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Unformatted text preview: oes not depend on r. Thus neither side can depend on either θ or r, and
hence both sides must be constant:
R dr r dR
dr = 1 d2 Θ
Θ dθ2 Thus the partial diﬀerential equation divides into two ordinary diﬀerential
= λΘ, Θ(θ + 2π ) = Θ(θ),
138 r d
dr r dR
dr = −λR. We have seen the ﬁrst of these equations before when we studied heat ﬂow in a
circular wire, and we recognize that with the periodic boundary conditions, the
only nontrivial solutions are
λ = 0, Θ= a0
2 where a0 is a constant, and
λ = −n2 , Θ = an cos nθ + bn sin nθ, where an and bn are constants, for n a positive integer. Substitution into the
second equation yields
− n2 R = 0.
If n = 0, the equation for R becomes
dr r dR
dr = 0, which is easily solved to yield
R(r) = A + B log r,
where A and B are constants of integration. In order for this equation to be wellbehaved as r → 0 we must have B = 0, and the solution u0 (r, θ) to Laplace’s
equation in this case is constant.
When n = 0, the equation fo...
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This document was uploaded on 01/12/2014.
- Winter '14