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]
