Page 465 - DMTH504_DIFFERENTIAL_AND_INTEGRAL_EQUATION

Differential and Integral Equation

Notes
Example: Consider the integral equation

b
x
x
t
y
t
x
y ( ) f ( ) K ( , ) ( )dt ...(1)
a
x
Find the solution when f(x) = e , K(x, t) = 2e x + t , a = 0, b = 1.
Substitute the value of f(x) and K(x, t) in (1) we have
1
t
y ( ) e x 2 e x e y ( )dt
t
x
0
1
t
e x 1 2 e y ( )dt
t
0
1
t
t
Let C e y ( )dt constant ...(2)
0
then y ( ) e x (1 2 C ) ...(3)
x
Substituting this value of y in (2) we have
1 (e 2 1)
t
C (1 2 C ) t . e e dt (1 2 C )
0 2
Solving for C i.e.
2C 2 C (e 2 1) (e 2 1)
(e 2 1)
C 2 ....(4)
2 1 (e 1)
Substituting in (3) we have
x
y ( ) e x [1 (e 2 1)] ...(5)
The denominator is non-zero.

Self Assessment

1. Solve the Fredholm integral equation

b
x
t
y ( ) f ( ) K y ( )dt
x
0
a
where K is a constant and show that for | | < 1/K (b a) the corresponding Neumann
0 0
Series is convergent.
30.3 Summary
 In case the parameter is small one gets the solution of Fredholm equation of the second
kind as a power series in called Neumann series.

 The Resolvent Kernel can also be expanded in powers of provided the Kernel K(x, t) is of
L -class. The resolvent Kernel is though an analytic function of but is not an entire
2
function in whole of complex -plane.

458 LOVELY PROFESSIONAL UNIVERSITY