Int. J. Contemp. Math. Sciences, Vol. 9, 2014, no. 9, 431 - 439
HIKARI Ltd, www.m-hikari.com
http://dx.doi.org/10.12988/ijcms.2014.4434
On Hamerstein-Volterra Integral Equation
R. O. Abd El Rahman
Department of Mathematics, Faculty of Science
Damanhur University, Egypt
Copyright © 2014 R. O. Abd El Rahman. This is an open access article distributed under the Creative
Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any
medium, provided the original work is properly cited.
Abstract
A numerical method is used to transform the Hamerstein–Volterra integral equation into
a system of Hamerstein integral equations. Then a classical method of the degenerate
kernel method is applied to solve the system of Hamerstein equations. Also, numerical
examples are solved.
Keywords: Integral equation, Hamerstein -Volterra (H-V), degenerate method
Introduction
Many problems of mathematical physics, contact problems in the theory of elasticity
and mixed problems of mechanics of continuous media are reduced to mixed type of
integral equations, (see [1-4]). Many different methods are used to solve the integral
equations analytically, see [5,6,7]. Also, for numerical methods, we refer to [8,9]. This
paper is concerned with the problem of finding numerical solution of the Hamerstein Volterra equation
1
t
( x, t ) k ( x , y ) (y, (y, t)) dy - F (t , ) ( x , ) d f ( x , t )
0
0
(1.1)
432
R. O. Abd El Rahman
where the kernel k ( x , y ) of Hamerstein integral term is considered in space and known
continuous function in L2 [ 0,1] while the kernel of Volterra term is considered in time
and known continuous functions in C [ 0, T ] .The known continuous function f ( x , t ) and
( y , ( y , t )) belong to the space L2 [0,1] C[0, T ]. While ( x, t ) is the unknown
function. In order to guarantee the existence of a unique solution we assume through
this work the following conditions.
(i) The kernel k ( x , y ) C ( [ 0,1] [ 0,1] ) , x , y [ 0,1] satisfies
k ( x , y ) M , M is a constant.
(ii) The positive continuous kernel F (t , ) C [0, T ] satisfies
F (t , ) M 2 , M 2 is a constant.
(iii) The given function f ( x , t ) with its partial derivatives are continuous in the space
L2 [ 0,1] C[ 0, T ] and its norm is defined as
t 1 2
1/ 2
f max { f ( x , ) dx}
d
0 T 0 0
(iv) The given continuous function ( x , ( x , )) satisfies the Lipshitz condition with
respect to the unknown argument
2.
( x, )
and it's norm is defined as
t 1 2
1/ 2
max { ( y , ( y , ) ) dy}
d .
0 T 0 0
Numerical Method
To represent the H-VIE of Eq. (1.1) to a system of HE. we divide the interval
[ 0, T ], 0 t T as 0 t 0 t1 ... t N , where t t n , n 0, 1, 2, ... N . Then
using the quadrature formula w j , j 0,1,...n, (see [8,9] ) we have
1
( x, t n ) k ( x, y ) ( y, ( y, t n )) dy )
0
n
p 1
w j F (tn , t j ) ( x, t j 0( n ) f ( x , t n )
j 0
(2.1)
Hamerstein-Volterra integral equation
where
n max h j ,
0 j n
433
h j t j 1 t j
The values of the weight formula w j and the constant p depend of the number of
3
derivatives of F (t , ) for all [ 0, T ] , for example, if F (t , ) C [ 0, T ], then we
have
p 3,
pn
and w0
h0
2
,
w3
h3
,
2
wj hj ,
j 1, 2 .
More information for the characteristic points and the quadrature coefficients are found
in [8,9]. Using the following notations
(2.2)
( x , t n ) n ( x ),
F ( t n , t j ) Fn , j ,
f ( x ,t k ) f k ( x ) ,
we can rewrite, after neglecting the error, the fomula (2.1) in the form
1
n 1
0
j 0
n n ( x ) k ( x , y ) ( y , n ( y ) ) dy f n ( x ) w j Fn, j j ( x )
( n 1 wn Fnn )
(2.3)
The integral equation (2.3) represents a system of Hamerstein integral equation of the
the first kind, when n 0 and for the second kind, for all values of n , wn Fnn 1 .
The solution of the system (2.3), when n 0 can be obtained using different
methods. In [10], a new collocation method is used to solve (2.3),when n 0 .
Also a variation of Nystrom method is presented in [11] to obtain the Hamerstein
integral equation of the second kind (i.e when n 0 )
3. Degenerate kernel method, see[6]
In this section we will apply the degenerate kernel method for this, assume
k l ( x , y ) is an approximation of the kernel k ( x , y ), that satisfies the condition
11
2
1/ 2
{ k ( x , y ) kl ( x , y ) dx dy}
0
as l
00
Also, assume
l
k l ( x , y ) Bi ( x ) Ci ( y )
i 1
where the set of the function Bi (x ) is assumed to be linear.
(3.1)
(3.2)
434
R. O. Abd El Rahman
It is natural to expect that the solution of the following equation associated with the
degenerate kernels k n ( x , t ) converges to the exact solution (1.1)
1
nl 1
0
j 0
n n ( x ) k l ( x, y ) ( y , n ( y )) dy f n ( x ) w j Fn , j n ( x )
l
l
l
l
(3.3)
Using (3.2) in (2.5), we get
l
n 1
i 1
j 0
n n ( x ) a n Bn ( x ) f n ( x ) w j Fn, j j ( x )
i i
l
l
(3.4)
where
1
a n Ci ( y ) ( y, n ( y )) dy .
i 0
l
Here an ' s represents the values of the constants that will be determined.
j
(3.5)
Using (3.3) in (3.5), we get
1
l
n1
1
an Cn ( y) ( y, ( f n ( y) an Bn ( y) w j Fn, j j ( y) ) dy
i 0 i
n
m1 m m
j0
l
(3.6)
Define the function
1
1
H n (a n , a n ,...a n ) C n ( y ) ( y,
( f n ( y)
1
2
l
0
n
l
n1
a nm B nm ( y ) w j Fn, j j ( y ) ) dy
m1
j 0
l
(3.7)
Here, the formula (3.6) represents a nonlinear system of algebraic equation in the form
an1
an2
.
=
.
.
an
l
H n1 ( an1 , an2 , ..., anl )
H n2 ( an1 , an2 ,..., anl )
.
.
.
.
.
.
.
. .
.
.
.
.
.
.
H n ( an , an , ..., an )
l 1 2
l
(3.8)
which can be written in the vector form
a H( a )
(3.9)
Hamerstein-Volterra integral equation
where a
T
435
( a n1 , a n2 , ..., a nl ) and H ( )T ( H n1 ( a ), H n2 ( a ),...,H nl ( a ) ) .
We shall show that the unique solution of Eq. (3.9) corresponds to the unique solution
of (3.4) for each nl under some mild assumptions.
Theorem 1. The system of integral equation (3.4) has a unique solution under the
following condition
11
1/ 2
{ k n ( x , y ) dxdy}
M1
(3.10)
00
To prove the theorm, we use the relation
kn ( x, y ) k ( x, y ) kn ( x, y ) k ( x, y )
Then with the aid of Eq. (3.1), we neglect the small constant where
k ( x, y ) kn ( x, y )
Assume, for n N , the integral operator
l
1
Tn ( x) k ( x, y ) ( y, n ( y ) )dy
l
l
0
and
~
n
(T nl ) ( x ) Tn ( x ) f n ( x ) w j Fn, j n ( x ).
j
l
j 0
(3.11)
It is straight forward to verify that Tnl M 1 A for all L2 [ 0 ,1 ] . Hence Tn is
l
a bounded nonlinear operator for all values of 0 n N , l is finite number. Also,we
have
~ (1) ~ (2)
(1)
( 2)
Tn Tn
Tn Tn
l
l
l
l
1
2 2
11 2
1
(1)
(2)
k ( x, y) dx dy ( y,n ( y) ) ( y,n ( y) ) dy .
l
00
0
(3.12)
Using the conditions (i) and (ii), we get
Tnl (1) Tnl ( 2 ) const . (1) ( 2 ) .
i.e (1) ( 2 )
hence the proof is completed.
Theorem 2. For the degenerate kernel k ( x, y ) , let
l
(3.13)
436
R. O. Abd El Rahman
1
2
l 1
M C { Bi ( x) dx}1l 2 { Ci ( x) dx}1/ 2
i 1 0
i 0 0
l
2
( M 1)
(3.14)
where, C is the constant of Lipschitz for the second argument of the function
( x, ( x, t ) ). Then the nonlinear algebraic equation (3.9) has a unique solution
*n *n1 , *n2 , ...*nl
(3.15)
and
l
nl f n ( x )
i 1
*i
Bi ( x )
l
w j F j ,nl n j
(3.16)
j 0
is the unique solution of (3.4).
Proof of this theorem depends on the definition of the discrete l 2 norm by
1
l
T
2
n l { n } 2 for n1 , n2 ,..., nl l 2 ( l ) .
2
j
j 1
For
(1)
(n1 ) , (n1 ) ,..., (n1 )
1
2
l
and
( 2 ) (n2 ) , (n2 ) ,..., (n2 )
1
2
l
we have
1
1
2
1
1
l
n
2
(1)
(2)
(1)
(2)
H ( ) H ( ) B{ Bi ( x) dx}2 { Ci ( x) dx}2
l
l2
i 10
i 10
Consequently F is a contraction operator in l( l ), M 1 .
It is not difficult for the reader to prove the following theorem
Theorem 3. For
11
2
1/ 2
k ( x , y ) kl ( x, y ) { k ( x , y ) k n ( x , y ) dx dy}
00
we have
l const. k k l . .
3.
Numerical results
Example 1:
In this example, we consider the following integral equation
1
t
3
( x, t ) x 2 y ( y, t ) dy ( x, ) d x 2 [t t t ] ,
3
3
0
0
Hamerstein-Volterra integral equation
437
the exact solution is ( x, t ) x 2 t , and the error is calculated in the table.
Example 2:
In this example, we consider the following integral equation
2
1
t
3
( x, t ) x y 2 ( y, t ) dy ( x, ) d x 2 [t t ] xt ,
3
6
0
0
Also, the exact solution of the previous integral equation is ( x, t ) x 2 t . Using the
degenerate kernel method, we obtained the error
At h=0.1, we have
E(x)
t
0
0
.2e 3 x .5e 5 x
2
0 .1
.3 e
0 .2
3 x 2
.2e 4 x
2
0 .3
.5e 3 x .4e 4 x
0 .4
.6e 3 x .7e 4 x
0,5
.8e 3 x .1e 4 x
0 .6
.9e 3 x .1e 4 x
0 .7
.1e 3 x .2e 4 x
2
2
2
2
0 .8
.1e
0 .9
.1e
1
.1e
At h 0.01, t 1 we have E ( x) .1e 4 x .2e 5 x
2
2x
2
2x
2
2x
2
.2 e
.2 e
.2 e
4x
4x
4x
438
R. O. Abd El Rahman
So when h decreases, the error is also decreased.
At h=0.1, we have
T
E(x)
0
0
0 .1
.1e 3 x
2
0 .2
.2 e 3 x
2
0 .3
.4e 3 x
2
0 .4
.5e 3 x
0,5
.6e 3 x
2
0 .6
.7e 3 x
2
0 .7
.9e 3 x
2
0 .8
.9e 2 x .
0 .9
.1e 2 x
2
2
1
2
.1e 2 x
2
At h 0.01, t 1 we have E ( x) .1e 4 x .
2
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Hamerstein-Volterra integral equation
439
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Received: April 15, 2014
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