Helmholtz's Equation

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Indroduction

This is a very well known equation given by

[math]\displaystyle{ \nabla^2 \phi + k^2 \phi = 0 }[/math].

It applies to a wide variety of situations that arise in electromagnetics and acoustics. It is also equivalent to the wave equation assuming a single frequency. In water waves, it arises when we Remove The Depth Dependence. Often there is then a cross over from the study of water waves to the study of scattering problems more generally. Also, if we perform a Cylindrical Eigenfunction Expansion we find that the modes all decay rapidly as distance goes to infinity except for the solutions which satisfy Helmholtz's equation. This means that many asymptotic results in linear water waves can be derived from results in acoustic or electromagnetic scattering.

Solution for a Circle

We can solve for the scattering by a circle using separation of variables. This is the basis of the method used in Bottom Mounted Cylinder

The Helmholtz equation in cylindrical coordinates is

[math]\displaystyle{ \frac{1}{r} \frac{\partial}{\partial r} \left( r \frac{\partial \phi}{\partial r} \right) + \frac{1}{r^2} \frac{\partial^2 \phi}{\partial \theta^2} = -k^2 \phi(r,\theta), }[/math]

we use the separation

[math]\displaystyle{ \phi(r,\theta) =: R(r) \Theta(\theta)\,. }[/math]

Substituting this into Laplace's equation yields

[math]\displaystyle{ \frac{r^2}{R(r)} \left[ \frac{1}{r} \frac{\mathrm{d}}{\mathrm{d}r} \left( r \frac{\mathrm{d} R}{\mathrm{d}r} \right) +k^2 R(r) \right] = - \frac{1}{\Theta (\theta)} \frac{\mathrm{d}^2 \Theta}{\mathrm{d} \theta^2} = \nu^2, }[/math]

[math]\displaystyle{ \Theta (\theta) }[/math] can therefore be expressed as

[math]\displaystyle{ \Theta (\theta) = A \, \mathrm{e}^{\mathrm{i} \nu \theta}, \quad \nu \in \mathbb{Z}. }[/math]

We also obtain the following expression

[math]\displaystyle{ r \frac{\mathrm{d}}{\mathrm{d}r} \left( r \frac{\mathrm{d} R}{\mathrm{d} r} \right) - (\nu^2 - k^2 r^2) R(r) = 0, \quad \nu \in \mathbb{Z}. }[/math]

Substituting [math]\displaystyle{ \tilde{r}:=k r }[/math] and writing [math]\displaystyle{ \tilde{R} (\tilde{r}) := R(\tilde{r}/k) = R(r) }[/math], this can be rewritten as

[math]\displaystyle{ \tilde{r}^2 \frac{\mathrm{d}^2 \tilde{R}}{\mathrm{d} \tilde{r}^2} + \tilde{r} \frac{\mathrm{d} \tilde{R}}{\mathrm{d} \tilde{r}} - (\nu^2 - \tilde{r}^2)\, \tilde{R} = 0, \quad \nu \in \mathbb{Z}, }[/math]

which is Bessel's equation. Substituting back, the general solution is given by

[math]\displaystyle{ R(r) = B \, J_\nu(k r) + C \, H^{(1)}_\nu(k r),\ \nu \in \mathbb{Z}, }[/math]

where [math]\displaystyle{ J_\nu \, }[/math] denotes a Bessel function of the first kind and [math]\displaystyle{ H^{(1)}_\nu \, }[/math] denotes a Hankel functions of order [math]\displaystyle{ \nu }[/math] (see Bessel functions for more information ). The choice of which Hankel function depends on whether we have positive or negative exponential time dependence. The potential outside the circle can therefore be written as

[math]\displaystyle{ \phi (r,\theta) = \sum_{\nu = - \infty}^{\infty} \left[ D_{\nu} J_\nu (k r) + E_{\nu} H^{(1)}_\nu (k r) \right] \mathrm{e}^{\mathrm{i} \nu \theta}, }[/math]

Note that the first term represents the incident wave (incoming wave) and the second term represents the scattered wave. In other words, we say that [math]\displaystyle{ \phi = \phi^{\mathrm{I}}+\phi^{\mathrm{S}} \, }[/math], where

[math]\displaystyle{ \phi^{\mathrm{I}} (r,\theta)= \sum_{\nu = - \infty}^{\infty} D_{\nu} J_\nu (k r) \mathrm{e}^{\mathrm{i} \nu \theta}, }[/math]

[math]\displaystyle{ \phi^{\mathrm{S}} (r,\theta)= \sum_{\nu = - \infty}^{\infty} E_{\nu} H^{(1)}_\nu (k r) \mathrm{e}^{\mathrm{i} \nu \theta}. }[/math]

We consider the case where we have Neumann boundary condition on the circle. Therefore we have [math]\displaystyle{ \partial_n\phi=0 }[/math] at [math]\displaystyle{ r=a \, }[/math]. This allows us to obtain

[math]\displaystyle{ E_{\nu} = - \frac{D_{\nu} J^{\prime}_\nu (k a)}{ H^{(1)\prime}_\nu (ka)}, }[/math]

which tells us that providing we know the form of the incident wave, we can compute the [math]\displaystyle{ D_\nu \, }[/math] coefficients and ultimately determine the potential throughout the circle. It is possible to expand a plane wave in terms of cylindrical waves using the Jacobi-Anger Identity.

Solution for an arbitrary scatterer

We can solve for an arbitrary scatterer by using Green's theorem. We express the potential as

[math]\displaystyle{ \epsilon\phi(\mathbf{x}) = \phi^{\mathrm{I}}(\mathbf{x}) + \frac{i}{4}\int_{\partial\Omega} \left( \partial_{n^{\prime}} H^{(1)}_0 (k |\mathbf{x} - \mathbf{x^{\prime}}|)\phi(\mathbf{x^{\prime}}) - H^{(1)}_0 (k |\mathbf{x} - \mathbf{x^{\prime}}|)\partial_{n^{\prime}}\phi(\mathbf{x^{\prime}}) \right) \mathrm{d} S^{\prime}, }[/math]

where [math]\displaystyle{ \epsilon = 1,1/2 \ \mbox{or} \ 0 }[/math], depending on whether we are exterior, on the boundary or in the interior of the domain (respectively), and the fundamental solution for the Helmholtz Equation (which incorporates Sommerfeld Radiation conditions) is given by [math]\displaystyle{ G(|\mathbf{x} - \mathbf{x}^\prime)|) = \frac{i}{4} H_{0}^{(1)}(k |\mathbf{x} - \mathbf{x}^\prime)|).\, }[/math]

If we consider again Neumann boundary conditions [math]\displaystyle{ \partial_{n^\prime}\phi(\mathbf{x}) = 0 }[/math] and restrict ourselves to the boundary we obtain the following integral equation

[math]\displaystyle{ \frac{1}{2}\phi(\mathbf{x}) = \phi^{\mathrm{I}}(\mathbf{x}) + \frac{i}{4} \int_{\partial\Omega} \partial_{n^{\prime}} H^{(1)}_0 (k|\mathbf{x} - \mathbf{x^{\prime}}|)\phi(\mathbf{x^{\prime}}) \mathrm{d} S^{\prime}. }[/math]

We solve this equation by the Galerkin method using a Fourier series as the basis. We parameterise the curve [math]\displaystyle{ \partial\Omega }[/math] by [math]\displaystyle{ \mathbf{s}(\gamma) }[/math] where [math]\displaystyle{ -\pi \leq \gamma \leq \pi }[/math]. We write the potential on the boundary as

[math]\displaystyle{ \phi(\mathbf{x}) = \sum_{n=-N}^{N} a_n e^{\mathrm{i} n \gamma}. }[/math]

We substitute this into the equation for the potential to obtain

[math]\displaystyle{ \frac{1}{2}\sum_{n=-N}^{N} a_n e^{\mathrm{i} n \gamma} = \phi^{\mathrm{I}}(\mathbf{x}) + \frac{i}{4} \int_{\partial\Omega} \partial_{n^{\prime}} H^{(1)}_0 (k|\mathbf{x} - \mathbf{x^{\prime}}|)\sum_{n=-N}^{N} a_n e^{\mathrm{i} n \gamma^{ \prime}} \mathrm{d} S^{\prime}. }[/math]

We now multiply by [math]\displaystyle{ e^{\mathrm{i} m \gamma} \, }[/math] and integrate to obtain

[math]\displaystyle{ \frac{1}{2} \sum_{n=-N}^{N} a_n \int_{\partial\Omega} e^{\mathrm{i} n \gamma} e^{\mathrm{i} m \gamma} \mathrm{d} S = \int_{\partial\Omega} \phi^{\mathrm{I}}(\mathbf{x})e^{\mathrm{i} m \gamma} \mathrm{d} S + \frac{i}{4} \sum_{n=-N}^{N} a_n \int_{\partial\Omega} \int_{\partial\Omega} \partial_{n^{\prime}} H^{(1)}_0 (k|\mathbf{x} - \mathbf{x^{\prime}}|)e^{\mathrm{i} n \gamma^{\prime}} e^{\mathrm{i} m \gamma} \mathrm{d} S^{\prime}\mathrm{d}S. }[/math]

External Links

Helmholtz Equation on Wikipedia