# Two Identical Docks using Symmetry

## Introduction

The problems consists three regions with a free surface and and two regions of identical length with a rigid surface through which not flow is possible. The solution method is an extension of Eigenfunction Matching for a Finite Dock using Symmetry in Two Dimensions. We begin with the simple problem when the waves are normally incident (so that the problem is truly two-dimensional). We then consider the case when the waves are incident at an angle. For the later we give the equations in slightly less detail.

## Governing Equations

We begin with the Frequency Domain Problem for a dock which occupies the region $x>0$ (we assume $e^{i\omega t}$ time dependence). 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

$\Delta\phi=0, \,\, -h

$\phi_{z}=0, \,\, z=-h,$

$\partial_z\phi=\alpha\phi, \,\, z=0,\,x<-L_2,\,-L_1L_2$

$\partial_z\phi=0, \,\, z=0,\,-L_2

where we require $L_1 and we define $L_2 - L_1 = 2L$ (so that the dock also has length $2L$).

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 begin by separating variables in the different regions.

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 not really the dispersion relation for a free surface, it would be better to refer to it as 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 floating dock

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

subject to the boundary conditions

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

and

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

The solution is $k=\kappa_{m}= \frac{m\pi}{h} \,$, $m\geq 0$ and

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

We note that

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

where

$C_{m} = \begin{cases} h,\quad m=0 \\ \frac{1}{2}h,\,\,\,m\neq 0 \end{cases}$

### Inner product between free surface and dock modes

$\int\nolimits_{-h}^{0}\phi_{n}(z)\psi_{m}(z) \mathrm{d} z=B_{mn}$

where

$B_{mn}=\frac{k_{n}\sin k_{n}h\cos\kappa_{m}h-\kappa_{m}\cos k_{n}h\sin \kappa_{m}h}{\left( \cos k_{n}h\right) \left( k_{n} ^{2}-\kappa_{m}^{2}\right) }$

### 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 $x < 0$ 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 $x < 0$.

The first term in the expansion of the diffracted potential for the domain $x < 0$ 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.

### Expansion of the potential

The solution method uses Symmetry in Two Dimensions and we write the potential as a symmetric and an anti-symmetric part and consider only the region $x<0$. We apply either Neuman (symmetric) or Dirichlet (anti-symmetric) boundary conditions at $x=0$. We separation of variables in the three regions, similar to as for the Eigenfunction Matching for a Finite Dock. We being with the symmetric potential which can be expanded as

$\phi^s(x,z)=e^{-k_{0}(x+L_2)}\phi_{0}\left( z\right) + \sum_{m=0}^{\infty}a^s_{m}e^{k_{m}(x+L_2)}\phi_{m}(z), \;\;x<-L_2$

$\phi(x,z)=b^s_0 \frac{x+L_1}{-2L}\psi_{0}(z) + \sum_{m=1}^{\infty}b^s_{m} e^{-\kappa_{m} (x+L_2)}\psi_{m}(z) +c^s_0 \frac{L_2+x}{2L}\psi_{0}(z) + \sum_{m=1}^{\infty}c^s_{m} e^{\kappa_{m} (x+L_1)}\psi_{m}(z) , \;\;-L_2

and

$\phi(x,z)=\sum_{m=0}^{\infty}d^s_{m} \frac{\cosh(k_{m}x)}{\cosh(k_m L_1)} \phi_{m}(z), \;\;-L_1

where $a^s_{m}$ and $d^s_{m}$ are the coefficients of the potential in the open water regions to the left and right and $b^s_m$ and $c^s_m$ are the coefficients 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 plate covered region. Therefore, the potentials and their derivatives at $x=- L_2$ and $x=-L_1$ have to be equal. We obtain

$\phi_{0}\left( z\right) + \sum_{m=0}^{\infty} a^s_{m} \phi_{m}\left( z\right) =\sum_{m=0}^{\infty}b^s_{m}\psi_{m}(z) + \sum_{m=1}^{\infty}c^s_{m}\psi_{m}(z)e^{-2L\kappa_m}$

$-k_{0}\phi_{0}\left( z\right) +\sum _{m=0}^{\infty} a^s_{m}k_{m}\phi_{m}\left( z\right) =-\frac{b^s_0}{2L}\psi_0(z) -\sum_{m=1}^{\infty}b^s_{m}\kappa_{m}\psi _{m}(z) + \frac{c^s_0}{2L}\psi_0(z)+\sum_{m=1}^{\infty}c^s_{m}\kappa_{m}\psi _{m}(z)e^{-2L\kappa_m}$

$\sum_{m=1}^{\infty}b^s_{m}\psi_{m}(z)e^{-2L\kappa_m} + \sum_{m=0}^{\infty}c^s_{m}\psi_{m}(z) =\sum_{m=0}^{\infty}d^s_{m} \phi_{m}\left( z\right)$

$-\frac{b^s_0}{2L}\psi_0(z) -\sum_{m=1}^{\infty}b^s_{m}\kappa_{m}\psi _{m}(z)e^{-2L\kappa_m} + \frac{c^s_0}{2L}\psi_0(z)+\sum_{m=1}^{\infty}c^s_{m}\kappa_{m}\psi _{m}(z) = \sum_{m=0}^{\infty}d^s_{m} \tanh(k_m L_1) k_m\phi_{m}\left( z\right)$

We solve these equations by multiplying both equations by $\phi_{l}(z)$ and integrating from $-h$ to $0$ to obtain:

$A_{0}\delta_{0l}+a^s_{l}A_{l} =\sum_{m=0}^{\infty}b^s_{m}B_{ml} + \sum_{m=1}^{\infty}c^s_{m}B_{ml}e^{-2L\kappa_m}$

$-k_{0}A_{0}\delta_{0l}+a^s_{l}k_{l}A_l = - b^s_0 \frac{B_{0l}}{2L} - \sum_{m=1}^{\infty}b^s_{m}\kappa_{m}B_{ml} + c^s_0 \frac{B_{0l}}{2L} + \sum_{m=1}^{\infty}c^s_{m}\kappa_{m}B_{ml} e^{-2L\kappa_m}$

$\sum_{m=1}^{\infty}b^s_{m}B_{ml}e^{-2L\kappa_m} + \sum_{m=0}^{\infty}c^s_{m}B_{ml} =d^s_l A_l$

$- b^s_0 \frac{B_{0l}}{2L} - \sum_{m=1}^{\infty}b^s_{m}\kappa_{m}B_{ml} e^{-2L\kappa_m} + c^s_0 \frac{B_{0l}}{2L} + \sum_{m=1}^{\infty}c^s_{m}\kappa_{m}B_{ml} = d^s_l \tanh(k_l L_1) k_l A_l$

## Numerical Solution

To solve the system of equations we set the upper limit of $l$ to be $M$. We then simply need to solve the linear system of equations.

## Anti-Symmetric Solution

The solution for the anti-symmetric potential proceeds in an almost identical manner.

$\phi^a(x,z)=e^{-k_{0}(x+L_2)}\phi_{0}\left( z\right) + \sum_{m=0}^{\infty}a^a_{m}e^{k_{m}(x+L_2)}\phi_{m}(z), \;\;x<-L_2$

$\phi(x,z)=b^a_0 \frac{x+L_1}{-2L}\psi_{0}(z) + \sum_{m=1}^{\infty}b^a_{m} e^{-\kappa_{m} (x+L_2)}\psi_{m}(z) +c^a_0 \frac{L_2+x}{2L}\psi_{0}(z) + \sum_{m=1}^{\infty}c^a_{m} e^{\kappa_{m} (x+L_1)}\psi_{m}(z) , \;\;-L_2

and

$\phi(x,z)=\sum_{m=0}^{\infty}d^a_{m} \frac{\sinh(k_{m}x)}{-\sinh(k_m L_1)} \phi_{m}(z), \;\;-L_1

where $a^a_{m}$ and $d^a_{m}$ are the coefficients of the potential in the open water regions to the left and right and $b^a_m$ and $c^a_m$ are the coefficients under the dock.

We solve these equations by multiplying both equations by $\phi_{l}(z)$ and integrating from $-h$ to $0$ to obtain:

$A_{0}\delta_{0l}+a^a_{l}A_{l} =\sum_{m=0}^{\infty}b^a_{m}B_{ml} + \sum_{m=1}^{\infty}c^a_{m}B_{ml}e^{-2L\kappa_m}$

$-k_{0}A_{0}\delta_{0l}+a^a_{l}k_{l}A_l = - b^a_0 \frac{B_{0l}}{2L} - \sum_{m=1}^{\infty}b^a_{m}\kappa_{m}B_{ml} + c^a_0 \frac{B_{0l}}{2L} + \sum_{m=1}^{\infty}c^a_{m}\kappa_{m}B_{ml} e^{-2L\kappa_m}$

$\sum_{m=1}^{\infty}b^a_{m}B_{ml}e^{-2L\kappa_m} + \sum_{m=0}^{\infty}c^a_{m}B_{ml} =d^a_l A_l$

$- b^a_0 \frac{B_{0l}}{2L} - \sum_{m=1}^{\infty}b^a_{m}\kappa_{m}B_{ml} e^{-2L\kappa_m} + c^a_0 \frac{B_{0l}}{2L} + \sum_{m=1}^{\infty}c^a_{m}\kappa_{m}B_{ml} = d^a_l \frac{k_l A_l}{\tanh(k_l L_1)}$

## Solution with Waves Incident at an Angle

We can consider the problem when the waves are incident at an angle $\theta$. In some ways the solution is now simpler because we do not need to write the zero term separately under the dock. When a wave in incident at an angle $\theta$ we have the wavenumber in the $y$ direction is $k_y = \sin\theta k_0$ where $k_0$ is as defined previously (note that $k_y$ is imaginary).

This means that the potential is now of the form $\phi(x,y,z)=e^{k_y y}\phi(x,z)$ so that when we separate variables we obtain

$k^2 = k_x^2 + k_y^2$

where $k$ is the separation constant calculated without an incident angle.

This means that the potential is now of the form $\phi(x,y,z)=e^{k_y y}\phi(x,z)$. Therefore the symmetric potential can be expanded as

$\phi^{s}_{0}\left( z\right) + \sum_{m=0}^{\infty} a^s_{m} \phi_{m}\left( z\right) =\sum_{m=0}^{\infty}b^s_{m}\psi_{m}(z) + \sum_{m=0}^{\infty}c^s_{m}\psi_{m}(z)e^{-2L\hat{\kappa}_m}$

$-\hat{k}_{0}\phi_{0}\left( z\right) +\sum _{m=0}^{\infty} a^s_{m}\hat{k}_{m}\phi_{m}\left( z\right) = -\sum_{m=0}^{\infty}b^s_{m}\hat{\kappa}_{m}\psi _{m}(z) +\sum_{m=0}^{\infty}c^s_{m}\hat{\kappa}_{m}\psi _{m}(z)e^{-2L\hat{\kappa}_m}$

$\sum_{m=1}^{\infty}b^s_{m}\psi_{m}(z)e^{-2L\hat{\kappa}_m} + \sum_{m=0}^{\infty}c^s_{m}\psi_{m}(z) =\sum_{m=0}^{\infty}d^s_{m} \phi_{m}\left( z\right)$

$-\sum_{m=0}^{\infty}b^s_{m}\hat{\kappa}_{m}\psi _{m}(z)e^{-2L\hat{\kappa}_m} + \sum_{m=0}^{\infty}c^s_{m}\hat{\kappa}_{m}\psi _{m}(z) = \sum_{m=0}^{\infty}d^s_{m} \tanh(\hat{k}_m L_1) \hat{k}_m\phi_{m}\left( z\right)$

where $\hat{k}_{m} = \sqrt{k_m^2 + k_y^2}$ and $\hat{\kappa}_{m} = \sqrt{\kappa_m^2 + k_y^2}$ and we always take the positive real root or the root with positive imaginary part.

We solve these equations by multiplying both equations by $\phi_{l}(z)$ and integrating from $-h$ to $0$ to obtain:

$A_{0}\delta_{0l}+a^s_{l}A_{l} =\sum_{m=0}^{\infty}b^s_{m}B_{ml} + \sum_{m=0}^{\infty}c^s_{m}B_{ml}e^{-2L\hat{\kappa}_m}$

$-\hat{k}_{0}A_{0}\delta_{0l}+a^s_{l}\hat{k}_{l}A_l = - \sum_{m=0}^{\infty}b^s_{m}\hat{\kappa}_{m}B_{ml} + \sum_{m=0}^{\infty}c^s_{m}\hat{\kappa}_{m}B_{ml} e^{-2L\hat{\kappa}}$

$\sum_{m=0}^{\infty}b^s_{m}B_{ml}e^{-2L\hat{\kappa}_m} + \sum_{m=0}^{\infty}c^s_{m}B_{ml} =d^s_l A_l$

$- \sum_{m=0}^{\infty}b^s_{m}\hat{\kappa}_{m}B_{ml} e^{-2L\hat{\kappa}_m} + \sum_{m=0}^{\infty}c^s_{m}\hat{\kappa}_{m}B_{ml} = d^s_l \tanh(\hat{k}_l L_1) \hat{k}_l A_l$

and these are solved exactly as before. The solution for the anti-symmetric potential is found in a similar fashion.

## Matlab Code

A program to calculate the coefficients for the finite dock problems can be found here two_finite_docks_symmetry.m