# Eigenfunction Matching for a Vertical Fixed Plate

## Introduction

We consider fixed vertical plate and determine scattering using Category:Symmetry in Two Dimensions

## Governing Equations

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 $z=-h$. We begin with the Frequency Domain Problem for a fixed vertical plate which occupies the region $x=0$ and $-a\gt z\gt -b$ where $0\lt a\lt b\lt h$. We assume $e^{i\omega t}$ time dependence.

The boundary value problem can therefore be expressed as

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

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

$\partial_z\phi=\alpha\phi, \,\, z=0,\,x\lt 0,$

$\partial_x\phi=0, \,\, -a\gt z\gt -b,\,x=0,$

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, $x\lt 0$ and $x\gt 0$.

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

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

### 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.

## Expansion of the Potential

Therefore the potential can be expanded as

$\phi(x,z)=e^{-{k}_0x}\phi_0(z)+\sum_{m=0}^{\infty}a_{m}e^{{k}_{m}x}\phi_{m}(z), \;\;x\lt 0$

and

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

## Solution using Symmetry

The problem is symmetric about the line $x=0$ and this allows us to solve the problem using symmetry. We decompose the solution into a symmetric and an anti-symmetric part as is described in Symmetry in Two Dimensions

### Symmetric solution

The symmetric potential can be expanded as

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

The boundary condition is that $\partial_x \phi = 0$ on $x=0$. The problem reduces to Waves reflecting off a vertical wall. $a_{0}^{s}=1$ $a_{m}^{s}=0,\, \,n \gt 0$

### Anti-symmetric solution

The anti-symmetric potential can be expanded as

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

For the anti-symmetric solution the potential satisfies $\partial_x \phi = 0, -a\gt z\gt -b$ on $x=0$ and $\phi = 0, 0\gt z\gt -a, -b\gt z\gt -h$. We impose this condition by integrating the following

$\int_{-h}^{0} \phi_m(z) \left\{ \begin{matrix} \phi^{a}(0,z),\,\,0\gt z\gt -a\,\,-b\gt z\gt -h\\ \partial_x\phi^{a}(0,z),\,\,-a\gt z\gt -b \end{matrix} \right\} dz = 0$

Therefore we have a system of equations of the form

$\sum_{n=0}^{N} A_{mn} a^{a}_n = f_m$

where

$A_{mn} = \int_{-h}^{-b} \phi_m(z)\phi_n(z) dz + \int_{-a}^{0} \phi_m(z)\phi_n(z) dz + \int_{-b}^{-a} k_n \phi_m(z)\phi_n(z) dz$

and

$f_m = \int_{-h}^{-b} \phi_m(z)\phi_0(z) dz + \int_{-a}^{0} \phi_m(z)\phi_0(z) dz - \int_{-b}^{-a} k_0 \phi_m(z)\phi_0(z) dz$

### Solution to the original problem

We can now reconstruct the potential for the finite dock from the two previous symmetric and anti-symmetric solution as explained in Symmetry in Two Dimensions. The amplitude in the left open-water region is simply obtained by the superposition principle

$a_{m} = \frac{1}{2}\left(a_{m}^{s}+a_{m}^{a}\right)$

and in the right open water region is just

$a_{m} = \frac{1}{2}\left(a_{m}^{s} - a_{m}^{a}\right)$

Therefore the scattered potential (without the incident wave, which will be added later) can be expanded as

$\phi(x,z)= e^{-k_{0}x}\phi_{0} + \sum_{m=0}^{\infty}\frac{1}{2}\left(a_{m}^{s} + a_{m}^{a}\right) e^{k_{m}x}\phi_{m}(z), \;\;x\lt 0$

and

$\phi(x,z)=\sum_{m=0}^{\infty}\frac{1}{2}\left(a_{m}^{s} - a_{m}^{a}\right) e^{-k_{m}x}\phi_{m}(z), \;\;x\gt 0$

## Solution with Waves Incident at an Angle

We can consider the problem when the waves are incident at an angle $\theta$.

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.

Therefore the potential can be expanded as

$\phi(x,z)=e^{-\hat{k}_0x}\phi_0(z)+\sum_{m=0}^{\infty}a_{m}e^{\hat{k}_{m}x}\phi_{m}(z), \;\;x\lt 0$

and

$\phi(x,z)=\sum_{m=0}^{\infty}b_{m} e^{-\hat{k}_{m}x}\phi_{m}(z), \;\;x\gt 0$

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

The equations are derived almost identically to those above and we obtain

## Matlab Code

A program to calculate the coefficients for the vertical fixed plate can be found here vertical_fixed_plate.m (note the solution uses symmetry but presents the full solution)