Page 434 - DMTH504_DIFFERENTIAL_AND_INTEGRAL_EQUATION

Unit 25: Volterra Equations and L Kernels and Functions
2

Hence if the solution exists, it should be given by letting n and given by Notes
x
x
u
x
f ( ) f ( ) H ( , , ) ( )du ...(11)
x
f
u
0
where H(x, u, ) is the resolvent Kernel given by the series
H ( , , ) n K n 1 ( , ) ...(12)
u
x
u
x
n 0
This method of successive approximation cannot only be applied to those of Volterra type
integral equations but a whole lot of other equations including the Fredholm integral equations.
Example: Let the Volterra integral equation be given by
x x
y
t
x
y ( ) X (x t ) ( )dt for 0 x 1
0
The interacted Kernels are
K 1 ( , ) x t
x
t
x (x ) t 3
K 2 ( , ) (x r )(r ) t dr
t
x
t 1.2.3.
3
(x r ) (r ) t dr (s t ) 5
t
x
K 3 ( , )
1.2.3 5
         
(x ) t 2n 1
t
K n ( , )
x
(2n 1)!
Hence
s (x ) t 3 (x ) t 5
y ( ) x (x ) t .... .dt
t
x
0 3! 5!
x
(x ) t 3 (x ) t 5 (x ) t 7
x .... sin x
3! 5! 7!
0
So the answer is
y(x) = sin x
25.3 L -Kernels and Functions
2
In the case of Volterra integral equation
x
x
u
u
f
y ( ) f ( ) K ( , ) ( )du ...(1)
x
x
0
The Kernel K(x, u) and the f(x) are supposed to be continuous and differentiable in the double
interval 0 x h and 0 u h. They are consequently bounded in the L -space. Namely the
2
Kernel and the function f(x) are quadratically integrable in the L -space i.e. 0 x h and
2
0 u h where h is constant i.e. the integrals
h h
K K 2 ( , )dxdu N 2 ...(2)
x
u
0 0
h
t f 2 ( )dx ...(3)
x
0
LOVELY PROFESSIONAL UNIVERSITY 427