WikiWaves - User contributions [en]

Helmholtz's Equation

2013-04-27T21:03:06Z

Mike smith: spelling fix

{{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 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 [[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 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
<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} = \nu^2,
</math>
</center>
<math>\Theta
(\theta)</math> can therefore be expressed as
<center>
<math>
\Theta (\theta) = A \, \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) = B \, J_\nu(k r) + C \, H^{(1)}_\nu(k r),\ \nu \in \mathbb{Z},
</math>
</center>
where <math>J_\nu \,</math> denotes a Bessel function
of the first kind and <math>H^{(1)}_\nu \,</math>
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
Hankel function depends on whether we have positive or negative exponential time dependence.
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>
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>\phi = \phi^{\mathrm{I}}+\phi^{\mathrm{S}} \,</math>, where
<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>
<math>
E_{\nu} = - \frac{D_{\nu} J^{\prime}_\nu (k a)}{ H^{(1)\prime}_\nu (ka)},
</math>
</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 ==

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}) + \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>
</center>
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>

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>
<math>
\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>
</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 as
<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 to obtain
<center>
<math>
\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>
</center>
We now multiply by <math>e^{\mathrm{i} m \gamma} \,</math> and integrate to obtain
<center>
<math>
\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>
</center>

==External Links==
[http://en.wikipedia.org/wiki/Helmholtz_equation Helmholtz Equation on Wikipedia]

[[Category:Linear Water-Wave Theory]]

Burgers Equation

2010-10-14T21:20:13Z

Mike smith: /* Exact Solution of Burgers equations */

{{nonlinear waves course
| chapter title = Burgers Equation
| next chapter =
| previous chapter = [[Reaction-Diffusion Systems]]
}}

==Introduction==

We have already met the conservation law for the traffic equations
<center><math>
\partial _{t}\rho +c\left( \rho \right) \partial _{x}\rho =0
</math></center>
and seen how this leads to shocks. We can smooth this equation by adding
dispersion to the equation to give us
<center><math>
\partial _{t}\rho +c\left( \rho \right) \partial _{x}\rho =\nu \partial
_{x}^{2}\rho
</math></center>
where <math>\nu >0.</math>

The simplest equation of this type is to write
<center><math>
\partial _{t}u+u\partial _{x}u=\nu \partial _{x}^{2}u
</math></center>
(changing variables to <math>u)\ </math>and this equation is known as Burgers equation.

==Travelling Wave Solution==

We can find a travelling wave solution by assuming that
<center><math>
u\left( x,t\right) =u\left( x-ct\right) =u\left( \xi \right)
</math></center>
This leads to the equations
<center><math>
-vu^{\prime }+u^{\prime }u-\nu u^{\prime \prime }=0
</math></center>
We begin by looking at the phase plane for this system, writing <math>w=u^{\prime
}</math> so that
<center><math>\begin{matrix}
\frac{du}{d\xi } &=&w \\
\frac{dw}{d\xi } &=&\frac{1}{\nu }\left( w\left( u-c\right) \right)
\end{matrix}</math></center>
This is a degenerate system with the entire <math>u</math> axis being equilibria.

We can also solve this equation exactly as follows.
<center><math>
-cu^{\prime }+u^{\prime }u-\nu u^{\prime \prime }=0
</math></center>
can be integrated to give
<center><math>
-cu+\frac{1}{2}\left( u\right) ^{2}-\nu u^{\prime }=c_{1}
</math></center>
which can be rearranged to give
<center><math>
u^{\prime }=\frac{1}{2\nu }\left( \left( u\right) ^{2}-2cu^{\prime
}-2c_{1}\right)
</math></center>
We define the two roots of the quadratic <math>\left( u\right) ^{2}-2\nu
u^{\prime }-2c_{1}=0</math> by <math>u_{1}</math> and <math>u_{2}</math>
and we assume that <math>u_{2} < u_{1}</math>. Note that there is only a bounded
solution if we have two real roots and for the bounded solution
<math>u_{2} < u < u_{1}</math>.
We note that the wave speed
is
<center><math>
v=\frac{1}{2}\left( u_{1}+u_{2}\right)
</math></center>
The equation can therefore be written as
<center><math>
2\nu u^{\prime }=\left( u-u_{1}\right) \left( u-u_{2}\right)
</math></center>
which has solution
<center><math>
u\left( \xi \right) =\frac{1}{2}\left( u_{1}+u_{2}\right) -\frac{1}{2}\left(
u_{1}-u_{2}\right) \tanh \left[ \left( \frac{\xi }{4\nu }\right) \left(
u_{1}-u_{2}\right) \right]
</math></center>

==Numerical Solution of Burgers equation==

We can solve the equation using our split step spectral method. The equation
can be written as
<center><math>
\partial _{t}u=-\frac{1}{2}\partial _{x}\left( u^{2}\right) +\nu \partial
_{x}^{2}u
</math></center>
We solve this by solving in Fourier space to give
<center><math>
\partial _{t}\hat{u}=-\frac{1}{2}ik\left( u^{2}\right) -\nu k^{2}\hat{u}
</math></center>
Then we solve each of the steps in turn to get
<center><math>
\partial _{t}u=\partial _{x}\left( u^{2}\right)
</math></center>
for a small time interval to give
<center><math>\begin{matrix}
\tilde{u}\left( k,t+\Delta t\right) &=&\hat{u}\left( k,t\right) -\frac{
\Delta t}{2}ik\mathcal{F}\left( \left[ \mathcal{F}^{-1}\hat{u}\left(
k,t\right) \right] ^{2}\right) \\
\hat{u}\left( k,t+\Delta t\right) &=&\tilde{u}\left( k,t+\Delta t\right)
\exp \left( -\nu k^{2}\Delta t\right)
\end{matrix}</math></center>

==Exact Solution of Burgers equations==

We can find an exact solution to Burgers equation. We want to solve
<center><math>\begin{matrix}
\partial _{t}u+u\partial _{x}u &=&\nu \partial _{x}^{2}u \\
u\left( x,0\right) &=&F\left( x\right)
\end{matrix}</math></center>
Frist we write the equation as
<center><math>
\partial _{t}u+\partial _{x}\left( \frac{u^{2}}{2}-\nu \partial _{x}u\right)
=0
</math></center>
We want to find a function <math>\psi \left( x,t\right) </math> such that
<center><math>
\partial _{x}\psi =u,\ \ \partial _{t}\psi =\nu \partial _{x}u-\frac{u^{2}}{2
}
</math></center>
Note that because <math>\partial _{x}\partial _{t}\psi =\partial _{t}\partial
_{x}\psi </math> we will satisfy Burgers equation. This gives us the following
equation for <math>\psi </math>
<center><math>
\partial _{t}\psi =\nu \partial _{x}^{2}\psi -\frac{1}{2}\left( \partial
_{x}\psi \right) ^{2}
</math></center>
We introduce the ''Cole-Hopf '' transformation
<center><math>
\psi =-2\nu \log \left( \phi \right)
</math></center>
From this we can obtain the three results:
<center><math>
\begin{align}
\partial _{x}\psi &=-2\nu \frac{\partial _{x}\phi }{\phi } \\
\partial _{x}^{2}\psi &=2\nu \left( \frac{\partial _{x}\phi }{\phi }\right)
^{2}-\frac{2\nu }{\phi }\partial _{x}^{2}\phi \\
\partial _{t}\psi &=-2\nu \frac{\partial _{t}\phi }{\phi }
\end{align}
</math></center>
Therefore
<center><math>
\partial _{t}\psi =\nu \partial _{x}^{2}\psi -\frac{1}{2}\left( \partial
_{x}\psi \right) ^{2}
</math></center>
becomes
<center><math>
-2\nu \frac{\partial _{t}\phi }{\phi }=2\nu ^{2}\left( \frac{\partial
_{x}\phi }{\phi }\right) ^{2}+\frac{1}{2}\left( 2\nu \frac{\partial _{x}\phi
}{\phi }\right) ^{2}
</math></center>
or
<center><math>
\partial _{t}\phi =\nu \partial _{x}^{2}\phi
</math></center>
which is just the diffusion equation. Note that we also have to transform the
boundary conditions. We have
<center><math>
F\left( x\right) =u\left( x,0\right) =-2\nu \frac{\partial _{x}\phi \left(
x,0\right) }{\phi \left( x,0\right) }
</math></center>
We can write this as
<center><math>
\frac{d}{dx}\left( \log \left( \phi \right) \right) =-\frac{1}{2\nu }F\left(
x\right)
</math></center>
which has solution
<center><math>
\phi \left( x,0\right) =\Phi \left( x\right) =\exp \left( -\frac{1}{2\nu }
\int_{0}^{x}F\left( s\right) ds\right)
</math></center>
We need to solve
<center><math>\begin{matrix}
\partial _{t}\phi &=&\nu \partial _{x}^{2}\phi \\
\phi \left( x,0\right) &=&\Phi \left( x\right)
\end{matrix}</math></center>
We take the Fourier transform and obtain
<center><math>\begin{matrix}
\partial _{t}\hat{\phi} &=&-k^{2}\nu \hat{\phi} \\
\hat{\phi}\left( k,0\right) &=&\hat{\Phi}\left( k\right)
\end{matrix}</math></center>
which has solution
<center><math>
\hat{\phi}\left( k,t\right) =\hat{\Phi}\left( k\right) e^{-k^{2}\nu t}
</math></center>
We can then use the convolution theorem to write
<center><math>\begin{matrix}
\phi \left( x,t\right) &=&\Phi \left( x\right) * \mathcal{F}^{-1}\left[
e^{-k^{2}\nu t}\right] \\
&=&\frac{1}{2\sqrt{\pi \nu t}}\int_{-\infty }^{\infty }\Phi \left( y\right)
\exp \left[ -\frac{\left( x-y\right) ^{2}}{4\nu t}\right] dy
\end{matrix}</math></center>
Which can be expressed as
<center><math>
\phi \left( x,t\right) =\frac{1}{2\sqrt{\pi \nu t}}\int_{-\infty }^{\infty
}\exp \left[ -\frac{f}{2\nu }\right] dy
</math></center>
where
<center><math>
f\left( x,y,t\right) =\frac{1}{2\nu }\int_{0}^{y}F\left( s\right) ds+\frac{
\left( x-y\right) ^{2}}{2t}
</math></center>
To find <math>u</math> we recall that
<center><math>\begin{matrix}
u\left( x,t\right) &=&-2\nu \frac{\partial _{x}\phi \left( x,t\right) }{\phi
\left( x,t\right) } \\
&=&\frac{\int_{-\infty }^{\infty }\left( \frac{x-y}{t}\right) \exp \left[ -
\frac{f}{2\nu }\right] dy}{\int_{-\infty }^{\infty }\exp \left[ -\frac{f}{
2\nu }\right] dy}
\end{matrix}</math></center>

Burgers Equation

2010-10-14T21:12:53Z

Mike smith: /* Exact Solution of Burgers equations */

{{nonlinear waves course
| chapter title = Burgers Equation
| next chapter =
| previous chapter = [[Reaction-Diffusion Systems]]
}}

==Introduction==

We have already met the conservation law for the traffic equations
<center><math>
\partial _{t}\rho +c\left( \rho \right) \partial _{x}\rho =0
</math></center>
and seen how this leads to shocks. We can smooth this equation by adding
dispersion to the equation to give us
<center><math>
\partial _{t}\rho +c\left( \rho \right) \partial _{x}\rho =\nu \partial
_{x}^{2}\rho
</math></center>
where <math>\nu >0.</math>

The simplest equation of this type is to write
<center><math>
\partial _{t}u+u\partial _{x}u=\nu \partial _{x}^{2}u
</math></center>
(changing variables to <math>u)\ </math>and this equation is known as Burgers equation.

==Travelling Wave Solution==

We can find a travelling wave solution by assuming that
<center><math>
u\left( x,t\right) =u\left( x-ct\right) =u\left( \xi \right)
</math></center>
This leads to the equations
<center><math>
-vu^{\prime }+u^{\prime }u-\nu u^{\prime \prime }=0
</math></center>
We begin by looking at the phase plane for this system, writing <math>w=u^{\prime
}</math> so that
<center><math>\begin{matrix}
\frac{du}{d\xi } &=&w \\
\frac{dw}{d\xi } &=&\frac{1}{\nu }\left( w\left( u-c\right) \right)
\end{matrix}</math></center>
This is a degenerate system with the entire <math>u</math> axis being equilibria.

We can also solve this equation exactly as follows.
<center><math>
-cu^{\prime }+u^{\prime }u-\nu u^{\prime \prime }=0
</math></center>
can be integrated to give
<center><math>
-cu+\frac{1}{2}\left( u\right) ^{2}-\nu u^{\prime }=c_{1}
</math></center>
which can be rearranged to give
<center><math>
u^{\prime }=\frac{1}{2\nu }\left( \left( u\right) ^{2}-2cu^{\prime
}-2c_{1}\right)
</math></center>
We define the two roots of the quadratic <math>\left( u\right) ^{2}-2\nu
u^{\prime }-2c_{1}=0</math> by <math>u_{1}</math> and <math>u_{2}</math>
and we assume that <math>u_{2} < u_{1}</math>. Note that there is only a bounded
solution if we have two real roots and for the bounded solution
<math>u_{2} < u < u_{1}</math>.
We note that the wave speed
is
<center><math>
v=\frac{1}{2}\left( u_{1}+u_{2}\right)
</math></center>
The equation can therefore be written as
<center><math>
2\nu u^{\prime }=\left( u-u_{1}\right) \left( u-u_{2}\right)
</math></center>
which has solution
<center><math>
u\left( \xi \right) =\frac{1}{2}\left( u_{1}+u_{2}\right) -\frac{1}{2}\left(
u_{1}-u_{2}\right) \tanh \left[ \left( \frac{\xi }{4\nu }\right) \left(
u_{1}-u_{2}\right) \right]
</math></center>

==Numerical Solution of Burgers equation==

We can solve the equation using our split step spectral method. The equation
can be written as
<center><math>
\partial _{t}u=-\frac{1}{2}\partial _{x}\left( u^{2}\right) +\nu \partial
_{x}^{2}u
</math></center>
We solve this by solving in Fourier space to give
<center><math>
\partial _{t}\hat{u}=-\frac{1}{2}ik\left( u^{2}\right) -\nu k^{2}\hat{u}
</math></center>
Then we solve each of the steps in turn to get
<center><math>
\partial _{t}u=\partial _{x}\left( u^{2}\right)
</math></center>
for a small time interval to give
<center><math>\begin{matrix}
\tilde{u}\left( k,t+\Delta t\right) &=&\hat{u}\left( k,t\right) -\frac{
\Delta t}{2}ik\mathcal{F}\left( \left[ \mathcal{F}^{-1}\hat{u}\left(
k,t\right) \right] ^{2}\right) \\
\hat{u}\left( k,t+\Delta t\right) &=&\tilde{u}\left( k,t+\Delta t\right)
\exp \left( -\nu k^{2}\Delta t\right)
\end{matrix}</math></center>

==Exact Solution of Burgers equations==

We can find an exact solution to Burgers equation. We want to solve
<center><math>\begin{matrix}
\partial _{t}u+u\partial _{x}u &=&\nu \partial _{x}^{2}u \\
u\left( x,0\right) &=&F\left( x\right)
\end{matrix}</math></center>
Frist we write the equation as
<center><math>
\partial _{t}u+\partial _{x}\left( \frac{u^{2}}{2}-\nu \partial _{x}u\right)
=0
</math></center>
We want to find a function <math>\psi \left( x,t\right) </math> such that
<center><math>
\partial _{x}\psi =u,\ \ \partial _{t}\psi =\nu \partial _{x}u-\frac{u^{2}}{2
}
</math></center>
Note that because <math>\partial _{x}\partial _{t}\psi =\partial _{t}\partial
_{x}\psi </math> we will satisfy Burgers equation. This gives us the following
equation for <math>\psi </math>
<center><math>
\partial _{t}\psi =\nu \partial _{x}^{2}\psi -\frac{1}{2}\left( \partial
_{x}\psi \right) ^{2}
</math></center>
We introduce the ''Cole-Hopf '' transformation
<center><math>
\psi =-2\nu \log \left( \phi \right)
</math></center>
From this we can obtain the three results:
<center><math>
\begin{align}
\partial _{x}\psi &=-2\nu \frac{\partial _{x}\phi }{\phi } \\
\partial _{x}^{2}\psi &=2\nu \left( \frac{\partial _{x}\phi }{\phi }\right)
^{2}-\frac{2\nu }{\phi }\partial _{x}^{2}\phi \\
\partial _{t}\psi &=-2\nu \frac{\partial _{t}\phi }{\phi }
\end{align}
</math></center>
Therefore
<center><math>
\partial _{t}\psi =\nu \partial _{x}^{2}\psi -\frac{1}{2}\left( \partial
_{x}\psi \right) ^{2}
</math></center>
becomes
<center><math>
-2\nu \frac{\partial _{t}\phi }{\phi }=2\nu ^{2}\left( \frac{\partial
_{x}\phi }{\phi }\right) ^{2}+\frac{1}{2}\left( 2\nu \frac{\partial _{x}\phi
}{\phi }\right) ^{2}
</math></center>
or
<center><math>
\partial _{t}\phi =\nu \partial _{x}^{2}\phi
</math></center>
which is just the diffusion equation. Note that we also have to transform the
boundary conditions. We have
<center><math>
F\left( x\right) =u\left( x,0\right) =-2\nu \frac{\partial _{x}\phi \left(
x,0\right) }{\phi \left( x,0\right) }
</math></center>
We can write this as
<center><math>
\frac{d}{dx}\left( \log \left( \phi \right) \right) =-\frac{1}{2\nu }F\left(
x\right)
</math></center>
which has solution
<center><math>
\phi \left( x,0\right) =\Phi \left( x\right) =\exp \left( -\frac{1}{2\nu }
\int_{0}^{x}F\left( s\right) ds\right)
</math></center>
We need to solve
<center><math>\begin{matrix}
\partial _{t}\phi &=&\nu \partial _{x}^{2}\phi \\
\phi \left( x,0\right) &=&\Phi \left( x\right)
\end{matrix}</math></center>
We take the Fourier transform and obtain
<center><math>\begin{matrix}
\partial _{t}\hat{\phi} &=&-k^{2}\nu \hat{\phi} \\
\hat{\phi}\left( k,0\right) &=&\hat{\Phi}\left( k\right)
\end{matrix}</math></center>
which has solution
<center><math>
\hat{\phi}\left( k,t\right) =\hat{\Phi}\left( k\right) e^{-k^{2}\nu t}
</math></center>
We can then use the convolution theorem to write
<center><math>\begin{matrix}
\phi \left( x,t\right) &=&\Phi \left( x\right) * \mathcal{F}^{-1}\left[
e^{-k^{2}\nu t}\right] \\
&=&\frac{1}{2\sqrt{\pi \nu t}}\int_{-\infty }^{\infty }\Phi \left( y\right)
\exp \left[ -\frac{\left( x-y\right) ^{2}}{4\nu t}\right] dy
\end{matrix}</math></center>
We can then write
<center><math>
\phi \left( x,t\right) =\frac{1}{2\sqrt{\pi \nu t}}\int_{-\infty }^{\infty
}\exp \left[ -\frac{f}{2\nu }\right] dy
</math></center>
where
<center><math>
f\left( x,y,t\right) =\frac{1}{2\nu }\int_{0}^{y}F\left( s\right) ds+\frac{
\left( x-y\right) ^{2}}{2t}
</math></center>
To find <math>u</math> we recall that
<center><math>\begin{matrix}
u\left( x,t\right) &=&-2\nu \frac{\partial _{x}\phi \left( x,t\right) }{\phi
\left( x,t\right) } \\
&=&\frac{\int_{-\infty }^{\infty }\left( \frac{x-y}{t}\right) \exp \left[ -
\frac{f}{2\nu }\right] dy}{\int_{-\infty }^{\infty }\exp \left[ -\frac{f}{
2\nu }\right] dy}
\end{matrix}</math></center>

Burgers Equation

2010-10-14T21:10:04Z

Mike smith: /* Exact Solution of Burgers equations */

{{nonlinear waves course
| chapter title = Burgers Equation
| next chapter =
| previous chapter = [[Reaction-Diffusion Systems]]
}}

==Introduction==

We have already met the conservation law for the traffic equations
<center><math>
\partial _{t}\rho +c\left( \rho \right) \partial _{x}\rho =0
</math></center>
and seen how this leads to shocks. We can smooth this equation by adding
dispersion to the equation to give us
<center><math>
\partial _{t}\rho +c\left( \rho \right) \partial _{x}\rho =\nu \partial
_{x}^{2}\rho
</math></center>
where <math>\nu >0.</math>

The simplest equation of this type is to write
<center><math>
\partial _{t}u+u\partial _{x}u=\nu \partial _{x}^{2}u
</math></center>
(changing variables to <math>u)\ </math>and this equation is known as Burgers equation.

==Travelling Wave Solution==

We can find a travelling wave solution by assuming that
<center><math>
u\left( x,t\right) =u\left( x-ct\right) =u\left( \xi \right)
</math></center>
This leads to the equations
<center><math>
-vu^{\prime }+u^{\prime }u-\nu u^{\prime \prime }=0
</math></center>
We begin by looking at the phase plane for this system, writing <math>w=u^{\prime
}</math> so that
<center><math>\begin{matrix}
\frac{du}{d\xi } &=&w \\
\frac{dw}{d\xi } &=&\frac{1}{\nu }\left( w\left( u-c\right) \right)
\end{matrix}</math></center>
This is a degenerate system with the entire <math>u</math> axis being equilibria.

We can also solve this equation exactly as follows.
<center><math>
-cu^{\prime }+u^{\prime }u-\nu u^{\prime \prime }=0
</math></center>
can be integrated to give
<center><math>
-cu+\frac{1}{2}\left( u\right) ^{2}-\nu u^{\prime }=c_{1}
</math></center>
which can be rearranged to give
<center><math>
u^{\prime }=\frac{1}{2\nu }\left( \left( u\right) ^{2}-2cu^{\prime
}-2c_{1}\right)
</math></center>
We define the two roots of the quadratic <math>\left( u\right) ^{2}-2\nu
u^{\prime }-2c_{1}=0</math> by <math>u_{1}</math> and <math>u_{2}</math>
and we assume that <math>u_{2} < u_{1}</math>. Note that there is only a bounded
solution if we have two real roots and for the bounded solution
<math>u_{2} < u < u_{1}</math>.
We note that the wave speed
is
<center><math>
v=\frac{1}{2}\left( u_{1}+u_{2}\right)
</math></center>
The equation can therefore be written as
<center><math>
2\nu u^{\prime }=\left( u-u_{1}\right) \left( u-u_{2}\right)
</math></center>
which has solution
<center><math>
u\left( \xi \right) =\frac{1}{2}\left( u_{1}+u_{2}\right) -\frac{1}{2}\left(
u_{1}-u_{2}\right) \tanh \left[ \left( \frac{\xi }{4\nu }\right) \left(
u_{1}-u_{2}\right) \right]
</math></center>

==Numerical Solution of Burgers equation==

We can solve the equation using our split step spectral method. The equation
can be written as
<center><math>
\partial _{t}u=-\frac{1}{2}\partial _{x}\left( u^{2}\right) +\nu \partial
_{x}^{2}u
</math></center>
We solve this by solving in Fourier space to give
<center><math>
\partial _{t}\hat{u}=-\frac{1}{2}ik\left( u^{2}\right) -\nu k^{2}\hat{u}
</math></center>
Then we solve each of the steps in turn to get
<center><math>
\partial _{t}u=\partial _{x}\left( u^{2}\right)
</math></center>
for a small time interval to give
<center><math>\begin{matrix}
\tilde{u}\left( k,t+\Delta t\right) &=&\hat{u}\left( k,t\right) -\frac{
\Delta t}{2}ik\mathcal{F}\left( \left[ \mathcal{F}^{-1}\hat{u}\left(
k,t\right) \right] ^{2}\right) \\
\hat{u}\left( k,t+\Delta t\right) &=&\tilde{u}\left( k,t+\Delta t\right)
\exp \left( -\nu k^{2}\Delta t\right)
\end{matrix}</math></center>

==Exact Solution of Burgers equations==

We can find an exact solution to Burgers equation. We want to solve
<center><math>\begin{matrix}
\partial _{t}u+u\partial _{x}u &=&\nu \partial _{x}^{2}u \\
u\left( x,0\right) &=&F\left( x\right)
\end{matrix}</math></center>
Frist we write the equation as
<center><math>
\partial _{t}u+\partial _{x}\left( \frac{u^{2}}{2}-\nu \partial _{x}u\right)
=0
</math></center>
We want to find a function <math>\psi \left( x,t\right) </math> such that
<center><math>
\partial _{x}\psi =u,\ \ \partial _{t}\psi =\nu \partial _{x}u-\frac{u^{2}}{2
}
</math></center>
Note that because <math>\partial _{x}\partial _{t}\psi =\partial _{t}\partial
_{x}\psi </math> we will satisfy Burgers equation. This gives us the following
equation for <math>\psi </math>
<center><math>
\partial _{t}\psi =\nu \partial _{x}^{2}\psi -\frac{1}{2}\left( \partial
_{x}\psi \right) ^{2}
</math></center>
We introduce the ''Cole-Hopf '' transformation
<center><math>
\psi =-2\nu \log \left( \phi \right)
</math></center>
From this we can obtain the three results:
<center><math>
\begin{align}
\partial _{x}\psi &=-2\nu \frac{\partial _{x}\phi }{\phi } \\
\partial _{x}^{2}\psi &=2\nu \left( \frac{\partial _{x}\phi }{\phi }\right)
^{2}-\frac{2\nu }{\phi }\partial _{x}^{2}\phi \\
\partial _{t}\psi &=-2\nu \frac{\partial _{t}\phi }{\phi }
\end{align}
</math></center>
Therefore
<center><math>
\partial _{t}\psi =\nu \partial _{x}^{2}\psi -\frac{1}{2}\left( \partial
_{x}\psi \right) ^{2}
</math></center>
becomes
<center><math>
-2\nu \frac{\partial _{t}\phi }{\phi }=2\nu ^{2}\left( \frac{\partial
_{x}\phi }{\phi }\right) ^{2}+\frac{1}{2}\left( 2\nu \frac{\partial _{x}\phi
}{\phi }\right) ^{2}
</math></center>
or
<center><math>
\partial _{t}\phi =\nu \partial _{x}^{2}\phi
</math></center>
which is just the diffusion equation. Note that we also have to transform the
boundary conditions. We have
<center><math>
F\left( x\right) =u\left( x,0\right) =-2\nu \frac{\partial _{x}\phi \left(
x,0\right) }{\phi \left( x,0\right) }
</math></center>
We can write this as
<center><math>
\frac{d}{dx}\left( \log \left( \phi \right) \right) =-\frac{1}{2\nu }F\left(
x\right)
</math></center>
which has solution
<center><math>
\phi \left( x,0\right) =\Phi \left( x\right) =\exp \left( -\frac{1}{2\nu }
\int_{0}^{x}F\left( x\right) ds\right)
</math></center>
We need to solve
<center><math>\begin{matrix}
\partial _{t}\phi &=&\nu \partial _{x}^{2}\phi \\
\phi \left( x,0\right) &=&\Phi \left( x\right)
\end{matrix}</math></center>
We take the Fourier transform and obtain
<center><math>\begin{matrix}
\partial _{t}\hat{\phi} &=&-k^{2}\nu \hat{\phi} \\
\hat{\phi}\left( k,0\right) &=&\hat{\Phi}\left( k\right)
\end{matrix}</math></center>
which has solution
<center><math>
\hat{\phi}\left( k,t\right) =\hat{\Phi}\left( k\right) e^{-k^{2}\nu t}
</math></center>
We can then use the convolution theorem to write
<center><math>\begin{matrix}
\phi \left( x,t\right) &=&\Phi \left( x\right) * \mathcal{F}^{-1}\left[
e^{-k^{2}\nu t}\right] \\
&=&\frac{1}{2\sqrt{\pi \nu t}}\int_{-\infty }^{\infty }\Phi \left( y\right)
\exp \left[ -\frac{\left( x-y\right) ^{2}}{4\nu t}\right] dy
\end{matrix}</math></center>
We can then write
<center><math>
\phi \left( x,t\right) =\frac{1}{2\sqrt{\pi \nu t}}\int_{-\infty }^{\infty
}\exp \left[ -\frac{f}{2\nu }\right] dy
</math></center>
where
<center><math>
f\left( x,y,t\right) =\frac{1}{2\nu }\int_{0}^{y}F\left( s\right) ds+\frac{
\left( x-y\right) ^{2}}{2t}
</math></center>
To find <math>u</math> we recall that
<center><math>\begin{matrix}
u\left( x,t\right) &=&-2\nu \frac{\partial _{x}\phi \left( x,t\right) }{\phi
\left( x,t\right) } \\
&=&\frac{\int_{-\infty }^{\infty }\left( \frac{x-y}{t}\right) \exp \left[ -
\frac{f}{2\nu }\right] dy}{\int_{-\infty }^{\infty }\exp \left[ -\frac{f}{
2\nu }\right] dy}
\end{matrix}</math></center>

Properties of the Linear Schrodinger Equation

2010-09-24T02:50:15Z

Mike smith: /* Example \delta function potential */

{{nonlinear waves course
| chapter title = Properties of the Linear Schrodinger Equation
| next chapter = [[Connection betwen KdV and the Schrodinger Equation]]
| previous chapter = [[Introduction to the Inverse Scattering Transform]]
}}

The linear Schrodinger equation
<center><math>
\partial_{x}^{2}w+uw=-\lambda w
</math></center>
has two kinds of solutions for <math>u\rightarrow0</math> as <math>x\rightarrow\pm\infty.</math> The
first are waves and the second are bound solutions. It is well known that
there are at most a finite number of bound solutions (provided <math>u\rightarrow0</math>
as <math>x\pm\infty</math> sufficiently rapidly) and a continum of solutions for the
incident waves. This is easiest seen through the following examples

==Example <math>\delta</math> function potential==

We consider here the case when <math>u\left( x,0\right) =\delta\left( x\right)
.</math> Note that this function can be thought of as the limit as of the potential
<center><math>
u\left( x\right) =\left\{
\begin{matrix}

0 & x\notin\left[ -\varepsilon,\varepsilon\right] \\
\frac{u_{0}}{2\varepsilon} & x\in\left[ -\varepsilon,\varepsilon\right]
\end{matrix}
\right.
</math></center>
In this case we need to solve
<center><math>
\partial_{x}^{2}w+\delta(x) w=-\lambda w
</math></center>
We consider the case of <math>\lambda<0</math> and <math>\lambda>0</math> separately. For the first
case we write <math>\lambda=-k^{2}</math> and we obtain
<center><math>
w\left( x\right) =\left\{
\begin{matrix}

ae^{kx}, & x<0\\
be^{-kx}, & x>0
\end{matrix}
\right.
</math></center>
We have two conditions at <math>x=0,</math> <math>w</math> must be continuous at <math>0</math> and
<math>\partial_{x}w\left( 0^{+}\right) -\partial_{x}w\left( 0^{-}\right)
+u_0 w\left( 0\right) =0.</math> This final condition is obtained by integrating `across' zero as follows
<center><math>\begin{align}
\int_{0^{-}}^{0^{+}} \partial_x^2 w +\delta(x) w + \lambda w \ dx = 0.
\end{align}
</math></center>

This gives the condition that <math>a=b</math> and
<math>k=u_{0}/2.</math> We need to normalise the eigenfunctions so that
<center><math>
\int_{-\infty}^{\infty}\left( w\left( x\right) \right) ^{2}dx=1.
</math></center>
Therefore
<center><math>
2\int_{0}^{\infty}\left( ae^{-u_{0}x/2}\right) ^{2}dx=1
</math></center>
which means that <math>a=\sqrt{u_{0}/2}.</math> Therefore, there is only one discrete
spectral point which we denote by <math>k_{1}=u_{0}/2</math>
<center><math>
w_{1}\left( x\right) =\left\{
\begin{matrix}

\sqrt{k_{1}}e^{k_{1}x}, & x<0\\
\sqrt{k_{1}}e^{-k_{1}x}, & x>0
\end{matrix}
\right.
</math></center>
The continuous eigenfunctions correspond to <math>\lambda=k^{2}>0</math> are of the form
<center><math>
w\left( x\right) =\left\{
\begin{matrix}

\mathrm{e}^{-\mathrm{i}kx}+r\mathrm{e}^{\mathrm{i}kx}, & x<0\\
a\mathrm{e}^{-\mathrm{i}kx}, & x>0
\end{matrix}
\right.
</math></center>
Again we have the conditions that <math>w</math> must be continuous at <math>0</math> and
<math>\partial_{x}w\left( 0^{+}\right) -\partial_{x}w\left( 0^{-}\right)
+u_{0}w\left( 0\right) =0.</math> This gives us
<center><math>\begin{matrix}
1+r & =a\\
-ika+ik-ikr & =-au_{0}
\end{matrix}</math></center>
which has solution
<center><math>\begin{matrix}
r & =\frac{u_{0}}{2ik-u_{0}}\\
a & =\frac{2ik}{2ik-u_{0}}
\end{matrix}</math></center>

==Example: Scattering by a Well==

The properties of the eigenfunction is perhaps seem most easily through the
following example
<center><math>
u\left( x\right) =\left\{
\begin{matrix}

0 & x\notin\left[ -\varsigma,\varsigma\right] \\
b & x\in\left[ -\varsigma,\varsigma\right]
\end{matrix}
\right.
</math></center>
where <math>b>0.</math>

===Case when <math>\lambda>0</math>===

If we solve this equation for the case when <math>\lambda<0,</math> <math>\lambda=-k^{2}</math> we
get
<center><math>
w\left( x\right) =\left\{
\begin{matrix}

a_{1}e^{kx}, & x<-\varsigma\\
b_{1}\cos\kappa x+b_{2}\sin\kappa x & -\varsigma<x<\varsigma\\
a_{2}e^{-kx} & x>\varsigma
\end{matrix}
\right.
</math></center>
where <math>\kappa=\sqrt{b-k^{2}}</math> which means that <math>0\leq k\leq\sqrt{b}</math> (there is
no solution for <math>k>\sqrt{b}).</math> We then match <math>w</math> and its derivative at
<math>x=\pm\varsigma</math> to solve for <math>a</math> and <math>b</math>. This leads to two system of
equation, one for the even (<math>a_{1}=a_{2}</math> and <math>b_{2}=0</math> ) and one for the odd
solutions (<math>a_{1}=-a_{2}</math> and <math>b_{1}=0)</math>. The solution for the even solutions
is
<center><math>
\left(
\begin{matrix}

e^{-k\varsigma} & -\cos\kappa\varsigma\\
ke^{-k\varsigma} & -\kappa\sin\kappa\varsigma
\end{matrix}
\right) \left(
\begin{matrix}

a_{1}\\
b_{1}
\end{matrix}
\right) =\left(
\begin{matrix}

0\\
0
\end{matrix}
\right)
</math></center>
This has non trivial solutions when
<center><math>
\det\left(
\begin{matrix}

e^{-k\varsigma} & -\cos\kappa\varsigma\\
ke^{-k\varsigma} & -\kappa\sin\kappa\varsigma
\end{matrix}
\right) =0
</math></center>
which gives us the equation
<center><math>
-\kappa e^{-k\varsigma}\sin\kappa\varsigma+k\cos\kappa\varsigma
e^{-k\varsigma}=0
</math></center>
or
<center><math>
\tan\kappa\varsigma=\frac{k}{\kappa}
</math></center>
We know that <math>0<\kappa<\sqrt{b}</math> and if we plot this we see that we obtain a
finite number of solutions.

The solution for the odd solutions is
<center><math>
\left(
\begin{matrix}

e^{-k\varsigma} & -\sin\kappa\varsigma\\
ke^{-k\varsigma} & \cos\kappa\varsigma
\end{matrix}
\right) \left(
\begin{matrix}

a_{1}\\
b_{1}
\end{matrix}
\right) =\left(
\begin{matrix}

0\\
0
\end{matrix}
\right)
</math></center>
This can non trivial solutions when
<center><math>
\det\left(
\begin{matrix}

e^{-k\varsigma} & -\sin\kappa\varsigma\\
ke^{-k\varsigma} & \kappa\cos\kappa\varsigma
\end{matrix}
\right) =0
</math></center>
which gives us the equation
<center><math>
\kappa e^{-k}a\cos\kappa\varsigma+k\sin\kappa\varsigma e^{-k}=0
</math></center>
or
<center><math>
\tan\kappa=-\frac{\kappa}{k}
</math></center>

=== Case when <math>\lambda>0</math> ===

When <math>\lambda>0</math> we write <math>\lambda=k^{2}</math> and we obtain solution
<center><math>
w\left( x\right) =\left\{
\begin{matrix}

\mathrm{e}^{-\mathrm{i}kx}+r\mathrm{e}^{\mathrm{i}kx}, & x<-\varsigma\\
b_{1}\cos\kappa x+b_{2}\sin\kappa x & -\varsigma<x<\varsigma\\
a\mathrm{e}^{-\mathrm{i}kx} & x>\varsigma
\end{matrix}
\right.
</math></center>
where <math>\kappa=\sqrt{b+k^{2}}.</math> Matching <math>w</math> and its derivaties at <math>x=\pm1</math> we
obtain
<center><math>
\left(
\begin{matrix}

-\mathrm{e}^{-\mathrm{i}k\varsigma} & \cos\kappa\varsigma & -\sin\kappa\varsigma & 0\\
ik\mathrm{e}^{-\mathrm{i}k\varsigma} & \kappa\sin\kappa\varsigma & \kappa\cos\kappa
\varsigma & 0\\
0 & \cos\kappa\varsigma & \sin\kappa\varsigma & -\mathrm{e}^{-\mathrm{i}k\varsigma}\\
0 & -\kappa\sin\kappa\varsigma & \kappa\cos\kappa\varsigma &
ik\mathrm{e}^{-\mathrm{i}k\varsigma}
\end{matrix}
\right) \left(
\begin{matrix}

r\\
b_{1}\\
b_{2}\\
a
\end{matrix}
\right) =\left(
\begin{matrix}

\mathrm{e}^{\mathrm{i}k}\\
ik\mathrm{e}^{-\mathrm{i}k}\\
0\\
0
\end{matrix}
\right)
</math></center>

Properties of the Linear Schrodinger Equation

2010-09-24T02:40:12Z

Mike smith: /* Example: Scattering by a Well */

{{nonlinear waves course
| chapter title = Properties of the Linear Schrodinger Equation
| next chapter = [[Connection betwen KdV and the Schrodinger Equation]]
| previous chapter = [[Introduction to the Inverse Scattering Transform]]
}}

The linear Schrodinger equation
<center><math>
\partial_{x}^{2}w+uw=-\lambda w
</math></center>
has two kinds of solutions for <math>u\rightarrow0</math> as <math>x\rightarrow\pm\infty.</math> The
first are waves and the second are bound solutions. It is well known that
there are at most a finite number of bound solutions (provided <math>u\rightarrow0</math>
as <math>x\pm\infty</math> sufficiently rapidly) and a continum of solutions for the
incident waves. This is easiest seen through the following examples

==Example <math>\delta</math> function potential==

We consider here the case when <math>u\left( x,0\right) =\delta\left( x\right)
.</math> Note that this function can be thought of as the limit as of the potential
<center><math>
u\left( x\right) =\left\{
\begin{matrix}

0 & x\notin\left[ -\varepsilon,\varepsilon\right] \\
\frac{u_{0}}{2\varepsilon} & x\in\left[ -\varepsilon,\varepsilon\right]
\end{matrix}
\right.
</math></center>
In this case we need to solve
<center><math>
\partial_{x}^{2}w+\delta(x) w=-\lambda w
</math></center>
We consider the case of <math>\lambda<0</math> and <math>\lambda>0</math> separately. For the first
case we write <math>\lambda=-k^{2}</math> and we obtain
<center><math>
w\left( x\right) =\left\{
\begin{matrix}

ae^{kx}, & x<0\\
be^{-kx}, & x>0
\end{matrix}
\right.
</math></center>
We have two conditions at <math>x=0,</math> <math>w</math> must be continuous at <math>0</math> and
<math>\partial_{x}w\left( 0^{+}\right) -\partial_{x}w\left( 0^{-}\right)
+u_{0}w\left( 0\right) =0.</math> This gives the condition that <math>a=b</math> and
<math>k=u_{0}/2.</math> We need to normalise the eigenfunctions so that
<center><math>
\int_{-\infty}^{\infty}\left( w\left( x\right) \right) ^{2}dx=1.
</math></center>
Therefore
<center><math>
2\int_{0}^{\infty}\left( ae^{-u_{0}x/2}\right) ^{2}dx=1
</math></center>
which means that <math>a=\sqrt{u_{0}/2}.</math> Therefore, there is only one discrete
spectral point which we denote by <math>k_{1}=u_{0}/2</math>
<center><math>
w_{1}\left( x\right) =\left\{
\begin{matrix}

\sqrt{k_{1}}e^{k_{1}x}, & x<0\\
\sqrt{k_{1}}e^{-k_{1}x}, & x>0
\end{matrix}
\right.
</math></center>
The continuous eigenfunctions correspond to <math>\lambda=k^{2}>0</math> are of the form
<center><math>
w\left( x\right) =\left\{
\begin{matrix}

\mathrm{e}^{-\mathrm{i}kx}+r\mathrm{e}^{\mathrm{i}kx}, & x<0\\
a\mathrm{e}^{-\mathrm{i}kx}, & x>0
\end{matrix}
\right.
</math></center>
Again we have the conditions that <math>w</math> must be continuous at <math>0</math> and
<math>\partial_{x}w\left( 0^{+}\right) -\partial_{x}w\left( 0^{-}\right)
+u_{0}w\left( 0\right) =0.</math> This gives us
<center><math>\begin{matrix}
1+r & =a\\
-ika+ik-ikr & =-au_{0}
\end{matrix}</math></center>
which has solution
<center><math>\begin{matrix}
r & =\frac{u_{0}}{2ik-u_{0}}\\
a & =\frac{2ik}{2ik-u_{0}}
\end{matrix}</math></center>

==Example: Scattering by a Well==

The properties of the eigenfunction is perhaps seem most easily through the
following example
<center><math>
u\left( x\right) =\left\{
\begin{matrix}

0 & x\notin\left[ -\varsigma,\varsigma\right] \\
b & x\in\left[ -\varsigma,\varsigma\right]
\end{matrix}
\right.
</math></center>
where <math>b>0.</math>

===Case when <math>\lambda>0</math>===

If we solve this equation for the case when <math>\lambda<0,</math> <math>\lambda=-k^{2}</math> we
get
<center><math>
w\left( x\right) =\left\{
\begin{matrix}

a_{1}e^{kx}, & x<-\varsigma\\
b_{1}\cos\kappa x+b_{2}\sin\kappa x & -\varsigma<x<\varsigma\\
a_{2}e^{-kx} & x>\varsigma
\end{matrix}
\right.
</math></center>
where <math>\kappa=\sqrt{b-k^{2}}</math> which means that <math>0\leq k\leq\sqrt{b}</math> (there is
no solution for <math>k>\sqrt{b}).</math> We then match <math>w</math> and its derivative at
<math>x=\pm\varsigma</math> to solve for <math>a</math> and <math>b</math>. This leads to two system of
equation, one for the even (<math>a_{1}=a_{2}</math> and <math>b_{2}=0</math> ) and one for the odd
solutions (<math>a_{1}=-a_{2}</math> and <math>b_{1}=0)</math>. The solution for the even solutions
is
<center><math>
\left(
\begin{matrix}

e^{-k\varsigma} & -\cos\kappa\varsigma\\
ke^{-k\varsigma} & -\kappa\sin\kappa\varsigma
\end{matrix}
\right) \left(
\begin{matrix}

a_{1}\\
b_{1}
\end{matrix}
\right) =\left(
\begin{matrix}

0\\
0
\end{matrix}
\right)
</math></center>
This has non trivial solutions when
<center><math>
\det\left(
\begin{matrix}

e^{-k\varsigma} & -\cos\kappa\varsigma\\
ke^{-k\varsigma} & -\kappa\sin\kappa\varsigma
\end{matrix}
\right) =0
</math></center>
which gives us the equation
<center><math>
-\kappa e^{-k\varsigma}\sin\kappa\varsigma+k\cos\kappa\varsigma
e^{-k\varsigma}=0
</math></center>
or
<center><math>
\tan\kappa\varsigma=\frac{k}{\kappa}
</math></center>
We know that <math>0<\kappa<\sqrt{b}</math> and if we plot this we see that we obtain a
finite number of solutions.

The solution for the odd solutions is
<center><math>
\left(
\begin{matrix}

e^{-k\varsigma} & -\sin\kappa\varsigma\\
ke^{-k\varsigma} & \cos\kappa\varsigma
\end{matrix}
\right) \left(
\begin{matrix}

a_{1}\\
b_{1}
\end{matrix}
\right) =\left(
\begin{matrix}

0\\
0
\end{matrix}
\right)
</math></center>
This can non trivial solutions when
<center><math>
\det\left(
\begin{matrix}

e^{-k\varsigma} & -\sin\kappa\varsigma\\
ke^{-k\varsigma} & \kappa\cos\kappa\varsigma
\end{matrix}
\right) =0
</math></center>
which gives us the equation
<center><math>
\kappa e^{-k}a\cos\kappa\varsigma+k\sin\kappa\varsigma e^{-k}=0
</math></center>
or
<center><math>
\tan\kappa=-\frac{\kappa}{k}
</math></center>

=== Case when <math>\lambda>0</math> ===

When <math>\lambda>0</math> we write <math>\lambda=k^{2}</math> and we obtain solution
<center><math>
w\left( x\right) =\left\{
\begin{matrix}

\mathrm{e}^{-\mathrm{i}kx}+r\mathrm{e}^{\mathrm{i}kx}, & x<-\varsigma\\
b_{1}\cos\kappa x+b_{2}\sin\kappa x & -\varsigma<x<\varsigma\\
a\mathrm{e}^{-\mathrm{i}kx} & x>\varsigma
\end{matrix}
\right.
</math></center>
where <math>\kappa=\sqrt{b+k^{2}}.</math> Matching <math>w</math> and its derivaties at <math>x=\pm1</math> we
obtain
<center><math>
\left(
\begin{matrix}

-\mathrm{e}^{-\mathrm{i}k\varsigma} & \cos\kappa\varsigma & -\sin\kappa\varsigma & 0\\
ik\mathrm{e}^{-\mathrm{i}k\varsigma} & \kappa\sin\kappa\varsigma & \kappa\cos\kappa
\varsigma & 0\\
0 & \cos\kappa\varsigma & \sin\kappa\varsigma & -\mathrm{e}^{-\mathrm{i}k\varsigma}\\
0 & -\kappa\sin\kappa\varsigma & \kappa\cos\kappa\varsigma &
ik\mathrm{e}^{-\mathrm{i}k\varsigma}
\end{matrix}
\right) \left(
\begin{matrix}

r\\
b_{1}\\
b_{2}\\
a
\end{matrix}
\right) =\left(
\begin{matrix}

\mathrm{e}^{\mathrm{i}k}\\
ik\mathrm{e}^{-\mathrm{i}k}\\
0\\
0
\end{matrix}
\right)
</math></center>

Properties of the Linear Schrodinger Equation

2010-09-24T02:38:40Z

Mike smith: /* Example \delta function potential */

{{nonlinear waves course
| chapter title = Properties of the Linear Schrodinger Equation
| next chapter = [[Connection betwen KdV and the Schrodinger Equation]]
| previous chapter = [[Introduction to the Inverse Scattering Transform]]
}}

The linear Schrodinger equation
<center><math>
\partial_{x}^{2}w+uw=-\lambda w
</math></center>
has two kinds of solutions for <math>u\rightarrow0</math> as <math>x\rightarrow\pm\infty.</math> The
first are waves and the second are bound solutions. It is well known that
there are at most a finite number of bound solutions (provided <math>u\rightarrow0</math>
as <math>x\pm\infty</math> sufficiently rapidly) and a continum of solutions for the
incident waves. This is easiest seen through the following examples

==Example <math>\delta</math> function potential==

We consider here the case when <math>u\left( x,0\right) =\delta\left( x\right)
.</math> Note that this function can be thought of as the limit as of the potential
<center><math>
u\left( x\right) =\left\{
\begin{matrix}

0 & x\notin\left[ -\varepsilon,\varepsilon\right] \\
\frac{u_{0}}{2\varepsilon} & x\in\left[ -\varepsilon,\varepsilon\right]
\end{matrix}
\right.
</math></center>
In this case we need to solve
<center><math>
\partial_{x}^{2}w+\delta(x) w=-\lambda w
</math></center>
We consider the case of <math>\lambda<0</math> and <math>\lambda>0</math> separately. For the first
case we write <math>\lambda=-k^{2}</math> and we obtain
<center><math>
w\left( x\right) =\left\{
\begin{matrix}

ae^{kx}, & x<0\\
be^{-kx}, & x>0
\end{matrix}
\right.
</math></center>
We have two conditions at <math>x=0,</math> <math>w</math> must be continuous at <math>0</math> and
<math>\partial_{x}w\left( 0^{+}\right) -\partial_{x}w\left( 0^{-}\right)
+u_{0}w\left( 0\right) =0.</math> This gives the condition that <math>a=b</math> and
<math>k=u_{0}/2.</math> We need to normalise the eigenfunctions so that
<center><math>
\int_{-\infty}^{\infty}\left( w\left( x\right) \right) ^{2}dx=1.
</math></center>
Therefore
<center><math>
2\int_{0}^{\infty}\left( ae^{-u_{0}x/2}\right) ^{2}dx=1
</math></center>
which means that <math>a=\sqrt{u_{0}/2}.</math> Therefore, there is only one discrete
spectral point which we denote by <math>k_{1}=u_{0}/2</math>
<center><math>
w_{1}\left( x\right) =\left\{
\begin{matrix}

\sqrt{k_{1}}e^{k_{1}x}, & x<0\\
\sqrt{k_{1}}e^{-k_{1}x}, & x>0
\end{matrix}
\right.
</math></center>
The continuous eigenfunctions correspond to <math>\lambda=k^{2}>0</math> are of the form
<center><math>
w\left( x\right) =\left\{
\begin{matrix}

\mathrm{e}^{-\mathrm{i}kx}+r\mathrm{e}^{\mathrm{i}kx}, & x<0\\
a\mathrm{e}^{-\mathrm{i}kx}, & x>0
\end{matrix}
\right.
</math></center>
Again we have the conditions that <math>w</math> must be continuous at <math>0</math> and
<math>\partial_{x}w\left( 0^{+}\right) -\partial_{x}w\left( 0^{-}\right)
+u_{0}w\left( 0\right) =0.</math> This gives us
<center><math>\begin{matrix}
1+r & =a\\
-ika+ik-ikr & =-au_{0}
\end{matrix}</math></center>
which has solution
<center><math>\begin{matrix}
r & =\frac{u_{0}}{2ik-u_{0}}\\
a & =\frac{2ik}{2ik-u_{0}}
\end{matrix}</math></center>

==Example: Scattering by a Well==

The properties of the eigenfunction is prehaps seem most easily through the
following example
<center><math>
u\left( x\right) =\left\{
\begin{matrix}

0 & x\notin\left[ -\varepsilon,\varepsilon\right] \\
b & x\in\left[ -\varepsilon,\varepsilon\right]
\end{matrix}
\right.
</math></center>
where <math>b>0.</math>

===Case when <math>\lambda>0</math>===

If we solve this equation for the case when <math>\lambda<0,</math> <math>\lambda=-k^{2}</math> we
get
<center><math>
w\left( x\right) =\left\{
\begin{matrix}

a_{1}e^{kx}, & x<-\varepsilon\\
b_{1}\cos\kappa x+b_{2}\sin\kappa x & -\varepsilon<x<\varepsilon\\
a_{2}e^{-kx} & x>\varepsilon
\end{matrix}
\right.
</math></center>
where <math>\kappa=\sqrt{b-k^{2}}</math> which means that <math>0\leq k\leq\sqrt{b}</math> (there is
no solution for <math>k>\sqrt{b}).</math> We then match <math>w</math> and its derivative at
<math>x=\pm\varepsilon</math> to solve for <math>a</math> and <math>b</math>. This leads to two system of
equation, one for the even (<math>a_{1}=a_{2}</math> and <math>b_{2}=0</math> ) and one for the odd
solutions (<math>a_{1}=-a_{2}</math> and <math>b_{1}=0)</math>. The solution for the even solutions
is
<center><math>
\left(
\begin{matrix}

e^{-k\varepsilon} & -\cos\kappa\varepsilon\\
ke^{-k\varepsilon} & -\kappa\sin\kappa\varepsilon
\end{matrix}
\right) \left(
\begin{matrix}

a_{1}\\
b_{1}
\end{matrix}
\right) =\left(
\begin{matrix}

0\\
0
\end{matrix}
\right)
</math></center>
This has non trivial solutions when
<center><math>
\det\left(
\begin{matrix}

e^{-k\varepsilon} & -\cos\kappa\varepsilon\\
ke^{-k\varepsilon} & -\kappa\sin\kappa\varepsilon
\end{matrix}
\right) =0
</math></center>
which gives us the equation
<center><math>
-\kappa e^{-k\varepsilon}\sin\kappa\varepsilon+k\cos\kappa\varepsilon
e^{-k\varepsilon}=0
</math></center>
or
<center><math>
\tan\kappa\varepsilon=\frac{k}{\kappa}
</math></center>
We know that <math>0<\kappa<\sqrt{b}</math> and if we plot this we see that we obtain a
finite number of solutions.

The solution for the odd solutions is
<center><math>
\left(
\begin{matrix}

e^{-k\varepsilon} & -\sin\kappa\varepsilon\\
ke^{-k\varepsilon} & \cos\kappa\varepsilon
\end{matrix}
\right) \left(
\begin{matrix}

a_{1}\\
b_{1}
\end{matrix}
\right) =\left(
\begin{matrix}

0\\
0
\end{matrix}
\right)
</math></center>
This can non trivial solutions when
<center><math>
\det\left(
\begin{matrix}

e^{-k\varepsilon} & -\sin\kappa\varepsilon\\
ke^{-k\varepsilon} & \kappa\cos\kappa\varepsilon
\end{matrix}
\right) =0
</math></center>
which gives us the equation
<center><math>
\kappa e^{-k}a\cos\kappa\varepsilon+k\sin\kappa\varepsilon e^{-k}=0
</math></center>
or
<center><math>
\tan\kappa=-\frac{\kappa}{k}
</math></center>

=== Case when <math>\lambda>0</math> ===

When <math>\lambda>0</math> we write <math>\lambda=k^{2}</math> and we obtain solution
<center><math>
w\left( x\right) =\left\{
\begin{matrix}

\mathrm{e}^{-\mathrm{i}kx}+r\mathrm{e}^{\mathrm{i}kx}, & x<-\varepsilon\\
b_{1}\cos\kappa x+b_{2}\sin\kappa x & -\varepsilon<x<\varepsilon\\
a\mathrm{e}^{-\mathrm{i}kx} & x>\varepsilon
\end{matrix}
\right.
</math></center>
where <math>\kappa=\sqrt{b+k^{2}}.</math> Matching <math>w</math> and its derivaties at <math>x=\pm1</math> we
obtain
<center><math>
\left(
\begin{matrix}

-\mathrm{e}^{-\mathrm{i}k\varepsilon} & \cos\kappa\varepsilon & -\sin\kappa\varepsilon & 0\\
ik\mathrm{e}^{-\mathrm{i}k\varepsilon} & \kappa\sin\kappa\varepsilon & \kappa\cos\kappa
\varepsilon & 0\\
0 & \cos\kappa\varepsilon & \sin\kappa\varepsilon & -\mathrm{e}^{-\mathrm{i}k\varepsilon}\\
0 & -\kappa\sin\kappa\varepsilon & \kappa\cos\kappa\varepsilon &
ik\mathrm{e}^{-\mathrm{i}k\varepsilon}
\end{matrix}
\right) \left(
\begin{matrix}

r\\
b_{1}\\
b_{2}\\
a
\end{matrix}
\right) =\left(
\begin{matrix}

\mathrm{e}^{\mathrm{i}k}\\
ik\mathrm{e}^{-\mathrm{i}k}\\
0\\
0
\end{matrix}
\right)
</math></center>

Properties of the Linear Schrodinger Equation

2010-09-24T02:37:43Z

Mike smith: /* Example \delta function potential */

{{nonlinear waves course
| chapter title = Properties of the Linear Schrodinger Equation
| next chapter = [[Connection betwen KdV and the Schrodinger Equation]]
| previous chapter = [[Introduction to the Inverse Scattering Transform]]
}}

The linear Schrodinger equation
<center><math>
\partial_{x}^{2}w+uw=-\lambda w
</math></center>
has two kinds of solutions for <math>u\rightarrow0</math> as <math>x\rightarrow\pm\infty.</math> The
first are waves and the second are bound solutions. It is well known that
there are at most a finite number of bound solutions (provided <math>u\rightarrow0</math>
as <math>x\pm\infty</math> sufficiently rapidly) and a continum of solutions for the
incident waves. This is easiest seen through the following examples

==Example <math>\delta</math> function potential==

We consider here the case when <math>u\left( x,0\right) =\delta\left( x\right)
.</math> Note that this function can be thought of as the limit as of the potential
<center><math>
u\left( x\right) =\left\{
\begin{matrix}

0 & x\notin\left[ -\varepsilon,\varepsilon\right] \\
\frac{u_{0}}{2\varepsilon} & x\in\left[ -\varepsilon,\varepsilon\right]
\end{matrix}
\right.
</math></center>
In this case we need to solve
<center><math>
\partial_{x}^{2}w+\delta w=-\lambda w
</math></center>
We consider the case of <math>\lambda<0</math> and <math>\lambda>0</math> separately. For the first
case we write <math>\lambda=-k^{2}</math> and we obtain
<center><math>
w\left( x\right) =\left\{
\begin{matrix}

ae^{kx}, & x<0\\
be^{-kx}, & x>0
\end{matrix}
\right.
</math></center>
We have two conditions at <math>x=0,</math> <math>w</math> must be continuous at <math>0</math> and
<math>\partial_{x}w\left( 0^{+}\right) -\partial_{x}w\left( 0^{-}\right)
+u_{0}w\left( 0\right) =0.</math> This gives the condition that <math>a=b</math> and
<math>k=u_{0}/2.</math> We need to normalise the eigenfunctions so that
<center><math>
\int_{-\infty}^{\infty}\left( w\left( x\right) \right) ^{2}dx=1.
</math></center>
Therefore
<center><math>
2\int_{0}^{\infty}\left( ae^{-u_{0}x/2}\right) ^{2}dx=1
</math></center>
which means that <math>a=\sqrt{u_{0}/2}.</math> Therefore, there is only one discrete
spectral point which we denote by <math>k_{1}=u_{0}/2</math>
<center><math>
w_{1}\left( x\right) =\left\{
\begin{matrix}

\sqrt{k_{1}}e^{k_{1}x}, & x<0\\
\sqrt{k_{1}}e^{-k_{1}x}, & x>0
\end{matrix}
\right.
</math></center>
The continuous eigenfunctions correspond to <math>\lambda=k^{2}>0</math> are of the form
<center><math>
w\left( x\right) =\left\{
\begin{matrix}

\mathrm{e}^{-\mathrm{i}kx}+r\mathrm{e}^{\mathrm{i}kx}, & x<0\\
a\mathrm{e}^{-\mathrm{i}kx}, & x>0
\end{matrix}
\right.
</math></center>
Again we have the conditions that <math>w</math> must be continuous at <math>0</math> and
<math>\partial_{x}w\left( 0^{+}\right) -\partial_{x}w\left( 0^{-}\right)
+u_{0}w\left( 0\right) =0.</math> This gives us
<center><math>\begin{matrix}
1+r & =a\\
-ika+ik-ikr & =-au_{0}
\end{matrix}</math></center>
which has solution
<center><math>\begin{matrix}
r & =\frac{u_{0}}{2ik-u_{0}}\\
a & =\frac{2ik}{2ik-u_{0}}
\end{matrix}</math></center>

==Example: Scattering by a Well==

The properties of the eigenfunction is prehaps seem most easily through the
following example
<center><math>
u\left( x\right) =\left\{
\begin{matrix}

0 & x\notin\left[ -\varepsilon,\varepsilon\right] \\
b & x\in\left[ -\varepsilon,\varepsilon\right]
\end{matrix}
\right.
</math></center>
where <math>b>0.</math>

===Case when <math>\lambda>0</math>===

If we solve this equation for the case when <math>\lambda<0,</math> <math>\lambda=-k^{2}</math> we
get
<center><math>
w\left( x\right) =\left\{
\begin{matrix}

a_{1}e^{kx}, & x<-\varepsilon\\
b_{1}\cos\kappa x+b_{2}\sin\kappa x & -\varepsilon<x<\varepsilon\\
a_{2}e^{-kx} & x>\varepsilon
\end{matrix}
\right.
</math></center>
where <math>\kappa=\sqrt{b-k^{2}}</math> which means that <math>0\leq k\leq\sqrt{b}</math> (there is
no solution for <math>k>\sqrt{b}).</math> We then match <math>w</math> and its derivative at
<math>x=\pm\varepsilon</math> to solve for <math>a</math> and <math>b</math>. This leads to two system of
equation, one for the even (<math>a_{1}=a_{2}</math> and <math>b_{2}=0</math> ) and one for the odd
solutions (<math>a_{1}=-a_{2}</math> and <math>b_{1}=0)</math>. The solution for the even solutions
is
<center><math>
\left(
\begin{matrix}

e^{-k\varepsilon} & -\cos\kappa\varepsilon\\
ke^{-k\varepsilon} & -\kappa\sin\kappa\varepsilon
\end{matrix}
\right) \left(
\begin{matrix}

a_{1}\\
b_{1}
\end{matrix}
\right) =\left(
\begin{matrix}

0\\
0
\end{matrix}
\right)
</math></center>
This has non trivial solutions when
<center><math>
\det\left(
\begin{matrix}

e^{-k\varepsilon} & -\cos\kappa\varepsilon\\
ke^{-k\varepsilon} & -\kappa\sin\kappa\varepsilon
\end{matrix}
\right) =0
</math></center>
which gives us the equation
<center><math>
-\kappa e^{-k\varepsilon}\sin\kappa\varepsilon+k\cos\kappa\varepsilon
e^{-k\varepsilon}=0
</math></center>
or
<center><math>
\tan\kappa\varepsilon=\frac{k}{\kappa}
</math></center>
We know that <math>0<\kappa<\sqrt{b}</math> and if we plot this we see that we obtain a
finite number of solutions.

The solution for the odd solutions is
<center><math>
\left(
\begin{matrix}

e^{-k\varepsilon} & -\sin\kappa\varepsilon\\
ke^{-k\varepsilon} & \cos\kappa\varepsilon
\end{matrix}
\right) \left(
\begin{matrix}

a_{1}\\
b_{1}
\end{matrix}
\right) =\left(
\begin{matrix}

0\\
0
\end{matrix}
\right)
</math></center>
This can non trivial solutions when
<center><math>
\det\left(
\begin{matrix}

e^{-k\varepsilon} & -\sin\kappa\varepsilon\\
ke^{-k\varepsilon} & \kappa\cos\kappa\varepsilon
\end{matrix}
\right) =0
</math></center>
which gives us the equation
<center><math>
\kappa e^{-k}a\cos\kappa\varepsilon+k\sin\kappa\varepsilon e^{-k}=0
</math></center>
or
<center><math>
\tan\kappa=-\frac{\kappa}{k}
</math></center>

=== Case when <math>\lambda>0</math> ===

When <math>\lambda>0</math> we write <math>\lambda=k^{2}</math> and we obtain solution
<center><math>
w\left( x\right) =\left\{
\begin{matrix}

\mathrm{e}^{-\mathrm{i}kx}+r\mathrm{e}^{\mathrm{i}kx}, & x<-\varepsilon\\
b_{1}\cos\kappa x+b_{2}\sin\kappa x & -\varepsilon<x<\varepsilon\\
a\mathrm{e}^{-\mathrm{i}kx} & x>\varepsilon
\end{matrix}
\right.
</math></center>
where <math>\kappa=\sqrt{b+k^{2}}.</math> Matching <math>w</math> and its derivaties at <math>x=\pm1</math> we
obtain
<center><math>
\left(
\begin{matrix}

-\mathrm{e}^{-\mathrm{i}k\varepsilon} & \cos\kappa\varepsilon & -\sin\kappa\varepsilon & 0\\
ik\mathrm{e}^{-\mathrm{i}k\varepsilon} & \kappa\sin\kappa\varepsilon & \kappa\cos\kappa
\varepsilon & 0\\
0 & \cos\kappa\varepsilon & \sin\kappa\varepsilon & -\mathrm{e}^{-\mathrm{i}k\varepsilon}\\
0 & -\kappa\sin\kappa\varepsilon & \kappa\cos\kappa\varepsilon &
ik\mathrm{e}^{-\mathrm{i}k\varepsilon}
\end{matrix}
\right) \left(
\begin{matrix}

r\\
b_{1}\\
b_{2}\\
a
\end{matrix}
\right) =\left(
\begin{matrix}

\mathrm{e}^{\mathrm{i}k}\\
ik\mathrm{e}^{-\mathrm{i}k}\\
0\\
0
\end{matrix}
\right)
</math></center>

Properties of the Linear Schrodinger Equation

2010-09-24T02:17:49Z

Mike smith: /* Example \delta function potential */

{{nonlinear waves course
| chapter title = Properties of the Linear Schrodinger Equation
| next chapter = [[Connection betwen KdV and the Schrodinger Equation]]
| previous chapter = [[Introduction to the Inverse Scattering Transform]]
}}

The linear Schrodinger equation
<center><math>
\partial_{x}^{2}w+uw=-\lambda w
</math></center>
has two kinds of solutions for <math>u\rightarrow0</math> as <math>x\rightarrow\pm\infty.</math> The
first are waves and the second are bound solutions. It is well known that
there are at most a finite number of bound solutions (provided <math>u\rightarrow0</math>
as <math>x\pm\infty</math> sufficiently rapidly) and a continum of solutions for the
incident waves. This is easiest seen through the following examples

==Example <math>\delta</math> function potential==

We consider here the case when <math>u\left( x,0\right) =\delta\left( x\right)
.</math> Note that this function can be thought of as the limit as of the potential
<center><math>
u\left( x\right) =\left\{
\begin{matrix}

0 & x\notin\left[ -\varepsilon,\varepsilon\right] \\
\frac{u_{0}}{2\varepsilon} & x\in\left[ -\varepsilon,\varepsilon\right]
\end{matrix}
\right.
</math></center>
In this case we need to solve
<center><math>
\partial_{x}^{2}w+\frac{u_{0}}{2 \epsilon} w=-\lambda w
</math></center>
We consider the case of <math>\lambda<0</math> and <math>\lambda>0</math> separately. For the first
case we write <math>\lambda=-k^{2}</math> and we obtain
<center><math>
w\left( x\right) =\left\{
\begin{matrix}

ae^{kx}, & x<0\\
be^{-kx}, & x>0
\end{matrix}
\right.
</math></center>
We have two conditions at <math>x=0,</math> <math>w</math> must be continuous at <math>0</math> and
<math>\partial_{x}w\left( 0^{+}\right) -\partial_{x}w\left( 0^{-}\right)
+u_{0}w\left( 0\right) =0.</math> This gives the condition that <math>a=b</math> and
<math>k=u_{0}/2.</math> We need to normalise the eigenfunctions so that
<center><math>
\int_{-\infty}^{\infty}\left( w\left( x\right) \right) ^{2}dx=1.
</math></center>
Therefore
<center><math>
2\int_{0}^{\infty}\left( ae^{-u_{0}x/2}\right) ^{2}dx=1
</math></center>
which means that <math>a=\sqrt{u_{0}/2}.</math> Therefore, there is only one discrete
spectral point which we denote by <math>k_{1}=u_{0}/2</math>
<center><math>
w_{1}\left( x\right) =\left\{
\begin{matrix}

\sqrt{k_{1}}e^{k_{1}x}, & x<0\\
\sqrt{k_{1}}e^{-k_{1}x}, & x>0
\end{matrix}
\right.
</math></center>
The continuous eigenfunctions correspond to <math>\lambda=k^{2}>0</math> are of the form
<center><math>
w\left( x\right) =\left\{
\begin{matrix}

\mathrm{e}^{-\mathrm{i}kx}+r\mathrm{e}^{\mathrm{i}kx}, & x<0\\
a\mathrm{e}^{-\mathrm{i}kx}, & x>0
\end{matrix}
\right.
</math></center>
Again we have the conditions that <math>w</math> must be continuous at <math>0</math> and
<math>\partial_{x}w\left( 0^{+}\right) -\partial_{x}w\left( 0^{-}\right)
+u_{0}w\left( 0\right) =0.</math> This gives us
<center><math>\begin{matrix}
1+r & =a\\
-ika+ik-ikr & =-au_{0}
\end{matrix}</math></center>
which has solution
<center><math>\begin{matrix}
r & =\frac{u_{0}}{2ik-u_{0}}\\
a & =\frac{2ik}{2ik-u_{0}}
\end{matrix}</math></center>

==Example: Scattering by a Well==

The properties of the eigenfunction is prehaps seem most easily through the
following example
<center><math>
u\left( x\right) =\left\{
\begin{matrix}

0 & x\notin\left[ -\varepsilon,\varepsilon\right] \\
b & x\in\left[ -\varepsilon,\varepsilon\right]
\end{matrix}
\right.
</math></center>
where <math>b>0.</math>

===Case when <math>\lambda>0</math>===

If we solve this equation for the case when <math>\lambda<0,</math> <math>\lambda=-k^{2}</math> we
get
<center><math>
w\left( x\right) =\left\{
\begin{matrix}

a_{1}e^{kx}, & x<-\varepsilon\\
b_{1}\cos\kappa x+b_{2}\sin\kappa x & -\varepsilon<x<\varepsilon\\
a_{2}e^{-kx} & x>\varepsilon
\end{matrix}
\right.
</math></center>
where <math>\kappa=\sqrt{b-k^{2}}</math> which means that <math>0\leq k\leq\sqrt{b}</math> (there is
no solution for <math>k>\sqrt{b}).</math> We then match <math>w</math> and its derivative at
<math>x=\pm\varepsilon</math> to solve for <math>a</math> and <math>b</math>. This leads to two system of
equation, one for the even (<math>a_{1}=a_{2}</math> and <math>b_{2}=0</math> ) and one for the odd
solutions (<math>a_{1}=-a_{2}</math> and <math>b_{1}=0)</math>. The solution for the even solutions
is
<center><math>
\left(
\begin{matrix}

e^{-k\varepsilon} & -\cos\kappa\varepsilon\\
ke^{-k\varepsilon} & -\kappa\sin\kappa\varepsilon
\end{matrix}
\right) \left(
\begin{matrix}

a_{1}\\
b_{1}
\end{matrix}
\right) =\left(
\begin{matrix}

0\\
0
\end{matrix}
\right)
</math></center>
This has non trivial solutions when
<center><math>
\det\left(
\begin{matrix}

e^{-k\varepsilon} & -\cos\kappa\varepsilon\\
ke^{-k\varepsilon} & -\kappa\sin\kappa\varepsilon
\end{matrix}
\right) =0
</math></center>
which gives us the equation
<center><math>
-\kappa e^{-k\varepsilon}\sin\kappa\varepsilon+k\cos\kappa\varepsilon
e^{-k\varepsilon}=0
</math></center>
or
<center><math>
\tan\kappa\varepsilon=\frac{k}{\kappa}
</math></center>
We know that <math>0<\kappa<\sqrt{b}</math> and if we plot this we see that we obtain a
finite number of solutions.

The solution for the odd solutions is
<center><math>
\left(
\begin{matrix}

e^{-k\varepsilon} & -\sin\kappa\varepsilon\\
ke^{-k\varepsilon} & \cos\kappa\varepsilon
\end{matrix}
\right) \left(
\begin{matrix}

a_{1}\\
b_{1}
\end{matrix}
\right) =\left(
\begin{matrix}

0\\
0
\end{matrix}
\right)
</math></center>
This can non trivial solutions when
<center><math>
\det\left(
\begin{matrix}

e^{-k\varepsilon} & -\sin\kappa\varepsilon\\
ke^{-k\varepsilon} & \kappa\cos\kappa\varepsilon
\end{matrix}
\right) =0
</math></center>
which gives us the equation
<center><math>
\kappa e^{-k}a\cos\kappa\varepsilon+k\sin\kappa\varepsilon e^{-k}=0
</math></center>
or
<center><math>
\tan\kappa=-\frac{\kappa}{k}
</math></center>

=== Case when <math>\lambda>0</math> ===

When <math>\lambda>0</math> we write <math>\lambda=k^{2}</math> and we obtain solution
<center><math>
w\left( x\right) =\left\{
\begin{matrix}

\mathrm{e}^{-\mathrm{i}kx}+r\mathrm{e}^{\mathrm{i}kx}, & x<-\varepsilon\\
b_{1}\cos\kappa x+b_{2}\sin\kappa x & -\varepsilon<x<\varepsilon\\
a\mathrm{e}^{-\mathrm{i}kx} & x>\varepsilon
\end{matrix}
\right.
</math></center>
where <math>\kappa=\sqrt{b+k^{2}}.</math> Matching <math>w</math> and its derivaties at <math>x=\pm1</math> we
obtain
<center><math>
\left(
\begin{matrix}

-\mathrm{e}^{-\mathrm{i}k\varepsilon} & \cos\kappa\varepsilon & -\sin\kappa\varepsilon & 0\\
ik\mathrm{e}^{-\mathrm{i}k\varepsilon} & \kappa\sin\kappa\varepsilon & \kappa\cos\kappa
\varepsilon & 0\\
0 & \cos\kappa\varepsilon & \sin\kappa\varepsilon & -\mathrm{e}^{-\mathrm{i}k\varepsilon}\\
0 & -\kappa\sin\kappa\varepsilon & \kappa\cos\kappa\varepsilon &
ik\mathrm{e}^{-\mathrm{i}k\varepsilon}
\end{matrix}
\right) \left(
\begin{matrix}

r\\
b_{1}\\
b_{2}\\
a
\end{matrix}
\right) =\left(
\begin{matrix}

\mathrm{e}^{\mathrm{i}k}\\
ik\mathrm{e}^{-\mathrm{i}k}\\
0\\
0
\end{matrix}
\right)
</math></center>

Properties of the Linear Schrodinger Equation

2010-09-24T02:17:30Z

Mike smith: /* Example \delta function potential */

{{nonlinear waves course
| chapter title = Properties of the Linear Schrodinger Equation
| next chapter = [[Connection betwen KdV and the Schrodinger Equation]]
| previous chapter = [[Introduction to the Inverse Scattering Transform]]
}}

The linear Schrodinger equation
<center><math>
\partial_{x}^{2}w+uw=-\lambda w
</math></center>
has two kinds of solutions for <math>u\rightarrow0</math> as <math>x\rightarrow\pm\infty.</math> The
first are waves and the second are bound solutions. It is well known that
there are at most a finite number of bound solutions (provided <math>u\rightarrow0</math>
as <math>x\pm\infty</math> sufficiently rapidly) and a continum of solutions for the
incident waves. This is easiest seen through the following examples

==Example <math>\delta</math> function potential==

We consider here the case when <math>u\left( x,0\right) =\delta\left( x\right)
.</math> Note that this function can be thought of as the limit as of the potential
<center><math>
u\left( x\right) =\left\{
\begin{matrix}

0 & x\notin\left[ -\varepsilon,\varepsilon\right] \\
\frac{u_{0}}{2\varepsilon} & x\in\left[ -\varepsilon,\varepsilon\right]
\end{matrix}
\right.
</math></center>
In this case we need to solve
<center><math>
\partial_{x}^{2}w+\frac{u_{0}}{2 \epsilon} \left( x\right) w=-\lambda w
</math></center>
We consider the case of <math>\lambda<0</math> and <math>\lambda>0</math> separately. For the first
case we write <math>\lambda=-k^{2}</math> and we obtain
<center><math>
w\left( x\right) =\left\{
\begin{matrix}

ae^{kx}, & x<0\\
be^{-kx}, & x>0
\end{matrix}
\right.
</math></center>
We have two conditions at <math>x=0,</math> <math>w</math> must be continuous at <math>0</math> and
<math>\partial_{x}w\left( 0^{+}\right) -\partial_{x}w\left( 0^{-}\right)
+u_{0}w\left( 0\right) =0.</math> This gives the condition that <math>a=b</math> and
<math>k=u_{0}/2.</math> We need to normalise the eigenfunctions so that
<center><math>
\int_{-\infty}^{\infty}\left( w\left( x\right) \right) ^{2}dx=1.
</math></center>
Therefore
<center><math>
2\int_{0}^{\infty}\left( ae^{-u_{0}x/2}\right) ^{2}dx=1
</math></center>
which means that <math>a=\sqrt{u_{0}/2}.</math> Therefore, there is only one discrete
spectral point which we denote by <math>k_{1}=u_{0}/2</math>
<center><math>
w_{1}\left( x\right) =\left\{
\begin{matrix}

\sqrt{k_{1}}e^{k_{1}x}, & x<0\\
\sqrt{k_{1}}e^{-k_{1}x}, & x>0
\end{matrix}
\right.
</math></center>
The continuous eigenfunctions correspond to <math>\lambda=k^{2}>0</math> are of the form
<center><math>
w\left( x\right) =\left\{
\begin{matrix}

\mathrm{e}^{-\mathrm{i}kx}+r\mathrm{e}^{\mathrm{i}kx}, & x<0\\
a\mathrm{e}^{-\mathrm{i}kx}, & x>0
\end{matrix}
\right.
</math></center>
Again we have the conditions that <math>w</math> must be continuous at <math>0</math> and
<math>\partial_{x}w\left( 0^{+}\right) -\partial_{x}w\left( 0^{-}\right)
+u_{0}w\left( 0\right) =0.</math> This gives us
<center><math>\begin{matrix}
1+r & =a\\
-ika+ik-ikr & =-au_{0}
\end{matrix}</math></center>
which has solution
<center><math>\begin{matrix}
r & =\frac{u_{0}}{2ik-u_{0}}\\
a & =\frac{2ik}{2ik-u_{0}}
\end{matrix}</math></center>

==Example: Scattering by a Well==

The properties of the eigenfunction is prehaps seem most easily through the
following example
<center><math>
u\left( x\right) =\left\{
\begin{matrix}

0 & x\notin\left[ -\varepsilon,\varepsilon\right] \\
b & x\in\left[ -\varepsilon,\varepsilon\right]
\end{matrix}
\right.
</math></center>
where <math>b>0.</math>

===Case when <math>\lambda>0</math>===

If we solve this equation for the case when <math>\lambda<0,</math> <math>\lambda=-k^{2}</math> we
get
<center><math>
w\left( x\right) =\left\{
\begin{matrix}

a_{1}e^{kx}, & x<-\varepsilon\\
b_{1}\cos\kappa x+b_{2}\sin\kappa x & -\varepsilon<x<\varepsilon\\
a_{2}e^{-kx} & x>\varepsilon
\end{matrix}
\right.
</math></center>
where <math>\kappa=\sqrt{b-k^{2}}</math> which means that <math>0\leq k\leq\sqrt{b}</math> (there is
no solution for <math>k>\sqrt{b}).</math> We then match <math>w</math> and its derivative at
<math>x=\pm\varepsilon</math> to solve for <math>a</math> and <math>b</math>. This leads to two system of
equation, one for the even (<math>a_{1}=a_{2}</math> and <math>b_{2}=0</math> ) and one for the odd
solutions (<math>a_{1}=-a_{2}</math> and <math>b_{1}=0)</math>. The solution for the even solutions
is
<center><math>
\left(
\begin{matrix}

e^{-k\varepsilon} & -\cos\kappa\varepsilon\\
ke^{-k\varepsilon} & -\kappa\sin\kappa\varepsilon
\end{matrix}
\right) \left(
\begin{matrix}

a_{1}\\
b_{1}
\end{matrix}
\right) =\left(
\begin{matrix}

0\\
0
\end{matrix}
\right)
</math></center>
This has non trivial solutions when
<center><math>
\det\left(
\begin{matrix}

e^{-k\varepsilon} & -\cos\kappa\varepsilon\\
ke^{-k\varepsilon} & -\kappa\sin\kappa\varepsilon
\end{matrix}
\right) =0
</math></center>
which gives us the equation
<center><math>
-\kappa e^{-k\varepsilon}\sin\kappa\varepsilon+k\cos\kappa\varepsilon
e^{-k\varepsilon}=0
</math></center>
or
<center><math>
\tan\kappa\varepsilon=\frac{k}{\kappa}
</math></center>
We know that <math>0<\kappa<\sqrt{b}</math> and if we plot this we see that we obtain a
finite number of solutions.

The solution for the odd solutions is
<center><math>
\left(
\begin{matrix}

e^{-k\varepsilon} & -\sin\kappa\varepsilon\\
ke^{-k\varepsilon} & \cos\kappa\varepsilon
\end{matrix}
\right) \left(
\begin{matrix}

a_{1}\\
b_{1}
\end{matrix}
\right) =\left(
\begin{matrix}

0\\
0
\end{matrix}
\right)
</math></center>
This can non trivial solutions when
<center><math>
\det\left(
\begin{matrix}

e^{-k\varepsilon} & -\sin\kappa\varepsilon\\
ke^{-k\varepsilon} & \kappa\cos\kappa\varepsilon
\end{matrix}
\right) =0
</math></center>
which gives us the equation
<center><math>
\kappa e^{-k}a\cos\kappa\varepsilon+k\sin\kappa\varepsilon e^{-k}=0
</math></center>
or
<center><math>
\tan\kappa=-\frac{\kappa}{k}
</math></center>

=== Case when <math>\lambda>0</math> ===

When <math>\lambda>0</math> we write <math>\lambda=k^{2}</math> and we obtain solution
<center><math>
w\left( x\right) =\left\{
\begin{matrix}

\mathrm{e}^{-\mathrm{i}kx}+r\mathrm{e}^{\mathrm{i}kx}, & x<-\varepsilon\\
b_{1}\cos\kappa x+b_{2}\sin\kappa x & -\varepsilon<x<\varepsilon\\
a\mathrm{e}^{-\mathrm{i}kx} & x>\varepsilon
\end{matrix}
\right.
</math></center>
where <math>\kappa=\sqrt{b+k^{2}}.</math> Matching <math>w</math> and its derivaties at <math>x=\pm1</math> we
obtain
<center><math>
\left(
\begin{matrix}

-\mathrm{e}^{-\mathrm{i}k\varepsilon} & \cos\kappa\varepsilon & -\sin\kappa\varepsilon & 0\\
ik\mathrm{e}^{-\mathrm{i}k\varepsilon} & \kappa\sin\kappa\varepsilon & \kappa\cos\kappa
\varepsilon & 0\\
0 & \cos\kappa\varepsilon & \sin\kappa\varepsilon & -\mathrm{e}^{-\mathrm{i}k\varepsilon}\\
0 & -\kappa\sin\kappa\varepsilon & \kappa\cos\kappa\varepsilon &
ik\mathrm{e}^{-\mathrm{i}k\varepsilon}
\end{matrix}
\right) \left(
\begin{matrix}

r\\
b_{1}\\
b_{2}\\
a
\end{matrix}
\right) =\left(
\begin{matrix}

\mathrm{e}^{\mathrm{i}k}\\
ik\mathrm{e}^{-\mathrm{i}k}\\
0\\
0
\end{matrix}
\right)
</math></center>

Properties of the Linear Schrodinger Equation

2010-09-24T02:13:45Z

Mike smith: /* Example \delta function potential */

{{nonlinear waves course
| chapter title = Properties of the Linear Schrodinger Equation
| next chapter = [[Connection betwen KdV and the Schrodinger Equation]]
| previous chapter = [[Introduction to the Inverse Scattering Transform]]
}}

The linear Schrodinger equation
<center><math>
\partial_{x}^{2}w+uw=-\lambda w
</math></center>
has two kinds of solutions for <math>u\rightarrow0</math> as <math>x\rightarrow\pm\infty.</math> The
first are waves and the second are bound solutions. It is well known that
there are at most a finite number of bound solutions (provided <math>u\rightarrow0</math>
as <math>x\pm\infty</math> sufficiently rapidly) and a continum of solutions for the
incident waves. This is easiest seen through the following examples

==Example <math>\delta</math> function potential==

We consider here the case when <math>u\left( x,0\right) =\delta\left( x\right)
.</math> Note that this function can be thought of as the limit as of the potential
<center><math>
u\left( x\right) =\left\{
\begin{matrix}

0 & x\notin\left[ -\varepsilon,\varepsilon\right] \\
\frac{u_{0}}{2\varepsilon} & x\in\left[ -\varepsilon,\varepsilon\right]
\end{matrix}
\right.
</math></center>
In this case we need to solve
<center><math>
\partial_{x}^{2}w+u_{0}\delta\left( x\right) w=-\lambda x
</math></center>
We consider the case of <math>\lambda<0</math> and <math>\lambda>0</math> separately. For the first
case we write <math>\lambda=-k^{2}</math> and we obtain
<center><math>
w\left( x\right) =\left\{
\begin{matrix}

ae^{kx}, & x<0\\
be^{-kx}, & x>0
\end{matrix}
\right.
</math></center>
We have two conditions at <math>x=0,</math> <math>w</math> must be continuous at <math>0</math> and
<math>\partial_{x}w\left( 0^{+}\right) -\partial_{x}w\left( 0^{-}\right)
+u_{0}w\left( 0\right) =0.</math> This gives the condition that <math>a=b</math> and
<math>k=u_{0}/2.</math> We need to normalise the eigenfunctions so that
<center><math>
\int_{-\infty}^{\infty}\left( w\left( x\right) \right) ^{2}dx=1.
</math></center>
Therefore
<center><math>
2\int_{0}^{\infty}\left( ae^{-u_{0}x/2}\right) ^{2}dx=1
</math></center>
which means that <math>a=\sqrt{u_{0}/2}.</math> Therefore, there is only one discrete
spectral point which we denote by <math>k_{1}=u_{0}/2</math>
<center><math>
w_{1}\left( x\right) =\left\{
\begin{matrix}

\sqrt{k_{1}}e^{k_{1}x}, & x<0\\
\sqrt{k_{1}}e^{-k_{1}x}, & x>0
\end{matrix}
\right.
</math></center>
The continuous eigenfunctions correspond to <math>\lambda=k^{2}>0</math> are of the form
<center><math>
w\left( x\right) =\left\{
\begin{matrix}

\mathrm{e}^{-\mathrm{i}kx}+r\mathrm{e}^{\mathrm{i}kx}, & x<0\\
a\mathrm{e}^{-\mathrm{i}kx}, & x>0
\end{matrix}
\right.
</math></center>
Again we have the conditions that <math>w</math> must be continuous at <math>0</math> and
<math>\partial_{x}w\left( 0^{+}\right) -\partial_{x}w\left( 0^{-}\right)
+u_{0}w\left( 0\right) =0.</math> This gives us
<center><math>\begin{matrix}
1+r & =a\\
-ika+ik-ikr & =-au_{0}
\end{matrix}</math></center>
which has solution
<center><math>\begin{matrix}
r & =\frac{u_{0}}{2ik-u_{0}}\\
a & =\frac{2ik}{2ik-u_{0}}
\end{matrix}</math></center>

==Example: Scattering by a Well==

The properties of the eigenfunction is prehaps seem most easily through the
following example
<center><math>
u\left( x\right) =\left\{
\begin{matrix}

0 & x\notin\left[ -\varepsilon,\varepsilon\right] \\
b & x\in\left[ -\varepsilon,\varepsilon\right]
\end{matrix}
\right.
</math></center>
where <math>b>0.</math>

===Case when <math>\lambda>0</math>===

If we solve this equation for the case when <math>\lambda<0,</math> <math>\lambda=-k^{2}</math> we
get
<center><math>
w\left( x\right) =\left\{
\begin{matrix}

a_{1}e^{kx}, & x<-\varepsilon\\
b_{1}\cos\kappa x+b_{2}\sin\kappa x & -\varepsilon<x<\varepsilon\\
a_{2}e^{-kx} & x>\varepsilon
\end{matrix}
\right.
</math></center>
where <math>\kappa=\sqrt{b-k^{2}}</math> which means that <math>0\leq k\leq\sqrt{b}</math> (there is
no solution for <math>k>\sqrt{b}).</math> We then match <math>w</math> and its derivative at
<math>x=\pm\varepsilon</math> to solve for <math>a</math> and <math>b</math>. This leads to two system of
equation, one for the even (<math>a_{1}=a_{2}</math> and <math>b_{2}=0</math> ) and one for the odd
solutions (<math>a_{1}=-a_{2}</math> and <math>b_{1}=0)</math>. The solution for the even solutions
is
<center><math>
\left(
\begin{matrix}

e^{-k\varepsilon} & -\cos\kappa\varepsilon\\
ke^{-k\varepsilon} & -\kappa\sin\kappa\varepsilon
\end{matrix}
\right) \left(
\begin{matrix}

a_{1}\\
b_{1}
\end{matrix}
\right) =\left(
\begin{matrix}

0\\
0
\end{matrix}
\right)
</math></center>
This has non trivial solutions when
<center><math>
\det\left(
\begin{matrix}

e^{-k\varepsilon} & -\cos\kappa\varepsilon\\
ke^{-k\varepsilon} & -\kappa\sin\kappa\varepsilon
\end{matrix}
\right) =0
</math></center>
which gives us the equation
<center><math>
-\kappa e^{-k\varepsilon}\sin\kappa\varepsilon+k\cos\kappa\varepsilon
e^{-k\varepsilon}=0
</math></center>
or
<center><math>
\tan\kappa\varepsilon=\frac{k}{\kappa}
</math></center>
We know that <math>0<\kappa<\sqrt{b}</math> and if we plot this we see that we obtain a
finite number of solutions.

The solution for the odd solutions is
<center><math>
\left(
\begin{matrix}

e^{-k\varepsilon} & -\sin\kappa\varepsilon\\
ke^{-k\varepsilon} & \cos\kappa\varepsilon
\end{matrix}
\right) \left(
\begin{matrix}

a_{1}\\
b_{1}
\end{matrix}
\right) =\left(
\begin{matrix}

0\\
0
\end{matrix}
\right)
</math></center>
This can non trivial solutions when
<center><math>
\det\left(
\begin{matrix}

e^{-k\varepsilon} & -\sin\kappa\varepsilon\\
ke^{-k\varepsilon} & \kappa\cos\kappa\varepsilon
\end{matrix}
\right) =0
</math></center>
which gives us the equation
<center><math>
\kappa e^{-k}a\cos\kappa\varepsilon+k\sin\kappa\varepsilon e^{-k}=0
</math></center>
or
<center><math>
\tan\kappa=-\frac{\kappa}{k}
</math></center>

=== Case when <math>\lambda>0</math> ===

When <math>\lambda>0</math> we write <math>\lambda=k^{2}</math> and we obtain solution
<center><math>
w\left( x\right) =\left\{
\begin{matrix}

\mathrm{e}^{-\mathrm{i}kx}+r\mathrm{e}^{\mathrm{i}kx}, & x<-\varepsilon\\
b_{1}\cos\kappa x+b_{2}\sin\kappa x & -\varepsilon<x<\varepsilon\\
a\mathrm{e}^{-\mathrm{i}kx} & x>\varepsilon
\end{matrix}
\right.
</math></center>
where <math>\kappa=\sqrt{b+k^{2}}.</math> Matching <math>w</math> and its derivaties at <math>x=\pm1</math> we
obtain
<center><math>
\left(
\begin{matrix}

-\mathrm{e}^{-\mathrm{i}k\varepsilon} & \cos\kappa\varepsilon & -\sin\kappa\varepsilon & 0\\
ik\mathrm{e}^{-\mathrm{i}k\varepsilon} & \kappa\sin\kappa\varepsilon & \kappa\cos\kappa
\varepsilon & 0\\
0 & \cos\kappa\varepsilon & \sin\kappa\varepsilon & -\mathrm{e}^{-\mathrm{i}k\varepsilon}\\
0 & -\kappa\sin\kappa\varepsilon & \kappa\cos\kappa\varepsilon &
ik\mathrm{e}^{-\mathrm{i}k\varepsilon}
\end{matrix}
\right) \left(
\begin{matrix}

r\\
b_{1}\\
b_{2}\\
a
\end{matrix}
\right) =\left(
\begin{matrix}

\mathrm{e}^{\mathrm{i}k}\\
ik\mathrm{e}^{-\mathrm{i}k}\\
0\\
0
\end{matrix}
\right)
</math></center>

Connection betwen KdV and the Schrodinger Equation

2010-09-24T01:59:10Z

Mike smith: /* Reflectionless Potential */

{{nonlinear waves course
| chapter title = Connection betwen KdV and the Schrodinger Equation
| next chapter = [[Example Calculations for the KdV and IST]]
| previous chapter = [[Properties of the Linear Schrodinger Equation]]
}}

If we substitute the relationship
<center><math>
\partial_{x}^{2}w+uw=-\lambda w
</math></center>
into the KdV after some manipulation we obtain
<center><math>
\partial_{t}\lambda w^{2}+\partial_{x}\left( w\partial_{x}Q-\partial
_{x}wQ\right) =0
</math></center>
where <math>Q=\partial_{t}w+\partial_{x}^{3}w-3\left( \lambda-u\right)
\partial_{x}w.</math> If we integrate this equation then we obtain the result that
<center><math>
\partial_{t}\lambda=0
</math></center>
provided that the eigenfunction <math>w</math> is bounded (which is true for the bound
state eigenfunctions). This shows that the discrete eigenvalues are unchanged
and <math>u\left( x,t\right) </math> evolves according to the KdV. Many other
properties can be found

==Scattering Data==

For the discrete spectrum the eigenfunctions behave like
<center><math>
w_{n}\left( x\right) =c_{n}\left( t\right) e^{-k_{n}x}
</math></center>
as <math>x\rightarrow\infty</math> with
<center><math>
\int_{-\infty}^{\infty}\left( w_{n}\left( x\right) \right) ^{2}dx=1
</math></center>
The continuous spectrum looks like
<center><math>
v\left( x,t\right) \approx \mathrm{e}^{-\mathrm{i}kx}+r\left( k,t\right) \mathrm{e}^{\mathrm{i}kx}
,\ \ \ x\rightarrow-\infty
</math></center>
<center><math>
v\left( x,t\right) \approx a\left( k,t\right) \mathrm{e}^{-\mathrm{i}kx},\ \ \ x\rightarrow
\infty
</math></center>
where <math>r</math> is the reflection coefficient and <math>a</math> is the transmission
coefficient. This gives us the scattering data at <math>t=0</math>
<center><math>
S\left( \lambda,0\right) =\left( \left\{ k_{n},c_{n}\left( 0\right)
\right\} _{n=1}^{N},r\left( k,0\right) ,a\left( k,0\right) \right)
</math></center>
The scattering data evolves as
<center><math>
k_{n}=k_{n}
</math></center>
<center><math>
c_{n}\left( t\right) =c_{n}\left( 0\right) e^{4k_{n}^{3}t}
</math></center>
<center><math>
r\left( k,t\right) =r\left( k,0\right) e^{8ik^{3}t}
</math></center>
<center><math>
a\left( k,t\right) =a\left( k,0\right)
</math></center>
We can recover <math>u</math> from scattering data. We write
<center><math>
F\left( x,t\right) =\sum_{n=1}^{N}c_{n}^{2}\left( t\right) e^{-k_{n}
x}+\frac{1}{2\pi}\int_{-\infty}^{\infty}r\left( k,t\right) \mathrm{e}^{\mathrm{i}kx}dk
</math></center>
Then solve
<center><math>
K\left( x,y;t\right) +F\left( x+y;t\right) +\int_{x}^{\infty}K\left(
x,z;t\right) F\left( z+y;t\right) dz=0
</math></center>
This is a linear integral equation called the \emph{Gelfand-Levitan-Marchenko
}equation. We then find <math>u</math> from
<center><math>
u\left( x,t\right) =2\partial_{x}K\left( x,x,t\right)
</math></center>

==Reflectionless Potential==

In general the IST is difficult to solve. However, there is a simplification
we can make when we have a reflectionless potential (which we will see gives
rise to the soliton solutions). The reflectionless potential is the case when
<math>r\left( k,0\right) =0</math> for all values of <math>k</math> for some <math>u.</math> In this case
<center><math>
F\left( x,t\right) =\sum_{n=1}^{N}c_{n}^{2}\left( t\right) e^{-k_{n}x}
</math></center>
then
<center><math>
K\left( x,y,t\right) +\sum_{n=1}^{N}c_{n}^{2}\left( t\right)
e^{-k_{n}\left( x+y\right) }+\int_{x}^{\infty}K\left( x,z,t\right)
\sum_{n=1}^{N}c_{n}^{2}\left( t\right) e^{-k_{n}\left( y+z\right) }dz=0
</math></center>
From the equation we can see that
<center><math>
K\left( x,y,t\right) =-\sum_{m=1}^{N}v_{m}\left( x,t\right) e^{-k_{m}y}
</math></center>
If we substitute this into the equation,
<center><math>
-\sum_{n=1}^{N}v_{n}\left( x,t\right) e^{-k_{n}y}+\sum_{n=1}^{N}c_{n}
^{2}\left( t\right) e^{-k_{n}\left( x+y\right) }+\int_{x}^{\infty}
-\sum_{m=1}^{N}v_{m}\left( x,t\right) e^{-k_{m}z}\sum_{n=1}^{N}c_{n}
^{2}\left( t\right) e^{-k_{n}\left( y+z\right) }dz=0
</math></center>
which leads to
<center><math>
-\sum_{n=1}^{N}c_{n}\left( t\right) v_{n}\left( x\right) e^{-k_{n}y}
+\sum_{n=1}^{N}c_{n}^{2}\left( t\right) e^{-k_{n}\left( x+y\right) }
-\sum_{n=1}^{N}\sum_{m=1}^{N}\frac{c_{m}\left( t\right) c_{n}^{2}\left(
t\right) }{k_{n}+k_{m}}v_{m}\left( x\right) e^{-k_{m}x}e^{-k_{n}\left(
y+x\right) }=0
</math></center>
and we can eliminate the sum over <math>n</math> , the <math>c_{n}\left( t\right) ,</math> and the
<math>e^{-k_{n}y}</math> to obtain
<center><math>
-v_{n}\left( x\right) +c_{n}\left( t\right) e^{-k_{n}x}-\sum_{m=1}
^{N}\frac{c_{n}\left( t\right) c_{m}\left( t\right) }{k_{n}+k_{m}}
v_{m}\left( x\right) e^{-\left( k_{m}+k_{n}\right) x}=0
</math></center>
which is an algebraic (finite dimensional system)\ for the unknows <math>v_{n}.</math> We
can write this as
<center><math>
\left( \mathbf{I}+\mathbf{C}\right) \vec{v}=\vec{f}
</math></center>
From the equation we can see that
<center><math>
K\left( x,y,t\right) =-\sum_{m=1}^{N}c_{m}\left( t\right) v_{m}\left(
x\right) e^{-k_{m}y}
</math></center>
If we substitute this into the equation
<center><math>
-\sum_{n=1}^{N}c_{n}\left( t\right) v_{n}\left( x\right) e^{-k_{n}y}
+\sum_{n=1}^{N}c_{n}^{2}\left( t\right) e^{-k_{n}\left( x+y\right) }
+\int_{x}^{\infty}-\sum_{m=1}^{N}c_{m}\left( t\right) v_{m}\left( x\right)
e^{-k_{m}z}\sum_{n=1}^{N}c_{n}^{2}\left( t\right) e^{-k_{n}\left(
y+z\right) }dz=0
</math></center>
which leads to
<center><math>
-\sum_{n=1}^{N}c_{n}\left( t\right) v_{n}\left( x\right) e^{-k_{n}y}
+\sum_{n=1}^{N}c_{n}^{2}\left( t\right) e^{-k_{n}\left( x+y\right) }
-\sum_{n=1}^{N}\sum_{m=1}^{N}\frac{c_{m}\left( t\right) c_{n}^{2}\left(
t\right) }{k_{n}+k_{m}}v_{m}\left( x\right) e^{-k_{m}x}e^{-k_{n}\left(
y+x\right) }=0
</math></center>
and we can eliminate the sum over <math>n</math> , the <math>c_{n}\left( t\right) ,</math> and the
<math>e^{-k_{n}y}</math> to obtain
<center><math>
-v_{n}\left( x\right) +c_{n}\left( t\right) e^{-k_{n}x}-\sum_{m=1}
^{N}\frac{c_{n}\left( t\right) c_{m}\left( t\right) }{k_{n}+k_{m}}
v_{m}\left( x\right) e^{-\left( k_{m}+k_{n}\right) x}=0
</math></center>
which is an algebraic (finite dimensional system)\ for the unknows <math>v_{n}.</math> We
can write this as
<center><math>
\left( \mathbf{I}+\mathbf{C}\right) \vec{v}=\vec{f}
</math></center>
where <math>f_{m}=c_{m}\left( t\right) e^{-k_{m}x}</math> and
<center><math>
c_{mn}=\sum_{m=1}^{N}\frac{c_{n}\left( t\right) c_{m}\left( t\right)
}{k_{n}+k_{m}}e^{-\left( k_{m}+k_{n}\right) x}
</math></center>
<center><math>
K\left( x,y,t\right) =-\sum_{m=1}^{N}c_{m}\left( t\right) \left(
\mathbf{I}+\mathbf{C}\right) ^{-1}\vec{f}e^{-k_{m}y}
</math></center>
This leads to
<center><math>
u\left( x,t\right) =2\partial_{x}^{2}\log\left[ \det\left( \mathbf{I}
+\mathbf{C}\right) \right]
</math></center>
Lets consider some simple examples. First of all if <math>n=1</math> (the single soliton
solution) we get
<center><math>\begin{matrix}
K\left( x,x,t\right) & =-\frac{c_{1}\left( t\right) c_{1}\left(
t\right) e^{-k_{1}x}e^{-k_{1}x}}{1+\frac{c_{1}\left( t\right) c_{1}\left(
t\right) }{k_{1}+k_{1}}e^{-\left( k_{1}+k_{1}\right) x}}\\
& =\frac{-1}{1+e^{2k_{1}x-8k_{1}^{3}t-\alpha}}
\end{matrix}</math></center>
where <math>e^{-\alpha}=2c_{0}^{2}\left( 0\right) .</math> Therefore
<center><math>\begin{align}
u\left( x,t\right) & =2\partial_{x}K\left( x,x,t\right) \\
& =\frac{4k_{1}e^{2k_{1}x-8k_{1}^{3}t-\alpha}}{\left( 1+e^{2k_{1}
x-8k_{1}^{3}t-\alpha}\right) ^{2}}\\
& =\frac{8k_{1}^{2}}{\left( \sqrt{2k_{1}}e^{\theta}+e^{-\theta}/\sqrt
{2k_{1}}\right) ^{2}}\\
& =2k_{1}^{2}\mbox{sech}^{2}\left\{ k_{1}\left( x-x_{0}\right) -4k_{1}
^{3}t\right\}
\end{align}</math></center>
where <math>\theta=k_{1}x-4k^{3}t-\alpha/2</math> and <math>\sqrt{2k}e^{-\alpha/2}=e^{-kx_{0}
}</math>. This is of course the single soliton solution.

Introduction to the Inverse Scattering Transform

2010-09-17T03:27:53Z

Mike smith: /* Case when \lambda */

{{nonlinear waves course
| chapter title = Introduction to the Inverse Scattering Transform
| next chapter = [[Reaction-Diffusion Systems]]
| previous chapter = [[Conservation Laws for the KdV]]
}}

== Introduction ==

The inverse scattering transformation gives a way to solve the KdV equation
exactly. You can think about is as being an analogous transformation to the
Fourier transformation, except it works for a non linear equation. We want to
be able to solve
<center><math>\begin{align}
\partial_{t}u+6u\partial_{x}u+\partial_{x}^{3}u & =0\\
u(x,0) & =f\left( x\right)
\end{align}</math></center>
with <math>\left\vert u\right\vert \rightarrow0</math> as <math>x\rightarrow\pm\infty.</math>

The Miura transformation is given by
<center><math>
u=v^{2}+v_{x} \,
</math></center>
and if <math>v</math> satisfies the mKdV
<center><math>
\partial_{t}v-6v^{2}\partial_{x}v+\partial_{x}^{3}v=0
</math></center>
then <math>u</math> satisfies the KdV (but not vice versa). We can think about the Miura
transformation as being a nonlinear ODE solving for <math>v</math> given <math>u.</math> This
nonlinear ODE is also known as the Riccati equation and there is a well know
transformation which linearises this equation. It we write
<center><math>
v=\frac{\left( \partial_{x}w\right) }{w}
</math></center>
then we obtain the equation
<center><math>
\partial_{x}^{2}w+uw=0
</math></center>
The KdV is invariant under the transformation <math>x\rightarrow x+6\lambda t,</math>
<math>u\rightarrow u+\lambda.</math> Therefore we consider the associated eigenvalue
problem
<center><math>
\partial_{x}^{2}w+uw=-\lambda w
</math></center>
The eigenfunctions and eigenvalues of this scattering problem play a key role
in the inverse scattering transformation. Note that this is Schrodinger's equation.

==Properties of the eigenfunctions==

The equation
<center><math>
\partial_{x}^{2}w+uw=-\lambda w
</math></center>
has two kinds of solutions for <math>u\rightarrow0</math> as <math>x\rightarrow\pm\infty.</math> The
first are waves and the second are bound solutions. It is well known that
there are at most a finite number of bound solutions (provided <math>u\rightarrow0</math>
as <math>x\pm\infty</math> sufficiently rapidly) and a continum of solutions for the
incident waves.

===Example: Scattering by a Well===

The properties of the eigenfunction is prehaps seem most easily through the
following example
<center><math>
u\left( x\right) =\left\{
\begin{matrix}

0, & x\notin\left[ -1,1\right] \\
b, & x\in\left[ -1,1\right]
\end{matrix}
\right.
</math></center>
where <math>b>0.</math>

==Case when <math>\lambda<0</math>==

If we solve this equation for the case when <math>\lambda<0,</math> <math>\lambda=-k^{2}</math> we
get
<center><math>
w\left( x\right) =\left\{
\begin{matrix}

a_{1}\mathrm{e}^{kx}, & x<-1\\
b_{1}\cos\kappa x+b_{2}\sin\kappa x & -1<x<1\\
a_{2}\mathrm{e}^{-kx} & x>1
\end{matrix}
\right.
</math></center>
where <math>\kappa=\sqrt{b-k^{2}}</math> where we have assumed that <math>b>k^{2}</math> (there is
no solution for <math>b<k^{2}).</math> We then match <math>w</math> and its derivative at <math>x=\pm1</math>
to solve for <math>a</math> and <math>b</math>. This leads to two system of equations, one for the
even (<math>a_{1}=a_{2}</math> and <math>b_{2}=0</math> ) and one for the odd solutions
(<math>a_{1}=-a_{2}</math> and <math>b_{1}=0)</math>. The solution for the even solutions is
<center><math>
\left(
\begin{matrix}

\mathrm{e}^{-k} & -\cos\kappa\\
k\mathrm{e}^{-k} & -\kappa \sin\kappa
\end{matrix}
\right) \left(
\begin{matrix}

a_{1}\\
b_{1}
\end{matrix}
\right) =\left(
\begin{matrix}

0\\
0
\end{matrix}
\right)
</math></center>
This has non trivial solutions when
<center><math>
\det\left(
\begin{matrix}

\mathrm{e}^{-k} & -\cos\kappa\\
k\mathrm{e}^{-k} & -\kappa \sin\kappa
\end{matrix}
\right) =0
</math></center>
which gives us the equation
<center><math>
- \kappa \sin\kappa \mathrm{e}^{-k}+\left( \cos\kappa\right) k\mathrm{e}^{-kx}=0
</math></center>
or
<center><math>
\kappa \tan\kappa=k=\sqrt{b-\kappa^{2}}
</math></center>
We know that <math>0<\kappa<\sqrt{b}</math> and if we plot this we see that we obtain a
finite number of solutions.

In other words, we solve the final equation above for <math>k</math> to obtain our eigenvalues corresponding to even solutions.
Similarly we repeat the above process for the odd solutions.

==Case when <math>\lambda>0</math>==

When <math>\lambda>0</math> we write <math>\lambda=k^{2}</math> and we obtain solution
<center><math>
w\left( x\right) =\left\{
\begin{matrix}

\mathrm{e}^{-\mathrm{i}kx}+r\mathrm{e}^{\mathrm{i}kx}, & x<-1\\
b_{1}\cos\kappa x+b_{2}\sin\kappa x & -1<x<1\\
a\mathrm{e}^{-\mathrm{i}kx} & x>1
\end{matrix}
\right.
</math></center>
where <math>\kappa=\sqrt{b+k^{2}}.</math> Matching <math>w</math> and its derivaties at <math>x=\pm1</math> we
obtain
<center><math>
\left(
\begin{matrix}

-\mathrm{e}^{-ik} & \cos\kappa & -\sin\kappa & 0\\
-ik\mathrm{e}^{-ik} & \kappa\sin\kappa & \kappa\cos\kappa & 0\\
0 & \cos\kappa & \sin\kappa & -\mathrm{e}^{-ik}\\
0 & -\kappa\sin\kappa & \kappa\cos\kappa & ik\mathrm{e}^{-ik}
\end{matrix}
\right) \left(
\begin{matrix}

r\\
b_{1}\\
b_{2}\\
a
\end{matrix}
\right) =\left(
\begin{matrix}

\mathrm{e}^{ik}\\
-ik\mathrm{e}^{-ik}\\
0\\
0
\end{matrix}
\right)
</math></center>

==Connection with the KdV==

If we substitute the relationship
<center><math>
\partial_{x}^{2}w+uw=-\lambda w
</math></center>
into the KdV after some manipulation we obtain
<center><math>
\partial_{t}\lambda w^{2}+\partial_{x}\left( w\partial_{x}Q-\partial
_{x}wQ\right) =0
</math></center>
where <math>Q=\partial_{t}w+\partial_{x}^{3}w-3\left( \lambda-u\right)
\partial_{x}w.</math> If we integrate this equation then we obtain the result that
<center><math>
\partial_{t}\lambda=0
</math></center>
provided that the eigenfunction <math>w</math> is bounded (which is true for the bound
state eigenfunctions). This shows that the discrete eigenvalues are unchanged
and <math>u\left( x,t\right) </math> evolves according to the KdV.

==Scattering Data==

For the discrete spectrum the eigenfunctions behave like
<center><math>
w_{n}\left( x\right) =c_{n}\left( t\right) \mathrm{e}^{-k_{n}x}
</math></center>
as <math>x\rightarrow\infty</math> with
<center><math>
\int_{-\infty}^{\infty}\left( w_{n}\left( x\right) \right) ^{2}dx=1
</math></center>
The continuous spectrum looks like
<center><math>
v\left( x,t\right) \approx \mathrm{e}^{-\mathrm{i}kx}+r\left( k,t\right) \mathrm{e}^{\mathrm{i}kx}
,\ \ \ x\rightarrow-\infty
</math></center>
<center><math>
v\left( x,t\right) \approx a\left( k,t\right) \mathrm{e}^{-\mathrm{i}kx},\ \ \ x\rightarrow
\infty
</math></center>
where <math>r</math> is the reflection coefficient and <math>a</math> is the transmission
coefficient. This gives us the scattering data at <math>t=0</math>
<center><math>
S\left( \lambda,0\right) =\left( \left\{ k_{n},c_{n}\left( 0\right)
\right\} _{n=1}^{N},r\left( k,0\right) ,a\left( k,0\right) \right)
</math></center>
The scattering data evolves as
<center><math>
k_{n}=k_{n}
</math></center>
<center><math>
c_{n}\left( t\right) =c_{n}\left( 0\right) \mathrm{e}^{4k_{n}^{3}t}
</math></center>
<center><math>
r\left( k,t\right) =r\left( k,0\right) \mathrm{e}^{8ik^{3}t}
</math></center>
<center><math>
a\left( k,t\right) =a\left( k,0\right)
</math></center>
We can recover <math>u</math> from scattering data. We write
<center><math>
F\left( x,t\right) =\sum_{n=1}^{N}c_{n}^{2}\left( t\right) \mathrm{e}^{-k_{n}
x}+\int_{-\infty}^{\infty}r\left( k,t\right) \mathrm{e}^{\mathrm{i}kx}\mathrm{d}k
</math></center>
Then solve
<center><math>
K\left( x,y;t\right) +F\left( x+y;t\right) +\int_{x}^{\infty}K\left(
x,z;t\right) F\left( z+y;t\right) \mathrm{d}z=0
</math></center>
This is a linear integral equation called the \emph{Gelfand-Levitan-Marchenko
}equation. We then find <math>u</math> from
<center><math>
u\left( x,t\right) =2\partial_{x}K\left( x,x,t\right)
</math></center>

==Reflectionless Potential==

In general the IST is difficult to solve. However, there is a simplification
we can make when we have a reflectionless potential (which we will see gives
rise to the soliton solutions). The reflectionless potential is the case when
<math>r\left( k,0\right) =0</math> for all values of <math>k</math> for some <math>u.</math> In this case
<center><math>
F\left( x,t\right) =\sum_{n=1}^{N}c_{n}^{2}\left( t\right) \mathrm{e}^{-k_{n}x}
</math></center>
then
<center><math>
K\left( x,y,t\right) +\sum_{n=1}^{N}c_{n}^{2}\left( t\right)
\mathrm{e}^{-k_{n}\left( x+y\right) }+\int_{x}^{\infty}K\left( x,z,t\right)
\sum_{n=1}^{N}c_{n}^{2}\left( t\right) \mathrm{e}^{-k_{n}\left( y+z\right) }dz=0
</math></center>
From the equation we can see that
<center><math>
K\left( x,y,t\right) =-\sum_{m=1}^{N}c_{m}\left( t\right) v_{m}\left(
x\right) \mathrm{e}^{-k_{m}y}
</math></center>
If we substitute this into the equation
<center><math>
-\sum_{n=1}^{N}c_{n}\left( t\right) v_{n}\left( x\right) \mathrm{e}^{-k_{n}y}
+\sum_{n=1}^{N}c_{n}^{2}\left( t\right) \mathrm{e}^{-k_{n}\left( x+y\right) }
+\int_{x}^{\infty}-\sum_{m=1}^{N}c_{m}\left( t\right) v_{m}\left( x\right)
\mathrm{e}^{-k_{m}y}\sum_{n=1}^{N}c_{n}^{2}\left( t\right) \mathrm{e}^{-k_{n}\left(
y+z\right) }dz=0
</math></center>
which leads to
<center><math>
-\sum_{n=1}^{N}c_{n}\left( t\right) v_{n}\left( x\right) \mathrm{e}^{-k_{n}y}
+\sum_{n=1}^{N}c_{n}^{2}\left( t\right) \mathrm{e}^{-k_{n}\left( x+y\right) }
-\sum_{n=1}^{N}\sum_{m=1}^{N}\frac{c_{m}\left( t\right) c_{n}^{2}\left(
t\right) }{k_{n}+k_{m}}v_{m}\left( x\right) \mathrm{e}^{-k_{m}x}\mathrm{e}^{-k_{n}\left(
y+x\right) }=0
</math></center>
and we can eliminate the sum over <math>n</math> , the <math>c_{n}\left( t\right) ,</math> and the
<math>\mathrm{e}^{-k_{n}y}</math> to obtain
<center><math>
-v_{n}\left( x\right) +c_{n}\left( t\right) \mathrm{e}^{-k_{n}x}-\sum_{m=1}
^{N}\frac{c_{n}\left( t\right) c_{m}\left( t\right) }{k_{n}+k_{m}}
v_{m}\left( x\right) \mathrm{e}^{-\left( k_{m}+k_{n}\right) x}=0
</math></center>
which is an algebraic (finite dimensional system)\ for the unknows <math>v_{n}.</math> We
can write this as
<center><math>
\left( \mathbf{I}+\mathbf{C}\right) \vec{v}=\vec{f}
</math></center>
where <math>f_{m}=c_{m}\left( t\right) \mathrm{e}^{-k_{m}x}</math> and
<center><math>
c_{mn}=\sum_{m=1}^{N}\frac{c_{n}\left( t\right) c_{m}\left( t\right)
}{k_{n}+k_{m}}\mathrm{e}^{-\left( k_{m}+k_{n}\right) x}
</math></center>
<center><math>
K\left( x,y,t\right) =-\sum_{m=1}^{N}c_{m}\left( t\right) \left(
\mathbf{I}+\mathbf{C}\right) ^{-1}\vec{f}\mathrm{e}^{-k_{m}y}
</math></center>
This leads to
<center><math>
u\left( x,t\right) =2\partial_{x}^{2}\log\left[ \det\left( \mathbf{I}
+\mathbf{C}\right) \right]
</math></center>
Lets consider some simple examples. First of all if <math>n=1</math> (the single soliton
solution) we get
<center><math>\begin{matrix}
K\left( x,x,t\right) & =-\frac{c_{1}\left( t\right) c_{1}\left(
t\right) \mathrm{e}^{-k_{1}x}\mathrm{e}^{-k_{1}x}}{1+\frac{c_{1}\left( t\right) c_{1}\left(
t\right) }{k_{1}+k_{1}}\mathrm{e}^{-\left( k_{1}+k_{1}\right) x}}\\
& =\frac{-1}{1+\mathrm{e}^{2k_{1}x-8k_{1}^{3}t-\alpha}}
\end{matrix}</math></center>
where <math>\mathrm{e}^{-\alpha}=2c_{0}^{2}\left( 0\right) .</math> Therefore
<center><math>\begin{matrix}
u\left( x,t\right) & =2\partial_{x}K\left( x,x,t\right) \\
& =\frac{4k_{1}\mathrm{e}^{2k_{1}x-8k_{1}^{3}t-\alpha}}{\left( 1+\mathrm{e}^{2k_{1}
x-8k_{1}^{3}t-\alpha}\right) ^{2}}\\
& =\frac{-8k_{1}^{2}}{\left( \sqrt{2k_{1}}\mathrm{e}^{\theta}+\mathrm{e}^{-\theta}
/\sqrt{2k_{1}}\right) ^{2}}\\
& =2k^{2}\sec^{2}\left\{ k_{1}\left( x-x_{0}\right) -4k_{1}^{3}t\right\}
\end{matrix}</math></center>
where <math>\theta=k_{1}x-4k^{3}t-\alpha/2</math> and <math>\sqrt{2k}\mathrm{e}^{-\alpha/2}=\mathrm{e}^{-kx_{0}
}</math>. This is of course the single soliton solution.

User:Mike smith

2010-09-17T01:13:44Z

Mike smith:

Is a PhD student at the University of Auckland Mathematics Dept (Supervisor: Dr Mike Meylan)

Introduction to the Inverse Scattering Transform

2010-09-17T01:09:05Z

Mike smith: /* Case when \lambda>0 */

{{nonlinear waves course
| chapter title = Introduction to the Inverse Scattering Transform
| next chapter = [[Reaction-Diffusion Systems]]
| previous chapter = [[Conservation Laws for the KdV]]
}}

== Introduction ==

The inverse scattering transformation gives a way to solve the KdV equation
exactly. You can think about is as being an analogous transformation to the
Fourier transformation, except it works for a non linear equation. We want to
be able to solve
<center><math>\begin{align}
\partial_{t}u+6u\partial_{x}u+\partial_{x}^{3}u & =0\\
u(x,0) & =f\left( x\right)
\end{align}</math></center>
with <math>\left\vert u\right\vert \rightarrow0</math> as <math>x\rightarrow\pm\infty.</math>

The Miura transformation is given by
<center><math>
u=v^{2}+v_{x} \,
</math></center>
and if <math>v</math> satisfies the mKdV
<center><math>
\partial_{t}v-6v^{2}\partial_{x}v+\partial_{x}^{3}v=0
</math></center>
then <math>u</math> satisfies the KdV (but not vice versa). We can think about the Miura
transformation as being a nonlinear ODE solving for <math>v</math> given <math>u.</math> This
nonlinear ODE is also known as the Riccati equation and there is a well know
transformation which linearises this equation. It we write
<center><math>
v=\frac{\left( \partial_{x}w\right) }{w}
</math></center>
then we obtain the equation
<center><math>
\partial_{x}^{2}w+uw=0
</math></center>
The KdV is invariant under the transformation <math>x\rightarrow x+6\lambda t,</math>
<math>u\rightarrow u+\lambda.</math> Therefore we consider the associated eigenvalue
problem
<center><math>
\partial_{x}^{2}w+uw=-\lambda w
</math></center>
The eigenfunctions and eigenvalues of this scattering problem play a key role
in the inverse scattering transformation. Note that this is Schrodinger's equation.

==Properties of the eigenfunctions==

The equation
<center><math>
\partial_{x}^{2}w+uw=-\lambda w
</math></center>
has two kinds of solutions for <math>u\rightarrow0</math> as <math>x\rightarrow\pm\infty.</math> The
first are waves and the second are bound solutions. It is well known that
there are at most a finite number of bound solutions (provided <math>u\rightarrow0</math>
as <math>x\pm\infty</math> sufficiently rapidly) and a continum of solutions for the
incident waves.

===Example: Scattering by a Well===

The properties of the eigenfunction is prehaps seem most easily through the
following example
<center><math>
u\left( x\right) =\left\{
\begin{matrix}

0, & x\notin\left[ -1,1\right] \\
b, & x\in\left[ -1,1\right]
\end{matrix}
\right.
</math></center>
where <math>b>0.</math>

==Case when <math>\lambda<0</math>==

If we solve this equation for the case when <math>\lambda<0,</math> <math>\lambda=-k^{2}</math> we
get
<center><math>
w\left( x\right) =\left\{
\begin{matrix}

a_{1}\mathrm{e}^{kx}, & x<-1\\
b_{1}\cos\kappa x+b_{2}\sin\kappa x & -1<x<1\\
a_{2}\mathrm{e}^{-kx} & x>1
\end{matrix}
\right.
</math></center>
where <math>\kappa=\sqrt{b-k^{2}}</math> where we have assumed that <math>b>k^{2}</math> (there is
no solution for <math>b<k^{2}).</math> We then match <math>w</math> and its derivative at <math>x=\pm1</math>
to solve for <math>a</math> and <math>b</math>. This leads to two system of equations, one for the
even (<math>a_{1}=a_{2}</math> and <math>b_{2}=0</math> ) and one for the odd solutions
(<math>a_{1}=-a_{2}</math> and <math>b_{1}=0)</math>. The solution for the even solutions is
<center><math>
\left(
\begin{matrix}

\mathrm{e}^{-kx} & -\cos\kappa\\
k\mathrm{e}^{-kx} & \sin\kappa
\end{matrix}
\right) \left(
\begin{matrix}

a_{1}\\
b_{1}
\end{matrix}
\right) =\left(
\begin{matrix}

0\\
0
\end{matrix}
\right)
</math></center>
This has non trivial solutions when
<center><math>
\det\left(
\begin{matrix}

\mathrm{e}^{-kx} & -\cos\kappa\\
k\mathrm{e}^{-kx} & - \kappa \sin\kappa
\end{matrix}
\right) =0
</math></center>
which gives us the equation
<center><math>
- \kappa \sin\kappa \mathrm{e}^{-kx}+\left( \cos\kappa\right) k\mathrm{e}^{-kx}=0
</math></center>
or
<center><math>
\kappa \tan\kappa=k=\sqrt{b-\kappa^{2}}
</math></center>
We know that <math>0<\kappa<\sqrt{b}</math> and if we plot this we see that we obtain a
finite number of solutions.

In other words, we solve the final equation above for <math>k</math> to obtain our eigenvalues corresponding to even solutions.
Similarly we repeat the above process for the odd solutions.

==Case when <math>\lambda>0</math>==

When <math>\lambda>0</math> we write <math>\lambda=k^{2}</math> and we obtain solution
<center><math>
w\left( x\right) =\left\{
\begin{matrix}

\mathrm{e}^{-\mathrm{i}kx}+r\mathrm{e}^{\mathrm{i}kx}, & x<-1\\
b_{1}\cos\kappa x+b_{2}\sin\kappa x & -1<x<1\\
a\mathrm{e}^{-\mathrm{i}kx} & x>1
\end{matrix}
\right.
</math></center>
where <math>\kappa=\sqrt{b+k^{2}}.</math> Matching <math>w</math> and its derivaties at <math>x=\pm1</math> we
obtain
<center><math>
\left(
\begin{matrix}

-\mathrm{e}^{-ik} & \cos\kappa & -\sin\kappa & 0\\
-ik\mathrm{e}^{-ik} & \kappa\sin\kappa & \kappa\cos\kappa & 0\\
0 & \cos\kappa & \sin\kappa & -\mathrm{e}^{-ik}\\
0 & -\kappa\sin\kappa & \kappa\cos\kappa & ik\mathrm{e}^{-ik}
\end{matrix}
\right) \left(
\begin{matrix}

r\\
b_{1}\\
b_{2}\\
a
\end{matrix}
\right) =\left(
\begin{matrix}

\mathrm{e}^{ik}\\
-ik\mathrm{e}^{-ik}\\
0\\
0
\end{matrix}
\right)
</math></center>

==Connection with the KdV==

If we substitute the relationship
<center><math>
\partial_{x}^{2}w+uw=-\lambda w
</math></center>
into the KdV after some manipulation we obtain
<center><math>
\partial_{t}\lambda w^{2}+\partial_{x}\left( w\partial_{x}Q-\partial
_{x}wQ\right) =0
</math></center>
where <math>Q=\partial_{t}w+\partial_{x}^{3}w-3\left( \lambda-u\right)
\partial_{x}w.</math> If we integrate this equation then we obtain the result that
<center><math>
\partial_{t}\lambda=0
</math></center>
provided that the eigenfunction <math>w</math> is bounded (which is true for the bound
state eigenfunctions). This shows that the discrete eigenvalues are unchanged
and <math>u\left( x,t\right) </math> evolves according to the KdV.

==Scattering Data==

For the discrete spectrum the eigenfunctions behave like
<center><math>
w_{n}\left( x\right) =c_{n}\left( t\right) \mathrm{e}^{-k_{n}x}
</math></center>
as <math>x\rightarrow\infty</math> with
<center><math>
\int_{-\infty}^{\infty}\left( w_{n}\left( x\right) \right) ^{2}dx=1
</math></center>
The continuous spectrum looks like
<center><math>
v\left( x,t\right) \approx \mathrm{e}^{-\mathrm{i}kx}+r\left( k,t\right) \mathrm{e}^{\mathrm{i}kx}
,\ \ \ x\rightarrow-\infty
</math></center>
<center><math>
v\left( x,t\right) \approx a\left( k,t\right) \mathrm{e}^{-\mathrm{i}kx},\ \ \ x\rightarrow
\infty
</math></center>
where <math>r</math> is the reflection coefficient and <math>a</math> is the transmission
coefficient. This gives us the scattering data at <math>t=0</math>
<center><math>
S\left( \lambda,0\right) =\left( \left\{ k_{n},c_{n}\left( 0\right)
\right\} _{n=1}^{N},r\left( k,0\right) ,a\left( k,0\right) \right)
</math></center>
The scattering data evolves as
<center><math>
k_{n}=k_{n}
</math></center>
<center><math>
c_{n}\left( t\right) =c_{n}\left( 0\right) \mathrm{e}^{4k_{n}^{3}t}
</math></center>
<center><math>
r\left( k,t\right) =r\left( k,0\right) \mathrm{e}^{8ik^{3}t}
</math></center>
<center><math>
a\left( k,t\right) =a\left( k,0\right)
</math></center>
We can recover <math>u</math> from scattering data. We write
<center><math>
F\left( x,t\right) =\sum_{n=1}^{N}c_{n}^{2}\left( t\right) \mathrm{e}^{-k_{n}
x}+\int_{-\infty}^{\infty}r\left( k,t\right) \mathrm{e}^{\mathrm{i}kx}\mathrm{d}k
</math></center>
Then solve
<center><math>
K\left( x,y;t\right) +F\left( x+y;t\right) +\int_{x}^{\infty}K\left(
x,z;t\right) F\left( z+y;t\right) \mathrm{d}z=0
</math></center>
This is a linear integral equation called the \emph{Gelfand-Levitan-Marchenko
}equation. We then find <math>u</math> from
<center><math>
u\left( x,t\right) =2\partial_{x}K\left( x,x,t\right)
</math></center>

==Reflectionless Potential==

In general the IST is difficult to solve. However, there is a simplification
we can make when we have a reflectionless potential (which we will see gives
rise to the soliton solutions). The reflectionless potential is the case when
<math>r\left( k,0\right) =0</math> for all values of <math>k</math> for some <math>u.</math> In this case
<center><math>
F\left( x,t\right) =\sum_{n=1}^{N}c_{n}^{2}\left( t\right) \mathrm{e}^{-k_{n}x}
</math></center>
then
<center><math>
K\left( x,y,t\right) +\sum_{n=1}^{N}c_{n}^{2}\left( t\right)
\mathrm{e}^{-k_{n}\left( x+y\right) }+\int_{x}^{\infty}K\left( x,z,t\right)
\sum_{n=1}^{N}c_{n}^{2}\left( t\right) \mathrm{e}^{-k_{n}\left( y+z\right) }dz=0
</math></center>
From the equation we can see that
<center><math>
K\left( x,y,t\right) =-\sum_{m=1}^{N}c_{m}\left( t\right) v_{m}\left(
x\right) \mathrm{e}^{-k_{m}y}
</math></center>
If we substitute this into the equation
<center><math>
-\sum_{n=1}^{N}c_{n}\left( t\right) v_{n}\left( x\right) \mathrm{e}^{-k_{n}y}
+\sum_{n=1}^{N}c_{n}^{2}\left( t\right) \mathrm{e}^{-k_{n}\left( x+y\right) }
+\int_{x}^{\infty}-\sum_{m=1}^{N}c_{m}\left( t\right) v_{m}\left( x\right)
\mathrm{e}^{-k_{m}y}\sum_{n=1}^{N}c_{n}^{2}\left( t\right) \mathrm{e}^{-k_{n}\left(
y+z\right) }dz=0
</math></center>
which leads to
<center><math>
-\sum_{n=1}^{N}c_{n}\left( t\right) v_{n}\left( x\right) \mathrm{e}^{-k_{n}y}
+\sum_{n=1}^{N}c_{n}^{2}\left( t\right) \mathrm{e}^{-k_{n}\left( x+y\right) }
-\sum_{n=1}^{N}\sum_{m=1}^{N}\frac{c_{m}\left( t\right) c_{n}^{2}\left(
t\right) }{k_{n}+k_{m}}v_{m}\left( x\right) \mathrm{e}^{-k_{m}x}\mathrm{e}^{-k_{n}\left(
y+x\right) }=0
</math></center>
and we can eliminate the sum over <math>n</math> , the <math>c_{n}\left( t\right) ,</math> and the
<math>\mathrm{e}^{-k_{n}y}</math> to obtain
<center><math>
-v_{n}\left( x\right) +c_{n}\left( t\right) \mathrm{e}^{-k_{n}x}-\sum_{m=1}
^{N}\frac{c_{n}\left( t\right) c_{m}\left( t\right) }{k_{n}+k_{m}}
v_{m}\left( x\right) \mathrm{e}^{-\left( k_{m}+k_{n}\right) x}=0
</math></center>
which is an algebraic (finite dimensional system)\ for the unknows <math>v_{n}.</math> We
can write this as
<center><math>
\left( \mathbf{I}+\mathbf{C}\right) \vec{v}=\vec{f}
</math></center>
where <math>f_{m}=c_{m}\left( t\right) \mathrm{e}^{-k_{m}x}</math> and
<center><math>
c_{mn}=\sum_{m=1}^{N}\frac{c_{n}\left( t\right) c_{m}\left( t\right)
}{k_{n}+k_{m}}\mathrm{e}^{-\left( k_{m}+k_{n}\right) x}
</math></center>
<center><math>
K\left( x,y,t\right) =-\sum_{m=1}^{N}c_{m}\left( t\right) \left(
\mathbf{I}+\mathbf{C}\right) ^{-1}\vec{f}\mathrm{e}^{-k_{m}y}
</math></center>
This leads to
<center><math>
u\left( x,t\right) =2\partial_{x}^{2}\log\left[ \det\left( \mathbf{I}
+\mathbf{C}\right) \right]
</math></center>
Lets consider some simple examples. First of all if <math>n=1</math> (the single soliton
solution) we get
<center><math>\begin{matrix}
K\left( x,x,t\right) & =-\frac{c_{1}\left( t\right) c_{1}\left(
t\right) \mathrm{e}^{-k_{1}x}\mathrm{e}^{-k_{1}x}}{1+\frac{c_{1}\left( t\right) c_{1}\left(
t\right) }{k_{1}+k_{1}}\mathrm{e}^{-\left( k_{1}+k_{1}\right) x}}\\
& =\frac{-1}{1+\mathrm{e}^{2k_{1}x-8k_{1}^{3}t-\alpha}}
\end{matrix}</math></center>
where <math>\mathrm{e}^{-\alpha}=2c_{0}^{2}\left( 0\right) .</math> Therefore
<center><math>\begin{matrix}
u\left( x,t\right) & =2\partial_{x}K\left( x,x,t\right) \\
& =\frac{4k_{1}\mathrm{e}^{2k_{1}x-8k_{1}^{3}t-\alpha}}{\left( 1+\mathrm{e}^{2k_{1}
x-8k_{1}^{3}t-\alpha}\right) ^{2}}\\
& =\frac{-8k_{1}^{2}}{\left( \sqrt{2k_{1}}\mathrm{e}^{\theta}+\mathrm{e}^{-\theta}
/\sqrt{2k_{1}}\right) ^{2}}\\
& =2k^{2}\sec^{2}\left\{ k_{1}\left( x-x_{0}\right) -4k_{1}^{3}t\right\}
\end{matrix}</math></center>
where <math>\theta=k_{1}x-4k^{3}t-\alpha/2</math> and <math>\sqrt{2k}\mathrm{e}^{-\alpha/2}=\mathrm{e}^{-kx_{0}
}</math>. This is of course the single soliton solution.

Introduction to the Inverse Scattering Transform

2010-09-17T01:01:26Z

Mike smith: /* Case when \lambda */

{{nonlinear waves course
| chapter title = Introduction to the Inverse Scattering Transform
| next chapter = [[Reaction-Diffusion Systems]]
| previous chapter = [[Conservation Laws for the KdV]]
}}

== Introduction ==

The inverse scattering transformation gives a way to solve the KdV equation
exactly. You can think about is as being an analogous transformation to the
Fourier transformation, except it works for a non linear equation. We want to
be able to solve
<center><math>\begin{align}
\partial_{t}u+6u\partial_{x}u+\partial_{x}^{3}u & =0\\
u(x,0) & =f\left( x\right)
\end{align}</math></center>
with <math>\left\vert u\right\vert \rightarrow0</math> as <math>x\rightarrow\pm\infty.</math>

The Miura transformation is given by
<center><math>
u=v^{2}+v_{x} \,
</math></center>
and if <math>v</math> satisfies the mKdV
<center><math>
\partial_{t}v-6v^{2}\partial_{x}v+\partial_{x}^{3}v=0
</math></center>
then <math>u</math> satisfies the KdV (but not vice versa). We can think about the Miura
transformation as being a nonlinear ODE solving for <math>v</math> given <math>u.</math> This
nonlinear ODE is also known as the Riccati equation and there is a well know
transformation which linearises this equation. It we write
<center><math>
v=\frac{\left( \partial_{x}w\right) }{w}
</math></center>
then we obtain the equation
<center><math>
\partial_{x}^{2}w+uw=0
</math></center>
The KdV is invariant under the transformation <math>x\rightarrow x+6\lambda t,</math>
<math>u\rightarrow u+\lambda.</math> Therefore we consider the associated eigenvalue
problem
<center><math>
\partial_{x}^{2}w+uw=-\lambda w
</math></center>
The eigenfunctions and eigenvalues of this scattering problem play a key role
in the inverse scattering transformation. Note that this is Schrodinger's equation.

==Properties of the eigenfunctions==

The equation
<center><math>
\partial_{x}^{2}w+uw=-\lambda w
</math></center>
has two kinds of solutions for <math>u\rightarrow0</math> as <math>x\rightarrow\pm\infty.</math> The
first are waves and the second are bound solutions. It is well known that
there are at most a finite number of bound solutions (provided <math>u\rightarrow0</math>
as <math>x\pm\infty</math> sufficiently rapidly) and a continum of solutions for the
incident waves.

===Example: Scattering by a Well===

The properties of the eigenfunction is prehaps seem most easily through the
following example
<center><math>
u\left( x\right) =\left\{
\begin{matrix}

0, & x\notin\left[ -1,1\right] \\
b, & x\in\left[ -1,1\right]
\end{matrix}
\right.
</math></center>
where <math>b>0.</math>

==Case when <math>\lambda<0</math>==

If we solve this equation for the case when <math>\lambda<0,</math> <math>\lambda=-k^{2}</math> we
get
<center><math>
w\left( x\right) =\left\{
\begin{matrix}

a_{1}\mathrm{e}^{kx}, & x<-1\\
b_{1}\cos\kappa x+b_{2}\sin\kappa x & -1<x<1\\
a_{2}\mathrm{e}^{-kx} & x>1
\end{matrix}
\right.
</math></center>
where <math>\kappa=\sqrt{b-k^{2}}</math> where we have assumed that <math>b>k^{2}</math> (there is
no solution for <math>b<k^{2}).</math> We then match <math>w</math> and its derivative at <math>x=\pm1</math>
to solve for <math>a</math> and <math>b</math>. This leads to two system of equations, one for the
even (<math>a_{1}=a_{2}</math> and <math>b_{2}=0</math> ) and one for the odd solutions
(<math>a_{1}=-a_{2}</math> and <math>b_{1}=0)</math>. The solution for the even solutions is
<center><math>
\left(
\begin{matrix}

\mathrm{e}^{-kx} & -\cos\kappa\\
k\mathrm{e}^{-kx} & \sin\kappa
\end{matrix}
\right) \left(
\begin{matrix}

a_{1}\\
b_{1}
\end{matrix}
\right) =\left(
\begin{matrix}

0\\
0
\end{matrix}
\right)
</math></center>
This has non trivial solutions when
<center><math>
\det\left(
\begin{matrix}

\mathrm{e}^{-kx} & -\cos\kappa\\
k\mathrm{e}^{-kx} & - \kappa \sin\kappa
\end{matrix}
\right) =0
</math></center>
which gives us the equation
<center><math>
- \kappa \sin\kappa \mathrm{e}^{-kx}+\left( \cos\kappa\right) k\mathrm{e}^{-kx}=0
</math></center>
or
<center><math>
\kappa \tan\kappa=k=\sqrt{b-\kappa^{2}}
</math></center>
We know that <math>0<\kappa<\sqrt{b}</math> and if we plot this we see that we obtain a
finite number of solutions.

In other words, we solve the final equation above for <math>k</math> to obtain our eigenvalues corresponding to even solutions.
Similarly we repeat the above process for the odd solutions.

==Case when <math>\lambda>0</math>==

When <math>\lambda>0</math> we write <math>\lambda=k^{2}</math> and we obtain solution
<center><math>
w\left( x\right) =\left\{
\begin{matrix}

\mathrm{e}^{-\mathrm{i}kx}+r\mathrm{e}^{\mathrm{i}kx}, & x<-1\\
b_{1}\cos\kappa x+b_{2}\sin\kappa x & -1<x<1\\
a\mathrm{e}^{-\mathrm{i}kx} & x>1
\end{matrix}
\right.
</math></center>
where <math>\kappa=\sqrt{b+k^{2}}.</math> Matching <math>w</math> and its derivaties at <math>x=\pm1</math> we
obtain
<center><math>
\left(
\begin{matrix}

-\mathrm{e}^{-ik} & \cos\kappa & -\sin\kappa & 0\\
ik\mathrm{e}^{-ik} & \kappa\sin\kappa & \kappa\cos\kappa & 0\\
0 & \cos\kappa & \sin\kappa & -\mathrm{e}^{-ik}\\
0 & -\kappa\sin\kappa & \kappa\cos\kappa & ik\mathrm{e}^{-ik}
\end{matrix}
\right) \left(
\begin{matrix}

r\\
b_{1}\\
b_{2}\\
a
\end{matrix}
\right) =\left(
\begin{matrix}

\mathrm{e}^{ik}\\
ik\mathrm{e}^{-ik}\\
0\\
0
\end{matrix}
\right)
</math></center>

==Connection with the KdV==

If we substitute the relationship
<center><math>
\partial_{x}^{2}w+uw=-\lambda w
</math></center>
into the KdV after some manipulation we obtain
<center><math>
\partial_{t}\lambda w^{2}+\partial_{x}\left( w\partial_{x}Q-\partial
_{x}wQ\right) =0
</math></center>
where <math>Q=\partial_{t}w+\partial_{x}^{3}w-3\left( \lambda-u\right)
\partial_{x}w.</math> If we integrate this equation then we obtain the result that
<center><math>
\partial_{t}\lambda=0
</math></center>
provided that the eigenfunction <math>w</math> is bounded (which is true for the bound
state eigenfunctions). This shows that the discrete eigenvalues are unchanged
and <math>u\left( x,t\right) </math> evolves according to the KdV.

==Scattering Data==

For the discrete spectrum the eigenfunctions behave like
<center><math>
w_{n}\left( x\right) =c_{n}\left( t\right) \mathrm{e}^{-k_{n}x}
</math></center>
as <math>x\rightarrow\infty</math> with
<center><math>
\int_{-\infty}^{\infty}\left( w_{n}\left( x\right) \right) ^{2}dx=1
</math></center>
The continuous spectrum looks like
<center><math>
v\left( x,t\right) \approx \mathrm{e}^{-\mathrm{i}kx}+r\left( k,t\right) \mathrm{e}^{\mathrm{i}kx}
,\ \ \ x\rightarrow-\infty
</math></center>
<center><math>
v\left( x,t\right) \approx a\left( k,t\right) \mathrm{e}^{-\mathrm{i}kx},\ \ \ x\rightarrow
\infty
</math></center>
where <math>r</math> is the reflection coefficient and <math>a</math> is the transmission
coefficient. This gives us the scattering data at <math>t=0</math>
<center><math>
S\left( \lambda,0\right) =\left( \left\{ k_{n},c_{n}\left( 0\right)
\right\} _{n=1}^{N},r\left( k,0\right) ,a\left( k,0\right) \right)
</math></center>
The scattering data evolves as
<center><math>
k_{n}=k_{n}
</math></center>
<center><math>
c_{n}\left( t\right) =c_{n}\left( 0\right) \mathrm{e}^{4k_{n}^{3}t}
</math></center>
<center><math>
r\left( k,t\right) =r\left( k,0\right) \mathrm{e}^{8ik^{3}t}
</math></center>
<center><math>
a\left( k,t\right) =a\left( k,0\right)
</math></center>
We can recover <math>u</math> from scattering data. We write
<center><math>
F\left( x,t\right) =\sum_{n=1}^{N}c_{n}^{2}\left( t\right) \mathrm{e}^{-k_{n}
x}+\int_{-\infty}^{\infty}r\left( k,t\right) \mathrm{e}^{\mathrm{i}kx}\mathrm{d}k
</math></center>
Then solve
<center><math>
K\left( x,y;t\right) +F\left( x+y;t\right) +\int_{x}^{\infty}K\left(
x,z;t\right) F\left( z+y;t\right) \mathrm{d}z=0
</math></center>
This is a linear integral equation called the \emph{Gelfand-Levitan-Marchenko
}equation. We then find <math>u</math> from
<center><math>
u\left( x,t\right) =2\partial_{x}K\left( x,x,t\right)
</math></center>

==Reflectionless Potential==

In general the IST is difficult to solve. However, there is a simplification
we can make when we have a reflectionless potential (which we will see gives
rise to the soliton solutions). The reflectionless potential is the case when
<math>r\left( k,0\right) =0</math> for all values of <math>k</math> for some <math>u.</math> In this case
<center><math>
F\left( x,t\right) =\sum_{n=1}^{N}c_{n}^{2}\left( t\right) \mathrm{e}^{-k_{n}x}
</math></center>
then
<center><math>
K\left( x,y,t\right) +\sum_{n=1}^{N}c_{n}^{2}\left( t\right)
\mathrm{e}^{-k_{n}\left( x+y\right) }+\int_{x}^{\infty}K\left( x,z,t\right)
\sum_{n=1}^{N}c_{n}^{2}\left( t\right) \mathrm{e}^{-k_{n}\left( y+z\right) }dz=0
</math></center>
From the equation we can see that
<center><math>
K\left( x,y,t\right) =-\sum_{m=1}^{N}c_{m}\left( t\right) v_{m}\left(
x\right) \mathrm{e}^{-k_{m}y}
</math></center>
If we substitute this into the equation
<center><math>
-\sum_{n=1}^{N}c_{n}\left( t\right) v_{n}\left( x\right) \mathrm{e}^{-k_{n}y}
+\sum_{n=1}^{N}c_{n}^{2}\left( t\right) \mathrm{e}^{-k_{n}\left( x+y\right) }
+\int_{x}^{\infty}-\sum_{m=1}^{N}c_{m}\left( t\right) v_{m}\left( x\right)
\mathrm{e}^{-k_{m}y}\sum_{n=1}^{N}c_{n}^{2}\left( t\right) \mathrm{e}^{-k_{n}\left(
y+z\right) }dz=0
</math></center>
which leads to
<center><math>
-\sum_{n=1}^{N}c_{n}\left( t\right) v_{n}\left( x\right) \mathrm{e}^{-k_{n}y}
+\sum_{n=1}^{N}c_{n}^{2}\left( t\right) \mathrm{e}^{-k_{n}\left( x+y\right) }
-\sum_{n=1}^{N}\sum_{m=1}^{N}\frac{c_{m}\left( t\right) c_{n}^{2}\left(
t\right) }{k_{n}+k_{m}}v_{m}\left( x\right) \mathrm{e}^{-k_{m}x}\mathrm{e}^{-k_{n}\left(
y+x\right) }=0
</math></center>
and we can eliminate the sum over <math>n</math> , the <math>c_{n}\left( t\right) ,</math> and the
<math>\mathrm{e}^{-k_{n}y}</math> to obtain
<center><math>
-v_{n}\left( x\right) +c_{n}\left( t\right) \mathrm{e}^{-k_{n}x}-\sum_{m=1}
^{N}\frac{c_{n}\left( t\right) c_{m}\left( t\right) }{k_{n}+k_{m}}
v_{m}\left( x\right) \mathrm{e}^{-\left( k_{m}+k_{n}\right) x}=0
</math></center>
which is an algebraic (finite dimensional system)\ for the unknows <math>v_{n}.</math> We
can write this as
<center><math>
\left( \mathbf{I}+\mathbf{C}\right) \vec{v}=\vec{f}
</math></center>
where <math>f_{m}=c_{m}\left( t\right) \mathrm{e}^{-k_{m}x}</math> and
<center><math>
c_{mn}=\sum_{m=1}^{N}\frac{c_{n}\left( t\right) c_{m}\left( t\right)
}{k_{n}+k_{m}}\mathrm{e}^{-\left( k_{m}+k_{n}\right) x}
</math></center>
<center><math>
K\left( x,y,t\right) =-\sum_{m=1}^{N}c_{m}\left( t\right) \left(
\mathbf{I}+\mathbf{C}\right) ^{-1}\vec{f}\mathrm{e}^{-k_{m}y}
</math></center>
This leads to
<center><math>
u\left( x,t\right) =2\partial_{x}^{2}\log\left[ \det\left( \mathbf{I}
+\mathbf{C}\right) \right]
</math></center>
Lets consider some simple examples. First of all if <math>n=1</math> (the single soliton
solution) we get
<center><math>\begin{matrix}
K\left( x,x,t\right) & =-\frac{c_{1}\left( t\right) c_{1}\left(
t\right) \mathrm{e}^{-k_{1}x}\mathrm{e}^{-k_{1}x}}{1+\frac{c_{1}\left( t\right) c_{1}\left(
t\right) }{k_{1}+k_{1}}\mathrm{e}^{-\left( k_{1}+k_{1}\right) x}}\\
& =\frac{-1}{1+\mathrm{e}^{2k_{1}x-8k_{1}^{3}t-\alpha}}
\end{matrix}</math></center>
where <math>\mathrm{e}^{-\alpha}=2c_{0}^{2}\left( 0\right) .</math> Therefore
<center><math>\begin{matrix}
u\left( x,t\right) & =2\partial_{x}K\left( x,x,t\right) \\
& =\frac{4k_{1}\mathrm{e}^{2k_{1}x-8k_{1}^{3}t-\alpha}}{\left( 1+\mathrm{e}^{2k_{1}
x-8k_{1}^{3}t-\alpha}\right) ^{2}}\\
& =\frac{-8k_{1}^{2}}{\left( \sqrt{2k_{1}}\mathrm{e}^{\theta}+\mathrm{e}^{-\theta}
/\sqrt{2k_{1}}\right) ^{2}}\\
& =2k^{2}\sec^{2}\left\{ k_{1}\left( x-x_{0}\right) -4k_{1}^{3}t\right\}
\end{matrix}</math></center>
where <math>\theta=k_{1}x-4k^{3}t-\alpha/2</math> and <math>\sqrt{2k}\mathrm{e}^{-\alpha/2}=\mathrm{e}^{-kx_{0}
}</math>. This is of course the single soliton solution.

Introduction to the Inverse Scattering Transform

2010-09-17T00:55:24Z

Mike smith: /* Example: Scattering by a Well */

{{nonlinear waves course
| chapter title = Introduction to the Inverse Scattering Transform
| next chapter = [[Reaction-Diffusion Systems]]
| previous chapter = [[Conservation Laws for the KdV]]
}}

== Introduction ==

The inverse scattering transformation gives a way to solve the KdV equation
exactly. You can think about is as being an analogous transformation to the
Fourier transformation, except it works for a non linear equation. We want to
be able to solve
<center><math>\begin{align}
\partial_{t}u+6u\partial_{x}u+\partial_{x}^{3}u & =0\\
u(x,0) & =f\left( x\right)
\end{align}</math></center>
with <math>\left\vert u\right\vert \rightarrow0</math> as <math>x\rightarrow\pm\infty.</math>

The Miura transformation is given by
<center><math>
u=v^{2}+v_{x} \,
</math></center>
and if <math>v</math> satisfies the mKdV
<center><math>
\partial_{t}v-6v^{2}\partial_{x}v+\partial_{x}^{3}v=0
</math></center>
then <math>u</math> satisfies the KdV (but not vice versa). We can think about the Miura
transformation as being a nonlinear ODE solving for <math>v</math> given <math>u.</math> This
nonlinear ODE is also known as the Riccati equation and there is a well know
transformation which linearises this equation. It we write
<center><math>
v=\frac{\left( \partial_{x}w\right) }{w}
</math></center>
then we obtain the equation
<center><math>
\partial_{x}^{2}w+uw=0
</math></center>
The KdV is invariant under the transformation <math>x\rightarrow x+6\lambda t,</math>
<math>u\rightarrow u+\lambda.</math> Therefore we consider the associated eigenvalue
problem
<center><math>
\partial_{x}^{2}w+uw=-\lambda w
</math></center>
The eigenfunctions and eigenvalues of this scattering problem play a key role
in the inverse scattering transformation. Note that this is Schrodinger's equation.

==Properties of the eigenfunctions==

The equation
<center><math>
\partial_{x}^{2}w+uw=-\lambda w
</math></center>
has two kinds of solutions for <math>u\rightarrow0</math> as <math>x\rightarrow\pm\infty.</math> The
first are waves and the second are bound solutions. It is well known that
there are at most a finite number of bound solutions (provided <math>u\rightarrow0</math>
as <math>x\pm\infty</math> sufficiently rapidly) and a continum of solutions for the
incident waves.

===Example: Scattering by a Well===

The properties of the eigenfunction is prehaps seem most easily through the
following example
<center><math>
u\left( x\right) =\left\{
\begin{matrix}

0, & x\notin\left[ -1,1\right] \\
b, & x\in\left[ -1,1\right]
\end{matrix}
\right.
</math></center>
where <math>b>0.</math>

==Case when <math>\lambda<0</math>==

If we solve this equation for the case when <math>\lambda<0,</math> <math>\lambda=-k^{2}</math> we
get
<center><math>
w\left( x\right) =\left\{
\begin{matrix}

a_{1}\mathrm{e}^{kx}, & x<-1\\
b_{1}\cos\kappa x+b_{2}\sin\kappa x & -1<x<1\\
a_{2}\mathrm{e}^{-kx} & x>1
\end{matrix}
\right.
</math></center>
where <math>\kappa=\sqrt{b-k^{2}}</math> where we have assumed that <math>b>k^{2}</math> (there is
no solution for <math>b<k^{2}).</math> We then match <math>w</math> and its derivative at <math>x=\pm1</math>
to solve for <math>a</math> and <math>b</math>. This leads to two system of equations, one for the
even (<math>a_{1}=a_{2}</math> and <math>b_{2}=0</math> ) and one for the odd solutions
(<math>a_{1}=-a_{2}</math> and <math>b_{1}=0)</math>. The solution for the even solutions is
<center><math>
\left(
\begin{matrix}

\mathrm{e}^{-kx} & -\cos\kappa\\
k\mathrm{e}^{-kx} & \sin\kappa
\end{matrix}
\right) \left(
\begin{matrix}

a_{1}\\
b_{1}
\end{matrix}
\right) =\left(
\begin{matrix}

0\\
0
\end{matrix}
\right)
</math></center>
This has non trivial solutions when
<center><math>
\det\left(
\begin{matrix}

\mathrm{e}^{-kx} & -\cos\kappa\\
k\mathrm{e}^{-kx} & - \kappa \sin\kappa
\end{matrix}
\right) =0
</math></center>
which gives us the equation
<center><math>
- \kappa \sin\kappa \mathrm{e}^{-kx}+\left( \cos\kappa\right) k\mathrm{e}^{-kx}=0
</math></center>
or
<center><math>
\kappa \tan\kappa=k=\sqrt{b-\kappa^{2}}
</math></center>
We know that <math>0<\kappa<\sqrt{b}</math> and if we plot this we see that we obtain a
finite number of solutions.

Similarly we repeat the above process for the odd solutions.

==Case when <math>\lambda>0</math>==

When <math>\lambda>0</math> we write <math>\lambda=k^{2}</math> and we obtain solution
<center><math>
w\left( x\right) =\left\{
\begin{matrix}

\mathrm{e}^{-\mathrm{i}kx}+r\mathrm{e}^{\mathrm{i}kx}, & x<-1\\
b_{1}\cos\kappa x+b_{2}\sin\kappa x & -1<x<1\\
a\mathrm{e}^{-\mathrm{i}kx} & x>1
\end{matrix}
\right.
</math></center>
where <math>\kappa=\sqrt{b+k^{2}}.</math> Matching <math>w</math> and its derivaties at <math>x=\pm1</math> we
obtain
<center><math>
\left(
\begin{matrix}

-\mathrm{e}^{-ik} & \cos\kappa & -\sin\kappa & 0\\
ik\mathrm{e}^{-ik} & \kappa\sin\kappa & \kappa\cos\kappa & 0\\
0 & \cos\kappa & \sin\kappa & -\mathrm{e}^{-ik}\\
0 & -\kappa\sin\kappa & \kappa\cos\kappa & ik\mathrm{e}^{-ik}
\end{matrix}
\right) \left(
\begin{matrix}

r\\
b_{1}\\
b_{2}\\
a
\end{matrix}
\right) =\left(
\begin{matrix}

\mathrm{e}^{ik}\\
ik\mathrm{e}^{-ik}\\
0\\
0
\end{matrix}
\right)
</math></center>

==Connection with the KdV==

If we substitute the relationship
<center><math>
\partial_{x}^{2}w+uw=-\lambda w
</math></center>
into the KdV after some manipulation we obtain
<center><math>
\partial_{t}\lambda w^{2}+\partial_{x}\left( w\partial_{x}Q-\partial
_{x}wQ\right) =0
</math></center>
where <math>Q=\partial_{t}w+\partial_{x}^{3}w-3\left( \lambda-u\right)
\partial_{x}w.</math> If we integrate this equation then we obtain the result that
<center><math>
\partial_{t}\lambda=0
</math></center>
provided that the eigenfunction <math>w</math> is bounded (which is true for the bound
state eigenfunctions). This shows that the discrete eigenvalues are unchanged
and <math>u\left( x,t\right) </math> evolves according to the KdV.

==Scattering Data==

For the discrete spectrum the eigenfunctions behave like
<center><math>
w_{n}\left( x\right) =c_{n}\left( t\right) \mathrm{e}^{-k_{n}x}
</math></center>
as <math>x\rightarrow\infty</math> with
<center><math>
\int_{-\infty}^{\infty}\left( w_{n}\left( x\right) \right) ^{2}dx=1
</math></center>
The continuous spectrum looks like
<center><math>
v\left( x,t\right) \approx \mathrm{e}^{-\mathrm{i}kx}+r\left( k,t\right) \mathrm{e}^{\mathrm{i}kx}
,\ \ \ x\rightarrow-\infty
</math></center>
<center><math>
v\left( x,t\right) \approx a\left( k,t\right) \mathrm{e}^{-\mathrm{i}kx},\ \ \ x\rightarrow
\infty
</math></center>
where <math>r</math> is the reflection coefficient and <math>a</math> is the transmission
coefficient. This gives us the scattering data at <math>t=0</math>
<center><math>
S\left( \lambda,0\right) =\left( \left\{ k_{n},c_{n}\left( 0\right)
\right\} _{n=1}^{N},r\left( k,0\right) ,a\left( k,0\right) \right)
</math></center>
The scattering data evolves as
<center><math>
k_{n}=k_{n}
</math></center>
<center><math>
c_{n}\left( t\right) =c_{n}\left( 0\right) \mathrm{e}^{4k_{n}^{3}t}
</math></center>
<center><math>
r\left( k,t\right) =r\left( k,0\right) \mathrm{e}^{8ik^{3}t}
</math></center>
<center><math>
a\left( k,t\right) =a\left( k,0\right)
</math></center>
We can recover <math>u</math> from scattering data. We write
<center><math>
F\left( x,t\right) =\sum_{n=1}^{N}c_{n}^{2}\left( t\right) \mathrm{e}^{-k_{n}
x}+\int_{-\infty}^{\infty}r\left( k,t\right) \mathrm{e}^{\mathrm{i}kx}\mathrm{d}k
</math></center>
Then solve
<center><math>
K\left( x,y;t\right) +F\left( x+y;t\right) +\int_{x}^{\infty}K\left(
x,z;t\right) F\left( z+y;t\right) \mathrm{d}z=0
</math></center>
This is a linear integral equation called the \emph{Gelfand-Levitan-Marchenko
}equation. We then find <math>u</math> from
<center><math>
u\left( x,t\right) =2\partial_{x}K\left( x,x,t\right)
</math></center>

==Reflectionless Potential==

In general the IST is difficult to solve. However, there is a simplification
we can make when we have a reflectionless potential (which we will see gives
rise to the soliton solutions). The reflectionless potential is the case when
<math>r\left( k,0\right) =0</math> for all values of <math>k</math> for some <math>u.</math> In this case
<center><math>
F\left( x,t\right) =\sum_{n=1}^{N}c_{n}^{2}\left( t\right) \mathrm{e}^{-k_{n}x}
</math></center>
then
<center><math>
K\left( x,y,t\right) +\sum_{n=1}^{N}c_{n}^{2}\left( t\right)
\mathrm{e}^{-k_{n}\left( x+y\right) }+\int_{x}^{\infty}K\left( x,z,t\right)
\sum_{n=1}^{N}c_{n}^{2}\left( t\right) \mathrm{e}^{-k_{n}\left( y+z\right) }dz=0
</math></center>
From the equation we can see that
<center><math>
K\left( x,y,t\right) =-\sum_{m=1}^{N}c_{m}\left( t\right) v_{m}\left(
x\right) \mathrm{e}^{-k_{m}y}
</math></center>
If we substitute this into the equation
<center><math>
-\sum_{n=1}^{N}c_{n}\left( t\right) v_{n}\left( x\right) \mathrm{e}^{-k_{n}y}
+\sum_{n=1}^{N}c_{n}^{2}\left( t\right) \mathrm{e}^{-k_{n}\left( x+y\right) }
+\int_{x}^{\infty}-\sum_{m=1}^{N}c_{m}\left( t\right) v_{m}\left( x\right)
\mathrm{e}^{-k_{m}y}\sum_{n=1}^{N}c_{n}^{2}\left( t\right) \mathrm{e}^{-k_{n}\left(
y+z\right) }dz=0
</math></center>
which leads to
<center><math>
-\sum_{n=1}^{N}c_{n}\left( t\right) v_{n}\left( x\right) \mathrm{e}^{-k_{n}y}
+\sum_{n=1}^{N}c_{n}^{2}\left( t\right) \mathrm{e}^{-k_{n}\left( x+y\right) }
-\sum_{n=1}^{N}\sum_{m=1}^{N}\frac{c_{m}\left( t\right) c_{n}^{2}\left(
t\right) }{k_{n}+k_{m}}v_{m}\left( x\right) \mathrm{e}^{-k_{m}x}\mathrm{e}^{-k_{n}\left(
y+x\right) }=0
</math></center>
and we can eliminate the sum over <math>n</math> , the <math>c_{n}\left( t\right) ,</math> and the
<math>\mathrm{e}^{-k_{n}y}</math> to obtain
<center><math>
-v_{n}\left( x\right) +c_{n}\left( t\right) \mathrm{e}^{-k_{n}x}-\sum_{m=1}
^{N}\frac{c_{n}\left( t\right) c_{m}\left( t\right) }{k_{n}+k_{m}}
v_{m}\left( x\right) \mathrm{e}^{-\left( k_{m}+k_{n}\right) x}=0
</math></center>
which is an algebraic (finite dimensional system)\ for the unknows <math>v_{n}.</math> We
can write this as
<center><math>
\left( \mathbf{I}+\mathbf{C}\right) \vec{v}=\vec{f}
</math></center>
where <math>f_{m}=c_{m}\left( t\right) \mathrm{e}^{-k_{m}x}</math> and
<center><math>
c_{mn}=\sum_{m=1}^{N}\frac{c_{n}\left( t\right) c_{m}\left( t\right)
}{k_{n}+k_{m}}\mathrm{e}^{-\left( k_{m}+k_{n}\right) x}
</math></center>
<center><math>
K\left( x,y,t\right) =-\sum_{m=1}^{N}c_{m}\left( t\right) \left(
\mathbf{I}+\mathbf{C}\right) ^{-1}\vec{f}\mathrm{e}^{-k_{m}y}
</math></center>
This leads to
<center><math>
u\left( x,t\right) =2\partial_{x}^{2}\log\left[ \det\left( \mathbf{I}
+\mathbf{C}\right) \right]
</math></center>
Lets consider some simple examples. First of all if <math>n=1</math> (the single soliton
solution) we get
<center><math>\begin{matrix}
K\left( x,x,t\right) & =-\frac{c_{1}\left( t\right) c_{1}\left(
t\right) \mathrm{e}^{-k_{1}x}\mathrm{e}^{-k_{1}x}}{1+\frac{c_{1}\left( t\right) c_{1}\left(
t\right) }{k_{1}+k_{1}}\mathrm{e}^{-\left( k_{1}+k_{1}\right) x}}\\
& =\frac{-1}{1+\mathrm{e}^{2k_{1}x-8k_{1}^{3}t-\alpha}}
\end{matrix}</math></center>
where <math>\mathrm{e}^{-\alpha}=2c_{0}^{2}\left( 0\right) .</math> Therefore
<center><math>\begin{matrix}
u\left( x,t\right) & =2\partial_{x}K\left( x,x,t\right) \\
& =\frac{4k_{1}\mathrm{e}^{2k_{1}x-8k_{1}^{3}t-\alpha}}{\left( 1+\mathrm{e}^{2k_{1}
x-8k_{1}^{3}t-\alpha}\right) ^{2}}\\
& =\frac{-8k_{1}^{2}}{\left( \sqrt{2k_{1}}\mathrm{e}^{\theta}+\mathrm{e}^{-\theta}
/\sqrt{2k_{1}}\right) ^{2}}\\
& =2k^{2}\sec^{2}\left\{ k_{1}\left( x-x_{0}\right) -4k_{1}^{3}t\right\}
\end{matrix}</math></center>
where <math>\theta=k_{1}x-4k^{3}t-\alpha/2</math> and <math>\sqrt{2k}\mathrm{e}^{-\alpha/2}=\mathrm{e}^{-kx_{0}
}</math>. This is of course the single soliton solution.

Introduction to the Inverse Scattering Transform

2010-09-17T00:49:57Z

Mike smith: /* Introduction */

{{nonlinear waves course
| chapter title = Introduction to the Inverse Scattering Transform
| next chapter = [[Reaction-Diffusion Systems]]
| previous chapter = [[Conservation Laws for the KdV]]
}}

== Introduction ==

The inverse scattering transformation gives a way to solve the KdV equation
exactly. You can think about is as being an analogous transformation to the
Fourier transformation, except it works for a non linear equation. We want to
be able to solve
<center><math>\begin{align}
\partial_{t}u+6u\partial_{x}u+\partial_{x}^{3}u & =0\\
u(x,0) & =f\left( x\right)
\end{align}</math></center>
with <math>\left\vert u\right\vert \rightarrow0</math> as <math>x\rightarrow\pm\infty.</math>

The Miura transformation is given by
<center><math>
u=v^{2}+v_{x} \,
</math></center>
and if <math>v</math> satisfies the mKdV
<center><math>
\partial_{t}v-6v^{2}\partial_{x}v+\partial_{x}^{3}v=0
</math></center>
then <math>u</math> satisfies the KdV (but not vice versa). We can think about the Miura
transformation as being a nonlinear ODE solving for <math>v</math> given <math>u.</math> This
nonlinear ODE is also known as the Riccati equation and there is a well know
transformation which linearises this equation. It we write
<center><math>
v=\frac{\left( \partial_{x}w\right) }{w}
</math></center>
then we obtain the equation
<center><math>
\partial_{x}^{2}w+uw=0
</math></center>
The KdV is invariant under the transformation <math>x\rightarrow x+6\lambda t,</math>
<math>u\rightarrow u+\lambda.</math> Therefore we consider the associated eigenvalue
problem
<center><math>
\partial_{x}^{2}w+uw=-\lambda w
</math></center>
The eigenfunctions and eigenvalues of this scattering problem play a key role
in the inverse scattering transformation. Note that this is Schrodinger's equation.

==Properties of the eigenfunctions==

The equation
<center><math>
\partial_{x}^{2}w+uw=-\lambda w
</math></center>
has two kinds of solutions for <math>u\rightarrow0</math> as <math>x\rightarrow\pm\infty.</math> The
first are waves and the second are bound solutions. It is well known that
there are at most a finite number of bound solutions (provided <math>u\rightarrow0</math>
as <math>x\pm\infty</math> sufficiently rapidly) and a continum of solutions for the
incident waves.

===Example: Scattering by a Well===

The properties of the eigenfunction is prehaps seem most easily through the
following example
<center><math>
u\left( x\right) =\left\{
\begin{matrix}

0, & x\notin\left[ -1,1\right] \\
b, & x\in\left[ -1,1\right]
\end{matrix}
\right.
</math></center>
where <math>b>0.</math>

\paragraph{Case when <math>\lambda>0</math>}

If we solve this equation for the case when <math>\lambda<0,</math> <math>\lambda=-k^{2}</math> we
get
<center><math>
w\left( x\right) =\left\{
\begin{matrix}

a_{1}\mathrm{e}^{kx}, & x<-1\\
b_{1}\cos\kappa x+b_{2}\sin\kappa x & -1<x<1\\
a_{2}\mathrm{e}^{-kx} & x>1
\end{matrix}
\right.
</math></center>
where <math>\kappa=\sqrt{b-k^{2}}</math> where we have assumed that <math>b>k^{2}</math> (there is
no solution for <math>b<k^{2}).</math> We then match <math>w</math> and its derivative at <math>x=\pm1</math>
to solve for <math>a</math> and <math>b</math>. This leads to two system of equation, one for the
even (<math>a_{1}=a_{2}</math> and <math>b_{2}=0</math> ) and one for the odd solutions
(<math>a_{1}=-a_{2}</math> and <math>b_{1}=0)</math>. The solution for the even solutions is
<center><math>
\left(
\begin{matrix}

\mathrm{e}^{-kx} & -\cos\kappa\\
k\mathrm{e}^{-kx} & \sin\kappa
\end{matrix}
\right) \left(
\begin{matrix}

a_{1}\\
b_{1}
\end{matrix}
\right) =\left(
\begin{matrix}

0\\
0
\end{matrix}
\right)
</math></center>
This can non trivial solutions when
<center><math>
\det\left(
\begin{matrix}

\mathrm{e}^{-kx} & -\cos\kappa\\
k\mathrm{e}^{-kx} & \sin\kappa
\end{matrix}
\right) =0
</math></center>
which gives us the equation
<center><math>
\mathrm{e}^{-kx}\sin\kappa+\left( \cos\kappa\right) k\mathrm{e}^{-kx}=0
</math></center>
or
<center><math>
\tan\kappa=-k=-\sqrt{b-\kappa^{2}}
</math></center>
We know that <math>0<\kappa<\sqrt{b}</math> and if we plot this we see that we obtain a
finite number of solutions.

\paragraph{Case when <math>\lambda>0</math>}

When <math>\lambda>0</math> we write <math>\lambda=k^{2}</math> and we obtain solution
<center><math>
w\left( x\right) =\left\{
\begin{matrix}

\mathrm{e}^{-\mathrm{i}kx}+r\mathrm{e}^{\mathrm{i}kx}, & x<-1\\
b_{1}\cos\kappa x+b_{2}\sin\kappa x & -1<x<1\\
a\mathrm{e}^{-\mathrm{i}kx} & x>1
\end{matrix}
\right.
</math></center>
where <math>\kappa=\sqrt{b+k^{2}}.</math> Matching <math>w</math> and its derivaties at <math>x=\pm1</math> we
obtain
<center><math>
\left(
\begin{matrix}

-\mathrm{e}^{-ik} & \cos\kappa & -\sin\kappa & 0\\
ik\mathrm{e}^{-ik} & \kappa\sin\kappa & \kappa\cos\kappa & 0\\
0 & \cos\kappa & \sin\kappa & -\mathrm{e}^{-ik}\\
0 & -\kappa\sin\kappa & \kappa\cos\kappa & ik\mathrm{e}^{-ik}
\end{matrix}
\right) \left(
\begin{matrix}

r\\
b_{1}\\
b_{2}\\
a
\end{matrix}
\right) =\left(
\begin{matrix}

\mathrm{e}^{ik}\\
ik\mathrm{e}^{-ik}\\
0\\
0
\end{matrix}
\right)
</math></center>

==Connection with the KdV==

If we substitute the relationship
<center><math>
\partial_{x}^{2}w+uw=-\lambda w
</math></center>
into the KdV after some manipulation we obtain
<center><math>
\partial_{t}\lambda w^{2}+\partial_{x}\left( w\partial_{x}Q-\partial
_{x}wQ\right) =0
</math></center>
where <math>Q=\partial_{t}w+\partial_{x}^{3}w-3\left( \lambda-u\right)
\partial_{x}w.</math> If we integrate this equation then we obtain the result that
<center><math>
\partial_{t}\lambda=0
</math></center>
provided that the eigenfunction <math>w</math> is bounded (which is true for the bound
state eigenfunctions). This shows that the discrete eigenvalues are unchanged
and <math>u\left( x,t\right) </math> evolves according to the KdV.

==Scattering Data==

For the discrete spectrum the eigenfunctions behave like
<center><math>
w_{n}\left( x\right) =c_{n}\left( t\right) \mathrm{e}^{-k_{n}x}
</math></center>
as <math>x\rightarrow\infty</math> with
<center><math>
\int_{-\infty}^{\infty}\left( w_{n}\left( x\right) \right) ^{2}dx=1
</math></center>
The continuous spectrum looks like
<center><math>
v\left( x,t\right) \approx \mathrm{e}^{-\mathrm{i}kx}+r\left( k,t\right) \mathrm{e}^{\mathrm{i}kx}
,\ \ \ x\rightarrow-\infty
</math></center>
<center><math>
v\left( x,t\right) \approx a\left( k,t\right) \mathrm{e}^{-\mathrm{i}kx},\ \ \ x\rightarrow
\infty
</math></center>
where <math>r</math> is the reflection coefficient and <math>a</math> is the transmission
coefficient. This gives us the scattering data at <math>t=0</math>
<center><math>
S\left( \lambda,0\right) =\left( \left\{ k_{n},c_{n}\left( 0\right)
\right\} _{n=1}^{N},r\left( k,0\right) ,a\left( k,0\right) \right)
</math></center>
The scattering data evolves as
<center><math>
k_{n}=k_{n}
</math></center>
<center><math>
c_{n}\left( t\right) =c_{n}\left( 0\right) \mathrm{e}^{4k_{n}^{3}t}
</math></center>
<center><math>
r\left( k,t\right) =r\left( k,0\right) \mathrm{e}^{8ik^{3}t}
</math></center>
<center><math>
a\left( k,t\right) =a\left( k,0\right)
</math></center>
We can recover <math>u</math> from scattering data. We write
<center><math>
F\left( x,t\right) =\sum_{n=1}^{N}c_{n}^{2}\left( t\right) \mathrm{e}^{-k_{n}
x}+\int_{-\infty}^{\infty}r\left( k,t\right) \mathrm{e}^{\mathrm{i}kx}\mathrm{d}k
</math></center>
Then solve
<center><math>
K\left( x,y;t\right) +F\left( x+y;t\right) +\int_{x}^{\infty}K\left(
x,z;t\right) F\left( z+y;t\right) \mathrm{d}z=0
</math></center>
This is a linear integral equation called the \emph{Gelfand-Levitan-Marchenko
}equation. We then find <math>u</math> from
<center><math>
u\left( x,t\right) =2\partial_{x}K\left( x,x,t\right)
</math></center>

==Reflectionless Potential==

In general the IST is difficult to solve. However, there is a simplification
we can make when we have a reflectionless potential (which we will see gives
rise to the soliton solutions). The reflectionless potential is the case when
<math>r\left( k,0\right) =0</math> for all values of <math>k</math> for some <math>u.</math> In this case
<center><math>
F\left( x,t\right) =\sum_{n=1}^{N}c_{n}^{2}\left( t\right) \mathrm{e}^{-k_{n}x}
</math></center>
then
<center><math>
K\left( x,y,t\right) +\sum_{n=1}^{N}c_{n}^{2}\left( t\right)
\mathrm{e}^{-k_{n}\left( x+y\right) }+\int_{x}^{\infty}K\left( x,z,t\right)
\sum_{n=1}^{N}c_{n}^{2}\left( t\right) \mathrm{e}^{-k_{n}\left( y+z\right) }dz=0
</math></center>
From the equation we can see that
<center><math>
K\left( x,y,t\right) =-\sum_{m=1}^{N}c_{m}\left( t\right) v_{m}\left(
x\right) \mathrm{e}^{-k_{m}y}
</math></center>
If we substitute this into the equation
<center><math>
-\sum_{n=1}^{N}c_{n}\left( t\right) v_{n}\left( x\right) \mathrm{e}^{-k_{n}y}
+\sum_{n=1}^{N}c_{n}^{2}\left( t\right) \mathrm{e}^{-k_{n}\left( x+y\right) }
+\int_{x}^{\infty}-\sum_{m=1}^{N}c_{m}\left( t\right) v_{m}\left( x\right)
\mathrm{e}^{-k_{m}y}\sum_{n=1}^{N}c_{n}^{2}\left( t\right) \mathrm{e}^{-k_{n}\left(
y+z\right) }dz=0
</math></center>
which leads to
<center><math>
-\sum_{n=1}^{N}c_{n}\left( t\right) v_{n}\left( x\right) \mathrm{e}^{-k_{n}y}
+\sum_{n=1}^{N}c_{n}^{2}\left( t\right) \mathrm{e}^{-k_{n}\left( x+y\right) }
-\sum_{n=1}^{N}\sum_{m=1}^{N}\frac{c_{m}\left( t\right) c_{n}^{2}\left(
t\right) }{k_{n}+k_{m}}v_{m}\left( x\right) \mathrm{e}^{-k_{m}x}\mathrm{e}^{-k_{n}\left(
y+x\right) }=0
</math></center>
and we can eliminate the sum over <math>n</math> , the <math>c_{n}\left( t\right) ,</math> and the
<math>\mathrm{e}^{-k_{n}y}</math> to obtain
<center><math>
-v_{n}\left( x\right) +c_{n}\left( t\right) \mathrm{e}^{-k_{n}x}-\sum_{m=1}
^{N}\frac{c_{n}\left( t\right) c_{m}\left( t\right) }{k_{n}+k_{m}}
v_{m}\left( x\right) \mathrm{e}^{-\left( k_{m}+k_{n}\right) x}=0
</math></center>
which is an algebraic (finite dimensional system)\ for the unknows <math>v_{n}.</math> We
can write this as
<center><math>
\left( \mathbf{I}+\mathbf{C}\right) \vec{v}=\vec{f}
</math></center>
where <math>f_{m}=c_{m}\left( t\right) \mathrm{e}^{-k_{m}x}</math> and
<center><math>
c_{mn}=\sum_{m=1}^{N}\frac{c_{n}\left( t\right) c_{m}\left( t\right)
}{k_{n}+k_{m}}\mathrm{e}^{-\left( k_{m}+k_{n}\right) x}
</math></center>
<center><math>
K\left( x,y,t\right) =-\sum_{m=1}^{N}c_{m}\left( t\right) \left(
\mathbf{I}+\mathbf{C}\right) ^{-1}\vec{f}\mathrm{e}^{-k_{m}y}
</math></center>
This leads to
<center><math>
u\left( x,t\right) =2\partial_{x}^{2}\log\left[ \det\left( \mathbf{I}
+\mathbf{C}\right) \right]
</math></center>
Lets consider some simple examples. First of all if <math>n=1</math> (the single soliton
solution) we get
<center><math>\begin{matrix}
K\left( x,x,t\right) & =-\frac{c_{1}\left( t\right) c_{1}\left(
t\right) \mathrm{e}^{-k_{1}x}\mathrm{e}^{-k_{1}x}}{1+\frac{c_{1}\left( t\right) c_{1}\left(
t\right) }{k_{1}+k_{1}}\mathrm{e}^{-\left( k_{1}+k_{1}\right) x}}\\
& =\frac{-1}{1+\mathrm{e}^{2k_{1}x-8k_{1}^{3}t-\alpha}}
\end{matrix}</math></center>
where <math>\mathrm{e}^{-\alpha}=2c_{0}^{2}\left( 0\right) .</math> Therefore
<center><math>\begin{matrix}
u\left( x,t\right) & =2\partial_{x}K\left( x,x,t\right) \\
& =\frac{4k_{1}\mathrm{e}^{2k_{1}x-8k_{1}^{3}t-\alpha}}{\left( 1+\mathrm{e}^{2k_{1}
x-8k_{1}^{3}t-\alpha}\right) ^{2}}\\
& =\frac{-8k_{1}^{2}}{\left( \sqrt{2k_{1}}\mathrm{e}^{\theta}+\mathrm{e}^{-\theta}
/\sqrt{2k_{1}}\right) ^{2}}\\
& =2k^{2}\sec^{2}\left\{ k_{1}\left( x-x_{0}\right) -4k_{1}^{3}t\right\}
\end{matrix}</math></center>
where <math>\theta=k_{1}x-4k^{3}t-\alpha/2</math> and <math>\sqrt{2k}\mathrm{e}^{-\alpha/2}=\mathrm{e}^{-kx_{0}
}</math>. This is of course the single soliton solution.

Helmholtz's Equation

2010-07-16T03:29:39Z

Mike smith: consistency

{{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 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 [[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 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
<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} = \nu^2,
</math>
</center>
<math>\Theta
(\theta)</math> can therefore be expressed as
<center>
<math>
\Theta (\theta) = A \, \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) = B \, J_\nu(k r) + C \, H^{(1)}_\nu(k r),\ \nu \in \mathbb{Z},
</math>
</center>
where <math>J_\nu \,</math> denotes a Bessel function
of the first kind and <math>H^{(1)}_\nu \,</math>
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
Hankel function depends on whether we have positive or negative exponential time dependence.
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>
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>\phi = \phi^{\mathrm{I}}+\phi^{\mathrm{S}} \,</math>, where
<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>
<math>
E_{\nu} = - \frac{D_{\nu} J^{\prime}_\nu (k a)}{ H^{(1)\prime}_\nu (ka)},
</math>
</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 ==

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}) + \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>
</center>
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 Helmholz 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>

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>
<math>
\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>
</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 as
<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 to obtain
<center>
<math>
\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>
</center>
We now multiply by <math>e^{\mathrm{i} m \gamma} \,</math> and integrate to obtain
<center>
<math>
\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>
</center>

==External Links==
[http://en.wikipedia.org/wiki/Helmholtz_equation Helmholtz Equation on Wikipedia]

[[Category:Linear Water-Wave Theory]]

Sommerfeld Radiation Condition

2010-07-08T21:17:45Z

Mike smith: correction to Sommerfeld Rad Cond. admins please doublecheck.

{{complete pages}}

This is a condition for the [[Frequency Domain Problem]] that the scattered wave is only
outgoing at infinity. It depends on the convention regarding whether the time dependence
is <math>\exp (i\omega t)\,</math> or <math>\exp (-i\omega t)\,</math>
Assuming the former (which is the standard convention on this wiki)
In two-dimensions the condition is
<center>
<math>
\sqrt{|{x}|}\left( \frac{\partial}{\partial|x|}+{i}k\right)
(\phi-\phi^{\mathrm{{In}}})=0,\;\mathrm{{as\;}}|x|\rightarrow\infty\mathrm{.}
</math>
</center>
where <math>\phi^{\mathrm{{In}}}</math> is the incident potential and <math>k</math>
is the wave number.

In three-dimensions the condition is
<center>
<math>
|\mathbf{r}|\left( \frac{\partial}{\partial|\mathbf{r}|}+{i}k\right)
(\phi-\phi^{\mathrm{{In}}})=0,\;\mathrm{{as\;}}|\mathbf{r}|\rightarrow\infty\mathrm{.}
</math>
</center>

If the time dependence is assumed to be <math>\exp (-i\omega t)\,</math> then we
have in two-dimensions
<center>
<math>
\sqrt{|{x}|} \left( \frac{\partial}{\partial|x|}-{i}k\right)
(\phi-\phi^{\mathrm{{In}}})=0,\;\mathrm{{as\;}}|x|\rightarrow\infty\mathrm{.}
</math>
</center>
and in three-dimensions
<center>
<math>
|\mathbf{r}|\left( \frac{\partial}{\partial|\mathbf{r}|}-{i}k\right)
(\phi-\phi^{\mathrm{{In}}})=0,\;\mathrm{{as\;}}|\mathbf{r}|\rightarrow\infty\mathrm{.}
</math>
</center>

[[Category:Linear Water-Wave Theory]]

Helmholtz's Equation

2010-06-21T00:26:30Z

Mike smith: missing i/4

{{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 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 [[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 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
<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} = \nu^2,
</math>
</center>
<math>\Theta
(\theta)</math> can therefore be expressed as
<center>
<math>
\Theta (\theta) = A \, \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) = B \, J_\nu(k_m r) + C \, H^{(1)}_\nu(k_m r),\ \nu \in \mathbb{Z},
</math>
</center>
where <math>J_\nu \,</math> denotes a Bessel function
of the first kind and <math>H^{(1)}_\nu \,</math>
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
Hankel function depends on whether we have positive or negative exponential time dependence.
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>
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>\phi = \phi^{\mathrm{I}}+\phi^{\mathrm{S}} \,</math>, where
<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>
<math>
E_{\nu} = - \frac{D_{\nu} J^{\prime}_\nu (k a)}{ H^{(1)\prime}_\nu (ka)},
</math>
</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 ==

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}) + \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>
</center>
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 Helmholz 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>

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>
<math>
\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>
</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 as
<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 to obtain
<center>
<math>
\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>
</center>
We now multiply by <math>e^{\mathrm{i} m \gamma} \,</math> and integrate to obtain
<center>
<math>
\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>
</center>

==External Links==
[http://en.wikipedia.org/wiki/Helmholtz_equation Helmholtz Equation on Wikipedia]

[[Category:Linear Water-Wave Theory]]

Helmholtz's Equation

2010-06-16T00:14:49Z

Mike smith: /* 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 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 [[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 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
<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} = \nu^2,
</math>
</center>
<math>\Theta
(\theta)</math> can therefore be expressed as
<center>
<math>
\Theta (\theta) = A \, \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) = B \, J_\nu(k_m r) + C \, H^{(1)}_\nu(k_m r),\ \nu \in \mathbb{Z},
</math>
</center>
where <math>J_\nu \,</math> denotes a Bessel function
of the first kind and <math>H^{(1)}_\nu \,</math>
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
Hankel function depends on whether we have positive or negative exponential time dependence.
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>
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>\phi = \phi^{\mathrm{I}}+\phi^{\mathrm{S}} \,</math>, where
<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>
<math>
E_{\nu} = - \frac{D_{\nu} J^{\prime}_\nu (k a)}{ H^{(1)\prime}_\nu (ka)},
</math>
</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 ==

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}) + \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>
</center>
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 Helmholz 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>

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>
<math>
\frac{1}{2}\phi(\mathbf{x}) = \phi^{\mathrm{I}}(\mathbf{x}) + \int_{\partial\Omega} \partial_{n^{\prime}} H^{(1)}_0
(k|\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 as
<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 to 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
(k|\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 to obtain
<center>
<math>
\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 +
\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>
</center>

==External Links==
[http://en.wikipedia.org/wiki/Helmholtz_equation Helmholtz Equation on Wikipedia]

[[Category:Linear Water-Wave Theory]]

Helmholtz's Equation

2010-06-16T00:13:19Z

Mike smith: /* 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 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 [[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 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
<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} = \nu^2,
</math>
</center>
<math>\Theta
(\theta)</math> can therefore be expressed as
<center>
<math>
\Theta (\theta) = A \, \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) = B \, J_\nu(k_m r) + C \, H^{(1)}_\nu(k_m r),\ \nu \in \mathbb{Z},
</math>
</center>
where <math>J_\nu \,</math> denotes a Bessel function
of the first kind and <math>H^{(1)}_\nu \,</math>
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
Hankel function depends on whether we have positive or negative exponential time dependence.
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>
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>\phi = \phi^{\mathrm{I}}+\phi^{\mathrm{S}} \,</math>, where
<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>
<math>
E_{\nu} = - \frac{D_{\nu} J^{\prime}_\nu (k a)}{ H^{(1)\prime}_\nu (ka)},
</math>
</center>
which tells us that providing we know the form of the incident wave, we can compute the <math>D_\nu \,</math> coefficients and compute the potential throughout the circle. It is possible to equate plane waves and cylindrical waves using the [http://en.wikipedia.org/wiki/Jacobi%E2%80%93Anger_expansion 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
<center>
<math>
\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>
</center>
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 Helmholz 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>

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>
<math>
\frac{1}{2}\phi(\mathbf{x}) = \phi^{\mathrm{I}}(\mathbf{x}) + \int_{\partial\Omega} \partial_{n^{\prime}} H^{(1)}_0
(k|\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 as
<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 to 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
(k|\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 to obtain
<center>
<math>
\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 +
\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>
</center>

==External Links==
[http://en.wikipedia.org/wiki/Helmholtz_equation Helmholtz Equation on Wikipedia]

[[Category:Linear Water-Wave Theory]]

User:Mike smith

2010-06-15T22:10:52Z

Mike smith:

Is an PhD student at the University of Auckland Mathematics Dept (Supervisor: Dr Mike Meylan)

Helmholtz's Equation

2010-06-15T22:06:18Z

Mike smith: /* 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 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 [[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 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
<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} = \nu^2,
</math>
</center>
<math>\Theta
(\theta)</math> can therefore be expressed as
<center>
<math>
\Theta (\theta) = A \, \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) = B \, J_\nu(k_m r) + C \, H^{(1)}_\nu(k_m r),\ \nu \in \mathbb{Z},
</math>
</center>
where <math>J_\nu \,</math> denotes a Bessel function
of the first kind and <math>H^{(1)}_\nu \,</math>
denotes a Hankel functions of order <math>\nu</math> (see [http://en.wikipedia.org/wiki/Bessel_function Bessel functions] for more information ). Note that the first term represents
incoming waves and the second term represents 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
<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 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>
<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}) + \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>
</center>
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 Helmholz 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>

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>
<math>
\frac{1}{2}\phi(\mathbf{x}) = \phi^{\mathrm{I}}(\mathbf{x}) + \int_{\partial\Omega} \partial_{n^{\prime}} H^{(1)}_0
(k|\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 as
<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 to 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
(k|\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 to obtain
<center>
<math>
\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 +
\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>
</center>

==External Links==
[http://en.wikipedia.org/wiki/Helmholtz_equation Helmholtz Equation on Wikipedia]

[[Category:Linear Water-Wave Theory]]

Helmholtz's Equation

2010-06-15T22:04:09Z

Mike smith: /* 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 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 [[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 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
<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} = \nu^2,
</math>
</center>
<math>\Theta
(\theta)</math> can therefore be expressed as
<center>
<math>
\Theta (\theta) = A \, \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) = B \, J_\nu(k_m r) + C \, H^{(1)}_\nu(k_m r),\ \nu \in \mathbb{Z},
</math>
</center>
where <math>J_\nu \,</math> denotes a Bessel function
of the first kind and <math>H^{(1)}_\nu \,</math>
denotes a Hankel functions of order <math>\nu</math> (see [http://en.wikipedia.org/wiki/Bessel_function Bessel functions] for more information ). Note that the first term represents
incoming waves and the second term represents 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
<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 Neumann 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}) + \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>
</center>
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 Helmholz 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>

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>
<math>
\frac{1}{2}\phi(\mathbf{x}) = \phi^{\mathrm{I}}(\mathbf{x}) + \int_{\partial\Omega} \partial_{n^{\prime}} H^{(1)}_0
(k|\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 as
<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 to 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
(k|\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 to obtain
<center>
<math>
\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 +
\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>
</center>

==External Links==
[http://en.wikipedia.org/wiki/Helmholtz_equation Helmholtz Equation on Wikipedia]

[[Category:Linear Water-Wave Theory]]

Helmholtz's Equation

2010-06-15T22:03:47Z

Mike smith: /* 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 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 [[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 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
<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} = \nu^2,
</math>
</center>
<math>\Theta
(\theta)</math> can therefore be expressed as
<center>
<math>
\Theta (\theta) = A \, \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) = B \, J_\nu(k_m r) + C \, H^{(1)}_\nu(k_m r),\ \nu \in \mathbb{Z},
</math>
</center>
where <math>J_\nu \,</math> denotes a Bessel function
of the first kind and <math>H^{(1)}_\nu \,</math>
denotes a Hankel functions of order <math>\nu</math> (see [http://en.wikipedia.org/wiki/Bessel_function Bessel functions] for more information ). Note that the first term represents
incoming waves and the second term represents 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
<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 Neumann 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}) + \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>
</center>
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 Helmholz 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>

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>
<math>
\frac{1}{2}\phi(\mathbf{x}) = \phi^{\mathrm{I}}(\mathbf{x}) + \int_{\partial\Omega} \partial_{n^{\prime}} H^{(1)}_0
(k|\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 to 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
(k|\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 to obtain
<center>
<math>
\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 +
\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>
</center>

==External Links==
[http://en.wikipedia.org/wiki/Helmholtz_equation Helmholtz Equation on Wikipedia]

[[Category:Linear Water-Wave Theory]]

Helmholtz's Equation

2010-06-10T00:29:31Z

Mike smith:

{{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 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 [[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 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
<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} = \nu^2,
</math>
</center>
<math>\Theta
(\theta)</math> can therefore be expressed as
<center>
<math>
\Theta (\theta) = A \, \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) = B \, J_\nu(k_m r) + C \, H^{(1)}_\nu(k_m r),\ \nu \in \mathbb{Z},
</math>
</center>
where <math>J_\nu \,</math> denotes a Bessel function
of the first kind and <math>H^{(1)}_\nu \,</math>
denotes a Hankel functions of order <math>\nu</math> (see [http://en.wikipedia.org/wiki/Bessel_function Bessel functions] for more information ). Note that the first term represents
incoming waves and the second term represents 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
<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 Neumann 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}) + \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>
</center>
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 Helmholz 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 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>
<math>
\frac{1}{2}\phi(\mathbf{x}) = \phi^{\mathrm{I}}(\mathbf{x}) + \int_{\partial\Omega} \partial_{n^{\prime}} H^{(1)}_0
(k|\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 to 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
(k|\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 to obtain
<center>
<math>
\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 +
\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>
</center>
==External Links==
[http://en.wikipedia.org/wiki/Helmholtz_equation Helmholtz Equation on Wikipedia]

[[Category:Linear Water-Wave Theory]]

Fundamental Solution for thin plates

2010-04-27T21:14:41Z

Mike smith:

{{complete pages}}

==Introduction==
On this page, we aim to derive the Green's function for a thin uniform plate. This derivation relies heavily on concepts discussed in [[Boyling 1996]]. We seek the fundamental solution for the Biharmonic equation in <math>\mathbb{R}^2 \,</math>, which is taken to be of the form
<center>
<math>
\left(\Delta^2 - k^2\right) u =0,
</math>
</center>
where <math>u \,</math> is the displacement, <math>k \,</math> is the wavenumber, and <math>\Delta = \partial_r^2 + \frac{1}{r}\partial_r, </math> in polar coordinates.

==Linear Operators==
We consider the operator <math>P(\Delta) = \left(\Delta^2 - k^4\right) = \left(\Delta - (ik)^2\right) \left(\Delta -k^2\right)</math>, whereas Boyling considers more general linear operators of the form

<center>
<math>
P(\lambda) = c(\lambda - \lambda_1)^{m_1}(\lambda - \lambda_2)^{m_2} \ldots (\lambda - \lambda_p)^{m_p},
</math>
</center>

which for our case, is identical when <math>c=1 \,</math>, <math>\lambda_1 = (ik)^2 \,</math>, <math>\lambda_2 = k^2 \,</math>, <math>m_1 = m_2 = 1 \,</math>, and <math>m_3=m_4=\ldots =0 \,</math>.

The reciprocal of <math>P(\lambda) \,</math> is taken to be of the form
<center>
<math>
\frac{1}{P(\lambda)} = \sum_{q=1}^{p} \sum_{n=1}^{m_q} \frac{c_{qn}}{(\lambda - \lambda_q)^n}.
</math>
</center>
where
<center>
<math>
c_{qn} = \lim_{\lambda \rightarrow \lambda_q} \frac{1}{(m_q - n)!} \frac{\partial^{m_q - n}}{\partial \lambda^{m_q-n}} \left[\frac{(\lambda - \lambda_q)^{m_q}}{P(\lambda)}\right].
</math>
</center>
It is straightforward to compute the <math>c_{qn} \,</math> coefficients
<center><math>
\begin{align}
c_{11} &= \lim_{\lambda \rightarrow \lambda_q} \frac{1}{(0)!} \frac{\partial^{0}}{\partial \lambda^{0}} \left[\frac{(\lambda - \lambda_1)}{P(\lambda)}\right] = \lim_{\lambda \rightarrow (ik)^2} \frac{1}{\lambda - \lambda_2} = - \frac{1}{2 k^2}, \\
c_{21} &= \frac{1}{2 k^2},
\end{align}
</math></center>
as well as determining a corresponding quantity <math>P_{qn} \,</math>
such that <math>\sum_{q=1}^{p} \sum_{n=1}^{m_q} c_{qn}P_{qn} = 1</math>, where
<center>
<math>
P_{qn}(\lambda) \left[\lambda - \lambda_q\right]^n = P(\lambda), \quad \mbox{for all q,n}.
</math>
</center>
That is, <math>P_{11} = (\lambda - k^2) \,</math> and <math>P_{21} = (\lambda - (ik)^2) \,</math>.

== Determining the Green's function==
In order to compute the Green's function for polynomials in the Laplacian, we express our unknown Green's function in terms of the fundamental solution of Helmholtz's equation:
<center>
<math>
G(r) = \sum_{q=1}^{p} \sum_{n=1}^{m_q} c_{qn} G_{n,\lambda_q}(r) = c_{11} G_{1,\lambda_1}+ c_{21} G_{1,\lambda_2}.
</math>
</center>

Using the Green's function for the [http://en.wikipedia.org/wiki/Helmholtz_equation Helmholtz Equation]
<center>
<math>
G_{1,\beta} = \frac{i}{4} H_0^{(1)} (\beta r),
</math>
</center>
(where <math>H_0^{(1)} \,</math> denotes Hankel functions of the first kind) which incorporates the Sommerfeld radiation condition
<center>
<math>
\lim_{r \rightarrow \infty} r^{1/2} \left(\frac{\partial}{\partial r} - ik \right)G = 0,
</math>
</center>
we obtain the final expression
<center><math>
\begin{align}
G(r) &= \left(-\frac{1}{2 k^2}\right)\left(\frac{i}{4}H_0^{(1)}(ikr)\right) + \left(\frac{1}{2 k^2}\right)\left(\frac{i}{4}H_0^{(1)}(kr)\right) \\
&= \frac{i}{8 k^2} \left[H_0^{(1)}(kr) - H_0^{(1)}(ikr) \right].
\end{align}
</math></center>
This is the fundamental solution for our governing equation above.

==Behaviour at <math>r=0</math>==
To determine the solution at <math>r \rightarrow 0</math> (where the above expression is singular), the following identities from [[Abramowitz and Stegun 1964]] are used
<center><math>
\begin{align}
H_0^{(1)}(z) &= J_0(z) + i Y_0(z), \\
K_0(z) &= \frac{\pi i}{2} H_0^{(1)}(iz), \quad -\pi < arg(z) \leq \pi/2, \\
K_0(z) &= \int_{0}^{\infty} \cos \left[z\sinh \theta \right]d \theta, \\
Y_0(z) &= \frac{1}{\pi}\int_0^\pi \sin \left[z \sin \theta \right] d \theta - \frac{1}{\pi} \int_0^\infty 2 e^{-z \sinh \theta} d \theta, \ | arg(z) | < \frac{\pi}{2}
\end{align}
</math></center>
from pages [http://www.math.auckland.ac.nz/~hcoh001/as/page_358.htm 358], [http://www.math.auckland.ac.nz/~hcoh001/as/page_375.htm 375], [http://www.math.auckland.ac.nz/~hcoh001/as/page_376.htm 376], and [http://www.math.auckland.ac.nz/~hcoh001/as/page_360.htm 360] respectively.

Let us consider
<center><math>
\begin{align}
&H_0^{(1)}(kr) - H_0^{(1)}(ikr) \\
=& J_0(kr) + i Y_0(kr) + \frac{2i}{\pi} K_0(kr) \\
=& J_0(kr) + \frac{i}{\pi}\int_0^\pi \sin \left(kr \sin \theta \right) d \theta - \frac{i}{\pi} \int_0^\infty 2 e^{-kr \sinh \theta} d \theta + \frac{2i}{\pi} \int_{0}^{\infty} \cos \left( kr \sinh \theta \right)d \theta
\end{align}
</math></center>
where introducing the limit yields
<center><math>
\begin{align}
\lim_{r \rightarrow 0}& J_0(kr) = 1, \\
\lim_{r \rightarrow 0}& \sin \left(kr \sin \theta \right) = 0, \\
\lim_{r \rightarrow 0}& e^{-kr \sinh \theta} =1, \\
\lim_{r \rightarrow 0}& \cos \left( kr \sinh \theta \right) = 1.
\end{align}
</math></center>
Consequently
<center><math>
\lim_{r \rightarrow 0} \left( H_0^{(1)}(kr) - H_0^{(1)}(ikr) \right) = \lim_{r \rightarrow 0} J_0(kr) -\frac{2i}{\pi}\int_{0}^{\infty} 1 d\theta + \frac{2i}{\pi}\int_{0}^{\infty} 1 d\theta = 1,
</math></center>
which then allows us to determine that
<center><math>
\lim_{r \rightarrow 0} G(r) = \frac{i}{8 k^2}.
</math></center>

==Solution==
The fundamental solution for a thin uniform plate in <math>\mathbb{R}^2 \,</math> governed by
<center>
<math>
\left(\Delta^2 - k^2\right) u =0,
</math>
</center>
is defined as follows
<center><math>
G(r) = \left\{
\begin{align}
&\frac{i}{8 k^2} \left[H_0^{(1)}(kr) - H_0^{(1)}(ikr) \right] &\mbox{for} \ r>0, \\
&\frac{i}{8 k^2} &\mbox{for } \ r=0.
\end{align}
\right.
</math></center>

Fundamental Solution for thin plates

2010-04-27T21:12:40Z

Mike smith: /* Introduction */

==Introduction==
On this page, we aim to derive the Green's function for a thin uniform plate. This derivation relies heavily on concepts discussed in [[Boyling 1996]]. We seek the fundamental solution for the Biharmonic equation in <math>\mathbb{R}^2 \,</math>, which is taken to be of the form
<center>
<math>
\left(\Delta^2 - k^2\right) u =0,
</math>
</center>
where <math>u \,</math> is the displacement, <math>k \,</math> is the wavenumber, and <math>\Delta = \partial_r^2 + \frac{1}{r}\partial_r, </math> in polar coordinates.

==Linear Operators==
We consider the operator <math>P(\Delta) = \left(\Delta^2 - k^4\right) = \left(\Delta - (ik)^2\right) \left(\Delta -k^2\right)</math>, whereas Boyling considers more general linear operators of the form

<center>
<math>
P(\lambda) = c(\lambda - \lambda_1)^{m_1}(\lambda - \lambda_2)^{m_2} \ldots (\lambda - \lambda_p)^{m_p},
</math>
</center>

which for our case, is identical when <math>c=1 \,</math>, <math>\lambda_1 = (ik)^2 \,</math>, <math>\lambda_2 = k^2 \,</math>, <math>m_1 = m_2 = 1 \,</math>, and <math>m_3=m_4=\ldots =0 \,</math>.

The reciprocal of <math>P(\lambda) \,</math> is taken to be of the form
<center>
<math>
\frac{1}{P(\lambda)} = \sum_{q=1}^{p} \sum_{n=1}^{m_q} \frac{c_{qn}}{(\lambda - \lambda_q)^n}.
</math>
</center>
where
<center>
<math>
c_{qn} = \lim_{\lambda \rightarrow \lambda_q} \frac{1}{(m_q - n)!} \frac{\partial^{m_q - n}}{\partial \lambda^{m_q-n}} \left[\frac{(\lambda - \lambda_q)^{m_q}}{P(\lambda)}\right].
</math>
</center>
It is straightforward to compute the <math>c_{qn} \,</math> coefficients
<center><math>
\begin{align}
c_{11} &= \lim_{\lambda \rightarrow \lambda_q} \frac{1}{(0)!} \frac{\partial^{0}}{\partial \lambda^{0}} \left[\frac{(\lambda - \lambda_1)}{P(\lambda)}\right] = \lim_{\lambda \rightarrow (ik)^2} \frac{1}{\lambda - \lambda_2} = - \frac{1}{2 k^2}, \\
c_{21} &= \frac{1}{2 k^2},
\end{align}
</math></center>
as well as determining a corresponding quantity <math>P_{qn} \,</math>
such that <math>\sum_{q=1}^{p} \sum_{n=1}^{m_q} c_{qn}P_{qn} = 1</math>, where
<center>
<math>
P_{qn}(\lambda) \left[\lambda - \lambda_q\right]^n = P(\lambda), \quad \mbox{for all q,n}.
</math>
</center>
That is, <math>P_{11} = (\lambda - k^2) \,</math> and <math>P_{21} = (\lambda - (ik)^2) \,</math>.

== Determining the Green's function==
In order to compute the Green's function for polynomials in the Laplacian, we express our unknown Green's function in terms of the fundamental solution of Helmholtz's equation:
<center>
<math>
G(r) = \sum_{q=1}^{p} \sum_{n=1}^{m_q} c_{qn} G_{n,\lambda_q}(r) = c_{11} G_{1,\lambda_1}+ c_{21} G_{1,\lambda_2}.
</math>
</center>

Using the Green's function for the [http://en.wikipedia.org/wiki/Helmholtz_equation Helmholtz Equation]
<center>
<math>
G_{1,\beta} = \frac{i}{4} H_0^{(1)} (\beta r),
</math>
</center>
where <math>H_0^{(1)} \,</math> denotes Hankel functions of the first kind, which incorporates the Sommerfeld radiation condition
<center>
<math>
\lim_{r \rightarrow \infty} r^{1/2} \left(\frac{\partial}{\partial r} - ik \right)G = 0,
</math>
</center>
we obtain the final expression
<center><math>
\begin{align}
G(r) &= \left(-\frac{1}{2 k^2}\right)\left(\frac{i}{4}H_0^{(1)}(ikr)\right) + \left(\frac{1}{2 k^2}\right)\left(\frac{i}{4}H_0^{(1)}(kr)\right) \\
&= \frac{i}{8 k^2} \left[H_0^{(1)}(kr) - H_0^{(1)}(ikr) \right].
\end{align}
</math></center>
This is the fundamental solution for our governing equation above.

==Behaviour at <math>r=0</math>==
To determine the solution at <math>r \rightarrow 0</math> (where the above expression is singular), the following identities from [[Abramowitz and Stegun 1964]] are used
<center><math>
\begin{align}
H_0^{(1)}(z) &= J_0(z) + i Y_0(z), \\
K_0(z) &= \frac{\pi i}{2} H_0^{(1)}(iz), \quad -\pi < arg(z) \leq \pi/2, \\
K_0(z) &= \int_{0}^{\infty} \cos \left[z\sinh \theta \right]d \theta, \\
Y_0(z) &= \frac{1}{\pi}\int_0^\pi \sin \left[z \sin \theta \right] d \theta - \frac{1}{\pi} \int_0^\infty 2 e^{-z \sinh \theta} d \theta, \ | arg(z) | < \frac{\pi}{2}
\end{align}
</math></center>
from pages [http://www.math.auckland.ac.nz/~hcoh001/as/page_358.htm 358], [http://www.math.auckland.ac.nz/~hcoh001/as/page_375.htm 375], [http://www.math.auckland.ac.nz/~hcoh001/as/page_376.htm 376], and [http://www.math.auckland.ac.nz/~hcoh001/as/page_360.htm 360] respectively.

Let us consider
<center><math>
\begin{align}
&H_0^{(1)}(kr) - H_0^{(1)}(ikr) \\
=& J_0(kr) + i Y_0(kr) + \frac{2i}{\pi} K_0(kr) \\
=& J_0(kr) + \frac{i}{\pi}\int_0^\pi \sin \left(kr \sin \theta \right) d \theta - \frac{i}{\pi} \int_0^\infty 2 e^{-kr \sinh \theta} d \theta + \frac{2i}{\pi} \int_{0}^{\infty} \cos \left( kr \sinh \theta \right)d \theta
\end{align}
</math></center>
where introducing the limit yields
<center><math>
\begin{align}
\lim_{r \rightarrow 0}& J_0(kr) = 1, \\
\lim_{r \rightarrow 0}& \sin \left(kr \sin \theta \right) = 0, \\
\lim_{r \rightarrow 0}& e^{-kr \sinh \theta} =1, \\
\lim_{r \rightarrow 0}& \cos \left( kr \sinh \theta \right) = 1.
\end{align}
</math></center>
Consequently
<center><math>
\lim_{r \rightarrow 0} \left( H_0^{(1)}(kr) - H_0^{(1)}(ikr) \right) = \lim_{r \rightarrow 0} J_0(kr) -\frac{2i}{\pi}\int_{0}^{\infty} 1 d\theta + \frac{2i}{\pi}\int_{0}^{\infty} 1 d\theta = 1,
</math></center>
which then allows us to determine that
<center><math>
\lim_{r \rightarrow 0} G(r) = \frac{i}{8 k^2}.
</math></center>

==Solution==
The fundamental solution for a thin uniform plate in <math>\mathbb{R}^2 \,</math> governed by
<center>
<math>
\left(\Delta^2 - k^2\right) u =0,
</math>
</center>
is defined as follows
<center><math>
G(r) = \left\{
\begin{align}
&\frac{i}{8 k^2} \left[H_0^{(1)}(kr) - H_0^{(1)}(ikr) \right] &\mbox{for} \ r>0, \\
&\frac{i}{8 k^2} &\mbox{for } \ r=0.
\end{align}
\right.
</math></center>

Fundamental Solution for thin plates

2010-04-27T21:10:14Z

Mike smith:

==Introduction==
On this page, we aim to derive the Green's function for a thin uniform plate. This derivation relies heavily on concepts discussed in [[Boyling 1996]]. We seek the fundamental solution for the Biharmonic equation in <math>\mathbb{R}^2 \,</math>, which is taken to be of the form
<center>
<math>
\left(\Delta^2 - k^2\right) u =0,
</math>
</center>
where <math>u \,</math> is the displacement, and <math>\Delta = \partial_r^2 + \frac{1}{r}\partial_r, </math> in polar coordinates.

==Linear Operators==
We consider the operator <math>P(\Delta) = \left(\Delta^2 - k^4\right) = \left(\Delta - (ik)^2\right) \left(\Delta -k^2\right)</math>, whereas Boyling considers more general linear operators of the form

<center>
<math>
P(\lambda) = c(\lambda - \lambda_1)^{m_1}(\lambda - \lambda_2)^{m_2} \ldots (\lambda - \lambda_p)^{m_p},
</math>
</center>

which for our case, is identical when <math>c=1 \,</math>, <math>\lambda_1 = (ik)^2 \,</math>, <math>\lambda_2 = k^2 \,</math>, <math>m_1 = m_2 = 1 \,</math>, and <math>m_3=m_4=\ldots =0 \,</math>.

The reciprocal of <math>P(\lambda) \,</math> is taken to be of the form
<center>
<math>
\frac{1}{P(\lambda)} = \sum_{q=1}^{p} \sum_{n=1}^{m_q} \frac{c_{qn}}{(\lambda - \lambda_q)^n}.
</math>
</center>
where
<center>
<math>
c_{qn} = \lim_{\lambda \rightarrow \lambda_q} \frac{1}{(m_q - n)!} \frac{\partial^{m_q - n}}{\partial \lambda^{m_q-n}} \left[\frac{(\lambda - \lambda_q)^{m_q}}{P(\lambda)}\right].
</math>
</center>
It is straightforward to compute the <math>c_{qn} \,</math> coefficients
<center><math>
\begin{align}
c_{11} &= \lim_{\lambda \rightarrow \lambda_q} \frac{1}{(0)!} \frac{\partial^{0}}{\partial \lambda^{0}} \left[\frac{(\lambda - \lambda_1)}{P(\lambda)}\right] = \lim_{\lambda \rightarrow (ik)^2} \frac{1}{\lambda - \lambda_2} = - \frac{1}{2 k^2}, \\
c_{21} &= \frac{1}{2 k^2},
\end{align}
</math></center>
as well as determining a corresponding quantity <math>P_{qn} \,</math>
such that <math>\sum_{q=1}^{p} \sum_{n=1}^{m_q} c_{qn}P_{qn} = 1</math>, where
<center>
<math>
P_{qn}(\lambda) \left[\lambda - \lambda_q\right]^n = P(\lambda), \quad \mbox{for all q,n}.
</math>
</center>
That is, <math>P_{11} = (\lambda - k^2) \,</math> and <math>P_{21} = (\lambda - (ik)^2) \,</math>.

== Determining the Green's function==
In order to compute the Green's function for polynomials in the Laplacian, we express our unknown Green's function in terms of the fundamental solution of Helmholtz's equation:
<center>
<math>
G(r) = \sum_{q=1}^{p} \sum_{n=1}^{m_q} c_{qn} G_{n,\lambda_q}(r) = c_{11} G_{1,\lambda_1}+ c_{21} G_{1,\lambda_2}.
</math>
</center>

Using the Green's function for the [http://en.wikipedia.org/wiki/Helmholtz_equation Helmholtz Equation]
<center>
<math>
G_{1,\beta} = \frac{i}{4} H_0^{(1)} (\beta r),
</math>
</center>
where <math>H_0^{(1)} \,</math> denotes Hankel functions of the first kind, which incorporates the Sommerfeld radiation condition
<center>
<math>
\lim_{r \rightarrow \infty} r^{1/2} \left(\frac{\partial}{\partial r} - ik \right)G = 0,
</math>
</center>
we obtain the final expression
<center><math>
\begin{align}
G(r) &= \left(-\frac{1}{2 k^2}\right)\left(\frac{i}{4}H_0^{(1)}(ikr)\right) + \left(\frac{1}{2 k^2}\right)\left(\frac{i}{4}H_0^{(1)}(kr)\right) \\
&= \frac{i}{8 k^2} \left[H_0^{(1)}(kr) - H_0^{(1)}(ikr) \right].
\end{align}
</math></center>
This is the fundamental solution for our governing equation above.

==Behaviour at <math>r=0</math>==
To determine the solution at <math>r \rightarrow 0</math> (where the above expression is singular), the following identities from [[Abramowitz and Stegun 1964]] are used
<center><math>
\begin{align}
H_0^{(1)}(z) &= J_0(z) + i Y_0(z), \\
K_0(z) &= \frac{\pi i}{2} H_0^{(1)}(iz), \quad -\pi < arg(z) \leq \pi/2, \\
K_0(z) &= \int_{0}^{\infty} \cos \left[z\sinh \theta \right]d \theta, \\
Y_0(z) &= \frac{1}{\pi}\int_0^\pi \sin \left[z \sin \theta \right] d \theta - \frac{1}{\pi} \int_0^\infty 2 e^{-z \sinh \theta} d \theta, \ | arg(z) | < \frac{\pi}{2}
\end{align}
</math></center>
from pages [http://www.math.auckland.ac.nz/~hcoh001/as/page_358.htm 358], [http://www.math.auckland.ac.nz/~hcoh001/as/page_375.htm 375], [http://www.math.auckland.ac.nz/~hcoh001/as/page_376.htm 376], and [http://www.math.auckland.ac.nz/~hcoh001/as/page_360.htm 360] respectively.

Let us consider
<center><math>
\begin{align}
&H_0^{(1)}(kr) - H_0^{(1)}(ikr) \\
=& J_0(kr) + i Y_0(kr) + \frac{2i}{\pi} K_0(kr) \\
=& J_0(kr) + \frac{i}{\pi}\int_0^\pi \sin \left(kr \sin \theta \right) d \theta - \frac{i}{\pi} \int_0^\infty 2 e^{-kr \sinh \theta} d \theta + \frac{2i}{\pi} \int_{0}^{\infty} \cos \left( kr \sinh \theta \right)d \theta
\end{align}
</math></center>
where introducing the limit yields
<center><math>
\begin{align}
\lim_{r \rightarrow 0}& J_0(kr) = 1, \\
\lim_{r \rightarrow 0}& \sin \left(kr \sin \theta \right) = 0, \\
\lim_{r \rightarrow 0}& e^{-kr \sinh \theta} =1, \\
\lim_{r \rightarrow 0}& \cos \left( kr \sinh \theta \right) = 1.
\end{align}
</math></center>
Consequently
<center><math>
\lim_{r \rightarrow 0} \left( H_0^{(1)}(kr) - H_0^{(1)}(ikr) \right) = \lim_{r \rightarrow 0} J_0(kr) -\frac{2i}{\pi}\int_{0}^{\infty} 1 d\theta + \frac{2i}{\pi}\int_{0}^{\infty} 1 d\theta = 1,
</math></center>
which then allows us to determine that
<center><math>
\lim_{r \rightarrow 0} G(r) = \frac{i}{8 k^2}.
</math></center>

==Solution==
The fundamental solution for a thin uniform plate in <math>\mathbb{R}^2 \,</math> governed by
<center>
<math>
\left(\Delta^2 - k^2\right) u =0,
</math>
</center>
is defined as follows
<center><math>
G(r) = \left\{
\begin{align}
&\frac{i}{8 k^2} \left[H_0^{(1)}(kr) - H_0^{(1)}(ikr) \right] &\mbox{for} \ r>0, \\
&\frac{i}{8 k^2} &\mbox{for } \ r=0.
\end{align}
\right.
</math></center>

Fundamental Solution for thin plates

2010-04-27T20:58:21Z

Mike smith: /* Behaviour at r=0 */

==Introduction==
On this page, we aim to derive the Green's function for a thin uniform plate. This derivation relies heavily on concepts discussed in [[Boyling 1996]]. We seek the fundamental solution for the Biharmonic equation in <math>\mathbb{R}^2 \,</math>, which is taken to be of the form
<center>
<math>
\left(\Delta^2 - k^2\right) u =0,
</math>
</center>
where <math>\Delta = \partial_r^2 + \frac{1}{r}\partial_r, </math> in polar coordinates.

==Linear Operators==
We consider the operator <math>P(\Delta) = \left(\Delta^2 - k^4\right) = \left(\Delta - (ik)^2\right) \left(\Delta -k^2\right)</math>, whereas Boyling considers more general linear operators of the form

<center>
<math>
P(\lambda) = c(\lambda - \lambda_1)^{m_1}(\lambda - \lambda_2)^{m_2} \ldots (\lambda - \lambda_p)^{m_p},
</math>
</center>

which for our case, is identical when <math>c=1 \,</math>, <math>\lambda_1 = (ik)^2 \,</math>, <math>\lambda_2 = k^2 \,</math>, <math>m_1 = m_2 = 1 \,</math>, and <math>m_3=m_4=\ldots =0 \,</math>.

The reciprocal of <math>P(\lambda) \,</math> is taken to be of the form
<center>
<math>
\frac{1}{P(\lambda)} = \sum_{q=1}^{p} \sum_{n=1}^{m_q} \frac{c_{qn}}{(\lambda - \lambda_q)^n}.
</math>
</center>
where
<center>
<math>
c_{qn} = \lim_{\lambda \rightarrow \lambda_q} \frac{1}{(m_q - n)!} \frac{\partial^{m_q - n}}{\partial \lambda^{m_q-n}} \left[\frac{(\lambda - \lambda_q)^{m_q}}{P(\lambda)}\right].
</math>
</center>
It is straightforward to compute the <math>c_{qn} \,</math> coefficients
<center><math>
\begin{align}
c_{11} &= \lim_{\lambda \rightarrow \lambda_q} \frac{1}{(0)!} \frac{\partial^{0}}{\partial \lambda^{0}} \left[\frac{(\lambda - \lambda_1)}{P(\lambda)}\right] = \lim_{\lambda \rightarrow (ik)^2} \frac{1}{\lambda - \lambda_2} = - \frac{1}{2 k^2}, \\
c_{21} &= \frac{1}{2 k^2},
\end{align}
</math></center>
as well as determining a corresponding quantity <math>P_{qn} \,</math>
such that <math>\sum_{q=1}^{p} \sum_{n=1}^{m_q} c_{qn}P_{qn} = 1</math>, where
<center>
<math>
P_{qn}(\lambda) \left[\lambda - \lambda_q\right]^n = P(\lambda), \quad \mbox{for all q,n}.
</math>
</center>
That is, <math>P_{11} = (\lambda - k^2) \,</math> and <math>P_{21} = (\lambda - (ik)^2) \,</math>.

== Determining the Green's function==
In order to compute the Green's function for polynomials in the Laplacian, we express our unknown Green's function in terms of the fundamental solution of Helmholtz's equation:
<center>
<math>
G(r) = \sum_{q=1}^{p} \sum_{n=1}^{m_q} c_{qn} G_{n,\lambda_q}(r) = c_{11} G_{1,\lambda_1}+ c_{21} G_{1,\lambda_2}.
</math>
</center>

Using the Green's function for the [http://en.wikipedia.org/wiki/Helmholtz_equation Helmholtz Equation]
<center>
<math>
G_{1,\beta} = \frac{i}{4} H_0^{(1)} (\beta r),
</math>
</center>
where <math>H_0^{(1)} \,</math> denotes Hankel functions of the first kind, which incorporates the Sommerfeld radiation condition
<center>
<math>
\lim_{r \rightarrow \infty} r^{1/2} \left(\frac{\partial}{\partial r} - ik \right)G = 0,
</math>
</center>
we obtain the final expression
<center><math>
\begin{align}
G(r) &= \left(-\frac{1}{2 k^2}\right)\left(\frac{i}{4}H_0^{(1)}(ikr)\right) + \left(\frac{1}{2 k^2}\right)\left(\frac{i}{4}H_0^{(1)}(kr)\right) \\
&= \frac{i}{8 k^2} \left[H_0^{(1)}(kr) - H_0^{(1)}(ikr) \right].
\end{align}
</math></center>
This is the fundamental solution for our governing equation above.

=Behaviour at <math>r=0</math>=
To determine the solution at <math>r \rightarrow 0</math> (where the above expression is singular), the following identities from [[Abramowitz and Stegun 1964]] are used
<center><math>
\begin{align}
H_0^{(1)}(z) &= J_0(z) + i Y_0(z), \\
K_0(z) &= \frac{\pi i}{2} H_0^{(1)}(iz), \quad -\pi < arg(z) \leq \pi/2, \\
K_0(z) &= \int_{0}^{\infty} \cos \left[z\sinh \theta \right]d \theta, \\
Y_0(z) &= \frac{1}{\pi}\int_0^\pi \sin \left[z \sin \theta \right] d \theta - \frac{1}{\pi} \int_0^\infty 2 e^{-z \sinh \theta} d \theta, \ | arg(z) | < \frac{\pi}{2}
\end{align}
</math></center>
from pages [http://www.math.auckland.ac.nz/~hcoh001/as/page_358.htm 358], [http://www.math.auckland.ac.nz/~hcoh001/as/page_375.htm 375], [http://www.math.auckland.ac.nz/~hcoh001/as/page_376.htm 376], and [http://www.math.auckland.ac.nz/~hcoh001/as/page_360.htm 360] respectively.

Let us consider
<center><math>
\begin{align}
&H_0^{(1)}(kr) - H_0^{(1)}(ikr) \\
=& J_0(kr) + i Y_0(kr) + \frac{2i}{\pi} K_0(kr) \\
=& J_0(kr) + \frac{i}{\pi}\int_0^\pi \sin \left(kr \sin \theta \right) d \theta - \frac{i}{\pi} \int_0^\infty 2 e^{-kr \sinh \theta} d \theta + \frac{2i}{\pi} \int_{0}^{\infty} \cos \left( kr \sinh \theta \right)d \theta
\end{align}
</math></center>
where introducing the limit yields
<center><math>
\begin{align}
\lim_{r \rightarrow 0}& J_0(kr) = 1, \\
\lim_{r \rightarrow 0}& \sin \left(kr \sin \theta \right) = 0, \\
\lim_{r \rightarrow 0}& e^{-kr \sinh \theta} =1, \\
\lim_{r \rightarrow 0}& \cos \left( kr \sinh \theta \right) = 1.
\end{align}
</math></center>
Consequently
<center><math>
\lim_{r \rightarrow 0} \left( H_0^{(1)}(kr) - H_0^{(1)}(ikr) \right) = \lim_{r \rightarrow 0} J_0(kr) -\frac{2i}{\pi}\int_{0}^{\infty} 1 d\theta + \frac{2i}{\pi}\int_{0}^{\infty} 1 d\theta = 1,
</math></center>
which then allows us to determine that
<center><math>
\lim_{r \rightarrow 0} G(r) = \frac{i}{8 k^2}.
</math></center>

Fundamental Solution for thin plates

2010-04-27T20:55:09Z

Mike smith:

==Introduction==
On this page, we aim to derive the Green's function for a thin uniform plate. This derivation relies heavily on concepts discussed in [[Boyling 1996]]. We seek the fundamental solution for the Biharmonic equation in <math>\mathbb{R}^2 \,</math>, which is taken to be of the form
<center>
<math>
\left(\Delta^2 - k^2\right) u =0,
</math>
</center>
where <math>\Delta = \partial_r^2 + \frac{1}{r}\partial_r, </math> in polar coordinates.

==Linear Operators==
We consider the operator <math>P(\Delta) = \left(\Delta^2 - k^4\right) = \left(\Delta - (ik)^2\right) \left(\Delta -k^2\right)</math>, whereas Boyling considers more general linear operators of the form

<center>
<math>
P(\lambda) = c(\lambda - \lambda_1)^{m_1}(\lambda - \lambda_2)^{m_2} \ldots (\lambda - \lambda_p)^{m_p},
</math>
</center>

which for our case, is identical when <math>c=1 \,</math>, <math>\lambda_1 = (ik)^2 \,</math>, <math>\lambda_2 = k^2 \,</math>, <math>m_1 = m_2 = 1 \,</math>, and <math>m_3=m_4=\ldots =0 \,</math>.

The reciprocal of <math>P(\lambda) \,</math> is taken to be of the form
<center>
<math>
\frac{1}{P(\lambda)} = \sum_{q=1}^{p} \sum_{n=1}^{m_q} \frac{c_{qn}}{(\lambda - \lambda_q)^n}.
</math>
</center>
where
<center>
<math>
c_{qn} = \lim_{\lambda \rightarrow \lambda_q} \frac{1}{(m_q - n)!} \frac{\partial^{m_q - n}}{\partial \lambda^{m_q-n}} \left[\frac{(\lambda - \lambda_q)^{m_q}}{P(\lambda)}\right].
</math>
</center>
It is straightforward to compute the <math>c_{qn} \,</math> coefficients
<center><math>
\begin{align}
c_{11} &= \lim_{\lambda \rightarrow \lambda_q} \frac{1}{(0)!} \frac{\partial^{0}}{\partial \lambda^{0}} \left[\frac{(\lambda - \lambda_1)}{P(\lambda)}\right] = \lim_{\lambda \rightarrow (ik)^2} \frac{1}{\lambda - \lambda_2} = - \frac{1}{2 k^2}, \\
c_{21} &= \frac{1}{2 k^2},
\end{align}
</math></center>
as well as determining a corresponding quantity <math>P_{qn} \,</math>
such that <math>\sum_{q=1}^{p} \sum_{n=1}^{m_q} c_{qn}P_{qn} = 1</math>, where
<center>
<math>
P_{qn}(\lambda) \left[\lambda - \lambda_q\right]^n = P(\lambda), \quad \mbox{for all q,n}.
</math>
</center>
That is, <math>P_{11} = (\lambda - k^2) \,</math> and <math>P_{21} = (\lambda - (ik)^2) \,</math>.

== Determining the Green's function==
In order to compute the Green's function for polynomials in the Laplacian, we express our unknown Green's function in terms of the fundamental solution of Helmholtz's equation:
<center>
<math>
G(r) = \sum_{q=1}^{p} \sum_{n=1}^{m_q} c_{qn} G_{n,\lambda_q}(r) = c_{11} G_{1,\lambda_1}+ c_{21} G_{1,\lambda_2}.
</math>
</center>

Using the Green's function for the [http://en.wikipedia.org/wiki/Helmholtz_equation Helmholtz Equation]
<center>
<math>
G_{1,\beta} = \frac{i}{4} H_0^{(1)} (\beta r),
</math>
</center>
where <math>H_0^{(1)} \,</math> denotes Hankel functions of the first kind, which incorporates the Sommerfeld radiation condition
<center>
<math>
\lim_{r \rightarrow \infty} r^{1/2} \left(\frac{\partial}{\partial r} - ik \right)G = 0,
</math>
</center>
we obtain the final expression
<center><math>
\begin{align}
G(r) &= \left(-\frac{1}{2 k^2}\right)\left(\frac{i}{4}H_0^{(1)}(ikr)\right) + \left(\frac{1}{2 k^2}\right)\left(\frac{i}{4}H_0^{(1)}(kr)\right) \\
&= \frac{i}{8 k^2} \left[H_0^{(1)}(kr) - H_0^{(1)}(ikr) \right].
\end{align}
</math></center>
This is the fundamental solution for our governing equation above.

=Behaviour at <math>r=0</math>=
To determine the solution at <math>r \rightarrow 0</math> (where the above expression is singular), the following identities from [[Abramowitz and Stegun 1964]] are used
<center><math>
\begin{align}
H_0^{(1)}(z) &= J_0(z) + i Y_0(z), \\
K_0(z) &= \frac{\pi i}{2} H_0^{(1)}(iz), \quad -\pi < arg(z) \leq \pi/2, \\
K_0(z) &= \int_{0}^{\infty} \cos \left[z\sinh \theta \right]d \theta, \\
Y_0(z) &= \frac{1}{\pi}\int_0^\pi \sin \left[z \sin \theta \right] d \theta - \frac{1}{\pi} \int_0^\infty 2 e^{-z \sinh \theta} d \theta, \ | arg(z) | < \frac{\pi}{2}
\end{align}
</math></center>
from pages [http://www.math.auckland.ac.nz/~hcoh001/as/page_358.htm 358], [http://www.math.auckland.ac.nz/~hcoh001/as/page_375.htm 375], [http://www.math.auckland.ac.nz/~hcoh001/as/page_376.htm 376], and [http://www.math.auckland.ac.nz/~hcoh001/as/page_360.htm 360] respectively.

Let us consider
<center><math>
\begin{align}
&H_0^{(1)}(kr) - H_0^{(1)}(ikr), \\
=& J_0(kr) + i Y_0(kr) + \frac{2i}{\pi} K_0(kr), \\
=& J_0(kr) + \frac{i}{\pi}\int_0^\pi \sin \left(kr \sin \theta \right) d \theta - \frac{i}{\pi} \int_0^\infty 2 e^{-kr \sinh \theta} d \theta + \frac{2i}{\pi} \int_{0}^{\infty} \cos \left( kr \sinh \theta \right)d \theta,
\end{align}
</math></center>
where introducing the limit yields
<center><math>
\begin{align}
\lim_{r \rightarrow 0}& J_0(kr) = 1 \\
\lim_{r \rightarrow 0}& \sin \left(kr \sin \theta \right) = 0 \\
\lim_{r \rightarrow 0}& e^{-kr \sinh \theta} =1 \\
\lim_{r \rightarrow 0}& \cos \left( kr \sinh \theta \right) = 1.
\end{align}
</math></center>
Consequently
<center><math>
\lim_{r \rightarrow 0} \left( H_0^{(1)}(kr) - H_0^{(1)}(ikr) \right) = \lim_{r \rightarrow 0} J_0(kr) -\frac{2i}{\pi}\int_{0}^{\infty} 1 d\theta + \frac{2i}{\pi}\int_{0}^{\infty} 1 d\theta = 1,
</math></center>
which then allows us to determine that
<center><math>
\lim_{r \rightarrow 0} G(r) = \frac{i}{8 k^2}
</math></center>

Fundamental Solution for thin plates

2010-04-27T20:45:28Z

Mike smith:

Fundamental Solution for thin plates

2010-04-27T20:37:48Z

Mike smith:

Boyling 1996

2010-04-27T20:22:59Z

Mike smith: Created page with 'Boyling, J. B. (1996) Green's functions for polynomials in the Laplacian, '' ZAMP'', '''47''' (3), pp. 485 - 492 Category:Reference'

Boyling, J. B. (1996) Green's functions for polynomials in the Laplacian, '' ZAMP'', '''47''' (3), pp. 485 - 492

[[Category:Reference]]

Fundamental Solution for thin plates

2010-04-27T06:21:51Z

Mike smith: Created page with 'On this page, we aim to derive the Green's function for a thin uniform plate.'

On this page, we aim to derive the Green's function for a thin uniform plate.

Eigenfunction Matching for a Semi-Infinite Floating Elastic Plate

2010-03-17T20:51:57Z

Mike smith: sign error (however is irrelevant as rhs = 0)

{{complete pages}}

== Introduction ==

We show here a solution for a semi-infinite [[:Category:Floating Elastic Plate|Floating Elastic Plate]] on [[Finite Depth]].
The problem was solved by [[Fox and Squire 1994]] but the solution method here is slightly different.
The simpler theory for a [[Eigenfunction Matching for a Semi-Infinite Dock|Dock]] describes
many of the ideas here in more detail.

[[Image:Semiinfinite plate.jpg|thumb|right|300px|Wave scattering by a submerged semi-infinite elastic plate]]

== Equations ==

We consider the problem of small-amplitude waves which are incident on a semi-infinite floating elastic
plate occupying water surface for <math>x>0</math>. The submergence of the plate is considered negligible.
We assume that the problem is invariant in the <math>y</math> direction, although we allow the waves to be
incident from an angle.
We also assume that the plate edges are free to move at
each boundary, although other boundary conditions could easily be considered using
the methods of solution presented here. We begin with the [[Frequency Domain Problem]] for a semi-infinite
[[Floating Elastic Plate|Floating Elastic Plates]]
in the non-dimensional form of [[Tayler_1986a|Tayler 1986]] ([[Dispersion Relation for a Floating Elastic Plate]]).
We also assume that the waves are normally incident (incidence at an angle will be discussed later).
<center><math>
\Delta \phi = 0, \;\;\; -h < z \leq 0,
</math></center>
<center><math>
\partial_z \phi = 0, \;\;\; z = - h,
</math></center>
<center><math>
\partial_z\phi=\alpha\phi, \,\, z=0,\,x<0,
</math></center>
<center><math>
\beta \partial_x^4\partial_z \phi
- \left( \gamma\alpha - 1 \right) \partial_z \phi = \alpha\phi, \;\;
z = 0, \;\;\; x \geq 0,
</math></center>
where <math>\alpha = \omega^2</math>, <math>\beta</math> and <math>\gamma</math>
are the stiffness and mass constant for the plate respectively. The free edge conditions
at the edge of the plate imply
<center><math>
\partial_x^3 \partial_z\phi = 0, \;\; z = 0, \;\;\; x = 0,
</math></center>
<center><math>
\partial_x^2 \partial_z\phi = 0, \;\; z = 0, \;\;\; x = 0,
</math></center>

== Method of solution ==

We use [http://en.wikipedia.org/wiki/Separation_of_Variables separation of variables] in the two regions, <math>x<0</math>
and <math>x>0</math>.

{{separation of variables in two dimensions}}

{{separation of variables for a free surface}}

{{separation of variables for a floating elastic plate}}

{{free surface floating elastic plate relations}}

{{incident potential for two dimensions}}

== An infinite dimensional system of equations ==

The potential and its derivative must be continuous across the
transition from open water to the plate covered region. Therefore, the
potentials and their derivatives at <math>x=0</math> have to be equal.
We also truncate the sum at <math>N</math> being careful that we have
two extra modes on the plate covered region to satisfy the edge conditions.
We obtain
<center>
<math>
\phi_{0}\left( z\right) + \sum_{m=0}^{N}
a_{m} \phi_{m}\left( z\right)
=\sum_{m=-2}^{N}b_{m}\psi_{m}(z)
</math>
</center>
and
<center>
<math>
-k_{0}\phi_{0}\left( z\right) +\sum
_{m=0}^{N} a_{m}k_{m}\phi_{m}\left( z\right)
=-\sum_{m=-2}^{N}b_{m}\kappa_{m}\psi
_{m}(z)
</math>
</center>
for each <math>N</math>.
We solve these equations by multiplying both equations by
<math>\phi_{l}(z)\,</math> and integrating from <math>-h</math> to <math>0</math> to obtain:
<center>
<math>
A_{0}\delta_{0l}+a_{l}A_{l}
=\sum_{m=-2}^{N}b_{m}B_{ml},\,\,0 \leq l \leq N
</math>
</center>
and
<center>
<math>
-k_{0}A_{0}\delta_{0l}+a_{l}k_{l}A_l
=-\sum_{m=-2}^{N}b_{m}\kappa_{m}B_{ml},\,\,0 \leq l \leq N
</math>
</center>
If we multiply the first equation by <math>k_l</math> and subtract the second equation
we obtain
<center>
<math>
(k_{0}+k_l)A_{0}\delta_{0l}
=\sum_{m=-2}^{N}b_{m}(k_l + \kappa_{m})B_{ml}
</math>
</center>
Finally, we need to apply the conditions at the edge of the plate to give us two further equations,
<center>
<math>
\partial_x^2\partial_z\phi = \sum_{m=-2}^{N}b_{m} \kappa_m^3 \tan\kappa_m h = 0
</math>
</center>
and
<center>
<math>
\partial_x^3\partial_z\phi = - \sum_{m=-2}^{N}b_{m} \kappa_m^4 \tan\kappa_m h = 0
</math>
</center>

== Numerical Solution ==

To solve the system of equations previously defined we set the upper limit of <math>l</math> to
be <math>N</math>, as stated before. In terms of matrix, we obtain
<center>
<math>
\begin{bmatrix}
\begin{bmatrix}
A_0&0 \quad \cdots&0\\
0&&\\
\vdots&A_l&\vdots\\
&&0\\
0&\cdots \quad 0 &A_N
\end{bmatrix}
&
\begin{bmatrix}
-B_{-2,0}&-B_{-1,0}\\
&\\
\vdots&\vdots\\
&\\
-B_{-2,N}&-B_{-1,N}
\end{bmatrix}
&
\begin{bmatrix}
-B_{0,0}&\cdots&-B_{0,N}\\
&&\\
\vdots&-B_{m,l}&\vdots\\
&&\\
-B_{N,0}&\cdots&-B_{N,N}
\end{bmatrix}
\\ \\
\begin{bmatrix}
0&&\cdots&&0\\
0&&\cdots&&0
\end{bmatrix}
&
\begin{bmatrix}
\kappa_{-2}^3\tan\kappa_{-2}h&\kappa_{-1}^3\tan\kappa_{-1}h\\
\kappa_{-2}^4\tan\kappa_{-2}h&\kappa_{-1}^4\tan\kappa_{-1}h
\end{bmatrix}
&
\begin{bmatrix}
\kappa_0^3\tan\kappa_0h&\cdots&\kappa_N^3\tan\kappa_Nh\\
\kappa_0^4\tan\kappa_0h&\cdots&\kappa_N^4\tan\kappa_Nh
\end{bmatrix}
\\ \\
\begin{bmatrix}
0&&\cdots&&0\\
&&&&\\
\vdots&&\ddots&&\vdots\\
&&&&\\
0&&\cdots&&0
\end{bmatrix}
&
\begin{bmatrix}
(k_0+\kappa_{-2})B_{-2,0}&(k_0+\kappa_{-1})B_{-1,0}\\
&\\
\vdots&\vdots\\
&\\
(k_N+\kappa_{-2})B_{-2,N}&(k_N+\kappa_{-1})B_{-1,N}
\end{bmatrix}
&
\begin{bmatrix}
(k_0 + \kappa_0) \, B_{0,0}&\cdots&(k_N + \kappa_{0}) \, B_{0,N}\\
&&\\
\vdots&(k_l + \kappa_{m}) \, B_{m,l}&\vdots\\
&&\\
(k_0 + \kappa_N) \, B_{N,0}&\cdots&(k_N + \kappa_{N}) \, B_{N,N}\\
\end{bmatrix}
\end{bmatrix}
\begin{bmatrix}

\begin{bmatrix}
a_{0} \\
\\
\vdots \\
\\
a_N
\end{bmatrix}
\\ \\
\begin{bmatrix}
b_{-2}\\
b_{-1}
\end{bmatrix}
\\ \\
\begin{bmatrix}
b_{0}\\
\\
\vdots \\
\\
b_N
\end{bmatrix}
\end{bmatrix}
=
\begin{bmatrix}
\begin{bmatrix}
- A_{0} \\
0 \\
\vdots \\
\\
0
\end{bmatrix}
\\ \\
\begin{bmatrix}
0 \\
0
\end{bmatrix}
\\ \\
\begin{bmatrix}
2k_{0}A_{0} \\
0 \\
\vdots \\
\\
0
\end{bmatrix}
\end{bmatrix}
</math>
</center>
We then simply need to solve the linear system of equations. Note that we can solve this equation for
<math>b_n</math> first and then solve for <math>a_n</math>

== Waves Incident at an Angle ==

We can consider the case of [[Waves Incident at an Angle]] <math>\theta</math>.
{{incident angle}}

It is shown that the potential can be expanded as
<center>
<math>
\phi(x,z)=e^{-\hat{k}_0x}\phi_0(z)+\sum_{m=0}^{\infty}a_{m}e^{\hat{k}_{m}x}\phi_{m}(z), \;\;x<0
</math>
</center>
and
<center>
<math>
\phi(x,z)=\sum_{m=-2}^{\infty}b_{m}
e^{-\hat{\kappa}_{m}x}\psi_{m}(z), \;\;x>0
</math>
</center>
where <math>\hat{k}_{m} = \sqrt{k_m^2 - k_y^2}</math> and <math>\hat{\kappa}_{m} = \sqrt{\kappa_m^2 - k_y^2}</math>
where we always take the positive real root or the root with positive imaginary part.

The edge conditions are also different and are
<center><math>
\left(\frac{\partial^3}{\partial x^3} + (2 - \nu)k^2_y\frac{\partial}{\partial x}\right) \frac{\partial\phi}{\partial z}= 0,
</math></center>
and
<center><math>
\left(\frac{\partial^2}{\partial x^2} + \nu k^2_y\right)\frac{\partial\phi}{\partial z} = 0,
</math></center>
where <math>\nu</math> is Poisons ratio.

We can expend these edge conditions, which respectively gives
<center>
<math>
-\sum_{m=-2}^{\infty}b_{m}\kappa_m (\hat{\kappa}_m^3+\hat{\kappa}_m k_y^2(2-\nu))\tan \kappa_m h=0
</math>
</center>
and
<center>
<math>
-\sum_{m=-2}^{\infty}b_{m}\kappa_m (\hat{\kappa}_m^2+k_y^2\nu))\tan \kappa_m h=0
</math>
</center>

The equations are derived almost identically to those above and we obtain
<center>
<math>
A_{0}\delta_{0l}+a_{l}A_{l}
=\sum_{m=-2}^{\infty}b_{m}B_{ml}
</math>
</center>
and
<center>
<math>
-\hat{k_{0}}A_{0}\delta_{0l}+a_{l}\hat{k}_{l}A_l
=-\sum_{m=-2}^{\infty}b_{m}\hat{\kappa}_{m}B_{ml}
</math>
</center>
and these are solved exactly as before.

== Energy Balance ==

We present a derivation of the energy balance here and also refer to the derivation
[[Energy Balance for Two Elastic Plates]]

{{energy contour and preliminaries}}

The contributions from the vertical ends are

<center><math>
\lim_{x\to-\infty} \Im\int_{-h}^{0}\phi^*\frac{\partial\phi}{\partial n} dz
= \lim_{x\to-\infty} \Im\int_{-h}^{0} \hat{k}_0 \left(e^{\hat{k}_0 x} + a_0^{*} e^{-\hat{k}_0 x} \right)
\left(e^{-\hat{k}_0 x} - a_0 e^{\hat{k}_0 x} \right)\phi_0(z)^2 dz
</math></center>
<center><math>
= \frac{\hat{k}_0}{i} A_0 \left(1 - |a_0^2| \right)
</math></center>
and
<center><math>
\lim_{x\to\infty} \Im\int_{-h}^{0}\phi^*\frac{\partial\phi}{\partial n} dz
= \lim_{x\to\infty} \Im\int_{-h}^{0} -\hat{\kappa}_0 \left(b_0^{*} e^{\hat{\kappa}_0 x} \right)
\left(b_0 e^{-\hat{\kappa}_0 x}\right)\psi_0(z)^2 dz
</math></center>
<center><math>
= -\frac{\hat{\kappa}_0}{i} |b_0^2| \int_{-h}^{0} \psi_0(z)^2 dz
</math></center>

The contribution from the surface integral under the plate is
<center><math>
\Im\int_{0}^{\infty}\phi^*\frac{\partial\phi}{\partial n} dx
= \Im\int_{0}^{\infty}
1/\alpha \left(\beta \partial_x^4 \frac{\partial\phi^{*}}{\partial n} - (\gamma\alpha - 1 ) \frac{\partial\phi^{*}}{\partial n} \right)
\frac{\partial\phi}{\partial n} dx
</math></center>
<center><math>
= \frac{2\beta}{\alpha} \frac{\hat{\kappa}_0^{3}}{i} |b_0^2| \partial_z\psi_0(0)^2
</math></center>
where we have integrated by parts and used the condition at the ends of the plate.

The energy balance is therefore
<center><math>
{\hat{k}_0} A_0 |a_0^2| + {\hat{\kappa}_0} |b_0^2| \int_{-h}^{0} \psi_0(z)^2 dz
- \frac{2\beta}{\alpha}{ \hat{\kappa}_0^{3}} |b_0^2| \partial_z\psi_0(0)^2 = {\hat{k}_0} A_0
</math></center>

Note that this formula is only valid for angles <math>\theta \in [-\theta_0, \theta_0]</math>, where <math>\theta_0</math>
is defined by <math>\sin \theta_0 =\frac{\kappa_0}{k_0}</math>. For large angles there is total reflection
and then <math>|R|^2=1</math>

== Matlab Code ==

A program to calculate the coefficients for the semi-infinite dock problems can be found here
[http://www.math.auckland.ac.nz/~meylan/code/eigenfunction_matching/semiinfinite_plate.m semiinfinite_plate.m]

=== Additional code ===

This program requires
* {{free surface dispersion equation code}}
* {{elastic plate dispersion equation code}}

[[Category:Floating Elastic Plate]]
[[Category:Eigenfunction Matching Method]]
[[Category:Complete Pages]]

Eigenfunction Matching for a Semi-Infinite Floating Elastic Plate

2010-03-17T20:36:49Z

Mike smith: /* Numerical Solution */

{{complete pages}}

== Introduction ==

We show here a solution for a semi-infinite [[:Category:Floating Elastic Plate|Floating Elastic Plate]] on [[Finite Depth]].
The problem was solved by [[Fox and Squire 1994]] but the solution method here is slightly different.
The simpler theory for a [[Eigenfunction Matching for a Semi-Infinite Dock|Dock]] describes
many of the ideas here in more detail.

[[Image:Semiinfinite plate.jpg|thumb|right|300px|Wave scattering by a submerged semi-infinite elastic plate]]

== Equations ==

We consider the problem of small-amplitude waves which are incident on a semi-infinite floating elastic
plate occupying water surface for <math>x>0</math>. The submergence of the plate is considered negligible.
We assume that the problem is invariant in the <math>y</math> direction, although we allow the waves to be
incident from an angle.
We also assume that the plate edges are free to move at
each boundary, although other boundary conditions could easily be considered using
the methods of solution presented here. We begin with the [[Frequency Domain Problem]] for a semi-infinite
[[Floating Elastic Plate|Floating Elastic Plates]]
in the non-dimensional form of [[Tayler_1986a|Tayler 1986]] ([[Dispersion Relation for a Floating Elastic Plate]]).
We also assume that the waves are normally incident (incidence at an angle will be discussed later).
<center><math>
\Delta \phi = 0, \;\;\; -h < z \leq 0,
</math></center>
<center><math>
\partial_z \phi = 0, \;\;\; z = - h,
</math></center>
<center><math>
\partial_z\phi=\alpha\phi, \,\, z=0,\,x<0,
</math></center>
<center><math>
\beta \partial_x^4\partial_z \phi
- \left( \gamma\alpha - 1 \right) \partial_z \phi = \alpha\phi, \;\;
z = 0, \;\;\; x \geq 0,
</math></center>
where <math>\alpha = \omega^2</math>, <math>\beta</math> and <math>\gamma</math>
are the stiffness and mass constant for the plate respectively. The free edge conditions
at the edge of the plate imply
<center><math>
\partial_x^3 \partial_z\phi = 0, \;\; z = 0, \;\;\; x = 0,
</math></center>
<center><math>
\partial_x^2 \partial_z\phi = 0, \;\; z = 0, \;\;\; x = 0,
</math></center>

== Method of solution ==

We use [http://en.wikipedia.org/wiki/Separation_of_Variables separation of variables] in the two regions, <math>x<0</math>
and <math>x>0</math>.

{{separation of variables in two dimensions}}

{{separation of variables for a free surface}}

{{separation of variables for a floating elastic plate}}

{{free surface floating elastic plate relations}}

{{incident potential for two dimensions}}

== An infinite dimensional system of equations ==

The potential and its derivative must be continuous across the
transition from open water to the plate covered region. Therefore, the
potentials and their derivatives at <math>x=0</math> have to be equal.
We also truncate the sum at <math>N</math> being careful that we have
two extra modes on the plate covered region to satisfy the edge conditions.
We obtain
<center>
<math>
\phi_{0}\left( z\right) + \sum_{m=0}^{N}
a_{m} \phi_{m}\left( z\right)
=\sum_{m=-2}^{N}b_{m}\psi_{m}(z)
</math>
</center>
and
<center>
<math>
-k_{0}\phi_{0}\left( z\right) +\sum
_{m=0}^{N} a_{m}k_{m}\phi_{m}\left( z\right)
=-\sum_{m=-2}^{N}b_{m}\kappa_{m}\psi
_{m}(z)
</math>
</center>
for each <math>N</math>.
We solve these equations by multiplying both equations by
<math>\phi_{l}(z)\,</math> and integrating from <math>-h</math> to <math>0</math> to obtain:
<center>
<math>
A_{0}\delta_{0l}+a_{l}A_{l}
=\sum_{m=-2}^{N}b_{m}B_{ml},\,\,0 \leq l \leq N
</math>
</center>
and
<center>
<math>
-k_{0}A_{0}\delta_{0l}+a_{l}k_{l}A_l
=-\sum_{m=-2}^{N}b_{m}\kappa_{m}B_{ml},\,\,0 \leq l \leq N
</math>
</center>
If we multiply the first equation by <math>k_l</math> and subtract the second equation
we obtain
<center>
<math>
(k_{0}+k_l)A_{0}\delta_{0l}
=\sum_{m=-2}^{N}b_{m}(k_l + \kappa_{m})B_{ml}
</math>
</center>
Finally, we need to apply the conditions at the edge of the plate to give us two further equations,
<center>
<math>
\partial_x^2\partial_z\phi = - \sum_{m=-2}^{N}b_{m} \kappa_m^3 \tan\kappa_m h = 0
</math>
</center>
and
<center>
<math>
\partial_x^3\partial_z\phi = \sum_{m=-2}^{N}b_{m} \kappa_m^4 \tan\kappa_m h = 0
</math>
</center>

== Numerical Solution ==

To solve the system of equations previously defined we set the upper limit of <math>l</math> to
be <math>N</math>, as stated before. In terms of matrix, we obtain
<center>
<math>
\begin{bmatrix}
\begin{bmatrix}
A_0&0 \quad \cdots&0\\
0&&\\
\vdots&A_l&\vdots\\
&&0\\
0&\cdots \quad 0 &A_N
\end{bmatrix}
&
\begin{bmatrix}
-B_{-2,0}&-B_{-1,0}\\
&\\
\vdots&\vdots\\
&\\
-B_{-2,N}&-B_{-1,N}
\end{bmatrix}
&
\begin{bmatrix}
-B_{0,0}&\cdots&-B_{0,N}\\
&&\\
\vdots&-B_{m,l}&\vdots\\
&&\\
-B_{N,0}&\cdots&-B_{N,N}
\end{bmatrix}
\\ \\
\begin{bmatrix}
0&&\cdots&&0\\
0&&\cdots&&0
\end{bmatrix}
&
\begin{bmatrix}
\kappa_{-2}^3\tan\kappa_{-2}h&\kappa_{-1}^3\tan\kappa_{-1}h\\
\kappa_{-2}^4\tan\kappa_{-2}h&\kappa_{-1}^4\tan\kappa_{-1}h
\end{bmatrix}
&
\begin{bmatrix}
\kappa_0^3\tan\kappa_0h&\cdots&\kappa_N^3\tan\kappa_Nh\\
\kappa_0^4\tan\kappa_0h&\cdots&\kappa_N^4\tan\kappa_Nh
\end{bmatrix}
\\ \\
\begin{bmatrix}
0&&\cdots&&0\\
&&&&\\
\vdots&&\ddots&&\vdots\\
&&&&\\
0&&\cdots&&0
\end{bmatrix}
&
\begin{bmatrix}
(k_0+\kappa_{-2})B_{-2,0}&(k_0+\kappa_{-1})B_{-1,0}\\
&\\
\vdots&\vdots\\
&\\
(k_N+\kappa_{-2})B_{-2,N}&(k_N+\kappa_{-1})B_{-1,N}
\end{bmatrix}
&
\begin{bmatrix}
(k_0 + \kappa_0) \, B_{0,0}&\cdots&(k_N + \kappa_{0}) \, B_{0,N}\\
&&\\
\vdots&(k_l + \kappa_{m}) \, B_{m,l}&\vdots\\
&&\\
(k_0 + \kappa_N) \, B_{N,0}&\cdots&(k_N + \kappa_{N}) \, B_{N,N}\\
\end{bmatrix}
\end{bmatrix}
\begin{bmatrix}

\begin{bmatrix}
a_{0} \\
\\
\vdots \\
\\
a_N
\end{bmatrix}
\\ \\
\begin{bmatrix}
b_{-2}\\
b_{-1}
\end{bmatrix}
\\ \\
\begin{bmatrix}
b_{0}\\
\\
\vdots \\
\\
b_N
\end{bmatrix}
\end{bmatrix}
=
\begin{bmatrix}
\begin{bmatrix}
- A_{0} \\
0 \\
\vdots \\
\\
0
\end{bmatrix}
\\ \\
\begin{bmatrix}
0 \\
0
\end{bmatrix}
\\ \\
\begin{bmatrix}
2k_{0}A_{0} \\
0 \\
\vdots \\
\\
0
\end{bmatrix}
\end{bmatrix}
</math>
</center>
We then simply need to solve the linear system of equations. Note that we can solve this equation for
<math>b_n</math> first and then solve for <math>a_n</math>

== Waves Incident at an Angle ==

We can consider the case of [[Waves Incident at an Angle]] <math>\theta</math>.
{{incident angle}}

It is shown that the potential can be expanded as
<center>
<math>
\phi(x,z)=e^{-\hat{k}_0x}\phi_0(z)+\sum_{m=0}^{\infty}a_{m}e^{\hat{k}_{m}x}\phi_{m}(z), \;\;x<0
</math>
</center>
and
<center>
<math>
\phi(x,z)=\sum_{m=-2}^{\infty}b_{m}
e^{-\hat{\kappa}_{m}x}\psi_{m}(z), \;\;x>0
</math>
</center>
where <math>\hat{k}_{m} = \sqrt{k_m^2 - k_y^2}</math> and <math>\hat{\kappa}_{m} = \sqrt{\kappa_m^2 - k_y^2}</math>
where we always take the positive real root or the root with positive imaginary part.

The edge conditions are also different and are
<center><math>
\left(\frac{\partial^3}{\partial x^3} + (2 - \nu)k^2_y\frac{\partial}{\partial x}\right) \frac{\partial\phi}{\partial z}= 0,
</math></center>
and
<center><math>
\left(\frac{\partial^2}{\partial x^2} + \nu k^2_y\right)\frac{\partial\phi}{\partial z} = 0,
</math></center>
where <math>\nu</math> is Poisons ratio.

We can expend these edge conditions, which respectively gives
<center>
<math>
-\sum_{m=-2}^{\infty}b_{m}\kappa_m (\hat{\kappa}_m^3+\hat{\kappa}_m k_y^2(2-\nu))\tan \kappa_m h=0
</math>
</center>
and
<center>
<math>
-\sum_{m=-2}^{\infty}b_{m}\kappa_m (\hat{\kappa}_m^2+k_y^2\nu))\tan \kappa_m h=0
</math>
</center>

The equations are derived almost identically to those above and we obtain
<center>
<math>
A_{0}\delta_{0l}+a_{l}A_{l}
=\sum_{m=-2}^{\infty}b_{m}B_{ml}
</math>
</center>
and
<center>
<math>
-\hat{k_{0}}A_{0}\delta_{0l}+a_{l}\hat{k}_{l}A_l
=-\sum_{m=-2}^{\infty}b_{m}\hat{\kappa}_{m}B_{ml}
</math>
</center>
and these are solved exactly as before.

== Energy Balance ==

We present a derivation of the energy balance here and also refer to the derivation
[[Energy Balance for Two Elastic Plates]]

{{energy contour and preliminaries}}

The contributions from the vertical ends are

<center><math>
\lim_{x\to-\infty} \Im\int_{-h}^{0}\phi^*\frac{\partial\phi}{\partial n} dz
= \lim_{x\to-\infty} \Im\int_{-h}^{0} \hat{k}_0 \left(e^{\hat{k}_0 x} + a_0^{*} e^{-\hat{k}_0 x} \right)
\left(e^{-\hat{k}_0 x} - a_0 e^{\hat{k}_0 x} \right)\phi_0(z)^2 dz
</math></center>
<center><math>
= \frac{\hat{k}_0}{i} A_0 \left(1 - |a_0^2| \right)
</math></center>
and
<center><math>
\lim_{x\to\infty} \Im\int_{-h}^{0}\phi^*\frac{\partial\phi}{\partial n} dz
= \lim_{x\to\infty} \Im\int_{-h}^{0} -\hat{\kappa}_0 \left(b_0^{*} e^{\hat{\kappa}_0 x} \right)
\left(b_0 e^{-\hat{\kappa}_0 x}\right)\psi_0(z)^2 dz
</math></center>
<center><math>
= -\frac{\hat{\kappa}_0}{i} |b_0^2| \int_{-h}^{0} \psi_0(z)^2 dz
</math></center>

The contribution from the surface integral under the plate is
<center><math>
\Im\int_{0}^{\infty}\phi^*\frac{\partial\phi}{\partial n} dx
= \Im\int_{0}^{\infty}
1/\alpha \left(\beta \partial_x^4 \frac{\partial\phi^{*}}{\partial n} - (\gamma\alpha - 1 ) \frac{\partial\phi^{*}}{\partial n} \right)
\frac{\partial\phi}{\partial n} dx
</math></center>
<center><math>
= \frac{2\beta}{\alpha} \frac{\hat{\kappa}_0^{3}}{i} |b_0^2| \partial_z\psi_0(0)^2
</math></center>
where we have integrated by parts and used the condition at the ends of the plate.

The energy balance is therefore
<center><math>
{\hat{k}_0} A_0 |a_0^2| + {\hat{\kappa}_0} |b_0^2| \int_{-h}^{0} \psi_0(z)^2 dz
- \frac{2\beta}{\alpha}{ \hat{\kappa}_0^{3}} |b_0^2| \partial_z\psi_0(0)^2 = {\hat{k}_0} A_0
</math></center>

Note that this formula is only valid for angles <math>\theta \in [-\theta_0, \theta_0]</math>, where <math>\theta_0</math>
is defined by <math>\sin \theta_0 =\frac{\kappa_0}{k_0}</math>. For large angles there is total reflection
and then <math>|R|^2=1</math>

== Matlab Code ==

A program to calculate the coefficients for the semi-infinite dock problems can be found here
[http://www.math.auckland.ac.nz/~meylan/code/eigenfunction_matching/semiinfinite_plate.m semiinfinite_plate.m]

=== Additional code ===

This program requires
* {{free surface dispersion equation code}}
* {{elastic plate dispersion equation code}}

[[Category:Floating Elastic Plate]]
[[Category:Eigenfunction Matching Method]]
[[Category:Complete Pages]]

Eigenfunction Matching for a Semi-Infinite Floating Elastic Plate

2010-03-17T20:36:27Z

Mike smith: block matrix more intuitive

{{complete pages}}

== Introduction ==

We show here a solution for a semi-infinite [[:Category:Floating Elastic Plate|Floating Elastic Plate]] on [[Finite Depth]].
The problem was solved by [[Fox and Squire 1994]] but the solution method here is slightly different.
The simpler theory for a [[Eigenfunction Matching for a Semi-Infinite Dock|Dock]] describes
many of the ideas here in more detail.

[[Image:Semiinfinite plate.jpg|thumb|right|300px|Wave scattering by a submerged semi-infinite elastic plate]]

== Equations ==

We consider the problem of small-amplitude waves which are incident on a semi-infinite floating elastic
plate occupying water surface for <math>x>0</math>. The submergence of the plate is considered negligible.
We assume that the problem is invariant in the <math>y</math> direction, although we allow the waves to be
incident from an angle.
We also assume that the plate edges are free to move at
each boundary, although other boundary conditions could easily be considered using
the methods of solution presented here. We begin with the [[Frequency Domain Problem]] for a semi-infinite
[[Floating Elastic Plate|Floating Elastic Plates]]
in the non-dimensional form of [[Tayler_1986a|Tayler 1986]] ([[Dispersion Relation for a Floating Elastic Plate]]).
We also assume that the waves are normally incident (incidence at an angle will be discussed later).
<center><math>
\Delta \phi = 0, \;\;\; -h < z \leq 0,
</math></center>
<center><math>
\partial_z \phi = 0, \;\;\; z = - h,
</math></center>
<center><math>
\partial_z\phi=\alpha\phi, \,\, z=0,\,x<0,
</math></center>
<center><math>
\beta \partial_x^4\partial_z \phi
- \left( \gamma\alpha - 1 \right) \partial_z \phi = \alpha\phi, \;\;
z = 0, \;\;\; x \geq 0,
</math></center>
where <math>\alpha = \omega^2</math>, <math>\beta</math> and <math>\gamma</math>
are the stiffness and mass constant for the plate respectively. The free edge conditions
at the edge of the plate imply
<center><math>
\partial_x^3 \partial_z\phi = 0, \;\; z = 0, \;\;\; x = 0,
</math></center>
<center><math>
\partial_x^2 \partial_z\phi = 0, \;\; z = 0, \;\;\; x = 0,
</math></center>

== Method of solution ==

We use [http://en.wikipedia.org/wiki/Separation_of_Variables separation of variables] in the two regions, <math>x<0</math>
and <math>x>0</math>.

{{separation of variables in two dimensions}}

{{separation of variables for a free surface}}

{{separation of variables for a floating elastic plate}}

{{free surface floating elastic plate relations}}

{{incident potential for two dimensions}}

== An infinite dimensional system of equations ==

The potential and its derivative must be continuous across the
transition from open water to the plate covered region. Therefore, the
potentials and their derivatives at <math>x=0</math> have to be equal.
We also truncate the sum at <math>N</math> being careful that we have
two extra modes on the plate covered region to satisfy the edge conditions.
We obtain
<center>
<math>
\phi_{0}\left( z\right) + \sum_{m=0}^{N}
a_{m} \phi_{m}\left( z\right)
=\sum_{m=-2}^{N}b_{m}\psi_{m}(z)
</math>
</center>
and
<center>
<math>
-k_{0}\phi_{0}\left( z\right) +\sum
_{m=0}^{N} a_{m}k_{m}\phi_{m}\left( z\right)
=-\sum_{m=-2}^{N}b_{m}\kappa_{m}\psi
_{m}(z)
</math>
</center>
for each <math>N</math>.
We solve these equations by multiplying both equations by
<math>\phi_{l}(z)\,</math> and integrating from <math>-h</math> to <math>0</math> to obtain:
<center>
<math>
A_{0}\delta_{0l}+a_{l}A_{l}
=\sum_{m=-2}^{N}b_{m}B_{ml},\,\,0 \leq l \leq N
</math>
</center>
and
<center>
<math>
-k_{0}A_{0}\delta_{0l}+a_{l}k_{l}A_l
=-\sum_{m=-2}^{N}b_{m}\kappa_{m}B_{ml},\,\,0 \leq l \leq N
</math>
</center>
If we multiply the first equation by <math>k_l</math> and subtract the second equation
we obtain
<center>
<math>
(k_{0}+k_l)A_{0}\delta_{0l}
=\sum_{m=-2}^{N}b_{m}(k_l + \kappa_{m})B_{ml}
</math>
</center>
Finally, we need to apply the conditions at the edge of the plate to give us two further equations,
<center>
<math>
\partial_x^2\partial_z\phi = - \sum_{m=-2}^{N}b_{m} \kappa_m^3 \tan\kappa_m h = 0
</math>
</center>
and
<center>
<math>
\partial_x^3\partial_z\phi = \sum_{m=-2}^{N}b_{m} \kappa_m^4 \tan\kappa_m h = 0
</math>
</center>

== Numerical Solution ==

To solve the system of equations previously defined we set the upper limit of <math>l</math> to
be <math>N</math>, as we already said before. In terms of matrix, we obtain
<center>
<math>
\begin{bmatrix}
\begin{bmatrix}
A_0&0 \quad \cdots&0\\
0&&\\
\vdots&A_l&\vdots\\
&&0\\
0&\cdots \quad 0 &A_N
\end{bmatrix}
&
\begin{bmatrix}
-B_{-2,0}&-B_{-1,0}\\
&\\
\vdots&\vdots\\
&\\
-B_{-2,N}&-B_{-1,N}
\end{bmatrix}
&
\begin{bmatrix}
-B_{0,0}&\cdots&-B_{0,N}\\
&&\\
\vdots&-B_{m,l}&\vdots\\
&&\\
-B_{N,0}&\cdots&-B_{N,N}
\end{bmatrix}
\\ \\
\begin{bmatrix}
0&&\cdots&&0\\
0&&\cdots&&0
\end{bmatrix}
&
\begin{bmatrix}
\kappa_{-2}^3\tan\kappa_{-2}h&\kappa_{-1}^3\tan\kappa_{-1}h\\
\kappa_{-2}^4\tan\kappa_{-2}h&\kappa_{-1}^4\tan\kappa_{-1}h
\end{bmatrix}
&
\begin{bmatrix}
\kappa_0^3\tan\kappa_0h&\cdots&\kappa_N^3\tan\kappa_Nh\\
\kappa_0^4\tan\kappa_0h&\cdots&\kappa_N^4\tan\kappa_Nh
\end{bmatrix}
\\ \\
\begin{bmatrix}
0&&\cdots&&0\\
&&&&\\
\vdots&&\ddots&&\vdots\\
&&&&\\
0&&\cdots&&0
\end{bmatrix}
&
\begin{bmatrix}
(k_0+\kappa_{-2})B_{-2,0}&(k_0+\kappa_{-1})B_{-1,0}\\
&\\
\vdots&\vdots\\
&\\
(k_N+\kappa_{-2})B_{-2,N}&(k_N+\kappa_{-1})B_{-1,N}
\end{bmatrix}
&
\begin{bmatrix}
(k_0 + \kappa_0) \, B_{0,0}&\cdots&(k_N + \kappa_{0}) \, B_{0,N}\\
&&\\
\vdots&(k_l + \kappa_{m}) \, B_{m,l}&\vdots\\
&&\\
(k_0 + \kappa_N) \, B_{N,0}&\cdots&(k_N + \kappa_{N}) \, B_{N,N}\\
\end{bmatrix}
\end{bmatrix}
\begin{bmatrix}

\begin{bmatrix}
a_{0} \\
\\
\vdots \\
\\
a_N
\end{bmatrix}
\\ \\
\begin{bmatrix}
b_{-2}\\
b_{-1}
\end{bmatrix}
\\ \\
\begin{bmatrix}
b_{0}\\
\\
\vdots \\
\\
b_N
\end{bmatrix}
\end{bmatrix}
=
\begin{bmatrix}
\begin{bmatrix}
- A_{0} \\
0 \\
\vdots \\
\\
0
\end{bmatrix}
\\ \\
\begin{bmatrix}
0 \\
0
\end{bmatrix}
\\ \\
\begin{bmatrix}
2k_{0}A_{0} \\
0 \\
\vdots \\
\\
0
\end{bmatrix}
\end{bmatrix}
</math>
</center>
We then simply need to solve the linear system of equations. Note that we can solve this equation for
<math>b_n</math> first and then solve for <math>a_n</math>

== Waves Incident at an Angle ==

We can consider the case of [[Waves Incident at an Angle]] <math>\theta</math>.
{{incident angle}}

It is shown that the potential can be expanded as
<center>
<math>
\phi(x,z)=e^{-\hat{k}_0x}\phi_0(z)+\sum_{m=0}^{\infty}a_{m}e^{\hat{k}_{m}x}\phi_{m}(z), \;\;x<0
</math>
</center>
and
<center>
<math>
\phi(x,z)=\sum_{m=-2}^{\infty}b_{m}
e^{-\hat{\kappa}_{m}x}\psi_{m}(z), \;\;x>0
</math>
</center>
where <math>\hat{k}_{m} = \sqrt{k_m^2 - k_y^2}</math> and <math>\hat{\kappa}_{m} = \sqrt{\kappa_m^2 - k_y^2}</math>
where we always take the positive real root or the root with positive imaginary part.

The edge conditions are also different and are
<center><math>
\left(\frac{\partial^3}{\partial x^3} + (2 - \nu)k^2_y\frac{\partial}{\partial x}\right) \frac{\partial\phi}{\partial z}= 0,
</math></center>
and
<center><math>
\left(\frac{\partial^2}{\partial x^2} + \nu k^2_y\right)\frac{\partial\phi}{\partial z} = 0,
</math></center>
where <math>\nu</math> is Poisons ratio.

We can expend these edge conditions, which respectively gives
<center>
<math>
-\sum_{m=-2}^{\infty}b_{m}\kappa_m (\hat{\kappa}_m^3+\hat{\kappa}_m k_y^2(2-\nu))\tan \kappa_m h=0
</math>
</center>
and
<center>
<math>
-\sum_{m=-2}^{\infty}b_{m}\kappa_m (\hat{\kappa}_m^2+k_y^2\nu))\tan \kappa_m h=0
</math>
</center>

The equations are derived almost identically to those above and we obtain
<center>
<math>
A_{0}\delta_{0l}+a_{l}A_{l}
=\sum_{m=-2}^{\infty}b_{m}B_{ml}
</math>
</center>
and
<center>
<math>
-\hat{k_{0}}A_{0}\delta_{0l}+a_{l}\hat{k}_{l}A_l
=-\sum_{m=-2}^{\infty}b_{m}\hat{\kappa}_{m}B_{ml}
</math>
</center>
and these are solved exactly as before.

== Energy Balance ==

We present a derivation of the energy balance here and also refer to the derivation
[[Energy Balance for Two Elastic Plates]]

{{energy contour and preliminaries}}

The contributions from the vertical ends are

<center><math>
\lim_{x\to-\infty} \Im\int_{-h}^{0}\phi^*\frac{\partial\phi}{\partial n} dz
= \lim_{x\to-\infty} \Im\int_{-h}^{0} \hat{k}_0 \left(e^{\hat{k}_0 x} + a_0^{*} e^{-\hat{k}_0 x} \right)
\left(e^{-\hat{k}_0 x} - a_0 e^{\hat{k}_0 x} \right)\phi_0(z)^2 dz
</math></center>
<center><math>
= \frac{\hat{k}_0}{i} A_0 \left(1 - |a_0^2| \right)
</math></center>
and
<center><math>
\lim_{x\to\infty} \Im\int_{-h}^{0}\phi^*\frac{\partial\phi}{\partial n} dz
= \lim_{x\to\infty} \Im\int_{-h}^{0} -\hat{\kappa}_0 \left(b_0^{*} e^{\hat{\kappa}_0 x} \right)
\left(b_0 e^{-\hat{\kappa}_0 x}\right)\psi_0(z)^2 dz
</math></center>
<center><math>
= -\frac{\hat{\kappa}_0}{i} |b_0^2| \int_{-h}^{0} \psi_0(z)^2 dz
</math></center>

The contribution from the surface integral under the plate is
<center><math>
\Im\int_{0}^{\infty}\phi^*\frac{\partial\phi}{\partial n} dx
= \Im\int_{0}^{\infty}
1/\alpha \left(\beta \partial_x^4 \frac{\partial\phi^{*}}{\partial n} - (\gamma\alpha - 1 ) \frac{\partial\phi^{*}}{\partial n} \right)
\frac{\partial\phi}{\partial n} dx
</math></center>
<center><math>
= \frac{2\beta}{\alpha} \frac{\hat{\kappa}_0^{3}}{i} |b_0^2| \partial_z\psi_0(0)^2
</math></center>
where we have integrated by parts and used the condition at the ends of the plate.

The energy balance is therefore
<center><math>
{\hat{k}_0} A_0 |a_0^2| + {\hat{\kappa}_0} |b_0^2| \int_{-h}^{0} \psi_0(z)^2 dz
- \frac{2\beta}{\alpha}{ \hat{\kappa}_0^{3}} |b_0^2| \partial_z\psi_0(0)^2 = {\hat{k}_0} A_0
</math></center>

Note that this formula is only valid for angles <math>\theta \in [-\theta_0, \theta_0]</math>, where <math>\theta_0</math>
is defined by <math>\sin \theta_0 =\frac{\kappa_0}{k_0}</math>. For large angles there is total reflection
and then <math>|R|^2=1</math>

== Matlab Code ==

A program to calculate the coefficients for the semi-infinite dock problems can be found here
[http://www.math.auckland.ac.nz/~meylan/code/eigenfunction_matching/semiinfinite_plate.m semiinfinite_plate.m]

=== Additional code ===

This program requires
* {{free surface dispersion equation code}}
* {{elastic plate dispersion equation code}}

[[Category:Floating Elastic Plate]]
[[Category:Eigenfunction Matching Method]]
[[Category:Complete Pages]]

Eigenfunction Matching for a Semi-Infinite Floating Elastic Plate

2010-03-17T20:11:01Z

Mike smith: either drop delta or keep k_l; matter of choice

{{complete pages}}

== Introduction ==

We show here a solution for a semi-infinite [[:Category:Floating Elastic Plate|Floating Elastic Plate]] on [[Finite Depth]].
The problem was solved by [[Fox and Squire 1994]] but the solution method here is slightly different.
The simpler theory for a [[Eigenfunction Matching for a Semi-Infinite Dock|Dock]] describes
many of the ideas here in more detail.

[[Image:Semiinfinite plate.jpg|thumb|right|300px|Wave scattering by a submerged semi-infinite elastic plate]]

== Equations ==

We consider the problem of small-amplitude waves which are incident on a semi-infinite floating elastic
plate occupying water surface for <math>x>0</math>. The submergence of the plate is considered negligible.
We assume that the problem is invariant in the <math>y</math> direction, although we allow the waves to be
incident from an angle.
We also assume that the plate edges are free to move at
each boundary, although other boundary conditions could easily be considered using
the methods of solution presented here. We begin with the [[Frequency Domain Problem]] for a semi-infinite
[[Floating Elastic Plate|Floating Elastic Plates]]
in the non-dimensional form of [[Tayler_1986a|Tayler 1986]] ([[Dispersion Relation for a Floating Elastic Plate]]).
We also assume that the waves are normally incident (incidence at an angle will be discussed later).
<center><math>
\Delta \phi = 0, \;\;\; -h < z \leq 0,
</math></center>
<center><math>
\partial_z \phi = 0, \;\;\; z = - h,
</math></center>
<center><math>
\partial_z\phi=\alpha\phi, \,\, z=0,\,x<0,
</math></center>
<center><math>
\beta \partial_x^4\partial_z \phi
- \left( \gamma\alpha - 1 \right) \partial_z \phi = \alpha\phi, \;\;
z = 0, \;\;\; x \geq 0,
</math></center>
where <math>\alpha = \omega^2</math>, <math>\beta</math> and <math>\gamma</math>
are the stiffness and mass constant for the plate respectively. The free edge conditions
at the edge of the plate imply
<center><math>
\partial_x^3 \partial_z\phi = 0, \;\; z = 0, \;\;\; x = 0,
</math></center>
<center><math>
\partial_x^2 \partial_z\phi = 0, \;\; z = 0, \;\;\; x = 0,
</math></center>

== Method of solution ==

We use [http://en.wikipedia.org/wiki/Separation_of_Variables separation of variables] in the two regions, <math>x<0</math>
and <math>x>0</math>.

{{separation of variables in two dimensions}}

{{separation of variables for a free surface}}

{{separation of variables for a floating elastic plate}}

{{free surface floating elastic plate relations}}

{{incident potential for two dimensions}}

== An infinite dimensional system of equations ==

The potential and its derivative must be continuous across the
transition from open water to the plate covered region. Therefore, the
potentials and their derivatives at <math>x=0</math> have to be equal.
We also truncate the sum at <math>N</math> being careful that we have
two extra modes on the plate covered region to satisfy the edge conditions.
We obtain
<center>
<math>
\phi_{0}\left( z\right) + \sum_{m=0}^{N}
a_{m} \phi_{m}\left( z\right)
=\sum_{m=-2}^{N}b_{m}\psi_{m}(z)
</math>
</center>
and
<center>
<math>
-k_{0}\phi_{0}\left( z\right) +\sum
_{m=0}^{N} a_{m}k_{m}\phi_{m}\left( z\right)
=-\sum_{m=-2}^{N}b_{m}\kappa_{m}\psi
_{m}(z)
</math>
</center>
for each <math>N</math>.
We solve these equations by multiplying both equations by
<math>\phi_{l}(z)\,</math> and integrating from <math>-h</math> to <math>0</math> to obtain:
<center>
<math>
A_{0}\delta_{0l}+a_{l}A_{l}
=\sum_{m=-2}^{N}b_{m}B_{ml},\,\,0 \leq l \leq N
</math>
</center>
and
<center>
<math>
-k_{0}A_{0}\delta_{0l}+a_{l}k_{l}A_l
=-\sum_{m=-2}^{N}b_{m}\kappa_{m}B_{ml},\,\,0 \leq l \leq N
</math>
</center>
If we multiply the first equation by <math>k_l</math> and subtract the second equation
we obtain
<center>
<math>
(k_{0}+k_l)A_{0}\delta_{0l}
=\sum_{m=-2}^{N}b_{m}(k_l + \kappa_{m})B_{ml}
</math>
</center>
Finally, we need to apply the conditions at the edge of the plate to give us two further equations,
<center>
<math>
\partial_x^2\partial_z\phi = - \sum_{m=-2}^{N}b_{m} \kappa_m^3 \tan\kappa_m h = 0
</math>
</center>
and
<center>
<math>
\partial_x^3\partial_z\phi = \sum_{m=-2}^{N}b_{m} \kappa_m^4 \tan\kappa_m h = 0
</math>
</center>

== Numerical Solution ==

To solve the system of equations previously defined we set the upper limit of <math>l</math> to
be <math>N</math>, as we already said before. In terms of matrix, we obtain
<center>
<math>
\begin{bmatrix}
\begin{bmatrix}
A_0&0 \quad \cdots&0\\
0&&\\
\vdots&A_l&\vdots\\
&&0\\
0&\cdots \quad 0 &A_N
\end{bmatrix}
&
\begin{bmatrix}
-B_{-20}&-B_{-10}\\
&\\
\vdots&\vdots\\
&\\
-B_{-2N}&-B_{-1N}
\end{bmatrix}
&
\begin{bmatrix}
-B_{00}&\cdots&-B_{0N}\\
&&\\
\vdots&-B_{ml}&\vdots\\
&&\\
-B_{N0}&\cdots&-B_{NN}
\end{bmatrix}
\\
\begin{bmatrix}
0&&\cdots&&0\\
0&&\cdots&&0
\end{bmatrix}
&
\begin{bmatrix}
\kappa_{-2}^3\tan\kappa_{-2}h&\kappa_{-1}^3\tan\kappa_{-1}h\\
\kappa_{-2}^4\tan\kappa_{-2}h&\kappa_{-1}^4\tan\kappa_{-1}h
\end{bmatrix}
&
\begin{bmatrix}
\kappa_0^3\tan\kappa_0h&\cdots&\kappa_N^3\tan\kappa_Nh\\
\kappa_0^4\tan\kappa_0h&\cdots&\kappa_N^4\tan\kappa_Nh
\end{bmatrix}
\\
\begin{bmatrix}
0&&\cdots&&0\\
&&&&\\
\vdots&&\ddots&&\vdots\\
&&&&\\
0&&\cdots&&0
\end{bmatrix}
&
\begin{bmatrix}
(k_0+\kappa_{-2})B_{-20}&(k_0+\kappa_{-1})B_{-10}\\
&\\
\vdots&\vdots\\
&\\
(k_N+\kappa_{-2})B_{-2N}&(k_N+\kappa_{-1})B_{-1N}
\end{bmatrix}
&
\begin{bmatrix}
(k_0 + \kappa_0) \, B_{00}&\cdots&(k_N + \kappa_{0}) \, B_{0N}\\
&&\\
\vdots&(k_l + \kappa_{m}) \, B_{ml}&\vdots\\
&&\\
(k_0 + \kappa_N) \, B_{N0}&\cdots&(k_N + \kappa_{N}) \, B_{NN}\\
\end{bmatrix}
\end{bmatrix}
\begin{bmatrix}
a_{0} \\
\\
\vdots \\
\\
a_N \\
b_{-2}\\
b_{-1}\\
b_{0}\\
\\
\vdots \\
\\
b_N
\end{bmatrix}
=
\begin{bmatrix}
- A_{0} \\
0 \\
\vdots \\
\\
0 \\
0 \\
0 \\
2k_{0}A_{0} \\
0 \\
\vdots \\
\\
0
\end{bmatrix}
</math>
</center>
We then simply need to solve the linear system of equations. Note that we can solve this equation for
<math>b_n</math> first and then solve for <math>a_n</math>

== Waves Incident at an Angle ==

We can consider the case of [[Waves Incident at an Angle]] <math>\theta</math>.
{{incident angle}}

It is shown that the potential can be expanded as
<center>
<math>
\phi(x,z)=e^{-\hat{k}_0x}\phi_0(z)+\sum_{m=0}^{\infty}a_{m}e^{\hat{k}_{m}x}\phi_{m}(z), \;\;x<0
</math>
</center>
and
<center>
<math>
\phi(x,z)=\sum_{m=-2}^{\infty}b_{m}
e^{-\hat{\kappa}_{m}x}\psi_{m}(z), \;\;x>0
</math>
</center>
where <math>\hat{k}_{m} = \sqrt{k_m^2 - k_y^2}</math> and <math>\hat{\kappa}_{m} = \sqrt{\kappa_m^2 - k_y^2}</math>
where we always take the positive real root or the root with positive imaginary part.

The edge conditions are also different and are
<center><math>
\left(\frac{\partial^3}{\partial x^3} + (2 - \nu)k^2_y\frac{\partial}{\partial x}\right) \frac{\partial\phi}{\partial z}= 0,
</math></center>
and
<center><math>
\left(\frac{\partial^2}{\partial x^2} + \nu k^2_y\right)\frac{\partial\phi}{\partial z} = 0,
</math></center>
where <math>\nu</math> is Poisons ratio.

We can expend these edge conditions, which respectively gives
<center>
<math>
-\sum_{m=-2}^{\infty}b_{m}\kappa_m (\hat{\kappa}_m^3+\hat{\kappa}_m k_y^2(2-\nu))\tan \kappa_m h=0
</math>
</center>
and
<center>
<math>
-\sum_{m=-2}^{\infty}b_{m}\kappa_m (\hat{\kappa}_m^2+k_y^2\nu))\tan \kappa_m h=0
</math>
</center>

The equations are derived almost identically to those above and we obtain
<center>
<math>
A_{0}\delta_{0l}+a_{l}A_{l}
=\sum_{m=-2}^{\infty}b_{m}B_{ml}
</math>
</center>
and
<center>
<math>
-\hat{k_{0}}A_{0}\delta_{0l}+a_{l}\hat{k}_{l}A_l
=-\sum_{m=-2}^{\infty}b_{m}\hat{\kappa}_{m}B_{ml}
</math>
</center>
and these are solved exactly as before.

== Energy Balance ==

We present a derivation of the energy balance here and also refer to the derivation
[[Energy Balance for Two Elastic Plates]]

{{energy contour and preliminaries}}

The contributions from the vertical ends are

<center><math>
\lim_{x\to-\infty} \Im\int_{-h}^{0}\phi^*\frac{\partial\phi}{\partial n} dz
= \lim_{x\to-\infty} \Im\int_{-h}^{0} \hat{k}_0 \left(e^{\hat{k}_0 x} + a_0^{*} e^{-\hat{k}_0 x} \right)
\left(e^{-\hat{k}_0 x} - a_0 e^{\hat{k}_0 x} \right)\phi_0(z)^2 dz
</math></center>
<center><math>
= \frac{\hat{k}_0}{i} A_0 \left(1 - |a_0^2| \right)
</math></center>
and
<center><math>
\lim_{x\to\infty} \Im\int_{-h}^{0}\phi^*\frac{\partial\phi}{\partial n} dz
= \lim_{x\to\infty} \Im\int_{-h}^{0} -\hat{\kappa}_0 \left(b_0^{*} e^{\hat{\kappa}_0 x} \right)
\left(b_0 e^{-\hat{\kappa}_0 x}\right)\psi_0(z)^2 dz
</math></center>
<center><math>
= -\frac{\hat{\kappa}_0}{i} |b_0^2| \int_{-h}^{0} \psi_0(z)^2 dz
</math></center>

The contribution from the surface integral under the plate is
<center><math>
\Im\int_{0}^{\infty}\phi^*\frac{\partial\phi}{\partial n} dx
= \Im\int_{0}^{\infty}
1/\alpha \left(\beta \partial_x^4 \frac{\partial\phi^{*}}{\partial n} - (\gamma\alpha - 1 ) \frac{\partial\phi^{*}}{\partial n} \right)
\frac{\partial\phi}{\partial n} dx
</math></center>
<center><math>
= \frac{2\beta}{\alpha} \frac{\hat{\kappa}_0^{3}}{i} |b_0^2| \partial_z\psi_0(0)^2
</math></center>
where we have integrated by parts and used the condition at the ends of the plate.

The energy balance is therefore
<center><math>
{\hat{k}_0} A_0 |a_0^2| + {\hat{\kappa}_0} |b_0^2| \int_{-h}^{0} \psi_0(z)^2 dz
- \frac{2\beta}{\alpha}{ \hat{\kappa}_0^{3}} |b_0^2| \partial_z\psi_0(0)^2 = {\hat{k}_0} A_0
</math></center>

Note that this formula is only valid for angles <math>\theta \in [-\theta_0, \theta_0]</math>, where <math>\theta_0</math>
is defined by <math>\sin \theta_0 =\frac{\kappa_0}{k_0}</math>. For large angles there is total reflection
and then <math>|R|^2=1</math>

== Matlab Code ==

A program to calculate the coefficients for the semi-infinite dock problems can be found here
[http://www.math.auckland.ac.nz/~meylan/code/eigenfunction_matching/semiinfinite_plate.m semiinfinite_plate.m]

=== Additional code ===

This program requires
* {{free surface dispersion equation code}}
* {{elastic plate dispersion equation code}}

[[Category:Floating Elastic Plate]]
[[Category:Eigenfunction Matching Method]]
[[Category:Complete Pages]]

Template:Separation of variables for a free surface

2010-03-17T19:36:46Z

Mike smith: /* Separation of variables for a free surface */

=== Separation of variables for a free surface ===

We use [http://en.wikipedia.org/wiki/Separation_of_Variables separation of variables]

{{separation of variables in two dimensions}}

{{separation of variables for a free surface first part}}

We denote the
positive imaginary solution of this equation by <math>k_{0} \,</math> and
the positive real solutions by <math>k_{m} \,</math>, <math>m\geq1</math>. We define
<center>
<math>
\phi_{m}\left( z\right) =\frac{\cos k_{m}(z+h)}{\cos k_{m}h},\quad m\geq0
</math>
</center>
as the vertical eigenfunction of the potential in the open
water region. From [http://en.wikipedia.org/wiki/Sturm-Liouville_theory Sturm-Liouville theory] the
vertical eigenfunctions are orthogonal. They can be normalised to be orthonormal, but
this has no advantages for a numerical implementation. It can be shown that
<center>
<math>
\int\nolimits_{-h}^{0}\phi_{m}(z)\phi_{n}(z) \mathrm{d} z=A_{n}\delta_{mn}
</math>
</center>
where
<center>
<math>
A_{n}=\frac{1}{2}\left( \frac{\cos k_{n}h\sin k_{n}h+k_{n}h}{k_{n}\cos
^{2}k_{n}h}\right).
</math>
</center>

Template:Separation of variables for a floating elastic plate

2010-03-17T19:33:48Z

Mike smith: /* Separation of variables under the Plate */

==Separation of variables under the Plate==
<center>
<math>
Z^{\prime\prime} + \kappa^2 Z =0.
</math>
</center>
subject to the boundary conditions
<center>
<math>
Z^{\prime}(-h) = 0
</math>
</center>
and
<center>
<math>
\left(\beta \kappa^4 + 1 - \alpha\gamma\right)Z^{\prime}(0) = \alpha Z(0)
</math>
</center>
(the first term comes from the beam eigenvalue problem, where <math>\partial_x^4 X = \kappa^4 X</math>). We then use the boundary condition at <math>z=-h \,</math> to write
<center>
<math>
Z = \frac{\cos \kappa(z+h)}{\cos \kappa h}
</math>
</center>
The boundary condition at the free surface (<math>z=0</math>) is
the [[Dispersion Relation for a Floating Elastic Plate]]
<center><math>
\kappa \tan{(\kappa h)}= -\frac{\alpha}{\beta \kappa^{4} + 1 - \alpha\gamma}
</math></center>
Solving for <math>\kappa \,</math> gives a pure imaginary root
with positive imaginary part, two complex roots (two complex conjugate paired roots
with positive imaginary part in most physical situations), an infinite number of positive real roots
which approach <math>{n\pi}/{h} \,</math> as <math>n</math> approaches infinity, and also the negative of all
these roots ([[Dispersion Relation for a Floating Elastic Plate]]) . We denote the two complex roots with positive imaginary part
by <math>\kappa_{-2} \,</math> and <math>\kappa_{-1} \,</math>, the purely imaginary
root with positive imaginary part by <math>\kappa_{0} \,</math> and the real roots with positive imaginary part
by <math>\kappa_{n} \,</math> for <math>n</math> a positive integer.
The imaginary root with positive imaginary part corresponds to a
reflected travelling mode propagating along the <math>x</math> axis.
The complex roots with positive imaginary parts correspond to damped reflected travelling modes and the real roots correspond to reflected evanescent modes.

Template:Free surface floating elastic plate relations

2010-03-17T19:30:36Z

Mike smith: requires additional scaling factor to be consistent with formulation and code

== Inner product between free surface and elastic plate modes ==

<center>
<math>
\int\nolimits_{-h}^{0}\phi_{n}(z)\psi_{m}(z) d z=B_{mn}
</math>
</center>
where
<center><math>
B_{mn}=\frac{k_{n}\sin k_{n}h\cos\kappa_{m}h-\kappa_{m}\cos k_{n}h\sin
\kappa_{m}h}{\left( \cos k_{n}h\right) \left( \cos \kappa_{m}h\right) \left( k_{n}
^{2}-\kappa_{m}^{2}\right) }
</math></center>

Template:Separation of variables for a floating elastic plate

2010-03-16T21:32:22Z

Mike smith: /* Separation of variables under the Plate */

==Separation of variables under the Plate==
<center>
<math>
Z^{\prime\prime} + k^2 Z =0.
</math>
</center>
subject to the boundary conditions
<center>
<math>
Z^{\prime}(-h) = 0
</math>
</center>
and
<center>
<math>
\left(\beta \kappa^4 + 1 - \alpha\gamma\right)Z^{\prime}(0) = \alpha Z(0)
</math>
</center>
(the first term comes from the beam eigenvalue problem, where <math>\partial_x^4 X = \kappa^4 X</math>). We then use the boundary condition at <math>z=-h \,</math> to write
<center>
<math>
Z = \frac{\cos \kappa(z+h)}{\cos \kappa h}
</math>
</center>
The boundary condition at the free surface (<math>z=0</math>) is
the [[Dispersion Relation for a Floating Elastic Plate]]
<center><math>
\kappa \tan{(\kappa h)}= -\frac{\alpha}{\beta \kappa^{4} + 1 - \alpha\gamma}
</math></center>
Solving for <math>\kappa \,</math> gives a pure imaginary root
with positive imaginary part, two complex roots (two complex conjugate paired roots
with positive imaginary part in most physical situations), an infinite number of positive real roots
which approach <math>{n\pi}/{h} \,</math> as <math>n</math> approaches infinity, and also the negative of all
these roots ([[Dispersion Relation for a Floating Elastic Plate]]) . We denote the two complex roots with positive imaginary part
by <math>\kappa_{-2} \,</math> and <math>\kappa_{-1} \,</math>, the purely imaginary
root with positive imaginary part by <math>\kappa_{0} \,</math> and the real roots with positive imaginary part
by <math>\kappa_{n} \,</math> for <math>n</math> a positive integer.
The imaginary root with positive imaginary part corresponds to a
reflected travelling mode propagating along the <math>x</math> axis.
The complex roots with positive imaginary parts correspond to damped reflected travelling modes and the real roots correspond to reflected evanescent modes.

Template:Separation of variables for a floating elastic plate

2010-03-16T21:28:58Z

Mike smith: consistent notation, correction for plate expression, subscript errors

==Separation of variables under the Plate==
<center>
<math>
Z^{\prime\prime} + k^2 Z =0.
</math>
</center>
subject to the boundary conditions
<center>
<math>
Z^{\prime}(-h) = 0
</math>
</center>
and
<center>
<math>
\left(\beta \kappa^4 + 1 - \alpha\gamma\right)Z^{\prime}(0) = \alpha Z(0)
</math>
</center>
(the first term comes from the beam eigenvalue problem, where <math>\partial_x^4 X_n = \kappa_n^4 X_n</math>). We then use the boundary condition at <math>z=-h \,</math> to write
<center>
<math>
Z = \frac{\cos \kappa(z+h)}{\cos \kappa h}
</math>
</center>
The boundary condition at the free surface (<math>z=0</math>) is
the [[Dispersion Relation for a Floating Elastic Plate]]
<center><math>
\kappa \tan{(\kappa h)}= -\frac{\alpha}{\beta \kappa^{4} + 1 - \alpha\gamma}
</math></center>
Solving for <math>\kappa</math> gives a pure imaginary root
with positive imaginary part, two complex roots (two complex conjugate paired roots
with positive imaginary part in most physical situations), an infinite number of positive real roots
which approach <math>{n\pi}/{h}</math> as <math>n</math> approaches infinity, and also the negative of all
these roots ([[Dispersion Relation for a Floating Elastic Plate]]) . We denote the two complex roots with positive imaginary part
by <math>\kappa_{-2} \,</math> and <math>\kappa_{-1} \,</math>, the purely imaginary
root with positive imaginary part by <math>\kappa_{0} \,</math> and the real roots with positive imaginary part
by <math>\kappa_{n} \,</math> for <math>n</math> a positive integer.
The imaginary root with positive imaginary part corresponds to a
reflected travelling mode propagating along the <math>x</math> axis.
The complex roots with positive imaginary parts correspond to damped reflected travelling modes and the real roots correspond to reflected evanescent modes.

Eigenfunction Matching for a Semi-Infinite Dock

2010-03-16T20:29:11Z

Mike smith: /* Numerical Solution */

{{complete pages}}

== Introduction ==

This is one of the simplest problem in eigenfunction matching. It also is an easy
problem to understand the [[:Category:Wiener-Hopf|Wiener-Hopf]] and
[[:Category:Residue Calculus|Residue Calculus]]. The problems consists of a region to the left
with a free surface and a region to the right with a rigid surface through which
not flow is possible.
We begin with the simply problem when the waves are normally incident (so that
the problem is truly two-dimensional. We then consider the case when the waves are incident
at an angle. For the later we give the equations in slightly less detail.
The case of a [[Finite Dock]] is treated very similarly. The problem can also
be generalised to a [[ Eigenfunction Matching for Submerged Semi-Infinite Dock|Submerged Semi-Infinite Dock]]

[[Image:semiinfinite_dock.jpg|thumb|right|300px|Wave scattering by a semi-infinite dock]]

== Governing Equations ==

We begin with the [[Frequency Domain Problem]] for a dock which occupies
the region <math>x>0</math> (we assume <math>e^{i\omega t}</math> time dependence).
The water is assumed to have
constant finite depth <math>h</math> and the <math>z</math>-direction points vertically
upward with the water surface at <math>z=0</math> and the sea floor at <math>z=-h</math>. The
boundary value problem can therefore be expressed as
<center>
<math>
\Delta\phi=0, \,\, -h<z<0,
</math>
</center>
<center>
<math>
\partial_z\phi=0, \,\, z=-h,
</math>
</center>
<center><math>
\partial_z\phi=\alpha\phi, \,\, z=0,\,x<0,
</math></center>
<center>
<math>
\partial_z\phi=0, \,\, z=0,\,x>0,
</math>
</center>
We
must also apply the [[Sommerfeld Radiation Condition]]
as <math>|x|\rightarrow\infty</math>. This essentially implies
that the only wave at infinity is propagating away and at negative infinity there is a unit incident wave
and a wave propagating away.

== Solution Method ==

We use [http://en.wikipedia.org/wiki/Separation_of_Variables separation of variables] in the two regions, <math>x<0</math>
and <math>x>0</math>.

{{separation of variables in two dimensions}}

{{separation of variables for a free surface}}

{{separation of variables for a dock}}

{{free surface dock relations}}

=== Expansion of the potential ===

We need to apply some boundary conditions at plus and minus infinity,
where are essentially the the solution cannot grow. This means that we
only have the positive (or negative) roots of the dispersion equation.
However, it does not help us with the purely imaginary root. Here we
must use a different condition, essentially identifying one solution
as the incoming wave and the other as the outgoing wave.

Therefore the scattered potential (without the incident wave, which will
be added later) can
be expanded as
<center>
<math>
\phi(x,z)=\sum_{m=0}^{\infty}a_{m}e^{k_{m}x}\phi_{m}(z), \;\;x<0
</math>
</center>
and
<center>
<math>
\phi(x,z)=\sum_{m=0}^{\infty}b_{m}
e^{-\kappa_{m}x}\psi_{m}(z), \;\;x>0
</math>
</center>
where <math>a_{m}</math> and <math>b_{m}</math>
are the coefficients of the potential in the open water and
the dock covered region respectively.

{{incident potential for two dimensions}}

=== An infinite dimensional system of equations ===

The potential and its derivative must be continuous across the
transition from open water to the plate covered region. Therefore, the
potentials and their derivatives at <math>x=0</math> have to be equal.
We obtain
<center>
<math>
\phi_{0}\left( z\right) + \sum_{m=0}^{\infty}
a_{m} \phi_{m}\left( z\right)
=\sum_{m=0}^{\infty}b_{m}\psi_{m}(z)
</math>
</center>
and
<center>
<math>
-k_{0}\phi_{0}\left( z\right) +\sum
_{m=0}^{\infty} a_{m}k_{m}\phi_{m}\left( z\right)
=-\sum_{m=0}^{\infty}b_{m}\kappa_{m}\psi
_{m}(z)
</math>
</center>
for each <math>m</math>.
We solve these equations by multiplying both equations by
<math>\phi_{l}(z) \,</math> and integrating from <math>-h</math> to <math>0</math> to obtain:
<center>
<math>
A_{0}\delta_{0l}+a_{l}A_{l}
=\sum_{m=0}^{\infty}b_{m}B_{ml}\,\,\,(3)
</math>
</center>
and
<center>
<math>
-k_{0}A_{0}\delta_{0l}+a_{l}k_{l}A_l
=-\sum_{m=0}^{\infty}b_{m}\kappa_{m}B_{ml} \,\,\,(4)
</math>
</center>
If we mutiply equation (3) by <math>k_l \,</math> and subtract equation (4)
we obtain
<center>
<math>
(k_{0}+k_l)A_{0}\delta_{0l}
=\sum_{m=0}^{\infty}b_{m}(k_l + \kappa_{m})B_{ml} \,\,\,(5)
</math>
</center>
This equation gives the required equations to solve for the
coefficients of the water velocity potential in the dock covered region.

== Numerical Solution ==

To solve the system of equations (3) and (5), we set the upper limit of <math>l</math> to
be <math>M</math>. This resulting system can be expressed in the block matrix form below,
<center>
<math>
\begin{bmatrix}
\begin{bmatrix}
A_0&0 \quad \cdots&0\\
0&&\\
\vdots&A_l&\vdots\\
&&0\\
0&\cdots \quad 0 &A_M
\end{bmatrix}
&
\begin{bmatrix}
-B_{00}&\cdots&-B_{0M}\\
&&\\
\vdots&-B_{ml}&\vdots\\
&&\\
-B_{M0}&\cdots&-B_{MM}
\end{bmatrix}
\\
\begin{bmatrix}
0&\cdots&0\\
\vdots&\ddots&\vdots\\
0&\cdots&0
\end{bmatrix}
&
\begin{bmatrix}
(k_0 + \kappa_0) \, B_{00}&\cdots&(k_M + \kappa_{0}) \, B_{0M}\\
&&\\
\vdots&(k_l + \kappa_{m}) \, B_{ml}&\vdots\\
&&\\
(k_0 + \kappa_M) \, B_{M0}&\cdots&(k_M + \kappa_{M}) \, B_{MM}\\
\end{bmatrix}
\end{bmatrix}
\begin{bmatrix}
a_{0} \\
a_{1} \\
\vdots \\
a_M \\
\\
b_{0}\\
b_1 \\
\vdots \\
\\
b_M
\end{bmatrix}
=
\begin{bmatrix}
- A_{0} \\
0 \\
\vdots \\
\\
0 \\
\\
2k_{0}A_{0} \\
0 \\
\vdots \\
\\
0
\end{bmatrix}
</math>
</center>
We then simply need to solve the linear system of equations.

== Solution with Waves Incident at an Angle ==

We can consider the problem when the waves are incident at an angle <math>\theta</math>.

{{incident angle}}

Therefore the potential can
be expanded as
<center>
<math>
\phi(x,z)=e^{-\hat{k}_0x}\phi_0(z)+\sum_{m=0}^{\infty}a_{m}e^{\hat{k}_{m}x}\phi_{m}(z), \;\;x<0
</math>
</center>
and
<center>
<math>
\phi(x,z)=\sum_{m=0}^{\infty}b_{m}
e^{-\hat{\kappa}_{m}x}\psi_{m}(z), \;\;x>0
</math>
</center>
where <math>\hat{k}_{m} = \sqrt{k_m^2 - k_y^2}</math> and <math>\hat{\kappa}_{m} = \sqrt{\kappa_m^2 - k_y^2}</math>
where we always take the positive real root or the root with positive imaginary part.

The equations are derived almost identically to those above and we obtain
<center>
<math>
A_{0}\delta_{0l}+a_{l}A_{l}
=\sum_{n=0}^{\infty}b_{m}B_{ml}
</math>
</center>
and
<center>
<math>
-\hat{k_{0}}A_{0}\delta_{0l}+a_{l}\hat{k}_{l}A_l
=-\sum_{m=0}^{\infty}b_{m}\hat{\kappa}_{m}B_{ml}
</math>
</center>
and these are solved exactly as before.

== Matlab Code ==

A program to calculate the coefficients for the semi-infinite dock problems can be found here
{{semiinfinite_dock code}}

=== Additional code ===

This program requires
* {{free surface dispersion equation code}}

[[Category:Eigenfunction Matching Method]]

Eigenfunction Matching for a Semi-Infinite Dock

2010-03-16T20:23:11Z

Mike smith: /* Numerical Solution */

{{complete pages}}

== Introduction ==

This is one of the simplest problem in eigenfunction matching. It also is an easy
problem to understand the [[:Category:Wiener-Hopf|Wiener-Hopf]] and
[[:Category:Residue Calculus|Residue Calculus]]. The problems consists of a region to the left
with a free surface and a region to the right with a rigid surface through which
not flow is possible.
We begin with the simply problem when the waves are normally incident (so that
the problem is truly two-dimensional. We then consider the case when the waves are incident
at an angle. For the later we give the equations in slightly less detail.
The case of a [[Finite Dock]] is treated very similarly. The problem can also
be generalised to a [[ Eigenfunction Matching for Submerged Semi-Infinite Dock|Submerged Semi-Infinite Dock]]

[[Image:semiinfinite_dock.jpg|thumb|right|300px|Wave scattering by a semi-infinite dock]]

== Governing Equations ==

We begin with the [[Frequency Domain Problem]] for a dock which occupies
the region <math>x>0</math> (we assume <math>e^{i\omega t}</math> time dependence).
The water is assumed to have
constant finite depth <math>h</math> and the <math>z</math>-direction points vertically
upward with the water surface at <math>z=0</math> and the sea floor at <math>z=-h</math>. The
boundary value problem can therefore be expressed as
<center>
<math>
\Delta\phi=0, \,\, -h<z<0,
</math>
</center>
<center>
<math>
\partial_z\phi=0, \,\, z=-h,
</math>
</center>
<center><math>
\partial_z\phi=\alpha\phi, \,\, z=0,\,x<0,
</math></center>
<center>
<math>
\partial_z\phi=0, \,\, z=0,\,x>0,
</math>
</center>
We
must also apply the [[Sommerfeld Radiation Condition]]
as <math>|x|\rightarrow\infty</math>. This essentially implies
that the only wave at infinity is propagating away and at negative infinity there is a unit incident wave
and a wave propagating away.

== Solution Method ==

We use [http://en.wikipedia.org/wiki/Separation_of_Variables separation of variables] in the two regions, <math>x<0</math>
and <math>x>0</math>.

{{separation of variables in two dimensions}}

{{separation of variables for a free surface}}

{{separation of variables for a dock}}

{{free surface dock relations}}

=== Expansion of the potential ===

We need to apply some boundary conditions at plus and minus infinity,
where are essentially the the solution cannot grow. This means that we
only have the positive (or negative) roots of the dispersion equation.
However, it does not help us with the purely imaginary root. Here we
must use a different condition, essentially identifying one solution
as the incoming wave and the other as the outgoing wave.

Therefore the scattered potential (without the incident wave, which will
be added later) can
be expanded as
<center>
<math>
\phi(x,z)=\sum_{m=0}^{\infty}a_{m}e^{k_{m}x}\phi_{m}(z), \;\;x<0
</math>
</center>
and
<center>
<math>
\phi(x,z)=\sum_{m=0}^{\infty}b_{m}
e^{-\kappa_{m}x}\psi_{m}(z), \;\;x>0
</math>
</center>
where <math>a_{m}</math> and <math>b_{m}</math>
are the coefficients of the potential in the open water and
the dock covered region respectively.

{{incident potential for two dimensions}}

=== An infinite dimensional system of equations ===

The potential and its derivative must be continuous across the
transition from open water to the plate covered region. Therefore, the
potentials and their derivatives at <math>x=0</math> have to be equal.
We obtain
<center>
<math>
\phi_{0}\left( z\right) + \sum_{m=0}^{\infty}
a_{m} \phi_{m}\left( z\right)
=\sum_{m=0}^{\infty}b_{m}\psi_{m}(z)
</math>
</center>
and
<center>
<math>
-k_{0}\phi_{0}\left( z\right) +\sum
_{m=0}^{\infty} a_{m}k_{m}\phi_{m}\left( z\right)
=-\sum_{m=0}^{\infty}b_{m}\kappa_{m}\psi
_{m}(z)
</math>
</center>
for each <math>m</math>.
We solve these equations by multiplying both equations by
<math>\phi_{l}(z) \,</math> and integrating from <math>-h</math> to <math>0</math> to obtain:
<center>
<math>
A_{0}\delta_{0l}+a_{l}A_{l}
=\sum_{m=0}^{\infty}b_{m}B_{ml}\,\,\,(3)
</math>
</center>
and
<center>
<math>
-k_{0}A_{0}\delta_{0l}+a_{l}k_{l}A_l
=-\sum_{m=0}^{\infty}b_{m}\kappa_{m}B_{ml} \,\,\,(4)
</math>
</center>
If we mutiply equation (3) by <math>k_l \,</math> and subtract equation (4)
we obtain
<center>
<math>
(k_{0}+k_l)A_{0}\delta_{0l}
=\sum_{m=0}^{\infty}b_{m}(k_l + \kappa_{m})B_{ml} \,\,\,(5)
</math>
</center>
This equation gives the required equations to solve for the
coefficients of the water velocity potential in the dock covered region.

== Numerical Solution ==

To solve the system of equations (3) and (5), we set the upper limit of <math>l</math> to
be <math>M</math>. In terms of matrix, we obtain
<center>
<math>
\begin{bmatrix}
\begin{bmatrix}
A_0&0 \quad \cdots&0\\
0&&\\
\vdots&A_l&\vdots\\
&&0\\
0&\cdots \quad 0 &A_M
\end{bmatrix}
&
\begin{bmatrix}
-B_{00}&\cdots&-B_{0M}\\
&&\\
\vdots&-B_{ml}&\vdots\\
&&\\
-B_{M0}&\cdots&-B_{MM}
\end{bmatrix}
\\
\begin{bmatrix}
0&\cdots&0\\
\vdots&\ddots&\vdots\\
0&\cdots&0
\end{bmatrix}
&
\begin{bmatrix}
(k_0 + \kappa_0) \, B_{00}&\cdots&(k_M + \kappa_{0}) \, B_{0M}\\
&&\\
\vdots&(k_l + \kappa_{m}) \, B_{ml}&\vdots\\
&&\\
(k_0 + \kappa_M) \, B_{M0}&\cdots&(k_M + \kappa_{M}) \, B_{MM}\\
\end{bmatrix}
\end{bmatrix}
\begin{bmatrix}
a_{0} \\
a_{1} \\
\vdots \\
a_M \\
\\
b_{0}\\
b_1 \\
\vdots \\
\\
b_M
\end{bmatrix}
=
\begin{bmatrix}
- A_{0} \\
0 \\
\vdots \\
\\
0 \\
\\
2k_{0}A_{0} \\
0 \\
\vdots \\
\\
0
\end{bmatrix}
</math>
</center>
We then simply need to solve the linear system of equations.

== Solution with Waves Incident at an Angle ==

We can consider the problem when the waves are incident at an angle <math>\theta</math>.

{{incident angle}}

Therefore the potential can
be expanded as
<center>
<math>
\phi(x,z)=e^{-\hat{k}_0x}\phi_0(z)+\sum_{m=0}^{\infty}a_{m}e^{\hat{k}_{m}x}\phi_{m}(z), \;\;x<0
</math>
</center>
and
<center>
<math>
\phi(x,z)=\sum_{m=0}^{\infty}b_{m}
e^{-\hat{\kappa}_{m}x}\psi_{m}(z), \;\;x>0
</math>
</center>
where <math>\hat{k}_{m} = \sqrt{k_m^2 - k_y^2}</math> and <math>\hat{\kappa}_{m} = \sqrt{\kappa_m^2 - k_y^2}</math>
where we always take the positive real root or the root with positive imaginary part.

The equations are derived almost identically to those above and we obtain
<center>
<math>
A_{0}\delta_{0l}+a_{l}A_{l}
=\sum_{n=0}^{\infty}b_{m}B_{ml}
</math>
</center>
and
<center>
<math>
-\hat{k_{0}}A_{0}\delta_{0l}+a_{l}\hat{k}_{l}A_l
=-\sum_{m=0}^{\infty}b_{m}\hat{\kappa}_{m}B_{ml}
</math>
</center>
and these are solved exactly as before.

== Matlab Code ==

A program to calculate the coefficients for the semi-infinite dock problems can be found here
{{semiinfinite_dock code}}

=== Additional code ===

This program requires
* {{free surface dispersion equation code}}

[[Category:Eigenfunction Matching Method]]

Eigenfunction Matching for a Semi-Infinite Dock

2010-03-16T20:21:34Z

Mike smith: seems more intuitive, previous result seen in block matrix below anyway

{{complete pages}}

== Introduction ==

This is one of the simplest problem in eigenfunction matching. It also is an easy
problem to understand the [[:Category:Wiener-Hopf|Wiener-Hopf]] and
[[:Category:Residue Calculus|Residue Calculus]]. The problems consists of a region to the left
with a free surface and a region to the right with a rigid surface through which
not flow is possible.
We begin with the simply problem when the waves are normally incident (so that
the problem is truly two-dimensional. We then consider the case when the waves are incident
at an angle. For the later we give the equations in slightly less detail.
The case of a [[Finite Dock]] is treated very similarly. The problem can also
be generalised to a [[ Eigenfunction Matching for Submerged Semi-Infinite Dock|Submerged Semi-Infinite Dock]]

[[Image:semiinfinite_dock.jpg|thumb|right|300px|Wave scattering by a semi-infinite dock]]

== Governing Equations ==

We begin with the [[Frequency Domain Problem]] for a dock which occupies
the region <math>x>0</math> (we assume <math>e^{i\omega t}</math> time dependence).
The water is assumed to have
constant finite depth <math>h</math> and the <math>z</math>-direction points vertically
upward with the water surface at <math>z=0</math> and the sea floor at <math>z=-h</math>. The
boundary value problem can therefore be expressed as
<center>
<math>
\Delta\phi=0, \,\, -h<z<0,
</math>
</center>
<center>
<math>
\partial_z\phi=0, \,\, z=-h,
</math>
</center>
<center><math>
\partial_z\phi=\alpha\phi, \,\, z=0,\,x<0,
</math></center>
<center>
<math>
\partial_z\phi=0, \,\, z=0,\,x>0,
</math>
</center>
We
must also apply the [[Sommerfeld Radiation Condition]]
as <math>|x|\rightarrow\infty</math>. This essentially implies
that the only wave at infinity is propagating away and at negative infinity there is a unit incident wave
and a wave propagating away.

== Solution Method ==

We use [http://en.wikipedia.org/wiki/Separation_of_Variables separation of variables] in the two regions, <math>x<0</math>
and <math>x>0</math>.

{{separation of variables in two dimensions}}

{{separation of variables for a free surface}}

{{separation of variables for a dock}}

{{free surface dock relations}}

=== Expansion of the potential ===

We need to apply some boundary conditions at plus and minus infinity,
where are essentially the the solution cannot grow. This means that we
only have the positive (or negative) roots of the dispersion equation.
However, it does not help us with the purely imaginary root. Here we
must use a different condition, essentially identifying one solution
as the incoming wave and the other as the outgoing wave.

Therefore the scattered potential (without the incident wave, which will
be added later) can
be expanded as
<center>
<math>
\phi(x,z)=\sum_{m=0}^{\infty}a_{m}e^{k_{m}x}\phi_{m}(z), \;\;x<0
</math>
</center>
and
<center>
<math>
\phi(x,z)=\sum_{m=0}^{\infty}b_{m}
e^{-\kappa_{m}x}\psi_{m}(z), \;\;x>0
</math>
</center>
where <math>a_{m}</math> and <math>b_{m}</math>
are the coefficients of the potential in the open water and
the dock covered region respectively.

{{incident potential for two dimensions}}

=== An infinite dimensional system of equations ===

The potential and its derivative must be continuous across the
transition from open water to the plate covered region. Therefore, the
potentials and their derivatives at <math>x=0</math> have to be equal.
We obtain
<center>
<math>
\phi_{0}\left( z\right) + \sum_{m=0}^{\infty}
a_{m} \phi_{m}\left( z\right)
=\sum_{m=0}^{\infty}b_{m}\psi_{m}(z)
</math>
</center>
and
<center>
<math>
-k_{0}\phi_{0}\left( z\right) +\sum
_{m=0}^{\infty} a_{m}k_{m}\phi_{m}\left( z\right)
=-\sum_{m=0}^{\infty}b_{m}\kappa_{m}\psi
_{m}(z)
</math>
</center>
for each <math>m</math>.
We solve these equations by multiplying both equations by
<math>\phi_{l}(z) \,</math> and integrating from <math>-h</math> to <math>0</math> to obtain:
<center>
<math>
A_{0}\delta_{0l}+a_{l}A_{l}
=\sum_{m=0}^{\infty}b_{m}B_{ml}\,\,\,(3)
</math>
</center>
and
<center>
<math>
-k_{0}A_{0}\delta_{0l}+a_{l}k_{l}A_l
=-\sum_{m=0}^{\infty}b_{m}\kappa_{m}B_{ml} \,\,\,(4)
</math>
</center>
If we mutiply equation (3) by <math>k_l \,</math> and subtract equation (4)
we obtain
<center>
<math>
(k_{0}+k_l)A_{0}\delta_{0l}
=\sum_{m=0}^{\infty}b_{m}(k_l + \kappa_{m})B_{ml} \,\,\,(5)
</math>
</center>
This equation gives the required equations to solve for the
coefficients of the water velocity potential in the dock covered region.

== Numerical Solution ==

To solve the system of equations (10) we set the upper limit of <math>l</math> to
be <math>M</math>. In terms of matrix, we obtain
<center>
<math>
\begin{bmatrix}
\begin{bmatrix}
A_0&0 \quad \cdots&0\\
0&&\\
\vdots&A_l\delta_{0l}&\vdots\\
&&0\\
0&\cdots \quad 0 &A_M
\end{bmatrix}
&
\begin{bmatrix}
-B_{00}&\cdots&-B_{0M}\\
&&\\
\vdots&-B_{ml}&\vdots\\
&&\\
-B_{M0}&\cdots&-B_{MM}
\end{bmatrix}
\\
\begin{bmatrix}
0&\cdots&0\\
\vdots&\ddots&\vdots\\
0&\cdots&0
\end{bmatrix}
&
\begin{bmatrix}
(k_0 + \kappa_0) \, B_{00}&\cdots&(k_M + \kappa_{0}) \, B_{0M}\\
&&\\
\vdots&(k_l + \kappa_{m}) \, B_{ml}&\vdots\\
&&\\
(k_0 + \kappa_M) \, B_{M0}&\cdots&(k_M + \kappa_{M}) \, B_{MM}\\
\end{bmatrix}
\end{bmatrix}
\begin{bmatrix}
a_{0} \\
a_{1} \\
\vdots \\
a_M \\
\\
b_{0}\\
b_1 \\
\vdots \\
\\
b_M
\end{bmatrix}
=
\begin{bmatrix}
- A_{0} \\
0 \\
\vdots \\
\\
0 \\
\\
2k_{0}A_{0} \\
0 \\
\vdots \\
\\
0
\end{bmatrix}
</math>
</center>
We then simply need to solve the linear system of equations.

== Solution with Waves Incident at an Angle ==

We can consider the problem when the waves are incident at an angle <math>\theta</math>.

{{incident angle}}

Therefore the potential can
be expanded as
<center>
<math>
\phi(x,z)=e^{-\hat{k}_0x}\phi_0(z)+\sum_{m=0}^{\infty}a_{m}e^{\hat{k}_{m}x}\phi_{m}(z), \;\;x<0
</math>
</center>
and
<center>
<math>
\phi(x,z)=\sum_{m=0}^{\infty}b_{m}
e^{-\hat{\kappa}_{m}x}\psi_{m}(z), \;\;x>0
</math>
</center>
where <math>\hat{k}_{m} = \sqrt{k_m^2 - k_y^2}</math> and <math>\hat{\kappa}_{m} = \sqrt{\kappa_m^2 - k_y^2}</math>
where we always take the positive real root or the root with positive imaginary part.

The equations are derived almost identically to those above and we obtain
<center>
<math>
A_{0}\delta_{0l}+a_{l}A_{l}
=\sum_{n=0}^{\infty}b_{m}B_{ml}
</math>
</center>
and
<center>
<math>
-\hat{k_{0}}A_{0}\delta_{0l}+a_{l}\hat{k}_{l}A_l
=-\sum_{m=0}^{\infty}b_{m}\hat{\kappa}_{m}B_{ml}
</math>
</center>
and these are solved exactly as before.

== Matlab Code ==

A program to calculate the coefficients for the semi-infinite dock problems can be found here
{{semiinfinite_dock code}}

=== Additional code ===

This program requires
* {{free surface dispersion equation code}}

[[Category:Eigenfunction Matching Method]]

Eigenfunction Matching for a Semi-Infinite Dock

2010-03-16T20:02:45Z

Mike smith: /* An infinite dimensional system of equations */

{{complete pages}}

== Introduction ==

This is one of the simplest problem in eigenfunction matching. It also is an easy
problem to understand the [[:Category:Wiener-Hopf|Wiener-Hopf]] and
[[:Category:Residue Calculus|Residue Calculus]]. The problems consists of a region to the left
with a free surface and a region to the right with a rigid surface through which
not flow is possible.
We begin with the simply problem when the waves are normally incident (so that
the problem is truly two-dimensional. We then consider the case when the waves are incident
at an angle. For the later we give the equations in slightly less detail.
The case of a [[Finite Dock]] is treated very similarly. The problem can also
be generalised to a [[ Eigenfunction Matching for Submerged Semi-Infinite Dock|Submerged Semi-Infinite Dock]]

[[Image:semiinfinite_dock.jpg|thumb|right|300px|Wave scattering by a semi-infinite dock]]

== Governing Equations ==

We begin with the [[Frequency Domain Problem]] for a dock which occupies
the region <math>x>0</math> (we assume <math>e^{i\omega t}</math> time dependence).
The water is assumed to have
constant finite depth <math>h</math> and the <math>z</math>-direction points vertically
upward with the water surface at <math>z=0</math> and the sea floor at <math>z=-h</math>. The
boundary value problem can therefore be expressed as
<center>
<math>
\Delta\phi=0, \,\, -h<z<0,
</math>
</center>
<center>
<math>
\partial_z\phi=0, \,\, z=-h,
</math>
</center>
<center><math>
\partial_z\phi=\alpha\phi, \,\, z=0,\,x<0,
</math></center>
<center>
<math>
\partial_z\phi=0, \,\, z=0,\,x>0,
</math>
</center>
We
must also apply the [[Sommerfeld Radiation Condition]]
as <math>|x|\rightarrow\infty</math>. This essentially implies
that the only wave at infinity is propagating away and at negative infinity there is a unit incident wave
and a wave propagating away.

== Solution Method ==

We use [http://en.wikipedia.org/wiki/Separation_of_Variables separation of variables] in the two regions, <math>x<0</math>
and <math>x>0</math>.

{{separation of variables in two dimensions}}

{{separation of variables for a free surface}}

{{separation of variables for a dock}}

{{free surface dock relations}}

=== Expansion of the potential ===

We need to apply some boundary conditions at plus and minus infinity,
where are essentially the the solution cannot grow. This means that we
only have the positive (or negative) roots of the dispersion equation.
However, it does not help us with the purely imaginary root. Here we
must use a different condition, essentially identifying one solution
as the incoming wave and the other as the outgoing wave.

Therefore the scattered potential (without the incident wave, which will
be added later) can
be expanded as
<center>
<math>
\phi(x,z)=\sum_{m=0}^{\infty}a_{m}e^{k_{m}x}\phi_{m}(z), \;\;x<0
</math>
</center>
and
<center>
<math>
\phi(x,z)=\sum_{m=0}^{\infty}b_{m}
e^{-\kappa_{m}x}\psi_{m}(z), \;\;x>0
</math>
</center>
where <math>a_{m}</math> and <math>b_{m}</math>
are the coefficients of the potential in the open water and
the dock covered region respectively.

{{incident potential for two dimensions}}

=== An infinite dimensional system of equations ===

The potential and its derivative must be continuous across the
transition from open water to the plate covered region. Therefore, the
potentials and their derivatives at <math>x=0</math> have to be equal.
We obtain
<center>
<math>
\phi_{0}\left( z\right) + \sum_{m=0}^{\infty}
a_{m} \phi_{m}\left( z\right)
=\sum_{m=0}^{\infty}b_{m}\psi_{m}(z)
</math>
</center>
and
<center>
<math>
-k_{0}\phi_{0}\left( z\right) +\sum
_{m=0}^{\infty} a_{m}k_{m}\phi_{m}\left( z\right)
=-\sum_{m=0}^{\infty}b_{m}\kappa_{m}\psi
_{m}(z)
</math>
</center>
for each <math>m</math>.
We solve these equations by multiplying both equations by
<math>\phi_{l}(z) \,</math> and integrating from <math>-h</math> to <math>0</math> to obtain:
<center>
<math>
A_{0}\delta_{0l}+a_{l}A_{l}
=\sum_{m=0}^{\infty}b_{m}B_{ml}\,\,\,(3)
</math>
</center>
and
<center>
<math>
-k_{0}A_{0}\delta_{0l}+a_{l}k_{l}A_l
=-\sum_{m=0}^{\infty}b_{m}\kappa_{m}B_{ml} \,\,\,(4)
</math>
</center>
If we mutiply equation (3) by <math>k_l \,</math> and subtract equation (4)
we obtain
<center>
<math>
2k_{0}A_{0}\delta_{0l}
=\sum_{m=0}^{\infty}b_{m}(k_l + \kappa_{m})B_{ml}
</math>
</center>
This equation gives the required equations to solve for the
coefficients of the water velocity potential in the dock covered region.

== Numerical Solution ==

To solve the system of equations (10) we set the upper limit of <math>l</math> to
be <math>M</math>. In terms of matrix, we obtain
<center>
<math>
\begin{bmatrix}
\begin{bmatrix}
A_0&0 \quad \cdots&0\\
0&&\\
\vdots&A_l\delta_{0l}&\vdots\\
&&0\\
0&\cdots \quad 0 &A_M
\end{bmatrix}
&
\begin{bmatrix}
-B_{00}&\cdots&-B_{0M}\\
&&\\
\vdots&-B_{ml}&\vdots\\
&&\\
-B_{M0}&\cdots&-B_{MM}
\end{bmatrix}
\\
\begin{bmatrix}
0&\cdots&0\\
\vdots&\ddots&\vdots\\
0&\cdots&0
\end{bmatrix}
&
\begin{bmatrix}
(k_0 + \kappa_0) \, B_{00}&\cdots&(k_M + \kappa_{0}) \, B_{0M}\\
&&\\
\vdots&(k_l + \kappa_{m}) \, B_{ml}&\vdots\\
&&\\
(k_0 + \kappa_M) \, B_{M0}&\cdots&(k_M + \kappa_{M}) \, B_{MM}\\
\end{bmatrix}
\end{bmatrix}
\begin{bmatrix}
a_{0} \\
a_{1} \\
\vdots \\
a_M \\
\\
b_{0}\\
b_1 \\
\vdots \\
\\
b_M
\end{bmatrix}
=
\begin{bmatrix}
- A_{0} \\
0 \\
\vdots \\
\\
0 \\
\\
2k_{0}A_{0} \\
0 \\
\vdots \\
\\
0
\end{bmatrix}
</math>
</center>
We then simply need to solve the linear system of equations.

== Solution with Waves Incident at an Angle ==

We can consider the problem when the waves are incident at an angle <math>\theta</math>.

{{incident angle}}

Therefore the potential can
be expanded as
<center>
<math>
\phi(x,z)=e^{-\hat{k}_0x}\phi_0(z)+\sum_{m=0}^{\infty}a_{m}e^{\hat{k}_{m}x}\phi_{m}(z), \;\;x<0
</math>
</center>
and
<center>
<math>
\phi(x,z)=\sum_{m=0}^{\infty}b_{m}
e^{-\hat{\kappa}_{m}x}\psi_{m}(z), \;\;x>0
</math>
</center>
where <math>\hat{k}_{m} = \sqrt{k_m^2 - k_y^2}</math> and <math>\hat{\kappa}_{m} = \sqrt{\kappa_m^2 - k_y^2}</math>
where we always take the positive real root or the root with positive imaginary part.

The equations are derived almost identically to those above and we obtain
<center>
<math>
A_{0}\delta_{0l}+a_{l}A_{l}
=\sum_{n=0}^{\infty}b_{m}B_{ml}
</math>
</center>
and
<center>
<math>
-\hat{k_{0}}A_{0}\delta_{0l}+a_{l}\hat{k}_{l}A_l
=-\sum_{m=0}^{\infty}b_{m}\hat{\kappa}_{m}B_{ml}
</math>
</center>
and these are solved exactly as before.

== Matlab Code ==

A program to calculate the coefficients for the semi-infinite dock problems can be found here
{{semiinfinite_dock code}}

=== Additional code ===

This program requires
* {{free surface dispersion equation code}}

[[Category:Eigenfunction Matching Method]]