Slide 1 - Mathematics at Dartmouth
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Singular Integral Equations arising in Water
Wave Problems
Aloknath Chakrabarti
Department of Mathematics
Indian Institute of Science
Bangalore-560012, India
Email:[email protected]
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.
2
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.
3
1.Solution of Abel’s Integral equation
and its generalization
Consider here the general form of Abel’s Integral equation
g t dt
x
h x h t
a
f x , 0 1 , a x b
(1.1)
where h(t) is a strictly monotonically increasing and differentiable function
for all t a, b
in (a, b) and h t 0,
Solution of (1.1) by using a very simple method :
Consider
x
I x
a
h u f u du
1
h x h u
(1.2)
By using (1.1), we can express (1.2), after interchanging the orders of
integration, as
4
x
h x dx
I x g t dt
1
a h u h t h x h u
a
x
Using the transformations
(1.3)
h u , 2 h x , 1 h t
(1.4)
we obtain, from (1.3) and (1.2), the following results
h u f u du
g t dt
1
sin a
a h x h u
x
x
(1.5)
giving, on differentiation :
g x
sin
u du
a h x h u 1
x
h u f
which solves the Abel’s integral equation (1.1) completely.
(1.6)
5
Examples of Abel’s integral equation equation (1.1)
t dt
0 x t
x
Example-1.1
Solution:
g
1
1 d
g x
dx
x
Example-1.2:
Solution:
g
0
x,
2
x
a
tdt
x t
t dt
cos x cos t
1
1
2
(1.7)
1
2x 2
f
2
x 1
(1.8)
x
(1.9)
a
g x
1 d
dx
x
a
sin t f t dt
cos t cos x
1
2
(1.10)
6
x
Example 1.3:
a
Solution:
g
x
g ( x)
Here we have chosen
b
Example 1.4 :
x
Solution:
2
t dt
t
2 sin
2
x2
1
2
f
2
x
d
dx
a
h x x
g t dt
t
2
,
f ( x)
f
x
2
t
t dt
t2
1
2
(1.11)
(1.12)
1/ 2
x b
2sin d b tf t dt
g ( x)
dx x t 2 x 2 1
(1.13)
(1.14)
7
A direct method of solution of Abel’s integral equation
x
t dt
x t
f x , 0 1 , x 0 , f (0) 0
(1.15)
0
Writing
, F , K u x , f x , k x eux dx
(1.16)
0
Then the integral equation (1.15) can be expressed as
F (u )
(u )
K (u )
where
K (u) k ( x)e
0
ux
(1.17)
dx x x eux dx
(1.18)
0
If we look at the given equation (1.15) as
(t )k x t dt
f ( x)
(1.19)
0
with k(x) = x-, and if we recall the convolution theorem in the following form
8
x
L t k x t dt u k (u )
0
(1.20)
Now, we easily find that
k u 1 u 1
(1.21)
Then, if we set
x
so that
d
,
dx
(u)
0
Then we obtain,
u
with 0 0
(1.22)
d ux
e dx u u
dx
(1.23)
F u u
1
(1.24)
By utilizing the convolution theorem (1.20) once more, in a clever manner,
we find that (1.24) gives:
x
1
1
x
x
t
f t dt
1 0
(1.25)
9
which, on using the identity
1
gives
sin
x
d sin d f t dt
x
dx
dx 0 x t 1
(1.26)
(1.27)
By using integration by parts we can also rewrite this as (since f(0) = 0)
x
sin
x
x t
1
f t dt
(1.28)
0
We shall next consider the general form of Abel’s integral equation which
is given by the relation
where
a( x)( A1 )( x) b( x)( A2 )( x) f ( x)
x
t dt
A
x
1
x t
x , 0 1
t dt
A2 x
t x
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.
(1.29)
(1.30)
10
Some Important Theorems and Results in Complex Function Theory
Theorem -1. If the function () satisfies the Hölder condition:
1 2 A 1 2
,
0 1
(A1)
where A is a positive constant, for all pairs of points 1 , 2 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
z
dz, where z
2 i z
(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).
11
Theorem -2: (The Basic Lemma)
If the density function () satisfies a Hölder condition, then the formula
1 t
z
d
2 i
z
(A3)
on passing through the point z = t, of the simple closed contour ,
behaves as a continuous function of z, i.e.,
(A4)
lim
z t
z
1
2 i
t
d
t
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 ].
12
Theorem-3 (Plemelj-Sokhotski Formulae) If
( z )
1
2 i
z dz,
(A5)
represents a sectionally analytic function, as in Theorem (*), then and exist, then
the following formulae hold good:
t t t ,
1
t t
d
,
i t
where t
(A6)
where lim
means that the points z approach the point t on from the
left of the positively oriented contour , and lim
means that z
approaches t from the right of .
z t
z t
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 :
13
t d ,
1
1
t t
d
.
2
2 i
t
1
1
t t
2
2 i
(A7)
(Plemelj formulae)
14
Generalized Abel Integral Equation and its Solution
The generalized Abel integral equation
x
a x
t dt
x t
b x
t dt
t x
f x , 0 1 x
(G1)
whereas the forcing term f(x) and the unknown function (x) belong to those
classes of functions which admit representations of the form
f
and
x
x
x
1
x
x
x x x
f
0
(G2)
where possesses a Hölder continuous derivative in and satisfies Hölder’s
condition in ,
15
z
With
so that
1
R z
t dt
,
t z
R z z z
1
z 0 , as
z
z ,
1
2
1
,
(G3)
(G4)
(G5)
and the associated Riemann – Hilbert problem is finally solved by utilizing
the Plemelj – Sokhotski formulae involving Cauchy-type singular
integrals.
16
Particular Example
Integral Equation:
t dt
x t
f x,
x
(G6)
Solution (by Gakhov [1996]):
(G7)
x
sin( ) d g t
x
,
1
dx x t
where
cot
2
1
dt
g ( x) f ( x)
R x f t
2
2
R t t x
(G8)
17
The Detailed Method
z
t dt
t z
As z tends to a point x
z x iy, y o i
2
x t dt t dt
,
0 1
t z
x t z
, from above z x iy , y o and below
(G9)
,
1 ,
Then the sectionally analytic function (z) ) tends to the following limiting values:
x e i A1 x A2 x
Where
A1 x
t dt
t z
,
and
A2 x
(G10)
t dt
t x
(G11)
x
The relation (D2) can also be expressed as
1
x
A1 x
x ,
2i sin
and
1
e i x e i x .
A2 x
2i sin
(G12)
18
By using the relations (G12) in the given integral equation (G1), we obtain
a x e i b x ( x) a x e ib x ( x ) 2i sin( ) f ( x ) x
(G13)
Relation (13) represents the special Riemann-Hilbert type problem
( x) G x ( x) g ( x),
with
and
x
a x e ib x
b x sin
G x
exp 2i arctan
i
a
x
e
b
x
a
x
b
x
cos
g x
2i sin f x
.
i
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:
z O
1
,
z
as z .
(G17)
19
First solve the homogeneous problem (14), satisfying the relation
0 ( x) G( x)0 ( x) 0,
Giving
where
(G18)
0 ( x) 0 ( x) G0 x ,
0 z exp 0 z
and
(G19)
G x exp G0 x
(G20)
Now we can express the function satisfying (19), as :
with
0 (t ) dt
t z
1
0 ( x) 2i sin A11G0 ( x),
0
where
z
(G21)
u
sin d
1
A
G
x
1 0
dx
(G22)
x
G0 t dt
x t
1
(G23)
Next, by utilizing (19) in (14), we obtain
g x
( x)
( x)
,
0
( x)
0
( x)
0
( x)
where
(G24)
20
0 ( x) exp 0 x
(G25)
with 0 x being obtainable by suing the relations (21)-(23).
Then, by utilizing the first of the formulae (G12), we can determine the
solution of the Riemann-Hilbert type problem (G24), as given by :
( z )
0 ( x)
where
(t ) dt
t z ,
g t dt
1
d
( x)
1
2 i dx
t
x
t
0
(G26)
.
(G27)
The relation (27) takes the equivalent form:
x
p t dt
1 p
( x)
,
1
1
2 i x
(t x)
with
g (t )
dp
p(t ) , p(t ) ,
0 (t )
dt
(G28)
(G29)
21
Next we obtain the following limiting values of the function (z),
as z approaches the point x , : [see (G10)]:
giving
where
( x) 0 x e i A1 x A2 x
(G30)
( x) ( x) h x (say),
(G31)
h( x) e i 0 ( x) e i 0 ( x) A1 x 0 ( x) 0 ( x) A2 x ,
(G32)
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
x
1 d
h(t )dt
x
.
1
2 i dx x t
(G33)
The result (G33) can also be expressed in the equivalent form:
1 h a
x h t dt
x
,
1
1
2 i x
x t
(G34)
22
2. Solution of Singular Integral Equations
of the Cauchy type
The general theory of a single linear singular integral equation of the type
c t t
dr f t , t ,
t
b
a
(2.1)
Defining a sectionally analytic function
t
1
z
dt , z ,
2 i
tz
(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 t
t
t
,
c t i
c t i
t ,
(2.3)
provided c i
23
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
p i
0 x
p i
0 x , x 0,1
(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:
x x
f x
, x 0,1
0 x 0 x p i 0 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 1
f t
z 0 z
0
dt E z
2 i 0 t p i t z
where E(z) is an arbitrary entire function of z.
(2.7)
24
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 an
arbitrary 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 0 z representing the solution of the homogeneous
problem (2.5).
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
z
0 z
,
z 1
i 2 i
e
,
i
so that we have
1
0 .
2
(2.9)
(2.10)
(2.11)
25
Then, observing that [by fixing the idea that 0 arg( z ) 2 ]
x i
(i) 0 x
e ,
1 x
x 3i
0 x
e , 0 x 1
1 x
(ii)
lim 0 z 1
z
(2.12)
(2.13)
as well as the fact that
lim z 0
z
(2.14)
We find that we must select
E z 0
giving
0 z 1
f t
z
0
dt , z
2 i
i 0 t t z
(2.15)
(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:
26
x x x
f x 0 x 0 x
0 x 0 x 1
f t
0
dt
2 i i
0 t t x
2 i 0 x
1
2
2
f
x 1 1 t
x
0
1
x
t
f t
dt
t
x
(2.17)
NOTE: The limiting case 0 of the integral equation (2.1) with
c t and 0,1 is the integral equation of the first kind as given by
1
0
t
tx
x ,
dt f
0 x 1
(2.18)
This limiting case gives 1/ 2 and the limit of the solution (2.17) is obtained as
1/ 2
1 x
x 2
1 x
1 t
t
1/ 2
1
0
f t
dt , 0 x 1]
tx
(2.19)
27
3. Hyper-singular Integral Equation
(singular integral equation having a higher order singularly in the integral)
H
1
t dt
t x
2
f
x , 1 x 1
(3A)
1
is considered for its solution for C1, 1,1 , C0, 1,1 (0 < <1), Cn, (-1,1)
The hypersingular integral Hf appearing in the equation (3A) is understood to be
equal to the Hadamard finite part (see Martin [1992]) of this divergent integral,
as given by the relation:
(3B)
1
x
t dt
t dt x x
H lim
.
2
2
0
1 t x
x t x
The equation (3A) has been solved by Martin [1992] and Chakrabarti and
Mandal [1998], under the circumstances when, 1 0 1 in the
following closed form:
x
1
2
1
1
x t
f (t ) log
1
1 xt 1 x 2 1 t 2 2
dt
(3C)
28
A Direct Function Theoretic Method
and The detailed analysis
Consider the sectionally analytic function
t dt
1
z
t z
2
z 1,1 .
,
(3.1)
1
Then if we utilize the following standard limiting values
1
1
lim
i x , x ,
y 0 x iy
x
and
1
1
lim
y 0
x iy
2
i x
x
2
, x ,
(3.2)
(3.3)
we obtain the following Plemelj – type formulae giving the limiting values of the
function (z), as z approaches a point on the cut (-1,1) from above y 0
1
and below y 0 respectively:
1
1
t dt
x i 0 x i t t x dt
2
1
1 t x
(3.4)
1
t dt
i x
,
for 1 x 1
29
2
1
t x
The limiting values (3.4) can also be derived by utilizing the standard
Plemelj formulae involving the limiting values of the Cauchy type integral
ˆ ( z)
giving
1
t dt
tz
1
, for z 1,1
ˆ x i 0
1
1
t dt
tx
(3.5)
(3.6)
and by the aid of the relation
ˆ
d
z
,
dz
(3.7)
along with the understanding that
H
d
T
dx
(3.8)
Now, the two relations (3.4) can also be viewed as the following two
equivalent relations
1
x x 2
t dt
t x
x x 2 i x ,
2
,
(3.9)
1
for 1 x 1.
30
By utilizing the first of the above two relations (3.9), we now rewrite the given
hypersingular integral equation as
x x 2 f x ,
1 x 1
(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
x x 0,
1 x 1
(3.11)
then we may rewrite the inhomogeneous problem (3.10) as
2 f x
( x) ( x)
,
0 ( x)
(3.12)
( z) ( z) 0 ( z).
(3.13)
with
Thus, then second of the relations (3.9) suggests that we can determine
the function (z) in the following form:
31
1
z
2 i
where
g t dt
1
t z
2
E0 z
(3.14)
1
dg
2 f ( x)
g x
dx
0 ( 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 g (t )dt
( x)
( x)
2 iE0 ( x)
2
2 2
1 t x
1
If we select
2
1
0
1 x 1 .
z 1
0 ( z )
z 1
giving
(3.16)
(3.17)
1
2
1 x
0 ( x) i
1 x
for 1 x 1,
(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:
32
1 1 x
x 2
1 x
1
2
1
g 0 (t )dt
t x
2
(3.19)
1
with
1
2
1 x
1
(3.20)
g 0 ( x) f x
1 x
Finally, by integrating the relation (3.19), we can determine the solution
of the given hypersingular integral equation, in the following form
x p( x) D0 ,
where
1 1 x
p x 2
1 x
1
2
(3.21)
1
g 0 (t )dt
1 (t x)2
(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).
33
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
1
1
2 g 0 (1)
1 1 x 2 g0 (t )
1 g 0 (1) g 0 (1)
p( x) 2
dt 2
1
2 12
2 2
1 x 1 (t x)
(1 x )
(1
x
)
1
x
Another integration gives, because of the relation (3.20):
p( x)
1 (1 x 2 ) 1 2
f t K x, t dt 4 g 0 (1)
2 1
1 x (1 x ) 2
1
1 t
g
(
1)
g
(1)
0
0
1 1 t
2
1
1
2
1
1
1
2
f (t )dt Sin 1 ( x ),
(3.23)
Ignoring an arbitrary constant [see(3.21)], when the following results are used:
1 t 1 1 t
1 x
1 t t x
1
and
2
2
1 t
K
2
x
1
x
2
1
2
1
2
1
x
t
1
1
t x 1 t
(3.24)
(3.25)
34
when (-1) = 0 = (1),
Special case:
the solution of the equation (3A), as given by the formulae (3.19) and
(3.21) is obtained in the form
x
1
2
1
f t K x, t dt ,
(3.26)
1
since we must have
1
1 t
g0 (1) 0 D0 , g 0 (1)
f t dt
1 t
1
1
2
(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.
35
4. Problems of Fluid Mechanics (Water Waves)
Mathematical Problem: Determination of the two-dimensional velocity
potentials j x, y , j 1,2 with i2 = -1, in the two-dimensional Cartesian
xy coordinates, in the half – plane y > 0, such that
2 j
x
with
2
j
y
j
x
2 j
y
2
0, x , y 0
k j 0, or
0,
on
y 0,
0 y t j
x
with
and
k 0, a known cons tan t
a1 =a, a 2 = 0, b2 = b, t1 = a- , t 2 = b ,
x 0 ,
1
(4.2)
+
j 0 , y j 0 , y ,
j
(4.1)
2
y L j (a j , b j )
for
,
as
y G j : 0, L j
x 0, y t
(4.3)
(4.4)
(4.5)
a1 =a, a 2 = 0, b2 = b, t1 = a- , t 2 = b+ ,
36
j
1 R e
j
eikx ky R j eikx ky ,
ikx ky
j
,
x
as
as
(4.6)
x
In which Rj’s are unknown constants to be determined, along with the
unknown functions j , and
j,
j 0, as
y
(4.7)
37
The methods of solution
j 1 R j e
Ky iKx
Aj L , y e x d ,
x0
0
(4.8)
j e Ky eiKx R j e iKx A j L , y e x d ,
x0
0
with j=1,2 and
L , y cos y K sin y
The unknown functions Aj and the unknown constants Rj are determined
form the following sets of dual integral equations:
Ky
A
L
,
y
d
R
e
, for y G j
j
j
0
(4.9)
and Aj L , y d iK (1 R j )e Ky , for y L j ( j 1, 2)
0
38
Existence of method of solutions
By Ursell [1947]
By Williams [1966]
39
(A). Ursell’s Method
The principal idea behind Ursell’s method involves setting
A ( cos y K sin y)d f ( y) iK (1 R )e
j
j
j
Ky
y (0, )( j 1, 2)
(4.10)
0
Then we observe that f j y 0, y Lj , because of the second of the
relations (4.9) and that the unknown functions, f j y , y G j , for are
singular at the turning points tj.
Utilizing Havelock’s expansion theorem we find that we must have
2
k 2 Aj
2
f t ( cos t K sin t )dt
j
j 1, 2
(4.11)
Gj
Substituting from relations (4.11) in to the first of the dual relations (4.9),
2 ( cos y K sin y )( cos t K sin t ) d
G f j t 0
2
2
K
j
R j e Ky for y G j , j 1, 2
dt (4.12)
40
The consistency of relation (4.10) demands that we must have
f y iK 1 R e
j
Ky
j
e
Ky
dy
(4.13)
0
Aj cos y K sin y e Ky dy d 0
0
0
Then
R j 1 2i f j y e Ky dy,
( j 1, 2)
(4.14)
Gj
Using Ursell’s approach, we next operate both sides of equation (4.12), for
each j by the operator d K formally and use the well-known identity
dy
sin y sin t
0
to obtain
Gj
d
1
y t
ln
,
2
y t
0 t, y
yt
1
1
f j (t ) K ln
dt 0
y t y t y t
y G
j,
j 1, 2
(4.15)
(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.
41
Here again, using Ursell’s idea, we first set
a
F1 ( y )
f1 (t ) dt
(4.17)
y
and
y
F2 ( y )
f 2 (t ) dt
b
Then obtain the following further reduced integral equations as given by
H j (u ) du
y2 u2
Gj
0,
j 1, 2
y Gj ,
(4.18)
For the two reduced functions H1(y) and H2 (y) as defined by the relations:
H1 y KF1 ( y ) f1 y
(4.19)
and
H2
y
KF2 ( y ) f 2
y
The singular integral equations (4.18) are best solved by using the results
available in Muskhelishvilli’s book and we easily deduce that
H1
y
and
H2
y
a
C1
2
y
C2 y
y
2
b
2
2
1
2
1
2
(4.21)
42
Then we find that
and
a
d Ky
e Ku du
f1 y C1
e
y a 2 u 2 12
dy
,
y 0, a
a
d Ky
e Ku du
f 2 y C2
e
y u 2 b 2 12
dy
,
y b,
(4.22)
(4.23)
Substituting from relations (4.22) and (4.23) into relations (4.11), after
integrating by parts we obtain
A1
and
C1 J 0 a
2 K2
(4.24)
C2bJ1 b
A2
2 K2
where
43
a
0
cos udu
a
2
u
2
1
2
2
J 0 a
and
1
b
1
u
cos
udu
0
b b a 2 u 2 1 2
b
1 cos udu
J1 b
2
(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
and
R1 1 iC1 I 0 Ka
R2 1 iC2bK1 bK
(4.26)
44
(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:
d
K Aj sin yd R j e Ky ,
dy
0
y Gj
(4.27)
and
d
K Aj sin yd iK 1 R j e Ky ,
dy
0
y L j j 1, 2
Then we must choose the constant Dj and Ej as follows:
0,
i (1 R2 )
2
R1
2K
E1 0,
D1
and
D2
E2
(4.28)
45
If we now set
A sin yd g y ,
j
j
for
y Gj
(4.29)
0
when the following identities are utilized
2
2 1/ 2
2
a y
and
(4.30)
Ky
ye
dy
b
bK1 bK
1/
2
y 2 b2
with In x and Kn x representing the standard modified Bessel functions.
0
cosh Kydy
I 0 ( Ka )
Finally we deduce that
and
R1 C1 K 0 Ka
R2 bC2 I1 Kb
(4.31)
after using the following identities:
46
and
cos y K sin y J a d
0
2
2
K 0 Ka e Ky , 0 y a
K
J
b
cos
y
K
sin
y
ky
0 1
d
I
Kb
e
,
y
b
1
2 K2
0
(4.32)
We can now easily determine the constants C1 , C2 , R1 and R2 by using
relations (4.26) and (4. 28), and we find that
and
R2
K 0 Ka
K 0 Ka i I 0
Ka
I1 Kb
I1 Kb iK1 Kb
R1
(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 j x, y , j 1,2 .
47
(C). A New Method
In this present approach, we start by rewriting the dual integral equations
(4.9) in the alternative forms
0 Aj cos y K sin y d R j e Ky , y G j
d Aj cos y K sin y d
and
0
i 1 R j e Ky D j ,
dy
y Lj
d
Operating both sides of the equations by
K produces
dy
0 Fj sin yd 0,
d
0
dy
where
Fj
(4C.1)
( j 1, 2)
for y G j
sin yd C j , for y L j
F j 2 K 2 A j
with C1 0, C2 arbitrary constants, so that for the case j=1 there is
no inconsistency as y .
(4C.2)
(4C.3)
48
We set
0 Fj sin yd
d b
dy y
1 d y t 1 t dt
a
,
y dy y 2 t 2 1/ 2
t 2 t dt
t
2
y
2 1/ 2
, j 2,
j 1, a y
0 y b
(4C.4)
Then we easily derive the following equations for the determination of the
two unknown functions 1 and 2 :
1 a
1 t dt
d
u
a
a
2
2 1/ 2
2
2 1/ 2
dy
u
a
u
t
and
y u
In
du 0, a y
y u
d b b t 2 t dt
y u
0
u
In
du C2 ,
dy u t 2 u 2 1/ 2
y u
0 y b
(4C.5)
(4C.6)
The above two equations (4C.5) and (4C.6) can easily be reduced to the
following two Abel type integral equations
49
y
t
and
y
0
1 t dt
2
y
t 2 t dt
y
2
t
0,
1/ 2
2
for y a
C2 y,
2 1/ 2
for 0 y b
(4C.7)
(4C.8)
by utilizing the following standard and elementary results:
a
t
u
u
udu
2
a
y
1/ 2
2
udu
2
t
y
2 1/ 2
2
2
u2
and
t
o
t 2 u 2
du
1/ 2
y
2
u
u2
2
0,
for y a
(4C.9)
for y t
0
, for t y
2 t 2 y 2 1/ 2
0
,
2 y y 2 t 2 1/ 2
for t y
(4C.10)
for t y
The solutions of the two Abel equations (4C.7) and (4C.8) are immediate
and we obtain
50
1 t 1
2 t C2
and
(4C.11)
(4C.12)
Then we obtain
and
F1 1J0 a
(4C.13)
F2 C2bJ1 b
(4C.14)
after utilizing the standard results that
cos y K sin y J a d K Ka e kv ,
0
0
2
2
0 y a
K
J1 b cos y K sin y
kv
0
d
I
Kb
e
,
y
b
1
2 K2
0
and
sin udu
J
a
0
2
2 1/ 2
2
u a
u
sin
udu
b0
J1 b
1/
2
b2 u 2 2
a
(4C.15)
(4C.16)
51
We finally obtain
1 J 0 a
A1 2
K2
C2bJ1 b
A2
2 K2
(4C.17)
where relations (4C.3) is utilized.
We thus observe that the principal unknown functions A1 and A2 are
determined in the same forms as those derived in relations (4.24), by
employing Ursell’s method.
We observe, as expected, that the final values of, A1 , A2 , R1 and R2
and obtained by this new method, agree completely with the ones obtained
earlier using Ursell’s method.
52
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).
53
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.
54
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.
55
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.
56
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.
57
THANK YOU
FOR YOUR
ATTENTION
58