Eigenfunctions of the Helmholtz equation in Toroidal geometry the Helmholtz equation 
$$\Delta \psi + k^2 \psi = 0$$
has a lot of fundamental applications in physics since it is a form of the wave equation $\Delta\phi - c^{-2}\partial_{tt}\phi = 0$ with an assumed harmonic time dependence $e^{\pm\mathrm{i}\omega t}$.
$k$ can be seen as some kind of potential - the equation is analogue to the stationary Schrödinger equation.  
The existance of solutions is to my knowledge linked to the separability of the Laplacian $\Delta$ in certain coordinate systems. Examples are cartesian, elliptical and cylindrical ones.
For now I am interested in a toroidal geometry, 
$$k(\mathbf{r}) = \begin{cases} k_{to} & \mathbf{r}\in T^2 \\ k_{out} & \text{else}\end{cases}$$
where $T^2 = \left\{ (x,y,z):\, r^2 \geq \left(  \sqrt{x^2 + y^2} - R\right)^2 + z^2  \right\}$
Hence the question:

Are there known solutions (in terms of eigenfunctions) of the Helmholtz equation for the given geometry?

Thank you in advance
Sincerely
Robert
Edit: As Hans pointed out, there might not be any solution according to a corresponding Wikipedia article. Unfortunately, there is no reference given - does anyone know where I could find the proof?
 A: *

*Normally $T^2$ means the Torus, which is a 2-manifold: $T^2 \cong [0,2\pi r]\times[0,2\pi R]$, the solution to 
$$
\Delta \psi + k^2\psi = 0\tag{1}
$$
bears the form: for $m\in \mathbb{Z}^2$, $\psi_k = e^{ i m\cdot x}$, with $|m| = \sqrt{m_1^2 +m_2^2} = k. $
The reason behind this is that $\mathbb{T}^2 \cong \mathbb{S}^1(r)\times \mathbb{S}^1(R) $, and for (1) on $\mathbb{S}^1$ has eigenvectors $e^{imx}$ where $|m| = k$, then the Fourier expansion on product spaces use basis $\prod e^{i m_i x_i}$.

*In your case it is actually a Toroid, according to the Field Theory Handbook the chapter about rotational system, the Helmholtz equation is not separable in toroidal geometry. Only Laplace equation is separable, please see section 6 in here.

*By that wikipedia article about Toroidal coordinates: we make the substitution for (1) as well:
$$\psi=u\sqrt{\cosh\tau-\cos\sigma},$$
then by the Laplacian in the toroidal geometry in that wiki entry:
\begin{align}
\Delta \psi =&
\frac{\left( \cosh \tau - \cos\sigma \right)^{3}}{a^{2}\sinh \tau} 
 \left[ 
\sinh \tau 
\frac{\partial}{\partial \sigma}
\left( \frac{1}{\cosh \tau - \cos\sigma}
\frac{\partial \Phi}{\partial \sigma}
\right) \right. \\[8pt]
& {}  + 
\left. \frac{\partial}{\partial \tau}
\left( \frac{\sinh \tau}{\cosh \tau - \cos\sigma}
\frac{\partial \Phi}{\partial \tau}
\right) + 
\frac{1}{\sinh \tau \left( \cosh \tau - \cos\sigma \right)}
\frac{\partial^2 \Phi}{\partial \phi^2}
\right].
\end{align}
(one extra thing to mention, the wiki entry failed to mention that $a^2 = R^2-r^2$) Equation (1) can be reduced as follows:
$$
\frac{\partial^2 u }{\partial \tau^2} + \frac{\cosh \tau}{\sinh\tau}\frac{\partial u }{\partial \tau} + \frac{1}{\sinh^2 \tau} \frac{\partial^2 u}{\partial \phi^2} + \frac{\partial^2 u}{\partial \sigma^2} + \left(\frac{ (R^2-r^2)k^2}{(\cosh\tau-\cos \sigma)^2} +\frac14\right)u= 0. 
$$
For above equation, though we separate it in three variables in toroidal coordinates, we can separate the $\phi$ variable:
$$
u = K(\tau,\sigma)\Phi(\phi).
$$
The equation becomes:
$$
\Delta_{\tau,\sigma} K + \frac{\cosh \tau}{\sinh\tau}\frac{\partial K }{\partial \tau} + \left(\frac{ (R^2-r^2)k^2}{(\cosh\tau-\cos \sigma)^2} +\frac14 -\frac{m^2}{\sinh^2 \tau}\right) K = 0,\tag{2}
$$
and 
$$
\Phi'' + m^2 \Phi = 0.
$$
Hence $u_m = K(\tau,\sigma)e^{im\theta}$, and $K$ satisfies (2). If someone knows how to proceed using analytical method for (2), I am interested in it as well.
