the depth slowly vary for water wave
p96-1 show
\[\beta_0=A_0 \cosh [\sigma(z-B)] \]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
\[\omega^2=\frac{\sigma}{\delta^2} \tanh \sigma \]p97-1show:
\[\nabla \cdot\left(k \int_B^1 \beta_0^2 d z+\left[\frac{\partial}{\partial T}\left(\omega \beta_0^2\right)\right]_{z=1}=0\right. \]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
\[\small \begin{aligned} &\nabla \cdot\left(\vec{k} \int_B^1 \beta_0^2 d z\right)=(\nabla \cdot \vec{k}) \int_B^1 \beta_0^2 d z+\vec{k} \cdot\left[\nabla \int_B^1 \beta_0^2 d z\right]\\ \end{aligned} \]\[\small \begin{split} \qquad \qquad \qquad \qquad &=(k_{\bar{x}}+l_{\bar{y}}) \int_B^1 \beta_0^2 d z+k \int_B^1 \frac{\partial (\beta_0^2)}{\partial \bar{x}} d z\\ &+l \int_B^1 \frac{\partial (\beta_0^2)}{\partial \bar{y}} d z-[(k B_{\bar{x}}+l B_{\bar{y}}) \beta_0^2]_{z=B}\\ \end{split} \]Q.E.D
consider
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\[\quad \frac{\partial \omega}{\partial k}=\frac{\delta^2 k \omega}{2 \sigma^2}\left(1+\frac{2 \sigma D}{\sinh \sigma D}\right) \]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
标签：right,partial,theory,water,beta,frac,left,sigma,Ray From： https://www.cnblogs.com/cicada-math/p/16993482.html