5.10 Some Applications of Legendre Polynomials 345
5.10 Some Applications of Legendre Polynomials
In this section we follow through the details involved in solving some problems
in which Legendre polynomials are used. First, we complete the problem stated
in the previous section.
A. Potential in a Sphere
We consider the axially symmetric potential equation — that is, with no vari-
ationinthelongitudinal-orθ-direction. The unknown functionumight rep-
resent an electrostatic potential, steady-state temperature, etc.
1
ρ^2{∂
∂ρ(
ρ^2∂u
∂ρ)
+
1
sin(φ)∂
∂φ(
sin(φ)∂u
∂φ)}
= 0 ,
0 <ρ<c, 0 <φ<π, (1)
u(c,φ)=f(φ), 0 <φ<π. (2)Of course, the functionuis to be bounded at the singular pointsφ=0,φ=π,
andρ=0. The assumption thatuhas the product form,u(ρ, φ)=(φ)R(ρ),
allows us to transform the partial differential equation into
(ρ^2 R′(r))′
R(r)+(sin(φ)′(φ))′
sin(φ)(φ)= 0.
FromhereweobtainequationsforRandindividually,
(
ρ^2 R′
)′
−μ^2 R= 0 , 0 <ρ<c, (3)
(
sin(φ)′)′
+μ^2 sin(φ)= 0 , 0 <φ<π. (4)
In Section 5.9 we found the eigenfunctions of Eq. (4), subject to the bound-
edness conditions atφ=0andπ,toben(φ)=Pn(cos(φ)), corresponding
to the eigenvaluesμ^2 n=n(n+ 1 ). We must still solve Eq. (3) forR.Afterthe
differentiation has been carried out, the problem forRbecomes
ρ^2 R′′n+ 2 ρR′n−n(n+ 1 )Rn= 0 , 0 <ρ<c,
Rnbounded atρ= 0.The equation is of the Cauchy–Euler type, solved by assumingR=ραand
determiningα. Two solutions,ρnandρ−(n+^1 ), are found, of which the sec-
ond is unbounded atρ=0. HenceRn=ρn, and our product solutions of the
potential equation have the form
un(ρ, φ)=Rn(ρ)n(φ)=ρnPn(
cos(φ)