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 Two Symmetric Bodies 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 [math]\displaystyle{ x\gt 0 }[/math]. The water is assumed to have constant finite depth [math]\displaystyle{ h }[/math] and the [math]\displaystyle{ z }[/math]-direction points vertically upward with the water surface at [math]\displaystyle{ z=0 }[/math] and the sea floor at [math]\displaystyle{ z=-h }[/math]. The boundary value problem can therefore be expressed as

[math]\displaystyle{ \Delta\phi=0, \,\, -h\lt z\lt 0, }[/math]

[math]\displaystyle{ \phi_{z}=0, \,\, z=-h, }[/math]

[math]\displaystyle{ \partial_z\phi=\alpha\phi, \,\, z=0,\,x\lt -L_2,\,-L_1\lt x\lt L_1, {\rm or} \, x\gt L_2 }[/math]

[math]\displaystyle{ \partial_z\phi=0, \,\, z=0,\,-L_2\lt x\lt -L_1, {\rm or} \, L_1\lt x\lt L_2, }[/math]

where we require [math]\displaystyle{ L_1\lt L_2 }[/math] and we define [math]\displaystyle{ L_2 - L_1 = 2L }[/math] (so that the dock also has length [math]\displaystyle{ 2L }[/math]).

We must also apply the Sommerfeld Radiation Condition as [math]\displaystyle{ |x|\rightarrow\infty }[/math]. 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

The solution method uses Two Symmetric Bodies in Two-Dimensions and we write the potential as a symmetric and an anti-symmetric part and consider only the region [math]\displaystyle{ x\lt 0 }[/math]. We apply either Neuman (symmetric) or Dirichlet (anti-symmetric) boundary conditions at [math]\displaystyle{ x=0 }[/math]. 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

[math]\displaystyle{ \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\lt -L_2 }[/math]

[math]\displaystyle{ \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\lt x\lt L_1 }[/math]

and

[math]\displaystyle{ \phi(x,z)=\sum_{m=0}^{\infty}d^s_{m} \frac{\cos(k_{m}x)}{\cos(k_m L_1)} \phi_{m}(z), \;\;-L_1\lt x\lt 0 }[/math]

where [math]\displaystyle{ a^s_{m} }[/math] and [math]\displaystyle{ d^s_{m} }[/math] are the coefficients of the potential in the open water regions to the left and right and [math]\displaystyle{ b^s_m }[/math] and [math]\displaystyle{ c^s_m }[/math] are the coefficients under the dock. [math]\displaystyle{ k_n }[/math] are the roots of the Dispersion Relation for a Free Surface. We denote the positive imaginary solutions by [math]\displaystyle{ k_{0} }[/math] and the positive real solutions by [math]\displaystyle{ k_{m} }[/math], [math]\displaystyle{ m\geq1 }[/math] (ordered with increasing imaginary part) and [math]\displaystyle{ \kappa_{m}=m\pi/h }[/math]. We define

[math]\displaystyle{ \phi_{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 and

[math]\displaystyle{ \psi_{m}\left( z\right) = \cos\kappa_{m}(z+h),\quad m\geq 0 }[/math]

as the vertical eigenfunction of the potential in the dock covered region. For later reference, we note that:

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

where

[math]\displaystyle{ A_{m}=\frac{1}{2}\left( \frac{\cos k_{m}h\sin k_{m}h+k_{m}h}{k_{m}\cos ^{2}k_{m}h}\right) }[/math]

and

[math]\displaystyle{ \int\nolimits_{-h}^{0}\phi_{n}(z)\psi_{m}(z) d z=B_{mn} }[/math]

where

[math]\displaystyle{ 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\cos\kappa_{m}h\right) \left( k_{n} ^{2}-\kappa_{m}^{2}\right) } }[/math]

and

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

where

[math]\displaystyle{ C_{m}=\frac{1}{2}h,\quad,m\neq 0 \quad \mathrm{and} \quad C_0 = h }[/math]

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 [math]\displaystyle{ x=- L_2 }[/math] and [math]\displaystyle{ x=-L_1 }[/math] have to be equal. We obtain

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

[math]\displaystyle{ -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} }[/math]

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

[math]\displaystyle{ -\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} \tan(k_m L_1) k_m\phi_{m}\left( z\right) }[/math]

for each [math]\displaystyle{ n }[/math]. We solve these equations by multiplying both equations by [math]\displaystyle{ \phi_{l}(z) }[/math] and integrating from [math]\displaystyle{ -h }[/math] to [math]\displaystyle{ 0 }[/math] to obtain:

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

[math]\displaystyle{ -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} }[/math]

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

[math]\displaystyle{ - 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 \tan(k_m L_1) k_l A_l }[/math]

Numerical Solution

To solve the system of equations we set the upper limit of [math]\displaystyle{ l }[/math] to be [math]\displaystyle{ M }[/math]. 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.

[math]\displaystyle{ \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\lt -L_2 }[/math]

[math]\displaystyle{ \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\lt x\lt L_1 }[/math]

and

[math]\displaystyle{ \phi(x,z)=\sum_{m=0}^{\infty}d^a_{m} \frac{\sin(k_{m}x)}{-\sin(k_m L_1)} \phi_{m}(z), \;\;-L_1\lt x\lt 0 }[/math]

where [math]\displaystyle{ a^a_{m} }[/math] and [math]\displaystyle{ d^a_{m} }[/math] are the coefficients of the potential in the open water regions to the left and right and [math]\displaystyle{ b^a_m }[/math] and [math]\displaystyle{ c^a_m }[/math] are the coefficients under the dock.

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

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

[math]\displaystyle{ -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} }[/math]

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

[math]\displaystyle{ - 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 \tan(k_m L_1) k_l A_l }[/math]

Solution with Waves Incident at an Angle

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

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

Therefore the potential can be expanded as

[math]\displaystyle{ \phi(x,z)=e^{-k_{0}(x+L)}\phi_{0}\left( z\right) + \sum_{m=0}^{\infty}a_{m}e^{\hat{k}_{m}(x+L)}\phi_{m}(z), \;\;x\lt -L }[/math]

[math]\displaystyle{ \phi(x,z)=\sum_{m=0}^{\infty}b_{m} e^{-\hat{\kappa}_{m} (x+L)}\psi_{m}(z) ++ \sum_{m=1}^{\infty}c_{m} e^{\hat{\kappa}_{m} (x-L)}\psi_{m}(z) , \;\;-L\lt x\lt L }[/math]

and

[math]\displaystyle{ \phi(x,z)=\sum_{m=0}^{\infty}d_{m} e^{-\hat{k}_{m}(x-L)}\phi_{m}(z), \;\;L\lt x }[/math]

where [math]\displaystyle{ \hat{k}_{m} = \sqrt{k_m^2 + k_y^2} }[/math] and [math]\displaystyle{ \hat{\kappa}_{m} = \sqrt{\kappa_m^2 + k_y^2} }[/math] 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

[math]\displaystyle{ A_{0}\delta_{0l}+a_{l}A_{l} =\sum_{m=0}^{\infty}b_{m}B_{ml} + \sum_{m=0}^{\infty}c_{m}B_{ml}e^{-2L\hat{\kappa}_m} }[/math]

[math]\displaystyle{ -\hat{k}_{0}A_{0}\delta_{0l}+a_{l}\hat{k}_{l}A_l = - \sum_{m=0}^{\infty}b_{m}\hat{\kappa}_{m}B_{ml} + \sum_{m=0}^{\infty}c_{m}\hat{\kappa}_{m}B_{ml} e^{-2L\hat{\kappa}_m} }[/math]

[math]\displaystyle{ \sum_{m=0}^{\infty}b_{m}B_{ml}e^{-2L\hat{\kappa}_m} + \sum_{m=0}^{\infty}c_{m}B_{ml} =d_l A_l }[/math]

[math]\displaystyle{ -\sum_{m=0}^{\infty}b_{m}\hat{\kappa}_{m}B_{ml} e^{-2L\hat{\kappa}_m} + \sum_{m=0}^{\infty}c_{m}\hat{\kappa}_{m}B_{ml} = -d_l \hat{k}_l A_l }[/math]

and these are solved exactly as before.

Matlab Code

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

Additional code

This program requires dispersion_free_surface.m to run