Page 484 - DMTH504_DIFFERENTIAL_AND_INTEGRAL_EQUATION

Unit 32: The Fredholm Theorem

Finally, we shall reduce the solvability condition (26 ) to a more readable and usual form as Notes
follows:
The term on the left side of (26) is, by (22) and (22 )

m b m b
t
r
f j j m ( ) 1 ( , ) ( )dr f ( )dt
t
r
t
a a
1 1
From (24), it is easily seen that the function in brackets on the right side is a solution of (2) with
g(t) 0, that is, of
b
( ) K ( , ) ( )ds 0 ...(30)
s
t
t
s
a
On the other hand, the general solution of (30) is given by linear combinations of the functions
in brackets. Therefore (26) is equivalent to the following for every solution (t) of (30).
b
t
t
f ( ) ( )dt 0 ...(31)
a
Similarly, we see that the condition (28) is equivalent to the following: for every solution (s) of
the equation
b
t
t
s
( ) K ( , ) ( )dt 0
s
a
b ...(32)
s
s
g ( ) ( )ds 0
a
Self Assessment
1. The Fredholm equation is given by
1
2 2
t
y
x
y ( ) f ( ) (xt x t ) ( )dt
x
0
3
solve for y(x) when f(x) = x .
32.3 Summary
 We have seen that Fredholm integral equation has solutions that depend on the nature of
the resolvent Kernel as well on the function f(s).
 If the parameter is not an eigenvalue then the non-homogeneous equation has one and
only one solution and the homogeneous equation has a solution Q(x) = 0.
 For to be one of the eigenvalues, the homogeneous equation admits a number of
independent solutions.
32.4 Keywords

The nature of the solution of the Fredholm integral equation of the second kind as well as on the
first kind depends upon the constant parameter as well as on the function f(x).
The eigenvalue problem puts certain conditions on the function f(s) for the solutions to exist.
Fredholm theorem elaborates on these points.

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