Polynomial solutions

SERIES SOLUTIONS OF ORDINARY DIFFERENTIAL EQUATIONS
derivative method. Let us first consider the case where the two solutions of the
indicial equation are equal. In this case a second solution is given by (16.28),
which may be written as
∂y(z, σ)
y2 (z) =
∂σ
σ=σ1
∞
∞ dan (σ)
= (ln z)z σ1
an (σ1 )z n + z σ1
zn
dσ σ=σ1
n=0
= y1 (z) ln z + z
n=1
σ1
∞
n
bn z ,
(16.31)
n=1
where bn = [dan (σ)/dσ]σ=σ1 . One could equally obtain the coefficients bn by direct
substitution of the form (16.31) into the original ODE.
In the case where the roots of the indicial equation differ by an integer (not
equal to zero), then from (16.30) a second solution is given by
∂
[(σ − σ2 )y(z, σ)]
y2 (z) =
∂σ
σ=σ2
∞
∞ d
σ
n
(σ − σ2 )an (σ)
= ln z (σ − σ2 )z
an (σ)z
+ z σ2
zn.
dσ
σ=σ2
n=0
σ=σ2
n=0
But, as we mentioned in the previous section, [(σ − σ2 )y(z, σ)] at σ = σ2 is just a
multiple of the first solution y(z, σ1 ). Therefore the second solution is of the form
y2 (z) = cy1 (z) ln z + z σ2
∞
bn z n ,
(16.32)
n=0
where c is a constant. In some cases, however, c might be zero, and so the second
solution would not contain the term in ln z and could be written simply as a
Frobenius series. Clearly this corresponds to the case in which the substitution
of a Frobenius series into the original ODE yields two solutions automatically.
In either case, the coefficients bn may also be found by direct substitution of the
form (16.32) into the original ODE.
16.5 Polynomial solutions
We have seen that the evaluation of successive terms of a series solution to a
differential equation is carried out by means of a recurrence relation. The form
of the relation for an depends upon n, the previous values of ar (r < n) and the
parameters of the equation. It may happen, as a result of this, that for some
value of n = N + 1 the computed value aN+1 is zero and that all higher ar also
vanish. If this is so, and the corresponding solution of the indicial equation σ
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16.5 POLYNOMIAL SOLUTIONS
is a positive integer or zero, then we are left with a finite polynomial of degree
N = N + σ as a solution of the ODE:
y(z) =
N
an z n+σ .
(16.33)
n=0
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 about z = 0 of
y − 2zy + λy = 0.
(16.34)
For what values of λ does the equation possess a polynomial solution? Find such a solution
for λ = 4.
Clearly
z = 0n is an ordinary point of (16.34) and so we look for solutions2 of the form
y= ∞
n=0 an z . Substituting this into the ODE and multiplying through by z we find
∞
[n(n − 1) − 2z 2 n + λz 2 ]an z n = 0.
n=0
By demanding that the coefficients of each power of z vanish 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
an−2
n(n − 1)
for n ≥ 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 set a1 = 0 and a0 = 1 to obtain
z2
z4
z6
− λ(4 − λ) − λ(4 − λ)(8 − λ) − · · ·
2!
4!
6!
or set a0 = 0 and a1 = 1 to obtain
y1 (z) = 1 − λ
(16.36)
z3
z5
z7
+ (2 − λ)(6 − λ) + (2 − λ)(6 − λ)(10 − λ) + · · · .
3!
5!
7!
Now, from the recurrence relation (16.35) (or in this case from the expressions for y1
and y2 themselves) we see that for the ODE to possess a polynomial solution we require
λ = 2(n − 2) for n ≥ 2 or, more simply, λ = 2n for n ≥ 0, i.e. λ must be an even positive
integer. If λ = 4 then from (16.36) the ODE has the polynomial solution
y2 (z) = z + (2 − λ)
y1 (z) = 1 −
4z 2
= 1 − 2z 2 . 2!
A simpler method of obtaining finite polynomial solutions is to assume a
solution of the form (16.33), where aN = 0. Instead of starting with the lowest
power of z, as we have done up to now, this time we start by considering the
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