Page 479 - DMTH504_DIFFERENTIAL_AND_INTEGRAL_EQUATION

Differential and Integral Equation

Notes For this reason, the Kernels (s, t) and (s, t) are called the resolvent Kernels of the equation (1)
and (2) respectively.
The foregoing theorem shows that is not an eigenvalue of either the Kernel K(s, t) or its
conjugate Kernel K (s, t),
K (s, t) = K(t, s) ...(15)
The General Case

We shall prove that there exist two sets of linearly independent continuous functions
s
s
s
( ), ( ), ..., ( )
1 2 m
...(15)
B 1 ( ), 2 ( ), ..., m ( )
t
t
t
defined on the interval [a, b], such that
2
b b m
K ( , ) v ( ) ( ) ds dt 1 ...(16)
s
s
t
t
v
a a
v 1
To prove this, let be an arbitrary positive number. Then we divide the interval (a, b) into a finite
number of sub-intervals I , I , ..., I , such that
1 2 n
t
K
s
s
sup | ( , ) K ( , t | )
a s v
for any pair of points t , t in each I . This is possible, because of the uniform continuity of
v
K(s, t) on the domain a s b, a t b. Let t be an inner point of I . Let I be an interval
v v v
contained in the interior of I and containing the point t . Then we define (t) as follows:
v v v
0 outside of I v
( )
t
v
1 on I v
such that the function (t) is continuous and 0 (t) 1 on the interval [a, b]. We now set
v v
s
s
( ) K ( , ) and
t
r
n
s
s
t
t
s
N ( , ) K ( , ) v ( ) ( )
t
v
v 1
Then we see that
t
s
s
s
t
| ( , )| | ( , ) K ( , t v )|
N
K
for t in I , and
v
s
t
N
s
t
s
t
| ( , )| | ( , ) K ( , t v ) ( )| 2 M
K
v
for t in I I where
v v
M = sup a s b a t b | ( , | ) t
K
s
Since and the sum of lengths of I I are both arbitrary, we can choose the values of them so
v v
small that
2
n
b b
t
K ( , ) v ( ) ( ) ds dt 1
s
t
s
v
a a
v 1
Clearly, the function (t), (t), ..., (t) are linearly independent. Hence, if (s), (s), ..., (s)
1 2 n 1 2 n
are linearly independent, then our proof is completed. If otherwise, say, (s) is written as a
n
linear combination of (s), (s), ..., – (s), then
1 2 n 1
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