CONTENTS

J. Duhok Univ., Vol.14, No.1 (Pure and Eng. Sciences), Pp 25-29, 2011
�
t
y ( t , x ) � G ( t , s ) g ( s, x ( s, x )) ds, m �0,1,2,...
m 0
��
m 0
and
�
bt ()
z ( t , x ) � g ( s, x ( s, x )) ds , m �0,1,2,..
m 0
at ()
m 0
Now, we claim that the sequence of functions
x (, t x)
is uniformly convergent on the domain
m
0
[ a, b] � Df
.
For m=1 in (9) using (3)-(5), we find that
t
x ( t , x ) � x ( t , x ) � � f ( s, x ( s, x ),
2 0 1 0
a
1 0
s
��� x 1( s, x 0
bs ( )
� as ( )
1 0
G ( s, � ) g ( �, )) d� , g ( �, x ( �, x )) d�
) �
� �
bs ( )
s
( , , ( , � ) ( � , ) �, g ( �, x ) d�)
f s x G s g x d ds
o
��
o
as ( )
t
( K x 1( s, x )
a
o � x o
�
� HL x 1(
s, x o ) � x o
�
�
�QhHx( s, x ) � x ) ds
� �
1
o o
t
�
� � ( KM ( b �a) � HLM ( b �a) �QhHM
( b � a)) ds
a
�
�
2
� ( K �H( L �Qh))
M ( b � a)
�
� �M ( b � a)
.
Suppose that the following inequality
k
x k �1( t , x 0) � x k ( t , x 0)
� � M ( b � a)
(14)
holds for some m=k, then we shall prove that
the inqualtiy
k �1
x ( t , x ) � x ( t , x ) � � M ( b � a)
k �20k �10
Is true for all t [ a, b]
� , xo � Df
From (9) when m=k+1 and the inequality
(14), we get
x ( t , x ) �x( t , x ) � ( K x ( s, x ) �x(
s, x ) �
k �20k �10
t
� a
k �1
0 k 0
�
HL x k �1( s, x 0) �xk( s, x 0) �QhH
x k �1(
s, x 0) �xk(
s, x 0)
) ds
�
t
k � k
� � ( K � M ( b �a) � HL � M ( b �a) �
a
�
k
QhH � M ( b �a))
ds
�
�
k
( K � H ( L �Qh )) � M ( b � a)
�
k �1
� M ( b �a)
�
By mathematical induction and by (9) and
(12) the following inequality holds:
m
x m�1( t , x 0) � x m(
t , x 0)
� � M ( b � a)
(15)
�
where � � ( K � H ( L � Qh))( b � a)
�
m �
for all 0,1,2,...
From (15) we conclude that for k � 1,
we
have the following inequality :
x ( t , x ) �x( t , x ) � x ( t , x ) �x(
t , x ) �
m �k 0 m 0 m �k 0 m �k �1
0
2
o
x ( t , x ) �x( t , x ) �... � x ( t , x ) �x(
t , x )
m �k �1 0 m �k �2 0 m �1
0 m 0
� � ) � � ) �... �
m �k �1 m �k �2
x 1( t , x 0 �x0x1( t , x 0 �x0
m
� x (, t x ) � x
1 0 0
m 2 k �2k�1 � � ( E � � � � �... � � � � ) x 1(, t x 0) � x 0
therefore
m
�1
x m �k ( t , x 0) � x m ( t , x 0)
� � ( E � �) M ( b � a)
(16)
where E is identity matrix, t � [ a, b]
and xo � D . f
By using the condition (8) and the inequality
(16), we find that
m
lim � � 0
(17)
m ��
The relations (16) and (17) prove the uniform
convergence of the sequence of function (9) in
the domain (10) as m ��.
Let
lim x ( t , x ) � x ( t , x )
(18)
m ��
m
0 � 0
Since the sequence of functions x m (, t x 0)
are
defined and continuous in the domain (10), then
the limiting function x � (, t x 0)
is also defined and
continuous in the domain (10).
Theorem 2. (Uniqueness Theorem)
With the hypotheses and all conditions of the
theorem 1 , the solution of Volterra integral
equation (1) is unique.
*
Proof. Let x (, t x 0)
be another solution of the
Volterra integral equation (1), i.e.
t s
* * *
( , ) ( ) ( , ( , ), ( , �) ( �, ( �, )) �,
o o
a
o
��
o
x t x F t f s x s x G s g x x d
� �� �
�
b( s)
as ( )
and then we have
g x d ds
*
( �, x ( �, o )) � ) ,
( , 0
*
( , 0
t
a
0
s
��� x( , x0
bs ( )
� as ( )
0
x t x ) � x t x ) � � f ( s, x ( s, x ),
G ( s, � ) g ( �, � )) d� , g ( �, x ( �, x )) d�
) �
� �
*
s
*
b ( s )
*
0
��
0
a( s)
0
f ( s, x ( s, x ), G ( s, � ) g ( �, x ( �, x )) d� , g ( �, x ( �, x )) d� ) ds
t
� � a
�
* �
*
( K x ( s, x o) � x ( s, x 0) � HL x ( s, x o)
� x ( s, x 0)
�
�
�QhHxsx� ds
*
( , o ) x (, s x 0)
)
�
�
*
( K � H ( L �Qh ))( b � a) x ( t , x o ) � x ( t , x 0)
so that
x ( t , x ) x ( t , x ) �� x ( t , x ) x ( t , x )
* *
0 � 0 0 � 0
By iteration we find
x ( t , x
*
) �x( t , x
m
) �� x ( t , x
*
) �x(
t , x )
0 0 0 0
But from the condition (8), we get
m
� � 0 when m ��, hence we obtain that
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