Mathematical Methods for Physics and Engineering : A Comprehensive Guide

(Darren Dugan) #1

21.3 SEPARATION OF VARIABLES IN POLAR COORDINATES


where the coefficientsR(r)andF(r) in the Legendre polynomial expansions


are functions ofr. Since in any particular problemρis given, we can find the


coefficientsF(r) in the expansion in the usual way (see subsection 18.1.2). It then


only remains to find the coefficientsR(r) in the expansion of the solutionu.


Writing∇^2 in spherical polars and substituting (21.64) and (21.65) into (21.63)

we obtain


∑∞


=0

[
P(cosθ)
r^2

d
dr

(
r^2

dR
dr

)
+

R
r^2 sinθ

d

(
sinθ

dP(cosθ)

)]
=

∑∞

=0

F(r)P(cosθ).

(21.66)

However, if, in equation (21.44) of our discussion of the angular part of the


solution to Laplace’s equation, we setm= 0 we conclude that


1
sinθ

d

(
sinθ

dP(cosθ)

)
=−(+1)P(cosθ).

Substituting this into (21.66), we find that the LHS is greatly simplified and we


obtain


∑∞

=0

[
1
r^2

d
dr

(
r^2

dR
dr

)

(+1)R
r^2

]
P(cosθ)=

∑∞

=0

F(r)P(cosθ).

This relation is most easily satisfied by equating terms on both sides for each


value ofseparately, so that for=0, 1 , 2 ,...we have


1
r^2

d
dr

(
r^2

dR
dr

)

(+1)R
r^2

=F(r). (21.67)

This is an ODE in whichF(r) is given, and it can therefore be solved for


R(r). The solution to Poisson’s equation,u, is then obtained by making the


superposition (21.64).


In a certain system, the electric charge densityρis distributed as follows:

ρ=

{


Arcosθ for 0 ≤r<a,
0 forr≥a.

Find the electrostatic potential insideand outside the charge distribution, given that both
the potential and its radial derivative are continuous everywhere.

The electrostatic potentialusatisfies


∇^2 u=

{


−(A/ 0 )rcosθ for 0≤r<a,
0forr≥a.

Forr<athe RHS can be written−(A/ 0 )rP 1 (cosθ), and the coefficients in (21.65) are
simplyF 1 (r)=−(Ar/ 0 )andF(r)=0for= 1. Therefore we need only calculateR 1 (r),
which satisfies (21.67) for=1:


1
r^2

d
dr

(


r^2

dR 1
dr

)



2 R 1


r^2

=−


Ar
 0

.

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