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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: − rd R dr r dR dr = 1 d2 Θ = λ. Θ dθ2 Thus the partial diﬀerential equation divides into two ordinary diﬀerential equations d2 Θ = λΘ, Θ(θ + 2π ) = Θ(θ), dθ2 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 d dR r r − n2 R = 0. dr dr If n = 0, the equation for R becomes d 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.

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