Riley hobson bence · Problem 18.24

Special functions

Mathematics — Methods · parabolic-cylinder-functions

Problem

The solutions

y(x, a)

of the equation

\frac{d^2y}{dx^2} - \left(\frac{1}{4}x^2 + a\right)y = 0 \qquad (*)

are known as parabolic cylinder functions. (a) If

y(x, a)

is a solution of

(*)

, determine which of the following are also solutions: (i)

y(a, -x)

, (ii)

y(-a, x)

, (iii)

y(a, ix)

and (iv)

y(-a, ix)

. (b) Show that one solution of

(*)

, even in

x

, is

y_1(x, a) = e^{-x^2/4} M\!\left(\tfrac{1}{2}a + \tfrac{1}{4},\, \tfrac{1}{2},\, \tfrac{1}{2}x^2\right),

where

M(\alpha, c, z)

is the confluent hypergeometric function satisfying

z\frac{d^2M}{dz^2} + (c - z)\frac{dM}{dz} - \alpha M = 0.

You may assume (or prove) that a second solution, odd in

x

, is given by

y_2(x, a) = xe^{-x^2/4}M\!\left(\tfrac{1}{2}a + \tfrac{3}{4},\, \tfrac{3}{2},\, \tfrac{1}{2}x^2\right)

. (c) Find, as an infinite series, an explicit expression for

e^{x^2/4}y_1(x, a)

. (d) Using the results from part (a), show that

y_1(x, a)

can also be written as

y_1(x, a) = e^{x^2/4} M\!\left(-\tfrac{1}{2}a + \tfrac{1}{4},\, \tfrac{1}{2},\, -\tfrac{1}{2}x^2\right).

(e) By making a suitable choice for

a

deduce that

1 + \sum_{n=1}^{\infty} \frac{b_n\, x^{2n}}{(2n)!} = e^{x^2/2}\left(1 + \sum_{n=1}^{\infty} \frac{(-1)^n b_n\, x^{2n}}{(2n)!}\right),

where

b_n = \prod_{r=1}^{n}(2r - \tfrac{3}{2})