Page 76 - DMTH504_DIFFERENTIAL_AND_INTEGRAL_EQUATION

Unit 3: Hermite Polynomials

Notes
x
x
or H n ( ) = 2n H n 1 ( )
(II) 2x H n ( ) = 2nH n 1 ( ) H n 1 ( )
x
x
x
H n ( )t n 2 t
x
we have ! n = e t 2 x
n 0
Differentiating both sides with respect to t, we get
x
H n ( ) n t n 1 e t 2 2tx ( 2t 2 )
x
! n =
n 0
x
H n ( ) n 1 2 2
or (n 1)! t = 2t e t 2tx 2x e t 2tx
n 1
x
H n ( ) n 1 H x H x
( ) n
( ) n
n
n
or (n 1)! t = 2t ! n t 2x ! n t
n 1 n 0 n 0
(Since term of L.H.S. Corresponding to n = 0 is zero)
H x H x H ( )
x
( ) n
n
or 2x ! n t = 2 n ( ) n 1 n t n 1
t
n 0 ! n (n 1)!
n 0 n 0
x
H x H n 1 ( ) n H n 1 ( ) n
x
( ) n
n
or 2x ! n t = 2 (n 1)! t ! n t
n 0 n 1 n 0
Equating the coefficient of t n , on both sides, we have
x
x
x
H ( ) H n 1 ( ) H n 1 ( )
2x n = 2
! n (n 1)! ! n
x
x
x
or 2x H n ( ) = 2n H n 1 ( ) H n 1 ( )
x
x
x
(III) H n ( ) = 2x H n ( ) H n 1 ( )
Writing recurrence formulae I and II, we have
x
x
H n ( ) = 2n H n 1 ( ) ...(i)
x
x
x
and 2x H n ( ) = 2n H n 1 ( ) H n 1 ( ) ...(ii)
Subtracting (ii) from (i), we have
x
x
x
H n ( ) = 2x H n ( ) H n 1 ( )
(IV) H n ( ) 2x H n ( ) 2nH n ( ) = 0
x
x
x
Hermite s differential equation is
2
d y 2x dy 2ny = 0
dx 2 dx
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