dynamics in strained rorating flow

the pdf of paper chaotic dynamics in a strained rotating flow: a precessing plane fluid layer

prove of (2.12)

paper have given the equation of motion in boundary (rotating) frame (2.4):

$$\frac{\partial \mathit{u}}{\partial t}+2[\hat{\mathit{z}}+\mathit{ϵ}(t)]\times \mathit{u}+\mathit{u}\cdot \mathrm{\nabla }\mathit{u}+\mathrm{\nabla }p=E{\mathrm{\nabla }}^2\mathit{u}+2z\mathit{ϵ}(t)$$

the boundary conditions

$$\mathit{u}(x,y,\pm \frac{1}{2})=\mathbf{0}$$

where

$$\mathit{ϵ}(t)=ϵ\hat{\mathit{X}}:=ϵ(\hat{\mathit{x}}\cos t-\hat{\mathit{y}}\sin t)$$

and a simple flow solution in the boundary frame is

$${\mathit{u}}_{\text{basic}}=2[\mathit{ϵ}(t)\times \hat{\mathit{z}}]\hat{\mathit{z}}\cdot \mathit{r}=-2ϵz(\hat{\mathit{x}}\sin t+\hat{\mathit{y}}\cos t)$$

${\mathit{u}}_{\text{basic}}$ satisfies

$$\frac{\partial {\mathit{u}}_{\text{basic}}}{\partial t}+2[\hat{\mathit{z}}+\mathit{ϵ}(t)]\times {\mathit{u}}_{\text{basic}}+{\mathit{u}}_{\text{basic}}\cdot \mathrm{\nabla }{\mathit{u}}_{\text{basic}}+\mathrm{\nabla }p=E{\mathrm{\nabla }}^2{\mathit{u}}_{\text{basic}}+2z\mathit{ϵ}(t)$$

let velocity perturbation $\tilde{\mathit{u}}=\mathit{u}-{\mathit{u}}_{\text{basic}}$

replacing the $\mathit{u}$ in (2.4) with $\tilde{\mathit{u}}+{\mathit{u}}_{\text{basic}}$ and applying the relation of basic flow, we are led to

$$\frac{\partial \tilde{\mathit{u}}}{\partial t}+2[\hat{\mathit{z}}+\mathit{ϵ}(t)]\times \tilde{\mathit{u}}+\tilde{\mathit{u}}\cdot \mathrm{\nabla }\tilde{\mathit{u}}+\tilde{\mathit{u}}\cdot \mathrm{\nabla }{\mathit{u}}_{\text{basic}}+{\mathit{u}}_{\text{basic}}\cdot \mathrm{\nabla }\tilde{\mathit{u}}+\mathrm{\nabla }\tilde{p}=E{\mathrm{\nabla }}^2\tilde{\mathit{u}}$$

here we have

$$\mathit{ϵ}(t)\times \tilde{\mathit{u}}=\left|\begin{array}{ccc}\hat{\mathit{x}}& \hat{\mathit{y}}& \hat{\mathit{z}}\\ ϵ\cos t& -ϵ\sin t& 0\\ {\tilde{u}}_x& {\tilde{u}}_y& {\tilde{u}}_z\end{array}\right|=\left[\begin{array}{c}-ϵ\sin t{\tilde{u}}_z\\ -ϵ\cos t{\tilde{u}}_z\\ ϵ\cos t{\tilde{u}}_y+ϵ\sin t{\tilde{u}}_x\end{array}\right]$$$$\tilde{\mathit{u}}\cdot \mathrm{\nabla }{\mathit{u}}_{\text{basic}}=({\tilde{u}}_x\frac{\partial }{\partial x}+{\tilde{u}}_y\frac{\partial }{\partial y}+{\tilde{u}}_z\frac{\partial }{\partial z})\left[\begin{array}{c}-2ϵz\sin t\\ -2ϵz\cos t\\ 0\end{array}\right]=\left[\begin{array}{c}-2ϵ\sin t{\tilde{u}}_z\\ -2ϵ\cos t{\tilde{u}}_z\\ 0\end{array}\right]$$$${\mathit{u}}_{\text{basic}}\cdot \mathrm{\nabla }\tilde{\mathit{u}}=(-2ϵz\sin t\frac{\partial }{\partial x}-2ϵz\cos t\frac{\partial }{\partial y})\left[\begin{array}{c}{\tilde{u}}_x\\ {\tilde{u}}_y\\ {\tilde{u}}_z\end{array}\right]=-2ϵz[\sin t\frac{\partial }{\partial x}+\cos t\frac{\partial }{\partial y}]\tilde{u}$$

so we have

$$\begin{array}{rl}& \frac{\partial \tilde{\mathit{u}}}{\partial t}+2\hat{\mathit{z}}\times \tilde{\mathit{u}}+\tilde{\mathit{u}}\cdot \mathrm{\nabla }\tilde{\mathit{u}}+\mathrm{\nabla }\tilde{p}-E{\mathrm{\nabla }}^2\tilde{\mathit{u}}=-2\mathit{ϵ}(t)\times \tilde{u}-\tilde{u}\cdot \mathrm{\nabla }{\mathit{u}}_{\text{basic}}-{\mathit{u}}_{\text{basic}}\cdot \mathrm{\nabla }\tilde{\mathit{u}}\\ & =\left[\begin{array}{c}2ϵ\sin t{\tilde{u}}_z\\ 2ϵ\cos t{\tilde{u}}_z\\ -2ϵ\cos t{\tilde{u}}_y-2ϵ\sin t{\tilde{u}}_x\end{array}\right]+\left[\begin{array}{c}2ϵ\sin t{\tilde{u}}_z\\ 2ϵ\cos t{\tilde{u}}_z\\ 0\end{array}\right]+2ϵz[\sin t\frac{\partial }{\partial x}+\cos t\frac{\partial }{\partial y}]\tilde{u}\\ & =\left[\begin{array}{c}4ϵ\sin t{\tilde{u}}_z\\ 4ϵ\cos t{\tilde{u}}_z\\ -2ϵ\cos t{\tilde{u}}_y-2ϵ\sin t{\tilde{u}}_x\end{array}\right]+2ϵz[\sin t\frac{\partial }{\partial x}+\cos t\frac{\partial }{\partial y}]\tilde{u}\\ & =ϵ\{2z(\sin t\frac{\partial }{\partial x}+\cos t\frac{\partial }{\partial y})+\left[\begin{array}{ccc}0& 0& 4\sin t\\ 0& 0& 4\cos t\\ -2\sin t& -2\cos t& 0\end{array}\right]\}\tilde{\mathit{u}}\end{array}$$

here we get (2.12)

and the boundary conditions

$$\frac{\partial \tilde{u}}{\partial z}=\frac{\partial \tilde{v}}{\partial z}=\tilde{w}=0\text{on}z=\pm \frac{1}{2}$$

where

$$\tilde{\mathit{u}}=\left[\begin{array}{c}\tilde{u}\\ \tilde{v}\\ \tilde{w}\end{array}\right]$$

prove of (4.2)

considering the linear part and ignoring viscosity in (2.12), we have

$$\begin{array}{c}\frac{\partial \tilde{\mathit{u}}}{\partial t}+2\hat{\mathit{z}}\times \tilde{\mathit{u}}+\mathrm{\nabla }\tilde{p}\hfill \\ \hfill =ϵ\{2z(\sin t\frac{\partial }{\partial x}+\cos t\frac{\partial }{\partial y})+\left[\begin{array}{ccc}0& 0& 4\sin t\\ 0& 0& 4\cos t\\ -2\sin t& -2\cos t& 0\end{array}\right]\}\tilde{\mathit{u}}\end{array}$$

here

$$\begin{array}{rl}& 2z(\sin t\frac{\partial }{\partial x}+\cos t\frac{\partial }{\partial y})+\left[\begin{array}{ccc}0& 0& 4\sin t\\ 0& 0& 4\cos t\\ -2\sin t& -2\cos t& 0\end{array}\right]\\ & =\left[\begin{array}{ccc}2z(\sin t{\partial }_x+\cos t{\partial }_y)& 0& 4\sin t\\ 0& 2z(\sin t{\partial }_x+\cos t{\partial }_y)& 4\cos t\\ -2\sin t& -2\cos t& 2z(\sin t{\partial }_x+\cos t{\partial }_y)\end{array}\right]\end{array}$$

applying $\sin t={\displaystyle \frac{{\mathrm{e}}^{it}-{\mathrm{e}}^{-it}}{2i}}$ and $\cos t={\displaystyle \frac{{\mathrm{e}}^{it}+{\mathrm{e}}^{-it}}{2}}$ yeilds

$$\begin{array}{rl}& =\left[\begin{array}{ccc}z({\displaystyle \frac{{\mathrm{e}}^{it}-{\mathrm{e}}^{-it}}{i}}{\partial }_x+{\displaystyle \frac{{\mathrm{e}}^{it}+{\mathrm{e}}^{-it}}{1}}{\partial }_y)& 0& 2{\displaystyle \frac{{\mathrm{e}}^{it}-{\mathrm{e}}^{-it}}{i}}\\ 0& z({\displaystyle \frac{{\mathrm{e}}^{it}-{\mathrm{e}}^{-it}}{i}}{\partial }_x+{\displaystyle \frac{{\mathrm{e}}^{it}+{\mathrm{e}}^{-it}}{1}}{\partial }_y)& 2{\displaystyle \frac{{\mathrm{e}}^{it}+{\mathrm{e}}^{-it}}{1}}\\ -{\displaystyle \frac{{\mathrm{e}}^{it}-{\mathrm{e}}^{-it}}{i}}& -{\displaystyle \frac{{\mathrm{e}}^{it}+{\mathrm{e}}^{-it}}{1}}& z({\displaystyle \frac{{\mathrm{e}}^{it}-{\mathrm{e}}^{-it}}{i}}{\partial }_x+{\displaystyle \frac{{\mathrm{e}}^{it}+{\mathrm{e}}^{-it}}{1}}{\partial }_y)\end{array}\right]\\ & ={\mathrm{e}}^{it}\left[\begin{array}{ccc}z({\partial }_y-i{\partial }_x)& 0& -2i\\ 0& z({\partial }_y-i{\partial }_x)& 2\\ i& -1& z({\partial }_y-i{\partial }_x)\end{array}\right]+{\mathrm{e}}^{-it}\left[\begin{array}{ccc}z({\partial }_y+i{\partial }_x)& 0& 2i\\ 0& z({\partial }_y+i{\partial }_x)& 2\\ -i& -1& z({\partial }_y+i{\partial }_x)\end{array}\right]\end{array}$$

the theory of 4.1 part

when $ϵ=0$

the inertial wave problem (Greenspan 1968)

$$\left[\begin{array}{c}u\\ v\\ w\\ p\end{array}\right]=\left[\begin{array}{c}{k}^2(k_x\lambda -2i{k}_y)\cos (n\pi [z+\frac{1}{2}])/4k_{\mathrm{\perp }}^2\\ k^2(k_y\lambda +2i{k}_x)\cos (n\pi [z+\frac{1}{2}])/4k_{\mathrm{\perp }}^2\\ -i{k}_z\sin (n\pi [z+\frac{1}{2}])/\lambda \\ \cos (n\pi [z+\frac{1}{2}])\end{array}\right]\exp (i(k_{x}x+k_{y}y+\lambda t))$$

with $k_{\mathrm{\perp }}^2:=k_x^2+k_y^2$ , $k_z:=n\pi$ , $k^2:=k_{\mathrm{\perp }}^2+k_z^2$ and the dispersion relation is

$$\lambda =\frac{\pm 2k_z}{k}$$

where $n=0,1,2,...$ and $k_x,k_y\in \mathbb{R}$ .

for $0<ϵ\ll 1$

consider a linear combination of two inertial waves (${\lambda }_A\ne {\lambda }_B$)

$$\tilde{\mathit{u}}=A(\tau ){\mathit{u}}_A(\mathit{x})\exp (i{\lambda }_{A}t)+B(\tau ){\mathit{u}}_B(\mathit{x})\exp (i{\lambda }_{B}t)+ϵ\tilde{\mathit{v}}(\mathit{x},t)+\cdots$$

to leading order in $ϵ$

using inner product to get the amplitude equations