Difference between revisions of "Eigenfunction Matching for a Circular Dock"

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=Introduction=
+
{{complete pages}}
  
We show here a solution for a dock on [[Finite Depth]] water.
+
== Introduction ==
  
=Governing Equations=
+
We show here a solution for a dock on [[Finite Depth]] water, which is circular. This is the three-dimensional
 +
analog of the [[Eigenfunction Matching for a Semi-Infinite Dock]].
 +
 
 +
== Governing Equations ==
  
 
We begin with the [[Frequency Domain Problem]].
 
We begin with the [[Frequency Domain Problem]].
Line 14: Line 17:
 
<center>
 
<center>
 
<math>
 
<math>
\Delta\phi=0, \,\, -H<z<0,
+
\Delta\phi=0, \,\, -h<z<0,
 
</math>
 
</math>
 
</center>
 
</center>
Line 34: Line 37:
 
denotes the derivative in <math>z</math>-direction.
 
denotes the derivative in <math>z</math>-direction.
  
=Solution Method=
+
== Solution Method ==
 +
 
 +
We use [http://en.wikipedia.org/wiki/Separation_of_Variables separation of variables] in the two regions, <math>r<a</math>
 +
and <math>r>a</math>.
 +
 
 +
{{separation of variables in cylindrical coordinates in finite depth}}
 +
 
 +
{{separation of variables for a free surface}}
 +
 
 +
{{separation of variables for a dock}}
 +
 
 +
{{free surface dock relations}}
  
== Separation of variables==
 
  
We now separate variables, noting that since the problem has
+
{{separation of variables in cylindrical coordinates}}
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>.
+
Since the solution must be bounded
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
 
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
 
<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
+
linear combination of <math>K_{n}(y)</math>.  
 +
The case <math>\kappa_0 =0</math> is a special case and the solution under
 +
the dock is <math>(r/a)^{|n|}</math>.
 +
Therefore the potential can
 
be expanded as
 
be expanded as
 
<center>
 
<center>
Line 156: Line 70:
 
<center>
 
<center>
 
<math>
 
<math>
\phi(r,\theta,z)=\sum_{n=-\infty}^{\infty}\sum_{m=-2}^{\infty}b_{mn}
+
\phi(r,\theta,z)=\sum_{n=-\infty}^{\infty}b_{0n}(r/a)^{|n|} e^{i n\theta}\psi_{0}(z)+
 +
\sum_{n=-\infty}^{\infty}\sum_{m=1}^{\infty}b_{mn}
 
I_{n}(\kappa_{m}r)e^{i n\theta}\psi_{m}(z), \;\;r<a
 
I_{n}(\kappa_{m}r)e^{i n\theta}\psi_{m}(z), \;\;r<a
 
</math>
 
</math>
Line 164: Line 79:
 
the plate covered region respectively.
 
the plate covered region respectively.
  
==Incident potential==
+
{{incident plane wave in cylindrical coordinates}}
  
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)
 
=\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==
+
The potential and its derivative must be continuous across the
 
+
transition from open water to the plate covered region. Therefore, the
The boundary conditions for the plate also have to be
+
potentials and their derivatives at <math>r=a</math> have to be equal
considered. The vertical force and bending moment must vanish, which can be
+
for each angle and we obtain
written as
 
 
<center>
 
<center>
 
<math>
 
<math>
\left[\bar{\Delta}-\frac{1-\nu}{r}\left(\frac{\partial}{\partial r}
+
I_{n}(k_{0}a)\phi_{0}\left( z\right) + \sum_{m=0}^{\infty}
+\frac{1}{r}\frac{\partial^{2}}{\partial\theta^{2}}\right)\right]
+
a_{mn} K_{n}(k_{m}a)\phi_{m}\left(  z\right)
w=0\,\,\,(3)
+
= b_{0n} \psi_{0}(z) +\sum_{m=1}^{\infty}b_{mn}I_{n}(\kappa_{m}a)\psi_{m}(z)
 
</math>
 
</math>
 
</center>
 
</center>
Line 196: Line 97:
 
<center>
 
<center>
 
<math>
 
<math>
\left[ \frac{\partial}{\partial r}\bar{\Delta}-\frac{1-\nu}{r^{2}}\left(
+
k_{0}I_{n}^{\prime}(k_{0}a)\phi_{0}\left( z\right)  +\sum
\frac{\partial}{\partial r}+\frac{1}{r}\right)  \frac{\partial^{2}}
+
_{m=0}^{\infty} a_{mn}k_{m}K_{n}^{\prime}(k_{m}a)\phi_{m}\left( z\right)
{\partial\theta^{2}}\right]  w=0 \,\,\,(4)
+
=b_{0n} \frac{|n|}{a} \psi_{0}(z) +\sum_{m=1}^{\infty}
 +
b_{mn}\kappa_{m}I_{n}^{\prime}(\kappa_{m}a)\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
+
I_{n}(k_{0}a)A_{0}\delta_{0l}+a_{ln}K_{n}(k_{l}a)A_{l}
}{\partial r}+\frac{1}{r^{2}}\frac{\partial^{2}}{\partial\theta^{2}}
+
=b_{0n}B_{0l} + \sum_{m=1}^{\infty}b_{mn}I_{n}(\kappa_{m}a)B_{ml}  
</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 229: Line 116:
 
<center>
 
<center>
 
<math>
 
<math>
(\beta\bar{\Delta}^{2}+1-\alpha\gamma)w=i\sqrt{\alpha}\phi,\quad r<a
+
k_{0}I_{n}^{\prime}(k_{0}a)A_{0}\delta_{0l}+a_{ln}k_{l}K_{n}^{\prime
</math>
+
}(k_{l}a)A_{l}  
</center>
+
= b_{0n}B_{0l}\frac{|n|}{a} +
The surface displacement can also be expanded in eigenfunctions
+
\sum_{m=1}^{\infty}b_{mn}\kappa_{m}I_{n}^{\prime}(\kappa_{m}a)B_{ml}  
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>
 
</math>
 
</center>
 
</center>
  
==An infinite dimensional system of equations==
+
== Numerical Solution ==
 +
 
 +
To solve the system of equations we set the upper limit of <math>l</math> to
 +
be <math>M</math>.
 +
 
 +
== A Simple Method To Calculate The [[Diffraction Transfer Matrix]] For The Case Of A Circular Dock ==
 +
 
 +
Let's consider an incident wave whose potential has the following expression
 +
<center><math>
 +
\phi^\mathrm{I} (r,\theta,z) = \sum_{n=0}^{\infty} \phi_n(z)
 +
\sum_{\nu = - \infty}^{\infty} D_{n\nu} I_\nu (k_n r) \mathrm{e}^{\mathrm{i}\nu \theta}.
 +
</math></center>
 +
Such an incident potential is found in the [[Kagemoto and Yue Interaction Theory]], where
 +
it can be written as the sum of an ambient incident potential and the scattered potentials
 +
of the other bodies, which are interpretated as incident potentials for the studied body.
 +
 
 +
We can apply the same eigenfunction matching that previously, considering the potential
 +
and its normal derivative continuous at <math>r=a</math>. Thus the potential and its normal
 +
derivative expressed at each side of this value of the radius have to be equal. We obtain
 +
the relationships
  
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
 
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>
 
<center>
 
<math>
 
<math>
e_{n}I_{n}(k_{0}a)\phi_{0}\left( z\right) + \sum_{m=0}^{\infty}
+
\sum_{m=0}^{\infty} D_{mn} I_n (k_m a) \phi_m(z) + \sum_{m=0}^{\infty}
 
a_{mn} K_{n}(k_{m}a)\phi_{m}\left(  z\right)  
 
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)
+
= b_{0n} \psi_{0}(z) +\sum_{m=1}^{\infty}b_{mn}I_{n}(\kappa_{m}a)\psi_{m}(z)
 
</math>
 
</math>
 
</center>
 
</center>
Line 294: Line 154:
 
<center>
 
<center>
 
<math>
 
<math>
e_{n}k_{0}I_{n}^{\prime}(k_{0}a)\phi_{0}\left( z\right) +\sum
+
\sum_{m=0}^{\infty} D_{mn} k_m I'_n (k_m a) \phi_m(z) +\sum
 
_{m=0}^{\infty} a_{mn}k_{m}K_{n}^{\prime}(k_{m}a)\phi_{m}\left(  z\right)  
 
_{m=0}^{\infty} a_{mn}k_{m}K_{n}^{\prime}(k_{m}a)\phi_{m}\left(  z\right)  
  =\sum_{m=-2}^{\infty}b_{mn}\kappa_{m}I_{n}^{\prime}(\kappa_{m}a)\psi
+
  =b_{0n} \frac{|n|}{a} \psi_{0}(z) +\sum_{m=1}^{\infty}
 +
b_{mn}\kappa_{m}I_{n}^{\prime}(\kappa_{m}a)\psi
 
_{m}(z)
 
_{m}(z)
 
</math>
 
</math>
 
</center>
 
</center>
 
for each <math>n</math>.
 
for each <math>n</math>.
We solve these equations by multiplying both equations by
+
We solve these equations with the same method that before, 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}
+
D_{ln} I_n (k_l a) A_l+a_{ln}K_{n}(k_{l}a)A_{l}
=\sum_{m=-2}^{\infty}b_{mn}I_{n}(\kappa_{m}a)B_{ml} \,\,\,(8)
+
=b_{0n}B_{0l} + \sum_{m=1}^{\infty}b_{mn}I_{n}(\kappa_{m}a)B_{ml}, \ \ \ (1)
 
</math>
 
</math>
 
</center>
 
</center>
Line 312: Line 174:
 
<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
+
D_{ln} k_l I'_n (k_l a) A_l+a_{ln}k_{l}K_{n}^{\prime
 
}(k_{l}a)A_{l}  
 
}(k_{l}a)A_{l}  
  =\sum_{m=-2}^{\infty}b_{mn}\kappa_{m}I_{n}^{\prime}(\kappa_{m}
+
  = b_{0n}B_{0l}\frac{|n|}{a} +
a)B_{ml} \,\,\,(9)
+
\sum_{m=1}^{\infty}b_{mn}\kappa_{m}I_{n}^{\prime}(\kappa_{m}a)B_{ml}, \ \ \ (2)
 
</math>
 
</math>
 
</center>
 
</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=
+
The [[Diffraction Transfer Matrix]]
 +
maps the coefficients of the incident wave with the coefficients of the scattered wave within
 +
the open water domain. The relation which links these two coefficients can be written as follows
 +
<center><math>
 +
a_{mn}=\sum_{l=0}^{\infty} T_{lmn} D_{ln}
 +
</math></center>
 +
 
 +
From the equations (1) and (2)
 +
we can write a linear system of equation, limiting the number of modes of the dispersion equation
 +
to <math>N</math> real ones
  
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
+
\begin{bmatrix}
n\theta }\phi_{m}(z), \;\;r>a  
+
\begin{bmatrix}
 +
-A_0 K_n(k_0 a)&0 \quad \cdots&0\\
 +
0&&\\
 +
\vdots&-A_l K_n(k_l a)&\vdots\\
 +
&&0\\
 +
0&\cdots \quad 0 & -A_N K_n(k_N a)
 +
\end{bmatrix}
 +
&
 +
\begin{bmatrix}
 +
B_{00}&I_n(\kappa_{1}a) B_{10}&\cdots&I_n(\kappa_{N}a) B_{N0}\\
 +
B_{01}&I_n(\kappa_{1}a) B_{11}&&\\
 +
\vdots&\vdots&&\vdots\\
 +
&&&\\
 +
B_{0N}&I_n(\kappa_{1}a) B_{1N}&\cdots&I_n(\kappa_{N}a) B_{NN}
 +
\end{bmatrix}
 +
\\
 +
\begin{bmatrix}
 +
-k_0 K'_n(k_0 a) A_0&0 \quad \cdots&0\\
 +
0&&\\
 +
\vdots&-k_l K'_n(k_l a) A_l&\vdots\\
 +
&&0\\
 +
0&\cdots \quad 0 & -k_N K'_n(k_N a) A_N
 +
\end{bmatrix}
 +
&
 +
\begin{bmatrix}
 +
B_{00}\frac{|n|}{a}&\kappa_{1} I'_n(\kappa_{1}a) B_{10}&\cdots&\kappa_{N} I'_n(\kappa_{N}a) B_{N0}\\
 +
B_{01}\frac{|n|}{a}&\kappa_{1} I'_n(\kappa_{1}a) B_{11}&&\\
 +
\vdots&\vdots&&\vdots\\
 +
&&&\\
 +
B_{0N}\frac{|n|}{a}&\kappa_{1} I'_n(\kappa_{1}a) B_{1N}&\cdots&\kappa_{N} I'_n(\kappa_{N}a) B_{NN}
 +
\end{bmatrix}
 +
\end{bmatrix}
 +
\begin{bmatrix}
 +
a_{0n} \\
 +
\\
 +
\vdots \\
 +
\\
 +
a_{Nn} \\
 +
b_{0n}\\
 +
\\
 +
\vdots \\
 +
\\
 +
b_{Nn}
 +
\end{bmatrix}
 +
=
 +
\begin{bmatrix}
 +
\begin{bmatrix}
 +
I_n(k_0 a) A_0&0 \quad \cdots&0\\
 +
0&&\\
 +
\vdots&I_n(k_l a) A_l&\vdots\\
 +
&&0\\
 +
0&\cdots \quad 0 & I_n(k_N a) A_N
 +
\end{bmatrix}
 +
\\
 +
\begin{bmatrix}
 +
k_0 I'_n(k_0 a) A_0&0 \quad \cdots&0\\
 +
0&&\\
 +
\vdots&k_l I'_n(k_l a) A_l&\vdots\\
 +
&&0\\
 +
0&\cdots \quad 0 &k_N I'_n(k_N a) A_N
 +
\end{bmatrix}
 +
\end{bmatrix}
 +
\begin{bmatrix}
 +
D_{0n} \\
 +
\\
 +
\vdots \\
 +
\\
 +
D_{Nn} \\
 +
D_{0n}\\
 +
\\
 +
\vdots \\
 +
\\
 +
D_{Nn}
 +
\end{bmatrix}
 
</math>
 
</math>
 
</center>
 
</center>
and
 
<center>
 
<math>
 
\phi(r,\theta,z)=\sum_{n=-N}^{N}\sum_{m=-2}^{M}b_{mn}I_{n}(\kappa
 
_{m}r)e^{i n\theta}\psi_{m}(z), \;\;r<a
 
</math>
 
</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 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>
 
<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}^{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>
 
</center>
 
and
 
<center>
 
<math>
 
\sum_{m=-2}^{M}(\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
 
</math>
 
</center>
 
It should be noted that the solutions for positive and negative
 
<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]]=
+
for each <math>n</math>.
  
The shallow water theory of [[Zilman_Miloh_2000a|Zilman and Miloh 2000]] can be recovered by
+
Therefore we can find a [[Diffraction Transfer Matrix]] for each <math>n</math>,
simply setting the depth shallow enough that the shallow water theory is valid
+
by setting
and setting <math>M=0</math>. If the shallow water theory is valid then
+
<center><math>
the first three roots of the dispersion equation for the ice will be exactly
+
\forall i \in [0, N], (D_{pn})_{p \in [0, N]} = \delta_{ip}
the same roots found in the shallow water theory by solving the polynomial
+
</math></center>
equation. The system of equations has four unknowns (three under the plate and
+
Then we solve the linear system defined previously, so that we can find the coefficients
one in the open water) exactly as for the theory of [[Zilman_Miloh_2000a|Zilman and Miloh 2000]].
+
<math>(a_{ln})_{l \in [0, N]}</math> for each <math>i</math>.
 
+
This vector represents exactly the <math>i^{th}</math> column of the [[Diffraction Transfer Matrix]],
=Numerical Results=
+
<math>n</math> being set.
 
 
[[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>
+
This method permits to obtain the matrix which links the coefficients of the incident and scattered
<math>-3.29 \!\times\!10^{-4}+1.43 \!\times\!10^{-3}i</math> </td><td>  <math>4.26 \!\times\!10^{-3}
+
potential in the free water domain. Applying this for each <math>n</math>, we finally obtain a 3-dimensional
-3.62 \!\times\!10^{-3}i</math></td><td>
+
matrix for the [[Diffraction Transfer Matrix]].
<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 circular 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/circle_dock_matching_one_n.m circle_dock_matching_one_n.m]
<math>-5.78 \!\times\!10^{-12}-6.21 \!\times\!10^{-12}i</math></td>
+
Note that this problem solves only for a single n.  
<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:Problems with Cylindrical Symmetry]]
 
[[Category:Problems with Cylindrical Symmetry]]
[[Category:Linear Hydroelasticity]]
 
 
[[Category:Eigenfunction Matching Method]]
 
[[Category:Eigenfunction Matching Method]]
 +
[[Category:Pages with Matlab Code]]
 +
[[Category:Complete Pages]]

Latest revision as of 23:59, 16 October 2009


Introduction

We show here a solution for a dock on Finite Depth water, which is circular. This is the three-dimensional analog of the Eigenfunction Matching for a Semi-Infinite Dock.

Governing Equations

We begin with the Frequency Domain Problem. We will use a cylindrical coordinate system, [math]\displaystyle{ (r,\theta,z) }[/math], assumed to have its origin at the centre of the circular plate which has radius [math]\displaystyle{ a }[/math]. The water is assumed to have constant finite depth [math]\displaystyle{ h }[/math] and the [math]\displaystyle{ z }[/math]-direction points vertically upward with the water surface at [math]\displaystyle{ z=0 }[/math] and the sea floor at [math]\displaystyle{ z=-h }[/math]. The boundary value problem can therefore be expressed as

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

[math]\displaystyle{ \phi_{z}=0, \,\, z=-h, }[/math]

[math]\displaystyle{ \phi_{z}=\alpha\phi, \,\, z=0,\,r\gt a, }[/math]

[math]\displaystyle{ \phi_{z}=0, \,\, z=0,\,r\lt a }[/math]

We must also apply the Sommerfeld Radiation Condition as [math]\displaystyle{ r\rightarrow\infty }[/math]. The subscript [math]\displaystyle{ z }[/math] denotes the derivative in [math]\displaystyle{ z }[/math]-direction.

Solution Method

We use separation of variables in the two regions, [math]\displaystyle{ r\lt a }[/math] and [math]\displaystyle{ r\gt a }[/math].

The solution of the problem for the potential in finite water depth can be found by a separation ansatz,

[math]\displaystyle{ \phi (r,\theta,z) =: Y(r,\theta) Z(z).\, }[/math]

Substituting this into the equation for [math]\displaystyle{ \phi }[/math] yields

[math]\displaystyle{ \frac{1}{Y(r,\theta)} \left[ \frac{1}{r} \frac{\partial}{\partial r} \left( r \frac{\partial Y}{\partial r} \right) + \frac{1}{r^2} \frac{\partial^2 Y}{\partial \theta^2} \right] = - \frac{1}{Z(z)} \frac{\mathrm{d}^2 Z}{\mathrm{d} z^2} = k^2. }[/math]

The possible separation constants [math]\displaystyle{ k }[/math] will be determined by the free surface condition and the bed condition.

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 for a Dock

The separation of variables equation for a floating dock

[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) = 0. }[/math]

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

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

We note that

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

where

[math]\displaystyle{ 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]\displaystyle{ \int\nolimits_{-h}^{0}\phi_{n}(z)\psi_{m}(z) \mathrm{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( k_{n} ^{2}-\kappa_{m}^{2}\right) } }[/math]


Separation for Cylindrical Coordinates

We now separate variables, noting that since the problem has circular symmetry we can write the potential as

[math]\displaystyle{ \phi(r,\theta,z)=\frac{\cos k(z+h)}{\cos kh}\sum_{n=-\infty}^{\infty}\rho_{n}(r)e^{i n \theta} }[/math]

We now solve for the function [math]\displaystyle{ \rho_{n}(r) }[/math]. Using Laplace's equation in polar coordinates we obtain

[math]\displaystyle{ \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}}+k^{2}\right) \rho_{n}=0. }[/math]

We can convert this equation to the standard form by substituting [math]\displaystyle{ y=k r }[/math] (provided that [math]\displaystyle{ \mu\neq 0 }[/math]to obtain

[math]\displaystyle{ 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]

The solution of this equation is a linear combination of the modified Bessel functions of order [math]\displaystyle{ n }[/math], [math]\displaystyle{ I_{n}(y) }[/math] and [math]\displaystyle{ K_{n}(y) }[/math] (Abramowitz and Stegun 1964).

Therefore

[math]\displaystyle{ \rho_n(r) = C_1 I_{n}(kr) + C_2 K_{n}(kr)\, }[/math]

for some constants [math]\displaystyle{ C_1 }[/math] and [math]\displaystyle{ C_2 }[/math]

Since the solution must be bounded we know that under the plate the solution will be a linear combination of [math]\displaystyle{ I_{n}(y) }[/math] while outside the plate the solution will be a linear combination of [math]\displaystyle{ K_{n}(y) }[/math]. The case [math]\displaystyle{ \kappa_0 =0 }[/math] is a special case and the solution under the dock is [math]\displaystyle{ (r/a)^{|n|} }[/math]. Therefore the potential can be expanded as

[math]\displaystyle{ \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\gt a }[/math]

and

[math]\displaystyle{ \phi(r,\theta,z)=\sum_{n=-\infty}^{\infty}b_{0n}(r/a)^{|n|} e^{i n\theta}\psi_{0}(z)+ \sum_{n=-\infty}^{\infty}\sum_{m=1}^{\infty}b_{mn} I_{n}(\kappa_{m}r)e^{i n\theta}\psi_{m}(z), \;\;r\lt a }[/math]

where [math]\displaystyle{ a_{mn} }[/math] and [math]\displaystyle{ b_{mn} }[/math] are the coefficients of the potential in the open water and the plate covered region respectively.

Incident potential

The incident potential is a wave of amplitude [math]\displaystyle{ A }[/math] in displacement travelling in the positive [math]\displaystyle{ x }[/math]-direction. The incident potential can therefore be written as

[math]\displaystyle{ \phi^{\mathrm{I}} =e^{k_{0}x}\phi_{0}\left( z\right) =\sum\limits_{n=-\infty}^{\infty} I_{n}(k_{0}r)\phi_{0}\left(z\right) e^{i n \theta} }[/math]

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{ r=a }[/math] have to be equal for each angle and we obtain

[math]\displaystyle{ 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) = b_{0n} \psi_{0}(z) +\sum_{m=1}^{\infty}b_{mn}I_{n}(\kappa_{m}a)\psi_{m}(z) }[/math]

and

[math]\displaystyle{ k_{0}I_{n}^{\prime}(k_{0}a)\phi_{0}\left( z\right) +\sum _{m=0}^{\infty} a_{mn}k_{m}K_{n}^{\prime}(k_{m}a)\phi_{m}\left( z\right) =b_{0n} \frac{|n|}{a} \psi_{0}(z) +\sum_{m=1}^{\infty} b_{mn}\kappa_{m}I_{n}^{\prime}(\kappa_{m}a)\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{ I_{n}(k_{0}a)A_{0}\delta_{0l}+a_{ln}K_{n}(k_{l}a)A_{l} =b_{0n}B_{0l} + \sum_{m=1}^{\infty}b_{mn}I_{n}(\kappa_{m}a)B_{ml} }[/math]

and

[math]\displaystyle{ k_{0}I_{n}^{\prime}(k_{0}a)A_{0}\delta_{0l}+a_{ln}k_{l}K_{n}^{\prime }(k_{l}a)A_{l} = b_{0n}B_{0l}\frac{|n|}{a} + \sum_{m=1}^{\infty}b_{mn}\kappa_{m}I_{n}^{\prime}(\kappa_{m}a)B_{ml} }[/math]

Numerical Solution

To solve the system of equations we set the upper limit of [math]\displaystyle{ l }[/math] to be [math]\displaystyle{ M }[/math].

A Simple Method To Calculate The Diffraction Transfer Matrix For The Case Of A Circular Dock

Let's consider an incident wave whose potential has the following expression

[math]\displaystyle{ \phi^\mathrm{I} (r,\theta,z) = \sum_{n=0}^{\infty} \phi_n(z) \sum_{\nu = - \infty}^{\infty} D_{n\nu} I_\nu (k_n r) \mathrm{e}^{\mathrm{i}\nu \theta}. }[/math]

Such an incident potential is found in the Kagemoto and Yue Interaction Theory, where it can be written as the sum of an ambient incident potential and the scattered potentials of the other bodies, which are interpretated as incident potentials for the studied body.

We can apply the same eigenfunction matching that previously, considering the potential and its normal derivative continuous at [math]\displaystyle{ r=a }[/math]. Thus the potential and its normal derivative expressed at each side of this value of the radius have to be equal. We obtain the relationships

[math]\displaystyle{ \sum_{m=0}^{\infty} D_{mn} I_n (k_m a) \phi_m(z) + \sum_{m=0}^{\infty} a_{mn} K_{n}(k_{m}a)\phi_{m}\left( z\right) = b_{0n} \psi_{0}(z) +\sum_{m=1}^{\infty}b_{mn}I_{n}(\kappa_{m}a)\psi_{m}(z) }[/math]

and

[math]\displaystyle{ \sum_{m=0}^{\infty} D_{mn} k_m I'_n (k_m a) \phi_m(z) +\sum _{m=0}^{\infty} a_{mn}k_{m}K_{n}^{\prime}(k_{m}a)\phi_{m}\left( z\right) =b_{0n} \frac{|n|}{a} \psi_{0}(z) +\sum_{m=1}^{\infty} b_{mn}\kappa_{m}I_{n}^{\prime}(\kappa_{m}a)\psi _{m}(z) }[/math]

for each [math]\displaystyle{ n }[/math]. We solve these equations with the same method that before, 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{ D_{ln} I_n (k_l a) A_l+a_{ln}K_{n}(k_{l}a)A_{l} =b_{0n}B_{0l} + \sum_{m=1}^{\infty}b_{mn}I_{n}(\kappa_{m}a)B_{ml}, \ \ \ (1) }[/math]

and

[math]\displaystyle{ D_{ln} k_l I'_n (k_l a) A_l+a_{ln}k_{l}K_{n}^{\prime }(k_{l}a)A_{l} = b_{0n}B_{0l}\frac{|n|}{a} + \sum_{m=1}^{\infty}b_{mn}\kappa_{m}I_{n}^{\prime}(\kappa_{m}a)B_{ml}, \ \ \ (2) }[/math]

The Diffraction Transfer Matrix maps the coefficients of the incident wave with the coefficients of the scattered wave within the open water domain. The relation which links these two coefficients can be written as follows

[math]\displaystyle{ a_{mn}=\sum_{l=0}^{\infty} T_{lmn} D_{ln} }[/math]

From the equations (1) and (2) we can write a linear system of equation, limiting the number of modes of the dispersion equation to [math]\displaystyle{ N }[/math] real ones

[math]\displaystyle{ \begin{bmatrix} \begin{bmatrix} -A_0 K_n(k_0 a)&0 \quad \cdots&0\\ 0&&\\ \vdots&-A_l K_n(k_l a)&\vdots\\ &&0\\ 0&\cdots \quad 0 & -A_N K_n(k_N a) \end{bmatrix} & \begin{bmatrix} B_{00}&I_n(\kappa_{1}a) B_{10}&\cdots&I_n(\kappa_{N}a) B_{N0}\\ B_{01}&I_n(\kappa_{1}a) B_{11}&&\\ \vdots&\vdots&&\vdots\\ &&&\\ B_{0N}&I_n(\kappa_{1}a) B_{1N}&\cdots&I_n(\kappa_{N}a) B_{NN} \end{bmatrix} \\ \begin{bmatrix} -k_0 K'_n(k_0 a) A_0&0 \quad \cdots&0\\ 0&&\\ \vdots&-k_l K'_n(k_l a) A_l&\vdots\\ &&0\\ 0&\cdots \quad 0 & -k_N K'_n(k_N a) A_N \end{bmatrix} & \begin{bmatrix} B_{00}\frac{|n|}{a}&\kappa_{1} I'_n(\kappa_{1}a) B_{10}&\cdots&\kappa_{N} I'_n(\kappa_{N}a) B_{N0}\\ B_{01}\frac{|n|}{a}&\kappa_{1} I'_n(\kappa_{1}a) B_{11}&&\\ \vdots&\vdots&&\vdots\\ &&&\\ B_{0N}\frac{|n|}{a}&\kappa_{1} I'_n(\kappa_{1}a) B_{1N}&\cdots&\kappa_{N} I'_n(\kappa_{N}a) B_{NN} \end{bmatrix} \end{bmatrix} \begin{bmatrix} a_{0n} \\ \\ \vdots \\ \\ a_{Nn} \\ b_{0n}\\ \\ \vdots \\ \\ b_{Nn} \end{bmatrix} = \begin{bmatrix} \begin{bmatrix} I_n(k_0 a) A_0&0 \quad \cdots&0\\ 0&&\\ \vdots&I_n(k_l a) A_l&\vdots\\ &&0\\ 0&\cdots \quad 0 & I_n(k_N a) A_N \end{bmatrix} \\ \begin{bmatrix} k_0 I'_n(k_0 a) A_0&0 \quad \cdots&0\\ 0&&\\ \vdots&k_l I'_n(k_l a) A_l&\vdots\\ &&0\\ 0&\cdots \quad 0 &k_N I'_n(k_N a) A_N \end{bmatrix} \end{bmatrix} \begin{bmatrix} D_{0n} \\ \\ \vdots \\ \\ D_{Nn} \\ D_{0n}\\ \\ \vdots \\ \\ D_{Nn} \end{bmatrix} }[/math]

for each [math]\displaystyle{ n }[/math].

Therefore we can find a Diffraction Transfer Matrix for each [math]\displaystyle{ n }[/math], by setting

[math]\displaystyle{ \forall i \in [0, N], (D_{pn})_{p \in [0, N]} = \delta_{ip} }[/math]

Then we solve the linear system defined previously, so that we can find the coefficients [math]\displaystyle{ (a_{ln})_{l \in [0, N]} }[/math] for each [math]\displaystyle{ i }[/math]. This vector represents exactly the [math]\displaystyle{ i^{th} }[/math] column of the Diffraction Transfer Matrix, [math]\displaystyle{ n }[/math] being set.

This method permits to obtain the matrix which links the coefficients of the incident and scattered potential in the free water domain. Applying this for each [math]\displaystyle{ n }[/math], we finally obtain a 3-dimensional matrix for the Diffraction Transfer Matrix.

Matlab Code

A program to calculate the coefficients for circular dock problems can be found here circle_dock_matching_one_n.m Note that this problem solves only for a single n.

Additional code

This program requires