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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?

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You probably mean $R$ instead of $R^2$ (if $R$ has its usual interpretation). –  Hans Lundmark Jan 14 '11 at 10:49
    
Anyway, for what it's worth: Wikipedia (en.wikipedia.org/wiki/Toroidal_coordinates#Standard_separation) claims (without pointing to any specific source) that the Helmholtz equation is not separable in toroidal coordinates. –  Hans Lundmark Jan 14 '11 at 10:50
    
@Hans: Thanks for pointing out to the error, I will correct it. And also thank you for the link. It really is a pitty that there is no reference given. Greets –  Robert Filter Jan 14 '11 at 14:32
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Another correction: existence of solutions is not linked to the coordinate system at all. The Laplace operator (or the Laplace-Beltrami operator) are geometric, so does not depend on the coordinates chosen. The existence of solutions however do depend on the global geometry of the manifold on which you are asking for the solution: it has to do with the spectrum of the Laplace-Beltrami operator. What you probably mean is "existence of nice, simple, closed-form expressions" of the solutions. –  Willie Wong Jan 14 '11 at 17:35
    
@Willie Wong: Thank you for the correction. Indeed, I mean it in this way. It seems to as if it is just a historic issue of applied mathematics that some "special" functions (trigonometric, Bessel etc) paved their way into standard textbooks. Nevertheless, do you know if there are implicit definitions of solutions available like the elliptic ones? If I remember correctly, the Greens function of the Helmholtz equation is normally constructed from eigenfunctions of the Laplace... Isn't an application of this procedure applicable here somehow? –  Robert Filter Jan 14 '11 at 19:05

1 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.

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Dear Shuhao, thanks for your answer, although my thanks is quite late. Thanks for your argumentation and calculation. Most valuable, though, seems to be the reference to Boyer et al., Nagoya Math. J. 60 (1976) you provided and therein, P. Morse and H. Feshbach, "Methods of Theoretical Physics", a must-have. –  Robert Filter Jan 2 at 12:17

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