2D Euler Equation:

\[ \begin{aligned} \rho \left( \partial _{t}+\boldsymbol{v}\cdot \nabla \right) \boldsymbol{v} +\nabla p &=\rho \boldsymbol{g}\\ \left( \partial _{t}+\boldsymbol{v}\cdot \nabla \right) \rho &=0\\ \nabla \cdot \boldsymbol{v} &=0 \end{aligned} \] \((x,y)\in \mathbb{T}\times \mathbb{R}\), \(\boldsymbol{v}=\left(v^{x}, v^{y}\right)=\nabla ^{\perp }\psi =\left(-\partial _{y},\partial _{x}\right)\psi\), \(\boldsymbol{g}=(0,-g)\).

Stationary solution: Shear flow \(\boldsymbol{v}_{0}=(U(y),0)\), \(\rho=\rho _{0}(y)\).

Consider Couette flow with exponential stratification: \(U(y)=Ry\), \(\rho _{0}(y)=Ae^{-\beta y}\), where \(A, \beta, R \in \mathbb{R}^+\).

After Linearization, \[ \begin{aligned} \boxed{-\beta \left[ \partial _{x}-\left( \partial _{t}+y\partial _{x}\right) \partial _{y}\right] \psi} +(\partial _{t}+y\partial _{x})\omega &=B^{2}\partial _{x}T\\ (\partial _{t}+y\partial _{x})T&=\partial _{x}\psi\\ \omega &=-\Delta \psi \end{aligned} \] If \(R \neq 0\), denote \(B^{2}=\frac{\beta g}{R^{2}}\) to be the Richardson number, \(T=\frac{R\rho }{\beta \rho _{0}}\) be the relative density perturbation, \(\omega =-\Delta \psi\) be the vorticity perturbation. A scaling in time is also applied here.

If \(\beta\) is small, by applying Boussinesq Approximation, the boxed term should be omitted.

Our goal is to determine the optimal decay rate of \(\|\boldsymbol{v}\|_{L^2}\) and \(\|T\|_{L^2}\). Results for Boussinesq Case: (For the original equation, an exponential weight is added, and the same decay rate is obtained)

where \(\nu^2 = {\frac{1}{4} - B^2}\).

It can be written in a Hamiltonian structure.

\[ \begin{align} (\partial _t + y \partial _x) \omega &= B^2 T_x,\\ (\partial _t + y \partial _x) T &= \psi _x. \end{align} \] \[ \begin{align} \partial _t \omega &= \partial _x \left(-y\omega + B^2 T \right),\\ \partial _t T &= \partial _x \left( -yT + \psi \right). \end{align} \] Define \[ \begin{equation} J = \partial _x \left( \begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array} \right) , L = \left( \begin{array}{cc} (-\Delta ) ^{-1} & -y \\ -y & B ^2 \end{array} \right) \end{equation} \] Then

\[ \frac{d}{dt} \left( \begin{array}{c} \omega \\ T \end{array} \right) = JL \left( \begin{array}{c} \omega \\ T \end{array} \right) \] Two invariants are \[ \begin{aligned} H \left( \omega, T \right) &= \frac{1}{2} \left\langle L \left( \begin{array}{c} \omega \\ T \end{array} \right), \left( \begin{array}{c} \omega \\ T \end{array} \right)\right\rangle \\&= \iint _{\mathbb{R} ^2 _+} \left( \frac{1}{2}\left| \nabla \psi \right| ^2 + \frac{1}{2} B ^2 T ^2 - y \omega T \right) \mathrm{d}x \mathrm{d}y\\ P(\omega, T) &= \frac{1}{2} \left\langle \left( \begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array} \right) \left( \begin{array}{c} \omega \\ T \end{array} \right), \left( \begin{array}{c} \omega \\ T \end{array} \right)\right\rangle\\ &= \iint _{\mathbb{R} ^2 _+} \omega T - \frac{1}{2} \beta T ^2 \mathrm{d}x \mathrm{d}y \end{aligned} \]

Shear Coordinates

Define \(\left( z,y\right) =\left(x-ty,y\right)\). Denote \[ \begin{aligned} f(t;z,y) &=\omega (t;z+ty,y)=\omega (t;x,y), \\ \phi (t;z,y) &=\psi (t;z+ty,y)=\psi (t;x,y), \\ \tau (t;z,y) &=T(t;z+ty,y)=T(t;x,y). \end{aligned} \] Then \[ \begin{aligned} \partial _{t}f =\left( \partial _{t}+y\partial _{x}\right) \omega =B^{2}\partial _{x}T =B^{2}\partial _{z}\tau, \\ \partial _{t}\tau =\left( \partial _{t}+y\partial _{x}\right) T=\partial _{x}\psi =\partial _{z}\phi ,\\ \left[ \partial _{zz}+(\partial _{y}-t\partial _{z})^{2}\right] \phi=\psi _{xx}+\psi _{yy}=-\omega =-f. \end{aligned} \]

Fourier Transform \((z,y)\rightarrow(k, \eta)\)

\[ \begin{aligned} \partial _{t}f &=B^{2}\partial _{z}\tau, \\ \partial _{t}\tau &=\partial _{z}\phi ,\\ \left[ \partial _{zz}+(\partial _{y}-t\partial _{z})^{2}\right] \phi &=-f \end{aligned} \] turn into \[ \begin{aligned} \hat{f}_{t}=B^{2}(ik)\hat{\tau},\\ \hat{\tau}_{t}=(ik)\hat{\phi},\\ \left[ (ik)^{2}+(i\eta -ikt)^{2}\right] \hat{\phi}=-\hat{f}. \end{aligned} \]

For fixed \(k\neq 0\) and \(\eta\), define \(s=t-\frac{\eta }{k}\) and \(s_{0}=-\frac{\eta }{k}\). Then we obtain a second order linear ODE for \(\hat{\phi}\): \[ (1+s^{2})\hat{\phi}_{tt}+4s\hat{\phi}_{t}+(2+B^{2})\hat{\phi}=0 \] It has two linearly independent solutions \[ g_{1}(s)=F\left( \frac{3}{4}-\frac{\nu }{2},\frac{3}{4}+\frac{\nu }{2};\frac{1}{2};-s^{2}\right)\\ g_{2}(s)=sF\left( \frac{5}{4}-\frac{\nu }{2},\frac{5}{4}+\frac{\nu }{2};\frac{3}{2};-s^{2}\right) . \] Recall that \(\nu^2 = {\frac{1}{4} - B^2}\).

Therefore, the general solutions to the equation can be written as \[ \hat{\phi}=C_{1}g_{1}(s)+C_{2}g_{2}(s) \] where \(C_{1},C_{2}\) are some constants depending only on \((k,\eta )\).

Assume \(c\) is not an integer. Euler's hypergeometric differential equation \[ \begin{aligned} z(1-z)f^{\prime \prime }(z)+\left[c-(a+b+1)z\right]f^{\prime }(z)-abf(z)=0 \end{aligned} \] has two linear independent solutions \[ \begin{aligned} f_1(z)&=F(a,b;c;z), \\ f_2(z)&=z^{1-c}F(1+a-c, 1+b-c; 2-c; z). \end{aligned} \] Here \(F\) is the Gauss's hypergeometric function. It is defined as a power series.

\[ \begin{equation*} F(a,b;c;z)=\sum_{n=0}^{\infty }\frac{(a)_{n}(b)_{n}}{(c)_{n}}\frac{z^{n}}{n!} \end{equation*} \] for \(|z|<1\), where \[ (x)_{n}=\left\{ \begin{array}{lc} 1 & n=0, \\ x(x+1)\cdots (x+n-1) & n>0. \end{array} \right. \] Its value \(F(z)\) for \(|z|\geq 1\) is defined by analytic continuation. Thus if \(c,z\in \mathbb{R}\), and \(a,b\) are conjugate, then \(F(a,b;c;z)\) is also real.

\[ \hat{\phi}=C_{1}g_{1}(s)+C_{2}g_{2}(s) \] The coefficients \(C_{1},C_{2}\) are determined by the initial conditions \(\psi (0;x,y)\) and \(T(0;x,y)\). Let \(\hat{\psi}^{0}(k,\eta ),\hat{T}^{0}(k,\eta )\) be Fourier transforms of \(\psi (0;x,y)\) and \(T(0;x,y)\).

It can be obtained that \[ \hat{\phi}(0;k,\eta )=\hat{\phi}^{0}(k,\eta )=\hat{\psi}^{0}(k,\eta ).\\ \hat{\phi}_{t}(0;k,\eta )=\frac{1}{1+s_{0}^{2}}\left( \frac{iB^{2}}{k}\hat{T}^{0}-2s_{0}\hat{\psi}^{0}\right) . \]

Hence the coefficients are determined by \[ \begin{aligned} C_{1}(k,\eta )=&\frac{1}{\Delta }\left[ g_{2}^{\prime }(s_{0})+\frac{2s_{0}}{ 1+s_{0}^{2}}g_{2}(s_{0})\right] \hat{\psi}^{0}(k,\eta )\\&+\frac{1}{\Delta } \left[ -\frac{iB^{2}}{1+s_{0}^{2}}g_{2}(s_{0})\right] \frac{\hat{T} ^{0}(k,\eta )}{k}, \\ C_{2}(k,\eta )=&\frac{1}{\Delta }\left[ -g_{1}^{\prime }(s_{0})-\frac{2s_{0}}{ 1+s_{0}^{2}}g_{1}(s_{0})\right] \hat{\psi}^{0}(k,\eta )\\&+\frac{1}{\Delta } \left[ \frac{iB^{2}}{1+s_{0}^{2}}g_{1}(s_{0})\right] \frac{\hat{T} ^{0}(k,\eta )}{k}, \end{aligned} \]

where \[ \Delta(s_0) =g_{1}(s_{0})g_{2}^{\prime }(s_{0})-g_{1}^{\prime }(s_{0})g_{2}(s_{0})=\frac{1}{\left( 1+s_{0}^{2}\right) ^{2}} \] is strictly positive for all \(s_{0}\in \mathbb{R}\). Thus the solution is given explicitly by \[ \hat{\phi}(t;k,\eta )=C_{1}(k,\eta )g_{1}(s)+C_{2}(k,\eta )g_{2}(s). \] Accordingly, \[ \begin{aligned} \hat{\tau}(t;k,\eta ) =&-\frac{ik}{B^{2}}\left( (1+s^{2})\hat{\phi}_{t}+2s\hat{\phi}\right) \end{aligned} \]

Now we will use the solution formula obtained in the last section to obtain the inviscid decay estimates, for solutions of the linearized equations under Boussinesq approximation.

By expanding \(g_{1}(s)\), \(g_{2}(s),\) \(g_{1}^{\prime }(s_{0})\), \(g_{2}^{\prime }(s_{0})\) at infinity, we obtain the following asymptotics. When \(B^2 \neq \frac{1}{4}\),

\[ \begin{aligned} g_{1}(s) =& \sqrt{\pi }\left[ \frac{\Gamma (\nu )}{\Gamma (-\frac{1}{4}+\frac{ \nu }{2})\Gamma (\frac{3}{4}+\frac{\nu }{2})}s^{-\frac{3}{2}+\nu } \right.\\&\left. +\frac{ \Gamma (-\nu )}{\Gamma (-\frac{1}{4}-\frac{\nu }{2})\Gamma (\frac{3}{4}- \frac{\nu }{2})}s^{-\frac{3}{2}-\nu }\right] +O\left( |s|^{-\frac{7}{2} +Re(\nu )}\right) , \\ g_{2}(s) =& \frac{\sqrt{\pi }}{2}\left[ \frac{\Gamma (\nu )}{\Gamma (\frac{1}{4} +\frac{\nu }{2})\Gamma (\frac{5}{4}+\frac{\nu }{2})}s^{-\frac{3}{2}+\nu } \right.\\&\left.+ \frac{\Gamma (-\nu )}{\Gamma (\frac{1}{4}-\frac{\nu }{2})\Gamma (\frac{5}{4}- \frac{\nu }{2})}s^{-\frac{3}{2}-\nu }\right] +O\left( |s|^{-\frac{5}{2} +Re(\nu )}\right) , \end{aligned} \]

In short, \[ \begin{aligned} g_{1,2}(s) &\sim s^{-\frac{3}{2}+Re(\nu) }\\ g_{1,2}'(s) &\sim s^{-\frac{5}{2}+Re(\nu) }\\ \end{aligned} \] Here "\(\sim\)" means of the same order as \(s \rightarrow \infty\). Thus for \(0 < B^2 < \frac{1}{4}\), \(\nu>0\), \[ \begin{aligned} |g_{1,2}(s)| \lesssim & \left\langle s\right\rangle ^{-\frac{3}{2}+\nu },\\ \left\vert C_{1,2}(k,\eta )\right\vert \lesssim &\left\langle s_{0}\right\rangle ^{\frac{3}{2}+\nu }\left( \left\vert \hat{\psi} ^{0}(k,\eta )\right\vert +\frac{\left\vert \hat{T}^{0}(k,\eta )\right\vert }{ \left\langle s_{0}\right\rangle |k|}\right) , \end{aligned} \]

Therefore \[ \begin{aligned} \left\vert \hat{\phi}(t;k,\eta )\right\vert =&\left\vert C_{1}(k,\eta)g_{1}(s)+C_{2}(k,\eta )g_{2}(s)\right\vert \\ \lesssim &\left\langle s\right\rangle ^{-\frac{3}{2}+\nu }\left\langle s_{0}\right\rangle ^{\frac{3}{2}+\nu }\left( \left\vert \hat{\psi} ^{0}(k,\eta )\right\vert +\frac{\left\vert \hat{T}^{0}(k,\eta )\right\vert }{ \left\langle s_{0}\right\rangle |k|}\right) . \end{aligned} \] And hence \[ \begin{aligned} v^{x}(t;x,y) =&-\partial _{y}\psi (t;x,y)=(-\partial _{y}+t\partial _{z})\phi (t;z,y), \\ v^{y}(t;x,y) =&\partial _{x}\psi (t;x,y)=\partial _{z}\phi (t;z,y), \end{aligned} \] By Fourier transform, we have

\[ \begin{aligned} \left\vert \hat{v}^{x}\left( t;k,\eta \right) \right\vert &=\left\vert iks \hat{\phi}(t;k,\eta )\right\vert \\&\leq \left\langle s\right\rangle ^{-\frac{1 }{2}+\nu }\left\langle s_{0}\right\rangle ^{\frac{3}{2}+\nu }\left( \left\vert k\right\vert \left\vert \hat{\psi}^{0}(k,\eta )\right\vert +\frac{ \left\vert \hat{T}^{0}(k,\eta )\right\vert }{\left\langle s_{0}\right\rangle }\right) ,\\ \left\vert \hat{v}^{y}\left( t;k,\eta \right) \right\vert &= \left\vert ik\hat{ \phi}(t;k,\eta )\right\vert \\&\leq \left\langle s\right\rangle ^{-\frac{3}{2} +\nu }\left\langle s_{0}\right\rangle ^{\frac{3}{2}+\nu }\left( \left\vert k\right\vert \left\vert \hat{\psi}^{0}(k,\eta )\right\vert +\frac{\left\vert \hat{T}^{0}(k,\eta )\right\vert }{\left\langle s_{0}\right\rangle }\right) ,\\ \left\vert \hat{\tau}(t;k,\eta )\right\vert & \lesssim \left\langle s\right\rangle ^{-\frac{1}{2}+\nu }\left\langle s_{0}\right\rangle ^{\frac{3}{2}+\nu }\left( |k|\left\vert \hat{\psi} ^{0}(k,\eta )\right\vert +\frac{\left\vert \hat{T}^{0}(k,\eta )\right\vert }{ \left\langle s_{0}\right\rangle }\right) . \end{aligned} \]

It has the form that \[ \begin{equation} \left\vert \hat{g}(t;k,\eta )\right\vert \lesssim \left\langle s\right\rangle ^{-a}\left\langle s_{0}\right\rangle ^{b}\left\vert k\right\vert ^{c}\left\vert \hat{h}(k,\eta )\right\vert ,\text{ }0\neq k\in \mathbb{Z}, \eta \in \mathbb{R}. \end{equation} \] We show that whenever this holds, provided \(g\) and \(h\) belongs to the proper space, \[ \begin{equation*} \left\Vert P_{\neq 0}g\left( t\right) \right\Vert _{L^{2}\left( \mathbb{ T\times R}\right) }\lesssim \left\langle t\right\rangle ^{-a}\left\Vert h\right\Vert _{H_{x}^{c}H_{y}^{b+a}\left( \mathbb{ T\times R}\right)}. \end{equation*} \]

Thus we have \[ \begin{aligned} \Vert P_{\neq 0}v^{x}\Vert _{L^{2}} \lesssim &\left\langle t\right\rangle ^{-\frac{1}{2}+\nu }\left( \Vert \psi ^{0}\Vert _{H_{x}^{1}H_{y}^{2}}+\Vert T^{0}\Vert _{L_{x}^{2}H_{y}^{1}}\right) , \\ \Vert v^{y}\Vert _{L^{2}} \lesssim& \left\langle t\right\rangle ^{-\frac{3}{ 2}+\nu }\left( \Vert \psi ^{0}\Vert _{H_{x}^{1}H_{y}^{3}}+\Vert T^{0}\Vert _{L_{x}^{2}H_{y}^{2}}\right) ,\\ \Vert P_{\neq 0}T\Vert _{L^{2}}=\Vert P_{\neq 0}\tau \Vert _{L^{2}}\lesssim& \left\langle t\right\rangle ^{-\frac{ 1}{2}+\nu }\left( \Vert \psi ^{0}\Vert _{H_{x}^{1}H_{y}^{2}}+\Vert T^{0}\Vert _{L_{x}^{2}H_{y}^{1}}\right) . \end{aligned}% \]

In the case \(B^{2}>\frac{1}{4}\), \(\nu =\sqrt{\frac{1}{4}-B^{2}}\) is pure imaginary. Then we have asymptotics \[ \begin{align*} |g_{1,2}(s)| \lesssim &\left\langle s\right\rangle ^{-\frac{3}{2}},\ |g_{1,2}^{\prime }(s_{0})|\lesssim \left\langle s_{0}\right\rangle ^{- \frac{5}{2}}. \end{align*} \] Therefore, \[ \begin{align*} \Vert P_{\neq 0}v^{x}\Vert _{L^{2}} \lesssim &\left\langle t\right\rangle ^{-\frac{1}{2}}\left( \Vert \psi ^{0}\Vert _{H_{x}^{1}H_{y}^{2}}+\Vert T^{0}\Vert _{L_{x}^{2}H_{y}^{1}}\right) , \\ \Vert v^{y}\Vert _{L^{2}} \lesssim &\left\langle t\right\rangle ^{-\frac{3}{ 2}}\left( \Vert \psi ^{0}\Vert _{H_{x}^{1}H_{y}^{3}}+\Vert T^{0}\Vert _{L_{x}^{2}H_{y}^{2}}\right) , \\ \Vert P_{\neq 0}T\Vert _{L^{2}} \lesssim &\left\langle t\right\rangle ^{- \frac{1}{2}}\left( \Vert \psi ^{0}\Vert _{H_{x}^{1}H_{y}^{2}}+\Vert T^{0}\Vert _{L_{x}^{2}H_{y}^{1}}\right) . \end{align*} \]

This critical case stands out because of its different expansion at infinity. When \(B^{2}=\frac{1}{4}\), \(\nu =0\), \[ \begin{align*} g_{1}(s) &= F\left( \frac{3}{4},\frac{3}{4};\frac{1}{2};-s^{2}\right) \\&= \frac{ 2\sqrt{\pi }}{\Gamma \left( -\frac{1}{4}\right) \Gamma \left( \frac{3}{4} \right) }s^{-\frac{3}{2}}\log \left( s\right) +O\left( |s|^{- \frac{3}{2}}\right) , \\ g_{2}(s) &= sF\left( \frac{5}{4},\frac{5}{4};\frac{3}{2};-s^{2}\right) \\&= \frac{\sqrt{\pi }}{\Gamma \left( \frac{1}{4}\right) \Gamma \left( \frac{5}{4} \right) }s^{-\frac{3}{2}}\log \left( s\right) +O\left( |s|^{- \frac{3}{2}}\right) , \end{align*} \]

Thus we obtain the following estimates \[ \begin{align*} \left\vert g_{1,2}(s)\right\vert \lesssim \left\langle s\right\rangle ^{- \frac{3}{2}}\left\langle \log \left\langle s\right\rangle \right\rangle ,\ \left\vert g_{1,2}^{\prime }(s_{0})\right\vert \lesssim &\left\langle s_{0}\right\rangle ^{-\frac{5}{2}}\left\langle \log \left\langle s_{0}\right\rangle \right\rangle \end{align*} \] and as a result \[ \begin{align*} \left\vert C_{1,2}(k,\eta )\right\vert \lesssim &\left\langle s_{0}\right\rangle ^{\frac{3}{2}}\left\langle \log \left\langle s_{0}\right\rangle \right\rangle \left( \left\vert \hat{\psi}^{0}(k,\eta )\right\vert +\frac{\left\vert \hat{T}^{0}(k,\eta )\right\vert }{ \left\langle s_{0}\right\rangle |k|}\right) \end{align*} \] Therefore, we have \[ \begin{align*} \left\vert \hat{\phi}(t;k,\eta )\right\vert \lesssim &\left\langle s\right\rangle ^{-\frac{3}{2}}\left\langle s_{0}\right\rangle ^{\frac{3}{2}}\left\langle \log \left\langle s\right\rangle \right\rangle \left\langle \log \left\langle s_{0}\right\rangle \right\rangle \left( \left\vert \hat{\psi}^{0}(k,\eta )\right\vert +\frac{\left\vert \hat{T}^{0}(k,\eta )\right\vert }{ \left\langle s_{0}\right\rangle |k|}\right) , \end{align*} \]

It can be shown by some kindergarden calculus (which is omitted here) that \[ \begin{equation*} \left\langle s\right\rangle ^{-a}\left\langle s_{0}\right\rangle ^{\frac{3}{2% }}\left\langle \log \left\langle s\right\rangle \right\rangle \left\langle \log \left\langle s_{0}\right\rangle \right\rangle \lesssim \left<t\right>^{-a}\left\langle \log \left\langle t\right\rangle \right\rangle \left\langle s_{0}\right\rangle ^{\frac{3}{2}+a }, \end{equation*}% \] for all \(a \geq \frac{1}{2}\). Thus we get \[ \begin{align*} \Vert P_{\neq 0}v^{x}\Vert _{L^{2}} \lesssim &\left\langle t\right\rangle ^{-\frac{1}{2}} \left\langle \log \left\langle t\right\rangle \right\rangle \left( \Vert \psi ^{0}\Vert _{H_{x}^{1}H_{y}^{2}}+\Vert T^{0}\Vert _{L_{x}^{2}H_{y}^{1}}\right) , \\ \Vert v^{y}\Vert _{L^{2}} \lesssim &\left\langle t\right\rangle ^{-\frac{3}{ 2}} \left\langle \log \left\langle t\right\rangle \right\rangle \left( \Vert \psi ^{0}\Vert _{H_{x}^{1}H_{y}^{3}}+\Vert T^{0}\Vert _{L_{x}^{2}H_{y}^{2}}\right) ,\\ \Vert P_{\neq 0}T\Vert _{L^{2}} \lesssim& \left\langle t\right\rangle ^{-\frac{ 1}{2}}\left\langle \log \left\langle t\right\rangle \right\rangle \left( \Vert \psi ^{0}\Vert _{H_{x}^{1}H_{y}^{2}}+\Vert T^{0}\Vert _{L_{x}^{2}H_{y}^{1}}\right) . \end{align*} \]

Recall \(B^2 = \frac{\beta g}{R^2}\), \(\nu = \sqrt{\frac{1}{4}-B^2}\).

\(B^2=0\) means \(\beta =0\), i.e. no stratification. We get \[ \begin{align*} \left( \partial _{t}+Ry\partial _{x}\right) \Delta \psi =&-\partial _{x}\left( \frac{\rho }{\rho _{0}}\right) g, \\ \left( \partial _{t}+Ry\partial _{x}\right) \left( \frac{\rho }{\rho _{0}} \right) =&0. \end{align*}% \] Hence the density is conserved. Similarly, by using a sheared coordinate, \[ \begin{align*} f(t;z,y) =&\omega (t;z+ty,y)=\omega (t;x,y), \\ \phi (t;z,y) =&\psi (t;z+ty,y)=\psi (t;x,y), \\ \tau (t;z,y) =&\frac{\rho }{\rho _{0}}(t;z+ty,y)=\frac{\rho }{\rho _{0}} (t;x,y). \end{align*} \]

Then \[ \begin{equation*} \partial _{t}f(t;z,y)=g\partial _{z}\tau (t;z,y),\ \partial _{t}\tau (t;z,y)=0. \end{equation*} \] So \[ \begin{align*} \hat{\tau}(t; k, \eta) =&\hat{\tau}(0; k, \eta), \\ \hat{f}\left( t; k, \eta\right) =&\hat{\omega}^{0}(k, \eta)+tikg\hat{\rho ^{0}}\left( k, \eta\right) ,\\ \hat{\phi}(t;k,\eta ) =&\frac{1}{k^{2}\left<s\right>^2 } \hat{f}\left( t;\eta ,k\right) \end{align*} \] where \(\omega (0;x,y)=\omega ^{0}(x,y),\ \frac{\rho }{\rho _{0}}(0;x,y)=\rho ^{0}\left( x,y\right)\).

Thus we have estimates, \[ \begin{align*} \left\vert \hat{\phi}(t;k,\eta )\right\vert \lesssim \left\langle s\right\rangle ^{-2}\left\langle s_{0}\right\rangle ^{2}\left\vert \hat{\psi}^{0}(k,\eta )\right\vert +|t|\frac{1}{\left\vert k\right\vert }\left\langle s\right\rangle ^{-2}\left\vert \hat{\rho} ^{0}(k,\eta )\right\vert.\\ \left\vert \hat{v}^{x}\left( t;k,\eta \right) \right\vert \lesssim \left\langle s\right\rangle ^{-1}\left\langle s_{0}\right\rangle ^{2}\left\vert k\right\vert \left\vert \hat{\psi}^{0}(k,\eta )\right\vert +|t|\left\langle s\right\rangle ^{-1}\left\vert \hat{\rho}^{0}(k,\eta )\right\vert ,\\ \left\vert \hat{v}^{y}\left( t;k,\eta \right) \right\vert \lesssim \left\langle s\right\rangle ^{-2}\left\langle s_{0}\right\rangle ^{2}\left\vert k\right\vert \left\vert \hat{\psi}^{0}(k,\eta )\right\vert +|t|\left\langle s\right\rangle ^{-2}\left\vert \hat{\rho}^{0}(k,\eta )\right\vert . \end{align*} \] By Lemma , we get \[ \begin{align*} \Vert P_{\neq 0}v^{x}\Vert _{L^{2}}\lesssim \Vert \rho ^{0}\Vert _{L_{x}^{2}H_{y}^{1}}+\left\langle t\right\rangle ^{-1}\Vert \psi ^{0}\Vert _{H_{x}^{1}H_{y}^{3}},\\ \Vert v^{y}\Vert _{L^{2}}\lesssim \left\langle t\right\rangle ^{-1}\Vert \rho ^{0}\Vert _{L_{x}^{2}H_{y}^{2}}+\left\langle t\right\rangle ^{-2}\Vert \psi ^{0}\Vert _{H_{x}^{1}H_{y}^{4}}. \end{align*} \] For \(\rho ^{0}=0\), i.e. constant density, this agrees with Lin and Zeng (2011) for the homogeneous fluids.

When there is no shear, i.e. \(R=0\), \(B^{2}=\infty\), the governing equations become \[ \begin{equation*} \partial _{t}\Delta \psi =-\partial _{x}\left( \frac{\rho }{\rho _{0}} \right) g,\ \ \ \ \ \ \partial _{t}\left( \frac{\rho }{\rho _{0}}\right) =\beta \partial _{x}\psi . \end{equation*} \] Denote \(T=\frac{\rho }{\beta \rho _{0}(y)},\) then \[ \begin{equation} \Delta \psi _{t}=-\partial _{x}T\beta g, \ \ \ \ \ \ \partial _{t}T=\partial _{x}\psi . \end{equation} \] Multiplying by \(\psi\) and then integrating by parts, we get the following invariant \[ \begin{equation*} \frac{\mathrm{d}}{\mathrm{d}t}\left( \beta g\iint T^{2}\mathrm{d}x \mathrm{d}y+\iint \left\vert \nabla \psi \right\vert ^{2}\mathrm{d}x \mathrm{d}y\right) =0. \end{equation*} \]

This shows that in the \(L^{2}\) norm, the perturbations of velocity and density are Liapunov stable but do not decay. However, the \(L^{\infty }\) norms decay due to the dispersive effects. Denote \(N^{2}=\beta g\) to be the squared Buoyancy frequency. By Fourier transform \((x,y)\rightarrow (k,\eta )\), \[ \begin{align} \left( (i\eta )^{2}+(ik)^{2}\right) \hat{\psi} _{t}&=-(ik)N^{2}\hat{T},\\ \hat{T}_{t}&=(ik)\hat{\psi }. \end{align}% \] Combining them, we get \[ \begin{equation*} \frac{d^{2}}{dt^{2}}\hat{\psi}=-\lambda ^{2}\hat{\psi}, \end{equation*} \] where \({\lambda ^{2}(k,\eta )=\frac{k^{2}N^{2}}{k^{2}+\eta ^{2}}}\).

This is a dispersive equation. By estimation on \(L^\infty\) norm, it can shown that \[ \begin{align*} \Vert P_{\neq 0}v^{x}\Vert _{L_{x}^{2}L_{y}^{\infty }} \lesssim &|t|^{- \frac{1}{3}}\left( \Vert \psi ^{0}\Vert _{H_{x}^{3/2}H_{y}^{9/2}}+\Vert \psi ^{0}\Vert _{H_{x}^{3/2}W_{y}^{1,1}} \right.\\&\left. +\Vert T^{0}\Vert _{H_{x}^{1/2}H_{y}^{9/2}}+\Vert T^{0}\Vert _{H_{x}^{1/2}W_{y}^{1,1}}\right) ,\\ \Vert v^{y}\Vert _{L_{x}^{2}L_{y}^{\infty }} \lesssim &|t|^{-\frac{1}{3} }\left( \Vert \psi ^{0}\Vert _{H_{x}^{5/2}H_{y}^{7/2}}+\Vert \psi ^{0}\Vert _{H_{x}^{5/2}L_{y}^{1}} \right.\\&\left. +\Vert T^{0}\Vert _{H_{x}^{3/2}H_{y}^{7/2}}+\Vert T^{0}\Vert _{H_{x}^{3/2}L_{y}^{1}}\right) .\\ \Vert P_{\neq 0}T\Vert _{L_{x}^{2}L_{y}^{\infty }} \lesssim& |t|^{-\frac{1}{3} }\left( \Vert \psi ^{0}\Vert _{H_{x}^{5/2}H_{y}^{9/2}}+\Vert \psi ^{0}\Vert _{H_{x}^{5/2}W_{y}^{1,1}} \right.\\&\left. +\Vert T^{0}\Vert _{H_{x}^{3/2}H_{y}^{7/2}}+\Vert T^{0}\Vert _{H_{x}^{3/2}L_{y}^{1}}\right) . \end{align*} \]

Now I am working on the equation in the \(y>0\) half space, with Dirichlet boundary condition, although I have not obtained any good results yet.

There are some results in the literature. For \(B^2 > \frac{1}{4}\) there are infinitely many pure imaginary eigen values. Hence no decay is possible in this case. It is still unclear about \(B^2 < \frac{1}{4}\), though it is shown that one or zero imaginary eigenvalue exists in this case.

Pointwise boundedness is proved by Dikki. He used Laplace transform to deal with the problem.

We are still looking for \(L^2\) stability or asymptotic stability.