= Solution
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:
$$
\boxed{r^2R''+2rR'=\lambda R,\qquad(\sin\theta\,\Theta')'=-\lambda\sin\theta\,\Theta.}
$$
The resulting differential identities extend across isolated zeros by <continuity>. Put $\rho=\cos\theta$. Then $\Theta'=-\sin\theta P'$ and
$$
(\sin\theta\,\Theta')'=\sin\theta\frac d{d\rho}\bigl((1-\rho^2)P'\bigr).
$$
This gives $((1-\rho^2)P')'+\lambda P=0$ on $(-1,1)$.
A nonzero angular factor must be regular at both poles: it must be bounded as $\rho\to\pm1$ 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 $\boxed{\lambda=n(n+1),\ n=0,1,2,\ldots}$ 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 verifies
$$
\begin{array}{c|c|c}
P&\lambda&R(r)\\\hline
1&0&A+B/r\\
\rho&2&Ar+Br^{-2}\\
(3\rho^2-1)/2&6&Ar^2+Br^{-3}
\end{array}
$$
Indeed $R=r^m$ 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 $\rho^2=\tfrac13P_0(\rho)+\tfrac23P_2(\rho)$. Normalize each <radial function>[radial solution] to vanish at $1$ and equal $1$ at $a$; <superposition> gives
$$
\boxed{u(r,\theta)=\frac13\frac{1-r^{-1}}{1-a^{-1}}+\frac23\frac{r^2-r^{-3}}{a^2-a^{-3}}P_2(\cos\theta).}
$$
Every 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 \b[the classical <Dirichlet problem>[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.
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