Past exam of the algebra, analysis and differential equations II course of the University of Oxford / 2010 / Differential Equations / 9

For an azimuth-independent product $u=R(r)\Theta (\theta )$, multiply Laplace's equation by $r^2/(R\Theta )$ where the factors are nonzero. The radial and angular expressions must equal opposite constants:The resulting differential identities extend across isolated zeros by continuity. Put $\rho =\cos \theta$. Then ${\Theta }^{\prime }=-\sin \theta P^{\prime }$ andThis gives $((1-{\rho }^2)P^{\prime }{)}^{\prime }+\lambda P=0$ on $(-1,1)$.

A nonzero angular factor must be regular at both poles: it must be bounded as $\rho \to \pm 1$ and extend smoothly as an axisymmetric function on the sphere. Equivalently one may use the regular endpoint solution at each end; the logarithmic endpoint solutions are excluded. These two endpoint conditions select $\boxed{\lambda =n(n+1),n=0,1,2,\dots }$ and the multiples of $P_n$. It is not sufficient to impose no condition at the endpoints, since local solutions exist for arbitrary $\lambda$. One way to see the quantization is the regular Legendre series about one endpoint: unless it terminates at $\lambda =n(n+1)$, continuation to the other endpoint has a logarithmic singularity.

Direct differentiation verifiesIndeed $R=r^m$ gives $m(m+1)=\lambda$, so the two powers for degree $n$ are $r^n$ and $r^{-n-1}$.

For the shell assume $a>1$. The outer datum decomposes as ${\rho }^2=\frac{1}{3}{P}_0(\rho )+\frac{2}{3}{P}_2(\rho )$. Normalize each radial solution to vanish at $1$ and equal $1$ at $a$; superposition givesEvery term is harmonic in the shell and regular at the angular poles; substitution at each boundary gives precisely the two prescribed data. For uniqueness among functions harmonic in the shell and continuous on its closure, subtract any two solutions. Their difference is harmonic, zero on both boundary spheres, and by the maximum principle is both at most zero and at least zero on the bounded shell. It vanishes identically. Thus the classical Dirichlet solution is unique. The usual continuity at the boundary is the interpretation of the stated boundary conditions; allowing singular boundary behavior would remove that hypothesis.