Page 56 - DMTH504_DIFFERENTIAL_AND_INTEGRAL_EQUATION

Unit 2: Legendre’s Polynomials

Self Assessment Notes

3. Prove

(n 1) ( ) = P n 1 ( ) xP n ( )
P
x
x
x
n
4. Prove that
2
x
x
)
x
(1 x P n ( ) = n P n 1 ( ) xP n ( )
2.5 Orthogonal Properties of Legendre Polynomials
Prove that
1
x
(i) P m ( )P n ( )dx 0 if m n and
x
1
1
2
(ii) [ ( )] dx 2
x
P
n
2n 1
1
Proof:
From Legendre equation P n ( ) being solution of it so we have
x
2
x
2 d P n ( ) dP n ( )
x
(1 x ) 2 2x ( n n 1) ( ) = 0
P
x
n
dx dt
x
d 2 dP n ( )
x
or (1 x ) ( n n 1)P n ( ) = 0 ...(A)
dx dx
In the same way, we have
d 2 dP m ( )
x
x
(1 x ) m (m 1)P m ( ) = 0 ...(B)
dx dx
Multiplying equation (A) by P m ( ) and (B) by P and subtracting
x
n
x
d 2 d P ( ) d 2 d P
x
P m (1 x ) n P n (1 x ) m ( ) [ (n 1) m (m 1)]P m ( )P n ( ) = 0 ...(C)
x
n
x
dx dx dx dx
Integrating equation (C) between the limits 1 to 1, we have
x
1 d 2 dP ( ) 1 d 2 dP
P m ( ) (1 x ) n dx P n ( ) (1 x ) m dx (n m )(n m 1)
x
x
1 dx dx 1 dx dx
1
P m ( ) ( )dx = 0
x
x
P
n
1
Integrating by parts we have
1 1 1
2 dP n dP m ( ) 2 dP n ( ) 2 dP m ( )
x
x
x
x
P ( )(1 x ) (1 x ) dx P ( )(1 x )
x
m n
dx 1 dx dx dx 1
1
1 1
d 2 dP m ( )
x
x
x
P n ( )(1 x ) dx (n m )(n m 1) P m ( ) ( )dx = 0
P
x
n
dx dx
1 1
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