Helmholtz's Equation
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 such as electromagnetics and acoustics. It is 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 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
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} = \eta^2, }[/math]
where the separation constant [math]\displaystyle{ \eta }[/math] must be an integer, say [math]\displaystyle{ \nu }[/math], in order for the potential to be continuous. [math]\displaystyle{ \Theta (\theta) }[/math] can therefore be expressed as
[math]\displaystyle{ \Theta (\theta) = C \, \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) = D_\nu \, J_\nu(k_m r) + E_\nu \, H^{(1)}_\nu(k_m r),\ \nu \in \mathbb{Z}, }[/math]
where Bessel functions of the first kind and Hankel functions of order [math]\displaystyle{ \nu }[/math]. Note that the first represent incoming waves and the second represent the scattered wave. 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]
We consider the case where we have Neuman boundary condition on the circle. Therefore we have [math]\displaystyle{ \partial_n\phi=0 }[/math] at [math]\displaystyle{ r=a }[/math]. We can therefore obtain
[math]\displaystyle{ E_{\nu} = - \frac{D_{\nu} J^{\prime}_\nu (k a)}{ H^{(1)\prime}_\nu (ka)} }[/math]
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}) + \int_{\partial\Omega} \left( \partial_{n^{\prime}} H^{1}_0 (|\mathbf{x} - \mathbf{x^{\prime}}|)\phi(\mathbf{x^{\prime}}) - H^{1}_0 (|\mathbf{x} - \mathbf{x^{\prime}}|)\partial_{n^{\prime}}\phi(\mathbf{x^{\prime}}) \right) \mathrm{d} S^{\prime} }[/math]
where [math]\displaystyle{ \epsilon }[/math] is one exterior to the domain, 1/2 on the boundary and zero inside.
It we consider again Neuman boundary conditions [math]\displaystyle{ \partial_n\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}) + \int_{\partial\Omega} \partial_{n^{\prime}} H^{1}_0 (|\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 are
[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 and we obtain
[math]\displaystyle{ \frac{1}{2}\sum_{n=-N}^{N} a_n e^{\mathrm{i} n \gamma} = \phi^{\mathrm{I}}(\mathbf{x}) + \int_{\partial\Omega} \partial_{n^{\prime}} H^{1}_0 (|\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 and obtain
[math]\displaystyle{ \frac{1}{2} a_n \sum_{n=-N}^{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 + \sum_{n=-N}^{N} a_n \int_{\partial\Omega} \int_{\partial\Omega} \partial_{n^{\prime}} H^{1}_0 (|\mathbf{x} - \mathbf{x^{\prime}}|)e^{\mathrm{i} n \gamma^{\prime}} e^{\mathrm{i} m \gamma} \mathrm{d} S^{\prime}\mathrm{d}S }[/math]
