# Eigenfunction Matching for a Semi-Infinite Rectangle

## 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 $d + ...$ freeboard with depth of submergence being $d$ 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 $x\gt 0$ (we assume $e^{i\omega t}$ time dependence). 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 &= 0, \,\,\, -h\leq z\leq 0, \,\, x \lt 0, \\ \Delta\phi &= 0, \,\,\, -h\leq z\leq -d, \,\, x \gt 0. \\ \partial_z\phi &= \alpha\phi, \,\,\, z=0, \,\, x\lt 0, \\ \partial_x\phi &= 0, \,\,\, -d\leq z\leq 0, \,\, x=0, \\ \partial_z\phi &= 0, \,\,\, z=-d, \,\, x\gt 0, \\ \partial_z\phi &= 0, \,\,\, z=-h. \end{align}

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

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

Before we can write down the expansion of the potentials we need to solve for the horizontal eigenfunctions. We start with the equation

$X^{\prime\prime} - k^2 X = 0,$

which has the general solution

$X(x) = a e^{kx} + b e^{-kx},$

where $a$ and $b$ are constants to be determined from the boundary conditions. The general solution, or eigenfunction expansion, for the potential $\phi$ is obtained by superimposing all the possible modes; for the region of the free surface we have

$\phi(x,z)=\sum_{m=0}^{\infty}\left(a_{m}e^{k_{m}x} + b_{m}e^{-k_{m}x}\right)\chi_{m}(z), \quad x\lt 0.$

In this expression the term involving $e^{ikx}$ (recall that $k_{0} = -ik$) corresponds to a wave propagating towards large positive $x$ while the term involving $e^{-ikx}$ corresponds to a wave propagating towards large negative $x$. Terms which decay as $x \rightarrow -\infty$ or as $x \rightarrow \infty$ are referred to as evanescent modes.

The expansion of the potential for the region under the dock is

$\phi(x,z)=\sum_{m=0}^{\infty}\left(a_{m}e^{\kappa_{m}x} + b_{m}e^{-\kappa_{m}x}\right)\psi_{m}(z), \quad x\gt 0.$

To obtain a unique solution, the diffracted field $\phi_{\mathrm{D}}$ 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

$\phi_{\mathrm{D}}(x,z)=\sum_{m=0}^{\infty}a_{m}e^{k_{m}x}\chi_{m}(z), \quad x\lt 0$

and

$\phi_{\mathrm{D}}(x,z)=\sum_{m=0}^{\infty}b_{m} e^{-\kappa_{m}x}\psi_{m}(z), \quad x\gt 0$

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 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 $x=0$ have to be equal. We obtain:

\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}

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

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

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

$-k_{0}A_{0}\delta_{0n} + k_{m}a_{m}A_{m}\delta_{mn} = -\sum^{\infty}_{m=1}\kappa_{m}b_{m}B_{mn}, \quad n\in\mathbb{N}\cup\left\{0\right\}$

Remember $\kappa_{0} = 0 \,$. 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 $m$ to be $N$. This resulting system can be expressed in the block matrix form below,

$\begin{bmatrix} B_{00} & B_{01} & \cdots & B_{0N} & -C_{0} & 0 & \cdots & 0 \\ B_{10} & B_{11} & \cdots & B_{1N} & 0 & -C_{1} & \cdots & \vdots \\ \vdots & \vdots & \ddots & \vdots & \vdots & \vdots & \ddots & 0 \\ B_{N0} & B_{N1} & \cdots & B_{NN} & 0 & \cdots & 0 & -C_{N} \\ k_{0}A_{0} & 0 & \cdots & 0 & 0 & \kappa_{1}B_{10} & \cdots & \kappa_{N}B_{N0} \\ 0 & k_{1}A_{1} & \cdots & \vdots & 0 & \kappa_{1}B_{11} & \cdots & \kappa_{N}B_{N1} \\ \vdots & \vdots & \ddots & 0 & \vdots & \vdots & \ddots & \vdots \\ 0 & \cdots & 0 & k_{N}A_{N} & 0 & \kappa_{1}B_{1N} & \cdots & \kappa_{N}B_{NN} \\ \end{bmatrix} \begin{bmatrix} a_{0} \\ a_{1} \\ \vdots \\ a_{N} \\ b_{0} \\ b_{1} \\ \vdots \\ b_{N} \\ \end{bmatrix} =\begin{bmatrix} -B_{00} \\ -B_{10} \\ \vdots \\ -B_{N0} \\ k_{0}A_{0} \\ 0 \\ \vdots \\ 0 \\ \end{bmatrix}$

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.m