Difference between revisions of "Eigenfunction Matching for a Semi-Infinite Floating Elastic Plate"

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=Introduction=
+
{{complete pages}}
 +
 
 +
== Introduction ==
  
 
We show here a solution for a semi-infinite [[:Category:Floating Elastic Plate|Floating Elastic Plate]] on [[Finite Depth]].  
 
We show here a solution for a semi-infinite [[:Category:Floating Elastic Plate|Floating Elastic Plate]] on [[Finite Depth]].  
The problem was solved by [[Fox and Squire 1994]] but the solution method here is slightly different.  
+
The problem was solved by [[Fox and Squire 1994]] but the solution method here is slightly different.
 
+
The simpler theory for a [[Eigenfunction Matching for a Semi-Infinite Dock|Dock]] describes
 +
many of the ideas here in more detail.
  
We show here a solution to the problem of wave propagation under many floating elastic plates of variable properties
+
[[Image:Semiinfinite plate.jpg|thumb|right|300px|Wave scattering by a submerged semi-infinite elastic plate]]
This work is based on [[Kohout et. al. 2006]]. This is a generalisation of the
 
[[Eigenfunction Matching Method for a Semi-Infinite Floating Elastic Plate]].
 
We assume that the first and last plate are semi-infinite. The presentation here does not
 
allow open water (it could be included but makes the formulation more complicated). In any case
 
open water can be considered by taking the limit as the
 
plate thickness tends to zero. The solution is derived using an extended eigenfunction matching method, in which
 
the plate boundary conditions are satisfied as auxiliary equations.
 
  
= Equations =
+
== Equations ==
  
 
We consider the problem of small-amplitude waves which are incident on a semi-infinite floating elastic
 
We consider the problem of small-amplitude waves which are incident on a semi-infinite floating elastic
Line 24: Line 20:
 
the methods of solution presented here. We begin with the [[Frequency Domain Problem]] for a semi-infinite
 
the methods of solution presented here. We begin with the [[Frequency Domain Problem]] for a semi-infinite
 
[[Floating Elastic Plate|Floating Elastic Plates]]
 
[[Floating Elastic Plate|Floating Elastic Plates]]
in the non-dimensional form of [[Tayler_1986a|Tayler 1986]] ([[Dispersion Relation for a Floating Elastic Plate]])
+
in the non-dimensional form of [[Tayler_1986a|Tayler 1986]] ([[Dispersion Relation for a Floating Elastic Plate]]).
 +
We also assume that the waves are normally incident (incidence at an angle will be discussed later).
 +
<center><math>
 +
\Delta \phi = 0, \;\;\; -h < z \leq 0,
 +
</math></center>
 +
<center><math>
 +
\partial_z \phi = 0, \;\;\; z = - h,
 +
</math></center>
 +
<center><math>
 +
\partial_z\phi=\alpha\phi, \,\, z=0,\,x<0,
 +
</math></center>
 +
<center><math>
 +
\beta \partial_x^4\partial_z \phi
 +
- \left( \gamma\alpha - 1 \right) \partial_z \phi = \alpha\phi, \;\;
 +
z = 0, \;\;\; x \geq 0,
 +
</math></center>
 +
where <math>\alpha = \omega^2</math>, <math>\beta</math> and <math>\gamma</math>
 +
are the stiffness and mass constant for the plate respectively. The free edge conditions
 +
at the edge of the plate imply
 +
<center><math>
 +
\partial_x^3 \partial_z\phi = 0, \;\; z = 0, \;\;\; x = 0,
 +
</math></center>
 +
<center><math>
 +
\partial_x^2 \partial_z\phi = 0, \;\; z = 0, \;\;\; x = 0,
 +
</math></center>
 +
 
 +
== Method of solution ==
 +
 
 +
We use [http://en.wikipedia.org/wiki/Separation_of_Variables separation of variables] in the two regions, <math>x<0</math>
 +
and <math>x>0</math>.
 +
 
 +
{{separation of variables in two dimensions}}
 +
 
 +
{{separation of variables for a free surface}}
 +
 
 +
{{separation of variables for a floating elastic plate}}
  
<center><math>\begin{matrix}
+
{{free surface floating elastic plate relations}}
\left(\frac{\partial^2}{\partial x^2} +
 
\frac{\partial^2}{\partial z^2} - k_y^2\right) \phi = 0, \;\;\;\; \mbox{ for } -h < z \leq 0,
 
\end{matrix}</math></center>
 
<center><math>\begin{matrix}
 
\frac{\partial \phi}{\partial z} = 0, \;\;\;\; \mbox{ at } z = - h,
 
\end{matrix}</math></center>
 
<center><math>\begin{matrix}
 
\left( \beta \left(\frac{\partial^2}{\partial x^2} - k^2_y\right)^2
 
- \gamma\alpha + 1\right)\frac{\partial \phi}{\partial z} - \alpha\phi = 0, \;\;\;\;
 
\mbox{ at } z = 0, \;\;\; x \geq 0,
 
\end{matrix}</math></center>
 
<center><math>\begin{matrix}
 
\frac{\partial \phi}{\partial z} - \alpha\phi = 0, \;\;\;\;
 
\mbox{ at } z = 0, \;\;\; x \leq 0,
 
\end{matrix}</math></center>
 
where <math>\alpha = \omega^2</math>, <math>\beta</math> and <math>\gamma</math>
 
and the stiffness and mass constant for the plate. The conditions
 
at the edge of the plate os
 
<center><math>\begin{matrix}
 
\left(\frac{\partial^3}{\partial x^3} - (2 - \nu)k^2_y\frac{\partial}{\partial x}\right) \frac{\partial\phi}{\partial z}= 0, \;\;\;\; \mbox{ at } z = 0, \;\;\; \mbox{ for } x = 0,
 
\end{matrix}</math></center>
 
<center><math>\begin{matrix}
 
\left(\frac{\partial^2}{\partial x^2} - \nu k^2_y\right)\frac{\partial\phi}{\partial z} = 0,
 
\;\;\;\;\mbox{ at } z = 0, \;\;\; \mbox{ for } x = 0.
 
\end{matrix}</math></center>
 
  
=Method of solution=
+
{{incident potential for two dimensions}}
  
==Eigenfunction expansion==
+
== An infinite dimensional system of equations ==
  
We will solve the system of equations using an [[:Category:Eigenfunction Matching Method|Eigenfunction Matching Method]].
+
The potential and its derivative must be continuous across the
The method was developed by [[Fox and Squire 1994]]. The separation of variables for the left hand region where
+
transition from open water to the plate covered region. Therefore, the
there is open water is described in [[Eigenfunction Matching Method for a Semi-Infinite Dock] and we consider
+
potentials and their derivatives at <math>x=0</math> have to be equal.
here only the separation of variables in the plate covered region.
+
We also truncate the sum at <math>N</math> being careful that we have
 +
two extra modes on the plate covered region to satisfy the edge conditions.
 +
We obtain
 +
<center>
 +
<math>
 +
\phi_{0}\left(  z\right) + \sum_{m=0}^{N}
 +
a_{m} \phi_{m}\left(  z\right)
 +
=\sum_{m=-2}^{N}b_{m}\psi_{m}(z)
 +
</math>
 +
</center>
 +
and
 +
<center>
 +
<math>
 +
-k_{0}\phi_{0}\left(  z\right)  +\sum
 +
_{m=0}^{N} a_{m}k_{m}\phi_{m}\left(  z\right)
 +
=-\sum_{m=-2}^{N}b_{m}\kappa_{m}\psi
 +
_{m}(z)
 +
</math>
 +
</center>
 +
for each <math>N</math>.
 +
We solve these equations by multiplying both equations by
 +
<math>\phi_{l}(z)\,</math> and integrating from <math>-h</math> to <math>0</math> to obtain:
 +
<center>
 +
<math>
 +
A_{0}\delta_{0l}+a_{l}A_{l}
 +
=\sum_{m=-2}^{N}b_{m}B_{ml},\,\,0 \leq l  \leq N
 +
</math>
 +
</center>
 +
and
 +
<center>
 +
<math>
 +
-k_{0}A_{0}\delta_{0l}+a_{l}k_{l}A_l
 +
=-\sum_{m=-2}^{N}b_{m}\kappa_{m}B_{ml},\,\,0 \leq l \leq N
 +
</math>
 +
</center>
 +
If we multiply the first equation by <math>k_l</math> and subtract the second equation
 +
we obtain
 +
<center>
 +
<math>
 +
(k_{0}+k_l)A_{0}\delta_{0l}
 +
=\sum_{m=-2}^{N}b_{m}(k_l + \kappa_{m})B_{ml}
 +
</math>
 +
</center>
 +
Finally, we need to apply the conditions at the edge of the plate to give us two further equations,
 +
<center>
 +
<math>
 +
\partial_x^2\partial_z\phi = \sum_{m=-2}^{N}b_{m} \kappa_m^3 \tan\kappa_m h = 0
 +
</math>
 +
</center>
 +
and
 +
<center>
 +
<math>
 +
\partial_x^3\partial_z\phi = - \sum_{m=-2}^{N}b_{m} \kappa_m^4 \tan\kappa_m h = 0
 +
</math>
 +
</center>
  
===Separation of variables under the Plate===
+
== Numerical Solution ==
The potential velocity can be written in terms of an infinite series of separated eigenfunctions under
 
each elastic plate, of the form
 
<math>\phi = e^{\kappa_\mu x} \cos(k_\mu(z+h)).</math>
 
If we apply the boundary conditions given
 
we obtain the [[Dispersion Relation for a Floating Elastic Plate]]
 
<center><math>\begin{matrix}
 
k_\mu\tan{(k_\mu h)}= & -\frac{\alpha}{\beta_\mu k_\mu^{4}  + 1 - \alpha\gamma_\mu}
 
\end{matrix}</math></center>
 
Solving for <math>k_\mu</math> gives a pure imaginary root
 
with positive imaginary part, two complex roots (two complex conjugate paired roots
 
with positive imaginary part in all physical situations), an infinite number of positive real roots
 
which approach <math>{n\pi}/{h}</math> as <math>n</math> approaches infinity, and also the negative of all
 
these roots ([[Dispersion Relation for a Floating Elastic Plate]]) . We denote the two complex roots with positive imaginary part
 
by <math>k_\mu(-2)</math> and <math>k_\mu(-1)</math>, the purely imaginary
 
root with positive imaginary part by <math>k_\mu(0)</math> and the real roots with positive imaginary part
 
by <math>k_\mu(n)</math> for <math>n</math> a positive integer.
 
The imaginary root with positive imaginary part corresponds to a
 
reflected travelling mode propagating along the <math>x</math> axis.
 
The complex roots with positive imaginary parts correspond to damped reflected travelling modes and the real roots correspond to reflected evanescent modes.
 
In a similar manner, the negative of these correspond to the transmitted travelling, damped and evanescent modes respectively.
 
The coefficient <math>\kappa_\mu</math> is
 
<center><math>\kappa_\mu(n) = \sqrt{k_\mu(n)^2 + k_y^2},</math></center>
 
where the root with positive real part is chosen or if the real part is negative with negative imaginary part.
 
  
==Expressions for the potential velocity==
+
To solve the system of equations previously defined we set the upper limit of <math>l</math> to
 +
be <math>N</math>, as stated before. In terms of matrix, we obtain
 +
<center>
 +
<math>
 +
\begin{bmatrix}
 +
\begin{bmatrix}
 +
A_0&0 \quad \cdots&0\\
 +
0&&\\
 +
\vdots&A_l&\vdots\\
 +
&&0\\
 +
0&\cdots \quad 0 &A_N
 +
\end{bmatrix}
 +
&
 +
\begin{bmatrix}
 +
-B_{-2,0}&-B_{-1,0}\\
 +
&\\
 +
\vdots&\vdots\\
 +
&\\
 +
-B_{-2,N}&-B_{-1,N}
 +
\end{bmatrix}
 +
&
 +
\begin{bmatrix}
 +
-B_{0,0}&\cdots&-B_{0,N}\\
 +
&&\\
 +
\vdots&-B_{m,l}&\vdots\\
 +
&&\\
 +
-B_{N,0}&\cdots&-B_{N,N}
 +
\end{bmatrix}
 +
\\ \\
 +
\begin{bmatrix}
 +
0&&\cdots&&0\\
 +
0&&\cdots&&0
 +
\end{bmatrix}
 +
&
 +
\begin{bmatrix}
 +
\kappa_{-2}^3\tan\kappa_{-2}h&\kappa_{-1}^3\tan\kappa_{-1}h\\
 +
\kappa_{-2}^4\tan\kappa_{-2}h&\kappa_{-1}^4\tan\kappa_{-1}h
 +
\end{bmatrix}
 +
&
 +
\begin{bmatrix}
 +
\kappa_0^3\tan\kappa_0h&\cdots&\kappa_N^3\tan\kappa_Nh\\
 +
\kappa_0^4\tan\kappa_0h&\cdots&\kappa_N^4\tan\kappa_Nh
 +
\end{bmatrix}
 +
\\ \\
 +
\begin{bmatrix}
 +
0&&\cdots&&0\\
 +
&&&&\\
 +
\vdots&&\ddots&&\vdots\\
 +
&&&&\\
 +
0&&\cdots&&0
 +
\end{bmatrix}
 +
&
 +
\begin{bmatrix}
 +
(k_0+\kappa_{-2})B_{-2,0}&(k_0+\kappa_{-1})B_{-1,0}\\
 +
&\\
 +
\vdots&\vdots\\
 +
&\\
 +
(k_N+\kappa_{-2})B_{-2,N}&(k_N+\kappa_{-1})B_{-1,N}
 +
\end{bmatrix}
 +
&
 +
\begin{bmatrix}
 +
(k_0 + \kappa_0) \, B_{0,0}&\cdots&(k_N + \kappa_{0}) \, B_{0,N}\\
 +
&&\\
 +
\vdots&(k_l + \kappa_{m}) \, B_{m,l}&\vdots\\
 +
&&\\
 +
(k_0 + \kappa_N) \, B_{N,0}&\cdots&(k_N + \kappa_{N}) \, B_{N,N}\\
 +
\end{bmatrix}
 +
\end{bmatrix}
 +
\begin{bmatrix}
  
We now expand the potential under each plate using the separation of variables solution.
+
\begin{bmatrix}
We always include the two complex and one imaginary root, and truncate the expansion
+
a_{0} \\
at <math>M</math> real roots of the dispersion equation.
+
\\
The potential <math>\phi</math> can now be expressed as the following sum of eigenfunctions:
+
\vdots \\
<center><math>
+
\\
\phi \approx \left\{
+
a_N
\begin{matrix}
+
\end{bmatrix}
{
+
\\ \\
Ie^{\kappa_{1}(0)(x-r_1)}\frac{\cos(k_1(0)(z+h))}{\cos(k_1(0)h)} }+\\
+
\begin{bmatrix}
{  
+
b_{-2}\\
        \qquad \qquad \sum_{n=-2}^{M}R_1(n) e^{\kappa_{1}(n)(x-r_1)} \frac{\cos(k_1(n)(z+h))}{\cos(k_1(n)h)} },&
+
b_{-1}
                \mbox{ for } x < r_1,\\
+
\end{bmatrix}
{
+
\\ \\
\sum_{n=-2}^{M}T_{\mu}(n) e^{-\kappa_\mu(n)(x-l_\mu)}
+
\begin{bmatrix}
        \frac{\cos(k_\mu(n)(z+h))}{\cos(k_\mu(n)h)} } + \\
+
b_{0}\\
{
+
\\
                \qquad \qquad \sum_{n=-2}^{M}R_{\mu}(n) e^{\kappa_\mu(n)(x-r_\mu)}
+
\vdots \\
                        \frac{\cos(k_\mu(n)(z+h))}{\cos(k_\mu(n)h)} },
+
\\
                                &\mbox{ for } l_\mu< x < r_\mu,\\
+
b_N
{
+
\end{bmatrix}
\sum_{n=-2}^{M}T_{\Lambda}(n)e^{-\kappa_{\Lambda}(n)(x-l_\Lambda)}
+
\end{bmatrix}
        \frac{\cos(k_\Lambda(n)(z+h))}{\cos(k_\Lambda(n)h)} }, &\mbox{ for } l_\Lambda<x,
+
=
\end{matrix} \right.
+
\begin{bmatrix}
</math></center>
+
\begin{bmatrix}
where <math>I</math> is the non-dimensional incident wave amplitude in potential,
+
- A_{0} \\
<math>\mu</math> is the <math>\mu^{th}</math> plate, <math>\Lambda</math> is the last plate,
+
0 \\
<math>r_\mu</math> represents the <math>x</math>-coordinate of the right edge of the <math>\mu^{th}</math> plate,
+
\vdots \\
<math>l_\mu</math> (<math>=r_{\mu-1}</math>) represents the <math>x</math>-coordinate of the left edge of the <math>\mu^{th}</math> plate,
+
\\
<math>R_\mu(n)</math> represents the reflected potential coefficient of the <math>n^{th}</math> mode under the <math>\mu^{th}</math> plate, and  
+
0
<math>T_\mu(n)</math> represents the transmitted potential coefficient of the <math>n^{th}</math> mode under the <math>\mu^{th}</math> plate.
+
\end{bmatrix}
Note that we have divided by  <math>\cos{(kh)}</math>,
+
\\ \\
so that the coefficients are normalised by the
+
\begin{bmatrix}
potential at the free surface rather than at the bottom surface.
+
0 \\
 +
0
 +
\end{bmatrix}
 +
\\ \\
 +
\begin{bmatrix}
 +
2k_{0}A_{0} \\
 +
0 \\
 +
\vdots \\
 +
\\
 +
0
 +
\end{bmatrix}
 +
\end{bmatrix}
 +
</math>
 +
</center>
 +
We then simply need to solve the linear system of equations. Note that we can solve this equation for
 +
<math>b_n</math> first and then solve for <math>a_n</math>
  
==Expressions for displacement==
+
== Waves Incident at an Angle ==  
The displacement is given by
 
<center><math>
 
\eta \approx \frac{i}{\omega}\left\{
 
\begin{matrix}
 
{
 
Ik_1(0)e^{\kappa_{1}(0)(x-r_1)}\tan{(k_1(0)h)} - } \\
 
{
 
        \qquad \sum_{n=-2}^{M}R_{1}(n)k_1(n)e^{\kappa_{1}(n)(x-r_1)} \tan{(k_1(n)h)} },
 
                & \mbox{ for } x < r_1, \\
 
{
 
-\sum_{n=-2}^{M}T_{\mu}(n)k_\mu(n)e^{-\kappa_\mu(n)(x-l_\mu)}\tan{(k_\mu(n)h)} - }\\
 
{
 
        \qquad \sum_{n=-2}^{M}R_{\mu}(n)k_\mu(n)e^{\kappa_\mu(n)(x-r_\mu)}
 
                \tan{(k_\mu(n)h)} },& \mbox{ for } l_\mu< x < r_\mu,\\
 
{
 
-\sum_{n=-2}^{M}T_{\Lambda}(n)k_\mu(n)e^{-\kappa_\mu(n)(x-l_\Lambda)}\tan{(k_\mu(n)h)} },
 
        &\mbox{ for } l_\Lambda<x.
 
\end{matrix}\right.
 
</math></center>
 
  
==Solving via eigenfunction matching==
+
We can consider the case of [[Waves Incident at an Angle]] <math>\theta</math>.
 +
{{incident angle}}
  
To solve for the coefficients, we require as many equations as we have unknowns.
+
It is shown that the potential can be expanded as
We derive the equations from the free edge conditions and from imposing conditions of
+
<center>
continuity of the potential and its derivative in the <math>x</math>-direction at each plate
+
<math>
boundary. We impose the latter condition by taking inner products with respect to
+
\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<0
the orthogonal functions <math>\cos \frac{m\pi}{h}(z+h)</math>, where <math>m</math> is a natural number. These functions are chosen for the following reasons.
+
</math>
The vertical eigenfunctions <math>\cos k_\mu(n)(z+h)</math> are not orthogonal (they are not even a basis) and could therefore lead to an ill-conditioned system of equations. Furthermore, by choosing  <math>\cos \frac{m\pi}{h}(z+h)</math> we can use the same functions to take the inner products under every plate. Finally, and most importantly, the plate eigenfunctions approach <math>\cos{(m\pi/h)(z + h)}</math> for large <math>m</math>, so that as we increase the number of modes the matrices become almost diagonal, leading to a very well-conditioned system of equations.  
+
</center>
 +
and
 +
<center>
 +
<math>
 +
\phi(x,z)=\sum_{m=-2}^{\infty}b_{m}
 +
e^{-\hat{\kappa}_{m}x}\psi_{m}(z), \;\;x>0
 +
</math>
 +
</center>
 +
where <math>\hat{k}_{m} = \sqrt{k_m^2 - k_y^2}</math> and <math>\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.  
  
Taking inner products leads to the following equations
+
The edge conditions are also different and are
 
<center><math>
 
<center><math>
\begin{matrix}
+
\left(\frac{\partial^3}{\partial x^3} + (2 - \nu)k^2_y\frac{\partial}{\partial x}\right) \frac{\partial\phi}{\partial z}= 0,
{
 
\int_{-h}^0 \phi_\mu(r_\mu,z)\cos \frac{m\pi}{h}(z+h) \, dz } &=&
 
{
 
\int_{-h}^0 \phi_{\mu+1}(l_{\mu+1},z)\cos \frac{m\pi}{h}(z+h) \, dz }\\
 
{
 
\int_{-h}^0 \frac{\partial\phi_\mu}{\partial x}(r_\mu,z) \cos \frac{m\pi}{h}(z+h) \, dz } &=&
 
{
 
\int_{-h}^0 \frac{\partial\phi_{\mu+1}}{\partial x}(l_{\mu+1},z) \cos \frac{m\pi}{h}(z+h) \, dz }
 
\end{matrix}
 
 
</math></center>
 
</math></center>
where <math>m\in[0,M]</math> and <math>\phi_\mu</math> denotes the potential under the <math>\mu</math>th plate, i.e. the expression
+
and
for <math>\phi</math> valid for <math>l_\mu <x<r_\mu</math>.
 
The remaining equations to be solved are given by the two edge conditions satisfied at both
 
edges of each plate
 
 
<center><math>
 
<center><math>
\begin{matrix}
+
\left(\frac{\partial^2}{\partial x^2} + \nu k^2_y\right)\frac{\partial\phi}{\partial z} = 0,
{
 
\left(\frac{\partial^3}{\partial x^3} - (2 - \nu)k_y^2\frac{\partial}{\partial x}\right)\frac{\partial\phi_\mu}{\partial z} }
 
&=&0, & \mbox{ for } z = 0 \mbox{ and } x = l_\mu,r_\mu,\\
 
{
 
\left(\frac{\partial^2}{\partial x^2} - \nu k_y^2\right)\frac{\partial\phi_\mu}{\partial z} }
 
&=&0, & \mbox{ for } z = 0 \mbox{ and } x = l_\mu,r_\mu.
 
\end{matrix}
 
 
</math></center>
 
</math></center>
 +
where <math>\nu</math> is Poisons ratio.
 +
 +
We can expend these edge conditions, which respectively gives
 +
<center>
 +
<math>
 +
-\sum_{m=-2}^{\infty}b_{m}\kappa_m (\hat{\kappa}_m^3+\hat{\kappa}_m k_y^2(2-\nu))\tan \kappa_m h=0
 +
</math>
 +
</center>
 +
and
 +
<center>
 +
<math>
 +
-\sum_{m=-2}^{\infty}b_{m}\kappa_m (\hat{\kappa}_m^2+k_y^2\nu))\tan \kappa_m h=0
 +
</math>
 +
</center>
 +
 +
The equations are derived almost identically to those above and we obtain
 +
<center>
 +
<math>
 +
A_{0}\delta_{0l}+a_{l}A_{l}
 +
=\sum_{m=-2}^{\infty}b_{m}B_{ml}
 +
</math>
 +
</center>
 +
and
 +
<center>
 +
<math>
 +
-\hat{k_{0}}A_{0}\delta_{0l}+a_{l}\hat{k}_{l}A_l
 +
=-\sum_{m=-2}^{\infty}b_{m}\hat{\kappa}_{m}B_{ml}
 +
</math>
 +
</center>
 +
and these are solved exactly as before.
  
We will show the explicit form of the linear system of equations which arise
+
== Energy Balance ==
when we solve these equations.
+
 
Let <math>{\mathbf T}_\mu</math> be a column vector given by
+
We present a derivation of the energy balance here and also refer to the derivation
<math>\left[T_{\mu}(-2), . . ., T_{\mu}(M)\right]^{{\mathbf T}}</math>
+
[[Energy Balance for Two Elastic Plates]]
and <math>{\mathbf R}_\mu</math> be a column vector given by 
+
 
<math>\left[R_{\mu}(-2) . . . R_{\mu}(M)\right]^{{\mathbf T}}</math>.
+
{{energy contour and preliminaries}}
 +
 
 +
The contributions from the vertical ends are
  
The equations which arise from matching at the boundary between the first
 
and second plate are
 
 
<center><math>
 
<center><math>
\begin{matrix}
+
\lim_{x\to-\infty} \Im\int_{-h}^{0}\phi^*\frac{\partial\phi}{\partial n} dz 
I{\mathbf C} + {\mathbf M}^{+}_{R_1} {\mathbf R}_1 ={\mathbf M}^{-}_{T_2}  {\mathbf T}_2
+
= \lim_{x\to-\infty} \Im\int_{-h}^{0} \hat{k}_0 \left(e^{\hat{k}_0 x} + a_0^{*} e^{-\hat{k}_0 x} \right)
+ {\mathbf M}^{-}_{R_2} {\mathbf R}_2,\\
+
\left(e^{-\hat{k}_0 x} - a_0 e^{\hat{k}_0 x} \right)\phi_0(z)^2 dz 
-\kappa_1(0)I\mathbf{C} + {\mathbf N}^{+}_{R_1} {\mathbf R}_1 = {\mathbf N}^{-}_{T_2} {\mathbf T}_2
 
+ {\mathbf N}^{-}_{R_2} {\mathbf R}_2.
 
\end{matrix}
 
 
</math></center>
 
</math></center>
The equations which arise from matching at the boundary of the <math>\mu</math>th and (<math>\mu+1</math>)th plate
 
boundary (<math>\mu>1</math>) are
 
 
<center><math>
 
<center><math>
\begin{matrix}
+
= \frac{\hat{k}_0}{i} A_0 \left(1 - |a_0^2| \right)
{\mathbf M}^{+}_{T_\mu} {\mathbf T}_\mu +{\mathbf M}^{+}_{R_\mu} {\mathbf R}_\mu 
 
={\mathbf M}^{-}_{T_{\mu+1}} {\mathbf T}_{\mu+1}
 
+ {\mathbf M}^{-}_{ R_{\mu+1}} {\mathbf R}_{\mu+1}, \\
 
{\mathbf N}^{+}_{T_\mu} {\mathbf T}_\mu + {\mathbf N}^{+}_{R_\mu}  {\mathbf R}_\mu
 
={\mathbf N}^{-}_{T_{\mu+1}}  {\mathbf T}_{\mu +1}
 
+{\mathbf N}^{-}_{ R_{\mu +1}}  {\mathbf R}_{\mu +1}.
 
\end{matrix}
 
 
</math></center>
 
</math></center>
The equations which arise from matching at the (<math>\Lambda-1</math>)th and <math>\Lambda</math>th boundary are
+
and  
 
<center><math>
 
<center><math>
\begin{matrix}
+
\lim_{x\to\infty} \Im\int_{-h}^{0}\phi^*\frac{\partial\phi}{\partial n} dz 
{\mathbf M}^{+}_{T_{\Lambda-1}} {\mathbf T}_{\Lambda-1}  
+
=  \lim_{x\to\infty} \Im\int_{-h}^{0} -\hat{\kappa}_0 \left(b_0^{*} e^{\hat{\kappa}_0 x} \right)
+ {\mathbf M}^{+}_{R_{\Lambda-1}}  {\mathbf R}_{\Lambda-1}  
+
\left(b_0 e^{-\hat{\kappa}_0 x}\right)\psi_0(z)^2 dz 
{\mathbf M}^{-}_{T_\Lambda } {\mathbf T}_{\Lambda}, \\
 
{\mathbf N}^{+}_{T_{\Lambda-1}} {\mathbf T}_{\Lambda-1}
 
+ {\mathbf N}^{+}_{R_{\Lambda-1}} {\mathbf R}_{\Lambda-1} =  {\mathbf N}^{-}_{T_\Lambda } {\mathbf T}_{\Lambda},
 
\end{matrix}
 
 
</math></center>
 
</math></center>
 
where <math>{\mathbf M}^{+}_{T_\mu}</math>, <math>{\mathbf M}^{+}_{R_\mu}</math>,
 
<math>{\mathbf M}^{-}_{T_\mu}</math>, and <math>{\mathbf M}^{-}_{R_\mu}</math>are <math>(M+1)</math> by <math>(M+3)</math> matrices given by
 
 
<center><math>
 
<center><math>
\begin{matrix}
+
= -\frac{\hat{\kappa}_0}{i} |b_0^2| \int_{-h}^{0} \psi_0(z)^2 dz   
{
 
{\mathbf M}^{+}_{T_\mu}(m,n) = \int_{-h}^0 e^{-\kappa_\mu(n) (r_\mu-l_\mu )} \frac{\cos \left(k_{\mu}(n) (z+h)\right)}{\cos \left(k_{\mu}(n) h\right)} \cos \left(\frac{m\pi}{h}(z+h)\right) \, dz}, \\
 
{
 
{\mathbf M}^{+}_{R_\mu}(m,n) = \int_{-h}^0 \frac{\cos \left(k_{\mu}(n) (z+h)\right)}{\cos \left(k_{\mu}(n) h\right)} \cos \left(\frac{m\pi}{h}(z+h)\right) \, dz },\\
 
{
 
{\mathbf M}^{-}_{T_\mu}(m,n) =  {\mathbf M}^{+}_{R_\mu}(m,n)  }\\
 
{
 
{\mathbf M}^{-}_{R_\mu}(m,n) = {\mathbf M}^{+}_{T_\mu}(m,n). }
 
\end{matrix}
 
</math></center>
 
<math>{\mathbf N}^{+}_{T_\mu}</math>, <math>{\mathbf N}^{+}_{R_\mu}</math>,
 
<math>{\mathbf N}^{-}_{T_\mu}</math>, and <math>{\mathbf N}^{-}_{R_\mu}</math> are given by
 
<center><math>
 
\begin{matrix}
 
{\mathbf N}^{\pm}_{T_\mu}(m,n)= -\kappa_\mu(n){\mathbf M}^{\pm}_{T_\mu}(m,n),\\
 
{\mathbf N}^{\pm}_{R_\mu}(m,n)= \kappa_\mu(n){\mathbf M}^{\pm}_{R_\mu}(m,n).
 
\end{matrix}
 
</math></center>
 
<math>\mathbf{C}</math> is a <math>(M+1)</math> vector which is given by
 
<center><math>
 
{\mathbf C}(m)=\int_{-h}^0 \frac{\cos (k_1(0)(z+h))}{\cos (k_1(0)h)} \cos \left(\frac{m\pi}{h}(z+h)\right)\, dz.
 
 
</math></center>
 
</math></center>
  
The integrals in the above equation are each solved analytically. Now, for all but the first and <math>\Lambda</math>th plate, the edge equation becomes
+
The contribution from the surface integral under the plate is
 
<center><math>
 
<center><math>
\begin{matrix}
+
\Im\int_{0}^{\infty}\phi^*\frac{\partial\phi}{\partial n} dx
{\mathbf E}^{+}_{T_\mu} {\mathbf T}_\mu + {\mathbf E}^{+}_{R_\mu} {\mathbf R}_\mu = 0,\\
+
\Im\int_{0}^{\infty}  
{\mathbf E}^{-}_{T_\mu} {\mathbf T}_\mu + {\mathbf E}^{-}_{R_\mu} {\mathbf R}_\mu = 0.
+
1/\alpha \left(\beta \partial_x^4 \frac{\partial\phi^{*}}{\partial n} - (\gamma\alpha - 1 ) \frac{\partial\phi^{*}}{\partial n} \right)
\end{matrix}
+
\frac{\partial\phi}{\partial n} dx
 
</math></center>
 
</math></center>
The first and last plates only require two equations, because each has only one plate
 
edge.
 
The equation for the first plate must be modified to include the effect of
 
the incident wave. This gives us
 
 
<center><math>
 
<center><math>
\begin{matrix}
+
= \frac{2\beta}{\alpha} \frac{\hat{\kappa}_0^{3}}{i} |b_0^2| \partial_z\psi_0(0)^2 
I \left(
 
\begin{matrix}
 
{\mathbf E}^{+}_{T_1}(1,0)\\
 
{\mathbf E}^{+}_{T_1}(2,0)
 
\end{matrix}
 
\right)
 
+ {\mathbf E}^{+}_{R_1} {\mathbf R}_1 = 0,\\
 
\end{matrix}
 
</math></center>
 
and for the <math>\Lambda</math>th plate we have no reflection so
 
<center><math>
 
\begin{matrix}
 
{\mathbf E}^{-}_{T_\mu} {\mathbf T}_\mu = 0.\\
 
\end{matrix}
 
 
</math></center>
 
</math></center>
 +
where we have integrated by parts and used the condition at the ends of the plate.
  
<math>{\mathbf E}^{+}_{T_\mu}</math>, <math>{\mathbf E}^{+}_{R_\mu}</math>, <math>{\mathbf E}^{-}_{T_\mu}</math> and <math>{\mathbf E}^{-}_{R_\mu}</math>  are 2 by M+3 matrices given by
+
The energy balance is therefore
 
<center><math>
 
<center><math>
\begin{matrix}
+
{\hat{k}_0} A_0 |a_0^2+ {\hat{\kappa}_0} |b_0^2| \int_{-h}^{0} \psi_0(z)^2 dz 
{\mathbf E}^{-}_{T_\mu}(1,n) = (\kappa_\mu(n)^2 - (2 - \nu)k_y^2)(k_{\mu}(n)\kappa_\mu(n)\tan{(k_{\mu}(n)h)}),\\
+
- \frac{2\beta}{\alpha}{ \hat{\kappa}_0^{3}} |b_0^2| \partial_z\psi_0(0)^2   = {\hat{k}_0} A_0
{\mathbf E}^{+}_{T_\mu}(1,n) = (\kappa_\mu(n)^2 - (2 - \nu)k_y^2)(k_{\mu}(n)\kappa_\mu(n)e^{-\kappa_\mu(n)(r_\mu - l_\mu)}\tan{(k_{\mu}(n)h)}),\\
 
{\mathbf E}^{-}_{R_\mu}(1,n) = (\kappa_\mu(n)^2 - (2 - \nu)k_y^2)(-k_{\mu}(n)\kappa_\mu(n)e^{\kappa_\mu(n)(l_\mu - r_\mu)}\tan{(k_{\mu}(n)h)}),\\
 
{\mathbf E}^{+}_{R_\mu}(1,n) = (\kappa_\mu(n)^2 - (2 - \nu)k_y^2)(-k_{\mu}(n)\kappa_\mu(n)\tan{(k_{\mu}(n)h)}),\\
 
 
 
{\mathbf E}^{-}_{T_\mu}(2,n) = (\kappa_\mu(n)^2 - \nu k_y^2)(-k_{\mu}(n)\tan{(k_{\mu}(n)h)}),\\
 
{\mathbf E}^{+}_{T_\mu}(2,n) = (\kappa_\mu(n)^2 - \nu k_y^2)(-k_{\mu}(n)e^{-\kappa_\mu(n)(r_\mu - l_\mu)}\tan{(k_{\mu}(n)h)}),\\
 
{\mathbf E}^{-}_{R_\mu}(2,n) = (\kappa_\mu(n)^2 - \nu k_y^2)(-k_{\mu}(n)e^{\kappa_\mu(n)(l_\mu - r_\mu)}\tan{(k_{\mu}(n)h)}),\\
 
{\mathbf E}^{+}_{R_\mu}(2,n) = (\kappa_\mu(n)^2 - \nu k_y^2)(-k_{\mu}(n)\tan{(k_{\mu}(n)h)}).\\
 
\end{matrix}
 
</math></center>
 
 
 
Now, the matching matrix is a <math>(2M+6)\times(\Lambda-1)</math> by
 
<math>(2M+1)\times(\Lambda -1)</math> matrix given by
 
<center><math>
 
{\mathbf M} =  
 
\left( \begin{matrix}
 
{\mathbf M}^{+}_{R_1} & -{\mathbf M}^{-}_{T_2} & -{\mathbf M}^{-}_{R_2} &    0    &    0    &        & 0 & 0 & 0 \\
 
{\mathbf N}^{+}_{R_1} & -{\mathbf N}^{-}_{T_2} & -{\mathbf N}^{-}_{R_2} &    0    &    0    &        & 0 & 0 & 0 \\
 
  0    &  {\mathbf M}^{+}_{T_2} &  {\mathbf M}^{+}_{R_2} & -{\mathbf M}^{-}_{T_3} & -{\mathbf M}^{-}_{R_3} &  & 0 & 0 & 0 \\
 
  0    &  {\mathbf N}^{+}_{T_2} &  {\mathbf N}^{+}_{R_2} & -{\mathbf N}^{-}_{T_3} & -{\mathbf N}^{-}_{R_3} &        & 0 & 0 & 0 \\
 
        &          &  \vdots  &          &          & \ddots & \\
 
  0    &    0    &    0    &    0    &    0    &        & {\mathbf M}^{+}_{T_{\Lambda - 1}} & {\mathbf M}^{+}_{R_{\Lambda - 1}} & -{\mathbf M}^{-}_{ T_{\Lambda}} \\
 
  0    &    0    &    0    &    0    &    0    &        & {\mathbf N}^{+}_{T_{\Lambda - 1}} & {\mathbf N}^{+}_{R_{\Lambda - 1}} & -{\mathbf N}^{-}_{T_{\Lambda }} \\
 
\end{matrix} \right),
 
 
</math></center>
 
</math></center>
  
the edge matrix is a <math>(2M+6)\times(\Lambda-1)</math> by <math>4(\Lambda-1)</math> matrix given by
 
 
<center><math>
 
{\mathbf E} =
 
\left( \begin{matrix}
 
  {\mathbf E}^{+}_{R_1} &    0  &          0        &    0    &    0    &        & 0 & 0 & 0 \\
 
  0    &  {\mathbf E}^{+}_{T_2} &  {\mathbf E}^{+}_{R_2} &    0    &    0    &        & 0 & 0 & 0 \\
 
  0    &  {\mathbf E}^{-}_{T_2} &  {\mathbf E}^{-}_{R_2} &    0    &    0    &        & 0 & 0 & 0 \\
 
  0    &    0    &    0    &  {\mathbf E}^{+}_{T_3} &  {\mathbf E}^{+}_{R_3} &  & 0 & 0 & 0 \\
 
  0    &    0    &    0    &  {\mathbf E}^{-}_{T_3} &  {\mathbf E}^{-}_{R_3} &        & 0 & 0 & 0 \\
 
        &          &  \vdots  &          &          & \ddots & \\
 
  0    &    0    &    0    &    0    &    0    &        & {\mathbf E}^{+}_{T_{\Lambda-1}} &  {\mathbf E}^{+}_{R_{\Lambda-1}} & 0 \\
 
  0    &    0    &    0    &    0    &    0    &        & {\mathbf E}^{-}_{T_{\Lambda-1}} &  {\mathbf E}^{-}_{R_{\Lambda-1}} & 0 \\
 
  0    &    0    &    0    &    0    &    0    &        & 0 & 0 & {\mathbf E}^{-}_{ T_\Lambda}
 
\end{matrix} \right),
 
</math></center>
 
  
and finally the complete system to be solved is given by
+
Note that this formula is only valid for angles <math>\theta \in [-\theta_0, \theta_0]</math>, where <math>\theta_0</math>
<center><math>
+
is defined by <math>\sin \theta_0 =\frac{\kappa_0}{k_0}</math>. For large angles there is total reflection
\left( \begin{matrix}
+
and then <math>|R|^2=1</math>
{\mathbf M}\\
 
{\mathbf E}\\
 
\end{matrix} \right)
 
\times
 
\left( \begin{matrix}
 
{\mathbf R}_1\\
 
{\mathbf T}_2\\
 
{\mathbf R}_2\\
 
{\mathbf T}_3\\
 
{\mathbf R}_3\\
 
\vdots\\
 
{\mathbf T}_{\Lambda-1}\\
 
{\mathbf R}_{\Lambda-1}\\
 
{\mathbf T}_{\Lambda}
 
  
\end{matrix} \right)
+
== Matlab Code ==
=
 
\left( \begin{matrix}
 
-I{\mathbf C}\\
 
\kappa_{1}(0)I{\mathbf C}\\
 
0\\
 
\vdots\\
 
-IE^{+}_{T_1}(1,0)\\
 
-IE^{+}_{T_1}(2,0)\\
 
0\\
 
\vdots
 
  
\end{matrix} \right).
+
A program to calculate the coefficients for the semi-infinite dock problems can be found here
</math></center>
+
[http://www.math.auckland.ac.nz/~meylan/code/eigenfunction_matching/semiinfinite_plate.m semiinfinite_plate.m]
  
The final system of equations has size <math>(2M+6)\times (\Lambda - 1)</math> by
+
=== Additional code ===
<math>(2M+6)\times (\Lambda - 1)</math>.
 
  
 +
This program requires
 +
* {{free surface dispersion equation code}}
 +
* {{elastic plate dispersion equation code}}
  
 
[[Category:Floating Elastic Plate]]
 
[[Category:Floating Elastic Plate]]
 
[[Category:Eigenfunction Matching Method]]
 
[[Category:Eigenfunction Matching Method]]
 
+
[[Category:Complete Pages]]
[[Category:Eigenfunction Matching Method]]
 
[[Category:Linear Hydroelasticity]]
 

Latest revision as of 20:51, 17 March 2010


Introduction

We show here a solution for a semi-infinite Floating Elastic Plate on Finite Depth. The problem was solved by Fox and Squire 1994 but the solution method here is slightly different. The simpler theory for a Dock describes many of the ideas here in more detail.

Wave scattering by a submerged semi-infinite elastic plate

Equations

We consider the problem of small-amplitude waves which are incident on a semi-infinite floating elastic plate occupying water surface for [math]\displaystyle{ x\gt 0 }[/math]. The submergence of the plate is considered negligible. We assume that the problem is invariant in the [math]\displaystyle{ y }[/math] direction, although we allow the waves to be incident from an angle. We also assume that the plate edges are free to move at each boundary, although other boundary conditions could easily be considered using the methods of solution presented here. We begin with the Frequency Domain Problem for a semi-infinite Floating Elastic Plates in the non-dimensional form of Tayler 1986 (Dispersion Relation for a Floating Elastic Plate). We also assume that the waves are normally incident (incidence at an angle will be discussed later).

[math]\displaystyle{ \Delta \phi = 0, \;\;\; -h \lt z \leq 0, }[/math]
[math]\displaystyle{ \partial_z \phi = 0, \;\;\; z = - h, }[/math]
[math]\displaystyle{ \partial_z\phi=\alpha\phi, \,\, z=0,\,x\lt 0, }[/math]
[math]\displaystyle{ \beta \partial_x^4\partial_z \phi - \left( \gamma\alpha - 1 \right) \partial_z \phi = \alpha\phi, \;\; z = 0, \;\;\; x \geq 0, }[/math]

where [math]\displaystyle{ \alpha = \omega^2 }[/math], [math]\displaystyle{ \beta }[/math] and [math]\displaystyle{ \gamma }[/math] are the stiffness and mass constant for the plate respectively. The free edge conditions at the edge of the plate imply

[math]\displaystyle{ \partial_x^3 \partial_z\phi = 0, \;\; z = 0, \;\;\; x = 0, }[/math]
[math]\displaystyle{ \partial_x^2 \partial_z\phi = 0, \;\; z = 0, \;\;\; x = 0, }[/math]

Method of solution

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:

[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]

Separation of variables under the Plate

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

subject to the boundary conditions

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

and

[math]\displaystyle{ \left(\beta \kappa^4 + 1 - \alpha\gamma\right)Z^{\prime}(0) = \alpha Z(0) }[/math]

(the first term comes from the beam eigenvalue problem, where [math]\displaystyle{ \partial_x^4 X = \kappa^4 X }[/math]). We then use the boundary condition at [math]\displaystyle{ z=-h \, }[/math] to write

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

The boundary condition at the free surface ([math]\displaystyle{ z=0 }[/math]) is the Dispersion Relation for a Floating Elastic Plate

[math]\displaystyle{ \kappa \tan{(\kappa h)}= -\frac{\alpha}{\beta \kappa^{4} + 1 - \alpha\gamma} }[/math]

Solving for [math]\displaystyle{ \kappa \, }[/math] gives a pure imaginary root with positive imaginary part, two complex roots (two complex conjugate paired roots with positive imaginary part in most physical situations), an infinite number of positive real roots which approach [math]\displaystyle{ {n\pi}/{h} \, }[/math] as [math]\displaystyle{ n }[/math] approaches infinity, and also the negative of all these roots (Dispersion Relation for a Floating Elastic Plate) . We denote the two complex roots with positive imaginary part by [math]\displaystyle{ \kappa_{-2} \, }[/math] and [math]\displaystyle{ \kappa_{-1} \, }[/math], the purely imaginary root with positive imaginary part by [math]\displaystyle{ \kappa_{0} \, }[/math] and the real roots with positive imaginary part by [math]\displaystyle{ \kappa_{n} \, }[/math] for [math]\displaystyle{ n }[/math] a positive integer. The imaginary root with positive imaginary part corresponds to a reflected travelling mode propagating along the [math]\displaystyle{ x }[/math] axis. The complex roots with positive imaginary parts correspond to damped reflected travelling modes and the real roots correspond to reflected evanescent modes.

Inner product between free surface and elastic plate modes

[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\right) \left( \cos \kappa_{m}h\right) \left( k_{n} ^{2}-\kappa_{m}^{2}\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.

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 also truncate the sum at [math]\displaystyle{ N }[/math] being careful that we have two extra modes on the plate covered region to satisfy the edge conditions. We obtain

[math]\displaystyle{ \phi_{0}\left( z\right) + \sum_{m=0}^{N} a_{m} \phi_{m}\left( z\right) =\sum_{m=-2}^{N}b_{m}\psi_{m}(z) }[/math]

and

[math]\displaystyle{ -k_{0}\phi_{0}\left( z\right) +\sum _{m=0}^{N} a_{m}k_{m}\phi_{m}\left( z\right) =-\sum_{m=-2}^{N}b_{m}\kappa_{m}\psi _{m}(z) }[/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_{l}A_{l} =\sum_{m=-2}^{N}b_{m}B_{ml},\,\,0 \leq l \leq N }[/math]

and

[math]\displaystyle{ -k_{0}A_{0}\delta_{0l}+a_{l}k_{l}A_l =-\sum_{m=-2}^{N}b_{m}\kappa_{m}B_{ml},\,\,0 \leq l \leq N }[/math]

If we multiply the first equation by [math]\displaystyle{ k_l }[/math] and subtract the second equation we obtain

[math]\displaystyle{ (k_{0}+k_l)A_{0}\delta_{0l} =\sum_{m=-2}^{N}b_{m}(k_l + \kappa_{m})B_{ml} }[/math]

Finally, we need to apply the conditions at the edge of the plate to give us two further equations,

[math]\displaystyle{ \partial_x^2\partial_z\phi = \sum_{m=-2}^{N}b_{m} \kappa_m^3 \tan\kappa_m h = 0 }[/math]

and

[math]\displaystyle{ \partial_x^3\partial_z\phi = - \sum_{m=-2}^{N}b_{m} \kappa_m^4 \tan\kappa_m h = 0 }[/math]

Numerical Solution

To solve the system of equations previously defined we set the upper limit of [math]\displaystyle{ l }[/math] to be [math]\displaystyle{ N }[/math], as stated before. In terms of matrix, we obtain

[math]\displaystyle{ \begin{bmatrix} \begin{bmatrix} A_0&0 \quad \cdots&0\\ 0&&\\ \vdots&A_l&\vdots\\ &&0\\ 0&\cdots \quad 0 &A_N \end{bmatrix} & \begin{bmatrix} -B_{-2,0}&-B_{-1,0}\\ &\\ \vdots&\vdots\\ &\\ -B_{-2,N}&-B_{-1,N} \end{bmatrix} & \begin{bmatrix} -B_{0,0}&\cdots&-B_{0,N}\\ &&\\ \vdots&-B_{m,l}&\vdots\\ &&\\ -B_{N,0}&\cdots&-B_{N,N} \end{bmatrix} \\ \\ \begin{bmatrix} 0&&\cdots&&0\\ 0&&\cdots&&0 \end{bmatrix} & \begin{bmatrix} \kappa_{-2}^3\tan\kappa_{-2}h&\kappa_{-1}^3\tan\kappa_{-1}h\\ \kappa_{-2}^4\tan\kappa_{-2}h&\kappa_{-1}^4\tan\kappa_{-1}h \end{bmatrix} & \begin{bmatrix} \kappa_0^3\tan\kappa_0h&\cdots&\kappa_N^3\tan\kappa_Nh\\ \kappa_0^4\tan\kappa_0h&\cdots&\kappa_N^4\tan\kappa_Nh \end{bmatrix} \\ \\ \begin{bmatrix} 0&&\cdots&&0\\ &&&&\\ \vdots&&\ddots&&\vdots\\ &&&&\\ 0&&\cdots&&0 \end{bmatrix} & \begin{bmatrix} (k_0+\kappa_{-2})B_{-2,0}&(k_0+\kappa_{-1})B_{-1,0}\\ &\\ \vdots&\vdots\\ &\\ (k_N+\kappa_{-2})B_{-2,N}&(k_N+\kappa_{-1})B_{-1,N} \end{bmatrix} & \begin{bmatrix} (k_0 + \kappa_0) \, B_{0,0}&\cdots&(k_N + \kappa_{0}) \, B_{0,N}\\ &&\\ \vdots&(k_l + \kappa_{m}) \, B_{m,l}&\vdots\\ &&\\ (k_0 + \kappa_N) \, B_{N,0}&\cdots&(k_N + \kappa_{N}) \, B_{N,N}\\ \end{bmatrix} \end{bmatrix} \begin{bmatrix} \begin{bmatrix} a_{0} \\ \\ \vdots \\ \\ a_N \end{bmatrix} \\ \\ \begin{bmatrix} b_{-2}\\ b_{-1} \end{bmatrix} \\ \\ \begin{bmatrix} b_{0}\\ \\ \vdots \\ \\ b_N \end{bmatrix} \end{bmatrix} = \begin{bmatrix} \begin{bmatrix} - A_{0} \\ 0 \\ \vdots \\ \\ 0 \end{bmatrix} \\ \\ \begin{bmatrix} 0 \\ 0 \end{bmatrix} \\ \\ \begin{bmatrix} 2k_{0}A_{0} \\ 0 \\ \vdots \\ \\ 0 \end{bmatrix} \end{bmatrix} }[/math]

We then simply need to solve the linear system of equations. Note that we can solve this equation for [math]\displaystyle{ b_n }[/math] first and then solve for [math]\displaystyle{ a_n }[/math]

Waves Incident at an Angle

We can consider the case of Waves 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.

It is shown that 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=-2}^{\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 edge conditions are also different and are

[math]\displaystyle{ \left(\frac{\partial^3}{\partial x^3} + (2 - \nu)k^2_y\frac{\partial}{\partial x}\right) \frac{\partial\phi}{\partial z}= 0, }[/math]

and

[math]\displaystyle{ \left(\frac{\partial^2}{\partial x^2} + \nu k^2_y\right)\frac{\partial\phi}{\partial z} = 0, }[/math]

where [math]\displaystyle{ \nu }[/math] is Poisons ratio.

We can expend these edge conditions, which respectively gives

[math]\displaystyle{ -\sum_{m=-2}^{\infty}b_{m}\kappa_m (\hat{\kappa}_m^3+\hat{\kappa}_m k_y^2(2-\nu))\tan \kappa_m h=0 }[/math]

and

[math]\displaystyle{ -\sum_{m=-2}^{\infty}b_{m}\kappa_m (\hat{\kappa}_m^2+k_y^2\nu))\tan \kappa_m h=0 }[/math]

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

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

and these are solved exactly as before.

Energy Balance

We present a derivation of the energy balance here and also refer to the derivation Energy Balance for Two Elastic Plates

Based on the method used in Evans and Davies 1968, a check can be made to ensure the solutions energy balance. The energy balance equation is derived by applying Green's theorem to [math]\displaystyle{ \phi }[/math] and its conjugate. The domain of integration is shown in the figure on the right. We assume that the angle is sufficiently small that we do not get total reflection.

A diagram depicting the area [math]\displaystyle{ \Omega }[/math] which is bounded by the rectangle [math]\displaystyle{ \partial\Omega }[/math]. The rectangle [math]\displaystyle{ \partial\Omega }[/math] is bounded by [math]\displaystyle{ -h\leq z \leq0 }[/math] and [math]\displaystyle{ -\infty\leq x \leq \infty }[/math]

Applying Green's theorem to [math]\displaystyle{ \phi }[/math] and its conjugate [math]\displaystyle{ \phi^* }[/math] gives

[math]\displaystyle{ { \iint_\Omega\left(\phi^*\nabla^2\phi - \phi\nabla^2\phi^* \right)\mathrm{d}x\mathrm{d}z = \int_{\partial\Omega}\left(\phi^*\frac{\partial\phi}{\partial n} - \phi\frac{\partial\phi^*}{\partial n} \right)\mathrm{d}l }, }[/math]

where [math]\displaystyle{ n }[/math] denotes the outward plane normal to the boundary and [math]\displaystyle{ l }[/math] denotes the plane parallel to the boundary. As [math]\displaystyle{ \phi }[/math] and [math]\displaystyle{ \phi^* }[/math] satisfy the Laplace's equation, the left hand side of the Green theorem equation vanishes so that it reduces to

[math]\displaystyle{ \Im\int_\mathcal{S}\phi^*\frac{\partial\phi}{\partial n} \mathrm{d}l = 0, }[/math]

The contributions from the vertical ends are

[math]\displaystyle{ \lim_{x\to-\infty} \Im\int_{-h}^{0}\phi^*\frac{\partial\phi}{\partial n} dz = \lim_{x\to-\infty} \Im\int_{-h}^{0} \hat{k}_0 \left(e^{\hat{k}_0 x} + a_0^{*} e^{-\hat{k}_0 x} \right) \left(e^{-\hat{k}_0 x} - a_0 e^{\hat{k}_0 x} \right)\phi_0(z)^2 dz }[/math]
[math]\displaystyle{ = \frac{\hat{k}_0}{i} A_0 \left(1 - |a_0^2| \right) }[/math]

and

[math]\displaystyle{ \lim_{x\to\infty} \Im\int_{-h}^{0}\phi^*\frac{\partial\phi}{\partial n} dz = \lim_{x\to\infty} \Im\int_{-h}^{0} -\hat{\kappa}_0 \left(b_0^{*} e^{\hat{\kappa}_0 x} \right) \left(b_0 e^{-\hat{\kappa}_0 x}\right)\psi_0(z)^2 dz }[/math]
[math]\displaystyle{ = -\frac{\hat{\kappa}_0}{i} |b_0^2| \int_{-h}^{0} \psi_0(z)^2 dz }[/math]

The contribution from the surface integral under the plate is

[math]\displaystyle{ \Im\int_{0}^{\infty}\phi^*\frac{\partial\phi}{\partial n} dx = \Im\int_{0}^{\infty} 1/\alpha \left(\beta \partial_x^4 \frac{\partial\phi^{*}}{\partial n} - (\gamma\alpha - 1 ) \frac{\partial\phi^{*}}{\partial n} \right) \frac{\partial\phi}{\partial n} dx }[/math]
[math]\displaystyle{ = \frac{2\beta}{\alpha} \frac{\hat{\kappa}_0^{3}}{i} |b_0^2| \partial_z\psi_0(0)^2 }[/math]

where we have integrated by parts and used the condition at the ends of the plate.

The energy balance is therefore

[math]\displaystyle{ {\hat{k}_0} A_0 |a_0^2| + {\hat{\kappa}_0} |b_0^2| \int_{-h}^{0} \psi_0(z)^2 dz - \frac{2\beta}{\alpha}{ \hat{\kappa}_0^{3}} |b_0^2| \partial_z\psi_0(0)^2 = {\hat{k}_0} A_0 }[/math]


Note that this formula is only valid for angles [math]\displaystyle{ \theta \in [-\theta_0, \theta_0] }[/math], where [math]\displaystyle{ \theta_0 }[/math] is defined by [math]\displaystyle{ \sin \theta_0 =\frac{\kappa_0}{k_0} }[/math]. For large angles there is total reflection and then [math]\displaystyle{ |R|^2=1 }[/math]

Matlab Code

A program to calculate the coefficients for the semi-infinite dock problems can be found here semiinfinite_plate.m

Additional code

This program requires