PDES: SEPARATION OF VARIABLES AND OTHER METHODS
Laplace’s equation in cylindrical polarsPassing to three dimensions, we now consider the solution of Laplace’s equation
in cylindrical polar coordinates,
1
ρ∂
∂ρ(
ρ∂u
∂ρ)
+1
ρ^2∂^2 u
∂φ^2+∂^2 u
∂z^2=0. (21.32)We note here that, even when considering a cylindrical physical system, if there
is no dependence of the physical variables onz(i.e. along the length of the
cylinder) then the problem may be treated using two-dimensional plane polars,
as discussed above.
For the more general case, however, we proceed as previously by trying asolution of the form
u(ρ, φ, z)=P(ρ)Φ(φ)Z(z),which, on substitution into (21.32) and division through byu=PΦZ, gives
1
Pρd
dρ(
ρdP
dρ)
+1
Φρ^2d^2 Φ
dφ^2+1
Zd^2 Z
dz^2=0.The last term depends only onz, and the first and second (taken together) depend
only onρandφ. Taking the separation constant to bek^2 , we find
1
Zd^2 Z
dz^2=k^2 ,1
Pρd
dρ(
ρdP
dρ)
+1
Φρ^2d^2 Φ
dφ^2+k^2 =0.The first of these equations has the straightforward solutionZ(z)=Eexp(−kz)+Fexpkz.Multiplying the second equation through byρ^2 ,weobtain
ρ
Pd
dρ(
ρdP
dρ)
+1
Φd^2 Φ
dφ^2+k^2 ρ^2 =0,in which the second term depends only on Φ and the other terms depend only
onρ. Taking the second separation constant to bem^2 , we find
1
Φd^2 Φ
dφ^2=−m^2 , (21.33)ρd
dρ(
ρdP
dρ)
+(k^2 ρ^2 −m^2 )P=0. (21.34)The equation in the azimuthal angleφhas the very familiar solutionΦ(φ)=Ccosmφ+Dsinmφ.