1540470959-Boundary_Value_Problems_and_Partial_Differential_Equations__Powers

4.5 Potential in a Disk 277

Knowing thatλ^2 n=n^2 ,wecaneasilyfindR(r). The equation forRbecomes

r^2 R′′+rR′−n^2 R= 0 , 0 <r<c,

when the indicated differentiations are carried out. This is a Cauchy–Euler
equation, whose solutions are known to have the formR(r)=rα,whereα
is constant. SubstitutingR=rα,R′=αrα−^1 ,andR′′=α(α− 1 )rα−^2 into it
leaves
(
α(α− 1 )+α−n^2

)

rα= 0 , 0 <r<c.

Becauserαis not zero, the constant factor in parentheses must be zero — that
is,α=±n. The general solution of the differential equation is any combina-
tion ofrnandr−n.Thelatter,however,isunboundedasrapproaches zero,
so we discard that solution, retainingRn(r)=rn.Inthespecialcasen=0, the
two solutions are the constant function 1 and ln(r). The logarithm is discarded
because of its behavior atr=0.
Now we reassemble our solution. The functions

r^0 · 1 = 1 , rncos(nθ), rnsin(nθ) (9)

are all solutions of the potential equation, so a general linear combination of
these solutions will also be a solution. Thusv(r,θ)may have the form

v(r,θ)=a 0 +

∑∞

n= 1

anrncos(nθ)+

∑∞

n= 1

bnrnsin(nθ). (10)

At the true boundaryr=c, the boundary condition reads

v(c,θ)=a 0 +

∑∞

n= 1

cn

(

ancos(nθ)+bnsin(nθ)

)

=f(θ ), −π<θ≤π.

This is a Fourier series problem, as in Section 1.1, solved by choosing

a 0 = 21 π

∫π

−π

f(θ )dθ,

an=^1

πcn

∫π

−π

f(θ )cos(nθ)dθ, bn=^1

πcn

∫π

−π

f(θ )sin(nθ)dθ. (11)

Example.
Consider the problem consisting of Eqs. (1)–(4) with

v(c,θ)=f(θ )=

{ 0 , −π<θ<−π/2,

1 , −π/ 2 <θ <π/2,

0 ,π/ 2 <θ <π.