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15.2 Legendre Polynomials 519

known as Legendre polynomials in the 1780s while continuing Laplace’s work on the potential
equation (Chapter 18).
2
Legendre’s differential equation is in Sturm-Liouville form, with r(x) = 1 − x , q(x) = 0
and p(x) = 1. Since r(−1) = r(1) = 0, Legendre’s equation forms a singular Sturm-Liouville
problem on [−1,1], with no boundary conditions. However, we seek solutions that are bounded
on this interval.
∞ n

To find such solutions, attempt a power series solution y = a n x expanded about 0 (see
n=0
Section 4.1). Substitute this series into Legendre’s equation to obtain
∞ ∞ ∞ ∞

n
n
n
n(n − 1)a n x n−2 − n(n − 1)a n x − 2na n x + λa n x = 0.
n=2 n=2 n=0 n=0
Rewrite the first summation as
∞ ∞
n−2 n
n(n − 1)a n x = (n + 2)(n + 1)a n+2 x
n=2 n=0
to obtain
∞ ∞ ∞ ∞

n n n n
(n + 2)(n + 1)a n+2 x − n(n − 1)a n x − 2na n x + λa n x = 0.
n=0 n=2 n=0 n=0
Write this as
2a 2 + 6a 3 x − 2a 1 x + λa 0 + λa 1 x
∞

2
n
+ [(n + 2)(n + 1)a n+2 − (n + n − λ)a n ]x = 0.
n=2
The constant term and the coefficient of each power of x on the left must be zero, so
2a 2 + λa 0 = 0
6a 3 − 2a 1 + λa 1 = 0, and
(n + 1)(n + 2)a n+2 −[n(n + 1) − λ]a n = 0for n = 2,3,··· .

From these equations, we obtain
λ
a 2 =− a 0 ,
2
2 − λ 2 − λ
a 3 = a 1 = a 1
6 3!
and
n(n + 1) − λ
a n+2 = a n for n = 2,3,··· . (15.5)
(n + 1)(n + 2)
Equation (15.5) is a recurrence relation which gives each a n+2 in terms of λ and a n . This enables
us to produce the coefficients in turn from previously found coefficients. We already have expres-
sions for a 2 and a 3 in terms of λ, a 0 and a 1 . From the recurrence relation of equation (15.5) with
n = 2,
6 − λ λ 6 − λ −λ(6 − λ)
a 4 = a 2 =− a 0 = a 0 .
(3)(4) 2 (3)(4) 4!

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October 14, 2010 15:20 THM/NEIL Page-519 27410_15_ch15_p505-562