Difference between revisions of "Free-Surface Green Function for a Floating Elastic Plate"

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 +
 
This is a special version of the free-surface Green function which applied when the [[Floating Elastic Plate]]
 
This is a special version of the free-surface Green function which applied when the [[Floating Elastic Plate]]
boundary condition applies at the free-surface
+
boundary condition applies at the free-surface. It reduced to the [[Free-Surface Green Function]] in the limit as the plate terms tend to zero and to the Green function for an infinite plate in the limit as the water terms tend to zero.
  
 
= Two Dimensions =
 
= Two Dimensions =
  
The Green function <math>G(x,z)</math> representing  
+
== Green function with singularity on the surface ==
outgoing waves as <math>|x|\rightarrow \infty</math> satisfies
+
For the case of a floating thin plate we almost always make the assumption of shallow draft and the Green function satisfies the following equations
 +
The Green function <math>G(x,x^{\prime},z)</math> representing outgoing waves as <math>|x|\rightarrow \infty</math> satisfies
 
<center><math>
 
<center><math>
 
\nabla^2 G = 0, -H<z<0,  
 
\nabla^2 G = 0, -H<z<0,  
Line 15: Line 18:
 
{\left( \beta \frac{\partial^4}{\partial x^4} -  
 
{\left( \beta \frac{\partial^4}{\partial x^4} -  
 
\gamma\alpha + 1\right)\frac{\partial G}{\partial z} - \alpha G}
 
\gamma\alpha + 1\right)\frac{\partial G}{\partial z} - \alpha G}
  = \delta(x), z=0,
+
  = \delta(x-x^{\prime}), z=0,
 
</math></center>
 
</math></center>
(note we are only considering singularities at the free surface).  
+
(note we are only considering singularities at the free surface). It is important to realise that these equations are based on the [[Frequency Domain Problem]] and that the exact form of the Green function is dependent on whether we have <math>e^{i\omega t}</math> or <math>e^{-i\omega t}</math> dependence. In what follows we assume <math>e^{i\omega t}</math> dependence.  
  
This problem can be solved (a solution is given below) to give
+
This problem can be solved (a solution is given below and also in [[Evans and Porter 2006]]) to give
 
<center><math>
 
<center><math>
G(x,z) = -i\sum_{n=-2}^\infty\frac{\sin{(k_n H)}\cos{(k_n(z+H))}}{2\alpha C(k_n)}e^{-k_n|x|},
+
G(x,x^{\prime},z) = -\sum_{n=-2}^\infty\frac{\sin{(k_n H)}\cos{(k_n(z+H))}}{2\alpha C(k_n)}e^{-k_n|x-x^{\prime}|},
 
</math></center>
 
</math></center>
 
where
 
where
 
<center><math>
 
<center><math>
C(k_n)=\frac{1}{2}\left(h + \frac{(5\beta k_n^4 + 1 - \alpha\gamma)\sin^2{(k_n H)}}{\alpha}\right),
+
C(k_n)=\frac{1}{2}\left(h - \frac{(5\beta k_n^4 + 1 - \alpha\gamma)\sin^2{(k_n H)}}{\alpha}\right),
 
</math></center>
 
</math></center>
 
and <math>k_n</math> are the solutions of the [[Dispersion Relation for a Floating Elastic Plate]],
 
and <math>k_n</math> are the solutions of the [[Dispersion Relation for a Floating Elastic Plate]],
Line 33: Line 36:
 
</math></center>
 
</math></center>
 
with <math>n=-1,-2</math> corresponding to the complex solutions with positive real part,  
 
with <math>n=-1,-2</math> corresponding to the complex solutions with positive real part,  
<math>n=0</math> corresponding to the imaginary solution with positive imaginary part and
+
<math>n=0</math> corresponding to the imaginary solution with positive imaginary part, since we have assumed <math>e^{i\omega t}</math> dependence (it would be of negative imaginary part if we had chosen <math>e^{-i\omega t}</math> dependence)
 
<math>n>0</math> corresponding to the real solutions with positive real part.
 
<math>n>0</math> corresponding to the real solutions with positive real part.
 
We can also write the Green function as
 
We can also write the Green function as
Line 41: Line 44:
 
\cosh q_n\left(  z+H\right)  .
 
\cosh q_n\left(  z+H\right)  .
 
</math></center>
 
</math></center>
where <math>r=|x|</math>, <math>q_n=ik_n</math> and we have used a different non-dimensionalisation so that
+
where <math>r=|x-x^{\prime}|</math>, <math>q_n=ik_n</math> and we have used a different non-dimensionalisation so that
 
the [[Dispersion Relation for a Floating Elastic Plate]] is written as
 
the [[Dispersion Relation for a Floating Elastic Plate]] is written as
 
<center><math>  
 
<center><math>  
Line 51: Line 54:
 
R\left(  q_n\right)  =\frac{\omega^{2}q_n}{\omega^{2}\left(  5q_n^{4}+u\right)
 
R\left(  q_n\right)  =\frac{\omega^{2}q_n}{\omega^{2}\left(  5q_n^{4}+u\right)
 
+H\left[  \left(  q_n^{5}+uq\right)  ^{2}-\omega^{4}\right]  }.  
 
+H\left[  \left(  q_n^{5}+uq\right)  ^{2}-\omega^{4}\right]  }.  
 +
</math></center>
 +
 +
== Green function with singularity in the water ==
 +
 +
We can define the Green function <math>G(x,x^{\prime},z)</math>, representing
 +
outgoing waves as <math>|x|\rightarrow \infty</math>, which satisfies
 +
<center><math>
 +
\nabla^2 G = \delta(x-x^{\prime})\delta(z-z^{\prime}), -H<z<0,
 +
</math></center>
 +
<center><math>
 +
\frac{\partial G}{\partial z} =0, z=-H,
 +
</math></center>
 +
<center><math>
 +
{\left( \beta \frac{\partial^4}{\partial x^4} -
 +
\gamma\alpha + 1\right)\frac{\partial G}{\partial z} - \alpha G}
 +
= 0, z=0.
 +
</math></center>
 +
Note that this Green function reduces the [[Free-Surface Green Function]] in the limit as <math>\beta</math> and <math>\gamma</math> tend to zero.
 +
 +
This problem can be solved to give [[Evans and Porter 2003]]
 +
<center><math>
 +
G(x,x^{\prime},z) = -\sum_{n=-2}^\infty\frac{\cos{(k_n(z^{\prime}+H))}\cos{(k_n(z+H))}}{2k_n C(k_n)}e^{-k_n|x-x^{\prime}|},
 +
</math></center>
 +
where <math>
 +
C(k_n)</math>
 +
and <math>k_n</math> are as before.
 +
 +
== Relationship between the two Green functions ==
 +
 +
If we denote the Green function with the singularity at the free-surface by <math>G_s</math> and the Green function with
 +
the singularity in the water by <math>G_w</math>then we have the following relationship between them
 +
<center><math>
 +
G_s = -\frac{1}{\alpha}\left. \frac{\partial G_w}{\partial z^{\prime}} \right|_{z^{\prime} =0}
 
</math></center>
 
</math></center>
  
Line 69: Line 105:
 
==Previous Work ==
 
==Previous Work ==
  
Radially symmetric waves in a thin plate without
+
In the following sections, the
the hydrodynamic support, have studied by [[Sneddon 1945]] using the two
 
dimensional radial Fourier transform. This integral transform is commonly
 
called the Hankel transform
 
<center><math>
 
\hat{w}\left(  \gamma\right)  =\frac{1}{2\pi}\int_{0}^{\infty}w\left(
 
r\right)  rJ_{0}\left(  \gamma r\right)  dr.
 
</math></center>
 
A complete analytical solution for deflection of an infinite plate due to a
 
localized static load was derived by [[Wyman 1950]] by choosing the
 
bounded solutions of
 
<center><math>
 
w_{rr}+\frac{1}{r}w_{r}+\mathrm{i}w=0
 
</math></center>
 
whose solutions satisfy the radial form of the (non-dimensional) thin plate
 
equation
 
<center><math>
 
\left[  \frac{\partial^{2}}{\partial r^{2}}+\frac{1}{r}\frac{\partial
 
}{\partial r}\right]  ^{2}w+w=0.
 
</math></center>
 
 
 
If we consider a plate floating on water we have to
 
to solve Laplace's equation, and thus the application of the Fourier
 
transform is not as straightforward as in the two examples cited (Sneddon
 
[[Sneddon 1945]] and Wyman [[Wyman 1950]]). Response of an infinite thin
 
elastic plate under fluid loading is studied in a series of papers by Crighton
 
[[Crighton 1972]], [[Crighton 1977]], [[Crighton 1979]]. In the following sections, the
 
 
deflection of a plate is expressed by the infinite number of modes that exist
 
deflection of a plate is expressed by the infinite number of modes that exist
 
in the vibrating plate floating on the water, instead of a finite number of
 
in the vibrating plate floating on the water, instead of a finite number of
Line 101: Line 111:
 
restraints. The fact that waves in an incompressible fluid can be expressed by
 
restraints. The fact that waves in an incompressible fluid can be expressed by
 
an infinite series of natural modes is shown by [[John 1949]] and [[John 1950]]
 
an infinite series of natural modes is shown by [[John 1949]] and [[John 1950]]
and discuss in [[Cylindrical Eigenfunction Expansion]]
+
and discussed in [[Cylindrical Eigenfunction Expansion]]
Each mode being a Hankel function of the first kind in the horizontal. It is known that the
+
Each mode being a Hankel function of the first kind in the horizontal  
oscillation of an infinite plate can also be expressed in terms of Hankel
+
(in the three-dimensional case) and a hyperbolic function
functions. [[Kheisin 1967]] chapter IV, studied the same
+
in the vertical. [[Kheisin 1967]] chapter IV, studied the same
problem solved in this chapter. Kheisin derived the same dispersion equation
+
problem solved in this chapter and derived the same dispersion equation
for the physical variables, and then studied the properties of the inverse
+
for the physical variables. He studied the properties of the inverse
transform for the simple shallow water and static load cases, which are
+
transform only for the simple shallow water and static load cases, which are
approximated versions of the solutions reported here.
+
approximated versions of the full solutions which is presented here.
  
 
==Mathematical model==
 
==Mathematical model==
Line 137: Line 147:
  
 
We solve the system of equations using the [http://en.wikipedia.org/wiki/Fourier_Transform Fourier Transform] in <math>\left(  x,y\right)  </math>-plane for three-dimensions and with
 
We solve the system of equations using the [http://en.wikipedia.org/wiki/Fourier_Transform Fourier Transform] in <math>\left(  x,y\right)  </math>-plane for three-dimensions and with
a Fourier Transform in
+
a [http://en.wikipedia.org/wiki/Fourier_Transform Fourier Transform] in
 
<math>x</math> for two dimensions. We choose the Fourier transform with respect to <math>x</math>
 
<math>x</math> for two dimensions. We choose the Fourier transform with respect to <math>x</math>
 
and <math>y</math> defined as
 
and <math>y</math> defined as
Line 143: Line 153:
 
\hat{\phi}\left(  \alpha,k,z\right)  =\int_{-\infty}^{\infty}\int_{-\infty
 
\hat{\phi}\left(  \alpha,k,z\right)  =\int_{-\infty}^{\infty}\int_{-\infty
 
}^{\infty}\phi\left(  x,y,z\right)  e^{\mathrm{i}\left(  \alpha
 
}^{\infty}\phi\left(  x,y,z\right)  e^{\mathrm{i}\left(  \alpha
x+ky\right)  }dxdy
+
x+ky\right)  }\mathrm{d}x\mathrm{d}y
 
</math></center>
 
</math></center>
and the inverse Fourier transform defined as
+
and the inverse [http://en.wikipedia.org/wiki/Fourier_Transform Fourier Transform] defined as
 
<center><math>
 
<center><math>
 
\phi\left(  x,y,z\right)  =\frac{1}{4\pi^{2}}\int_{-\infty}^{\infty}
 
\phi\left(  x,y,z\right)  =\frac{1}{4\pi^{2}}\int_{-\infty}^{\infty}
 
\int_{-\infty}^{\infty}\hat{\phi}\left(  \alpha,k,z\right)
 
\int_{-\infty}^{\infty}\hat{\phi}\left(  \alpha,k,z\right)
e^{-\mathrm{i}\left(  \alpha x+ky\right)  }d\alpha dk.
+
e^{-\mathrm{i}\left(  \alpha x+ky\right)  }\mathrm{d}\alpha \mathrm{d}k.
 
</math></center>
 
</math></center>
 
For the one-dimensional case, the definitions are
 
For the one-dimensional case, the definitions are
 
<center><math>
 
<center><math>
 
\hat{\phi}\left(  \alpha,z\right)    =\int_{-\infty}^{\infty}\phi\left(
 
\hat{\phi}\left(  \alpha,z\right)    =\int_{-\infty}^{\infty}\phi\left(
x,z\right)  e^{\mathrm{i}\alpha x}dx,  
+
x,z\right)  e^{\mathrm{i}\alpha x}\mathrm{d}x,  
 
</math></center>
 
</math></center>
 
<center><math>
 
<center><math>
 
\phi\left(  x,z\right)    =\frac{1}{2\pi}\int_{-\infty}^{\infty}\hat{\phi
 
\phi\left(  x,z\right)    =\frac{1}{2\pi}\int_{-\infty}^{\infty}\hat{\phi
}\left(  \alpha,z\right)  e^{-\mathrm{i}\alpha x}d\alpha.  
+
}\left(  \alpha,z\right)  e^{-\mathrm{i}\alpha x}\mathrm{d}\alpha.  
 
</math></center>
 
</math></center>
We denote the spatial Fourier transform by using a hat over <math>w</math> and <math>\phi</math>.
+
We denote the spatial [http://en.wikipedia.org/wiki/Fourier_Transform Fourier Transform]
 +
by using a hat over <math>w</math> and <math>\phi</math>.
  
The Fourier transform of both sides of the Laplace's equation  
+
The [http://en.wikipedia.org/wiki/Fourier_Transform Fourier Transform] of both sides of the Laplace's equation  
 
becomes an ordinary differential equation with respect to <math>z</math>,
 
becomes an ordinary differential equation with respect to <math>z</math>,
 
<center><math>
 
<center><math>
Line 252: Line 263:
 
<center><math>
 
<center><math>
 
w\left(  r\right)  =\frac{1}{2\pi}\int_{0}^{\infty}\hat{w}\left(
 
w\left(  r\right)  =\frac{1}{2\pi}\int_{0}^{\infty}\hat{w}\left(
\gamma\right)  \gamma J_{0}\left(  \gamma r\right)  d\gamma  
+
\gamma\right)  \gamma J_{0}\left(  \gamma r\right)  \mathrm{d}\gamma  
 
</math></center>
 
</math></center>
 
for the three dimensional problem,
 
for the three dimensional problem,
Line 261: Line 272:
 
<center><math>
 
<center><math>
 
w\left(  r\right)  =\frac{1}{\pi}\int_{0}^{\infty}\hat{w}\left(
 
w\left(  r\right)  =\frac{1}{\pi}\int_{0}^{\infty}\hat{w}\left(
\gamma\right)  \cos\left(  \gamma r\right)  d\gamma  
+
\gamma\right)  \cos\left(  \gamma r\right)  \mathrm{d}\gamma  
 
</math></center>
 
</math></center>
 
where again <math>r=\left|  x\right|  </math>. Note that the factors <math>1/2\pi</math> and <math>1/\pi</math> result from the form of the Fourier transform in two and one dimensional spaces.
 
where again <math>r=\left|  x\right|  </math>. Note that the factors <math>1/2\pi</math> and <math>1/\pi</math> result from the form of the Fourier transform in two and one dimensional spaces.
  
The integrals are calculated using the [[Mittag-Leffler Expansion for the Floating Elastic Plate Dispersion Relation]].
+
The integrals are calculated using the [[Mittag-Leffler Expansion for the Floating Elastic Plate Dispersion Relation]]
Pairing each pole <math>q_n</math> with its negative counterpart <math>-q_n</math>, gives
+
where we show that
 
<center><math>
 
<center><math>
 
\hat{w}\left(  \gamma\right)  =\sum_{n=-2}^{\infty}\frac{2q_nR\left(
 
\hat{w}\left(  \gamma\right)  =\sum_{n=-2}^{\infty}\frac{2q_nR\left(
Line 281: Line 292:
 
<center><math>
 
<center><math>
 
\int_{0}^{\infty}\frac{\gamma}{\gamma^{2}-q^{2}}J_{0}\left(  \gamma r\right)
 
\int_{0}^{\infty}\frac{\gamma}{\gamma^{2}-q^{2}}J_{0}\left(  \gamma r\right)
d\gamma=K_{0}\left(  -\mathrm{i}qr\right)   
+
\mathrm{d}\gamma=K_{0}\left(  -\mathrm{i}qr\right)   
 
</math></center>
 
</math></center>
 
for <math>Im q>0</math>, <math>r>0</math>, where <math>K_{0}</math> is a modified Bessel function and an identity between the modified Bessel function.
 
for <math>Im q>0</math>, <math>r>0</math>, where <math>K_{0}</math> is a modified Bessel function and an identity between the modified Bessel function.
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q_n R\left(  q_n\right)  K_{0}\left(  -\mathrm{i}q_n r\right)
 
q_n R\left(  q_n\right)  K_{0}\left(  -\mathrm{i}q_n r\right)
 
</math></center>
 
</math></center>
</math></center>
+
<center><math>
 
=\frac{\mathrm{i}}{2}\sum_{n=-2}^{\infty}q_n R\left(  q_n\right)
 
=\frac{\mathrm{i}}{2}\sum_{n=-2}^{\infty}q_n R\left(  q_n\right)
 
H_{0}^{\left(  1\right)  }\left(  q_n r\right).
 
H_{0}^{\left(  1\right)  }\left(  q_n r\right).
 
</math></center>
 
</math></center>
  
In two-dimensions the identity ([[Erdelyi et. al.]] formula 1.2 (11) with <math>x=\gamma</math>, <math>y=r</math>, and <math>a=-\mathrm{i}q</math>)
+
In two-dimensions the identity ([[Erdelyi et. al.1954]] formula 1.2 (11) with <math>x=\gamma</math>, <math>y=r</math>, and <math>a=-\mathrm{i}q</math>)
 
<center><math>
 
<center><math>
 
\int_{0}^{\infty}\frac{\cos\left(  \gamma r\right)  }{\gamma^{2}-q^{2}}
 
\int_{0}^{\infty}\frac{\cos\left(  \gamma r\right)  }{\gamma^{2}-q^{2}}
d\gamma=-\frac{\pi}{\mathrm{i}2q}\exp\left(  \mathrm{i}qr\right)
+
\mathrm{d}\gamma=-\frac{\pi}{\mathrm{i}2q}\exp\left(  \mathrm{i}qr\right)
 
</math></center>
 
</math></center>
 
for <math>Im q>0</math>, <math>r>0</math> can be used. This gives us
 
for <math>Im q>0</math>, <math>r>0</math> can be used. This gives us
Line 324: Line 335:
 
\cosh q_n\left(  z+H\right).
 
\cosh q_n\left(  z+H\right).
 
</math></center>
 
</math></center>
 +
 +
 +
[[Category:Floating Elastic Plate]]

Latest revision as of 16:40, 8 December 2009


This is a special version of the free-surface Green function which applied when the Floating Elastic Plate boundary condition applies at the free-surface. It reduced to the Free-Surface Green Function in the limit as the plate terms tend to zero and to the Green function for an infinite plate in the limit as the water terms tend to zero.

Two Dimensions

Green function with singularity on the surface

For the case of a floating thin plate we almost always make the assumption of shallow draft and the Green function satisfies the following equations The Green function [math]\displaystyle{ G(x,x^{\prime},z) }[/math] representing outgoing waves as [math]\displaystyle{ |x|\rightarrow \infty }[/math] satisfies

[math]\displaystyle{ \nabla^2 G = 0, -H\lt z\lt 0, }[/math]
[math]\displaystyle{ \frac{\partial G}{\partial z} =0, z=-H, }[/math]
[math]\displaystyle{ {\left( \beta \frac{\partial^4}{\partial x^4} - \gamma\alpha + 1\right)\frac{\partial G}{\partial z} - \alpha G} = \delta(x-x^{\prime}), z=0, }[/math]

(note we are only considering singularities at the free surface). It is important to realise that these equations are based on the Frequency Domain Problem and that the exact form of the Green function is dependent on whether we have [math]\displaystyle{ e^{i\omega t} }[/math] or [math]\displaystyle{ e^{-i\omega t} }[/math] dependence. In what follows we assume [math]\displaystyle{ e^{i\omega t} }[/math] dependence.

This problem can be solved (a solution is given below and also in Evans and Porter 2006) to give

[math]\displaystyle{ G(x,x^{\prime},z) = -\sum_{n=-2}^\infty\frac{\sin{(k_n H)}\cos{(k_n(z+H))}}{2\alpha C(k_n)}e^{-k_n|x-x^{\prime}|}, }[/math]

where

[math]\displaystyle{ C(k_n)=\frac{1}{2}\left(h - \frac{(5\beta k_n^4 + 1 - \alpha\gamma)\sin^2{(k_n H)}}{\alpha}\right), }[/math]

and [math]\displaystyle{ k_n }[/math] are the solutions of the Dispersion Relation for a Floating Elastic Plate,

[math]\displaystyle{ \beta k^5 \sin(kH) - k \left(1 - \alpha \gamma \right) \sin(kH) = -\alpha \cos(kH) \, }[/math]

with [math]\displaystyle{ n=-1,-2 }[/math] corresponding to the complex solutions with positive real part, [math]\displaystyle{ n=0 }[/math] corresponding to the imaginary solution with positive imaginary part, since we have assumed [math]\displaystyle{ e^{i\omega t} }[/math] dependence (it would be of negative imaginary part if we had chosen [math]\displaystyle{ e^{-i\omega t} }[/math] dependence) [math]\displaystyle{ n\gt 0 }[/math] corresponding to the real solutions with positive real part. We can also write the Green function as

[math]\displaystyle{ G\left( r,z\right) = -\omega\sum_{n=-2}^{\infty} \frac{R\left( q_n\right) }{q_n\sinh q_nH}\exp\left( \mathrm{i}q_nr\right) \cosh q_n\left( z+H\right) . }[/math]

where [math]\displaystyle{ r=|x-x^{\prime}| }[/math], [math]\displaystyle{ q_n=ik_n }[/math] and we have used a different non-dimensionalisation so that the Dispersion Relation for a Floating Elastic Plate is written as

[math]\displaystyle{ -q^5 \sinh(kH) - k \left(1 - m\omega^2 \right) \sinh(kH) = -\omega^2 \cosh(kH) \, }[/math]

and

[math]\displaystyle{ R\left( q_n\right) =\frac{\omega^{2}q_n}{\omega^{2}\left( 5q_n^{4}+u\right) +H\left[ \left( q_n^{5}+uq\right) ^{2}-\omega^{4}\right] }. }[/math]

Green function with singularity in the water

We can define the Green function [math]\displaystyle{ G(x,x^{\prime},z) }[/math], representing outgoing waves as [math]\displaystyle{ |x|\rightarrow \infty }[/math], which satisfies

[math]\displaystyle{ \nabla^2 G = \delta(x-x^{\prime})\delta(z-z^{\prime}), -H\lt z\lt 0, }[/math]
[math]\displaystyle{ \frac{\partial G}{\partial z} =0, z=-H, }[/math]
[math]\displaystyle{ {\left( \beta \frac{\partial^4}{\partial x^4} - \gamma\alpha + 1\right)\frac{\partial G}{\partial z} - \alpha G} = 0, z=0. }[/math]

Note that this Green function reduces the Free-Surface Green Function in the limit as [math]\displaystyle{ \beta }[/math] and [math]\displaystyle{ \gamma }[/math] tend to zero.

This problem can be solved to give Evans and Porter 2003

[math]\displaystyle{ G(x,x^{\prime},z) = -\sum_{n=-2}^\infty\frac{\cos{(k_n(z^{\prime}+H))}\cos{(k_n(z+H))}}{2k_n C(k_n)}e^{-k_n|x-x^{\prime}|}, }[/math]

where [math]\displaystyle{ C(k_n) }[/math] and [math]\displaystyle{ k_n }[/math] are as before.

Relationship between the two Green functions

If we denote the Green function with the singularity at the free-surface by [math]\displaystyle{ G_s }[/math] and the Green function with the singularity in the water by [math]\displaystyle{ G_w }[/math]then we have the following relationship between them

[math]\displaystyle{ G_s = -\frac{1}{\alpha}\left. \frac{\partial G_w}{\partial z^{\prime}} \right|_{z^{\prime} =0} }[/math]

Three Dimensions

[math]\displaystyle{ G\left( r,z\right) =-\frac{\omega}{2}\sum_{n=-2}^{\infty} \frac{R\left( q_n\right) }{\sinh q_nH}H_{0}^{\left( 1\right) }\left( q_n r\right) \cosh q_n\left( z+H\right) }[/math]

Derivation of the Green function

We present here a derivation of the Green function for both the two and three dimensional problem. The devivation uses the Mittag-Leffler Expansion for the Floating Elastic Plate Dispersion Equation. This derivation is based on the PhD thesis of Hyuck Chung.

Previous Work

In the following sections, the deflection of a plate is expressed by the infinite number of modes that exist in the vibrating plate floating on the water, instead of a finite number of modes that represent the free oscillation of an elastic plate without restraints. The fact that waves in an incompressible fluid can be expressed by an infinite series of natural modes is shown by John 1949 and John 1950 and discussed in Cylindrical Eigenfunction Expansion Each mode being a Hankel function of the first kind in the horizontal (in the three-dimensional case) and a hyperbolic function in the vertical. Kheisin 1967 chapter IV, studied the same problem solved in this chapter and derived the same dispersion equation for the physical variables. He studied the properties of the inverse transform only for the simple shallow water and static load cases, which are approximated versions of the full solutions which is presented here.

Mathematical model

In terms of non-dimensional variables the Floating Elastic Plate equation becomes

[math]\displaystyle{ \left[ \nabla^{4}-m\omega^{2}+1\right] w+\mathrm{i}\omega \phi=p_n. }[/math]

where [math]\displaystyle{ p_n }[/math] is the pressure and [math]\displaystyle{ n }[/math] is the dimension. We assume that [math]\displaystyle{ p_2 = \delta(x) }[/math] and that [math]\displaystyle{ p_3 = \delta(x) \delta(y) }[/math]. [math]\displaystyle{ w }[/math] is the displacement of the surface which is related to the velocity potential by the equation [math]\displaystyle{ \frac{\partial\phi}{\partial z} = \frac{\partial w}{\partial t} }[/math] We also have Laplace's equation for the velocity potential and the bottom condition

[math]\displaystyle{ \nabla^{2}\phi=0 }[/math]

for [math]\displaystyle{ -\infty\lt x,y\lt \infty, }[/math] [math]\displaystyle{ -H\lt z\lt 0 }[/math] and

[math]\displaystyle{ \phi_{z}=0\quad \mathrm{at} \quad \;z=-H. }[/math]

The system of equations from together with the Sommerfeld Radiation Condition form the BVP, which we will solve for [math]\displaystyle{ w }[/math] and [math]\displaystyle{ \phi }[/math] for a given [math]\displaystyle{ \omega }[/math].

Spatial Fourier transform

We solve the system of equations using the Fourier Transform in [math]\displaystyle{ \left( x,y\right) }[/math]-plane for three-dimensions and with a Fourier Transform in [math]\displaystyle{ x }[/math] for two dimensions. We choose the Fourier transform with respect to [math]\displaystyle{ x }[/math] and [math]\displaystyle{ y }[/math] defined as

[math]\displaystyle{ \hat{\phi}\left( \alpha,k,z\right) =\int_{-\infty}^{\infty}\int_{-\infty }^{\infty}\phi\left( x,y,z\right) e^{\mathrm{i}\left( \alpha x+ky\right) }\mathrm{d}x\mathrm{d}y }[/math]

and the inverse Fourier Transform defined as

[math]\displaystyle{ \phi\left( x,y,z\right) =\frac{1}{4\pi^{2}}\int_{-\infty}^{\infty} \int_{-\infty}^{\infty}\hat{\phi}\left( \alpha,k,z\right) e^{-\mathrm{i}\left( \alpha x+ky\right) }\mathrm{d}\alpha \mathrm{d}k. }[/math]

For the one-dimensional case, the definitions are

[math]\displaystyle{ \hat{\phi}\left( \alpha,z\right) =\int_{-\infty}^{\infty}\phi\left( x,z\right) e^{\mathrm{i}\alpha x}\mathrm{d}x, }[/math]
[math]\displaystyle{ \phi\left( x,z\right) =\frac{1}{2\pi}\int_{-\infty}^{\infty}\hat{\phi }\left( \alpha,z\right) e^{-\mathrm{i}\alpha x}\mathrm{d}\alpha. }[/math]

We denote the spatial Fourier Transform by using a hat over [math]\displaystyle{ w }[/math] and [math]\displaystyle{ \phi }[/math].

The Fourier Transform of both sides of the Laplace's equation becomes an ordinary differential equation with respect to [math]\displaystyle{ z }[/math],

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

with a solution

[math]\displaystyle{ \hat{\phi}\left( \alpha,k,z\right) =A\left( \gamma\right) e^{\gamma z}+B\left( \gamma\right) e^{-\gamma z} }[/math]

where [math]\displaystyle{ \gamma^{2}=\alpha^{2}+k^{2} }[/math] and [math]\displaystyle{ A }[/math], [math]\displaystyle{ B }[/math] are unknown coefficients to be determined by the boundary condition. We find that [math]\displaystyle{ \hat{\phi} }[/math] is a function only of the magnitude of the Fourier transform variables, [math]\displaystyle{ \gamma=\left\| \mathbf{\gamma}\right\| =\left| \left| \left( \alpha,k\right) \right| \right| }[/math], hence we may now denote [math]\displaystyle{ \hat{\phi }\left( \alpha,k,z\right) }[/math] by [math]\displaystyle{ \hat{\phi}\left( \gamma,z\right) }[/math]. We can reach the same conclusion using the fact that [math]\displaystyle{ w\left( x,y\right) }[/math] and [math]\displaystyle{ \phi\left( x,y,z\right) }[/math] are functions of [math]\displaystyle{ r=\sqrt{x^{2}+y^{2}} }[/math] thus the Fourier transforms must be functions of [math]\displaystyle{ \gamma=\sqrt{\alpha^{2}+k^{2}} }[/math].

We find the unknown coefficients [math]\displaystyle{ A }[/math] and [math]\displaystyle{ B }[/math] from the Fourier transformed ocean floor condition that [math]\displaystyle{ \left. \hat{\phi}_{z}\right| _{z=-H}=0 }[/math] to be [math]\displaystyle{ A\left( \gamma\right) =Ce^{\gamma H} }[/math], and [math]\displaystyle{ B\left( \gamma\right) =Ce^{-\gamma H} }[/math]. Thus, we obtain the depth dependence of the Fourier transform of the potential

[math]\displaystyle{ \hat{\phi}\left( \gamma,z\right) =\hat{\phi}\left( \gamma,0\right) \frac{\cosh\gamma\left( z+H\right) }{\cosh\gamma H}. }[/math]

At the surface, [math]\displaystyle{ z=0 }[/math], differentiating both sides with respect to [math]\displaystyle{ z }[/math] the vertical component of the velocity is

[math]\displaystyle{ \hat{\phi}_{z}\left( \gamma,0\right) =\hat{\phi}\left( \gamma,0\right) \gamma\tanh\gamma H. }[/math]

Using this relationship to substitute for [math]\displaystyle{ \hat{\phi}_{z} }[/math] in the non-dimensional form of the kinematic condition, we find that

[math]\displaystyle{ \hat{\phi}\left( \gamma,0\right) =\frac{\mathrm{i}\omega\hat {w}\left( \gamma\right) }{\gamma\tanh\gamma H}. }[/math]

Thus, once we find [math]\displaystyle{ \hat{w}\left( \gamma\right) }[/math] then we can find [math]\displaystyle{ \hat{\phi}\left( \gamma,z\right) }[/math].

The Fourier transform of the plate equation is also an algebraic equation in the parameter [math]\displaystyle{ \gamma }[/math]

[math]\displaystyle{ \left( \gamma^{4}+1-m\omega^{2}\right) \hat{w}+\mathrm{i}\omega \hat{\phi}=1. }[/math]

Hence, we have the single algebraic equation for each spatial Fourier component of [math]\displaystyle{ w }[/math]

[math]\displaystyle{ \left( \gamma^{4}+1-m\omega^{2}-\frac{\omega^{2}}{\gamma\tanh\gamma H}\right) \hat{w}=1. }[/math]

Notice that we have used the fact that the Fourier transform of the delta function is [math]\displaystyle{ 1 }[/math].

We find that the spatial Fourier transform of the displacement of the ice sheet for the localized forcing, both point and line, is

[math]\displaystyle{ \hat{w}\left( \gamma\right) =\frac{1}{d\left( \gamma,\omega\right) } }[/math]

where

[math]\displaystyle{ d\left( \gamma,\omega\right) =\gamma^{4}+1-m\omega^{2}-\frac{\omega^{2} }{\gamma\tanh\gamma H}. }[/math]

Where [math]\displaystyle{ d\left( \gamma,\omega\right) }[/math] in the Dispersion Relation for a Floating Elastic Plate function, and the associated algebraic equation [math]\displaystyle{ d\left( \gamma ,\omega\right) =0 }[/math] the dispersion equation for waves propagating through an ice sheet. This dispersion relation (and the Fourier transform of it) was derived by Kheisin 1967 chapter IV, and Fox and Squire 1994.

Inverse Fourier Transform

Our task now is to derive the inverse Fourier transform). We notice that the roots of the Dispersion Relation for a Floating Elastic Plate for a fixed [math]\displaystyle{ \omega }[/math], are the poles of the function [math]\displaystyle{ \hat{w}\left( \gamma\right) }[/math], which are necessary for calculation of integrals involved in the inverse Fourier transform. There are a pair of two real roots [math]\displaystyle{ \pm q_0 }[/math], an infinite number of pure imaginary roots [math]\displaystyle{ \pm q_{n},\,n\in\mathbb{N} }[/math], ([math]\displaystyle{ \mathrm{Im} q_{n}\gt 0 }[/math]) plus four complex roots (in all physical situations) occur at plus and minus complex-conjugate pairs [math]\displaystyle{ \pm q_{-1} }[/math] and [math]\displaystyle{ \pm q_{-2} }[/math] with [math]\displaystyle{ \mathrm{Im} \left( q_{-1,-2}\right) \gt 0 }[/math]).

We note that the dispersion equation represents a relationship between the spatial wavenumbers [math]\displaystyle{ \gamma }[/math] and the radial frequency [math]\displaystyle{ \omega }[/math], which is how the name "dispersion" came about.

We may also solve for the velocity potential at the surface of the water

[math]\displaystyle{ \hat{\phi}\left( \gamma,0\right) =\frac{\mathrm{i}\omega w} {\gamma\tanh\gamma H}=\frac{\mathrm{i}\omega}{\gamma\tanh\left( \gamma H\right) d\left( \gamma,\omega\right) }, (3-10) }[/math]

which is also a function of [math]\displaystyle{ \gamma }[/math] only and has the same poles as [math]\displaystyle{ \hat{w}\left( \gamma\right) }[/math] does.

We calculate the displacement [math]\displaystyle{ w\left( x,y\right) }[/math] using the inverse Fourier transform of [math]\displaystyle{ \hat{w}\left( \gamma\right) }[/math]. Since [math]\displaystyle{ \hat{w}\left( \gamma\right) }[/math] is radially symmetric, the inverse transform may be written (Bracewell 1965)

[math]\displaystyle{ w\left( r\right) =\frac{1}{2\pi}\int_{0}^{\infty}\hat{w}\left( \gamma\right) \gamma J_{0}\left( \gamma r\right) \mathrm{d}\gamma }[/math]

for the three dimensional problem, where [math]\displaystyle{ J_{0} }[/math] is Bessel function and [math]\displaystyle{ r }[/math] is the distance from the point of forcing. For the two-dimensional problem the inverse Fourier transform of [math]\displaystyle{ \hat{w}\left( \gamma\right) }[/math] in the [math]\displaystyle{ x }[/math]-axis and since [math]\displaystyle{ \hat{w}\left( \gamma\right) }[/math] is an even function, this is

[math]\displaystyle{ w\left( r\right) =\frac{1}{\pi}\int_{0}^{\infty}\hat{w}\left( \gamma\right) \cos\left( \gamma r\right) \mathrm{d}\gamma }[/math]

where again [math]\displaystyle{ r=\left| x\right| }[/math]. Note that the factors [math]\displaystyle{ 1/2\pi }[/math] and [math]\displaystyle{ 1/\pi }[/math] result from the form of the Fourier transform in two and one dimensional spaces.

The integrals are calculated using the Mittag-Leffler Expansion for the Floating Elastic Plate Dispersion Relation where we show that

[math]\displaystyle{ \hat{w}\left( \gamma\right) =\sum_{n=-2}^{\infty}\frac{2q_nR\left( q_n\right) }{\gamma^{2}-q_n^{2}} }[/math]

where [math]\displaystyle{ R\left( q_n\right) }[/math] is the residue of [math]\displaystyle{ \hat{w}\left( \gamma\right) }[/math] at [math]\displaystyle{ \gamma=q_n }[/math] given by

[math]\displaystyle{ R\left( q_n\right) =\frac{\omega^{2}q_n}{\omega^{2}\left( 5q_n^{4}+u\right) +H\left[ \left( q_n^{5}+uq_n\right) ^{2}-\omega^{4}\right] }. }[/math]

.

Using the identities Abramowitz and Stegun 1964 formula 11.4.44 with [math]\displaystyle{ \nu=0 }[/math], [math]\displaystyle{ \mu=0 }[/math], [math]\displaystyle{ z=-\mathrm{i}q }[/math] and [math]\displaystyle{ a=r }[/math])

[math]\displaystyle{ \int_{0}^{\infty}\frac{\gamma}{\gamma^{2}-q^{2}}J_{0}\left( \gamma r\right) \mathrm{d}\gamma=K_{0}\left( -\mathrm{i}qr\right) }[/math]

for [math]\displaystyle{ Im q\gt 0 }[/math], [math]\displaystyle{ r\gt 0 }[/math], where [math]\displaystyle{ K_{0} }[/math] is a modified Bessel function and an identity between the modified Bessel function.

We can calculate the integral of the inverse Fourier transform of [math]\displaystyle{ \hat{w} }[/math] and, using the identity between [math]\displaystyle{ K_{0} }[/math] and Hankel function of the first kind (Abramowitz and Stegun 1964 formula 9.6.4),

[math]\displaystyle{ K_{0}\left( \zeta\right) =\frac{\mathrm{i}\pi}{2}H_{0}^{\left( 1\right) }\left( \mathrm{i}\zeta\right) }[/math]

for [math]\displaystyle{ Re \zeta\geq0 }[/math], the displacement for three-dimensions may be written in the equivalent forms

[math]\displaystyle{ w\left( r\right) =\frac{1}{\pi}\sum_{n=-2}^{\infty} q_n R\left( q_n\right) K_{0}\left( -\mathrm{i}q_n r\right) }[/math]
[math]\displaystyle{ =\frac{\mathrm{i}}{2}\sum_{n=-2}^{\infty}q_n R\left( q_n\right) H_{0}^{\left( 1\right) }\left( q_n r\right). }[/math]

In two-dimensions the identity (Erdelyi et. al.1954 formula 1.2 (11) with [math]\displaystyle{ x=\gamma }[/math], [math]\displaystyle{ y=r }[/math], and [math]\displaystyle{ a=-\mathrm{i}q }[/math])

[math]\displaystyle{ \int_{0}^{\infty}\frac{\cos\left( \gamma r\right) }{\gamma^{2}-q^{2}} \mathrm{d}\gamma=-\frac{\pi}{\mathrm{i}2q}\exp\left( \mathrm{i}qr\right) }[/math]

for [math]\displaystyle{ Im q\gt 0 }[/math], [math]\displaystyle{ r\gt 0 }[/math] can be used. This gives us

[math]\displaystyle{ w\left( r\right) =\mathrm{i}\sum_{n=-2}^{\infty} R\left( q\right) \exp\left( \mathrm{i}qr\right). }[/math]

Solution for the Velocity Potential

The velocity potential in the water can be found in two-dimensions

[math]\displaystyle{ \phi\left( r,z\right) =-\frac{\omega}{2}\sum_{n=-2}^{\infty} \frac{R\left( q_n\right) }{\sinh q_nH}H_{0}^{\left( 1\right) }\left( q_nr\right) \cosh q_n\left( z+H\right) }[/math]

and for three-dimensions

[math]\displaystyle{ \phi\left( r,z\right) =-\omega\sum_{n=-2}^{\infty} \frac{R\left( q_n\right) }{q_n\sinh q_nH}\exp\left( \mathrm{i}q_nr\right) \cosh q_n\left( z+H\right). }[/math]