Page 68 - DMTH504_DIFFERENTIAL_AND_INTEGRAL_EQUATION

Unit 3: Hermite Polynomials

Notes
2
2( 1) 3 2 ( 1)( 3) 5
= a x x x
0
3! 5!
m
( 2) ( 1)( 3) ( 2m 1) 2m 1
x
(2m 1)!
= y (say) ...(x)
2
From (viii) and (x) it is obvious that (x) is the part of solution, given by (viii). But as the two are
the solutions of the same equations so (x) must not be the part of solution (viii).
a = 0 and the solution in the case k = 0 must be given by (ix).
1
Hence the general solution of Hermite s equation is

y Ay By ,
1 2
where A and B are arbitrary constants and y y are given by (ix) and (x).
,
1
2
Hermite s Polynomials
When is an even integer, equation (ix) gives an even polynomial of degree n.
Let = n, n being an even integer and let

a = ( 1) n /2 n !
0
n
!
2
n
Coefficient of x in (ix) is
n n n
2 . 1 1
n
/2
n
( 1) n /2 n ! ( 2) ( n n 2) (n n 2) = 2 2 2 .
n
n ! n ( /2)!
!
2
Similarly coefficient of x n 2
( 1) n /2 ! n ( 2) (n 2)/2 ( n n 2) (n n 2 2)
n
( /2)! (n 2)!
2 n 2 ( n n 1) /2( /2 1) 2
n
n
( /2)!
n
( n n 1) n 2
2
1!
and so on.
So value of y is given by

x
x
y = (2 )x n ( n n 1) (2 ) n 2 ( n n 1)(n 2)(n 3) (2 ) n 4 ( 1) n /2 ! n
n 1! 2! ( /2)!
n

LOVELY PROFESSIONAL UNIVERSITY 61