abstract: the depth slowly vary for water wave
set-1
show(p96-1 ):
\[ \beta_0=A_0 \cosh [\sigma(z-B)] \] We have: (note that \(\omega \rightarrow\) omega) \[ \beta_{0 z z}-\sigma^2 \beta_0=0, \quad \sigma=\sqrt{\delta^2\left(k^2+l^2\right)}>0 \] general solution: \[ \beta_0=c_1 e^{-\sigma z}+c_2 e^{\sigma z} \] boundary condition: \[ \beta_{0 z}=\delta^2 \omega^2 \beta_0, \ z=1 ; \quad \beta_{0 z}=0 \text { on }\ z=B \] then: \[ -\sigma c_1 e^{-\sigma}+\sigma c_2 e^\sigma=\delta^2 \omega^2\left(c_1 e^{-\sigma}+c_2 e^\sigma\right) \] \[ -\sigma c_1 e^{-\sigma B}+\sigma c_2 e^{\sigma B}=0 \tag{1} \]
\[ \rightarrow c_1\left(\sigma e^{-\sigma}+\delta^2 \omega^2 e^{-\sigma}\right)+c_2\left(\delta^2 \omega^2 e^\sigma-\sigma e^\sigma\right)= 0 \]
then substitute (1) to it: \[ c_1=e^{2 \sigma B} c_2 \tag{2} \]
\[ \small c_2\left(\sigma e^{2 \sigma B-\sigma}+\delta^2 \omega^2 e^{2 \sigma B-\sigma}\right)+c_2\left(\delta^2 \omega^2 e^\sigma-\sigma e^\sigma\right)=0 \tag{3} \]
therefore, from (2) \[ \begin{aligned} & \beta_0=c_2 e^{2 \sigma B-\sigma z}+c_2 e^{\sigma z} \\ & =c_2 e^{-\sigma B}\left(e^{\sigma B-\sigma z}+e^{\sigma z-\sigma B}\right) \\ \end{aligned} \] Let \(c_2 e^{-\sigma B}=A_0\) \[ \beta_0=A_0 \cosh [\sigma(z-B)] \] From (3): \[ \begin{aligned} & \sigma e^{2 \sigma B-\sigma}+\delta^2 \omega^2 e^{2 \sigma B-\sigma}+\delta^2 \omega^2 e^\sigma-\sigma e^\sigma=0 \\ \\ & \delta^2 \omega^2\left(e^{2 \sigma B-\sigma}+e^\sigma\right)=\sigma\left(e^\sigma-e^{2 \sigma B-\sigma}\right) \\ \\ & \omega^2=\frac{\sigma}{\delta^2} \frac{e^\sigma-e^{2 \sigma B-\sigma} }{e^\sigma+e^{2 \sigma B-\sigma} }=\frac{\sigma}{\delta^2} \frac{e^{\sigma-\sigma B}-e^{\sigma B-\sigma} }{e^{\sigma-\sigma B}+e^{\sigma B-\sigma} } \\ \end{aligned} \] when \(B=0\) then the dispersion relation \[ \omega^2=\frac{\sigma}{\delta^2} \tanh \sigma \]
set-2
show(p97-1): \[ \nabla \cdot\left(k \int_B^1 \beta_0^2 d z+\left[\frac{\partial}{\partial T}\left(\omega \beta_0^2\right)\right]_{z=1}=0\right. \] We have: \[ \small \begin{aligned} & {\left[i \delta^2 \frac{\partial}{\partial T}\left(\omega \beta_0^2\right)\right]_{z=1}-\left[i \delta^2\left(k B_{\bar{x} }+l B_{\bar{y} }\right) \beta_0^2\right]_{z=B} } \\ &=-i \delta^2\left[k \int_B^1 \frac{\partial\left(\beta_0^2\right)}{\partial \bar{x} } d z+l \int_B^1 \frac{\partial}{\partial \bar{y} }\left(\beta_0^2\right) d z+\left(k_{\bar{x} }+l_\bar{y}\right) \int_B^1 \beta_0^2 d z\right] \end{aligned} \] consider that: \[ \small \frac{d}{d x} \int_{b(x)}^{a(x)} f(x, y) d y=\int_a^b f_x(x, y) d y+f(x, b) b_x-f(x, a) a_x \] we write: \(\quad(\nabla=(\partial / \partial x, \partial / \partial y) \quad \vec{k}=(k, \nu))\) \[ \small \begin{aligned} \nabla \cdot \int_{B(\bar{x}, \bar{y})}^1 \beta_0^2 d z &=\left\{\int_B^1 \frac{\partial\left(\beta_0^2\right)}{\partial \bar{x} } d z+\left[\beta_0^2\right]_{z=1} \cdot 0-\left[\beta_0^2\right]_{z=B} \cdot B_\bar{x}\right\} \vec{i} \\ & +\left\{\int_B^1 \frac{\partial\left(\beta_0^2\right)}{\partial \bar{y} } d z+\left[\beta_0^2\right]_{z=1} \cdot 0-\left[\beta_0^2\right]_{z=B} \cdot B_{\bar{y} }\right\} \vec{j} \end{aligned} \] \[ \int_B^1 \frac{\partial\left(\beta_0^2\right)}{\partial \bar{x} } \vec{i}+\frac{\partial\left(\beta_0^2\right)}{\partial \bar{y} } \vec{j} d z-\left[\beta_0^2\right]_{z=B} \cdot B_{\bar{x} } \vec{i}-\left[\beta_0^2\right]_{z=B} \cdot B_{\bar{y} } \vec{j} \] then: \[ \begin{aligned} \vec{k} \cdot \nabla \int_B^1 \beta_0^2 d z&=k \int_B^1 \frac{\partial\left(\beta_0^2\right)}{\partial \bar{x} } d z-k\left[\beta_0^2\right]_{z=B} \cdot B_\bar{x} \\ & +l \int_B^1 \frac{\partial\left(\beta_0^2\right)}{\partial \bar{y} } d z-l\left[\beta_0^2\right]_{z=B} \cdot B_{\bar{y} } \\ & \end{aligned} \] we see that: \[ (\nabla \vec{k}) \int_B^1 \beta_0^2 d z=\left(k_{\bar{x} }+l_{\bar{y} }\right) \int_B^1 \beta_0^2 d z \] and
\[ \small\nabla \cdot\left(\vec{k} \int_B^1 \beta_0^2 d z\right)=(\nabla \cdot \vec{k}) \int_B^1 \beta_0^2 d z+\vec{k} \cdot\left[\nabla \int_B^1 \beta_0^2 d z\right] \]
\[ \small \begin{split} \qquad \qquad \qquad \qquad &=(k_{\bar{x} }+l_{\bar{y} }) \int_B^1 \beta_0^2 d z+k \int_B^1 \frac{\partial (\beta_0^2)}{\partial \bar{x} } d z\\ &+l \int_B^1 \frac{\partial (\beta_0^2)}{\partial \bar{y} } d z-[(k B_{\bar{x} }+l B_{\bar{y} }) \beta_0^2]_{z=B}\\ \end{split} \] Q.E.D
set-3
consider \[ \omega^2=\frac{\sigma}{\delta^2} \tanh [\sigma(1-B)],\ \sigma=\delta \sqrt{k^2+l^2}\ \& \ D=1-B \]
Show \[ \quad \frac{\partial \omega}{\partial k}=\frac{\delta^2 k \omega}{2 \sigma^2}\left(1+\frac{2 \sigma D}{\sinh \sigma D}\right) \] write derivatives of LHD with respect to \(k\) \[ \frac{\partial\left(\omega^2\right)}{\partial K}=2 \omega \frac{\partial \omega}{\partial K} \] Similarly the RHD \[ \begin{aligned} \frac{\partial(R H D)}{\partial k} & =\frac{1}{\delta} \tanh \sigma D+\frac{\sigma}{\delta^2} \frac{D}{\cosh ^2 \sigma D} \frac{\partial \sigma}{\partial k} \\ & = \end{aligned} \] To be continued