Dual Integral Equations

7/29/2019 Dual Integral Equations

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IJMMS 2003:17, 10931100PII. S0161171203007488

http://ijmms.hindawi.com Hindawi Publishing Corp.

DUAL INTEGRAL EQUATIONSREVISITED

SUDESHNA BANERJEA and C. C. KAR

Received 30 April 2001

Dual integral equations with trigonometric kernel are reinvestigated here for a

solution. The behaviour of one of the integrals at the end points of the interval

complementary to the one in which it is defined plays the key role in determin-

ing the solution of the dual integral equations. The solution of the dual integralequations is then applied to find an exact solution of the water wave scattering

problems.

2000 Mathematics Subject Classification: 45F10.

1. Introduction. Boundary value problems with mixed boundary conditions

arising in different branches of mathematical physics can be reduced to dual

integral equations. A mixed boundary condition is the one in which one con-

dition is prescribed at one part of the boundary while some other condition is

prescribed at the remaining part of the boundary. The solution of the dual in-tegral equations essentially depends on the behaviour of one of the integrals at

the end points of the interval complementary to the one in which it is defined

[1, 4]. This behaviour is dictated by the physics of the problem.

In the present paper, we consider the following dual integral equations:

0Aj (k)L(k,y)dk =Rj exp(Ky), y Gj ,

0kAj (k)L(k,y)dk = iK

1Rj

exp(Ky ), y Bj ,

(1.1)

where

L(k,y)= k cos kyKsinky,

Gj = (0,)Bj ,(1.2)

A(k) is an unknown function, and R is an unknown constant. This integral

equation arises in the well-known problem of scattering water waves by a ver-

tical barrier under the assumption of linearised theory [5, 6, 7, 8]. The vertical

barrier may be (i) partially immersed in deep water, (ii) completely submerged

and extending infinitely downwards in deep water, (iii) a vertical wall with a

gap, or (iv) a submerged plate. The solution of (1.1) has been obtained here by

noting the behaviour of the second equation of (1.1) at the end points of the

interval Gj , which can be determined from physical consideration. Equation
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1094 S. BANERJEA AND C. C. KAR

(1.1) was then reduced to a singular integral equation whose kernel involves

Cauchy and logarithmic type singularity. The solution of this singular integral

equation is known (cf. [3, 4, 6, 8]). The solution of (1.1) was then obtained

by utilizing the solution of aforesaid singular integral equation. Knowing the

solution of (1.1), the solution of the corresponding scattering problems was

obtained in a closed form. In Section 2, we consider the genesis of dual inte-

gral equation (1.1), and in Section 3, we find the solution of (1.1) and hence the

solution of the corresponding scattering problems.

2. Genesis of the dual integral equations. The two-dimensional problem of

the scattering of surface waves by a vertical barrier present in deep water under

the assumption of linearised theory consists in solving mixed two-dimensional

boundary value problem given as follows: j satisfies

2j = 0 in < x < , y 0, (2.1)

the free surface condition

Kj+jy= 0 on y= 0, K=2

g, a constant, (2.2)

the condition on the barrier,

j

x= 0 on x = 0 y Bj , j = 1, 2, 3, 4. (2.3)

Here, Bj represents the vertical barrier. (i) For j = 1, the barrier is partially

immersed to a depth a1 below the mean free surface y= 0 so that B1 = (0, a1).

(ii) For j = 2, the vertical barrier is completely submerged and extends infinitely

downwards, so B2 = (a2,). (iii) For j = 3, the vertical barrier is in the form

of a wall with a gap, so B3 = (0, a3)+ (a4,). (iv) For j = 4, the barrier is in

the form of a plate submerged in deep water, so B4 = (a5, a6). The bottom

condition is given by

j 0 as y . (2.4)

At the sharp edges of the barrier, we must have

r1/2j bounded as r 0, (2.5)

where r denotes the distance from sharp edges aj of the barrier, j = 1, . . . , 6

j

Rj expKyiKx

+exp(Ky+iKx) as x ,

Tj exp(Ky+iKx) as x ,(2.6)


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DUAL INTEGRAL EQUATIONSREVISITED 1095

where Tj , Rj are unknown complex constant. The function j , j = 1, 2, 3, 4, rep-

resents the velocity potential for two-dimensional irrotational motion corre-

sponding to various scattering problems. The function exp(Ky+iKx) (drop-

ping the time dependent factor exp(it) where is the circular frequency

K= 2/g, g being acceleration due to gravity) represents the wave propagat-

ing from the negative x-direction incident upon the barrier Bj . The complex

constants Rj and Tj are the reflection and transmission coefficients, respec-

tively.

By Havelock expansion of water wave potential, a suitable representation of

j satisfying (2.1), (2.2), (2.4), and (2.6) is

j =

Rj exp(KyiKx)+exp(Ky+iKx)

+

0Bj (k)L(k,y)exp(kx)dk, x < 0,

Tj exp(Ky+iKx)+

0Aj (k)L(k,y)exp(kx)dk, x > 0,

(2.7)

where (cf. [8])

Tj+Rj = 1, Aj (k)=Bj (k). (2.8)

By condition (2.3), using (2.7) we have

0kAj (k)L(k,y)dk = iK

1Rj

exp(ky ), y Bj . (2.9)

Also, j is continuous across the gap Gj below/above/between the barrier so

that

j (+0, y)=j (0, y), y Gj . (2.10)

Using (2.7), we have

0Aj (k)L(k,y)dk = Rj exp(ky ), y Gj . (2.11)

Here, G1 = (a1,), G2 = (0, a2), G3 = (a3, a4), and G4 = (0, a5)+(a6,). Equa-

tions (2.9) and (2.11) give the required integral equations. In the following

section, we determine the solution of (1.1).

3. The solution of (1.1). Let

iK

1Rj

exp(Ky )

0kAj (k)L(k,y)dk =

0, y Bj ,

hj (y ), y Gj ,(3.1)


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where hj (y) is the unknown function. In view of(2.9), (2.3), and (2.4),

h1(y)

O

ya11/2

as y a1,

0 as y ,(3.2)

h2(y)

Oya21/2

as y a2,

bounded as y 0,(3.3)

h3(y)

Oyai1/2

as y ai, i= 3, 4, (3.4)

h4(y)

O

yai1/2

as y ai, i= 5, 6,

0 as y

,

bounded as y 0.

(3.5)

By Havelocks expansion theorem [8], we have from (3.1)

i

1Rj= 2

Gj

hj (t) exp(Kt)dt, (3.6)

kAj (k)=2

1

K2+k2

Gj

hj (t)L(k,t)dt. (3.7)

Substituting Aj (k) from (3.7) into (2.11), we have

2

Gj

hj (t)

0

L(k,t)L(k,y)

k

K2+k2 dkdt = Rj exp(Ky), y Gj . (3.8)

Simplifying (3.8) and applying (d/dy+K), we have

Gj

hj (t)

Kln

yty+t+ 1y+t +

1

yt

dt = 0, yGj . (3.9)

This is a singular integral equation in hj (t), whose kernel involves a combi-

nation of Cauchy type and logarithmic singularity. An appropriate solution of

(3.9) can be obtained by considering the behaviour of hj (t) at the end points

ofGj , which is given in (3.2), (3.3), (3.4), and (3.5) for various configurations of

the barrier. Hence (3.6) and (3.7) show that the behaviour of hj (t) at the end

points of Gj plays the key role in determining the solution of (1.1).

Now, considering (3.2), (3.3), (3.4), and (3.5), we find hj (t) for j = 1, 2, 3, 4

and hence Aj (k) and Rj for j = 1, 2, 3, 4.


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(1) Knowing (3.2), h1(t) is given by (cf. [8])

h1

(t) = C1

d

dyexp(ky)

y

a

t exp(Kt)t2a2

1/2

dt, yG1

, (3.10)

where C1 is a constant. Substituting h1(t) in (3.6) and (3.7), we have

A1(k)=a1C1

K2+k2J1(ka), R1 = 1+ia1C1K1(Ka). (3.11)

To find C1, A1(k) and R1 are substituted in the first equation of (1.1) to get

C1 =1

a11 , 1 =I1

Ka1

iK1

Ka1

. (3.12)

So that

A1(k)=J1

ka1

1

K2+k2

, R = I1

ka1

1. (3.13)

(2) For j = 2,

h2(y)= C2 ddy

exp(ky)y

bexp(Kv)

b2v21/2 dv (cf. [6]), (3.14)

where C2 is a constant. Substituting in (3.6) and (3.7)

A2(k)=C2

K2+k2J0

ka2

, R2 = 1+iC2I0

Ka2

. (3.15)

The constant C2 is determined by substituting A2(k), R2 in first equation of

(1.1). On simplification, this gives

C2 =1

K0

Ka2+iI0

Ka2

. (3.16)

(3) For j = 3 (cf. [3]),

h3(y)=d

dyexp(Ky )

ya4

C3 exp(Ku)(u)du, (3.17)

where

(u)=u

R(u)

2

F1

a3, a4, u

, (3.18)


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C3 is a constant,

F1a3, a4, u= a3

0

R(v)

v2

u2

dv,

R(u)=a23u2

1/2a24u21/2,

=K1 exp(Ka)+(2/)2

K, F1

2(K)

,

i(K) =i(K, 1), i

K, F1=

ti

uF1

a3, a4, u

R(u)du,

ti =

a3, a3

, i= 1,

a3, a4

, i= 2,

a4,

, i= 3,

(3.19)

and hence (3.6) and (3.7) give

A3(k)=2

C3

k

K2+k2sinka+k

a4a3

(u) cos kudu

,

R3 = C3I,

I=

1(K)3(K)

2

1

K, F13

K, F1.

(3.20)

To find C3, substitute A3(k) and R3 in the first equation of (1.1) to get

C3 =i

J+iI, (3.21)

where

J=K1 exp(ka)+2(K)2

K, F1

. (3.22)

(4) For j = 4 (cf. [2]),

h4(y)=

d

dy

exp(Ky )

ya5

exp(Ku)P(u)du

, y < a5,

d

dy

exp(Ky )

ya6

exp(Ku)P(u)du

, y < a6,

(3.23)

where

P(u)=C4

R0(u)

d20u

2

, (3.24)

C4 and d20 are constants,

R0(u)=u2a251/2u2a261/2, (3.25)


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and (3.6) and (3.7) give

A4(k)=J(k)

K2

+k2

C4, J (k)= b

a

d20u2

R0(u)

sinkudu, (3.26)

R4 = 1iC4

00

. (3.27)

To determine C4 and d20, we substitute A4(k) in the first equation of (1.1) to

get the relations

R4 = C40, (3.28)

R4 = C4

0

a6a5

d20x

2

R0(x)exp(Kx)dx

, (3.29)

which yield

a6a5

d20x

2

R0(x)exp(Kx)dx = 0. (3.30)

This determines d20. Equating (3.26) and (3.28), we have

C4 =i

4, 4 =00i0, (3.31)

where

0 =

a5a5

d20x

2

R0(x)exp(Kx)dx,

0 =

a6

d20x

2

R0(x)exp(Kx)dx,

0 =

a6a5

d20x

2

R0(x)exp(Kx)dx.

(3.32)

Thus, knowing Aj (k) and Rj , the corresponding j (x,y) for j = 1, 2, 3, 4 are

known from (2.7).

Acknowledgment. This work was supported by the National Board of

Higher Mathematics (NBHM) through research project No. 48/3/99-R&DII/611,

given to S. Banerjea.

References

[1] S. Banerjea and C. C. Kar, A note on some dual integral equations, ZAMM Z. Angew.

Math. Mech. 80 (2000), no. 3, 205210.

[2] S. Banerjea and B. N. Mandal, Solution of a singular integral equation in a double

interval arising in the theory of water waves, Appl. Math. Lett. 6 (1993), no. 3,8184.

[3] , On a singular integral equation with logarithmic and Cauchy kernel, Int.

J. Math. Educ. Sci. Technol. 43 (1995), 267313.


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[4] A. Chakrabarti and N. Mandal, Solutions of some dual integral equations, ZAMM Z.

Angew. Math. Mech. 78 (1998), no. 2, 141144.

[5] D. V. Evans, Diffraction of water waves by a submerged vertical plate, J. Fluid Mech.

40 (1970), 433451.[6] B. N Mandal and P. K. Kundu, Scattering of water waves by a vertical barrier and

associated mathematical methods, Proc. Indian Nat. Sci. Acad. Part A 53

(1987), 514530.

[7] D. Porter, The transmission of waves through a gap in a vertical barrier, Proc.

Cambridge Philos. Soc. 71 (1972), 411421.

[8] F. Ursell, The effect of a fixed vertical barrier on surface waves in deep water, Proc.

Cambridge Philos. Soc. 43 (1947), 374382.

Sudeshna Banerjea: Department of Applied Mathematics, University of Calcutta, 92

APC Road, Calcutta-700009, India

E-mail address: [email protected]

C. C. Kar: Mathurapur High School, Mathurapur, 24 Parganas, West Bengal, India
mailto:[email protected]:[email protected]


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Dual Integral Equations