WikiWaves - User contributions [en]

Helmholtz's Equation

2010-06-09T01:23:20Z

Administrator: /* Solution for an arbitrary scatterer */

{{complete pages}}

== Indroduction ==

This is a very well known equation given by
<center>
<math>\nabla^2 \phi + k^2 \phi = 0 </math>.
</center>
It applies to a wide variety of situations such as electromagnetics and acoustics.
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.
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
<center>
<math>
\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>
</center>
we use the separation
<center>
<math>
\phi(r,\theta) =: R(r) \Theta(\theta)\,.
</math>
</center>
Substituting this into Laplace's equation yields
<center>
<math>
\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>
</center>
where the separation constant <math>\eta</math> must be an integer, say <math>\nu</math>,
in order for the potential to be continuous. <math>\Theta
(\theta)</math> can therefore be expressed as
<center>
<math>
\Theta (\theta) = C \, \mathrm{e}^{\mathrm{i} \nu \theta}, \quad \nu \in \mathbb{Z}.
</math>
</center>
We also obtain the following expression
<center>
<math>
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>
</center>
Substituting <math>\tilde{r}:=k r</math> and writing <math>\tilde{R} (\tilde{r}) :=
R(\tilde{r}/k) = R(r)</math>, this can be rewritten as
<center>
<math>
\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>
</center>
which is Bessel's equation. Substituting back,
the general solution is given by
<center>
<math>
R(r) = D_\nu \, J_\nu(k_m r) + E_\nu \, H^{(1)}_\nu(k_m r),\ \nu \in \mathbb{Z},
</math>
</center>
where
[http://en.wikipedia.org/wiki/Bessel_function Bessel functions] of the first kind
and Hankel functions of order <math>\nu</math>.
The potential outside the circle can therefore be written as
<center>
<math>
\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>
</center>

We consider the case where we have Neuman boundary condition on the circle. Therefore
we have <math>\partial_n\phi=0</math> at <math>r=a</math>. We can therefore obtain
<center>
<math>
E_{\nu} = - \frac{D_{\nu} J^{\prime}_\nu (k a)}{ H^{(1)\prime}_\nu (ka)}
</math>
</center>

== Solution for an arbitrary scatterer ==

We can solve for an arbitrary scatterer by using Green's theorem. We express the potential as
<center>
<math>
\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>
</center>
where <math>\epsilon</math> is one exterior to the domain, 1/2 on the boundary and zero inside.

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
<center>
<math>
\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>
</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
<center>
<math>
\phi(\mathbf{x}) = \sum_{n=-N}^{N} a_n e^{\mathrm{i} n \gamma}
</math>
</center>
We substitute this into the equation for the potential and we obtain
<center>
<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
(|\mathbf{x} - \mathbf{x^{\prime}}|)\sum_{n=-N}^{N} a_n e^{\mathrm{i} n \gamma^{\prime}}
\mathrm{d} S^{\prime}
</math>
</center>
We now multiply by <math>e^{\mathrm{i} m \gamma}</math> and integrate and obtain
<center>
<math>
\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>
</center>

[http://en.wikipedia.org/wiki/Helmholtz_equation External link]

[[Category:Linear Water-Wave Theory]]

Helmholtz's Equation

2010-06-09T01:08:25Z

Administrator: /* Solution for an arbitrary scatterer */

{{complete pages}}

== Indroduction ==

This is a very well known equation given by
<center>
<math>\nabla^2 \phi + k^2 \phi = 0 </math>.
</center>
It applies to a wide variety of situations such as electromagnetics and acoustics.
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.
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
<center>
<math>
\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>
</center>
we use the separation
<center>
<math>
\phi(r,\theta) =: R(r) \Theta(\theta)\,.
</math>
</center>
Substituting this into Laplace's equation yields
<center>
<math>
\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>
</center>
where the separation constant <math>\eta</math> must be an integer, say <math>\nu</math>,
in order for the potential to be continuous. <math>\Theta
(\theta)</math> can therefore be expressed as
<center>
<math>
\Theta (\theta) = C \, \mathrm{e}^{\mathrm{i} \nu \theta}, \quad \nu \in \mathbb{Z}.
</math>
</center>
We also obtain the following expression
<center>
<math>
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>
</center>
Substituting <math>\tilde{r}:=k r</math> and writing <math>\tilde{R} (\tilde{r}) :=
R(\tilde{r}/k) = R(r)</math>, this can be rewritten as
<center>
<math>
\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>
</center>
which is Bessel's equation. Substituting back,
the general solution is given by
<center>
<math>
R(r) = D_\nu \, J_\nu(k_m r) + E_\nu \, H^{(1)}_\nu(k_m r),\ \nu \in \mathbb{Z},
</math>
</center>
where
[http://en.wikipedia.org/wiki/Bessel_function Bessel functions] of the first kind
and Hankel functions of order <math>\nu</math>.
The potential outside the circle can therefore be written as
<center>
<math>
\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>
</center>

We consider the case where we have Neuman boundary condition on the circle. Therefore
we have <math>\partial_n\phi=0</math> at <math>r=a</math>. We can therefore obtain
<center>
<math>
E_{\nu} = - \frac{D_{\nu} J^{\prime}_\nu (k a)}{ H^{(1)\prime}_\nu (ka)}
</math>
</center>

== Solution for an arbitrary scatterer ==

We can solve for an arbitrary scatterer by using Green's theorem. We express the potential as
<center>
<math>
\epsilon\phi(\mathbf{x}) = \phi^{\mathrm{I}}(\mathbf{x}) + \int_{\partial\Omega} \left( \partial_n H^{1}_0
(|\mathbf{x} - \mathbf{x^{\prime}}|)\phi(\mathbf{x^{\prime}}) -
H^{1}_0 (|\mathbf{x} - \mathbf{x^{\prime}}|)\partial_n\phi(\mathbf{x^{\prime}}) \right)
\mathrm{d} S
</math>
</center>
where <math>\epsilon</math> is one exterior to the domain, 1/2 on the boundary and zero inside.

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
<center>
<math>
\frac{1}{2}\phi(\mathbf{x}) = \phi^{\mathrm{I}}(\mathbf{x}) + \int_{\partial\Omega} \partial_n H^{1}_0
(|\mathbf{x} - \mathbf{x^{\prime}}|)\phi(\mathbf{x^{\prime}})
\mathrm{d} S
</math>
</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
<center>
<math>
\phi(\mathbf{x}) = \sum_{n=-N}^{N} a_n e^{\mathrm{i} n \gamma}
</math>
</center>
We substitute this into the equation for the potential and we obtain
<center>
<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 H^{1}_0
(|\mathbf{x} - \mathbf{x^{\prime}}|)\sum_{n=-N}^{N} a_n e^{\mathrm{i} n \gamma}
\mathrm{d} S
</math>
</center>
We now multiply by <math>e^{\mathrm{i} m \gamma}</math> and integrate and obtain
<center>
<math>
\frac{1}{2} a_n \sum_{n=-N}^{N} \int_{\partial\Omega} e^{\mathrm{i} n \gamma} e^{\mathrm{i} m \gamma^{\prime}}
\mathrm{d} S^{\prime}
= \int_{\partial\Omega} \phi^{\mathrm{I}}(\mathbf{x})e^{\mathrm{i} m \gamma^{\prime}}
\mathrm{d} S^{\prime} +
\sum_{n=-N}^{N} a_n \int_{\partial\Omega} \int_{\partial\Omega} \partial_n H^{1}_0
(|\mathbf{x} - \mathbf{x^{\prime}}|)e^{\mathrm{i} n \gamma}
e^{\mathrm{i} m \gamma^{\prime}} \mathrm{d} S \mathrm{d} S^{\prime}
</math>
</center>

[http://en.wikipedia.org/wiki/Helmholtz_equation External link]

[[Category:Linear Water-Wave Theory]]

Helmholtz's Equation

2010-06-09T00:48:44Z

Administrator: /* Solution for a Circle */

{{complete pages}}

== Indroduction ==

This is a very well known equation given by
<center>
<math>\nabla^2 \phi + k^2 \phi = 0 </math>.
</center>
It applies to a wide variety of situations such as electromagnetics and acoustics.
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.
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
<center>
<math>
\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>
</center>
we use the separation
<center>
<math>
\phi(r,\theta) =: R(r) \Theta(\theta)\,.
</math>
</center>
Substituting this into Laplace's equation yields
<center>
<math>
\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>
</center>
where the separation constant <math>\eta</math> must be an integer, say <math>\nu</math>,
in order for the potential to be continuous. <math>\Theta
(\theta)</math> can therefore be expressed as
<center>
<math>
\Theta (\theta) = C \, \mathrm{e}^{\mathrm{i} \nu \theta}, \quad \nu \in \mathbb{Z}.
</math>
</center>
We also obtain the following expression
<center>
<math>
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>
</center>
Substituting <math>\tilde{r}:=k r</math> and writing <math>\tilde{R} (\tilde{r}) :=
R(\tilde{r}/k) = R(r)</math>, this can be rewritten as
<center>
<math>
\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>
</center>
which is Bessel's equation. Substituting back,
the general solution is given by
<center>
<math>
R(r) = D_\nu \, J_\nu(k_m r) + E_\nu \, H^{(1)}_\nu(k_m r),\ \nu \in \mathbb{Z},
</math>
</center>
where
[http://en.wikipedia.org/wiki/Bessel_function Bessel functions] of the first kind
and Hankel functions of order <math>\nu</math>.
The potential outside the circle can therefore be written as
<center>
<math>
\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>
</center>

We consider the case where we have Neuman boundary condition on the circle. Therefore
we have <math>\partial_n\phi=0</math> at <math>r=a</math>. We can therefore obtain
<center>
<math>
E_{\nu} = - \frac{D_{\nu} J^{\prime}_\nu (k a)}{ H^{(1)\prime}_\nu (ka)}
</math>
</center>

== Solution for an arbitrary scatterer ==

We can solve for an arbitrary scatterer by using Green's theorem. We express the potential as
<center>
<math>
\epsilon\phi(\mathbf{x}) = \phi^{\mathrm{I}}(\mathbf{x}) + \int_{\partial\Omega} \left( \partial_n H^{1}_0
(|\mathbf{x} - \mathbf{x^{\prime}}|)\phi(\mathbf{x}) -
H^{1}_0 (|\mathbf{x} - \mathbf{x^{\prime}}|)\partial_n\phi(\mathbf{x}) \right)
\mathrm{d} S
</math>
</center>
where <math>\epsilon</math> is one exterior to the domain, 1/2 on the boundary and zero inside.

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
<center>
<math>
\frac{1}{2}\phi(\mathbf{x}) = \phi^{\mathrm{I}}(\mathbf{x}) + \int_{\partial\Omega} \partial_n H^{1}_0
(|\mathbf{x} - \mathbf{x^{\prime}}|)\phi(\mathbf{x})
\mathrm{d} S
</math>
</center>

[http://en.wikipedia.org/wiki/Helmholtz_equation External link]

[[Category:Linear Water-Wave Theory]]

Helmholtz's Equation

2010-06-09T00:33:23Z

Administrator: /* Solution for a Circle */

Helmholtz's Equation

2010-06-09T00:26:17Z

Administrator: /* Solution for a Circle */

Helmholtz's Equation

2010-06-09T00:20:18Z

Administrator:

Template:Separation of variables for the r and theta coordinates

2010-06-09T00:12:39Z

Administrator: /* Separation of Variable for the r and \theta coordinates */

== Separation of Variable for the <math>r</math> and <math>\theta</math> coordinates ==

For the solution of
<center>
<math>
\frac{1}{r} \frac{\partial}{\partial r} \left( r \frac{\partial
Y}{\partial r} \right) + \frac{1}{r^2} \frac{\partial^2 Y}{\partial
\theta^2} = k_m^2 Y(r,\theta),
</math>
</center>
we use the separation
<center>
<math>
\,\!Y(r,\theta) =: R(r) \Theta(\theta).
</math>
</center>
Substituting this into Laplace's equation yields
<center>
<math>
\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_m^2 R(r) \right] = -
\frac{1}{\Theta (\theta)} \frac{\mathrm{d}^2 \Theta}{\mathrm{d}
\theta^2} = \eta^2,
</math>
</center>
where the separation constant <math>\eta</math> must be an integer, say <math>\nu</math>,
in order for the potential to be continuous. <math>\Theta
(\theta)</math> can therefore be expressed as
<center>
<math>
\Theta (\theta) = C \, \mathrm{e}^{\mathrm{i} \nu \theta}, \quad \nu \in \mathbb{Z}.
</math>
</center>
We also obtain the following expression
<center>
<math>
r \frac{\mathrm{d}}{\mathrm{d}r} \left( r \frac{\mathrm{d}
R}{\mathrm{d} r} \right) - (\nu^2 + k_m^2 r^2) R(r) = 0, \quad \nu \in
\mathbb{Z}.
</math>
</center>
Substituting <math>\tilde{r}:=k_m r</math> and writing <math>\tilde{R} (\tilde{r}) :=
R(\tilde{r}/k_m) = R(r)</math>, this can be rewritten as
<center>
<math>
\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>
</center>
which is the modified version of Bessel's equation. Substituting back,
the general solution is given by
<center>
<math>
R(r) = D_\nu \, I_\nu(k_m r) + E_\nu \, K_\nu(k_m r),\ \nu \in \mathbb{Z},
</math>
</center>
where <math>I_\nu</math> and <math>K_\nu</math> are the modified
[http://en.wikipedia.org/wiki/Bessel_function Bessel functions] of the first
and second kind, respectively, of order <math>\nu</math>.

Note that <math>K_\nu (\mathrm{i} x) = \pi / 2\,\,
\mathrm{i}^{\nu+1} H_\nu^{(2)}(x)</math> with <math>H_\nu^{(2)}</math> denoting
the Hankel function of the second kind of order <math>\nu</math>.
Also, <math>I_\nu</math> does not satisfy the [[Sommerfeld Radiation Condition]]
since it becomes unbounded for increasing real argument and it
represents incoming waves.

Helmholtz's Equation

2010-06-09T00:09:16Z

Administrator: /* Solution for a Circle */

Helmholtz's Equation

2010-06-09T00:08:25Z

Administrator: /* Solution for a Circle */

Helmholtz's Equation

2010-06-09T00:07:55Z

Administrator:

Template:Removing the depth dependence

2010-06-09T00:04:09Z

Administrator:

If we have a problem in which all the scatterers are of constant cross sections so
that
<center>
<math>\partial\Omega = \partial\hat{\Omega} \times z\in[-h,0]
</math>
</center>
where <math>\partial\hat{\Omega} </math> is a function only of <math>x,y</math>
i.e. the boundary of the scattering bodies is uniform with respect to depth.
We can remove the depth dependence [http://en.wikipedia.org/wiki/Separation_of_Variables separation of variables]
and obtain that the dependence on depth is given by
<center>
<math>
\phi(x,y,z) = \frac{\cosh \big( k (z+h) \big)}{\cosh(k h)} \bar{\phi}(x,y)
</math>
</center>
Since <math>\phi</math> satisfies [[Laplace's Equation]], then <math>\Phi</math> satisfies [[Helmholtz's Equation]]
<center>
<math>\nabla^2 \bar{\phi} + k^2 \bar{\phi} = 0 </math>
</center>
in the region not occupied by the scatterers.