View source for Energy Balance for Two Elastic Plates

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 [[Evans and Davies 1968]].
We set up the problem as given in Figure 256 
\begin{figure}[H]
\begin{center}
\includegraphics[width=.8\textwidth]{Figures/EnergyBalanceTricks}
\end{center}
\caption[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>.} 
(256)
\end{figure}

Applying Green's theorem to <math>\phi</math> and its conjugate <math>\phi^*</math> gives
<center><math>(257)
{  \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_{-\infty}^{\infty}(\phi\frac{\partial\phi^*}{\partial z})\big|_{z=0}dx }, 
</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_{-h}^0(\phi\frac{\partial\phi^*}{\partial x})\big|_{x=-\infty}dz } = 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)}\frac{\cos{(k_1(0)(z+h))}}{\cos{(k_1(0)h)}} + 
	R_1(0)e^{\kappa_{1}(0)(x)}\frac{\cos{(k_{1}(0)(z+h))}}{\cos{(k_{1}(0)h)}}.
</math></center>
To simplify the derivation, we re-express the previous expression as
<center><math>
\phi \approx \left(e^{-i\kappa^I_1(x)}+ R_1(0)e^{i\kappa^I_1(x)}\right)\frac{\cosh{(k^I_1(z+h))}}{\cosh{(k^I_1h)}},  
</math></center>
where <math>k^I_1 = -\Im k_{1}(0)</math> and <math>\kappa^I_1=-\Im \kappa_1(0)</math>, so that
<center><math>
\frac{\partial\phi}{\partial x} \approx \left(-i\kappa^{I}_1e^{-i\kappa^{I}_1(x)} 
	+ i\kappa^{I}_1R_1(0)e^{i\kappa^{I}_1(x)}\right)\frac{\cosh{(k^I_1(z+h))}}{\cosh{(k^I_1h)}}.
</math></center>
Therefore,
<center><math>
\begin{matrix}
\xi_1 & = & {\Im\int_{-h}^0 \left[\left(e^{-i\kappa^{I}_1(x-r_1)}+ R_1(0)e^{i\kappa^{I}_1(x-r_1)}\right)\right.}
 {   \left(i\kappa^{I}_1e^{i\kappa^{I}_1(x-r_1)} - i\kappa^{I}_1R_1(0)^*e^{-i\kappa^{I}_1(x-r_1)}\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 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>

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>

\begin{multline*}
\frac{\partial^2}{\partial x^2}\frac{\partial\phi(x_2,0)}{\partial z}\centerdot
		\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-l_\Lambda)} \tanh{(k^I_\Lambda h)}\right]\\,
		\left[i\kappa^I_{\Lambda} k^I_\Lambda T_\Lambda(0)^*e^{i\kappa^I_{\Lambda}(x-l_\Lambda)} \tanh{(k^I_\Lambda h)}\right],
\end{multline*}
\begin{multline*}
 = -i(\kappa^I_{\Lambda})^3(k^I_\Lambda)^2\tanh^2{(k^I_\Lambda h)}|T_\Lambda(0)|^2,
\end{multline*}
and finally,
\begin{multline*}
\frac{\partial^2}{\partial x^1}\frac{\partial\phi(x_1,0)}{\partial z}\centerdot
		\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)} + R_1(0)e^{i\kappa^I_{1}(x-r_1)}\right)
			\tanh{(k^I_1 h)}\right]\\
		\left[i\kappa^I_{1} k^I_1\left(e^{i\kappa^I_{1}(x-r_1)} - R_1(0)e^{-i\kappa^I_{1}(x-r_1)}\right)
			\tanh{(k^I_1 h)}\right],
\end{multline*}
<center><math>
= -i(\kappa^I_{1})^3(k^I_1)^2\tanh^2{(k^I_1h)}\left(1 - |R_1(0)|^2\right).
</math></center>

We can now express \eqref{Eqn:EnergyTerm1} as 
\begin{multline*}
\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] \right.\\
	-\left.\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] \right.\\
	-\left.\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]\right.\\
	+\left.\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]
\end{multline*}
\begin{multline*}
= \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{multline*}

= Solving the Energy Balance Equation =

Pulling it all together, \eqref{Eqn:EnergyPhi} becomes
<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.
\end{multline*}
Re-arranging gives
\begin{multline*}
\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)}\right. \\
	\left.+ \frac{1}{2k^I_\Lambda}+\frac{h}{2\sinh{(k^I_\Lambda h)}\cosh{(k^I_\Lambda h)}}\right)|T_\Lambda(0)|^2 
\end{multline*}
\begin{multline*}
	\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)}\right. \\
	\left.+ \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]]