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[water wave] Ray theory-3

开发技术 开发技术 2022-12-20 次浏览
the depth slowly vary for water wave

[water wave] Ray theory-3

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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