Page 57 - DMTH504_DIFFERENTIAL_AND_INTEGRAL_EQUATION

Differential and Integral Equation

Notes or
1 1
x
dP ( ) dP ( ) 2 dP ( ) dP ( ) 2
x
x
x
0 m n (1 x )dx 0 n m (1 x )dx
dx dx dx dx
1 1
1
x
(n m )(n m 1) P ( )P ( )dx = 0
x
m n
1
1
or (n m )(n m 1) P m ( ) P n ( )dx = 0
x
x
1
Thus
1
P m ( ) ( )dx = 0 for m n ...(D)
P
x
x
n
1
This proves the first part.
To prove (ii)
We have
(1 2xh h 2 ) 1 = (1 2xh h 2 ) 1/2 .(1 2x h 2 ) 1
n
m
x
x
= h P n ( ) h P m ( )
n 0 m 0
x
x
x
= h 2n P n 2 ( ) 2 h m n P n ( ) P m ( ) ...(E)
n 0 m 0
n 0
m n
Integrating between the limits 1 to +1, we have
1 1 1
2n
x
h P 2 ( )dx 2 h m n P ( )P ( )dx = dx
x
x
n n m 2
1 n 0 1 m 0 1 (1 2hx h )
n 0
m n
1 1
x
Thus h 2n P n 2 ( )dx = dx 2 1/2 ...(F)
n 0 1 1 (1 2xh h )
1 1 h 2 1 1 h
= log log
2h 1 h h 1 h
Expanding the R.H.S. in powers of h, we have
1 2 3 4 2 3 4
h 2n P n 2 ( )dx = 1 h h h h ... h h h h ...
x
n 0 1 h 2 3 4 2 3 4
2 h 3 h 5 h 7
= h h 3 5 7 ...

1
2n
= 2 h 2n 1
n 0

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