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 [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]. We begin with the Frequency Domain Problem for a fixed vertical plate which occupies the region [math]\displaystyle{ x=0 }[/math] and [math]\displaystyle{ -a\gt z\gt -b }[/math] where [math]\displaystyle{ 0\lt a\lt b\lt h }[/math]. We assume [math]\displaystyle{ e^{i\omega t} }[/math] time dependence.
The boundary value problem can therefore be expressed as
[math]\displaystyle{ \Delta\phi=0, \,\, -h\lt z\lt 0, }[/math]
[math]\displaystyle{ \partial_z\phi=0, \,\, z=-h, }[/math]
[math]\displaystyle{ \partial_x\phi=0, \,\, -a\gt z\gt -b,\,x=0, }[/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
We use separation of variables in the two regions, [math]\displaystyle{ x\lt 0 }[/math] and [math]\displaystyle{ x\gt 0 }[/math].
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]
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:
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]
Incident potential
To create meaningful solutions of the velocity potential [math]\displaystyle{ \phi }[/math] in the specified domains we add an incident wave term to the expansion for the domain of [math]\displaystyle{ x \lt 0 }[/math] above. The incident potential is a wave of amplitude [math]\displaystyle{ A }[/math] in displacement travelling in the positive [math]\displaystyle{ x }[/math]-direction. We would only see this in the time domain [math]\displaystyle{ \Phi(x,z,t) }[/math] however, in the frequency domain the incident potential can be written as
[math]\displaystyle{ \phi_{\mathrm{I}}(x,z) =e^{-k_{0}x}\chi_{0}\left( z\right). }[/math]
The total velocity (scattered) potential now becomes [math]\displaystyle{ \phi = \phi_{\mathrm{I}} + \phi_{\mathrm{D}} }[/math] for the domain of [math]\displaystyle{ x \lt 0 }[/math].
The first term in the expansion of the diffracted potential for the domain [math]\displaystyle{ x \lt 0 }[/math] is given by
[math]\displaystyle{ a_{0}e^{k_{0}x}\chi_{0}\left( z\right) }[/math]
which represents the reflected wave.
In any scattering problem [math]\displaystyle{ |R|^2 + |T|^2 = 1 }[/math] where [math]\displaystyle{ R }[/math] and [math]\displaystyle{ T }[/math] are the reflection and transmission coefficients respectively. In our case of the semi-infinite dock [math]\displaystyle{ |a_{0}| = |R| = 1 }[/math] and [math]\displaystyle{ |T| = 0 }[/math] as there are no transmitted waves in the region under the dock.
Solution using Symmetry
The problem is symmetric about the line [math]\displaystyle{ x=0 }[/math] 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
[math]\displaystyle{ \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 }[/math]
The boundary condition is that [math]\displaystyle{ \partial_x \phi = 0 }[/math] on [math]\displaystyle{ x=0 }[/math]. The problem reduces to Waves reflecting off a vertical wall. [math]\displaystyle{ a_{0}^{s}=1 }[/math] [math]\displaystyle{ a_{m}^{s}=0,\, \,n \gt 0 }[/math]
Anti-symmetric solution
The anti-symmetric potential can be expanded as
[math]\displaystyle{ \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 }[/math]
For the anti-symmetric solution the potential satisfies [math]\displaystyle{ \partial_x \phi = 0, -a\gt z\gt -b }[/math] on [math]\displaystyle{ x=0 }[/math] and [math]\displaystyle{ \phi = 0, 0\gt z\gt -a, -b\gt z\gt -h }[/math]. We impose this condition by integrating the following
[math]\displaystyle{ \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 }[/math]
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
[math]\displaystyle{ a_{m} = \frac{1}{2}\left(a_{m}^{s}+a_{m}^{a}\right) }[/math]
and in the right open water region is just
[math]\displaystyle{ a_{m} = \frac{1}{2}\left(a_{m}^{s} - a_{m}^{a}\right) }[/math]
Therefore the scattered potential (without the incident wave, which will be added later) can be expanded as
[math]\displaystyle{ \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 }[/math]
and
[math]\displaystyle{ \phi(x,z)=\sum_{m=0}^{\infty}\frac{1}{2}\left(a_{m}^{s} - a_{m}^{a}\right) e^{-\kappa_{m}x}\phi_{m}(z), \;\;x\gt 0 }[/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].
When a wave in incident at an angle [math]\displaystyle{ \theta }[/math] 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).
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
[math]\displaystyle{ k^2 = k_x^2 + k_y^2 }[/math]
where [math]\displaystyle{ k }[/math] is the separation constant calculated without an incident angle.
Therefore the potential can be expanded as
[math]\displaystyle{ \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 }[/math]
and
[math]\displaystyle{ \phi(x,z)=\sum_{m=0}^{\infty}b_{m} e^{-\hat{\kappa}_{m}x}\psi_{m}(z), \;\;x\gt 0 }[/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_{n=0}^{\infty}b_{m}B_{ml} }[/math]
and
[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} }[/math]
and these are solved exactly as before.
Matlab Code
A program to calculate the coefficients for the semi-infinite dock problems can be found here semiinfinite_dock.m
Additional code
This program requires
