Solving a Laplacian in polar coordinates I came across the following boundary value problem that I can't solve.  It's the Laplacian on the upper half of an annulus with radius $1 \leq r \leq 2$ in polar coordinates:
$u_{rr} + \frac{1}{r} u_r + \frac{1}{r^2} u_{\theta \theta} =0$
$u(1,\theta)=u(2,\theta)=0, ~~~ 0<\theta<\pi$
$u(r,0)=0, ~~ u(r,\pi)=r, ~~~ 1<r<2.$
The problem mentions that one must show in this case the choice of separation constant $\lambda=-\alpha^2<0$ leads to eigenvalues and eigenfunctions.  The problem is that in this case, after separating variables and solving
$ \Theta'' + \lambda \Theta = 0$
$ r^2R'' + rR' - \lambda R =0$
we would get $\Theta(\theta)=c_1 \cosh(\alpha \theta) + c_2 \sinh(\alpha \theta)$ which is not periodic in $\theta$.  Usually we always have $\alpha^2>0$ which gives us the periodic sine and cosine solution for $\Theta$.  Help would be appreciated!
 A: There's no problem with peridocity here. Periodicity is required when the domain consists of full circles, in which case a lack of periodicity would imply a lack of continuity. If the function is only defined for $0\lt\theta\lt\pi$, there's no reason for imposing periodicity.
A: Notice that to satisfy the boundary conditions, $R(r)$ must be oscillatory. 
This forces $\Theta(\theta)$ to have exponential behavior. 
Assume $\Theta'' - \lambda\Theta = 0$, where $\lambda > 0$, to see this directly. 
After some work we find we are forced by the boundary conditions on $R(r)$ to take $\lambda < 0$.
It is a good exercise to work this out.${}^\dagger$
As joriki notes, we need not (and should not) impose the constraint that $\Theta$ be periodic.
Solve the ODE for $R(r)$ and use the boundary conditions to eliminate one of the solutions and find the eigenvalues $\lambda_n$. 
Use the boundary condition $\Theta(0) = 0$ to eliminate one of the angular solutions. 
This gives a solution of the form $u_n(r,\theta) = R_n(r)\Theta_n(\theta)$
that satisfies $u_n(1,\theta) = u_n(2,\theta) = u_n(r,0) = 0$. 
The solution satisfying the final boundary condition, $u(r,\pi) = r$, 
will be a sum
$$u(r,\theta) = \sum_{n=1}^\infty a_n u_n(r,\theta).$$
We need
$$\sum_{n=1}^\infty a_n u_n(r,\pi) = r.$$
Sturm-Liouville theory tells us the eigenfunctions $R_n$ are orthogonal,
$$\int_1^2 \frac{dr}{r} R_m R_n \propto \delta_{mn},$$
where $\delta_{mn}$ is the Kronecker delta.
(Note that the weight function for the ODE for $R(r)$ is $1/r$.)
To find the $a_n$s, multiply by $u_m(r,\pi)/r$ and integrate from $1$ to $2$.
This is totally analogous to the process of finding the coefficients in a Fourier series. 
There's a nice closed form for the $a_n$. 
Below we plot $r$ and the partial sum $\sum_{n=1}^{25}a_n u_n(r,\pi)$.

Figure 1. The sum $\sum_{n=1}^{25}a_n u_n(r,\pi)$. 


${}^\dagger$Think of solving Laplace's equation in Cartesian coordinates. 
After separating variables, we can't have both $X(x)$ and $Y(y)$ oscillatory since we need
$\frac{X''}{X} + \frac{Y''}{Y} = 0$. 

