Page 481 - DMTH504_DIFFERENTIAL_AND_INTEGRAL_EQUATION

Differential and Integral Equation

Notes From this follows the fact that to solve the equation (1) is equivalent to finding a solution (s) of
the equation (18) with the term in brackets as the Kernel and for the right side the given function
b
s
f ( ) 1 ( , ) ( ) dr
s
r
f
r
a
We shall prove incidentally that
b
r
r
s
( ) ( , ) ( )dr (v = 1, 2, ..., m) ...(19)
s
v 1 v
a
are linearly independent. To prove this, suppose
m b
s
r
s
0 c ( ) ( , ) ( )dr
r
v v 1 v
a
v 1
m m
b
r
r
s
c v v ( ) 1 ( , ) c v v ( ) dr
s
a
v 1 v 1
m
c
and | | 0. Then, by the properties of the resolvent Kernel (s, t), we have
v
v 1 1
m b
r
s
c v v ( ) 0 K 1 ( , )0.dr 0
s
a
v 1
This contradicts the linear independence of (s).
v
The equation (18) is reduced to the system of equations
m
b b
r
s
r
r
( ) f ( ) 1 ( , ) ( ) v v ( ) 1 ( , ) v ( )dr
s
s
s
s
f
r
a a ...(20)
v 1
b
t
( ) ( )dt ( = 1, 2, ..., m) ...(21)
t
a
Hence, substituting (20) in (21), we have a system of linear equations in unknowns, , , ... ,
1 2 m
m b b
s
r
s
r
s
s
( ) ( )ds ( , ) ( ) ( )dr ds
v v 1 v
v 1 a a
b b
t
r
r
t
t
( ) 1 ( , ) ( )dr f ( )dt ( = 1, 2, ..., m) ...(22)
a a
Accordingly to solve the equation (1) is equivalent to finding the solutions of (22); indeed,
substituting the solution in (20), we obtain the solution of (1).
Similarly, we see that to solve the equation (2) is equivalent to solving the following system of
linear equations in the unknowns
, ,..., ,
1 2 m
m b b
t
r
t
( ) ( )dt ( , ) ( ) ( )dr dt
t
r
t
v v 1 v ...(23)
a a
v 1
b b
s
s
s
r
r
a ( ) 1 ( , ) ( )dr g ( )ds ( = 1, 2, ..., m)
a a
and the solution (t) of (2) is given by
m
b b
t
t
t
( ) g ( ) 1 ( , ) ( )dr v v ( ) 1 ( , ) ( )dr ...(24)
r
r
g
r
t
r
t
v
a a
v 1
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