Difference between revisions of "Green Function Method for a Floating Body on the Surface"

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{{incomplete pages}}
+
{{complete pages}}
  
 
== Introduction ==
 
== Introduction ==
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{{finite floating body on the surface frequency domain}}
 
{{finite floating body on the surface frequency domain}}
  
== Transformation using [[Eigenfunctions for a Uniform Free Beam]] ==
 
  
{{equations for a eigenfunction of a free beam}}
 
  
This solution is discussed further in [[Eigenfunctions for a Free Beam]]. <br /><br />
+
== Equation in Terms of the Modes ==
Using the expression <math>\partial_n \phi =\partial_t w</math>, we can form
 
<center>
 
<math>
 
\frac{\partial \phi}{\partial z} = i\omega \sum_{n=0}^{\infty} \zeta_n X_n
 
</math>
 
</center>
 
where <math>\zeta_n \,</math> are coefficients to be evaluated.
 
  
== Equation in Terms of the Modes of the Plate ==
+
The equations are
 
+
{{general dock type body equations}}
Under these assumptions, the equations become
 
 
<center><math>
 
<center><math>
\Delta\phi  =0,\,\,-h<z<0,
+
\mathrm{i}\omega\sum_{n=0,1}\zeta_{n}X_{n}   =\partial_{z}\phi,\,\,x\in
</math></center>
+
(-L,L),\,\, z=0,
<center><math>
 
\partial_{z}\phi  =0,\,\,z=-h,
 
</math></center>
 
<center><math>
 
\alpha\phi  =\partial_{z}\phi,\,\,x\notin(-L,L),\ \ z=0,
 
 
</math></center>
 
</math></center>
 
<center><math>
 
<center><math>
i\omega\sum_{n=0}^{\infty}\zeta_{n}X_{n}   =\partial_{z}\phi,\,\,x\in
+
\left(- \alpha\gamma + 1\right)  \zeta_{n}=-i\omega
 +
\int_{-L}^{L}\phi X_{n}\mathrm{d}x, \,\,x\in
 
(-L,L),\,\, z=0,
 
(-L,L),\,\, z=0,
</math></center>
 
<center><math>
 
\sum_{n=0}^{\infty}\zeta_{n}\left(  1+\beta\lambda_{n}^{4}\right)
 
X_{n}-\alpha\gamma\sum_{n=0}^{\infty}\zeta_{n}X_{n}    = -i\omega
 
\phi,\,\,x\in(-L,L),\,\, z=0.
 
 
</math></center>
 
</math></center>
 
We solve for the potential (and displacement) as the sum of
 
We solve for the potential (and displacement) as the sum of
Line 57: Line 38:
 
{{diffraction potential equations for a dock}}
 
{{diffraction potential equations for a dock}}
 
{{radiation condition for diffracted potential}}
 
{{radiation condition for diffracted potential}}
 
As the plate is floating on the surface, we can denote it as follows:
 
<center><math>
 
\phi^{\rm I}|_{z=0}  =  e^{-i kx} \,
 
</math></center>
 
 
 
We now consider the scattered potentials <math>\phi^{\mathrm{S}}</math>.  The relationship between scattered potentials, diffracted potentials and the incident wave are as follows:
 
<center><math>
 
\phi^{\mathrm{D}}=\phi^{\mathrm{I}}+\phi^{\mathrm{S}} \,
 
</math></center>
 
from this, we can construct the following conditions:
 
<center><math>
 
\Delta\phi^{\mathrm{S}}    =0,\,\,-h<z<0,
 
</math></center>
 
<center><math>
 
\partial_{z}\phi^{\mathrm{S}}    =0,\,\,z=-h,
 
</math></center>
 
<center><math>
 
\partial_{z}\phi^{\mathrm{S}}    =\alpha\phi^{\mathrm{S}},\,\,x\notin(-L,L),\, \,
 
z=0
 
</math></center>
 
<center><math>
 
\partial_{z}\phi^{\mathrm{S}}    = -\partial_{z}\phi^{\mathrm{I}},\,\,x\in(-L,L),\,\,z=0.
 
</math></center> <br /><br />
 
  
 
We now consider the radiation potentials <math>\phi^{\mathrm{R}}</math>.  We can express the radiation potential as:
 
We now consider the radiation potentials <math>\phi^{\mathrm{R}}</math>.  We can express the radiation potential as:
 
<center><math>
 
<center><math>
\phi^{\mathrm{R}}=\sum_{n=0}^{\infty}\zeta_n \phi_n^{\mathrm{R}}
+
\phi^{\mathrm{R}}=\sum_{n=0,1}\zeta_n \phi_n^{\mathrm{R}}
 
</math></center>
 
</math></center>
 
which satisfy the following equations
 
which satisfy the following equations
<center><math>
+
{{radiation potential equations for a dock like structure}}
\Delta\phi_n^{\mathrm{R}}    =0,\,\,-h<z<0,
 
</math></center>
 
<center><math>
 
\partial_{z}\phi_n^{\mathrm{R}}    =0,\,\,z=-h,
 
</math></center>
 
<center><math>
 
\partial_{z}\phi_n^{\mathrm{R}}    =\alpha\phi_n^{\mathrm{R}},\,\,x\notin(-L,L),\, \,
 
z=0
 
</math></center>
 
<center><math>
 
\partial_{z}\phi_n^{\mathrm{R}}    = i\omega X_{n},\,\,x\in(-L,L),\,\,z=0.
 
</math></center>
 
The radiation condition for the radiation potential is
 
<center><math>
 
\frac{\partial\phi_n^{\mathrm{R}}}{\partial x}\pm ik\phi_n^{\mathrm{R}}=0,\,\,\mathrm{as}
 
\,\,x\rightarrow\pm\infty.
 
</math></center>
 
 
Therefore we find the potential as
 
Therefore we find the potential as
 
<center><math>
 
<center><math>
\phi=\phi^{\mathrm{D}} +\sum_{n=0}^{\infty}\zeta_{n}\phi_n^{\mathrm{R}},
+
\left( - \alpha\gamma + 1\right)  \zeta_{n}=-i\omega
</math></center>
 
so that
 
<center><math>
 
\sum_{n=0}^{\infty}\left( 1+\beta\lambda_{n}^{4} - \alpha\gamma\right)
 
\zeta_{n}X_{n}=-i\omega \left( \phi^{\mathrm{D}}+\sum_{n=0}^{\infty}\zeta_{n}\phi_n^{\mathrm{R}} \right).
 
</math></center>
 
If we multiply by <math>X_m</math> and take an inner product over the plate we obtain
 
<center><math>
 
\left(  1+\beta\lambda_{n}^{4} - \alpha\gamma\right)  \zeta_{n}=-i\omega
 
 
\int_{-L}^{L}\phi^{\mathrm{D}} X_{n}\mathrm{d}x +
 
\int_{-L}^{L}\phi^{\mathrm{D}} X_{n}\mathrm{d}x +
\sum_{m=0}^{\infty}\left(\omega^2 a_{mn}(\omega) - i\omega b_{mn}(\omega)\right)
+
\sum_{m=0,1}\left(\omega^2 a_{mn}(\omega) - i\omega b_{mn}(\omega)\right)
 
\zeta_{m},
 
\zeta_{m},
 
</math></center>
 
</math></center>
Line 126: Line 56:
 
\omega^2 a_{mn}(\omega) -i\omega b_{mn}(\omega) = - i\omega\int_{-L}^{L}\phi_m^{\mathrm{R}}X_{n}\mathrm{d}x,
 
\omega^2 a_{mn}(\omega) -i\omega b_{mn}(\omega) = - i\omega\int_{-L}^{L}\phi_m^{\mathrm{R}}X_{n}\mathrm{d}x,
 
</math></center>
 
</math></center>
and they are referred to as the added mass and damping coefficients (see [[Added-Mass, Damping Coefficients And Exciting Forces]]
+
and they are referred to as the added mass and damping coefficients (see [[Added-Mass, Damping Coefficients And Exciting Forces]])
for the equivalent definition for a rigid body).
 
 
respectively.
 
respectively.
This equation is solved by truncating the number of modes.
+
 
 +
Note that for this simple example the added mass and damping matrices are diagonal.
  
 
== Solution for the Radiation and Diffracted Potential ==
 
== Solution for the Radiation and Diffracted Potential ==
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and  
 
and  
<center><math>
+
 
\phi_n^{\mathrm{R}}(x) =  \int_{-L}^{L}G(x,\xi)
+
{{Green's function equations for the radiation potential for a dock like structure}}
\left(
 
\alpha\phi_n^{\mathrm{R}}(\xi) - i\omega X_n(\xi)
 
\right)\mathrm{d} \xi
 
</math></center>
 
  
 
== Reflection and Transmission Coefficients ==
 
== Reflection and Transmission Coefficients ==
Line 149: Line 75:
 
== Matlab Code ==
 
== Matlab Code ==
  
{{elastic plate modes code}}
+
{{rigid plate modes code}}
 
 
== Alternative Solution Method using Green Functions for a Uniform Plate ==
 
 
 
We can also solve the equation by a closely related method which was given in
 
[[Meylan and Squire 1994]].
 
We can transform the equations to
 
<center><math>
 
\phi(x) = \phi^{\rm I}(x) + \int_{-L}^{L}G(x,\xi)
 
\left(
 
\alpha\phi(\xi) - \partial_z\phi(\xi)
 
\right)\mathrm{d} \xi
 
</math></center>
 
 
 
Expanding as before
 
<center>
 
<math>
 
\partial_z \phi = i\omega \sum \xi_n X_n
 
</math>
 
</center>
 
we obtain
 
<center><math>
 
-i\omega \phi = \sum \left(\beta\lambda_n^4 - \gamma\alpha + 1\right)\xi_n X_n
 
</math>
 
</center>
 
This leads to the following equation
 
<center>
 
<math>
 
\partial_z\phi(x) = \frac{1}{\alpha} \int_{-L}^{L} \frac{X_n(x)X_n(\xi)}{\beta\lambda_n^4 - \gamma\alpha + 1} \phi(\xi)\mathrm{d}\xi
 
</math>
 
</center>
 
or
 
<center>
 
<math>
 
\partial_z\phi(x) = \frac{1}{\alpha} \int_{-L}^{L} g(x,\xi) \phi(\xi)\mathrm{d}\xi
 
</math>
 
</center>
 
where
 
<center>
 
<math>
 
g(x,\xi) = \frac{X_n(x)X_n(\xi)}{\beta\lambda_n^4 - \gamma\alpha + 1}
 
</math>
 
</center>
 
which is the Green function for the plate.
 
 
 
[[Category:Floating Elastic Plate]]
 

Latest revision as of 00:31, 24 September 2009


Introduction

The problem of a two-dimensional floating body which has negligible submergence is solved using a green function. The problem of a dock is solved in Green Function Method for a Finite Dock and for a floating elastic plate is solved in Green Function Methods for Floating Elastic Plates

Equations for a Finite Plate in Frequency Domain

We consider the problem of small-amplitude waves which are incident on finite floating body occupying water surface for [math]\displaystyle{ -L\lt x\lt L }[/math]. The submergence of the body is considered negligible. We assume that the problem is invariant in the [math]\displaystyle{ y }[/math] direction.

[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 -L,\,\,{\rm or}\,\,x\gt L }[/math]

where [math]\displaystyle{ \alpha = \omega^2 }[/math]. The equation under the body consists of the kinematic condition

[math]\displaystyle{ \mathrm{i}\omega w = \partial_z \phi,\,\,\, z=0,\,\,-L\leq x\leq L }[/math]

plus the kinematic condition. The body motion is expanded using the modes for heave and pitch. Using the expression [math]\displaystyle{ \partial_n \phi =\partial_t w }[/math], we can form

[math]\displaystyle{ \frac{\partial \phi}{\partial z} = i\omega \sum_{n=0,1} \xi_n X_n(x) }[/math]

where [math]\displaystyle{ \xi_n \, }[/math] are coefficients to be evaluated. The functions [math]\displaystyle{ X_n(x) }[/math] are given by

[math]\displaystyle{ X_0 = \frac{1}{\sqrt{2L}} }[/math]

and

[math]\displaystyle{ X_1 = \sqrt{\frac{3}{2L^3}} x }[/math]

Note that this numbering is non-standard for a floating body and comes from Eigenfunctions for a Uniform Free Beam.


Equation in Terms of the Modes

The equations are

[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 -L,\,\,{\rm or}\,\,x\gt L }[/math]
[math]\displaystyle{ \mathrm{i}\omega\sum_{n=0,1}\zeta_{n}X_{n} =\partial_{z}\phi,\,\,x\in (-L,L),\,\, z=0, }[/math]
[math]\displaystyle{ \left(- \alpha\gamma + 1\right) \zeta_{n}=-i\omega \int_{-L}^{L}\phi X_{n}\mathrm{d}x, \,\,x\in (-L,L),\,\, z=0, }[/math]

We solve for the potential (and displacement) as the sum of the diffracted and radiation potentials in the standard way, as for a rigid body.

[math]\displaystyle{ \phi=\phi^{\mathrm{D}}+\phi^{\mathrm{R}} ,\, }[/math]

We begin with the diffraction potential [math]\displaystyle{ \phi^{\mathrm{D}} }[/math] which satisfies the following equations

[math]\displaystyle{ \Delta\phi^{\mathrm{D}} =0,\,\,-h\lt z\lt 0, }[/math]
[math]\displaystyle{ \partial_{z}\phi^{\mathrm{D}} =0,\,\,z=-h, }[/math]
[math]\displaystyle{ \partial_{z}\phi^{\mathrm{D}} =\alpha \phi^{\mathrm{D}},\,\,x\notin(-L,L),\,\, z=0, }[/math]
[math]\displaystyle{ \partial_{z}\phi^{\mathrm{D}} =0,\,\,x\in(-L,L),\,\,z=0. }[/math]

[math]\displaystyle{ \phi^{\mathrm{D}} }[/math] satisfies the Sommerfeld Radiation Condition

[math]\displaystyle{ \frac{\partial}{\partial x} \left(\phi^{\mathrm{D}}-\phi^{\rm I} \right) \pm k_0\left( \phi^{\mathrm{D}}-\phi^{\rm I}\right) = 0 ,\,\,\mathrm{as} \,\,x\rightarrow\infty. }[/math]

[math]\displaystyle{ \phi^{\mathrm{I}}\, }[/math] is a plane wave travelling in the [math]\displaystyle{ x }[/math] direction,

[math]\displaystyle{ \phi^{\mathrm{I}}(x,z)=A \phi_0(z) e^{\mathrm{i} k x} \, }[/math]

where [math]\displaystyle{ A }[/math] is the wave amplitude (in potential) [math]\displaystyle{ \mathrm{i} k }[/math] is the positive imaginary solution of the Dispersion Relation for a Free Surface (note we are assuming that the time dependence is of the form [math]\displaystyle{ \exp(-\mathrm{i}\omega t) }[/math]) and

[math]\displaystyle{ \phi_0(z) =\frac{\cosh k(z+h)}{\cosh k h} }[/math]

We now consider the radiation potentials [math]\displaystyle{ \phi^{\mathrm{R}} }[/math]. We can express the radiation potential as:

[math]\displaystyle{ \phi^{\mathrm{R}}=\sum_{n=0,1}\zeta_n \phi_n^{\mathrm{R}} }[/math]

which satisfy the following equations

[math]\displaystyle{ \Delta\phi_n^{\mathrm{R}} =0,\,\,-h\lt z\lt 0, }[/math]
[math]\displaystyle{ \partial_{z}\phi_n^{\mathrm{R}} =0,\,\,z=-h, }[/math]
[math]\displaystyle{ \partial_{z}\phi_n^{\mathrm{R}} =\alpha\phi_n^{\mathrm{R}},\,\,x\notin(-L,L),\, \, z=0 }[/math]
[math]\displaystyle{ \partial_{z}\phi_n^{\mathrm{R}} = i\omega X_{n},\,\,x\in(-L,L),\,\,z=0. }[/math]

The radiation condition for the radiation potential is

[math]\displaystyle{ \frac{\partial\phi_n^{\mathrm{R}}}{\partial x}\pm ik\phi_n^{\mathrm{R}}=0,\,\,\mathrm{as} \,\,x\rightarrow\pm\infty. }[/math]

Therefore we find the potential as

[math]\displaystyle{ \left( - \alpha\gamma + 1\right) \zeta_{n}=-i\omega \int_{-L}^{L}\phi^{\mathrm{D}} X_{n}\mathrm{d}x + \sum_{m=0,1}\left(\omega^2 a_{mn}(\omega) - i\omega b_{mn}(\omega)\right) \zeta_{m}, }[/math]

where the functions [math]\displaystyle{ a_{mn}(\omega) }[/math] and [math]\displaystyle{ b_{mn}(\omega) }[/math] are given by

[math]\displaystyle{ \omega^2 a_{mn}(\omega) -i\omega b_{mn}(\omega) = - i\omega\int_{-L}^{L}\phi_m^{\mathrm{R}}X_{n}\mathrm{d}x, }[/math]

and they are referred to as the added mass and damping coefficients (see Added-Mass, Damping Coefficients And Exciting Forces) respectively.

Note that for this simple example the added mass and damping matrices are diagonal.

Solution for the Radiation and Diffracted Potential

We use the Free-Surface Green Function for two-dimensional waves, with singularity at the water surface since we are only interested in its value at [math]\displaystyle{ z=0 }[/math] (details about this method can be found in Integral Equation for the Finite Depth Green Function at Surface). Using this we can transform the system of equations to

[math]\displaystyle{ \phi^{\mathrm{D}}(x) = \phi^{\mathrm{I}}(x) + \int_{-L}^{L}G(x,\xi) \alpha\phi^{\mathrm{D}}(\xi) \mathrm{d} \xi }[/math]

and

[math]\displaystyle{ \phi_n^{\mathrm{R}}(x) = \int_{-L}^{L}G(x,\xi) \left( \alpha\phi_n^{\mathrm{R}}(\xi) - i\omega X_n(\xi) \right)\mathrm{d} \xi }[/math]

Reflection and Transmission Coefficients

The Reflection and Transmission Coefficients represent the ratio of the amplitude of the reflected or transmitted wave to the amplitude of the incident wave. Conservation of energy means that [math]\displaystyle{ |R|^2+|T|^2=1\, }[/math].

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 \leq 0 \, }[/math] and [math]\displaystyle{ -\infty \leq x \leq \infty \, }[/math] or [math]\displaystyle{ -N \leq x \leq N\, }[/math]

We can calculate the Reflection and Transmission coefficients by applying Green's theorem to [math]\displaystyle{ \phi\, }[/math] and [math]\displaystyle{ \phi^{\mathrm{I}}\, }[/math] [math]\displaystyle{ \phi^{\mathrm{I}}\, }[/math] is a plane wave travelling in the [math]\displaystyle{ x }[/math] direction,

[math]\displaystyle{ \phi^{\mathrm{I}}(x,z)=A \phi_0(z) e^{\mathrm{i} k x} \, }[/math]

where [math]\displaystyle{ A }[/math] is the wave amplitude (in potential) [math]\displaystyle{ \mathrm{i} k }[/math] is the positive imaginary solution of the Dispersion Relation for a Free Surface (note we are assuming that the time dependence is of the form [math]\displaystyle{ \exp(-\mathrm{i}\omega t) }[/math]) and

[math]\displaystyle{ \phi_0(z) =\frac{\cosh k(z+h)}{\cosh k h} }[/math]

We assume that [math]\displaystyle{ A=1 }[/math]. This gives us

[math]\displaystyle{ \iint_{\Omega}(\phi\Delta\phi^{\mathrm{I}} - \phi^{\mathrm{I}}\Delta\phi)\mathrm{d}x\mathrm{d}z = \int_{\partial\Omega}(\phi \partial_n \phi^{\rm I} - \phi^{\rm I}\partial_n\phi)\mathrm{d}s = 0, }[/math]

This means that (using the far field behaviour of the potential [math]\displaystyle{ \phi }[/math])

[math]\displaystyle{ \int_{\partial\Omega_{B}} (\phi \partial_n \phi^{\rm I} - \phi^{\rm I}\partial_n\phi)\mathrm{d}s + 2k_0 R \int_{-h}^{0} \left(\phi_0(z)\right)^2 \mathrm{d}z = 0, }[/math]

For the present case the body is present only on the surface and we therefore have

[math]\displaystyle{ \int_{-L}^{L} e^{-k_0 x} \left(\alpha \phi(x) - \partial_n \phi(x)\right)\mathrm{d}x + 2k_0 R \int_{-h}^{0} \left(\phi_0(z)\right)^2 \mathrm{d}z = 0 }[/math]

Therefore

[math]\displaystyle{ R = -\frac{\int_{-L}^{L} e^{-k_0 x} \left(\alpha \phi(x) - \partial_n \phi(x)\right)\mathrm{d}x } {2 k_0 \int_{-h}^{0} \left(\phi_0(z)\right)^2 \mathrm{d}z}. }[/math]

and using a wave incident from the right we obtain

[math]\displaystyle{ T = 1 - \frac{\int_{-L}^{L} e^{k_0 x} \left(\alpha \phi(x) - \partial_n \phi(x)\right)\mathrm{d}x } {2 k_0 \int_{-h}^{0} \left(\phi_0(z)\right)^2 \mathrm{d}z}. }[/math]

Note that an expression for the integral in the denominator can be found in Eigenfunction Matching for a Semi-Infinite Dock

Matlab Code

A program to calculate the solution in elastic modes can be found here

rigid_plate_modes.m

Additional code

This program requires