Difference between revisions of "Helmholtz's Equation"

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<math>\nabla^2 \phi + k^2 \phi = 0 </math>.
 
<math>\nabla^2 \phi + k^2 \phi = 0 </math>.
 
</center>
 
</center>
It applies to a wide variety of situations such as electromagnetics and acoustics. It is the wave equation
+
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.  
 
assuming a single frequency.  
In water waves it arises when we [[Removing The Depth Dependence|Remove The Depth Dependence]]. Often there is then a cross
+
In water waves, it arises when we [[Removing The Depth Dependence|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.
 
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  
 
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
+
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
 
satisfy Helmholtz's equation. This means that many asymptotic results in linear water waves can be
 
derived from results in acoustic or electromagnetic scattering.
 
derived from results in acoustic or electromagnetic scattering.
Line 21: Line 21:
 
of the method used in [[Bottom Mounted Cylinder]]
 
of the method used in [[Bottom Mounted Cylinder]]
  
Helmholtz equation in cylindrical coordinates is   
+
The Helmholtz equation in cylindrical coordinates is   
 
<center>
 
<center>
 
<math>
 
<math>
Line 41: Line 41:
 
\frac{\mathrm{d} R}{\mathrm{d}r} \right) +k^2 R(r) \right] = -
 
\frac{\mathrm{d} R}{\mathrm{d}r} \right) +k^2 R(r) \right] = -
 
\frac{1}{\Theta (\theta)} \frac{\mathrm{d}^2 \Theta}{\mathrm{d}
 
\frac{1}{\Theta (\theta)} \frac{\mathrm{d}^2 \Theta}{\mathrm{d}
\theta^2} = \eta^2,  
+
\theta^2} = \nu^2,  
 
</math>
 
</math>
 
</center>
 
</center>
where the separation constant <math>\eta</math> must be an integer, say <math>\nu</math>,
+
<math>\Theta
in order for the potential to be continuous. <math>\Theta
 
 
(\theta)</math> can therefore be expressed as  
 
(\theta)</math> can therefore be expressed as  
 
<center>
 
<center>
 
<math>
 
<math>
\Theta (\theta) = C \, \mathrm{e}^{\mathrm{i} \nu \theta}, \quad \nu \in \mathbb{Z}.
+
\Theta (\theta) = A \, \mathrm{e}^{\mathrm{i} \nu \theta}, \quad \nu \in \mathbb{Z}.
 
</math>
 
</math>
 
</center>
 
</center>
Line 73: Line 72:
 
<center>
 
<center>
 
<math>
 
<math>
R(r) = D_\nu \, J_\nu(k_m r) + E_\nu \, H^{(1)}_\nu(k_m r),\ \nu \in \mathbb{Z},
+
R(r) = B \, J_\nu(k r) + C \, H^{(1)}_\nu(k r),\ \nu \in \mathbb{Z},
 
</math>
 
</math>
 
</center>
 
</center>
where
+
where <math>J_\nu \,</math> denotes a Bessel function
[http://en.wikipedia.org/wiki/Bessel_function Bessel functions] of the first kind
+
of the first kind and <math>H^{(1)}_\nu \,</math>
and Hankel functions of order <math>\nu</math>. Note that the first represent
+
denotes a Hankel functions of order <math>\nu</math> (see [http://en.wikipedia.org/wiki/Bessel_function Bessel functions] for more information ). The choice of which
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.  
 
Hankel function depends on whether we have positive or negative exponential time dependence.  
 
The potential outside the circle can therefore be written as
 
The potential outside the circle can therefore be written as
Line 89: Line 87:
 
</math>
 
</math>
 
</center>
 
</center>
 
+
Note that the first term represents the incident wave
We consider the case where we have Neuman boundary condition on the circle. Therefore
+
(incoming wave) and the second term represents the scattered wave.  In other words, we say that <math>\phi = \phi^{\mathrm{I}}+\phi^{\mathrm{S}} \,</math>, where
we have <math>\partial_n\phi=0</math> at <math>r=a</math>. We can therefore obtain
+
<center>
 +
<math>
 +
\phi^{\mathrm{I}} (r,\theta)= \sum_{\nu = -
 +
\infty}^{\infty}  D_{\nu} J_\nu (k r) \mathrm{e}^{\mathrm{i} \nu \theta},
 +
</math>
 +
</center>
 +
<center>
 +
<math>
 +
\phi^{\mathrm{S}} (r,\theta)= \sum_{\nu = -
 +
\infty}^{\infty} E_{\nu} H^{(1)}_\nu (k
 +
r)  \mathrm{e}^{\mathrm{i} \nu \theta}.
 +
</math>
 +
</center>
 +
We consider the case where we have Neumann boundary condition on the circle. Therefore
 +
we have <math>\partial_n\phi=0</math> at <math>r=a \,</math>. This allows us to obtain
 
<center>
 
<center>
 
<math>
 
<math>
  E_{\nu} = - \frac{D_{\nu} J^{\prime}_\nu (k a)}{ H^{(1)\prime}_\nu (ka)}
+
  E_{\nu} = - \frac{D_{\nu} J^{\prime}_\nu (k a)}{ H^{(1)\prime}_\nu (ka)},
 
</math>
 
</math>
 
</center>
 
</center>
 +
which tells us that providing we know the form of the incident wave, we can compute the <math>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 [http://en.wikipedia.org/wiki/Jacobi%E2%80%93Anger_expansion Jacobi-Anger Identity].
  
 
== Solution for an arbitrary scatterer ==
 
== Solution for an arbitrary scatterer ==
Line 103: Line 116:
 
<center>
 
<center>
 
<math>
 
<math>
\epsilon\phi(\mathbf{x}) = \phi^{\mathrm{I}}(\mathbf{x}) + \int_{\partial\Omega} \left( \partial_{n^{\prime}} H^{1}_0  
+
\epsilon\phi(\mathbf{x}) = \phi^{\mathrm{I}}(\mathbf{x}) + \frac{i}{4}\int_{\partial\Omega} \left( \partial_{n^{\prime}} H^{(1)}_0  
(|\mathbf{x} - \mathbf{x^{\prime}}|)\phi(\mathbf{x^{\prime}}) -  
+
(k |\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)
+
H^{(1)}_0 (k |\mathbf{x} - \mathbf{x^{\prime}}|)\partial_{n^{\prime}}\phi(\mathbf{x^{\prime}}) \right)
\mathrm{d} S^{\prime}
+
\mathrm{d} S^{\prime},
 
</math>
 
</math>
 
</center>
 
</center>
where <math>\epsilon</math> is one exterior to the domain, 1/2 on the boundary and zero inside.  
+
where <math>\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>G(|\mathbf{x} - \mathbf{x}^\prime)|) = \frac{i}{4} H_{0}^{(1)}(k |\mathbf{x} - \mathbf{x}^\prime)|).\,</math>
  
It we consider again Neuman boundary conditions <math>\partial_n\phi(\mathbf{x}) = 0</math> and restrict ourselves to the boundary we obtain the following integral equation
+
If we consider again Neumann boundary conditions <math>\partial_{n^\prime}\phi(\mathbf{x}) = 0</math> and restrict ourselves to the boundary we obtain the following integral equation
 
<center>
 
<center>
 
<math>
 
<math>
\frac{1}{2}\phi(\mathbf{x}) = \phi^{\mathrm{I}}(\mathbf{x}) + \int_{\partial\Omega} \partial_{n^{\prime}} H^{1}_0  
+
\frac{1}{2}\phi(\mathbf{x}) = \phi^{\mathrm{I}}(\mathbf{x}) + \frac{i}{4} \int_{\partial\Omega} \partial_{n^{\prime}} H^{(1)}_0  
(|\mathbf{x} - \mathbf{x^{\prime}}|)\phi(\mathbf{x^{\prime}})  
+
(k|\mathbf{x} - \mathbf{x^{\prime}}|)\phi(\mathbf{x^{\prime}})  
\mathrm{d} S^{\prime}
+
\mathrm{d} S^{\prime}.
 
</math>
 
</math>
 
</center>
 
</center>
  
We solve this equation by the Galerkin method using a Fourier series as the basis. We parameterise the curve <math>\partial\Omega</math> by <math>\mathbf{s}(\gamma)</math> where <math>-\pi \leq \gamma \leq \pi</math>. We write the potential on the boundary are
+
We solve this equation by the Galerkin method using a Fourier series as the basis. We parameterise the curve <math>\partial\Omega</math> by <math>\mathbf{s}(\gamma)</math> where <math>-\pi \leq \gamma \leq \pi</math>. We write the potential on the boundary as
 
<center>
 
<center>
 
<math>
 
<math>
\phi(\mathbf{x}) = \sum_{n=-N}^{N} a_n e^{\mathrm{i} n \gamma}
+
\phi(\mathbf{x}) = \sum_{n=-N}^{N} a_n e^{\mathrm{i} n \gamma}.
 
</math>
 
</math>
 
</center>
 
</center>
We substitute this into the equation for the potential and we obtain
+
We substitute this into the equation for the potential to obtain
 
<center>
 
<center>
 
<math>
 
<math>
\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  
+
\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  
(|\mathbf{x} - \mathbf{x^{\prime}}|)\sum_{n=-N}^{N} a_n e^{\mathrm{i} n \gamma^{\prime}}
+
(k|\mathbf{x} - \mathbf{x^{\prime}}|)\sum_{n=-N}^{N} a_n e^{\mathrm{i} n \gamma^{ \prime}}
\mathrm{d} S^{\prime}
+
\mathrm{d} S^{\prime}.
 
</math>
 
</math>
 
</center>
 
</center>
We now multiply by <math>e^{\mathrm{i} m \gamma}</math> and integrate and obtain
+
We now multiply by <math>e^{\mathrm{i} m \gamma} \,</math> and integrate to obtain
 
<center>
 
<center>
 
<math>
 
<math>
\frac{1}{2} a_n \sum_{n=-N}^{N} \int_{\partial\Omega} e^{\mathrm{i} n \gamma} e^{\mathrm{i} m \gamma}  
+
\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
 
\mathrm{d} S
 
= \int_{\partial\Omega} \phi^{\mathrm{I}}(\mathbf{x})e^{\mathrm{i} m \gamma}  
 
= \int_{\partial\Omega} \phi^{\mathrm{I}}(\mathbf{x})e^{\mathrm{i} m \gamma}  
\mathrm{d} S +  
+
\mathrm{d} S + \frac{i}{4}
\sum_{n=-N}^{N} a_n \int_{\partial\Omega} \int_{\partial\Omega} \partial_{n^{\prime}} H^{1}_0  
+
\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}}
+
(k|\mathbf{x} - \mathbf{x^{\prime}}|)e^{\mathrm{i} n \gamma^{\prime}}
e^{\mathrm{i} m \gamma} \mathrm{d}  S^{\prime}\mathrm{d}S  
+
e^{\mathrm{i} m \gamma} \mathrm{d}  S^{\prime}\mathrm{d}S.
 
</math>
 
</math>
 
</center>
 
</center>
  
[http://en.wikipedia.org/wiki/Helmholtz_equation External link]
+
==External Links==
 +
[http://en.wikipedia.org/wiki/Helmholtz_equation Helmholtz Equation on Wikipedia]
  
 
[[Category:Linear Water-Wave Theory]]
 
[[Category:Linear Water-Wave Theory]]

Latest revision as of 21:03, 27 April 2013


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