Page 438 - DMTH504_DIFFERENTIAL_AND_INTEGRAL_EQUATION

Unit 25: Volterra Equations and L Kernels and Functions
2

With the help of (4), it is easy to prove that the function y(x) given by (2) satisfies (1). Also Notes
x
x
x
u
f
u
y 0 ( ) f ( ) H ( , , ) ( )du ...(9)
x
0
certainly belongs to L proved that f(x) belongs to the same class. But then we have
2
x
x
y
y 0 ( ) K ( , ) ( )du
x
u
u
0
0
x x x z
x
x
u
x
z
f
u
z
u
u
f
u
f ( ) H ( , , ) ( )du K ( , ) ( )du r K ( , )dz H ( , , ) ( )du
u
f
x
0 0 0 0
x x
x
u
z
z
u
H
x
u
u
x
x
f ( ) K ( , ) H ( , , ) K ( , ) ( , , )dz f ( )du
0 0
f ( ) 0 f ( )
x
x
So the function y (x) from (9) is the only function of class L of the given equation, neglecting the
0 2
function y(x) given by
x
u
y ( ) K ( , ) ( )du ...(10)
x
u
x
0
known as a zero function in L -space. For this we observe that let v be the norm of y(x) in the
2
basic interval (0, h)
h
v 2 y 2 ( )dx
x
0
then from (10) using Schwarz inequality, it follows that
x x
2
u
x
x
z
y 2 ( ) | | 2 K 2 ( , )du y 2 ( )dz | | A 2 ( )v 2
x
0 0
and successively
x x x
4
2
4
y 2 ( ) | | v 2 K 2 ( , )du A 2 ( )dz | | v A 2 ( ) A 2 ( )du
z
u
x
x
x
u
0 0 0
x x u
6
u
u
z
y 2 ( ) | | v 2 K 2 ( , )du A 2 ( )du A 2 ( )dz
x
x
0 0 0
x y
6
2
x
u
z
| | v A 2 ( ) A 2 ( )du A 2 ( )dz
0 0
By analogy to (7) we have
n 2n
x u n 1 1 x N
u
A 2 (u 1 )du 1 ...... A 2 (u n )du n A 2 ( )du ...(11)
0 0 0 ! n 0 ! n
hence we can write
2
(| | N 2 n
)
2 2 2 2
x
( ) v A ( ) , (n 0, 1, 2, ...)
x
! n
and this shows that y(x) = 0 at any point where A(x) is finite. So we have shown that y (x) is a
0
unique solution of (1).
An alternative approx of proving the existence and uniqueness of the solution of the Volterra
integral equation is by Picard’s process of successive approximation method. It is advisable to
try it as an alternative as given in Yosida book.
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