Difference between revisions of "Eigenfunction Matching for a Semi-Infinite Dock"

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= Introduction =
+
{{complete pages}}
  
This is the simplest problem in eigenfunction matching. It also is an easy
+
== Introduction ==
problem to understand the [[Wiener-Hopf]] and [[Residue Calculus]]
 
  
=Introduction=
+
This is one of the simplest problem in eigenfunction matching. It also is an easy
 +
problem to understand the [[:Category:Wiener-Hopf|Wiener-Hopf]] and
 +
[[:Category:Residue Calculus|Residue Calculus]].  The problems consists of a region to the left
 +
with a free surface and a region to the right with a rigid surface through which
 +
not flow is possible.
 +
We begin with the simply problem when the waves are normally incident (so that
 +
the problem is truly two-dimensional.  We then consider the case when the waves are incident
 +
at an angle. For the later we give the equations in slightly less detail.
 +
The case of a [[Finite Dock]] is treated very similarly. The problem can also
 +
be generalised to a [[ Eigenfunction Matching for Submerged Semi-Infinite Dock|Submerged Semi-Infinite Dock]]
  
We show here a solution for a [[Floating Elastic Plate]] on [[Finite Depth]] water
+
[[Image:semiinfinite_dock.jpg|thumb|right|300px|Wave scattering by a semi-infinite dock]]
based on [[Peter_Meylan_Chung_2004a|Peter, Meylan and Chung 2004]]. A solution
 
for [[Shallow Depth]] was given in [[Zilman_Miloh_2000a|Zilman and Miloh 2000]] and we will also show this.
 
  
=Governing Equations=
+
== Governing Equations ==
  
 
We begin with the [[Frequency Domain Problem]] for a dock which occupies
 
We begin with the [[Frequency Domain Problem]] for a dock which occupies
the region <math>x>0</math>.
+
the region <math>x>0</math> (we assume <math>e^{i\omega t}</math> time dependence).
 
The water is assumed to have
 
The water is assumed to have
constant finite depth <math>H</math> and the <math>z</math>-direction points vertically
+
constant finite depth <math>h</math> and the <math>z</math>-direction points vertically
upward with the water surface at <math>z=0</math> and the sea floor at <math>z=-H</math>. The
+
upward with the water surface at <math>z=0</math> and the sea floor at <math>z=-h</math>. The
 
boundary value problem can therefore be expressed as
 
boundary value problem can therefore be expressed as
 
<center>
 
<center>
 
<math>
 
<math>
\Delta\phi=0, \,\, -H<z<0,
+
\Delta\phi=0, \,\, -h < z < 0,
 
</math>
 
</math>
 
</center>
 
</center>
 
<center>
 
<center>
 
<math>
 
<math>
\phi_{z}=0, \,\, z=-H,
+
\partial_z\phi=0, \,\, z=-h,
 
</math>
 
</math>
 
</center>
 
</center>
 
<center><math>
 
<center><math>
\phi_{z}=\alpha\phi, \,\, z=0,\,x<0,
+
\partial_z\phi=\alpha\phi, \,\, z=0,\,x<0,
 
</math></center>
 
</math></center>
 
<center>
 
<center>
 
<math>
 
<math>
\phi_{z}=0, \,\, z=0,\,x>0,
+
\partial_z\phi=0, \,\, z=0,\,x>0,
 
</math>
 
</math>
 
</center>
 
</center>
 
We
 
We
 
must also apply the [[Sommerfeld Radiation Condition]]
 
must also apply the [[Sommerfeld Radiation Condition]]
as <math>|x|\rightarrow\infty</math>. The subscript <math>z</math>
+
as <math>|x|\rightarrow\infty</math>. This essentially implies
denotes the derivative in <math>z</math>-direction.
+
that the only wave at infinity is propagating away and at negative infinity there is a unit incident wave
 +
and a wave propagating away.
  
=Solution Method=
+
== Solution Method ==
  
== Separation of variables==
+
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>.
  
We now separate variables and write the potential as
+
{{separation of variables in two dimensions}}
<center>
+
 
<math>
+
{{separation of variables for a free surface}}
\phi(x,z)=\zeta(z)\rho(x)
+
 
</math>
+
{{separation of variables for a dock}}
</center>
+
 
Applying Laplace's equation we obtain
+
{{free surface dock relations}}
<center>
+
 
<math>
+
=== Expansion of the potential ===
\zeta_{zz}+\mu^{2}\zeta=0
+
 
</math>
+
We need to apply some boundary conditions at plus and minus infinity,  
</center>
+
where are essentially the the solution cannot grow. This means that we
so that:
+
only have the positive (or negative) roots of the dispersion equation.
<center>
+
However, it does not help us with the purely imaginary root. Here we
<math>
+
must use a different condition, essentially identifying one solution
\zeta=\cos\mu(z+H)
+
as the incoming wave and the other as the outgoing wave.
</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 x<0\,\,\,(1)
 
</math>
 
</center>
 
and
 
<center>
 
<math>
 
\kappa\tan(\kappa H)=0,\quad
 
x>0 \,\,\,(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)
 
<math>\kappa_{m}=m\pi/H</math>, <math>m\geq 0</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 0
 
</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>
 
  
Therefore the potential can
+
Therefore the scattered potential (without the incident wave, which will
 +
be added later) can
 
be expanded as
 
be expanded as
 
<center>
 
<center>
Line 139: Line 87:
 
where <math>a_{m}</math> and <math>b_{m}</math>
 
where <math>a_{m}</math> and <math>b_{m}</math>
 
are the coefficients of the potential in the open water and
 
are the coefficients of the potential in the open water and
the plate covered region respectively.
+
the dock covered region respectively.
  
==Incident potential==
+
{{incident potential for two dimensions}}
  
The incident potential is a wave of amplitude <math>A</math>
+
=== An infinite dimensional system of equations ===
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)
 
</math>
 
</center>
 
where <math>e_{n}=A/\left(i\sqrt{\alpha}\right)</math>.
 
  
==Boundary conditions==
 
  
The boundary conditions for the plate also have to be
 
considered. The vertical force and bending moment must vanish, which can be
 
written as
 
<center>
 
<math>
 
\left[\bar{\Delta}-\frac{1-\nu}{r}\left(\frac{\partial}{\partial r}
 
+\frac{1}{r}\frac{\partial^{2}}{\partial\theta^{2}}\right)\right]
 
w=0\,\,\,(3)
 
</math>
 
</center>
 
and
 
<center>
 
<math>
 
\left[  \frac{\partial}{\partial r}\bar{\Delta}-\frac{1-\nu}{r^{2}}\left(
 
\frac{\partial}{\partial r}+\frac{1}{r}\right)  \frac{\partial^{2}}
 
{\partial\theta^{2}}\right]  w=0 \,\,\,(4)
 
</math>
 
</center>
 
where <math>w</math> is the time-independent surface
 
displacement, <math>\nu</math> is Poisson's ratio, and <math>\bar{\Delta}</math> is the
 
polar coordinate Laplacian
 
<center>
 
<math>
 
\bar{\Delta}=\frac{\partial^{2}}{\partial r^{2}}+\frac{1}{r}\frac{\partial
 
}{\partial r}+\frac{1}{r^{2}}\frac{\partial^{2}}{\partial\theta^{2}}
 
</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>
 
</center>
 
and
 
<center>
 
<math>
 
(\beta\bar{\Delta}^{2}+1-\alpha\gamma)w=i\sqrt{\alpha}\phi,\quad r<a
 
</math>
 
</center>
 
The surface displacement can also be expanded in eigenfunctions
 
as
 
<center>
 
<math>
 
w(r,\theta)=\sum_{n=-\infty}^{\infty}\sum_{m=0}^{\infty}i\sqrt{\alpha}
 
a_{mn}K_{n}(k_{m}r)e^{i n\theta},\;\;r>a
 
</math>
 
</center>
 
and:
 
<center>
 
<math>
 
w(r,\theta)=
 
\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>
 
</center>
 
using the fact that
 
<center>
 
<math>
 
\bar{\Delta}\left(  I_{n}(\kappa_{m}r)e^{i n\theta}\right)  =\kappa_{m}
 
^{2}I_{n}(\kappa_{m}r)e^{i n\theta}\,\,\,(5)
 
</math>
 
</center>
 
 
==An infinite dimensional system of equations==
 
 
The boundary conditions (3) and
 
(4) can be expressed in terms of the potential
 
using (5). Since the angular modes are uncoupled the
 
conditions apply to each mode, giving
 
<center>
 
<math>
 
\sum_{m=-2}^{\infty}(\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\,\,\,(6)
 
</math>
 
</center>
 
<center>
 
<math>
 
\sum_{m=-2}^{\infty}(\beta\kappa_{m}^{4}+1-\alpha\gamma)^{-1}b_{mn}
 
\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
 
The potential and its derivative must be continuous across the
 
transition from open water to the plate covered region. Therefore, 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.
+
potentials and their derivatives at <math>x=0</math> have to be equal.
Again we know that this must be true for each angle and we obtain
+
We obtain
 
<center>
 
<center>
 
<math>
 
<math>
e_{n}I_{n}(k_{0}a)\phi_{0}\left(  z\right) + \sum_{m=0}^{\infty}
+
\phi_{0}\left(  z\right) + \sum_{m=0}^{\infty}
a_{mn} K_{n}(k_{m}a)\phi_{m}\left(  z\right)  
+
a_{m} \phi_{m}\left(  z\right)  
=\sum_{m=-2}^{\infty}b_{mn}I_{n}(\kappa_{m}a)\psi_{m}(z)
+
=\sum_{m=0}^{\infty}b_{m}\psi_{m}(z)
 
</math>
 
</math>
 
</center>
 
</center>
Line 267: Line 108:
 
<center>
 
<center>
 
<math>
 
<math>
  e_{n}k_{0}I_{n}^{\prime}(k_{0}a)\phi_{0}\left(  z\right)  +\sum
+
  -k_{0}\phi_{0}\left(  z\right)  +\sum
_{m=0}^{\infty} a_{mn}k_{m}K_{n}^{\prime}(k_{m}a)\phi_{m}\left(  z\right)  
+
_{m=0}^{\infty} a_{m}k_{m}\phi_{m}\left(  z\right)  
  =\sum_{m=-2}^{\infty}b_{mn}\kappa_{m}I_{n}^{\prime}(\kappa_{m}a)\psi
+
  =-\sum_{m=0}^{\infty}b_{m}\kappa_{m}\psi
 
_{m}(z)
 
_{m}(z)
 
</math>
 
</math>
 
</center>
 
</center>
for each <math>n</math>.
+
for each <math>m</math>.
 
We solve these equations by multiplying both equations by
 
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:
+
<math>\phi_{l}(z) \,</math> and integrating from <math>-h</math> to <math>0</math> to obtain:
 
<center>
 
<center>
 
<math>
 
<math>
e_{n}I_{n}(k_{0}a)A_{0}\delta_{0l}+a_{ln}K_{n}(k_{l}a)A_{l}
+
A_{0}\delta_{0l}+a_{l}A_{l}
=\sum_{m=-2}^{\infty}b_{mn}I_{n}(\kappa_{m}a)B_{ml} \,\,\,(8)
+
=\sum_{m=0}^{\infty}b_{m}B_{ml}\,\,\,(3)
 
</math>
 
</math>
 
</center>
 
</center>
Line 285: Line 126:
 
<center>
 
<center>
 
<math>
 
<math>
e_{n}k_{0}I_{n}^{\prime}(k_{0}a)A_{0}\delta_{0l}+a_{ln}k_{l}K_{n}^{\prime
+
-k_{0}A_{0}\delta_{0l}+a_{l}k_{l}A_l
}(k_{l}a)A_{l}
+
  =-\sum_{m=0}^{\infty}b_{m}\kappa_{m}B_{ml} \,\,\,(4)
  =\sum_{m=-2}^{\infty}b_{mn}\kappa_{m}I_{n}^{\prime}(\kappa_{m}
 
a)B_{ml} \,\,\,(9)
 
 
</math>
 
</math>
 
</center>
 
</center>
Equation (8) can be solved for the open water
+
If we mutiply equation (3) by <math>k_l \,</math> and subtract equation (4)
coefficients <math>a_{mn}</math>
+
we obtain
 
<center>
 
<center>
 
<math>
 
<math>
a_{ln}=-e_{n}\frac{I_{n}(k_{0}a)}{K_{n}(k_{0}a)}\delta_{0l}+\sum
+
(k_{0}+k_l)A_{0}\delta_{0l}
_{m=-2}^{\infty}b_{mn}\frac{I_{n}(\kappa_{m}a)B_{ml}}{K_{n}(k_{l}a)A_{l}}
+
=\sum_{m=0}^{\infty}b_{m}(k_l + \kappa_{m})B_{ml} \,\,\,(5)
 
</math>
 
</math>
 
</center>
 
</center>
which can then be substituted into equation
+
This equation gives the required equations to solve for the
(9) to give us
+
coefficients of the water velocity potential in the dock covered region.
 +
 
 +
== Numerical Solution ==
 +
 
 +
To solve the system of equations (3) and (5), we set the upper limit of <math>l</math> to
 +
be <math>M</math>. This resulting system can be expressed in the block matrix form below,
 
<center>
 
<center>
 
<math>
 
<math>
\left(  k_{0}I_{n}^{\prime}(k_{0}a)-k_{0}\frac{K_{n}^{\prime}(k_{0}
+
\begin{bmatrix}
a)}{K_{n}(k_{0}a)}I_{n}(k_{0}a)\right)  e_{n}A_{0}\delta_{0l}
+
\begin{bmatrix}
=\sum_{m=-2}^{\infty}\left( \kappa_{m}I_{n}^{\prime}(\kappa_{m}
+
A_0&0 \quad \cdots&0\\
a)-k_{l}\frac{K_{n}^{\prime}(k_{l}a)}{K_{n}(k_{l}a)}I_{n}(\kappa
+
0&&\\
_{m}a)\right)  B_{ml}b_{mn}\,\,\,(10)
+
\vdots&A_l&\vdots\\
 +
&&0\\
 +
0&\cdots \quad 0 &A_M
 +
\end{bmatrix}
 +
&
 +
\begin{bmatrix}
 +
-B_{00}&\cdots&-B_{0M}\\
 +
&&\\
 +
\vdots&-B_{ml}&\vdots\\
 +
&&\\
 +
-B_{M0}&\cdots&-B_{MM}
 +
\end{bmatrix}
 +
\\
 +
\begin{bmatrix}
 +
0&\cdots&0\\
 +
\vdots&\ddots&\vdots\\
 +
0&\cdots&0
 +
\end{bmatrix}
 +
&
 +
\begin{bmatrix}
 +
(k_0 + \kappa_0) \, B_{00}&\cdots&(k_M + \kappa_{0}) \, B_{0M}\\
 +
&&\\
 +
\vdots&(k_l + \kappa_{m}) \, B_{ml}&\vdots\\
 +
&&\\
 +
(k_0 + \kappa_M) \, B_{M0}&\cdots&(k_M + \kappa_{M}) \, B_{MM}\\
 +
\end{bmatrix}
 +
\end{bmatrix}
 +
\begin{bmatrix}
 +
a_{0} \\
 +
a_{1} \\
 +
\vdots \\
 +
a_M \\
 +
\\
 +
b_{0}\\
 +
b_1 \\
 +
\vdots \\
 +
\\
 +
b_M
 +
\end{bmatrix}
 +
=
 +
\begin{bmatrix}
 +
- A_{0} \\
 +
0 \\
 +
\vdots \\
 +
\\
 +
0 \\
 +
\\
 +
2k_{0}A_{0} \\
 +
0 \\
 +
\vdots \\
 +
\\
 +
0
 +
\end{bmatrix}
 
</math>
 
</math>
 
</center>
 
</center>
for each <math>n</math>.
+
We then simply need to solve the linear system of equations.
Together with equations (6) and (7)
+
 
equation (10) gives the required equations to solve for the
+
== Solution with Waves Incident at an Angle ==
coefficients of the water velocity potential in the plate covered region.
+
 
 +
We can consider the problem when the waves are incident at an angle <math>\theta</math>.  
  
=Numerical Solution=
+
{{incident angle}}
  
To solve the system of equations (10) together
+
Therefore the potential can
with the boundary conditions (6 and 7) we set the upper limit of <math>l</math> to
+
be expanded as
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
+
\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
n\theta }\phi_{m}(z), \;\;r>a
 
 
</math>
 
</math>
 
</center>
 
</center>
Line 330: Line 224:
 
<center>
 
<center>
 
<math>
 
<math>
\phi(r,\theta,z)=\sum_{n=-N}^{N}\sum_{m=-2}^{M}b_{mn}I_{n}(\kappa
+
\phi(x,z)=\sum_{m=0}^{\infty}b_{m}
_{m}r)e^{i n\theta}\psi_{m}(z), \;\;r<a
+
e^{-\hat{\kappa}_{m}x}\psi_{m}(z), \;\;x>0
 
</math>
 
</math>
 
</center>
 
</center>
Since <math>l</math> is an integer with <math>0\leq l\leq
+
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>
M</math> this leads to a system of <math>M+1</math> equations.
+
where we always take the positive real root or the root with positive imaginary part.
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)
+
The equations are derived almost identically to those above and we obtain
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_{n=0}^{\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 359: Line 241:
 
<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=0}^{\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]]=
 
 
 
The shallow water theory of [[Zilman_Miloh_2000a|Zilman and Miloh 2000]] can be recovered by
 
simply setting the depth shallow enough that the shallow water theory is valid
 
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=
 
 
 
[[Image:ComparisionH25.jpg|thumb|right|300px|Figure 1]]
 
 
 
We present solutions for a plate of radius <math>a=100</math>. The wavelength is
 
<math>\lambda=50</math> (recall that <math>\alpha=2\pi/\lambda\tanh\left(  2\pi
 
H/\lambda\right)</math>), <math>\beta=10^{5}</math> and
 
<math>\gamma=0</math>.
 
We compare with the method
 
presented in [[Meylan_2002a|Meylan 2002]] for an arbitrary shaped plate modified to compute
 
the solution for finite depth. The circle is represented in this scheme by
 
square panels which are arranged to, as nearly as possible, form a circular
 
shape.
 
 
 
Figure 1 shows the
 
real part (a and c) and imaginary part (b and d) of the displacement for depth
 
<math>H=25</math>. The number
 
of points in the angular expansion is <math>N=16</math>. The number of
 
roots of the dispersion equation is <math>M=8</math>. Plots (a) and
 
(b) are calculated using the circular plate method described here. Plots (c)
 
and (d) are calculated using an arbitrary shaped plate method, with the
 
panels shown being the actual panels used in the calculation. We see the
 
expected agreement between the two methods.
 
 
 
 
 
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>
 
<math>-3.29 \!\times\!10^{-4}+1.43 \!\times\!10^{-3}i</math> </td><td>  <math>4.26 \!\times\!10^{-3}
 
-3.62 \!\times\!10^{-3}i</math></td><td>
 
<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>
+
{{semiinfinite_dock code}}
<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}}
  
 
[[Category:Eigenfunction Matching Method]]
 
[[Category:Eigenfunction Matching Method]]

Latest revision as of 00:43, 25 April 2017


Introduction

This is one of the simplest problem in eigenfunction matching. It also is an easy problem to understand the Wiener-Hopf and Residue Calculus. The problems consists of a region to the left with a free surface and a region to the right with a rigid surface through which not flow is possible. We begin with the simply problem when the waves are normally incident (so that the problem is truly two-dimensional. We then consider the case when the waves are incident at an angle. For the later we give the equations in slightly less detail. The case of a Finite Dock is treated very similarly. The problem can also be generalised to a Submerged Semi-Infinite Dock

Wave scattering by a semi-infinite dock

Governing Equations

We begin with the Frequency Domain Problem for a dock which occupies the region [math]x\gt 0[/math] (we assume [math]e^{i\omega t}[/math] time dependence). The water is assumed to have constant finite depth [math]h[/math] and the [math]z[/math]-direction points vertically upward with the water surface at [math]z=0[/math] and the sea floor at [math]z=-h[/math]. The boundary value problem can therefore be expressed as

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

[math] \partial_z\phi=0, \,\, z=-h, [/math]

[math] \partial_z\phi=\alpha\phi, \,\, z=0,\,x\lt 0, [/math]

[math] \partial_z\phi=0, \,\, z=0,\,x\gt 0, [/math]

We must also apply the Sommerfeld Radiation Condition as [math]|x|\rightarrow\infty[/math]. This essentially implies that the only wave at infinity is propagating away and at negative infinity there is a unit incident wave and a wave propagating away.

Solution Method

We use separation of variables in the two regions, [math]x\lt 0[/math] and [math]x\gt 0[/math].

We express the potential as

[math] \phi(x,z) = X(x)Z(z)\, [/math]

and then Laplace's equation becomes

[math] \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] \phi(x,z) = X(x)Z(z)\, [/math]

and then Laplace's equation becomes

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

subject to the boundary conditions

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

and

[math] Z^{\prime}(0) = \alpha Z(0) [/math]

We can then use the boundary condition at [math]z=-h \, [/math] to write

[math] 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]z=0[/math]. The boundary condition at the free surface ([math]z=0 \,[/math]) gives rise to:

[math] 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]k_{0}=\pm ik \,[/math] and the positive real solutions by [math]k_{m} \,[/math], [math]m\geq1[/math]. The [math]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] \cos ix = \cosh x, \quad \sin ix = i\sinh x, [/math]

to arrive at the dispersion relation

[math] \alpha = k\tanh kh. [/math]

We note that for a specified frequency [math]\omega \,[/math] the equation determines the wavenumber [math]k \,[/math].

Finally we define the function [math]Z(z) \,[/math] as

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

where

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


Separation of Variables for a Dock

The separation of variables equation for a floating dock

[math] Z^{\prime\prime} + k^2 Z =0, [/math]

subject to the boundary conditions

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

and

[math] Z^{\prime} (0) = 0. [/math]

The solution is [math]k=\kappa_{m}= \frac{m\pi}{h} \,[/math], [math]m\geq 0[/math] and

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

We note that

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

where

[math] C_{m} = \begin{cases} h,\quad m=0 \\ \frac{1}{2}h,\,\,\,m\neq 0 \end{cases} [/math]

Inner product between free surface and dock modes

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

where

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

Expansion of the potential

We need to apply some boundary conditions at plus and minus infinity, where are essentially the the solution cannot grow. This means that we only have the positive (or negative) roots of the dispersion equation. However, it does not help us with the purely imaginary root. Here we must use a different condition, essentially identifying one solution as the incoming wave and the other as the outgoing wave.

Therefore the scattered potential (without the incident wave, which will be added later) can be expanded as

[math] \phi(x,z)=\sum_{m=0}^{\infty}a_{m}e^{k_{m}x}\phi_{m}(z), \;\;x\lt 0 [/math]

and

[math] \phi(x,z)=\sum_{m=0}^{\infty}b_{m} e^{-\kappa_{m}x}\psi_{m}(z), \;\;x\gt 0 [/math]

where [math]a_{m}[/math] and [math]b_{m}[/math] are the coefficients of the potential in the open water and the dock covered region respectively.

Incident potential

To create meaningful solutions of the velocity potential [math]\phi[/math] in the specified domains we add an incident wave term to the expansion for the domain of [math]x \lt 0[/math] above. The incident potential is a wave of amplitude [math]A[/math] in displacement travelling in the positive [math]x[/math]-direction. We would only see this in the time domain [math]\Phi(x,z,t)[/math] however, in the frequency domain the incident potential can be written as

[math] \phi_{\mathrm{I}}(x,z) =e^{-k_{0}x}\chi_{0}\left( z\right). [/math]

The total velocity (scattered) potential now becomes [math]\phi = \phi_{\mathrm{I}} + \phi_{\mathrm{D}}[/math] for the domain of [math]x \lt 0[/math].

The first term in the expansion of the diffracted potential for the domain [math]x \lt 0[/math] is given by

[math] a_{0}e^{k_{0}x}\chi_{0}\left( z\right) [/math]

which represents the reflected wave.

In any scattering problem [math]|R|^2 + |T|^2 = 1[/math] where [math]R[/math] and [math]T[/math] are the reflection and transmission coefficients respectively. In our case of the semi-infinite dock [math]|a_{0}| = |R| = 1[/math] and [math]|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]x=0[/math] have to be equal. We obtain

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

and

[math] -k_{0}\phi_{0}\left( z\right) +\sum _{m=0}^{\infty} a_{m}k_{m}\phi_{m}\left( z\right) =-\sum_{m=0}^{\infty}b_{m}\kappa_{m}\psi _{m}(z) [/math]

for each [math]m[/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:

[math] A_{0}\delta_{0l}+a_{l}A_{l} =\sum_{m=0}^{\infty}b_{m}B_{ml}\,\,\,(3) [/math]

and

[math] -k_{0}A_{0}\delta_{0l}+a_{l}k_{l}A_l =-\sum_{m=0}^{\infty}b_{m}\kappa_{m}B_{ml} \,\,\,(4) [/math]

If we mutiply equation (3) by [math]k_l \,[/math] and subtract equation (4) we obtain

[math] (k_{0}+k_l)A_{0}\delta_{0l} =\sum_{m=0}^{\infty}b_{m}(k_l + \kappa_{m})B_{ml} \,\,\,(5) [/math]

This equation gives the required equations to solve for the coefficients of the water velocity potential in the dock covered region.

Numerical Solution

To solve the system of equations (3) and (5), we set the upper limit of [math]l[/math] to be [math]M[/math]. This resulting system can be expressed in the block matrix form below,

[math] \begin{bmatrix} \begin{bmatrix} A_0&0 \quad \cdots&0\\ 0&&\\ \vdots&A_l&\vdots\\ &&0\\ 0&\cdots \quad 0 &A_M \end{bmatrix} & \begin{bmatrix} -B_{00}&\cdots&-B_{0M}\\ &&\\ \vdots&-B_{ml}&\vdots\\ &&\\ -B_{M0}&\cdots&-B_{MM} \end{bmatrix} \\ \begin{bmatrix} 0&\cdots&0\\ \vdots&\ddots&\vdots\\ 0&\cdots&0 \end{bmatrix} & \begin{bmatrix} (k_0 + \kappa_0) \, B_{00}&\cdots&(k_M + \kappa_{0}) \, B_{0M}\\ &&\\ \vdots&(k_l + \kappa_{m}) \, B_{ml}&\vdots\\ &&\\ (k_0 + \kappa_M) \, B_{M0}&\cdots&(k_M + \kappa_{M}) \, B_{MM}\\ \end{bmatrix} \end{bmatrix} \begin{bmatrix} a_{0} \\ a_{1} \\ \vdots \\ a_M \\ \\ b_{0}\\ b_1 \\ \vdots \\ \\ b_M \end{bmatrix} = \begin{bmatrix} - A_{0} \\ 0 \\ \vdots \\ \\ 0 \\ \\ 2k_{0}A_{0} \\ 0 \\ \vdots \\ \\ 0 \end{bmatrix} [/math]

We then simply need to solve the linear system of equations.

Solution with Waves Incident at an Angle

We can consider the problem when the waves are incident at an angle [math]\theta[/math].

When a wave in incident at an angle [math]\theta [/math] we have the wavenumber in the [math]y[/math] direction is [math]k_y = \sin\theta k_0[/math] where [math]k_0[/math] is as defined previously (note that [math]k_y[/math] is imaginary).

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

[math] k^2 = k_x^2 + k_y^2 [/math]

where [math]k[/math] is the separation constant calculated without an incident angle.

Therefore the potential can be expanded as

[math] \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] \phi(x,z)=\sum_{m=0}^{\infty}b_{m} e^{-\hat{\kappa}_{m}x}\psi_{m}(z), \;\;x\gt 0 [/math]

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 equations are derived almost identically to those above and we obtain

[math] A_{0}\delta_{0l}+a_{l}A_{l} =\sum_{n=0}^{\infty}b_{m}B_{ml} [/math]

and

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

and these are solved exactly as before.

Matlab Code

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

Additional code

This program requires