Mathematical Methods for Physics and Engineering : A Comprehensive Guide

PDES: SEPARATION OF VARIABLES AND OTHER METHODS

This can be rearranged to give

r^2 R′′ 1 +2rR 1 ′− 2 R 1 =−

Ar^3

 0

,

where the prime denotes differentiation with respect tor. The LHS is homogeneous and
the equation can be reduced by the substitutionr=expt, and writingR 1 (r)=S(t), to

̈S+S ̇− 2 S=−A

 0

exp 3t, (21.68)

where the dots indicate differentiation with respect tot.
This is an inhomogeneous second-order ODE with constant coefficients and can be
straightforwardly solved by the methods of subsection 15.2.1 to give

S(t)=c 1 expt+c 2 exp(− 2 t)−

A

10  0

exp 3t.

Recalling thatr=exptwe find

R 1 (r)=c 1 r+c 2 r−^2 −

A

10  0

r^3.

Since we are interested in the regionr<awe must havec 2 = 0 for the solution to remain
finite. Thus inside the charge distribution the electrostatic potential has the form

u 1 (r, θ, φ)=

(

c 1 r−

A

10  0

r^3

)

P 1 (cosθ). (21.69)

Outside the charge distribution (forr≥a), however, the electrostatic potential obeys
Laplace’s equation,∇^2 u= 0, and so given the symmetry of the problem and the requirement
thatu→∞asr→∞the solution must take the form

u 2 (r, θ, φ)=

∑∞

=0

B

r+1

P(cosθ). (21.70)

We can now use the boundary conditions atr=ato fix the constants in (21.69) and
(21.70). The requirement of continuity of the potential and its radial derivative atr=a
imply that

u 1 (a, θ, φ)=u 2 (a, θ, φ),

∂u 1

∂r

(a, θ, φ)=

∂u 2

∂r

(a, θ, φ).

ClearlyB=0for= 1; carrying out the necessary differentiations and settingr=ain
(21.69) and (21.70) we obtain the simultaneous equations

c 1 a−

A

10  0

a^3 =

B 1

a^2

,

c 1 −

3 A

10  0

a^2 =−

2 B 1

a^3

,

which may be solved to givec 1 =Aa^2 /(6 0 )andB 1 =Aa^5 /(15 0 ). SinceP 1 (cosθ)=cosθ,
the electrostatic potentials inside and outside the charge distribution are given, respectively,
by

u 1 (r, θ, φ)=

A

 0

(

a^2 r

6

−

r^3

10

)

cosθ, u 2 (r, θ, φ)=

Aa^5

15  0

cosθ

r^2

.