The Chemistry Maths Book, Second Edition

382 Chapter 13Second-order differential equations. Some special functions

The Hermite polynomials satisfy the recurrence relation

H

n+ 1

(x) 1 − 12 xH

n

(x) 1 + 12 nH

n− 1

(x) 1 = 10 (13.33)

so that, givenH

0

1 = 11 andH

1

1 = 12 x, all higher polynomials can be derived.

EXAMPLE 13.10

(i) Use the series expansion (13.31) to findH

4

(x).

(ii) Verify by substitution in (13.30) thatH

4

(x)is a solution of the Hermite equation.

H

4

′(x) 1 = 164 x

3

1 − 196 x, H

4

′′(x) 1 = 1192 x

2

1 − 196

Therefore, forn 1 = 14 , the Hermite equation is

(iii) Use the recurrence relation (13.33) to findH

5

(x).

H

5

1 = 12 xH

4

1 − 18 H

3

1 = 12 x[16x

4

1 − 148 x

2

1 + 1 12] 1 − 1 8[8x

3

1 − 112 x] 1 = 132 x

5

1 − 1160 x

3

1 + 1120 x

0 Exercise 21

Hermite functions

A differential equation that is related to the Hermite equation is

y′′ 1 + 1 (1 1 − 1 x

2

1 + 12 n)y 1 = 10 (13.34)

and this is solved by making the substitution

y(x) 1 = 1 e

−x

2

22

u(x) (13.35)

with second derivative

y′′ 1 = 1 e

−x

2

22

[u′′ 1 − 12 xu′ 1 − 1 (1 1 − 1 x

2

)u]

Substitution of yand its second derivative in (13.34) gives

e

−x

2

22

[u′′ 1 − 12 xu′ 1 + 12 nu] 1 = 10

′′− ′+= −













−−













yxyy x 2 8 192 96 2 64 96xx x

23

++−+













816 48 12 0=

42

xx

Hx x x x x

4

42 04

2

43

1

2

432

2

()()=−() () 21648

!

• 
  

!

=−

×××

xx

2

• 12