Mathematical Methods for Physics and Engineering : A Comprehensive Guide

(Darren Dugan) #1

21.3 SEPARATION OF VARIABLES IN POLAR COORDINATES


and the coefficientsAnmay be expressed as


An=

2 T 0


a^2 J^21 (kna)

[


aJ 1 (kna)
kn

]


=


2 T 0


knaJ 1 (kna)

.


The steady-state temperature in the cylinder is then given by

u(ρ, φ, z)=

∑∞


n=1

2 T 0


knaJ 1 (kna)

J 0 (knρ)exp(−knz).

We note that if, in the above example, the base of the cylinder were not kept at

a uniform temperatureT 0 , but instead had some fixed temperature distribution


T(ρ, φ), then the solution of the problem would become more complicated. In


such a case, the required temperature distributionu(ρ, φ, z)isingeneralnotaxially


symmetric, and so the separation constantmis not restricted to be zero but may


take any integer value. The solution will then take the form


u(ρ, φ, z)=

∑∞

m=0

∑∞

n=1

Jm(knmρ)(Cnmcosmφ+Dnmsinmφ) exp(−knmz),

where the separation constantsknmare such thatJm(knma) = 0, i.e.knmais thenth


zero of themth-order Bessel function. At the base of the cylinder we would then


require


u(ρ, φ,0) =

∑∞

m=0

∑∞

n=1

Jm(knmρ)(Cnmcosmφ+Dnmsinmφ)=T(ρ, φ).
(21.37)

The coefficientsCnmcould be found by multiplying (21.37) byJq(krqρ)cosqφ,


integrating with respect toρandφover the base of the cylinder and exploiting


the orthogonality of the Bessel functions and of the trigonometric functions. The


Dnmcould be found in a similar way by multiplying (21.37) byJq(krqρ)sinqφ.


Laplace’s equation in spherical polars

We now come to an equation that is very widely applicable in physical science,


namely∇^2 u= 0 in spherical polar coordinates:


1
r^2


∂r

(
r^2

∂u
∂r

)
+

1
r^2 sinθ


∂θ

(
sinθ

∂u
∂θ

)
+

1
r^2 sin^2 θ

∂^2 u
∂φ^2

=0. (21.38)

Our method of procedure will be as before; we try a solution of the form

u(r, θ, φ)=R(r)Θ(θ)Φ(φ).

Substituting this in (21.38), dividing through byu=RΘΦ and multiplying byr^2 ,


we obtain


1
R

d
dr

(
r^2

dR
dr

)
+

1
Θsinθ

d

(
sinθ



)
+

1
Φsin^2 θ

d^2 Φ
dφ^2

=0. (21.39)
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