Today, we consider those PDEs with Laplacian in spherical coordinate.

$$ \Delta u = \frac{1}{r^2} \left( \frac{\d}{\d r} r^2 \frac{\d u}{\d r} \right) + \frac{1}{r^2} \frac{1}{\sin \theta} \frac{\d }{\d \theta} ( \sin \theta \frac{\d u}{\d \theta}) + \frac{1}{r^2 \sin^2 \theta} \frac{\d^2 u}{\d \phi^2}. $$

Notice that $\theta$ and $\phi$ varies in a bounded domain, hence the eigenvalue problems for these variables have discrete eigenvalues.

We look for eigenfunctions of the form $$ u (r, \theta, \phi) = R® \Theta(\theta) \Phi(\phi). $$ Then $ \Delta u = \lambda u$ is equivalent to $$ \lambda = \frac{1}{u} \Delta u = \frac{1}{R} \frac{1}{r^2} \left( \frac{\d}{\d r} r^2 \frac{\d R}{\d r} \right) + \frac{1}{\Theta} \frac{1}{r^2} \frac{1}{\sin \theta} \frac{\d }{\d \theta} ( \sin \theta \frac{\d \Theta }{\d \theta}) + \frac{1}{\Phi } \frac{1}{r^2 \sin^2 \theta} \frac{\d^2 \Phi}{\d \phi^2}. $$ which in turns breaks into equations $$ \frac{\d^2 \Phi}{\d \phi^2} = \lambda_\phi \Phi $$ $$ \frac{1}{\sin \theta} \frac{\d }{\d \theta} ( \sin \theta \frac{\d \Theta}{\d \theta}) + \frac{\lambda_\phi}{\sin^2 \theta} \Theta = \lambda_\theta \Theta $$ $$ \frac{1}{R} \frac{1}{r^2} \left( \frac{\d}{\d r} r^2 \frac{\d R}{\d r} \right) + \frac{\lambda_\theta}{r^2} = \lambda_r \quad \Rightarrow \left( \frac{\d}{\d r} r^2 \frac{\d R}{\d r} \right) + (\lambda_\theta - \lambda_r r^2) R = 0 $$ The total eigenvalue $\lambda = \lambda_r$.

The equation about $\Phi$ has $\sin(m\phi), \cos(m\phi)$ as eigenfunctions, with eigenvalues $\lambda_\phi = -m^2$.

The equation about $\Theta$ has associated Legendre polynomial as solution, recall that $y(\theta) = P_l^m (\cos \theta)$ solves equation (Problem 12.10.2) $$ \frac{1}{\sin \theta} \frac{\d }{\d \theta} ( \sin \theta \frac{\d y}{\d \theta}) + \left( l(l+1) - \frac{m^2}{\sin^2 \theta} \right) y = 0 $$ By comparison, we get eigenvalues $\lambda_\theta = -l (l+1)$.

The equation about $R$ is related to the spherical Bessel function's equation. Recall $y® = j_n®, y_n®$ solves $$ \left( \frac{\d}{\d r} r^2 \frac{\d y}{\d r} \right) + (r^2 - n(n+1) ) y = 0 $$ Hence, for $\lambda_\theta = l(l+1)$, we can solve $R®$ to get $$ R® = \begin{cases} j_l( k r), y_l(kr) & \lambda_r = -k^2 \cr j_l( i k r), y_l(i kr) & \lambda_r = k^2 \cr r^l, r^{-l-1} & \lambda_r=0 \end{cases} $$ Usually, we require $R(0)$ to be finite, and $y_l® \sim r^{-l-1}$ as $r \to 0$, hence we only keep the $j_l(kr), j_l(ikr), r^l$ solution above. The case $\lambda_r$ can be obtained as the leading order behavior of $j_l(kr)$ after rescaling by a constant $$ r^l = \lim_{k \to 0} k^{-l} j_l( k r). $$

In some cases, suppose we solve the equation in the exterior of a ball and require the solution to decay as $r \to \infty$, then we will not ask $R(0)$ to be finite, instead we require $R(\infty) = 0$. Then the $y_l(kr), y_l(ikr), r^{-l-1}$ solutions will come into play.

The general eigenfunction is $$u(r,\theta,\phi) = \begin{cases} j_l(kr) \cr y_l(kr) \end{cases} P_l^\cos \theta) \begin{cases} \cos(m \phi) \cr \sin(m \phi) \end{cases}, \quad \lambda = -k^2 $$

We solve $ \Delta u = 0$ for $r \leq 1$ with $u(r=1, \theta, \phi) = f(\theta,\phi)$. We may first decompose $f(\theta, \phi)$ as $$ f(\theta, \phi) = \sum_{l=0}^\infty \sum_{m=0}^l [a_{l m} \cos(m\phi) + b_{l m} \sin(m \phi) ] P_l^\cos \theta) $$ with $b_{l 0}=0$. Then the solulution for $u$ is obtained by setting $\lambda = \lambda_r=0$ $$ u(r, \theta, \phi) = \sum_{l=0}^\infty \sum_{m=0}^l r^l [a_{l m} \cos(m\phi) + b_{l m} \sin(m \phi) ] P_l^\cos \theta). $$

To obtain the coefficients $a_{lm}, b_{lm}$ from $f(\theta, \phi)$, we uses orthogonality of these functions $P_l^\cos \theta) \cos(m \phi), P_l^\cos \theta) \sin(m \phi)$ on the two sphere with volume form $\sin \theta d\theta d\phi$. For example, we claim that, if $(l, m) \neq (l', m')$, then $$ \int_{\phi = 0}^{2\pi} \int_{\theta=0}^\pi P_l^\cos \theta) \cos(m \phi) P_{l'}^{m'}(\cos \theta) \cos(m' \phi) \sin \theta d\theta d\phi = 0$$ Indeed, integrating $d\phi$, we see that if $m \neq m'$ the result is zero; if $m=m'$ but $l \neq l'$, then we use the orthogonality of associated Legendre functions (see section 12.10 of Boas), to show that the integral is zero.