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18.7 Steady-State Equation for a Sphere 657

Then
d d
2

[
(ϕ)sin(ϕ) =− [(1 − x )G (x)]
dϕ dϕ
d dx
2

=− [(1 − x )G (x)]
dx dϕ
d
2

=− [(1 − x )G (x)][−sin(ϕ)].
dx
We conclude that
1 d d
2
[
(ϕ)sin(ϕ)]= [(1 − x )G (x)].
sin(ϕ) dϕ dx
The point to this calculation is that equation (18.5) transforms to
2

[(1 − x )G (x)] + λG(x) = 0,
which is Legendre’s differential equation, considered on the interval [−1,1]. In Section 15.2, we
found the eigenvalues λ n =n(n +1) for n =0,1,2,···. The eigenfunctions are constant multiples
of the Legendre polynomials P n (x). For nonnegative integers n, we therefore have a solution of
the differential equation for
:

n (ϕ) = G(cos(ϕ)) = P n (cos(ϕ)).
Now that we know the eigenvalues, the differential equation for X is
2

ρ X + 2ρX − n(n + 1)X = 0.
This is an Euler equation with general solution
n
X(ρ) = aρ + bρ −n−1 .
Choose b = 0 to have a solution that is bounded as ρ → 0+, which is the center of the sphere.
For each nonnegative integer n, we now have a function
n
u n (ρ,ϕ) = a n ρ P n (cos(ϕ))
that satisfies the steady-state heat equation. To satisfy the boundary condition, write a superposi-
tion
∞
n
u(ρ,ϕ) = a n ρ P n (cos(ϕ)).
n=0
We must choose the coefficients to satisfy

∞
n
u(R,ϕ) = a n R P n cos(ϕ).
n=0
To put this into the context of an eigenfunction expansion in terms of Legendre polynomials,
recall that ϕ(x) = arccos(x) to obtain
∞
n
a n R P n (x) = f (arccos(x)).
n=0
Then

2n + 1 1
n
a n R = f (arccos(x))P n (x)dx
2 −1

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October 14, 2010 15:27 THM/NEIL Page-657 27410_18_ch18_p641-666