Page 44 - DMTH504_DIFFERENTIAL_AND_INTEGRAL_EQUATION

Unit 2: Legendre’s Polynomials

We now apply Leibnitz theorem to differentiate equation r times. Here Leibnitz theorem states Notes
th
that the r differentiation of product of two functions is given by
2
r
d r ( fg ) = d g df d r 1 ( r r 1) d f d r 2 g
2
dx r f dx r r dx dx r 1 g 2 dx dx n 2 .... ...(C)
So differentiating equation (B) r times we get
r
r
r
2 d r 2 u d r 1 u ( r r 1) d u d r 1 u d u d u
(1 x ) r .( 2 ) ( 2) 2(n 1) x . r 2n r = 0
x
dx r 2 dx r 1 2 dx r dx r 1 dx r dx
or rearranging terms
r
2 d r 2 u d r 1 u d u
r
(1 x ) r 2 2 (n 1 r ) r 1 r ( r r 1) 2 (n 1) 2n = 0 ...(D)
x
dx dx dx
Simplifying the above equation and putting
r
d u
u r = r , ...(E)
dx
We get

2
2 d u r du r
x
(1 x ) 2 (n 1 r ) (r 1)(2n r )u r = 0
dx 2 dx r
We now put r = n and get
2
2 d u n du n
(1 x ) 2 ( 1) (n 1)( )u n = 0
n
x
dx 2 dx
This is Legendre’s equation. Hence for r = x, u satisfies Legendre’s equation. Thus the Legendre’s
n
polynomial are given by
d n 2 n
C
P ( ) = (x 1) ( ) ...(F)
x
n n
dx
n
Where C is a constant. To evaluate C we compare the coefficients of x on both sides of (F) i.e.
n
(2 )! x n d n 2n 2n
n
= C x C (2 )(2n 1)...(n 1)x
n
2
n
2 ( !) dx n
(2 )! n
n
= C x
! n
Thus
1
( !)2 n = C
n
Thus
1 d n 2 n
x
P n ( ) = n n (x 1)
!
2 n dx
This is Rodrigue’s formula for the Legendre’s polynomials. We can again find a few Legendre
polynomials from this formula.
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