Mathematical Methods for Physics and Engineering : A Comprehensive Guide

16.5 POLYNOMIAL SOLUTIONS

is a positive integer or zero, then we are left with a finite polynomial of degree

N′=N+σas a solution of the ODE:

y(z)=

∑N

n=0

anzn+σ. (16.33)

In many applications in theoretical physics (particularly in quantum mechanics)

the termination of a potentially infinite series after a finite number of terms

is of crucial importance in establishing physically acceptable descriptions and

properties of systems. The condition under which such a termination occurs is

therefore of considerable importance.

Find power series solutions aboutz=0of

y′′− 2 zy′+λy=0. (16.34)

For what values ofλdoes the equation possess a polynomial solution?Find such a solution

forλ=4.

Clearlyz= 0 is an ordinary point of (16.34) and so we look for solutions of the form
y=

∑∞

n=0anz

n. Substituting this into the ODE and multiplying through byz (^2) we find
∑∞
n=0
[n(n−1)− 2 z^2 n+λz^2 ]anzn=0.
By demanding that the coefficients of each power ofzvanish separately we derive the
recurrence relation
n(n−1)an−2(n−2)an− 2 +λan− 2 =0,
which may be rearranged to give
an=
2(n−2)−λ
n(n−1)
an− 2 forn≥ 2. (16.35)
The odd and even coefficients are therefore independent of one another, and two solutions
to (16.34) may be derived. We either seta 1 =0anda 0 =1toobtain
y 1 (z)=1−λ
z^2
2!
−λ(4−λ)
z^4
4!
−λ(4−λ)(8−λ)
z^6
6!

−··· (16.36)

or seta 0 =0anda 1 =1toobtain

y 2 (z)=z+(2−λ)

z^3

3!

+(2−λ)(6−λ)

z^5

5!

+(2−λ)(6−λ)(10−λ)

z^7

7!

+···.

Now, from the recurrence relation (16.35) (or in this case from the expressions fory 1
andy 2 themselves) we see that for the ODE to possess a polynomial solution we require
λ=2(n−2) forn≥2 or, more simply,λ=2nforn≥0, i.e.λmust be an even positive
integer. Ifλ= 4 then from (16.36) the ODE has the polynomial solution

y 1 (z)=1−

4 z^2

2!

=1− 2 z^2 .

A simpler method of obtaining finite polynomial solutions is toassumea

solution of the form (16.33), whereaN= 0. Instead of starting with the lowest

power ofz, as we have done up to now, this time we start by considering the