Methods | Part IB, 2002

The potential Φ(r,ϑ)\Phi(r, \vartheta), satisfies Laplace's equation everywhere except on a sphere of unit radius and Φ0\Phi \rightarrow 0 as rr \rightarrow \infty. The potential is continuous at r=1r=1, but the derivative of the potential satisfies

limr1+Φrlimr1Φr=Vcos2ϑ\lim _{r \rightarrow 1^{+}} \frac{\partial \Phi}{\partial r}-\lim _{r \rightarrow 1^{-}} \frac{\partial \Phi}{\partial r}=V \cos ^{2} \vartheta

where VV is a constant. Use the method of separation of variables to find Φ\Phi for both r>1r>1 and r<1r<1.

[The Laplacian in spherical polar coordinates for axisymmetric systems is

21r2(rr2r)+1r2sinϑ(ϑsinϑϑ)\nabla^{2} \equiv \frac{1}{r^{2}}\left(\frac{\partial}{\partial r} r^{2} \frac{\partial}{\partial r}\right)+\frac{1}{r^{2} \sin \vartheta}\left(\frac{\partial}{\partial \vartheta} \sin \vartheta \frac{\partial}{\partial \vartheta}\right)

You may assume that the equation

((1x2)y)+λy=0\left(\left(1-x^{2}\right) y^{\prime}\right)^{\prime}+\lambda y=0

has polynomial solutions of degree nn, which are regular at x=±1x=\pm 1, if and only if λ=n(n+1).]\lambda=n(n+1) .]

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