TITLE: Eigenfunctions of the Laplacian
QUESTION [8 upvotes]: I am willing to offer a bounty for this one, so I will give you an exact idea of what I need:
I am looking for solutions of $$\Delta \Psi(r,\theta)=k^2\Psi(r,\theta)$$
where $k\in \mathbb{R}$.
Such that: $\Psi(r,\theta)=0$ for r approaching infinity. $\partial_r \Psi(r,\theta)=0$ for r approaching infinity and at some $R \in \mathbb{R}_{>0}$ we want to have $$\Psi(R,\theta)=\sum_{l=0}^{\infty} f(l)\frac{1}{R} P_l(\cos(\theta))$$
where $P_l$ is the l-th Legendre Polynomial and $f(l)=\frac{1}{2l+1}(P_{l+1}(\cos(\alpha)-P_{l-1}(\cos(\alpha))$. 
I am only interested in solutions on $$\mathbb{R}^3\backslash B(0,R)$$.
Actually, you do not need to proof that you have found a solution. It would be totally sufficient to find a solution. It might be interesting to know that when we only have $r$ dependence, then the solution is given by something like $\frac{A e^{-r}}{r}$, where A is some constant.
This is the big task, now if you have troubles with the form of $\Psi(R,\theta)$, I am also interested in a solution, where you can replace the one given above by: $\Psi(R,\theta)=CR\cos(\theta)$, where $C$ is a negative constant. (The other conditions (like $Psi$ goes to zero when r approaching infinity, remain all the same).
The first one with an answer to any of both boundary conditions will get the full bounty.

REPLY [5 votes]: The eigenfunctions can be found by separation of variables, 
and are of the form 
$${\psi}_{l m}(r,\theta,\phi) =
    \left\{
        \begin{array}[c]{l}
        {i_l}(k r) \\
        {k_l}(k r)
        \end{array}
    \right\}
    \left\{
        \begin{array}[c]{l}
        P_{l}^{m}(\cos \theta) \\
        Q_{l}^{m}(\cos \theta)
        \end{array}
    \right\}
    \{ e^{\pm i m\phi} \},$$
where the $i$s and $k$s are modified spherical Bessel functions.
($i_l(z) = \sqrt{\pi/(2z)}I_{l+1/2}(z)$ and similarly for $k$, where $I$ and $K$ are the modified Bessel functions.)
The $P$s and $Q$s are Legendre functions.
$\{e^{\pm i m\phi}\}$ should be read, for example, as "a linear combination of $e^{i m\phi}$ and $e^{-i m\phi}$."
We are interested in solutions with the following properties: They 

vanish at infinity; 
are well-behaved functions of $\theta$; 
have no $\phi$ dependence. 

Asymptotically, $i_l(z) \sim e^z/(2z)$ and $k_l(z)\sim \pi e^{-z}/(2z)$. 
In addition, $Q_l(\cos\theta)$ diverges for $\theta = n\pi$, $n\in\mathbb{N}$. 
Thus, the solutions are of the form 
$$\psi_l(r,\theta) = a_l k_l(k r) P_l(\cos\theta),$$
where $a_l$ is some constant dependent only on $l$. 
It can be shown that 
$\lim_{r\to\infty}\partial \psi_l(r,\theta)/\partial r = 0$, 
so we need only match the remaining boundary condition. 
The function of interest is a linear combination of the $\psi_l$s, 
$$\Psi(r,\theta) = \sum_{l=0}^\infty a_l k_l(k r) P_l(\cos\theta).$$
Then 
\begin{eqnarray*}
\Psi(R,\theta) &=& \sum_{l=0}^\infty a_l k_l(k R) P_l(\cos\theta) \\
&=& \sum_{l=0}^\infty \frac{f(l)}{R}P_l(\cos\theta)
\end{eqnarray*}
and so 
$$a_l = \frac{f(l)}{R \, k_l(k R)}.$$
The second boundary condition
To satisfy the boundary condition $\Psi(R,\theta) = C R\cos\theta$ we need only the $l=1$ term, since $P_1(\cos\theta)=\cos\theta$. 
We find 
$$\Psi(r,\theta) = \frac{C R k_1(k r)\cos\theta}{k_1(k R)}  
= \frac{C R^3(1+k r)e^{-k(r-R)}\cos\theta}{r^2(1+k R)}.$$
Derivation of eigenfunctions
In spherical coordinates we have 
$$\frac{1}{r^2}(r^2 \psi_r)_r + \frac{1}{r^2\sin\theta}(\psi_\theta \sin\theta)_\theta + \frac{1}{r^2\sin^2\theta}\psi_{\phi\phi} - k^2\psi = 0,$$
where subscripts denote partial differentiation. 
Letting $\psi(r,\theta,\phi) = R(r)\Theta(\theta)\Phi(\phi)$ 
and multiplying by $r^2\sin^2\theta/(R\Theta\Phi)$ we find 
$$\frac{\sin^2\theta}{R}(r^2 R')' + \frac{\sin\theta}{\Theta}(\Theta'\sin\theta)' + \frac{1}{\Phi}\Phi'' - k^2r^2\sin^2\theta = 0.$$
Since $\Phi''/\Phi$ depends only on $\phi$ and the other terms on $r$ and $\theta$, $\Phi''/\Phi$ must be constant. 
We are looking for oscillatory solutions in $\phi$. 
Thus, 
$$\begin{equation*}
\Phi''/\Phi = -m^2,\tag{1}
\end{equation*}$$
so 
$$\Phi(\phi) = \left\{e^{\pm i m\phi}\right\}.$$
Now we must solve 
$$\frac{1}{R}(r^2 R')' + \frac{1}{\Theta\sin\theta}(\Theta'\sin\theta)' -\frac{m^2}{\sin^2\theta} - k^2r^2 = 0.$$
The first and fourth term depend on $r$ and the rest on $\theta$, 
therefore, 
$$\frac{1}{R}(r^2 R')' - k^2r^2 = l(l+1)$$
for some constant $l$. 
Rewriting this we find 
$$\begin{equation*}
\frac{1}{r^2}(r^2 R')' - \left(k^2 + \frac{l(l+1)}{r^2}\right)R = 0.\tag{2}
\end{equation*}$$
This is the modified spherical Bessel differential equation with solutions
$$R(r) =     
    \left\{
        \begin{array}[c]{l}
        {i_l}(k r) \\
        {k_l}(k r)
        \end{array}
    \right\}.$$
Lastly we must solve 
$$\frac{1}{\Theta\sin\theta}(\Theta'\sin\theta)' -\frac{m^2}{\sin^2\theta} +l(l+1) = 0.$$
Letting $x = \cos\theta$ we find the associated Legendre differential equation, 
$$\begin{equation*}
(1-x^2)\Theta_{xx} - 2x \Theta_x + \left(l(l+1)-\frac{m^2}{1-x^2}\right)\Theta = 0,\tag{3}
\end{equation*}$$
with solutions 
$P_l^m(x)$ and $Q_l^m(x)$, the associated Legendre functions. 
We have 
$$\Theta(\theta) =    
    \left\{
        \begin{array}[c]{l}
        P_{l}^{m}(\cos \theta) \\
        Q_{l}^{m}(\cos \theta)
        \end{array}
    \right\}.$$