An eigenvalue problem on a half circle 
Solve the eigenvalue problem for the half-circle $x^2+y^2\le R^2$ $y\ge 0$ with homogeneous Dirichlet conditions as boundary conditions.

This is what I did:
Let $u=u(r,\phi)$
$$\Delta u=-\lambda u$$
$$R_{rr}\Phi+\frac{1}{r}R_r\Phi+\frac{1}{r^2}R\Phi_{\phi\phi}=-\lambda R\Phi$$
This gives $$\frac{r^2R_{rr}}{R}+\frac{rR_r}{R}+\frac{\Phi_{\phi\phi}}{\Phi}=-\lambda$$
Rearranging and setting both equations equal to some constant $\mu$:
$$\frac{r^2R_{rr}}{R}+\frac{rR_r}{R}+\lambda=-\frac{\Phi_{\phi\phi}}{\Phi}=\mu$$
Gives the two ODEs:
$$\frac{r^2R_{rr}}{R}+\frac{rR_r}{R}+\lambda=\mu$$
$$-\frac{\Phi_{\phi\phi}}{\Phi}=\mu$$
The second gives with Dirichlet conditions:
$$-\frac{\Phi_{\phi\phi}}{\Phi}=\mu \longrightarrow \Phi(\phi)=C_1\sin\sqrt{\mu}\phi$$
The former gives the Bessel equation,:
$$R_{rr}+rR_r+(\lambda-\mu)\frac{R}{r^2}=0 $$
But this should rather be in the form:
$$R_{rr}+rR_r+(\lambda-\frac{\mu}{r^2})R=0 $$
What went wrong, and what about the fact that with Dirichlet conditions we should also include $J(R)=0$? That gives a trivial solution.
Is this procedure incomplete?
Thanks
 A: You forget to multiply $\lambda$ by $r^2$. It should be
$$ \frac{r^2R_{rr}}{R} + \frac{rR_r}{R} + \frac{\Phi_{\phi\phi}}{\Phi} = -\lambda r^2. $$
Following your notation, you would arrive at
$$ r^2R_{rr} + rR_r + (\lambda r^2 - \mu)R = 0. $$
This is a Bessel differential equation after you do a change of variable $s = \sqrt{\lambda} r$, which you can write down the solution.
A: 
[...]
$$R_{rr}\Phi+\frac{1}{r}R_r\Phi+\frac{1}{r^2}R\Phi_{\phi\phi}=-\lambda R\Phi$$
This gives $$\frac{r^2R_{rr}}{R}+\frac{rR_r}{R}+\frac{\Phi_{\phi\phi}}{\Phi}=-\lambda$$
[...]
What went wrong[?]

A factor of $r^2$ has been dropped from the right-hand side at this step.

But this should rather be in the form:
$$R_{rr}+rR_r+(\lambda-\frac{\mu}{r^2})R=0 $$

The coefficient of the $R_r$ term here should be $1/r$.

what about the fact that with Dirichlet conditions we should also include $J(R)=0$? That gives a trivial solution.

For the azimuthal equation, the zero-value condition at $\phi = \pi$ implies that $\sin (\sqrt{\mu} \pi) = 0$, and hence $\sqrt{\mu} = n$, where $n = 1, 2, 3, \text{etc}$.
Eventually you should get
$$R_{rr} + \frac{R_r}{r} + \left( \lambda-\frac{n^2}{r^2} \right) R = 0, $$
whose fundamental solutions are Bessel $J_n (r \sqrt{\lambda})$ and $Y_n (r \sqrt{\lambda})$.

*

*To get the solution to vanish at $r = 0$, you only want $J_n (r \sqrt{\lambda})$ (since Bessel-Y has a singularity).

*To get the solution to vanish at $r = r_\text{max}$ (I've avoided $R$ which clashes with the radial dependent variable), you need $r_\text{max} \sqrt{\lambda} = \alpha_{n,k}$, where $\alpha_{n,k}$ is the $k$th root of $J_n$, with $k = 1, 2, 3, \text{etc}$.

