Series solutions about an ordinary point

16.2 SERIES SOLUTIONS ABOUT AN ORDINARY POINT
Equation
Hypergeometric
z(1 − z)y + [c − (a + b + 1)z]y − aby = 0
Regular
singularities
Essential
singularities
0, 1, ∞
—
Legendre
(1 − z 2 )y − 2zy + ( + 1)y = 0
−1, 1, ∞
—
Associated Legendre (1 − z 2 )y − 2zy + ( + 1) −
−1, 1, ∞
—
m2
y=0
2
1−z
Chebyshev
(1 − z 2 )y − zy + ν 2 y = 0
−1, 1, ∞
—
Confluent hypergeometric
zy + (c − z)y − ay = 0
0
∞
Bessel
z 2 y + zy + (z 2 − ν 2 )y = 0
0
∞
Laguerre
zy + (1 − z)y + νy = 0
0
∞
Associated Laguerre
zy + (m + 1 − z)y + (ν − m)y = 0
0
∞
Hermite
y − 2zy + 2νy = 0
—
∞
Simple harmonic oscillator
y + ω 2 y = 0
—
∞
Table 16.1 Important second-order linear ODEs in the physical sciences and
engineering.
Table 16.1 lists the singular points of several second-order linear ODEs that
play important roles in the analysis of many problems in physics and engineering.
A full discussion of the solutions to each of the equations in table 16.1 and their
properties is left until chapter 18. We now discuss the general methods by which
series solutions may be obtained.
16.2 Series solutions about an ordinary point
If z = z0 is an ordinary point of (16.7) then it may be shown that every solution
y(z) of the equation is also analytic at z = z0 . From now on we will take z0 as the
origin, i.e. z0 = 0. If this is not already the case, then a substitution Z = z − z0
will make it so. Since every solution is analytic, y(z) can be represented by a
535
SERIES SOLUTIONS OF ORDINARY DIFFERENTIAL EQUATIONS
power series of the form (see section 24.11)
y(z) =
∞
an z n .
(16.9)
n=0
Moreover, it may be shown that such a power series converges for |z| < R, where
R is the radius of convergence and is equal to the distance from z = 0 to the
nearest singular point of the ODE (see chapter 24). At the radius of convergence,
however, the series may or may not converge (as shown in section 4.5).
Since every solution of (16.7) is analytic at an ordinary point, it is always
possible to obtain two independent solutions (from which the general solution
(16.2) can be constructed) of the form (16.9). The derivatives of y with respect to
z are given by
y =
y =
∞
n=0
∞
nan z n−1 =
∞
(n + 1)an+1 z n ,
(16.10)
n=0
n(n − 1)an z n−2 =
∞
n=0
(n + 2)(n + 1)an+2 z n .
(16.11)
n=0
Note that, in each case, in the first equality the sum can still start at n = 0 since
the first term in (16.10) and the first two terms in (16.11) are automatically zero.
The second equality in each case is obtained by shifting the summation index
so that the sum can be written in terms of coefficients of z n . By substituting
(16.9)–(16.11) into the ODE (16.7), and requiring that the coefficients of each
power of z sum to zero, we obtain a recurrence relation expressing each an in
terms of the previous ar (0 ≤ r ≤ n − 1).
Find the series solutions, about z = 0, of
y + y = 0.
By inspection, z = 0 is an ordinary point of the equation,
and so we may obtain two
n
independent solutions by making the substitution y = ∞
n=0 an z . Using (16.9) and (16.11)
we find
∞
∞
(n + 2)(n + 1)an+2 z n +
an z n = 0,
n=0
n=0
which may be written as
∞
[(n + 2)(n + 1)an+2 + an ]z n = 0.
n=0
For this equation to be satisfied we require that the coefficient of each power of z vanishes
separately, and so we obtain the two-term recurrence relation
an
an+2 = −
for n ≥ 0.
(n + 2)(n + 1)
Using this relation, we can calculate, say, the even coefficients a2 , a4 , a6 and so on, for
536
16.2 SERIES SOLUTIONS ABOUT AN ORDINARY POINT
a given a0 . Alternatively, starting with a1 , we obtain the odd coefficients a3 , a5 , etc. Two
independent solutions of the ODE can be obtained by setting either a0 = 0 or a1 = 0.
Firstly, if we set a1 = 0 and choose a0 = 1 then we obtain the solution
(−1)n
z2
z4
+
− ··· =
z 2n .
2! 4!
(2n)!
n=0
∞
y1 (z) = 1 −
Secondly, if we set a0 = 0 and choose a1 = 1 then we obtain a second, independent, solution
(−1)n
z3
z5
+
− ··· =
z 2n+1 .
3! 5!
(2n + 1)!
n=0
∞
y2 (z) = z −
Recognising these two series as cos z and sin z, we can write the general solution as
y(z) = c1 cos z + c2 sin z,
where c1 and c2 are arbitrary constants that are fixed by boundary conditions (if supplied).
We note that both solutions converge for all z, as might be expected since the ODE
possesses no singular points (except |z| → ∞). Solving the above example was quite straightforward and the resulting series
were easily recognised and written in closed form (i.e. in terms of elementary
functions); this is not usually the case. Another simplifying feature of the previous
example was that we obtained a two-term recurrence relation relating an+2 and
an , so that the odd- and even-numbered coefficients were independent of one
another. In general, the recurrence relation expresses an in terms of any number
of the previous ar (0 ≤ r ≤ n − 1).
Find the series solutions, about z = 0, of
y −
2
y = 0.
(1 − z)2
By inspection, z = 0 is an ordinary
point, and therefore we may find two independent
n
solutions by substituting y = ∞
n=0 an z . Using (16.10) and (16.11), and multiplying through
2
by (1 − z) , we find
(1 − 2z + z 2 )
∞
n(n − 1)an z n−2 − 2
∞
n=0
an z n = 0,
n=0
which leads to
∞
∞
∞
∞
n(n − 1)an z n−2 − 2
n(n − 1)an z n−1 +
n(n − 1)an z n − 2
an z n = 0.
n=0
n=0
n=0
n=0
In order to write all these series in terms of the coefficients of z n , we must shift the
summation index in the first two sums, obtaining
∞
(n + 2)(n + 1)an+2 z n − 2
n=0
∞
(n + 1)nan+1 z n +
n=0
∞
(n2 − n − 2)an z n = 0,
n=0
which can be written as
∞
(n + 1)[(n + 2)an+2 − 2nan+1 + (n − 2)an ]z n = 0.
n=0
537