 of thederivatrive of a function. . . . . . . . . . . . . . . . . . . . 265

7.6 Finite Hankel Transform over 0 � x � 1 of order n ofd2fdx2 + 1

xdfdx , when p is the root of Jn(p) = 0. . . . . . . 266

7.7 Finite Hankel Transform of f ′′(x) + 1x f ′(x) − n2

x2 f(x), wherep is the root of Jn(p) = 0 in 0 � x � 1 . . . . . . . . . . 266

7.8 Other forms of finite Hankel Transforms. . . . . . . . . . . 267

7.9 Illustrations. . . . . . . . . . . . . . . . . . . . . . . . . . 268

7.10 Application of finite Hankel Transforms. . . . . . . . . . . 269

8 The Mellin Transform 277

8.1 Introduction. . . . . . . . . . . . . . . . . . . . . . . . . . 277

8.2 Definition of Mellin Transform. . . . . . . . . . . . . . . . 278

8.3 Mellin Transform of derivative of a function. . . . . . . . . 281

8.4 Mellin Transform of Integral of a function. . . . . . . . . . 283

8.5 Mellin Inversion theorem. . . . . . . . . . . . . . . . . . . 285

8.6 Convolution theorem of Mellin Transform. . . . . . . . . . 286

8.7 Illustrative solved Examples. . . . . . . . . . . . . . . . . 287

8.8 Solution of Integral equations. . . . . . . . . . . . . . . . . 292

8.9 Application to Summation of Series. . . . . . . . . . . . . 293

8.10 The Generalised Mellin Transform. . . . . . . . . . . . . . 295

8.11 Convolution of generalised Mellin Transform. . . . . . . . 297

8.12 Finite Mellin Transform. . . . . . . . . . . . . . . . . . . . 297

9 Finite Laplace Transforms 302

9.1 Introduction. . . . . . . . . . . . . . . . . . . . . . . . . . 302

9.2 Definition of Finite Laplace Transform. . . . . . . . . . . . 302

9.3 Finite Laplace Transform of elementary functions. . . . . 304

9.4 Operational Properties. . . . . . . . . . . . . . . . . . . . 307

9.5 The Initial Value and the Final Value Theorem . . . . . . 311

9.6 Applications . . . . . . . . . . . . . . . . . . . . . . . . . . 312

10 Legendre Transforms 317

10.1 Introduction. . . . . . . . . . . . . . . . . . . . . . . . . . 317

10.2 Definition of Legendre Transform. . . . . . . . . . . . . . 317

10.3 Elementary properties of Legendre Transforms. . . . . . . 318

10.4 Operational Properties of Legendre Transforms . . . . . . 323

10.5 Application to Boundary Value Problems. . . . . . . . . . 325

11 The Kontorovich-Lebedev Transform 328

11.1 Introduction. . . . . . . . . . . . . . . . . . . . . . . . . . 328

11.2 Definition of Kontorovich - Lebedev Transform. . . . . . . 328

11.3 Parseval Relation for Kontorovich-Lebedev Transforms. . 329

11.4 Illustrative Examples. . . . . . . . . . . . . . . . . . . . . 330

11.5 Boundary Value Problem in a wedge of finite thickness. . 332

12 The Mehler-Fock Transform 335

12.1 Introduction. . . . . . . . . . . . . . . . . . . . . . . . . . 335

12.2 Fock’s Theorem (with weaker restriction). . . . . . . . . . 335

12.3 Mehler-Fock Transform of zero order and its properties. . 337

12.4 Parseval type relation. . . . . . . . . . . . . . . . . . . . . 339

12.5 Mehler-Fock Transform of order m . . . . . . . . . . . . . 341

12.6 Application to Boundary Value Problems. . . . . . . . . . 342

12.6.1 First Example . . . . . . . . . . . . . . . . . . . . 342

12.6.2 Second Example . . . . . . . . . . . . . . . . . . . 344

12.6.3 Third Example . . . . . . . . . . . . . . . . . . . . 345

12.6.4 Fourth Example . . . . . . . . . . . . . . . . . . . 347

12.7 Application of Mehler-Fock Transform for solving dualintegral equation. . . . . . . . . . . . . . . . . . . . . . . . 348

13 Jacobi, Gegenbauer, Laguerre and Hermite Transforms351

13.1 Introduction. . . . . . . . . . . . . . . . . . . . . . . . . . 351

13.2 Definition of Jacobi Transform. . . . . . . . . . . . . . . . 351

13.3 The Gegenbauer Transform. . . . . . . . . . . . . . . . . . 355

13.4 Convolution Theorem . . . . . . . . . . . . . . . . . . . . 356

13.5 Application of the Transforms . . . . . . . . . . . . . . . . 357

13.6 The Laguerre Transform . . . . . . . . . . . . . . . . . . . 359

13.7 Operational properties . . . . . . . . . . . . . . . . . . . . 361

13.8 Hermite Transform. . . . . . . . . . . . . . . . . . . . . . 364

13.9 Operational Properties. . . . . . . . . . . . . . . . . . . . 366

13.10Hermite Transform of derivative of a function. . . . . . . . 367

14 The Z-Transform 372

14.1 Introduction. . . . . . . . . . . . . . . . . . . . . . . . . . 372

14.2 Z - Transform : Definition. . . . . . . . . . . . . . . . . . 372

14.3 Some Operational Properties of Z-Transform. . . . . . . . 376

14.4 Application of Z-Transforms. . . . . . . . . . . . . . . . . 383

Appendix 390Bibliography 405Index 407

Chapter 1

FOURIER TRANSFORM

1.1 Introduction.

The method of integral transforms is one of the most easy and effectivemethods for solving problems arising in Mathematical Physics, AppliedMathematics and Engineering Science which are defined by differentialequations, difference equations and integral equations. The main ideain the application of the method is to transform the unknown function,say, f(t) of some variable t to a different function, say, F (p) of a complexvariable p. Then the associated differential equation can be directly re-duced to either a differential equation of lower dimension or an algebraicequation in the variable p. There are several forms of integral transformsand one form may be obtained from the other by a transformation of thecoordinates and the functions. The choice of integral transform dependson the structure of the equation and on the geometry of the domainunder consideration. This method of integral transform simplifies thecomputational techniques considerably.

Suppose that there exists a known function K(p, t) of two variablesp and t and that ∫ b

aK(p, t) f(t) dt = F (p) (1.1)

is convergent. Then F (p) is called the integral transform of the functionf(t) by the function K(p, t), which is called the kernel of the transform.Here, the variable p is a parameter, which may be real or complex. Dif-ferent forms of kernels will generate different form of transforms. Belowwe discuss a variety of integral transforms with their applications afterusing different forms of kernels in variety of domains. To begin with, weconsider Fourier Transform.

2 An Introduction to Integral Transforms

1.2 Classes of functions

A singlevalued function f(x) of the independent variable x which iscontinuous in an interval [a,b], is said to belong to a class denoted byf ∈ C [a, b].

A function f(x) is said to be piecewise continuous in an interval (a, b)if the interval can be partitioned into finite number of non-intersectingsubintervals (a, a1), (a1, a2), ..., (an−1, b), in each of which the function iscontinuous and has finite limits as x approaches the end points of eachof the sub-intervals. Such a function is said to belong to a class denotedby f ∈ P (a, b).

A piecewise continuous function f in (a, b), whose first order deriv-ative is also a piecewise continuous function in (a, b) and belongs to aclass denoted by f ∈ P 1(a, b).

The set of functions f(x) is said to be absolutely integrable over Ω,if∫Ω |f(x)|dx is finite. Then we say that this function belongs to a class

denoted by f(x) ∈ A1(Ω). Similarly, the statement f ∈ Am(Ω) implies∫Ω |f(x)|mdx is finite.

Finally, we introduce a class of functions f(x), which satisfies thefollowing conditions

(i) f(x) is defined in c < x < c + 2 l

(ii) f(x) is periodic function of period 2 l and(iii) f(x) and f ′(x) are piecewise continuous in c < x < c+2l denoting

by f(x) ∈ P 1(c, c + 2l).

This class of function f(x) is said to satisfy Dirichlet’s conditions.

A function f(x) is said to be of exponential order σ as x → ∞, ifconstants σ, m(> 0) can be so found that |e−σxf(x)| < m implying|f(x)| < m emσ for x > x0. Equivalently, we also write f(x) = 0 (emσ)as x → ∞.

1.3 Fourier Series and Fourier Integral Formula

Suppose that a function f(x) satisfies Dirichlet’s conditions over theinterval (−l, l) and it belongs to the class P 1(R) and also to the class

Fourier Transform 3

A1(R). Let the principal period of f(x) be 2 l. Then f(x) admits of theFourier series

f(x) = a0 +∞∑

n=1

[an cos

nπx

l+ bn sin

nπx

l

]

where a0 =12l

∫ l

−lf(x)dx

(an, bn) =1l

∫ l

−lf(x)

[cos

nπx

l, sin

nπx

l

]dx

so that an − i bn =1l

∫ l

−lf(x) e

−inπxl dx

Let us now set a0 = c0

an = cn + c−n

i bn = cn − c−n

Then, f(x) = c0 +∞∑

n=1

[cne

−inπxl + c−ne

inπxl

]

=+∞∑−∞

cne−inπx

l (1.2)

where, cn =12l

∫ l

−lf(x) e

inπxl dx =

12l

∫ l

−lf(t) e

inπtl dt

(1.3)

This form of Fourier series is called the complex form of Fourier seriesin (−l, l). Thus, from (1.2) and (1.3), one gets

f(x) =+∞∑−∞

[12l

∫ l

−lf(t) e

inπtl

dt

]e−

inπxl (1.4)

Let us now put πl = δξ. We note that δξ → 0 as l → ∞ such that

l.δξ = π = a finite number. Then the series (1.4), before taking thelimit as δξ → 0, becomes

f(x) =12π

+∞∑−∞

δξ

[∫ πδξ

−πδξ

f(t) eint δξ dt

]e−inx δξ

=12π

∫ πδξ

−πδξ

f(t)

[+∞∑−∞

δξ.ein(t−x)δξ

]dt


after interchanging formally the summation and integration signs. Since,by assumption f(x) ∈ P 1(R) and also f(x) ∈ A1(R), making δξ → 0and using the definition of Riemann definite integral as limit of a sumwe get

f(x) =12π

∫ +∞

−∞f(t)

[∫ +∞

−∞ei(t−x)ydy

]dt (1.5)

⇒ f(x) =1√2π

∫ +∞

−∞e−iyx[F (y)] dy, (1.6)

where F (y) =1√2π

∫ +∞

−∞f(t)eiyt dt (1.7)

Again, from eqn. (1.5) one gets

f(x) =1π

∫ ∞

0dy

∫ +∞

−∞f(t) cos y(−t + x) dt (1.8)

Eqn. (1.8) is known as the Fourier Integral formula. At a point offinite discontinuity of f(x), the left hand side of (1.8) is replaced by12 [f(x + 0) + f(x − 0)] in the sense of limiting values.

A detailed proof of this statement is included in the following corol-laries.

Coro 1.1 Fourier’s Integral Theorem.

If f(t) ∈ P 1(R) and also f(t) ∈ A1(R), then for all x ∈ R,

1π

∫ ∞

0dy

∫ +∞

−∞f(t) cos[y(x − t)]dt =

12[f(x + 0) + f(x − 0)]

Proof. If k > 0,∫ ∞

0f(t)dt

∫ λ

0cos[y(t − x)]dy −

∫ λ

0dy

∫ ∞

0f(t) cos[y(t − x)]dt

=∫ k

0f(t)dt

∫ λ

0cos[y(t − x)]dy +

∫ ∞

kf(t)dt

∫ λ

0cos[y(t − x)]dy

−∫ λ

0dy

∫ k

0f(t) cos[y(t − x)]dt −

∫ λ

0dy

∫ ∞

kf(t) cos[y(t − x)]dt

=∫ ∞

kf(t)dt

∫ λ


∫ λ

0dy

∫ ∞


Since f(t) ∈ A1(R), for every arbitrary positive number ∈, we canfind a number K1 such that∫ ∞

k|f(t)| dt <

∈2λ

, k > K1

Fourier Transform 5

Thus, ∣∣∣∣∫ λ

0dy

∫ ∞


∣∣∣∣ <∫ λ

0dy

∫ ∞

k|f(t)|dt <

∈2

.

Also, since∫ λ

0cos[y(t − x)]dy =

sin[λ(t − x)]t − x

it follows that∣∣∣∣∫ ∞

kf(t)dt

∫ ∞

0cos[y(t − x)]dy

∣∣∣∣ =∣∣∣∣∫ ∞

k−xf(t + x)

sinλ t

tdt

∣∣∣∣Thus, there exists a number K2 such that if k > K2 + x,∣∣∣∣∫ ∞

k−xf(t + x)[sin λt/t]dt

∣∣∣∣ < ∈2

.∣∣∣∣∫ ∞

k−xf(t)dt

∫ λ


∫ λ

0dy

∫ ∞


∣∣∣∣ <∈,

for k > K = max(K1,K2) even for large values of λ

In other words,

limλ→∞

∫ ∞

0f(t)dt

∫ λ

0cos[y(t − x)]dy

= limλ→∞

∫ λ

0dy

∫ ∞

0f(t) cos[y(t − x)]dt (1.9)

In an exactly similar way we can show that

limλ→∞

∫ 0

−∞f(t)dt

∫ λ

0cos[y(t − x)]dy

= limλ→∞

∫ λ

0dy

∫ 0

−∞f(t) cos[y(t − x)]dt (1.10)

Adding (1.9) and (1.10) we get

limλ→∞

∫ ∞

−∞f(t)dt

∫ λ

0cos[y(t − x)]dy

= limλ→∞

∫ λ

0dy

∫ +∞

−∞f(t) cos[y(t − x)]dt

Again since,∫ 0

−∞f(x + t)

sin λ t

tdt =

∫ ∞

0f(x − t)

sin λ t

tdt

and for all x ∈ R,

∫ +∞

−∞f(x + t)

sin λ t

tdt → π

2[f(x + 0) + f(x − 0)].


We have,

12[f(x + 0) + f(x − 0)] = lim

λ→∞1π

∫ +∞

−∞f(x + u)

sin λ u

udu

= limλ→∞

1π

∫ +∞

−∞f(t)

sin[λ(t − x)]t − x

dt

= limλ→∞

1π

∫ +∞

−∞f(t)dt

∫ λ

0cos[y(t − x)] dy

= limλ→∞

1π

∫ λ

0dy

∫ +∞

−∞f(t) cos[y(t − x)] dt

Thus, the result of the Fourier Integral Theorem follows immediately.

In addition, if f(t) is continuous at the point t = x, we have

12π

∫ +∞

−∞e−iyxdy

∫ +∞

−∞f(t) eiytdt = f(x) (1.11)

From eqn. (1.11) the result in eqn. (1.8) of article 1.3 will follow.

1.4 Fourier Transforms

We are now ready to define Fourier transform of a piecewise continuousfunction f(x) ∈ P 1(R) and f(x) ∈ A1(R). It is defined by

F [f(x);x → ξ] = F [f(x)] = F (ξ) = f(ξ) =1√2π

∫ +∞

−∞f(x)eiξxdx

(1.12)

and f(x) = F−1[F (ξ)] =1√2π

∫ +∞

−∞F (ξ) e−iξx dξ (1.13)

at a point of continuity of f(x).

F (ξ) in (1.12) is called Fourier transform of f(x) and f(x) in (1.13) iscalled the inverse Fourier transform of F (ξ).

Some authors define Fourier transform (1.12) of f(x) and inversionformula in (1.13) by

F [f(x) ; x → ξ] = F (ξ) =∫ +∞

−∞f(x) eiξxdx (1.14)

and f(x) =12π

∫ +∞

−∞F (ξ) e−iξxdξ (1.15)

respectively.

Fourier Transform 7

At a point of discontinuity of x ∈ R, the left hand sides of eqns.(1.13) and (1.15) take the form

12

[f(x + 0) + f(x − 0)]

instead of f(x).

From the definitions of Fourier transform and Inverse Fourier trans-form we can, therefore, write from (1.12)

F (ξ) = F [f(x)] = F [F−1(F (ξ))] ⇒ F F−1[F (ξ)] ≡ I[F (ξ)]

⇒ F F−1 = I, an identity operator, F and F−1 being Fourier andits inverse operators respectively and from (1.13)

f(x) = F−1[F (ξ)] = F−1[F (f(x))] = F−1F [f(x)] = I[f(x)]

⇒ F−1F = I (1.16)

Thus in operator notation FF−1 = F−1 F = I . It shows that theseoperators F and F−1 are commutative.

1.4.1 Fourier sine and cosine Transforms.

Let in addition belonging to the classes P 1(R) and A1(R), the functionf(x) be an odd function of x ∈ R. Then, clearly f(−x) = −f(x), Now,by (1.12)

F [f(x);x → ξ] = F (ξ) =1√2π

∫ ∞

0f(x)

[eiξx − e−iξx

]dx

= i

√2π

∫ ∞

0f(x) sin ξx dx

≡ i Fs(ξ), say (1.17)

where, Fs(ξ) =

√2π

∫ ∞

0f(x) sin ξx dx (1.18)

Fs(ξ) in (1.18) is defined to be Fourier sine transform of the functionf(x). Its connection with the Fourier transform is given by (1.17) pro-vided f(x) in an odd function of x ∈ R. The inversion formula of (1.18)is then defined by

f(x) =

√2π

∫ ∞

0Fs(ξ) sin ξx dξ (1.19)


Let f(x) be an even function of x ∈ R in addition belonging to classesP 1(R) and A1(R). Then, f(−x) = f(x) and therefore, by (1.12) above

F [f(x);x → ξ] = F (ξ) =1√2π

∫ ∞

0f(x)

[eiξx + e−iξx

]dx

=

√2π

∫ ∞

0f(x) cos ξx dx

≡ Fc(ξ), say, (1.20)

where Fc(ξ) =

√2π

∫ ∞

0f(x) cos ξx dx (1.21)

Fc(ξ) in (1.21) is defined to be Fourier cosine transform of the func-tion f(x). Its connection with the Fourier transform is given by eqn.(1.20), provided f(x) is an even function of x ∈ R. The inversion for-mula of Fourier cosine transform defined by (1.21) is then defined as

f(x) =

√2π

∫ ∞

0Fc(ξ) cos ξx dξ (1.22)

1.5 Linearity property of Fourier Transforms.

If c1 and c2 are constants, then

(i) F [c1 f1(x) + c2 f2(x)] = c1 F [f1(x)] + c2 F [f2(x)]

(ii) Fs[c1 f1(x) + c2 f2(x)] = c1 Fs[f1(x)] + c2 Fs[f2(x)]

(iii) Fc[c1 f1(x) + c2 f2(x)] = c1 Fc[f1(x)] + c2 Fc[f2(x)]

Proof.

(i) By definition of Fourier Transform in eqn.(1.12) of article 1.4, wehave

F [c1 f1(x) + c2 f2(x);x → ξ]

=1√2π

∫ +∞

−∞[c1 f1(x) + c2 f2(x)] · eiξx dx

= c1 · 1√2π

∫ +∞

−∞f1(x)eiξx dx + c2 · 1√

2π

∫ +∞

−∞f2(x)eiξx dx

= c1 F [f1(x);x → ξ] + c2 F [f2(x);x → ξ]

Fourier Transform 9

(ii) By definition of Fourier sine transform in eqn. (1.18) of article1.4.1 we get

Fs[c1 f1(x) + c2 f2(x);x → ξ] =

√2π

∫ ∞

0c1 f1(x) sin ξx dx

+

√2π

∫ ∞

0c2 f2(x) sin ξx dx

= c1 Fs[f1(x);x → ξ] + c2 Fs[f2(x);x → ξ].

(iii) By definition of Fourier cosine transform in eqn.(1.21) of article1.4.1 proceeding similarly as above, we have

Fc[c1 f1(x) + c2 f2(x)] = c1 Fc[f1(x)] + c2 Fc[f2(x)] .

1.6 Change of Scale property.

(i) If F [f(x);x → ξ] = F (ξ), then F [f(ax) ; x → ξ] = 1a F

(ξa

)

Proof. Since F [f(x);x → ξ] =1√2π

∫ +∞

−∞f(x)eiξx dx = F (ξ),

we have F [f(ax);x → ξ] =1√2π

∫ +∞

−∞f(ax)eiξx dx

=1√2π

∫ +∞

−∞f(t)ei ξ

at 1a

dt

=1a· 1√

2π

∫ +∞

−∞f(t)ei( ξ

a)t dt =1aF

(ξ

a

).

(ii) If Fs[f(x)] = Fs(ξ), then Fs[f(ax)] = 1a Fs

(ξa

)Proof. By definition

Fs[f(x);x → ξ] =

√2π

∫ ∞

0f(x) sin ξx dx = Fs(ξ).

Therefore,

Fs[f(ax);x → ξ] =

√2π

∫ ∞

0f(ax) sin ξx dx

=1a

√2π

∫ ∞

0f(t) sin

(ξ

at

)dt

=1aFs

(ξ

a

)


(iii) Similarly, it is easy to prove that if

Fc[f(x);x → ξ] = Fc(ξ), then Fc[f(ax);x → ξ] = 1a Fc

(ξa

).

1.6 The shifting property of Fourier Transform.

If F (ξ) is Fourier transform of f(x) in R, then Fourier transform off(x − a) is eiξaF (ξ).

Proof. By definition, we have

F (ξ) =1√2π

∫ +∞

−∞f(x) eiξx dx = F [f(x);x → ξ]

Therefore,

F [f(x − a);x → ξ] =1√2π

∫ +∞

−∞f(x − a) eiξx dx

=1√2π

∫ +∞

−∞f(t) eiξ(t+a) dt

= eiξa · 1√2π

∫ +∞

−∞f(t) eiξt dt

= eiξaF (ξ).

1.7 The Modulation theorem.

Theorem : If F [f(x);x → ξ] = F (ξ), then F [f(x) cos ax ;x → ξ]= 1

2 [F (ξ + a) + F (ξ − a)].

Proof. Since it is given that

F [f(x);x → ξ] = F (ξ) =1√2π

∫ +∞

−∞f(x)eiξx dx

we have

F [f(x) cos ax ;x → ξ] =1√2π

∫ +∞

−∞f(x) · eiax + e−iax

2· eiξx dx

=12

[1√2π

∫ +∞

−∞f(x) ei(ξ+a)x dx

+1√2π

∫ +∞

−∞f(x) ei(ξ−a)x dx

]= [F (ξ + a) + F (ξ − a)].

Other three parts of the theorem are

(a) if Fs[f(x);x → ξ] = Fs(ξ) then

Fourier Transform 11

Fs[f(x) cos ax ;x → ξ] =12[Fs(ξ + a) + Fs(ξ − a)]

(b) if Fs[f(x);x → ξ] = Fs(ξ) then

Fc[f(x) sin ax ;x → ξ] =12[Fs(ξ + a) − Fs(ξ − a)]

and (c) if Fc[f(x);x → ξ] = Fc(ξ) then

Fs[f(x) sin ax ;x → ξ] =12[Fc(ξ − a) − Fc(ξ + a)]

These results can similarly be proved as was done in the first case onreplacing cos ax and sin ax by their exponential forms in the definitionsof the corresponding transforms of the left hand side.

1.8 Evaluation of integrals by means of inversion theorems.

By definitions of Fourier transform and its inversion formula in eqns.(1.12), (1.13), (1.17), (1.19), (1.21) and (1.22) of article 1.4, we mayemploy these results to evaluate certain integrals involving trigonometricfunctions. For example, first of all we consider

I1 =∫ ∞

0e−bx cos ax dx , I2 =

∫ ∞

0e−bx sin ax dx

Integrating by parts we can now evaluate I1 and I2 as

I1 =b

a2 + b2, I2 =

a

a2 + b2

These results means that if we set f(x) = e−bx, then its cosine andsine transforms are given by

Fc

[e−bx;x → ξ

]=

√2π

b

ξ2 + b2

Fs

[e−bx;x → ξ

]=

√2π

ξ

ξ2 + b2

Substituting these expressions in eqns. (1.19) and (1.22) of article1.4.1, we get ∫ ∞

0

cos ξx

ξ2 + b2dξ =

π

2be−bx

and∫ ∞

0

ξ sin ξx

ξ2 + b2dξ =

π

2e−bx


As a second example, if we take f(x) =

{1 , 0 < x < a

0 , x > a,

then

Fc(ξ) =

√2π

∫ a

01 cos ξx dx =

√2π

sin ξa

ξ

and so it gives

2π

∫ ∞

0

sin ξa

ξcos ξx dξ =

{1, if 0 < x < a

0, if x � a

Let us consider a third example when f(x) =

⎧⎪⎨⎪⎩

0 , 0 < x < a

x , a ≤ x ≤ b

0 , x > b

Then,

Fs(ξ) =

√2π

∫ b

ax sin ξx dx

=

√2π

[a cos ξ a − b cos ξ b

ξ+

sin ξ b − sin ξ a

ξ2

]

and this result gives

2π

∫ ∞

0sin ξ x [Fs(ξ)] dξ =

⎧⎪⎨⎪⎩

0 , 0 < x < a

x , a ≤ x ≤ b

0 , x > b

In the fourth case if we choose f(x) =

{1, 0 < x < a

0, x > a

Then

Fs(ξ) =1 − cos ξ a

ξ

and it gives

2π

∫ ∞

0sin ξ x

[1 − cos ξ a

ξ

]dξ = f(x) =

{1 , 0 < x < a

0 , x > a

Similarly taking f(x) =

{1, a < x b

Combining these above results, we get

2π

∫ ∞

0sin ξ x

[a − b

ξ+

cos a ξ − cos b ξ

ξ2

]dξ =

⎧⎪⎨⎪⎩

a − b , 0 < x < a

x − b , a < x b

1.9 Fourier Transform of some particular functions.

We consider the following step function f(x) defined by

f(x) =

{0 , x < 01 , x > 0

This function is called Heaviside unit step function and it is denotedby the symbol H(x). Thus, we have

H(x) =

{0 , x < 0 (or sometimes for x ≤ 0)1 , x > 0

Almost all step functions can be expressed as a combination of Heavisideunit step functions. For example,

f(x) =

{1 , |x| < a

0 , otherwise

is equivalent to f(x) = H(x + a) − H(x − a) ≡ H(a − |x|)

and the step function f(x) =

{k , x > a

0 , otherwise

is equivalent to f(x) = k H(x − a), for all x

Example 1.1. Evaluate Fourier transform of H(x + a) − H(x − a)

Solution. By definition,

F [H(x + a) − H(x − a);x → ξ] =1√2π

∫ ∞

0eiξx[H(x + a) − H(x − a)]dx


=1√2π

∫ a

−aeiξx dx

=1√2π

[eiξx

iξ

]a

−a

=1√2πiξ

[eiξa − e−iξa

]

=

√2π

sin a ξ

ξ

Therefore, using the formula for inverse F T we get

1√2π

∫ +∞

−∞

√2π

sin a ξ

ξe−iξx dξ = H(x + a) − H(x − a)

This result implies,

1π

∫ +∞

−∞

sin a ξ

ξ· [cos ξ x − i sin ξ x] dξ = H(x + a) − H(x − a)

i.e∫ +∞

−∞

sin a ξ

ξcos ξ x dξ = π[H(x + a) − H(x − a)]

=

{π , for |x| < a

0 , for |x| > a

Putting x = 0 and a = 1, in the above we get∫ +∞

−∞

sin ξ

ξdξ = π

implying,

∫ ∞

0

sin ξ

ξdξ =

π

2

and∫ +∞

−∞

sin a ξ sin ξ x

ξdξ = 0

Example 1.2. Evalute FT of

f(x) =

{x , |x| < a

0 , |x| > a


F [f(x) ; x → ξ] =1√2π

∫ a

−ax eiξx dx

=1√2π

{[x

eiξx

iξ

]a

−a

− 1iξ

∫ a

−aeiξx dx

}


=1√2π

[a eiξa + a e−iξa

iξ+

1ξ2

{eiaξ − e−iaξ

}]

=1√2π

[2 a cos a ξ

iξ+

2 i sin a ξ

ξ2

]

=

√2π

i

[sin a ξ − a ξ cos a ξ

ξ2

].

Example 1.3. If F T of f(x) is F (ξ), calculate F T of(i) eiλxf(x) and (ii) f(−x) in terms of F (ξ).

Solution. We know that,

F [f(x) ; x → ξ] =1√2π

∫ +∞

−∞f(x) eiξx dx = F (ξ).

Therefore,

(i) F[eiλx f(x) ; x → ξ

]=

1√2π

∫ +∞

−∞f(x) · ei(ξ+λ)x dx = F (ξ + λ).

(ii) F [f(−x) ; x → ξ] =1√2π

∫ +∞

−∞f(−x) eiξx dx

=1√2π

∫ +∞

−∞f(y) e−iξy dy = F (−ξ) = F−1(ξ)

Example 1.4. Find the function whose cosine transform is

√2π

sin a ξ

ξ.

Solution. Let f(x) be the required function such that

Fc[f(x) ; x → ξ] =

√2π

sin a ξ

ξ

f(x) =2π

∫ ∞

0

sin a ξ

ξ· cos ξ x d ξ

=1π

∫ ∞

0

1ξ[sin ξ(a + x) + sin ξ(a − x)] dξ

=1π

[∫ ∞

0

sin ξ(a + x)ξ

dξ +∫ ∞

0

sin ξ(a − x)ξ

dξ

]

=1π

[π2

+π

2

], when x < a and

1π

[π2− π

2

], when x > a

So, f(x) =

{1 , when x < a

0 , when x > a


Example 1.5. Find the F T of f(x) =

{1 − x2 , |x| ≤ 10 , |x| > 1

and hence evaluate∫ ∞

0

x cos x − sin x

x3cos

x

2dx

Solution. From definition, we have

F (ξ) = F [f(x) ; x → ξ] =1√2π

∫ 1

−1(1 − x2) eiξx dx

=[− 2

ξ2(eiξ + e−iξ) +

2iξ3

(eiξ − e−iξ)]· 1√

2π

Therefore,

F (ξ) =−4√2π

[(ξ cos ξ − sin ξ) /ξ3]

Now, the inversion formula gives

1√2π

∫ +∞

−∞

−4√2π

[ξ cos ξ − sin ξ

ξ3

]· e−iξx dx = f(x)

Hence,∫ +∞

−∞

[−4(ξ cos ξ − sin ξ)ξ3

][cos ξ x − i sin ξ x] dξ = 2πf(x)

Equating real parts, we have∫ ∞

−∞

sin ξ − ξ cos ξ

ξ3cos ξ x dξ =

π

2f(x)

Or, 2∫ ∞

0


ξ3cos ξ x dξ =

{π2 (1 − x2) , |x| ≤ 10 , |x| > 1

Now, putting x = 12 in above we get∫ ∞

0


ξ3cos

ξ

2dξ =

12· π

2· 34

=3π16

This gives,∫ ∞

0

x cos x − sin x

x3cos

x

2dx = −3π

16

Example 1.6. Find FT of f(x) = [1 − |x|]H[1 − |x|]Solution.

Here f(x) =

{1 − |x| , when − 1 < x < 10 , otherwise


Here, f(x) is an even function of x and so

F [f(x) ; x → ξ] = Fc[f(x) ; x → ξ]

=

√2π

∫ 1

0(1 − x) cos ξ x dx

=

√2π

1 − cos ξ

ξ2

So, F−1c [Fc{f(x) ; x → ξ} ; ξ → x] = [1 − |x|] H [1 − |x|]

Or,2π

∫ ∞

0

1 − cos ξ

ξ2cos ξ x dξ =

{1 − x , 0 < x < 10 , x > 1

(i)

Some Deductions :

(a) Putting x = 0 in the final result (i) of the example 1.6 we get∫ ∞

0

1 − cos ξ

ξ2dξ =

π

2

Or,∫ ∞

0

sin2 x

x2dx =

π

2

(b) Integrating the result (i) in the example 1.6 with respect to x from0 to X, we get∫ ∞

0

1 − cos ξ

ξ3

sin ξ X

ξdξ =

π

2

[X − X2

2

](ii)

Now, putting X = 12 , one gets∫ ∞

0

2 sin3 ξ2

8(

ξ2

)3 · 2d(

ξ

2

)=

π

2

[38

]

⇒∫ ∞

0

sin3 x

x3dx =

3π8

(c) We have seen in Example 1.1 that∫ ∞

0

sin x

xdx =

π

2

Again∫ ∞

0

sin2 x

x2dx =

π

2, from (a)

Also in (b),∫ ∞

0

sin3 x

x3dx =

3π8

<π

2


These results show that

sin x

x�[sin x

x

]2�[sin x

x

]3

(d) Again integrating the result (ii) in (b) above with respect to X

from X = 0 to X = 1, we get

∫ ∞

0

1 − cos ξ

ξ2

[− cos ξ X

ξ2

]10

dξ =π

2

[X2

2− X3

6

]10

Or,∫ ∞

0

sin4 x

x4dx =

π

3

Example 1.7. Find the F T of (i) f(x) = e−a|x|, a > 0 and(ii) f(x) = |x|e−a|x|, a > 0.

Solutions (i) F [f(x) ; x → ξ] =1√2π

∫ +∞

−∞e−a|x| · eiξx dx

=1√2π

[∫ 0

−∞eax · eiξx dx +

∫ ∞

0e−ax · eiξx dx

]

=1√2π

[1

a + iξ+

1a − iξ

]

=

√2π

a

a2 + ξ2

(ii) F [ |x|e−a|x| ; x → ξ ]

=1√2π

∫ +∞

−∞|x| e−a|x| · eiξx dx

=1√2π

[− d

da

∫ +∞

−∞e−a|x| eiξx dx

]

= − d

da

[√2π

a

a2 + ξ2

]

=

√2π

[a2 − ξ2

(a2 + ξ2)2

]

Example 1.8. (a) Find the F T of f(x) =

{e−ax, x > 0, a > 0−eax, x < 0, a > 0

Solution. Clearly, the given function is an odd function of x in R.


Therefore

F [f(x) ; x → ξ] = i

√2π

∫ ∞

0e−ax sin ξ x dx

= i

√2π· ξ

ξ2 + a2

(b) Find F T of f(x) = x e−a|x| , a > 0.

Solution. The given function f(x) is an odd function of x ∈ R. Hence,

F [f(x) ; x → ξ] = i

√2π

∫ ∞

0x e−ax · sin ξ x dx

= i

√2π

(− d

da

)∫ ∞

0e−ax sin ξ x dx = −i

√2π

d

da

[ξ

a2 + ξ2

]

= i

√2π

2 a ξ

(ξ2 + a2)2

Example 1.9. Find the F T of f(x) = e−αx2, α > 0.

Solution. We know from definition that

F [f(x) ; x → ξ] =1√2π

∫ +∞

−∞e−αx2

eiξx dx

=1√2π

∫ ∞

−∞e−�√

αx− iξ2√

α

�2− ξ2

4α dx

=e−

ξ2

4α√2π

∫ +∞

−∞e−z2 · dz√

α

=2 · e− ξ2

4α√2πα

∫ ∞

0e−z2

dz

F [e−αx2] =

2 · e− ξ2

4α√2πα

·√

π

2=

1√2α

e−ξ2

4α (i)

Some important deductions.

(i) Let α = 1 . Then F [e−x2; x → ξ] = 1√

2e

−ξ2

4

(ii) Let α = 12 . Then F [e

−x2

2 ; x → ξ] = e−ξ2

2

This result shows that F−1 [e−ξ2

2 ; ξ → x] = e−x2

2


Such a function f(x) = e−x2

2 having the property that

F [f(x) ; x → ξ] = f(ξ)

is called self reciprocal.(iii) Differentiating (i) with respect to ξ we get

d

dξ

[√2π

∫ ∞

0e−αx2

cos ξx dx

]=

d

dξ

1√2α

· e−ξ2

4α

⇒ Fs[xe−αx2; x → ξ] =

1√8α3

ξ e−ξ2

4α

Putting , α =12

we get Fs

[x e

−x2

2 ; x → ξ

]= ξ e

−ξ2

2

Thus f(x) = x e−x2

2 is self reciprocal with regard to Fourier sine trans-form.

Putting, α =i

2we get Fc

[e

ix2

2 ; x → ξ

]=

1 − i√2

eiξ2

2

Therefore, Fc

[cos

x2

2; x → ξ

]=

1√2

Re

[(1 − i)e

iξ2

2

]

= cos(

ξ2

2− π

4

)(ii)

Also, Fc

[sin

x2

2; x → ξ

]=

1√2

[cos

ξ2

2− sin

ξ2

2

]

= − sin(

ξ2

2− π

4

)(iii)

after equating real and imaginary parts of both sides. Combining resultsin (ii) and (iii) above one gets

Fc

[cos(

x2

2− π

8

); x → ξ

]= cos

[ξ2

2− π

8

]

1.10 Convolution or Faltung of two integrable functions.

The convolution or Falting or Faltung of two integrable functions f(x)and g(x), where −∞ < x < ∞ is denoted and defined as

f ∗ g =1√2π

∫ +∞

−∞f(x − u) g(u) du (1.23)


It possesses many formal properties such as

f ∗ (λg) = (λf) ∗ g = λ(f ∗ g), λ = constant

f ∗ g = g ∗ f , (the commutative property)

f ∗ (g + h) = f ∗ g + f ∗ h ,

which can be verified easily directly from definition (1.23). Further, ifboth f(x) and g(x) belong to C1(R) and A1(R) classes, then so doestheir convolution h(x) = f ∗ g, since

√2π∫ ∞

−∞|h(x)|dx =

∫ +∞

−∞dx

∣∣∣∣∫ +∞

−∞f(u)g(x − u)du

∣∣∣∣<

∫ +∞

−∞dx

∫ +∞

−∞|f(u)g(x − u)| du

=∫ +∞

−∞|f(u)|du

∫ +∞

−∞|g(x − u)| dx

⇒∫ +∞

−∞|h(x)|dx <

1√2π

∫ +∞

−∞|f(u)|du

∫ +∞

−∞|g(υ)|dυ

and the result follows from the fact that both f(x) and g(x) belong tothe class A1(R).

Again, (f ∗g)∗h = f ∗(g∗h) is true by direct verification. Therefore,the convolution property is also associative.

We shall now discuss Fourier transform of the convolution of a pairof functions as detailed below, by the name Convolution or Falting orFaltung theorem for F T .

1.11 Convolution or Falting or Faltung Theorem for FT.

Theorem.

The FT of the convolution of f(x) and g(x), both belonging to theclasses C1(R) and A1(R), is the product of the F T of f(x) and g(x).That means that,

F [f ∗ g;x → ξ] = F (ξ) G(ξ)

where F (ξ) =1√2π

∫ +∞

−∞f(x) eiξx dx and

G(ξ) =1√2π

∫ +∞

−∞g(x) eiξxdx


Proof. Since f(x) and g(x) both belong to the classes C1(R) andA1(R), f ∗g also belongs to the classes C1(R) and A1(R) and hence F T

of f ∗ g also exists. Therefore,

F [f ∗ g ; x → ξ] =1√2π

∫ +∞

−∞eiξx dx√

2π

∫ +∞

−∞f(x − u)g(u)du

=12π

∫ +∞

−∞

∫ +∞

−∞f(x − u)g(u)eiξx dx du

=12π

∫ +∞

−∞g(u)

[∫ +∞

−∞eiξxf(x − u)dx

]du

=1√2π

∫ ∞

−∞g(u)eiξu

[1√2π

∫ +∞

−∞eiξυf(υ) dυ

]du

=1√2π

∫ +∞

−∞g(u)eiξu · F (ξ) du

F [f ∗ g ; x → ξ] = F (ξ) · G(ξ) (1.24)

This proves the theorem.

Coro 1.2 Another interpretation of convolution theorem from eqn.(1.24) is that

F−1 [F (ξ)G(ξ) ; ξ → x] =1√2π

∫ +∞

−∞f(x − u)g(u)du

Or,1√2π

∫ +∞

−∞F (ξ)G(ξ)e−iξxdξ =

1√2π

∫ +∞

−∞f(υ) g(x − υ)dυ

= f ∗ g (1.25)

Coro 1.3 From the result (1.25) above, we have

1√2π

∫ +∞

−∞F (ξ) G(ξ) e−iξxdξ = f ∗ g =

1√2π

∫ +∞

−∞f(υ)g(x − υ)dυ

Putting x = 0 in the above equation, we get∫ +∞

−∞F (ξ) G(ξ) dξ =

∫ +∞

−∞f(υ) g(−υ) dυ (1.26)

Coro 1.4 Let f(x) ∈ R. Then F (ξ) = F [f(x) ; x → ξ] is a complexvalued function and its conjugate, denoted by F (ξ) is then given by

F (ξ) =1√2π

∫ +∞

−∞f(x)e−iξxdx = F−1[f(x);x → ξ]


i.e, F{f(x);x → ξ} = F−1[f(x) ; x → ξ]

Therefore, F ≡ F−1 (1.27)

Coro 1.5 Similar results for sine and cosine transforms of the func-tions f(x) and g(x) ∈ R may also be established in regard to theirconvolution. For example, if f(x), g(x) are even functions and

Fc [f(x) ; x → ξ] =

√2π

∫ ∞

0f(x) cos ξ x dx

Fc [g(x) ; x → ξ] =

√2π

∫ ∞

0g(x) cos ξ x dξ ,

then Fc [f ∗ g ; x → ξ] = Fc(ξ) Gc(ξ) (1.28)

Also, if f(x), g(x) are odd functions of x in R and if

Fs[f(x) ; x → ξ] = Fs(ξ) and Fs[g(x) ; x → ξ] = Gs(ξ)

then Fs [f ∗ g ; x → ξ] = Fs(ξ) Gs(ξ) (1.29)

It may be noted that the convolution had already been defined inarticle 1.10 and keeping this together with evenness and oddness char-acter of the associated functions one can very easily deduce the resultsin equations (1.28) and (1.29) respectively.

1.12 Parseval’s relations for Fourier Transforms.

Theorem.

If F (ξ) and G(ξ) are complex F.T of f(x) and g(x) respectively, then

(i)1√2π

∫ +∞

−∞F (ξ)G(ξ) dξ =

1√2π

∫ +∞

−∞f(x)g(x) dx (1.30)

(ii)1√2π

∫ +∞

−∞|F (ξ)|2dξ =

1√2π

∫ +∞

−∞|f(x)|2dx (1.31)

where bar sign over function signifies complex conjugate of the complexfunctions or absolute value for real functions

Proof (i) By Fourier Inverse transform formula, we have

g(x) =1√2π

∫ +∞

−∞G(ξ)e−iξxdξ


Taking complex conjugates on both sides, the above equation gives

g(x) =1√2π

∫ +∞

−∞G(ξ) eiξx dξ

Therefore,1√2π

∫ +∞

−∞f(x) g (x)dx

=1√2π

∫ +∞

−∞f(x)

[1√2π

∫ +∞

−∞G(ξ) eiξxdξ

]dx

=1√2π

∫ +∞

−∞G(ξ)

[1√2π

∫ +∞

−∞f(x)eiξxdx

]dξ

=1√2π

∫ +∞

−∞G(ξ)F (ξ) dξ , which proves the result (i).

(ii) In the result (i) if we take g(x) = f(x), we get

1√2π

∫ +∞

−∞f(x) f(x) dx =

1√2π

∫ +∞

−∞F (ξ) F (ξ) dξ

i.e,1√2π

∫ +∞

−∞|f(x)|2dx =

1√2π

∫ +∞

−∞|F (ξ)|2 dξ

This proves the part (ii) of the Parseval’s relations.

Note : In the above parseval’s relations or identities one may drop thefactors 1√

2πfrom either sides of (1.30) and (1.31)

There exists other four Parseval’s identities given by

(iii)

√2π

∫ ∞

0Fc(ξ)Gc(ξ)dξ =

√2π

∫ ∞

0f(x)g(x)dx (1.32)

(iv)

√2π

∫ ∞

0Fs(ξ)Gs(ξ)dξ =

√2π

∫ ∞

0f(x)g(x)dx (1.33)

(v)

√2π

∫ ∞

0|Fc(ξ)|2dξ =

√2π

∫ ∞

0|f(x)|2dx (1.34)

and (vi)

√2π

∫ ∞

0|Fs(ξ)2dξ =

√2π

∫ ∞

0|f(x)|2dx (1.35)

connecting to Fourier cosine and sine transforms. Where again the con-

stant terms√

2π may be dropped from either sides in the above results.

Their proofs can similarly be deduced as were done in cases (i) or (ii)above.


Example 1.10. Let f(x) = e−bx , g(x) = e−ax. Use parseval’s relationof Fourier cosine transform to evaluate the integral to prove∫ ∞

0

dx

(a2 + x2)(b2 + x2)=

π

2ab(a + b)

Solution.

We have Fc [f(x) ; x → ξ] =

√2π

∫ ∞

0e−bx cos ξ x dx

=

√2π

b

b2 + ξ2≡ Fc(ξ)

Similarly, Gc(ξ) =

√2π

a

a2 + ξ2

Therefore, by the Parseval’s relation (iii) above, we get√2π

∫ ∞

0

2π

ab

(ξ2 + a2)(ξ2 + b2)(ξ2 + b2)dξ =

√2π

∫ ∞

0e−(a+b)xdx

i.e,∫ ∞

0

dξ

(ξ2 + a2)(ξ2 + b2)=

π

2ab(a + b).

Example 1.11. Let f(x) =

{1 , 0 < x < a

0 , x > a

Use Parseval′s identity to evaluate∫ ∞

0

sin2 ax

x2dx .

Solution. We have from example 4 of article 1.9 that

Fc[f(x) ; x → ∞] =

√2π

sin a ξ

ξ.

Then by the Parseval’s identity (1.34) of the corollary above√2π

∫ ∞

0

2π

∣∣∣∣sin a ξ

ξ

∣∣∣∣2

dξ =

√2π

∫ ∞

0|f(x)|2dx

i.e,∫ ∞

0

sin2 a ξ

ξ2dξ =

π

2

∫ a

0dx =

πa

2

Example 1.12. Taking g(x) = e−ax and f(x) =

{1 , 0 < x b


and using proper Parseval’s relation prove that

∫ ∞

0

sin at

t(a2 + t2)dt =

π

21 − e−a2

a2

Solution. By Fourier cosine transform

Gc(ξ) = Fc[g(x);x → ξ] =a

ξ2 + a2and Fc[f(x);x → ξ] =

sin aξ

ξ= Fc(ξ)

Then by the Parseval’s relation connecting Fourier cosine transform weget

∫ ∞

0

a sin a ξ

ξ(ξ2 + a2)dξ =

∫ a

0e−ax · 1 dx =

1 − e−a2

a

⇒∫ ∞

0

sin at

t(t2 + a2)dt =

1 − e−a2

a2

1.13 Fourier Transform of the derivative of a function.

In many applications of the theory of Fourier transform to boundaryvalue problems of Mathematical Physics it is necessary to express FT

of the derivative of a function f(x) in terms of FT of the function f(x).To this direction we now denote FT of dnf(x)

dxn by F (n)(ξ), say. Then, weget

F (n)(ξ) =−1√2π

∫ +∞

−∞(iξ)

dn−1f(x)d xn−1

eiξxdx +[

1√2π

dn−1f(x)d xn−1

· eiξx

]+∞

−∞,

which is obtained after integrating by parts the right hand side of thedefinition of

F (n)(ξ) =1√2π

∫ +∞

−∞

dnf(x)dxn

eiξx dx (1.36)

If we assume that dn−1f(x)dxn−1 tends to zero as |x| → ∞, the above result

takes the formF (n) = −iξF (n−1)

By repeated application of this rule and by the assumption

lim|x|→∞

[drf(x)

dxr

]= 0 , r = 1, 2, · · · n − 1,


we have finallyF (n) = (−iξ)n F

This implies that FT of the nth derivative of a function f(x) is (−iξ)n

times FT of the function, provided that the first (n − 1) derivative ofthe function vanish as |x| → ∞.

The corresponding results for Fourier cosine and Fourier sine trans-form of a function are not so simple. For discussion towards this direc-tion, we define F

(n)s and F

(n)c by the equations

F (n)s =

√2π

∫ ∞

0

dnf(x)dxn

sin ξx dx ,

F (n)c =

√2π

∫ ∞

0

dnf(x)dxn

cos ξ x dx (1.37)

Then, integrating by parts the right hand side of the second of the aboveequations in (1.37) we get

F (n)c =

√2π

[dn−1f(x)

dxn−1cos ξ x

]∞0

+ ξ

√2π

∫ ∞

0

dn−1f(x)dxn−1

sin ξ x dx

= −an−1 + ξ F (n−1)s , (1.38)

on the assumption that limx→∞

dn−1f(x)dxn−1

= 0,

limx→0

√2π

dn−1f(x)dxn−1

= an−1

Similarly, from the first of the equation in (1.37) we derive

F (n)s = −ξ F (n−1)

c (1.39)

Using the result (1.39) in (1.38) we have

F (n)c = −an−1 − ξ2 F (n−2)

c (1.40)

Thus, by repeated application of equation (1.40) one can reduce F(n)c to

a sum of ξ’s and either F(1)c of Fc according as n is odd or even integer

respectively. Therefore, we obtain the following formulae :

F (2m)c = −

m−1∑s=0

(−1)sa2m−2s−1 ξ2s + (−1)m ξ2m Fc (1.41)

F (2m+1)c = −

m∑s=0

(−1)sa2m−2s ξ2s + (−1)m ξ2m+1 Fs (1.42)


Similar results hold for sine transforms too. From (1.39) and (1.40) weobtain,

F (n)s = ξ an−2 − ξ2 F (n−2)

s

From this formula we can derive the formulas

F (2n)s = −

n∑k=1

(−1)kξ2k−1a2n−2k + (−1)n+1ξ2n Fs (1.43)

F (2n+1)s = −

n∑k=1

(−1)k ξ2k−1a2n−2k+1 + (−1)n+1 ξ2n+1Fc (1.44)

Certain special cases of the above formulae arise frequently. For example,if df

dx = d3fdx3 = 0, when x = 0, then

√2π

∫ ∞

0

d2f

dx2cos ξ x dx = −ξ2 Fc

and

√2π

∫ ∞

0

d4f

dx4cos ξ x dx = ξ4 Fc

On the other hand, if f(x) = d2f(x)dx2 = 0, when x = 0, then

√2π

∫ ∞

0

d2f

dx2sin ξ x dx = −ξ2 Fs

and

√2π

∫ ∞

0

d4f

dx4sin ξ x dx = ξ4 Fs

Corollary 1.6. If a function f(x) has a finite discontinuity at a singlepoint x = a ∈ R for simplicity, the above formula for F T of itsderivative has to be modified in the following manner :

We write

F [f ′(x) ; x → ξ] =1√2π

∫ a−0

−∞f ′(x) eiξxdx +

1√2π

∫ ∞

a+0f ′(x) eiξxdx

As before, integrating the right hand side integrals by parts, one gets

F [f ′(x) ; x → ξ] =1√2π

[f(x) eiξx

]a−0

−∞+

1√2π

[f(x)eiξx

]∞a+0

− iξ√2π

∫ a−0

−∞f(x)eiξxdx − iξ√

2π

∫ ∞

a+0f(x)eiξxdx,


which can be written in the form

F [f ′(x) ; x → ξ] = −iξ F [f(x) ; x → ξ] − 1√2π

eiξa[f(x)]a (1.45)

where [f(x)]a = f(a + 0) − f(a − 0), called the jump of f(x) at x = a .

As a generalisation if we now suppose that f(x) has n points of finitediscontinuities at points x = ai, i = 1, 2, · · · n, the modified form ofeqn.(1.45) is

F [f ′(x);x → ξ] = −iξ F [f(x);x → ξ] − 1√2π

n∑i=1

[f(x)]ai eiξai (1.46)

Corollary 1.7.

We know that f(x) ∈ C1(R), then regarding the function f(x) eiξx

as a function of x and ξ, we see that f(x) eiξx ∈ C1(R×R) and thereforeF (ξ) = 1√

2π

∫ +∞−∞ f(x)eiξxdx is convergent and the integral

i√2π

∫ +∞

−∞xf(x) eiξxdx =

1√2π

∫ +∞

−∞

∂

∂ξ

[f(x)eiξx

]dx

is uniformly convergent. Then

F ′(ξ) = iF [xf(x) ; x → ξ] (1.47)

We can continue the above process any finite number of times to obtain

dr

dξrF (ξ) = irF [xrf(x) ; x → ξ], r = 0, 1, · · · (1.48)

Corollary 1.8.

If a > 0, Fc [f(ax) ; x → ξ] =

√2π

∫ ∞

0f(ax) cos ξ x dx

= a−1

√2π

∫ ∞

0f(x) cos

(ξx

a

)dx

Therefore, Fc[f(ax) ; x → ξ] =1aFc[f(x) ; x → ξ

a], a > 0, (1.49)

Then, Fc[f(x) cos(ωx) ; x → ξ] =12[Fc(ξ + ω) + Fc(ξ − ω)]

(1.50)

and Fc[f(x) sin ω x ; x → ξ] =12[Fs(ω + ξ) + Fs(ω − ξ)]

(1.51)


Corollary 1.9.

As in corollary 1.8 above, we can easily deduce that

Fs[f(ax) ; x → ξ] = a−1Fs

[f(x) ; x → ξ

a

], a > 0 (1.52)

Fs[f(x) cos ωx ; x → ξ] =12[Fs(ξ + ω) + Fs(ξ − ω)]

and Fs[f(x) sin ωx ; x → ξ] =12[Fc(ξ − ω) − Fc(ξ + ω)]

Corollary 1.10.

In actual calculation of Fourier sine transform the following resultmay be found useful. Let us denote

Fs [f(x) ; x → ξ] = ϕ(ξ) , for ξ > 0

Then, for ξ < 0 we have

−ϕ(η) ≡ −ϕ(−ξ) = Fs [f(x) ; x → ξ] , ξ = −η , η > 0

Therefore, we have in general

Fs[f(x) ; x → ξ] = ϕ(|ξ|) · sgn ξ

where sgn ξ =

{+1 , for ξ > 0−1 , for ξ < 0

1.14 Fourier Transform of some more useful functions.

(a) We require to find Fc[e−axxn−1;x → ξ] and Fs[e−ax xn−1; x → ξ]where a > 0 and n > 0. For this purpose let us assume√

2π

∫ ∞

0e−axxn−1 cos ξ x dx = C

and

√2π

∫ ∞

0e−axxn−1 sin ξ x dx = S

Then, C − iS =

√2π

∫ ∞

0xn−1 · e−∝xdx, where ∝= a + iξ = reiθ, say

=

√2π

∫ ∞

0

1∝n

e−zzn−1dz


=

√2π

1∝n

Γ(n) ,

=

√2π

Γ(n)rn

· e−inθ

where r =√

a2 + ξ2 and θ = tan−1 ξ

a.

Thus, Fc [xn−1e−ax ; x → ξ] =

√2π

cos n θ · Γ(n)(a2 + ξ2)

n2

and Fs [xn−1e−ax ; x → ξ] =

√2π

Γ(n) · sinnθ

(a2 + ξ2)n2

Deductions :

Since Fc F−1c = I and Fs F−1

s = I, we get

xn−1 e−ax = Fc

[√2π

Γ(n)r−n cos n θ ; ξ → x

]

=2π

Γ(n)∫ ∞

0r−n cos nθ cos ξx dξ

Putting a = 1 in the above equation we get

π

2· xn−1 e−x

Γ(n)=∫ ∞

0(1 + ξ2)−

n2 cos(n tan−1 ξ) cos ξx dξ

As tan θ = ξ ⇒ sec2 θ dθ = dξ , we have

π

2xn−1 e−x

Γ(n)=∫ π

2

0cosn−2 θ cos nθ cos(x tan θ) dθ (1.53)

Similarly,

xn−1 e−x = Fs

[√2π

sin nθ Γ(n)(1 + ξ2)

n2

; ξ → x

]

=2π

∫ ∞

0

sin nθ · Γ(n)(1 + ξ2)

n2

sin ξx dξ

This result gives,

π

2· xn−1e−x

Γ(n)=∫ π/2

0cosn−2 θ · sin nθ · sin(x tan θ) dθ (1.54)


If we multiply (1.53) by e−x xm−1 and integrate the result withrespect to x from 0 to ∞, we get

π

2Γ(n)

∫ ∞

0e−2xxm+n−2dx =

∫ π/2

0cosn−2 θ · cos nθ · A dθ

where A =∫ ∞

0e−xxm−1 · cos(x tan θ) dx (1.55)

Similarly, multiplying (1.54) by e−x xm−1 and integrating theresult with respect to x from 0 to ∞, we get

π

2Γ(n)

∫ ∞

0e−2xxm+n−2 dx =

∫ π/2

0cosn−2 θ · sin nθ · B dθ

where B =∫ ∞

0e−xxm−1 sin(x tan θ) dx (1.56)

Therefore, from (1.55) and (1.56)

A − iB =∫ ∞

0e−x xm−1 e−ix tan θ dx

=∫ ∞

0e−∝x xm−1 dx, where ∝= 1 + i tan θ =

eiθ

cos θ

=1

∝mΓ(m) = Γ(m) cosm θ e−imθ

This result implies, A = Γ(m) cosm θ cos mθ,B = Γ(m) cosm θ sinmθ.

Again,π

2Γ(n)

∫ ∞

0e−2xxm+n−2dx =

π

2m+n

Γ(m + n − 1)Γ(n)

Therefore,π Γ(m + n − 1)

2m+nΓ(n)=∫ π/2

0cosn−2 θ cos nθ

[Γ(m) cosm θ cos mθ]dθ

⇒ π Γ(m + n − 1)2m+nΓ(m)Γ(n)

=∫ π/2

0cosm+n−2 θ cos mθ cos nθ dθ

Also,π Γ(m + n − 1)2m+nΓ(m)Γ(n)

=∫ π/2

0cosm+n−2 θ sin nθ sinmθ dθ

Adding these two results, we get

π Γ(m + n − 1)2m+n−1 Γ(m)Γ(n)

=∫ π/2

0cosm+n−2 θ cos(m − n)θ dθ


If we now set m + n = 2, then we get

π

2Γ(n)(1 − n)Γ(1 − n)=

sin nπ

2(1 − n)

Therefore, we get Γ(n)Γ(1 − n) = πsinnπ , the duplication formula

for the Gamma function.(b) We now evaluate Fc

[(a2 − x2)ν−

12 H(a − x) ; x → ξ

]Solution. Following the definition, we have

Fc[(a2 − x2)ν−12 H(a − x) ; x → ξ] =

√2π

∫ a

0(a2 − x2)ν−

12 cos ξxdx

=

√2π· I, say

Now, from the series expansion of cos ξx, we can write

I =∫ a

0(a2 − x2)ν−

12

∞∑r=0

(−1)r(ξx)2r

(2r)!dx

=∞∑

r=0

(−1)rξ2r

(2r)!

∫ a

0(a2 − x2)ν−

12 x2r dx

=∞∑

r=0

(−1)rξ2r

(2r)!· a2ν+2r

2

∫ 1

0zr− 1

2 (1 − z)ν−12 dz,where x2 = a2z

=∞∑

r=0

(−1)rξ2ra2ν+2r

2(2r)!Γ(ν + 1

2 )Γ(r + 12)

Γ(ν + r + 1)

=∞∑

r=0

(−1)rΓ(ν + 12 )

Γ(ν + r + 1)·(

ξ

2

)2r

· a2ν+2r

r!

√π

2

Thus,

Fc

[(a2 − x2)ν−

12 H(a − x) ; x → ξ

]

=

√2π

√π

2

[ ∞∑r=0

(−1)r

Γ(r + 1)Γ(ν + r + 1)

](aξ

2

)2r+ν (ξ

2

)−ν

aνΓ(

ν +12

)

=1√2

Jν (aξ) aν Γ(

ν +12

)(ξ

2

)−ν

=(

a

ξ

)ν

2ν− 12 Γ(

ν +12

)Jν(aξ)


⇒ Fc

[(a

ξ

)ν

2ν− 12 Jν(aξ) ; ξ → x

]= (a2 − x2)ν−

12 H(a − x)

=

{(a2 − x2)ν−

12 , 0 < x < a

0 , otherwise

Deduction. Putting ν = 0, a = 1 in the above, we get

Fc

[√π

2J0 (ξ); ξ → x

]=

{1√

1−x2, 0 < x < 1

0 , otherwise .

(c) Find Fourier cosine transforms of e−ax cos ax and e−ax sin ax andhence evaluate that of 1

x4+k4

Solution.

Let C = Fc[e−ax cos ax ; x → ξ]

and S = Fc[e−ax sin ax ; x → ξ]

Then, C − iS =

√2π

∫ ∞

0e−αx cos ξ x dx, where α = a(1 + i)

=

√2π

I, say.

So, I =α

ξ2− α

ξ2I

⇒ I =α

α2 + ξ2

Therefore, C − iS =

√2π

a(1 + i)ξ2 + a2(1 + i)2

=

√2π· a(1 + i)(ξ2 − 2a2i)(ξ2 + 2a2i)(ξ2 − 2a2i)

=

√2π

a[(ξ2 + 2a2) + i(ξ2 − 2a2)]ξ4 + 4a4

So, Fc[e−ax cos ax;x → ξ] =

√2π

a(ξ2 + 2a2)ξ4 + 4a4

and Fc[e−ax sin ax;x → ξ] =

√2π

a(−ξ2 + 2a2)ξ4 + 4a4

.

Then, Fc[e−ax(λ cos ax + μ sin ax) ; x → ξ]

=

√2π

a(λ − μ)ξ2 + 2a2(λ + μ)

ξ4 + 4a4


Now choosing, 1 = λ = μ �= 0, we get

Fc[e−ax(cos ax + sin ax);x → ξ] =

√2π

4a3

ξ4 + 4a4

∴ Fc

[√2π

4a3

ξ4 + 4a4; ξ → x

]= e−ax(cos ax + sin ax)

Thus Fc

[1

ξ4 + 4a4; ξ → x

]=√

π

2e−ax

4a3(cos ax + sin ax)

∴ Fc

[1

ξ4 + k4; ξ → x

]=√

π

2· e

−kx√2√

2k3

(cos

kx√2

+ sinkx√

2

)

Or, Fc

[1

x4 + k4; x → ξ

]=√

π

2· e

−kξ√2√

2k3

(cos

kξ√2

+ sinkξ√

2

).

(d) From Pn(x) =1

2n n!dn

dxn(x2 − 1)n evaluate

F [Pn(x) H(1 − |x|) ; x → ξ]

Solution.

F [Pn(x)H(1 − |x|);x → ξ]

=1√2π

· 12n n!

∫ 1

−1

dn

dxn(x2 − 1)neiξx dx

=(−iξ)√2π2n n!

∫ 1

−1

dn−1

dxn−1(x2 − 1)n · eiξx dx , on integration by parts

=(−iξ)2√2π2n n!

∫ 1

−1

dn−2

dxn−2(x2 − 1)n · eiξx dx

Similarly proceeding after n-th step we get

=1√2π

(−iξ)n

2n n!

∫ 1

−1(x2 − 1)neiξxdx

=inξn

√2π2n n!

∫ 1

−1(1 − x2)neiξxdx

=inξn

√2π2nn!

· 2 ·[∫ 1

0(1 − x2)n cos ξ x dx

]

But from the result of (b) above, after putting a = 1 and ν − 12 = n, we

get

F [Pn(x) H(1 − |x|) ; x → ξ]


=

√2π

inξn

2n n!·√

π

2ξ−n− 1

2 2nJn+ 12

(ξ) · Γ(n + 1)

=in√ξ

Jn+ 12

(ξ)

1.15 Fourier Transforms of Rational Functions.

Using calculus of residues FT of rational functions can be calculated.Let us assume that the function of a complex variable z admits of thefollowing properties :

(i) f(z) has finite number of singularities at z = a1, a2 · · · an in thehalf plane Im z > 0

(ii) f(z) is analytic everywhere on real z axis except at the pointsb1, b2, · · · bm, which are simple poles and

(iii)∣∣zf(z) eiξz

∣∣→ 0 as |z| → ∞, Im z � 0.

Let us now integrate eiξzf(z), ξ > 0 round the contour shown in theadjoining figure and making use of Cauchy’s residue theorem to obtain

Fig. 1.1

12πi

∫ +∞

−∞f(x)eiξxdx =

12

m∑s=1

res [f(bs)] · eibsξ +n∑

s=1

res [f(as)] · eiξas

This formula yeilds for ξ > 0

F [f(x);x → ξ] =√

2πi

[n∑

s=1

res {f(as)}eiξas +12

m∑s=1

res {f(bs)}eiξbs

]


To provide an application of the above result we first consider thesimplest example when f(z) = 1

z .

This function has a simple pole at z = 0 and it has no other singu-larities inside and on the contour shown above. Then, for ξ > 0

F [1x

; x → ξ] =√

2πi · 12

res [f(0)] · eiξ.0

= i

√π

2

Again f(x) = 1x is an odd function of x in R. This gives

Fs

[1x

; x → ξ

]=√

π

2, ξ > 0 .

To provide another example, consider f(z) = 1z(z2+a2)

where a is areal number. This function f(z) has simple poles inside the contour asdescribed above at z = 0 and at z = ia having residues a−2 and −1

2a−2

respectively, when ξ > 0. Then, we get

F

[1

x(x2 + a2); x → ξ

]=√

π

21a2

i(1 − e−ξ|a|

)

and from it we have

Fs

[1

x(x2 + a2); x → ξ

]=√

π

21a2

(1 − e−ξ|a|

)sgn(ξ) .

1.16 Other important examples concerning derivative of

FT .

The following example will depict different methods for finding F T ofgiven function using derivative of the transformed function whenevernecessary.

Example 1.13. Find Fourier sine and cosine transform of f(x) = x.

Solution.

Let Fc(ξ) =

√2π

∫ ∞

0x cos ξ x dx

and Fs(ξ) =

√2π

∫ ∞

0x sin ξ x dx


Thus, Fc(ξ) − i Fs(ξ) =

√2π

∫ ∞

0x e−iξxdx

=

√2π

∫ ∞

0

t e−t

iξ

dt

iξ, where iξx = t

= −Γ(2)ξ2

= − 1ξ2

Equating real and imaginary parts, we get the required result.

Example 1.14. Find f(x), if Fs[f(x);x → ξ] = e−aξ

ξ . Hence, evaluate

F−1s

[1ξ ; ξ → x

].

Solution. we know that

f(x) = F−1s [Fs(ξ)] =

2π

∫ ∞

0

e−aξ

ξsin ξ x dξ (1.57)

Therefore,df (x)dx

=2π

∫ ∞

0e−aξ cos ξ x dξ =

2π

a

a2 + x2

Integrating, we get

f(x) =2aπ

1a

tan−1 x

a+ A , A is an arbitrary constant

Putting x = 0 in the above, we get A = f(0) = 0, from (1.57).

So, f(x) =2π

tan−1 x

a= F−1

s [Fs(ξ)] = F−1s

[1ξ

e−aξ

]

Putting a = 0 in the above, we get

2π

tan−1(x

0

)=

2π· π

2= 1 = F−1

s

(1ξ· 1)

⇒ F−1s

(1ξ

)= 1

Example 1.15. Find Fourier sine transform of f(x) = 1x(x2+a2)

[An alternative method].

Solution. Let I(ξ) = Fs

[1

x(x2 + a2); x → ξ

]

=

√2π

∫ ∞

0

sin ξ x

x(x2 + a2)dx ,

dI

dξ=

√2π

∫ ∞

0

cos ξ x

x2 + a2dx


Therefore,d2I

dξ2= −

√2π

∫ ∞

0

[1 − a2

x2 + a2

]sin ξ x

xdx

= −√

2π

∫ ∞

0

sin ξ x

xdx + a2 I(ξ)

= −√

π

2+ a2 I(ξ)

Solving this differential equation, we get

I(ξ) = A eaξ + B e−aξ +√

π

21a2

⇒ dI(ξ)dξ

= Aa eaξ − Ba e−aξ

Taking ξ = 0, Aa − Ba =

√2π

π

2· 1a

=√

π

2· 1a

⇒ A − B =√

π

2· 1a2

Also, A + B = −√

π

2· 1a2

Solving we get, A = 0 and B = −√

π

2· 1a2

∴ Fs

[1

x(x2 + a2)

]= −

√π

21a2

e−aξ +√

π

2· 1a2

=√

π

21a2

(1 − e−aξ

)

i.e

√2π

∫ ∞

0

sin ξ x

x(x2 + a2)dx =

√π

2· 1a2

(1 − e−aξ

)

i.e∫ ∞

0

sin ξ x

x(x2 + a2)dx =

π

2a2

(1 − e−aξ

).

Example 1.16. Find Fourier cosine inverse of 11+ξ2 .

Solution.

Solution. Let f(x) = F−1c

[1

1 + ξ2; ξ → x

]

=

√2π

∫ ∞

0

cos ξ x

1 + ξ2dξ

∴ df (x)dx

= −√

2π· π

2+

√2π

∫ ∞

0

sin ξ x dξ

ξ(1 + ξ2)


= −√

π

2e−x .

Thus, f(x) =√

π2 e−x, after evaluating the constant of integration by

putting x = 0 there.

Example 1.17. Evaluate Fc[f(x) ; x → ξ] and Fc[g(x) ; x → ξ] where

f(x) = x−s, 0 < s < 1, and g(x) =(1 − x2

)ν− 12 H(1 − x),

(ν > −1

2

)

Solution. We find Fc[f(x) ; x → ξ] by evaluating the contour integral∫Γ f(z) eiξz dz, where Γ is a positively described quarter circle |z| = R

with identation at z = 0 and as depicted in the adjoining figure.

Fig. 1.2

It may be noted here that

z · f(z) eiξz −→ 0 as ∈→ 0 ⇒ z → 0

and z · f(z) eiξz −→ 0 as R → ∞, 0 < θ <π

2,where z = Reiθ on C2

Therefore,∫

c1

f(z) eiξz dz −→ 0 as |z| → 0

and∫

c2

f(z) eiξz dz −→ 0 as |z| → ∞

Then by Cauchy’s residue theorem we have∫ ∞

0x−s eiξx dx = e

−πsi2

∫ ∞

0y−se−ξy i dy

since the integrand has no singularity inside the contour depicted aboveand is regular there.


This gives,∫ ∞

0x−s eiξx dx = i

(cos

πs

2− i sin

πs

2

)ξs−1Γ(1 − s), which implies

Fc[f(x) ; x → ξ] =√

2π

∫∞0 x−s cos ξ x dx

=√

2π sin sπ

2 Γ(1 − s)ξs−1 and

Fs[f(x) ; x → ξ] =√

2π

∫∞0 x−s sin ξ x dx

=√

2π cos sπ

2 · Γ(1 − s) ξs−1

⎫⎪⎪⎪⎪⎪⎪⎬⎪⎪⎪⎪⎪⎪⎭(1.58)

Also, from the result in (b) of article 1.14 we get

Fc [g(x) ; x → ξ]

= Gc(ξ)

= 2ν− 12 Γ(

ν +12

)ξ−ν Jν(ξ)

Then by the result (iii) of article 1.12 we have∫ ∞

0Fc(ξ) Gc(ξ) dξ =

∫ ∞

0f(x) g(x) dx

implying that

2νπ− 12 sin

sπ

2Γ(1 − s) Γ

(ν +

12

)∫ ∞

0ξs−ν−1 Jν(ξ) dξ

=∫ 1

0x−s(1 − x2)ν−

12 dx , ν > −1

2

=12

∫ 1

0y−

s2− 1

2 (1 − y)ν−12 dy

= Γ(

ν +12

)Γ(

12− s

2

)/[2Γ(ν − s

2+ 1)]

Since, π cosec(sπ

2

)= Γ

(s

2

)Γ(1 − s

2

)and Γ

(12

)Γ(1 − s) = 2−s Γ

(12− s

2

)Γ(1 − s

2

)

we have∫ ∞

0ξs−ν−1 Jν(ξ) dξ =

2s−ν−1 Γ(

s2

)Γ(ν − s

2 + 1) , 0 < s < 1, ν > −1

2.


Similarly another pair of formulae∫ ∞

0Fs(ξ)Gc(ξ) sin(ξx) dξ

=12

∫ ∞

0f(u)[g(|u − x|) − g(u + x)] du

=12

∫ ∞

0g(u)[f(x + u) − f(u − x)] du

can also be used for the derivation of∫ ∞

0Fs(ξ)Gc(ξ) dξ =

∫ ∞

0f(x)g(x) dx

in this case.

Example 1.18. Find Fourier sine and cosine transforms of the function

f(x) =(eax + e−ax

)/(eπx − e−πx

)=

ch ax

sh πx.

Solution. Fs [f(x) ; x → ξ] =

√2π

∫ ∞

0

ch ax

sh πxsin ξ x dx

=

√2π

∫ ∞

0

ch ax

sh πx· eiξx − e−iξx

2idx

=

√2π· 12i

∫ ∞

0

[{e(a+iξ)x − e−(a+iξ)x

}eπx − e−πx

−{e(a−iξ)x − e−(a−iξ)x

}eπx − e−πx

]dx

=

√2π· 12i

[12

tana + iξ

2− 1

2tan

a − iξ

2

],by use of tables of integrals

=

√12π

sh ξ

cos a + ch ξ, on simplification.

Similarly, Fc [f(x) ; x → ξ] =

√2π

∫ ∞

0

ch ax

sh πxcos ξ x dx

=

√2π

12

∫ ∞

0

[{e(a+iξ)x + e−(a+iξ)x

}eπx − e−πx

+

{e(a−iξ)x + e−(a−iξ)x

}eπx − e−πx

]dx

=1√2π

[sec

a + iξ

2+ sec

a − iξ

2

], from tables of integrals

=1√2π

cos a2 ch ξ

2

cos a + ch ξ.


Example 1.19. Show that∫ ∞

0

cos px

1 + p2dp =

π

2e−x , x � 0 ,

Solution. We know that∫ ∞

0e−x cos px dx =

11 + p2

This means that

Fc[e−x ; x → ξ] =

√2π

11 + ξ2

Therefore, by inverse Fourier cosine transform formula we get

F−1c

[√2π

11 + ξ2

; ξ → x

]= e−x

⇒√

2π

∫ ∞

0

√2π

11 + ξ2

cos ξ x dξ = e−x

Or,∫ ∞

0

cos ξ x

1 + ξ2dξ =

π

2· e−x .

Example 1.20. Find Fourier sine and Fourier cosine transforms of thefunction

f(x) =

{sin x , 0 < x < a

0 , x > a

Solution. Fs [ f(x) ; x → ξ ]

=

√2π

∫ a

0sin x sin ξx dx

=1√2π

∫ a

0[cos(1 − ξ)x − cos(1 + ξ)x]dx

=1√2π

[sin(1 − ξ)a

1 − ξ− sin(1 + ξ)a

1 + ξ

]

Fc [f(x) ; x → ξ]

=

√2π

∫ a

0sin x cos ξ x dx

=1√2π

∫ a

0[sin(1 + ξ)x + sin(1 − ξ)x]dx


=1√2π

[−cos(1 + ξ)x

1 + ξ− cos(1 − ξ)x

1 − ξ

]a

0

=1√2π

[− cos(1 + ξ)a + 11 + ξ

+− cos(1 − ξ)a + 1

1 − ξ

]

=

√2π

[sin2(1 + ξ)a

2

1 + ξ+

sin2(1 − ξ)a2

1 − ξ

].

Example 1.21. Find f(x), if its Fourier sine transform is√

2π

ξ1+ξ2

Solution. By question, f(x) = F−1s

[√2π

ξ1+ξ2 ; ξ → x

]

Therefore, f(x) =2π

∫ ∞

0

ξ

1 + ξ2sin ξ x dξ

=2π

∫ ∞

0

[sin ξx

ξ− sin ξx

ξ(1 + ξ2)

]dξ

= 1 − 2π

∫ ∞

0

sin ξx

ξ(1 + ξ2)dξ (i)

∴ df

dx= − 2

π

∫ ∞

0

cos ξx

1 + ξ2dξ (ii)

⇒ d2f

dx2=

2π

∫ ∞

0

ξ sin ξx

1 + ξ2dξ = f(x)

⇒ f(x) = A ex + B e−x (iii)

But f(0) = 1 = A + B , by (i) and (ii) above.

Also since[

df

dx

]x=0

= −1 = A − B , by (ii) and (iii)

∴ A = 0 and B = 1

Hence, f(x) = e−x .

Example 1.22. Using Fourier transform evaluate the following integralsto prove that

(a)∫ +∞

−∞

dx

(x2 + a2)(x2 + b2)=

π

ab(a + b), a > 0 , b > 0

(b)∫ ∞

0

x−pdx

a2 + x2=

π

2a−(p+1) sec

(πp

2

)

(c)∫ ∞

0

x2 dx

(x2 + a2)(x2 + b2)=

π

2(a + b), a > 0 , b > 0

(d)∫ ∞

0

x2

(x2 + a2)4dx =

π

(2a)5, a > 0


Solution.

(a) Let us take f(x) = e−a|x| and g(x) = e−b|x|

Then F (ξ) =

√2π

a

ξ2 + a2, G(ξ) =

√2π

b

ξ2 + b2

Therefore,∫ +∞

−∞F (ξ) G(ξ) eiξxdξ =

∫ +∞

−∞f(t) g(x − t) dt

Hence, by putting x = 0 on both sides we get∫ +∞

−∞F (ξ) G(ξ) dξ =

∫ +∞

−∞f(t) g(−t) dt

⇒∫ +∞

−∞

dξ

(ξ2 + a2)(ξ2 + b2)=

π

2ab

∫ +∞

−∞e−|t|(a+b) dt

=π

ab

∫ ∞

0e−(a+b)tdt

=π

ab(a + b)

(b) Taking f(x) = e−ax, g(x) = xp−1 we have

Fc(ξ) = Fc

[e−ax ; x → ξ

]=

√2π

a

a2 + ξ2

and Fc

[xp−1 ; x → ξ

]=

√2π

ξ−p Γ(p) cosπp

2= Gc(ξ)

By using the Parseval’s relation for Fourier cosine transform weget ∫ ∞

0Fc(ξ) Gc(ξ) dξ =

∫ ∞

0f(x) g(x) dx

Hence,2aπ

cosπp

2Γ(p)

∫ ∞

0

ξ−p dξ

ξ2 + a2

=∫ ∞

0xp−1 · e−ax dx

=Γ(p)ap

, after putting ax = t

Therefore,∫ ∞

0

ξ−p

a2 + ξ2dξ =

π

2ap+1sec

πp

2.


(c) We know that

Fs

[e−ax ; x → ξ

]=

√2π

ξ

ξ2 + a2= Fs(ξ), say

Fs

[e−bx ; x → ξ

]=

√2π

ξ

ξ2 + b2= Gs(ξ), say,

if we choose f(x) = e−ax and g(x) = e−bx.

Then by convolution theorem we have∫ ∞

0Fs(ξ) Gs(ξ) cos ξ x dξ =

12

∫ ∞

0g(ξ) [f(ξ + x) + f(ξ − x)] dξ

Putting x = 0, we get

2π

∫ ∞

0

ξ2 dξ

(ξ2 + a2)(ξ2 + b2)

=∫ ∞

0e−(a+b)ξ dξ

=1

a + b

⇒∫ ∞

0

ξ2 dξ

(ξ2 + a2)(ξ2 + b2)=

π

2(a + b).

(d) By convolution theorem we have∫ +∞

−∞f(x) f(x) dx =

∫ +∞

−∞F (ξ) F (ξ) dξ as the Parseval′s relation,

where F (ξ) is the Fourier transform of f(x).

Let us now write f(x) =1

2(x2 + a2)⇒ f ′(x) = − x

(x2 + a2)4

and F [f(x) ; x → ξ] = F (ξ) =√

π

212a

exp (−a|ξ|) .

Hence,∫ +∞

−∞|f ′(x)|2 dx =

∫ +∞

−∞|F{f ′(x) ; x → ξ}|2 dξ

=∫ +∞

−∞|(iξ) · 1

2a

√π

2exp (−a|ξ|)|2 dξ

⇒∫ +∞

−∞

x2 dx

(x2 + a2)4=

π

22

4a2

∫ ∞

0ξ2 exp (−2aξ) dξ

=2π

(2a)5.


1.17 The solution of Integral Equations of Convolution

Type.

An integral equation is an equation in which the unknown function ap-pears under the integral sign. The most general type of integral equationcan, in general, be broadly classified into two types - Fredholm integralequation where limits of integral are constants and Volterra integralequation in which one of the limits of integration is variable.

The simplest integral equation that can be tackled by means ofFourier transform technique is the integral equation of convolution type

1√2π

∫ +∞

−∞f(t) k(x − t) dt = g(x) , −∞ < x < ∞ (1.59)

Here k(x) , g(x) are prescribed functions for all real values of x and areassumed to possess Fourier transforms K(ξ) and G(ξ) respectively.

Taking F T of each side of (1.59) and making use of the convolutionproperty we obtain formally the relation

F (ξ) K(ξ) = G(ξ) ⇒ F (ξ) = G(ξ) L(ξ), (1.60)

where L(ξ) =1

K(ξ)·

Now, if F−1[L(ξ) ; ξ → x] = l(x)

exists, then applying the convolution theorem we would obtain the so-lution of (1.60) and hence from (1.59) that

f(x) =1√2π

∫ +∞

−∞g(t) l(x − t) dt (1.61)

The main difficulty in the above procedure will arise if l(x), as definedabove, does not exist. However if it so happens that for some positiveinteger n, the inverse transform

F−1

[1

(−iξ)n K(ξ); ξ → x

]= m(x) (1.62)

exists, then (1.60) can be expressed as

F (ξ) = (−iξ)n G(ξ) · M(ξ) ,where M(ξ) = F [m(x) ; x → ξ]


which finally yeilds

f(x) =1√2π

∫ +∞

−∞m(x − t) g(n)(t) dt , (1.63)

where, g(n)(t) =dn

dtng(t) .

As an example, we consider the integral equation

1√2π

∫ +∞

−∞|x − t|− 1

2 f(t) dt = g(x) (1.64)

Here, k(x) = |x|− 12 and hence K(ξ) = |ξ|− 1

2 , from (1.58) of Example1.17 of article 1.16. Now, since

F−1

[1

(−iξ) K(ξ); ξ → x

]

= F−1[i sgn(ξ) |ξ|− 1

2 ; ξ → x]

= |x|− 12 sgn(x)

the solution of the integral equation in (1.63) is given by

f(x) =1√2π

∫ x

−∞

g′(t) dt√x − t

− 1√2π

∫ ∞

x

g′(t) dt√t − x

.

In cases in which the inverse transform (1.62) does not exist it is some-times possible to have a solution of eqn. (1.59), provided that the func-tion g(x) has derivatives of all orders. The corresponding method dueto Eddington consists essentially in assuming the solution of eqn. (1.59)in the form

f(x) =∞∑

n=1

Cn g(n)(x) (1.65)

where Cn (n = 0, 1, 2, · · ·) are constants to be determined. For such atype of solution we have

F [f(x) ; x → ξ] = F (ξ) =∞∑

n=0

Cn(−iξ)n G(ξ),

provided F T of all the derivatives of g(x) exist. Substituting this formof F (ξ) in eqn. (1.60) we get

[K(ξ)]−1 =∞∑

n=0

Cn(−iξ)n ,


so that if 1/K(ξ) can be expanded as a Maclaurin series in the neighbour-hood of ξ = 0, then (−i)nCn is the coefficient of ξn in that expansion.

If there is an Eddington solution of the form (1.65), then Ci’s areuniquely determined from k(x) ; but there exists still some kernels k(x)which exclude, for any infinitely differentiable function g(x), all possi-bility that an Eddington solution exists.

For the case in which Eddington solution exists, we consider theintegral equation

1σ√

2π

∫ +∞

−∞f(t)e−

12(x−t)2/σ2

dt = g(x),

where gn(x) exists for all n ∈ I. Here

k(x) = σ−1e−12x2/σ2 ⇒ K(ξ) = e−

12ξ2σ2

.

Assuming the solution due to Eddington we have∞∑

n=0

Cn(−iξ)n = e12ξ2σ2

.

This equation uniquely determines Cn. In fact, here

C2m−1 = 0

and C2m = (−1)m12σ2m

m!.

Then the solution of the integral equation is

f(x) =∞∑

m=0

(−1)m[

12 σ2m

m!

]· g2m (x) .

Example 1.23. Solve the integral equation∫ ∞

0f(x) cos ξ x dx =

{(1 − ξ) , 0 ≤ ξ ≤ 10 , ξ > 1

Solution. By definition, we can have

Fc[f(x) ; x → ξ] = Fc(ξ)

=

√2π

∫ ∞

0f(x) cos ξ x dx

=

{ √2π (1 − ξ) , 0 ≤ ξ ≤ 1

0 , ξ > 1


Therefore, taking inverse Fourier cosine transform we have√2π

∫ ∞

0Fc [f(x) ; x → ξ] cos ξ x dξ = f(x)

⇒ f(x) =

√2π·√

2π

∫ 1

0(1 − ξ) cos ξ x dξ

=2π

∫ 1

0(1 − ξ) cos ξ x dξ

=2π

[(1 − ξ)

sin ξ x

x

∣∣∣∣1

0

+∫ 1

0

sin ξx

xdξ

]

=2π

[−cos ξ x

x2

∣∣∣∣1

0

]

=2(1 − cos x)

πx2.

Example 1.24. Solve the integral equations

(i)∫ ∞

0f(x) cos λx dx = e−λ

and (ii)∫ ∞

0f(x) sin λx dx = e−λ

Solution. (i) By definition of F.C. transform, we have

Fc[f(x) ; x → ξ] = Fc(ξ) =

√2π

∫ ∞

0f(x) cos ξx dx

=

√2π

e−ξ , by the problem (i) .

Thus by Inverse transform, we get

f(x) =2π

∫ ∞

0e−ξ cos ξ x dξ

=2π

[e−ξ

1 + x2(− cos ξ x + x sin ξ x)

∣∣∣∣∞

0

]

=2π

[1

1 + x2

]

(ii) Fs[f(x) ; x → ξ] = Fs(ξ) =

√2π

∫ ∞

0f(x) sin ξ x dx

=

√2π

e−ξ , by the given question


Then on taking the inverse Fourier sine transform, we get

f(x) =2π

∫ ∞

0e−ξ sin ξ x dξ

=2π

[x

1 + x2

], by evaluating the integral.

Example 1.25. Solve the integral equation

∫ ∞

0f(x) sin ξ x dx = F (ξ) =

⎧⎪⎨⎪⎩

1 , 0 � ξ < 12 , 1 � ξ < 20 , ξ � 2 .

Solution. As in example (1.23),we can have

f(x) =2π

∫ ∞

0F (ξ) sin ξ x dξ

=2π

[∫ 1

0sin ξ x dξ +

∫ 2

12 sin ξ x dξ

]

=2π

[− cos ξ x

x

∣∣∣∣1

0

− 2 cos ξ x

x

∣∣∣∣2

1

]

=2π

[1 − cos x

x+

2x

(cos x − cos 2x)]

=2

πx[1 + cos x − 2 cos 2x]

Example 1.26. Solve the following Fredholm integral equation withconvolution type kernel∫ +∞

−∞f(t) k(x − t) dt + λ f(x) = u(x)

for the particular cases

(i) k(x) =1x

(ii) k(x) =12

x

|x| , λ = 1

(iii) k(x) = f(x) , λ = 0 and u(x) =1

x2 + a2

(iv) k(x) =1

x2 + a2, λ = 0 and u(x) =

1x2 + b2

(v) k(x) = 4 e−a(x) , λ = 1 and u(x) = e−|x|


Solution. Fourier transform of the integral equation gives√

2π F (ξ) K(ξ) + λ F (ξ) = U(ξ)

Or, F (ξ) = U(ξ)/[√

2π K(ξ) + λ]

Inverting the Fourier transformed function we get

f(x) =1√2π

∫ +∞

−∞

U(ξ) eiξx

[√

2πK(ξ) + λ]dξ

(i) Since k(x) = 1x , we have K(ξ) = i

√π2 sgn ξ .

Therefore, the solution of the particular integral equation is

f(x) =1√2π

∫ +∞

−∞

U(ξ) eiξx

λ + i π sgn (ξ)dξ

(ii) If k(x) = 12

x|x| , K(ξ) = 1√

2π1iξ , then the solution of the particular

integral equation is given by

f(x) =1√2π

∫ +∞

−∞iξ

U(ξ) eiξx

(1 + iξ)dξ

=1√2π

∫ +∞

−∞F [u′(x);x → ξ] F [

√2π e−x;x → ξ] eiξx dξ

= u′(x) ∗√

2π e−x =∫ +∞

−∞u′(ξ) exp (ξ − x) dξ

(iii) In this particular case we have

F (ξ) =1√2a

exp{−1

2a |ξ|

}

Therefore, f(x) =1√2π

1√2a

∫ +∞

−∞exp

{iξx − 1

2a |ξ|

}dξ

=1

2√

πa

[∫ ∞

0exp{−ξ(a

2+ ix

)}dξ

+∫ ∞

0exp{−ξ(a

2− ix

)}dξ

]

=√

a

π

24x2 + a2


(iv) For this particular case Fourier transform of the integral equationis

√2π F (ξ) F

[1

x2 + a2; x → ξ

]=√

π

2e−b|ξ|

b

Or√

2π F (ξ)√

π

2e−a|ξ|

a=√

π

2e−b|ξ|

b

Thus, F (ξ) =1√2π

a

bexp {−|ξ|(b − a)}

Inverting the Fourier transformed function we get

f(x) =a

2πb

[∫ ∞

0exp[−ξ{(b − a) + ix}] dξ

+∫ ∞

0exp[−ξ{(b − a) − ix}]dξ

]

=a

2πb

[1

(b − a) + ix+

1(b − a) − ix

]

=a

πb

[b − a

(b − a)2 + x2

]

(v) In this case Fourier transform of the integral equation is

F (ξ) =a2 + ξ2

a2 + ξ2 + 8aU(ξ)

Inverting, f(x) =1√2π

∫ +∞

−∞(a2 + ξ2)U(ξ)a2 + ξ2 + 8a

eiξx dξ

If u(x) = e−|x|, a = 1, U(ξ) =

√2π

11 + ξ2

Therefore, f(x) =1π

∫ +∞

−∞

eiξx

ξ2 + 32dξ

=13

exp {−3|x|} ,

after separately considering the cases x > 0 and x < 0 and thencombining them.

1.18 Fourier Transform of Functions of several variables.

Similar results like those in article 1.13, FT of the functions of severalvariables do hold. For simplicity, we state them for a function f(x, y)


of two independent variables. The extension to a higher number ofvariables is obvious. These extensions are sometimes called multipleFourier transforms.

Let us define

F [f(x, y) ; x → ξ] = F [ξ, y] =1√2π

∫ +∞

−∞f(x, y) eiξx dx

Fc[f(x, y) ; x → ξ] = Fc[ξ, y] =

√2π

∫ ∞

0f(x, y) cos ξ x dx

and Fs[f(x, y) ; x → ξ] = Fs[ξ, y] =

√2π

∫ ∞

0f(x, y) sin ξ x dx

The corresponding inversion formulae are then given by

F−1[F (ξ, y) ; ξ → x] = F−1(ξ, y) =1√2π

∫ +∞

−∞F (ξ, y) · e−iξx dξ,

F−1[Fc(ξ, y) ; ξ → x] = F−1c (ξ, y) =

√2π

∫ ∞

0Fc(ξ, y) cos ξ x dξ

and F−1[Fs(ξ, y) ; ξ → x] = F−1s (ξ, y) =

√2π

∫ ∞

0Fs(ξ, y) sin ξ x dξ

respectively.

The analogues of the results in 1.13 for continuous functions are

F

[∂f

∂x; x → ξ

]= −iξ F (ξ, y)

Fc

[∂f

∂x; x → ξ

]= −f(0, y) + ξ Fs(ξ, y)

Fs

[∂f

∂x; x → ξ

]= −ξ Fc(ξ, y)

Applying these results in succession we find that

F

[∂2f

∂x2; x → ξ

]= −ξ2 F (ξ, y)

Fc

[∂2f

∂x2; x → ξ

]= −ξ2 Fc(ξ, y) − ∂f

∂x(0, y)

Fs

[∂2f

∂x2; x → ξ

]= −ξ2 Fs(ξ, y) + ξ f(0, y)

Similarly, FT of other higher order partial derivatives with respect tocorresponding variables can be deduced. These results will help us in


reducing partial differential equation to a lower dimensional equation ofvariables. Thus, FT can be used in solving the boundary value problemsin two or higher dimensions.

The above ideas can be extended to several variables. Let f(x, y)be a function of two independent variables x, y. Temporarily treatingf(x, y) as a function of x, we have

F [f(x, y) ; x → ξ] = f(ξ, y) =1√2π

∫ +∞

−∞f(x, y) eiξx dx

Then treating f(ξ, y) as a function of the independent variable y, its FT

is given by

f=

(ξ, η) =1√2π

∫ +∞

−∞f(ξ, y) eiηy dy = F [f(ξ, y) ; y → η]

Thus, finally we have

f=

(ξ, η) =12π

∫ +∞

−∞

∫ +∞

−∞f(x, y) ei( ξx+ηy ) dx dy

and the corresponding inversion formula is

f(x, y) =12π

∫ +∞

−∞

∫ +∞

−∞f=

(ξ, η) e−i(ξx+ηy) dξ dη .

Such extensions to Fourier sine and Fourier cosine transforms in func-tions of several variables do exist and the corresponding results can bededuced easily.

Using the above definitions of FT to functions of several variables,we can apply FT to mixed partial derivatives too arising in partial dif-ferential equations in solving boundary value problems.

1.19 Application of Fourier Transform to Boundary Value

Problems

We first discuss the use of sine and cosine transform and finally discussthe use of complex Fourier transform arising in boundary value problems.

The sine and cosine transforms can be applied when the range ofvariable selected for exclusion is from 0 to ∞. The choice of sine and


cosine transform is decided by the form of the boundary conditions atthe lower limit of the variable selected for exclusion.

For example,

Fs

[∂2u(x, y)

∂x2;x → ξ

]=

√2π

∫ ∞

0

∂2u

∂x2sin ξ x dx

=

√2π

ξ

∫ ∞

0

∂u

∂xcos ξ x dx

=

√2π

ξ u(0, y) −√

2π

ξ2 us(ξ, y), (1.66)

provided u(x, y) is known at x = 0 and∂u

∂x→ 0 as x → ∞

Similarly,

Fc

[∂2u(x, y)

∂x2; x → ξ

]=

√2π

∫ ∞

0

∂2u

∂x2cos ξ x dx

= −√

2π

[∂u(x, y)

∂x

]x=0

−√

2π

ξ2

∫ ∞

0u(x, y) cos ξ x dx, (1.67)

provided∂u(0, y)

∂xis known and u,

∂u

∂x→ 0 as x → ∞.

Noting carefully the results in (1.66) and (1.67) it may be seen thatremoving a term ∂2u(x,y)

∂x2 from a partial differential equation requires theknowledge of u(0, y) for use of a sine transform while the use of a cosinetransform for the same purpose requires the knowledge of ux(0, y) to beknown.

It may be noted that a term ∂u∂x or any odd order partial derivative

can not be removed with the help of sine or cosine transforms.

Again, complex Fourier transform will be found useful for the samepurpose as above, if the range of variable is from −∞ to +∞ of thepartial differential equation.

Example 1.27. The temperature u(x, t) of a semi-infinite rod is deter-mined by the partial differential equation ∂u

∂t = ∂2u∂x2 , x > 0 , t > 0

subject to the initial condition u(x, 0) = 1 , 0 < x < 1

= 0 , x > 1

and the boundary condition u(0, t) = 0. Find the temperature at anytime t at any distance x from x = 0.


Solution. Since the variable x varies from 0 to ∞ and since thevalue of u(x, t) at x = 0 is prescribed, Fourier sine transform of bothsides of the partial differential equation is to be taken on the variablex which is to be excluded in the transformed equation. Then the givenequation becomes

d

dtus(ξ, t) =

√2π

∫ ∞

0

∂2u

∂x2sin ξx dx

=

√2π

[ξ u(0, t)] − ξ2 us(ξ, t)

= −ξ2 us(ξ, t)

∴ us(ξ, t) = c e−ξ2t, where c is an arbitrary constant . (i)

It is given that initially

u(x, 0) = 1 , 0 < x < 1

= 0 , x > 1

∴ us(ξ, 0) =

√2π

∫ ∞

0u(x, 0) · sin ξ x dx

=

√2π

∫ 1

0sin ξ x dx =

√2π

[− cos ξ x

ξ

]10

=

√2π

[1 − cos ξ

ξ

](ii)

Therefore, from (i) and (ii) we get

c =

√2π

1 − cos ξ

ξ

Using this value of c eqn(i) gives

u(x, t) =2π

∫ ∞

0

1 − cos ξ

ξe−ξ2t sin ξ x dξ.

This is the required temperature function u(x, t).

Example 1.28. Solve the diffusion equation ∂u∂t = ∂2u

∂x2 , x > 0, t > 0subject to the boundary condition ux(0, t) = 0 and u(x, t) is bounded.The initial condition is being

u(x, 0) =

{x , 0 ≤ x ≤ 10 , x > 1


Solution. Since the range of x is from 0 to ∞ and the value ofux(0, t) is prescribed, it is useful to apply Fourier cosine transform toremove the variable x from the PDE.

Thus we get,

d

dtuc(ξ, t) =

√2π

∫ ∞

0

∂2u

∂x2cos ξ x dx

=

√2π

[cos ξ x

∂u

∂x

∣∣∣∣∞

0

+ ξ

∫ ∞

0

∂u

∂xsin ξ x dx

]

= −√

2π

ξ2

∫ ∞

0u(x, t) cos ξ x dx = −ξ2 uc(ξ, t)

∴ uc(ξ, t) = A e−ξ2t ⇒ uc(ξ, 0) = A, arbitrary constant.

Now, uc(ξ, 0) =

√2π

∫ ∞

0u(x, 0) cos ξ x dx =

√2π

∫ 1

0x cos ξ x dx,

using given initial condition

i.e uc(ξ, 0) =

√2π

[sin ξ

ξ+

cos ξ − 1ξ2

]= A

This result gives

uc(ξ, t) =

√2π

[sin ξ

ξ− 1 − cos ξ

ξ2

]. e−ξ2t

Taking inverse Fourier cosine transform the required solution is

u(x, t) =2π

∫ ∞

0

ξ sin ξ − 1 + cos ξ

ξ2e−ξ2t cos ξ x dξ .

Example 1.29. Use Fourier transform to determine the displacementy(x, t) of an infinite vibrating string, given that the string is initially atrest and that the initial displacement is f(x),−∞ < x < ∞. Show thatthe required solution can also be put in the form

y(x, t) =12[f(x + ct) + f(x − ct)]

when the one dimensional wave equation of the vibrating string is givenby

∂2y

∂t2= c2 ∂2y

∂x2, −∞ < x < ∞ , t > 0 .


Solution. Since the variable x varies from −∞ to ∞, to remove thevariable from the PDE, we apply complex FT to the given equation toobtain

d2y(ξ, t)dt2

=c2

√2π

∫ +∞

−∞

∂2y(x, t)∂x2

eiξxdx

= −c2ξ2y(ξ, t)

Therefore, y(ξ, t) = A cos(cξt) + B sin(cξt) (i)

and yt(ξ, t) = A c ξ sin(cξt) + B c ξ cos(cξt) (ii)

where A and B are two arbitrary constants. Now, the given initialconditions are

y(x, 0) = initial displacement = f(x) (iii)

and∂y

∂t(x, 0) = 0, since initial velocity of vibrating string is zero.

(iv)

From (ii) and (iii) we get

y(ξ, 0) =1√2π

∫ +∞

−∞f(x) eiξx dx = f(ξ) , say (v)

and∂y

∂t(ξ, 0) = yt(ξ, 0) = 0 . (vi)

Then from eqn. (i) we have

y(ξ, 0) = A = f(ξ) , by eqn. (v)

Also, from eqn. (ii) we have

yt(ξ, 0) = B c ξ = 0, by eqn. (vi)

Thus, B = 0 .

Theny(ξ, t) = f(ξ) cos (c ξ t)

Taking inverse FT to the above equation we get

y(x, t) =1√2π

∫ +∞

−∞f(ξ)

[eicξt + e−icξt

2

]e−iξx dξ


=12

1√2π

∫ +∞

−∞f(ξ) ·

[e−iξ(x−ct) + e−iξ(x+ct)

]dξ

=12

[1√2π

∫ +∞

−∞f(ξ) e−ξ(x−ct)dξ +

1√2π

∫ +∞

−∞f(ξ) · e−iξ(x+ct)dξ

]

⇒ y(x, t) =12

[f(x − ct) + f(x + ct)] , by inverse F T formula.

Thus, the required result is proved.

Example 1.30. The steady state temperature distribution in a semi-infinite solid y > 0 is represented by the two-dimensional Laplace equa-tion

∂2u

∂x2+

∂2u

∂y2= 0, 0 < y < ∞ , −∞ < x < ∞

subject to the boundary conditions

u(x, 0) = 1 , |x| < a

and u(x, 0) = 0 , for |x| > a

Thus the temperature on the insulated surface y = 0 is kept at unityover the strip |x| < a and at zero outside the strip. Then using cosinetransform show that

u(x, y) =1π

[tan−1 a + x

y+ tan−1 a − x

y

],

after assuming the result that∫ ∞

0e−sx sin rx

xdx = tan−1 r

s, (r, s > 0).

Solution. Since the surface y = 0 is kept at a fixed temperature,there is no flow of heat normal to it and hence ∂u

∂x → 0 as x → 0 and asx → ∞. Therefore, taking Fourier cosine transform of the given partialdifferential equation in variable x we get√

2π

∫ ∞

0

∂2u

∂x2cos ξ x dx +

√2π

∫ ∞

0

∂2u

∂y2cos ξx dx = 0

Or, − ξ2 uc(ξ, y) +d2

dy2uc(ξ, y) = 0

Its general solution is

uc(ξ, y) = A eξy + B e−ξy


But u(x, y) is finite as y → ∞, uc(ξ, y) is also finite and hence we musthave A = 0. Thus

uc(ξ, y) = B e−ξy (i)

Again,

√2π

∫ ∞

0u(x, 0) cos ξ x dx =

√2π

∫ 1

0cos ξ x dx

=sin ξa

ξ·√

2π

Thus uc(ξ, 0) =

√2π

sin ξ a

ξ= B , from eqn. (i)

Hence, we have

uc(ξ, y) =

√2π

sin ξ a

ξe−ξy

Therefore, u(x, y) =2π

∫ ∞

0

sin ξ a

ξ· e−ξy cos ξ x dξ

=1π

∫ ∞

0

e−ξy

ξ[sin(a + x)ξ + sin(a − x)ξ] dξ

=1π

[tan−1 a + x

y+ tan−1 a − x

y

].

Example 1.31. Solve the one dimensional heat conduction problemwhere the temperature of a semi-infinite bar satisfies the PDE

∂u

∂t= k

∂2u

∂x2, t > 0, 0 < x < ∞,

the boundary and initial prescribed conditions are being

u(0, t) = f(t), u(x, 0) = 0 and u(x, t) → 0 as x → ∞.

Solution. Applying Fourier sine transform over the variable x thePDE transforms to

d

dtus(ξ, t) = k ξ

√2π

u(0, t) − k ξ2 us(ξ, t)

Or,d us

dt(ξ, t) = −k ξ2 us(ξ, t) +

√2π

k ξ f(t)

This linear ODE in us(ξ, t) has the solution given by

us(ξ, t) · ekξ2t =

√2π

kξ

∫f(t) ekξ2t dt + A

=

√2π

kξ

∫ t

f(ς) ekξ2ς dς + A (i)


Again, u(x, 0) = 0 , by given condition and therefore us(ξ, 0) = 0 .

Putting t = 0 in eqn. (i) we get

us(ξ, 0) · 1 =

√2π

kξ

∫ ς=0

f(ς) ekξ2ςdς + A (ii)

Therefore, ekξ2tus(ξ, t) =

√2π

kξ

∫ t

0f(ς) ekξ2ςdς

Or, us(ξ, t) =

√2π

kξ

∫ t

0f(ς) e−kξ2(t−ς)dς

So, u(x, t) = F−1s

[√2π

kξ

∫ t

0f(ς) e−kξ2(t−ς)dς ; ξ → x

]

=

√2π

k

∫ t

0f(ς)F−1

s [ξ e−lξ2; ξ → x] ,where l = k(t − ς)

From the deduction (iii) of example 1.9 we can have the result

F−1s

[ξe−lξ2

; ξ → x]

=1√8l3

x e−x2

4l = g(x, ς), say

Therefore, u(x, t) =

√2π

k

∫ t

0f(ς) g(x, ς) dς,

where g(x, ς) =x(t − ς)−3

√8k3

· e−x2

[4k(t−ς)]

gives the required solution of the problem.

Example 1.32. Find the solution of the Laplace’s equation

∂2u

∂x2+

∂2u

∂y2= 0

in the half-plane y � 0, subject to the boundary conditionu(x, 0) = f(x),−∞ < x < ∞ and the limiting condition u(x, y) → 0 asρ =

√x2 + y2 → ∞.

Solution. Introducing Fourier transform to remove the variable x,we get

d2 u (ξ, y)dy2

− ξ2 u(ξ, y) = 0 (i)

From the given boundary condition we get

u(ξ, 0) =1√2π

∫ +∞

−∞f(x) eiξxdx = F (ξ), say (ii)


The limiting condition gives u(ξ, y) → 0 , y → ∞ . (iii)

Using (ii) and (iii), the solution of (i) is given by

u(ξ, y) = F (ξ) e−|ξ|y with F (ξ) = F [f(x) ; x → ξ]

Then using the convolution theoem we can invert the above equation toobtain

u(x, y) =1√2π

∫ +∞

−∞f(t) g(x − t) dt (iv)

where g(x) = F−1[e−|ξ|y ; ξ → x

]

=

√2π

y

x2 + y2

Substituting this result in (iv) we get

u(x, y) =y

π

∫ +∞

−∞

f(t)dt

(x − t)2 + y2, y > 0 .

Example 1.33. Derive the solution of Laplace equation in the infinitestrip −∞ < x < ∞, 0 � y � a subject to the boundary conditionsu(x.0) = f(x), u(x, a) = g(x) and limiting condition u(x, y) → 0 asx → ±∞ .

Solution. To remove the variable x, we introduce Fourier transformover the variable x. Then the Laplace equation transforms to

d2

dy2u(ξ, y) − ξ2 u(ξ, y) = 0 (i)

Also, the boundary conditions under F T transform to

u(ξ, 0) = F (ξ) = F [f(x) ; x → ξ] , u(ξ, a) = G(ξ) = F [g(x) ; x → ξ] .

Therefore, the solution of eqn. (i) under transformed boundary conditionis

u(ξ, y) = F (ξ)sh ξ(a − y)

sh ξa+ G(ξ)

sh ξy

sh ξa, 0 < y < a .

By convolution theorem, the required solution of the problem is givenby

u(x, y) =12a

∫ +∞

−∞f(t)K1(x − t , a − y) dt


+12a

∫ +∞

−∞g(t)K1(x − t , y) dt (ii)

where K1(x, y) =

√2π

a F−1

[sh ξ y

sh ξ a; ξ → x

]

=

√2π

a Fc

[sh ξ y

sh ξ a; ξ → x

],

By calculus of residues, above Fourier cosine transform is evaluated as

K1 (x, y) =sin(πy

a

)ch(

πxa

)+ cos

(πya

)Substituting this result in eqn. (ii), we get

u(x, y) =12a

sin(πy

a

) ∫ +∞

−∞

f(t)dt

ch π (x−t)a − cos πy

a

+12a

sin(πy

a

)∫ +∞

−∞

g(t) dt

ch π (x−t)a + cos πy

a

Example 1.34. Consider the problem as in 1.33 subject to the bound-ary conditions u(x, 0) = f(x) and uy(x, a) = 0. Find the temperaturedistribution u(x, y) of the problem defined by the Laplace equation

∂2u

∂x2+

∂2u

∂y2= 0 , −∞ < x < ∞ , 0 � y � a

Solution. We apply F T over variable x to obtain

u(ξ, y) = F (ξ)chξ(a − y)

ch ξa, 0 � y � a (i)

where F (ξ) =1√2π

∫ +∞

−∞f(x) eiξx dx (ii)

after using the transformed value of the boundary conditionsu(ξ, 0) = F (ξ) and uy(ξ, 0) = 0. Then by convolution theorem we get

u(x, y) =∫ +∞

−∞f(t) K2(x − t , a − y) dt

where K2(x, y) =1√2π

Fc

[ch (ξy)ch (ξa)

; ξ → x

]


This K2(x, y) may be evaluated by method of contour integral. But thisevaluation is being quite involved, we stop here in persuing it. We onlymention the result for K2(x, y) as

K2 (x, y) =ch(

πx2a

)cos(πy

2a

)ch(

πxa

)+ cos

(πya

)which can also be found from the tables of Integral transform. Then thesolution of the problem is given by

u(x, y) = sin(πy

2a

) ∫ +∞

−∞

f(t) chπ(x−t)2a dt

chπ(x−t)a − cos πy

a

.

Example 1.35. Illustrate the method of solution of the boundary valueproblem defined by the Laplace’s equation

∂2u

∂x2+

∂2u

∂y2= 0 , x � 0 , 0 � y � a

in the semi-infinite strip subjected to the boundary conditions

(i) u(0, y) = 0, 0 � y � a ; u(x, 0) = f(x) , x � 0 ;

u(x, a) = g(x) , x � 0

and (ii) u(x, 0) = f(x), x � 0;ux(0, y) = 0, 0 � y � a;uy(x, a) = 0, x � 0.

Solution. (i) Physical interpretation of this boundary value prob-lem is the distribution of the temperature u(x, y) in the semi-infinitestrip where the faces are maintained at prescribed temperatures e.g.f(x) on y = 0 and g(x) on y = a and zero on x = 0 , 0 � y � a.These conditions tells us that Fourier sine transform over the variablex is found useful in solving the problem as below. Applying the saidtransform to the PDE we get

[d2

dy2− ξ2

]us(ξ, y) = 0 , where us(ξ, y) =

√2π

∫ ∞

0u(x, y) sin ξ x dx

This equation satisfies the boundary conditions on y = 0 and y = a.

Therefore, in 0 ≤ y ≤ a we have

us(ξ, y) = Fs(ξ)sh ξ(a − y)

sh ξa+ Gs(ξ)

sh ξy

sh ξa


where Fs(ξ) and Gs(ξ) are Fourier sine transforms of f(x) and g(x)respectively. Then proceeding as in example 1.34 we get

us(ξ, y) =1a

√π

2[Fs(ξ)Fc{K1(x, a−y) ; x → ξ}+Gs(ξ)Fc{K1(x, y) ; x → ξ}]

Now, using the convolution theorem the solution of the problem is givenby

u(x, y) =12a

∫ ∞

0f(t){K1(|t − x| , a − y) − K1(t + x , a − y)} dt

+12a

∫ ∞

0g(t){K1(|t − x| , y) − K1(t + x , y)} dt

Solution. (ii) In this case, the given boundary conditions suggestthat use of Fourier cosine transform is found useful over variable x.Hence, the PDE transforms to[

d2

dy2− ξ2

]uc(ξ, y) = 0 ,

where uc(ξ, y) = Fc[u(x, y) ; x → ξ] (i)

The transformed form of the boundary conditions are then

uc(ξ, 0) = Fc(ξ) = Fc[f(x) ; x → ξ]

and ddy uc(ξ, a) = 0

⎫⎪⎬⎪⎭ (ii)

The appropriate solution of eqn.(i) under eqns. (ii) is given by

uc(ξ, y) = Fc(ξ)ch ξ(a − y)

ch ξa=

√2π Fc (ξ) Fc [K2(x , a − y) ; x → ξ]

where K2(x, y) is same as defined in the solution of the example in 1.34.Now making use of convolution theorem we get

u(x, y) =∫ ∞

0f(t)[K2(x + t , a − y) + K2(|x − t| , a − y)] dt

A more explicit expression as in examples 1.34 and 1.35 can similarlybe obtained. But we do not persue it due to lack of space.

Example 1.36. Solve the PDE

∂u

∂t= 2

∂2u

∂x2


subject to the conditions u(0, t) = 0, u(x, 0) = e−x , x > 0 and u(x, t)is bounded when x > 0 , t > 0 .

Solution. Taking Fourier sine transform on x, we get the PDE as

d

dtus(ξ, t) = 2 ·

√2π

∫ ∞

0

∂2u

∂x2sin ξ x dx

= −2ξ2us(ξ, t)

Integrating, us(ξ, t) = A e−2ξ2t (i)

Now, from the condition u(x, 0) = e−x, we have

us(ξ, 0) =

√2π

ξ

(1 + ξ2)(ii)

Using (ii) in (i) we have A = ξ1+ξ2

√2π and therefore,

us(ξ, t) =

√2π

ξ

1 + ξ2e−2ξ2t

Now, taking inverse Fourier sine transform of the above result the solu-tion of the PDE under given conditions is

u(x, t) =2π

∫ ∞

0

ξ

1 + ξ2e−2ξ2t sin ξ x dξ

Example 1.37. Prove that the steady state temperature distributionu(x, y, ) inside a semi-infinite strip x > 0 , 0 < y < b under the boundaryconditions u(x, 0) = f(x) , 0 < x < ∞ ; u(x, b) = 0 , 0 < x < ∞ andu(0, y) = 0 , 0 < y < b is given by

u(x, y) =2π

∫ ∞

0f(u)du

∫ ∞

0

sh(b − y)ξsh bξ

sin ξ x sin ξ u dξ

Solution. Since u(x, y) is the steady-state temperature, it must satisfythe PDE

∂2u

∂x2+

∂2u

∂y2= 0 (i)

Taking Fourier sine transform over the variable x, eqn.(i) becomes

d2us

dy2− ξ2 us(ξ, y) = 0 , after using the value u(0, y) = 0 .


Its solution is given by, us(ξ, y) = A ch ξy + B sh(b − y)ξ, (ii)where A, B are arbitrary constants.

Taking Fourier sine transforms of u(x, 0) = f(x) and u(x, b) = 0 weget

us(ξ, 0) =

√2π

∫ ∞

0f(u) sin ξ u dx and us(ξ, b) = 0 .

Using these results in (ii) we obtain, A = 0 and

B =1

sh bξ[us(ξ, 0)] =

√2π

∫ ∞

0

f(u) sin ξ u du

sh bξ

Therefore,

us(ξ, y) =

√2π

sh(b − y)ξsh bξ

∫ ∞

0f(u) sin ξ u du

Then taking inverse Fourier sine transform we get

u(x, y) =2π

∫ ∞

0sin ξ x

[sinh(b − y)ξ

sh bξ

∫ ∞

0f(u) sin ξ u du

]dξ

=2π

∫ ∞

0f(u)du

[∫ ∞

0

sh(b − y)ξsh bξ

sin ξ x sin ξ u dξ

]

Example 1.38. The temperature u(x, t) of a semi-infinite rod0 � x < ∞ satisfies the PDE ∂u

∂t = κ∂2u∂x2 , subject to conditions

u(x, 0) = 0 , x � 0 and ∂u∂x = −λ, a constant, when x = 0 , t > 0.

Solution. Using Fourier cosine transform over the variable x, thePDE transforms to

d uc(ξ, t)dt

=

√2π

κ λ − κ ξ2 uc(ξ, t)

⇒ d uc

dt+ κ ξ2 uc(t) = κ λ

√2π

(i)

using given conditions on the boundary.

Now solving the eqn.(i) we have

uc(ξ, t) =λ

ξ2

√2π

+ A e−κξ2t (ii)

Again, u(x, 0) = 0 ⇒ us(ξ, 0) = 0 (iii)


Using (iii) in (ii), we get

uc(ξ, t) =

√2π

λ

ξ2

[1 − e−κξ2t

]

Therefore, by inverse cosine transform we get the required temperature

u(x, t) =2λπ

∫ ∞

0

cos ξx

ξ2(1 − e−κξ2t) dξ .

Example 1.39. Solve the PDE ∂2u∂x2 − ∂u

∂y = 0 in −∞ < x < ∞ andy > 0 under the conditions u(x, 0) = f(x) if u(x, y) is bounded when|x| → ∞ .

Solution. Taking F T over the variable x, the given PDE gives

1√2π

∫ +∞

−∞

∂2u

∂x2eiξx dx =

1√2π

∫ +∞

−∞

∂u(x, y)∂y

· eiξx dx

Thus, − ξ2 u(ξ, y) =d

dyu(ξ, y)

Solving this simple ordinary differential equation we get

u(ξ, y) = A e−ξ2y (i)

The condition u(x, 0) = f(x) gives rise to

u(ξ, 0) =1√2π

∫ +∞

−∞f(x) eiξx dx (ii)

Also, from (i), putting y = 0 we get

u(ξ, 0) = A (iii)

Comparing (ii) and (iii) and using the result in (i) we finally get

u(ξ, y) = e−ξ2y 1√2π

∫ +∞

−∞f(u) eiξu du

Taking inverse Fourier transform, we get

u(x, y) =12π

∫ +∞

−∞e−iξx

[e−ξ2y

∫ +∞

−∞f(u) · eiξudu

]dξ

=12π

∫ +∞

−∞f(u)

[∫ +∞

−∞e−ξ2y · e−iξ(x−u)dξ

]du


=12π

∫ +∞

−∞f(u)

[∫ +∞

−∞e−y{ξ+ i(x−u)

2y}2− (x−u)2

4y dξ

]du

=12π

∫ +∞

−∞f(u) · e

−(x−u)2

4y

[∫ +∞

−∞e−y{ξ+ i(x−u)

2y}2

dξ

]du

Now, putting[ξ +

i(x − u)2y

]=

η√y

, we get

dξ =dη√

y

so that the inner integral becomes∫ +∞

−∞e−η2 · dη =

√π , a standard result.

Thus we get,

u(x, y) =12π

∫ +∞

−∞

f(u)√y

e−(x−u)2

4y√

π du

=1

2√

πy

∫ +∞

−∞f(u) exp

[−(x − u)2

4y

]du .

Example 1.40. Solve the boundary value problem for the Laplaceequation uxx + uyy = 0 in the quarter plane 0 < x < ∞ , 0 < y < ∞with the boundary conditions

u(0, y) = a , u(x, 0) = 0

∇u → 0 as r =√

x2 + y2 → ∞

where a is a constant. Using Fourier transform solve it.

Solution. Applying Fourier sine transform over x the Laplace equa-tion takes the form

d2Us

dy2− ξ2 Us +

√2π

ξa = 0,

where Us =

√2π

∫ ∞

0u(x, y) sin ξx dx

Its solution is

Us(ξ, y) = A e−ξy +

√2π

a

ξ,


where A is a constant. Now, sine transform of u(x, 0) = 0 is

Us(ξ, 0) = 0 .

Hence, Us(ξ, y) =a

ξ

√2ξ

[1 − e−ξy

].

Inverting, the solution of the boundary value problem is

u(x, y) =2aπ

∫ ∞

0

1ξ

(1 − e−ξy) sin ξ x dξ

= a − 2aπ

[π2− tan−1 y

x

]=

2aπ

tan−1

(x

y

).

Example 1.41. Solve the Laplace equation uxx + uyy = 0 in a semi-infinite strip in the xy-plane defined by 0 < x < ∞ , 0 < y < b underthe boundary conditions

u(0, y) = 0 , u(x, y) → 0 as x → ∞ for 0 < y < b

u(x, b) = 0 , u(x, 0) = f(x) , for 0 < x < ∞ .

Solution. Sine transform of the Laplace equation with respect to x

under boundary conditions u(0, y) = 0 and u(x, y) → 0 as x → ∞ givesrise to

d2Us

dy2− ξ2Us = 0

Also sine transforms of u(x, b) = 0 , u(x, 0) = f(x) become

Us(ξ, b) = 0

Us(ξ, 0) = Fs(ξ)

Hence, the solution of the transformed differential equation under trans-formed boundary conditions is

Us(ξ, y) = Fs(ξ)sinh [ξ(b − y)]

sinh ξb

Now, inverting this equation we get

u(x, y) =

√2π

∫ ∞

0Fs(ξ)

sinh[ξ(b − y)]sinh ξb

sin ξ x dξ

=2π

∫ ∞

0

[∫ ∞

0f(t) sin ξt dt

]sinh ξ(b − y)

sinh ξbsin ξ x dξ


When ξb → ∞ ,[

sh ξ(b−y)sh ξb

]∼ exp (−ξb) and so the solution becomes

u(x, y) =1π

∫ ∞

0f(t) dt

∫ ∞

0[cos ξ(x − t) − cos ξ(x + t)] e−ξy dξ

Or, u(x, y) =1π

∫ ∞

0f(t)

[y

(x − t)2 + y2− y

(x + t)2 + y2

]dt

as the final solution of the corresponding quarter plane problem0 < x < ∞ , 0 < y < ∞ of the Laplace equation under the givenboundary conditions when ξb → ∞.

Exercises

(1) Find Fourier Transform of the function f(x), if

f(x) =sin ax

x, a > 0 .

[Ans. π , when |ξ| < a

0 , when |ξ| > a

]

(2) Find Fourier Transform of

f(x) =

{eiωx , 0 < x < a

0 , otherwise

to prove that F [f(x) ; x → ξ] =i√2π

[ei(ξ+ω)a − ei(ξ+ω)b

ξ + ω

]

(3) Prove that F [H(a − |x|) ; x → ξ] =√

2π

sin aξξ , where H(x) is the

Heaviside’s unit step function.

(4) Prove that Fc F−1c ≡ I and Fc = F−1

c , where Fc and F−1c are co-

sine transform and inverse cosine transform operators respectively.

(5) Find Fourier transform of each of the following functions :

(a) f(x) = 11+x2

(b) f(x) = ex exp (−ex)

(c) f(x) = x exp(−ax2

2

), a > 0

[Ans. (a)√

π2 e−|ξ| (b) Γ(1−iξ)/

√2π (c) Take f(x) = − 1

addx

(e

−ax2

2

)etc].


(6) (a) Prove that F [δ(x − ct) + δ(x + ct) ; x → ξ] =√

π2 cos ξ ct

(b) Prove that F [H(ct − |x|) ; x → ξ] =√

π2

sin ξ ctξ

(c) Prove that F [xnf(x) ; x → ξ] = in dn

dξn F (ξ)

(d) If f(x) has a finite discontinuity at x = a, show thatF [f ′(x) ; x → ξ] = iξ F (ξ)− 1√

2πexp (−iξa)[f(a+0)−f(a−0)] .

(7) (a) Prove that√

2πδ(x) ∗ f(x) = f(x)

(b) Prove that ddx [f(x) ∗ g(x)] = f ′(x) ∗ g(x) = f(x) ∗ g′(x).

(8) (a) Prove that Fs

[1x ; x → ξ

]=√

π2

(b) Prove that Fs [e−ax ; x → ξ] =√

2π

ξa2+ξ2

(c) Fc[e−ax ; x → ξ] =√

2π

aa2+ξ2 , a > 0

(9) If Fc[f(x) ; x → ξ] = Fc(ξ) , prove that

Fc[f(x) cos ω x ; x → ξ] = 12 [Fc(ξ + ω) + Fc(ξ − ω)]

(10) Prove the following :

(a) F [e−x2; x → ξ] = 1√

2e

−ξ2

2

(b) Fs

[e−ax

x ; x → ξ]

=√

2π tan−1 ξ

a ,

(11) Prove that F−1s

[e−πξ ; ξ → x

]=√

2π

xx2+π2

(12) Use shift theorem to prove that

F[e−|x−t| ; t→ξ

]= eiξxF

[e−|t| ; t → ξ

]

=

√2π

eiξx

1 + ξ2

Hence, show that eiξx =12(1 + ξ2)

∫ +∞

−∞e−|x−t|+iξtdt .


(13) Use Parseval’s theorem to prove that

∫ ∞

0

sin at

t(a2 + t2)dt =

π

21 − e−a2

a2

by choosing f(x) = e−ax and g(x) =

{1 , 0 < x < a

0 , x > a

(14) Prove that∫ ∞

0

[a − b

ξ+

cos aξ − cos bξ

ξ2

]sin ξ x dξ

=

⎧⎪⎨⎪⎩

π2 (a − b) , 0 < x < aπ2 (x − b) , a < x b

(15) From the identity

f(x) = − d

dx

∫ ∞

xf(u)du

prove that Fs[f(x);x → ξ] = ξ Fc

[∫ ∞

xf(u)du;x → ξ

]by using the result Fs [f ′(x) ; x → ξ] = −ξ Fc [f(t) ; t → ξ] .

(16) If δn(x) = n√π

e−n2x2, prove that

F [δn(x) ; x → ξ] =1√2π

e−ξ2

4n2

Hence, prove that∫ +∞

−∞δn(x) dx = 1 .

(17) Show that

F

[cosh ax

cosh πx; x → ξ

]=

1√2π

2 cos a2 · ch ξ

2

cos a + ch ξ

by using contour integration method round a rectangular contourdescribed by x = ± R , y = 0 , y = 2π, after suitably indented atpoints if needed.


(18) Putting f(x) = x− 12 in the sine and cosine forms of Fourier’s inte-

gral theorem show that∫ ∞

0

sin x√x

dx =∫ ∞

0

cos x√x

dx =√

π

2

and deduce that Fs

[t−

12 ; t → ξ

]= Fc

[t−

12 ; t → ξ

]= ξ−

12

(19) From Fourier cosine transforms of e−ax cos ax and of e−ax sin ax

deduce F. C. transforms of (x4+k4)−1 and x2(x4+k4)−1 , (k > 0).

(20) From Fourier sine transforms of e−ax cos ax and of e−ax sin ax

deduce F.S. transforms of (a) x(x4 +k4)−1 (b) x3(x4 +k4)−1

(c) x(x4 + k4)−2 (d) x3(x4 + k4)−2

(21) Show that if b and ξ are real,

Fs

[x2 − b2

x(x2 + b2); x → ξ

]=

√2π[e−|bξ| − 1

2

]sgn (ξ)

(22) Show that if ξ > 0∫ ∞

0

sin x cos ξx

xdx =

π

2H(1 − ξ)) .

(23) If f(x) = (1−|x|)H(1−|x|), find F [f(x) ; x → ξ] and deduce that

(a)∫ ∞

0

(sin x

x

)2

dx =π

2

(b)∫ ∞

0

(sin x

x

)3

dx =3π8

(c)∫ ∞

0

(sin x

x

)4

dx =π

3

(24) Show that F T of 34a3 (a2 − x2)H

[1 − |x|

a

], a > 0 is 1√

2πϕ(ξa),

where ϕ(t) = 3 sin tt3

− cos tt2

(25) If a > 0, b > 0 show that

(a) Fc

[e−bx−e−ax

x ; x → ξ]

= 1√2π

log a2+ξ2

b2+ξ2 , (a > b)

(b) Fs

[e−bx−e−ax

x ; x → ξ]

=√

2π tan−1 (a−b)ξ

ab+ξ2 , (a > b)


(26) If F (ξ) = Fc

[e−

x2

2 ; x → ξ

], show that dF

dξ = −ξF , F (0) = 1 and

hence duduce that F (ξ) = e−ξ2

2 .

(27) Using Parseval’s theorem show that if a and b are positive con-stants ∫ ∞

0sin ax sin bx

dx

x2=

π

2min (a, b).

(28) Making use of Fourier cosine transform and the Parseval’s relationprove that ∫ ∞

0

x−sdx

1 + x2=

π

2sec(π

2s)

, 0 < s < 1

(29) Use Fourier transform to solve the ODE given below :

(a) y′′(x)−y(x)+2 f(x) = 0 , where f(x) = 0 when x < −a andwhen x > a and y(x) and its derivatives vanish at x → ±∞(b) 2y′′(x) + xy′(x) + y(x) = 0

[Ans. (a) y(x) = e−x∫ x−a etf(t)dt + ex

∫ ax e−tf(t)dt

(b) y(x) = A exp(−x2

4

), A is a constant]

(30) Solve the following integral equation for f(x) :

(a)∫ +∞

−∞e−at2f(x − t)dt = e−bx2

, a, b > 0

(b) f(x) +∫ +∞

−∞f(x − t)e−atdt =

1x2 + b2

(c)1π

∫ +∞

−∞

f(t)x − t

dt = ϕ(x) ,

the integral is being in Cauchy principle value sense

[Ans. (a) f(x) =a√

π(a − b)exp

(−abx2

a − b

)

(b) f(x) =

√2π

{b

a(a − b)

}1

(a − b)2 + x2

(c) f(x) = − 1π

∫ +∞

−∞

ϕ(t)x − t

dt when F (ξ) =1π· iπΦ(ξ)

sgn ξ]


(31) Solve the PDE utt + uxxx = 0 , −∞ < x < ∞ , t > 0 under theconditions u(x, 0) = f(x) , ut(x, 0) = 0 for −∞ < x < ∞.

[Use F−1[cos(ξ2t) ; ξ → x] =1√2t

cos(

x2

4t− π

4

)for solution.]

(32) A function y(x, t) satisfies the diffusion equation k ∂2y∂x2 = ∂y

∂t

on the half line x � 0, for t > 0, the initial condition

y(x, 0) = Cx e−x2

4a2 , C and a are being constants, and boundaryconditions y(0, t) = 0 and y(x, t) → 0 as x → ∞. Prove thaty(x, t) = Cx(1 + kt/a2)

−32 exp

[− x2

4(a2+kt)

].

(33) A harmonic function u(x, y) in x � 0 , y � 0 satisfies the boundaryconditions ux(0, y) = f(y) , u(x, 0) = 0. Show that

uy(x, 0) = − 2π

∫ ∞

0

tf(t)dt

x2 + t2

Find uy(x, 0) if f(y) = q H(b − y) , where q, b > 0 .

(34) Show that θ(x, t) satisfying ∂2θ∂x2 = ∂θ

θt , x � 0, t � 0, θ(x, 0) = f(x)and ∂θ

∂x − hθ = 0 for x = 0 is given by

θ(x, t) =1√2π

∫ ∞

0ω(u, t)[f(|x − u|) + f(x + u) + h

∫ x+u

|x−u|f(υ)dυ]du,

where ω(x, t) =1√2t

e−x2

4t −√

π

2hehx+h2t Erfc

(12xt−

12 + ht−

12

).

(35) Prove that Fourier cosine transform of f(x) = 1x does not exist but

Fourier sine transform of it exists.

(36) Find Fourier sine transform of

(a) f(x) = x e−ax , a > 0

(b) f(x) = e−ax

x , a > 0

[Ans. (a) , (b) : Differentiate with respect to a and integrate w.r.ta both sides the integral

∫∞0 e−ax sin ξ x dx = ξ

ξ2+a2 to obtain theresults respectively.]

(37) If Fc(ξ) = Fc[e−ax2; x → ξ] , prove that Fc(ξ) satisfies the ODE

dFcdξ + ξ

2a Fc = 0 , when Fc(0) = 1


(38) Apply Fourier cosine transform to solve uxx+uyy = 0, 0< x, y< ∞subject to the conditions u(x, 0) = H(a − x) ,

a > x ; ux(0, y) = 0 ; 0 < x, y < ∞.

[Ans. u(x, y) = − 2π

∫ ∞

0

sin aξ

ξe−ξy cos ξ x dξ]

(39) Use Fourier sine or cosine transform properly to solve the followingintegral equations :

(a)∫ ∞

0f(x) cos ξ x dx =

√π

2ξ

(b)∫ ∞

0f(x) sin ξ x dx =

a

a2 + ξ2

(c)2π

∫ ∞

0f(x) sin ξ x dx = J0(aξ)

(d)∫ ∞

0f(x) cos ξ x dx =

sin aξ

ξ[Ans. (a) f(x) =

1√x

(b) f(x) = e−ax

(c) f(x) =H(x − a)√

x2 − a2(d) f(x) = H(a − x)

]

(40) Use Parseval formula to evaluate the following integral whena, b > 0 :

(a)∫ +∞

−∞

dx

(x2 + a2)2(b)

∫ +∞

−∞

sin ax dx

x(x2 + b2)(c)∫ +∞

−∞

sin2 ax

x2dx[

Ans. (a)πa3

2(b)

π

b2

(1 − e−ab

)(c) πa

]

(41) Solve the following free vibration problem of a semi-inifinite stringdefined by utt = c2 uxx , 0 < x < ∞ , t > 0 subject to conditionu(0, t) = 0 , u(x, 0) = f(x) , ut(x, 0) = g(x).

[ For x > ct , u(x, t) =12[f(x − ct) + f(x + ct)] +

12c

∫ x+ct

x−ctg(ξ)dξ.

Similarly result for x < ct can also be obtained ] .

Chapter 2

FINITE FOURIER TRANSFORM

2.1 Introduction.

Integral transform applied to solve the boundary value and initial valueproblems in mathematical physics in which the range of values of one ofthe independent variables say x, is finite for reducing the order of theassociated partial differential is of the type∫ b

af(x, · · ·)K(x, ξ)dx,

where a, b are finite numbers. Such a transform is termed as a finiteIntegral transform. Further, if the kernel K(x, ξ) is a Fourier kernellike sin xξ or cos xξ then the corresponding transforms are called finiteFourier transforms.

2.2 Finite Fourier cosine and sine transforms.

It is well-known result in the theory of Fourier series that, if f(x) satisfiesDirichlet’s conditions in 0 ≤ x ≤ a, a being a finite real number andhence f(x) ε ℘1 (0, a), finite Fourier cosine transform fc(n) of f(x) isdefined by

fc[f(x) ; n] = fc(n) =∫ a

0f(x) cos

nπx

adx (2.1)

and the corresponding inverse finite cosine transform is given by

f−1c [fc (n); x] = f(x) =

1a

fc(0) +2a

∞∑n=1

fc(n) cosnπx

adx (2.2)

at a point of continuity of f(x) in 0 ≤ x ≤ a. But at a pointing ofdiscontinuity x = c, 0 < c < a, the right hand side series in (2.2)


takes the value 12 [f(c + 0) + f (c − 0)] and the series takes the value

12 [f(0) + f(a)] both at x = 0 or at x = a. In the sequel, (2.2) will bereferred to as the inversion formula for finite Fourier cosine transform.

Similar results hold for finite sine transform. For a function f(x)ε ℘1(0, a),and thereby satisfying the Dirichlet’s conditions, finite Fourier sine trans-form fs (n) of f(x) in (0, a) is defined by

fs[f(x);n] = fs (n) =∫ a

0f(x) sin

nπx

adx (2.3)

and its corresponding inversion formula is given by

f−1s [fs (n); x] = f(x) =

2a

∞∑n=1

fs (n) sinnπx

a(2.4)

at a point of continuity x of f(x) in (0, a) and at a point of discontinuityx = c, 0 < c < a of f(x), the right hand side series in eqn (2.4) takes thevalue 1

2 [f(c + 0) + f(c− 0)] and the series takes the value 12 [f(0) + f(a)]

both at x = 0 or x = a .

Corollary 2.1. In particular, if a = π and that if f(x) ∈ ℘1(0, π),finite Fourier cosine transform of f(x) and the corresponding inversetransform are given by

fc(n) =∫ π

0f(x) cos nx dx (2.5)

and f−1c [fc(n)] = f−1

c [fc(n);x] = f(x) =1π

fc(0) +2π

∞∑n=1

fc(n) cos nx

(2.6)

Similarly, finite Fourier sine transform of f(x) and corresponding Inversetransform are given by

fs(n) =∫ π

0f(x) sin nx dx (2.7)

and f−1s [fs(n) ; x] = f(x) =

2π

∞∑n=1

fs(n) sin nx (2.8)

The right hand series in eqns (2.6) and (2.8) converge to 12 [f(x + 0) +

f(x−0)] at a point of discontinuity x and to f(x) at a point of continuityx and to 1

2 [f(0) + f(π)] at points x = 0 or x = π.

Finite Fourier Transform 81

2.3 Relation between finite Fourier transforms of the deriv-

atives of a function.

In applications finite Fourier transforms of the derivatives of a functionare found useful to the solution of boundary value and initial valueproblems. Accordingly, the following theorems are important.

Theorem 2.1. Let f(x) be continuous and f ′(x) be sectionally contin-uous on the interval 0 ≤ x ≤ a, then

(i) fc[f ′(x) ; x → n] = (−1)nf(a) − f(0) + nπa fs(n), n ∈ Z∗

(ii) fs[f ′(x) ; x → n] = −nπa fc(n), n ∈ N

Proof.(i) From the definition of finite Fourier cosine transform we have

fc[f ′(x) ; n] =∫ a

0

df

dx· cos nπx

adx

=[f(x) · cos nπx

a

]a0

+nπ

a

∫ a

0f(x) · sin nπx

adx

= (−1)nf(a) − f(0) +nπ

afs(n), n = 0, 1, 2, ..... = Z∗

(2.9)

(ii) Again

fs[f ′(x) ; n] =∫ a

0

df

dxsin

nπx

adx

= −nπ

a

∫ a

0f(x) cos

nπx

adx

= −nπ

afc(n) , n = N (2.10)

Theorem 2.2. Let f(x) and f ′(x) be continuous and f ′′(x) be section-ally continuous in 0 ≤ x ≤ a . Then

(i) fc[f ′′(x) ; n] = −f ′(0) + (−1)n f ′(a) − n2π2

a2fc (n)

(ii) fs[f ′′(x) ; n] =nπ

af(0) − (−1)n

nπ

af(a) − n2π2

a2fs (n)

Proof.(i) From the definition we have

fc [f ′′(x);n] =∫ a

0

d2f

dx2cos

nπx

adx =

[df

dx· cos

nπx

a

]a

0

+nπ

afs[f ′(x);n]

= (−1)n f ′(a) − f ′(0) − n2π2

a2fc(n), by (2.10) of Th. 2.1. (2.11)


(ii) Again, fs[f ′′(x) ; n] =∫ a

0

d2f

dx2sin

nπx

adx

= −nπ

a

∫ a

0

df

dx· cos

nπx

adx

= −nπ

a(−1)n f(a) +

nπ

af(0) +

n2π2

a2fc(n).

(2.12)

For finite Fourier transform for the derivatives of order greater than 2of a function may be obtained by induction. In the analysis of bound-ary value problems we often need to know the results for fourth-orderderivatives. It follows that if d2f

dx2 is zero at x = 0 and at x = a, then

∫ a

0

d4f(x)dx4

sinnπx

adx = −n2π2

a2

∫ a

0

d2f

dx2sin

nπx

adx

=n4π4

a4fs (n) (2.13)

Similarly, if d3fdx3 = df

dx = 0 when x = 0 and when x = a, it is seen that

∫ a

0

d4f(x)dx4

cosnπx

adx =

n4π4

a4fc (n) (2.14)

2.4 Faltung or convolution theorems for finite Fourier

transform.

Like ordinary Fourier transform, the convolution theorems for finiteFourier transform can not be put into such simple forms. Accordingly,we introduce two new functions f1(x) and f2(x) instead of f(x), whichare defined as odd and even periodic extensions respectively of f(x) withperiod 2π in the sense

f1(x) =

{f(x) , 0 � x � π

−f(−x) , −π < x < 0

f1(x + 2 r π) = f1(x) , r = ± 1 , ± 2, · · ·

and f2(x) =

{f(x) , 0 ≤ x ≤ π

f(−x) , −π < x < 0

f2(x + 2 r π) = f2(x) , r = ± 1 , ± 2, · · ·


Thus with the above understanding if f1(x) and g1(x) are the oddperiodic extensions of f(x) and g(x) respectively, then∫ π

−πf1(x − u)g1(u)du =

∫ π

0f(x − u)g(u)du −

∫ π

0f(u − x)g(u)du

(2.15)

Again since f1(x) is periodic∫ π

−πf1(x − u)g1(u)du =

∫ π

0f(−x − u)g(u)du −

∫ π

0f(x + u)g(u)du

(2.16)

We call ∫ π

−πf1(x − u) g1(u)du ≡ f1(x) ∗ g1(x),

as the convolution of the functions f1(x) and g1(x), introducing thenotation introduced by Churchill.

If finite Fourier sine transforms of f(x) and g(x) in (0, π) be fs(n)and gs(n) respectively, then

2π

m∑n=1

fs(n)gs(n) cos nx =2π

∫ π

0dξ

∫ π

0f(ξ)g(η)dη

m∑n=1

cos nx sin nξ sinnη

Making use of

4 sin nξ sin nη cos nx = cos(ξ − η + x)n + cos(ξ − η − x)n

− cos(ξ + η + x)n − cos(ξ + η − x)n

andn∑

r=1

cos ry = −12

+sin(m − 1

2

)y

2 sin(

12y)

we get2π

m∑r=1

fs(n)gs(n) cos nx =14π

∫ π

0f(ξ)dξ

∫ π

0g(η)dη

[sin (m − 1

2)(ξ − η + x)sin 1

2(ξ − η + x)+

sin (m − 12)(ξ − η − x)

sin 12(ξ − η − x)

−sin (m − 12)(ξ + η + x)

sin 12(ξ + η + x)

− sin (m − 12)(ξ + η − x)

sin 12(ξ + η − x)

]

Introducing the transformations

λ = ξ − η + x, μ = η


in the above four integrals, we get the first integral

14π

∫ π

0f(ξ)dξ

∫ π

0g(η)

sin(m − 1

2

)(ξ − η + x)

sin 12(ξ − η + x)

dη

=12π

∫ π

0g(μ)dμ

∫ x+π−μ

x−μf(λ + μ − x)

sin(m − 1

2

)λ

λ

12λ

sin 12λ

dλ

→ 14

∫ π

0g(μ)f(μ − x)dμ as m → ∞ .

Similarly, for m → ∞ the other three integrals can be evaluated to give

2π

∞∑n=1

fs(n)gs(n) cos nx = −14

∫ π

0g(μ)[f(x − μ) + f(−x − μ)

−f(μ − x) − f(μ + x)]dμ

Thus, by (2.15) and (2.16) we get from the above equation that

2π

∞∑n=1

fs(n) gs(n) cos nx = −12

∫ π

−πf1(x − μ)g1(μ)dμ

= −12f1(x) ∗ g1(x)

Then by inverse finite cosine transform we express the above relationas

fc[f1(x) ∗ g1(x)] = −2 fs(n) gs(n) (2.17)

Similarly, by considering the limit as m → ∞ of the sum

2π

m∑n=1

fs(n) gc(n) sin nx

we can have the result

fs[f1(x) ∗ g1(x)] = 2 fs(n) gc(n) (2.18)

We can also show that

fc[f2(x) ∗ g2(x)] = 2 fc(n) gc(n) (2.19)


2.5 Multiple Finite Fourier Transform.

Let f(x, y) be a function of two independent variables x and y definedin the rectangle 0 � x � a , 0 � y � b. For the time being, regard-ing f(x, y) as a function of x satisfying certain necessary conditions, itpossess a finite Fourier sine transform. We denote this transform byfs(m, y) and it is defined by

fs(m, y) = fs[f(x, y);x → m] =∫ a

0f(x, y) sin

mπx

adx (2.20)

This function will itself have a finite Fourier sine transform over variabley which we may now be denote d by fss(m,n) and defined by

fss(m,n) = f=

ss[f(x, y) ; (x, y) → (m,n)]

=∫ b

0

∫ a

0f(x, y) sin

mπx

asin

nπy

bdx dy (2.21)

We regard this equation as a double finite sine transform of the functionf(x, y) defined over the rectangular domain 0 � x � a , 0 � y � b.

The corresponding inversion formula giving f(x, y) in terms of fss(m,n),are easily given by

fs(m, y) =2b

∞∑n=1

fss(m,n) sinnπy

b(2.22)

and f(x, y) =2a

∞∑m=1

fs(m, y) sinmπx

a(2.23)

and therefore, by (2.22) and (2.23) we finally have

f(x, y) =4ab

∞∑m=1

∞∑n=1

fss(m,n) sinmπx

asin

nπy

b(2.24)

This is the inversion formula for the double finite Fourier transformfss(m,n) of f(x, y).

Proceeding as above, the double finite Fourier cosine transform fcc(m,n)of f(x, y) in the region 0 � x � a , 0 � y � b can be defined as

fcc(m,n) =∫ a

0

∫ b

0f(x, y) cos

mπx

acos

nπy

bdx dy (2.25)


In a similar way we can define double finite Fourier transforms of thetypes

fcs(m,n) = fc[fs{f(x, y); y → n};x → m]

=∫ a

0

∫ b

0f(x, y) cos

mπx

asin

nπy

bdx dy

fsc(m,n) =∫ a

0

∫ b

0f(x, y) sin

mπx

acos

nπy

bdx dy

Inversion formulas for double transforms of these types can easilybe deduced by successive applications of the inversion formulas for thefinite sine and cosine transforms. For example, in the case, in whichf(x, y) is defined on the region 0 � x � a , 0 � y � b , we get

f−1cs [fcs(m,n) ; (m,n) → (x, y)] ≡ f(x, y)

=2ab

∞∑n=1

fcs(0, n) sinnyπ

b+

4ab

∞∑m=1

∞∑n=1

fcs(m,n) cosmπx

asin

nπy

b

2.6 Double Transforms of partial derivatives of functions.

The double finite Fourier sine transforms of the partial derivatives offunction f(x, y) ∈ C2(D), where

D = {(x, y) : 0 < x < a, 0 < y < b}is given by

fs

[∂2f

∂x2; x → m]

]=∫ a

0

∂2f

∂x2sin

mπx

adx

=mπ

a

[f(0, y) + (−1)m+1 f(a, y)

]− m2π2

a2fs [f(x, y) ;x → m]

Therefore, fss

[∂2f

∂x2; (x, y) → (m,n)

]= Θ1(m,n) − m2π2

a2fss(m,n)

(2.26)

where Θ1(m,n) =mπ

a

[fs{f(0, y) ; y → n}

+ (−1)m+1fs{f(a, y); y → n}] (2.27)

Thus, fss

[∂2f

∂x2; (x, y) → (m,n)

]= −m2π2

a2fss(m,n),

if f(0, y) = f(a, y) = 0, 0 < y < b, in particular .


A similar result holds for double finite sine transform of ∂2f∂y2 as

fss

[∂2f

∂y2; (x, y) → (m,n)

]=

nπ

b

[fs{f(x, 0) ; x → m}]

+nπ

b

[(−1)n+1 fs{f(x, b) ; x → m}]− n2π2

b2fss(m,n)

= Θ2(m,n) − n2π2

b2fss(m,n), say. (2.28)

In particular, if f(x, 0) = f(x, b) = 0, 0 < x < a, then

fss

[∂2f

∂y2; (x, y) → (m,n)

]= −n2π2

b2fss (m,n) (2.29)

From (2.26) and (2.28) we immediately get

fss

[∂2f

∂x2+

∂2f

∂y2; (x, y) → (m,n)

]

= Θ1(m,n) + Θ2(m,n) − π2

[m2

a2+

n2

b2

]fss(m,n) (2.30)

If, in particular, f(x, y) vanishes along the boundary of D, then

fss

[∇2f(x, y) ; (x, y) → (m,n)]

= −π2

[m2

a2+

n2

b2

]fss(m,n) (2.31)

The above deductions may be extended for functions of more than twovariables similarly in its corresponding region of definitions.

2.7 Application of finite Fourier transforms to boundary

value problems.

When the range of one of the variables is finite in the associated PDEof a given boundary value problem, finite Fourier transform over thatvariable may be applied to remove it. Further, the choice of sine andcosine transform is decided by the given form of the boundary / initialconditions of the problem. We shall now briefly indicate below someapplications of finite Fourier transforms to the solution of boundaryvalue and initial value problems.

Example 2.1. The temperature u(x, t) at any point x at any time t ina solid bounded by planes x = 0 and x = 4 satisfies the heat conduction


equation ∂2u∂x2 = ∂u

∂t , when the end faces x = 0 and x = 4 are kept at zerotemperature. Initially the temperature at x is 2x. It is required to findu(x, t) for all x and t.

Solution. The initial and the boundary value problem is defined by

∂2u

∂x2=

∂u

∂t, 0 ≤ x ≤ 4 , t � 0 (i)

subject to the initial condition u(x, 0) = 2x and boundary conditionsu(0, t) = u(4, t) = 0.

Since u(0, t) and u(4, t) are prescribed, we introduce finite Fouriertransform over x and then eqn. (i) becomes

∫ 4

0

∂u(x, t)∂t

sinnπx

4dx =

∫ 4

0

∂2u(x, t)∂x2

sinnπx

4dx

i.e,d us(n, t)

dt= −nπ

4[u(4, t) cos nπ − u(0, t)] − n2π2

16us(n, t)

= −n2π2

16us(n, t)

Its solution is us(n, t) = c e−n2π2

16t (ii)

Putting t = 0 , us(n, 0) = c

Now, us(n, 0) =∫ 4

0u(x, 0) sin

nπx

4dx

=32(−1)n+1

nπ(iii)

Therefore, from (ii) and (iii), we get

us(n, t) =32(−1)n+1

nπe−

n2π2

16t (iv)

Taking inverse finite Fourier transform of eqn (iv), we get

u(x, t) =24

∞∑n=1

us(n, t) sinnπx

4

=16π

∞∑n=1

(−1)n+1

ne

−n2π2

16t sin

nπx

4(v)

as the expression of the temperature of the solid at any point at anytime.


Example 2.2. Solve the wave equation

∂2u

∂x2=

1c2

∂2u

∂t2, 0 ≤ x ≤ a , t > 0 (i)

satisfying the boundary conditions u(0, t) = u(a, t) = 0 , t > 0 and theinitial conditions u(x, 0) = 4b

a2 x(a − x), ∂u(x,0)∂t = 0 , 0 � x � a to

determine the displacement u(x, t).

Solution. Since the boundary condition u(0, t) and u(a.t) are prescribedwe may apply finite Fourier sine transform over the variable x. Thenthe given PDE (i) becomes

[d2

dt2+

c2n2π2

a2

]us(n, t) = 0, where us(n, t) = fs(u(x, t) ; x → n)

Its solution is given by

us(n, t) = A cosnπc

at + B sin

nπc

at (ii)

Further, taking finite Fourier sine transform over x on initial conditionswe get

us(n, 0) =8baπ3

1n3

[1 + (−1)n+1

], (iii)

d

dtus(n, 0) = 0. (iv)

Using (iii), (iv) in eqn. (ii), we get

us(n, t) =8ab

π3

[1 + (−1)n+1

n3

]cos

nπct

a(v)

Making use of inversion formula for finite Fourier transform, we get therequired displacement as

u(x, t) =∞∑

r=0

16b(2r + 1)−3

π3cos

(2r + 1)πct

asin

(2r + 1)πx

a.

Example 2.3. Consider the diffusion equation

∂2u

∂x2=

1k

∂u

∂t, 0 < x < a , t > 0


for its solution satisfying the boundary conditions

∂u(0, t)∂x

=∂u(a, t)

∂x= 0 , t > 0

and the initial condition u(x, 0) = f(x) , 0 � x � a , using proper finiteFourier transform.

Solution. Since ∂u∂x on the boundaries x = 0 , a are prescribed the

proper finite Fourier transform to be applied here is finite cosine trans-form over x. Hence, the diffusion equation and the initial conditiontransform to

duc

dt+

kn2π2

a2uc(n, t) = 0 (i)

and uc(n, 0) = fc(n) = fc[f(x) ; x → n] (ii)

Solving eqn.(i) under condition (ii), one gets

uc(n, t) = fc(n) e−k n2 π2

a2 t

Therefore, by inversion theorem for finite Fourier cosine transform, weget

u(x, t) =1afc(0) +

2a

∞∑n=1

fc(n) e−kn2π2

a2 t cosnπx

a.

Example 2.4. The displacement u(x, t) describing the vibration of abeam of length a with freely hinged end points satisfies the equation

∂4u

∂x4+

1c2

∂2u

∂t2=

p (x, t)EI

, 0 ≤ x ≤ a , t > 0

The displacement u(x, t) satisfies the initial conditionsu(x, 0) = ∂u

∂t (x, 0) = 0 , 0 ≤ x ≤ a and the boundary conditionsu(0, t) = ∂2u

∂x2 (0, t) = 0 and u(a, t) = ∂2u∂x2 (a, t) = 0 , t > 0

It is necessary to determine the displacement of the beam anywhere andat any time using finite Fourier transform.

Solution. Taking Fourier sine transform over x, the PDE transformsto

n4π4

a4us(n, t) +

d2

dt2us(n, t) =

c2

EIps(n, t), (i)

where ps(n, t) = fs[ p (x, t) ; x → n ] ,


after using the four boundary conditions. Again, the transformed formof the two given initial conditions become

us(n, 0) =d us(n, 0)

dt= 0 (ii)

Therefore, the solution of the new initial value problem defined byeqn. (i) and under the initial conditions in eqn. (ii) is given by

us(n, t) =a2

n2π2c

∫ t

0

c2

EIps(n, t) sin

n2π2c(t − ς)a2

dς

Inverting the above equation we get

u(x, t) =2ac

π2EI

∞∑n=1

1n2

sinnπx

a

∫ t

0ps(n, t) sin

n2π2c

a2(t − ς) dς

Example 2.5. Find the solution of the one dimensional heat conductionequation in a bar with the temperature distribution satisfying ∂u

∂t = k ∂2u∂x2

under the boundary conditions ∂u∂x(x, t) = 0 at x = 0 and at x = π and

the given initial condition u(x, 0) = f(x) , 0 � x � π.

Solution. Taking Fourier cosine transform over x, the given PDE underthe boundary conditions becomes

d uc

dt= −kn2 uc(n, t)

Its solution is, clearly , uc(n, t) = uc(n, 0) e−kn2t

Now, since u(x, 0) = f(x), we have uc(n, 0) =∫ π0 f(x) cos nx dx

Therefore, uc(n, t) = e−kn2t∫ π0 f(x) cos nx dx

Now, taking inverse finite Fourier transform, the above eqn. gives

u(x, t) =1π

∫ π

0f(y)dy +

2π

∞∑n=1

[e−kn2t cos nx

∫ π

0f(y) cos ny dy

].

Example 2.6. The end points of a solid bounded by x = 0 and x = π

are maintained at temperatures u(0, t) = 1, u(π, t) = 3, where u(x, t)represents its temperature at any point of it at any time t. Initially,the solid was held at 1 unit temperature with its surfaces were insu-lated. Find the temperature distribution u(x, t) of the solid, given thatuxx(x, t) = ut(x, t).


Solution. The heat conduction equation of the solid is

uxx = ut (i)

with boundary conditions u(0, t) = 1, u(π, t) = 3 and the initial conditionu(x, 0) = 1.

Taking finite Fourier transform over x, the eqn.(i) becomes

dus

dt= −n [u(π, t) cos nπ − u(0, t) + nus(n, t)]

i.e,dus

dt+ n2us(n, t) = n [1 − 3 cos nπ] .

Solving this linear ODE we get

us(n, t) = c e−n2t + (1 − 3 cos nπ)/n (ii)

Putting t = 0, (ii) gives

us(n, 0) = c + (1 − 3 cos nπ)/n

Also, since us(n, 0) =∫ π

0u(x, 0) sin nx dx =

1 − cos nπ

n

we get c =2 cos nπ

nand hence eqn. (ii) takes the form

us(n, t) =2 cos nπ

ne−n2t +

1 − 3 cos nπ

n

Therefore, after inversion

u(x, t) =2π

∞∑n=1

[2 cos nπ

ne−n2t +

1 − 3 cos nπ

n

]sin nx

=4π

∞∑n=1

cos nπ

ne−n2t sin nx +

2π

∞∑n=1

1 − 3 cos nπ

nsin nx

(iii)

But fs [1 ; x → n] =∫ π

0sin nx dx =

1 − cos nπ

nand therefore

1 =2π

∞∑n=1

1 − cos nπ

nsin px (iv)

Also, fs[x ; x → n] =∫ π

0x sin nx dx = −π cos nπ

nimplying

x =2π

∞∑n=1

[−π cos nπ

n

]sin nx (v)


Thus from (iv) and (v) one gets

1 +2xπ

=2π

∞∑n=1

1 − cos nπ

nsin nx (vi)

Using (vi) in (iii), it gives

u(x, t) = 1 +2xπ

+4π

∞∑n=1

cos nπ

ne−n2t sin nx

Example 2.7. A uniform string of length π is stretched between itsends and initially one end, say x = 0, is given a small oscillation a sin ωt

while the other end is kept fixed. Using finite Fourier transform provethat the displacement of the point x of the string at time t is given by

a sin wt sinw(π − x)

ccosec

wπ

c+

2acw

π

∞∑n=1

1w2 − n2c2

sin nx sin nct,

if the displacement u(x, t) of the string satisfies the PDE

∂2u

∂t2= c2 ∂2u

∂x2(i)

Solution. The boundary and the initial conditions of the problem aregiven by

u(0, t) = a sin wt (ii)

u(π, t) = 0 (iii)

u(x, 0) = 0 (iv)

and∂u

∂t(x, 0) = 0 (v)

The form of the boundary conditions in (ii) and (iii) imply that finiteFourier sine transform need to be applied over the variable x to reduceeqn (i) to

d2 us(n, t)(dt2)

= −c2n [u(π, t) cos nπ − u(0, t) + nus(n, t)]

Using (ii) and (iii), this equation becomes[d2

dt2+ c2n2

]us(n, t) = c2 a n sin wt


The general solution of this equation is given by

us(n, t) = A cos nct + B sin nct +c2a n sinwt

c2n2 − w2(vi)

The transformed form of (iv) and (v) becomes

us(n, 0) = 0 (vii)

and[d us(n, t)

dt

]t=0

= 0 (viii)

Using (vii) and (viii), the solution becomes

us(n, t) =acw

w2 − c2n2sin cnt − c2an sinwt

w2 − c2n2

Inverting the above transformed solution, we get

u(x, t) =2π

∞∑n=1

[acw

w2 − c2n2sin cnt − c2an sinwt

w2 − c2n2

]sin nt (ix)

To write the above solution in the required form we note that

fs

[sin

w(π − x)c

; x → n

]=∫ π

0sin

(π − x)wc

sinnx dx

=12

∫ π

0

[cos{

w(π − x)c

− nx

}− cos

{w(π − x)

c+ nx

}]dx

= sin(wπ

c

)· n

n2 − w2

c2

=nc2

n2c2 − w2sin

wπ

c.

This result gives

sinw(π − x)

c=

2π

∞∑n=1

nc2

n2c2 − w2sin

wπ

csin nx

Thus,2ac2 sin wt

π

∞∑n=1

n sin nx

n2c2 − w2= a sin wt sin

w(π − x)c

cosecwπ

c

(x)

The solution (ix), after use of eqn. (x) becomes

u(x, t) = a sin wt sinw(π − x)

ccosec

wπ

c

+2acw

π

∞∑n=1

1w2 − c2n2

sin nx sin nct


Example 2.8. Solve the three dimensional Laplace equation

∂2V

∂x2+

∂2V

∂y2+

∂2V

∂z2= 0, 0 � x � π, 0 � y � π, 0 � z � π

with the boundary conditions

V = V0 , when y = π ; V = 0 , when y = 0 and

V = 0 , when x = 0, π and V = 0 , when z = 0, π .

Solution. The boundary conditions can be expressed as

V (0, y, z) = V (π, y, z) = 0 (i)

V (x, 0, z) = 0 (ii)

V (x, π, z) = V0 (iii)

and V (x, y, 0) = V (x, y, π) = 0 (iv)

The form of the boundary conditions in (i) and (iv) show that finiteFourier sine transforms over variables x and z successively reduce thegiven Laplace equation in the following forms :

Firstly, taking finite Fourier sine transform over the variable x weget

−m2Vs(m, y, z) +[

∂2

∂y2+

∂2

∂z2

]Vs(m, y, z) = 0, by (i). (v)

Again, taking finite Fourier transform over the variable z, the eqn.(v) gives

−m2V s(m, y, n) − n2V s(m, y, n) +d2

dy2V s(m, y, n) = 0, · · · (vi)

since V s(m, y, 0) = V s(m, y, π) = 0

Again, since∫ π

0V (x, π, z) sinmx dx =

∫ π

0V0 sin mx dx

= V0(1 − cos mπ)/m

V s(m,π, z) =

{0 , m = even2V0m , m = odd

· · · (vii)

and V s(m, 0, z) = 0

The solution of the eqn (vi) is

V s(m, y, n) = c1 ch μy + c2 sh μ y


where, μ =√

m2 + n2

Using the transformed boundary conditions (vii), the above generalsolution of eqn(vi) becomes,

V s(m, y, n) = c2 sh μy,

where c2 is given by

c2 =V s(m,π, n)

sh μ π

But since,∫ π

0V s(m,π, z) sin nz dz = 0,when m = even

and∫ π

0V s(m,π, z) sin nz dz

=∫ π

0

2 V0

msin nz dz =

2 V0

mn[1 − cos nπ]

=

{0, if m = odd, n = even4V0mn , if m = odd, n = odd.

V s(m,π, n) =4V0

mn, when m,n are both odd.

Hence, V s(m, y, n) =4V0

mn

sin μy

sin μπ, after substituting the value of c2 .

Thus, V s(m, y, n) =4V0

(2m + 1)(2n + 1)sh μ y

sh μ π,

where, μ =[√

m2 + n2]m,n=odd

=√

(2m + 1)2 + (2n + 1)2;m,n = 0, 1, 2, · · ·

Inverting firstly over z, we get

V s(m, y, z) =2π

∞∑n=0

V s(m, y, n) sin(2n + 1)z

Again, inverting over the variable x.

V (x, y, z) =16π2

∞∑m=0

∞∑n=0

sh μy

sh μπ· sin(2m + 1)x

2m + 1· sin (2n + 1)z

(2n + 1),

where μ =√ [

(2m + 1)2 + (2n + 1)2].

Example 2.9. The deflection y(x) of a uniform beam of finite lengthl satisfies the well known ODE d4y

dx4 = w(x)EI = W (x) , 0 � x � l. Find


y(x) if the beam is freely hinged at x = 0 and at x = l, implying

y(x) = y′′(x) = 0 at x = 0, l.

Solution. Application of finite Fourier transform the given equationunder the given boundary condition gives y(n) =

(l

nπ

)4Ws(n). Inverting

the result we get

y(x) =2l3

π4

∞∑n=1

1n4

sinnπx

lW s(n)

Or y(x) =2l3

π4

∞∑n=1

1n4

sinnπx

l

∫ l

0W (ξ) sin

nπξ

ldξ.

In particular, if W (x) = W0 δ(x − α), where W0 is a constant, then

y(x) =2l3W0

π4

∞∑n=1

1n4

sinnπx

lsin

nπα

l, where 0 < α < l.

Example 2.10. The transverse displacement of an elastic membraneu(x, y, t) satisfies the PDE ∂2u

∂x2 + ∂2u∂y2 = 1

c2∂2u∂t2

, under the boundaryconditions u = 0 on the boundary, u = f(x, y), ut = g(x, y) at t = 0.Find the displacement u(x, y, t) after utilising finite Fourier transform.

Solution. Applying double finite Fourier sine transform we get

us(m,n) =∫ a

0

∫ b

0u(x, y) sin

mπx

asin

nπy

bdx dy .

Then the PDE givesd2us

dt2+ c2π2

[m2

a2+

n2

b2

]us = 0

with transformed boundary conditions

us(m,n, 0) = fs(m,n),[dus

dt

]t=0

= gs(m,n).

The solution of this transformed problem is

us(m,n, t) = fs(m,n) cos (cπwmn t) + gs(m,n) sin (cπwmn t)

where wmn =[m2

a2+

n2

b2

]1/2


Inverting this equation we get the solution of the problem as

u(x, y, t) =4ab

∞∑m=1

∞∑n=1

sinmπx

a· sin

nπx

b

[fs(m,n)

cos (cπwmn t) + gs(m,n) sin (cπwmn t)]

where[fs(m,n), gs (m,n)

]=∫ a

o

∫ b

0sin

mπξ

asin

nπη

b[f(ξ, η), g(ξ, η)] dξ dη

Exercises.

(1) Find finite sine and cosine transforms of f(x) = x2, 0 < x < π.

[Ans. fs(n) =2n3

(cos nπ − 1) − π2

ncos nπ, n = 1, 2, · · ·

=π2

3, n = 0 ;

fc(n) = 2π(cos nπ − 1)/n2]

(2) Find f(x) if (i) fs(n) = (1 − cos nπ)/n2π2 and 0 < x < 1[Ans.

2π3

∞∑n=1

1 − cos nπ

n2sin nx

]

(3) Calculate fs(n) , if f(x) =

{x , 0 � x � π

2

π − x , π2 � x � π[

Ans.2n2

sinnπ

2

]

(4) Find fc(n), if f(x) = ch c(π−x)sh πc .

[Ans. c/(c2 + n2)

]

(5) Find fs(n), if f(x) = 2π tan−1 2b sin x

1−b2

[Ans. bn [1 − (−1)n] /n, |b| ≤ 1]

(6) Prove that, if f(x) is defined in (0, a)

(a) fs [1 ; n] =a

nπ

[1 + (−1)n+1

](b) fs [x(a − x) ; n] =

2a3

n3π3

[1 + (−1)n+1

](c) f−1

s

[1n

; x

]=

π

a

[1 − x

a

]


(7) By finding finite Fourier sine and cosine transform of

f(x) = ekx, 0 � x � a, deduce that

f−1s

[n

k2 + n2 π2

a2

; n → x

]=

a

π

sh {k(a − x)}sh{ka} , 0 � x � a

(8) Show that the solution u(x, t) of the diffusion equation∂2u∂x2 = 1

k∂u∂t , 0 � x � a , t > 0 satisfying boundary conditions

u(0, t) = u(a, t) = 0, t > 0

and the initial condition

u(x, 0) = f(x), 0 � x � a

can be expressed as

u(x, t) =2π

∞∑n=1

fs(n) e−k n2π2t

a2 sin(nπx

a

).

(9) The function u(x, y) satisfying ∂2u∂x2 + ∂2u

dy2 = 0 in the semi-infinite

strip { (x,y) : 0 ≤ x < ∞ , 0 ≤ y ≤ b}, and the boundary

conditions

u(x, 0) = 0 , u(x, b) = 0 , x � 0

u(0, y) = f(y) , 0 ≤ y ≤ b

u(x, y) → 0 as x → ∞ , 0 ≤ y ≤ b

can be expressed as

u(x, y) =2b

∞∑n=1

fs(n) · e−nπx/b cosnπy

b.

(10) u(x, y) satisfies the PDE

∂2u

∂x2+

∂2u

∂y2= 0

in the rectangle R = {(x, y) : 0 ≤ x ≤ a , 0 ≤ y ≤ b}and the boundary conditions

∂u(0, y)∂x

=∂u(a, y)

∂x= 0, 0 ≤ y ≤ b

u(x, 0) = f(x) , u(x, b) = 0 , 0 ≤ x ≤ b


Then prove that taking Fourier cosine transform over the variable x

given by fc [u(x, y) ; x → m] = fc(m), the solution of the boundaryvalue problem is given by

u(x, y) = f−1c

[fc(m)

sinh{mπ(b − y)/a}sinh(mπb/a)

; m → x

]

(11) Find the steady state temperature in a long square bar of side π

when one face of it is kept at constant temperature u0 and the otherface at zero temperature. Assume that the temperature function u(x, y)is bounded and satisfies the PDE uxx + uyy = 0.

[Hint : Take the boundary conditions as u(0, y) = u(π, y) = 0 , u(x, 0) = 0and u(x, π) = u0. Also take finite Fourier transform over x] .[

Ans. u(x, y) =4u0

π

∞∑n=0

sh(2n + 1)y sin(2n + 1)x(2n + 1)sh(2n + 1)π

]

(12) Prove that

(a) Fc [x2 ; n] =

{a3

3 , if n = 02(

anπ

)2a(−1)n, if n = 1, 2, 3, · · ·

(b) Fs

[x2;n

]= a3

nπ (−1)n+1 − 2(

anπ

)3 [1 + (−1)n+1]

when 0 < x < a

(13) Prove that the initial-boundary value problem defined by

ut = k uxx, 0 ≤ x ≤ a, t > 0 under the conditions u(x, 0) = 0,

u(0, t) = f(t), u(a, t) = 0 is given by

u(x, t) =2πk

a

∞∑n=1

n sinnπx

a

∫ t

0f(ς)exp

[−k(t − ζ)

(nπ

a

)2]

dς

(14) Prove the following results defined over (0, π)

(i) Fs

[x2 (π − x) ; x → n

]= 1

n3

[1 + (−1)n+1

](ii) Fs

[sh a(π−x)

sh πa ; x → n]

= nn2+a2 , a �= 0

(iii) Fc

[(π − x)2 ; x → n

]= 2π

n2 for n �= 0 and π3

3 for n = 0


(iv) Fc [ch a(π − x) ; x → n] = a sh (aπ)n2+a2 , for a �= 0.

(v) Fs

[ddx{f(x) ∗ g(x)} ; x → n

]= 2nfs(n) gs(n)

(vi) Fs

[∫ x0 f(t) ∗ g(t)dt ; x → n

]= 2

n fs(n) gs(n)

(15) If fc(p) =∫ π0 f(x) cos px dx , fs(p) =

∫ π0 f(x) sin px dx, where

p is not necessarily an integer, show for any constant α that

(i) Fc [{2f(x) cos αx} ; x → n] = fc(n − α) + fc(n + α)

(ii) Fc [{2f(x) sin αx} ; x → n] = fs(n + α) − fs(n − α)

(iii) Fs [{2f(x) cos αx} ; x → n] = fs(n + α) + fs(n − α)

(iv) Fs [{2f(x) sin αx} ; x → n] = fc(n − α) − fc(n + α) .

Chapter 3

THE LAPLACE TRANSFORM

3.1 Introduction.

In article 1.3 of chapter 1 it is noted that if∫ +∞−∞ |f(t)|dt is not conver-

gent, Fourier transform F (ξ) of the function f(t) need not exist for allreal ξ. For example, when f(t) = sin ωt, ω = real, F (ξ) does not ex-ist. But such situations do arise occasionally in practice. To handle thissituation, we consider a new function f1(t) connected to f(t) defined by

f1(t) = e−γtf(t)H(t),

where γ is an arbitrary real positive constant and H(t) is the Heavisideunit step function. Clearly, here f1(t) ∈ P 1(R) and f1(t) ∈ A1(R) forall t ∈ R and therefore, Fourier transform of f1(t) exists, since∫ +∞

−∞f1(t)dt =

∫ ∞

0e−γtf(t) dt

is convergent. In fact, in this case, by Fourier Integral theorem

f1(t) =12π

∫ +∞

−∞e−iξt dξ

∫ +∞

−∞f1(u) eiξu du

implying

f(t) =12π

∫ +∞

−∞e−iξt dξ · eγt

∫ ∞

0f(u) · e−(γ−iξ)u du

If we write p = γ − iξ, the above relation can be expressed as

f(t) =1

2πi

∫ γ+i∞

γ−i∞ept

[∫ ∞

0f(u) e−pu du

]dp

=1

2πi

∫ γ+i∞

γ−i∞ept f (p) dp (3.1)

where f(p) =∫ ∞

0f(u)e−pu du, Re p = γ > 0 (3.2)

Eqns. (3.1) and (3.2) constitute a transform with K(p, u) = e−pu as thekernel of it.

The Laplace Transform 103

3.2 Definitions

We now define Laplace transform of a piecewise continuous function f(t)of the real variable t defined on the semi-axis t � 0 by

L [f(t) ; t → p] = L [f(t)] = F (p) = f(p) =∫ ∞

0f(t)e−ptdt (3.3)

and inverse Laplace transform by

f(t) = L−1 [f(p), p → t] = L−1 [F (p)] =1

2πi

∫ γ+i∞

γ−i∞eptf(p) dp (3.4)

where γ = Re p > 0.

Note 1. Some authors use variable s in place of p.

Note 2. The function f(t) must be of exponential order for the existenceof its Laplace transform.

Note 3. In operator notation eqn.(3.3) is expressed as

L [f(t) ; t → p ] = f(p) (3.5)

and relation in eqn.(3.4) is expressed as

L−1 [f(p) ; p → t] = f(t) (3.6)

Thus, L−1[L(f(t))] = f(t) = I [f(t)]

⇒ L−1 L ≡ I

Also, L[L−1(f(p))] = f(p) = I[f(p)]

⇒ LL−1 ≡ I

This means that L−1L ≡ LL−1 ≡ I, showing that these operators L andL−1 are commutative.

3.3 Sufficient conditions for existence of Laplace Trans-

form.

Theorem 3.1. If f(t) is of some exponential order for large t and ispiecewise continuous over o � t � ∞, then Laplace transform of f(t)exists.


Proof. Let f(t) be of exponential order σ such that

|f(t)| < Meσt , for t � t0

Then we have

L [f(t); t → p] =∫ ∞

0e−ptf(t) dt

=∫ t0

0e−ptf(t) dt +

∫ ∞

t0

e−ptf(t) dt

= I1 + I2 , say

Since f(t) is a piecewise continuous function on every finite interval0 � t � to , I1 exists and it is convergent Also, we have

| I2 | = |∫ ∞

t0

e−ptf(t) dt |

�∫ ∞

t0

e−pt|f(t)| dt

< M

∫ ∞

t0

e−(p−σ)tdt =Me−(p−σ)t0

(p − σ), if p > σ

Therefore, |I2| is finite for all t0 > 0 and p > σ and hence, I2 is conver-gent. Thus, L[f(t)] exists for all p > σ.

Note 3.1 Though the conditions stated in the theorem above are suf-ficient for the existence of Laplace transform, but these are not . Thismeans that even if the above conditions are not satisfied by a function,Laplace transform of that function may or may not exist. This can beshown by considering the example f(t) = 1√

t.

Here, f(t) → ∞ as t → 0.Hence, f(t) is not a piecewise continuousfunction on every finite interval for t � 0.

Now, L[ 1√t] =

∫∞0 e−pt 1√

tdt = 2√

p

∫∞0 e−x2

dx =√

πp , p > 0. This

proves the existence of Laplace transform of f(t).

3.4 Linearity property of Laplace Transform.

Theorem 3.2. If L[f1(t) ; t → p ] and L[f2(t) ; t → p ] both exist andc1, c2 are constants, then

L[c1f1(t) + c2f2(t) ; t → p ] = c1L[f1(t) ; t → p ] + c2[f2(t) ; t → p ]


Proof. By definition,

L[c1f1(t) + c2f2(t) ; t → p ] =∫ ∞

0e−pt[c1f1(t) + c2f2(t)]dt

= c1

∫ ∞

0e−ptf1(t)dt + c2

∫ ∞

0e−ptf2(t)dt

= c1 L[f1(t); t → p] + c2 L [f2(t); t → p ]

3.5 Laplace Transforms of some elementary functions

We calculate below Laplace transforms of some elementary functionsfrom definition.

Example 3.1. L[H(t) ; t → p]

=∫ ∞

0e−ptH(t)dt =

∫ ∞

0e−ptdt =

1p.

Therefore, L[H(t − a) ; t → p] =∫ ∞

0H(t − a)e−ptdt

=∫ ∞

ae−ptdt =

e−ap

p

Example 3.2. L[tν ; t → p ]

=∫ ∞

0tνe−ptdt = p−ν−1

∫ ∞

0uνe−udu,when u = p t

= p−ν−1 Γ(ν + 1), which exists when Re ν > −1.

Thus in particular,

L[tn ; t → p ] =n!

pn+1, n = 1, 2, 3, · · · , p > 0

so that L[t ; t → p] =1p2

, L[t2 ; t → p] =2p3

, · · · etc., p > 0

Example 3.3. L[eat ; t → p ]

=∫ ∞

0e−pteat dt =

1p − a

, p > a.

Example 3.4.

L [sin at ; t → p ] =∫ ∞

0e−pt sin at dt


=∫ ∞

0e−pt

[eait − e−ait

2i

]dt

=

[∫ ∞

0

e−(p−ai)t − e−(p+ai)t

2i

]dt

=12i

[1

p − ai− 1

p + ai

], by linearity property

=a

p2 + a2, p > 0,

Example 3.5. L [cos at ; t → p ]

=∫ ∞

0e−pt eait + e−ait

2dt

=12

[∫ ∞

0e−(p−ai)t dt

]+

12

[∫ ∞

0e−(p+ai)t dt

], by linearity property

=12

[1

p − ai+

1p + ai

]

=p

p2 + a2, p > 0

Example 3.6. L [cosh at ; t → p ]

=∫ ∞

0e−pt

[eat + e−at

2

]dt

=12

∫ ∞

0e−(p−a)t dt +

12

∫ ∞

0e−(p+a)t dt, by linearity property

=12

[1

p − a+

1p + a

]

=p

p2 − a2, p > a > 0

Example 3.7. L [sinh at ; t → p ]

=∫ ∞

0e−pt

[eat − e−at

2

]dt

=12

∫ ∞

0e−(p−a)t dt − 1

2

∫ ∞

0e−(p+a)t dt, by linearity property

=12

[1

p − a− 1

p + a

]

=a

p2 − a2, p > a > 0


3.6 First shift theorem.

Theorem 3.3. If Laplace transform of f(t) is f(p), then Laplacetransform of eatf(t) is f(p − a).

Proof. Given that

L [ f(t) ; t → p ] =∫ ∞

0f(t) e−pt dt

= f(p)

Therefore, L[

eatf(t)]

=∫ ∞

0eatf(t) e−pt dt

=∫ ∞

0f(t) · e−(p−a)t dt

= f(p − a). (3.7)

3.7 Second shift theorem.

Theorem 3.4. If Laplace transform of f(t) is f(p), then Laplace trans-form of f(t − a)H(t − a) is e−apf(p).

Proof. Given that

L [f(t) ; t → p] =∫ ∞

0f(t) e−pt dt

= f(p)

Therefore , L [ f(t − a)H(t − a) ; t → p ]

=∫ ∞

0f(t − a)H(t − a)e−ptdt

=∫ ∞

af(t − a) e−pt dt

=∫ ∞

0f(x) e−px · e−pa dx

= e−pa · f(p). (3.8)

3.8 The change of scale property.

Theorem 3.5. If L [f(t); t → p] = f(p), then L [f(at); t → p] = 1a f( p

a

)


Proof. Given that,∫ ∞

0f(t) e−pt dt = f(p)

Therefore, L [f(at) ; t → p ] =∫ ∞

0f(at) e−pt dt

=1a

∫ ∞

0f(x) · e−P

ax dx

=1a

∫ ∞

0f(x) · e−( p

a)x dx

=1a

f(p

a

). (3.9)

3.9 Examples

Example 3.8. Evaluate

L[f(t)] , where f(t) =

{t/a , 0 < t < a

1 , t > a

Solution.

L [f(t)] =∫ ∞

0e−ptf(t) dt

=∫ a

0e−pt · t

adt +

∫ ∞

ae−pt dt

=1 − e−ap

ap2

Example 3.9. Evaluate L[

1√πt

]

Solution. L

[1√πt

]=∫ ∞

0e−pt 1√

πtdt

=√

p√π

∫ ∞

0e−x x

12−1 dx

p

=1√πp

Γ(

12

)

=√

π√πp

=1√p

.



tνe−at ; t → p], if Re ν + 1 > 0

Solution. It is known that L [ tν ; t → p ] = Γ(ν+1)pν+1 .

Therefore, by shift theorem, we get

L[

tνe−at; t → p]

=Γ(ν + 1)

(p + a)ν+1

Example 3.11. Evaluate L [sin(t − a)H(t − a) ; t → p] .Hence, evaluate L

[e(t−a)k sin(t − a) H(t − a) ; t → p

].

Solution. We know that L [sin t ; t → p] = 1p2+1

Hence, by the second shift theorem, we get

L [sin(t − a) H(t − a)] = e−pa/(1 + p2)

Thus, L[ek(t−a) sin(t − a)H(t − a) ; t → p

]= e−(p−k)/

[1 + (p − k)2

], after using the first shift theorem.

Example 3.12. Evaluate L [ sinh bt ; t → p ]. Hence evaluateL[

eat sinh bt ; t → p]

Solution. Since, L [ sinh t ; t → p ] = 1p2−1

, we have by the change ofscale property that

L [ sinh bt ; t → p ] =b

p2 − b2

Hence, L[eat sh bt ; t → p

]=

b

(p − a)2 + b2


eat cos bt ; t → p]. Hence, evaluate

L[ea(t−c) cos b(t − c)H(t − c) ; t → p

]Solution. We know that L [ cos bt ; t → p ] = p

p2+b2

So, L[

eat cos bt ; t → p]

=p − a

(p − a)2 + b2,by shift theorem.

Hence , L[

ea(t−c) cos b (t − c)H (t − c) ; t → p]

= e−pc (p − a)(p − a)2 + b2

Example 3.14. Evaluate L[(πt)−

12 e

t2

].


Solution. We know that L[

1√πt

]= 1√

p

Hence, L[

1√πt

et2 ; t → p

]= 1�

p− 12

.

Example 3.15. Evaluate(i) L [t H (t − 1) ; t → p ]and (ii) L [ H(t − 1) cos at ; t → p ]

Solution. From definition L [ f(t)H(t − a) ; t → p ]

=∫ ∞

ae−pt f(t) dt

=∫ ∞

0e−p(u+a) f(u + a) du

= e−pa

∫ ∞

0e−pu f(u + a) du

= e−paL [f(t + a) ; t → p]

(i) L [t H (t − 1) ; t → p]

=∫ ∞

1e−pt t dt , a = 1

= e−p

∫ ∞

0e−pu (t + 1) dt

= e−p

[1p2

+1p

]

=p + 1p2

e−p

(ii) L [ H(t − 1) cos at ]

= e−p L [ cos a(t + 1)]

= e−p L [ cos at. cos a − sin at. sin a]

=e−p

p2 + 1[ p cos a − a sin a ] .

3.10 Laplace Transform of derivatives of a function.

Theorem 3.9. Let f(t) be a continuous function of t � 0 and it isof exponential order for large t and if f ′(t) is a piecewise continuousfunction for t � 0, then Laplace transform of the derivative f ′(t) existswhen p > σ and is given by

L[

f ′(t)]

= p L [ f(t) ] − f(0)


Proof. By definition L [ f ′(t) ]

=∫ ∞

0e−pt f ′(t) dt

=[

e−pt. f(t)]∞0

−∫ ∞

0(−p) e−pt f(t) dt

= limt→∞ e−ptf(t) − f(o) + p L [ f(t) ] (3.10)

Since, f(t) is of exponential order σ for t → ∞, we have

|f(t)| � M eσt , t � 0

Hence, |f(t) e−pt| = e−pt |f(t)| � M e−pt · eσt = M e−(p−σ)t

So, limt→∞ e−pt f(t) = 0, as p − σ > 0

Thus from (3.10), we get

L[

f ′(t)]

= p L [ f(t) ] − f(0). (3.11)

Corollary 3.1. If f ′′(t) exists for t � 0 and is a piecewise continuousfunction, then proceeding as above we can extend the result of the abovetheorem as

L[

f ′′(t)]

= p L[

f ′(t)]− f ′(0).

= p [ p{L(f(t))} − f(0)] − f ′(0)

= p2 L [(f(t)] ] − pf(0) − f ′(0) (3.12)

Corollary 3.2. In general, if fn(t) exists for t � 0 and is a piecewisecontinuous function of t, then

L [fn(t)] = pn L [f(t)] − pn−1 f (0) − pn−2f ′(0) − · · · − f (n−1)(0)

(3.13)

Theorem 3.10. Suppose f(t) is not continuous at t = a, 0 < a < ∞and f(a − 0), f(a + 0) exist and f(t) is of requisite order for existenceof L [f(t)]. Further if f ′(t) exists, for t �= a > 0 then

L [f ′(t) ] = p [ f(p) ] − f(0) − e−ap [f(a + 0) − f(a − 0)]

Proof. By definition

L[

f ′(t)]

=∫ ∞

0e−pt. f ′(t) dt


=∫ a−0

0e−pt. f ′(t) dt +

∫ ∞

a+0e−pt f ′(t) dt

=[e−pt f(t)

]a−0

0+ p

∫ a−0

0e−ptf(t) dt

+[e−pt f(t)

]∞a+0

+ p

∫ ∞

a+0e−ptf(t) dt

=[e−p(a−0) f(a − 0) − f(0)

]+[0 − e−p(a+0) f(a + 0)

]+ p

[∫ a−0

0e−ptf(t) dt +

∫ ∞

a+0e−ptf(t) dt

]= p L [f(t)] − f(0) − e−ap [f(a + 0) − f(a − 0)] (3.14)

Corollary 3.3. If f(t) is discontinuous at t = a1, a2, · · · ak, theabove theorem 3.10 can easily be extended as

L[

f ′(t)]

= p L [ f(t) ] − f(0) −k∑

i=1

e−ai p{f(ai + 0) − f(ai − 0)}

(3.15)

Remark. 3.1 In the light of the above Theorem 3.9 and Theorem3.10, one can easily get the extension of the theorems for the Laplacetransforms of f ′′(t), f ′′′(t). · · · etc, provided that the functions satisfy therequisite properties for the existence of their Laplace transforms.

3.11 Laplace Transform of Integral of a function

. Theorem 3.11. If Laplace transform of a function f(t) is f(p). thenLaplace transform of

∫ t0 f(τ)dτ is f(p)/p.

Proof. Given that

L [f(t) ; t → p] =∫ ∞

0e−pt f(t) dt = f(p)

Now, L

[∫ t

0f(τ)dτ

]

=∫ ∞

0e−p t

[∫ t

0f(τ) d τ

]dt

=∫ ∞

0f(τ)

[∫ ∞

τe−pt dt

]dτ,


by changing order of integration

=∫ ∞

0f (τ).

1p

e−p τ dτ

=1p

f(p). (3.16)

An alternative proof.

Let∫ t

0f(τ)dτ = g(t). Then, g′(t) = f(t) and g(0) = 0.

Now, L

[ ∫ t

0f(τ) dτ

]. = L [g(t)] .

But, by the theorem 3.9 of article 3.10, we know that

L[g′ (t)

]= p L [g(t)] − g(0) = p L [g(t)] .

Therefore, L [g (t) ] =1p

L[

g′ (t)].

=1p

L [ f (t) ] =1p

f(p)

Hence, L

[ ∫ t

0f (τ) dτ

]=

1p

f (p).

3.12 Laplace Transform of tnf(t)

Theorem 3.12. If Laplace transform of a piecewise continuous functionf(t) is f(p), then Laplace transform of tnf(t) is (−1)n dnf(p)

dpn , when n isa positive integer.

Proof. Before proving the above result, we may restate it as thederivative of the Laplace transform of a given function satisfying requisiteconditions for existence of the necessary results.

Given that, f(p) =∫ ∞

0e−pt f(t) dt

Differentiating both sides of the above equation with respect to p, weget


d f(p)dp

=d

dp

∫ ∞

0e−pt f(t)dt

= −∫ ∞

0t e−pt f(t)dt ,

valid for assumed nature of the function f(t)

= −L [ t f(t) ]

Thus, L [tf(t)] = (−1)1d

dpf(p) , (3.17)

which is the case of the theorem, when n = 1.

To prove the general case for any positive integral value of n, let usassume that for n = k

L [ tn f(t)] = (−1)ndn

dpnf(p) (3.18)

Hence, we have

L[

tk f(t)]

= (−1)kdk

dpkf(p)

implying,

∫ ∞

0e−pt{tkf(t)} dt = (−1)k

dk

dpkf(p) (3.19)

Differentiating both sides of (3.19) with respect to p once, we get∫ ∞

0e−pt {tk+1f(t)} dt = (−1)k+1 dk+1

dpk+1f(p)

implying, L[tk+1f(t)

]= (−1)k+1 dk+1

dpk+1f(p) (3.20)

Equation (3.20) shows that the result of the theorem is true for n = k+1,if it is true for n = k. But it is seen in equation (3.17) that it is true forn = 1. Hence, it is true for n = 1 + 1 = 2, n = 2 + 1 = 3, · · ·. Therefore,by the method of mathematical induction the required general result ofthe theorem is true for all positive integral value of n.

This result is sometimes known as the derivative of the transformedfunction.

3.13 Laplace Transform of f(t)t

Theorem 3.13. If Laplace transform of a piecewise continuous functionf(t) be f(p), then Laplace transform of f(t)/t is

∫∞p f(p)dp.


Proof. Given that ∫ ∞

0f(t) e−pt dt = f(p)

Integrating both sides of the above equation with respect to p from p to∞, we get∫ ∞

pf(p1)dp1 =

∫ ∞

p

[ ∫ ∞

0e−p1tf(t) dt

]dp1,

=∫ ∞

0f(t)

{∫ ∞

pe−p1t dp1

}dt, by changing order of integration.

=∫ ∞

0

f(t)t

e−pt dt

= L [ f(t)/t ] . (3.21)

Thus, the result of the theorem is proved to be true. This result issometimes known as the integral of Laplace transformed function.

3.14 Laplace Transform of a periodic function.

Theorem 3.14. Let f(t) be a periodic function of period τ , so thatf(t + nτ) = f(t), for n = 1, 2, 3 · · · . If f(t) is a piecewise continuousfunction for t > 0, then

L [ f(t) ] =1

1 − e−pτ

∫ τ

0e−pt f(t) dt

Proof.

L [ f(t) ] =∫ ∞

0e−pt f(t) dt

=∫ τ

0e−pt f(t)dt +

∫ 2τ

τe−pt f(t) +

∫ 3τ

2τe−pt f(t) dt + · · ·

=∞∑

n=0

∫ (n+1)τ

nτe−pt f(t) dt

=∞∑

n=0

∫ τ

0e−p(x+nτ) f(x + nτ)dx, putting t = x + nτ

=∞∑

n=0

e−pnτ

∫ τ

0e−px f(x) dx


=[ ∫ τ

0e−px f(x)dx

] ∞∑n=0

e−pnτ

=∫ τ

0e−px f(x)dx .

11 − e−pτ

, since p > 0, e−pτ < 1.

=1

1 − e−pτ

∫ τ

0e−pt f(t)dt, changing x to t. (3.22)

3.15 The initial-value theorem and the final-value theo-

rem of Laplace Transform.

Theorem 3.15.

Let f(t) be a continuous function for all t � 0 and is of exponentialorder for large t and f ′(t) be a piecewise continuous function of t .Then,

limt→0

f(t) = limp→∞ p L [f(t); t → p]

This result is called the initial-value theorem.

Proof. We know by theorem 3.9 of article 3.10 that

L[f ′(t)

]=∫ ∞

0e−ptf ′(t) dt = p L [f(t)] − f(0)

Taking p → ∞, the above equation gives

limp→∞

∫ ∞

0e−ptf ′(t)dt =

∫ ∞

0lim

p→∞[e−ptf ′(t)

]dt = lim

p→∞ [p L (f(t) − f(0)]

Since f ′(t) is sectionally continuous and of exponential order, wehave from above that

limp→∞ [pL{f(t)} − f(0)] = 0

which implies f(0) = limp→∞ p [L{f(t)}]

Thus, limt→0

f(t) = limp→∞ p [L{ f(t)}] (3.23)

Theorem 3.16. Let f(t) be a continuous function for all t � 0 and isof exponential order and f ′(t) be a piecewise continuous function of t.Then the result of the final-value theorem states that

limt→∞ f(t) = lim

p→0p L [ f(t) ]


Proof. As in theorem 3.15 above we have

limp→0

∫ ∞

0e−ptf ′ (t) dt = lim

p→0[ pL{f(t)} − f(0) ]

This result gives∫ ∞

0

[limp→0

e−pt

]f ′(t) dt = lim

p→0p L [f(t)] − f(0)

Or∫ ∞

0f ′(t) dt = [ f(t) ]∞0 = lim

p→0p L [f(t)] − f(0)

Or limt→∞ f(t) − f(0) = lim

p→0p L [ f(t) ] − f(0)

Or limt→∞ f(t) = lim

p→0p L [ f(t) ] (3.24)

Thus the result of the final value theorem of Laplace transform isproved.

3.16 Examples

Example 3.16 Find the Laplace transform of f(t) = sin at − at cos at

Solution. By linearity property of Laplace transform, we have

L [sin at − at cos at] = L [sin at] − L [ at cos at ]

=a

p2 + a2+ a

d

dpL [ cos at ] , by article 2.12

=a

p2 + a2+ a

d

dp

[p

p2 + a2

]

=a

p2 + a2+ a

{1

p2 + a2− 2p2

(p2 + a2)2

}

=2a

(p2 + a2)2[

a2]

=2a3

(p2 + a2)2


sh tt

].

Solution. From article 3.13 that L[

f(t)t

]=∫∞p f(p) dp, where

f(p) = L [ f(t)] . Now choosing f(t) = sh t , we have

L

[sh t

t

]=∫ ∞

pL [sht ; t → p] dp

=∫ ∞

p

1p2 − 1

dp =12

logp − 1p + 1


Example 3.18. Evaluate L [sin at/t]. Does L [ cos at/t ] exist ?

Solution. As in the above example 2, we have

L [ sin at/t ] =∫ ∞

pL [ sin at ; t → p ] dp

=∫ ∞

p

a

p2 + a2dp =

[tan−1 p

a

]∞p

= cot−1 p

a

Thus,∫ ∞

0e−pt sin at

t= cot−1 (p/a).

Putting p = 0 and a = 1 we have,∫ ∞

0

sin t

tdt = lim

p→0

( π

2− tan−1 p

)=

π

2, an important result

Again, L [ cos at/t ] =∫ ∞

p

p

p2 + a2dp

=12[

log(

p2 + a2) ]∞

p

But, limp→∞ log (p2 + a2) is unbounded and therefore L

[cos at

t

]does not

exist.


t1n

]and then evaluate L

[sin

√t]

and

finally evaluate L[

cos√

t / i.e cos√

t√t

√t].

Solution. From definition, we have

L[

t1n

]=∫ ∞

0e−pt. t

1n dt =

Γ(

1n + 1

)p

1n

+1

Again, since sin√

t = t12 − t

32

3!+

t52

5!− t

72

7!· · · · · · ,

we have

L[

sin√

t]

=Γ(3

2)

p32

− 13!

Γ(52 )

p5/2+

15!

Γ(72 )

p7/2· · · · · ·

=√

π

2p32

[1 − 1

4p+

12!

(14p

)2

− 13!

(14p

)3

+ · · · · · ·]

=√

π

2p32

e−14p .

Now considering f(t) = sin√

t , we get

f ′(t) =cos

√t

2√

t


and also L[f ′(t)

]= p L [ f(t) ] − f(0) implying

L

[cos

√t

2√

t

]= p

√π

2p32

e− 1

4p

Thus, L

[cos

√t√

t

]=√

π

pe−

14p

Example 3.20. Let f(t) be a periodic function of period 4, where

f(t) =

{3t , 0 < t < 26 , 2 < t < 4

Evaluate. L[f(t)]

Solution. By the result of article 3.14 we have here

L[f(t) ; t → p] =1

1 − e−4p

∫ 4

0f(t) · e−ptdt

=1

1 − e−4p

[∫ 2

03t e−ptdt +

∫ 4

26 e−ptdt

]

=1

1 − e−4p

[3(1 − e−2p − 2p e−4p2

)p2

], after simplification.

Example 3.21. Evaluate L[cos at ; t → p] and hence deduce L[sin at].

Solution. From the definition

L[cos at] =∫ ∞

0e−pt cos at dt =

p

p2 + a2, by actual integration.

Again,

∫ t

0cos at dt =

sin at

a·

Hence, L

[sin at

a

]= L

[∫ t

0cos at dt

]

=1p

L[cos at] =1p· p

p2 + a2

This result implies,L[sin at] =

a

p2 + a2.

Example 3.22. Verify the result of (a) initial-value theorem and(b) final-value theorem for the function f(t) = e−at cos bt, where a > 0.


Solution. (a) It is seen that f(t) satisfy the necessary condition forapplication of these theorems. We have now

limt→0

f(t) = limt→0

e−at cos bt = 1

Also, L[f(t)] =(p + a)

(p + a)2 + b2· So, pL[f(t)] =

p(p + a)(p + a)2 + b2

Therefore, limp→∞ pL [ f(t) ; t → p ] = lim

p→∞p(p + a)

(p + a)2 + b2= 1

Thus, the initial value theorem is verified.

(b) Again, limt→∞ f(t) = lim

t→∞ e−at cos bt = 0

Also, limp→0

pL [ f(t) ; t → p ]

= limp→0

p(p + a)(p + a)2 + b2

= 0

Therefore, the final-value theorem of the above function is verified.

Example 3.23. Is it possible to use the initial-value or the final-valuetheorem to evaluate lim

t→0f(t) or to evaluate lim

t→∞ f(t), when f(t) =et sin t ? Give reason for your answer. Also comment on your answer.

Solution.

L[f(t)] = L[et sin t] =1

(p − 1)2 + 1Now, lim

t→0f(t) = lim

t→0et sin t = 0

limp→∞ pL[f(t)] = lim

p→∞p

(p − 1)2 + 1= 0

For this function, IVT is applicable.

Again, limt→∞ f(t) = lim

t→∞ et sin t = does not exist

Also, limp→0

pL [f(t)] = limp→0

p

(p − 1)2 + 1= 0

This shows that FVT is not applicable here.

The reason for it is that f(t) does not satisfy the condition of expo-nential order of f(t)for large t.

The above two results show that the conditions in the theorems arethough sufficient but not necessary.


Example 3.24. Verify the result L[f ′′(t)] = p2f(p) − pf(0) − f ′(0) forthe case f(t) = et sin t .

Solution. We have f ′(t) = et(sin t + cos t) , f ′′(t) = 2et cos t

Now, L[f ′′(t)] = L[2et cos t] =2(p − 1)

(p − 1)2 + 1

Also, f(p) = L[f(t)] =1

(p − 1)2 + 1, f(0) = 0 , f ′(0) = 1

Therefore, p2f(p) − p f(0) − f ′(0)

=p2

(p − 1)2 + 1− p × 0 − 1 =

p2 − (p − 1)2 − 1(p − 1)2 + 1

=2(p − 1)

(p − 1)2 + 1. Thus, the result is verified.

Example 3.25. If y(t) = 12k [ekt

∫ t0 e−kudu− e−kt

∫ t0 ekudu] , and noting

y′′(t) − k2y(t) = 1 , verify that y(p) = 1p(p2−k2)

when k �= 0, after usingLaplace transform in article 2.10.

Solution. We have,

L[y′′(t) − k2y(t)] = L[y′′(t)] − k2L[y(t)]

= p2y(p) − py(0) − y′(0) − k2y(p)

= (p2 − k2) y(p) ≡ L[1] =1p

⇒ y(p) =1

p(p2 − k2)Also, y(p) = L [y(t)]

=1

2k2

[−2

p+

2pp2 − k2

]

=1

p (p2 − k2)

Thus, value of y(p) is verified.

3.17 Laplace Transform of some special functions.

I. The Heaviside functionWe have already introduced the unit step function H(t) defined by

H(t) =

{0 , t < 01 , t > 0,


This is usually known as “Heaviside unit function”. It may further benoted that any step function f(t) can always be expressed as a combi-nation of different Heaviside functions. For example, let a step functionbe defined by the relation

f(t) =

⎧⎪⎨⎪⎩

g(t) , 0 < t < a

k , a < t b

Then we can express it as

f(t) = g(t)H(a − t) + k[H(t − a) − H(t − b)]

Further, L[H(t)] =∫ ∞

0H(t)e−ptdt

=∫ ∞

01 · e−pt =

1p

implying, L−1

[1p

]= H(t)

II. The Dirac delta function.

A practical problem that arises in application in finding the Laplaceinversion of unity or H(t). Also, similar situation arises in handling unitimpulse function. Dirac introduced it as, δ(t) which acts for a very shorttime duration τ with a finite impulse 1 satisfying the relations

δ(t) = 0 , for t �= 0 (3.25)

and∫ ∞

−∞δ(t)dt =

∫ τ

0δ(t)dt = 1 (3.26)

In mathematical analysis, such function is unlike ordinary one and intro-duction of such a function sometimes leads to inconsistency in analysis.For this reason, Dirac called it as an “improper function” . It may beused only when no inconsistency do follow afterwards in analysis. Thus,entirely formally such function may be used in classical analysis withoutsearching for rigour in treatment. The equations (3.25) and (3.26) abovedo not depict any clear picture for δ(t). Since, the precise variation ofδ(t) with t in the interval is not an important issue but its effect is thepoint of study, we can use it properly with assumption that it does nothave any unnecessary violent oscillations in the neighbourhood of t = 0.


For example, if we define a step function δ∈(t) by

δ∈(t) =

⎧⎪⎨⎪⎩

0 , t < 01∈ , 0 < t <∈0 , t >∈

(3.27)

then since∫ +∞

−∞δ∈ (t) dt =

∫ ∈

0

1∈ dt = 1 , (3.28)

it follows that lim∈→0

δ∈ (t) = δ(t) (3.29)

as defined in (3.25) and (3.26)

If f(t) be a continuous function in the neighbourhood of t = 0, then∫ +∞

−∞δ∈(t)f(t)dt =

∫ ∈

0δ∈(t)f(t)dt =

1∈∫ ∈

0f(t)dt

=f(θ ∈)

∈∫ ∈

0dt = f(θ ∈) , 0 < θ < 1 , by mean value theorem.

Thus,∫ +∞

−∞f(t)δ∈(t) dt = f(θ ∈) , 0 < θ < 1

Now, letting ∈→ 0 we get∫ +∞

−∞f(t) δ(t) dt = f(0) (3.30)

In general, putting∫ +∞

−∞f(ξ) δ(ξ − a) dξ

=∫ +∞

−∞f(t + a) δ(t) dt , putting ξ − a = t

= f(0 + a) = f(a) (3.31)

This is an important result which can be used formally in analysis when-ever no confusion do arise.

The Laplace transform of δ(t) is then given by definition that

L[δ(t) ; t → p] =∫ ∞

0e−ptδ(t)dt

= e−p.0 = 1 = H(t) , t > 0 (3.32)

Thus, L[δ(t)] = H(t)

and in turn L−1[H(t)] = δ(t) (3.33)


Hence, (3.32) and (3.33) are the required result obtained as proposed inthe beginning of this analysis.

III. Laplace Transform of the sine and the cosine integrals.The sine integral Si(t) is denoted and defined by

Si(t) =∫ t

0

sin u

udu

Now, L [ Si(t) ; t → p ] =∫ ∞

0e−pt

[∫ t

0

sinu

udu

]dt

=1p

∫ ∞

0

sin t

te−ptdt

=1p

[∫ ∞

pL [ sin t ] dp

], by 3.13

=1p

[∫ ∞

p

1p2 + 1

dp

]=

1p

[π2− tan−1 p

]

=1p

cot−1 p. (3.34)

Again, the cosine integral Ci(t) is denoted by

Ci(t) = −∫ ∞

t

cos u

udu

Therefore, L [ Ci(t) ; t → p ]

= −∫ ∞

0e−pt

[∫ ∞

t

cos u

udu

]dt

= −∫ ∞

0

cos u

u

[∫ u

0e−ptdt

]du ,

by changing the order of integration

= −1p

∫ ∞

0

cos u

u(1 − e−pu)du (3.35)

Again, f(p1) =∫ ∞

0f(u) e−p1udp1

Integrating this result with respect to p1 between the limits 0 to p, weget ∫ p

0f(p1) dp1 =

∫ p

0

[∫ ∞

0f(u) e−p1u du

]dp1


Therefore,∫ p

0f(p1)dp1 =

∫ ∞

0f(u)

[∫ p

0e−p1udp1

]du ,

by changing the order of integration.

=∫ ∞

0

f(u)u

(1 − e−pu)du

Assuming, f(u) = cos u , we have f(p) = L[f(u) ; u → p] =p

p2 + 1

and hence, ∫ p

0

p1

p21 + 1

dp1 =∫ ∞

0

cos u

u(1 − e−pu) du (3.36)

Using (3.36) in (3.35) above, we get

L[Ci (t) ; t → p] = −1p

∫ p

0

p1 dp1

p21 + 1

= − 12p

log (1 + p2)

An alternative method.

Since, Ci (t) = −∫ ∞

t

cos u

udu

t Ci′ (t) = cos t

So, L[t Ci′ (t)] =p

p2 + 1⇒ − d

dp[L{C ′

i (t)}] =p

p2 + 1

or,d

dp[p L{Ci (t)} − Ci (0)] = − p

p2 + 1

or,d

dp[p L{Ci (t)}] = − p

p2 + 1

Integrating both sides with respect to p,we get

p L[Ci (t)] = −12

log (p2 + 1) + c where c is a constant

Now, by the final-value theorem

limp→0

p L[Ci (t)] = limt→∞Ci(t)

we get, limp→0

[−1

2log (p2 + 1) + c

]= − lim

t→∞

∫ ∞

t

cos u

udu = 0

Thus c = 0 ⇒ p L[Ci (t)] = −12

log (p2 + 1)

⇒ L[Ci (t)] = − 12p

(p2 + 1) .


IV. Laplace Transform of the Exponential integral.

The exponential integral is defined by

Ei (t) =∫ ∞

t

e−u

udu

Then, Ei′ (t) = −e−t

t⇒ t E′

i (t) = −e−t

Thus as before,

L[t Ei′ (t)] = − 1p + 1

⇒ − d

dp[ L{Ei′ (t)} ] = − 1

p + 1

Or,d

dp[ p L{Ei (t)} − Ei (0) ] =

1p + 1

Or,d

dp[ p L{Ei (t)} ] =

1p + 1

Integrating both sides with respect to p , we get

p L[Ei (t)] = log (p + 1) + c , a constant

Now, by final-value theorem, we get

limp→0

[ log (p + 1) + c ] = limt→∞Ei(t) = 0 ⇒ c = 0

Thus, L[Ei (t)] =1p

log (p + 1) (3.37)

V. Laplace Transform of the Error function.

The error function Erf(t) is defined by

Erf(t) =2√π

∫ t

0e−x2

dx

and the error complementary function Erfc(t) is also defined by

Erfc(t) =2√π

∫ ∞

te−x2

dx = 1 − 2√π

∫ t

0e−x2

dx (3.38)

= 1 − Erf(t),

since,2√π

∫ ∞

0e−x2

dx = 1 , by use of Gamma function.

Thus, by evaluating Laplace transform of Erf(t), the same for Erfc(t)can easily be evaluated.


To make the evaluation steps relatively simpler, we evaluate Laplacetransform of Erf(

√t) instead of Erf(t) in the following steps. We know

that

Erf(√

t) =2√π

∫ √t

0e−x2

dx

=2√π

∫ √t

0

[1 − x2 +

x4

2!− x6

3!+ · · ·

]dx,

=2√π

[t1/2 − t3/2

3+

t5/2

5.2!− t7/2

7.3!+ · · ·

]

Thus,

L[Erf(√

t)] =2√π

[Γ(3/2)p3/2

− Γ(5/2)3 · p5/2

+Γ(7/2)

5.2! · p7/2− Γ(9/2)

7.3!p9/2+ · · ·

]

=1

p3/2

[1 − 1

2· 1p

+1.32.4

1p2

− 1.3.52.4.6

1p3

+ · · ·]

=1

p3/2

(1 +

1p

)− 12

=1

p(p + 1)1/2(3.39)

Therefore,

L[Erfc(√

t)] = L[1 − Erf(√

t)]

=1p− 1

p(p + 1)1/2=

(p + 1)1/2 − 1p(p + 1)1/2

(3.40)

VI. Laplace Transforms of Bessel functions.

The Bessel function of first kind of order ν, Jν(t) is defined by

Jν(t) =∞∑

r=0

(−1)r(

12 t)ν+2r

r! Γ(ν + r + 1).

Its Laplace transform is given by the equation

L [tνJν (t) ; t → p] =∞∑

r=0

(−1)ν2−ν−2r

r! Γ(r + ν + 1)L[t2ν+2r ; t → p

](3.41)

The duplication formula for the Gamma function is

22z−1 Γ(z) Γ(

z +12

)=

√π Γ(2z) (3.42)

Using (3.42) we have the relation

1Γ(r + ν + 1)

L[t2ν+2r; t → p

]=

Γ(2ν + 2r + 1)Γ(ν + r + 1)

p−2ν−2r−1


= π− 12 22ν+2r Γ

(ν + r +

12

)p−2ν−2r−1,(

Re p > 0 , ν > −12

)(3.43)

Therefore, results in eqns.(3.41) and (3.43) give

L[tν Jν(t) ; t → p] = 2ν π− 12 p−2ν−1

∞∑r=0

Γ(ν + 1

2 + r)

r!(−p−2

)rIf |p| > 1, the above series is convergent and it gives

L[tν Jν(t) ; t → p] =2ν Γ

(ν + 1

2

)Γ(

12

) (p2 + 1)−ν− 12 , Re p > 1 (3.44)

after using the Binomial theorem

(1 − x)−α =∞∑

r=0

Γ(α + r)xr

r! Γ(α), |x| < 1.

In particular, putting ν = 0 in (3.44), we get

L[J0(t) ; t → p] = (p2 + 1)−12 , Re p > 1

Letting p → 0 , it follows that∫∞0 J0(t)dt = 1.

Again by the result in eqn. (3.44), we have

L [ tν Jν(at) ; t → p ]

=(2a)ν Γ

(ν + 1

2

)√

π(p2 + a2)ν+ 12

, ν > −12

, Re p > a > 0

and L[J0(at) ; t → p] = (p2 + a2)−12 , Re p > a > 0

Also, L[e−atJ0(at) ; t → p] =1√

p2 + 2ap + 2a2

Deduction 3.1.

If Re p > 1 , like eqn. (3.41) one can write

L [ tν+1 Jν(at) ; t → p ] =∞∑

r=0

(−1)r 2−ν−2r Γ(2ν + 2r + 2)r! Γ(ν + r + 1) p2ν+2r+2


At this stage, using the duplication formula of the Gamma function , itis found that

2−ν−2r Γ(2ν + 2r + 2)Γ(ν + r + 1)

= 2ν+1 1√π

Γ(

ν + r +32

),

and so,

L[tν+1Jν(t) ; t → p

]= 2ν+1 · 1√

π· 1p2ν+2

∞∑r=0

(− 1

p2

)ν Γ(ν + r + 3

2

)r!

= 2ν+1 1√π

p−2ν−2(1 + p−2

)−ν− 32 Γ

(ν +

32

),

after using Binomial theorem. Thus,

L[tν+1Jν(t) ; t → p

]=

2ν+1p Γ(ν + 3

2

)√

π(p2 + 1)ν+ 32

, Re p > 1

Therefore, L[tν+1Jν(at) ; t → p

]=

2ν+1 aνp Γ(ν + 3

2

)√

π(p2 + a2)ν+ 32

, Re p > a > 0

(3.45)

In particular, if ν = 0,

L [tJ0(at)] =p

(p2 + a2)32

.

Deductions 3.2. By principle of analytic continuation we can extendthe result of the definition of Bessel function of first kind for the per-missible range of argument of the variable as

tν2 Jν(2

√t) =

∞∑r=0

(−1)rtν+r

r! Γ(ν + r + 1)

and hence

L[t

ν2 Jν(2

√t) ; t → p

]=

∞∑r=0

(−1)r

r! Γ(r + ν + 1)L[tν+r ; t → p

]

= p−ν−1∞∑

r=0

(−1

p

)r

r!, if Re p > 0

= p−ν−1 e− 1

p , Re p > 0

Also, if a > 0 , Re p > 0. (3.46)

L[t

ν2 Jν(2

√at) ; t → p

]= a

ν2 p−ν−1e

− ap (3.47)

and L[J0(2

√at) ; t → p

]=

1pe− a

p .


Deduction 3.3. Modified Bessel function of the first kind Iν(t) ofthe variable t is given by

Iν(t) =∞∑

r=0

(12 t)ν+2r

r! Γ(ν + r + 1)

Therefore, by the same procedure as above, we get

L [tν Iν(t) ; t → p] = π− 12 2νp−2ν−1

∞∑r=0

Γ(ν + r + 1

2

)r!

p−2r

This result implies that if Re p > 1 , ν > −12 ,

L [tν Iν(t) ; t → p] =1√π

2ν Γ(

ν +12

)(p2 − 1)−ν− 1

2

Also, L [tν Iν(at) ; t → p] =(2a)ν Γ

(ν + 1

2

)√

π(p2 − a2)ν+ 12

, for Re p > a > 0

(3.48)

In a similar way one can prove that if ν > −12

L[tν+1Iν(at) ; t → p

]=

2ν+1aνp Γ(ν + 3

2

)√

π(p2 − a2)ν+ 32

, for Re p > a > 0 (3.49)

Again,

L[t

ν2 Iν(2

√at) ; t → p

]= a

ν2 p−ν−1e

ap , for Re p > 0 , a > 0 , ν > −1

2(3.50)

by similar arguments made above.

The special cases for ν = 0 are given by

L [I0(at) ; t → p] = (p2 − a2)−12 , Re p > a > 0 (3.51)

L [t I0(at) ; t → p] = p(p2 − a2)−12 , Re p > a > 0 (3.52)

and L[I0(2

√at) ; t → p

]= p−1 e

ap , Re p > 0 , a > 0 (3.53)

An important result due to Tricomi [cf Tricomi (1935)] can be statedas

L

[t

ν2

∫ ∞

0x− ν

2 Jν

(2√

xt)

f(x)dx ; t → p

]= p−ν−1 f

(1p

)(3.54)

We do not persue the derivation of this last result due to lack ofscope of this book.


3.18 The Convolution of two functions.

Let f(t) and g(t) be two piecewise continuous functions and are of someexponential order for large t and for every t � 0. Then the convolutionof these functions is denoted by f ∗ g(t) and is defined by

f ∗ g(t) =∫ t

0f(u)g(t − u)du (3.55)

The above relation in eqn. (3.55) is also known as faltung or resultantof f(t) and g(t). By definition,

f ∗ g(t) =∫ t

0f(u)g(t − u) du

=∫ t

0f(t − τ) g(τ) dτ

= g ∗ f(t) (3.56)

Therefore, convolution of two functions satisfies the commutative law .It also satisfies the distributive law

f ∗ [g + h](t) = f ∗ g(t) + f ∗ h(t) (3.57)

and also the associative law

[f ∗ (g ∗ h)](t) = [(f ∗ g) ∗ h](t) , (3.58)

The two results in eqns. (3.57) and (3.58) can be directly verified fromthe definition (3.55). For example,

[(f ∗ g) ∗ h](t) =∫ t

0h(t − τ) dτ

[∫ τ

0f(ξ) g(τ − ξ) dξ

]

=∫ t

0f(ξ)dξ

∫ t

ξg(τ − ξ)h(t − τ)dτ, Changing the order of integration

=∫ t

0f(ξ) dξ

∫ t−ξ

0g(η) h(t − ξ − η) dη

= f ∗ (g ∗ h) (t)

We now turn our attention to evaluate Laplace transform of the convo-lution of two functions f(t) and g(t).

L[(f ∗ g)(t) ; t → p] = L

[∫ t

0f(τ) g(t − τ) dτ ; t → p

]


=∫ ∞

0e−pt

[∫ t

0f(τ) g(t − τ) dτ

]dt

=∫ ∞

0f(τ)

[∫ ∞

τe−pt g(t − τ) dt

]dτ

=∫ ∞

0f(τ) e−pτ

[∫ ∞

0e−pη g(η) dη

]dτ

= f(p) g(p) , (3.59)

where f(p) and g(p) are Laplace transforms of f(t) and g(t) respectively.

In particular, L[f ∗ f(t) ; t → p] =[f(p)

]2As an illustration, if f(t) = ta−1, a > 0, g(t) = tb−1, b > 0

L[ta−1; t → p

]=

Γ(a)pa

, L[tb−1 ; t → p

]=

Γ(b)pb

then L[(ta−1 ∗ tb−1) ; t → p

]=

Γ(a) Γ(b)pa+b

(3.60)

Also, it is known that

L[ta+b−1 ; t → p

]=

Γ(a + b)pa+b

(3.61)

Dividing the results of (3.60) by that of (3.61) one gets

β(a, b) =Γ(a) Γ(b)Γ(a + b)

=L[(

ta−1 ∗ tb−1)

; t → p]

L [ta+b−1 ; t → p]

3.19 Applications

Example 3.26. By considering Si(t), find its Laplace transform andhence prove that ∫ ∞

0

sin u

udu =

π

2

Solution. We have, from III of article 3.17 that

L[Si(t) ; t → p] =∫ ∞

0e−pt

[∫ t

0

sin u

udu

]dt

=1p

∫ ∞

0

sin t

te−pt dt

=1p

[∫ ∞

pL{sin t ; t → p}dp

]


=1p

∫ ∞

p

1p2 + 1

dp =1p

[π2− tan−1 p

]

Thus,∫ ∞

0

sin t

te−pt dt =

π

2− tan−1 p (3.62)

In eqn.(3.62) making p → 0, we get∫ ∞

0

sin t

tdt =

π

2.

Example 3.27. Prove that L

[e−

t2

4 ; t → p

]=

√π ep2

Erfc (p) .

Solution. We know that

L [ Erfc{g(t)} ; t → p ] = L [ 1 − Erf{g(t)} ; t → p ]

=1p− L [Erf{g(t)} ; t → p]

Now, L

[e−

t2

4 ; t → p

]=∫ ∞

0e−pt e−

t2

4 dt

=∫ ∞

0e−�( t

2+p)2−p2

�dt = ep2

∫ ∞

0e−( t

2+p)2

dt

= ep2

∫ ∞

pe−u2 · 2 du , putting u = p +

t

2

= 2 ep2 × Erfc(p) ×√

π

2, by eqn. (3.38) of article 3.17 (iv)

=√

π ep2Erfc(p) .

Example 3.28. Find L [ Erf (t) ; t → p ] to show that its value is

expressed as 1p e

p2

4 Erfc (12 p).

Solution. We have from definition

L [ Erf(t) ; t → p ] =∫ ∞

0e−pt

[2√π

∫ t

0e−u2

du

]dt

=2√π

∫ ∞

0e−u2

[ ∫ ∞

ue−pt dt

]du

=2

p√

π

∫ ∞

0e−u2−pu du

=2

p√

π

∫ ∞

0e−( u+ p

2 )2

· e p2

4 du

=2ep2/4

p√

π

∫ ∞

p/2e−v2

dv


=2ep2/4

p√

π· Erfc

( p

2

)·√

π

2

=1p

ep2/4 Erfc

(12p

)

Deduction.

L [ Erf (bt) ; t → p ] =1b

L[

Erf (t); t → p

b

]=

b

p· 1

be

p2

4b2 · Erfc( p

2b

), by the above result.

Example 3.29. If f(t) = | sin t|, prove that L [ | sin t| ; t → p ] is equalto coth (pπ

2 )/(1 + p2).

Solution. Clearly, f(t) = | sin t| is a periodic function of principalperiod τ = π. Hence

L [ f(t) ; t → p ] = L [ | sin t | ; t → p ]

=1

1 − e−pπ

∫ π

0e−pt | sin t| dt

=1

1 − e−pπ

∫ π

0e−pt sin t dt =

11 − e−pn

· 1 + e−pπ

1 + p2

=1

1 + p2coth

(pπ

2

)

Example 3.30. Find the Laplace transform of f(t) =∫∞0 cos tx2 dx

Solution. We have, by definition

L

[ ∫ ∞

0cos tx2dx

]=∫ ∞

0e−pt

[ ∫ ∞

0cos tx2 dx

]dt

=∫ ∞

0

[ ∫ ∞

0cos t x2 .e−pt dt

]dx ,

after changing order of integration.

=∫ ∞

0

p

p2 + x4dx

=1

2√

p

∫ π2

0

√cot θ dθ, where x2 = p tan θ

=π

2√

2p


Example 3.31. Find (i) L [ f(t) ] and (ii) L [ f ′(t) ] where f(t) isdefined by

f(t) =

{2t , 0 � t < 1t , t > 1

Solution.

(i) L[f(t) ; t → p]

=∫ 1

02t · e−ptdt +

∫ ∞

1t e−ptdt

=2p2

−(

1p

+1p2

)e−p , after evaluation of integrals

(ii) Since f(t) =

{2t , 0 � t < 1t , t > 1

, we have f ′(t) =

{2, 0 � t < 11, t > 1

Therefore, L[f ′(t) ; t → p] =∫ 1

02e−pt dt +

∫ ∞

1e−pt dt

=2 − e−p

p.

Example 3.32. The Laguerre polynomial Ln(t) is defined by

Ln(t) =et

n!dn

dtn[

e−t tn]

, n = 0, 1, 2, . . .

Prove that

L [Ln(t) ; t → p ] =(p − 1)n

pn+1, n = 0, 1, 2, . . .


L [ Ln(t) ;→ p ] =∫ ∞

0

e−t(p−1)

n!dn

dtn(e−t tn) dt

=1n!

(p − 1)∫ ∞

0e−t(p−1) dn−1

dtn−1

(e−t · tn

)dt

= · · · · · ·=

(p − 1)n

n!

∫ ∞

0e−pt · tn dt

=(p − 1)n

n!Γ(n + 1)

pn+1

=(p − 1)n

pn+1.


Example 3.33. If f(t) = | sin t| , prove that f(t) is a periodic functionof t of principal period τ = π. Find the Laplace transform of f(t).

Solution. Here

f(t) = | sin(t)|, f(t + τ) = f(t + π) = | sin(π + t)|= | − sin t| = | sin t| = f(t)

f(t + nτ) = | sin t| = f(t), n = 0, 1, 2, · · ·Hence, f(t) is a periodic function of period τ = π. Therefore,

L [ f(t) ; t → p ]

= L [ | sin t| ; t → p ]

=1

1 − e−pπ

∫ π

0f(t)e−pt dt

=1

1 − e−pπ

∫ π

0| sin t| e−pt dt

=1

1 − e−pπ

∫ π

0sin t · e−pt dt

=e−pπ + 11 − e−pπ

· 1p2 + 1

, Re p > 0

= cothp π

2/(p2 + 1) , Re p > 0

Example 3.34.

Let f(t) =

{1 , 0 � t < τ/2

−1 , τ/2 < t < τ

and f(t) is a periodic function of principal period τ . Prove that

L [ f(t) ; t → p] =1p

tanhpτ

2·

Solution. We can express f(t) = H(t) − 2 H(

t − τ2

), 0 < t < τ ,

and f(t + rτ) = f(t), for all t > 0.

L [ f(t) ; t → p ] =1

1 − e−pτ

∫ τ

0f(t)e−pt dt

=1

1 − e−pτ

[∫ τ2

0e−pt dt −

∫ τ

τ/2e−pt dt

]

=1

p ( 1 − e−pτ )

[1 − e−pτ/2

]2=

1p

tanhpτ

2, Re p > 0


In particular, if τ = π we get

L [ f(t) ; t → p ]τ=π =1p

tanhpπ

2, Re p > 0

Example 3.35. By considering the Fourier series of a periodic functionf(t) of principal period 2π, find Laplace transform of f(t) to deduce

a0

2p+

∞∑n=1

pan + n bn

p2 + n2=

11 − e−2pπ

∫ 2π

0f(u) e−pu du

Hence, deduce the Mittag-Leffler identity

12

+2pπ

∞∑r=1

[p2 + (2r − 1)2

]−1 =(1 + e−pπ

)−1

for f(t) =

{1, 0 < t < π

0, π < t < 2π

Solution. The Fourier series corresponding to f(t) is given by

f(t) � a0

2+

∞∑n=1

[ an cos nt + bn sin nt ] (1)

where an = 1π

∫ 2π0 f(u) cos nu du

bn = 1π

∫ 2π0 f(u) sin nu du

⎫⎪⎬⎪⎭ (2)

Taking Laplace transform of both sides of eqn.(1) one gets

11 − e−2pπ

∫ 2π

0f(u)e−pudu =

a0

2p+

∞∑n=1

pan + nbn

p2 + n2(3)

where an and bn are given in eqns.(2).

As an example let f(t) =

{1 , 0 < t < π

0 , π < t < 2π

be a periodic function of period 2π. Then,

a0 = 1, an = 0 for n = 1, 2, · · ·

and b2n = 0, b2n−1 =(

n − 12

)−1

π, n = 1, 2, 3, · · ·

Also,∫ 2π

0f(u)e−pudu =

∫ π

0e−pudu =

1p

(1 − e−pπ

),


Substituting these results in eqn. (3), we get

12

+2pπ

∞∑n=1

1p2 + (2n − 1)2

=1

1 + e−pπ,

which is the required result.

Example 3.36. Using the definition of convolution of two functions,find

L

[∫ t

0J0(τ) J0(t − τ) dτ ; t → p

]

in its simplest form and hence evaluate∫ t0 J0(τ)J0(t − τ)dτ .

Solution. Let f(t) = J0(t) and g(t) = J0(t). Then

f(t) ∗ g(t) =∫ t

0f(τ) g(t − τ) dτ

=∫ t

0J0(τ) J0(t − τ) dτ

∴ L

[∫ t

0J0(τ) J0(t − τ)dτ

]= L [f ∗ g(t)]

= f(p) g(p)

=[f(p)

]2, since f(t) = g(t) ⇒ f(p) = g(p)

∴ L

[∫ t

0J0(τ) J0(t − τ)dτ ; t → p

]=

1p2 + 1

This result leads to

L−1

[L

{∫ t

0J0(τ) J0(t − τ) dτ

}]= L−1

[1

p2 + 1

]= sin t

Or∫ t

0J0(τ) J0(t − τ) dτ = sin t

Excercises

(1) Find the Laplace transform of

f(t) =

{0 , 0 < t < 1

(t − 1)2 , t > 1[Ans. 2e−p

p3

]


(2) Evaluate

L [ f(t) ; t → p ] where

(i) f(t) =

{sin t , 0 < t < π

0 , t > π

(ii) f(t) = sin√

t[Ans. (i)(1 + e−pπ) / (p2 + 1) (ii)

√π e−1/4p / (2p

32 )]

(3) Find Laplace transform of f(t), where

(i) f(t) = sh at cos at

(ii) f(t) = t e−t(3 sh 2t − 5 ch 2t)

(iii) f(t) = [ g(t) cos wt ] if L[g(t)] = g(p)[Ans. (i) a(p2 − 2a2)/(p4 + 4a4) (ii) − d

dp

[1−5p

p2+2p−3

]

(iii) 12 [ g(p − iw) + g(p + iw) ]

](4) Evaluate L [(sin at − at cos at) ; t → p] and hence find

limt→0

L [ sin at − at cos at ; t → p ].

[Ans. 2a3/(p2 + a2)2 , p > 0 ; 0]

(5) Using Laplace transform evaluate I =∫∞0 t e−2tdt to prove

that I = 325 .

(6) Prove that∫∞0 (e−at − e−bt)/t dt = log(b/a) .Hence, find the value

of∫∞0 (e−t − e−3t)dt

t

[Ans. log 3]

(7) Prove that L [ J1(t) ; t → p ] = 1 − p√p2+1

and L [ t J1(t) ; t → p ] = 1(p2+1)3/2

(8) Prove that(i) L[t erf(2

√t) ; t → p] = (3p+8)

p2(p2+4)3/2


(ii) L[e3terf√

t ; t → p] = 1(p−3)

√p−2

(9) Prove that∫∞0

f(t)t dt =

∫∞0 f(p)dp , provided that the integrals

converge.

(10) If L[f(t) ; t → p] = f(p) , show that L[∫∞

0f(u)

u du ; t → p]

= 1p

∫∞p f(x)dx . Hence, show that L

[∫ t0

sinuu du ; t → p

]= 1

p cot−1 p

(11) Prove that L[Erf (t) ; t → p] = 2p√

π

∫∞0 e−u2−pu du

(12) Prove that L[tn eiat ; t → p] = n!(p+ia)n+1(p2 +a2)−n−1 , wheren is a positive integer.

Chapter 4

THE INVERSE LAPLACE TRANSFORM

AND

APPLICATION OF LAPLACE TRANSFORM

4.1 Introduction.

If f(t) belongs to a class A (meaning that it is a piecewise continuousfunction over 0 ≤ t < ∞ and is of some exponential order), its Laplacetransform f(p) exists. This is denoted symbolically by

L [f(t); t → p] =∫ ∞

0e−pt f(t) dt ≡ f(p) (4.1)

Then, in turn f(t), the object function is the inverse Laplace transformof the image function f(p) and is given symbolically by

L−1[f(p); p → t

] ≡ f(t) (4.2)

Thus by eqn. (4.1), one can evaluate f(p) for a given f(t), since f(p)exists. We shall now consider the inverse problem - deriving informationfrom the prescribed f(p) that enable us to derive the original functionf(t) through some formula, called the Laplace inversion formula.

Before answering this basic problem raised above, we shall at firstpresent heuristic approach of determining the inverse Laplace transformof some given f(p). Also as per need we define a new function, calledthe null function in this connection together with a connected theorem,known as Lerch’s theorem.

Definition.Null function. A function N(t) satisfying the condition

∫ t0 N (τ) dτ = 0,

for all t > 0 is called a null function.

Theorem 4.1.Lerch’s Theorem. If L [ f1(t) ; t → p ] , Re p > c1 and


L [ f2(t) ; t → p ] , Re p > c2 both exists and if

L [ f1(t) ; t → p ] = L [ f2(t) ; t → p ]

for Rep > c = max(c1, c2), then f2(t) − f1(t) = N(t)

In addition, if f1(t) and f2(t) are continuous on the whole real line, thenf1(t) = f2(t) , t > 0.

Proof. We shall not persue here the proof of this theorem. But thecomment that the inverse Laplace transform of a given f(p) is uniqueexcept for a null function, should be kept in mind.

In the proposed heuristic approach, we reduce the given function f(p)as a combination of some functions which are the Laplace transformsof known functions. Then applying the idea of the Lerch’s theorem,the inverse Laplace transform of the given f(p) can be obtained byusing the inverse results already discussed in the previous article of thischapter. This approach will be just like evaluation of the integrationas an inverse process of differentiation of a given function, called theintegrand, after reducing it to sum of functions which are the derivativesof known standard functions. Sometimes for reduction of f(p) into sumof standard expressions for manipulation of Laplace inversion, partialfraction rules are found useful. Here, again the linearity property ofLaplace inversion holds.

Theorem 4.2. If f1(p) = L [f1(t); t → p] and f2(p) = L [f2(t); t → p]and c1, c2 be two arbitrary constants, then

L−1[

c1 f1(p) + c2 f2(p) ; p → t]

= c1 L−1[f1(p) ; p → t

]+ c2 L−1

[f2(p) ; p → t

]= c1 f1(t) + c2 f2(t).

Proof. Since, L−1[f1(p) ; p → t

]= f1(t) and

L−1[f2(p) ; p → t

]= f2(t)

under the given condition, we have by the linearity property of theLaplace transform

L [c1 f1(t) + c2 f2(t); t → p] = c1L [f1(t); t → p] + c2L [f2(t) ; t → p]

= c1f1(p) + c2f2(p).

The Inverse Laplace Transform 143

Hence, L−1 L [c1 f1(t) + c2 f2(t) ; t → p]

= L−1[c1f1(p) +c2 f2(p) ; p → t

]

implying, L−1[c1 f1(p) + c2 f2(p) ; p → t

]= c1 f1 (t) + c2 f2 (t)

= c1L−1[

f1(p) ; p → t]+ c2L

−1[

f2(p) ; p → t]

Thus, the theorem is proved.

4.2 Calculation of Laplace inversion of some elementary

functions.

Before finding Laplace inversion of elementary functions which are a bitcomplex in nature, we state below some results obtained directly fromthe previous articles in 3.5 and 3.9. These results may be considered asformulae for further applications.

(i) L [ H(t ) ; t → p ] = 1p implies L−1

[1p ; p → t

]= H(t)

(ii) L [tν ; t → p] = Γ(ν+1)pν+1 implies L−1

[Γ(ν+1)pν+1 ; p → t

]= tν , Re ν > −1

(iii) L[eat; t → p

]= 1

p−a implies L−1[

1p−a ; p → t

]= eat, Re p > a

(iv) L [sinat; t → p] = ap2+a2 implies L−1

[a

p2+a2 ; p → t]

= sinata , p > 0

(v) L [cos at; t → p] = pp2+a2 implies L−1

[p

p2+a2 ; p → t]

= cos at, p > 0

(vi) L [sh at ; t → p] = ap2−a2 implies L−1

[1

p2−a2 ; p → t]

= sh ata ,

p > a > 0

(vii) L [ch at ; t → p] = pp2−a2 ; implies L−1

[p

p2−a2 ; p → t]

= ch at,p > a > 0

(viii) L[tνe−at; t → p

]= Γ(ν+1)

(p+a)ν+1 implies

L−1[

Γ(ν+1)

(p+a)ν+1 ; p → t]

= tνe−at, Re ν + 1 > 0


(ix) L [ sin(t − a)H(t − a); t → p] = e−pa

1+p2 implies

L−1[

e−pa

1+p2 ; p → t]

= sin(t − a) H(t − a)

(x) L [cos(t − a)H(t − a); t → p] = pe−pa

1+p2 implies

L−1[

pe−pa

1+p2 ; p → t]

= cos(t − a) H(t − a)

(xi) L[sh(t − a)H(t − a); t → p] = e−pa

p2−1implies

L−1[ e−pa

p2−1; p → t] = sh(t − a)H(t − a)

(xii) L [ch(t − a)H(t − a); t → p] = pe−pa

p2−1implies

L−1[

pe−pa

p2−1 ; p → t]

= ch(t − a) H(t − a)

We now turn our attention in discussing some rules of manipulationof Laplace inversion of some combinations of elementary functions of p

through the following examples.

Example 4.1. Evaluate L−1[

1(p+1)(p2+1)

; p → t]

Solution. Resolving into partial fractions under usual method we have

1(p2 + 1)(p + 1)

≡ 12

1p + 1

− 12

p

p2 + 1+

12

1p2 + 1

Therefore,

L−1

[1

(p + 1)(p2 + 1)

]=

12L−1

[1

p + 1

]− 1

2L−1

[p

p2 + 1

]+

12L−1

[1

p2 + 1

]

=12[e−t − cos t + sin t

]


6p2+22p+18p3+6p2+11p+6

]Solution.

L−1

[6p2 + 22p + 18

p3 + 6p2 + 11p + 6

]

= L−1

[6p2 + 22p + 18

(p + 1)(p + 2)(p + 3)

]

= L−1

[1

p + 1+

2p + 2

+3

p + 3

]= e−t + 2e−2t + 3e−3t


Example 4.3. Evaluate L−1

[ (√p−1p

)2]

Solution.

L−1

[ (√p − 1p

)2]

= L−1

[p − 2

√p + 1

p2

]

= L−1

[1p− 2

p3/2+

1p2

]

= H(t) − 4

√t

π+ t

4.3 Method of expansion into partial fractions of the ratio

of two polynomials

Let f(p) and g(p) be two polynomials in p such that the degree of f(p)is less than that of g(p). Then f(p)/g(p) can be expressed as

f(p)g(p)

=n∑

r=1

Ar

p − ar, where p − ar are the factors of g(p)

with Ar = limp→ar

(p − ar) f(p)g(p)

=f(ar)g′(ar)

, for r = 1, 2, · · · n.

To prove the above result, we write

f(p)g(p)

≡ A1

p − a1+

A2

p − a2+ · · · An

p − an(4.3)

Then, limp→ar

(p − ar) f(p))g(p)

= Ar, the other terms being zero in the limit.

Therefore

Ar = limp→ar

f(p)g(p)/p − ar

=f(ar)g′(ar)

, provided g′(ar) �= 0

meaning that p − ar is a non-repeated factor of g(p). Thus, the resultstated above is proved.


p+5(p+1)(p2+1)

]


Solution. We express by partial fraction that[p + 5

(p + 1)(p2 + 1)

]≡ A1

p + 1+

A2

p + i+

A3

p − i

=2

p − 1+

i−52(1+i)

p + i+

i+52(i−1)

p − i

=2

p + 1+

−2pp2 + 1

+3

p2 + 1,

Therefore, L−1

[p + 5

(p + 1)(p2 + 1); p → t

]= 2e−t − 2 cos t + 3 sin t


p2−6p3+4p2+3p

]Solution. We express, by partial fraction that

p2 − 6p3 + 4p2 + 3p

≡ A1

p+

A2

p + 1+

A3

p + 3=

f(p)g(p)

, say

where f(p) = p2 − 6, g(p) = p3 + 4p2 + 3p, g′(p) = 3p2 + 8p + 3

Then we have

A1 =f(0)g′(0)

= −2, A2 =f(−1)g′(−1)

=52, A3 =

f(−3)g′(−3)

=12

Hence,

L−1

[p2 − 6

p3 + 4p2 + 3p

]=

52

e−t +12

e−3t − 2H(t)


4p+5(p−1)2(p+2)

]Solution. By method of partial fraction we express

4p + 5(p − 1)2(p + 2)

≡ A

p − 1+

B

(p − 1)2+

C

p + 2Then, 4p + 5 ≡ A(p − 1)(p + 2) + B(p + 2) + C(p − 1)2

From this identity we get

9 = 3B, ( by putting p = 1) =⇒ B = 3

−3 = 9C, ( by putting p = −2) =⇒ C = −13

and 5=−2A + 2B + C, (by equating term independent of p) =⇒ A =13


Therefore , L−1

[4p + 5

(p − 1)2(p + 2)

]=

13et + 3L−1

[1

(p − 1)2

]− 1

3e−2t

=13et − 1

3e−2t + 3 t et, by article 4.2


1√2p+3

], after proving

L−1[f (ap + b)

]=(

1a

)e−

bta f(

ta

),where L−1

[f(p)

]= f(t)

Solution. Given that

f(p) =∫ ∞

0e−ptf(t) dt , by definition.

So, f(ap + b) =∫ ∞

0e−pat . e−bt f(t) dt

=1a

∫ ∞

0e−px e−

bxa f

(x

a

)dx , putting at = x

= L

[1a

e−bta f

(t

a

) ]

⇒ L−1[

f(ap + b)]

=1a

e−bta f

(t

a

).

Now putting a = 2, b = 3 we get.

f(p) =1√p

=√

π√π√

p=

1√π

Γ(

12

)p

12

so that f(t) =1√πt

. Therefore, f

(t

a

)= f

(t

2

)=

√2√

π t

These results give,

L−1

[1√

2p + 3

]=

12

e−32t

√2√

π t=

1√2πt

e−3t2

Example. 4.8. Evaluate

L−1

[(p + 1) e−πp

p2 + p + 1

]

Solution. We have

L−1

[p + 1

p2 + p + 1

]


= L−1

⎡⎢⎣ p + 1

2 + 12

(p + 12)2 +

( √3

2

)2

⎤⎥⎦

= L−1

⎡⎢⎣ p + 1

2

(p + 12)2 +

( √3

2

)2

⎤⎥⎦+

1√3

L−1

⎡⎢⎣

√3

2

(p + 12)2 +

( √3

2

)2

⎤⎥⎦

= e−t2 L−1

⎡⎢⎣ p

p2 +( √

32

)2

⎤⎥⎦+

e−t2√3

L−1

⎡⎢⎣

√3

2

p2 +( √

32

)2

⎤⎥⎦

= e−t2 cos

( √3

2t

)+ e−

t2

1√3

sin

( √3

2t

)

∴ L−1

[(p + 1)e−πp

p2 + p + 1

]

= e−(t−π)

2

[cos

{√3

2(t − π)

}+

1√3

sin

{√3

2(t − π)

}]· H(t − π),

by the second shift theorem of Laplace transform

Example. 4.9. Evaluate L−1[

3(1+e−pπ)p2+9

]Solution. We have

L−1

[3(1 + e−pπ)

p2 + 9

]

= L−1

[3

p2 + 32+ e−pπ 3

p2 + 32

]

= L−1

[3

p2 + 32

]+ L−1

[e−pπ 3

p2 + 32

]= sin 3t + H(t − π) sin 3 (t − π)

= sin 3t − sin 3 t. H(t − π)

Example 4.10. Using the result of inverse Laplace transform evaluateL−1

[p+1

(p2+2p+1)2

].

Solution. We know that L [tn f(t)] = (−1)n dn

dpn f(p), for n = 1, 2, 3, · · · .This result implies

L−1

[dn

dpnf(p)

]= (−1)tnf(t) , where L[f(t)] = f(p)


Now, L−1

[p + 1

{ (p + 1)2 + 1 }2

]

= e−t L−1

[p

(p2 + 1)2

]

= −e−t L−1

[12

d

dp.

(1

(p2 + 1)

) ]

= −e−t . (−1)1t

2sin t

= e−t t

2sin t


log p+3p+2

]Solution. Let f(p) =

[log p+3

p+2

]. We wish to calculate f(t).

Now,d

dpf(p) =

1p + 3

− 1p + 2

L−1

[d

dpf(p)

]= e−3t − e−2t

Therefore, −L−1

[d

dpf(p)

]= tf(t) = e−2t − e−3t

implying, f(t) =e−2t − e−3t

t


log(

1 − 1p2

) ]Solution. Let

f(p) = log(

1 − 1p2

)⇒ − d

dp

[f(p)

]=

2p− 2p

p2 − 1

∴ −L−1

[d

dpf(p)

]= L−1

[2p− 2p

p2 − 1

]⇒ tf(t) = 2H(t) − 2 cosh t

Or, f(t) = 2 [ H(t) − cosh t ] /t.

Example 4.13. Assuming that limt→∞

f(t)t do exist, prove that

L−1[

f(p)p

]=∫ t0 f(x) dx. Hence, evaluate L−1

[1

p3(p2+1)

]Solution. First part is discussed in article 2.10.

By the above result, L−1[

1p

(1

p2+1

) ]=∫ t0 sin x dx = 1 − cos t

L−1

[1p

{1

p(p2 + 1)

} ]=∫ t

0(1 − cos x) dx = t − sin t.


Hence, L−1

[1p

{1

p2(1 + p2)

}]=∫ t

0(x − sin x) dx =

t2

2+ cos t − 1.

Example 4.14. Evaluate

L−1

[1

p(p + 1)3

]

Solution. We have[1

p(p + 1)3

]= L−1

[1

{(p + 1) − 1 }(p + 1)3

]

= e−tL−1

[1

(p − 1)p3

]

Now, L−1

[1

p − 1

]= et

∴ L−1

[1

p3(p − 1)

]=∫ t

0

[ ∫ t

0

{ ∫ t

0etdt

}dt

]dt

=∫ t

0

[ ∫ t

0

(et − 1

)dt

]dt =

∫ t

0

(et − 1 − t

)dt

= et − 1 − t − t2

2

∴ L−1

[1

p(p + 1)3

]= e−t

[et −

(1 + t +

t2

2

)]

= 1 − e−t

(1 + t +

t2

2

)

Example 4.15. If L−1[

p(p2+1)2

]= 1

2 t sin t, evaluate L−1[

1(p2+1)2

]Solution. We have

L−1

[1

(p2 + 1)2

]= L−1

[1p

p

(p2 + 1)2

]

=∫ t

0

(12

x sin x

)dx

=12

[sin t − t cos t]

Example 4.16. By using the convolution theorem evaluate L−1[

1p2(p+1)2

]to show that its value is t e−t + 2 e−t + t − 2.

Solution. We have by convolution theorem

L−1[f(p) · g(p)] = f ∗ g =∫ t

0f(τ)g(t − τ) dτ


Now, L−1

[1p2

]= t and L−1

[1

(p + 1)2

]= (t)e−t.

Therefore, L−1

[1p2

.1

(p + 1)2

]=∫ t

0e−u . u(t − u)du

= te−t + 2e−t + t − 2 , on simplification.


p(p2+4)3

]by using covolution theorem.

Solution. We know that L−1[

1(p2+4)

]= sin 2t

2

and L−1

[p

(p2 + 4)2

]= L−1

[−1

2d

dp

1(p2 + 4)

]

= −12.t(−1)1 L−1

[1

p2 + 4

]

=12t

sin 2t2

. =14

t sin 2t

Therefore, L−1

[p

(p2 + 4)2.

1p2 + 4

]=∫ t

0

u sin 2u4

.12

sin 2(t − u)du

=18

∫ t

0u sin 2u( sin 2t cos 2u − cos 2t sin 2u) du

=sin 2t

8

∫ t

0u sin 2u cos 2udu − cos 2t

8

∫ t

0u sin2 2u du

=164

[ sin 2t − 2t cos 2t ] ,

after evalution of the integrals and on simplification.

Example 4.18. Prove that L−1[

1p√

p+4

]= 1

2 erf (2√

t).


L−1

[1p

]= H(t)

and L−1

[1√

p + 4

]= e−4t L−1

[1√p

]=

e−4t

√πt

Therefore, L−1

[1p

1√p + 4

]=∫ t

0

e−4τ

√πτ

H(t − τ)dτ

=∫ t

0

e−4τ

√4τ

dτ =1√π

∫ 2√

t

0e−u2

du, on putting 4τ = u2

=12

erf (2√

t) , by definition of erf(x) =2√π

∫ x

0e−α2

dα



1√p(p−a)

]by the convolution theorem.

Hence deduce the value of L−1[

1p√

p+a

]Solution. We know that L−1

[1√p

]= 1√

πtand L−1

[1

(p−a)

]= eat

Therefore, by the convolution theorem

L−1

[1√p

· 1(p − a)

]=∫ t

0

1√πu

ea(t−u)du

=eat

√π

∫ t

0

1√u

e−au du , putting au = ς2

=eat

√a

· 2√π

∫ √at

0e−ς2dς

=eat

√a

erf(√

at).

We have now

L−1

[1

p√

p + a

]= L−1

[1

(p + a) − a· 1√

p + a

]

e−at L−1

[1

(p − a)· 1√

p

]

= e−at eat

√a

erf(√

at) =1√a

erf(√

at).

Example 4.20. A periodic function f(t) of period 2π having a finitediscontinuity at t = π is given by

f(t) =

{sin t , 0 � t < π

cos t , π < t � 2π

Evaluate its Laplace transform

Solution The Laplace transform of f(t) is given by

f(p) =1

1 − e−2πp

[∫ 2π

0f(t)e−pt dt

]

=1

1 − e−2πp

[∫ π−0

0e−pt sin t dt +

∫ 2π

π+0e−pt sin t dt

]

=1

1 − e−2pπ

[1 + e−pπ

1 + p2− pe−pπ(1 + e−pπ)

1 + p2

]

=1

1 − e−pπ

[1 − pe−pπ

1 + p2

]


4.4 The general evaluation technique of inverse Laplace

Transform.

The inverse Laplace transform has already been discussed in articles 3.3and 3.4 and is given by

f(t) = L−1[f(p) ; p → t] =1

2πiP.V.

∫ c+i∞

c−i∞eptf(p)dp , t > 0 (4.4)

where Re p = c > 0 and an arbitrary constant, provided f(t) = 0(eγt),Re p > γ with

f(p) = L [ f(t); t → p ] =∫ ∞

0f(t)e−pt dt (4.5)

where f(p) is being the function of a complex variable p in the half-planeRe p > γ. The integral in (4.4) is called a Bromwich integral.

The working rule for determining the inverse Laplace transform f(t)of f(p) with the help of the contour integral of a complex variable isshown below:

Since, f(t) =1

2πiP.V.

∫ c+i∞

c−i∞eptf(p) dp, t > 0,

to evaluate f(t) by (4.4), we evaluate the contour integral

12πi

∫C

eptf(p) dp, (4.6)

where C is the contour ABCDEA described in the counter-clockwisesense and consists of the line Re p = c > γ and the sectorial part of thecircle of large radius R with centre at the origin of the complex p-plane.All the singularities of f(p) are lying to the left of the line segment fromc − iR to c + iR and therefore they are all enclosed inside the contourC. This contour C is known as Bromwich contour in turn and is shownin FIG 1 below.


For actual evaluation of the integral in (4.6) let us assume thatpn, (n = 1, 2, · · · , N) denote the singularities of f(p) inside C. Let R0

denotes the largest of the moduli of all these singular points. The para-metric representation of the semicircular part, say, CR is given by

p = c + Reiθ ,π

2� θ � 3π

2(4.7)

where R > R0 + c. Note that for each pn,

|pn − c| � |pn| + c � R0 + c < R.

Then by Cauchy’s residue theorem, theorem we get

∫Γ

eptf(p) dp = 2πiN∑

n=1

Resp=pn

[eptf(p)

]− ∫CR

eptf(p) dp (4.8)

It may be noted at this point that if f(p) has any branch point singu-larity, corresponding branch-cut must be given suitably for making f(p)analytic except for the singular points inside the modified contour C.

Suppose now that, for all point p on CR, there is a positive constantMR such that |f(p)| � MR, where MR → 0 as R → ∞. We may use theparametric representation (4.7) for CR to write

∫CR

ept f(p) dp =∫ 3π

2

π2

exp(ct + Rteiθ

)f(c + Reiθ

)· Ri eiθdθ

Then since,∣∣∣ exp (ct + Rteiθ)

∣∣∣ = ect · eRt cos θ


and∣∣∣ f(c + Reiθ)

∣∣∣ � MR, we find that∣∣∣∣∫

CR

eptf)(p)dp

∣∣∣∣ � ect MR R

∫ 3π/2

π/2eRt cos θ dθ (4.9)

But the substitution φ = θ− π2 together with Jordan’s inequality reveals

that ∫ 3π/2

π/2eRt cos θ dθ =

∫ π

0e−Rt sin φ dφ <

π

Rt

Thus inequality (4.9) becomes∣∣∣∣∫

CR

eptf(p) dp

∣∣∣∣ � ect MR π

t(4.10)

and hence limR→∞

∫CR

eptf(p) dp = 0 (4.11)

Therefore, letting R → ∞ in eqn. (4.4), we see that f(t), defined bythe equation

f(t) =1

2πilim

R→∞

∫ c+i∞

c−i∞eptf(p) dp , t > 0 (4.12)

can be written as

f(t) =N∑

n=1

Resp=pn

[eptf(p)

], t > 0. (4.13)

Further it is to be noted in case that when f(p) has countable infi-nite number of singular points, the evaluation of f(t) in (4.13) may bereplaced by

f(t) =∞∑

n=1

Resp=pn

[eptf(p)

], t > 0. (4.14)

whenever it is possible to show that

limR,N→∞

∫CR

eptf(p) dp = 0 (4.15)

In many applications of Laplace transforms, such as the solution of ODE

and PDE arising in initial value problem or boundary-value problem,the form of f(p) oblained there may not be that much simple to put


it in a combination of the forms discussed in the heuristic approachwith elementary functions. Thus, in those cases the necessity of thisgeneral evaluation technique of the inverse Laplace transform may befound useful.

It may also be noted that both heuristic method and the general con-tour integration method inversion of Laplace transform lead to identicalresults.

Example 4.21. Using heuristic as well as complex Inverse formula eval-uate the inverse Laplace transform of the image function f(p) = p

(p2+1)2.

Solution. By heuristic approach, we get

L−1[

f(p)]

= L−1

[p

(p2 + 1)2

]= L−1

[−1

2d

dp

{1

(p2 + 1)

}]

=12

t L−1

[1

p2 + 1

]

=12

t sin t = f(t)

Also, by complex inversion rule we have

f(t) =1

2πi.

∫ c+i∞

c−i∞eptf(p) dp

with f(p) = p(p2+1)2

, which has singularities only at points ±i , both ofthem being double poles. Therefore, we choose the arbitrary constantc > 0 and the corresponding Bromwhich contour C as shown below.

In this case, f(t) = Sum of the residues of ept p(p2+1)2 at p = ±i


= limp→i

d

dp

[(p − i)2

eptp

(p + i)2(p − i)2

]+ lim

p→−i

d

dp

[(p + i)2

eptp

(p + i)2(p − i)2

]

= − it

4eit +

14it e−it

=t

2sin t , after simplification

Example 4.22. Use complex inversion formula to evaluate inverseLaplace transform of f(p) = 2p2−4

(p+1)(p−2)(p−3)

Solution. If inverse transform of f(p) be f(t), then by complex Inversionformula for Laplace transform it is given by

f(t) =1

2πi.

∫ c+i∞

c−i∞f(p)ept dp

=1

2πi

∫ c+i∞

c−i∞

(2p2 − 4) ept dp

(p + 1)(p − 2)(p − 3)=

12πi

∫C

(2p2 − 4) ept dp

(p + 1)(p − 2)(p − 3)

where C is a positively oriented closed contour with the Bramwich linefrom c − i∞ to c + i∞ as a part and enclosing all the poles of f(p).

Since, f(p) has only three simple poles at points p = −1, 2, 3 we choosehere the arbitrary constant c > 3 for the Bromwich integral. Thus byarticle 2.21

f(t) = sum of the residues at the poles p = −1, p = 2 and p = 3 ofthe integrand

[f(p)ept

].

Now, residue of f(p)ept at p = −1

= limp→−1

[(p + 1)/ (2p2 − 4)ept

(p + 1)/ (p + 2)(p − 3)

]=

16

e−t

Residue of f(p) ept at p = 2 = limp→2

[(p − 2)(2p2 − 4)ept

(p + 1)(p − 2)(p − 3)

]

= −43e2t

and residue of f(p)ept at p = 3 = limp→3

[(p − 3)(2p2 − 4)ept

(p + 1)(p − 2)(p − 3)

]

=72e3t

Therefore, f(t) = −16

e−t − 43

e2t +72

e3t


4.5 Inversion Formula from a different stand point : The

Tricomi’s method.

The Laguerre polynomial Ln(t) is defined by

Ln(t) =n∑

r=o

cn

r

(−t)r

r!(4.16)

Therefore, L [Ln(t); t → p ] =n∑

r=0

(−1)r cn

r

1pr+1

=1p

(1 − 1

p

)2

(4.17)

Making use of change of scale and the first shift theorem of Laplacetransform, we get

L[

e−bt Ln(2bt); t → p]

=1

p + b

(p − b

p + b

)n

(4.18)

Therefore, if f(t) =∞∑

n=0

an e−bt Ln(2bt) (4.19)

then L [ f(t); t → p ] = f(p)

=∞∑

n=0

an1

p + b

(p − b

p + b

)n

(4.20)

by Lerch’s theorem.

Thus the equation (4.19) defines the Tricomi’s method of determiningLaplace inversion L−1

[f(p); p → t

], when f(p) is expressed in the

form (4.20). For this purpose we substitute p = b(

1+σ1−σ

)and then

express the given f(p) in the form

f(p) =1 − σ

2b

∞∑n=0

anσn

Thus, f(t) = L−1[

f(p); p → t]

is interpreted as

f(t) =∞∑

n=0

an Bn(bt) , (4.21)

where Bn(bt) = e−btLn(2bt) (4.22)

To verify the accuracy of the method we discuss the following specialcase as discussed by Ward [(1954) Proc. camb. Phil. Soc.,50,49 ] , where

f(p) =1

(p + 1)(p − 2)and f(t) = e−t − e−2t


To evaluate f(t) by the Tricomi’s method, we substitutep = (1 + σ)/(1 − σ) , with b = 1. Then we get

f(p) =12(1 − σ) .

13(1 − σ

3)−1(1 − σ)

=1 − σ

2

[13− 2

3

∞∑n=1

( σ

3

)n]

So, f(t) =13

B0(t) − 23

∞∑n=1

3−nBn(t), (4.23)

where Bn(t) is defined in (4.22) above.

Estimated accuracy can be verified by keeping seven terms in (4.23),for fixed values of t.

A similar result can be derived involving Legendre Polynomials too.

Example 4.23. If f(t) = L−1[f(s); s → t

], then prove that

L−1

[1sf(s) ; s → t

]=∫ t

0f(x) dx

Solution. Consider g(t) = 1 ⇒ g(s) = 1s . Therefore, by convolution

theorem L−1[

1s f(s); s → t

]=∫ t0 f(t − τ) dτ if L [ f(t); t → s ] = f(s)

This equation implies,

L−1

[1sf(s); s → t

]=∫ t

0f(x)dx , where t − τ = x

Example 4.24 Evaluate

L−1

[ch x

√sa

s ch(√

sa l) ; s → t

]

Solution. By complex inversion formula

f(t) =1

2πi

∫ c+i∞

c−i∞est ch(αx)

ch(αl)ds

s, where α =

√s

a

Clearly, the integrand has poles at s = 0, s = sn = −(2n + 1)2 aπ2

4l2,

n = 0 or integer.

Residue at s = 0 is 1.


Residue at s = sn isexp(−snt)ch

{i(2n + 1) πx

2l

}[

s dds

{ch l

√sa

} ]s=sn

=4(−1)n+1

(2n + 1) πexp

[−{

(2n + 1)π2l

}2

at

]ch{

(2n + 1)πx

2l

}

Thus, f(t) = Sum of the residues at the poles of the integrand

= 1 +4π

∞∑n=0

(−1)n+1

(2n + 1)exp

[−(2n + 1)2

π2at

4l2

]ch{

(2n + 1)πx

2l

}.

Example 4.25. Using power series expansion in s due to Heavisideevaluate the Laplace inversion of f(s) = 1

ssh x

√s

sh√

s, 0 < x < 1, s > 0.

Solution. We have

f(s) =1s

ex√

s − e−x√

s

e√

s − e−√

s

=1s

e−(1−x)√

s − e−(1+x)√

s

1 − e−2√

s

=1s

[e−(1−x)

√s − e−(1+x)

√s].

∞∑n=0

exp(−2n√

s)

Or f(s) =1s

∞∑n=0

[exp{−(1 − x + 2n)

√s}− exp

{−(1 + x + 2n)√

s}]

Therefore, f(t) =∞∑

n=0

[erfc

(1 − x + 2n

2√

t

)− erfc

(1 + x + 2n

2√

t

)]


1(s2+a2)2

; s → t]

Solution. We know that L−1[

1(s2+a2) ; s → t

]= sin at

a

Therefore, L−1

[1

(s2 + a2)2; s → t

]=

sin at

a∗ sin at

a

=1a2

∫ t

0sin a τ sin a(t − τ) dτ, by convolution theorem

=1

2a3(sin at − at cos at).

Example 4.27. Prove that L−1[

1(√

s+a); s → t

]= 1√

π t−aea2t erfc (a

√t)


Solution.1

(√

s + a)=

1√s− a√

s(√

s + a)

∴ L−1

[1

(√

s + a); s → t

]

= L−1

[1√s

; s → t

]− a L−1

[(√

s − a)(s − a2)

√s

; s → t

]

= L−1

[1√s

; s → t

]− a L−1

[1

s − a2; s → t

]

+a2L−1

[1√

s(s − a2); s → t

]

=1√πt

− a exp (a2t) + a exp (a2t) erf (a√

t)

=1√πt

− a exp (a2t) erfc (a√

t).

4.6 The Double Laplace Transform

The double Laplace transform of a function of two variables f(x, y) inthe positive quadrant of the xy- plane is defined by the equation

f(p, q) = L2 [f(x, y) ; (x, y) → (p, q)]

= L [L {f(x, y) ; x → p} ; y → q] (4.24)

and hence by the double integral

f(p, q) =∫ ∞

0

∫ ∞

0f(x, y) e−(px+qy) dx dy (4.25)

whenever the integral exists .

It therefore, follows that

L2 [f(ax, by); (x, y) → (p, q)] =1ab

f(p

a,q

b

), a > 0, b > 0

(4.26)

and L2 [f(ax) g(by); (x, y) → (p, q) ] =1ab

f(p

a

)g(q

b

), a > 0, b > 0

(4.27)

In particular, we have therefore

L2 [ f(x); (x, y) → (p, q)] =1qf(p) (4.28)


The formulae (4.26)- (4.28) follow from the definitions (4.24) and(4.25) above.

To evaluate double Laplace transform of a function f(x + y), wechange the independent variables (x, y) to (ξ, η) where we definex = 1

2(ξ − η), y = 12 (ξ + η) implying dxdy = 1

2 dξdη

Thus,

L2 [f(x + y); (x, y) → (p, q)] =12

∫ ∞

0f(ξ)dξ

[∫ ξ

−ξe−

12(p+q)ξ− 1

2(p−q)ηdη

]

=12

∫ ∞

0f(ξ)

[2

p − q(e−qξ − e−pξ)

]dξ

=1

p − q

[f(q) − f(p)

](4.29)

after introducing the new region of double integration in the ξη - planeand evaluating the double integral.

To evaluate the double Laplace transform of f(x−y) no such simpleformula as above exists. For the case when f(t) is an even function oft, we have from definition that

L2 [f(x − y) ; (x, y) → (p, q)] =∫ ∫

Q1

f(x − y) e−px−qy dx dy

+∫ ∫

Q2

f(y − x) e−px−qy dx dy

(4.30)

while if f(t) is an odd function of t

L2 [f(x − y) ; (x, y) → (p, q)] =∫ ∫

Q1

f(x − y) e−px−qy dx dy

−∫ ∫

Q2

f(y − x) e−px−qy dx dy (4.31)

where Q1 and Q2 are infinite regions shown below in Fig. 1.4.4


After making the transformation x = 12(ξ + η), y = 1

2(η − ξ) we findthat∫ ∫

Q1

f(x − y)e−px−qydxdy =12

∫ ∞

0f(ξ)dξ

[∫ ∞

ξe−

12(p−q)ξ− 1

2(p+q)ηdη

]= f(p)/(p + q) (4.32)

On the other hand making the transformation

x =12(η − ς), y =

12(η + ς)

we get∫ ∫

Q2

f(y − x) e−px−qy dx dy

=12

∫ ∞

0f(ς)dς

[∫ ∞

ςe−

12(q−p) ς− 1

2(p+q)η dη

]

=f(q)p + q

(4.33)

Therefore, from above when f(t) is an even function of t

L2 [ f(x − y); (x, y) → (p, q)] =1

p + q

[f(p) + f(q)

]. (4.34)


But, if f(t) is an odd function of t, we have

L2 [ f(x − y); (x, y) → (p, q)] =1

p + q

[f(p) − f(q)

]. (4.35)

Also, L2 [ f(x) H (x − y); (x, y) → (p, q) ]

=∫ ∞

0f(x)e−px

( ∫ x

0e−qydy

)dx

=1q

[f(p) − f(p + q)

], (4.36)

L2 [ f(x) H (y − x); (x, y) → (p, q) ]

=∫ ∞

0f(x)e−px

[ ∫ ∞

xe−qydy

]dx

=1q

[f(p + q)

], (4.37)

and L2 [ f(x) H (x + y) ; (x, y) → (p, q) ] =1qf(p) (4.38)

As a special case if f(x) = 1, f(p) = 1p so that

L2 [H (x − y); (x, y) → (p, q) ] =1

p(p + q)(4.39)

The formulae for double Laplace transform of the partial derivativesof a function of two variables can be obtained from the definition asgiven below.

L2

[∂u(x, y)

∂x; (x, y) → (p, q)

]=∫ ∞

0e−qydy

[∫ ∞

0

∂u

∂xe−pxdx

]

= −∫ ∞

0e−qydy

[u(0, y) − p

∫ ∞

0e−pxu(x, y)dx

]= p u (p, q) − u1(q),

where u1(q) = L [u(0, y); y → q]

and similarly,

L2

[∂u(x, y)

∂y; (x, y) → (p, q)

]= q u (p, q) − u2(p), (4.40)

where u2(p) = L [u(x, 0);x → p]


Laplace transform of the higher order partial derivatives of a functionu(x, y). of the variables x, y can similarly be derived as

L2

[∂2u

∂x ∂y; (x, y) → (p, q)

]= pqu(p, q) − qu1(q) − pu2(p) + u(0, 0)

(4.41)

L2

[∂2u

∂x2; (x, y) → (p, q)

]= p2u(p, q) − pu1(q) − u3(q) (4.42)

L2

[∂2u

∂y2; (x, y) → (p, q)

]= q2u(p, q) − qu2(p) − u4(p) (4.43)

where, u3(q) = L

[∂u

∂x(0, y); y → q

]and u4(p) = L

[∂u

∂y(x, 0);x → p

]

The convolution theorem in this case is generalised as below.

The convolution of two double variable functions f(x, y) and g(x, y)is defined by

f ∗ ∗ g =∫ x

0ds

∫ y

0f(s, t) g (x − s, y − t)dt (4.44)

and its double Laplace transform is given by

L2 [ f ∗ ∗ g ; (x, y) → (p, q) ] = f(p, q) g(p, q) (4.45)

The proof of (4.45) follows directly from the definition of double Laplacetransform.

From (4.45) it therefore follows that

L−12

[f(p, q) g (p, q) ; (p, q) → (x, y)

]= f ∗ ∗g.

Example 4.28 Solve the PDE ∂u∂x = ∂u

∂y satisfying the boundaryconditions u(x, 0) = a(x), u(0, y) = b(y), for x > 0, y > 0.

Solution. Taking double Laplace transform, the PDE gives

L2

[∂u

∂x; (x, y) → (p, q)

]= L2

[∂u

∂y; (x, y) → (p, q)

]implying, pu(p, q) − b(q) = qu(p, q) − a(p).

Thus, u(p, q) =b(q) − a(p)

p − q

Or, u(x, y) = L−12

[b(q) − a(p)

p − q; (p, q) → (x, y)

]


4.7 The iterative Laplace Transform.

In case of double Laplace transform of a function of two variables andthose for its partial derivatives if the transformed variables p and q areidential, then such a double Laplace transform, in particular, is calledan iterated Laplace transform which is denoted and defined as

I [ f(x, y) ; p ] = L2 [ f(x, y); (x, y) → (p, p) ]

=∫ ∞

0

∫ ∞

0f(x, y)e−p(x+y) dx dy (4.46)

Thus, by substituting q = p in the last section 4.4 all the propertiesof iterated Laplace transform can be derived.

In this case the generalised convolution of a function f(x, y) is de-noted and defined by

f (2)(x) =∫ x

0f(x − y, y) dy (4.47)

and hence

L[f (2)(x); x → p

]=∫ ∞

0e−px dx

[∫ x

0(x − y, y) dy

](4.48)

Using the variable transformation x = u + v , y = v ⇒ ∂(x,y)∂(u,v) = 1

L[f (2)(x); x → p

]=∫ ∞

0

∫ ∞

0f(u, v)e−p(u+v)du dv , from (4.48)

= I [ f(x, y) ; p ] , [ by (4.46) ] (4.49)

4.8 The Bilateral Laplace Transform.

Here, we shall now define a transform on functions which are definedfor all real values of the independent variables with the kernel e−px asdenoted and defined by the equation

f+(p) = L+ [ f(x) ; x → p ] =∫ +∞

−∞f(x)e−pxdx (4.50)

provided that f(x) is such that integral on the right hand side of eqn.(4.50) is convergent for some values of p. f+(p) is sometimes called thetwo-sided Laplace transform of f(x) by some authors.


Writing (4.50) in the form

L+ [ f(x) ; x → p ] =∫ ∞

0f(x)e−pxdx +

∫ ∞

0f(−x)epxdx

we have

L+[f(x) ; x → p] = L [ f(x) ; x → p ] + L [ f(−x) ; x → −p ]

(4.51)

Let the first integral on the r.h.s converges for Re p > σ1 . and thesecond for Re(−p) > −σ2 ⇒ Re p < σ2 . Therefore, for the bilateralLaplace transform of f(x) is to exist these two half-planes must overlapimplying that σ2 > σ1. Hence, the condition for existance of the bilateralLaplace transform defined above in eqn. (4.50) is when Re p satisfies

σ1 < Rep < σ2 (4.52)

If σ1 = σ2, the strip contracts to a line Re p = σ1 and the region ofconvergence is the line Re p = σ1. But, if σ2 < σ1, the bilateral Laplacetransform for f(x) as defined above does not exist.

For example, if f(x) = e−x2, the bilateral Laplace transform is

defined for all Re p. While if f(x) = e−a|x|, the bilateral transformexists for −a < Re p < +a. But the function f(x) = e−ax does not havebilateral Laplace transform.

Thus , if f+(p) =∫ +∞

−∞f(x)e−pxdx, σ1 < Re p < σ2,

its inversion formula is given by

f(x) =1

2πi

∫ c+i∞

c−i∞f+(p)epxdp, σ1 < c < σ2 (4.53)

Here of course the region of convergence must always be specifiedsince two distinct object functions may have the same image function.For example, L+ [ H(t); t → p ] = 1, Re p > 0 and L+ [ −H(−t) ; t → p ]= 1, Re p < 0


4.9 Application of Laplace Transforms.

A. Solution of ODE with constant coeffiecients.

Suppose that we wish to solve the nth order linear ODE

dny

dtn+ c1

dn−1y

dtn−1+ · · · + cn−1

dy

dt+ cn y = f(t) (4.54)

where ci, i = 1, 2, · · · n are given constants, subject to the initial condi-tions y(0) = k1, y

′(0) = k2, · · · yn=1(0) = kn. Taking Laplace transformof both sides of the ODE using the results of article 3.9, and the initialconditions the ODE gives,[

y(p)pn − pn−1k1 − · · · − p kn−1 − kn

]+ c1

[y(p)pn−1 − pn−2k1 − · · · kn−1

]+ · · · + cn−1 [py(p) − k1] + cn.y(p) = f(p).

Thus,

y(p)[pn + c1p

n−1 + · · · + cn

]= f(p) + k1

[pn−1 + c1p

n−2 · · · + cn−1 p]

+k2

[pn−2 + c1p

n−3 + . . .]+ · · · + kn−1 [p + c1] + kn

=⇒ y(p) =[f(p) + g(p)

]/[pn + c1p

n−1 + · · · + cn

]y(p) =

f(p)h(p)

+g(p)h(p)

, where f(p) = L [f(t)] , (4.55)

g(p) is a polynomial of degree n − 1 in p and h(p) is also a polynomialof degree n in p.

Thus, the solution of the ODE under the given initial conditions canbe obtained after taking inverse Laplace transform of the last equationin(4.55). Since, the right side is a known function, we have

y(t) = L−1

[f(p)h(p)

]+ L−1

[g(p)h(p)

](4.56)

If eqn.(4.54) is a homogeneous one, then f(t) ≡ 0 and so, the requiredsolution of the homogeneous ODE under given initial conditions is

y(t) = L−1 [ g(p)/h(p) ] . (4.57)

Example 4.29 Solve by using Laplace transform the initial -valueproblem y′′′(t) + 2y′′(t) − y′(t) − 2y(t) = 0 with initial conditionsy(0) = y′(0) = 0 and y′′(0) = 6.


Solution. Taking Laplace transform of both sides of the given ODEunder the initial conditions, we get[

p3y(p) − p2y(0) − py′(0) − y′′(0)]+ 2[p2y(p) − py(0) − y′(0)

]− [py(p) − y(0)] − 2y(p) = 0

=⇒ y(p) = 6/[p3 − 2p2 − p − 2

]=⇒ y(p) =

6(p − 1)(p + 1)(p + 2)

≡ 1p − 1

− 3p + 1

+2

p + 2

Thus inverting both sides , we get

y(t) = et − 3e−t + 2e−2t .

Example 4.30. Solve the initial-value problem defined by[d2

dt2+ n2

]x(t) = a sin (nt + α) ; x(0) = x′(o) = 0 .

Solution. Taking Laplace transform of both sides of the given ODEunder the initial conditions, we get(

p2 + n2)x(p) = a cos α

n

p2 + n2+ a sin α

p

p2 + n2

⇒ x(p) = an cos α1

(p2 + n2)2+ a sin α

p

(p2 + n2)2

On inversion, the above eqn. gives

x(t) = an cos α1

2n3[ sinnt − nt cos nt ] + a sin α

t

2nsin nt

= a [ sinnt cos α − nt cos(nt + α) ] /2n2

Example 4.31. Solve the initial-value problem

dy

dt+ 2y +

∫ t

0ydt = sin t , y(0) = 1.

Solution. Taking Laplace transform of the ODE we get under the giveninitial condition that

py(p) − 1 + 2y(p) +1py(p) =

1p2 + 1

⇒ y(p) =p(2 + p2)

(p2 + 1)(p + 1)2


By partial fraction method we can express

y(p) =1

(p + 1)− 3

21

(p + 1)2+

12.

1p2 + 1

On inverting, we get

y(t) = e−t − 32t e−t +

12

sin t

as the required solution of the given problem.

Example 4.32. Solve the BVP d2xdt2

+ 9x = cos 2t, under the boundaryconditions x(0) = 1, x

(π2

)= −1

Solution. Taking Laplace transform, the given ODE leads to

(p2 + 9) x(p) =p

p2 + 4+ p + A ; where A = x′(0), say

Therefore,

x(p) =p

(p2 + 32)(p2 + 22)+

p

p2 + 32+

3A3(p2 + 32)

=15

[p

p2 + 22− p

p2 + 32

]+

p

p2 + 32+

A

3· 3p2 + 32

Inverting,

x(t) =15

[ cos 2t − cos 3t ] + cos 3t +A

3· sin 3t

=15

cos 2t +45

cos 3t +A

3sin 3t

Since, x(

π2

)= −1 is being given, the above equation gives A = 4

5

Thus the solution of the BVP is

x(t) =15

[ cos 2t + 4cos 3t + 4 sin 3t ]

Example 4.33. A voltage Ee−at is applied at t = 0 to a circuit ofinductance L and resistance R connected in series where a,E,L and R

are constants. Using Laplace transform show that the current i(t) atany time t is given by

i(t) =E

R − aL

[e−at − e

−RtL

]Solution. In an electrical network with given voltage E(t), resistanceR and inductance L, the current i(t) builds at the rate given by

Ldi

dt+ Ri(t) = E(t)


Here it is given that E(t) = Ee−at and i(0) = 0. Therefore, takingLaplace transform of the above equation under given conditions, we get

L [ p i(p) − 0 ] + R i(p) = E1

p + a

⇒ i(p) =E

(Lp + R)(p + a)≡ E

L(

a − RL

)[

1p + R

L

− 1p + a

]

=E

R − aL

[1

p + a− 1

p + RL

]

Inverting, i(t) =E

R − aL

[e−at − e−

RtL

].

B. Solution of simultaneous ODE with constant co-efficeents

The ODE involving more than one dependent variables but witha single independent variable give rise to simultaneous equations. Theprocedure for solving such simultaneous equations is almost same as thatdiscussed in section A. Here also, we have to take Laplace transform ofthe simultaneous equations to reduce them to corresponding number ofalgebraic equations which can then be solved for the Laplace transformeddependent variables. Finally inverting these relations we can recoverthe dependent variables forming the required solutions. The method isillustrated through the following examples.

Example 4.34. Solve the following initial value problem defined by thesimultaneous ODE

dx

dt− y = et dy

dt+ x = sin t, given that x(0) = 1, y(0) = 0

Solution. Taking Laplace transform the two given ODE under assignedinitial conditions, we get

px(p) − 1 − y(p) =1

p − 1

py(p) + x(p) =1

p2 + 1

Solving these two equations for x(p) and y(p), we get

x(p) =p

p2 + 1+

p

(p − 1)(p2 + 1)+

1(p2 + 1)2


=⇒ x(t) = 12

[cos t + 2 sin t + et − t cos t

], on inversion of Laplace

transform.

Also, y(p) = −1 − 1p−1 + px(p)

This expression after Laplace inversion and on simplification givesrise to

y(t) =12[

t sin t − et + cos t − sin t]

Example 4.35. Solve the IVP defined by

x′(t)+y′(t) = t, x′′(t)−y(t) = e−t if x(0) = 3, x′(0) = −2 and y(0) = 0.

Solution. Taking Laplace transform of the ODE under given initialconditions we get,

px(p) − 3 + py(p) =1p2

and p2x(p) − 3p + 2 − y(p) =1

p + 1

Solving these equations for x(p) and y(p), we get

x(p) =3p2 + 1

p3(p2 + 1)+

3p2 + p − 1(p + 1)(p2 + 1)

and y(p) =3p2 + 1

p2(p2 + 1)− 3p2 + p − 1

(p + 1)(p2 + 1)

Expressing into partial fractions, we get

x(p) =2p

+1p3

− 32(p2 + 1)

+p

2(p2 + 1)+

12(p + 1)

and y(p) =1p

+1

2(p + 1)+

32(p2 + 1)

− p

2(p2 + 1)

Taking inverse Laplace transform, it is found that

x(t) = 2 +12t2 +

12et − 3

2sin t +

12

cos t

y(t) = 1 − 12e−t − 1

2cos t +

32

sin t,

as solutions of the given problem.


C. Solution of ODE with variable co-efficients.

In this case let the differenial equation be a combination of terms ofthe form tnyn(t) and the Laplace transform of which are given by theformulae

L [tmyn(t)] = (−1)mdm

dpm

[L{

y(n)(t)} ]

,m = 1, 2, . . .

Then the usual method of solution, as in the case A, may be employedfor its solution.

Example 4.36. Solve the ODE with variable coefficients

ty′′(t) + y′(t) + ty(t) = 0

under the initial condition y(0) = 1

Solution. Taking Laplace transform, the differential equation gives

− d

dp

[p2 y(p) − p.1 − y′(0)

]+ [ p y(p) − 1 ] − d

dpy(p) = 0

under the given initial condition. On simplification, it gives

− d

dp

[(p2 + 1) y (p)

]+ py(p) = −1 + 1 = 0

⇒ −(p2 + 1)d y(p)

dp− py(p) = 0

or,dy(p)y(p)

= − p dp

p2 + 1

Integrating, log y(p) = −12

log(p2 + 1) + log c

or, y(p) =c√

1 + p2

or, y(t) = c J0(t)

But, by given condition y(0) = 1 ⇒ 1 = c.1 ⇒ c = 1. Therefore, therequired particular solution is given by

y(t) = J0(t).

Example 4.37. Solve the initial value problem defined by the ODEwith variable coefficients given by

ty′′(t) + 2y′(t) + ty(t) = 0, y(0) = 1


Solution. Taking Laplace transform under the given initial condition,the ODE leads to

− d

dp

[p2y(p) − py(0) − y′(0)

]+ 2 [ py(p) − y(0) ] − d

dpy(p) = 0

⇒ (p2 + 1)dy(p)dp

= −1

Or,∫

dy(p) = −∫

dp

1 + p2

Or, y(p) = cot−1 p +(c − π

2

), c is being an arbitrary constant.

After inverting, the above relation gives

y(t) = L−1(

−π

2+ c

)+

sin t

tAs t → 0, y(0) = 1 and therefore we get

1 = L−1[

c − π

2

]+ 1 ⇒ L−1

[c − π

2

]= 0

Hence, y(t) = sin tt is the required solution of the IVP.

[Note here that this last result can also be obtained by considering thelimit of y(p) as t → 0 or as p → ∞ leading to c = π

2 ].

Example 4.38. Solve the ODE

ty′′(t) + (t − 1)y′(t) − y(t) = 0 under the conditions y(0) = 5 andy′(0) = −5 , if y(∞) is finite.

Solution. Taking Laplace transform, the ODE under given initial con-ditions leads to

− d

dp

[(p2 + p)y(p) − 5p − y′(0) − 5

]= py(p) + y(p) − 5

or,dy(p)dp

+3p + 2

p(p + 1)y(p) =

10p(p + 1)

The I.F of this linear ODE in y(p) is e� (3p+2)dp

p(p+1) = p2(p + 1). Hence itssolution is given by

y(p) · p2(p + 1) =∫ [

I.F × 10p(p + 1)

]dp + A, A = arbitrary constant

⇒ (p3 + p2)y (p) = 5p2 + A


Now by the final value theorem of Laplace transform as t → ∞implying p → 0 we get

limp→0

(p + p2) · limp→0

p y(p) = A

⇒ 0 · y(∞) = A

∴ A = 0 and hence y(p) =5

p + 1=⇒ y(t) = 5e−t, as the required

solution.

D. Ordinary differential equations in application to Electricalnetworks.

We consider an electrical network consisting of (i) a power source orsupplier of electromotive force E(volts) (ii) a resistor having rasistanceR(ohms) (iii) an inductor having inductance L(henrys) and (iv) a ca-pacitor having capacitance C(farads). When the circuit is switched onat time t = 0 by Kirchhoff law the circuit usually known as an L-C-Rcircuit satisfies the ODE

Ld2Q

dt2+ R

dQ

dt+

Q(t)C

= E

with the current i(t) at time t is given by i(t) = dQdt , where Q is the

charge measured in coulombs. To discuss such a problem we considerthe following examples.

Example 4.39. At time t = 0, a constant voltage E is applied to aL-C-R circuit. The initial current i and the charge Q are zero. Find thecurrent at any time t > 0, when R2 < or = or > 4L

C .

Solution. By Kirchhoff law, the ODE in Q at time t for the circuit isgiven by Ld2Q

dt2+ R dQ

dt + QC = E , i(t) = dQ

dt under i(0) = 0, Q(0) = 0.The above pair of simultaneous ODE can also expressed as

Ldi

dt+ Ri +

Q

C= E ,

dQ

dt= i

Taking Laplace transform, these equations under given initial conditionsbecome

L [ p i (p) − i(0) ] + R i (p) +1C

Q (p) =E

p

and p Q (p) − Q(0) = i(p)


Since i(0) = 0, Q(0) = 0, these equations become

i(p) [ Lp + R ] +1C

Q (p) =E

p, i (p) = p Q(p) · · · (i)

Solving for i(p), we get[Lp + R +

1Cp

]i (p) =

E

p⇒ i (p) =

E

L[

(p + R2L)2 + n2

] · · · (ii)

where we assume1

CL− R2

4L2= n2

We shall now discuss three different cases individually.

Case I. If R2 < 4LC , R2

4L2 < 1CL ⇒ n2 > 0.

Then inverting, the equation (ii) gives

i(t) = L−1

[E

L{

(p + R2L)2 + n2

}]

=E

Le−

Rt2L L−1

{1

p2 + n2

}=

E

Lne−

Rt2L sin nt

Case II. If R2 = 4LC , R2

4L2 = 1CL ⇒ n = 0

Then after inversion of Laplace transform, the eqn.(ii) gives

i(t) =E

LL−1

[1(

p + R2L

)2]

=E

Le−

Rt2L L−1

(1p2

)=

Et

Le−

Rt2L .

Case III. If R2 > 4LC , n2 < 0 ⇒ n2 = −m2, say

Then after inversion of Laplace transform, the equation(ii) gives

i(t) =E

LL−1

[1(

p + R2L

)2 − m2

]

=E

Le−

Rt2L L−1

(1

p2 − m2

)

=E

Lme−

Rt2L sinh mt

Example 4.40. In an electrical network with e.m.f E(t), rasistance R

and inductance L in series, the current i builds up at the rate given byL di

dt + Ri = E(t). If the switch of the circuit is connected at t = 0 anddisconnected at t = a, find the current i at any time.


Solution. In this case, it is given that i = 0 at t = 0 and

E(t) =

{E , for 0 < t < a

0 , for t > a .

Taking Laplace transform, the given ODE in i(t) gives

(Lp + R) i(p) =∫ ∞

0e−ptE(t) dt =

∫ a

0e−pt E dt =

E

p

(1 − e−ap

)This gives i(p) =

E

p (Lp + R)− Ee−ap

p (Lp + R)

To find i(t) we have to invert the above equation.

Now, L−1

[E

p (Lp + R)

]= L−1

[E

R

{1p− 1

p + RL

}]

=E

R

(1 − e−

RtL

)

and L−1

[E e−ap

p(Lp + R)

]= L−1

[E e−ap

R

{1p− 1

p + RL

}]

=E

R

(1 − e−

R(t−a)L

)H(t − a), by 2nd shift theorem.

Therfore, we get

i(t) =E

R

(1 − e−

RtL

)− E

R

(1 − e

−R(t−a)L

)H(t − a).

where H(t − a) =

{0 , 0 < t < a

1 , t > a

E. Application to Mechanics.

In the vibration problem the displacement of a particle measuredfrom given fixed point satisfies some ODE. In case of loaded beams thetransverse deflection of the beam also satisfies some ODE. Under initialand boundary conditions for these respective cases these displacementswill be evaluated in the following examples. It may be noted that in caseof beam problems we assume the boundary conditions as (i) y = y′ = 0for clamped or built in fixed end (ii) y = y′′ = 0 for hinged or simplysupported ends and (iii) y′′ = y′′′ = 0 for free ends, y being the deflectionof the beam.


Example 4.41 A mass m moves along the x-axis under the influenceof a force which varies to its instantaneous speed and acts in a directionopposite to the direction of motion of the mass. Assuming that themass is located at x = a initially and is moving to the right side oralong positive x-axis with the speed v0 , find the position where the masscomes to rest if the position of the mass at any time is x and it satisfiesthe ODE m d2x

dt2= −μ dx

dt .

Solution. The initial conditions of the problem are x(0) = a, dxdt (0) = v0 .

Now taking Laplace transform of the ODE under the given initial con-ditions, we get

m[p2 x(p) − pa − v0

]= −μ [ px(p) − a]

This implies , x(p)[

mp2 + μp]

= mpa + mv0 + μa

or x(p) =mv0 + μa

μp− mv0

μ(p + μm)

Inverting Laplace transform, the above equation gives the positionx(t) of the mass as

x(t) =mv0 + μa

μ− mv0

μe−

μm t

Thus the velocity of the mass at any time is given by

dx

dt= v0 e−

μtm

When the mass comes to rest, dxdt = 0 ⇒ e−

μtm = 0

Thus for such a time t when e−μtm = 0, we get

x =mv0 + μa

μ.

Hence, the mass comes to rest when it is at a distance a + mv0μ from

the origin.

Example 4.42. Obtain the deflection y at a distance x from one end ofa weightless beam of length l and freely supported at both ends, whena concentrated load W acts at x = a. The differential equation fordeflection y is being

EId4y

dx4= Wδ(x − a),


where E is the Young’s modulus of elasticity of the beam and I is themoment of inertia of a cross-section of the beam about its axis. Itmay be noted that for freely supported ends, the following conditionsy(0) = y′′(0) = 0 and y(l) = y′′(l) = 0 are satisfied

Solution. Taking Laplace transform in variable x the given ODE afterusing the initial conditions y(0) = y′′(0) = 0 and assuming y′(0) = c1

, y′′′′(0) = c2, we get

p4y(p) = c1p2 + c2 +

W

EIe−ap

Or, y(p) =c1

p2+

c2

p4+

W

EI

e−ap

p4

Taking inverse Laplace transform the last equation gives

y(x) = c1x + c2x3

3!+

W

EI

(x − a)3

3!H(x − a)

Therefore, y′(x) = c1 +12c2x

2 +W

2EI(x − a)2 H(x − a)

y′′(x) = c2x +W

EI(x − a) H(x − a)

Using the conditions y(l) = 0, y′′(l) = 0 we get

c2 =−W

EI

l − a

l

c1 =[

Wa(l − a)6EIl

][ 2l − a ]

Therefore, the required deflection of the beam at a distance x from theend x = 0 is

y(x) =W

6EI

[a(l − a)(2l − a)

lx − l − a

lx3 + (x − a)3 H (x − a)

]

In particular, if 0 < x < a , H(x − a) = 0

∴ y(x) =W

6(EI)

[a(l − a)(2l − a)

lx − (l − a)

lx3

]Also , if a < x < l , H(x − a) = 1

∴ y(x) =W

6EI

[a(l − a)(2l − a)

lx − l − a

lx3 + (x − a)3

]


From these above two results we have at x = a

y(a) =13

W

EI

a2(l − a)2

l.

F. Application to partial differential equation : Boundaryvalue Problem.

It is known that in case of ordinary differential equation the Laplacetransform reduces it to an algebraic equation of the transformed func-tion. But it may now be noted that by the use of Laplace transform oneof the independent variable of the equation will be removed by the trans-formed parameter p or s. Thus the PDE in two independent variableswill be reduced to an ODE and for PDE in more than two indepen-dent variables, one of the variable will be removed by the transformedparameter p , after taking account of the boundary conditions on thatvariable.

To start with the matter under discussion, let u(x, t) be a functionof two independent variables defined for a � x � b and t � 0. Alsolet u(x, t), regarded as a function t , be of some exponential order forlarge t and is a piecewise continuous function over t � 0. Then Laplacetransform over the variable t of u(x, t) is defined as

L [ u(x, t); t → p ] =∫ ∞

0e−ptu(x, t) dt = u(x, p) (4.58)

Therefore , L

[∂u

∂t

]= pu(x, p) − u(x, 0)

L

[∂2u

∂t2

]= p2u(x, p) − pu(x, 0) − ∂u

∂t(x, 0) , etc.

The proof of the above results will follow from the definition in (4.58).Similarly, for functions of more than two independent variables of whichone of them is t, say u(x, y, t), or u(x, y, z, t) etc, let it be defined fort � 0 and is of some exponential order for large t and is piecewisecontinuous function in t � 0. Then Laplace transform of such functionand of their partial derivatives can be defined as an extension of theabove case of two or more variables.

Example 4.43. Find the solution of the PDE ∂2y∂x2 − ∂2y

∂t2= xt subject

to y(x, 0) = ∂y∂t (x, 0) = 0 and y(0, t) = 0. It is given that the required

solution is bounded both for x � 0 and t � 0.


Solution. Taking Laplace transform on the variable t the given PDEunder first two initial conditions gives

d2y(x, p)dx2

− [ p2 y(x, p) − p.0 − 0]

=x

p2

Or,d2y(x, p)

dx2− p2 y(x, p) =

x

p2

The general solution of the last equation is

y (x, p) = Aepx + Be−px − x

p4(i)

Since y (x, t) is bounded for all x and t, so also y (x, p) for all x andp. Hence for large x , y (x, p) is bounded and therefore, the arbitraryconstant A = 0 in the above form (i) of y (x, p). So,

y (x, p) = B e−px +−x

p4(ii)

But, by the given condition y(0, t) = 0 we get

y(0, p) = 0 (iii)

Therefore (ii) implies under (iii) that B = 0. Then eqn. (ii) implies

y (x, p) = − x

p4

Taking inverse Laplace transform the above solution reduces to

y (x, t) = −x t3

6

as the required solution of the PDE under given conditions.

Example 4.44. A bar of length a is at zero temperature. At t = 0, theend x = a is suddenly raised to a temperature u0 and the end x = 0 isinsulated. Find the temperature at any point x of the bar at any timet > 0 assuming that the surface of bar is insulated.

Solution. Let u(x, t) be the temperature at any point x and at anytime t of the bar so that it satisfies the heat conduction equation

∂u

∂t= c2 ∂2u

∂x2, (0 � x � a , t � 0) (i)


subject to the conditions

u(x, 0) = 0 (ii)

ux(0, t) = 0 (insulated end means heat flow from that end is nil) (iii)

and u(a, t) = u0 (iv)

Taking Laplace transform of eqn. (i) above over the variable t weget

pu(x, p) − u(x, 0) = c2 d2

dx2u(x, p) (v)

where u (x, p) = L [ u(x, t) ; t → p ] =∫ ∞

0e−ptu(x, t) dt

Using condition given in eqn. (ii) , the eqn (v) becomes

d2u(x, p)dx2

− p

c2u(x, p) = 0 (vi)

Also, Laplace transform of eqns. (iii) and (iv) become

ux(0, p) = 0 (vii)

and u(a, p) =u0

p(viii)

Solving eqn. (vi), we have

u(x, p) = A e√

px

c + B e−√

px

c

Under the condition (vii) we get A = B and therefore,

u(x, p) = 2A cosh√

px

c

Now, using condition in eqn. (viii) we have finally

u(x, p) =uo cosh

(√pxc

)p cosh

(√pac

) (ix)

To obtain the temperature u(x, t) we have to invert Laplace trans-form of u(x, p) given in eqn. (ix). But the inversion can not be obtainedby heuristic method. Therefore, we take recourse to general inversion


formula of Laplace transform as discussed in article 4.4. By such aformula we have here,

u(x, t) =1

2πi. P.V

∫ λ+i∞

λ−i∞eptu(x, p) dp , (x)

where the Bromwich line contour Γ from λ− i∞ to λ+ i∞ , (λ is a realconstant) is such that all the singularities of u(x, p) in p-plane lie to theleft of Re p = λ. Then if C as shown in the figure below is a positivelyoriented closed contour consisting of Γ and a large (infinite) circle CR

enclosing all the singularities of u(x, p) after giving branch-cut in thep-plane required to make the integrand single valued eqn. (x) can bewritten equivalently as

u(x, t) =1

2πi

∫C

eptu(x, p) dp (xi)

=N∑

n=1

Resp=pn

[eptu(x, p)

], (xii)

where N= number of poles of u(x, p) inside C. Here also

limR→∞

∫CR

eptu(x, p) dp = 0

.


Also,∫

L1+L2

ept u (x, p) dp = 0 ,

limr→0

∫Cr

ept u (x, p) dp = 0

Thus eqn. (xii) follows from eqns. (x) and (xi). Now the poles ofthe integrand in (xi) inside C are given by

p = pn = −(2n − 1)2

4a2c2π2 , n = 0, 1, 2, · · ·

Residue at p = pn

= limp→pn

⎡⎣uo(p − pn) ept cosh

(√pxc

)p cosh

( √p a

c

)⎤⎦

= u0 limp→pn

[ {p − pn

ch√

p ac

} ]. limp→pn

[eptch(

√p x

c )p

]

= u0 limp→pn

{1

sh(√

p ac )

.2√

pc

a

}. limp→pn

[eptch(

√p x

c )p

]

=4u0(−1)n

(2n − 1)π. e

−(2n−1)2π2 c2 t

4a2 . cos(2n − 1)πx

2a

Thus, we get

u(x, t) =4u0

π

∞∑n=1

(−1)n

(2n − 1)e

−(2n−1)2π2 c2 t

4a2 cos(2n − 1)πx

2a,

as the required solution of the given heat conduction problem.

Example 4.45. A semi-infinite solid x > 0 has its initial temperaturezero. A constant heat flux H is applied at the face x = 0 and hence−kux(0, t) = A, where k is the thermal diffusivity of the material of thesolid and u(x, t) is the temperature of the solid at any point of it at timet. If the heat conduction equation of the solid is given by

∂u

∂t= k

∂2u

∂x2, 0 < x < ∞, t > 0 (i)

with the boundary condition ∂u(0,t)∂x = −H

τ and the initial conditionu(x, 0) = 0, find the temperature of the solid at any point x > 0 at anyinstant, given that


L−1

[e−x

√p

p32

; p → t

]=√

tπ e

−x2

4t2 − x2 erfc

{x

(2√

t )

}Solution. Taking Laplace transform over the variable t the eqn. (i)under the initial condition takes the form

pu(x, p) = kd2

dx2u (x, p)

Its general solution is given by

u(x, p) = Aex√

p/k + Be−x√

p/k (ii)

Since the temperature u(x, t) is bounded for large x, so is u(x, p) andtherefore eqn. (ii) gives A = 0. Thus eqn.(ii) becomes

u(x, p) = Be−x√

pk (iii)

Also, Laplace transform of the boundary condition becomes

∂u

∂x(0, p) = −H

τp, (τ = thermal conductivity ) (iv)

Using (iv) in (iii), we get B = H

τ√

kp32

and hence eqn. (iii) implies

u(x, p) =√

k H e−x√

pk

τ p32

Now, inverting Laplace transform the required solution of theproblem is given by

u(x, t) = L−1

[H√

k

τ

e−x√

pk

p32

; p → t

]

Therefore, by given condition

u(x, t) =H√

k

τ

[√t

πe−

x2

4t2k − x

2√

kerfc

{x

2√

kt

}]

Example 4.46. The temperature u(x, t) of a slab of material mediumsatisfies the heat conduction equation ∂u

∂t = k ∂2u∂x2 , 0 < x < l, t > 0. If

the initial temperature at the bounding planes x = 0 and x = l of the


slab be u0, find the temperature in this solid after the face at x = 0 isinsulated and the temperature of the face at x = l is reduced to zero.

Solution. The temperature distribution u(x, t) of the slab satisfies thePDE

∂u

∂t= k

∂2u

dx2(i)

subject to the boundary conditions du(o,t)dx = 0 and u(l, t) = 0 and the

initial condition u(x, 0) = u0, 0 < x < l. Taking Laplace transformequation (i) under the initial condition gives

pu(x, p) − u0 = kd2

dx2u(x, p)

Solving the above equation we get

u(x, p) = A cosh(√

p

kx

)+ B sinh

(√p

kx

)+

u0

p(ii)

The Laplace transform of the given boundary conditions become

d

dxu(0, p) = 0 and u(l, p) = 0 (iii)

Using (iii) in (ii) we get B = 0 and A = − u0

p cosh(√

pk l)

Therefore, eqn. (ii) becomes

u(x, p) =u0

p− u0

p

cosh(√

pk x)

cosh(√

pk l)

Inverting Laplace transform the above relation gives

u(x, t) = u0 − u0L−1

⎡⎣ cosh

(√pk x)

p cosh(√

pk l)⎤⎦ (iv)

The second term in the above is almost equivalent to eqn. (ix) ofexample-4.44. Hence, following the same technique (iv) can be evaluatedto get the solution u(x, t) of the problem in terms of x and t.


Note: As an alternative procedure for evaluation of Laplace inversionthe expression 1

p . ch(√

pk x)

/ch(√

pk l)

may be expanded in an infi-nite series and then term by term inversion may be obtained for solution,noting that L−1

[1p e−c

√p]

= erfc(

c2√

t

).

Example 4.47. Consider the diffusion equation in a semi-infinite line0 < x < ∞ defined by ∂2θ

∂x2 = 1k

∂θ∂t , t � 0 and subject to the boundary

condition θ(0, t) = f(t) and the initial condition θ(x, 0) = 0 togetherwith the limiting condition θ (x, t) → 0 as x → ∞. Obtain the temper-ature θ (x, t) of the line under this general case and of the particularcases (i) θ (0, t) = θ0 (T/t )

12 , θ0 and T are positive constants

(ii) θ(0, t) = θ0, a constant and (iii) ∂∂x θ(0, t) = f(t) instead of the

condition θ(0, t) = f(t) by using Laplace transform.

Solution. Taking Laplace transform the PDE under the given initialcondition reduces to

d2θ (x, p)dx2

=1k

[p θ (x, p)

]We solve the above second order ODE to obtain

θ (x, p) = Ae−√

pkx + Be

√pkx (i)

Now, taking Laplace transform to the given boundary condition

θ (0, t) = f(t) we getθ (0, p) = f(p)

Under Laplace transformed the limiting condition becomes θ(x, p) → 0as x → ∞. Using these two results, we get

θ (x, p) = f(p) e−√

pkx , since B = 0 (ii)

Therefore, inverting this equation we have by the convolution theoremthat

θ (x, t) =∫ t

0f (τ) g (x, t − τ) dτ (iii)

where g (x, t) = L−1[

e−√

pkx ; p → t

](iv)

But since , L[

t−12 e−c/t ; t → p

]=√

π

pe−2

√cp, differentiating both


sides with respect to c we obtain

L[

t−32 e− c/t ; t → p

]=√

π

ce−2

√cp (v)

Therefore, g (x, t) =1√

4kπt3e−x2/4kt

and hence the solution of this general problem is given by

θ (x, t) =1√4πk

∫ t

0f(τ)exp

[x2

4k (t − τ)

]dτ

(t − τ)32

We shall now take up the particular cases shown below.

Case-I

θ(0, t) = θ0

(T

t

) 12

So, f(p) = θ0 T12 L

[t−

12 ; t → p

]= θ0

(πT

p

)12

and hence

θ(x, p) = θ0 (πT )12 p−

12 e−x

√pk

Inverting , θ(x, t) = θ0

(T

t

)12

e−x2

4kt

Case-II

θ(0, t) = θ0 , a constant .

In this case , f(p) =θ0

pand so θ(x, p) =

θ0

pe−x

√pk

After Laplace inversion we get

θ (x, t) = θ0 L−1

[1p

e−x√

pk ; p → t

]

= θ0 Erfc

(12

x√kt

).

Case - III

In this case∂

∂x

[θ(0, p)

]= f(p) , θ(x, p) → 0 as x → ∞

Therefore, θ(x, p) = −√

kf(p)1√p

e−x√

pk


Hence, θ (x, t) =∫ t

0f (τ) g (x, t − τ) dτ

where g(x, t) = −k12 L−1

[1√p

e−x

√pk ; p → t

]

= −(

k

πt

) 12

e−x2

4kt

Therefore, the required solution for this general case is given by

u(x, t) = −√

k

π

∫ t

0f(τ) exp

[− x2

4k(t − τ)

]d τ√t − τ

.

Example 4.48. An infinitely long string of negligible mass having oneend at x = 0, is initially at rest along the x-axis. The end x = 0 isgiven a transverse displacement f(t), t > 0. Find the displacement ofany point of the string at any time by using Laplace transform.

Solution. Let y(x, t) be the transverse displacement of any point x

of the string at any time t. Then, y(x, t) satisfies the wave equation∂2y∂t2 = c2 ∂2y

∂x2 , x > 0, t > 0 ; subject to the initial conditions y(x, 0) = 0,∂y∂t (x, 0) = 0 and the boundary condition y(0, t) = f(t) and y(x, t) isbounded. Laplace transform of the PDE over the variable t gives

p2y (x, p) − p y (x, 0) − ∂y (x, 0)∂t

= c2 d2y(x, p)dx2

Using the initial conditions we rewrite the above equation as

d2y

dx2=( p

c

)2y (x, p) (i)

Again, Laplace transform of the given boundary conditions are

y(0, p) = f(p) and y(x, p) is bounded.

Solving (i), we get

y (x, p) = A epxc + B e−

pxc

Since y(x, t) is bounded so is y(x, p) for all x. Then we have

y (x, p) = Be−pxc , because A = 0.


By use of Laplace transformed boundary condition y(0, p) = f(p) wehave

y (x, p) = f(p) . e−pxc

Therefore, by complex inversion formula of Laplace transform we get

y(x, t) =1

2πi

∫ λ+i∞

λ−i∞e(t−x

c)p f(p) dp

= f(t − x

c

).

Example 4.49 A light flexible string of length a is stretched betweentwo fixed points at x = 0 and at x = a. If the string is displaced intothe curve y = b sin π x

a and released from rest in that position initially,find its displacement at any time t and at any point x.

Solution. The transverse vibration y(x, t) of the string satisfies thePDE ∂2y

∂t2= c2 ∂2y

∂x2 subject to boundary conditions y(0, t) = 0, y(a, t) =0 and the initial conditions y(x, 0) = b sin πx

a , ∂y∂t (x, 0) = 0. Taking

Laplace transform of the PDE over the variable t under the initial con-ditions we get

d2y(x, p)dx2

− p2

c2y (x, p) = −bp

c2sin

πx

a

The general solution of this second order ODE is

y(x, p) = A epxc + B e−

pxc +

bp sin(

πxa

)p2 +

(πca

)2 (i)

Now, Laplace transform of the given boundary conditions are

u(0, p) = 0 and u(a, p) = 0

Using these results in (i) we get

A + B = 0 and A epac + B e−

pac = 0.

Solving these equations for A and B, we get A = B = 0, then

y(x, p) = b sinπx

a.

p

p2 +(

πca

)2


Therefore, inverting this equation we get the required solution of thegiven problem as

y(x, t) = b sinπx

aL−1

[p

p2 +(

πca

)2 ; p → t

]

= b sinπx

acos

πct

a

Example 4.50. A tightly stretched flexible string has its end points atx = 0 and at x = l. At time t = 0 the string is given the shape definedby y = μ

∑∞n=1 sin nπ

l x, where μ is a constant, and then released fromrest. Find the displacement y = (x, t) of the string.

Solution. The partial differential equation satisfying the transversedisplacement y(x, t) of the string and satisfying the initial conditionsy(x, 0) = μ

∑∞n=1 sin nπx

l and ∂y (x,0)∂t = 0 and boundary conditions

y(0, t) = y(l, t) = 0 is given by

∂2y

∂t2= c2 ∂2y

∂x2,

c being a material constant of the string.

Taking Laplace transform of the PDE under the initial conditions,we have

p2y(x, p) − py(x, 0) − ∂y(x, 0)∂t

= c2 d2

dx2y(x, p)

Or[

d2

dx2− p2

c2

]y (x, p) = −pμ

c2

∞∑n=1

sinnπx

l

This equation has the solution

y(x, p) = A chp

c(l − x) + B sh

p

c(l − x) + μ

∞∑n=1

p sin nπxl

p2 +(

nπcl

)2Laplace transform of the given boundary conditions are

y(0, p) = y(l, p) = 0

Using these results, the above form of y(x, p) gives

A chpl

c+ B sh

pl

c= 0

and A = 0. Thus , B = 0 and hence

y(x, p) = μ

∞∑n=1

p sin nπxl

p2 +(

nπcl

)2


Inverting Laplace transform we get the solution of the problem as

y(x, t) = μ

∞∑n=1

cosnπct

lsin

nπx

l.

Example 4.51. An infinite string lying along the x-axis is vibratingby giving a displacement f(t) to the end x = 0. The other end of thestring at large distance from the origin remains fixed. If initially thedisplacement and the velocity of every point of the string were zero,find the displacement of every point of the string at any time, when thedisplacement satisfies the PDE

∂2y

∂t2= c2 ∂2y

∂x2(i)

Solution. As per statement of the problem the initial and the boundaryconditions are

y(x, 0) = 0,∂y (x, 0)

dt= 0 and y(0, t) = f(t), lim

x→∞ y(x, t) = 0

respectively. Taking Laplace transform over t, eqn. (i) becomes

p2y(x, p) − py(x, 0) − ∂y

∂t(x, 0) = c2 d2

dx2y(x, p)

Using the initial conditions, we have then

d2y

dx2(x, p) − p2

c2y(x, p) = 0 , p > 0.

The general solution of this ODE is

y(x, p) = Aepc

x + Be−pc

x (ii)

Since, limx→∞ y(x, 0) = 0, we must have from the above solution that A = 0.

Therefore, the solution (ii) becomes

y(x, p) = Be−pc

x (iii)

But, again from the boundary condition y(0, t) = f(t) under Laplacetransform, we get y(0, p) = f(p) = L [ f(t) ]. Hence, by (iii) we get

y(0, p) = B = f(p)


Therefore, y(x, p) = f(p) e−pcx. Inverting the above transformed solu-

tion y(x, p), we have then finally

L−1 [ y(x, p) ; p → t ] = L−1[

f(p) . e−pcx ; p → t

]Or y(x, t) = f

(t − x

c

)H(

t − x

c

).

by the second shift theorem of Laplace transform. Thus the requireddisplacement is

y(x, t) = 0, t <x

c

= f(t − x

c

), t >

x

c

G. Application to solution of Integral equation and Integro-differntial equation.

An integral equation is an equation in which the unknown functionappears under integral sign. For example,

f(s) =∫ b

ak (s, t) g(t) dt ,

g(s) = f(s) +∫ b

ak (s, t) g (t) dt

and g(s) =∫ b

ak(s, t) [g (t)]2 dt ,

where b, the upper limit of the integral may be either a variable or aconstant. The function g(s) is the unknown function while all otherfunctions are known. These functions may be complex valued functionsof the real values of s and t. If the limits a and b are both constants,the integral equation is known as Fredholm integral equation. If theupper limit b is a variable and a is a constant, the corresponding in-tegral equation is called Volterra equation. The kernel k(s, t) of theintegral equation is a known function. If k(s, t) = K(s − t), a functionof the difference s− t only, the corresponding integral equation is calleda convolution type integral equation. Many interesting problems of me-chanics and physics lead to integral equation in which the kernel k(s, t)is a function of the difference (s− t) only. Thus, when k(s, t) = K(s− t),where K is a function of one variable, though k(s, t) is a function of twovariables, Laplace transform is one of the necessary tools towards thesolution of such convolution type Volterra integral equation


g(s) = f(s) +∫ s

0K(s − t)g(t) dt (4.59)

as shown below.

On applying Laplace transform to both sides of the eqn.(4.59) and usingthe convolution formula, we have

g(p) = f(p) + K(p) . g (p)

Or g(p) = f(p)/[1 − K(p)

], (4.60)

and then the inversion of eqn. (4.60) yields the solution of (4.59)

A special type convolution integral equation∫ s

0

f(u)(s − u)α

du = g(s), 0 < α < 1 (4.61)

is called an Abel’s integral equation. Here f(u) is an unknown and g(s)is a known function.

Laplace transform method may also be effectively used to solve suchan equation in (4.61) too.

The following examples may be considered for much more details inapplications of the method.


s =∫ s

0es−t g(t) dt

Solution. Taking Laplace transform of both sides, we obtain

1p2

= K(p) g (p) =1

p − 1g(p)

Or , g (p) =1p− 1

p2

Inverting , g (s) = 1 − s , as the solution of the integral equation


f(s) = s2 +∫ s

0f(u) sin (s − u) du


Solution. Taking Laplace transform on both sides, we get

f(p) =2p3

+ L [ f(s) ∗ sin s ] =2p3

+ f(p)/(p2 + 1)

⇒ f(p) =2(p2 + 1)

p5=

2p3

+2p5

.

Inverting , f(s) = t2 +t4

12.


g(t) = 1 −∫ t

o( t − τ) g (τ) dτ

Solution. Taking Laplace transform, the given integral equationbecomes

g(p) =1p− L

[ ∫ t

o( t − τ) g (τ) dτ ; t → p

]

=1p− 1

p2. g (p) , by convolution theorem .

Therefore , g(p)[

1 + p2

p2

]=

1p

⇒ g(p) =p

p2 + 1

Inverting the last equation, the required solution of the integralequation is

g(t) = L−1

[p

p2 + 1; p → t

]= cos t


sin t =∫ t

0J0 (t − τ ) g (τ) dτ

Solution. Taking Laplace transform, the given integral equationbecomes

1p2 + 1

= L [ Jo (t) ; t → p] . g (p) =1√

p2 + 1g (p)

Thus , g (p) =1√

p2 + 1.


Inverting, we get the solution of the problem as

g(t) = L−1

[1√

p2 + 1

]= J0 (t) .


cos t =∫ t

0J0 (t − τ ) g ( τ ) dτ

Solution. Taking Laplace transform the integral equation gives

p

p2 + 1=

1√p2 + 1

. g(p) ⇒ g (p) =p√

p2 + 1

Inverting, g(t) = L−1

[p.

1√p2 + 1

]=∫ t

0H (τ)J0 (t − τ)dτ

=∫ t

0J0 (t − τ) dτ , t > τ > 0

since it is known that , L

[ ∫ t

0f (τ) g (t − τ)dτ

]= f(p) . g (p) .

Example 4.57. Solve the Abel’s integral equation∫ t

0

f (τ)dτ

(t − τ)13

= t (1 + t)

Solution. It is known that f(t) ∗ t−13 = t + t2. Taking the Laplace

transform of both sides we get

L [ f(t) ; t → p ] . L[t−

13

]= L [t] + L

[t2]

Or , f(p) .Γ (2

3 )

p23

=1p2

+2p3

Or , f(p) =1

Γ (23)

[1

p43

+2

p73

]

Taking inverse Laplace transform, the above equation gives the solutionof the Abel’s integral equation as

f(t) =1

Γ(

23

)[

t43−1

Γ (43 )

+ 2t

73−1

Γ(73 )

]


=1

Γ (23 )

[t

13

13 Γ (1

3 )+

2 t1+13

43 . 1

3 Γ (13)

]

=t

13

Γ (13 )Γ (1 − 1

3)

[3 +

92t

]

=3t

13 sin π

3

π[1 +

32

t]

=3√

3 t13

4π(2 + 3t)

Example 4.58. Solve the integro-differential equation

f ′(t) = t +∫ t

0f (t − τ) cos τ dτ , if f(0) = 4.

Solution. Taking Laplace transform of both sides the given equationbecomes

pf(p) − 4 =1p2

+ f(p) · p

p2 + 1

⇒ pf(p)[1 − 1

1 + p2

]=

1p2

+ 4

Or,p3

1 + p2f(p) =

1p2

+ 4 =1 + 4p2

p2

Or, f(p) =1p5

+5p3

+4p

.

Inverting Laplace transform, the solution of the equation is

f(t) =t4

24+

5t2

2+ 4.

G. Application to solution of Difference and Differential-differenceeqations.

Difference equation arise in a natural way in physics and engineering.Let a, a + 1, a + 2, . . . be consequitive finite values of the argument x ofa function denoted by u(x) or ux at discrete points. Here the constant1 is called the interval of difference. The nth order finite difference ofur is defined by

Δnur ≡ Δn−1 [Δur] = Δn−1 [ur+1 − ur] =n∑

k=0

(−1)k

(n

k

)ur+n−k

(4.62)


Any equation containing the differences of different orders of the functionur is called a difference equation. The highest order finite difference ap-pearing in a difference equation is called its order. Again any differenceequation containing the derivatives of the unknown function is called adifferential-difference equation. Thus a differential - difference equationpossesses two orders - one due to the largest order difference and theother associated with the highest order derivatives. For example,

Δ2un − Δun + un = 0 (4.63)

and u′(t) + u (t − 1) = f(t) (4.64)

are second order difference equation and first order differential-differenceequation respectively. To solve difference equation it facilitates to intro-duce a step function Sn(t) defined by

Sn(t) = H (t − n) − H (t − n − 1) , n � t � n + 1 (4.65)

where H(t) is the Heaviside unit step function

H(t) =

{1 , t > 00 , otherwise

(4.66)

Then Laplace transform of Sn(t) is given by

L [Sn(t) ; t → p] =∫ ∞

0e−pt [H (t − n) − H(t − n − 1)] dt (4.67)

=∫ n+1

ne−pt dt =

1p

(1 − e−p) e−np

= S0 (p) e−np (4.68)

where S0(p) =1p

(1 − e−p)

Let us now define a function u(t) by a series

u(t) =∞∑

n=0

un Sn(t) (4.69)

where {un} is a given sequence of finite numbers. Then it follows from(4.69) that


u(t) = un for n � t < n + 1

and u(t + 1) =∞∑

n=0

un Sn (t + 1)

=∞∑

n=0

un [H(t + 1 − n) − H(t − n)]

=∞∑

n=1

unSn−1(t)

=∞∑

n=0

un+1Sn(t) (4.70)

Similarly , u (t + 2) =∞∑

n=0

un+2 Sn(t) (4.71)

and more generally , u (t + j) =∞∑

n=0

un+j Sn(t) (4.72)

when j = 1, 2, 3, · · · · · · · · ·

Then from (4.69) Laplace transform of u(t) is given by

L [u(t) ; t → p ] = u (p) =∫ ∞

0e−ptu (t) dt

=∫ ∞

0e−pt [

∞∑n=0

unSn (t) ] dt

=∞∑

n=0

un

∫ ∞

0e−ptSn (t) dt

=∞∑

n=0

un Sn (p) (4.73)

Now this eqn. (4.73) gives

u (p) =∞∑

n=0

S0 (p) e−np

=1p

(1 − e−p)∞∑

n=0

un e−np

= S (p) ζ (p) , (4.74)


where ζ (p) =∞∑

n=0

un e−np (4.75)

Then , u(t) = L−1 [ S0 (p) ζ (p) ; p → t ] (4.76)

In particular, if un = an then in this case

ζ (p) =∞∑

n=0

an e−np =∞∑

n=0

(a e−p)n

=1

1 − ae−p=

ep

ep − a(4.77)

L [an] ≡ L [an (t) ] = S0 (p) ζ (p) implying , n � t < n + 1

L−1 [S0 (p)ep

ep − a] = an, by eqn. (4.77) (4.78)

Also, if un = (n + 1) an , then correspondingly

ζ (p) =∞∑

n=0

(n + 1) an e−np

=∞∑

n=0

(n + 1) (ae−p)n = (1 − ae−p)−2 (4.79)

Hence,

L [ (n + 1) an] = L [(n + 1) an (t)] = S0 (p)ζ(p), n � t < n + 1

=1

(1 − ae−p)2S0 (p)

=e2p

(ep − a)2S0 (p) (4.80)

This equation therefore implies

L−1

[e2p S0 (p)(ep − a)2

; p → t

]= (n + 1) an (4.81)

Again,

∞∑n=0

nan e−np =aep

(1 − ae−p)2

and so

L[nan] ≡ L[nan (t)] = S0 (p)ζ (p) , n � t < n + 1

=S0 (p) aep

(ep − a)2

Therefore, L−1

[aS0 (p) ep

(ep − a)2

]= n an (4.82)


Theorem 4.3.

If u (p) = L[u (t) ; t → p ] , then

L [u (t + 1) ; t → p ] = ep [u (p) − u0 S0 (p) ] (4.83)

where u0 = u(0).Proof. By definition

L[u (t + 1) ; t → p ] =∫ ∞

0e−pt u (t + 1) dt

=∫ ∞

1e−pτ ep u(τ) dτ

= ep

[ ∫ ∞

0e−pτu (τ) dτ −

∫ 1

0e−pτ u(τ) dτ

]

= ep[u(p) −∫ 1

0e−pτu(0) dτ ] , since u (τ) = u0 in (0, 1)

= ep u(p) − u0 ep S0 (p)

= ep[L{u(t) ; t → p} − u0S0(p)

]Also in view of the above theorem, we get

L[u (t + 2) ; t → p ] = ep [ L{ u (t + 1) ; t → p } − u(1)S0 (p) ]

= e2p [u (p) − u0 S0(p) ] − ep u(1)S0 (p)

= e2p [ u (p) − { u0 + u1 e−p } S0 (p) ] (4.84)

More generally,

L [ u (t + j) ; t → p ] = ejp [u(p) − S0 (p)j−1∑i=0

ui e−ip ] (4.85)

Example 4.59. Solve the difference equation

yk+2 − 5y k+1 + 6yk = 0 when y0 = 0, y1 = 1.

Solution. Taking Laplace transform the given equation results in

e2p [y (p)−{ y0 + y1e−p } S0 (p)]− 5ep [ y (p)− y0 S0 (p) ] + 6y (p) = 0

We use the given initial condition to obtain[e2p − 5ep + 6

]y (p) = ep S0 (p)

Thus , y(p) =ep

ep − 3S0 (p) − ep

ep − 2S0 (p)

⇒ yk = 3k − 2k


Example 4.60. Solve the difference equation

yn+2 − 2λ yn+1 + λ2yn = 0 , when y0 = 0 and y1 = 1

Solution. Taking Laplace transform the given difference equation underthe initial condition becomes

(ep − λ)2 y (p) = ep S0 (p)]

⇒ y (p) =ep

(ep − λ )2S0(p)

=1λ

λ ep

(ep − λ)2S0(p)

Inverting this Laplace transformed equation we get

yn =1λ

n λn , by (4.83)

= n λn−1

Example 4.61. Solve the differential-difference equation y′(t) = y(t−1)under the initial condition y(0) = 1

Solution. Laplace transform of the given equation we get

p y (p) − 1 = e−p [ y (p) − S0 (p)]

⇒ y(p) [ p − e−p ] = 1 +e−p − 1

p. e−p

Therefore,y (p) =

1p − e−p

− e−p

p (p − e−p)+

e−2p

p(p − e−p)

=1p

+e−2p

p2 (1 − e−p

p )

=1p

+(

e−p

p

)2 [1 − e−p

p

]−1

=1p

+e−2p

p2+

e−3p

p3+

e−4p

p4+ · · · · · ·

On inversion of Laplace transform, the above result gives

y(t) = 1 +t − 21!

+(t − 3)2

2!+ · · · · · · + (t − n)n−1

(n − 1)!, t > n.

Example 4.62. Solve the non-homogeneous difference equationyk+1 − 3yk = 1 under the condition y0 = 1

2


Solution. Laplace transform of the given difference equation results in

ep [ y (p) − y0 S (p) ] − 3y (p) =1p

⇒ (ep − 3) y (p) =12

ep S0 (p) +1p

⇒ y (p) =12

ep

ep − 3S0 (p) +

1p(ep − 3)

=12

ep

ep − 3S0 (p) +

1p

1 − e−p

(1 − e−p)(ep − 3)

=12

ep

ep − 3S0 (p) + S0 (p)

ep

e2p − 4ep + 3

=[

12

ep

ep − 3+

ep

(ep − 3) (ep − 1)

]S0 (p)

=[

12

ep

ep − 3+

12

ep

ep − 3− 1

2ep

ep − 1

]S0 (p)

=ep

ep − 3S0 (p) − 1

2ep

ep − 1S0 (p)

Inverting Laplace transform, we get

yk = 3k − 12

. 1k

Or, yk = 3k − 12.

Example 4.63. Show that the solution of the difference equation

un+2 + 4 un+1 + un = 0 with u0 = 0 and u1 = 1 is given by

un =1

2√

3

[(√

3 − 2)n + (−1)n+1 (2 +√

3)n]

Solution. Application of Laplace transform yields

e2p[u (p) − (e−pu1 + u0) S0(p) ] + 4ep[u (p) − u0 S0(p)] + u (p) = 0

Or [e2p + 4 ep + 1 ]u (p) = ep S0(p)

Or, u(p) =ep

(ep − α)(ep − β)S0(p),where α =

√3 − 2 and β = −(

√3 + 2)

=1

−β + α

[ −ep

ep − β+

ep

ep − α

]S0 (p)

=1

2√

3

[ep

ep − αS0 (p) − ep

ep − βS0 (p)

]


Inverting Laplace transform we get

un =1

2√

3

[(√

3 − 2)n − { −(√

3 + 2)}n]

=1

2√

3

[(√

3 − 2)n + (−1)n+1(√

3 + 2)n]

as the solution of the given difference equation.

Exercises

(1) Find inverse Laplace transforms of :

(a)p

(p + 1)2(p2 + 1)(b)

p2 + 6(p2 + 1)(p2 + 4)

(c)p + 2

(p2 + 4p + 5)2(d)

p

p4 + p2 + 1

(e)p

p4 + 4a4(f)

1p(p + 1)3

[Ans. (a)

12

(sin t − t e−t) (b)13

(5 sin t − sin 2 t)

(c)12

t e−2t sin t (d)2√3

sint

2· sin

√3

2t

(e)1

2a2sin at sh at (f) 1 − e−t

(1 + t +

t2

2

)]


(a) L [ (1 − 2t) e−2t ; t → s ] = s (s + 2)−2

(b) L [t H (t − a) ; t → s ] = (1 + sa) s−2 exp (−as)

(c) L [(1 + 2at) t−12 exp (at) ; t → s] = s

√π (s − a)−

32

(d) L

[1t

{e−at − e−bt

}; t → s

]= log

(s + b

s + a

)

(3) Prove that

(i) L−1

[1p

sin1p

]= t − t3

(3!)2+

t5

(5!)2− t7

(7!)2+ · · · and

(ii) L−1

[1p

cos1p

]= 1 − t2

(2!)2+

t4

(4!)2− t6

(6!)2+ · · ·


(4) Find inverse Laplace transform of :

(a) logp + 1p − 1

(b) logp − 1

p(c)

3p + 1(p + 1)4

(d)12

logp2 + b2

p2 + a2(e) cos−1

(p

2

)(f) log

(1 +

1p2

)

[Ans. (a)

2 sh t

t(b)

1 − et

t(c) e−t

(32

t2 − 13

t3)

(d)cos at − cos bt

t(e)

sin (2t)t

(f)2(1 − cos t)

t

]

(5) Prove that

L−1

[1pf(p)

]=∫ t

0f(x)dx and hence obtain

(i) L−1

[1

p(p + 1)

]and (ii) L−1

[1

p2(p + a)

][Ans. (i)

1a

(1 − e−at) (ii)1a2

(at + e−at − 1)]

(6) Prove that

(a) L−1

[e−as

s2; s → t

]= (t − a)H(t − a)

(b) L−1

[w

s2 + w2f(s) ; s → t

]=∫ t

0f(t − τ) sin wτ dτ

(c) L−1

[1s

e−a√

s ; s → t

]= erfc

(a

2√

t

)

(d) L−1[tan−1

(a

s

); s → t

]=

1t

sh(at)

(e) L−1

[f(s)s2

; s → t

]=∫ t

0(t − τ) f(τ) dτ

(7) Show that

(i)∫ ∞

0cos x2dx =

√π

2√

2=∫ ∞

0sin x2dx and

(ii)∫ ∞

0e−x2

dx =√

π

2,

by using Laplace transform and its inversion and considering func-tions f(t) =

∫∞0 cos(t x2)dx and f(t) =

∫∞0 e−tx2

dx in cases (i)and (ii) respectively.


(8) Evaluate inverse Laplace transform of :

(i)3(1 + e−pπ)

(p2 + 9)(ii)

32p(16 p2 + 1)2

(iii)1p

logp + 2p + 2

(iv) log(1 − 1/p2)[Ans. (i) − sin 3t H(t − π) + sin 3t (ii)

t

4sin

t

4

(iii)∫ t

0

e−x − e−2x

xdx (iv)

2(1 − ch t)t

]

(9) Prove that

(i) L−1

[1p

log(

1 +1p2

)]=∫ x

∂

2x

(1 − cos x)dx and

(ii) L−1

[1

p2(1 + p2)

]= t e−t + 2 e−t + t − 2

(10) Apply convolution theorem to evaluate

(a) L−1

[p2

(p2 + a2)(p2 + b2)

](b) L−1

[1

p(p2 + 4)

]

(c) L−1

[p

(p2 + 4)3

](d) L−1

[8

(p2 + 1)3

]

(e) L−1

[1

(p + 2)2(p − 2)

]

[ Ans. (a)(a sin at − b sin bt)

(a2 − b2)(b)

14(1 − cos 2t)

(c)t

64[sin 2t − 2t cos 2t] (d) (3 − t2) sin t − 3t cos t

(e)116[e2t − (4t + 1)e−2t

](11) Using Laplace transform properly prove that∫ ∞

0

e−t − e−3t

tdt = log 3

(12) Find inverse Laplace transform of

(a)p e−p/2 + π e−p

p2 + 4(b)

p e−ap

p2 − w2, a > 0

(c)w

(1 − e−πp

w )(p2 + w2)(d)

e−p

(p − 1)(p − 2)


[Ans. (a) sin πt

[H

(t − 1

2

)− H(t − 1)

](b) ch w(t − a) H(t − a)

(c) f(t),where f(t) =

{sin wt , 0 < t < π

w

0 , πw < t < 2π

w

and f(t) is a periodic function of period2πw

(d) [e2(t−1) − e(t−1)H(t − 1)]]

(13) Using L[J0(t)] prove that L[J1(t)] = 1− p√p2+1

. Also prove that

L[t J1(t)] = 1

(1+p2)32

and hence show that

L−1

[1√

p2 + 2ap + 2a2

]= e−atJ0(at)

(14) By expanding tan−1(

1p

)in ascending powers of p, and taking term

by term inverse Laplace transform show that

L−1

[tan−1

(1p

); p → t

]=

sin t

t.

(15) Evaluate L[et erf√

t] and hence evaluate L−1

[3p+2

2p2(p+1)32

][Ans.

1(p − 1)

√p

, t erf√

t

]

(16) Evaluate L−1[

1p3(1+p2)

]by using

(a) partial fractions,

(b) the convolution theorem.

[Ans. −1 + t2

2 − cos t](17) Apply convolution theorem to prove that

(a) L−1

[1

p2(p2 − a2)

]=

(sh at − at)a3

(b) L−1

[1

(p − 1)√

p

]= et erf(

√t)

(c) L−1

[1p

p

(p2 + 1)2

]=

12(sin t − t cos t),

using L−1

[p

(p2 + 1)2

]=

12

t sin t


(18) Prove that

L−1[e−

√p]

=e−

14t

2√

π t32

(19) Use convolution theorem to prove that∫ t

0J0 (τ) J0 (t − τ) dτ = sin t

(20) (a) If f(t) = H(t − π

2

)sin t, prove that L[f(t) ; t → s] = s

s2+1e−

πs2

(b) Prove that L[ | sin at| ; t → s] =a

s2 + a2coth

πs

2a, s > 0

(c) Prove that L

[d

dt(f ∗ g) ; t → s

]= g(0)f (s) + L[f ∗ g′(t) ; t → s] = sf(s) g(s)

(d) Show that f(t) = sin(a√

t) satisfies the ODE

4t f ′′(t) + 2f ′(t) + a2f(t) = 0

(e) Establish that L

[∫ ∞

t

f(x)x

dx ; t → s

]=

1s

∫ s

0f(x) dx

(f) Prove that L

[cos at − cos bt

t; t → s

]=

12

log(

s2 + a2

s2 + b2

)

(21) Find Laplace transform of the triangular wave function defined in(0, 2a) by

f(t) =

{t , 0 < t < a

2a − t , a < t < 2a[Ans. s−2 th

(as

2

)]

(22) Apply convolution theorem to show that

(a)∫ t

0sin u cos(t − u)du =

12

t sin t

(b) L−1

[1√

p(p − a)

]=

eat

√a

erf√

at and deduce that

L−1

[1p

1√p + a

]=

erf√

at√a


(23) Evaluate, by the method of residues that

(a) L−1

[p

(p + 1)3(p − 1)2

]

(b) L−1

[1

(p2 + 1)2

]

(c) L−1

[2p2 − 4

(p + 1)(p − 2)(p − 3)

]

(d) L−1[e−

√p]

(e) L−1

[ch x

√p

p cosh√

p

], 0 < x < 1

(f) L−1

[1p2

thπp

2

]

[Ans. (a)e−t

16(1 − 2t2) +

et

16(2t − 1)

(b)12t cos t +

12

sin t

(c)72

e3t − 43

e2t − 16

e−t

(d) t−32 e−

14t /(2

√π)

(e)4π

∞∑n=1

(−1)n

2n − 1e−(2n−1)2 π2t

4 cos(

n − 12

)π x + 1

(f) A periodic function f(t) =

{t , 0 < t < π

2π − t , π < t < 2πof period 2π]

(24) Solve the following equations by using Laplace transform :

(a) x′′(t) − 2x′(t) + x(t) = et , when x(0) = 2 , x′(0) = −1

(b) y′′(t) + 4y′(t) + 3y(t) = e−t , when y(0) = y′(0) = 1

(c) y′′(t) − 3y′(t) + 2y(t) = 4t + e3t , when y(0) = 1 , y′(0) = −1

(d) y′′′(t) + 2y′′(t) − y′(t) − 2y(t) = 0 , when y(0) = 1, y′(0) = 2,

y′′(0) = 2

(e) y′′(t) + 2y′(t) + 5y(t) = et sin t , when y(0) = 0, y′(0) = 1

(f) x′′(t) + 9x(t) = cos 2t , when x(0) = 1, x(π

2

)= −1

(g) (D3 − 3D2 + 3D − 1)y(t) = t2e2t,where y(0) = 1, y′(0) = 0,

y′′(0) = −2


(h) (D4 + 2D2 + 1)y(t) = 0,where y(0) = 0, y′(0) = 1, y′′(0) = 2,

y′′′(0) = −3

(i) (D3 − 2D2 + 5D)x(t) = 0, where x(0) = 0, x′(0) = 1, x(π

8

)= 1

(j) (D2 − 1) x (t) = a ch nt , where x(0) = x′(0) = 0,

(k) (D3 + 1) x (t) = 1 , x(0) = x′(0) = x′′(0) = 0

(l) x′′(t) − k2x(t) = f(t) , where x(0) = x′(0) = 0, k �= 0

(m) (D2 + 1)x(t) = f(t) , x(0) = x′(0) = 0

[ Ans. (a) x = 2et − 3t et +12

t2 et (b) y =74

e−t − 34

e−3t − 12t e−t

(c) y = 2t + 3 +13(e3t − et

)− 2e2t (d) y =13(5et + e−2t) − e−t

(e) y =113

e−t(sin t + sin 2t) (f) x =15(4 cos 3t + 4 sin 3t + cos 2t)

(g) y = (t2 − 6t + 12) e2t −(

32t2 + 7t + 11

)et

(h) y = t(sin t + cos t) (i) x = 1 − et(cos at − sin 2t)

(j) x =a(ch nt − ch t)

(n2 − 1)(k) x = 1 − 1

3e−t − 2

3e

t2 cos

√3t

2

(l) x =12k

[ekt

∫ t

0e−kuf(u) du − e−kt

∫ t

0ekuf(u) du

]

(m) x =∫ t

0sin(t − u)f(u) du ]

(25) Solve the following simultaneous ODE by using Laplace transform:

(a) x′(t) − 2x(t) + 3y(t) = 0, y′(t) + 2x(t) − y(t) = 0,

when x(0) = 8 , y(0) = 3

(b) (D2 − D) x (t) + y(t) = 0,(

D − 12

)x (t) + D y (t) = 0,

when x(0) = 0, y(0) = 1, x′(0) = 0

(c) D2 x (t) + y(t) = −5 cos 2t, D2 y (t) + x(t) = 5 cos 2t,

when x(0) = x′(0) = y′(0) = y′(0) = 1 , y(0) = −1

(d) x′(t) − y′(t) − 2x(t) + 2y(t) = 1 − 2t, x′′ (t) + 2y′(t)

+x(t) = 0, x(0) = x′(0) = y(0) = 0

(e) x′(t) + 5x(t) − 2y(t) = t, y′(t) + 2x(t) + y(t) = 0,


when x(0) = y(0) = 0

(f) (D2 − 1)x(t) + 5Dy(t) = t, −2Dx(t) + (D2 − 4) y (t) = 2,

if x(0) = y(0) = x′(0) = y′(0) = 0

(g) Dx(t) + Dy(t) = t, D2x(t) − y(t) = e−t,

if x(0) = 3, x′(0) = −2, y(0) = 0

(h) (D2 + 2) x (t) − Dy(t) = 1, Dx(t) + (D2 + 2) y (t) = 0,

if x(0) = x′(0) = y(0) = y′(0) = 0

[ Ans. (a) x(t) = 5e−t + 3e4t , y(t) = 5e−t − 2e4t

(b) x(t) =12(sh t − tet

), y(t) = ch t

(c) x(t) = sin t + cos 2t, y(t) = sin t − cos 2t

(d) x(t) = 2(1 − e−t − te−t

), y(t) = −t(1 + 2e−t)

+2(1 − e−t)

(e) x(t) = − 127

(1 + 6t)e−3t +127

(1 + 3t) ,

y(t) = −19(2 + 3t)e−3t +

127

(2 − 3t)

(f) x(t) = 5 sin t − 2 sin 2t − t, y(t) = 1 − 2 cos t + cos 2t

(g) x(t) = 2 +t2

2+

et

2− 3 sin t

2+

cos t

2,

y(t) = 1 − 12e−t − cos t

2+

3 sin t

2

(h) x(t) =16[3 − 2 cos t − cos 2t], y(t) =

16[−2 sin t + sin 2t] ]

(26) Solve (D2 + tD − 1)x(t) = 0, when x(0) = 0 , x′(0) = 1 using

Laplace transform to prove that x(t) = t + cL−1

[1p2 e

p2

2

]. Also

discuss for the particular solution when x(0) = 0.

[ Ans. x = t ](27) Using Laplace transform prove that x(t) = 1 + 2t is the particular

solution of the ODE x′′(t) − tx′(t) + x(t) = 1, when x(0) = 1,x′(0) = 2. Assume here that L−1(pn) = 0, for n = 0, 1, 2, 3, · · ·

(28) Using Laplace transform prove that x(t) = sin t/t is the solutionof the ODE with variable coefficients tx′′(t) + 2x′(t) + tx(t) = 0,if x(0) = 0.


(29) Prove, by using Laplace transform that y = (1 − t) e−t + 12 sin t is

the solution of the integro-differential equation given by

dy

dt+ 2y(t) +

∫ t

0y(u)du = sin t , y(0) = 1

(30) Solve the following integral/integro-differential equation :

(a) f(t) = e−t − 2∫ t

0cos(t − u)f(u)du

(b) 16 sin 4t =∫ t

0f(u)f(t − u)du

(c) f ′(t) = sin t +∫ t

0f(t − u) cos u du , if f(0) = 0

(d) f ′(t) = t +∫ t

0f(t − u) cos u du , if f(0) = 4

(e)∫ t

0f ′(t)f(t − u)du = 24t3 , if f(0) = 0

(f)∫ t

0f(u) cos(t − u)du = f ′(t), if f(0) = 1

[Ans. (a) f(t) = e−t(1 − t)2 (b) f(t) = ±8 J0(4t)

(c) f(t) =t2

2(d) f(t) = 4 +

52

t2 +t4

24

(e) f(t) = ±16 t32 /√

π (f) f(t) = 1 +t2

2]

(31) A mechanical system, with two degrees of freedom, satisfies theequations 2x′′(t) + 3y′(t) = 4 , 2y′′(t) − 3x′(t) = 0. Use Laplacetransform to determine x(t) and y(t), given that x(t), y(t), x′(t), y′(t)all vanish at t = 0.[Ans. x =

89

(1 − cos

32

t

), y =

89

(32

t − sin32

t

)]

(32) The co-ordinates (x, y) of a particle moving along a plane curve atany time t are given by

y′(t) + 2x(t) = sin 2t , x′(t) − 2y(t) = cos 2t , (t > 0)

If x(t = 0) = 1, y(t = 0) = 0, using Laplace transform show thatthe particle moves on the curve 4x2 + 4xy + 5y2 = 4


(33) Solve the differential system

dx

dt= Ax , x(0) =

[01

]where x =

[x1

x2

], A =

[0 1

−2 3

][Ans. x(t) =

[e2t − et

2e2t − et

] ]

(34) Solve the system

d2x1

dt2− 3x1 − 4x2 = 0 ,

d2x2

dt2+ x1 + x2 = 0 , t > 0 with

x1 = x2 = 0 ,dx1

dt= 2 ,

dx2

dt= 0 at t = 0

[Ans. x1 = 2t ch t, x2 = sh t − t ch t]

(35) Obtain the solution of tx′′(t) + x′(t) + a2x(t) = 0 , x(0) = 1in the form

x(t) = AL−1

[1√

s2 + a2; s → t

]= A J0(at)

(36) Solve the following differential equations under the given initialconditions :

(a)dX

dt= AX,X(0) =

[10

],where X =

[x

y

], A =

[1 −2

−2 1

]

(b)dx1

dt= x1 + 2x2 + t,

dx2

dt= x2 + 2x1 + t, x1(0) = 2, x2(0) = 4

(c)dx

dt= 2x − 3y,

dy

dt= y − 2x ; x(0) = 2, y(0) = 1

[Ans. (a) x(t) =12(e3t + e−t

), y(t) =

12(e3t − e−t

)(b) x1 =

289

e3t − e−t − t

3− 1

9, x2 =

289

e3t + e−t − t

3− 1

9

(c) x =15(7e−t + 3e4t

), y =

15(7e−t − 2e4t

)]

(37) The zero-order chemical reaction satisfies the initial value problem

dC(t)dt

+ k0 = 0 , t > 0 with C(0) = c0 at t = 0

where k0 is a positive constant and C(t) is the concentration of areacting substance at time t. Show that

C(t) = c0 − k0 t


(38) Using Laplace transform investigate the motion of a particlegoverned by the equations of motion

d2x

dt2−w

dy

dt= 0 ,

d2y

dt2+w

dx

dt= w2a under initial conditions

x(0) = y(0) =d

dt(x(0)) =

d

dty(0) = 0

[Ans. x(t) = a(wt − sin wt) , y(t) = a(1 − cos wt)]

(39) A weightless beam of length l has its ends at x = 0 and x = l.The beam is hinged at these two points. If a concentrated load W

acts at point x = l3 , show that the deflection of the beam at any

point x of it is given by

EI y(x) =W

81x (5l2 − 9x2) +

W

6

(x − l

3

)3

H

(x − l

3

),

on the assumption that the deflection y(x) satisfies the ODE

EI y(iv)(x) = Wδ

(x − l

3

), E, I being the elastic constants of the

beam.(40) The deflection of a beam of length l, clamped horizontally at both

ends and loaded at x = l4 by a weight W is given by

EI y(iv)(x) = Wδ

(x − l

4

). Find the deflection curve of the beam,

given that y(x) = y′(x) = 0, when x = 0, l.

[Ans. EI y(x) =W

6

(x − l

4

)3

H

(x − l

4

)+

9128

WIx2− 964

W 2x3. ]

(41) Find Laplace transform of(i) a square-wave periodic function of period a given by

f(t) =

{1, 0 < t < a

2

−1, a2 < t < a

(ii) a rectified semi-wave periodic function of period 2πw given by

f(t) =

{sin wt, 0 < t < π

w

0, πw < t < 2π

w⎡⎣Ans. (i)

1p

th(ap

4

)(ii)

w[(1 − e

−πpw

)(p2 + w2)

]⎤⎦


(42) Using Laplace transform solve the initial value problem defined by

(D2 + a2)2 x (t) = cos at, if x(0), x′(0), x′′(0), x′′′(0) all vanish.[Ans. x(t) =

18a3

[t sin at − at2 cos at

]]

(43) Solve the following ODE :

(a) ty′′(t) + (2t + 3)y′(t) + (t + 3)y(t) = a e−t , y(0) = 0 y′(0) =a

3(b) y′′(t) + at y′(t) − 2ay(t) = 1, y(0) = y′(0) = 0 , a > 0

[Ans. (a) y(t) =at e−t

3(b) y(t) =

t2

2]

(44) Solve the following initial value problems defined by

(a) x′′(t) + y′(t) + 3x(t) = 15e−t, y′′(t) − 4x′(t) + 3y(t) = 15 sin 2t,

x(0) = 35, x′(0) = −48, y(0) = 27 and y′(0) = −55

(b) x′(t) + 2y′′(t) = e−t , x′(t) + 2x(t) − y(t) = 1, x(0) = y(0) = 0

[Ans. (a) x(t) = 30 cos t − 15 sin 3t − 3e−t + 2cos 2t

y(t) = 30 cos 3t − 60 sin t − 3e−t + sin 2t

(b) x(t) = 1 + e−t − e−at − e−bt, y(t) = 1 + e−t − be−at − ae−bt

where a = 1 − 1√2

and b = 1 +1√2

]

(45) An alternating e.m.f. E sin wt is applied to an electrial networkconsisting of a constant inductance L and capacitance C connectedin series. If Q(t) be the charge at any time t and i(t) be the currentat that instant and they satisfy the ODE

LdQdt + Q

C = E sin wt , i(t) = dQdt with initial charge Q(0) = 0 and

initial current i(0) = 0, show that the current i(t) is given by

i(t) =Ew(cos wt − cos nt)

[ (n2 − w2) L ]where LC =

1n2

(46) (a) A beam which is clamped at the ends x = 0 and x = l, carriesa uniform load w0 per unit length. Show that the deflection y(x)at any point is given by

y(x) =[w0 x2(l − x)2]

(24 E I)


(b) If the beam in (a) be clamped at x = 0 but is free at x = l andcarries a uniform load w0 per unit length, show that its deflectionat any point x is given by

y(x) =(

w0 x2

24 E I

)(x2 − 4lx + 6l2)

(47) An impulsive voltage Eδ(t) is applied to a circuit consisting ofL,R,C in series with zero initial conditions. If I(t) be the currentat any subsequent time t, find the limit lim

t→0I(t) and justify your

answer. [ Ans. EL ]

(48) A semi-infinite transmission line of negligible inductance and leak-age per unit length satisfy the PDE ∂v

∂t = −Ri, ∂i∂x = −C ∂v

∂t wherev is the voltage and i is the current, R is the constant resistanceand C is the constant capacitance. If v0, the constant voltage, isapplied at the sending end x = 0 at t = 0, prove that the voltage

and current at any point are given by v(x, t) = v0 erfc

(x2

√RCt

)and i(x, t) = v0

√x

2

√CR t−3/2 e−(RCx2/4t).

(49) Using Laplace transform prove that the bounded solution of thePDE

(a) ∂y∂t = ∂2y

∂x2 − 4y under conditions y(0, t) = y(π, t) = 0 andy(x, 0) = 6 sin x−4 sin 2x is y(x, t) = 6e−5t sinx−4e−8t sin 2x

(b) ∂y∂x = y + 2∂y

∂t under condition y(x, 0) = 6e−3x is given byy(x, t) = 6 e−(3x+2t)

(50) A semi-infinite solid x > 0 is initially at temperature zero. At timet = 0, a constant temperature u0 > 0 is applied and maintained atthe face x = 0. Prove that at any point of the solid at a later timeis given by u0 erfc[x/2

√kt] provided the heat conduction equation

of the solid is given by ∂u∂t = k ∂2u

∂x2 , 0 < x < ∞ , t > 0 , k is thediffusivity of the material.

(51) The faces x = 0 and x = l of a slab of material for which k = 1 arekept at temperature zero and until the temperature distributionbecomes u(x, 0) = x. Prove that the temperature u(x, t) at asubsequent time is given by

u(x, t) =2π

∞∑n=1

(−1)n

ne−n2π2t sin nπx .


[ The following result may be used for inversion of Laplace trans-formation :

L−1

[sh x

√p

p sh√

p; p → t

]= x +

2π

∞∑n=1

(−1)n

ne−n2π2t sin(nπx) ] .

(52) An infinite string having one end at x = 0 is initially at reston the x-axis. The end x = 0 undergoes a periodic transversedisplacement A0 sin nt. Find the displacement of any point on thestring at t > 0. Assume that the displacement u(x, t) satisfies thewave equation ∂2u

∂t2 = c2 ∂2u∂x2 .

[Ans. u(x, t) = A0 sin n(t − x

c

)H(t − x

c

). ]

(53) Solve the BVP ∂2u∂t2

= a2 ∂2u∂x2 − g , x > 0 , t > 0 under conditions

u(x, 0) = ut(x, 0) = 0 , u(0, t) = 0 , limx→∞

ux(x, t) = 0 , t � 0 to

show that u(x, t) = 12 g(t − x

a

)2H(t − x

a

)− 12 g t2 .

(54) Prove that, the temperature u(x, t) in the semi-infinite mediumx > 0, when the end x = 0 is maintained at zero temperature andthe initial distribution of temperature is f(x), is given by

u(x, t) =2π

∫ ∞

0f(p) e−c2p2t sin xp dp .

where f(p) is Laplace transform of f(x).

(55) If the initial temperature of an infinite bar is given by

θ(x, 0) =

{θ0 , |x| < a

0 , |x| > a, determine the temperature at

any point x and at any instant t to prove that

θ(x, t) =θ0

2

[erf

(a + x

2c√

t

)+ erf

(a − x

2c√

t

)]

(56) Prove that y(x, t) = f(t − xc ) is the displacement of an infinitely

long string having one end at x = 0 and is initially at rest alongx-axis, when the end x = 0 is given a transverse displacementf(t) , t > 0 .


(57) Show that u(x, y) = L−12

[f(p,q)p+q ; (p, q) → (x, y)

]is the solution of

the BVP defined by

∂u

∂x+

∂u

∂y= f(x, y) in x � 0 , y � 0 and which vanishes on the

co-ordinate axes.

(58) Use the method of double Laplace transform to find the solutionin the positive quadrant (x � 0 , y � 0) of the PDE

∂2u(x, y)∂x∂y

+ u(x, y) = 0

if u(0, y) = u(x, 0) = 1.[Ans. u(x, y) = L−1

2

[2

p + q; (p, q) → (x, y)

]]

(59) Use eqn. (4.26) of article 4.4 to evaluate the double Laplace trans-form of (i) sin(ax + by) and of (ii) cos(ax + by).

(60) Solve the following difference equations with prescribed initial con-ditions :

(a) un+1 − 2 un = 0 , u0 = 1 [Ans. un = 2n]

(b) yk+2−5 yk+1+6 yk = 2, y0 = 0, y1 = 2 [Ans. yk = 1−2k+3k]

(c) 2 yk+1−yk−4 = 0 , y0 = 3[Ans. yk = 4 − (1

2

)k](d) u′(t) − α u(t − 1) = β, u(0) = 0[

Ans. u(t) = β

{t +

α(t − 1)2

Γ(3)+ α2 (t − 2)3

Γ(4)+ · · · + αn(t − n)n+1

Γ(n + 2)

}]

(e) Δ2un + 3un = 0 , u0 = 0 , u1 = 1 [Ans. un = n 2n . ]

(61) Prove that the general solution of the difference equation

un+2 − 4 un+1 + 3 un = 0

is un = A . 3n + B . 2n , where A = u1 − 2u0 and B = 3u0 − u1.

If u0 = 2 and u1 = −2 , find the general solution un.


(62) Solve the following integral equations :

(a) f(t) = sin 2t +∫ t

0f(t − τ) sin τ dτ

(b) f(t) =t

2sin t +

∫ t

0f ′′(τ) sin(t − τ)dτ , f (0) = f ′(0) = 0

(c)∫ t

0f(τ) J0{a(t − τ)} dτ = sin at

(d) f(t) = sin t +∫ t

0f(τ) sin(2t − 2τ)dτ

(e) f(t) = t2 +∫ t

0f ′(t − τ)e−aτdτ , f (0) = 0

[Ans. (a)12

(t +

32

sin 2t)

(b) (t − cos t) (c) a J0(at)

(d) 3 sin t −√

2 sin(√

2t) (e) t2 +2ta

]

(63) Using Laplace transform evaluate the following :

(a)∫ ∞

0

sin tx

x(x2 + a2)dx, (a, t > 0) (b)

∫ +∞

−∞

cos tx dx

x2 + a2, (a, t > 0)

(c)∫ +∞

−∞

x sin xt

x2 + a2dx , (a, t > 0) (d)

∫ ∞

0exp (−tx2)dx , (t > 0)

[Ans. (a)

π

2a2(1 − e−at) (b)

π

ae−at (c) 2π e−at (d)

√π

4t

]

(64) Solve the following difference equations using Laplace transform :

(a) Δun − 2un = 0 , u0 = 1

(b) Δ2un − 2un+1 + 3un = 0 , u0 = 0 , u1 = 1

(c) un+2 − 4un+1 + 4un = 0 , u0 = 1 , u1 = 4

(d) un+2 − 5un+1 + 6un = 0 , u0 = 1 , u1 = 4

(e) Δ2un + 3un = 0 , u0 = 0 , u1 = 0

[Ans. (a) un = 3n (b) un = n 2n−1 (c) un = (n + 1)2n

(d)un = 2(3n − 2n−1) (e) un = n2n]

Chapter 5

Hilbert and Stieltjes Transforms

5.1 Introduction.

Both Hilbert and Stieltjes transforms appear in many problems ofApplied Mathematics. Specially, Hilbert transform plays important rolein solving problems of fluid mechanics, electronics and signal processingetc. Also while solving problems of fracture mechanics through integralequation technique use of Stieltjes transform may be found useful dueto its simple inversion formula. At first this chapter deals with Hilberttransform.

5.2 Definition of Hilbert Transform

Let f(t) be defined in the real line −∞ < t < ∞. The Hilbert transformof f(t) denoted by fH(x), is defined by

fH(x) = H[f(t) ; t → x] =1π

∫ +∞

−∞

f(t)dt

t − x(5.1)

The integral on the right hand side of eqn. (5.1) is defined in the Cauchyprincipal value sense.

The inverse Hilbert transform is derived formally by use of Fouriertransform in the following steps.

Let us rewrite the formula in eqn. (5.1) as an integral equation ofthe convolution type for the determination of the function f(t) given by

fH(x) =1π

∫ +∞

−∞

f(t)dt

t − x≡ 1√

2π

∫ +∞

−∞f(t) g(x − t) dt (5.2)

where g(x) =√

2π

(− 1x

)and on the assumption that fH(x) is a known

function of x ∈ (−∞ , ∞).

Hilbert and Stieltjes Transforms 221

Since g(x) = −√

2π

x−1

its Fourier transform g(ξ) is given by

g (ξ) = − i sgn (ξ)

Now, taking Fourier transform of eqn.(5.2), it is found that

f=

H (ξ) = − i sgn (ξ) · f (ξ) , (5.3)

where f=

H(ξ) and f(ξ) are the Fourier transforms of fH(x) and f(x)respectively. Now, from eqn. (5.3) one gets

f(ξ) =−1

i sgn(ξ)f=

H(ξ) = i sgn(ξ) f=

H(ξ)

Taking Inverse Fourier transform we finally get

H−1[fH(x)

] ≡ f(x) = − 1π

∫ +∞

−∞

fH(t)dt

t − x, (5.4)

as the required inversion formula for Hilbert transform after reversingall the arguments used above.

For a rigorous proof of this formula one may be referred to the textbooks of Titchmarsh (Oxford Univ. Press , 1948) or Integral equationsby Tricomi (Q.J.Maths, Oxford, 2, 199, 1951).

5.3 Some Important properties of Hilbert Transforms.

From the definition of Hilbert transform in eqn. (5.1) and its inversionformula in eqn. (5.4) it can be derived in operator forms that

H−1 = −H (5.5)

Also, H[ f(t + a) ; t → x ] =1π

∫ +∞

−∞

f(t + a)dt

t − x

=1π

∫ +∞

−∞

f(y)dy

y − a − x=

1π

∫ ∞

−∞

f(y)dy

y − (a + x)

= H[ f(t) ; t → x + a ] = fH (x + a) (5.6)


H[ f(at) ; t → x ] =1π

∫ +∞

−∞

f(at)dt

t − x

=1π

∫ +∞

−∞

f(y)dy

a[ya − x]=

1π

∫ +∞

−∞

f(y)dy

y − (ax), (a > 0)

= H[ f(t) ; t → ax ] ≡ fH(ax) , (a > 0) (5.7)

Also, H[ f(−at) ; t → x ] =1π

∫ +∞

−∞

f(−at)dt

t − x

=1π

∫ +∞

−∞

f(y)dy

a[−y

a − x] , (a > 0)

=−1π

∫ +∞

−∞

f(y)dy

y − (−ax)= −H[ f(t) ; t → −ax ] , (a > 0)

≡ −fH(−ax) (5.8)

H[ f ′(t) ; t → x ] =1π

∫ +∞

−∞

f ′(t)dt

t − x

=1π

[ {1

t − xf(t)}t→+∞

t→−∞+∫ +∞

−∞

f(t)dt

(t − x)2

], integrating by parts

=1π

∫ +∞

−∞

f(t)dt

(t − x)2≡ d

dxfH(x), (5.9)

sinced

dx[ fH(x) ] =

d

dx

{1π

∫ +∞

−∞

f(t)dt

t − x

}

=1π

∫ +∞

−∞

f(t)(t − x)2

dt

and assuming the existence of all the integrals involved here in theCauchy principal value sense.

Finally, H[tf(t); t → x] =1π

∫ +∞

−∞

tf(t)t − x

dt =1π

∫ +∞

−∞

t − x + x

t − xf(t)dt

=1π

∫ +∞

−∞f(t)dt +

x

π

∫ +∞

−∞

f(t)dt

t − x= x fH(x) + (π)−1

∫ +∞

−∞f(t)dt

(5.10)


Theorem 5.1. If f(t) is an even function of t, then

fH (x) =x

π

∫ +∞

−∞

f(t) − f(x)t2 − x2

dt (5.11)

Proof. Since by Cauchy principal value sense∫ +∞

−∞

dt

t − x= 0,

we have fH (x) =1π

∫ +∞

−∞f(t)dt

t − x

=1π

∫ +∞

−∞

f(t) − f(x)t − x

dt

=1π

∫ +∞

−∞

(t + x)[ f(t) − f(x) ]t2 − x2

dt

=1π

∫ +∞

−∞

x[ f(t) − f(x) ]t2 − x2

dt +1π

∫ +∞

−∞

t[ f(t) − f(x) ]t2 − x2

dt

So, fH (x) =x

π

∫ +∞

−∞

f(t) − f(x)t2 − x2

dt

as because the second integral of the right-hand side is zero, theintegrand is being an odd function of t.

Example 5.1.

Find the Hilbert transform of

f(t) =

{1 , for |t| < a

0 , for |t| > a

Solution. We have by definition

fH (x) =1π

∫ +a

−a

dt

t − x

If |t| < a, the integrand has a singularity at t = x and hence

fH (x) =1π

lim∈→0

[∫ x−∈

−a

dt

t − x+∫ a

x+∈

dt

t − x

]


=1π

lim∈→0

[ log | ∈ | − log |(a + x)| + log |(a − x)| − log | ∈ | ], for |x| < a

=1π

log∣∣∣∣a − x

a + x

∣∣∣∣ , for |x| < a

If |t| > a, the integrand has no singularity in −a < t < a andtherefore,

fH (x) =1π

∫ a

−a

dt

t − x=

1π

[ log |t − x| ]a−a

=1π


a + x

∣∣∣∣ , for |x| > a

Thus for all x, in this case

fH (x) =1π


a + x

∣∣∣∣Example 5.2. Find the Hilbert transform of

f(t) =t

t2 + a2, a > 0


fH (x) =1π

∫ +∞

−∞

t dt

(t2 + a2)(t − x)

=1π

∫ +∞

−∞

1a2 + x2

[a2

t2 + a2+

x

t − x− xt

t2 + a2

]dt

=1

π(a2 + x2)

[a2

∫ +∞

−∞

dt

t2 + a2+ x

∫ +∞

−∞

dt

t − x− x

∫ +∞

−∞

t dt

t2 + a2

]

=1

π(a2 + x2)· (aπ) + 0 + 0 , by Cauchy principal value of the

second and the third integrals

=a

x2 + a2·

Example 5.3. Find the Hilbert transforms of

(i) f(t) = cos wt and

(ii) f(t) = sinwt.


Solution.

(i) From the definition of Hilbert transform, we get

fH (x) =1π

∫ +∞

−∞

cos wt

t − xdt

=1π

∫ +∞

−∞

cos{w(t − x) + wx}t − x

dt

=cos wx

π

∫ +∞

−∞

cos w(t − x)t − x

dt − sin wx

π

∫ +∞

−∞

sin w(t − x)t − x

dt

=cos wx

π

∫ +∞

−∞

cos wα

αdα − sin wx

π· (π)

= 0 − sin wx = − sinwx

(ii) It can similarly be shown that

fH (x) = cos wx.

5.4 Relation between Hilbert Transform and Fourier Trans-

form.

If we write

a(t) =1π

∫ +∞

−∞f(α) cos (αt) dα

b(t) =1π

∫ +∞

−∞f(α) sin (αt) dα (5.12)

Fourier integral can be expressed as

f(x) =12

∫ +∞

−∞[ a(t) + i b(t) ] e−ixt dt

=∫ ∞

0[ a(t) cos (xt) + b(t) sin (xt) ] dt

Let us now define a function of the complex variable z as

φ (z) =∫ ∞

0[ a(t) − i b(t) ] eizt dt , where z = x + iy

≡ U(x, y) + i V (x, y) , say,


We have then

limy→0

U(x, y) = f(x)

and also − limy→0

V (x, y) =∫ ∞

0[ b(t) cos (xt) − a(t) sin (xt) ] dt

Substituting the values of a(t) and b(t) from eqn. (5.12), we find that

− limy→0

V (x, y) =1π

∫ ∞

0dt

[∫ +∞

−∞f(α) sin(α − x)t dα

]

= limλ→∞

1π

∫ +∞

−∞

1 − cos λ(α − x)α − x

f(α) dα

= limλ→∞

1π

∫ ∞

0

1 − cos λ t

t{ f(x + t) − f(x − t)} dt

If f(t) is sufficiently smooth, we have

limλ→∞

∫ ∞

0cos λ t

f(x + t) − f(x − t)t

dt = 0

by the Riemann-Lebesgue lemma. Therefore,

− limy→0

V (x, y) =1π

∫ ∞

0

f(x + t) − f(x − t)t

dt (5.13)

≡ H [ f(t) ; t → x ] = fH (x) ·

This last step is obtained by changing the variable of integration in thesecond integral in eqn. (5.13) and combining it with the first integral.

5.5 Finite Hilbert Transform.

The finite Hilbert transform is defined by Tricomi (1951) as

H[ f(t) , a, b ] = fH (x, a, b) =1π

∫ b

a

f(t)t − x

dt (5.12)

when x ∈ (a < x < b). In studying the airfoil theory such transformdo arise in Aerodynamics. Also this transform is used in finding thesolution of singular integral equation of the type

1π

∫ 1

−1

f(t)t − x

dt = fH (x) , (5.13)


where fH (x) and f(t) are respectively known and unknown functionssatisfying Holder conditions on (−1, 1) of the variables. Eqn. (5.13) arisein boundary value problems in solid and in fracture mechanics. In thisrespect works of the authors like Muskhelishvili (1963), Gakhov (1966),Peters (1972), Chakraborty & Williams (1980), Comninou (1977), Das,Patra and Debnath (2004) and others may be referred to.

Several authors including Okikiolu (1965) and Kober (1967)introduced the modified Hilbert transform defined by

Hα [f(t)] = fHα (x) =cosec

(πα2

)2 Γ(α)

∫ +∞

−∞(t − x)α−1 f(t) dt ,

(5.14)

where 0 < α < 1 and x ∈ (−∞ , ∞) ·

The Parseval’s relation of this modified Hilbert transform is given inthe following theorem.

Theorem 5.2. If Hα [f(t)] = fHα (x) , then∫ +∞

−∞Hα [ f(t), x ] h(x) dx = −

∫ +∞

−∞Hα [ h(t), x ] f(x) dx (5.15)

Proof.

We do not persue with the proof of this theorem due to lack of spaceand scope of this treatise.

5.6 One-sided Hilbert Transform.

The two-sided Hilbert transform of f(t) in −∞ < t < x, defined ineqn (5.1), can be written as

fH (x) =1π

∫ +∞

−∞

f(t)dt

t − x

=1π

∫ ∞

0

f(t)t − x

dt − 1π

∫ ∞

0

f(−t)t + x

dt (5.16)

when x > 0 , where the second integral is actually the Stieltjes transformof [−f(−t)] to be defined in this chapter. A similar expression for fH (x)can also be written for x < 0. We now define one-sided Hilbert transform


of f(t) as

H+ [f(t)] = f+H (x) =

1π

∫ ∞

0

f(t)t − x

dt (5.17)

Further, Mellin transform of H+{f(t)}, to be defined in the next chapter,is given by

M [ H+{f(t)} ] =∫ ∞

0xp−1

[1π

∫ ∞

0

f(t)dt

t − x

]dx

=1π

∫ ∞

0f(t)

[∫ ∞

0

xp−1 dx

t − x

]dt

= cot (πp)∫ ∞

0xp−1 f(t) dt

= cot (πp) M{f(t)}

Taking inverse Mellin transform, we obtain

H+ [ f(t) ] = f+ (x) =1

2πi

∫ c+i∞

c−i∞x−p cot(πp) f(p) dp (5.18)

where f(x) is defined on 0 < x < ∞ and p may be a complex number.

5.7 Asymptotic Expansions of one-sided Hilbert Trans-

form.

The one-sided Hilbert transform of a function f(t) that is analytic for0 < t < ∞ and having the asymptotic expansion with known an, An, Bn

and w > 0 in the form

f(t) =∞∑

n=1

an

tn+ cos wt

∞∑n=1

An

tn+ sinwt

∞∑n=1

Bn

tn, as t → ∞, (5.19)

is defined as

H+ [ f(t) ; t → x ] = f+H (x) =

∫ ∞

0

f(t)t − x

dt (5.20)

Therefore, to find the asymptotic expansion of f+H (x) for such f given

in (5.19), we need to find the expansions of

P

∫ ∞

0

cos wt

t − xdt and P

∫ ∞

0

sinwt

t − xdt


Now, since P

∫ ∞

0

eiwt

t − xdt = πi (exp iwx) +

∫ ∞

0

eiwt

t − xdt ,

(5.21)

where the contour in the right hand side integral is being indented aboveat t = x. Deforming this contour into the positive imaginary axis bysetting t = iu , u > 0 we get

∫ ∞

0

eiwt

t − xdt = − i

∫ ∞

0

e−wu

x − iudu

= −∞∑

n=0

in+1

xn+1

∫ ∞

0un e−wu du , by Watson′s lemma

= −∞∑

n=0

n!(wx)n+1

exp{

(n + 1) iπ

2

}(5.22)

From (5.17) and (5.18), separating the real and imaginary parts we get

P

∫ ∞

0

cos wt

t − xdt ∼ −π sin wx −

∞∑n=0

n!(wx)n+1

cos{

(n + 1)π

2

}as x → ∞

and

P

∫ ∞

0

sinwt

t − xdt ∼ −π cos wx −

∞∑n=0

n!(wx)n+1

sin{

(n + 1)π

2

}as x → ∞

(5.23)

Then the asymptotic expansion of one-sided Hilbert transform of f(t)satisfying (5.19) is given by

f+H (x) =

∫ ∞

0

f(t)dt

t − x∼ −

∞∑n=1

dn

xn− log x

∞∑n=1

an

xn

−(π sin wx)∞∑

n=1

An

xn+ (π cos wx)

∞∑n=1

Bn

xnas x → ∞

(5.24)

where dn = M{f(t), n} =∫ ∞

0tn−1f(t) dt (5.25)

The detailed of the derivation of (5.24) and (5.25) is not taken up here.For interested readers, the paper entilled “Integrals with a large para-meter : Hilbert transforms (1983), Math Proc. Camb, Phil. Soc., 93,141-149” may be referred to.


5.8 The Stieltjes Transform.

From the definition of Laplace Transform we have

f(p) = L[ f(t) ; t → p ] =∫ ∞

0e−ptf(t) dt (5.26)

and so Laplace Transform f(p) in variable p to y is given by

L[ f(p) ; p → y ] =∫ ∞

0e−py f(p) dp

=∫ ∞

0e−py dp

∫ ∞

0f(t) e−pt dt , (5.27)

provided that the right hand side integral exists.

Now since,∫ ∞

0e−pt−py dp =

1t + y

we see that

L [ L{f(t) ; t → p} ; p → y ] =∫ ∞

0

f(t)t + y

dt (5.28)

If the right hand side integral of eqn. (5.28) is convergent for complexvalues of y belonging to some region Ω of y-plane then we representeqn. (5.28) as

f2 (y) =∫ ∞

0

f(t)t + y

dt , for y ∈ Ω

≡ S[ f(t) ; t → y ] (5.29)

This is called Stieltjes transform of f(t) for y ∈ Ω. In particular, f2 (y)is analytic in the complex y-plane with a branch cut along the negativey-axis.


5.9 Some Deductions.

Since,t

t + y= 1 − y

t + y, we have formally

S [ t f(t) ; t → y ] =∫ ∞

0f(t) dt − y f2 (y) (5.30)

Also since,1

(t + a)(t + y)=

1a − y

[1

t + y− 1

t + a

],we have

S

[f(t)t + a

; t → y

]=

1a − y

[ f2 (y) − f2 (a) ] (5.31)

Again, from the change of scale we can easily deduce that

S [ f(ax) ; x → y ] = f2 (y) , for a > 0 (5.32)

Also, putting x = 1u we can show that∫ ∞

0

1x f(

1x

)x + y

dx =∫ ∞

0

f(u)du

1 + uy

This result implies that

S[

x−1 f(x−1)

; x → y]

= y−1 f2

(y−1)

(5.33)

and using (5.32) in (5.33) it is obtained that

S[

x−1 f(a

x

); x → y

]= y f2

(a

y

)(5.34)

Substituting t = u2 it is deduced that

S [f(t

12

); t → y] = 2

∫ ∞

0

uf (u)u2 + y

du

Again since,2u

u2 + y=

1u + i

√y

+1

u − i√

y, we obtain

S [f(t

12

); t → y] = f2(i

√y) + f2(−i

√y) (5.35)

Stieltjes transform of the derivative of f(t) is given by

S [f ′(t) ; t → y] =∫ ∞

0

f ′(t)dt

t + y

=[

f(t)t + y

]∞0

+∫ ∞

0

f(t) dt

(t + y)2

= −1y

f(0) − d

dyf2(y) (5.36)


If |Arg a| < π and |Arg y| < π, we have from definition

S

[1

t + a; t → y

]=

1a − y

log∣∣∣∣ay∣∣∣∣ (5.37)

If S [ f(t) ; t → y ] =∫ ∞

0

f(t) dt

t + y, then

dn

dyn[ S{f(t) ; t → y} ] = (−1)n n!

∫ ∞

0

f(t) dt

(t + y)n+1, n = 1, 2, 3, · · ·

As a matter of fact Stieltjes transform can be looked upon as a repeatedLaplace transform and the calculation of Stieltjes transform of functionscan be found easily. For example, we know that

L

[1√t

e−at ; t → p

]=

√π (p + a)−

12

L[√

π (p + a)−12 ; p → y

]=

π√y

eay Erfc (√

ay)

These results imply that

S

[1√t

e−at ; t → y

]=

π√y

eay Erfc (√

ay) (5.38)

5.10 The Inverse Stieltjes Transform.


We describe below an informal proof of inverse Stieltjes transform.A rigorous proof of it may be found in the book of Widder (The LaplaceTransform, Princeton University Press, 1941).

Let us consider the contour integral of g(z) epz over an anticlockwisecontour C in the z-plane, consisting of a straight line γ from c − iR

to c + iR, the circular arc AB of large radius R, a branch cut at z = 0along the negative z-axis from B to B1, with a small circle γ1 : |z| = δ

and then from C1 to C and the circular arc CD of large radius R asshown in the above figure 1 where the function g(z) is assumed to beanalytic except for its branch singularities on the negative real z-axis.The function is also assumed to be such that

lim|z|→∞

g(z)epz = 0 , Re z � c , ( c is arbitrary ) (5.39)

lim|z|→0

z g(z) = 0 (5.40)

Then clearly, ∫C

g(z) epz dz = 0

This result implies that∫ c+iR

c−iRg(z) epz dz +

∫arcAB

g(z) epz dz +∫ δ

Rg(x eiπ) e−px(−dx)

+∮

γ1

g(z) epz dz +∫ R

δg(x e−iπ) e−px(−dx) +

∫arc CD

g(z) epz dz = 0

(5.41)

Making R → ∞ and δ → 0 when on γ1 : z = δ eiθ, we get from abovethat

limδ→0

∫γ1

g(z) epz dz = 0 , by eqn. (5.40) above

limR→∞

∫arcAB

g(z) epz dz = 0 and limR→∞

∫arc CD

g(z) epz dz = 0, by (5.39)

above

Therefore, eqn. (5.41) reduces to∫ c+i∞

c−i∞g(z) epz dz = −

∫ ∞

0g(x eiπ

)e−pxdx +

∫ ∞

0g(x e−iπ

)e−pxdx

=∫ ∞

0

[g(x e−iπ

)− g(x eiπ

) ]e−px dx


The last equation can be rewritten as

2πi L−1 [g(z) ; z → p]

= L[{

g(x e−iπ

)− g(x eiπ

)}; x → p

]Therefore,

g(z) = L[[

L (2πi)−1 {g (x e−iπ)− g

(x eiπ

)}; x → p

]; p → z

](5.42)

Again, it is known from double Laplace transform that

L [ L { f(x) ; x → p } ; p → y ] = S [ f(x) ; x → y ] (5.43)

Therefore, comparing eqns. (5.42) and (5.43), we can write

S−1 [ g(y) ; y → x ]

=1

2πi

{g(x e−πi

)− g(x eiπ

)}(5.44)

as the formula for inverse stieltjes transform

5.11 Relation between Hilbert Transform and Stieltjes

Transform.

Since Hilbert transform and Stieltjes transform are closely related weshall find a relation between them as below.

We know that

fH (x) = H [ f(t) ; t → x ] =1π

∫ +∞

−∞

f(t) dt

t − x,

where x is real and the integral is in the sense of Cauchy principal value.

If x > 0, we may write

fH (x) =1π

∫ ∞

0

f(t) dt

t − x+

1π

∫ 0

−∞

f(t) dt

t − x

=12π

∫ ∞

0

f(t) dt

t + x eiπ+

12π

∫ ∞

0

f(t) dt

t + x e−iπ− 1

π

∫ ∞

0

f(−t)(t + x)

dt


∴ fH (x) = (2π)−1 S[

f(t) ; t → x eiπ]

+ (2π)−1 S[

f(t) ; t → x e−iπ]− 1

πS [ f(−t) ; t → x ]

for x > 0 (5.45)

Similarly, when x < 0, we have x = −|x| and hence we have

fH (x) =1π

∫ ∞

0

f(t) dt

t + x+

1π

∫ 0

−∞

f(t) dt

t − x

= (π)−1 S [ f(t) ; t → |x| ] − 1π

∫ ∞

0

f(−t)t + x

dt

= (π)−1 S [ f(t) ; t → −x ] − 12π

∫ ∞

0

f(−t) dt

t + (−x eiπ)

− 12π

∫ ∞

0

f(−t) dt

t + (−x e−iπ)for x < 0 (5.46)

Excercises

(1) Find Hilbert transform of the following functions :

(a) f(t) =1

(a2 + t2), Re a > 0

(b) f(t) =tα

(t + a), | Re α | < 1

(c) f(t) = exp (−at)

(d) f(t) = t−α exp (−at) , Re a > 0 , Re α < 1

(e) f(t) =1t

sin (a√

t) , a > 0

(f) f(t) =sin t

t

[ Ans. (a) (a2 + x2)−1[πx

2a− log

(x

a

)](b) (a − x)−1(aα − xα) π cosec πα

(c) − exp (ax) Ei (−ax)

(d) Γ(1 − α) x−a exp (ax) Γ (α , ax)

(e)π

x[1 − exp(−a

√x)]

(f)(cos x − 1)

x]


(2) Find Stieltjes transform of the following functions.

(a) f(t) =tα−1

(t + a)

(b) f(t) =1

(t2 + a2)

(c) f(t) =t

(t2 + a2)

[ Ans. (a)(a − x)−1(xα−1 − aα−1

)π cosec (απ)

(b) (a2 + x2)−1[πx

2a− log

(x

a

)]

(c) (a2 + x2)−1[πa

2+ x log

(x

a

)]]

(3) Show that

(a) S

[1√t

cos a√

t ; t → y

]=

π√y

exp (−a√

x) , x > 0

(b) S[sin (k

√t) ; t → y

]= π exp (−k

√y) , k > 0

(4) Prove that

lim∈ → 0+

1π

∫ ∞

∈

[f(x + u) − f(x − u)

u

]du =

1π

P

∫ ∞

−∞

f(t)t − x

dt

(5) Prove that if f(t) = H(t), then

(a) [H{f(t − a) − f(t − b)} ; t → x] =1π

log∣∣∣∣ b − x

a − x

∣∣∣∣ , b > a > 0

(b) H

[f(t − a)

t; t → x

]=

1πx

loga

|a − x| , a > 0

(c) H

[f(t)

at + b; t → x

]=

1π(ax + b)

logb

a|x| ,

(a, b > 0 , x �= b

a

)

(d) H

[αt + βa

t2 + a2; t → x

]=

αa − βx

x2 + a2, a > 0


(6) Prove that

(a) S

[1

(a + x)(b + x); x → y

]=

⎧⎪⎪⎨⎪⎪⎩

(a−b) log(ay)−(a−y) log(a

b)

(a−y)(b−y)(a−b) , b �= a

a log( ya)−(a−y)

(a−y)2a , b = a

(b) S[xν−1 ; x → y

]= π

yν−1

sin ν π, 0 < ν < 1

(c) S[(a + x)−1 ; x → y

]= (a − y)−1 log

(a

y

)

(7) Show that f(t) = A

[Γ′( 1

2)√πt

− 2 log t√t

]is the solution of the integral

equation f(s) = − 1√π

∫∞0 e−st f(t) dt .

Chapter 6

Hankel Transforms

6.1 Introduction.

Hankel transform arises in the discussion of axisymmetric problems inthe cylindrical coordinates involving Bessel functions. This chapter dealswith the definitions, elementary properties, the inversion theorem andparseval relations of this transform and its relation with other trans-forms. Applications of this transform to various axisymmetric problemsin applied Mathematics are also discussed in this chapter.

6.2 The Hankel Transform.

The (infinite) Hankel transform of order ν of f(r) is formally definedand denoted by ∫∞

0 r f(r) Jν(ξr) dr ≡ Hν [ f(r); ξ ] ,

= fν(ξ),

⎫⎪⎬⎪⎭ (6.1)

if f(r) is integrable over (0,∞)

Theoretically, ν may be zero, integer or a non-integral number. Butin physical application ν is usually either zero or a positive integer.Accordingly, in discussions of elementary properties of Hankel transformν is taken as ν = n , n = 0 or a positive integer.

6.3 Elementary properties

Theorem 6.1. If f1(r) and f2(r) be two integrable functions in (0,∞)and c1, c2 are constants, then

Hn [ c1 f1(r) + c2 f2(r) ] = c1 Hn [ f1(r) ] + c2 Hn [ f2(r) ] (6.2)

Hankel Transforms 239

This property is called the linearity property of Hankel transform.

Proof. We have by definition,

Hn [ c1 f1(r) + c2 f2(r) ] =∫ ∞

0r [ c1 f1(r) + c2 f2(r) ] Jn (ξr) dr

= c1

∫ ∞

0f1(r) Jn (ξr) dr + c2

∫ ∞

0f2(r) Jn (ξr) dr

= c1 Hn [ f1(r) ; ξ ] + c2 Hn [ f2(r) ; ξ ] (6.2)

Theorem 6.2. If a is a constant and H [ f(r) ; ξ ] is Hilbert transformof f(r) of order ν, then

Hν [ f(ar) ; ξ ] =1a2

Hν

[f(r) ;

ξ

a

]=

1a2

fν

(ξ

a

)

Proof. By definition of Hankel transform in (6.1), we have

Hν [ f(ar) ; ξ ] =∫ ∞

0f(ar) · r Jν(ξr) dr

=∫ ∞

0f(x)

x

aJν

(ξ

ax

)1a

dx , taking ar = x

=1a2

∫ ∞

0x f(x) Jν

(ξ

ax

)dx

=1a2

Hν

[f(x) ;

ξ

a

]=

1a2

fν

(ξ

a

)(6.3)

This property of Hankel transform is known as the change of scaleproperty of the transform.

Theorem 6.3. If fn(ξ) be Hankel transform of f(r) of order n, thenHankel transform of f ′(r) of order n is given by

Hn

[f ′(r) ; ξ

]=

ξ

2n[

(n − 1)fn+1(ξ) − (n + 1)fn−1(ξ)], n � 1

(6.4)

H1

[f ′(r) ; ξ

]= −ξ f0 (ξ) (6.5)

provided [rf(r)] vanishes as r → ∞ and as r → 0.

Proof. We have

Hn

[f ′(r) ; ξ

]=∫ ∞

0rJn(ξr)f ′(r)dr

= [ rJn(ξr)f(r) ]∞0 −∫ ∞

0

d

dr{ rJn(ξr) } f(r) dr


Using the properties of Bessel function we have

d

dr[ rJn(ξr) ] = (1 − n)Jn(ξr) + ξrJn−1(ξr)

Therefore, if { rJn(ξr)f(r) } tends to zero as r → ∞ and also as r → 0,then

Hn

[f ′(r) ; ξ

]= −

∫ ∞

0{ (1 − n)Jn(ξr) + ξrJn−1(ξr) } f(r)dr

= (n − 1)∫ ∞

0f(r)Jn(ξr)dr − ξ

∫ ∞

0rf(r)Jn−1(ξr)dr

Again, using the recurrence relation

Jn(ξr) =ξr

2n[Jn−1(ξr) + Jn+1(ξr)]

of Bessel function, we get

Hn

[f ′(r) ; ξ

]= −ξfn−1(ξ) +

ξ(n − 1)2n

[∫ ∞

0rf(r)Jn−1(ξr) dr

+∫ ∞

0rf(r)Jn+1(ξr)dr

]

= −ξfn−1(ξ) + ξ(n − 1)

2nfn−1(ξ) + ξ

(n − 1)2n

fn+1(ξ)

= −(n + 1)2n

ξfn−1 (ξ) +(n − 1)

2nξ fn+1 (ξ)

Hn

[f ′(r) ; ξ

]=

ξ

2n[

(n − 1)fn+1(ξ) − (n + 1)fn−1(ξ)]

When n = 1 , it follows that

H1

[f ′(r) ; ξ

]= −ξ f0 (ξ)

Thus, the eqns. (6.4) and (6.5) are proved.

Repeated application of (6.4) gives

Hn

[f ′′(r) ; ξ

]=

ξ

2n[

(n − 1)f ′n+1(ξ) − (n + 1)f ′

n−1(ξ)]

=ξ2

4

[n + 1n − 1

fn−2(ξ) − 2n2 − 3n2 − 1

fn (ξ) +n − 1n + 1

fn+2 (ξ)]

(6.6)


Theorem 6.4. If fn(ξ) be Hankel transform of order n of f(r), thenHankel transform of order n of f ′′(r) + 1

r f ′(r) − n2

r2 f(r) is −ξ2fn(ξ),provided that rf ′(r) → 0 and rf(r) → 0 when r → 0 and when r → ∞Proof. From the definition of Hankel transform we have

Hn

[f ′′(r) +

1rf ′(r) − n2

r2f(r) ; ξ

]

=∫ ∞

0

{f ′′(r) +

1rf ′(r) − n2

r2f(r)

}r Jn (ξr) dr

Now, integrating by parts the first integral gives∫ ∞

0r

d2

dr2f(r)Jn (ξr) dr = −

∫ ∞

0

df (r)dr

Jn (ξr) dr

−∫ ∞

0rdf (r)dr

J ′n (ξr) dr

and therefore,∫ ∞

0

{d2f

dr2+

1r

df

dr

}r Jn (ξr) dr = −

∫ ∞

0

df

drr J ′

n (ξr) dr

=∫ ∞

0f(r)

d

dr

{r J ′

n(ξr)}

dr, on integrating by parts

Since, Jn(r) is Bessel function of order n, we have

d

dr

[r

d

drJn(r)

]+(

1 − n2

r2

)rJn(r) = 0

Here replacing r by ξr we get,

d

dr

[rJ ′

n(ξr)]

= −(

ξ2 − n2

r2

)rJn(ξr)

Using this result we get∫ ∞

0

{f ′′(r) +

1rf ′(r) − n2

r2f(r)

}rJn(ξr) dr = −ξ2

∫ ∞

0rf(r)Jn(ξr)dr

Or∫ ∞

0

{f ′′(r) +

1rf ′(r) − n2

r2f(r)

}rJn(ξr) dr = −ξ2fn(ξ) (6.7)

The last equation proves the required result under the given condition.


6.4 Inversion formula for Hankel Transform.

The valid proof for Hankel inversion theorem for general order ν is com-plicated. In application of Hankel transform to the solution of boundaryvalue problems in mathematical physics it is usually only the transformsof order zero and unity are involved. So we present here a proof of Hankelinversion theorem of non-negative order n.

Theorem 6.5. If fn(ξ) is Hankel transform of the function f(r)

implying fn(ξ) = H [ f(r) ; n ] =∫ ∞

0r f(r)Jn(ξr) dr

then f(r) =∫ ∞

0ξfn(ξ)Jn(ξr) dξ (6.8)

is called Hankel inversion formula of fn(ξ) and we express it as

f(r) = H−1[

fn(ξ) ; r]

(6.9)

Proof. Extending the results of general Fourier transform and its in-version formula of a function of one variable for function of two variables,we get

f(s, t) =∫ +∞

−∞

∫ +∞

−∞f(x, y) ei(sx+ty)dx dy (6.10)

f(x, y) =1

4π2

∫ +∞

−∞

∫ +∞

−∞f(s, t) e−i(sx+ty) ds dt (6.11)

Taking the following change of variables

x = r cos θ, y = r sin θ ; s = ξ cos α, t = ξ sin α

in eqns. (6.10) and (6.11) respectively, we get the transform pair as

f(ξ, α) =∫ ∞

0rdr

∫ 2π

0f(r, θ) eiξr cos(θ−α) dθ (6.12)

and f(r, θ) =1

4π2

∫ ∞

0ξdξ

∫ 2π

0f(ξ, α) e−iξr cos(θ−α) dα (6.13)

Let us now choose

f(r, θ) = f(r)e−inθ (6.14)

Then f(ξ, α) =∫ ∞

0f(r)r dr

∫ 2π

0ei{−nθ+ξr cos(θ−α) }dθ (6.15)


Now putting φ = α − θ − π

2, we get∫ 2π

0ei{−nθ+ξr cos(θ−α)} dθ = ein(π

2−α)∫ 2π

0ei(nφ−ξr sinφ) dφ

= ein(π2−α) · 2π Jn(ξr), since Jn(ξr)

=12π

∫ 2π

0ei(nφ−ξr sinφ) dφ

Therefore, eqn. (6.15) reduces to

f(ξ, α) =∫ ∞

0r f(r) ein(π

2−α) · 2π Jn(ξr) dr (6.16)

= 2π ein(π2−α) ·

∫ ∞

0r f(r) Jn(ξr) dr

= 2π ein(π2−α) · f(ξ) (6.17)

Thus using eqns, (6.14) and (6.17) in eqn. (6.13), we get

f(r) e−inθ =12π

∫ ∞

0

[ξ f(ξ) e−inθ · 2π Jn(ξr)

]dξ

= e−inθ

∫ ∞

0ξf(ξ)Jn(ξr) dξ

Or f(r) =∫ ∞

0ξ f(ξ) Jn(ξr) dr (6.18)

This result is the required inversion formula for the nth order Hankeltransformed function fn(ξ).

Note 6.1. For the case of general order ν > −12 if

√rf(r) is piecewise

continuous and absolutely integrable on the positive real line, Hankelinversion theorem is given by∫ ∞

0ξ fν(ξ) Jν(ξr)dξ =

12[f(r + 0) + f(r − 0)], (6.19)

where f ν(ξ) = Hν [f(r) ; ξ] .


6.5 The Parseval Relation for Hankel Transforms.

Theorem 6.6. The Parseval theorem.

Let f(ξ) and g(ξ) be Hankel transforms of the functions f(r) andg(r) respectively. Then∫ ∞

0r f(r) g(r) dr =

∫ ∞

0ξ f(ξ) g(ξ) dξ

Proof. It is known from eqn. (6.19) that

f(r) =∫ ∞

0f(ξ) ξ Jn(ξr) dξ

and also that g(ξ) =∫∞0 r g(r) Jn(ξr) dr from the definition of Hankel

transform.

Therefore,∫ ∞

0ξ f(ξ) g(ξ)dξ

=∫ ∞

0ξf(ξ)

[ ∫ ∞

0rg(r)Jn(ξr)dr

]dξ

=∫ ∞

0rg(r)

[ ∫ ∞

0ξf(ξ)Jn(ξr)dξ

]dr,

by interchanging the order of integration.

=∫ ∞

0rg(r)f(r)dr (6.20)

Thus the Parseval relation is formally established for the order n ofHankel transform.

Note 6.2. For the case of general order ν of Hankel transform, thecorresponding parseval relation is given by∫ ∞

0ξ f(ξ) g(ξ)dξ =

∫ ∞

0rf(r)g(r)dr (6.21)

where f(ξ) =∫ ∞

0rf(r)Jν(ξr)dr

and g(ξ) =∫ ∞

0rg(r)Jν(ξr)dr


6.6 Illustrative Examples.

Example 6.1.

Find Hankel transform of order zero of f(x) =

{1 , 0 < x < a

0 , x > a

Solution. We know from definition of Hankel transform that

H0 [f(x)] =∫ ∞

0xf(x)J0 (ξx) dx

=∫ a

0xJ0 (ξx) dx

=[

x

ξJ1(ξx)

]a

0

by the recurrence relationd

dx[ xnJn(x)] = xJn−1(x)

=a

ξJ1(ξa)

Example 6.2. Evaluate (a)H1

[e−ax

x ; ξ]

(b)H0

[1x

; ξ

]

Solution.

(a) We have, H1

[e−ax

x; ξ

]=∫ ∞

ox.

e−ax

xJ1 (ξx) dx

=∫ ∞

oe−axJ1 (ξx) dx

=(a2 + ξ2)

12 − a

ξ(a2 + ξ2)12

(b) H0

[1x

; ξ

]=∫ ∞

oJ0 (ξx) dx =

∫ ∞

oe−0xJ0 (ξx) dx

= (ξ2)−12 =

1ξ, since

∫ ∞

oe−axJ0(bx)dx =

1√a2 + ξ2

Example 6.3. Find Hankel transform of order zero of e−iax

x . and hence

evaluate H0

[cos ax

x , sin (ax)x

]



H0

[e−iax

x; ξ

]=∫ ∞

0(cos ax − i sin ax)J0(ξx) dx

=∫ ∞

0cos axJ0(ξx)dx − i

∫ ∞

0sin ax J0(ξx) dx

= Re(−a2 + ξ2)−12 − Im(ξ2 − a2)−

12

∴ H0

[ cos ax

x; ξ]

=

{(ξ2 − a2)−

12 , ξ > a

0 , 0 < ξ < a

and H0

[sin ax

x; ξ

]=

{0 , ξ > a

(a2 − ξ2)−12 , ξ < a

Example 6.4. If

f(x) =

{xn, 0 < x < a,

0, x > a

find Hankel transform of order n of f(x).

Solution. We have, by definition

Hn [ f(x) ] =∫ ∞

0xJn(ξx)f(x) dx

=∫ a

0xn. x Jn(ξx) dx

By recurrence relation of Bessel function we have

d

dx

[xn+1Jn+1(x)

]= xn+1Jn(x)

⇒ 1ξ

d

dx

[xn+1Jn+1(ξx)

]= xn+1Jn(ξx)

Therefore,

Hn [f(x)] =∫ a

0

1ξ

d

dx

[xn+1Jn+1(ξx)

]dx

=an+1

ξJn+1(ξa).


Example. 6.5. Find Hankel transform of order zero of[d2

dx2+

1x

d

dx

](e−ax

x

).

Solution. We know from (6.7) for n = 0 that∫ ∞

0

[f ′′(r) +

1rf ′(r)

]rJ0(ξr)dr = −ξ2f0(ξ)

Therefore,∫ ∞

0

[d2

dx2+

1x

d

dx

] (e−ax

x

). xJ0(ξx) dx

= −ξ2

∫ ∞

0e−axJ0 ( ξx ) dx

= − ξ2

(a2 + ξ2)12

.

Example. 6.6. Deduce that

Hν [xν(a2 − x2)μ−ν−1H(a − x) ; ξ] = 2μ−ν−1 Γ(μ − ν)aμξν−μJμ(ξa)

and hence prove that

Hν

[x ν−μ Jμ(ax); ξ

]=

ξν(a2 − ξ2) μ−ν−1

2 μ−ν−1 Γ( μ − ν) aμH(a − ξ).

Solution. From definition of Hankel transform we have

Hν

[xν(

a2 − x2)μ−ν−1

H(a − x) ; ξ]

=∫ a

0xν+1(a2 − x2)μ−ν−1Jν(xξ) dx

=∞∑

r=0

(−1)r( ξ2 )ν+2r

r! Γ(r + ν + 1)

∫ a

0x2ν+2r+1(a2 − x2)μ−ν−1 dx,

for μ > ν � 0 , a > 0 . (6.22)

The integral on the right hand side of this last equation has the value

12

a2ν+2r .Γ(r + ν + 1) Γ(μ − ν)

Γ (r + μ + 1)

and therefore,

Hν

[xν(a2 − x2)μ−ν−1 H(a − x) ; ξ

]


= 2μ−ν−1Γ(μ − ν)aμξν−μ∞∑

r=0

(−1)r(12ξa)μ+2r

r! Γ (r + μ + 1)

= 2μ−ν−1Γ(μ − ν)aμξν−μJμ(ξa)

(6.23)

Now, using Hankel inversion theorem we have if μ > ν � 0

2μ−ν−1 Γ(μ − ν)aμ Hν

[ξν−μJμ(ξa)

]= xν(a2 − x2)μ−ν−1 H(a − x)

This result in an alternative form can be expressed as

Hν [xν−μJμ(ax) ; ξ] = ξν(a2−ξ2)μ−ν−1

2μ−ν−1 Γ (μ−ν) aμ H(a − ξ)

⇒ ∫∞0 x1−μ+νJμ(ax)Jν (ξx) dx = ξν(a2−ξ2)μ−ν−1

2μ−ν−1 Γ (μ−ν) aμ H(a − ξ)

⎫⎪⎪⎬⎪⎪⎭ (6.24)

Example 6.7. Using eqns.(6.23) and (6.24) in example 6.6, deduce thecorresponding results for the following particular cases

(a) μ = ν + 1 , (b) μ = ν +12

, (c) ν = 0 , (d) ν =12

Solution. (a) Puttintg μ = ν + 1 in eqns. (6.23) and (6.24) we obtain

Hν [ xνH (a − x) ; ξ ] =aν+1

ξJν+1(ξa) , a > 0

Hν

[x−1Jν+1(ax) ; ξ

]=

ξν

aν+1H (a − ξ) , a > 0

(b) Putting μ = ν + 12 in the pair of eqns. (6.23) and (6.24) we get,

Hν

[xν(a2 − x2)−

12 H (a − x) ; ξ

]=√

π

2ξaν+ 1

2 Jν+ 12

(ξa)

Hν

[x− 1

2 Jν+ 12

(ax) ; ξ]

=

√2π

ξν H (a − ξ)

aν+ 12

√a2 − ξ2

, a > 0, ν � 0

(6.25)

(c) Taking ν = 0, the above two equations become

H0

[J1(ax)

x; ξ

]=

H (a − ξ)a

, a > 0 (6.26)

H0

[sin(ax)

x; ξ

]=

H (a − ξ)√a2 − ξ2

, a > 0, ξ < a.


(d) Similarly taking ν = 12 , the above pair of equations give∫ ∞

0

√x J1 (ax) J 1

2(xξ) dx =

√2π

√ξ

a2 − ξ2

H(a − ξ)a

Interchanging a and ξ, we have

H1

[sin(ax)

x; ξ

]=

aH (ξ − a)

ξ√

ξ2 − a2

Example 6.8. Prove that H0

[1 − J0(ax)

x2; ξ

]= log

a

ξ. H (a − ξ)

Solution. From eqn. (6.26) we can write∫ ∞

0J1 (ax) J0 (ξx) dx =

H (a − ξ)a

(6.27)

Here, using the results∫ a

0J1 ( ax ) da =

1 − J0 (ax)x

and∫ a

0

H (a − ξ)a

da = H (a − ξ) log(

a

ξ

),

we find on integrating (6.27) with respect to a from 0 to a that∫ ∞

0

1 − J1 (ax )x

J0 (ξx)dx = H(a − ξ) log(

a

ξ

)

⇒ H0

[x−2 { 1 − J0 (ax) } ; ξ

]= H(a − ξ) log

(a

ξ

).

Example 6.9. Using the result

L [ xνJν (xa) ; x → p ] =2 νaν Γ (ν + 1

2)√

π(p2 + a2)ν+ 12

,

prove that Hν

[xν−1 e−px ; ξ

]=

2 νξν Γ (ν + 12)

√π(p2 + ξ2)ν+ 1

2

Solution. From the definition of Hankel transform we know that

Hν

[xν−1 e−px ; ξ

]=∫ ∞

0x. xν−1 e−px Jν (ξx) dx

=∫ ∞

0{ xν . Jν (ξx)} e−px dx

= L [ xνJν(ξx) ; x → p ]

=2νξν Γ ( ν + 1

2)√

π (p2 + ξ2)ν+ 12

, by using the given condition.


Example 6.10. Evaluate Hν

[x−2Jν(ax) ; ξ

]when ν > −1

2

Solution. Consider the following two functions

f(x) = xνH(a − x) , g(x) = Hxν H(b − x) , ν = −12

Then they give, fν(ξ) = aν+1

ξ Jν+1(ξa), gν(ξ) = bν+1

ξ Jν+1 (ξb),

by using the result of Example 6.7(a). Therefore, using Parseval relationof f(x) and g(x) we have

(ab)ν+1

∫ ∞

0ξ−1 Jν+1(ξa)Jν+1(ξb) dξ =

∫ min (a,b)

0x2ν+1 dx

Now, suppose 0 < a < b. Then the last equation becomes∫ ∞

0ξ−1 Jν+1 (ξa) Jν+1 (ξb) d ξ =

12(ν + 1)

( a

b

)ν+1, ν > −1

2

⇒ Hν

[x−2 Jν(ax) ; ξ

]=

{12ν ( ξ

a)ν , 0 < ξ < a12ν (a

ξ )ν , ξ > a , ν > −12

Example 6.11. Find Hankel inversion of order 1 for the followingfunctions

(a)1ξ

e−aξ (b)1ξ2

e−aξ

Solution.

(a) H−1

[1ξ

e−aξ ; ξ → x

]=∫ ∞

0

e−aξ

ξ. ξ J1 (ξx) dξ

=∫ ∞

0e−aξ J1 (ξx) dξ

=1x− a

x (a2 + x2)12

, by table of integrals.

(b) H−1

[1ξ2

e−aξ ; ξ → x

]=∫ ∞

0

e−aξ

ξJ1 (aξ) dξ

=(a2 + x2)

12 − a

x, by example (a) above

Example 6.12. Find Hankel transform of order zero of 1x . Then apply

the inversion formula to get the original functions.


Solution.

H0

[1x

; ξ

]=∫ ∞

0

1x

. x J0 (ξx) dx

=∫ ∞

0J0 (ξx) dx =

1ξ, by table of integrals

Thus, H−10

[1ξ

; x

]=∫ ∞

0

1ξ

. ξ J0 (ξx) dξ

=∫ ∞

0J0 (ξx) dx =

1x

.

Example 6.13. Obtain zero-order Hankel transforms of

(a) r−1e−ar (b)δ(r)r

(c) H(a − r)

Solution.

(a) H0

[r−1e−ar ; ξ

]=∫ ∞

0e−ar J0 (ξr) dr =

1√ξ2 + a2

(b) H0

[δ(r)r

; r → ξ

]=∫ ∞

0δ(r) J0 (ξr) dr = 1

(c) H0 [H(a − r) ; ξ ] =∫ ∞

0rH(a − r)J0(ξr) dr =

∫ a

0rJ0 (ξr) dr

=1ξ2

∫ ar

0p J0 (p) dp =

1ξ2

[p J1(p)]aξ0 =

a

ξJ1(aξ).

Example 6.14. Find the nth order Hankel transforms for the functions

(a) rn H(a − r) (b) rne−ar2

Solution.

(a) Hn [ rn H(a − r) ; ξ ] =∫ ∞

0rn+1 H(a − r) Jn (ξr) dr

=∫ a

0rn+1 Jn(ξr) dr =

an+1

ξJn+1 (aξ)

(b) Hn

[rn e−ar2

; ξ]

=∫ ∞

0rn+1 e−ar2

Jn(ξr) dr

=ξn

(2a)n+1exp(− ξ2

2a

), from the table of integrals


Example 6.15. Solve the differential equation

∂2u

∂r2+

1r

∂u

∂r+

∂2u

∂z2= 0 , r � 0 , z � 0

satisfying the conditions (i) u → 0 as z → ∞ and as r → ∞,

(ii) u = f(r), on z = 0, r � 0.

Solution. Let u0(ξ, z) be Hankel transform of zero order of u(r, z).Then taking the zeroth order Hankel transform to

∂2u

∂r2+

1r

∂u

∂r+

∂2u

∂z2= 0

we get it as

−ξ2 u0 (ξ, z) +d2 u0

∂z2(ξ, z) = 0

Solving the last ODE we have

u0 (ξ, z) = A eξz + B e−ξz (6.28)

where A, B are constants to be calculated. Now Hankel transform oforder zero of given condition (i) becomes u0 (ξ , z) = 0 , as z → ∞ andhence (6.28) gives A = 0 implying

u0 (ξ , z) = B e−ξz (6.29)

Again, taking Hankel transform of order zero to the given condition(ii), it becomes

u0 (ξ , 0) = f0 (ξ) ≡ H0 [ f(r) ; r → ξ ] (6.30)

Then, eqn. (6.29.) by virtue of eqn. (6.30.) gives

B = f0 (ξ)

Thus, u (ξ , z) = f0 (ξ) · e−ξz

Applying Hankel inversion formula we have, therefore,

u(r , z) =∫ ∞

0f0 (ξ) e−ξz ξ J0 (ξr) dξ

as the required solution of the B V P.


Example 6.16. The vibration u(r, t) of a large membrane is given by

∂2u

∂r2+

1r

∂u

∂r=

1c2

∂2u

∂t2, r � 0 , t � 0

with the conditions

u(r, 0) = f(r) ,∂u(r, 0)

∂t= g(r).

Find u(r, t) for t > 0 .

Solution. Taking zeroth order Hankel transform, the given PDEgoverning the vibration becomes

− ξ2 u0 (ξ, t) =1c2

d2 u0

dt2(ξ, t)


u0 (ξ, t) = A cos (c ξ t) + B sin (c ξ t)

Let f0 (ξ) and g0 (ξ) be Hankel transform of order zero of the givenfunctions f(r) and g(r) respectively.

Then from the given conditions

u0 (ξ, 0) = f0 (ξ) =∫ ∞

0f(r) r J0 (ξr) dr (6.31)

and∂

∂ tu0 (ξ, 0) = g0 (ξ) =

∫ ∞

0g (r) rJ0 (ξr) dr (6.32)

From the general solution and (6.31) we get

f0(ξ) = A

and hence the general solution and (6.32) finally give

g0(ξ) = Bcξ

Therefore, the solution u0(ξ, t) ultimately becomes

u0 (ξ, t) = f0 (ξ) cos (c ξt) +g0 (ξ)c ξ

sin (cξt)

Now, applying the inversion formula of Hankel transform of order zerowe get

u (r, t) =∫ ∞

0

[f0 (ξ) cos (c ξt) +

g0 (ξ)c ξ

sin (c ξ t)]

ξ J0 ( ξ r) dr


Example 6.17. Consider the axisymmetric Dirichlet problem for a

thick plate |z| � b determined by the function u(ρ, z) which is harmonicand satisfies the boundary conditions.

u(ρ, b) = f(ρ) , u(ρ,−b) = g(ρ) (6.33)

Find the solution u(ρ, z). If f(ρ) = g(ρ), evaluate u(ρ, 0) and∂

∂zu(ρ, 0)

Solution. Since u(ρ, z) satisfies the PDE[∂2

∂ρ2+

1ρ

∂

∂ρ+

∂2

∂z2

][u(ρ, z) = 0 (6.34)

applying zeroth order Hankel transform to (6.34), we get

d2

dz2uo (ξ, z) − ξ2 u0 (ξ, z) = 0

The appropriate solution of this equation is

u0 (ξ, z) = A (ξ) sinh {ξ (b + z)} + B (ξ) sinh { ξ (b − z)} (6.35)

Now, Hankel transformed boundary conditions in eqn. (6.33) are

u0(ξ, b) = f(ξ) and u0(ξ,−b) = g0 (ξ)

Using these conditions, the solution results in

A (ξ) sinh (2 ξ b) = f0 (ξ)

and B (ξ) sinh (2ξb) = g0 (ξ) .

Thus the solution in (6.35) takes the form

u0 (ξ, z) = cosech (2 ξ b)[f0 (ξ) sinh {ξ(z + b)} + g0 (ξ) sinh { ξ (b − z)}]

Applying Hankel inversion theorem we obtain the solution as

u(ρ , z) = H0

[f0 (ξ)

sinh ξ(b + z)sinh (2 ξ b)

+ g0 (ξ)sinh ξ(b − z)sinh (2 ξ b)

; ξ → ρ

](6.36)

If, in particular, f(ρ) = g(ρ) ⇒ f0(ξ) = g0(ξ) , then

u(ξ , z) = f0(ξ)cosh (ξz)cosh (ξb)


and the solution is

u(ρ , z) = H0

[f0(ξ)

cosh(ξz)cosh(ξb)

; ξ → ρ

](6.37)

From (6.36) we can derive the general formulae

u(ρ, 0) = H0

[12{f0 (ξ) + g0 (ξ)} sech (ξb) ; ξ → ρ

]∂u

∂z(ρ, 0) = H0

[12ξ{f0 (ξ) − g0 (ξ)}cosech (ξb) ; ξ → ρ

]

Also, from (6.37) we can derive the particular solution

u(ρ, 0) = H0 [f0 (ξ) sech (ξb) ; ξ → ρ]∂u(ρ, 0)

∂z= 0

Example 6.18. Consider the transverse displacement z(ρ, t) of a largemembrane which satisfies the non-homogeneous wave equation

∂2z

∂ρ2+

1ρ

∂z

∂ρ=

1c2

∂2z

∂t2− p (ρ, t)

T(6.38)

where T is the tension in the membrane, p (ρ, t) is the symmetricalpressure applied normally to the membrane and c2 = T/σ, where σ isthe density per unit area of the membrane.

If the initial conditions are

z (ρ , 0) = f(ρ) ,∂z(ρ, 0)

∂t= g(ρ), (6.39)

find the solution of the problem in the Hankel transformed domain.

Discuss the particular case of the problem if there is no appliedpressure and if the initial displacement is ∈ (1 + ρ2/a2)−1/2 to themembrane.

Solution. Introducing Hankel transform of order zero, the equation(6.38) becomes

d2z (ξ, t)dt2

+ c2 ξ2 z (ξ, t) =p0(ξ, t)

σ(6.40)

where, p0 (ξ, t) = H0 [ p (ρ, t) ; ρ → ξ ]


Let Hankel transform of order zero of the initial conditions in (6.39)be given by

z (ξ, 0) = f0 (ξ) ,d

dtz(ξ, 0) = g0 (ξ) (6.41)

Then, the solution of (6.40) under the initial conditions (6.41) is givenby

z (ξ, t) = f0 (ξ) cos (c t ξ) + (c ξ)−1 g0 (ξ) sin c t ξ

+(c σ ξ)−1

∫ t

0p0 (ξ, τ) sin {c ξ (t − τ)} dτ (6.42)

Particular case.

In this case p0 (ξ, t) = 0 and g0 (ξ) = 0. Also f0 (ξ) =∈ a ξ−1 e−ξa.Then the solution in eqn. (6.42) becomes

z (ξ, t) = ∈ a ξ−1 e−ξa cos (c t ξ)

Now, using the inversion theorem of Hankel transform of order zero tothe above transformed solution, we get

z (ρ, t) = ∈ a H0

[ξ−1 e−ξa cos (c t ξ) ; ξ → ρ

]= ∈ a Re

[∫ ∞

0e−ξ(a+ict) J0 (ξρ) dξ

]

= ∈ a Re[ρ2 + (a + ict)2

]− 12

= ∈ a R−1 cos ϕ

where R and ϕ are given by

R4 = (a2 + ρ2 − c2t2)2 + 4a2c2t2

and tan 2ϕ =(2 act)

(a2 + ρ2 − c2t2).

Example 6.19. Find the potential V (r, z) of a field due to a flatcircular disc of unit radius with its center at the origin, axis along thez-axis and satisfying the differential equation

∂2V

∂r2+

1r

∂V

∂r+

∂2V

∂z2= 0 , 0 � r < ∞ , z � 0 (6.43)

and the boundary conditions

V (r, z) = V0, when z = 0, 0 � r < 1 (6.44)

and∂V (r, z)

∂z= 0, when z = 0, r > 1 (6.45)


Solution. Introducing Hankel transform of order zero, eqn. (6.43)becomes

d2V 0

dz2− ξ2 V 0 = 0


V 0 (ξ, z) = A(ξ) eξz + B(ξ) e−ξz

where A(ξ) , B(ξ) are arbitrary functions of ξ .

On physical grounds, V 0 (ξ, z) → 0 as ξ → ∞, since V (r, z) → 0as r → ∞ . Therefore, A(ξ) = 0 implying that

V 0 (ξ, z) = B(ξ) e−ξz

Applying Hankel inversion formula, we get

V (r, z) =∫ ∞

0B(ξ) ξ e−ξz J0(ξr) dξ (6.46)

and∂V (r, z)

∂z= −

∫ ∞

0B(ξ) ξ2 e−ξz J0(ξr) dξ (6.47)

Now, the boundary conditions in eqns. (6.44) and (6.45) can beexpressed as∫∞

0 ξ B(ξ) J0 (ξr) dξ = V0 , 0 � r < 1

and∫∞0 ξ2 B(ξ) J0 (ξr) dξ = 0 , r > 1

⎫⎪⎬⎪⎭ (6.48)

after utilising eqns. (6.46) and (6.47).

But it is known that∫ ∞

0J0 (ξr)

sin ξ

ξd ξ =

π

2, 0 � r < 1

and∫ ∞

0J0 (ξr) sin ξ d ξ = 0 , r > 1

Therefore, comparing the last two equations with the eqns. in (6.48),we get

B(ξ) =2V0 sin ξ

πξ2

So, V (r, z) =2V0

π

∫ ∞

0e−ξz sin ξ

ξJ0(ξr) dξ.


Exercises.

(1) Find Hankel transform of order 1 of x−2 e−x.[Ans.

�(1+ξ2)

12 −1

�

ξ

]

(2) If f(x) =

{a2 − x2 , 0 � x < a,

0 , x > a

calculate Hankel transform of f(x) of order zero.

[Ans. 4a

ξ3 J1(aξ) − 2a2

ξ2 J0(ξa)]

.

(3) (a) Prove that H0

[dfdx ; x → ξ

]= −ξ

√a2 + ξ2, if f(x) = 1

x e−ax.

(b) Prove that H0

[δ(r)r ; r → ξ

]= 1 .

(c) Prove that H1 [e−ar ; r → ξ] = ξ(a2+ξ2)

.

(d) Prove that Hn

[rn e−ar2

; r → ξ]

= ξn

(2a)n+1 e−r2

4a .

(4) Show that the solution of the boundary value problem defined by

∂2u

∂r2+

1r

∂u

∂r+

∂2u

∂z2= 0 , 0 � r < ∞ , z > 0 ,

u(r, 0) = u0 for 0 � r < a , where u0 is a constant andu(r, z) → 0 as z → ∞ is

u(r, z) = a u0

∫ ∞

0J1 (aξ) J0 (ξr) e−zξ dξ

(5) Solve the Neumann problem for the Laplace equation

urr +1r

ur + uzz = 0 , 0 < r < ∞ , 0 < z < ∞ when

uz(r, 0) = − 1πa2

H(a − r) , 0 < r < ∞ , u(r, z) → 0 as z → ∞

to prove that u(r, z) =1πa

∫ ∞

0

1ξ

J1(aξ) J0(rξ) e−ξz dξ

(6) Heat is supplied at a constant rate Q per unit area per unit timeover a circular area of radius a in the plane z = 0 to an infinite solid


of thermal conductivity k, the rest of the plane is being kept atzero temperature. Solve the steady state heat conduction problemto find the temperature field u(r, z) satisfying the equation

urr + 1rur + uzz = 0 , 0 < r < ∞ , −∞ < z < ∞

with the boundary conditions u → 0 as r → ∞, u → 0 as|z| → ∞ and −k uz = 1

2 Q H(a − r) when z = 0. Finally, provethat

u(r, z) = Qa2k

∫∞0 ξ−1 exp (−|z|ξ) J1(aξ) J0(rξ) dξ .

Chapter 7

Finite Hankel Transforms

7.1 Introduction.

Finite Hankel transform arises in discussing the solution of certain specialtype of boundary value problems. Sneddon (Phil. Mag., 37, 1946) wasthe first author who introduced this transform. Later application of thistransform was found in the works of several other authors in discussingsolutions of axisymmetric physical problems in long circular cylindersand membranes.

7.2 Expansion of some functions in series involving cylin-

der functions : Fourier-Bessel Series.

Suppose f(r) is a real piecewise continuous function in (0, a) and is ofbounded variation in every subinterval [r1, r2] where 0 < r1 < r2 < a.

Then, if the integral∫ a0

√r |f(r)|dr is finite, a Fourier series like series

expansion to f(r) represented by

f(r) =∞∑

m=1

cm Jν

(xνm

r

a

), ν � −1

2(7.1)

where xνm is given by Jν (xνm ) = 0,m = 1, 2, 3, · · · do exist at everypoint of continuity of f(r) and to

12

[ f(r + 0) + f(r − 0) ] (7.2)

at every point of discontinuity of f(r).

The co-efficient cm of the expansion of f(r) in (7.1) can be deter-mined by using the orthogonality property of the system of functions

Jν

(xνm

r

a,)

m = 1, 2, . . . . (7.3)

Finite Hankel Transforms 261

To prove this assertion, let there exist two distinct non zero real numbersα, β such that Jν (α r) , Jν (β r ) are the Bessel functions of first kind.

Then we would have[d 2

dr2+

1r

d

dr+(

α2 − ν2

r2

) ]Jν ( α r) = 0[

d 2

dr2+

1r

d

dr+(

β2 − ν2

r2

) ]Jν ( β r) = 0

Now, subtracting the second equation multiplied by r Jν (α r) fromthe first equation multiplied by r Jν (β r) , and integrating the resultwith respect to r from r = 0 to r = a, we get∫ a

0r Jν (αr) Jν (βr) dr

=[aβ Jν (aα) J ′

ν ( aβ) − aα Jν (βa) Jν′ (αa)]

(α2 − β2)(7.4)

Setting, aα = xνm and aβ = xνn we get from equation (7.4) that∫ a

0r Jν (αr) Jν (βr) dr = 0, if m �= n.

When m = n, taking the limit of eqn. (7.4) as β → α we get afterelimination of J ′′

ν that∫ a

0r J2

ν (αr) dr =a2

2

[J ′ 2

ν (αa) +(

1 − ν2

a2α2

)J2

ν (αa)]

(7.5)

Or,∫ a

0r J2

ν (xmr

a) dr =

a2

2J

′2ν (xνn) (7.6)

=a2

2J2

ν+1 (xνn) (7.7)

Thus assuming the possibility of an expansion of f(r) of the form (7.1),multiplying it by r Jν(xνn

ra) and integrating term by term from r = 0

to r = a , we obtain

cm =2

a2 J2ν+1 (xνm)

∫ a

0rf(r) Jν

(xνm

r

a

)dr, m = 1, 2, · · · (7.8)

The series (7.1) with co-efficients calculated from eqn.(7.8), is called theFourier-Bessel series of f(r).

We are now ready to define finite Hankel transform in the followingsection.


7.3 The Finite Hankel Transform.

Form I If f(r) satisfies the conditions laid down in article 7.1 overthe finite interval 0 � r � a, then finite Hankel transform of order n isdenoted by Hn[f(r)] and is defined by

Hn [ f(r) ] =∫ a

0r f(r) Jn (rpi) dr ≡ fn (pi) (7.9)

where pi is a positive root of the equation

Jn (api) = 0 (7.10)

The function f(r) can be considered as inverse finite Hankel transformof fn(pi) and it is denoted by

f(r) = H−1n

[fn (pi)

]and is defined by

f(r) =2a2

∞∑i=1

fn (pi)Jn (rpi)

J2n+1 (api)

(7.11)

where pi (0 < p1 < p2 < · · ·) are the roots of the equation Jn(api) = 0.This means that

J ′n (api) = Jn−1 (api) = −Jn+1 (api)

due to standard recurrence relation of Besel function. Here, the abbre-viation api is used instead of xνi as was used in article 7.2.

As a particular case, zeroth order finite Hankel transform of f(r)and its inversion are then defined by

H0 [ f(r) ] ≡ f0 (pi) =∫ a

0r f(r) J0 (rpi) dr (7.12)

H−10 [ f0 (pi) ] ≡ f(r) =

2a2

∞∑i=1

f0 (pi)J0 (rpi)J2

1 (api)(7.13)

where the summation is taken over all the positive roots of J0(ap) = 0.Similarly, finite Hankel transforms of other different orders of f(r) canbe defined as shown above.



Example 7.1. Find finite Hankel transform of order n of f(r) = rn,for 0 � r � a.


d

dx[xn Jn (x)] = xn Jn−1 (x)

Therefore,d

dr

[rn+1 Jn+1 (ki r)

]= ki rn+1 Jn (ki r)

and hence Hn[rn] =∫ a

0rn · r Jn (ki r) dr

=1ki

∫ a

0

d

dr

[rn+1 Jn+1 (ki r)

]dr

=an+1

kiJn+1 (ki a)

In particular if n = 0, then

Hn[ 1 ] =a

kiJ1 (aki) .

Example 7.2. Find finite Hankel transform of (a2 − r2) of order zerofor 0 � r � a.

Solution. We have

H0

[(a2 − r2)

]=∫ a

0r (a2 − r2) J0 (r ki) dr

=∫ a

0a2r J0 (rki) dr −

∫ a

0r3 J0 (rki) dr

= a2

∫ a

0

1ki

.d

dr{ rJ1 (kir) } dr

−∫ a

0r2 1

ki.d

dr{ rJ1 (kir) } dr

=a2

ki[ rJ1 ( kir ) ]a0 −

[r2

kir J1 (kir)

]a

0

+∫ a

0

2rki

rJ1 (ki r) dr

=a3

kiJ1 (ki a) − a3

kiJ1 (ki a) +

2k2

i

[r2 J 2 (ki r)

]a0

=2a2

k2i

J2(ki a)


=2a2

k2i

[2

ki aJ1(ki a) − J0 (ki a)

],by recurrence relation

=4ak3

i

J1 (ki a) − 2a2

k2i

J0 (ki a) .

Example 7.3. Prove that∫ 10

Jn (α x)Jn (α) . x Jn (ki x) dx = ki

α2−k2i

J ′n (ki),

where ki is the positive root of Jn (ki) = 0.

Solution. The Bessel function Jn(x) satisfies the ODE[x2 d2

dx2+ x

d

dx+ (x2 − n2)

]Jn(x) = 0 (i)

Replacing x by ki x and α x successively in (i), we get

x2 d2

dx2Jn (ki x) + x

d

dxJn (ki x) + (k2

i x2 − n2) Jn (ki x) = 0

and x2 d2

dx2Jn (α x) + x

d

dxJn (α x) + (α2x2 − n2) Jn (α x) = 0

Multiplying the first equation by Jn(α x) and the second by Jn (ki x)and then subtracting, we get

xd

dx

[x

{Jn (α x)

d

dxJn (ki x) − Jn (kix)

d

dxJn (α x)

}]−x2 (α2 − k2

i ) Jn (ki x) Jn (αx) = 0.

Or xJn (kix) Jn (α x)

=1

α2 − k2i

d

dx[x

{Jn (α x)

d

dxJn (ki x) − Jn (ki x)

d

dxJn (αx)

} ]

∴∫ 1

0

Jn (α x)Jn (α)

xJn (ki x) dx

=1

α2 − k2i

1Jn (α)[

ki Jn (α)J ′n (ki) − α Jn (ki) J ′

n (α)]

=ki

α2 − k2i

J ′n (ki) , since Jn (ki) = 0

Example. 7.4. Prove that finite Hankel transform of order n of

f(x) =21+n−m

Γ (m − n).xn (1 − x2)m−n−1 , 0 < x < 1,

is pn−m Jn (p).


Solution. By definition

Hn [f (x)] =∫ 1

0

21+n−m

Γ (m − n)xn (1 − x2)m−n−1 xJn (px) dx

=∞∑

r=0

∫ 1

0

21+n−m

Γ (m − n)xn+1 (1 − x2)m−n−1 (−1)r

r!Γ (n + r + 1)(p x)n+2r

2n+2rdx

∴ Hn [f(x)] =1

Γ(m − n)

∞∑r=0

(−1)r pn+2r

Γ(n + r + 1)2m+2r

∫ 1

0y(n+r+1)−1

(1 − y)m−n−1 dy

=1

Γ(m − n)

∞∑r=0

(−1)r pn+2r

Γ(n + r + 1)2m+2r.Γ (n + r + 1) Γ (m − n)Γ (n + r + 1 + m − n)

= pn−m∞∑

r=0

(−1)r

Γ(m + r + 1)

(p

2

)m+2r

= pn−m Jm (p)

Example. 7.5. Find H−1[

cp J1 (ap)

], where p is a positive root of

J0 (ap) = 0.Solution. If f(r) be the required function such thatH [f(r)] = fH (p) = c

p J1 (ap), then from equation (7.13) we have

[2a2

∑p

{c

pJ1 (ap)

}J0 (p r)J2

1 (p a)

]= f(r)

Or2ca2

∑p

J0 (pr)J1(p a)

= f(r)

7.5 Finite Hankel Transform of order n in 0 � x � 1 of

the derivatrive of a function.

For 0 � x � 1. Hankel transform of first order derivative df(x)dx of order

n is given by

Hn

[df

dx

]=∫ 1

0

df

dx.x Jn (px) dx , where p is a root ofJn(p) = 0

= [ f(x) . xJn (px) ]10 −∫ 1

0f(x)

d

dx{ x Jn (px) } dx


= −∫ 1

0f(x).

[px

2n{(n + 1) Jn−1(px) − (n − 1)Jn+1(px)}

]dx

=p

2n

[(n − 1)

∫ 1

0f(x) x Jn+1(px) dx

−(n + 1)∫ 1

0f(x) x Jn−1(px) dx

]

=p

2n[ (n − 1) Hn+1{f(x)} − (n + 1) Hn−1{f (x)} ] (7.14)

Corollary. 7.1. Putting n = 1 in eqn. (7.14), we get

H1

[df

dx

]= − p Ho [f(x)] (7.15)

7.6 Finite Hankel Transform over 0 � x � 1 of order n ofd2fdx2 + 1

xdfdx

, when p is the root of Jn(p) = 0.

By definition of Hankel transform of order n in 0 � x � 1, we have

Hn

[d2f

dx2+

1r

df

dx

]=∫ 1

0

[d2f

dx2+

1x

df

dx

]x Jn (px) dx

=∫ 1

0x f ′′(x) Jn (px) dx +

∫ 1

0f ′(x) Jn (px) dx

=[x Jn(px) f ′(x)

]10−∫ 1

0

d

dx{x Jn(px)}f ′(x) dx +

∫ 1

0f ′(x) J ′

n(px)dx

= −∫ 1

0px J ′

n(px) f ′(x) dx

= −p

2

∫ 1

0x f ′(x)[Jn−1(px) − Jn+1(px)] dx ,

byd

dxJn(px) =

p

2[Jn−1(px) − Jn+1(px)]

=p

2Hn+1 [f ′(x)] − p

2Hn−1 [f ′(x)] . (7.16)

7.7 Finite Hankel Transform of f ′′(x) + 1xf ′(x) − n2

x2 f(x),

where p is the root of Jn(p) = 0 in 0 � x � 1 .

We have, by the result of article 7.4 that

Hn

[f ′′(x) +

1x

f ′(x) − n2

x2f(x)

]


=∫ 1

0

[f ′′(x) +

1x

f ′(x)]

x Jn (px) dx −∫ 1

0

n2

x2f(x). xJn (px) dx

= −p

∫ 1

0x Jn

′ (px)f ′ (x) dx − n2

∫ 1

0

1x

f(x) Jn (px) dx

= −p

[ {xJ ′

n(px) f(x)}1

0−∫ 1

0

{J ′

n(px) + px J ′′n(px)

}f(x) dx

]

−n2

∫ 1

0

1x

f(x)Jn(px)dx

= −p J ′n(p) f(1) + p

[∫ 1

0

{J ′

n(px) + px

(n2

p2x2− 1)

Jn(px) − J ′n(px)

}

f(x) dx]− n2

∫ 1

0

1x

f(x) Jn (px) dx,

by px J ′′n (px) + J ′

n(px) = px

(n2

p2x2− 1

)Jn (px).

Therefore,

Hn

[f ′′(x) +

1x

f ′(x) − n2

x2f(x)

]

= −p Jn′(p)f(1) − p2

∫ 1

0x Jn(x) f(x) dx, after simplification.

(7.17)

Corollary 7.2 Putting n = 0 and using the result J ′0(p) = −J1(p) we

get from the above result that

H0

[f ′′(x) +

1x

f ′(x)]

= p J1 (p) f(1) − p2 H0 [f(x)] (7.18)

7.8 Other forms of finite Hankel Transforms.

For utility purpose in applications other two different forms of finiteHankel transforms of order n are introduced below.

Form II If f(r) is a piecewise continuous function in 0 � r � 1. thenits finite Hankel transform of order n is denoted by f(p) and is definedby

Hn [ f(r) ] = f(p) =∫ 1

0f(r). r Jn (pr) dr, (7.19)

where p is a positive root of

p Jn′ (p) + h Jn (p) = 0 (7.20)


Correspondingly, inverse finite Hankel transform of f(p) is given by

H−1n

[f(p)

] ≡ f(r) = 2∑

p

p2f(p)h2 + p2 − n2

.Jn (pr){Jn (p)}2 (7.21)

where the summation is being taken over all the positive roots of theequation (7.20).

Form III If the field of variation of the variable r ranges finitely as0 < a � r � b, a third form of finite Hankel transform of order n of afunction f(r), which is piecewise continuous in the range, is defined by

Hn [f(r) ] = f(p) =∫ b

arf(r) Bn (pr) dr, (7.22)

where Bn (p r) = Jn(p r) Yn(p a) − Yn(pr) Jn(p a) (7.23)

and Yn(p r) is the Bessel function of second kind of order n. Here p is apositive root of the equation

Bn (p b) = 0 ⇒ Jn(p b) Yn(p a) = Yn(p b) Jn(p a) (7.24)

Then inverse of f(p) in this case is defined and denoted by

H−1n

[f (p)

] ≡ f(r) =π2

2

∑p

p2 { Jn(pb)}2 f (p){ Jn(pa) }2 − { Jn(pb) }2 Bn (pr),

(7.25)

where the summation is being taken over all the positive roots of theequation (7.25).

7.9 Illustrations.

(i) As an illustration we evaluate finite Hankel transform of order zeroin (0, 1) of f ′′(x) + 1

xf ′(x), where, p is a positive root ofp J ′

0(p) + h J0(p) = 0 below.

By definition, we have

H0

[f ′′(x) +

1x

f ′(x)]

=∫ 1

0

[f ′′(x) +

1x

f ′(x)]

xJ0(px)dx


= [ f ′(x) · x J0(px) ]10 −∫ 1

0f ′(x) [ x p J ′

0(px) + J0(px) ] dx

+∫ 1

0f ′(x)J0(px)dx

= [f ′(1)J0(p)] − p

∫ 1

0f ′(x) · x J ′

0(px)dx

= f ′(1)J0(p) − p[{f(x) · x J ′0(px)}1

0 −∫ 1

0f(x)

d

dx{x J ′

0(px)}dx]

= f ′(1)J0(p) − p f(1)J ′0(p) + p

∫ 1

0f(x){J ′

0(px) + pxJ ′′0 (px)}dx

= f ′(1)J0(p) − p f(1)J ′0(p) + p

∫ 1

0(−px) J0(px) f(x)dx

= f ′(1)J0(p) − p f(1) J ′0(p) − p2 H0[f(x)]

But since p is a root of p J ′0(p) + h J0(p) = 0, we have

J ′0(p) = −h

J0(p)p

and therefore,

H0

[f ′′(x) +

1x

f ′(x)]

= [f ′(1) + h f(1)]J0(p) − p2 H0[f(x)] (7.26)

(ii) For illustration of the form-III of finite Hankel transform of ordern in 0 < a � r � b, of f ′′(r) + 1

r f ′(r) − n2f(r)r2 , where p is a root of

Jn(pb) Yn(pa) = Yn(pb) Jn(pa), is stated without proof as

Hn

[f ′′(r) +

1rf ′(r) − n2

r2f(r)

]

=2π

[f(b)

Jn(pa)Jn(pb)

− f(a)]− p2 Hn [f(r)] (7.27)

7.10 Application of finite Hankel Transforms.

We discuss below the application of finite Hankel transform to solvesome initial-boundary value problems.

Example 7.6 Solve the following initial-boundary value problemdefined by

∂2υ

∂r2+

1r

∂υ

∂r=

1κ

∂υ

dt, 0 � r < 1 , t � 0


under the conditions υ(1, t) = υ0, a costant, when t > 0 and

υ(r, 0) = 0 , 0 � r < 1, by using finite Hankel transform in 0 � r < 1.

Solution. Let υ0 (p, t) be Hankel transform of υ(r, t) over 0 � r < 1of order zero. Then, taking Hankel transform of order zero, the givenPDE under the given boundary condition υ(1.t) = υ0 becomes

p υ(1, t) J1(p) − p2 H0[υ] =1κ

∫ 1

0

∂υ

∂tr J0(pr) dr

Hence, p υ0 J1(p) − p2 υ0 =1κ

d

dtυ0

Thus, υ0(p, t) is a solution of the linear ODE

d υ0

dt+ κ p2 υ0 = κ p υ0 J1(p)

Its solutions is given by

υ0(p, t) =(

υ0

p

)J1(p) + c e−κp2t, where c is a constant.

Now, taking zero order Hankel transform of υ(r, 0) = 0 we getυ0(p, 0) = 0 and therefore, the above solution gives c = −

(υ0p

)J1(p).

Hence,υ0(p, t) =

υ0

pJ1(p)

(1 − e−κp2t

)Applying inversion formula of Hankel transform we finally get the solu-tion as

υ(r, t) = 2∑

p

υ0 (p, t)J0(pr)[J ′

0(p)]2

= 2 υ0

∑p

(1 − e−κp2t

){ J0(pr)p J1(p)

}, since J ′

0(p) = −J1(p),

where summation is taken over all p satisfying J0(p) = 0.

Example 7.7. Find the temperature distribution in a long circularcylinder defined by the axisymmetric heat conduction equation

1κ

∂u

∂t=

∂2u

∂r2+

1r

∂u

∂r, 0 � r � a , t > 0


under the boundary and initial conditions

u(r, t) = f(t) , on r = a , t > 0

u(r, 0) = 0 , 0 � r � a

Solution. Under finite Hankel transform of order zero, the given PDEbecomes

1κ

ut(p, t) = −p2u (p, t) + ap J1(ap) f(t) , by (7.18)

Also, u(r, 0) = 0 , 0 � r � a under the transformed domain results in

u (p , 0) = 0

The solution of the above first order ODE gives

u (p, t) = κ ap J1 (ap)∫ t

0f(τ) exp

{−κp2(t − τ)}

dτ

Thus, the formal solution of the boundary value problem is given by

u(r, t) =2κa

∑p

p J0(pr)J1(pa)

∫ t

0f(τ) exp

{−κp2(t − τ)}

dτ,

where the summation is taken over all positive root p satisfying J0(ap) = 0.In particular, if f(t) = constant = T0 , say, then

u(r, t) =2κ T0

a

∑p

p J0(pr)J1(pa)

∫ t

0exp

{−κp2(t − τ)}

dτ

=2T0

a

∑p

J0(pr)p J1(pa)

[1 − exp (−κp2t)

]

Thus, the temperature distribution consists of a transient term whichdecays to zero as t → ∞ together with a steady-state term.

Example 7.8. The axisymmetric unsteady motion of a viscous fluid inan infinitely long circular cylinder of radius a is governed by the PDE

∂u

∂t= ν

[∂2u

∂r2+

1r

∂u

∂r− u(r, t)

r2

], 0 � r � a , t > 0 ,

where u = u(r, t) is the tangential fluid velocity and ν is the constantkinematic viscosity of the fluid. The cylinder is initially at rest at t = 0,


and it is then allowed to rotate with constant angular velocity Ω. Thusthe respective boundary and initial conditions are

u(r, t) = a Ω , t > 0

and u(r, t) = 0 , t = 0

Solution. Taking Laplace transform in t and finite Hankel transformof order one over 0 < r < a jointly defined by

¯u (p, s) =∫ ∞

0e−st dt

[∫ a

0r J1(pr) u(r, t) dr

],

where p is the positive root of the equation J1(ap) = 0 and using (7.17)the given PDE under the boundary and initial conditions becomes

s¯u (p, s) = −ν[p2 ¯u(p, s)

]− ν a2 p Ωs

J ′1(ap)

Or ¯u (p, s) = −ν a2 Ω p J ′1(ap)

s(s + νp2)

The inverse Laplace transform gives

u(p, t) = −a2 Ωp

J ′1(ap)

[1 − exp (−νtp2)

]Finally, using inverse Hankel transform with J ′

1(ap) = −J2(ap) the aboveequation gives

u(r, t) = 2Ω∑

p

J1(rp)p J2(ap)

[1 − exp(−νtp2)]

Since, Hn [rn] =an+1

pJn+1(ap) , for n = 0, 1, 2, · · ·

we have r = H−11

[a2

pJ2(ap)

]= 2∑

p

J1(rp)p J2(ap)

Using this result we get the final solution of the initial-boundary valueproblem as

u(r, t) = rΩ − 2Ω∑

p

J1(pr)p J2(pa)

exp (−νtp2)

This solution consists of a steady part rΩ together with a transient part−2Ω

∑p

J1(pr)p J2(pa) exp (−νp2t) which tends to zero as t → ∞.


Example 7.9. A viscous fluid of kinematic viscosity ν is containedbetween two infinitely long concentric circular cylinders of radi a andb. The inner cylinder is kept at rest and the outer cylinder suddenlystarted rotating with uniform angular velocity ω. Find the tangentialvelocity u(r, t) of the fluid if the equation of motion is

∂2u

∂r2+

1r

∂u

∂r− u

r2=

1ν

du

dt, a < r 0,

given that u = b ω when r = b , u = 0 when r = a and also u = 0 whent = 0 .

Solution. Given the boundary values and the initial condition of theinitial boundary value problem as

u(b, t) = b ω , u(a, t) = 0 (i)

and u(r, 0) = 0 (ii)

Now, taking finite Hankel transform of order 1 over a � r � b on r, letu(p, t) be given by

u(p, t) =∫ b

ar B1(pr) u(r, t) dr (iii)

where p is a positive root of the equation

J1(pb) Y1(pa) = Y1(pb) J1(pa) (iv)

Then, the given constitutive PDE becomes

H1

[∂2u

∂r2+

1r

∂u

∂r− u2

r2

]=

1ν

d

dtu (p , t)

Simplifying, we get

2π

[u(b, t)

J1(pa)J1(pb)

− u(a, t)]− p2u(p, t) =

1ν

du

dt, by (7.27)

Ordu

dt+ p2ν u(p, t) =

2bνω

π

J1(pa)J1(pb)

This linear ODE has the solution given by

u(p, t) =2bωπp2

J1(pa)J1(pb)

+ C e−νp2t (v)


But taking Hankel transform of order 1 on r, the initial condition in (ii)gives

u(p , 0) = 0 (vi)

Employing (vi) in (v), we get

C = −2bωπp2

J1(pa)J1(pb)

and thus (v) becomes

u(p, t) =2bωπp2

J1(pa)J1(pb)

[1 − e−νp2t

](vii)

Now, inverting Hankel transform of order 1, the above equation leads to

u(r, t) = π ω b∑

p

(1 − e−νp2t) J1(pa) J1(pb)[J1(pa)]2 − [J1(pb)]2

· B1(pr)

as the final solution of the initial-boundary value problem.

Example 7.10. The free symmetric vibration of a thin circular mem-brane of radius a is governed by the wave equation

∂2u

∂r2+

1r

∂u

∂r=

1c2

∂2u

∂t2, 0 < r < a , t > 0

with boundary and initial condition

u(a, t) = 0 , t � 0

u(r, 0) = f(r) ,∂u(r, 0)

∂t= g(r) , 0 < r < a

Determine the solution of the problem

Solution. Applying zeroth order finite Hankel transform over r in0 < r < a the given constitutive equation becomes

d2u

dt2+ c2 p2 u = 0 (i)

and the given boundary conditions become

u(p, 0) = f(p) andd

dt[ u (p , 0) ] = g(p) (ii)


The solution of this system of second order ODE in (i) under (ii) is then

u(p, t) = f(p) cos (p c t) +g(p)pc

sin (p c t) (iii)

Finally the inverse Hankel transform yeilds the solution as

u(r, t) =2a2

∑p

f(p) cos (p c t)· J0(rp)J2

1 (ap)+

2ca2

∑p

g(p) sin (p c t)J0(rp)

p J21 (ap)

,

where the summation is taken over all positive roots p of J0(pa) = 0 .

Exercises.

(1) Find the zeroth order Hankel transform of(a) f(r) = r2 (b) f(r) = J0(α r)

[Ans. (a)

a2

p2

(ap − 4

ap

)J1(ap) (b)

ap

(α2 − p2)J0(aα) J1(ap)

]

(2) Prove that H0

[1r f ′(r)

]= p H1

{1r f(r)

}− f(a) , where p is apositive root of J0(ap) = 0

(3) Find the solution of the forced symmetric vibrations of a thinelastic membrane that satisfy the initial boundary value problem

∂2u

∂r2+

1r

∂u

∂r− 1

c2

∂2u

∂t2= −p (r, t)

T0,

where p(r, t) is the applied pressure for producing the vibrationand the membrane is stretched by a constant tension T0. Themembrane is set into motion from rest from its equilibrium positionso that

u(r, 0) = 0 =[∂u

∂t

]t=0

(4) Find finite Hankel transform of (r2−b2)r with the kernel

J1(pr) Y1(pb) − J1(pb) Y1(pr).


[Hint: Start with evalvating∫ ba (r2−b2){J1(pr)Y1(pb)−Y1(pr)J1(pb)}dr]

[Ans. H1

{(r2 − b2)

r

}=

b2 − a2

ap2

J1(pb)J1(pa)

, where p is the positive

root of Y1(pb) J1(ap) − J1(pb) Y1(pa) = 0

and Y1(pb) J ′1(pa) − J1(pb) Y ′

1(pa) =1pa

Jn (pb)Jn (pa)

] .

Chapter 8

The Mellin Transform

8.1 Introduction.

This chapter deals with the origin, theory and applications of Mellintransforms. We derive the transform from its natural way in connectionwith the solution of a problem of potential theory in an infinite wedge.Riemann was the first author to discuss this transform in 1876 and manyothers too have shown its applications later.

Let us consider a problem of potential theory in an infinite wedgedefined by ρ > 0, |ϕ| < α where ρ, ϕ are plane polar co-ordinates.We wish to find the solutions u(ρ, ϕ) of the two-dimensional Laplaceequation

∂2u

∂ρ2+

1ρ

∂u

∂ρ+

1ρ2

∂2u

∂ϕ2= 0 (8.1)

under the conditions that

u (ρ , ± α) = f(ρ) , (a given function) (8.2)

and ρs u (ρ , ϕ) , ρs+1 ∂u

∂ρ(8.3)

tend to zero as ρ → 0 and also as ρ → ∞ for some specified complexnumber s. From (8.1), we can write

∂

∂ρ

[ρ

∂u

∂ρ

]+

1ρ

∂2u

∂ϕ2= 0

Now multiplying the above equation by ρs and integrating the resultwith respect to ρ from ρ = 0 to ρ → ∞ we get∫ ∞

0ρs

[∂

∂ρ

{ρ

∂u

∂ρ

}]dρ +

∫ ∞

0ρs−1 ∂2

∂ϕ2u(ρ, ϕ) dρ = 0


If we use the formula for integration by parts in the first term we getthe above equation as[

ρs+1 ∂u

∂ρ− s ρs u(ρ, ϕ)

]∞0

+ s2

∫ ∞

0ρs−1 u(ρ, ϕ) dρ

+d2

dφ2

∫ ∞

0ρs−1 u(ρ, ϕ) dρ = 0

Now, using the conditions (8.3) we get[d2

dϕ2+ s2

] ∫ ∞

0ρs−1 u(ρ, ϕ) dρ = 0 (8.4)

Defining Mellin transform of the function u(ρ, ϕ) over 0 < ρ < ∞ as

u∗(s, ϕ) ≡ M [ u(ρ, ϕ) ; ρ → s ] =∫ ∞

0ρs−1 u(ρ, ϕ) dρ

the equation (8.4) can be expressed as

d2u∗

dϕ2(s, ϕ) + s2 u∗(s, ϕ) = 0 (8.5)

which is an ordinary differential equation.

If the conditions on the faces of wedge are as specified in (8.2) beused, we have

u∗(s ,± α) = f∗(s)

after taking Mellin transforms. Then the solution of the boundary valueproblem under prescribed conditions is

u∗(s, ϕ) = f∗(s) · cos sϕ

cos sα(8.6)

Therefore, the solution u∗(s, ϕ) of the boundary-value problem satis-fying the prescribed condition is a function whose Mellin transform isgiven by

f∗(s)cos sϕ

cos sα.

8.2 Definition of Mellin Transform.

Let f(x) be defined for 0 < x < ∞. Mellin transform of this function isa new function of the complex variable s defined by

f∗(s) = M [ f(x) ; x → s ] =∫ ∞

0f(x) xs−1 dx (8.7)

The Mellin Transform 279

for those values of s for which the right hand side improper integral ineqn. (8.7) converges.

We shall now discuss some of the basic important properties of Mellintransform.

Property I. Linearity property.

Let c1, c2 be constants and let f1(x) and f2(x) be two given functions.Then the linearity property of Mellin transform states that

M [c1f1(x)+c2f2(x) ;x → s] = c1M [f1(x) ; x → s]+c2 M [f2(x) ; x → s]

Proof.

The proof of this result is very simple and can be seen in the followingsteps.

M [c1f1(x) + c2f2(x) ; x → s]

=∫ ∞

0xs−1 [c1f1(x) + c2f2(x)] dx

= c1

∫ ∞

0f1(x) xs−1 dx + c2

∫ ∞

0f2(x) xs−1 dx

= c1f∗1 (s) + c2f

∗2 (s) , (8.8)

for those s for which both the improper integrals do exist.

Property II. Change of scale property.

If M [f(x) ; x → s] = f∗(s) , then

M [f(ax) ; x → s] = a−s f∗(s)

Proof.

The proof of this result is deduced by a transformation of variablein the integral form of the transform.

M [f(ax) ; x → s] =∫ ∞

0f(ax)xs−1dx

=∫ ∞

0f(t) · ts−1

asdt , putting ax = t

= a−s f∗(s) (8.9)

Property III.

If M [f(x) ; x → s] = f∗(s) , then

M [xaf(x) ; x → s] = f∗(s + a)


Proof.

To prove the above result, we get

M [xaf(x) ; x → s] =∫ ∞

0xaf(x) · xs−1 dx =

∫ ∞

0f(x) · x(s+a)−1 dx

= f∗(s + a) , by use of (8.7) (8.10)

Property IV.

If M [f(x) ; x → s] = f∗(s) , then

M

[1x

f

(1x

); x → s

]= f∗(1 − s) .

Proof.

To prove the above result, we have by definition of Mellin transform

M

[1x

f

(1x

); x → s

]

=∫ ∞

0

1x

f

(1x

)xs−1 dx

=∫ ∞

0t−sf(t) dt , putting x =

1t

=∫ ∞

0t(1−s)−1 f(t) dt

= f∗(1 − s) , by (8.7) (8.11)

Property V.

If M [f(x) ; x → s] = f∗(s) , then

M [f(xa) ; x → s] =1a

f∗(s

a

), a > 0

Proof. To prove the above result, we have

M [f(xa) ; x → s] =∫ ∞

0xs−1 f(xa) dx

=1a

∫ ∞

0t(

sa−1) f(t) dt , where xa = t

=1a

f∗(s

a

), by (8.7) (8.12)


Property VI. Derivative of Mellin Transform.

If M [f(x) ; x → s] = f∗(s), then M [log x.f(x) ; x → s] is equalto d

ds f∗(s).

Proof. We have, by definition of Mellin transform

M [log x.f(x) ; x → s] =∫ ∞

0xs−1 log x.f(x) dx

But, from the given conditiond

ds[f∗(s)] =

∫ ∞

0

d

ds

[xs−1 f(x)

]dx

=∫ ∞

0f(x) .

1x

. xs log x dx =∫ ∞

0xs−1 . f(x) log x.dx

Thus, M [ log x.f(x) ; x → s ] =d

dsf∗(s) (8.13)

8.3 Mellin Transform of derivative of a function.

The following two sets of results are found to be true in these cases.

Results I (a) M [f ′(x) ; x → s] = −(s − 1)f∗(s − 1)

(b) M [f ′′(x) ; x → s] = (s − 1)(s − 2)f∗(s − 2)

(c) M [fn(x);x → s] = (−1)nΓ(s)

Γ(s − n)f∗(s − n), n = 1, 2, 3, · · ·

where M [f(x) ; x → s] = f∗(s) and fn(x) =dn

dxnf(x)

Proof.

(a) By definition of Mellin transform

M [f ′(x) ; x → s] =∫ ∞

0xs−1f ′(x) dx

=[xs−1 f(x)

]∞0

− (s − 1)∫ ∞

0xs−2 f(x) dx

= −(s − 1)∫ ∞

0x(s−1)−1 f(x) dx ,

for the existence of real numbers λ1 and λ2 such that

lims→0

xs−1 f(x) = 0 , lims→∞ xs−1 f(x) = 0

for λ1 < Re s < λ2 and f∗(s − 1) exists.

Thus, M [ f ′(x) ; x → s ] = −(s − 1) f∗(s − 1)

(8.14)


(b) Again application of the result in (8.14) will give

M [f ′′(x) ; x → s] = M [ϕ′(x) ; x → s] , where ϕ(x) = f ′(x)

= −(s − 1) ϕ∗(s − 1) = −(s − 1) M [f ′(x) ; x → s]

= −{(s − 1)}{−(s − 2)} M [f(x) ; x → s]

= (s − 1)(s − 2) f∗(x) (8.15)

(c) For the last case (c), if n is a positive integer, by using theMathematical induction n times, we get

M [f (n)(x) ; x → s] = (−1)n (s − 1)(s − 2) · · · (s − n + 1)f∗(s − n)

= (−1)nΓ(s)

Γ(s − n)f∗(s − n) (8.16)

Results II If M [f(x) ; x → s] = f∗(s) , then

(a) M

[(x

d

dx

)f(x) ; x → s

]= −s f∗(s)

(b) M

[(x

d

dx

)2

f(x) ; x → s

]= s2 f∗(s)

(c) M

[(x

d

dx

)n

f(x) ; x → s

]= (−1)n sn f∗(s) , n = 2, 3, · · ·

Proof.

(a) By (8.10) we know that

M [ x φ (x) ; x → s ] = ϕ∗ (s + 1)

Replacing ϕ(x) ≡ f ′(x) , we get

M [ x f ′(x) ; x → s ] = M [ f ′(x) ; x → s + 1 ] (8.17)

Also, since M [ f ′(x) ; x → s ] = −(s− 1) f∗(s− 1) , replacing s

by s + 1 here we get

M [ f ′(x) ; x → s + 1 ] = −s f∗(s) (8.18)

Therefore, from eqn. (8.17) and (8.18) we get

M [ x f ′(x) ; x → s ] = −s f∗(s) , which is the result in (a).(8.19)


(b) Let us assume here that(x

d

dx

)f(x) = g(x)

Then,

M [x g′(x) ; x → s] = M [g′(x) ; x → s + 1] , by (8.17)

= −s g∗(s)

= −s

{M

[x

d

dxf(x)

]}= −s {−s f∗(s)} , by (8.18)

= s2 f∗(s) (8.20)

(c) To prove this result, we use the method of Mathematical inductionto obtain

M

[(x

d

dx

)n

f(x) ; x → s

]= (−1)n sn f∗(s) (8.21)

after the nth stage.

8.4 Mellin Transform of Integral of a function.

Result I. If M [f(x) ; x → s] = f∗(s), then

(a) M

[∫ x

0f(t)dt ; x → s

]= −1

sf∗(s + 1)

(b) M

[∫ x

0dy

{∫ y

0f(t)dt

}; x → s

]=

1s(s + 1)

f∗(s + 2)

(c) M [In f(x) ; x → s] = (−1)nΓ(s)

Γ(s + n)f∗(s + n) , where by

In f(x) we mean In f(x) =∫ x

0In−1 f(t) dt

Proof. The proof of these results are given below :

(a) Let∫ x0 f(t) dt = G(x) ⇒ G′(x) = f(x)

Now, since M [G′(x) ; x → s] = −(s − 1)G∗(s − 1)

= −(s − 1)M [G(x) ; x → s − 1]

= −(s − 1)M[∫ x

0f(t)dt ; x → s − 1

]


we get M [f(x) ; x → s + 1] = −s M

[∫ x

0f(t)dt ; x → s

]

∴ M

[∫ x

0f(t)dt ; x → s

]= −1

sM [f(x) ; x → s + 1]

= −1s

f∗(s + 1) (8.22)

(b) Repeating the result in (a) we get

M

[∫ x

0dy

{∫ y

0f(t)dt

};x → s

]= −1

sM

[∫ x

0f(t)dt;x → s + 1

]

= −1s

{− 1

s + 1f∗(s + 2)

}

=1

s(s + 1)f∗(s + 2) (8.23)

(c) For proving the result in this part the method of Mathematicalinduction will be applied on the index n.

Result II. If M [f(x) ; x → s] = f∗(s) , then

(a) M

[∫ ∞

xf(t)dt ; x → s

]=

1s

f∗(s + 1)

(b) M

[∫ ∞

xdy

{∫ ∞

yf(t)dt

};x → s

]=

1s(s + 1)

f∗(s + 2)

(c) M [I∞n f(x) ; x → s] =Γ(s)

Γ(s + n)f∗(s + n) ,

where I∞n f(x) =∫ ∞

xI∞n−1 f(t)dt .

Proof. As in result I, the result of these parts is shown below.

(a) Let∫ ∞

xf(t)dt = g(x) ⇒ −f(x) = g′(x)

Therefore, M [−f(x) ; x → s] = M [g′(x) ; x → s]

= −(s − 1)M[∫ ∞

xf(t)dt ; x → s − 1

]

⇒ M [f(x) ; x → s] = (s − 1) M

[∫ ∞

xf(t)dt ; x → s − 1

]

⇒ M [f(x) ; x → s + 1] = s M

[∫ ∞

xf(t)dt ; x → s

],

(8.24)


which is the proof of the result (a).

(b) Part (b) can be proved by using the result of part (a) onceagain as in Result I.

(c) To prove this part once again the method of Mathematicalinduction on index n will be used.

8.5 Mellin Inversion theorem.

Fourier Transform of g(t) , −∞ < t < ∞ is defined as

G(ξ) =1√2π

∫ +∞

−∞g(t)eiξt dt

Substituting x = et , s = c + iξ in the above we get

G (ic − is) =1√2π

∫ ∞

0x−c g(log x) . xs−1 dx (8.25)

Again in the corresponding inversion formula of Fourier transform

g(t) =1√2π

∫ +∞

−∞G(ξ) e−iξt dξ ,

if the above transform be applied we get

g(log x) =1√2π

i

∫ c+i∞

c−i∞G(ic − is) xc−s ds (8.26)

Now, writing

f(x) = (2π)−12 x−c g(log x) , f∗(s) = G(ic − is)

we have from eqns (8.25) and (8.26)

f∗(s) =∫ ∞

0xs−1 f(x) dx , (8.27)

f(x) =1

2πi

∫ c+i∞

c−i∞f∗(s) x−s ds (8.28)

as the definition of Mellin transform of f(x) and its correspondinginversion formula respectively.

It may be noted here that the left hand side of eqn. (8.28) can alsobe expressed as

f(x) = M−1 [f∗(s) ; s → x] =1

2πi

∫ c+i∞

c−i∞f∗(s) x−s ds .


8.6 Convolution theorem of Mellin Transform.

Theorem 8.1. If f∗(s) and g∗(s) be Mellin Transforms of f(x) andg(x) respectively, then

M [f(x) g(x) ; x → s] =1

2πi

∫ c+i∞

c−i∞f∗(z) · g∗(s − z)dz .

Proof.

M [f(x) g(x) ; x → s]

=∫ ∞

0f(x) g(x) · xs−1 dx

=∫ ∞

0f(x) xs−1

[1

2πi

∫ c+i∞

c−i∞g∗(z) x−z dz

]dx

=1

2πi

∫ c+i∞

c−i∞g∗(z)

[∫ ∞

0f(x) xs−z−1 dx

]dz

=1

2πi

∫ c+i∞

c−i∞g∗(z) f∗(s − z) dz

Similarly, we would get

M [f(x) g(x) ; x → s] =∫ ∞

0f(x) g(x) xs−1 dx

=1

2πi

∫ c+i∞

c−i∞f∗(z) g∗(s − z) dz

Thus we get

M [f(x) g(x) ; x → s]

= 12πi

∫ c+i∞c−i∞ f∗(z) g∗(s − z) dz

= 12πi

∫ c+i∞c−i∞ g∗(z) f∗(s − z) dz

⎫⎪⎬⎪⎭ (8.29)

as the result of the convolution theorem.

Corollary 8.1. Another form of convolution theorem of Mellin trans-form is given below :


Theorem 8.2. If f∗(s) and g∗(s) are Mellin transforms of the functionsf(x) and g(x) respectively, then

M−1 [f∗(s) g∗(s) ; s → x] =∫ ∞

0f(x

t

)g(t)

dt

t

Proof. We know that

M−1 [ f∗(s) g∗(s) ; s → x ]

=1

2πi

∫ c+i∞

c−i∞x−s f∗(s) g∗(s) ds , by (8.28)

=1

2πi

∫ c+i∞

c−i∞x−s f∗(s)

{∫ ∞

0ts−1 · g(t)dt

}ds

=∫ ∞

0

g(t)t

{1

2πi

∫ c+i∞

c−i∞f∗(s)

(x

t

)−sds

}dt

=∫ ∞

0

g(t)t

f(x

t

)dt , (8.30)

since f(x

t

)=

12πi

∫ c+i∞

c−i∞

(x

t

)−sf∗(s) ds (8.31)

8.7 Illustrative solved Examples.

Example 8.1. Evaluate Mellin transform of

(a) f(x) = e−nx

(b) f(x) = (1 + x)−a

(c) f(x) = (1 + xa)−b, where 0 < Re s < Re (ab)

(d) f(x) = cos kx and f(x) = sin kx

(e) f(x) = 2(e2x−1)

Solutions. By definition of Mellin transform, we have

(a) M[e−nx ; x → s

]=∫ ∞

0xs−1 e−nx dx

=1ns

∫ ∞

0ts−1 e−t dt, where nx = t

=Γ(p)ns


(b) M[(1 + x)−a ; x → s

]=∫ ∞

0

xs−1

(1 + x)s+(a−s)dx

= B (s , a − s) , for Re s > 0 , Re (a − s) > 0

=Γ(s) Γ(a − s)

Γ(a), for Re a > Re s > 0

(c) M

[1

(1 + xa)b; x → s

]=∫ ∞

0

xs−1

(1 + xa)bdx

=∫ ∞

0

(y)sa−1dy

(1 + y)b, where xa = y,

a

xdx =

dy

y

=1a

∫ ∞

0

y( sa−1)

(1 + y)(b− sa)+ s

a

dy

=1a

B(b − s

a,s

a

)

=Γ(b − s

a

)Γ(

sa

)a Γ(b)

, where 0 < Re s < Re (ab)

(d) M [cos kx ; x → s] =∫ ∞

0xs−1 · eikx + e−ikx

2dx

=12

∫ ∞

0e−ikx · xs−1 dx +

12

∫ ∞

0eikx xs−1 dx

We know from (a) above that∫ ∞

0e−nx · xs−1 dx =

Γ(s)ns

Hence,∫ ∞

0e−ikx · xs−1 dx =

Γ(s)(ik)s

and∫ ∞

0eikx · xs−1 dx =

Γ(s)(−ik)s

Therefore,∫ ∞

0cos kx · xs−1 dx =

Γ(s)ks

cosπs

2

and similarly∫ ∞

0xs−1 sin kx dx =

Γ(s)ks

sinπs

2


(e) we have M

[2

e2x − 1; x → s

]

= 2∫ ∞

0xs−1 dx

e2x − 1dx

= 2∞∑

n=1

∫ ∞

0xs−1 · e−2nx dx

= 2∞∑

n=1

Γ(s)(2n)s

= 21−s Γ(s)∞∑

n=1

1ns

= 21−s Γ(s) ζ(s) ,

where ζ(s) =∞∑

n=1

1ns

, Re s > 1 is Riemann zeta function.

Example 8.2. If f∗(s) and g∗(s) are Mellin transforms of f(x) andg(x) respectively, prove that∫ ∞

0f(x) g(x) dx =

12πi

∫ c+i∞

c−i∞f∗(z) g∗(1 − z) dz .

Solution. We know, by convolution theorem of Mellin transform that∫ ∞

0xs−1 f(x) g(x) dx =

12πi

∫ c+i∞

c−i∞f∗(z) g∗(s − z) dz

Now, putting s = 1 in this relation we get∫ ∞

0f(x) g(x) dx =

12πi

∫ c+i∞

c−i∞f∗(z) g∗(1 − z) dz .

Example 8.3. Find Mellin inversion of Γ(s) .

Solution. Let f(x) be the required Mellin inversion of Γ(s) . Then

f(x) =1

2πi

∫ c+i∞

c−i∞x−s Γ(s) ds

=1

2πi

∫ c+i∞

c−i∞x−s · π ds

Γ(1 − s) sin sπ

=12i

∫ c+i∞

c−i∞x−s ds

Γ(1 − s) sin sπ

=12i

∫C

x−s ds

Γ(1 − s) sin sπ,


where C is an anticlockwise closed large semi-circular contourwith centre at s = 0 lying to the left of the imaginary s-axis. Then theintegrand has singularities at its poles at s = 0,−1,−2, · · · . Thus, thevalue of the integral is 2πi times sum of the residues of the integrandat its poles. Hence,

f(x) =12i

2πi∞∑

n=0

xn

Γ(1 + n)[

dds (sin sπ)

]s=−n

=∞∑

n=0

(−1)nxn

n!

= e−x .

Example 8.4. Solve the boundary value problem defined by

x2 uxx + x ux + uyy = 0 , 0 � x < ∞ , 0 < y < 1

subject to boundary conditions u(x, 0) = 0 , u(x, 1) =

{A, 0 � x � 10, x > 1

where A is a constant, by applying Mellin transform over x .

Solution. Applying Mellin transform over x we get the PDE as

u∗yy + s2 u∗ = 0 , 0 < y < 1

and the boundary conditions transform to

u∗(s, 0) = 0 , u∗(s, 1) = A

∫ 1

0xs−1 dx =

A

s

Then the solution of the transformed ODE is

u∗(s, y) =A

s

sin sy

sin s, 0 < Re s < 1 .

Taking inverse Mellin transform we get

u(x, y) =A

2πi

∫ c+i∞

c−i∞

x−s

s

sin sy

sin sds .

The function u∗(s, y) is analytic in 0 < Re s < π and thus 0 < c < π.The above integrand has simple poles at s = nπ , n = 1, 2, 3, · · · and all


those lie inside a large semi-circular contour in the right-half of s-plane.Therefore,

u(x, y) =A

π

∞∑n=1

1n

(−1)n x−nπ sin (nπy) .

Example 8.5. Consider the problem of determining the potentialϕ(r, θ) that satisfies the Laplace equation

r2 ϕrr + r ϕr + ϕθθ = 0

in an infinite wedge 0 < r < ∞ , −α < θ < α and the boundaryconditions

ϕ(r, α) = f(r) , ϕ(r,−α) = g(r) , 0 � r < ∞ϕ(r, θ) → 0 as r → ∞ for all θ in − α < θ < α .

Solution. Applying Mellin transform over r, the Laplace equationbecomes

s2 ϕ∗(s, θ) +d2 ϕ∗(s, θ)

dθ2= 0

and the boundary conditions transform to

ϕ∗(s, α) = f∗(s) , ϕ∗(s,−α) = g∗(s) and ϕ∗(s, θ) → 0 as s → ∞

The general solution of the transformed ODE is

ϕ∗(s, θ) = A cos sθ + B sin sθ ,

where A and B are arbitrary constants to be determined under thetransformed boundary conditions. These give

A cos s α + B sin s α = f∗(s)

A cos s α − B sin s α = g∗(s)

Solving for A and B , we get

A =f∗(s) + g∗(s)

2 cos s α, B =

f∗(s) − g∗(s)2 sin s α

Thus the solution becomes

ϕ∗(s, θ) = f∗(s)sin s(α + θ)

sin (2sα)+ g∗(s)

sin s(α − θ)sin (2sα)


Inverting the Mellin transformed equation, the solution of the problemcan be expressed as

ϕ(r, θ) = M−1 [f∗(s) h∗(s, α + θ)] + M−1 [g∗(s) h∗(s, α − θ)]

where h∗(s, θ) =sin (sθ)sin (2sα)

Now, h(r, θ) = M−1 [h∗(s, θ)] = M−1

[sin sθ

sin 2sα

]

=12α

rn sin (nθ)1 + 2rn cos nθ + r2n

,

after simplification where n =π

2αTherefore, the above form of solution of the problem is represented as

ϕ(r, θ) =n rn cos nθ

π

[∫ ∞

0

un−1 f(u) du

u2n − 2rnun sin nθ + r2n

]

+nrn cos nθ

π

[∫ ∞

0

un−1 g(u) du

u2n + 2rnun sin nθ + r2n

],− π

2n< θ <

π

2n

8.8 Solution of Integral equations.

We begin with the solution of the following integral equation of generaltype ∫ ∞

0f(x) K(xt) dx = g(t) , t > 0 (8.32)

Taking Mellin transform of both sides of eqn. (8.32) we get

f∗(1 − s) K∗(s) = g∗(s)

by using eqn. (8.30).

Now replacing s by 1 − s we find

f∗(s) = g∗(1 − s) L∗(s) , where L∗(s) =1

K∗(1 − s)

This solution can be written in the form

f(x) =∫ ∞

0g(t) L(xt) dt, (8.33)


provided that the inverse Mellin transform in

L(x) = M−1

[1

K∗(1 − s); s → x

]exists.

In particular, the equation (8.32) will have the solution

f(x) =∫ ∞

0g(t) K(xt) dt , x > 0

if K∗(s) K∗(1 − s) = 1

Example 8.6.

Solve the integral equation∫ ∞

0f(ξ) g

(x

ξ

)dξ

ξ= h(x) , x > 0

Solution. Applying Mellin transform the given integral equation re-duces to

f∗(s) = h∗(s) K∗(s) , where K∗(s) =1

g∗(s)

Then inverting the Mellin transform we get the required solution of thegiven integral equation as

f(x) = M−1 [h∗(s) K∗(s)]

=∫ ∞

0h(ξ) K

(x

ξ

)dξ

ξ, by use of (8.30)

8.9 Application to Summation of Series.

A useful method of summing slowly convergent series has been given byMacfarlane (Phil Mag., 40,188,(1950)).

Theorem 8.3. If f∗(s) is Mellin transform of f(x), then∞∑

n=0

f(n + a) =1

2πi

∫ c+i∞

c−i∞f∗(s) ζ(s, a) ds

where ζ(s, a) is the generalised zeta function of Riemann, defined whenRe s > 1, by the equation

ζ(s, a) =∞∑

n=0

(s + a)−n , 0 < a � 1 , Re s > 1


Proof. It follows from inverse Mellin transform that

f(n + a) =1

2πi

∫ c+i∞

c−i∞f∗(s) (n + a)−s ds (8.34)

Summing this result over all n from n = 0 to n → ∞, it gives

∞∑n=0

f(n + a) =1

2πi

∫ c+i∞

c−i∞f∗(s) ζ(s, a) ds (8.35)

Similarly proceeding, we can have

f(nx) = M−1[n−s f∗(s)

]=

12πi

∫ c+i∞

c−i∞x−s n−s f∗(s) ds

Thus,∞∑

n=0

f(nx) =1

2πi

∫ c+i∞

c−i∞x−s f∗(s) ζ(s) ds = M−1 [f∗(s) ζ(s)]

(8.36)

where ζ(s) =∞∑

n=1

n−s

When x = 1 , eqn. (8.36) reduces to∞∑

n=1

f(n) =1

2πi

∫ c+i∞

c−i∞f∗(s) ζ(s) ds (8.37)

Illustrative Examples.

Example 8.7. Show that

(a)∞∑

n=1

(−1)n−1 n−s = (1 − 21−s) ζ(s)

(b)∞∑

n=1

sin an

n=

12

(π − a) .

Solutions.

(a) We know that

∞∑n=1

(−1)n−1 n−s · tn =∞∑

n=1

(−1)n−1 tn · 1Γ(s)

∫ ∞

0xs−1 e−nx dx

=1

Γ(s)

∫ ∞

0xs−1 dx

∞∑n=1

(−1)n−1 tn e−nx


=1

Γ(s)

∫ ∞

0xs−1 t e−x

1 + t e−xdx

=1

Γ(s)

∫ ∞

0xs−1 t

ex + tdx

Thus in the limit as t → 1, we finally get

∞∑n=1

(−1)n−1 n−s =1

Γ(s)M

[1

1 + ex; x → s

]

= (1 − 21−s) ζ(s)

(b) Mellin transform of f(x) = sin axx is given by

M

[sin ax

x; x → s

]=∫ ∞

0xs−2 sin ax dx

= −Γ(s − 1)as−1

cosπs

2, using Fourier sine transform

Thus,∞∑

n=1

sin an

n=

12πi

∫ c+i∞

c−i∞

Γ(s − 1)as−1

ζ(s) cosπs

2ds .

After using the functional equation for the zeta function

(2π)s ζ(1 − s) = 2 Γ(s) ζ(s) cos(πs

2

)the above equation gives

∞∑n=1

sin an

n= −a

2· 12πi

∫ c+i∞

c−i∞

(2πa

)s ζ(1 − s)(s − 1)

ds

=12

(π − a) ,

since the integrand has two simple poles at s = 0 and at s = 1inside the large semicircular contour closed on the left side of thes-plane.

8.10 The Generalised Mellin Transform.

In order to extend the applicability of Mellin transform D. Naylor(J. Math. Mech. 12, 265 - 274, 1963) has introduced the generalised


Mellin transform in a � r < ∞ when the unknown function in theconstitutive equation is prescribed at r = a. It is defined as

M− [f(r) ; r → s] =∫ ∞

a

[rs−1 − a2s

rs+1

]f(r) dr ≡ f∗

−(s) (8.38)

together with its inverse transform

M−1− [f∗(s) ; s → r] = f(r) ≡ 1

2πi

∫ c+i∞

c−i∞r−s f∗

−(s) ds , r � a, (8.39)

where f∗−(s) is analytic in |Re s| < γ with c < γ .

Again, on the other hand if the derivative of the unknown function inthe constitutive equation is prescribed, then the corresponding transfrompair is given by

M+ [f(r) ; r → s] = f∗+(s) ≡

∫ ∞

a

[rs−1 +

a2s

rs+1

]f(r) dr, (8.40)

and M−1+ [f∗

+(s) ; s → r] = f(r) ≡ 12πi

∫ c+i∞

c−i∞r−s f∗

+(s) ds , r � a

(8.41)

where c is defined as in eqn. (8.39)

Under (8.38), the differential operator expression over f(r) becomes

M−[r2 ∂2f

∂r2+ r

∂f

∂r

]= s2 f∗

−(s) + 2sas f(a) , (8.42)

provided f(r) is appropriately behaved at infinity. More precisely

limr→∞

[(rs − a2s r−s) r f∗

−(s) − s (rs + a2s r−s) f(r)]

= 0 .

Also under the eqn. (8.40) the same expression over f(r) becomes

M+

[r2 ∂2f

∂r2+ r

∂f

∂r

]= s2 f∗

+(s) − 2as+1 f ′(a) , (8.43)

where f ′(r) exists at r = a and f(r) is appropriately behaved at infinity.

These types of generalised Mellin transforms have been applied insolving physical problems on truncated conical region by B. Patra [ Bull.De L’Academic Pol. des. Sci., 24, 10, 1976] and many others.


8.11 Convolution of generalised Mellin Transform.

Theorem 8.4. If M+[f(r) ; r → s] = f∗+(s)

and M+[g(r) ; r → s] = g∗+(s), then

M+ [ g(r) f(r) ; r → s ] =1

2πi

∫L

f∗+(ξ) g∗+(s − ξ) dξ .

Proof. Assuming that f∗+(s) and g∗+(s) are analytic in |Re s| < γ , we

have

M+ [f(r) g(r) ; r → s] =∫ ∞

a

[rs−1 +

a2s

rs+1

]f(r) g(r) dr

=1

2πi

∫L

f∗+(ξ) dξ

(∫ ∞

ars−ξ−1 g(r) dr

)

+1

2πi

∫ ∞

a

a2s

rs+1g(r) dr

∫L

r−ξ f∗+(ξ) dξ

Replacing ξ by −ξ and using f∗+ (ξ) = a2ξ f∗

+ (−ξ) we obtain∫L

r−ξ f∗+ (ξ) dξ =

∫L

rξ a−2ξ f∗+ (ξ) dξ

The path of integration L : Re ξ = c becomes Re ξ = −c, but thesepaths can be reconciled if f∗

+(ξ) → 0 as [Im ξ] >> 1. Then∫ ∞

a

a2s

rs+1f(r) g(r) dr =

12πi

∫L

f∗+(ξ) dξ

∫ ∞

a

a2s−2ξ

rs−ξ+1g(r) dr

Therefore, using these results we get

M+ [f(r) g(r) ; r → s] =1

2πi

∫L

f∗+ (ξ) dξ

∫ ∞

ars−ξ−1 g(r) dr

+1

2πi

∫L

f∗+(ξ) dξ

∫ ∞

a

a2s−2ξ

rs−ξ+1g(r) dr (8.44)

=1

2πi

∫L

f∗+(ξ) g∗+(s − ξ) dξ .

Thus the proof is complete.

8.12 Finite Mellin Transform.

We conclude this chapter with an account of a finite Mellin transformdue to D. Naylor (1963) as defined below :


The finite Mellin transform of first kind over the range 0 � r � a isdefined as

M1 [ f(r); r → s ] =∫ a

0

[a2s

rs+1− rs−1

]f(r) dr ≡ f∗

1 (s) (8.45)

and its corresponding inversion formula is defined as

M−11 [ f∗

1 (s) ; s → r ] =1

2πi

∫L

r−sf∗+(s) ds ≡ f(r), (8.46)

where the contour L : (c − i∞ , c + i∞) in the s-plane in some strip|Re s| � γ and 0 < c < γ.

Also for the finite range of r in 0 � r � a , Mellin transform ofsecond kind is defined as

M2 [ f(r) ; r → s ] =∫ a

0

[a2s

rs+1+ rs−1

]f(r) dr ≡ f∗

2 (s) (8.47)

and its corresponding inversion formula is

M−12 [ f∗

2 (s) ; s → r ] =1

2πi

∫L

r−sf∗2 (s) ds ≡ f(r) (8.48)

The transform pairs shown above in eqns. (8.45) - (8.48) are useful insolving boundary value problems in finite conical regions.

Other finite Mellin transform pair, the third kind, suitable foranalysis of boundary value problems in spherical polar co ordinates aresimilarly defined.

For example, in 0 � r � a, the third kind Mellin transform and itsinversion are defined as

M3 [ f(r) ; r → s] =∫ a

0

[a2s+1

rs+1− rs

]f(r) dr ≡ f∗

3 (s) (8.49)

M−13 [ f∗

3 (s) ; s → r] =1

2πi

∫L

r−s−1 f∗3 (s) ds ≡ f(r) (8.50)

The finite Mellin transform of fourth kind and its inversion formula aredefined as

M4[f(r) ; r → s] =∫ a

0

[a2s+2

rs+1+ rs+1

]f(r) dr ≡ f∗

4 (s) (8.51)

M−14 [f∗

4 (s) ; s → r] =1

2πi

∫L

r−s−1 f∗4 (s) ds ≡ f(r) (8.52)


Illustrations.

(a) It is easy to verify by application of “integration by parts” that

M1

[r2frr + r fr ; r → s

]=∫ a

0

[a2s

rs+1− rs−1

] [r2 frr + r fr

]dr

= 2s as f(a) − s2 f∗1 (s) (8.53)

Clearly, this transform is useful if f(a) is prescribed as one of theboundary conditions in the corresponding boundary value problems.

(b) Similarly it may be verified that

M2

[r2 frr + r fr ; r → s

]=∫ a

0

[a2s

rs+1+ rs−1

] [r2 frr + r fr

]dr

= 2as−1 f ′(a) + s2 f∗2 (s) . (8.54)

This transform is useful when f ′(a) is prescribed as one of theboundary conditions in the corresponding boundary value problems.

(c) When there is an axisymmetric differential operator

L =[r2 d2

dr2+ 2r

d

dr

]

in the constitutive Laplace equation of the boundary value problemwith 0 � r � a, we can similarly, as in (a) and (b), verify that

M3 [Lf(r) ; r → s] = s(s + 1) f∗3 (s) + 2as+1 f(a) (8.55)

and M4 [Lf(r) ; r → s] = s(s + 1)f∗4 (s) + 2as+3 f ′(a) (8.56)

These formulae in eqns. (8.55) or (8.56) are useful in transformingthe portion Lf(r) of the associated axisymmetric Laplace equationof the physical problem when either f(a) or f ′(a) are prescribedas one of the respective boundary conditions in the correspondingboundary value problems.


Exercises.

(1) Show that

(a) M [ cos x ; x → s ] = Γ(s) cos πs2 , 0 < Re s < 1

(b) M [ sin x ; x → s ] = Γ(s) sin πs2 , 0 < Re s < 1

(c) M[(1 + x2)−1 ; x → s

]= π

2 cosec(

πs2

), 0 < Re s < 2

(d) M[12 sech x ; x → s

]= Γ(s) L[s] , where L(s) =

∑∞n=1

1(2n−1)s

(2) Show that

M[e−x cos ϕ cos(x sin ϕ) ; x → s

]= Γ(s) cos s ϕ, Re s > 0 , |ϕ| <

π

2M[e−x cos ϕ sin(x sin ϕ) ; x → s

]= Γ(s) sin s ϕ, Re s > −1 , |ϕ| <

π

2

Deduce that if |ϕ| < π2 ,

M−1[ cos s ϕ

sin s π; s → x

]=

1π

· 1 + x cos ϕ

1 + 2x cos ϕ + x2

M−1

[sin s ϕ

sin s π; s → x

]=

1π

· x sin ϕ

1 + 2x cos ϕ + x2

and hence that if |ϕ| < π2n , n > 0

M−1

[cos s ϕ

sin sπn

; s → x

]=

n

π· 1 + xn cos (nϕ)

1 + 2xn cos (nϕ) + x2n

M−1

[sin s ϕ

sin sπn

; s → x

]=

n

π· xn sin (nϕ)

1 + 2xn cos (nϕ) + x2n

(3) Prove that

(a) M

[2√π

∫ x

0

f(t) dt√x2 − t2

; x → s

]=

Γ(−s+1

2

)Γ(1 − s

2

) f∗(s)

(b) M

[2√π

∫ ∞

x

f(t) dt√t2 − x2

; x → s

]=

Γ(

s2

)Γ(

s+12

) f∗(s)

(4) Show that the integral equation

f(x) = h(x) +∫ ∞

0g(xξ) f(ξ) dξ


has the formal solution

f(x) =1

2πi

∫ c+i∞

c−i∞

[h∗(s) + g∗(s) h∗(1 − s)

1 − g∗(s) g∗(1 − s)

]x−s ds

(5) Show that∞∑

n=1

cos kn

n2=

k2

4− πk

2+

π2

6

Hence deduce that ∞∑n=1

1n2

=π2

6

(6) If f(x) =∑∞n=1

an e−nx , show that

M [f(x) ; x → s] = f∗(s) = Γ(s) g∗(s)

where g∗(s) =∞∑

n=1

an n−s .

If an = 1, for all n , derive

f∗(s) = Γ(s) ζ(s)

Show also that M

[exp (−ax)1 − e−x

; x → s

]= Γ(s) ζ(s, a) .

(7) Show that

∞∑n=1

cos an

n3=

112[a3 − 3πa2 + 2π2a

]

(8) Prove that

(a) M

[∫ ∞

0ξn f(xξ) g(ξ) dξ ; x → s

]= f∗(s) g∗(1 + n − s)

(b) M

[∫ ∞

0ξn f

(x

ξ

)g(ξ) dξ ; x → s

]= f∗(s) g∗(s + n + 1)

(9) Show that the solution of the boundary value problem defined by

r2 ϕrr + r ϕr + ϕθθ

= 0 , 0 < r < ∞ , 0 < θ < π

ϕ(r, 0) = ϕ(r, π) = f(r)

is ϕ(r, θ) =1

2πi

∫ c+i∞

c−i∞r−s f∗(s) cos

{s(θ − π

2

)}ds

cos(

πs2

)

Chapter 9

Finite Laplace Transforms

9.1 Introduction.

Laplace transform of a piecewise continuous function f(t) for 0 � t < ∞,where |f(t)| < Meσ t,M, σ being real constants defined in section 3.3by

F (s) =∫ ∞

0f(t) e−st dt

exists when Re s > σ. But from the above definition it can not beconcluded that Laplace transform of any general form of f(t) do exist.For example, if f(t) = eat2 , a > 0, its Laplace transform does notexist. Again from the physical point of view there are situations wheredetermining the response of a linear system with disturbance f(t) of anygeneral form in 0 � t � T, T is finite is also a pertinent question fordiscussion.

To study all such situations raised above the power and elegance ofFinite Laplace transform defined below may be found very useful. In thisdirection the papers of H.S. Dunn (Proc. Camb. Phil. Soc. 63, 1967)and L. Debnath and J. Thomas (ZAMM, 56, 1976) may be consulted asbasic and pioneering works.

9.2 Definition of Finite Laplace Transform.

Finite Laplace transform of a continuous function or of a piecewise con-tinuous function f(t) in 0 � t � T satisfying the Lipschitz condition isdenoted by

LT [ f (t) ; t → s ] = f(s, T )

and is defined by

LT [f(t) ; t → s] = f(s, T ) =∫ T

0f(t) e−st dt, (9.1)

Finite Laplace Transforms 303

where s is any real or complex number and T is a finite real number.

The inverse finite Laplace transform is defined by the complexintegral

f(t) = L−1T [ f (s, T ) ; s → t ] =

12πi

∫ c+i∞

c−i∞f (s, T ) est ds, (9.2)

where the contour of integration denoted by Γ is a straight line joiningpoints c − iR and c + iR as R → ∞ and when the function f(t) iscontinuous in 0 � t � T . If f(t) is a piecewise continuous functionwith a finite number of finite discontinuities in 0 � t � T , then thecorresponding inverse finite transform of f (s, T ) at t is given by

12πi

∫Γ

f (s, T ) e−st ds =12[ f (t − 0) + f (t + 0)] (9.3)

as R → ∞. This formula (9.3) is obtained due to the fact that f (s, T )is an entire function of the complex variable s.

Note. 9.1. Let∫

f (t) e−stdt = −F (s, t) e−st.

Then f (s, t) =∫ T

0f(t) e−st dt = F (s, 0) − F (s, T ) e−st, (9.4)

where F (s, 0) = f (s) =∫ ∞

0f (t). e−st dt (9.5)

Therefore, from (9.2) and (9.4) we can write

f(t) =1

2πi

∫Γ

F (s, 0) est ds − 12πi

∫Γ

F (s, T ) es(t−T ) dt, (9.6)

where Γ is a line contour from c − iR to c + iR as R → ∞.

The first integral in (9.6) may be closed on the left half plane. Fort < T, the contour of the second integral is closed in the right half plane.So, selectintg Γ such that all the poles of F (s, 0) lie to the left of Γ andhence the first integral is the solution of the initial value problem, thesecond integral being zero in this case. When t > T , the second integralis closed in the left half of the complex s-plane so that f(t) = 0. Thus,there is no need to consider the second integral and this is identical withusual Laplace transform.

Note. 9.2. Following the definitions of usual Laplace transform andFinite Laplace transform it may be noted that if the usual Laplace trans-form of a function f(t) exists, then its Finite Laplace transform must


exist. This is so because

f(s) = L [f(t); t → s] =∫ ∞

0f(t)e−stdt =

∫ T

0e−stf(t)dt +

∫ ∞

Te−stf(t)dt

and the existence of the left hand side integral implies the existence ofthe second integral of the right hand side of the above equation. Butthe converse of this statement is not true. Since, for example

L [eat2 ; t → s ] =∫∞0 eat2e−st dt , for a > 0 does not exist though∫ T

0 eat2 e−stdt does exist for a > 0.

9.3 Finite Laplace Transform of elementary functions.

Determine finite Laplace transforms of f(t) in the following examplessupplied in 9.1-9.10

Example 9.1. f(t) = 1

Solution.

LT [1; t → s ] =∫ T

01e−st dt =

1s

[1 − e−sT ] (9.7)

Example 9.2. f(t) = eat

Solution.

LT [eat ; t → s] =∫ T

0e−(s−a)t dt =

1 − e−(s−a)T

s − a(9.8)

Example 9.3. f(t) = sin at

Solution.

LT [sin at ; t → s ] =∫ T

0sin at. e−st dt

=a

s2 + a2− e−sT

s2 + a2(s sin aT + a cos aT ) (9.9)

Example 9.4. f(t) = cos at

Solution.

LT [cos at ; t → s ] =∫ T

0cos at. e−st dt

=s

s2 + a2+

e−sT

s2 + a2[a sin aT − s cos aT ] (9.10)


Example 9.5. f(t) = t

Solution.

LT [ t; t → s ] =∫ T

0t e−st dt

=1s2

− e−sT

s[1s

+ T ] (9.11)

Example 9.6. f(t) = tn , n = positive integer.

Solution.

LT [ tn; t → s ] =∫ T

0tn e−st dt

=n!

sn+1− e−sT

s[ T n +

n

sT n−1 +

n(n − 1)s2

T n−2

+ · · · + n! T

sn−1+

n!sn

] , (9.12)

after using integration by parts successively.

Example 9.7. f(t) = ta , a > −1

Solution.

LT [ ta; t → s ] =∫ T

0ta. e−st dt , putting u = st

= s−(a+1)

∫ sT

0ua e−u du

= s−(a+1)γ (a + 1 , sT ) , (9.13)

by use of incomplete gamma function defined by

γ (α, x) =∫ x

0e−u uα−1 du

Example 9.8. f(t) = a periodic function of t of period w.

Solution. We have f(t + nw) = f(t), for all integer value of n.

Now,

LT [f(t) ; t → s] =∫ T

0f(t) e−st dt

=∫ nw

0f(t) e−st dt


=∫ w

0f(t)e−stdt +

∫ 2w

wf(t) e−stdt + · · · +

∫ nw

(n−1) wf(t) e−st dt

=∫ w

0f(u) e−su du + e−sw

∫ w

0f(u) e−su du + · · ·

+e−s(n−1)w

∫ w

0f(u)e−su du

=1 − e−nsw

1 − e−sw

∫ w

0e−su f(u) du

=1 − e−nsw

1 − e−swf (s,w) (9.14)

When n → ∞, the above result reduces to

f(s) =

∫ w0 f(u)e−sudu

(1 − e−sw),

which is a known result when T → ∞(i, e n → ∞) , for usual LaplaceTransform.

Example 9.9. f(t) = erf(at)

Solution. LT [ erf(at) ; t → s ] =∫ T0 erf (at) e−st dt

Defining the Error function by erf(x) = 2√π

∫ x0 e−α2

dα, we have

∫ T

0erf (at).e−st dt

= −[

e−st

serf (at)

]T

0

+1s

2√π

∫ T

0exp[−{ st + a2t2 }] dt

= −e−st

serf(aT ) +

1s

exp

(s2

4a2

)2√π

∫ aT+ 12a

12a

e−u2du

= −e−sT

serf(aT ) +

1sexp (

s2

4a2)[erf(aT +

s

2a

)− erf

( s

2a

)].

(9.15)

Example 9.10. f(t) = H(t − a)

Solution.

LT [ H(t − a) ; t → s ] =∫ T

0H(t − a) e−st dt

=∫ T

ae−st dt =

1s

(e−as − e−Ta) . (9.16)


Example 9.11. Find finite Laplace Transform of

(i) f(t) = cosh at, a > 0 (ii) f(t) = t4 e−at, a > 0

Solutions.

(i) LT [ cosh at ; t → s ] =∫ T

0

12

(eat + e−at) . e−st dt

=12

1 − e−(s−a) T

s − a+

12

1 − e−(s+a)T

s + a

=s

s2 − a2− e−st

[eaT

s − a+

e−aT

s + a

]

(ii) LT [ t4 e−at ; t → s ] =∫ T

0t4 e−at e−st dt

=∫ T

0t4 e−(s+a)t dt

=4!

(s + a)5− e−(s+a)T

s + a

[T 4 +

4 T 3

s + a+

12 T 2

(s + a)2+

24 T

(s + a)3+

24(s + a)4

]

Example 9.12. If f(x) has a discontinuity at t = a where 0 < a < T ,prove that

LT [ f ′(t) ; t → s ] = sf (s, T )+e−sT f(T )−f(0)−e−sa [f(a + 0) − f(a − 0)]

Solution.

LT [ f ′(t) ; t → s]

=∫ a−0

0f ′(t)e−st dt +

∫ T

a+0f ′(t) e−st dt

=[e−stf(t)

]a−0

0+ s

∫ a−0

0f(t)e−stdt +

[e−stf(t)

]Ta+0

+ s

∫ T

a+0e−stf(t)dt

= sf(s, T ) + e−sT f(T ) − f(0) − e−sa [f(a + 0) − f(a − 0)] .

When f(t) has a finite number k of finite discontinuities at t = a1, a2, · · · ak

this result can be generalised after application of the above method ina similar manner.

9.4 Operational Properties.

Theorem 9.1. If LT [ f(t) ; t → s ] = f(s, T ), then

LT { e−atf(t) ; t → s } = f(s + a , T )


This property of finite Laplace transform is called its shifting property.

Proof. It is given that f(s, T ) =∫ T0 f(t)e−st dt.

Therefore,

LT

[e−at f(t) ; t → s

]=∫ T

0f(t) . e−at.e−st dt

=∫ T

0f(t) . e−(s+a)t. dt

= f (s + a, T ) (9.17)

Theorem 9.2. If LT [ f(t) ; t → s ] = f(s, T ),then LT [ f(at ; t → s) ] = 1

a f ( sa , aT )

This property of finite Laplace transform is called its scaling property.

Proof. Since, LT [ f(t) ; t → s ] = f(s, T ), we have

LT [f(at) ; t → s ] =∫ T

0f(at) . e−st dt

=1a

∫ aT

0f(x) . e−

sax. dx, where at = x

=1af(

s

a, aT ). (9.18)

Theorem 9.3. (Finite Laplace transform of Derivatives)

If LT [f(t) ; t → s ] = f (s, T ) ,

then LT [f ′(t) ; t → s ] = sf (s, T ) − f(0) + e−sT f(T )

LT [f ′′(t) ; t → s ] = s2f (s, T ) − sf(0) − f ′(0) + sf(T )e−sT + f ′(T )e−sT

and generally, LT [ f (n)(t) ; t → s ] = snf(s, T ) −n∑

k=1

sn−k f (k−1)(0)

+e−sTn∑

k=1

sn−k f (k−1) (T ) .

Proof. Applying integration by parts, we have

LT [f ′ (t) ; t → s ] =∫ T

0f ′ (t) e−st dt

= sf (s, T ) − f(0) + e−sT f(T ) (9.19)

Let f ′(t) ≡ ϕ(t) . Then f ′′(t) = ϕ′ (t) and hence


LT [f ′′(t) ; t → s] = LT [ϕ′(t) ; t → s]

= s LT [ϕ(t) ; t → s] − ϕ(0) + e−sT ϕ(T )

= s[sf(s, T ) − f(0) + e−sT f(T )] − f ′(0) + e−sT f ′(T )

= s2 f(s, T ) − s f(0) − f ′(0) + e−sT [s f(T ) + f ′(T )] (9.20)

Repeating the above process (n−1) times, the last formula can be provedeasily.

Theorem 9.4. (Finite Laplace transform of Integrals).If LT [f(t) ; t → s] = f(s, T ), then

LT

[∫ t

0f(u) du

]=

1s

[f(s, T ) − e−sT

∫ T

0f(u) du

]

Proof. Let the integral∫ t

0f(u) du = ϕ(t), so that ϕ′(t) = f(t)

Now, LT [ϕ′(t) ; t → s] = s LT [ϕ(t) ; t → s] − ϕ(0) + e−sT ϕ(T )

⇒ LT [f(t) ; t → s] = s LT

[∫ t

0f(u) du ; t → s

]+ e−sT ϕ(T )

⇒ LT

[∫ t

0f(u) du ; t → s

]=

1s

[f(s, T ) − e−sT

∫ T

0f(u) du

](9.21)

Theorem 9.5. (Derivative of finite Laplace transform).

If LT [f(t) ; t → s] = f(s, T ), thend

ds[f(s, T )] = LT [(−t)f(t) ; t → s]

d2

ds2[f(s, T )] = LT [(−t)2f(t) ; t → s]

and generallydn

dsn[f(s, T )] = LT [(−t)n f(t) ; t → s] .

Proof.

Since, f(s, T ) =∫ T

0e−st f(t) dt

we haved

dsf(s, T ) =

∫ T

0

∂

∂se−st · f(t) dt


=∫ T

0e−st (−t) f(t) dt

= LT [(−t) f(t) ; t → s] (9.22)d2

ds2f(s, t) =

∫ T

0

∂2

∂s2e−st · f(t) dt

=∫ T

0e−st · (−t)2f(t)dt

= LT [(−t)2f(t) ; t → s] . (9.23)

In general, similarly we get

dn

dsnf(s, T ) =

∫ T

0

∂n

∂sne−st · f(t) dt

=∫ T

0e−st (−t)n f(t) dt

= LT [(−t)nf(t) ; t → s] . (9.24)

Theorem 9.6. Finite Laplace transform of the product of tn and m thderivatives of f(t) is

(−1)ndn

dsn

[smf(s, t) −

m∑k=1

sm−kfk−1(0) + e−sTm∑

k=1

sm−kf (k−1)(T )

].

Proof.

We have LT [ tnf ′(t) ; t → s ] = (−1)ndn

dsn[ LT{f ′(t) ; t → s} ]

= (−1)ndn

dsn[s f(s, T ) − f(0) + e−sT f(T )]

= (−1)n[

dn

dsn{s f(s, T )}

]+ (T )n e−sT f(T ) (9.25)

Similarly, LT [tnf (m)(t) ; t → s]

= (−1)ndn

dsn

[smf(s, T ) −

m∑k=1

sm−k f (k−1) (0) + e−sTm∑

k=1

sm−kf (k−1)(T )

]

(9.26)

Theorem 9.7. Integral of finite Laplace transform of a function is givenby ∫ T

sf(s, T )ds =

∫ T

0

f(t)t

e−stdt −∫ T

0

f(t)t

e−tT dt.


Proof.

Let f(s, T ) = LT [f(t) ; t → s] =∫ T

0f(t) · e−st dt

Then∫ T

sf(s, T ) ds =

∫ T

sds

∫ T

0f(t) e−st dt

=∫ T

0f(t) dt

∫ T

se−st ds

=∫ T

0

f(t)t

e−st dt −∫ T

0

f(t)t

e−tT dt , (9.27)

provided that both the integrals on the right hand side of eqn. (9.27)exist.

9.5 The Initial Value and the Final Value Theorem .

These theorems give the behavior of the object functions in terms of thebehavior of the transformed functions

Theorem 9.8. (The Initial Value Theorem).

If f(t) be at most a piecewise continuous function, for 0 ≤ t ≤ T andif LT [f(t) ; t → s] = f(s, T ) then,

lims→∞[sf (s, T ) ] = lim

t→0f(t) (9.28)

Proof. The proof of the above theorem can be developed like that inthe theorem of section 3.15.

Theorem 9.9. (The Final value Theorem)

If f(t) be at most a piecewise continuous function for 0 � t � T andif LT [f(t) ; t → s ] = f(s, T ) then,

limt→∞ f(t) = lim

s→0sf (s, T ) (9.29)

Proof. The proof of this theorem can also be developed as that in theTheorem of section 3.16

Theorem 9.10. The solution of an initial value problem is identicalwith that of the final value problem.

Proof. Suppose fin(t) is the solution of the initial value problem, andit is given by

fin(t) =1

2πi

∫Γ

F (s, 0) est ds (9.30)


where Γ is the Bromwich contour extending from c − iR to c + iR asR → ∞.

Again suppose ffi(t) is the solution of the finial value problem andis defined by

ffi(t) =1

2πi

∫Γ

F (s, T ) es(t−T )ds (9.31)

where Γ lies to the left of the singularities of F (s, t) or of F (s, 0). Thenfrom eqns. (9.27) and (9.28) we get

fin(t) − ffi (t) =1

2πi

∫C

[ F (s, 0) − F (s, T ) e−sT ] est ds

=1

2πi

∫C

f(s, T ) est ds (9.32)

where C is a closed contour which contains all the singularities of F (s, 0)or of F (s, T ) and F (s, t) is defined by

−F (s, t)e−st =∫

f(t) e−st dt

The integrand in eqn. (9.32) is an entire function of s and hence byCauchy’s theorem, the integral must vanish.

Hence.fin (t) = ffi (t) = f(t) (9.33)

9.6 Applications

Example 9.13. Verify Theorems 9.8 and 9.9 when f(t) = exp (−t).

Solution.

Let LT [f(t) ; t → s] = f(s, T ). Then for f(t) = exp (−t),

we have f(s, T ) =1 − e−(1+s)T

(1 + s). Therefore,

lims→∞ f(s, T ) = 0, by L′Hospital′s rule. Again,

lims→∞ s f(s, T ) = lim

s→∞(1 + s − 1)(1 − e−(s+1)T )

1 + s

= lims→∞[1 − e−(s+1)T ] − lim

s→∞1 − e−(s+1)T

1 + s


= 1 − 0 = 1

Also, limt→0

f(t) = limt→0

e−t = 1

Thus lims→∞ s f(s, T ) = lim

t→0f(t) is verified.

Also, lims→0

f(s, T ) = lims→0

1 − e−(1+s)T

1 + s= 1 − e−T

and∫ T

0f(t) dt = 1 − e−T

∴ lims→0

f(s, T ) =∫ T

0f(t) dt is verified

Further, lims→0

s f(s, T ) = lims→0

s

[1 − e−(1+s)T

]s + 1

= 0

Thus both the theorems 9.8 and 9.9 are verified.

Example 9.14. Use finite Laplace transform to solve the initial valueproblem

di

dt= A , 0 ≤ t ≤ T ; i(0) = 0 .

Solution. Finite Laplace transform of the ODE under the initial con-dition x(0) = 0 gives

s i (s, T ) + i(T ) e−sT =A

s

[1 − e−sT

]or i (s, T ) +

e−sT

si(T ) = A

[1 − e−sT

s2

]

Here i (s, T ) is not an entire function unless we choose i(T ) = AT .Then we have

i (s, T ) =−AT

s[ e−sT ] +

A

s2(1 − e−sT )

= A

[1s2

− e−sT

s

(1s

+ T

)]

Inverting finite Laplace transform the above equation gives

i(t) = At.

Example. 9.15. The equation for vertical displacement of horizontaltaut string between two fixed points x = 0 and x = L caused by a


concentrated or distributed load W (x) normalised with the tension ofthe string is

d2y

dx2= W (x) , 0 ≤ x ≤ L

Find y(x).

Solution. The above boundary value problem under the above state-ment is subjected to the boundary conditions

y(0) = y(L) = 0

Let us assume here that the given load is a concentrated load of unitmagnitude at x = a and therefore.

W (x) = δ(x − a) , 0 < a < L

Taking finite Laplace transform over x ∈ (0, L) the above differentialequation under the loading W (x) = δ(x − a) gives

y (s, L) =1s2

[e−sa + y′(0) − e−sL y′(L)

]This function y(s, L) is not an entire function of s unless we choosey′(0) = y′(L) − 1. Using this condition the above equation can beexpressed as

y(s, L) =y′(0)s2

[1 − e−sL − sL e−sL

]+

e−sa

s2

[1 − e−s(L−a)

+ y′(0) sL e−s(L−a)]

To complete the inversion of finite Laplace transform in the above equa-tion we assume Ly′(0) = a − L so that we get

y(x) = xy′(0) + (x − a) H (x − a)

Example 9.16. In an L − R electrical circuit the current I(t) satisfiedthe ODE

LdI

dt+ RI = E0 cos wt , 0 � t � T,

under the emf E0 cos wt and I(t) = 0 at t = 0 ; L,R,E0 being constants.Find I(t) using finite Laplace transform.


Solution.

Applying finite Laplace transform the given ODE becomes

sI(s, T ) + e−sT I(T ) +R

LI(s, T ) =

E0

L

[s

s2 + w2

+e−sT

s2 + w2(w sin wT − s cos wT )

]

I(s, T ) = −e−sT I(T )s + R

L

+E0

L(s + RL )

[s

s2 + w2

+e−sT


]

Since I(s, T ) is not an entire function of s, we make it entire by choosing

I(T ) =E0

L

w

w2 + R2

L2

[R

wLcos wT + sin wT − R e−

RTL

wL

]

Using this result we get

I(s, T ) =w E0

L(w2 + R2

L2

) [ R

wL

{s

s2 + w2+

e−sT


}]

+w E0

L(w2 + R2

L2

) [{ w

s2 + w2− e−sT

s2 + w2(s sin wT + w cos wT )

}]

− w E0

L(w2 + R2

L2

)[

R

wL(s + R

L

) {1 − e−(s+ RL )T}]

Inverting finite Laplace transformed equation, the above relation gives

I(t) =w E0

L(w2 + R2

L2

) { R

wLcos wt + sin wt − R

wLe−

RtL

}

It may be noted that the first two terms on the right hand side of theabove equation represent the steady-state current and the third termthere represents the transient part of the current field I(t) atany time t.


Exercises.

(1) Find finite Laplace transform of

f(t) = sinh at[Ans. a

s2−a2 + e−sT

2

(eaT

a−s + e−aT

a+s

)]

(2) Use finite Laplace transform to solve the initial value problem

dx

dt+ α x = At , 0 � t � T ; x(0) = a

to prove that

x(t) =(

a +A

α2

)e−αt +

At

α− A

α2

(3) Solve the simple harmonic oscillator governed by

d2x

dt2+ w2x = F , x(0) = a , x(0) = u

where F, a and u are constants, to prove that

x(t) =(

a − F

w2

)cos wt +

u

w2sin wt +

F

w2.

Chapter 10

Legendre Transforms

10.1 Introduction.

As an aid to solve some special type of boundary value problems thischapter deals with the Legendre transform whose kernel is a Legendrepolynomial. Initially, definition and basic operational properties of thetransform are discussed based on the papers of Churchill (J.Math. andPhysics, 33. 165, 1954) and Churchill and Dolph (Proc. Amer. Math,Sec, 5, 93, 1954).

10.2 Definition of Legendre Transform.

Legendre transform of a function f(x) over the interval −1 � x � 1 isdefined by

fl (n) ≡ l [ f(x) ; x → n ] =∫ 1

−1f(x) Pn(x) dx , (10.1)

where Pn(x) is the Legendre polynomial of integral degree n(� 0) pro-vided that the integral in (10.1) exists.

The orthogonal property of Legendre polynomial is well-known as∫ 1

−1Pm(x) Pn(x) dx =

22n + 1

δmn (10.2)

and this can also be expressed as

l [ Pm(x) ; x → n ] =2

2n + 1δmn (10.3)

From eqns. (10.2) and (10.3) Fourier-Legendre expansion theorem ofa piecewise continuous function f(x) for −1 � x � 1 is given by


∞∑n=0

(n +

12

)Pn(x)fl(n) = f(x) ,

if f(x) ∈ C (−1, 1) =12

[f(x + 0) + f(x − 0)] , if f(x) ∈ P 1(−1, 1)

(10.4)

We, therefore, have the inversion formula for Legendre transform is

l−1[

fl (n) ; n → x]

=∞∑

n=0

(n +

12

)fl(n) Pn(x) (10.5)

10.3 Elementary properties of Legendre Transforms.

(1) Since P0(x) = 1 and P1(x) = x we get

fl(0) =∫ 1

−1f(x) dx (10.6)

and fl(1) =∫ 1

−1xf(x) dx (10.7)

Again, l[f ′(x) ; x → 0] =∫ 1

−1f ′(x)dx = f(1) − f(−1)

(10.8)

l[f ′(x) ; x → 1] =∫ 1

−1xf ′(x) dx = f(1) + f(−1)

−∫ 1

−1f(x) dx (10.9)

Also, l[f ′′(x) ; x → 1] = f ′(1) + f ′(−1) −∫ 1

−1f ′(x)dx

= f ′(1) + f ′(−1) − f(1) + f(−1) (10.10)

and l[C ; x → 0] =∫ 1

−1C dx = 2C (10.11)

l[C ; x → n] =∫ 1

−1C Pn(x)dx

= C

∫ 1

−1Pn(x)dx = C

∫ 1

−1P0(x) Pn(x) dx

⇒ l[C ; x → n] = 0 , when n = 1, 2, 3, · · ·(by putting m = 0 in eqn. (10.3)). (10.12)

Legendre Transforms 319

(2) Again since, Pn(−x) = (−1)n Pn(x) we have

l[f(−x) ; x → n] =∫ 1

−1f(−x) Pn(x) dx

=∫ 1

−1f(x) Pn(−x) dx

= (−1)n∫ 1

−1f(x) Pn(x) dx

= (−1)n l[f(x) ; x → n]

From this result we can write

l−1[(−1)n fl(n) ; n → x

]= f(−x) (10.13)

(3) Legendre transform of eiαx is given by

ln [eiαx ; x → n] =∫ 1

−1eiαx Pn(x) dx =

√2πα

in Jn+ 12

(α),

(10.14)from table of integrals (Copson, 1935). Similarly,

ln [eαx ; x → n] =

√2πα

In+ 12

(α) (10.15)

(4) Also

l[(1 − x2)−

12 ; x → n

]=∫ 1

−1

Pn(x)dx√1 − x2

= π P 2n(0) (10.16)

and l

[1

2(t − x); x → n

]=

12

∫ 1

−1

Pn(x) dx

t − x= Qn(t) , |t| > 1

(10.17)

These results are also obtained from “Tables of integral Trans-forms” by Harry Batemann (1954).

(5) From the generating relation of Legendre polynomial we have

(1 − 2rx + r2)−12 =

∞∑n=0

rn Pn(x) , |r| < 1


Multiplying both sides by Pn(x) and integrating the result withrespect to x from −1 to 1, we get∫ 1

−1(1 − 2rx + r2)−

12 Pn(x) dx = rn

∫ 1

−1P 2

n(x) dx

=2rn

(2n + 1)(10.18)

When r = 1, we get∫ 1

−1[2(1 − x)]−

12 Pn(x) dx =

22n + 1

⇒∫ 1

−1(1 − x)−

12 Pn(x) dx =

2√

22n + 1

Or l [(1 − x)−12 ; x → n ] =

2√

22n + 1

(10.19)

Defferentiating eqn. (10.18) with respect to r and after multiplyingthe result by r, we get

12

∫ 1

−1(1 − 2rx + r2)−

32 (2rx − 2r2) Pn(x) dx =

2nrn

2n + 1

⇒ 12

∫ 1

−1(1 − 2rx + r2)−

32[−{1 − 2rx + r2 + r2 − 1}]Pn(x)dx

=2nrn

2n + 1

⇒ −12

∫ 1

−1(1 − 2rx + r2)−

12 Pn (x)

+(1 − r2)

2

∫ 1

−1(1 − 2rx + r2)−

32 Pn (x) dx

=2nrn

2n + 1

⇒ −l[

(1 − 2rx + r2)−12 ; x → n

]+(1 − r2) l

[(1 − 2rx + r2)−

32 ; x → n

]=

4nrn

2n + 1

Now using eqn. (10.18), the above result reduces to

l[(1 − 2rx + r2)−32 ; x → n]

=[

4nrn

2n + 1+

2rn

2n + 1

]1

1 − r2


=2rn

1 − r2(10.20)

(6) Legendre transform of H(x) is given by

l [ H(x) ; x → n ] =∫ 1

−1H (x) Pn(x) dx

=∫ 1

0H(x) Pn(x) dx

=∫ 1

0Pn(x) dx

If n = 0, Pn(x) = P0(x) = 1 and hence l[ H(x) ; x → n ] = 1.

If n > 1, we know from the recurrence relation of Legendre poly-nomial that

(2n + 1) Pn(x) = P′

n+1 (x) − P′

n−1 (x) (10.21)

Therefore, in this case∫ 1

0Pn (x) dx

=1

2n + 1

∫ 1

0[ P

′n+1(x) − P

′n−1 (x) ] dx

=1

2n + 1[Pn−1 (0) − Pn+1 (0)]

Thus, l[ H(x) ; x → n ] =

{1 , for n = 0

12n+1 [Pn−1 (0) − Pn+1 (0)] , for n � 1

(10.22)

(7) From eqn. (10.21) we have∫ 1

−1g(x) Pn(x) dx =

12n + 1

[ ∫ 1

−1g(x)P

′n+1 (x) dx

−∫ 1

−1g(x) P

′n−1 (x) dx

]

=1

2n + 1

[g(x) Pn+1 (x)|1−1 −

∫ 1

−1g′(x) Pn+1 (x) dx

−g(x) P′

n−1 (x)|1−1 +∫ 1

−1g′(x) Pn−1(x) dx

]


=1

2n + 1[ g(x) { Pn+1(x) − Pn−1 (x)}1

−1

−∫ 1

−1g′(x) {Pn+1(x) − Pn−1 (x)} dx ]

Thus if g(x) =∫ x−1 f(t)dt is a continuous function in (−1, 1)

l [ g(x) ; x → n ] =1

2n + 1[

l{ g′(x) ; x → (n − 1) } ]− 1

2n + 1[l{ g′ (x) ; x → n + 1 } ] (10.23)

provided the expression [· · ·]1−1 vanishes. Thus, in particular, wehave the special case

l

[ ∫ x

−1f(t) dt ; x → n

]=

12n + 1

[fl(n − 1) − fl(n + 1)

](10.24)

Therefore, by repeated application of (10.8) and (10.9) and by(10.23) we get

l[

f ′ (x) ; x → 2m]

= f(1) − f(−1) −m−1∑r=0

(4m − 4r − 1) fl (2m − 2r − 1)

(10.25)

l[

f ′ (x) ; x → 2m + 1]

= f(1) − f(−1) −m−1∑r=0

(4m − 4r + 1) fl (2m − 2r)

(10.26)

Also from the recurrence relation

(n + 1) Pn+1 (x) − (2n + 1) x Pn (x) + n Pn−1 (x) = 0

we can easily deduce that

l [ xf (x) ; x → n ] =1

2n + 1[

(n + 1) fl (n + 1) + n fl (n)]

(10.27)


10.4 Operational Properties of Legendre Transforms

Theorem 10.1 If f ′(x) is continuous and f ′′(x) is bounded and inte-grable in each subinterval of −1 � x � 1 and if Legendre transform off(x) exists and

lim|x|→1

(1 − x2)[

f(x) , f ′(x)]

= 0

then l

[d

dx{ (1 − x2)

d

dxf(x) } ; x → n

]= −(n + 1) n fl (n).

Proof. We have

l

[d

dx

{(1 − x2)

d

dxf(x)

}; x → n

]

=∫ 1

−1

d

dx

{(1 − x2)

d

dxf(x)

}Pn (x) dx.

= −∫ 1

−1(1 − x2)

d

dxf(x) P ′

n(x) dx ,

on integration by parts and using the given condition.

Again on integration by part the above result becomes

= −[(1 − x2) P ′

n (x) f(x)|1−1 −∫ 1

−1

d

dx

{(1 − x2) P ′

n (x)}

f(x) dx

]

=∫ 1

−1

d

dx

{(1 − x2)

d

dxPn (x)

}f(x) dx , by using given condition.

But it is known that

d

dx

[(1 − x2)

d

dxPn (x)

]= −n(n + 1) Pn(x)

and therefore,

l

[d

dx

{(1 − x2)

df (x)dx

}; x → n

]= −n (n + 1)

∫ 1

−1f(x) Pn(x) dx

= −n(n + 1) fl(n) (10.28)

Note 10.1. If we denote the operator R by

Rf(x) ≡ d

dx

[(1 − x2)

d

dxf (x)

]


we can extend the result of theorem 10.1 as

l[

R2 f (x) ; x → n]

= (−1)2 n2 (n + 1)2fl (n)

l[

R3 f (x) ; x → n]

= (−1)3 n3 (n + 1)3fl (n)

· · · · · ·l[

Rk f(x) ; x → n]

= (−1)k nk (n + 1)k fl (n) (10.29)

Corollary 10.1. It is seen that n(n + 1) =(n + 1

2

)2 − 14 . Therefore

we have

l [R f(x) ; x → n] = −n(n + 1) fl(n)

= −[(

n +12

)2

− 14

]fl(n)

= −(

n +12

)2

fl(n) +14

fl(n)

Thus, l

[14f(x) − R(f(x)) ; x → n

]=(

n +12

)2

fl(n) (10.30)

Extending this result, we get

(−1)k l [Rk(f(x))−4k ; x → n] =k−1∑r=0

(−1)r kcr

[4−r

(n +

12

)2k−2r]

fl(n)

Theorem 10.2. (Convolution Theorem).

If l [f(x) ; x → n] = fl(n) and l[g(x) ; x → n] = gl(n)

then l−1 [fl(n)gl(n) ; n → x] = f(x) ∗ g(x) (10.31)

where f(x)∗g(x) is called the convolution of f(x) and g(x) and is definedby

f(x) ∗ g(x) =1π

∫ π

0f(cos λ) sin λ dλ

∫ π

0g (cos (η)) dβ (10.32)

with x = cos μ and cos η = cos λ cos μ + sinλ sin μ cos β

Proof. We have

fl(n) gl(n) =∫ π

0f(cos λ)Pn(cos λ)(sin λ)dλ

∫ π

0g(cos η)Pn(cos η) sin η dη

=∫ π

0f(cos λ) sin λ

[∫ π

0g(cos η)Pn(cos η)Pn(cos λ) sin η

]dλ

where f(x) = f(cos λ) and g(x) = g (cos η)


Further, since

Pn(cos η) Pn(cos λ) =1π

∫ π

0Pn(cos μ) dα

where cos μ = cos η cos λ + sin η sin λ cos α , we can rewrite

fl(n)gl(n) =1π

∫ π

0f(cos λ) sin λ

[∫ π

0

∫ π

0g(cos λ)Pn(cos λ) sin η dα dη

]dλ

Now the double integral on the right hand of above equation is given by∫ π

0

∫ π

0g(cos λ cos μ + sinλ sin μ cos β) Pn(cos μ) sin μ dμ

Using this result and changing the order of integration, we get

fl(n)gl(n) =1π

∫ π

0Pn(cos μ) sin μ

[∫ π

0

∫ π

0f(cos λ) sin λ g(cos η)dλ dβ

]dμ

=∫ π

0h(cos μ) Pn(cos μ) sinμ dμ,

where cos η = cos λ cos μ + sin λ sin μ cos β

and h(cos μ) =1π

∫ π

0f(cos λ) sinλ dλ

∫ π

0g(cos η) dβ

Thus the theorem is proved.

In particular, if μ = 0 we get from (10.32) that

h(1) =∫ 1

−1f(t) g(t) dt (10.33)

and when μ = π, h(−1) =∫ 1

−1f(t) g(−t) dt (10.34)

10.5 Application to Boundary Value Problems.

As an illustration we consider application of finite Legendre transform inthe solution of interior Dirichlet problem for the potential u(r, θ) insidea unit sphere (r = 1) satisfying the partial differential equation

∂

∂r

[r2 ∂u

∂r

]+

∂

∂x

[(1 − x2)

∂u

∂x

]= 0 , 0 < r � 1 (10.35)

with boundary condition

u(1, x) = f(x) , −1 < x < 1 (10.36)


where x = cos θ .

Applying finite Legendre transform over the variable x, the PDE(10.35) and the boundary condition (10.36) reduce to

r2 d2ul(r, n)dr2

+ 2rdul(r, n)

dr− n(n + 1) ul(r, n) = 0 (10.37)

and ul(1, n) = fl(n) (10.38)

respectively, where ul(r, n) is a continuous function of r in 0 < r � 1

The bounded solution of (10.37) and (10.38) is

ul(r, n) = fl(n) rn , 0 < r � 1 , where n is a non − negative integer.

On inversion the solution of the problem is given by

u(r, x) =∞∑

n=0

(n +

12

)fl(n) rn Pn(x) , 0 < r � 1 , |x| < 1 (10.39)

Another representation of this solution can also be obtained throughapplication of convolution theorem and also the formula in equation(10.20) as

u(r, cos θ) = l−1 [ fl(n) rn ; n → x = cos θ ]

=12π

∫ π


∫ π

0

(1 − r2)dη

(1 − 2r cos μ + r2)32

(10.40)

where cos μ = cos λ cos θ + sinλ sin θ cos η

Integral in (10.40) is called the Poisson integral formula of the potentialinside the unit sphere.

The solution of exterior Dirichlet problem under boundary conditionw(1, x) = f(x) , x = cos θ of the unit sphere r = 1 is then given by

wl (r, n) =1r

fl(n) r−n , n = 0, 1, 2, · · ·On inversion it gives

w (r, cos θ) =1r

w

(1r

, cos θ

), r > 1

=12π

∫ π


∫ π

0

(r2 − 1) dη

(1 − 2r cos μ + r2)32

(10.41)


where cos μ = cos λ cos θ + sin λ sin θ cos η and w(r , cos θ) is thesolution of the exterior problem.

Exercises.

(1) (a) Show that

l [ xn ; x → n ] =2n+1 (n!)2

(2n + 1)!

(b) Using the result in eqn. (10.27) find l [ x2f(x) ; x → n ]

(2) Using the definition of odd Legendre transform as

f2n+1 ≡ l0 [ f(x) ; x → n ] =∫ 1

0f(x) P2n+1 (x) dx

and the even Legendre transform as

f2n ≡ le [ f(x) ; x → n ] =∫ 1

0f(x) P2n (x) dx

and their respective inversion formulae

l0 [ f2n+1 ; n → x ] ≡ f(x) =∞∑

n=0

(4n + 3) f2n+1 P2n+1(x), 0 � x � 1,

l−1e [ f2n ; n → x ] ≡ f(x) =

∞∑n=0

(4n + 1) f2n P2n(x) , 0 � x � 1

and denoting the operator L = ∂∂x

[(1 − x2) ∂

∂x

]prove that

l0 [ L f(x) ; x → n ] = (2n + 1) P2n(0) f(0)

−(2n + 1)(2n + 2) f2n+1

and le [ L f(x) ; x → n ] = −P2n(0) f ′(0) − 2n(2n + 1) f2n

(3) Prove that the Legendre transformed solution of the Dirichletboundary value problem for u(r, θ) defined by

∂2u

∂r2+

1r

∂u

∂r+ (1 − x2)

∂2u

∂x2− 2x

∂u

∂x= 0 , 0 � r � a , 0 � θ � π

u (a, θ) = f(x) , 0 � θ � π

where x = cos θ, is given by

ul (r, n) =(r

a

)nfl(n) .

Chapter 11

The Kontorovich-Lebedev Transform

11.1 Introduction.

In discussing the variable separable solution of Boundary value Problemsin cylindrical co-ordinates for wage type region Kontorovich-Lebedevtransform has been found application. To understand the applicabilityof the transform some elementary properties of the modified Besselfunction of the second kind, also called Macdonald function, is needed.We present below only operational proof of the results at various stagesof required theorems for easy understanding of the readers in engineeringand physics.

11.2 Definition of Kontorovich - Lebedev Transform.

Kontorovich-Lebedev transform of a function f(x) over the positive realline is denoted by f(τ) and is defined by

K[f(x) ; x → τ ] ≡ f(τ) =∫ ∞

0x−1 f(x)Kiτ (x) dx, (x � 0) (11.1)

The corresponding inversion theorem of Kontorovich-Lebedev transformis given by

K−1[f (τ) ; τ → x

] ≡ f(x) =2π2

∫ ∞

0Kiτ (x) τ sinh (πτ)f (τ) dτ,

(11.2)if f(x) is a function defined on the positive real line such that 1

xf(x) iscontinuously differentiable and xf(x) and x d

dx [ 1x f(x)] are absolutely

integrable over the positive real line.

If f(x) is a piecewise continuous function over (0 � x < ∞] havinga finite discontinuity at x, then (11.2) becomes

K−1 [f (τ) ; τ → x] =2π2

∫ ∞

0Kiτ (x)τ sinh (πτ)f (τ) dτ

=12

[f(x + 0) + f(x − 0)] (11.3)

The Kontorovich-Lebedev Transform 329

We present below an operational proof of the above inversion theorem,since the rigorous proof is a bit complicated.

We know that the cosine transform of Kiτ (x) is given by

Fc[ Kiτ (x) ; τ → t ] =√

π

2. e−x cosh t (11.4)

Therefore, we have

Fc [K {f(x) ; x → τ} ; τ → t]

=√

π

2

∫ ∞

0e−xch t f(x)

xdx (11.5)

This result can also be expressed as Laplace transform of 1x f(x) as

L

[f (x)

x; x → p

]=

2π

∫ ∞

0f (τ) cos (τ cosh−1 p ) dτ (11.6)

But since, L [ f(x) ; x → p ] =−∂

∂pL

[f(x)

x; x → p

]

we have from eqn. (11.6) that

L[f(x) ; x → p ] =2π

∫ ∞

0τ f(τ)

sin (τ cosh−1 p)√p2 − 1

dτ (11.7)

Taking Laplace inversion of both sides of the above equation we obtain

K−1[f(τ) ; τ → x ] ≡ f(x) =2π2

∫ ∞

0τ sinh (πτ) Kiτ (x) f (τ) dτ

(11.8)as the inversion formula for Kontorovich-Lebedev transform.

11.3 Parseval Relation for Kontorovich-Lebedev Trans-

forms.

Theorem 11.1. If f(τ) and g(τ) are respectively Kontorovich-Labedevtransforms of f(x) and g(x) respectively, x � 0 then

2π2

∫ ∞

0τ sinh (πτ) f(τ)g(τ) dτ =

∫ ∞

0x−1f(x) g(x) dx (11.9)


Proof. We have

2π2

∫ ∞

0τ sinh (πτ) f (τ)g (τ) dτ

=2π2

∫ ∞

0τ sinh (πτ) f (τ) dτ

∫ ∞

0x−1 g(x) Kiτ (x) dx

=∫ ∞

0x−1 g(x) dx.

2π2

∫ ∞

0τ sinh (πτ) f (τ) Kiτ (x) dτ

=∫ ∞

0x−1g(x) f(x) dx , by (11.8)

This proves the Parseval relation of Kontorovich-Lebedev transform.


Example 11.1. Prove that

K[te−t cos x ; t → τ

]=

π sinh(xτ)sinh(πτ) sin x

and hence deduce the values of K[t ; t → τ ] and K[te−t ; t → τ ].

Solution. By definition in (11.1) we have

K[

t e−t cos x ; t → τ]

=∫ ∞

0

1t

. t e−t cos x Kiτ (t) dt

=∫ ∞

0Kiτ (t) e−pt dt , where p = cos x

= L [ Kiτ (t) ; t → p ] =π sinh (xτ)

sinh (πτ) sin x,

using table of integral transforms.

Thus, in particular, if x = π2 and x = 0 we get respectively

K[t ; t → τ ] = π2 sech πτ

2 and K[t e−t ; t → τ ] = πτ cosech (πτ)

Example 11.2. Prove that

K[x sin α e−x cos α ; x → τ

]=

πsh(ατ)sh(πτ)

and K[e−xcht ; t → τ

]=

π cos (τt)τ sh (πτ)

Solution. Proceeding as the solution of Example 11.1, the above resultscan also be proved easily.


Example 11.3. Using the results of Example 11.2 deduce that

(a)∫ ∞

0e−x f(x) dx =

2π

∫ ∞

0τ2 f(τ) dτ

(b)∫ ∞

0e−x(1+cht) dx = 2

∫ ∞

0

cos (τt) dτ

sh (πτ)

and hence deduce that

Fc [ τ cosech (πτ) ; τ → t ] =1

2√

2πsech2

(t

2

)

Solutions.

(a) Taking g(x) = x sin α e−x cos α , we have from the table ofintegral transform of Copson that

g(τ) = K[ g(x) ; x → τ ] =π sh (ατ)sh (πτ)

Therefore, Parseval relation for Kontorovich-Lebedev transforms∫ ∞

0x−1 f(x) g(x) dx =

2π2

∫ ∞

0τ sh(πτ) f(τ) g(τ) dτ ,

for this g(x), gives∫ ∞

0x−1f(x) · x sin α e−x cos α dx =

2π2

∫ ∞

0τ sinh(πτ) f(τ) · πsh(ατ)

sh(πτ)dτ

⇒∫ ∞

0e−x cos α f(x) dx =

2π

∫ ∞

0τsh(ατ)sinα

f(τ) dτ

Now letting α → 0 , we get∫ ∞

0e−x f(x) dx =

2π

∫ ∞

0τ2f(τ) dτ

(b) Now taking f(x) = e−xcht we get

f(τ) =π

τ

cos(τt)sh (πτ)

Then using these relations in the final result of (a), we get∫ ∞

0e−x(1+cht)dx =

2π

∫ ∞

0τ2 π

τ

cos(τt)sh(πτ)

dτ


⇒ 12

sech2 t

2= 2∫ ∞

0

τ cos (τt)sh (πτ)

dτ

=√

2π Fc [ τ cosech (πτ) ; τ → t ]

⇒ Fc [ τ cosech (πτ) ; τ → t ] =1

2√

2πsech2

(t

2

).

11.5 Boundary Value Problem in a wedge of finite thickness.

Consider the boundary value problem of determining a harmonic func-tion u(ρ, ϕ, z) in a finite wedge defined by

ρ � 0 , 0 � ϕ � α , 0 � z � a (11.10)

when the function u satisfies the boundary conditions

u (ρ , α , z) = f (ρ , z) ,

u (ρ , ϕ , 0) = u (ρ , ϕ , a) = 0and u (ρ , 0 , z) = 0 .

⎫⎪⎬⎪⎭ (11.11)

Since u (ρ , ϕ , z) is a harmonic function, it satisfies the PDE

∂2u

∂ρ2+

1ρ

∂u

∂ρ+

1ρ2

∂2u

∂ϕ2+

∂2u

∂z2= 0 (11.12)

We seek a variable separable non-zero solution of (11.12) in the form

u (ρ , ϕ , z) = R(ρ) Φ(ϕ) Z(z) �= 0

so that after substituting the above value of u, differential equation(11.12) becomes

1R

d2R

dρ2+

1ρR

dR

dρ+

1Φ ρ2

d2Φdϕ2

= − 1Z

d2Z

dz2= σ2, say (11.13)

Therefore, Z(z) = A(σ) cos σz + B(σ) sin σz (11.14)

andρ2

R

d2R

dρ2+

ρ

R

dR

dρ− σ2ρ2 =

−1Φ

d2Φdφ2

= − τ2, say (11.15)

Therefore, Φ (φ) = C(τ) ch (τφ) + D(τ) sh(τφ) (11.16)

Also from (11.15) we have

ρ2 d2R

dρ2+ ρ

dR

dρ− (σ2ρ2 − τ2)R = 0 (11.17)


Putting τ = −iλ ⇒ λ = iτ eqn. (11.17) becomes

ρ2 d2R

dρ2+ ρ

dR

dρ− (λ2 + σ2ρ2)R = 0

One solution of this equation is Kλ (ρσ) = Kiτ (ρσ)

Hence,

u(ρ, φ, z) = L (τ) Kiτ (ρσ) [A(σ) cos σz + B(σ) sin σz]

[C (τ) ch τφ + D (τ) sh τϕ ].

From (11.11) since u (ρ, φ, 0) = u (φ, φ, a) = 0, we have

A(σ) = 0 and sin σa = 0 = sinnπ, n = 1, 2, · · ·

Also, since u(ρ, 0, z) = 0 we have C (τ) = 0. Therefore, by method oflinear superposition the solution under the last three boundary condi-tions in (11.11) of the problem can be expressed as

u(ρ, φ, z) =∞∑

n=1

L(τ) B(n) D(τ) sinnπz

ash τφ.Kiτ

(nπρ

a

)

for any variable value of the positive constant τ . After little adjustmentof the constants L(τ)B(n)D(τ), on linear superposition of solutions overτ the above solution u(ρ, φ, z) can also be expressed as

u(ρ, φ, z) =2π2

∞∑n=1

sinnπz

a

∫ ∞

0τFn(τ)

sh (τφ)sh (τ a)

Kiτ (nπρ

a) sh πτ dτ

(11.18)

This final form of u (ρ, φ, z) in (11.18) will provide a solution of theboundary value problem satisfying all the boundary conditions in (11.11),provided that we can find Fn(τ) such that

f(ρ, z) =∞∑

n=1

sinnπz

a

[K−1

{Fn(τ) ; τ → nπρ

a

}]

Inverting as the finite Fourier transform we get

K−1[

Fn (τ) ; τ → nπρ

a

]=

2a

∫ a

0f (ρ, y) sin

nπy

ady


Now replacing ρ by arnπ , we see that the above equation is equivalent to

K−1 [ Fn (τ) ; τ → r ] =2a

∫ a

0f( ar

nπ, y)

sinnπy

ady

and hence

Fn(τ) =2a

∫ a

0sin

nπy

aK[

f( ar

nπ, y)

; r → τ]

dy

=2a

∫ a

0sin

nπy

ady

∫ ∞

0r−1 f

( ar

nπ, y)

Kiτ (r) dr

Changing the variable of integration in last integration we can expressit as

Fn(τ) =2a

∫ a

0sin

nπy

ady

{∫ ∞

0x−1f(x, y) Kiτ

(nπx

a

)dx

}(11.19)

Substituting this value of Fn(τ) in eqn. (11.18), the solution of theboundary value is obtained.

Exercises.

(1) Prove that

K [ 1 ; τ → x ] =π

21

x sinh(

π2 x)

(2) Prove that

(a)∫ ∞

0

{τ2 Kit (x) dt

}=

π

2x e−x

(b)∫ ∞

0Kit (x) dt =

π

2e−x

(3) Prove that

K[

e−x cos t ; x → τ]

=π ch (tτ)τ sh (πτ)

Hence deduce that

(a) K [ e−x ; x → τ ] =π

τ

1sh(πτ)

(b) K [ ex ; x → τ ] =π

τcoth (πτ)

(c) K [ e−xcht ; x → τ ] =π cos (tτ)τ sh (πτ)

Chapter 12

The Mehler-Fock Transform

12.1 Introduction.

In solving some of the boundary value problems associated with themathematical theory of elasticity application of the Mehler-Fock trans-form has been found. Specially in problems concerning analysis of stressfield in the vicinity of external crack in elastic region these transformshave found significant applications . In this chapter our main objectiveis to develop some basic concepts of this transform in an elegant mannertogether with some useful basic properties of the associated Legendrefunctions of first kind.

12.2 Fock’s Theorem (with weaker restriction).

Theorem 12.1 If a function g(x) defined on 1 � x < ∞ is such thatthe integral

∫ ∞

1

|g(x)| dx√(x + 1)

exists, then at every point x in whose neighbourhood g(x) has a boundedvariation∫ ∞

0P− 1

2+iτ (x)g∗0(τ) dτ =

12

[ g(x + 0) + g(x − 0) ] (12.1)

where g∗0(τ) = τ tanh (πτ)∫ ∞

1P− 1

2+iτ (x) g(x) dx (12.2)

However, if g(x) is continuous at x ∈ [1,∞) we have (12.1) is equivalentto ∫ ∞

0P− 1

2+iτ (x)g∗0(τ) dτ = g(x) (12.3)


Proof. To supply a formal proof of this form of Fock’s theorem wereplace g (cosh α) = f(α) and proceed as below.

Let us introduce a pair of operators Φ0 and Φ−10 such that

Φ0 [ f(α) ; α → τ ] = f∗0 (τ)

=∫ ∞

0f(α)P− 1

2+iτ (cosh α) sinhα dα (12.4)

and Φ−10 [ f∗

0 (τ) ; τ → α ] = f(α)

=∫ ∞

0τ tanh (π τ) P− 1

2+iτ (cosh α) f∗

0 (τ) dτ (12.5)

We begin by finding the forms of Φ−10 and Φ0. Determining these

operators is equivalent to finding a solution

f∗0 (τ) = Φ0 [ f(α) ; α → τ ]

of the integral equation (12.4). Now introducing the formula

P− 12+iτ (cosh α) =

√2 cth (πτ)

π

∫ ∞

α

sin (τt)√cht − chα

dt (12.6)

in (12.5) we find

f(α) =√

2π

∫ ∞

0τf∗

0 (τ) dτ

∫ ∞

α

sin (τt)√(cht − chα)

dt

=1√π

∫ ∞

α

dt√ch t − ch α

√2π

∫ ∞

0τ f∗

0 (τ) sin (τ t) dτ

=1√π

∫ ∞

αFs [ τf∗

0 (τ) ; τ → t ]dt√

(ch t − ch α)(12.7)

where Fs [· · · ] is the Fourier sine transform of τf∗0 (τ). Treating (12.7)

as an Abel integral equation, its solution is given by

Fs[ τf∗0 (τ) ; τ → t ] = − 1√

π

d

dt

∫ ∞

t

f(α) dα√ch α − cht

(12.8)

Again inverting (12.8) by means of Fourier sine inversion theorem weget

τ f∗0 (τ) = − 1√

πFs

[d

dt

∫ ∞

t

f(α) sh α dα√ch α − ch t

; t → τ

]

The Mehler-Fock Transform 337

=τ√π

Fc

[ ∫ ∞

t

f (α) sh α dα√ch α − ch t

; t → τ

]Therefore,

f∗0 (τ) =

√2

π

∫ ∞

0cos(τt) dτ

∫ ∞

t

f(α)sh α dα√(ch α − ch t)

=∫ ∞

0f(α) sh α dα

√2

π

∫ α

0

cos(τ t)√(ch α − ch t)

dt

⇒ f∗0 (τ) = Φ0[f(α);α → τ ] =

∫ ∞

0f(α)P− 1

2+iτ(chα)shα dα

(12.9)

since, P− 12 +iτ

(chα) =√

2π

∫ α

0

cos(τ t)√(ch α − ch t)

dt

Thus (12.4) is obtained after starting from (12.5).

If we replace f(α) by g(chα) and if

g(x) =∫ ∞

0P− 1

2 +iτ(x) g0 (τ) dτ , x ∈ [1,∞] (12.10)

then eqn. (12.9) implies

g∗0(τ) = τ th(πτ)

∫ ∞

1P− 1

2+iτ(x) g(x) dx (12.11)

This result proves the theorem for the case when g(x) is a continuousfunction.

There are other two forms of the Fock’s theorem which we need notpersue here.

12.3 Mehler-Fock Transform of zero order and its prop-

erties.

The Mehler-Fock integral transform of order zero of a function f(α),α ∈ [0,∞) is denoted by f∗

0(τ) and is defined by

f∗0(τ) = Φ0 [f(α);α → τ ] =

∫ ∞

0f(α) P− 1

2 +iτ(chα)shα dα (12.12)

The inverse of this transform is given by

f(α) = Φ−1

0[f∗

0(τ) ; τ → α] =

∫ ∞

0τ f∗

0(τ) th(πτ) P− 1

2+iτ (chα)dτ

(12.13)


Properties.

Considering the special case ν = −12 + iτ , θ = iα and m = 0 in the

F.G. Mehler’s formula [Math.Ann. 18,161,1881] we find that

P− 12+iτ (chα) =

√2

π

∫ α

0

cos(τt) dt√(ch α − ch t)

(12.14)

=1√π

√2π

∫ ∞

0

cos(τt)H(α − t)√(ch α − ch t)

dt (12.15)

and in the Lagrange formula [Mem. des. Sci. Math. Fasic 47, 1939]that

P− 12+iτ (chα) =

√2

πcth (πτ)

∫ ∞

α

sin(τt) dt√(ch t − ch α)

(12.16)

=1√π

√2π

cth (πτ)∫ ∞

0

sin(τt)H(t − α)√(ch t − ch α)

dt

(12.17)

Treating eqn. (12.14) as an Abel integral equation, we get after itsinversion that

d

dt

[sin τt

τ

]= cos(τt) =

1√2

d

dt

∫ t

0P− 1

2+iτ (chα)

sh α · d α√(ch t − ch α)

⇒ sin(τt)τ

=1√2

∫ t

0P− 1

2+iτ (chα)

sinhα dα√(ch t − ch α)

(12.18)

Also treating (12.16) as another Abel integral equation we get after itsinversion that√

2π

sin(τt) cth(πτ) = − 1π

d

dt

∫ ∞

t

P− 12+iτ (chα)shα dα√{(ch α − ch t)}

⇒ − d

dt

[cos τt

τ

]= sin τt = − 1√

2d

dt

∫ ∞

tP− 1

2+iτ (chα) · th(πτ)

shα dα√(ch α − ch t)

⇒ cos τt

τ=

1√2

th πτ

∫ ∞

t

P− 12+iτ (chα) · shα dα√

(ch α − ch t)(12.19)

Thus from (12.18) we can have

1√2

∫ ∞

0H(t − α) P− 1

2+iτ (chα)

shα dα√(ch t − ch α)

=sin(τt)

τ


The last equation can be expressed as

Φ0

[H(t − α)√

(ch t − ch α); α → τ

]=

√2 sin τt

τ(12.20)

Again (12.19) gives∫ ∞

0H(α − t)P− 1

2+iτ (chα)

shα dα√(ch α − ch t)

=√

2 cos(τt)τ

cth (πτ)

This equation can also be expressed as

Φ0

[H(α − t)√

(ch α − ch t); α → τ

]=

√2

τcth (πτ) cos(τt) (12.21)

12.4 Parseval type relation.

From Fourier transform the corresponding parseval type relation of Mehler-Fock transform of order zero can be deduced easily [N.N. Lebedev, Dokl.Akad. Nauk. SSSR, 68,445, 1949] if we define alternatively that

φ0 [f(x) ; x → τ ] = f0(τ) =∫ ∞

1f(x)P− 1

2+iτ (x) dx (12.22)

and f(x) =∫ ∞

0τ th (πτ) f0(τ) dτ ≡ φ−1

0[f0(τ) ; τ → x]

(12.23)

For convenience if we define

f0(τ) = φ0 [f(x) ; x → τ ]

g0(τ) = φ0[g(x) ; x → τ ]

where f0(τ) =∫ ∞

1f(x) P− 1

2+iτ (x) dx, (12.24)

and g0(τ) =∫ ∞

1g(x) P− 1

2+iτ (x) dx (12.25)

then∫ ∞

0τ th(πτ) f0(τ) g0(τ) dτ

=∫ ∞

0τ th (πτ) g0 (τ) dτ

∫ ∞

1f(x)P− 1

2+iτ (x) dx

=∫ ∞

1f(x)dx

∫ ∞

0τ th (πτ) g0(τ)P− 1

2+iτ (x) dτ

=∫ ∞

1f(x) g(x) dx, (12.26)


by using inversion theorem in the inner integral.

This is the required Parseval type relation.

For example, if

f(x) =1

x + s, g(x) =

1x + t

we have f0(τ) =∫ ∞

1

1x + s

P− 12+iτ (x) dx

= πsech (πτ) P− 12+iτ (s) , |s| < 1

and g0(τ) = πsech (πτ) P− 12+iτ (t) , |t < 1

If s �= t ,

∫ ∞

1

dx

(x + s)(x + t)=

1t − s

log1 + t

1 + s

while∫ ∞

1

dx

(x + s)2=

1s + 1

Thus from (12.26) we have

∫ ∞

0

τ sh πτ

ch3 πτP− 1

2+iτ (s)P− 1

2+iτ (t) dτ =

{1

π2(t−s) log 1+t1+s , t �= s

1π2(1+s)

, t = s

where |s| < 1 , |t| < 1. (12.27)

As another example, let

f(x) = e−ax , g(x) = e−bx

Then f0(τ) =

√2πa

Kiτ (a) and g0(τ) =

√2πb

Kiτ (b)

Also,∫ ∞

1e−(a+b)x dx =

e−(a+b)

(a + b).

Therefore, by (12.26) we get

∫ ∞

0τ th (πτ) Kiτ (a)Kiτ (b) dτ =

π√

ab

2(a + b)e−(a+b) (12.28)

As a third example, let

f(x) = x− 32 , g(x) = x− 5

2

and hence f0(τ) =√

2 sech(πτ

2

)and


g0(τ) =2√

23

τ cosechπτ

2

Again, since∫ ∞

1f(x) g(x) dx =

∫ ∞

1x−4 dx =

13

,

we have from (12.26) that∫ ∞

0τ2 sech π τ d τ =

18

(12.29)

12.5 Mehler-Fock Transform of order m

The Mehler-Fock transform of order m is denoted and defined by

f∗m(τ) = Φm [f(α) ; α → τ ] =

∫ ∞

0f(α)P

m

− 12+iτ (chα) dα (12.30)

The corresponding inversion formula for the transform is

f(α) = (−1)m∫ ∞

0τ th(πτ)P

m

− 12+iτ (chα)f∗

m(τ) dτ

≡ Φ−1

m [f∗m

(τ) ; τ → α] (12.31)

where Pmν (x) = (x2 − 1)

m2

dm

dxmPν(x) , m > 0

Let chα = x. Then eqn. (12.31) can be expressed as

f(ch−1x) = (−1)m(x2 − 1)m2

dm

dxm

∫ ∞

0τ th (πτ)P− 1

2+iτ (x) f∗

m (τ) dτ

(12.32)Also, if we define the Mehler-Fock transform of order m by the equation

Φm [f(α) ; α → τ ] = f∗m (τ) =

∫ ∞

0f(α) P

−m

− 12+iτ (chα) shα dα (12.33)

then the corresponding inversion formula will be

Φ−1

m[f∗

m(τ) ; τ → α] = f(α) = (−1)m∫ ∞

0τ th (πτ) f∗

m(τ)Pm

− 12+iτ (chα) dτ,

(12.34)since the associated Legendre equation does not change after replacingm by −m and therefore the corresponding Legendre functions.


Also, for some class of functions f for convenience, if we definealtarnatively

φm [f(x) ; x → τ ] ≡ fm(τ) =∫ ∞

1f(x) P

m

− 12+iτ (x) dx (12.35)

the corresponding inversion formula will be

φ−1

m[fm(τ) ; τ → x] ≡ f(x) = (−1)m

∫ ∞

0τ th (πτ) f∗

m(τ) P

m

− 12+iτ (x) dx

(12.36)

12.6 Application to Boundary Value Problems.

12.6.1 First Example

At first we consider to determine a harmonic function, say, ψ in cylin-drical co-ordinates (ρ, φ, z) under certain given boundary conditions like

ψ = f(ρ) , 0 � ρ � a (12.37)

and∂ψ

∂z= 0 , ρ > a on the boundary plane z = 0 (12.38)

If we introduce a toroidal co-ordinate system in the same region as(α, β, φ) the equations of transformations are

ρ =a shα

chα + cos β, z =

a sin β

chα + cos β, φ = φ (12.39)

implying ρ + iz = a th12(α + iβ) ≡ f(α + iβ), say , (12.40)

where f is a holomorphic function of (α + iβ). It may be noticed from(12.39) that

ρ2 + (z + a cot β)2 = a2 cosec2 β (12.41)

and (ρ − a th α)2 + z2 = a2 cosech2 α (12.42)

representing surfaces β = β1, say and α = α1, say, as toruses respectivelyabout z-axis .

In (12.41), if 0 < β1 < π , β = β1 is the part of the sphere withcentre at (0, 0,−α cot β1) and radius a cosec β1 lying above the xy-planei.e, z > 0. Again if −π < β1 < 0, the surface is that part of the above


sphere which lies below the xy-plane i.e, where z < 0. In cylindricalsystem the Laplace equation in ψ

∂2ψ

∂ρ2+

1ρ

∂ψ

∂ρ+

1ρ2

∂2ψ

∂φ2+

∂2ψ

∂z2= 0 (12.43)

then takes the form

∂

∂α

[ρ∂ψ

∂α

]+

1∂β

[ρ∂ψ

∂β

]+

|f ′|2ρ

∂2ψ

∂φ2= 0 (12.44)

Again from (12.40) we have

|f ′|2 = a2(chα + cos β)−2

and so the above PDE in (12.44) can be expressed as

∂

∂α

[shα

chα + cos β

∂ψ

∂α

]+

∂

∂β

[shα

chα + cos β

∂ψ

∂β

]

+1

shα(chα + cos β)∂2ψ

∂φ2= 0 (12.45)

Let us seek a solution of (12.45) in toroidal co-ordinates in the form

ψ(α, β, φ) =√

(chα + cos β) · υ(α, β, φ) (12.46)

Then (12.45) leads to a PDE in υ(α, β, φ) as

shα∂2υ

∂α2+ chα

∂υ

∂α+ shα

∂2υ

∂β2+

14

υ shα +1

shα

∂2υ

∂φ2= 0 (12.47)

Eqn. (12.47) has the separable solutions of the form

υ(α, β, φ) = A(α) · e±τβ · e±imφ ,

if A(α) satisfy the equation

(1 − μ2)d2A

dμ2− 2μ

dA

dμ−[

m2

1 − μ2+(

14

+ τ2

)]A = 0 (12.48)

where μ = chα .

Now representing ν = −12 + iτ , we have ν(ν + 1) = − (1

4 + τ2)

andhence eqn, (12.48) can be expressed as

(1 − μ2)d2A

dμ2− 2μ

dA

dμ+[ν(ν + 1) − m2

1 − μ2

]A = 0 (12.49)


which is an associated Legendre differential equation. This equation(12.49) has a solution of the form

A(μ) = c P−mν (μ) = c P

−m

− 12+iτ (chα)

implying ν(α, β, φ) = c P−m

− 12+iτ (chα) e±τβ · e±imφ

Therefore, ψ(α, β, φ) =√

(chα + cos β) · P−m

− 12+iτ (chα) e±τβ · e±imφ

(12.50)

are the forms of the general solution of eqn. (12.44)

If we are interested to find the separable solutions of the Laplaceequation in the axisymmetric case that is in the φ-independent case, wehave

ψ(α, β) =√

(chα + cos β) · P− 12+iτ (chα) e±τβ (12.51)

Thus by the principle of superposition the general solution of the Laplaceequation in axisymmetric case is expressed as

ψ(α, β) =√

(chα + cos β) · Φ−1

0[A(τ) ch β τ + B(τ) sh β τ ; τ → α]

(12.52)

where Φ−1

0[χ(τ) ; τ → α] =

∫ ∞

0τ th (πτ) P− 1

2+iτ (chα) χ (τ) dτ

(12.53)

For the general case this solution similarly can be expressed as

ψ(α, β, φ) =√

(chα + cos β)∞∑

m=−∞eimφ Φ

−1

m[Am(τ)ch(βτ)

+Bm(τ) sh(βτ) ; τ → α] , (12.54)

where Φ−1

m[χm(τ); τ → α] = (−1)m

∫ ∞

0τ th(πτ) P

−m

− 12+iτ (chα)χm(τ)dτ

(12.55)

12.6.2 Second Example

Secondly for example, let us consider the particular problem of deter-mination of the solutions ψ of the Laplace equation in a half-spacez � 0 under the prescribed boundary conditions on ψ(ρ, z, φ) on a disk0 < ρ < a , z = 0 and ∂ψ

∂z = 0 in the region of the boundary outside


that disk in cylindrical co-ordinate system. In toroidal co-ordinate sys-tem (α, β, φ), as discussed above in 12.6.1, the given boundary conditionsare transformed to

ψ (α, 0, φ) = f(α, φ) , 0 < φ < 2π , α > 0 (12.56)

and∂ψ

∂β(α, π, φ) = 0 . (12.57)

Following the result in eqn. (12.50), we take a separable solution of theLaplace equation as

ψ(α, β, φ) =√

(chα + cos β)·∞∑

m=−∞Φ

−1

m

[Am(τ)

ch(πτ − βτ)ch(πτ)

; τ → α

]eimφ

(12.58)Here we may note that this form of solution identically satisfied thecondition ∂ψ

∂β (α, π, φ) = 0 . Now, to satisfy the boundary condition ineqn. (12.56) we have to find the functions Am(τ) for m ranging fromm → −∞ to m → +∞ through integral values such that

√2 sh

(α

2

)m=+∞∑m=−∞

Φ−1

m[Am(τ) ; τ → α]eimφ = f(α, φ) (12.59)

implying that

Φ−1

m[Am(τ) ; τ → α] =

2− 3

2

πcosech

(α

2

)∫ 2π

0f(α, φ)e−imφ dφ

Therefore, Am(τ) = 2− 3

2 π−1

∫ 2π

0e−imφΦm

[cosech

(α

2

)f(α, φ);α → τ

]dφ

(12.60)

12.6.3 Third Example

Thirdly another important problem of interest is to find the solutionψ(ρ, z) of Laplace equation in a half space z � 0 for the axisymmetriccase under the boundary conditions

ψ(ρ, 0) = F (ρ) for 0 � ρ � a (12.61)

and[∂ψ

∂z

]= 0, for ρ > a (12.62)


Then in the toroidal co-ordinate system the φ independent solution isobtained from eqn. (12.58) with a little variation after putting φ = 0and m = 0 in the form

ψ(α, β) =√

(chα + cos β) Φ−1

0

[A0(τ)

ch(πτ − βτ)ch(πτ)

; τ → α

](12.63)

for identically satisfying the condition in eqn. (12.62). Also the condi-tion in eqn. (12.61) will be satisfied if we have found out A0(τ) suchthat

A0(τ) = Φ0

[1√2

sech(α

2

)F(a th

α

2

); α → τ

](12.64)

= Mehler − Fock transform of1√2

sech(α

2

)F(a tan

α

2

)

Thus eqns. (12.63) together with (12.64) will constitute the requiredsolution of the given boundary value problem in the axi-symmetric case.

This boundary value problem corresponds to the axisymmetric con-tact problem for a half-space in the theory of elasticity.

As a particular case of the above problem, let us suppose that F (α) = k,

a constant

implying1√2

sech(α

2

)F(a th

α

2

)=

k√2

sech(α

2

),

Now, A0(τ) = Φ0

[k√2

sechα

2; α → τ

]=

k√2· 2 τ−1 cosech (πτ)

=√

2 k τ−1 cosech (πτ)

∴ ψ(α, β) = k√

[2(chα + cos β)] Φ−1

0

[τ−1 2 ch(πτ − βτ)

sh(2πτ); τ → α

]

=2kπ

tan−1

[chα + cos β

1 − cos β

] 12

, (12.65)

from table of integrals.

A quantity of physical interest is the number

P = −2π∫ a

0ρ

(∂ψ

∂z

)z=0

dρ

to be evaluated now.


We have, a∂ψ

∂z= −sh α sinβ

∂ψ

∂α+ (1 + chα cos β)

∂ψ

∂β

∴[∂ψ

∂z

]z=0

=1 + chα

a

(∂ψ

∂β

)β=0

Now,∂ψ

∂β=

√2 k

πcos(

β

2

)(chα + cos β)−

12

So,[∂ψ

∂z

]z=0

= − 2kπa

ch(α

2

)

∴ P = 2 k a

∫ ∞

0th

α

2sech

α

2dα , since ρ = a th

α

2= 4 a k (12.66)

12.6.4 Fourth Example

Finally we consider the problem of determining the axisymmetric stressesin an elastic body weakened by a penny-shaped crack . This problem isreduced to determining a harmonic function ψ in the half-space z � 0satisfying the conditions[

∂ψ

∂z

]z=0

= p (ρ) , 0 � ρ � a (12.67)

[Ψ]z=0 = 0 , ρ > a (12.68)

As before, introducing toroidal co-ordinates (α, β, φ) under transforma-tion

ρ =a sh α

chα + cos β, z = a

sin β

chα + cos β, φ = φ

the boundary conditions in eqns. (12.67) and (12.68) reduce to

∂ψ (α, 0)∂β

= −2a sech2(α

2

)p(a th

α

2

)(12.69)

and ψ(α, π) = 0 respectively. (12.70)

So, we choose ψ(α, β) satisfying (12.70) identically as

ψ(α, β) =√

(chα + cos β) Φ−1

0

[τ−1 f∗(τ)

sh(πτ − βτ)ch(πτ)

; τ → α

]

To satisfy (12.69) we take

f∗(τ) = Φ0 [f(α) ; α → τ ]

where f(α) =√

2 a p(a tanh

α

2

)sech3 α

2(12.71)


For a particular case, if p(ρ) = k, a constant, then

f(α) =√

2 k a sech3(α

2

)and hence f∗(τ) = 8

√2 k a τ cosech (πτ) , from table of integrals.

Therefore, ψ(α, β) = 16

√(ch2

α

2− sin2 β

2

)k a

Φ−1

0

[2 sh(πτ − βτ)

sh(2πτ); τ → α

]

12.7 Application of Mehler-Fock Transform for solving

dual integral equation.

Consider the dual integral equations for determination of f(τ) as∫ ∞

0f(τ) cos x τ dτ = g1(x) , 0 � x � a (12.72)∫ ∞

0f(τ) cth (πτ) sin x τ dτ = h2(x) , x > a , (12.73)

where g1(x) and h2(x) are two prescribed functions in their given ranges.

Multiplying both sides of the eqn. (12.72) by√

2π√

(chα − chx)and

integrating the result with respect to x from 0 to α and finally usingthe result in eqn. (12.14) we get∫ α

0

√2

π√

(chα − ch x)

[∫ ∞

0f(τ) cos xτ dτ

]dx =

∫ α

0

√2

π

g1(x)dx√(chα − ch x)

Thus∫ ∞

0f(τ) P− 1

2+iτ (chα) dτ = Ω1(α), say

=√

2π

∫ α

0

g1(x)dx√(chα − ch x)

, 0 < α < a

(12.74)

Similarly, multiplying eqn. (12.73) by√

2π (chx− chα)−

12 and integrating

the result with respect to x from α to ∞ and finally using the result ineqn (12.16) we get∫ ∞

0f(τ) P− 1

2+iτ (chα) dτ =

√2

π

∫ ∞

α

h2(x)dx√(chx − chα)

≡ Ω2(α), say, α > a

(12.75)


Let us now define

Ω (α) =

{Ω1 (α) , 0 � α � a

Ω2 (α) , α > a

so that the eqns in (12.74) and (12.75) are equivalent to∫ ∞

0f(τ) P− 1

2+iτ (chα) dτ = Ω (x)

⇒ Φ−1

0

[τ−1 cth (πτ) f(τ) ; τ → α

]= Ω (α)

After Mehler-Fock inversion the above equation yields

f(τ) = τ th (πτ) Φ0 [ Ω(α) ; α → τ ]

= τ th (πτ)∫ a

0Ω1(α) P− 1

2+iτ (chα) sh α dα

+ τ th (πτ)∫ ∞

aΩ2(x) P− 1

2+iτ (chα) sh α dα (12.76)

Thus the dual integral equation is solved in closed form.

Exercises

(1) Considering the alternative notation of Mehler-Fock transform as

f0(τ) = ϕ0 [f(x) ; x → τ ] =∫ ∞

1f(x) P− 1

2+iτ (x) dx

and the inverse

f(x) = ϕ−1

0[f0(τ) ; τ → x] =

∫ ∞

0τ th (πτ)f0(τ) P− 1

2+iτ (x) dτ

prove that

(a) ϕ0 [e−ax ; x → τ ] =

√2πa

Kiτ (a) , |arg a| <π

2

(b) ϕ0 [x− 32 ; x → τ ] =

√2 sech

(πτ

2

)(c) ϕ0 [(chα + cos β)−

32 ; α → τ ] = 2

√2 cosec β

sh(βτ)sh(πτ)

,−π < β < π

(d) ϕ0

[sech3

(α

2

); α → τ

]= 8 τ cosech (πτ)


(2) Following the dual integral equations discussed in section 12.7prove that∫ ∞

0f(τ) cos(x τ) dτ =

1√2

d

dx

∫ a

0

Ω1(α)sh α dα√(ch x − chα)

+1√2

d

dx

∫ x

a

Ω2(α) sh α dα√(ch x − chα)

, for x > a

Ω1(α) , Ω2(α) are functions defined in the section.

(3) Reduce the following dual integral equations∫ ∞

0τ−1 f(τ) sin xτ dτ = j1(x) , 0 � x � a∫ ∞

0cth (πτ) f(τ) sin xτ dτ = h2(x) , x > a

as a pair of dual integral equations discussed in section 12.7.

(4) Prove that

(a) Φ0

[sech

(α

2

); α → τ

]= 2τ−1 cosech (πτ)

(b) Φ0 [√

(sech (α)) ; α → τ ] =1√2 τ

cosech(πτ

2

)

Chapter 13

Jacobi, Gegenbauer, Laguerre and Hermite

Transforms

13.1 Introduction.

In this section we present below a brief account of some important in-tegral transforms like Jacobi, Gegenbauer, Laguerre and Hermite trans-forms. These transforms are applicable to a limited class of problems ofphysical science and therefore these are sparingly discussed in literature.We only persue here their definitions, basic operational properties anda few important applications.

13.2 Definition of Jacobi Transform.

In view of the orthogonal relation of Jacobi polynomials of ordersα(> −1) and β(> −1) and degrees m and n given by

∫ 1

−1(1 − x)α (1 + x)β P (α,β)

n (x) P (α,β)m (x) dx = δn δmn (13.1)

where δnm is the Kronecker delta symbol and

δn =2α+β+1 Γ(n + α + 1) Γ(n + β + 1)

n! (α + β + 2n + 1) Γ(n + α + β + 1)(13.2)

there exist under certain restrictions a series expansion of f(x) of theform

f(x) =∞∑

n=1

an P (α,β)n (x) ,−1 < x < 1 (13.3)

with an =1δn

∫ 1

−1f(x) P (α,β)

n (x) dx =1δn

f (α,β)(n) (13.4)


Accordingly, the Jacobi transform of degree n of a function f(x) in−1 < x < 1 is defined by Debnath (Bull Cal Math Soe., 55, 1963) as

J [f(x);x → n] ≡ f (α,β)(n) =∫ 1

−1(1 − x)α (1 + x)β P (α,β)

n (x) f(x) dx

(13.5)and the inverse Jocobi transform is given by

J−1[f (α,β) (n); n → x

]≡ f(x) =

∞∑n=1

(δn)−1 f (α,β) (n) P (α,β)n (x)

(13.6)

Example 13.1. Prove that the Jacobi Transform of degree n of

(a) f(x), a polynomial of degree m < n is zero.

(b) P(α,β)m (x) is δmn

Solution.

(a) Since f(x) is a polynomial of degree m < n, it admits of a seriesexpansion in Jacobi polynomials as

f(x) =m<n∑r=1

ar P (α,β)r (x) , −1 < x < 1

and therefore,

J [f(x) ; x → n] =∫ 1

−1(1 − x)α(1 + x)βP (α,β)

n (x)

{m<n∑r=1

ar P (α,β)r (x)

}dx

=m∑

r=1

ar

∫ 1

−1(1 − x)α(a + x)βP (α,β)

n (x) P (α,β)r (x) dx

= δn δnr = 0 , since n �= r(= 1, 2, · · ·m < n)

with δn is as defined in (13.2).

(b) J[P (α,β)

m (x) ; x → n]

=∫ 1

−1(1 − x)α (1 + x)β P (α,β)

n (x) P (α,β)m (x) dx

= δnm.

[Tables of Integral Transforms, Erdelyi et al,Mc Graw Hill, NY]

Jacobi, Gegenbauer, Laguerre and Hermite Transforms 353

Theorem 13.1

Defining a differential operator R over a function f(x) by

R [f(x)] = (1 − x)−α (1 + x)−β d

dx

[(1 − x)α+1 (1 + x)β+1 d

dxf(x)

]

and assuming

lim|x|→1

[(1 − x)α+1 (1 + x)β+1 f(x)] = 0 = lim|x|→1

[(1 − x)α+1 (1 + x)β+1f ′(x)]

(13.7)if J [f(x) ; x → n] = f (α,β)(n) , then

J {R[f(x)] ; x → n} = −n(n + α + β + 1) f (α,β)(n), n = 0, 1, 2, · · ·(13.8)


J{R[f(x)]} =∫ 1

−1(1 − x)α(1 + x)β(1 − x)−α(1 + x)−β

d

dx

[(1 − x)1+α(1 + x)1+β df (x)

dx

]P (α,β)

n (x) dx

=∫ 1

−1

d

dx

[(1 − x)1+α (1 + x)1+β df (x)

dx

]P (α,β)

n (x) dx

=[P (α,β)

n (x)(1 − x)1+α(1 + x)1+βf ′(x)]1−1

−∫ 1

−1(1 − x)1+α(1 + x)1+βf ′(x)

d

dxP (α,β)

n (x) dx

= −∫ 1

−1(1 − x)1+α(1 + x)1+βf ′(x)

d

dxP (α,β)

n (x)dx, after integrating by parts

Again integrating by parts and using the orthogonal relation (13.1), theabove relation further reduces to

J{R[f(x)]} = −n(n + α + β + 1)∫ 1

−1(1 − x)α(1 + x)βP (α,β)

n (x)f(x)dx

= −n(n + α + β + 1) f (α,β) (n)

This completes the proof.

Corollary 13.1 If f(x) and R[f(x)] satisfy the conditions in (13.7) thenJ{R{R[f(x)]}} ≡ J{R2[f(x)]} exists and is given by

J{R2 [f(x)]} = (−1)2 n2(n + α + β + 1)2 f (α,β) (n)


More generally, if f(x) and Rj[f(x)] satisfy the conditions in(13.7) forj = 1, 2, · · · then J{Rk[f(x)]} = (−1)k nk(n + α + β + 1)k f (α,β) (n).

Example 13.2

Find the Jacobi Transform of the following functions :

(a) f(x) = xn

(b) f(x) = (1 + x)p−β

(c) f(x) = (1 − x)q−α

Solution.

(a) Since P (α,β)n (x) =

(1 + α + β)2n xn

2n n!(1 + α + β)n+ a polynomial of degree(n − 1),

therefore

xn =2nn!(1 + α + β)n

(1 + α + β)2nP (α,β)

n (x) + a polynomial of(n − 1)thdegree

So,

J [xn; x → n] =∫ 1

−1(1 − x)α (1 + x)βPα,β

n (x) xn dx

=∫ 1

−1(1 − x)α (1 + x)β .

[Pα,β

n (x)]2

.2nn!(1 + α + β)n

(1 + α + β)2ndx

+∫ 1

−1(1 − x)α (1 + x)βPα,β

n .(x). [a polynomial of (n − 1)th degree] dx

= 2n+α+β+1 Γ(n + α + 1) Γ(n + β + 1)Γ(n + α + β + 1)

, as the second integral vanishes

Here, (α)n is the factorial function defined by

(α)n = α(α + 1)(α + 2) · · · (α + n − 1), n � 1

=Γ(α + n)

Γ(α).

(b) J[(1 + x)p−β ; x → n

]=∫ 1

−1(1 − x)α(1 + x)p P (α,β)

n (x) dx

=

(n + α

n

)2n+p+1 Γ(p + 1) Γ(α + 1) Γ(p − β + 1)

Γ(α + p + n + 2) Γ(p − β + n + 1),


follows from the Tables of Integrals [ Erdelyi et al, McGraw Hill, NY.]

In particular, if α = β = 0, then

J[(1 + x)p−β ; x → n

]=∫ 1

−1P (0,0)

n (x)(1 + x)p dx

=∫ 1

−1(1 + x)p Pn(x)dx

=2p+1{Γ(1 + p)}2

Γ(p + n + 2) Γ(p + n + 1).

This result again follows from the Tables of Integrals [Erdelyi et al,Mc Graw Hill, NY]

(c) J[(1 − x)q−α; x → n

]=∫ 1

−1(1 − x)q (1 + x)βPα,β

n (x)dx

=2 q+β+1

n!Γ(α − q)Γ(q + 1) Γ(n + β + 1)Γ(n + α − q)

Γ(n + β + q + 2)

This result also follows from the tables of integrals [ Erdelyi et al,Mc Graw Hill, NY.]

13.3 The Gegenbauer Transform.

The Gegenbauer Transform and its inversion is denoted and defined by

G [f(x);x → n] ≡ f (ν)(n) =∫ 1

−1(1 − x2)(ν−

12)Cν

n(x)f(x) dx

and G−1[f (ν)(n);n → x

]= f(x) =

∞∑n=0

δ−1n Cν

n (x)f (ν)(n)

with δn =21−2ν πΓ(n + 2ν)n!(ν + n) [Γ(ν)]2

(13.9)

When α = β = ν − 12 , the Jacobi Polynomial P

(α,β)n (x) becomes the

Gegenbauer polynomial Cνn(x). The corresponding orthogonal relation

of Gegenbauer polynomial is∫ 1

−1(1 − x2)ν−

12 Cν

m(x)Cνn(x)dx = δn δnm (13.10)


where δn =21−2νπΓ(n + 2ν)n!(n + ν)[Γ(ν)]2

Also the differential form R defined by

R[f(x)] =(1 − x2

) d2f(x)dx2

− (2ν + 1) xdf (x)dx

can be expressed as

R[f(x)] = (1 − x2)12−ν d

dx

[(1 − x2

)ν+ 12

df

dx

](13.11)

Then under the Gegenbauer transform it reduces to

G[R{f(x)};x → n] = −n(n + 2ν) f (ν)(n) (13.12)

Similarly,

G[Rk{f(x)}; x → n] = (−1)k nk(n + 2ν)k f (ν)(n),

where k = 1, 2, 3, · · · (13.13)

13.4 Convolution Theorem

If G[ f(x) ; x → n] = f (ν)(n) and G [ g(x) ; x → n] = g(ν)(n), then

f ν(n)g(ν)(n) = G [ h(x) ; x → n ] = h(ν)(n),

where h(x) is given by

h(cos ψ) = A(sin ψ)1−2ν

∫ π

0

∫ π

0f(cos θ) g (cos φ) (sin θ)2ν

(sin φ)2ν−1 (sin λ)2ν−1 dθ dα

and α is defined by

cos φ = cos θ cos ψ + sin θ sin ψ cos α

Proof. From definition

f (ν)(n) g(ν)(n)

=∫ 1

−1f(x) (1 − x2)ν−

12 Cν

n (x) dx

∫ 1

−1g(x) (1 − x2)ν−

12 Cν

n (x) dx

=∫ π

0f(cos θ)(sin θ)2ν Cν

n(cos θ)dθ.

∫ π

0g(cos φ)(sin φ)2ν Cν

n(cos φ) dφ

=∫ π

0f(cos θ)(sin θ)2ν

[∫ π

0g(cos φ) Cν

n(cos θ) Cνn(cos φ)(sin φ)2νdφ

]dθ

(13.14)


By the additional formula of Gegenbauer polynomial, we have

Cνn(cos θ) Cν

n(cos φ) = A

∫ π

0Cν

n(cos ψ)(sin λ)2ν−1 dλ, (13.15)

where A ={Γ(n + 2ν)/ n! 22ν−1 Γ2(ν)

}(13.16)

and cos ψ = cos θ cos φ + sin θ sin φ cos λ (13.17)

Accordingly we get

f (ν)(n) g(ν)(n) =∫ π

0f(cos θ) (sin θ)2ν

[ ∫ π

0

∫ π

0g (cos φ) Cν

n (cos ψ)

(sin φ)2ν (sin λ)2ν−1 . dλ dφ]dθ (13.18)

If we make the variable transformation

cos φ = cos θ cos ψ + sin θ sin ψ cos α (13.19)

then from eqns (13.17) and (13.19) we get

dλ dφ = (sin ψ/ sin φ) dψ dα

and so the double integral in eqn. (13.18) becomes∫ π

0

∫ π

0(cos φ) Cν

n (cos ψ) (sin φ)2ν−1(sin λ)2ν−1 sin ψ dψ dα (13.20)

Then we get from (13.18) as

f (ν)(n) g(ν)(n) =∫ π

0(sin ψ)2ν Cν

n(cos ψ) h(cos ψ) dψ

= G[h (cos ψ)],

where h [cos ψ] = A(sin ψ)1−2ν

∫ π

0

∫ π

0f (cos θ) g (cos θ) (sin θ)2ν .

(sin φ)2ν−1 (sin λ)2ν−1 dθ dα (13.21)

Thus the convolution theorem is proved.

13.5 Application of the Transforms

Case I We consider below the one-dimensional generalised heat con-duction equation defined by

∂

∂x

[(1 − x2)

∂u

∂x

]+ [(1 − x) β − (1 + x)α]

∂u

∂x= ρc

∂u

∂t(13.22)


under the initial condition

u(x, 0) = g(x) , for − 1 < x < 1 (13.23)

This equation can be expressed as

(1 − x)−α (1 + x)−β

[∂

∂x

{(1 − x)α+1 (1 + x)β+1 ∂u

∂x

}]= ρα

∂u

∂t

(13.24)

Or R[u(x, t)] =1c

∂u

∂t,

where c =α

ρdis a positive constant depending on the density, specific

heat and thermal conductivity of the material.

We solve this boundary value problem by using the Jacobi transform.Taking Jacobi transform, the eqns. (13.24) and (13.23) give rise to

d

dtu(α,β)(n, t) = −cn(n + α + β + 1) u(α,β)(n, t) (13.25)

and u(α,β)(n, 0) = g(α,β)(n) (13.26)

respectively. Solving these equations we get

u(α,β)(n, t) = g(α,β) (n) exp [−n(n + α + β + 1) ct]

Therefore, the formal solution of the problem after inversion of Jacobitransform then becomes

u(x, t) =∞∑

n=0

δ−1n g(α,β) (n) P (α,β)

n (x) exp [−n(n + α + β + 1) ct ]

(13.27)

Case II If the heat conduction equation for a homogeneous solid beambe

∂

∂x

[(1 − x2)

∂u

∂x

]− (2ν + 1) x

∂u

∂x=

1c

∂u

∂t(13.28)

under the initial condition u (x, 0) = g(x), −1 < x < 1 (13.29)

we can apply Gegenbauer transform to solve it. After application of theGegenbauer transform the PDE (13.28) becomes

d

dtu(ν) (n, t) = −dn(n + 2ν) u(ν) (n, t) (13.30)


and the initial condition (13.29) becomes

u(ν)(n, 0) = g(ν)(n) (13.31)

The solution of equation (13.30) under (13.31) is then given by

u(ν)(n, t) = g(ν)(n) exp [−n(n + 2ν) ct ] (13.32)

On inversion of (13.32) we get the formal solution of the given boundaryvalue problem as

u(x, t) =∞∑

n=0

δ−1n Cν

n (x) exp [−n(n + 2ν) ct] (13.33)

13.6 The Laguerre Transform

Debnath and McCully introduced this transform through their pioneer-ing works. Laguerre transform of a function f(x) over 0 � x < ∞ isdenoted and defined by the integral

L[f(x); x → n] ≡ fa(n) =∫ ∞

0e−xxα Lα

n (x) f(x) dx (13.34)

where Lαn(x) is the Laguerre polynomial of degree n(� 0) and order

α (> −1) which satisfy the ODE

d

dx

[e−x xα+1 d

dxLα

n (x)]

+ ne−x xα Lαn(x) = 0 (13.35)

These Laguerre polynomials satisfy the orthogonal property∫ ∞

0e−x xα Lα

n (x) Lαm (x) dx =

(n + α

n

)Γ (n + 1) δn = δn δmn

(13.36)

where δmn is the Kronecker delta function and δn is given by

δn =

(n + α

n

)Γ (n + 1) (13.37)

Accordingly, the inverse Laguerre transform is defined by

f(x) ≡ L−1[fα(n) ; n → x

]=

∞∑n=0

(δn)−1 fα (n) Lαn (x) (13.38)


If α = 0, according to McCully, Laguerre transform pair takes the form

L[f(x); x → n] ≡ f0(n) =∫ ∞

0e−xLn(x) f(x) dx (13.39)

L−1[f0(n); n → x] ≡ f(x) =∞∑

n=0

f0 (n) Ln (x) (13.40)

where Ln(x) is the Laguerre polynomial of degree n and order zero.

Example 13.3. If f(x) = Lαm(x) then prove that L[f(x)] = δnδmn


L [f(x); x → n ] = L[Lαm(x); x → n ]

=∫ ∞

0e−xxαLα

n (x) Lαm (x) dx

=

(n + α

n

)Γ (α + 1) δmn ,

by the orthogonal property of the Laguerre polynomial

= δn δmn

Example 13.4 Find the Laguerre transform of f(x) = e−αx, α > −1

Solution. We have L[e−αx ; x → n ]

=∫ ∞

0e−(α+1)x xα Lα

n (x) dx

=Γ(n + α + 1) αn

n!(α − 1)n+α+1,

from the tables of integrals [Erdelyi et al (1954).]

Example 13.5. Prove that∫ ∞

0e−x xk Ln(x) dx =

{0, k �= n

(−1)n n!, k = n

where Ln(x) is a Laguerre polynomial due to McCully.

Solution. We can express xk as linear combination of polynomialsL0(x), L1 (x), · · ·Lk(x). Let it be

xk =k∑

r=0

CrLr(x)


Then∫ ∞

0e−xxkLn (x) dx =

k∑r=0

Cr

∫ ∞

0e−x Lr(x) Ln(x) dx

=k∑

r=0

Cr × 0 ,

by the orthogonality relation since k = r �= n.

But, if k = n , then xk = xn =n∑

r=0

Cr Lr (x)

=n−1∑r=0

Cr Lr (x) + Cn Ln (x)

Therefore,∫ ∞

0e−x xnLn(x)dx

=∫ ∞

0e−xL2

n (x).Cn dx

= Cn.1

Now, in the expansion of xn

xn = C0 L0(x) + C1L1(x) + · · · + Cn Ln(x)

= [C0L0(x) + C1 L1(x) + · · · + Cn−1 Ln−1]

+Cn

[(−1)nxn

n!+ a polynomial of degree (n − 1)

]

as xn occurs only on the single term Cn(−1)n

n! xn on the right hand sideof the above expansion we have by comparing the coefficients of xn onboth sides that

1 = Cn (−1)n/n!

This implies that Cn = (−1)nn! and therefore∫ ∞

0e−x xn Ln (x) dx = (−1)n n!

13.7 Operational properties

(a) Laguerre transform of the derivative of a function.

Let L[f(x) ; x → n] = fα(n), then

L[f ′(x) ; x → n] =∫ ∞

0e−x xαLα

n(x)f ′(x) dx


=[e−x xαLα

n (x) f(x)]∞0

+∫ ∞

0e−xxαLα

n (x)f(x) dx

−α

∫ ∞

0e−x xα−1 Lα

n(x) f(x) dx −∫ ∞

0e−xxα

{d

dxLα

n(x)}

f(x) dx

= fα(n) − α

n∑k=0

fn−1(k) +n∑

k=0

fα(k) , (13.41)

from Higher Transandental function, [Erdelyi, 1953.]

(b) If L [f(x) ; x → n] = fa(n), then

L [R {f(x)} ;x → n] = −nfa(n) (13.42)

where R is the differential operator defined by

R[f(x)] = ex x−α d

dx

[e−x xα+1 df (x)

dx

].


L [R{f(x)} ; x → n] =∫ ∞

0Lα

n (x)d

dx

[e−x xα+1 df (x)

dx

]dx

=[Lα

n(x){

e−xxα+1 df (x)dx

}]∞0

−∫ ∞

0e−xxα+1 dLα

n(x)dx

.df (x)dx

dx

= −[e−xxα+1 d

dxLα

n (x) f(x)]∞

0

+∫ ∞

0f(x)

d

dx

{e−xxα+1 d

dxLα

n(x)}

dx

=∫ ∞

0f(x).

{−ne−x xα Lαn (x)

}dx ,by eqn. (13.35)

= −n

∫ ∞

0e−x xα Lα

n(x) f(x)dx

= −nfα(n) (13.43)

On extendingthis proof, we can have

L[R2{f(x) ; x → n}] = (−1)2 n2fα(n)

More generally,

L[Rk {f(x)} ; x → n

]= (−1)k nkfα(n), k = 1, 2, 3, · · · (13.44)

(c) The convolution theorem of Laguerre transform of order zero(α = 0) is given by

L−1[f0(n) g0(n)] = h(x) (13.45)


where h(x) is given by

h(x) =1π

∫ ∞

0e−t f(t) dt

∫ π

0exp (

√xt cos θ) cos (

√xt sin θ)

g(x + t − 2

√xt cos θ

)dθ, (13.46)

if L[f(x) ; x → n] = f0(n) and L[g(x) ; x → n] = g0(n).

Proof. We have by definition

f0(n) g0(n) =∫ ∞

0e−x Ln (x) f(x)dx

∫ ∞

0e−yLy(y)g(y)dy

=∫ ∞

0e−x f(x) dx

[∫ ∞

0e−y Ln (x) Ly(y) g(y) dy

]

Now from a formula of the product of two Laguerre functions wehave

Ln(x)Ly(y) =1π

∫ π

0e√

xy cos θ cos(√

xy sin θ) Ln(x+y−2√

xy cos θ) dθ

[Bateman, Partial differential equation, 1944]. Therefore we canexpress

πf0(n)g0(n) =∫ ∞

0e−x f(x) dx

[∫ ∞

0e−y g(y){∫ π

0exp(

√xy cos θ) cos(

√xy sin θ). Ln(x + y − 2

√xy cos θ) dθ

}dy

]

The integral within the square bracket is

∫ ∞

0e−tLn (t) dt

∫ π

0exp (

√xt cos φ) cos(

√xt sinφ)

g (x + t − 2√

xt cos φ) dφ

So,we can write, f0(n)g0(n) = L[h(t); t → n]

=∫ ∞

0e−tLn(t)h(t)dt

where, h(x) =1π

∫ ∞

0e−tf(t)dt

[∫ π

0exp (

√xt cos θ) cos(

√xt sin θ)

g(x + t − 2√

xt cos θ)dθ]

This is the convolution theorem of the Laguere transform and theproof of it is complete.


13.8 Hermite Transform.

In 1964 Debnath introduced Hermite transform through his paperentitled “on Hermite transform”, Mathematicki Vesnik, 1,(16)[].He denoted and defined it by

H[f(x); x → n] ≡ fH(n) =∫ +∞

−∞exp (−x2) Hn(x) f(x) dx,

(13.47)where Hn(x) is the Hermite polynomial of degree n defined throughthe Rodrigues formula

Hn(x) = (−1)n exp (x2)dn

dxn

{exp (−x2)

}(13.48)

The inverse Hermite transform denoted by H−1[fH(n);n → x] andis defined by

H−1[fH(n) ; n → x] ≡ f(x) =∞∑

n=0

δ−1n fH(n) Hn(x) (13.49)

where δn is given byδn =

√π n! 2n, (13.50)

Since any function f(x) can be expanded in a series of Hermitepolynomials like

f(x) =∞∑

n=0

an Hn(x)

where the coefficients an of the (n + 1)th term is given by

an = δ−1n fH(n)

after using the orthogonality relation of Hermite polynomials∫ ∞

−∞exp (−x2) Hn(x) Hm(x) dx = δnmδn (13.51)

Example 13.6. Find Hermite transform of each of the followingfunctions :

(a) f(x) = a polynomial of degree m < n

(b) f(x) = Hm(x)

(c) f(x) = exp (2xt − t2)

(d) f(x) = exp (ax)


Solutions.

(a) Let f(x) = Pr(x), r < n

Then

H[f(x) ; x → n] =∫ +∞

−∞exp (−x2)

r∑k=0

ak Hk(x).

Hn(x) dx, r < n

=∞∑

k=0

ak {0} = 0 ,

from the orthogonality relation of Hermite polynomials.

(b) H [Hm (x) ; x → n]

=∫ +∞

−∞exp(−x2)Hm (x)Hn(x) dx

= δn δmn

(c) To find H[exp(2xt − t2) ; x → n

]we consider the generating

function of Hermite polynomial as

exp (2xt − t2) =∞∑

n=0

tn

n!Hn(x)

Then H[exp(2xt − t2) ; x → n

]=∫ +∞

−∞exp(−x2) Hn(x).

∞∑n=0

tn

n!Hn (x) dx

=∞∑

n=0

tn

n!

∫ +∞

−∞exp(−x2) H2

n (x) dx

=∞∑

n=0

tn

n!δn =

√π

∞∑n=0

(2t)n , |t| <12.

(d) We have from definition

H[exp(ax) ; x → n]

=∫ +∞

−∞exp(−x2) exp (ax) Hn(x) dx

Now since ,∫ +∞

−∞exp(−x2 + 2bx) Hn (x) dx


=√

π

∞∑n=0

(2b)n exp (b2),

the above relation gives

H [exp (ax) ; x → n] =√

π

∞∑n=0

an exp

(a2

4

)

13.9 Operational Properties.

(a) Theorem 13.2. If f ′(x) is continious and f ′′(x) is integrable in−∞ < x < ∞ and if H[f(x); x → n] = fH(n), then

H[R{f(x)} ; x → n] = −2nfH(n),

where R[f(x)] is defined as

R[f(x)] = exp(x2)d

dx

[exp(−x2)

df

dx

].

Proof. We have by definition,

H [R{f(x)} ; x → n] =∫ +∞

−∞Hn (x).

d

dx

[exp (−x2)

df

dx

]dx

= −∫ +∞

−∞df

dx(x) e−x2

H ′n(x) dx

=∫ +∞

−∞f(x)

[e−x2

Hn”(x) − 2xe−x2H ′

n(x)]dx

=∫ +∞

−∞e−x2

f(x) [−2n Hn (x)] dx

= −2n fH (n) (13.52)

If R [f(x)] also satisfies conditions of the theorem, then

H[R2{f(x)} ; x → n

]= (−1)2 (2n)2fH(n).

In general, when Rk−1satisfies the conditions of the theorem, then

H[Rk{f(x)} ; x → n

]= (−1)k (2n)kfH(n),

where k = 1, 2, · · ·

Theorem 13.3. If f(x) is bounded and integrable and fH(0) = 0, thenfor each constant C

H−1

[−f

H(n)

2n

]= R−1[f(x)] =

∫ x

exp (s2)∫ s

−∞exp (−t2) f(t) dt + C,


where R−1 is the inverse of the differential operator R and n is a positiveinteger.

Proof.Let R−1[f(x)] = y(x) ⇒ R [y(x)] = f(x).

Thus y(x) is the solution of a second order ODE. Since,

fH (0) = 0 and H0(x) = 1, we have∫ +∞

−∞e−x2

. R [y (x)] dx =∫ +∞

−∞e−x2

R [y (x)] Ho(x) dx

=∫ +∞

−∞e−x2

f(x) H0(x) dx = fH

(0) = 0 (13.53)

Therefore, the first integral of R [y(x)] = f(x) is

exp(−x2)dy

dx=∫ x

−∞exp(−t2) f (t) dt

As |x| → ∞, the right hand side integral in above tends to zero byvirtue of eqn. (13.53). Then the second integral is

y(x) =∫ x

0exp(s2){∫ s

−∞

[exp (−t2)f(t)

]dt

}ds + C (13.54)

Therefore ,

lim|x|→∞

exp (−x2) y(x) = 0 and H [y(x)] exists.

So, H [R{y (x)} ; x → n] = −2n H [y(x)] implying

H[f (x) ; x → n] = −2n H [y (x)]

⇒ fH

(n) = −2n H[R−1{f(x)}]

⇒ H[R−1{f (x)}] = −f

H(n)

2n(13.55)

13.10 Hermite Transform of derivative of a function.

Theorem 13.4. If H[f (x) ; x → n ] = fH

(n) and m is a positiveinteger, then H[f (m) (x) ; x → n ] = fH (n + m)

Proof. we have by definition

H[f ′ (x) ; x → n ] =∫ +∞

−∞exp (−x2) Hn (x) f ′ (x) dx


=[exp (−x2) f(x) Hn (x)

]+∞−∞ −

∫ +∞

−∞f(x)

d

dx

{e−x2

Hn(x)}

dx

= 2∫ +∞

−∞xe−x2

Hn (x) f(x) dx −∫ +∞

−∞)f(x) e−x2

H ′n (x) dx

=∫ +∞

−∞e−x2

[Hn+1 (x) + 2n Hn−1 x] f(x) dx

−2n∫ +∞

−∞e−x2

Hn−1 (x) f(x) dx,

since Hn+1(x) = 2x Hn (x) − 2n Hn−1 (x) (13.56)

and H ′n (x) = 2n Hn−1 (x) (13.57)

=∫ +∞

−∞exp (−x2) Hn+1 (x) f(x) dx

= fH

(n + 1). (13.58)

Proceeding in the similar manner we can arrive at

H [f (m)(x) ; x → n] = fH

(n + m)

Thoerem 13.5. If Harmite transforms of f(x) and of x fm−1(x) exists,then

H [x f (m)(x) ; x → n] = n fH

(m + n − 1) +12fH

(m + n + 1)


H[x f (m)(x) ; x → n] =∫ +∞

−∞exp (−x2)Hn(x)

{x f (m)(x)

}dx

=[x exp (−x2) Hn(x) f (m−1)(x)

]+∞−∞

−∫ +∞

−∞d

dx

{x exp(−x2) Hn(x)

}fm−1(x) dx

=∫ +∞

−∞2x2 exp(−x2) Hn(x)fm−1(x)dx

−∫ +∞

−∞exp(−x2)Hn(x)f (m−1)(x)dx

−n

∫ +∞

−∞2x exp (−x2)Hn−1(x)f (m−1)(x) dx

Using eqns. (13.56) and (13.57), we therefore have

H[xfm(x);x → n] =∫ +∞

−∞x e−x2

[Hn+1(x) + 2n Hn−1(x)]f (m−1)(x)dx


−n

∫ +∞

−∞e−x2

[Hn(x) + 2(n − 1)Hn−2(x)]f (m−1)(x)dx − fH

(n + m + 1)

=12

∫ +∞

−∞e−x2

[Hn+2(x) + 2(n + 1)Hn(x)] f (m−1)(x) dx

+n

∫ +∞

−∞e−x2

[Hn(x) + 2(n − 1)Hn−2(x)] f (m−1)(x) dx

−n fH

(n + m − 1) − 2n(n − 1)fH

(n + m − 3) − fH

(n + m + 1) ,

since H[f (m)(x)] = fH

(n + m)

=12

fH

(n + m + 1) + (n + 1)fH

(n + m − 1) + n[fH

(n + m − 1)

+2(n − 1)fH

(n + m − 3)] − nfH

(n + m − 1) − 2n(n − 1)fH

(n + m − 3)

−fH

(n + m + 1)

= n fH

(n + m − 1) +12

fH

(n + m + 1)

Accordingly, for m = 1 and m = 2, we get

H [x f ′(x) ; x → n] = n fH

(n) +12

fH

(n + 2)

H [x f ′′(x) ; x → n] = n fH

(n + 1) +12

fH

(n + 3)

Exercises

(1) Find Jacobi Transform of the following functions to prove that

(a) J(1 ; n) = 0

(b) J [xm ; x → n] =

{0 , if m �= n

1 , if m = n

(c) J [(1 − x)−1P (α,β)n (x) ; x → n]

=2α+β Γ(α + n + 1)Γ(β + n + 1)

n! αΓ(α + β + n + 1), Re α > 0 , Re β > −1

(d) J [(1 − x)α P (α,β)n (x) ; x → n]

=24α+β+1 Γ

(α + 1

2

) {Γ(α + n + 1)}2 Γ(β + 2n + 1)√π(n!)2 Γ(α + 1) Γ(2α + β + 2n + 2)

, Re α > −12

Re β > −1

after using the Tables of integrals, if necessary.


(2) Express∂

∂x

[(1 − x2)

∂u

∂x

]+ [β(1 − x) − α(1 + x)]

∂u

∂x

as R [u(x, t)], if R is the differential operator defined by

(1 − x)−α(1 + x)−β

[∂

∂x

{(1 − x)α+1 (1 + x)β+1 ∂

∂x

}]

(3) Prove that∫ 1

−1

[(1 − x) P (α+1,β)

n (x) + (1 + x) P (α,β+1)n (x)

](1 − x)−α(1 + x)−β P (α,β)

n (x) dx

= 2∫ 1

−1(1 − x)−α (1 + x)−β

[P (α,β)

n (x)]2

dx

= 2.

(4) Expressing xβ , β > 0 as a series using the result from Erdelyi inthe form

xβ = Γ(α + β + 1)∞∑

n=0

(−β)nΓ(n + α + 1)

Lα

n(x) , x > 0, α > −1.

prove that L[xβ ; x → n] = Γ(α + β + 1)∞∑

n=0

(−β)nΓ(n + α + 1)

δn

where δn =

(n + α

n

)Γ(α + 1)

(5) Prove that

fα+1 (n) = (n + α + 1) fα(n) − (n + 1) fα(n + 1)

by using the recurrence relation of Laguerre polynomial

x Lα+1

n(x) = (n + α + 1)L

α

n(x) − (n + 1) L

α

n+1 (x)

(6) Prove that the zero order Laguerre transfrom of

(a) f(x) = H(x − a), a � 0 is exp (−a) if n = 0 and is

e−a[Ln(a) − Ln−1(a)], if n � 1

(b) f(x) = e−ax , a > −1 is an(1 + a)n+1

(c) f(x) = xm is 0 if n > m and is (−1)n(

m

n

), if m � n

(d) f(x) = Ln(x) is 1, if n is a non zero positive integer.


(7) If L [f(x) ; x → n] = f0(n) =∫ ∞

0e−x Ln(x) f(x) dx,

prove that L

[ex d

dx

{x e−x f ′(x)

}; x → n

]= −n f0 (n)

(8) Prove that H[xm ; x → n] = 0 , for m = 0, 1, 2, · · · (n − 1)

(9) Show that H[xn ; x → n] =√

π(n!) Pn(1) ,where Pn(x)

is a Legendre Polynomial.

(10) Prove that H[eax(1 + ax) ; x → n] = H[x eax ; x → n].

Chapter 14

The Z-Transform

14.1 Introduction.

Analysing the behavior of functions whose values are known on a finiteor infinite set of discrete points in a given domain cannot be achievedby using the existing Fourier or Laplace Transform techniques. Accord-ingly, some efficient procedure for the numerical evaluation of the systemwill be aimed at. For example, such necessity do arise for the determi-nation of the output relation when the discrete input function in knownin a signal processing system. The Z-transform technique is one suchmethods to answer the problem proposed above.

14.2 Z - Transform : Definition.

We have learnt that, under certain restrictions, the Laplace transformof a function f(t) is defined as

F (s) =∫ ∞

0f(t)e−st dt (14.1)

After writing the integral on the right hand side of eqn. (14.1) as aRiemann sum we have

F (s) = T

∞∑k=0

f(kT )exp(−kTs), (14.2)

where T is the length of the subinterval chosen to get a proper repre-sentation of f(t) with its sampled values at the discrete set of pointstk = kT, given by f(kT ), k = 0, 1, 2, . . . and the value of the integralunder (14.1) with desired accuracy.

The Z-Transform 373

Then putting z = exp(sT ), we have F (s) in (14.2) is expressed as

F (s) = T∞∑

k=0

f(kT )z−k (14.3)

Here, it shows that F (s) is expanded in a series in power of z. Thisrelation in (14.3) can further be generalized by considering a sequence{xk} for k = 0, 1, 2, · · · and write a series in powers of z correspondingto the above as

∞∑k=0

xkz−k = x(z) (14.4)

after conveniently setting T = 1 as the interval between samples. Thissetting and representations have no effect on the properties of this trans-form.

This relation in (14.4) defines the Z - transform of the sequence {xk}.Symbolically, we write

Z [{xk}] = X(z) =+∞∑k=0

xkz−k (14.5)

as the Z-transform of {xk} and

Z−1[X(z)] = {xk} (14.6)

as the inverse Z-transform of X(z). It means that with a given sequencewe can associate a series in a domain of complex z-plane and vice versa.

To obtain the inversion of Z-transform in terms of a complex integralin the z-plane, we consider eqn. (14.5) as

X(z) =∞∑

k=0

xk z−k ≡∞∑

k=0

x(k) z−k , say (14.7)

Or X(z) = x(0) + x(1)z−1 + x(2)z−2 + · · · + x(k)z−k + · · · (14.8)

Multiplying both sides by zk−1 and integrating with respect to z

along the closed positive contour C, which encloses all singularities ofX(z), we get


12πi

∮C

X(z) zk−1 dz =1

2πi

[∮C

x(0) zk−1 dz +∮

Cx(1)zk−2dz + · · ·+

+∮

Cx(k)z−1 dz +

∫C

x(k + 1)z−2 dz + · · · · · ·]

By Canchy’s integral formula, the above relation gives

12πi

∮C

X(z) zk−1 dz =1

2πi

∮C

x(k)z

dz

= x(k) ,

since all other integrals on the right hand side vanish.

Thus we get,

12πi

∮C

X(z) zk−1 dz = x(k) (14.9)

implying that

z−1[X(z)] = x(k) = {xk} =1

2πi

∮C

X(z) zk−1 dz (14.10)

Example 14.1.

If x(n) = H(n) =

{1, n � 00, n < 0

,find the Z − transform of x(n).

Solution.

Z[x(n)] =∞∑

n=0

[x(0) + x(1)z−1 + · · · + x(n)z−n + · · ·]

=α∑

n=0

(1z

)n

=1

1 − 1z

=z

z − 1, |z| > 1.

Example 14.2. If x(n) = an, then find its Z-transform.

Solution.

X(z) = Z[x(n)] =∞∑

n=0

(a

z

)n=

z

z − a, |z| > a.

The Z-Transform 375

Example 14.3. Find the Z-transform of the following functions

(a) x(n) = n

(b) x(n) = n2

(c) x(n) = 1n!

(d) x(n) = einx

(e) x(n) = cosh nx

(f) x(n) = H(n), where H(n) =

{0, if n < 01, if n > 0

and n is an integer.

Solution.

(a) Z[n] =∞∑

n=0

n z−n = z∞∑

n=0

n z−n−1 = −zd

dz

[ ∞∑n=0

z−n

]

= z/(z − 1)2, |z| > 1

(b) Z[n2] = Z[n.n] =∞∑

n=0

n.nz−n

= z∞∑

n=0

n.n z−(n+1)

= z

∞∑n=0

n.

[− d

dzz−n

]

= −zd

dz

∞∑n=0

n z−n

= −zd

dz

[z

(z − 1)2

], |z| > 1,by using result in (a)

=z(z + 1)(z − 1)3

, |z| > 1

(c) Z

[1n!

]=

∞∑n=0

1n!

z−n = exp(

1z

), for all z

(d) Z[einx]

=∞∑

n=0

einx

zn=

∞∑n=0

(eix)n

z−n =∞∑

n=0

(eix

z

)n

=1

1 − eix

z

=z

z − eix

Similarly Z[e−inx

]=

z

z − e−ix


Therefore, Z [cos n x] =z(z − cos x)

z2 − 2z cos x + 1

and Z [sin nx] =z sin x

z2 − 2z cos x + 1

(e) Z [cosh nx] =12Z[enx + e−nx

]=

12

[z

z − ex+

z

z − e−x

]=

z(z − chx)z2 − 2z chx + 1

(f) [H(n)] =∞∑

n=0

H(n) z−n =∞∑

n=0

z−n =z

z − 1, |z| > 1

Example 14.4. If f(n) is a periodic sequence of integral period N ,then prove that

F (z) = Z[f(n)] =zN

zN − 1F1(z)

where F1(z) =N−1∑k=0

f(k)z−k


F (z) = Z[f(n)] =∞∑

n=0

f(n)z−n

=[f(0) + f(1)z−1 + f(2)z−2 + · · · + f(N − 1)z−N+1

]+[f(0)z−Nz0 + f(1)z−N .z−1 + · · · + f(N − 1) z−N z−N+1

]+ · · ·

=N−1∑k=0

f(k)z−k.1 + z−NN−1∑k=0

f(k)z−k + z−2NN−1∑k=0

f(k)z−k + · · ·

=N−1∑k=0

f(k)z−k

[1 +

1zN

+1

z2N+ · · ·

]

=1

1 − 1zN

F1(z) =zN

zN − 1F1(Z)

14.3 Some Operational Properties of Z-Transform.

Theorem 14.1 : (Translation). If Z[f(n)] = F (z) and m � 0, then

Z [f(n − m)] = z−m

[F (z) +

−1∑k=−m

f(k)z−k

], (14.11)

The Z-Transform 377

Z [f(n + m)] = zm

[F (z) −

m−1∑k=0

f(k)z−k

]. (14.12)

In particular, if m = 1, 2, 3, · · · ., then from the first relation

Z[f(n − 1)] = z−1 F (z)

Z[f(n − 2)] = z−2 F (z) +∑−1

k=−2

f(k)z−k

⎫⎬⎭ (14.13)

Similarly, it follows from the second relation (14.12) that

Z[f(n + 1)] = z[F (z) − f(0)]Z[f(n + 2)] = z2[F (z) − f(0)] − zf(1)Z[f(n + 3)] = z3[F (z) − f(0)] − z2f(1) − zf(2)

⎫⎪⎬⎪⎭ (14.14)

Result in (14.12) will be useful in solving difference and differential equa-tion for initial value problems.


Z[f(n − m)] =∞∑

m=0

f(n − m) z−n

=∞∑

k=−m

f(k)z−m−k

= z−m∞∑

k=−m

f(k) z−k

= z−m

[ −1∑k=−m

f(k)z−k +∞∑

k=0

f(k) z−k

]

= z−m

[ −1∑k=−m

f(k)z−k + F (z)

]

If m = 1, then Z [f(n − 1)] = z−1 F (z),when f(k) = 0, k < 0

Also, Z [f(n + m)] =∞∑

m=0

f(n + m)z−n

= zm∞∑

k=m

f(k)z−k


= zm

[ ∞∑k=0

f(k)z−k −m−1∑k=0

f(k) z−k

]

= zm

[F (z) −

m−1∑k=0

f(k) z−k

]

For m = 1, 2, 3, · · · the results in (14.14) will follow immediately.

Theorem 14.2 : (Multiplication)

If Z [f(n)] = F (z), then Z [anf(n)] = F(z

a

), |z| > |a|,

Z[e−bnf(n)

]= F (zeb) and Z [nf(n)] = −z

d

dzF (z).

Proof. From definition Z [anf(n)] =∞∑

n=0

an f(n) z−n

=∞∑

n=0

f(n)(z

a

)−n= F

(z

a

),|z||a| > 1 (14.15)

Again, Z[e−bn f(n)

]=

∞∑n=0

f(n)z−ne−bn

=∞∑

n=0

f(n)(zeb)−n

= F(zeb)

(14.16)

Also, Z [nf(n)] =∞∑

n=0

nf(n) z−n

= z∞∑

n=0

f(n){− d

dzz−n

}

= −zd

dz

[ ∞∑n=0

f(n)z−n

]

= −zd

dz[F (z)] (14.17)

Thus, all the results of the theorem is proved.

Theorem 14.3 : (Division)

If Z [f(n)] = F (z), then Z

[f(n)

n + m

]= −zm

∫ z

0

F (ξ)ξm+1

dξ

The Z-Transform 379

Proof. By definition, we have

Z

[f(n)

n + m

]=

∞∑n=0

f(n)n + m

z−n, m � 0

=∞∑

n=0

f(n)[−∫ z

0ξ−(m+n+1)dξ

]

= −zm

∫ z

0ξ−(m+1)

[ ∞∑n=0

f(n)ξ−n

]dξ

= −zm

∫ z

0ξ−(m+1)F (ξ) dξ (14.18)

Theorem 14.4 : (The convolution Theorem).

If Z [f(n)] = F (z) and Z [g(n)] = G(z) , then Z-transform of theconvolution f(n) ∗ g(n) defined by

f(n) ∗ g(n) =∞∑

m=0

f(n − m)g(m) (14.19)

is given byZ [f(n) ∗ g(n)]

= F (z) G(z)

= Z [f(n)]Z [g(n)]

Equivalently, Z−1 [F (z) G(z)] = f(n) ∗ g(n) =∞∑

m=0

f(n − m) g(m)

Proof. From definition

Z [f(n) ∗ g(n)] =∞∑

n=0

z−n

[ ∞∑m=0

f(n − m) g(m)

]

=∞∑

m=0

g(m)∞∑

n=0

f(n − m)z−n

=∞∑

m=0

g(m)z−m∞∑

k=−m

f(k)z−k

=∞∑

m=0

g(m)z−m∞∑

k=0

f(k)z−k , on the assumption that f(k) = 0, k < 0

= Z [f(n)] Z [g(n)] (14.20)

Thus the theorem is proved.


Theorem 14.5 : (The initial value Theorem).

If Z [f(n)] = F (z), then f(0) = limz→∞F (z).

Also if f(0) = 0, then f(1) = limz→∞ zF (z).

Proof. we have by definition

F (z) =∞∑

n=0

f(n)z−n = f(0) +f(1)

z+

f(2)z2

+ · · ·

Then letting z → ∞, we get from this relation that

limz→∞F (z) = f(0) (14.21)

If f(0) = 0, then we get

F (z) =f(1)

z+

f(2)z2

+ · · ·Therefore, lim

z→∞ z F (z) = f(1) (14.22)

Theorem 14.6 : (The Final Value Theorem)

If Z [f(n)] = F (z), then limn→∞ f(n) = lim

z→1[(z − 1)F (z)] ,

provided the limits exist.

Proof. We have from (14.14)

Z [f(n + 1) − f(n)] = z[F (z) − f(0)] − F (z)

This implies that

∞∑n=0

[f(n + 1) − f(n)] z−n = (z − 1)F (z) − zf(0)

In the limit as z → 1, we obtain

limz→1

∞∑n=0

[f(n + 1) − f(n)] z−n = limz→1

(z − 1) F (z) − f(0)

Or f(∞) − f(0) = limz→1

(z − 1) F (z) − f(0)

Thus , limz→∞f(n) = lim

z→1(z − 1) F (z), (14.23)

The Z-Transform 381

provided the limits exist. Hence the result of this theorem is proved.

Theorem 14.7 : The Z-transform of the partial derivatives of afunction.

The Z-transform of ∂∂a [f(n, a)] is given by

Z

[∂

∂af(n, a)

]=

∂

∂a[Z{f(n, a)}]


Z

[∂

∂af (n, a)

]=

∞∑n=0

[∂

∂af(n, a)

]z−n

=∂

∂a

[ ∞∑n=0

f(n, a)z−n

]

=∂

∂aZ[f(n, a)] (14.24)

For example , if f(n, a) = nena then

Z [nena] = Z

[∂

∂aena

]

=∂

∂a[Z{ena}]

=∂

∂a

[z

z − ea

]=

zea

(z − ea)2

Example 14.5 If F (z) = z(z−a)(z−b) , find f(0) and f(1) by using the

initial value theorem.

Solution. By initial value theorem we get from eqns. (14.21) and(14.22) that

f(0) = limz→∞F (z) = lim

z→∞z

(z − a)(z − b)= 0

Again,

f(1) = limz→∞ zF (z)

= limz→∞

z2

(z − a)(z − b)= 1

Example. 14.6

Find the inverses of the following Z-transformed functions:

(a) F (z) =z

z − 2


(b) F (z) = exp (1z)

(c) F (z) =3z2 − z

(z − 1)(z − 2)2

(d) F (z) =z(z + 1)(z − 1)3

Solution.

(a) Here F (z) = zz−2

=1

1 − 2z

=(

1 − 2z

)−1

= 1 +2z

+22

z2+

23

z3+ · · ·

Therefore, Z−1[F (z)] = Z−1

[1 +

2z

+22

z2+ · · ·

]Here f(0) = 1, f(1) = 2, f(2) = 22, f(3) = 23, · · ·

Therefore, f(n) = 2n

So, Z−1[F (z)] = f(n) = 2n

(b) We have in this case

F (z) = 1 +1z

+12!

1z2

+ · · ·

Therefore, f(0) = 1, f(1) = 1, f(2) =12!

, · · ·

∴ f(n) =1n!

∴ Z−1[F (z)] = f(n) =1n!

Thus Z−1[e

1z

]=

1n!

(c) Here, by partial fraction method

3z2 − z

(z − 1)(z − 2)2≡ 2

z

z − 1− 2

z

z − 2+

52

2z(z − 2)2

Then, Z−1

[3z2 − z

(z − 1)(z − 2)2

]

= Z−1

[2z

z − 1

]− Z−1

[2z

z − 2

]+

52

Z−1

[2z

(z − 2)2

]

The Z-Transform 383

= 2.1 − (2)n+1 +52

. n 2n, by theorem (14.7)

= 2 − 2n+1 + 5n 2n−1

(d) We can express

z(z + 1)(z − 1)3

≡ z

(z − 1)2

[z

z − 1+

1z − 1

]

Let F (z) =z

(z − 1)2and G(z) =

z

z − 1+

1z − 1

Then by inversion of Z-transform we have

f(n) = n and g(n) = H(n) + H(n − 1)

where H(n) is the unit function .

Therefore, by convolution theorem

Z−1

[z(z + 1)(z − 1)3

]= f(n) ∗ g(n)

=n∑

m=0

m[H(n − m) + H(n − m − 1)]

= n2.

14.4 Application of Z-Transforms.

(a) Solution of difference equations.

Example 14.7

Solve the following initial value problems :

(i) f(n + 1) + 3f(n) = n, f(0) = 1

(ii) f(n + 2) − f(n + 1) − 6f(n) = 0, f(0) = 0, f(1) = 3

(iii) un+1 = un + un−1, u1 = u(0) = 1

(iv) u(n + 2) − u(n + 1) + u(n) = 0, u(0) = 1, u(1) = 2


Solutions.

(i) f(n + 1) + 3 f(n) = n , f(0) = 1The Z-transform of this difference equation yields

(z + 3) F (z) = z +z

(z − 1)2

Or F (z) =z

z + 3

(1 +

116

)+

14

z

(z − 1)2− 1

16z

z − 1

Upon inverting this transformed equation by Z-transform weget

f(n) =1716

(−3)n − 116

(1)n +14n.

(ii) f(n + 2) − f(n + 1) − 6 f(n) = 0 , f(0) = 0 , f(1) = 3.The Z-transform of the difference equation under the initialconditions yeilds

F (z)[z2 − z − 6] = 3z

Or F (z) =3z5

[1

z − 3− 1

z + 2

]

=35

z

z − 3− 3

5z

z + 2Upon inversion the Z-transformed equation yeilds

f(n) =353n − 3

5(−2)n.

(iii) un+1 = un + un−1 , u1 = u(0) = 1This equation defines a sequence which is known as theFibonacci Sequence.The Z-transform of the above difference equation yeilds

U(z) =z2

z2 − z − 1Upon inversion of the above transformed equation we get

un = Z−1

[z2

z2 − z − 1

]

≡ Z−1

[z2

(z − a)(z − b)

], say

where (a, b) =

(1 +

√5

2,

1 −√5

2

)

The Z-Transform 385

Therefore,

un = Z−1

[az

z − a

1a − b

]− Z−1

[bz

z − b

1a − b

]

=a

a − b(a)n − b

a − b(b)n

=an+1 − bn+1

a − b

(iv) u(n + 2) − u(n + 1) + u(n) = 0, u(0) = 1, u(1) = 2The Z-transform of this difference equation under the initialconditions yeilds

U(z)[z2 − z + 1

]= (z2 + z)

Or U(z) =z2 − z

2

z2 − z + 1+

√3(√

32 z)

z2 − z + 1

Upon inversion of the Z-transformed equation we get

u(n) = cosnπ

3+

√3 sin

(nπ

3

)as the solution of the given initial value problem.

(b) Summation of Infinite Series

Theorem 14.8 If Z[f(n)] = F (z), then

n∑k=1

f(k) = Z−1

[z

z − 1F (z)

]

and∞∑

k=1

f(k) = limz→1

F (z) = F (1).

Proof. Let h(n) =∑nk=1

f(k). Then clearly,

h(n) − h(n − 1) = f(n)

Applying Z-transform this relation gives

H(z) − z−1 H(z) = F (z) , by (14.13).

Therefore,H(z) =

z

z − 1F (z)


Inverting Z-transform we get

h(n) =n∑

k=1

f(k)

= Z−1

[z

z − 1F (z)

]

Also, if z → 1 we have from the Final Value Theorem that

limn→∞h(n) = lim

n→∞

n∑k=1

f(k)

= limz→1

[(z − 1).z

z − 1F (z)

]

Or∞∑

k=1

f(k) = limz→1

[zF (z)] = F (1).

Thus the proof is complete.

Example 14.8 Find the sum of the following series using Z-transform:

(a)∞∑

n=0

aneinx

(b)∞∑

n=0

e−(2n+1)x

(c)∞∑

n=0

(−1)nxn+1

n + 1

(d)∞∑

n=0

an cos nx

Solutions.

(a) Let f(n) = aneinx

We know that Z[einx]

=z

z − eix

Therefore , Z[f(n)] = Z[aneinx

]=

z

z − aeix

So,∞∑

n=0

aneinx = limz→1

z

z − aeix

=(1 − aeix

)−1

The Z-Transform 387

(b)∞∑

n=0

e−(2n+1)x = e−x∞∑

n=0

e−2nx

So, Z[e−(2n+1)x

]= Z

[e−xe−2nx

]=

e−x

1 − ze−2x

Therefore ,

∞∑n=0

e−(2n+1)x

= limz→1

e−x

1 − ze−2x

=1

ex − e−x= (2 sinh x)−1

(c) We know that Z[xn+1

]=

zx

z − x

Hence , Z

[xn+1

n + 1

]= z

∫ ∞

z

zx

z − x

dz

z2

= xz

∫ ∞

z

dz

z(z − x)

= xz

[1x

logz − x

z

]∞z

= −z log[z − x

z

]Now replacing x by − x, this above result gives

Z

[(−1)n

xn+1

n + 1

]= z log

[z + x

z

]Therefore ,

∞∑n=0

(−1)nxn+1

n + 1= lim

z→1z log

[z + x

z

]

= log(1 + x)

(d) We know that

Z[cos nx] =z(z − cos x)

z2 − 2z cos x + 1

∴ Z [an cos nx] =za

(za − cos x

)z2

a2 − 2za cos x + 1

=z(z − a cos x)

z2 − 2az cos x + a2


So ,

∞∑n=0

an cos nx = limz→1

[z(z − a cos x)

z2 − 2az cos x + a2

]

=1 − a cos x

1 − 2a cos x + a2

Exercises.


(a) Z[n3]

= (z3 + 4z2 + z)/(z − 1)4

(b) Z [H(n) − H(n − 2)] = 1 +1z

(c) Z[n2an

]=

z(z + a)(z − a)2

(d) Z [sinhna] =z sinh a

z2 − 2z cosh a + 1

(e) Z [nanf(n)] = −zd

dz

[F(z

a

)](f) Z

[f(n)

n + m

]= zm

∫ ∞

z

F (ξ)dξ

ξm+1,

Hence duduce from the result of (f) that

Z

[1

n + 1

]= z log

(z

z − 1

)

(2) Find the inverse Z transform of the following functions :

(a)1

(z − a)2(b)

z3

(z2 − 1)(z − 2)(c)

z2

(z − 1)(z − 12 )

(d)z

(z − a)m+1(e) z log

z + 1z

(f)z

z − 2[Ans. (a) (n − 1) an−2 H(n − 1) (b)

16[(−1)n + 2n+3 − 3(1)n

](c) (2 − 2−n) (d) n(n − 1) · · · (n − m + 1) an−m /m!

(e)(−1)n

(n + 1)(f) 2n

]

(3) Solve the following difference equations :

(a) f(n + 1) + 2f(n) = n, f(0) = 1

The Z-Transform 389

(b) f(n + 2) − 3f(n + 1) + 2f(n) = 0, f(0) = 1, f(1) = 2

(c) f(n + 2) − 5f(n + 1) + 6f(n) = 2n, f(0) = 1, f(1) = 0

(d) f(n + 2) − 2xf(n + 1) + f(n) = 0 , |x| ≤ 1 ,

f(0) = f0, f(1) = 0, n � 2

(e) f(n + 3) − 3f(n + 2) + 3f(n + 1) − f(n) = 0,

f(0) = 1, f(1) = 0 ; f(2) = 1

Ans. (a)19

[10(−2)n + 3n − 1] (b) 2n (c) 2n+1 − 3n − n 2n−1

(d)f0 z[(z − x) − 2x(1 − x2)−

12 ]

(z2 − 2xz + 1)(e) (n − 1)2

(4) Show that∞∑

n=0

(−1)n e−n

n + 1= e log (1 + e−1)

(5) (a) If F (z) = zz−α , prove by using initial value theorem that

f(0) = 1

(b) Use final value theorem to evaluate

limn→∞ f(n) if F (z) =

(3z2 − z)[(z − 1)(z − 2)2]

[Ans.2](6) Solve the difference equation

yk+2 − 3yk+1 + 2yk = 0

to prove that Y (z) =1

1 − 2z− 1

1 − z

and hence deduce that yk

= 2k − 1 , if y(0) = 0 , y(1) = 1.

Appendix : Tables of Integral Transforms

We collect here some important formulae derived in this text or elsewherewhich are of common use. For an exhaustive list of transforms the readershould also consult G.A. Campbell and R.M. Foster (Bell Telephone-system, Technical Publication, 1931), W. Magnus and F. Oberhettinger(Springer, Berlin, 1948), R.V. Churchill (Modern Operational Mathe-matics in Engineering, Mc Graw-Hill, NY, 1944), Erdelyi et al. (Tablesof Integral Transforms, Mc Graw-Hill, NY, 1954), Oberhettinger andHiggens (Boeing Sc. Res. Lab. Maths Note 246, Seattle, 1961), Ian N.Sneddon (Mc Graw-Hill Book Company, 1951), etc. Unless otherwisestated here a, t are assumed to be positive constants.

A1 : Fourier Transforms.f(t) F [f(t) ; t → ξ]

1√

2π δ(ξ)

tkH(t)√

2π ik[

12δ(k)(ξ) + (−1)kk!

2πiξk+1

]e−a|t|

√2π

aa2+ξ2

te−α|t|√

2π2aiξ (a2 + ξ2)−2

|t|e−a|t|√

2π (a2 − ξ2) (a2 + ξ2)−2

e−a2t2 1√2a

e−14

ξ2

a2

δ(t) 1√2π

1t −i

√π2 sgn ξ

Appendix : Tables of Integral Transforms 391

A1 : Fourier Transforms. (Continued)

f(t) F [f(t) ; t → ξ]

tk sgn t√

2k (−iξ)−k−1k!

sin att

{ √π2 , if |ξ| < a

0 , if |ξ| > a

chatchπt ,−π < a < π

√2π cos a

2 ch ξ2/[chξ + cos a]

shatshπt ,−π < a < π 1√

2πsin a

chξ+cos a

{1√

a2−t2, |t| < a

0 , |t| > aπ2 J0(aξ)

|t|−1 1|ξ|

H(a − |t|)√

2π

sin aξξ

t−1 i√

π2 sgn ξ

(1 − t2)(1 + t2)−2√

π2 ξ e−ξ

A2 : Fourier Cosine Transforms.

f(t) fc [f(t) ; t → ξ]

{1 , 0 < t < a

0 , t > a

√2π

sin aξξ

a(t2 + a2)−1, a > 0√

π2 e−aξ

e−at√

2π

aa2+ξ2

tp−1, 0 < p < 1√

2π Γ(p)ξ−p cos pπ

2

392 An Introduction to Integral Transform

A2 : Fourier Cosine Transforms. (Continued)

f(t) fc [f(t) ; t → ξ]

e−a2t2 1√2|a| e−

14

ξ2

a2

cos t2

21√2

[cos ξ2

2 + sin ξ2

2 ]

sin t2

21√2

[cos ξ2

2 − sin ξ2

2 ]

sh(αt)sh(βt) , 0 < α < β

√π2

1β

sin( παβ

)

ch(πξβ

)+cos�

παβ

�

ch(αt)ch(βt) , 0 < α < β

√2π 1

β

cos�

πα2β

�ch�

πξ2β

�

ch�

πξβ

�+cos

�παβ

�

t−νJν(at), ν > −12

(a2−ξ2)ν− 12 H(a−ξ)

2ν− 12 aν Γ(ν+ 1

2)

K0(at)√

π2

(a2 + ξ2

) 12

2e−t sin tt

√2π tan−1

(2ξ2

)(1−t2)(1+t2)2

√π2 ξ e−ξ

A3 : Fourier Sine Transforms.

f(t) Fs [f(t) ; t → ξ]

1t

√π2 sgn ξ

e−at√

2π

ξξ2+a2

t e−at√

2π

2aξ(ξ2+a2)2


A3: Fourier Sine Transforms. (Continued)

f(t) Fs [f(t) ; t → ξ]

t−1 e−at√

2π tan−1

(ξa

)

tp−1 , 0 a

J0(aξ)

1t(t2+a2)

√π2

1a2 (1 − e−|ξa|) sgn ξ

H(a − t) , a > 0√

2π

1−cos aξξ

cosech t√

π2 tanh(1

2πξ)

cosech(αt)sinh(βt) , 0 < α < β

√π2

1β

sinh�

πξβ

�

cosh�

πξβ

�+ cos

�παβ

�

sinh(αt)cosh(βt) , 0 < α < β

√2πβ

sin�

πα2β

�sinh

�πξ2β

�

cos�

παβ

�+ cosh

�πξβ

�

J0(at)t

{ √π2 , a < ξ < ∞√2π sin−1

(ξa

), 0 < ξ < a

t J0(at)k2+t2

√π2 e−kξ I0 (ak) , ξ > a


A4 : Laplace Transforms.

f(t) L [f(t) ; t → p]

1 1p , p > 0

tν , ν > −1 p−ν−1 Γ (ν + 1)

eat 1p−a , Re p > a

cosh (at) p(p2−a2)

, Re p > a

sin (at) a(p2−a2)

, Re p > a

t cosh (at) (p2 + a2)(p2 − a2)−2 , Re p > a

t sinh (at) 2ap((p2 − a2)−2 , Re p > a

cos (at) p (p2 + a2)−1

sin (at) a (p2 + a2)−1

t cos (at) (p2 − a2) (p2 + a2)−2

t sin (at) 2ap (p2 + a2)−2

t−1 sin (at) cot−1 ( pa)

e−14b2t2 √

π b−1ep2

b2 Erfc (pb ) , b > 0

Erf(√

t) p−1(p + 1)−12

Erf( a√t) p−1e−2a

√p

tνJν(at) (2a)ν Γ (ν+ 12)

√π(p2+a2)ν+1

2, ν > −1

2

tν+1Jν(at) (2a)ν+1aν p Γ (ν+ 32)

√π(p2+a2)ν+3

2, R ep > a > 0

tνI0(at) p (p2 − a2)−32 , R ep > a > 0

eat−1a

1p(p−a)


A4 : Laplace Transforms. (Continued)

f(t) L [f(t) ; t → p]

sin(a√

t) 12 a√

πp3 e

−a2

4p , Re p > a > 0

sin(a√

t)t Erf(1

2a√p) , Re p > a > 0

{0 , 0 < t be−bp

∑nm=0

n! bm

m!pn−m+1 , Re p > 0

{0 , 0 < t bbK1(bp) , Re p > 0.

e−14

at

√a p−

12 K1(

√pa) , Re a � 0.

log t −p−1 log(γp)

2t sinh(at) log p+a

p−a , Re p > |Re a|

t cos at (p2−a2)(p2+a2)2

t cosh at (p2 + a2)/(p2 − a2)−2

sin(at)t tan−1 (a

p )

J0 (at) (p2 + a2)−12

I0(at) (p2 − a2)−12

δ(t − a) exp (−ap) , a > 0

| sin at| , a > 0 a(p2+a2)

coth (πp2a )


A5 : Hilbert Transforms.f(t) H [f(t) ; t → x]

tt2+a2

ax2+a2

H(t − a) − H(t − b) 1π log | b−x

a−x | , b > a > 0

H(t−a)t

1πx log a

|a−x| , a > 0

(at + b)−1 H(t) 1π(ax+b) log b

a|x| , a, b > 0, x �= ba

|t|ν−1 , 0 < Re ν < 1 −ctn (νπ2 ) |x|ν−1 sgn x

eiat i eiax

sin(at)/t {cos(ax)−1}x

sin atJ1(at) cos(ax) J1 (a x)

cos atJ1(at) − sin(ax) J1 (a x)

A6 : Stieltjes Transforms.

f(x) S [ f(x) ; x → y ] =∫∞

0

f(x)x+y dx

1x+a

1a−y log |ay |

e−ax√x

π√y eay Erfc (

√ay).

1a2+x2

1a2+y2

[ny2a − log y

a

]xν , −1 < Re ν < 0 −πyν cosec (πν)

xν

a+x , −1 < Re ν < 1 π(aν−yν)(a−y) sin ν π

x−νe−ax , Re α > 0, Re ν < 1 Γ(1 − ν) y−ν eay Γ (ν, ay)


A6 : Stieltjes Transforms. (Continued)

f(x) S [ f(x) ; x → y ] =∫∞

0

f(x)x+y dx

log x(a+x) , |arg a| < π 1

2 (y − a)−1[(log y)2 − (log a)2

]sin(a

√x) πexp(−a

√y)

x−1 sin(a√

x) πy−1[1 − exp (−a

√y)]

1√x

cos(a√

x) π√y exp (−a

√y )

A7 : Hankel Transforms.

f(x) Hν [ f(x) ; x → ξ ]

{xν , 0 < x < a

0 , x > a(ν > −1) aν+1

ξ Jν+1 (ξa)

(a2 − x2) H (a − x) 4aξ3 J1(ξa) − 2a2

ξ2 J0(ξa)

e−px , (ν = 0) p (ξ2 + p2)−12

e−px

x , (ν = 0) (ξ2 + p2)−12

e−px , (ν = 1) ξ (ξ2 + p2)−12

e−px

x , (ν = 1) 1ξ − p

ξ√

ξ2+p2

a√a2+x2

, (ν = 0) e−aξ

sin(ax)x , (ν = 0) (a2 − ξ2)−

12 H(a − ξ)

sin(ax)x , (ν = 1) a

ξ√

ξ2−a2H (ξ − a)


A7: Hankel Transforms. (Continued)

f(x) Hν [ f(x) ; x → ξ ]

xνe−x2

a2(

12a2)ν+1

ξν e−ξ2a2

4

x e−ax , (ν = 1) ξ(a2 + ξ2

)− 32

1x , (ν = 0) 1

ξ

√x (a2 + x2)−

12 , (ν = 0) 1√

ξe−aξ

√x (a2 − x2)−

12 H(a − x) , (ν = 0) 1√

ξsin (−aξ)

√x (x2 − a2)−

12 H (x − a) , (ν = 0) 1√

ξcos (−aξ)

1√x

log x , (ν = 0) − 1√ξ

log (2ξ)

1√x

sin (ax) , (ν = 0) ξ−12

(a2 − ξ2

)− 12 H (a − ξ)

1√x

cos (ax) , (ν = 0) ξ12

√(ξ2 − a2) H (ξ − a)

√x cos

(a2x2

2

), (ν = 0)

√ξ

2a2

H(a − x) , (ν = 0) aξ J1(aξ)

1x2 (1 − cos ax) , (ν = 0) cosh−1 (a

ξ ) H (a − ξ)

(x2 + a2)−12 , (ν = 0) 1

ξ e−aξ{√

ξ2 + a2 − a}2

e−ax

x , (ν = 1) 1ξ

[1 − a

(ξ2+a2)12

]


A8 : Finite Hankel Transforms.

f(r)∫ a0 f(r) r Jn (rξ)dr = Hν [ f(r) ; r → ξ ]

c, (n=0) caξ J1(ξa)

(a2 − r2) , (n = 0) 4aξ3 J1(ξa) − 2a2

ξ2 J0(ξa)

rn , (n > −1) an+1

ξ Jn+1 (ξa)

J0(ar)J0 (a) − 1 , (n = 0) −J1(ξ)

ξ�1− ξ2

a2

�

1r , (n = 1) 1

ξ{1 − J0(aξ)}

(a2 − r2)−12 , (n = 0) 1

ξ sin aξ

1r (a2 − r2)−

12 , (n = 0) (1−cos aξ)

(aξ)

A9 : Mellin Transforms.

f(x) M [f(x) ; x → s] =∫∞

0f(x) xs−1dx

e−px p−s Γ (s) , Re s > 0

e−x2 12 Γ

(s2

)(1 + x)−a Γ (s) Γ(a − s)/Γ(a) , 0 < Re s < Re a

(1 + xa)−1 πa cosec πs

a , 0 < Re s < Re a

sin x Γ (s) sin(

sπ2

), 0 < Re s < 1

cos x Γ (s) cos(

sπ2

), 0 < Re s < 1

H(a − x) as

s


A9 : Mellin Transforms. (Continued)

f(x) M [f(x) ; x → s] =∫∞

0f(x) xs−1dx

(1 − x)p−1 H (1 − x) Γ (s)Γ(p)Γ(s+p) , Re p > 0

log (1 + x) πs cosec sπ

(1 − x)−1 π cot (sπ) , 0 < Re s < 1

xν H (1 − x) (s + ν)−1 , Re s > −Re ν

log x . H (a − x) 1sas

(log a − 1

s

), Re s > 0

{log x}(a+x) , |arg a| < π πas−1 cosec πs [log a − π cot πs] ,

0 < Re s < 1

xν log x H (1 − x) −(s + ν)−2, Re s > −Re ν

e−x(log x)n dn

dsn [Γ (s)] , Re s > 0

log |1 − x| πs−1 cot (πs), −1 < Re s < 0

log |1+x1−x | π

s tan πs2 , −1 < Re s < 1

Jν (ax) , (a > 0) 2s−1 Γ(

s2 + ν

2

)as Γ

(ν2 − s

2 + 1)

,

Re ν < Re s < 32

log axH(a − x) as

s2

x−1 log (1 + x) π (1 − s)−1 cosec πs


A9 : Mellin Transforms. (Continued)

f(x) M [f(x) ; x → s] =∫∞

0f(x) xs−1dx

xa (1 + x)−b Γ(a + s) Γ (b − a − s)Γ(b)

(1 + ax)−n Γ(s) Γ(n−s)asΓ (x) , 0 < Re s < n

A10 : Kontorovich - Lebedev Transforms.

f(x) K [f(x) ; τ ] ≡ f(τ) =∫∞

0x−1 f(x) Kiτ (x) dx

1 π2τ cosech

(πτ2

)x π

2 sech(

πτ2

)e−x π

τ cosech (τπ)

ex πτ coth (τπ)

xe−x πτ cosech (τπ)

e−x cosh t π cos(τt)τ sinh(πτ)

x e−x cos t π sinh(τt)cosec tsinh(πτ)

x e−x cosh t π sin(τt)sinh(πτ) sinh t


A11 : Mehler - Fock Transforms (of zero order)f(α) f∗

0 (τ) = Φ[f(α) ; α → τ ]

H(t − α) (ch t − ch α)−12

√2 τ−1 sin tτ

H(α − t) (ch α − ch t)−12

√2 τ−1 cth (πτ) cos(τ t)

(ch α + cos β)−1 π sech (τπ) P− 12+iτ (cos β), −π < β < π

(ch α + cos β)−12

√2

τch (βτ)sh (πτ) , π < β < π

sech (α2 ) 2 τ−1 cosech (πτ)

√sech (α) 1√

2τcosech

(π τ2

)

A11 : Another form of the Mehler-Fock Transforms.

f(x) f0(τ) = Φ0[f(x) ; x → τ ] =∫∞1 f(x) P− 1

2+iτ(x) dx

e−ax√

2πa Kiτ (a) , |arg a| < π

2

x− 12

√2 sech (π τ

2 )

x− 32

2√

23 τ cosech(πτ

2 )

1x+t πsech (πτ)P− 1

2+iτ (t), |t| < 1

1√x+t

√2 ch(τ cos−1 t)

τ sh (πτ)


A12 : The Z - Transformsyn

∑ynzn

{1 , n = k0 , n �= k zk

c c1−z

n z(1−z)2

n2 z(1+z)(1−z)3

n3 z(z2+4z+1)(1−z)4

cn 1(1−cz)

ncn cz(1−cz)

(n + k

k

)cn 1

(1−cz)k+1

(b

n

)cn ab−n (a + cz)b

cn

n , (n = 1, 2, 3, · · ·) −ln(1 − cz)

cn

n , (n = 1, 3, 5, · · ·) 12 ln

(1+cz1−cz

)= tanh−1 (cz)

cn

n , (n = 2, 4, 6, · · ·) −12 ln

(1 − c2z2

)cn/n! ecz

cn/n! , (n = 1, 3, 5, · · ·) sinh (cz)

cn/n! , (n = 0, 2, 4, · · ·) cosh (cz)

(lnc)n/n! cn


A12 : The Z - Transforms (Continued)

yn∑

ynzn

sin (cn) z sin cz2+1−2z cos c

cos(cn) 1−z cos cz2+1−2z cos c

b−an sin(cn) z sin cba+b−az2−2z cos c

b−an cos (cn) ba−z cos cb−az2+ba−2z cos c

Bibliography

The following is a sketch of the bibliography which can not be consideredas comprehensive and exhaustive. It consists of books and researchpapers to which reference is made in the text or otherwise. Many otherrelated materials from other books or papers are also included in thetext. Acknowledgement is due to all the authors.

(1) Bateman, Harry (1954). Tables of Integral Transforms, Mc Graw-Hill Book Company Inc.

(2) Chakraborty, A and Williams, W.E. (1980). A note on a singularIntegral Equations, J. Inst. Maths. Applics, 26, 321-323.

(3) Churchill, R. V (1951). The Operational calculus of LegendreTransforms, J. Math. and Phy, 33, 165-178.

(4) Churchill, R.V. and Dolph, C.L. (1954). Inverse Transform ofproduct of Legendre Transforms, Proc. Amer. Math. Soc., 5,93-100.

(5) Comninou, M (1977). The interface Crack, J. Appl. Mech., 44,631-636.

(6) Das, S, Patra, B and Debnath, L (2004). On elasto-dynamicalproblem of interfacial Griffith cracks in composite media, Int. J.Engng. Sci., 42, 735-752.

(7) Debnath, L (1995). Integral Transforms and their applications, CR C Press Inc.

(8) Debnath, L and Thomas, J (1976). On finite Laplace Transforma-tion with Applications, ZAMM, 56, 559-563

(9) Dunn, H.S. (1967). A generalisation of Laplace Transform, Proc.Camb. Phil. Soc., 63, 155-161.

(10) Gakhov, F. D. (1966). Boundary Value Problems, Pergamon Press,London.

(11) Kober, H. (1967). A modification of Hilbert Transform, the Weylintegral and functional equations, J. Lond. Math. Soc., 42, 42-50.

406 Introduction to Integral Transform

(12) Muskhelishvili, N.I. (1953). Singular Integral Equations, Noord-hoff, Groningen

(13) Magnus, W and Oberhettinger, F. (1943). Formulas and Theoremsfor the functions of Mathematical Physics. Chelsea Publishing Co.,NY

(14) Naylor, D. (1963). On a Mellin Type Integral Transform, J. Math.Mech., 12, 265-274

(15) Okikiolu, G. O. (1965). A generalisation of Hilbert Transform.(16) Peters, A.S. (1972). Pair of Cauchy Singular Integral Equations,

Comm. Pure. Appl. Maths, 25, 369-402(17) Riemann, B. (1976). Uber die Anzahi der Primzahlen under eine

Gegebenen Grosse, Gesammelte Math. Werde, 136-144.(18) Sneddon, I. N. (1946). Finite Hankel Transform, Phil Mag., 37,

17.(19) Sneddon, I. N. (1974). The use of Integral Transform, Tata Mc

Graw-Hill Publishing Co. Ltd.(20) Senddon, I.N. (1951). Fourier Transforms, Mc Graw-Hill Book

Company, NY.(21) Tricomi, F.G. (1951). On the finite Hilbert Transformation, Q.J.

Maths., Oxford, 2, 199-211(22) Tricomi, F.G. (1955). Integral Equations, Dover Publications, Inc,

NY.(23) Titchmarsh, E.C. (1959). An Introduction to the Theory of Fourier

Integrals, Second Edition, Oxford Univ. Press, Oxford.(24) Ursell, F. (1983). Integrals with large Parameters : Hilbert Trans-

forms, Math. Proc. Comb. Phil. Soc., 93, 141-149.(25) Gradshteyn, I.S. and Ryzhik, I.M. (1980). Table of Integrals, Se-

ries and Products, Academic Press, NY.(26) Oberhettinger, F(1974). Tables of Mellin Transforms and Appli-

cation, Springer-Verlag, N.Y.(27) Watson, E. J. (1981). Laplace Transforms and Applications, Van

Nostrand Reinhold, N.Y.

Index

AAbel’s integral equation, 196absolutely integrable, 2analytic, 154analytic continuation, 129Application to Mechanics, 177associated Legendre equation, 341associative law, 131asymptotic expansion, 228axisymmetric contact problem, 346

BBessel function, 127Bilateral Laplace Transform, 166Binomial theorem, 128Boundary value Problem, 55, 180Boundary Value Problem in a

wedge, 332branch point singularity, 154branch-cut, 154Bromwich integral, 153BVP, 170

CCanchy’s integral formula, 374Cauchy principal value sense, 220,

222Cauchy’s residue theorem, 36, 154Cauchy’s theorem, 312Change of Scale, 9, 107, 239, 279circular disc, 256Classes of functions, 2commutative law, 131complex form of Fourier series, 3complex Fourier transform, 56

complex Inverse formula, 156, 159concentrated load, 178contour, 37contour integral, 153Convolution, 20, 131, 324Convolution of generalised Mellin

transform, 297convolution theorem, 22, 150, 286,

324, 356convolution theorems for finite Fourier

transform, 82cosine integral, 124cosine transforms, 7

DDerivative of finite Laplace trans-

form, 309derivative of FT, 37Derivative of Mellin Transform, 281Difference, 197difference equations, 1, 198differential - difference equation, 197,

198differential equations, 1diffusion equation, 57, 89, 187Dirac delta function, 122Dirichlet’s conditions, 2distributive law, 131double finite Fourier cosine trans-

form, 85double finite Fourier transforms, 86double finite sine transform, 85Double Laplace Transform, 161Double Transforms of partial

derivatives, 86

408 book name

duplication formula, 127, 129

EEddington solution, 49elastic membrane, 97electrical network, 170, 175entire function, 314error complementary function, 126Error function, 126Estimated accuracy, 159even function, 8, 163existence, 103exponential, 104Exponential integral, 126exponential order, 2exterior Dirichlet problem, 326external crack, 335

FFalting, 20, 21Faltung, 20, 82faltung or resultant, 131Faltung Theorem, 21Fibonacci Sequence, 384final-value theorem, 116, 311finite Hankel transform, 268finite difference, 197finite Fourier cosine transform, 79,

80finite Fourier sine transform, 80finite Fourier transforms, 79Finite Hankel transform, 260, 262,

269finite Hankel transform of order n,

267finite Hilbert transform, 226Finite Laplace transform, 302Finite Laplace transform of Deriv-

atives, 308

Finite Laplace transform of Inte-grals, 309

Finite Mellin transform, 297finite sine transform, 80First shift theorem, 107flexible string, 190Fourier Integral Formula, 2, 4Fourier kernel, 79Fourier Series, 2Fourier sine transform, 7Fourier transform, 6, 7Fourier’s integral theorem, 75Fourier-Bessel Series, 260, 261Fredholm integral equation, 47, 51,

193freely hinged, 97freely supported ends, 179Functions of several variables, 53

GGamma function, 33, 127, 129Gegenbauer polynomial, 355Gegenbauer Transform, 355general evaluation technique of in-

verse Laplace transform, 153,156

Generalised Mellin transform, 295

HHankel inversion formula, 242Hankel transform, 238, 243harmonic, 254heat conduction equation, 88 91,

92, 181, 184heat conduction problem, 61Heaviside function, 121Heaviside unit step function, 13, 121,

198Hermite transform, 364

Index 409

Hermite transform of derivative ofa function, 367

heuristic approach, 141Hilbert transform, 220hinged end, 90

Iidentity operator, 7image function, 141improper function, 122infinite strip, 63infinite wedge, 277Initial Value, 311initial-boundary value

problems, 269initial-value problem, 169initial-value theorem, 116insulated surface, 60integral equation, 1, 47, 193integral equation of convolution

type, 47Integral of a function, 112Integral of finite Laplace

transform, 310integral of Laplace transformed

function, 115integral transforms, 1Integro-differntial equation, 193, 197interior Dirichlet problem, 325inverse finite cosine transform, 79inverse finite Hankel

transform, 262, 268inverse Fourier transform, 6, 7inverse Hilbert transform, 220, 221inverse Laplace transform, 103, 141Inverse Stieltjes

Transform, 232, 233, 234

inversion formula for, 243inversion formula for Legendre

transform, 318inversion of Z-transform, 373iterative Laplace transform, 166

JJacobi polynomials, 351Jacobi Transform, 351

Kkernel, 1, 193kinematic viscosity, 271Kontorovich-Lebedev transform, 328

LL’ Hospital’s rule, 312L-C-R circuit, 175Laguerre polynomial, 135, 158Laguerre polynomial due to McCully

Mittag-Leffler, 137, 360Laguerre transform, 359Laguerre transform of order zero,

362Laguerre transform of the deriva-

tive, 361Laplace equation, 291Laplace equation in a half space,

345Laplace inversion formula, 141Laplace transform, 103Legendre Transform, 317Lerch’s theorem, 141, 158linear superposition, 333Linearity property, 8, 104, 142, 279linearity property of Hankel trans-

form, 239Lipschitz condition, 302

410 book name

MMaclaurin series, 49mathematical induction, 114, 284mean value theorem, 123Mehler-Fock inversion, 349Mehler-Fock transform of

order m, 341Mehler-Fock transform of zero or-

der, 337Mellin Inversion theorem, 285Mellin transform, 278Mellin transform of derivative, 281Mellin transform of Integral, 283Modified Bessel function, 130modified Hilbert transform, 227Modulation theorem, 10moment of inertia, 179Multiple Finite Fourier

Transform, 85multiple Fourier transforms, 54

NNull function, 141

Oodd and even periodic extensions,

82odd function, 7, 164ODE with variable co-efficients, 173one-sided Hilbert transform, 228Operational Properties of

Z-transform, 376Operational Properties, 307, 323operators, 7, 103original function, 141orthogonality property, 260

PParseval relation, 23, 244, 329

Parseval type relation, 339Parseval’s identities, 24partial differential equation, 180partial fractions, 145penny-shaped crack, 347periodic extensions, 83periodic function, 115, 136, 152piecewise continuous, 2Poisson integral formula, 326power series expansion, 160principal period, 3problem of potential theory, 277properties of Legendre

transforms, 318

Qquarter plane, 70

RRational Functions, 36Riemann sum, 372Rodrigues formula, 364

Sscaling property, 308Second shift theorem, 107semi-infinite strip, 99shift theorem, 158shifting property, 10, 308simple poles, 3, 36simultaneous ODE, 171sine integral Si(t), 124singlevalued function, 2singular integral equation, 226singular points, 154singularities, 36, 37slowly convergent series, 293Solution of difference equations, 383Solution of Integral equations, 292

Index 411

Solution of ODE, 168special functions, 121steady state, 60steady state temperature, 67, 100step function, 13Stieltjes Transform, 230strip, 60Sufficient conditions, 103, 104Summation of Infinite Series, 385Summation of Series, 293

TThe convolution Theorem, 82, 379The Final value Theorem, 311, 380The Initial Value Theorem, 311, 380The inverse finite Laplace

transform, 303The Z-transform of the partial

derivatives of a function, 381thermal diffusivity, 184three dimensional Laplace

equation, 95to Eddington, 48toroidal co-ordinate system, 342transverse displacement, 189, 255Tricomi’s method, 158two-sided Hilbert transform, 227two-sided Laplace transform, 166

Uunit step function, 102

Vvibrating string, 58vibration, 90viscous fluid, 271Volterra integral equation, 47, 193

Wwave equation, 89, 255, 274

YYoung’s modulus, 179

ZZ - Transform , 372zeta function, 293

Documents

An Introduction to Integral Transforms