WikiWaves - User contributions [en]

Method of Characteristics for Linear Equations

2017-11-03T00:47:38Z

Fmontiel: /* Non-homogeneous Example */

{{nonlinear waves course
| chapter title = Method of Characteristics for Linear Equations
| next chapter = [[Traffic Waves]]
| previous chapter =
}}

{{complete pages}}

We present here a brief account of the method of characteristic for linear waves.

== Introduction ==

The method of characteristics is an important method for hyperbolic PDE's which
applies to both linear and nonlinear equations.

We begin with the simplest wave equation
<center>
<math>
\partial_t u + \partial_x u = 0,\,\,-\infty < x < \infty,\,\,t>0,
</math></center>
subject to the initial conditions
<center>
<math>
\left. u \right|_{t=0} = f(x)
</math></center>

We consider the solution along the curve <math>(x,t) = (X(t),t)</math>. We then have
<center>
<math>
\frac{\mathrm{d} }{\mathrm{d} t} u(X(t),t) = \partial_t u + \frac{\mathrm{d} X}{\mathrm{d}t}\partial_x u = \partial_x u \left(\frac{\mathrm{d} X}{\mathrm{d}t} - 1 \right)
</math></center>

Therefore along the curve <math>\frac{\mathrm{d} X}{\mathrm{d}t} = 1</math> <math>u(x,t)</math> must be a constant.
These are nothing but the straight lines <math>x = t+c</math>
This means that we have
<center>
<math>
u(x,t) = u(t+c,t) = u(c,0) = f(c) = f(x-t)\,
</math></center>
Therefore the solution is <math>u(x,t) = f(x-t)\,</math>.

== General Form ==

If we consider the equation
<center>
<math>
\partial_t u + a(x,t)\partial_x u = 0,\,\,-\infty < x < \infty,\,\,t>0,
</math></center>
then we can apply the method of characteristics.
We consider the solution along the curve <math>(x,t) = (X(t),t)</math>. We then have
<center>
<math>
\frac{\mathrm{d} }{\mathrm{d} t} u(X(t),t) = \partial_t u + \frac{\mathrm{d} X}{\mathrm{d}t}\partial_x u = \partial_x u \left(\frac{\mathrm{d} X}{\mathrm{d}t} - a(x,t) \right).
</math>
</center>
This gives us the following o.d.e. for the characteristic curves (along which the solution is a constant)
<center>
<math>
\frac{\mathrm{d} X}{\mathrm{d} t} = a(x,t) .
</math>
</center>

== Example 1 ==

[[Image:Characteristic_linear1.jpg|thumb|right|350px|Characteristic for Example 1]]
Consider the equation
<center>
<math>
\partial_t u + x \partial_x u = 0,\,\,-\infty < x < \infty,\,\,t>0,
</math></center>
subject to the initial conditions
<center>
<math>
\left. u \right|_{t=0} = f(x)
</math></center>

We consider the solution along the curve <math>(x,t) = (X(t),t)</math>. We then have
<center>
<math>
\frac{\mathrm{d} }{\mathrm{d} t} u(X(t),t) = \partial_t u + \frac{\mathrm{d}X}{\mathrm{d}t}\partial_x u = \partial_x u \left(\frac{\mathrm{d} X}{\mathrm{d}t} - x \right)
</math></center>

Therefore along the curve <math>\frac{\mathrm{d} X}{\mathrm{d}t} = x</math> <math>u(x,t)</math> must be a constant.
These are the curves <math>x = ce^t</math>
This means that we have
<center>
<math>
u(x,t) = u(ce^t,t) = u(c,0) = f(c) = f(xe^{-t})\,
</math></center>
Therefore the solution is given <math>u(x,t) = f(xe^{-t})\,</math>.

[[Image:Waterfall_linear1.jpg|thumb|right|350px|Solution for Example 1 with <math>f(x) = e^{-x^2}</math>]]

== Example 2 ==

Consider the equation
<center>
<math>
\partial_t u + t \partial_x u = 0,\,\,-\infty < x < \infty,\,\,t>0,
</math></center>
subject to the initial conditions
<center>
<math>
\left. u \right|_{t=0} = f(x)
</math></center>

We consider the solution along the curve <math>(x,t) = (X(t),t)</math>. We then have
<center>
<math>
\frac{\mathrm{d} }{\mathrm{d} t} u(X(t),t) = \partial_t u + \frac{\mathrm{d} X}{\mathrm{d}t}\partial_x u = \partial_x u \left(\frac{\mathrm{d} X}{\mathrm{d}t} - t \right)
</math></center>

Therefore along the curve <math>\frac{\mathrm{d} X}{\mathrm{d}t} = t</math> <math>u(x,t)</math> must be a constant.
These are the curves <math>x = t^2/2+c</math>
This means that we have
<center>
<math>
u(x,t) = u(t^2/2 + c,t) = u(c,0) = f(c) = f(x - t^2/2)\,
</math></center>
Therefore the solution is given <math>u(x,t) = f(x - t^2/2)\,</math>.

== Non-homogeneous Example ==

We can also use the method of characteristics in the non-homogeneous case. We show this through an example
Consider the equation
<center>
<math>
\partial_t u + t \partial_x u = xt,\,\,-\infty < x < \infty,\,\,t>0,
</math></center>
subject to the initial conditions
<center>
<math>
\left. u \right|_{t=0} = f(x)
</math></center>

We consider the solution along the curve <math>(x,t) = (X(t),t)</math>. We then have
<center>
<math>
\frac{\mathrm{d} u}{\mathrm{d} t} = \partial_t u + \frac{\mathrm{d} X}{\mathrm{d}t}\partial_x u = \partial_x u \left(\frac{\mathrm{d}X}{\mathrm{d}t} - t \right) + xt
</math></center>

Therefore along the curve <math>\frac{\mathrm{d} X}{\mathrm{d}t} = t</math> which are the curves <math>x = t^2/2+c</math>
<center>
<math>
\frac{\mathrm{d}}{\mathrm{d}t}u(x,t) = xt = t^3/2 + c t
</math>
</center>
Therefore
<center>
<math>
u(t^2/2+c,t) = t^4/8 + c t^2/2 + c_2\,
</math>
</center>
Now
<center>
<math>
u(c,0) = c_2 = f(c)
</math></center>

Therefore the solution is given <math>u(x,t) = t^4/8 + (x -t^2/2) t^2/2 + f(x-t^2/2)\,</math> or
<math>u(x,t) = -t^4/8 + x t^2/2 + f(x-t^2/2)\,</math>

Method of Characteristics for Linear Equations

2017-11-03T00:47:21Z

Fmontiel: /* Example 2 */

{{nonlinear waves course
| chapter title = Method of Characteristics for Linear Equations
| next chapter = [[Traffic Waves]]
| previous chapter =
}}

{{complete pages}}

We present here a brief account of the method of characteristic for linear waves.

== Introduction ==

The method of characteristics is an important method for hyperbolic PDE's which
applies to both linear and nonlinear equations.

We begin with the simplest wave equation
<center>
<math>
\partial_t u + \partial_x u = 0,\,\,-\infty < x < \infty,\,\,t>0,
</math></center>
subject to the initial conditions
<center>
<math>
\left. u \right|_{t=0} = f(x)
</math></center>

We consider the solution along the curve <math>(x,t) = (X(t),t)</math>. We then have
<center>
<math>
\frac{\mathrm{d} }{\mathrm{d} t} u(X(t),t) = \partial_t u + \frac{\mathrm{d} X}{\mathrm{d}t}\partial_x u = \partial_x u \left(\frac{\mathrm{d} X}{\mathrm{d}t} - 1 \right)
</math></center>

Therefore along the curve <math>\frac{\mathrm{d} X}{\mathrm{d}t} = 1</math> <math>u(x,t)</math> must be a constant.
These are nothing but the straight lines <math>x = t+c</math>
This means that we have
<center>
<math>
u(x,t) = u(t+c,t) = u(c,0) = f(c) = f(x-t)\,
</math></center>
Therefore the solution is <math>u(x,t) = f(x-t)\,</math>.

== General Form ==

If we consider the equation
<center>
<math>
\partial_t u + a(x,t)\partial_x u = 0,\,\,-\infty < x < \infty,\,\,t>0,
</math></center>
then we can apply the method of characteristics.
We consider the solution along the curve <math>(x,t) = (X(t),t)</math>. We then have
<center>
<math>
\frac{\mathrm{d} }{\mathrm{d} t} u(X(t),t) = \partial_t u + \frac{\mathrm{d} X}{\mathrm{d}t}\partial_x u = \partial_x u \left(\frac{\mathrm{d} X}{\mathrm{d}t} - a(x,t) \right).
</math>
</center>
This gives us the following o.d.e. for the characteristic curves (along which the solution is a constant)
<center>
<math>
\frac{\mathrm{d} X}{\mathrm{d} t} = a(x,t) .
</math>
</center>

== Example 1 ==

[[Image:Characteristic_linear1.jpg|thumb|right|350px|Characteristic for Example 1]]
Consider the equation
<center>
<math>
\partial_t u + x \partial_x u = 0,\,\,-\infty < x < \infty,\,\,t>0,
</math></center>
subject to the initial conditions
<center>
<math>
\left. u \right|_{t=0} = f(x)
</math></center>

We consider the solution along the curve <math>(x,t) = (X(t),t)</math>. We then have
<center>
<math>
\frac{\mathrm{d} }{\mathrm{d} t} u(X(t),t) = \partial_t u + \frac{\mathrm{d}X}{\mathrm{d}t}\partial_x u = \partial_x u \left(\frac{\mathrm{d} X}{\mathrm{d}t} - x \right)
</math></center>

Therefore along the curve <math>\frac{\mathrm{d} X}{\mathrm{d}t} = x</math> <math>u(x,t)</math> must be a constant.
These are the curves <math>x = ce^t</math>
This means that we have
<center>
<math>
u(x,t) = u(ce^t,t) = u(c,0) = f(c) = f(xe^{-t})\,
</math></center>
Therefore the solution is given <math>u(x,t) = f(xe^{-t})\,</math>.

[[Image:Waterfall_linear1.jpg|thumb|right|350px|Solution for Example 1 with <math>f(x) = e^{-x^2}</math>]]

== Example 2 ==

Consider the equation
<center>
<math>
\partial_t u + t \partial_x u = 0,\,\,-\infty < x < \infty,\,\,t>0,
</math></center>
subject to the initial conditions
<center>
<math>
\left. u \right|_{t=0} = f(x)
</math></center>

We consider the solution along the curve <math>(x,t) = (X(t),t)</math>. We then have
<center>
<math>
\frac{\mathrm{d} }{\mathrm{d} t} u(X(t),t) = \partial_t u + \frac{\mathrm{d} X}{\mathrm{d}t}\partial_x u = \partial_x u \left(\frac{\mathrm{d} X}{\mathrm{d}t} - t \right)
</math></center>

Therefore along the curve <math>\frac{\mathrm{d} X}{\mathrm{d}t} = t</math> <math>u(x,t)</math> must be a constant.
These are the curves <math>x = t^2/2+c</math>
This means that we have
<center>
<math>
u(x,t) = u(t^2/2 + c,t) = u(c,0) = f(c) = f(x - t^2/2)\,
</math></center>
Therefore the solution is given <math>u(x,t) = f(x - t^2/2)\,</math>.

== Non-homogeneous Example ==

We can also use the method of characteristics in the non-homogeneous case. We show this through an example
Consider the equation
<center>
<math>
\partial_t u + t \partial_x u = xt,\,\,-\infty<x<\infty,\,\,t>0,
</math></center>
subject to the initial conditions
<center>
<math>
\left. u \right|_{t=0} = f(x)
</math></center>

We consider the solution along the curve <math>(x,t) = (X(t),t)</math>. We then have
<center>
<math>
\frac{\mathrm{d} u}{\mathrm{d} t} = \partial_t u + \frac{\mathrm{d} X}{\mathrm{d}t}\partial_x u = \partial_x u \left(\frac{\mathrm{d}X}{\mathrm{d}t} - t \right) + xt
</math></center>

Therefore along the curve <math>\frac{\mathrm{d} X}{\mathrm{d}t} = t</math> which are the curves <math>x = t^2/2+c</math>
<center>
<math>
\frac{\mathrm{d}}{\mathrm{d}t}u(x,t) = xt = t^3/2 + c t
</math>
</center>
Therefore
<center>
<math>
u(t^2/2+c,t) = t^4/8 + c t^2/2 + c_2\,
</math>
</center>
Now
<center>
<math>
u(c,0) = c_2 = f(c)
</math></center>

Therefore the solution is given <math>u(x,t) = t^4/8 + (x -t^2/2) t^2/2 + f(x-t^2/2)\,</math> or
<math>u(x,t) = -t^4/8 + x t^2/2 + f(x-t^2/2)\,</math>

Method of Characteristics for Linear Equations

2017-11-03T00:47:01Z

Fmontiel: /* Example 1 */

{{nonlinear waves course
| chapter title = Method of Characteristics for Linear Equations
| next chapter = [[Traffic Waves]]
| previous chapter =
}}

{{complete pages}}

We present here a brief account of the method of characteristic for linear waves.

== Introduction ==

The method of characteristics is an important method for hyperbolic PDE's which
applies to both linear and nonlinear equations.

We begin with the simplest wave equation
<center>
<math>
\partial_t u + \partial_x u = 0,\,\,-\infty < x < \infty,\,\,t>0,
</math></center>
subject to the initial conditions
<center>
<math>
\left. u \right|_{t=0} = f(x)
</math></center>

We consider the solution along the curve <math>(x,t) = (X(t),t)</math>. We then have
<center>
<math>
\frac{\mathrm{d} }{\mathrm{d} t} u(X(t),t) = \partial_t u + \frac{\mathrm{d} X}{\mathrm{d}t}\partial_x u = \partial_x u \left(\frac{\mathrm{d} X}{\mathrm{d}t} - 1 \right)
</math></center>

Therefore along the curve <math>\frac{\mathrm{d} X}{\mathrm{d}t} = 1</math> <math>u(x,t)</math> must be a constant.
These are nothing but the straight lines <math>x = t+c</math>
This means that we have
<center>
<math>
u(x,t) = u(t+c,t) = u(c,0) = f(c) = f(x-t)\,
</math></center>
Therefore the solution is <math>u(x,t) = f(x-t)\,</math>.

== General Form ==

If we consider the equation
<center>
<math>
\partial_t u + a(x,t)\partial_x u = 0,\,\,-\infty < x < \infty,\,\,t>0,
</math></center>
then we can apply the method of characteristics.
We consider the solution along the curve <math>(x,t) = (X(t),t)</math>. We then have
<center>
<math>
\frac{\mathrm{d} }{\mathrm{d} t} u(X(t),t) = \partial_t u + \frac{\mathrm{d} X}{\mathrm{d}t}\partial_x u = \partial_x u \left(\frac{\mathrm{d} X}{\mathrm{d}t} - a(x,t) \right).
</math>
</center>
This gives us the following o.d.e. for the characteristic curves (along which the solution is a constant)
<center>
<math>
\frac{\mathrm{d} X}{\mathrm{d} t} = a(x,t) .
</math>
</center>

== Example 1 ==

[[Image:Characteristic_linear1.jpg|thumb|right|350px|Characteristic for Example 1]]
Consider the equation
<center>
<math>
\partial_t u + x \partial_x u = 0,\,\,-\infty < x < \infty,\,\,t>0,
</math></center>
subject to the initial conditions
<center>
<math>
\left. u \right|_{t=0} = f(x)
</math></center>

We consider the solution along the curve <math>(x,t) = (X(t),t)</math>. We then have
<center>
<math>
\frac{\mathrm{d} }{\mathrm{d} t} u(X(t),t) = \partial_t u + \frac{\mathrm{d}X}{\mathrm{d}t}\partial_x u = \partial_x u \left(\frac{\mathrm{d} X}{\mathrm{d}t} - x \right)
</math></center>

Therefore along the curve <math>\frac{\mathrm{d} X}{\mathrm{d}t} = x</math> <math>u(x,t)</math> must be a constant.
These are the curves <math>x = ce^t</math>
This means that we have
<center>
<math>
u(x,t) = u(ce^t,t) = u(c,0) = f(c) = f(xe^{-t})\,
</math></center>
Therefore the solution is given <math>u(x,t) = f(xe^{-t})\,</math>.

[[Image:Waterfall_linear1.jpg|thumb|right|350px|Solution for Example 1 with <math>f(x) = e^{-x^2}</math>]]

== Example 2 ==

Consider the equation
<center>
<math>
\partial_t u + t \partial_x u = 0,\,\,-\infty<x<\infty,\,\,t>0,
</math></center>
subject to the initial conditions
<center>
<math>
\left. u \right|_{t=0} = f(x)
</math></center>

We consider the solution along the curve <math>(x,t) = (X(t),t)</math>. We then have
<center>
<math>
\frac{\mathrm{d} }{\mathrm{d} t} u(X(t),t) = \partial_t u + \frac{\mathrm{d} X}{\mathrm{d}t}\partial_x u = \partial_x u \left(\frac{\mathrm{d} X}{\mathrm{d}t} - t \right)
</math></center>

Therefore along the curve <math>\frac{\mathrm{d} X}{\mathrm{d}t} = t</math> <math>u(x,t)</math> must be a constant.
These are the curves <math>x = t^2/2+c</math>
This means that we have
<center>
<math>
u(x,t) = u(t^2/2 + c,t) = u(c,0) = f(c) = f(x - t^2/2)\,
</math></center>
Therefore the solution is given <math>u(x,t) = f(x - t^2/2)\,</math>.

== Non-homogeneous Example ==

We can also use the method of characteristics in the non-homogeneous case. We show this through an example
Consider the equation
<center>
<math>
\partial_t u + t \partial_x u = xt,\,\,-\infty<x<\infty,\,\,t>0,
</math></center>
subject to the initial conditions
<center>
<math>
\left. u \right|_{t=0} = f(x)
</math></center>

We consider the solution along the curve <math>(x,t) = (X(t),t)</math>. We then have
<center>
<math>
\frac{\mathrm{d} u}{\mathrm{d} t} = \partial_t u + \frac{\mathrm{d} X}{\mathrm{d}t}\partial_x u = \partial_x u \left(\frac{\mathrm{d}X}{\mathrm{d}t} - t \right) + xt
</math></center>

Therefore along the curve <math>\frac{\mathrm{d} X}{\mathrm{d}t} = t</math> which are the curves <math>x = t^2/2+c</math>
<center>
<math>
\frac{\mathrm{d}}{\mathrm{d}t}u(x,t) = xt = t^3/2 + c t
</math>
</center>
Therefore
<center>
<math>
u(t^2/2+c,t) = t^4/8 + c t^2/2 + c_2\,
</math>
</center>
Now
<center>
<math>
u(c,0) = c_2 = f(c)
</math></center>

Therefore the solution is given <math>u(x,t) = t^4/8 + (x -t^2/2) t^2/2 + f(x-t^2/2)\,</math> or
<math>u(x,t) = -t^4/8 + x t^2/2 + f(x-t^2/2)\,</math>

Method of Characteristics for Linear Equations

2017-11-03T00:46:44Z

Fmontiel: /* General Form */

{{nonlinear waves course
| chapter title = Method of Characteristics for Linear Equations
| next chapter = [[Traffic Waves]]
| previous chapter =
}}

{{complete pages}}

We present here a brief account of the method of characteristic for linear waves.

== Introduction ==

The method of characteristics is an important method for hyperbolic PDE's which
applies to both linear and nonlinear equations.

We begin with the simplest wave equation
<center>
<math>
\partial_t u + \partial_x u = 0,\,\,-\infty < x < \infty,\,\,t>0,
</math></center>
subject to the initial conditions
<center>
<math>
\left. u \right|_{t=0} = f(x)
</math></center>

We consider the solution along the curve <math>(x,t) = (X(t),t)</math>. We then have
<center>
<math>
\frac{\mathrm{d} }{\mathrm{d} t} u(X(t),t) = \partial_t u + \frac{\mathrm{d} X}{\mathrm{d}t}\partial_x u = \partial_x u \left(\frac{\mathrm{d} X}{\mathrm{d}t} - 1 \right)
</math></center>

Therefore along the curve <math>\frac{\mathrm{d} X}{\mathrm{d}t} = 1</math> <math>u(x,t)</math> must be a constant.
These are nothing but the straight lines <math>x = t+c</math>
This means that we have
<center>
<math>
u(x,t) = u(t+c,t) = u(c,0) = f(c) = f(x-t)\,
</math></center>
Therefore the solution is <math>u(x,t) = f(x-t)\,</math>.

== General Form ==

If we consider the equation
<center>
<math>
\partial_t u + a(x,t)\partial_x u = 0,\,\,-\infty < x < \infty,\,\,t>0,
</math></center>
then we can apply the method of characteristics.
We consider the solution along the curve <math>(x,t) = (X(t),t)</math>. We then have
<center>
<math>
\frac{\mathrm{d} }{\mathrm{d} t} u(X(t),t) = \partial_t u + \frac{\mathrm{d} X}{\mathrm{d}t}\partial_x u = \partial_x u \left(\frac{\mathrm{d} X}{\mathrm{d}t} - a(x,t) \right).
</math>
</center>
This gives us the following o.d.e. for the characteristic curves (along which the solution is a constant)
<center>
<math>
\frac{\mathrm{d} X}{\mathrm{d} t} = a(x,t) .
</math>
</center>

== Example 1 ==

[[Image:Characteristic_linear1.jpg|thumb|right|350px|Characteristic for Example 1]]
Consider the equation
<center>
<math>
\partial_t u + x \partial_x u = 0,\,\,-\infty<x<\infty,\,\,t>0,
</math></center>
subject to the initial conditions
<center>
<math>
\left. u \right|_{t=0} = f(x)
</math></center>

We consider the solution along the curve <math>(x,t) = (X(t),t)</math>. We then have
<center>
<math>
\frac{\mathrm{d} }{\mathrm{d} t} u(X(t),t) = \partial_t u + \frac{\mathrm{d}X}{\mathrm{d}t}\partial_x u = \partial_x u \left(\frac{\mathrm{d} X}{\mathrm{d}t} - x \right)
</math></center>

Therefore along the curve <math>\frac{\mathrm{d} X}{\mathrm{d}t} = x</math> <math>u(x,t)</math> must be a constant.
These are the curves <math>x = ce^t</math>
This means that we have
<center>
<math>
u(x,t) = u(ce^t,t) = u(c,0) = f(c) = f(xe^{-t})\,
</math></center>
Therefore the solution is given <math>u(x,t) = f(xe^{-t})\,</math>.

[[Image:Waterfall_linear1.jpg|thumb|right|350px|Solution for Example 1 with <math>f(x) = e^{-x^2}</math>]]

== Example 2 ==

Consider the equation
<center>
<math>
\partial_t u + t \partial_x u = 0,\,\,-\infty<x<\infty,\,\,t>0,
</math></center>
subject to the initial conditions
<center>
<math>
\left. u \right|_{t=0} = f(x)
</math></center>

We consider the solution along the curve <math>(x,t) = (X(t),t)</math>. We then have
<center>
<math>
\frac{\mathrm{d} }{\mathrm{d} t} u(X(t),t) = \partial_t u + \frac{\mathrm{d} X}{\mathrm{d}t}\partial_x u = \partial_x u \left(\frac{\mathrm{d} X}{\mathrm{d}t} - t \right)
</math></center>

Therefore along the curve <math>\frac{\mathrm{d} X}{\mathrm{d}t} = t</math> <math>u(x,t)</math> must be a constant.
These are the curves <math>x = t^2/2+c</math>
This means that we have
<center>
<math>
u(x,t) = u(t^2/2 + c,t) = u(c,0) = f(c) = f(x - t^2/2)\,
</math></center>
Therefore the solution is given <math>u(x,t) = f(x - t^2/2)\,</math>.

== Non-homogeneous Example ==

We can also use the method of characteristics in the non-homogeneous case. We show this through an example
Consider the equation
<center>
<math>
\partial_t u + t \partial_x u = xt,\,\,-\infty<x<\infty,\,\,t>0,
</math></center>
subject to the initial conditions
<center>
<math>
\left. u \right|_{t=0} = f(x)
</math></center>

We consider the solution along the curve <math>(x,t) = (X(t),t)</math>. We then have
<center>
<math>
\frac{\mathrm{d} u}{\mathrm{d} t} = \partial_t u + \frac{\mathrm{d} X}{\mathrm{d}t}\partial_x u = \partial_x u \left(\frac{\mathrm{d}X}{\mathrm{d}t} - t \right) + xt
</math></center>

Therefore along the curve <math>\frac{\mathrm{d} X}{\mathrm{d}t} = t</math> which are the curves <math>x = t^2/2+c</math>
<center>
<math>
\frac{\mathrm{d}}{\mathrm{d}t}u(x,t) = xt = t^3/2 + c t
</math>
</center>
Therefore
<center>
<math>
u(t^2/2+c,t) = t^4/8 + c t^2/2 + c_2\,
</math>
</center>
Now
<center>
<math>
u(c,0) = c_2 = f(c)
</math></center>

Therefore the solution is given <math>u(x,t) = t^4/8 + (x -t^2/2) t^2/2 + f(x-t^2/2)\,</math> or
<math>u(x,t) = -t^4/8 + x t^2/2 + f(x-t^2/2)\,</math>

Method of Characteristics for Linear Equations

2017-11-03T00:46:23Z

Fmontiel: /* Introduction */

{{nonlinear waves course
| chapter title = Method of Characteristics for Linear Equations
| next chapter = [[Traffic Waves]]
| previous chapter =
}}

{{complete pages}}

We present here a brief account of the method of characteristic for linear waves.

== Introduction ==

The method of characteristics is an important method for hyperbolic PDE's which
applies to both linear and nonlinear equations.

We begin with the simplest wave equation
<center>
<math>
\partial_t u + \partial_x u = 0,\,\,-\infty < x < \infty,\,\,t>0,
</math></center>
subject to the initial conditions
<center>
<math>
\left. u \right|_{t=0} = f(x)
</math></center>

We consider the solution along the curve <math>(x,t) = (X(t),t)</math>. We then have
<center>
<math>
\frac{\mathrm{d} }{\mathrm{d} t} u(X(t),t) = \partial_t u + \frac{\mathrm{d} X}{\mathrm{d}t}\partial_x u = \partial_x u \left(\frac{\mathrm{d} X}{\mathrm{d}t} - 1 \right)
</math></center>

Therefore along the curve <math>\frac{\mathrm{d} X}{\mathrm{d}t} = 1</math> <math>u(x,t)</math> must be a constant.
These are nothing but the straight lines <math>x = t+c</math>
This means that we have
<center>
<math>
u(x,t) = u(t+c,t) = u(c,0) = f(c) = f(x-t)\,
</math></center>
Therefore the solution is <math>u(x,t) = f(x-t)\,</math>.

== General Form ==

If we consider the equation
<center>
<math>
\partial_t u + a(x,t)\partial_x u = 0,\,\,-\infty<x<\infty,\,\,t>0,
</math></center>
then we can apply the method of characteristics.
We consider the solution along the curve <math>(x,t) = (X(t),t)</math>. We then have
<center>
<math>
\frac{\mathrm{d} }{\mathrm{d} t} u(X(t),t) = \partial_t u + \frac{\mathrm{d} X}{\mathrm{d}t}\partial_x u = \partial_x u \left(\frac{\mathrm{d} X}{\mathrm{d}t} - a(x,t) \right).
</math>
</center>
This gives us the following o.d.e. for the characteristic curves (along which the solution is a constant)
<center>
<math>
\frac{\mathrm{d} X}{\mathrm{d} t} = a(x,t) .
</math>
</center>

== Example 1 ==

[[Image:Characteristic_linear1.jpg|thumb|right|350px|Characteristic for Example 1]]
Consider the equation
<center>
<math>
\partial_t u + x \partial_x u = 0,\,\,-\infty<x<\infty,\,\,t>0,
</math></center>
subject to the initial conditions
<center>
<math>
\left. u \right|_{t=0} = f(x)
</math></center>

We consider the solution along the curve <math>(x,t) = (X(t),t)</math>. We then have
<center>
<math>
\frac{\mathrm{d} }{\mathrm{d} t} u(X(t),t) = \partial_t u + \frac{\mathrm{d}X}{\mathrm{d}t}\partial_x u = \partial_x u \left(\frac{\mathrm{d} X}{\mathrm{d}t} - x \right)
</math></center>

Therefore along the curve <math>\frac{\mathrm{d} X}{\mathrm{d}t} = x</math> <math>u(x,t)</math> must be a constant.
These are the curves <math>x = ce^t</math>
This means that we have
<center>
<math>
u(x,t) = u(ce^t,t) = u(c,0) = f(c) = f(xe^{-t})\,
</math></center>
Therefore the solution is given <math>u(x,t) = f(xe^{-t})\,</math>.

[[Image:Waterfall_linear1.jpg|thumb|right|350px|Solution for Example 1 with <math>f(x) = e^{-x^2}</math>]]

== Example 2 ==

Consider the equation
<center>
<math>
\partial_t u + t \partial_x u = 0,\,\,-\infty<x<\infty,\,\,t>0,
</math></center>
subject to the initial conditions
<center>
<math>
\left. u \right|_{t=0} = f(x)
</math></center>

We consider the solution along the curve <math>(x,t) = (X(t),t)</math>. We then have
<center>
<math>
\frac{\mathrm{d} }{\mathrm{d} t} u(X(t),t) = \partial_t u + \frac{\mathrm{d} X}{\mathrm{d}t}\partial_x u = \partial_x u \left(\frac{\mathrm{d} X}{\mathrm{d}t} - t \right)
</math></center>

Therefore along the curve <math>\frac{\mathrm{d} X}{\mathrm{d}t} = t</math> <math>u(x,t)</math> must be a constant.
These are the curves <math>x = t^2/2+c</math>
This means that we have
<center>
<math>
u(x,t) = u(t^2/2 + c,t) = u(c,0) = f(c) = f(x - t^2/2)\,
</math></center>
Therefore the solution is given <math>u(x,t) = f(x - t^2/2)\,</math>.

== Non-homogeneous Example ==

We can also use the method of characteristics in the non-homogeneous case. We show this through an example
Consider the equation
<center>
<math>
\partial_t u + t \partial_x u = xt,\,\,-\infty<x<\infty,\,\,t>0,
</math></center>
subject to the initial conditions
<center>
<math>
\left. u \right|_{t=0} = f(x)
</math></center>

We consider the solution along the curve <math>(x,t) = (X(t),t)</math>. We then have
<center>
<math>
\frac{\mathrm{d} u}{\mathrm{d} t} = \partial_t u + \frac{\mathrm{d} X}{\mathrm{d}t}\partial_x u = \partial_x u \left(\frac{\mathrm{d}X}{\mathrm{d}t} - t \right) + xt
</math></center>

Therefore along the curve <math>\frac{\mathrm{d} X}{\mathrm{d}t} = t</math> which are the curves <math>x = t^2/2+c</math>
<center>
<math>
\frac{\mathrm{d}}{\mathrm{d}t}u(x,t) = xt = t^3/2 + c t
</math>
</center>
Therefore
<center>
<math>
u(t^2/2+c,t) = t^4/8 + c t^2/2 + c_2\,
</math>
</center>
Now
<center>
<math>
u(c,0) = c_2 = f(c)
</math></center>

Therefore the solution is given <math>u(x,t) = t^4/8 + (x -t^2/2) t^2/2 + f(x-t^2/2)\,</math> or
<math>u(x,t) = -t^4/8 + x t^2/2 + f(x-t^2/2)\,</math>

Sommerfeld Radiation Condition

2012-09-04T04:55:38Z

Fmontiel:

{{complete pages}}

This is a condition for the [[Frequency Domain Problem]] that the scattered wave is only
outgoing at infinity. It depends on the convention regarding whether the time dependence
is <math>\exp (i\omega t)\,</math> or <math>\exp (-i\omega t)\,</math>.
Assuming the former (which is the standard convention on this wiki).
In two dimensions the condition is
<center>
<math>
\left( \frac{\partial}{\partial|x|}+\mathrm{i}k\right)
(\phi-\phi^{\mathrm{{In}}})\rightarrow0,\;\mathrm{{as\;}}|x|\rightarrow\infty\mathrm{.}
</math>
</center>
where <math>\phi^{\mathrm{{In}}}</math> is the incident potential and <math>k</math>
is the wave number.

In three dimensions the condition is
<center>
<math>
r^{1/2}\left( \frac{\partial}{\partial r}+\mathrm{i}k\right)
(\phi-\phi^{\mathrm{{In}}})\rightarrow0,\;\mathrm{{as\;}}r\rightarrow\infty\mathrm{.}
</math>
</center>

If the time dependence is assumed to be <math>\exp (-i\omega t)\,</math>, then we
have in two dimensions
<center>
<math>
\left( \frac{\partial}{\partial|x|}-\mathrm{i}k\right)
(\phi-\phi^{\mathrm{{In}}})\rightarrow0,\;\mathrm{{as\;}}|x|\rightarrow\infty\mathrm{.}
</math>
</center>
and in three dimensions
<center>
<math>
r^{1/2}\left( \frac{\partial}{\partial r}-\mathrm{i}k\right)
(\phi-\phi^{\mathrm{{In}}})\rightarrow0,\;\mathrm{{as\;}}r\rightarrow\infty\mathrm{.}
</math>
</center>

[[Category:Linear Water-Wave Theory]]

Energy Balance for Two Elastic Plates

2008-08-27T00:15:59Z

Fmontiel: /* Expanding <math>\mathbf{\xi_2}</math> */

= Introduction =

We derive here a condition which must be satisfied for energy balance for two semi-infinite elastic plates. The solution applies
to the problem of [[Eigenfunction Matching Method for Floating Elastic Plates|multiple elastic plates]] and from here we know we can write the potential as
<center><math>
\phi \approx \left\{
\begin{matrix}
{
Ie^{\kappa_{1}(0)(x-r_1)}\frac{\cos(k_1(0)(z+h))}{\cos(k_1(0)h)} }+\\
{
\qquad \qquad \sum_{n=-2}^{M}R_1(n) e^{\kappa_{1}(n)(x-r_1)} \frac{\cos(k_1(n)(z+h))}{\cos(k_1(n)h)} },&
\mbox{ for } x < r_1,\\
{
\sum_{n=-2}^{M}T_{\mu}(n) e^{-\kappa_\mu(n)(x-l_\mu)}
\frac{\cos(k_\mu(n)(z+h))}{\cos(k_\mu(n)h)} } + \\
{
\qquad \qquad \sum_{n=-2}^{M}R_{\mu}(n) e^{\kappa_\mu(n)(x-r_\mu)}
\frac{\cos(k_\mu(n)(z+h))}{\cos(k_\mu(n)h)} },
&\mbox{ for } l_\mu< x < r_\mu,\\
{
\sum_{n=-2}^{M}T_{\Lambda}(n)e^{-\kappa_{\Lambda}(n)(x-l_\Lambda)}
\frac{\cos(k_\Lambda(n)(z+h))}{\cos(k_\Lambda(n)h)} }, &\mbox{ for } l_\Lambda<x,
\end{matrix} \right.
</math></center>
(for details of this notation see [[Eigenfunction Matching Method for Floating Elastic Plates]]).

= Equations =

Based on the method used in [[Evans and Davies 1968]], a check can be made to ensure the solutions of the floating plate problem are in energy balance.
This is simply a condition that the incident energy is equal to the sum of the radiated energy.
When the first and final plates have different properties, the energy balance equation is derived by applying Green's theorem to <math>\phi</math> and its conjugate.
The domain of integration is shown in the figure on the right.

[[Image:energy_schematic.jpg|thumb|right|300px|A diagram depicting the area <math>\mathcal{U}</math> which is bounded by the rectangle <math>\mathcal{S}</math>.]
{A diagram depicting the area <math>\mathcal{U}</math> which is bounded by the rectangle <math>\mathcal{S}</math>.
The rectangle <math>\mathcal{S}</math> is bounded by <math> -h\leq z \leq0</math> and <math>-\infty\leq x \leq \infty</math>]]

Applying Green's theorem to <math>\phi</math> and its conjugate <math>\phi^*</math> gives
<center><math>
{ \int\int_\mathcal{U}(\phi\nabla^2\phi^* - \phi^*\nabla^2\phi)dxdz
= \int_\mathcal{S}(\phi\frac{\partial\phi^*}{\partial n} - \phi^*\frac{\partial\phi}{\partial n})dl },
</math></center>
where <math>n</math> denotes the outward plane normal to the boundary and <math>l</math> denotes the plane parallel to the boundary.
As <math>\phi</math> and <math>\phi^*</math> satisfy the Laplace's equation, the left hand side of the Green theorem equation vanishes so that it reduces to
<center><math>
\Im\int_\mathcal{S}\phi\frac{\partial\phi^*}{\partial n} dl = 0,
</math></center>
Expanding gives
<center><math>
\xi_1 +\xi_2 + \xi_3 = 0,
</math></center>
where
<center><math>
\xi_1 =
-{ \Im\int_{-h}^0(\phi\frac{\partial\phi^*}{\partial x})\big|_{x=-\infty}dz },
</math></center>
<center><math>
\xi_2 =
{ \Im\int_{-h}^0(\phi\frac{\partial\phi^*}{\partial x})\big|_{x=\infty}dz },
</math></center>
and
<center><math>
\xi_3 =
{ \Im\int_{-\infty}^{\infty}(\phi\frac{\partial\phi^*}{\partial z})\big|_{z=0}dx } ,
</math></center>
where <math>\Im</math> denotes the imaginary part. The bottom domain composant is obviously equal to zero because of the no-influx condition over the seabed (<math>\frac{\partial\phi^*}{\partial z}\big|_{z=-h}=0</math>).

= Expanding <math>\mathbf{\xi_1}</math> =

Near <math>x=-\infty</math>, we approximate <math>\phi</math> by
<center><math>
\phi \approx e^{-\kappa_{1}(0)(x-r_1)}\frac{\cos{(k_1(0)(z+h))}}{\cos{(k_1(0)h)}} +
R_1(0)e^{\kappa_{1}(0)(x-r_1)}\frac{\cos{(k_{1}(0)(z+h))}}{\cos{(k_{1}(0)h)}}
</math></center>
Therefore,
<center><math>
\begin{matrix}
\xi_1 & = & {-\Im\int_{-h}^0 \left[\left(e^{-i\kappa^{I}_1(x)}+ R_1(0)e^{i\kappa^{I}_1(x)}\right)\right.}
{ \left(i\kappa^{I}_1e^{i\kappa^{I}_1(x)} - i\kappa^{I}_1R_1(0)^*e^{-i\kappa^{I}_1(x)}\right) }
{ \left.\left(\frac{\cosh^2{(k^I_1(z+h))}}{\cosh^2{(k^I_1h)}}\right)\right]dz, }\\ \\
& = & { - \Im\left[\frac{i\kappa^{I}_1\left(1 - |R_1(0)|^2\right)}{2\cosh^2{(k^I_1h)}}
\int_{-h}^0 \left(\cosh{(2k^I_1(z+h))+1}\right)dz\right] ,}\\ \\
& = & { - \Im\left[\frac{i\kappa^{I}_1\left(1 - |R_1(0)|^2\right)}{2\cosh^2{(k^I_1h)}}
\left[\frac{1}{2k^I_1}\sinh{(2k^I_1(z+h))} + z\right]_{-h}^0\right], }\\ \\
& = & { - \frac{\kappa^{I}_1\left(1 - |R_1(0)|^2\right)}{2\cosh^2{(k^I_1h)}} \left(\frac{1}{2k^I_1}\sinh{(2k^I_1h))} + h\right) ,}\\ \\
& = & { - \frac{\kappa^{I}_1\left(1 - |R_1(0)|^2\right)}{2k^I_1}\left(\tanh{(k^I_1h)}+\frac{hk^I_1}{\cosh^2{(k^I_1h)}}\right) , }
\end{matrix}
</math></center>
where <math>R_1(0)^*</math> is the conjugate of <math>R_1(0)</math>.

= Expanding <math>\mathbf{\xi_2}</math> =

Near <math>x=\infty</math>, we approximate <math>\phi</math> by
<center><math>
\begin{matrix}
{ \phi \approx T_\Lambda(0)e^{-\kappa_{\Lambda}(0)(x)}\frac{\cos{(k_\Lambda(0)(z+h))}}{\cos{(k_\Lambda(0)h)}}, }
\end{matrix}
</math></center>
and re-express as
<center><math>
\phi \approx T_\Lambda(0)e^{-i\kappa^I_\Lambda(x)}\frac{\cosh{(k^I_\Lambda(z+h))}}{\cosh{(k^I_\Lambda h)}},
</math></center>
where <math>k^I_\Lambda = \Im k_{\Lambda}(0)</math> and <math>\kappa^I_\Lambda=\Im \kappa_\Lambda(0)</math>, so that
<center><math>
\frac{\partial\phi}{\partial x} \approx -i\kappa^I_{\Lambda}T_\Lambda(0)e^{-i\kappa^I_{\Lambda}(x)}\frac{\cosh{(k^I_\Lambda(z+h))}}{\cosh{(k^I_\Lambda h)}}.
</math></center>
We know that <math>\kappa_\Lambda(0)=\sqrt{k_{\Lambda}(0)^2-(ik_y)^2}</math>, where <math>k_y</math> is real and depends on the incident angle <math>\theta</math>. When <math>\theta</math> becomes greater than a certain angle <math>\theta_0</math> defined by <math>\sin \theta_0=\frac{k_{\Lambda}(0)}{k_{1}(0)}</math>, <math>\kappa_\Lambda(0)</math> becomes real so the potential becomes real as well. Thus the imaginary part is equal to 0. In that case, we have <math>\xi_2=0</math>.
Therefore, if <math>\theta \in [-\theta_0, \theta_0]</math>,
<center><math>
\begin{matrix}
\xi_2 & = & \Im\int_{-h}^0 \left[\right.(T_\Lambda(0)e^{-i\kappa^I_{\Lambda}(x)})(i\kappa^I_{\Lambda}T_\Lambda(0)^*e^{i\kappa^I_{\Lambda}(x)})
\frac{\cosh^2{(k^I_\Lambda(z+h))}}{\cosh^2{(k^I_\Lambda h)}}\left.\right]dz, \\ \\
& = & { \frac{\kappa^I_{\Lambda}|T_\Lambda(0)|^2}{2\cosh^2{(k^I_\Lambda h)}}
\left(\frac{1}{2k^I_\Lambda}\sinh{(2k^I_\Lambda h)} + h\right)},\\ \\
& = & { \frac{\kappa^I_{\Lambda}|T_\Lambda(0)|^2}{2k^I_\Lambda}\left(\tanh{(k^I_\Lambda h)}+\frac{k^I_\Lambda h}{\cosh^2{(k^I_\Lambda h)}}\right)},
\end{matrix}
</math></center>
otherwise <math>\xi_2=0</math>.

= Expanding <math>\mathbf{\xi_3}</math> =

The ice-covered boundary condition for the [[Floating Elastic Plate]] gives
<center><math>
\xi_3 = { \Im\int_{-\infty}^{\infty}\left(\frac{\beta}{\alpha}
\left(\frac{\partial^2}{\partial x^2} - k_y^2\right)^2 - \gamma + \frac{1}{\alpha}\right)
\frac{\partial\phi}{\partial z}\cdot\frac{\partial\phi^*}{\partial z}\bigg|_{z=0}dx.}
</math></center>
Since <math>{\frac{\partial\phi}{\partial z}\cdot\frac{\partial\phi^*}{\partial z}}</math> is real,
<center><math>
\xi_3 = \Im\int_{-\infty}^{\infty}\left(\frac{\beta}{\alpha}\frac{\partial^2}{\partial x^2}
\left(\frac{\partial^2}{\partial x^2} - 2k_y^2\right) \right)
\frac{\partial\phi}{\partial z}\cdot\frac{\partial\phi^*}{\partial z}\bigg|_{z=0}dx.
</math></center>
Integration by parts gives
<center><math>
\xi_3 = \Im\left\{\left[\frac{\beta}{\alpha}\frac{\partial}{\partial x}
\left(\frac{\partial^2}{\partial x^2}-2k_y^2\right)
\frac{\partial\phi}{\partial z}\cdot\frac{\partial\phi^*}{\partial z}\right]_{-\infty}^{\infty}
-\int_{-\infty}^{\infty}\frac{\beta}{\alpha}\frac{\partial}{\partial x}
\left(\frac{\partial^2}{\partial x^2} - 2k_y^2\right)
\frac{\partial\phi}{\partial z}\cdot\frac
{\partial}{\partial x}\frac{\partial\phi^*}{\partial z}dx
\right\}.
</math></center>
As <math>{2k_y^2\frac{\partial}{\partial x}\frac{\partial\phi}{\partial z}\cdot \frac{\partial}{\partial x}\frac{\partial\phi^*}{\partial z}}</math> is real and by integration by parts, the expression of <math>\xi_3</math> becomes,
<center><math>
\xi_3 =\Im\left\{
\left[\frac{\beta}{\alpha}\frac{\partial}{\partial x}
\left(\frac{\partial^2}{\partial x^2}-2k_y^2\right)
\frac{\partial\phi}{\partial z}\cdot\frac{\partial\phi^*}{\partial z}\right]_{-\infty}^{\infty}
- \left[\frac{\beta}{\alpha}\frac{\partial^2}{\partial x^2}\frac{\partial\phi}{\partial z}\cdot
\frac{\partial}{\partial x}\frac{\partial\phi^*}{\partial z}\right]_{-\infty}^{\infty}
+ \int_{-\infty}^\infty\frac{\beta}{\alpha}\frac{\partial^2}{\partial x^2}\frac{\partial\phi}{\partial z}\cdot
\frac{\partial^2}{\partial x^2}\frac{\partial\phi^*}{\partial z}dx\right\}.
</math></center>
As <math>{\frac{\partial^2}{\partial x^2}\frac{\partial\phi}{\partial z}\cdot \frac{\partial^2}{\partial x^2}\frac{\partial\phi^*}{\partial z}}</math> is real, we obtain the new expression of <math>\xi_3</math>
<center><math>
\xi_3 = \Im\left\{
\left[\frac{\beta}{\alpha}\frac{\partial}{\partial x}
\left(\frac{\partial^2}{\partial x^2}-2k_y^2\right)
\frac{\partial\phi}{\partial z}\cdot\frac{\partial\phi^*}{\partial z}\right]_{-\infty}^{\infty}-
\left[\frac{\beta}{\alpha}\frac{\partial^2}{\partial x^2}\frac{\partial\phi}{\partial z}\cdot
\frac{\partial}{\partial x}\frac{\partial\phi^*}{\partial z}\right]_{-\infty}^{\infty}\right\}.
</math></center>

Now breaking <math>\xi_3</math> down, we can simplify the left hand term for <math>x>0</math>,
if <math>\theta \in [-\theta_0, \theta_0]</math>,

<center><math>
\begin{matrix}
\frac{\partial}{\partial x}\left(\frac{\partial^2}{\partial x^2}-2k_y^2\right)
\frac{\partial\phi(x_2,0)}{\partial z}\cdot\frac{\partial\phi(x_2,0)^*}{\partial z} \\ \\
= \left((-i\kappa^I_{\Lambda})^3k^I_\Lambda T_\Lambda(0)e^{-i\kappa^I_{\Lambda}(x)}\tanh{(k^I_\Lambda h)}\right.
\left.- 2k_y^2(-i\kappa_{\Lambda}(0))k^I_\Lambda T_\Lambda(0)e^{-i\kappa^I_{\Lambda}(x)} \tanh{(k^I_\Lambda h)}\right)
\left(k^I_\Lambda T_\Lambda(0)^*e^{i\kappa^I_{\Lambda}(x)}\tanh{(k^I_\Lambda h)}\right), \\ \\
= i\kappa^I_{\Lambda} (k^I_\Lambda)^2\left((\kappa^I_{\Lambda})^2 + 2k_y^2 \right) \tanh^2{(k^I_\Lambda h)}|T_\Lambda(0)|^2,
\end{matrix}
</math></center>
else this term is equal to 0.

and for <math>x<0</math>

<center><math>
\begin{matrix}
\frac{\partial}{\partial x}\left(\frac{\partial^2}{\partial x^2}-2k_y^2\right)\frac{\partial\phi(x_1,0)}{\partial z}\cdot
\frac{\partial\phi(x_1,0)^*}{\partial z} \\ \\
= \left[i(\kappa^I_{1})^3k^I_1\left(e^{-i\kappa^I_{1}(x)}-R_1(0)e^{i\kappa^I_{1}(x)}\right) \tanh{(k^I_1 h)}
- 2ik_y^2\kappa^I_{1}k^I_1\left(-e^{-i\kappa^I_{1}(x)} + R_1(0)e^{i\kappa^I_{1}(x)}\right) \tanh{(k^I_1h)}\right]
\left[k^I_1\left(e^{i\kappa^I_{1}(x)} + R_1(0)^*e^{-i\kappa^I_{1}(x))}\right) \tanh{(k^I_1h)}\right], \\ \\
= \left[i\kappa^I_{1}k_1(0)((\kappa^I_{1})^2 + 2k_y^2)\left(e^{-i\kappa^I_{1}(x)} - R_1(0)e^{i\kappa^I_{1}(x)}\right)
\tanh{(k^I_1h)}\right]
\left[k^I_1\left(e^{i\kappa^I_{1}(x)} + R_1(0)^*e^{-i\kappa^I_{1}(x))}\right) \tanh{(k^I_1h)}\right], \\ \\
= i\kappa^I_{1}(k^I_1)^2\left((\kappa^I_{1})^2 + 2k_y^2\right)\tanh^2{(k^I_1h)}\left(1 - |R_1(0)|^2\right),
\end{matrix}
</math></center>

Likewise we expand the right hand term for <math>x>0</math>, if <math>\theta \in [-\theta_0, \theta_0]</math>,

<center><math>
\begin{matrix}
\frac{\partial^2}{\partial x^2}\frac{\partial\phi(x_2,0)}{\partial z}\cdot
\frac{\partial}{\partial x}\frac{\partial\phi^*}{\partial z} \\ \\
= \left[-(\kappa^I_{\Lambda})^2k^I_\Lambda T_\Lambda(0)e^{-i\kappa^I_{\Lambda}(x)} \tanh{(k^I_\Lambda h)}\right]
\left[i\kappa^I_{\Lambda} k^I_\Lambda T_\Lambda(0)^*e^{i\kappa^I_{\Lambda}(x)} \tanh{(k^I_\Lambda h)}\right], \\ \\
= -i(\kappa^I_{\Lambda})^3(k^I_\Lambda)^2\tanh^2{(k^I_\Lambda h)}|T_\Lambda(0)|^2,
\end{matrix}
</math></center>
else this term is equal to 0 as well,
and finally for <math>x<0</math>,

<center><math>
\begin{matrix}
\frac{\partial^2}{\partial x^1}\frac{\partial\phi(x_1,0)}{\partial z}\cdot
\frac{\partial}{\partial x}\frac{\partial\phi^*}{\partial z}\\ \\
= \left[-(\kappa^I_{1})^2k^I_1\left(e^{-i\kappa^I_{1}(x)} + R_1(0)e^{i\kappa^I_{1}(x)}\right)\tanh{(k^I_1 h)}\right]
\left[i\kappa^I_{1} k^I_1\left(e^{i\kappa^I_{1}(x)} - R_1(0)e^{-i\kappa^I_{1}(x)}\right)\tanh{(k^I_1 h)}\right], \\ \\
= -i(\kappa^I_{1})^3(k^I_1)^2\tanh^2{(k^I_1h)}\left(1 - |R_1(0)|^2\right).
\end{matrix}
</math></center>

We can now express <math>\xi_3</math> as

<center><math>
\begin{matrix}
\xi_3 & = & \Im
\left\{\frac{\beta_\Lambda}{\alpha}\left[i\kappa^I_{\Lambda} (k^I_\Lambda)^2\left((\kappa^I_{\Lambda})^2 + 2k_y^2 \right)
\tanh^2{(k^I_\Lambda h)}|T_\Lambda(0)|^2\right]
-\frac{\beta_1}{\alpha}\left[i\kappa^I_{1}(k^I_1)^2\left((\kappa^I_{1})^2 + 2k_y^2\right)\tanh^2{(k^I_1h)}
\left(1 - |R_1(0)|^2\right)\right]
-\frac{\beta_\Lambda}{\alpha}\left[-i(\kappa^I_{\Lambda})^3(k^I_\Lambda)^2\tanh^2{(k^I_\Lambda h)}|T_\Lambda(0)|^2\right]
+\frac{\beta_1}{\alpha}\left[-i(\kappa^I_{1})^3(k^I_1)^2\tanh^2{(k^I_1h)} \left(1 - |R_1(0)|^2\right)\right]\right\}, \\ \\
& = & \frac{\beta_\Lambda}{\alpha}\left[2\kappa^I_{\Lambda} (k^I_\Lambda)^2((\kappa^I_{\Lambda})^2 + k_y^2)\tanh^2{(k^I_\Lambda h)}|T_\Lambda(0)|^2\right]
-\frac{\beta_1}{\alpha}\left[2\kappa^I_{1}(k^I_1)^2((\kappa^I_{1})^2 + k_y^2)\tanh^2{(k^I_1h)}(1-|R_1(0)|^2)\right].
\end{matrix}
</math></center>
if <math>\theta \in [-\theta_0, \theta_0]</math>, and if not, we obtain:
<center>
<math>
\xi_3 = -\frac{\beta_1}{\alpha}\left[2\kappa^I_{1}(k^I_1)^2((\kappa^I_{1})^2 + k_y^2)\tanh^2{(k^I_1h)}(1-|R_1(0)|^2)\right].
</math>
</center>

= Solving the Energy Balance Equation =

Pulling it all together for the case <math>\theta \in [-\theta_0, \theta_0]</math>, we finally obtain
<center><math>
\frac{\beta_\Lambda}{\alpha}\left[2\kappa^I_{\Lambda}(k^I_\Lambda)^2((\kappa^I_{\Lambda})^2 +k_y^2)
\tanh^2{(k^I_\Lambda h)}|T_\Lambda(0)|^2\right]
-\frac{\beta_1}{\alpha}\left[2\kappa^I_{1}(k^I_1)^2((\kappa^I_{1})^2 +k_y^2)\tanh^2{(k^I_1h)}(1-|R_1(0)|^2)\right]
+ \frac{\kappa^I_{\Lambda}|T_\Lambda(0)|^2}{2k^I_\Lambda}\left(\tanh{(k^I_\Lambda h)}+
\frac{k^I_\Lambda h}{\cosh^2{(k^I_\Lambda h)}}\right)
- \frac{\kappa^I_{1}\left(1 - |R_1(0)|^2\right)}{2k^I_1}\left(\tanh{(k^I_1h)}+\frac{hk^I_1}{\cosh^2{(k^I_1h)}}\right)=0.
</math></center>

Re-arranging gives

<center><math>
\kappa^I_{\Lambda}\tanh{(k^I_\Lambda h)}
\left(\frac{\beta_\Lambda}{\alpha}2(k^I_\Lambda)^2((\kappa^I_{\Lambda})^2 +k_y^2)\tanh{(k^I_\Lambda h)}
+ \frac{1}{2k^I_\Lambda}+\frac{h}{2\sinh{(k^I_\Lambda h)}\cosh{(k^I_\Lambda h)}}\right)|T_\Lambda(0)|^2
\quad - \kappa^I_{1}\tanh{(k^I_1 h)}
\left(\frac{\beta_1}{\alpha}2(k^I_1)^2((\kappa^I_{1})^2 +k_y^2)\tanh{(k^I_1 h)}
+ \frac{1}{2it}+\frac{h}{2\sinh{(k^I_1 h)}\cosh{(k^I_1 h)}}\right)(1-|R_1(0)|^2)=0
</math></center>

which can be expressed as
<center><math>
D |T_{\Lambda}(0)|^2 + |R_{1}(0)|^2 = 1,
</math></center>
where <math>D</math> is given by
<center><math>
D = \left(\frac{\kappa^I_{\Lambda} k^I_1\cosh^2{(k^I_1h)}}{\kappa^I_{1}k^I_\Lambda\cosh^2{(k^I_\Lambda h)}}\right)
\left(\frac{-\frac{\beta_\Lambda}{\alpha}4(k^I_\Lambda)^5\sinh^2{(k^I_\Lambda h)} +
\frac{1}{2}{\sinh{(2k^I_\Lambda h)}}+k^I_\Lambda h}
{-\frac{\beta_1}{\alpha}4(k^I_1)^5\sinh^2{(k^I_1h)} +
\frac{1}{2}{\sinh{(2k^I_1h)}}+k^I_1h}\right)
</math></center>
For the case of greater angles, there are no terms depending on <math>|T_{\Lambda}(0)|^2</math>, so we obtain <math>D=0</math> and <math>|R_{1}(0)|^2 = 1</math>, which is the case of the total reflexion. No energy is transmitted in the <math>x>0</math> region.

[[Category:Floating Elastic Plate]]

Eigenfunction Matching for a Semi-Infinite Floating Elastic Plate

2008-08-27T00:15:12Z

Fmontiel: /* Energy Balance */

Eigenfunction Matching for a Semi-Infinite Floating Elastic Plate

2008-08-26T23:46:20Z

Fmontiel: /* Energy Balance */

Energy Balance for Two Elastic Plates

2008-08-26T23:43:11Z

Fmontiel: /* Solving the Energy Balance Equation */

= Introduction =

We derive here a condition which must be satisfied for energy balance for two semi-infinite elastic plates. The solution applies
to the problem of [[Eigenfunction Matching Method for Floating Elastic Plates|multiple elastic plates]] and from here we know we can write the potential as
<center><math>
\phi \approx \left\{
\begin{matrix}
{
Ie^{\kappa_{1}(0)(x-r_1)}\frac{\cos(k_1(0)(z+h))}{\cos(k_1(0)h)} }+\\
{
\qquad \qquad \sum_{n=-2}^{M}R_1(n) e^{\kappa_{1}(n)(x-r_1)} \frac{\cos(k_1(n)(z+h))}{\cos(k_1(n)h)} },&
\mbox{ for } x < r_1,\\
{
\sum_{n=-2}^{M}T_{\mu}(n) e^{-\kappa_\mu(n)(x-l_\mu)}
\frac{\cos(k_\mu(n)(z+h))}{\cos(k_\mu(n)h)} } + \\
{
\qquad \qquad \sum_{n=-2}^{M}R_{\mu}(n) e^{\kappa_\mu(n)(x-r_\mu)}
\frac{\cos(k_\mu(n)(z+h))}{\cos(k_\mu(n)h)} },
&\mbox{ for } l_\mu< x < r_\mu,\\
{
\sum_{n=-2}^{M}T_{\Lambda}(n)e^{-\kappa_{\Lambda}(n)(x-l_\Lambda)}
\frac{\cos(k_\Lambda(n)(z+h))}{\cos(k_\Lambda(n)h)} }, &\mbox{ for } l_\Lambda<x,
\end{matrix} \right.
</math></center>
(for details of this notation see [[Eigenfunction Matching Method for Floating Elastic Plates]]).

= Equations =

Based on the method used in [[Evans and Davies 1968]], a check can be made to ensure the solutions of the floating plate problem are in energy balance.
This is simply a condition that the incident energy is equal to the sum of the radiated energy.
When the first and final plates have different properties, the energy balance equation is derived by applying Green's theorem to <math>\phi</math> and its conjugate.
The domain of integration is shown in the figure on the right.

[[Image:energy_schematic.jpg|thumb|right|300px|A diagram depicting the area <math>\mathcal{U}</math> which is bounded by the rectangle <math>\mathcal{S}</math>.]
{A diagram depicting the area <math>\mathcal{U}</math> which is bounded by the rectangle <math>\mathcal{S}</math>.
The rectangle <math>\mathcal{S}</math> is bounded by <math> -h\leq z \leq0</math> and <math>-\infty\leq x \leq \infty</math>]]

Applying Green's theorem to <math>\phi</math> and its conjugate <math>\phi^*</math> gives
<center><math>
{ \int\int_\mathcal{U}(\phi\nabla^2\phi^* - \phi^*\nabla^2\phi)dxdz
= \int_\mathcal{S}(\phi\frac{\partial\phi^*}{\partial n} - \phi^*\frac{\partial\phi}{\partial n})dl },
</math></center>
where <math>n</math> denotes the outward plane normal to the boundary and <math>l</math> denotes the plane parallel to the boundary.
As <math>\phi</math> and <math>\phi^*</math> satisfy the Laplace's equation, the left hand side of the Green theorem equation vanishes so that it reduces to
<center><math>
\Im\int_\mathcal{S}\phi\frac{\partial\phi^*}{\partial n} dl = 0,
</math></center>
Expanding gives
<center><math>
\xi_1 +\xi_2 + \xi_3 = 0,
</math></center>
where
<center><math>
\xi_1 =
-{ \Im\int_{-h}^0(\phi\frac{\partial\phi^*}{\partial x})\big|_{x=-\infty}dz },
</math></center>
<center><math>
\xi_2 =
{ \Im\int_{-h}^0(\phi\frac{\partial\phi^*}{\partial x})\big|_{x=\infty}dz },
</math></center>
and
<center><math>
\xi_3 =
{ \Im\int_{-\infty}^{\infty}(\phi\frac{\partial\phi^*}{\partial z})\big|_{z=0}dx } ,
</math></center>
where <math>\Im</math> denotes the imaginary part. The bottom domain composant is obviously equal to zero because of the no-influx condition over the seabed (<math>\frac{\partial\phi^*}{\partial z}\big|_{z=-h}=0</math>).

= Expanding <math>\mathbf{\xi_1}</math> =

Near <math>x=-\infty</math>, we approximate <math>\phi</math> by
<center><math>
\phi \approx e^{-\kappa_{1}(0)(x-r_1)}\frac{\cos{(k_1(0)(z+h))}}{\cos{(k_1(0)h)}} +
R_1(0)e^{\kappa_{1}(0)(x-r_1)}\frac{\cos{(k_{1}(0)(z+h))}}{\cos{(k_{1}(0)h)}}
</math></center>
Therefore,
<center><math>
\begin{matrix}
\xi_1 & = & {-\Im\int_{-h}^0 \left[\left(e^{-i\kappa^{I}_1(x)}+ R_1(0)e^{i\kappa^{I}_1(x)}\right)\right.}
{ \left(i\kappa^{I}_1e^{i\kappa^{I}_1(x)} - i\kappa^{I}_1R_1(0)^*e^{-i\kappa^{I}_1(x)}\right) }
{ \left.\left(\frac{\cosh^2{(k^I_1(z+h))}}{\cosh^2{(k^I_1h)}}\right)\right]dz, }\\ \\
& = & { - \Im\left[\frac{i\kappa^{I}_1\left(1 - |R_1(0)|^2\right)}{2\cosh^2{(k^I_1h)}}
\int_{-h}^0 \left(\cosh{(2k^I_1(z+h))+1}\right)dz\right] ,}\\ \\
& = & { - \Im\left[\frac{i\kappa^{I}_1\left(1 - |R_1(0)|^2\right)}{2\cosh^2{(k^I_1h)}}
\left[\frac{1}{2k^I_1}\sinh{(2k^I_1(z+h))} + z\right]_{-h}^0\right], }\\ \\
& = & { - \frac{\kappa^{I}_1\left(1 - |R_1(0)|^2\right)}{2\cosh^2{(k^I_1h)}} \left(\frac{1}{2k^I_1}\sinh{(2k^I_1h))} + h\right) ,}\\ \\
& = & { - \frac{\kappa^{I}_1\left(1 - |R_1(0)|^2\right)}{2k^I_1}\left(\tanh{(k^I_1h)}+\frac{hk^I_1}{\cosh^2{(k^I_1h)}}\right) , }
\end{matrix}
</math></center>
where <math>R_1(0)^*</math> is the conjugate of <math>R_1(0)</math>.

= Expanding <math>\mathbf{\xi_2}</math> =

Near <math>x=\infty</math>, we approximate <math>\phi</math> by
<center><math>
\begin{matrix}
{ \phi \approx T_\Lambda(0)e^{-\kappa_{\Lambda}(0)(x)}\frac{\cos{(k_\Lambda(0)(z+h))}}{\cos{(k_\Lambda(0)h)}}, }
\end{matrix}
</math></center>
and re-express as
<center><math>
\phi \approx T_\Lambda(0)e^{-i\kappa^I_\Lambda(x)}\frac{\cosh{(k^I_\Lambda(z+h))}}{\cosh{(k^I_\Lambda h)}},
</math></center>
where <math>k^I_\Lambda = \Im k_{\Lambda}(0)</math> and <math>\kappa^I_\Lambda=\Im \kappa_\Lambda(0)</math>, so that
<center><math>
\frac{\partial\phi}{\partial x} \approx -i\kappa^I_{\Lambda}T_\Lambda(0)e^{-i\kappa^I_{\Lambda}(x)}\frac{\cosh{(k^I_\Lambda(z+h))}}{\cosh{(k^I_\Lambda h)}}.
</math></center>
We know that <math>\kappa_\Lambda(0)=\sqrt{k_{\Lambda}(0)^2-(ik_y)^2}</math>, where <math>k_y</math> is real and depends on the incident angle <math>\theta</math>. When <math>\theta</math> becomes greater than a certain angle <math>\theta_0</math> defined by <math>\sin \theta_0=\sqrt{\frac{k_{\Lambda}(0)}{k_{1}(0)}}</math>, <math>\kappa_\Lambda(0)</math> becomes real so the potential becomes real as well. Thus the imaginary part is equal to 0. In that case, we have <math>\xi_2=0</math>.
Therefore, if <math>\theta \in [-\theta_0, \theta_0]</math>,
<center><math>
\begin{matrix}
\xi_2 & = & \Im\int_{-h}^0 \left[\right.(T_\Lambda(0)e^{-i\kappa^I_{\Lambda}(x)})(i\kappa^I_{\Lambda}T_\Lambda(0)^*e^{i\kappa^I_{\Lambda}(x)})
\frac{\cosh^2{(k^I_\Lambda(z+h))}}{\cosh^2{(k^I_\Lambda h)}}\left.\right]dz, \\ \\
& = & { \frac{\kappa^I_{\Lambda}|T_\Lambda(0)|^2}{2\cosh^2{(k^I_\Lambda h)}}
\left(\frac{1}{2k^I_\Lambda}\sinh{(2k^I_\Lambda h)} + h\right)},\\ \\
& = & { \frac{\kappa^I_{\Lambda}|T_\Lambda(0)|^2}{2k^I_\Lambda}\left(\tanh{(k^I_\Lambda h)}+\frac{k^I_\Lambda h}{\cosh^2{(k^I_\Lambda h)}}\right)},
\end{matrix}
</math></center>
otherwise <math>\xi_2=0</math>.

= Expanding <math>\mathbf{\xi_3}</math> =

The ice-covered boundary condition for the [[Floating Elastic Plate]] gives
<center><math>
\xi_3 = { \Im\int_{-\infty}^{\infty}\left(\frac{\beta}{\alpha}
\left(\frac{\partial^2}{\partial x^2} - k_y^2\right)^2 - \gamma + \frac{1}{\alpha}\right)
\frac{\partial\phi}{\partial z}\cdot\frac{\partial\phi^*}{\partial z}\bigg|_{z=0}dx.}
</math></center>
Since <math>{\frac{\partial\phi}{\partial z}\cdot\frac{\partial\phi^*}{\partial z}}</math> is real,
<center><math>
\xi_3 = \Im\int_{-\infty}^{\infty}\left(\frac{\beta}{\alpha}\frac{\partial^2}{\partial x^2}
\left(\frac{\partial^2}{\partial x^2} - 2k_y^2\right) \right)
\frac{\partial\phi}{\partial z}\cdot\frac{\partial\phi^*}{\partial z}\bigg|_{z=0}dx.
</math></center>
Integration by parts gives
<center><math>
\xi_3 = \Im\left\{\left[\frac{\beta}{\alpha}\frac{\partial}{\partial x}
\left(\frac{\partial^2}{\partial x^2}-2k_y^2\right)
\frac{\partial\phi}{\partial z}\cdot\frac{\partial\phi^*}{\partial z}\right]_{-\infty}^{\infty}
-\int_{-\infty}^{\infty}\frac{\beta}{\alpha}\frac{\partial}{\partial x}
\left(\frac{\partial^2}{\partial x^2} - 2k_y^2\right)
\frac{\partial\phi}{\partial z}\cdot\frac
{\partial}{\partial x}\frac{\partial\phi^*}{\partial z}dx
\right\}.
</math></center>
As <math>{2k_y^2\frac{\partial}{\partial x}\frac{\partial\phi}{\partial z}\cdot \frac{\partial}{\partial x}\frac{\partial\phi^*}{\partial z}}</math> is real and by integration by parts, the expression of <math>\xi_3</math> becomes,
<center><math>
\xi_3 =\Im\left\{
\left[\frac{\beta}{\alpha}\frac{\partial}{\partial x}
\left(\frac{\partial^2}{\partial x^2}-2k_y^2\right)
\frac{\partial\phi}{\partial z}\cdot\frac{\partial\phi^*}{\partial z}\right]_{-\infty}^{\infty}
- \left[\frac{\beta}{\alpha}\frac{\partial^2}{\partial x^2}\frac{\partial\phi}{\partial z}\cdot
\frac{\partial}{\partial x}\frac{\partial\phi^*}{\partial z}\right]_{-\infty}^{\infty}
+ \int_{-\infty}^\infty\frac{\beta}{\alpha}\frac{\partial^2}{\partial x^2}\frac{\partial\phi}{\partial z}\cdot
\frac{\partial^2}{\partial x^2}\frac{\partial\phi^*}{\partial z}dx\right\}.
</math></center>
As <math>{\frac{\partial^2}{\partial x^2}\frac{\partial\phi}{\partial z}\cdot \frac{\partial^2}{\partial x^2}\frac{\partial\phi^*}{\partial z}}</math> is real, we obtain the new expression of <math>\xi_3</math>
<center><math>
\xi_3 = \Im\left\{
\left[\frac{\beta}{\alpha}\frac{\partial}{\partial x}
\left(\frac{\partial^2}{\partial x^2}-2k_y^2\right)
\frac{\partial\phi}{\partial z}\cdot\frac{\partial\phi^*}{\partial z}\right]_{-\infty}^{\infty}-
\left[\frac{\beta}{\alpha}\frac{\partial^2}{\partial x^2}\frac{\partial\phi}{\partial z}\cdot
\frac{\partial}{\partial x}\frac{\partial\phi^*}{\partial z}\right]_{-\infty}^{\infty}\right\}.
</math></center>

Now breaking <math>\xi_3</math> down, we can simplify the left hand term for <math>x>0</math>,
if <math>\theta \in [-\theta_0, \theta_0]</math>,

<center><math>
\begin{matrix}
\frac{\partial}{\partial x}\left(\frac{\partial^2}{\partial x^2}-2k_y^2\right)
\frac{\partial\phi(x_2,0)}{\partial z}\cdot\frac{\partial\phi(x_2,0)^*}{\partial z} \\ \\
= \left((-i\kappa^I_{\Lambda})^3k^I_\Lambda T_\Lambda(0)e^{-i\kappa^I_{\Lambda}(x)}\tanh{(k^I_\Lambda h)}\right.
\left.- 2k_y^2(-i\kappa_{\Lambda}(0))k^I_\Lambda T_\Lambda(0)e^{-i\kappa^I_{\Lambda}(x)} \tanh{(k^I_\Lambda h)}\right)
\left(k^I_\Lambda T_\Lambda(0)^*e^{i\kappa^I_{\Lambda}(x)}\tanh{(k^I_\Lambda h)}\right), \\ \\
= i\kappa^I_{\Lambda} (k^I_\Lambda)^2\left((\kappa^I_{\Lambda})^2 + 2k_y^2 \right) \tanh^2{(k^I_\Lambda h)}|T_\Lambda(0)|^2,
\end{matrix}
</math></center>
else this term is equal to 0.

and for <math>x<0</math>

<center><math>
\begin{matrix}
\frac{\partial}{\partial x}\left(\frac{\partial^2}{\partial x^2}-2k_y^2\right)\frac{\partial\phi(x_1,0)}{\partial z}\cdot
\frac{\partial\phi(x_1,0)^*}{\partial z} \\ \\
= \left[i(\kappa^I_{1})^3k^I_1\left(e^{-i\kappa^I_{1}(x)}-R_1(0)e^{i\kappa^I_{1}(x)}\right) \tanh{(k^I_1 h)}
- 2ik_y^2\kappa^I_{1}k^I_1\left(-e^{-i\kappa^I_{1}(x)} + R_1(0)e^{i\kappa^I_{1}(x)}\right) \tanh{(k^I_1h)}\right]
\left[k^I_1\left(e^{i\kappa^I_{1}(x)} + R_1(0)^*e^{-i\kappa^I_{1}(x))}\right) \tanh{(k^I_1h)}\right], \\ \\
= \left[i\kappa^I_{1}k_1(0)((\kappa^I_{1})^2 + 2k_y^2)\left(e^{-i\kappa^I_{1}(x)} - R_1(0)e^{i\kappa^I_{1}(x)}\right)
\tanh{(k^I_1h)}\right]
\left[k^I_1\left(e^{i\kappa^I_{1}(x)} + R_1(0)^*e^{-i\kappa^I_{1}(x))}\right) \tanh{(k^I_1h)}\right], \\ \\
= i\kappa^I_{1}(k^I_1)^2\left((\kappa^I_{1})^2 + 2k_y^2\right)\tanh^2{(k^I_1h)}\left(1 - |R_1(0)|^2\right),
\end{matrix}
</math></center>

Likewise we expand the right hand term for <math>x>0</math>, if <math>\theta \in [-\theta_0, \theta_0]</math>,

<center><math>
\begin{matrix}
\frac{\partial^2}{\partial x^2}\frac{\partial\phi(x_2,0)}{\partial z}\cdot
\frac{\partial}{\partial x}\frac{\partial\phi^*}{\partial z} \\ \\
= \left[-(\kappa^I_{\Lambda})^2k^I_\Lambda T_\Lambda(0)e^{-i\kappa^I_{\Lambda}(x)} \tanh{(k^I_\Lambda h)}\right]
\left[i\kappa^I_{\Lambda} k^I_\Lambda T_\Lambda(0)^*e^{i\kappa^I_{\Lambda}(x)} \tanh{(k^I_\Lambda h)}\right], \\ \\
= -i(\kappa^I_{\Lambda})^3(k^I_\Lambda)^2\tanh^2{(k^I_\Lambda h)}|T_\Lambda(0)|^2,
\end{matrix}
</math></center>
else this term is equal to 0 as well,
and finally for <math>x<0</math>,

<center><math>
\begin{matrix}
\frac{\partial^2}{\partial x^1}\frac{\partial\phi(x_1,0)}{\partial z}\cdot
\frac{\partial}{\partial x}\frac{\partial\phi^*}{\partial z}\\ \\
= \left[-(\kappa^I_{1})^2k^I_1\left(e^{-i\kappa^I_{1}(x)} + R_1(0)e^{i\kappa^I_{1}(x)}\right)\tanh{(k^I_1 h)}\right]
\left[i\kappa^I_{1} k^I_1\left(e^{i\kappa^I_{1}(x)} - R_1(0)e^{-i\kappa^I_{1}(x)}\right)\tanh{(k^I_1 h)}\right], \\ \\
= -i(\kappa^I_{1})^3(k^I_1)^2\tanh^2{(k^I_1h)}\left(1 - |R_1(0)|^2\right).
\end{matrix}
</math></center>

We can now express <math>\xi_3</math> as

<center><math>
\begin{matrix}
\xi_3 & = & \Im
\left\{\frac{\beta_\Lambda}{\alpha}\left[i\kappa^I_{\Lambda} (k^I_\Lambda)^2\left((\kappa^I_{\Lambda})^2 + 2k_y^2 \right)
\tanh^2{(k^I_\Lambda h)}|T_\Lambda(0)|^2\right]
-\frac{\beta_1}{\alpha}\left[i\kappa^I_{1}(k^I_1)^2\left((\kappa^I_{1})^2 + 2k_y^2\right)\tanh^2{(k^I_1h)}
\left(1 - |R_1(0)|^2\right)\right]
-\frac{\beta_\Lambda}{\alpha}\left[-i(\kappa^I_{\Lambda})^3(k^I_\Lambda)^2\tanh^2{(k^I_\Lambda h)}|T_\Lambda(0)|^2\right]
+\frac{\beta_1}{\alpha}\left[-i(\kappa^I_{1})^3(k^I_1)^2\tanh^2{(k^I_1h)} \left(1 - |R_1(0)|^2\right)\right]\right\}, \\ \\
& = & \frac{\beta_\Lambda}{\alpha}\left[2\kappa^I_{\Lambda} (k^I_\Lambda)^2((\kappa^I_{\Lambda})^2 + k_y^2)\tanh^2{(k^I_\Lambda h)}|T_\Lambda(0)|^2\right]
-\frac{\beta_1}{\alpha}\left[2\kappa^I_{1}(k^I_1)^2((\kappa^I_{1})^2 + k_y^2)\tanh^2{(k^I_1h)}(1-|R_1(0)|^2)\right].
\end{matrix}
</math></center>
if <math>\theta \in [-\theta_0, \theta_0]</math>, and if not, we obtain:
<center>
<math>
\xi_3 = -\frac{\beta_1}{\alpha}\left[2\kappa^I_{1}(k^I_1)^2((\kappa^I_{1})^2 + k_y^2)\tanh^2{(k^I_1h)}(1-|R_1(0)|^2)\right].
</math>
</center>

= Solving the Energy Balance Equation =

Pulling it all together for the case <math>\theta \in [-\theta_0, \theta_0]</math>, we finally obtain
<center><math>
\frac{\beta_\Lambda}{\alpha}\left[2\kappa^I_{\Lambda}(k^I_\Lambda)^2((\kappa^I_{\Lambda})^2 +k_y^2)
\tanh^2{(k^I_\Lambda h)}|T_\Lambda(0)|^2\right]
-\frac{\beta_1}{\alpha}\left[2\kappa^I_{1}(k^I_1)^2((\kappa^I_{1})^2 +k_y^2)\tanh^2{(k^I_1h)}(1-|R_1(0)|^2)\right]
+ \frac{\kappa^I_{\Lambda}|T_\Lambda(0)|^2}{2k^I_\Lambda}\left(\tanh{(k^I_\Lambda h)}+
\frac{k^I_\Lambda h}{\cosh^2{(k^I_\Lambda h)}}\right)
- \frac{\kappa^I_{1}\left(1 - |R_1(0)|^2\right)}{2k^I_1}\left(\tanh{(k^I_1h)}+\frac{hk^I_1}{\cosh^2{(k^I_1h)}}\right)=0.
</math></center>

Re-arranging gives

<center><math>
\kappa^I_{\Lambda}\tanh{(k^I_\Lambda h)}
\left(\frac{\beta_\Lambda}{\alpha}2(k^I_\Lambda)^2((\kappa^I_{\Lambda})^2 +k_y^2)\tanh{(k^I_\Lambda h)}
+ \frac{1}{2k^I_\Lambda}+\frac{h}{2\sinh{(k^I_\Lambda h)}\cosh{(k^I_\Lambda h)}}\right)|T_\Lambda(0)|^2
\quad - \kappa^I_{1}\tanh{(k^I_1 h)}
\left(\frac{\beta_1}{\alpha}2(k^I_1)^2((\kappa^I_{1})^2 +k_y^2)\tanh{(k^I_1 h)}
+ \frac{1}{2it}+\frac{h}{2\sinh{(k^I_1 h)}\cosh{(k^I_1 h)}}\right)(1-|R_1(0)|^2)=0
</math></center>

which can be expressed as
<center><math>
D |T_{\Lambda}(0)|^2 + |R_{1}(0)|^2 = 1,
</math></center>
where <math>D</math> is given by
<center><math>
D = \left(\frac{\kappa^I_{\Lambda} k^I_1\cosh^2{(k^I_1h)}}{\kappa^I_{1}k^I_\Lambda\cosh^2{(k^I_\Lambda h)}}\right)
\left(\frac{-\frac{\beta_\Lambda}{\alpha}4(k^I_\Lambda)^5\sinh^2{(k^I_\Lambda h)} +
\frac{1}{2}{\sinh{(2k^I_\Lambda h)}}+k^I_\Lambda h}
{-\frac{\beta_1}{\alpha}4(k^I_1)^5\sinh^2{(k^I_1h)} +
\frac{1}{2}{\sinh{(2k^I_1h)}}+k^I_1h}\right)
</math></center>
For the case of greater angles, there are no terms depending on <math>|T_{\Lambda}(0)|^2</math>, so we obtain <math>D=0</math> and <math>|R_{1}(0)|^2 = 1</math>, which is the case of the total reflexion. No energy is transmitted in the <math>x>0</math> region.

[[Category:Floating Elastic Plate]]

Energy Balance for Two Elastic Plates

2008-08-26T23:42:33Z

Fmontiel: /* Solving the Energy Balance Equation */

= Introduction =

We derive here a condition which must be satisfied for energy balance for two semi-infinite elastic plates. The solution applies
to the problem of [[Eigenfunction Matching Method for Floating Elastic Plates|multiple elastic plates]] and from here we know we can write the potential as
<center><math>
\phi \approx \left\{
\begin{matrix}
{
Ie^{\kappa_{1}(0)(x-r_1)}\frac{\cos(k_1(0)(z+h))}{\cos(k_1(0)h)} }+\\
{
\qquad \qquad \sum_{n=-2}^{M}R_1(n) e^{\kappa_{1}(n)(x-r_1)} \frac{\cos(k_1(n)(z+h))}{\cos(k_1(n)h)} },&
\mbox{ for } x < r_1,\\
{
\sum_{n=-2}^{M}T_{\mu}(n) e^{-\kappa_\mu(n)(x-l_\mu)}
\frac{\cos(k_\mu(n)(z+h))}{\cos(k_\mu(n)h)} } + \\
{
\qquad \qquad \sum_{n=-2}^{M}R_{\mu}(n) e^{\kappa_\mu(n)(x-r_\mu)}
\frac{\cos(k_\mu(n)(z+h))}{\cos(k_\mu(n)h)} },
&\mbox{ for } l_\mu< x < r_\mu,\\
{
\sum_{n=-2}^{M}T_{\Lambda}(n)e^{-\kappa_{\Lambda}(n)(x-l_\Lambda)}
\frac{\cos(k_\Lambda(n)(z+h))}{\cos(k_\Lambda(n)h)} }, &\mbox{ for } l_\Lambda<x,
\end{matrix} \right.
</math></center>
(for details of this notation see [[Eigenfunction Matching Method for Floating Elastic Plates]]).

= Equations =

Based on the method used in [[Evans and Davies 1968]], a check can be made to ensure the solutions of the floating plate problem are in energy balance.
This is simply a condition that the incident energy is equal to the sum of the radiated energy.
When the first and final plates have different properties, the energy balance equation is derived by applying Green's theorem to <math>\phi</math> and its conjugate.
The domain of integration is shown in the figure on the right.

[[Image:energy_schematic.jpg|thumb|right|300px|A diagram depicting the area <math>\mathcal{U}</math> which is bounded by the rectangle <math>\mathcal{S}</math>.]
{A diagram depicting the area <math>\mathcal{U}</math> which is bounded by the rectangle <math>\mathcal{S}</math>.
The rectangle <math>\mathcal{S}</math> is bounded by <math> -h\leq z \leq0</math> and <math>-\infty\leq x \leq \infty</math>]]

Applying Green's theorem to <math>\phi</math> and its conjugate <math>\phi^*</math> gives
<center><math>
{ \int\int_\mathcal{U}(\phi\nabla^2\phi^* - \phi^*\nabla^2\phi)dxdz
= \int_\mathcal{S}(\phi\frac{\partial\phi^*}{\partial n} - \phi^*\frac{\partial\phi}{\partial n})dl },
</math></center>
where <math>n</math> denotes the outward plane normal to the boundary and <math>l</math> denotes the plane parallel to the boundary.
As <math>\phi</math> and <math>\phi^*</math> satisfy the Laplace's equation, the left hand side of the Green theorem equation vanishes so that it reduces to
<center><math>
\Im\int_\mathcal{S}\phi\frac{\partial\phi^*}{\partial n} dl = 0,
</math></center>
Expanding gives
<center><math>
\xi_1 +\xi_2 + \xi_3 = 0,
</math></center>
where
<center><math>
\xi_1 =
-{ \Im\int_{-h}^0(\phi\frac{\partial\phi^*}{\partial x})\big|_{x=-\infty}dz },
</math></center>
<center><math>
\xi_2 =
{ \Im\int_{-h}^0(\phi\frac{\partial\phi^*}{\partial x})\big|_{x=\infty}dz },
</math></center>
and
<center><math>
\xi_3 =
{ \Im\int_{-\infty}^{\infty}(\phi\frac{\partial\phi^*}{\partial z})\big|_{z=0}dx } ,
</math></center>
where <math>\Im</math> denotes the imaginary part. The bottom domain composant is obviously equal to zero because of the no-influx condition over the seabed (<math>\frac{\partial\phi^*}{\partial z}\big|_{z=-h}=0</math>).

= Expanding <math>\mathbf{\xi_1}</math> =

Near <math>x=-\infty</math>, we approximate <math>\phi</math> by
<center><math>
\phi \approx e^{-\kappa_{1}(0)(x-r_1)}\frac{\cos{(k_1(0)(z+h))}}{\cos{(k_1(0)h)}} +
R_1(0)e^{\kappa_{1}(0)(x-r_1)}\frac{\cos{(k_{1}(0)(z+h))}}{\cos{(k_{1}(0)h)}}
</math></center>
Therefore,
<center><math>
\begin{matrix}
\xi_1 & = & {-\Im\int_{-h}^0 \left[\left(e^{-i\kappa^{I}_1(x)}+ R_1(0)e^{i\kappa^{I}_1(x)}\right)\right.}
{ \left(i\kappa^{I}_1e^{i\kappa^{I}_1(x)} - i\kappa^{I}_1R_1(0)^*e^{-i\kappa^{I}_1(x)}\right) }
{ \left.\left(\frac{\cosh^2{(k^I_1(z+h))}}{\cosh^2{(k^I_1h)}}\right)\right]dz, }\\ \\
& = & { - \Im\left[\frac{i\kappa^{I}_1\left(1 - |R_1(0)|^2\right)}{2\cosh^2{(k^I_1h)}}
\int_{-h}^0 \left(\cosh{(2k^I_1(z+h))+1}\right)dz\right] ,}\\ \\
& = & { - \Im\left[\frac{i\kappa^{I}_1\left(1 - |R_1(0)|^2\right)}{2\cosh^2{(k^I_1h)}}
\left[\frac{1}{2k^I_1}\sinh{(2k^I_1(z+h))} + z\right]_{-h}^0\right], }\\ \\
& = & { - \frac{\kappa^{I}_1\left(1 - |R_1(0)|^2\right)}{2\cosh^2{(k^I_1h)}} \left(\frac{1}{2k^I_1}\sinh{(2k^I_1h))} + h\right) ,}\\ \\
& = & { - \frac{\kappa^{I}_1\left(1 - |R_1(0)|^2\right)}{2k^I_1}\left(\tanh{(k^I_1h)}+\frac{hk^I_1}{\cosh^2{(k^I_1h)}}\right) , }
\end{matrix}
</math></center>
where <math>R_1(0)^*</math> is the conjugate of <math>R_1(0)</math>.

= Expanding <math>\mathbf{\xi_2}</math> =

Near <math>x=\infty</math>, we approximate <math>\phi</math> by
<center><math>
\begin{matrix}
{ \phi \approx T_\Lambda(0)e^{-\kappa_{\Lambda}(0)(x)}\frac{\cos{(k_\Lambda(0)(z+h))}}{\cos{(k_\Lambda(0)h)}}, }
\end{matrix}
</math></center>
and re-express as
<center><math>
\phi \approx T_\Lambda(0)e^{-i\kappa^I_\Lambda(x)}\frac{\cosh{(k^I_\Lambda(z+h))}}{\cosh{(k^I_\Lambda h)}},
</math></center>
where <math>k^I_\Lambda = \Im k_{\Lambda}(0)</math> and <math>\kappa^I_\Lambda=\Im \kappa_\Lambda(0)</math>, so that
<center><math>
\frac{\partial\phi}{\partial x} \approx -i\kappa^I_{\Lambda}T_\Lambda(0)e^{-i\kappa^I_{\Lambda}(x)}\frac{\cosh{(k^I_\Lambda(z+h))}}{\cosh{(k^I_\Lambda h)}}.
</math></center>
We know that <math>\kappa_\Lambda(0)=\sqrt{k_{\Lambda}(0)^2-(ik_y)^2}</math>, where <math>k_y</math> is real and depends on the incident angle <math>\theta</math>. When <math>\theta</math> becomes greater than a certain angle <math>\theta_0</math> defined by <math>\sin \theta_0=\sqrt{\frac{k_{\Lambda}(0)}{k_{1}(0)}}</math>, <math>\kappa_\Lambda(0)</math> becomes real so the potential becomes real as well. Thus the imaginary part is equal to 0. In that case, we have <math>\xi_2=0</math>.
Therefore, if <math>\theta \in [-\theta_0, \theta_0]</math>,
<center><math>
\begin{matrix}
\xi_2 & = & \Im\int_{-h}^0 \left[\right.(T_\Lambda(0)e^{-i\kappa^I_{\Lambda}(x)})(i\kappa^I_{\Lambda}T_\Lambda(0)^*e^{i\kappa^I_{\Lambda}(x)})
\frac{\cosh^2{(k^I_\Lambda(z+h))}}{\cosh^2{(k^I_\Lambda h)}}\left.\right]dz, \\ \\
& = & { \frac{\kappa^I_{\Lambda}|T_\Lambda(0)|^2}{2\cosh^2{(k^I_\Lambda h)}}
\left(\frac{1}{2k^I_\Lambda}\sinh{(2k^I_\Lambda h)} + h\right)},\\ \\
& = & { \frac{\kappa^I_{\Lambda}|T_\Lambda(0)|^2}{2k^I_\Lambda}\left(\tanh{(k^I_\Lambda h)}+\frac{k^I_\Lambda h}{\cosh^2{(k^I_\Lambda h)}}\right)},
\end{matrix}
</math></center>
otherwise <math>\xi_2=0</math>.

= Expanding <math>\mathbf{\xi_3}</math> =

The ice-covered boundary condition for the [[Floating Elastic Plate]] gives
<center><math>
\xi_3 = { \Im\int_{-\infty}^{\infty}\left(\frac{\beta}{\alpha}
\left(\frac{\partial^2}{\partial x^2} - k_y^2\right)^2 - \gamma + \frac{1}{\alpha}\right)
\frac{\partial\phi}{\partial z}\cdot\frac{\partial\phi^*}{\partial z}\bigg|_{z=0}dx.}
</math></center>
Since <math>{\frac{\partial\phi}{\partial z}\cdot\frac{\partial\phi^*}{\partial z}}</math> is real,
<center><math>
\xi_3 = \Im\int_{-\infty}^{\infty}\left(\frac{\beta}{\alpha}\frac{\partial^2}{\partial x^2}
\left(\frac{\partial^2}{\partial x^2} - 2k_y^2\right) \right)
\frac{\partial\phi}{\partial z}\cdot\frac{\partial\phi^*}{\partial z}\bigg|_{z=0}dx.
</math></center>
Integration by parts gives
<center><math>
\xi_3 = \Im\left\{\left[\frac{\beta}{\alpha}\frac{\partial}{\partial x}
\left(\frac{\partial^2}{\partial x^2}-2k_y^2\right)
\frac{\partial\phi}{\partial z}\cdot\frac{\partial\phi^*}{\partial z}\right]_{-\infty}^{\infty}
-\int_{-\infty}^{\infty}\frac{\beta}{\alpha}\frac{\partial}{\partial x}
\left(\frac{\partial^2}{\partial x^2} - 2k_y^2\right)
\frac{\partial\phi}{\partial z}\cdot\frac
{\partial}{\partial x}\frac{\partial\phi^*}{\partial z}dx
\right\}.
</math></center>
As <math>{2k_y^2\frac{\partial}{\partial x}\frac{\partial\phi}{\partial z}\cdot \frac{\partial}{\partial x}\frac{\partial\phi^*}{\partial z}}</math> is real and by integration by parts, the expression of <math>\xi_3</math> becomes,
<center><math>
\xi_3 =\Im\left\{
\left[\frac{\beta}{\alpha}\frac{\partial}{\partial x}
\left(\frac{\partial^2}{\partial x^2}-2k_y^2\right)
\frac{\partial\phi}{\partial z}\cdot\frac{\partial\phi^*}{\partial z}\right]_{-\infty}^{\infty}
- \left[\frac{\beta}{\alpha}\frac{\partial^2}{\partial x^2}\frac{\partial\phi}{\partial z}\cdot
\frac{\partial}{\partial x}\frac{\partial\phi^*}{\partial z}\right]_{-\infty}^{\infty}
+ \int_{-\infty}^\infty\frac{\beta}{\alpha}\frac{\partial^2}{\partial x^2}\frac{\partial\phi}{\partial z}\cdot
\frac{\partial^2}{\partial x^2}\frac{\partial\phi^*}{\partial z}dx\right\}.
</math></center>
As <math>{\frac{\partial^2}{\partial x^2}\frac{\partial\phi}{\partial z}\cdot \frac{\partial^2}{\partial x^2}\frac{\partial\phi^*}{\partial z}}</math> is real, we obtain the new expression of <math>\xi_3</math>
<center><math>
\xi_3 = \Im\left\{
\left[\frac{\beta}{\alpha}\frac{\partial}{\partial x}
\left(\frac{\partial^2}{\partial x^2}-2k_y^2\right)
\frac{\partial\phi}{\partial z}\cdot\frac{\partial\phi^*}{\partial z}\right]_{-\infty}^{\infty}-
\left[\frac{\beta}{\alpha}\frac{\partial^2}{\partial x^2}\frac{\partial\phi}{\partial z}\cdot
\frac{\partial}{\partial x}\frac{\partial\phi^*}{\partial z}\right]_{-\infty}^{\infty}\right\}.
</math></center>

Now breaking <math>\xi_3</math> down, we can simplify the left hand term for <math>x>0</math>,
if <math>\theta \in [-\theta_0, \theta_0]</math>,

<center><math>
\begin{matrix}
\frac{\partial}{\partial x}\left(\frac{\partial^2}{\partial x^2}-2k_y^2\right)
\frac{\partial\phi(x_2,0)}{\partial z}\cdot\frac{\partial\phi(x_2,0)^*}{\partial z} \\ \\
= \left((-i\kappa^I_{\Lambda})^3k^I_\Lambda T_\Lambda(0)e^{-i\kappa^I_{\Lambda}(x)}\tanh{(k^I_\Lambda h)}\right.
\left.- 2k_y^2(-i\kappa_{\Lambda}(0))k^I_\Lambda T_\Lambda(0)e^{-i\kappa^I_{\Lambda}(x)} \tanh{(k^I_\Lambda h)}\right)
\left(k^I_\Lambda T_\Lambda(0)^*e^{i\kappa^I_{\Lambda}(x)}\tanh{(k^I_\Lambda h)}\right), \\ \\
= i\kappa^I_{\Lambda} (k^I_\Lambda)^2\left((\kappa^I_{\Lambda})^2 + 2k_y^2 \right) \tanh^2{(k^I_\Lambda h)}|T_\Lambda(0)|^2,
\end{matrix}
</math></center>
else this term is equal to 0.

and for <math>x<0</math>

<center><math>
\begin{matrix}
\frac{\partial}{\partial x}\left(\frac{\partial^2}{\partial x^2}-2k_y^2\right)\frac{\partial\phi(x_1,0)}{\partial z}\cdot
\frac{\partial\phi(x_1,0)^*}{\partial z} \\ \\
= \left[i(\kappa^I_{1})^3k^I_1\left(e^{-i\kappa^I_{1}(x)}-R_1(0)e^{i\kappa^I_{1}(x)}\right) \tanh{(k^I_1 h)}
- 2ik_y^2\kappa^I_{1}k^I_1\left(-e^{-i\kappa^I_{1}(x)} + R_1(0)e^{i\kappa^I_{1}(x)}\right) \tanh{(k^I_1h)}\right]
\left[k^I_1\left(e^{i\kappa^I_{1}(x)} + R_1(0)^*e^{-i\kappa^I_{1}(x))}\right) \tanh{(k^I_1h)}\right], \\ \\
= \left[i\kappa^I_{1}k_1(0)((\kappa^I_{1})^2 + 2k_y^2)\left(e^{-i\kappa^I_{1}(x)} - R_1(0)e^{i\kappa^I_{1}(x)}\right)
\tanh{(k^I_1h)}\right]
\left[k^I_1\left(e^{i\kappa^I_{1}(x)} + R_1(0)^*e^{-i\kappa^I_{1}(x))}\right) \tanh{(k^I_1h)}\right], \\ \\
= i\kappa^I_{1}(k^I_1)^2\left((\kappa^I_{1})^2 + 2k_y^2\right)\tanh^2{(k^I_1h)}\left(1 - |R_1(0)|^2\right),
\end{matrix}
</math></center>

Likewise we expand the right hand term for <math>x>0</math>, if <math>\theta \in [-\theta_0, \theta_0]</math>,

<center><math>
\begin{matrix}
\frac{\partial^2}{\partial x^2}\frac{\partial\phi(x_2,0)}{\partial z}\cdot
\frac{\partial}{\partial x}\frac{\partial\phi^*}{\partial z} \\ \\
= \left[-(\kappa^I_{\Lambda})^2k^I_\Lambda T_\Lambda(0)e^{-i\kappa^I_{\Lambda}(x)} \tanh{(k^I_\Lambda h)}\right]
\left[i\kappa^I_{\Lambda} k^I_\Lambda T_\Lambda(0)^*e^{i\kappa^I_{\Lambda}(x)} \tanh{(k^I_\Lambda h)}\right], \\ \\
= -i(\kappa^I_{\Lambda})^3(k^I_\Lambda)^2\tanh^2{(k^I_\Lambda h)}|T_\Lambda(0)|^2,
\end{matrix}
</math></center>
else this term is equal to 0 as well,
and finally for <math>x<0</math>,

<center><math>
\begin{matrix}
\frac{\partial^2}{\partial x^1}\frac{\partial\phi(x_1,0)}{\partial z}\cdot
\frac{\partial}{\partial x}\frac{\partial\phi^*}{\partial z}\\ \\
= \left[-(\kappa^I_{1})^2k^I_1\left(e^{-i\kappa^I_{1}(x)} + R_1(0)e^{i\kappa^I_{1}(x)}\right)\tanh{(k^I_1 h)}\right]
\left[i\kappa^I_{1} k^I_1\left(e^{i\kappa^I_{1}(x)} - R_1(0)e^{-i\kappa^I_{1}(x)}\right)\tanh{(k^I_1 h)}\right], \\ \\
= -i(\kappa^I_{1})^3(k^I_1)^2\tanh^2{(k^I_1h)}\left(1 - |R_1(0)|^2\right).
\end{matrix}
</math></center>

We can now express <math>\xi_3</math> as

<center><math>
\begin{matrix}
\xi_3 & = & \Im
\left\{\frac{\beta_\Lambda}{\alpha}\left[i\kappa^I_{\Lambda} (k^I_\Lambda)^2\left((\kappa^I_{\Lambda})^2 + 2k_y^2 \right)
\tanh^2{(k^I_\Lambda h)}|T_\Lambda(0)|^2\right]
-\frac{\beta_1}{\alpha}\left[i\kappa^I_{1}(k^I_1)^2\left((\kappa^I_{1})^2 + 2k_y^2\right)\tanh^2{(k^I_1h)}
\left(1 - |R_1(0)|^2\right)\right]
-\frac{\beta_\Lambda}{\alpha}\left[-i(\kappa^I_{\Lambda})^3(k^I_\Lambda)^2\tanh^2{(k^I_\Lambda h)}|T_\Lambda(0)|^2\right]
+\frac{\beta_1}{\alpha}\left[-i(\kappa^I_{1})^3(k^I_1)^2\tanh^2{(k^I_1h)} \left(1 - |R_1(0)|^2\right)\right]\right\}, \\ \\
& = & \frac{\beta_\Lambda}{\alpha}\left[2\kappa^I_{\Lambda} (k^I_\Lambda)^2((\kappa^I_{\Lambda})^2 + k_y^2)\tanh^2{(k^I_\Lambda h)}|T_\Lambda(0)|^2\right]
-\frac{\beta_1}{\alpha}\left[2\kappa^I_{1}(k^I_1)^2((\kappa^I_{1})^2 + k_y^2)\tanh^2{(k^I_1h)}(1-|R_1(0)|^2)\right].
\end{matrix}
</math></center>
if <math>\theta \in [-\theta_0, \theta_0]</math>, and if not, we obtain:
<center>
<math>
\xi_3 = -\frac{\beta_1}{\alpha}\left[2\kappa^I_{1}(k^I_1)^2((\kappa^I_{1})^2 + k_y^2)\tanh^2{(k^I_1h)}(1-|R_1(0)|^2)\right].
</math>
</center>

= Solving the Energy Balance Equation =

Pulling it all together for the case <math>\theta \in [-\theta_0, \theta_0]</math>, we finally obtain
<center><math>
\frac{\beta_\Lambda}{\alpha}\left[2\kappa^I_{\Lambda}(k^I_\Lambda)^2((\kappa^I_{\Lambda})^2 +k_y^2)
\tanh^2{(k^I_\Lambda h)}|T_\Lambda(0)|^2\right]
-\frac{\beta_1}{\alpha}\left[2\kappa^I_{1}(k^I_1)^2((\kappa^I_{1})^2 +k_y^2)\tanh^2{(k^I_1h)}(1-|R_1(0)|^2)\right]
+ \frac{\kappa^I_{\Lambda}|T_\Lambda(0)|^2}{2k^I_\Lambda}\left(\tanh{(k^I_\Lambda h)}+
\frac{k^I_\Lambda h}{\cosh^2{(k^I_\Lambda h)}}\right)
- \frac{\kappa^I_{1}\left(1 - |R_1(0)|^2\right)}{2k^I_1}\left(\tanh{(k^I_1h)}+\frac{hk^I_1}{\cosh^2{(k^I_1h)}}\right)=0.
</math></center>

Re-arranging gives

<center><math>
\kappa^I_{\Lambda}\tanh{(k^I_\Lambda h)}
\left(\frac{\beta_\Lambda}{\alpha}2(k^I_\Lambda)^2((\kappa^I_{\Lambda})^2 +k_y^2)\tanh{(k^I_\Lambda h)}
+ \frac{1}{2k^I_\Lambda}+\frac{h}{2\sinh{(k^I_\Lambda h)}\cosh{(k^I_\Lambda h)}}\right)|T_\Lambda(0)|^2
\quad - \kappa^I_{1}\tanh{(k^I_1 h)}
\left(\frac{\beta_1}{\alpha}2(k^I_1)^2((\kappa^I_{1})^2 +k_y^2)\tanh{(k^I_1 h)}
+ \frac{1}{2it}+\frac{h}{2\sinh{(k^I_1 h)}\cosh{(k^I_1 h)}}\right)(1-|R_1(0)|^2)=0
</math></center>

which can be expressed as
<center><math>
D |T_{\Lambda}(0)|^2 + |R_{1}(0)|^2 = 1,
</math></center>
where <math>D</math> is given by
<center><math>
D = \left(\frac{\kappa^I_{\Lambda} k^I_1\cosh^2{(k^I_1h)}}{\kappa^I_{1}k^I_\Lambda\cosh^2{(k^I_\Lambda h)}}\right)
\left(\frac{-\frac{\beta_\Lambda}{\alpha}4(k^I_\Lambda)^5\sinh^2{(k^I_\Lambda h)} +
\frac{1}{2}{\sinh{(2k^I_\Lambda h)}}+k^I_\Lambda h}
{-\frac{\beta_1}{\alpha}4(k^I_1)^5\sinh^2{(k^I_1h)} +
\frac{1}{2}{\sinh{(2k^I_1h)}}+k^I_1h}\right)
</math></center>
For the case of greater angles, there are no terms depending on |T_{\Lambda}(0)|^2, so we obtain <math>D=0</math> and <math>|R_{1}(0)|^2 = 1</math>, which is the case of the total reflexion. No energy is transmitted in the <math>x>0</math> region.

[[Category:Floating Elastic Plate]]

Energy Balance for Two Elastic Plates

2008-08-26T23:37:03Z

Fmontiel: /* Expanding <math>\mathbf{\xi_3}</math> */

= Introduction =

We derive here a condition which must be satisfied for energy balance for two semi-infinite elastic plates. The solution applies
to the problem of [[Eigenfunction Matching Method for Floating Elastic Plates|multiple elastic plates]] and from here we know we can write the potential as
<center><math>
\phi \approx \left\{
\begin{matrix}
{
Ie^{\kappa_{1}(0)(x-r_1)}\frac{\cos(k_1(0)(z+h))}{\cos(k_1(0)h)} }+\\
{
\qquad \qquad \sum_{n=-2}^{M}R_1(n) e^{\kappa_{1}(n)(x-r_1)} \frac{\cos(k_1(n)(z+h))}{\cos(k_1(n)h)} },&
\mbox{ for } x < r_1,\\
{
\sum_{n=-2}^{M}T_{\mu}(n) e^{-\kappa_\mu(n)(x-l_\mu)}
\frac{\cos(k_\mu(n)(z+h))}{\cos(k_\mu(n)h)} } + \\
{
\qquad \qquad \sum_{n=-2}^{M}R_{\mu}(n) e^{\kappa_\mu(n)(x-r_\mu)}
\frac{\cos(k_\mu(n)(z+h))}{\cos(k_\mu(n)h)} },
&\mbox{ for } l_\mu< x < r_\mu,\\
{
\sum_{n=-2}^{M}T_{\Lambda}(n)e^{-\kappa_{\Lambda}(n)(x-l_\Lambda)}
\frac{\cos(k_\Lambda(n)(z+h))}{\cos(k_\Lambda(n)h)} }, &\mbox{ for } l_\Lambda<x,
\end{matrix} \right.
</math></center>
(for details of this notation see [[Eigenfunction Matching Method for Floating Elastic Plates]]).

= Equations =

Based on the method used in [[Evans and Davies 1968]], a check can be made to ensure the solutions of the floating plate problem are in energy balance.
This is simply a condition that the incident energy is equal to the sum of the radiated energy.
When the first and final plates have different properties, the energy balance equation is derived by applying Green's theorem to <math>\phi</math> and its conjugate.
The domain of integration is shown in the figure on the right.

[[Image:energy_schematic.jpg|thumb|right|300px|A diagram depicting the area <math>\mathcal{U}</math> which is bounded by the rectangle <math>\mathcal{S}</math>.]
{A diagram depicting the area <math>\mathcal{U}</math> which is bounded by the rectangle <math>\mathcal{S}</math>.
The rectangle <math>\mathcal{S}</math> is bounded by <math> -h\leq z \leq0</math> and <math>-\infty\leq x \leq \infty</math>]]

Applying Green's theorem to <math>\phi</math> and its conjugate <math>\phi^*</math> gives
<center><math>
{ \int\int_\mathcal{U}(\phi\nabla^2\phi^* - \phi^*\nabla^2\phi)dxdz
= \int_\mathcal{S}(\phi\frac{\partial\phi^*}{\partial n} - \phi^*\frac{\partial\phi}{\partial n})dl },
</math></center>
where <math>n</math> denotes the outward plane normal to the boundary and <math>l</math> denotes the plane parallel to the boundary.
As <math>\phi</math> and <math>\phi^*</math> satisfy the Laplace's equation, the left hand side of the Green theorem equation vanishes so that it reduces to
<center><math>
\Im\int_\mathcal{S}\phi\frac{\partial\phi^*}{\partial n} dl = 0,
</math></center>
Expanding gives
<center><math>
\xi_1 +\xi_2 + \xi_3 = 0,
</math></center>
where
<center><math>
\xi_1 =
-{ \Im\int_{-h}^0(\phi\frac{\partial\phi^*}{\partial x})\big|_{x=-\infty}dz },
</math></center>
<center><math>
\xi_2 =
{ \Im\int_{-h}^0(\phi\frac{\partial\phi^*}{\partial x})\big|_{x=\infty}dz },
</math></center>
and
<center><math>
\xi_3 =
{ \Im\int_{-\infty}^{\infty}(\phi\frac{\partial\phi^*}{\partial z})\big|_{z=0}dx } ,
</math></center>
where <math>\Im</math> denotes the imaginary part. The bottom domain composant is obviously equal to zero because of the no-influx condition over the seabed (<math>\frac{\partial\phi^*}{\partial z}\big|_{z=-h}=0</math>).

= Expanding <math>\mathbf{\xi_1}</math> =

Near <math>x=-\infty</math>, we approximate <math>\phi</math> by
<center><math>
\phi \approx e^{-\kappa_{1}(0)(x-r_1)}\frac{\cos{(k_1(0)(z+h))}}{\cos{(k_1(0)h)}} +
R_1(0)e^{\kappa_{1}(0)(x-r_1)}\frac{\cos{(k_{1}(0)(z+h))}}{\cos{(k_{1}(0)h)}}
</math></center>
Therefore,
<center><math>
\begin{matrix}
\xi_1 & = & {-\Im\int_{-h}^0 \left[\left(e^{-i\kappa^{I}_1(x)}+ R_1(0)e^{i\kappa^{I}_1(x)}\right)\right.}
{ \left(i\kappa^{I}_1e^{i\kappa^{I}_1(x)} - i\kappa^{I}_1R_1(0)^*e^{-i\kappa^{I}_1(x)}\right) }
{ \left.\left(\frac{\cosh^2{(k^I_1(z+h))}}{\cosh^2{(k^I_1h)}}\right)\right]dz, }\\ \\
& = & { - \Im\left[\frac{i\kappa^{I}_1\left(1 - |R_1(0)|^2\right)}{2\cosh^2{(k^I_1h)}}
\int_{-h}^0 \left(\cosh{(2k^I_1(z+h))+1}\right)dz\right] ,}\\ \\
& = & { - \Im\left[\frac{i\kappa^{I}_1\left(1 - |R_1(0)|^2\right)}{2\cosh^2{(k^I_1h)}}
\left[\frac{1}{2k^I_1}\sinh{(2k^I_1(z+h))} + z\right]_{-h}^0\right], }\\ \\
& = & { - \frac{\kappa^{I}_1\left(1 - |R_1(0)|^2\right)}{2\cosh^2{(k^I_1h)}} \left(\frac{1}{2k^I_1}\sinh{(2k^I_1h))} + h\right) ,}\\ \\
& = & { - \frac{\kappa^{I}_1\left(1 - |R_1(0)|^2\right)}{2k^I_1}\left(\tanh{(k^I_1h)}+\frac{hk^I_1}{\cosh^2{(k^I_1h)}}\right) , }
\end{matrix}
</math></center>
where <math>R_1(0)^*</math> is the conjugate of <math>R_1(0)</math>.

= Expanding <math>\mathbf{\xi_2}</math> =

Near <math>x=\infty</math>, we approximate <math>\phi</math> by
<center><math>
\begin{matrix}
{ \phi \approx T_\Lambda(0)e^{-\kappa_{\Lambda}(0)(x)}\frac{\cos{(k_\Lambda(0)(z+h))}}{\cos{(k_\Lambda(0)h)}}, }
\end{matrix}
</math></center>
and re-express as
<center><math>
\phi \approx T_\Lambda(0)e^{-i\kappa^I_\Lambda(x)}\frac{\cosh{(k^I_\Lambda(z+h))}}{\cosh{(k^I_\Lambda h)}},
</math></center>
where <math>k^I_\Lambda = \Im k_{\Lambda}(0)</math> and <math>\kappa^I_\Lambda=\Im \kappa_\Lambda(0)</math>, so that
<center><math>
\frac{\partial\phi}{\partial x} \approx -i\kappa^I_{\Lambda}T_\Lambda(0)e^{-i\kappa^I_{\Lambda}(x)}\frac{\cosh{(k^I_\Lambda(z+h))}}{\cosh{(k^I_\Lambda h)}}.
</math></center>
We know that <math>\kappa_\Lambda(0)=\sqrt{k_{\Lambda}(0)^2-(ik_y)^2}</math>, where <math>k_y</math> is real and depends on the incident angle <math>\theta</math>. When <math>\theta</math> becomes greater than a certain angle <math>\theta_0</math> defined by <math>\sin \theta_0=\sqrt{\frac{k_{\Lambda}(0)}{k_{1}(0)}}</math>, <math>\kappa_\Lambda(0)</math> becomes real so the potential becomes real as well. Thus the imaginary part is equal to 0. In that case, we have <math>\xi_2=0</math>.
Therefore, if <math>\theta \in [-\theta_0, \theta_0]</math>,
<center><math>
\begin{matrix}
\xi_2 & = & \Im\int_{-h}^0 \left[\right.(T_\Lambda(0)e^{-i\kappa^I_{\Lambda}(x)})(i\kappa^I_{\Lambda}T_\Lambda(0)^*e^{i\kappa^I_{\Lambda}(x)})
\frac{\cosh^2{(k^I_\Lambda(z+h))}}{\cosh^2{(k^I_\Lambda h)}}\left.\right]dz, \\ \\
& = & { \frac{\kappa^I_{\Lambda}|T_\Lambda(0)|^2}{2\cosh^2{(k^I_\Lambda h)}}
\left(\frac{1}{2k^I_\Lambda}\sinh{(2k^I_\Lambda h)} + h\right)},\\ \\
& = & { \frac{\kappa^I_{\Lambda}|T_\Lambda(0)|^2}{2k^I_\Lambda}\left(\tanh{(k^I_\Lambda h)}+\frac{k^I_\Lambda h}{\cosh^2{(k^I_\Lambda h)}}\right)},
\end{matrix}
</math></center>
otherwise <math>\xi_2=0</math>.

= Expanding <math>\mathbf{\xi_3}</math> =

The ice-covered boundary condition for the [[Floating Elastic Plate]] gives
<center><math>
\xi_3 = { \Im\int_{-\infty}^{\infty}\left(\frac{\beta}{\alpha}
\left(\frac{\partial^2}{\partial x^2} - k_y^2\right)^2 - \gamma + \frac{1}{\alpha}\right)
\frac{\partial\phi}{\partial z}\cdot\frac{\partial\phi^*}{\partial z}\bigg|_{z=0}dx.}
</math></center>
Since <math>{\frac{\partial\phi}{\partial z}\cdot\frac{\partial\phi^*}{\partial z}}</math> is real,
<center><math>
\xi_3 = \Im\int_{-\infty}^{\infty}\left(\frac{\beta}{\alpha}\frac{\partial^2}{\partial x^2}
\left(\frac{\partial^2}{\partial x^2} - 2k_y^2\right) \right)
\frac{\partial\phi}{\partial z}\cdot\frac{\partial\phi^*}{\partial z}\bigg|_{z=0}dx.
</math></center>
Integration by parts gives
<center><math>
\xi_3 = \Im\left\{\left[\frac{\beta}{\alpha}\frac{\partial}{\partial x}
\left(\frac{\partial^2}{\partial x^2}-2k_y^2\right)
\frac{\partial\phi}{\partial z}\cdot\frac{\partial\phi^*}{\partial z}\right]_{-\infty}^{\infty}
-\int_{-\infty}^{\infty}\frac{\beta}{\alpha}\frac{\partial}{\partial x}
\left(\frac{\partial^2}{\partial x^2} - 2k_y^2\right)
\frac{\partial\phi}{\partial z}\cdot\frac
{\partial}{\partial x}\frac{\partial\phi^*}{\partial z}dx
\right\}.
</math></center>
As <math>{2k_y^2\frac{\partial}{\partial x}\frac{\partial\phi}{\partial z}\cdot \frac{\partial}{\partial x}\frac{\partial\phi^*}{\partial z}}</math> is real and by integration by parts, the expression of <math>\xi_3</math> becomes,
<center><math>
\xi_3 =\Im\left\{
\left[\frac{\beta}{\alpha}\frac{\partial}{\partial x}
\left(\frac{\partial^2}{\partial x^2}-2k_y^2\right)
\frac{\partial\phi}{\partial z}\cdot\frac{\partial\phi^*}{\partial z}\right]_{-\infty}^{\infty}
- \left[\frac{\beta}{\alpha}\frac{\partial^2}{\partial x^2}\frac{\partial\phi}{\partial z}\cdot
\frac{\partial}{\partial x}\frac{\partial\phi^*}{\partial z}\right]_{-\infty}^{\infty}
+ \int_{-\infty}^\infty\frac{\beta}{\alpha}\frac{\partial^2}{\partial x^2}\frac{\partial\phi}{\partial z}\cdot
\frac{\partial^2}{\partial x^2}\frac{\partial\phi^*}{\partial z}dx\right\}.
</math></center>
As <math>{\frac{\partial^2}{\partial x^2}\frac{\partial\phi}{\partial z}\cdot \frac{\partial^2}{\partial x^2}\frac{\partial\phi^*}{\partial z}}</math> is real, we obtain the new expression of <math>\xi_3</math>
<center><math>
\xi_3 = \Im\left\{
\left[\frac{\beta}{\alpha}\frac{\partial}{\partial x}
\left(\frac{\partial^2}{\partial x^2}-2k_y^2\right)
\frac{\partial\phi}{\partial z}\cdot\frac{\partial\phi^*}{\partial z}\right]_{-\infty}^{\infty}-
\left[\frac{\beta}{\alpha}\frac{\partial^2}{\partial x^2}\frac{\partial\phi}{\partial z}\cdot
\frac{\partial}{\partial x}\frac{\partial\phi^*}{\partial z}\right]_{-\infty}^{\infty}\right\}.
</math></center>

Now breaking <math>\xi_3</math> down, we can simplify the left hand term for <math>x>0</math>,
if <math>\theta \in [-\theta_0, \theta_0]</math>,

<center><math>
\begin{matrix}
\frac{\partial}{\partial x}\left(\frac{\partial^2}{\partial x^2}-2k_y^2\right)
\frac{\partial\phi(x_2,0)}{\partial z}\cdot\frac{\partial\phi(x_2,0)^*}{\partial z} \\ \\
= \left((-i\kappa^I_{\Lambda})^3k^I_\Lambda T_\Lambda(0)e^{-i\kappa^I_{\Lambda}(x)}\tanh{(k^I_\Lambda h)}\right.
\left.- 2k_y^2(-i\kappa_{\Lambda}(0))k^I_\Lambda T_\Lambda(0)e^{-i\kappa^I_{\Lambda}(x)} \tanh{(k^I_\Lambda h)}\right)
\left(k^I_\Lambda T_\Lambda(0)^*e^{i\kappa^I_{\Lambda}(x)}\tanh{(k^I_\Lambda h)}\right), \\ \\
= i\kappa^I_{\Lambda} (k^I_\Lambda)^2\left((\kappa^I_{\Lambda})^2 + 2k_y^2 \right) \tanh^2{(k^I_\Lambda h)}|T_\Lambda(0)|^2,
\end{matrix}
</math></center>
else this term is equal to 0.

and for <math>x<0</math>

<center><math>
\begin{matrix}
\frac{\partial}{\partial x}\left(\frac{\partial^2}{\partial x^2}-2k_y^2\right)\frac{\partial\phi(x_1,0)}{\partial z}\cdot
\frac{\partial\phi(x_1,0)^*}{\partial z} \\ \\
= \left[i(\kappa^I_{1})^3k^I_1\left(e^{-i\kappa^I_{1}(x)}-R_1(0)e^{i\kappa^I_{1}(x)}\right) \tanh{(k^I_1 h)}
- 2ik_y^2\kappa^I_{1}k^I_1\left(-e^{-i\kappa^I_{1}(x)} + R_1(0)e^{i\kappa^I_{1}(x)}\right) \tanh{(k^I_1h)}\right]
\left[k^I_1\left(e^{i\kappa^I_{1}(x)} + R_1(0)^*e^{-i\kappa^I_{1}(x))}\right) \tanh{(k^I_1h)}\right], \\ \\
= \left[i\kappa^I_{1}k_1(0)((\kappa^I_{1})^2 + 2k_y^2)\left(e^{-i\kappa^I_{1}(x)} - R_1(0)e^{i\kappa^I_{1}(x)}\right)
\tanh{(k^I_1h)}\right]
\left[k^I_1\left(e^{i\kappa^I_{1}(x)} + R_1(0)^*e^{-i\kappa^I_{1}(x))}\right) \tanh{(k^I_1h)}\right], \\ \\
= i\kappa^I_{1}(k^I_1)^2\left((\kappa^I_{1})^2 + 2k_y^2\right)\tanh^2{(k^I_1h)}\left(1 - |R_1(0)|^2\right),
\end{matrix}
</math></center>

Likewise we expand the right hand term for <math>x>0</math>, if <math>\theta \in [-\theta_0, \theta_0]</math>,

<center><math>
\begin{matrix}
\frac{\partial^2}{\partial x^2}\frac{\partial\phi(x_2,0)}{\partial z}\cdot
\frac{\partial}{\partial x}\frac{\partial\phi^*}{\partial z} \\ \\
= \left[-(\kappa^I_{\Lambda})^2k^I_\Lambda T_\Lambda(0)e^{-i\kappa^I_{\Lambda}(x)} \tanh{(k^I_\Lambda h)}\right]
\left[i\kappa^I_{\Lambda} k^I_\Lambda T_\Lambda(0)^*e^{i\kappa^I_{\Lambda}(x)} \tanh{(k^I_\Lambda h)}\right], \\ \\
= -i(\kappa^I_{\Lambda})^3(k^I_\Lambda)^2\tanh^2{(k^I_\Lambda h)}|T_\Lambda(0)|^2,
\end{matrix}
</math></center>
else this term is equal to 0 as well,
and finally for <math>x<0</math>,

<center><math>
\begin{matrix}
\frac{\partial^2}{\partial x^1}\frac{\partial\phi(x_1,0)}{\partial z}\cdot
\frac{\partial}{\partial x}\frac{\partial\phi^*}{\partial z}\\ \\
= \left[-(\kappa^I_{1})^2k^I_1\left(e^{-i\kappa^I_{1}(x)} + R_1(0)e^{i\kappa^I_{1}(x)}\right)\tanh{(k^I_1 h)}\right]
\left[i\kappa^I_{1} k^I_1\left(e^{i\kappa^I_{1}(x)} - R_1(0)e^{-i\kappa^I_{1}(x)}\right)\tanh{(k^I_1 h)}\right], \\ \\
= -i(\kappa^I_{1})^3(k^I_1)^2\tanh^2{(k^I_1h)}\left(1 - |R_1(0)|^2\right).
\end{matrix}
</math></center>

We can now express <math>\xi_3</math> as

<center><math>
\begin{matrix}
\xi_3 & = & \Im
\left\{\frac{\beta_\Lambda}{\alpha}\left[i\kappa^I_{\Lambda} (k^I_\Lambda)^2\left((\kappa^I_{\Lambda})^2 + 2k_y^2 \right)
\tanh^2{(k^I_\Lambda h)}|T_\Lambda(0)|^2\right]
-\frac{\beta_1}{\alpha}\left[i\kappa^I_{1}(k^I_1)^2\left((\kappa^I_{1})^2 + 2k_y^2\right)\tanh^2{(k^I_1h)}
\left(1 - |R_1(0)|^2\right)\right]
-\frac{\beta_\Lambda}{\alpha}\left[-i(\kappa^I_{\Lambda})^3(k^I_\Lambda)^2\tanh^2{(k^I_\Lambda h)}|T_\Lambda(0)|^2\right]
+\frac{\beta_1}{\alpha}\left[-i(\kappa^I_{1})^3(k^I_1)^2\tanh^2{(k^I_1h)} \left(1 - |R_1(0)|^2\right)\right]\right\}, \\ \\
& = & \frac{\beta_\Lambda}{\alpha}\left[2\kappa^I_{\Lambda} (k^I_\Lambda)^2((\kappa^I_{\Lambda})^2 + k_y^2)\tanh^2{(k^I_\Lambda h)}|T_\Lambda(0)|^2\right]
-\frac{\beta_1}{\alpha}\left[2\kappa^I_{1}(k^I_1)^2((\kappa^I_{1})^2 + k_y^2)\tanh^2{(k^I_1h)}(1-|R_1(0)|^2)\right].
\end{matrix}
</math></center>
if <math>\theta \in [-\theta_0, \theta_0]</math>, and if not, we obtain:
<center>
<math>
\xi_3 = -\frac{\beta_1}{\alpha}\left[2\kappa^I_{1}(k^I_1)^2((\kappa^I_{1})^2 + k_y^2)\tanh^2{(k^I_1h)}(1-|R_1(0)|^2)\right].
</math>
</center>

= Solving the Energy Balance Equation =

Pulling it all together, we finally obtain
<center><math>
\frac{\beta_\Lambda}{\alpha}\left[2\kappa^I_{\Lambda}(k^I_\Lambda)^2((\kappa^I_{\Lambda})^2 +k_y^2)
\tanh^2{(k^I_\Lambda h)}|T_\Lambda(0)|^2\right]
-\frac{\beta_1}{\alpha}\left[2\kappa^I_{1}(k^I_1)^2((\kappa^I_{1})^2 +k_y^2)\tanh^2{(k^I_1h)}(1-|R_1(0)|^2)\right]
+ \frac{\kappa^I_{\Lambda}|T_\Lambda(0)|^2}{2k^I_\Lambda}\left(\tanh{(k^I_\Lambda h)}+
\frac{k^I_\Lambda h}{\cosh^2{(k^I_\Lambda h)}}\right)
- \frac{\kappa^I_{1}\left(1 - |R_1(0)|^2\right)}{2k^I_1}\left(\tanh{(k^I_1h)}+\frac{hk^I_1}{\cosh^2{(k^I_1h)}}\right)=0.
</math></center>

Re-arranging gives

<center><math>
\kappa^I_{\Lambda}\tanh{(k^I_\Lambda h)}
\left(\frac{\beta_\Lambda}{\alpha}2(k^I_\Lambda)^2((\kappa^I_{\Lambda})^2 +k_y^2)\tanh{(k^I_\Lambda h)}
+ \frac{1}{2k^I_\Lambda}+\frac{h}{2\sinh{(k^I_\Lambda h)}\cosh{(k^I_\Lambda h)}}\right)|T_\Lambda(0)|^2
\quad - \kappa^I_{1}\tanh{(k^I_1 h)}
\left(\frac{\beta_1}{\alpha}2(k^I_1)^2((\kappa^I_{1})^2 +k_y^2)\tanh{(k^I_1 h)}
+ \frac{1}{2it}+\frac{h}{2\sinh{(k^I_1 h)}\cosh{(k^I_1 h)}}\right)(1-|R_1(0)|^2)=0
</math></center>

which can be expressed as
<center><math>
D |T_{\Lambda}(0)|^2 + |R_{1}(0)|^2 = 1,
</math></center>
where <math>D</math> is given by
<center><math>
D = \left(\frac{\kappa^I_{\Lambda} k^I_1\cosh^2{(k^I_1h)}}{\kappa^I_{1}k^I_\Lambda\cosh^2{(k^I_\Lambda h)}}\right)
\left(\frac{-\frac{\beta_\Lambda}{\alpha}4(k^I_\Lambda)^5\sinh^2{(k^I_\Lambda h)} +
\frac{1}{2}{\sinh{(2k^I_\Lambda h)}}+k^I_\Lambda h}
{-\frac{\beta_1}{\alpha}4(k^I_1)^5\sinh^2{(k^I_1h)} +
\frac{1}{2}{\sinh{(2k^I_1h)}}+k^I_1h}\right)
</math></center>

[[Category:Floating Elastic Plate]]

Energy Balance for Two Elastic Plates

2008-08-26T23:35:12Z

Fmontiel: /* Expanding <math>\mathbf{\xi_3}</math> */

= Introduction =

We derive here a condition which must be satisfied for energy balance for two semi-infinite elastic plates. The solution applies
to the problem of [[Eigenfunction Matching Method for Floating Elastic Plates|multiple elastic plates]] and from here we know we can write the potential as
<center><math>
\phi \approx \left\{
\begin{matrix}
{
Ie^{\kappa_{1}(0)(x-r_1)}\frac{\cos(k_1(0)(z+h))}{\cos(k_1(0)h)} }+\\
{
\qquad \qquad \sum_{n=-2}^{M}R_1(n) e^{\kappa_{1}(n)(x-r_1)} \frac{\cos(k_1(n)(z+h))}{\cos(k_1(n)h)} },&
\mbox{ for } x < r_1,\\
{
\sum_{n=-2}^{M}T_{\mu}(n) e^{-\kappa_\mu(n)(x-l_\mu)}
\frac{\cos(k_\mu(n)(z+h))}{\cos(k_\mu(n)h)} } + \\
{
\qquad \qquad \sum_{n=-2}^{M}R_{\mu}(n) e^{\kappa_\mu(n)(x-r_\mu)}
\frac{\cos(k_\mu(n)(z+h))}{\cos(k_\mu(n)h)} },
&\mbox{ for } l_\mu< x < r_\mu,\\
{
\sum_{n=-2}^{M}T_{\Lambda}(n)e^{-\kappa_{\Lambda}(n)(x-l_\Lambda)}
\frac{\cos(k_\Lambda(n)(z+h))}{\cos(k_\Lambda(n)h)} }, &\mbox{ for } l_\Lambda<x,
\end{matrix} \right.
</math></center>
(for details of this notation see [[Eigenfunction Matching Method for Floating Elastic Plates]]).

= Equations =

Based on the method used in [[Evans and Davies 1968]], a check can be made to ensure the solutions of the floating plate problem are in energy balance.
This is simply a condition that the incident energy is equal to the sum of the radiated energy.
When the first and final plates have different properties, the energy balance equation is derived by applying Green's theorem to <math>\phi</math> and its conjugate.
The domain of integration is shown in the figure on the right.

[[Image:energy_schematic.jpg|thumb|right|300px|A diagram depicting the area <math>\mathcal{U}</math> which is bounded by the rectangle <math>\mathcal{S}</math>.]
{A diagram depicting the area <math>\mathcal{U}</math> which is bounded by the rectangle <math>\mathcal{S}</math>.
The rectangle <math>\mathcal{S}</math> is bounded by <math> -h\leq z \leq0</math> and <math>-\infty\leq x \leq \infty</math>]]

Applying Green's theorem to <math>\phi</math> and its conjugate <math>\phi^*</math> gives
<center><math>
{ \int\int_\mathcal{U}(\phi\nabla^2\phi^* - \phi^*\nabla^2\phi)dxdz
= \int_\mathcal{S}(\phi\frac{\partial\phi^*}{\partial n} - \phi^*\frac{\partial\phi}{\partial n})dl },
</math></center>
where <math>n</math> denotes the outward plane normal to the boundary and <math>l</math> denotes the plane parallel to the boundary.
As <math>\phi</math> and <math>\phi^*</math> satisfy the Laplace's equation, the left hand side of the Green theorem equation vanishes so that it reduces to
<center><math>
\Im\int_\mathcal{S}\phi\frac{\partial\phi^*}{\partial n} dl = 0,
</math></center>
Expanding gives
<center><math>
\xi_1 +\xi_2 + \xi_3 = 0,
</math></center>
where
<center><math>
\xi_1 =
-{ \Im\int_{-h}^0(\phi\frac{\partial\phi^*}{\partial x})\big|_{x=-\infty}dz },
</math></center>
<center><math>
\xi_2 =
{ \Im\int_{-h}^0(\phi\frac{\partial\phi^*}{\partial x})\big|_{x=\infty}dz },
</math></center>
and
<center><math>
\xi_3 =
{ \Im\int_{-\infty}^{\infty}(\phi\frac{\partial\phi^*}{\partial z})\big|_{z=0}dx } ,
</math></center>
where <math>\Im</math> denotes the imaginary part. The bottom domain composant is obviously equal to zero because of the no-influx condition over the seabed (<math>\frac{\partial\phi^*}{\partial z}\big|_{z=-h}=0</math>).

= Expanding <math>\mathbf{\xi_1}</math> =

Near <math>x=-\infty</math>, we approximate <math>\phi</math> by
<center><math>
\phi \approx e^{-\kappa_{1}(0)(x-r_1)}\frac{\cos{(k_1(0)(z+h))}}{\cos{(k_1(0)h)}} +
R_1(0)e^{\kappa_{1}(0)(x-r_1)}\frac{\cos{(k_{1}(0)(z+h))}}{\cos{(k_{1}(0)h)}}
</math></center>
Therefore,
<center><math>
\begin{matrix}
\xi_1 & = & {-\Im\int_{-h}^0 \left[\left(e^{-i\kappa^{I}_1(x)}+ R_1(0)e^{i\kappa^{I}_1(x)}\right)\right.}
{ \left(i\kappa^{I}_1e^{i\kappa^{I}_1(x)} - i\kappa^{I}_1R_1(0)^*e^{-i\kappa^{I}_1(x)}\right) }
{ \left.\left(\frac{\cosh^2{(k^I_1(z+h))}}{\cosh^2{(k^I_1h)}}\right)\right]dz, }\\ \\
& = & { - \Im\left[\frac{i\kappa^{I}_1\left(1 - |R_1(0)|^2\right)}{2\cosh^2{(k^I_1h)}}
\int_{-h}^0 \left(\cosh{(2k^I_1(z+h))+1}\right)dz\right] ,}\\ \\
& = & { - \Im\left[\frac{i\kappa^{I}_1\left(1 - |R_1(0)|^2\right)}{2\cosh^2{(k^I_1h)}}
\left[\frac{1}{2k^I_1}\sinh{(2k^I_1(z+h))} + z\right]_{-h}^0\right], }\\ \\
& = & { - \frac{\kappa^{I}_1\left(1 - |R_1(0)|^2\right)}{2\cosh^2{(k^I_1h)}} \left(\frac{1}{2k^I_1}\sinh{(2k^I_1h))} + h\right) ,}\\ \\
& = & { - \frac{\kappa^{I}_1\left(1 - |R_1(0)|^2\right)}{2k^I_1}\left(\tanh{(k^I_1h)}+\frac{hk^I_1}{\cosh^2{(k^I_1h)}}\right) , }
\end{matrix}
</math></center>
where <math>R_1(0)^*</math> is the conjugate of <math>R_1(0)</math>.

= Expanding <math>\mathbf{\xi_2}</math> =

Near <math>x=\infty</math>, we approximate <math>\phi</math> by
<center><math>
\begin{matrix}
{ \phi \approx T_\Lambda(0)e^{-\kappa_{\Lambda}(0)(x)}\frac{\cos{(k_\Lambda(0)(z+h))}}{\cos{(k_\Lambda(0)h)}}, }
\end{matrix}
</math></center>
and re-express as
<center><math>
\phi \approx T_\Lambda(0)e^{-i\kappa^I_\Lambda(x)}\frac{\cosh{(k^I_\Lambda(z+h))}}{\cosh{(k^I_\Lambda h)}},
</math></center>
where <math>k^I_\Lambda = \Im k_{\Lambda}(0)</math> and <math>\kappa^I_\Lambda=\Im \kappa_\Lambda(0)</math>, so that
<center><math>
\frac{\partial\phi}{\partial x} \approx -i\kappa^I_{\Lambda}T_\Lambda(0)e^{-i\kappa^I_{\Lambda}(x)}\frac{\cosh{(k^I_\Lambda(z+h))}}{\cosh{(k^I_\Lambda h)}}.
</math></center>
We know that <math>\kappa_\Lambda(0)=\sqrt{k_{\Lambda}(0)^2-(ik_y)^2}</math>, where <math>k_y</math> is real and depends on the incident angle <math>\theta</math>. When <math>\theta</math> becomes greater than a certain angle <math>\theta_0</math> defined by <math>\sin \theta_0=\sqrt{\frac{k_{\Lambda}(0)}{k_{1}(0)}}</math>, <math>\kappa_\Lambda(0)</math> becomes real so the potential becomes real as well. Thus the imaginary part is equal to 0. In that case, we have <math>\xi_2=0</math>.
Therefore, if <math>\theta \in [-\theta_0, \theta_0]</math>,
<center><math>
\begin{matrix}
\xi_2 & = & \Im\int_{-h}^0 \left[\right.(T_\Lambda(0)e^{-i\kappa^I_{\Lambda}(x)})(i\kappa^I_{\Lambda}T_\Lambda(0)^*e^{i\kappa^I_{\Lambda}(x)})
\frac{\cosh^2{(k^I_\Lambda(z+h))}}{\cosh^2{(k^I_\Lambda h)}}\left.\right]dz, \\ \\
& = & { \frac{\kappa^I_{\Lambda}|T_\Lambda(0)|^2}{2\cosh^2{(k^I_\Lambda h)}}
\left(\frac{1}{2k^I_\Lambda}\sinh{(2k^I_\Lambda h)} + h\right)},\\ \\
& = & { \frac{\kappa^I_{\Lambda}|T_\Lambda(0)|^2}{2k^I_\Lambda}\left(\tanh{(k^I_\Lambda h)}+\frac{k^I_\Lambda h}{\cosh^2{(k^I_\Lambda h)}}\right)},
\end{matrix}
</math></center>
otherwise <math>\xi_2=0</math>.

= Expanding <math>\mathbf{\xi_3}</math> =

The ice-covered boundary condition for the [[Floating Elastic Plate]] gives
<center><math>
\xi_3 = { \Im\int_{-\infty}^{\infty}\left(\frac{\beta}{\alpha}
\left(\frac{\partial^2}{\partial x^2} - k_y^2\right)^2 - \gamma + \frac{1}{\alpha}\right)
\frac{\partial\phi}{\partial z}\cdot\frac{\partial\phi^*}{\partial z}\bigg|_{z=0}dx.}
</math></center>
Since <math>{\frac{\partial\phi}{\partial z}\cdot\frac{\partial\phi^*}{\partial z}}</math> is real,
<center><math>
\xi_3 = \Im\int_{-\infty}^{\infty}\left(\frac{\beta}{\alpha}\frac{\partial^2}{\partial x^2}
\left(\frac{\partial^2}{\partial x^2} - 2k_y^2\right) \right)
\frac{\partial\phi}{\partial z}\cdot\frac{\partial\phi^*}{\partial z}\bigg|_{z=0}dx.
</math></center>
Integration by parts gives
<center><math>
\xi_3 = \Im\left\{\left[\frac{\beta}{\alpha}\frac{\partial}{\partial x}
\left(\frac{\partial^2}{\partial x^2}-2k_y^2\right)
\frac{\partial\phi}{\partial z}\cdot\frac{\partial\phi^*}{\partial z}\right]_{-\infty}^{\infty}
-\int_{-\infty}^{\infty}\frac{\beta}{\alpha}\frac{\partial}{\partial x}
\left(\frac{\partial^2}{\partial x^2} - 2k_y^2\right)
\frac{\partial\phi}{\partial z}\cdot\frac
{\partial}{\partial x}\frac{\partial\phi^*}{\partial z}dx
\right\}.
</math></center>
As <math>{2k_y^2\frac{\partial}{\partial x}\frac{\partial\phi}{\partial z}\cdot \frac{\partial}{\partial x}\frac{\partial\phi^*}{\partial z}}</math> is real and by integration by parts, the expression of <math>\xi_3</math> becomes,
<center><math>
\xi_3 =\Im\left\{
\left[\frac{\beta}{\alpha}\frac{\partial}{\partial x}
\left(\frac{\partial^2}{\partial x^2}-2k_y^2\right)
\frac{\partial\phi}{\partial z}\cdot\frac{\partial\phi^*}{\partial z}\right]_{-\infty}^{\infty}
- \left[\frac{\beta}{\alpha}\frac{\partial^2}{\partial x^2}\frac{\partial\phi}{\partial z}\cdot
\frac{\partial}{\partial x}\frac{\partial\phi^*}{\partial z}\right]_{-\infty}^{\infty}
+ \int_{-\infty}^\infty\frac{\beta}{\alpha}\frac{\partial^2}{\partial x^2}\frac{\partial\phi}{\partial z}\cdot
\frac{\partial^2}{\partial x^2}\frac{\partial\phi^*}{\partial z}dx\right\}.
</math></center>
As <math>{\frac{\partial^2}{\partial x^2}\frac{\partial\phi}{\partial z}\cdot \frac{\partial^2}{\partial x^2}\frac{\partial\phi^*}{\partial z}}</math> is real, we obtain the new expression of <math>\xi_3</math>
<center><math>
\xi_3 = \Im\left\{
\left[\frac{\beta}{\alpha}\frac{\partial}{\partial x}
\left(\frac{\partial^2}{\partial x^2}-2k_y^2\right)
\frac{\partial\phi}{\partial z}\cdot\frac{\partial\phi^*}{\partial z}\right]_{-\infty}^{\infty}-
\left[\frac{\beta}{\alpha}\frac{\partial^2}{\partial x^2}\frac{\partial\phi}{\partial z}\cdot
\frac{\partial}{\partial x}\frac{\partial\phi^*}{\partial z}\right]_{-\infty}^{\infty}\right\}.
</math></center>

Now breaking <math>\xi_3</math> down, we can simplify the left hand term for <math>x>0</math>,
if <math>\theta \in [-\theta_0, \theta_0]</math>,

<center><math>
\begin{matrix}
\frac{\partial}{\partial x}\left(\frac{\partial^2}{\partial x^2}-2k_y^2\right)
\frac{\partial\phi(x_2,0)}{\partial z}\cdot\frac{\partial\phi(x_2,0)^*}{\partial z} \\ \\
= \left((-i\kappa^I_{\Lambda})^3k^I_\Lambda T_\Lambda(0)e^{-i\kappa^I_{\Lambda}(x)}\tanh{(k^I_\Lambda h)}\right.
\left.- 2k_y^2(-i\kappa_{\Lambda}(0))k^I_\Lambda T_\Lambda(0)e^{-i\kappa^I_{\Lambda}(x)} \tanh{(k^I_\Lambda h)}\right)
\left(k^I_\Lambda T_\Lambda(0)^*e^{i\kappa^I_{\Lambda}(x)}\tanh{(k^I_\Lambda h)}\right), \\ \\
= i\kappa^I_{\Lambda} (k^I_\Lambda)^2\left((\kappa^I_{\Lambda})^2 + 2k_y^2 \right) \tanh^2{(k^I_\Lambda h)}|T_\Lambda(0)|^2,
\end{matrix}
</math></center>
else this term is equal to 0.

and for <math>x<0</math>

<center><math>
\begin{matrix}
\frac{\partial}{\partial x}\left(\frac{\partial^2}{\partial x^2}-2k_y^2\right)\frac{\partial\phi(x_1,0)}{\partial z}\cdot
\frac{\partial\phi(x_1,0)^*}{\partial z} \\ \\
= \left[i(\kappa^I_{1})^3k^I_1\left(e^{-i\kappa^I_{1}(x)}-R_1(0)e^{i\kappa^I_{1}(x)}\right) \tanh{(k^I_1 h)}
- 2ik_y^2\kappa^I_{1}k^I_1\left(-e^{-i\kappa^I_{1}(x)} + R_1(0)e^{i\kappa^I_{1}(x)}\right) \tanh{(k^I_1h)}\right]
\left[k^I_1\left(e^{i\kappa^I_{1}(x)} + R_1(0)^*e^{-i\kappa^I_{1}(x))}\right) \tanh{(k^I_1h)}\right], \\ \\
= \left[i\kappa^I_{1}k_1(0)((\kappa^I_{1})^2 + 2k_y^2)\left(e^{-i\kappa^I_{1}(x)} - R_1(0)e^{i\kappa^I_{1}(x)}\right)
\tanh{(k^I_1h)}\right]
\left[k^I_1\left(e^{i\kappa^I_{1}(x)} + R_1(0)^*e^{-i\kappa^I_{1}(x))}\right) \tanh{(k^I_1h)}\right], \\ \\
= i\kappa^I_{1}(k^I_1)^2\left((\kappa^I_{1})^2 + 2k_y^2\right)\tanh^2{(k^I_1h)}\left(1 - |R_1(0)|^2\right),
\end{matrix}
</math></center>

Likewise we expand the right hand term for <math>x>0</math>, if <math>\theta \in [-\theta_0, \theta_0]</math>,

<center><math>
\begin{matrix}
\frac{\partial^2}{\partial x^2}\frac{\partial\phi(x_2,0)}{\partial z}\cdot
\frac{\partial}{\partial x}\frac{\partial\phi^*}{\partial z} \\ \\
= \left[-(\kappa^I_{\Lambda})^2k^I_\Lambda T_\Lambda(0)e^{-i\kappa^I_{\Lambda}(x)} \tanh{(k^I_\Lambda h)}\right]
\left[i\kappa^I_{\Lambda} k^I_\Lambda T_\Lambda(0)^*e^{i\kappa^I_{\Lambda}(x)} \tanh{(k^I_\Lambda h)}\right], \\ \\
= -i(\kappa^I_{\Lambda})^3(k^I_\Lambda)^2\tanh^2{(k^I_\Lambda h)}|T_\Lambda(0)|^2,
\end{matrix}
</math></center>
else this term is equal to 0 as well,
and finally for <math>x<0</math>,

<center><math>
\begin{matrix}
\frac{\partial^2}{\partial x^1}\frac{\partial\phi(x_1,0)}{\partial z}\cdot
\frac{\partial}{\partial x}\frac{\partial\phi^*}{\partial z}\\ \\
= \left[-(\kappa^I_{1})^2k^I_1\left(e^{-i\kappa^I_{1}(x)} + R_1(0)e^{i\kappa^I_{1}(x)}\right)\tanh{(k^I_1 h)}\right]
\left[i\kappa^I_{1} k^I_1\left(e^{i\kappa^I_{1}(x)} - R_1(0)e^{-i\kappa^I_{1}(x)}\right)\tanh{(k^I_1 h)}\right], \\ \\
= -i(\kappa^I_{1})^3(k^I_1)^2\tanh^2{(k^I_1h)}\left(1 - |R_1(0)|^2\right).
\end{matrix}
</math></center>

We can now express <math>\xi_3</math> as

<center><math>
\begin{matrix}
\xi_3 & = & \Im
\left\{\frac{\beta_\Lambda}{\alpha}\left[i\kappa^I_{\Lambda} (k^I_\Lambda)^2\left((\kappa^I_{\Lambda})^2 + 2k_y^2 \right)
\tanh^2{(k^I_\Lambda h)}|T_\Lambda(0)|^2\right]
-\frac{\beta_1}{\alpha}\left[i\kappa^I_{1}(k^I_1)^2\left((\kappa^I_{1})^2 + 2k_y^2\right)\tanh^2{(k^I_1h)}
\left(1 - |R_1(0)|^2\right)\right]
-\frac{\beta_\Lambda}{\alpha}\left[-i(\kappa^I_{\Lambda})^3(k^I_\Lambda)^2\tanh^2{(k^I_\Lambda h)}|T_\Lambda(0)|^2\right]
+\frac{\beta_1}{\alpha}\left[-i(\kappa^I_{1})^3(k^I_1)^2\tanh^2{(k^I_1h)} \left(1 - |R_1(0)|^2\right)\right]\right\}, \\ \\
& = & \frac{\beta_\Lambda}{\alpha}\left[2\kappa^I_{\Lambda} (k^I_\Lambda)^2((\kappa^I_{\Lambda})^2 + k_y^2)\tanh^2{(k^I_\Lambda h)}|T_\Lambda(0)|^2\right]
-\frac{\beta_1}{\alpha}\left[2\kappa^I_{1}(k^I_1)^2((\kappa^I_{1})^2 + k_y^2)\tanh^2{(k^I_1h)}(1-|R_1(0)|^2)\right].
\end{matrix}
</math></center>
if <math>\theta \in [-\theta_0, \theta_0]</math>, and if not, we obtain:
<center>
<math>
\xi_3 = -\frac{\beta_1}{\alpha}\left[2\kappa^I_{1}(k^I_1)^2((\kappa^I_{1})^2 + k_y^2)\tanh^2{(k^I_1h)}(1-|R_1(0)|^2)\right].

= Solving the Energy Balance Equation =

Pulling it all together, we finally obtain
<center><math>
\frac{\beta_\Lambda}{\alpha}\left[2\kappa^I_{\Lambda}(k^I_\Lambda)^2((\kappa^I_{\Lambda})^2 +k_y^2)
\tanh^2{(k^I_\Lambda h)}|T_\Lambda(0)|^2\right]
-\frac{\beta_1}{\alpha}\left[2\kappa^I_{1}(k^I_1)^2((\kappa^I_{1})^2 +k_y^2)\tanh^2{(k^I_1h)}(1-|R_1(0)|^2)\right]
+ \frac{\kappa^I_{\Lambda}|T_\Lambda(0)|^2}{2k^I_\Lambda}\left(\tanh{(k^I_\Lambda h)}+
\frac{k^I_\Lambda h}{\cosh^2{(k^I_\Lambda h)}}\right)
- \frac{\kappa^I_{1}\left(1 - |R_1(0)|^2\right)}{2k^I_1}\left(\tanh{(k^I_1h)}+\frac{hk^I_1}{\cosh^2{(k^I_1h)}}\right)=0.
</math></center>

Re-arranging gives

<center><math>
\kappa^I_{\Lambda}\tanh{(k^I_\Lambda h)}
\left(\frac{\beta_\Lambda}{\alpha}2(k^I_\Lambda)^2((\kappa^I_{\Lambda})^2 +k_y^2)\tanh{(k^I_\Lambda h)}
+ \frac{1}{2k^I_\Lambda}+\frac{h}{2\sinh{(k^I_\Lambda h)}\cosh{(k^I_\Lambda h)}}\right)|T_\Lambda(0)|^2
\quad - \kappa^I_{1}\tanh{(k^I_1 h)}
\left(\frac{\beta_1}{\alpha}2(k^I_1)^2((\kappa^I_{1})^2 +k_y^2)\tanh{(k^I_1 h)}
+ \frac{1}{2it}+\frac{h}{2\sinh{(k^I_1 h)}\cosh{(k^I_1 h)}}\right)(1-|R_1(0)|^2)=0
</math></center>

which can be expressed as
<center><math>
D |T_{\Lambda}(0)|^2 + |R_{1}(0)|^2 = 1,
</math></center>
where <math>D</math> is given by
<center><math>
D = \left(\frac{\kappa^I_{\Lambda} k^I_1\cosh^2{(k^I_1h)}}{\kappa^I_{1}k^I_\Lambda\cosh^2{(k^I_\Lambda h)}}\right)
\left(\frac{-\frac{\beta_\Lambda}{\alpha}4(k^I_\Lambda)^5\sinh^2{(k^I_\Lambda h)} +
\frac{1}{2}{\sinh{(2k^I_\Lambda h)}}+k^I_\Lambda h}
{-\frac{\beta_1}{\alpha}4(k^I_1)^5\sinh^2{(k^I_1h)} +
\frac{1}{2}{\sinh{(2k^I_1h)}}+k^I_1h}\right)
</math></center>

[[Category:Floating Elastic Plate]]

Energy Balance for Two Elastic Plates

2008-08-26T23:23:41Z

Fmontiel:

= Introduction =

We derive here a condition which must be satisfied for energy balance for two semi-infinite elastic plates. The solution applies
to the problem of [[Eigenfunction Matching Method for Floating Elastic Plates|multiple elastic plates]] and from here we know we can write the potential as
<center><math>
\phi \approx \left\{
\begin{matrix}
{
Ie^{\kappa_{1}(0)(x-r_1)}\frac{\cos(k_1(0)(z+h))}{\cos(k_1(0)h)} }+\\
{
\qquad \qquad \sum_{n=-2}^{M}R_1(n) e^{\kappa_{1}(n)(x-r_1)} \frac{\cos(k_1(n)(z+h))}{\cos(k_1(n)h)} },&
\mbox{ for } x < r_1,\\
{
\sum_{n=-2}^{M}T_{\mu}(n) e^{-\kappa_\mu(n)(x-l_\mu)}
\frac{\cos(k_\mu(n)(z+h))}{\cos(k_\mu(n)h)} } + \\
{
\qquad \qquad \sum_{n=-2}^{M}R_{\mu}(n) e^{\kappa_\mu(n)(x-r_\mu)}
\frac{\cos(k_\mu(n)(z+h))}{\cos(k_\mu(n)h)} },
&\mbox{ for } l_\mu< x < r_\mu,\\
{
\sum_{n=-2}^{M}T_{\Lambda}(n)e^{-\kappa_{\Lambda}(n)(x-l_\Lambda)}
\frac{\cos(k_\Lambda(n)(z+h))}{\cos(k_\Lambda(n)h)} }, &\mbox{ for } l_\Lambda<x,
\end{matrix} \right.
</math></center>
(for details of this notation see [[Eigenfunction Matching Method for Floating Elastic Plates]]).

= Equations =

Based on the method used in [[Evans and Davies 1968]], a check can be made to ensure the solutions of the floating plate problem are in energy balance.
This is simply a condition that the incident energy is equal to the sum of the radiated energy.
When the first and final plates have different properties, the energy balance equation is derived by applying Green's theorem to <math>\phi</math> and its conjugate.
The domain of integration is shown in the figure on the right.

[[Image:energy_schematic.jpg|thumb|right|300px|A diagram depicting the area <math>\mathcal{U}</math> which is bounded by the rectangle <math>\mathcal{S}</math>.]
{A diagram depicting the area <math>\mathcal{U}</math> which is bounded by the rectangle <math>\mathcal{S}</math>.
The rectangle <math>\mathcal{S}</math> is bounded by <math> -h\leq z \leq0</math> and <math>-\infty\leq x \leq \infty</math>]]

Applying Green's theorem to <math>\phi</math> and its conjugate <math>\phi^*</math> gives
<center><math>
{ \int\int_\mathcal{U}(\phi\nabla^2\phi^* - \phi^*\nabla^2\phi)dxdz
= \int_\mathcal{S}(\phi\frac{\partial\phi^*}{\partial n} - \phi^*\frac{\partial\phi}{\partial n})dl },
</math></center>
where <math>n</math> denotes the outward plane normal to the boundary and <math>l</math> denotes the plane parallel to the boundary.
As <math>\phi</math> and <math>\phi^*</math> satisfy the Laplace's equation, the left hand side of the Green theorem equation vanishes so that it reduces to
<center><math>
\Im\int_\mathcal{S}\phi\frac{\partial\phi^*}{\partial n} dl = 0,
</math></center>
Expanding gives
<center><math>
\xi_1 +\xi_2 + \xi_3 = 0,
</math></center>
where
<center><math>
\xi_1 =
-{ \Im\int_{-h}^0(\phi\frac{\partial\phi^*}{\partial x})\big|_{x=-\infty}dz },
</math></center>
<center><math>
\xi_2 =
{ \Im\int_{-h}^0(\phi\frac{\partial\phi^*}{\partial x})\big|_{x=\infty}dz },
</math></center>
and
<center><math>
\xi_3 =
{ \Im\int_{-\infty}^{\infty}(\phi\frac{\partial\phi^*}{\partial z})\big|_{z=0}dx } ,
</math></center>
where <math>\Im</math> denotes the imaginary part. The bottom domain composant is obviously equal to zero because of the no-influx condition over the seabed (<math>\frac{\partial\phi^*}{\partial z}\big|_{z=-h}=0</math>).

= Expanding <math>\mathbf{\xi_1}</math> =

Near <math>x=-\infty</math>, we approximate <math>\phi</math> by
<center><math>
\phi \approx e^{-\kappa_{1}(0)(x-r_1)}\frac{\cos{(k_1(0)(z+h))}}{\cos{(k_1(0)h)}} +
R_1(0)e^{\kappa_{1}(0)(x-r_1)}\frac{\cos{(k_{1}(0)(z+h))}}{\cos{(k_{1}(0)h)}}
</math></center>
Therefore,
<center><math>
\begin{matrix}
\xi_1 & = & {-\Im\int_{-h}^0 \left[\left(e^{-i\kappa^{I}_1(x)}+ R_1(0)e^{i\kappa^{I}_1(x)}\right)\right.}
{ \left(i\kappa^{I}_1e^{i\kappa^{I}_1(x)} - i\kappa^{I}_1R_1(0)^*e^{-i\kappa^{I}_1(x)}\right) }
{ \left.\left(\frac{\cosh^2{(k^I_1(z+h))}}{\cosh^2{(k^I_1h)}}\right)\right]dz, }\\ \\
& = & { - \Im\left[\frac{i\kappa^{I}_1\left(1 - |R_1(0)|^2\right)}{2\cosh^2{(k^I_1h)}}
\int_{-h}^0 \left(\cosh{(2k^I_1(z+h))+1}\right)dz\right] ,}\\ \\
& = & { - \Im\left[\frac{i\kappa^{I}_1\left(1 - |R_1(0)|^2\right)}{2\cosh^2{(k^I_1h)}}
\left[\frac{1}{2k^I_1}\sinh{(2k^I_1(z+h))} + z\right]_{-h}^0\right], }\\ \\
& = & { - \frac{\kappa^{I}_1\left(1 - |R_1(0)|^2\right)}{2\cosh^2{(k^I_1h)}} \left(\frac{1}{2k^I_1}\sinh{(2k^I_1h))} + h\right) ,}\\ \\
& = & { - \frac{\kappa^{I}_1\left(1 - |R_1(0)|^2\right)}{2k^I_1}\left(\tanh{(k^I_1h)}+\frac{hk^I_1}{\cosh^2{(k^I_1h)}}\right) , }
\end{matrix}
</math></center>
where <math>R_1(0)^*</math> is the conjugate of <math>R_1(0)</math>.

= Expanding <math>\mathbf{\xi_2}</math> =

Near <math>x=\infty</math>, we approximate <math>\phi</math> by
<center><math>
\begin{matrix}
{ \phi \approx T_\Lambda(0)e^{-\kappa_{\Lambda}(0)(x)}\frac{\cos{(k_\Lambda(0)(z+h))}}{\cos{(k_\Lambda(0)h)}}, }
\end{matrix}
</math></center>
and re-express as
<center><math>
\phi \approx T_\Lambda(0)e^{-i\kappa^I_\Lambda(x)}\frac{\cosh{(k^I_\Lambda(z+h))}}{\cosh{(k^I_\Lambda h)}},
</math></center>
where <math>k^I_\Lambda = \Im k_{\Lambda}(0)</math> and <math>\kappa^I_\Lambda=\Im \kappa_\Lambda(0)</math>, so that
<center><math>
\frac{\partial\phi}{\partial x} \approx -i\kappa^I_{\Lambda}T_\Lambda(0)e^{-i\kappa^I_{\Lambda}(x)}\frac{\cosh{(k^I_\Lambda(z+h))}}{\cosh{(k^I_\Lambda h)}}.
</math></center>
We know that <math>\kappa_\Lambda(0)=\sqrt{k_{\Lambda}(0)^2-(ik_y)^2}</math>, where <math>k_y</math> is real and depends on the incident angle <math>\theta</math>. When <math>\theta</math> becomes greater than a certain angle <math>\theta_0</math> defined by <math>\sin \theta_0=\sqrt{\frac{k_{\Lambda}(0)}{k_{1}(0)}}</math>, <math>\kappa_\Lambda(0)</math> becomes real so the potential becomes real as well. Thus the imaginary part is equal to 0. In that case, we have <math>\xi_2=0</math>.
Therefore, if <math>\theta \in [-\theta_0, \theta_0]</math>,
<center><math>
\begin{matrix}
\xi_2 & = & \Im\int_{-h}^0 \left[\right.(T_\Lambda(0)e^{-i\kappa^I_{\Lambda}(x)})(i\kappa^I_{\Lambda}T_\Lambda(0)^*e^{i\kappa^I_{\Lambda}(x)})
\frac{\cosh^2{(k^I_\Lambda(z+h))}}{\cosh^2{(k^I_\Lambda h)}}\left.\right]dz, \\ \\
& = & { \frac{\kappa^I_{\Lambda}|T_\Lambda(0)|^2}{2\cosh^2{(k^I_\Lambda h)}}
\left(\frac{1}{2k^I_\Lambda}\sinh{(2k^I_\Lambda h)} + h\right)},\\ \\
& = & { \frac{\kappa^I_{\Lambda}|T_\Lambda(0)|^2}{2k^I_\Lambda}\left(\tanh{(k^I_\Lambda h)}+\frac{k^I_\Lambda h}{\cosh^2{(k^I_\Lambda h)}}\right)},
\end{matrix}
</math></center>
otherwise <math>\xi_2=0</math>.

= Expanding <math>\mathbf{\xi_3}</math> =

The ice-covered boundary condition for the [[Floating Elastic Plate]] gives
<center><math>
\xi_3 = { \Im\int_{-\infty}^{\infty}\left(\frac{\beta}{\alpha}
\left(\frac{\partial^2}{\partial x^2} - k_y^2\right)^2 - \gamma + \frac{1}{\alpha}\right)
\frac{\partial\phi}{\partial z}\cdot\frac{\partial\phi^*}{\partial z}\bigg|_{z=0}dx.}
</math></center>
Since <math>{\frac{\partial\phi}{\partial z}\cdot\frac{\partial\phi^*}{\partial z}}</math> is real,
<center><math>
\xi_3 = \Im\int_{-\infty}^{\infty}\left(\frac{\beta}{\alpha}\frac{\partial^2}{\partial x^2}
\left(\frac{\partial^2}{\partial x^2} - 2k_y^2\right) \right)
\frac{\partial\phi}{\partial z}\cdot\frac{\partial\phi^*}{\partial z}\bigg|_{z=0}dx.
</math></center>
Integration by parts gives
<center><math>
\xi_3 = \Im\left\{\left[\frac{\beta}{\alpha}\frac{\partial}{\partial x}
\left(\frac{\partial^2}{\partial x^2}-2k_y^2\right)
\frac{\partial\phi}{\partial z}\cdot\frac{\partial\phi^*}{\partial z}\right]_{-\infty}^{\infty}
-\int_{-\infty}^{\infty}\frac{\beta}{\alpha}\frac{\partial}{\partial x}
\left(\frac{\partial^2}{\partial x^2} - 2k_y^2\right)
\frac{\partial\phi}{\partial z}\cdot\frac
{\partial}{\partial x}\frac{\partial\phi^*}{\partial z}dx
\right\}.
</math></center>
As <math>{2k_y^2\frac{\partial}{\partial x}\frac{\partial\phi}{\partial z}\cdot \frac{\partial}{\partial x}\frac{\partial\phi^*}{\partial z}}</math> is real and by integration by parts, the expression of <math>\xi_3</math> becomes,
<center><math>
\xi_3 =\Im\left\{
\left[\frac{\beta}{\alpha}\frac{\partial}{\partial x}
\left(\frac{\partial^2}{\partial x^2}-2k_y^2\right)
\frac{\partial\phi}{\partial z}\cdot\frac{\partial\phi^*}{\partial z}\right]_{-\infty}^{\infty}
- \left[\frac{\beta}{\alpha}\frac{\partial^2}{\partial x^2}\frac{\partial\phi}{\partial z}\cdot
\frac{\partial}{\partial x}\frac{\partial\phi^*}{\partial z}\right]_{-\infty}^{\infty}
+ \int_{-\infty}^\infty\frac{\beta}{\alpha}\frac{\partial^2}{\partial x^2}\frac{\partial\phi}{\partial z}\cdot
\frac{\partial^2}{\partial x^2}\frac{\partial\phi^*}{\partial z}dx\right\}.
</math></center>
As <math>{\frac{\partial^2}{\partial x^2}\frac{\partial\phi}{\partial z}\cdot \frac{\partial^2}{\partial x^2}\frac{\partial\phi^*}{\partial z}}</math> is real, we obtain the new expression of <math>\xi_3</math>
<center><math>
\xi_3 = \Im\left\{
\left[\frac{\beta}{\alpha}\frac{\partial}{\partial x}
\left(\frac{\partial^2}{\partial x^2}-2k_y^2\right)
\frac{\partial\phi}{\partial z}\cdot\frac{\partial\phi^*}{\partial z}\right]_{-\infty}^{\infty}-
\left[\frac{\beta}{\alpha}\frac{\partial^2}{\partial x^2}\frac{\partial\phi}{\partial z}\cdot
\frac{\partial}{\partial x}\frac{\partial\phi^*}{\partial z}\right]_{-\infty}^{\infty}\right\}.
</math></center>

Now breaking <math>\xi_3</math> down, we can simplify the left hend term for <math>x>0</math>

<center><math>
\begin{matrix}
\frac{\partial}{\partial x}\left(\frac{\partial^2}{\partial x^2}-2k_y^2\right)
\frac{\partial\phi(x_2,0)}{\partial z}\cdot\frac{\partial\phi(x_2,0)^*}{\partial z} \\ \\
= \left((-i\kappa^I_{\Lambda})^3k^I_\Lambda T_\Lambda(0)e^{-i\kappa^I_{\Lambda}(x)}\tanh{(k^I_\Lambda h)}\right.
\left.- 2k_y^2(-i\kappa_{\Lambda}(0))k^I_\Lambda T_\Lambda(0)e^{-i\kappa^I_{\Lambda}(x)} \tanh{(k^I_\Lambda h)}\right)
\left(k^I_\Lambda T_\Lambda(0)^*e^{i\kappa^I_{\Lambda}(x)}\tanh{(k^I_\Lambda h)}\right), \\ \\
= i\kappa^I_{\Lambda} (k^I_\Lambda)^2\left((\kappa^I_{\Lambda})^2 + 2k_y^2 \right) \tanh^2{(k^I_\Lambda h)}|T_\Lambda(0)|^2,
\end{matrix}
</math></center>

and for <math>x<0</math>

<center><math>
\begin{matrix}
\frac{\partial}{\partial x}\left(\frac{\partial^2}{\partial x^2}-2k_y^2\right)\frac{\partial\phi(x_1,0)}{\partial z}\cdot
\frac{\partial\phi(x_1,0)^*}{\partial z} \\ \\
= \left[i(\kappa^I_{1})^3k^I_1\left(e^{-i\kappa^I_{1}(x)}-R_1(0)e^{i\kappa^I_{1}(x)}\right) \tanh{(k^I_1 h)}
- 2ik_y^2\kappa^I_{1}k^I_1\left(-e^{-i\kappa^I_{1}(x)} + R_1(0)e^{i\kappa^I_{1}(x)}\right) \tanh{(k^I_1h)}\right]
\left[k^I_1\left(e^{i\kappa^I_{1}(x)} + R_1(0)^*e^{-i\kappa^I_{1}(x))}\right) \tanh{(k^I_1h)}\right], \\ \\
= \left[i\kappa^I_{1}k_1(0)((\kappa^I_{1})^2 + 2k_y^2)\left(e^{-i\kappa^I_{1}(x)} - R_1(0)e^{i\kappa^I_{1}(x)}\right)
\tanh{(k^I_1h)}\right]
\left[k^I_1\left(e^{i\kappa^I_{1}(x)} + R_1(0)^*e^{-i\kappa^I_{1}(x))}\right) \tanh{(k^I_1h)}\right], \\ \\
= i\kappa^I_{1}(k^I_1)^2\left((\kappa^I_{1})^2 + 2k_y^2\right)\tanh^2{(k^I_1h)}\left(1 - |R_1(0)|^2\right),
\end{matrix}
</math></center>

Likewise we expand the right hend term for <math>x>0</math>

<center><math>
\begin{matrix}
\frac{\partial^2}{\partial x^2}\frac{\partial\phi(x_2,0)}{\partial z}\cdot
\frac{\partial}{\partial x}\frac{\partial\phi^*}{\partial z} \\ \\
= \left[-(\kappa^I_{\Lambda})^2k^I_\Lambda T_\Lambda(0)e^{-i\kappa^I_{\Lambda}(x)} \tanh{(k^I_\Lambda h)}\right]
\left[i\kappa^I_{\Lambda} k^I_\Lambda T_\Lambda(0)^*e^{i\kappa^I_{\Lambda}(x)} \tanh{(k^I_\Lambda h)}\right], \\ \\
= -i(\kappa^I_{\Lambda})^3(k^I_\Lambda)^2\tanh^2{(k^I_\Lambda h)}|T_\Lambda(0)|^2,
\end{matrix}
</math></center>

and finally for <math>x<0</math>,

<center><math>
\begin{matrix}
\frac{\partial^2}{\partial x^1}\frac{\partial\phi(x_1,0)}{\partial z}\cdot
\frac{\partial}{\partial x}\frac{\partial\phi^*}{\partial z}\\ \\
= \left[-(\kappa^I_{1})^2k^I_1\left(e^{-i\kappa^I_{1}(x)} + R_1(0)e^{i\kappa^I_{1}(x)}\right)\tanh{(k^I_1 h)}\right]
\left[i\kappa^I_{1} k^I_1\left(e^{i\kappa^I_{1}(x)} - R_1(0)e^{-i\kappa^I_{1}(x)}\right)\tanh{(k^I_1 h)}\right], \\ \\
= -i(\kappa^I_{1})^3(k^I_1)^2\tanh^2{(k^I_1h)}\left(1 - |R_1(0)|^2\right).
\end{matrix}
</math></center>

We can now express <math>\xi_3</math> as

<center><math>
\begin{matrix}
\xi_3 & = & \Im
\left\{\frac{\beta_\Lambda}{\alpha}\left[i\kappa^I_{\Lambda} (k^I_\Lambda)^2\left((\kappa^I_{\Lambda})^2 + 2k_y^2 \right)
\tanh^2{(k^I_\Lambda h)}|T_\Lambda(0)|^2\right]
-\frac{\beta_1}{\alpha}\left[i\kappa^I_{1}(k^I_1)^2\left((\kappa^I_{1})^2 + 2k_y^2\right)\tanh^2{(k^I_1h)}
\left(1 - |R_1(0)|^2\right)\right]
-\frac{\beta_\Lambda}{\alpha}\left[-i(\kappa^I_{\Lambda})^3(k^I_\Lambda)^2\tanh^2{(k^I_\Lambda h)}|T_\Lambda(0)|^2\right]
+\frac{\beta_1}{\alpha}\left[-i(\kappa^I_{1})^3(k^I_1)^2\tanh^2{(k^I_1h)} \left(1 - |R_1(0)|^2\right)\right]\right\}, \\ \\
& = & \frac{\beta_\Lambda}{\alpha}\left[2\kappa^I_{\Lambda} (k^I_\Lambda)^2((\kappa^I_{\Lambda})^2 + k_y^2)\tanh^2{(k^I_\Lambda h)}|T_\Lambda(0)|^2\right]
-\frac{\beta_1}{\alpha}\left[2\kappa^I_{1}(k^I_1)^2((\kappa^I_{1})^2 + k_y^2)\tanh^2{(k^I_1h)}(1-|R_1(0)|^2)\right].
\end{matrix}
</math></center>

= Solving the Energy Balance Equation =

Pulling it all together, we finally obtain
<center><math>
\frac{\beta_\Lambda}{\alpha}\left[2\kappa^I_{\Lambda}(k^I_\Lambda)^2((\kappa^I_{\Lambda})^2 +k_y^2)
\tanh^2{(k^I_\Lambda h)}|T_\Lambda(0)|^2\right]
-\frac{\beta_1}{\alpha}\left[2\kappa^I_{1}(k^I_1)^2((\kappa^I_{1})^2 +k_y^2)\tanh^2{(k^I_1h)}(1-|R_1(0)|^2)\right]
+ \frac{\kappa^I_{\Lambda}|T_\Lambda(0)|^2}{2k^I_\Lambda}\left(\tanh{(k^I_\Lambda h)}+
\frac{k^I_\Lambda h}{\cosh^2{(k^I_\Lambda h)}}\right)
- \frac{\kappa^I_{1}\left(1 - |R_1(0)|^2\right)}{2k^I_1}\left(\tanh{(k^I_1h)}+\frac{hk^I_1}{\cosh^2{(k^I_1h)}}\right)=0.
</math></center>

Re-arranging gives

<center><math>
\kappa^I_{\Lambda}\tanh{(k^I_\Lambda h)}
\left(\frac{\beta_\Lambda}{\alpha}2(k^I_\Lambda)^2((\kappa^I_{\Lambda})^2 +k_y^2)\tanh{(k^I_\Lambda h)}
+ \frac{1}{2k^I_\Lambda}+\frac{h}{2\sinh{(k^I_\Lambda h)}\cosh{(k^I_\Lambda h)}}\right)|T_\Lambda(0)|^2
\quad - \kappa^I_{1}\tanh{(k^I_1 h)}
\left(\frac{\beta_1}{\alpha}2(k^I_1)^2((\kappa^I_{1})^2 +k_y^2)\tanh{(k^I_1 h)}
+ \frac{1}{2it}+\frac{h}{2\sinh{(k^I_1 h)}\cosh{(k^I_1 h)}}\right)(1-|R_1(0)|^2)=0
</math></center>

which can be expressed as
<center><math>
D |T_{\Lambda}(0)|^2 + |R_{1}(0)|^2 = 1,
</math></center>
where <math>D</math> is given by
<center><math>
D = \left(\frac{\kappa^I_{\Lambda} k^I_1\cosh^2{(k^I_1h)}}{\kappa^I_{1}k^I_\Lambda\cosh^2{(k^I_\Lambda h)}}\right)
\left(\frac{-\frac{\beta_\Lambda}{\alpha}4(k^I_\Lambda)^5\sinh^2{(k^I_\Lambda h)} +
\frac{1}{2}{\sinh{(2k^I_\Lambda h)}}+k^I_\Lambda h}
{-\frac{\beta_1}{\alpha}4(k^I_1)^5\sinh^2{(k^I_1h)} +
\frac{1}{2}{\sinh{(2k^I_1h)}}+k^I_1h}\right)
</math></center>

[[Category:Floating Elastic Plate]]

Energy Balance for Two Elastic Plates

2008-08-26T05:12:44Z

Fmontiel: /* Solving the Energy Balance Equation */

= Introduction =

We derive here a condition which must be satisfied for energy balance for two semi-infinite elastic plates. The solution applies
to the problem of [[Eigenfunction Matching Method for Floating Elastic Plates|multiple elastic plates]] and from here we know we can write the potential as
<center><math>
\phi \approx \left\{
\begin{matrix}
{
Ie^{\kappa_{1}(0)(x-r_1)}\frac{\cos(k_1(0)(z+h))}{\cos(k_1(0)h)} }+\\
{
\qquad \qquad \sum_{n=-2}^{M}R_1(n) e^{\kappa_{1}(n)(x-r_1)} \frac{\cos(k_1(n)(z+h))}{\cos(k_1(n)h)} },&
\mbox{ for } x < r_1,\\
{
\sum_{n=-2}^{M}T_{\mu}(n) e^{-\kappa_\mu(n)(x-l_\mu)}
\frac{\cos(k_\mu(n)(z+h))}{\cos(k_\mu(n)h)} } + \\
{
\qquad \qquad \sum_{n=-2}^{M}R_{\mu}(n) e^{\kappa_\mu(n)(x-r_\mu)}
\frac{\cos(k_\mu(n)(z+h))}{\cos(k_\mu(n)h)} },
&\mbox{ for } l_\mu< x < r_\mu,\\
{
\sum_{n=-2}^{M}T_{\Lambda}(n)e^{-\kappa_{\Lambda}(n)(x-l_\Lambda)}
\frac{\cos(k_\Lambda(n)(z+h))}{\cos(k_\Lambda(n)h)} }, &\mbox{ for } l_\Lambda<x,
\end{matrix} \right.
</math></center>
(for details of this notation see [[Eigenfunction Matching Method for Floating Elastic Plates]]).

= Equations =

Based on the method used in [[Evans and Davies 1968]], a check can be made to ensure the solutions of the floating plate problem are in energy balance.
This is simply a condition that the incident energy is equal to the sum of the radiated energy.
When the first and final plates have different properties, the energy balance equation is derived by applying Green's theorem to <math>\phi</math> and its conjugate.
The domain of integration is shown in the figure on the right.

[[Image:energy_schematic.jpg|thumb|right|300px|A diagram depicting the area <math>\mathcal{U}</math> which is bounded by the rectangle <math>\mathcal{S}</math>.]
{A diagram depicting the area <math>\mathcal{U}</math> which is bounded by the rectangle <math>\mathcal{S}</math>.
The rectangle <math>\mathcal{S}</math> is bounded by <math> -h\leq z \leq0</math> and <math>-\infty\leq x \leq \infty</math>]]

Applying Green's theorem to <math>\phi</math> and its conjugate <math>\phi^*</math> gives
<center><math>
{ \int\int_\mathcal{U}(\phi\nabla^2\phi^* - \phi^*\nabla^2\phi)dxdz
= \int_\mathcal{S}(\phi\frac{\partial\phi^*}{\partial n} - \phi^*\frac{\partial\phi}{\partial n})dl },
</math></center>
where <math>n</math> denotes the outward plane normal to the boundary and <math>l</math> denotes the plane parallel to the boundary.
As <math>\phi</math> and <math>\phi^*</math> satisfy the Laplace's equation, the left hand side of the Green theorem equation vanishes so that it reduces to
<center><math>
\Im\int_\mathcal{S}\phi\frac{\partial\phi^*}{\partial n} dl = 0,
</math></center>
Expanding gives
<center><math>
\xi_1 +\xi_2 + \xi_3 = 0,
</math></center>
where
<center><math>
\xi_1 =
-{ \Im\int_{-h}^0(\phi\frac{\partial\phi^*}{\partial x})\big|_{x=-\infty}dz },
</math></center>
<center><math>
\xi_2 =
{ \Im\int_{-h}^0(\phi\frac{\partial\phi^*}{\partial x})\big|_{x=\infty}dz },
</math></center>
and
<center><math>
\xi_3 =
{ \Im\int_{-\infty}^{\infty}(\phi\frac{\partial\phi^*}{\partial z})\big|_{z=0}dx } ,
</math></center>
where <math>\Im</math> denotes the imaginary part. The bottom domain composant is obviously equal to zero because of the no-influx condition over the seabed (<math>\frac{\partial\phi^*}{\partial z}\big|_{z=-h}=0</math>).

= Expanding <math>\mathbf{\xi_1}</math> =

Near <math>x=-\infty</math>, we approximate <math>\phi</math> by
<center><math>
\phi \approx e^{-\kappa_{1}(0)(x-r_1)}\frac{\cos{(k_1(0)(z+h))}}{\cos{(k_1(0)h)}} +
R_1(0)e^{\kappa_{1}(0)(x-r_1)}\frac{\cos{(k_{1}(0)(z+h))}}{\cos{(k_{1}(0)h)}}
</math></center>
Therefore,
<center><math>
\begin{matrix}
\xi_1 & = & {-\Im\int_{-h}^0 \left[\left(e^{-i\kappa^{I}_1(x)}+ R_1(0)e^{i\kappa^{I}_1(x)}\right)\right.}
{ \left(i\kappa^{I}_1e^{i\kappa^{I}_1(x)} - i\kappa^{I}_1R_1(0)^*e^{-i\kappa^{I}_1(x)}\right) }
{ \left.\left(\frac{\cosh^2{(k^I_1(z+h))}}{\cosh^2{(k^I_1h)}}\right)\right]dz, }\\ \\
& = & { - \Im\left[\frac{i\kappa^{I}_1\left(1 - |R_1(0)|^2\right)}{2\cosh^2{(k^I_1h)}}
\int_{-h}^0 \left(\cosh{(2k^I_1(z+h))+1}\right)dz\right] ,}\\ \\
& = & { - \Im\left[\frac{i\kappa^{I}_1\left(1 - |R_1(0)|^2\right)}{2\cosh^2{(k^I_1h)}}
\left[\frac{1}{2k^I_1}\sinh{(2k^I_1(z+h))} + z\right]_{-h}^0\right], }\\ \\
& = & { - \frac{\kappa^{I}_1\left(1 - |R_1(0)|^2\right)}{2\cosh^2{(k^I_1h)}} \left(\frac{1}{2k^I_1}\sinh{(2k^I_1h))} + h\right) ,}\\ \\
& = & { - \frac{\kappa^{I}_1\left(1 - |R_1(0)|^2\right)}{2k^I_1}\left(\tanh{(k^I_1h)}+\frac{hk^I_1}{\cosh^2{(k^I_1h)}}\right) , }
\end{matrix}
</math></center>
where <math>R_1(0)^*</math> is the conjugate of <math>R_1(0)</math>.

= Expanding <math>\mathbf{\xi_2}</math> =

Near <math>x=\infty</math>, we approximate <math>\phi</math> by
<center><math>
\begin{matrix}
{ \phi \approx T_\Lambda(0)e^{-\kappa_{\Lambda}(0)(x)}\frac{\cos{(k_\Lambda(0)(z+h))}}{\cos{(k_\Lambda(0)h)}}, }
\end{matrix}
</math></center>
and re-express as
<center><math>
\phi \approx T_\Lambda(0)e^{-i\kappa^I_\Lambda(x)}\frac{\cosh{(k^I_\Lambda(z+h))}}{\cosh{(k^I_\Lambda h)}},
</math></center>
where <math>k^I_\Lambda = \Im k_{\Lambda}(0)</math> and <math>\kappa^I_\Lambda=\Im \kappa_\Lambda(0)</math>, so that
<center><math>
\frac{\partial\phi}{\partial x} \approx -i\kappa^I_{\Lambda}T_\Lambda(0)e^{-i\kappa^I_{\Lambda}(x)}\frac{\cosh{(k^I_\Lambda(z+h))}}{\cosh{(k^I_\Lambda h)}}.
</math></center>
Therefore,
<center><math>
\begin{matrix}
\xi_2 & = & \Im\int_{-h}^0 \left[\right.(T_\Lambda(0)e^{-i\kappa^I_{\Lambda}(x)})(i\kappa^I_{\Lambda}T_\Lambda(0)^*e^{i\kappa^I_{\Lambda}(x)})
\frac{\cosh^2{(k^I_\Lambda(z+h))}}{\cosh^2{(k^I_\Lambda h)}}\left.\right]dz, \\ \\
& = & { \frac{\kappa^I_{\Lambda}|T_\Lambda(0)|^2}{2\cosh^2{(k^I_\Lambda h)}}
\left(\frac{1}{2k^I_\Lambda}\sinh{(2k^I_\Lambda h)} + h\right)},\\ \\
& = & { \frac{\kappa^I_{\Lambda}|T_\Lambda(0)|^2}{2k^I_\Lambda}\left(\tanh{(k^I_\Lambda h)}+\frac{k^I_\Lambda h}{\cosh^2{(k^I_\Lambda h)}}\right)}.
\end{matrix}
</math></center>

= Expanding <math>\mathbf{\xi_3}</math> =

The ice-covered boundary condition for the [[Floating Elastic Plate]] gives
<center><math>
\xi_3 = { \Im\int_{-\infty}^{\infty}\left(\frac{\beta}{\alpha}
\left(\frac{\partial^2}{\partial x^2} - k_y^2\right)^2 - \gamma + \frac{1}{\alpha}\right)
\frac{\partial\phi}{\partial z}\cdot\frac{\partial\phi^*}{\partial z}\bigg|_{z=0}dx.}
</math></center>
Since <math>{\frac{\partial\phi}{\partial z}\cdot\frac{\partial\phi^*}{\partial z}}</math> is real,
<center><math>
\xi_3 = \Im\int_{-\infty}^{\infty}\left(\frac{\beta}{\alpha}\frac{\partial^2}{\partial x^2}
\left(\frac{\partial^2}{\partial x^2} - 2k_y^2\right) \right)
\frac{\partial\phi}{\partial z}\cdot\frac{\partial\phi^*}{\partial z}\bigg|_{z=0}dx.
</math></center>
Integration by parts gives
<center><math>
\xi_3 = \Im\left\{\left[\frac{\beta}{\alpha}\frac{\partial}{\partial x}
\left(\frac{\partial^2}{\partial x^2}-2k_y^2\right)
\frac{\partial\phi}{\partial z}\cdot\frac{\partial\phi^*}{\partial z}\right]_{-\infty}^{\infty}
-\int_{-\infty}^{\infty}\frac{\beta}{\alpha}\frac{\partial}{\partial x}
\left(\frac{\partial^2}{\partial x^2} - 2k_y^2\right)
\frac{\partial\phi}{\partial z}\cdot\frac
{\partial}{\partial x}\frac{\partial\phi^*}{\partial z}dx
\right\}.
</math></center>
As <math>{2k_y^2\frac{\partial}{\partial x}\frac{\partial\phi}{\partial z}\cdot \frac{\partial}{\partial x}\frac{\partial\phi^*}{\partial z}}</math> is real and by integration by parts, the expression of <math>\xi_3</math> becomes,
<center><math>
\xi_3 =\Im\left\{
\left[\frac{\beta}{\alpha}\frac{\partial}{\partial x}
\left(\frac{\partial^2}{\partial x^2}-2k_y^2\right)
\frac{\partial\phi}{\partial z}\cdot\frac{\partial\phi^*}{\partial z}\right]_{-\infty}^{\infty}
- \left[\frac{\beta}{\alpha}\frac{\partial^2}{\partial x^2}\frac{\partial\phi}{\partial z}\cdot
\frac{\partial}{\partial x}\frac{\partial\phi^*}{\partial z}\right]_{-\infty}^{\infty}
+ \int_{-\infty}^\infty\frac{\beta}{\alpha}\frac{\partial^2}{\partial x^2}\frac{\partial\phi}{\partial z}\cdot
\frac{\partial^2}{\partial x^2}\frac{\partial\phi^*}{\partial z}dx\right\}.
</math></center>
As <math>{\frac{\partial^2}{\partial x^2}\frac{\partial\phi}{\partial z}\cdot \frac{\partial^2}{\partial x^2}\frac{\partial\phi^*}{\partial z}}</math> is real, we obtain the new expression of <math>\xi_3</math>
<center><math>
\xi_3 = \Im\left\{
\left[\frac{\beta}{\alpha}\frac{\partial}{\partial x}
\left(\frac{\partial^2}{\partial x^2}-2k_y^2\right)
\frac{\partial\phi}{\partial z}\cdot\frac{\partial\phi^*}{\partial z}\right]_{-\infty}^{\infty}-
\left[\frac{\beta}{\alpha}\frac{\partial^2}{\partial x^2}\frac{\partial\phi}{\partial z}\cdot
\frac{\partial}{\partial x}\frac{\partial\phi^*}{\partial z}\right]_{-\infty}^{\infty}\right\}.
</math></center>

Now breaking <math>\xi_3</math> down, we can simplify the left hend term for <math>x>0</math>

<center><math>
\begin{matrix}
\frac{\partial}{\partial x}\left(\frac{\partial^2}{\partial x^2}-2k_y^2\right)
\frac{\partial\phi(x_2,0)}{\partial z}\cdot\frac{\partial\phi(x_2,0)^*}{\partial z} \\ \\
= \left((-i\kappa^I_{\Lambda})^3k^I_\Lambda T_\Lambda(0)e^{-i\kappa^I_{\Lambda}(x)}\tanh{(k^I_\Lambda h)}\right.
\left.- 2k_y^2(-i\kappa_{\Lambda}(0))k^I_\Lambda T_\Lambda(0)e^{-i\kappa^I_{\Lambda}(x)} \tanh{(k^I_\Lambda h)}\right)
\left(k^I_\Lambda T_\Lambda(0)^*e^{i\kappa^I_{\Lambda}(x)}\tanh{(k^I_\Lambda h)}\right), \\ \\
= i\kappa^I_{\Lambda} (k^I_\Lambda)^2\left((\kappa^I_{\Lambda})^2 + 2k_y^2 \right) \tanh^2{(k^I_\Lambda h)}|T_\Lambda(0)|^2,
\end{matrix}
</math></center>

and for <math>x<0</math>

<center><math>
\begin{matrix}
\frac{\partial}{\partial x}\left(\frac{\partial^2}{\partial x^2}-2k_y^2\right)\frac{\partial\phi(x_1,0)}{\partial z}\cdot
\frac{\partial\phi(x_1,0)^*}{\partial z} \\ \\
= \left[i(\kappa^I_{1})^3k^I_1\left(e^{-i\kappa^I_{1}(x)}-R_1(0)e^{i\kappa^I_{1}(x)}\right) \tanh{(k^I_1 h)}
- 2ik_y^2\kappa^I_{1}k^I_1\left(-e^{-i\kappa^I_{1}(x)} + R_1(0)e^{i\kappa^I_{1}(x)}\right) \tanh{(k^I_1h)}\right]
\left[k^I_1\left(e^{i\kappa^I_{1}(x)} + R_1(0)^*e^{-i\kappa^I_{1}(x))}\right) \tanh{(k^I_1h)}\right], \\ \\
= \left[i\kappa^I_{1}k_1(0)((\kappa^I_{1})^2 + 2k_y^2)\left(e^{-i\kappa^I_{1}(x)} - R_1(0)e^{i\kappa^I_{1}(x)}\right)
\tanh{(k^I_1h)}\right]
\left[k^I_1\left(e^{i\kappa^I_{1}(x)} + R_1(0)^*e^{-i\kappa^I_{1}(x))}\right) \tanh{(k^I_1h)}\right], \\ \\
= i\kappa^I_{1}(k^I_1)^2\left((\kappa^I_{1})^2 + 2k_y^2\right)\tanh^2{(k^I_1h)}\left(1 - |R_1(0)|^2\right),
\end{matrix}
</math></center>

Likewise we expand the right hend term for <math>x>0</math>

<center><math>
\begin{matrix}
\frac{\partial^2}{\partial x^2}\frac{\partial\phi(x_2,0)}{\partial z}\cdot
\frac{\partial}{\partial x}\frac{\partial\phi^*}{\partial z} \\ \\
= \left[-(\kappa^I_{\Lambda})^2k^I_\Lambda T_\Lambda(0)e^{-i\kappa^I_{\Lambda}(x)} \tanh{(k^I_\Lambda h)}\right]
\left[i\kappa^I_{\Lambda} k^I_\Lambda T_\Lambda(0)^*e^{i\kappa^I_{\Lambda}(x)} \tanh{(k^I_\Lambda h)}\right], \\ \\
= -i(\kappa^I_{\Lambda})^3(k^I_\Lambda)^2\tanh^2{(k^I_\Lambda h)}|T_\Lambda(0)|^2,
\end{matrix}
</math></center>

and finally for <math>x<0</math>,

<center><math>
\begin{matrix}
\frac{\partial^2}{\partial x^1}\frac{\partial\phi(x_1,0)}{\partial z}\cdot
\frac{\partial}{\partial x}\frac{\partial\phi^*}{\partial z}\\ \\
= \left[-(\kappa^I_{1})^2k^I_1\left(e^{-i\kappa^I_{1}(x)} + R_1(0)e^{i\kappa^I_{1}(x)}\right)\tanh{(k^I_1 h)}\right]
\left[i\kappa^I_{1} k^I_1\left(e^{i\kappa^I_{1}(x)} - R_1(0)e^{-i\kappa^I_{1}(x)}\right)\tanh{(k^I_1 h)}\right], \\ \\
= -i(\kappa^I_{1})^3(k^I_1)^2\tanh^2{(k^I_1h)}\left(1 - |R_1(0)|^2\right).
\end{matrix}
</math></center>

We can now express <math>\xi_3</math> as

<center><math>
\begin{matrix}
\xi_3 & = & \Im
\left\{\frac{\beta_\Lambda}{\alpha}\left[i\kappa^I_{\Lambda} (k^I_\Lambda)^2\left((\kappa^I_{\Lambda})^2 + 2k_y^2 \right)
\tanh^2{(k^I_\Lambda h)}|T_\Lambda(0)|^2\right]
-\frac{\beta_1}{\alpha}\left[i\kappa^I_{1}(k^I_1)^2\left((\kappa^I_{1})^2 + 2k_y^2\right)\tanh^2{(k^I_1h)}
\left(1 - |R_1(0)|^2\right)\right]
-\frac{\beta_\Lambda}{\alpha}\left[-i(\kappa^I_{\Lambda})^3(k^I_\Lambda)^2\tanh^2{(k^I_\Lambda h)}|T_\Lambda(0)|^2\right]
+\frac{\beta_1}{\alpha}\left[-i(\kappa^I_{1})^3(k^I_1)^2\tanh^2{(k^I_1h)} \left(1 - |R_1(0)|^2\right)\right]\right\}, \\ \\
& = & \frac{\beta_\Lambda}{\alpha}\left[2\kappa^I_{\Lambda} (k^I_\Lambda)^2((\kappa^I_{\Lambda})^2 + k_y^2)\tanh^2{(k^I_\Lambda h)}|T_\Lambda(0)|^2\right]
-\frac{\beta_1}{\alpha}\left[2\kappa^I_{1}(k^I_1)^2((\kappa^I_{1})^2 + k_y^2)\tanh^2{(k^I_1h)}(1-|R_1(0)|^2)\right].
\end{matrix}
</math></center>

= Solving the Energy Balance Equation =

Pulling it all together, we finally obtain
<center><math>
\frac{\beta_\Lambda}{\alpha}\left[2\kappa^I_{\Lambda}(k^I_\Lambda)^2((\kappa^I_{\Lambda})^2 +k_y^2)
\tanh^2{(k^I_\Lambda h)}|T_\Lambda(0)|^2\right]
-\frac{\beta_1}{\alpha}\left[2\kappa^I_{1}(k^I_1)^2((\kappa^I_{1})^2 +k_y^2)\tanh^2{(k^I_1h)}(1-|R_1(0)|^2)\right]
+ \frac{\kappa^I_{\Lambda}|T_\Lambda(0)|^2}{2k^I_\Lambda}\left(\tanh{(k^I_\Lambda h)}+
\frac{k^I_\Lambda h}{\cosh^2{(k^I_\Lambda h)}}\right)
- \frac{\kappa^I_{1}\left(1 - |R_1(0)|^2\right)}{2k^I_1}\left(\tanh{(k^I_1h)}+\frac{hk^I_1}{\cosh^2{(k^I_1h)}}\right)=0.
</math></center>

Re-arranging gives

<center><math>
\kappa^I_{\Lambda}\tanh{(k^I_\Lambda h)}
\left(\frac{\beta_\Lambda}{\alpha}2(k^I_\Lambda)^2((\kappa^I_{\Lambda})^2 +k_y^2)\tanh{(k^I_\Lambda h)}
+ \frac{1}{2k^I_\Lambda}+\frac{h}{2\sinh{(k^I_\Lambda h)}\cosh{(k^I_\Lambda h)}}\right)|T_\Lambda(0)|^2
\quad - \kappa^I_{1}\tanh{(k^I_1 h)}
\left(\frac{\beta_1}{\alpha}2(k^I_1)^2((\kappa^I_{1})^2 +k_y^2)\tanh{(k^I_1 h)}
+ \frac{1}{2it}+\frac{h}{2\sinh{(k^I_1 h)}\cosh{(k^I_1 h)}}\right)(1-|R_1(0)|^2)=0
</math></center>

which can be expressed as
<center><math>
D |T_{\Lambda}(0)|^2 + |R_{1}(0)|^2 = 1,
</math></center>
where <math>D</math> is given by
<center><math>
D = \left(\frac{\kappa^I_{\Lambda} k^I_1\cosh^2{(k^I_1h)}}{\kappa^I_{1}k^I_\Lambda\cosh^2{(k^I_\Lambda h)}}\right)
\left(\frac{-\frac{\beta_\Lambda}{\alpha}4(k^I_\Lambda)^5\sinh^2{(k^I_\Lambda h)} +
\frac{1}{2}{\sinh{(2k^I_\Lambda h)}}+k^I_\Lambda h}
{-\frac{\beta_1}{\alpha}4(k^I_1)^5\sinh^2{(k^I_1h)} +
\frac{1}{2}{\sinh{(2k^I_1h)}}+k^I_1h}\right)
</math></center>

[[Category:Floating Elastic Plate]]

Energy Balance for Two Elastic Plates

2008-08-26T05:11:09Z

Fmontiel: /* Expanding <math>\mathbf{\xi_1}</math> */

= Introduction =

We derive here a condition which must be satisfied for energy balance for two semi-infinite elastic plates. The solution applies
to the problem of [[Eigenfunction Matching Method for Floating Elastic Plates|multiple elastic plates]] and from here we know we can write the potential as
<center><math>
\phi \approx \left\{
\begin{matrix}
{
Ie^{\kappa_{1}(0)(x-r_1)}\frac{\cos(k_1(0)(z+h))}{\cos(k_1(0)h)} }+\\
{
\qquad \qquad \sum_{n=-2}^{M}R_1(n) e^{\kappa_{1}(n)(x-r_1)} \frac{\cos(k_1(n)(z+h))}{\cos(k_1(n)h)} },&
\mbox{ for } x < r_1,\\
{
\sum_{n=-2}^{M}T_{\mu}(n) e^{-\kappa_\mu(n)(x-l_\mu)}
\frac{\cos(k_\mu(n)(z+h))}{\cos(k_\mu(n)h)} } + \\
{
\qquad \qquad \sum_{n=-2}^{M}R_{\mu}(n) e^{\kappa_\mu(n)(x-r_\mu)}
\frac{\cos(k_\mu(n)(z+h))}{\cos(k_\mu(n)h)} },
&\mbox{ for } l_\mu< x < r_\mu,\\
{
\sum_{n=-2}^{M}T_{\Lambda}(n)e^{-\kappa_{\Lambda}(n)(x-l_\Lambda)}
\frac{\cos(k_\Lambda(n)(z+h))}{\cos(k_\Lambda(n)h)} }, &\mbox{ for } l_\Lambda<x,
\end{matrix} \right.
</math></center>
(for details of this notation see [[Eigenfunction Matching Method for Floating Elastic Plates]]).

= Equations =

Based on the method used in [[Evans and Davies 1968]], a check can be made to ensure the solutions of the floating plate problem are in energy balance.
This is simply a condition that the incident energy is equal to the sum of the radiated energy.
When the first and final plates have different properties, the energy balance equation is derived by applying Green's theorem to <math>\phi</math> and its conjugate.
The domain of integration is shown in the figure on the right.

[[Image:energy_schematic.jpg|thumb|right|300px|A diagram depicting the area <math>\mathcal{U}</math> which is bounded by the rectangle <math>\mathcal{S}</math>.]
{A diagram depicting the area <math>\mathcal{U}</math> which is bounded by the rectangle <math>\mathcal{S}</math>.
The rectangle <math>\mathcal{S}</math> is bounded by <math> -h\leq z \leq0</math> and <math>-\infty\leq x \leq \infty</math>]]

Applying Green's theorem to <math>\phi</math> and its conjugate <math>\phi^*</math> gives
<center><math>
{ \int\int_\mathcal{U}(\phi\nabla^2\phi^* - \phi^*\nabla^2\phi)dxdz
= \int_\mathcal{S}(\phi\frac{\partial\phi^*}{\partial n} - \phi^*\frac{\partial\phi}{\partial n})dl },
</math></center>
where <math>n</math> denotes the outward plane normal to the boundary and <math>l</math> denotes the plane parallel to the boundary.
As <math>\phi</math> and <math>\phi^*</math> satisfy the Laplace's equation, the left hand side of the Green theorem equation vanishes so that it reduces to
<center><math>
\Im\int_\mathcal{S}\phi\frac{\partial\phi^*}{\partial n} dl = 0,
</math></center>
Expanding gives
<center><math>
\xi_1 +\xi_2 + \xi_3 = 0,
</math></center>
where
<center><math>
\xi_1 =
-{ \Im\int_{-h}^0(\phi\frac{\partial\phi^*}{\partial x})\big|_{x=-\infty}dz },
</math></center>
<center><math>
\xi_2 =
{ \Im\int_{-h}^0(\phi\frac{\partial\phi^*}{\partial x})\big|_{x=\infty}dz },
</math></center>
and
<center><math>
\xi_3 =
{ \Im\int_{-\infty}^{\infty}(\phi\frac{\partial\phi^*}{\partial z})\big|_{z=0}dx } ,
</math></center>
where <math>\Im</math> denotes the imaginary part. The bottom domain composant is obviously equal to zero because of the no-influx condition over the seabed (<math>\frac{\partial\phi^*}{\partial z}\big|_{z=-h}=0</math>).

= Expanding <math>\mathbf{\xi_1}</math> =

Near <math>x=-\infty</math>, we approximate <math>\phi</math> by
<center><math>
\phi \approx e^{-\kappa_{1}(0)(x-r_1)}\frac{\cos{(k_1(0)(z+h))}}{\cos{(k_1(0)h)}} +
R_1(0)e^{\kappa_{1}(0)(x-r_1)}\frac{\cos{(k_{1}(0)(z+h))}}{\cos{(k_{1}(0)h)}}
</math></center>
Therefore,
<center><math>
\begin{matrix}
\xi_1 & = & {-\Im\int_{-h}^0 \left[\left(e^{-i\kappa^{I}_1(x)}+ R_1(0)e^{i\kappa^{I}_1(x)}\right)\right.}
{ \left(i\kappa^{I}_1e^{i\kappa^{I}_1(x)} - i\kappa^{I}_1R_1(0)^*e^{-i\kappa^{I}_1(x)}\right) }
{ \left.\left(\frac{\cosh^2{(k^I_1(z+h))}}{\cosh^2{(k^I_1h)}}\right)\right]dz, }\\ \\
& = & { - \Im\left[\frac{i\kappa^{I}_1\left(1 - |R_1(0)|^2\right)}{2\cosh^2{(k^I_1h)}}
\int_{-h}^0 \left(\cosh{(2k^I_1(z+h))+1}\right)dz\right] ,}\\ \\
& = & { - \Im\left[\frac{i\kappa^{I}_1\left(1 - |R_1(0)|^2\right)}{2\cosh^2{(k^I_1h)}}
\left[\frac{1}{2k^I_1}\sinh{(2k^I_1(z+h))} + z\right]_{-h}^0\right], }\\ \\
& = & { - \frac{\kappa^{I}_1\left(1 - |R_1(0)|^2\right)}{2\cosh^2{(k^I_1h)}} \left(\frac{1}{2k^I_1}\sinh{(2k^I_1h))} + h\right) ,}\\ \\
& = & { - \frac{\kappa^{I}_1\left(1 - |R_1(0)|^2\right)}{2k^I_1}\left(\tanh{(k^I_1h)}+\frac{hk^I_1}{\cosh^2{(k^I_1h)}}\right) , }
\end{matrix}
</math></center>
where <math>R_1(0)^*</math> is the conjugate of <math>R_1(0)</math>.

= Expanding <math>\mathbf{\xi_2}</math> =

Near <math>x=\infty</math>, we approximate <math>\phi</math> by
<center><math>
\begin{matrix}
{ \phi \approx T_\Lambda(0)e^{-\kappa_{\Lambda}(0)(x)}\frac{\cos{(k_\Lambda(0)(z+h))}}{\cos{(k_\Lambda(0)h)}}, }
\end{matrix}
</math></center>
and re-express as
<center><math>
\phi \approx T_\Lambda(0)e^{-i\kappa^I_\Lambda(x)}\frac{\cosh{(k^I_\Lambda(z+h))}}{\cosh{(k^I_\Lambda h)}},
</math></center>
where <math>k^I_\Lambda = \Im k_{\Lambda}(0)</math> and <math>\kappa^I_\Lambda=\Im \kappa_\Lambda(0)</math>, so that
<center><math>
\frac{\partial\phi}{\partial x} \approx -i\kappa^I_{\Lambda}T_\Lambda(0)e^{-i\kappa^I_{\Lambda}(x)}\frac{\cosh{(k^I_\Lambda(z+h))}}{\cosh{(k^I_\Lambda h)}}.
</math></center>
Therefore,
<center><math>
\begin{matrix}
\xi_2 & = & \Im\int_{-h}^0 \left[\right.(T_\Lambda(0)e^{-i\kappa^I_{\Lambda}(x)})(i\kappa^I_{\Lambda}T_\Lambda(0)^*e^{i\kappa^I_{\Lambda}(x)})
\frac{\cosh^2{(k^I_\Lambda(z+h))}}{\cosh^2{(k^I_\Lambda h)}}\left.\right]dz, \\ \\
& = & { \frac{\kappa^I_{\Lambda}|T_\Lambda(0)|^2}{2\cosh^2{(k^I_\Lambda h)}}
\left(\frac{1}{2k^I_\Lambda}\sinh{(2k^I_\Lambda h)} + h\right)},\\ \\
& = & { \frac{\kappa^I_{\Lambda}|T_\Lambda(0)|^2}{2k^I_\Lambda}\left(\tanh{(k^I_\Lambda h)}+\frac{k^I_\Lambda h}{\cosh^2{(k^I_\Lambda h)}}\right)}.
\end{matrix}
</math></center>

= Expanding <math>\mathbf{\xi_3}</math> =

The ice-covered boundary condition for the [[Floating Elastic Plate]] gives
<center><math>
\xi_3 = { \Im\int_{-\infty}^{\infty}\left(\frac{\beta}{\alpha}
\left(\frac{\partial^2}{\partial x^2} - k_y^2\right)^2 - \gamma + \frac{1}{\alpha}\right)
\frac{\partial\phi}{\partial z}\cdot\frac{\partial\phi^*}{\partial z}\bigg|_{z=0}dx.}
</math></center>
Since <math>{\frac{\partial\phi}{\partial z}\cdot\frac{\partial\phi^*}{\partial z}}</math> is real,
<center><math>
\xi_3 = \Im\int_{-\infty}^{\infty}\left(\frac{\beta}{\alpha}\frac{\partial^2}{\partial x^2}
\left(\frac{\partial^2}{\partial x^2} - 2k_y^2\right) \right)
\frac{\partial\phi}{\partial z}\cdot\frac{\partial\phi^*}{\partial z}\bigg|_{z=0}dx.
</math></center>
Integration by parts gives
<center><math>
\xi_3 = \Im\left\{\left[\frac{\beta}{\alpha}\frac{\partial}{\partial x}
\left(\frac{\partial^2}{\partial x^2}-2k_y^2\right)
\frac{\partial\phi}{\partial z}\cdot\frac{\partial\phi^*}{\partial z}\right]_{-\infty}^{\infty}
-\int_{-\infty}^{\infty}\frac{\beta}{\alpha}\frac{\partial}{\partial x}
\left(\frac{\partial^2}{\partial x^2} - 2k_y^2\right)
\frac{\partial\phi}{\partial z}\cdot\frac
{\partial}{\partial x}\frac{\partial\phi^*}{\partial z}dx
\right\}.
</math></center>
As <math>{2k_y^2\frac{\partial}{\partial x}\frac{\partial\phi}{\partial z}\cdot \frac{\partial}{\partial x}\frac{\partial\phi^*}{\partial z}}</math> is real and by integration by parts, the expression of <math>\xi_3</math> becomes,
<center><math>
\xi_3 =\Im\left\{
\left[\frac{\beta}{\alpha}\frac{\partial}{\partial x}
\left(\frac{\partial^2}{\partial x^2}-2k_y^2\right)
\frac{\partial\phi}{\partial z}\cdot\frac{\partial\phi^*}{\partial z}\right]_{-\infty}^{\infty}
- \left[\frac{\beta}{\alpha}\frac{\partial^2}{\partial x^2}\frac{\partial\phi}{\partial z}\cdot
\frac{\partial}{\partial x}\frac{\partial\phi^*}{\partial z}\right]_{-\infty}^{\infty}
+ \int_{-\infty}^\infty\frac{\beta}{\alpha}\frac{\partial^2}{\partial x^2}\frac{\partial\phi}{\partial z}\cdot
\frac{\partial^2}{\partial x^2}\frac{\partial\phi^*}{\partial z}dx\right\}.
</math></center>
As <math>{\frac{\partial^2}{\partial x^2}\frac{\partial\phi}{\partial z}\cdot \frac{\partial^2}{\partial x^2}\frac{\partial\phi^*}{\partial z}}</math> is real, we obtain the new expression of <math>\xi_3</math>
<center><math>
\xi_3 = \Im\left\{
\left[\frac{\beta}{\alpha}\frac{\partial}{\partial x}
\left(\frac{\partial^2}{\partial x^2}-2k_y^2\right)
\frac{\partial\phi}{\partial z}\cdot\frac{\partial\phi^*}{\partial z}\right]_{-\infty}^{\infty}-
\left[\frac{\beta}{\alpha}\frac{\partial^2}{\partial x^2}\frac{\partial\phi}{\partial z}\cdot
\frac{\partial}{\partial x}\frac{\partial\phi^*}{\partial z}\right]_{-\infty}^{\infty}\right\}.
</math></center>

Now breaking <math>\xi_3</math> down, we can simplify the left hend term for <math>x>0</math>

<center><math>
\begin{matrix}
\frac{\partial}{\partial x}\left(\frac{\partial^2}{\partial x^2}-2k_y^2\right)
\frac{\partial\phi(x_2,0)}{\partial z}\cdot\frac{\partial\phi(x_2,0)^*}{\partial z} \\ \\
= \left((-i\kappa^I_{\Lambda})^3k^I_\Lambda T_\Lambda(0)e^{-i\kappa^I_{\Lambda}(x)}\tanh{(k^I_\Lambda h)}\right.
\left.- 2k_y^2(-i\kappa_{\Lambda}(0))k^I_\Lambda T_\Lambda(0)e^{-i\kappa^I_{\Lambda}(x)} \tanh{(k^I_\Lambda h)}\right)
\left(k^I_\Lambda T_\Lambda(0)^*e^{i\kappa^I_{\Lambda}(x)}\tanh{(k^I_\Lambda h)}\right), \\ \\
= i\kappa^I_{\Lambda} (k^I_\Lambda)^2\left((\kappa^I_{\Lambda})^2 + 2k_y^2 \right) \tanh^2{(k^I_\Lambda h)}|T_\Lambda(0)|^2,
\end{matrix}
</math></center>

and for <math>x<0</math>

<center><math>
\begin{matrix}
\frac{\partial}{\partial x}\left(\frac{\partial^2}{\partial x^2}-2k_y^2\right)\frac{\partial\phi(x_1,0)}{\partial z}\cdot
\frac{\partial\phi(x_1,0)^*}{\partial z} \\ \\
= \left[i(\kappa^I_{1})^3k^I_1\left(e^{-i\kappa^I_{1}(x)}-R_1(0)e^{i\kappa^I_{1}(x)}\right) \tanh{(k^I_1 h)}
- 2ik_y^2\kappa^I_{1}k^I_1\left(-e^{-i\kappa^I_{1}(x)} + R_1(0)e^{i\kappa^I_{1}(x)}\right) \tanh{(k^I_1h)}\right]
\left[k^I_1\left(e^{i\kappa^I_{1}(x)} + R_1(0)^*e^{-i\kappa^I_{1}(x))}\right) \tanh{(k^I_1h)}\right], \\ \\
= \left[i\kappa^I_{1}k_1(0)((\kappa^I_{1})^2 + 2k_y^2)\left(e^{-i\kappa^I_{1}(x)} - R_1(0)e^{i\kappa^I_{1}(x)}\right)
\tanh{(k^I_1h)}\right]
\left[k^I_1\left(e^{i\kappa^I_{1}(x)} + R_1(0)^*e^{-i\kappa^I_{1}(x))}\right) \tanh{(k^I_1h)}\right], \\ \\
= i\kappa^I_{1}(k^I_1)^2\left((\kappa^I_{1})^2 + 2k_y^2\right)\tanh^2{(k^I_1h)}\left(1 - |R_1(0)|^2\right),
\end{matrix}
</math></center>

Likewise we expand the right hend term for <math>x>0</math>

<center><math>
\begin{matrix}
\frac{\partial^2}{\partial x^2}\frac{\partial\phi(x_2,0)}{\partial z}\cdot
\frac{\partial}{\partial x}\frac{\partial\phi^*}{\partial z} \\ \\
= \left[-(\kappa^I_{\Lambda})^2k^I_\Lambda T_\Lambda(0)e^{-i\kappa^I_{\Lambda}(x)} \tanh{(k^I_\Lambda h)}\right]
\left[i\kappa^I_{\Lambda} k^I_\Lambda T_\Lambda(0)^*e^{i\kappa^I_{\Lambda}(x)} \tanh{(k^I_\Lambda h)}\right], \\ \\
= -i(\kappa^I_{\Lambda})^3(k^I_\Lambda)^2\tanh^2{(k^I_\Lambda h)}|T_\Lambda(0)|^2,
\end{matrix}
</math></center>

and finally for <math>x<0</math>,

<center><math>
\begin{matrix}
\frac{\partial^2}{\partial x^1}\frac{\partial\phi(x_1,0)}{\partial z}\cdot
\frac{\partial}{\partial x}\frac{\partial\phi^*}{\partial z}\\ \\
= \left[-(\kappa^I_{1})^2k^I_1\left(e^{-i\kappa^I_{1}(x)} + R_1(0)e^{i\kappa^I_{1}(x)}\right)\tanh{(k^I_1 h)}\right]
\left[i\kappa^I_{1} k^I_1\left(e^{i\kappa^I_{1}(x)} - R_1(0)e^{-i\kappa^I_{1}(x)}\right)\tanh{(k^I_1 h)}\right], \\ \\
= -i(\kappa^I_{1})^3(k^I_1)^2\tanh^2{(k^I_1h)}\left(1 - |R_1(0)|^2\right).
\end{matrix}
</math></center>

We can now express <math>\xi_3</math> as

<center><math>
\begin{matrix}
\xi_3 & = & \Im
\left\{\frac{\beta_\Lambda}{\alpha}\left[i\kappa^I_{\Lambda} (k^I_\Lambda)^2\left((\kappa^I_{\Lambda})^2 + 2k_y^2 \right)
\tanh^2{(k^I_\Lambda h)}|T_\Lambda(0)|^2\right]
-\frac{\beta_1}{\alpha}\left[i\kappa^I_{1}(k^I_1)^2\left((\kappa^I_{1})^2 + 2k_y^2\right)\tanh^2{(k^I_1h)}
\left(1 - |R_1(0)|^2\right)\right]
-\frac{\beta_\Lambda}{\alpha}\left[-i(\kappa^I_{\Lambda})^3(k^I_\Lambda)^2\tanh^2{(k^I_\Lambda h)}|T_\Lambda(0)|^2\right]
+\frac{\beta_1}{\alpha}\left[-i(\kappa^I_{1})^3(k^I_1)^2\tanh^2{(k^I_1h)} \left(1 - |R_1(0)|^2\right)\right]\right\}, \\ \\
& = & \frac{\beta_\Lambda}{\alpha}\left[2\kappa^I_{\Lambda} (k^I_\Lambda)^2((\kappa^I_{\Lambda})^2 + k_y^2)\tanh^2{(k^I_\Lambda h)}|T_\Lambda(0)|^2\right]
-\frac{\beta_1}{\alpha}\left[2\kappa^I_{1}(k^I_1)^2((\kappa^I_{1})^2 + k_y^2)\tanh^2{(k^I_1h)}(1-|R_1(0)|^2)\right].
\end{matrix}
</math></center>

= Solving the Energy Balance Equation =

Pulling it all together, we finally obtain
<center><math>
\frac{\beta_\Lambda}{\alpha}\left[2\kappa^I_{\Lambda}(k^I_\Lambda)^2((\kappa^I_{\Lambda})^2 +k_y^2)
\tanh^2{(k^I_\Lambda h)}|T_\Lambda(0)|^2\right]
-\frac{\beta_1}{\alpha}\left[2\kappa^I_{1}(k^I_1)^2((\kappa^I_{1})^2 +k_y^2)\tanh^2{(k^I_1h)}(1-|R_1(0)|^2)\right]
+ \frac{\kappa^I_{\Lambda}|T_\Lambda(0)|^2}{2k^I_\Lambda}\left(\tanh{(k^I_\Lambda h)}+
\frac{k^I_\Lambda h}{\cosh^2{(k^I_\Lambda h)}}\right)
- \frac{\kappa^I_{1}\left(1 - |R_1(0)|^2\right)}{2k^I_1}\left(\tanh{(k^I_1h)}+\frac{hk^I_1}{\cosh^2{(k^I_1h)}}\right)=0.
</math></center>

Re-arranging gives

<center><math>
\kappa^I_{\Lambda}\tanh{(k^I_\Lambda h)}
\left(\frac{\beta_\Lambda}{\alpha}2(k^I_\Lambda)^2((\kappa^I_{\Lambda})^2 +k_y^2)\tanh{(k^I_\Lambda h)}
+ \frac{1}{2k^I_\Lambda}+\frac{h}{2\sinh{(k^I_\Lambda h)}\cosh{(k^I_\Lambda h)}}\right)|T_\Lambda(0)|^2
\quad - \kappa^I_{1}\tanh{(k^I_1 h)}
\left(\frac{\beta_1}{\alpha}2(k^I_1)^2((\kappa^I_{1})^2 +k_y^2)\tanh{(k^I_1 h)}
+ \frac{1}{2it}+\frac{h}{2\sinh{(k^I_1 h)}\cosh{(k^I_1 h)}}\right)(1-|R_1(0)|^2)=0
</math></center>

which can be expressed as
<center><math>
D |T_{\Lambda}(0)|^2 + |R_{1}(0)|^2 = 1,
</math></center>
where <math>D</math> is given by
<center><math>
D = \left(\frac{\kappa^I_{\Lambda} k^I_1\cosh^2{(k^I_1h)}}{\kappa^I_{1}k^I_\Lambda\cosh^2{(k^I_\Lambda h)}}\right)
\left(\frac{\frac{\beta_\Lambda}{\alpha}4(k^I_\Lambda)^2\kappa^I_{\Lambda}((\kappa^I_{\Lambda})^2 -k_y^2)\sinh^2{(k^I_\Lambda h)} +
\frac{1}{2}{\sinh{(2k^I_\Lambda h)}}+k^I_\Lambda h}
{\frac{\beta_1}{\alpha}4(k^I_1)^2\kappa^I_1((\kappa^I_{1})^2 -k_y^2)\sinh^2{(k^I_1h)} +
\frac{1}{2}{\sinh{(2k^I_1h)}}+k^I_1h}\right)
</math></center>

[[Category:Floating Elastic Plate]]

Eigenfunction Matching for a Semi-Infinite Floating Elastic Plate

2008-08-26T04:12:03Z

Fmontiel: /* Energy Balance */

=Introduction=

We show here a solution for a semi-infinite [[:Category:Floating Elastic Plate|Floating Elastic Plate]] on [[Finite Depth]].
The problem was solved by [[Fox and Squire 1994]] but the solution method here is slightly different.
The simpler theory for a [[Eigenfunction Matching for a Semi-Infinite Dock|Dock]] describes
many of the ideas here in more detail.

[[Image:Semiinfinite plate.jpg|thumb|right|300px|Wave scattering by a submerged semi-infinite elastic plate]]

= Equations =

We consider the problem of small-amplitude waves which are incident on a semi-infinite floating elastic
plate occupying water surface for <math>x>0</math>. The submergence of the plate is considered negligible.
We assume that the problem is invariant in the <math>y</math> direction, although we allow the waves to be
incident from an angle.
We also assume that the plate edges are free to move at
each boundary, although other boundary conditions could easily be considered using
the methods of solution presented here. We begin with the [[Frequency Domain Problem]] for a semi-infinite
[[Floating Elastic Plate|Floating Elastic Plates]]
in the non-dimensional form of [[Tayler_1986a|Tayler 1986]] ([[Dispersion Relation for a Floating Elastic Plate]]).
We also assume that the waves are normally incident (incidence at an angle will be discussed later).
<center><math>\begin{matrix}
\Delta \phi = 0, \;\;\; -h < z \leq 0,
\end{matrix}</math></center>
<center><math>
\partial_z \phi = 0, \;\;\; z = - h,
</math></center>
<center><math>
\partial_z\phi=\alpha\phi, \,\, z=0,\,x<0,
</math></center>
<center><math>
\beta \partial_x^4\partial_z \phi
- \left( \gamma\alpha - 1 \right) \partial_z \phi = \alpha\phi, \;\;
z = 0, \;\;\; x \geq 0,
</math></center>
where <math>\alpha = \omega^2</math>, <math>\beta</math> and <math>\gamma</math>
are the stiffness and mass constant for the plate respectively. The free edge conditions
at the edge of the plate imply
<center><math>
\partial_x^3 \partial_z\phi = 0, \;\; z = 0, \;\;\; x = 0,
</math></center>
<center><math>
\partial_x^2 \partial_z\phi = 0, \;\; z = 0, \;\;\; x = 0,
</math></center>

=Method of solution=

==Eigenfunction expansion==

We will solve the system of equations using an [[:Category:Eigenfunction Matching Method|Eigenfunction Matching Method]].
The method was developed by [[Fox and Squire 1994]]. The separation of variables for the left hand region where
there is open water, and the incident wave, is described in [[Eigenfunction Matching for a Semi-Infinite Dock]] and we consider
here only the separation of variables in the plate covered region.

===Separation of variables under the Plate===
The potential velocity can be written in terms of an infinite series of separated eigenfunctions under
each elastic plate, of the form
<center><math>
\psi = \frac{\cos(\kappa (z+h))}{\cos\kappa h}.\,
</math></center>
If we apply the boundary conditions given
we obtain the [[Dispersion Relation for a Floating Elastic Plate]]
<center><math>
\kappa \tan{(\kappa h)}= -\frac{\alpha}{\beta \kappa^{4} + 1 - \alpha\gamma}
</math></center>
Solving for <math>\kappa</math> gives a pure imaginary root
with positive imaginary part, two complex roots (two complex conjugate paired roots
with positive imaginary part in most physical situations), an infinite number of positive real roots
which approach <math>{n\pi}/{h}</math> as <math>n</math> approaches infinity, and also the negative of all
these roots ([[Dispersion Relation for a Floating Elastic Plate]]) . We denote the two complex roots with positive imaginary part
by <math>\kappa_(-2)</math> and <math>\kappa_(-1)</math>, the purely imaginary
root with positive imaginary part by <math>\kappa_{0}</math> and the real roots with positive imaginary part
by <math>\kappa_(n)</math> for <math>n</math> a positive integer.
The imaginary root with positive imaginary part corresponds to a
reflected travelling mode propagating along the <math>x</math> axis.
The complex roots with positive imaginary parts correspond to damped reflected travelling modes and the real roots correspond to reflected evanescent modes.

==Expressions for the potential velocity==

We now expand the potential in the two regions using the separation of variables solution.
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.
The potential <math>\phi</math> can now be expressed as the following sum of eigenfunctions:
be expanded as
<center>
<math>
\phi(x,z)= e^{-k_{0}x}\phi_{0}(z) + \sum_{m=0}^{\infty}a_{m}e^{k_{m}x}\phi_{m}(z), \;\;x<0
</math>
</center>
and
<center>
<math>
\phi(x,z)=\sum_{m=-2}^{\infty}b_{m}
e^{-\kappa_{m}x}\psi_{m}(z), \;\;x>0
</math>
</center>
where <math>a_{m}</math> and <math>b_{m}</math> are the coefficients to be determined.

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>

==An infinite dimensional system of equations==

The potential and its derivative must be continuous across the
transition from open water to the plate covered region. Therefore, the
potentials and their derivatives at <math>x=0</math> have to be equal.
We also truncate the sum at <math>N</math> being careful that we have
two extra modes on the plate covered region to satisfy the edge conditions.
We obtain
<center>
<math>
\phi_{0}\left( z\right) + \sum_{m=0}^{N}
a_{m} \phi_{m}\left( z\right)
=\sum_{m=-2}^{N}b_{m}\psi_{m}(z)
</math>
</center>
and
<center>
<math>
-k_{0}\phi_{0}\left( z\right) +\sum
_{m=0}^{N} a_{m}k_{m}\phi_{m}\left( z\right)
=-\sum_{m=-2}^{N}b_{m}\kappa_{m}\psi
_{m}(z)
</math>
</center>
for each <math>N</math>.
We solve these equations by multiplying both equations by
<math>\phi_{l}(z)\,</math> and integrating from <math>-h</math> to <math>0</math> to obtain:
<center>
<math>
A_{0}\delta_{0l}+a_{l}A_{l}
=\sum_{m=-2}^{N}b_{m}B_{ml},\,\,0 \leq l \leq N
</math>
</center>
and
<center>
<math>
-k_{0}A_{0}\delta_{0l}+a_{l}k_{l}A_l
=-\sum_{m=-2}^{N}b_{m}\kappa_{m}B_{ml},\,\,0 \leq l \leq N
</math>
</center>
If we multiply the first equation by <math>k_l</math> and subtract the second equation
we obtain
<center>
<math>
2k_{0}A_{0}\delta_{0l}
=\sum_{m=-2}^{N}b_{m}(k_l + \kappa_{m})B_{ml}
</math>
</center>
Finally, we need to apply the conditions at the edge of the plate to give us two further equations,
<center>
<math>
\partial_x^2\partial_z\phi = - \sum_{m=-2}^{N}b_{m} \kappa_m^3 \tan\kappa_m h = 0
</math>
</center>
and
<center>
<math>
\partial_x^3\partial_z\phi = \sum_{m=-2}^{N}b_{m} \kappa_m^4 \tan\kappa_m h = 0
</math>
</center>

=Numerical Solution=

To solve the system of equations previously defined we set the upper limit of <math>l</math> to
be <math>N</math>, as we already said before. In terms of matrix, we obtain
<center>
<math>
\begin{bmatrix}
\begin{bmatrix}
A_0&0 \quad \cdots&0\\
0&&\\
\vdots&A_l&\vdots\\
&&0\\
0&\cdots \quad 0 &A_N
\end{bmatrix}
&
\begin{bmatrix}
-B_{-20}&-B_{-10}\\
&\\
\vdots&\vdots\\
&\\
-B_{-2N}&-B_{-1N}
\end{bmatrix}
&
\begin{bmatrix}
-B_{00}&\cdots&-B_{0N}\\
&&\\
\vdots&-B_{ml}&\vdots\\
&&\\
-B_{N0}&\cdots&-B_{NN}
\end{bmatrix}
\\
\begin{bmatrix}
0&&\cdots&&0\\
0&&\cdots&&0
\end{bmatrix}
&
\begin{bmatrix}
\kappa_{-2}^3\tan\kappa_{-2}h&\kappa_{-1}^3\tan\kappa_{-1}h\\
\kappa_{-2}^4\tan\kappa_{-2}h&\kappa_{-1}^4\tan\kappa_{-1}h
\end{bmatrix}
&
\begin{bmatrix}
\kappa_0^3\tan\kappa_0h&\cdots&\kappa_N^3\tan\kappa_Nh\\
\kappa_0^4\tan\kappa_0h&\cdots&\kappa_N^4\tan\kappa_Nh
\end{bmatrix}
\\
\begin{bmatrix}
0&&\cdots&&0\\
&&&&\\
\vdots&&\ddots&&\vdots\\
&&&&\\
0&&\cdots&&0
\end{bmatrix}
&
\begin{bmatrix}
(k_0+\kappa_{-2})B_{-20}&(k_0+\kappa_{-1})B_{-10}\\
&\\
\vdots&\vdots\\
&\\
(k_N+\kappa_{-2})B_{-2N}&(k_N+\kappa_{-1})B_{-1N}
\end{bmatrix}
&
\begin{bmatrix}
(k_0 + \kappa_0) \, B_{00}&\cdots&(k_N + \kappa_{0}) \, B_{0N}\\
&&\\
\vdots&(k_l + \kappa_{m}) \, B_{ml}&\vdots\\
&&\\
(k_0 + \kappa_N) \, B_{N0}&\cdots&(k_N + \kappa_{N}) \, B_{NN}\\
\end{bmatrix}
\end{bmatrix}
\begin{bmatrix}
a_{0} \\
\\
\vdots \\
\\
a_N \\
b_{-2}\\
b_{-1}\\
b_{0}\\
\\
\vdots \\
\\
b_N
\end{bmatrix}
=
\begin{bmatrix}
- A_{0} \\
0 \\
\vdots \\
\\
0 \\
0 \\
0 \\
2k_{0}A_{0} \\
0 \\
\vdots \\
\\
0
\end{bmatrix}
</math>
</center>
We then simply need to solve the linear system of equations. Note that we can solve this equation for
<math>b_n</math> first and then solve for <math>a_n</math>

= Waves Incident at an Angle =

We can consider the case of [[Waves Incident at an Angle]] <math>\theta</math>. The theory shows that we can introduce
a wavenumber in the <math>y</math>-direction which express <math>k_y=k_0 \sin \theta</math>.

It is shown that the potential can be expanded as
<center>
<math>
\phi(x,z)=e^{-\hat{k}_0x}\phi_0(z)+\sum_{m=0}^{\infty}a_{m}e^{\hat{k}_{m}x}\phi_{m}(z), \;\;x<0
</math>
</center>
and
<center>
<math>
\phi(x,z)=\sum_{m=-2}^{\infty}b_{m}
e^{-\hat{\kappa}_{m}x}\psi_{m}(z), \;\;x>0
</math>
</center>
where <math>\hat{k}_{m} = \sqrt{k_m^2 - k_y^2}</math> and <math>\hat{\kappa}_{m} = \sqrt{\kappa_m^2 - k_y^2}</math>
where we always take the positive real root or the root with positive imaginary part.

The edge conditions are also different and are
<center><math>
\left(\frac{\partial^3}{\partial x^3} + (2 - \nu)k^2_y\frac{\partial}{\partial x}\right) \frac{\partial\phi}{\partial z}= 0,
</math></center>
and
<center><math>
\left(\frac{\partial^2}{\partial x^2} + \nu k^2_y\right)\frac{\partial\phi}{\partial z} = 0,
</math></center>
where <math>\nu</math> is Poisons ratio.

We can expend these edge conditions, which respectively gives
<center>
<math>
-\sum_{m=-2}^{\infty}b_{m}\kappa_m (\hat{\kappa}_m^3+\hat{\kappa}_m k_y^2(2-\nu))\tan \kappa_m h=0
</math>
</center>
and
<center>
<math>
-\sum_{m=-2}^{\infty}b_{m}\kappa_m (\hat{\kappa}_m^2+k_y^2\nu))\tan \kappa_m h=0
</math>
</center>

The equations are derived almost identically to those above and we obtain
<center>
<math>
A_{0}\delta_{0l}+a_{l}A_{l}
=\sum_{m=-2}^{\infty}b_{m}B_{ml}
</math>
</center>
and
<center>
<math>
-\hat{k_{0}}A_{0}\delta_{0l}+a_{l}\hat{k}_{l}A_l
=-\sum_{m=-2}^{\infty}b_{m}\hat{\kappa}_{m}B_{ml}
</math>
</center>
and these are solved exactly as before.

= Energy Balance =

The energy balance equation is ([[Energy Balance for Two Elastic Plates]] where the left-hand side plate
is open water for our problem, so we need to take the stiffness of this plate equal to <math>0</math>, so as to
model water) given by
<center><math>
D |T|^2 + |R|^2 = 1, \,
</math></center>
where <math> T=b_0 </math>, <math>R=a_0</math>, and <math>D</math> is given by
<center><math>
D = \left(\frac{\hat{\kappa}_0 k_0\cos^2{(k_0 h)}}{\hat{k}_0 \kappa_0\cos^2{(\kappa_0 h)}}\right)
\left(\frac{-\frac{4\beta}{\alpha}(\kappa_0)^3(\hat{\kappa}_0^2 +k_y^2)\sin^2{(\kappa_0 h)} +
\frac{1}{2}{\sin{(2 \kappa_0 h)}}+\kappa_0 h}
{
\frac{1}{2}{\sin{(2 k_0 h)}}+k_0h}\right)
</math></center>
where the index <math>0</math> refers to the travelling wave (imaginary root of the dispersion equation).

= Matlab Code =

A program to calculate the coefficients for the semi-infinite dock problems can be found here
[http://www.math.auckland.ac.nz/~meylan/code/eigenfunction_matching/semiinfinite_plate.m semiinfinite_plate.m]

== Additional code ==

This program requires
* {{free surface dispersion equation code}}
* {{elastic plate dispersion equation code}}

[[Category:Floating Elastic Plate]]
[[Category:Eigenfunction Matching Method]]
[[Category:Complete Pages]]

Eigenfunction Matching for a Semi-Infinite Floating Elastic Plate

2008-08-25T02:58:32Z

Fmontiel: /* Energy Balance */

=Introduction=

We show here a solution for a semi-infinite [[:Category:Floating Elastic Plate|Floating Elastic Plate]] on [[Finite Depth]].
The problem was solved by [[Fox and Squire 1994]] but the solution method here is slightly different.
The simpler theory for a [[Eigenfunction Matching for a Semi-Infinite Dock|Dock]] describes
many of the ideas here in more detail.

[[Image:Semiinfinite plate.jpg|thumb|right|300px|Wave scattering by a submerged semi-infinite elastic plate]]

= Equations =

We consider the problem of small-amplitude waves which are incident on a semi-infinite floating elastic
plate occupying water surface for <math>x>0</math>. The submergence of the plate is considered negligible.
We assume that the problem is invariant in the <math>y</math> direction, although we allow the waves to be
incident from an angle.
We also assume that the plate edges are free to move at
each boundary, although other boundary conditions could easily be considered using
the methods of solution presented here. We begin with the [[Frequency Domain Problem]] for a semi-infinite
[[Floating Elastic Plate|Floating Elastic Plates]]
in the non-dimensional form of [[Tayler_1986a|Tayler 1986]] ([[Dispersion Relation for a Floating Elastic Plate]]).
We also assume that the waves are normally incident (incidence at an angle will be discussed later).
<center><math>\begin{matrix}
\Delta \phi = 0, \;\;\; -h < z \leq 0,
\end{matrix}</math></center>
<center><math>
\partial_z \phi = 0, \;\;\; z = - h,
</math></center>
<center><math>
\partial_z\phi=\alpha\phi, \,\, z=0,\,x<0,
</math></center>
<center><math>
\beta \partial_x^4\partial_z \phi
- \left( \gamma\alpha - 1 \right) \partial_z \phi = \alpha\phi, \;\;
z = 0, \;\;\; x \geq 0,
</math></center>
where <math>\alpha = \omega^2</math>, <math>\beta</math> and <math>\gamma</math>
are the stiffness and mass constant for the plate respectively. The free edge conditions
at the edge of the plate imply
<center><math>
\partial_x^3 \partial_z\phi = 0, \;\; z = 0, \;\;\; x = 0,
</math></center>
<center><math>
\partial_x^2 \partial_z\phi = 0, \;\; z = 0, \;\;\; x = 0,
</math></center>

=Method of solution=

==Eigenfunction expansion==

We will solve the system of equations using an [[:Category:Eigenfunction Matching Method|Eigenfunction Matching Method]].
The method was developed by [[Fox and Squire 1994]]. The separation of variables for the left hand region where
there is open water, and the incident wave, is described in [[Eigenfunction Matching for a Semi-Infinite Dock]] and we consider
here only the separation of variables in the plate covered region.

===Separation of variables under the Plate===
The potential velocity can be written in terms of an infinite series of separated eigenfunctions under
each elastic plate, of the form
<center><math>
\psi = \frac{\cos(\kappa (z+h))}{\cos\kappa h}.\,
</math></center>
If we apply the boundary conditions given
we obtain the [[Dispersion Relation for a Floating Elastic Plate]]
<center><math>
\kappa \tan{(\kappa h)}= -\frac{\alpha}{\beta \kappa^{4} + 1 - \alpha\gamma}
</math></center>
Solving for <math>\kappa</math> gives a pure imaginary root
with positive imaginary part, two complex roots (two complex conjugate paired roots
with positive imaginary part in most physical situations), an infinite number of positive real roots
which approach <math>{n\pi}/{h}</math> as <math>n</math> approaches infinity, and also the negative of all
these roots ([[Dispersion Relation for a Floating Elastic Plate]]) . We denote the two complex roots with positive imaginary part
by <math>\kappa_(-2)</math> and <math>\kappa_(-1)</math>, the purely imaginary
root with positive imaginary part by <math>\kappa_{0}</math> and the real roots with positive imaginary part
by <math>\kappa_(n)</math> for <math>n</math> a positive integer.
The imaginary root with positive imaginary part corresponds to a
reflected travelling mode propagating along the <math>x</math> axis.
The complex roots with positive imaginary parts correspond to damped reflected travelling modes and the real roots correspond to reflected evanescent modes.

==Expressions for the potential velocity==

We now expand the potential in the two regions using the separation of variables solution.
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.
The potential <math>\phi</math> can now be expressed as the following sum of eigenfunctions:
be expanded as
<center>
<math>
\phi(x,z)= e^{-k_{0}x}\phi_{0}(z) + \sum_{m=0}^{\infty}a_{m}e^{k_{m}x}\phi_{m}(z), \;\;x<0
</math>
</center>
and
<center>
<math>
\phi(x,z)=\sum_{m=-2}^{\infty}b_{m}
e^{-\kappa_{m}x}\psi_{m}(z), \;\;x>0
</math>
</center>
where <math>a_{m}</math> and <math>b_{m}</math> are the coefficients to be determined.

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>

==An infinite dimensional system of equations==

The potential and its derivative must be continuous across the
transition from open water to the plate covered region. Therefore, the
potentials and their derivatives at <math>x=0</math> have to be equal.
We also truncate the sum at <math>N</math> being careful that we have
two extra modes on the plate covered region to satisfy the edge conditions.
We obtain
<center>
<math>
\phi_{0}\left( z\right) + \sum_{m=0}^{N}
a_{m} \phi_{m}\left( z\right)
=\sum_{m=-2}^{N}b_{m}\psi_{m}(z)
</math>
</center>
and
<center>
<math>
-k_{0}\phi_{0}\left( z\right) +\sum
_{m=0}^{N} a_{m}k_{m}\phi_{m}\left( z\right)
=-\sum_{m=-2}^{N}b_{m}\kappa_{m}\psi
_{m}(z)
</math>
</center>
for each <math>N</math>.
We solve these equations by multiplying both equations by
<math>\phi_{l}(z)\,</math> and integrating from <math>-h</math> to <math>0</math> to obtain:
<center>
<math>
A_{0}\delta_{0l}+a_{l}A_{l}
=\sum_{m=-2}^{N}b_{m}B_{ml},\,\,0 \leq l \leq N
</math>
</center>
and
<center>
<math>
-k_{0}A_{0}\delta_{0l}+a_{l}k_{l}A_l
=-\sum_{m=-2}^{N}b_{m}\kappa_{m}B_{ml},\,\,0 \leq l \leq N
</math>
</center>
If we multiply the first equation by <math>k_l</math> and subtract the second equation
we obtain
<center>
<math>
2k_{0}A_{0}\delta_{0l}
=\sum_{m=-2}^{N}b_{m}(k_l + \kappa_{m})B_{ml}
</math>
</center>
Finally, we need to apply the conditions at the edge of the plate to give us two further equations,
<center>
<math>
\partial_x^2\partial_z\phi = - \sum_{m=-2}^{N}b_{m} \kappa_m^3 \tan\kappa_m h = 0
</math>
</center>
and
<center>
<math>
\partial_x^3\partial_z\phi = \sum_{m=-2}^{N}b_{m} \kappa_m^4 \tan\kappa_m h = 0
</math>
</center>

=Numerical Solution=

To solve the system of equations previously defined we set the upper limit of <math>l</math> to
be <math>N</math>, as we already said before. In terms of matrix, we obtain
<center>
<math>
\begin{bmatrix}
\begin{bmatrix}
A_0&0 \quad \cdots&0\\
0&&\\
\vdots&A_l&\vdots\\
&&0\\
0&\cdots \quad 0 &A_N
\end{bmatrix}
&
\begin{bmatrix}
-B_{-20}&-B_{-10}\\
&\\
\vdots&\vdots\\
&\\
-B_{-2N}&-B_{-1N}
\end{bmatrix}
&
\begin{bmatrix}
-B_{00}&\cdots&-B_{0N}\\
&&\\
\vdots&-B_{ml}&\vdots\\
&&\\
-B_{N0}&\cdots&-B_{NN}
\end{bmatrix}
\\
\begin{bmatrix}
0&&\cdots&&0\\
0&&\cdots&&0
\end{bmatrix}
&
\begin{bmatrix}
\kappa_{-2}^3\tan\kappa_{-2}h&\kappa_{-1}^3\tan\kappa_{-1}h\\
\kappa_{-2}^4\tan\kappa_{-2}h&\kappa_{-1}^4\tan\kappa_{-1}h
\end{bmatrix}
&
\begin{bmatrix}
\kappa_0^3\tan\kappa_0h&\cdots&\kappa_N^3\tan\kappa_Nh\\
\kappa_0^4\tan\kappa_0h&\cdots&\kappa_N^4\tan\kappa_Nh
\end{bmatrix}
\\
\begin{bmatrix}
0&&\cdots&&0\\
&&&&\\
\vdots&&\ddots&&\vdots\\
&&&&\\
0&&\cdots&&0
\end{bmatrix}
&
\begin{bmatrix}
(k_0+\kappa_{-2})B_{-20}&(k_0+\kappa_{-1})B_{-10}\\
&\\
\vdots&\vdots\\
&\\
(k_N+\kappa_{-2})B_{-2N}&(k_N+\kappa_{-1})B_{-1N}
\end{bmatrix}
&
\begin{bmatrix}
(k_0 + \kappa_0) \, B_{00}&\cdots&(k_N + \kappa_{0}) \, B_{0N}\\
&&\\
\vdots&(k_l + \kappa_{m}) \, B_{ml}&\vdots\\
&&\\
(k_0 + \kappa_N) \, B_{N0}&\cdots&(k_N + \kappa_{N}) \, B_{NN}\\
\end{bmatrix}
\end{bmatrix}
\begin{bmatrix}
a_{0} \\
\\
\vdots \\
\\
a_N \\
b_{-2}\\
b_{-1}\\
b_{0}\\
\\
\vdots \\
\\
b_N
\end{bmatrix}
=
\begin{bmatrix}
- A_{0} \\
0 \\
\vdots \\
\\
0 \\
0 \\
0 \\
2k_{0}A_{0} \\
0 \\
\vdots \\
\\
0
\end{bmatrix}
</math>
</center>
We then simply need to solve the linear system of equations. Note that we can solve this equation for
<math>b_n</math> first and then solve for <math>a_n</math>

= Waves Incident at an Angle =

We can consider the case of [[Waves Incident at an Angle]] <math>\theta</math>. The theory shows that we can introduce
a wavenumber in the <math>y</math>-direction which express <math>k_y=k_0 \sin \theta</math>.

It is shown that the potential can be expanded as
<center>
<math>
\phi(x,z)=e^{-\hat{k}_0x}\phi_0(z)+\sum_{m=0}^{\infty}a_{m}e^{\hat{k}_{m}x}\phi_{m}(z), \;\;x<0
</math>
</center>
and
<center>
<math>
\phi(x,z)=\sum_{m=-2}^{\infty}b_{m}
e^{-\hat{\kappa}_{m}x}\psi_{m}(z), \;\;x>0
</math>
</center>
where <math>\hat{k}_{m} = \sqrt{k_m^2 - k_y^2}</math> and <math>\hat{\kappa}_{m} = \sqrt{\kappa_m^2 - k_y^2}</math>
where we always take the positive real root or the root with positive imaginary part.

The edge conditions are also different and are
<center><math>
\left(\frac{\partial^3}{\partial x^3} - (2 - \nu)k^2_y\frac{\partial}{\partial x}\right) \frac{\partial\phi}{\partial z}= 0,
</math></center>
and
<center><math>
\left(\frac{\partial^2}{\partial x^2} - \nu k^2_y\right)\frac{\partial\phi}{\partial z} = 0,
</math></center>
where <math>\nu</math> is Poisons ratio.

We can expend these edge conditions, which respectively gives
<center>
<math>
-\sum_{m=-2}^{\infty}b_{m}\kappa_m (\hat{\kappa}_m^3-\hat{\kappa}_m k_y^2(2-\nu))\tan \kappa_m h=0
</math>
</center>
and
<center>
<math>
-\sum_{m=-2}^{\infty}b_{m}\kappa_m (\hat{\kappa}_m^2-k_y^2\nu))\tan \kappa_m h=0
</math>
</center>

The equations are derived almost identically to those above and we obtain
<center>
<math>
A_{0}\delta_{0l}+a_{l}A_{l}
=\sum_{m=-2}^{\infty}b_{m}B_{ml}
</math>
</center>
and
<center>
<math>
-\hat{k_{0}}A_{0}\delta_{0l}+a_{l}\hat{k}_{l}A_l
=-\sum_{m=-2}^{\infty}b_{m}\hat{\kappa}_{m}B_{ml}
</math>
</center>
and these are solved exactly as before.

= Energy Balance =

The energy balance equation is ([[Energy Balance for Two Elastic Plates]] where the left-hand side plate
is open water for our problem, so we need to take the stiffness of this plate equal to <math>0</math>, so as to
model water) given by
<center><math>
D |T|^2 + |R|^2 = 1, \,
</math></center>
where <math> T=b_0 </math>, <math>R=a_0</math>, and <math>D</math> is given by
<center><math>
D = \left(\frac{\hat{\kappa}_0 k_0\cos^2{(k_0 h)}}{\hat{k}_0 \kappa_0\cos^2{(\kappa_0 h)}}\right)
\left(\frac{\frac{4\beta}{\alpha}(\kappa_0)^2\hat{\kappa}_0(\hat{\kappa}_0^2 -k_y^2)\sin^2{(\kappa_0 h)} +
\frac{1}{2}{\sin{(2 \kappa_0 h)}}+\kappa_0 h}
{
\frac{1}{2}{\sin{(2 k_0 h)}}+k_0h}\right)
</math></center>
where the index <math>0</math> refers to the travelling wave (imaginary root of the dispersion equation).

= Matlab Code =

A program to calculate the coefficients for the semi-infinite dock problems can be found here
[http://www.math.auckland.ac.nz/~meylan/code/eigenfunction_matching/semiinfinite_plate.m semiinfinite_plate.m]

== Additional code ==

This program requires
* {{free surface dispersion equation code}}
* {{elastic plate dispersion equation code}}

[[Category:Floating Elastic Plate]]
[[Category:Eigenfunction Matching Method]]
[[Category:Complete Pages]]

Eigenfunction Matching for a Semi-Infinite Floating Elastic Plate

2008-08-25T02:57:05Z

Fmontiel: /* Energy Balance */

=Introduction=

We show here a solution for a semi-infinite [[:Category:Floating Elastic Plate|Floating Elastic Plate]] on [[Finite Depth]].
The problem was solved by [[Fox and Squire 1994]] but the solution method here is slightly different.
The simpler theory for a [[Eigenfunction Matching for a Semi-Infinite Dock|Dock]] describes
many of the ideas here in more detail.

[[Image:Semiinfinite plate.jpg|thumb|right|300px|Wave scattering by a submerged semi-infinite elastic plate]]

= Equations =

We consider the problem of small-amplitude waves which are incident on a semi-infinite floating elastic
plate occupying water surface for <math>x>0</math>. The submergence of the plate is considered negligible.
We assume that the problem is invariant in the <math>y</math> direction, although we allow the waves to be
incident from an angle.
We also assume that the plate edges are free to move at
each boundary, although other boundary conditions could easily be considered using
the methods of solution presented here. We begin with the [[Frequency Domain Problem]] for a semi-infinite
[[Floating Elastic Plate|Floating Elastic Plates]]
in the non-dimensional form of [[Tayler_1986a|Tayler 1986]] ([[Dispersion Relation for a Floating Elastic Plate]]).
We also assume that the waves are normally incident (incidence at an angle will be discussed later).
<center><math>\begin{matrix}
\Delta \phi = 0, \;\;\; -h < z \leq 0,
\end{matrix}</math></center>
<center><math>
\partial_z \phi = 0, \;\;\; z = - h,
</math></center>
<center><math>
\partial_z\phi=\alpha\phi, \,\, z=0,\,x<0,
</math></center>
<center><math>
\beta \partial_x^4\partial_z \phi
- \left( \gamma\alpha - 1 \right) \partial_z \phi = \alpha\phi, \;\;
z = 0, \;\;\; x \geq 0,
</math></center>
where <math>\alpha = \omega^2</math>, <math>\beta</math> and <math>\gamma</math>
are the stiffness and mass constant for the plate respectively. The free edge conditions
at the edge of the plate imply
<center><math>
\partial_x^3 \partial_z\phi = 0, \;\; z = 0, \;\;\; x = 0,
</math></center>
<center><math>
\partial_x^2 \partial_z\phi = 0, \;\; z = 0, \;\;\; x = 0,
</math></center>

=Method of solution=

==Eigenfunction expansion==

We will solve the system of equations using an [[:Category:Eigenfunction Matching Method|Eigenfunction Matching Method]].
The method was developed by [[Fox and Squire 1994]]. The separation of variables for the left hand region where
there is open water, and the incident wave, is described in [[Eigenfunction Matching for a Semi-Infinite Dock]] and we consider
here only the separation of variables in the plate covered region.

===Separation of variables under the Plate===
The potential velocity can be written in terms of an infinite series of separated eigenfunctions under
each elastic plate, of the form
<center><math>
\psi = \frac{\cos(\kappa (z+h))}{\cos\kappa h}.\,
</math></center>
If we apply the boundary conditions given
we obtain the [[Dispersion Relation for a Floating Elastic Plate]]
<center><math>
\kappa \tan{(\kappa h)}= -\frac{\alpha}{\beta \kappa^{4} + 1 - \alpha\gamma}
</math></center>
Solving for <math>\kappa</math> gives a pure imaginary root
with positive imaginary part, two complex roots (two complex conjugate paired roots
with positive imaginary part in most physical situations), an infinite number of positive real roots
which approach <math>{n\pi}/{h}</math> as <math>n</math> approaches infinity, and also the negative of all
these roots ([[Dispersion Relation for a Floating Elastic Plate]]) . We denote the two complex roots with positive imaginary part
by <math>\kappa_(-2)</math> and <math>\kappa_(-1)</math>, the purely imaginary
root with positive imaginary part by <math>\kappa_{0}</math> and the real roots with positive imaginary part
by <math>\kappa_(n)</math> for <math>n</math> a positive integer.
The imaginary root with positive imaginary part corresponds to a
reflected travelling mode propagating along the <math>x</math> axis.
The complex roots with positive imaginary parts correspond to damped reflected travelling modes and the real roots correspond to reflected evanescent modes.

==Expressions for the potential velocity==

We now expand the potential in the two regions using the separation of variables solution.
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.
The potential <math>\phi</math> can now be expressed as the following sum of eigenfunctions:
be expanded as
<center>
<math>
\phi(x,z)= e^{-k_{0}x}\phi_{0}(z) + \sum_{m=0}^{\infty}a_{m}e^{k_{m}x}\phi_{m}(z), \;\;x<0
</math>
</center>
and
<center>
<math>
\phi(x,z)=\sum_{m=-2}^{\infty}b_{m}
e^{-\kappa_{m}x}\psi_{m}(z), \;\;x>0
</math>
</center>
where <math>a_{m}</math> and <math>b_{m}</math> are the coefficients to be determined.

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>

==An infinite dimensional system of equations==

The potential and its derivative must be continuous across the
transition from open water to the plate covered region. Therefore, the
potentials and their derivatives at <math>x=0</math> have to be equal.
We also truncate the sum at <math>N</math> being careful that we have
two extra modes on the plate covered region to satisfy the edge conditions.
We obtain
<center>
<math>
\phi_{0}\left( z\right) + \sum_{m=0}^{N}
a_{m} \phi_{m}\left( z\right)
=\sum_{m=-2}^{N}b_{m}\psi_{m}(z)
</math>
</center>
and
<center>
<math>
-k_{0}\phi_{0}\left( z\right) +\sum
_{m=0}^{N} a_{m}k_{m}\phi_{m}\left( z\right)
=-\sum_{m=-2}^{N}b_{m}\kappa_{m}\psi
_{m}(z)
</math>
</center>
for each <math>N</math>.
We solve these equations by multiplying both equations by
<math>\phi_{l}(z)\,</math> and integrating from <math>-h</math> to <math>0</math> to obtain:
<center>
<math>
A_{0}\delta_{0l}+a_{l}A_{l}
=\sum_{m=-2}^{N}b_{m}B_{ml},\,\,0 \leq l \leq N
</math>
</center>
and
<center>
<math>
-k_{0}A_{0}\delta_{0l}+a_{l}k_{l}A_l
=-\sum_{m=-2}^{N}b_{m}\kappa_{m}B_{ml},\,\,0 \leq l \leq N
</math>
</center>
If we multiply the first equation by <math>k_l</math> and subtract the second equation
we obtain
<center>
<math>
2k_{0}A_{0}\delta_{0l}
=\sum_{m=-2}^{N}b_{m}(k_l + \kappa_{m})B_{ml}
</math>
</center>
Finally, we need to apply the conditions at the edge of the plate to give us two further equations,
<center>
<math>
\partial_x^2\partial_z\phi = - \sum_{m=-2}^{N}b_{m} \kappa_m^3 \tan\kappa_m h = 0
</math>
</center>
and
<center>
<math>
\partial_x^3\partial_z\phi = \sum_{m=-2}^{N}b_{m} \kappa_m^4 \tan\kappa_m h = 0
</math>
</center>

=Numerical Solution=

To solve the system of equations previously defined we set the upper limit of <math>l</math> to
be <math>N</math>, as we already said before. In terms of matrix, we obtain
<center>
<math>
\begin{bmatrix}
\begin{bmatrix}
A_0&0 \quad \cdots&0\\
0&&\\
\vdots&A_l&\vdots\\
&&0\\
0&\cdots \quad 0 &A_N
\end{bmatrix}
&
\begin{bmatrix}
-B_{-20}&-B_{-10}\\
&\\
\vdots&\vdots\\
&\\
-B_{-2N}&-B_{-1N}
\end{bmatrix}
&
\begin{bmatrix}
-B_{00}&\cdots&-B_{0N}\\
&&\\
\vdots&-B_{ml}&\vdots\\
&&\\
-B_{N0}&\cdots&-B_{NN}
\end{bmatrix}
\\
\begin{bmatrix}
0&&\cdots&&0\\
0&&\cdots&&0
\end{bmatrix}
&
\begin{bmatrix}
\kappa_{-2}^3\tan\kappa_{-2}h&\kappa_{-1}^3\tan\kappa_{-1}h\\
\kappa_{-2}^4\tan\kappa_{-2}h&\kappa_{-1}^4\tan\kappa_{-1}h
\end{bmatrix}
&
\begin{bmatrix}
\kappa_0^3\tan\kappa_0h&\cdots&\kappa_N^3\tan\kappa_Nh\\
\kappa_0^4\tan\kappa_0h&\cdots&\kappa_N^4\tan\kappa_Nh
\end{bmatrix}
\\
\begin{bmatrix}
0&&\cdots&&0\\
&&&&\\
\vdots&&\ddots&&\vdots\\
&&&&\\
0&&\cdots&&0
\end{bmatrix}
&
\begin{bmatrix}
(k_0+\kappa_{-2})B_{-20}&(k_0+\kappa_{-1})B_{-10}\\
&\\
\vdots&\vdots\\
&\\
(k_N+\kappa_{-2})B_{-2N}&(k_N+\kappa_{-1})B_{-1N}
\end{bmatrix}
&
\begin{bmatrix}
(k_0 + \kappa_0) \, B_{00}&\cdots&(k_N + \kappa_{0}) \, B_{0N}\\
&&\\
\vdots&(k_l + \kappa_{m}) \, B_{ml}&\vdots\\
&&\\
(k_0 + \kappa_N) \, B_{N0}&\cdots&(k_N + \kappa_{N}) \, B_{NN}\\
\end{bmatrix}
\end{bmatrix}
\begin{bmatrix}
a_{0} \\
\\
\vdots \\
\\
a_N \\
b_{-2}\\
b_{-1}\\
b_{0}\\
\\
\vdots \\
\\
b_N
\end{bmatrix}
=
\begin{bmatrix}
- A_{0} \\
0 \\
\vdots \\
\\
0 \\
0 \\
0 \\
2k_{0}A_{0} \\
0 \\
\vdots \\
\\
0
\end{bmatrix}
</math>
</center>
We then simply need to solve the linear system of equations. Note that we can solve this equation for
<math>b_n</math> first and then solve for <math>a_n</math>

= Waves Incident at an Angle =

We can consider the case of [[Waves Incident at an Angle]] <math>\theta</math>. The theory shows that we can introduce
a wavenumber in the <math>y</math>-direction which express <math>k_y=k_0 \sin \theta</math>.

It is shown that the potential can be expanded as
<center>
<math>
\phi(x,z)=e^{-\hat{k}_0x}\phi_0(z)+\sum_{m=0}^{\infty}a_{m}e^{\hat{k}_{m}x}\phi_{m}(z), \;\;x<0
</math>
</center>
and
<center>
<math>
\phi(x,z)=\sum_{m=-2}^{\infty}b_{m}
e^{-\hat{\kappa}_{m}x}\psi_{m}(z), \;\;x>0
</math>
</center>
where <math>\hat{k}_{m} = \sqrt{k_m^2 - k_y^2}</math> and <math>\hat{\kappa}_{m} = \sqrt{\kappa_m^2 - k_y^2}</math>
where we always take the positive real root or the root with positive imaginary part.

The edge conditions are also different and are
<center><math>
\left(\frac{\partial^3}{\partial x^3} - (2 - \nu)k^2_y\frac{\partial}{\partial x}\right) \frac{\partial\phi}{\partial z}= 0,
</math></center>
and
<center><math>
\left(\frac{\partial^2}{\partial x^2} - \nu k^2_y\right)\frac{\partial\phi}{\partial z} = 0,
</math></center>
where <math>\nu</math> is Poisons ratio.

We can expend these edge conditions, which respectively gives
<center>
<math>
-\sum_{m=-2}^{\infty}b_{m}\kappa_m (\hat{\kappa}_m^3-\hat{\kappa}_m k_y^2(2-\nu))\tan \kappa_m h=0
</math>
</center>
and
<center>
<math>
-\sum_{m=-2}^{\infty}b_{m}\kappa_m (\hat{\kappa}_m^2-k_y^2\nu))\tan \kappa_m h=0
</math>
</center>

The equations are derived almost identically to those above and we obtain
<center>
<math>
A_{0}\delta_{0l}+a_{l}A_{l}
=\sum_{m=-2}^{\infty}b_{m}B_{ml}
</math>
</center>
and
<center>
<math>
-\hat{k_{0}}A_{0}\delta_{0l}+a_{l}\hat{k}_{l}A_l
=-\sum_{m=-2}^{\infty}b_{m}\hat{\kappa}_{m}B_{ml}
</math>
</center>
and these are solved exactly as before.

= Energy Balance =

The energy balance equation is ([[Energy Balance for Two Elastic Plates]] where the left-hand side plate
is open water for our problem, so we need to take the stiffness of this plate equal to <math>0</math>, so as to
modeling water) given by
<center><math>
D |T|^2 + |R|^2 = 1, \,
</math></center>
where <math> T=b_0 </math>, <math>R=a_0</math>, and <math>D</math> is given by
<center><math>
D = \left(\frac{\hat{\kappa}_0 k_0\cos^2{(k_0 h)}}{\hat{k}_0 \kappa_0\cos^2{(\kappa_0 h)}}\right)
\left(\frac{\frac{4\beta}{\alpha}(\kappa_0)^2\hat{\kappa}_0(\hat{\kappa}_0^2 -k_y^2)\sin^2{(\kappa_0 h)} +
\frac{1}{2}{\sin{(2 \kappa_0 h)}}+\kappa_0 h}
{
\frac{1}{2}{\sin{(2 k_0 h)}}+k_0h}\right)
</math></center>
where the index <math>0</math> refers to the travelling wave (imaginary root of the dispersion equation).

= Matlab Code =

A program to calculate the coefficients for the semi-infinite dock problems can be found here
[http://www.math.auckland.ac.nz/~meylan/code/eigenfunction_matching/semiinfinite_plate.m semiinfinite_plate.m]

== Additional code ==

This program requires
* {{free surface dispersion equation code}}
* {{elastic plate dispersion equation code}}

[[Category:Floating Elastic Plate]]
[[Category:Eigenfunction Matching Method]]
[[Category:Complete Pages]]

Energy Balance for Two Elastic Plates

2008-08-25T02:30:05Z

Fmontiel: /* Solving the Energy Balance Equation */

= Introduction =

We derive here a condition which must be satisfied for energy balance for two semi-infinite elastic plates. The solution applies
to the problem of [[Eigenfunction Matching Method for Floating Elastic Plates|multiple elastic plates]] and from here we know we can write the potential as
<center><math>
\phi \approx \left\{
\begin{matrix}
{
Ie^{\kappa_{1}(0)(x-r_1)}\frac{\cos(k_1(0)(z+h))}{\cos(k_1(0)h)} }+\\
{
\qquad \qquad \sum_{n=-2}^{M}R_1(n) e^{\kappa_{1}(n)(x-r_1)} \frac{\cos(k_1(n)(z+h))}{\cos(k_1(n)h)} },&
\mbox{ for } x < r_1,\\
{
\sum_{n=-2}^{M}T_{\mu}(n) e^{-\kappa_\mu(n)(x-l_\mu)}
\frac{\cos(k_\mu(n)(z+h))}{\cos(k_\mu(n)h)} } + \\
{
\qquad \qquad \sum_{n=-2}^{M}R_{\mu}(n) e^{\kappa_\mu(n)(x-r_\mu)}
\frac{\cos(k_\mu(n)(z+h))}{\cos(k_\mu(n)h)} },
&\mbox{ for } l_\mu< x < r_\mu,\\
{
\sum_{n=-2}^{M}T_{\Lambda}(n)e^{-\kappa_{\Lambda}(n)(x-l_\Lambda)}
\frac{\cos(k_\Lambda(n)(z+h))}{\cos(k_\Lambda(n)h)} }, &\mbox{ for } l_\Lambda<x,
\end{matrix} \right.
</math></center>
(for details of this notation see [[Eigenfunction Matching Method for Floating Elastic Plates]]).

= Equations =

Based on the method used in [[Evans and Davies 1968]], a check can be made to ensure the solutions of the floating plate problem are in energy balance.
This is simply a condition that the incident energy is equal to the sum of the radiated energy.
When the first and final plates have different properties, the energy balance equation is derived by applying Green's theorem to <math>\phi</math> and its conjugate.
The domain of integration is shown in the figure on the right.

[[Image:energy_schematic.jpg|thumb|right|300px|A diagram depicting the area <math>\mathcal{U}</math> which is bounded by the rectangle <math>\mathcal{S}</math>.]
{A diagram depicting the area <math>\mathcal{U}</math> which is bounded by the rectangle <math>\mathcal{S}</math>.
The rectangle <math>\mathcal{S}</math> is bounded by <math> -h\leq z \leq0</math> and <math>-\infty\leq x \leq \infty</math>]]

Applying Green's theorem to <math>\phi</math> and its conjugate <math>\phi^*</math> gives
<center><math>
{ \int\int_\mathcal{U}(\phi\nabla^2\phi^* - \phi^*\nabla^2\phi)dxdz
= \int_\mathcal{S}(\phi\frac{\partial\phi^*}{\partial n} - \phi^*\frac{\partial\phi}{\partial n})dl },
</math></center>
where <math>n</math> denotes the outward plane normal to the boundary and <math>l</math> denotes the plane parallel to the boundary.
As <math>\phi</math> and <math>\phi^*</math> satisfy the Laplace's equation, the left hand side of the Green theorem equation vanishes so that it reduces to
<center><math>
\Im\int_\mathcal{S}\phi\frac{\partial\phi^*}{\partial n} dl = 0,
</math></center>
Expanding gives
<center><math>
\xi_1 +\xi_2 + \xi_3 = 0,
</math></center>
where
<center><math>
\xi_1 =
-{ \Im\int_{-h}^0(\phi\frac{\partial\phi^*}{\partial x})\big|_{x=-\infty}dz },
</math></center>
<center><math>
\xi_2 =
{ \Im\int_{-h}^0(\phi\frac{\partial\phi^*}{\partial x})\big|_{x=\infty}dz },
</math></center>
and
<center><math>
\xi_3 =
{ \Im\int_{-\infty}^{\infty}(\phi\frac{\partial\phi^*}{\partial z})\big|_{z=0}dx } ,
</math></center>
where <math>\Im</math> denotes the imaginary part. The bottom domain composant is obviously equal to zero because of the no-influx condition over the seabed (<math>\frac{\partial\phi^*}{\partial z}\big|_{z=-h}=0</math>).

= Expanding <math>\mathbf{\xi_1}</math> =

Near <math>x=-\infty</math>, we approximate <math>\phi</math> by
<center><math>
\phi \approx e^{-\kappa_{1}(0)(x-r_1)}\frac{\cos{(k_1(0)(z+h))}}{\cos{(k_1(0)h)}} +
R_1(0)e^{\kappa_{1}(0)(x-r_1)}\frac{\cos{(k_{1}(0)(z+h))}}{\cos{(k_{1}(0)h)}}
</math></center>
Therefore,
<center><math>
\begin{matrix}
\xi_1 & = & {\Im\int_{-h}^0 \left[\left(e^{-i\kappa^{I}_1(x)}+ R_1(0)e^{i\kappa^{I}_1(x)}\right)\right.}
{ \left(i\kappa^{I}_1e^{i\kappa^{I}_1(x)} - i\kappa^{I}_1R_1(0)^*e^{-i\kappa^{I}_1(x)}\right) }
{ \left.\left(\frac{\cosh^2{(k^I_1(z+h))}}{\cosh^2{(k^I_1h)}}\right)\right]dz, }\\ \\
& = & { \Im\left[\frac{i\kappa^{I}_1\left(1 - |R_1(0)|^2\right)}{2\cosh^2{(k^I_1h)}}
\int_{-h}^0 \left(\cosh{(2k^I_1(z+h))+1}\right)dz\right] ,}\\ \\
& = & { \Im\left[\frac{i\kappa^{I}_1\left(1 - |R_1(0)|^2\right)}{2\cosh^2{(k^I_1h)}}
\left[\frac{1}{2k^I_1}\sinh{(2k^I_1(z+h))} + z\right]_{-h}^0\right], }\\ \\
& = & { \frac{\kappa^{I}_1\left(1 - |R_1(0)|^2\right)}{2\cosh^2{(k^I_1h)}} \left(\frac{1}{2k^I_1}\sinh{(2k^I_1h))} + h\right) ,}\\ \\
& = & { \frac{\kappa^{I}_1\left(1 - |R_1(0)|^2\right)}{2k^I_1}\left(\tanh{(k^I_1h)}+\frac{hk^I_1}{\cosh^2{(k^I_1h)}}\right) , }
\end{matrix}
</math></center>
where <math>R_1(0)^*</math> is the conjugate of <math>R_1(0)</math>.

= Expanding <math>\mathbf{\xi_2}</math> =

Near <math>x=\infty</math>, we approximate <math>\phi</math> by
<center><math>
\begin{matrix}
{ \phi \approx T_\Lambda(0)e^{-\kappa_{\Lambda}(0)(x)}\frac{\cos{(k_\Lambda(0)(z+h))}}{\cos{(k_\Lambda(0)h)}}, }
\end{matrix}
</math></center>
and re-express as
<center><math>
\phi \approx T_\Lambda(0)e^{-i\kappa^I_\Lambda(x)}\frac{\cosh{(k^I_\Lambda(z+h))}}{\cosh{(k^I_\Lambda h)}},
</math></center>
where <math>k^I_\Lambda = \Im k_{\Lambda}(0)</math> and <math>\kappa^I_\Lambda=\Im \kappa_\Lambda(0)</math>, so that
<center><math>
\frac{\partial\phi}{\partial x} \approx -i\kappa^I_{\Lambda}T_\Lambda(0)e^{-i\kappa^I_{\Lambda}(x)}\frac{\cosh{(k^I_\Lambda(z+h))}}{\cosh{(k^I_\Lambda h)}}.
</math></center>
Therefore,
<center><math>
\begin{matrix}
\xi_2 & = & \Im\int_{-h}^0 \left[\right.(T_\Lambda(0)e^{-i\kappa^I_{\Lambda}(x)})(i\kappa^I_{\Lambda}T_\Lambda(0)^*e^{i\kappa^I_{\Lambda}(x)})
\frac{\cosh^2{(k^I_\Lambda(z+h))}}{\cosh^2{(k^I_\Lambda h)}}\left.\right]dz, \\ \\
& = & { \frac{\kappa^I_{\Lambda}|T_\Lambda(0)|^2}{2\cosh^2{(k^I_\Lambda h)}}
\left(\frac{1}{2k^I_\Lambda}\sinh{(2k^I_\Lambda h)} + h\right)},\\ \\
& = & { \frac{\kappa^I_{\Lambda}|T_\Lambda(0)|^2}{2k^I_\Lambda}\left(\tanh{(k^I_\Lambda h)}+\frac{k^I_\Lambda h}{\cosh^2{(k^I_\Lambda h)}}\right)}.
\end{matrix}
</math></center>

= Expanding <math>\mathbf{\xi_3}</math> =

The ice-covered boundary condition for the [[Floating Elastic Plate]] gives
<center><math>
\xi_3 = { \Im\int_{-\infty}^{\infty}\left(\frac{\beta}{\alpha}
\left(\frac{\partial^2}{\partial x^2} - k_y^2\right)^2 - \gamma + \frac{1}{\alpha}\right)
\frac{\partial\phi}{\partial z}\cdot\frac{\partial\phi^*}{\partial z}\bigg|_{z=0}dx.}
</math></center>
Since <math>{\frac{\partial\phi}{\partial z}\cdot\frac{\partial\phi^*}{\partial z}}</math> is real,
<center><math>
\xi_3 = \Im\int_{-\infty}^{\infty}\left(\frac{\beta}{\alpha}\frac{\partial^2}{\partial x^2}
\left(\frac{\partial^2}{\partial x^2} - 2k_y^2\right) \right)
\frac{\partial\phi}{\partial z}\cdot\frac{\partial\phi^*}{\partial z}\bigg|_{z=0}dx.
</math></center>
Integration by parts gives
<center><math>
\xi_3 = \Im\left\{\left[\frac{\beta}{\alpha}\frac{\partial}{\partial x}
\left(\frac{\partial^2}{\partial x^2}-2k_y^2\right)
\frac{\partial\phi}{\partial z}\cdot\frac{\partial\phi^*}{\partial z}\right]_{-\infty}^{\infty}
-\int_{-\infty}^{\infty}\frac{\beta}{\alpha}\frac{\partial}{\partial x}
\left(\frac{\partial^2}{\partial x^2} - 2k_y^2\right)
\frac{\partial\phi}{\partial z}\cdot\frac
{\partial}{\partial x}\frac{\partial\phi^*}{\partial z}dx
\right\}.
</math></center>
As <math>{2k_y^2\frac{\partial}{\partial x}\frac{\partial\phi}{\partial z}\cdot \frac{\partial}{\partial x}\frac{\partial\phi^*}{\partial z}}</math> is real and by integration by parts, the expression of <math>\xi_3</math> becomes,
<center><math>
\xi_3 =\Im\left\{
\left[\frac{\beta}{\alpha}\frac{\partial}{\partial x}
\left(\frac{\partial^2}{\partial x^2}-2k_y^2\right)
\frac{\partial\phi}{\partial z}\cdot\frac{\partial\phi^*}{\partial z}\right]_{-\infty}^{\infty}
- \left[\frac{\beta}{\alpha}\frac{\partial^2}{\partial x^2}\frac{\partial\phi}{\partial z}\cdot
\frac{\partial}{\partial x}\frac{\partial\phi^*}{\partial z}\right]_{-\infty}^{\infty}
+ \int_{-\infty}^\infty\frac{\beta}{\alpha}\frac{\partial^2}{\partial x^2}\frac{\partial\phi}{\partial z}\cdot
\frac{\partial^2}{\partial x^2}\frac{\partial\phi^*}{\partial z}dx\right\}.
</math></center>
As <math>{\frac{\partial^2}{\partial x^2}\frac{\partial\phi}{\partial z}\cdot \frac{\partial^2}{\partial x^2}\frac{\partial\phi^*}{\partial z}}</math> is real, we obtain the new expression of <math>\xi_3</math>
<center><math>
\xi_3 = \Im\left\{
\left[\frac{\beta}{\alpha}\frac{\partial}{\partial x}
\left(\frac{\partial^2}{\partial x^2}-2k_y^2\right)
\frac{\partial\phi}{\partial z}\cdot\frac{\partial\phi^*}{\partial z}\right]_{-\infty}^{\infty}-
\left[\frac{\beta}{\alpha}\frac{\partial^2}{\partial x^2}\frac{\partial\phi}{\partial z}\cdot
\frac{\partial}{\partial x}\frac{\partial\phi^*}{\partial z}\right]_{-\infty}^{\infty}\right\}.
</math></center>

Now breaking <math>\xi_3</math> down, we can simplify the left hend term for <math>x>0</math>

<center><math>
\begin{matrix}
\frac{\partial}{\partial x}\left(\frac{\partial^2}{\partial x^2}-2k_y^2\right)
\frac{\partial\phi(x_2,0)}{\partial z}\cdot\frac{\partial\phi(x_2,0)^*}{\partial z} \\ \\
= \left((-i\kappa^I_{\Lambda})^3k^I_\Lambda T_\Lambda(0)e^{-i\kappa^I_{\Lambda}(x)}\tanh{(k^I_\Lambda h)}\right.
\left.- 2k_y^2(-i\kappa_{\Lambda}(0))k^I_\Lambda T_\Lambda(0)e^{-i\kappa^I_{\Lambda}(x)} \tanh{(k^I_\Lambda h)}\right)
\left(k^I_\Lambda T_\Lambda(0)^*e^{i\kappa^I_{\Lambda}(x)}\tanh{(k^I_\Lambda h)}\right), \\ \\
= i\kappa^I_{\Lambda} (k^I_\Lambda)^2\left((\kappa^I_{\Lambda})^2 + 2k_y^2 \right) \tanh^2{(k^I_\Lambda h)}|T_\Lambda(0)|^2,
\end{matrix}
</math></center>

and for <math>x<0</math>

<center><math>
\begin{matrix}
\frac{\partial}{\partial x}\left(\frac{\partial^2}{\partial x^2}-2k_y^2\right)\frac{\partial\phi(x_1,0)}{\partial z}\cdot
\frac{\partial\phi(x_1,0)^*}{\partial z} \\ \\
= \left[i(\kappa^I_{1})^3k^I_1\left(e^{-i\kappa^I_{1}(x)}-R_1(0)e^{i\kappa^I_{1}(x)}\right) \tanh{(k^I_1 h)}
- 2ik_y^2\kappa^I_{1}k^I_1\left(-e^{-i\kappa^I_{1}(x)} + R_1(0)e^{i\kappa^I_{1}(x)}\right) \tanh{(k^I_1h)}\right]
\left[k^I_1\left(e^{i\kappa^I_{1}(x)} + R_1(0)^*e^{-i\kappa^I_{1}(x))}\right) \tanh{(k^I_1h)}\right], \\ \\
= \left[i\kappa^I_{1}k_1(0)((\kappa^I_{1})^2 + 2k_y^2)\left(e^{-i\kappa^I_{1}(x)} - R_1(0)e^{i\kappa^I_{1}(x)}\right)
\tanh{(k^I_1h)}\right]
\left[k^I_1\left(e^{i\kappa^I_{1}(x)} + R_1(0)^*e^{-i\kappa^I_{1}(x))}\right) \tanh{(k^I_1h)}\right], \\ \\
= i\kappa^I_{1}(k^I_1)^2\left((\kappa^I_{1})^2 + 2k_y^2\right)\tanh^2{(k^I_1h)}\left(1 - |R_1(0)|^2\right),
\end{matrix}
</math></center>

Likewise we expand the right hend term for <math>x>0</math>

<center><math>
\begin{matrix}
\frac{\partial^2}{\partial x^2}\frac{\partial\phi(x_2,0)}{\partial z}\cdot
\frac{\partial}{\partial x}\frac{\partial\phi^*}{\partial z} \\ \\
= \left[-(\kappa^I_{\Lambda})^2k^I_\Lambda T_\Lambda(0)e^{-i\kappa^I_{\Lambda}(x)} \tanh{(k^I_\Lambda h)}\right]
\left[i\kappa^I_{\Lambda} k^I_\Lambda T_\Lambda(0)^*e^{i\kappa^I_{\Lambda}(x)} \tanh{(k^I_\Lambda h)}\right], \\ \\
= -i(\kappa^I_{\Lambda})^3(k^I_\Lambda)^2\tanh^2{(k^I_\Lambda h)}|T_\Lambda(0)|^2,
\end{matrix}
</math></center>

and finally for <math>x<0</math>,

<center><math>
\begin{matrix}
\frac{\partial^2}{\partial x^1}\frac{\partial\phi(x_1,0)}{\partial z}\cdot
\frac{\partial}{\partial x}\frac{\partial\phi^*}{\partial z}\\ \\
= \left[-(\kappa^I_{1})^2k^I_1\left(e^{-i\kappa^I_{1}(x)} + R_1(0)e^{i\kappa^I_{1}(x)}\right)\tanh{(k^I_1 h)}\right]
\left[i\kappa^I_{1} k^I_1\left(e^{i\kappa^I_{1}(x)} - R_1(0)e^{-i\kappa^I_{1}(x)}\right)\tanh{(k^I_1 h)}\right], \\ \\
= -i(\kappa^I_{1})^3(k^I_1)^2\tanh^2{(k^I_1h)}\left(1 - |R_1(0)|^2\right).
\end{matrix}
</math></center>

We can now express <math>\xi_3</math> as

<center><math>
\begin{matrix}
\xi_3 & = & \Im
\left\{\frac{\beta_\Lambda}{\alpha}\left[i\kappa^I_{\Lambda} (k^I_\Lambda)^2\left((\kappa^I_{\Lambda})^2 + 2k_y^2 \right)
\tanh^2{(k^I_\Lambda h)}|T_\Lambda(0)|^2\right]
-\frac{\beta_1}{\alpha}\left[i\kappa^I_{1}(k^I_1)^2\left((\kappa^I_{1})^2 + 2k_y^2\right)\tanh^2{(k^I_1h)}
\left(1 - |R_1(0)|^2\right)\right]
-\frac{\beta_\Lambda}{\alpha}\left[-i(\kappa^I_{\Lambda})^3(k^I_\Lambda)^2\tanh^2{(k^I_\Lambda h)}|T_\Lambda(0)|^2\right]
+\frac{\beta_1}{\alpha}\left[-i(\kappa^I_{1})^3(k^I_1)^2\tanh^2{(k^I_1h)} \left(1 - |R_1(0)|^2\right)\right]\right\}, \\ \\
& = & \frac{\beta_\Lambda}{\alpha}\left[2\kappa^I_{\Lambda} (k^I_\Lambda)^2((\kappa^I_{\Lambda})^2 + k_y^2)\tanh^2{(k^I_\Lambda h)}|T_\Lambda(0)|^2\right]
-\frac{\beta_1}{\alpha}\left[2\kappa^I_{1}(k^I_1)^2((\kappa^I_{1})^2 + k_y^2)\tanh^2{(k^I_1h)}(1-|R_1(0)|^2)\right].
\end{matrix}
</math></center>

= Solving the Energy Balance Equation =

Pulling it all together, we finally obtain
<center><math>
\frac{\beta_\Lambda}{\alpha}\left[2\kappa^I_{\Lambda}(k^I_\Lambda)^2((\kappa^I_{\Lambda})^2 +k_y^2)
\tanh^2{(k^I_\Lambda h)}|T_\Lambda(0)|^2\right]
-\frac{\beta_1}{\alpha}\left[2\kappa^I_{1}(k^I_1)^2((\kappa^I_{1})^2 +k_y^2)\tanh^2{(k^I_1h)}(1-|R_1(0)|^2)\right]
+ \frac{\kappa^I_{\Lambda}|T_\Lambda(0)|^2}{2k^I_\Lambda}\left(\tanh{(k^I_\Lambda h)}+
\frac{k^I_\Lambda h}{\cosh^2{(k^I_\Lambda h)}}\right)
- \frac{\kappa^I_{1}\left(1 - |R_1(0)|^2\right)}{2k^I_1}\left(\tanh{(k^I_1h)}+\frac{hk^I_1}{\cosh^2{(k^I_1h)}}\right)=0.
</math></center>

Re-arranging gives

<center><math>
\kappa^I_{\Lambda}\tanh{(k^I_\Lambda h)}
\left(\frac{\beta_\Lambda}{\alpha}2(k^I_\Lambda)^2((\kappa^I_{\Lambda})^2 +k_y^2)\tanh{(k^I_\Lambda h)}
+ \frac{1}{2k^I_\Lambda}+\frac{h}{2\sinh{(k^I_\Lambda h)}\cosh{(k^I_\Lambda h)}}\right)|T_\Lambda(0)|^2
\quad - \kappa^I_{1}\tanh{(k^I_1 h)}
\left(\frac{\beta_1}{\alpha}2(k^I_1)^2((\kappa^I_{1})^2 +k_y^2)\tanh{(k^I_1 h)}
+ \frac{1}{2it}+\frac{h}{2\sinh{(k^I_1 h)}\cosh{(k^I_1 h)}}\right)(1-|R_1(0)|^2)=0
</math></center>

which can be expressed as
<center><math>
D |T_{\Lambda}(0)|^2 + |R_{1}(0)|^2 = 1,
</math></center>
where <math>D</math> is given by
<center><math>
D = \left(\frac{\kappa^I_{\Lambda} k^I_1\cosh^2{(k^I_1h)}}{\kappa^I_{1}k^I_\Lambda\cosh^2{(k^I_\Lambda h)}}\right)
\left(\frac{\frac{\beta_\Lambda}{\alpha}4(k^I_\Lambda)^2\kappa^I_{\Lambda}((\kappa^I_{\Lambda})^2 -k_y^2)\sinh^2{(k^I_\Lambda h)} +
\frac{1}{2}{\sinh{(2k^I_\Lambda h)}}+k^I_\Lambda h}
{\frac{\beta_1}{\alpha}4(k^I_1)^2\kappa^I_1((\kappa^I_{1})^2 -k_y^2)\sinh^2{(k^I_1h)} +
\frac{1}{2}{\sinh{(2k^I_1h)}}+k^I_1h}\right)
</math></center>

[[Category:Floating Elastic Plate]]

Eigenfunction Matching for a Semi-Infinite Floating Elastic Plate

2008-08-25T02:26:22Z

Fmontiel: /* Energy Balance */

=Introduction=

We show here a solution for a semi-infinite [[:Category:Floating Elastic Plate|Floating Elastic Plate]] on [[Finite Depth]].
The problem was solved by [[Fox and Squire 1994]] but the solution method here is slightly different.
The simpler theory for a [[Eigenfunction Matching for a Semi-Infinite Dock|Dock]] describes
many of the ideas here in more detail.

[[Image:Semiinfinite plate.jpg|thumb|right|300px|Wave scattering by a submerged semi-infinite elastic plate]]

= Equations =

We consider the problem of small-amplitude waves which are incident on a semi-infinite floating elastic
plate occupying water surface for <math>x>0</math>. The submergence of the plate is considered negligible.
We assume that the problem is invariant in the <math>y</math> direction, although we allow the waves to be
incident from an angle.
We also assume that the plate edges are free to move at
each boundary, although other boundary conditions could easily be considered using
the methods of solution presented here. We begin with the [[Frequency Domain Problem]] for a semi-infinite
[[Floating Elastic Plate|Floating Elastic Plates]]
in the non-dimensional form of [[Tayler_1986a|Tayler 1986]] ([[Dispersion Relation for a Floating Elastic Plate]]).
We also assume that the waves are normally incident (incidence at an angle will be discussed later).
<center><math>\begin{matrix}
\Delta \phi = 0, \;\;\; -h < z \leq 0,
\end{matrix}</math></center>
<center><math>
\partial_z \phi = 0, \;\;\; z = - h,
</math></center>
<center><math>
\partial_z\phi=\alpha\phi, \,\, z=0,\,x<0,
</math></center>
<center><math>
\beta \partial_x^4\partial_z \phi
- \left( \gamma\alpha - 1 \right) \partial_z \phi = \alpha\phi, \;\;
z = 0, \;\;\; x \geq 0,
</math></center>
where <math>\alpha = \omega^2</math>, <math>\beta</math> and <math>\gamma</math>
are the stiffness and mass constant for the plate respectively. The free edge conditions
at the edge of the plate imply
<center><math>
\partial_x^3 \partial_z\phi = 0, \;\; z = 0, \;\;\; x = 0,
</math></center>
<center><math>
\partial_x^2 \partial_z\phi = 0, \;\; z = 0, \;\;\; x = 0,
</math></center>

=Method of solution=

==Eigenfunction expansion==

We will solve the system of equations using an [[:Category:Eigenfunction Matching Method|Eigenfunction Matching Method]].
The method was developed by [[Fox and Squire 1994]]. The separation of variables for the left hand region where
there is open water, and the incident wave, is described in [[Eigenfunction Matching for a Semi-Infinite Dock]] and we consider
here only the separation of variables in the plate covered region.

===Separation of variables under the Plate===
The potential velocity can be written in terms of an infinite series of separated eigenfunctions under
each elastic plate, of the form
<center><math>
\psi = \frac{\cos(\kappa (z+h))}{\cos\kappa h}.\,
</math></center>
If we apply the boundary conditions given
we obtain the [[Dispersion Relation for a Floating Elastic Plate]]
<center><math>
\kappa \tan{(\kappa h)}= -\frac{\alpha}{\beta \kappa^{4} + 1 - \alpha\gamma}
</math></center>
Solving for <math>\kappa</math> gives a pure imaginary root
with positive imaginary part, two complex roots (two complex conjugate paired roots
with positive imaginary part in most physical situations), an infinite number of positive real roots
which approach <math>{n\pi}/{h}</math> as <math>n</math> approaches infinity, and also the negative of all
these roots ([[Dispersion Relation for a Floating Elastic Plate]]) . We denote the two complex roots with positive imaginary part
by <math>\kappa_(-2)</math> and <math>\kappa_(-1)</math>, the purely imaginary
root with positive imaginary part by <math>\kappa_{0}</math> and the real roots with positive imaginary part
by <math>\kappa_(n)</math> for <math>n</math> a positive integer.
The imaginary root with positive imaginary part corresponds to a
reflected travelling mode propagating along the <math>x</math> axis.
The complex roots with positive imaginary parts correspond to damped reflected travelling modes and the real roots correspond to reflected evanescent modes.

==Expressions for the potential velocity==

We now expand the potential in the two regions using the separation of variables solution.
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.
The potential <math>\phi</math> can now be expressed as the following sum of eigenfunctions:
be expanded as
<center>
<math>
\phi(x,z)= e^{-k_{0}x}\phi_{0}(z) + \sum_{m=0}^{\infty}a_{m}e^{k_{m}x}\phi_{m}(z), \;\;x<0
</math>
</center>
and
<center>
<math>
\phi(x,z)=\sum_{m=-2}^{\infty}b_{m}
e^{-\kappa_{m}x}\psi_{m}(z), \;\;x>0
</math>
</center>
where <math>a_{m}</math> and <math>b_{m}</math> are the coefficients to be determined.

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>

==An infinite dimensional system of equations==

The potential and its derivative must be continuous across the
transition from open water to the plate covered region. Therefore, the
potentials and their derivatives at <math>x=0</math> have to be equal.
We also truncate the sum at <math>N</math> being careful that we have
two extra modes on the plate covered region to satisfy the edge conditions.
We obtain
<center>
<math>
\phi_{0}\left( z\right) + \sum_{m=0}^{N}
a_{m} \phi_{m}\left( z\right)
=\sum_{m=-2}^{N}b_{m}\psi_{m}(z)
</math>
</center>
and
<center>
<math>
-k_{0}\phi_{0}\left( z\right) +\sum
_{m=0}^{N} a_{m}k_{m}\phi_{m}\left( z\right)
=-\sum_{m=-2}^{N}b_{m}\kappa_{m}\psi
_{m}(z)
</math>
</center>
for each <math>N</math>.
We solve these equations by multiplying both equations by
<math>\phi_{l}(z)\,</math> and integrating from <math>-h</math> to <math>0</math> to obtain:
<center>
<math>
A_{0}\delta_{0l}+a_{l}A_{l}
=\sum_{m=-2}^{N}b_{m}B_{ml},\,\,0 \leq l \leq N
</math>
</center>
and
<center>
<math>
-k_{0}A_{0}\delta_{0l}+a_{l}k_{l}A_l
=-\sum_{m=-2}^{N}b_{m}\kappa_{m}B_{ml},\,\,0 \leq l \leq N
</math>
</center>
If we multiply the first equation by <math>k_l</math> and subtract the second equation
we obtain
<center>
<math>
2k_{0}A_{0}\delta_{0l}
=\sum_{m=-2}^{N}b_{m}(k_l + \kappa_{m})B_{ml}
</math>
</center>
Finally, we need to apply the conditions at the edge of the plate to give us two further equations,
<center>
<math>
\partial_x^2\partial_z\phi = - \sum_{m=-2}^{N}b_{m} \kappa_m^3 \tan\kappa_m h = 0
</math>
</center>
and
<center>
<math>
\partial_x^3\partial_z\phi = \sum_{m=-2}^{N}b_{m} \kappa_m^4 \tan\kappa_m h = 0
</math>
</center>

=Numerical Solution=

To solve the system of equations previously defined we set the upper limit of <math>l</math> to
be <math>N</math>, as we already said before. In terms of matrix, we obtain
<center>
<math>
\begin{bmatrix}
\begin{bmatrix}
A_0&0 \quad \cdots&0\\
0&&\\
\vdots&A_l&\vdots\\
&&0\\
0&\cdots \quad 0 &A_N
\end{bmatrix}
&
\begin{bmatrix}
-B_{-20}&-B_{-10}\\
&\\
\vdots&\vdots\\
&\\
-B_{-2N}&-B_{-1N}
\end{bmatrix}
&
\begin{bmatrix}
-B_{00}&\cdots&-B_{0N}\\
&&\\
\vdots&-B_{ml}&\vdots\\
&&\\
-B_{N0}&\cdots&-B_{NN}
\end{bmatrix}
\\
\begin{bmatrix}
0&&\cdots&&0\\
0&&\cdots&&0
\end{bmatrix}
&
\begin{bmatrix}
\kappa_{-2}^3\tan\kappa_{-2}h&\kappa_{-1}^3\tan\kappa_{-1}h\\
\kappa_{-2}^4\tan\kappa_{-2}h&\kappa_{-1}^4\tan\kappa_{-1}h
\end{bmatrix}
&
\begin{bmatrix}
\kappa_0^3\tan\kappa_0h&\cdots&\kappa_N^3\tan\kappa_Nh\\
\kappa_0^4\tan\kappa_0h&\cdots&\kappa_N^4\tan\kappa_Nh
\end{bmatrix}
\\
\begin{bmatrix}
0&&\cdots&&0\\
&&&&\\
\vdots&&\ddots&&\vdots\\
&&&&\\
0&&\cdots&&0
\end{bmatrix}
&
\begin{bmatrix}
(k_0+\kappa_{-2})B_{-20}&(k_0+\kappa_{-1})B_{-10}\\
&\\
\vdots&\vdots\\
&\\
(k_N+\kappa_{-2})B_{-2N}&(k_N+\kappa_{-1})B_{-1N}
\end{bmatrix}
&
\begin{bmatrix}
(k_0 + \kappa_0) \, B_{00}&\cdots&(k_N + \kappa_{0}) \, B_{0N}\\
&&\\
\vdots&(k_l + \kappa_{m}) \, B_{ml}&\vdots\\
&&\\
(k_0 + \kappa_N) \, B_{N0}&\cdots&(k_N + \kappa_{N}) \, B_{NN}\\
\end{bmatrix}
\end{bmatrix}
\begin{bmatrix}
a_{0} \\
\\
\vdots \\
\\
a_N \\
b_{-2}\\
b_{-1}\\
b_{0}\\
\\
\vdots \\
\\
b_N
\end{bmatrix}
=
\begin{bmatrix}
- A_{0} \\
0 \\
\vdots \\
\\
0 \\
0 \\
0 \\
2k_{0}A_{0} \\
0 \\
\vdots \\
\\
0
\end{bmatrix}
</math>
</center>
We then simply need to solve the linear system of equations. Note that we can solve this equation for
<math>b_n</math> first and then solve for <math>a_n</math>

= Waves Incident at an Angle =

We can consider the case of [[Waves Incident at an Angle]] <math>\theta</math>. The theory shows that we can introduce
a wavenumber in the <math>y</math>-direction which express <math>k_y=k_0 \sin \theta</math>.

It is shown that the potential can be expanded as
<center>
<math>
\phi(x,z)=e^{-\hat{k}_0x}\phi_0(z)+\sum_{m=0}^{\infty}a_{m}e^{\hat{k}_{m}x}\phi_{m}(z), \;\;x<0
</math>
</center>
and
<center>
<math>
\phi(x,z)=\sum_{m=-2}^{\infty}b_{m}
e^{-\hat{\kappa}_{m}x}\psi_{m}(z), \;\;x>0
</math>
</center>
where <math>\hat{k}_{m} = \sqrt{k_m^2 - k_y^2}</math> and <math>\hat{\kappa}_{m} = \sqrt{\kappa_m^2 - k_y^2}</math>
where we always take the positive real root or the root with positive imaginary part.

The edge conditions are also different and are
<center><math>
\left(\frac{\partial^3}{\partial x^3} - (2 - \nu)k^2_y\frac{\partial}{\partial x}\right) \frac{\partial\phi}{\partial z}= 0,
</math></center>
and
<center><math>
\left(\frac{\partial^2}{\partial x^2} - \nu k^2_y\right)\frac{\partial\phi}{\partial z} = 0,
</math></center>
where <math>\nu</math> is Poisons ratio.

We can expend these edge conditions, which respectively gives
<center>
<math>
-\sum_{m=-2}^{\infty}b_{m}\kappa_m (\hat{\kappa}_m^3-\hat{\kappa}_m k_y^2(2-\nu))\tan \kappa_m h=0
</math>
</center>
and
<center>
<math>
-\sum_{m=-2}^{\infty}b_{m}\kappa_m (\hat{\kappa}_m^2-k_y^2\nu))\tan \kappa_m h=0
</math>
</center>

The equations are derived almost identically to those above and we obtain
<center>
<math>
A_{0}\delta_{0l}+a_{l}A_{l}
=\sum_{m=-2}^{\infty}b_{m}B_{ml}
</math>
</center>
and
<center>
<math>
-\hat{k_{0}}A_{0}\delta_{0l}+a_{l}\hat{k}_{l}A_l
=-\sum_{m=-2}^{\infty}b_{m}\hat{\kappa}_{m}B_{ml}
</math>
</center>
and these are solved exactly as before.

= Energy Balance =

The energy balance equation is ([[Energy Balance for Two Elastic Plates]]) given by
<center><math>
D |T|^2 + |R|^2 = 1, \,
</math></center>
where <math> T=b_0 </math>, <math>R=a_0</math>, and <math>D</math> is given by
<center><math>
D = \left(\frac{\hat{\kappa}_0 k_0\cos^2{(k_0 h)}}{\hat{k}_0 \kappa_0\cos^2{(\kappa_0 h)}}\right)
\left(\frac{\frac{4\beta}{\alpha}(\kappa_0)^2\hat{\kappa}_0(\hat{\kappa}_0^2 -k_y^2)\sin^2{(\kappa_0 h)} +
\frac{1}{2}{\sin{(2 \kappa_0 h)}}+\kappa_0 h}
{
\frac{1}{2}{\sin{(2 k_0 h)}}+k_0h}\right)
</math></center>
where the index <math>0</math> refers to the travelling wave (imaginary root of the dispersion equation).

= Matlab Code =

A program to calculate the coefficients for the semi-infinite dock problems can be found here
[http://www.math.auckland.ac.nz/~meylan/code/eigenfunction_matching/semiinfinite_plate.m semiinfinite_plate.m]

== Additional code ==

This program requires
* {{free surface dispersion equation code}}
* {{elastic plate dispersion equation code}}

[[Category:Floating Elastic Plate]]
[[Category:Eigenfunction Matching Method]]
[[Category:Complete Pages]]

Energy Balance for Two Elastic Plates

2008-08-25T01:31:36Z

Fmontiel: /* Equations */

= Introduction =

We derive here a condition which must be satisfied for energy balance for two semi-infinite elastic plates. The solution applies
to the problem of [[Eigenfunction Matching Method for Floating Elastic Plates|multiple elastic plates]] and from here we know we can write the potential as
<center><math>
\phi \approx \left\{
\begin{matrix}
{
Ie^{\kappa_{1}(0)(x-r_1)}\frac{\cos(k_1(0)(z+h))}{\cos(k_1(0)h)} }+\\
{
\qquad \qquad \sum_{n=-2}^{M}R_1(n) e^{\kappa_{1}(n)(x-r_1)} \frac{\cos(k_1(n)(z+h))}{\cos(k_1(n)h)} },&
\mbox{ for } x < r_1,\\
{
\sum_{n=-2}^{M}T_{\mu}(n) e^{-\kappa_\mu(n)(x-l_\mu)}
\frac{\cos(k_\mu(n)(z+h))}{\cos(k_\mu(n)h)} } + \\
{
\qquad \qquad \sum_{n=-2}^{M}R_{\mu}(n) e^{\kappa_\mu(n)(x-r_\mu)}
\frac{\cos(k_\mu(n)(z+h))}{\cos(k_\mu(n)h)} },
&\mbox{ for } l_\mu< x < r_\mu,\\
{
\sum_{n=-2}^{M}T_{\Lambda}(n)e^{-\kappa_{\Lambda}(n)(x-l_\Lambda)}
\frac{\cos(k_\Lambda(n)(z+h))}{\cos(k_\Lambda(n)h)} }, &\mbox{ for } l_\Lambda<x,
\end{matrix} \right.
</math></center>
(for details of this notation see [[Eigenfunction Matching Method for Floating Elastic Plates]]).

= Equations =

Based on the method used in [[Evans and Davies 1968]], a check can be made to ensure the solutions of the floating plate problem are in energy balance.
This is simply a condition that the incident energy is equal to the sum of the radiated energy.
When the first and final plates have different properties, the energy balance equation is derived by applying Green's theorem to <math>\phi</math> and its conjugate.
The domain of integration is shown in the figure on the right.

[[Image:energy_schematic.jpg|thumb|right|300px|A diagram depicting the area <math>\mathcal{U}</math> which is bounded by the rectangle <math>\mathcal{S}</math>.]
{A diagram depicting the area <math>\mathcal{U}</math> which is bounded by the rectangle <math>\mathcal{S}</math>.
The rectangle <math>\mathcal{S}</math> is bounded by <math> -h\leq z \leq0</math> and <math>-\infty\leq x \leq \infty</math>]]

Applying Green's theorem to <math>\phi</math> and its conjugate <math>\phi^*</math> gives
<center><math>
{ \int\int_\mathcal{U}(\phi\nabla^2\phi^* - \phi^*\nabla^2\phi)dxdz
= \int_\mathcal{S}(\phi\frac{\partial\phi^*}{\partial n} - \phi^*\frac{\partial\phi}{\partial n})dl },
</math></center>
where <math>n</math> denotes the outward plane normal to the boundary and <math>l</math> denotes the plane parallel to the boundary.
As <math>\phi</math> and <math>\phi^*</math> satisfy the Laplace's equation, the left hand side of the Green theorem equation vanishes so that it reduces to
<center><math>
\Im\int_\mathcal{S}\phi\frac{\partial\phi^*}{\partial n} dl = 0,
</math></center>
Expanding gives
<center><math>
\xi_1 +\xi_2 + \xi_3 = 0,
</math></center>
where
<center><math>
\xi_1 =
-{ \Im\int_{-h}^0(\phi\frac{\partial\phi^*}{\partial x})\big|_{x=-\infty}dz },
</math></center>
<center><math>
\xi_2 =
{ \Im\int_{-h}^0(\phi\frac{\partial\phi^*}{\partial x})\big|_{x=\infty}dz },
</math></center>
and
<center><math>
\xi_3 =
{ \Im\int_{-\infty}^{\infty}(\phi\frac{\partial\phi^*}{\partial z})\big|_{z=0}dx } ,
</math></center>
where <math>\Im</math> denotes the imaginary part. The bottom domain composant is obviously equal to zero because of the no-influx condition over the seabed (<math>\frac{\partial\phi^*}{\partial z}\big|_{z=-h}=0</math>).

= Expanding <math>\mathbf{\xi_1}</math> =

Near <math>x=-\infty</math>, we approximate <math>\phi</math> by
<center><math>
\phi \approx e^{-\kappa_{1}(0)(x-r_1)}\frac{\cos{(k_1(0)(z+h))}}{\cos{(k_1(0)h)}} +
R_1(0)e^{\kappa_{1}(0)(x-r_1)}\frac{\cos{(k_{1}(0)(z+h))}}{\cos{(k_{1}(0)h)}}
</math></center>
Therefore,
<center><math>
\begin{matrix}
\xi_1 & = & {\Im\int_{-h}^0 \left[\left(e^{-i\kappa^{I}_1(x)}+ R_1(0)e^{i\kappa^{I}_1(x)}\right)\right.}
{ \left(i\kappa^{I}_1e^{i\kappa^{I}_1(x)} - i\kappa^{I}_1R_1(0)^*e^{-i\kappa^{I}_1(x)}\right) }
{ \left.\left(\frac{\cosh^2{(k^I_1(z+h))}}{\cosh^2{(k^I_1h)}}\right)\right]dz, }\\ \\
& = & { \Im\left[\frac{i\kappa^{I}_1\left(1 - |R_1(0)|^2\right)}{2\cosh^2{(k^I_1h)}}
\int_{-h}^0 \left(\cosh{(2k^I_1(z+h))+1}\right)dz\right] ,}\\ \\
& = & { \Im\left[\frac{i\kappa^{I}_1\left(1 - |R_1(0)|^2\right)}{2\cosh^2{(k^I_1h)}}
\left[\frac{1}{2k^I_1}\sinh{(2k^I_1(z+h))} + z\right]_{-h}^0\right], }\\ \\
& = & { \frac{\kappa^{I}_1\left(1 - |R_1(0)|^2\right)}{2\cosh^2{(k^I_1h)}} \left(\frac{1}{2k^I_1}\sinh{(2k^I_1h))} + h\right) ,}\\ \\
& = & { \frac{\kappa^{I}_1\left(1 - |R_1(0)|^2\right)}{2k^I_1}\left(\tanh{(k^I_1h)}+\frac{hk^I_1}{\cosh^2{(k^I_1h)}}\right) , }
\end{matrix}
</math></center>
where <math>R_1(0)^*</math> is the conjugate of <math>R_1(0)</math>.

= Expanding <math>\mathbf{\xi_2}</math> =

Near <math>x=\infty</math>, we approximate <math>\phi</math> by
<center><math>
\begin{matrix}
{ \phi \approx T_\Lambda(0)e^{-\kappa_{\Lambda}(0)(x)}\frac{\cos{(k_\Lambda(0)(z+h))}}{\cos{(k_\Lambda(0)h)}}, }
\end{matrix}
</math></center>
and re-express as
<center><math>
\phi \approx T_\Lambda(0)e^{-i\kappa^I_\Lambda(x)}\frac{\cosh{(k^I_\Lambda(z+h))}}{\cosh{(k^I_\Lambda h)}},
</math></center>
where <math>k^I_\Lambda = \Im k_{\Lambda}(0)</math> and <math>\kappa^I_\Lambda=\Im \kappa_\Lambda(0)</math>, so that
<center><math>
\frac{\partial\phi}{\partial x} \approx -i\kappa^I_{\Lambda}T_\Lambda(0)e^{-i\kappa^I_{\Lambda}(x)}\frac{\cosh{(k^I_\Lambda(z+h))}}{\cosh{(k^I_\Lambda h)}}.
</math></center>
Therefore,
<center><math>
\begin{matrix}
\xi_2 & = & \Im\int_{-h}^0 \left[\right.(T_\Lambda(0)e^{-i\kappa^I_{\Lambda}(x)})(i\kappa^I_{\Lambda}T_\Lambda(0)^*e^{i\kappa^I_{\Lambda}(x)})
\frac{\cosh^2{(k^I_\Lambda(z+h))}}{\cosh^2{(k^I_\Lambda h)}}\left.\right]dz, \\ \\
& = & { \frac{\kappa^I_{\Lambda}|T_\Lambda(0)|^2}{2\cosh^2{(k^I_\Lambda h)}}
\left(\frac{1}{2k^I_\Lambda}\sinh{(2k^I_\Lambda h)} + h\right)},\\ \\
& = & { \frac{\kappa^I_{\Lambda}|T_\Lambda(0)|^2}{2k^I_\Lambda}\left(\tanh{(k^I_\Lambda h)}+\frac{k^I_\Lambda h}{\cosh^2{(k^I_\Lambda h)}}\right)}.
\end{matrix}
</math></center>

= Expanding <math>\mathbf{\xi_3}</math> =

The ice-covered boundary condition for the [[Floating Elastic Plate]] gives
<center><math>
\xi_3 = { \Im\int_{-\infty}^{\infty}\left(\frac{\beta}{\alpha}
\left(\frac{\partial^2}{\partial x^2} - k_y^2\right)^2 - \gamma + \frac{1}{\alpha}\right)
\frac{\partial\phi}{\partial z}\cdot\frac{\partial\phi^*}{\partial z}\bigg|_{z=0}dx.}
</math></center>
Since <math>{\frac{\partial\phi}{\partial z}\cdot\frac{\partial\phi^*}{\partial z}}</math> is real,
<center><math>
\xi_3 = \Im\int_{-\infty}^{\infty}\left(\frac{\beta}{\alpha}\frac{\partial^2}{\partial x^2}
\left(\frac{\partial^2}{\partial x^2} - 2k_y^2\right) \right)
\frac{\partial\phi}{\partial z}\cdot\frac{\partial\phi^*}{\partial z}\bigg|_{z=0}dx.
</math></center>
Integration by parts gives
<center><math>
\xi_3 = \Im\left\{\left[\frac{\beta}{\alpha}\frac{\partial}{\partial x}
\left(\frac{\partial^2}{\partial x^2}-2k_y^2\right)
\frac{\partial\phi}{\partial z}\cdot\frac{\partial\phi^*}{\partial z}\right]_{-\infty}^{\infty}
-\int_{-\infty}^{\infty}\frac{\beta}{\alpha}\frac{\partial}{\partial x}
\left(\frac{\partial^2}{\partial x^2} - 2k_y^2\right)
\frac{\partial\phi}{\partial z}\cdot\frac
{\partial}{\partial x}\frac{\partial\phi^*}{\partial z}dx
\right\}.
</math></center>
As <math>{2k_y^2\frac{\partial}{\partial x}\frac{\partial\phi}{\partial z}\cdot \frac{\partial}{\partial x}\frac{\partial\phi^*}{\partial z}}</math> is real and by integration by parts, the expression of <math>\xi_3</math> becomes,
<center><math>
\xi_3 =\Im\left\{
\left[\frac{\beta}{\alpha}\frac{\partial}{\partial x}
\left(\frac{\partial^2}{\partial x^2}-2k_y^2\right)
\frac{\partial\phi}{\partial z}\cdot\frac{\partial\phi^*}{\partial z}\right]_{-\infty}^{\infty}
- \left[\frac{\beta}{\alpha}\frac{\partial^2}{\partial x^2}\frac{\partial\phi}{\partial z}\cdot
\frac{\partial}{\partial x}\frac{\partial\phi^*}{\partial z}\right]_{-\infty}^{\infty}
+ \int_{-\infty}^\infty\frac{\beta}{\alpha}\frac{\partial^2}{\partial x^2}\frac{\partial\phi}{\partial z}\cdot
\frac{\partial^2}{\partial x^2}\frac{\partial\phi^*}{\partial z}dx\right\}.
</math></center>
As <math>{\frac{\partial^2}{\partial x^2}\frac{\partial\phi}{\partial z}\cdot \frac{\partial^2}{\partial x^2}\frac{\partial\phi^*}{\partial z}}</math> is real, we obtain the new expression of <math>\xi_3</math>
<center><math>
\xi_3 = \Im\left\{
\left[\frac{\beta}{\alpha}\frac{\partial}{\partial x}
\left(\frac{\partial^2}{\partial x^2}-2k_y^2\right)
\frac{\partial\phi}{\partial z}\cdot\frac{\partial\phi^*}{\partial z}\right]_{-\infty}^{\infty}-
\left[\frac{\beta}{\alpha}\frac{\partial^2}{\partial x^2}\frac{\partial\phi}{\partial z}\cdot
\frac{\partial}{\partial x}\frac{\partial\phi^*}{\partial z}\right]_{-\infty}^{\infty}\right\}.
</math></center>

Now breaking <math>\xi_3</math> down, we can simplify the left hend term for <math>x>0</math>

<center><math>
\begin{matrix}
\frac{\partial}{\partial x}\left(\frac{\partial^2}{\partial x^2}-2k_y^2\right)
\frac{\partial\phi(x_2,0)}{\partial z}\cdot\frac{\partial\phi(x_2,0)^*}{\partial z} \\ \\
= \left((-i\kappa^I_{\Lambda})^3k^I_\Lambda T_\Lambda(0)e^{-i\kappa^I_{\Lambda}(x)}\tanh{(k^I_\Lambda h)}\right.
\left.- 2k_y^2(-i\kappa_{\Lambda}(0))k^I_\Lambda T_\Lambda(0)e^{-i\kappa^I_{\Lambda}(x)} \tanh{(k^I_\Lambda h)}\right)
\left(k^I_\Lambda T_\Lambda(0)^*e^{i\kappa^I_{\Lambda}(x)}\tanh{(k^I_\Lambda h)}\right), \\ \\
= i\kappa^I_{\Lambda} (k^I_\Lambda)^2\left((\kappa^I_{\Lambda})^2 + 2k_y^2 \right) \tanh^2{(k^I_\Lambda h)}|T_\Lambda(0)|^2,
\end{matrix}
</math></center>

and for <math>x<0</math>

<center><math>
\begin{matrix}
\frac{\partial}{\partial x}\left(\frac{\partial^2}{\partial x^2}-2k_y^2\right)\frac{\partial\phi(x_1,0)}{\partial z}\cdot
\frac{\partial\phi(x_1,0)^*}{\partial z} \\ \\
= \left[i(\kappa^I_{1})^3k^I_1\left(e^{-i\kappa^I_{1}(x)}-R_1(0)e^{i\kappa^I_{1}(x)}\right) \tanh{(k^I_1 h)}
- 2ik_y^2\kappa^I_{1}k^I_1\left(-e^{-i\kappa^I_{1}(x)} + R_1(0)e^{i\kappa^I_{1}(x)}\right) \tanh{(k^I_1h)}\right]
\left[k^I_1\left(e^{i\kappa^I_{1}(x)} + R_1(0)^*e^{-i\kappa^I_{1}(x))}\right) \tanh{(k^I_1h)}\right], \\ \\
= \left[i\kappa^I_{1}k_1(0)((\kappa^I_{1})^2 + 2k_y^2)\left(e^{-i\kappa^I_{1}(x)} - R_1(0)e^{i\kappa^I_{1}(x)}\right)
\tanh{(k^I_1h)}\right]
\left[k^I_1\left(e^{i\kappa^I_{1}(x)} + R_1(0)^*e^{-i\kappa^I_{1}(x))}\right) \tanh{(k^I_1h)}\right], \\ \\
= i\kappa^I_{1}(k^I_1)^2\left((\kappa^I_{1})^2 + 2k_y^2\right)\tanh^2{(k^I_1h)}\left(1 - |R_1(0)|^2\right),
\end{matrix}
</math></center>

Likewise we expand the right hend term for <math>x>0</math>

<center><math>
\begin{matrix}
\frac{\partial^2}{\partial x^2}\frac{\partial\phi(x_2,0)}{\partial z}\cdot
\frac{\partial}{\partial x}\frac{\partial\phi^*}{\partial z} \\ \\
= \left[-(\kappa^I_{\Lambda})^2k^I_\Lambda T_\Lambda(0)e^{-i\kappa^I_{\Lambda}(x)} \tanh{(k^I_\Lambda h)}\right]
\left[i\kappa^I_{\Lambda} k^I_\Lambda T_\Lambda(0)^*e^{i\kappa^I_{\Lambda}(x)} \tanh{(k^I_\Lambda h)}\right], \\ \\
= -i(\kappa^I_{\Lambda})^3(k^I_\Lambda)^2\tanh^2{(k^I_\Lambda h)}|T_\Lambda(0)|^2,
\end{matrix}
</math></center>

and finally for <math>x<0</math>,

<center><math>
\begin{matrix}
\frac{\partial^2}{\partial x^1}\frac{\partial\phi(x_1,0)}{\partial z}\cdot
\frac{\partial}{\partial x}\frac{\partial\phi^*}{\partial z}\\ \\
= \left[-(\kappa^I_{1})^2k^I_1\left(e^{-i\kappa^I_{1}(x)} + R_1(0)e^{i\kappa^I_{1}(x)}\right)\tanh{(k^I_1 h)}\right]
\left[i\kappa^I_{1} k^I_1\left(e^{i\kappa^I_{1}(x)} - R_1(0)e^{-i\kappa^I_{1}(x)}\right)\tanh{(k^I_1 h)}\right], \\ \\
= -i(\kappa^I_{1})^3(k^I_1)^2\tanh^2{(k^I_1h)}\left(1 - |R_1(0)|^2\right).
\end{matrix}
</math></center>

We can now express <math>\xi_3</math> as

<center><math>
\begin{matrix}
\xi_3 & = & \Im
\left\{\frac{\beta_\Lambda}{\alpha}\left[i\kappa^I_{\Lambda} (k^I_\Lambda)^2\left((\kappa^I_{\Lambda})^2 + 2k_y^2 \right)
\tanh^2{(k^I_\Lambda h)}|T_\Lambda(0)|^2\right]
-\frac{\beta_1}{\alpha}\left[i\kappa^I_{1}(k^I_1)^2\left((\kappa^I_{1})^2 + 2k_y^2\right)\tanh^2{(k^I_1h)}
\left(1 - |R_1(0)|^2\right)\right]
-\frac{\beta_\Lambda}{\alpha}\left[-i(\kappa^I_{\Lambda})^3(k^I_\Lambda)^2\tanh^2{(k^I_\Lambda h)}|T_\Lambda(0)|^2\right]
+\frac{\beta_1}{\alpha}\left[-i(\kappa^I_{1})^3(k^I_1)^2\tanh^2{(k^I_1h)} \left(1 - |R_1(0)|^2\right)\right]\right\}, \\ \\
& = & \frac{\beta_\Lambda}{\alpha}\left[2\kappa^I_{\Lambda} (k^I_\Lambda)^2((\kappa^I_{\Lambda})^2 + k_y^2)\tanh^2{(k^I_\Lambda h)}|T_\Lambda(0)|^2\right]
-\frac{\beta_1}{\alpha}\left[2\kappa^I_{1}(k^I_1)^2((\kappa^I_{1})^2 + k_y^2)\tanh^2{(k^I_1h)}(1-|R_1(0)|^2)\right].
\end{matrix}
</math></center>

= Solving the Energy Balance Equation =

Pulling it all together, we finally obtain
<center><math>
\frac{\beta_\Lambda}{\alpha}\left[2\kappa^I_{\Lambda}(k^I_\Lambda)^2((\kappa^I_{\Lambda})^2 +k_y^2)
\tanh^2{(k^I_\Lambda h)}|T_\Lambda(0)|^2\right]
-\frac{\beta_1}{\alpha}\left[2\kappa^I_{1}(k^I_1)^2((\kappa^I_{1})^2 +k_y^2)\tanh^2{(k^I_1h)}(1-|R_1(0)|^2)\right]
+ \frac{\kappa^I_{\Lambda}|T_\Lambda(0)|^2}{2k^I_\Lambda}\left(\tanh{(k^I_\Lambda h)}+
\frac{k^I_\Lambda h}{\cosh^2{(k^I_\Lambda h)}}\right)
- \frac{\kappa^I_{1}\left(1 - |R_1(0)|^2\right)}{2k^I_1}\left(\tanh{(k^I_1h)}+\frac{hk^I_1}{\cosh^2{(k^I_1h)}}\right)=0.
</math></center>

Re-arranging gives

<center><math>
\kappa^I_{\Lambda}\tanh{(k^I_\Lambda h)}
\left(\frac{\beta_\Lambda}{\alpha}2(k^I_\Lambda)^2((\kappa^I_{\Lambda})^2 +k_y^2)\tanh{(k^I_\Lambda h)}
+ \frac{1}{2k^I_\Lambda}+\frac{h}{2\sinh{(k^I_\Lambda h)}\cosh{(k^I_\Lambda h)}}\right)|T_\Lambda(0)|^2
\quad - \kappa^I_{1}\tanh{(k^I_1 h)}
\left(\frac{\beta_1}{\alpha}2(k^I_1)^2((\kappa^I_{1})^2 +k_y^2)\tanh{(k^I_1 h)}
+ \frac{1}{2it}+\frac{h}{2\sinh{(k^I_1 h)}\cosh{(k^I_1 h)}}\right)(1-|R_1(0)|^2)=0
</math></center>

which can be expressed as
<center><math>
D |T_{\Lambda}(0)|^2 + |R_{1}(0)|^2 = 1,
</math></center>
where <math>D</math> is given by
<center><math>
D = \left(\frac{\kappa^I_{\Lambda} k^I_1\cosh^2{(k^I_1h)}}{\kappa^I_{1}k^I_\Lambda\cosh^2{(k^I_\Lambda h)}}\right)
\left(\frac{\frac{\beta_\Lambda}{\alpha}4(k^I_\Lambda)^3((\kappa^I_{\Lambda})^2 +k_y^2)\sinh^2{(k^I_\Lambda h)} +
\frac{1}{2}{\sinh{(2k^I_\Lambda h)}}+k^I_\Lambda h}
{\frac{\beta_1}{\alpha}4(k^I_1)^3((\kappa^I_{1})^2 +k_y^2)\sinh^2{(k^I_1h)} +
\frac{1}{2}{\sinh{(2k^I_1h)}}+k^I_1h}\right)
</math></center>

[[Category:Floating Elastic Plate]]

Energy Balance for Two Elastic Plates

2008-08-23T07:06:58Z

Fmontiel: /* Expanding <math>\mathbf{\xi_3}</math> */

= Introduction =

We derive here a condition which must be satisfied for energy balance for two semi-infinite elastic plates. The solution applies
to the problem of [[Eigenfunction Matching Method for Floating Elastic Plates|multiple elastic plates]] and from here we know we can write the potential as
<center><math>
\phi \approx \left\{
\begin{matrix}
{
Ie^{\kappa_{1}(0)(x-r_1)}\frac{\cos(k_1(0)(z+h))}{\cos(k_1(0)h)} }+\\
{
\qquad \qquad \sum_{n=-2}^{M}R_1(n) e^{\kappa_{1}(n)(x-r_1)} \frac{\cos(k_1(n)(z+h))}{\cos(k_1(n)h)} },&
\mbox{ for } x < r_1,\\
{
\sum_{n=-2}^{M}T_{\mu}(n) e^{-\kappa_\mu(n)(x-l_\mu)}
\frac{\cos(k_\mu(n)(z+h))}{\cos(k_\mu(n)h)} } + \\
{
\qquad \qquad \sum_{n=-2}^{M}R_{\mu}(n) e^{\kappa_\mu(n)(x-r_\mu)}
\frac{\cos(k_\mu(n)(z+h))}{\cos(k_\mu(n)h)} },
&\mbox{ for } l_\mu< x < r_\mu,\\
{
\sum_{n=-2}^{M}T_{\Lambda}(n)e^{-\kappa_{\Lambda}(n)(x-l_\Lambda)}
\frac{\cos(k_\Lambda(n)(z+h))}{\cos(k_\Lambda(n)h)} }, &\mbox{ for } l_\Lambda<x,
\end{matrix} \right.
</math></center>
(for details of this notation see [[Eigenfunction Matching Method for Floating Elastic Plates]]).

= Equations =

Based on the method used in [[Evans and Davies 1968]], a check can be made to ensure the solutions of the floating plate problem are in energy balance.
This is simply a condition that the incident energy is equal to the sum of the radiated energy.
When the first and final plates have different properties, the energy balance equation is derived by applying Green's theorem to <math>\phi</math> and its conjugate.
The domain of integration is shown in the figure on the right.

[[Image:energy_schematic.jpg|thumb|right|300px|A diagram depicting the area <math>\mathcal{U}</math> which is bounded by the rectangle <math>\mathcal{S}</math>.]
{A diagram depicting the area <math>\mathcal{U}</math> which is bounded by the rectangle <math>\mathcal{S}</math>.
The rectangle <math>\mathcal{S}</math> is bounded by <math> -h\leq z \leq0</math> and <math>-\infty\leq x \leq \infty</math>]]

Applying Green's theorem to <math>\phi</math> and its conjugate <math>\phi^*</math> gives
<center><math>
{ \int\int_\mathcal{U}(\phi\nabla^2\phi^* - \phi^*\nabla^2\phi)dxdz
= \int_\mathcal{S}(\phi\frac{\partial\phi^*}{\partial n} - \phi^*\frac{\partial\phi}{\partial n})dl },
</math></center>
where <math>n</math> denotes the outward plane normal to the boundary and <math>l</math> denotes the plane parallel to the boundary.
As <math>\phi</math> and <math>\phi^*</math> satisfy the Laplace's equation, the left hand side of the Green theorem equation vanishes so that it reduces to
<center><math>
\Im\int_\mathcal{S}\phi\frac{\partial\phi^*}{\partial n} dl = 0,
</math></center>
Expanding gives
<center><math>
\xi_1 +\xi_2 + \xi_3 = 0,
</math></center>
where
<center><math>
\xi_1 =
{ \Im\int_{-h}^0(\phi\frac{\partial\phi^*}{\partial x})\big|_{x=-\infty}dz },
</math></center>
<center><math>
\xi_2 =
{ \Im\int_{-h}^0(\phi\frac{\partial\phi^*}{\partial x})\big|_{x=\infty}dz },
</math></center>
and
<center><math>
\xi_3 =
{ \Im\int_{-\infty}^{\infty}(\phi\frac{\partial\phi^*}{\partial z})\big|_{z=0}dx } ,
</math></center>
where <math>\Im</math> denotes the imaginary part. The bottom domain composant is obviously equal to zero because of the no-influx condition over the seabed (<math>\frac{\partial\phi^*}{\partial z}\big|_{z=-h}=0</math>).

= Expanding <math>\mathbf{\xi_1}</math> =

Near <math>x=-\infty</math>, we approximate <math>\phi</math> by
<center><math>
\phi \approx e^{-\kappa_{1}(0)(x-r_1)}\frac{\cos{(k_1(0)(z+h))}}{\cos{(k_1(0)h)}} +
R_1(0)e^{\kappa_{1}(0)(x-r_1)}\frac{\cos{(k_{1}(0)(z+h))}}{\cos{(k_{1}(0)h)}}
</math></center>
Therefore,
<center><math>
\begin{matrix}
\xi_1 & = & {\Im\int_{-h}^0 \left[\left(e^{-i\kappa^{I}_1(x)}+ R_1(0)e^{i\kappa^{I}_1(x)}\right)\right.}
{ \left(i\kappa^{I}_1e^{i\kappa^{I}_1(x)} - i\kappa^{I}_1R_1(0)^*e^{-i\kappa^{I}_1(x)}\right) }
{ \left.\left(\frac{\cosh^2{(k^I_1(z+h))}}{\cosh^2{(k^I_1h)}}\right)\right]dz, }\\ \\
& = & { \Im\left[\frac{i\kappa^{I}_1\left(1 - |R_1(0)|^2\right)}{2\cosh^2{(k^I_1h)}}
\int_{-h}^0 \left(\cosh{(2k^I_1(z+h))+1}\right)dz\right] ,}\\ \\
& = & { \Im\left[\frac{i\kappa^{I}_1\left(1 - |R_1(0)|^2\right)}{2\cosh^2{(k^I_1h)}}
\left[\frac{1}{2k^I_1}\sinh{(2k^I_1(z+h))} + z\right]_{-h}^0\right], }\\ \\
& = & { \frac{\kappa^{I}_1\left(1 - |R_1(0)|^2\right)}{2\cosh^2{(k^I_1h)}} \left(\frac{1}{2k^I_1}\sinh{(2k^I_1h))} + h\right) ,}\\ \\
& = & { \frac{\kappa^{I}_1\left(1 - |R_1(0)|^2\right)}{2k^I_1}\left(\tanh{(k^I_1h)}+\frac{hk^I_1}{\cosh^2{(k^I_1h)}}\right) , }
\end{matrix}
</math></center>
where <math>R_1(0)^*</math> is the conjugate of <math>R_1(0)</math>.

= Expanding <math>\mathbf{\xi_2}</math> =

Near <math>x=\infty</math>, we approximate <math>\phi</math> by
<center><math>
\begin{matrix}
{ \phi \approx T_\Lambda(0)e^{-\kappa_{\Lambda}(0)(x)}\frac{\cos{(k_\Lambda(0)(z+h))}}{\cos{(k_\Lambda(0)h)}}, }
\end{matrix}
</math></center>
and re-express as
<center><math>
\phi \approx T_\Lambda(0)e^{-i\kappa^I_\Lambda(x)}\frac{\cosh{(k^I_\Lambda(z+h))}}{\cosh{(k^I_\Lambda h)}},
</math></center>
where <math>k^I_\Lambda = \Im k_{\Lambda}(0)</math> and <math>\kappa^I_\Lambda=\Im \kappa_\Lambda(0)</math>, so that
<center><math>
\frac{\partial\phi}{\partial x} \approx -i\kappa^I_{\Lambda}T_\Lambda(0)e^{-i\kappa^I_{\Lambda}(x)}\frac{\cosh{(k^I_\Lambda(z+h))}}{\cosh{(k^I_\Lambda h)}}.
</math></center>
Therefore,
<center><math>
\begin{matrix}
\xi_2 & = & \Im\int_{-h}^0 \left[\right.(T_\Lambda(0)e^{-i\kappa^I_{\Lambda}(x)})(i\kappa^I_{\Lambda}T_\Lambda(0)^*e^{i\kappa^I_{\Lambda}(x)})
\frac{\cosh^2{(k^I_\Lambda(z+h))}}{\cosh^2{(k^I_\Lambda h)}}\left.\right]dz, \\ \\
& = & { \frac{\kappa^I_{\Lambda}|T_\Lambda(0)|^2}{2\cosh^2{(k^I_\Lambda h)}}
\left(\frac{1}{2k^I_\Lambda}\sinh{(2k^I_\Lambda h)} + h\right)},\\ \\
& = & { \frac{\kappa^I_{\Lambda}|T_\Lambda(0)|^2}{2k^I_\Lambda}\left(\tanh{(k^I_\Lambda h)}+\frac{k^I_\Lambda h}{\cosh^2{(k^I_\Lambda h)}}\right)}.
\end{matrix}
</math></center>

= Expanding <math>\mathbf{\xi_3}</math> =

The ice-covered boundary condition for the [[Floating Elastic Plate]] gives
<center><math>
\xi_3 = { \Im\int_{-\infty}^{\infty}\left(\frac{\beta}{\alpha}
\left(\frac{\partial^2}{\partial x^2} - k_y^2\right)^2 - \gamma + \frac{1}{\alpha}\right)
\frac{\partial\phi}{\partial z}\cdot\frac{\partial\phi^*}{\partial z}\bigg|_{z=0}dx.}
</math></center>
Since <math>{\frac{\partial\phi}{\partial z}\cdot\frac{\partial\phi^*}{\partial z}}</math> is real,
<center><math>
\xi_3 = \Im\int_{-\infty}^{\infty}\left(\frac{\beta}{\alpha}\frac{\partial^2}{\partial x^2}
\left(\frac{\partial^2}{\partial x^2} - 2k_y^2\right) \right)
\frac{\partial\phi}{\partial z}\cdot\frac{\partial\phi^*}{\partial z}\bigg|_{z=0}dx.
</math></center>
Integration by parts gives
<center><math>
\xi_3 = \Im\left\{\left[\frac{\beta}{\alpha}\frac{\partial}{\partial x}
\left(\frac{\partial^2}{\partial x^2}-2k_y^2\right)
\frac{\partial\phi}{\partial z}\cdot\frac{\partial\phi^*}{\partial z}\right]_{-\infty}^{\infty}
-\int_{-\infty}^{\infty}\frac{\beta}{\alpha}\frac{\partial}{\partial x}
\left(\frac{\partial^2}{\partial x^2} - 2k_y^2\right)
\frac{\partial\phi}{\partial z}\cdot\frac
{\partial}{\partial x}\frac{\partial\phi^*}{\partial z}dx
\right\}.
</math></center>
As <math>{2k_y^2\frac{\partial}{\partial x}\frac{\partial\phi}{\partial z}\cdot \frac{\partial}{\partial x}\frac{\partial\phi^*}{\partial z}}</math> is real and by integration by parts, the expression of <math>\xi_3</math> becomes,
<center><math>
\xi_3 =\Im\left\{
\left[\frac{\beta}{\alpha}\frac{\partial}{\partial x}
\left(\frac{\partial^2}{\partial x^2}-2k_y^2\right)
\frac{\partial\phi}{\partial z}\cdot\frac{\partial\phi^*}{\partial z}\right]_{-\infty}^{\infty}
- \left[\frac{\beta}{\alpha}\frac{\partial^2}{\partial x^2}\frac{\partial\phi}{\partial z}\cdot
\frac{\partial}{\partial x}\frac{\partial\phi^*}{\partial z}\right]_{-\infty}^{\infty}
+ \int_{-\infty}^\infty\frac{\beta}{\alpha}\frac{\partial^2}{\partial x^2}\frac{\partial\phi}{\partial z}\cdot
\frac{\partial^2}{\partial x^2}\frac{\partial\phi^*}{\partial z}dx\right\}.
</math></center>
As <math>{\frac{\partial^2}{\partial x^2}\frac{\partial\phi}{\partial z}\cdot \frac{\partial^2}{\partial x^2}\frac{\partial\phi^*}{\partial z}}</math> is real, we obtain the new expression of <math>\xi_3</math>
<center><math>
\xi_3 = \Im\left\{
\left[\frac{\beta}{\alpha}\frac{\partial}{\partial x}
\left(\frac{\partial^2}{\partial x^2}-2k_y^2\right)
\frac{\partial\phi}{\partial z}\cdot\frac{\partial\phi^*}{\partial z}\right]_{-\infty}^{\infty}-
\left[\frac{\beta}{\alpha}\frac{\partial^2}{\partial x^2}\frac{\partial\phi}{\partial z}\cdot
\frac{\partial}{\partial x}\frac{\partial\phi^*}{\partial z}\right]_{-\infty}^{\infty}\right\}.
</math></center>

Now breaking <math>\xi_3</math> down, we can simplify the left hend term for <math>x>0</math>

<center><math>
\begin{matrix}
\frac{\partial}{\partial x}\left(\frac{\partial^2}{\partial x^2}-2k_y^2\right)
\frac{\partial\phi(x_2,0)}{\partial z}\cdot\frac{\partial\phi(x_2,0)^*}{\partial z} \\ \\
= \left((-i\kappa^I_{\Lambda})^3k^I_\Lambda T_\Lambda(0)e^{-i\kappa^I_{\Lambda}(x)}\tanh{(k^I_\Lambda h)}\right.
\left.- 2k_y^2(-i\kappa_{\Lambda}(0))k^I_\Lambda T_\Lambda(0)e^{-i\kappa^I_{\Lambda}(x)} \tanh{(k^I_\Lambda h)}\right)
\left(k^I_\Lambda T_\Lambda(0)^*e^{i\kappa^I_{\Lambda}(x)}\tanh{(k^I_\Lambda h)}\right), \\ \\
= i\kappa^I_{\Lambda} (k^I_\Lambda)^2\left((\kappa^I_{\Lambda})^2 + 2k_y^2 \right) \tanh^2{(k^I_\Lambda h)}|T_\Lambda(0)|^2,
\end{matrix}
</math></center>

and for <math>x<0</math>

<center><math>
\begin{matrix}
\frac{\partial}{\partial x}\left(\frac{\partial^2}{\partial x^2}-2k_y^2\right)\frac{\partial\phi(x_1,0)}{\partial z}\cdot
\frac{\partial\phi(x_1,0)^*}{\partial z} \\ \\
= \left[i(\kappa^I_{1})^3k^I_1\left(e^{-i\kappa^I_{1}(x)}-R_1(0)e^{i\kappa^I_{1}(x)}\right) \tanh{(k^I_1 h)}
- 2ik_y^2\kappa^I_{1}k^I_1\left(-e^{-i\kappa^I_{1}(x)} + R_1(0)e^{i\kappa^I_{1}(x)}\right) \tanh{(k^I_1h)}\right]
\left[k^I_1\left(e^{i\kappa^I_{1}(x)} + R_1(0)^*e^{-i\kappa^I_{1}(x))}\right) \tanh{(k^I_1h)}\right], \\ \\
= \left[i\kappa^I_{1}k_1(0)((\kappa^I_{1})^2 + 2k_y^2)\left(e^{-i\kappa^I_{1}(x)} - R_1(0)e^{i\kappa^I_{1}(x)}\right)
\tanh{(k^I_1h)}\right]
\left[k^I_1\left(e^{i\kappa^I_{1}(x)} + R_1(0)^*e^{-i\kappa^I_{1}(x))}\right) \tanh{(k^I_1h)}\right], \\ \\
= i\kappa^I_{1}(k^I_1)^2\left((\kappa^I_{1})^2 + 2k_y^2\right)\tanh^2{(k^I_1h)}\left(1 - |R_1(0)|^2\right),
\end{matrix}
</math></center>

Likewise we expand the right hend term for <math>x>0</math>

<center><math>
\begin{matrix}
\frac{\partial^2}{\partial x^2}\frac{\partial\phi(x_2,0)}{\partial z}\cdot
\frac{\partial}{\partial x}\frac{\partial\phi^*}{\partial z} \\ \\
= \left[-(\kappa^I_{\Lambda})^2k^I_\Lambda T_\Lambda(0)e^{-i\kappa^I_{\Lambda}(x)} \tanh{(k^I_\Lambda h)}\right]
\left[i\kappa^I_{\Lambda} k^I_\Lambda T_\Lambda(0)^*e^{i\kappa^I_{\Lambda}(x)} \tanh{(k^I_\Lambda h)}\right], \\ \\
= -i(\kappa^I_{\Lambda})^3(k^I_\Lambda)^2\tanh^2{(k^I_\Lambda h)}|T_\Lambda(0)|^2,
\end{matrix}
</math></center>

and finally for <math>x<0</math>,

<center><math>
\begin{matrix}
\frac{\partial^2}{\partial x^1}\frac{\partial\phi(x_1,0)}{\partial z}\cdot
\frac{\partial}{\partial x}\frac{\partial\phi^*}{\partial z}\\ \\
= \left[-(\kappa^I_{1})^2k^I_1\left(e^{-i\kappa^I_{1}(x)} + R_1(0)e^{i\kappa^I_{1}(x)}\right)\tanh{(k^I_1 h)}\right]
\left[i\kappa^I_{1} k^I_1\left(e^{i\kappa^I_{1}(x)} - R_1(0)e^{-i\kappa^I_{1}(x)}\right)\tanh{(k^I_1 h)}\right], \\ \\
= -i(\kappa^I_{1})^3(k^I_1)^2\tanh^2{(k^I_1h)}\left(1 - |R_1(0)|^2\right).
\end{matrix}
</math></center>

We can now express <math>\xi_3</math> as

<center><math>
\begin{matrix}
\xi_3 & = & \Im
\left\{\frac{\beta_\Lambda}{\alpha}\left[i\kappa^I_{\Lambda} (k^I_\Lambda)^2\left((\kappa^I_{\Lambda})^2 + 2k_y^2 \right)
\tanh^2{(k^I_\Lambda h)}|T_\Lambda(0)|^2\right]
-\frac{\beta_1}{\alpha}\left[i\kappa^I_{1}(k^I_1)^2\left((\kappa^I_{1})^2 + 2k_y^2\right)\tanh^2{(k^I_1h)}
\left(1 - |R_1(0)|^2\right)\right]
-\frac{\beta_\Lambda}{\alpha}\left[-i(\kappa^I_{\Lambda})^3(k^I_\Lambda)^2\tanh^2{(k^I_\Lambda h)}|T_\Lambda(0)|^2\right]
+\frac{\beta_1}{\alpha}\left[-i(\kappa^I_{1})^3(k^I_1)^2\tanh^2{(k^I_1h)} \left(1 - |R_1(0)|^2\right)\right]\right\}, \\ \\
& = & \frac{\beta_\Lambda}{\alpha}\left[2\kappa^I_{\Lambda} (k^I_\Lambda)^2((\kappa^I_{\Lambda})^2 + k_y^2)\tanh^2{(k^I_\Lambda h)}|T_\Lambda(0)|^2\right]
-\frac{\beta_1}{\alpha}\left[2\kappa^I_{1}(k^I_1)^2((\kappa^I_{1})^2 + k_y^2)\tanh^2{(k^I_1h)}(1-|R_1(0)|^2)\right].
\end{matrix}
</math></center>

= Solving the Energy Balance Equation =

Pulling it all together, we finally obtain
<center><math>
\frac{\beta_\Lambda}{\alpha}\left[2\kappa^I_{\Lambda}(k^I_\Lambda)^2((\kappa^I_{\Lambda})^2 +k_y^2)
\tanh^2{(k^I_\Lambda h)}|T_\Lambda(0)|^2\right]
-\frac{\beta_1}{\alpha}\left[2\kappa^I_{1}(k^I_1)^2((\kappa^I_{1})^2 +k_y^2)\tanh^2{(k^I_1h)}(1-|R_1(0)|^2)\right]
+ \frac{\kappa^I_{\Lambda}|T_\Lambda(0)|^2}{2k^I_\Lambda}\left(\tanh{(k^I_\Lambda h)}+
\frac{k^I_\Lambda h}{\cosh^2{(k^I_\Lambda h)}}\right)
- \frac{\kappa^I_{1}\left(1 - |R_1(0)|^2\right)}{2k^I_1}\left(\tanh{(k^I_1h)}+\frac{hk^I_1}{\cosh^2{(k^I_1h)}}\right)=0.
</math></center>

Re-arranging gives

<center><math>
\kappa^I_{\Lambda}\tanh{(k^I_\Lambda h)}
\left(\frac{\beta_\Lambda}{\alpha}2(k^I_\Lambda)^2((\kappa^I_{\Lambda})^2 +k_y^2)\tanh{(k^I_\Lambda h)}
+ \frac{1}{2k^I_\Lambda}+\frac{h}{2\sinh{(k^I_\Lambda h)}\cosh{(k^I_\Lambda h)}}\right)|T_\Lambda(0)|^2
\quad - \kappa^I_{1}\tanh{(k^I_1 h)}
\left(\frac{\beta_1}{\alpha}2(k^I_1)^2((\kappa^I_{1})^2 +k_y^2)\tanh{(k^I_1 h)}
+ \frac{1}{2it}+\frac{h}{2\sinh{(k^I_1 h)}\cosh{(k^I_1 h)}}\right)(1-|R_1(0)|^2)=0
</math></center>

which can be expressed as
<center><math>
D |T_{\Lambda}(0)|^2 + |R_{1}(0)|^2 = 1,
</math></center>
where <math>D</math> is given by
<center><math>
D = \left(\frac{\kappa^I_{\Lambda} k^I_1\cosh^2{(k^I_1h)}}{\kappa^I_{1}k^I_\Lambda\cosh^2{(k^I_\Lambda h)}}\right)
\left(\frac{\frac{\beta_\Lambda}{\alpha}4(k^I_\Lambda)^3((\kappa^I_{\Lambda})^2 +k_y^2)\sinh^2{(k^I_\Lambda h)} +
\frac{1}{2}{\sinh{(2k^I_\Lambda h)}}+k^I_\Lambda h}
{\frac{\beta_1}{\alpha}4(k^I_1)^3((\kappa^I_{1})^2 +k_y^2)\sinh^2{(k^I_1h)} +
\frac{1}{2}{\sinh{(2k^I_1h)}}+k^I_1h}\right)
</math></center>

[[Category:Floating Elastic Plate]]

Energy Balance for Two Elastic Plates

2008-08-23T06:55:00Z

Fmontiel: /* Equations */

= Introduction =

We derive here a condition which must be satisfied for energy balance for two semi-infinite elastic plates. The solution applies
to the problem of [[Eigenfunction Matching Method for Floating Elastic Plates|multiple elastic plates]] and from here we know we can write the potential as
<center><math>
\phi \approx \left\{
\begin{matrix}
{
Ie^{\kappa_{1}(0)(x-r_1)}\frac{\cos(k_1(0)(z+h))}{\cos(k_1(0)h)} }+\\
{
\qquad \qquad \sum_{n=-2}^{M}R_1(n) e^{\kappa_{1}(n)(x-r_1)} \frac{\cos(k_1(n)(z+h))}{\cos(k_1(n)h)} },&
\mbox{ for } x < r_1,\\
{
\sum_{n=-2}^{M}T_{\mu}(n) e^{-\kappa_\mu(n)(x-l_\mu)}
\frac{\cos(k_\mu(n)(z+h))}{\cos(k_\mu(n)h)} } + \\
{
\qquad \qquad \sum_{n=-2}^{M}R_{\mu}(n) e^{\kappa_\mu(n)(x-r_\mu)}
\frac{\cos(k_\mu(n)(z+h))}{\cos(k_\mu(n)h)} },
&\mbox{ for } l_\mu< x < r_\mu,\\
{
\sum_{n=-2}^{M}T_{\Lambda}(n)e^{-\kappa_{\Lambda}(n)(x-l_\Lambda)}
\frac{\cos(k_\Lambda(n)(z+h))}{\cos(k_\Lambda(n)h)} }, &\mbox{ for } l_\Lambda<x,
\end{matrix} \right.
</math></center>
(for details of this notation see [[Eigenfunction Matching Method for Floating Elastic Plates]]).

= Equations =

Based on the method used in [[Evans and Davies 1968]], a check can be made to ensure the solutions of the floating plate problem are in energy balance.
This is simply a condition that the incident energy is equal to the sum of the radiated energy.
When the first and final plates have different properties, the energy balance equation is derived by applying Green's theorem to <math>\phi</math> and its conjugate.
The domain of integration is shown in the figure on the right.

[[Image:energy_schematic.jpg|thumb|right|300px|A diagram depicting the area <math>\mathcal{U}</math> which is bounded by the rectangle <math>\mathcal{S}</math>.]
{A diagram depicting the area <math>\mathcal{U}</math> which is bounded by the rectangle <math>\mathcal{S}</math>.
The rectangle <math>\mathcal{S}</math> is bounded by <math> -h\leq z \leq0</math> and <math>-\infty\leq x \leq \infty</math>]]

Applying Green's theorem to <math>\phi</math> and its conjugate <math>\phi^*</math> gives
<center><math>
{ \int\int_\mathcal{U}(\phi\nabla^2\phi^* - \phi^*\nabla^2\phi)dxdz
= \int_\mathcal{S}(\phi\frac{\partial\phi^*}{\partial n} - \phi^*\frac{\partial\phi}{\partial n})dl },
</math></center>
where <math>n</math> denotes the outward plane normal to the boundary and <math>l</math> denotes the plane parallel to the boundary.
As <math>\phi</math> and <math>\phi^*</math> satisfy the Laplace's equation, the left hand side of the Green theorem equation vanishes so that it reduces to
<center><math>
\Im\int_\mathcal{S}\phi\frac{\partial\phi^*}{\partial n} dl = 0,
</math></center>
Expanding gives
<center><math>
\xi_1 +\xi_2 + \xi_3 = 0,
</math></center>
where
<center><math>
\xi_1 =
{ \Im\int_{-h}^0(\phi\frac{\partial\phi^*}{\partial x})\big|_{x=-\infty}dz },
</math></center>
<center><math>
\xi_2 =
{ \Im\int_{-h}^0(\phi\frac{\partial\phi^*}{\partial x})\big|_{x=\infty}dz },
</math></center>
and
<center><math>
\xi_3 =
{ \Im\int_{-\infty}^{\infty}(\phi\frac{\partial\phi^*}{\partial z})\big|_{z=0}dx } ,
</math></center>
where <math>\Im</math> denotes the imaginary part. The bottom domain composant is obviously equal to zero because of the no-influx condition over the seabed (<math>\frac{\partial\phi^*}{\partial z}\big|_{z=-h}=0</math>).

= Expanding <math>\mathbf{\xi_1}</math> =

Near <math>x=-\infty</math>, we approximate <math>\phi</math> by
<center><math>
\phi \approx e^{-\kappa_{1}(0)(x-r_1)}\frac{\cos{(k_1(0)(z+h))}}{\cos{(k_1(0)h)}} +
R_1(0)e^{\kappa_{1}(0)(x-r_1)}\frac{\cos{(k_{1}(0)(z+h))}}{\cos{(k_{1}(0)h)}}
</math></center>
Therefore,
<center><math>
\begin{matrix}
\xi_1 & = & {\Im\int_{-h}^0 \left[\left(e^{-i\kappa^{I}_1(x)}+ R_1(0)e^{i\kappa^{I}_1(x)}\right)\right.}
{ \left(i\kappa^{I}_1e^{i\kappa^{I}_1(x)} - i\kappa^{I}_1R_1(0)^*e^{-i\kappa^{I}_1(x)}\right) }
{ \left.\left(\frac{\cosh^2{(k^I_1(z+h))}}{\cosh^2{(k^I_1h)}}\right)\right]dz, }\\ \\
& = & { \Im\left[\frac{i\kappa^{I}_1\left(1 - |R_1(0)|^2\right)}{2\cosh^2{(k^I_1h)}}
\int_{-h}^0 \left(\cosh{(2k^I_1(z+h))+1}\right)dz\right] ,}\\ \\
& = & { \Im\left[\frac{i\kappa^{I}_1\left(1 - |R_1(0)|^2\right)}{2\cosh^2{(k^I_1h)}}
\left[\frac{1}{2k^I_1}\sinh{(2k^I_1(z+h))} + z\right]_{-h}^0\right], }\\ \\
& = & { \frac{\kappa^{I}_1\left(1 - |R_1(0)|^2\right)}{2\cosh^2{(k^I_1h)}} \left(\frac{1}{2k^I_1}\sinh{(2k^I_1h))} + h\right) ,}\\ \\
& = & { \frac{\kappa^{I}_1\left(1 - |R_1(0)|^2\right)}{2k^I_1}\left(\tanh{(k^I_1h)}+\frac{hk^I_1}{\cosh^2{(k^I_1h)}}\right) , }
\end{matrix}
</math></center>
where <math>R_1(0)^*</math> is the conjugate of <math>R_1(0)</math>.

= Expanding <math>\mathbf{\xi_2}</math> =

Near <math>x=\infty</math>, we approximate <math>\phi</math> by
<center><math>
\begin{matrix}
{ \phi \approx T_\Lambda(0)e^{-\kappa_{\Lambda}(0)(x)}\frac{\cos{(k_\Lambda(0)(z+h))}}{\cos{(k_\Lambda(0)h)}}, }
\end{matrix}
</math></center>
and re-express as
<center><math>
\phi \approx T_\Lambda(0)e^{-i\kappa^I_\Lambda(x)}\frac{\cosh{(k^I_\Lambda(z+h))}}{\cosh{(k^I_\Lambda h)}},
</math></center>
where <math>k^I_\Lambda = \Im k_{\Lambda}(0)</math> and <math>\kappa^I_\Lambda=\Im \kappa_\Lambda(0)</math>, so that
<center><math>
\frac{\partial\phi}{\partial x} \approx -i\kappa^I_{\Lambda}T_\Lambda(0)e^{-i\kappa^I_{\Lambda}(x)}\frac{\cosh{(k^I_\Lambda(z+h))}}{\cosh{(k^I_\Lambda h)}}.
</math></center>
Therefore,
<center><math>
\begin{matrix}
\xi_2 & = & \Im\int_{-h}^0 \left[\right.(T_\Lambda(0)e^{-i\kappa^I_{\Lambda}(x)})(i\kappa^I_{\Lambda}T_\Lambda(0)^*e^{i\kappa^I_{\Lambda}(x)})
\frac{\cosh^2{(k^I_\Lambda(z+h))}}{\cosh^2{(k^I_\Lambda h)}}\left.\right]dz, \\ \\
& = & { \frac{\kappa^I_{\Lambda}|T_\Lambda(0)|^2}{2\cosh^2{(k^I_\Lambda h)}}
\left(\frac{1}{2k^I_\Lambda}\sinh{(2k^I_\Lambda h)} + h\right)},\\ \\
& = & { \frac{\kappa^I_{\Lambda}|T_\Lambda(0)|^2}{2k^I_\Lambda}\left(\tanh{(k^I_\Lambda h)}+\frac{k^I_\Lambda h}{\cosh^2{(k^I_\Lambda h)}}\right)}.
\end{matrix}
</math></center>

= Expanding <math>\mathbf{\xi_3}</math> =

The ice-covered boundary condition for the [[Floating Elastic Plate]] gives
<center><math>
\xi_3 = { \Im\int_{-\infty}^{\infty}\left(\frac{\beta}{\alpha}
\left(\frac{\partial^2}{\partial x^2} - k_y^2\right)^2 - \gamma + \frac{1}{\alpha}\right)
\frac{\partial\phi}{\partial z}\cdot\frac{\partial\phi^*}{\partial z}\bigg|_{z=0}dx.}
</math></center>
Since <math>{\frac{\partial\phi}{\partial z}\cdot\frac{\partial\phi^*}{\partial z}}</math> is real,
<center><math>
\xi_3 = \Im\int_{-\infty}^{\infty}\left(\frac{\beta}{\alpha}\frac{\partial^2}{\partial x^2}
\left(\frac{\partial^2}{\partial x^2} - 2k_y^2\right)^2 \right)
\frac{\partial\phi}{\partial z}\cdot\frac{\partial\phi^*}{\partial z}\bigg|_{z=0}dx.
</math></center>
Integration by parts gives
<center><math>
\xi_3 = \Im\left\{\left[\frac{\beta}{\alpha}\frac{\partial}{\partial x}
\left(\frac{\partial^2}{\partial x^2}-2k_y^2\right)
\frac{\partial\phi}{\partial z}\cdot\frac{\partial\phi^*}{\partial z}\right]_{-\infty}^{\infty}
-\int_{-\infty}^{\infty}\frac{\beta}{\alpha}\frac{\partial}{\partial x}
\left(\frac{\partial^2}{\partial x^2} - 2k_y^2\right)
\frac{\partial\phi}{\partial z}\cdot\frac
{\partial}{\partial x}\frac{\partial\phi^*}{\partial z}dx
\right\}.
</math></center>
As <math>{2k_y^2\frac{\partial}{\partial x}\frac{\partial\phi}{\partial z}\cdot \frac{\partial}{\partial x}\frac{\partial\phi^*}{\partial z}}</math> is real and by integration by parts, the expression of <math>\xi_3</math> becomes,
<center><math>
\xi_3 =\Im\left\{
\left[\frac{\beta}{\alpha}\frac{\partial}{\partial x}
\left(\frac{\partial^2}{\partial x^2}-2k_y^2\right)
\frac{\partial\phi}{\partial z}\cdot\frac{\partial\phi^*}{\partial z}\right]_{-\infty}^{\infty}
- \left[\frac{\beta}{\alpha}\frac{\partial^2}{\partial x^2}\frac{\partial\phi}{\partial z}\cdot
\frac{\partial}{\partial x}\frac{\partial\phi^*}{\partial z}\right]_{-\infty}^{\infty}
+ \int_{-\infty}^\infty\frac{\beta}{\alpha}\frac{\partial^2}{\partial x^2}\frac{\partial\phi}{\partial z}\cdot
\frac{\partial^2}{\partial x^2}\frac{\partial\phi^*}{\partial z}dx\right\}.
</math></center>
As <math>{\frac{\partial^2}{\partial x^2}\frac{\partial\phi}{\partial z}\cdot \frac{\partial^2}{\partial x^2}\frac{\partial\phi^*}{\partial z}}</math> is real, we obtain the new expression of <math>\xi_3</math>
<center><math>
\xi_3 = \Im\left\{
\left[\frac{\beta}{\alpha}\frac{\partial}{\partial x}
\left(\frac{\partial^2}{\partial x^2}-2k_y^2\right)
\frac{\partial\phi}{\partial z}\cdot\frac{\partial\phi^*}{\partial z}\right]_{-\infty}^{\infty}-
\left[\frac{\beta}{\alpha}\frac{\partial^2}{\partial x^2}\frac{\partial\phi}{\partial z}\cdot
\frac{\partial}{\partial x}\frac{\partial\phi^*}{\partial z}\right]_{-\infty}^{\infty}\right\}.
</math></center>

Now breaking <math>\xi_3</math> down, we can simplify the left hend term for <math>x>0</math>

<center><math>
\begin{matrix}
\frac{\partial}{\partial x}\left(\frac{\partial^2}{\partial x^2}-2k_y^2\right)
\frac{\partial\phi(x_2,0)}{\partial z}\cdot\frac{\partial\phi(x_2,0)^*}{\partial z} \\ \\
= \left((-i\kappa^I_{\Lambda})^3k^I_\Lambda T_\Lambda(0)e^{-i\kappa^I_{\Lambda}(x)}\tanh{(k^I_\Lambda h)}\right.
\left.- 2k_y^2(-i\kappa_{\Lambda}(0))k^I_\Lambda T_\Lambda(0)e^{-i\kappa^I_{\Lambda}(x)} \tanh{(k^I_\Lambda h)}\right)
\left(k^I_\Lambda T_\Lambda(0)^*e^{i\kappa^I_{\Lambda}(x)}\tanh{(k^I_\Lambda h)}\right), \\ \\
= i\kappa^I_{\Lambda} (k^I_\Lambda)^2\left((\kappa^I_{\Lambda})^2 + 2k_y^2 \right) \tanh^2{(k^I_\Lambda h)}|T_\Lambda(0)|^2,
\end{matrix}
</math></center>

and for <math>x<0</math>

<center><math>
\begin{matrix}
\frac{\partial}{\partial x}\left(\frac{\partial^2}{\partial x^2}-2k_y^2\right)\frac{\partial\phi(x_1,0)}{\partial z}\cdot
\frac{\partial\phi(x_1,0)^*}{\partial z} \\ \\
= \left[i(\kappa^I_{1})^3k^I_1\left(e^{-i\kappa^I_{1}(x)}-R_1(0)e^{i\kappa^I_{1}(x)}\right) \tanh{(k^I_1 h)}
- 2ik_y^2\kappa^I_{1}k^I_1\left(-e^{-i\kappa^I_{1}(x)} + R_1(0)e^{i\kappa^I_{1}(x)}\right) \tanh{(k^I_1h)}\right]
\left[k^I_1\left(e^{i\kappa^I_{1}(x)} + R_1(0)^*e^{-i\kappa^I_{1}(x))}\right) \tanh{(k^I_1h)}\right], \\ \\
= \left[i\kappa^I_{1}k_1(0)((\kappa^I_{1})^2 + 2k_y^2)\left(e^{-i\kappa^I_{1}(x)} - R_1(0)e^{i\kappa^I_{1}(x)}\right)
\tanh{(k^I_1h)}\right]
\left[k^I_1\left(e^{i\kappa^I_{1}(x)} + R_1(0)^*e^{-i\kappa^I_{1}(x))}\right) \tanh{(k^I_1h)}\right], \\ \\
= i\kappa^I_{1}(k^I_1)^2\left((\kappa^I_{1})^2 + 2k_y^2\right)\tanh^2{(k^I_1h)}\left(1 - |R_1(0)|^2\right),
\end{matrix}
</math></center>

Likewise we expand the right hend term for <math>x>0</math>

<center><math>
\begin{matrix}
\frac{\partial^2}{\partial x^2}\frac{\partial\phi(x_2,0)}{\partial z}\cdot
\frac{\partial}{\partial x}\frac{\partial\phi^*}{\partial z} \\ \\
= \left[-(\kappa^I_{\Lambda})^2k^I_\Lambda T_\Lambda(0)e^{-i\kappa^I_{\Lambda}(x)} \tanh{(k^I_\Lambda h)}\right]
\left[i\kappa^I_{\Lambda} k^I_\Lambda T_\Lambda(0)^*e^{i\kappa^I_{\Lambda}(x)} \tanh{(k^I_\Lambda h)}\right], \\ \\
= -i(\kappa^I_{\Lambda})^3(k^I_\Lambda)^2\tanh^2{(k^I_\Lambda h)}|T_\Lambda(0)|^2,
\end{matrix}
</math></center>

and finally for <math>x<0</math>,

<center><math>
\begin{matrix}
\frac{\partial^2}{\partial x^1}\frac{\partial\phi(x_1,0)}{\partial z}\cdot
\frac{\partial}{\partial x}\frac{\partial\phi^*}{\partial z}\\ \\
= \left[-(\kappa^I_{1})^2k^I_1\left(e^{-i\kappa^I_{1}(x)} + R_1(0)e^{i\kappa^I_{1}(x)}\right)\tanh{(k^I_1 h)}\right]
\left[i\kappa^I_{1} k^I_1\left(e^{i\kappa^I_{1}(x)} - R_1(0)e^{-i\kappa^I_{1}(x)}\right)\tanh{(k^I_1 h)}\right], \\ \\
= -i(\kappa^I_{1})^3(k^I_1)^2\tanh^2{(k^I_1h)}\left(1 - |R_1(0)|^2\right).
\end{matrix}
</math></center>

We can now express <math>\xi_3</math> as

<center><math>
\begin{matrix}
\xi_3 & = & \Im
\left\{\frac{\beta_\Lambda}{\alpha}\left[i\kappa^I_{\Lambda} (k^I_\Lambda)^2\left((\kappa^I_{\Lambda})^2 + 2k_y^2 \right)
\tanh^2{(k^I_\Lambda h)}|T_\Lambda(0)|^2\right]
-\frac{\beta_1}{\alpha}\left[i\kappa^I_{1}(k^I_1)^2\left((\kappa^I_{1})^2 + 2k_y^2\right)\tanh^2{(k^I_1h)}
\left(1 - |R_1(0)|^2\right)\right]
-\frac{\beta_\Lambda}{\alpha}\left[-i(\kappa^I_{\Lambda})^3(k^I_\Lambda)^2\tanh^2{(k^I_\Lambda h)}|T_\Lambda(0)|^2\right]
+\frac{\beta_1}{\alpha}\left[-i(\kappa^I_{1})^3(k^I_1)^2\tanh^2{(k^I_1h)} \left(1 - |R_1(0)|^2\right)\right]\right\}, \\ \\
& = & \frac{\beta_\Lambda}{\alpha}\left[2\kappa^I_{\Lambda} (k^I_\Lambda)^2((\kappa^I_{\Lambda})^2 + k_y^2)\tanh^2{(k^I_\Lambda h)}|T_\Lambda(0)|^2\right]
-\frac{\beta_1}{\alpha}\left[2\kappa^I_{1}(k^I_1)^2((\kappa^I_{1})^2 + k_y^2)\tanh^2{(k^I_1h)}(1-|R_1(0)|^2)\right].
\end{matrix}
</math></center>

= Solving the Energy Balance Equation =

Pulling it all together, we finally obtain
<center><math>
\frac{\beta_\Lambda}{\alpha}\left[2\kappa^I_{\Lambda}(k^I_\Lambda)^2((\kappa^I_{\Lambda})^2 +k_y^2)
\tanh^2{(k^I_\Lambda h)}|T_\Lambda(0)|^2\right]
-\frac{\beta_1}{\alpha}\left[2\kappa^I_{1}(k^I_1)^2((\kappa^I_{1})^2 +k_y^2)\tanh^2{(k^I_1h)}(1-|R_1(0)|^2)\right]
+ \frac{\kappa^I_{\Lambda}|T_\Lambda(0)|^2}{2k^I_\Lambda}\left(\tanh{(k^I_\Lambda h)}+
\frac{k^I_\Lambda h}{\cosh^2{(k^I_\Lambda h)}}\right)
- \frac{\kappa^I_{1}\left(1 - |R_1(0)|^2\right)}{2k^I_1}\left(\tanh{(k^I_1h)}+\frac{hk^I_1}{\cosh^2{(k^I_1h)}}\right)=0.
</math></center>

Re-arranging gives

<center><math>
\kappa^I_{\Lambda}\tanh{(k^I_\Lambda h)}
\left(\frac{\beta_\Lambda}{\alpha}2(k^I_\Lambda)^2((\kappa^I_{\Lambda})^2 +k_y^2)\tanh{(k^I_\Lambda h)}
+ \frac{1}{2k^I_\Lambda}+\frac{h}{2\sinh{(k^I_\Lambda h)}\cosh{(k^I_\Lambda h)}}\right)|T_\Lambda(0)|^2
\quad - \kappa^I_{1}\tanh{(k^I_1 h)}
\left(\frac{\beta_1}{\alpha}2(k^I_1)^2((\kappa^I_{1})^2 +k_y^2)\tanh{(k^I_1 h)}
+ \frac{1}{2it}+\frac{h}{2\sinh{(k^I_1 h)}\cosh{(k^I_1 h)}}\right)(1-|R_1(0)|^2)=0
</math></center>

which can be expressed as
<center><math>
D |T_{\Lambda}(0)|^2 + |R_{1}(0)|^2 = 1,
</math></center>
where <math>D</math> is given by
<center><math>
D = \left(\frac{\kappa^I_{\Lambda} k^I_1\cosh^2{(k^I_1h)}}{\kappa^I_{1}k^I_\Lambda\cosh^2{(k^I_\Lambda h)}}\right)
\left(\frac{\frac{\beta_\Lambda}{\alpha}4(k^I_\Lambda)^3((\kappa^I_{\Lambda})^2 +k_y^2)\sinh^2{(k^I_\Lambda h)} +
\frac{1}{2}{\sinh{(2k^I_\Lambda h)}}+k^I_\Lambda h}
{\frac{\beta_1}{\alpha}4(k^I_1)^3((\kappa^I_{1})^2 +k_y^2)\sinh^2{(k^I_1h)} +
\frac{1}{2}{\sinh{(2k^I_1h)}}+k^I_1h}\right)
</math></center>

[[Category:Floating Elastic Plate]]

Energy Balance for Two Elastic Plates

2008-08-23T06:37:37Z

Fmontiel: /* Expanding <math>\mathbf{\xi_2}</math> */

= Introduction =

We derive here a condition which must be satisfied for energy balance for two semi-infinite elastic plates. The solution applies
to the problem of [[Eigenfunction Matching Method for Floating Elastic Plates|multiple elastic plates]] and from here we know we can write the potential as
<center><math>
\phi \approx \left\{
\begin{matrix}
{
Ie^{\kappa_{1}(0)(x-r_1)}\frac{\cos(k_1(0)(z+h))}{\cos(k_1(0)h)} }+\\
{
\qquad \qquad \sum_{n=-2}^{M}R_1(n) e^{\kappa_{1}(n)(x-r_1)} \frac{\cos(k_1(n)(z+h))}{\cos(k_1(n)h)} },&
\mbox{ for } x < r_1,\\
{
\sum_{n=-2}^{M}T_{\mu}(n) e^{-\kappa_\mu(n)(x-l_\mu)}
\frac{\cos(k_\mu(n)(z+h))}{\cos(k_\mu(n)h)} } + \\
{
\qquad \qquad \sum_{n=-2}^{M}R_{\mu}(n) e^{\kappa_\mu(n)(x-r_\mu)}
\frac{\cos(k_\mu(n)(z+h))}{\cos(k_\mu(n)h)} },
&\mbox{ for } l_\mu< x < r_\mu,\\
{
\sum_{n=-2}^{M}T_{\Lambda}(n)e^{-\kappa_{\Lambda}(n)(x-l_\Lambda)}
\frac{\cos(k_\Lambda(n)(z+h))}{\cos(k_\Lambda(n)h)} }, &\mbox{ for } l_\Lambda<x,
\end{matrix} \right.
</math></center>
(for details of this notation see [[Eigenfunction Matching Method for Floating Elastic Plates]]).

= Equations =

Based on the method used in [[Evans and Davies 1968]], a check can be made to ensure the solutions of the floating plate problem are in energy balance.
This is simply a condition that the incident energy is equal to the sum of the radiated energy.
When the first and final plates have different properties, the energy balance equation is derived by applying Green's theorem to <math>\phi</math> and its conjugate.
The domain of integration is shown in the figure on the right.

[[Image:energy_schematic.jpg|thumb|right|300px|A diagram depicting the area <math>\mathcal{U}</math> which is bounded by the rectangle <math>\mathcal{S}</math>.]
{A diagram depicting the area <math>\mathcal{U}</math> which is bounded by the rectangle <math>\mathcal{S}</math>.
The rectangle <math>\mathcal{S}</math> is bounded by <math> -h\leq z \leq0</math> and <math>-\infty\leq x \leq \infty</math>]]

Applying Green's theorem to <math>\phi</math> and its conjugate <math>\phi^*</math> gives
<center><math>
{ \int\int_\mathcal{U}(\phi\nabla^2\phi^* - \phi^*\nabla^2\phi)dxdz
= \int_\mathcal{S}(\phi\frac{\partial\phi^*}{\partial n} - \phi^*\frac{\partial\phi}{\partial n})dl },
</math></center>
where <math>n</math> denotes the outward plane normal to the boundary and <math>l</math> denotes the plane parallel to the boundary.
As <math>\phi</math> and <math>\phi^*</math> satisfy the Laplace's equation, the left hand side of the Green theorem equation vanishes so that it reduces to
<center><math>
\Im\int_\mathcal{S}\phi\frac{\partial\phi^*}{\partial n} dl = 0,
</math></center>
Expanding gives
<center><math>
\xi_1 +\xi_2 + \xi_3 = 0,
</math></center>
where
<center><math>
\xi_1 =
{ \Im\int_{-h}^0(\phi\frac{\partial\phi^*}{\partial x})\big|_{x=-\infty}dz },
</math></center>
<center><math>
\xi_2 =
{ \Im\int_{-h}^0(\phi\frac{\partial\phi^*}{\partial x})\big|_{x=\infty}dz },
</math></center>
and
<center><math>
\xi_3 =
{ \Im\int_{-\infty}^{\infty}(\phi\frac{\partial\phi^*}{\partial z})\big|_{z=0}dx }= 0 ,
</math></center>
where <math>\Im</math> denotes the imaginary part. The bottom domain composant is obviously equal to zero because of the no-influx condition over the seabed (<math>\frac{\partial\phi^*}{\partial z}\big|_{z=-h}=0</math>).

= Expanding <math>\mathbf{\xi_1}</math> =

Near <math>x=-\infty</math>, we approximate <math>\phi</math> by
<center><math>
\phi \approx e^{-\kappa_{1}(0)(x-r_1)}\frac{\cos{(k_1(0)(z+h))}}{\cos{(k_1(0)h)}} +
R_1(0)e^{\kappa_{1}(0)(x-r_1)}\frac{\cos{(k_{1}(0)(z+h))}}{\cos{(k_{1}(0)h)}}
</math></center>
Therefore,
<center><math>
\begin{matrix}
\xi_1 & = & {\Im\int_{-h}^0 \left[\left(e^{-i\kappa^{I}_1(x)}+ R_1(0)e^{i\kappa^{I}_1(x)}\right)\right.}
{ \left(i\kappa^{I}_1e^{i\kappa^{I}_1(x)} - i\kappa^{I}_1R_1(0)^*e^{-i\kappa^{I}_1(x)}\right) }
{ \left.\left(\frac{\cosh^2{(k^I_1(z+h))}}{\cosh^2{(k^I_1h)}}\right)\right]dz, }\\ \\
& = & { \Im\left[\frac{i\kappa^{I}_1\left(1 - |R_1(0)|^2\right)}{2\cosh^2{(k^I_1h)}}
\int_{-h}^0 \left(\cosh{(2k^I_1(z+h))+1}\right)dz\right] ,}\\ \\
& = & { \Im\left[\frac{i\kappa^{I}_1\left(1 - |R_1(0)|^2\right)}{2\cosh^2{(k^I_1h)}}
\left[\frac{1}{2k^I_1}\sinh{(2k^I_1(z+h))} + z\right]_{-h}^0\right], }\\ \\
& = & { \frac{\kappa^{I}_1\left(1 - |R_1(0)|^2\right)}{2\cosh^2{(k^I_1h)}} \left(\frac{1}{2k^I_1}\sinh{(2k^I_1h))} + h\right) ,}\\ \\
& = & { \frac{\kappa^{I}_1\left(1 - |R_1(0)|^2\right)}{2k^I_1}\left(\tanh{(k^I_1h)}+\frac{hk^I_1}{\cosh^2{(k^I_1h)}}\right) , }
\end{matrix}
</math></center>
where <math>R_1(0)^*</math> is the conjugate of <math>R_1(0)</math>.

= Expanding <math>\mathbf{\xi_2}</math> =

Near <math>x=\infty</math>, we approximate <math>\phi</math> by
<center><math>
\begin{matrix}
{ \phi \approx T_\Lambda(0)e^{-\kappa_{\Lambda}(0)(x)}\frac{\cos{(k_\Lambda(0)(z+h))}}{\cos{(k_\Lambda(0)h)}}, }
\end{matrix}
</math></center>
and re-express as
<center><math>
\phi \approx T_\Lambda(0)e^{-i\kappa^I_\Lambda(x)}\frac{\cosh{(k^I_\Lambda(z+h))}}{\cosh{(k^I_\Lambda h)}},
</math></center>
where <math>k^I_\Lambda = \Im k_{\Lambda}(0)</math> and <math>\kappa^I_\Lambda=\Im \kappa_\Lambda(0)</math>, so that
<center><math>
\frac{\partial\phi}{\partial x} \approx -i\kappa^I_{\Lambda}T_\Lambda(0)e^{-i\kappa^I_{\Lambda}(x)}\frac{\cosh{(k^I_\Lambda(z+h))}}{\cosh{(k^I_\Lambda h)}}.
</math></center>
Therefore,
<center><math>
\begin{matrix}
\xi_2 & = & \Im\int_{-h}^0 \left[\right.(T_\Lambda(0)e^{-i\kappa^I_{\Lambda}(x)})(i\kappa^I_{\Lambda}T_\Lambda(0)^*e^{i\kappa^I_{\Lambda}(x)})
\frac{\cosh^2{(k^I_\Lambda(z+h))}}{\cosh^2{(k^I_\Lambda h)}}\left.\right]dz, \\ \\
& = & { \frac{\kappa^I_{\Lambda}|T_\Lambda(0)|^2}{2\cosh^2{(k^I_\Lambda h)}}
\left(\frac{1}{2k^I_\Lambda}\sinh{(2k^I_\Lambda h)} + h\right)},\\ \\
& = & { \frac{\kappa^I_{\Lambda}|T_\Lambda(0)|^2}{2k^I_\Lambda}\left(\tanh{(k^I_\Lambda h)}+\frac{k^I_\Lambda h}{\cosh^2{(k^I_\Lambda h)}}\right)}.
\end{matrix}
</math></center>

= Expanding <math>\mathbf{\xi_3}</math> =

The ice-covered boundary condition for the [[Floating Elastic Plate]] gives
<center><math>
\xi_3 = { \Im\int_{-\infty}^{\infty}\left(\frac{\beta}{\alpha}
\left(\frac{\partial^2}{\partial x^2} - k_y^2\right)^2 - \gamma + \frac{1}{\alpha}\right)
\frac{\partial\phi}{\partial z}\cdot\frac{\partial\phi^*}{\partial z}\bigg|_{z=0}dx.}
</math></center>
Since <math>{\frac{\partial\phi}{\partial z}\cdot\frac{\partial\phi^*}{\partial z}}</math> is real,
<center><math>
\xi_3 = \Im\int_{-\infty}^{\infty}\left(\frac{\beta}{\alpha}\frac{\partial^2}{\partial x^2}
\left(\frac{\partial^2}{\partial x^2} - 2k_y^2\right)^2 \right)
\frac{\partial\phi}{\partial z}\cdot\frac{\partial\phi^*}{\partial z}\bigg|_{z=0}dx.
</math></center>
Integration by parts gives
<center><math>
\xi_3 = \Im\left\{\left[\frac{\beta}{\alpha}\frac{\partial}{\partial x}
\left(\frac{\partial^2}{\partial x^2}-2k_y^2\right)
\frac{\partial\phi}{\partial z}\cdot\frac{\partial\phi^*}{\partial z}\right]_{-\infty}^{\infty}
-\int_{-\infty}^{\infty}\frac{\beta}{\alpha}\frac{\partial}{\partial x}
\left(\frac{\partial^2}{\partial x^2} - 2k_y^2\right)
\frac{\partial\phi}{\partial z}\cdot\frac
{\partial}{\partial x}\frac{\partial\phi^*}{\partial z}dx
\right\}.
</math></center>
As <math>{2k_y^2\frac{\partial}{\partial x}\frac{\partial\phi}{\partial z}\cdot \frac{\partial}{\partial x}\frac{\partial\phi^*}{\partial z}}</math> is real and by integration by parts, the expression of <math>\xi_3</math> becomes,
<center><math>
\xi_3 =\Im\left\{
\left[\frac{\beta}{\alpha}\frac{\partial}{\partial x}
\left(\frac{\partial^2}{\partial x^2}-2k_y^2\right)
\frac{\partial\phi}{\partial z}\cdot\frac{\partial\phi^*}{\partial z}\right]_{-\infty}^{\infty}
- \left[\frac{\beta}{\alpha}\frac{\partial^2}{\partial x^2}\frac{\partial\phi}{\partial z}\cdot
\frac{\partial}{\partial x}\frac{\partial\phi^*}{\partial z}\right]_{-\infty}^{\infty}
+ \int_{-\infty}^\infty\frac{\beta}{\alpha}\frac{\partial^2}{\partial x^2}\frac{\partial\phi}{\partial z}\cdot
\frac{\partial^2}{\partial x^2}\frac{\partial\phi^*}{\partial z}dx\right\}.
</math></center>
As <math>{\frac{\partial^2}{\partial x^2}\frac{\partial\phi}{\partial z}\cdot \frac{\partial^2}{\partial x^2}\frac{\partial\phi^*}{\partial z}}</math> is real, we obtain the new expression of <math>\xi_3</math>
<center><math>
\xi_3 = \Im\left\{
\left[\frac{\beta}{\alpha}\frac{\partial}{\partial x}
\left(\frac{\partial^2}{\partial x^2}-2k_y^2\right)
\frac{\partial\phi}{\partial z}\cdot\frac{\partial\phi^*}{\partial z}\right]_{-\infty}^{\infty}-
\left[\frac{\beta}{\alpha}\frac{\partial^2}{\partial x^2}\frac{\partial\phi}{\partial z}\cdot
\frac{\partial}{\partial x}\frac{\partial\phi^*}{\partial z}\right]_{-\infty}^{\infty}\right\}.
</math></center>

Now breaking <math>\xi_3</math> down, we can simplify the left hend term for <math>x>0</math>

<center><math>
\begin{matrix}
\frac{\partial}{\partial x}\left(\frac{\partial^2}{\partial x^2}-2k_y^2\right)
\frac{\partial\phi(x_2,0)}{\partial z}\cdot\frac{\partial\phi(x_2,0)^*}{\partial z} \\ \\
= \left((-i\kappa^I_{\Lambda})^3k^I_\Lambda T_\Lambda(0)e^{-i\kappa^I_{\Lambda}(x)}\tanh{(k^I_\Lambda h)}\right.
\left.- 2k_y^2(-i\kappa_{\Lambda}(0))k^I_\Lambda T_\Lambda(0)e^{-i\kappa^I_{\Lambda}(x)} \tanh{(k^I_\Lambda h)}\right)
\left(k^I_\Lambda T_\Lambda(0)^*e^{i\kappa^I_{\Lambda}(x)}\tanh{(k^I_\Lambda h)}\right), \\ \\
= i\kappa^I_{\Lambda} (k^I_\Lambda)^2\left((\kappa^I_{\Lambda})^2 + 2k_y^2 \right) \tanh^2{(k^I_\Lambda h)}|T_\Lambda(0)|^2,
\end{matrix}
</math></center>

and for <math>x<0</math>

<center><math>
\begin{matrix}
\frac{\partial}{\partial x}\left(\frac{\partial^2}{\partial x^2}-2k_y^2\right)\frac{\partial\phi(x_1,0)}{\partial z}\cdot
\frac{\partial\phi(x_1,0)^*}{\partial z} \\ \\
= \left[i(\kappa^I_{1})^3k^I_1\left(e^{-i\kappa^I_{1}(x)}-R_1(0)e^{i\kappa^I_{1}(x)}\right) \tanh{(k^I_1 h)}
- 2ik_y^2\kappa^I_{1}k^I_1\left(-e^{-i\kappa^I_{1}(x)} + R_1(0)e^{i\kappa^I_{1}(x)}\right) \tanh{(k^I_1h)}\right]
\left[k^I_1\left(e^{i\kappa^I_{1}(x)} + R_1(0)^*e^{-i\kappa^I_{1}(x))}\right) \tanh{(k^I_1h)}\right], \\ \\
= \left[i\kappa^I_{1}k_1(0)((\kappa^I_{1})^2 + 2k_y^2)\left(e^{-i\kappa^I_{1}(x)} - R_1(0)e^{i\kappa^I_{1}(x)}\right)
\tanh{(k^I_1h)}\right]
\left[k^I_1\left(e^{i\kappa^I_{1}(x)} + R_1(0)^*e^{-i\kappa^I_{1}(x))}\right) \tanh{(k^I_1h)}\right], \\ \\
= i\kappa^I_{1}(k^I_1)^2\left((\kappa^I_{1})^2 + 2k_y^2\right)\tanh^2{(k^I_1h)}\left(1 - |R_1(0)|^2\right),
\end{matrix}
</math></center>

Likewise we expand the right hend term for <math>x>0</math>

<center><math>
\begin{matrix}
\frac{\partial^2}{\partial x^2}\frac{\partial\phi(x_2,0)}{\partial z}\cdot
\frac{\partial}{\partial x}\frac{\partial\phi^*}{\partial z} \\ \\
= \left[-(\kappa^I_{\Lambda})^2k^I_\Lambda T_\Lambda(0)e^{-i\kappa^I_{\Lambda}(x)} \tanh{(k^I_\Lambda h)}\right]
\left[i\kappa^I_{\Lambda} k^I_\Lambda T_\Lambda(0)^*e^{i\kappa^I_{\Lambda}(x)} \tanh{(k^I_\Lambda h)}\right], \\ \\
= -i(\kappa^I_{\Lambda})^3(k^I_\Lambda)^2\tanh^2{(k^I_\Lambda h)}|T_\Lambda(0)|^2,
\end{matrix}
</math></center>

and finally for <math>x<0</math>,

<center><math>
\begin{matrix}
\frac{\partial^2}{\partial x^1}\frac{\partial\phi(x_1,0)}{\partial z}\cdot
\frac{\partial}{\partial x}\frac{\partial\phi^*}{\partial z}\\ \\
= \left[-(\kappa^I_{1})^2k^I_1\left(e^{-i\kappa^I_{1}(x)} + R_1(0)e^{i\kappa^I_{1}(x)}\right)\tanh{(k^I_1 h)}\right]
\left[i\kappa^I_{1} k^I_1\left(e^{i\kappa^I_{1}(x)} - R_1(0)e^{-i\kappa^I_{1}(x)}\right)\tanh{(k^I_1 h)}\right], \\ \\
= -i(\kappa^I_{1})^3(k^I_1)^2\tanh^2{(k^I_1h)}\left(1 - |R_1(0)|^2\right).
\end{matrix}
</math></center>

We can now express <math>\xi_3</math> as

<center><math>
\begin{matrix}
\xi_3 & = & \Im
\left\{\frac{\beta_\Lambda}{\alpha}\left[i\kappa^I_{\Lambda} (k^I_\Lambda)^2\left((\kappa^I_{\Lambda})^2 + 2k_y^2 \right)
\tanh^2{(k^I_\Lambda h)}|T_\Lambda(0)|^2\right]
-\frac{\beta_1}{\alpha}\left[i\kappa^I_{1}(k^I_1)^2\left((\kappa^I_{1})^2 + 2k_y^2\right)\tanh^2{(k^I_1h)}
\left(1 - |R_1(0)|^2\right)\right]
-\frac{\beta_\Lambda}{\alpha}\left[-i(\kappa^I_{\Lambda})^3(k^I_\Lambda)^2\tanh^2{(k^I_\Lambda h)}|T_\Lambda(0)|^2\right]
+\frac{\beta_1}{\alpha}\left[-i(\kappa^I_{1})^3(k^I_1)^2\tanh^2{(k^I_1h)} \left(1 - |R_1(0)|^2\right)\right]\right\}, \\ \\
& = & \frac{\beta_\Lambda}{\alpha}\left[2\kappa^I_{\Lambda} (k^I_\Lambda)^2((\kappa^I_{\Lambda})^2 + k_y^2)\tanh^2{(k^I_\Lambda h)}|T_\Lambda(0)|^2\right]
-\frac{\beta_1}{\alpha}\left[2\kappa^I_{1}(k^I_1)^2((\kappa^I_{1})^2 + k_y^2)\tanh^2{(k^I_1h)}(1-|R_1(0)|^2)\right].
\end{matrix}
</math></center>

= Solving the Energy Balance Equation =

Pulling it all together, we finally obtain
<center><math>
\frac{\beta_\Lambda}{\alpha}\left[2\kappa^I_{\Lambda}(k^I_\Lambda)^2((\kappa^I_{\Lambda})^2 +k_y^2)
\tanh^2{(k^I_\Lambda h)}|T_\Lambda(0)|^2\right]
-\frac{\beta_1}{\alpha}\left[2\kappa^I_{1}(k^I_1)^2((\kappa^I_{1})^2 +k_y^2)\tanh^2{(k^I_1h)}(1-|R_1(0)|^2)\right]
+ \frac{\kappa^I_{\Lambda}|T_\Lambda(0)|^2}{2k^I_\Lambda}\left(\tanh{(k^I_\Lambda h)}+
\frac{k^I_\Lambda h}{\cosh^2{(k^I_\Lambda h)}}\right)
- \frac{\kappa^I_{1}\left(1 - |R_1(0)|^2\right)}{2k^I_1}\left(\tanh{(k^I_1h)}+\frac{hk^I_1}{\cosh^2{(k^I_1h)}}\right)=0.
</math></center>

Re-arranging gives

<center><math>
\kappa^I_{\Lambda}\tanh{(k^I_\Lambda h)}
\left(\frac{\beta_\Lambda}{\alpha}2(k^I_\Lambda)^2((\kappa^I_{\Lambda})^2 +k_y^2)\tanh{(k^I_\Lambda h)}
+ \frac{1}{2k^I_\Lambda}+\frac{h}{2\sinh{(k^I_\Lambda h)}\cosh{(k^I_\Lambda h)}}\right)|T_\Lambda(0)|^2
\quad - \kappa^I_{1}\tanh{(k^I_1 h)}
\left(\frac{\beta_1}{\alpha}2(k^I_1)^2((\kappa^I_{1})^2 +k_y^2)\tanh{(k^I_1 h)}
+ \frac{1}{2it}+\frac{h}{2\sinh{(k^I_1 h)}\cosh{(k^I_1 h)}}\right)(1-|R_1(0)|^2)=0
</math></center>

which can be expressed as
<center><math>
D |T_{\Lambda}(0)|^2 + |R_{1}(0)|^2 = 1,
</math></center>
where <math>D</math> is given by
<center><math>
D = \left(\frac{\kappa^I_{\Lambda} k^I_1\cosh^2{(k^I_1h)}}{\kappa^I_{1}k^I_\Lambda\cosh^2{(k^I_\Lambda h)}}\right)
\left(\frac{\frac{\beta_\Lambda}{\alpha}4(k^I_\Lambda)^3((\kappa^I_{\Lambda})^2 +k_y^2)\sinh^2{(k^I_\Lambda h)} +
\frac{1}{2}{\sinh{(2k^I_\Lambda h)}}+k^I_\Lambda h}
{\frac{\beta_1}{\alpha}4(k^I_1)^3((\kappa^I_{1})^2 +k_y^2)\sinh^2{(k^I_1h)} +
\frac{1}{2}{\sinh{(2k^I_1h)}}+k^I_1h}\right)
</math></center>

[[Category:Floating Elastic Plate]]

Energy Balance for Two Elastic Plates

2008-08-23T06:31:58Z

Fmontiel: /* Expanding <math>\mathbf{\xi_2}</math> */

= Introduction =

We derive here a condition which must be satisfied for energy balance for two semi-infinite elastic plates. The solution applies
to the problem of [[Eigenfunction Matching Method for Floating Elastic Plates|multiple elastic plates]] and from here we know we can write the potential as
<center><math>
\phi \approx \left\{
\begin{matrix}
{
Ie^{\kappa_{1}(0)(x-r_1)}\frac{\cos(k_1(0)(z+h))}{\cos(k_1(0)h)} }+\\
{
\qquad \qquad \sum_{n=-2}^{M}R_1(n) e^{\kappa_{1}(n)(x-r_1)} \frac{\cos(k_1(n)(z+h))}{\cos(k_1(n)h)} },&
\mbox{ for } x < r_1,\\
{
\sum_{n=-2}^{M}T_{\mu}(n) e^{-\kappa_\mu(n)(x-l_\mu)}
\frac{\cos(k_\mu(n)(z+h))}{\cos(k_\mu(n)h)} } + \\
{
\qquad \qquad \sum_{n=-2}^{M}R_{\mu}(n) e^{\kappa_\mu(n)(x-r_\mu)}
\frac{\cos(k_\mu(n)(z+h))}{\cos(k_\mu(n)h)} },
&\mbox{ for } l_\mu< x < r_\mu,\\
{
\sum_{n=-2}^{M}T_{\Lambda}(n)e^{-\kappa_{\Lambda}(n)(x-l_\Lambda)}
\frac{\cos(k_\Lambda(n)(z+h))}{\cos(k_\Lambda(n)h)} }, &\mbox{ for } l_\Lambda<x,
\end{matrix} \right.
</math></center>
(for details of this notation see [[Eigenfunction Matching Method for Floating Elastic Plates]]).

= Equations =

Based on the method used in [[Evans and Davies 1968]], a check can be made to ensure the solutions of the floating plate problem are in energy balance.
This is simply a condition that the incident energy is equal to the sum of the radiated energy.
When the first and final plates have different properties, the energy balance equation is derived by applying Green's theorem to <math>\phi</math> and its conjugate.
The domain of integration is shown in the figure on the right.

[[Image:energy_schematic.jpg|thumb|right|300px|A diagram depicting the area <math>\mathcal{U}</math> which is bounded by the rectangle <math>\mathcal{S}</math>.]
{A diagram depicting the area <math>\mathcal{U}</math> which is bounded by the rectangle <math>\mathcal{S}</math>.
The rectangle <math>\mathcal{S}</math> is bounded by <math> -h\leq z \leq0</math> and <math>-\infty\leq x \leq \infty</math>]]

Applying Green's theorem to <math>\phi</math> and its conjugate <math>\phi^*</math> gives
<center><math>
{ \int\int_\mathcal{U}(\phi\nabla^2\phi^* - \phi^*\nabla^2\phi)dxdz
= \int_\mathcal{S}(\phi\frac{\partial\phi^*}{\partial n} - \phi^*\frac{\partial\phi}{\partial n})dl },
</math></center>
where <math>n</math> denotes the outward plane normal to the boundary and <math>l</math> denotes the plane parallel to the boundary.
As <math>\phi</math> and <math>\phi^*</math> satisfy the Laplace's equation, the left hand side of the Green theorem equation vanishes so that it reduces to
<center><math>
\Im\int_\mathcal{S}\phi\frac{\partial\phi^*}{\partial n} dl = 0,
</math></center>
Expanding gives
<center><math>
\xi_1 +\xi_2 + \xi_3 = 0,
</math></center>
where
<center><math>
\xi_1 =
{ \Im\int_{-h}^0(\phi\frac{\partial\phi^*}{\partial x})\big|_{x=-\infty}dz },
</math></center>
<center><math>
\xi_2 =
{ \Im\int_{-h}^0(\phi\frac{\partial\phi^*}{\partial x})\big|_{x=\infty}dz },
</math></center>
and
<center><math>
\xi_3 =
{ \Im\int_{-\infty}^{\infty}(\phi\frac{\partial\phi^*}{\partial z})\big|_{z=0}dx }= 0 ,
</math></center>
where <math>\Im</math> denotes the imaginary part. The bottom domain composant is obviously equal to zero because of the no-influx condition over the seabed (<math>\frac{\partial\phi^*}{\partial z}\big|_{z=-h}=0</math>).

= Expanding <math>\mathbf{\xi_1}</math> =

Near <math>x=-\infty</math>, we approximate <math>\phi</math> by
<center><math>
\phi \approx e^{-\kappa_{1}(0)(x-r_1)}\frac{\cos{(k_1(0)(z+h))}}{\cos{(k_1(0)h)}} +
R_1(0)e^{\kappa_{1}(0)(x-r_1)}\frac{\cos{(k_{1}(0)(z+h))}}{\cos{(k_{1}(0)h)}}
</math></center>
Therefore,
<center><math>
\begin{matrix}
\xi_1 & = & {\Im\int_{-h}^0 \left[\left(e^{-i\kappa^{I}_1(x)}+ R_1(0)e^{i\kappa^{I}_1(x)}\right)\right.}
{ \left(i\kappa^{I}_1e^{i\kappa^{I}_1(x)} - i\kappa^{I}_1R_1(0)^*e^{-i\kappa^{I}_1(x)}\right) }
{ \left.\left(\frac{\cosh^2{(k^I_1(z+h))}}{\cosh^2{(k^I_1h)}}\right)\right]dz, }\\ \\
& = & { \Im\left[\frac{i\kappa^{I}_1\left(1 - |R_1(0)|^2\right)}{2\cosh^2{(k^I_1h)}}
\int_{-h}^0 \left(\cosh{(2k^I_1(z+h))+1}\right)dz\right] ,}\\ \\
& = & { \Im\left[\frac{i\kappa^{I}_1\left(1 - |R_1(0)|^2\right)}{2\cosh^2{(k^I_1h)}}
\left[\frac{1}{2k^I_1}\sinh{(2k^I_1(z+h))} + z\right]_{-h}^0\right], }\\ \\
& = & { \frac{\kappa^{I}_1\left(1 - |R_1(0)|^2\right)}{2\cosh^2{(k^I_1h)}} \left(\frac{1}{2k^I_1}\sinh{(2k^I_1h))} + h\right) ,}\\ \\
& = & { \frac{\kappa^{I}_1\left(1 - |R_1(0)|^2\right)}{2k^I_1}\left(\tanh{(k^I_1h)}+\frac{hk^I_1}{\cosh^2{(k^I_1h)}}\right) , }
\end{matrix}
</math></center>
where <math>R_1(0)^*</math> is the conjugate of <math>R_1(0)</math>.

= Expanding <math>\mathbf{\xi_2}</math> =

Near <math>x=\infty</math>, we approximate <math>\phi</math> by
<center><math>
\begin{matrix}
{ \phi \approx T_\Lambda(0)e^{-\kappa_{\Lambda}(0)(x)}\frac{\cos{(k_\Lambda(0)(z+h))}}{\cos{(k_\Lambda(0)h)}}, }
\end{matrix}
</math></center>
and re-express as
<center><math>
\phi \approx T_\Lambda(0)e^{-i\kappa^I_\Lambda(x)}\frac{\cosh{(k^I_\Lambda(z+h))}}{\cosh{(k^I_\Lambda h)}},
</math></center>
where <math>k^I_\Lambda = \Im k_{\Lambda}(0)</math> and <math>\kappa^I_\Lambda=\Im \kappa_\Lambda(0)</math>, so that
<center><math>
\frac{\partial\phi}{\partial x} \approx -i\kappa^I_{\Lambda}T_\Lambda(0)e^{-i\kappa^I_{\Lambda}(x)}\frac{\cosh{(k^I_\Lambda(z+h))}}{\cosh{(k^I_\Lambda h)}}.
</math></center>
Therefore,
<center><math>
\begin{matrix}
\xi_2 & = & \Im\int_{-h}^0 \left[\right.(T_\Lambda(0)e^{-i\kappa^I_{\Lambda}(x)})(i\kappa^I_{\Lambda}T_\Lambda(0)^*e^{-i\kappa^I_{\Lambda}(x)})
\frac{\cosh^2{(k^I_\Lambda(z+h))}}{\cosh^2{(k^I_\Lambda h)}}\left.\right]dz, \\ \\
& = & { \frac{\kappa^I_{\Lambda}|T_\Lambda(0)|^2}{2\cosh^2{(k^I_\Lambda h)}}
\left(\frac{1}{2k^I_\Lambda}\sinh{(2k^I_\Lambda h)} + h\right)},\\ \\
& = & { \frac{\kappa^I_{\Lambda}|T_\Lambda(0)|^2}{2k^I_\Lambda}\left(\tanh{(k^I_\Lambda h)}+\frac{k^I_\Lambda h}{\cosh^2{(k^I_\Lambda h)}}\right)}.
\end{matrix}
</math></center>

= Expanding <math>\mathbf{\xi_3}</math> =

The ice-covered boundary condition for the [[Floating Elastic Plate]] gives
<center><math>
\xi_3 = { \Im\int_{-\infty}^{\infty}\left(\frac{\beta}{\alpha}
\left(\frac{\partial^2}{\partial x^2} - k_y^2\right)^2 - \gamma + \frac{1}{\alpha}\right)
\frac{\partial\phi}{\partial z}\cdot\frac{\partial\phi^*}{\partial z}\bigg|_{z=0}dx.}
</math></center>
Since <math>{\frac{\partial\phi}{\partial z}\cdot\frac{\partial\phi^*}{\partial z}}</math> is real,
<center><math>
\xi_3 = \Im\int_{-\infty}^{\infty}\left(\frac{\beta}{\alpha}\frac{\partial^2}{\partial x^2}
\left(\frac{\partial^2}{\partial x^2} - 2k_y^2\right)^2 \right)
\frac{\partial\phi}{\partial z}\cdot\frac{\partial\phi^*}{\partial z}\bigg|_{z=0}dx.
</math></center>
Integration by parts gives
<center><math>
\xi_3 = \Im\left\{\left[\frac{\beta}{\alpha}\frac{\partial}{\partial x}
\left(\frac{\partial^2}{\partial x^2}-2k_y^2\right)
\frac{\partial\phi}{\partial z}\cdot\frac{\partial\phi^*}{\partial z}\right]_{-\infty}^{\infty}
-\int_{-\infty}^{\infty}\frac{\beta}{\alpha}\frac{\partial}{\partial x}
\left(\frac{\partial^2}{\partial x^2} - 2k_y^2\right)
\frac{\partial\phi}{\partial z}\cdot\frac
{\partial}{\partial x}\frac{\partial\phi^*}{\partial z}dx
\right\}.
</math></center>
As <math>{2k_y^2\frac{\partial}{\partial x}\frac{\partial\phi}{\partial z}\cdot \frac{\partial}{\partial x}\frac{\partial\phi^*}{\partial z}}</math> is real and by integration by parts, the expression of <math>\xi_3</math> becomes,
<center><math>
\xi_3 =\Im\left\{
\left[\frac{\beta}{\alpha}\frac{\partial}{\partial x}
\left(\frac{\partial^2}{\partial x^2}-2k_y^2\right)
\frac{\partial\phi}{\partial z}\cdot\frac{\partial\phi^*}{\partial z}\right]_{-\infty}^{\infty}
- \left[\frac{\beta}{\alpha}\frac{\partial^2}{\partial x^2}\frac{\partial\phi}{\partial z}\cdot
\frac{\partial}{\partial x}\frac{\partial\phi^*}{\partial z}\right]_{-\infty}^{\infty}
+ \int_{-\infty}^\infty\frac{\beta}{\alpha}\frac{\partial^2}{\partial x^2}\frac{\partial\phi}{\partial z}\cdot
\frac{\partial^2}{\partial x^2}\frac{\partial\phi^*}{\partial z}dx\right\}.
</math></center>
As <math>{\frac{\partial^2}{\partial x^2}\frac{\partial\phi}{\partial z}\cdot \frac{\partial^2}{\partial x^2}\frac{\partial\phi^*}{\partial z}}</math> is real, we obtain the new expression of <math>\xi_3</math>
<center><math>
\xi_3 = \Im\left\{
\left[\frac{\beta}{\alpha}\frac{\partial}{\partial x}
\left(\frac{\partial^2}{\partial x^2}-2k_y^2\right)
\frac{\partial\phi}{\partial z}\cdot\frac{\partial\phi^*}{\partial z}\right]_{-\infty}^{\infty}-
\left[\frac{\beta}{\alpha}\frac{\partial^2}{\partial x^2}\frac{\partial\phi}{\partial z}\cdot
\frac{\partial}{\partial x}\frac{\partial\phi^*}{\partial z}\right]_{-\infty}^{\infty}\right\}.
</math></center>

Now breaking <math>\xi_3</math> down, we can simplify the left hend term for <math>x>0</math>

<center><math>
\begin{matrix}
\frac{\partial}{\partial x}\left(\frac{\partial^2}{\partial x^2}-2k_y^2\right)
\frac{\partial\phi(x_2,0)}{\partial z}\cdot\frac{\partial\phi(x_2,0)^*}{\partial z} \\ \\
= \left((-i\kappa^I_{\Lambda})^3k^I_\Lambda T_\Lambda(0)e^{-i\kappa^I_{\Lambda}(x)}\tanh{(k^I_\Lambda h)}\right.
\left.- 2k_y^2(-i\kappa_{\Lambda}(0))k^I_\Lambda T_\Lambda(0)e^{-i\kappa^I_{\Lambda}(x)} \tanh{(k^I_\Lambda h)}\right)
\left(k^I_\Lambda T_\Lambda(0)^*e^{i\kappa^I_{\Lambda}(x)}\tanh{(k^I_\Lambda h)}\right), \\ \\
= i\kappa^I_{\Lambda} (k^I_\Lambda)^2\left((\kappa^I_{\Lambda})^2 + 2k_y^2 \right) \tanh^2{(k^I_\Lambda h)}|T_\Lambda(0)|^2,
\end{matrix}
</math></center>

and for <math>x<0</math>

<center><math>
\begin{matrix}
\frac{\partial}{\partial x}\left(\frac{\partial^2}{\partial x^2}-2k_y^2\right)\frac{\partial\phi(x_1,0)}{\partial z}\cdot
\frac{\partial\phi(x_1,0)^*}{\partial z} \\ \\
= \left[i(\kappa^I_{1})^3k^I_1\left(e^{-i\kappa^I_{1}(x)}-R_1(0)e^{i\kappa^I_{1}(x)}\right) \tanh{(k^I_1 h)}
- 2ik_y^2\kappa^I_{1}k^I_1\left(-e^{-i\kappa^I_{1}(x)} + R_1(0)e^{i\kappa^I_{1}(x)}\right) \tanh{(k^I_1h)}\right]
\left[k^I_1\left(e^{i\kappa^I_{1}(x)} + R_1(0)^*e^{-i\kappa^I_{1}(x))}\right) \tanh{(k^I_1h)}\right], \\ \\
= \left[i\kappa^I_{1}k_1(0)((\kappa^I_{1})^2 + 2k_y^2)\left(e^{-i\kappa^I_{1}(x)} - R_1(0)e^{i\kappa^I_{1}(x)}\right)
\tanh{(k^I_1h)}\right]
\left[k^I_1\left(e^{i\kappa^I_{1}(x)} + R_1(0)^*e^{-i\kappa^I_{1}(x))}\right) \tanh{(k^I_1h)}\right], \\ \\
= i\kappa^I_{1}(k^I_1)^2\left((\kappa^I_{1})^2 + 2k_y^2\right)\tanh^2{(k^I_1h)}\left(1 - |R_1(0)|^2\right),
\end{matrix}
</math></center>

Likewise we expand the right hend term for <math>x>0</math>

<center><math>
\begin{matrix}
\frac{\partial^2}{\partial x^2}\frac{\partial\phi(x_2,0)}{\partial z}\cdot
\frac{\partial}{\partial x}\frac{\partial\phi^*}{\partial z} \\ \\
= \left[-(\kappa^I_{\Lambda})^2k^I_\Lambda T_\Lambda(0)e^{-i\kappa^I_{\Lambda}(x)} \tanh{(k^I_\Lambda h)}\right]
\left[i\kappa^I_{\Lambda} k^I_\Lambda T_\Lambda(0)^*e^{i\kappa^I_{\Lambda}(x)} \tanh{(k^I_\Lambda h)}\right], \\ \\
= -i(\kappa^I_{\Lambda})^3(k^I_\Lambda)^2\tanh^2{(k^I_\Lambda h)}|T_\Lambda(0)|^2,
\end{matrix}
</math></center>

and finally for <math>x<0</math>,

<center><math>
\begin{matrix}
\frac{\partial^2}{\partial x^1}\frac{\partial\phi(x_1,0)}{\partial z}\cdot
\frac{\partial}{\partial x}\frac{\partial\phi^*}{\partial z}\\ \\
= \left[-(\kappa^I_{1})^2k^I_1\left(e^{-i\kappa^I_{1}(x)} + R_1(0)e^{i\kappa^I_{1}(x)}\right)\tanh{(k^I_1 h)}\right]
\left[i\kappa^I_{1} k^I_1\left(e^{i\kappa^I_{1}(x)} - R_1(0)e^{-i\kappa^I_{1}(x)}\right)\tanh{(k^I_1 h)}\right], \\ \\
= -i(\kappa^I_{1})^3(k^I_1)^2\tanh^2{(k^I_1h)}\left(1 - |R_1(0)|^2\right).
\end{matrix}
</math></center>

We can now express <math>\xi_3</math> as

<center><math>
\begin{matrix}
\xi_3 & = & \Im
\left\{\frac{\beta_\Lambda}{\alpha}\left[i\kappa^I_{\Lambda} (k^I_\Lambda)^2\left((\kappa^I_{\Lambda})^2 + 2k_y^2 \right)
\tanh^2{(k^I_\Lambda h)}|T_\Lambda(0)|^2\right]
-\frac{\beta_1}{\alpha}\left[i\kappa^I_{1}(k^I_1)^2\left((\kappa^I_{1})^2 + 2k_y^2\right)\tanh^2{(k^I_1h)}
\left(1 - |R_1(0)|^2\right)\right]
-\frac{\beta_\Lambda}{\alpha}\left[-i(\kappa^I_{\Lambda})^3(k^I_\Lambda)^2\tanh^2{(k^I_\Lambda h)}|T_\Lambda(0)|^2\right]
+\frac{\beta_1}{\alpha}\left[-i(\kappa^I_{1})^3(k^I_1)^2\tanh^2{(k^I_1h)} \left(1 - |R_1(0)|^2\right)\right]\right\}, \\ \\
& = & \frac{\beta_\Lambda}{\alpha}\left[2\kappa^I_{\Lambda} (k^I_\Lambda)^2((\kappa^I_{\Lambda})^2 + k_y^2)\tanh^2{(k^I_\Lambda h)}|T_\Lambda(0)|^2\right]
-\frac{\beta_1}{\alpha}\left[2\kappa^I_{1}(k^I_1)^2((\kappa^I_{1})^2 + k_y^2)\tanh^2{(k^I_1h)}(1-|R_1(0)|^2)\right].
\end{matrix}
</math></center>

= Solving the Energy Balance Equation =

Pulling it all together, we finally obtain
<center><math>
\frac{\beta_\Lambda}{\alpha}\left[2\kappa^I_{\Lambda}(k^I_\Lambda)^2((\kappa^I_{\Lambda})^2 +k_y^2)
\tanh^2{(k^I_\Lambda h)}|T_\Lambda(0)|^2\right]
-\frac{\beta_1}{\alpha}\left[2\kappa^I_{1}(k^I_1)^2((\kappa^I_{1})^2 +k_y^2)\tanh^2{(k^I_1h)}(1-|R_1(0)|^2)\right]
+ \frac{\kappa^I_{\Lambda}|T_\Lambda(0)|^2}{2k^I_\Lambda}\left(\tanh{(k^I_\Lambda h)}+
\frac{k^I_\Lambda h}{\cosh^2{(k^I_\Lambda h)}}\right)
- \frac{\kappa^I_{1}\left(1 - |R_1(0)|^2\right)}{2k^I_1}\left(\tanh{(k^I_1h)}+\frac{hk^I_1}{\cosh^2{(k^I_1h)}}\right)=0.
</math></center>

Re-arranging gives

<center><math>
\kappa^I_{\Lambda}\tanh{(k^I_\Lambda h)}
\left(\frac{\beta_\Lambda}{\alpha}2(k^I_\Lambda)^2((\kappa^I_{\Lambda})^2 +k_y^2)\tanh{(k^I_\Lambda h)}
+ \frac{1}{2k^I_\Lambda}+\frac{h}{2\sinh{(k^I_\Lambda h)}\cosh{(k^I_\Lambda h)}}\right)|T_\Lambda(0)|^2
\quad - \kappa^I_{1}\tanh{(k^I_1 h)}
\left(\frac{\beta_1}{\alpha}2(k^I_1)^2((\kappa^I_{1})^2 +k_y^2)\tanh{(k^I_1 h)}
+ \frac{1}{2it}+\frac{h}{2\sinh{(k^I_1 h)}\cosh{(k^I_1 h)}}\right)(1-|R_1(0)|^2)=0
</math></center>

which can be expressed as
<center><math>
D |T_{\Lambda}(0)|^2 + |R_{1}(0)|^2 = 1,
</math></center>
where <math>D</math> is given by
<center><math>
D = \left(\frac{\kappa^I_{\Lambda} k^I_1\cosh^2{(k^I_1h)}}{\kappa^I_{1}k^I_\Lambda\cosh^2{(k^I_\Lambda h)}}\right)
\left(\frac{\frac{\beta_\Lambda}{\alpha}4(k^I_\Lambda)^3((\kappa^I_{\Lambda})^2 +k_y^2)\sinh^2{(k^I_\Lambda h)} +
\frac{1}{2}{\sinh{(2k^I_\Lambda h)}}+k^I_\Lambda h}
{\frac{\beta_1}{\alpha}4(k^I_1)^3((\kappa^I_{1})^2 +k_y^2)\sinh^2{(k^I_1h)} +
\frac{1}{2}{\sinh{(2k^I_1h)}}+k^I_1h}\right)
</math></center>

[[Category:Floating Elastic Plate]]

Energy Balance for Two Elastic Plates

2008-08-23T06:29:00Z

Fmontiel: /* Equations */

= Introduction =

We derive here a condition which must be satisfied for energy balance for two semi-infinite elastic plates. The solution applies
to the problem of [[Eigenfunction Matching Method for Floating Elastic Plates|multiple elastic plates]] and from here we know we can write the potential as
<center><math>
\phi \approx \left\{
\begin{matrix}
{
Ie^{\kappa_{1}(0)(x-r_1)}\frac{\cos(k_1(0)(z+h))}{\cos(k_1(0)h)} }+\\
{
\qquad \qquad \sum_{n=-2}^{M}R_1(n) e^{\kappa_{1}(n)(x-r_1)} \frac{\cos(k_1(n)(z+h))}{\cos(k_1(n)h)} },&
\mbox{ for } x < r_1,\\
{
\sum_{n=-2}^{M}T_{\mu}(n) e^{-\kappa_\mu(n)(x-l_\mu)}
\frac{\cos(k_\mu(n)(z+h))}{\cos(k_\mu(n)h)} } + \\
{
\qquad \qquad \sum_{n=-2}^{M}R_{\mu}(n) e^{\kappa_\mu(n)(x-r_\mu)}
\frac{\cos(k_\mu(n)(z+h))}{\cos(k_\mu(n)h)} },
&\mbox{ for } l_\mu< x < r_\mu,\\
{
\sum_{n=-2}^{M}T_{\Lambda}(n)e^{-\kappa_{\Lambda}(n)(x-l_\Lambda)}
\frac{\cos(k_\Lambda(n)(z+h))}{\cos(k_\Lambda(n)h)} }, &\mbox{ for } l_\Lambda<x,
\end{matrix} \right.
</math></center>
(for details of this notation see [[Eigenfunction Matching Method for Floating Elastic Plates]]).

= Equations =

Based on the method used in [[Evans and Davies 1968]], a check can be made to ensure the solutions of the floating plate problem are in energy balance.
This is simply a condition that the incident energy is equal to the sum of the radiated energy.
When the first and final plates have different properties, the energy balance equation is derived by applying Green's theorem to <math>\phi</math> and its conjugate.
The domain of integration is shown in the figure on the right.

[[Image:energy_schematic.jpg|thumb|right|300px|A diagram depicting the area <math>\mathcal{U}</math> which is bounded by the rectangle <math>\mathcal{S}</math>.]
{A diagram depicting the area <math>\mathcal{U}</math> which is bounded by the rectangle <math>\mathcal{S}</math>.
The rectangle <math>\mathcal{S}</math> is bounded by <math> -h\leq z \leq0</math> and <math>-\infty\leq x \leq \infty</math>]]

Applying Green's theorem to <math>\phi</math> and its conjugate <math>\phi^*</math> gives
<center><math>
{ \int\int_\mathcal{U}(\phi\nabla^2\phi^* - \phi^*\nabla^2\phi)dxdz
= \int_\mathcal{S}(\phi\frac{\partial\phi^*}{\partial n} - \phi^*\frac{\partial\phi}{\partial n})dl },
</math></center>
where <math>n</math> denotes the outward plane normal to the boundary and <math>l</math> denotes the plane parallel to the boundary.
As <math>\phi</math> and <math>\phi^*</math> satisfy the Laplace's equation, the left hand side of the Green theorem equation vanishes so that it reduces to
<center><math>
\Im\int_\mathcal{S}\phi\frac{\partial\phi^*}{\partial n} dl = 0,
</math></center>
Expanding gives
<center><math>
\xi_1 +\xi_2 + \xi_3 = 0,
</math></center>
where
<center><math>
\xi_1 =
{ \Im\int_{-h}^0(\phi\frac{\partial\phi^*}{\partial x})\big|_{x=-\infty}dz },
</math></center>
<center><math>
\xi_2 =
{ \Im\int_{-h}^0(\phi\frac{\partial\phi^*}{\partial x})\big|_{x=\infty}dz },
</math></center>
and
<center><math>
\xi_3 =
{ \Im\int_{-\infty}^{\infty}(\phi\frac{\partial\phi^*}{\partial z})\big|_{z=0}dx }= 0 ,
</math></center>
where <math>\Im</math> denotes the imaginary part. The bottom domain composant is obviously equal to zero because of the no-influx condition over the seabed (<math>\frac{\partial\phi^*}{\partial z}\big|_{z=-h}=0</math>).

= Expanding <math>\mathbf{\xi_1}</math> =

Near <math>x=-\infty</math>, we approximate <math>\phi</math> by
<center><math>
\phi \approx e^{-\kappa_{1}(0)(x-r_1)}\frac{\cos{(k_1(0)(z+h))}}{\cos{(k_1(0)h)}} +
R_1(0)e^{\kappa_{1}(0)(x-r_1)}\frac{\cos{(k_{1}(0)(z+h))}}{\cos{(k_{1}(0)h)}}
</math></center>
Therefore,
<center><math>
\begin{matrix}
\xi_1 & = & {\Im\int_{-h}^0 \left[\left(e^{-i\kappa^{I}_1(x)}+ R_1(0)e^{i\kappa^{I}_1(x)}\right)\right.}
{ \left(i\kappa^{I}_1e^{i\kappa^{I}_1(x)} - i\kappa^{I}_1R_1(0)^*e^{-i\kappa^{I}_1(x)}\right) }
{ \left.\left(\frac{\cosh^2{(k^I_1(z+h))}}{\cosh^2{(k^I_1h)}}\right)\right]dz, }\\ \\
& = & { \Im\left[\frac{i\kappa^{I}_1\left(1 - |R_1(0)|^2\right)}{2\cosh^2{(k^I_1h)}}
\int_{-h}^0 \left(\cosh{(2k^I_1(z+h))+1}\right)dz\right] ,}\\ \\
& = & { \Im\left[\frac{i\kappa^{I}_1\left(1 - |R_1(0)|^2\right)}{2\cosh^2{(k^I_1h)}}
\left[\frac{1}{2k^I_1}\sinh{(2k^I_1(z+h))} + z\right]_{-h}^0\right], }\\ \\
& = & { \frac{\kappa^{I}_1\left(1 - |R_1(0)|^2\right)}{2\cosh^2{(k^I_1h)}} \left(\frac{1}{2k^I_1}\sinh{(2k^I_1h))} + h\right) ,}\\ \\
& = & { \frac{\kappa^{I}_1\left(1 - |R_1(0)|^2\right)}{2k^I_1}\left(\tanh{(k^I_1h)}+\frac{hk^I_1}{\cosh^2{(k^I_1h)}}\right) , }
\end{matrix}
</math></center>
where <math>R_1(0)^*</math> is the conjugate of <math>R_1(0)</math>.

= Expanding <math>\mathbf{\xi_2}</math> =

Near <math>x=\infty</math>, we approximate <math>\phi</math> by
<center><math>
\begin{matrix}
{ \phi \approx T_\Lambda(0)e^{-\kappa_{\Lambda}(0)(x)}\frac{\cos{(k_\Lambda(0)(z+h))}}{\cos{(k_\Lambda(0)h)}}, }
\end{matrix}
</math></center>
and re-express as
<center><math>
\phi \approx T_\Lambda(0)e^{-i\kappa^I_\Lambda(x)}\frac{\cosh{(k^I_\Lambda(z+h))}}{\cosh{(k^I_\Lambda h)}},
</math></center>
where <math>k^I_\Lambda = -\Im k_{\Lambda}(0)</math> and <math>\kappa^I_\Lambda=-\Im \kappa_\Lambda(0)</math>, so that
<center><math>
\frac{\partial\phi}{\partial x} \approx -i\kappa^I_{\Lambda}T_\Lambda(0)e^{-i\kappa^I_{\Lambda}(x)}\frac{\cosh{(k^I_\Lambda(z+h))}}{\cosh{(k^I_\Lambda h)}}.
</math></center>
Therefore,
<center><math>
\begin{matrix}
\xi_2 & = & \Im\int_{-h}^0 \left[\right.(T_\Lambda(0)e^{-i\kappa^I_{\Lambda}(x)})(i\kappa^I_{\Lambda}T_\Lambda(0)^*e^{-i\kappa^I_{\Lambda}(x)})
\frac{\cosh^2{(k^I_\Lambda(z+h))}}{\cosh^2{(k^I_\Lambda h)}}\left.\right]dz, \\ \\
& = & { \frac{\kappa^I_{\Lambda}|T_\Lambda(0)|^2}{2\cosh^2{(k^I_\Lambda h)}}
\left(\frac{1}{2k^I_\Lambda}\sinh{(2k^I_\Lambda h)} + h\right)},\\ \\
& = & { \frac{\kappa^I_{\Lambda}|T_\Lambda(0)|^2}{2k^I_\Lambda}\left(\tanh{(k^I_\Lambda h)}+\frac{k^I_\Lambda h}{\cosh^2{(k^I_\Lambda h)}}\right)}.
\end{matrix}
</math></center>

= Expanding <math>\mathbf{\xi_3}</math> =

The ice-covered boundary condition for the [[Floating Elastic Plate]] gives
<center><math>
\xi_3 = { \Im\int_{-\infty}^{\infty}\left(\frac{\beta}{\alpha}
\left(\frac{\partial^2}{\partial x^2} - k_y^2\right)^2 - \gamma + \frac{1}{\alpha}\right)
\frac{\partial\phi}{\partial z}\cdot\frac{\partial\phi^*}{\partial z}\bigg|_{z=0}dx.}
</math></center>
Since <math>{\frac{\partial\phi}{\partial z}\cdot\frac{\partial\phi^*}{\partial z}}</math> is real,
<center><math>
\xi_3 = \Im\int_{-\infty}^{\infty}\left(\frac{\beta}{\alpha}\frac{\partial^2}{\partial x^2}
\left(\frac{\partial^2}{\partial x^2} - 2k_y^2\right)^2 \right)
\frac{\partial\phi}{\partial z}\cdot\frac{\partial\phi^*}{\partial z}\bigg|_{z=0}dx.
</math></center>
Integration by parts gives
<center><math>
\xi_3 = \Im\left\{\left[\frac{\beta}{\alpha}\frac{\partial}{\partial x}
\left(\frac{\partial^2}{\partial x^2}-2k_y^2\right)
\frac{\partial\phi}{\partial z}\cdot\frac{\partial\phi^*}{\partial z}\right]_{-\infty}^{\infty}
-\int_{-\infty}^{\infty}\frac{\beta}{\alpha}\frac{\partial}{\partial x}
\left(\frac{\partial^2}{\partial x^2} - 2k_y^2\right)
\frac{\partial\phi}{\partial z}\cdot\frac
{\partial}{\partial x}\frac{\partial\phi^*}{\partial z}dx
\right\}.
</math></center>
As <math>{2k_y^2\frac{\partial}{\partial x}\frac{\partial\phi}{\partial z}\cdot \frac{\partial}{\partial x}\frac{\partial\phi^*}{\partial z}}</math> is real and by integration by parts, the expression of <math>\xi_3</math> becomes,
<center><math>
\xi_3 =\Im\left\{
\left[\frac{\beta}{\alpha}\frac{\partial}{\partial x}
\left(\frac{\partial^2}{\partial x^2}-2k_y^2\right)
\frac{\partial\phi}{\partial z}\cdot\frac{\partial\phi^*}{\partial z}\right]_{-\infty}^{\infty}
- \left[\frac{\beta}{\alpha}\frac{\partial^2}{\partial x^2}\frac{\partial\phi}{\partial z}\cdot
\frac{\partial}{\partial x}\frac{\partial\phi^*}{\partial z}\right]_{-\infty}^{\infty}
+ \int_{-\infty}^\infty\frac{\beta}{\alpha}\frac{\partial^2}{\partial x^2}\frac{\partial\phi}{\partial z}\cdot
\frac{\partial^2}{\partial x^2}\frac{\partial\phi^*}{\partial z}dx\right\}.
</math></center>
As <math>{\frac{\partial^2}{\partial x^2}\frac{\partial\phi}{\partial z}\cdot \frac{\partial^2}{\partial x^2}\frac{\partial\phi^*}{\partial z}}</math> is real, we obtain the new expression of <math>\xi_3</math>
<center><math>
\xi_3 = \Im\left\{
\left[\frac{\beta}{\alpha}\frac{\partial}{\partial x}
\left(\frac{\partial^2}{\partial x^2}-2k_y^2\right)
\frac{\partial\phi}{\partial z}\cdot\frac{\partial\phi^*}{\partial z}\right]_{-\infty}^{\infty}-
\left[\frac{\beta}{\alpha}\frac{\partial^2}{\partial x^2}\frac{\partial\phi}{\partial z}\cdot
\frac{\partial}{\partial x}\frac{\partial\phi^*}{\partial z}\right]_{-\infty}^{\infty}\right\}.
</math></center>

Now breaking <math>\xi_3</math> down, we can simplify the left hend term for <math>x>0</math>

<center><math>
\begin{matrix}
\frac{\partial}{\partial x}\left(\frac{\partial^2}{\partial x^2}-2k_y^2\right)
\frac{\partial\phi(x_2,0)}{\partial z}\cdot\frac{\partial\phi(x_2,0)^*}{\partial z} \\ \\
= \left((-i\kappa^I_{\Lambda})^3k^I_\Lambda T_\Lambda(0)e^{-i\kappa^I_{\Lambda}(x)}\tanh{(k^I_\Lambda h)}\right.
\left.- 2k_y^2(-i\kappa_{\Lambda}(0))k^I_\Lambda T_\Lambda(0)e^{-i\kappa^I_{\Lambda}(x)} \tanh{(k^I_\Lambda h)}\right)
\left(k^I_\Lambda T_\Lambda(0)^*e^{i\kappa^I_{\Lambda}(x)}\tanh{(k^I_\Lambda h)}\right), \\ \\
= i\kappa^I_{\Lambda} (k^I_\Lambda)^2\left((\kappa^I_{\Lambda})^2 + 2k_y^2 \right) \tanh^2{(k^I_\Lambda h)}|T_\Lambda(0)|^2,
\end{matrix}
</math></center>

and for <math>x<0</math>

<center><math>
\begin{matrix}
\frac{\partial}{\partial x}\left(\frac{\partial^2}{\partial x^2}-2k_y^2\right)\frac{\partial\phi(x_1,0)}{\partial z}\cdot
\frac{\partial\phi(x_1,0)^*}{\partial z} \\ \\
= \left[i(\kappa^I_{1})^3k^I_1\left(e^{-i\kappa^I_{1}(x)}-R_1(0)e^{i\kappa^I_{1}(x)}\right) \tanh{(k^I_1 h)}
- 2ik_y^2\kappa^I_{1}k^I_1\left(-e^{-i\kappa^I_{1}(x)} + R_1(0)e^{i\kappa^I_{1}(x)}\right) \tanh{(k^I_1h)}\right]
\left[k^I_1\left(e^{i\kappa^I_{1}(x)} + R_1(0)^*e^{-i\kappa^I_{1}(x))}\right) \tanh{(k^I_1h)}\right], \\ \\
= \left[i\kappa^I_{1}k_1(0)((\kappa^I_{1})^2 + 2k_y^2)\left(e^{-i\kappa^I_{1}(x)} - R_1(0)e^{i\kappa^I_{1}(x)}\right)
\tanh{(k^I_1h)}\right]
\left[k^I_1\left(e^{i\kappa^I_{1}(x)} + R_1(0)^*e^{-i\kappa^I_{1}(x))}\right) \tanh{(k^I_1h)}\right], \\ \\
= i\kappa^I_{1}(k^I_1)^2\left((\kappa^I_{1})^2 + 2k_y^2\right)\tanh^2{(k^I_1h)}\left(1 - |R_1(0)|^2\right),
\end{matrix}
</math></center>

Likewise we expand the right hend term for <math>x>0</math>

<center><math>
\begin{matrix}
\frac{\partial^2}{\partial x^2}\frac{\partial\phi(x_2,0)}{\partial z}\cdot
\frac{\partial}{\partial x}\frac{\partial\phi^*}{\partial z} \\ \\
= \left[-(\kappa^I_{\Lambda})^2k^I_\Lambda T_\Lambda(0)e^{-i\kappa^I_{\Lambda}(x)} \tanh{(k^I_\Lambda h)}\right]
\left[i\kappa^I_{\Lambda} k^I_\Lambda T_\Lambda(0)^*e^{i\kappa^I_{\Lambda}(x)} \tanh{(k^I_\Lambda h)}\right], \\ \\
= -i(\kappa^I_{\Lambda})^3(k^I_\Lambda)^2\tanh^2{(k^I_\Lambda h)}|T_\Lambda(0)|^2,
\end{matrix}
</math></center>

and finally for <math>x<0</math>,

<center><math>
\begin{matrix}
\frac{\partial^2}{\partial x^1}\frac{\partial\phi(x_1,0)}{\partial z}\cdot
\frac{\partial}{\partial x}\frac{\partial\phi^*}{\partial z}\\ \\
= \left[-(\kappa^I_{1})^2k^I_1\left(e^{-i\kappa^I_{1}(x)} + R_1(0)e^{i\kappa^I_{1}(x)}\right)\tanh{(k^I_1 h)}\right]
\left[i\kappa^I_{1} k^I_1\left(e^{i\kappa^I_{1}(x)} - R_1(0)e^{-i\kappa^I_{1}(x)}\right)\tanh{(k^I_1 h)}\right], \\ \\
= -i(\kappa^I_{1})^3(k^I_1)^2\tanh^2{(k^I_1h)}\left(1 - |R_1(0)|^2\right).
\end{matrix}
</math></center>

We can now express <math>\xi_3</math> as

<center><math>
\begin{matrix}
\xi_3 & = & \Im
\left\{\frac{\beta_\Lambda}{\alpha}\left[i\kappa^I_{\Lambda} (k^I_\Lambda)^2\left((\kappa^I_{\Lambda})^2 + 2k_y^2 \right)
\tanh^2{(k^I_\Lambda h)}|T_\Lambda(0)|^2\right]
-\frac{\beta_1}{\alpha}\left[i\kappa^I_{1}(k^I_1)^2\left((\kappa^I_{1})^2 + 2k_y^2\right)\tanh^2{(k^I_1h)}
\left(1 - |R_1(0)|^2\right)\right]
-\frac{\beta_\Lambda}{\alpha}\left[-i(\kappa^I_{\Lambda})^3(k^I_\Lambda)^2\tanh^2{(k^I_\Lambda h)}|T_\Lambda(0)|^2\right]
+\frac{\beta_1}{\alpha}\left[-i(\kappa^I_{1})^3(k^I_1)^2\tanh^2{(k^I_1h)} \left(1 - |R_1(0)|^2\right)\right]\right\}, \\ \\
& = & \frac{\beta_\Lambda}{\alpha}\left[2\kappa^I_{\Lambda} (k^I_\Lambda)^2((\kappa^I_{\Lambda})^2 + k_y^2)\tanh^2{(k^I_\Lambda h)}|T_\Lambda(0)|^2\right]
-\frac{\beta_1}{\alpha}\left[2\kappa^I_{1}(k^I_1)^2((\kappa^I_{1})^2 + k_y^2)\tanh^2{(k^I_1h)}(1-|R_1(0)|^2)\right].
\end{matrix}
</math></center>

= Solving the Energy Balance Equation =

Pulling it all together, we finally obtain
<center><math>
\frac{\beta_\Lambda}{\alpha}\left[2\kappa^I_{\Lambda}(k^I_\Lambda)^2((\kappa^I_{\Lambda})^2 +k_y^2)
\tanh^2{(k^I_\Lambda h)}|T_\Lambda(0)|^2\right]
-\frac{\beta_1}{\alpha}\left[2\kappa^I_{1}(k^I_1)^2((\kappa^I_{1})^2 +k_y^2)\tanh^2{(k^I_1h)}(1-|R_1(0)|^2)\right]
+ \frac{\kappa^I_{\Lambda}|T_\Lambda(0)|^2}{2k^I_\Lambda}\left(\tanh{(k^I_\Lambda h)}+
\frac{k^I_\Lambda h}{\cosh^2{(k^I_\Lambda h)}}\right)
- \frac{\kappa^I_{1}\left(1 - |R_1(0)|^2\right)}{2k^I_1}\left(\tanh{(k^I_1h)}+\frac{hk^I_1}{\cosh^2{(k^I_1h)}}\right)=0.
</math></center>

Re-arranging gives

<center><math>
\kappa^I_{\Lambda}\tanh{(k^I_\Lambda h)}
\left(\frac{\beta_\Lambda}{\alpha}2(k^I_\Lambda)^2((\kappa^I_{\Lambda})^2 +k_y^2)\tanh{(k^I_\Lambda h)}
+ \frac{1}{2k^I_\Lambda}+\frac{h}{2\sinh{(k^I_\Lambda h)}\cosh{(k^I_\Lambda h)}}\right)|T_\Lambda(0)|^2
\quad - \kappa^I_{1}\tanh{(k^I_1 h)}
\left(\frac{\beta_1}{\alpha}2(k^I_1)^2((\kappa^I_{1})^2 +k_y^2)\tanh{(k^I_1 h)}
+ \frac{1}{2it}+\frac{h}{2\sinh{(k^I_1 h)}\cosh{(k^I_1 h)}}\right)(1-|R_1(0)|^2)=0
</math></center>

which can be expressed as
<center><math>
D |T_{\Lambda}(0)|^2 + |R_{1}(0)|^2 = 1,
</math></center>
where <math>D</math> is given by
<center><math>
D = \left(\frac{\kappa^I_{\Lambda} k^I_1\cosh^2{(k^I_1h)}}{\kappa^I_{1}k^I_\Lambda\cosh^2{(k^I_\Lambda h)}}\right)
\left(\frac{\frac{\beta_\Lambda}{\alpha}4(k^I_\Lambda)^3((\kappa^I_{\Lambda})^2 +k_y^2)\sinh^2{(k^I_\Lambda h)} +
\frac{1}{2}{\sinh{(2k^I_\Lambda h)}}+k^I_\Lambda h}
{\frac{\beta_1}{\alpha}4(k^I_1)^3((\kappa^I_{1})^2 +k_y^2)\sinh^2{(k^I_1h)} +
\frac{1}{2}{\sinh{(2k^I_1h)}}+k^I_1h}\right)
</math></center>

[[Category:Floating Elastic Plate]]

Eigenfunction Matching for a Semi-Infinite Floating Elastic Plate

2008-08-22T02:50:03Z

Fmontiel: /* Equations */

=Introduction=

We show here a solution for a semi-infinite [[:Category:Floating Elastic Plate|Floating Elastic Plate]] on [[Finite Depth]].
The problem was solved by [[Fox and Squire 1994]] but the solution method here is slightly different.
The simpler theory for a [[Eigenfunction Matching for a Semi-Infinite Dock|Dock]] describes
many of the ideas here in more detail.

[[Image:Semiinfinite plate.jpg|thumb|right|300px|Wave scattering by a submerged semi-infinite elastic plate]]

= Equations =

We consider the problem of small-amplitude waves which are incident on a semi-infinite floating elastic
plate occupying water surface for <math>x>0</math>. The submergence of the plate is considered negligible.
We assume that the problem is invariant in the <math>y</math> direction, although we allow the waves to be
incident from an angle.
We also assume that the plate edges are free to move at
each boundary, although other boundary conditions could easily be considered using
the methods of solution presented here. We begin with the [[Frequency Domain Problem]] for a semi-infinite
[[Floating Elastic Plate|Floating Elastic Plates]]
in the non-dimensional form of [[Tayler_1986a|Tayler 1986]] ([[Dispersion Relation for a Floating Elastic Plate]]).
We also assume that the waves are normally incident (incidence at an angle will be discussed later).
<center><math>\begin{matrix}
\Delta \phi = 0, \;\;\; -h < z \leq 0,
\end{matrix}</math></center>
<center><math>
\partial_z \phi = 0, \;\;\; z = - h,
</math></center>
<center><math>
\partial_z\phi=\alpha\phi, \,\, z=0,\,x<0,
</math></center>
<center><math>
\beta \partial_x^4\partial_z \phi
- \left( \gamma\alpha - 1 \right) \partial_z \phi = \alpha\phi, \;\;
z = 0, \;\;\; x \geq 0,
</math></center>
where <math>\alpha = \omega^2</math>, <math>\beta</math> and <math>\gamma</math>
are the stiffness and mass constant for the plate respectively. The free edge conditions
at the edge of the plate imply
<center><math>
\partial_x^3 \partial_z\phi = 0, \;\; z = 0, \;\;\; x = 0,
</math></center>
<center><math>
\partial_x^2 \partial_z\phi = 0, \;\; z = 0, \;\;\; x = 0,
</math></center>

=Method of solution=

==Eigenfunction expansion==

We will solve the system of equations using an [[:Category:Eigenfunction Matching Method|Eigenfunction Matching Method]].
The method was developed by [[Fox and Squire 1994]]. The separation of variables for the left hand region where
there is open water, and the incident wave, is described in [[Eigenfunction Matching for a Semi-Infinite Dock]] and we consider
here only the separation of variables in the plate covered region.

===Separation of variables under the Plate===
The potential velocity can be written in terms of an infinite series of separated eigenfunctions under
each elastic plate, of the form
<center><math>
\psi = \frac{\cos(\kappa (z+h))}{\cos\kappa h}.\,
</math></center>
If we apply the boundary conditions given
we obtain the [[Dispersion Relation for a Floating Elastic Plate]]
<center><math>
\kappa \tan{(\kappa h)}= -\frac{\alpha}{\beta \kappa^{4} + 1 - \alpha\gamma}
</math></center>
Solving for <math>\kappa</math> gives a pure imaginary root
with positive imaginary part, two complex roots (two complex conjugate paired roots
with positive imaginary part in most physical situations), an infinite number of positive real roots
which approach <math>{n\pi}/{h}</math> as <math>n</math> approaches infinity, and also the negative of all
these roots ([[Dispersion Relation for a Floating Elastic Plate]]) . We denote the two complex roots with positive imaginary part
by <math>\kappa_(-2)</math> and <math>\kappa_(-1)</math>, the purely imaginary
root with positive imaginary part by <math>\kappa_{0}</math> and the real roots with positive imaginary part
by <math>\kappa_(n)</math> for <math>n</math> a positive integer.
The imaginary root with positive imaginary part corresponds to a
reflected travelling mode propagating along the <math>x</math> axis.
The complex roots with positive imaginary parts correspond to damped reflected travelling modes and the real roots correspond to reflected evanescent modes.

==Expressions for the potential velocity==

We now expand the potential in the two regions using the separation of variables solution.
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.
The potential <math>\phi</math> can now be expressed as the following sum of eigenfunctions:
be expanded as
<center>
<math>
\phi(x,z)= e^{-k_{0}x}\phi_{0}(z) + \sum_{m=0}^{\infty}a_{m}e^{k_{m}x}\phi_{m}(z), \;\;x<0
</math>
</center>
and
<center>
<math>
\phi(x,z)=\sum_{m=-2}^{\infty}b_{m}
e^{-\kappa_{m}x}\psi_{m}(z), \;\;x>0
</math>
</center>
where <math>a_{m}</math> and <math>b_{m}</math> are the coefficients to be determined.

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>

==An infinite dimensional system of equations==

The potential and its derivative must be continuous across the
transition from open water to the plate covered region. Therefore, the
potentials and their derivatives at <math>x=0</math> have to be equal.
We also truncate the sum at <math>N</math> being careful that we have
two extra modes on the plate covered region to satisfy the edge conditions.
We obtain
<center>
<math>
\phi_{0}\left( z\right) + \sum_{m=0}^{N}
a_{m} \phi_{m}\left( z\right)
=\sum_{m=-2}^{N}b_{m}\psi_{m}(z)
</math>
</center>
and
<center>
<math>
-k_{0}\phi_{0}\left( z\right) +\sum
_{m=0}^{N} a_{m}k_{m}\phi_{m}\left( z\right)
=-\sum_{m=-2}^{N}b_{m}\kappa_{m}\psi
_{m}(z)
</math>
</center>
for each <math>N</math>.
We solve these equations by multiplying both equations by
<math>\phi_{l}(z)\,</math> and integrating from <math>-h</math> to <math>0</math> to obtain:
<center>
<math>
A_{0}\delta_{0l}+a_{l}A_{l}
=\sum_{m=-2}^{N}b_{m}B_{ml},\,\,0 \leq l \leq N
</math>
</center>
and
<center>
<math>
-k_{0}A_{0}\delta_{0l}+a_{l}k_{l}A_l
=-\sum_{m=-2}^{N}b_{m}\kappa_{m}B_{ml},\,\,0 \leq l \leq N
</math>
</center>
If we multiply the first equation by <math>k_l</math> and subtract the second equation
we obtain
<center>
<math>
2k_{0}A_{0}\delta_{0l}
=\sum_{m=-2}^{N}b_{m}(k_l + \kappa_{m})B_{ml}
</math>
</center>
Finally, we need to apply the conditions at the edge of the plate to give us two further equations,
<center>
<math>
\partial_x^2\partial_z\phi = - \sum_{m=-2}^{N}b_{m} \kappa_m^3 \tan\kappa_m h = 0
</math>
</center>
and
<center>
<math>
\partial_x^3\partial_z\phi = \sum_{m=-2}^{N}b_{m} \kappa_m^4 \tan\kappa_m h = 0
</math>
</center>

=Numerical Solution=

To solve the system of equations previously defined we set the upper limit of <math>l</math> to
be <math>N</math>, as we already said before. In terms of matrix, we obtain
<center>
<math>
\begin{bmatrix}
\begin{bmatrix}
A_0&0 \quad \cdots&0\\
0&&\\
\vdots&A_l&\vdots\\
&&0\\
0&\cdots \quad 0 &A_N
\end{bmatrix}
&
\begin{bmatrix}
-B_{-20}&-B_{-10}\\
&\\
\vdots&\vdots\\
&\\
-B_{-2N}&-B_{-1N}
\end{bmatrix}
&
\begin{bmatrix}
-B_{00}&\cdots&-B_{0N}\\
&&\\
\vdots&-B_{ml}&\vdots\\
&&\\
-B_{N0}&\cdots&-B_{NN}
\end{bmatrix}
\\
\begin{bmatrix}
0&&\cdots&&0\\
0&&\cdots&&0
\end{bmatrix}
&
\begin{bmatrix}
\kappa_{-2}^3\tan\kappa_{-2}h&\kappa_{-1}^3\tan\kappa_{-1}h\\
\kappa_{-2}^4\tan\kappa_{-2}h&\kappa_{-1}^4\tan\kappa_{-1}h
\end{bmatrix}
&
\begin{bmatrix}
\kappa_0^3\tan\kappa_0h&\cdots&\kappa_N^3\tan\kappa_Nh\\
\kappa_0^4\tan\kappa_0h&\cdots&\kappa_N^4\tan\kappa_Nh
\end{bmatrix}
\\
\begin{bmatrix}
0&&\cdots&&0\\
&&&&\\
\vdots&&\ddots&&\vdots\\
&&&&\\
0&&\cdots&&0
\end{bmatrix}
&
\begin{bmatrix}
(k_0+\kappa_{-2})B_{-20}&(k_0+\kappa_{-1})B_{-10}\\
&\\
\vdots&\vdots\\
&\\
(k_N+\kappa_{-2})B_{-2N}&(k_N+\kappa_{-1})B_{-1N}
\end{bmatrix}
&
\begin{bmatrix}
(k_0 + \kappa_0) \, B_{00}&\cdots&(k_N + \kappa_{0}) \, B_{0N}\\
&&\\
\vdots&(k_l + \kappa_{m}) \, B_{ml}&\vdots\\
&&\\
(k_0 + \kappa_N) \, B_{N0}&\cdots&(k_N + \kappa_{N}) \, B_{NN}\\
\end{bmatrix}
\end{bmatrix}
\begin{bmatrix}
a_{0} \\
\\
\vdots \\
\\
a_N \\
b_{-2}\\
b_{-1}\\
b_{0}\\
\\
\vdots \\
\\
b_N
\end{bmatrix}
=
\begin{bmatrix}
- A_{0} \\
0 \\
\vdots \\
\\
0 \\
0 \\
0 \\
2k_{0}A_{0} \\
0 \\
\vdots \\
\\
0
\end{bmatrix}
</math>
</center>
We then simply need to solve the linear system of equations. Note that we can solve this equation for
<math>b_n</math> first and then solve for <math>a_n</math>

= Waves Incident at an Angle =

We can consider the case of [[Waves Incident at an Angle]] <math>\theta</math>. The theory shows that we can introduce
a wavenumber in the <math>y</math>-direction which express <math>k_y=k_0 \sin \theta</math>.

It is shown that the potential can be expanded as
<center>
<math>
\phi(x,z)=e^{-\hat{k}_0x}\phi_0(z)+\sum_{m=0}^{\infty}a_{m}e^{\hat{k}_{m}x}\phi_{m}(z), \;\;x<0
</math>
</center>
and
<center>
<math>
\phi(x,z)=\sum_{m=-2}^{\infty}b_{m}
e^{-\hat{\kappa}_{m}x}\psi_{m}(z), \;\;x>0
</math>
</center>
where <math>\hat{k}_{m} = \sqrt{k_m^2 - k_y^2}</math> and <math>\hat{\kappa}_{m} = \sqrt{\kappa_m^2 - k_y^2}</math>
where we always take the positive real root or the root with positive imaginary part.

The edge conditions are also different and are
<center><math>
\left(\frac{\partial^3}{\partial x^3} - (2 - \nu)k^2_y\frac{\partial}{\partial x}\right) \frac{\partial\phi}{\partial z}= 0,
</math></center>
and
<center><math>
\left(\frac{\partial^2}{\partial x^2} - \nu k^2_y\right)\frac{\partial\phi}{\partial z} = 0,
</math></center>
where <math>\nu</math> is Poisons ratio.

We can expend these edge conditions, which respectively gives
<center>
<math>
-\sum_{m=-2}^{\infty}b_{m}\kappa_m (\hat{\kappa}_m^3-\hat{\kappa}_m k_y^2(2-\nu))\tan \kappa_m h=0
</math>
</center>
and
<center>
<math>
-\sum_{m=-2}^{\infty}b_{m}\kappa_m (\hat{\kappa}_m^2-k_y^2\nu))\tan \kappa_m h=0
</math>
</center>

The equations are derived almost identically to those above and we obtain
<center>
<math>
A_{0}\delta_{0l}+a_{l}A_{l}
=\sum_{m=-2}^{\infty}b_{m}B_{ml}
</math>
</center>
and
<center>
<math>
-\hat{k_{0}}A_{0}\delta_{0l}+a_{l}\hat{k}_{l}A_l
=-\sum_{m=-2}^{\infty}b_{m}\hat{\kappa}_{m}B_{ml}
</math>
</center>
and these are solved exactly as before.

= Energy Balance =

The energy balance equation is ([[Energy Balance for Two Elastic Plates]]) given by
<center><math>
D |T|^2 + |R|^2 = 1, \,
</math></center>
where <math> T=b_0 </math>, <math>R=a_0</math>, and <math>D</math> is given by
<center><math>
D = \left(\frac{\hat{\kappa}_0 k_0\cos^2{(k_0 h)}}{\hat{k}_0 \kappa_0\cos^2{(\kappa_0 h)}}\right)
\left(\frac{\frac{4\beta}{\alpha}(\kappa_0)^3((-\hat{\kappa}_0)^2 +k_y^2)\sin^2{(\kappa_0 h)} +
\frac{1}{2}{\sin{(2 \kappa_0 h)}}+\kappa_0 h}
{
\frac{1}{2}{\sin{(2 k_0 h)}}+k_0h}\right)
</math></center>
where the index <math>0</math> refers to the travelling wave (imaginary root of the dispersion equation).

= Matlab Code =

A program to calculate the coefficients for the semi-infinite dock problems can be found here
[http://www.math.auckland.ac.nz/~meylan/code/eigenfunction_matching/semiinfinite_plate.m semiinfinite_plate.m]

== Additional code ==

This program requires
* {{free surface dispersion equation code}}
* {{elastic plate dispersion equation code}}

[[Category:Floating Elastic Plate]]
[[Category:Eigenfunction Matching Method]]
[[Category:Complete Pages]]

Waves Incident at an Angle

2008-08-22T01:51:16Z

Fmontiel:

We can consider the problem when the waves are incident at an angle <math>\theta</math>. Thus the incident
potential can be expressed as follow:
<center>
<math>
\phi^I=e^{-k_0(\cos \theta x + \sin \theta y)}
</math>
</center>
In this
case the wavenumber in the <math>y</math>-direction is <math>k_y = \sin\theta k_0</math>
where <math>k_0</math> is the travelling mode, the pure imaginary root of the
[[Dispersion Relation for a Free Surface]] (note that <math>k_y</math> is imaginary).

We assume here that the object has infinite length in the <math>y</math>-direction, so the solution
does not vary in that direction except over a period. This means that the potential is now of the
form <math>\phi(x,y,z)=e^{k_y y}X(x)Z(z)</math>. The separation of variables and the application of
Laplace's equation clearly shows that the dependance of <math>z</math> remains unchanged. However,
we obtain
<center>
<math>
\frac{1}{e^{k_yy}X(x)}(k_y^2e^{k_yy}X(x)+e^{k_yy}\frac{d^2X}{dx^2})=k_n^2=-\frac{1}{Z(z)}\frac{d^2Z}{dz^2}
</math>
</center>
where <math>k_n</math> is a root of the dispersion equation. This simplifies as
<center>
<math>
k_y^2+\frac{1}{X}\frac{d^2X}{dx^2}=k_n^2
</math>
</center>
which permits to obtain the following differential equation for <math>X</math>
<center>
<math>
\frac{d^2X}{dx^2}-(k_n^2-k_y^2)X=0
</math>
</center>
If we introduce the new variable <math>k_x^2=k_n^2-k_y^2</math>, we obtain a similar differential equation to solve
in the <math>x</math>-direction as for the null incident angle problem, so that we can develop a solution for
<math>X</math> of the same form that the one with no angle, taking care of replacing the wave number <math>k_n</math>
by <math>\hat{k}_{n} = \sqrt{k_n^2 - k_y^2}</math>.

Waves Incident at an Angle

2008-08-21T01:10:05Z

Fmontiel:

We can consider the problem when the waves are incident at an angle <math>\theta</math>. Thus the incident
potential can be expressed as follow:
<center>
<math>
\phi^I=e^{-k_0(\cos \theta x + \sin \theta y)}
</math>
</center>
In this
case the wavenumber in the <math>y</math>-direction is <math>k_y = \sin\theta k_0</math>
where <math>k_0</math> is the travelling mode, the pure imaginary root of the
[[Dispersion Relation for a Free Surface]] (note that <math>k_y</math> is imaginary).

We assume here that the object has infinite length in the <math>y</math>-direction, so the solution
does not vary in that direction except over a period. This means that the potential is now of the
form <math>\phi(x,y,z)=e^{k_y y}W(x)Z(z)</math>. The separation of variables and the application of
Laplace's equation clearly shows that the dependance of <math>z</math> remains unchanged. However,
we obtain
<center>
<math>
\frac{1}{e^{k_yy}X(x)}(k_y^2e^{k_yy}X(x)+e^{k_yy}\frac{d^2X}{dx^2})=k_n^2=-\frac{1}{Z(z)}\frac{d^2Z}{dz^2}
</math>
</center>
where <math>k_n</math> is a root of the dispersion equation. This simplifies as
<center>
<math>
k_y^2+\frac{1}{X}\frac{d^2X}{dx^2}=k_n^2
</math>
</center>
which permits to obtain the following differential equation for <math>X</math>
<center>
<math>
\frac{d^2X}{dx^2}-(k_n^2-k_y^2)X=0
</math>
</center>
If we introduce the new variable <math>k_x^2=k_n^2-k_y^2</math>, we obtain a similar differential equation to solve
in the <math>x</math>-direction as for the null incident angle problem, so that we can develop a solution for
<math>X</math> of the same form that the one with no angle, taking care of replacing the wave number <math>k_n</math>
by <math>\hat{k}_{n} = \sqrt{k_n^2 - k_y^2}</math>.

Eigenfunction Matching for a Semi-Infinite Floating Elastic Plate

2008-08-21T00:45:35Z

Fmontiel: /* Waves Incident at an Angle */

=Introduction=

We show here a solution for a semi-infinite [[:Category:Floating Elastic Plate|Floating Elastic Plate]] on [[Finite Depth]].
The problem was solved by [[Fox and Squire 1994]] but the solution method here is slightly different.
The simpler theory for a [[Eigenfunction Matching for a Semi-Infinite Dock|Dock]] describes
many of the ideas here in more detail.

[[Image:Semiinfinite plate.jpg|thumb|right|300px|Wave scattering by a submerged semi-infinite elastic plate]]

= Equations =

We consider the problem of small-amplitude waves which are incident on a semi-infinite floating elastic
plate occupying water surface for <math>x>0</math>. The submergence of the plate is considered negligible.
We assume that the problem is invariant in the <math>y</math> direction, although we allow the waves to be
incident from an angle.
We also assume that the plate edges are free to move at
each boundary, although other boundary conditions could easily be considered using
the methods of solution presented here. We begin with the [[Frequency Domain Problem]] for a semi-infinite
[[Floating Elastic Plate|Floating Elastic Plates]]
in the non-dimensional form of [[Tayler_1986a|Tayler 1986]] ([[Dispersion Relation for a Floating Elastic Plate]]).
We also assume that the waves are normally incident (incidence at an angle will be discussed later).
<center><math>\begin{matrix}
\Delta \phi = 0, \;\;\; -h < z \leq 0,
\end{matrix}</math></center>
<center><math>
\partial_z \phi = 0, \;\;\; z = - h,
</math></center>
<center><math>
\partial_z\phi=\alpha\phi, \,\, z=0,\,x<0,
</math></center>
<center><math>
\beta \partial_x^4\partial_z \phi
- \left( \gamma\alpha - 1 \right) \partial_z \phi + \alpha\phi = 0, \;\;
z = 0, \;\;\; x \geq 0,
</math></center>
where <math>\alpha = \omega^2</math>, <math>\beta</math> and <math>\gamma</math>
are the stiffness and mass constant for the plate respectively. The free edge conditions
at the edge of the plate imply
<center><math>
\partial_x^3 \partial_z\phi = 0, \;\; z = 0, \;\;\; x = 0,
</math></center>
<center><math>
\partial_x^2 \partial_z\phi = 0, \;\; z = 0, \;\;\; x = 0,
</math></center>

=Method of solution=

==Eigenfunction expansion==

We will solve the system of equations using an [[:Category:Eigenfunction Matching Method|Eigenfunction Matching Method]].
The method was developed by [[Fox and Squire 1994]]. The separation of variables for the left hand region where
there is open water, and the incident wave, is described in [[Eigenfunction Matching for a Semi-Infinite Dock]] and we consider
here only the separation of variables in the plate covered region.

===Separation of variables under the Plate===
The potential velocity can be written in terms of an infinite series of separated eigenfunctions under
each elastic plate, of the form
<center><math>
\psi = \frac{\cos(\kappa (z+h))}{\cos\kappa h}.\,
</math></center>
If we apply the boundary conditions given
we obtain the [[Dispersion Relation for a Floating Elastic Plate]]
<center><math>
\kappa \tan{(\kappa h)}= -\frac{\alpha}{\beta \kappa^{4} + 1 - \alpha\gamma}
</math></center>
Solving for <math>\kappa</math> gives a pure imaginary root
with positive imaginary part, two complex roots (two complex conjugate paired roots
with positive imaginary part in most physical situations), an infinite number of positive real roots
which approach <math>{n\pi}/{h}</math> as <math>n</math> approaches infinity, and also the negative of all
these roots ([[Dispersion Relation for a Floating Elastic Plate]]) . We denote the two complex roots with positive imaginary part
by <math>\kappa_(-2)</math> and <math>\kappa_(-1)</math>, the purely imaginary
root with positive imaginary part by <math>\kappa_{0}</math> and the real roots with positive imaginary part
by <math>\kappa_(n)</math> for <math>n</math> a positive integer.
The imaginary root with positive imaginary part corresponds to a
reflected travelling mode propagating along the <math>x</math> axis.
The complex roots with positive imaginary parts correspond to damped reflected travelling modes and the real roots correspond to reflected evanescent modes.

==Expressions for the potential velocity==

We now expand the potential in the two regions using the separation of variables solution.
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.
The potential <math>\phi</math> can now be expressed as the following sum of eigenfunctions:
be expanded as
<center>
<math>
\phi(x,z)= e^{-k_{0}x}\phi_{0}(z) + \sum_{m=0}^{\infty}a_{m}e^{k_{m}x}\phi_{m}(z), \;\;x<0
</math>
</center>
and
<center>
<math>
\phi(x,z)=\sum_{m=-2}^{\infty}b_{m}
e^{-\kappa_{m}x}\psi_{m}(z), \;\;x>0
</math>
</center>
where <math>a_{m}</math> and <math>b_{m}</math> are the coefficients to be determined.

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>

==An infinite dimensional system of equations==

The potential and its derivative must be continuous across the
transition from open water to the plate covered region. Therefore, the
potentials and their derivatives at <math>x=0</math> have to be equal.
We also truncate the sum at <math>N</math> being careful that we have
two extra modes on the plate covered region to satisfy the edge conditions.
We obtain
<center>
<math>
\phi_{0}\left( z\right) + \sum_{m=0}^{N}
a_{m} \phi_{m}\left( z\right)
=\sum_{m=-2}^{N}b_{m}\psi_{m}(z)
</math>
</center>
and
<center>
<math>
-k_{0}\phi_{0}\left( z\right) +\sum
_{m=0}^{N} a_{m}k_{m}\phi_{m}\left( z\right)
=-\sum_{m=-2}^{N}b_{m}\kappa_{m}\psi
_{m}(z)
</math>
</center>
for each <math>N</math>.
We solve these equations by multiplying both equations by
<math>\phi_{l}(z)\,</math> and integrating from <math>-h</math> to <math>0</math> to obtain:
<center>
<math>
A_{0}\delta_{0l}+a_{l}A_{l}
=\sum_{m=-2}^{N}b_{m}B_{ml},\,\,0 \leq l \leq N
</math>
</center>
and
<center>
<math>
-k_{0}A_{0}\delta_{0l}+a_{l}k_{l}A_l
=-\sum_{m=-2}^{N}b_{m}\kappa_{m}B_{ml},\,\,0 \leq l \leq N
</math>
</center>
If we multiply the first equation by <math>k_l</math> and subtract the second equation
we obtain
<center>
<math>
2k_{0}A_{0}\delta_{0l}
=\sum_{m=-2}^{N}b_{m}(k_l + \kappa_{m})B_{ml}
</math>
</center>
Finally, we need to apply the conditions at the edge of the plate to give us two further equations,
<center>
<math>
\partial_x^2\partial_z\phi = - \sum_{m=-2}^{N}b_{m} \kappa_m^3 \tan\kappa_m h = 0
</math>
</center>
and
<center>
<math>
\partial_x^3\partial_z\phi = \sum_{m=-2}^{N}b_{m} \kappa_m^4 \tan\kappa_m h = 0
</math>
</center>

=Numerical Solution=

To solve the system of equations previously defined we set the upper limit of <math>l</math> to
be <math>N</math>, as we already said before. In terms of matrix, we obtain
<center>
<math>
\begin{bmatrix}
\begin{bmatrix}
A_0&0 \quad \cdots&0\\
0&&\\
\vdots&A_l&\vdots\\
&&0\\
0&\cdots \quad 0 &A_N
\end{bmatrix}
&
\begin{bmatrix}
-B_{-20}&-B_{-10}\\
&\\
\vdots&\vdots\\
&\\
-B_{-2N}&-B_{-1N}
\end{bmatrix}
&
\begin{bmatrix}
-B_{00}&\cdots&-B_{0N}\\
&&\\
\vdots&-B_{ml}&\vdots\\
&&\\
-B_{N0}&\cdots&-B_{NN}
\end{bmatrix}
\\
\begin{bmatrix}
0&&\cdots&&0\\
0&&\cdots&&0
\end{bmatrix}
&
\begin{bmatrix}
\kappa_{-2}^3\tan\kappa_{-2}h&\kappa_{-1}^3\tan\kappa_{-1}h\\
\kappa_{-2}^4\tan\kappa_{-2}h&\kappa_{-1}^4\tan\kappa_{-1}h
\end{bmatrix}
&
\begin{bmatrix}
\kappa_0^3\tan\kappa_0h&\cdots&\kappa_N^3\tan\kappa_Nh\\
\kappa_0^4\tan\kappa_0h&\cdots&\kappa_N^4\tan\kappa_Nh
\end{bmatrix}
\\
\begin{bmatrix}
0&&\cdots&&0\\
&&&&\\
\vdots&&\ddots&&\vdots\\
&&&&\\
0&&\cdots&&0
\end{bmatrix}
&
\begin{bmatrix}
(k_0+\kappa_{-2})B_{-20}&(k_0+\kappa_{-1})B_{-10}\\
&\\
\vdots&\vdots\\
&\\
(k_N+\kappa_{-2})B_{-2N}&(k_N+\kappa_{-1})B_{-1N}
\end{bmatrix}
&
\begin{bmatrix}
(k_0 + \kappa_0) \, B_{00}&\cdots&(k_N + \kappa_{0}) \, B_{0N}\\
&&\\
\vdots&(k_l + \kappa_{m}) \, B_{ml}&\vdots\\
&&\\
(k_0 + \kappa_N) \, B_{N0}&\cdots&(k_N + \kappa_{N}) \, B_{NN}\\
\end{bmatrix}
\end{bmatrix}
\begin{bmatrix}
a_{0} \\
\\
\vdots \\
\\
a_N \\
b_{-2}\\
b_{-1}\\
b_{0}\\
\\
\vdots \\
\\
b_N
\end{bmatrix}
=
\begin{bmatrix}
- A_{0} \\
0 \\
\vdots \\
\\
0 \\
0 \\
0 \\
2k_{0}A_{0} \\
0 \\
\vdots \\
\\
0
\end{bmatrix}
</math>
</center>
We then simply need to solve the linear system of equations. Note that we can solve this equation for
<math>b_n</math> first and then solve for <math>a_n</math>

= Waves Incident at an Angle =

We can consider the case of [[Waves Incident at an Angle]] <math>\theta</math>. The theory shows that we can introduce
a wavenumber in the <math>y</math>-direction which express <math>k_y=k_0 \sin \theta</math>.

It is shown that the potential can be expanded as
<center>
<math>
\phi(x,z)=e^{-\hat{k}_0x}\phi_0(z)+\sum_{m=0}^{\infty}a_{m}e^{\hat{k}_{m}x}\phi_{m}(z), \;\;x<0
</math>
</center>
and
<center>
<math>
\phi(x,z)=\sum_{m=-2}^{\infty}b_{m}
e^{-\hat{\kappa}_{m}x}\psi_{m}(z), \;\;x>0
</math>
</center>
where <math>\hat{k}_{m} = \sqrt{k_m^2 - k_y^2}</math> and <math>\hat{\kappa}_{m} = \sqrt{\kappa_m^2 - k_y^2}</math>
where we always take the positive real root or the root with positive imaginary part.

The edge conditions are also different and are
<center><math>
\left(\frac{\partial^3}{\partial x^3} - (2 - \nu)k^2_y\frac{\partial}{\partial x}\right) \frac{\partial\phi}{\partial z}= 0,
</math></center>
and
<center><math>
\left(\frac{\partial^2}{\partial x^2} - \nu k^2_y\right)\frac{\partial\phi}{\partial z} = 0,
</math></center>
where <math>\nu</math> is Poisons ratio.

We can expend these edge conditions, which respectively gives
<center>
<math>
-\sum_{m=-2}^{\infty}b_{m}\kappa_m (\hat{\kappa}_m^3-\hat{\kappa}_m k_y^2(2-\nu))\tan \kappa_m h=0
</math>
</center>
and
<center>
<math>
-\sum_{m=-2}^{\infty}b_{m}\kappa_m (\hat{\kappa}_m^2-k_y^2\nu))\tan \kappa_m h=0
</math>
</center>

The equations are derived almost identically to those above and we obtain
<center>
<math>
A_{0}\delta_{0l}+a_{l}A_{l}
=\sum_{m=-2}^{\infty}b_{m}B_{ml}
</math>
</center>
and
<center>
<math>
-\hat{k_{0}}A_{0}\delta_{0l}+a_{l}\hat{k}_{l}A_l
=-\sum_{m=-2}^{\infty}b_{m}\hat{\kappa}_{m}B_{ml}
</math>
</center>
and these are solved exactly as before.

= Energy Balance =

The energy balance equation is ([[Energy Balance for Two Elastic Plates]]) given by
<center><math>
D |T|^2 + |R|^2 = 1, \,
</math></center>
where <math> T=b_0 </math>, <math>R=a_0</math>, and <math>D</math> is given by
<center><math>
D = \left(\frac{\hat{\kappa}_0 k_0\cos^2{(k_0 h)}}{\hat{k}_0 \kappa_0\cos^2{(\kappa_0 h)}}\right)
\left(\frac{\frac{4\beta}{\alpha}(\kappa_0)^3((-\hat{\kappa}_0)^2 +k_y^2)\sin^2{(\kappa_0 h)} +
\frac{1}{2}{\sin{(2 \kappa_0 h)}}+\kappa_0 h}
{
\frac{1}{2}{\sin{(2 k_0 h)}}+k_0h}\right)
</math></center>
where the index <math>0</math> refers to the travelling wave (imaginary root of the dispersion equation).

= Matlab Code =

A program to calculate the coefficients for the semi-infinite dock problems can be found here
[http://www.math.auckland.ac.nz/~meylan/code/eigenfunction_matching/semiinfinite_plate.m semiinfinite_plate.m]

== Additional code ==

This program requires
* {{free surface dispersion equation code}}
* {{elastic plate dispersion equation code}}

[[Category:Floating Elastic Plate]]
[[Category:Eigenfunction Matching Method]]
[[Category:Complete Pages]]

Waves Incident at an Angle

2008-08-21T00:43:47Z

Fmontiel:

We can consider the problem when the waves are incident at an angle <math>\theta</math>. Thus the incident
potential can be expressed as follow:
<center>
<math>
\phi^I=e^{-k_0(\cos \theta x + \sin \theta y)}
</math>
</center>
In this
case the wavenumber in the <math>y</math>-direction is <math>k_y = \sin\theta k_0</math>
where <math>k_0</math> is the travelling mode, the pure imaginary root of the
[[Dispersion Relation for a Free Surface]] (note that <math>k_y</math> is imaginary).

We assume here that the object has infinite length in the <math>y</math>-direction, so the solution
does not vary in that direction except over a period. This means that the potential is now of the
form <math>\phi(x,y,z)=e^{k_y y}W(x)Z(z)</math>. The separation of variables and the application of
Laplace's equation clearly shows that the dependance of <math>z</math> remains unchanged. However,
we obtain
<center>
<math>
\frac{1}{e^{k_yy}X(x)}(k_y^2e^{k_yy}X(x)+e^{k_yy}\frac{d^2X}{dx^2})=k_n^2=-\frac{1}{Z(z)}\frac{d^2Z}{dz^2}
</math>
</center>
where <math>k_n</math> is a root of the dispersion equation. This simplifies as
<center>
<math>
k_y^2+\frac{1}{X}\frac{d^2X}{dx^2}=k_n^2
</math>
</center>
which permits to obtain the followinf differential equation for <math>X</math>
<center>
<math>
\frac{d^2X}{dx^2}-(k_n^2-k_y^2)X=0
</math>
</center>
If we introduce the new variable <math>k_x^2=k_n^2-k_y^2</math>, we obtain a similar differential equation to solve
in the <math>x</math>-direction as for the null incident angle problem, so that we can develop a solution for
<math>X</math> of the same form that the one with no angle, taking care of replacing the wave number <math>k_n</math>
by <math>\hat{k}_{n} = \sqrt{k_n^2 - k_y^2}</math>.

Waves Incident at an Angle

2008-08-21T00:43:10Z

Fmontiel:

We can consider the problem when the waves are incident at an angle <math>\theta</math>. Thus the incident
potential can be expressed as follow:
<center>
<math>
\phi^I=e^{-k_0(\cos \theta x + \sin \theta y)}
</math>
</center>
In this
case the wavenumber in the <math>y</math>-direction is <math>k_y = \sin\theta k_0</math>
where <math>k_0</math> is the travelling mode, the pure imaginary root of the
[[Dispersion Relation For A Free Surface]] (note that <math>k_y</math> is imaginary).

We assume here that the object has infinite length in the <math>y</math>-direction, so the solution
does not vary in that direction except over a period. This means that the potential is now of the
form <math>\phi(x,y,z)=e^{k_y y}W(x)Z(z)</math>. The separation of variables and the application of
Laplace's equation clearly shows that the dependance of <math>z</math> remains unchanged. However,
we obtain
<center>
<math>
\frac{1}{e^{k_yy}X(x)}(k_y^2e^{k_yy}X(x)+e^{k_yy}\frac{d^2X}{dx^2})=k_n^2=-\frac{1}{Z(z)}\frac{d^2Z}{dz^2}
</math>
</center>
where <math>k_n</math> is a root of the dispersion equation. This simplifies as
<center>
<math>
k_y^2+\frac{1}{X}\frac{d^2X}{dx^2}=k_n^2
</math>
</center>
which permits to obtain the followinf differential equation for <math>X</math>
<center>
<math>
\frac{d^2X}{dx^2}-(k_n^2-k_y^2)X=0
</math>
</center>
If we introduce the new variable <math>k_x^2=k_n^2-k_y^2</math>, we obtain a similar differential equation to solve
in the <math>x</math>-direction as for the null incident angle problem, so that we can develop a solution for
<math>X</math> of the same form that the one with no angle, taking care of replacing the wave number <math>k_n</math>
by <math>\hat{k}_{n} = \sqrt{k_n^2 - k_y^2}</math>.

Waves Incident at an Angle

2008-08-21T00:35:24Z

Fmontiel:

We can consider the problem when the waves are incident at an angle <math>\theta</math>. Thus the incident
potential can be expressed as follow:
<center>
<math>
\phi^I=e^{-k_0(\cos \theta x + \sin \theta y)}
</math>
</center>
In this
case the wavenumber in the <math>y</math>-direction is <math>k_y = \sin\theta k_0</math>
where <math>k_0</math> is as defined previously (note that <math>k_y</math> is imaginary).

We assume here that the object has infinite length in the <math>y</math>-direction, so the solution
does not vary in that direction except over a period. This means that the potential is now of the
form <math>\phi(x,y,z)=e^{k_y y}W(x)Z(z)</math>. The separation of variables and the application of
Laplace's equation clearly shows that the dependance of <math>z</math> remains unchanged. However,
we obtain
<center>
<math>
\frac{1}{e^{k_yy}X(x)}(k_y^2e^{k_yy}X(x)+e^{k_yy}\frac{d^2X}{dx^2})=k_n^2=-\frac{1}{Z(z)}\frac{d^2Z}{dz^2}
</math>
</center>
where <math>k_n</math> is a root of the dispersion equation. This simplifies as
<center>
<math>
k_y^2+\frac{1}{X}\frac{d^2X}{dx^2}=k_n^2
</math>
</center>
which permits to obtain the followinf differential equation for <math>X</math>
<center>
<math>
\frac{d^2X}{dx^2}-(k_n^2-k_y^2)X=0
</math>
</center>
If we introduce the new variable <math>k_x^2=k_n^2-k_y^2</math>, we obtain a similar differential equation to solve
in the <math>x</math>-direction as for the null incident angle problem, so that we can develop a solution for
<math>X</math> of the same form that the one with no angle, taking care of replacing the wave number <math>k_n</math>
by <math>\hat{k}_{n} = \sqrt{k_n^2 - k_y^2}</math>.

Eigenfunction Matching for a Semi-Infinite Floating Elastic Plate

2008-08-20T23:45:01Z

Fmontiel: /* Waves Incident at an Angle */

=Introduction=

We show here a solution for a semi-infinite [[:Category:Floating Elastic Plate|Floating Elastic Plate]] on [[Finite Depth]].
The problem was solved by [[Fox and Squire 1994]] but the solution method here is slightly different.
The simpler theory for a [[Eigenfunction Matching for a Semi-Infinite Dock|Dock]] describes
many of the ideas here in more detail.

[[Image:Semiinfinite plate.jpg|thumb|right|300px|Wave scattering by a submerged semi-infinite elastic plate]]

= Equations =

We consider the problem of small-amplitude waves which are incident on a semi-infinite floating elastic
plate occupying water surface for <math>x>0</math>. The submergence of the plate is considered negligible.
We assume that the problem is invariant in the <math>y</math> direction, although we allow the waves to be
incident from an angle.
We also assume that the plate edges are free to move at
each boundary, although other boundary conditions could easily be considered using
the methods of solution presented here. We begin with the [[Frequency Domain Problem]] for a semi-infinite
[[Floating Elastic Plate|Floating Elastic Plates]]
in the non-dimensional form of [[Tayler_1986a|Tayler 1986]] ([[Dispersion Relation for a Floating Elastic Plate]]).
We also assume that the waves are normally incident (incidence at an angle will be discussed later).
<center><math>\begin{matrix}
\Delta \phi = 0, \;\;\; -h < z \leq 0,
\end{matrix}</math></center>
<center><math>
\partial_z \phi = 0, \;\;\; z = - h,
</math></center>
<center><math>
\partial_z\phi=\alpha\phi, \,\, z=0,\,x<0,
</math></center>
<center><math>
\beta \partial_x^4\partial_z \phi
- \left( \gamma\alpha - 1 \right) \partial_z \phi + \alpha\phi = 0, \;\;
z = 0, \;\;\; x \geq 0,
</math></center>
where <math>\alpha = \omega^2</math>, <math>\beta</math> and <math>\gamma</math>
are the stiffness and mass constant for the plate respectively. The free edge conditions
at the edge of the plate imply
<center><math>
\partial_x^3 \partial_z\phi = 0, \;\; z = 0, \;\;\; x = 0,
</math></center>
<center><math>
\partial_x^2 \partial_z\phi = 0, \;\; z = 0, \;\;\; x = 0,
</math></center>

=Method of solution=

==Eigenfunction expansion==

We will solve the system of equations using an [[:Category:Eigenfunction Matching Method|Eigenfunction Matching Method]].
The method was developed by [[Fox and Squire 1994]]. The separation of variables for the left hand region where
there is open water, and the incident wave, is described in [[Eigenfunction Matching for a Semi-Infinite Dock]] and we consider
here only the separation of variables in the plate covered region.

===Separation of variables under the Plate===
The potential velocity can be written in terms of an infinite series of separated eigenfunctions under
each elastic plate, of the form
<center><math>
\psi = \frac{\cos(\kappa (z+h))}{\cos\kappa h}.\,
</math></center>
If we apply the boundary conditions given
we obtain the [[Dispersion Relation for a Floating Elastic Plate]]
<center><math>
\kappa \tan{(\kappa h)}= -\frac{\alpha}{\beta \kappa^{4} + 1 - \alpha\gamma}
</math></center>
Solving for <math>\kappa</math> gives a pure imaginary root
with positive imaginary part, two complex roots (two complex conjugate paired roots
with positive imaginary part in most physical situations), an infinite number of positive real roots
which approach <math>{n\pi}/{h}</math> as <math>n</math> approaches infinity, and also the negative of all
these roots ([[Dispersion Relation for a Floating Elastic Plate]]) . We denote the two complex roots with positive imaginary part
by <math>\kappa_(-2)</math> and <math>\kappa_(-1)</math>, the purely imaginary
root with positive imaginary part by <math>\kappa_{0}</math> and the real roots with positive imaginary part
by <math>\kappa_(n)</math> for <math>n</math> a positive integer.
The imaginary root with positive imaginary part corresponds to a
reflected travelling mode propagating along the <math>x</math> axis.
The complex roots with positive imaginary parts correspond to damped reflected travelling modes and the real roots correspond to reflected evanescent modes.

==Expressions for the potential velocity==

We now expand the potential in the two regions using the separation of variables solution.
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.
The potential <math>\phi</math> can now be expressed as the following sum of eigenfunctions:
be expanded as
<center>
<math>
\phi(x,z)= e^{-k_{0}x}\phi_{0}(z) + \sum_{m=0}^{\infty}a_{m}e^{k_{m}x}\phi_{m}(z), \;\;x<0
</math>
</center>
and
<center>
<math>
\phi(x,z)=\sum_{m=-2}^{\infty}b_{m}
e^{-\kappa_{m}x}\psi_{m}(z), \;\;x>0
</math>
</center>
where <math>a_{m}</math> and <math>b_{m}</math> are the coefficients to be determined.

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>

==An infinite dimensional system of equations==

The potential and its derivative must be continuous across the
transition from open water to the plate covered region. Therefore, the
potentials and their derivatives at <math>x=0</math> have to be equal.
We also truncate the sum at <math>N</math> being careful that we have
two extra modes on the plate covered region to satisfy the edge conditions.
We obtain
<center>
<math>
\phi_{0}\left( z\right) + \sum_{m=0}^{N}
a_{m} \phi_{m}\left( z\right)
=\sum_{m=-2}^{N}b_{m}\psi_{m}(z)
</math>
</center>
and
<center>
<math>
-k_{0}\phi_{0}\left( z\right) +\sum
_{m=0}^{N} a_{m}k_{m}\phi_{m}\left( z\right)
=-\sum_{m=-2}^{N}b_{m}\kappa_{m}\psi
_{m}(z)
</math>
</center>
for each <math>N</math>.
We solve these equations by multiplying both equations by
<math>\phi_{l}(z)\,</math> and integrating from <math>-h</math> to <math>0</math> to obtain:
<center>
<math>
A_{0}\delta_{0l}+a_{l}A_{l}
=\sum_{m=-2}^{N}b_{m}B_{ml},\,\,0 \leq l \leq N
</math>
</center>
and
<center>
<math>
-k_{0}A_{0}\delta_{0l}+a_{l}k_{l}A_l
=-\sum_{m=-2}^{N}b_{m}\kappa_{m}B_{ml},\,\,0 \leq l \leq N
</math>
</center>
If we multiply the first equation by <math>k_l</math> and subtract the second equation
we obtain
<center>
<math>
2k_{0}A_{0}\delta_{0l}
=\sum_{m=-2}^{N}b_{m}(k_l + \kappa_{m})B_{ml}
</math>
</center>
Finally, we need to apply the conditions at the edge of the plate to give us two further equations,
<center>
<math>
\partial_x^2\partial_z\phi = - \sum_{m=-2}^{N}b_{m} \kappa_m^3 \tan\kappa_m h = 0
</math>
</center>
and
<center>
<math>
\partial_x^3\partial_z\phi = \sum_{m=-2}^{N}b_{m} \kappa_m^4 \tan\kappa_m h = 0
</math>
</center>

=Numerical Solution=

To solve the system of equations previously defined we set the upper limit of <math>l</math> to
be <math>N</math>, as we already said before. In terms of matrix, we obtain
<center>
<math>
\begin{bmatrix}
\begin{bmatrix}
A_0&0 \quad \cdots&0\\
0&&\\
\vdots&A_l&\vdots\\
&&0\\
0&\cdots \quad 0 &A_N
\end{bmatrix}
&
\begin{bmatrix}
-B_{-20}&-B_{-10}\\
&\\
\vdots&\vdots\\
&\\
-B_{-2N}&-B_{-1N}
\end{bmatrix}
&
\begin{bmatrix}
-B_{00}&\cdots&-B_{0N}\\
&&\\
\vdots&-B_{ml}&\vdots\\
&&\\
-B_{N0}&\cdots&-B_{NN}
\end{bmatrix}
\\
\begin{bmatrix}
0&&\cdots&&0\\
0&&\cdots&&0
\end{bmatrix}
&
\begin{bmatrix}
\kappa_{-2}^3\tan\kappa_{-2}h&\kappa_{-1}^3\tan\kappa_{-1}h\\
\kappa_{-2}^4\tan\kappa_{-2}h&\kappa_{-1}^4\tan\kappa_{-1}h
\end{bmatrix}
&
\begin{bmatrix}
\kappa_0^3\tan\kappa_0h&\cdots&\kappa_N^3\tan\kappa_Nh\\
\kappa_0^4\tan\kappa_0h&\cdots&\kappa_N^4\tan\kappa_Nh
\end{bmatrix}
\\
\begin{bmatrix}
0&&\cdots&&0\\
&&&&\\
\vdots&&\ddots&&\vdots\\
&&&&\\
0&&\cdots&&0
\end{bmatrix}
&
\begin{bmatrix}
(k_0+\kappa_{-2})B_{-20}&(k_0+\kappa_{-1})B_{-10}\\
&\\
\vdots&\vdots\\
&\\
(k_N+\kappa_{-2})B_{-2N}&(k_N+\kappa_{-1})B_{-1N}
\end{bmatrix}
&
\begin{bmatrix}
(k_0 + \kappa_0) \, B_{00}&\cdots&(k_N + \kappa_{0}) \, B_{0N}\\
&&\\
\vdots&(k_l + \kappa_{m}) \, B_{ml}&\vdots\\
&&\\
(k_0 + \kappa_N) \, B_{N0}&\cdots&(k_N + \kappa_{N}) \, B_{NN}\\
\end{bmatrix}
\end{bmatrix}
\begin{bmatrix}
a_{0} \\
\\
\vdots \\
\\
a_N \\
b_{-2}\\
b_{-1}\\
b_{0}\\
\\
\vdots \\
\\
b_N
\end{bmatrix}
=
\begin{bmatrix}
- A_{0} \\
0 \\
\vdots \\
\\
0 \\
0 \\
0 \\
2k_{0}A_{0} \\
0 \\
\vdots \\
\\
0
\end{bmatrix}
</math>
</center>
We then simply need to solve the linear system of equations. Note that we can solve this equation for
<math>b_n</math> first and then solve for <math>a_n</math>

= Waves Incident at an Angle =

We can consider the problem when the waves are incident at an angle <math>\theta</math>. Thus the incident
potential can be expressed as follow:
<center>
<math>
\phi^I=e^{-k_0(\cos \theta x + \sin \theta y)}
</math>
</center>
In this
case we have the wavenumber in the <math>y</math> direction is <math>k_y = \sin\theta k_0</math>
where <math>k_0</math> is as defined previously (note that <math>k_y</math> is imaginary).

We assume here that the plate has infinite length in the <math>y</math>-direction, so the solution
does not vary in that direction except over a period. This means that the potential is now of the
form <math>\phi(x,y,z)=e^{k_y y}W(x,z)</math>. The separation of variables and the application of
Laplace's equation clearly shows that the dependance of <math>z</math> remains unchanged. However,
we obtain
<center>
<math>
\frac{1}{e^{k_yy}X(x)}(k_y^2e^{k_yy}X(x)+e^{k_yy}\frac{d^2X}{dx^2})=k_n^2=-\frac{1}{Z(z)}\frac{d^2Z}{dz^2}
</math>
</center>
where <math>k_n</math> is a root of the dispersion equation. This simplifies as
<center>
<math>
k_y^2+\frac{1}{X}\frac{d^2X}{dx^2}=k_n^2
</math>
</center>
which permits to obtain the followinf differential equation for <math>X</math>
<center>
<math>
\frac{d^2X}{dx^2}-(k_n^2-k_y^2)X=0
</math>
</center>
If we introduce the new variable <math>k_x^2=k_n^2-k_y^2</math>, we obtain a similar differential equation to solve
in the <math>x</math>-direction as for the null incident angle problem. Here we did not clarify the fact that we are
either in the open water region or the plate covered region, because the development is exactly the same in both cases.

Therefore the potential can be expanded as
<center>
<math>
\phi(x,z)=e^{-\hat{k}_0x}\phi_0(z)+\sum_{m=0}^{\infty}a_{m}e^{\hat{k}_{m}x}\phi_{m}(z), \;\;x<0
</math>
</center>
and
<center>
<math>
\phi(x,z)=\sum_{m=-2}^{\infty}b_{m}
e^{-\hat{\kappa}_{m}x}\psi_{m}(z), \;\;x>0
</math>
</center>
where <math>\hat{k}_{m} = \sqrt{k_m^2 - k_y^2}</math> and <math>\hat{\kappa}_{m} = \sqrt{\kappa_m^2 - k_y^2}</math>
where we always take the positive real root or the root with positive imaginary part.

The edge conditions are also different and are
<center><math>
\left(\frac{\partial^3}{\partial x^3} - (2 - \nu)k^2_y\frac{\partial}{\partial x}\right) \frac{\partial\phi}{\partial z}= 0,
</math></center>
and
<center><math>
\left(\frac{\partial^2}{\partial x^2} - \nu k^2_y\right)\frac{\partial\phi}{\partial z} = 0,
</math></center>
where <math>\nu</math> is Poisons ratio.

We can expend these edge conditions, which respectively gives
<center>
<math>
-\sum_{m=-2}^{\infty}b_{m}\kappa_m (\hat{\kappa}_m^3-\hat{\kappa}_m k_y^2(2-\nu))\tan \kappa_m h=0
</math>
</center>
and
<center>
<math>
-\sum_{m=-2}^{\infty}b_{m}\kappa_m (\hat{\kappa}_m^2-k_y^2\nu))\tan \kappa_m h=0
</math>
</center>

The equations are derived almost identically to those above and we obtain
<center>
<math>
A_{0}\delta_{0l}+a_{l}A_{l}
=\sum_{m=-2}^{\infty}b_{m}B_{ml}
</math>
</center>
and
<center>
<math>
-\hat{k_{0}}A_{0}\delta_{0l}+a_{l}\hat{k}_{l}A_l
=-\sum_{m=-2}^{\infty}b_{m}\hat{\kappa}_{m}B_{ml}
</math>
</center>
and these are solved exactly as before.

= Energy Balance =

The energy balance equation is ([[Energy Balance for Two Elastic Plates]]) given by
<center><math>
D |T|^2 + |R|^2 = 1, \,
</math></center>
where <math> T=b_0 </math>, <math>R=a_0</math>, and <math>D</math> is given by
<center><math>
D = \left(\frac{\hat{\kappa}_0 k_0\cos^2{(k_0 h)}}{\hat{k}_0 \kappa_0\cos^2{(\kappa_0 h)}}\right)
\left(\frac{\frac{4\beta}{\alpha}(\kappa_0)^3((-\hat{\kappa}_0)^2 +k_y^2)\sin^2{(\kappa_0 h)} +
\frac{1}{2}{\sin{(2 \kappa_0 h)}}+\kappa_0 h}
{
\frac{1}{2}{\sin{(2 k_0 h)}}+k_0h}\right)
</math></center>
where the index <math>0</math> refers to the travelling wave (imaginary root of the dispersion equation).

= Matlab Code =

A program to calculate the coefficients for the semi-infinite dock problems can be found here
[http://www.math.auckland.ac.nz/~meylan/code/eigenfunction_matching/semiinfinite_plate.m semiinfinite_plate.m]

== Additional code ==

This program requires
* {{free surface dispersion equation code}}
* {{elastic plate dispersion equation code}}

[[Category:Floating Elastic Plate]]
[[Category:Eigenfunction Matching Method]]
[[Category:Complete Pages]]

Eigenfunction Matching for a Semi-Infinite Floating Elastic Plate

2008-08-20T23:40:40Z

Fmontiel: /* Waves Incident at an Angle */

=Introduction=

We show here a solution for a semi-infinite [[:Category:Floating Elastic Plate|Floating Elastic Plate]] on [[Finite Depth]].
The problem was solved by [[Fox and Squire 1994]] but the solution method here is slightly different.
The simpler theory for a [[Eigenfunction Matching for a Semi-Infinite Dock|Dock]] describes
many of the ideas here in more detail.

[[Image:Semiinfinite plate.jpg|thumb|right|300px|Wave scattering by a submerged semi-infinite elastic plate]]

= Equations =

We consider the problem of small-amplitude waves which are incident on a semi-infinite floating elastic
plate occupying water surface for <math>x>0</math>. The submergence of the plate is considered negligible.
We assume that the problem is invariant in the <math>y</math> direction, although we allow the waves to be
incident from an angle.
We also assume that the plate edges are free to move at
each boundary, although other boundary conditions could easily be considered using
the methods of solution presented here. We begin with the [[Frequency Domain Problem]] for a semi-infinite
[[Floating Elastic Plate|Floating Elastic Plates]]
in the non-dimensional form of [[Tayler_1986a|Tayler 1986]] ([[Dispersion Relation for a Floating Elastic Plate]]).
We also assume that the waves are normally incident (incidence at an angle will be discussed later).
<center><math>\begin{matrix}
\Delta \phi = 0, \;\;\; -h < z \leq 0,
\end{matrix}</math></center>
<center><math>
\partial_z \phi = 0, \;\;\; z = - h,
</math></center>
<center><math>
\partial_z\phi=\alpha\phi, \,\, z=0,\,x<0,
</math></center>
<center><math>
\beta \partial_x^4\partial_z \phi
- \left( \gamma\alpha - 1 \right) \partial_z \phi + \alpha\phi = 0, \;\;
z = 0, \;\;\; x \geq 0,
</math></center>
where <math>\alpha = \omega^2</math>, <math>\beta</math> and <math>\gamma</math>
are the stiffness and mass constant for the plate respectively. The free edge conditions
at the edge of the plate imply
<center><math>
\partial_x^3 \partial_z\phi = 0, \;\; z = 0, \;\;\; x = 0,
</math></center>
<center><math>
\partial_x^2 \partial_z\phi = 0, \;\; z = 0, \;\;\; x = 0,
</math></center>

=Method of solution=

==Eigenfunction expansion==

We will solve the system of equations using an [[:Category:Eigenfunction Matching Method|Eigenfunction Matching Method]].
The method was developed by [[Fox and Squire 1994]]. The separation of variables for the left hand region where
there is open water, and the incident wave, is described in [[Eigenfunction Matching for a Semi-Infinite Dock]] and we consider
here only the separation of variables in the plate covered region.

===Separation of variables under the Plate===
The potential velocity can be written in terms of an infinite series of separated eigenfunctions under
each elastic plate, of the form
<center><math>
\psi = \frac{\cos(\kappa (z+h))}{\cos\kappa h}.\,
</math></center>
If we apply the boundary conditions given
we obtain the [[Dispersion Relation for a Floating Elastic Plate]]
<center><math>
\kappa \tan{(\kappa h)}= -\frac{\alpha}{\beta \kappa^{4} + 1 - \alpha\gamma}
</math></center>
Solving for <math>\kappa</math> gives a pure imaginary root
with positive imaginary part, two complex roots (two complex conjugate paired roots
with positive imaginary part in most physical situations), an infinite number of positive real roots
which approach <math>{n\pi}/{h}</math> as <math>n</math> approaches infinity, and also the negative of all
these roots ([[Dispersion Relation for a Floating Elastic Plate]]) . We denote the two complex roots with positive imaginary part
by <math>\kappa_(-2)</math> and <math>\kappa_(-1)</math>, the purely imaginary
root with positive imaginary part by <math>\kappa_{0}</math> and the real roots with positive imaginary part
by <math>\kappa_(n)</math> for <math>n</math> a positive integer.
The imaginary root with positive imaginary part corresponds to a
reflected travelling mode propagating along the <math>x</math> axis.
The complex roots with positive imaginary parts correspond to damped reflected travelling modes and the real roots correspond to reflected evanescent modes.

==Expressions for the potential velocity==

We now expand the potential in the two regions using the separation of variables solution.
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.
The potential <math>\phi</math> can now be expressed as the following sum of eigenfunctions:
be expanded as
<center>
<math>
\phi(x,z)= e^{-k_{0}x}\phi_{0}(z) + \sum_{m=0}^{\infty}a_{m}e^{k_{m}x}\phi_{m}(z), \;\;x<0
</math>
</center>
and
<center>
<math>
\phi(x,z)=\sum_{m=-2}^{\infty}b_{m}
e^{-\kappa_{m}x}\psi_{m}(z), \;\;x>0
</math>
</center>
where <math>a_{m}</math> and <math>b_{m}</math> are the coefficients to be determined.

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>

==An infinite dimensional system of equations==

The potential and its derivative must be continuous across the
transition from open water to the plate covered region. Therefore, the
potentials and their derivatives at <math>x=0</math> have to be equal.
We also truncate the sum at <math>N</math> being careful that we have
two extra modes on the plate covered region to satisfy the edge conditions.
We obtain
<center>
<math>
\phi_{0}\left( z\right) + \sum_{m=0}^{N}
a_{m} \phi_{m}\left( z\right)
=\sum_{m=-2}^{N}b_{m}\psi_{m}(z)
</math>
</center>
and
<center>
<math>
-k_{0}\phi_{0}\left( z\right) +\sum
_{m=0}^{N} a_{m}k_{m}\phi_{m}\left( z\right)
=-\sum_{m=-2}^{N}b_{m}\kappa_{m}\psi
_{m}(z)
</math>
</center>
for each <math>N</math>.
We solve these equations by multiplying both equations by
<math>\phi_{l}(z)\,</math> and integrating from <math>-h</math> to <math>0</math> to obtain:
<center>
<math>
A_{0}\delta_{0l}+a_{l}A_{l}
=\sum_{m=-2}^{N}b_{m}B_{ml},\,\,0 \leq l \leq N
</math>
</center>
and
<center>
<math>
-k_{0}A_{0}\delta_{0l}+a_{l}k_{l}A_l
=-\sum_{m=-2}^{N}b_{m}\kappa_{m}B_{ml},\,\,0 \leq l \leq N
</math>
</center>
If we multiply the first equation by <math>k_l</math> and subtract the second equation
we obtain
<center>
<math>
2k_{0}A_{0}\delta_{0l}
=\sum_{m=-2}^{N}b_{m}(k_l + \kappa_{m})B_{ml}
</math>
</center>
Finally, we need to apply the conditions at the edge of the plate to give us two further equations,
<center>
<math>
\partial_x^2\partial_z\phi = - \sum_{m=-2}^{N}b_{m} \kappa_m^3 \tan\kappa_m h = 0
</math>
</center>
and
<center>
<math>
\partial_x^3\partial_z\phi = \sum_{m=-2}^{N}b_{m} \kappa_m^4 \tan\kappa_m h = 0
</math>
</center>

=Numerical Solution=

To solve the system of equations previously defined we set the upper limit of <math>l</math> to
be <math>N</math>, as we already said before. In terms of matrix, we obtain
<center>
<math>
\begin{bmatrix}
\begin{bmatrix}
A_0&0 \quad \cdots&0\\
0&&\\
\vdots&A_l&\vdots\\
&&0\\
0&\cdots \quad 0 &A_N
\end{bmatrix}
&
\begin{bmatrix}
-B_{-20}&-B_{-10}\\
&\\
\vdots&\vdots\\
&\\
-B_{-2N}&-B_{-1N}
\end{bmatrix}
&
\begin{bmatrix}
-B_{00}&\cdots&-B_{0N}\\
&&\\
\vdots&-B_{ml}&\vdots\\
&&\\
-B_{N0}&\cdots&-B_{NN}
\end{bmatrix}
\\
\begin{bmatrix}
0&&\cdots&&0\\
0&&\cdots&&0
\end{bmatrix}
&
\begin{bmatrix}
\kappa_{-2}^3\tan\kappa_{-2}h&\kappa_{-1}^3\tan\kappa_{-1}h\\
\kappa_{-2}^4\tan\kappa_{-2}h&\kappa_{-1}^4\tan\kappa_{-1}h
\end{bmatrix}
&
\begin{bmatrix}
\kappa_0^3\tan\kappa_0h&\cdots&\kappa_N^3\tan\kappa_Nh\\
\kappa_0^4\tan\kappa_0h&\cdots&\kappa_N^4\tan\kappa_Nh
\end{bmatrix}
\\
\begin{bmatrix}
0&&\cdots&&0\\
&&&&\\
\vdots&&\ddots&&\vdots\\
&&&&\\
0&&\cdots&&0
\end{bmatrix}
&
\begin{bmatrix}
(k_0+\kappa_{-2})B_{-20}&(k_0+\kappa_{-1})B_{-10}\\
&\\
\vdots&\vdots\\
&\\
(k_N+\kappa_{-2})B_{-2N}&(k_N+\kappa_{-1})B_{-1N}
\end{bmatrix}
&
\begin{bmatrix}
(k_0 + \kappa_0) \, B_{00}&\cdots&(k_N + \kappa_{0}) \, B_{0N}\\
&&\\
\vdots&(k_l + \kappa_{m}) \, B_{ml}&\vdots\\
&&\\
(k_0 + \kappa_N) \, B_{N0}&\cdots&(k_N + \kappa_{N}) \, B_{NN}\\
\end{bmatrix}
\end{bmatrix}
\begin{bmatrix}
a_{0} \\
\\
\vdots \\
\\
a_N \\
b_{-2}\\
b_{-1}\\
b_{0}\\
\\
\vdots \\
\\
b_N
\end{bmatrix}
=
\begin{bmatrix}
- A_{0} \\
0 \\
\vdots \\
\\
0 \\
0 \\
0 \\
2k_{0}A_{0} \\
0 \\
\vdots \\
\\
0
\end{bmatrix}
</math>
</center>
We then simply need to solve the linear system of equations. Note that we can solve this equation for
<math>b_n</math> first and then solve for <math>a_n</math>

= Waves Incident at an Angle =

We can consider the problem when the waves are incident at an angle <math>\theta</math>. Thus the incident
potential can be expressed as follow:
<center>
<math>
\phi^I=e^{-k_0(\cos \theta x + \sin \theta y)}
</math>
</center>
In this
case we have the wavenumber in the <math>y</math> direction is <math>k_y = \sin\theta k_0</math>
where <math>k_0</math> is as defined previously (note that <math>k_y</math> is imaginary).

We assume here that the plate has infinite length in the <math>y</math>-direction, so the solution
does not vary in that direction except over a period. This means that the potential is now of the
form <math>\phi(x,y,z)=e^{k_y y}W(x,z)</math>. The separation of variables and the application of
Laplace's equation clearly shows that the dependance of <math>z</math> remains unchanged. However,
we obtain
<center>
<math>
\frac{1}{e^{k_yy}X(x)}(k_y^2e^{k_yy}X(x)+e^{k_yy}\frac{d^2X}{dx^2})=k_n^2=-\frac{1}{Z(z)}\frac{d^2Z}{dz^2}
</math>
</center>
where <math>k_n</math> is a root of the dispersion equation. This simplifies as
<center>
<math>
k_y^2+\frac{1}{X}\frac{d^2X}{dx^2}=k_n^2
</math>
</center>
which permits to obtain the followinf differential equation for <math>X</math>
<center>
<math>
\frac{d^2X}{dx^2}-(k_n^2-k_y^2)X=0
</math>
</center>
If we introduce the new variable <math>k_x^2=k_n^2-k_y^2</math>, we obtain a similar differential equation to solve
in the <math>x</math>-direction as for the null incident angle problem. Here we did not clarify the fact that we are
in the open water region or the plate covered region, because the development is exactly the same in both cases.

Therefore the potential can be expanded as
<center>
<math>
\phi(x,z)=e^{-\hat{k}_0x}\phi_0(z)+\sum_{m=0}^{\infty}a_{m}e^{\hat{k}_{m}x}\phi_{m}(z), \;\;x<0
</math>
</center>
and
<center>
<math>
\phi(x,z)=\sum_{m=-2}^{\infty}b_{m}
e^{-\hat{\kappa}_{m}x}\psi_{m}(z), \;\;x>0
</math>
</center>
where <math>\hat{k}_{m} = \sqrt{k_m^2 - k_y^2}</math> and <math>\hat{\kappa}_{m} = \sqrt{\kappa_m^2 - k_y^2}</math>
where we always take the positive real root or the root with positive imaginary part.

The edge conditions are also different and are
<center><math>
\left(\frac{\partial^3}{\partial x^3} - (2 - \nu)k^2_y\frac{\partial}{\partial x}\right) \frac{\partial\phi}{\partial z}= 0,
</math></center>
and
<center><math>
\left(\frac{\partial^2}{\partial x^2} - \nu k^2_y\right)\frac{\partial\phi}{\partial z} = 0,
</math></center>
where <math>\nu</math> is Poisons ratio.

We can expend these edge conditions, which respectively gives
<center>
<math>
-\sum_{m=-2}^{\infty}b_{m}\kappa_m (\hat{\kappa}_m^3-\hat{\kappa}_m k_y^2(2-\nu))\tan \kappa_m h=0
</math>
</center>
and
<center>
<math>
-\sum_{m=-2}^{\infty}b_{m}\kappa_m (\hat{\kappa}_m^2-k_y^2\nu))\tan \kappa_m h=0
</math>
</center>

The equations are derived almost identically to those above and we obtain
<center>
<math>
A_{0}\delta_{0l}+a_{l}A_{l}
=\sum_{m=-2}^{\infty}b_{m}B_{ml}
</math>
</center>
and
<center>
<math>
-\hat{k_{0}}A_{0}\delta_{0l}+a_{l}\hat{k}_{l}A_l
=-\sum_{m=-2}^{\infty}b_{m}\hat{\kappa}_{m}B_{ml}
</math>
</center>
and these are solved exactly as before.

= Energy Balance =

The energy balance equation is ([[Energy Balance for Two Elastic Plates]]) given by
<center><math>
D |T|^2 + |R|^2 = 1, \,
</math></center>
where <math> T=b_0 </math>, <math>R=a_0</math>, and <math>D</math> is given by
<center><math>
D = \left(\frac{\hat{\kappa}_0 k_0\cos^2{(k_0 h)}}{\hat{k}_0 \kappa_0\cos^2{(\kappa_0 h)}}\right)
\left(\frac{\frac{4\beta}{\alpha}(\kappa_0)^3((-\hat{\kappa}_0)^2 +k_y^2)\sin^2{(\kappa_0 h)} +
\frac{1}{2}{\sin{(2 \kappa_0 h)}}+\kappa_0 h}
{
\frac{1}{2}{\sin{(2 k_0 h)}}+k_0h}\right)
</math></center>
where the index <math>0</math> refers to the travelling wave (imaginary root of the dispersion equation).

= Matlab Code =

A program to calculate the coefficients for the semi-infinite dock problems can be found here
[http://www.math.auckland.ac.nz/~meylan/code/eigenfunction_matching/semiinfinite_plate.m semiinfinite_plate.m]

== Additional code ==

This program requires
* {{free surface dispersion equation code}}
* {{elastic plate dispersion equation code}}

[[Category:Floating Elastic Plate]]
[[Category:Eigenfunction Matching Method]]
[[Category:Complete Pages]]

Eigenfunction Matching for a Semi-Infinite Floating Elastic Plate

2008-08-20T03:50:25Z

Fmontiel: /* Waves Incident at an Angle */

=Introduction=

We show here a solution for a semi-infinite [[:Category:Floating Elastic Plate|Floating Elastic Plate]] on [[Finite Depth]].
The problem was solved by [[Fox and Squire 1994]] but the solution method here is slightly different.
The simpler theory for a [[Eigenfunction Matching for a Semi-Infinite Dock|Dock]] describes
many of the ideas here in more detail.

[[Image:Semiinfinite plate.jpg|thumb|right|300px|Wave scattering by a submerged semi-infinite elastic plate]]

= Equations =

We consider the problem of small-amplitude waves which are incident on a semi-infinite floating elastic
plate occupying water surface for <math>x>0</math>. The submergence of the plate is considered negligible.
We assume that the problem is invariant in the <math>y</math> direction, although we allow the waves to be
incident from an angle.
We also assume that the plate edges are free to move at
each boundary, although other boundary conditions could easily be considered using
the methods of solution presented here. We begin with the [[Frequency Domain Problem]] for a semi-infinite
[[Floating Elastic Plate|Floating Elastic Plates]]
in the non-dimensional form of [[Tayler_1986a|Tayler 1986]] ([[Dispersion Relation for a Floating Elastic Plate]]).
We also assume that the waves are normally incident (incidence at an angle will be discussed later).
<center><math>\begin{matrix}
\Delta \phi = 0, \;\;\; -h < z \leq 0,
\end{matrix}</math></center>
<center><math>
\partial_z \phi = 0, \;\;\; z = - h,
</math></center>
<center><math>
\partial_z\phi=\alpha\phi, \,\, z=0,\,x<0,
</math></center>
<center><math>
\beta \partial_x^4\partial_z \phi
- \left( \gamma\alpha - 1 \right) \partial_z \phi + \alpha\phi = 0, \;\;
z = 0, \;\;\; x \geq 0,
</math></center>
where <math>\alpha = \omega^2</math>, <math>\beta</math> and <math>\gamma</math>
are the stiffness and mass constant for the plate respectively. The free edge conditions
at the edge of the plate imply
<center><math>
\partial_x^3 \partial_z\phi = 0, \;\; z = 0, \;\;\; x = 0,
</math></center>
<center><math>
\partial_x^2 \partial_z\phi = 0, \;\; z = 0, \;\;\; x = 0,
</math></center>

=Method of solution=

==Eigenfunction expansion==

We will solve the system of equations using an [[:Category:Eigenfunction Matching Method|Eigenfunction Matching Method]].
The method was developed by [[Fox and Squire 1994]]. The separation of variables for the left hand region where
there is open water, and the incident wave, is described in [[Eigenfunction Matching for a Semi-Infinite Dock]] and we consider
here only the separation of variables in the plate covered region.

===Separation of variables under the Plate===
The potential velocity can be written in terms of an infinite series of separated eigenfunctions under
each elastic plate, of the form
<center><math>
\psi = \frac{\cos(\kappa (z+h))}{\cos\kappa h}.\,
</math></center>
If we apply the boundary conditions given
we obtain the [[Dispersion Relation for a Floating Elastic Plate]]
<center><math>
\kappa \tan{(\kappa h)}= -\frac{\alpha}{\beta \kappa^{4} + 1 - \alpha\gamma}
</math></center>
Solving for <math>\kappa</math> gives a pure imaginary root
with positive imaginary part, two complex roots (two complex conjugate paired roots
with positive imaginary part in most physical situations), an infinite number of positive real roots
which approach <math>{n\pi}/{h}</math> as <math>n</math> approaches infinity, and also the negative of all
these roots ([[Dispersion Relation for a Floating Elastic Plate]]) . We denote the two complex roots with positive imaginary part
by <math>\kappa_(-2)</math> and <math>\kappa_(-1)</math>, the purely imaginary
root with positive imaginary part by <math>\kappa_{0}</math> and the real roots with positive imaginary part
by <math>\kappa_(n)</math> for <math>n</math> a positive integer.
The imaginary root with positive imaginary part corresponds to a
reflected travelling mode propagating along the <math>x</math> axis.
The complex roots with positive imaginary parts correspond to damped reflected travelling modes and the real roots correspond to reflected evanescent modes.

==Expressions for the potential velocity==

We now expand the potential in the two regions using the separation of variables solution.
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.
The potential <math>\phi</math> can now be expressed as the following sum of eigenfunctions:
be expanded as
<center>
<math>
\phi(x,z)= e^{-k_{0}x}\phi_{0}(z) + \sum_{m=0}^{\infty}a_{m}e^{k_{m}x}\phi_{m}(z), \;\;x<0
</math>
</center>
and
<center>
<math>
\phi(x,z)=\sum_{m=-2}^{\infty}b_{m}
e^{-\kappa_{m}x}\psi_{m}(z), \;\;x>0
</math>
</center>
where <math>a_{m}</math> and <math>b_{m}</math> are the coefficients to be determined.

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>

==An infinite dimensional system of equations==

The potential and its derivative must be continuous across the
transition from open water to the plate covered region. Therefore, the
potentials and their derivatives at <math>x=0</math> have to be equal.
We also truncate the sum at <math>N</math> being careful that we have
two extra modes on the plate covered region to satisfy the edge conditions.
We obtain
<center>
<math>
\phi_{0}\left( z\right) + \sum_{m=0}^{N}
a_{m} \phi_{m}\left( z\right)
=\sum_{m=-2}^{N}b_{m}\psi_{m}(z)
</math>
</center>
and
<center>
<math>
-k_{0}\phi_{0}\left( z\right) +\sum
_{m=0}^{N} a_{m}k_{m}\phi_{m}\left( z\right)
=-\sum_{m=-2}^{N}b_{m}\kappa_{m}\psi
_{m}(z)
</math>
</center>
for each <math>N</math>.
We solve these equations by multiplying both equations by
<math>\phi_{l}(z)\,</math> and integrating from <math>-h</math> to <math>0</math> to obtain:
<center>
<math>
A_{0}\delta_{0l}+a_{l}A_{l}
=\sum_{m=-2}^{N}b_{m}B_{ml},\,\,0 \leq l \leq N
</math>
</center>
and
<center>
<math>
-k_{0}A_{0}\delta_{0l}+a_{l}k_{l}A_l
=-\sum_{m=-2}^{N}b_{m}\kappa_{m}B_{ml},\,\,0 \leq l \leq N
</math>
</center>
If we multiply the first equation by <math>k_l</math> and subtract the second equation
we obtain
<center>
<math>
2k_{0}A_{0}\delta_{0l}
=\sum_{m=-2}^{N}b_{m}(k_l + \kappa_{m})B_{ml}
</math>
</center>
Finally, we need to apply the conditions at the edge of the plate to give us two further equations,
<center>
<math>
\partial_x^2\partial_z\phi = - \sum_{m=-2}^{N}b_{m} \kappa_m^3 \tan\kappa_m h = 0
</math>
</center>
and
<center>
<math>
\partial_x^3\partial_z\phi = \sum_{m=-2}^{N}b_{m} \kappa_m^4 \tan\kappa_m h = 0
</math>
</center>

=Numerical Solution=

To solve the system of equations previously defined we set the upper limit of <math>l</math> to
be <math>N</math>, as we already said before. In terms of matrix, we obtain
<center>
<math>
\begin{bmatrix}
\begin{bmatrix}
A_0&0 \quad \cdots&0\\
0&&\\
\vdots&A_l&\vdots\\
&&0\\
0&\cdots \quad 0 &A_N
\end{bmatrix}
&
\begin{bmatrix}
-B_{-20}&-B_{-10}\\
&\\
\vdots&\vdots\\
&\\
-B_{-2N}&-B_{-1N}
\end{bmatrix}
&
\begin{bmatrix}
-B_{00}&\cdots&-B_{0N}\\
&&\\
\vdots&-B_{ml}&\vdots\\
&&\\
-B_{N0}&\cdots&-B_{NN}
\end{bmatrix}
\\
\begin{bmatrix}
0&&\cdots&&0\\
0&&\cdots&&0
\end{bmatrix}
&
\begin{bmatrix}
\kappa_{-2}^3\tan\kappa_{-2}h&\kappa_{-1}^3\tan\kappa_{-1}h\\
\kappa_{-2}^4\tan\kappa_{-2}h&\kappa_{-1}^4\tan\kappa_{-1}h
\end{bmatrix}
&
\begin{bmatrix}
\kappa_0^3\tan\kappa_0h&\cdots&\kappa_N^3\tan\kappa_Nh\\
\kappa_0^4\tan\kappa_0h&\cdots&\kappa_N^4\tan\kappa_Nh
\end{bmatrix}
\\
\begin{bmatrix}
0&&\cdots&&0\\
&&&&\\
\vdots&&\ddots&&\vdots\\
&&&&\\
0&&\cdots&&0
\end{bmatrix}
&
\begin{bmatrix}
(k_0+\kappa_{-2})B_{-20}&(k_0+\kappa_{-1})B_{-10}\\
&\\
\vdots&\vdots\\
&\\
(k_N+\kappa_{-2})B_{-2N}&(k_N+\kappa_{-1})B_{-1N}
\end{bmatrix}
&
\begin{bmatrix}
(k_0 + \kappa_0) \, B_{00}&\cdots&(k_N + \kappa_{0}) \, B_{0N}\\
&&\\
\vdots&(k_l + \kappa_{m}) \, B_{ml}&\vdots\\
&&\\
(k_0 + \kappa_N) \, B_{N0}&\cdots&(k_N + \kappa_{N}) \, B_{NN}\\
\end{bmatrix}
\end{bmatrix}
\begin{bmatrix}
a_{0} \\
\\
\vdots \\
\\
a_N \\
b_{-2}\\
b_{-1}\\
b_{0}\\
\\
\vdots \\
\\
b_N
\end{bmatrix}
=
\begin{bmatrix}
- A_{0} \\
0 \\
\vdots \\
\\
0 \\
0 \\
0 \\
2k_{0}A_{0} \\
0 \\
\vdots \\
\\
0
\end{bmatrix}
</math>
</center>
We then simply need to solve the linear system of equations. Note that we can solve this equation for
<math>b_n</math> first and then solve for <math>a_n</math>

= Waves Incident at an Angle =

We can consider the problem when the waves are incident at an angle <math>\theta</math>. Thus the incident
potential can be expressed as follow:
<center>
<math>
\phi^I=e^{-k_0(\cos \theta x + \sin \theta y)}
</math>
</center>
In this
case we have the wavenumber in the <math>y</math> direction is <math>k_y = \sin\theta k_0</math>
where <math>k_0</math> is as defined previously (note that <math>k_y</math> is imaginary).

We assume here that the plate has infinite length in the <math>y</math>-direction, so the solution
does not vary in that direction except over a period. This means that the potential is now of the
form <math>\phi(x,y,z)=e^{k_y y}W(x,z)</math>. The separation of variables and the application of
Laplace's equation clearly shows that the dependance of <math>z</math> remains unchanged. However,
we obtain

Therefore the potential can
be expanded as
<center>
<math>
\phi(x,z)=e^{-\hat{k}_0x}\phi_0(z)+\sum_{m=0}^{\infty}a_{m}e^{\hat{k}_{m}x}\phi_{m}(z), \;\;x<0
</math>
</center>
and
<center>
<math>
\phi(x,z)=\sum_{m=-2}^{\infty}b_{m}
e^{-\hat{\kappa}_{m}x}\psi_{m}(z), \;\;x>0
</math>
</center>
where <math>\hat{k}_{m} = \sqrt{k_m^2 + k_y^2}</math> and <math>\hat{\kappa}_{m} = \sqrt{\kappa_m^2 + k_y^2}</math>
where we always take the positive real root or the root with positive imaginary part.

The edge conditions are also different and are
<center><math>
\left(\frac{\partial^3}{\partial x^3} - (2 - \nu)k^2_y\frac{\partial}{\partial x}\right) \frac{\partial\phi}{\partial z}= 0,
</math></center>
and
<center><math>
\left(\frac{\partial^2}{\partial x^2} - \nu k^2_y\right)\frac{\partial\phi}{\partial z} = 0,
</math></center>
where <math>\nu</math> is Poisons ratio.

We can expend these edge conditions, which respectively gives
<center>
<math>
-\sum_{m=-2}^{\infty}b_{m}\kappa_m (\hat{\kappa}_m^3-\hat{\kappa}_m k_y^2(2-\nu))\tan \kappa_m h=0
</math>
</center>
and
<center>
<math>
-\sum_{m=-2}^{\infty}b_{m}\kappa_m (\hat{\kappa}_m^2-k_y^2\nu))\tan \kappa_m h=0
</math>
</center>

The equations are derived almost identically to those above and we obtain
<center>
<math>
A_{0}\delta_{0l}+a_{l}A_{l}
=\sum_{m=-2}^{\infty}b_{m}B_{ml}
</math>
</center>
and
<center>
<math>
-\hat{k_{0}}A_{0}\delta_{0l}+a_{l}\hat{k}_{l}A_l
=-\sum_{m=-2}^{\infty}b_{m}\hat{\kappa}_{m}B_{ml}
</math>
</center>
and these are solved exactly as before.

= Energy Balance =

The energy balance equation is ([[Energy Balance for Two Elastic Plates]]) given by
<center><math>
D |T|^2 + |R|^2 = 1, \,
</math></center>
where <math> T=b_0 </math>, <math>R=a_0</math>, and <math>D</math> is given by
<center><math>
D = \left(\frac{\hat{\kappa}_0 k_0\cos^2{(k_0 h)}}{\hat{k}_0 \kappa_0\cos^2{(\kappa_0 h)}}\right)
\left(\frac{\frac{4\beta}{\alpha}(\kappa_0)^3((-\hat{\kappa}_0)^2 +k_y^2)\sin^2{(\kappa_0 h)} +
\frac{1}{2}{\sin{(2 \kappa_0 h)}}+\kappa_0 h}
{
\frac{1}{2}{\sin{(2 k_0 h)}}+k_0h}\right)
</math></center>
where the index <math>0</math> refers to the travelling wave (imaginary root of the dispersion equation).

= Matlab Code =

A program to calculate the coefficients for the semi-infinite dock problems can be found here
[http://www.math.auckland.ac.nz/~meylan/code/eigenfunction_matching/semiinfinite_plate.m semiinfinite_plate.m]

== Additional code ==

This program requires
* {{free surface dispersion equation code}}
* {{elastic plate dispersion equation code}}

[[Category:Floating Elastic Plate]]
[[Category:Eigenfunction Matching Method]]
[[Category:Complete Pages]]

Eigenfunction Matching for a Semi-Infinite Floating Elastic Plate

2008-08-20T03:46:22Z

Fmontiel: /* Waves Incident at an Angle */

=Introduction=

We show here a solution for a semi-infinite [[:Category:Floating Elastic Plate|Floating Elastic Plate]] on [[Finite Depth]].
The problem was solved by [[Fox and Squire 1994]] but the solution method here is slightly different.
The simpler theory for a [[Eigenfunction Matching for a Semi-Infinite Dock|Dock]] describes
many of the ideas here in more detail.

[[Image:Semiinfinite plate.jpg|thumb|right|300px|Wave scattering by a submerged semi-infinite elastic plate]]

= Equations =

We consider the problem of small-amplitude waves which are incident on a semi-infinite floating elastic
plate occupying water surface for <math>x>0</math>. The submergence of the plate is considered negligible.
We assume that the problem is invariant in the <math>y</math> direction, although we allow the waves to be
incident from an angle.
We also assume that the plate edges are free to move at
each boundary, although other boundary conditions could easily be considered using
the methods of solution presented here. We begin with the [[Frequency Domain Problem]] for a semi-infinite
[[Floating Elastic Plate|Floating Elastic Plates]]
in the non-dimensional form of [[Tayler_1986a|Tayler 1986]] ([[Dispersion Relation for a Floating Elastic Plate]]).
We also assume that the waves are normally incident (incidence at an angle will be discussed later).
<center><math>\begin{matrix}
\Delta \phi = 0, \;\;\; -h < z \leq 0,
\end{matrix}</math></center>
<center><math>
\partial_z \phi = 0, \;\;\; z = - h,
</math></center>
<center><math>
\partial_z\phi=\alpha\phi, \,\, z=0,\,x<0,
</math></center>
<center><math>
\beta \partial_x^4\partial_z \phi
- \left( \gamma\alpha - 1 \right) \partial_z \phi + \alpha\phi = 0, \;\;
z = 0, \;\;\; x \geq 0,
</math></center>
where <math>\alpha = \omega^2</math>, <math>\beta</math> and <math>\gamma</math>
are the stiffness and mass constant for the plate respectively. The free edge conditions
at the edge of the plate imply
<center><math>
\partial_x^3 \partial_z\phi = 0, \;\; z = 0, \;\;\; x = 0,
</math></center>
<center><math>
\partial_x^2 \partial_z\phi = 0, \;\; z = 0, \;\;\; x = 0,
</math></center>

=Method of solution=

==Eigenfunction expansion==

We will solve the system of equations using an [[:Category:Eigenfunction Matching Method|Eigenfunction Matching Method]].
The method was developed by [[Fox and Squire 1994]]. The separation of variables for the left hand region where
there is open water, and the incident wave, is described in [[Eigenfunction Matching for a Semi-Infinite Dock]] and we consider
here only the separation of variables in the plate covered region.

===Separation of variables under the Plate===
The potential velocity can be written in terms of an infinite series of separated eigenfunctions under
each elastic plate, of the form
<center><math>
\psi = \frac{\cos(\kappa (z+h))}{\cos\kappa h}.\,
</math></center>
If we apply the boundary conditions given
we obtain the [[Dispersion Relation for a Floating Elastic Plate]]
<center><math>
\kappa \tan{(\kappa h)}= -\frac{\alpha}{\beta \kappa^{4} + 1 - \alpha\gamma}
</math></center>
Solving for <math>\kappa</math> gives a pure imaginary root
with positive imaginary part, two complex roots (two complex conjugate paired roots
with positive imaginary part in most physical situations), an infinite number of positive real roots
which approach <math>{n\pi}/{h}</math> as <math>n</math> approaches infinity, and also the negative of all
these roots ([[Dispersion Relation for a Floating Elastic Plate]]) . We denote the two complex roots with positive imaginary part
by <math>\kappa_(-2)</math> and <math>\kappa_(-1)</math>, the purely imaginary
root with positive imaginary part by <math>\kappa_{0}</math> and the real roots with positive imaginary part
by <math>\kappa_(n)</math> for <math>n</math> a positive integer.
The imaginary root with positive imaginary part corresponds to a
reflected travelling mode propagating along the <math>x</math> axis.
The complex roots with positive imaginary parts correspond to damped reflected travelling modes and the real roots correspond to reflected evanescent modes.

==Expressions for the potential velocity==

We now expand the potential in the two regions using the separation of variables solution.
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.
The potential <math>\phi</math> can now be expressed as the following sum of eigenfunctions:
be expanded as
<center>
<math>
\phi(x,z)= e^{-k_{0}x}\phi_{0}(z) + \sum_{m=0}^{\infty}a_{m}e^{k_{m}x}\phi_{m}(z), \;\;x<0
</math>
</center>
and
<center>
<math>
\phi(x,z)=\sum_{m=-2}^{\infty}b_{m}
e^{-\kappa_{m}x}\psi_{m}(z), \;\;x>0
</math>
</center>
where <math>a_{m}</math> and <math>b_{m}</math> are the coefficients to be determined.

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>

==An infinite dimensional system of equations==

The potential and its derivative must be continuous across the
transition from open water to the plate covered region. Therefore, the
potentials and their derivatives at <math>x=0</math> have to be equal.
We also truncate the sum at <math>N</math> being careful that we have
two extra modes on the plate covered region to satisfy the edge conditions.
We obtain
<center>
<math>
\phi_{0}\left( z\right) + \sum_{m=0}^{N}
a_{m} \phi_{m}\left( z\right)
=\sum_{m=-2}^{N}b_{m}\psi_{m}(z)
</math>
</center>
and
<center>
<math>
-k_{0}\phi_{0}\left( z\right) +\sum
_{m=0}^{N} a_{m}k_{m}\phi_{m}\left( z\right)
=-\sum_{m=-2}^{N}b_{m}\kappa_{m}\psi
_{m}(z)
</math>
</center>
for each <math>N</math>.
We solve these equations by multiplying both equations by
<math>\phi_{l}(z)\,</math> and integrating from <math>-h</math> to <math>0</math> to obtain:
<center>
<math>
A_{0}\delta_{0l}+a_{l}A_{l}
=\sum_{m=-2}^{N}b_{m}B_{ml},\,\,0 \leq l \leq N
</math>
</center>
and
<center>
<math>
-k_{0}A_{0}\delta_{0l}+a_{l}k_{l}A_l
=-\sum_{m=-2}^{N}b_{m}\kappa_{m}B_{ml},\,\,0 \leq l \leq N
</math>
</center>
If we multiply the first equation by <math>k_l</math> and subtract the second equation
we obtain
<center>
<math>
2k_{0}A_{0}\delta_{0l}
=\sum_{m=-2}^{N}b_{m}(k_l + \kappa_{m})B_{ml}
</math>
</center>
Finally, we need to apply the conditions at the edge of the plate to give us two further equations,
<center>
<math>
\partial_x^2\partial_z\phi = - \sum_{m=-2}^{N}b_{m} \kappa_m^3 \tan\kappa_m h = 0
</math>
</center>
and
<center>
<math>
\partial_x^3\partial_z\phi = \sum_{m=-2}^{N}b_{m} \kappa_m^4 \tan\kappa_m h = 0
</math>
</center>

=Numerical Solution=

To solve the system of equations previously defined we set the upper limit of <math>l</math> to
be <math>N</math>, as we already said before. In terms of matrix, we obtain
<center>
<math>
\begin{bmatrix}
\begin{bmatrix}
A_0&0 \quad \cdots&0\\
0&&\\
\vdots&A_l&\vdots\\
&&0\\
0&\cdots \quad 0 &A_N
\end{bmatrix}
&
\begin{bmatrix}
-B_{-20}&-B_{-10}\\
&\\
\vdots&\vdots\\
&\\
-B_{-2N}&-B_{-1N}
\end{bmatrix}
&
\begin{bmatrix}
-B_{00}&\cdots&-B_{0N}\\
&&\\
\vdots&-B_{ml}&\vdots\\
&&\\
-B_{N0}&\cdots&-B_{NN}
\end{bmatrix}
\\
\begin{bmatrix}
0&&\cdots&&0\\
0&&\cdots&&0
\end{bmatrix}
&
\begin{bmatrix}
\kappa_{-2}^3\tan\kappa_{-2}h&\kappa_{-1}^3\tan\kappa_{-1}h\\
\kappa_{-2}^4\tan\kappa_{-2}h&\kappa_{-1}^4\tan\kappa_{-1}h
\end{bmatrix}
&
\begin{bmatrix}
\kappa_0^3\tan\kappa_0h&\cdots&\kappa_N^3\tan\kappa_Nh\\
\kappa_0^4\tan\kappa_0h&\cdots&\kappa_N^4\tan\kappa_Nh
\end{bmatrix}
\\
\begin{bmatrix}
0&&\cdots&&0\\
&&&&\\
\vdots&&\ddots&&\vdots\\
&&&&\\
0&&\cdots&&0
\end{bmatrix}
&
\begin{bmatrix}
(k_0+\kappa_{-2})B_{-20}&(k_0+\kappa_{-1})B_{-10}\\
&\\
\vdots&\vdots\\
&\\
(k_N+\kappa_{-2})B_{-2N}&(k_N+\kappa_{-1})B_{-1N}
\end{bmatrix}
&
\begin{bmatrix}
(k_0 + \kappa_0) \, B_{00}&\cdots&(k_N + \kappa_{0}) \, B_{0N}\\
&&\\
\vdots&(k_l + \kappa_{m}) \, B_{ml}&\vdots\\
&&\\
(k_0 + \kappa_N) \, B_{N0}&\cdots&(k_N + \kappa_{N}) \, B_{NN}\\
\end{bmatrix}
\end{bmatrix}
\begin{bmatrix}
a_{0} \\
\\
\vdots \\
\\
a_N \\
b_{-2}\\
b_{-1}\\
b_{0}\\
\\
\vdots \\
\\
b_N
\end{bmatrix}
=
\begin{bmatrix}
- A_{0} \\
0 \\
\vdots \\
\\
0 \\
0 \\
0 \\
2k_{0}A_{0} \\
0 \\
\vdots \\
\\
0
\end{bmatrix}
</math>
</center>
We then simply need to solve the linear system of equations. Note that we can solve this equation for
<math>b_n</math> first and then solve for <math>a_n</math>

= Waves Incident at an Angle =

We can consider the problem when the waves are incident at an angle <math>\theta</math>. Thus the incident
potential can be expressed as follow:
<center>
<math>
\phi^I=e^{-k_0(\cos \theta x + \sin \theta y)}
</math>
</center>
In this
case we have the wavenumber in the <math>y</math> direction is <math>k_y = \sin\theta k_0</math>
where <math>k_0</math> is as defined previously (note that <math>k_y</math> is imaginary).

We assume here that the plate has infinite length in the <math>y</math>-direction, so the solution
does not vary in that direction except over a period. This means that the potential is now of the
form <math>\phi(x,y,z)=e^{k_y y}W(x,z)</math>

Therefore the potential can
be expanded as
<center>
<math>
\phi(x,z)=e^{-\hat{k}_0x}\phi_0(z)+\sum_{m=0}^{\infty}a_{m}e^{\hat{k}_{m}x}\phi_{m}(z), \;\;x<0
</math>
</center>
and
<center>
<math>
\phi(x,z)=\sum_{m=-2}^{\infty}b_{m}
e^{-\hat{\kappa}_{m}x}\psi_{m}(z), \;\;x>0
</math>
</center>
where <math>\hat{k}_{m} = \sqrt{k_m^2 + k_y^2}</math> and <math>\hat{\kappa}_{m} = \sqrt{\kappa_m^2 + k_y^2}</math>
where we always take the positive real root or the root with positive imaginary part.

The edge conditions are also different and are
<center><math>
\left(\frac{\partial^3}{\partial x^3} - (2 - \nu)k^2_y\frac{\partial}{\partial x}\right) \frac{\partial\phi}{\partial z}= 0,
</math></center>
and
<center><math>
\left(\frac{\partial^2}{\partial x^2} - \nu k^2_y\right)\frac{\partial\phi}{\partial z} = 0,
</math></center>
where <math>\nu</math> is Poisons ratio.

We can expend these edge conditions, which respectively gives
<center>
<math>
-\sum_{m=-2}^{\infty}b_{m}\kappa_m (\hat{\kappa}_m^3-\hat{\kappa}_m k_y^2(2-\nu))\tan \kappa_m h=0
</math>
</center>
and
<center>
<math>
-\sum_{m=-2}^{\infty}b_{m}\kappa_m (\hat{\kappa}_m^2-k_y^2\nu))\tan \kappa_m h=0
</math>
</center>

The equations are derived almost identically to those above and we obtain
<center>
<math>
A_{0}\delta_{0l}+a_{l}A_{l}
=\sum_{m=-2}^{\infty}b_{m}B_{ml}
</math>
</center>
and
<center>
<math>
-\hat{k_{0}}A_{0}\delta_{0l}+a_{l}\hat{k}_{l}A_l
=-\sum_{m=-2}^{\infty}b_{m}\hat{\kappa}_{m}B_{ml}
</math>
</center>
and these are solved exactly as before.

= Energy Balance =

The energy balance equation is ([[Energy Balance for Two Elastic Plates]]) given by
<center><math>
D |T|^2 + |R|^2 = 1, \,
</math></center>
where <math> T=b_0 </math>, <math>R=a_0</math>, and <math>D</math> is given by
<center><math>
D = \left(\frac{\hat{\kappa}_0 k_0\cos^2{(k_0 h)}}{\hat{k}_0 \kappa_0\cos^2{(\kappa_0 h)}}\right)
\left(\frac{\frac{4\beta}{\alpha}(\kappa_0)^3((-\hat{\kappa}_0)^2 +k_y^2)\sin^2{(\kappa_0 h)} +
\frac{1}{2}{\sin{(2 \kappa_0 h)}}+\kappa_0 h}
{
\frac{1}{2}{\sin{(2 k_0 h)}}+k_0h}\right)
</math></center>
where the index <math>0</math> refers to the travelling wave (imaginary root of the dispersion equation).

= Matlab Code =

A program to calculate the coefficients for the semi-infinite dock problems can be found here
[http://www.math.auckland.ac.nz/~meylan/code/eigenfunction_matching/semiinfinite_plate.m semiinfinite_plate.m]

== Additional code ==

This program requires
* {{free surface dispersion equation code}}
* {{elastic plate dispersion equation code}}

[[Category:Floating Elastic Plate]]
[[Category:Eigenfunction Matching Method]]
[[Category:Complete Pages]]

Eigenfunction Matching for a Circular Floating Elastic Plate

2008-08-04T02:04:54Z

Fmontiel: /* A Simple Method To Calculate The Diffraction Transfer Matrix For The Case Of A Circular Plate */

=Introduction=

We show here a solution for a [[Floating Elastic Plate]] on [[Finite Depth]] water
based on [[Peter_Meylan_Chung_2004a|Peter, Meylan and Chung 2004]]. A solution
for [[Shallow Depth]] was given in [[Zilman_Miloh_2000a|Zilman and Miloh 2000]] and we will also show this.
The solution is an extension of the [[Eigenfunction Matching for a Circular Dock]].

=Governing Equations=

We begin with the [[Frequency Domain Problem]] for a [[Floating Elastic Plate]]
in the non-dimensional form of [[Tayler_1986a|Tayler 1986]] ([[Dispersion Relation for a Floating Elastic Plate]])
We will use a cylindrical coordinate system, <math>(r,\theta,z)</math>,
assumed to have its origin at the centre of the circular
plate which has radius <math>a</math>. The water is assumed to have
constant finite depth <math>H</math> and the <math>z</math>-direction points vertically
upward with the water surface at <math>z=0</math> and the sea floor at <math>z=-H</math>. The
boundary value problem can therefore be expressed as
<center>
<math>
\Delta\phi=0, \,\, -H<z<0,
</math>
</center>
<center>
<math>
\phi_{z}=0, \,\, z=-H,
</math>
</center>
<center><math>
\phi_{z}=\alpha\phi, \,\, z=0,\,r>a,
</math></center>
<center>
<math>
(\beta\Delta^{2}+1-\alpha\gamma)\phi_{z}=\alpha\phi, \,\, z=0,\,r<a
</math>
</center>
where the constants <math>\beta</math> and <math>\gamma</math> are given by
<center>
<math>
\beta=\frac{D}{\rho\,L^{4}g}, \gamma=\frac{\rho_{i}h}{\rho\,L}
</math>
</center>
and <math>\rho_{i}</math> is the density of the plate. We
must also apply the edge conditions for the plate and the [[Sommerfeld Radiation Condition]]
as <math>r\rightarrow\infty</math>. The subscript <math>z</math>
denotes the derivative in <math>z</math>-direction.

=Solution Method=

== Separation of variables==

We now separate variables, noting that since the problem has
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>.
Using Laplace's equation in polar coordinates we obtain
<center>
<math>
\frac{\mathrm{d}^{2}\rho_{n}}{\mathrm{d}r^{2}}+\frac{1}{r}
\frac{\mathrm{d}\rho_{n}}{\mathrm{d}r}-\left(
\frac{n^{2}}{r^{2}}+\mu^{2}\right) \rho_{n}=0
</math>
</center>
where <math>\mu</math> is <math>k_{m}</math> or
<math>\kappa_{m},</math> depending on whether <math>r</math> is
greater or less than <math>a</math>. We can convert this equation to the
standard form by substituting <math>y=\mu r</math> to obtain
<center>
<math>
y^{2}\frac{\mathrm{d}^{2}\rho_{n}}{\mathrm{d}y^{2}}+y\frac{\mathrm{d}\rho_{n}
}{\rm{d}y}-(n^{2}+y^{2})\rho_{n}=0
</math>
</center>
The solution of this equation is a linear combination of the
modified Bessel functions of order <math>n</math>, <math>I_{n}(y)</math> and
<math>K_{n}(y)</math> [[Abramowitz and Stegun 1964]]. Since the solution must be bounded
we know that under the plate the solution will be a linear combination of
<math>I_{n}(y)</math> while outside the plate the solution will be a
linear combination of <math>K_{n}(y)</math>. Therefore the potential can
be expanded as
<center>
<math>
\phi(r,\theta,z)=\sum_{n=-\infty}^{\infty}\sum_{m=0}^{\infty}a_{mn}K_{n}
(k_{m}r)e^{i n\theta}\phi_{m}(z), \;\;r>a
</math>
</center>
and
<center>
<math>
\phi(r,\theta,z)=\sum_{n=-\infty}^{\infty}\sum_{m=-2}^{\infty}b_{mn}
I_{n}(\kappa_{m}r)e^{i n\theta}\psi_{m}(z), \;\;r<a
</math>
</center>
where <math>a_{mn}</math> and <math>b_{mn}</math>
are the coefficients of the potential in the open water and
the plate covered region respectively.

==Incident potential==

The incident potential is a wave of amplitude <math>A</math>
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 boundary conditions for the plate also have to be
considered. The vertical force and bending moment must vanish, which can be
written as
<center>
<math>
\left[\bar{\Delta}-\frac{1-\nu}{r}\left(\frac{\partial}{\partial r}
+\frac{1}{r}\frac{\partial^{2}}{\partial\theta^{2}}\right)\right]
w=0\,\,\,(3)
</math>
</center>
and
<center>
<math>
\left[ \frac{\partial}{\partial r}\bar{\Delta}-\frac{1-\nu}{r^{2}}\left(
-\frac{\partial}{\partial r}+\frac{1}{r}\right) \frac{\partial^{2}}
{\partial\theta^{2}}\right] w=0 \,\,\,(4)
</math>
</center>
where <math>w</math> is the time-independent surface
displacement, <math>\nu</math> is Poisson's ratio, and <math>\bar{\Delta}</math> is the
polar coordinate Laplacian
<center>
<math>
\bar{\Delta}=\frac{\partial^{2}}{\partial r^{2}}+\frac{1}{r}\frac{\partial
}{\partial r}+\frac{1}{r^{2}}\frac{\partial^{2}}{\partial\theta^{2}}
</math>
</center>

== Displacement of the plate ==

The surface displacement and the water velocity potential at
the water surface are linked through the kinematic boundary condition
<center>
<math>
\phi_{z}=-i\sqrt{\alpha}w,\,\,\,z=0
</math>
</center>
From the equations the potential and the surface
displacement are therefore related by
<center>
<math>
w=i\sqrt{\alpha}\phi,\quad r>a
</math>
</center>
and
<center>
<math>
(\beta\bar{\Delta}^{2}+1-\alpha\gamma)w=i\sqrt{\alpha}\phi,\quad r<a
</math>
</center>
The surface displacement can also be expanded in eigenfunctions
as
<center>
<math>
w(r,\theta)=\sum_{n=-\infty}^{\infty}\sum_{m=0}^{\infty}i\sqrt{\alpha}
a_{mn}K_{n}(k_{m}r)e^{i n\theta},\;\;r>a
</math>
</center>
and:
<center>
<math>
w(r,\theta)=
\sum_{n=-\infty}^{\infty}\sum_{m=-2}^{\infty}i\sqrt{\alpha}(\beta\kappa
_{m}^{4}+1-\alpha\gamma)^{-1}b_{mn}I_{n}(\kappa_{m}r)e^{i
n\theta},\; r<a
</math>
</center>
using the fact that
<center>
<math>
\bar{\Delta}\left( I_{n}(\kappa_{m}r)e^{i n\theta}\right) =\kappa_{m}
^{2}I_{n}(\kappa_{m}r)e^{i n\theta}\,\,\,(5)
</math>
</center>

==An infinite dimensional system of equations==

The boundary conditions (3) and
(4) can be expressed in terms of the potential
using (5). Since the angular modes are uncoupled the
conditions apply to each mode, giving
<center>
<math>
\sum_{m=-2}^{\infty}(\beta\kappa_{m}^{4}+1-\alpha\gamma)^{-1}b_{mn}
\left(\kappa_{m}^{2}I_{n}(\kappa_{m}a)-\frac{1-\nu}{a}\left(\kappa
_{m}I_{n}^{\prime}(\kappa_{m}a)-\frac{n^{2}}{a}I_{n}(\kappa_{m}a)\right)
\right) =0\,\,\,(6)
</math>
</center>
<center>
<math>
\sum_{m=-2}^{\infty}(\beta\kappa_{m}^{4}+1-\alpha\gamma)^{-1}b_{mn}
\left( \kappa_{m}^{3}I_{n}^{\prime}(\kappa_{m}a)+n^{2}\frac{1-\nu}{a^{2}
}\left( -\kappa_{m}I_{n}^{\prime}(\kappa_{m}a)+\frac{1}{a}I_{n}(\kappa
_{m}a)\right) \right)
=0\,\,\,(7)
</math>
</center>
The potential and its derivative must be continuous across the
transition from open water to the plate covered region. Therefore, the
potentials and their derivatives at <math>r=a</math> have to be equal.
Again we know that this must be true for each angle and we obtain
<center>
<math>
e_{n}I_{n}(k_{0}a)\phi_{0}\left( z\right) + \sum_{m=0}^{\infty}
a_{mn} K_{n}(k_{m}a)\phi_{m}\left( z\right)
=\sum_{m=-2}^{\infty}b_{mn}I_{n}(\kappa_{m}a)\psi_{m}(z)
</math>
</center>
and
<center>
<math>
e_{n}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)
=\sum_{m=-2}^{\infty}b_{mn}\kappa_{m}I_{n}^{\prime}(\kappa_{m}a)\psi
_{m}(z)
</math>
</center>
for each <math>n</math>.
We solve these equations by multiplying both equations by
<math>\phi_{l}(z)</math> and integrating from <math>-H</math> to <math>0</math> to obtain:
<center>
<math>
e_{n}I_{n}(k_{0}a)A_{0}\delta_{0l}+a_{ln}K_{n}(k_{l}a)A_{l}
=\sum_{m=-2}^{\infty}b_{mn}I_{n}(\kappa_{m}a)B_{ml} \,\,\,(8)
</math>
</center>
and
<center>
<math>
e_{n}k_{0}I_{n}^{\prime}(k_{0}a)A_{0}\delta_{0l}+a_{ln}k_{l}K_{n}^{\prime
}(k_{l}a)A_{l}
=\sum_{m=-2}^{\infty}b_{mn}\kappa_{m}I_{n}^{\prime}(\kappa_{m}
a)B_{ml} \,\,\,(9)
</math>
</center>
Equation (8) can be solved for the open water
coefficients <math>a_{mn}</math>
<center>
<math>
a_{ln}=-e_{n}\frac{I_{n}(k_{0}a)}{K_{n}(k_{0}a)}\delta_{0l}+\sum
_{m=-2}^{\infty}b_{mn}\frac{I_{n}(\kappa_{m}a)B_{ml}}{K_{n}(k_{l}a)A_{l}}
</math>
</center>
which can then be substituted into equation
(9) to give us
<center>
<math>
\left( k_{0}I_{n}^{\prime}(k_{0}a)-k_{0}\frac{K_{n}^{\prime}(k_{0}
a)}{K_{n}(k_{0}a)}I_{n}(k_{0}a)\right) e_{n}A_{0}\delta_{0l}
=\sum_{m=-2}^{\infty}\left( \kappa_{m}I_{n}^{\prime}(\kappa_{m}
a)-k_{l}\frac{K_{n}^{\prime}(k_{l}a)}{K_{n}(k_{l}a)}I_{n}(\kappa
_{m}a)\right) B_{ml}b_{mn}\,\,\,(10)
</math>
</center>
for each <math>n</math>.
Together with equations (6) and (7)
equation (10) gives the required equations to solve for the
coefficients of the water velocity potential in the plate covered region.

=Numerical Solution=

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>
<math>
\phi(r,\theta,z)=\sum_{n=-N}^{N}\sum_{m=0}^{M}a_{mn}K_{n}(k_{m}r)e^{i
n\theta }\phi_{m}(z), \;\;r>a
</math>
</center>
and
<center>
<math>
\phi(r,\theta,z)=\sum_{n=-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]]=

The shallow water theory of [[Zilman_Miloh_2000a|Zilman and Miloh 2000]] can be recovered by
simply setting the depth shallow enough that the shallow water theory is valid
and setting <math>M=0</math>. If the shallow water theory is valid then
the first three roots of the dispersion equation for the ice will be exactly
the same roots found in the shallow water theory by solving the polynomial
equation. The system of equations has four unknowns (three under the plate and
one in the open water) exactly as for the theory of [[Zilman_Miloh_2000a|Zilman and Miloh 2000]].

= A Simple Method To Calculate The [[Diffraction Transfer Matrix]] For The Case Of A Circular Plate =

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
<center><math>
\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)
=\sum_{m=-2}^{\infty}b_{mn}I_{n}(\kappa_{m}a)\psi_{m}(z)
</math></center>
and
<center><math>
\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}(k_{m}a)\phi_{m}\left( z\right)
=\sum_{m=-2}^{\infty}b_{mn} \kappa_m I'_{n}(\kappa_{m}a)\psi_{m}(z)
</math></center>
for each <math>n</math>.
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:
<center><math>
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},\ \ \ (11)
</math></center>
and
<center><math>
D_{ln} k_l I'_n (k_l a) A_l + a_{ln} k_l K'_{n}(k_{l} a) A_l
=\sum_{m=-2}^{\infty}b_{mn} \kappa_m I'_{n}(\kappa_{m}a) B_{ml} \ \ \ (12)
</math></center>

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>

Furthermore the boundary conditions are exactly the same that before, namely
<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
</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 .
</math>
</center>
For the further study, let's call
<center>
<math>
L^1_{mn}=(\beta\kappa_{m}^{4}+1-\alpha\gamma)^{-1}
\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)
</math>
</center>
and
<center>
<math>
L^2_{mn}=(\beta\kappa_{m}^{4}+1-\alpha\gamma)^{-1}
\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)
</math>
</center>
From the equations (11), (12) and the boundary conditions over the edges of the plate,
we can write a linear system of equation, limiting the number of modes of the dispersion equation
to <math>N</math> real ones
<center>
<math>
\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}
I_n(\kappa_{-2}a) B_{-20}&\cdots&I_n(\kappa_{N}a) B_{N0}\\
&&\\
\vdots&&\vdots\\
&&\\
I_n(\kappa_{-2}a) B_{-2N}&\cdots&I_n(\kappa_{N}a) B_{NN}
\end{bmatrix}
\\
\begin{bmatrix}
0&&\cdots&&0\\
0&&\cdots&&0
\end{bmatrix}
&
\begin{bmatrix}
L^1_{-2n}&\cdots&L^1_{Nn}\\
L^2_{-2n}&\cdots&L^2_{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}
\kappa_{-2} I'_n(\kappa_{-2}a) B_{-20}&\cdots&\kappa_{N} I'_n(\kappa_{N}a) B_{N0}\\
&&\\
\vdots&&\vdots\\
&&\\
\kappa_{-2} I'_n(\kappa_{-2}a) B_{-2N}&\cdots&\kappa_{N} I'_n(\kappa_{N}a) B_{NN}
\end{bmatrix}
\end{bmatrix}
\begin{bmatrix}
a_{0n} \\
\\
\vdots \\
\\
a_{Nn} \\
b_{-2n}\\
b_{-1n}\\
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}
0&&\cdots&&0\\
0&&\cdots&&0
\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} \\
0\\
0\\
D_{0n}\\
\\
\vdots \\
\\
D_{Nn}
\end{bmatrix}
</math>
</center>
for each <math>n</math>.

Therefore we can find a [[Diffraction Transfer Matrix]] for each <math>n</math>,
by setting
<center><math>
\forall i \in [0, N], (D_{pn})_{p \in [0, N]} = \delta_{ip}
</math></center>
Then we solve the linear system defined previously, so that we can find the coefficients
<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]],
<math>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>n</math>, we finally obtain a 3-dimensional
matrix for the [[Diffraction Transfer Matrix]].

= Matlab Code =

A program to calculate the coefficients for circular plate problems can be found here
[http://www.math.auckland.ac.nz/~meylan/code/eigenfunction_matching/circle_plate_matching_one_n.m circle_plate_matching_one_n.m]
Note that this problem solves only for a single n.

== Additional code ==

This program requires
[http://www.math.auckland.ac.nz/~meylan/code/dispersion/dispersion_free_surface.m dispersion_free_surface.m]
and [http://www.math.auckland.ac.nz/~meylan/code/dispersion/dispersion_elastic_surface.m dispersion_elastic_surface.m]
to run

[[Category:Problems with Cylindrical Symmetry]]
[[Category:Linear Hydroelasticity]]
[[Category:Eigenfunction Matching Method]]
[[Category:Complete Pages]]

Eigenfunction Matching for a Circular Dock

2008-08-04T02:04:28Z

Fmontiel: /* A Simple Method To Calculate The Diffraction Transfer Matrix For The Case Of A Circular Dock */

=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>(r,\theta,z)</math>,
assumed to have its origin at the centre of the circular
plate which has radius <math>a</math>. The water is assumed to have
constant finite depth <math>h</math> and the <math>z</math>-direction points vertically
upward with the water surface at <math>z=0</math> and the sea floor at <math>z=-h</math>. The
boundary value problem can therefore be expressed as
<center>
<math>
\Delta\phi=0, \,\, -h<z<0,
</math>
</center>
<center>
<math>
\phi_{z}=0, \,\, z=-h,
</math>
</center>
<center><math>
\phi_{z}=\alpha\phi, \,\, z=0,\,r>a,
</math></center>
<center>
<math>
\phi_{z}=0, \,\, z=0,\,r<a
</math>
</center>
We must also apply the [[Sommerfeld Radiation Condition]]
as <math>r\rightarrow\infty</math>. The subscript <math>z</math>
denotes the derivative in <math>z</math>-direction.

=Solution Method=

== Separation of variables==

We now separate variables, noting that since the problem has
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
</math>
</center>
and
<center>
<math>
\kappa\tan(\kappa h)=0,\quad
r<a
</math>
</center>
Note that we have set <math>\mu=k</math> under the free
surface and <math>\mu=\kappa</math> under the dock. We denote the
positive imaginary solution of the [[Dispersion Relation for a Free Surface]]
by <math>k_{0}</math> and
the positive real solutions by <math>k_{m}</math>, <math>m\geq1</math>. The solutions of
second equation will be denoted by
<math>\kappa_{m} = m\pi/h</math>, <math>m\geq 0</math>.

We define
<center>
<math>
\phi_{m}\left( z\right) =\frac{\cos k_{m}(z+h)}{\cos k_{m}h},\quad m\geq0
</math>
</center>
as the vertical eigenfunction of the potential in the open
water region and
<center>
<math>
\psi_{m}\left( z\right) = \cos\kappa_{m}(z+h),\quad
m\geq 0
</math>
</center>
as the vertical eigenfunction of the potential in the dock
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\right) \left( k_{n}
^{2}-\kappa_{m}^{2}\right) }
</math></center>
and
<center>
<math>
\int\nolimits_{-h}^{0}\psi_{m}(z)\psi_{n}(z) d z=C_{m}\delta_{mn}
</math>
</center>
where
<center>
<math>
C_{m}=\frac{1}{2}h,\quad,m\neq 0 \quad \mathrm{and} \quad C_0 = h
</math></center>

We now solve for the function <math>\rho_{n}(r)</math>.
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> (provided that
<math>\mu\neq 0</math>to obtain
<center>
<math>
y^{2}\frac{\mathrm{d}^{2}\rho_{n}}{\mathrm{d}y^{2}}+y\frac{\mathrm{d}\rho_{n}
}{\rm{d}y}-(n^{2}+y^{2})\rho_{n}=0
</math>
</center>
The solution of this equation is a linear combination of the
modified Bessel functions of order <math>n</math>, <math>I_{n}(y)</math> and
<math>K_{n}(y)</math> ([[Abramowitz and Stegun 1964]]). Since the solution must be bounded
we know that under the plate the solution will be a linear combination of
<math>I_{n}(y)</math> while outside the plate the solution will be a
linear combination of <math>K_{n}(y)</math>.
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
<center>
<math>
\phi(r,\theta,z)=\sum_{n=-\infty}^{\infty}\sum_{m=0}^{\infty}a_{mn}K_{n}
(k_{m}r)e^{i n\theta}\phi_{m}(z), \;\;r>a
</math>
</center>
and
<center>
<math>
\phi(r,\theta,z)=\sum_{n=-\infty}^{\infty}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
</math>
</center>
where <math>a_{mn}</math> and <math>b_{mn}</math>
are the coefficients of the potential in the open water and
the plate covered region respectively.

==Incident potential==

The incident potential is a wave of amplitude <math>A</math>
in displacement travelling in the positive <math>x</math>-direction.
The incident potential can therefore be written as
<center>
<math>
\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>
</center>

==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>r=a</math> have to be equal
for each angle and we obtain
<center>
<math>
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>
</center>
and
<center>
<math>
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>
</center>
for each <math>n</math>.
We solve these equations by multiplying both equations by
<math>\phi_{l}(z)</math> and integrating from <math>-H</math> to <math>0</math> to obtain:
<center>
<math>
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>
</center>
and
<center>
<math>
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>
</center>

= 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

<center>
<math>
\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>
</center>
and
<center>
<math>
\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>
</center>
for each <math>n</math>.
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:

<center>
<math>
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>
</center>
and
<center>
<math>
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>
</center>

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

<center>
<math>
\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>
</center>

for each <math>n</math>.

Therefore we can find a [[Diffraction Transfer Matrix]] for each <math>n</math>,
by setting
<center><math>
\forall i \in [0, N], (D_{pn})_{p \in [0, N]} = \delta_{ip}
</math></center>
Then we solve the linear system defined previously, so that we can find the coefficients
<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]],
<math>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>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
[http://www.math.auckland.ac.nz/~meylan/code/eigenfunction_matching/circle_dock_matching_one_n.m circle_dock_matching_one_n.m]
Note that this problem solves only for a single n.

== Additional code ==

This program requires
[http://www.math.auckland.ac.nz/~meylan/code/dispersion/dispersion_free_surface.m dispersion_free_surface.m]
to run

[[Category:Problems with Cylindrical Symmetry]]
[[Category:Eigenfunction Matching Method]]
[[Category:Pages with Matlab Code]]
[[Category:Complete Pages]]

Eigenfunction Matching for a Circular Dock

2008-08-04T01:55:27Z

Fmontiel:

=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>(r,\theta,z)</math>,
assumed to have its origin at the centre of the circular
plate which has radius <math>a</math>. The water is assumed to have
constant finite depth <math>h</math> and the <math>z</math>-direction points vertically
upward with the water surface at <math>z=0</math> and the sea floor at <math>z=-h</math>. The
boundary value problem can therefore be expressed as
<center>
<math>
\Delta\phi=0, \,\, -h<z<0,
</math>
</center>
<center>
<math>
\phi_{z}=0, \,\, z=-h,
</math>
</center>
<center><math>
\phi_{z}=\alpha\phi, \,\, z=0,\,r>a,
</math></center>
<center>
<math>
\phi_{z}=0, \,\, z=0,\,r<a
</math>
</center>
We must also apply the [[Sommerfeld Radiation Condition]]
as <math>r\rightarrow\infty</math>. The subscript <math>z</math>
denotes the derivative in <math>z</math>-direction.

=Solution Method=

== Separation of variables==

We now separate variables, noting that since the problem has
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
</math>
</center>
and
<center>
<math>
\kappa\tan(\kappa h)=0,\quad
r<a
</math>
</center>
Note that we have set <math>\mu=k</math> under the free
surface and <math>\mu=\kappa</math> under the dock. We denote the
positive imaginary solution of the [[Dispersion Relation for a Free Surface]]
by <math>k_{0}</math> and
the positive real solutions by <math>k_{m}</math>, <math>m\geq1</math>. The solutions of
second equation will be denoted by
<math>\kappa_{m} = m\pi/h</math>, <math>m\geq 0</math>.

We define
<center>
<math>
\phi_{m}\left( z\right) =\frac{\cos k_{m}(z+h)}{\cos k_{m}h},\quad m\geq0
</math>
</center>
as the vertical eigenfunction of the potential in the open
water region and
<center>
<math>
\psi_{m}\left( z\right) = \cos\kappa_{m}(z+h),\quad
m\geq 0
</math>
</center>
as the vertical eigenfunction of the potential in the dock
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\right) \left( k_{n}
^{2}-\kappa_{m}^{2}\right) }
</math></center>
and
<center>
<math>
\int\nolimits_{-h}^{0}\psi_{m}(z)\psi_{n}(z) d z=C_{m}\delta_{mn}
</math>
</center>
where
<center>
<math>
C_{m}=\frac{1}{2}h,\quad,m\neq 0 \quad \mathrm{and} \quad C_0 = h
</math></center>

We now solve for the function <math>\rho_{n}(r)</math>.
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> (provided that
<math>\mu\neq 0</math>to obtain
<center>
<math>
y^{2}\frac{\mathrm{d}^{2}\rho_{n}}{\mathrm{d}y^{2}}+y\frac{\mathrm{d}\rho_{n}
}{\rm{d}y}-(n^{2}+y^{2})\rho_{n}=0
</math>
</center>
The solution of this equation is a linear combination of the
modified Bessel functions of order <math>n</math>, <math>I_{n}(y)</math> and
<math>K_{n}(y)</math> ([[Abramowitz and Stegun 1964]]). Since the solution must be bounded
we know that under the plate the solution will be a linear combination of
<math>I_{n}(y)</math> while outside the plate the solution will be a
linear combination of <math>K_{n}(y)</math>.
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
<center>
<math>
\phi(r,\theta,z)=\sum_{n=-\infty}^{\infty}\sum_{m=0}^{\infty}a_{mn}K_{n}
(k_{m}r)e^{i n\theta}\phi_{m}(z), \;\;r>a
</math>
</center>
and
<center>
<math>
\phi(r,\theta,z)=\sum_{n=-\infty}^{\infty}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
</math>
</center>
where <math>a_{mn}</math> and <math>b_{mn}</math>
are the coefficients of the potential in the open water and
the plate covered region respectively.

==Incident potential==

The incident potential is a wave of amplitude <math>A</math>
in displacement travelling in the positive <math>x</math>-direction.
The incident potential can therefore be written as
<center>
<math>
\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>
</center>

==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>r=a</math> have to be equal
for each angle and we obtain
<center>
<math>
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>
</center>
and
<center>
<math>
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>
</center>
for each <math>n</math>.
We solve these equations by multiplying both equations by
<math>\phi_{l}(z)</math> and integrating from <math>-H</math> to <math>0</math> to obtain:
<center>
<math>
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>
</center>
and
<center>
<math>
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>
</center>

= 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

<center>
<math>
\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>
</center>
and
<center>
<math>
\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>
</center>
for each <math>n</math>.
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:

<center>
<math>
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>
</center>
and
<center>
<math>
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>
</center>

The [[Diffraction Transfer Matrix]]
maps the coefficients of the incident wave with the coefficients of the scattered wave within
the free 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

<center>
<math>
\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>
</center>

for each <math>n</math>.

Therefore we can find a [[Diffraction Transfer Matrix]] for each <math>n</math>,
by setting
<center><math>
\forall i \in [0, N], (D_{pn})_{p \in [0, N]} = \delta_{ip}
</math></center>
Then we solve the linear system defined previously, so that we can find the coefficients
<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]],
<math>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>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
[http://www.math.auckland.ac.nz/~meylan/code/eigenfunction_matching/circle_dock_matching_one_n.m circle_dock_matching_one_n.m]
Note that this problem solves only for a single n.

== Additional code ==

This program requires
[http://www.math.auckland.ac.nz/~meylan/code/dispersion/dispersion_free_surface.m dispersion_free_surface.m]
to run

[[Category:Problems with Cylindrical Symmetry]]
[[Category:Eigenfunction Matching Method]]
[[Category:Pages with Matlab Code]]
[[Category:Complete Pages]]

Eigenfunction Matching for a Circular Dock

2008-08-04T01:54:32Z

Fmontiel: /* A Simple Method To Calculate The Diffraction Transfer Matrix For The Case Of A Circular Plate */

=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>(r,\theta,z)</math>,
assumed to have its origin at the centre of the circular
plate which has radius <math>a</math>. The water is assumed to have
constant finite depth <math>h</math> and the <math>z</math>-direction points vertically
upward with the water surface at <math>z=0</math> and the sea floor at <math>z=-h</math>. The
boundary value problem can therefore be expressed as
<center>
<math>
\Delta\phi=0, \,\, -h<z<0,
</math>
</center>
<center>
<math>
\phi_{z}=0, \,\, z=-h,
</math>
</center>
<center><math>
\phi_{z}=\alpha\phi, \,\, z=0,\,r>a,
</math></center>
<center>
<math>
\phi_{z}=0, \,\, z=0,\,r<a
</math>
</center>
We must also apply the [[Sommerfeld Radiation Condition]]
as <math>r\rightarrow\infty</math>. The subscript <math>z</math>
denotes the derivative in <math>z</math>-direction.

=Solution Method=

== Separation of variables==

We now separate variables, noting that since the problem has
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
</math>
</center>
and
<center>
<math>
\kappa\tan(\kappa h)=0,\quad
r<a
</math>
</center>
Note that we have set <math>\mu=k</math> under the free
surface and <math>\mu=\kappa</math> under the dock. We denote the
positive imaginary solution of the [[Dispersion Relation for a Free Surface]]
by <math>k_{0}</math> and
the positive real solutions by <math>k_{m}</math>, <math>m\geq1</math>. The solutions of
second equation will be denoted by
<math>\kappa_{m} = m\pi/h</math>, <math>m\geq 0</math>.

We define
<center>
<math>
\phi_{m}\left( z\right) =\frac{\cos k_{m}(z+h)}{\cos k_{m}h},\quad m\geq0
</math>
</center>
as the vertical eigenfunction of the potential in the open
water region and
<center>
<math>
\psi_{m}\left( z\right) = \cos\kappa_{m}(z+h),\quad
m\geq 0
</math>
</center>
as the vertical eigenfunction of the potential in the dock
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\right) \left( k_{n}
^{2}-\kappa_{m}^{2}\right) }
</math></center>
and
<center>
<math>
\int\nolimits_{-h}^{0}\psi_{m}(z)\psi_{n}(z) d z=C_{m}\delta_{mn}
</math>
</center>
where
<center>
<math>
C_{m}=\frac{1}{2}h,\quad,m\neq 0 \quad \mathrm{and} \quad C_0 = h
</math></center>

We now solve for the function <math>\rho_{n}(r)</math>.
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> (provided that
<math>\mu\neq 0</math>to obtain
<center>
<math>
y^{2}\frac{\mathrm{d}^{2}\rho_{n}}{\mathrm{d}y^{2}}+y\frac{\mathrm{d}\rho_{n}
}{\rm{d}y}-(n^{2}+y^{2})\rho_{n}=0
</math>
</center>
The solution of this equation is a linear combination of the
modified Bessel functions of order <math>n</math>, <math>I_{n}(y)</math> and
<math>K_{n}(y)</math> ([[Abramowitz and Stegun 1964]]). Since the solution must be bounded
we know that under the plate the solution will be a linear combination of
<math>I_{n}(y)</math> while outside the plate the solution will be a
linear combination of <math>K_{n}(y)</math>.
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
<center>
<math>
\phi(r,\theta,z)=\sum_{n=-\infty}^{\infty}\sum_{m=0}^{\infty}a_{mn}K_{n}
(k_{m}r)e^{i n\theta}\phi_{m}(z), \;\;r>a
</math>
</center>
and
<center>
<math>
\phi(r,\theta,z)=\sum_{n=-\infty}^{\infty}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
</math>
</center>
where <math>a_{mn}</math> and <math>b_{mn}</math>
are the coefficients of the potential in the open water and
the plate covered region respectively.

==Incident potential==

The incident potential is a wave of amplitude <math>A</math>
in displacement travelling in the positive <math>x</math>-direction.
The incident potential can therefore be written as
<center>
<math>
\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>
</center>

==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>r=a</math> have to be equal
for each angle and we obtain
<center>
<math>
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>
</center>
and
<center>
<math>
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>
</center>
for each <math>n</math>.
We solve these equations by multiplying both equations by
<math>\phi_{l}(z)</math> and integrating from <math>-H</math> to <math>0</math> to obtain:
<center>
<math>
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>
</center>
and
<center>
<math>
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>
</center>

= 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

<center>
<math>
\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>
</center>
and
<center>
<math>
\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>
</center>
for each <math>n</math>.
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:

<center>
<math>
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>
</center>
and
<center>
<math>
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>
</center>

The [[Diffraction Transfer Matrix]]
maps the coefficients of the incident wave with the coefficients of the scattered wave within
the free 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

<center>
<math>
\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>
</center>

for each <math>n</math>.

Therefore we can find a [[Diffraction Transfer Matrix]] for each <math>n</math>,
by setting
<center><math>
\forall i \in [0, N], (D_{pn})_{p \in [0, N]} = \delta_{ip}
</math></center>
Then we solve the linear system defined previously, so that we can find the coefficients
<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]],
<math>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>n</math>, we finally obtain a 3-dimensional
matrix for the [[Diffraction Transfer Matrix]].

= Numerical Solution =

To solve the system of equations we set the upper limit of <math>l</math> to
be <math>M</math>.

= Matlab Code =

A program to calculate the coefficients for circular dock problems can be found here
[http://www.math.auckland.ac.nz/~meylan/code/eigenfunction_matching/circle_dock_matching_one_n.m circle_dock_matching_one_n.m]
Note that this problem solves only for a single n.

== Additional code ==

This program requires
[http://www.math.auckland.ac.nz/~meylan/code/dispersion/dispersion_free_surface.m dispersion_free_surface.m]
to run

[[Category:Problems with Cylindrical Symmetry]]
[[Category:Eigenfunction Matching Method]]
[[Category:Pages with Matlab Code]]
[[Category:Complete Pages]]

Eigenfunction Matching for a Circular Floating Elastic Plate

2008-08-04T01:52:48Z

Fmontiel: /* A Simple Method To Calculate The Diffraction Transfer Matrix For The Case Of A Circular Plate */

=Introduction=

We show here a solution for a [[Floating Elastic Plate]] on [[Finite Depth]] water
based on [[Peter_Meylan_Chung_2004a|Peter, Meylan and Chung 2004]]. A solution
for [[Shallow Depth]] was given in [[Zilman_Miloh_2000a|Zilman and Miloh 2000]] and we will also show this.
The solution is an extension of the [[Eigenfunction Matching for a Circular Dock]].

=Governing Equations=

We begin with the [[Frequency Domain Problem]] for a [[Floating Elastic Plate]]
in the non-dimensional form of [[Tayler_1986a|Tayler 1986]] ([[Dispersion Relation for a Floating Elastic Plate]])
We will use a cylindrical coordinate system, <math>(r,\theta,z)</math>,
assumed to have its origin at the centre of the circular
plate which has radius <math>a</math>. The water is assumed to have
constant finite depth <math>H</math> and the <math>z</math>-direction points vertically
upward with the water surface at <math>z=0</math> and the sea floor at <math>z=-H</math>. The
boundary value problem can therefore be expressed as
<center>
<math>
\Delta\phi=0, \,\, -H<z<0,
</math>
</center>
<center>
<math>
\phi_{z}=0, \,\, z=-H,
</math>
</center>
<center><math>
\phi_{z}=\alpha\phi, \,\, z=0,\,r>a,
</math></center>
<center>
<math>
(\beta\Delta^{2}+1-\alpha\gamma)\phi_{z}=\alpha\phi, \,\, z=0,\,r<a
</math>
</center>
where the constants <math>\beta</math> and <math>\gamma</math> are given by
<center>
<math>
\beta=\frac{D}{\rho\,L^{4}g}, \gamma=\frac{\rho_{i}h}{\rho\,L}
</math>
</center>
and <math>\rho_{i}</math> is the density of the plate. We
must also apply the edge conditions for the plate and the [[Sommerfeld Radiation Condition]]
as <math>r\rightarrow\infty</math>. The subscript <math>z</math>
denotes the derivative in <math>z</math>-direction.

=Solution Method=

== Separation of variables==

We now separate variables, noting that since the problem has
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>.
Using Laplace's equation in polar coordinates we obtain
<center>
<math>
\frac{\mathrm{d}^{2}\rho_{n}}{\mathrm{d}r^{2}}+\frac{1}{r}
\frac{\mathrm{d}\rho_{n}}{\mathrm{d}r}-\left(
\frac{n^{2}}{r^{2}}+\mu^{2}\right) \rho_{n}=0
</math>
</center>
where <math>\mu</math> is <math>k_{m}</math> or
<math>\kappa_{m},</math> depending on whether <math>r</math> is
greater or less than <math>a</math>. We can convert this equation to the
standard form by substituting <math>y=\mu r</math> to obtain
<center>
<math>
y^{2}\frac{\mathrm{d}^{2}\rho_{n}}{\mathrm{d}y^{2}}+y\frac{\mathrm{d}\rho_{n}
}{\rm{d}y}-(n^{2}+y^{2})\rho_{n}=0
</math>
</center>
The solution of this equation is a linear combination of the
modified Bessel functions of order <math>n</math>, <math>I_{n}(y)</math> and
<math>K_{n}(y)</math> [[Abramowitz and Stegun 1964]]. Since the solution must be bounded
we know that under the plate the solution will be a linear combination of
<math>I_{n}(y)</math> while outside the plate the solution will be a
linear combination of <math>K_{n}(y)</math>. Therefore the potential can
be expanded as
<center>
<math>
\phi(r,\theta,z)=\sum_{n=-\infty}^{\infty}\sum_{m=0}^{\infty}a_{mn}K_{n}
(k_{m}r)e^{i n\theta}\phi_{m}(z), \;\;r>a
</math>
</center>
and
<center>
<math>
\phi(r,\theta,z)=\sum_{n=-\infty}^{\infty}\sum_{m=-2}^{\infty}b_{mn}
I_{n}(\kappa_{m}r)e^{i n\theta}\psi_{m}(z), \;\;r<a
</math>
</center>
where <math>a_{mn}</math> and <math>b_{mn}</math>
are the coefficients of the potential in the open water and
the plate covered region respectively.

==Incident potential==

The incident potential is a wave of amplitude <math>A</math>
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 boundary conditions for the plate also have to be
considered. The vertical force and bending moment must vanish, which can be
written as
<center>
<math>
\left[\bar{\Delta}-\frac{1-\nu}{r}\left(\frac{\partial}{\partial r}
+\frac{1}{r}\frac{\partial^{2}}{\partial\theta^{2}}\right)\right]
w=0\,\,\,(3)
</math>
</center>
and
<center>
<math>
\left[ \frac{\partial}{\partial r}\bar{\Delta}-\frac{1-\nu}{r^{2}}\left(
-\frac{\partial}{\partial r}+\frac{1}{r}\right) \frac{\partial^{2}}
{\partial\theta^{2}}\right] w=0 \,\,\,(4)
</math>
</center>
where <math>w</math> is the time-independent surface
displacement, <math>\nu</math> is Poisson's ratio, and <math>\bar{\Delta}</math> is the
polar coordinate Laplacian
<center>
<math>
\bar{\Delta}=\frac{\partial^{2}}{\partial r^{2}}+\frac{1}{r}\frac{\partial
}{\partial r}+\frac{1}{r^{2}}\frac{\partial^{2}}{\partial\theta^{2}}
</math>
</center>

== Displacement of the plate ==

The surface displacement and the water velocity potential at
the water surface are linked through the kinematic boundary condition
<center>
<math>
\phi_{z}=-i\sqrt{\alpha}w,\,\,\,z=0
</math>
</center>
From the equations the potential and the surface
displacement are therefore related by
<center>
<math>
w=i\sqrt{\alpha}\phi,\quad r>a
</math>
</center>
and
<center>
<math>
(\beta\bar{\Delta}^{2}+1-\alpha\gamma)w=i\sqrt{\alpha}\phi,\quad r<a
</math>
</center>
The surface displacement can also be expanded in eigenfunctions
as
<center>
<math>
w(r,\theta)=\sum_{n=-\infty}^{\infty}\sum_{m=0}^{\infty}i\sqrt{\alpha}
a_{mn}K_{n}(k_{m}r)e^{i n\theta},\;\;r>a
</math>
</center>
and:
<center>
<math>
w(r,\theta)=
\sum_{n=-\infty}^{\infty}\sum_{m=-2}^{\infty}i\sqrt{\alpha}(\beta\kappa
_{m}^{4}+1-\alpha\gamma)^{-1}b_{mn}I_{n}(\kappa_{m}r)e^{i
n\theta},\; r<a
</math>
</center>
using the fact that
<center>
<math>
\bar{\Delta}\left( I_{n}(\kappa_{m}r)e^{i n\theta}\right) =\kappa_{m}
^{2}I_{n}(\kappa_{m}r)e^{i n\theta}\,\,\,(5)
</math>
</center>

==An infinite dimensional system of equations==

The boundary conditions (3) and
(4) can be expressed in terms of the potential
using (5). Since the angular modes are uncoupled the
conditions apply to each mode, giving
<center>
<math>
\sum_{m=-2}^{\infty}(\beta\kappa_{m}^{4}+1-\alpha\gamma)^{-1}b_{mn}
\left(\kappa_{m}^{2}I_{n}(\kappa_{m}a)-\frac{1-\nu}{a}\left(\kappa
_{m}I_{n}^{\prime}(\kappa_{m}a)-\frac{n^{2}}{a}I_{n}(\kappa_{m}a)\right)
\right) =0\,\,\,(6)
</math>
</center>
<center>
<math>
\sum_{m=-2}^{\infty}(\beta\kappa_{m}^{4}+1-\alpha\gamma)^{-1}b_{mn}
\left( \kappa_{m}^{3}I_{n}^{\prime}(\kappa_{m}a)+n^{2}\frac{1-\nu}{a^{2}
}\left( -\kappa_{m}I_{n}^{\prime}(\kappa_{m}a)+\frac{1}{a}I_{n}(\kappa
_{m}a)\right) \right)
=0\,\,\,(7)
</math>
</center>
The potential and its derivative must be continuous across the
transition from open water to the plate covered region. Therefore, the
potentials and their derivatives at <math>r=a</math> have to be equal.
Again we know that this must be true for each angle and we obtain
<center>
<math>
e_{n}I_{n}(k_{0}a)\phi_{0}\left( z\right) + \sum_{m=0}^{\infty}
a_{mn} K_{n}(k_{m}a)\phi_{m}\left( z\right)
=\sum_{m=-2}^{\infty}b_{mn}I_{n}(\kappa_{m}a)\psi_{m}(z)
</math>
</center>
and
<center>
<math>
e_{n}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)
=\sum_{m=-2}^{\infty}b_{mn}\kappa_{m}I_{n}^{\prime}(\kappa_{m}a)\psi
_{m}(z)
</math>
</center>
for each <math>n</math>.
We solve these equations by multiplying both equations by
<math>\phi_{l}(z)</math> and integrating from <math>-H</math> to <math>0</math> to obtain:
<center>
<math>
e_{n}I_{n}(k_{0}a)A_{0}\delta_{0l}+a_{ln}K_{n}(k_{l}a)A_{l}
=\sum_{m=-2}^{\infty}b_{mn}I_{n}(\kappa_{m}a)B_{ml} \,\,\,(8)
</math>
</center>
and
<center>
<math>
e_{n}k_{0}I_{n}^{\prime}(k_{0}a)A_{0}\delta_{0l}+a_{ln}k_{l}K_{n}^{\prime
}(k_{l}a)A_{l}
=\sum_{m=-2}^{\infty}b_{mn}\kappa_{m}I_{n}^{\prime}(\kappa_{m}
a)B_{ml} \,\,\,(9)
</math>
</center>
Equation (8) can be solved for the open water
coefficients <math>a_{mn}</math>
<center>
<math>
a_{ln}=-e_{n}\frac{I_{n}(k_{0}a)}{K_{n}(k_{0}a)}\delta_{0l}+\sum
_{m=-2}^{\infty}b_{mn}\frac{I_{n}(\kappa_{m}a)B_{ml}}{K_{n}(k_{l}a)A_{l}}
</math>
</center>
which can then be substituted into equation
(9) to give us
<center>
<math>
\left( k_{0}I_{n}^{\prime}(k_{0}a)-k_{0}\frac{K_{n}^{\prime}(k_{0}
a)}{K_{n}(k_{0}a)}I_{n}(k_{0}a)\right) e_{n}A_{0}\delta_{0l}
=\sum_{m=-2}^{\infty}\left( \kappa_{m}I_{n}^{\prime}(\kappa_{m}
a)-k_{l}\frac{K_{n}^{\prime}(k_{l}a)}{K_{n}(k_{l}a)}I_{n}(\kappa
_{m}a)\right) B_{ml}b_{mn}\,\,\,(10)
</math>
</center>
for each <math>n</math>.
Together with equations (6) and (7)
equation (10) gives the required equations to solve for the
coefficients of the water velocity potential in the plate covered region.

=Numerical Solution=

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>
<math>
\phi(r,\theta,z)=\sum_{n=-N}^{N}\sum_{m=0}^{M}a_{mn}K_{n}(k_{m}r)e^{i
n\theta }\phi_{m}(z), \;\;r>a
</math>
</center>
and
<center>
<math>
\phi(r,\theta,z)=\sum_{n=-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]]=

The shallow water theory of [[Zilman_Miloh_2000a|Zilman and Miloh 2000]] can be recovered by
simply setting the depth shallow enough that the shallow water theory is valid
and setting <math>M=0</math>. If the shallow water theory is valid then
the first three roots of the dispersion equation for the ice will be exactly
the same roots found in the shallow water theory by solving the polynomial
equation. The system of equations has four unknowns (three under the plate and
one in the open water) exactly as for the theory of [[Zilman_Miloh_2000a|Zilman and Miloh 2000]].

= A Simple Method To Calculate The [[Diffraction Transfer Matrix]] For The Case Of A Circular Plate =

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
<center><math>
\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)
=\sum_{m=-2}^{\infty}b_{mn}I_{n}(\kappa_{m}a)\psi_{m}(z)
</math></center>
and
<center><math>
\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}(k_{m}a)\phi_{m}\left( z\right)
=\sum_{m=-2}^{\infty}b_{mn} \kappa_m I'_{n}(\kappa_{m}a)\psi_{m}(z)
</math></center>
for each <math>n</math>.
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:
<center><math>
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},\ \ \ (11)
</math></center>
and
<center><math>
D_{ln} k_l I'_n (k_l a) A_l + a_{ln} k_l K'_{n}(k_{l} a) A_l
=\sum_{m=-2}^{\infty}b_{mn} \kappa_m I'_{n}(\kappa_{m}a) B_{ml} \ \ \ (12)
</math></center>

The [[Diffraction Transfer Matrix]]
maps the coefficients of the incident wave with the coefficients of the scattered wave within
the free 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>

Furthermore the boundary conditions are exactly the same that before, namely
<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
</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 .
</math>
</center>
For the further study, let's call
<center>
<math>
L^1_{mn}=(\beta\kappa_{m}^{4}+1-\alpha\gamma)^{-1}
\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)
</math>
</center>
and
<center>
<math>
L^2_{mn}=(\beta\kappa_{m}^{4}+1-\alpha\gamma)^{-1}
\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)
</math>
</center>
From the equations (11), (12) and the boundary conditions over the edges of the plate,
we can write a linear system of equation, limiting the number of modes of the dispersion equation
to <math>N</math> real ones
<center>
<math>
\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}
I_n(\kappa_{-2}a) B_{-20}&\cdots&I_n(\kappa_{N}a) B_{N0}\\
&&\\
\vdots&&\vdots\\
&&\\
I_n(\kappa_{-2}a) B_{-2N}&\cdots&I_n(\kappa_{N}a) B_{NN}
\end{bmatrix}
\\
\begin{bmatrix}
0&&\cdots&&0\\
0&&\cdots&&0
\end{bmatrix}
&
\begin{bmatrix}
L^1_{-2n}&\cdots&L^1_{Nn}\\
L^2_{-2n}&\cdots&L^2_{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}
\kappa_{-2} I'_n(\kappa_{-2}a) B_{-20}&\cdots&\kappa_{N} I'_n(\kappa_{N}a) B_{N0}\\
&&\\
\vdots&&\vdots\\
&&\\
\kappa_{-2} I'_n(\kappa_{-2}a) B_{-2N}&\cdots&\kappa_{N} I'_n(\kappa_{N}a) B_{NN}
\end{bmatrix}
\end{bmatrix}
\begin{bmatrix}
a_{0n} \\
\\
\vdots \\
\\
a_{Nn} \\
b_{-2n}\\
b_{-1n}\\
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}
0&&\cdots&&0\\
0&&\cdots&&0
\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} \\
0\\
0\\
D_{0n}\\
\\
\vdots \\
\\
D_{Nn}
\end{bmatrix}
</math>
</center>
for each <math>n</math>.

Therefore we can find a [[Diffraction Transfer Matrix]] for each <math>n</math>,
by setting
<center><math>
\forall i \in [0, N], (D_{pn})_{p \in [0, N]} = \delta_{ip}
</math></center>
Then we solve the linear system defined previously, so that we can find the coefficients
<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]],
<math>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>n</math>, we finally obtain a 3-dimensional
matrix for the [[Diffraction Transfer Matrix]].

= Matlab Code =

A program to calculate the coefficients for circular plate problems can be found here
[http://www.math.auckland.ac.nz/~meylan/code/eigenfunction_matching/circle_plate_matching_one_n.m circle_plate_matching_one_n.m]
Note that this problem solves only for a single n.

== Additional code ==

This program requires
[http://www.math.auckland.ac.nz/~meylan/code/dispersion/dispersion_free_surface.m dispersion_free_surface.m]
and [http://www.math.auckland.ac.nz/~meylan/code/dispersion/dispersion_elastic_surface.m dispersion_elastic_surface.m]
to run

[[Category:Problems with Cylindrical Symmetry]]
[[Category:Linear Hydroelasticity]]
[[Category:Eigenfunction Matching Method]]
[[Category:Complete Pages]]

Eigenfunction Matching for a Circular Dock

2008-08-04T01:20:50Z

Fmontiel: /* An infinite dimensional system of equations */

Eigenfunction Matching for a Circular Floating Elastic Plate

2008-08-04T01:18:59Z

Fmontiel: /* A Simple Method To Calculate The Diffraction Transfer Matrix For The Case Of A Circular Plate */

=Introduction=

We show here a solution for a [[Floating Elastic Plate]] on [[Finite Depth]] water
based on [[Peter_Meylan_Chung_2004a|Peter, Meylan and Chung 2004]]. A solution
for [[Shallow Depth]] was given in [[Zilman_Miloh_2000a|Zilman and Miloh 2000]] and we will also show this.
The solution is an extension of the [[Eigenfunction Matching for a Circular Dock]].

=Governing Equations=

We begin with the [[Frequency Domain Problem]] for a [[Floating Elastic Plate]]
in the non-dimensional form of [[Tayler_1986a|Tayler 1986]] ([[Dispersion Relation for a Floating Elastic Plate]])
We will use a cylindrical coordinate system, <math>(r,\theta,z)</math>,
assumed to have its origin at the centre of the circular
plate which has radius <math>a</math>. The water is assumed to have
constant finite depth <math>H</math> and the <math>z</math>-direction points vertically
upward with the water surface at <math>z=0</math> and the sea floor at <math>z=-H</math>. The
boundary value problem can therefore be expressed as
<center>
<math>
\Delta\phi=0, \,\, -H<z<0,
</math>
</center>
<center>
<math>
\phi_{z}=0, \,\, z=-H,
</math>
</center>
<center><math>
\phi_{z}=\alpha\phi, \,\, z=0,\,r>a,
</math></center>
<center>
<math>
(\beta\Delta^{2}+1-\alpha\gamma)\phi_{z}=\alpha\phi, \,\, z=0,\,r<a
</math>
</center>
where the constants <math>\beta</math> and <math>\gamma</math> are given by
<center>
<math>
\beta=\frac{D}{\rho\,L^{4}g}, \gamma=\frac{\rho_{i}h}{\rho\,L}
</math>
</center>
and <math>\rho_{i}</math> is the density of the plate. We
must also apply the edge conditions for the plate and the [[Sommerfeld Radiation Condition]]
as <math>r\rightarrow\infty</math>. The subscript <math>z</math>
denotes the derivative in <math>z</math>-direction.

=Solution Method=

== Separation of variables==

We now separate variables, noting that since the problem has
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>.
Using Laplace's equation in polar coordinates we obtain
<center>
<math>
\frac{\mathrm{d}^{2}\rho_{n}}{\mathrm{d}r^{2}}+\frac{1}{r}
\frac{\mathrm{d}\rho_{n}}{\mathrm{d}r}-\left(
\frac{n^{2}}{r^{2}}+\mu^{2}\right) \rho_{n}=0
</math>
</center>
where <math>\mu</math> is <math>k_{m}</math> or
<math>\kappa_{m},</math> depending on whether <math>r</math> is
greater or less than <math>a</math>. We can convert this equation to the
standard form by substituting <math>y=\mu r</math> to obtain
<center>
<math>
y^{2}\frac{\mathrm{d}^{2}\rho_{n}}{\mathrm{d}y^{2}}+y\frac{\mathrm{d}\rho_{n}
}{\rm{d}y}-(n^{2}+y^{2})\rho_{n}=0
</math>
</center>
The solution of this equation is a linear combination of the
modified Bessel functions of order <math>n</math>, <math>I_{n}(y)</math> and
<math>K_{n}(y)</math> [[Abramowitz and Stegun 1964]]. Since the solution must be bounded
we know that under the plate the solution will be a linear combination of
<math>I_{n}(y)</math> while outside the plate the solution will be a
linear combination of <math>K_{n}(y)</math>. Therefore the potential can
be expanded as
<center>
<math>
\phi(r,\theta,z)=\sum_{n=-\infty}^{\infty}\sum_{m=0}^{\infty}a_{mn}K_{n}
(k_{m}r)e^{i n\theta}\phi_{m}(z), \;\;r>a
</math>
</center>
and
<center>
<math>
\phi(r,\theta,z)=\sum_{n=-\infty}^{\infty}\sum_{m=-2}^{\infty}b_{mn}
I_{n}(\kappa_{m}r)e^{i n\theta}\psi_{m}(z), \;\;r<a
</math>
</center>
where <math>a_{mn}</math> and <math>b_{mn}</math>
are the coefficients of the potential in the open water and
the plate covered region respectively.

==Incident potential==

The incident potential is a wave of amplitude <math>A</math>
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 boundary conditions for the plate also have to be
considered. The vertical force and bending moment must vanish, which can be
written as
<center>
<math>
\left[\bar{\Delta}-\frac{1-\nu}{r}\left(\frac{\partial}{\partial r}
+\frac{1}{r}\frac{\partial^{2}}{\partial\theta^{2}}\right)\right]
w=0\,\,\,(3)
</math>
</center>
and
<center>
<math>
\left[ \frac{\partial}{\partial r}\bar{\Delta}-\frac{1-\nu}{r^{2}}\left(
-\frac{\partial}{\partial r}+\frac{1}{r}\right) \frac{\partial^{2}}
{\partial\theta^{2}}\right] w=0 \,\,\,(4)
</math>
</center>
where <math>w</math> is the time-independent surface
displacement, <math>\nu</math> is Poisson's ratio, and <math>\bar{\Delta}</math> is the
polar coordinate Laplacian
<center>
<math>
\bar{\Delta}=\frac{\partial^{2}}{\partial r^{2}}+\frac{1}{r}\frac{\partial
}{\partial r}+\frac{1}{r^{2}}\frac{\partial^{2}}{\partial\theta^{2}}
</math>
</center>

== Displacement of the plate ==

The surface displacement and the water velocity potential at
the water surface are linked through the kinematic boundary condition
<center>
<math>
\phi_{z}=-i\sqrt{\alpha}w,\,\,\,z=0
</math>
</center>
From the equations the potential and the surface
displacement are therefore related by
<center>
<math>
w=i\sqrt{\alpha}\phi,\quad r>a
</math>
</center>
and
<center>
<math>
(\beta\bar{\Delta}^{2}+1-\alpha\gamma)w=i\sqrt{\alpha}\phi,\quad r<a
</math>
</center>
The surface displacement can also be expanded in eigenfunctions
as
<center>
<math>
w(r,\theta)=\sum_{n=-\infty}^{\infty}\sum_{m=0}^{\infty}i\sqrt{\alpha}
a_{mn}K_{n}(k_{m}r)e^{i n\theta},\;\;r>a
</math>
</center>
and:
<center>
<math>
w(r,\theta)=
\sum_{n=-\infty}^{\infty}\sum_{m=-2}^{\infty}i\sqrt{\alpha}(\beta\kappa
_{m}^{4}+1-\alpha\gamma)^{-1}b_{mn}I_{n}(\kappa_{m}r)e^{i
n\theta},\; r<a
</math>
</center>
using the fact that
<center>
<math>
\bar{\Delta}\left( I_{n}(\kappa_{m}r)e^{i n\theta}\right) =\kappa_{m}
^{2}I_{n}(\kappa_{m}r)e^{i n\theta}\,\,\,(5)
</math>
</center>

==An infinite dimensional system of equations==

The boundary conditions (3) and
(4) can be expressed in terms of the potential
using (5). Since the angular modes are uncoupled the
conditions apply to each mode, giving
<center>
<math>
\sum_{m=-2}^{\infty}(\beta\kappa_{m}^{4}+1-\alpha\gamma)^{-1}b_{mn}
\left(\kappa_{m}^{2}I_{n}(\kappa_{m}a)-\frac{1-\nu}{a}\left(\kappa
_{m}I_{n}^{\prime}(\kappa_{m}a)-\frac{n^{2}}{a}I_{n}(\kappa_{m}a)\right)
\right) =0\,\,\,(6)
</math>
</center>
<center>
<math>
\sum_{m=-2}^{\infty}(\beta\kappa_{m}^{4}+1-\alpha\gamma)^{-1}b_{mn}
\left( \kappa_{m}^{3}I_{n}^{\prime}(\kappa_{m}a)+n^{2}\frac{1-\nu}{a^{2}
}\left( -\kappa_{m}I_{n}^{\prime}(\kappa_{m}a)+\frac{1}{a}I_{n}(\kappa
_{m}a)\right) \right)
=0\,\,\,(7)
</math>
</center>
The potential and its derivative must be continuous across the
transition from open water to the plate covered region. Therefore, the
potentials and their derivatives at <math>r=a</math> have to be equal.
Again we know that this must be true for each angle and we obtain
<center>
<math>
e_{n}I_{n}(k_{0}a)\phi_{0}\left( z\right) + \sum_{m=0}^{\infty}
a_{mn} K_{n}(k_{m}a)\phi_{m}\left( z\right)
=\sum_{m=-2}^{\infty}b_{mn}I_{n}(\kappa_{m}a)\psi_{m}(z)
</math>
</center>
and
<center>
<math>
e_{n}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)
=\sum_{m=-2}^{\infty}b_{mn}\kappa_{m}I_{n}^{\prime}(\kappa_{m}a)\psi
_{m}(z)
</math>
</center>
for each <math>n</math>.
We solve these equations by multiplying both equations by
<math>\phi_{l}(z)</math> and integrating from <math>-H</math> to <math>0</math> to obtain:
<center>
<math>
e_{n}I_{n}(k_{0}a)A_{0}\delta_{0l}+a_{ln}K_{n}(k_{l}a)A_{l}
=\sum_{m=-2}^{\infty}b_{mn}I_{n}(\kappa_{m}a)B_{ml} \,\,\,(8)
</math>
</center>
and
<center>
<math>
e_{n}k_{0}I_{n}^{\prime}(k_{0}a)A_{0}\delta_{0l}+a_{ln}k_{l}K_{n}^{\prime
}(k_{l}a)A_{l}
=\sum_{m=-2}^{\infty}b_{mn}\kappa_{m}I_{n}^{\prime}(\kappa_{m}
a)B_{ml} \,\,\,(9)
</math>
</center>
Equation (8) can be solved for the open water
coefficients <math>a_{mn}</math>
<center>
<math>
a_{ln}=-e_{n}\frac{I_{n}(k_{0}a)}{K_{n}(k_{0}a)}\delta_{0l}+\sum
_{m=-2}^{\infty}b_{mn}\frac{I_{n}(\kappa_{m}a)B_{ml}}{K_{n}(k_{l}a)A_{l}}
</math>
</center>
which can then be substituted into equation
(9) to give us
<center>
<math>
\left( k_{0}I_{n}^{\prime}(k_{0}a)-k_{0}\frac{K_{n}^{\prime}(k_{0}
a)}{K_{n}(k_{0}a)}I_{n}(k_{0}a)\right) e_{n}A_{0}\delta_{0l}
=\sum_{m=-2}^{\infty}\left( \kappa_{m}I_{n}^{\prime}(\kappa_{m}
a)-k_{l}\frac{K_{n}^{\prime}(k_{l}a)}{K_{n}(k_{l}a)}I_{n}(\kappa
_{m}a)\right) B_{ml}b_{mn}\,\,\,(10)
</math>
</center>
for each <math>n</math>.
Together with equations (6) and (7)
equation (10) gives the required equations to solve for the
coefficients of the water velocity potential in the plate covered region.

=Numerical Solution=

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>
<math>
\phi(r,\theta,z)=\sum_{n=-N}^{N}\sum_{m=0}^{M}a_{mn}K_{n}(k_{m}r)e^{i
n\theta }\phi_{m}(z), \;\;r>a
</math>
</center>
and
<center>
<math>
\phi(r,\theta,z)=\sum_{n=-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]]=

The shallow water theory of [[Zilman_Miloh_2000a|Zilman and Miloh 2000]] can be recovered by
simply setting the depth shallow enough that the shallow water theory is valid
and setting <math>M=0</math>. If the shallow water theory is valid then
the first three roots of the dispersion equation for the ice will be exactly
the same roots found in the shallow water theory by solving the polynomial
equation. The system of equations has four unknowns (three under the plate and
one in the open water) exactly as for the theory of [[Zilman_Miloh_2000a|Zilman and Miloh 2000]].

= A Simple Method To Calculate The [[Diffraction Transfer Matrix]] For The Case Of A Circular Plate =

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
<center><math>
\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)
=\sum_{m=-2}^{\infty}b_{mn}I_{n}(\kappa_{m}a)\psi_{m}(z)
</math></center>
and
<center><math>
\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}(k_{m}a)\phi_{m}\left( z\right)
=\sum_{m=-2}^{\infty}b_{mn} \kappa_m I'_{n}(\kappa_{m}a)\psi_{m}(z)
</math></center>
for each <math>n</math>.
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:
<center><math>
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},\ \ \ (11)
</math></center>
and
<center><math>
D_{ln} k_l I'_n (k_l a) A_l + a_{ln} k_l K'_{n}(k_{l} a) A_l
=\sum_{m=-2}^{\infty}b_{mn} \kappa_m I'_{n}(\kappa_{m}a) B_{ml} \ \ \ (12)
</math></center>

The [[Diffraction Transfer Matrix]]
maps the coefficients of the incident wave with the coefficients of the scattered wave within
the free 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>

Furthermore the boundary conditions are exactly the same that before, namely
<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
</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 .
</math>
</center>
For the further study, let's call
<center>
<math>
L^1_{mn}=(\beta\kappa_{m}^{4}+1-\alpha\gamma)^{-1}
\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)
</math>
</center>
and
<center>
<math>
L^2_{mn}=(\beta\kappa_{m}^{4}+1-\alpha\gamma)^{-1}
\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)
</math>
</center>
From the equations (11), (12) and the boundary conditions over the edges of the plate,
we can write a linear system of equation, limiting the number of modes of the dispersion equation
to <math>N</math> real ones
<center>
<math>
\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}
I_n(\kappa_{-2}a) B_{-20}&\cdots&I_n(\kappa_{N}a) B_{N0}\\
&&\\
\vdots&&\vdots\\
&&\\
I_n(\kappa_{-2}a) B_{-2N}&\cdots&I_n(\kappa_{N}a) B_{NN}
\end{bmatrix}
\\
\begin{bmatrix}
0&&\cdots&&0\\
0&&\cdots&&0
\end{bmatrix}
&
\begin{bmatrix}
L^1_{-2n}&\cdots&L^1_{Nn}\\
L^2_{-2n}&\cdots&L^2_{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}
\kappa_{-2} I'_n(\kappa_{-2}a) B_{-20}&\cdots&\kappa_{N} I'_n(\kappa_{N}a) B_{N0}\\
&&\\
\vdots&&\vdots\\
&&\\
\kappa_{-2} I'_n(\kappa_{-2}a) B_{-2N}&\cdots&\kappa_{N} I'_n(\kappa_{N}a) B_{NN}
\end{bmatrix}
\end{bmatrix}
\begin{bmatrix}
a_{0} \\
\\
\vdots \\
\\
a_N \\
b_{-2}\\
b_{-1}\\
b_{0}\\
\\
\vdots \\
\\
b_N
\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}
0&&\cdots&&0\\
0&&\cdots&&0
\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} \\
0\\
0\\
D_{0n}\\
\\
\vdots \\
\\
D_{Nn}
\end{bmatrix}
</math>
</center>
for each <math>n</math>.

Therefore we can find a [[Diffraction Transfer Matrix]] for each <math>n</math>,
by setting
<center><math>
\forall i \in [0, N], (D_{pn})_{p \in [0, N]} = \delta_{ip}
</math></center>
Then we solve the linear system defined previously, so that we can find the coefficients
<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]],
<math>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>n</math>, we finally obtain a 3-dimensional
matrix for the [[Diffraction Transfer Matrix]].

= Matlab Code =

A program to calculate the coefficients for circular plate problems can be found here
[http://www.math.auckland.ac.nz/~meylan/code/eigenfunction_matching/circle_plate_matching_one_n.m circle_plate_matching_one_n.m]
Note that this problem solves only for a single n.

== Additional code ==

This program requires
[http://www.math.auckland.ac.nz/~meylan/code/dispersion/dispersion_free_surface.m dispersion_free_surface.m]
and [http://www.math.auckland.ac.nz/~meylan/code/dispersion/dispersion_elastic_surface.m dispersion_elastic_surface.m]
to run

[[Category:Problems with Cylindrical Symmetry]]
[[Category:Linear Hydroelasticity]]
[[Category:Eigenfunction Matching Method]]
[[Category:Complete Pages]]

Eigenfunction Matching for a Circular Floating Elastic Plate

2008-08-04T01:16:33Z

Fmontiel: /* Solving for the case of a more complicated incident wave */

=Introduction=

We show here a solution for a [[Floating Elastic Plate]] on [[Finite Depth]] water
based on [[Peter_Meylan_Chung_2004a|Peter, Meylan and Chung 2004]]. A solution
for [[Shallow Depth]] was given in [[Zilman_Miloh_2000a|Zilman and Miloh 2000]] and we will also show this.
The solution is an extension of the [[Eigenfunction Matching for a Circular Dock]].

=Governing Equations=

We begin with the [[Frequency Domain Problem]] for a [[Floating Elastic Plate]]
in the non-dimensional form of [[Tayler_1986a|Tayler 1986]] ([[Dispersion Relation for a Floating Elastic Plate]])
We will use a cylindrical coordinate system, <math>(r,\theta,z)</math>,
assumed to have its origin at the centre of the circular
plate which has radius <math>a</math>. The water is assumed to have
constant finite depth <math>H</math> and the <math>z</math>-direction points vertically
upward with the water surface at <math>z=0</math> and the sea floor at <math>z=-H</math>. The
boundary value problem can therefore be expressed as
<center>
<math>
\Delta\phi=0, \,\, -H<z<0,
</math>
</center>
<center>
<math>
\phi_{z}=0, \,\, z=-H,
</math>
</center>
<center><math>
\phi_{z}=\alpha\phi, \,\, z=0,\,r>a,
</math></center>
<center>
<math>
(\beta\Delta^{2}+1-\alpha\gamma)\phi_{z}=\alpha\phi, \,\, z=0,\,r<a
</math>
</center>
where the constants <math>\beta</math> and <math>\gamma</math> are given by
<center>
<math>
\beta=\frac{D}{\rho\,L^{4}g}, \gamma=\frac{\rho_{i}h}{\rho\,L}
</math>
</center>
and <math>\rho_{i}</math> is the density of the plate. We
must also apply the edge conditions for the plate and the [[Sommerfeld Radiation Condition]]
as <math>r\rightarrow\infty</math>. The subscript <math>z</math>
denotes the derivative in <math>z</math>-direction.

=Solution Method=

== Separation of variables==

We now separate variables, noting that since the problem has
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>.
Using Laplace's equation in polar coordinates we obtain
<center>
<math>
\frac{\mathrm{d}^{2}\rho_{n}}{\mathrm{d}r^{2}}+\frac{1}{r}
\frac{\mathrm{d}\rho_{n}}{\mathrm{d}r}-\left(
\frac{n^{2}}{r^{2}}+\mu^{2}\right) \rho_{n}=0
</math>
</center>
where <math>\mu</math> is <math>k_{m}</math> or
<math>\kappa_{m},</math> depending on whether <math>r</math> is
greater or less than <math>a</math>. We can convert this equation to the
standard form by substituting <math>y=\mu r</math> to obtain
<center>
<math>
y^{2}\frac{\mathrm{d}^{2}\rho_{n}}{\mathrm{d}y^{2}}+y\frac{\mathrm{d}\rho_{n}
}{\rm{d}y}-(n^{2}+y^{2})\rho_{n}=0
</math>
</center>
The solution of this equation is a linear combination of the
modified Bessel functions of order <math>n</math>, <math>I_{n}(y)</math> and
<math>K_{n}(y)</math> [[Abramowitz and Stegun 1964]]. Since the solution must be bounded
we know that under the plate the solution will be a linear combination of
<math>I_{n}(y)</math> while outside the plate the solution will be a
linear combination of <math>K_{n}(y)</math>. Therefore the potential can
be expanded as
<center>
<math>
\phi(r,\theta,z)=\sum_{n=-\infty}^{\infty}\sum_{m=0}^{\infty}a_{mn}K_{n}
(k_{m}r)e^{i n\theta}\phi_{m}(z), \;\;r>a
</math>
</center>
and
<center>
<math>
\phi(r,\theta,z)=\sum_{n=-\infty}^{\infty}\sum_{m=-2}^{\infty}b_{mn}
I_{n}(\kappa_{m}r)e^{i n\theta}\psi_{m}(z), \;\;r<a
</math>
</center>
where <math>a_{mn}</math> and <math>b_{mn}</math>
are the coefficients of the potential in the open water and
the plate covered region respectively.

==Incident potential==

The incident potential is a wave of amplitude <math>A</math>
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 boundary conditions for the plate also have to be
considered. The vertical force and bending moment must vanish, which can be
written as
<center>
<math>
\left[\bar{\Delta}-\frac{1-\nu}{r}\left(\frac{\partial}{\partial r}
+\frac{1}{r}\frac{\partial^{2}}{\partial\theta^{2}}\right)\right]
w=0\,\,\,(3)
</math>
</center>
and
<center>
<math>
\left[ \frac{\partial}{\partial r}\bar{\Delta}-\frac{1-\nu}{r^{2}}\left(
-\frac{\partial}{\partial r}+\frac{1}{r}\right) \frac{\partial^{2}}
{\partial\theta^{2}}\right] w=0 \,\,\,(4)
</math>
</center>
where <math>w</math> is the time-independent surface
displacement, <math>\nu</math> is Poisson's ratio, and <math>\bar{\Delta}</math> is the
polar coordinate Laplacian
<center>
<math>
\bar{\Delta}=\frac{\partial^{2}}{\partial r^{2}}+\frac{1}{r}\frac{\partial
}{\partial r}+\frac{1}{r^{2}}\frac{\partial^{2}}{\partial\theta^{2}}
</math>
</center>

== Displacement of the plate ==

The surface displacement and the water velocity potential at
the water surface are linked through the kinematic boundary condition
<center>
<math>
\phi_{z}=-i\sqrt{\alpha}w,\,\,\,z=0
</math>
</center>
From the equations the potential and the surface
displacement are therefore related by
<center>
<math>
w=i\sqrt{\alpha}\phi,\quad r>a
</math>
</center>
and
<center>
<math>
(\beta\bar{\Delta}^{2}+1-\alpha\gamma)w=i\sqrt{\alpha}\phi,\quad r<a
</math>
</center>
The surface displacement can also be expanded in eigenfunctions
as
<center>
<math>
w(r,\theta)=\sum_{n=-\infty}^{\infty}\sum_{m=0}^{\infty}i\sqrt{\alpha}
a_{mn}K_{n}(k_{m}r)e^{i n\theta},\;\;r>a
</math>
</center>
and:
<center>
<math>
w(r,\theta)=
\sum_{n=-\infty}^{\infty}\sum_{m=-2}^{\infty}i\sqrt{\alpha}(\beta\kappa
_{m}^{4}+1-\alpha\gamma)^{-1}b_{mn}I_{n}(\kappa_{m}r)e^{i
n\theta},\; r<a
</math>
</center>
using the fact that
<center>
<math>
\bar{\Delta}\left( I_{n}(\kappa_{m}r)e^{i n\theta}\right) =\kappa_{m}
^{2}I_{n}(\kappa_{m}r)e^{i n\theta}\,\,\,(5)
</math>
</center>

==An infinite dimensional system of equations==

The boundary conditions (3) and
(4) can be expressed in terms of the potential
using (5). Since the angular modes are uncoupled the
conditions apply to each mode, giving
<center>
<math>
\sum_{m=-2}^{\infty}(\beta\kappa_{m}^{4}+1-\alpha\gamma)^{-1}b_{mn}
\left(\kappa_{m}^{2}I_{n}(\kappa_{m}a)-\frac{1-\nu}{a}\left(\kappa
_{m}I_{n}^{\prime}(\kappa_{m}a)-\frac{n^{2}}{a}I_{n}(\kappa_{m}a)\right)
\right) =0\,\,\,(6)
</math>
</center>
<center>
<math>
\sum_{m=-2}^{\infty}(\beta\kappa_{m}^{4}+1-\alpha\gamma)^{-1}b_{mn}
\left( \kappa_{m}^{3}I_{n}^{\prime}(\kappa_{m}a)+n^{2}\frac{1-\nu}{a^{2}
}\left( -\kappa_{m}I_{n}^{\prime}(\kappa_{m}a)+\frac{1}{a}I_{n}(\kappa
_{m}a)\right) \right)
=0\,\,\,(7)
</math>
</center>
The potential and its derivative must be continuous across the
transition from open water to the plate covered region. Therefore, the
potentials and their derivatives at <math>r=a</math> have to be equal.
Again we know that this must be true for each angle and we obtain
<center>
<math>
e_{n}I_{n}(k_{0}a)\phi_{0}\left( z\right) + \sum_{m=0}^{\infty}
a_{mn} K_{n}(k_{m}a)\phi_{m}\left( z\right)
=\sum_{m=-2}^{\infty}b_{mn}I_{n}(\kappa_{m}a)\psi_{m}(z)
</math>
</center>
and
<center>
<math>
e_{n}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)
=\sum_{m=-2}^{\infty}b_{mn}\kappa_{m}I_{n}^{\prime}(\kappa_{m}a)\psi
_{m}(z)
</math>
</center>
for each <math>n</math>.
We solve these equations by multiplying both equations by
<math>\phi_{l}(z)</math> and integrating from <math>-H</math> to <math>0</math> to obtain:
<center>
<math>
e_{n}I_{n}(k_{0}a)A_{0}\delta_{0l}+a_{ln}K_{n}(k_{l}a)A_{l}
=\sum_{m=-2}^{\infty}b_{mn}I_{n}(\kappa_{m}a)B_{ml} \,\,\,(8)
</math>
</center>
and
<center>
<math>
e_{n}k_{0}I_{n}^{\prime}(k_{0}a)A_{0}\delta_{0l}+a_{ln}k_{l}K_{n}^{\prime
}(k_{l}a)A_{l}
=\sum_{m=-2}^{\infty}b_{mn}\kappa_{m}I_{n}^{\prime}(\kappa_{m}
a)B_{ml} \,\,\,(9)
</math>
</center>
Equation (8) can be solved for the open water
coefficients <math>a_{mn}</math>
<center>
<math>
a_{ln}=-e_{n}\frac{I_{n}(k_{0}a)}{K_{n}(k_{0}a)}\delta_{0l}+\sum
_{m=-2}^{\infty}b_{mn}\frac{I_{n}(\kappa_{m}a)B_{ml}}{K_{n}(k_{l}a)A_{l}}
</math>
</center>
which can then be substituted into equation
(9) to give us
<center>
<math>
\left( k_{0}I_{n}^{\prime}(k_{0}a)-k_{0}\frac{K_{n}^{\prime}(k_{0}
a)}{K_{n}(k_{0}a)}I_{n}(k_{0}a)\right) e_{n}A_{0}\delta_{0l}
=\sum_{m=-2}^{\infty}\left( \kappa_{m}I_{n}^{\prime}(\kappa_{m}
a)-k_{l}\frac{K_{n}^{\prime}(k_{l}a)}{K_{n}(k_{l}a)}I_{n}(\kappa
_{m}a)\right) B_{ml}b_{mn}\,\,\,(10)
</math>
</center>
for each <math>n</math>.
Together with equations (6) and (7)
equation (10) gives the required equations to solve for the
coefficients of the water velocity potential in the plate covered region.

=Numerical Solution=

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>
<math>
\phi(r,\theta,z)=\sum_{n=-N}^{N}\sum_{m=0}^{M}a_{mn}K_{n}(k_{m}r)e^{i
n\theta }\phi_{m}(z), \;\;r>a
</math>
</center>
and
<center>
<math>
\phi(r,\theta,z)=\sum_{n=-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]]=

The shallow water theory of [[Zilman_Miloh_2000a|Zilman and Miloh 2000]] can be recovered by
simply setting the depth shallow enough that the shallow water theory is valid
and setting <math>M=0</math>. If the shallow water theory is valid then
the first three roots of the dispersion equation for the ice will be exactly
the same roots found in the shallow water theory by solving the polynomial
equation. The system of equations has four unknowns (three under the plate and
one in the open water) exactly as for the theory of [[Zilman_Miloh_2000a|Zilman and Miloh 2000]].

= A Simple Method To Calculate The [[Diffraction Transfer Matrix]] For The Case Of A Circular Plate =

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
<center><math>
\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)
=\sum_{m=-2}^{\infty}b_{mn}I_{n}(\kappa_{m}a)\psi_{m}(z)
</math></center>
and
<center><math>
\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}(k_{m}a)\phi_{m}\left( z\right)
=\sum_{m=-2}^{\infty}b_{mn} \kappa_m I'_{n}(\kappa_{m}a)\psi_{m}(z)
</math></center>
for each <math>n</math>.
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:
<center><math>
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},\ \ \ (11)
</math></center>
and
<center><math>
D_{ln} k_l I'_n (k_l a) A_l + a_{ln} k_l K'_{n}(k_{l} a) A_l
=\sum_{m=-2}^{\infty}b_{mn} \kappa_m I'_{n}(\kappa_{m}a) B_{ml} \ \ \ (12)
</math></center>
As a consequence we can find an expression of the coefficients <math>a_{ln}</math>
<center><math>
a_{ln} = -\frac{K_{n}(k_l a)}{I_n (k_l a)} D_{ln} +
\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 the equation for the normal derivative
of the potential, to give us
<center><math>
k_l A_l \left( I'_n (k_l a) - \frac{K'_{n}(k_{l} a) K_{n}(k_l a)}{I_n (k_l a)} \right) D_{ln}
= \sum_{m=-2}^{\infty} \left( \kappa_m I'_{n}(\kappa_{m}a) -
k_l \frac{K'_{n}(k_{l} a) I_{n}(\kappa_m a)}{K_n (k_l a)} \right) B_{ml} b_{mn}
</math></center>
for each <math>n</math>.

We notice here that there is still a dependence between the coefficients <math>a_{mn}</math>
and <math>D_{ln}</math>. We can get rid of this dependence, using a [[Diffraction Transfer Matrix]],
which maps the coefficients of the incident wave with the coefficients of the scattered wave within
the free water domain. Then 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>

Furthermore the boundary conditions are exactly the same that before, namely
<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
</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 .
</math>
</center>
For the further study, let's call
<center>
<math>
L^1_{mn}=(\beta\kappa_{m}^{4}+1-\alpha\gamma)^{-1}
\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)
</math>
</center>
and
<center>
<math>
L^2_{mn}=(\beta\kappa_{m}^{4}+1-\alpha\gamma)^{-1}
\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)
</math>
</center>
From the equations (11), (12) and the boundary conditions over the edges of the plate,
we can write a linear system of equation, limiting the number of modes of the dispersion equation
to <math>N</math> real ones
<center>
<math>
\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}
I_n(\kappa_{-2}a) B_{-20}&\cdots&I_n(\kappa_{N}a) B_{N0}\\
&&\\
\vdots&&\vdots\\
&&\\
I_n(\kappa_{-2}a) B_{-2N}&\cdots&I_n(\kappa_{N}a) B_{NN}
\end{bmatrix}
\\
\begin{bmatrix}
0&&\cdots&&0\\
0&&\cdots&&0
\end{bmatrix}
&
\begin{bmatrix}
L^1_{-2n}&\cdots&L^1_{Nn}\\
L^2_{-2n}&\cdots&L^2_{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}
\kappa_{-2} I'_n(\kappa_{-2}a) B_{-20}&\cdots&\kappa_{N} I'_n(\kappa_{N}a) B_{N0}\\
&&\\
\vdots&&\vdots\\
&&\\
\kappa_{-2} I'_n(\kappa_{-2}a) B_{-2N}&\cdots&\kappa_{N} I'_n(\kappa_{N}a) B_{NN}
\end{bmatrix}
\end{bmatrix}
\begin{bmatrix}
a_{0} \\
\\
\vdots \\
\\
a_N \\
b_{-2}\\
b_{-1}\\
b_{0}\\
\\
\vdots \\
\\
b_N
\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}
0&&\cdots&&0\\
0&&\cdots&&0
\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} \\
0\\
0\\
D_{0n}\\
\\
\vdots \\
\\
D_{Nn}
\end{bmatrix}
</math>
</center>
for each <math>n</math>.

Therefore we can find a [[Diffraction Transfer Matrix]] for each <math>n</math>,
by setting
<center><math>
\forall i \in [0, N], (D_{pn})_{p \in [0, N]} = \delta_{ip}
</math></center>
Then we solve the linear system defined previously, so that we can find the coefficients
<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]],
<math>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>n</math>, we finally obtain a 3-dimensional
matrix for the [[Diffraction Transfer Matrix]].

= Matlab Code =

A program to calculate the coefficients for circular plate problems can be found here
[http://www.math.auckland.ac.nz/~meylan/code/eigenfunction_matching/circle_plate_matching_one_n.m circle_plate_matching_one_n.m]
Note that this problem solves only for a single n.

== Additional code ==

This program requires
[http://www.math.auckland.ac.nz/~meylan/code/dispersion/dispersion_free_surface.m dispersion_free_surface.m]
and [http://www.math.auckland.ac.nz/~meylan/code/dispersion/dispersion_elastic_surface.m dispersion_elastic_surface.m]
to run

[[Category:Problems with Cylindrical Symmetry]]
[[Category:Linear Hydroelasticity]]
[[Category:Eigenfunction Matching Method]]
[[Category:Complete Pages]]

Kagemoto and Yue Interaction Theory

2008-08-02T03:24:49Z

Fmontiel: /* Final Equations */

= Introduction =

This is an interaction theory which provides the exact solution (i.e. it is not based on a [[Wide Spacing Approximation]]).
The theory uses the [[Cylindrical Eigenfunction Expansion]] and [[Graf's Addition Theorem]] to represent the potential in local coordinates. The incident and scattered potential of each body are then related by the associated [[Diffraction Transfer Matrix]]. [[Interaction Theory for Cylinders]] presents a simplified version for cylinders.

The basic idea is as follows: The scattered potential of each body is represented in the [[Cylindrical Eigenfunction Expansion]] associated with the local coordinates centred at the mean centre position of the body. Using [[Graf's Addition Theorem]], the scattered potential of all bodies (given in their local coordinates) can be mapped to an incident potential associated with the coordinates of all other bodies. Doing this, the incident potential of each body (which is given by the ambient incident potential plus the scattered potentials of all other bodies) is given in the [[Cylindrical Eigenfunction Expansion]] associated with its local coordinates. Using the [[Diffraction Transfer Matrix]], which relates the incident and scattered potential of each body in isolation, a system of equations for the coefficients of the scattered potentials of all bodies is obtained.

The theory is described in [[Kagemoto and Yue 1986]] and in
[[Peter and Meylan 2004]].

The derivation of the theory in [[Infinite Depth]] is also presented, see
[[Kagemoto and Yue Interaction Theory for Infinite Depth]].

[[Category:Interaction Theory]]

= Equations of Motion =

The problem consists of <math>n</math> bodies
<math>\Delta_j</math> with immersed body
surface <math>\Gamma_j</math>. Each body is subject to
the [[Standard Linear Wave Scattering Problem]] and the particluar
equations of motion for each body (e.g. rigid, or freely floating)
can be different for each body.
It is a [[Frequency Domain Problem]] with frequency <math>\omega</math>.
The solution is exact, up to the
restriction that the escribed cylinder of each body may not contain any
other body.
To simplify notation, <math>\mathbf{y} = (x,y,z)</math> always denotes a point
in the water, which is assumed to be of [[Finite Depth]] <math>H</math>,
while <math>\mathbf{x}</math> always denotes a point of the undisturbed water
surface assumed at <math>z=0</math>.

Writing <math>\alpha = \omega^2/g</math> where <math>g</math> is the acceleration due to
gravity, the potential <math>\phi</math> has to
satisfy the standard boundary-value problem
<center><math>
\nabla^2 \phi = 0, \; \mathbf{y} \in D
</math></center>
<center><math>
\frac{\partial \phi}{\partial z} = \alpha \phi, \;
{\mathbf{x}} \in \Gamma^\mathrm{f},
</math></center>
<center><math>
\frac{\partial \phi}{\partial z} = 0, \; \mathbf{y} \in D, \ z=-H,
</math></center>
where <math>D</math> is the
is the domain occupied by the water and
<math>\Gamma^\mathrm{f}</math> is the free water surface. At the immersed body
surface <math>\Gamma_j</math> of body <math>\Delta_j</math>, <math>j=1,\dots,N</math>, the water velocity potential has to
equal the normal velocity of the body <math>\mathbf{v}_j</math>,
<center><math>
\frac{\partial \phi}{\partial n} = \mathbf{v}_j, \; {\mathbf{y}}
\in \Gamma_j.
</math></center>
where the normal derivative is given by the particaluar equations of motion of the body.
Moreover, the [[Sommerfeld Radiation Condition]] is imposed.

=Eigenfunction expansion of the potential=

Each body is subject to an incident potential and moves in response to this
incident potential to produce a scattered potential. Each of these is
expanded using the [[Cylindrical Eigenfunction Expansion]]
The scattered potential of a body
<math>\Delta_j</math> can be expressed as
<center><math>
\phi_j^\mathrm{S} (r_j,\theta_j,z) =
\sum_{m=0}^{\infty} f_m(z) \sum_{\mu = -
\infty}^{\infty} A_{m \mu}^j K_\mu (k_m r_j) \mathrm{e}^{\mathrm{i}\mu \theta_j},
</math></center>
with discrete coefficients <math>A_{m \mu}^j</math>, where <math>(r_j,\theta_j,z)</math>
are cylindrical polar coordinates centered at each body
<center><math>
f_m(z) = \frac{\cos k_m (z+H)}{\cos k_m H}.
</math></center>
where <math>k_m</math> are found from <math>\alpha</math> by the [[Dispersion Relation for a Free Surface]]
<center><math>
\alpha + k_m \tan k_m H = 0\,.
</math></center>
where <math>k_0</math> is the
imaginary root with positive imaginary part
and <math>k_m</math>, <math>m>0</math>, are given the positive real roots ordered
with increasing size.

The incident potential upon body <math>\Delta_j</math> can be also be expanded in
regular cylindrical eigenfunctions,
<center><math>
\phi_j^\mathrm{I} (r_j,\theta_j,z) = \sum_{n=0}^{\infty} f_n(z)
\sum_{\nu = - \infty}^{\infty} D_{n\nu}^j I_\nu (k_n r_j) \mathrm{e}^{\mathrm{i}\nu \theta_j},
</math></center>
with discrete coefficients <math>D_{n\nu}^j</math>. In these expansions, <math>I_\nu</math>
and <math>K_\nu</math> denote the modified [http://en.wikipedia.org/wiki/Bessel_function : Bessel functions]
of the first and second kind, respectively, both of order <math>\nu</math>.

Note that the term for <math>m =0</math> or
<math>n=0</math> corresponds to the propagating modes while the
terms for <math>m\geq 1</math> (<math>n\geq 1</math>) correspond to the evanescent modes.

=Derivation of the system of equations=

A system of equations for the unknown
coefficients of the
scattered wavefields of all bodies is developed. This system of
equations is based on transforming the
scattered potential of <math>\Delta_j</math> into an incident potential upon
<math>\Delta_l</math> (<math>j \neq l</math>). Doing this for all bodies simultaneously,
and relating the incident and scattered potential for each body, a system
of equations for the unknown coefficients is developed.
Making use of the periodicity of the geometry and of the ambient incident
wave, this system of equations can then be simplified.

The scattered potential <math>\phi_j^{\mathrm{S}}</math> of body <math>\Delta_j</math> needs to be
represented in terms of the incident potential <math>\phi_l^{\mathrm{I}}</math>
upon <math>\Delta_l</math>, <math>j \neq l</math>. This can be accomplished by using
[[Graf's Addition Theorem]]
<center><math>
K_\tau(k_m r_j) \mathrm{e}^{\mathrm{i}\tau (\theta_j-\varphi_{jl})} =
\sum_{\nu = - \infty}^{\infty} K_{\tau + \nu} (k_m R_{jl}) \,
I_\nu (k_m r_l) \mathrm{e}^{\mathrm{i}\nu (\pi - \theta_l + \varphi_{jl})}, \quad j \neq l,
</math></center>
which is valid provided that <math>r_l < R_{jl}</math>. Here, <math>(R_{jl},\varphi_{jl})</math> are the polar coordinates of the mean centre position of <math>\Delta_{l}</math> in the local coordinates of <math>\Delta_{j}</math>.

The limitation <math>r_l < R_{jl}</math> only requires that the escribed cylinder of each body
<math>\Delta_l</math> does not enclose any other origin <math>O_j</math> (<math>j \neq l</math>). However, the
expansion of the scattered and incident potential in cylindrical
eigenfunctions is only valid outside the escribed cylinder of each
body. Therefore the condition that the
escribed cylinder of each body <math>\Delta_l</math> does not enclose any other
origin <math>O_j</math> (<math>j \neq l</math>) is superseded by the more rigorous
restriction that the escribed cylinder of each body may not contain any
other body.

Making use of the eigenfunction expansion as well as [[Graf's Addition Theorem]], the scattered potential
of <math>\Delta_j</math> can be expressed in terms of the
incident potential upon <math>\Delta_l</math> as
<center><math>
\phi_j^{\mathrm{S}} (r_l,\theta_l,z)
= \sum_{m=0}^\infty f_m(z) \sum_{\tau = -
\infty}^{\infty} A_{m\tau}^j \sum_{\nu = -\infty}^{\infty}
(-1)^\nu K_{\tau-\nu} (k_m R_{jl}) I_\nu (k_m r_l) \mathrm{e}^{\mathrm{i}\nu
\theta_l} \mathrm{e}^{\mathrm{i}(\tau-\nu) \varphi_{jl}}
</math></center>
<center><math>
= \sum_{m=0}^\infty f_m(z) \sum_{\nu =
-\infty}^{\infty} \Big[ \sum_{\tau = - \infty}^{\infty} A_{m\tau}^j
(-1)^\nu K_{\tau-\nu} (k_m R_{jl}) \mathrm{e}^{\mathrm{i}(\tau - \nu)
\varphi_{jl}} \Big] I_\nu (k_m r_l) \mathrm{e}^{\mathrm{i}\nu \theta_l}.
</math></center>
The ambient incident wavefield <math>\phi^{\mathrm{In}}</math> can also be
expanded in the eigenfunctions corresponding to the incident wavefield upon
<math>\Delta_l</math>. Let <math>\tilde{D}_{n\nu}^{l}</math> denote the coefficients of this
ambient incident wavefield in the incoming eigenfunction expansion for
<math>\Delta_l</math> (cf. the example in [[Cylindrical Eigenfunction Expansion]]).
<center><math>
\phi^{\mathrm{In}}(r_l,\theta_l,z)= \sum_{n=0}^\infty f_n(z) \sum_{\nu = -\infty}^{\infty}
\tilde{D}_{n\nu}^{l} I_\nu (k_n
r_l) \mathrm{e}^{\mathrm{i}\nu \theta_l}.
</math></center>
The total
incident wavefield upon body <math>\Delta_j</math> can now be expressed as
<center><math>
\phi_l^{\mathrm{I}}(r_l,\theta_l,z) = \phi^{\mathrm{In}}(r_l,\theta_l,z) +
\sum_{j=1,j \neq l}^{N} \, \phi_j^{\mathrm{S}}
(r_l,\theta_l,z)
</math></center>
This allows us to write
<center><math>
\sum_{n=0}^{\infty} f_n(z)
\sum_{\nu = - \infty}^{\infty} D_{n\nu}^l I_\nu (k_n r_l) \mathrm{e}^{\mathrm{i}\nu \theta_l}
</math></center>
<center><math>
= \sum_{n=0}^\infty f_n(z) \sum_{\nu = -\infty}^{\infty}
\Big[ \tilde{D}_{n\nu}^{l} +
\sum_{j=1,j \neq l}^{N} \sum_{\tau =
-\infty}^{\infty} A_{n\tau}^j (-1)^\nu K_{\tau - \nu} (k_n
R_{jl}) \mathrm{e}^{\mathrm{i}(\tau - \nu) \varphi_{jl}} \Big] I_\nu (k_n
r_l) \mathrm{e}^{\mathrm{i}\nu \theta_l}.
</math></center>
It therefore follows that
<center><math>
D_{n\nu}^l =
\tilde{D}_{n\nu}^{l} +
\sum_{j=1,j \neq l}^{N} \sum_{\tau =
-\infty}^{\infty} A_{n\tau}^j (-1)^\nu K_{\tau - \nu} (k_n
R_{jl}) \mathrm{e}^{\mathrm{i}(\tau - \nu) \varphi_{jl}}
</math></center>

= Final Equations =

The scattered and incident potential of each body <math>\Delta_l</math> can be related by the
[[Diffraction Transfer Matrix]] acting in the following way,
<center><math>
A_{m \mu}^l = \sum_{n=0}^{\infty} \sum_{\nu = -\infty}^{\infty} B_{m n
\mu \nu}^l D_{n\nu}^l.
</math></center>

The substitution of this into the equation for relating
the coefficients <math>D_{n\nu}^l</math> and
<math>A_{m \mu}^l</math>gives the
required equations to determine the coefficients of the scattered
wavefields of all bodies,
<center><math>
A_{m\mu}^l = \sum_{n=0}^{\infty}
\sum_{\nu = -\infty}^{\infty} B_{mn\mu\nu}^l
\Big[ \tilde{D}_{n\nu}^{l} +
\sum_{j=1,j \neq l}^{N} \sum_{\tau =
-\infty}^{\infty} A_{n\tau}^j (-1)^\nu K_{\tau - \nu} (k_n
R_{jl}) \mathrm{e}^{\mathrm{i}(\tau -\nu) \varphi_{jl}} \Big],
</math></center>
<math>m \in \mathbb{N}</math>, <math>\mu \in \mathbb{Z}</math>, <math>l=1,\dots,N</math>.

= Interaction for the case of circular plates =

Eigenfunction Matching for a Circular Floating Elastic Plate

2008-08-02T02:23:29Z

Fmontiel: /* Solving for the case of a more complicated incident wave */

=Introduction=

We show here a solution for a [[Floating Elastic Plate]] on [[Finite Depth]] water
based on [[Peter_Meylan_Chung_2004a|Peter, Meylan and Chung 2004]]. A solution
for [[Shallow Depth]] was given in [[Zilman_Miloh_2000a|Zilman and Miloh 2000]] and we will also show this.
The solution is an extension of the [[Eigenfunction Matching for a Circular Dock]].

=Governing Equations=

We begin with the [[Frequency Domain Problem]] for a [[Floating Elastic Plate]]
in the non-dimensional form of [[Tayler_1986a|Tayler 1986]] ([[Dispersion Relation for a Floating Elastic Plate]])
We will use a cylindrical coordinate system, <math>(r,\theta,z)</math>,
assumed to have its origin at the centre of the circular
plate which has radius <math>a</math>. The water is assumed to have
constant finite depth <math>H</math> and the <math>z</math>-direction points vertically
upward with the water surface at <math>z=0</math> and the sea floor at <math>z=-H</math>. The
boundary value problem can therefore be expressed as
<center>
<math>
\Delta\phi=0, \,\, -H<z<0,
</math>
</center>
<center>
<math>
\phi_{z}=0, \,\, z=-H,
</math>
</center>
<center><math>
\phi_{z}=\alpha\phi, \,\, z=0,\,r>a,
</math></center>
<center>
<math>
(\beta\Delta^{2}+1-\alpha\gamma)\phi_{z}=\alpha\phi, \,\, z=0,\,r<a
</math>
</center>
where the constants <math>\beta</math> and <math>\gamma</math> are given by
<center>
<math>
\beta=\frac{D}{\rho\,L^{4}g}, \gamma=\frac{\rho_{i}h}{\rho\,L}
</math>
</center>
and <math>\rho_{i}</math> is the density of the plate. We
must also apply the edge conditions for the plate and the [[Sommerfeld Radiation Condition]]
as <math>r\rightarrow\infty</math>. The subscript <math>z</math>
denotes the derivative in <math>z</math>-direction.

=Solution Method=

== Separation of variables==

We now separate variables, noting that since the problem has
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>.
Using Laplace's equation in polar coordinates we obtain
<center>
<math>
\frac{\mathrm{d}^{2}\rho_{n}}{\mathrm{d}r^{2}}+\frac{1}{r}
\frac{\mathrm{d}\rho_{n}}{\mathrm{d}r}-\left(
\frac{n^{2}}{r^{2}}+\mu^{2}\right) \rho_{n}=0
</math>
</center>
where <math>\mu</math> is <math>k_{m}</math> or
<math>\kappa_{m},</math> depending on whether <math>r</math> is
greater or less than <math>a</math>. We can convert this equation to the
standard form by substituting <math>y=\mu r</math> to obtain
<center>
<math>
y^{2}\frac{\mathrm{d}^{2}\rho_{n}}{\mathrm{d}y^{2}}+y\frac{\mathrm{d}\rho_{n}
}{\rm{d}y}-(n^{2}+y^{2})\rho_{n}=0
</math>
</center>
The solution of this equation is a linear combination of the
modified Bessel functions of order <math>n</math>, <math>I_{n}(y)</math> and
<math>K_{n}(y)</math> [[Abramowitz and Stegun 1964]]. Since the solution must be bounded
we know that under the plate the solution will be a linear combination of
<math>I_{n}(y)</math> while outside the plate the solution will be a
linear combination of <math>K_{n}(y)</math>. Therefore the potential can
be expanded as
<center>
<math>
\phi(r,\theta,z)=\sum_{n=-\infty}^{\infty}\sum_{m=0}^{\infty}a_{mn}K_{n}
(k_{m}r)e^{i n\theta}\phi_{m}(z), \;\;r>a
</math>
</center>
and
<center>
<math>
\phi(r,\theta,z)=\sum_{n=-\infty}^{\infty}\sum_{m=-2}^{\infty}b_{mn}
I_{n}(\kappa_{m}r)e^{i n\theta}\psi_{m}(z), \;\;r<a
</math>
</center>
where <math>a_{mn}</math> and <math>b_{mn}</math>
are the coefficients of the potential in the open water and
the plate covered region respectively.

==Incident potential==

The incident potential is a wave of amplitude <math>A</math>
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 boundary conditions for the plate also have to be
considered. The vertical force and bending moment must vanish, which can be
written as
<center>
<math>
\left[\bar{\Delta}-\frac{1-\nu}{r}\left(\frac{\partial}{\partial r}
+\frac{1}{r}\frac{\partial^{2}}{\partial\theta^{2}}\right)\right]
w=0\,\,\,(3)
</math>
</center>
and
<center>
<math>
\left[ \frac{\partial}{\partial r}\bar{\Delta}-\frac{1-\nu}{r^{2}}\left(
-\frac{\partial}{\partial r}+\frac{1}{r}\right) \frac{\partial^{2}}
{\partial\theta^{2}}\right] w=0 \,\,\,(4)
</math>
</center>
where <math>w</math> is the time-independent surface
displacement, <math>\nu</math> is Poisson's ratio, and <math>\bar{\Delta}</math> is the
polar coordinate Laplacian
<center>
<math>
\bar{\Delta}=\frac{\partial^{2}}{\partial r^{2}}+\frac{1}{r}\frac{\partial
}{\partial r}+\frac{1}{r^{2}}\frac{\partial^{2}}{\partial\theta^{2}}
</math>
</center>

== Displacement of the plate ==

The surface displacement and the water velocity potential at
the water surface are linked through the kinematic boundary condition
<center>
<math>
\phi_{z}=-i\sqrt{\alpha}w,\,\,\,z=0
</math>
</center>
From the equations the potential and the surface
displacement are therefore related by
<center>
<math>
w=i\sqrt{\alpha}\phi,\quad r>a
</math>
</center>
and
<center>
<math>
(\beta\bar{\Delta}^{2}+1-\alpha\gamma)w=i\sqrt{\alpha}\phi,\quad r<a
</math>
</center>
The surface displacement can also be expanded in eigenfunctions
as
<center>
<math>
w(r,\theta)=\sum_{n=-\infty}^{\infty}\sum_{m=0}^{\infty}i\sqrt{\alpha}
a_{mn}K_{n}(k_{m}r)e^{i n\theta},\;\;r>a
</math>
</center>
and:
<center>
<math>
w(r,\theta)=
\sum_{n=-\infty}^{\infty}\sum_{m=-2}^{\infty}i\sqrt{\alpha}(\beta\kappa
_{m}^{4}+1-\alpha\gamma)^{-1}b_{mn}I_{n}(\kappa_{m}r)e^{i
n\theta},\; r<a
</math>
</center>
using the fact that
<center>
<math>
\bar{\Delta}\left( I_{n}(\kappa_{m}r)e^{i n\theta}\right) =\kappa_{m}
^{2}I_{n}(\kappa_{m}r)e^{i n\theta}\,\,\,(5)
</math>
</center>

==An infinite dimensional system of equations==

The boundary conditions (3) and
(4) can be expressed in terms of the potential
using (5). Since the angular modes are uncoupled the
conditions apply to each mode, giving
<center>
<math>
\sum_{m=-2}^{\infty}(\beta\kappa_{m}^{4}+1-\alpha\gamma)^{-1}b_{mn}
\left(\kappa_{m}^{2}I_{n}(\kappa_{m}a)-\frac{1-\nu}{a}\left(\kappa
_{m}I_{n}^{\prime}(\kappa_{m}a)-\frac{n^{2}}{a}I_{n}(\kappa_{m}a)\right)
\right) =0\,\,\,(6)
</math>
</center>
<center>
<math>
\sum_{m=-2}^{\infty}(\beta\kappa_{m}^{4}+1-\alpha\gamma)^{-1}b_{mn}
\left( \kappa_{m}^{3}I_{n}^{\prime}(\kappa_{m}a)+n^{2}\frac{1-\nu}{a^{2}
}\left( -\kappa_{m}I_{n}^{\prime}(\kappa_{m}a)+\frac{1}{a}I_{n}(\kappa
_{m}a)\right) \right)
=0\,\,\,(7)
</math>
</center>
The potential and its derivative must be continuous across the
transition from open water to the plate covered region. Therefore, the
potentials and their derivatives at <math>r=a</math> have to be equal.
Again we know that this must be true for each angle and we obtain
<center>
<math>
e_{n}I_{n}(k_{0}a)\phi_{0}\left( z\right) + \sum_{m=0}^{\infty}
a_{mn} K_{n}(k_{m}a)\phi_{m}\left( z\right)
=\sum_{m=-2}^{\infty}b_{mn}I_{n}(\kappa_{m}a)\psi_{m}(z)
</math>
</center>
and
<center>
<math>
e_{n}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)
=\sum_{m=-2}^{\infty}b_{mn}\kappa_{m}I_{n}^{\prime}(\kappa_{m}a)\psi
_{m}(z)
</math>
</center>
for each <math>n</math>.
We solve these equations by multiplying both equations by
<math>\phi_{l}(z)</math> and integrating from <math>-H</math> to <math>0</math> to obtain:
<center>
<math>
e_{n}I_{n}(k_{0}a)A_{0}\delta_{0l}+a_{ln}K_{n}(k_{l}a)A_{l}
=\sum_{m=-2}^{\infty}b_{mn}I_{n}(\kappa_{m}a)B_{ml} \,\,\,(8)
</math>
</center>
and
<center>
<math>
e_{n}k_{0}I_{n}^{\prime}(k_{0}a)A_{0}\delta_{0l}+a_{ln}k_{l}K_{n}^{\prime
}(k_{l}a)A_{l}
=\sum_{m=-2}^{\infty}b_{mn}\kappa_{m}I_{n}^{\prime}(\kappa_{m}
a)B_{ml} \,\,\,(9)
</math>
</center>
Equation (8) can be solved for the open water
coefficients <math>a_{mn}</math>
<center>
<math>
a_{ln}=-e_{n}\frac{I_{n}(k_{0}a)}{K_{n}(k_{0}a)}\delta_{0l}+\sum
_{m=-2}^{\infty}b_{mn}\frac{I_{n}(\kappa_{m}a)B_{ml}}{K_{n}(k_{l}a)A_{l}}
</math>
</center>
which can then be substituted into equation
(9) to give us
<center>
<math>
\left( k_{0}I_{n}^{\prime}(k_{0}a)-k_{0}\frac{K_{n}^{\prime}(k_{0}
a)}{K_{n}(k_{0}a)}I_{n}(k_{0}a)\right) e_{n}A_{0}\delta_{0l}
=\sum_{m=-2}^{\infty}\left( \kappa_{m}I_{n}^{\prime}(\kappa_{m}
a)-k_{l}\frac{K_{n}^{\prime}(k_{l}a)}{K_{n}(k_{l}a)}I_{n}(\kappa
_{m}a)\right) B_{ml}b_{mn}\,\,\,(10)
</math>
</center>
for each <math>n</math>.
Together with equations (6) and (7)
equation (10) gives the required equations to solve for the
coefficients of the water velocity potential in the plate covered region.

=Numerical Solution=

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>
<math>
\phi(r,\theta,z)=\sum_{n=-N}^{N}\sum_{m=0}^{M}a_{mn}K_{n}(k_{m}r)e^{i
n\theta }\phi_{m}(z), \;\;r>a
</math>
</center>
and
<center>
<math>
\phi(r,\theta,z)=\sum_{n=-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]]=

The shallow water theory of [[Zilman_Miloh_2000a|Zilman and Miloh 2000]] can be recovered by
simply setting the depth shallow enough that the shallow water theory is valid
and setting <math>M=0</math>. If the shallow water theory is valid then
the first three roots of the dispersion equation for the ice will be exactly
the same roots found in the shallow water theory by solving the polynomial
equation. The system of equations has four unknowns (three under the plate and
one in the open water) exactly as for the theory of [[Zilman_Miloh_2000a|Zilman and Miloh 2000]].

= Solving for the case of a more complicated incident wave =

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
<center><math>
\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)
=\sum_{m=-2}^{\infty}b_{mn}I_{n}(\kappa_{m}a)\psi_{m}(z)
</math></center>
and
<center><math>
\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}(k_{m}a)\phi_{m}\left( z\right)
=\sum_{m=-2}^{\infty}b_{mn} \kappa_m I'_{n}(\kappa_{m}a)\psi_{m}(z)
</math></center>
for each <math>n</math>.
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:
<center><math>
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},\ \ \ (11)
</math></center>
and
<center><math>
D_{ln} k_l I'_n (k_l a) A_l + a_{ln} k_l K'_{n}(k_{l} a) A_l
=\sum_{m=-2}^{\infty}b_{mn} \kappa_m I'_{n}(\kappa_{m}a) B_{ml} \ \ \ (12)
</math></center>
As a consequence we can find an expression of the coefficients <math>a_{ln}</math>
<center><math>
a_{ln} = -\frac{K_{n}(k_l a)}{I_n (k_l a)} D_{ln} +
\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 the equation for the normal derivative
of the potential, to give us
<center><math>
k_l A_l \left( I'_n (k_l a) - \frac{K'_{n}(k_{l} a) K_{n}(k_l a)}{I_n (k_l a)} \right) D_{ln}
= \sum_{m=-2}^{\infty} \left( \kappa_m I'_{n}(\kappa_{m}a) -
k_l \frac{K'_{n}(k_{l} a) I_{n}(\kappa_m a)}{K_n (k_l a)} \right) B_{ml} b_{mn}
</math></center>
for each <math>n</math>.

We notice here that there is still a dependence between the coefficients <math>a_{mn}</math>
and <math>D_{ln}</math>. We can get rid of this dependence, using a [[Diffraction Transfer Matrix]],
which maps the coefficients of the incident wave with the coefficients of the scattered wave within
the free water domain. Then 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>

Furthermore the boundary conditions are exactly the same that before, namely
<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
</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 .
</math>
</center>
For the further study, let's call
<center>
<math>
L^1_{mn}=(\beta\kappa_{m}^{4}+1-\alpha\gamma)^{-1}
\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)
</math>
</center>
and
<center>
<math>
L^2_{mn}=(\beta\kappa_{m}^{4}+1-\alpha\gamma)^{-1}
\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)
</math>
</center>
From the equations (11), (12) and the boundary conditions over the edges of the plate,
we can write a linear system of equation, limiting the number of modes of the dispersion equation
to <math>N</math> real ones
<center>
<math>
\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}
I_n(\kappa_{-2}a) B_{-20}&\cdots&I_n(\kappa_{N}a) B_{N0}\\
&&\\
\vdots&&\vdots\\
&&\\
I_n(\kappa_{-2}a) B_{-2N}&\cdots&I_n(\kappa_{N}a) B_{NN}
\end{bmatrix}
\\
\begin{bmatrix}
0&&\cdots&&0\\
0&&\cdots&&0
\end{bmatrix}
&
\begin{bmatrix}
L^1_{-2n}&\cdots&L^1_{Nn}\\
L^2_{-2n}&\cdots&L^2_{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}
\kappa_{-2} I'_n(\kappa_{-2}a) B_{-20}&\cdots&\kappa_{N} I'_n(\kappa_{N}a) B_{N0}\\
&&\\
\vdots&&\vdots\\
&&\\
\kappa_{-2} I'_n(\kappa_{-2}a) B_{-2N}&\cdots&\kappa_{N} I'_n(\kappa_{N}a) B_{NN}
\end{bmatrix}
\end{bmatrix}
\begin{bmatrix}
a_{0} \\
\\
\vdots \\
\\
a_N \\
b_{-2}\\
b_{-1}\\
b_{0}\\
\\
\vdots \\
\\
b_N
\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}
0&&\cdots&&0\\
0&&\cdots&&0
\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} \\
0\\
0\\
D_{0n}\\
\\
\vdots \\
\\
D_{Nn}
\end{bmatrix}
</math>
</center>
for each <math>n</math>.

Therefore we can find a [[Diffraction Transfer Matrix]] for each <math>n</math>,
by setting
<center><math>
\forall i \in [0, N], (D_{pn})_{p \in [0, N]} = \delta_{ip}
</math></center>
Then we solve the linear system defined previously, so that we can find the coefficients
<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]],
<math>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>n</math>, we finally obtain a 3-dimensional
matrix for the [[Diffraction Transfer Matrix]].

= Matlab Code =

A program to calculate the coefficients for circular plate problems can be found here
[http://www.math.auckland.ac.nz/~meylan/code/eigenfunction_matching/circle_plate_matching_one_n.m circle_plate_matching_one_n.m]
Note that this problem solves only for a single n.

== Additional code ==

This program requires
[http://www.math.auckland.ac.nz/~meylan/code/dispersion/dispersion_free_surface.m dispersion_free_surface.m]
and [http://www.math.auckland.ac.nz/~meylan/code/dispersion/dispersion_elastic_surface.m dispersion_elastic_surface.m]
to run

[[Category:Problems with Cylindrical Symmetry]]
[[Category:Linear Hydroelasticity]]
[[Category:Eigenfunction Matching Method]]
[[Category:Complete Pages]]

Eigenfunction Matching for a Circular Floating Elastic Plate

2008-08-02T02:00:40Z

Fmontiel: /* Solving for the case of a more complicated incident wave */

=Introduction=

We show here a solution for a [[Floating Elastic Plate]] on [[Finite Depth]] water
based on [[Peter_Meylan_Chung_2004a|Peter, Meylan and Chung 2004]]. A solution
for [[Shallow Depth]] was given in [[Zilman_Miloh_2000a|Zilman and Miloh 2000]] and we will also show this.
The solution is an extension of the [[Eigenfunction Matching for a Circular Dock]].

=Governing Equations=

We begin with the [[Frequency Domain Problem]] for a [[Floating Elastic Plate]]
in the non-dimensional form of [[Tayler_1986a|Tayler 1986]] ([[Dispersion Relation for a Floating Elastic Plate]])
We will use a cylindrical coordinate system, <math>(r,\theta,z)</math>,
assumed to have its origin at the centre of the circular
plate which has radius <math>a</math>. The water is assumed to have
constant finite depth <math>H</math> and the <math>z</math>-direction points vertically
upward with the water surface at <math>z=0</math> and the sea floor at <math>z=-H</math>. The
boundary value problem can therefore be expressed as
<center>
<math>
\Delta\phi=0, \,\, -H<z<0,
</math>
</center>
<center>
<math>
\phi_{z}=0, \,\, z=-H,
</math>
</center>
<center><math>
\phi_{z}=\alpha\phi, \,\, z=0,\,r>a,
</math></center>
<center>
<math>
(\beta\Delta^{2}+1-\alpha\gamma)\phi_{z}=\alpha\phi, \,\, z=0,\,r<a
</math>
</center>
where the constants <math>\beta</math> and <math>\gamma</math> are given by
<center>
<math>
\beta=\frac{D}{\rho\,L^{4}g}, \gamma=\frac{\rho_{i}h}{\rho\,L}
</math>
</center>
and <math>\rho_{i}</math> is the density of the plate. We
must also apply the edge conditions for the plate and the [[Sommerfeld Radiation Condition]]
as <math>r\rightarrow\infty</math>. The subscript <math>z</math>
denotes the derivative in <math>z</math>-direction.

=Solution Method=

== Separation of variables==

We now separate variables, noting that since the problem has
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>.
Using Laplace's equation in polar coordinates we obtain
<center>
<math>
\frac{\mathrm{d}^{2}\rho_{n}}{\mathrm{d}r^{2}}+\frac{1}{r}
\frac{\mathrm{d}\rho_{n}}{\mathrm{d}r}-\left(
\frac{n^{2}}{r^{2}}+\mu^{2}\right) \rho_{n}=0
</math>
</center>
where <math>\mu</math> is <math>k_{m}</math> or
<math>\kappa_{m},</math> depending on whether <math>r</math> is
greater or less than <math>a</math>. We can convert this equation to the
standard form by substituting <math>y=\mu r</math> to obtain
<center>
<math>
y^{2}\frac{\mathrm{d}^{2}\rho_{n}}{\mathrm{d}y^{2}}+y\frac{\mathrm{d}\rho_{n}
}{\rm{d}y}-(n^{2}+y^{2})\rho_{n}=0
</math>
</center>
The solution of this equation is a linear combination of the
modified Bessel functions of order <math>n</math>, <math>I_{n}(y)</math> and
<math>K_{n}(y)</math> [[Abramowitz and Stegun 1964]]. Since the solution must be bounded
we know that under the plate the solution will be a linear combination of
<math>I_{n}(y)</math> while outside the plate the solution will be a
linear combination of <math>K_{n}(y)</math>. Therefore the potential can
be expanded as
<center>
<math>
\phi(r,\theta,z)=\sum_{n=-\infty}^{\infty}\sum_{m=0}^{\infty}a_{mn}K_{n}
(k_{m}r)e^{i n\theta}\phi_{m}(z), \;\;r>a
</math>
</center>
and
<center>
<math>
\phi(r,\theta,z)=\sum_{n=-\infty}^{\infty}\sum_{m=-2}^{\infty}b_{mn}
I_{n}(\kappa_{m}r)e^{i n\theta}\psi_{m}(z), \;\;r<a
</math>
</center>
where <math>a_{mn}</math> and <math>b_{mn}</math>
are the coefficients of the potential in the open water and
the plate covered region respectively.

==Incident potential==

The incident potential is a wave of amplitude <math>A</math>
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 boundary conditions for the plate also have to be
considered. The vertical force and bending moment must vanish, which can be
written as
<center>
<math>
\left[\bar{\Delta}-\frac{1-\nu}{r}\left(\frac{\partial}{\partial r}
+\frac{1}{r}\frac{\partial^{2}}{\partial\theta^{2}}\right)\right]
w=0\,\,\,(3)
</math>
</center>
and
<center>
<math>
\left[ \frac{\partial}{\partial r}\bar{\Delta}-\frac{1-\nu}{r^{2}}\left(
-\frac{\partial}{\partial r}+\frac{1}{r}\right) \frac{\partial^{2}}
{\partial\theta^{2}}\right] w=0 \,\,\,(4)
</math>
</center>
where <math>w</math> is the time-independent surface
displacement, <math>\nu</math> is Poisson's ratio, and <math>\bar{\Delta}</math> is the
polar coordinate Laplacian
<center>
<math>
\bar{\Delta}=\frac{\partial^{2}}{\partial r^{2}}+\frac{1}{r}\frac{\partial
}{\partial r}+\frac{1}{r^{2}}\frac{\partial^{2}}{\partial\theta^{2}}
</math>
</center>

== Displacement of the plate ==

The surface displacement and the water velocity potential at
the water surface are linked through the kinematic boundary condition
<center>
<math>
\phi_{z}=-i\sqrt{\alpha}w,\,\,\,z=0
</math>
</center>
From the equations the potential and the surface
displacement are therefore related by
<center>
<math>
w=i\sqrt{\alpha}\phi,\quad r>a
</math>
</center>
and
<center>
<math>
(\beta\bar{\Delta}^{2}+1-\alpha\gamma)w=i\sqrt{\alpha}\phi,\quad r<a
</math>
</center>
The surface displacement can also be expanded in eigenfunctions
as
<center>
<math>
w(r,\theta)=\sum_{n=-\infty}^{\infty}\sum_{m=0}^{\infty}i\sqrt{\alpha}
a_{mn}K_{n}(k_{m}r)e^{i n\theta},\;\;r>a
</math>
</center>
and:
<center>
<math>
w(r,\theta)=
\sum_{n=-\infty}^{\infty}\sum_{m=-2}^{\infty}i\sqrt{\alpha}(\beta\kappa
_{m}^{4}+1-\alpha\gamma)^{-1}b_{mn}I_{n}(\kappa_{m}r)e^{i
n\theta},\; r<a
</math>
</center>
using the fact that
<center>
<math>
\bar{\Delta}\left( I_{n}(\kappa_{m}r)e^{i n\theta}\right) =\kappa_{m}
^{2}I_{n}(\kappa_{m}r)e^{i n\theta}\,\,\,(5)
</math>
</center>

==An infinite dimensional system of equations==

The boundary conditions (3) and
(4) can be expressed in terms of the potential
using (5). Since the angular modes are uncoupled the
conditions apply to each mode, giving
<center>
<math>
\sum_{m=-2}^{\infty}(\beta\kappa_{m}^{4}+1-\alpha\gamma)^{-1}b_{mn}
\left(\kappa_{m}^{2}I_{n}(\kappa_{m}a)-\frac{1-\nu}{a}\left(\kappa
_{m}I_{n}^{\prime}(\kappa_{m}a)-\frac{n^{2}}{a}I_{n}(\kappa_{m}a)\right)
\right) =0\,\,\,(6)
</math>
</center>
<center>
<math>
\sum_{m=-2}^{\infty}(\beta\kappa_{m}^{4}+1-\alpha\gamma)^{-1}b_{mn}
\left( \kappa_{m}^{3}I_{n}^{\prime}(\kappa_{m}a)+n^{2}\frac{1-\nu}{a^{2}
}\left( -\kappa_{m}I_{n}^{\prime}(\kappa_{m}a)+\frac{1}{a}I_{n}(\kappa
_{m}a)\right) \right)
=0\,\,\,(7)
</math>
</center>
The potential and its derivative must be continuous across the
transition from open water to the plate covered region. Therefore, the
potentials and their derivatives at <math>r=a</math> have to be equal.
Again we know that this must be true for each angle and we obtain
<center>
<math>
e_{n}I_{n}(k_{0}a)\phi_{0}\left( z\right) + \sum_{m=0}^{\infty}
a_{mn} K_{n}(k_{m}a)\phi_{m}\left( z\right)
=\sum_{m=-2}^{\infty}b_{mn}I_{n}(\kappa_{m}a)\psi_{m}(z)
</math>
</center>
and
<center>
<math>
e_{n}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)
=\sum_{m=-2}^{\infty}b_{mn}\kappa_{m}I_{n}^{\prime}(\kappa_{m}a)\psi
_{m}(z)
</math>
</center>
for each <math>n</math>.
We solve these equations by multiplying both equations by
<math>\phi_{l}(z)</math> and integrating from <math>-H</math> to <math>0</math> to obtain:
<center>
<math>
e_{n}I_{n}(k_{0}a)A_{0}\delta_{0l}+a_{ln}K_{n}(k_{l}a)A_{l}
=\sum_{m=-2}^{\infty}b_{mn}I_{n}(\kappa_{m}a)B_{ml} \,\,\,(8)
</math>
</center>
and
<center>
<math>
e_{n}k_{0}I_{n}^{\prime}(k_{0}a)A_{0}\delta_{0l}+a_{ln}k_{l}K_{n}^{\prime
}(k_{l}a)A_{l}
=\sum_{m=-2}^{\infty}b_{mn}\kappa_{m}I_{n}^{\prime}(\kappa_{m}
a)B_{ml} \,\,\,(9)
</math>
</center>
Equation (8) can be solved for the open water
coefficients <math>a_{mn}</math>
<center>
<math>
a_{ln}=-e_{n}\frac{I_{n}(k_{0}a)}{K_{n}(k_{0}a)}\delta_{0l}+\sum
_{m=-2}^{\infty}b_{mn}\frac{I_{n}(\kappa_{m}a)B_{ml}}{K_{n}(k_{l}a)A_{l}}
</math>
</center>
which can then be substituted into equation
(9) to give us
<center>
<math>
\left( k_{0}I_{n}^{\prime}(k_{0}a)-k_{0}\frac{K_{n}^{\prime}(k_{0}
a)}{K_{n}(k_{0}a)}I_{n}(k_{0}a)\right) e_{n}A_{0}\delta_{0l}
=\sum_{m=-2}^{\infty}\left( \kappa_{m}I_{n}^{\prime}(\kappa_{m}
a)-k_{l}\frac{K_{n}^{\prime}(k_{l}a)}{K_{n}(k_{l}a)}I_{n}(\kappa
_{m}a)\right) B_{ml}b_{mn}\,\,\,(10)
</math>
</center>
for each <math>n</math>.
Together with equations (6) and (7)
equation (10) gives the required equations to solve for the
coefficients of the water velocity potential in the plate covered region.

=Numerical Solution=

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>
<math>
\phi(r,\theta,z)=\sum_{n=-N}^{N}\sum_{m=0}^{M}a_{mn}K_{n}(k_{m}r)e^{i
n\theta }\phi_{m}(z), \;\;r>a
</math>
</center>
and
<center>
<math>
\phi(r,\theta,z)=\sum_{n=-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]]=

The shallow water theory of [[Zilman_Miloh_2000a|Zilman and Miloh 2000]] can be recovered by
simply setting the depth shallow enough that the shallow water theory is valid
and setting <math>M=0</math>. If the shallow water theory is valid then
the first three roots of the dispersion equation for the ice will be exactly
the same roots found in the shallow water theory by solving the polynomial
equation. The system of equations has four unknowns (three under the plate and
one in the open water) exactly as for the theory of [[Zilman_Miloh_2000a|Zilman and Miloh 2000]].

= Solving for the case of a more complicated incident wave =

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
<center><math>
\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)
=\sum_{m=-2}^{\infty}b_{mn}I_{n}(\kappa_{m}a)\psi_{m}(z)
</math></center>
and
<center><math>
\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}(k_{m}a)\phi_{m}\left( z\right)
=\sum_{m=-2}^{\infty}b_{mn} \kappa_m I'_{n}(\kappa_{m}a)\psi_{m}(z)
</math></center>
for each <math>n</math>.
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:
<center><math>
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},\ \ \ (11)
</math></center>
and
<center><math>
D_{ln} k_l I'_n (k_l a) A_l + a_{ln} k_l K'_{n}(k_{l} a) A_l
=\sum_{m=-2}^{\infty}b_{mn} \kappa_m I'_{n}(\kappa_{m}a) B_{ml} \ \ \ (12)
</math></center>
As a consequence we can find an expression of the coefficients <math>a_{ln}</math>
<center><math>
a_{ln} = -\frac{K_{n}(k_l a)}{I_n (k_l a)} D_{ln} +
\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 the equation for the normal derivative
of the potential, to give us
<center><math>
k_l A_l \left( I'_n (k_l a) - \frac{K'_{n}(k_{l} a) K_{n}(k_l a)}{I_n (k_l a)} \right) D_{ln}
= \sum_{m=-2}^{\infty} \left( \kappa_m I'_{n}(\kappa_{m}a) -
k_l \frac{K'_{n}(k_{l} a) I_{n}(\kappa_m a)}{K_n (k_l a)} \right) B_{ml} b_{mn}
</math></center>
for each <math>n</math>.

We notice here that there is still a dependence between the coefficients <math>a_{mn}</math>
and <math>D_{ln}</math>. We can get rid of this dependence, using a [[Diffraction Transfer Matrix]],
which maps the coefficients of the incident wave with the coefficients of the scattered wave within
the free water domain. Then 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>

Furthermore the boundary conditions are exactly the same that before, namely
<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
</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 .
</math>
</center>
For the further study, let's call
<center>
<math>
L^1_{mn}=(\beta\kappa_{m}^{4}+1-\alpha\gamma)^{-1}
\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)
</math>
</center>
and
<center>
<math>
L^2_{mn}=(\beta\kappa_{m}^{4}+1-\alpha\gamma)^{-1}
\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)
</math>
</center>
From the equations (11), (12) and the boundary conditions over the edges of the plate,
we can write a linear system of equation, limiting the number of modes of the dispersion equation
to <math>N</math> real ones
<center>
<math>
\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}
I_n(\kappa_{-2}a) B_{-20}&\cdots&I_n(\kappa_{N}a) B_{N0}\\
&&\\
\vdots&&\vdots\\
&&\\
I_n(\kappa_{-2}a) B_{-2N}&\cdots&I_n(\kappa_{N}a) B_{NN}
\end{bmatrix}
\\
\begin{bmatrix}
0&&\cdots&&0\\
0&&\cdots&&0
\end{bmatrix}
&
\begin{bmatrix}
L^1_{-2n}&\cdots&L^1_{Nn}\\
L^2_{-2n}&\cdots&L^2_{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}
\kappa_{-2} I'_n(\kappa_{-2}a) B_{-20}&\cdots&\kappa_{N} I'_n(\kappa_{N}a) B_{N0}\\
&&\\
\vdots&&\vdots\\
&&\\
\kappa_{-2} I'_n(\kappa_{-2}a) B_{-2N}&\cdots&\kappa_{N} I'_n(\kappa_{N}a) B_{NN}
\end{bmatrix}
\end{bmatrix}
\begin{bmatrix}
a_{0} \\
\\
\vdots \\
\\
a_N \\
b_{-2}\\
b_{-1}\\
b_{0}\\
\\
\vdots \\
\\
b_N
\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}
0&&\cdots&&0\\
0&&\cdots&&0
\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} \\
0\\
0\\
D_{0n}\\
\\
\vdots \\
\\
D_{Nn}
\end{bmatrix}
</math>
</center>
for each <math>n</math>.

Therefore we can find a diffraction transfer matrix for each <math>n</math>,
by setting
<center><math>
\forall i \in [0, N], (D_{pn})_{p \in [0, N]} = \delta_{ip}
</math></center>
Then we solve the linear system defined previously, so that we can find the coefficients
<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]],
<math>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>n</math>, we finally obtain a 3-dimensional
matrix for the [[Diffraction Transfer Matrix]].

= Matlab Code =

A program to calculate the coefficients for circular plate problems can be found here
[http://www.math.auckland.ac.nz/~meylan/code/eigenfunction_matching/circle_plate_matching_one_n.m circle_plate_matching_one_n.m]
Note that this problem solves only for a single n.

== Additional code ==

This program requires
[http://www.math.auckland.ac.nz/~meylan/code/dispersion/dispersion_free_surface.m dispersion_free_surface.m]
and [http://www.math.auckland.ac.nz/~meylan/code/dispersion/dispersion_elastic_surface.m dispersion_elastic_surface.m]
to run

[[Category:Problems with Cylindrical Symmetry]]
[[Category:Linear Hydroelasticity]]
[[Category:Eigenfunction Matching Method]]
[[Category:Complete Pages]]

Eigenfunction Matching for a Circular Floating Elastic Plate

2008-08-01T05:10:20Z

Fmontiel: /* Solving for the case of a more complicated incident wave */

=Introduction=

We show here a solution for a [[Floating Elastic Plate]] on [[Finite Depth]] water
based on [[Peter_Meylan_Chung_2004a|Peter, Meylan and Chung 2004]]. A solution
for [[Shallow Depth]] was given in [[Zilman_Miloh_2000a|Zilman and Miloh 2000]] and we will also show this.
The solution is an extension of the [[Eigenfunction Matching for a Circular Dock]].

=Governing Equations=

We begin with the [[Frequency Domain Problem]] for a [[Floating Elastic Plate]]
in the non-dimensional form of [[Tayler_1986a|Tayler 1986]] ([[Dispersion Relation for a Floating Elastic Plate]])
We will use a cylindrical coordinate system, <math>(r,\theta,z)</math>,
assumed to have its origin at the centre of the circular
plate which has radius <math>a</math>. The water is assumed to have
constant finite depth <math>H</math> and the <math>z</math>-direction points vertically
upward with the water surface at <math>z=0</math> and the sea floor at <math>z=-H</math>. The
boundary value problem can therefore be expressed as
<center>
<math>
\Delta\phi=0, \,\, -H<z<0,
</math>
</center>
<center>
<math>
\phi_{z}=0, \,\, z=-H,
</math>
</center>
<center><math>
\phi_{z}=\alpha\phi, \,\, z=0,\,r>a,
</math></center>
<center>
<math>
(\beta\Delta^{2}+1-\alpha\gamma)\phi_{z}=\alpha\phi, \,\, z=0,\,r<a
</math>
</center>
where the constants <math>\beta</math> and <math>\gamma</math> are given by
<center>
<math>
\beta=\frac{D}{\rho\,L^{4}g}, \gamma=\frac{\rho_{i}h}{\rho\,L}
</math>
</center>
and <math>\rho_{i}</math> is the density of the plate. We
must also apply the edge conditions for the plate and the [[Sommerfeld Radiation Condition]]
as <math>r\rightarrow\infty</math>. The subscript <math>z</math>
denotes the derivative in <math>z</math>-direction.

=Solution Method=

== Separation of variables==

We now separate variables, noting that since the problem has
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>.
Using Laplace's equation in polar coordinates we obtain
<center>
<math>
\frac{\mathrm{d}^{2}\rho_{n}}{\mathrm{d}r^{2}}+\frac{1}{r}
\frac{\mathrm{d}\rho_{n}}{\mathrm{d}r}-\left(
\frac{n^{2}}{r^{2}}+\mu^{2}\right) \rho_{n}=0
</math>
</center>
where <math>\mu</math> is <math>k_{m}</math> or
<math>\kappa_{m},</math> depending on whether <math>r</math> is
greater or less than <math>a</math>. We can convert this equation to the
standard form by substituting <math>y=\mu r</math> to obtain
<center>
<math>
y^{2}\frac{\mathrm{d}^{2}\rho_{n}}{\mathrm{d}y^{2}}+y\frac{\mathrm{d}\rho_{n}
}{\rm{d}y}-(n^{2}+y^{2})\rho_{n}=0
</math>
</center>
The solution of this equation is a linear combination of the
modified Bessel functions of order <math>n</math>, <math>I_{n}(y)</math> and
<math>K_{n}(y)</math> [[Abramowitz and Stegun 1964]]. Since the solution must be bounded
we know that under the plate the solution will be a linear combination of
<math>I_{n}(y)</math> while outside the plate the solution will be a
linear combination of <math>K_{n}(y)</math>. Therefore the potential can
be expanded as
<center>
<math>
\phi(r,\theta,z)=\sum_{n=-\infty}^{\infty}\sum_{m=0}^{\infty}a_{mn}K_{n}
(k_{m}r)e^{i n\theta}\phi_{m}(z), \;\;r>a
</math>
</center>
and
<center>
<math>
\phi(r,\theta,z)=\sum_{n=-\infty}^{\infty}\sum_{m=-2}^{\infty}b_{mn}
I_{n}(\kappa_{m}r)e^{i n\theta}\psi_{m}(z), \;\;r<a
</math>
</center>
where <math>a_{mn}</math> and <math>b_{mn}</math>
are the coefficients of the potential in the open water and
the plate covered region respectively.

==Incident potential==

The incident potential is a wave of amplitude <math>A</math>
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 boundary conditions for the plate also have to be
considered. The vertical force and bending moment must vanish, which can be
written as
<center>
<math>
\left[\bar{\Delta}-\frac{1-\nu}{r}\left(\frac{\partial}{\partial r}
+\frac{1}{r}\frac{\partial^{2}}{\partial\theta^{2}}\right)\right]
w=0\,\,\,(3)
</math>
</center>
and
<center>
<math>
\left[ \frac{\partial}{\partial r}\bar{\Delta}-\frac{1-\nu}{r^{2}}\left(
-\frac{\partial}{\partial r}+\frac{1}{r}\right) \frac{\partial^{2}}
{\partial\theta^{2}}\right] w=0 \,\,\,(4)
</math>
</center>
where <math>w</math> is the time-independent surface
displacement, <math>\nu</math> is Poisson's ratio, and <math>\bar{\Delta}</math> is the
polar coordinate Laplacian
<center>
<math>
\bar{\Delta}=\frac{\partial^{2}}{\partial r^{2}}+\frac{1}{r}\frac{\partial
}{\partial r}+\frac{1}{r^{2}}\frac{\partial^{2}}{\partial\theta^{2}}
</math>
</center>

== Displacement of the plate ==

The surface displacement and the water velocity potential at
the water surface are linked through the kinematic boundary condition
<center>
<math>
\phi_{z}=-i\sqrt{\alpha}w,\,\,\,z=0
</math>
</center>
From the equations the potential and the surface
displacement are therefore related by
<center>
<math>
w=i\sqrt{\alpha}\phi,\quad r>a
</math>
</center>
and
<center>
<math>
(\beta\bar{\Delta}^{2}+1-\alpha\gamma)w=i\sqrt{\alpha}\phi,\quad r<a
</math>
</center>
The surface displacement can also be expanded in eigenfunctions
as
<center>
<math>
w(r,\theta)=\sum_{n=-\infty}^{\infty}\sum_{m=0}^{\infty}i\sqrt{\alpha}
a_{mn}K_{n}(k_{m}r)e^{i n\theta},\;\;r>a
</math>
</center>
and:
<center>
<math>
w(r,\theta)=
\sum_{n=-\infty}^{\infty}\sum_{m=-2}^{\infty}i\sqrt{\alpha}(\beta\kappa
_{m}^{4}+1-\alpha\gamma)^{-1}b_{mn}I_{n}(\kappa_{m}r)e^{i
n\theta},\; r<a
</math>
</center>
using the fact that
<center>
<math>
\bar{\Delta}\left( I_{n}(\kappa_{m}r)e^{i n\theta}\right) =\kappa_{m}
^{2}I_{n}(\kappa_{m}r)e^{i n\theta}\,\,\,(5)
</math>
</center>

==An infinite dimensional system of equations==

The boundary conditions (3) and
(4) can be expressed in terms of the potential
using (5). Since the angular modes are uncoupled the
conditions apply to each mode, giving
<center>
<math>
\sum_{m=-2}^{\infty}(\beta\kappa_{m}^{4}+1-\alpha\gamma)^{-1}b_{mn}
\left(\kappa_{m}^{2}I_{n}(\kappa_{m}a)-\frac{1-\nu}{a}\left(\kappa
_{m}I_{n}^{\prime}(\kappa_{m}a)-\frac{n^{2}}{a}I_{n}(\kappa_{m}a)\right)
\right) =0\,\,\,(6)
</math>
</center>
<center>
<math>
\sum_{m=-2}^{\infty}(\beta\kappa_{m}^{4}+1-\alpha\gamma)^{-1}b_{mn}
\left( \kappa_{m}^{3}I_{n}^{\prime}(\kappa_{m}a)+n^{2}\frac{1-\nu}{a^{2}
}\left( -\kappa_{m}I_{n}^{\prime}(\kappa_{m}a)+\frac{1}{a}I_{n}(\kappa
_{m}a)\right) \right)
=0\,\,\,(7)
</math>
</center>
The potential and its derivative must be continuous across the
transition from open water to the plate covered region. Therefore, the
potentials and their derivatives at <math>r=a</math> have to be equal.
Again we know that this must be true for each angle and we obtain
<center>
<math>
e_{n}I_{n}(k_{0}a)\phi_{0}\left( z\right) + \sum_{m=0}^{\infty}
a_{mn} K_{n}(k_{m}a)\phi_{m}\left( z\right)
=\sum_{m=-2}^{\infty}b_{mn}I_{n}(\kappa_{m}a)\psi_{m}(z)
</math>
</center>
and
<center>
<math>
e_{n}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)
=\sum_{m=-2}^{\infty}b_{mn}\kappa_{m}I_{n}^{\prime}(\kappa_{m}a)\psi
_{m}(z)
</math>
</center>
for each <math>n</math>.
We solve these equations by multiplying both equations by
<math>\phi_{l}(z)</math> and integrating from <math>-H</math> to <math>0</math> to obtain:
<center>
<math>
e_{n}I_{n}(k_{0}a)A_{0}\delta_{0l}+a_{ln}K_{n}(k_{l}a)A_{l}
=\sum_{m=-2}^{\infty}b_{mn}I_{n}(\kappa_{m}a)B_{ml} \,\,\,(8)
</math>
</center>
and
<center>
<math>
e_{n}k_{0}I_{n}^{\prime}(k_{0}a)A_{0}\delta_{0l}+a_{ln}k_{l}K_{n}^{\prime
}(k_{l}a)A_{l}
=\sum_{m=-2}^{\infty}b_{mn}\kappa_{m}I_{n}^{\prime}(\kappa_{m}
a)B_{ml} \,\,\,(9)
</math>
</center>
Equation (8) can be solved for the open water
coefficients <math>a_{mn}</math>
<center>
<math>
a_{ln}=-e_{n}\frac{I_{n}(k_{0}a)}{K_{n}(k_{0}a)}\delta_{0l}+\sum
_{m=-2}^{\infty}b_{mn}\frac{I_{n}(\kappa_{m}a)B_{ml}}{K_{n}(k_{l}a)A_{l}}
</math>
</center>
which can then be substituted into equation
(9) to give us
<center>
<math>
\left( k_{0}I_{n}^{\prime}(k_{0}a)-k_{0}\frac{K_{n}^{\prime}(k_{0}
a)}{K_{n}(k_{0}a)}I_{n}(k_{0}a)\right) e_{n}A_{0}\delta_{0l}
=\sum_{m=-2}^{\infty}\left( \kappa_{m}I_{n}^{\prime}(\kappa_{m}
a)-k_{l}\frac{K_{n}^{\prime}(k_{l}a)}{K_{n}(k_{l}a)}I_{n}(\kappa
_{m}a)\right) B_{ml}b_{mn}\,\,\,(10)
</math>
</center>
for each <math>n</math>.
Together with equations (6) and (7)
equation (10) gives the required equations to solve for the
coefficients of the water velocity potential in the plate covered region.

=Numerical Solution=

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>
<math>
\phi(r,\theta,z)=\sum_{n=-N}^{N}\sum_{m=0}^{M}a_{mn}K_{n}(k_{m}r)e^{i
n\theta }\phi_{m}(z), \;\;r>a
</math>
</center>
and
<center>
<math>
\phi(r,\theta,z)=\sum_{n=-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]]=

The shallow water theory of [[Zilman_Miloh_2000a|Zilman and Miloh 2000]] can be recovered by
simply setting the depth shallow enough that the shallow water theory is valid
and setting <math>M=0</math>. If the shallow water theory is valid then
the first three roots of the dispersion equation for the ice will be exactly
the same roots found in the shallow water theory by solving the polynomial
equation. The system of equations has four unknowns (three under the plate and
one in the open water) exactly as for the theory of [[Zilman_Miloh_2000a|Zilman and Miloh 2000]].

= Solving for the case of a more complicated incident wave =

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
<center><math>
\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)
=\sum_{m=-2}^{\infty}b_{mn}I_{n}(\kappa_{m}a)\psi_{m}(z)
</math></center>
and
<center><math>
\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}(k_{m}a)\phi_{m}\left( z\right)
=\sum_{m=-2}^{\infty}b_{mn} \kappa_m I'_{n}(\kappa_{m}a)\psi_{m}(z)
</math></center>
for each <math>n</math>.
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:
<center><math>
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},\ \ \ (11)
</math></center>
and
<center><math>
D_{ln} k_l I'_n (k_l a) A_l + a_{ln} k_l K'_{n}(k_{l} a) A_l
=\sum_{m=-2}^{\infty}b_{mn} \kappa_m I'_{n}(\kappa_{m}a) B_{ml} \ \ \ (12)
</math></center>
As a consequence we can find an expression of the coefficients <math>a_{ln}</math>
<center><math>
a_{ln} = -\frac{K_{n}(k_l a)}{I_n (k_l a)} D_{ln} +
\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 the equation for the normal derivative
of the potential, to give us
<center><math>
k_l A_l \left( I'_n (k_l a) - \frac{K'_{n}(k_{l} a) K_{n}(k_l a)}{I_n (k_l a)} \right) D_{ln}
= \sum_{m=-2}^{\infty} \left( \kappa_m I'_{n}(\kappa_{m}a) -
k_l \frac{K'_{n}(k_{l} a) I_{n}(\kappa_m a)}{K_n (k_l a)} \right) B_{ml} b_{mn}
</math></center>
for each <math>n</math>.

We notice here that there is still a dependence between the coefficients <math>a_{mn}</math>
and <math>D_{ln}</math>. We can get rid of this dependence, using a [[Diffraction Transfer Matrix]],
which maps the coefficients of the incident wave with the coefficients of the scattered wave within
the free water domain.

Furthermore the boundary conditions are exactly the same that before, namely
<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
</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 .
</math>
</center>
For the further study, let's call
<center>
<math>
L^1_{mn}=(\beta\kappa_{m}^{4}+1-\alpha\gamma)^{-1}
\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)
</math>
</center>
and
<center>
<math>
L^2_{mn}=(\beta\kappa_{m}^{4}+1-\alpha\gamma)^{-1}
\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)
</math>
</center>
From the equations (11), (12) and the boundary conditions over the edges of the plate,
we can write a linear system of equation, limiting the number of modes of the dispersion equation
to <math>N</math> real ones
<center>
<math>
\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}
I_n(\kappa_{-2}a) B_{-20}&\cdots&I_n(\kappa_{N}a) B_{N0}\\
&&\\
\vdots&&\vdots\\
&&\\
I_n(\kappa_{-2}a) B_{-2N}&\cdots&I_n(\kappa_{N}a) B_{NN}
\end{bmatrix}
\\
\begin{bmatrix}
0&&\cdots&&0\\
0&&\cdots&&0
\end{bmatrix}
&
\begin{bmatrix}
L^1_{-2n}&\cdots&L^1_{Nn}\\
L^2_{-2n}&\cdots&L^2_{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}
\kappa{-2} I'_n(\kappa_{-2}a) B_{-20}&\cdots&\kappa_{N} I'_n(\kappa_{N}a) B_{N0}\\
&&\\
\vdots&&\vdots\\
&&\\
\kappa{-2} I'_n(\kappa_{-2}a) B_{-2N}&\cdots&\kappa_{N} I'_n(\kappa_{N}a) B_{NN}
\end{bmatrix}
\end{bmatrix}
\begin{bmatrix}
a_{0} \\
\\
\vdots \\
\\
a_N \\
b_{-2}\\
b_{-1}\\
b_{0}\\
\\
\vdots \\
\\
b_N
\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}
0&&\cdots&&0\\
0&&\cdots&&0
\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} \\
0\\
0\\
D_{0n}\\
\\
\vdots \\
\\
D_{Nn}
\end{bmatrix}
</math>
</center>

= Matlab Code =

A program to calculate the coefficients for circular plate problems can be found here
[http://www.math.auckland.ac.nz/~meylan/code/eigenfunction_matching/circle_plate_matching_one_n.m circle_plate_matching_one_n.m]
Note that this problem solves only for a single n.

== Additional code ==

This program requires
[http://www.math.auckland.ac.nz/~meylan/code/dispersion/dispersion_free_surface.m dispersion_free_surface.m]
and [http://www.math.auckland.ac.nz/~meylan/code/dispersion/dispersion_elastic_surface.m dispersion_elastic_surface.m]
to run

[[Category:Problems with Cylindrical Symmetry]]
[[Category:Linear Hydroelasticity]]
[[Category:Eigenfunction Matching Method]]
[[Category:Complete Pages]]