Stability of the 2D Boussinesq equations with a velocity damping term in the strip domain

Abstract

We prove the global well-posedness for the 2D Boussinesq equations with a velocity damping term around the equilibrium state $(0,x_2)$ in the strip domain $\mathbb R\times (0, 1)$ with Navier-type slip boundary condition. It is worth mentioning that the results of low regularity are obtained using only the energy estimate and the structure of the equations.

Appendix

This section is devoted to give a sketch proof of global well-posedness for system (1.3) in $\mathbb R^2$. Indeed, we have the following Proposition similar as Section 4.

Proposition 5.1

Assume that the solution $(u,\theta )$ of system (1.3) satisfies

$$\begin{aligned} \displaystyle \sup \limits _{0\le t\le T}\big (\Vert u(t)\Vert _{H^3(\mathbb R^2)}^2+\Vert \theta (t)\Vert _{H^3(\mathbb R^2)}^2\big )\le c_0^2. \end{aligned}$$

If $c_0$ is suitable small, then for some $C_0>0$, it holds that

$$\begin{aligned} \displaystyle {\mathcal E}(t)+ \int \limits _0^t{\mathcal F}(s)ds\le C_0\Big (\Vert u_0\Vert ^2_{H^3(\mathbb R^2)}+\Vert \theta _0\Vert ^2_{H^3(\mathbb R^2)}\Big ) \end{aligned}$$

for any $t\in [0,T]$.

Proof

The proof of Proposition 5.1 is based on several steps of careful energy estimates, which is similar to the proof of Proposition 4.1.

Step 1. $L^2$ estimate of $u, \theta$.

Due to $\langle u,\nabla p \rangle = -\langle \mathop {\textrm{div}}\nolimits {u}, p \rangle = 0$ in $\mathbb R^2$, we take the $L^2$ inner product of equations (1.3)$_1$ and (1.3)$_2$ with u and $\theta$, respectively, and integrate by parts to obtain

$$\begin{aligned} \frac{1}{2} \frac{d}{dt} (\Vert u\Vert _{L^2}^2+\Vert \theta \Vert _{L^2}^2)+\Vert u\Vert _{L^2}^2=0 \end{aligned}$$

(5.1)

for any $t\in [0, T]$.

Step 2. $\dot{H}^1$ estimate of $u, \theta$.

We apply $\nabla$ to equations (1.3)$_1$ and (1.3)$_2$ and then take the $L^2$ inner product of the resulting equations with $\nabla u$ and $\nabla \theta$, respectively, to obtain

$$\begin{aligned}&\displaystyle \frac{1}{2}\frac{d}{dt}(\Vert \nabla u\Vert ^2_{L^2} + \Vert \nabla \theta \Vert _{L^2}^2)+\Vert \nabla u\Vert ^2_{L^2} = -\langle \nabla u, \nabla (u\cdot \nabla u) \rangle - \langle \nabla \theta , \nabla (u\cdot \nabla \theta ) \rangle \nonumber \\&\quad =-\langle \nabla u, \nabla u\cdot \nabla u \rangle -\langle \nabla \theta , \nabla u_1\partial _1\theta \rangle - \langle \partial _1\theta \nonumber \\&\quad \lesssim (\Vert u\Vert _{H^3}+\Vert \theta \Vert _{H^3})(\Vert u\Vert _{H^3}^2+\Vert \partial _1\theta \Vert _{H^2}^2) \le Cc_0{\mathcal F}(t), \end{aligned}$$

(5.2)

which gives the $\dot{H}^1$ estimate of $(u,\theta )$.

Step 3. Dissipation estimate of $\partial _1 \theta$.

Thanks to expression (1.5), we apply $\nabla$ to equations (1.3)$_1$ and (1.3)$_2$ and take the $L^2$ inner product of the resulting equations with $-\nabla (\theta e_2)$ and $-\nabla u_2$, respectively, to obtain

$$\begin{aligned}&\displaystyle -\frac{d}{dt} \langle \nabla u_2, \nabla \theta \rangle + \Vert \partial _1\theta \Vert _{L^2}^2- \langle \nabla (\theta e_2), \nabla u \rangle - \Vert \nabla u_2\Vert _{L^2}^2 \nonumber \\&\quad = \Big (\langle \nabla (\theta e_2), \nabla (u\cdot \nabla u) \rangle + \langle \nabla u_2, \nabla (u\cdot \nabla \theta ) \rangle \Big ) + \langle \nabla (\theta e_2), \nabla \nabla (-\Delta )^{-1}\mathop {\textrm{div}}\nolimits (u\cdot \nabla u) \rangle \end{aligned}$$

(5.3)

$$\begin{aligned}&\quad = \langle \nabla \theta , \nabla u\cdot \nabla u_2 \rangle + \langle \nabla u_2, \nabla u\cdot \nabla \theta \rangle - \langle \partial _2\theta , \partial _1 u\cdot \nabla u_1 + \partial _2 u\cdot \nabla u_2 \rangle + \int \limits _{\Omega }u\cdot \nabla (\nabla \theta \nabla u_2)\textrm{d}x\nonumber \\&\quad \lesssim \Vert \nabla \theta \Vert _{L^2}\Vert \nabla u\Vert _{L^4}^2 \lesssim \Vert \theta \Vert _{H^3}\Vert u\Vert _{H^3}^2\lesssim Cc_0{\mathcal F}(t), \end{aligned}$$

(5.4)

here we use

$$\begin{aligned} -\langle \nabla (\theta e_2), \nabla u \rangle = -\langle \partial _1\theta ,\partial _1 u_2 \rangle + \langle \partial _1\theta , \partial _2 u_1 \rangle , \end{aligned}$$

and

$$\begin{aligned} \displaystyle -\langle \nabla (\theta e_2), \nabla (-\theta e_2 + \nabla p_2) \rangle = \Vert \nabla \theta \Vert _{L^2}^2 + \langle \nabla \theta , \nabla \partial _2^2(-\Delta )^{-1}\theta \rangle = \Vert \nabla \theta \Vert _{L^2}^2 - \Vert \partial _2\theta \Vert _{L^2}^2 = \Vert \partial _1\theta \Vert _{L^2}^2. \end{aligned}$$

Then, we obtain the dissipation estimate of $\partial _1\theta$.

Step 4. $\dot{H}^2$ estimate of $u,\theta$.

We apply $\Delta$ to equations (1.3)$_1$ and (1.3)$_2$ and then take the $L^2$ inner product with $\Delta u$ and $\Delta \theta$, respectively, to obtain

$$\begin{aligned} \displaystyle \frac{1}{2}\frac{d}{dt}(\Vert \Delta u\Vert _{L^2}^2 + \Vert \Delta \theta \Vert _{L^2}^2) + \Vert \Delta u\Vert _{L^2}^2&= - \langle \Delta u, \Delta (u\cdot \nabla u) \rangle - \langle \Delta \theta , \Delta (u\cdot \nabla \theta ) \rangle \nonumber \\&\lesssim (\Vert u\Vert _{H^3}+\Vert \theta \Vert _{H^3})(\Vert u\Vert _{H^3}^2+\Vert \partial _1\theta \Vert _{H^2}^2) \le Cc_0{\mathcal F}(t), \end{aligned}$$

(5.5)

which gives the $\dot{H}^2$ estimate of $(u,\theta )$.

Step 5. Dissipation estimate of $\nabla \partial _1 \theta$.

Thanks to expression (1.5), we apply $\Delta$ to equation (1.3)$_1$ and (1.3)$_2$, and take the $L^2$ inner product of the resulting equations with $-\Delta (\theta e_2)$ and $-\Delta u_2$, respectively, to obtain

$$\begin{aligned}&\displaystyle -\frac{d}{dt}\langle \Delta u_2, \Delta \theta \rangle + \Vert \nabla \partial _1\theta \Vert _{L^2}^2 - \langle \Delta \theta , \Delta u_2 \rangle - \Vert \Delta u_2\Vert _{L^2}^2 \nonumber \\&\quad = \Big (\langle \Delta (\theta e_2), \Delta (u\cdot \nabla u) \rangle + \langle \Delta u_2,\Delta (u\cdot \nabla \theta ) \rangle \Big ) - \langle \Delta \theta , \Delta \partial _2 (-\Delta )^{-1} \mathop {\textrm{div}}\nolimits (u\cdot \nabla u) \rangle \nonumber \\&\quad = \langle \Delta \theta , \Delta u\cdot \nabla u_2 + 2\partial _1 u\cdot \nabla \partial _1 u_2 + 2\partial _2 u\cdot \nabla \partial _2 u_2 \rangle \nonumber \\&\qquad + \langle \Delta u_2, \Delta u\cdot \nabla \theta + 2\partial _1 u\cdot \nabla \partial _1\theta + 2\partial _2 u\cdot \nabla \partial _2\theta \rangle + \int \limits _{\Omega }u\cdot \nabla (\Delta \theta \Delta u_2)\textrm{d}x\nonumber \\&\qquad - \langle \Delta \theta , \partial _1\partial _2u\cdot \nabla u_1 + \partial _1u\cdot \nabla \partial _2u_1 + \partial _2^2u\cdot \nabla u_2 + \partial _2u\cdot \nabla \partial _2u_2 \rangle \nonumber \\&\quad \lesssim \Vert \nabla ^2\theta \Vert _{L^2}\Vert \nabla ^2 u\Vert _{L^4}\Vert \nabla u\Vert _{L^4} + \Vert \nabla \theta \Vert _{L^4}\Vert \nabla ^2 u\Vert _{L^4}\Vert \nabla ^2 u\Vert _{L^2} \lesssim \Vert \theta \Vert _{H^3}\Vert u\Vert _{H^3}^2\le Cc_0{\mathcal F}(t). \end{aligned}$$

(5.6)

Here we use

$$\begin{aligned} \displaystyle - \langle \Delta \theta , \Delta u_2 \rangle = -\langle \partial _1^2\theta , \Delta u_2 \rangle - \langle \partial _1\partial _2\theta , \partial _1\partial _2u_2 \rangle + \langle \partial _1\partial _2\theta , \partial _2^2u_1 \rangle , \end{aligned}$$

and

$$\begin{aligned} \displaystyle \langle -\Delta (\theta e_2), \Delta (-\theta e_2 + \nabla p_2) \rangle = \Vert \Delta \theta \Vert _{L^2}^2 + \langle \Delta \theta , \Delta \partial _2^2(-\Delta )^{-1}\theta \rangle = \langle \Delta \theta , \partial _1^2\theta \rangle = \Vert \nabla \partial _1\theta \Vert _{L^2}^2. \end{aligned}$$

Thus we obtain the dissipation estimate of $\nabla \partial _1\theta$.

Step 6. $\dot{H}^3$ estimate of $u,\theta$.

We apply $\nabla \Delta$ to equations (1.3)$_1$ and (1.3)$_2$ and then take the $L^2$ inner product of the resulting equations with $\nabla \Delta u$ and $\nabla \Delta \theta$ to obtain

$$\begin{aligned}&\displaystyle \frac{1}{2}\frac{d}{dt}(\Vert \nabla \Delta u\Vert _{L^2}^2 + \Vert \nabla \Delta \theta \Vert _{L^2}^2) + \Vert \nabla \Delta u\Vert _{L^2}^2 \nonumber \\&\quad = - \langle \nabla \Delta u, \nabla \Delta (u\cdot \nabla u) \rangle - \langle \nabla \Delta \theta , \nabla \Delta (u\cdot \nabla \theta ) \rangle := VI_{1} + VI_{2}\nonumber \\&\quad = - \langle \nabla \Delta u, \nabla \Delta u\cdot \nabla u \rangle - \langle \nabla \Delta u, \Delta u\cdot \nabla \nabla u \rangle - \langle \nabla \Delta u, \nabla u\cdot \nabla \Delta u \rangle \nonumber \\&\qquad - \langle \nabla \Delta u, 2\nabla \partial _1 u\cdot \nabla \partial _1 u + 2\nabla \partial _2 u\cdot \nabla \partial _2 u + 2\partial _1u\cdot \nabla \nabla \partial _1 u + 2\partial _2u\cdot \nabla \nabla \partial _2 u \rangle \nonumber \\&\qquad -\langle \nabla \Delta \theta , \nabla \Delta u\cdot \nabla \theta \rangle - \langle \nabla \Delta \theta , \Delta u\cdot \nabla \nabla \theta \rangle - \langle \nabla \Delta \theta , 2\partial _1\nabla u\cdot \nabla \partial _1\theta + 2\partial _2\nabla u\cdot \nabla \partial _2\theta \rangle \nonumber \\&\qquad -\langle \nabla \Delta \theta , 2\partial _1u\cdot \nabla \nabla \partial _1\theta + 2\partial _2u\cdot \nabla \nabla \partial _2\theta \rangle - \langle \nabla \Delta \theta , \nabla u\cdot \nabla \Delta \theta \rangle \nonumber \\&\quad \lesssim (\Vert u\Vert _{H^3} + \Vert \theta \Vert _{H^3}+ \Vert \theta \Vert _{H^3}^2+ \Vert \theta \Vert _{H^3}^3)(\Vert u\Vert _{H^3}^2 + \Vert \partial _1\theta \Vert _{H^2}^2) \le Cc_0{\mathcal F}(t), \end{aligned}$$

(5.7)

which gives the $\dot{H}^3$ estimate of $(u,\theta )$.

Step 7. Dissipation estimate of $\nabla ^2\partial _1 \theta$.

Thanks to expression (1.5), we apply $\nabla \Delta$ to equations (1.3)$_1$ and (1.3)$_2$ and take the $L^2$ inner product of the resulting equations with $-\nabla \Delta (\theta e_2)$ and $-\nabla \Delta u_2$, respectively, to obtain

$$\begin{aligned}&\displaystyle -\frac{d}{dt}\langle \nabla \Delta u_2, \nabla \Delta \theta \rangle + \Vert \Delta \partial _1\theta \Vert _{L^2}^2 - \langle \nabla \Delta \theta , \nabla \Delta u_2 \rangle - \Vert \nabla \Delta u_2\Vert _{L^2}^2 \nonumber \\&\quad = \langle \nabla \Delta (\theta e_2), \nabla \Delta (u\cdot \nabla u) \rangle + \langle \nabla \Delta u_2, \nabla \Delta (u\cdot \nabla \theta ) \rangle + \langle \nabla \Delta (\theta e_2), \nabla \Delta \nabla (-\Delta )^{-1}\mathop {\textrm{div}}\nolimits (u\cdot \nabla u) \rangle \nonumber \\&\quad = \langle \nabla \Delta \theta , u\cdot \nabla \nabla \Delta u_2 \rangle + \langle \nabla \Delta \theta , \nabla \Delta u\cdot \nabla u_2 + \nabla u\cdot \nabla \Delta u_2 + \Delta u\cdot \nabla \nabla u_2 \rangle \nonumber \\&\qquad + 2\langle \nabla \Delta \theta , \nabla \partial _1u\cdot \nabla \partial _1u_2 + \nabla \partial _2u\cdot \nabla \partial _2u_2 + \partial _1u\cdot \nabla \nabla \partial _1u_2 + \partial _2u\cdot \nabla \nabla \partial _2u_2 \rangle \nonumber \\&\qquad -\langle \nabla \Delta \theta , \nabla \partial _1\partial _2 u\cdot \nabla u_1 + \nabla \partial _1u\cdot \nabla \partial _2u_1 + \nabla \partial _2^2u\cdot \nabla u_2 + \nabla \partial _2u\cdot \nabla \partial _2u_2 \rangle \nonumber \\&\qquad - \langle \nabla \Delta \theta , \partial _1\partial _2 u\cdot \nabla \nabla u_1 + \partial _1u\cdot \nabla \nabla \partial _2u_1 + \partial _2^2u\cdot \nabla \nabla u_2 + \partial _2u\cdot \nabla \nabla \partial _2u_2 \rangle \nonumber \\&\quad \lesssim (\Vert u\Vert _{H^3} + \Vert \theta \Vert _{H^3}+ \Vert \theta \Vert _{H^3}^2+ \Vert \theta \Vert _{H^3}^3)(\Vert u\Vert _{H^3}^2 + \Vert \partial _1\theta \Vert _{H^2}^2) \le Cc_0{\mathcal F}(t), \end{aligned}$$

(5.8)

Here we use

$$\begin{aligned} \displaystyle -\langle \nabla \Delta \theta , \nabla \Delta u_2 \rangle = -\langle \partial _1\Delta \theta , \partial _1\Delta u_2 \rangle - \langle \partial _2\partial _1^2\theta , \partial _2\Delta u_2 \rangle + \langle \partial _1\partial _2^2\theta , \partial _1^2\partial _2u_1 \rangle + \langle \partial _1\partial _2^2\theta , \partial _2^3u_1 \rangle , \end{aligned}$$

and

$$\begin{aligned} \displaystyle \langle -\nabla \Delta (\theta e_2), \nabla \Delta (-\theta e_2 - \nabla p_2) \rangle = \Vert \nabla \Delta \theta \Vert _{L^2}^2 - \langle \nabla \Delta \theta , \nabla \partial _2^2\theta \rangle = \langle \nabla \Delta \theta , \nabla \partial _1^2\theta \rangle = \Vert \Delta \partial _1\theta \Vert _{L^2}^2. \end{aligned}$$

Thus, we obtain the dissipation estimate of $\nabla ^2\partial _1\theta$.

Step 8. Closing of the a priori estimates.

With the estimates in step 1–7, we can take a similar procedure as that of Proposition 4.1 to conclude that

$$\begin{aligned} \displaystyle {\mathcal E}(t) + \int \limits _{0}^{t} {\mathcal F}(s)\textrm{d}s \le C_0(\Vert u_0\Vert _{H^3}^2 + \Vert \theta _0\Vert _{H^3}^2) \end{aligned}$$

for some positive constant $C_0$ provided that $c_0$ is suitable small. This completes the proof of Proposition 5.1. $\square$