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]\displaystyle{ \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]\displaystyle{ \mathcal{U} }[/math] which is bounded by the rectangle [math]\displaystyle{ \mathcal{S} }[/math].] {A diagram depicting the area [math]\displaystyle{ \mathcal{U} }[/math] which is bounded by the rectangle [math]\displaystyle{ \mathcal{S} }[/math]. The rectangle [math]\displaystyle{ \mathcal{S} }[/math] is bounded by [math]\displaystyle{ -h\leq z \leq0 }[/math] and [math]\displaystyle{ -\infty\leq x \leq \infty }[/math].} (256) \end{figure}

Applying Green's theorem to [math]\displaystyle{ \phi }[/math] and its conjugate [math]\displaystyle{ \phi^* }[/math] gives

[math]\displaystyle{ (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]

where [math]\displaystyle{ n }[/math] denotes the outward plane normal to the boundary and [math]\displaystyle{ l }[/math] denotes the plane parallel to the boundary. As [math]\displaystyle{ \phi }[/math] and [math]\displaystyle{ \phi^* }[/math] satisfy the Laplace's equation, the left hand side of the Green theorem equation vanishes so that it reduces to

[math]\displaystyle{ \Im\int_\mathcal{S}\phi\frac{\partial\phi^*}{\partial n} dl = 0, }[/math]

Expanding gives

[math]\displaystyle{ \xi_1 +\xi_2 + \xi_3 = 0, }[/math]

where

[math]\displaystyle{ \xi_1 = { \Im\int_{-\infty}^{\infty}(\phi\frac{\partial\phi^*}{\partial z})\big|_{z=0}dx }, }[/math] [math]\displaystyle{ \xi_2 = { \Im\int_{-h}^0(\phi\frac{\partial\phi^*}{\partial x})\big|_{x=\infty}dz }, }[/math]

and

[math]\displaystyle{ \xi_3 = { - \Im\int_{-h}^0(\phi\frac{\partial\phi^*}{\partial x})\big|_{x=-\infty}dz } = 0 , }[/math]

where [math]\displaystyle{ \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]\displaystyle{ \frac{\partial\phi^*}{\partial z}\big|_{z=-h}=0 }[/math]).

Expanding [math]\displaystyle{ \mathbf{\xi_1} }[/math]

Near [math]\displaystyle{ x=-\infty }[/math], we approximate [math]\displaystyle{ \phi }[/math] by

[math]\displaystyle{ \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]

To simplify the derivation, we re-express the previous expression as

[math]\displaystyle{ \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]

where [math]\displaystyle{ k^I_1 = -\Im k_{1}(0) }[/math] and [math]\displaystyle{ \kappa^I_1=-\Im \kappa_1(0) }[/math], so that

[math]\displaystyle{ \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]

Therefore,

[math]\displaystyle{ \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]

where [math]\displaystyle{ R_1(0)^* }[/math] is the conjugate of [math]\displaystyle{ R_1(0) }[/math].

Expanding [math]\displaystyle{ \mathbf{\xi_2} }[/math]

Near [math]\displaystyle{ x=\infty }[/math], we approximate [math]\displaystyle{ \phi }[/math] by

[math]\displaystyle{ \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]

and re-express as

[math]\displaystyle{ \phi \approx T_\Lambda(0)e^{-i\kappa^I_\Lambda(x)})\frac{\cosh{(k^I_\Lambda(z+h))}}{\cosh{(k^I_\Lambda h)}}, }[/math]

where [math]\displaystyle{ k^I_\Lambda = -\Im k_{\Lambda}(0) }[/math] and [math]\displaystyle{ \kappa^I_\Lambda=-\Im \kappa_\Lambda(0) }[/math], so that

[math]\displaystyle{ \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]

Therefore,

[math]\displaystyle{ \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]

Expanding [math]\displaystyle{ \mathbf{\xi_3} }[/math]

The ice-covered boundary condition, \eqref{Eq:IceCoveredCondition}, gives

[math]\displaystyle{ (261) \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}\centerdot\frac{\partial\phi^*}{\partial z}\bigg|_{z=0}dx.} }[/math]

Since [math]\displaystyle{ {\frac{\partial\phi}{\partial z}\centerdot{\partial\phi^*}{\partial z}} }[/math] is real, [math]\displaystyle{ }[/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}\centerdot\frac{\partial\phi^*}{\partial z}\bigg|_{z=0}dx.[math]\displaystyle{ }[/math] Integration by parts gives \begin{multline}(262) \xi_3 = \Im\biggl[\biggr.\left[\frac{\beta}{\alpha}\frac{\partial}{\partial x} \left(\frac{\partial^2}{\partial x^2}-2k_y^2\right) \frac{\partial\phi}{\partial z}\centerdot\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}\centerdot\frac {\partial}{\partial x}\frac{\partial\phi^*}{\partial z}dx \biggl.\biggr]. \end{multline} As [math]\displaystyle{ {2k_y^2\frac{\partial}{\partial x}\frac{\partial\phi}{\partial z}\centerdot \frac{\partial}{\partial x}\frac{\partial\phi^*}{\partial z}} }[/math] is real and by integration by parts, \eqref{Eqn:EnergyIceCover2} becomes, \begin{multline}(263) \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}\centerdot\frac{\partial\phi^*}{\partial z}\right]_{-\infty}^{\infty}\right.\\ - \left[\frac{\beta}{\alpha}\frac{\partial^2}{\partial x^2}\frac{\partial\phi}{\partial z}\centerdot \frac{\partial}{\partial x}\frac{\partial\phi^*}{\partial z}\right]_{-\infty}^{\infty} \\ + \left.\int_{-\infty}^\infty\frac{\beta}{\alpha}\frac{\partial^2}{\partial x^2}\frac{\partial\phi}{\partial z}\centerdot \frac{\partial^2}{\partial x^2}\frac{\partial\phi^*}{\partial z}dx\right]. \end{multline} As [math]\displaystyle{ {\frac{\partial^2}{\partial x^2}\frac{\partial\phi}{\partial z}\centerdot \frac{\partial^2}{\partial x^2}\frac{\partial\phi^*}{\partial z}} }[/math] is real, \eqref{Eqn:Energy4} becomes

[math]\displaystyle{ (264) \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}\centerdot\frac{\partial\phi^*}{\partial z}\right]_{-\infty}^{\infty}- \left[\frac{\beta}{\alpha}\frac{\partial^2}{\partial x^2}\frac{\partial\phi}{\partial z}\centerdot \frac{\partial}{\partial x}\frac{\partial\phi^*}{\partial z}\right]_{-\infty}^{\infty}\right]. }[/math]

Now breaking [math]\displaystyle{ \xi_3 }[/math] down, \begin{multline*} \frac{\partial}{\partial x}\left(\frac{\partial^2}{\partial x^2}-2k_y^2\right) \frac{\partial\phi(x_2,0)}{\partial z}\centerdot \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-l_\Lambda)}\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-l_\Lambda)} \tanh{(k^I_\Lambda h)}\right)\\ \left(k^I_\Lambda T_\Lambda(0)^*e^{i\kappa^I_{\Lambda}(x-l_\Lambda)}\tanh{(k^I_\Lambda h)}\right), \end{multline*} \begin{flalign*} = 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{flalign*} \begin{multline*} \frac{\partial}{\partial x}\left(\frac{\partial^2}{\partial x^2}-2k_y^2\right) \frac{\partial\phi(x_1,0)}{\partial z}\centerdot \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)}-R_1(0)e^{i\kappa^I_{1}(x-r_1)}\right) \tanh{(k^I_1 h)}\right.\\ \left.- 2ik_y^2\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_1h)}\right]\\ \left[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_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)} - R_1(0)e^{i\kappa^I_{1}(x-r_1)}\right) \tanh{(k^I_1h)}\right]\\ \left[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_1h)}\right], \end{multline*}

[math]\displaystyle{ = 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). }[/math]

\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*}

[math]\displaystyle{ = -i(\kappa^I_{1})^3(k^I_1)^2\tanh^2{(k^I_1h)}\left(1 - |R_1(0)|^2\right). }[/math]

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

[math]\displaystyle{ \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]

which can be expressed as

[math]\displaystyle{ D |T_{\Lambda}(0)|^2 + |R_{1}(0)|^2 = 1, }[/math]

where [math]\displaystyle{ D }[/math] is given by

[math]\displaystyle{ 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]