boundary conditions for Schr$\ddot{\textrm{o}}$dinger equation in 2D polars? What are the boundary conditions at $r=0$ for Schr$\ddot{\textrm{o}}$dinger's equation for a quantum particle in 2D polars $(r,\theta)$, with potential $U=0$ for $r<a$ and $U=\infty$ for $r>a$?
I've shown that the angular part of the wave-function is $e^{im\theta}$ where $m\in\mathbb{Z}$, and I've got an ODE for the radial part $R(r)$, with boundary condition $R(a)=0$.
Since $e^{im\theta}$ is not well-defined as $(r,\theta)$ approaches the origin, it looks as though we should have $R(0)=0\;$ to get a well-defined wave-function. But according to the question, the series expansion for $R(r)$ has arbitrary constant term and all odd-order terms zero. What sort of boundary conditions give this??
Many thanks for any help with this!
 A: Schrödinger's equation takes the form 
$$(\nabla^2 + k^2)\psi = 0,$$
where $k = \sqrt{2 m E/\hbar^2}$ inside the well. 
This is the Helmholtz equation. 
The solutions in polar coordinates are well-known, 
$${\psi}_{n}(r,\theta) =
    \left\{
        \begin{array}[c]{l}
        {J_n}(k r) \\
        {Y_n}(k r)
        \end{array}
    \right\}
    \{ e^{\pm i n\theta} \},$$
where $J_n$ and $Y_n$ are the Bessel functions of the first and second kind. 
(The braces are shorthand for "linear combinations of.")
The boundary conditions are that $\psi$ be finite, vanish at $r=a$, and be single-valued.
Therefore, the eigenfunctions are 
$$\psi_{n i}(r,\theta) =  J_n\left(\alpha_{n i} \frac{r}{a}\right) (a_{n i}\cos n\theta + b_{n i}\sin n\theta),$$
where
$n = 0,1,2,\ldots$ and where
$\alpha_{n i}$ is the $i$th zero of $J_n$. 
(Energy quantization follows since 
$k = k_{n i} = \alpha_{n i}/a$.)
Notice that $J_n(0) = 0$ for $n\ge 1$. 
Furthermore, $\psi_0$ has no angular dependence.
Thus, we need not impose any conditions at $r=0$. 
The solutions already "know" about this possible trouble and have taken care of it! 
