the depth slowly vary for water wave
![[water wave] Ray theory-3](https://pic.wangt.cc/download/pic_router.php?path=https://img2023.cnblogs.com/blog/2894364/202212/2894364-20221218091211389-1134867362.jpg)
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^2left(k^2+l^2right)}>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^2left(c_1 e^{-sigma}+c_2 e^sigmaright)
]
[-sigma c_1 e^{-sigma B}+sigma c_2 e^{sigma B}=0 tag{1}
]
[rightarrow c_1left(sigma e^{-sigma}+delta^2 omega^2 e^{-sigma}right)+c_2left(delta^2 omega^2 e^sigma-sigma e^sigmaright)= 0
]
then substitute (1) to it:
[c_1=e^{2 sigma B} c_2 tag{2}
]
[small
c_2left(sigma e^{2 sigma B-sigma}+delta^2 omega^2 e^{2 sigma B-sigma}right)+c_2left(delta^2 omega^2 e^sigma-sigma e^sigmaright)=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^2left(e^{2 sigma B-sigma}+e^sigmaright)=sigmaleft(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 cdotleft(k int_B^1 beta_0^2 d z+left[frac{partial}{partial T}left(omega beta_0^2right)right]_{z=1}=0right.
]
We have:
[small
begin{aligned}
& {left[i delta^2 frac{partial}{partial T}left(omega beta_0^2right)right]_{z=1}-left[i delta^2left(k B_{bar{x}}+l B_{bar{y}}right) beta_0^2right]_{z=B}} \
&=-i delta^2left[k int_B^1 frac{partialleft(beta_0^2right)}{partial bar{x}} d z+l int_B^1 frac{partial}{partial bar{y}}left(beta_0^2right) d z+left(k_{bar{x}}+l_bar{y}right) int_B^1 beta_0^2 d zright]
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{partialleft(beta_0^2right)}{partial bar{x}} d z+left[beta_0^2right]_{z=1} cdot 0-left[beta_0^2right]_{z=B} cdot B_bar{x}right} vec{i} \
& +left{int_B^1 frac{partialleft(beta_0^2right)}{partial bar{y}} d z+left[beta_0^2right]_{z=1} cdot 0-left[beta_0^2right]_{z=B} cdot B_{bar{y}}right} vec{j}
end{aligned}
]
[int_B^1 frac{partialleft(beta_0^2right)}{partial bar{x}} vec{i}+frac{partialleft(beta_0^2right)}{partial bar{y}} vec{j} d z-left[beta_0^2right]_{z=B} cdot B_{bar{x}} vec{i}-left[beta_0^2right]_{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{partialleft(beta_0^2right)}{partial bar{x}} d z-kleft[beta_0^2right]_{z=B} cdot B_bar{x} \
& +l int_B^1 frac{partialleft(beta_0^2right)}{partial bar{y}} d z-lleft[beta_0^2right]_{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 cdotleft(vec{k} int_B^1 beta_0^2 d zright)=(nabla cdot vec{k}) int_B^1 beta_0^2 d z+vec{k} cdotleft[nabla int_B^1 beta_0^2 d zright]\
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
[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{partialleft(omega^2right)}{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
程序员灯塔
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