Singular Integral Equations arising in Water Wave Problems

Department of Mathematics Indian Institute of Science Bangalore-560012, India Email:alok@math.iisc.ernet.in

Aloknath Chakrabarti

2

Abstract

Mixed Boundary Value Problems occur, in a natural way, in varieties of branches of Physics and Engineering and several mathematical methods have been developed to solve this class of problems of Applied Mathematics.

While understanding applications of such boundary value problems are of immense value to Physicists and Engineers, analyzing these problems mathematically and determining their solutions by utilizing the most appropriate analytical or numerical methods are the concerns of Applied Mathematicians.

Of the various analytical methods, which are useful to solve certain mixed boundary value problems arising in the theory of Scattering of Surface Water Waves, the methods involving complex function theory and singular integral equations will be examined in detail along with some recent developments of such methods.

3

Literature Review

Sneddon [1972]: Varieties of mixed boundary value problems of mathematical physics can be solved by reducing them to integral equations of one type or the other.

Muskhelishvili [1953], Gakhov [1966], Mikhlin [1964] : Certain singular integral equations and their methods of solution in detail.

Chakrabarti [2006] : Development of above recently.

Chakrabarti [1997] and Mandal and Chakrabarti [2000: Book]: Occurrences of such singular integral equations in studies on problems of scattering of surface water waves by barriers, present in the fluid medium.

4

1.Solution of Abel’s Integral equation and its generalization

Consider here the general form of Abel’s Integral equation

, 0 1 ,x

a

g t dtf x a x b

h x h t

where h(t) is a strictly monotonically increasing and differentiable function in (a, b) and

Solution of (1.1) by using a very simple method :

Consider 1

x

a

h u f u duI x

h x h u

By using (1.1), we can express (1.2), after interchanging the orders of integration, as

0, ,h t for all t a b

(1.1)

(1.2)

5

1

x x

a a

h x dxI x g t dt

h u h t h x h u

(1.3)

Using the transformations 2 1, ,h u h x h t

we obtain, from (1.3) and (1.2), the following results

1sin

x x

a a

h u f u dug t dt

h x h u

giving, on differentiation :

1

sin x

a

h u f u dug x

h x h u

which solves the Abel’s integral equation (1.1) completely.

(1.4)

(1.5)

(1.6)

6

Examples of Abel’s integral equation equation (1.1)

Example-1.1

12

0

, 0 1x g t dt

x xx t

Solution:

12

12

1 2x

a

d tdt xg x

dx x t

Example-1.2:

1

2cos cos

x

a

g t dtf x

x t

Solution:

12

sin1

cos cos

x

a

t f t dtdg x

dx t x

(1.7)

(1.8)

(1.9)

(1.10)

7

Example 1.3:

1

22 2

x

a

g t dtf t

x t

Solution:

122 2

2sin( )

x

a

f t dtdg x

dx x t

Here we have chosen 2 , 1/ 2h x x

Example 1.4 :

2 2( )

b

x

g t dtf x x b

t x

Solution: 12 2

2sin( )

b

x

tf t dtdg x

dx t x

(1.11)

(1.12)

(1.13)

(1.14)

8

A direct method of solution of Abel’s integral equation

0

, 0 1 , 0 , (0) 0x t dt

f x x fx t

(1.15)

Writing

0

, , , , uxF K u x f x k x e dx

(1.16)

Then the integral equation (1.15) can be expressed as

)(

)()(

uK

uFu (1.17)

where

0 0

( ) ( ) ux x uxK u k x e dx x e dx

(1.18)

If we look at the given equation (1.15) as

0

( ) ( )t k x t dt f x

(1.19)

with k(x) = x-, and if we recall the convolution theorem in the following form

9

0

( )x

L t k x t dt u k u

(1.20)

Now, we easily find that

11k u u (1.21)

Then, if we set

, 0 0d

x withdx

so that

0

( ) uxdu e dx u u

dx

(1.22)

(1.23)

Then we obtain,

1

F u uu

(1.24)

By utilizing the convolution theorem (1.20) once more, in a clever manner, we find that (1.24) gives:

1

0

1

1

x

x x t f t dt

(1.25)

10

which, on using the identity

1

sin

(1.26)gives

1

0

sin x f t dtd dx

dx dx x t

(1.27)

By using integration by parts we can also rewrite this as (since f(0) = 0)

1

0

sin x

x x t f t dt

(1.28)

We shall next consider the general form of Abel’s integral equation which is given by the relation

1 2( )( )( ) ( )( )( ) ( )a x A x b x A x f x

1

2

, 0 1

x t dtA x

x tx

t dtA x

t x

(1.29)

(1.30)

where

The method of solution of the general Abel’s integral equation (1.29) involves the theory of functions of a complex variable leading to Rieman Hilbert type boundary value problems.

11

Some Important Theorems and Results in Complex Function Theory

Theorem -1. If the function () satisfies the Hölder condition:

1 2 1 2 , 0 1A

where A is a positive constant, for all pairs of points on a simple closed, positively oriented contour of the complex z – plane (z = x+iy, i2 = -1), then the Cauchy-type as given by the relation:

1 2,

1,

2z dz where z

i z

(A1)

(A2)

represents a “sectionally analytic” (analytic except for points z lying on ) function of the complex variable z. The function (), in the relation (A2) is called the “density function” of the Cauchy-type integral (z).

12

Theorem -2: (The Basic Lemma)

If the density function () satisfies a Hölder condition, then the formula

on passing through the point z = t, of the simple closed contour , behaves as a continuous function of z, i.e.,

exists and is equal to (t).

[Note : Theorem also holds even if is an arc in the z- plane, provided that the point g does not coincide with any end point of ].

1

2

tz d

i z

(A3)

1lim

2z t

tz d

i t

(A4)

13

Theorem-3 (Plemelj-Sokhotski Formulae) If

represents a sectionally analytic function, as in Theorem (*), then and exist, then the following formulae hold good:

1( ) ,

2z dz

i z

,

1,

t t t

where tt t d

i t

where means that the points z approach the point t on from the left of the positively oriented contour , and means that z approaches t from the right of .

The formulae (A5) are known as the Plemelj-Sokhotski formulae (also referred to as just the Plemelj formulae) involving the Cauchy-type integrals (z). The formulae (A5) can also be expressed as :

limz t

limz t

(A5)

(A6)

14

1 1,

2 2

1 1.

2 2

t t di t

t t di t

(A7)

(Plemelj formulae)

15

Generalized Abel Integral Equation and its Solution

The generalized Abel integral equation

, 0 1x t dt t dt

a x b x f x xx t t x

whereas the forcing term f(x) and the unknown function (x) belong to those classes of functions which admit representations of the form

1

0

f x x x f x

xand x

x x

where possesses a Hölder continuous derivative in and satisfies Hölder’s condition in ,

(G1)

(G2)

16

1

,t dt

zR z t z

With 12 1

,R z z z

so that 10 , ,z as zz

and the associated Riemann – Hilbert problem is finally solved by utilizing the Plemelj – Sokhotski formulae involving Cauchy-type singular integrals.

(G3)

(G4)

(G5)

17

Particular Example

Integral Equation: ,t dt

f x xx t

(G6)

Solution (by Gakhov [1996]):

1

sin( ),

x g tdx

dx x t

(G7)

where cot 21

( ) ( )2 2

dtg x f x R x f t

R t t x

(G8)

18

The Detailed Method

0 1 ,x

x

t dt t dt t dtz

t z t z t z

(G9)

As z tends to a point x , from above ,

2, 1 ,z x iy y o i

Then the sectionally analytic function (z) ) tends to the following limiting values:

1 2ix e A x A x (G10)

Where 1 ,t dt

A xt z

and 2

x

t dtA x

t x

(G11)

The relation (D2) can also be expressed as

1

1,

2 sinA x x x

i

and

2

1.

2 sini iA x e x e x

i

(G12)

,z x iy y o and below

19

By using the relations (G12) in the given integral equation (G1), we obtain

( ) ( ) 2 sin( ) ( )i ia x e b x x a x e b x x i f x x (G13)

Relation (13) represents the special Riemann-Hilbert type problem

( ) ( ) ( ),x G x x g x x with

sinexp 2 arctan

cos

i

i

a x e b x b xG x i

a x e b x a x b x

and

2 sin.

i

i f xg x

a x e b x

(G14)

(G15)

(G16)

Method of solution of the new Riemann-Hilbert type problem (14):

(z), given by equation (G9), satisfies the following condition at infinity:

1, .z O as z

z (G17)

20

First solve the homogeneous problem (14), satisfying the relation

0 0( ) ( ) ( ) 0,x G x x Giving 0 0 0( ) ( ) ,x x G x

where 0 0expz z and 0expG x G x

(G18)

(G19)

(G20)

Now we can express the function satisfying (19), as :

00

( )u

t dtz

t z

where 1 1

0 1 0( ) 2 sin ( ),x i A G x

with

011 0 1

sin x G t dtdA G x

dx x t

(G21)

(G22)

(G23)

Next, by utilizing (19) in (14), we obtain

0 0 0

( ) ( ),

( ) ( ) ( )

g xx x

x x x

(G24)

where

21

0 0( ) expx x with being obtainable by suing the relations (21)-(23). 0 x

Then, by utilizing the first of the formulae (G12), we can determine the solution of the Riemann-Hilbert type problem (G24), as given by :

0

( ) ( ),

( )

z t dt

x t z

where 1

0

1( ) .

2

g t dtdx

i dx t x t

(G25)

(G26)

(G27)

The relation (27) takes the equivalent form:

1 1

1( ) ,

2 ( )

xp p t dtx

i t xx

with

0

( )( ) , ( ) ,

( )

g t dpp t p t

t dt

(G28)

(G29)

22

Next we obtain the following limiting values of the function (z), as z approaches the point , :x [see (G10)]:

0 1 2( ) ix x e A x A x giving

( ) ( ) ( ),x x h x say where 0 0 1 0 0 2( ) ( ) ( ) ( ) ( ) ,i ih x e x e x A x x x A x

Finally, by utilizing the first formula in (G12), once again, we obtain the required solution of the given generalized Abel integral equation (G1) in the form

1

1 ( ).

2

xd h t dtx

i dx x t

The result (G33) can also be expressed in the equivalent form:

1 1

1,

2xh a h t dt

xi x x t

(G30)

(G31)

(G32)

(G33)

(G34)

23

2. Solution of Singular Integral Equations of the Cauchy type

The general theory of a single linear singular integral equation of the type

, ,bac t t dr f t t

t

(2.1)

Defining a sectionally analytic function

1, ,

2

tz dt z

i t z

(2.2)

Utilizing the Plemelj-Sokhotski formulae we can rewrite (2.1) as

,c t t t i t t f t i.e.

, ,

c t i f tt t t

c t i c t i

provided c i

(2.3)

24

The relation (2.3) is a particular case of the most general such relation, as given by

,t G t t h t t (2.4)

Consider the case c = p in equation (2.1) and is the open interval 0 < x < 1:

We first solve the homogeneous Riemann-Hilbert problem (2.3) in this particular case.

Here 0 0 , 0,1

p ix x x

p i

(2.5)

Then, by the aid of any suitable solution 0 ,z of the homogeneous

problem (2.5), we can cast the original Riemann-Hilbert problem as:

0 0 0

, 0,1x x f x

xx x p i x

(2.6)

The general solution of the Riemann-Hilbert problem (2.6) can be written down by the aid of the Plemelj-Sokhotski formulae. The general solution is given by

1

0 00

1

2

f tz z dt E z

i t p i t z

where E(z) is an arbitrary entire function of z.

(2.7)

25

Then we find that the general solution of our integral equation (2.1) can be determined by means of the relation

,x x x x (2.8)

We thus find that the general solution of the integral equation (2.1) depends on anarbitrary choice of an entire function E(z) appearing in the relation (2.7). A special choice of E(z) can be made depending on the class of the forcing functions f(x) and the selection of the function representing the solution of the homogeneous problem (2.5).

0 z

To illustrate the above procedure we take up the special case such that

x and f(x) are bounded at x=0 but unbounded at x=1 , with an integrable

singularity there. We select

0 ,1

zz

z

2 ,iie

i

so that we have 1

0 .2

(2.9)

(2.10)

(2.11)

26

Then, observing that [by fixing the idea that 0 arg( ) 2 ]z

0 ,1

ixx e

x

30 , 0 1

1ix

x e xx

(i)

(ii) 0lim 1z

z

(2.12)

(2.13)

as well as the fact that

lim 0z

z

(2.14)

We find that we must select

0E z (2.15)

giving

0 1

00

,2

z f tz dt z

i i t t z

(2.16)

Using the Plemeji-Sokhotski formulae on the relation (2.16), together with the results (2.12), we find that the relation (2.8) produces the unique solution of our integral equation (2.1) in this special circumstance. It is given by:

27

x x x

0 00 0 1

00 02 2

f x x xx x f tdt

i i t t x i x

102 2

1 1

1

f tx tf x dt

x t t x

(2.17)

NOTE: The limiting case of the integral equation (2.1) with is the integral equation of the first kind as given by

0

0,1c t and

10 , 0 1

tdt f x x

t x

(2.18)

This limiting case gives and the limit of the solution (2.17) is obtained as1/ 2

1/ 2 1/ 2102

1 1, 0 1]

1

f tx tx dt x

x t t x

(2.19)

28

3. Hyper-singular Integral Equation (singular integral equation having a higher order singularly in the integral)

1

21

, 1 1t dt

H f x xt x

(3A)

is considered for its solution for 1, 0, n,1,1 , 1,1 (0 < <1), C (-1,1) C C

The equation (3A) has been solved by Martin [1992] and Chakrabarti and Mandal [1998], under the circumstances when, in the following closed form:

1 0 1

1

122 21 2

1( ) log

1 1 1

x tx f t dt

xt x t

The hypersingular integral Hf appearing in the equation (3A) is understood to beequal to the Hadamard finite part (see Martin [1992]) of this divergent integral, as given by the relation:

1

2 20

1

lim .x

x

t dt t dt x xH

t x t x

(3B)

(3C)

29

A Direct Function Theoretic Method and The detailed analysis

Consider the sectionally analytic function

1

21

, 1,1 .t dt

z zt z

Then if we utilize the following standard limiting values

(3.1)

0

1 1lim , ,y

i x xx iy x

and

2 20

1 1lim , ,y

i x xxx iy

(3.2)

(3.3)

we obtain the following Plemelj – type formulae giving the limiting values of thefunction (z), as z approaches a point on the cut (-1,1) from above and below respectively:

0y 10y

1 1

21 1

1

21

0

, 1 1

t dtx i x i t t x dt

t x

t dti x for x

t x

(3.4)

30

The limiting values (3.4) can also be derived by utilizing the standard Plemelj formulae involving the limiting values of the Cauchy type integral

1

1

ˆ ( ) , 1,1t dt

z for zt z

giving

1

1

ˆ 0t dt

x it x

(3.5)

(3.6)

and by the aid of the relation

ˆ

,d

zdz

along with the understanding that

dH T

dx

(3.7)

(3.8)

Now, the two relations (3.4) can also be viewed as the following two equivalent relations

1

21

2 ,

2 , 1 1.

t dtx x

t x

x x i x for x

(3.9)

31

By utilizing the first of the above two relations (3.9), we now rewrite the givenhypersingular integral equation as

2 , 1 1x x f x x (3.10)

which represents a special Riemann-Hilbert type boundary value problem for the determination of the unknown function (z).

If 0(z) represents a nontrivial solution of the homogeneous problem (3.10), satisfying

0, 1 1x x x (3.11)

then we may rewrite the inhomogeneous problem (3.10) as

0

2( ) ( ) ,

( )

f xx x

x

(3.12)

with0( ) ( ) ( ).z z z (3.13)

Thus, then second of the relations (3.9) suggests that we can determine the function (z) in the following form:

32

1

021

1

2

g t dtz E z

i t z

(3.14)

where

0

2 ( )

( )

dg f xg x

dx x

(3.15)

Next, by utilizing the form (3.14) of the function (z), along with the relation (3.13)and the second of the Plemelj-type formulae (3.9), we obtain the following result:

1

0 0221

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

2

g t dtx x iE x x

t x

(3.16)

If we select 12

0

1( )

1

zz

z

(3.17)

giving 12

0

1( ) 1 1,

1

xx i for x

x

(3.18)

we find that, because of the relations (3.13) and (3.14), we must select E0(z) to be equal to zero.

Then, using the relation (3.18), along with the relation (3.15), we obtain from the relation (3.16), the following result:

33

12 1

022

1

( )1 1

1

g t dtxx

x t x

with

1

2

10

1( )

1

xg x f x

x

(3.19)

(3.20)

Finally, by integrating the relation (3.19), we can determine the solution of the given hypersingular integral equation, in the following form

0( ) ,x p x D

where

1

2 10

2 21

( )1 1

1 ( )

g t dtxp x

x t x

(3.21)

(3.22)

This completes the method of solution f the hypersingular integral equation (3A), in principle, once the hypersingular integral occurring in the relation (3.20) is evaluated, for a given forcing function f(x).

34

We can derive the known form (3C) of the solution of the equation (3A), as obtained by Martin [1992], by using a procedure as described below:

By integrating by parts, we obtain form the relation (3.22), that

12

1 12 2

10 0 0 0

2 2 2 21

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

1 ( ) (1 ) (1 ) 1

g t g g gxp x dt

x t x x x x

Another integration gives, because of the relation (3.20):

12

12

12

1 2

02 21

11

0 021

1 1 (1 )( ) , 4 (1)

1 (1 )

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

1

xp x f t K x t dt g

x x

tg g f t dt Sin x

t

(3.23)

Ignoring an arbitrary constant [see(3.21)], when the following results are used:

1

212

211 1 1 1

1 1 1

tt

t t x x t x t

(3.24)

and 122

2

1 1

1

K t

x x x t

(3.25)

35

Special case:

the solution of the equation (3A), as given by the formulae (3.19) and (3.21) is obtained in the form

1

21

1, ,x f t K x t dt

since we must have

1

21

0 0 0

1

1(1) 0 , ( 1)

1

tg D g f t dt

t

(3.26)

(3.27)

The result (3.26) agrees with the form (3C), involving a weakly singular integral.

The analysis presented above is believed to be self-contained and straightforward.

when (-1) = 0 = (1),

36

4. Problems of Fluid Mechanics (Water Waves)

Mathematical Problem: Determination of the two-dimensional velocity potentials with i2 = -1, in the two-dimensional Cartesian xy coordinates, in the half – plane y > 0, such that

, , 1, 2j x y j

2 2

2 20, , 0j j x y

x y

with

0, 0, 0, tanjjk or y k a known cons t

y

0, 0 , ( , )jj j jon x y L a b

x

(4.1)

(4.2)

(4.3)

0 , 0 , , : 0,j j j jy y for y G L

120 , 0,j

jy t as x y tx

with

+1 2 2 1 - 2a =a, a = 0, b = b, t = a , t = b ,

+1 2 2 1 - 2a =a, a = 0, b = b, t = a , t = b ,

(4.4)

(4.5)

and

37

1 ,

,

ikx kyj j

ikx ky ikx kyj j

R e as x

e R e as x

(4.6)

In which Rj’s are unknown constants to be determined, along with the unknown functions , and j

, 0,j j as y (4.7)

38

The methods of solution

0

0

1 , , 0

, , 0

Ky iKx xj j j

Ky iKx iKx xj j j

R e A L y e d x

e e R e A L y e d x

(4.8)

with j=1,2 and

, sinL y cos y K y

The unknown functions and the unknown constants Rj are determined form the following sets of dual integral equations:

jA

0

0

, ,

, (1 ) , ( 1, 2)

Kyj j j

Kyj j j

A L y d R e for y G

and A L y d iK R e for y L j

(4.9)

39

Existence of method of solutions

By Ursell [1947]

By Williams [1966]

40

(A). Ursell’s Method

The principal idea behind Ursell’s method involves setting

0

( cos sin ) ( ) (1 ) (0, )( 1,2)Kyj j jA y K y d f y iK R e y j

(4.10)

Then we observe that because of the second of the relations (4.9) and that the unknown functions, for are singular at the turning points tj.

0, ,j jf y y L , ,j jf y y G

Utilizing Havelock’s expansion theorem we find that we must have

2 2 2( cos sin ) 1,2

j

j j

G

k A f t t K t dt j

(4.11)

Substituting from relations (4.11) in to the first of the dual relations (4.9),

2 20

2 ( cos sin )( cos sin )

, 1, 2

j

j

G

Kyj j

y K y t K t df t dt

K

R e for y G j

(4.12)

41

The consistency of relation (4.10) demands that we must have

0

0 0

1

cos sin 0

Ky Kyj j

Kyj

f y iK R e e dy

A y K y e dy d

(4.13)

Then 1 2 , ( 1,2)

j

Kyj j

G

R i f y e dy j (4.14)

Using Ursell’s approach, we next operate both sides of equation (4.12), for each j by the operator formally and use the well-known identity d

Kdy

0

sin sin 1ln , 0 ,

2

y t y td t y

y t

(4.15)

to obtain

,

1 1( ) ln 0 1,2j j

Gj

y tf t K dt y G j

y t y t y t

(4.16)

Many researchers, including Ursell (1947), have studied the singular integral equations (4.16). The employment of various methods and solutions of such integral equations have become central in many important and interesting studies involving singular integral equations.

42

Here again, using Ursell’s idea, we first set

1 1

2 2

( ) ( )

( ) ( )

a

y

y

b

F y f t dt

F y f t dt

(4.17)

Then obtain the following further reduced integral equations as given by

2 2

( )0, , 1, 2

j

jj

G

H u duy G j

y u

(4.18)

For the two reduced functions H1(y) and H2 (y) as defined by the relations:

1 1 1

2 2 2

( )

( )

H y KF y f y

and

H y KF y f y

(4.19)

The singular integral equations (4.18) are best solved by using the results available in Muskhelishvilli’s book and we easily deduce that

12

12

11

2 2

22

2 2

CH y

a y

C yH y

y b

(4.21)and

and

43

Then we find that

12

1 12 2

, 0,a Ku

Ky

y

d e duf y C e y a

dy a u

and

12

2 22 2

, ,a Ku

Ky

y

d e duf y C e y b

dy u b

(4.22)

(4.23)

Substituting from relations (4.22) and (4.23) into relations (4.11), after integrating by parts we obtain

1 01 2 2

C J aA

K

2 12 2 2

C bJ bA

K

and(4.24)

where

44

012 2 2

0

112 2 2

0

cos

2

1 1 cos 1 cos

2

a

b

b

uduJ a

a u

and

uudu udu J b

b b a u

(4.25)

where J0 (x) and J1 (x) represent the standard Bessel functions of the first kind.

Then, by using relations (4.22) and (4.23) in relations (4.14) and integrating by parts, we derive that

1 1 0

2 2 1

1

1

R iC I Ka

and

R iC bK bK

(4.26)

45

(B). Williams’s Method

The major deviation in Williams’s method from Ursell’s method lies in rewriting the basic dual integral equations (4.9) in the following alternative forms:

0

0

sin ,

sin 1 , 1,2

Kyj j j

Kyj j j

dK A yd R e y G

dy

and

dK A yd iK R e y L j

dy

(4.27)

Then we must choose the constant Dj and Ej as follows:

11

1

2

22

2

0,

0,

(1 ) E

2

RD

KE

D

i R

(4.28)

and

46

0

sin ,j j jA yd g y for y G

If we now set

(4.29)

when the following identities are utilized

0 01/ 22 2

11/ 22 2

cosh( )

2

Ky

b

KydyI Ka

a y

ye dybK bK

y b

and (4.30)

with representing the standard modified Bessel functions. n nI x and K x

Finally we deduce that

1 1 0

2 2 1

R C K Ka

R bC I Kb

and (4.31)

after using the following identities:

47

0 0 02 2

10 12 2

cos sin, 0

cos sin,

Ky

ky

y K yJ a d K Ka e y a

K

J b y K yd I Kb e y b

K

and (4.32)

We can now easily determine the constants by using relations (4.26) and (4. 28), and we find that

1 2 1 2, ,C C R and R

01

0 0

12

1 1

K KaR

K Ka i I Ka

I KbR

I Kb iK Kb

and (4.33)

which are the most familiar results derived by Ursell [1947].

The full solutions of the two boundary value problems are thus completed when the relations (4.24) are substituted, in conjunction with relations (4.28) and (4.30), into the expressions (8), for the potentials . , , 1, 2j x y j

48

(C). A New Method

In this present approach, we start by rewriting the dual integral equations (4.9) in the alternative forms

0

0 ,

cos sin ,

cos sin1

Kyj j j

j Kyj j

j

A y K y d R e y G

A y K y ddi R e D

dy

y L

and (4C.1)

Operating both sides of the equations by producesdK

dy

0

0

sin 0,

sin , ( 1,2)

j j

jj j

F yd for y G

Fdyd C for y L j

dy

(4C.2)

where

2 2j jF K A (4C.3)

with arbitrary constants, so that for the case j=1 there is no inconsistency as .

1 20,C Cy

49

We set

11/ 22 2

0

21/ 22 2

1, 1,

sin

, 2, 0

ya

j

by

t t dtdj a y

y dy y tF yd

t t dtdj y b

dy t y

(4C.4)

Then we easily derive the following equations for the determination of the two unknown functions and : 1 2

1 11/ 2 1/ 22 2 2 2

0,ua a

a t dtd y uIn du a y

dy y uu a u t

(4C.5)

and

20 21/ 22 2

, 0b bu

t t dtd y uIn du C y b

dy u y ut u

(4C.6)

The above two equations (4C.5) and (4C.6) can easily be reduced to the following two Abel type integral equations

50

11/ 22 2

0,y

t dtfor y a

t y

and

2

0 21/ 22 2, 0y t t dt

C y for y by t

by utilizing the following standard and elementary results:

1/ 22 2 2 20,a

udufor y a

u a y u

1/ 22 2 2 2 1/ 22 2

0

,2

t

for y tudu

for t yu t y u t y

1/ 22 2 2 2 1/ 22 2

0

,2

to

for t ydu

for t yt u y u y y t

and

(4C.8)

(4C.9)

(4C.10)

(4C.7)

The solutions of the two Abel equations (4C.7) and (4C.8) are immediate and we obtain

51

1 1t and

2 2t C (4C.11)

(4C.12)

Then we obtain

1 1 0F J a

2 2 1F C bJ b and

(4C.13)

(4C.14)

after utilizing the standard results that

0 0 02 2

10 12 2

cos sin, 0

cos sin,

kv

kv

y K yJ a d K Ka e y a

K

J b y K yd I Kb e y b

K

01/ 22 2

0 11/ 22 2

sin

2

sin

2

a

b

uduJ a

u a

u uduJ b

b u

and

(4C.15)

(4C.16)

52

We finally obtain

1 01 2 2

2 12 2 2

J aA

K

C bJ bA

K

where relations (4C.3) is utilized.

We thus observe that the principal unknown functions are determined in the same forms as those derived in relations (4.24), by employing Ursell’s method.

1 2A and A

We observe, as expected, that the final values of, and obtained by this new method, agree completely with the ones obtained earlier using Ursell’s method.

1 2 1 2, ,A A R and R

(4C.17)

53

References

1. Chakrabarti, A and Mandal, B. N., “ Derivation of the solution of a simple Hypersingular integral equation”, 1998, Int. J. Math. Edcu. Sci. Technol., 29, 47 – 53.

2. Chakrabarti, A. – “A Survey on Two International Methods used in Scattering of Surface Water Waves”, Advances in Fluid Mechanics, WIT Press, Edited by B. N. Mandal, 1997, pp 232-253.

3. Chakrabarti, A. – “ Solution of the Generalized Abel Integral Equations”, Jl. Int. Eqns and Appl., 2006 (accepted).

4. Chakrabarti, A. – “Solution of a Simple Hyper – Singular Integral Equation”, Jl. Int. Eqns and Appl., 2006 (accepted).

5. Grakhov, F.D.- “On new types of integral equations, soluble ion closed form”, Problems of Continuum Mechanics, pp. 118-132, Published by the Society of Industrial and Applied Mathematics, (SIAM), Philadelphia, Pennsylvania, (1961).

54

References

6. Grakhov, F. D. – “Boundary Value Problems”, Perganon, Oxford, 1966.

7. Grakhov, F.D.-Boundary Value Problems, pp. 531-535, Oxford Press, London, Edinburgh, New York, Paris, Frankfurt (1996).

8. Jones, D. S., 1982, “ The theory of Generalized functions”, Cambridge University Press, Cambridge.

9. Lundgren, T. and Chiang, D.- “Solution of a class of singular integral equations”, Quart. Appl. Math. Vol. 24, No. 4. (1967), pp. 301-313.

10. Mandal, B. N. and Chakrabarti, A. “Water Wave Scattering by Bariers” WIT Press, 2000.

11. Martin, P. A., 1992, Exact solution of a simple hypersigular integral equation, “J. Integral Equation Appic.” 4, 197 – 204.

12. Mikhlin, S. G – “ Integral Equations”, Pergamon, New York, 1964.

55

References

13. Muskhelishvili, N. I. – “Singular Integral equations”, Noordhiff, Graringen, 1953

14. Muskhelishvili N. I., 1977, “Singular Integral Equations”, (Groningen : Noordhoff).

15. Sakalyuk, K.D.- “Abel’s generalized integral equation”, Dokl. Akad. Nauk., SSSR, Vol. 131, No.4. (1960), pp. 748-751.

16. Sneddon, I. N. – “The use of Integral Transforms”, McGraw – Hill, New York, 1972.

17. Ursell, F. – “The Effect of a fixed Vertical Barier on Surface Waves in Deep Water”, Proc. Camb. Phil. Soc., 43, (1947), pp 374 – 382.

18. Williams, W. E. – “A note on scattering of water waves by a vertical barier” Proc. Camb. Phil. Soc., 62, (1966), pp 507 – 509.

56

Acknowledgement

I take this opportunity to thank the

University Grants Commission (UGC), New Delhi, India

for awarding me an Emeritus Fellowship

to carry out the research involving this lecture.

57

Whilst most mathematicians like to enjoy handling mathematical problems for their complete solution, there exists a class of mathematicians who enjoy creating mathematical problems which can not be solved completely by the aid of existing mathematical ideas.

58

THANK YOU

FOR YOUR

ATTENTION