Eigenfunctions of the Helmholtz equation in Toroidal geometry

Question

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?

Answer 1

• 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.