Page 460 - DMTH504_DIFFERENTIAL_AND_INTEGRAL_EQUATION

Unit 29: Fredholm Equations Solution by the Method of Successive Approximation

Notes
Example: Let us consider the following integral equation

1
y ( ) e x u y ( )du f ( ) ...(1)
x
x
u
0
we now have
1 1
K 2 ( , ) e x 1 u 1 u u du 1 e x u du 1 e x u K ( , )
x
u
u
x
0 0
with this consequence that all the iterated Kernels K coincide with the given Kernel K(x, u) and
n
the series (6) becomes
x
x
u
H ( , , ) K ( , ) (1 2 ....) ...(2)
u
Hence we have
x
u
K ( , )
u
x
H ( , , ) ...(3)
( 1)
and we see that the resolvent Kernel is analytic function of . So we have one and only one
solution for 1.
e x 1 u
u
y ( ) f ( ) e f ( )du ...(4)
x
x
( 1) 0
We started with the integral equation (1)
1
x
u
y ( ) K ( , ) ( ) f ( ) ...(1)
y
x
x
u
0
and arrived at the equation (13)
1
x
y ( ) f ( ) H ( , u , ) ( )du ...(13)
f
x
x
u
0
where the resolvent Kernel satisfies the equation (12)
u
K ( , ) H ( , , ) H ( , u 1 , ) (u 1 , )du ...(11)
u
x
x
u
x
Substituting (13) into L.H.S. of (1)
1 1 1
f
u
f ( ) H ( , , ) ( )du K ( , ) f ( ) H ( , u 1 , ) (u 1 )du 1 du
x
f
u
x
u
u
u
x
0 0 0
1 1
1
u
u
f
x
x
u
x
= ( )f x du f ( ) K ( , ) H ( , , ) H ( , u 1 , ) (u 1 )du K ( , )du
u
u
1
0
0 0
1 1
u
u
u
x
= ( )f x du K ( , ) H ( , , ) H ( , u 1 , ) (u 1 , ) f ( )
x
x
K
u
0 0
= f(x) + 0 = R.H.S.
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