continue
\[ \frac{\partial(\omega)}{\partial k}=\frac{\partial}{\partial k}\left[\frac{\sigma}{\delta^2} \tanh [\sigma(1-B)]]\right. \]
the RHD : \[ \begin{aligned} R H D & =\frac{k+2 D k \sqrt{k^2+l^2} \delta \operatorname{csch}\left[2 D \sqrt{k^2+l^2} \delta\right] \omega}{2\left(k^2+l^2\right)} \\ & =\frac{k \omega(1+2 \sigma D \operatorname{csch}(2 \sigma D)}{2 \sigma^2 / \delta^2} \\ & =\frac{k \omega \delta^2}{2 \sigma^2}\left[1+\frac{2 \sigma D}{\sinh (2 \sigma D)}\right] \end{aligned} \] Q.E.D
set-1
show(p98-1 ):
\[ \frac{\partial}{\partial T}\left(\frac{E}{\omega}\right)+\nabla \cdot\left(\frac{E}{\omega} \vec c_g\right)=0, \vec c_g=\left(\frac{\partial \omega}{\partial k}, \frac{\partial \omega}{\partial l}\right) \]
Since \[ E=\frac{1}{2} \omega^2 A_0^2 \cosh ^2 \sigma D \]
\[ \nabla \cdot\left(\frac{E}{\omega} \vec c_g\right)=\nabla \cdot\left(\frac{1}{2} \omega A_0^2 \cosh ^2 (\sigma D)\vec c_g\right) \]
we have: \[ \begin{gathered} \vec{k} \int_B^1 \beta_0^2 d z=\left(\omega A_0^2 \cosh ^2 \sigma D\right) \vec c_g \\ \nabla \cdot\left(\vec{k} \int_B^1 \beta_0^2 d z\right)+\left[\frac{\partial}{\partial T}\left(\omega \beta_0^2\right)\right]_{z=1}=0 \quad(1) \end{gathered} \] hence: \[ \nabla \cdot\left(\vec{k} \int_B^1 \beta_0^2 d z\right)=\nabla \cdot\left(\frac{2 E}{\omega} \vec c_g\right)=2 \nabla \cdot\left(\frac{E}{\omega} \vec c_g\right) \] from equ 2.77 \[ \beta_0=A_0 \cosh [\sigma(z-B)] \] then: \[ \begin{aligned} \frac{\partial}{\partial T}\left[\omega \beta_0^2\right]_{z=1} & =\frac{\partial}{\partial T}\left[\omega A_0^2 \cosh ^2 \sigma D\right] \\ & =\frac{\partial}{\partial T}\left[\frac{2 E}{\omega}\right] \end{aligned} \] Substitute (2) (3) to (1) \[ 2 \nabla \cdot\left(\frac{E}{\omega} \vec c_g\right)+2 \frac{\partial}{\partial T}\left(\frac{E}{\omega}\right)=0 \] \(Q \cdot E \cdot D\)