Eigenfunction Matching for a Semi-Infinite Rectangle
Introduction
The problem consists of a region to the left with a free water surface and a region to the right with a rigid semi infinite rectangular block with height [math]\displaystyle{ d }[/math] fully submerged through which not flow is possible. We begin with the simple problem when the waves are normally incident (so that the problem is truly two-dimensional) and later we also consider the case when the waves are incident at an angle.
Governing Equations
We begin with the Frequency Domain Problem for a submerged dock which occupies the region [math]\displaystyle{ x\gt 0 }[/math] (we assume [math]\displaystyle{ e^{i\omega t} }[/math] time dependence). The depth of submergence is [math]\displaystyle{ d }[/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
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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]
Separation of Variables for a Dock
The separation of variables equation for the fully submerged dock is given by:
subject to the boundary conditions
and
The solution is [math]\displaystyle{ k=\kappa_{m}= \frac{m\pi}{h-d} \, }[/math], [math]\displaystyle{ m\in\mathbb{N}\cup\left\{0\right\} }[/math] and
We note that
where
[math]\displaystyle{ C_{m} = \frac{1}{2} \left( \frac{\cos(\kappa_{m}(h-d))\sin(\kappa_{m}(h-d))+\kappa_{m}(h-d)}{\kappa_{m}} \right) }[/math]
Inner product between free surface and dock modes
where
Expansion of the potential
We need to apply some boundary conditions at plus and minus infinity, where are essentially the the solution cannot grow. This means that we only have the positive (or negative) roots of the dispersion equation. However, it does not help us with the purely imaginary root. Here we must use a different condition, essentially identifying one solution as the incoming wave and the other as the outgoing wave.
Therefore the scattered potential (without the incident wave, which will be added later) can be expanded as
[math]\displaystyle{ \phi(x,z)=\sum_{m=0}^{\infty}a_{m}e^{k_{m}x}\phi_{m}(z), \;\;x\lt 0 }[/math]
and
[math]\displaystyle{ \phi(x,z)=\sum_{m=0}^{\infty}b_{m} e^{-\kappa_{m}x}\psi_{m}(z), \;\;x\gt 0 }[/math]
where [math]\displaystyle{ a_{m} }[/math] and [math]\displaystyle{ b_{m} }[/math] are the coefficients of the potential in the open water and the dock covered region respectively.
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.
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=0 }[/math] have to be equal. We obtain:
For the first equation we multiply both sides by [math]\displaystyle{ \psi_{n}(z) \, }[/math] and integrating from [math]\displaystyle{ -h }[/math] to [math]\displaystyle{ -d }[/math] to obtain:
and for the second equation we multiply both sides by [math]\displaystyle{ \phi_{n}(z) \, }[/math] and integrating from [math]\displaystyle{ -h }[/math] to [math]\displaystyle{ 0 }[/math] to obtain:
Solving the equations above will yield the coefficients of the water velocity potential in the dock covered region.
Numerical Solution
To solve the system of equations, we set the upper limit of [math]\displaystyle{ n }[/math] to be [math]\displaystyle{ N }[/math]. This resulting system can be expressed in the block matrix form below,
[math]\displaystyle{ \begin{bmatrix} B_{00} & B_{01} & \cdots & B_{0N} & -C_{0} & 0 & \cdots & 0 \\ B_{10} & B_{11} & \cdots & B_{1N} & 0 & -C_{1} & \cdots & \vdots \\ \vdots & \vdots & \ddots & \vdots & \vdots & \vdots & \ddots & 0 \\ B_{N0} & B_{N1} & \cdots & B_{NN} & 0 & \cdots & 0 & -C_{N} \\ k_{0}A_{0} & 0 & \cdots & 0 & \kappa_{0}B_{00} & \kappa_{1}B_{10} & \cdots & \kappa_{N}B_{N0} \\ 0 & k_{1}A_{1} & \cdots & \vdots & \kappa_{0}B_{01} & \kappa_{1}B_{11} & \cdots & \kappa_{N}B_{N1} \\ \vdots & \vdots & \ddots & 0 & \vdots & \vdots & \ddots & \vdots \\ 0 & \cdots & 0 & k_{N}A_{N} & \kappa_{0}B_{0N} & \kappa_{1}B_{1N} & \cdots & \kappa_{N}B_{NN} \\ \end{bmatrix} \begin{bmatrix} a_{0} \\ a_{1} \\ \vdots \\ a_{N} \\ b_{0} \\ b_{1} \\ \vdots \\ b_{N} \\ \end{bmatrix} =\begin{bmatrix} -B_{00} \\ -B_{10} \\ \vdots \\ -B_{N0} \\ k_{0}A_{0} \\ 0 \\ \vdots \\ 0 \\ \end{bmatrix} }[/math]
We then simply need to solve the linear system of equations.
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
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
