Boundary Element Method for a Fixed Body in Finite Depth

Introduction

We show how to solve for the wave scattering by a rigid body in constant Finite Depth using the Boundary Element Method.

Equations

The Standard Linear Wave Scattering Problem in Finite Depth for a fixed body is

[math]\displaystyle{ \begin{align} \Delta\phi &=0, &-h\lt z\lt 0,\,\,\mathbf{x} \in \Omega \\ \partial_z\phi &= 0, &z=-h, \\ \partial_z \phi &= \alpha \phi, &z=0,\,\,\mathbf{x} \in \partial \Omega_{\mathrm{F}}, \end{align} }[/math]

(note that the last expression can be obtained from combining the expressions:

[math]\displaystyle{ \begin{align} \partial_z \phi &= -\mathrm{i} \omega \zeta, &z=0,\,\,\mathbf{x} \in \partial \Omega_{\mathrm{F}}, \\ \mathrm{i} \omega \phi &= g\zeta, &z=0,\,\,\mathbf{x} \in \partial \Omega_{\mathrm{F}}, \end{align} }[/math]

where [math]\displaystyle{ \alpha = \omega^2/g \, }[/math]) The body boundary condition for a rigid body is just

[math]\displaystyle{ \partial_{n}\phi=0,\ \ \mathbf{x}\in\partial\Omega_{\mathrm{B}}, }[/math]

The equation is subject to some radiation conditions at infinity. We assume the following. [math]\displaystyle{ \phi^{\mathrm{I}}\, }[/math] is a plane wave travelling in the [math]\displaystyle{ x }[/math] direction,

[math]\displaystyle{ \phi^{\mathrm{I}}(x,z)=A \phi_0(z) e^{\mathrm{i} k x} \, }[/math]

where [math]\displaystyle{ A }[/math] is the wave amplitude (in potential) [math]\displaystyle{ \mathrm{i} k }[/math] is the positive imaginary solution of the Dispersion Relation for a Free Surface (note we are assuming that the time dependence is of the form [math]\displaystyle{ \exp(-\mathrm{i}\omega t) }[/math]) and

[math]\displaystyle{ \phi_0(z) =\frac{\cosh k(z+h)}{\cosh k h} }[/math]

In two-dimensions the Sommerfeld Radiation Condition is

[math]\displaystyle{ \left( \frac{\partial}{\partial|x|} - \mathrm{i} k \right) (\phi-\phi^{\mathrm{{I}}})=0,\;\mathrm{{as\;}}|x|\rightarrow\infty\mathrm{.} }[/math]

where [math]\displaystyle{ \phi^{\mathrm{{I}}} }[/math] is the incident potential.

Solution Method

We divide the domain into three regions, [math]\displaystyle{ x\lt -l }[/math], [math]\displaystyle{ x\gt r }[/math], and [math]\displaystyle{ -l\lt x\lt r }[/math] so that the body surface is entirely in the finite region.

Solution in the finite region

We use the Boundary Element Method in the finite region.

Solution in the Semi-infinite Domains

We now solve Laplace's equation in the semi-infinite domains [math]\displaystyle{ \Omega ^{-}=\left\{ x\lt -l,\;-h\leq z\leq 0\right\} }[/math] and [math]\displaystyle{ \Omega ^{+}=\left\{ x\gt r,\;-h\leq z\leq 0\right\} }[/math]. Since the water depth is constant in these regions we can solve Laplace's equation by separation of variables. The potential in the region [math]\displaystyle{ \Omega ^{-} }[/math] satisfies the following equation

[math]\displaystyle{ \left. \begin{matrix} \nabla ^{2}\phi =0,\;\;\mathbf{x}\in \Omega ^{-}, \\ \phi _{n} = \alpha \phi =0,\;\;z=0, \\ \phi _{n}=0,\;\;z=-h, \\ \phi =\tilde{\phi}\left( z\right) ,\;\;x=-\,l, \\ \lim\limits_{x\rightarrow -\infty }\phi \left( x,z\right) =\cosh \left( k_{t}^{\left( 1\right) }\left( z+1\right) \right) e^{ik_{t}^{\left( 1\right) }x} \\ +R\,\cosh \left( k_{t}^{\left( 1\right) }\left( z+1\right) \right) e^{-ik_{t}^{\left( 1\right) }x}, \end{matrix} \right\} }[/math]

where [math]\displaystyle{ \mathbf{x}=\left( x,z\right) \ }[/math]and [math]\displaystyle{ \tilde{\phi}\left( z\right) }[/math] is an arbitrary continuous function[math]\displaystyle{ . }[/math] Our aim is to find the outward normal derivative of the potential on [math]\displaystyle{ x=-l }[/math] as a function of [math]\displaystyle{ \tilde{\phi}\left( z\right) }[/math].

Separation of variables for a free surface

We use separation of variables

We express the potential as

[math]\displaystyle{ \phi(x,z) = X(x)Z(z)\, }[/math]

and then Laplace's equation becomes

[math]\displaystyle{ \frac{X^{\prime\prime}}{X} = - \frac{Z^{\prime\prime}}{Z} = k^2 }[/math]

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

[math]\displaystyle{ Z^{\prime\prime} + k^2 Z =0. }[/math]

subject to the boundary conditions

[math]\displaystyle{ Z^{\prime}(-h) = 0 }[/math]

and

[math]\displaystyle{ Z^{\prime}(0) = \alpha Z(0) }[/math]

We can then use the boundary condition at [math]\displaystyle{ z=-h \, }[/math] to write

[math]\displaystyle{ Z = \frac{\cos k(z+h)}{\cos kh} }[/math]

where we have chosen the value of the coefficent so we have unit value at [math]\displaystyle{ z=0 }[/math]. The boundary condition at the free surface ([math]\displaystyle{ z=0 \, }[/math]) gives rise to:

[math]\displaystyle{ k\tan\left( kh\right) =-\alpha \, }[/math]

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 [math]\displaystyle{ k_{0}=\pm ik \, }[/math] and the positive real solutions by [math]\displaystyle{ k_{m} \, }[/math], [math]\displaystyle{ m\geq1 }[/math]. The [math]\displaystyle{ k \, }[/math] of the imaginary solution is the wavenumber. We put the imaginary roots back into the equation above and use the hyperbolic relations

[math]\displaystyle{ \cos ix = \cosh x, \quad \sin ix = i\sinh x, }[/math]

to arrive at the dispersion relation

[math]\displaystyle{ \alpha = k\tanh kh. }[/math]

We note that for a specified frequency [math]\displaystyle{ \omega \, }[/math] the equation determines the wavenumber [math]\displaystyle{ k \, }[/math].

Finally we define the function [math]\displaystyle{ Z(z) \, }[/math] as

[math]\displaystyle{ \chi_{m}\left( z\right) =\frac{\cos k_{m}(z+h)}{\cos k_{m}h},\quad m\geq0 }[/math]

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

[math]\displaystyle{ \int\nolimits_{-h}^{0}\chi_{m}(z)\chi_{n}(z) \mathrm{d} z=A_{n}\delta_{mn} }[/math]

where

[math]\displaystyle{ 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). }[/math]

Expansion of the potential

This gives us the following expression for the potential in the region [math]\displaystyle{ \Omega ^{-}, }[/math]

[math]\displaystyle{ \phi \left( x,z\right) = \phi_0(z)e^{k_0 x} +\sum_{m=0}^{\infty }\left\langle \tilde{\phi}\left( z\right) ,\phi _{m}\left( z\right) \right\rangle \frac{\phi _{m}\left( z\right)}{A_m} e^{k_{m}^{\left( 1\right) }\left( x+l\right) } }[/math]

We define [math]\displaystyle{ \mathbf{Q}_{1} }[/math] as

[math]\displaystyle{ \mathbf{Q}_{1}\tilde{\phi}\left( z\right) = \sum_{m=0}^{\infty } k_m \left\langle \tilde{\phi}\left( z\right) ,\phi _{m}\left( z\right) \right\rangle \frac{\phi _{m}\left( z\right)}{A_m} e^{k_{m}^{\left( 1\right) }\left( x+l\right) } }[/math]

Similarly, we now consider the potential in the region [math]\displaystyle{ \Omega ^{+} }[/math] which satisfies

[math]\displaystyle{ \left. \begin{matrix} \nabla ^{2}\phi =0,\;\;\mathbf{x}\in \Omega ^{+}, \\ \phi _{n}-\nu \phi =0,\;\;z=0, \\ \phi _{n}=0, \;\;z=-H, \\ \phi =\tilde{\phi}\left( z\right) ,\;\;x=\,l, \\ \lim\limits_{x\rightarrow -\infty }\phi \left( x,z\right) =T\cosh \left( k_{t}^{\left( 2\right) }\left( z+H_{2}\right) \right) e^{ik_{t}^{\left( 2\right) }x}. \end{matrix} \right\} }[/math]

Solving by separation of variables as before we obtain

[math]\displaystyle{ \left. \phi _{n}\right| _{x=l}=\mathbf{Q}_{2}\tilde{\phi}\left( z\right) , }[/math]

where the outward normal is with respect to the [math]\displaystyle{ \Omega }[/math] domain. The integral operator [math]\displaystyle{ \mathbf{Q}_{2} }[/math] is given by

[math]\displaystyle{ \mathbf{Q}_{2}\tilde{\phi}\left( z\right) =\sum_{m=0}^{\infty }k_{m}^{\left( 2\right) }\left\langle \tilde{\phi}\left( z\right) ,\tau _{m}^{\left( 2\right) }\left( z\right) \right\rangle \tau _{m}^{\left( 2\right) }\left( z\right) . (21) }[/math]