# Eigenfunction Matching for a Finite Rectangle using Symmetry

## Introduction

The problem consists of a finite rigid fixed rectangle of length $2L$ centered at $x = 0$ and having a height of $d + ...$ freeboard with depth of submergence being $d$ 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 $x = 0$ we can decompose the velocity potential $\phi$ into a symmetric $\phi^s$ and antisymmetric $\phi^a$ potential. We then solve for these two potentials in the left domain only $x \leq 0$ and subsequently recompose the solution $\phi$ 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 $2L$ we are only interested in the dock of $-L \leq x \leq 0$ together with the fluid domain $x \leq 0$. We assume $e^{i\omega t}$ time dependence and the depth of submergence is $d$. The water is assumed to have constant finite depth $h$ and the $z$-direction points vertically upward with the water surface at $z=0$ and the sea floor at $z=-h$. The boundary value problem can therefore be expressed as

\begin{align} \Delta\phi^{s,a} &= 0, \,\,\, -h\leq z\leq 0, \,\, x \lt -L, \\ \Delta\phi^{s,a} &= 0, \,\,\, -h\leq z\leq -d, \,\, -L \lt x \lt 0. \\ \partial_z\phi^{s,a} &= \alpha\phi^{s,a}, \,\,\, z=0, \,\, x \lt -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 \lt x \lt 0, \\ z = -h, \,\, x \lt 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}

Where $\alpha = \omega^2/g$.

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

We must also apply the Sommerfeld Radiation Condition as $|x|\rightarrow\infty$. 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 separation of variables in the two regions, $x\leq 0$ and $x\geq 0$.

We express the potential as

$\phi(x,z) = X(x)Z(z)\,$

and then Laplace's equation becomes

$\frac{X^{\prime\prime}}{X} = - \frac{Z^{\prime\prime}}{Z} = k^2$

### Separation of variables for a free surface

We use separation of variables

We express the potential as

$\phi(x,z) = X(x)Z(z)\,$

and then Laplace's equation becomes

$\frac{X^{\prime\prime}}{X} = - \frac{Z^{\prime\prime}}{Z} = k^2$

The separation of variables equation for deriving free surface eigenfunctions is as follows:

$Z^{\prime\prime} + k^2 Z =0.$

subject to the boundary conditions

$Z^{\prime}(-h) = 0$

and

$Z^{\prime}(0) = \alpha Z(0)$

We can then use the boundary condition at $z=-h \,$ to write

$Z = \frac{\cos k(z+h)}{\cos kh}$

where we have chosen the value of the coefficent so we have unit value at $z=0$. The boundary condition at the free surface ($z=0 \,$) gives rise to:

$k\tan\left( kh\right) =-\alpha \,$

which is the Dispersion Relation for a Free Surface

The above equation is a transcendental equation. If we solve for all roots in the complex plane we find that the first root is a pair of imaginary roots. We denote the imaginary solutions of this equation by $k_{0}=\pm ik \,$ and the positive real solutions by $k_{m} \,$, $m\geq1$. The $k \,$ of the imaginary solution is the wavenumber. We put the imaginary roots back into the equation above and use the hyperbolic relations

$\cos ix = \cosh x, \quad \sin ix = i\sinh x,$

to arrive at the dispersion relation

$\alpha = k\tanh kh.$

We note that for a specified frequency $\omega \,$ the equation determines the wavenumber $k \,$.

Finally we define the function $Z(z) \,$ as

$\chi_{m}\left( z\right) =\frac{\cos k_{m}(z+h)}{\cos k_{m}h},\quad m\geq0$

as the vertical eigenfunction of the potential in the open water region. From Sturm-Liouville theory the vertical eigenfunctions are orthogonal. They can be normalised to be orthonormal, but this has no advantages for a numerical implementation. It can be shown that

$\int\nolimits_{-h}^{0}\chi_{m}(z)\chi_{n}(z) \mathrm{d} z=A_{n}\delta_{mn}$

where

$A_{n}=\frac{1}{2}\left( \frac{\cos k_{n}h\sin k_{n}h+k_{n}h}{k_{n}\cos ^{2}k_{n}h}\right).$

### Separation of Variables for a Dock

The separation of variables equation for a partially submerged dock is given by:

$Z^{\prime\prime} + k^2 Z =0,$

subject to the boundary conditions

$Z^{\prime} (-h) = 0,$

and

$Z^{\prime} (-d) = 0.$

The solution is $k=\kappa_{m}= \frac{m\pi}{h-d} \,$ , $m\in\mathbb{N}\cup\left\{0\right\}$ and

$Z = \psi_{m}\left( z\right) = \cos\kappa_{m}(z+h),\quad m\in\mathbb{N}\cup\left\{0\right\}.$

We note that

$\int\nolimits_{-h}^{-d}\psi_{m}(z)\psi_{n}(z) \mathrm{d} z=C_{m}\delta_{mn},$

where

$C_{m} = \frac{1}{2} \left( \frac{\cos\kappa_{m}(h-d)\sin\kappa_{m}(h-d)+\kappa_{m}(h-d)}{\kappa_{m}} \right).$

In addition there is the special case of when $m=n=0$ ,

$C_{0} = h-d,$

where $\delta_{00}=1$ .

### Inner product between free surface and dock modes

$B_{mn} = \int\nolimits_{-h}^{-d}\psi_{m}(z)\chi_{n}(z) \mathrm{d} z , \quad m\geq1, \quad n\geq0,$

where

$B_{mn}= \int\nolimits_{-h}^{-d} \frac{\cos\kappa_{m}(z+h)\cos k_{n}(z+h)}{\cos(k_{n}h)} \mathrm{d} z=\frac{-(-1)^m k_{n}\sin k_{n}(h-d)} {\cos(k_{n}h)(\kappa_{m}^{2}-k_{n}^{2})}.$

The evaluation of this inner product works on the assumption that the two roots $\kappa_{m}$ and $k_{n}$ 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.

$B_{nm} = \int\nolimits_{-h}^{-d}\chi_{m}(z)\psi_{n}(z) \mathrm{d} z , \quad m\geq0, \quad n\geq0.$

For $m \geq 0$ and $n \geq 1$ the evaluation is above with the exception that the indices of the expression are switched.

Finally there is the special case of when $m \geq 0$ and $n = 0$.

$B_{0m} = \frac{\sin k_{m}(h-d)}{k_{m}\cos(k_{m}h)}.$

### 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

$\phi(x,z)=\sum_{m=0}^{\infty}a_{m}e^{k_{m}x}\chi_{m}(z), \;\;x\lt 0$

and

$\phi(x,z)=\sum_{m=0}^{\infty}b_{m} e^{-\kappa_{m}x}\psi_{m}(z), \;\;x\gt 0$

where $a_{m}$ and $b_{m}$ are the coefficients of the potential in the open water and the dock covered region respectively.

### Incident potential

To create meaningful solutions of the velocity potential $\phi$ in the specified domains we add an incident wave term to the expansion for the domain of $x \lt 0$ above. The incident potential is a wave of amplitude $A$ in displacement travelling in the positive $x$-direction. We would only see this in the time domain $\Phi(x,z,t)$ however, in the frequency domain the incident potential can be written as

$\phi_{\mathrm{I}}(x,z) =e^{-k_{0}x}\chi_{0}\left( z\right).$

The total velocity (scattered) potential now becomes $\phi = \phi_{\mathrm{I}} + \phi_{\mathrm{D}}$ for the domain of $x \lt 0$.

The first term in the expansion of the diffracted potential for the domain $x \lt 0$ is given by

$a_{0}e^{k_{0}x}\chi_{0}\left( z\right)$

which represents the reflected wave.

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

### An infinite dimensional system of equations

The symmetric and antisymmetric potentials $\phi^{s,a}(x,z) \,$ must be continuous across the transition from open water to the dock covered region. Therefore, the potentials at $x=-L$ have to be equal. For these cases we obtain:

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

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

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

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

\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}

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

\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}

For the first equation we multiply both sides by $\chi_{n}(z) \,$ and integrating from $-h$ to $0$ to obtain:

$-k_{0}A_{0}\delta_{0n} + k_{m}a_{m}A_{m} = -\sum^{\infty}_{m=0}\kappa_{m}b_{m}B_{mn}, \quad n\in\mathbb{N}\cup\left\{0\right\}$

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

$D_{n} = \int\nolimits_{-h}^{-d}\chi_{n}(z) \mathrm{d} z = \int\nolimits_{-h}^{-d} \frac{\cos k_{n}(z+h)}{\cos k_{n}h} \mathrm{d} z=\frac{\sin k_{n}(h-d)} {k_{n}\cos k_{n}h}.$

## Numerical Solution

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

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

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

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

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

## Matlab Code

A program to calculate the coefficients for the single finite dock problem using symmetry can be found here semiinfinite_dock.m