21.3 SEPARATION OF VARIABLES IN POLAR COORDINATES
As in the two-dimensional case, single-valuedness ofurequires thatmis an
integer. However, in the particular casem= 0 the solution is
Φ(φ)=Cφ+D.
This form is appropriate to a solution with axial symmetry (C= 0) or one that is
multivalued, but manageably so, such as the magnetic scalar potential associated
with a currentI(in which caseC=I/(2π)andDis arbitrary).
Finally, theρ-equation (21.34) may be transformed into Bessel’s equation of
ordermby writingμ=kρ. This has the solution
P(ρ)=AJm(kρ)+BYm(kρ).
The properties of these functions were investigated in chapter 16 and will not
be pursued here. We merely note thatYm(kρ) is singular atρ= 0, and so, when
seeking solutions to Laplace’s equation in cylindrical coordinates within some
region containing theρ= 0 axis, we requireB=0.
The complete separated-variable solution in cylindrical polars of Laplace’s
equation∇^2 u= 0 is thus given by
u(ρ, φ, z)=[AJm(kρ)+BYm(kρ)][Ccosmφ+Dsinmφ][Eexp(−kz)+Fexpkz].
(21.35)
Of course we may use the principle of superposition to build up more general
solutions by adding together solutions of the form (21.35) for all allowed values
of the separation constantskandm.
A semi-infinite solid cylinder of radiusahas its curved surface held at 0 ◦Cand its base
held at a temperatureT 0. Find the steady-state temperature distribution in the cylinder.
The physical situation is shown in figure 21.5. The steady-state temperature distribution
u(ρ, φ, z) must satisfy Laplace’s equation subject to the imposed boundary conditions. Let
us take the cylinder to have its base in thez= 0 plane and to extend along the positive
z-axis. From (21.35), in order thatuis finite everywhere in the cylinder we immediately
requireB=0andF= 0. Furthermore, since the boundary conditions, and hence the
temperature distribution, are axially symmetric, we requirem=0,andsothegeneral
solution must be a superposition of solutions of the formJ 0 (kρ)exp(−kz) for all allowed
values of the separation constantk.
The boundary conditionu(a, φ, z) = 0 restricts the allowed values ofk,sincewemust
haveJ 0 (ka) = 0. The zeros of Bessel functions are given in most books of mathematical
tables, and we find that, to two decimal places,
J 0 (x)=0 forx=2. 40 , 5. 52 , 8. 65 ,....
Writing the allowed values ofkasknforn=1, 2 , 3 ,...(so, for example,k 1 =2. 40 /a), the
required solution takes the form
u(ρ, φ, z)=
∑∞
n=1
AnJ 0 (knρ)exp(−knz).