WikiWaves - User contributions [en]

Template:Incident potential for two dimensions

2012-03-21T21:24:10Z

Tertius: /* Incident potential */

===Incident potential===

To create meaningful solutions of the velocity potential <math>\phi</math> in the specified domains we add an incident wave term to the expansion for the domain of <math>x < 0</math> above. The incident potential is a wave of amplitude <math>A</math>
in displacement travelling in the positive <math>x</math>-direction. We would only see this in the time domain <math>\Phi(x,z,t)</math> however, in the frequency domain the incident potential can be written as
<center>
<math>
\phi_{\mathrm{I}}(x,z) =e^{-k_{0}x}\chi_{0}\left(
z\right).
</math>
</center>
The total velocity (scattered) potential now becomes <math>\phi = \phi_{\mathrm{I}} + \phi_{\mathrm{D}}</math> for the domain of <math>x < 0</math>.

The first term in the expansion of the diffracted potential for the domain <math>x < 0</math> is given by
<center>
<math>
a_{0}e^{k_{0}x}\chi_{0}\left(
z\right)
</math>
</center>
which represents the reflected wave.

In any scattering problem <math>|R|^2 + |T|^2 = 1</math> where <math>R</math> and <math>T</math> are the reflection and transmission coefficients respectively. In our case of the semi-infinite dock <math>|a_{0}| = |R| = 1</math> and <math>|T| = 0</math> as there are no transmitted waves in the region under the dock.

Eigenfunction Matching for a Finite Rectangle using Symmetry

2012-03-19T01:18:48Z

Tertius: /* An infinite dimensional system of equations */

{{incomplete pages}}

== Introduction ==

The problem consists of a finite rigid fixed rectangle of length <math>2L</math> centered at <math>x = 0</math> and having a height of <math>d + ...</math> freeboard with depth of submergence being <math>d</math> through which no flow is possible. The regions to the left and right of the rectangle are the free water surface. Since the geometry of the rectangle is symmetric about <math>x = 0</math> we can decompose the velocity potential <math>\phi</math> into a symmetric <math>\phi^s</math> and antisymmetric <math>\phi^a</math> potential. We then solve for these two potentials in the left domain only <math>x \leq 0</math> and subsequently recompose the solution <math>\phi</math> for the entire domain. We begin with the simple problem when the waves are normally incident on the rectangle (dock) (so that the problem is truly two-dimensional). We can extend this solution to a pair of partially submerged finite rigid fixed rectangles using symmetry
[[Eigenfunction Matching for a Finite Rectangle using Symmetry]]

== Governing Equations ==

We begin with the [[Frequency Domain Problem]] for a partially submerged finite dock. Although the dock is of length <math>2L</math> we are only interested in the dock of <math>-L \leq x \leq 0</math> together with the fluid domain <math>x \leq 0</math>. We assume <math>e^{i\omega t}</math> time dependence and
the depth of submergence is <math>d</math>.
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>
\begin{align}
\Delta\phi^{s,a} &= 0, \,\,\, -h\leq z\leq 0, \,\, x < -L, \\

\Delta\phi^{s,a} &= 0, \,\,\, -h\leq z\leq -d, \,\, -L < x < 0. \\

\partial_z\phi^{s,a} &= \alpha\phi^{s,a}, \,\,\, z=0, \,\, x < -L, \\

\partial_x\phi^{s,a} &= 0, \,\,\, -d\leq z\leq 0, \,\, x = -L, \\

\partial_z\phi^{s,a} &= 0, \,\,\,
\begin{cases}
z = -d, \,\, -L < x < 0, \\
z = -h, \,\, x < 0.
\end{cases} \\

\partial_x\phi^s &= 0, \,\,\, x = 0, \,\, -h \leq z \leq -d, \\

\phi^a &= 0, \,\,\, x = 0, \,\, -h \leq z \leq -d.
\end{align}
</math></center>

Where <math>\alpha = \omega^2/g</math>.

<math>\phi^s(x,z)</math> is an even function and at <math>x = 0</math> the derivative of this function is equal to zero. <math>\phi^a(x,z)</math> is an odd function and at <math>x = 0</math> this function is equal to zero. These boundary conditions are important and enable the recomposed solution <math>\phi(x,z)</math> to be reflected into the domain of <math>x \geq 0</math>.

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\leq 0</math>
and <math>x\geq 0</math>.

{{separation of variables in two dimensions}}

{{separation of variables for a free surface}}

=== Separation of Variables for a Dock ===

The separation of variables equation for a partially submerged dock is given by:
<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>
Z^{\prime} (-d) = 0.
</math></center>
The solution is
<math>k=\kappa_{m}= \frac{m\pi}{h-d} \,</math> , <math>m\in\mathbb{N}\cup\left\{0\right\}</math> and
<center><math>
Z = \psi_{m}\left( z\right) = \cos\kappa_{m}(z+h),\quad
m\in\mathbb{N}\cup\left\{0\right\}.</math></center>
We note that
<center><math>
\int\nolimits_{-h}^{-d}\psi_{m}(z)\psi_{n}(z) \mathrm{d} z=C_{m}\delta_{mn},
</math></center>
where
<center>
<math>
C_{m} = \frac{1}{2} \left( \frac{\cos\kappa_{m}(h-d)\sin\kappa_{m}(h-d)+\kappa_{m}(h-d)}{\kappa_{m}} \right).
</math>
</center>
In addition there is the special case of when <math>m=n=0</math> ,
<center>
<math>
C_{0} = h-d,
</math>
</center>
where <math>\delta_{00}=1</math> .

=== Inner product between free surface and dock modes ===

<center><math>
B_{mn} = \int\nolimits_{-h}^{-d}\psi_{m}(z)\chi_{n}(z) \mathrm{d} z , \quad m\geq1, \quad n\geq0,
</math></center>
where
<center><math>
B_{mn}= \int\nolimits_{-h}^{-d} \frac{\cos\kappa_{m}(z+h)\cos k_{n}(z+h)}{\cos(k_{n}h)} \mathrm{d} z=\frac{-(-1)^m k_{n}\sin k_{n}(h-d)}
{\cos(k_{n}h)(\kappa_{m}^{2}-k_{n}^{2})}.
</math></center>
The evaluation of this inner product works on the assumption that the two roots <math>\kappa_{m}</math> and <math>k_{n}</math> do not coincide. The Matlab code given below has a routine to check for this case and make necessary adjustments.

We also have the case where the functions of the inner product are switched.
<center><math>
B_{nm} = \int\nolimits_{-h}^{-d}\chi_{m}(z)\psi_{n}(z) \mathrm{d} z , \quad m\geq0, \quad n\geq0.
</math></center>
For <math>m \geq 0</math> and <math>n \geq 1</math> the evaluation is above with the exception that the indices of the expression are switched.

Finally there is the special case of when <math>m \geq 0</math> and <math>n = 0</math>.
<center><math>
B_{0m} = \frac{\sin k_{m}(h-d)}{k_{m}\cos(k_{m}h)}.
</math></center>

=== Expansion of the potential ===

We need to apply some boundary conditions at plus and minus infinity,
where essentially 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 roots. Here we
must use a different condition, essentially identifying one solution
as the incoming wave and the other as the outgoing wave.

Therefore the diffracted potentials can
be expanded as
<center>
<math>
\phi(x,z)=\sum_{m=0}^{\infty}a_{m}e^{k_{m}x}\chi_{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 '''symmetric''' and '''antisymmetric''' potentials <math>\phi^{s,a}(x,z) \,</math> must be continuous across the
transition from open water to the dock covered region. Therefore, the
potentials at <math>x=-L</math> have to be equal.
For these cases we obtain:

<center><math>
\chi_{0}\left( z\right) + \sum_{m=0}^{\infty}a^{s,a}_{m} \chi_{m}\left(z\right)
=\sum_{m=0}^{\infty}b^{s,a}_{m}\psi_{m}(z), \quad -h \leq z\leq -d.
</math></center>

We multiply both sides by <math>\psi_{n}(z) \,</math> and integrate from <math>-h</math> to <math>-d</math> to obtain:

<center><math>
B_{n0} + \sum^{\infty}_{m=0}a^{s,a}_{m}B_{nm} = b^{s,a}_{m}C_{m}\delta_{mn}, \quad n\in\mathbb{N}\cup\left\{0\right\}.
</math></center>

The derivative of the '''symmetric''' potential <math>\partial_x\phi^{s}(x,z) \,</math> must be continuous across the
transition from open water to the dock covered region. Therefore, the derivatives of the potentials at <math>x=-L</math> have to be equal.
For this case we obtain:

<center><math>
\begin{align}
-k_{0}\chi_{0}\left( z\right) +\sum_{m=0}^{\infty} a^{s}_{m}k_{m}\chi_{m}\left(z\right) =
\begin{cases}
\quad \quad \quad \quad \quad 0,\quad \quad \quad \quad \quad -d\leq z\leq 0 \\
-\sum_{m=1}^{\infty}b^{s}_{m}\kappa_{m}\tanh(\kappa_{m}L)\psi_{m}(z),\quad -h\leq z\leq -d.
\end{cases}
\end{align}
</math></center>

The derivative of the '''antisymmetric''' potential <math>\partial_x\phi^{a}(x,z) \,</math> must be continuous across the
transition from open water to the dock covered region. Therefore, the derivatives of the potentials at <math>x=-L</math> have to be equal.
For this case we obtain:

<center><math>
\begin{align}
-k_{0}\chi_{0}\left( z\right) +\sum_{m=0}^{\infty} a^{a}_{m}k_{m}\chi_{m}\left(z\right) =
\begin{cases}
\quad \quad \quad \quad \quad 0,\quad \quad \quad \quad \quad \quad \quad \quad -d\leq z\leq 0 \\
-\frac{1}{L}b^a_{0} -\sum_{m=1}^{\infty}b^{a}_{m}\kappa_{m}\coth(\kappa_{m}L)\psi_{m}(z),\quad -h\leq z\leq -d.
\end{cases}
\end{align}
</math></center>

For the first equation we multiply both sides by <math>\chi_{n}(z) \,</math> and integrating from <math>-h</math> to <math>0</math> to obtain:
<center><math>
-k_{0}A_{0}\delta_{0n} + k_{m}a_{m}A_{m} = -\sum^{\infty}_{m=0}\kappa_{m}b_{m}B_{mn}, \quad n\in\mathbb{N}\cup\left\{0\right\}
</math></center>

Solving the equations above will yield the coefficients of the water velocity potential in the dock covered region.

<center><math>
D_{n} = \int\nolimits_{-h}^{-d}\chi_{n}(z) \mathrm{d} z = \int\nolimits_{-h}^{-d} \frac{\cos k_{n}(z+h)}{\cos k_{n}h} \mathrm{d} z=\frac{\sin k_{n}(h-d)}
{k_{n}\cos k_{n}h}.
</math></center>

== Numerical Solution ==

To solve the system of equations for the '''symmetric''' case, we set the upper limit of <math>m</math> to be <math>N</math>. This resulting system can be expressed in the block matrix form below,

<center>
<math>
\begin{bmatrix}
B_{00} & B_{01} & \cdots & B_{0N} & -C_{0} & 0 & \cdots & 0 \\
B_{10} & B_{11} & \cdots & B_{1N} & 0 & -C_{1} & \cdots & \vdots \\
\vdots & \vdots & \ddots & \vdots & \vdots & \vdots & \ddots & 0 \\
B_{N0} & B_{N1} & \cdots & B_{NN} & 0 & \cdots & 0 & -C_{N} \\
k_{0}A_{0} & 0 & \cdots & 0 & 0 & \kappa_{1}\tanh(\kappa_{1}L)B_{10} & \cdots & \kappa_{N}\tanh(\kappa_{N}L)B_{N0} \\
0 & k_{1}A_{1} & \cdots & \vdots & 0 & \kappa_{1}\tanh(\kappa_{1}L)B_{11} & \cdots & \kappa_{N}\tanh(\kappa_{N}L)B_{N1} \\
\vdots & \vdots & \ddots & 0 & \vdots & \vdots & \ddots & \vdots \\
0 & \cdots & 0 & k_{N}A_{N} & 0 & \kappa_{1}\tanh(\kappa_{1}L)B_{1N} & \cdots & \kappa_{N}\tanh(\kappa_{N}L)B_{NN} \\
\end{bmatrix}
\begin{bmatrix}
a^s_{0} \\ a^s_{1} \\ \vdots \\ a^s_{N} \\ b^s_{0} \\ b^s_{1} \\ \vdots \\ b^s_{N} \\
\end{bmatrix}
=\begin{bmatrix}
-B_{00} \\ -B_{10} \\ \vdots \\ -B_{N0} \\ k_{0}A_{0} \\ 0 \\ \vdots \\ 0 \\
\end{bmatrix}
</math>
</center>

To solve the system of equations for the '''antisymmetric''' case, we set the upper limit of <math>m</math> to be <math>N</math>. This resulting system can be expressed in the block matrix form below,

<center>
<math>
\begin{bmatrix}
B_{00} & B_{01} & \cdots & B_{0N} & -C_{0} & 0 & \cdots & 0 \\
B_{10} & B_{11} & \cdots & B_{1N} & 0 & -C_{1} & \cdots & \vdots \\
\vdots & \vdots & \ddots & \vdots & \vdots & \vdots & \ddots & 0 \\
B_{N0} & B_{N1} & \cdots & B_{NN} & 0 & \cdots & 0 & -C_{N} \\
k_{0}A_{0} & 0 & \cdots & 0 & \frac{1}{L}D_{0} & \kappa_{1}\coth(\kappa_{1}L)B_{10} & \cdots & \kappa_{N}\coth(\kappa_{N}L)B_{N0} \\
0 & k_{1}A_{1} & \cdots & \vdots & \frac{1}{L}D_{1} & \kappa_{1}\coth(\kappa_{1}L)B_{11} & \cdots & \kappa_{N}\coth(\kappa_{N}L)B_{N1} \\
\vdots & \vdots & \ddots & 0 & \vdots & \vdots & \ddots & \vdots \\
0 & \cdots & 0 & k_{N}A_{N} & \frac{1}{L}D_{N} & \kappa_{1}\coth(\kappa_{1}L)B_{1N} & \cdots & \kappa_{N}\coth(\kappa_{N}L)B_{NN} \\
\end{bmatrix}
\begin{bmatrix}
a^a_{0} \\ a^a_{1} \\ \vdots \\ a^a_{N} \\ b^a_{0} \\ b^a_{1} \\ \vdots \\ b^a_{N} \\
\end{bmatrix}
=\begin{bmatrix}
-B_{00} \\ -B_{10} \\ \vdots \\ -B_{N0} \\ k_{0}A_{0} \\ 0 \\ \vdots \\ 0 \\
\end{bmatrix}
</math>
</center>

We then simply need to solve the two sets of <math>2(N+1) \mbox{ x } 2(N+1)</math> linear systems of equations. If we had not taken advantage of the symmetry we would need to solve a <math>4(N+1) \mbox{ x } 4(N+1)</math> linear system of equations for four coefficients <math>a_{m}</math>, <math>b_{m}</math>, <math>c_{m}</math> and <math>d_{m}</math>.

== Matlab Code ==

A program to calculate the coefficients for the single finite dock problem using symmetry 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 Finite Rectangle using Symmetry

2012-03-16T06:37:03Z

Tertius: /* An infinite dimensional system of equations */

{{incomplete pages}}

== Introduction ==

The problem consists of a finite rigid fixed rectangle of length <math>2L</math> centered at <math>x = 0</math> and having a height of <math>d + ...</math> freeboard with depth of submergence being <math>d</math> through which no flow is possible. The regions to the left and right of the rectangle are the free water surface. Since the geometry of the rectangle is symmetric about <math>x = 0</math> we can decompose the velocity potential <math>\phi</math> into a symmetric <math>\phi^s</math> and antisymmetric <math>\phi^a</math> potential. We then solve for these two potentials in the left domain only <math>x \leq 0</math> and subsequently recompose the solution <math>\phi</math> for the entire domain. We begin with the simple problem when the waves are normally incident on the rectangle (dock) (so that the problem is truly two-dimensional). We can extend this solution to a pair of partially submerged finite rigid fixed rectangles using symmetry
[[Eigenfunction Matching for a Finite Rectangle using Symmetry]]

== Governing Equations ==

We begin with the [[Frequency Domain Problem]] for a partially submerged finite dock. Although the dock is of length <math>2L</math> we are only interested in the dock of <math>-L \leq x \leq 0</math> together with the fluid domain <math>x \leq 0</math>. We assume <math>e^{i\omega t}</math> time dependence and
the depth of submergence is <math>d</math>.
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>
\begin{align}
\Delta\phi^{s,a} &= 0, \,\,\, -h\leq z\leq 0, \,\, x < -L, \\

\Delta\phi^{s,a} &= 0, \,\,\, -h\leq z\leq -d, \,\, -L < x < 0. \\

\partial_z\phi^{s,a} &= \alpha\phi^{s,a}, \,\,\, z=0, \,\, x < -L, \\

\partial_x\phi^{s,a} &= 0, \,\,\, -d\leq z\leq 0, \,\, x = -L, \\

\partial_z\phi^{s,a} &= 0, \,\,\,
\begin{cases}
z = -d, \,\, -L < x < 0, \\
z = -h, \,\, x < 0.
\end{cases} \\

\partial_x\phi^s &= 0, \,\,\, x = 0, \,\, -h \leq z \leq -d, \\

\phi^a &= 0, \,\,\, x = 0, \,\, -h \leq z \leq -d.
\end{align}
</math></center>

Where <math>\alpha = \omega^2/g</math>.

<math>\phi^s(x,z)</math> is an even function and at <math>x = 0</math> the derivative of this function is equal to zero. <math>\phi^a(x,z)</math> is an odd function and at <math>x = 0</math> this function is equal to zero. These boundary conditions are important and enable the recomposed solution <math>\phi(x,z)</math> to be reflected into the domain of <math>x \geq 0</math>.

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\leq 0</math>
and <math>x\geq 0</math>.

{{separation of variables in two dimensions}}

{{separation of variables for a free surface}}

=== Separation of Variables for a Dock ===

The separation of variables equation for a partially submerged dock is given by:
<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>
Z^{\prime} (-d) = 0.
</math></center>
The solution is
<math>k=\kappa_{m}= \frac{m\pi}{h-d} \,</math> , <math>m\in\mathbb{N}\cup\left\{0\right\}</math> and
<center><math>
Z = \psi_{m}\left( z\right) = \cos\kappa_{m}(z+h),\quad
m\in\mathbb{N}\cup\left\{0\right\}.</math></center>
We note that
<center><math>
\int\nolimits_{-h}^{-d}\psi_{m}(z)\psi_{n}(z) \mathrm{d} z=C_{m}\delta_{mn},
</math></center>
where
<center>
<math>
C_{m} = \frac{1}{2} \left( \frac{\cos\kappa_{m}(h-d)\sin\kappa_{m}(h-d)+\kappa_{m}(h-d)}{\kappa_{m}} \right).
</math>
</center>
In addition there is the special case of when <math>m=n=0</math> ,
<center>
<math>
C_{0} = h-d,
</math>
</center>
where <math>\delta_{00}=1</math> .

=== Inner product between free surface and dock modes ===

<center><math>
B_{mn} = \int\nolimits_{-h}^{-d}\psi_{m}(z)\chi_{n}(z) \mathrm{d} z , \quad m\geq1, \quad n\geq0,
</math></center>
where
<center><math>
B_{mn}= \int\nolimits_{-h}^{-d} \frac{\cos\kappa_{m}(z+h)\cos k_{n}(z+h)}{\cos(k_{n}h)} \mathrm{d} z=\frac{-(-1)^m k_{n}\sin k_{n}(h-d)}
{\cos(k_{n}h)(\kappa_{m}^{2}-k_{n}^{2})}.
</math></center>
The evaluation of this inner product works on the assumption that the two roots <math>\kappa_{m}</math> and <math>k_{n}</math> do not coincide. The Matlab code given below has a routine to check for this case and make necessary adjustments.

We also have the case where the functions of the inner product are switched.
<center><math>
B_{nm} = \int\nolimits_{-h}^{-d}\chi_{m}(z)\psi_{n}(z) \mathrm{d} z , \quad m\geq0, \quad n\geq0.
</math></center>
For <math>m \geq 0</math> and <math>n \geq 1</math> the evaluation is above with the exception that the indices of the expression are switched.

Finally there is the special case of when <math>m \geq 0</math> and <math>n = 0</math>.
<center><math>
B_{0m} = \frac{\sin k_{m}(h-d)}{k_{m}\cos(k_{m}h)}.
</math></center>

=== Expansion of the potential ===

We need to apply some boundary conditions at plus and minus infinity,
where essentially 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 roots. Here we
must use a different condition, essentially identifying one solution
as the incoming wave and the other as the outgoing wave.

Therefore the diffracted potentials can
be expanded as
<center>
<math>
\phi(x,z)=\sum_{m=0}^{\infty}a_{m}e^{k_{m}x}\chi_{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 '''symmetric''' and '''antisymmetric''' potentials <math>\phi^{s,a}(x,z) \,</math> must be continuous across the
transition from open water to the dock covered region. Therefore, the
potentials at <math>x=-L</math> have to be equal.
For these cases we obtain:

<center><math>
\chi_{0}\left( z\right) + \sum_{m=0}^{\infty}a^{s,a}_{m} \chi_{m}\left(z\right)
=\sum_{m=0}^{\infty}b^{s,a}_{m}\psi_{m}(z), \quad -h \leq z\leq -d.
</math></center>

We multiply both sides by <math>\psi_{n}(z) \,</math> and integrate from <math>-h</math> to <math>-d</math> to obtain:

<center><math>
B_{n0} + \sum^{\infty}_{m=0}a^{s,a}_{m}B_{nm} = b^{s,a}_{m}C_{m}\delta_{mn}, \quad n\in\mathbb{N}\cup\left\{0\right\}.
</math></center>

The derivative of the '''symmetric''' potential <math>\partial_x\phi^{s}(x,z) \,</math> must be continuous across the
transition from open water to the dock covered region. Therefore, the derivatives of the potentials at <math>x=-L</math> have to be equal.
For this case we obtain:

<center><math>
\begin{align}
-k_{0}\chi_{0}\left( z\right) +\sum_{m=0}^{\infty} a^{s}_{m}k_{m}\chi_{m}\left(z\right) =
\begin{cases}
\quad \quad \quad \quad \quad 0,\quad \quad \quad \quad \quad \quad -d\leq z\leq 0 \\
-\sum_{m=1}^{\infty}b^{s}_{m}\kappa_{m}\tanh(\kappa_{m}L)\psi_{m}(z),\quad -h\leq z\leq -d.
\end{cases}
\end{align}
</math></center>

The derivative of the '''antisymmetric''' potential <math>\partial_x\phi^{a}(x,z) \,</math> must be continuous across the
transition from open water to the dock covered region. Therefore, the derivatives of the potentials at <math>x=-L</math> have to be equal.
For this case we obtain:

<center><math>
\begin{align}
-k_{0}\chi_{0}\left( z\right) +\sum_{m=0}^{\infty} a^{a}_{m}k_{m}\chi_{m}\left(z\right) =
\begin{cases}
\quad \quad \quad \quad \quad 0,\quad \quad \quad \quad \quad \quad \quad \quad -d\leq z\leq 0 \\
-\frac{1}{L}b^a_{0} -\sum_{m=1}^{\infty}b^{a}_{m}\kappa_{m}\coth(\kappa_{m}L)\psi_{m}(z),\quad -h\leq z\leq -d.
\end{cases}
\end{align}
</math></center>

For the first equation we multiply both sides by <math>\chi_{n}(z) \,</math> and integrating from <math>-h</math> to <math>0</math> to obtain:
<center><math>
-k_{0}A_{0}\delta_{0n} + k_{m}a_{m}A_{m} = -\sum^{\infty}_{m=0}\kappa_{m}b_{m}B_{mn}, \quad n\in\mathbb{N}\cup\left\{0\right\}
</math></center>

Solving the equations above will yield the coefficients of the water velocity potential in the dock covered region.

<center><math>
D_{n} = \int\nolimits_{-h}^{-d}\chi_{n}(z) \mathrm{d} z = \int\nolimits_{-h}^{-d} \frac{\cos k_{n}(z+h)}{\cos k_{n}h} \mathrm{d} z=\frac{\sin k_{n}(h-d)}
{k_{n}\cos k_{n}h}.
</math></center>

== Numerical Solution ==

To solve the system of equations for the '''symmetric''' case, we set the upper limit of <math>m</math> to be <math>N</math>. This resulting system can be expressed in the block matrix form below,

<center>
<math>
\begin{bmatrix}
B_{00} & B_{01} & \cdots & B_{0N} & -C_{0} & 0 & \cdots & 0 \\
B_{10} & B_{11} & \cdots & B_{1N} & 0 & -C_{1} & \cdots & \vdots \\
\vdots & \vdots & \ddots & \vdots & \vdots & \vdots & \ddots & 0 \\
B_{N0} & B_{N1} & \cdots & B_{NN} & 0 & \cdots & 0 & -C_{N} \\
k_{0}A_{0} & 0 & \cdots & 0 & 0 & \kappa_{1}\tanh(\kappa_{1}L)B_{10} & \cdots & \kappa_{N}\tanh(\kappa_{N}L)B_{N0} \\
0 & k_{1}A_{1} & \cdots & \vdots & 0 & \kappa_{1}\tanh(\kappa_{1}L)B_{11} & \cdots & \kappa_{N}\tanh(\kappa_{N}L)B_{N1} \\
\vdots & \vdots & \ddots & 0 & \vdots & \vdots & \ddots & \vdots \\
0 & \cdots & 0 & k_{N}A_{N} & 0 & \kappa_{1}\tanh(\kappa_{1}L)B_{1N} & \cdots & \kappa_{N}\tanh(\kappa_{N}L)B_{NN} \\
\end{bmatrix}
\begin{bmatrix}
a^s_{0} \\ a^s_{1} \\ \vdots \\ a^s_{N} \\ b^s_{0} \\ b^s_{1} \\ \vdots \\ b^s_{N} \\
\end{bmatrix}
=\begin{bmatrix}
-B_{00} \\ -B_{10} \\ \vdots \\ -B_{N0} \\ k_{0}A_{0} \\ 0 \\ \vdots \\ 0 \\
\end{bmatrix}
</math>
</center>

To solve the system of equations for the '''antisymmetric''' case, we set the upper limit of <math>m</math> to be <math>N</math>. This resulting system can be expressed in the block matrix form below,

<center>
<math>
\begin{bmatrix}
B_{00} & B_{01} & \cdots & B_{0N} & -C_{0} & 0 & \cdots & 0 \\
B_{10} & B_{11} & \cdots & B_{1N} & 0 & -C_{1} & \cdots & \vdots \\
\vdots & \vdots & \ddots & \vdots & \vdots & \vdots & \ddots & 0 \\
B_{N0} & B_{N1} & \cdots & B_{NN} & 0 & \cdots & 0 & -C_{N} \\
k_{0}A_{0} & 0 & \cdots & 0 & \frac{1}{L}D_{0} & \kappa_{1}\coth(\kappa_{1}L)B_{10} & \cdots & \kappa_{N}\coth(\kappa_{N}L)B_{N0} \\
0 & k_{1}A_{1} & \cdots & \vdots & \frac{1}{L}D_{1} & \kappa_{1}\coth(\kappa_{1}L)B_{11} & \cdots & \kappa_{N}\coth(\kappa_{N}L)B_{N1} \\
\vdots & \vdots & \ddots & 0 & \vdots & \vdots & \ddots & \vdots \\
0 & \cdots & 0 & k_{N}A_{N} & \frac{1}{L}D_{N} & \kappa_{1}\coth(\kappa_{1}L)B_{1N} & \cdots & \kappa_{N}\coth(\kappa_{N}L)B_{NN} \\
\end{bmatrix}
\begin{bmatrix}
a^a_{0} \\ a^a_{1} \\ \vdots \\ a^a_{N} \\ b^a_{0} \\ b^a_{1} \\ \vdots \\ b^a_{N} \\
\end{bmatrix}
=\begin{bmatrix}
-B_{00} \\ -B_{10} \\ \vdots \\ -B_{N0} \\ k_{0}A_{0} \\ 0 \\ \vdots \\ 0 \\
\end{bmatrix}
</math>
</center>

We then simply need to solve the two sets of <math>2(N+1) \mbox{ x } 2(N+1)</math> linear systems of equations. If we had not taken advantage of the symmetry we would need to solve a <math>4(N+1) \mbox{ x } 4(N+1)</math> linear system of equations for four coefficients <math>a_{m}</math>, <math>b_{m}</math>, <math>c_{m}</math> and <math>d_{m}</math>.

== Matlab Code ==

A program to calculate the coefficients for the single finite dock problem using symmetry 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 Finite Rectangle using Symmetry

2012-03-16T06:36:19Z

Tertius: /* An infinite dimensional system of equations */

{{incomplete pages}}

== Introduction ==

The problem consists of a finite rigid fixed rectangle of length <math>2L</math> centered at <math>x = 0</math> and having a height of <math>d + ...</math> freeboard with depth of submergence being <math>d</math> through which no flow is possible. The regions to the left and right of the rectangle are the free water surface. Since the geometry of the rectangle is symmetric about <math>x = 0</math> we can decompose the velocity potential <math>\phi</math> into a symmetric <math>\phi^s</math> and antisymmetric <math>\phi^a</math> potential. We then solve for these two potentials in the left domain only <math>x \leq 0</math> and subsequently recompose the solution <math>\phi</math> for the entire domain. We begin with the simple problem when the waves are normally incident on the rectangle (dock) (so that the problem is truly two-dimensional). We can extend this solution to a pair of partially submerged finite rigid fixed rectangles using symmetry
[[Eigenfunction Matching for a Finite Rectangle using Symmetry]]

== Governing Equations ==

We begin with the [[Frequency Domain Problem]] for a partially submerged finite dock. Although the dock is of length <math>2L</math> we are only interested in the dock of <math>-L \leq x \leq 0</math> together with the fluid domain <math>x \leq 0</math>. We assume <math>e^{i\omega t}</math> time dependence and
the depth of submergence is <math>d</math>.
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>
\begin{align}
\Delta\phi^{s,a} &= 0, \,\,\, -h\leq z\leq 0, \,\, x < -L, \\

\Delta\phi^{s,a} &= 0, \,\,\, -h\leq z\leq -d, \,\, -L < x < 0. \\

\partial_z\phi^{s,a} &= \alpha\phi^{s,a}, \,\,\, z=0, \,\, x < -L, \\

\partial_x\phi^{s,a} &= 0, \,\,\, -d\leq z\leq 0, \,\, x = -L, \\

\partial_z\phi^{s,a} &= 0, \,\,\,
\begin{cases}
z = -d, \,\, -L < x < 0, \\
z = -h, \,\, x < 0.
\end{cases} \\

\partial_x\phi^s &= 0, \,\,\, x = 0, \,\, -h \leq z \leq -d, \\

\phi^a &= 0, \,\,\, x = 0, \,\, -h \leq z \leq -d.
\end{align}
</math></center>

Where <math>\alpha = \omega^2/g</math>.

<math>\phi^s(x,z)</math> is an even function and at <math>x = 0</math> the derivative of this function is equal to zero. <math>\phi^a(x,z)</math> is an odd function and at <math>x = 0</math> this function is equal to zero. These boundary conditions are important and enable the recomposed solution <math>\phi(x,z)</math> to be reflected into the domain of <math>x \geq 0</math>.

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\leq 0</math>
and <math>x\geq 0</math>.

{{separation of variables in two dimensions}}

{{separation of variables for a free surface}}

=== Separation of Variables for a Dock ===

The separation of variables equation for a partially submerged dock is given by:
<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>
Z^{\prime} (-d) = 0.
</math></center>
The solution is
<math>k=\kappa_{m}= \frac{m\pi}{h-d} \,</math> , <math>m\in\mathbb{N}\cup\left\{0\right\}</math> and
<center><math>
Z = \psi_{m}\left( z\right) = \cos\kappa_{m}(z+h),\quad
m\in\mathbb{N}\cup\left\{0\right\}.</math></center>
We note that
<center><math>
\int\nolimits_{-h}^{-d}\psi_{m}(z)\psi_{n}(z) \mathrm{d} z=C_{m}\delta_{mn},
</math></center>
where
<center>
<math>
C_{m} = \frac{1}{2} \left( \frac{\cos\kappa_{m}(h-d)\sin\kappa_{m}(h-d)+\kappa_{m}(h-d)}{\kappa_{m}} \right).
</math>
</center>
In addition there is the special case of when <math>m=n=0</math> ,
<center>
<math>
C_{0} = h-d,
</math>
</center>
where <math>\delta_{00}=1</math> .

=== Inner product between free surface and dock modes ===

<center><math>
B_{mn} = \int\nolimits_{-h}^{-d}\psi_{m}(z)\chi_{n}(z) \mathrm{d} z , \quad m\geq1, \quad n\geq0,
</math></center>
where
<center><math>
B_{mn}= \int\nolimits_{-h}^{-d} \frac{\cos\kappa_{m}(z+h)\cos k_{n}(z+h)}{\cos(k_{n}h)} \mathrm{d} z=\frac{-(-1)^m k_{n}\sin k_{n}(h-d)}
{\cos(k_{n}h)(\kappa_{m}^{2}-k_{n}^{2})}.
</math></center>
The evaluation of this inner product works on the assumption that the two roots <math>\kappa_{m}</math> and <math>k_{n}</math> do not coincide. The Matlab code given below has a routine to check for this case and make necessary adjustments.

We also have the case where the functions of the inner product are switched.
<center><math>
B_{nm} = \int\nolimits_{-h}^{-d}\chi_{m}(z)\psi_{n}(z) \mathrm{d} z , \quad m\geq0, \quad n\geq0.
</math></center>
For <math>m \geq 0</math> and <math>n \geq 1</math> the evaluation is above with the exception that the indices of the expression are switched.

Finally there is the special case of when <math>m \geq 0</math> and <math>n = 0</math>.
<center><math>
B_{0m} = \frac{\sin k_{m}(h-d)}{k_{m}\cos(k_{m}h)}.
</math></center>

=== Expansion of the potential ===

We need to apply some boundary conditions at plus and minus infinity,
where essentially 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 roots. Here we
must use a different condition, essentially identifying one solution
as the incoming wave and the other as the outgoing wave.

Therefore the diffracted potentials can
be expanded as
<center>
<math>
\phi(x,z)=\sum_{m=0}^{\infty}a_{m}e^{k_{m}x}\chi_{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 '''symmetric''' and '''antisymmetric''' potentials <math>\phi^{s,a}(x,z) \,</math> must be continuous across the
transition from open water to the dock covered region. Therefore, the
potentials at <math>x=-L</math> have to be equal.
For these cases we obtain:

<center><math>
\chi_{0}\left( z\right) + \sum_{m=0}^{\infty}a^{s,a}_{m} \chi_{m}\left(z\right)
=\sum_{m=0}^{\infty}b^{s,a}_{m}\psi_{m}(z), \quad -h \leq z\leq -d.
</math></center>

We multiply both sides by <math>\psi_{n}(z) \,</math> and integrate from <math>-h</math> to <math>-d</math> to obtain:

<center><math>
B_{n0} + \sum^{\infty}_{m=0}a^{s,a}_{m}B_{nm} = b^{s,a}_{m}C_{m}\delta_{mn}, \quad n\in\mathbb{N}\cup\left\{0\right\}.
</math></center>

The derivative of the '''symmetric''' potential <math>\partial_x\phi^{s}(x,z) \,</math> must be continuous across the
transition from open water to the dock covered region. Therefore, the derivatives of the potentials at <math>x=-L</math> have to be equal.
For this case we obtain:

<center><math>
\begin{align}
-k_{0}\chi_{0}\left( z\right) +\sum_{m=0}^{\infty} a^{s}_{m}k_{m}\chi_{m}\left(z\right) =
\begin{cases}
\quad \quad \quad \quad \quad 0,\quad \quad \quad \quad \quad \quad -d\leq z\leq 0 \\
-\sum_{m=1}^{\infty}b^{s}_{m}\kappa_{m}\tanh(\kappa_{m}L)\psi_{m}(z),\quad -h\leq z\leq -d.
\end{cases}
\end{align}
</math></center>

The derivative of the '''antisymmetric''' potential <math>\partial_x\phi^{a}(x,z) \,</math> must be continuous across the
transition from open water to the dock covered region. Therefore, the derivatives of the potentials at <math>x=-L</math> have to be equal.
For this case we obtain:

<center><math>
\begin{align}
-k_{0}\chi_{0}\left( z\right) +\sum_{m=0}^{\infty} a^{a}_{m}k_{m}\chi_{m}\left(z\right) =
\begin{cases}
\quad \quad \quad \quad \quad 0,\quad \quad \quad \quad \quad \quad \quad \quad -d\leq z\leq 0 \\
-\frac{1}{L}b^a_{0} -\sum_{m=1}^{\infty}b^{a}_{m}\kappa_{m}\coth(\kappa_{m}L)\psi_{m}(z),\quad -h\leq z\leq -d.
\end{cases}
\end{align}
</math></center>

For the first equation we multiply both sides by <math>\chi_{n}(z) \,</math> and integrating from <math>-h</math> to <math>0</math> to obtain:
<center><math>
-k_{0}A_{0}\delta_{0n} + k_{m}a_{m}A_{m} = -\sum^{\infty}_{m=0}\kappa_{m}b_{m}B_{mn}, \quad n\in\mathbb{N}\cup\left\{0\right\}
</math></center>

Solving the equations above will yield the coefficients of the water velocity potential in the dock covered region.

<center><math>
D_{n} = \int\nolimits_{-h}^{-d}\chi_{n}(z) \mathrm{d} z = \int\nolimits_{-h}^{-d} \frac{\cos k_{n}(z+h)}{\cos k_{n}h} \mathrm{d} z=\frac{\sin k_{n}(h-d)}
{k_{n}\cos k_{n}}.
</math></center>

== Numerical Solution ==

To solve the system of equations for the '''symmetric''' case, we set the upper limit of <math>m</math> to be <math>N</math>. This resulting system can be expressed in the block matrix form below,

<center>
<math>
\begin{bmatrix}
B_{00} & B_{01} & \cdots & B_{0N} & -C_{0} & 0 & \cdots & 0 \\
B_{10} & B_{11} & \cdots & B_{1N} & 0 & -C_{1} & \cdots & \vdots \\
\vdots & \vdots & \ddots & \vdots & \vdots & \vdots & \ddots & 0 \\
B_{N0} & B_{N1} & \cdots & B_{NN} & 0 & \cdots & 0 & -C_{N} \\
k_{0}A_{0} & 0 & \cdots & 0 & 0 & \kappa_{1}\tanh(\kappa_{1}L)B_{10} & \cdots & \kappa_{N}\tanh(\kappa_{N}L)B_{N0} \\
0 & k_{1}A_{1} & \cdots & \vdots & 0 & \kappa_{1}\tanh(\kappa_{1}L)B_{11} & \cdots & \kappa_{N}\tanh(\kappa_{N}L)B_{N1} \\
\vdots & \vdots & \ddots & 0 & \vdots & \vdots & \ddots & \vdots \\
0 & \cdots & 0 & k_{N}A_{N} & 0 & \kappa_{1}\tanh(\kappa_{1}L)B_{1N} & \cdots & \kappa_{N}\tanh(\kappa_{N}L)B_{NN} \\
\end{bmatrix}
\begin{bmatrix}
a^s_{0} \\ a^s_{1} \\ \vdots \\ a^s_{N} \\ b^s_{0} \\ b^s_{1} \\ \vdots \\ b^s_{N} \\
\end{bmatrix}
=\begin{bmatrix}
-B_{00} \\ -B_{10} \\ \vdots \\ -B_{N0} \\ k_{0}A_{0} \\ 0 \\ \vdots \\ 0 \\
\end{bmatrix}
</math>
</center>

To solve the system of equations for the '''antisymmetric''' case, we set the upper limit of <math>m</math> to be <math>N</math>. This resulting system can be expressed in the block matrix form below,

<center>
<math>
\begin{bmatrix}
B_{00} & B_{01} & \cdots & B_{0N} & -C_{0} & 0 & \cdots & 0 \\
B_{10} & B_{11} & \cdots & B_{1N} & 0 & -C_{1} & \cdots & \vdots \\
\vdots & \vdots & \ddots & \vdots & \vdots & \vdots & \ddots & 0 \\
B_{N0} & B_{N1} & \cdots & B_{NN} & 0 & \cdots & 0 & -C_{N} \\
k_{0}A_{0} & 0 & \cdots & 0 & \frac{1}{L}D_{0} & \kappa_{1}\coth(\kappa_{1}L)B_{10} & \cdots & \kappa_{N}\coth(\kappa_{N}L)B_{N0} \\
0 & k_{1}A_{1} & \cdots & \vdots & \frac{1}{L}D_{1} & \kappa_{1}\coth(\kappa_{1}L)B_{11} & \cdots & \kappa_{N}\coth(\kappa_{N}L)B_{N1} \\
\vdots & \vdots & \ddots & 0 & \vdots & \vdots & \ddots & \vdots \\
0 & \cdots & 0 & k_{N}A_{N} & \frac{1}{L}D_{N} & \kappa_{1}\coth(\kappa_{1}L)B_{1N} & \cdots & \kappa_{N}\coth(\kappa_{N}L)B_{NN} \\
\end{bmatrix}
\begin{bmatrix}
a^a_{0} \\ a^a_{1} \\ \vdots \\ a^a_{N} \\ b^a_{0} \\ b^a_{1} \\ \vdots \\ b^a_{N} \\
\end{bmatrix}
=\begin{bmatrix}
-B_{00} \\ -B_{10} \\ \vdots \\ -B_{N0} \\ k_{0}A_{0} \\ 0 \\ \vdots \\ 0 \\
\end{bmatrix}
</math>
</center>

We then simply need to solve the two sets of <math>2(N+1) \mbox{ x } 2(N+1)</math> linear systems of equations. If we had not taken advantage of the symmetry we would need to solve a <math>4(N+1) \mbox{ x } 4(N+1)</math> linear system of equations for four coefficients <math>a_{m}</math>, <math>b_{m}</math>, <math>c_{m}</math> and <math>d_{m}</math>.

== Matlab Code ==

A program to calculate the coefficients for the single finite dock problem using symmetry 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 Finite Rectangle using Symmetry

2012-03-16T06:33:51Z

Tertius: /* An infinite dimensional system of equations */

{{incomplete pages}}

== Introduction ==

The problem consists of a finite rigid fixed rectangle of length <math>2L</math> centered at <math>x = 0</math> and having a height of <math>d + ...</math> freeboard with depth of submergence being <math>d</math> through which no flow is possible. The regions to the left and right of the rectangle are the free water surface. Since the geometry of the rectangle is symmetric about <math>x = 0</math> we can decompose the velocity potential <math>\phi</math> into a symmetric <math>\phi^s</math> and antisymmetric <math>\phi^a</math> potential. We then solve for these two potentials in the left domain only <math>x \leq 0</math> and subsequently recompose the solution <math>\phi</math> for the entire domain. We begin with the simple problem when the waves are normally incident on the rectangle (dock) (so that the problem is truly two-dimensional). We can extend this solution to a pair of partially submerged finite rigid fixed rectangles using symmetry
[[Eigenfunction Matching for a Finite Rectangle using Symmetry]]

== Governing Equations ==

We begin with the [[Frequency Domain Problem]] for a partially submerged finite dock. Although the dock is of length <math>2L</math> we are only interested in the dock of <math>-L \leq x \leq 0</math> together with the fluid domain <math>x \leq 0</math>. We assume <math>e^{i\omega t}</math> time dependence and
the depth of submergence is <math>d</math>.
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>
\begin{align}
\Delta\phi^{s,a} &= 0, \,\,\, -h\leq z\leq 0, \,\, x < -L, \\

\Delta\phi^{s,a} &= 0, \,\,\, -h\leq z\leq -d, \,\, -L < x < 0. \\

\partial_z\phi^{s,a} &= \alpha\phi^{s,a}, \,\,\, z=0, \,\, x < -L, \\

\partial_x\phi^{s,a} &= 0, \,\,\, -d\leq z\leq 0, \,\, x = -L, \\

\partial_z\phi^{s,a} &= 0, \,\,\,
\begin{cases}
z = -d, \,\, -L < x < 0, \\
z = -h, \,\, x < 0.
\end{cases} \\

\partial_x\phi^s &= 0, \,\,\, x = 0, \,\, -h \leq z \leq -d, \\

\phi^a &= 0, \,\,\, x = 0, \,\, -h \leq z \leq -d.
\end{align}
</math></center>

Where <math>\alpha = \omega^2/g</math>.

<math>\phi^s(x,z)</math> is an even function and at <math>x = 0</math> the derivative of this function is equal to zero. <math>\phi^a(x,z)</math> is an odd function and at <math>x = 0</math> this function is equal to zero. These boundary conditions are important and enable the recomposed solution <math>\phi(x,z)</math> to be reflected into the domain of <math>x \geq 0</math>.

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\leq 0</math>
and <math>x\geq 0</math>.

{{separation of variables in two dimensions}}

{{separation of variables for a free surface}}

=== Separation of Variables for a Dock ===

The separation of variables equation for a partially submerged dock is given by:
<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>
Z^{\prime} (-d) = 0.
</math></center>
The solution is
<math>k=\kappa_{m}= \frac{m\pi}{h-d} \,</math> , <math>m\in\mathbb{N}\cup\left\{0\right\}</math> and
<center><math>
Z = \psi_{m}\left( z\right) = \cos\kappa_{m}(z+h),\quad
m\in\mathbb{N}\cup\left\{0\right\}.</math></center>
We note that
<center><math>
\int\nolimits_{-h}^{-d}\psi_{m}(z)\psi_{n}(z) \mathrm{d} z=C_{m}\delta_{mn},
</math></center>
where
<center>
<math>
C_{m} = \frac{1}{2} \left( \frac{\cos\kappa_{m}(h-d)\sin\kappa_{m}(h-d)+\kappa_{m}(h-d)}{\kappa_{m}} \right).
</math>
</center>
In addition there is the special case of when <math>m=n=0</math> ,
<center>
<math>
C_{0} = h-d,
</math>
</center>
where <math>\delta_{00}=1</math> .

=== Inner product between free surface and dock modes ===

<center><math>
B_{mn} = \int\nolimits_{-h}^{-d}\psi_{m}(z)\chi_{n}(z) \mathrm{d} z , \quad m\geq1, \quad n\geq0,
</math></center>
where
<center><math>
B_{mn}= \int\nolimits_{-h}^{-d} \frac{\cos\kappa_{m}(z+h)\cos k_{n}(z+h)}{\cos(k_{n}h)} \mathrm{d} z=\frac{-(-1)^m k_{n}\sin k_{n}(h-d)}
{\cos(k_{n}h)(\kappa_{m}^{2}-k_{n}^{2})}.
</math></center>
The evaluation of this inner product works on the assumption that the two roots <math>\kappa_{m}</math> and <math>k_{n}</math> do not coincide. The Matlab code given below has a routine to check for this case and make necessary adjustments.

We also have the case where the functions of the inner product are switched.
<center><math>
B_{nm} = \int\nolimits_{-h}^{-d}\chi_{m}(z)\psi_{n}(z) \mathrm{d} z , \quad m\geq0, \quad n\geq0.
</math></center>
For <math>m \geq 0</math> and <math>n \geq 1</math> the evaluation is above with the exception that the indices of the expression are switched.

Finally there is the special case of when <math>m \geq 0</math> and <math>n = 0</math>.
<center><math>
B_{0m} = \frac{\sin k_{m}(h-d)}{k_{m}\cos(k_{m}h)}.
</math></center>

=== Expansion of the potential ===

We need to apply some boundary conditions at plus and minus infinity,
where essentially 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 roots. Here we
must use a different condition, essentially identifying one solution
as the incoming wave and the other as the outgoing wave.

Therefore the diffracted potentials can
be expanded as
<center>
<math>
\phi(x,z)=\sum_{m=0}^{\infty}a_{m}e^{k_{m}x}\chi_{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 '''symmetric''' and '''antisymmetric''' potentials <math>\phi^{s,a}(x,z) \,</math> must be continuous across the
transition from open water to the dock covered region. Therefore, the
potentials at <math>x=-L</math> have to be equal.
For these cases we obtain:

<center><math>
\chi_{0}\left( z\right) + \sum_{m=0}^{\infty}a^{s,a}_{m} \chi_{m}\left(z\right)
=\sum_{m=0}^{\infty}b^{s,a}_{m}\psi_{m}(z), \quad -h \leq z\leq -d.
</math></center>

We multiply both sides by <math>\psi_{n}(z) \,</math> and integrate from <math>-h</math> to <math>-d</math> to obtain:

<center><math>
B_{n0} + \sum^{\infty}_{m=0}a^{s,a}_{m}B_{nm} = b^{s,a}_{m}C_{m}\delta_{mn}, \quad n\in\mathbb{N}\cup\left\{0\right\}.
</math></center>

The derivative of the '''symmetric''' potential <math>\partial_x\phi^{s}(x,z) \,</math> must be continuous across the
transition from open water to the dock covered region. Therefore, the derivatives of the potentials at <math>x=-L</math> have to be equal.
For this case we obtain:

<center><math>
\begin{align}
-k_{0}\chi_{0}\left( z\right) +\sum_{m=0}^{\infty} a^{s}_{m}k_{m}\chi_{m}\left(z\right) =
\begin{cases}
\quad \quad \quad \quad \quad 0,\quad \quad \quad \quad \quad \quad -d\leq z\leq 0 \\
-\sum_{m=1}^{\infty}b^{s}_{m}\kappa_{m}\tanh(\kappa_{m}L)\psi_{m}(z),\quad -h\leq z\leq -d.
\end{cases}
\end{align}
</math></center>

The derivative of the '''antisymmetric''' potential <math>\partial_x\phi^{a}(x,z) \,</math> must be continuous across the
transition from open water to the dock covered region. Therefore, the derivatives of the potentials at <math>x=-L</math> have to be equal.
For this case we obtain:

<center><math>
\begin{align}
-k_{0}\chi_{0}\left( z\right) +\sum_{m=0}^{\infty} a^{a}_{m}k_{m}\chi_{m}\left(z\right) =
\begin{cases}
\quad \quad \quad \quad \quad 0,\quad \quad \quad \quad \quad \quad \quad \quad -d\leq z\leq 0 \\
-\frac{1}{L}b^a_{0} -\sum_{m=1}^{\infty}b^{a}_{m}\kappa_{m}\coth(\kappa_{m}L)\psi_{m}(z),\quad -h\leq z\leq -d.
\end{cases}
\end{align}
</math></center>

For the first equation we multiply both sides by <math>\chi_{n}(z) \,</math> and integrating from <math>-h</math> to <math>0</math> to obtain:
<center><math>
-k_{0}A_{0}\delta_{0n} + k_{m}a_{m}A_{m} = -\sum^{\infty}_{m=0}\kappa_{m}b_{m}B_{mn}, \quad n\in\mathbb{N}\cup\left\{0\right\}
</math></center>

Solving the equations above will yield the coefficients of the water velocity potential in the dock covered region.

<center><math>
D_{n} = \int\nolimits_{-h}^{-d}\chi_{n}(z) \mathrm{d} z = \int\nolimits_{-h}^{-d} \frac{\cos k_{n}(z+h)}{\cos k_{n}h} \mathrm{d} z=\frac{-(-1)^m k_{n}\sin k_{n}(h-d)}
{\cos(k_{n}h)(\kappa_{m}^{2}-k_{n}^{2})}.
</math></center>

== Numerical Solution ==

To solve the system of equations for the '''symmetric''' case, we set the upper limit of <math>m</math> to be <math>N</math>. This resulting system can be expressed in the block matrix form below,

<center>
<math>
\begin{bmatrix}
B_{00} & B_{01} & \cdots & B_{0N} & -C_{0} & 0 & \cdots & 0 \\
B_{10} & B_{11} & \cdots & B_{1N} & 0 & -C_{1} & \cdots & \vdots \\
\vdots & \vdots & \ddots & \vdots & \vdots & \vdots & \ddots & 0 \\
B_{N0} & B_{N1} & \cdots & B_{NN} & 0 & \cdots & 0 & -C_{N} \\
k_{0}A_{0} & 0 & \cdots & 0 & 0 & \kappa_{1}\tanh(\kappa_{1}L)B_{10} & \cdots & \kappa_{N}\tanh(\kappa_{N}L)B_{N0} \\
0 & k_{1}A_{1} & \cdots & \vdots & 0 & \kappa_{1}\tanh(\kappa_{1}L)B_{11} & \cdots & \kappa_{N}\tanh(\kappa_{N}L)B_{N1} \\
\vdots & \vdots & \ddots & 0 & \vdots & \vdots & \ddots & \vdots \\
0 & \cdots & 0 & k_{N}A_{N} & 0 & \kappa_{1}\tanh(\kappa_{1}L)B_{1N} & \cdots & \kappa_{N}\tanh(\kappa_{N}L)B_{NN} \\
\end{bmatrix}
\begin{bmatrix}
a^s_{0} \\ a^s_{1} \\ \vdots \\ a^s_{N} \\ b^s_{0} \\ b^s_{1} \\ \vdots \\ b^s_{N} \\
\end{bmatrix}
=\begin{bmatrix}
-B_{00} \\ -B_{10} \\ \vdots \\ -B_{N0} \\ k_{0}A_{0} \\ 0 \\ \vdots \\ 0 \\
\end{bmatrix}
</math>
</center>

To solve the system of equations for the '''antisymmetric''' case, we set the upper limit of <math>m</math> to be <math>N</math>. This resulting system can be expressed in the block matrix form below,

<center>
<math>
\begin{bmatrix}
B_{00} & B_{01} & \cdots & B_{0N} & -C_{0} & 0 & \cdots & 0 \\
B_{10} & B_{11} & \cdots & B_{1N} & 0 & -C_{1} & \cdots & \vdots \\
\vdots & \vdots & \ddots & \vdots & \vdots & \vdots & \ddots & 0 \\
B_{N0} & B_{N1} & \cdots & B_{NN} & 0 & \cdots & 0 & -C_{N} \\
k_{0}A_{0} & 0 & \cdots & 0 & \frac{1}{L}D_{0} & \kappa_{1}\coth(\kappa_{1}L)B_{10} & \cdots & \kappa_{N}\coth(\kappa_{N}L)B_{N0} \\
0 & k_{1}A_{1} & \cdots & \vdots & \frac{1}{L}D_{1} & \kappa_{1}\coth(\kappa_{1}L)B_{11} & \cdots & \kappa_{N}\coth(\kappa_{N}L)B_{N1} \\
\vdots & \vdots & \ddots & 0 & \vdots & \vdots & \ddots & \vdots \\
0 & \cdots & 0 & k_{N}A_{N} & \frac{1}{L}D_{N} & \kappa_{1}\coth(\kappa_{1}L)B_{1N} & \cdots & \kappa_{N}\coth(\kappa_{N}L)B_{NN} \\
\end{bmatrix}
\begin{bmatrix}
a^a_{0} \\ a^a_{1} \\ \vdots \\ a^a_{N} \\ b^a_{0} \\ b^a_{1} \\ \vdots \\ b^a_{N} \\
\end{bmatrix}
=\begin{bmatrix}
-B_{00} \\ -B_{10} \\ \vdots \\ -B_{N0} \\ k_{0}A_{0} \\ 0 \\ \vdots \\ 0 \\
\end{bmatrix}
</math>
</center>

We then simply need to solve the two sets of <math>2(N+1) \mbox{ x } 2(N+1)</math> linear systems of equations. If we had not taken advantage of the symmetry we would need to solve a <math>4(N+1) \mbox{ x } 4(N+1)</math> linear system of equations for four coefficients <math>a_{m}</math>, <math>b_{m}</math>, <math>c_{m}</math> and <math>d_{m}</math>.

== Matlab Code ==

A program to calculate the coefficients for the single finite dock problem using symmetry 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 Finite Rectangle using Symmetry

2012-03-16T06:22:13Z

Tertius: /* An infinite dimensional system of equations */

{{incomplete pages}}

== Introduction ==

The problem consists of a finite rigid fixed rectangle of length <math>2L</math> centered at <math>x = 0</math> and having a height of <math>d + ...</math> freeboard with depth of submergence being <math>d</math> through which no flow is possible. The regions to the left and right of the rectangle are the free water surface. Since the geometry of the rectangle is symmetric about <math>x = 0</math> we can decompose the velocity potential <math>\phi</math> into a symmetric <math>\phi^s</math> and antisymmetric <math>\phi^a</math> potential. We then solve for these two potentials in the left domain only <math>x \leq 0</math> and subsequently recompose the solution <math>\phi</math> for the entire domain. We begin with the simple problem when the waves are normally incident on the rectangle (dock) (so that the problem is truly two-dimensional). We can extend this solution to a pair of partially submerged finite rigid fixed rectangles using symmetry
[[Eigenfunction Matching for a Finite Rectangle using Symmetry]]

== Governing Equations ==

We begin with the [[Frequency Domain Problem]] for a partially submerged finite dock. Although the dock is of length <math>2L</math> we are only interested in the dock of <math>-L \leq x \leq 0</math> together with the fluid domain <math>x \leq 0</math>. We assume <math>e^{i\omega t}</math> time dependence and
the depth of submergence is <math>d</math>.
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>
\begin{align}
\Delta\phi^{s,a} &= 0, \,\,\, -h\leq z\leq 0, \,\, x < -L, \\

\Delta\phi^{s,a} &= 0, \,\,\, -h\leq z\leq -d, \,\, -L < x < 0. \\

\partial_z\phi^{s,a} &= \alpha\phi^{s,a}, \,\,\, z=0, \,\, x < -L, \\

\partial_x\phi^{s,a} &= 0, \,\,\, -d\leq z\leq 0, \,\, x = -L, \\

\partial_z\phi^{s,a} &= 0, \,\,\,
\begin{cases}
z = -d, \,\, -L < x < 0, \\
z = -h, \,\, x < 0.
\end{cases} \\

\partial_x\phi^s &= 0, \,\,\, x = 0, \,\, -h \leq z \leq -d, \\

\phi^a &= 0, \,\,\, x = 0, \,\, -h \leq z \leq -d.
\end{align}
</math></center>

Where <math>\alpha = \omega^2/g</math>.

<math>\phi^s(x,z)</math> is an even function and at <math>x = 0</math> the derivative of this function is equal to zero. <math>\phi^a(x,z)</math> is an odd function and at <math>x = 0</math> this function is equal to zero. These boundary conditions are important and enable the recomposed solution <math>\phi(x,z)</math> to be reflected into the domain of <math>x \geq 0</math>.

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\leq 0</math>
and <math>x\geq 0</math>.

{{separation of variables in two dimensions}}

{{separation of variables for a free surface}}

=== Separation of Variables for a Dock ===

The separation of variables equation for a partially submerged dock is given by:
<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>
Z^{\prime} (-d) = 0.
</math></center>
The solution is
<math>k=\kappa_{m}= \frac{m\pi}{h-d} \,</math> , <math>m\in\mathbb{N}\cup\left\{0\right\}</math> and
<center><math>
Z = \psi_{m}\left( z\right) = \cos\kappa_{m}(z+h),\quad
m\in\mathbb{N}\cup\left\{0\right\}.</math></center>
We note that
<center><math>
\int\nolimits_{-h}^{-d}\psi_{m}(z)\psi_{n}(z) \mathrm{d} z=C_{m}\delta_{mn},
</math></center>
where
<center>
<math>
C_{m} = \frac{1}{2} \left( \frac{\cos\kappa_{m}(h-d)\sin\kappa_{m}(h-d)+\kappa_{m}(h-d)}{\kappa_{m}} \right).
</math>
</center>
In addition there is the special case of when <math>m=n=0</math> ,
<center>
<math>
C_{0} = h-d,
</math>
</center>
where <math>\delta_{00}=1</math> .

=== Inner product between free surface and dock modes ===

<center><math>
B_{mn} = \int\nolimits_{-h}^{-d}\psi_{m}(z)\chi_{n}(z) \mathrm{d} z , \quad m\geq1, \quad n\geq0,
</math></center>
where
<center><math>
B_{mn}= \int\nolimits_{-h}^{-d} \frac{\cos\kappa_{m}(z+h)\cos k_{n}(z+h)}{\cos(k_{n}h)} \mathrm{d} z=\frac{-(-1)^m k_{n}\sin k_{n}(h-d)}
{\cos(k_{n}h)(\kappa_{m}^{2}-k_{n}^{2})}.
</math></center>
The evaluation of this inner product works on the assumption that the two roots <math>\kappa_{m}</math> and <math>k_{n}</math> do not coincide. The Matlab code given below has a routine to check for this case and make necessary adjustments.

We also have the case where the functions of the inner product are switched.
<center><math>
B_{nm} = \int\nolimits_{-h}^{-d}\chi_{m}(z)\psi_{n}(z) \mathrm{d} z , \quad m\geq0, \quad n\geq0.
</math></center>
For <math>m \geq 0</math> and <math>n \geq 1</math> the evaluation is above with the exception that the indices of the expression are switched.

Finally there is the special case of when <math>m \geq 0</math> and <math>n = 0</math>.
<center><math>
B_{0m} = \frac{\sin k_{m}(h-d)}{k_{m}\cos(k_{m}h)}.
</math></center>

=== Expansion of the potential ===

We need to apply some boundary conditions at plus and minus infinity,
where essentially 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 roots. Here we
must use a different condition, essentially identifying one solution
as the incoming wave and the other as the outgoing wave.

Therefore the diffracted potentials can
be expanded as
<center>
<math>
\phi(x,z)=\sum_{m=0}^{\infty}a_{m}e^{k_{m}x}\chi_{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 '''symmetric''' and '''antisymmetric''' potentials <math>\phi^{s,a}(x,z) \,</math> must be continuous across the
transition from open water to the dock covered region. Therefore, the
potentials at <math>x=-L</math> have to be equal.
For these cases we obtain:

<center><math>
\chi_{0}\left( z\right) + \sum_{m=0}^{\infty}a^{s,a}_{m} \chi_{m}\left(z\right)
=\sum_{m=0}^{\infty}b^{s,a}_{m}\psi_{m}(z), \quad -h \leq z\leq -d.
</math></center>

We multiply both sides by <math>\psi_{n}(z) \,</math> and integrate from <math>-h</math> to <math>-d</math> to obtain:

<center><math>
B_{n0} + \sum^{\infty}_{m=0}a^{s,a}_{m}B_{nm} = b^{s,a}_{m}C_{m}\delta_{mn}, \quad n\in\mathbb{N}\cup\left\{0\right\}.
</math></center>

The derivative of the '''symmetric''' potential <math>\partial_x\phi^{s}(x,z) \,</math> must be continuous across the
transition from open water to the dock covered region. Therefore, the derivatives of the potentials at <math>x=-L</math> have to be equal.
For this case we obtain:

<center><math>
\begin{align}
-k_{0}\chi_{0}\left( z\right) +\sum_{m=0}^{\infty} a^{s}_{m}k_{m}\chi_{m}\left(z\right) =
\begin{cases}
\quad \quad \quad \quad \quad 0,\quad \quad \quad \quad \quad \quad -d\leq z\leq 0 \\
-\sum_{m=1}^{\infty}b^{s}_{m}\kappa_{m}\tanh(\kappa_{m}L)\psi_{m}(z),\quad -h\leq z\leq -d.
\end{cases}
\end{align}
</math></center>

The derivative of the '''antisymmetric''' potential <math>\partial_x\phi^{a}(x,z) \,</math> must be continuous across the
transition from open water to the dock covered region. Therefore, the derivatives of the potentials at <math>x=-L</math> have to be equal.
For this case we obtain:

<center><math>
\begin{align}
-k_{0}\chi_{0}\left( z\right) +\sum_{m=0}^{\infty} a^{a}_{m}k_{m}\chi_{m}\left(z\right) =
\begin{cases}
\quad \quad \quad \quad \quad 0,\quad \quad \quad \quad \quad \quad \quad \quad -d\leq z\leq 0 \\
-\frac{1}{L}b^a_{0} -\sum_{m=1}^{\infty}b^{a}_{m}\kappa_{m}\coth(\kappa_{m}L)\psi_{m}(z),\quad -h\leq z\leq -d.
\end{cases}
\end{align}
</math></center>

For the first equation we multiply both sides by <math>\chi_{n}(z) \,</math> and integrating from <math>-h</math> to <math>0</math> to obtain:
<center><math>
-k_{0}A_{0}\delta_{0n} + k_{m}a_{m}A_{m} = -\sum^{\infty}_{m=0}\kappa_{m}b_{m}B_{mn}, \quad n\in\mathbb{N}\cup\left\{0\right\}
</math></center>

Solving the equations above will yield the coefficients of the water velocity potential in the dock covered region.

== Numerical Solution ==

To solve the system of equations for the '''symmetric''' case, we set the upper limit of <math>m</math> to be <math>N</math>. This resulting system can be expressed in the block matrix form below,

<center>
<math>
\begin{bmatrix}
B_{00} & B_{01} & \cdots & B_{0N} & -C_{0} & 0 & \cdots & 0 \\
B_{10} & B_{11} & \cdots & B_{1N} & 0 & -C_{1} & \cdots & \vdots \\
\vdots & \vdots & \ddots & \vdots & \vdots & \vdots & \ddots & 0 \\
B_{N0} & B_{N1} & \cdots & B_{NN} & 0 & \cdots & 0 & -C_{N} \\
k_{0}A_{0} & 0 & \cdots & 0 & 0 & \kappa_{1}\tanh(\kappa_{1}L)B_{10} & \cdots & \kappa_{N}\tanh(\kappa_{N}L)B_{N0} \\
0 & k_{1}A_{1} & \cdots & \vdots & 0 & \kappa_{1}\tanh(\kappa_{1}L)B_{11} & \cdots & \kappa_{N}\tanh(\kappa_{N}L)B_{N1} \\
\vdots & \vdots & \ddots & 0 & \vdots & \vdots & \ddots & \vdots \\
0 & \cdots & 0 & k_{N}A_{N} & 0 & \kappa_{1}\tanh(\kappa_{1}L)B_{1N} & \cdots & \kappa_{N}\tanh(\kappa_{N}L)B_{NN} \\
\end{bmatrix}
\begin{bmatrix}
a^s_{0} \\ a^s_{1} \\ \vdots \\ a^s_{N} \\ b^s_{0} \\ b^s_{1} \\ \vdots \\ b^s_{N} \\
\end{bmatrix}
=\begin{bmatrix}
-B_{00} \\ -B_{10} \\ \vdots \\ -B_{N0} \\ k_{0}A_{0} \\ 0 \\ \vdots \\ 0 \\
\end{bmatrix}
</math>
</center>

To solve the system of equations for the '''antisymmetric''' case, we set the upper limit of <math>m</math> to be <math>N</math>. This resulting system can be expressed in the block matrix form below,

<center>
<math>
\begin{bmatrix}
B_{00} & B_{01} & \cdots & B_{0N} & -C_{0} & 0 & \cdots & 0 \\
B_{10} & B_{11} & \cdots & B_{1N} & 0 & -C_{1} & \cdots & \vdots \\
\vdots & \vdots & \ddots & \vdots & \vdots & \vdots & \ddots & 0 \\
B_{N0} & B_{N1} & \cdots & B_{NN} & 0 & \cdots & 0 & -C_{N} \\
k_{0}A_{0} & 0 & \cdots & 0 & \frac{1}{L}D_{0} & \kappa_{1}\coth(\kappa_{1}L)B_{10} & \cdots & \kappa_{N}\coth(\kappa_{N}L)B_{N0} \\
0 & k_{1}A_{1} & \cdots & \vdots & \frac{1}{L}D_{1} & \kappa_{1}\coth(\kappa_{1}L)B_{11} & \cdots & \kappa_{N}\coth(\kappa_{N}L)B_{N1} \\
\vdots & \vdots & \ddots & 0 & \vdots & \vdots & \ddots & \vdots \\
0 & \cdots & 0 & k_{N}A_{N} & \frac{1}{L}D_{N} & \kappa_{1}\coth(\kappa_{1}L)B_{1N} & \cdots & \kappa_{N}\coth(\kappa_{N}L)B_{NN} \\
\end{bmatrix}
\begin{bmatrix}
a^a_{0} \\ a^a_{1} \\ \vdots \\ a^a_{N} \\ b^a_{0} \\ b^a_{1} \\ \vdots \\ b^a_{N} \\
\end{bmatrix}
=\begin{bmatrix}
-B_{00} \\ -B_{10} \\ \vdots \\ -B_{N0} \\ k_{0}A_{0} \\ 0 \\ \vdots \\ 0 \\
\end{bmatrix}
</math>
</center>

We then simply need to solve the two sets of <math>2(N+1) \mbox{ x } 2(N+1)</math> linear systems of equations. If we had not taken advantage of the symmetry we would need to solve a <math>4(N+1) \mbox{ x } 4(N+1)</math> linear system of equations for four coefficients <math>a_{m}</math>, <math>b_{m}</math>, <math>c_{m}</math> and <math>d_{m}</math>.

== Matlab Code ==

A program to calculate the coefficients for the single finite dock problem using symmetry 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 Finite Rectangle using Symmetry

2012-03-16T06:21:19Z

Tertius: /* An infinite dimensional system of equations */

{{incomplete pages}}

== Introduction ==

The problem consists of a finite rigid fixed rectangle of length <math>2L</math> centered at <math>x = 0</math> and having a height of <math>d + ...</math> freeboard with depth of submergence being <math>d</math> through which no flow is possible. The regions to the left and right of the rectangle are the free water surface. Since the geometry of the rectangle is symmetric about <math>x = 0</math> we can decompose the velocity potential <math>\phi</math> into a symmetric <math>\phi^s</math> and antisymmetric <math>\phi^a</math> potential. We then solve for these two potentials in the left domain only <math>x \leq 0</math> and subsequently recompose the solution <math>\phi</math> for the entire domain. We begin with the simple problem when the waves are normally incident on the rectangle (dock) (so that the problem is truly two-dimensional). We can extend this solution to a pair of partially submerged finite rigid fixed rectangles using symmetry
[[Eigenfunction Matching for a Finite Rectangle using Symmetry]]

== Governing Equations ==

We begin with the [[Frequency Domain Problem]] for a partially submerged finite dock. Although the dock is of length <math>2L</math> we are only interested in the dock of <math>-L \leq x \leq 0</math> together with the fluid domain <math>x \leq 0</math>. We assume <math>e^{i\omega t}</math> time dependence and
the depth of submergence is <math>d</math>.
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>
\begin{align}
\Delta\phi^{s,a} &= 0, \,\,\, -h\leq z\leq 0, \,\, x < -L, \\

\Delta\phi^{s,a} &= 0, \,\,\, -h\leq z\leq -d, \,\, -L < x < 0. \\

\partial_z\phi^{s,a} &= \alpha\phi^{s,a}, \,\,\, z=0, \,\, x < -L, \\

\partial_x\phi^{s,a} &= 0, \,\,\, -d\leq z\leq 0, \,\, x = -L, \\

\partial_z\phi^{s,a} &= 0, \,\,\,
\begin{cases}
z = -d, \,\, -L < x < 0, \\
z = -h, \,\, x < 0.
\end{cases} \\

\partial_x\phi^s &= 0, \,\,\, x = 0, \,\, -h \leq z \leq -d, \\

\phi^a &= 0, \,\,\, x = 0, \,\, -h \leq z \leq -d.
\end{align}
</math></center>

Where <math>\alpha = \omega^2/g</math>.

<math>\phi^s(x,z)</math> is an even function and at <math>x = 0</math> the derivative of this function is equal to zero. <math>\phi^a(x,z)</math> is an odd function and at <math>x = 0</math> this function is equal to zero. These boundary conditions are important and enable the recomposed solution <math>\phi(x,z)</math> to be reflected into the domain of <math>x \geq 0</math>.

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\leq 0</math>
and <math>x\geq 0</math>.

{{separation of variables in two dimensions}}

{{separation of variables for a free surface}}

=== Separation of Variables for a Dock ===

The separation of variables equation for a partially submerged dock is given by:
<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>
Z^{\prime} (-d) = 0.
</math></center>
The solution is
<math>k=\kappa_{m}= \frac{m\pi}{h-d} \,</math> , <math>m\in\mathbb{N}\cup\left\{0\right\}</math> and
<center><math>
Z = \psi_{m}\left( z\right) = \cos\kappa_{m}(z+h),\quad
m\in\mathbb{N}\cup\left\{0\right\}.</math></center>
We note that
<center><math>
\int\nolimits_{-h}^{-d}\psi_{m}(z)\psi_{n}(z) \mathrm{d} z=C_{m}\delta_{mn},
</math></center>
where
<center>
<math>
C_{m} = \frac{1}{2} \left( \frac{\cos\kappa_{m}(h-d)\sin\kappa_{m}(h-d)+\kappa_{m}(h-d)}{\kappa_{m}} \right).
</math>
</center>
In addition there is the special case of when <math>m=n=0</math> ,
<center>
<math>
C_{0} = h-d,
</math>
</center>
where <math>\delta_{00}=1</math> .

=== Inner product between free surface and dock modes ===

<center><math>
B_{mn} = \int\nolimits_{-h}^{-d}\psi_{m}(z)\chi_{n}(z) \mathrm{d} z , \quad m\geq1, \quad n\geq0,
</math></center>
where
<center><math>
B_{mn}= \int\nolimits_{-h}^{-d} \frac{\cos\kappa_{m}(z+h)\cos k_{n}(z+h)}{\cos(k_{n}h)} \mathrm{d} z=\frac{-(-1)^m k_{n}\sin k_{n}(h-d)}
{\cos(k_{n}h)(\kappa_{m}^{2}-k_{n}^{2})}.
</math></center>
The evaluation of this inner product works on the assumption that the two roots <math>\kappa_{m}</math> and <math>k_{n}</math> do not coincide. The Matlab code given below has a routine to check for this case and make necessary adjustments.

We also have the case where the functions of the inner product are switched.
<center><math>
B_{nm} = \int\nolimits_{-h}^{-d}\chi_{m}(z)\psi_{n}(z) \mathrm{d} z , \quad m\geq0, \quad n\geq0.
</math></center>
For <math>m \geq 0</math> and <math>n \geq 1</math> the evaluation is above with the exception that the indices of the expression are switched.

Finally there is the special case of when <math>m \geq 0</math> and <math>n = 0</math>.
<center><math>
B_{0m} = \frac{\sin k_{m}(h-d)}{k_{m}\cos(k_{m}h)}.
</math></center>

=== Expansion of the potential ===

We need to apply some boundary conditions at plus and minus infinity,
where essentially 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 roots. Here we
must use a different condition, essentially identifying one solution
as the incoming wave and the other as the outgoing wave.

Therefore the diffracted potentials can
be expanded as
<center>
<math>
\phi(x,z)=\sum_{m=0}^{\infty}a_{m}e^{k_{m}x}\chi_{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 '''symmetric''' and '''antisymmetric''' potentials <math>\phi^{s,a}(x,z) \,</math> must be continuous across the
transition from open water to the dock covered region. Therefore, the
potentials at <math>x=-L</math> have to be equal.
For these cases we obtain:

<center><math>
\chi_{0}\left( z\right) + \sum_{m=0}^{\infty}a^{s,a}_{m} \chi_{m}\left(z\right)
=\sum_{m=0}^{\infty}b^{s,a}_{m}\psi_{m}(z), \quad -h \leq z\leq -d.
</math></center>

We multiply both sides by <math>\psi_{n}(z) \,</math> and integrate from <math>-h</math> to <math>-d</math> to obtain:

<center><math>
B_{n0} + \sum^{\infty}_{m=0}a^{s,a}_{m}B_{nm} = b^{s,a}_{m}C_{m}\delta_{mn}, \quad n\in\mathbb{N}\cup\left\{0\right\}.
</math></center>

The derivative of the '''symmetric''' potential <math>\partial_x\phi^{s}(x,z) \,</math> must be continuous across the
transition from open water to the dock covered region. Therefore, the derivatives of the potentials at <math>x=-L</math> have to be equal.
For this case we obtain:

<center><math>
\begin{align}
-k_{0}\chi_{0}\left( z\right) +\sum_{m=0}^{\infty} a^{s}_{m}k_{m}\chi_{m}\left(z\right) =
\begin{cases}
\quad \quad \quad \quad \quad 0,\quad \quad \quad \quad \quad \quad -d\leq z\leq 0 \\
-\sum_{m=1}^{\infty}b^{s}_{m}\kappa_{m}\tanh(\kappa_{m}L)\psi_{m}(z),\quad -h\leq z\leq -d
\end{cases}
\end{align}
</math></center>

The derivative of the '''antisymmetric''' potential <math>\partial_x\phi^{a}(x,z) \,</math> must be continuous across the
transition from open water to the dock covered region. Therefore, the derivatives of the potentials at <math>x=-L</math> have to be equal.
For this case we obtain:

<center><math>
\begin{align}
-k_{0}\chi_{0}\left( z\right) +\sum_{m=0}^{\infty} a^{a}_{m}k_{m}\chi_{m}\left(z\right) =
\begin{cases}
\quad \quad \quad \quad \quad 0,\quad \quad \quad \quad \quad \quad \quad \quad -d\leq z\leq 0 \\
-\frac{1}{L}b^a_{0} -\sum_{m=1}^{\infty}b^{a}_{m}\kappa_{m}\coth(\kappa_{m}L)\psi_{m}(z),\quad -h\leq z\leq -d
\end{cases}
\end{align}
</math></center>

For the first equation we multiply both sides by <math>\chi_{n}(z) \,</math> and integrating from <math>-h</math> to <math>0</math> to obtain:
<center><math>
-k_{0}A_{0}\delta_{0n} + k_{m}a_{m}A_{m} = -\sum^{\infty}_{m=0}\kappa_{m}b_{m}B_{mn}, \quad n\in\mathbb{N}\cup\left\{0\right\}
</math></center>

Solving the equations above will yield the coefficients of the water velocity potential in the dock covered region.

== Numerical Solution ==

To solve the system of equations for the '''symmetric''' case, we set the upper limit of <math>m</math> to be <math>N</math>. This resulting system can be expressed in the block matrix form below,

<center>
<math>
\begin{bmatrix}
B_{00} & B_{01} & \cdots & B_{0N} & -C_{0} & 0 & \cdots & 0 \\
B_{10} & B_{11} & \cdots & B_{1N} & 0 & -C_{1} & \cdots & \vdots \\
\vdots & \vdots & \ddots & \vdots & \vdots & \vdots & \ddots & 0 \\
B_{N0} & B_{N1} & \cdots & B_{NN} & 0 & \cdots & 0 & -C_{N} \\
k_{0}A_{0} & 0 & \cdots & 0 & 0 & \kappa_{1}\tanh(\kappa_{1}L)B_{10} & \cdots & \kappa_{N}\tanh(\kappa_{N}L)B_{N0} \\
0 & k_{1}A_{1} & \cdots & \vdots & 0 & \kappa_{1}\tanh(\kappa_{1}L)B_{11} & \cdots & \kappa_{N}\tanh(\kappa_{N}L)B_{N1} \\
\vdots & \vdots & \ddots & 0 & \vdots & \vdots & \ddots & \vdots \\
0 & \cdots & 0 & k_{N}A_{N} & 0 & \kappa_{1}\tanh(\kappa_{1}L)B_{1N} & \cdots & \kappa_{N}\tanh(\kappa_{N}L)B_{NN} \\
\end{bmatrix}
\begin{bmatrix}
a^s_{0} \\ a^s_{1} \\ \vdots \\ a^s_{N} \\ b^s_{0} \\ b^s_{1} \\ \vdots \\ b^s_{N} \\
\end{bmatrix}
=\begin{bmatrix}
-B_{00} \\ -B_{10} \\ \vdots \\ -B_{N0} \\ k_{0}A_{0} \\ 0 \\ \vdots \\ 0 \\
\end{bmatrix}
</math>
</center>

To solve the system of equations for the '''antisymmetric''' case, we set the upper limit of <math>m</math> to be <math>N</math>. This resulting system can be expressed in the block matrix form below,

<center>
<math>
\begin{bmatrix}
B_{00} & B_{01} & \cdots & B_{0N} & -C_{0} & 0 & \cdots & 0 \\
B_{10} & B_{11} & \cdots & B_{1N} & 0 & -C_{1} & \cdots & \vdots \\
\vdots & \vdots & \ddots & \vdots & \vdots & \vdots & \ddots & 0 \\
B_{N0} & B_{N1} & \cdots & B_{NN} & 0 & \cdots & 0 & -C_{N} \\
k_{0}A_{0} & 0 & \cdots & 0 & \frac{1}{L}D_{0} & \kappa_{1}\coth(\kappa_{1}L)B_{10} & \cdots & \kappa_{N}\coth(\kappa_{N}L)B_{N0} \\
0 & k_{1}A_{1} & \cdots & \vdots & \frac{1}{L}D_{1} & \kappa_{1}\coth(\kappa_{1}L)B_{11} & \cdots & \kappa_{N}\coth(\kappa_{N}L)B_{N1} \\
\vdots & \vdots & \ddots & 0 & \vdots & \vdots & \ddots & \vdots \\
0 & \cdots & 0 & k_{N}A_{N} & \frac{1}{L}D_{N} & \kappa_{1}\coth(\kappa_{1}L)B_{1N} & \cdots & \kappa_{N}\coth(\kappa_{N}L)B_{NN} \\
\end{bmatrix}
\begin{bmatrix}
a^a_{0} \\ a^a_{1} \\ \vdots \\ a^a_{N} \\ b^a_{0} \\ b^a_{1} \\ \vdots \\ b^a_{N} \\
\end{bmatrix}
=\begin{bmatrix}
-B_{00} \\ -B_{10} \\ \vdots \\ -B_{N0} \\ k_{0}A_{0} \\ 0 \\ \vdots \\ 0 \\
\end{bmatrix}
</math>
</center>

We then simply need to solve the two sets of <math>2(N+1) \mbox{ x } 2(N+1)</math> linear systems of equations. If we had not taken advantage of the symmetry we would need to solve a <math>4(N+1) \mbox{ x } 4(N+1)</math> linear system of equations for four coefficients <math>a_{m}</math>, <math>b_{m}</math>, <math>c_{m}</math> and <math>d_{m}</math>.

== Matlab Code ==

A program to calculate the coefficients for the single finite dock problem using symmetry 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 Finite Rectangle using Symmetry

2012-03-16T06:17:03Z

Tertius: /* An infinite dimensional system of equations */

{{incomplete pages}}

== Introduction ==

The problem consists of a finite rigid fixed rectangle of length <math>2L</math> centered at <math>x = 0</math> and having a height of <math>d + ...</math> freeboard with depth of submergence being <math>d</math> through which no flow is possible. The regions to the left and right of the rectangle are the free water surface. Since the geometry of the rectangle is symmetric about <math>x = 0</math> we can decompose the velocity potential <math>\phi</math> into a symmetric <math>\phi^s</math> and antisymmetric <math>\phi^a</math> potential. We then solve for these two potentials in the left domain only <math>x \leq 0</math> and subsequently recompose the solution <math>\phi</math> for the entire domain. We begin with the simple problem when the waves are normally incident on the rectangle (dock) (so that the problem is truly two-dimensional). We can extend this solution to a pair of partially submerged finite rigid fixed rectangles using symmetry
[[Eigenfunction Matching for a Finite Rectangle using Symmetry]]

== Governing Equations ==

We begin with the [[Frequency Domain Problem]] for a partially submerged finite dock. Although the dock is of length <math>2L</math> we are only interested in the dock of <math>-L \leq x \leq 0</math> together with the fluid domain <math>x \leq 0</math>. We assume <math>e^{i\omega t}</math> time dependence and
the depth of submergence is <math>d</math>.
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>
\begin{align}
\Delta\phi^{s,a} &= 0, \,\,\, -h\leq z\leq 0, \,\, x < -L, \\

\Delta\phi^{s,a} &= 0, \,\,\, -h\leq z\leq -d, \,\, -L < x < 0. \\

\partial_z\phi^{s,a} &= \alpha\phi^{s,a}, \,\,\, z=0, \,\, x < -L, \\

\partial_x\phi^{s,a} &= 0, \,\,\, -d\leq z\leq 0, \,\, x = -L, \\

\partial_z\phi^{s,a} &= 0, \,\,\,
\begin{cases}
z = -d, \,\, -L < x < 0, \\
z = -h, \,\, x < 0.
\end{cases} \\

\partial_x\phi^s &= 0, \,\,\, x = 0, \,\, -h \leq z \leq -d, \\

\phi^a &= 0, \,\,\, x = 0, \,\, -h \leq z \leq -d.
\end{align}
</math></center>

Where <math>\alpha = \omega^2/g</math>.

<math>\phi^s(x,z)</math> is an even function and at <math>x = 0</math> the derivative of this function is equal to zero. <math>\phi^a(x,z)</math> is an odd function and at <math>x = 0</math> this function is equal to zero. These boundary conditions are important and enable the recomposed solution <math>\phi(x,z)</math> to be reflected into the domain of <math>x \geq 0</math>.

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\leq 0</math>
and <math>x\geq 0</math>.

{{separation of variables in two dimensions}}

{{separation of variables for a free surface}}

=== Separation of Variables for a Dock ===

The separation of variables equation for a partially submerged dock is given by:
<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>
Z^{\prime} (-d) = 0.
</math></center>
The solution is
<math>k=\kappa_{m}= \frac{m\pi}{h-d} \,</math> , <math>m\in\mathbb{N}\cup\left\{0\right\}</math> and
<center><math>
Z = \psi_{m}\left( z\right) = \cos\kappa_{m}(z+h),\quad
m\in\mathbb{N}\cup\left\{0\right\}.</math></center>
We note that
<center><math>
\int\nolimits_{-h}^{-d}\psi_{m}(z)\psi_{n}(z) \mathrm{d} z=C_{m}\delta_{mn},
</math></center>
where
<center>
<math>
C_{m} = \frac{1}{2} \left( \frac{\cos\kappa_{m}(h-d)\sin\kappa_{m}(h-d)+\kappa_{m}(h-d)}{\kappa_{m}} \right).
</math>
</center>
In addition there is the special case of when <math>m=n=0</math> ,
<center>
<math>
C_{0} = h-d,
</math>
</center>
where <math>\delta_{00}=1</math> .

=== Inner product between free surface and dock modes ===

<center><math>
B_{mn} = \int\nolimits_{-h}^{-d}\psi_{m}(z)\chi_{n}(z) \mathrm{d} z , \quad m\geq1, \quad n\geq0,
</math></center>
where
<center><math>
B_{mn}= \int\nolimits_{-h}^{-d} \frac{\cos\kappa_{m}(z+h)\cos k_{n}(z+h)}{\cos(k_{n}h)} \mathrm{d} z=\frac{-(-1)^m k_{n}\sin k_{n}(h-d)}
{\cos(k_{n}h)(\kappa_{m}^{2}-k_{n}^{2})}.
</math></center>
The evaluation of this inner product works on the assumption that the two roots <math>\kappa_{m}</math> and <math>k_{n}</math> do not coincide. The Matlab code given below has a routine to check for this case and make necessary adjustments.

We also have the case where the functions of the inner product are switched.
<center><math>
B_{nm} = \int\nolimits_{-h}^{-d}\chi_{m}(z)\psi_{n}(z) \mathrm{d} z , \quad m\geq0, \quad n\geq0.
</math></center>
For <math>m \geq 0</math> and <math>n \geq 1</math> the evaluation is above with the exception that the indices of the expression are switched.

Finally there is the special case of when <math>m \geq 0</math> and <math>n = 0</math>.
<center><math>
B_{0m} = \frac{\sin k_{m}(h-d)}{k_{m}\cos(k_{m}h)}.
</math></center>

=== Expansion of the potential ===

We need to apply some boundary conditions at plus and minus infinity,
where essentially 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 roots. Here we
must use a different condition, essentially identifying one solution
as the incoming wave and the other as the outgoing wave.

Therefore the diffracted potentials can
be expanded as
<center>
<math>
\phi(x,z)=\sum_{m=0}^{\infty}a_{m}e^{k_{m}x}\chi_{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 '''symmetric''' and '''antisymmetric''' potentials <math>\phi^{s,a}(x,z) \,</math> must be continuous across the
transition from open water to the dock covered region. Therefore, the
potentials at <math>x=-L</math> have to be equal.
For these cases we obtain:

<center><math>
\chi_{0}\left( z\right) + \sum_{m=0}^{\infty}a^{s,a}_{m} \chi_{m}\left(z\right)
=\sum_{m=0}^{\infty}b^{s,a}_{m}\psi_{m}(z), \quad -h \leq z\leq -d.
</math></center>

We multiply both sides by <math>\psi_{n}(z) \,</math> and integrate from <math>-h</math> to <math>-d</math> to obtain:

<center><math>
B_{n0} + \sum^{\infty}_{m=0}a^{s,a}_{m}B_{nm} = b^{s,a}_{m}C_{m}\delta_{mn}, \quad n\in\mathbb{N}\cup\left\{0\right\}.
</math></center>

The derivative of the '''symmetric''' potential <math>\partial_x\phi^{s}(x,z) \,</math> must be continuous across the
transition from open water to the dock covered region. Therefore, the derivatives of the potentials at <math>x=-L</math> have to be equal.
For this case we obtain:

<center><math>
\begin{align}
-k_{0}\chi_{0}\left( z\right) +\sum_{m=0}^{\infty} a^{s}_{m}k_{m}\chi_{m}\left(z\right) =
\begin{cases}
\quad \quad \quad \quad \quad 0,\quad \quad \quad \quad \quad \quad -d\leq z\leq 0 \\
-\sum_{m=1}^{\infty}b^{s}_{m}\kappa_{m}\tanh(\kappa_{m}L)\psi_{m}(z),\quad -h\leq z\leq -d
\end{cases}
\end{align}
</math></center>

The derivative of the '''antisymmetric''' potential <math>\partial_x\phi^{a}(x,z) \,</math> must be continuous across the
transition from open water to the dock covered region. Therefore, the derivatives of the potentials at <math>x=-L</math> have to be equal.
For this case we obtain:

<center><math>
\begin{align}
-k_{0}\chi_{0}\left( z\right) +\sum_{m=0}^{\infty} a^{a}_{m}k_{m}\chi_{m}\left(z\right) =
\begin{cases}
\quad \quad \quad \quad \quad 0,\quad \quad \quad \quad \quad \quad -d\leq z\leq 0 \\
-\frac{1}{L}b^a_{0} -\sum_{m=1}^{\infty}b^{a}_{m}\kappa_{m}\coth(\kappa_{m}L)\psi_{m}(z),\quad -h\leq z\leq -d
\end{cases}
\end{align}
</math></center>

For the first equation we multiply both sides by <math>\chi_{n}(z) \,</math> and integrating from <math>-h</math> to <math>0</math> to obtain:
<center><math>
-k_{0}A_{0}\delta_{0n} + k_{m}a_{m}A_{m} = -\sum^{\infty}_{m=0}\kappa_{m}b_{m}B_{mn}, \quad n\in\mathbb{N}\cup\left\{0\right\}
</math></center>

Solving the equations above will yield the coefficients of the water velocity potential in the dock covered region.

== Numerical Solution ==

To solve the system of equations for the '''symmetric''' case, we set the upper limit of <math>m</math> to be <math>N</math>. This resulting system can be expressed in the block matrix form below,

<center>
<math>
\begin{bmatrix}
B_{00} & B_{01} & \cdots & B_{0N} & -C_{0} & 0 & \cdots & 0 \\
B_{10} & B_{11} & \cdots & B_{1N} & 0 & -C_{1} & \cdots & \vdots \\
\vdots & \vdots & \ddots & \vdots & \vdots & \vdots & \ddots & 0 \\
B_{N0} & B_{N1} & \cdots & B_{NN} & 0 & \cdots & 0 & -C_{N} \\
k_{0}A_{0} & 0 & \cdots & 0 & 0 & \kappa_{1}\tanh(\kappa_{1}L)B_{10} & \cdots & \kappa_{N}\tanh(\kappa_{N}L)B_{N0} \\
0 & k_{1}A_{1} & \cdots & \vdots & 0 & \kappa_{1}\tanh(\kappa_{1}L)B_{11} & \cdots & \kappa_{N}\tanh(\kappa_{N}L)B_{N1} \\
\vdots & \vdots & \ddots & 0 & \vdots & \vdots & \ddots & \vdots \\
0 & \cdots & 0 & k_{N}A_{N} & 0 & \kappa_{1}\tanh(\kappa_{1}L)B_{1N} & \cdots & \kappa_{N}\tanh(\kappa_{N}L)B_{NN} \\
\end{bmatrix}
\begin{bmatrix}
a^s_{0} \\ a^s_{1} \\ \vdots \\ a^s_{N} \\ b^s_{0} \\ b^s_{1} \\ \vdots \\ b^s_{N} \\
\end{bmatrix}
=\begin{bmatrix}
-B_{00} \\ -B_{10} \\ \vdots \\ -B_{N0} \\ k_{0}A_{0} \\ 0 \\ \vdots \\ 0 \\
\end{bmatrix}
</math>
</center>

To solve the system of equations for the '''antisymmetric''' case, we set the upper limit of <math>m</math> to be <math>N</math>. This resulting system can be expressed in the block matrix form below,

<center>
<math>
\begin{bmatrix}
B_{00} & B_{01} & \cdots & B_{0N} & -C_{0} & 0 & \cdots & 0 \\
B_{10} & B_{11} & \cdots & B_{1N} & 0 & -C_{1} & \cdots & \vdots \\
\vdots & \vdots & \ddots & \vdots & \vdots & \vdots & \ddots & 0 \\
B_{N0} & B_{N1} & \cdots & B_{NN} & 0 & \cdots & 0 & -C_{N} \\
k_{0}A_{0} & 0 & \cdots & 0 & \frac{1}{L}D_{0} & \kappa_{1}\coth(\kappa_{1}L)B_{10} & \cdots & \kappa_{N}\coth(\kappa_{N}L)B_{N0} \\
0 & k_{1}A_{1} & \cdots & \vdots & \frac{1}{L}D_{1} & \kappa_{1}\coth(\kappa_{1}L)B_{11} & \cdots & \kappa_{N}\coth(\kappa_{N}L)B_{N1} \\
\vdots & \vdots & \ddots & 0 & \vdots & \vdots & \ddots & \vdots \\
0 & \cdots & 0 & k_{N}A_{N} & \frac{1}{L}D_{N} & \kappa_{1}\coth(\kappa_{1}L)B_{1N} & \cdots & \kappa_{N}\coth(\kappa_{N}L)B_{NN} \\
\end{bmatrix}
\begin{bmatrix}
a^a_{0} \\ a^a_{1} \\ \vdots \\ a^a_{N} \\ b^a_{0} \\ b^a_{1} \\ \vdots \\ b^a_{N} \\
\end{bmatrix}
=\begin{bmatrix}
-B_{00} \\ -B_{10} \\ \vdots \\ -B_{N0} \\ k_{0}A_{0} \\ 0 \\ \vdots \\ 0 \\
\end{bmatrix}
</math>
</center>

We then simply need to solve the two sets of <math>2(N+1) \mbox{ x } 2(N+1)</math> linear systems of equations. If we had not taken advantage of the symmetry we would need to solve a <math>4(N+1) \mbox{ x } 4(N+1)</math> linear system of equations for four coefficients <math>a_{m}</math>, <math>b_{m}</math>, <math>c_{m}</math> and <math>d_{m}</math>.

== Matlab Code ==

A program to calculate the coefficients for the single finite dock problem using symmetry 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 Finite Rectangle using Symmetry

2012-03-16T06:09:03Z

Tertius: /* An infinite dimensional system of equations */

{{incomplete pages}}

== Introduction ==

The problem consists of a finite rigid fixed rectangle of length <math>2L</math> centered at <math>x = 0</math> and having a height of <math>d + ...</math> freeboard with depth of submergence being <math>d</math> through which no flow is possible. The regions to the left and right of the rectangle are the free water surface. Since the geometry of the rectangle is symmetric about <math>x = 0</math> we can decompose the velocity potential <math>\phi</math> into a symmetric <math>\phi^s</math> and antisymmetric <math>\phi^a</math> potential. We then solve for these two potentials in the left domain only <math>x \leq 0</math> and subsequently recompose the solution <math>\phi</math> for the entire domain. We begin with the simple problem when the waves are normally incident on the rectangle (dock) (so that the problem is truly two-dimensional). We can extend this solution to a pair of partially submerged finite rigid fixed rectangles using symmetry
[[Eigenfunction Matching for a Finite Rectangle using Symmetry]]

== Governing Equations ==

We begin with the [[Frequency Domain Problem]] for a partially submerged finite dock. Although the dock is of length <math>2L</math> we are only interested in the dock of <math>-L \leq x \leq 0</math> together with the fluid domain <math>x \leq 0</math>. We assume <math>e^{i\omega t}</math> time dependence and
the depth of submergence is <math>d</math>.
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>
\begin{align}
\Delta\phi^{s,a} &= 0, \,\,\, -h\leq z\leq 0, \,\, x < -L, \\

\Delta\phi^{s,a} &= 0, \,\,\, -h\leq z\leq -d, \,\, -L < x < 0. \\

\partial_z\phi^{s,a} &= \alpha\phi^{s,a}, \,\,\, z=0, \,\, x < -L, \\

\partial_x\phi^{s,a} &= 0, \,\,\, -d\leq z\leq 0, \,\, x = -L, \\

\partial_z\phi^{s,a} &= 0, \,\,\,
\begin{cases}
z = -d, \,\, -L < x < 0, \\
z = -h, \,\, x < 0.
\end{cases} \\

\partial_x\phi^s &= 0, \,\,\, x = 0, \,\, -h \leq z \leq -d, \\

\phi^a &= 0, \,\,\, x = 0, \,\, -h \leq z \leq -d.
\end{align}
</math></center>

Where <math>\alpha = \omega^2/g</math>.

<math>\phi^s(x,z)</math> is an even function and at <math>x = 0</math> the derivative of this function is equal to zero. <math>\phi^a(x,z)</math> is an odd function and at <math>x = 0</math> this function is equal to zero. These boundary conditions are important and enable the recomposed solution <math>\phi(x,z)</math> to be reflected into the domain of <math>x \geq 0</math>.

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\leq 0</math>
and <math>x\geq 0</math>.

{{separation of variables in two dimensions}}

{{separation of variables for a free surface}}

=== Separation of Variables for a Dock ===

The separation of variables equation for a partially submerged dock is given by:
<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>
Z^{\prime} (-d) = 0.
</math></center>
The solution is
<math>k=\kappa_{m}= \frac{m\pi}{h-d} \,</math> , <math>m\in\mathbb{N}\cup\left\{0\right\}</math> and
<center><math>
Z = \psi_{m}\left( z\right) = \cos\kappa_{m}(z+h),\quad
m\in\mathbb{N}\cup\left\{0\right\}.</math></center>
We note that
<center><math>
\int\nolimits_{-h}^{-d}\psi_{m}(z)\psi_{n}(z) \mathrm{d} z=C_{m}\delta_{mn},
</math></center>
where
<center>
<math>
C_{m} = \frac{1}{2} \left( \frac{\cos\kappa_{m}(h-d)\sin\kappa_{m}(h-d)+\kappa_{m}(h-d)}{\kappa_{m}} \right).
</math>
</center>
In addition there is the special case of when <math>m=n=0</math> ,
<center>
<math>
C_{0} = h-d,
</math>
</center>
where <math>\delta_{00}=1</math> .

=== Inner product between free surface and dock modes ===

<center><math>
B_{mn} = \int\nolimits_{-h}^{-d}\psi_{m}(z)\chi_{n}(z) \mathrm{d} z , \quad m\geq1, \quad n\geq0,
</math></center>
where
<center><math>
B_{mn}= \int\nolimits_{-h}^{-d} \frac{\cos\kappa_{m}(z+h)\cos k_{n}(z+h)}{\cos(k_{n}h)} \mathrm{d} z=\frac{-(-1)^m k_{n}\sin k_{n}(h-d)}
{\cos(k_{n}h)(\kappa_{m}^{2}-k_{n}^{2})}.
</math></center>
The evaluation of this inner product works on the assumption that the two roots <math>\kappa_{m}</math> and <math>k_{n}</math> do not coincide. The Matlab code given below has a routine to check for this case and make necessary adjustments.

We also have the case where the functions of the inner product are switched.
<center><math>
B_{nm} = \int\nolimits_{-h}^{-d}\chi_{m}(z)\psi_{n}(z) \mathrm{d} z , \quad m\geq0, \quad n\geq0.
</math></center>
For <math>m \geq 0</math> and <math>n \geq 1</math> the evaluation is above with the exception that the indices of the expression are switched.

Finally there is the special case of when <math>m \geq 0</math> and <math>n = 0</math>.
<center><math>
B_{0m} = \frac{\sin k_{m}(h-d)}{k_{m}\cos(k_{m}h)}.
</math></center>

=== Expansion of the potential ===

We need to apply some boundary conditions at plus and minus infinity,
where essentially 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 roots. Here we
must use a different condition, essentially identifying one solution
as the incoming wave and the other as the outgoing wave.

Therefore the diffracted potentials can
be expanded as
<center>
<math>
\phi(x,z)=\sum_{m=0}^{\infty}a_{m}e^{k_{m}x}\chi_{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 '''symmetric''' and '''antisymmetric''' potentials <math>\phi^{s,a}(x,z) \,</math> must be continuous across the
transition from open water to the dock covered region. Therefore, the
potentials at <math>x=-L</math> have to be equal.
For these cases we obtain:

<center><math>
\chi_{0}\left( z\right) + \sum_{m=0}^{\infty}a^{s,a}_{m} \chi_{m}\left(z\right)
=\sum_{m=0}^{\infty}b^{s,a}_{m}\psi_{m}(z), \quad -h \leq z\leq -d.
</math></center>

We multiply both sides by <math>\psi_{n}(z) \,</math> and integrate from <math>-h</math> to <math>-d</math> to obtain:

<center><math>
B_{n0} + \sum^{\infty}_{m=0}a^{s,a}_{m}B_{nm} = b^{s,a}_{m}C_{m}\delta_{mn}, \quad n\in\mathbb{N}\cup\left\{0\right\}.
</math></center>

The derivative of the '''symmetric''' potential <math>\partial_x\phi^{s}(x,z) \,</math> must be continuous across the
transition from open water to the dock covered region. Therefore, the derivatives of the potentials at <math>x=-L</math> have to be equal.
For this case we obtain:

<center><math>
\begin{align}
-k_{0}\chi_{0}\left( z\right) +\sum_{m=0}^{\infty} a^{s}_{m}k_{m}\chi_{m}\left(z\right) =
\begin{cases}
\quad \quad \quad \quad \quad 0,\quad \quad \quad \quad \quad \quad -d\leq z\leq 0 \\
-\sum_{m=1}^{\infty}b^{s}_{m}\kappa_{m}\tanh(\kappa_{m}L)\psi_{m}(z),\quad -h\leq z\leq -d
\end{cases}
\end{align}
</math></center>

The derivative of the '''antisymmetric''' potential <math>\partial_x\phi^{a}(x,z) \,</math> must be continuous across the
transition from open water to the dock covered region. Therefore, the derivatives of the potentials at <math>x=-L</math> have to be equal.
For this case we obtain:

<center><math>
\begin{align}
-k_{0}\chi_{0}\left( z\right) +\sum_{m=0}^{\infty} a^{s}_{m}k_{m}\chi_{m}\left(z\right) =
\begin{cases}
\quad \quad \quad \quad \quad 0,\quad \quad \quad \quad \quad \quad -d\leq z\leq 0 \\
-\sum_{m=1}^{\infty}b^{s}_{m}\kappa_{m}\tanh(\kappa_{m}L)\psi_{m}(z),\quad -h\leq z\leq -d
\end{cases}
\end{align}
</math></center>

For the first equation we multiply both sides by <math>\chi_{n}(z) \,</math> and integrating from <math>-h</math> to <math>0</math> to obtain:
<center><math>
-k_{0}A_{0}\delta_{0n} + k_{m}a_{m}A_{m} = -\sum^{\infty}_{m=0}\kappa_{m}b_{m}B_{mn}, \quad n\in\mathbb{N}\cup\left\{0\right\}
</math></center>

Solving the equations above will yield the coefficients of the water velocity potential in the dock covered region.

== Numerical Solution ==

To solve the system of equations for the '''symmetric''' case, we set the upper limit of <math>m</math> to be <math>N</math>. This resulting system can be expressed in the block matrix form below,

<center>
<math>
\begin{bmatrix}
B_{00} & B_{01} & \cdots & B_{0N} & -C_{0} & 0 & \cdots & 0 \\
B_{10} & B_{11} & \cdots & B_{1N} & 0 & -C_{1} & \cdots & \vdots \\
\vdots & \vdots & \ddots & \vdots & \vdots & \vdots & \ddots & 0 \\
B_{N0} & B_{N1} & \cdots & B_{NN} & 0 & \cdots & 0 & -C_{N} \\
k_{0}A_{0} & 0 & \cdots & 0 & 0 & \kappa_{1}\tanh(\kappa_{1}L)B_{10} & \cdots & \kappa_{N}\tanh(\kappa_{N}L)B_{N0} \\
0 & k_{1}A_{1} & \cdots & \vdots & 0 & \kappa_{1}\tanh(\kappa_{1}L)B_{11} & \cdots & \kappa_{N}\tanh(\kappa_{N}L)B_{N1} \\
\vdots & \vdots & \ddots & 0 & \vdots & \vdots & \ddots & \vdots \\
0 & \cdots & 0 & k_{N}A_{N} & 0 & \kappa_{1}\tanh(\kappa_{1}L)B_{1N} & \cdots & \kappa_{N}\tanh(\kappa_{N}L)B_{NN} \\
\end{bmatrix}
\begin{bmatrix}
a^s_{0} \\ a^s_{1} \\ \vdots \\ a^s_{N} \\ b^s_{0} \\ b^s_{1} \\ \vdots \\ b^s_{N} \\
\end{bmatrix}
=\begin{bmatrix}
-B_{00} \\ -B_{10} \\ \vdots \\ -B_{N0} \\ k_{0}A_{0} \\ 0 \\ \vdots \\ 0 \\
\end{bmatrix}
</math>
</center>

To solve the system of equations for the '''antisymmetric''' case, we set the upper limit of <math>m</math> to be <math>N</math>. This resulting system can be expressed in the block matrix form below,

<center>
<math>
\begin{bmatrix}
B_{00} & B_{01} & \cdots & B_{0N} & -C_{0} & 0 & \cdots & 0 \\
B_{10} & B_{11} & \cdots & B_{1N} & 0 & -C_{1} & \cdots & \vdots \\
\vdots & \vdots & \ddots & \vdots & \vdots & \vdots & \ddots & 0 \\
B_{N0} & B_{N1} & \cdots & B_{NN} & 0 & \cdots & 0 & -C_{N} \\
k_{0}A_{0} & 0 & \cdots & 0 & \frac{1}{L}D_{0} & \kappa_{1}\coth(\kappa_{1}L)B_{10} & \cdots & \kappa_{N}\coth(\kappa_{N}L)B_{N0} \\
0 & k_{1}A_{1} & \cdots & \vdots & \frac{1}{L}D_{1} & \kappa_{1}\coth(\kappa_{1}L)B_{11} & \cdots & \kappa_{N}\coth(\kappa_{N}L)B_{N1} \\
\vdots & \vdots & \ddots & 0 & \vdots & \vdots & \ddots & \vdots \\
0 & \cdots & 0 & k_{N}A_{N} & \frac{1}{L}D_{N} & \kappa_{1}\coth(\kappa_{1}L)B_{1N} & \cdots & \kappa_{N}\coth(\kappa_{N}L)B_{NN} \\
\end{bmatrix}
\begin{bmatrix}
a^a_{0} \\ a^a_{1} \\ \vdots \\ a^a_{N} \\ b^a_{0} \\ b^a_{1} \\ \vdots \\ b^a_{N} \\
\end{bmatrix}
=\begin{bmatrix}
-B_{00} \\ -B_{10} \\ \vdots \\ -B_{N0} \\ k_{0}A_{0} \\ 0 \\ \vdots \\ 0 \\
\end{bmatrix}
</math>
</center>

We then simply need to solve the two sets of <math>2(N+1) \mbox{ x } 2(N+1)</math> linear systems of equations. If we had not taken advantage of the symmetry we would need to solve a <math>4(N+1) \mbox{ x } 4(N+1)</math> linear system of equations for four coefficients <math>a_{m}</math>, <math>b_{m}</math>, <math>c_{m}</math> and <math>d_{m}</math>.

== Matlab Code ==

A program to calculate the coefficients for the single finite dock problem using symmetry 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 Finite Rectangle using Symmetry

2012-03-16T06:04:20Z

Tertius: /* An infinite dimensional system of equations */

{{incomplete pages}}

== Introduction ==

The problem consists of a finite rigid fixed rectangle of length <math>2L</math> centered at <math>x = 0</math> and having a height of <math>d + ...</math> freeboard with depth of submergence being <math>d</math> through which no flow is possible. The regions to the left and right of the rectangle are the free water surface. Since the geometry of the rectangle is symmetric about <math>x = 0</math> we can decompose the velocity potential <math>\phi</math> into a symmetric <math>\phi^s</math> and antisymmetric <math>\phi^a</math> potential. We then solve for these two potentials in the left domain only <math>x \leq 0</math> and subsequently recompose the solution <math>\phi</math> for the entire domain. We begin with the simple problem when the waves are normally incident on the rectangle (dock) (so that the problem is truly two-dimensional). We can extend this solution to a pair of partially submerged finite rigid fixed rectangles using symmetry
[[Eigenfunction Matching for a Finite Rectangle using Symmetry]]

== Governing Equations ==

We begin with the [[Frequency Domain Problem]] for a partially submerged finite dock. Although the dock is of length <math>2L</math> we are only interested in the dock of <math>-L \leq x \leq 0</math> together with the fluid domain <math>x \leq 0</math>. We assume <math>e^{i\omega t}</math> time dependence and
the depth of submergence is <math>d</math>.
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>
\begin{align}
\Delta\phi^{s,a} &= 0, \,\,\, -h\leq z\leq 0, \,\, x < -L, \\

\Delta\phi^{s,a} &= 0, \,\,\, -h\leq z\leq -d, \,\, -L < x < 0. \\

\partial_z\phi^{s,a} &= \alpha\phi^{s,a}, \,\,\, z=0, \,\, x < -L, \\

\partial_x\phi^{s,a} &= 0, \,\,\, -d\leq z\leq 0, \,\, x = -L, \\

\partial_z\phi^{s,a} &= 0, \,\,\,
\begin{cases}
z = -d, \,\, -L < x < 0, \\
z = -h, \,\, x < 0.
\end{cases} \\

\partial_x\phi^s &= 0, \,\,\, x = 0, \,\, -h \leq z \leq -d, \\

\phi^a &= 0, \,\,\, x = 0, \,\, -h \leq z \leq -d.
\end{align}
</math></center>

Where <math>\alpha = \omega^2/g</math>.

<math>\phi^s(x,z)</math> is an even function and at <math>x = 0</math> the derivative of this function is equal to zero. <math>\phi^a(x,z)</math> is an odd function and at <math>x = 0</math> this function is equal to zero. These boundary conditions are important and enable the recomposed solution <math>\phi(x,z)</math> to be reflected into the domain of <math>x \geq 0</math>.

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\leq 0</math>
and <math>x\geq 0</math>.

{{separation of variables in two dimensions}}

{{separation of variables for a free surface}}

=== Separation of Variables for a Dock ===

The separation of variables equation for a partially submerged dock is given by:
<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>
Z^{\prime} (-d) = 0.
</math></center>
The solution is
<math>k=\kappa_{m}= \frac{m\pi}{h-d} \,</math> , <math>m\in\mathbb{N}\cup\left\{0\right\}</math> and
<center><math>
Z = \psi_{m}\left( z\right) = \cos\kappa_{m}(z+h),\quad
m\in\mathbb{N}\cup\left\{0\right\}.</math></center>
We note that
<center><math>
\int\nolimits_{-h}^{-d}\psi_{m}(z)\psi_{n}(z) \mathrm{d} z=C_{m}\delta_{mn},
</math></center>
where
<center>
<math>
C_{m} = \frac{1}{2} \left( \frac{\cos\kappa_{m}(h-d)\sin\kappa_{m}(h-d)+\kappa_{m}(h-d)}{\kappa_{m}} \right).
</math>
</center>
In addition there is the special case of when <math>m=n=0</math> ,
<center>
<math>
C_{0} = h-d,
</math>
</center>
where <math>\delta_{00}=1</math> .

=== Inner product between free surface and dock modes ===

<center><math>
B_{mn} = \int\nolimits_{-h}^{-d}\psi_{m}(z)\chi_{n}(z) \mathrm{d} z , \quad m\geq1, \quad n\geq0,
</math></center>
where
<center><math>
B_{mn}= \int\nolimits_{-h}^{-d} \frac{\cos\kappa_{m}(z+h)\cos k_{n}(z+h)}{\cos(k_{n}h)} \mathrm{d} z=\frac{-(-1)^m k_{n}\sin k_{n}(h-d)}
{\cos(k_{n}h)(\kappa_{m}^{2}-k_{n}^{2})}.
</math></center>
The evaluation of this inner product works on the assumption that the two roots <math>\kappa_{m}</math> and <math>k_{n}</math> do not coincide. The Matlab code given below has a routine to check for this case and make necessary adjustments.

We also have the case where the functions of the inner product are switched.
<center><math>
B_{nm} = \int\nolimits_{-h}^{-d}\chi_{m}(z)\psi_{n}(z) \mathrm{d} z , \quad m\geq0, \quad n\geq0.
</math></center>
For <math>m \geq 0</math> and <math>n \geq 1</math> the evaluation is above with the exception that the indices of the expression are switched.

Finally there is the special case of when <math>m \geq 0</math> and <math>n = 0</math>.
<center><math>
B_{0m} = \frac{\sin k_{m}(h-d)}{k_{m}\cos(k_{m}h)}.
</math></center>

=== Expansion of the potential ===

We need to apply some boundary conditions at plus and minus infinity,
where essentially 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 roots. Here we
must use a different condition, essentially identifying one solution
as the incoming wave and the other as the outgoing wave.

Therefore the diffracted potentials can
be expanded as
<center>
<math>
\phi(x,z)=\sum_{m=0}^{\infty}a_{m}e^{k_{m}x}\chi_{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 '''symmetric''' and '''antisymmetric''' potentials <math>\phi^{s,a}(x,z) \,</math> must be continuous across the
transition from open water to the dock covered region. Therefore, the
potentials at <math>x=-L</math> have to be equal.
For these cases we obtain:

<center><math>
\chi_{0}\left( z\right) + \sum_{m=0}^{\infty}a^{s,a}_{m} \chi_{m}\left(z\right)
=\sum_{m=0}^{\infty}b^{s,a}_{m}\psi_{m}(z), \quad -h \leq z\leq -d.
</math></center>

We multiply both sides by <math>\psi_{n}(z) \,</math> and integrate from <math>-h</math> to <math>-d</math> to obtain:

<center><math>
B_{n0} + \sum^{\infty}_{m=0}a^{s,a}_{m}B_{nm} = b^{s,a}_{m}C_{m}\delta_{mn}, \quad n\in\mathbb{N}\cup\left\{0\right\}.
</math></center>

The derivative of the '''symmetric''' potential <math>\partial_x\phi^{s}(x,z) \,</math> must be continuous across the
transition from open water to the dock covered region. Therefore, the derivatives of the potentials at <math>x=-L</math> have to be equal.
For this case we obtain:

<center><math>
\begin{align}
-k_{0}\chi_{0}\left( z\right) +\sum_{m=0}^{\infty} a^{s}_{m}k_{m}\chi_{m}\left(z\right) =
\begin{cases}
\quad \quad \quad \quad \quad 0,\quad \quad \quad \quad \quad \quad -d\leq z\leq 0 \\
-\sum_{m=1}^{\infty}b^{s}_{m}\kappa_{m}\tanh(\kappa_{m}L)\psi_{m}(z),\quad -h\leq z\leq -d
\end{cases}
\end{align}
</math></center>

and for the second equation we multiply both sides by <math>\chi_{n}(z) \,</math> and integrating from <math>-h</math> to <math>0</math> to obtain:
<center><math>
-k_{0}A_{0}\delta_{0n} + k_{m}a_{m}A_{m} = -\sum^{\infty}_{m=0}\kappa_{m}b_{m}B_{mn}, \quad n\in\mathbb{N}\cup\left\{0\right\}
</math></center>

Solving the equations above will yield the coefficients of the water velocity potential in the dock covered region.

== Numerical Solution ==

To solve the system of equations for the '''symmetric''' case, we set the upper limit of <math>m</math> to be <math>N</math>. This resulting system can be expressed in the block matrix form below,

<center>
<math>
\begin{bmatrix}
B_{00} & B_{01} & \cdots & B_{0N} & -C_{0} & 0 & \cdots & 0 \\
B_{10} & B_{11} & \cdots & B_{1N} & 0 & -C_{1} & \cdots & \vdots \\
\vdots & \vdots & \ddots & \vdots & \vdots & \vdots & \ddots & 0 \\
B_{N0} & B_{N1} & \cdots & B_{NN} & 0 & \cdots & 0 & -C_{N} \\
k_{0}A_{0} & 0 & \cdots & 0 & 0 & \kappa_{1}\tanh(\kappa_{1}L)B_{10} & \cdots & \kappa_{N}\tanh(\kappa_{N}L)B_{N0} \\
0 & k_{1}A_{1} & \cdots & \vdots & 0 & \kappa_{1}\tanh(\kappa_{1}L)B_{11} & \cdots & \kappa_{N}\tanh(\kappa_{N}L)B_{N1} \\
\vdots & \vdots & \ddots & 0 & \vdots & \vdots & \ddots & \vdots \\
0 & \cdots & 0 & k_{N}A_{N} & 0 & \kappa_{1}\tanh(\kappa_{1}L)B_{1N} & \cdots & \kappa_{N}\tanh(\kappa_{N}L)B_{NN} \\
\end{bmatrix}
\begin{bmatrix}
a^s_{0} \\ a^s_{1} \\ \vdots \\ a^s_{N} \\ b^s_{0} \\ b^s_{1} \\ \vdots \\ b^s_{N} \\
\end{bmatrix}
=\begin{bmatrix}
-B_{00} \\ -B_{10} \\ \vdots \\ -B_{N0} \\ k_{0}A_{0} \\ 0 \\ \vdots \\ 0 \\
\end{bmatrix}
</math>
</center>

To solve the system of equations for the '''antisymmetric''' case, we set the upper limit of <math>m</math> to be <math>N</math>. This resulting system can be expressed in the block matrix form below,

<center>
<math>
\begin{bmatrix}
B_{00} & B_{01} & \cdots & B_{0N} & -C_{0} & 0 & \cdots & 0 \\
B_{10} & B_{11} & \cdots & B_{1N} & 0 & -C_{1} & \cdots & \vdots \\
\vdots & \vdots & \ddots & \vdots & \vdots & \vdots & \ddots & 0 \\
B_{N0} & B_{N1} & \cdots & B_{NN} & 0 & \cdots & 0 & -C_{N} \\
k_{0}A_{0} & 0 & \cdots & 0 & \frac{1}{L}D_{0} & \kappa_{1}\coth(\kappa_{1}L)B_{10} & \cdots & \kappa_{N}\coth(\kappa_{N}L)B_{N0} \\
0 & k_{1}A_{1} & \cdots & \vdots & \frac{1}{L}D_{1} & \kappa_{1}\coth(\kappa_{1}L)B_{11} & \cdots & \kappa_{N}\coth(\kappa_{N}L)B_{N1} \\
\vdots & \vdots & \ddots & 0 & \vdots & \vdots & \ddots & \vdots \\
0 & \cdots & 0 & k_{N}A_{N} & \frac{1}{L}D_{N} & \kappa_{1}\coth(\kappa_{1}L)B_{1N} & \cdots & \kappa_{N}\coth(\kappa_{N}L)B_{NN} \\
\end{bmatrix}
\begin{bmatrix}
a^a_{0} \\ a^a_{1} \\ \vdots \\ a^a_{N} \\ b^a_{0} \\ b^a_{1} \\ \vdots \\ b^a_{N} \\
\end{bmatrix}
=\begin{bmatrix}
-B_{00} \\ -B_{10} \\ \vdots \\ -B_{N0} \\ k_{0}A_{0} \\ 0 \\ \vdots \\ 0 \\
\end{bmatrix}
</math>
</center>

We then simply need to solve the two sets of <math>2(N+1) \mbox{ x } 2(N+1)</math> linear systems of equations. If we had not taken advantage of the symmetry we would need to solve a <math>4(N+1) \mbox{ x } 4(N+1)</math> linear system of equations for four coefficients <math>a_{m}</math>, <math>b_{m}</math>, <math>c_{m}</math> and <math>d_{m}</math>.

== Matlab Code ==

A program to calculate the coefficients for the single finite dock problem using symmetry 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 Finite Rectangle using Symmetry

2012-03-16T05:57:42Z

Tertius: /* An infinite dimensional system of equations */

{{incomplete pages}}

== Introduction ==

The problem consists of a finite rigid fixed rectangle of length <math>2L</math> centered at <math>x = 0</math> and having a height of <math>d + ...</math> freeboard with depth of submergence being <math>d</math> through which no flow is possible. The regions to the left and right of the rectangle are the free water surface. Since the geometry of the rectangle is symmetric about <math>x = 0</math> we can decompose the velocity potential <math>\phi</math> into a symmetric <math>\phi^s</math> and antisymmetric <math>\phi^a</math> potential. We then solve for these two potentials in the left domain only <math>x \leq 0</math> and subsequently recompose the solution <math>\phi</math> for the entire domain. We begin with the simple problem when the waves are normally incident on the rectangle (dock) (so that the problem is truly two-dimensional). We can extend this solution to a pair of partially submerged finite rigid fixed rectangles using symmetry
[[Eigenfunction Matching for a Finite Rectangle using Symmetry]]

== Governing Equations ==

We begin with the [[Frequency Domain Problem]] for a partially submerged finite dock. Although the dock is of length <math>2L</math> we are only interested in the dock of <math>-L \leq x \leq 0</math> together with the fluid domain <math>x \leq 0</math>. We assume <math>e^{i\omega t}</math> time dependence and
the depth of submergence is <math>d</math>.
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>
\begin{align}
\Delta\phi^{s,a} &= 0, \,\,\, -h\leq z\leq 0, \,\, x < -L, \\

\Delta\phi^{s,a} &= 0, \,\,\, -h\leq z\leq -d, \,\, -L < x < 0. \\

\partial_z\phi^{s,a} &= \alpha\phi^{s,a}, \,\,\, z=0, \,\, x < -L, \\

\partial_x\phi^{s,a} &= 0, \,\,\, -d\leq z\leq 0, \,\, x = -L, \\

\partial_z\phi^{s,a} &= 0, \,\,\,
\begin{cases}
z = -d, \,\, -L < x < 0, \\
z = -h, \,\, x < 0.
\end{cases} \\

\partial_x\phi^s &= 0, \,\,\, x = 0, \,\, -h \leq z \leq -d, \\

\phi^a &= 0, \,\,\, x = 0, \,\, -h \leq z \leq -d.
\end{align}
</math></center>

Where <math>\alpha = \omega^2/g</math>.

<math>\phi^s(x,z)</math> is an even function and at <math>x = 0</math> the derivative of this function is equal to zero. <math>\phi^a(x,z)</math> is an odd function and at <math>x = 0</math> this function is equal to zero. These boundary conditions are important and enable the recomposed solution <math>\phi(x,z)</math> to be reflected into the domain of <math>x \geq 0</math>.

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\leq 0</math>
and <math>x\geq 0</math>.

{{separation of variables in two dimensions}}

{{separation of variables for a free surface}}

=== Separation of Variables for a Dock ===

The separation of variables equation for a partially submerged dock is given by:
<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>
Z^{\prime} (-d) = 0.
</math></center>
The solution is
<math>k=\kappa_{m}= \frac{m\pi}{h-d} \,</math> , <math>m\in\mathbb{N}\cup\left\{0\right\}</math> and
<center><math>
Z = \psi_{m}\left( z\right) = \cos\kappa_{m}(z+h),\quad
m\in\mathbb{N}\cup\left\{0\right\}.</math></center>
We note that
<center><math>
\int\nolimits_{-h}^{-d}\psi_{m}(z)\psi_{n}(z) \mathrm{d} z=C_{m}\delta_{mn},
</math></center>
where
<center>
<math>
C_{m} = \frac{1}{2} \left( \frac{\cos\kappa_{m}(h-d)\sin\kappa_{m}(h-d)+\kappa_{m}(h-d)}{\kappa_{m}} \right).
</math>
</center>
In addition there is the special case of when <math>m=n=0</math> ,
<center>
<math>
C_{0} = h-d,
</math>
</center>
where <math>\delta_{00}=1</math> .

=== Inner product between free surface and dock modes ===

<center><math>
B_{mn} = \int\nolimits_{-h}^{-d}\psi_{m}(z)\chi_{n}(z) \mathrm{d} z , \quad m\geq1, \quad n\geq0,
</math></center>
where
<center><math>
B_{mn}= \int\nolimits_{-h}^{-d} \frac{\cos\kappa_{m}(z+h)\cos k_{n}(z+h)}{\cos(k_{n}h)} \mathrm{d} z=\frac{-(-1)^m k_{n}\sin k_{n}(h-d)}
{\cos(k_{n}h)(\kappa_{m}^{2}-k_{n}^{2})}.
</math></center>
The evaluation of this inner product works on the assumption that the two roots <math>\kappa_{m}</math> and <math>k_{n}</math> do not coincide. The Matlab code given below has a routine to check for this case and make necessary adjustments.

We also have the case where the functions of the inner product are switched.
<center><math>
B_{nm} = \int\nolimits_{-h}^{-d}\chi_{m}(z)\psi_{n}(z) \mathrm{d} z , \quad m\geq0, \quad n\geq0.
</math></center>
For <math>m \geq 0</math> and <math>n \geq 1</math> the evaluation is above with the exception that the indices of the expression are switched.

Finally there is the special case of when <math>m \geq 0</math> and <math>n = 0</math>.
<center><math>
B_{0m} = \frac{\sin k_{m}(h-d)}{k_{m}\cos(k_{m}h)}.
</math></center>

=== Expansion of the potential ===

We need to apply some boundary conditions at plus and minus infinity,
where essentially 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 roots. Here we
must use a different condition, essentially identifying one solution
as the incoming wave and the other as the outgoing wave.

Therefore the diffracted potentials can
be expanded as
<center>
<math>
\phi(x,z)=\sum_{m=0}^{\infty}a_{m}e^{k_{m}x}\chi_{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 '''symmetric''' and '''antisymmetric''' potentials <math>\phi^{s,a}(x,z) \,</math> must be continuous across the
transition from open water to the dock covered region. Therefore, the
potentials at <math>x=-L</math> have to be equal.
For these cases we obtain:

<center><math>
\chi_{0}\left( z\right) + \sum_{m=0}^{\infty}a^{s,a}_{m} \chi_{m}\left(z\right)
=\sum_{m=0}^{\infty}b^{s,a}_{m}\psi_{m}(z), \quad -h \leq z\leq -d.
</math></center>

We multiply both sides by <math>\psi_{n}(z) \,</math> and integrate from <math>-h</math> to <math>-d</math> to obtain:

<center><math>
B_{n0} + \sum^{\infty}_{m=0}a^{s,a}_{m}B_{nm} = b^{s,a}_{m}C_{m}\delta_{mn}, \quad n\in\mathbb{N}\cup\left\{0\right\}.
</math></center>

The derivative of the '''symmetric''' potential <math>\phi^{s}(x,z) \,</math> must be continuous across the
transition from open water to the dock covered region. Therefore, the derivatives of the potentials at <math>x=-L</math> have to be equal.
For this case we obtain:

<center><math>
\begin{align}
-k_{0}\chi_{0}\left( z\right) +\sum_{m=0}^{\infty} a^{s}_{m}k_{m}\chi_{m}\left(z\right) =
\begin{cases}
\quad \quad \quad \quad \quad 0,\quad \quad \quad \quad \quad \quad -d\leq z\leq 0 \\
-\sum_{m=1}^{\infty}b^{s}_{m}\kappa_{m}\tanh(\kappa_{m}L)\psi_{m}(z),\quad -h\leq z\leq -d
\end{cases}
\end{align}
</math></center>

and for the second equation we multiply both sides by <math>\chi_{n}(z) \,</math> and integrating from <math>-h</math> to <math>0</math> to obtain:
<center><math>
-k_{0}A_{0}\delta_{0n} + k_{m}a_{m}A_{m} = -\sum^{\infty}_{m=0}\kappa_{m}b_{m}B_{mn}, \quad n\in\mathbb{N}\cup\left\{0\right\}
</math></center>

Solving the equations above will yield the coefficients of the water velocity potential in the dock covered region.

== Numerical Solution ==

To solve the system of equations for the '''symmetric''' case, we set the upper limit of <math>m</math> to be <math>N</math>. This resulting system can be expressed in the block matrix form below,

<center>
<math>
\begin{bmatrix}
B_{00} & B_{01} & \cdots & B_{0N} & -C_{0} & 0 & \cdots & 0 \\
B_{10} & B_{11} & \cdots & B_{1N} & 0 & -C_{1} & \cdots & \vdots \\
\vdots & \vdots & \ddots & \vdots & \vdots & \vdots & \ddots & 0 \\
B_{N0} & B_{N1} & \cdots & B_{NN} & 0 & \cdots & 0 & -C_{N} \\
k_{0}A_{0} & 0 & \cdots & 0 & 0 & \kappa_{1}\tanh(\kappa_{1}L)B_{10} & \cdots & \kappa_{N}\tanh(\kappa_{N}L)B_{N0} \\
0 & k_{1}A_{1} & \cdots & \vdots & 0 & \kappa_{1}\tanh(\kappa_{1}L)B_{11} & \cdots & \kappa_{N}\tanh(\kappa_{N}L)B_{N1} \\
\vdots & \vdots & \ddots & 0 & \vdots & \vdots & \ddots & \vdots \\
0 & \cdots & 0 & k_{N}A_{N} & 0 & \kappa_{1}\tanh(\kappa_{1}L)B_{1N} & \cdots & \kappa_{N}\tanh(\kappa_{N}L)B_{NN} \\
\end{bmatrix}
\begin{bmatrix}
a^s_{0} \\ a^s_{1} \\ \vdots \\ a^s_{N} \\ b^s_{0} \\ b^s_{1} \\ \vdots \\ b^s_{N} \\
\end{bmatrix}
=\begin{bmatrix}
-B_{00} \\ -B_{10} \\ \vdots \\ -B_{N0} \\ k_{0}A_{0} \\ 0 \\ \vdots \\ 0 \\
\end{bmatrix}
</math>
</center>

To solve the system of equations for the '''antisymmetric''' case, we set the upper limit of <math>m</math> to be <math>N</math>. This resulting system can be expressed in the block matrix form below,

<center>
<math>
\begin{bmatrix}
B_{00} & B_{01} & \cdots & B_{0N} & -C_{0} & 0 & \cdots & 0 \\
B_{10} & B_{11} & \cdots & B_{1N} & 0 & -C_{1} & \cdots & \vdots \\
\vdots & \vdots & \ddots & \vdots & \vdots & \vdots & \ddots & 0 \\
B_{N0} & B_{N1} & \cdots & B_{NN} & 0 & \cdots & 0 & -C_{N} \\
k_{0}A_{0} & 0 & \cdots & 0 & \frac{1}{L}D_{0} & \kappa_{1}\coth(\kappa_{1}L)B_{10} & \cdots & \kappa_{N}\coth(\kappa_{N}L)B_{N0} \\
0 & k_{1}A_{1} & \cdots & \vdots & \frac{1}{L}D_{1} & \kappa_{1}\coth(\kappa_{1}L)B_{11} & \cdots & \kappa_{N}\coth(\kappa_{N}L)B_{N1} \\
\vdots & \vdots & \ddots & 0 & \vdots & \vdots & \ddots & \vdots \\
0 & \cdots & 0 & k_{N}A_{N} & \frac{1}{L}D_{N} & \kappa_{1}\coth(\kappa_{1}L)B_{1N} & \cdots & \kappa_{N}\coth(\kappa_{N}L)B_{NN} \\
\end{bmatrix}
\begin{bmatrix}
a^a_{0} \\ a^a_{1} \\ \vdots \\ a^a_{N} \\ b^a_{0} \\ b^a_{1} \\ \vdots \\ b^a_{N} \\
\end{bmatrix}
=\begin{bmatrix}
-B_{00} \\ -B_{10} \\ \vdots \\ -B_{N0} \\ k_{0}A_{0} \\ 0 \\ \vdots \\ 0 \\
\end{bmatrix}
</math>
</center>

We then simply need to solve the two sets of <math>2(N+1) \mbox{ x } 2(N+1)</math> linear systems of equations. If we had not taken advantage of the symmetry we would need to solve a <math>4(N+1) \mbox{ x } 4(N+1)</math> linear system of equations for four coefficients <math>a_{m}</math>, <math>b_{m}</math>, <math>c_{m}</math> and <math>d_{m}</math>.

== Matlab Code ==

A program to calculate the coefficients for the single finite dock problem using symmetry 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 Finite Rectangle using Symmetry

2012-03-16T05:47:40Z

Tertius: /* An infinite dimensional system of equations */

{{incomplete pages}}

== Introduction ==

The problem consists of a finite rigid fixed rectangle of length <math>2L</math> centered at <math>x = 0</math> and having a height of <math>d + ...</math> freeboard with depth of submergence being <math>d</math> through which no flow is possible. The regions to the left and right of the rectangle are the free water surface. Since the geometry of the rectangle is symmetric about <math>x = 0</math> we can decompose the velocity potential <math>\phi</math> into a symmetric <math>\phi^s</math> and antisymmetric <math>\phi^a</math> potential. We then solve for these two potentials in the left domain only <math>x \leq 0</math> and subsequently recompose the solution <math>\phi</math> for the entire domain. We begin with the simple problem when the waves are normally incident on the rectangle (dock) (so that the problem is truly two-dimensional). We can extend this solution to a pair of partially submerged finite rigid fixed rectangles using symmetry
[[Eigenfunction Matching for a Finite Rectangle using Symmetry]]

== Governing Equations ==

We begin with the [[Frequency Domain Problem]] for a partially submerged finite dock. Although the dock is of length <math>2L</math> we are only interested in the dock of <math>-L \leq x \leq 0</math> together with the fluid domain <math>x \leq 0</math>. We assume <math>e^{i\omega t}</math> time dependence and
the depth of submergence is <math>d</math>.
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>
\begin{align}
\Delta\phi^{s,a} &= 0, \,\,\, -h\leq z\leq 0, \,\, x < -L, \\

\Delta\phi^{s,a} &= 0, \,\,\, -h\leq z\leq -d, \,\, -L < x < 0. \\

\partial_z\phi^{s,a} &= \alpha\phi^{s,a}, \,\,\, z=0, \,\, x < -L, \\

\partial_x\phi^{s,a} &= 0, \,\,\, -d\leq z\leq 0, \,\, x = -L, \\

\partial_z\phi^{s,a} &= 0, \,\,\,
\begin{cases}
z = -d, \,\, -L < x < 0, \\
z = -h, \,\, x < 0.
\end{cases} \\

\partial_x\phi^s &= 0, \,\,\, x = 0, \,\, -h \leq z \leq -d, \\

\phi^a &= 0, \,\,\, x = 0, \,\, -h \leq z \leq -d.
\end{align}
</math></center>

Where <math>\alpha = \omega^2/g</math>.

<math>\phi^s(x,z)</math> is an even function and at <math>x = 0</math> the derivative of this function is equal to zero. <math>\phi^a(x,z)</math> is an odd function and at <math>x = 0</math> this function is equal to zero. These boundary conditions are important and enable the recomposed solution <math>\phi(x,z)</math> to be reflected into the domain of <math>x \geq 0</math>.

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\leq 0</math>
and <math>x\geq 0</math>.

{{separation of variables in two dimensions}}

{{separation of variables for a free surface}}

=== Separation of Variables for a Dock ===

The separation of variables equation for a partially submerged dock is given by:
<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>
Z^{\prime} (-d) = 0.
</math></center>
The solution is
<math>k=\kappa_{m}= \frac{m\pi}{h-d} \,</math> , <math>m\in\mathbb{N}\cup\left\{0\right\}</math> and
<center><math>
Z = \psi_{m}\left( z\right) = \cos\kappa_{m}(z+h),\quad
m\in\mathbb{N}\cup\left\{0\right\}.</math></center>
We note that
<center><math>
\int\nolimits_{-h}^{-d}\psi_{m}(z)\psi_{n}(z) \mathrm{d} z=C_{m}\delta_{mn},
</math></center>
where
<center>
<math>
C_{m} = \frac{1}{2} \left( \frac{\cos\kappa_{m}(h-d)\sin\kappa_{m}(h-d)+\kappa_{m}(h-d)}{\kappa_{m}} \right).
</math>
</center>
In addition there is the special case of when <math>m=n=0</math> ,
<center>
<math>
C_{0} = h-d,
</math>
</center>
where <math>\delta_{00}=1</math> .

=== Inner product between free surface and dock modes ===

<center><math>
B_{mn} = \int\nolimits_{-h}^{-d}\psi_{m}(z)\chi_{n}(z) \mathrm{d} z , \quad m\geq1, \quad n\geq0,
</math></center>
where
<center><math>
B_{mn}= \int\nolimits_{-h}^{-d} \frac{\cos\kappa_{m}(z+h)\cos k_{n}(z+h)}{\cos(k_{n}h)} \mathrm{d} z=\frac{-(-1)^m k_{n}\sin k_{n}(h-d)}
{\cos(k_{n}h)(\kappa_{m}^{2}-k_{n}^{2})}.
</math></center>
The evaluation of this inner product works on the assumption that the two roots <math>\kappa_{m}</math> and <math>k_{n}</math> do not coincide. The Matlab code given below has a routine to check for this case and make necessary adjustments.

We also have the case where the functions of the inner product are switched.
<center><math>
B_{nm} = \int\nolimits_{-h}^{-d}\chi_{m}(z)\psi_{n}(z) \mathrm{d} z , \quad m\geq0, \quad n\geq0.
</math></center>
For <math>m \geq 0</math> and <math>n \geq 1</math> the evaluation is above with the exception that the indices of the expression are switched.

Finally there is the special case of when <math>m \geq 0</math> and <math>n = 0</math>.
<center><math>
B_{0m} = \frac{\sin k_{m}(h-d)}{k_{m}\cos(k_{m}h)}.
</math></center>

=== Expansion of the potential ===

We need to apply some boundary conditions at plus and minus infinity,
where essentially 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 roots. Here we
must use a different condition, essentially identifying one solution
as the incoming wave and the other as the outgoing wave.

Therefore the diffracted potentials can
be expanded as
<center>
<math>
\phi(x,z)=\sum_{m=0}^{\infty}a_{m}e^{k_{m}x}\chi_{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 must be continuous across the
transition from open water to the dock covered region. Therefore, the
potentials at <math>x=-L</math> have to be equal.
We obtain for the '''symmetric''' and '''antisymmetric''' cases:

<center><math>
\chi_{0}\left( z\right) + \sum_{m=0}^{\infty}a^{s,a}_{m} \chi_{m}\left(z\right)
=\sum_{m=0}^{\infty}b^{s,a}_{m}\psi_{m}(z), \quad -h \leq z\leq -d.
</math></center>

We multiply both sides by <math>\psi_{n}(z) \,</math> and integrate from <math>-h</math> to <math>-d</math> to obtain:

<center><math>
B_{n0} + \sum^{\infty}_{m=0}a^{s,a}_{m}B_{nm} = b^{s,a}_{m}C_{m}\delta_{mn}, \quad n\in\mathbb{N}\cup\left\{0\right\}.
</math></center>

The derivative of the '''symmetric''' potential <math>\phi^{s}(x,z) \,</math> must be continuous across the
transition from open water to the dock covered region. Therefore, the derivatives of the potentials at <math>x=-L</math> have to be equal.
For this case we obtain:

<center><math>
\begin{align}
-k_{0}\chi_{0}\left( z\right) +\sum_{m=0}^{\infty} a^{s}_{m}k_{m}\chi_{m}\left(z\right) =
\begin{cases}
\quad \quad \quad \quad \quad 0,\quad \quad \quad \quad \quad \quad -d\leq z\leq 0 \\
-\sum_{m=1}^{\infty}b^{s}_{m}\kappa_{m}\tanh(\kappa_{m}L)\psi_{m}(z),\quad -h\leq z\leq -d
\end{cases}
\end{align}
</math></center>

and for the second equation we multiply both sides by <math>\chi_{n}(z) \,</math> and integrating from <math>-h</math> to <math>0</math> to obtain:
<center><math>
-k_{0}A_{0}\delta_{0n} + k_{m}a_{m}A_{m} = -\sum^{\infty}_{m=0}\kappa_{m}b_{m}B_{mn}, \quad n\in\mathbb{N}\cup\left\{0\right\}
</math></center>

Solving the equations above will yield the coefficients of the water velocity potential in the dock covered region.

== Numerical Solution ==

To solve the system of equations for the '''symmetric''' case, we set the upper limit of <math>m</math> to be <math>N</math>. This resulting system can be expressed in the block matrix form below,

<center>
<math>
\begin{bmatrix}
B_{00} & B_{01} & \cdots & B_{0N} & -C_{0} & 0 & \cdots & 0 \\
B_{10} & B_{11} & \cdots & B_{1N} & 0 & -C_{1} & \cdots & \vdots \\
\vdots & \vdots & \ddots & \vdots & \vdots & \vdots & \ddots & 0 \\
B_{N0} & B_{N1} & \cdots & B_{NN} & 0 & \cdots & 0 & -C_{N} \\
k_{0}A_{0} & 0 & \cdots & 0 & 0 & \kappa_{1}\tanh(\kappa_{1}L)B_{10} & \cdots & \kappa_{N}\tanh(\kappa_{N}L)B_{N0} \\
0 & k_{1}A_{1} & \cdots & \vdots & 0 & \kappa_{1}\tanh(\kappa_{1}L)B_{11} & \cdots & \kappa_{N}\tanh(\kappa_{N}L)B_{N1} \\
\vdots & \vdots & \ddots & 0 & \vdots & \vdots & \ddots & \vdots \\
0 & \cdots & 0 & k_{N}A_{N} & 0 & \kappa_{1}\tanh(\kappa_{1}L)B_{1N} & \cdots & \kappa_{N}\tanh(\kappa_{N}L)B_{NN} \\
\end{bmatrix}
\begin{bmatrix}
a^s_{0} \\ a^s_{1} \\ \vdots \\ a^s_{N} \\ b^s_{0} \\ b^s_{1} \\ \vdots \\ b^s_{N} \\
\end{bmatrix}
=\begin{bmatrix}
-B_{00} \\ -B_{10} \\ \vdots \\ -B_{N0} \\ k_{0}A_{0} \\ 0 \\ \vdots \\ 0 \\
\end{bmatrix}
</math>
</center>

To solve the system of equations for the '''antisymmetric''' case, we set the upper limit of <math>m</math> to be <math>N</math>. This resulting system can be expressed in the block matrix form below,

<center>
<math>
\begin{bmatrix}
B_{00} & B_{01} & \cdots & B_{0N} & -C_{0} & 0 & \cdots & 0 \\
B_{10} & B_{11} & \cdots & B_{1N} & 0 & -C_{1} & \cdots & \vdots \\
\vdots & \vdots & \ddots & \vdots & \vdots & \vdots & \ddots & 0 \\
B_{N0} & B_{N1} & \cdots & B_{NN} & 0 & \cdots & 0 & -C_{N} \\
k_{0}A_{0} & 0 & \cdots & 0 & \frac{1}{L}D_{0} & \kappa_{1}\coth(\kappa_{1}L)B_{10} & \cdots & \kappa_{N}\coth(\kappa_{N}L)B_{N0} \\
0 & k_{1}A_{1} & \cdots & \vdots & \frac{1}{L}D_{1} & \kappa_{1}\coth(\kappa_{1}L)B_{11} & \cdots & \kappa_{N}\coth(\kappa_{N}L)B_{N1} \\
\vdots & \vdots & \ddots & 0 & \vdots & \vdots & \ddots & \vdots \\
0 & \cdots & 0 & k_{N}A_{N} & \frac{1}{L}D_{N} & \kappa_{1}\coth(\kappa_{1}L)B_{1N} & \cdots & \kappa_{N}\coth(\kappa_{N}L)B_{NN} \\
\end{bmatrix}
\begin{bmatrix}
a^a_{0} \\ a^a_{1} \\ \vdots \\ a^a_{N} \\ b^a_{0} \\ b^a_{1} \\ \vdots \\ b^a_{N} \\
\end{bmatrix}
=\begin{bmatrix}
-B_{00} \\ -B_{10} \\ \vdots \\ -B_{N0} \\ k_{0}A_{0} \\ 0 \\ \vdots \\ 0 \\
\end{bmatrix}
</math>
</center>

We then simply need to solve the two sets of <math>2(N+1) \mbox{ x } 2(N+1)</math> linear systems of equations. If we had not taken advantage of the symmetry we would need to solve a <math>4(N+1) \mbox{ x } 4(N+1)</math> linear system of equations for four coefficients <math>a_{m}</math>, <math>b_{m}</math>, <math>c_{m}</math> and <math>d_{m}</math>.

== Matlab Code ==

A program to calculate the coefficients for the single finite dock problem using symmetry 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 Finite Rectangle using Symmetry

2012-03-16T05:41:49Z

Tertius: /* An infinite dimensional system of equations */

{{incomplete pages}}

== Introduction ==

The problem consists of a finite rigid fixed rectangle of length <math>2L</math> centered at <math>x = 0</math> and having a height of <math>d + ...</math> freeboard with depth of submergence being <math>d</math> through which no flow is possible. The regions to the left and right of the rectangle are the free water surface. Since the geometry of the rectangle is symmetric about <math>x = 0</math> we can decompose the velocity potential <math>\phi</math> into a symmetric <math>\phi^s</math> and antisymmetric <math>\phi^a</math> potential. We then solve for these two potentials in the left domain only <math>x \leq 0</math> and subsequently recompose the solution <math>\phi</math> for the entire domain. We begin with the simple problem when the waves are normally incident on the rectangle (dock) (so that the problem is truly two-dimensional). We can extend this solution to a pair of partially submerged finite rigid fixed rectangles using symmetry
[[Eigenfunction Matching for a Finite Rectangle using Symmetry]]

== Governing Equations ==

We begin with the [[Frequency Domain Problem]] for a partially submerged finite dock. Although the dock is of length <math>2L</math> we are only interested in the dock of <math>-L \leq x \leq 0</math> together with the fluid domain <math>x \leq 0</math>. We assume <math>e^{i\omega t}</math> time dependence and
the depth of submergence is <math>d</math>.
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>
\begin{align}
\Delta\phi^{s,a} &= 0, \,\,\, -h\leq z\leq 0, \,\, x < -L, \\

\Delta\phi^{s,a} &= 0, \,\,\, -h\leq z\leq -d, \,\, -L < x < 0. \\

\partial_z\phi^{s,a} &= \alpha\phi^{s,a}, \,\,\, z=0, \,\, x < -L, \\

\partial_x\phi^{s,a} &= 0, \,\,\, -d\leq z\leq 0, \,\, x = -L, \\

\partial_z\phi^{s,a} &= 0, \,\,\,
\begin{cases}
z = -d, \,\, -L < x < 0, \\
z = -h, \,\, x < 0.
\end{cases} \\

\partial_x\phi^s &= 0, \,\,\, x = 0, \,\, -h \leq z \leq -d, \\

\phi^a &= 0, \,\,\, x = 0, \,\, -h \leq z \leq -d.
\end{align}
</math></center>

Where <math>\alpha = \omega^2/g</math>.

<math>\phi^s(x,z)</math> is an even function and at <math>x = 0</math> the derivative of this function is equal to zero. <math>\phi^a(x,z)</math> is an odd function and at <math>x = 0</math> this function is equal to zero. These boundary conditions are important and enable the recomposed solution <math>\phi(x,z)</math> to be reflected into the domain of <math>x \geq 0</math>.

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\leq 0</math>
and <math>x\geq 0</math>.

{{separation of variables in two dimensions}}

{{separation of variables for a free surface}}

=== Separation of Variables for a Dock ===

The separation of variables equation for a partially submerged dock is given by:
<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>
Z^{\prime} (-d) = 0.
</math></center>
The solution is
<math>k=\kappa_{m}= \frac{m\pi}{h-d} \,</math> , <math>m\in\mathbb{N}\cup\left\{0\right\}</math> and
<center><math>
Z = \psi_{m}\left( z\right) = \cos\kappa_{m}(z+h),\quad
m\in\mathbb{N}\cup\left\{0\right\}.</math></center>
We note that
<center><math>
\int\nolimits_{-h}^{-d}\psi_{m}(z)\psi_{n}(z) \mathrm{d} z=C_{m}\delta_{mn},
</math></center>
where
<center>
<math>
C_{m} = \frac{1}{2} \left( \frac{\cos\kappa_{m}(h-d)\sin\kappa_{m}(h-d)+\kappa_{m}(h-d)}{\kappa_{m}} \right).
</math>
</center>
In addition there is the special case of when <math>m=n=0</math> ,
<center>
<math>
C_{0} = h-d,
</math>
</center>
where <math>\delta_{00}=1</math> .

=== Inner product between free surface and dock modes ===

<center><math>
B_{mn} = \int\nolimits_{-h}^{-d}\psi_{m}(z)\chi_{n}(z) \mathrm{d} z , \quad m\geq1, \quad n\geq0,
</math></center>
where
<center><math>
B_{mn}= \int\nolimits_{-h}^{-d} \frac{\cos\kappa_{m}(z+h)\cos k_{n}(z+h)}{\cos(k_{n}h)} \mathrm{d} z=\frac{-(-1)^m k_{n}\sin k_{n}(h-d)}
{\cos(k_{n}h)(\kappa_{m}^{2}-k_{n}^{2})}.
</math></center>
The evaluation of this inner product works on the assumption that the two roots <math>\kappa_{m}</math> and <math>k_{n}</math> do not coincide. The Matlab code given below has a routine to check for this case and make necessary adjustments.

We also have the case where the functions of the inner product are switched.
<center><math>
B_{nm} = \int\nolimits_{-h}^{-d}\chi_{m}(z)\psi_{n}(z) \mathrm{d} z , \quad m\geq0, \quad n\geq0.
</math></center>
For <math>m \geq 0</math> and <math>n \geq 1</math> the evaluation is above with the exception that the indices of the expression are switched.

Finally there is the special case of when <math>m \geq 0</math> and <math>n = 0</math>.
<center><math>
B_{0m} = \frac{\sin k_{m}(h-d)}{k_{m}\cos(k_{m}h)}.
</math></center>

=== Expansion of the potential ===

We need to apply some boundary conditions at plus and minus infinity,
where essentially 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 roots. Here we
must use a different condition, essentially identifying one solution
as the incoming wave and the other as the outgoing wave.

Therefore the diffracted potentials can
be expanded as
<center>
<math>
\phi(x,z)=\sum_{m=0}^{\infty}a_{m}e^{k_{m}x}\chi_{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 must be continuous across the
transition from open water to the dock covered region. Therefore, the
potentials at <math>x=-L</math> have to be equal.
We obtain for the '''symmetric''' and '''antisymmetric''' cases:

<center><math>
\chi_{0}\left( z\right) + \sum_{m=0}^{\infty}a^{s,a}_{m} \chi_{m}\left(z\right)
=\sum_{m=0}^{\infty}b^{s,a}_{m}\psi_{m}(z), \quad -h \leq z\leq -d.
</math></center>

We multiply both sides by <math>\psi_{n}(z) \,</math> and integrate from <math>-h</math> to <math>-d</math> to obtain:

<center><math>
B_{n0} + \sum^{\infty}_{m=0}a^{s,a}_{m}B_{nm} = b^{s,a}_{m}C_{m}\delta_{mn}, \quad n\in\mathbb{N}\cup\left\{0\right\}.
</math></center>

The derivative of the potential must be continuous across the
transition from open water to the dock covered region. Therefore, the derivatives of the potentials at <math>x=-L</math> have to be equal.
We obtain for the '''symmetric''' <math>\phi^{s}(x,z) \,</math> case:

<center><math>
\begin{align}
-k_{0}\chi_{0}\left( z\right) +\sum_{m=0}^{\infty} a^{s}_{m}k_{m}\chi_{m}\left(z\right) =
\begin{cases}
\quad \quad \quad \quad \quad 0,\quad \quad \quad \quad \quad \quad -d\leq z\leq 0 \\
-\sum_{m=1}^{\infty}b^{s}_{m}\kappa_{m}\tanh(\kappa_{m}L)\psi_{m}(z),\quad -h\leq z\leq -d
\end{cases}
\end{align}
</math></center>

and for the second equation we multiply both sides by <math>\chi_{n}(z) \,</math> and integrating from <math>-h</math> to <math>0</math> to obtain:
<center><math>
-k_{0}A_{0}\delta_{0n} + k_{m}a_{m}A_{m} = -\sum^{\infty}_{m=0}\kappa_{m}b_{m}B_{mn}, \quad n\in\mathbb{N}\cup\left\{0\right\}
</math></center>

Solving the equations above will yield the coefficients of the water velocity potential in the dock covered region.

== Numerical Solution ==

To solve the system of equations for the '''symmetric''' case, we set the upper limit of <math>m</math> to be <math>N</math>. This resulting system can be expressed in the block matrix form below,

<center>
<math>
\begin{bmatrix}
B_{00} & B_{01} & \cdots & B_{0N} & -C_{0} & 0 & \cdots & 0 \\
B_{10} & B_{11} & \cdots & B_{1N} & 0 & -C_{1} & \cdots & \vdots \\
\vdots & \vdots & \ddots & \vdots & \vdots & \vdots & \ddots & 0 \\
B_{N0} & B_{N1} & \cdots & B_{NN} & 0 & \cdots & 0 & -C_{N} \\
k_{0}A_{0} & 0 & \cdots & 0 & 0 & \kappa_{1}\tanh(\kappa_{1}L)B_{10} & \cdots & \kappa_{N}\tanh(\kappa_{N}L)B_{N0} \\
0 & k_{1}A_{1} & \cdots & \vdots & 0 & \kappa_{1}\tanh(\kappa_{1}L)B_{11} & \cdots & \kappa_{N}\tanh(\kappa_{N}L)B_{N1} \\
\vdots & \vdots & \ddots & 0 & \vdots & \vdots & \ddots & \vdots \\
0 & \cdots & 0 & k_{N}A_{N} & 0 & \kappa_{1}\tanh(\kappa_{1}L)B_{1N} & \cdots & \kappa_{N}\tanh(\kappa_{N}L)B_{NN} \\
\end{bmatrix}
\begin{bmatrix}
a^s_{0} \\ a^s_{1} \\ \vdots \\ a^s_{N} \\ b^s_{0} \\ b^s_{1} \\ \vdots \\ b^s_{N} \\
\end{bmatrix}
=\begin{bmatrix}
-B_{00} \\ -B_{10} \\ \vdots \\ -B_{N0} \\ k_{0}A_{0} \\ 0 \\ \vdots \\ 0 \\
\end{bmatrix}
</math>
</center>

To solve the system of equations for the '''antisymmetric''' case, we set the upper limit of <math>m</math> to be <math>N</math>. This resulting system can be expressed in the block matrix form below,

<center>
<math>
\begin{bmatrix}
B_{00} & B_{01} & \cdots & B_{0N} & -C_{0} & 0 & \cdots & 0 \\
B_{10} & B_{11} & \cdots & B_{1N} & 0 & -C_{1} & \cdots & \vdots \\
\vdots & \vdots & \ddots & \vdots & \vdots & \vdots & \ddots & 0 \\
B_{N0} & B_{N1} & \cdots & B_{NN} & 0 & \cdots & 0 & -C_{N} \\
k_{0}A_{0} & 0 & \cdots & 0 & \frac{1}{L}D_{0} & \kappa_{1}\coth(\kappa_{1}L)B_{10} & \cdots & \kappa_{N}\coth(\kappa_{N}L)B_{N0} \\
0 & k_{1}A_{1} & \cdots & \vdots & \frac{1}{L}D_{1} & \kappa_{1}\coth(\kappa_{1}L)B_{11} & \cdots & \kappa_{N}\coth(\kappa_{N}L)B_{N1} \\
\vdots & \vdots & \ddots & 0 & \vdots & \vdots & \ddots & \vdots \\
0 & \cdots & 0 & k_{N}A_{N} & \frac{1}{L}D_{N} & \kappa_{1}\coth(\kappa_{1}L)B_{1N} & \cdots & \kappa_{N}\coth(\kappa_{N}L)B_{NN} \\
\end{bmatrix}
\begin{bmatrix}
a^a_{0} \\ a^a_{1} \\ \vdots \\ a^a_{N} \\ b^a_{0} \\ b^a_{1} \\ \vdots \\ b^a_{N} \\
\end{bmatrix}
=\begin{bmatrix}
-B_{00} \\ -B_{10} \\ \vdots \\ -B_{N0} \\ k_{0}A_{0} \\ 0 \\ \vdots \\ 0 \\
\end{bmatrix}
</math>
</center>

We then simply need to solve the two sets of <math>2(N+1) \mbox{ x } 2(N+1)</math> linear systems of equations. If we had not taken advantage of the symmetry we would need to solve a <math>4(N+1) \mbox{ x } 4(N+1)</math> linear system of equations for four coefficients <math>a_{m}</math>, <math>b_{m}</math>, <math>c_{m}</math> and <math>d_{m}</math>.

== Matlab Code ==

A program to calculate the coefficients for the single finite dock problem using symmetry 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 Finite Rectangle using Symmetry

2012-03-16T05:39:29Z

Tertius: /* An infinite dimensional system of equations */

{{incomplete pages}}

== Introduction ==

The problem consists of a finite rigid fixed rectangle of length <math>2L</math> centered at <math>x = 0</math> and having a height of <math>d + ...</math> freeboard with depth of submergence being <math>d</math> through which no flow is possible. The regions to the left and right of the rectangle are the free water surface. Since the geometry of the rectangle is symmetric about <math>x = 0</math> we can decompose the velocity potential <math>\phi</math> into a symmetric <math>\phi^s</math> and antisymmetric <math>\phi^a</math> potential. We then solve for these two potentials in the left domain only <math>x \leq 0</math> and subsequently recompose the solution <math>\phi</math> for the entire domain. We begin with the simple problem when the waves are normally incident on the rectangle (dock) (so that the problem is truly two-dimensional). We can extend this solution to a pair of partially submerged finite rigid fixed rectangles using symmetry
[[Eigenfunction Matching for a Finite Rectangle using Symmetry]]

== Governing Equations ==

We begin with the [[Frequency Domain Problem]] for a partially submerged finite dock. Although the dock is of length <math>2L</math> we are only interested in the dock of <math>-L \leq x \leq 0</math> together with the fluid domain <math>x \leq 0</math>. We assume <math>e^{i\omega t}</math> time dependence and
the depth of submergence is <math>d</math>.
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>
\begin{align}
\Delta\phi^{s,a} &= 0, \,\,\, -h\leq z\leq 0, \,\, x < -L, \\

\Delta\phi^{s,a} &= 0, \,\,\, -h\leq z\leq -d, \,\, -L < x < 0. \\

\partial_z\phi^{s,a} &= \alpha\phi^{s,a}, \,\,\, z=0, \,\, x < -L, \\

\partial_x\phi^{s,a} &= 0, \,\,\, -d\leq z\leq 0, \,\, x = -L, \\

\partial_z\phi^{s,a} &= 0, \,\,\,
\begin{cases}
z = -d, \,\, -L < x < 0, \\
z = -h, \,\, x < 0.
\end{cases} \\

\partial_x\phi^s &= 0, \,\,\, x = 0, \,\, -h \leq z \leq -d, \\

\phi^a &= 0, \,\,\, x = 0, \,\, -h \leq z \leq -d.
\end{align}
</math></center>

Where <math>\alpha = \omega^2/g</math>.

<math>\phi^s(x,z)</math> is an even function and at <math>x = 0</math> the derivative of this function is equal to zero. <math>\phi^a(x,z)</math> is an odd function and at <math>x = 0</math> this function is equal to zero. These boundary conditions are important and enable the recomposed solution <math>\phi(x,z)</math> to be reflected into the domain of <math>x \geq 0</math>.

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\leq 0</math>
and <math>x\geq 0</math>.

{{separation of variables in two dimensions}}

{{separation of variables for a free surface}}

=== Separation of Variables for a Dock ===

The separation of variables equation for a partially submerged dock is given by:
<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>
Z^{\prime} (-d) = 0.
</math></center>
The solution is
<math>k=\kappa_{m}= \frac{m\pi}{h-d} \,</math> , <math>m\in\mathbb{N}\cup\left\{0\right\}</math> and
<center><math>
Z = \psi_{m}\left( z\right) = \cos\kappa_{m}(z+h),\quad
m\in\mathbb{N}\cup\left\{0\right\}.</math></center>
We note that
<center><math>
\int\nolimits_{-h}^{-d}\psi_{m}(z)\psi_{n}(z) \mathrm{d} z=C_{m}\delta_{mn},
</math></center>
where
<center>
<math>
C_{m} = \frac{1}{2} \left( \frac{\cos\kappa_{m}(h-d)\sin\kappa_{m}(h-d)+\kappa_{m}(h-d)}{\kappa_{m}} \right).
</math>
</center>
In addition there is the special case of when <math>m=n=0</math> ,
<center>
<math>
C_{0} = h-d,
</math>
</center>
where <math>\delta_{00}=1</math> .

=== Inner product between free surface and dock modes ===

<center><math>
B_{mn} = \int\nolimits_{-h}^{-d}\psi_{m}(z)\chi_{n}(z) \mathrm{d} z , \quad m\geq1, \quad n\geq0,
</math></center>
where
<center><math>
B_{mn}= \int\nolimits_{-h}^{-d} \frac{\cos\kappa_{m}(z+h)\cos k_{n}(z+h)}{\cos(k_{n}h)} \mathrm{d} z=\frac{-(-1)^m k_{n}\sin k_{n}(h-d)}
{\cos(k_{n}h)(\kappa_{m}^{2}-k_{n}^{2})}.
</math></center>
The evaluation of this inner product works on the assumption that the two roots <math>\kappa_{m}</math> and <math>k_{n}</math> do not coincide. The Matlab code given below has a routine to check for this case and make necessary adjustments.

We also have the case where the functions of the inner product are switched.
<center><math>
B_{nm} = \int\nolimits_{-h}^{-d}\chi_{m}(z)\psi_{n}(z) \mathrm{d} z , \quad m\geq0, \quad n\geq0.
</math></center>
For <math>m \geq 0</math> and <math>n \geq 1</math> the evaluation is above with the exception that the indices of the expression are switched.

Finally there is the special case of when <math>m \geq 0</math> and <math>n = 0</math>.
<center><math>
B_{0m} = \frac{\sin k_{m}(h-d)}{k_{m}\cos(k_{m}h)}.
</math></center>

=== Expansion of the potential ===

We need to apply some boundary conditions at plus and minus infinity,
where essentially 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 roots. Here we
must use a different condition, essentially identifying one solution
as the incoming wave and the other as the outgoing wave.

Therefore the diffracted potentials can
be expanded as
<center>
<math>
\phi(x,z)=\sum_{m=0}^{\infty}a_{m}e^{k_{m}x}\chi_{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 must be continuous across the
transition from open water to the dock covered region. Therefore, the
potentials at <math>x=-L</math> have to be equal.
We obtain for the '''symmetric''' and '''antisymmetric''' cases:

<center><math>
\chi_{0}\left( z\right) + \sum_{m=0}^{\infty}a^{s,a}_{m} \chi_{m}\left(z\right)
=\sum_{m=0}^{\infty}b^{s,a}_{m}\psi_{m}(z), \quad -h \leq z\leq -d.
</math></center>

We multiply both sides by <math>\psi_{n}(z) \,</math> and integrate from <math>-h</math> to <math>-d</math> to obtain:

<center><math>
B_{n0} + \sum^{\infty}_{m=0}a^{s,a}_{m}B_{nm} = b^{s,a}_{m}C_{m}\delta_{mn}, \quad n\in\mathbb{N}\cup\left\{0\right\}.
</math></center>

The derivative of the potential must be continuous across the
transition from open water to the dock covered region. Therefore, the derivatives of the potentials at <math>x=-L</math> have to be equal.
We obtain for the '''symmetric''' <math>\phi_{s}(x,z) \,</math> case:

<center><math>
\begin{align}
-k_{0}\chi_{0}\left( z\right) +\sum_{m=0}^{\infty} a^{s}_{m}k_{m}\chi_{m}\left(z\right) =
\begin{cases}
\quad \quad \quad \quad 0,\quad \quad \quad \quad \quad \quad -d\leq z\leq 0 \\
-\sum_{m=1}^{\infty}b^{s}_{m}\kappa_{m}\tanh(\kappa_{m}L)\psi_{m}(z),\quad -h\leq z\leq -d
\end{cases}
\end{align}
</math></center>

and for the second equation we multiply both sides by <math>\chi_{n}(z) \,</math> and integrating from <math>-h</math> to <math>0</math> to obtain:
<center><math>
-k_{0}A_{0}\delta_{0n} + k_{m}a_{m}A_{m} = -\sum^{\infty}_{m=0}\kappa_{m}b_{m}B_{mn}, \quad n\in\mathbb{N}\cup\left\{0\right\}
</math></center>

Solving the equations above will yield the coefficients of the water velocity potential in the dock covered region.

== Numerical Solution ==

To solve the system of equations for the '''symmetric''' case, we set the upper limit of <math>m</math> to be <math>N</math>. This resulting system can be expressed in the block matrix form below,

<center>
<math>
\begin{bmatrix}
B_{00} & B_{01} & \cdots & B_{0N} & -C_{0} & 0 & \cdots & 0 \\
B_{10} & B_{11} & \cdots & B_{1N} & 0 & -C_{1} & \cdots & \vdots \\
\vdots & \vdots & \ddots & \vdots & \vdots & \vdots & \ddots & 0 \\
B_{N0} & B_{N1} & \cdots & B_{NN} & 0 & \cdots & 0 & -C_{N} \\
k_{0}A_{0} & 0 & \cdots & 0 & 0 & \kappa_{1}\tanh(\kappa_{1}L)B_{10} & \cdots & \kappa_{N}\tanh(\kappa_{N}L)B_{N0} \\
0 & k_{1}A_{1} & \cdots & \vdots & 0 & \kappa_{1}\tanh(\kappa_{1}L)B_{11} & \cdots & \kappa_{N}\tanh(\kappa_{N}L)B_{N1} \\
\vdots & \vdots & \ddots & 0 & \vdots & \vdots & \ddots & \vdots \\
0 & \cdots & 0 & k_{N}A_{N} & 0 & \kappa_{1}\tanh(\kappa_{1}L)B_{1N} & \cdots & \kappa_{N}\tanh(\kappa_{N}L)B_{NN} \\
\end{bmatrix}
\begin{bmatrix}
a^s_{0} \\ a^s_{1} \\ \vdots \\ a^s_{N} \\ b^s_{0} \\ b^s_{1} \\ \vdots \\ b^s_{N} \\
\end{bmatrix}
=\begin{bmatrix}
-B_{00} \\ -B_{10} \\ \vdots \\ -B_{N0} \\ k_{0}A_{0} \\ 0 \\ \vdots \\ 0 \\
\end{bmatrix}
</math>
</center>

To solve the system of equations for the '''antisymmetric''' case, we set the upper limit of <math>m</math> to be <math>N</math>. This resulting system can be expressed in the block matrix form below,

<center>
<math>
\begin{bmatrix}
B_{00} & B_{01} & \cdots & B_{0N} & -C_{0} & 0 & \cdots & 0 \\
B_{10} & B_{11} & \cdots & B_{1N} & 0 & -C_{1} & \cdots & \vdots \\
\vdots & \vdots & \ddots & \vdots & \vdots & \vdots & \ddots & 0 \\
B_{N0} & B_{N1} & \cdots & B_{NN} & 0 & \cdots & 0 & -C_{N} \\
k_{0}A_{0} & 0 & \cdots & 0 & \frac{1}{L}D_{0} & \kappa_{1}\coth(\kappa_{1}L)B_{10} & \cdots & \kappa_{N}\coth(\kappa_{N}L)B_{N0} \\
0 & k_{1}A_{1} & \cdots & \vdots & \frac{1}{L}D_{1} & \kappa_{1}\coth(\kappa_{1}L)B_{11} & \cdots & \kappa_{N}\coth(\kappa_{N}L)B_{N1} \\
\vdots & \vdots & \ddots & 0 & \vdots & \vdots & \ddots & \vdots \\
0 & \cdots & 0 & k_{N}A_{N} & \frac{1}{L}D_{N} & \kappa_{1}\coth(\kappa_{1}L)B_{1N} & \cdots & \kappa_{N}\coth(\kappa_{N}L)B_{NN} \\
\end{bmatrix}
\begin{bmatrix}
a^a_{0} \\ a^a_{1} \\ \vdots \\ a^a_{N} \\ b^a_{0} \\ b^a_{1} \\ \vdots \\ b^a_{N} \\
\end{bmatrix}
=\begin{bmatrix}
-B_{00} \\ -B_{10} \\ \vdots \\ -B_{N0} \\ k_{0}A_{0} \\ 0 \\ \vdots \\ 0 \\
\end{bmatrix}
</math>
</center>

We then simply need to solve the two sets of <math>2(N+1) \mbox{ x } 2(N+1)</math> linear systems of equations. If we had not taken advantage of the symmetry we would need to solve a <math>4(N+1) \mbox{ x } 4(N+1)</math> linear system of equations for four coefficients <math>a_{m}</math>, <math>b_{m}</math>, <math>c_{m}</math> and <math>d_{m}</math>.

== Matlab Code ==

A program to calculate the coefficients for the single finite dock problem using symmetry 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 Finite Rectangle using Symmetry

2012-03-16T05:37:20Z

Tertius: /* An infinite dimensional system of equations */

{{incomplete pages}}

== Introduction ==

The problem consists of a finite rigid fixed rectangle of length <math>2L</math> centered at <math>x = 0</math> and having a height of <math>d + ...</math> freeboard with depth of submergence being <math>d</math> through which no flow is possible. The regions to the left and right of the rectangle are the free water surface. Since the geometry of the rectangle is symmetric about <math>x = 0</math> we can decompose the velocity potential <math>\phi</math> into a symmetric <math>\phi^s</math> and antisymmetric <math>\phi^a</math> potential. We then solve for these two potentials in the left domain only <math>x \leq 0</math> and subsequently recompose the solution <math>\phi</math> for the entire domain. We begin with the simple problem when the waves are normally incident on the rectangle (dock) (so that the problem is truly two-dimensional). We can extend this solution to a pair of partially submerged finite rigid fixed rectangles using symmetry
[[Eigenfunction Matching for a Finite Rectangle using Symmetry]]

== Governing Equations ==

We begin with the [[Frequency Domain Problem]] for a partially submerged finite dock. Although the dock is of length <math>2L</math> we are only interested in the dock of <math>-L \leq x \leq 0</math> together with the fluid domain <math>x \leq 0</math>. We assume <math>e^{i\omega t}</math> time dependence and
the depth of submergence is <math>d</math>.
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>
\begin{align}
\Delta\phi^{s,a} &= 0, \,\,\, -h\leq z\leq 0, \,\, x < -L, \\

\Delta\phi^{s,a} &= 0, \,\,\, -h\leq z\leq -d, \,\, -L < x < 0. \\

\partial_z\phi^{s,a} &= \alpha\phi^{s,a}, \,\,\, z=0, \,\, x < -L, \\

\partial_x\phi^{s,a} &= 0, \,\,\, -d\leq z\leq 0, \,\, x = -L, \\

\partial_z\phi^{s,a} &= 0, \,\,\,
\begin{cases}
z = -d, \,\, -L < x < 0, \\
z = -h, \,\, x < 0.
\end{cases} \\

\partial_x\phi^s &= 0, \,\,\, x = 0, \,\, -h \leq z \leq -d, \\

\phi^a &= 0, \,\,\, x = 0, \,\, -h \leq z \leq -d.
\end{align}
</math></center>

Where <math>\alpha = \omega^2/g</math>.

<math>\phi^s(x,z)</math> is an even function and at <math>x = 0</math> the derivative of this function is equal to zero. <math>\phi^a(x,z)</math> is an odd function and at <math>x = 0</math> this function is equal to zero. These boundary conditions are important and enable the recomposed solution <math>\phi(x,z)</math> to be reflected into the domain of <math>x \geq 0</math>.

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\leq 0</math>
and <math>x\geq 0</math>.

{{separation of variables in two dimensions}}

{{separation of variables for a free surface}}

=== Separation of Variables for a Dock ===

The separation of variables equation for a partially submerged dock is given by:
<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>
Z^{\prime} (-d) = 0.
</math></center>
The solution is
<math>k=\kappa_{m}= \frac{m\pi}{h-d} \,</math> , <math>m\in\mathbb{N}\cup\left\{0\right\}</math> and
<center><math>
Z = \psi_{m}\left( z\right) = \cos\kappa_{m}(z+h),\quad
m\in\mathbb{N}\cup\left\{0\right\}.</math></center>
We note that
<center><math>
\int\nolimits_{-h}^{-d}\psi_{m}(z)\psi_{n}(z) \mathrm{d} z=C_{m}\delta_{mn},
</math></center>
where
<center>
<math>
C_{m} = \frac{1}{2} \left( \frac{\cos\kappa_{m}(h-d)\sin\kappa_{m}(h-d)+\kappa_{m}(h-d)}{\kappa_{m}} \right).
</math>
</center>
In addition there is the special case of when <math>m=n=0</math> ,
<center>
<math>
C_{0} = h-d,
</math>
</center>
where <math>\delta_{00}=1</math> .

=== Inner product between free surface and dock modes ===

<center><math>
B_{mn} = \int\nolimits_{-h}^{-d}\psi_{m}(z)\chi_{n}(z) \mathrm{d} z , \quad m\geq1, \quad n\geq0,
</math></center>
where
<center><math>
B_{mn}= \int\nolimits_{-h}^{-d} \frac{\cos\kappa_{m}(z+h)\cos k_{n}(z+h)}{\cos(k_{n}h)} \mathrm{d} z=\frac{-(-1)^m k_{n}\sin k_{n}(h-d)}
{\cos(k_{n}h)(\kappa_{m}^{2}-k_{n}^{2})}.
</math></center>
The evaluation of this inner product works on the assumption that the two roots <math>\kappa_{m}</math> and <math>k_{n}</math> do not coincide. The Matlab code given below has a routine to check for this case and make necessary adjustments.

We also have the case where the functions of the inner product are switched.
<center><math>
B_{nm} = \int\nolimits_{-h}^{-d}\chi_{m}(z)\psi_{n}(z) \mathrm{d} z , \quad m\geq0, \quad n\geq0.
</math></center>
For <math>m \geq 0</math> and <math>n \geq 1</math> the evaluation is above with the exception that the indices of the expression are switched.

Finally there is the special case of when <math>m \geq 0</math> and <math>n = 0</math>.
<center><math>
B_{0m} = \frac{\sin k_{m}(h-d)}{k_{m}\cos(k_{m}h)}.
</math></center>

=== Expansion of the potential ===

We need to apply some boundary conditions at plus and minus infinity,
where essentially 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 roots. Here we
must use a different condition, essentially identifying one solution
as the incoming wave and the other as the outgoing wave.

Therefore the diffracted potentials can
be expanded as
<center>
<math>
\phi(x,z)=\sum_{m=0}^{\infty}a_{m}e^{k_{m}x}\chi_{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 must be continuous across the
transition from open water to the dock covered region. Therefore, the
potentials at <math>x=-L</math> have to be equal.
We obtain for the '''symmetric''' and '''antisymmetric''' cases:

<center><math>
\chi_{0}\left( z\right) + \sum_{m=0}^{\infty}a^{s,a}_{m} \chi_{m}\left(z\right)
=\sum_{m=0}^{\infty}b^{s,a}_{m}\psi_{m}(z), \quad -h \leq z\leq -d.
</math></center>

We multiply both sides by <math>\psi_{n}(z) \,</math> and integrate from <math>-h</math> to <math>-d</math> to obtain:

<center><math>
B_{n0} + \sum^{\infty}_{m=0}a^{s,a}_{m}B_{nm} = b^{s,a}_{m}C_{m}\delta_{mn}, \quad n\in\mathbb{N}\cup\left\{0\right\}.
</math></center>

The derivative of the potential must be continuous across the
transition from open water to the dock covered region. Therefore, the derivatives of the potentials at <math>x=-L</math> have to be equal.
We obtain for the '''symmetric''' case:

<center><math>
\begin{align}
-k_{0}\chi_{0}\left( z\right) +\sum_{m=0}^{\infty} a^{s}_{m}k_{m}\chi_{m}\left(z\right) =
\begin{cases}
\quad \quad \quad \quad 0,\quad \quad \quad \quad \quad -d\leq z\leq 0 \\
-\sum_{m=1}^{\infty}b^{s}_{m}\kappa_{m}\tanh(\kappa_{m}L)\psi_{m}(z),\quad -h\leq z\leq -d
\end{cases}
\end{align}
</math></center>

and for the second equation we multiply both sides by <math>\chi_{n}(z) \,</math> and integrating from <math>-h</math> to <math>0</math> to obtain:
<center><math>
-k_{0}A_{0}\delta_{0n} + k_{m}a_{m}A_{m} = -\sum^{\infty}_{m=0}\kappa_{m}b_{m}B_{mn}, \quad n\in\mathbb{N}\cup\left\{0\right\}
</math></center>

Solving the equations above will yield the coefficients of the water velocity potential in the dock covered region.

== Numerical Solution ==

To solve the system of equations for the '''symmetric''' case, we set the upper limit of <math>m</math> to be <math>N</math>. This resulting system can be expressed in the block matrix form below,

<center>
<math>
\begin{bmatrix}
B_{00} & B_{01} & \cdots & B_{0N} & -C_{0} & 0 & \cdots & 0 \\
B_{10} & B_{11} & \cdots & B_{1N} & 0 & -C_{1} & \cdots & \vdots \\
\vdots & \vdots & \ddots & \vdots & \vdots & \vdots & \ddots & 0 \\
B_{N0} & B_{N1} & \cdots & B_{NN} & 0 & \cdots & 0 & -C_{N} \\
k_{0}A_{0} & 0 & \cdots & 0 & 0 & \kappa_{1}\tanh(\kappa_{1}L)B_{10} & \cdots & \kappa_{N}\tanh(\kappa_{N}L)B_{N0} \\
0 & k_{1}A_{1} & \cdots & \vdots & 0 & \kappa_{1}\tanh(\kappa_{1}L)B_{11} & \cdots & \kappa_{N}\tanh(\kappa_{N}L)B_{N1} \\
\vdots & \vdots & \ddots & 0 & \vdots & \vdots & \ddots & \vdots \\
0 & \cdots & 0 & k_{N}A_{N} & 0 & \kappa_{1}\tanh(\kappa_{1}L)B_{1N} & \cdots & \kappa_{N}\tanh(\kappa_{N}L)B_{NN} \\
\end{bmatrix}
\begin{bmatrix}
a^s_{0} \\ a^s_{1} \\ \vdots \\ a^s_{N} \\ b^s_{0} \\ b^s_{1} \\ \vdots \\ b^s_{N} \\
\end{bmatrix}
=\begin{bmatrix}
-B_{00} \\ -B_{10} \\ \vdots \\ -B_{N0} \\ k_{0}A_{0} \\ 0 \\ \vdots \\ 0 \\
\end{bmatrix}
</math>
</center>

To solve the system of equations for the '''antisymmetric''' case, we set the upper limit of <math>m</math> to be <math>N</math>. This resulting system can be expressed in the block matrix form below,

<center>
<math>
\begin{bmatrix}
B_{00} & B_{01} & \cdots & B_{0N} & -C_{0} & 0 & \cdots & 0 \\
B_{10} & B_{11} & \cdots & B_{1N} & 0 & -C_{1} & \cdots & \vdots \\
\vdots & \vdots & \ddots & \vdots & \vdots & \vdots & \ddots & 0 \\
B_{N0} & B_{N1} & \cdots & B_{NN} & 0 & \cdots & 0 & -C_{N} \\
k_{0}A_{0} & 0 & \cdots & 0 & \frac{1}{L}D_{0} & \kappa_{1}\coth(\kappa_{1}L)B_{10} & \cdots & \kappa_{N}\coth(\kappa_{N}L)B_{N0} \\
0 & k_{1}A_{1} & \cdots & \vdots & \frac{1}{L}D_{1} & \kappa_{1}\coth(\kappa_{1}L)B_{11} & \cdots & \kappa_{N}\coth(\kappa_{N}L)B_{N1} \\
\vdots & \vdots & \ddots & 0 & \vdots & \vdots & \ddots & \vdots \\
0 & \cdots & 0 & k_{N}A_{N} & \frac{1}{L}D_{N} & \kappa_{1}\coth(\kappa_{1}L)B_{1N} & \cdots & \kappa_{N}\coth(\kappa_{N}L)B_{NN} \\
\end{bmatrix}
\begin{bmatrix}
a^a_{0} \\ a^a_{1} \\ \vdots \\ a^a_{N} \\ b^a_{0} \\ b^a_{1} \\ \vdots \\ b^a_{N} \\
\end{bmatrix}
=\begin{bmatrix}
-B_{00} \\ -B_{10} \\ \vdots \\ -B_{N0} \\ k_{0}A_{0} \\ 0 \\ \vdots \\ 0 \\
\end{bmatrix}
</math>
</center>

We then simply need to solve the two sets of <math>2(N+1) \mbox{ x } 2(N+1)</math> linear systems of equations. If we had not taken advantage of the symmetry we would need to solve a <math>4(N+1) \mbox{ x } 4(N+1)</math> linear system of equations for four coefficients <math>a_{m}</math>, <math>b_{m}</math>, <math>c_{m}</math> and <math>d_{m}</math>.

== Matlab Code ==

A program to calculate the coefficients for the single finite dock problem using symmetry 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 Finite Rectangle using Symmetry

2012-03-16T05:35:43Z

Tertius: /* An infinite dimensional system of equations */

{{incomplete pages}}

== Introduction ==

The problem consists of a finite rigid fixed rectangle of length <math>2L</math> centered at <math>x = 0</math> and having a height of <math>d + ...</math> freeboard with depth of submergence being <math>d</math> through which no flow is possible. The regions to the left and right of the rectangle are the free water surface. Since the geometry of the rectangle is symmetric about <math>x = 0</math> we can decompose the velocity potential <math>\phi</math> into a symmetric <math>\phi^s</math> and antisymmetric <math>\phi^a</math> potential. We then solve for these two potentials in the left domain only <math>x \leq 0</math> and subsequently recompose the solution <math>\phi</math> for the entire domain. We begin with the simple problem when the waves are normally incident on the rectangle (dock) (so that the problem is truly two-dimensional). We can extend this solution to a pair of partially submerged finite rigid fixed rectangles using symmetry
[[Eigenfunction Matching for a Finite Rectangle using Symmetry]]

== Governing Equations ==

We begin with the [[Frequency Domain Problem]] for a partially submerged finite dock. Although the dock is of length <math>2L</math> we are only interested in the dock of <math>-L \leq x \leq 0</math> together with the fluid domain <math>x \leq 0</math>. We assume <math>e^{i\omega t}</math> time dependence and
the depth of submergence is <math>d</math>.
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>
\begin{align}
\Delta\phi^{s,a} &= 0, \,\,\, -h\leq z\leq 0, \,\, x < -L, \\

\Delta\phi^{s,a} &= 0, \,\,\, -h\leq z\leq -d, \,\, -L < x < 0. \\

\partial_z\phi^{s,a} &= \alpha\phi^{s,a}, \,\,\, z=0, \,\, x < -L, \\

\partial_x\phi^{s,a} &= 0, \,\,\, -d\leq z\leq 0, \,\, x = -L, \\

\partial_z\phi^{s,a} &= 0, \,\,\,
\begin{cases}
z = -d, \,\, -L < x < 0, \\
z = -h, \,\, x < 0.
\end{cases} \\

\partial_x\phi^s &= 0, \,\,\, x = 0, \,\, -h \leq z \leq -d, \\

\phi^a &= 0, \,\,\, x = 0, \,\, -h \leq z \leq -d.
\end{align}
</math></center>

Where <math>\alpha = \omega^2/g</math>.

<math>\phi^s(x,z)</math> is an even function and at <math>x = 0</math> the derivative of this function is equal to zero. <math>\phi^a(x,z)</math> is an odd function and at <math>x = 0</math> this function is equal to zero. These boundary conditions are important and enable the recomposed solution <math>\phi(x,z)</math> to be reflected into the domain of <math>x \geq 0</math>.

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\leq 0</math>
and <math>x\geq 0</math>.

{{separation of variables in two dimensions}}

{{separation of variables for a free surface}}

=== Separation of Variables for a Dock ===

The separation of variables equation for a partially submerged dock is given by:
<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>
Z^{\prime} (-d) = 0.
</math></center>
The solution is
<math>k=\kappa_{m}= \frac{m\pi}{h-d} \,</math> , <math>m\in\mathbb{N}\cup\left\{0\right\}</math> and
<center><math>
Z = \psi_{m}\left( z\right) = \cos\kappa_{m}(z+h),\quad
m\in\mathbb{N}\cup\left\{0\right\}.</math></center>
We note that
<center><math>
\int\nolimits_{-h}^{-d}\psi_{m}(z)\psi_{n}(z) \mathrm{d} z=C_{m}\delta_{mn},
</math></center>
where
<center>
<math>
C_{m} = \frac{1}{2} \left( \frac{\cos\kappa_{m}(h-d)\sin\kappa_{m}(h-d)+\kappa_{m}(h-d)}{\kappa_{m}} \right).
</math>
</center>
In addition there is the special case of when <math>m=n=0</math> ,
<center>
<math>
C_{0} = h-d,
</math>
</center>
where <math>\delta_{00}=1</math> .

=== Inner product between free surface and dock modes ===

<center><math>
B_{mn} = \int\nolimits_{-h}^{-d}\psi_{m}(z)\chi_{n}(z) \mathrm{d} z , \quad m\geq1, \quad n\geq0,
</math></center>
where
<center><math>
B_{mn}= \int\nolimits_{-h}^{-d} \frac{\cos\kappa_{m}(z+h)\cos k_{n}(z+h)}{\cos(k_{n}h)} \mathrm{d} z=\frac{-(-1)^m k_{n}\sin k_{n}(h-d)}
{\cos(k_{n}h)(\kappa_{m}^{2}-k_{n}^{2})}.
</math></center>
The evaluation of this inner product works on the assumption that the two roots <math>\kappa_{m}</math> and <math>k_{n}</math> do not coincide. The Matlab code given below has a routine to check for this case and make necessary adjustments.

We also have the case where the functions of the inner product are switched.
<center><math>
B_{nm} = \int\nolimits_{-h}^{-d}\chi_{m}(z)\psi_{n}(z) \mathrm{d} z , \quad m\geq0, \quad n\geq0.
</math></center>
For <math>m \geq 0</math> and <math>n \geq 1</math> the evaluation is above with the exception that the indices of the expression are switched.

Finally there is the special case of when <math>m \geq 0</math> and <math>n = 0</math>.
<center><math>
B_{0m} = \frac{\sin k_{m}(h-d)}{k_{m}\cos(k_{m}h)}.
</math></center>

=== Expansion of the potential ===

We need to apply some boundary conditions at plus and minus infinity,
where essentially 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 roots. Here we
must use a different condition, essentially identifying one solution
as the incoming wave and the other as the outgoing wave.

Therefore the diffracted potentials can
be expanded as
<center>
<math>
\phi(x,z)=\sum_{m=0}^{\infty}a_{m}e^{k_{m}x}\chi_{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 must be continuous across the
transition from open water to the dock covered region. Therefore, the
potentials at <math>x=-L</math> have to be equal.
We obtain for the '''symmetric''' and '''antisymmetric''' cases:

<center><math>
\chi_{0}\left( z\right) + \sum_{m=0}^{\infty}a^{s,a}_{m} \chi_{m}\left(z\right)
=\sum_{m=0}^{\infty}b^{s,a}_{m}\psi_{m}(z), \quad -h \leq z\leq -d.
</math></center>

We multiply both sides by <math>\psi_{n}(z) \,</math> and integrate from <math>-h</math> to <math>-d</math> to obtain:

<center><math>
B_{n0} + \sum^{\infty}_{m=0}a^{s,a}_{m}B_{nm} = b^{s,a}_{m}C_{m}\delta_{mn}, \quad n\in\mathbb{N}\cup\left\{0\right\}.
</math></center>

The derivative of the potential must be continuous across the
transition from open water to the dock covered region. Therefore, the derivatives of the potentials at <math>x=-L</math> have to be equal.
We obtain for the '''symmetric''' case:

<center><math>
\begin{align}
-k_{0}\chi_{0}\left( z\right) +\sum_{m=0}^{\infty} a^{s}_{m}k_{m}\chi_{m}\left(z\right) =
\begin{cases}
\quad \quad \quad \quad 0,\quad \quad \quad \quad -d\leq z\leq 0 \\
-\sum_{m=1}^{\infty}b^{s}_{m}\kappa_{m}\tanh(\kappa_{m}L)\psi_{m}(z),\quad -h\leq z\leq -d
\end{cases}
\end{align}
</math></center>

and for the second equation we multiply both sides by <math>\chi_{n}(z) \,</math> and integrating from <math>-h</math> to <math>0</math> to obtain:
<center><math>
-k_{0}A_{0}\delta_{0n} + k_{m}a_{m}A_{m} = -\sum^{\infty}_{m=0}\kappa_{m}b_{m}B_{mn}, \quad n\in\mathbb{N}\cup\left\{0\right\}
</math></center>

Solving the equations above will yield the coefficients of the water velocity potential in the dock covered region.

== Numerical Solution ==

To solve the system of equations for the '''symmetric''' case, we set the upper limit of <math>m</math> to be <math>N</math>. This resulting system can be expressed in the block matrix form below,

<center>
<math>
\begin{bmatrix}
B_{00} & B_{01} & \cdots & B_{0N} & -C_{0} & 0 & \cdots & 0 \\
B_{10} & B_{11} & \cdots & B_{1N} & 0 & -C_{1} & \cdots & \vdots \\
\vdots & \vdots & \ddots & \vdots & \vdots & \vdots & \ddots & 0 \\
B_{N0} & B_{N1} & \cdots & B_{NN} & 0 & \cdots & 0 & -C_{N} \\
k_{0}A_{0} & 0 & \cdots & 0 & 0 & \kappa_{1}\tanh(\kappa_{1}L)B_{10} & \cdots & \kappa_{N}\tanh(\kappa_{N}L)B_{N0} \\
0 & k_{1}A_{1} & \cdots & \vdots & 0 & \kappa_{1}\tanh(\kappa_{1}L)B_{11} & \cdots & \kappa_{N}\tanh(\kappa_{N}L)B_{N1} \\
\vdots & \vdots & \ddots & 0 & \vdots & \vdots & \ddots & \vdots \\
0 & \cdots & 0 & k_{N}A_{N} & 0 & \kappa_{1}\tanh(\kappa_{1}L)B_{1N} & \cdots & \kappa_{N}\tanh(\kappa_{N}L)B_{NN} \\
\end{bmatrix}
\begin{bmatrix}
a^s_{0} \\ a^s_{1} \\ \vdots \\ a^s_{N} \\ b^s_{0} \\ b^s_{1} \\ \vdots \\ b^s_{N} \\
\end{bmatrix}
=\begin{bmatrix}
-B_{00} \\ -B_{10} \\ \vdots \\ -B_{N0} \\ k_{0}A_{0} \\ 0 \\ \vdots \\ 0 \\
\end{bmatrix}
</math>
</center>

To solve the system of equations for the '''antisymmetric''' case, we set the upper limit of <math>m</math> to be <math>N</math>. This resulting system can be expressed in the block matrix form below,

<center>
<math>
\begin{bmatrix}
B_{00} & B_{01} & \cdots & B_{0N} & -C_{0} & 0 & \cdots & 0 \\
B_{10} & B_{11} & \cdots & B_{1N} & 0 & -C_{1} & \cdots & \vdots \\
\vdots & \vdots & \ddots & \vdots & \vdots & \vdots & \ddots & 0 \\
B_{N0} & B_{N1} & \cdots & B_{NN} & 0 & \cdots & 0 & -C_{N} \\
k_{0}A_{0} & 0 & \cdots & 0 & \frac{1}{L}D_{0} & \kappa_{1}\coth(\kappa_{1}L)B_{10} & \cdots & \kappa_{N}\coth(\kappa_{N}L)B_{N0} \\
0 & k_{1}A_{1} & \cdots & \vdots & \frac{1}{L}D_{1} & \kappa_{1}\coth(\kappa_{1}L)B_{11} & \cdots & \kappa_{N}\coth(\kappa_{N}L)B_{N1} \\
\vdots & \vdots & \ddots & 0 & \vdots & \vdots & \ddots & \vdots \\
0 & \cdots & 0 & k_{N}A_{N} & \frac{1}{L}D_{N} & \kappa_{1}\coth(\kappa_{1}L)B_{1N} & \cdots & \kappa_{N}\coth(\kappa_{N}L)B_{NN} \\
\end{bmatrix}
\begin{bmatrix}
a^a_{0} \\ a^a_{1} \\ \vdots \\ a^a_{N} \\ b^a_{0} \\ b^a_{1} \\ \vdots \\ b^a_{N} \\
\end{bmatrix}
=\begin{bmatrix}
-B_{00} \\ -B_{10} \\ \vdots \\ -B_{N0} \\ k_{0}A_{0} \\ 0 \\ \vdots \\ 0 \\
\end{bmatrix}
</math>
</center>

We then simply need to solve the two sets of <math>2(N+1) \mbox{ x } 2(N+1)</math> linear systems of equations. If we had not taken advantage of the symmetry we would need to solve a <math>4(N+1) \mbox{ x } 4(N+1)</math> linear system of equations for four coefficients <math>a_{m}</math>, <math>b_{m}</math>, <math>c_{m}</math> and <math>d_{m}</math>.

== Matlab Code ==

A program to calculate the coefficients for the single finite dock problem using symmetry 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 Finite Rectangle using Symmetry

2012-03-16T05:21:34Z

Tertius: /* An infinite dimensional system of equations */

{{incomplete pages}}

== Introduction ==

The problem consists of a finite rigid fixed rectangle of length <math>2L</math> centered at <math>x = 0</math> and having a height of <math>d + ...</math> freeboard with depth of submergence being <math>d</math> through which no flow is possible. The regions to the left and right of the rectangle are the free water surface. Since the geometry of the rectangle is symmetric about <math>x = 0</math> we can decompose the velocity potential <math>\phi</math> into a symmetric <math>\phi^s</math> and antisymmetric <math>\phi^a</math> potential. We then solve for these two potentials in the left domain only <math>x \leq 0</math> and subsequently recompose the solution <math>\phi</math> for the entire domain. We begin with the simple problem when the waves are normally incident on the rectangle (dock) (so that the problem is truly two-dimensional). We can extend this solution to a pair of partially submerged finite rigid fixed rectangles using symmetry
[[Eigenfunction Matching for a Finite Rectangle using Symmetry]]

== Governing Equations ==

We begin with the [[Frequency Domain Problem]] for a partially submerged finite dock. Although the dock is of length <math>2L</math> we are only interested in the dock of <math>-L \leq x \leq 0</math> together with the fluid domain <math>x \leq 0</math>. We assume <math>e^{i\omega t}</math> time dependence and
the depth of submergence is <math>d</math>.
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>
\begin{align}
\Delta\phi^{s,a} &= 0, \,\,\, -h\leq z\leq 0, \,\, x < -L, \\

\Delta\phi^{s,a} &= 0, \,\,\, -h\leq z\leq -d, \,\, -L < x < 0. \\

\partial_z\phi^{s,a} &= \alpha\phi^{s,a}, \,\,\, z=0, \,\, x < -L, \\

\partial_x\phi^{s,a} &= 0, \,\,\, -d\leq z\leq 0, \,\, x = -L, \\

\partial_z\phi^{s,a} &= 0, \,\,\,
\begin{cases}
z = -d, \,\, -L < x < 0, \\
z = -h, \,\, x < 0.
\end{cases} \\

\partial_x\phi^s &= 0, \,\,\, x = 0, \,\, -h \leq z \leq -d, \\

\phi^a &= 0, \,\,\, x = 0, \,\, -h \leq z \leq -d.
\end{align}
</math></center>

Where <math>\alpha = \omega^2/g</math>.

<math>\phi^s(x,z)</math> is an even function and at <math>x = 0</math> the derivative of this function is equal to zero. <math>\phi^a(x,z)</math> is an odd function and at <math>x = 0</math> this function is equal to zero. These boundary conditions are important and enable the recomposed solution <math>\phi(x,z)</math> to be reflected into the domain of <math>x \geq 0</math>.

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\leq 0</math>
and <math>x\geq 0</math>.

{{separation of variables in two dimensions}}

{{separation of variables for a free surface}}

=== Separation of Variables for a Dock ===

The separation of variables equation for a partially submerged dock is given by:
<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>
Z^{\prime} (-d) = 0.
</math></center>
The solution is
<math>k=\kappa_{m}= \frac{m\pi}{h-d} \,</math> , <math>m\in\mathbb{N}\cup\left\{0\right\}</math> and
<center><math>
Z = \psi_{m}\left( z\right) = \cos\kappa_{m}(z+h),\quad
m\in\mathbb{N}\cup\left\{0\right\}.</math></center>
We note that
<center><math>
\int\nolimits_{-h}^{-d}\psi_{m}(z)\psi_{n}(z) \mathrm{d} z=C_{m}\delta_{mn},
</math></center>
where
<center>
<math>
C_{m} = \frac{1}{2} \left( \frac{\cos\kappa_{m}(h-d)\sin\kappa_{m}(h-d)+\kappa_{m}(h-d)}{\kappa_{m}} \right).
</math>
</center>
In addition there is the special case of when <math>m=n=0</math> ,
<center>
<math>
C_{0} = h-d,
</math>
</center>
where <math>\delta_{00}=1</math> .

=== Inner product between free surface and dock modes ===

<center><math>
B_{mn} = \int\nolimits_{-h}^{-d}\psi_{m}(z)\chi_{n}(z) \mathrm{d} z , \quad m\geq1, \quad n\geq0,
</math></center>
where
<center><math>
B_{mn}= \int\nolimits_{-h}^{-d} \frac{\cos\kappa_{m}(z+h)\cos k_{n}(z+h)}{\cos(k_{n}h)} \mathrm{d} z=\frac{-(-1)^m k_{n}\sin k_{n}(h-d)}
{\cos(k_{n}h)(\kappa_{m}^{2}-k_{n}^{2})}.
</math></center>
The evaluation of this inner product works on the assumption that the two roots <math>\kappa_{m}</math> and <math>k_{n}</math> do not coincide. The Matlab code given below has a routine to check for this case and make necessary adjustments.

We also have the case where the functions of the inner product are switched.
<center><math>
B_{nm} = \int\nolimits_{-h}^{-d}\chi_{m}(z)\psi_{n}(z) \mathrm{d} z , \quad m\geq0, \quad n\geq0.
</math></center>
For <math>m \geq 0</math> and <math>n \geq 1</math> the evaluation is above with the exception that the indices of the expression are switched.

Finally there is the special case of when <math>m \geq 0</math> and <math>n = 0</math>.
<center><math>
B_{0m} = \frac{\sin k_{m}(h-d)}{k_{m}\cos(k_{m}h)}.
</math></center>

=== Expansion of the potential ===

We need to apply some boundary conditions at plus and minus infinity,
where essentially 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 roots. Here we
must use a different condition, essentially identifying one solution
as the incoming wave and the other as the outgoing wave.

Therefore the diffracted potentials can
be expanded as
<center>
<math>
\phi(x,z)=\sum_{m=0}^{\infty}a_{m}e^{k_{m}x}\chi_{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 must be continuous across the
transition from open water to the dock covered region. Therefore, the
potentials at <math>x=-L</math> have to be equal.
We obtain for the '''symmetric''' and '''antisymmetric''' cases:

<center><math>
\chi_{0}\left( z\right) + \sum_{m=0}^{\infty}a^{s,a}_{m} \chi_{m}\left(z\right)
=\sum_{m=0}^{\infty}b^{s,a}_{m}\psi_{m}(z), \quad -h \leq z\leq -d.
</math></center>

We multiply both sides by <math>\psi_{n}(z) \,</math> and integrate from <math>-h</math> to <math>-d</math> to obtain:

<center><math>
B_{n0} + \sum^{\infty}_{m=0}a^{s,a}_{m}B_{nm} = b^{s,a}_{m}C_{m}\delta_{mn}, \quad n\in\mathbb{N}\cup\left\{0\right\}.
</math></center>

The derivative of the potential must be continuous across the
transition from open water to the dock covered region. Therefore, the derivatives of the potentials at <math>x=-L</math> have to be equal.
We obtain for the '''symmetric''' case:

<center><math>
\begin{align}
-k_{0}\chi_{0}\left( z\right) +\sum_{m=0}^{\infty} a^{s}_{m}k_{m}\chi_{m}\left(z\right) =
\begin{cases}
\quad \quad \quad 0,\quad \quad \quad -d\leq z\leq 0 \\
-\sum_{m=0}^{\infty}b^{s}_{m}\kappa_{m}\psi_{m}(z),\quad -h\leq z\leq -d
\end{cases}
\end{align}
</math></center>

and for the second equation we multiply both sides by <math>\chi_{n}(z) \,</math> and integrating from <math>-h</math> to <math>0</math> to obtain:
<center><math>
-k_{0}A_{0}\delta_{0n} + k_{m}a_{m}A_{m} = -\sum^{\infty}_{m=0}\kappa_{m}b_{m}B_{mn}, \quad n\in\mathbb{N}\cup\left\{0\right\}
</math></center>

Solving the equations above will yield the coefficients of the water velocity potential in the dock covered region.

== Numerical Solution ==

To solve the system of equations for the '''symmetric''' case, we set the upper limit of <math>m</math> to be <math>N</math>. This resulting system can be expressed in the block matrix form below,

<center>
<math>
\begin{bmatrix}
B_{00} & B_{01} & \cdots & B_{0N} & -C_{0} & 0 & \cdots & 0 \\
B_{10} & B_{11} & \cdots & B_{1N} & 0 & -C_{1} & \cdots & \vdots \\
\vdots & \vdots & \ddots & \vdots & \vdots & \vdots & \ddots & 0 \\
B_{N0} & B_{N1} & \cdots & B_{NN} & 0 & \cdots & 0 & -C_{N} \\
k_{0}A_{0} & 0 & \cdots & 0 & 0 & \kappa_{1}\tanh(\kappa_{1}L)B_{10} & \cdots & \kappa_{N}\tanh(\kappa_{N}L)B_{N0} \\
0 & k_{1}A_{1} & \cdots & \vdots & 0 & \kappa_{1}\tanh(\kappa_{1}L)B_{11} & \cdots & \kappa_{N}\tanh(\kappa_{N}L)B_{N1} \\
\vdots & \vdots & \ddots & 0 & \vdots & \vdots & \ddots & \vdots \\
0 & \cdots & 0 & k_{N}A_{N} & 0 & \kappa_{1}\tanh(\kappa_{1}L)B_{1N} & \cdots & \kappa_{N}\tanh(\kappa_{N}L)B_{NN} \\
\end{bmatrix}
\begin{bmatrix}
a^s_{0} \\ a^s_{1} \\ \vdots \\ a^s_{N} \\ b^s_{0} \\ b^s_{1} \\ \vdots \\ b^s_{N} \\
\end{bmatrix}
=\begin{bmatrix}
-B_{00} \\ -B_{10} \\ \vdots \\ -B_{N0} \\ k_{0}A_{0} \\ 0 \\ \vdots \\ 0 \\
\end{bmatrix}
</math>
</center>

To solve the system of equations for the '''antisymmetric''' case, we set the upper limit of <math>m</math> to be <math>N</math>. This resulting system can be expressed in the block matrix form below,

<center>
<math>
\begin{bmatrix}
B_{00} & B_{01} & \cdots & B_{0N} & -C_{0} & 0 & \cdots & 0 \\
B_{10} & B_{11} & \cdots & B_{1N} & 0 & -C_{1} & \cdots & \vdots \\
\vdots & \vdots & \ddots & \vdots & \vdots & \vdots & \ddots & 0 \\
B_{N0} & B_{N1} & \cdots & B_{NN} & 0 & \cdots & 0 & -C_{N} \\
k_{0}A_{0} & 0 & \cdots & 0 & \frac{1}{L}D_{0} & \kappa_{1}\coth(\kappa_{1}L)B_{10} & \cdots & \kappa_{N}\coth(\kappa_{N}L)B_{N0} \\
0 & k_{1}A_{1} & \cdots & \vdots & \frac{1}{L}D_{1} & \kappa_{1}\coth(\kappa_{1}L)B_{11} & \cdots & \kappa_{N}\coth(\kappa_{N}L)B_{N1} \\
\vdots & \vdots & \ddots & 0 & \vdots & \vdots & \ddots & \vdots \\
0 & \cdots & 0 & k_{N}A_{N} & \frac{1}{L}D_{N} & \kappa_{1}\coth(\kappa_{1}L)B_{1N} & \cdots & \kappa_{N}\coth(\kappa_{N}L)B_{NN} \\
\end{bmatrix}
\begin{bmatrix}
a^a_{0} \\ a^a_{1} \\ \vdots \\ a^a_{N} \\ b^a_{0} \\ b^a_{1} \\ \vdots \\ b^a_{N} \\
\end{bmatrix}
=\begin{bmatrix}
-B_{00} \\ -B_{10} \\ \vdots \\ -B_{N0} \\ k_{0}A_{0} \\ 0 \\ \vdots \\ 0 \\
\end{bmatrix}
</math>
</center>

We then simply need to solve the two sets of <math>2(N+1) \mbox{ x } 2(N+1)</math> linear systems of equations. If we had not taken advantage of the symmetry we would need to solve a <math>4(N+1) \mbox{ x } 4(N+1)</math> linear system of equations for four coefficients <math>a_{m}</math>, <math>b_{m}</math>, <math>c_{m}</math> and <math>d_{m}</math>.

== Matlab Code ==

A program to calculate the coefficients for the single finite dock problem using symmetry 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 Finite Rectangle using Symmetry

2012-03-16T05:20:56Z

Tertius: /* An infinite dimensional system of equations */

{{incomplete pages}}

== Introduction ==

The problem consists of a finite rigid fixed rectangle of length <math>2L</math> centered at <math>x = 0</math> and having a height of <math>d + ...</math> freeboard with depth of submergence being <math>d</math> through which no flow is possible. The regions to the left and right of the rectangle are the free water surface. Since the geometry of the rectangle is symmetric about <math>x = 0</math> we can decompose the velocity potential <math>\phi</math> into a symmetric <math>\phi^s</math> and antisymmetric <math>\phi^a</math> potential. We then solve for these two potentials in the left domain only <math>x \leq 0</math> and subsequently recompose the solution <math>\phi</math> for the entire domain. We begin with the simple problem when the waves are normally incident on the rectangle (dock) (so that the problem is truly two-dimensional). We can extend this solution to a pair of partially submerged finite rigid fixed rectangles using symmetry
[[Eigenfunction Matching for a Finite Rectangle using Symmetry]]

== Governing Equations ==

We begin with the [[Frequency Domain Problem]] for a partially submerged finite dock. Although the dock is of length <math>2L</math> we are only interested in the dock of <math>-L \leq x \leq 0</math> together with the fluid domain <math>x \leq 0</math>. We assume <math>e^{i\omega t}</math> time dependence and
the depth of submergence is <math>d</math>.
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>
\begin{align}
\Delta\phi^{s,a} &= 0, \,\,\, -h\leq z\leq 0, \,\, x < -L, \\

\Delta\phi^{s,a} &= 0, \,\,\, -h\leq z\leq -d, \,\, -L < x < 0. \\

\partial_z\phi^{s,a} &= \alpha\phi^{s,a}, \,\,\, z=0, \,\, x < -L, \\

\partial_x\phi^{s,a} &= 0, \,\,\, -d\leq z\leq 0, \,\, x = -L, \\

\partial_z\phi^{s,a} &= 0, \,\,\,
\begin{cases}
z = -d, \,\, -L < x < 0, \\
z = -h, \,\, x < 0.
\end{cases} \\

\partial_x\phi^s &= 0, \,\,\, x = 0, \,\, -h \leq z \leq -d, \\

\phi^a &= 0, \,\,\, x = 0, \,\, -h \leq z \leq -d.
\end{align}
</math></center>

Where <math>\alpha = \omega^2/g</math>.

<math>\phi^s(x,z)</math> is an even function and at <math>x = 0</math> the derivative of this function is equal to zero. <math>\phi^a(x,z)</math> is an odd function and at <math>x = 0</math> this function is equal to zero. These boundary conditions are important and enable the recomposed solution <math>\phi(x,z)</math> to be reflected into the domain of <math>x \geq 0</math>.

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\leq 0</math>
and <math>x\geq 0</math>.

{{separation of variables in two dimensions}}

{{separation of variables for a free surface}}

=== Separation of Variables for a Dock ===

The separation of variables equation for a partially submerged dock is given by:
<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>
Z^{\prime} (-d) = 0.
</math></center>
The solution is
<math>k=\kappa_{m}= \frac{m\pi}{h-d} \,</math> , <math>m\in\mathbb{N}\cup\left\{0\right\}</math> and
<center><math>
Z = \psi_{m}\left( z\right) = \cos\kappa_{m}(z+h),\quad
m\in\mathbb{N}\cup\left\{0\right\}.</math></center>
We note that
<center><math>
\int\nolimits_{-h}^{-d}\psi_{m}(z)\psi_{n}(z) \mathrm{d} z=C_{m}\delta_{mn},
</math></center>
where
<center>
<math>
C_{m} = \frac{1}{2} \left( \frac{\cos\kappa_{m}(h-d)\sin\kappa_{m}(h-d)+\kappa_{m}(h-d)}{\kappa_{m}} \right).
</math>
</center>
In addition there is the special case of when <math>m=n=0</math> ,
<center>
<math>
C_{0} = h-d,
</math>
</center>
where <math>\delta_{00}=1</math> .

=== Inner product between free surface and dock modes ===

<center><math>
B_{mn} = \int\nolimits_{-h}^{-d}\psi_{m}(z)\chi_{n}(z) \mathrm{d} z , \quad m\geq1, \quad n\geq0,
</math></center>
where
<center><math>
B_{mn}= \int\nolimits_{-h}^{-d} \frac{\cos\kappa_{m}(z+h)\cos k_{n}(z+h)}{\cos(k_{n}h)} \mathrm{d} z=\frac{-(-1)^m k_{n}\sin k_{n}(h-d)}
{\cos(k_{n}h)(\kappa_{m}^{2}-k_{n}^{2})}.
</math></center>
The evaluation of this inner product works on the assumption that the two roots <math>\kappa_{m}</math> and <math>k_{n}</math> do not coincide. The Matlab code given below has a routine to check for this case and make necessary adjustments.

We also have the case where the functions of the inner product are switched.
<center><math>
B_{nm} = \int\nolimits_{-h}^{-d}\chi_{m}(z)\psi_{n}(z) \mathrm{d} z , \quad m\geq0, \quad n\geq0.
</math></center>
For <math>m \geq 0</math> and <math>n \geq 1</math> the evaluation is above with the exception that the indices of the expression are switched.

Finally there is the special case of when <math>m \geq 0</math> and <math>n = 0</math>.
<center><math>
B_{0m} = \frac{\sin k_{m}(h-d)}{k_{m}\cos(k_{m}h)}.
</math></center>

=== Expansion of the potential ===

We need to apply some boundary conditions at plus and minus infinity,
where essentially 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 roots. Here we
must use a different condition, essentially identifying one solution
as the incoming wave and the other as the outgoing wave.

Therefore the diffracted potentials can
be expanded as
<center>
<math>
\phi(x,z)=\sum_{m=0}^{\infty}a_{m}e^{k_{m}x}\chi_{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 must be continuous across the
transition from open water to the dock covered region. Therefore, the
potentials at <math>x=-L</math> have to be equal.
We obtain for the '''symmetric''' and '''antisymmetric''' cases:

<center><math>
\chi_{0}\left( z\right) + \sum_{m=0}^{\infty}a^{s,a}_{m} \chi_{m}\left(z\right)
=\sum_{m=0}^{\infty}b^{s,a}_{m}\psi_{m}(z), \quad -h \leq z\leq -d.
</math></center>

We multiply both sides by <math>\psi_{n}(z) \,</math> and integrate from <math>-h</math> to <math>-d</math> to obtain:
<center><math>
B_{n0} + \sum^{\infty}_{m=0}a^{s,a}_{m}B_{nm} = b^{s,a}_{m}C_{m}\delta_{mn}, \quad n\in\mathbb{N}\cup\left\{0\right\}.
</math></center>

The derivative of the potential must be continuous across the
transition from open water to the dock covered region. Therefore, the derivatives of the potentials at <math>x=-L</math> have to be equal.
We obtain for the '''symmetric''' case:

<center><math>
\begin{align}
-k_{0}\chi_{0}\left( z\right) +\sum_{m=0}^{\infty} a^{s}_{m}k_{m}\chi_{m}\left(z\right) =
\begin{cases}
\quad \quad \quad 0,\quad \quad \quad -d\leq z\leq 0 \\
-\sum_{m=0}^{\infty}b^{s}_{m}\kappa_{m}\psi_{m}(z),\quad -h\leq z\leq -d
\end{cases}
\end{align}
</math></center>

and for the second equation we multiply both sides by <math>\chi_{n}(z) \,</math> and integrating from <math>-h</math> to <math>0</math> to obtain:
<center><math>
-k_{0}A_{0}\delta_{0n} + k_{m}a_{m}A_{m} = -\sum^{\infty}_{m=0}\kappa_{m}b_{m}B_{mn}, \quad n\in\mathbb{N}\cup\left\{0\right\}
</math></center>

Solving the equations above will yield the coefficients of the water velocity potential in the dock covered region.

== Numerical Solution ==

To solve the system of equations for the '''symmetric''' case, we set the upper limit of <math>m</math> to be <math>N</math>. This resulting system can be expressed in the block matrix form below,

<center>
<math>
\begin{bmatrix}
B_{00} & B_{01} & \cdots & B_{0N} & -C_{0} & 0 & \cdots & 0 \\
B_{10} & B_{11} & \cdots & B_{1N} & 0 & -C_{1} & \cdots & \vdots \\
\vdots & \vdots & \ddots & \vdots & \vdots & \vdots & \ddots & 0 \\
B_{N0} & B_{N1} & \cdots & B_{NN} & 0 & \cdots & 0 & -C_{N} \\
k_{0}A_{0} & 0 & \cdots & 0 & 0 & \kappa_{1}\tanh(\kappa_{1}L)B_{10} & \cdots & \kappa_{N}\tanh(\kappa_{N}L)B_{N0} \\
0 & k_{1}A_{1} & \cdots & \vdots & 0 & \kappa_{1}\tanh(\kappa_{1}L)B_{11} & \cdots & \kappa_{N}\tanh(\kappa_{N}L)B_{N1} \\
\vdots & \vdots & \ddots & 0 & \vdots & \vdots & \ddots & \vdots \\
0 & \cdots & 0 & k_{N}A_{N} & 0 & \kappa_{1}\tanh(\kappa_{1}L)B_{1N} & \cdots & \kappa_{N}\tanh(\kappa_{N}L)B_{NN} \\
\end{bmatrix}
\begin{bmatrix}
a^s_{0} \\ a^s_{1} \\ \vdots \\ a^s_{N} \\ b^s_{0} \\ b^s_{1} \\ \vdots \\ b^s_{N} \\
\end{bmatrix}
=\begin{bmatrix}
-B_{00} \\ -B_{10} \\ \vdots \\ -B_{N0} \\ k_{0}A_{0} \\ 0 \\ \vdots \\ 0 \\
\end{bmatrix}
</math>
</center>

To solve the system of equations for the '''antisymmetric''' case, we set the upper limit of <math>m</math> to be <math>N</math>. This resulting system can be expressed in the block matrix form below,

<center>
<math>
\begin{bmatrix}
B_{00} & B_{01} & \cdots & B_{0N} & -C_{0} & 0 & \cdots & 0 \\
B_{10} & B_{11} & \cdots & B_{1N} & 0 & -C_{1} & \cdots & \vdots \\
\vdots & \vdots & \ddots & \vdots & \vdots & \vdots & \ddots & 0 \\
B_{N0} & B_{N1} & \cdots & B_{NN} & 0 & \cdots & 0 & -C_{N} \\
k_{0}A_{0} & 0 & \cdots & 0 & \frac{1}{L}D_{0} & \kappa_{1}\coth(\kappa_{1}L)B_{10} & \cdots & \kappa_{N}\coth(\kappa_{N}L)B_{N0} \\
0 & k_{1}A_{1} & \cdots & \vdots & \frac{1}{L}D_{1} & \kappa_{1}\coth(\kappa_{1}L)B_{11} & \cdots & \kappa_{N}\coth(\kappa_{N}L)B_{N1} \\
\vdots & \vdots & \ddots & 0 & \vdots & \vdots & \ddots & \vdots \\
0 & \cdots & 0 & k_{N}A_{N} & \frac{1}{L}D_{N} & \kappa_{1}\coth(\kappa_{1}L)B_{1N} & \cdots & \kappa_{N}\coth(\kappa_{N}L)B_{NN} \\
\end{bmatrix}
\begin{bmatrix}
a^a_{0} \\ a^a_{1} \\ \vdots \\ a^a_{N} \\ b^a_{0} \\ b^a_{1} \\ \vdots \\ b^a_{N} \\
\end{bmatrix}
=\begin{bmatrix}
-B_{00} \\ -B_{10} \\ \vdots \\ -B_{N0} \\ k_{0}A_{0} \\ 0 \\ \vdots \\ 0 \\
\end{bmatrix}
</math>
</center>

We then simply need to solve the two sets of <math>2(N+1) \mbox{ x } 2(N+1)</math> linear systems of equations. If we had not taken advantage of the symmetry we would need to solve a <math>4(N+1) \mbox{ x } 4(N+1)</math> linear system of equations for four coefficients <math>a_{m}</math>, <math>b_{m}</math>, <math>c_{m}</math> and <math>d_{m}</math>.

== Matlab Code ==

A program to calculate the coefficients for the single finite dock problem using symmetry 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 Finite Rectangle using Symmetry

2012-03-16T05:18:42Z

Tertius: /* An infinite dimensional system of equations */

{{incomplete pages}}

== Introduction ==

The problem consists of a finite rigid fixed rectangle of length <math>2L</math> centered at <math>x = 0</math> and having a height of <math>d + ...</math> freeboard with depth of submergence being <math>d</math> through which no flow is possible. The regions to the left and right of the rectangle are the free water surface. Since the geometry of the rectangle is symmetric about <math>x = 0</math> we can decompose the velocity potential <math>\phi</math> into a symmetric <math>\phi^s</math> and antisymmetric <math>\phi^a</math> potential. We then solve for these two potentials in the left domain only <math>x \leq 0</math> and subsequently recompose the solution <math>\phi</math> for the entire domain. We begin with the simple problem when the waves are normally incident on the rectangle (dock) (so that the problem is truly two-dimensional). We can extend this solution to a pair of partially submerged finite rigid fixed rectangles using symmetry
[[Eigenfunction Matching for a Finite Rectangle using Symmetry]]

== Governing Equations ==

We begin with the [[Frequency Domain Problem]] for a partially submerged finite dock. Although the dock is of length <math>2L</math> we are only interested in the dock of <math>-L \leq x \leq 0</math> together with the fluid domain <math>x \leq 0</math>. We assume <math>e^{i\omega t}</math> time dependence and
the depth of submergence is <math>d</math>.
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>
\begin{align}
\Delta\phi^{s,a} &= 0, \,\,\, -h\leq z\leq 0, \,\, x < -L, \\

\Delta\phi^{s,a} &= 0, \,\,\, -h\leq z\leq -d, \,\, -L < x < 0. \\

\partial_z\phi^{s,a} &= \alpha\phi^{s,a}, \,\,\, z=0, \,\, x < -L, \\

\partial_x\phi^{s,a} &= 0, \,\,\, -d\leq z\leq 0, \,\, x = -L, \\

\partial_z\phi^{s,a} &= 0, \,\,\,
\begin{cases}
z = -d, \,\, -L < x < 0, \\
z = -h, \,\, x < 0.
\end{cases} \\

\partial_x\phi^s &= 0, \,\,\, x = 0, \,\, -h \leq z \leq -d, \\

\phi^a &= 0, \,\,\, x = 0, \,\, -h \leq z \leq -d.
\end{align}
</math></center>

Where <math>\alpha = \omega^2/g</math>.

<math>\phi^s(x,z)</math> is an even function and at <math>x = 0</math> the derivative of this function is equal to zero. <math>\phi^a(x,z)</math> is an odd function and at <math>x = 0</math> this function is equal to zero. These boundary conditions are important and enable the recomposed solution <math>\phi(x,z)</math> to be reflected into the domain of <math>x \geq 0</math>.

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\leq 0</math>
and <math>x\geq 0</math>.

{{separation of variables in two dimensions}}

{{separation of variables for a free surface}}

=== Separation of Variables for a Dock ===

The separation of variables equation for a partially submerged dock is given by:
<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>
Z^{\prime} (-d) = 0.
</math></center>
The solution is
<math>k=\kappa_{m}= \frac{m\pi}{h-d} \,</math> , <math>m\in\mathbb{N}\cup\left\{0\right\}</math> and
<center><math>
Z = \psi_{m}\left( z\right) = \cos\kappa_{m}(z+h),\quad
m\in\mathbb{N}\cup\left\{0\right\}.</math></center>
We note that
<center><math>
\int\nolimits_{-h}^{-d}\psi_{m}(z)\psi_{n}(z) \mathrm{d} z=C_{m}\delta_{mn},
</math></center>
where
<center>
<math>
C_{m} = \frac{1}{2} \left( \frac{\cos\kappa_{m}(h-d)\sin\kappa_{m}(h-d)+\kappa_{m}(h-d)}{\kappa_{m}} \right).
</math>
</center>
In addition there is the special case of when <math>m=n=0</math> ,
<center>
<math>
C_{0} = h-d,
</math>
</center>
where <math>\delta_{00}=1</math> .

=== Inner product between free surface and dock modes ===

<center><math>
B_{mn} = \int\nolimits_{-h}^{-d}\psi_{m}(z)\chi_{n}(z) \mathrm{d} z , \quad m\geq1, \quad n\geq0,
</math></center>
where
<center><math>
B_{mn}= \int\nolimits_{-h}^{-d} \frac{\cos\kappa_{m}(z+h)\cos k_{n}(z+h)}{\cos(k_{n}h)} \mathrm{d} z=\frac{-(-1)^m k_{n}\sin k_{n}(h-d)}
{\cos(k_{n}h)(\kappa_{m}^{2}-k_{n}^{2})}.
</math></center>
The evaluation of this inner product works on the assumption that the two roots <math>\kappa_{m}</math> and <math>k_{n}</math> do not coincide. The Matlab code given below has a routine to check for this case and make necessary adjustments.

We also have the case where the functions of the inner product are switched.
<center><math>
B_{nm} = \int\nolimits_{-h}^{-d}\chi_{m}(z)\psi_{n}(z) \mathrm{d} z , \quad m\geq0, \quad n\geq0.
</math></center>
For <math>m \geq 0</math> and <math>n \geq 1</math> the evaluation is above with the exception that the indices of the expression are switched.

Finally there is the special case of when <math>m \geq 0</math> and <math>n = 0</math>.
<center><math>
B_{0m} = \frac{\sin k_{m}(h-d)}{k_{m}\cos(k_{m}h)}.
</math></center>

=== Expansion of the potential ===

We need to apply some boundary conditions at plus and minus infinity,
where essentially 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 roots. Here we
must use a different condition, essentially identifying one solution
as the incoming wave and the other as the outgoing wave.

Therefore the diffracted potentials can
be expanded as
<center>
<math>
\phi(x,z)=\sum_{m=0}^{\infty}a_{m}e^{k_{m}x}\chi_{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 must be continuous across the
transition from open water to the dock covered region. Therefore, the
potentials at <math>x=-L</math> have to be equal.
We obtain for the '''symmetric''' and '''antisymmetric''' cases:

<center><math>
\chi_{0}\left( z\right) + \sum_{m=0}^{\infty}a^{s,a}_{m} \chi_{m}\left(z\right)
=\sum_{m=0}^{\infty}b^{s,a}_{m}\psi_{m}(z), \quad -h \leq z\leq -d.
</math></center>

We multiply both sides by <math>\psi_{n}(z) \,</math> and integrating from <math>-h</math> to <math>-d</math> to obtain:
<center><math>
B_{n0} + \sum^{\infty}_{m=0}a^{s,a}_{m}B_{nm} = b^{s,a}_{m}C_{m}\delta_{mn}, \quad n\in\mathbb{N}\cup\left\{0\right\}.
</math></center>

The derivative of the potential must be continuous across the
transition from open water to the dock covered region. Therefore, the derivatives of the potentials at <math>x=-L</math> have to be equal.
We obtain for the '''symmetric''' case:

<center><math>
\begin{align}
-k_{0}\chi_{0}\left( z\right) +\sum_{m=0}^{\infty} a^{s}_{m}k_{m}\chi_{m}\left(z\right) =
\begin{cases}
\quad \quad \quad 0,\quad \quad \quad -d\leq z\leq 0 \\
-\sum_{m=0}^{\infty}b^{s}_{m}\kappa_{m}\psi_{m}(z),\quad -h\leq z\leq -d
\end{cases}
\end{align}
</math></center>

and for the second equation we multiply both sides by <math>\chi_{n}(z) \,</math> and integrating from <math>-h</math> to <math>0</math> to obtain:
<center><math>
-k_{0}A_{0}\delta_{0n} + k_{m}a_{m}A_{m} = -\sum^{\infty}_{m=0}\kappa_{m}b_{m}B_{mn}, \quad n\in\mathbb{N}\cup\left\{0\right\}
</math></center>

Solving the equations above will yield the coefficients of the water velocity potential in the dock covered region.

== Numerical Solution ==

To solve the system of equations for the '''symmetric''' case, we set the upper limit of <math>m</math> to be <math>N</math>. This resulting system can be expressed in the block matrix form below,

<center>
<math>
\begin{bmatrix}
B_{00} & B_{01} & \cdots & B_{0N} & -C_{0} & 0 & \cdots & 0 \\
B_{10} & B_{11} & \cdots & B_{1N} & 0 & -C_{1} & \cdots & \vdots \\
\vdots & \vdots & \ddots & \vdots & \vdots & \vdots & \ddots & 0 \\
B_{N0} & B_{N1} & \cdots & B_{NN} & 0 & \cdots & 0 & -C_{N} \\
k_{0}A_{0} & 0 & \cdots & 0 & 0 & \kappa_{1}\tanh(\kappa_{1}L)B_{10} & \cdots & \kappa_{N}\tanh(\kappa_{N}L)B_{N0} \\
0 & k_{1}A_{1} & \cdots & \vdots & 0 & \kappa_{1}\tanh(\kappa_{1}L)B_{11} & \cdots & \kappa_{N}\tanh(\kappa_{N}L)B_{N1} \\
\vdots & \vdots & \ddots & 0 & \vdots & \vdots & \ddots & \vdots \\
0 & \cdots & 0 & k_{N}A_{N} & 0 & \kappa_{1}\tanh(\kappa_{1}L)B_{1N} & \cdots & \kappa_{N}\tanh(\kappa_{N}L)B_{NN} \\
\end{bmatrix}
\begin{bmatrix}
a^s_{0} \\ a^s_{1} \\ \vdots \\ a^s_{N} \\ b^s_{0} \\ b^s_{1} \\ \vdots \\ b^s_{N} \\
\end{bmatrix}
=\begin{bmatrix}
-B_{00} \\ -B_{10} \\ \vdots \\ -B_{N0} \\ k_{0}A_{0} \\ 0 \\ \vdots \\ 0 \\
\end{bmatrix}
</math>
</center>

To solve the system of equations for the '''antisymmetric''' case, we set the upper limit of <math>m</math> to be <math>N</math>. This resulting system can be expressed in the block matrix form below,

<center>
<math>
\begin{bmatrix}
B_{00} & B_{01} & \cdots & B_{0N} & -C_{0} & 0 & \cdots & 0 \\
B_{10} & B_{11} & \cdots & B_{1N} & 0 & -C_{1} & \cdots & \vdots \\
\vdots & \vdots & \ddots & \vdots & \vdots & \vdots & \ddots & 0 \\
B_{N0} & B_{N1} & \cdots & B_{NN} & 0 & \cdots & 0 & -C_{N} \\
k_{0}A_{0} & 0 & \cdots & 0 & \frac{1}{L}D_{0} & \kappa_{1}\coth(\kappa_{1}L)B_{10} & \cdots & \kappa_{N}\coth(\kappa_{N}L)B_{N0} \\
0 & k_{1}A_{1} & \cdots & \vdots & \frac{1}{L}D_{1} & \kappa_{1}\coth(\kappa_{1}L)B_{11} & \cdots & \kappa_{N}\coth(\kappa_{N}L)B_{N1} \\
\vdots & \vdots & \ddots & 0 & \vdots & \vdots & \ddots & \vdots \\
0 & \cdots & 0 & k_{N}A_{N} & \frac{1}{L}D_{N} & \kappa_{1}\coth(\kappa_{1}L)B_{1N} & \cdots & \kappa_{N}\coth(\kappa_{N}L)B_{NN} \\
\end{bmatrix}
\begin{bmatrix}
a^a_{0} \\ a^a_{1} \\ \vdots \\ a^a_{N} \\ b^a_{0} \\ b^a_{1} \\ \vdots \\ b^a_{N} \\
\end{bmatrix}
=\begin{bmatrix}
-B_{00} \\ -B_{10} \\ \vdots \\ -B_{N0} \\ k_{0}A_{0} \\ 0 \\ \vdots \\ 0 \\
\end{bmatrix}
</math>
</center>

We then simply need to solve the two sets of <math>2(N+1) \mbox{ x } 2(N+1)</math> linear systems of equations. If we had not taken advantage of the symmetry we would need to solve a <math>4(N+1) \mbox{ x } 4(N+1)</math> linear system of equations for four coefficients <math>a_{m}</math>, <math>b_{m}</math>, <math>c_{m}</math> and <math>d_{m}</math>.

== Matlab Code ==

A program to calculate the coefficients for the single finite dock problem using symmetry 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 Finite Rectangle using Symmetry

2012-03-16T05:16:56Z

Tertius: /* An infinite dimensional system of equations */

{{incomplete pages}}

== Introduction ==

The problem consists of a finite rigid fixed rectangle of length <math>2L</math> centered at <math>x = 0</math> and having a height of <math>d + ...</math> freeboard with depth of submergence being <math>d</math> through which no flow is possible. The regions to the left and right of the rectangle are the free water surface. Since the geometry of the rectangle is symmetric about <math>x = 0</math> we can decompose the velocity potential <math>\phi</math> into a symmetric <math>\phi^s</math> and antisymmetric <math>\phi^a</math> potential. We then solve for these two potentials in the left domain only <math>x \leq 0</math> and subsequently recompose the solution <math>\phi</math> for the entire domain. We begin with the simple problem when the waves are normally incident on the rectangle (dock) (so that the problem is truly two-dimensional). We can extend this solution to a pair of partially submerged finite rigid fixed rectangles using symmetry
[[Eigenfunction Matching for a Finite Rectangle using Symmetry]]

== Governing Equations ==

We begin with the [[Frequency Domain Problem]] for a partially submerged finite dock. Although the dock is of length <math>2L</math> we are only interested in the dock of <math>-L \leq x \leq 0</math> together with the fluid domain <math>x \leq 0</math>. We assume <math>e^{i\omega t}</math> time dependence and
the depth of submergence is <math>d</math>.
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>
\begin{align}
\Delta\phi^{s,a} &= 0, \,\,\, -h\leq z\leq 0, \,\, x < -L, \\

\Delta\phi^{s,a} &= 0, \,\,\, -h\leq z\leq -d, \,\, -L < x < 0. \\

\partial_z\phi^{s,a} &= \alpha\phi^{s,a}, \,\,\, z=0, \,\, x < -L, \\

\partial_x\phi^{s,a} &= 0, \,\,\, -d\leq z\leq 0, \,\, x = -L, \\

\partial_z\phi^{s,a} &= 0, \,\,\,
\begin{cases}
z = -d, \,\, -L < x < 0, \\
z = -h, \,\, x < 0.
\end{cases} \\

\partial_x\phi^s &= 0, \,\,\, x = 0, \,\, -h \leq z \leq -d, \\

\phi^a &= 0, \,\,\, x = 0, \,\, -h \leq z \leq -d.
\end{align}
</math></center>

Where <math>\alpha = \omega^2/g</math>.

<math>\phi^s(x,z)</math> is an even function and at <math>x = 0</math> the derivative of this function is equal to zero. <math>\phi^a(x,z)</math> is an odd function and at <math>x = 0</math> this function is equal to zero. These boundary conditions are important and enable the recomposed solution <math>\phi(x,z)</math> to be reflected into the domain of <math>x \geq 0</math>.

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\leq 0</math>
and <math>x\geq 0</math>.

{{separation of variables in two dimensions}}

{{separation of variables for a free surface}}

=== Separation of Variables for a Dock ===

The separation of variables equation for a partially submerged dock is given by:
<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>
Z^{\prime} (-d) = 0.
</math></center>
The solution is
<math>k=\kappa_{m}= \frac{m\pi}{h-d} \,</math> , <math>m\in\mathbb{N}\cup\left\{0\right\}</math> and
<center><math>
Z = \psi_{m}\left( z\right) = \cos\kappa_{m}(z+h),\quad
m\in\mathbb{N}\cup\left\{0\right\}.</math></center>
We note that
<center><math>
\int\nolimits_{-h}^{-d}\psi_{m}(z)\psi_{n}(z) \mathrm{d} z=C_{m}\delta_{mn},
</math></center>
where
<center>
<math>
C_{m} = \frac{1}{2} \left( \frac{\cos\kappa_{m}(h-d)\sin\kappa_{m}(h-d)+\kappa_{m}(h-d)}{\kappa_{m}} \right).
</math>
</center>
In addition there is the special case of when <math>m=n=0</math> ,
<center>
<math>
C_{0} = h-d,
</math>
</center>
where <math>\delta_{00}=1</math> .

=== Inner product between free surface and dock modes ===

<center><math>
B_{mn} = \int\nolimits_{-h}^{-d}\psi_{m}(z)\chi_{n}(z) \mathrm{d} z , \quad m\geq1, \quad n\geq0,
</math></center>
where
<center><math>
B_{mn}= \int\nolimits_{-h}^{-d} \frac{\cos\kappa_{m}(z+h)\cos k_{n}(z+h)}{\cos(k_{n}h)} \mathrm{d} z=\frac{-(-1)^m k_{n}\sin k_{n}(h-d)}
{\cos(k_{n}h)(\kappa_{m}^{2}-k_{n}^{2})}.
</math></center>
The evaluation of this inner product works on the assumption that the two roots <math>\kappa_{m}</math> and <math>k_{n}</math> do not coincide. The Matlab code given below has a routine to check for this case and make necessary adjustments.

We also have the case where the functions of the inner product are switched.
<center><math>
B_{nm} = \int\nolimits_{-h}^{-d}\chi_{m}(z)\psi_{n}(z) \mathrm{d} z , \quad m\geq0, \quad n\geq0.
</math></center>
For <math>m \geq 0</math> and <math>n \geq 1</math> the evaluation is above with the exception that the indices of the expression are switched.

Finally there is the special case of when <math>m \geq 0</math> and <math>n = 0</math>.
<center><math>
B_{0m} = \frac{\sin k_{m}(h-d)}{k_{m}\cos(k_{m}h)}.
</math></center>

=== Expansion of the potential ===

We need to apply some boundary conditions at plus and minus infinity,
where essentially 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 roots. Here we
must use a different condition, essentially identifying one solution
as the incoming wave and the other as the outgoing wave.

Therefore the diffracted potentials can
be expanded as
<center>
<math>
\phi(x,z)=\sum_{m=0}^{\infty}a_{m}e^{k_{m}x}\chi_{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 must be continuous across the
transition from open water to the dock covered region. Therefore, the
potentials at <math>x=-L</math> have to be equal.
We obtain for the '''symmetric''' and '''antisymmetric''' cases:

<center><math>
\chi_{0}\left( z\right) + \sum_{m=0}^{\infty}a^{s,a}_{m} \chi_{m}\left(z\right)
=\sum_{m=0}^{\infty}b^{s,a}_{m}\psi_{m}(z), \quad -h \leq z\leq -d.
</math></center>

We multiply both sides by <math>\psi_{n}(z) \,</math> and integrating from <math>-h</math> to <math>-d</math> to obtain:
<center><math>
B_{n0} + \sum^{\infty}_{m=0}a^{s,a}_{m}B_{nm} = b^{s,a}_{m}C_{m}\delta_{mn}, \quad n\in\mathbb{N}\cup\left\{0\right\}
</math></center>

The derivative of the potential must be continuous across the
transition from open water to the dock covered region. Therefore, the derivatives of the potentials at <math>x=-L</math> have to be equal.
We obtain for the '''symmetric''' case:

<center><math>
\begin{align}
-k_{0}\chi_{0}\left( z\right) +\sum_{m=0}^{\infty} a^{s}_{m}k_{m}\chi_{m}\left(z\right) =
\begin{cases}
\quad \quad \quad 0,\quad \quad \quad -d\leq z\leq 0 \\
-\sum_{m=0}^{\infty}b^{s}_{m}\kappa_{m}\psi_{m}(z),\quad -h\leq z\leq -d
\end{cases}
\end{align}
</math></center>

and for the second equation we multiply both sides by <math>\chi_{n}(z) \,</math> and integrating from <math>-h</math> to <math>0</math> to obtain:
<center><math>
-k_{0}A_{0}\delta_{0n} + k_{m}a_{m}A_{m} = -\sum^{\infty}_{m=0}\kappa_{m}b_{m}B_{mn}, \quad n\in\mathbb{N}\cup\left\{0\right\}
</math></center>

Solving the equations above will yield the coefficients of the water velocity potential in the dock covered region.

== Numerical Solution ==

To solve the system of equations for the '''symmetric''' case, we set the upper limit of <math>m</math> to be <math>N</math>. This resulting system can be expressed in the block matrix form below,

<center>
<math>
\begin{bmatrix}
B_{00} & B_{01} & \cdots & B_{0N} & -C_{0} & 0 & \cdots & 0 \\
B_{10} & B_{11} & \cdots & B_{1N} & 0 & -C_{1} & \cdots & \vdots \\
\vdots & \vdots & \ddots & \vdots & \vdots & \vdots & \ddots & 0 \\
B_{N0} & B_{N1} & \cdots & B_{NN} & 0 & \cdots & 0 & -C_{N} \\
k_{0}A_{0} & 0 & \cdots & 0 & 0 & \kappa_{1}\tanh(\kappa_{1}L)B_{10} & \cdots & \kappa_{N}\tanh(\kappa_{N}L)B_{N0} \\
0 & k_{1}A_{1} & \cdots & \vdots & 0 & \kappa_{1}\tanh(\kappa_{1}L)B_{11} & \cdots & \kappa_{N}\tanh(\kappa_{N}L)B_{N1} \\
\vdots & \vdots & \ddots & 0 & \vdots & \vdots & \ddots & \vdots \\
0 & \cdots & 0 & k_{N}A_{N} & 0 & \kappa_{1}\tanh(\kappa_{1}L)B_{1N} & \cdots & \kappa_{N}\tanh(\kappa_{N}L)B_{NN} \\
\end{bmatrix}
\begin{bmatrix}
a^s_{0} \\ a^s_{1} \\ \vdots \\ a^s_{N} \\ b^s_{0} \\ b^s_{1} \\ \vdots \\ b^s_{N} \\
\end{bmatrix}
=\begin{bmatrix}
-B_{00} \\ -B_{10} \\ \vdots \\ -B_{N0} \\ k_{0}A_{0} \\ 0 \\ \vdots \\ 0 \\
\end{bmatrix}
</math>
</center>

To solve the system of equations for the '''antisymmetric''' case, we set the upper limit of <math>m</math> to be <math>N</math>. This resulting system can be expressed in the block matrix form below,

<center>
<math>
\begin{bmatrix}
B_{00} & B_{01} & \cdots & B_{0N} & -C_{0} & 0 & \cdots & 0 \\
B_{10} & B_{11} & \cdots & B_{1N} & 0 & -C_{1} & \cdots & \vdots \\
\vdots & \vdots & \ddots & \vdots & \vdots & \vdots & \ddots & 0 \\
B_{N0} & B_{N1} & \cdots & B_{NN} & 0 & \cdots & 0 & -C_{N} \\
k_{0}A_{0} & 0 & \cdots & 0 & \frac{1}{L}D_{0} & \kappa_{1}\coth(\kappa_{1}L)B_{10} & \cdots & \kappa_{N}\coth(\kappa_{N}L)B_{N0} \\
0 & k_{1}A_{1} & \cdots & \vdots & \frac{1}{L}D_{1} & \kappa_{1}\coth(\kappa_{1}L)B_{11} & \cdots & \kappa_{N}\coth(\kappa_{N}L)B_{N1} \\
\vdots & \vdots & \ddots & 0 & \vdots & \vdots & \ddots & \vdots \\
0 & \cdots & 0 & k_{N}A_{N} & \frac{1}{L}D_{N} & \kappa_{1}\coth(\kappa_{1}L)B_{1N} & \cdots & \kappa_{N}\coth(\kappa_{N}L)B_{NN} \\
\end{bmatrix}
\begin{bmatrix}
a^a_{0} \\ a^a_{1} \\ \vdots \\ a^a_{N} \\ b^a_{0} \\ b^a_{1} \\ \vdots \\ b^a_{N} \\
\end{bmatrix}
=\begin{bmatrix}
-B_{00} \\ -B_{10} \\ \vdots \\ -B_{N0} \\ k_{0}A_{0} \\ 0 \\ \vdots \\ 0 \\
\end{bmatrix}
</math>
</center>

We then simply need to solve the two sets of <math>2(N+1) \mbox{ x } 2(N+1)</math> linear systems of equations. If we had not taken advantage of the symmetry we would need to solve a <math>4(N+1) \mbox{ x } 4(N+1)</math> linear system of equations for four coefficients <math>a_{m}</math>, <math>b_{m}</math>, <math>c_{m}</math> and <math>d_{m}</math>.

== Matlab Code ==

A program to calculate the coefficients for the single finite dock problem using symmetry 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 Finite Rectangle using Symmetry

2012-03-16T05:15:14Z

Tertius: /* An infinite dimensional system of equations */

{{incomplete pages}}

== Introduction ==

The problem consists of a finite rigid fixed rectangle of length <math>2L</math> centered at <math>x = 0</math> and having a height of <math>d + ...</math> freeboard with depth of submergence being <math>d</math> through which no flow is possible. The regions to the left and right of the rectangle are the free water surface. Since the geometry of the rectangle is symmetric about <math>x = 0</math> we can decompose the velocity potential <math>\phi</math> into a symmetric <math>\phi^s</math> and antisymmetric <math>\phi^a</math> potential. We then solve for these two potentials in the left domain only <math>x \leq 0</math> and subsequently recompose the solution <math>\phi</math> for the entire domain. We begin with the simple problem when the waves are normally incident on the rectangle (dock) (so that the problem is truly two-dimensional). We can extend this solution to a pair of partially submerged finite rigid fixed rectangles using symmetry
[[Eigenfunction Matching for a Finite Rectangle using Symmetry]]

== Governing Equations ==

We begin with the [[Frequency Domain Problem]] for a partially submerged finite dock. Although the dock is of length <math>2L</math> we are only interested in the dock of <math>-L \leq x \leq 0</math> together with the fluid domain <math>x \leq 0</math>. We assume <math>e^{i\omega t}</math> time dependence and
the depth of submergence is <math>d</math>.
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>
\begin{align}
\Delta\phi^{s,a} &= 0, \,\,\, -h\leq z\leq 0, \,\, x < -L, \\

\Delta\phi^{s,a} &= 0, \,\,\, -h\leq z\leq -d, \,\, -L < x < 0. \\

\partial_z\phi^{s,a} &= \alpha\phi^{s,a}, \,\,\, z=0, \,\, x < -L, \\

\partial_x\phi^{s,a} &= 0, \,\,\, -d\leq z\leq 0, \,\, x = -L, \\

\partial_z\phi^{s,a} &= 0, \,\,\,
\begin{cases}
z = -d, \,\, -L < x < 0, \\
z = -h, \,\, x < 0.
\end{cases} \\

\partial_x\phi^s &= 0, \,\,\, x = 0, \,\, -h \leq z \leq -d, \\

\phi^a &= 0, \,\,\, x = 0, \,\, -h \leq z \leq -d.
\end{align}
</math></center>

Where <math>\alpha = \omega^2/g</math>.

<math>\phi^s(x,z)</math> is an even function and at <math>x = 0</math> the derivative of this function is equal to zero. <math>\phi^a(x,z)</math> is an odd function and at <math>x = 0</math> this function is equal to zero. These boundary conditions are important and enable the recomposed solution <math>\phi(x,z)</math> to be reflected into the domain of <math>x \geq 0</math>.

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\leq 0</math>
and <math>x\geq 0</math>.

{{separation of variables in two dimensions}}

{{separation of variables for a free surface}}

=== Separation of Variables for a Dock ===

The separation of variables equation for a partially submerged dock is given by:
<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>
Z^{\prime} (-d) = 0.
</math></center>
The solution is
<math>k=\kappa_{m}= \frac{m\pi}{h-d} \,</math> , <math>m\in\mathbb{N}\cup\left\{0\right\}</math> and
<center><math>
Z = \psi_{m}\left( z\right) = \cos\kappa_{m}(z+h),\quad
m\in\mathbb{N}\cup\left\{0\right\}.</math></center>
We note that
<center><math>
\int\nolimits_{-h}^{-d}\psi_{m}(z)\psi_{n}(z) \mathrm{d} z=C_{m}\delta_{mn},
</math></center>
where
<center>
<math>
C_{m} = \frac{1}{2} \left( \frac{\cos\kappa_{m}(h-d)\sin\kappa_{m}(h-d)+\kappa_{m}(h-d)}{\kappa_{m}} \right).
</math>
</center>
In addition there is the special case of when <math>m=n=0</math> ,
<center>
<math>
C_{0} = h-d,
</math>
</center>
where <math>\delta_{00}=1</math> .

=== Inner product between free surface and dock modes ===

<center><math>
B_{mn} = \int\nolimits_{-h}^{-d}\psi_{m}(z)\chi_{n}(z) \mathrm{d} z , \quad m\geq1, \quad n\geq0,
</math></center>
where
<center><math>
B_{mn}= \int\nolimits_{-h}^{-d} \frac{\cos\kappa_{m}(z+h)\cos k_{n}(z+h)}{\cos(k_{n}h)} \mathrm{d} z=\frac{-(-1)^m k_{n}\sin k_{n}(h-d)}
{\cos(k_{n}h)(\kappa_{m}^{2}-k_{n}^{2})}.
</math></center>
The evaluation of this inner product works on the assumption that the two roots <math>\kappa_{m}</math> and <math>k_{n}</math> do not coincide. The Matlab code given below has a routine to check for this case and make necessary adjustments.

We also have the case where the functions of the inner product are switched.
<center><math>
B_{nm} = \int\nolimits_{-h}^{-d}\chi_{m}(z)\psi_{n}(z) \mathrm{d} z , \quad m\geq0, \quad n\geq0.
</math></center>
For <math>m \geq 0</math> and <math>n \geq 1</math> the evaluation is above with the exception that the indices of the expression are switched.

Finally there is the special case of when <math>m \geq 0</math> and <math>n = 0</math>.
<center><math>
B_{0m} = \frac{\sin k_{m}(h-d)}{k_{m}\cos(k_{m}h)}.
</math></center>

=== Expansion of the potential ===

We need to apply some boundary conditions at plus and minus infinity,
where essentially 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 roots. Here we
must use a different condition, essentially identifying one solution
as the incoming wave and the other as the outgoing wave.

Therefore the diffracted potentials can
be expanded as
<center>
<math>
\phi(x,z)=\sum_{m=0}^{\infty}a_{m}e^{k_{m}x}\chi_{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 must be continuous across the
transition from open water to the dock covered region. Therefore, the
potentials at <math>x=-L</math> have to be equal.
We obtain for the '''symmetric''' and '''antisymmetric''' cases:

<center><math>
\chi_{0}\left( z\right) + \sum_{m=0}^{\infty}a^{s,a}_{m} \chi_{m}\left(z\right)
=\sum_{m=0}^{\infty}b^{s,a}_{m}\psi_{m}(z), \quad -h \leq z\leq -d.
</math></center>

We multiply both sides by <math>\psi_{n}(z) \,</math> and integrating from <math>-h</math> to <math>-d</math> to obtain:
<center><math>
B_{n0} + \sum^{\infty}_{m=0}a^{s,a}_{m}B_{nm} = b^{s,a}_{m}C_{m}\delta_{mn}, n\in\mathbb{N}\cup\left\{0\right\}
</math></center>

The derivative of the potential must be continuous across the
transition from open water to the dock covered region. Therefore, the derivatives of the potentials at <math>x=-L</math> have to be equal.
We obtain for the '''symmetric''' case:

<center><math>
\begin{align}
-k_{0}\chi_{0}\left( z\right) +\sum_{m=0}^{\infty} a^{s}_{m}k_{m}\chi_{m}\left(z\right) =
\begin{cases}
\quad \quad \quad 0,\quad \quad \quad -d\leq z\leq 0 \\
-\sum_{m=0}^{\infty}b^{s}_{m}\kappa_{m}\psi_{m}(z),\quad -h\leq z\leq -d
\end{cases}
\end{align}
</math></center>

and for the second equation we multiply both sides by <math>\chi_{n}(z) \,</math> and integrating from <math>-h</math> to <math>0</math> to obtain:
<center><math>
-k_{0}A_{0}\delta_{0n} + k_{m}a_{m}A_{m} = -\sum^{\infty}_{m=0}\kappa_{m}b_{m}B_{mn}, n\in\mathbb{N}\cup\left\{0\right\}
</math></center>

Solving the equations above will yield the coefficients of the water velocity potential in the dock covered region.

== Numerical Solution ==

To solve the system of equations for the '''symmetric''' case, we set the upper limit of <math>m</math> to be <math>N</math>. This resulting system can be expressed in the block matrix form below,

<center>
<math>
\begin{bmatrix}
B_{00} & B_{01} & \cdots & B_{0N} & -C_{0} & 0 & \cdots & 0 \\
B_{10} & B_{11} & \cdots & B_{1N} & 0 & -C_{1} & \cdots & \vdots \\
\vdots & \vdots & \ddots & \vdots & \vdots & \vdots & \ddots & 0 \\
B_{N0} & B_{N1} & \cdots & B_{NN} & 0 & \cdots & 0 & -C_{N} \\
k_{0}A_{0} & 0 & \cdots & 0 & 0 & \kappa_{1}\tanh(\kappa_{1}L)B_{10} & \cdots & \kappa_{N}\tanh(\kappa_{N}L)B_{N0} \\
0 & k_{1}A_{1} & \cdots & \vdots & 0 & \kappa_{1}\tanh(\kappa_{1}L)B_{11} & \cdots & \kappa_{N}\tanh(\kappa_{N}L)B_{N1} \\
\vdots & \vdots & \ddots & 0 & \vdots & \vdots & \ddots & \vdots \\
0 & \cdots & 0 & k_{N}A_{N} & 0 & \kappa_{1}\tanh(\kappa_{1}L)B_{1N} & \cdots & \kappa_{N}\tanh(\kappa_{N}L)B_{NN} \\
\end{bmatrix}
\begin{bmatrix}
a^s_{0} \\ a^s_{1} \\ \vdots \\ a^s_{N} \\ b^s_{0} \\ b^s_{1} \\ \vdots \\ b^s_{N} \\
\end{bmatrix}
=\begin{bmatrix}
-B_{00} \\ -B_{10} \\ \vdots \\ -B_{N0} \\ k_{0}A_{0} \\ 0 \\ \vdots \\ 0 \\
\end{bmatrix}
</math>
</center>

To solve the system of equations for the '''antisymmetric''' case, we set the upper limit of <math>m</math> to be <math>N</math>. This resulting system can be expressed in the block matrix form below,

<center>
<math>
\begin{bmatrix}
B_{00} & B_{01} & \cdots & B_{0N} & -C_{0} & 0 & \cdots & 0 \\
B_{10} & B_{11} & \cdots & B_{1N} & 0 & -C_{1} & \cdots & \vdots \\
\vdots & \vdots & \ddots & \vdots & \vdots & \vdots & \ddots & 0 \\
B_{N0} & B_{N1} & \cdots & B_{NN} & 0 & \cdots & 0 & -C_{N} \\
k_{0}A_{0} & 0 & \cdots & 0 & \frac{1}{L}D_{0} & \kappa_{1}\coth(\kappa_{1}L)B_{10} & \cdots & \kappa_{N}\coth(\kappa_{N}L)B_{N0} \\
0 & k_{1}A_{1} & \cdots & \vdots & \frac{1}{L}D_{1} & \kappa_{1}\coth(\kappa_{1}L)B_{11} & \cdots & \kappa_{N}\coth(\kappa_{N}L)B_{N1} \\
\vdots & \vdots & \ddots & 0 & \vdots & \vdots & \ddots & \vdots \\
0 & \cdots & 0 & k_{N}A_{N} & \frac{1}{L}D_{N} & \kappa_{1}\coth(\kappa_{1}L)B_{1N} & \cdots & \kappa_{N}\coth(\kappa_{N}L)B_{NN} \\
\end{bmatrix}
\begin{bmatrix}
a^a_{0} \\ a^a_{1} \\ \vdots \\ a^a_{N} \\ b^a_{0} \\ b^a_{1} \\ \vdots \\ b^a_{N} \\
\end{bmatrix}
=\begin{bmatrix}
-B_{00} \\ -B_{10} \\ \vdots \\ -B_{N0} \\ k_{0}A_{0} \\ 0 \\ \vdots \\ 0 \\
\end{bmatrix}
</math>
</center>

We then simply need to solve the two sets of <math>2(N+1) \mbox{ x } 2(N+1)</math> linear systems of equations. If we had not taken advantage of the symmetry we would need to solve a <math>4(N+1) \mbox{ x } 4(N+1)</math> linear system of equations for four coefficients <math>a_{m}</math>, <math>b_{m}</math>, <math>c_{m}</math> and <math>d_{m}</math>.

== Matlab Code ==

A program to calculate the coefficients for the single finite dock problem using symmetry 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 Finite Rectangle using Symmetry

2012-03-16T05:01:35Z

Tertius: /* An infinite dimensional system of equations */

{{incomplete pages}}

== Introduction ==

The problem consists of a finite rigid fixed rectangle of length <math>2L</math> centered at <math>x = 0</math> and having a height of <math>d + ...</math> freeboard with depth of submergence being <math>d</math> through which no flow is possible. The regions to the left and right of the rectangle are the free water surface. Since the geometry of the rectangle is symmetric about <math>x = 0</math> we can decompose the velocity potential <math>\phi</math> into a symmetric <math>\phi^s</math> and antisymmetric <math>\phi^a</math> potential. We then solve for these two potentials in the left domain only <math>x \leq 0</math> and subsequently recompose the solution <math>\phi</math> for the entire domain. We begin with the simple problem when the waves are normally incident on the rectangle (dock) (so that the problem is truly two-dimensional). We can extend this solution to a pair of partially submerged finite rigid fixed rectangles using symmetry
[[Eigenfunction Matching for a Finite Rectangle using Symmetry]]

== Governing Equations ==

We begin with the [[Frequency Domain Problem]] for a partially submerged finite dock. Although the dock is of length <math>2L</math> we are only interested in the dock of <math>-L \leq x \leq 0</math> together with the fluid domain <math>x \leq 0</math>. We assume <math>e^{i\omega t}</math> time dependence and
the depth of submergence is <math>d</math>.
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>
\begin{align}
\Delta\phi^{s,a} &= 0, \,\,\, -h\leq z\leq 0, \,\, x < -L, \\

\Delta\phi^{s,a} &= 0, \,\,\, -h\leq z\leq -d, \,\, -L < x < 0. \\

\partial_z\phi^{s,a} &= \alpha\phi^{s,a}, \,\,\, z=0, \,\, x < -L, \\

\partial_x\phi^{s,a} &= 0, \,\,\, -d\leq z\leq 0, \,\, x = -L, \\

\partial_z\phi^{s,a} &= 0, \,\,\,
\begin{cases}
z = -d, \,\, -L < x < 0, \\
z = -h, \,\, x < 0.
\end{cases} \\

\partial_x\phi^s &= 0, \,\,\, x = 0, \,\, -h \leq z \leq -d, \\

\phi^a &= 0, \,\,\, x = 0, \,\, -h \leq z \leq -d.
\end{align}
</math></center>

Where <math>\alpha = \omega^2/g</math>.

<math>\phi^s(x,z)</math> is an even function and at <math>x = 0</math> the derivative of this function is equal to zero. <math>\phi^a(x,z)</math> is an odd function and at <math>x = 0</math> this function is equal to zero. These boundary conditions are important and enable the recomposed solution <math>\phi(x,z)</math> to be reflected into the domain of <math>x \geq 0</math>.

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\leq 0</math>
and <math>x\geq 0</math>.

{{separation of variables in two dimensions}}

{{separation of variables for a free surface}}

=== Separation of Variables for a Dock ===

The separation of variables equation for a partially submerged dock is given by:
<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>
Z^{\prime} (-d) = 0.
</math></center>
The solution is
<math>k=\kappa_{m}= \frac{m\pi}{h-d} \,</math> , <math>m\in\mathbb{N}\cup\left\{0\right\}</math> and
<center><math>
Z = \psi_{m}\left( z\right) = \cos\kappa_{m}(z+h),\quad
m\in\mathbb{N}\cup\left\{0\right\}.</math></center>
We note that
<center><math>
\int\nolimits_{-h}^{-d}\psi_{m}(z)\psi_{n}(z) \mathrm{d} z=C_{m}\delta_{mn},
</math></center>
where
<center>
<math>
C_{m} = \frac{1}{2} \left( \frac{\cos\kappa_{m}(h-d)\sin\kappa_{m}(h-d)+\kappa_{m}(h-d)}{\kappa_{m}} \right).
</math>
</center>
In addition there is the special case of when <math>m=n=0</math> ,
<center>
<math>
C_{0} = h-d,
</math>
</center>
where <math>\delta_{00}=1</math> .

=== Inner product between free surface and dock modes ===

<center><math>
B_{mn} = \int\nolimits_{-h}^{-d}\psi_{m}(z)\chi_{n}(z) \mathrm{d} z , \quad m\geq1, \quad n\geq0,
</math></center>
where
<center><math>
B_{mn}= \int\nolimits_{-h}^{-d} \frac{\cos\kappa_{m}(z+h)\cos k_{n}(z+h)}{\cos(k_{n}h)} \mathrm{d} z=\frac{-(-1)^m k_{n}\sin k_{n}(h-d)}
{\cos(k_{n}h)(\kappa_{m}^{2}-k_{n}^{2})}.
</math></center>
The evaluation of this inner product works on the assumption that the two roots <math>\kappa_{m}</math> and <math>k_{n}</math> do not coincide. The Matlab code given below has a routine to check for this case and make necessary adjustments.

We also have the case where the functions of the inner product are switched.
<center><math>
B_{nm} = \int\nolimits_{-h}^{-d}\chi_{m}(z)\psi_{n}(z) \mathrm{d} z , \quad m\geq0, \quad n\geq0.
</math></center>
For <math>m \geq 0</math> and <math>n \geq 1</math> the evaluation is above with the exception that the indices of the expression are switched.

Finally there is the special case of when <math>m \geq 0</math> and <math>n = 0</math>.
<center><math>
B_{0m} = \frac{\sin k_{m}(h-d)}{k_{m}\cos(k_{m}h)}.
</math></center>

=== Expansion of the potential ===

We need to apply some boundary conditions at plus and minus infinity,
where essentially 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 roots. Here we
must use a different condition, essentially identifying one solution
as the incoming wave and the other as the outgoing wave.

Therefore the diffracted potentials can
be expanded as
<center>
<math>
\phi(x,z)=\sum_{m=0}^{\infty}a_{m}e^{k_{m}x}\chi_{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 must be continuous across the
transition from open water to the dock covered region. Therefore, the
potentials at <math>x=-L</math> have to be equal.
We obtain for the '''symmetric''' and '''antisymmetric''' cases:

<center><math>
\chi_{0}\left( z\right) + \sum_{m=0}^{\infty}a^{s,a}_{m} \chi_{m}\left(z\right)
=\sum_{m=0}^{\infty}b^{s,a}_{m}\psi_{m}(z), \quad -h \leq z\leq -d.
</math></center>

The derivative of the potential must be continuous across the
transition from open water to the dock covered region. Therefore, the derivatives of the potentials at <math>x=-L</math> have to be equal.
We obtain for the '''symmetric''' case:

<center><math>
\begin{align}
-k_{0}\chi_{0}\left( z\right) +\sum_{m=0}^{\infty} a^{s}_{m}k_{m}\chi_{m}\left(z\right) =
\begin{cases}
\quad \quad \quad 0,\quad \quad \quad -d\leq z\leq 0 \\
-\sum_{m=0}^{\infty}b^{s}_{m}\kappa_{m}\psi_{m}(z),\quad -h\leq z\leq -d
\end{cases}
\end{align}
</math></center>

For the first equation we multiply both sides by <math>\psi_{n}(z) \,</math> and integrating from <math>-h</math> to <math>-d</math> to obtain:
<center><math>
B_{n0} + \sum^{\infty}_{m=0}a_{m}B_{nm} = b_{m}C_{m}\delta_{mn}, n\in\mathbb{N}\cup\left\{0\right\}
</math></center>
and for the second equation we multiply both sides by <math>\chi_{n}(z) \,</math> and integrating from <math>-h</math> to <math>0</math> to obtain:
<center><math>
-k_{0}A_{0}\delta_{0n} + k_{m}a_{m}A_{m} = -\sum^{\infty}_{m=0}\kappa_{m}b_{m}B_{mn}, n\in\mathbb{N}\cup\left\{0\right\}
</math></center>

Solving the equations above will yield the coefficients of the water velocity potential in the dock covered region.

== Numerical Solution ==

To solve the system of equations for the '''symmetric''' case, we set the upper limit of <math>m</math> to be <math>N</math>. This resulting system can be expressed in the block matrix form below,

<center>
<math>
\begin{bmatrix}
B_{00} & B_{01} & \cdots & B_{0N} & -C_{0} & 0 & \cdots & 0 \\
B_{10} & B_{11} & \cdots & B_{1N} & 0 & -C_{1} & \cdots & \vdots \\
\vdots & \vdots & \ddots & \vdots & \vdots & \vdots & \ddots & 0 \\
B_{N0} & B_{N1} & \cdots & B_{NN} & 0 & \cdots & 0 & -C_{N} \\
k_{0}A_{0} & 0 & \cdots & 0 & 0 & \kappa_{1}\tanh(\kappa_{1}L)B_{10} & \cdots & \kappa_{N}\tanh(\kappa_{N}L)B_{N0} \\
0 & k_{1}A_{1} & \cdots & \vdots & 0 & \kappa_{1}\tanh(\kappa_{1}L)B_{11} & \cdots & \kappa_{N}\tanh(\kappa_{N}L)B_{N1} \\
\vdots & \vdots & \ddots & 0 & \vdots & \vdots & \ddots & \vdots \\
0 & \cdots & 0 & k_{N}A_{N} & 0 & \kappa_{1}\tanh(\kappa_{1}L)B_{1N} & \cdots & \kappa_{N}\tanh(\kappa_{N}L)B_{NN} \\
\end{bmatrix}
\begin{bmatrix}
a^s_{0} \\ a^s_{1} \\ \vdots \\ a^s_{N} \\ b^s_{0} \\ b^s_{1} \\ \vdots \\ b^s_{N} \\
\end{bmatrix}
=\begin{bmatrix}
-B_{00} \\ -B_{10} \\ \vdots \\ -B_{N0} \\ k_{0}A_{0} \\ 0 \\ \vdots \\ 0 \\
\end{bmatrix}
</math>
</center>

To solve the system of equations for the '''antisymmetric''' case, we set the upper limit of <math>m</math> to be <math>N</math>. This resulting system can be expressed in the block matrix form below,

<center>
<math>
\begin{bmatrix}
B_{00} & B_{01} & \cdots & B_{0N} & -C_{0} & 0 & \cdots & 0 \\
B_{10} & B_{11} & \cdots & B_{1N} & 0 & -C_{1} & \cdots & \vdots \\
\vdots & \vdots & \ddots & \vdots & \vdots & \vdots & \ddots & 0 \\
B_{N0} & B_{N1} & \cdots & B_{NN} & 0 & \cdots & 0 & -C_{N} \\
k_{0}A_{0} & 0 & \cdots & 0 & \frac{1}{L}D_{0} & \kappa_{1}\coth(\kappa_{1}L)B_{10} & \cdots & \kappa_{N}\coth(\kappa_{N}L)B_{N0} \\
0 & k_{1}A_{1} & \cdots & \vdots & \frac{1}{L}D_{1} & \kappa_{1}\coth(\kappa_{1}L)B_{11} & \cdots & \kappa_{N}\coth(\kappa_{N}L)B_{N1} \\
\vdots & \vdots & \ddots & 0 & \vdots & \vdots & \ddots & \vdots \\
0 & \cdots & 0 & k_{N}A_{N} & \frac{1}{L}D_{N} & \kappa_{1}\coth(\kappa_{1}L)B_{1N} & \cdots & \kappa_{N}\coth(\kappa_{N}L)B_{NN} \\
\end{bmatrix}
\begin{bmatrix}
a^a_{0} \\ a^a_{1} \\ \vdots \\ a^a_{N} \\ b^a_{0} \\ b^a_{1} \\ \vdots \\ b^a_{N} \\
\end{bmatrix}
=\begin{bmatrix}
-B_{00} \\ -B_{10} \\ \vdots \\ -B_{N0} \\ k_{0}A_{0} \\ 0 \\ \vdots \\ 0 \\
\end{bmatrix}
</math>
</center>

We then simply need to solve the two sets of <math>2(N+1) \mbox{ x } 2(N+1)</math> linear systems of equations. If we had not taken advantage of the symmetry we would need to solve a <math>4(N+1) \mbox{ x } 4(N+1)</math> linear system of equations for four coefficients <math>a_{m}</math>, <math>b_{m}</math>, <math>c_{m}</math> and <math>d_{m}</math>.

== Matlab Code ==

A program to calculate the coefficients for the single finite dock problem using symmetry 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 Finite Rectangle using Symmetry

2012-03-16T05:00:12Z

Tertius: /* An infinite dimensional system of equations */

{{incomplete pages}}

== Introduction ==

The problem consists of a finite rigid fixed rectangle of length <math>2L</math> centered at <math>x = 0</math> and having a height of <math>d + ...</math> freeboard with depth of submergence being <math>d</math> through which no flow is possible. The regions to the left and right of the rectangle are the free water surface. Since the geometry of the rectangle is symmetric about <math>x = 0</math> we can decompose the velocity potential <math>\phi</math> into a symmetric <math>\phi^s</math> and antisymmetric <math>\phi^a</math> potential. We then solve for these two potentials in the left domain only <math>x \leq 0</math> and subsequently recompose the solution <math>\phi</math> for the entire domain. We begin with the simple problem when the waves are normally incident on the rectangle (dock) (so that the problem is truly two-dimensional). We can extend this solution to a pair of partially submerged finite rigid fixed rectangles using symmetry
[[Eigenfunction Matching for a Finite Rectangle using Symmetry]]

== Governing Equations ==

We begin with the [[Frequency Domain Problem]] for a partially submerged finite dock. Although the dock is of length <math>2L</math> we are only interested in the dock of <math>-L \leq x \leq 0</math> together with the fluid domain <math>x \leq 0</math>. We assume <math>e^{i\omega t}</math> time dependence and
the depth of submergence is <math>d</math>.
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>
\begin{align}
\Delta\phi^{s,a} &= 0, \,\,\, -h\leq z\leq 0, \,\, x < -L, \\

\Delta\phi^{s,a} &= 0, \,\,\, -h\leq z\leq -d, \,\, -L < x < 0. \\

\partial_z\phi^{s,a} &= \alpha\phi^{s,a}, \,\,\, z=0, \,\, x < -L, \\

\partial_x\phi^{s,a} &= 0, \,\,\, -d\leq z\leq 0, \,\, x = -L, \\

\partial_z\phi^{s,a} &= 0, \,\,\,
\begin{cases}
z = -d, \,\, -L < x < 0, \\
z = -h, \,\, x < 0.
\end{cases} \\

\partial_x\phi^s &= 0, \,\,\, x = 0, \,\, -h \leq z \leq -d, \\

\phi^a &= 0, \,\,\, x = 0, \,\, -h \leq z \leq -d.
\end{align}
</math></center>

Where <math>\alpha = \omega^2/g</math>.

<math>\phi^s(x,z)</math> is an even function and at <math>x = 0</math> the derivative of this function is equal to zero. <math>\phi^a(x,z)</math> is an odd function and at <math>x = 0</math> this function is equal to zero. These boundary conditions are important and enable the recomposed solution <math>\phi(x,z)</math> to be reflected into the domain of <math>x \geq 0</math>.

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\leq 0</math>
and <math>x\geq 0</math>.

{{separation of variables in two dimensions}}

{{separation of variables for a free surface}}

=== Separation of Variables for a Dock ===

The separation of variables equation for a partially submerged dock is given by:
<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>
Z^{\prime} (-d) = 0.
</math></center>
The solution is
<math>k=\kappa_{m}= \frac{m\pi}{h-d} \,</math> , <math>m\in\mathbb{N}\cup\left\{0\right\}</math> and
<center><math>
Z = \psi_{m}\left( z\right) = \cos\kappa_{m}(z+h),\quad
m\in\mathbb{N}\cup\left\{0\right\}.</math></center>
We note that
<center><math>
\int\nolimits_{-h}^{-d}\psi_{m}(z)\psi_{n}(z) \mathrm{d} z=C_{m}\delta_{mn},
</math></center>
where
<center>
<math>
C_{m} = \frac{1}{2} \left( \frac{\cos\kappa_{m}(h-d)\sin\kappa_{m}(h-d)+\kappa_{m}(h-d)}{\kappa_{m}} \right).
</math>
</center>
In addition there is the special case of when <math>m=n=0</math> ,
<center>
<math>
C_{0} = h-d,
</math>
</center>
where <math>\delta_{00}=1</math> .

=== Inner product between free surface and dock modes ===

<center><math>
B_{mn} = \int\nolimits_{-h}^{-d}\psi_{m}(z)\chi_{n}(z) \mathrm{d} z , \quad m\geq1, \quad n\geq0,
</math></center>
where
<center><math>
B_{mn}= \int\nolimits_{-h}^{-d} \frac{\cos\kappa_{m}(z+h)\cos k_{n}(z+h)}{\cos(k_{n}h)} \mathrm{d} z=\frac{-(-1)^m k_{n}\sin k_{n}(h-d)}
{\cos(k_{n}h)(\kappa_{m}^{2}-k_{n}^{2})}.
</math></center>
The evaluation of this inner product works on the assumption that the two roots <math>\kappa_{m}</math> and <math>k_{n}</math> do not coincide. The Matlab code given below has a routine to check for this case and make necessary adjustments.

We also have the case where the functions of the inner product are switched.
<center><math>
B_{nm} = \int\nolimits_{-h}^{-d}\chi_{m}(z)\psi_{n}(z) \mathrm{d} z , \quad m\geq0, \quad n\geq0.
</math></center>
For <math>m \geq 0</math> and <math>n \geq 1</math> the evaluation is above with the exception that the indices of the expression are switched.

Finally there is the special case of when <math>m \geq 0</math> and <math>n = 0</math>.
<center><math>
B_{0m} = \frac{\sin k_{m}(h-d)}{k_{m}\cos(k_{m}h)}.
</math></center>

=== Expansion of the potential ===

We need to apply some boundary conditions at plus and minus infinity,
where essentially 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 roots. Here we
must use a different condition, essentially identifying one solution
as the incoming wave and the other as the outgoing wave.

Therefore the diffracted potentials can
be expanded as
<center>
<math>
\phi(x,z)=\sum_{m=0}^{\infty}a_{m}e^{k_{m}x}\chi_{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 must be continuous across the
transition from open water to the dock covered region. Therefore, the
potentials at <math>x=-L</math> have to be equal.
We obtain for the '''symmetric''' and '''antisymmetric''' cases:

<center><math>
\chi_{0}\left( z\right) + \sum_{m=0}^{\infty}a^{s,a}_{m} \chi_{m}\left(z\right)
=\sum_{m=0}^{\infty}b^{s,a}_{m}\psi_{m}(z), \quad -h \leq z\leq -d.
</math></center>

The derivative of the potential must be continuous across the
transition from open water to the dock covered region. Therefore, the derivatives of the potentials at <math>x=-L</math> have to be equal.
We obtain for the '''symmetric''' case:

<center><math>
\begin{align}
-k_{0}\chi_{0}\left( z\right) +\sum_{m=0}^{\infty} a^{s,a}_{m}k_{m}\chi_{m}\left(z\right) =
\begin{cases}
\quad \quad \quad 0,\quad \quad \quad -d\leq z\leq 0 \\
-\sum_{m=0}^{\infty}b^{s,a}_{m}\kappa_{m}\psi_{m}(z),\quad -h\leq z\leq -d
\end{cases}
\end{align}
</math></center>

For the first equation we multiply both sides by <math>\psi_{n}(z) \,</math> and integrating from <math>-h</math> to <math>-d</math> to obtain:
<center><math>
B_{n0} + \sum^{\infty}_{m=0}a_{m}B_{nm} = b_{m}C_{m}\delta_{mn}, n\in\mathbb{N}\cup\left\{0\right\}
</math></center>
and for the second equation we multiply both sides by <math>\chi_{n}(z) \,</math> and integrating from <math>-h</math> to <math>0</math> to obtain:
<center><math>
-k_{0}A_{0}\delta_{0n} + k_{m}a_{m}A_{m} = -\sum^{\infty}_{m=0}\kappa_{m}b_{m}B_{mn}, n\in\mathbb{N}\cup\left\{0\right\}
</math></center>

Solving the equations above will yield the coefficients of the water velocity potential in the dock covered region.

== Numerical Solution ==

To solve the system of equations for the '''symmetric''' case, we set the upper limit of <math>m</math> to be <math>N</math>. This resulting system can be expressed in the block matrix form below,

<center>
<math>
\begin{bmatrix}
B_{00} & B_{01} & \cdots & B_{0N} & -C_{0} & 0 & \cdots & 0 \\
B_{10} & B_{11} & \cdots & B_{1N} & 0 & -C_{1} & \cdots & \vdots \\
\vdots & \vdots & \ddots & \vdots & \vdots & \vdots & \ddots & 0 \\
B_{N0} & B_{N1} & \cdots & B_{NN} & 0 & \cdots & 0 & -C_{N} \\
k_{0}A_{0} & 0 & \cdots & 0 & 0 & \kappa_{1}\tanh(\kappa_{1}L)B_{10} & \cdots & \kappa_{N}\tanh(\kappa_{N}L)B_{N0} \\
0 & k_{1}A_{1} & \cdots & \vdots & 0 & \kappa_{1}\tanh(\kappa_{1}L)B_{11} & \cdots & \kappa_{N}\tanh(\kappa_{N}L)B_{N1} \\
\vdots & \vdots & \ddots & 0 & \vdots & \vdots & \ddots & \vdots \\
0 & \cdots & 0 & k_{N}A_{N} & 0 & \kappa_{1}\tanh(\kappa_{1}L)B_{1N} & \cdots & \kappa_{N}\tanh(\kappa_{N}L)B_{NN} \\
\end{bmatrix}
\begin{bmatrix}
a^s_{0} \\ a^s_{1} \\ \vdots \\ a^s_{N} \\ b^s_{0} \\ b^s_{1} \\ \vdots \\ b^s_{N} \\
\end{bmatrix}
=\begin{bmatrix}
-B_{00} \\ -B_{10} \\ \vdots \\ -B_{N0} \\ k_{0}A_{0} \\ 0 \\ \vdots \\ 0 \\
\end{bmatrix}
</math>
</center>

To solve the system of equations for the '''antisymmetric''' case, we set the upper limit of <math>m</math> to be <math>N</math>. This resulting system can be expressed in the block matrix form below,

<center>
<math>
\begin{bmatrix}
B_{00} & B_{01} & \cdots & B_{0N} & -C_{0} & 0 & \cdots & 0 \\
B_{10} & B_{11} & \cdots & B_{1N} & 0 & -C_{1} & \cdots & \vdots \\
\vdots & \vdots & \ddots & \vdots & \vdots & \vdots & \ddots & 0 \\
B_{N0} & B_{N1} & \cdots & B_{NN} & 0 & \cdots & 0 & -C_{N} \\
k_{0}A_{0} & 0 & \cdots & 0 & \frac{1}{L}D_{0} & \kappa_{1}\coth(\kappa_{1}L)B_{10} & \cdots & \kappa_{N}\coth(\kappa_{N}L)B_{N0} \\
0 & k_{1}A_{1} & \cdots & \vdots & \frac{1}{L}D_{1} & \kappa_{1}\coth(\kappa_{1}L)B_{11} & \cdots & \kappa_{N}\coth(\kappa_{N}L)B_{N1} \\
\vdots & \vdots & \ddots & 0 & \vdots & \vdots & \ddots & \vdots \\
0 & \cdots & 0 & k_{N}A_{N} & \frac{1}{L}D_{N} & \kappa_{1}\coth(\kappa_{1}L)B_{1N} & \cdots & \kappa_{N}\coth(\kappa_{N}L)B_{NN} \\
\end{bmatrix}
\begin{bmatrix}
a^a_{0} \\ a^a_{1} \\ \vdots \\ a^a_{N} \\ b^a_{0} \\ b^a_{1} \\ \vdots \\ b^a_{N} \\
\end{bmatrix}
=\begin{bmatrix}
-B_{00} \\ -B_{10} \\ \vdots \\ -B_{N0} \\ k_{0}A_{0} \\ 0 \\ \vdots \\ 0 \\
\end{bmatrix}
</math>
</center>

We then simply need to solve the two sets of <math>2(N+1) \mbox{ x } 2(N+1)</math> linear systems of equations. If we had not taken advantage of the symmetry we would need to solve a <math>4(N+1) \mbox{ x } 4(N+1)</math> linear system of equations for four coefficients <math>a_{m}</math>, <math>b_{m}</math>, <math>c_{m}</math> and <math>d_{m}</math>.

== Matlab Code ==

A program to calculate the coefficients for the single finite dock problem using symmetry 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 Finite Rectangle using Symmetry

2012-03-16T04:35:33Z

Tertius: /* An infinite dimensional system of equations */

{{incomplete pages}}

== Introduction ==

The problem consists of a finite rigid fixed rectangle of length <math>2L</math> centered at <math>x = 0</math> and having a height of <math>d + ...</math> freeboard with depth of submergence being <math>d</math> through which no flow is possible. The regions to the left and right of the rectangle are the free water surface. Since the geometry of the rectangle is symmetric about <math>x = 0</math> we can decompose the velocity potential <math>\phi</math> into a symmetric <math>\phi^s</math> and antisymmetric <math>\phi^a</math> potential. We then solve for these two potentials in the left domain only <math>x \leq 0</math> and subsequently recompose the solution <math>\phi</math> for the entire domain. We begin with the simple problem when the waves are normally incident on the rectangle (dock) (so that the problem is truly two-dimensional). We can extend this solution to a pair of partially submerged finite rigid fixed rectangles using symmetry
[[Eigenfunction Matching for a Finite Rectangle using Symmetry]]

== Governing Equations ==

We begin with the [[Frequency Domain Problem]] for a partially submerged finite dock. Although the dock is of length <math>2L</math> we are only interested in the dock of <math>-L \leq x \leq 0</math> together with the fluid domain <math>x \leq 0</math>. We assume <math>e^{i\omega t}</math> time dependence and
the depth of submergence is <math>d</math>.
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>
\begin{align}
\Delta\phi^{s,a} &= 0, \,\,\, -h\leq z\leq 0, \,\, x < -L, \\

\Delta\phi^{s,a} &= 0, \,\,\, -h\leq z\leq -d, \,\, -L < x < 0. \\

\partial_z\phi^{s,a} &= \alpha\phi^{s,a}, \,\,\, z=0, \,\, x < -L, \\

\partial_x\phi^{s,a} &= 0, \,\,\, -d\leq z\leq 0, \,\, x = -L, \\

\partial_z\phi^{s,a} &= 0, \,\,\,
\begin{cases}
z = -d, \,\, -L < x < 0, \\
z = -h, \,\, x < 0.
\end{cases} \\

\partial_x\phi^s &= 0, \,\,\, x = 0, \,\, -h \leq z \leq -d, \\

\phi^a &= 0, \,\,\, x = 0, \,\, -h \leq z \leq -d.
\end{align}
</math></center>

Where <math>\alpha = \omega^2/g</math>.

<math>\phi^s(x,z)</math> is an even function and at <math>x = 0</math> the derivative of this function is equal to zero. <math>\phi^a(x,z)</math> is an odd function and at <math>x = 0</math> this function is equal to zero. These boundary conditions are important and enable the recomposed solution <math>\phi(x,z)</math> to be reflected into the domain of <math>x \geq 0</math>.

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\leq 0</math>
and <math>x\geq 0</math>.

{{separation of variables in two dimensions}}

{{separation of variables for a free surface}}

=== Separation of Variables for a Dock ===

The separation of variables equation for a partially submerged dock is given by:
<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>
Z^{\prime} (-d) = 0.
</math></center>
The solution is
<math>k=\kappa_{m}= \frac{m\pi}{h-d} \,</math> , <math>m\in\mathbb{N}\cup\left\{0\right\}</math> and
<center><math>
Z = \psi_{m}\left( z\right) = \cos\kappa_{m}(z+h),\quad
m\in\mathbb{N}\cup\left\{0\right\}.</math></center>
We note that
<center><math>
\int\nolimits_{-h}^{-d}\psi_{m}(z)\psi_{n}(z) \mathrm{d} z=C_{m}\delta_{mn},
</math></center>
where
<center>
<math>
C_{m} = \frac{1}{2} \left( \frac{\cos\kappa_{m}(h-d)\sin\kappa_{m}(h-d)+\kappa_{m}(h-d)}{\kappa_{m}} \right).
</math>
</center>
In addition there is the special case of when <math>m=n=0</math> ,
<center>
<math>
C_{0} = h-d,
</math>
</center>
where <math>\delta_{00}=1</math> .

=== Inner product between free surface and dock modes ===

<center><math>
B_{mn} = \int\nolimits_{-h}^{-d}\psi_{m}(z)\chi_{n}(z) \mathrm{d} z , \quad m\geq1, \quad n\geq0,
</math></center>
where
<center><math>
B_{mn}= \int\nolimits_{-h}^{-d} \frac{\cos\kappa_{m}(z+h)\cos k_{n}(z+h)}{\cos(k_{n}h)} \mathrm{d} z=\frac{-(-1)^m k_{n}\sin k_{n}(h-d)}
{\cos(k_{n}h)(\kappa_{m}^{2}-k_{n}^{2})}.
</math></center>
The evaluation of this inner product works on the assumption that the two roots <math>\kappa_{m}</math> and <math>k_{n}</math> do not coincide. The Matlab code given below has a routine to check for this case and make necessary adjustments.

We also have the case where the functions of the inner product are switched.
<center><math>
B_{nm} = \int\nolimits_{-h}^{-d}\chi_{m}(z)\psi_{n}(z) \mathrm{d} z , \quad m\geq0, \quad n\geq0.
</math></center>
For <math>m \geq 0</math> and <math>n \geq 1</math> the evaluation is above with the exception that the indices of the expression are switched.

Finally there is the special case of when <math>m \geq 0</math> and <math>n = 0</math>.
<center><math>
B_{0m} = \frac{\sin k_{m}(h-d)}{k_{m}\cos(k_{m}h)}.
</math></center>

=== Expansion of the potential ===

We need to apply some boundary conditions at plus and minus infinity,
where essentially 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 roots. Here we
must use a different condition, essentially identifying one solution
as the incoming wave and the other as the outgoing wave.

Therefore the diffracted potentials can
be expanded as
<center>
<math>
\phi(x,z)=\sum_{m=0}^{\infty}a_{m}e^{k_{m}x}\chi_{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 dock covered region. Therefore, the
potentials and their derivatives at <math>x=-L</math> have to be equal.
We obtain for the '''symmetric''' and '''antisymmetric''' cases:

<center><math>
\chi_{0}\left( z\right) + \sum_{m=0}^{\infty}a^{s,a}_{m} \chi_{m}\left(z\right)
=\sum_{m=0}^{\infty}b^{s,a}_{m}\psi_{m}(z), \quad -h \leq z\leq -d.
</math></center>

The derivative of the potential must be continuous across the
transition from open water to the dock covered region. Therefore, the derivatives of the potentials at <math>x=-L</math> have to be equal.
We obtain for the '''symmetric''' case:

<center><math>
\begin{align}
-k_{0}\chi_{0}\left( z\right) +\sum_{m=0}^{\infty} a^{s,a}_{m}k_{m}\chi_{m}\left(z\right) =
\begin{cases}
\quad \quad \quad 0,\quad \quad \quad -d\leq z\leq 0 \\
-\sum_{m=0}^{\infty}b^{s,a}_{m}\kappa_{m}\psi_{m}(z),\quad -h\leq z\leq -d
\end{cases}
\end{align}
</math></center>

For the first equation we multiply both sides by <math>\psi_{n}(z) \,</math> and integrating from <math>-h</math> to <math>-d</math> to obtain:
<center><math>
B_{n0} + \sum^{\infty}_{m=0}a_{m}B_{nm} = b_{m}C_{m}\delta_{mn}, n\in\mathbb{N}\cup\left\{0\right\}
</math></center>
and for the second equation we multiply both sides by <math>\chi_{n}(z) \,</math> and integrating from <math>-h</math> to <math>0</math> to obtain:
<center><math>
-k_{0}A_{0}\delta_{0n} + k_{m}a_{m}A_{m} = -\sum^{\infty}_{m=0}\kappa_{m}b_{m}B_{mn}, n\in\mathbb{N}\cup\left\{0\right\}
</math></center>

Solving the equations above will yield the coefficients of the water velocity potential in the dock covered region.

== Numerical Solution ==

To solve the system of equations for the '''symmetric''' case, we set the upper limit of <math>m</math> to be <math>N</math>. This resulting system can be expressed in the block matrix form below,

<center>
<math>
\begin{bmatrix}
B_{00} & B_{01} & \cdots & B_{0N} & -C_{0} & 0 & \cdots & 0 \\
B_{10} & B_{11} & \cdots & B_{1N} & 0 & -C_{1} & \cdots & \vdots \\
\vdots & \vdots & \ddots & \vdots & \vdots & \vdots & \ddots & 0 \\
B_{N0} & B_{N1} & \cdots & B_{NN} & 0 & \cdots & 0 & -C_{N} \\
k_{0}A_{0} & 0 & \cdots & 0 & 0 & \kappa_{1}\tanh(\kappa_{1}L)B_{10} & \cdots & \kappa_{N}\tanh(\kappa_{N}L)B_{N0} \\
0 & k_{1}A_{1} & \cdots & \vdots & 0 & \kappa_{1}\tanh(\kappa_{1}L)B_{11} & \cdots & \kappa_{N}\tanh(\kappa_{N}L)B_{N1} \\
\vdots & \vdots & \ddots & 0 & \vdots & \vdots & \ddots & \vdots \\
0 & \cdots & 0 & k_{N}A_{N} & 0 & \kappa_{1}\tanh(\kappa_{1}L)B_{1N} & \cdots & \kappa_{N}\tanh(\kappa_{N}L)B_{NN} \\
\end{bmatrix}
\begin{bmatrix}
a^s_{0} \\ a^s_{1} \\ \vdots \\ a^s_{N} \\ b^s_{0} \\ b^s_{1} \\ \vdots \\ b^s_{N} \\
\end{bmatrix}
=\begin{bmatrix}
-B_{00} \\ -B_{10} \\ \vdots \\ -B_{N0} \\ k_{0}A_{0} \\ 0 \\ \vdots \\ 0 \\
\end{bmatrix}
</math>
</center>

To solve the system of equations for the '''antisymmetric''' case, we set the upper limit of <math>m</math> to be <math>N</math>. This resulting system can be expressed in the block matrix form below,

<center>
<math>
\begin{bmatrix}
B_{00} & B_{01} & \cdots & B_{0N} & -C_{0} & 0 & \cdots & 0 \\
B_{10} & B_{11} & \cdots & B_{1N} & 0 & -C_{1} & \cdots & \vdots \\
\vdots & \vdots & \ddots & \vdots & \vdots & \vdots & \ddots & 0 \\
B_{N0} & B_{N1} & \cdots & B_{NN} & 0 & \cdots & 0 & -C_{N} \\
k_{0}A_{0} & 0 & \cdots & 0 & \frac{1}{L}D_{0} & \kappa_{1}\coth(\kappa_{1}L)B_{10} & \cdots & \kappa_{N}\coth(\kappa_{N}L)B_{N0} \\
0 & k_{1}A_{1} & \cdots & \vdots & \frac{1}{L}D_{1} & \kappa_{1}\coth(\kappa_{1}L)B_{11} & \cdots & \kappa_{N}\coth(\kappa_{N}L)B_{N1} \\
\vdots & \vdots & \ddots & 0 & \vdots & \vdots & \ddots & \vdots \\
0 & \cdots & 0 & k_{N}A_{N} & \frac{1}{L}D_{N} & \kappa_{1}\coth(\kappa_{1}L)B_{1N} & \cdots & \kappa_{N}\coth(\kappa_{N}L)B_{NN} \\
\end{bmatrix}
\begin{bmatrix}
a^a_{0} \\ a^a_{1} \\ \vdots \\ a^a_{N} \\ b^a_{0} \\ b^a_{1} \\ \vdots \\ b^a_{N} \\
\end{bmatrix}
=\begin{bmatrix}
-B_{00} \\ -B_{10} \\ \vdots \\ -B_{N0} \\ k_{0}A_{0} \\ 0 \\ \vdots \\ 0 \\
\end{bmatrix}
</math>
</center>

We then simply need to solve the two sets of <math>2(N+1) \mbox{ x } 2(N+1)</math> linear systems of equations. If we had not taken advantage of the symmetry we would need to solve a <math>4(N+1) \mbox{ x } 4(N+1)</math> linear system of equations for four coefficients <math>a_{m}</math>, <math>b_{m}</math>, <math>c_{m}</math> and <math>d_{m}</math>.

== Matlab Code ==

A program to calculate the coefficients for the single finite dock problem using symmetry 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 Finite Rectangle using Symmetry

2012-03-16T04:34:08Z

Tertius: /* An infinite dimensional system of equations */

{{incomplete pages}}

== Introduction ==

The problem consists of a finite rigid fixed rectangle of length <math>2L</math> centered at <math>x = 0</math> and having a height of <math>d + ...</math> freeboard with depth of submergence being <math>d</math> through which no flow is possible. The regions to the left and right of the rectangle are the free water surface. Since the geometry of the rectangle is symmetric about <math>x = 0</math> we can decompose the velocity potential <math>\phi</math> into a symmetric <math>\phi^s</math> and antisymmetric <math>\phi^a</math> potential. We then solve for these two potentials in the left domain only <math>x \leq 0</math> and subsequently recompose the solution <math>\phi</math> for the entire domain. We begin with the simple problem when the waves are normally incident on the rectangle (dock) (so that the problem is truly two-dimensional). We can extend this solution to a pair of partially submerged finite rigid fixed rectangles using symmetry
[[Eigenfunction Matching for a Finite Rectangle using Symmetry]]

== Governing Equations ==

We begin with the [[Frequency Domain Problem]] for a partially submerged finite dock. Although the dock is of length <math>2L</math> we are only interested in the dock of <math>-L \leq x \leq 0</math> together with the fluid domain <math>x \leq 0</math>. We assume <math>e^{i\omega t}</math> time dependence and
the depth of submergence is <math>d</math>.
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>
\begin{align}
\Delta\phi^{s,a} &= 0, \,\,\, -h\leq z\leq 0, \,\, x < -L, \\

\Delta\phi^{s,a} &= 0, \,\,\, -h\leq z\leq -d, \,\, -L < x < 0. \\

\partial_z\phi^{s,a} &= \alpha\phi^{s,a}, \,\,\, z=0, \,\, x < -L, \\

\partial_x\phi^{s,a} &= 0, \,\,\, -d\leq z\leq 0, \,\, x = -L, \\

\partial_z\phi^{s,a} &= 0, \,\,\,
\begin{cases}
z = -d, \,\, -L < x < 0, \\
z = -h, \,\, x < 0.
\end{cases} \\

\partial_x\phi^s &= 0, \,\,\, x = 0, \,\, -h \leq z \leq -d, \\

\phi^a &= 0, \,\,\, x = 0, \,\, -h \leq z \leq -d.
\end{align}
</math></center>

Where <math>\alpha = \omega^2/g</math>.

<math>\phi^s(x,z)</math> is an even function and at <math>x = 0</math> the derivative of this function is equal to zero. <math>\phi^a(x,z)</math> is an odd function and at <math>x = 0</math> this function is equal to zero. These boundary conditions are important and enable the recomposed solution <math>\phi(x,z)</math> to be reflected into the domain of <math>x \geq 0</math>.

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\leq 0</math>
and <math>x\geq 0</math>.

{{separation of variables in two dimensions}}

{{separation of variables for a free surface}}

=== Separation of Variables for a Dock ===

The separation of variables equation for a partially submerged dock is given by:
<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>
Z^{\prime} (-d) = 0.
</math></center>
The solution is
<math>k=\kappa_{m}= \frac{m\pi}{h-d} \,</math> , <math>m\in\mathbb{N}\cup\left\{0\right\}</math> and
<center><math>
Z = \psi_{m}\left( z\right) = \cos\kappa_{m}(z+h),\quad
m\in\mathbb{N}\cup\left\{0\right\}.</math></center>
We note that
<center><math>
\int\nolimits_{-h}^{-d}\psi_{m}(z)\psi_{n}(z) \mathrm{d} z=C_{m}\delta_{mn},
</math></center>
where
<center>
<math>
C_{m} = \frac{1}{2} \left( \frac{\cos\kappa_{m}(h-d)\sin\kappa_{m}(h-d)+\kappa_{m}(h-d)}{\kappa_{m}} \right).
</math>
</center>
In addition there is the special case of when <math>m=n=0</math> ,
<center>
<math>
C_{0} = h-d,
</math>
</center>
where <math>\delta_{00}=1</math> .

=== Inner product between free surface and dock modes ===

<center><math>
B_{mn} = \int\nolimits_{-h}^{-d}\psi_{m}(z)\chi_{n}(z) \mathrm{d} z , \quad m\geq1, \quad n\geq0,
</math></center>
where
<center><math>
B_{mn}= \int\nolimits_{-h}^{-d} \frac{\cos\kappa_{m}(z+h)\cos k_{n}(z+h)}{\cos(k_{n}h)} \mathrm{d} z=\frac{-(-1)^m k_{n}\sin k_{n}(h-d)}
{\cos(k_{n}h)(\kappa_{m}^{2}-k_{n}^{2})}.
</math></center>
The evaluation of this inner product works on the assumption that the two roots <math>\kappa_{m}</math> and <math>k_{n}</math> do not coincide. The Matlab code given below has a routine to check for this case and make necessary adjustments.

We also have the case where the functions of the inner product are switched.
<center><math>
B_{nm} = \int\nolimits_{-h}^{-d}\chi_{m}(z)\psi_{n}(z) \mathrm{d} z , \quad m\geq0, \quad n\geq0.
</math></center>
For <math>m \geq 0</math> and <math>n \geq 1</math> the evaluation is above with the exception that the indices of the expression are switched.

Finally there is the special case of when <math>m \geq 0</math> and <math>n = 0</math>.
<center><math>
B_{0m} = \frac{\sin k_{m}(h-d)}{k_{m}\cos(k_{m}h)}.
</math></center>

=== Expansion of the potential ===

We need to apply some boundary conditions at plus and minus infinity,
where essentially 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 roots. Here we
must use a different condition, essentially identifying one solution
as the incoming wave and the other as the outgoing wave.

Therefore the diffracted potentials can
be expanded as
<center>
<math>
\phi(x,z)=\sum_{m=0}^{\infty}a_{m}e^{k_{m}x}\chi_{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 dock covered region. Therefore, the
potentials and their derivatives at <math>x=-L</math> have to be equal.
We obtain for the '''symmetric''' and '''antisymmetric''' cases:

<center><math>
\chi_{0}\left( z\right) + \sum_{m=0}^{\infty}a^{s,a}_{m} \chi_{m}\left(z\right)
=\sum_{m=0}^{\infty}b^{s,a}_{m}\psi_{m}(z), \quad -h \leq z\leq -d.
</math></center>

The derivative of the potential must be continuous across the
transition from open water to the dock covered region. Therefore, the derivative of the potential at <math>x=-L</math> have to be equal.
We obtain for the '''symmetric''' case:

<center><math>
\begin{align}
-k_{0}\chi_{0}\left( z\right) +\sum_{m=0}^{\infty} a^{s,a}_{m}k_{m}\chi_{m}\left(z\right) =
\begin{cases}
\quad \quad \quad 0,\quad \quad \quad -d\leq z\leq 0 \\
-\sum_{m=0}^{\infty}b^{s,a}_{m}\kappa_{m}\psi_{m}(z),\quad -h\leq z\leq -d
\end{cases}
\end{align}
</math></center>

For the first equation we multiply both sides by <math>\psi_{n}(z) \,</math> and integrating from <math>-h</math> to <math>-d</math> to obtain:
<center><math>
B_{n0} + \sum^{\infty}_{m=0}a_{m}B_{nm} = b_{m}C_{m}\delta_{mn}, n\in\mathbb{N}\cup\left\{0\right\}
</math></center>
and for the second equation we multiply both sides by <math>\chi_{n}(z) \,</math> and integrating from <math>-h</math> to <math>0</math> to obtain:
<center><math>
-k_{0}A_{0}\delta_{0n} + k_{m}a_{m}A_{m} = -\sum^{\infty}_{m=0}\kappa_{m}b_{m}B_{mn}, n\in\mathbb{N}\cup\left\{0\right\}
</math></center>

Solving the equations above will yield the coefficients of the water velocity potential in the dock covered region.

== Numerical Solution ==

To solve the system of equations for the '''symmetric''' case, we set the upper limit of <math>m</math> to be <math>N</math>. This resulting system can be expressed in the block matrix form below,

<center>
<math>
\begin{bmatrix}
B_{00} & B_{01} & \cdots & B_{0N} & -C_{0} & 0 & \cdots & 0 \\
B_{10} & B_{11} & \cdots & B_{1N} & 0 & -C_{1} & \cdots & \vdots \\
\vdots & \vdots & \ddots & \vdots & \vdots & \vdots & \ddots & 0 \\
B_{N0} & B_{N1} & \cdots & B_{NN} & 0 & \cdots & 0 & -C_{N} \\
k_{0}A_{0} & 0 & \cdots & 0 & 0 & \kappa_{1}\tanh(\kappa_{1}L)B_{10} & \cdots & \kappa_{N}\tanh(\kappa_{N}L)B_{N0} \\
0 & k_{1}A_{1} & \cdots & \vdots & 0 & \kappa_{1}\tanh(\kappa_{1}L)B_{11} & \cdots & \kappa_{N}\tanh(\kappa_{N}L)B_{N1} \\
\vdots & \vdots & \ddots & 0 & \vdots & \vdots & \ddots & \vdots \\
0 & \cdots & 0 & k_{N}A_{N} & 0 & \kappa_{1}\tanh(\kappa_{1}L)B_{1N} & \cdots & \kappa_{N}\tanh(\kappa_{N}L)B_{NN} \\
\end{bmatrix}
\begin{bmatrix}
a^s_{0} \\ a^s_{1} \\ \vdots \\ a^s_{N} \\ b^s_{0} \\ b^s_{1} \\ \vdots \\ b^s_{N} \\
\end{bmatrix}
=\begin{bmatrix}
-B_{00} \\ -B_{10} \\ \vdots \\ -B_{N0} \\ k_{0}A_{0} \\ 0 \\ \vdots \\ 0 \\
\end{bmatrix}
</math>
</center>

To solve the system of equations for the '''antisymmetric''' case, we set the upper limit of <math>m</math> to be <math>N</math>. This resulting system can be expressed in the block matrix form below,

<center>
<math>
\begin{bmatrix}
B_{00} & B_{01} & \cdots & B_{0N} & -C_{0} & 0 & \cdots & 0 \\
B_{10} & B_{11} & \cdots & B_{1N} & 0 & -C_{1} & \cdots & \vdots \\
\vdots & \vdots & \ddots & \vdots & \vdots & \vdots & \ddots & 0 \\
B_{N0} & B_{N1} & \cdots & B_{NN} & 0 & \cdots & 0 & -C_{N} \\
k_{0}A_{0} & 0 & \cdots & 0 & \frac{1}{L}D_{0} & \kappa_{1}\coth(\kappa_{1}L)B_{10} & \cdots & \kappa_{N}\coth(\kappa_{N}L)B_{N0} \\
0 & k_{1}A_{1} & \cdots & \vdots & \frac{1}{L}D_{1} & \kappa_{1}\coth(\kappa_{1}L)B_{11} & \cdots & \kappa_{N}\coth(\kappa_{N}L)B_{N1} \\
\vdots & \vdots & \ddots & 0 & \vdots & \vdots & \ddots & \vdots \\
0 & \cdots & 0 & k_{N}A_{N} & \frac{1}{L}D_{N} & \kappa_{1}\coth(\kappa_{1}L)B_{1N} & \cdots & \kappa_{N}\coth(\kappa_{N}L)B_{NN} \\
\end{bmatrix}
\begin{bmatrix}
a^a_{0} \\ a^a_{1} \\ \vdots \\ a^a_{N} \\ b^a_{0} \\ b^a_{1} \\ \vdots \\ b^a_{N} \\
\end{bmatrix}
=\begin{bmatrix}
-B_{00} \\ -B_{10} \\ \vdots \\ -B_{N0} \\ k_{0}A_{0} \\ 0 \\ \vdots \\ 0 \\
\end{bmatrix}
</math>
</center>

We then simply need to solve the two sets of <math>2(N+1) \mbox{ x } 2(N+1)</math> linear systems of equations. If we had not taken advantage of the symmetry we would need to solve a <math>4(N+1) \mbox{ x } 4(N+1)</math> linear system of equations for four coefficients <math>a_{m}</math>, <math>b_{m}</math>, <math>c_{m}</math> and <math>d_{m}</math>.

== Matlab Code ==

A program to calculate the coefficients for the single finite dock problem using symmetry 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 Finite Rectangle using Symmetry

2012-03-16T04:31:54Z

Tertius: /* An infinite dimensional system of equations */

{{incomplete pages}}

== Introduction ==

The problem consists of a finite rigid fixed rectangle of length <math>2L</math> centered at <math>x = 0</math> and having a height of <math>d + ...</math> freeboard with depth of submergence being <math>d</math> through which no flow is possible. The regions to the left and right of the rectangle are the free water surface. Since the geometry of the rectangle is symmetric about <math>x = 0</math> we can decompose the velocity potential <math>\phi</math> into a symmetric <math>\phi^s</math> and antisymmetric <math>\phi^a</math> potential. We then solve for these two potentials in the left domain only <math>x \leq 0</math> and subsequently recompose the solution <math>\phi</math> for the entire domain. We begin with the simple problem when the waves are normally incident on the rectangle (dock) (so that the problem is truly two-dimensional). We can extend this solution to a pair of partially submerged finite rigid fixed rectangles using symmetry
[[Eigenfunction Matching for a Finite Rectangle using Symmetry]]

== Governing Equations ==

We begin with the [[Frequency Domain Problem]] for a partially submerged finite dock. Although the dock is of length <math>2L</math> we are only interested in the dock of <math>-L \leq x \leq 0</math> together with the fluid domain <math>x \leq 0</math>. We assume <math>e^{i\omega t}</math> time dependence and
the depth of submergence is <math>d</math>.
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>
\begin{align}
\Delta\phi^{s,a} &= 0, \,\,\, -h\leq z\leq 0, \,\, x < -L, \\

\Delta\phi^{s,a} &= 0, \,\,\, -h\leq z\leq -d, \,\, -L < x < 0. \\

\partial_z\phi^{s,a} &= \alpha\phi^{s,a}, \,\,\, z=0, \,\, x < -L, \\

\partial_x\phi^{s,a} &= 0, \,\,\, -d\leq z\leq 0, \,\, x = -L, \\

\partial_z\phi^{s,a} &= 0, \,\,\,
\begin{cases}
z = -d, \,\, -L < x < 0, \\
z = -h, \,\, x < 0.
\end{cases} \\

\partial_x\phi^s &= 0, \,\,\, x = 0, \,\, -h \leq z \leq -d, \\

\phi^a &= 0, \,\,\, x = 0, \,\, -h \leq z \leq -d.
\end{align}
</math></center>

Where <math>\alpha = \omega^2/g</math>.

<math>\phi^s(x,z)</math> is an even function and at <math>x = 0</math> the derivative of this function is equal to zero. <math>\phi^a(x,z)</math> is an odd function and at <math>x = 0</math> this function is equal to zero. These boundary conditions are important and enable the recomposed solution <math>\phi(x,z)</math> to be reflected into the domain of <math>x \geq 0</math>.

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\leq 0</math>
and <math>x\geq 0</math>.

{{separation of variables in two dimensions}}

{{separation of variables for a free surface}}

=== Separation of Variables for a Dock ===

The separation of variables equation for a partially submerged dock is given by:
<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>
Z^{\prime} (-d) = 0.
</math></center>
The solution is
<math>k=\kappa_{m}= \frac{m\pi}{h-d} \,</math> , <math>m\in\mathbb{N}\cup\left\{0\right\}</math> and
<center><math>
Z = \psi_{m}\left( z\right) = \cos\kappa_{m}(z+h),\quad
m\in\mathbb{N}\cup\left\{0\right\}.</math></center>
We note that
<center><math>
\int\nolimits_{-h}^{-d}\psi_{m}(z)\psi_{n}(z) \mathrm{d} z=C_{m}\delta_{mn},
</math></center>
where
<center>
<math>
C_{m} = \frac{1}{2} \left( \frac{\cos\kappa_{m}(h-d)\sin\kappa_{m}(h-d)+\kappa_{m}(h-d)}{\kappa_{m}} \right).
</math>
</center>
In addition there is the special case of when <math>m=n=0</math> ,
<center>
<math>
C_{0} = h-d,
</math>
</center>
where <math>\delta_{00}=1</math> .

=== Inner product between free surface and dock modes ===

<center><math>
B_{mn} = \int\nolimits_{-h}^{-d}\psi_{m}(z)\chi_{n}(z) \mathrm{d} z , \quad m\geq1, \quad n\geq0,
</math></center>
where
<center><math>
B_{mn}= \int\nolimits_{-h}^{-d} \frac{\cos\kappa_{m}(z+h)\cos k_{n}(z+h)}{\cos(k_{n}h)} \mathrm{d} z=\frac{-(-1)^m k_{n}\sin k_{n}(h-d)}
{\cos(k_{n}h)(\kappa_{m}^{2}-k_{n}^{2})}.
</math></center>
The evaluation of this inner product works on the assumption that the two roots <math>\kappa_{m}</math> and <math>k_{n}</math> do not coincide. The Matlab code given below has a routine to check for this case and make necessary adjustments.

We also have the case where the functions of the inner product are switched.
<center><math>
B_{nm} = \int\nolimits_{-h}^{-d}\chi_{m}(z)\psi_{n}(z) \mathrm{d} z , \quad m\geq0, \quad n\geq0.
</math></center>
For <math>m \geq 0</math> and <math>n \geq 1</math> the evaluation is above with the exception that the indices of the expression are switched.

Finally there is the special case of when <math>m \geq 0</math> and <math>n = 0</math>.
<center><math>
B_{0m} = \frac{\sin k_{m}(h-d)}{k_{m}\cos(k_{m}h)}.
</math></center>

=== Expansion of the potential ===

We need to apply some boundary conditions at plus and minus infinity,
where essentially 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 roots. Here we
must use a different condition, essentially identifying one solution
as the incoming wave and the other as the outgoing wave.

Therefore the diffracted potentials can
be expanded as
<center>
<math>
\phi(x,z)=\sum_{m=0}^{\infty}a_{m}e^{k_{m}x}\chi_{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 dock covered region. Therefore, the
potentials and their derivatives at <math>x=-L</math> have to be equal.
We obtain for the '''symmetric''' and '''antisymmetric''' cases:

<center><math>
\chi_{0}\left( z\right) + \sum_{m=0}^{\infty}a^{s,a}_{m} \chi_{m}\left(z\right)
=\sum_{m=0}^{\infty}b^{s,a}_{m}\psi_{m}(z), \quad -h \leq z\leq -d
</math></center>

The derivative of the potential must be continuous across the
transition from open water to the dock covered region. Therefore, the derivative of the potential at <math>x=-L</math> have to be equal.
We obtain for the '''symmetric''' case:

<center><math>
\begin{align}
-k_{0}\chi_{0}\left( z\right) +\sum_{m=0}^{\infty} a^{s,a}_{m}k_{m}\chi_{m}\left(z\right) =
\begin{cases}
\quad \quad \quad 0,\quad \quad \quad -d\leq z\leq 0 \\
-\sum_{m=0}^{\infty}b^{s,a}_{m}\kappa_{m}\psi_{m}(z),\quad -h\leq z\leq -d
\end{cases}
\end{align}
</math></center>

For the first equation we multiply both sides by <math>\psi_{n}(z) \,</math> and integrating from <math>-h</math> to <math>-d</math> to obtain:
<center><math>
B_{n0} + \sum^{\infty}_{m=0}a_{m}B_{nm} = b_{m}C_{m}\delta_{mn}, n\in\mathbb{N}\cup\left\{0\right\}
</math></center>
and for the second equation we multiply both sides by <math>\chi_{n}(z) \,</math> and integrating from <math>-h</math> to <math>0</math> to obtain:
<center><math>
-k_{0}A_{0}\delta_{0n} + k_{m}a_{m}A_{m} = -\sum^{\infty}_{m=0}\kappa_{m}b_{m}B_{mn}, n\in\mathbb{N}\cup\left\{0\right\}
</math></center>

Solving the equations above will yield the coefficients of the water velocity potential in the dock covered region.

== Numerical Solution ==

To solve the system of equations for the '''symmetric''' case, we set the upper limit of <math>m</math> to be <math>N</math>. This resulting system can be expressed in the block matrix form below,

<center>
<math>
\begin{bmatrix}
B_{00} & B_{01} & \cdots & B_{0N} & -C_{0} & 0 & \cdots & 0 \\
B_{10} & B_{11} & \cdots & B_{1N} & 0 & -C_{1} & \cdots & \vdots \\
\vdots & \vdots & \ddots & \vdots & \vdots & \vdots & \ddots & 0 \\
B_{N0} & B_{N1} & \cdots & B_{NN} & 0 & \cdots & 0 & -C_{N} \\
k_{0}A_{0} & 0 & \cdots & 0 & 0 & \kappa_{1}\tanh(\kappa_{1}L)B_{10} & \cdots & \kappa_{N}\tanh(\kappa_{N}L)B_{N0} \\
0 & k_{1}A_{1} & \cdots & \vdots & 0 & \kappa_{1}\tanh(\kappa_{1}L)B_{11} & \cdots & \kappa_{N}\tanh(\kappa_{N}L)B_{N1} \\
\vdots & \vdots & \ddots & 0 & \vdots & \vdots & \ddots & \vdots \\
0 & \cdots & 0 & k_{N}A_{N} & 0 & \kappa_{1}\tanh(\kappa_{1}L)B_{1N} & \cdots & \kappa_{N}\tanh(\kappa_{N}L)B_{NN} \\
\end{bmatrix}
\begin{bmatrix}
a^s_{0} \\ a^s_{1} \\ \vdots \\ a^s_{N} \\ b^s_{0} \\ b^s_{1} \\ \vdots \\ b^s_{N} \\
\end{bmatrix}
=\begin{bmatrix}
-B_{00} \\ -B_{10} \\ \vdots \\ -B_{N0} \\ k_{0}A_{0} \\ 0 \\ \vdots \\ 0 \\
\end{bmatrix}
</math>
</center>

To solve the system of equations for the '''antisymmetric''' case, we set the upper limit of <math>m</math> to be <math>N</math>. This resulting system can be expressed in the block matrix form below,

<center>
<math>
\begin{bmatrix}
B_{00} & B_{01} & \cdots & B_{0N} & -C_{0} & 0 & \cdots & 0 \\
B_{10} & B_{11} & \cdots & B_{1N} & 0 & -C_{1} & \cdots & \vdots \\
\vdots & \vdots & \ddots & \vdots & \vdots & \vdots & \ddots & 0 \\
B_{N0} & B_{N1} & \cdots & B_{NN} & 0 & \cdots & 0 & -C_{N} \\
k_{0}A_{0} & 0 & \cdots & 0 & \frac{1}{L}D_{0} & \kappa_{1}\coth(\kappa_{1}L)B_{10} & \cdots & \kappa_{N}\coth(\kappa_{N}L)B_{N0} \\
0 & k_{1}A_{1} & \cdots & \vdots & \frac{1}{L}D_{1} & \kappa_{1}\coth(\kappa_{1}L)B_{11} & \cdots & \kappa_{N}\coth(\kappa_{N}L)B_{N1} \\
\vdots & \vdots & \ddots & 0 & \vdots & \vdots & \ddots & \vdots \\
0 & \cdots & 0 & k_{N}A_{N} & \frac{1}{L}D_{N} & \kappa_{1}\coth(\kappa_{1}L)B_{1N} & \cdots & \kappa_{N}\coth(\kappa_{N}L)B_{NN} \\
\end{bmatrix}
\begin{bmatrix}
a^a_{0} \\ a^a_{1} \\ \vdots \\ a^a_{N} \\ b^a_{0} \\ b^a_{1} \\ \vdots \\ b^a_{N} \\
\end{bmatrix}
=\begin{bmatrix}
-B_{00} \\ -B_{10} \\ \vdots \\ -B_{N0} \\ k_{0}A_{0} \\ 0 \\ \vdots \\ 0 \\
\end{bmatrix}
</math>
</center>

We then simply need to solve the two sets of <math>2(N+1) \mbox{ x } 2(N+1)</math> linear systems of equations. If we had not taken advantage of the symmetry we would need to solve a <math>4(N+1) \mbox{ x } 4(N+1)</math> linear system of equations for four coefficients <math>a_{m}</math>, <math>b_{m}</math>, <math>c_{m}</math> and <math>d_{m}</math>.

== Matlab Code ==

A program to calculate the coefficients for the single finite dock problem using symmetry 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 Finite Rectangle using Symmetry

2012-03-16T04:21:25Z

Tertius: /* An infinite dimensional system of equations */

{{incomplete pages}}

== Introduction ==

The problem consists of a finite rigid fixed rectangle of length <math>2L</math> centered at <math>x = 0</math> and having a height of <math>d + ...</math> freeboard with depth of submergence being <math>d</math> through which no flow is possible. The regions to the left and right of the rectangle are the free water surface. Since the geometry of the rectangle is symmetric about <math>x = 0</math> we can decompose the velocity potential <math>\phi</math> into a symmetric <math>\phi^s</math> and antisymmetric <math>\phi^a</math> potential. We then solve for these two potentials in the left domain only <math>x \leq 0</math> and subsequently recompose the solution <math>\phi</math> for the entire domain. We begin with the simple problem when the waves are normally incident on the rectangle (dock) (so that the problem is truly two-dimensional). We can extend this solution to a pair of partially submerged finite rigid fixed rectangles using symmetry
[[Eigenfunction Matching for a Finite Rectangle using Symmetry]]

== Governing Equations ==

We begin with the [[Frequency Domain Problem]] for a partially submerged finite dock. Although the dock is of length <math>2L</math> we are only interested in the dock of <math>-L \leq x \leq 0</math> together with the fluid domain <math>x \leq 0</math>. We assume <math>e^{i\omega t}</math> time dependence and
the depth of submergence is <math>d</math>.
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>
\begin{align}
\Delta\phi^{s,a} &= 0, \,\,\, -h\leq z\leq 0, \,\, x < -L, \\

\Delta\phi^{s,a} &= 0, \,\,\, -h\leq z\leq -d, \,\, -L < x < 0. \\

\partial_z\phi^{s,a} &= \alpha\phi^{s,a}, \,\,\, z=0, \,\, x < -L, \\

\partial_x\phi^{s,a} &= 0, \,\,\, -d\leq z\leq 0, \,\, x = -L, \\

\partial_z\phi^{s,a} &= 0, \,\,\,
\begin{cases}
z = -d, \,\, -L < x < 0, \\
z = -h, \,\, x < 0.
\end{cases} \\

\partial_x\phi^s &= 0, \,\,\, x = 0, \,\, -h \leq z \leq -d, \\

\phi^a &= 0, \,\,\, x = 0, \,\, -h \leq z \leq -d.
\end{align}
</math></center>

Where <math>\alpha = \omega^2/g</math>.

<math>\phi^s(x,z)</math> is an even function and at <math>x = 0</math> the derivative of this function is equal to zero. <math>\phi^a(x,z)</math> is an odd function and at <math>x = 0</math> this function is equal to zero. These boundary conditions are important and enable the recomposed solution <math>\phi(x,z)</math> to be reflected into the domain of <math>x \geq 0</math>.

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\leq 0</math>
and <math>x\geq 0</math>.

{{separation of variables in two dimensions}}

{{separation of variables for a free surface}}

=== Separation of Variables for a Dock ===

The separation of variables equation for a partially submerged dock is given by:
<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>
Z^{\prime} (-d) = 0.
</math></center>
The solution is
<math>k=\kappa_{m}= \frac{m\pi}{h-d} \,</math> , <math>m\in\mathbb{N}\cup\left\{0\right\}</math> and
<center><math>
Z = \psi_{m}\left( z\right) = \cos\kappa_{m}(z+h),\quad
m\in\mathbb{N}\cup\left\{0\right\}.</math></center>
We note that
<center><math>
\int\nolimits_{-h}^{-d}\psi_{m}(z)\psi_{n}(z) \mathrm{d} z=C_{m}\delta_{mn},
</math></center>
where
<center>
<math>
C_{m} = \frac{1}{2} \left( \frac{\cos\kappa_{m}(h-d)\sin\kappa_{m}(h-d)+\kappa_{m}(h-d)}{\kappa_{m}} \right).
</math>
</center>
In addition there is the special case of when <math>m=n=0</math> ,
<center>
<math>
C_{0} = h-d,
</math>
</center>
where <math>\delta_{00}=1</math> .

=== Inner product between free surface and dock modes ===

<center><math>
B_{mn} = \int\nolimits_{-h}^{-d}\psi_{m}(z)\chi_{n}(z) \mathrm{d} z , \quad m\geq1, \quad n\geq0,
</math></center>
where
<center><math>
B_{mn}= \int\nolimits_{-h}^{-d} \frac{\cos\kappa_{m}(z+h)\cos k_{n}(z+h)}{\cos(k_{n}h)} \mathrm{d} z=\frac{-(-1)^m k_{n}\sin k_{n}(h-d)}
{\cos(k_{n}h)(\kappa_{m}^{2}-k_{n}^{2})}.
</math></center>
The evaluation of this inner product works on the assumption that the two roots <math>\kappa_{m}</math> and <math>k_{n}</math> do not coincide. The Matlab code given below has a routine to check for this case and make necessary adjustments.

We also have the case where the functions of the inner product are switched.
<center><math>
B_{nm} = \int\nolimits_{-h}^{-d}\chi_{m}(z)\psi_{n}(z) \mathrm{d} z , \quad m\geq0, \quad n\geq0.
</math></center>
For <math>m \geq 0</math> and <math>n \geq 1</math> the evaluation is above with the exception that the indices of the expression are switched.

Finally there is the special case of when <math>m \geq 0</math> and <math>n = 0</math>.
<center><math>
B_{0m} = \frac{\sin k_{m}(h-d)}{k_{m}\cos(k_{m}h)}.
</math></center>

=== Expansion of the potential ===

We need to apply some boundary conditions at plus and minus infinity,
where essentially 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 roots. Here we
must use a different condition, essentially identifying one solution
as the incoming wave and the other as the outgoing wave.

Therefore the diffracted potentials can
be expanded as
<center>
<math>
\phi(x,z)=\sum_{m=0}^{\infty}a_{m}e^{k_{m}x}\chi_{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 dock covered region. Therefore, the
potentials and their derivatives at <math>x=-L</math> have to be equal.
We obtain for the '''symmetric''' and '''antisymmetric''' cases:

<center><math>
\chi_{0}\left( z\right) + \sum_{m=0}^{\infty}a^{s,a}_{m} \chi_{m}\left(z\right)
=\sum_{m=0}^{\infty}b^{s,a}_{m}\psi_{m}(z), \quad -h \leq z\leq -d
</math></center>

<center><math>
\begin{align}
-k_{0}\chi_{0}\left( z\right) +\sum_{m=0}^{\infty} a^{s,a}_{m}k_{m}\chi_{m}\left(z\right) =
\begin{cases}
\quad \quad \quad 0,\quad \quad \quad -d\leq z\leq 0 \\
-\sum_{m=0}^{\infty}b^{s,a}_{m}\kappa_{m}\psi_{m}(z),\quad -h\leq z\leq -d
\end{cases}
\end{align}
</math></center>

For the first equation we multiply both sides by <math>\psi_{n}(z) \,</math> and integrating from <math>-h</math> to <math>-d</math> to obtain:
<center><math>
B_{n0} + \sum^{\infty}_{m=0}a_{m}B_{nm} = b_{m}C_{m}\delta_{mn}, n\in\mathbb{N}\cup\left\{0\right\}
</math></center>
and for the second equation we multiply both sides by <math>\chi_{n}(z) \,</math> and integrating from <math>-h</math> to <math>0</math> to obtain:
<center><math>
-k_{0}A_{0}\delta_{0n} + k_{m}a_{m}A_{m} = -\sum^{\infty}_{m=0}\kappa_{m}b_{m}B_{mn}, n\in\mathbb{N}\cup\left\{0\right\}
</math></center>

Solving the equations above will yield the coefficients of the water velocity potential in the dock covered region.

== Numerical Solution ==

To solve the system of equations for the '''symmetric''' case, we set the upper limit of <math>m</math> to be <math>N</math>. This resulting system can be expressed in the block matrix form below,

<center>
<math>
\begin{bmatrix}
B_{00} & B_{01} & \cdots & B_{0N} & -C_{0} & 0 & \cdots & 0 \\
B_{10} & B_{11} & \cdots & B_{1N} & 0 & -C_{1} & \cdots & \vdots \\
\vdots & \vdots & \ddots & \vdots & \vdots & \vdots & \ddots & 0 \\
B_{N0} & B_{N1} & \cdots & B_{NN} & 0 & \cdots & 0 & -C_{N} \\
k_{0}A_{0} & 0 & \cdots & 0 & 0 & \kappa_{1}\tanh(\kappa_{1}L)B_{10} & \cdots & \kappa_{N}\tanh(\kappa_{N}L)B_{N0} \\
0 & k_{1}A_{1} & \cdots & \vdots & 0 & \kappa_{1}\tanh(\kappa_{1}L)B_{11} & \cdots & \kappa_{N}\tanh(\kappa_{N}L)B_{N1} \\
\vdots & \vdots & \ddots & 0 & \vdots & \vdots & \ddots & \vdots \\
0 & \cdots & 0 & k_{N}A_{N} & 0 & \kappa_{1}\tanh(\kappa_{1}L)B_{1N} & \cdots & \kappa_{N}\tanh(\kappa_{N}L)B_{NN} \\
\end{bmatrix}
\begin{bmatrix}
a^s_{0} \\ a^s_{1} \\ \vdots \\ a^s_{N} \\ b^s_{0} \\ b^s_{1} \\ \vdots \\ b^s_{N} \\
\end{bmatrix}
=\begin{bmatrix}
-B_{00} \\ -B_{10} \\ \vdots \\ -B_{N0} \\ k_{0}A_{0} \\ 0 \\ \vdots \\ 0 \\
\end{bmatrix}
</math>
</center>

To solve the system of equations for the '''antisymmetric''' case, we set the upper limit of <math>m</math> to be <math>N</math>. This resulting system can be expressed in the block matrix form below,

<center>
<math>
\begin{bmatrix}
B_{00} & B_{01} & \cdots & B_{0N} & -C_{0} & 0 & \cdots & 0 \\
B_{10} & B_{11} & \cdots & B_{1N} & 0 & -C_{1} & \cdots & \vdots \\
\vdots & \vdots & \ddots & \vdots & \vdots & \vdots & \ddots & 0 \\
B_{N0} & B_{N1} & \cdots & B_{NN} & 0 & \cdots & 0 & -C_{N} \\
k_{0}A_{0} & 0 & \cdots & 0 & \frac{1}{L}D_{0} & \kappa_{1}\coth(\kappa_{1}L)B_{10} & \cdots & \kappa_{N}\coth(\kappa_{N}L)B_{N0} \\
0 & k_{1}A_{1} & \cdots & \vdots & \frac{1}{L}D_{1} & \kappa_{1}\coth(\kappa_{1}L)B_{11} & \cdots & \kappa_{N}\coth(\kappa_{N}L)B_{N1} \\
\vdots & \vdots & \ddots & 0 & \vdots & \vdots & \ddots & \vdots \\
0 & \cdots & 0 & k_{N}A_{N} & \frac{1}{L}D_{N} & \kappa_{1}\coth(\kappa_{1}L)B_{1N} & \cdots & \kappa_{N}\coth(\kappa_{N}L)B_{NN} \\
\end{bmatrix}
\begin{bmatrix}
a^a_{0} \\ a^a_{1} \\ \vdots \\ a^a_{N} \\ b^a_{0} \\ b^a_{1} \\ \vdots \\ b^a_{N} \\
\end{bmatrix}
=\begin{bmatrix}
-B_{00} \\ -B_{10} \\ \vdots \\ -B_{N0} \\ k_{0}A_{0} \\ 0 \\ \vdots \\ 0 \\
\end{bmatrix}
</math>
</center>

We then simply need to solve the two sets of <math>2(N+1) \mbox{ x } 2(N+1)</math> linear systems of equations. If we had not taken advantage of the symmetry we would need to solve a <math>4(N+1) \mbox{ x } 4(N+1)</math> linear system of equations for four coefficients <math>a_{m}</math>, <math>b_{m}</math>, <math>c_{m}</math> and <math>d_{m}</math>.

== Matlab Code ==

A program to calculate the coefficients for the single finite dock problem using symmetry 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 Finite Rectangle using Symmetry

2012-03-16T04:20:25Z

Tertius: /* An infinite dimensional system of equations */

{{incomplete pages}}

== Introduction ==

The problem consists of a finite rigid fixed rectangle of length <math>2L</math> centered at <math>x = 0</math> and having a height of <math>d + ...</math> freeboard with depth of submergence being <math>d</math> through which no flow is possible. The regions to the left and right of the rectangle are the free water surface. Since the geometry of the rectangle is symmetric about <math>x = 0</math> we can decompose the velocity potential <math>\phi</math> into a symmetric <math>\phi^s</math> and antisymmetric <math>\phi^a</math> potential. We then solve for these two potentials in the left domain only <math>x \leq 0</math> and subsequently recompose the solution <math>\phi</math> for the entire domain. We begin with the simple problem when the waves are normally incident on the rectangle (dock) (so that the problem is truly two-dimensional). We can extend this solution to a pair of partially submerged finite rigid fixed rectangles using symmetry
[[Eigenfunction Matching for a Finite Rectangle using Symmetry]]

== Governing Equations ==

We begin with the [[Frequency Domain Problem]] for a partially submerged finite dock. Although the dock is of length <math>2L</math> we are only interested in the dock of <math>-L \leq x \leq 0</math> together with the fluid domain <math>x \leq 0</math>. We assume <math>e^{i\omega t}</math> time dependence and
the depth of submergence is <math>d</math>.
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>
\begin{align}
\Delta\phi^{s,a} &= 0, \,\,\, -h\leq z\leq 0, \,\, x < -L, \\

\Delta\phi^{s,a} &= 0, \,\,\, -h\leq z\leq -d, \,\, -L < x < 0. \\

\partial_z\phi^{s,a} &= \alpha\phi^{s,a}, \,\,\, z=0, \,\, x < -L, \\

\partial_x\phi^{s,a} &= 0, \,\,\, -d\leq z\leq 0, \,\, x = -L, \\

\partial_z\phi^{s,a} &= 0, \,\,\,
\begin{cases}
z = -d, \,\, -L < x < 0, \\
z = -h, \,\, x < 0.
\end{cases} \\

\partial_x\phi^s &= 0, \,\,\, x = 0, \,\, -h \leq z \leq -d, \\

\phi^a &= 0, \,\,\, x = 0, \,\, -h \leq z \leq -d.
\end{align}
</math></center>

Where <math>\alpha = \omega^2/g</math>.

<math>\phi^s(x,z)</math> is an even function and at <math>x = 0</math> the derivative of this function is equal to zero. <math>\phi^a(x,z)</math> is an odd function and at <math>x = 0</math> this function is equal to zero. These boundary conditions are important and enable the recomposed solution <math>\phi(x,z)</math> to be reflected into the domain of <math>x \geq 0</math>.

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\leq 0</math>
and <math>x\geq 0</math>.

{{separation of variables in two dimensions}}

{{separation of variables for a free surface}}

=== Separation of Variables for a Dock ===

The separation of variables equation for a partially submerged dock is given by:
<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>
Z^{\prime} (-d) = 0.
</math></center>
The solution is
<math>k=\kappa_{m}= \frac{m\pi}{h-d} \,</math> , <math>m\in\mathbb{N}\cup\left\{0\right\}</math> and
<center><math>
Z = \psi_{m}\left( z\right) = \cos\kappa_{m}(z+h),\quad
m\in\mathbb{N}\cup\left\{0\right\}.</math></center>
We note that
<center><math>
\int\nolimits_{-h}^{-d}\psi_{m}(z)\psi_{n}(z) \mathrm{d} z=C_{m}\delta_{mn},
</math></center>
where
<center>
<math>
C_{m} = \frac{1}{2} \left( \frac{\cos\kappa_{m}(h-d)\sin\kappa_{m}(h-d)+\kappa_{m}(h-d)}{\kappa_{m}} \right).
</math>
</center>
In addition there is the special case of when <math>m=n=0</math> ,
<center>
<math>
C_{0} = h-d,
</math>
</center>
where <math>\delta_{00}=1</math> .

=== Inner product between free surface and dock modes ===

<center><math>
B_{mn} = \int\nolimits_{-h}^{-d}\psi_{m}(z)\chi_{n}(z) \mathrm{d} z , \quad m\geq1, \quad n\geq0,
</math></center>
where
<center><math>
B_{mn}= \int\nolimits_{-h}^{-d} \frac{\cos\kappa_{m}(z+h)\cos k_{n}(z+h)}{\cos(k_{n}h)} \mathrm{d} z=\frac{-(-1)^m k_{n}\sin k_{n}(h-d)}
{\cos(k_{n}h)(\kappa_{m}^{2}-k_{n}^{2})}.
</math></center>
The evaluation of this inner product works on the assumption that the two roots <math>\kappa_{m}</math> and <math>k_{n}</math> do not coincide. The Matlab code given below has a routine to check for this case and make necessary adjustments.

We also have the case where the functions of the inner product are switched.
<center><math>
B_{nm} = \int\nolimits_{-h}^{-d}\chi_{m}(z)\psi_{n}(z) \mathrm{d} z , \quad m\geq0, \quad n\geq0.
</math></center>
For <math>m \geq 0</math> and <math>n \geq 1</math> the evaluation is above with the exception that the indices of the expression are switched.

Finally there is the special case of when <math>m \geq 0</math> and <math>n = 0</math>.
<center><math>
B_{0m} = \frac{\sin k_{m}(h-d)}{k_{m}\cos(k_{m}h)}.
</math></center>

=== Expansion of the potential ===

We need to apply some boundary conditions at plus and minus infinity,
where essentially 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 roots. Here we
must use a different condition, essentially identifying one solution
as the incoming wave and the other as the outgoing wave.

Therefore the diffracted potentials can
be expanded as
<center>
<math>
\phi(x,z)=\sum_{m=0}^{\infty}a_{m}e^{k_{m}x}\chi_{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 dock covered region. Therefore, the
potentials and their derivatives at <math>x=-L</math> have to be equal.
We obtain for the '''symmetric''' and '''antisymmetric''' cases:

<center><math>
\chi_{0}\left( z\right) + \sum_{m=0}^{\infty}a^{s,a}_{m} \chi_{m}\left(z\right)
&=\sum_{m=0}^{\infty}b^{s,a}_{m}\psi_{m}(z), \quad -h \leq z\leq -d
</math></center>

<center><math>
\begin{align}
-k_{0}\chi_{0}\left( z\right) +\sum_{m=0}^{\infty} a^{s,a}_{m}k_{m}\chi_{m}\left(z\right) &=
\begin{cases}
\quad \quad \quad 0,\quad \quad \quad -d\leq z\leq 0 \\
-\sum_{m=0}^{\infty}b^{s,a}_{m}\kappa_{m}\psi_{m}(z),\quad -h\leq z\leq -d
\end{cases}
\end{align}
</math></center>

For the first equation we multiply both sides by <math>\psi_{n}(z) \,</math> and integrating from <math>-h</math> to <math>-d</math> to obtain:
<center><math>
B_{n0} + \sum^{\infty}_{m=0}a_{m}B_{nm} = b_{m}C_{m}\delta_{mn}, n\in\mathbb{N}\cup\left\{0\right\}
</math></center>
and for the second equation we multiply both sides by <math>\chi_{n}(z) \,</math> and integrating from <math>-h</math> to <math>0</math> to obtain:
<center><math>
-k_{0}A_{0}\delta_{0n} + k_{m}a_{m}A_{m} = -\sum^{\infty}_{m=0}\kappa_{m}b_{m}B_{mn}, n\in\mathbb{N}\cup\left\{0\right\}
</math></center>

Solving the equations above will yield the coefficients of the water velocity potential in the dock covered region.

== Numerical Solution ==

To solve the system of equations for the '''symmetric''' case, we set the upper limit of <math>m</math> to be <math>N</math>. This resulting system can be expressed in the block matrix form below,

<center>
<math>
\begin{bmatrix}
B_{00} & B_{01} & \cdots & B_{0N} & -C_{0} & 0 & \cdots & 0 \\
B_{10} & B_{11} & \cdots & B_{1N} & 0 & -C_{1} & \cdots & \vdots \\
\vdots & \vdots & \ddots & \vdots & \vdots & \vdots & \ddots & 0 \\
B_{N0} & B_{N1} & \cdots & B_{NN} & 0 & \cdots & 0 & -C_{N} \\
k_{0}A_{0} & 0 & \cdots & 0 & 0 & \kappa_{1}\tanh(\kappa_{1}L)B_{10} & \cdots & \kappa_{N}\tanh(\kappa_{N}L)B_{N0} \\
0 & k_{1}A_{1} & \cdots & \vdots & 0 & \kappa_{1}\tanh(\kappa_{1}L)B_{11} & \cdots & \kappa_{N}\tanh(\kappa_{N}L)B_{N1} \\
\vdots & \vdots & \ddots & 0 & \vdots & \vdots & \ddots & \vdots \\
0 & \cdots & 0 & k_{N}A_{N} & 0 & \kappa_{1}\tanh(\kappa_{1}L)B_{1N} & \cdots & \kappa_{N}\tanh(\kappa_{N}L)B_{NN} \\
\end{bmatrix}
\begin{bmatrix}
a^s_{0} \\ a^s_{1} \\ \vdots \\ a^s_{N} \\ b^s_{0} \\ b^s_{1} \\ \vdots \\ b^s_{N} \\
\end{bmatrix}
=\begin{bmatrix}
-B_{00} \\ -B_{10} \\ \vdots \\ -B_{N0} \\ k_{0}A_{0} \\ 0 \\ \vdots \\ 0 \\
\end{bmatrix}
</math>
</center>

To solve the system of equations for the '''antisymmetric''' case, we set the upper limit of <math>m</math> to be <math>N</math>. This resulting system can be expressed in the block matrix form below,

<center>
<math>
\begin{bmatrix}
B_{00} & B_{01} & \cdots & B_{0N} & -C_{0} & 0 & \cdots & 0 \\
B_{10} & B_{11} & \cdots & B_{1N} & 0 & -C_{1} & \cdots & \vdots \\
\vdots & \vdots & \ddots & \vdots & \vdots & \vdots & \ddots & 0 \\
B_{N0} & B_{N1} & \cdots & B_{NN} & 0 & \cdots & 0 & -C_{N} \\
k_{0}A_{0} & 0 & \cdots & 0 & \frac{1}{L}D_{0} & \kappa_{1}\coth(\kappa_{1}L)B_{10} & \cdots & \kappa_{N}\coth(\kappa_{N}L)B_{N0} \\
0 & k_{1}A_{1} & \cdots & \vdots & \frac{1}{L}D_{1} & \kappa_{1}\coth(\kappa_{1}L)B_{11} & \cdots & \kappa_{N}\coth(\kappa_{N}L)B_{N1} \\
\vdots & \vdots & \ddots & 0 & \vdots & \vdots & \ddots & \vdots \\
0 & \cdots & 0 & k_{N}A_{N} & \frac{1}{L}D_{N} & \kappa_{1}\coth(\kappa_{1}L)B_{1N} & \cdots & \kappa_{N}\coth(\kappa_{N}L)B_{NN} \\
\end{bmatrix}
\begin{bmatrix}
a^a_{0} \\ a^a_{1} \\ \vdots \\ a^a_{N} \\ b^a_{0} \\ b^a_{1} \\ \vdots \\ b^a_{N} \\
\end{bmatrix}
=\begin{bmatrix}
-B_{00} \\ -B_{10} \\ \vdots \\ -B_{N0} \\ k_{0}A_{0} \\ 0 \\ \vdots \\ 0 \\
\end{bmatrix}
</math>
</center>

We then simply need to solve the two sets of <math>2(N+1) \mbox{ x } 2(N+1)</math> linear systems of equations. If we had not taken advantage of the symmetry we would need to solve a <math>4(N+1) \mbox{ x } 4(N+1)</math> linear system of equations for four coefficients <math>a_{m}</math>, <math>b_{m}</math>, <math>c_{m}</math> and <math>d_{m}</math>.

== Matlab Code ==

A program to calculate the coefficients for the single finite dock problem using symmetry 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 Finite Rectangle using Symmetry

2012-03-16T03:28:30Z

Tertius: /* An infinite dimensional system of equations */

{{incomplete pages}}

== Introduction ==

The problem consists of a finite rigid fixed rectangle of length <math>2L</math> centered at <math>x = 0</math> and having a height of <math>d + ...</math> freeboard with depth of submergence being <math>d</math> through which no flow is possible. The regions to the left and right of the rectangle are the free water surface. Since the geometry of the rectangle is symmetric about <math>x = 0</math> we can decompose the velocity potential <math>\phi</math> into a symmetric <math>\phi^s</math> and antisymmetric <math>\phi^a</math> potential. We then solve for these two potentials in the left domain only <math>x \leq 0</math> and subsequently recompose the solution <math>\phi</math> for the entire domain. We begin with the simple problem when the waves are normally incident on the rectangle (dock) (so that the problem is truly two-dimensional). We can extend this solution to a pair of partially submerged finite rigid fixed rectangles using symmetry
[[Eigenfunction Matching for a Finite Rectangle using Symmetry]]

== Governing Equations ==

We begin with the [[Frequency Domain Problem]] for a partially submerged finite dock. Although the dock is of length <math>2L</math> we are only interested in the dock of <math>-L \leq x \leq 0</math> together with the fluid domain <math>x \leq 0</math>. We assume <math>e^{i\omega t}</math> time dependence and
the depth of submergence is <math>d</math>.
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>
\begin{align}
\Delta\phi^{s,a} &= 0, \,\,\, -h\leq z\leq 0, \,\, x < -L, \\

\Delta\phi^{s,a} &= 0, \,\,\, -h\leq z\leq -d, \,\, -L < x < 0. \\

\partial_z\phi^{s,a} &= \alpha\phi^{s,a}, \,\,\, z=0, \,\, x < -L, \\

\partial_x\phi^{s,a} &= 0, \,\,\, -d\leq z\leq 0, \,\, x = -L, \\

\partial_z\phi^{s,a} &= 0, \,\,\,
\begin{cases}
z = -d, \,\, -L < x < 0, \\
z = -h, \,\, x < 0.
\end{cases} \\

\partial_x\phi^s &= 0, \,\,\, x = 0, \,\, -h \leq z \leq -d, \\

\phi^a &= 0, \,\,\, x = 0, \,\, -h \leq z \leq -d.
\end{align}
</math></center>

Where <math>\alpha = \omega^2/g</math>.

<math>\phi^s(x,z)</math> is an even function and at <math>x = 0</math> the derivative of this function is equal to zero. <math>\phi^a(x,z)</math> is an odd function and at <math>x = 0</math> this function is equal to zero. These boundary conditions are important and enable the recomposed solution <math>\phi(x,z)</math> to be reflected into the domain of <math>x \geq 0</math>.

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\leq 0</math>
and <math>x\geq 0</math>.

{{separation of variables in two dimensions}}

{{separation of variables for a free surface}}

=== Separation of Variables for a Dock ===

The separation of variables equation for a partially submerged dock is given by:
<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>
Z^{\prime} (-d) = 0.
</math></center>
The solution is
<math>k=\kappa_{m}= \frac{m\pi}{h-d} \,</math> , <math>m\in\mathbb{N}\cup\left\{0\right\}</math> and
<center><math>
Z = \psi_{m}\left( z\right) = \cos\kappa_{m}(z+h),\quad
m\in\mathbb{N}\cup\left\{0\right\}.</math></center>
We note that
<center><math>
\int\nolimits_{-h}^{-d}\psi_{m}(z)\psi_{n}(z) \mathrm{d} z=C_{m}\delta_{mn},
</math></center>
where
<center>
<math>
C_{m} = \frac{1}{2} \left( \frac{\cos\kappa_{m}(h-d)\sin\kappa_{m}(h-d)+\kappa_{m}(h-d)}{\kappa_{m}} \right).
</math>
</center>
In addition there is the special case of when <math>m=n=0</math> ,
<center>
<math>
C_{0} = h-d,
</math>
</center>
where <math>\delta_{00}=1</math> .

=== Inner product between free surface and dock modes ===

<center><math>
B_{mn} = \int\nolimits_{-h}^{-d}\psi_{m}(z)\chi_{n}(z) \mathrm{d} z , \quad m\geq1, \quad n\geq0,
</math></center>
where
<center><math>
B_{mn}= \int\nolimits_{-h}^{-d} \frac{\cos\kappa_{m}(z+h)\cos k_{n}(z+h)}{\cos(k_{n}h)} \mathrm{d} z=\frac{-(-1)^m k_{n}\sin k_{n}(h-d)}
{\cos(k_{n}h)(\kappa_{m}^{2}-k_{n}^{2})}.
</math></center>
The evaluation of this inner product works on the assumption that the two roots <math>\kappa_{m}</math> and <math>k_{n}</math> do not coincide. The Matlab code given below has a routine to check for this case and make necessary adjustments.

We also have the case where the functions of the inner product are switched.
<center><math>
B_{nm} = \int\nolimits_{-h}^{-d}\chi_{m}(z)\psi_{n}(z) \mathrm{d} z , \quad m\geq0, \quad n\geq0.
</math></center>
For <math>m \geq 0</math> and <math>n \geq 1</math> the evaluation is above with the exception that the indices of the expression are switched.

Finally there is the special case of when <math>m \geq 0</math> and <math>n = 0</math>.
<center><math>
B_{0m} = \frac{\sin k_{m}(h-d)}{k_{m}\cos(k_{m}h)}.
</math></center>

=== Expansion of the potential ===

We need to apply some boundary conditions at plus and minus infinity,
where essentially 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 roots. Here we
must use a different condition, essentially identifying one solution
as the incoming wave and the other as the outgoing wave.

Therefore the diffracted potentials can
be expanded as
<center>
<math>
\phi(x,z)=\sum_{m=0}^{\infty}a_{m}e^{k_{m}x}\chi_{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 dock covered region. Therefore, the
potentials and their derivatives at <math>x=-L</math> have to be equal.
We obtain for the '''symmetric''' and '''antisymmetric''' cases:
<center><math>
\begin{align}
\chi_{0}\left( z\right) + \sum_{m=0}^{\infty}a^{s,a}_{m} \chi_{m}\left( z\right)
&=\sum_{m=0}^{\infty}b^{s,a}_{m}\psi_{m}(z), \quad -h \leq z\leq -d \\

-k_{0}\chi_{0}\left( z\right) +\sum_{m=0}^{\infty} a^{s,a}_{m}k_{m}\chi_{m}\left( z\right) &=
\begin{cases}
\quad \quad \quad 0,\quad \quad \quad -d\leq z\leq 0 \\
-\sum_{m=0}^{\infty}b^{s,a}_{m}\kappa_{m}\psi_{m}(z),\quad -h\leq z\leq -d
\end{cases}
\end{align}
</math></center>

For the first equation we multiply both sides by <math>\psi_{n}(z) \,</math> and integrating from <math>-h</math> to <math>-d</math> to obtain:
<center><math>
B_{n0} + \sum^{\infty}_{m=0}a_{m}B_{nm} = b_{m}C_{m}\delta_{mn}, n\in\mathbb{N}\cup\left\{0\right\}
</math></center>
and for the second equation we multiply both sides by <math>\chi_{n}(z) \,</math> and integrating from <math>-h</math> to <math>0</math> to obtain:
<center><math>
-k_{0}A_{0}\delta_{0n} + k_{m}a_{m}A_{m} = -\sum^{\infty}_{m=0}\kappa_{m}b_{m}B_{mn}, n\in\mathbb{N}\cup\left\{0\right\}
</math></center>

Solving the equations above will yield the coefficients of the water velocity potential in the dock covered region.

== Numerical Solution ==

To solve the system of equations for the '''symmetric''' case, we set the upper limit of <math>m</math> to be <math>N</math>. This resulting system can be expressed in the block matrix form below,

<center>
<math>
\begin{bmatrix}
B_{00} & B_{01} & \cdots & B_{0N} & -C_{0} & 0 & \cdots & 0 \\
B_{10} & B_{11} & \cdots & B_{1N} & 0 & -C_{1} & \cdots & \vdots \\
\vdots & \vdots & \ddots & \vdots & \vdots & \vdots & \ddots & 0 \\
B_{N0} & B_{N1} & \cdots & B_{NN} & 0 & \cdots & 0 & -C_{N} \\
k_{0}A_{0} & 0 & \cdots & 0 & 0 & \kappa_{1}\tanh(\kappa_{1}L)B_{10} & \cdots & \kappa_{N}\tanh(\kappa_{N}L)B_{N0} \\
0 & k_{1}A_{1} & \cdots & \vdots & 0 & \kappa_{1}\tanh(\kappa_{1}L)B_{11} & \cdots & \kappa_{N}\tanh(\kappa_{N}L)B_{N1} \\
\vdots & \vdots & \ddots & 0 & \vdots & \vdots & \ddots & \vdots \\
0 & \cdots & 0 & k_{N}A_{N} & 0 & \kappa_{1}\tanh(\kappa_{1}L)B_{1N} & \cdots & \kappa_{N}\tanh(\kappa_{N}L)B_{NN} \\
\end{bmatrix}
\begin{bmatrix}
a^s_{0} \\ a^s_{1} \\ \vdots \\ a^s_{N} \\ b^s_{0} \\ b^s_{1} \\ \vdots \\ b^s_{N} \\
\end{bmatrix}
=\begin{bmatrix}
-B_{00} \\ -B_{10} \\ \vdots \\ -B_{N0} \\ k_{0}A_{0} \\ 0 \\ \vdots \\ 0 \\
\end{bmatrix}
</math>
</center>

To solve the system of equations for the '''antisymmetric''' case, we set the upper limit of <math>m</math> to be <math>N</math>. This resulting system can be expressed in the block matrix form below,

<center>
<math>
\begin{bmatrix}
B_{00} & B_{01} & \cdots & B_{0N} & -C_{0} & 0 & \cdots & 0 \\
B_{10} & B_{11} & \cdots & B_{1N} & 0 & -C_{1} & \cdots & \vdots \\
\vdots & \vdots & \ddots & \vdots & \vdots & \vdots & \ddots & 0 \\
B_{N0} & B_{N1} & \cdots & B_{NN} & 0 & \cdots & 0 & -C_{N} \\
k_{0}A_{0} & 0 & \cdots & 0 & \frac{1}{L}D_{0} & \kappa_{1}\coth(\kappa_{1}L)B_{10} & \cdots & \kappa_{N}\coth(\kappa_{N}L)B_{N0} \\
0 & k_{1}A_{1} & \cdots & \vdots & \frac{1}{L}D_{1} & \kappa_{1}\coth(\kappa_{1}L)B_{11} & \cdots & \kappa_{N}\coth(\kappa_{N}L)B_{N1} \\
\vdots & \vdots & \ddots & 0 & \vdots & \vdots & \ddots & \vdots \\
0 & \cdots & 0 & k_{N}A_{N} & \frac{1}{L}D_{N} & \kappa_{1}\coth(\kappa_{1}L)B_{1N} & \cdots & \kappa_{N}\coth(\kappa_{N}L)B_{NN} \\
\end{bmatrix}
\begin{bmatrix}
a^a_{0} \\ a^a_{1} \\ \vdots \\ a^a_{N} \\ b^a_{0} \\ b^a_{1} \\ \vdots \\ b^a_{N} \\
\end{bmatrix}
=\begin{bmatrix}
-B_{00} \\ -B_{10} \\ \vdots \\ -B_{N0} \\ k_{0}A_{0} \\ 0 \\ \vdots \\ 0 \\
\end{bmatrix}
</math>
</center>

We then simply need to solve the two sets of <math>2(N+1) \mbox{ x } 2(N+1)</math> linear systems of equations. If we had not taken advantage of the symmetry we would need to solve a <math>4(N+1) \mbox{ x } 4(N+1)</math> linear system of equations for four coefficients <math>a_{m}</math>, <math>b_{m}</math>, <math>c_{m}</math> and <math>d_{m}</math>.

== Matlab Code ==

A program to calculate the coefficients for the single finite dock problem using symmetry 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 Finite Rectangle using Symmetry

2012-03-16T03:17:20Z

Tertius: /* An infinite dimensional system of equations */

{{incomplete pages}}

== Introduction ==

The problem consists of a finite rigid fixed rectangle of length <math>2L</math> centered at <math>x = 0</math> and having a height of <math>d + ...</math> freeboard with depth of submergence being <math>d</math> through which no flow is possible. The regions to the left and right of the rectangle are the free water surface. Since the geometry of the rectangle is symmetric about <math>x = 0</math> we can decompose the velocity potential <math>\phi</math> into a symmetric <math>\phi^s</math> and antisymmetric <math>\phi^a</math> potential. We then solve for these two potentials in the left domain only <math>x \leq 0</math> and subsequently recompose the solution <math>\phi</math> for the entire domain. We begin with the simple problem when the waves are normally incident on the rectangle (dock) (so that the problem is truly two-dimensional). We can extend this solution to a pair of partially submerged finite rigid fixed rectangles using symmetry
[[Eigenfunction Matching for a Finite Rectangle using Symmetry]]

== Governing Equations ==

We begin with the [[Frequency Domain Problem]] for a partially submerged finite dock. Although the dock is of length <math>2L</math> we are only interested in the dock of <math>-L \leq x \leq 0</math> together with the fluid domain <math>x \leq 0</math>. We assume <math>e^{i\omega t}</math> time dependence and
the depth of submergence is <math>d</math>.
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>
\begin{align}
\Delta\phi^{s,a} &= 0, \,\,\, -h\leq z\leq 0, \,\, x < -L, \\

\Delta\phi^{s,a} &= 0, \,\,\, -h\leq z\leq -d, \,\, -L < x < 0. \\

\partial_z\phi^{s,a} &= \alpha\phi^{s,a}, \,\,\, z=0, \,\, x < -L, \\

\partial_x\phi^{s,a} &= 0, \,\,\, -d\leq z\leq 0, \,\, x = -L, \\

\partial_z\phi^{s,a} &= 0, \,\,\,
\begin{cases}
z = -d, \,\, -L < x < 0, \\
z = -h, \,\, x < 0.
\end{cases} \\

\partial_x\phi^s &= 0, \,\,\, x = 0, \,\, -h \leq z \leq -d, \\

\phi^a &= 0, \,\,\, x = 0, \,\, -h \leq z \leq -d.
\end{align}
</math></center>

Where <math>\alpha = \omega^2/g</math>.

<math>\phi^s(x,z)</math> is an even function and at <math>x = 0</math> the derivative of this function is equal to zero. <math>\phi^a(x,z)</math> is an odd function and at <math>x = 0</math> this function is equal to zero. These boundary conditions are important and enable the recomposed solution <math>\phi(x,z)</math> to be reflected into the domain of <math>x \geq 0</math>.

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\leq 0</math>
and <math>x\geq 0</math>.

{{separation of variables in two dimensions}}

{{separation of variables for a free surface}}

=== Separation of Variables for a Dock ===

The separation of variables equation for a partially submerged dock is given by:
<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>
Z^{\prime} (-d) = 0.
</math></center>
The solution is
<math>k=\kappa_{m}= \frac{m\pi}{h-d} \,</math> , <math>m\in\mathbb{N}\cup\left\{0\right\}</math> and
<center><math>
Z = \psi_{m}\left( z\right) = \cos\kappa_{m}(z+h),\quad
m\in\mathbb{N}\cup\left\{0\right\}.</math></center>
We note that
<center><math>
\int\nolimits_{-h}^{-d}\psi_{m}(z)\psi_{n}(z) \mathrm{d} z=C_{m}\delta_{mn},
</math></center>
where
<center>
<math>
C_{m} = \frac{1}{2} \left( \frac{\cos\kappa_{m}(h-d)\sin\kappa_{m}(h-d)+\kappa_{m}(h-d)}{\kappa_{m}} \right).
</math>
</center>
In addition there is the special case of when <math>m=n=0</math> ,
<center>
<math>
C_{0} = h-d,
</math>
</center>
where <math>\delta_{00}=1</math> .

=== Inner product between free surface and dock modes ===

<center><math>
B_{mn} = \int\nolimits_{-h}^{-d}\psi_{m}(z)\chi_{n}(z) \mathrm{d} z , \quad m\geq1, \quad n\geq0,
</math></center>
where
<center><math>
B_{mn}= \int\nolimits_{-h}^{-d} \frac{\cos\kappa_{m}(z+h)\cos k_{n}(z+h)}{\cos(k_{n}h)} \mathrm{d} z=\frac{-(-1)^m k_{n}\sin k_{n}(h-d)}
{\cos(k_{n}h)(\kappa_{m}^{2}-k_{n}^{2})}.
</math></center>
The evaluation of this inner product works on the assumption that the two roots <math>\kappa_{m}</math> and <math>k_{n}</math> do not coincide. The Matlab code given below has a routine to check for this case and make necessary adjustments.

We also have the case where the functions of the inner product are switched.
<center><math>
B_{nm} = \int\nolimits_{-h}^{-d}\chi_{m}(z)\psi_{n}(z) \mathrm{d} z , \quad m\geq0, \quad n\geq0.
</math></center>
For <math>m \geq 0</math> and <math>n \geq 1</math> the evaluation is above with the exception that the indices of the expression are switched.

Finally there is the special case of when <math>m \geq 0</math> and <math>n = 0</math>.
<center><math>
B_{0m} = \frac{\sin k_{m}(h-d)}{k_{m}\cos(k_{m}h)}.
</math></center>

=== Expansion of the potential ===

We need to apply some boundary conditions at plus and minus infinity,
where essentially 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 roots. Here we
must use a different condition, essentially identifying one solution
as the incoming wave and the other as the outgoing wave.

Therefore the diffracted potentials can
be expanded as
<center>
<math>
\phi(x,z)=\sum_{m=0}^{\infty}a_{m}e^{k_{m}x}\chi_{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 dock covered region. Therefore, the
potentials and their derivatives at <math>x=-L</math> have to be equal.
We obtain for the '''symmetric''' case:
<center><math>
\begin{align}
\chi_{0}\left( z\right) + \sum_{m=0}^{\infty}a^{s,a}_{m} \chi_{m}\left( z\right)
&=\sum_{m=0}^{\infty}b^{s,a}_{m}\psi_{m}(z), \quad -h \leq z\leq -d \\

-k_{0}\chi_{0}\left( z\right) +\sum_{m=0}^{\infty} a^{s,a}_{m}k_{m}\chi_{m}\left( z\right) &=
\begin{cases}
\quad \quad \quad 0,\quad \quad \quad -d\leq z\leq 0 \\
-\sum_{m=0}^{\infty}b^{s,a}_{m}\kappa_{m}\psi_{m}(z),\quad -h\leq z\leq -d
\end{cases}
\end{align}
</math></center>

For the first equation we multiply both sides by <math>\psi_{n}(z) \,</math> and integrating from <math>-h</math> to <math>-d</math> to obtain:
<center><math>
B_{n0} + \sum^{\infty}_{m=0}a_{m}B_{nm} = b_{m}C_{m}\delta_{mn}, n\in\mathbb{N}\cup\left\{0\right\}
</math></center>
and for the second equation we multiply both sides by <math>\chi_{n}(z) \,</math> and integrating from <math>-h</math> to <math>0</math> to obtain:
<center><math>
-k_{0}A_{0}\delta_{0n} + k_{m}a_{m}A_{m} = -\sum^{\infty}_{m=0}\kappa_{m}b_{m}B_{mn}, n\in\mathbb{N}\cup\left\{0\right\}
</math></center>

Solving the equations above will yield the coefficients of the water velocity potential in the dock covered region.

== Numerical Solution ==

To solve the system of equations for the '''symmetric''' case, we set the upper limit of <math>m</math> to be <math>N</math>. This resulting system can be expressed in the block matrix form below,

<center>
<math>
\begin{bmatrix}
B_{00} & B_{01} & \cdots & B_{0N} & -C_{0} & 0 & \cdots & 0 \\
B_{10} & B_{11} & \cdots & B_{1N} & 0 & -C_{1} & \cdots & \vdots \\
\vdots & \vdots & \ddots & \vdots & \vdots & \vdots & \ddots & 0 \\
B_{N0} & B_{N1} & \cdots & B_{NN} & 0 & \cdots & 0 & -C_{N} \\
k_{0}A_{0} & 0 & \cdots & 0 & 0 & \kappa_{1}\tanh(\kappa_{1}L)B_{10} & \cdots & \kappa_{N}\tanh(\kappa_{N}L)B_{N0} \\
0 & k_{1}A_{1} & \cdots & \vdots & 0 & \kappa_{1}\tanh(\kappa_{1}L)B_{11} & \cdots & \kappa_{N}\tanh(\kappa_{N}L)B_{N1} \\
\vdots & \vdots & \ddots & 0 & \vdots & \vdots & \ddots & \vdots \\
0 & \cdots & 0 & k_{N}A_{N} & 0 & \kappa_{1}\tanh(\kappa_{1}L)B_{1N} & \cdots & \kappa_{N}\tanh(\kappa_{N}L)B_{NN} \\
\end{bmatrix}
\begin{bmatrix}
a^s_{0} \\ a^s_{1} \\ \vdots \\ a^s_{N} \\ b^s_{0} \\ b^s_{1} \\ \vdots \\ b^s_{N} \\
\end{bmatrix}
=\begin{bmatrix}
-B_{00} \\ -B_{10} \\ \vdots \\ -B_{N0} \\ k_{0}A_{0} \\ 0 \\ \vdots \\ 0 \\
\end{bmatrix}
</math>
</center>

To solve the system of equations for the '''antisymmetric''' case, we set the upper limit of <math>m</math> to be <math>N</math>. This resulting system can be expressed in the block matrix form below,

<center>
<math>
\begin{bmatrix}
B_{00} & B_{01} & \cdots & B_{0N} & -C_{0} & 0 & \cdots & 0 \\
B_{10} & B_{11} & \cdots & B_{1N} & 0 & -C_{1} & \cdots & \vdots \\
\vdots & \vdots & \ddots & \vdots & \vdots & \vdots & \ddots & 0 \\
B_{N0} & B_{N1} & \cdots & B_{NN} & 0 & \cdots & 0 & -C_{N} \\
k_{0}A_{0} & 0 & \cdots & 0 & \frac{1}{L}D_{0} & \kappa_{1}\coth(\kappa_{1}L)B_{10} & \cdots & \kappa_{N}\coth(\kappa_{N}L)B_{N0} \\
0 & k_{1}A_{1} & \cdots & \vdots & \frac{1}{L}D_{1} & \kappa_{1}\coth(\kappa_{1}L)B_{11} & \cdots & \kappa_{N}\coth(\kappa_{N}L)B_{N1} \\
\vdots & \vdots & \ddots & 0 & \vdots & \vdots & \ddots & \vdots \\
0 & \cdots & 0 & k_{N}A_{N} & \frac{1}{L}D_{N} & \kappa_{1}\coth(\kappa_{1}L)B_{1N} & \cdots & \kappa_{N}\coth(\kappa_{N}L)B_{NN} \\
\end{bmatrix}
\begin{bmatrix}
a^a_{0} \\ a^a_{1} \\ \vdots \\ a^a_{N} \\ b^a_{0} \\ b^a_{1} \\ \vdots \\ b^a_{N} \\
\end{bmatrix}
=\begin{bmatrix}
-B_{00} \\ -B_{10} \\ \vdots \\ -B_{N0} \\ k_{0}A_{0} \\ 0 \\ \vdots \\ 0 \\
\end{bmatrix}
</math>
</center>

We then simply need to solve the two sets of <math>2(N+1) \mbox{ x } 2(N+1)</math> linear systems of equations. If we had not taken advantage of the symmetry we would need to solve a <math>4(N+1) \mbox{ x } 4(N+1)</math> linear system of equations for four coefficients <math>a_{m}</math>, <math>b_{m}</math>, <math>c_{m}</math> and <math>d_{m}</math>.

== Matlab Code ==

A program to calculate the coefficients for the single finite dock problem using symmetry 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 Finite Rectangle using Symmetry

2012-03-16T03:16:37Z

Tertius: /* An infinite dimensional system of equations */

{{incomplete pages}}

== Introduction ==

The problem consists of a finite rigid fixed rectangle of length <math>2L</math> centered at <math>x = 0</math> and having a height of <math>d + ...</math> freeboard with depth of submergence being <math>d</math> through which no flow is possible. The regions to the left and right of the rectangle are the free water surface. Since the geometry of the rectangle is symmetric about <math>x = 0</math> we can decompose the velocity potential <math>\phi</math> into a symmetric <math>\phi^s</math> and antisymmetric <math>\phi^a</math> potential. We then solve for these two potentials in the left domain only <math>x \leq 0</math> and subsequently recompose the solution <math>\phi</math> for the entire domain. We begin with the simple problem when the waves are normally incident on the rectangle (dock) (so that the problem is truly two-dimensional). We can extend this solution to a pair of partially submerged finite rigid fixed rectangles using symmetry
[[Eigenfunction Matching for a Finite Rectangle using Symmetry]]

== Governing Equations ==

We begin with the [[Frequency Domain Problem]] for a partially submerged finite dock. Although the dock is of length <math>2L</math> we are only interested in the dock of <math>-L \leq x \leq 0</math> together with the fluid domain <math>x \leq 0</math>. We assume <math>e^{i\omega t}</math> time dependence and
the depth of submergence is <math>d</math>.
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>
\begin{align}
\Delta\phi^{s,a} &= 0, \,\,\, -h\leq z\leq 0, \,\, x < -L, \\

\Delta\phi^{s,a} &= 0, \,\,\, -h\leq z\leq -d, \,\, -L < x < 0. \\

\partial_z\phi^{s,a} &= \alpha\phi^{s,a}, \,\,\, z=0, \,\, x < -L, \\

\partial_x\phi^{s,a} &= 0, \,\,\, -d\leq z\leq 0, \,\, x = -L, \\

\partial_z\phi^{s,a} &= 0, \,\,\,
\begin{cases}
z = -d, \,\, -L < x < 0, \\
z = -h, \,\, x < 0.
\end{cases} \\

\partial_x\phi^s &= 0, \,\,\, x = 0, \,\, -h \leq z \leq -d, \\

\phi^a &= 0, \,\,\, x = 0, \,\, -h \leq z \leq -d.
\end{align}
</math></center>

Where <math>\alpha = \omega^2/g</math>.

<math>\phi^s(x,z)</math> is an even function and at <math>x = 0</math> the derivative of this function is equal to zero. <math>\phi^a(x,z)</math> is an odd function and at <math>x = 0</math> this function is equal to zero. These boundary conditions are important and enable the recomposed solution <math>\phi(x,z)</math> to be reflected into the domain of <math>x \geq 0</math>.

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\leq 0</math>
and <math>x\geq 0</math>.

{{separation of variables in two dimensions}}

{{separation of variables for a free surface}}

=== Separation of Variables for a Dock ===

The separation of variables equation for a partially submerged dock is given by:
<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>
Z^{\prime} (-d) = 0.
</math></center>
The solution is
<math>k=\kappa_{m}= \frac{m\pi}{h-d} \,</math> , <math>m\in\mathbb{N}\cup\left\{0\right\}</math> and
<center><math>
Z = \psi_{m}\left( z\right) = \cos\kappa_{m}(z+h),\quad
m\in\mathbb{N}\cup\left\{0\right\}.</math></center>
We note that
<center><math>
\int\nolimits_{-h}^{-d}\psi_{m}(z)\psi_{n}(z) \mathrm{d} z=C_{m}\delta_{mn},
</math></center>
where
<center>
<math>
C_{m} = \frac{1}{2} \left( \frac{\cos\kappa_{m}(h-d)\sin\kappa_{m}(h-d)+\kappa_{m}(h-d)}{\kappa_{m}} \right).
</math>
</center>
In addition there is the special case of when <math>m=n=0</math> ,
<center>
<math>
C_{0} = h-d,
</math>
</center>
where <math>\delta_{00}=1</math> .

=== Inner product between free surface and dock modes ===

<center><math>
B_{mn} = \int\nolimits_{-h}^{-d}\psi_{m}(z)\chi_{n}(z) \mathrm{d} z , \quad m\geq1, \quad n\geq0,
</math></center>
where
<center><math>
B_{mn}= \int\nolimits_{-h}^{-d} \frac{\cos\kappa_{m}(z+h)\cos k_{n}(z+h)}{\cos(k_{n}h)} \mathrm{d} z=\frac{-(-1)^m k_{n}\sin k_{n}(h-d)}
{\cos(k_{n}h)(\kappa_{m}^{2}-k_{n}^{2})}.
</math></center>
The evaluation of this inner product works on the assumption that the two roots <math>\kappa_{m}</math> and <math>k_{n}</math> do not coincide. The Matlab code given below has a routine to check for this case and make necessary adjustments.

We also have the case where the functions of the inner product are switched.
<center><math>
B_{nm} = \int\nolimits_{-h}^{-d}\chi_{m}(z)\psi_{n}(z) \mathrm{d} z , \quad m\geq0, \quad n\geq0.
</math></center>
For <math>m \geq 0</math> and <math>n \geq 1</math> the evaluation is above with the exception that the indices of the expression are switched.

Finally there is the special case of when <math>m \geq 0</math> and <math>n = 0</math>.
<center><math>
B_{0m} = \frac{\sin k_{m}(h-d)}{k_{m}\cos(k_{m}h)}.
</math></center>

=== Expansion of the potential ===

We need to apply some boundary conditions at plus and minus infinity,
where essentially 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 roots. Here we
must use a different condition, essentially identifying one solution
as the incoming wave and the other as the outgoing wave.

Therefore the diffracted potentials can
be expanded as
<center>
<math>
\phi(x,z)=\sum_{m=0}^{\infty}a_{m}e^{k_{m}x}\chi_{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 dock covered region. Therefore, the
potentials and their derivatives at <math>x=-L</math> have to be equal.
We obtain for the '''symmetric''' case:
<center><math>
\begin{align}
\chi_{0}\left( z\right) + \sum_{m=0}^{\infty}a^{s,a}_{m} \chi_{m}\left( z\right)
&=\sum_{m=0}^{\infty}b^{s,a}_{m}\psi_{m}(z), \quad -h \leq z\leq -d \\

-k_{0}\chi_{0}\left( z\right) +\sum_{m=0}^{\infty} a_{m}k_{m}\chi_{m}\left( z\right) &=
\begin{cases}
\quad \quad \quad 0,\quad \quad \quad -d\leq z\leq 0 \\
-\sum_{m=0}^{\infty}b_{m}\kappa_{m}\psi_{m}(z),\quad -h\leq z\leq -d
\end{cases}
\end{align}
</math></center>

For the first equation we multiply both sides by <math>\psi_{n}(z) \,</math> and integrating from <math>-h</math> to <math>-d</math> to obtain:
<center><math>
B_{n0} + \sum^{\infty}_{m=0}a_{m}B_{nm} = b_{m}C_{m}\delta_{mn}, n\in\mathbb{N}\cup\left\{0\right\}
</math></center>
and for the second equation we multiply both sides by <math>\chi_{n}(z) \,</math> and integrating from <math>-h</math> to <math>0</math> to obtain:
<center><math>
-k_{0}A_{0}\delta_{0n} + k_{m}a_{m}A_{m} = -\sum^{\infty}_{m=0}\kappa_{m}b_{m}B_{mn}, n\in\mathbb{N}\cup\left\{0\right\}
</math></center>

Solving the equations above will yield the coefficients of the water velocity potential in the dock covered region.

== Numerical Solution ==

To solve the system of equations for the '''symmetric''' case, we set the upper limit of <math>m</math> to be <math>N</math>. This resulting system can be expressed in the block matrix form below,

<center>
<math>
\begin{bmatrix}
B_{00} & B_{01} & \cdots & B_{0N} & -C_{0} & 0 & \cdots & 0 \\
B_{10} & B_{11} & \cdots & B_{1N} & 0 & -C_{1} & \cdots & \vdots \\
\vdots & \vdots & \ddots & \vdots & \vdots & \vdots & \ddots & 0 \\
B_{N0} & B_{N1} & \cdots & B_{NN} & 0 & \cdots & 0 & -C_{N} \\
k_{0}A_{0} & 0 & \cdots & 0 & 0 & \kappa_{1}\tanh(\kappa_{1}L)B_{10} & \cdots & \kappa_{N}\tanh(\kappa_{N}L)B_{N0} \\
0 & k_{1}A_{1} & \cdots & \vdots & 0 & \kappa_{1}\tanh(\kappa_{1}L)B_{11} & \cdots & \kappa_{N}\tanh(\kappa_{N}L)B_{N1} \\
\vdots & \vdots & \ddots & 0 & \vdots & \vdots & \ddots & \vdots \\
0 & \cdots & 0 & k_{N}A_{N} & 0 & \kappa_{1}\tanh(\kappa_{1}L)B_{1N} & \cdots & \kappa_{N}\tanh(\kappa_{N}L)B_{NN} \\
\end{bmatrix}
\begin{bmatrix}
a^s_{0} \\ a^s_{1} \\ \vdots \\ a^s_{N} \\ b^s_{0} \\ b^s_{1} \\ \vdots \\ b^s_{N} \\
\end{bmatrix}
=\begin{bmatrix}
-B_{00} \\ -B_{10} \\ \vdots \\ -B_{N0} \\ k_{0}A_{0} \\ 0 \\ \vdots \\ 0 \\
\end{bmatrix}
</math>
</center>

To solve the system of equations for the '''antisymmetric''' case, we set the upper limit of <math>m</math> to be <math>N</math>. This resulting system can be expressed in the block matrix form below,

<center>
<math>
\begin{bmatrix}
B_{00} & B_{01} & \cdots & B_{0N} & -C_{0} & 0 & \cdots & 0 \\
B_{10} & B_{11} & \cdots & B_{1N} & 0 & -C_{1} & \cdots & \vdots \\
\vdots & \vdots & \ddots & \vdots & \vdots & \vdots & \ddots & 0 \\
B_{N0} & B_{N1} & \cdots & B_{NN} & 0 & \cdots & 0 & -C_{N} \\
k_{0}A_{0} & 0 & \cdots & 0 & \frac{1}{L}D_{0} & \kappa_{1}\coth(\kappa_{1}L)B_{10} & \cdots & \kappa_{N}\coth(\kappa_{N}L)B_{N0} \\
0 & k_{1}A_{1} & \cdots & \vdots & \frac{1}{L}D_{1} & \kappa_{1}\coth(\kappa_{1}L)B_{11} & \cdots & \kappa_{N}\coth(\kappa_{N}L)B_{N1} \\
\vdots & \vdots & \ddots & 0 & \vdots & \vdots & \ddots & \vdots \\
0 & \cdots & 0 & k_{N}A_{N} & \frac{1}{L}D_{N} & \kappa_{1}\coth(\kappa_{1}L)B_{1N} & \cdots & \kappa_{N}\coth(\kappa_{N}L)B_{NN} \\
\end{bmatrix}
\begin{bmatrix}
a^a_{0} \\ a^a_{1} \\ \vdots \\ a^a_{N} \\ b^a_{0} \\ b^a_{1} \\ \vdots \\ b^a_{N} \\
\end{bmatrix}
=\begin{bmatrix}
-B_{00} \\ -B_{10} \\ \vdots \\ -B_{N0} \\ k_{0}A_{0} \\ 0 \\ \vdots \\ 0 \\
\end{bmatrix}
</math>
</center>

We then simply need to solve the two sets of <math>2(N+1) \mbox{ x } 2(N+1)</math> linear systems of equations. If we had not taken advantage of the symmetry we would need to solve a <math>4(N+1) \mbox{ x } 4(N+1)</math> linear system of equations for four coefficients <math>a_{m}</math>, <math>b_{m}</math>, <math>c_{m}</math> and <math>d_{m}</math>.

== Matlab Code ==

A program to calculate the coefficients for the single finite dock problem using symmetry 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 Finite Rectangle using Symmetry

2012-03-16T03:13:05Z

Tertius: /* An infinite dimensional system of equations */

{{incomplete pages}}

== Introduction ==

The problem consists of a finite rigid fixed rectangle of length <math>2L</math> centered at <math>x = 0</math> and having a height of <math>d + ...</math> freeboard with depth of submergence being <math>d</math> through which no flow is possible. The regions to the left and right of the rectangle are the free water surface. Since the geometry of the rectangle is symmetric about <math>x = 0</math> we can decompose the velocity potential <math>\phi</math> into a symmetric <math>\phi^s</math> and antisymmetric <math>\phi^a</math> potential. We then solve for these two potentials in the left domain only <math>x \leq 0</math> and subsequently recompose the solution <math>\phi</math> for the entire domain. We begin with the simple problem when the waves are normally incident on the rectangle (dock) (so that the problem is truly two-dimensional). We can extend this solution to a pair of partially submerged finite rigid fixed rectangles using symmetry
[[Eigenfunction Matching for a Finite Rectangle using Symmetry]]

== Governing Equations ==

We begin with the [[Frequency Domain Problem]] for a partially submerged finite dock. Although the dock is of length <math>2L</math> we are only interested in the dock of <math>-L \leq x \leq 0</math> together with the fluid domain <math>x \leq 0</math>. We assume <math>e^{i\omega t}</math> time dependence and
the depth of submergence is <math>d</math>.
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>
\begin{align}
\Delta\phi^{s,a} &= 0, \,\,\, -h\leq z\leq 0, \,\, x < -L, \\

\Delta\phi^{s,a} &= 0, \,\,\, -h\leq z\leq -d, \,\, -L < x < 0. \\

\partial_z\phi^{s,a} &= \alpha\phi^{s,a}, \,\,\, z=0, \,\, x < -L, \\

\partial_x\phi^{s,a} &= 0, \,\,\, -d\leq z\leq 0, \,\, x = -L, \\

\partial_z\phi^{s,a} &= 0, \,\,\,
\begin{cases}
z = -d, \,\, -L < x < 0, \\
z = -h, \,\, x < 0.
\end{cases} \\

\partial_x\phi^s &= 0, \,\,\, x = 0, \,\, -h \leq z \leq -d, \\

\phi^a &= 0, \,\,\, x = 0, \,\, -h \leq z \leq -d.
\end{align}
</math></center>

Where <math>\alpha = \omega^2/g</math>.

<math>\phi^s(x,z)</math> is an even function and at <math>x = 0</math> the derivative of this function is equal to zero. <math>\phi^a(x,z)</math> is an odd function and at <math>x = 0</math> this function is equal to zero. These boundary conditions are important and enable the recomposed solution <math>\phi(x,z)</math> to be reflected into the domain of <math>x \geq 0</math>.

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\leq 0</math>
and <math>x\geq 0</math>.

{{separation of variables in two dimensions}}

{{separation of variables for a free surface}}

=== Separation of Variables for a Dock ===

The separation of variables equation for a partially submerged dock is given by:
<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>
Z^{\prime} (-d) = 0.
</math></center>
The solution is
<math>k=\kappa_{m}= \frac{m\pi}{h-d} \,</math> , <math>m\in\mathbb{N}\cup\left\{0\right\}</math> and
<center><math>
Z = \psi_{m}\left( z\right) = \cos\kappa_{m}(z+h),\quad
m\in\mathbb{N}\cup\left\{0\right\}.</math></center>
We note that
<center><math>
\int\nolimits_{-h}^{-d}\psi_{m}(z)\psi_{n}(z) \mathrm{d} z=C_{m}\delta_{mn},
</math></center>
where
<center>
<math>
C_{m} = \frac{1}{2} \left( \frac{\cos\kappa_{m}(h-d)\sin\kappa_{m}(h-d)+\kappa_{m}(h-d)}{\kappa_{m}} \right).
</math>
</center>
In addition there is the special case of when <math>m=n=0</math> ,
<center>
<math>
C_{0} = h-d,
</math>
</center>
where <math>\delta_{00}=1</math> .

=== Inner product between free surface and dock modes ===

<center><math>
B_{mn} = \int\nolimits_{-h}^{-d}\psi_{m}(z)\chi_{n}(z) \mathrm{d} z , \quad m\geq1, \quad n\geq0,
</math></center>
where
<center><math>
B_{mn}= \int\nolimits_{-h}^{-d} \frac{\cos\kappa_{m}(z+h)\cos k_{n}(z+h)}{\cos(k_{n}h)} \mathrm{d} z=\frac{-(-1)^m k_{n}\sin k_{n}(h-d)}
{\cos(k_{n}h)(\kappa_{m}^{2}-k_{n}^{2})}.
</math></center>
The evaluation of this inner product works on the assumption that the two roots <math>\kappa_{m}</math> and <math>k_{n}</math> do not coincide. The Matlab code given below has a routine to check for this case and make necessary adjustments.

We also have the case where the functions of the inner product are switched.
<center><math>
B_{nm} = \int\nolimits_{-h}^{-d}\chi_{m}(z)\psi_{n}(z) \mathrm{d} z , \quad m\geq0, \quad n\geq0.
</math></center>
For <math>m \geq 0</math> and <math>n \geq 1</math> the evaluation is above with the exception that the indices of the expression are switched.

Finally there is the special case of when <math>m \geq 0</math> and <math>n = 0</math>.
<center><math>
B_{0m} = \frac{\sin k_{m}(h-d)}{k_{m}\cos(k_{m}h)}.
</math></center>

=== Expansion of the potential ===

We need to apply some boundary conditions at plus and minus infinity,
where essentially 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 roots. Here we
must use a different condition, essentially identifying one solution
as the incoming wave and the other as the outgoing wave.

Therefore the diffracted potentials can
be expanded as
<center>
<math>
\phi(x,z)=\sum_{m=0}^{\infty}a_{m}e^{k_{m}x}\chi_{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 dock covered region. Therefore, the
potentials and their derivatives at <math>x=-L</math> have to be equal.
We obtain for the '''symmetric''' case:
<center><math>
\begin{align}
\chi_{0}\left( z\right) + \sum_{m=0}^{\infty}a_{m} \chi_{m}\left( z\right)
&=\sum_{m=0}^{\infty}b_{m}\psi_{m}(z), \quad -h \leq z\leq -d \\

-k_{0}\chi_{0}\left( z\right) +\sum_{m=0}^{\infty} a_{m}k_{m}\chi_{m}\left( z\right) &=
\begin{cases}
\quad \quad \quad 0,\quad \quad \quad -d\leq z\leq 0 \\
-\sum_{m=0}^{\infty}b_{m}\kappa_{m}\psi_{m}(z),\quad -h\leq z\leq -d
\end{cases}
\end{align}
</math></center>

For the first equation we multiply both sides by <math>\psi_{n}(z) \,</math> and integrating from <math>-h</math> to <math>-d</math> to obtain:
<center><math>
B_{n0} + \sum^{\infty}_{m=0}a_{m}B_{nm} = b_{m}C_{m}\delta_{mn}, n\in\mathbb{N}\cup\left\{0\right\}
</math></center>
and for the second equation we multiply both sides by <math>\chi_{n}(z) \,</math> and integrating from <math>-h</math> to <math>0</math> to obtain:
<center><math>
-k_{0}A_{0}\delta_{0n} + k_{m}a_{m}A_{m} = -\sum^{\infty}_{m=0}\kappa_{m}b_{m}B_{mn}, n\in\mathbb{N}\cup\left\{0\right\}
</math></center>

Solving the equations above will yield the coefficients of the water velocity potential in the dock covered region.

== Numerical Solution ==

To solve the system of equations for the '''symmetric''' case, we set the upper limit of <math>m</math> to be <math>N</math>. This resulting system can be expressed in the block matrix form below,

<center>
<math>
\begin{bmatrix}
B_{00} & B_{01} & \cdots & B_{0N} & -C_{0} & 0 & \cdots & 0 \\
B_{10} & B_{11} & \cdots & B_{1N} & 0 & -C_{1} & \cdots & \vdots \\
\vdots & \vdots & \ddots & \vdots & \vdots & \vdots & \ddots & 0 \\
B_{N0} & B_{N1} & \cdots & B_{NN} & 0 & \cdots & 0 & -C_{N} \\
k_{0}A_{0} & 0 & \cdots & 0 & 0 & \kappa_{1}\tanh(\kappa_{1}L)B_{10} & \cdots & \kappa_{N}\tanh(\kappa_{N}L)B_{N0} \\
0 & k_{1}A_{1} & \cdots & \vdots & 0 & \kappa_{1}\tanh(\kappa_{1}L)B_{11} & \cdots & \kappa_{N}\tanh(\kappa_{N}L)B_{N1} \\
\vdots & \vdots & \ddots & 0 & \vdots & \vdots & \ddots & \vdots \\
0 & \cdots & 0 & k_{N}A_{N} & 0 & \kappa_{1}\tanh(\kappa_{1}L)B_{1N} & \cdots & \kappa_{N}\tanh(\kappa_{N}L)B_{NN} \\
\end{bmatrix}
\begin{bmatrix}
a^s_{0} \\ a^s_{1} \\ \vdots \\ a^s_{N} \\ b^s_{0} \\ b^s_{1} \\ \vdots \\ b^s_{N} \\
\end{bmatrix}
=\begin{bmatrix}
-B_{00} \\ -B_{10} \\ \vdots \\ -B_{N0} \\ k_{0}A_{0} \\ 0 \\ \vdots \\ 0 \\
\end{bmatrix}
</math>
</center>

To solve the system of equations for the '''antisymmetric''' case, we set the upper limit of <math>m</math> to be <math>N</math>. This resulting system can be expressed in the block matrix form below,

<center>
<math>
\begin{bmatrix}
B_{00} & B_{01} & \cdots & B_{0N} & -C_{0} & 0 & \cdots & 0 \\
B_{10} & B_{11} & \cdots & B_{1N} & 0 & -C_{1} & \cdots & \vdots \\
\vdots & \vdots & \ddots & \vdots & \vdots & \vdots & \ddots & 0 \\
B_{N0} & B_{N1} & \cdots & B_{NN} & 0 & \cdots & 0 & -C_{N} \\
k_{0}A_{0} & 0 & \cdots & 0 & \frac{1}{L}D_{0} & \kappa_{1}\coth(\kappa_{1}L)B_{10} & \cdots & \kappa_{N}\coth(\kappa_{N}L)B_{N0} \\
0 & k_{1}A_{1} & \cdots & \vdots & \frac{1}{L}D_{1} & \kappa_{1}\coth(\kappa_{1}L)B_{11} & \cdots & \kappa_{N}\coth(\kappa_{N}L)B_{N1} \\
\vdots & \vdots & \ddots & 0 & \vdots & \vdots & \ddots & \vdots \\
0 & \cdots & 0 & k_{N}A_{N} & \frac{1}{L}D_{N} & \kappa_{1}\coth(\kappa_{1}L)B_{1N} & \cdots & \kappa_{N}\coth(\kappa_{N}L)B_{NN} \\
\end{bmatrix}
\begin{bmatrix}
a^a_{0} \\ a^a_{1} \\ \vdots \\ a^a_{N} \\ b^a_{0} \\ b^a_{1} \\ \vdots \\ b^a_{N} \\
\end{bmatrix}
=\begin{bmatrix}
-B_{00} \\ -B_{10} \\ \vdots \\ -B_{N0} \\ k_{0}A_{0} \\ 0 \\ \vdots \\ 0 \\
\end{bmatrix}
</math>
</center>

We then simply need to solve the two sets of <math>2(N+1) \mbox{ x } 2(N+1)</math> linear systems of equations. If we had not taken advantage of the symmetry we would need to solve a <math>4(N+1) \mbox{ x } 4(N+1)</math> linear system of equations for four coefficients <math>a_{m}</math>, <math>b_{m}</math>, <math>c_{m}</math> and <math>d_{m}</math>.

== Matlab Code ==

A program to calculate the coefficients for the single finite dock problem using symmetry 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 Finite Rectangle using Symmetry

2012-03-16T03:08:07Z

Tertius: /* Numerical Solution */

{{incomplete pages}}

== Introduction ==

The problem consists of a finite rigid fixed rectangle of length <math>2L</math> centered at <math>x = 0</math> and having a height of <math>d + ...</math> freeboard with depth of submergence being <math>d</math> through which no flow is possible. The regions to the left and right of the rectangle are the free water surface. Since the geometry of the rectangle is symmetric about <math>x = 0</math> we can decompose the velocity potential <math>\phi</math> into a symmetric <math>\phi^s</math> and antisymmetric <math>\phi^a</math> potential. We then solve for these two potentials in the left domain only <math>x \leq 0</math> and subsequently recompose the solution <math>\phi</math> for the entire domain. We begin with the simple problem when the waves are normally incident on the rectangle (dock) (so that the problem is truly two-dimensional). We can extend this solution to a pair of partially submerged finite rigid fixed rectangles using symmetry
[[Eigenfunction Matching for a Finite Rectangle using Symmetry]]

== Governing Equations ==

We begin with the [[Frequency Domain Problem]] for a partially submerged finite dock. Although the dock is of length <math>2L</math> we are only interested in the dock of <math>-L \leq x \leq 0</math> together with the fluid domain <math>x \leq 0</math>. We assume <math>e^{i\omega t}</math> time dependence and
the depth of submergence is <math>d</math>.
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>
\begin{align}
\Delta\phi^{s,a} &= 0, \,\,\, -h\leq z\leq 0, \,\, x < -L, \\

\Delta\phi^{s,a} &= 0, \,\,\, -h\leq z\leq -d, \,\, -L < x < 0. \\

\partial_z\phi^{s,a} &= \alpha\phi^{s,a}, \,\,\, z=0, \,\, x < -L, \\

\partial_x\phi^{s,a} &= 0, \,\,\, -d\leq z\leq 0, \,\, x = -L, \\

\partial_z\phi^{s,a} &= 0, \,\,\,
\begin{cases}
z = -d, \,\, -L < x < 0, \\
z = -h, \,\, x < 0.
\end{cases} \\

\partial_x\phi^s &= 0, \,\,\, x = 0, \,\, -h \leq z \leq -d, \\

\phi^a &= 0, \,\,\, x = 0, \,\, -h \leq z \leq -d.
\end{align}
</math></center>

Where <math>\alpha = \omega^2/g</math>.

<math>\phi^s(x,z)</math> is an even function and at <math>x = 0</math> the derivative of this function is equal to zero. <math>\phi^a(x,z)</math> is an odd function and at <math>x = 0</math> this function is equal to zero. These boundary conditions are important and enable the recomposed solution <math>\phi(x,z)</math> to be reflected into the domain of <math>x \geq 0</math>.

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\leq 0</math>
and <math>x\geq 0</math>.

{{separation of variables in two dimensions}}

{{separation of variables for a free surface}}

=== Separation of Variables for a Dock ===

The separation of variables equation for a partially submerged dock is given by:
<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>
Z^{\prime} (-d) = 0.
</math></center>
The solution is
<math>k=\kappa_{m}= \frac{m\pi}{h-d} \,</math> , <math>m\in\mathbb{N}\cup\left\{0\right\}</math> and
<center><math>
Z = \psi_{m}\left( z\right) = \cos\kappa_{m}(z+h),\quad
m\in\mathbb{N}\cup\left\{0\right\}.</math></center>
We note that
<center><math>
\int\nolimits_{-h}^{-d}\psi_{m}(z)\psi_{n}(z) \mathrm{d} z=C_{m}\delta_{mn},
</math></center>
where
<center>
<math>
C_{m} = \frac{1}{2} \left( \frac{\cos\kappa_{m}(h-d)\sin\kappa_{m}(h-d)+\kappa_{m}(h-d)}{\kappa_{m}} \right).
</math>
</center>
In addition there is the special case of when <math>m=n=0</math> ,
<center>
<math>
C_{0} = h-d,
</math>
</center>
where <math>\delta_{00}=1</math> .

=== Inner product between free surface and dock modes ===

<center><math>
B_{mn} = \int\nolimits_{-h}^{-d}\psi_{m}(z)\chi_{n}(z) \mathrm{d} z , \quad m\geq1, \quad n\geq0,
</math></center>
where
<center><math>
B_{mn}= \int\nolimits_{-h}^{-d} \frac{\cos\kappa_{m}(z+h)\cos k_{n}(z+h)}{\cos(k_{n}h)} \mathrm{d} z=\frac{-(-1)^m k_{n}\sin k_{n}(h-d)}
{\cos(k_{n}h)(\kappa_{m}^{2}-k_{n}^{2})}.
</math></center>
The evaluation of this inner product works on the assumption that the two roots <math>\kappa_{m}</math> and <math>k_{n}</math> do not coincide. The Matlab code given below has a routine to check for this case and make necessary adjustments.

We also have the case where the functions of the inner product are switched.
<center><math>
B_{nm} = \int\nolimits_{-h}^{-d}\chi_{m}(z)\psi_{n}(z) \mathrm{d} z , \quad m\geq0, \quad n\geq0.
</math></center>
For <math>m \geq 0</math> and <math>n \geq 1</math> the evaluation is above with the exception that the indices of the expression are switched.

Finally there is the special case of when <math>m \geq 0</math> and <math>n = 0</math>.
<center><math>
B_{0m} = \frac{\sin k_{m}(h-d)}{k_{m}\cos(k_{m}h)}.
</math></center>

=== Expansion of the potential ===

We need to apply some boundary conditions at plus and minus infinity,
where essentially 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 roots. Here we
must use a different condition, essentially identifying one solution
as the incoming wave and the other as the outgoing wave.

Therefore the diffracted potentials can
be expanded as
<center>
<math>
\phi(x,z)=\sum_{m=0}^{\infty}a_{m}e^{k_{m}x}\chi_{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 dock covered region. Therefore, the
potentials and their derivatives at <math>x=0</math> have to be equal.
We obtain:
<center><math>
\begin{align}
\chi_{0}\left( z\right) + \sum_{m=0}^{\infty}a_{m} \chi_{m}\left( z\right)
&=\sum_{m=0}^{\infty}b_{m}\psi_{m}(z), \quad -h \leq z\leq -d \\

-k_{0}\chi_{0}\left( z\right) +\sum_{m=0}^{\infty} a_{m}k_{m}\chi_{m}\left( z\right) &=
\begin{cases}
\quad \quad \quad 0,\quad \quad \quad -d\leq z\leq 0 \\
-\sum_{m=0}^{\infty}b_{m}\kappa_{m}\psi_{m}(z),\quad -h\leq z\leq -d
\end{cases}
\end{align}
</math></center>

For the first equation we multiply both sides by <math>\psi_{n}(z) \,</math> and integrating from <math>-h</math> to <math>-d</math> to obtain:
<center><math>
B_{n0} + \sum^{\infty}_{m=0}a_{m}B_{nm} = b_{m}C_{m}\delta_{mn}, n\in\mathbb{N}\cup\left\{0\right\}
</math></center>
and for the second equation we multiply both sides by <math>\chi_{n}(z) \,</math> and integrating from <math>-h</math> to <math>0</math> to obtain:
<center><math>
-k_{0}A_{0}\delta_{0n} + k_{m}a_{m}A_{m} = -\sum^{\infty}_{m=0}\kappa_{m}b_{m}B_{mn}, n\in\mathbb{N}\cup\left\{0\right\}
</math></center>

Solving the equations above will yield the coefficients of the water velocity potential in the dock covered region.

== Numerical Solution ==

To solve the system of equations for the '''symmetric''' case, we set the upper limit of <math>m</math> to be <math>N</math>. This resulting system can be expressed in the block matrix form below,

<center>
<math>
\begin{bmatrix}
B_{00} & B_{01} & \cdots & B_{0N} & -C_{0} & 0 & \cdots & 0 \\
B_{10} & B_{11} & \cdots & B_{1N} & 0 & -C_{1} & \cdots & \vdots \\
\vdots & \vdots & \ddots & \vdots & \vdots & \vdots & \ddots & 0 \\
B_{N0} & B_{N1} & \cdots & B_{NN} & 0 & \cdots & 0 & -C_{N} \\
k_{0}A_{0} & 0 & \cdots & 0 & 0 & \kappa_{1}\tanh(\kappa_{1}L)B_{10} & \cdots & \kappa_{N}\tanh(\kappa_{N}L)B_{N0} \\
0 & k_{1}A_{1} & \cdots & \vdots & 0 & \kappa_{1}\tanh(\kappa_{1}L)B_{11} & \cdots & \kappa_{N}\tanh(\kappa_{N}L)B_{N1} \\
\vdots & \vdots & \ddots & 0 & \vdots & \vdots & \ddots & \vdots \\
0 & \cdots & 0 & k_{N}A_{N} & 0 & \kappa_{1}\tanh(\kappa_{1}L)B_{1N} & \cdots & \kappa_{N}\tanh(\kappa_{N}L)B_{NN} \\
\end{bmatrix}
\begin{bmatrix}
a^s_{0} \\ a^s_{1} \\ \vdots \\ a^s_{N} \\ b^s_{0} \\ b^s_{1} \\ \vdots \\ b^s_{N} \\
\end{bmatrix}
=\begin{bmatrix}
-B_{00} \\ -B_{10} \\ \vdots \\ -B_{N0} \\ k_{0}A_{0} \\ 0 \\ \vdots \\ 0 \\
\end{bmatrix}
</math>
</center>

To solve the system of equations for the '''antisymmetric''' case, we set the upper limit of <math>m</math> to be <math>N</math>. This resulting system can be expressed in the block matrix form below,

<center>
<math>
\begin{bmatrix}
B_{00} & B_{01} & \cdots & B_{0N} & -C_{0} & 0 & \cdots & 0 \\
B_{10} & B_{11} & \cdots & B_{1N} & 0 & -C_{1} & \cdots & \vdots \\
\vdots & \vdots & \ddots & \vdots & \vdots & \vdots & \ddots & 0 \\
B_{N0} & B_{N1} & \cdots & B_{NN} & 0 & \cdots & 0 & -C_{N} \\
k_{0}A_{0} & 0 & \cdots & 0 & \frac{1}{L}D_{0} & \kappa_{1}\coth(\kappa_{1}L)B_{10} & \cdots & \kappa_{N}\coth(\kappa_{N}L)B_{N0} \\
0 & k_{1}A_{1} & \cdots & \vdots & \frac{1}{L}D_{1} & \kappa_{1}\coth(\kappa_{1}L)B_{11} & \cdots & \kappa_{N}\coth(\kappa_{N}L)B_{N1} \\
\vdots & \vdots & \ddots & 0 & \vdots & \vdots & \ddots & \vdots \\
0 & \cdots & 0 & k_{N}A_{N} & \frac{1}{L}D_{N} & \kappa_{1}\coth(\kappa_{1}L)B_{1N} & \cdots & \kappa_{N}\coth(\kappa_{N}L)B_{NN} \\
\end{bmatrix}
\begin{bmatrix}
a^a_{0} \\ a^a_{1} \\ \vdots \\ a^a_{N} \\ b^a_{0} \\ b^a_{1} \\ \vdots \\ b^a_{N} \\
\end{bmatrix}
=\begin{bmatrix}
-B_{00} \\ -B_{10} \\ \vdots \\ -B_{N0} \\ k_{0}A_{0} \\ 0 \\ \vdots \\ 0 \\
\end{bmatrix}
</math>
</center>

We then simply need to solve the two sets of <math>2(N+1) \mbox{ x } 2(N+1)</math> linear systems of equations. If we had not taken advantage of the symmetry we would need to solve a <math>4(N+1) \mbox{ x } 4(N+1)</math> linear system of equations for four coefficients <math>a_{m}</math>, <math>b_{m}</math>, <math>c_{m}</math> and <math>d_{m}</math>.

== Matlab Code ==

A program to calculate the coefficients for the single finite dock problem using symmetry 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 Finite Rectangle using Symmetry

2012-03-16T03:01:56Z

Tertius: /* Numerical Solution */

{{incomplete pages}}

== Introduction ==

The problem consists of a finite rigid fixed rectangle of length <math>2L</math> centered at <math>x = 0</math> and having a height of <math>d + ...</math> freeboard with depth of submergence being <math>d</math> through which no flow is possible. The regions to the left and right of the rectangle are the free water surface. Since the geometry of the rectangle is symmetric about <math>x = 0</math> we can decompose the velocity potential <math>\phi</math> into a symmetric <math>\phi^s</math> and antisymmetric <math>\phi^a</math> potential. We then solve for these two potentials in the left domain only <math>x \leq 0</math> and subsequently recompose the solution <math>\phi</math> for the entire domain. We begin with the simple problem when the waves are normally incident on the rectangle (dock) (so that the problem is truly two-dimensional). We can extend this solution to a pair of partially submerged finite rigid fixed rectangles using symmetry
[[Eigenfunction Matching for a Finite Rectangle using Symmetry]]

== Governing Equations ==

We begin with the [[Frequency Domain Problem]] for a partially submerged finite dock. Although the dock is of length <math>2L</math> we are only interested in the dock of <math>-L \leq x \leq 0</math> together with the fluid domain <math>x \leq 0</math>. We assume <math>e^{i\omega t}</math> time dependence and
the depth of submergence is <math>d</math>.
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>
\begin{align}
\Delta\phi^{s,a} &= 0, \,\,\, -h\leq z\leq 0, \,\, x < -L, \\

\Delta\phi^{s,a} &= 0, \,\,\, -h\leq z\leq -d, \,\, -L < x < 0. \\

\partial_z\phi^{s,a} &= \alpha\phi^{s,a}, \,\,\, z=0, \,\, x < -L, \\

\partial_x\phi^{s,a} &= 0, \,\,\, -d\leq z\leq 0, \,\, x = -L, \\

\partial_z\phi^{s,a} &= 0, \,\,\,
\begin{cases}
z = -d, \,\, -L < x < 0, \\
z = -h, \,\, x < 0.
\end{cases} \\

\partial_x\phi^s &= 0, \,\,\, x = 0, \,\, -h \leq z \leq -d, \\

\phi^a &= 0, \,\,\, x = 0, \,\, -h \leq z \leq -d.
\end{align}
</math></center>

Where <math>\alpha = \omega^2/g</math>.

<math>\phi^s(x,z)</math> is an even function and at <math>x = 0</math> the derivative of this function is equal to zero. <math>\phi^a(x,z)</math> is an odd function and at <math>x = 0</math> this function is equal to zero. These boundary conditions are important and enable the recomposed solution <math>\phi(x,z)</math> to be reflected into the domain of <math>x \geq 0</math>.

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\leq 0</math>
and <math>x\geq 0</math>.

{{separation of variables in two dimensions}}

{{separation of variables for a free surface}}

=== Separation of Variables for a Dock ===

The separation of variables equation for a partially submerged dock is given by:
<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>
Z^{\prime} (-d) = 0.
</math></center>
The solution is
<math>k=\kappa_{m}= \frac{m\pi}{h-d} \,</math> , <math>m\in\mathbb{N}\cup\left\{0\right\}</math> and
<center><math>
Z = \psi_{m}\left( z\right) = \cos\kappa_{m}(z+h),\quad
m\in\mathbb{N}\cup\left\{0\right\}.</math></center>
We note that
<center><math>
\int\nolimits_{-h}^{-d}\psi_{m}(z)\psi_{n}(z) \mathrm{d} z=C_{m}\delta_{mn},
</math></center>
where
<center>
<math>
C_{m} = \frac{1}{2} \left( \frac{\cos\kappa_{m}(h-d)\sin\kappa_{m}(h-d)+\kappa_{m}(h-d)}{\kappa_{m}} \right).
</math>
</center>
In addition there is the special case of when <math>m=n=0</math> ,
<center>
<math>
C_{0} = h-d,
</math>
</center>
where <math>\delta_{00}=1</math> .

=== Inner product between free surface and dock modes ===

<center><math>
B_{mn} = \int\nolimits_{-h}^{-d}\psi_{m}(z)\chi_{n}(z) \mathrm{d} z , \quad m\geq1, \quad n\geq0,
</math></center>
where
<center><math>
B_{mn}= \int\nolimits_{-h}^{-d} \frac{\cos\kappa_{m}(z+h)\cos k_{n}(z+h)}{\cos(k_{n}h)} \mathrm{d} z=\frac{-(-1)^m k_{n}\sin k_{n}(h-d)}
{\cos(k_{n}h)(\kappa_{m}^{2}-k_{n}^{2})}.
</math></center>
The evaluation of this inner product works on the assumption that the two roots <math>\kappa_{m}</math> and <math>k_{n}</math> do not coincide. The Matlab code given below has a routine to check for this case and make necessary adjustments.

We also have the case where the functions of the inner product are switched.
<center><math>
B_{nm} = \int\nolimits_{-h}^{-d}\chi_{m}(z)\psi_{n}(z) \mathrm{d} z , \quad m\geq0, \quad n\geq0.
</math></center>
For <math>m \geq 0</math> and <math>n \geq 1</math> the evaluation is above with the exception that the indices of the expression are switched.

Finally there is the special case of when <math>m \geq 0</math> and <math>n = 0</math>.
<center><math>
B_{0m} = \frac{\sin k_{m}(h-d)}{k_{m}\cos(k_{m}h)}.
</math></center>

=== Expansion of the potential ===

We need to apply some boundary conditions at plus and minus infinity,
where essentially 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 roots. Here we
must use a different condition, essentially identifying one solution
as the incoming wave and the other as the outgoing wave.

Therefore the diffracted potentials can
be expanded as
<center>
<math>
\phi(x,z)=\sum_{m=0}^{\infty}a_{m}e^{k_{m}x}\chi_{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 dock covered region. Therefore, the
potentials and their derivatives at <math>x=0</math> have to be equal.
We obtain:
<center><math>
\begin{align}
\chi_{0}\left( z\right) + \sum_{m=0}^{\infty}a_{m} \chi_{m}\left( z\right)
&=\sum_{m=0}^{\infty}b_{m}\psi_{m}(z), \quad -h \leq z\leq -d \\

-k_{0}\chi_{0}\left( z\right) +\sum_{m=0}^{\infty} a_{m}k_{m}\chi_{m}\left( z\right) &=
\begin{cases}
\quad \quad \quad 0,\quad \quad \quad -d\leq z\leq 0 \\
-\sum_{m=0}^{\infty}b_{m}\kappa_{m}\psi_{m}(z),\quad -h\leq z\leq -d
\end{cases}
\end{align}
</math></center>

For the first equation we multiply both sides by <math>\psi_{n}(z) \,</math> and integrating from <math>-h</math> to <math>-d</math> to obtain:
<center><math>
B_{n0} + \sum^{\infty}_{m=0}a_{m}B_{nm} = b_{m}C_{m}\delta_{mn}, n\in\mathbb{N}\cup\left\{0\right\}
</math></center>
and for the second equation we multiply both sides by <math>\chi_{n}(z) \,</math> and integrating from <math>-h</math> to <math>0</math> to obtain:
<center><math>
-k_{0}A_{0}\delta_{0n} + k_{m}a_{m}A_{m} = -\sum^{\infty}_{m=0}\kappa_{m}b_{m}B_{mn}, n\in\mathbb{N}\cup\left\{0\right\}
</math></center>

Solving the equations above will yield the coefficients of the water velocity potential in the dock covered region.

== Numerical Solution ==

To solve the system of equations for the '''symmetric''' case, we set the upper limit of <math>m</math> to be <math>N</math>. This resulting system can be expressed in the block matrix form below,

<center>
<math>
\begin{bmatrix}
B_{00} & B_{01} & \cdots & B_{0N} & -C_{0} & 0 & \cdots & 0 \\
B_{10} & B_{11} & \cdots & B_{1N} & 0 & -C_{1} & \cdots & \vdots \\
\vdots & \vdots & \ddots & \vdots & \vdots & \vdots & \ddots & 0 \\
B_{N0} & B_{N1} & \cdots & B_{NN} & 0 & \cdots & 0 & -C_{N} \\
k_{0}A_{0} & 0 & \cdots & 0 & 0 & \kappa_{1}\tanh(\kappa_{1}L)B_{10} & \cdots & \kappa_{N}\tanh(\kappa_{N}L)B_{N0} \\
0 & k_{1}A_{1} & \cdots & \vdots & 0 & \kappa_{1}\tanh(\kappa_{1}L)B_{11} & \cdots & \kappa_{N}\tanh(\kappa_{N}L)B_{N1} \\
\vdots & \vdots & \ddots & 0 & \vdots & \vdots & \ddots & \vdots \\
0 & \cdots & 0 & k_{N}A_{N} & 0 & \kappa_{1}\tanh(\kappa_{1}L)B_{1N} & \cdots & \kappa_{N}\tanh(\kappa_{N}L)B_{NN} \\
\end{bmatrix}
\begin{bmatrix}
a^s_{0} \\ a^s_{1} \\ \vdots \\ a^s_{N} \\ b^s_{0} \\ b^s_{1} \\ \vdots \\ b^s_{N} \\
\end{bmatrix}
=\begin{bmatrix}
-B_{00} \\ -B_{10} \\ \vdots \\ -B_{N0} \\ k_{0}A_{0} \\ 0 \\ \vdots \\ 0 \\
\end{bmatrix}
</math>
</center>

To solve the system of equations for the '''antisymmetric''' case, we set the upper limit of <math>m</math> to be <math>N</math>. This resulting system can be expressed in the block matrix form below,

<center>
<math>
\begin{bmatrix}
B_{00} & B_{01} & \cdots & B_{0N} & -C_{0} & 0 & \cdots & 0 \\
B_{10} & B_{11} & \cdots & B_{1N} & 0 & -C_{1} & \cdots & \vdots \\
\vdots & \vdots & \ddots & \vdots & \vdots & \vdots & \ddots & 0 \\
B_{N0} & B_{N1} & \cdots & B_{NN} & 0 & \cdots & 0 & -C_{N} \\
k_{0}A_{0} & 0 & \cdots & 0 & \frac{1}{L}D_{0} & \kappa_{1}\coth(\kappa_{1}L)B_{10} & \cdots & \kappa_{N}\coth(\kappa_{N}L)B_{N0} \\
0 & k_{1}A_{1} & \cdots & \vdots & \frac{1}{L}D_{1} & \kappa_{1}\coth(\kappa_{1}L)B_{11} & \cdots & \kappa_{N}\coth(\kappa_{N}L)B_{N1} \\
\vdots & \vdots & \ddots & 0 & \vdots & \vdots & \ddots & \vdots \\
0 & \cdots & 0 & k_{N}A_{N} & \frac{1}{L}D_{N} & \kappa_{1}\coth(\kappa_{1}L)B_{1N} & \cdots & \kappa_{N}\coth(\kappa_{N}L)B_{NN} \\
\end{bmatrix}
\begin{bmatrix}
a^a_{0} \\ a^a_{1} \\ \vdots \\ a^a_{N} \\ b^a_{0} \\ b^a_{1} \\ \vdots \\ b^a_{N} \\
\end{bmatrix}
=\begin{bmatrix}
-B_{00} \\ -B_{10} \\ \vdots \\ -B_{N0} \\ k_{0}A_{0} \\ 0 \\ \vdots \\ 0 \\
\end{bmatrix}
</math>
</center>

We then simply need to solve the two sets of <math>2(N+1) \mbox{ x } 2(N+1)</math> linear systems of equations. If we had not taken advantage of the symmetry we would need to solve a <math>4(N+1) \mbox{ x } 4(N+1)</math> linear system of equations for four coefficients <math>a_{m}</math>, <math>b_{m}</math>, <math>c_{m}</math>, <math>d_{m}</math>.

== Matlab Code ==

A program to calculate the coefficients for the single finite dock problem using symmetry 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 Finite Rectangle using Symmetry

2012-03-16T02:00:19Z

Tertius: /* Numerical Solution */

{{incomplete pages}}

== Introduction ==

The problem consists of a finite rigid fixed rectangle of length <math>2L</math> centered at <math>x = 0</math> and having a height of <math>d + ...</math> freeboard with depth of submergence being <math>d</math> through which no flow is possible. The regions to the left and right of the rectangle are the free water surface. Since the geometry of the rectangle is symmetric about <math>x = 0</math> we can decompose the velocity potential <math>\phi</math> into a symmetric <math>\phi^s</math> and antisymmetric <math>\phi^a</math> potential. We then solve for these two potentials in the left domain only <math>x \leq 0</math> and subsequently recompose the solution <math>\phi</math> for the entire domain. We begin with the simple problem when the waves are normally incident on the rectangle (dock) (so that the problem is truly two-dimensional). We can extend this solution to a pair of partially submerged finite rigid fixed rectangles using symmetry
[[Eigenfunction Matching for a Finite Rectangle using Symmetry]]

== Governing Equations ==

We begin with the [[Frequency Domain Problem]] for a partially submerged finite dock. Although the dock is of length <math>2L</math> we are only interested in the dock of <math>-L \leq x \leq 0</math> together with the fluid domain <math>x \leq 0</math>. We assume <math>e^{i\omega t}</math> time dependence and
the depth of submergence is <math>d</math>.
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>
\begin{align}
\Delta\phi^{s,a} &= 0, \,\,\, -h\leq z\leq 0, \,\, x < -L, \\

\Delta\phi^{s,a} &= 0, \,\,\, -h\leq z\leq -d, \,\, -L < x < 0. \\

\partial_z\phi^{s,a} &= \alpha\phi^{s,a}, \,\,\, z=0, \,\, x < -L, \\

\partial_x\phi^{s,a} &= 0, \,\,\, -d\leq z\leq 0, \,\, x = -L, \\

\partial_z\phi^{s,a} &= 0, \,\,\,
\begin{cases}
z = -d, \,\, -L < x < 0, \\
z = -h, \,\, x < 0.
\end{cases} \\

\partial_x\phi^s &= 0, \,\,\, x = 0, \,\, -h \leq z \leq -d, \\

\phi^a &= 0, \,\,\, x = 0, \,\, -h \leq z \leq -d.
\end{align}
</math></center>

Where <math>\alpha = \omega^2/g</math>.

<math>\phi^s(x,z)</math> is an even function and at <math>x = 0</math> the derivative of this function is equal to zero. <math>\phi^a(x,z)</math> is an odd function and at <math>x = 0</math> this function is equal to zero. These boundary conditions are important and enable the recomposed solution <math>\phi(x,z)</math> to be reflected into the domain of <math>x \geq 0</math>.

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\leq 0</math>
and <math>x\geq 0</math>.

{{separation of variables in two dimensions}}

{{separation of variables for a free surface}}

=== Separation of Variables for a Dock ===

The separation of variables equation for a partially submerged dock is given by:
<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>
Z^{\prime} (-d) = 0.
</math></center>
The solution is
<math>k=\kappa_{m}= \frac{m\pi}{h-d} \,</math> , <math>m\in\mathbb{N}\cup\left\{0\right\}</math> and
<center><math>
Z = \psi_{m}\left( z\right) = \cos\kappa_{m}(z+h),\quad
m\in\mathbb{N}\cup\left\{0\right\}.</math></center>
We note that
<center><math>
\int\nolimits_{-h}^{-d}\psi_{m}(z)\psi_{n}(z) \mathrm{d} z=C_{m}\delta_{mn},
</math></center>
where
<center>
<math>
C_{m} = \frac{1}{2} \left( \frac{\cos\kappa_{m}(h-d)\sin\kappa_{m}(h-d)+\kappa_{m}(h-d)}{\kappa_{m}} \right).
</math>
</center>
In addition there is the special case of when <math>m=n=0</math> ,
<center>
<math>
C_{0} = h-d,
</math>
</center>
where <math>\delta_{00}=1</math> .

=== Inner product between free surface and dock modes ===

<center><math>
B_{mn} = \int\nolimits_{-h}^{-d}\psi_{m}(z)\chi_{n}(z) \mathrm{d} z , \quad m\geq1, \quad n\geq0,
</math></center>
where
<center><math>
B_{mn}= \int\nolimits_{-h}^{-d} \frac{\cos\kappa_{m}(z+h)\cos k_{n}(z+h)}{\cos(k_{n}h)} \mathrm{d} z=\frac{-(-1)^m k_{n}\sin k_{n}(h-d)}
{\cos(k_{n}h)(\kappa_{m}^{2}-k_{n}^{2})}.
</math></center>
The evaluation of this inner product works on the assumption that the two roots <math>\kappa_{m}</math> and <math>k_{n}</math> do not coincide. The Matlab code given below has a routine to check for this case and make necessary adjustments.

We also have the case where the functions of the inner product are switched.
<center><math>
B_{nm} = \int\nolimits_{-h}^{-d}\chi_{m}(z)\psi_{n}(z) \mathrm{d} z , \quad m\geq0, \quad n\geq0.
</math></center>
For <math>m \geq 0</math> and <math>n \geq 1</math> the evaluation is above with the exception that the indices of the expression are switched.

Finally there is the special case of when <math>m \geq 0</math> and <math>n = 0</math>.
<center><math>
B_{0m} = \frac{\sin k_{m}(h-d)}{k_{m}\cos(k_{m}h)}.
</math></center>

=== Expansion of the potential ===

We need to apply some boundary conditions at plus and minus infinity,
where essentially 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 roots. Here we
must use a different condition, essentially identifying one solution
as the incoming wave and the other as the outgoing wave.

Therefore the diffracted potentials can
be expanded as
<center>
<math>
\phi(x,z)=\sum_{m=0}^{\infty}a_{m}e^{k_{m}x}\chi_{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 dock covered region. Therefore, the
potentials and their derivatives at <math>x=0</math> have to be equal.
We obtain:
<center><math>
\begin{align}
\chi_{0}\left( z\right) + \sum_{m=0}^{\infty}a_{m} \chi_{m}\left( z\right)
&=\sum_{m=0}^{\infty}b_{m}\psi_{m}(z), \quad -h \leq z\leq -d \\

-k_{0}\chi_{0}\left( z\right) +\sum_{m=0}^{\infty} a_{m}k_{m}\chi_{m}\left( z\right) &=
\begin{cases}
\quad \quad \quad 0,\quad \quad \quad -d\leq z\leq 0 \\
-\sum_{m=0}^{\infty}b_{m}\kappa_{m}\psi_{m}(z),\quad -h\leq z\leq -d
\end{cases}
\end{align}
</math></center>

For the first equation we multiply both sides by <math>\psi_{n}(z) \,</math> and integrating from <math>-h</math> to <math>-d</math> to obtain:
<center><math>
B_{n0} + \sum^{\infty}_{m=0}a_{m}B_{nm} = b_{m}C_{m}\delta_{mn}, n\in\mathbb{N}\cup\left\{0\right\}
</math></center>
and for the second equation we multiply both sides by <math>\chi_{n}(z) \,</math> and integrating from <math>-h</math> to <math>0</math> to obtain:
<center><math>
-k_{0}A_{0}\delta_{0n} + k_{m}a_{m}A_{m} = -\sum^{\infty}_{m=0}\kappa_{m}b_{m}B_{mn}, n\in\mathbb{N}\cup\left\{0\right\}
</math></center>

Solving the equations above will yield the coefficients of the water velocity potential in the dock covered region.

== Numerical Solution ==

To solve the system of equations for the '''symmetric''' case, we set the upper limit of <math>m</math> to be <math>N</math>. This resulting system can be expressed in the block matrix form below,

<center>
<math>
\begin{bmatrix}
B_{00} & B_{01} & \cdots & B_{0N} & -C_{0} & 0 & \cdots & 0 \\
B_{10} & B_{11} & \cdots & B_{1N} & 0 & -C_{1} & \cdots & \vdots \\
\vdots & \vdots & \ddots & \vdots & \vdots & \vdots & \ddots & 0 \\
B_{N0} & B_{N1} & \cdots & B_{NN} & 0 & \cdots & 0 & -C_{N} \\
k_{0}A_{0} & 0 & \cdots & 0 & 0 & \kappa_{1}\tanh(\kappa_{1}L)B_{10} & \cdots & \kappa_{N}\tanh(\kappa_{N}L)B_{N0} \\
0 & k_{1}A_{1} & \cdots & \vdots & 0 & \kappa_{1}\tanh(\kappa_{1}L)B_{11} & \cdots & \kappa_{N}\tanh(\kappa_{N}L)B_{N1} \\
\vdots & \vdots & \ddots & 0 & \vdots & \vdots & \ddots & \vdots \\
0 & \cdots & 0 & k_{N}A_{N} & 0 & \kappa_{1}\tanh(\kappa_{1}L)B_{1N} & \cdots & \kappa_{N}\tanh(\kappa_{N}L)B_{NN} \\
\end{bmatrix}
\begin{bmatrix}
a^s_{0} \\ a^s_{1} \\ \vdots \\ a^s_{N} \\ b^s_{0} \\ b^s_{1} \\ \vdots \\ b^s_{N} \\
\end{bmatrix}
=\begin{bmatrix}
-B_{00} \\ -B_{10} \\ \vdots \\ -B_{N0} \\ k_{0}A_{0} \\ 0 \\ \vdots \\ 0 \\
\end{bmatrix}
</math>
</center>

To solve the system of equations for the '''antisymmetric''' case, we set the upper limit of <math>m</math> to be <math>N</math>. This resulting system can be expressed in the block matrix form below,

<center>
<math>
\begin{bmatrix}
B_{00} & B_{01} & \cdots & B_{0N} & -C_{0} & 0 & \cdots & 0 \\
B_{10} & B_{11} & \cdots & B_{1N} & 0 & -C_{1} & \cdots & \vdots \\
\vdots & \vdots & \ddots & \vdots & \vdots & \vdots & \ddots & 0 \\
B_{N0} & B_{N1} & \cdots & B_{NN} & 0 & \cdots & 0 & -C_{N} \\
k_{0}A_{0} & 0 & \cdots & 0 & \frac{1}{L}D_{0} & \kappa_{1}\coth(\kappa_{1}L)B_{10} & \cdots & \kappa_{N}\coth(\kappa_{N}L)B_{N0} \\
0 & k_{1}A_{1} & \cdots & \vdots & \frac{1}{L}D_{1} & \kappa_{1}\coth(\kappa_{1}L)B_{11} & \cdots & \kappa_{N}\coth(\kappa_{N}L)B_{N1} \\
\vdots & \vdots & \ddots & 0 & \vdots & \vdots & \ddots & \vdots \\
0 & \cdots & 0 & k_{N}A_{N} & \frac{1}{L}D_{N} & \kappa_{1}\coth(\kappa_{1}L)B_{1N} & \cdots & \kappa_{N}\coth(\kappa_{N}L)B_{NN} \\
\end{bmatrix}
\begin{bmatrix}
a^a_{0} \\ a^a_{1} \\ \vdots \\ a^a_{N} \\ b^a_{0} \\ b^a_{1} \\ \vdots \\ b^a_{N} \\
\end{bmatrix}
=\begin{bmatrix}
-B_{00} \\ -B_{10} \\ \vdots \\ -B_{N0} \\ k_{0}A_{0} \\ 0 \\ \vdots \\ 0 \\
\end{bmatrix}
</math>
</center>

We then simply need to solve the two sets of <math>2(N+1) \mbox{ x } 2(N+1)</math> linear system of equations.

== Matlab Code ==

A program to calculate the coefficients for the single finite dock problem using symmetry 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 Finite Rectangle using Symmetry

2012-03-16T01:58:48Z

Tertius: /* Numerical Solution */

{{incomplete pages}}

== Introduction ==

The problem consists of a finite rigid fixed rectangle of length <math>2L</math> centered at <math>x = 0</math> and having a height of <math>d + ...</math> freeboard with depth of submergence being <math>d</math> through which no flow is possible. The regions to the left and right of the rectangle are the free water surface. Since the geometry of the rectangle is symmetric about <math>x = 0</math> we can decompose the velocity potential <math>\phi</math> into a symmetric <math>\phi^s</math> and antisymmetric <math>\phi^a</math> potential. We then solve for these two potentials in the left domain only <math>x \leq 0</math> and subsequently recompose the solution <math>\phi</math> for the entire domain. We begin with the simple problem when the waves are normally incident on the rectangle (dock) (so that the problem is truly two-dimensional). We can extend this solution to a pair of partially submerged finite rigid fixed rectangles using symmetry
[[Eigenfunction Matching for a Finite Rectangle using Symmetry]]

== Governing Equations ==

We begin with the [[Frequency Domain Problem]] for a partially submerged finite dock. Although the dock is of length <math>2L</math> we are only interested in the dock of <math>-L \leq x \leq 0</math> together with the fluid domain <math>x \leq 0</math>. We assume <math>e^{i\omega t}</math> time dependence and
the depth of submergence is <math>d</math>.
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>
\begin{align}
\Delta\phi^{s,a} &= 0, \,\,\, -h\leq z\leq 0, \,\, x < -L, \\

\Delta\phi^{s,a} &= 0, \,\,\, -h\leq z\leq -d, \,\, -L < x < 0. \\

\partial_z\phi^{s,a} &= \alpha\phi^{s,a}, \,\,\, z=0, \,\, x < -L, \\

\partial_x\phi^{s,a} &= 0, \,\,\, -d\leq z\leq 0, \,\, x = -L, \\

\partial_z\phi^{s,a} &= 0, \,\,\,
\begin{cases}
z = -d, \,\, -L < x < 0, \\
z = -h, \,\, x < 0.
\end{cases} \\

\partial_x\phi^s &= 0, \,\,\, x = 0, \,\, -h \leq z \leq -d, \\

\phi^a &= 0, \,\,\, x = 0, \,\, -h \leq z \leq -d.
\end{align}
</math></center>

Where <math>\alpha = \omega^2/g</math>.

<math>\phi^s(x,z)</math> is an even function and at <math>x = 0</math> the derivative of this function is equal to zero. <math>\phi^a(x,z)</math> is an odd function and at <math>x = 0</math> this function is equal to zero. These boundary conditions are important and enable the recomposed solution <math>\phi(x,z)</math> to be reflected into the domain of <math>x \geq 0</math>.

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\leq 0</math>
and <math>x\geq 0</math>.

{{separation of variables in two dimensions}}

{{separation of variables for a free surface}}

=== Separation of Variables for a Dock ===

The separation of variables equation for a partially submerged dock is given by:
<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>
Z^{\prime} (-d) = 0.
</math></center>
The solution is
<math>k=\kappa_{m}= \frac{m\pi}{h-d} \,</math> , <math>m\in\mathbb{N}\cup\left\{0\right\}</math> and
<center><math>
Z = \psi_{m}\left( z\right) = \cos\kappa_{m}(z+h),\quad
m\in\mathbb{N}\cup\left\{0\right\}.</math></center>
We note that
<center><math>
\int\nolimits_{-h}^{-d}\psi_{m}(z)\psi_{n}(z) \mathrm{d} z=C_{m}\delta_{mn},
</math></center>
where
<center>
<math>
C_{m} = \frac{1}{2} \left( \frac{\cos\kappa_{m}(h-d)\sin\kappa_{m}(h-d)+\kappa_{m}(h-d)}{\kappa_{m}} \right).
</math>
</center>
In addition there is the special case of when <math>m=n=0</math> ,
<center>
<math>
C_{0} = h-d,
</math>
</center>
where <math>\delta_{00}=1</math> .

=== Inner product between free surface and dock modes ===

<center><math>
B_{mn} = \int\nolimits_{-h}^{-d}\psi_{m}(z)\chi_{n}(z) \mathrm{d} z , \quad m\geq1, \quad n\geq0,
</math></center>
where
<center><math>
B_{mn}= \int\nolimits_{-h}^{-d} \frac{\cos\kappa_{m}(z+h)\cos k_{n}(z+h)}{\cos(k_{n}h)} \mathrm{d} z=\frac{-(-1)^m k_{n}\sin k_{n}(h-d)}
{\cos(k_{n}h)(\kappa_{m}^{2}-k_{n}^{2})}.
</math></center>
The evaluation of this inner product works on the assumption that the two roots <math>\kappa_{m}</math> and <math>k_{n}</math> do not coincide. The Matlab code given below has a routine to check for this case and make necessary adjustments.

We also have the case where the functions of the inner product are switched.
<center><math>
B_{nm} = \int\nolimits_{-h}^{-d}\chi_{m}(z)\psi_{n}(z) \mathrm{d} z , \quad m\geq0, \quad n\geq0.
</math></center>
For <math>m \geq 0</math> and <math>n \geq 1</math> the evaluation is above with the exception that the indices of the expression are switched.

Finally there is the special case of when <math>m \geq 0</math> and <math>n = 0</math>.
<center><math>
B_{0m} = \frac{\sin k_{m}(h-d)}{k_{m}\cos(k_{m}h)}.
</math></center>

=== Expansion of the potential ===

We need to apply some boundary conditions at plus and minus infinity,
where essentially 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 roots. Here we
must use a different condition, essentially identifying one solution
as the incoming wave and the other as the outgoing wave.

Therefore the diffracted potentials can
be expanded as
<center>
<math>
\phi(x,z)=\sum_{m=0}^{\infty}a_{m}e^{k_{m}x}\chi_{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 dock covered region. Therefore, the
potentials and their derivatives at <math>x=0</math> have to be equal.
We obtain:
<center><math>
\begin{align}
\chi_{0}\left( z\right) + \sum_{m=0}^{\infty}a_{m} \chi_{m}\left( z\right)
&=\sum_{m=0}^{\infty}b_{m}\psi_{m}(z), \quad -h \leq z\leq -d \\

-k_{0}\chi_{0}\left( z\right) +\sum_{m=0}^{\infty} a_{m}k_{m}\chi_{m}\left( z\right) &=
\begin{cases}
\quad \quad \quad 0,\quad \quad \quad -d\leq z\leq 0 \\
-\sum_{m=0}^{\infty}b_{m}\kappa_{m}\psi_{m}(z),\quad -h\leq z\leq -d
\end{cases}
\end{align}
</math></center>

For the first equation we multiply both sides by <math>\psi_{n}(z) \,</math> and integrating from <math>-h</math> to <math>-d</math> to obtain:
<center><math>
B_{n0} + \sum^{\infty}_{m=0}a_{m}B_{nm} = b_{m}C_{m}\delta_{mn}, n\in\mathbb{N}\cup\left\{0\right\}
</math></center>
and for the second equation we multiply both sides by <math>\chi_{n}(z) \,</math> and integrating from <math>-h</math> to <math>0</math> to obtain:
<center><math>
-k_{0}A_{0}\delta_{0n} + k_{m}a_{m}A_{m} = -\sum^{\infty}_{m=0}\kappa_{m}b_{m}B_{mn}, n\in\mathbb{N}\cup\left\{0\right\}
</math></center>

Solving the equations above will yield the coefficients of the water velocity potential in the dock covered region.

== Numerical Solution ==

To solve the system of equations for the '''symmetric''' case, we set the upper limit of <math>m</math> to be <math>N</math>. This resulting system can be expressed in the block matrix form below,

<center>
<math>
\begin{bmatrix}
B_{00} & B_{01} & \cdots & B_{0N} & -C_{0} & 0 & \cdots & 0 \\
B_{10} & B_{11} & \cdots & B_{1N} & 0 & -C_{1} & \cdots & \vdots \\
\vdots & \vdots & \ddots & \vdots & \vdots & \vdots & \ddots & 0 \\
B_{N0} & B_{N1} & \cdots & B_{NN} & 0 & \cdots & 0 & -C_{N} \\
k_{0}A_{0} & 0 & \cdots & 0 & 0 & \kappa_{1}\tanh(\kappa_{1}L)B_{10} & \cdots & \kappa_{N}\tanh(\kappa_{N}L)B_{N0} \\
0 & k_{1}A_{1} & \cdots & \vdots & 0 & \kappa_{1}\tanh(\kappa_{1}L)B_{11} & \cdots & \kappa_{N}\tanh(\kappa_{N}L)B_{N1} \\
\vdots & \vdots & \ddots & 0 & \vdots & \vdots & \ddots & \vdots \\
0 & \cdots & 0 & k_{N}A_{N} & 0 & \kappa_{1}\tanh(\kappa_{1}L)B_{1N} & \cdots & \kappa_{N}\tanh(\kappa_{N}L)B_{NN} \\
\end{bmatrix}
\begin{bmatrix}
a^s_{0} \\ a^s_{1} \\ \vdots \\ a^s_{N} \\ b^s_{0} \\ b^s_{1} \\ \vdots \\ b^s_{N} \\
\end{bmatrix}
=\begin{bmatrix}
-B_{00} \\ -B_{10} \\ \vdots \\ -B_{N0} \\ k_{0}A_{0} \\ 0 \\ \vdots \\ 0 \\
\end{bmatrix}
</math>
</center>

To solve the system of equations for the '''antisymmetric''' case, we set the upper limit of <math>m</math> to be <math>N</math>. This resulting system can be expressed in the block matrix form below,

<center>
<math>
\begin{bmatrix}
B_{00} & B_{01} & \cdots & B_{0N} & -C_{0} & 0 & \cdots & 0 \\
B_{10} & B_{11} & \cdots & B_{1N} & 0 & -C_{1} & \cdots & \vdots \\
\vdots & \vdots & \ddots & \vdots & \vdots & \vdots & \ddots & 0 \\
B_{N0} & B_{N1} & \cdots & B_{NN} & 0 & \cdots & 0 & -C_{N} \\
k_{0}A_{0} & 0 & \cdots & 0 & \frac{1}{L}D_{0} & \kappa_{1}\coth(\kappa_{1}L)B_{10} & \cdots & \kappa_{N}\coth(\kappa_{N}L)B_{N0} \\
0 & k_{1}A_{1} & \cdots & \vdots & \frac{1}{L}D_{1} & \kappa_{1}\coth(\kappa_{1}L)B_{11} & \cdots & \kappa_{N}\coth(\kappa_{N}L)B_{N1} \\
\vdots & \vdots & \ddots & 0 & \vdots & \vdots & \ddots & \vdots \\
0 & \cdots & 0 & k_{N}A_{N} & \frac{1}{L}D_{N} & \kappa_{1}\coth(\kappa_{1}L)B_{1N} & \cdots & \kappa_{N}\coth(\kappa_{N}L)B_{NN} \\
\end{bmatrix}
\begin{bmatrix}
a^a_{0} \\ a^a_{1} \\ \vdots \\ a^a_{N} \\ b^a_{0} \\ b^a_{1} \\ \vdots \\ b^a_{N} \\
\end{bmatrix}
=\begin{bmatrix}
-B_{00} \\ -B_{10} \\ \vdots \\ -B_{N0} \\ k_{0}A_{0} \\ 0 \\ \vdots \\ 0 \\
\end{bmatrix}
</math>
</center>

We then simply need to solve the two sets of <math>2(N+1) \mbox{x} 2(N+1)</math> linear system of equations.

== Matlab Code ==

A program to calculate the coefficients for the single finite dock problem using symmetry 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 Finite Rectangle using Symmetry

2012-03-16T01:50:33Z

Tertius: /* Numerical Solution */

{{incomplete pages}}

== Introduction ==

The problem consists of a finite rigid fixed rectangle of length <math>2L</math> centered at <math>x = 0</math> and having a height of <math>d + ...</math> freeboard with depth of submergence being <math>d</math> through which no flow is possible. The regions to the left and right of the rectangle are the free water surface. Since the geometry of the rectangle is symmetric about <math>x = 0</math> we can decompose the velocity potential <math>\phi</math> into a symmetric <math>\phi^s</math> and antisymmetric <math>\phi^a</math> potential. We then solve for these two potentials in the left domain only <math>x \leq 0</math> and subsequently recompose the solution <math>\phi</math> for the entire domain. We begin with the simple problem when the waves are normally incident on the rectangle (dock) (so that the problem is truly two-dimensional). We can extend this solution to a pair of partially submerged finite rigid fixed rectangles using symmetry
[[Eigenfunction Matching for a Finite Rectangle using Symmetry]]

== Governing Equations ==

We begin with the [[Frequency Domain Problem]] for a partially submerged finite dock. Although the dock is of length <math>2L</math> we are only interested in the dock of <math>-L \leq x \leq 0</math> together with the fluid domain <math>x \leq 0</math>. We assume <math>e^{i\omega t}</math> time dependence and
the depth of submergence is <math>d</math>.
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>
\begin{align}
\Delta\phi^{s,a} &= 0, \,\,\, -h\leq z\leq 0, \,\, x < -L, \\

\Delta\phi^{s,a} &= 0, \,\,\, -h\leq z\leq -d, \,\, -L < x < 0. \\

\partial_z\phi^{s,a} &= \alpha\phi^{s,a}, \,\,\, z=0, \,\, x < -L, \\

\partial_x\phi^{s,a} &= 0, \,\,\, -d\leq z\leq 0, \,\, x = -L, \\

\partial_z\phi^{s,a} &= 0, \,\,\,
\begin{cases}
z = -d, \,\, -L < x < 0, \\
z = -h, \,\, x < 0.
\end{cases} \\

\partial_x\phi^s &= 0, \,\,\, x = 0, \,\, -h \leq z \leq -d, \\

\phi^a &= 0, \,\,\, x = 0, \,\, -h \leq z \leq -d.
\end{align}
</math></center>

Where <math>\alpha = \omega^2/g</math>.

<math>\phi^s(x,z)</math> is an even function and at <math>x = 0</math> the derivative of this function is equal to zero. <math>\phi^a(x,z)</math> is an odd function and at <math>x = 0</math> this function is equal to zero. These boundary conditions are important and enable the recomposed solution <math>\phi(x,z)</math> to be reflected into the domain of <math>x \geq 0</math>.

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\leq 0</math>
and <math>x\geq 0</math>.

{{separation of variables in two dimensions}}

{{separation of variables for a free surface}}

=== Separation of Variables for a Dock ===

The separation of variables equation for a partially submerged dock is given by:
<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>
Z^{\prime} (-d) = 0.
</math></center>
The solution is
<math>k=\kappa_{m}= \frac{m\pi}{h-d} \,</math> , <math>m\in\mathbb{N}\cup\left\{0\right\}</math> and
<center><math>
Z = \psi_{m}\left( z\right) = \cos\kappa_{m}(z+h),\quad
m\in\mathbb{N}\cup\left\{0\right\}.</math></center>
We note that
<center><math>
\int\nolimits_{-h}^{-d}\psi_{m}(z)\psi_{n}(z) \mathrm{d} z=C_{m}\delta_{mn},
</math></center>
where
<center>
<math>
C_{m} = \frac{1}{2} \left( \frac{\cos\kappa_{m}(h-d)\sin\kappa_{m}(h-d)+\kappa_{m}(h-d)}{\kappa_{m}} \right).
</math>
</center>
In addition there is the special case of when <math>m=n=0</math> ,
<center>
<math>
C_{0} = h-d,
</math>
</center>
where <math>\delta_{00}=1</math> .

=== Inner product between free surface and dock modes ===

<center><math>
B_{mn} = \int\nolimits_{-h}^{-d}\psi_{m}(z)\chi_{n}(z) \mathrm{d} z , \quad m\geq1, \quad n\geq0,
</math></center>
where
<center><math>
B_{mn}= \int\nolimits_{-h}^{-d} \frac{\cos\kappa_{m}(z+h)\cos k_{n}(z+h)}{\cos(k_{n}h)} \mathrm{d} z=\frac{-(-1)^m k_{n}\sin k_{n}(h-d)}
{\cos(k_{n}h)(\kappa_{m}^{2}-k_{n}^{2})}.
</math></center>
The evaluation of this inner product works on the assumption that the two roots <math>\kappa_{m}</math> and <math>k_{n}</math> do not coincide. The Matlab code given below has a routine to check for this case and make necessary adjustments.

We also have the case where the functions of the inner product are switched.
<center><math>
B_{nm} = \int\nolimits_{-h}^{-d}\chi_{m}(z)\psi_{n}(z) \mathrm{d} z , \quad m\geq0, \quad n\geq0.
</math></center>
For <math>m \geq 0</math> and <math>n \geq 1</math> the evaluation is above with the exception that the indices of the expression are switched.

Finally there is the special case of when <math>m \geq 0</math> and <math>n = 0</math>.
<center><math>
B_{0m} = \frac{\sin k_{m}(h-d)}{k_{m}\cos(k_{m}h)}.
</math></center>

=== Expansion of the potential ===

We need to apply some boundary conditions at plus and minus infinity,
where essentially 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 roots. Here we
must use a different condition, essentially identifying one solution
as the incoming wave and the other as the outgoing wave.

Therefore the diffracted potentials can
be expanded as
<center>
<math>
\phi(x,z)=\sum_{m=0}^{\infty}a_{m}e^{k_{m}x}\chi_{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 dock covered region. Therefore, the
potentials and their derivatives at <math>x=0</math> have to be equal.
We obtain:
<center><math>
\begin{align}
\chi_{0}\left( z\right) + \sum_{m=0}^{\infty}a_{m} \chi_{m}\left( z\right)
&=\sum_{m=0}^{\infty}b_{m}\psi_{m}(z), \quad -h \leq z\leq -d \\

-k_{0}\chi_{0}\left( z\right) +\sum_{m=0}^{\infty} a_{m}k_{m}\chi_{m}\left( z\right) &=
\begin{cases}
\quad \quad \quad 0,\quad \quad \quad -d\leq z\leq 0 \\
-\sum_{m=0}^{\infty}b_{m}\kappa_{m}\psi_{m}(z),\quad -h\leq z\leq -d
\end{cases}
\end{align}
</math></center>

For the first equation we multiply both sides by <math>\psi_{n}(z) \,</math> and integrating from <math>-h</math> to <math>-d</math> to obtain:
<center><math>
B_{n0} + \sum^{\infty}_{m=0}a_{m}B_{nm} = b_{m}C_{m}\delta_{mn}, n\in\mathbb{N}\cup\left\{0\right\}
</math></center>
and for the second equation we multiply both sides by <math>\chi_{n}(z) \,</math> and integrating from <math>-h</math> to <math>0</math> to obtain:
<center><math>
-k_{0}A_{0}\delta_{0n} + k_{m}a_{m}A_{m} = -\sum^{\infty}_{m=0}\kappa_{m}b_{m}B_{mn}, n\in\mathbb{N}\cup\left\{0\right\}
</math></center>

Solving the equations above will yield the coefficients of the water velocity potential in the dock covered region.

== Numerical Solution ==

To solve the system of equations for the '''symmetric''' case, we set the upper limit of <math>m</math> to be <math>N</math>. This resulting system can be expressed in the block matrix form below,

<center>
<math>
\begin{bmatrix}
B_{00} & B_{01} & \cdots & B_{0N} & -C_{0} & 0 & \cdots & 0 \\
B_{10} & B_{11} & \cdots & B_{1N} & 0 & -C_{1} & \cdots & \vdots \\
\vdots & \vdots & \ddots & \vdots & \vdots & \vdots & \ddots & 0 \\
B_{N0} & B_{N1} & \cdots & B_{NN} & 0 & \cdots & 0 & -C_{N} \\
k_{0}A_{0} & 0 & \cdots & 0 & 0 & \kappa_{1}\tanh(\kappa_{1}L)B_{10} & \cdots & \kappa_{N}\tanh(\kappa_{N}L)B_{N0} \\
0 & k_{1}A_{1} & \cdots & \vdots & 0 & \kappa_{1}\tanh(\kappa_{1}L)B_{11} & \cdots & \kappa_{N}\tanh(\kappa_{N}L)B_{N1} \\
\vdots & \vdots & \ddots & 0 & \vdots & \vdots & \ddots & \vdots \\
0 & \cdots & 0 & k_{N}A_{N} & 0 & \kappa_{1}\tanh(\kappa_{1}L)B_{1N} & \cdots & \kappa_{N}\tanh(\kappa_{N}L)B_{NN} \\
\end{bmatrix}
\begin{bmatrix}
a^s_{0} \\ a^s_{1} \\ \vdots \\ a^s_{N} \\ b^s_{0} \\ b^s_{1} \\ \vdots \\ b^s_{N} \\
\end{bmatrix}
=\begin{bmatrix}
-B_{00} \\ -B_{10} \\ \vdots \\ -B_{N0} \\ k_{0}A_{0} \\ 0 \\ \vdots \\ 0 \\
\end{bmatrix}
</math>
</center>

To solve the system of equations for the '''antisymmetric''' case, we set the upper limit of <math>m</math> to be <math>N</math>. This resulting system can be expressed in the block matrix form below,

<center>
<math>
\begin{bmatrix}
B_{00} & B_{01} & \cdots & B_{0N} & -C_{0} & 0 & \cdots & 0 \\
B_{10} & B_{11} & \cdots & B_{1N} & 0 & -C_{1} & \cdots & \vdots \\
\vdots & \vdots & \ddots & \vdots & \vdots & \vdots & \ddots & 0 \\
B_{N0} & B_{N1} & \cdots & B_{NN} & 0 & \cdots & 0 & -C_{N} \\
k_{0}A_{0} & 0 & \cdots & 0 & \frac{1}{L}D_{0} & \kappa_{1}\coth(\kappa_{1}L)B_{10} & \cdots & \kappa_{N}\coth(\kappa_{N}L)B_{N0} \\
0 & k_{1}A_{1} & \cdots & \vdots & \frac{1}{L}D_{1} & \kappa_{1}\coth(\kappa_{1}L)B_{11} & \cdots & \kappa_{N}\coth(\kappa_{N}L)B_{N1} \\
\vdots & \vdots & \ddots & 0 & \vdots & \vdots & \ddots & \vdots \\
0 & \cdots & 0 & k_{N}A_{N} & \frac{1}{L}D_{N} & \kappa_{1}\coth(\kappa_{1}L)B_{1N} & \cdots & \kappa_{N}\coth(\kappa_{N}L)B_{NN} \\
\end{bmatrix}
\begin{bmatrix}
a^a_{0} \\ a^a_{1} \\ \vdots \\ a^a_{N} \\ b^a_{0} \\ b^a_{1} \\ \vdots \\ b^a_{N} \\
\end{bmatrix}
=\begin{bmatrix}
-B_{00} \\ -B_{10} \\ \vdots \\ -B_{N0} \\ k_{0}A_{0} \\ 0 \\ \vdots \\ 0 \\
\end{bmatrix}
</math>
</center>

We then simply need to solve the two sets of <math>2(N+1) 2(N+1)</math> linear system of equations.

== Matlab Code ==

A program to calculate the coefficients for the single finite dock problem using symmetry 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 Finite Rectangle using Symmetry

2012-03-16T01:41:13Z

Tertius: /* Numerical Solution */

{{incomplete pages}}

== Introduction ==

The problem consists of a finite rigid fixed rectangle of length <math>2L</math> centered at <math>x = 0</math> and having a height of <math>d + ...</math> freeboard with depth of submergence being <math>d</math> through which no flow is possible. The regions to the left and right of the rectangle are the free water surface. Since the geometry of the rectangle is symmetric about <math>x = 0</math> we can decompose the velocity potential <math>\phi</math> into a symmetric <math>\phi^s</math> and antisymmetric <math>\phi^a</math> potential. We then solve for these two potentials in the left domain only <math>x \leq 0</math> and subsequently recompose the solution <math>\phi</math> for the entire domain. We begin with the simple problem when the waves are normally incident on the rectangle (dock) (so that the problem is truly two-dimensional). We can extend this solution to a pair of partially submerged finite rigid fixed rectangles using symmetry
[[Eigenfunction Matching for a Finite Rectangle using Symmetry]]

== Governing Equations ==

We begin with the [[Frequency Domain Problem]] for a partially submerged finite dock. Although the dock is of length <math>2L</math> we are only interested in the dock of <math>-L \leq x \leq 0</math> together with the fluid domain <math>x \leq 0</math>. We assume <math>e^{i\omega t}</math> time dependence and
the depth of submergence is <math>d</math>.
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>
\begin{align}
\Delta\phi^{s,a} &= 0, \,\,\, -h\leq z\leq 0, \,\, x < -L, \\

\Delta\phi^{s,a} &= 0, \,\,\, -h\leq z\leq -d, \,\, -L < x < 0. \\

\partial_z\phi^{s,a} &= \alpha\phi^{s,a}, \,\,\, z=0, \,\, x < -L, \\

\partial_x\phi^{s,a} &= 0, \,\,\, -d\leq z\leq 0, \,\, x = -L, \\

\partial_z\phi^{s,a} &= 0, \,\,\,
\begin{cases}
z = -d, \,\, -L < x < 0, \\
z = -h, \,\, x < 0.
\end{cases} \\

\partial_x\phi^s &= 0, \,\,\, x = 0, \,\, -h \leq z \leq -d, \\

\phi^a &= 0, \,\,\, x = 0, \,\, -h \leq z \leq -d.
\end{align}
</math></center>

Where <math>\alpha = \omega^2/g</math>.

<math>\phi^s(x,z)</math> is an even function and at <math>x = 0</math> the derivative of this function is equal to zero. <math>\phi^a(x,z)</math> is an odd function and at <math>x = 0</math> this function is equal to zero. These boundary conditions are important and enable the recomposed solution <math>\phi(x,z)</math> to be reflected into the domain of <math>x \geq 0</math>.

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\leq 0</math>
and <math>x\geq 0</math>.

{{separation of variables in two dimensions}}

{{separation of variables for a free surface}}

=== Separation of Variables for a Dock ===

The separation of variables equation for a partially submerged dock is given by:
<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>
Z^{\prime} (-d) = 0.
</math></center>
The solution is
<math>k=\kappa_{m}= \frac{m\pi}{h-d} \,</math> , <math>m\in\mathbb{N}\cup\left\{0\right\}</math> and
<center><math>
Z = \psi_{m}\left( z\right) = \cos\kappa_{m}(z+h),\quad
m\in\mathbb{N}\cup\left\{0\right\}.</math></center>
We note that
<center><math>
\int\nolimits_{-h}^{-d}\psi_{m}(z)\psi_{n}(z) \mathrm{d} z=C_{m}\delta_{mn},
</math></center>
where
<center>
<math>
C_{m} = \frac{1}{2} \left( \frac{\cos\kappa_{m}(h-d)\sin\kappa_{m}(h-d)+\kappa_{m}(h-d)}{\kappa_{m}} \right).
</math>
</center>
In addition there is the special case of when <math>m=n=0</math> ,
<center>
<math>
C_{0} = h-d,
</math>
</center>
where <math>\delta_{00}=1</math> .

=== Inner product between free surface and dock modes ===

<center><math>
B_{mn} = \int\nolimits_{-h}^{-d}\psi_{m}(z)\chi_{n}(z) \mathrm{d} z , \quad m\geq1, \quad n\geq0,
</math></center>
where
<center><math>
B_{mn}= \int\nolimits_{-h}^{-d} \frac{\cos\kappa_{m}(z+h)\cos k_{n}(z+h)}{\cos(k_{n}h)} \mathrm{d} z=\frac{-(-1)^m k_{n}\sin k_{n}(h-d)}
{\cos(k_{n}h)(\kappa_{m}^{2}-k_{n}^{2})}.
</math></center>
The evaluation of this inner product works on the assumption that the two roots <math>\kappa_{m}</math> and <math>k_{n}</math> do not coincide. The Matlab code given below has a routine to check for this case and make necessary adjustments.

We also have the case where the functions of the inner product are switched.
<center><math>
B_{nm} = \int\nolimits_{-h}^{-d}\chi_{m}(z)\psi_{n}(z) \mathrm{d} z , \quad m\geq0, \quad n\geq0.
</math></center>
For <math>m \geq 0</math> and <math>n \geq 1</math> the evaluation is above with the exception that the indices of the expression are switched.

Finally there is the special case of when <math>m \geq 0</math> and <math>n = 0</math>.
<center><math>
B_{0m} = \frac{\sin k_{m}(h-d)}{k_{m}\cos(k_{m}h)}.
</math></center>

=== Expansion of the potential ===

We need to apply some boundary conditions at plus and minus infinity,
where essentially 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 roots. Here we
must use a different condition, essentially identifying one solution
as the incoming wave and the other as the outgoing wave.

Therefore the diffracted potentials can
be expanded as
<center>
<math>
\phi(x,z)=\sum_{m=0}^{\infty}a_{m}e^{k_{m}x}\chi_{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 dock covered region. Therefore, the
potentials and their derivatives at <math>x=0</math> have to be equal.
We obtain:
<center><math>
\begin{align}
\chi_{0}\left( z\right) + \sum_{m=0}^{\infty}a_{m} \chi_{m}\left( z\right)
&=\sum_{m=0}^{\infty}b_{m}\psi_{m}(z), \quad -h \leq z\leq -d \\

-k_{0}\chi_{0}\left( z\right) +\sum_{m=0}^{\infty} a_{m}k_{m}\chi_{m}\left( z\right) &=
\begin{cases}
\quad \quad \quad 0,\quad \quad \quad -d\leq z\leq 0 \\
-\sum_{m=0}^{\infty}b_{m}\kappa_{m}\psi_{m}(z),\quad -h\leq z\leq -d
\end{cases}
\end{align}
</math></center>

For the first equation we multiply both sides by <math>\psi_{n}(z) \,</math> and integrating from <math>-h</math> to <math>-d</math> to obtain:
<center><math>
B_{n0} + \sum^{\infty}_{m=0}a_{m}B_{nm} = b_{m}C_{m}\delta_{mn}, n\in\mathbb{N}\cup\left\{0\right\}
</math></center>
and for the second equation we multiply both sides by <math>\chi_{n}(z) \,</math> and integrating from <math>-h</math> to <math>0</math> to obtain:
<center><math>
-k_{0}A_{0}\delta_{0n} + k_{m}a_{m}A_{m} = -\sum^{\infty}_{m=0}\kappa_{m}b_{m}B_{mn}, n\in\mathbb{N}\cup\left\{0\right\}
</math></center>

Solving the equations above will yield the coefficients of the water velocity potential in the dock covered region.

== Numerical Solution ==

To solve the system of equations for the '''symmetric''' case, we set the upper limit of <math>m</math> to be <math>N</math>. This resulting system can be expressed in the block matrix form below,

<center>
<math>
\begin{bmatrix}
B_{00} & B_{01} & \cdots & B_{0N} & -C_{0} & 0 & \cdots & 0 \\
B_{10} & B_{11} & \cdots & B_{1N} & 0 & -C_{1} & \cdots & \vdots \\
\vdots & \vdots & \ddots & \vdots & \vdots & \vdots & \ddots & 0 \\
B_{N0} & B_{N1} & \cdots & B_{NN} & 0 & \cdots & 0 & -C_{N} \\
k_{0}A_{0} & 0 & \cdots & 0 & 0 & \kappa_{1}\tanh(\kappa_{1}L)B_{10} & \cdots & \kappa_{N}\tanh(\kappa_{N}L)B_{N0} \\
0 & k_{1}A_{1} & \cdots & \vdots & 0 & \kappa_{1}\tanh(\kappa_{1}L)B_{11} & \cdots & \kappa_{N}\tanh(\kappa_{N}L)B_{N1} \\
\vdots & \vdots & \ddots & 0 & \vdots & \vdots & \ddots & \vdots \\
0 & \cdots & 0 & k_{N}A_{N} & 0 & \kappa_{1}\tanh(\kappa_{1}L)B_{1N} & \cdots & \kappa_{N}\tanh(\kappa_{N}L)B_{NN} \\
\end{bmatrix}
\begin{bmatrix}
a^s_{0} \\ a^s_{1} \\ \vdots \\ a^s_{N} \\ b^s_{0} \\ b^s_{1} \\ \vdots \\ b^s_{N} \\
\end{bmatrix}
=\begin{bmatrix}
-B_{00} \\ -B_{10} \\ \vdots \\ -B_{N0} \\ k_{0}A_{0} \\ 0 \\ \vdots \\ 0 \\
\end{bmatrix}
</math>
</center>

To solve the system of equations for the '''antisymmetric''' case, we set the upper limit of <math>m</math> to be <math>N</math>. This resulting system can be expressed in the block matrix form below,

<center>
<math>
\begin{bmatrix}
B_{00} & B_{01} & \cdots & B_{0N} & -C_{0} & 0 & \cdots & 0 \\
B_{10} & B_{11} & \cdots & B_{1N} & 0 & -C_{1} & \cdots & \vdots \\
\vdots & \vdots & \ddots & \vdots & \vdots & \vdots & \ddots & 0 \\
B_{N0} & B_{N1} & \cdots & B_{NN} & 0 & \cdots & 0 & -C_{N} \\
k_{0}A_{0} & 0 & \cdots & 0 & \frac{1}{L}D_{0} & \kappa_{1}\coth(\kappa_{1}L)B_{10} & \cdots & \kappa_{N}\coth(\kappa_{N}L)B_{N0} \\
0 & k_{1}A_{1} & \cdots & \vdots & \frac{1}{L}D_{1} & \kappa_{1}\coth(\kappa_{1}L)B_{11} & \cdots & \kappa_{N}\coth(\kappa_{N}L)B_{N1} \\
\vdots & \vdots & \ddots & 0 & \vdots & \vdots & \ddots & \vdots \\
0 & \cdots & 0 & k_{N}A_{N} & \frac{1}{L}D_{N} & \kappa_{1}\coth(\kappa_{1}L)B_{1N} & \cdots & \kappa_{N}\coth(\kappa_{N}L)B_{NN} \\
\end{bmatrix}
\begin{bmatrix}
a^a_{0} \\ a^a_{1} \\ \vdots \\ a^a_{N} \\ b^a_{0} \\ b^a_{1} \\ \vdots \\ b^a_{N} \\
\end{bmatrix}
=\begin{bmatrix}
-B_{00} \\ -B_{10} \\ \vdots \\ -B_{N0} \\ k_{0}A_{0} \\ 0 \\ \vdots \\ 0 \\
\end{bmatrix}
</math>
</center>

We then simply need to solve the linear system of equations.

== Matlab Code ==

A program to calculate the coefficients for the single finite dock problem using symmetry 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 Finite Rectangle using Symmetry

2012-03-16T01:32:21Z

Tertius: /* Matlab Code */

{{incomplete pages}}

== Introduction ==

The problem consists of a finite rigid fixed rectangle of length <math>2L</math> centered at <math>x = 0</math> and having a height of <math>d + ...</math> freeboard with depth of submergence being <math>d</math> through which no flow is possible. The regions to the left and right of the rectangle are the free water surface. Since the geometry of the rectangle is symmetric about <math>x = 0</math> we can decompose the velocity potential <math>\phi</math> into a symmetric <math>\phi^s</math> and antisymmetric <math>\phi^a</math> potential. We then solve for these two potentials in the left domain only <math>x \leq 0</math> and subsequently recompose the solution <math>\phi</math> for the entire domain. We begin with the simple problem when the waves are normally incident on the rectangle (dock) (so that the problem is truly two-dimensional). We can extend this solution to a pair of partially submerged finite rigid fixed rectangles using symmetry
[[Eigenfunction Matching for a Finite Rectangle using Symmetry]]

== Governing Equations ==

We begin with the [[Frequency Domain Problem]] for a partially submerged finite dock. Although the dock is of length <math>2L</math> we are only interested in the dock of <math>-L \leq x \leq 0</math> together with the fluid domain <math>x \leq 0</math>. We assume <math>e^{i\omega t}</math> time dependence and
the depth of submergence is <math>d</math>.
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>
\begin{align}
\Delta\phi^{s,a} &= 0, \,\,\, -h\leq z\leq 0, \,\, x < -L, \\

\Delta\phi^{s,a} &= 0, \,\,\, -h\leq z\leq -d, \,\, -L < x < 0. \\

\partial_z\phi^{s,a} &= \alpha\phi^{s,a}, \,\,\, z=0, \,\, x < -L, \\

\partial_x\phi^{s,a} &= 0, \,\,\, -d\leq z\leq 0, \,\, x = -L, \\

\partial_z\phi^{s,a} &= 0, \,\,\,
\begin{cases}
z = -d, \,\, -L < x < 0, \\
z = -h, \,\, x < 0.
\end{cases} \\

\partial_x\phi^s &= 0, \,\,\, x = 0, \,\, -h \leq z \leq -d, \\

\phi^a &= 0, \,\,\, x = 0, \,\, -h \leq z \leq -d.
\end{align}
</math></center>

Where <math>\alpha = \omega^2/g</math>.

<math>\phi^s(x,z)</math> is an even function and at <math>x = 0</math> the derivative of this function is equal to zero. <math>\phi^a(x,z)</math> is an odd function and at <math>x = 0</math> this function is equal to zero. These boundary conditions are important and enable the recomposed solution <math>\phi(x,z)</math> to be reflected into the domain of <math>x \geq 0</math>.

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\leq 0</math>
and <math>x\geq 0</math>.

{{separation of variables in two dimensions}}

{{separation of variables for a free surface}}

=== Separation of Variables for a Dock ===

The separation of variables equation for a partially submerged dock is given by:
<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>
Z^{\prime} (-d) = 0.
</math></center>
The solution is
<math>k=\kappa_{m}= \frac{m\pi}{h-d} \,</math> , <math>m\in\mathbb{N}\cup\left\{0\right\}</math> and
<center><math>
Z = \psi_{m}\left( z\right) = \cos\kappa_{m}(z+h),\quad
m\in\mathbb{N}\cup\left\{0\right\}.</math></center>
We note that
<center><math>
\int\nolimits_{-h}^{-d}\psi_{m}(z)\psi_{n}(z) \mathrm{d} z=C_{m}\delta_{mn},
</math></center>
where
<center>
<math>
C_{m} = \frac{1}{2} \left( \frac{\cos\kappa_{m}(h-d)\sin\kappa_{m}(h-d)+\kappa_{m}(h-d)}{\kappa_{m}} \right).
</math>
</center>
In addition there is the special case of when <math>m=n=0</math> ,
<center>
<math>
C_{0} = h-d,
</math>
</center>
where <math>\delta_{00}=1</math> .

=== Inner product between free surface and dock modes ===

<center><math>
B_{mn} = \int\nolimits_{-h}^{-d}\psi_{m}(z)\chi_{n}(z) \mathrm{d} z , \quad m\geq1, \quad n\geq0,
</math></center>
where
<center><math>
B_{mn}= \int\nolimits_{-h}^{-d} \frac{\cos\kappa_{m}(z+h)\cos k_{n}(z+h)}{\cos(k_{n}h)} \mathrm{d} z=\frac{-(-1)^m k_{n}\sin k_{n}(h-d)}
{\cos(k_{n}h)(\kappa_{m}^{2}-k_{n}^{2})}.
</math></center>
The evaluation of this inner product works on the assumption that the two roots <math>\kappa_{m}</math> and <math>k_{n}</math> do not coincide. The Matlab code given below has a routine to check for this case and make necessary adjustments.

We also have the case where the functions of the inner product are switched.
<center><math>
B_{nm} = \int\nolimits_{-h}^{-d}\chi_{m}(z)\psi_{n}(z) \mathrm{d} z , \quad m\geq0, \quad n\geq0.
</math></center>
For <math>m \geq 0</math> and <math>n \geq 1</math> the evaluation is above with the exception that the indices of the expression are switched.

Finally there is the special case of when <math>m \geq 0</math> and <math>n = 0</math>.
<center><math>
B_{0m} = \frac{\sin k_{m}(h-d)}{k_{m}\cos(k_{m}h)}.
</math></center>

=== Expansion of the potential ===

We need to apply some boundary conditions at plus and minus infinity,
where essentially 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 roots. Here we
must use a different condition, essentially identifying one solution
as the incoming wave and the other as the outgoing wave.

Therefore the diffracted potentials can
be expanded as
<center>
<math>
\phi(x,z)=\sum_{m=0}^{\infty}a_{m}e^{k_{m}x}\chi_{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 dock covered region. Therefore, the
potentials and their derivatives at <math>x=0</math> have to be equal.
We obtain:
<center><math>
\begin{align}
\chi_{0}\left( z\right) + \sum_{m=0}^{\infty}a_{m} \chi_{m}\left( z\right)
&=\sum_{m=0}^{\infty}b_{m}\psi_{m}(z), \quad -h \leq z\leq -d \\

-k_{0}\chi_{0}\left( z\right) +\sum_{m=0}^{\infty} a_{m}k_{m}\chi_{m}\left( z\right) &=
\begin{cases}
\quad \quad \quad 0,\quad \quad \quad -d\leq z\leq 0 \\
-\sum_{m=0}^{\infty}b_{m}\kappa_{m}\psi_{m}(z),\quad -h\leq z\leq -d
\end{cases}
\end{align}
</math></center>

For the first equation we multiply both sides by <math>\psi_{n}(z) \,</math> and integrating from <math>-h</math> to <math>-d</math> to obtain:
<center><math>
B_{n0} + \sum^{\infty}_{m=0}a_{m}B_{nm} = b_{m}C_{m}\delta_{mn}, n\in\mathbb{N}\cup\left\{0\right\}
</math></center>
and for the second equation we multiply both sides by <math>\chi_{n}(z) \,</math> and integrating from <math>-h</math> to <math>0</math> to obtain:
<center><math>
-k_{0}A_{0}\delta_{0n} + k_{m}a_{m}A_{m} = -\sum^{\infty}_{m=0}\kappa_{m}b_{m}B_{mn}, n\in\mathbb{N}\cup\left\{0\right\}
</math></center>

Solving the equations above will yield the coefficients of the water velocity potential in the dock covered region.

== Numerical Solution ==

To solve the system of equations for the '''symmetric''' case, we set the upper limit of <math>m</math> to be <math>N</math>. This resulting system can be expressed in the block matrix form below,

<center>
<math>
\begin{bmatrix}
B_{00} & B_{01} & \cdots & B_{0N} & -C_{0} & 0 & \cdots & 0 \\
B_{10} & B_{11} & \cdots & B_{1N} & 0 & -C_{1} & \cdots & \vdots \\
\vdots & \vdots & \ddots & \vdots & \vdots & \vdots & \ddots & 0 \\
B_{N0} & B_{N1} & \cdots & B_{NN} & 0 & \cdots & 0 & -C_{N} \\
k_{0}A_{0} & 0 & \cdots & 0 & 0 & \kappa_{1}\tanh(\kappa_{1}L)B_{10} & \cdots & \kappa_{N}\tanh(\kappa_{N}L)B_{N0} \\
0 & k_{1}A_{1} & \cdots & \vdots & 0 & \kappa_{1}\tanh(\kappa_{1}L)B_{11} & \cdots & \kappa_{N}\tanh(\kappa_{N}L)B_{N1} \\
\vdots & \vdots & \ddots & 0 & \vdots & \vdots & \ddots & \vdots \\
0 & \cdots & 0 & k_{N}A_{N} & 0 & \kappa_{1}\tanh(\kappa_{1}L)B_{1N} & \cdots & \kappa_{N}\tanh(\kappa_{N}L)B_{NN} \\
\end{bmatrix}
\begin{bmatrix}
a^s_{0} \\ a^s_{1} \\ \vdots \\ a^s_{N} \\ b^s_{0} \\ b^s_{1} \\ \vdots \\ b^s_{N} \\
\end{bmatrix}
=\begin{bmatrix}
-B_{00} \\ -B_{10} \\ \vdots \\ -B_{N0} \\ k_{0}A_{0} \\ 0 \\ \vdots \\ 0 \\
\end{bmatrix}
</math>
</center>

To solve the system of equations for the '''antisymmetric''' case, we set the upper limit of <math>m</math> to be <math>N</math>. This resulting system can be expressed in the block matrix form below,

<center>
<math>
\begin{bmatrix}
B_{00} & B_{01} & \cdots & B_{0N} & -C_{0} & 0 & \cdots & 0 \\
B_{10} & B_{11} & \cdots & B_{1N} & 0 & -C_{1} & \cdots & \vdots \\
\vdots & \vdots & \ddots & \vdots & \vdots & \vdots & \ddots & 0 \\
B_{N0} & B_{N1} & \cdots & B_{NN} & 0 & \cdots & 0 & -C_{N} \\
k_{0}A_{0} & 0 & \cdots & 0 & 0 & \kappa_{1}\coth(\kappa_{1}L)B_{10} & \cdots & \kappa_{N}\coth(\kappa_{N}L)B_{N0} \\
0 & k_{1}A_{1} & \cdots & \vdots & 0 & \kappa_{1}\coth(\kappa_{1}L)B_{11} & \cdots & \kappa_{N}\coth(\kappa_{N}L)B_{N1} \\
\vdots & \vdots & \ddots & 0 & \vdots & \vdots & \ddots & \vdots \\
0 & \cdots & 0 & k_{N}A_{N} & 0 & \kappa_{1}\coth(\kappa_{1}L)B_{1N} & \cdots & \kappa_{N}\coth(\kappa_{N}L)B_{NN} \\
\end{bmatrix}
\begin{bmatrix}
a^a_{0} \\ a^a_{1} \\ \vdots \\ a^a_{N} \\ b^a_{0} \\ b^a_{1} \\ \vdots \\ b^a_{N} \\
\end{bmatrix}
=\begin{bmatrix}
-B_{00} \\ -B_{10} \\ \vdots \\ -B_{N0} \\ k_{0}A_{0} \\ 0 \\ \vdots \\ 0 \\
\end{bmatrix}
</math>
</center>

We then simply need to solve the linear system of equations.

== Matlab Code ==

A program to calculate the coefficients for the single finite dock problem using symmetry 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 Finite Rectangle using Symmetry

2012-03-16T01:30:47Z

Tertius: /* Numerical Solution */

{{incomplete pages}}

== Introduction ==

The problem consists of a finite rigid fixed rectangle of length <math>2L</math> centered at <math>x = 0</math> and having a height of <math>d + ...</math> freeboard with depth of submergence being <math>d</math> through which no flow is possible. The regions to the left and right of the rectangle are the free water surface. Since the geometry of the rectangle is symmetric about <math>x = 0</math> we can decompose the velocity potential <math>\phi</math> into a symmetric <math>\phi^s</math> and antisymmetric <math>\phi^a</math> potential. We then solve for these two potentials in the left domain only <math>x \leq 0</math> and subsequently recompose the solution <math>\phi</math> for the entire domain. We begin with the simple problem when the waves are normally incident on the rectangle (dock) (so that the problem is truly two-dimensional). We can extend this solution to a pair of partially submerged finite rigid fixed rectangles using symmetry
[[Eigenfunction Matching for a Finite Rectangle using Symmetry]]

== Governing Equations ==

We begin with the [[Frequency Domain Problem]] for a partially submerged finite dock. Although the dock is of length <math>2L</math> we are only interested in the dock of <math>-L \leq x \leq 0</math> together with the fluid domain <math>x \leq 0</math>. We assume <math>e^{i\omega t}</math> time dependence and
the depth of submergence is <math>d</math>.
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>
\begin{align}
\Delta\phi^{s,a} &= 0, \,\,\, -h\leq z\leq 0, \,\, x < -L, \\

\Delta\phi^{s,a} &= 0, \,\,\, -h\leq z\leq -d, \,\, -L < x < 0. \\

\partial_z\phi^{s,a} &= \alpha\phi^{s,a}, \,\,\, z=0, \,\, x < -L, \\

\partial_x\phi^{s,a} &= 0, \,\,\, -d\leq z\leq 0, \,\, x = -L, \\

\partial_z\phi^{s,a} &= 0, \,\,\,
\begin{cases}
z = -d, \,\, -L < x < 0, \\
z = -h, \,\, x < 0.
\end{cases} \\

\partial_x\phi^s &= 0, \,\,\, x = 0, \,\, -h \leq z \leq -d, \\

\phi^a &= 0, \,\,\, x = 0, \,\, -h \leq z \leq -d.
\end{align}
</math></center>

Where <math>\alpha = \omega^2/g</math>.

<math>\phi^s(x,z)</math> is an even function and at <math>x = 0</math> the derivative of this function is equal to zero. <math>\phi^a(x,z)</math> is an odd function and at <math>x = 0</math> this function is equal to zero. These boundary conditions are important and enable the recomposed solution <math>\phi(x,z)</math> to be reflected into the domain of <math>x \geq 0</math>.

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\leq 0</math>
and <math>x\geq 0</math>.

{{separation of variables in two dimensions}}

{{separation of variables for a free surface}}

=== Separation of Variables for a Dock ===

The separation of variables equation for a partially submerged dock is given by:
<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>
Z^{\prime} (-d) = 0.
</math></center>
The solution is
<math>k=\kappa_{m}= \frac{m\pi}{h-d} \,</math> , <math>m\in\mathbb{N}\cup\left\{0\right\}</math> and
<center><math>
Z = \psi_{m}\left( z\right) = \cos\kappa_{m}(z+h),\quad
m\in\mathbb{N}\cup\left\{0\right\}.</math></center>
We note that
<center><math>
\int\nolimits_{-h}^{-d}\psi_{m}(z)\psi_{n}(z) \mathrm{d} z=C_{m}\delta_{mn},
</math></center>
where
<center>
<math>
C_{m} = \frac{1}{2} \left( \frac{\cos\kappa_{m}(h-d)\sin\kappa_{m}(h-d)+\kappa_{m}(h-d)}{\kappa_{m}} \right).
</math>
</center>
In addition there is the special case of when <math>m=n=0</math> ,
<center>
<math>
C_{0} = h-d,
</math>
</center>
where <math>\delta_{00}=1</math> .

=== Inner product between free surface and dock modes ===

<center><math>
B_{mn} = \int\nolimits_{-h}^{-d}\psi_{m}(z)\chi_{n}(z) \mathrm{d} z , \quad m\geq1, \quad n\geq0,
</math></center>
where
<center><math>
B_{mn}= \int\nolimits_{-h}^{-d} \frac{\cos\kappa_{m}(z+h)\cos k_{n}(z+h)}{\cos(k_{n}h)} \mathrm{d} z=\frac{-(-1)^m k_{n}\sin k_{n}(h-d)}
{\cos(k_{n}h)(\kappa_{m}^{2}-k_{n}^{2})}.
</math></center>
The evaluation of this inner product works on the assumption that the two roots <math>\kappa_{m}</math> and <math>k_{n}</math> do not coincide. The Matlab code given below has a routine to check for this case and make necessary adjustments.

We also have the case where the functions of the inner product are switched.
<center><math>
B_{nm} = \int\nolimits_{-h}^{-d}\chi_{m}(z)\psi_{n}(z) \mathrm{d} z , \quad m\geq0, \quad n\geq0.
</math></center>
For <math>m \geq 0</math> and <math>n \geq 1</math> the evaluation is above with the exception that the indices of the expression are switched.

Finally there is the special case of when <math>m \geq 0</math> and <math>n = 0</math>.
<center><math>
B_{0m} = \frac{\sin k_{m}(h-d)}{k_{m}\cos(k_{m}h)}.
</math></center>

=== Expansion of the potential ===

We need to apply some boundary conditions at plus and minus infinity,
where essentially 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 roots. Here we
must use a different condition, essentially identifying one solution
as the incoming wave and the other as the outgoing wave.

Therefore the diffracted potentials can
be expanded as
<center>
<math>
\phi(x,z)=\sum_{m=0}^{\infty}a_{m}e^{k_{m}x}\chi_{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 dock covered region. Therefore, the
potentials and their derivatives at <math>x=0</math> have to be equal.
We obtain:
<center><math>
\begin{align}
\chi_{0}\left( z\right) + \sum_{m=0}^{\infty}a_{m} \chi_{m}\left( z\right)
&=\sum_{m=0}^{\infty}b_{m}\psi_{m}(z), \quad -h \leq z\leq -d \\

-k_{0}\chi_{0}\left( z\right) +\sum_{m=0}^{\infty} a_{m}k_{m}\chi_{m}\left( z\right) &=
\begin{cases}
\quad \quad \quad 0,\quad \quad \quad -d\leq z\leq 0 \\
-\sum_{m=0}^{\infty}b_{m}\kappa_{m}\psi_{m}(z),\quad -h\leq z\leq -d
\end{cases}
\end{align}
</math></center>

For the first equation we multiply both sides by <math>\psi_{n}(z) \,</math> and integrating from <math>-h</math> to <math>-d</math> to obtain:
<center><math>
B_{n0} + \sum^{\infty}_{m=0}a_{m}B_{nm} = b_{m}C_{m}\delta_{mn}, n\in\mathbb{N}\cup\left\{0\right\}
</math></center>
and for the second equation we multiply both sides by <math>\chi_{n}(z) \,</math> and integrating from <math>-h</math> to <math>0</math> to obtain:
<center><math>
-k_{0}A_{0}\delta_{0n} + k_{m}a_{m}A_{m} = -\sum^{\infty}_{m=0}\kappa_{m}b_{m}B_{mn}, n\in\mathbb{N}\cup\left\{0\right\}
</math></center>

Solving the equations above will yield the coefficients of the water velocity potential in the dock covered region.

== Numerical Solution ==

To solve the system of equations for the '''symmetric''' case, we set the upper limit of <math>m</math> to be <math>N</math>. This resulting system can be expressed in the block matrix form below,

<center>
<math>
\begin{bmatrix}
B_{00} & B_{01} & \cdots & B_{0N} & -C_{0} & 0 & \cdots & 0 \\
B_{10} & B_{11} & \cdots & B_{1N} & 0 & -C_{1} & \cdots & \vdots \\
\vdots & \vdots & \ddots & \vdots & \vdots & \vdots & \ddots & 0 \\
B_{N0} & B_{N1} & \cdots & B_{NN} & 0 & \cdots & 0 & -C_{N} \\
k_{0}A_{0} & 0 & \cdots & 0 & 0 & \kappa_{1}\tanh(\kappa_{1}L)B_{10} & \cdots & \kappa_{N}\tanh(\kappa_{N}L)B_{N0} \\
0 & k_{1}A_{1} & \cdots & \vdots & 0 & \kappa_{1}\tanh(\kappa_{1}L)B_{11} & \cdots & \kappa_{N}\tanh(\kappa_{N}L)B_{N1} \\
\vdots & \vdots & \ddots & 0 & \vdots & \vdots & \ddots & \vdots \\
0 & \cdots & 0 & k_{N}A_{N} & 0 & \kappa_{1}\tanh(\kappa_{1}L)B_{1N} & \cdots & \kappa_{N}\tanh(\kappa_{N}L)B_{NN} \\
\end{bmatrix}
\begin{bmatrix}
a^s_{0} \\ a^s_{1} \\ \vdots \\ a^s_{N} \\ b^s_{0} \\ b^s_{1} \\ \vdots \\ b^s_{N} \\
\end{bmatrix}
=\begin{bmatrix}
-B_{00} \\ -B_{10} \\ \vdots \\ -B_{N0} \\ k_{0}A_{0} \\ 0 \\ \vdots \\ 0 \\
\end{bmatrix}
</math>
</center>

To solve the system of equations for the '''antisymmetric''' case, we set the upper limit of <math>m</math> to be <math>N</math>. This resulting system can be expressed in the block matrix form below,

<center>
<math>
\begin{bmatrix}
B_{00} & B_{01} & \cdots & B_{0N} & -C_{0} & 0 & \cdots & 0 \\
B_{10} & B_{11} & \cdots & B_{1N} & 0 & -C_{1} & \cdots & \vdots \\
\vdots & \vdots & \ddots & \vdots & \vdots & \vdots & \ddots & 0 \\
B_{N0} & B_{N1} & \cdots & B_{NN} & 0 & \cdots & 0 & -C_{N} \\
k_{0}A_{0} & 0 & \cdots & 0 & 0 & \kappa_{1}\coth(\kappa_{1}L)B_{10} & \cdots & \kappa_{N}\coth(\kappa_{N}L)B_{N0} \\
0 & k_{1}A_{1} & \cdots & \vdots & 0 & \kappa_{1}\coth(\kappa_{1}L)B_{11} & \cdots & \kappa_{N}\coth(\kappa_{N}L)B_{N1} \\
\vdots & \vdots & \ddots & 0 & \vdots & \vdots & \ddots & \vdots \\
0 & \cdots & 0 & k_{N}A_{N} & 0 & \kappa_{1}\coth(\kappa_{1}L)B_{1N} & \cdots & \kappa_{N}\coth(\kappa_{N}L)B_{NN} \\
\end{bmatrix}
\begin{bmatrix}
a^a_{0} \\ a^a_{1} \\ \vdots \\ a^a_{N} \\ b^a_{0} \\ b^a_{1} \\ \vdots \\ b^a_{N} \\
\end{bmatrix}
=\begin{bmatrix}
-B_{00} \\ -B_{10} \\ \vdots \\ -B_{N0} \\ k_{0}A_{0} \\ 0 \\ \vdots \\ 0 \\
\end{bmatrix}
</math>
</center>

We then simply need to solve the linear system of equations.

== 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 Rectangle

2012-03-16T01:28:57Z

Tertius: /* Numerical Solution */

{{incomplete pages}}

== Introduction ==

The problem consists of a region to the left with a free water surface and a region to the right with a rigid fixed semi infinite rectangular block with height <math>d + ...</math> freeboard with depth of submergence being <math>d</math> through which no flow is possible.
We begin with the simple problem when the waves are normally incident (so that the problem is truly two-dimensional). We can extend this solution to a finite dock using symmetry
[[Eigenfunction Matching for a Finite Rectangle using Symmetry]]

== Governing Equations ==

We begin with the [[Frequency Domain Problem]] for a partially submerged dock which occupies
the region <math>x>0</math> (we assume <math>e^{i\omega t}</math> time dependence).
The depth of submergence is <math>d</math>.
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>
\begin{align}
\Delta\phi &= 0, \,\,\, -h\leq z\leq 0, \,\, x < 0, \\

\Delta\phi &= 0, \,\,\, -h\leq z\leq -d, \,\, x > 0. \\

\partial_z\phi &= \alpha\phi, \,\,\, z=0, \,\, x<0, \\

\partial_x\phi &= 0, \,\,\, -d\leq z\leq 0, \,\, x=0, \\

\partial_z\phi &= 0, \,\,\, z=-d, \,\, x>0, \\

\partial_z\phi &= 0, \,\,\, z=-h.
\end{align}
</math></center>

Where <math>\alpha = \omega^2/g</math>.

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\leq 0</math>
and <math>x\geq 0</math>.

{{separation of variables in two dimensions}}

{{separation of variables for a free surface}}

=== Separation of Variables for a Dock ===

The separation of variables equation for a partially submerged dock is given by:
<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>
Z^{\prime} (-d) = 0.
</math></center>
The solution is
<math>k=\kappa_{m}= \frac{m\pi}{h-d} \,</math> , <math>m\in\mathbb{N}\cup\left\{0\right\}</math> and
<center><math>
Z = \psi_{m}\left( z\right) = \cos\kappa_{m}(z+h),\quad
m\in\mathbb{N}\cup\left\{0\right\}.</math></center>
We note that
<center><math>
\int\nolimits_{-h}^{-d}\psi_{m}(z)\psi_{n}(z) \mathrm{d} z=C_{m}\delta_{mn},
</math></center>
where
<center>
<math>
C_{m} = \frac{1}{2} \left( \frac{\cos\kappa_{m}(h-d)\sin\kappa_{m}(h-d)+\kappa_{m}(h-d)}{\kappa_{m}} \right).
</math>
</center>
In addition there is the special case of when <math>m=n=0</math> ,
<center>
<math>
C_{0} = h-d,
</math>
</center>
where <math>\delta_{00}=1</math> .

=== Inner product between free surface and dock modes ===

<center><math>
B_{mn} = \int\nolimits_{-h}^{-d}\psi_{m}(z)\chi_{n}(z) \mathrm{d} z , \quad m\geq1, \quad n\geq0,
</math></center>
where
<center><math>
B_{mn}= \int\nolimits_{-h}^{-d} \frac{\cos\kappa_{m}(z+h)\cos k_{n}(z+h)}{\cos(k_{n}h)} \mathrm{d} z=\frac{-(-1)^m k_{n}\sin k_{n}(h-d)}
{\cos(k_{n}h)(\kappa_{m}^{2}-k_{n}^{2})}.
</math></center>
The evaluation of this inner product works on the assumption that the two roots <math>\kappa_{m}</math> and <math>k_{n}</math> do not coincide. The Matlab code given below has a routine to check for this case and make necessary adjustments.

We also have the case where the functions of the inner product are switched.
<center><math>
B_{nm} = \int\nolimits_{-h}^{-d}\chi_{m}(z)\psi_{n}(z) \mathrm{d} z , \quad m\geq0, \quad n\geq0.
</math></center>
For <math>m \geq 0</math> and <math>n \geq 1</math> the evaluation is above with the exception that the indices of the expression are switched.

Finally there is the special case of when <math>m \geq 0</math> and <math>n = 0</math>.
<center><math>
B_{0m} = \frac{\sin k_{m}(h-d)}{k_{m}\cos(k_{m}h)}.
</math></center>

=== Expansion of the potential ===

Before we can write down the expansion of the potentials we need to solve for the horizontal eigenfunctions. We start with the equation
<center>
<math>
X^{\prime\prime} - k^2 X = 0,
</math>
</center>
which has the general solution
<center>
<math>
X(x) = a e^{kx} + b e^{-kx},
</math>
</center>
where <math>a</math> and <math>b</math> are constants to be determined from the boundary conditions. The general solution, or eigenfunction expansion, for the potential <math>\phi</math> is obtained by superimposing all the possible modes; for the region of the '''free surface''' we have
<center>
<math>
\phi(x,z)=\sum_{m=0}^{\infty}\left(a_{m}e^{k_{m}x} + b_{m}e^{-k_{m}x}\right)\chi_{m}(z), \quad x<0.
</math>
</center>
In this expression the term involving <math>e^{ikx}</math> (recall that <math>k_{0} = -ik</math>) corresponds to a wave propagating towards large positive <math>x</math> while the term involving <math>e^{-ikx}</math> corresponds to a wave propagating towards large negative <math>x</math>. Terms which decay as <math>x \rightarrow -\infty</math> or as <math>x \rightarrow \infty</math> are referred to as evanescent modes.

The expansion of the potential for the region under the '''dock''' is
<center>
<math>
\phi(x,z)=\sum_{m=0}^{\infty}\left(a_{m}e^{\kappa_{m}x} + b_{m}e^{-\kappa_{m}x}\right)\psi_{m}(z), \quad x>0.
</math>
</center>
To obtain a unique solution, the diffracted field <math>\phi_{\mathrm{D}}</math> must satisfy the Sommerfeld radiation condition (see further above) specifying that the waves of this potential propagate away from the structure and the solution be bounded.

Application of the radiation condition to both the general solutions immediately above give diffracted potentials of
<center>
<math>
\phi_{\mathrm{D}}(x,z)=\sum_{m=0}^{\infty}a_{m}e^{k_{m}x}\chi_{m}(z), \quad x<0
</math>
</center>
and
<center>
<math>
\phi_{\mathrm{D}}(x,z)=\sum_{m=0}^{\infty}b_{m}
e^{-\kappa_{m}x}\psi_{m}(z), \quad 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 dock covered region. Therefore, the
potentials and their derivatives at <math>x=0</math> have to be equal.
We obtain:
<center><math>
\begin{align}
\chi_{0}\left( z\right) + \sum_{m=0}^{\infty}a_{m} \chi_{m}\left( z\right)
&=\sum_{m=0}^{\infty}b_{m}\psi_{m}(z), \quad -h \leq z\leq -d \\

-k_{0}\chi_{0}\left( z\right) +\sum_{m=0}^{\infty} a_{m}k_{m}\chi_{m}\left( z\right) &=
\begin{cases}
\quad \quad \quad 0,\quad \quad \quad \quad -d\leq z\leq 0 \\
-\sum_{m=1}^{\infty}b_{m}\kappa_{m}\psi_{m}(z),\quad -h\leq z\leq -d
\end{cases}
\end{align}
</math></center>

For the first equation we multiply both sides by <math>\psi_{n}(z) \,</math> and integrating from <math>-h</math> to <math>-d</math> to obtain:
<center><math>
B_{n0} + \sum^{\infty}_{m=0}a_{m}B_{nm} = b_{m}C_{m}\delta_{mn}, \quad n\in\mathbb{N}\cup\left\{0\right\}
</math></center>
and for the second equation we multiply both sides by <math>\chi_{n}(z) \,</math> and integrating from <math>-h</math> to <math>0</math> to obtain:
<center><math>
-k_{0}A_{0}\delta_{0n} + k_{m}a_{m}A_{m}\delta_{mn} = -\sum^{\infty}_{m=1}\kappa_{m}b_{m}B_{mn}, \quad n\in\mathbb{N}\cup\left\{0\right\}
</math></center>

Remember <math>\kappa_{0} = 0 \,</math>. Solving the equations above will yield the coefficients of the water velocity potential in the free surface and dock covered regions.

== Numerical Solution ==

To solve the system of equations, we set the upper limit of <math>m</math> to be <math>N</math>. This resulting system can be expressed in the block matrix form below,

<center>
<math>
\begin{bmatrix}
B_{00} & B_{01} & \cdots & B_{0N} & -C_{0} & 0 & \cdots & 0 \\
B_{10} & B_{11} & \cdots & B_{1N} & 0 & -C_{1} & \cdots & \vdots \\
\vdots & \vdots & \ddots & \vdots & \vdots & \vdots & \ddots & 0 \\
B_{N0} & B_{N1} & \cdots & B_{NN} & 0 & \cdots & 0 & -C_{N} \\
k_{0}A_{0} & 0 & \cdots & 0 & 0 & \kappa_{1}B_{10} & \cdots & \kappa_{N}B_{N0} \\
0 & k_{1}A_{1} & \cdots & \vdots & 0 & \kappa_{1}B_{11} & \cdots & \kappa_{N}B_{N1} \\
\vdots & \vdots & \ddots & 0 & \vdots & \vdots & \ddots & \vdots \\
0 & \cdots & 0 & k_{N}A_{N} & 0 & \kappa_{1}B_{1N} & \cdots & \kappa_{N}B_{NN} \\
\end{bmatrix}
\begin{bmatrix}
a_{0} \\ a_{1} \\ \vdots \\ a_{N} \\ b_{0} \\ b_{1} \\ \vdots \\ b_{N} \\
\end{bmatrix}
=\begin{bmatrix}
-B_{00} \\ -B_{10} \\ \vdots \\ -B_{N0} \\ k_{0}A_{0} \\ 0 \\ \vdots \\ 0 \\
\end{bmatrix}
</math>
</center>

We then simply need to solve the linear system of equations.

== Matlab Code ==

A program to calculate the coefficients for the semi-infinite dock problem 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 Rectangle

2012-03-16T01:28:34Z

Tertius: /* Numerical Solution */

Eigenfunction Matching for a Semi-Infinite Rectangle

2012-03-16T01:28:16Z

Tertius: /* Numerical Solution */

Eigenfunction Matching for a Finite Rectangle using Symmetry

2012-03-16T01:23:34Z

Tertius: /* Numerical Solution */

{{incomplete pages}}

== Introduction ==

The problem consists of a finite rigid fixed rectangle of length <math>2L</math> centered at <math>x = 0</math> and having a height of <math>d + ...</math> freeboard with depth of submergence being <math>d</math> through which no flow is possible. The regions to the left and right of the rectangle are the free water surface. Since the geometry of the rectangle is symmetric about <math>x = 0</math> we can decompose the velocity potential <math>\phi</math> into a symmetric <math>\phi^s</math> and antisymmetric <math>\phi^a</math> potential. We then solve for these two potentials in the left domain only <math>x \leq 0</math> and subsequently recompose the solution <math>\phi</math> for the entire domain. We begin with the simple problem when the waves are normally incident on the rectangle (dock) (so that the problem is truly two-dimensional). We can extend this solution to a pair of partially submerged finite rigid fixed rectangles using symmetry
[[Eigenfunction Matching for a Finite Rectangle using Symmetry]]

== Governing Equations ==

We begin with the [[Frequency Domain Problem]] for a partially submerged finite dock. Although the dock is of length <math>2L</math> we are only interested in the dock of <math>-L \leq x \leq 0</math> together with the fluid domain <math>x \leq 0</math>. We assume <math>e^{i\omega t}</math> time dependence and
the depth of submergence is <math>d</math>.
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>
\begin{align}
\Delta\phi^{s,a} &= 0, \,\,\, -h\leq z\leq 0, \,\, x < -L, \\

\Delta\phi^{s,a} &= 0, \,\,\, -h\leq z\leq -d, \,\, -L < x < 0. \\

\partial_z\phi^{s,a} &= \alpha\phi^{s,a}, \,\,\, z=0, \,\, x < -L, \\

\partial_x\phi^{s,a} &= 0, \,\,\, -d\leq z\leq 0, \,\, x = -L, \\

\partial_z\phi^{s,a} &= 0, \,\,\,
\begin{cases}
z = -d, \,\, -L < x < 0, \\
z = -h, \,\, x < 0.
\end{cases} \\

\partial_x\phi^s &= 0, \,\,\, x = 0, \,\, -h \leq z \leq -d, \\

\phi^a &= 0, \,\,\, x = 0, \,\, -h \leq z \leq -d.
\end{align}
</math></center>

Where <math>\alpha = \omega^2/g</math>.

<math>\phi^s(x,z)</math> is an even function and at <math>x = 0</math> the derivative of this function is equal to zero. <math>\phi^a(x,z)</math> is an odd function and at <math>x = 0</math> this function is equal to zero. These boundary conditions are important and enable the recomposed solution <math>\phi(x,z)</math> to be reflected into the domain of <math>x \geq 0</math>.

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\leq 0</math>
and <math>x\geq 0</math>.

{{separation of variables in two dimensions}}

{{separation of variables for a free surface}}

=== Separation of Variables for a Dock ===

The separation of variables equation for a partially submerged dock is given by:
<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>
Z^{\prime} (-d) = 0.
</math></center>
The solution is
<math>k=\kappa_{m}= \frac{m\pi}{h-d} \,</math> , <math>m\in\mathbb{N}\cup\left\{0\right\}</math> and
<center><math>
Z = \psi_{m}\left( z\right) = \cos\kappa_{m}(z+h),\quad
m\in\mathbb{N}\cup\left\{0\right\}.</math></center>
We note that
<center><math>
\int\nolimits_{-h}^{-d}\psi_{m}(z)\psi_{n}(z) \mathrm{d} z=C_{m}\delta_{mn},
</math></center>
where
<center>
<math>
C_{m} = \frac{1}{2} \left( \frac{\cos\kappa_{m}(h-d)\sin\kappa_{m}(h-d)+\kappa_{m}(h-d)}{\kappa_{m}} \right).
</math>
</center>
In addition there is the special case of when <math>m=n=0</math> ,
<center>
<math>
C_{0} = h-d,
</math>
</center>
where <math>\delta_{00}=1</math> .

=== Inner product between free surface and dock modes ===

<center><math>
B_{mn} = \int\nolimits_{-h}^{-d}\psi_{m}(z)\chi_{n}(z) \mathrm{d} z , \quad m\geq1, \quad n\geq0,
</math></center>
where
<center><math>
B_{mn}= \int\nolimits_{-h}^{-d} \frac{\cos\kappa_{m}(z+h)\cos k_{n}(z+h)}{\cos(k_{n}h)} \mathrm{d} z=\frac{-(-1)^m k_{n}\sin k_{n}(h-d)}
{\cos(k_{n}h)(\kappa_{m}^{2}-k_{n}^{2})}.
</math></center>
The evaluation of this inner product works on the assumption that the two roots <math>\kappa_{m}</math> and <math>k_{n}</math> do not coincide. The Matlab code given below has a routine to check for this case and make necessary adjustments.

We also have the case where the functions of the inner product are switched.
<center><math>
B_{nm} = \int\nolimits_{-h}^{-d}\chi_{m}(z)\psi_{n}(z) \mathrm{d} z , \quad m\geq0, \quad n\geq0.
</math></center>
For <math>m \geq 0</math> and <math>n \geq 1</math> the evaluation is above with the exception that the indices of the expression are switched.

Finally there is the special case of when <math>m \geq 0</math> and <math>n = 0</math>.
<center><math>
B_{0m} = \frac{\sin k_{m}(h-d)}{k_{m}\cos(k_{m}h)}.
</math></center>

=== Expansion of the potential ===

We need to apply some boundary conditions at plus and minus infinity,
where essentially 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 roots. Here we
must use a different condition, essentially identifying one solution
as the incoming wave and the other as the outgoing wave.

Therefore the diffracted potentials can
be expanded as
<center>
<math>
\phi(x,z)=\sum_{m=0}^{\infty}a_{m}e^{k_{m}x}\chi_{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 dock covered region. Therefore, the
potentials and their derivatives at <math>x=0</math> have to be equal.
We obtain:
<center><math>
\begin{align}
\chi_{0}\left( z\right) + \sum_{m=0}^{\infty}a_{m} \chi_{m}\left( z\right)
&=\sum_{m=0}^{\infty}b_{m}\psi_{m}(z), \quad -h \leq z\leq -d \\

-k_{0}\chi_{0}\left( z\right) +\sum_{m=0}^{\infty} a_{m}k_{m}\chi_{m}\left( z\right) &=
\begin{cases}
\quad \quad \quad 0,\quad \quad \quad -d\leq z\leq 0 \\
-\sum_{m=0}^{\infty}b_{m}\kappa_{m}\psi_{m}(z),\quad -h\leq z\leq -d
\end{cases}
\end{align}
</math></center>

For the first equation we multiply both sides by <math>\psi_{n}(z) \,</math> and integrating from <math>-h</math> to <math>-d</math> to obtain:
<center><math>
B_{n0} + \sum^{\infty}_{m=0}a_{m}B_{nm} = b_{m}C_{m}\delta_{mn}, n\in\mathbb{N}\cup\left\{0\right\}
</math></center>
and for the second equation we multiply both sides by <math>\chi_{n}(z) \,</math> and integrating from <math>-h</math> to <math>0</math> to obtain:
<center><math>
-k_{0}A_{0}\delta_{0n} + k_{m}a_{m}A_{m} = -\sum^{\infty}_{m=0}\kappa_{m}b_{m}B_{mn}, n\in\mathbb{N}\cup\left\{0\right\}
</math></center>

Solving the equations above will yield the coefficients of the water velocity potential in the dock covered region.

== Numerical Solution ==

To solve the system of equations for the '''symmetric''' case, we set the upper limit of <math>m</math> to be <math>N</math>. This resulting system can be expressed in the block matrix form below,

<center>
<math>
\begin{bmatrix}
B_{00} & B_{01} & \cdots & B_{0N} & -C_{0} & 0 & \cdots & 0 \\
B_{10} & B_{11} & \cdots & B_{1N} & 0 & -C_{1} & \cdots & \vdots \\
\vdots & \vdots & \ddots & \vdots & \vdots & \vdots & \ddots & 0 \\
B_{N0} & B_{N1} & \cdots & B_{NN} & 0 & \cdots & 0 & -C_{N} \\
k_{0}A_{0} & 0 & \cdots & 0 & 0 & \kappa_{1}\tanh(\kappa_{1}L)B_{10} & \cdots & \kappa_{N}\tanh(\kappa_{N}L)B_{N0} \\
0 & k_{1}A_{1} & \cdots & \vdots & 0 & \kappa_{1}\tanh(\kappa_{1}L)B_{11} & \cdots & \kappa_{N}\tanh(\kappa_{N}L)B_{N1} \\
\vdots & \vdots & \ddots & 0 & \vdots & \vdots & \ddots & \vdots \\
0 & \cdots & 0 & k_{N}A_{N} & 0 & \kappa_{1}\tanh(\kappa_{1}L)B_{1N} & \cdots & \kappa_{N}\tanh(\kappa_{N}L)B_{NN} \\
\end{bmatrix}
\begin{bmatrix}
a^s_{0} \\ a^s_{1} \\ \vdots \\ a^s_{N} \\ b^s_{0} \\ b^s_{1} \\ \vdots \\ b^s_{N} \\
\end{bmatrix}
=\begin{bmatrix}
-B_{00} \\ -B_{10} \\ \vdots \\ -B_{N0} \\ k_{0}A_{0} \\ 0 \\ \vdots \\ 0 \\
\end{bmatrix}
</math>
</center>

To solve the system of equations for the '''antisymmetric''' case, we set the upper limit of <math>m</math> to be <math>N</math>. This resulting system can be expressed in the block matrix form below,

<center>
<math>
\begin{bmatrix}
B_{00} & B_{01} & \cdots & B_{0N} & -C_{0} & 0 & \cdots & 0 \\
B_{10} & B_{11} & \cdots & B_{1N} & 0 & -C_{1} & \cdots & \vdots \\
\vdots & \vdots & \ddots & \vdots & \vdots & \vdots & \ddots & 0 \\
B_{N0} & B_{N1} & \cdots & B_{NN} & 0 & \cdots & 0 & -C_{N} \\
k_{0}A_{0} & 0 & \cdots & 0 & 0 & \kappa_{1}\coth(\kappa_{1}L)B_{10} & \cdots & \kappa_{N}\coth(\kappa_{N}L)B_{N0} \\
0 & k_{1}A_{1} & \cdots & \vdots & 0 & \kappa_{1}\coth(\kappa_{1}L)B_{11} & \cdots & \kappa_{N}\coth(\kappa_{N}L)B_{N1} \\
\vdots & \vdots & \ddots & 0 & \vdots & \vdots & \ddots & \vdots \\
0 & \cdots & 0 & k_{N}A_{N} & 0 & \kappa_{1}\coth(\kappa_{1}L)B_{1N} & \cdots & \kappa_{N}\coth(\kappa_{N}L)B_{NN} \\
\end{bmatrix}
\begin{bmatrix}
a^a_{0} \\ a^a_{1} \\ \vdots \\ a^a_{N} \\ b^a_{0} \\ b^a_{1} \\ \vdots \\ b^a_{N} \\
\end{bmatrix}
=\begin{bmatrix}
-B_{00} \\ -B_{10} \\ \vdots \\ -B_{N0} \\ k_{0}A_{0} \\ 0 \\ \vdots \\ 0 \\
\end{bmatrix}
</math>
</center>

We then simply need to solve the linear system of equations.

== 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 Finite Rectangle using Symmetry

2012-03-16T01:20:12Z

Tertius: /* Numerical Solution */

{{incomplete pages}}

== Introduction ==

The problem consists of a finite rigid fixed rectangle of length <math>2L</math> centered at <math>x = 0</math> and having a height of <math>d + ...</math> freeboard with depth of submergence being <math>d</math> through which no flow is possible. The regions to the left and right of the rectangle are the free water surface. Since the geometry of the rectangle is symmetric about <math>x = 0</math> we can decompose the velocity potential <math>\phi</math> into a symmetric <math>\phi^s</math> and antisymmetric <math>\phi^a</math> potential. We then solve for these two potentials in the left domain only <math>x \leq 0</math> and subsequently recompose the solution <math>\phi</math> for the entire domain. We begin with the simple problem when the waves are normally incident on the rectangle (dock) (so that the problem is truly two-dimensional). We can extend this solution to a pair of partially submerged finite rigid fixed rectangles using symmetry
[[Eigenfunction Matching for a Finite Rectangle using Symmetry]]

== Governing Equations ==

We begin with the [[Frequency Domain Problem]] for a partially submerged finite dock. Although the dock is of length <math>2L</math> we are only interested in the dock of <math>-L \leq x \leq 0</math> together with the fluid domain <math>x \leq 0</math>. We assume <math>e^{i\omega t}</math> time dependence and
the depth of submergence is <math>d</math>.
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>
\begin{align}
\Delta\phi^{s,a} &= 0, \,\,\, -h\leq z\leq 0, \,\, x < -L, \\

\Delta\phi^{s,a} &= 0, \,\,\, -h\leq z\leq -d, \,\, -L < x < 0. \\

\partial_z\phi^{s,a} &= \alpha\phi^{s,a}, \,\,\, z=0, \,\, x < -L, \\

\partial_x\phi^{s,a} &= 0, \,\,\, -d\leq z\leq 0, \,\, x = -L, \\

\partial_z\phi^{s,a} &= 0, \,\,\,
\begin{cases}
z = -d, \,\, -L < x < 0, \\
z = -h, \,\, x < 0.
\end{cases} \\

\partial_x\phi^s &= 0, \,\,\, x = 0, \,\, -h \leq z \leq -d, \\

\phi^a &= 0, \,\,\, x = 0, \,\, -h \leq z \leq -d.
\end{align}
</math></center>

Where <math>\alpha = \omega^2/g</math>.

<math>\phi^s(x,z)</math> is an even function and at <math>x = 0</math> the derivative of this function is equal to zero. <math>\phi^a(x,z)</math> is an odd function and at <math>x = 0</math> this function is equal to zero. These boundary conditions are important and enable the recomposed solution <math>\phi(x,z)</math> to be reflected into the domain of <math>x \geq 0</math>.

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\leq 0</math>
and <math>x\geq 0</math>.

{{separation of variables in two dimensions}}

{{separation of variables for a free surface}}

=== Separation of Variables for a Dock ===

The separation of variables equation for a partially submerged dock is given by:
<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>
Z^{\prime} (-d) = 0.
</math></center>
The solution is
<math>k=\kappa_{m}= \frac{m\pi}{h-d} \,</math> , <math>m\in\mathbb{N}\cup\left\{0\right\}</math> and
<center><math>
Z = \psi_{m}\left( z\right) = \cos\kappa_{m}(z+h),\quad
m\in\mathbb{N}\cup\left\{0\right\}.</math></center>
We note that
<center><math>
\int\nolimits_{-h}^{-d}\psi_{m}(z)\psi_{n}(z) \mathrm{d} z=C_{m}\delta_{mn},
</math></center>
where
<center>
<math>
C_{m} = \frac{1}{2} \left( \frac{\cos\kappa_{m}(h-d)\sin\kappa_{m}(h-d)+\kappa_{m}(h-d)}{\kappa_{m}} \right).
</math>
</center>
In addition there is the special case of when <math>m=n=0</math> ,
<center>
<math>
C_{0} = h-d,
</math>
</center>
where <math>\delta_{00}=1</math> .

=== Inner product between free surface and dock modes ===

<center><math>
B_{mn} = \int\nolimits_{-h}^{-d}\psi_{m}(z)\chi_{n}(z) \mathrm{d} z , \quad m\geq1, \quad n\geq0,
</math></center>
where
<center><math>
B_{mn}= \int\nolimits_{-h}^{-d} \frac{\cos\kappa_{m}(z+h)\cos k_{n}(z+h)}{\cos(k_{n}h)} \mathrm{d} z=\frac{-(-1)^m k_{n}\sin k_{n}(h-d)}
{\cos(k_{n}h)(\kappa_{m}^{2}-k_{n}^{2})}.
</math></center>
The evaluation of this inner product works on the assumption that the two roots <math>\kappa_{m}</math> and <math>k_{n}</math> do not coincide. The Matlab code given below has a routine to check for this case and make necessary adjustments.

We also have the case where the functions of the inner product are switched.
<center><math>
B_{nm} = \int\nolimits_{-h}^{-d}\chi_{m}(z)\psi_{n}(z) \mathrm{d} z , \quad m\geq0, \quad n\geq0.
</math></center>
For <math>m \geq 0</math> and <math>n \geq 1</math> the evaluation is above with the exception that the indices of the expression are switched.

Finally there is the special case of when <math>m \geq 0</math> and <math>n = 0</math>.
<center><math>
B_{0m} = \frac{\sin k_{m}(h-d)}{k_{m}\cos(k_{m}h)}.
</math></center>

=== Expansion of the potential ===

We need to apply some boundary conditions at plus and minus infinity,
where essentially 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 roots. Here we
must use a different condition, essentially identifying one solution
as the incoming wave and the other as the outgoing wave.

Therefore the diffracted potentials can
be expanded as
<center>
<math>
\phi(x,z)=\sum_{m=0}^{\infty}a_{m}e^{k_{m}x}\chi_{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 dock covered region. Therefore, the
potentials and their derivatives at <math>x=0</math> have to be equal.
We obtain:
<center><math>
\begin{align}
\chi_{0}\left( z\right) + \sum_{m=0}^{\infty}a_{m} \chi_{m}\left( z\right)
&=\sum_{m=0}^{\infty}b_{m}\psi_{m}(z), \quad -h \leq z\leq -d \\

-k_{0}\chi_{0}\left( z\right) +\sum_{m=0}^{\infty} a_{m}k_{m}\chi_{m}\left( z\right) &=
\begin{cases}
\quad \quad \quad 0,\quad \quad \quad -d\leq z\leq 0 \\
-\sum_{m=0}^{\infty}b_{m}\kappa_{m}\psi_{m}(z),\quad -h\leq z\leq -d
\end{cases}
\end{align}
</math></center>

For the first equation we multiply both sides by <math>\psi_{n}(z) \,</math> and integrating from <math>-h</math> to <math>-d</math> to obtain:
<center><math>
B_{n0} + \sum^{\infty}_{m=0}a_{m}B_{nm} = b_{m}C_{m}\delta_{mn}, n\in\mathbb{N}\cup\left\{0\right\}
</math></center>
and for the second equation we multiply both sides by <math>\chi_{n}(z) \,</math> and integrating from <math>-h</math> to <math>0</math> to obtain:
<center><math>
-k_{0}A_{0}\delta_{0n} + k_{m}a_{m}A_{m} = -\sum^{\infty}_{m=0}\kappa_{m}b_{m}B_{mn}, n\in\mathbb{N}\cup\left\{0\right\}
</math></center>

Solving the equations above will yield the coefficients of the water velocity potential in the dock covered region.

== Numerical Solution ==

To solve the system of equations for the '''symmetric''' case, we set the upper limit of <math>m</math> to be <math>N</math>. This resulting system can be expressed in the block matrix form below,

<center>
<math>
\begin{bmatrix}
B_{00} & B_{01} & \cdots & B_{0N} & -C_{0} & 0 & \cdots & 0 \\
B_{10} & B_{11} & \cdots & B_{1N} & 0 & -C_{1} & \cdots & \vdots \\
\vdots & \vdots & \ddots & \vdots & \vdots & \vdots & \ddots & 0 \\
B_{N0} & B_{N1} & \cdots & B_{NN} & 0 & \cdots & 0 & -C_{N} \\
k_{0}A_{0} & 0 & \cdots & 0 & 0 & \kappa_{1}\tanh(\kappa_{1}L)B_{10} & \cdots & \kappa_{N}\tanh(\kappa_{N}L)B_{N0} \\
0 & k_{1}A_{1} & \cdots & \vdots & 0 & \kappa_{1}\tanh(\kappa_{1}L)B_{11} & \cdots & \kappa_{N}\tanh(\kappa_{N}L)B_{N1} \\
\vdots & \vdots & \ddots & 0 & \vdots & \vdots & \ddots & \vdots \\
0 & \cdots & 0 & k_{N}A_{N} & 0 & \kappa_{1}\tanh(\kappa_{1}L)B_{1N} & \cdots & \kappa_{N}\tanh(\kappa_{N}L)B_{NN} \\
\end{bmatrix}
\begin{bmatrix}
a^s_{0} \\ a^s_{1} \\ \vdots \\ a^s_{N} \\ b^s_{0} \\ b^s_{1} \\ \vdots \\ b^s_{N} \\
\end{bmatrix}
=\begin{bmatrix}
-B_{00} \\ -B_{10} \\ \vdots \\ -B_{N0} \\ k_{0}A_{0} \\ 0 \\ \vdots \\ 0 \\
\end{bmatrix}
</math>
</center>

To solve the system of equations for the '''antisymmetric''' case, we set the upper limit of <math>m</math> to be <math>N</math>. This resulting system can be expressed in the block matrix form below,

<center>
<math>
\begin{bmatrix}
B_{00} & B_{01} & \cdots & B_{0N} & -C_{0} & 0 & \cdots & 0 \\
B_{10} & B_{11} & \cdots & B_{1N} & 0 & -C_{1} & \cdots & \vdots \\
\vdots & \vdots & \ddots & \vdots & \vdots & \vdots & \ddots & 0 \\
B_{N0} & B_{N1} & \cdots & B_{NN} & 0 & \cdots & 0 & -C_{N} \\
k_{0}A_{0} & 0 & \cdots & 0 & 0 & \kappa_{1}\tanh(\kappa_{1}L)B_{10} & \cdots & \kappa_{N}\tanh(\kappa_{N}L)B_{N0} \\
0 & k_{1}A_{1} & \cdots & \vdots & 0 & \kappa_{1}\tanh(\kappa_{1}L)B_{11} & \cdots & \kappa_{N}\tanh(\kappa_{N}L)B_{N1} \\
\vdots & \vdots & \ddots & 0 & \vdots & \vdots & \ddots & \vdots \\
0 & \cdots & 0 & k_{N}A_{N} & 0 & \kappa_{1}\tanh(\kappa_{1}L)B_{1N} & \cdots & \kappa_{N}\tanh(\kappa_{N}L)B_{NN} \\
\end{bmatrix}
\begin{bmatrix}
a^s_{0} \\ a^s_{1} \\ \vdots \\ a^s_{N} \\ b^s_{0} \\ b^s_{1} \\ \vdots \\ b^s_{N} \\
\end{bmatrix}
=\begin{bmatrix}
-B_{00} \\ -B_{10} \\ \vdots \\ -B_{N0} \\ k_{0}A_{0} \\ 0 \\ \vdots \\ 0 \\
\end{bmatrix}
</math>
</center>

We then simply need to solve the linear system of equations.

== 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 Finite Rectangle using Symmetry

2012-03-16T01:18:03Z

Tertius: /* Numerical Solution */

{{incomplete pages}}

== Introduction ==

The problem consists of a finite rigid fixed rectangle of length <math>2L</math> centered at <math>x = 0</math> and having a height of <math>d + ...</math> freeboard with depth of submergence being <math>d</math> through which no flow is possible. The regions to the left and right of the rectangle are the free water surface. Since the geometry of the rectangle is symmetric about <math>x = 0</math> we can decompose the velocity potential <math>\phi</math> into a symmetric <math>\phi^s</math> and antisymmetric <math>\phi^a</math> potential. We then solve for these two potentials in the left domain only <math>x \leq 0</math> and subsequently recompose the solution <math>\phi</math> for the entire domain. We begin with the simple problem when the waves are normally incident on the rectangle (dock) (so that the problem is truly two-dimensional). We can extend this solution to a pair of partially submerged finite rigid fixed rectangles using symmetry
[[Eigenfunction Matching for a Finite Rectangle using Symmetry]]

== Governing Equations ==

We begin with the [[Frequency Domain Problem]] for a partially submerged finite dock. Although the dock is of length <math>2L</math> we are only interested in the dock of <math>-L \leq x \leq 0</math> together with the fluid domain <math>x \leq 0</math>. We assume <math>e^{i\omega t}</math> time dependence and
the depth of submergence is <math>d</math>.
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>
\begin{align}
\Delta\phi^{s,a} &= 0, \,\,\, -h\leq z\leq 0, \,\, x < -L, \\

\Delta\phi^{s,a} &= 0, \,\,\, -h\leq z\leq -d, \,\, -L < x < 0. \\

\partial_z\phi^{s,a} &= \alpha\phi^{s,a}, \,\,\, z=0, \,\, x < -L, \\

\partial_x\phi^{s,a} &= 0, \,\,\, -d\leq z\leq 0, \,\, x = -L, \\

\partial_z\phi^{s,a} &= 0, \,\,\,
\begin{cases}
z = -d, \,\, -L < x < 0, \\
z = -h, \,\, x < 0.
\end{cases} \\

\partial_x\phi^s &= 0, \,\,\, x = 0, \,\, -h \leq z \leq -d, \\

\phi^a &= 0, \,\,\, x = 0, \,\, -h \leq z \leq -d.
\end{align}
</math></center>

Where <math>\alpha = \omega^2/g</math>.

<math>\phi^s(x,z)</math> is an even function and at <math>x = 0</math> the derivative of this function is equal to zero. <math>\phi^a(x,z)</math> is an odd function and at <math>x = 0</math> this function is equal to zero. These boundary conditions are important and enable the recomposed solution <math>\phi(x,z)</math> to be reflected into the domain of <math>x \geq 0</math>.

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\leq 0</math>
and <math>x\geq 0</math>.

{{separation of variables in two dimensions}}

{{separation of variables for a free surface}}

=== Separation of Variables for a Dock ===

The separation of variables equation for a partially submerged dock is given by:
<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>
Z^{\prime} (-d) = 0.
</math></center>
The solution is
<math>k=\kappa_{m}= \frac{m\pi}{h-d} \,</math> , <math>m\in\mathbb{N}\cup\left\{0\right\}</math> and
<center><math>
Z = \psi_{m}\left( z\right) = \cos\kappa_{m}(z+h),\quad
m\in\mathbb{N}\cup\left\{0\right\}.</math></center>
We note that
<center><math>
\int\nolimits_{-h}^{-d}\psi_{m}(z)\psi_{n}(z) \mathrm{d} z=C_{m}\delta_{mn},
</math></center>
where
<center>
<math>
C_{m} = \frac{1}{2} \left( \frac{\cos\kappa_{m}(h-d)\sin\kappa_{m}(h-d)+\kappa_{m}(h-d)}{\kappa_{m}} \right).
</math>
</center>
In addition there is the special case of when <math>m=n=0</math> ,
<center>
<math>
C_{0} = h-d,
</math>
</center>
where <math>\delta_{00}=1</math> .

=== Inner product between free surface and dock modes ===

<center><math>
B_{mn} = \int\nolimits_{-h}^{-d}\psi_{m}(z)\chi_{n}(z) \mathrm{d} z , \quad m\geq1, \quad n\geq0,
</math></center>
where
<center><math>
B_{mn}= \int\nolimits_{-h}^{-d} \frac{\cos\kappa_{m}(z+h)\cos k_{n}(z+h)}{\cos(k_{n}h)} \mathrm{d} z=\frac{-(-1)^m k_{n}\sin k_{n}(h-d)}
{\cos(k_{n}h)(\kappa_{m}^{2}-k_{n}^{2})}.
</math></center>
The evaluation of this inner product works on the assumption that the two roots <math>\kappa_{m}</math> and <math>k_{n}</math> do not coincide. The Matlab code given below has a routine to check for this case and make necessary adjustments.

We also have the case where the functions of the inner product are switched.
<center><math>
B_{nm} = \int\nolimits_{-h}^{-d}\chi_{m}(z)\psi_{n}(z) \mathrm{d} z , \quad m\geq0, \quad n\geq0.
</math></center>
For <math>m \geq 0</math> and <math>n \geq 1</math> the evaluation is above with the exception that the indices of the expression are switched.

Finally there is the special case of when <math>m \geq 0</math> and <math>n = 0</math>.
<center><math>
B_{0m} = \frac{\sin k_{m}(h-d)}{k_{m}\cos(k_{m}h)}.
</math></center>

=== Expansion of the potential ===

We need to apply some boundary conditions at plus and minus infinity,
where essentially 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 roots. Here we
must use a different condition, essentially identifying one solution
as the incoming wave and the other as the outgoing wave.

Therefore the diffracted potentials can
be expanded as
<center>
<math>
\phi(x,z)=\sum_{m=0}^{\infty}a_{m}e^{k_{m}x}\chi_{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 dock covered region. Therefore, the
potentials and their derivatives at <math>x=0</math> have to be equal.
We obtain:
<center><math>
\begin{align}
\chi_{0}\left( z\right) + \sum_{m=0}^{\infty}a_{m} \chi_{m}\left( z\right)
&=\sum_{m=0}^{\infty}b_{m}\psi_{m}(z), \quad -h \leq z\leq -d \\

-k_{0}\chi_{0}\left( z\right) +\sum_{m=0}^{\infty} a_{m}k_{m}\chi_{m}\left( z\right) &=
\begin{cases}
\quad \quad \quad 0,\quad \quad \quad -d\leq z\leq 0 \\
-\sum_{m=0}^{\infty}b_{m}\kappa_{m}\psi_{m}(z),\quad -h\leq z\leq -d
\end{cases}
\end{align}
</math></center>

For the first equation we multiply both sides by <math>\psi_{n}(z) \,</math> and integrating from <math>-h</math> to <math>-d</math> to obtain:
<center><math>
B_{n0} + \sum^{\infty}_{m=0}a_{m}B_{nm} = b_{m}C_{m}\delta_{mn}, n\in\mathbb{N}\cup\left\{0\right\}
</math></center>
and for the second equation we multiply both sides by <math>\chi_{n}(z) \,</math> and integrating from <math>-h</math> to <math>0</math> to obtain:
<center><math>
-k_{0}A_{0}\delta_{0n} + k_{m}a_{m}A_{m} = -\sum^{\infty}_{m=0}\kappa_{m}b_{m}B_{mn}, n\in\mathbb{N}\cup\left\{0\right\}
</math></center>

Solving the equations above will yield the coefficients of the water velocity potential in the dock covered region.

== Numerical Solution ==

To solve the system of equations for the symmetric case, we set the upper limit of <math>m</math> to be <math>N</math>. This resulting system can be expressed in the block matrix form below,

<center>
<math>
\begin{bmatrix}
B_{00} & B_{01} & \cdots & B_{0N} & -C_{0} & 0 & \cdots & 0 \\
B_{10} & B_{11} & \cdots & B_{1N} & 0 & -C_{1} & \cdots & \vdots \\
\vdots & \vdots & \ddots & \vdots & \vdots & \vdots & \ddots & 0 \\
B_{N0} & B_{N1} & \cdots & B_{NN} & 0 & \cdots & 0 & -C_{N} \\
k_{0}A_{0} & 0 & \cdots & 0 & 0 & \kappa_{1}\tanh(\kappa_{1}L)B_{10} & \cdots & \kappa_{N}\tanh(\kappa_{N}L)B_{N0} \\
0 & k_{1}A_{1} & \cdots & \vdots & 0 & \kappa_{1}\tanh(\kappa_{1}L)B_{11} & \cdots & \kappa_{N}\tanh(\kappa_{N}L)B_{N1} \\
\vdots & \vdots & \ddots & 0 & \vdots & \vdots & \ddots & \vdots \\
0 & \cdots & 0 & k_{N}A_{N} & 0 & \kappa_{1}\tanh(\kappa_{1}L)B_{1N} & \cdots & \kappa_{N}\tanh(\kappa_{N}L)B_{NN} \\
\end{bmatrix}
\begin{bmatrix}
a^s_{0} \\ a^s_{1} \\ \vdots \\ a^s_{N} \\ b^s_{0} \\ b^s_{1} \\ \vdots \\ b^s_{N} \\
\end{bmatrix}
=\begin{bmatrix}
-B_{00} \\ -B_{10} \\ \vdots \\ -B_{N0} \\ k_{0}A_{0} \\ 0 \\ \vdots \\ 0 \\
\end{bmatrix}
</math>
</center>

To solve the system of equations for the antisymmetric case, we set the upper limit of <math>m</math> to be <math>N</math>. This resulting system can be expressed in the block matrix form below,

<center>
<math>
\begin{bmatrix}
B_{00} & B_{01} & \cdots & B_{0N} & -C_{0} & 0 & \cdots & 0 \\
B_{10} & B_{11} & \cdots & B_{1N} & 0 & -C_{1} & \cdots & \vdots \\
\vdots & \vdots & \ddots & \vdots & \vdots & \vdots & \ddots & 0 \\
B_{N0} & B_{N1} & \cdots & B_{NN} & 0 & \cdots & 0 & -C_{N} \\
k_{0}A_{0} & 0 & \cdots & 0 & 0 & \kappa_{1}\tanh(\kappa_{1}L)B_{10} & \cdots & \kappa_{N}\tanh(\kappa_{N}L)B_{N0} \\
0 & k_{1}A_{1} & \cdots & \vdots & 0 & \kappa_{1}\tanh(\kappa_{1}L)B_{11} & \cdots & \kappa_{N}\tanh(\kappa_{N}L)B_{N1} \\
\vdots & \vdots & \ddots & 0 & \vdots & \vdots & \ddots & \vdots \\
0 & \cdots & 0 & k_{N}A_{N} & 0 & \kappa_{1}\tanh(\kappa_{1}L)B_{1N} & \cdots & \kappa_{N}\tanh(\kappa_{N}L)B_{NN} \\
\end{bmatrix}
\begin{bmatrix}
a^s_{0} \\ a^s_{1} \\ \vdots \\ a^s_{N} \\ b^s_{0} \\ b^s_{1} \\ \vdots \\ b^s_{N} \\
\end{bmatrix}
=\begin{bmatrix}
-B_{00} \\ -B_{10} \\ \vdots \\ -B_{N0} \\ k_{0}A_{0} \\ 0 \\ \vdots \\ 0 \\
\end{bmatrix}
</math>
</center>

We then simply need to solve the linear system of equations.

== 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 Finite Rectangle using Symmetry

2012-03-16T01:17:27Z

Tertius: /* Numerical Solution */

{{incomplete pages}}

== Introduction ==

The problem consists of a finite rigid fixed rectangle of length <math>2L</math> centered at <math>x = 0</math> and having a height of <math>d + ...</math> freeboard with depth of submergence being <math>d</math> through which no flow is possible. The regions to the left and right of the rectangle are the free water surface. Since the geometry of the rectangle is symmetric about <math>x = 0</math> we can decompose the velocity potential <math>\phi</math> into a symmetric <math>\phi^s</math> and antisymmetric <math>\phi^a</math> potential. We then solve for these two potentials in the left domain only <math>x \leq 0</math> and subsequently recompose the solution <math>\phi</math> for the entire domain. We begin with the simple problem when the waves are normally incident on the rectangle (dock) (so that the problem is truly two-dimensional). We can extend this solution to a pair of partially submerged finite rigid fixed rectangles using symmetry
[[Eigenfunction Matching for a Finite Rectangle using Symmetry]]

== Governing Equations ==

We begin with the [[Frequency Domain Problem]] for a partially submerged finite dock. Although the dock is of length <math>2L</math> we are only interested in the dock of <math>-L \leq x \leq 0</math> together with the fluid domain <math>x \leq 0</math>. We assume <math>e^{i\omega t}</math> time dependence and
the depth of submergence is <math>d</math>.
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>
\begin{align}
\Delta\phi^{s,a} &= 0, \,\,\, -h\leq z\leq 0, \,\, x < -L, \\

\Delta\phi^{s,a} &= 0, \,\,\, -h\leq z\leq -d, \,\, -L < x < 0. \\

\partial_z\phi^{s,a} &= \alpha\phi^{s,a}, \,\,\, z=0, \,\, x < -L, \\

\partial_x\phi^{s,a} &= 0, \,\,\, -d\leq z\leq 0, \,\, x = -L, \\

\partial_z\phi^{s,a} &= 0, \,\,\,
\begin{cases}
z = -d, \,\, -L < x < 0, \\
z = -h, \,\, x < 0.
\end{cases} \\

\partial_x\phi^s &= 0, \,\,\, x = 0, \,\, -h \leq z \leq -d, \\

\phi^a &= 0, \,\,\, x = 0, \,\, -h \leq z \leq -d.
\end{align}
</math></center>

Where <math>\alpha = \omega^2/g</math>.

<math>\phi^s(x,z)</math> is an even function and at <math>x = 0</math> the derivative of this function is equal to zero. <math>\phi^a(x,z)</math> is an odd function and at <math>x = 0</math> this function is equal to zero. These boundary conditions are important and enable the recomposed solution <math>\phi(x,z)</math> to be reflected into the domain of <math>x \geq 0</math>.

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\leq 0</math>
and <math>x\geq 0</math>.

{{separation of variables in two dimensions}}

{{separation of variables for a free surface}}

=== Separation of Variables for a Dock ===

The separation of variables equation for a partially submerged dock is given by:
<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>
Z^{\prime} (-d) = 0.
</math></center>
The solution is
<math>k=\kappa_{m}= \frac{m\pi}{h-d} \,</math> , <math>m\in\mathbb{N}\cup\left\{0\right\}</math> and
<center><math>
Z = \psi_{m}\left( z\right) = \cos\kappa_{m}(z+h),\quad
m\in\mathbb{N}\cup\left\{0\right\}.</math></center>
We note that
<center><math>
\int\nolimits_{-h}^{-d}\psi_{m}(z)\psi_{n}(z) \mathrm{d} z=C_{m}\delta_{mn},
</math></center>
where
<center>
<math>
C_{m} = \frac{1}{2} \left( \frac{\cos\kappa_{m}(h-d)\sin\kappa_{m}(h-d)+\kappa_{m}(h-d)}{\kappa_{m}} \right).
</math>
</center>
In addition there is the special case of when <math>m=n=0</math> ,
<center>
<math>
C_{0} = h-d,
</math>
</center>
where <math>\delta_{00}=1</math> .

=== Inner product between free surface and dock modes ===

<center><math>
B_{mn} = \int\nolimits_{-h}^{-d}\psi_{m}(z)\chi_{n}(z) \mathrm{d} z , \quad m\geq1, \quad n\geq0,
</math></center>
where
<center><math>
B_{mn}= \int\nolimits_{-h}^{-d} \frac{\cos\kappa_{m}(z+h)\cos k_{n}(z+h)}{\cos(k_{n}h)} \mathrm{d} z=\frac{-(-1)^m k_{n}\sin k_{n}(h-d)}
{\cos(k_{n}h)(\kappa_{m}^{2}-k_{n}^{2})}.
</math></center>
The evaluation of this inner product works on the assumption that the two roots <math>\kappa_{m}</math> and <math>k_{n}</math> do not coincide. The Matlab code given below has a routine to check for this case and make necessary adjustments.

We also have the case where the functions of the inner product are switched.
<center><math>
B_{nm} = \int\nolimits_{-h}^{-d}\chi_{m}(z)\psi_{n}(z) \mathrm{d} z , \quad m\geq0, \quad n\geq0.
</math></center>
For <math>m \geq 0</math> and <math>n \geq 1</math> the evaluation is above with the exception that the indices of the expression are switched.

Finally there is the special case of when <math>m \geq 0</math> and <math>n = 0</math>.
<center><math>
B_{0m} = \frac{\sin k_{m}(h-d)}{k_{m}\cos(k_{m}h)}.
</math></center>

=== Expansion of the potential ===

We need to apply some boundary conditions at plus and minus infinity,
where essentially 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 roots. Here we
must use a different condition, essentially identifying one solution
as the incoming wave and the other as the outgoing wave.

Therefore the diffracted potentials can
be expanded as
<center>
<math>
\phi(x,z)=\sum_{m=0}^{\infty}a_{m}e^{k_{m}x}\chi_{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 dock covered region. Therefore, the
potentials and their derivatives at <math>x=0</math> have to be equal.
We obtain:
<center><math>
\begin{align}
\chi_{0}\left( z\right) + \sum_{m=0}^{\infty}a_{m} \chi_{m}\left( z\right)
&=\sum_{m=0}^{\infty}b_{m}\psi_{m}(z), \quad -h \leq z\leq -d \\

-k_{0}\chi_{0}\left( z\right) +\sum_{m=0}^{\infty} a_{m}k_{m}\chi_{m}\left( z\right) &=
\begin{cases}
\quad \quad \quad 0,\quad \quad \quad -d\leq z\leq 0 \\
-\sum_{m=0}^{\infty}b_{m}\kappa_{m}\psi_{m}(z),\quad -h\leq z\leq -d
\end{cases}
\end{align}
</math></center>

For the first equation we multiply both sides by <math>\psi_{n}(z) \,</math> and integrating from <math>-h</math> to <math>-d</math> to obtain:
<center><math>
B_{n0} + \sum^{\infty}_{m=0}a_{m}B_{nm} = b_{m}C_{m}\delta_{mn}, n\in\mathbb{N}\cup\left\{0\right\}
</math></center>
and for the second equation we multiply both sides by <math>\chi_{n}(z) \,</math> and integrating from <math>-h</math> to <math>0</math> to obtain:
<center><math>
-k_{0}A_{0}\delta_{0n} + k_{m}a_{m}A_{m} = -\sum^{\infty}_{m=0}\kappa_{m}b_{m}B_{mn}, n\in\mathbb{N}\cup\left\{0\right\}
</math></center>

Solving the equations above will yield the coefficients of the water velocity potential in the dock covered region.

== Numerical Solution ==

To solve the system of equations for the symmetric case, we set the upper limit of <math>m</math> to be <math>N</math>. This resulting system can be expressed in the block matrix form below,

<center>
<math>
\begin{bmatrix}
B_{00} & B_{01} & \cdots & B_{0N} & -C_{0} & 0 & \cdots & 0 \\
B_{10} & B_{11} & \cdots & B_{1N} & 0 & -C_{1} & \cdots & \vdots \\
\vdots & \vdots & \ddots & \vdots & \vdots & \vdots & \ddots & 0 \\
B_{N0} & B_{N1} & \cdots & B_{NN} & 0 & \cdots & 0 & -C_{N} \\
k_{0}A_{0} & 0 & \cdots & 0 & 0 & \kappa_{1}\tanh(\kappa_{1}L)B_{10} & \cdots & \kappa_{N}\tanh(\kappa_{N}L)B_{N0} \\
0 & k_{1}A_{1} & \cdots & \vdots & 0 & \kappa_{1}\tanh(\kappa_{1}L)B_{11} & \cdots & \kappa_{N}\tanh(\kappa_{N}L)B_{N1} \\
\vdots & \vdots & \ddots & 0 & \vdots & \vdots & \ddots & \vdots \\
0 & \cdots & 0 & k_{N}A_{N} & 0 & \kappa_{1}\tanh(\kappa_{1}L)B_{1N} & \cdots & \kappa_{N}\tanh(\kappa_{N}L)B_{NN} \\
\end{bmatrix}
\begin{bmatrix}
a^s_{0} \\ a^s_{1} \\ \vdots \\ a^s_{N} \\ b^s_{0} \\ b^s_{1} \\ \vdots \\ b^s_{N} \\
\end{bmatrix}
=\begin{bmatrix}
-B_{00} \\ -B_{10} \\ \vdots \\ -B_{N0} \\ k_{0}A_{0} \\ 0 \\ \vdots \\ 0 \\
\end{bmatrix}
</math>
</center>

To solve the system of equations for the antisymmetric case, we set the upper limit of <math>m</math> to be <math>N</math>. This resulting system can be expressed in the block matrix form below,

<center>
<math>
\begin{bmatrix}
B_{00} & B_{01} & \cdots & B_{0N} & -C_{0} & 0 & \cdots & 0 \\
B_{10} & B_{11} & \cdots & B_{1N} & 0 & -C_{1} & \cdots & \vdots \\
\vdots & \vdots & \ddots & \vdots & \vdots & \vdots & \ddots & 0 \\
B_{N0} & B_{N1} & \cdots & B_{NN} & 0 & \cdots & 0 & -C_{N} \\
k_{0}A_{0} & 0 & \cdots & 0 & 0 & \kappa_{1}\tanh(\kappa_{1}L)B_{10} & \cdots & \kappa_{N}\tanh(\kappa_{N}L)B_{N0} \\
0 & k_{1}A_{1} & \cdots & \vdots & 0 & \kappa_{1}\tanh(\kappa_{1}L)B_{11} & \cdots & \kappa_{N}\tanh(\kappa_{N}L)B_{N1} \\
\vdots & \vdots & \ddots & 0 & \vdots & \vdots & \ddots & \vdots \\
0 & \cdots & 0 & k_{N}A_{N} & 0 & \kappa_{1}\tanh(\kappa_{1}L)B_{1N} & \cdots & \kappa_{N}\tanh(\kappa_{N}L)B_{NN} \\
\end{bmatrix}
\begin{bmatrix}
a^s_{0} \\ a^s_{1} \\ \vdots \\ a^s_{N} \\ b^s_{0} \\ b^s_{1} \\ \vdots \\ b^s_{N} \\
\end{bmatrix}
=\begin{bmatrix}
-B_{00} \\ -B_{10} \\ \vdots \\ -B_{N0} \\ k_{0}A_{0} \\ 0 \\ \vdots \\ 0 \\
\end{bmatrix}
</math>
</center>

We then simply need to solve the linear system of equations.

== 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 Finite Rectangle using Symmetry

2012-03-16T01:16:37Z

Tertius: /* Numerical Solution */

{{incomplete pages}}

== Introduction ==

The problem consists of a finite rigid fixed rectangle of length <math>2L</math> centered at <math>x = 0</math> and having a height of <math>d + ...</math> freeboard with depth of submergence being <math>d</math> through which no flow is possible. The regions to the left and right of the rectangle are the free water surface. Since the geometry of the rectangle is symmetric about <math>x = 0</math> we can decompose the velocity potential <math>\phi</math> into a symmetric <math>\phi^s</math> and antisymmetric <math>\phi^a</math> potential. We then solve for these two potentials in the left domain only <math>x \leq 0</math> and subsequently recompose the solution <math>\phi</math> for the entire domain. We begin with the simple problem when the waves are normally incident on the rectangle (dock) (so that the problem is truly two-dimensional). We can extend this solution to a pair of partially submerged finite rigid fixed rectangles using symmetry
[[Eigenfunction Matching for a Finite Rectangle using Symmetry]]

== Governing Equations ==

We begin with the [[Frequency Domain Problem]] for a partially submerged finite dock. Although the dock is of length <math>2L</math> we are only interested in the dock of <math>-L \leq x \leq 0</math> together with the fluid domain <math>x \leq 0</math>. We assume <math>e^{i\omega t}</math> time dependence and
the depth of submergence is <math>d</math>.
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>
\begin{align}
\Delta\phi^{s,a} &= 0, \,\,\, -h\leq z\leq 0, \,\, x < -L, \\

\Delta\phi^{s,a} &= 0, \,\,\, -h\leq z\leq -d, \,\, -L < x < 0. \\

\partial_z\phi^{s,a} &= \alpha\phi^{s,a}, \,\,\, z=0, \,\, x < -L, \\

\partial_x\phi^{s,a} &= 0, \,\,\, -d\leq z\leq 0, \,\, x = -L, \\

\partial_z\phi^{s,a} &= 0, \,\,\,
\begin{cases}
z = -d, \,\, -L < x < 0, \\
z = -h, \,\, x < 0.
\end{cases} \\

\partial_x\phi^s &= 0, \,\,\, x = 0, \,\, -h \leq z \leq -d, \\

\phi^a &= 0, \,\,\, x = 0, \,\, -h \leq z \leq -d.
\end{align}
</math></center>

Where <math>\alpha = \omega^2/g</math>.

<math>\phi^s(x,z)</math> is an even function and at <math>x = 0</math> the derivative of this function is equal to zero. <math>\phi^a(x,z)</math> is an odd function and at <math>x = 0</math> this function is equal to zero. These boundary conditions are important and enable the recomposed solution <math>\phi(x,z)</math> to be reflected into the domain of <math>x \geq 0</math>.

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\leq 0</math>
and <math>x\geq 0</math>.

{{separation of variables in two dimensions}}

{{separation of variables for a free surface}}

=== Separation of Variables for a Dock ===

The separation of variables equation for a partially submerged dock is given by:
<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>
Z^{\prime} (-d) = 0.
</math></center>
The solution is
<math>k=\kappa_{m}= \frac{m\pi}{h-d} \,</math> , <math>m\in\mathbb{N}\cup\left\{0\right\}</math> and
<center><math>
Z = \psi_{m}\left( z\right) = \cos\kappa_{m}(z+h),\quad
m\in\mathbb{N}\cup\left\{0\right\}.</math></center>
We note that
<center><math>
\int\nolimits_{-h}^{-d}\psi_{m}(z)\psi_{n}(z) \mathrm{d} z=C_{m}\delta_{mn},
</math></center>
where
<center>
<math>
C_{m} = \frac{1}{2} \left( \frac{\cos\kappa_{m}(h-d)\sin\kappa_{m}(h-d)+\kappa_{m}(h-d)}{\kappa_{m}} \right).
</math>
</center>
In addition there is the special case of when <math>m=n=0</math> ,
<center>
<math>
C_{0} = h-d,
</math>
</center>
where <math>\delta_{00}=1</math> .

=== Inner product between free surface and dock modes ===

<center><math>
B_{mn} = \int\nolimits_{-h}^{-d}\psi_{m}(z)\chi_{n}(z) \mathrm{d} z , \quad m\geq1, \quad n\geq0,
</math></center>
where
<center><math>
B_{mn}= \int\nolimits_{-h}^{-d} \frac{\cos\kappa_{m}(z+h)\cos k_{n}(z+h)}{\cos(k_{n}h)} \mathrm{d} z=\frac{-(-1)^m k_{n}\sin k_{n}(h-d)}
{\cos(k_{n}h)(\kappa_{m}^{2}-k_{n}^{2})}.
</math></center>
The evaluation of this inner product works on the assumption that the two roots <math>\kappa_{m}</math> and <math>k_{n}</math> do not coincide. The Matlab code given below has a routine to check for this case and make necessary adjustments.

We also have the case where the functions of the inner product are switched.
<center><math>
B_{nm} = \int\nolimits_{-h}^{-d}\chi_{m}(z)\psi_{n}(z) \mathrm{d} z , \quad m\geq0, \quad n\geq0.
</math></center>
For <math>m \geq 0</math> and <math>n \geq 1</math> the evaluation is above with the exception that the indices of the expression are switched.

Finally there is the special case of when <math>m \geq 0</math> and <math>n = 0</math>.
<center><math>
B_{0m} = \frac{\sin k_{m}(h-d)}{k_{m}\cos(k_{m}h)}.
</math></center>

=== Expansion of the potential ===

We need to apply some boundary conditions at plus and minus infinity,
where essentially 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 roots. Here we
must use a different condition, essentially identifying one solution
as the incoming wave and the other as the outgoing wave.

Therefore the diffracted potentials can
be expanded as
<center>
<math>
\phi(x,z)=\sum_{m=0}^{\infty}a_{m}e^{k_{m}x}\chi_{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 dock covered region. Therefore, the
potentials and their derivatives at <math>x=0</math> have to be equal.
We obtain:
<center><math>
\begin{align}
\chi_{0}\left( z\right) + \sum_{m=0}^{\infty}a_{m} \chi_{m}\left( z\right)
&=\sum_{m=0}^{\infty}b_{m}\psi_{m}(z), \quad -h \leq z\leq -d \\

-k_{0}\chi_{0}\left( z\right) +\sum_{m=0}^{\infty} a_{m}k_{m}\chi_{m}\left( z\right) &=
\begin{cases}
\quad \quad \quad 0,\quad \quad \quad -d\leq z\leq 0 \\
-\sum_{m=0}^{\infty}b_{m}\kappa_{m}\psi_{m}(z),\quad -h\leq z\leq -d
\end{cases}
\end{align}
</math></center>

For the first equation we multiply both sides by <math>\psi_{n}(z) \,</math> and integrating from <math>-h</math> to <math>-d</math> to obtain:
<center><math>
B_{n0} + \sum^{\infty}_{m=0}a_{m}B_{nm} = b_{m}C_{m}\delta_{mn}, n\in\mathbb{N}\cup\left\{0\right\}
</math></center>
and for the second equation we multiply both sides by <math>\chi_{n}(z) \,</math> and integrating from <math>-h</math> to <math>0</math> to obtain:
<center><math>
-k_{0}A_{0}\delta_{0n} + k_{m}a_{m}A_{m} = -\sum^{\infty}_{m=0}\kappa_{m}b_{m}B_{mn}, n\in\mathbb{N}\cup\left\{0\right\}
</math></center>

Solving the equations above will yield the coefficients of the water velocity potential in the dock covered region.

== Numerical Solution ==

To solve the system of equations for the symmetric case, we set the upper limit of <math>m</math> to be <math>N</math>. This resulting system can be expressed in the block matrix form below,

<center>
<math>
\begin{bmatrix}
B_{00} & B_{01} & \cdots & B_{0N} & -C_{0} & 0 & \cdots & 0 \\
B_{10} & B_{11} & \cdots & B_{1N} & 0 & -C_{1} & \cdots & \vdots \\
\vdots & \vdots & \ddots & \vdots & \vdots & \vdots & \ddots & 0 \\
B_{N0} & B_{N1} & \cdots & B_{NN} & 0 & \cdots & 0 & -C_{N} \\
k_{0}A_{0} & 0 & \cdots & 0 & 0 & \kappa_{1}\tanh(\kappa_{1}L)B_{10} & \cdots & \kappa_{N}\tanh(\kappa_{N}L)B_{N0} \\
0 & k_{1}A_{1} & \cdots & \vdots & 0 & \kappa_{1}\tanh(\kappa_{1}L)B_{11} & \cdots & \kappa_{N}\tanh(\kappa_{N}L)B_{N1} \\
\vdots & \vdots & \ddots & 0 & \vdots & \vdots & \ddots & \vdots \\
0 & \cdots & 0 & k_{N}A_{N} & 0 & \kappa_{1}\tanh(\kappa_{1}L)B_{1N} & \cdots & \kappa_{N}\tanh(\kappa_{N}L)B_{NN} \\
\end{bmatrix}
\begin{bmatrix}
a^s_{0} \\ a^s_{1} \\ \vdots \\ a^s_{N} \\ b^s_{0} \\ b^s_{1} \\ \vdots \\ b^s_{N} \\
\end{bmatrix}
=\begin{bmatrix}
-B_{00} \\ -B_{10} \\ \vdots \\ -B_{N0} \\ k_{0}A_{0} \\ 0 \\ \vdots \\ 0 \\
\end{bmatrix}
</math>
</center>

To solve the system of equations for the antisymmetric case, we set the upper limit of <math>m</math> to be <math>N</math>. This resulting system can be expressed in the block matrix form below,

<center>
<math>
\begin{bmatrix}
B_{00} & B_{01} & \cdots & B_{0N} & -C_{0} & 0 & \cdots & 0 \\
B_{10} & B_{11} & \cdots & B_{1N} & 0 & -C_{1} & \cdots & \vdots \\
\vdots & \vdots & \ddots & \vdots & \vdots & \vdots & \ddots & 0 \\
B_{N0} & B_{N1} & \cdots & B_{NN} & 0 & \cdots & 0 & -C_{N} \\
k_{0}A_{0} & 0 & \cdots & 0 & 0 & \kappa_{1}\tanh(\kappa_{1}L)B_{10} & \cdots & \kappa_{N}\tanh(\kappa_{N}L)B_{N0} \\
0 & k_{1}A_{1} & \cdots & \vdots & 0 & \kappa_{1}\tanh(\kappa_{1}L)B_{11} & \cdots & \kappa_{N}\tanh(\kappa_{N}L)B_{N1} \\
\vdots & \vdots & \ddots & 0 & \vdots & \vdots & \ddots & \vdots \\
0 & \cdots & 0 & k_{N}A_{N} & 0 & \kappa_{1}\tanh(\kappa_{1}L)B_{1N} & \cdots & \kappa_{N}\tanh(\kappa_{N}L)B_{NN} \\
\end{bmatrix}
\begin{bmatrix}
a^s_{0} \\ a^s_{1} \\ \vdots \\ a^s_{N} \\ b^s_{0} \\ b^s_{1} \\ \vdots \\ b^s_{N} \\
\end{bmatrix}
=\begin{bmatrix}
-B_{00} \\ -B_{10} \\ \vdots \\ -B_{N0} \\ k_{0}A_{0} \\ 0 \\ \vdots \\ 0 \\
\end{bmatrix}
</math>
</center>

We then simply need to solve the linear system of equations.

== 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 Finite Rectangle using Symmetry

2012-03-16T01:12:07Z

Tertius: /* Numerical Solution */

{{incomplete pages}}

== Introduction ==

The problem consists of a finite rigid fixed rectangle of length <math>2L</math> centered at <math>x = 0</math> and having a height of <math>d + ...</math> freeboard with depth of submergence being <math>d</math> through which no flow is possible. The regions to the left and right of the rectangle are the free water surface. Since the geometry of the rectangle is symmetric about <math>x = 0</math> we can decompose the velocity potential <math>\phi</math> into a symmetric <math>\phi^s</math> and antisymmetric <math>\phi^a</math> potential. We then solve for these two potentials in the left domain only <math>x \leq 0</math> and subsequently recompose the solution <math>\phi</math> for the entire domain. We begin with the simple problem when the waves are normally incident on the rectangle (dock) (so that the problem is truly two-dimensional). We can extend this solution to a pair of partially submerged finite rigid fixed rectangles using symmetry
[[Eigenfunction Matching for a Finite Rectangle using Symmetry]]

== Governing Equations ==

We begin with the [[Frequency Domain Problem]] for a partially submerged finite dock. Although the dock is of length <math>2L</math> we are only interested in the dock of <math>-L \leq x \leq 0</math> together with the fluid domain <math>x \leq 0</math>. We assume <math>e^{i\omega t}</math> time dependence and
the depth of submergence is <math>d</math>.
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>
\begin{align}
\Delta\phi^{s,a} &= 0, \,\,\, -h\leq z\leq 0, \,\, x < -L, \\

\Delta\phi^{s,a} &= 0, \,\,\, -h\leq z\leq -d, \,\, -L < x < 0. \\

\partial_z\phi^{s,a} &= \alpha\phi^{s,a}, \,\,\, z=0, \,\, x < -L, \\

\partial_x\phi^{s,a} &= 0, \,\,\, -d\leq z\leq 0, \,\, x = -L, \\

\partial_z\phi^{s,a} &= 0, \,\,\,
\begin{cases}
z = -d, \,\, -L < x < 0, \\
z = -h, \,\, x < 0.
\end{cases} \\

\partial_x\phi^s &= 0, \,\,\, x = 0, \,\, -h \leq z \leq -d, \\

\phi^a &= 0, \,\,\, x = 0, \,\, -h \leq z \leq -d.
\end{align}
</math></center>

Where <math>\alpha = \omega^2/g</math>.

<math>\phi^s(x,z)</math> is an even function and at <math>x = 0</math> the derivative of this function is equal to zero. <math>\phi^a(x,z)</math> is an odd function and at <math>x = 0</math> this function is equal to zero. These boundary conditions are important and enable the recomposed solution <math>\phi(x,z)</math> to be reflected into the domain of <math>x \geq 0</math>.

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\leq 0</math>
and <math>x\geq 0</math>.

{{separation of variables in two dimensions}}

{{separation of variables for a free surface}}

=== Separation of Variables for a Dock ===

The separation of variables equation for a partially submerged dock is given by:
<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>
Z^{\prime} (-d) = 0.
</math></center>
The solution is
<math>k=\kappa_{m}= \frac{m\pi}{h-d} \,</math> , <math>m\in\mathbb{N}\cup\left\{0\right\}</math> and
<center><math>
Z = \psi_{m}\left( z\right) = \cos\kappa_{m}(z+h),\quad
m\in\mathbb{N}\cup\left\{0\right\}.</math></center>
We note that
<center><math>
\int\nolimits_{-h}^{-d}\psi_{m}(z)\psi_{n}(z) \mathrm{d} z=C_{m}\delta_{mn},
</math></center>
where
<center>
<math>
C_{m} = \frac{1}{2} \left( \frac{\cos\kappa_{m}(h-d)\sin\kappa_{m}(h-d)+\kappa_{m}(h-d)}{\kappa_{m}} \right).
</math>
</center>
In addition there is the special case of when <math>m=n=0</math> ,
<center>
<math>
C_{0} = h-d,
</math>
</center>
where <math>\delta_{00}=1</math> .

=== Inner product between free surface and dock modes ===

<center><math>
B_{mn} = \int\nolimits_{-h}^{-d}\psi_{m}(z)\chi_{n}(z) \mathrm{d} z , \quad m\geq1, \quad n\geq0,
</math></center>
where
<center><math>
B_{mn}= \int\nolimits_{-h}^{-d} \frac{\cos\kappa_{m}(z+h)\cos k_{n}(z+h)}{\cos(k_{n}h)} \mathrm{d} z=\frac{-(-1)^m k_{n}\sin k_{n}(h-d)}
{\cos(k_{n}h)(\kappa_{m}^{2}-k_{n}^{2})}.
</math></center>
The evaluation of this inner product works on the assumption that the two roots <math>\kappa_{m}</math> and <math>k_{n}</math> do not coincide. The Matlab code given below has a routine to check for this case and make necessary adjustments.

We also have the case where the functions of the inner product are switched.
<center><math>
B_{nm} = \int\nolimits_{-h}^{-d}\chi_{m}(z)\psi_{n}(z) \mathrm{d} z , \quad m\geq0, \quad n\geq0.
</math></center>
For <math>m \geq 0</math> and <math>n \geq 1</math> the evaluation is above with the exception that the indices of the expression are switched.

Finally there is the special case of when <math>m \geq 0</math> and <math>n = 0</math>.
<center><math>
B_{0m} = \frac{\sin k_{m}(h-d)}{k_{m}\cos(k_{m}h)}.
</math></center>

=== Expansion of the potential ===

We need to apply some boundary conditions at plus and minus infinity,
where essentially 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 roots. Here we
must use a different condition, essentially identifying one solution
as the incoming wave and the other as the outgoing wave.

Therefore the diffracted potentials can
be expanded as
<center>
<math>
\phi(x,z)=\sum_{m=0}^{\infty}a_{m}e^{k_{m}x}\chi_{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 dock covered region. Therefore, the
potentials and their derivatives at <math>x=0</math> have to be equal.
We obtain:
<center><math>
\begin{align}
\chi_{0}\left( z\right) + \sum_{m=0}^{\infty}a_{m} \chi_{m}\left( z\right)
&=\sum_{m=0}^{\infty}b_{m}\psi_{m}(z), \quad -h \leq z\leq -d \\

-k_{0}\chi_{0}\left( z\right) +\sum_{m=0}^{\infty} a_{m}k_{m}\chi_{m}\left( z\right) &=
\begin{cases}
\quad \quad \quad 0,\quad \quad \quad -d\leq z\leq 0 \\
-\sum_{m=0}^{\infty}b_{m}\kappa_{m}\psi_{m}(z),\quad -h\leq z\leq -d
\end{cases}
\end{align}
</math></center>

For the first equation we multiply both sides by <math>\psi_{n}(z) \,</math> and integrating from <math>-h</math> to <math>-d</math> to obtain:
<center><math>
B_{n0} + \sum^{\infty}_{m=0}a_{m}B_{nm} = b_{m}C_{m}\delta_{mn}, n\in\mathbb{N}\cup\left\{0\right\}
</math></center>
and for the second equation we multiply both sides by <math>\chi_{n}(z) \,</math> and integrating from <math>-h</math> to <math>0</math> to obtain:
<center><math>
-k_{0}A_{0}\delta_{0n} + k_{m}a_{m}A_{m} = -\sum^{\infty}_{m=0}\kappa_{m}b_{m}B_{mn}, n\in\mathbb{N}\cup\left\{0\right\}
</math></center>

Solving the equations above will yield the coefficients of the water velocity potential in the dock covered region.

== Numerical Solution ==

To solve the system of equations for the symmetric case, we set the upper limit of <math>m</math> to be <math>N</math>. This resulting system can be expressed in the block matrix form below,
<center>
<math>
\begin{bmatrix}
B_{00} & B_{01} & \cdots & B_{0N} & -C_{0} & 0 & \cdots & 0 \\
B_{10} & B_{11} & \cdots & B_{1N} & 0 & -C_{1} & \cdots & \vdots \\
\vdots & \vdots & \ddots & \vdots & \vdots & \vdots & \ddots & 0 \\
B_{N0} & B_{N1} & \cdots & B_{NN} & 0 & \cdots & 0 & -C_{N} \\
k_{0}A_{0} & 0 & \cdots & 0 & 0 & \kappa_{1}\tanh(\kappa_{1}L)B_{10} & \cdots & \kappa_{N}\tanh(\kappa_{N}L)B_{N0} \\
0 & k_{1}A_{1} & \cdots & \vdots & 0 & \kappa_{1}\tanh(\kappa_{1}L)B_{11} & \cdots & \kappa_{N}\tanh(\kappa_{N}L)B_{N1} \\
\vdots & \vdots & \ddots & 0 & \vdots & \vdots & \ddots & \vdots \\
0 & \cdots & 0 & k_{N}A_{N} & 0 & \kappa_{1}\tanh(\kappa_{1}L)B_{1N} & \cdots & \kappa_{N}\tanh(\kappa_{N}L)B_{NN} \\
\end{bmatrix}
\begin{bmatrix}
a^s_{0} \\ a^s_{1} \\ \vdots \\ a^s_{N} \\ b^s_{0} \\ b^s_{1} \\ \vdots \\ b^s_{N} \\
\end{bmatrix}
=\begin{bmatrix}
-B_{00} \\ -B_{10} \\ \vdots \\ -B_{N0} \\ k_{0}A_{0} \\ 0 \\ \vdots \\ 0 \\
\end{bmatrix}
</math>
</center>
We then simply need to solve the linear system of equations.

== 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 Finite Rectangle using Symmetry

2012-03-16T01:07:24Z

Tertius: /* Numerical Solution */

{{incomplete pages}}

== Introduction ==

The problem consists of a finite rigid fixed rectangle of length <math>2L</math> centered at <math>x = 0</math> and having a height of <math>d + ...</math> freeboard with depth of submergence being <math>d</math> through which no flow is possible. The regions to the left and right of the rectangle are the free water surface. Since the geometry of the rectangle is symmetric about <math>x = 0</math> we can decompose the velocity potential <math>\phi</math> into a symmetric <math>\phi^s</math> and antisymmetric <math>\phi^a</math> potential. We then solve for these two potentials in the left domain only <math>x \leq 0</math> and subsequently recompose the solution <math>\phi</math> for the entire domain. We begin with the simple problem when the waves are normally incident on the rectangle (dock) (so that the problem is truly two-dimensional). We can extend this solution to a pair of partially submerged finite rigid fixed rectangles using symmetry
[[Eigenfunction Matching for a Finite Rectangle using Symmetry]]

== Governing Equations ==

We begin with the [[Frequency Domain Problem]] for a partially submerged finite dock. Although the dock is of length <math>2L</math> we are only interested in the dock of <math>-L \leq x \leq 0</math> together with the fluid domain <math>x \leq 0</math>. We assume <math>e^{i\omega t}</math> time dependence and
the depth of submergence is <math>d</math>.
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>
\begin{align}
\Delta\phi^{s,a} &= 0, \,\,\, -h\leq z\leq 0, \,\, x < -L, \\

\Delta\phi^{s,a} &= 0, \,\,\, -h\leq z\leq -d, \,\, -L < x < 0. \\

\partial_z\phi^{s,a} &= \alpha\phi^{s,a}, \,\,\, z=0, \,\, x < -L, \\

\partial_x\phi^{s,a} &= 0, \,\,\, -d\leq z\leq 0, \,\, x = -L, \\

\partial_z\phi^{s,a} &= 0, \,\,\,
\begin{cases}
z = -d, \,\, -L < x < 0, \\
z = -h, \,\, x < 0.
\end{cases} \\

\partial_x\phi^s &= 0, \,\,\, x = 0, \,\, -h \leq z \leq -d, \\

\phi^a &= 0, \,\,\, x = 0, \,\, -h \leq z \leq -d.
\end{align}
</math></center>

Where <math>\alpha = \omega^2/g</math>.

<math>\phi^s(x,z)</math> is an even function and at <math>x = 0</math> the derivative of this function is equal to zero. <math>\phi^a(x,z)</math> is an odd function and at <math>x = 0</math> this function is equal to zero. These boundary conditions are important and enable the recomposed solution <math>\phi(x,z)</math> to be reflected into the domain of <math>x \geq 0</math>.

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\leq 0</math>
and <math>x\geq 0</math>.

{{separation of variables in two dimensions}}

{{separation of variables for a free surface}}

=== Separation of Variables for a Dock ===

The separation of variables equation for a partially submerged dock is given by:
<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>
Z^{\prime} (-d) = 0.
</math></center>
The solution is
<math>k=\kappa_{m}= \frac{m\pi}{h-d} \,</math> , <math>m\in\mathbb{N}\cup\left\{0\right\}</math> and
<center><math>
Z = \psi_{m}\left( z\right) = \cos\kappa_{m}(z+h),\quad
m\in\mathbb{N}\cup\left\{0\right\}.</math></center>
We note that
<center><math>
\int\nolimits_{-h}^{-d}\psi_{m}(z)\psi_{n}(z) \mathrm{d} z=C_{m}\delta_{mn},
</math></center>
where
<center>
<math>
C_{m} = \frac{1}{2} \left( \frac{\cos\kappa_{m}(h-d)\sin\kappa_{m}(h-d)+\kappa_{m}(h-d)}{\kappa_{m}} \right).
</math>
</center>
In addition there is the special case of when <math>m=n=0</math> ,
<center>
<math>
C_{0} = h-d,
</math>
</center>
where <math>\delta_{00}=1</math> .

=== Inner product between free surface and dock modes ===

<center><math>
B_{mn} = \int\nolimits_{-h}^{-d}\psi_{m}(z)\chi_{n}(z) \mathrm{d} z , \quad m\geq1, \quad n\geq0,
</math></center>
where
<center><math>
B_{mn}= \int\nolimits_{-h}^{-d} \frac{\cos\kappa_{m}(z+h)\cos k_{n}(z+h)}{\cos(k_{n}h)} \mathrm{d} z=\frac{-(-1)^m k_{n}\sin k_{n}(h-d)}
{\cos(k_{n}h)(\kappa_{m}^{2}-k_{n}^{2})}.
</math></center>
The evaluation of this inner product works on the assumption that the two roots <math>\kappa_{m}</math> and <math>k_{n}</math> do not coincide. The Matlab code given below has a routine to check for this case and make necessary adjustments.

We also have the case where the functions of the inner product are switched.
<center><math>
B_{nm} = \int\nolimits_{-h}^{-d}\chi_{m}(z)\psi_{n}(z) \mathrm{d} z , \quad m\geq0, \quad n\geq0.
</math></center>
For <math>m \geq 0</math> and <math>n \geq 1</math> the evaluation is above with the exception that the indices of the expression are switched.

Finally there is the special case of when <math>m \geq 0</math> and <math>n = 0</math>.
<center><math>
B_{0m} = \frac{\sin k_{m}(h-d)}{k_{m}\cos(k_{m}h)}.
</math></center>

=== Expansion of the potential ===

We need to apply some boundary conditions at plus and minus infinity,
where essentially 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 roots. Here we
must use a different condition, essentially identifying one solution
as the incoming wave and the other as the outgoing wave.

Therefore the diffracted potentials can
be expanded as
<center>
<math>
\phi(x,z)=\sum_{m=0}^{\infty}a_{m}e^{k_{m}x}\chi_{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 dock covered region. Therefore, the
potentials and their derivatives at <math>x=0</math> have to be equal.
We obtain:
<center><math>
\begin{align}
\chi_{0}\left( z\right) + \sum_{m=0}^{\infty}a_{m} \chi_{m}\left( z\right)
&=\sum_{m=0}^{\infty}b_{m}\psi_{m}(z), \quad -h \leq z\leq -d \\

-k_{0}\chi_{0}\left( z\right) +\sum_{m=0}^{\infty} a_{m}k_{m}\chi_{m}\left( z\right) &=
\begin{cases}
\quad \quad \quad 0,\quad \quad \quad -d\leq z\leq 0 \\
-\sum_{m=0}^{\infty}b_{m}\kappa_{m}\psi_{m}(z),\quad -h\leq z\leq -d
\end{cases}
\end{align}
</math></center>

For the first equation we multiply both sides by <math>\psi_{n}(z) \,</math> and integrating from <math>-h</math> to <math>-d</math> to obtain:
<center><math>
B_{n0} + \sum^{\infty}_{m=0}a_{m}B_{nm} = b_{m}C_{m}\delta_{mn}, n\in\mathbb{N}\cup\left\{0\right\}
</math></center>
and for the second equation we multiply both sides by <math>\chi_{n}(z) \,</math> and integrating from <math>-h</math> to <math>0</math> to obtain:
<center><math>
-k_{0}A_{0}\delta_{0n} + k_{m}a_{m}A_{m} = -\sum^{\infty}_{m=0}\kappa_{m}b_{m}B_{mn}, n\in\mathbb{N}\cup\left\{0\right\}
</math></center>

Solving the equations above will yield the coefficients of the water velocity potential in the dock covered region.

== Numerical Solution ==

To solve the system of equations for the symmetric case, we set the upper limit of <math>m</math> to be <math>N</math>. This resulting system can be expressed in the block matrix form below,
<center>
<math>
\begin{bmatrix}
B_{00} & B_{01} & \cdots & B_{0N} & -C_{0} & 0 & \cdots & 0 \\
B_{10} & B_{11} & \cdots & B_{1N} & 0 & -C_{1} & \cdots & \vdots \\
\vdots & \vdots & \ddots & \vdots & \vdots & \vdots & \ddots & 0 \\
B_{N0} & B_{N1} & \cdots & B_{NN} & 0 & \cdots & 0 & -C_{N} \\
k_{0}A_{0} & 0 & \cdots & 0 & \kappa_{0}B_{00} & \kappa_{1}\tanh(\kappa_{1}L)B_{10} & \cdots & \kappa_{N}\tanh(\kappa_{N}L)B_{N0} \\
0 & k_{1}A_{1} & \cdots & \vdots & \kappa_{0}B_{01} & \kappa_{1}\tanh(\kappa_{1}L)B_{11} & \cdots & \kappa_{N}\tanh(\kappa_{N}L)B_{N1} \\
\vdots & \vdots & \ddots & 0 & \vdots & \vdots & \ddots & \vdots \\
0 & \cdots & 0 & k_{N}A_{N} & \kappa_{0}B_{0N} & \kappa_{1}\tanh(\kappa_{1}L)B_{1N} & \cdots & \kappa_{N}\tanh(\kappa_{N}L)B_{NN} \\
\end{bmatrix}
\begin{bmatrix}
a^s_{0} \\ a^s_{1} \\ \vdots \\ a^s_{N} \\ b^s_{0} \\ b^s_{1} \\ \vdots \\ b^s_{N} \\
\end{bmatrix}
=\begin{bmatrix}
-B_{00} \\ -B_{10} \\ \vdots \\ -B_{N0} \\ k_{0}A_{0} \\ 0 \\ \vdots \\ 0 \\
\end{bmatrix}
</math>
</center>
We then simply need to solve the linear system of equations.

== 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]]