Difference between revisions of "Energy Balance for Two Elastic Plates"

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= Introduction =
 
= Introduction =
  
We derive here a condition which must be satisfied for energy balance for two semi-infinite elastic plates.
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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)} }+\\
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{
 +
        \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)} },&
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                \mbox{ for } x < r_1,\\
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{
 +
\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)} } + \\
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{
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                \qquad \qquad \sum_{n=-2}^{M}R_{\mu}(n) e^{\kappa_\mu(n)(x-r_\mu)}
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                        \frac{\cos(k_\mu(n)(z+h))}{\cos(k_\mu(n)h)} },
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                                &\mbox{ for } l_\mu< x < r_\mu,\\
 +
{
 +
\sum_{n=-2}^{M}T_{\Lambda}(n)e^{-\kappa_{\Lambda}(n)(x-l_\Lambda)}
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        \frac{\cos(k_\Lambda(n)(z+h))}{\cos(k_\Lambda(n)h)} }, &\mbox{ for } l_\Lambda<x,
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\end{matrix} \right.
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</math></center>
 +
(for details of this notation see [[Eigenfunction Matching Method for Floating Elastic Plates]]).
  
 
= Equations =  
 
= Equations =  

Revision as of 09:08, 11 July 2008

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 multiple elastic plates and from here we know we can write the potential as

[math]\displaystyle{ \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 \lt 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\lt x \lt 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\lt x, \end{matrix} \right. }[/math]

(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]\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{ { \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_{-h}^0(\phi\frac{\partial\phi^*}{\partial x})\big|_{x=-\infty}dz }, }[/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_{-\infty}^{\infty}(\phi\frac{\partial\phi^*}{\partial z})\big|_{z=0}dx }= 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^{-\hat{k_0}x}\frac{\cos{(k_0(z+h))}}{\cos{(k_0h)}} + R e^{\kappa_0x}\frac{\cos{(k_0(z+h))}}{\cos{(k_0h)}}. }[/math]

We also need the derivative of the potential with respect to [math]\displaystyle{ x }[/math]

[math]\displaystyle{ \frac{\partial\phi}{\partial x} \approx \left(-k_0e^{-k_0x} + \kappa_0a_0e^{\kappa_0x}\right)\frac{\cos{(k_0(z+h))}}{\cos{(k_0h)}}. }[/math]

Therefore,

[math]\displaystyle{ \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]

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 for the Floating Elastic Plate gives

[math]\displaystyle{ \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]

Since [math]\displaystyle{ {\frac{\partial\phi}{\partial z}\cdot\frac{\partial\phi^*}{\partial z}} }[/math] is real,

[math]\displaystyle{ \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]

Integration by parts gives

[math]\displaystyle{ \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]

As [math]\displaystyle{ {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]\displaystyle{ \xi_3 }[/math] becomes,

[math]\displaystyle{ \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]

As [math]\displaystyle{ {\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]\displaystyle{ \xi_3 }[/math]

[math]\displaystyle{ \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]

Now breaking [math]\displaystyle{ \xi_3 }[/math] down, we can simplify the left hend term for [math]\displaystyle{ x\gt 0 }[/math]

[math]\displaystyle{ \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]

and for [math]\displaystyle{ x\lt 0 }[/math]

[math]\displaystyle{ \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]

Likewise we expand the right hend term for [math]\displaystyle{ x\gt 0 }[/math]

[math]\displaystyle{ \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]

and finally for [math]\displaystyle{ x\lt 0 }[/math],

[math]\displaystyle{ \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]

We can now express [math]\displaystyle{ \xi_3 }[/math] as

[math]\displaystyle{ \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]

Solving the Energy Balance Equation

Pulling it all together, we finally obtain

[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. }[/math]

Re-arranging gives

[math]\displaystyle{ \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]

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]