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 A solution
+
The problem was solved by [[Fox and Squire 1994]] but the solution method here is slightly different.
for [[Shallow Depth]] was given in [[Zilman_Miloh_2000a|Zilman and Miloh 2000]] and we will also show this.
+
The simpler theory for a [[Eigenfunction Matching for a Semi-Infinite Dock|Dock]] describes
 +
many of the ideas here in more detail.
  
=Governing Equations=
+
[[Image:Semiinfinite plate.jpg|thumb|right|300px|Wave scattering by a submerged semi-infinite elastic plate]]
  
We begin with the [[Frequency Domain Problem]] for a [[Floating Elastic Plate]]
+
== Equations ==
in the non-dimensional form of [[Tayler_1986a|Tayler 1986]] ([[Dispersion Relation for a Floating Elastic Plate]])
+
 
We will use a cylindrical coordinate system, <math>(r,\theta,z)</math>,
+
We consider the problem of small-amplitude waves which are incident on a semi-infinite floating elastic
assumed to have its origin at the centre of the circular
+
plate occupying water surface for <math>x>0</math>. The submergence of the plate is considered negligible.
plate which has radius <math>a</math>. The water is assumed to have
+
We assume that the problem is invariant in the <math>y</math> direction, although we allow the waves to be
constant finite depth <math>H</math> and the <math>z</math>-direction points vertically
+
incident from an angle.
upward with the water surface at <math>z=0</math> and the sea floor at <math>z=-H</math>. The
+
We also assume that the plate edges are free to move at
boundary value problem can therefore be expressed as
+
each boundary, although other boundary conditions could easily be considered using
<center>
+
the methods of solution presented here. We begin with the [[Frequency Domain Problem]] for a semi-infinite
<math>
+
[[Floating Elastic Plate|Floating Elastic Plates]]
\Delta\phi=0, \,\, -H<z<0,
+
in the non-dimensional form of [[Tayler_1986a|Tayler 1986]] ([[Dispersion Relation for a Floating Elastic Plate]]).
</math>
+
We also assume that the waves are normally incident (incidence at an angle will be discussed later).
</center>
+
<center><math>
<center>
+
\Delta \phi = 0, \;\;\; -h < z \leq 0,  
<math>
+
</math></center>
\phi_{z}=0, \,\, z=-H,
+
<center><math>
</math>
+
\partial_z \phi = 0, \;\;\; z = - h,
</center>
+
</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>
 
<center><math>
\phi_{z}=\alpha\phi, \,\, z=0,\,r>a,
+
\partial_x^2 \partial_z\phi = 0, \;\; z = 0, \;\;\; x = 0,
 
</math></center>
 
</math></center>
<center>
 
<math>
 
(\beta\Delta^{2}+1-\alpha\gamma)\phi_{z}=\alpha\phi, \,\, z=0,\,r<a
 
</math>
 
</center>
 
where the constants <math>\beta</math> and <math>\gamma</math> are given by
 
<center>
 
<math>
 
\beta=\frac{D}{\rho\,L^{4}g}, \gamma=\frac{\rho_{i}h}{\rho\,L}
 
</math>
 
</center>
 
and <math>\rho_{i}</math> is the density of the plate. We
 
must also apply the edge conditions for the plate and the [[Sommerfeld Radiation Condition]]
 
as <math>r\rightarrow\infty</math>. The subscript <math>z</math>
 
denotes the derivative in <math>z</math>-direction.
 
  
=Solution Method=
+
== 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==
+
{{separation of variables in two dimensions}}
  
We now separate variables, noting that since the problem has
+
{{separation of variables for a free surface}}
circular symmetry we can write the potential as
 
<center>
 
<math>
 
\phi(r,\theta,z)=\zeta(z)\sum_{n=-\infty}^{\infty}\rho_{n}(r)e^{i n \theta}
 
</math>
 
</center>
 
Applying Laplace's equation we obtain
 
<center>
 
<math>
 
\zeta_{zz}+\mu^{2}\zeta=0
 
</math>
 
</center>
 
so that:
 
<center>
 
<math>
 
\zeta=\cos\mu(z+H)
 
</math>
 
</center>
 
where the separation constant <math>\mu^{2}</math> must
 
satisfy the [[Dispersion Relation for a Free Surface]]
 
<center>
 
<math>
 
k\tan\left(  kH\right)  =-\alpha,\quad r>a\,\,\,(1)
 
</math>
 
</center>
 
and the [[Dispersion Relation for a Floating Elastic Plate]]
 
<center>
 
<math>
 
\kappa\tan(\kappa H)=\frac{-\alpha}{\beta\kappa^{4}+1-\alpha\gamma},\quad
 
r<a \,\,\,(2)
 
</math>
 
</center>
 
Note that we have set <math>\mu=k</math> under the free
 
surface and <math>\mu=\kappa</math> under the plate. We denote the
 
positive imaginary solution of (1) by <math>k_{0}</math> and
 
the positive real solutions by <math>k_{m}</math>, <math>m\geq1</math>. The solutions of
 
(2) will be denoted by 
 
<math>\kappa_{m}</math>, <math>m\geq-2</math>. The fully complex
 
solutions with positive imaginary part are <math>\kappa_{-2}</math> and
 
<math>\kappa_{-1}</math> (where <math>\kappa_{-1}=\overline{\kappa_{-2}}</math>),
 
the negative imaginary solution is <math>\kappa_{0}</math> and the positive real
 
solutions are <math>\kappa_{m}</math>, <math>m\geq1</math>. We define
 
<center>
 
<math>
 
\phi_{m}\left(  z\right)  =\frac{\cos k_{m}(z+H)}{\cos k_{m}H},\quad m\geq0
 
</math>
 
</center>
 
as the vertical eigenfunction of the potential in the open
 
water region and
 
<center>
 
<math>
 
\psi_{m}\left(  z\right)  =\frac{\cos\kappa_{m}(z+H)}{\cos\kappa_{m}H},\quad
 
m\geq-2
 
</math>
 
</center>
 
as the vertical eigenfunction of the potential in the plate
 
covered region. For later reference, we note that:
 
<center>
 
<math>
 
\int\nolimits_{-H}^{0}\phi_{m}(z)\phi_{n}(z) d z=A_{m}\delta_{mn}
 
</math>
 
</center>
 
where
 
<center>
 
<math>
 
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>
 
</center>
 
and
 
<center>
 
<math>
 
\int\nolimits_{-H}^{0}\phi_{n}(z)\psi_{m}(z) d z=B_{mn}
 
</math>
 
</center>
 
where
 
<center><math>
 
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></center>
 
  
We now solve for the function <math>\rho_{n}(r)</math>.
+
{{separation of variables for a floating elastic plate}}
Using Laplace's equation in polar coordinates we obtain
 
<center>
 
<math>
 
\frac{\mathrm{d}^{2}\rho_{n}}{\mathrm{d}r^{2}}+\frac{1}{r}
 
\frac{\mathrm{d}\rho_{n}}{\mathrm{d}r}-\left(
 
\frac{n^{2}}{r^{2}}+\mu^{2}\right)  \rho_{n}=0
 
</math>
 
</center>
 
where <math>\mu</math> is <math>k_{m}</math> or
 
<math>\kappa_{m},</math> depending on whether <math>r</math> is
 
greater or less than <math>a</math>. We can convert this equation to the
 
standard form by substituting <math>y=\mu r</math> to obtain
 
<center>
 
<math>
 
y^{2}\frac{\mathrm{d}^{2}\rho_{n}}{\mathrm{d}y^{2}}+y\frac{\mathrm{d}\rho_{n}
 
}{\rm{d}y}-(n^{2}+y^{2})\rho_{n}=0
 
</math>
 
</center>
 
The solution of this equation is a linear combination of the
 
modified Bessel functions of order <math>n</math>, <math>I_{n}(y)</math> and
 
<math>K_{n}(y)</math> [[Abramowitz and Stegun 1964]]. Since the solution must be bounded
 
we know that under the plate the solution will be a linear combination of
 
<math>I_{n}(y)</math> while outside the plate the solution will be a
 
linear combination of <math>K_{n}(y)</math>. Therefore the potential can
 
be expanded as
 
<center>
 
<math>
 
\phi(r,\theta,z)=\sum_{n=-\infty}^{\infty}\sum_{m=0}^{\infty}a_{mn}K_{n}
 
(k_{m}r)e^{i n\theta}\phi_{m}(z), \;\;r>a
 
</math>
 
</center>
 
and
 
<center>
 
<math>
 
\phi(r,\theta,z)=\sum_{n=-\infty}^{\infty}\sum_{m=-2}^{\infty}b_{mn}
 
I_{n}(\kappa_{m}r)e^{i n\theta}\psi_{m}(z), \;\;r<a
 
</math>
 
</center>
 
where <math>a_{mn}</math> and <math>b_{mn}</math>
 
are the coefficients of the potential in the open water and
 
the plate covered region respectively.
 
  
==Incident potential==
+
{{free surface floating elastic plate relations}}
  
The incident potential is a wave of amplitude <math>A</math>
+
{{incident potential for two dimensions}}
in displacement travelling in the positive <math>x</math>-direction.
 
The incident potential can therefore be written as
 
<center>
 
<math>
 
\phi^{\mathrm{I}}  =\frac{A}{i\sqrt{\alpha}}e^{k_{0}x}\phi_{0}\left(
 
z\right)
 
=\sum\limits_{n=-\infty}^{\infty}e_{n}I_{n}(k_{0}r)\phi_{0}\left(z\right)
 
e^{i n \theta}
 
</math>
 
</center>
 
where <math>e_{n}=A/\left(i\sqrt{\alpha}\right)</math>
 
(we retain the dependence on <math>n</math> for situations
 
where the incident potential might take another form).
 
  
==Boundary conditions==
+
== An infinite dimensional system of equations ==
  
The boundary conditions for the plate also have to be
+
The potential and its derivative must be continuous across the
considered. The vertical force and bending moment must vanish, which can be
+
transition from open water to the plate covered region. Therefore, the
written as
+
potentials and their derivatives at <math>x=0</math> have to be equal.
 +
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>
 
<center>
 
<math>
 
<math>
\left[\bar{\Delta}-\frac{1-\nu}{r}\left(\frac{\partial}{\partial r}
+
\phi_{0}\left( z\right) + \sum_{m=0}^{N}
+\frac{1}{r}\frac{\partial^{2}}{\partial\theta^{2}}\right)\right]
+
a_{m} \phi_{m}\left(  z\right)
w=0\,\,\,(3)
+
=\sum_{m=-2}^{N}b_{m}\psi_{m}(z)
 
</math>
 
</math>
 
</center>
 
</center>
Line 206: Line 79:
 
<center>
 
<center>
 
<math>
 
<math>
\left[ \frac{\partial}{\partial r}\bar{\Delta}-\frac{1-\nu}{r^{2}}\left(
+
  -k_{0}\phi_{0}\left(  z\right)  +\sum
\frac{\partial}{\partial r}+\frac{1}{r}\right)  \frac{\partial^{2}}
+
_{m=0}^{N} a_{m}k_{m}\phi_{m}\left(  z\right)  
{\partial\theta^{2}}\right]  w=0 \,\,\,(4)
+
  =-\sum_{m=-2}^{N}b_{m}\kappa_{m}\psi
 +
_{m}(z)
 
</math>
 
</math>
 
</center>
 
</center>
where <math>w</math> is the time-independent surface
+
for each <math>N</math>.
displacement, <math>\nu</math> is Poisson's ratio, and <math>\bar{\Delta}</math> is the
+
We solve these equations by multiplying both equations by
polar coordinate Laplacian
+
<math>\phi_{l}(z)\,</math> and integrating from <math>-h</math> to <math>0</math> to obtain:
 
<center>
 
<center>
 
<math>
 
<math>
\bar{\Delta}=\frac{\partial^{2}}{\partial r^{2}}+\frac{1}{r}\frac{\partial
+
A_{0}\delta_{0l}+a_{l}A_{l}
}{\partial r}+\frac{1}{r^{2}}\frac{\partial^{2}}{\partial\theta^{2}}
+
=\sum_{m=-2}^{N}b_{m}B_{ml},\,\,0 \leq l  \leq N
</math>
 
</center>
 
== Displacement of the plate ==
 
 
 
The surface displacement and the water velocity potential at
 
the water surface are linked through the kinematic boundary condition
 
<center>
 
<math>
 
\phi_{z}=-i\sqrt{\alpha}w,\,\,\,z=0
 
</math>
 
</center>
 
From equations (\ref{bvp_plate}) the potential and the surface
 
displacement are therefore related by
 
<center>
 
<math>
 
w=i\sqrt{\alpha}\phi,\quad r>a
 
 
</math>
 
</math>
 
</center>
 
</center>
Line 239: Line 97:
 
<center>
 
<center>
 
<math>
 
<math>
(\beta\bar{\Delta}^{2}+1-\alpha\gamma)w=i\sqrt{\alpha}\phi,\quad r<a
+
-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>
 
</math>
 
</center>
 
</center>
The surface displacement can also be expanded in eigenfunctions
+
If we multiply the first equation by <math>k_l</math> and subtract the second equation
as
+
we obtain
 
<center>
 
<center>
 
<math>
 
<math>
w(r,\theta)=\sum_{n=-\infty}^{\infty}\sum_{m=0}^{\infty}i\sqrt{\alpha}
+
(k_{0}+k_l)A_{0}\delta_{0l}
a_{mn}K_{n}(k_{m}r)e^{i n\theta},\;\;r>a
+
=\sum_{m=-2}^{N}b_{m}(k_l + \kappa_{m})B_{ml}
 
</math>
 
</math>
 
</center>
 
</center>
and:
+
Finally, we need to apply the conditions at the edge of the plate to give us two further equations,
 
<center>
 
<center>
 
<math>
 
<math>
w(r,\theta)=
+
\partial_x^2\partial_z\phi = \sum_{m=-2}^{N}b_{m} \kappa_m^3 \tan\kappa_m h = 0
\sum_{n=-\infty}^{\infty}\sum_{m=-2}^{\infty}i\sqrt{\alpha}(\beta\kappa
 
_{m}^{4}+1-\alpha\gamma)^{-1}b_{mn}I_{n}(\kappa_{m}r)e^{i
 
n\theta},\; r<a
 
 
</math>
 
</math>
 
</center>
 
</center>
using the fact that
+
and
 
<center>
 
<center>
 
<math>
 
<math>
\bar{\Delta}\left(  I_{n}(\kappa_{m}r)e^{i n\theta}\right)  =\kappa_{m}
+
\partial_x^3\partial_z\phi = - \sum_{m=-2}^{N}b_{m} \kappa_m^4 \tan\kappa_m h = 0
^{2}I_{n}(\kappa_{m}r)e^{i n\theta}\,\,\,(5)
 
 
</math>
 
</math>
 
</center>
 
</center>
  
==An infinite dimensional system of equations==
+
== Numerical Solution ==
  
The boundary conditions (3) and
+
To solve the system of equations previously defined we set the upper limit of <math>l</math> to
(4) can be expressed in terms of the potential
+
be <math>N</math>, as stated before. In terms of matrix, we obtain
using (5). Since the angular modes are uncoupled the
 
conditions apply to each mode, giving
 
 
<center>
 
<center>
 
<math>
 
<math>
\sum_{m=-2}^{\infty}(\beta\kappa_{m}^{4}+1-\alpha\gamma)^{-1}b_{mn}
+
\begin{bmatrix}
\left(\kappa_{m}^{2}I_{n}(\kappa_{m}a)-\frac{1-\nu}{a}\left(\kappa
+
\begin{bmatrix}
_{m}I_{n}^{\prime}(\kappa_{m}a)-\frac{n^{2}}{a}I_{n}(\kappa_{m}a)\right)
+
A_0&0 \quad \cdots&0\\
\right) =0\,\,\,(6)
+
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>
 
</math>
 
</center>
 
</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>
 +
 +
== Waves Incident at an Angle ==
 +
 +
We can consider the case of [[Waves Incident at an Angle]] <math>\theta</math>.
 +
{{incident angle}}
 +
 +
It is shown that the potential can be expanded as
 
<center>
 
<center>
 
<math>
 
<math>
\sum_{m=-2}^{\infty}(\beta\kappa_{m}^{4}+1-\alpha\gamma)^{-1}b_{mn}
+
\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
\left(  \kappa_{m}^{3}I_{n}^{\prime}(\kappa_{m}a)+n^{2}\frac{1-\nu}{a^{2}
 
}\left(  \kappa_{m}I_{n}^{\prime}(\kappa_{m}a)+\frac{1}{a}I_{n}(\kappa
 
_{m}a)\right)  \right)
 
=0\,\,\,(7)
 
</math>
 
</center>
 
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>r=a</math> have to be equal.
 
Again we know that this must be true for each angle and we obtain
 
<center>
 
<math>
 
e_{n}I_{n}(k_{0}a)\phi_{0}\left(  z\right) + \sum_{m=0}^{\infty}
 
a_{mn} K_{n}(k_{m}a)\phi_{m}\left(  z\right)
 
=\sum_{m=-2}^{\infty}b_{mn}I_{n}(\kappa_{m}a)\psi_{m}(z)
 
 
</math>
 
</math>
 
</center>
 
</center>
Line 304: Line 257:
 
<center>
 
<center>
 
<math>
 
<math>
e_{n}k_{0}I_{n}^{\prime}(k_{0}a)\phi_{0}\left(  z\right) +\sum
+
\phi(x,z)=\sum_{m=-2}^{\infty}b_{m}
_{m=0}^{\infty} a_{mn}k_{m}K_{n}^{\prime}(k_{m}a)\phi_{m}\left(  z\right)
+
e^{-\hat{\kappa}_{m}x}\psi_{m}(z), \;\;x>0
=\sum_{m=-2}^{\infty}b_{mn}\kappa_{m}I_{n}^{\prime}(\kappa_{m}a)\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>
 
e_{n}I_{n}(k_{0}a)A_{0}\delta_{0l}+a_{ln}K_{n}(k_{l}a)A_{l}
 
=\sum_{m=-2}^{\infty}b_{mn}I_{n}(\kappa_{m}a)B_{ml} \,\,\,(8)
 
 
</math>
 
</math>
 
</center>
 
</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.
 +
 +
The edge conditions are also different and are
 +
<center><math>
 +
\left(\frac{\partial^3}{\partial x^3} + (2 - \nu)k^2_y\frac{\partial}{\partial x}\right) \frac{\partial\phi}{\partial z}= 0, 
 +
</math></center>
 
and
 
and
<center>
+
<center><math>
<math>
+
\left(\frac{\partial^2}{\partial x^2} + \nu k^2_y\right)\frac{\partial\phi}{\partial z} = 0,
e_{n}k_{0}I_{n}^{\prime}(k_{0}a)A_{0}\delta_{0l}+a_{ln}k_{l}K_{n}^{\prime
+
</math></center>
}(k_{l}a)A_{l}
+
where <math>\nu</math> is Poisons ratio.  
=\sum_{m=-2}^{\infty}b_{mn}\kappa_{m}I_{n}^{\prime}(\kappa_{m}
 
a)B_{ml} \,\,\,(9)
 
</math>
 
</center>
 
Equation (8) can be solved for the open water
 
coefficients <math>a_{mn}</math>
 
<center>
 
<math>
 
a_{ln}=-e_{n}\frac{I_{n}(k_{0}a)}{K_{n}(k_{0}a)}\delta_{0l}+\sum
 
_{m=-2}^{\infty}b_{mn}\frac{I_{n}(\kappa_{m}a)B_{ml}}{K_{n}(k_{l}a)A_{l}}
 
</math>
 
</center>
 
which can then be substituted into equation
 
(9) to give us
 
<center>
 
<math>
 
\left(  k_{0}I_{n}^{\prime}(k_{0}a)-k_{0}\frac{K_{n}^{\prime}(k_{0}
 
a)}{K_{n}(k_{0}a)}I_{n}(k_{0}a)\right)  e_{n}A_{0}\delta_{0l}
 
=\sum_{m=-2}^{\infty}\left(  \kappa_{m}I_{n}^{\prime}(\kappa_{m}
 
a)-k_{l}\frac{K_{n}^{\prime}(k_{l}a)}{K_{n}(k_{l}a)}I_{n}(\kappa
 
_{m}a)\right)  B_{ml}b_{mn}\,\,\,(10)
 
</math>
 
</center>
 
for each <math>n</math>.
 
Together with equations (6) and (7)
 
equation (10) gives the required equations to solve for the
 
coefficients of the water velocity potential in the plate covered region.
 
  
=Numerical Solution=
+
We can expend these edge conditions, which respectively gives  
 
 
To solve the system of equations (10) together
 
with the boundary conditions (6 and 7) we set the upper limit of <math>l</math> to
 
be <math>M</math>. We also set the angular expansion to be from
 
<math>n=-N</math> to <math>N</math>. This gives us
 
 
<center>
 
<center>
 
<math>
 
<math>
\phi(r,\theta,z)=\sum_{n=-N}^{N}\sum_{m=0}^{M}a_{mn}K_{n}(k_{m}r)e^{i
+
-\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
n\theta }\phi_{m}(z), \;\;r>a
 
 
</math>
 
</math>
 
</center>
 
</center>
Line 367: Line 283:
 
<center>
 
<center>
 
<math>
 
<math>
\phi(r,\theta,z)=\sum_{n=-N}^{N}\sum_{m=-2}^{M}b_{mn}I_{n}(\kappa
+
-\sum_{m=-2}^{\infty}b_{m}\kappa_m (\hat{\kappa}_m^2+k_y^2\nu))\tan \kappa_m h=0
_{m}r)e^{i n\theta}\psi_{m}(z), \;\;r<a
 
 
</math>
 
</math>
 
</center>
 
</center>
Since <math>l</math> is an integer with <math>0\leq l\leq
+
 
M</math> this leads to a system of <math>M+1</math> equations.
+
The equations are derived almost identically to those above and we obtain
The number of unknowns is <math>M+3</math> and the two extra equations
 
are obtained from the boundary conditions for the free plate (6)
 
and (7). The equations to be solved for each <math>n</math> are
 
 
<center>
 
<center>
 
<math>
 
<math>
\left(  k_{0}I_{n}^{\prime}(k_{0}a)-k_{0}\frac{K_{n}^{\prime}(k_{0}
+
A_{0}\delta_{0l}+a_{l}A_{l}
a)}{K_{n}(k_{0}a)}I_{n}(k_{0}a)\right)  e_{n}A_{0}\delta_{0l}
+
=\sum_{m=-2}^{\infty}b_{m}B_{ml}
=\sum_{m=-2}^{M}\left(  \kappa_{m}I_{n}^{\prime}(\kappa_{m}a)-k_{l}
 
\frac{K_{n}^{\prime}(k_{l}a)}{K_{n}(k_{l}a)}I_{n}(\kappa_{m}a)\right)
 
B_{ml}b_{mn}
 
</math>
 
</center>
 
<center>
 
<math>
 
\sum_{m=-2}^{M}(\beta\kappa_{m}^{4}+1-\alpha\gamma)^{-1}b_{mn}
 
\left(  \kappa_{m}^{2}I_{n}(\kappa_{m}a)-\frac{1-\nu}{a}\left(  \kappa
 
_{m}I_{n}^{\prime}(\kappa_{m}a)-\frac{n^{2}}{a}I_{n}(\kappa_{m}a)\right)
 
\right) =0
 
 
</math>
 
</math>
 
</center>
 
</center>
Line 396: Line 297:
 
<center>
 
<center>
 
<math>
 
<math>
\sum_{m=-2}^{M}(\beta\kappa_{m}^{4}+1-\alpha\gamma)^{-1}b_{mn}
+
-\hat{k_{0}}A_{0}\delta_{0l}+a_{l}\hat{k}_{l}A_l
\left(  \kappa_{m}^{3}I_{n}^{\prime}(\kappa_{m}a)+n^{2}\frac{1-\nu}{a^{2}
+
  =-\sum_{m=-2}^{\infty}b_{m}\hat{\kappa}_{m}B_{ml}
}\left( \kappa_{m}I_{n}^{\prime}(\kappa_{m}a)+\frac{1}{a}I_{n}(\kappa
 
_{m}a)\right)  \right) =0
 
 
</math>
 
</math>
 
</center>
 
</center>
It should be noted that the solutions for positive and negative
+
and these are solved exactly as before.
<math>n</math> are identical so that they do not both need to be
 
calculated. There are some minor simplifications which are a consequence of
 
this which are discussed in more detail in [[Zilman_Miloh 2000a|Zilman and Miloh 2000]].
 
  
=The [[Shallow Depth]] Theory of [[Zilman_Miloh_2000a|Zilman and Miloh 2000]]=
+
== Energy Balance ==
  
The shallow water theory of [[Zilman_Miloh_2000a|Zilman and Miloh 2000]] can be recovered by
+
We present a derivation of the energy balance here and also refer to the derivation
simply setting the depth shallow enough that the shallow water theory is valid
+
[[Energy Balance for Two Elastic Plates]]  
and setting <math>M=0</math>. If the shallow water theory is valid then
 
the first three roots of the dispersion equation for the ice will be exactly
 
the same roots found in the shallow water theory by solving the polynomial
 
equation. The system of equations has four unknowns (three under the plate and
 
one in the open water) exactly as for the theory of [[Zilman_Miloh_2000a|Zilman and Miloh 2000]].
 
  
=Numerical Results=
+
{{energy contour and preliminaries}}
  
[[Image:ComparisionH25.jpg|thumb|right|300px|Figure 1]]
+
The contributions from the vertical ends are
  
We present solutions for a plate of radius <math>a=100</math>. The wavelength is
+
<center><math>
<math>\lambda=50</math> (recall that <math>\alpha=2\pi/\lambda\tanh\left( 2\pi
+
\lim_{x\to-\infty} \Im\int_{-h}^{0}\phi^*\frac{\partial\phi}{\partial n} dz 
H/\lambda\right)</math>), <math>\beta=10^{5}</math> and
+
= \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)
<math>\gamma=0</math>.
+
\left(e^{-\hat{k}_0 x} - a_0 e^{\hat{k}_0 x} \right)\phi_0(z)^2 dz 
We compare with the method
+
</math></center>
presented in [[Meylan_2002a|Meylan 2002]] for an arbitrary shaped plate modified to compute
+
<center><math>
the solution for finite depth. The circle is represented in this scheme by
+
= \frac{\hat{k}_0}{i} A_0 \left(1 - |a_0^2| \right)
square panels which are arranged to, as nearly as possible, form a circular
+
</math></center>
shape.
+
and
 +
<center><math>
 +
\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></center>
 +
<center><math>
 +
= -\frac{\hat{\kappa}_0}{i} |b_0^2| \int_{-h}^{0} \psi_0(z)^2 dz 
 +
</math></center>
  
Figure 1 shows the
+
The contribution from the surface integral under the plate is
real part (a and c) and imaginary part (b and d) of the displacement for depth
+
<center><math>
<math>H=25</math>. The number
+
\Im\int_{0}^{\infty}\phi^*\frac{\partial\phi}{\partial n} dx
of points in the angular expansion is <math>N=16</math>. The number of
+
=  \Im\int_{0}^{\infty}
roots of the dispersion equation is <math>M=8</math>. Plots (a) and
+
1/\alpha \left(\beta \partial_x^4 \frac{\partial\phi^{*}}{\partial n} - (\gamma\alpha - 1 ) \frac{\partial\phi^{*}}{\partial n} \right)
(b) are calculated using the circular plate method described here. Plots (c)
+
\frac{\partial\phi}{\partial n} dx
and (d) are calculated using an arbitrary shaped plate method, with the
+
</math></center>
panels shown being the actual panels used in the calculation. We see the
+
<center><math>
expected agreement between the two methods.
+
= \frac{2\beta}{\alpha} \frac{\hat{\kappa}_0^{3}}{i} |b_0^2| \partial_z\psi_0(0)^2 
 +
</math></center>
 +
where we have integrated by parts and used the condition at the ends of the plate.  
  
 +
The energy balance is therefore
 +
<center><math>
 +
{\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></center>
  
The table below shows the values of the coefficients
 
<math>b_{mn}</math> for the case for previous case (<math>\lambda=50</math>,
 
<math>a=100</math>, <math>\beta=10^5</math>, <math>\gamma=0</math>, and <math>H=25</math>). The very rapid
 
decay of the higher evanescent modes is apparent. This shows how efficient this method of
 
solution is since only a small number of modes are required.
 
 
<blockquote style="background: white;  padding: 0em;">
 
<table border="1">
 
<tr>
 
<td> <math>b_{mn}</math> </td><td> <math>n=0</math> </td>
 
<td> <math>n=1</math> </td><td> <math>n=2</math> </td><td> <math>n=3</math> </td>
 
</tr>
 
<tr>
 
<td><math>m=-2</math></td> <td><math>1.32 \!\times\!10^{-1}-9.71 \!\times\!10^{-1}i</math> </td>
 
<td> <math>6.85 \!\times\!10^{-1} -6.37 \!\times\!10^{-1}i</math></td>
 
<td>  <math>2.95 \!\times\!10^{-1}-1.12 \!\times\!10^{0}i</math> </td><td>  <math>6.09 \!\times\!10^{-1}
 
-4.95 \!\times\!10^{-1}i</math> </td></tr>
 
 
<tr><td><math>m=-1</math> </td><td>  <math>-6.38 \!\times\!10^{-5}+1.47 \!\times\!10^{-3}i</math> </td><td>
 
<math>-3.92 \!\times\!10^{-3} + 3.99 \!\times\!10^{-3}i</math></td>
 
<td> <math>1.41 \!\times\!10^{-3}+2.82 \!\times\!10^{-3}i</math> </td><td>
 
<math>-4.28 \!\times\!10^{-3} +3.89 \!\times\!10^{-3}i</math></td></tr>
 
  
<tr><td><math>m=0</math> </td><td>
+
Note that this formula is only valid for angles <math>\theta \in [-\theta_0, \theta_0]</math>, where <math>\theta_0</math>
<math>-3.29 \!\times\!10^{-4}+1.43 \!\times\!10^{-3}i</math> </td><td>  <math>4.26 \!\times\!10^{-3}
+
is defined by <math>\sin \theta_0 =\frac{\kappa_0}{k_0}</math>. For large angles there is total reflection
-3.62 \!\times\!10^{-3}i</math></td><td>  
+
and then <math>|R|^2=1</math>
<math>-2.62 \!\times\!10^{-3}+1.76 \!\times\!10^{-3}i</math> </td><td>  <math>4.68 \!\times\!10^{-3}
 
-3.39 \!\times\!10^{-3}i</math></td></tr>
 
  
<tr><td><math>m=1</math> </td><td>  <math>4.31 \!\times\!10^{-7}-3.18 \!\times\!10^{-6}i</math> </td><td>
+
== Matlab Code ==
<math>-6.64 \!\times\!10^{-6} -7.14 \!\times\!10^{-6}i</math></td>
 
<td> <math>2.07 \!\times\!10^{-7}-7.89 \!\times\!10^{-7}i</math> </td><td>
 
<math>-6.30 \!\times\!10^{-6} -7.74 \!\times\!10^{-6}i</math></td></tr>
 
  
<tr><td><math>m=2</math> </td><td>
+
A program to calculate the coefficients for the semi-infinite dock problems can be found here
<math>6.79 \!\times\!10^{-13}-5.01 \!\times\!10^{-12}i</math> </td><td>
+
[http://www.math.auckland.ac.nz/~meylan/code/eigenfunction_matching/semiinfinite_plate.m semiinfinite_plate.m]
<math>-5.78 \!\times\!10^{-12}-6.21 \!\times\!10^{-12}i</math></td>
 
<td> <math>8.87 \!\times\!10^{-13}-3.38 \!\times\!10^{-12}i</math> </td><td>
 
<math>-5.54 \!\times\!10^{-12}-6.81 \!\times\!10^{-12}i</math></td></tr>
 
  
<tr><td><math>m=3</math> </td><td>  <math>1.35 \!\times\!10^{-18}-9.95 \!\times\!10^{-18}i</math> </td><td>
+
=== Additional code ===
<math>-9.69 \!\times\!10^{-18}-1.04 \!\times\!10^{-17}i</math></td>
 
<td> <math>1.94 \!\times\!10^{-18}-7.39 \!\times\!10^{-18}i</math>  </td><td>
 
<math>-9.37 \!\times\!10^{-18}-1.15 \!\times\!10^{-17}i</math></td>
 
</tr>
 
</table>
 
</blockquote>
 
  
 +
This program requires
 +
* {{free surface dispersion equation code}}
 +
* {{elastic plate dispersion equation code}}
  
[[Category:Problems with Cylindrical Symmetry]]
+
[[Category:Floating Elastic Plate]]
[[Category:Linear Hydroelasticity]]
+
[[Category:Eigenfunction Matching Method]]
 +
[[Category:Complete Pages]]

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