Stability and Large Time Behavior of the 2D Boussinesq Equations with Mixed Partial Dissipation Near Hydrostatic Equilibrium

Abstract

This paper establishes the large time behavior of the solution to two dimensional Boussinesq equations with mixed partial dissipation. Our main result is achieved in terms of the global $H^{2}$-stability. Finally, we also obtain the decay estimates of linearized Boussinesq equations.

1 Introduction

This note is concerned with the following two dimensional Boussinesq equations with mixed partial dissipation:

$$ \left \{ \begin{aligned} &\partial _{t} U_{1}+U\cdot \nabla U_{1}-\mu \partial _{22}U_{1}+ \partial _{1}\overline{\pi}=0, &(x,t)\in \mathbf {R}^{2}\times (0,\infty ), \\ &\partial _{t} U_{2}+U\cdot \nabla U_{2}-\kappa \partial _{11}U_{2}+ \partial _{2}\overline{\pi}=\Theta , &(x,t)\in \mathbf {R}^{2}\times (0, \infty ), \\ &\partial _{t}\Theta +U\cdot \nabla \Theta -\eta \partial _{11}\Theta =0, &(x,t)\in \mathbf {R}^{2}\times (0,\infty ), \\ &\nabla \cdot U=0, & (x,t)\in \mathbf {R}^{2}\times (0,\infty ), \\ &U|_{t=0}= U_{0},\Theta |_{t=0}=\Theta _{0}, &x\in \mathbf {R}^{2}, \end{aligned} \right . $$

(1.1)

where the unknown $U=(U_{1},U_{2})$ denotes the velocity field, $\overline{\pi}$ is the pressure, $\Theta$ is the temperature, $\mu$ and $\kappa$ are the velocity viscosity, $\eta$ is the thermal diffusivity. For the term $\Theta$ in the second equation of (1.1) represents the buoyancy forcing generated due to the temperature variation.

Let

$$\begin{aligned} u=U-U^{0},\theta =\Theta -\Theta ^{0},\pi =\overline{\pi}- \overline{\pi}^{0}. \end{aligned}$$

Then $(u,\theta , \pi )$ obeys

$$ \left \{ \begin{aligned} &\partial _{t} u_{1}+u\cdot \nabla u_{1}-\mu \partial _{22}u_{1}+ \partial _{1}\pi =0, &(x,t)\in \mathbf {R}^{2}\times (0,\infty ), \\ &\partial _{t} u_{2}+u\cdot \nabla u_{2}-\kappa \partial _{11}u_{2}+ \partial _{2}\pi =\theta , &(x,t)\in \mathbf {R}^{2}\times (0,\infty ), \\ &\partial _{t}\theta +u\cdot \nabla \theta -\eta \partial _{11}\theta =-u_{2}, &(x,t)\in \mathbf {R}^{2}\times (0,\infty ), \\ &\nabla \cdot u=0, & (x,t)\in \mathbf {R}^{2}\times (0,\infty ), \\ &u|_{t=0}= u_{0},\theta |_{t=0}=\theta _{0}, &x\in \mathbf {R}^{2}, \end{aligned} \right . $$

(1.2)

where

$$\begin{aligned} U^{0}=0,\quad \Theta ^{0}=x_{2}, \quad \overline{\pi}^{0}=\frac{1}{2}x^{2}_{2} \end{aligned}$$

(1.3)

is the steady solution of (1.1). Many geophysical flows such as atmospheric fronts and ocean circulations can be modeled by the Boussinesq equations. Recently, the stability and large time behavior issues on the Boussinesq equations have gained more and more interests and become the center of mathematic investigation. In the last thirty years, a considerable amount of literature has been published on the stability problem concerning the Boussinesq equations. Some of them focus on the stability of 2D Boussinesq equations with various partial dissipation (see e. g. [1], [2], [4], [5], [8], [9], [10], [11]). In 2019, Ji, Li, Wei and Wu [6] obtained the stability of the 2D Boussinesq eqution (1.2) under the assumption that $H^{1}$-norm of initial data is small. However, they didn’t give the large time behavior of the system (1.2). Very recently, Lai, Wu and Zhong [7] have established the global existence and stability of 2D Boussinesq equations with partial dissipation and temperature damping in the Sobolev space $H^{2}(\mathbf {R}^{2})$. In addition, the large-time behavior of $\|\nabla u\|_{L^{2}}$ and $\|\nabla \theta \|_{L^{2}}$ is also obtained via energy methods. Motivated by [1], [9] and [7], the purpose of this paper is to address large time behavior of the solution to the system (1.2) and decay estimates of linearized equation of system (1.2). Our results are stated as follows.

Theorem 1.1

Let $(u_{0},\theta _{0})\in H^{2}(\mathbf {R}^{2})$ and $\nabla \cdot u_{0}=0$. If

$$\begin{aligned} &\|u_{0}\|_{H^{2}}+\|\theta _{0}\|_{H^{2}}\leq \varepsilon , \end{aligned}$$

(1.4)

holds for sufficiently small $\varepsilon >0$, then, the system (1.2) admits a unique global smooth solution satisfying

$$\begin{aligned} &\|u(t)\|_{H^{2}}^{2}+\|\theta (t)\|_{H^{2}}^{2}+2\int _{0}^{t}\mu \|\partial _{2}u_{1}(\tau )\|_{H^{2}}^{2}+\kappa \|\partial _{1}u_{2}( \tau )\|_{H^{2}}^{2} +\eta \|\partial _{1}\theta (\tau )\|_{H^{2}}^{2}d \tau \leq C\varepsilon ^{2} \end{aligned}$$

(1.5)

for all $t>0$ and $C=C(\mu ,\kappa ,\eta )$ is a positive constant. Moreover,

$$\begin{aligned} &\|\partial _{1}u_{2}(t)\|_{L^{2}}\rightarrow 0,\quad \|\partial _{2}u_{1}(t) \|_{L^{2}}\rightarrow 0,\quad \|\partial _{1}\theta (t)\|_{L^{2}} \rightarrow 0,\quad \textit{as}\quad t\rightarrow \infty . \end{aligned}$$

(1.6)

Remark 1.2

Compared with Theorem 1.1 in [6], we obtain the stability under the $H^{2}$-norm of the initial data $(u_{0},\theta _{0})$ is small because the achievement of large time behavior of the solution $(u,\theta )$ to system (1.2) is heavily dependent on the uniform estimate (1.5).

Applying the $\partial _{1}$ and $\partial _{2}$ to $\text{(1.2)}_{1}$ and $\text{(1.2)}_{2}$, respectively, to conclude

$$\begin{aligned} \pi =\Delta ^{-1}\partial _{2}\theta -\Delta ^{-1}\nabla \cdot \nabla \cdot (u\otimes u)+\mu \Delta ^{-1}\partial _{1}\partial _{22}u_{1}+ \kappa \Delta ^{-1}\partial _{2}\partial _{11}u_{2}. \end{aligned}$$

(1.7)

Then, the equation (1.2) can be rewritten as

$$ \left \{ \begin{aligned} \partial _{t}&u_{1}+u\cdot \nabla u_{1}-\mu \partial _{22}u_{1}+ \partial _{1}\Delta ^{-1}\partial _{2}\theta -\partial _{1}\Delta ^{-1} \nabla \cdot \nabla \cdot (u\otimes u) \\ &+\mu \partial _{1}\Delta ^{-1}\partial _{1}\partial _{22}u_{1}+\kappa \partial _{1}\Delta ^{-1}\partial _{2}\partial _{11}u_{2}=0, \\ \partial _{t}&u_{2}+u\cdot \nabla u_{2}-\kappa \partial _{11}u_{2}- \partial _{1}\partial _{1}\Delta ^{-1}\theta -\partial _{2}\Delta ^{-1} \nabla \cdot \nabla \cdot (u\otimes u) \\ &+\mu \partial _{2}\Delta ^{-1}\partial _{1}\partial _{22}u_{1}+\kappa \partial _{2}\Delta ^{-1}\partial _{2}\partial _{11}u_{2}=0, \\ \partial _{t}&\theta +u\cdot \nabla \theta -\eta \partial _{11}\theta =-u_{2}. \end{aligned} \right . $$

(1.8)

The linearized equations of (1.8) is

$$ \left \{ \begin{aligned} \partial _{t}&u_{1}-\Delta ^{-1}(\mu \partial _{2}^{4}+\kappa \partial _{1}^{4})u_{1}+ \partial _{1}\partial _{2}\Delta ^{-1}\theta =0, \\ \partial _{t}&u_{2}-\Delta ^{-1}(\mu \partial _{2}^{4}+\kappa \partial _{1}^{4})u_{2}- \partial _{1}\partial _{1}\Delta ^{-1}\theta =0, \\ \partial _{t}&\theta -\eta \partial _{11}\theta =-u_{2}. \end{aligned} \right . $$

(1.9)

The following theorem gives the explicit decay rates of the solution of (1.9).

Theorem 1.3

Let $(u,\theta )$ be the corresponding solution of (1.9). Then we have the following two conclusions:

(i) Let $\sigma >0$. Assume initial data $(u_{0},\theta _{0})$ with $\nabla \cdot u_{0}=0$ satisfying

$$\begin{aligned} \|\Lambda _{1}^{-\sigma}u_{0}\|_{L^{2}}+\|\Lambda _{1}^{-\sigma} \theta _{0}\|_{L^{2}}+\|\Lambda _{1}^{-(\sigma +2)}\theta _{0}\|_{L^{2}} \leq \varepsilon \end{aligned}$$

(1.10)

for some $\varepsilon$ small enough. Then $(u,\theta )$ obeys the following decay estimate

$$\begin{aligned} \|u(t)\|_{L^{2}}+\|\theta (t)\|_{L^{2}}\leq C\varepsilon t^{- \frac {\sigma}{2}}, \end{aligned}$$

(1.11)

where $C>0$ is a constant independent of $\varepsilon$ and t.

(ii)Let $m>0$. Assume initial data $(u_{0},\theta _{0})$ with $\nabla \cdot u_{0}=0$ satisfying

$$\begin{aligned} \|u_{0}\|_{L^{2}}+\|\theta _{0}\|_{L^{2}}+\|\Lambda _{1}^{-2}\theta _{0} \|_{L^{2}}\leq \varepsilon \end{aligned}$$

(1.12)

for some $\varepsilon$ small enough. Then $(u,\theta )$ obeys the following decay estimate

$$\begin{aligned} \|\partial _{1}^{m}u(t)\|_{L^{2}}+\|\partial _{1}^{m}\theta (t)\|_{L^{2}} \leq C\varepsilon t^{-\frac {m}{2}}, \end{aligned}$$

(1.13)

where $C>0$ is a constant independent of $\varepsilon$ and $t$.

Remark 1.4

By taking the time derivative on (1.9) and making several substitutions, the system (1.9) turns into the following degenerate wave equations with damping:

$$\{\begin{array}{rl}{\partial }_{\mathit{tt}}& u_1+(\mu R_2^2{\partial }_2^2+\kappa R_1^2{\partial }_1^2-\eta {\partial }_{11}){\partial }_t{u}_1-(R_1^2+\mu \eta R_1^2{\partial }_2^4+\kappa \eta R_1^2{\partial }_1^4)u_1=0,\\ {\partial }_{\mathit{tt}}& u_2+(\mu R_2^2{\partial }_2^2+\kappa R_1^2{\partial }_1^2-\eta {\partial }_{11}){\partial }_t{u}_2-(R_1^2+\mu \eta R_1^2{\partial }_2^4+\kappa \eta R_1^2{\partial }_1^4)u_2=0,\\ {\partial }_{\mathit{tt}}& \theta +(\mu R_2^2{\partial }_2^2+\kappa R_1^2{\partial }_1^2-\eta {\partial }_{11}){\partial }_t\theta -(R_1^2+\mu \eta R_1^2{\partial }_2^4+\kappa \eta R_1^2{\partial }_1^4)\theta =0,\end{array}$$

(1.14)

where $R_{i}=\partial _{i}(-\Delta )^{-\frac{1}{2}}$ with $i=1,2$ denotes the standard Resiz transform. Compared with the wave equations in [1], this system is more complex. The upper bounds for the kernel function $G_{1}$ and $G_{2}$, which is presented in Sect. 4, are more sophisticated to handle than that in [1]. These upper bounds play a crucial role in achieving the decay estimate in Theorem 1.3.

Remark 1.5

Now we explain why we cannot obtain the decay rate of the system (1.8). The methods of proving Theorem 1.3 heavily depends on the spectral analysis of the wave equations (1.14). Unfortunately, it is very difficult for us to decouple the system (1.8). Consequently, we cannot build the decay estimates of system (1.8) via spectral methods. It is of great interest to address this problem.

The rest of this paper is organized as follows. Some crucial lemmas are presented in Sect. 2. We first build a priori estimates and exploy the bootstrap argument to establish $H^{2}$-stability in Sect. 3. The large time behavior of the solution to system (1.1) is also obtained in Sect. 3. The proof of Theorem 1.3 can be found in Sect. 4.

Notation 1

We recall the definition of the fractional Laplacian, $\widehat{\Lambda _{i}^{\beta}f}(\xi )=|\xi _{i}|^{\beta}\hat{f}(\xi )$, for any real number $\beta$ and $i=1,2$, $\xi =(\xi _{1},\xi _{2})$.

2 Several Useful Lemmas

For the convenience, we first recall the following version of the two dimensional anisotropic inequalities in the whole space $\mathbf {R}^{2}$. Lemma 2.1 is due to Cao and Wu [3].

Lemma 2.1

Assume $f$, $g$, $h$, $\partial _{1}g$ and $\partial _{2}h$ are in $L^{2}(\mathbf {R}^{2})$, then, for a constant $C$,

$${\int }_{{\mathbf{R}}^2}|\mathit{fgh}|dx\le C{\parallel f\parallel }_{L^2({\mathbf{R}}^2)}{\parallel g\parallel }_{L^2({\mathbf{R}}^2)}^{\frac{1}{2}}{\parallel {\partial }_{1}g\parallel }_{L^2({\mathbf{R}}^2)}^{\frac{1}{2}}{\parallel h\parallel }_{L^2({\mathbf{R}}^2)}^{\frac{1}{2}}{\parallel {\partial }_{2}h\parallel }_{L^2({\mathbf{R}}^2)}^{\frac{1}{2}}.$$

(2.1)

Lemma 2.2

Let $f=f(t)$, with $t\in [0,\infty )$ be nonnegative continuous function. Assume $f$ is integrable on $[0,\infty )$,

$$ \int _{0}^{\infty }f(t)dt< \infty . $$

Assume that for any $\delta >0$, there is $\rho >0$ such that, for any $0\leq t_{1}< t_{2}$ with $t_{2}-t_{1}\leq \rho$, either $f(t_{2})\leq f(t_{1})$ or $f(t_{2})\geq f(t_{1})$ and $f(t_{2})-f(t_{1})\leq \delta$. Then

$$\begin{aligned} &f(t)\rightarrow 0, \quad \textit{as} \quad t\rightarrow \infty . \end{aligned}$$

(2.2)

This Lemma can be found in [7].

3 Proofs of Theorem 1.1

3.1 $H^{2}$-Stability

For the sake of conciseness, we construct a suitable energy functional:

$$\begin{aligned} \begin{aligned}[b] E(t)=&\displaystyle \sup _{0\leq \tau \leq t}(\|u(\tau )\|_{H^{2}}^{2}+ \|\theta (\tau )\|_{H^{2}}^{2}) \\ &+2\int _{0}^{t}\mu \|\partial _{2}u_{1}(\tau )\|_{H^{2}}^{2}+\kappa \|\partial _{1}u_{2}(\tau )\|_{H^{2}}^{2}+\eta \|\partial _{1}\theta ( \tau )\|_{H^{2}}^{2}d\tau . \end{aligned} \end{aligned}$$

(3.1)

Step 1 $L^{2}$-energy estimate. A standard energy method yields

$$\begin{aligned} \begin{aligned}[b] \|u(t)\|_{L^{2}}^{2}+\|\theta (t)\|_{L^{2}}^{2}&+2\int _{0}^{t}\mu \| \partial _{2}u_{1}(\tau )\|_{L^{2}}^{2} +\kappa \|\partial _{1}u_{2}( \tau )\|_{L^{2}}^{2}+\eta \|\partial _{1}\theta (\tau )\|_{L^{2}}^{2}d \tau \\ &\le \|u_{0}\|_{L^{2}}^{2}+\|\theta _{0}\|_{L^{2}}^{2}. \end{aligned} \end{aligned}$$

(3.2)

Step 2 $\dot{H}^{2}$-energy estimate. Applying $\Delta$ to both sides of the first, the second and the third equation of (1.2), respectively, then taking the $L^{2}$-inner product with $(\Delta u_{1},\Delta u_{2},\Delta \theta )$ to obtain

$$\begin{aligned} \begin{aligned}[b] \frac{1}{2}\frac {d}{dt}&(\|\Delta u(t)\|_{L^{2}}^{2}+\|\Delta \theta (t) \|_{L^{2}}^{2})+\mu \|\partial _{2}\Delta u_{1}\|_{L^{2}}^{2} +\kappa \|\partial _{1}\Delta u_{2}\|_{L^{2}}^{2}+\eta \|\partial _{1}\Delta \theta \|_{L^{2}}^{2} \\ =&-\int _{\mathbf {R}^{2}}\Delta (u\cdot \nabla u_{1})\Delta u_{1}+ \Delta (u\cdot \nabla u_{2})\Delta u_{2}dx+\int _{\mathbf {R}^{2}}( \Delta \theta \Delta u_{2}-\Delta u_{2}\Delta \theta )dx \\ &-\int _{\mathbf {R}^{2}}\Delta (u\cdot \nabla \theta )\Delta \theta dx:=I_{1}+I_{2}+I_{3}. \end{aligned} \end{aligned}$$

(3.3)

It is not difficult to check that $I_{2}=0$. To estimate $I_{1}$, we decompose $I_{1}$ into the following form:

$$\begin{aligned} \begin{aligned}[b] I_{1} =&-\int _{\mathbf {R}^{2}}\partial _{11}u\cdot \nabla u_{1}\partial _{11}u_{1}+2 \partial _{1}u\cdot \nabla \partial _{1} u_{1}\partial _{11}u_{1}dx \\ &-\int _{\mathbf {R}^{2}}\partial _{22}u\cdot \nabla u_{1}\partial _{22}u_{1}+2 \partial _{2}u\cdot \nabla \partial _{2} u_{1}\partial _{22}u_{1}dx \\ &-\int _{\mathbf {R}^{2}}\partial _{11}u\cdot \nabla u_{2}\partial _{11}u_{2}+2 \partial _{1}u\cdot \nabla \partial _{1} u_{2}\partial _{11}u_{2}dx \\ &-\int _{\mathbf {R}^{2}}\partial _{22}u\cdot \nabla u_{2}\partial _{22}u_{2}+2 \partial _{2}u\cdot \nabla \partial _{2} u_{2}\partial _{22}u_{2}dx \\ &:=I_{11}+I_{12}+I_{13}+I_{14}. \end{aligned} \end{aligned}$$

(3.4)

Thanks to the fact that $\nabla \cdot u=0$ and the Sobolev embedding, one gets

$$\begin{aligned} \begin{aligned}[b] I_{11}=&-3\int _{\mathbf {R}^{2}}(\partial _{11}u_{1})^{2}\partial _{1}u_{1}dx- \int _{\mathbf {R}^{2}}\partial _{11}u_{2}\partial _{2}u_{1}\partial _{11}u_{1}dx \\ &-2\int _{\mathbf {R}^{2}}\partial _{1}u_{2}\partial _{2}\partial _{1} u_{1} \partial _{11}u_{1}dx \\ \leq &C\|\partial _{1}u_{1}\|_{L^{2}}\|\partial _{11}u_{1}\|_{L^{4}}^{2}+C \|\partial _{2}u_{1}\|_{L^{\infty}}\|\partial _{11}u_{2}\|_{L^{2}}\| \partial _{11}u_{1}\|_{L^{2}} \\ &+C\|\partial _{1}u_{2}\|_{L^{\infty}}\|\partial _{2}\partial _{1}u_{1} \|_{L^{2}}\|\partial _{11}u_{1}\|_{L^{2}} \\ \leq &C\|u\|_{H^{2}}(\|\partial _{2}u_{1}\|_{H^{2}}^{2}+\|\partial _{1}u_{2} \|_{H^{2}}^{2}). \end{aligned} \end{aligned}$$

(3.5)

Similarly,

$$\begin{aligned} I_{12}, I_{13}, I_{14}\leq C\|u\|_{H^{2}}(\|\partial _{2}u_{1}\|_{H^{2}}^{2}+ \|\partial _{1}u_{2}\|_{H^{2}}^{2}). \end{aligned}$$

(3.6)

Next, we split $I_{3}$ into the following two parts:

$$\begin{aligned} I_{3}=&-\int _{\mathbf {R}^{2}}\partial _{11}(u\cdot \nabla \theta ) \partial _{11}\theta +\partial _{22}(u\cdot \nabla \theta )\partial _{22} \theta dx \\ =&-\int _{\mathbf {R}^{2}}\partial _{11}u\cdot \nabla \theta \partial _{11} \theta +2\partial _{1}u\cdot \nabla \partial _{1}\theta \partial _{11} \theta dx \\ &-\int _{\mathbf {R}^{2}}\partial _{22}u\cdot \nabla \theta \partial _{22} \theta +2\partial _{2}u\cdot \nabla \partial _{2}\theta \partial _{22} \theta dx \\ :=&I_{31}+I_{32}. \end{aligned}$$

We can infer from the Hölder inequality and the Sobolev inequality

$$\begin{aligned} \begin{aligned}[b] I_{31}=&-\int _{\mathbf {R}^{2}}\partial _{11}u_{1}\partial _{1}\theta \partial _{11}\theta dx-\int _{\mathbf {R}^{2}}\partial _{11}u_{2}\partial _{2} \theta \partial _{11}\theta dx-2\int _{\mathbf {R}^{2}}\partial _{1}u\cdot \nabla \partial _{1}\theta \partial _{11}\theta dx \\ \leq &C\|\partial _{11}u_{1}\|_{L^{2}}\|\partial _{1}\theta \|_{L^{4}} \|\partial _{11}\theta \|_{L^{4}}+C\|\partial _{11}u_{2}\|_{L^{4}}\| \partial _{2}\theta \|_{L^{2}}\|\partial _{11}\theta \|_{L^{4}} \\ &+C\|\partial _{1}u\|_{L^{2}}\|\nabla \partial _{1}\theta \|_{L^{4}}\| \partial _{11}\theta \|_{L^{4}} \\ \leq &C\|u\|_{H^{2}}\|\partial _{1}\theta \|_{H^{2}}^{2}+C\|\theta \|_{H^{2}}( \|\partial _{1}u_{2}\|_{H^{2}}^{2}+\|\partial _{1}\theta \|_{H^{2}}^{2}) \\ \leq &C(\|u\|_{H^{2}}+\|\theta \|_{H^{2}})(\|\partial _{1}u_{2}\|_{H^{2}}^{2}+ \|\partial _{1}\theta \|_{H^{2}}^{2}). \end{aligned} \end{aligned}$$

(3.7)

To handle $I_{32}$, we write

$$\begin{aligned} I_{32}\leq &C|\int _{\mathbf {R}^{2}}\partial _{22}u_{1}\partial _{1} \theta \partial _{22}\theta +\partial _{2}u_{1}\partial _{1}\partial _{2} \theta \partial _{22}\theta dx| \\ &+C|\int _{\mathbf {R}^{2}}\partial _{22}u_{2}\partial _{2}\theta \partial _{22} \theta dx|+C|\int _{\mathbf {R}^{2}}\partial _{2}u_{2}(\partial _{22} \theta )^{2}dx| \\ :=&I_{321}+I_{322}+I_{323}. \end{aligned}$$

According to the Hölder inequality, it deduces

$$\begin{aligned} \begin{aligned}[b] I_{321}\leq &C(\|\partial _{22}u_{1}\|_{L^{4}}\|\partial _{1}\theta \|_{L^{4}}+ \|\partial _{2}u_{1}\|_{L^{4}}\|\partial _{1}\partial _{2}\theta \|_{L^{4}}) \|\partial _{22}\theta \|_{L^{2}} \\ \leq &C\|\theta \|_{H^{2}}(\|\partial _{2}u_{1}\|_{H^{2}}^{2}+\| \partial _{1}\theta \|_{H^{2}}^{2}). \end{aligned} \end{aligned}$$

(3.8)

Integrating by parts and the Hölder inequality give rise to

$$\begin{aligned} \begin{aligned}[b] I_{322}\leq &C|\int _{\mathbf {R}^{2}}\partial _{1}\partial _{2}u_{1} \partial _{2}\theta \partial _{22}\theta dx| \\ \leq &C|\int _{\mathbf {R}^{2}}\partial _{2}u_{1}(\partial _{1}\partial _{2} \theta \partial _{22}\theta +\partial _{2}\theta \partial _{22}\partial _{1} \theta ) dx| \\ \leq &C\|\partial _{2}u_{1}\|_{L^{4}}(\|\partial _{1}\partial _{2} \theta \|_{L^{4}}\|\partial _{22}\theta \|_{L^{2}}+\|\partial _{2} \theta \|_{L^{4}}\|\partial _{1}\partial _{22}\theta \|_{L^{2}}) \\ \leq &C\|\theta \|_{H^{2}}(\|\partial _{2}u_{1}\|_{H^{2}}^{2}+\| \partial _{1}\theta \|_{H^{2}}^{2}). \end{aligned} \end{aligned}$$

(3.9)

Form Lemma 2.1 and the Young inequality, one can follow that

$$\begin{aligned} \begin{aligned}[b] I_{323}\leq &C|\int _{\mathbf {R}^{2}}\partial _{1}u_{1}(\partial _{22} \theta )^{2}dx| \\ \leq &C|\int _{\mathbf {R}^{2}}u_{1}\partial _{1}\partial _{22}\theta \partial _{22}\theta dx| \\ \leq &C\|\partial _{1}\partial _{22}\theta \|_{L^{2}}\|\partial _{22} \theta \|_{L^{2}}^{\frac{1}{2}}\|\partial _{1}\partial _{22}\theta \|_{L^{2}}^{ \frac{1}{2}} \|u_{1}\|_{L^{2}}^{\frac{1}{2}}\|\partial _{2}u_{1}\|_{L^{2}}^{ \frac{1}{2}} \\ \leq &C(\|u\|_{H^{2}}+\|\theta \|_{H^{2}})(\|\partial _{2}u_{1}\|_{H^{2}}^{2}+ \|\partial _{1}\theta \|_{H^{2}}^{2}). \end{aligned} \end{aligned}$$

(3.10)

Combining the estimates from (3.5) to (3.10) and integrating over $[0,t]$, we can obtain

$$\begin{aligned} \|\Delta & u(t)\|_{L^{2}}^{2}+\|\Delta \theta (t)\|_{L^{2}}^{2}+2 \int _{0}^{t}\mu \|\partial _{2}\Delta u_{1}\|_{L^{2}}^{2} +\kappa \| \partial _{1}\Delta u_{2}\|_{L^{2}}^{2}+\eta \|\partial _{1}\Delta \theta \|_{L^{2}}^{2}d\tau \\ \leq &\|\Delta u_{0}\|_{L^{2}}^{2}+\|\Delta \theta _{0}\|_{L^{2}}^{2}+C \int _{0}^{t}(\|u\|_{H^{2}}+\|\theta \|_{H^{2}})(\|\partial _{2}u_{1} \|_{H^{2}}^{2}+\|\partial _{1}u_{2}\|_{H^{2}}^{2}+\|\partial _{1} \theta \|_{H^{2}}^{2})d\tau . \end{aligned}$$

(3.11)

Adding (3.2) and (3.11) leads to

$$\begin{aligned} \begin{aligned}[b] \| u(t)&\|_{H^{2}}^{2}+\|\theta (t)\|_{H^{2}}^{2}+2\int _{0}^{t}\mu \|\partial _{2}u_{1}\|_{H^{2}}^{2} +\kappa \|\partial _{1}u_{2}\|_{H^{2}}^{2}+ \eta \|\partial _{1}\theta \|_{H^{2}}^{2}d\tau \\ \leq &C_{0}(\|u_{0}\|_{H^{2}}^{2}+\|\theta _{0}\|_{H^{2}}^{2}) +C_{1} \displaystyle \sup _{0\leq \tau \leq t}(\|u(\tau )\|_{H^{2}}+\| \theta (\tau )\|_{H^{2}}) \\ &\times \int _{0}^{t}(\|\partial _{2}u_{1}\|_{H^{2}}^{2}+\|\partial _{1}u_{2} \|_{H^{2}}^{2}+\|\partial _{1}\theta \|_{H^{2}}^{2})d\tau , \end{aligned} \end{aligned}$$

(3.12)

which along with the definition of $E(t)$ ensures

$$\begin{aligned} E(t)\leq C_{0}E(0)+C_{1}E^{\frac{3}{2}}(t). \end{aligned}$$

(3.13)

To apply the bootstrapping argument, we make the ansatz

$$\begin{aligned} E(t)\leq \frac {1}{4C_{1}^{2}}. \end{aligned}$$

(3.14)

We choose $\varepsilon$ suitable small such that the initial $H^{2}$-norm $E(0)$ sufficiently small, namely,

$$\begin{aligned} E(0):=\|u_{0}\|_{H^{2}}^{2}+\|\theta _{0}\|_{H^{2}}^{2}\leq \varepsilon ^{2}:=\frac {1}{4C_{1}^{2}C_{0}}. \end{aligned}$$

(3.15)

In fact, when (3.14) and (3.15) holds, (3.13) implies

$$ E(t)\leq \frac {1}{4C_{1}^{2}}+\frac{1}{2}E(t).$$

Therefore, the bootstrapping argument then concludes that, for all $t>0$

$$ E(t)\leq \frac {1}{8C_{1}^{2}}\leq \frac {C_{0}}{2}\varepsilon ^{2},$$

which gives the desired inequality (1.5).

3.2 Large Time Behavior of the Boussinesq equation (1.2)

Now we pay our attention to show the inequality (1.6). Applying $\partial _{2}$ to $\text{(1.8)}_{1}$ and $\partial _{1}$ to $\text{(1.8)}_{2}$, then taking the $L^{2}$-inner product with $\partial _{2}u_{1}$ and $\partial _{1}u_{2}$, respectively. After performing $L^{2}$-inner product on both side of $\text{(1.8)}_{3}$ with $\partial _{1}\theta$, we add them to get

$$\begin{aligned} \frac{1}{2}&\frac {d}{dt}(\|\partial _{2}u_{1}(t)\|_{L^{2}}^{2}+\| \partial _{1}u_{2}(t)\|_{L^{2}}^{2}+\|\partial _{1}\theta (t)\|_{L^{2}}^{2})+ \mu \|\partial _{22}u_{1}\|_{L^{2}}^{2} +\kappa \|\partial _{11}u_{2}\|_{L^{2}}^{2}+ \eta \|\partial _{11}\theta \|_{L^{2}}^{2} \\ =&-\int _{\mathbf {R}^{2}}\partial _{2}u\cdot \nabla u_{1}\partial _{2}u_{1}+ \partial _{1}u\cdot \nabla u_{2}\partial _{1}u_{2}dx \\ &-\int _{\mathbf {R}^{2}}\partial _{2}\partial _{1}\partial _{2}\Delta ^{-1} \theta \partial _{2}u_{1}-\partial _{1}\partial _{1}\partial _{1}\Delta ^{-1} \theta \partial _{1}u_{2}dx \\ &+\int _{\mathbf {R}^{2}}\partial _{2}\partial _{1}\Delta ^{-1}\nabla \cdot \nabla \cdot (u\otimes u)\partial _{2}u_{1}dx+\int _{\mathbf {R}^{2}} \partial _{1}\partial _{2}\Delta ^{-1}\nabla \cdot \nabla \cdot (u \otimes u)\partial _{1}u_{2}dx \\ &-\int _{\mathbf {R}^{2}}\partial _{1}u\cdot \nabla \theta \partial _{1} \theta dx-\int _{\mathbf {R}^{2}}\partial _{1}u_{2}\partial _{1}\theta dx \\ &+\int _{\mathbf {R}^{2}}(\mu \partial _{1}\Delta ^{-1}\partial _{1} \partial _{22}u_{1}+\kappa \partial _{1}\Delta ^{-1}\partial _{2} \partial _{11}u_{2})\partial _{2}u_{1}dx \\ &+\int _{\mathbf {R}^{2}}(\mu \partial _{2}\Delta ^{-1}\partial _{1} \partial _{22}u_{1}+\kappa \partial _{2}\Delta ^{-1}\partial _{2} \partial _{11}u_{2})\partial _{1}u_{2}dx \\ :=&J_{1}+J_{2}+J_{3}+J_{4}+J_{5}+J_{6}++J_{7}+J_{8}. \end{aligned}$$

(3.16)

Thanks to the fact $\nabla \cdot u=0$, it’s not hard to see that

$$\begin{aligned} J_{1}&=-\int _{\mathbf {R}^{2}}(\partial _{2}u_{1})^{2}\partial _{1}u_{1}+ \partial _{2}u_{2}(\partial _{2}u_{1})^{2}+\partial _{1}u_{1}(\partial _{1}u_{2})^{2}+( \partial _{1}u_{2})^{2}\partial _{2}u_{2}dx \\ &=0. \end{aligned}$$

By $L^{p}$- boundedness of the Riesz transform and the Hölder inequality, we get

$$\begin{aligned} J_{2}\le &C\|R_{22}\partial _{1}\theta \|_{L^{2}}\|\partial _{2}u_{1}\|_{L^{2}}+C \|R_{11}\partial _{1}\theta \|_{L^{2}}\|\partial _{1}u_{2}\|_{L^{2}} \\ \le &C\|\partial _{1}\theta \|_{L^{2}}(\|\partial _{2}u_{1}\|_{L^{2}}+ \|\partial _{1}u_{2}\|_{L^{2}}) \\ \le &C\|u\|_{H^{2}}\|\theta \|_{H^{2}}. \end{aligned}$$

Thanks to $L^{p}$- boundedness of the Riesz transform and Sobolev’s embedding, one arrives at

$$\begin{aligned} J_{3}\le &C\|R_{2}R_{1}(\partial _{1}(u\cdot \nabla u_{1})+\partial _{2}(u \cdot \nabla u_{2}))\|_{L^{2}}\|\partial _{2}u_{1}\|_{L^{2}} \\ \le &C\|\partial _{1}(u\cdot \nabla u_{1})+\partial _{2}(u\cdot \nabla u_{2}) \|_{L^{2}}\|\partial _{2}u_{1}\|_{L^{2}} \\ \le &C\|\partial _{1}u\cdot \nabla u_{1}+u\cdot \nabla \partial _{1}u_{1}+ \partial _{2}u\cdot \nabla u_{2}+u\cdot \nabla \partial _{2} u_{2}\|_{L^{2}} \|\partial _{2}u_{1}\|_{L^{2}} \\ \le &C(\|\partial _{1}u\|_{L^{4}}\|\nabla u_{1}\|_{L^{4}}+\|u\|_{L^{ \infty}}\|\nabla \partial _{1}u_{1}\|_{L^{2}}+C\|\partial _{2}u\|_{L^{4}} \|\nabla u_{2}\|_{L^{4}})\|\partial _{2}u_{1}\|_{L^{2}} \\ \le &C\|u\|_{H^{2}}^{3}. \end{aligned}$$

Similarly,

$$\begin{aligned} J_{4}\le &C\|u\|_{H^{2}}^{3}. \end{aligned}$$

Applying the Hölder inequality to get

$$\begin{aligned} J_{5}\le &C\|\partial _{1}u\|_{L^{4}}\|\nabla \theta \|_{L^{4}}\| \partial _{1}\theta \|_{L^{2}}\le C\|u\|_{H^{2}}\|\theta \|_{H^{2}}^{2}, \end{aligned}$$

and

$$\begin{aligned} J_{6}\le &C\|\partial _{1}u_{2}\|_{L^{2}}\|\partial _{1}\theta \|_{L^{2}} \le C\|u\|_{H^{2}}\|\theta \|_{H^{2}}. \end{aligned}$$

Integrating by parts and the $L^{p}$- boundedness of the Riesz transform give rise to

$$\begin{aligned} J_{7}+J_{8}\le &C\|u\|_{H^{2}}^{2}. \end{aligned}$$

Inserting the estimate from $J_{1}$ to $J_{6}$ into (3.16) and integrating over $[s,t]$ with $0< s< t<\infty$ to obtain

$$\begin{aligned} (\|&\partial _{2}u_{1}(t)\|_{L^{2}}^{2}+\|\partial _{1}u_{2}(t)\|_{L^{2}}^{2}+ \|\partial _{1}\theta (t)\|_{L^{2}}^{2}) \\ &-(\|\partial _{2}u_{1}(s)\|_{L^{2}}^{2}+\|\partial _{1}u_{2}(s)\|_{L^{2}}^{2}+ \|\partial _{1}\theta (s)\|_{L^{2}}^{2})\leq C(\varepsilon ^{2}+ \varepsilon ^{3})(t-s). \end{aligned}$$

(3.17)

Thanks to (1.5), one has

$$\begin{aligned} \int _{0}^{\infty}\|\partial _{2}u_{1}(\tau )\|_{L^{2}}^{2}+\|\partial _{1}u_{2}( \tau )\|_{L^{2}}^{2}+\|\partial _{1}\theta (\tau )\|_{L^{2}}^{2}d\tau \leq C\varepsilon ^{2}. \end{aligned}$$

Therefore, as a result of Lemma 2.2, we conclude that

$$\begin{aligned} &\|\partial _{1}u_{2}(t)\|_{L^{2}}\rightarrow 0, \quad \|\partial _{2}u_{1}(t)\|_{L^{2}}\rightarrow 0, \quad \|\partial _{1}\theta (t)\|_{L^{2}}\rightarrow 0, \quad \text{as} \quad t\rightarrow \infty . \end{aligned}$$

This helps us to complete the proof of Theorem 1.1.

4 Proofs of Theorem 1.3

Lemma 4.1

Assume that $\phi$ satisfies the follow equation in $\mathbf {R}^{2}$,

$${\partial }_{\mathit{tt}}\varphi +(\mu R_2^2{\partial }_2^2+\kappa R_1^2{\partial }_1^2-\eta {\partial }_{11}){\partial }_t\varphi -(R_1^2+\mu \eta R_1^2{\partial }_2^4+\kappa \eta R_1^2{\partial }_1^4)\varphi =0,$$

(4.1)

with the initial conditions

$$\begin{aligned} \phi (x,0)=\phi _{0}(x), \quad \partial _{t}\phi (x,0)=\phi _{1}(x). \end{aligned}$$

Then the solution $\phi$ to (4.1) can be explicitly represented as

$$\begin{aligned} \phi (x,t)=G_{1}\bigg(\phi _{1}-\frac{1}{2}(\Delta ^{-1}(\mu \partial _{2}^{4}+ \kappa \partial _{1}^{4})+\eta \partial _{11})\phi _{0}\bigg)+G_{2} \phi _{0}, \end{aligned}$$

(4.2)

where $G_{1}$ and $G_{2}$ are given as follows,

$$\begin{aligned} \widehat{G}_{1}(\xi ,t)= \frac {e^{\lambda _{2}t}-e^{\lambda _{1}t}}{\lambda _{2}-\lambda _{1}}, \quad \widehat{G}_{2}(\xi ,t)=\frac{1}{2}(e^{\lambda _{1}t}+e^{\lambda _{2}t}), \end{aligned}$$

(4.3)

with $\lambda _{1}$ and $\lambda _{2}$ being the roots of the characteristic equation

$$\begin{aligned} \lambda ^{2}+\bigg( \frac {\mu \xi _{2}^{4}+\kappa \xi _{1}^{4}}{|\xi |^{2}}+\eta \xi _{1}^{2} \bigg)\lambda + \frac {\xi _{1}^{2}+\eta \xi _{1}^{2}(\mu \xi _{2}^{4}+\kappa \xi _{1}^{4})}{|\xi |^{2}}=0, \end{aligned}$$

or

$$\begin{aligned} \lambda _{1}=-\frac{1}{2}\bigg( \frac {\mu \xi _{2}^{4}+\kappa \xi _{1}^{4}}{|\xi |^{2}}+\eta \xi _{1}^{2} \bigg)-\frac{1}{2}\sqrt{\Gamma}, \end{aligned}$$

(4.4)

$$\begin{aligned} \lambda _{2}=-\frac{1}{2}\bigg( \frac {\mu \xi _{2}^{4}+\kappa \xi _{1}^{4}}{|\xi |^{2}}+\eta \xi _{1}^{2} \bigg)+\frac{1}{2}\sqrt{\Gamma}, \end{aligned}$$

(4.5)

here

$$\begin{aligned} \Gamma =\bigg( \frac {\mu \xi _{2}^{4}+\kappa \xi _{1}^{4}}{|\xi |^{2}}+\eta \xi _{1}^{2} \bigg)^{2}- \frac {4\xi _{1}^{2}+4\eta \xi _{1}^{2}(\mu \xi _{2}^{4}+\kappa \xi _{1}^{4})}{|\xi |^{2}}. \end{aligned}$$

(4.6)

Proof

Applying the Fourier transform on the space variable $x$ to both sides of (4.1), we obtain

$${\partial }_{\mathit{tt}}\hat{\varphi }+(\frac{\mu {\xi }_2^4+\kappa {\xi }_1^4}{{|\xi |}^2}+\eta {\xi }_1^2){\partial }_t\hat{\varphi }+\frac{{\xi }_1^2+\eta {\xi }_1^2(\mu {\xi }_2^4+\kappa {\xi }_1^4)}{{|\xi |}^2}\hat{\varphi }=0,$$

namely,

$$\begin{aligned} (\partial _{t}-\lambda _{2})(\partial _{t}-\lambda _{1})\widehat{\phi}=0 \quad \text{or} \quad (\partial _{t}-\lambda _{1})(\partial _{t}-\lambda _{2})\widehat{\phi}=0. \end{aligned}$$

It is not difficult to rewrite the wave equation into two different systems,

$$\begin{aligned} (\partial _{t}-\lambda _{2})\widehat{\phi}=\widehat{f}, \end{aligned}$$

(4.7)

$$\begin{aligned} (\partial _{t}-\lambda _{1})\widehat{f}=0, \end{aligned}$$

(4.8)

or

$$\begin{aligned} (\partial _{t}-\lambda _{1})\widehat{\phi}=\widehat{g}, \end{aligned}$$

(4.9)

$$\begin{aligned} (\partial _{t}-\lambda _{2})\widehat{g}=0. \end{aligned}$$

(4.10)

By taking the difference of (4.9) and (4.7), it deduces

$$\begin{aligned} \widehat{\phi}(\xi ,t)=(\lambda _{2}-\lambda _{1})^{-1}(\widehat{g}- \widehat{f}). \end{aligned}$$

(4.11)

Then, (4.8) and (4.10) yield

$$\begin{aligned} \widehat{f}(\xi ,t)=e^{\lambda _{1}t}\widehat{f}(\xi ,0)=e^{\lambda _{1}t}( \widehat{\phi}_{1}-\lambda _{2}\widehat{\phi}_{0}), \end{aligned}$$

(4.12)

$$\begin{aligned} \widehat{g}(\xi ,t)=e^{\lambda _{2}t}\widehat{g}(\xi ,0)=e^{\lambda _{2}t}( \widehat{\phi}_{1}-\lambda _{1}\widehat{\phi}_{0}). \end{aligned}$$

(4.13)

Inserting (4.12) into (4.11) leads to

$$\begin{aligned} \begin{aligned}[b] \widehat{\phi}(\xi ,t)&=(\lambda _{2}-\lambda _{1})^{-1}\bigg((e^{ \lambda _{2}t}-e^{\lambda _{1}t})\widehat{\phi}_{1}+ (\lambda _{2}e^{ \lambda _{1}t}-\lambda _{1}e^{\lambda _{2}t})\widehat{\phi}_{0}\bigg) \\ &= \frac {e^{\lambda _{2}t}-e^{\lambda _{1}t}}{\lambda _{2}-\lambda _{1}}( \widehat{\phi}_{1}-\lambda _{2}\widehat{\phi}_{0})+e^{\lambda _{2}t} \widehat{\phi}_{0} \\ &= \frac {e^{\lambda _{2}t}-e^{\lambda _{1}t}}{\lambda _{2}-\lambda _{1}} \bigg(\widehat{\phi}_{1}+\frac{1}{2}( \frac {\mu \xi _{2}^{4}+\kappa \xi _{1}^{4}}{|\xi |^{2}}+\eta \xi _{1}^{2}) \widehat{\phi}_{0}\bigg) +\frac{1}{2}(e^{\lambda _{1}t}+e^{\lambda _{2}t}) \widehat{\phi}_{0} \\ &=\widehat{G}_{1}\bigg(\widehat{\phi}_{1}+\frac{1}{2}( \frac {\mu \xi _{2}^{4}+\kappa \xi _{1}^{4}}{|\xi |^{2}}+\eta \xi _{1}^{2}) \widehat{\phi}_{0}\bigg)+\widehat{G}_{2}\widehat{\phi}_{0}, \end{aligned} \end{aligned}$$

(4.14)

where we used the definition of $\lambda _{2}$ in the third inequality. This completes the proof of Lemma 4.1. □

Due to the fact that $\widehat{G}_{1}(\xi ,t)$ and $\widehat{G}_{2}(\xi ,t)$ have a strong dependence on frequency, we need to be divided frequency space into several subdomains to obtain the optimal upper bound of $\widehat{G}_{1}(\xi ,t)$ and $\widehat{G}_{2}(\xi ,t)$.

Lemma 4.2

Let $\mathbf {R}^{2}=S_{1}\cup S_{2}$. Here

$$\begin{aligned} \begin{aligned}[b] S_{1}=\biggl\{ \xi \in \mathbf {R}^{2}: \Gamma &= \bigg( \frac {\mu \xi _{2}^{4}+\kappa \xi _{1}^{4}}{|\xi |^{2}}+\eta \xi _{1}^{2} \bigg)^{2}- \frac {4\xi _{1}^{2}+4\eta \xi _{1}^{2}(\mu \xi _{2}^{4}+\kappa \xi _{1}^{4})}{|\xi |^{2}} \\ &\leq \frac{1}{4}\bigg( \frac {\mu \xi _{2}^{4}+\kappa \xi _{1}^{4}}{|\xi |^{2}}+\eta \xi _{1}^{2} \bigg)^{2}\biggr\} , \end{aligned} \end{aligned}$$

$S_{2}=\mathbf {R}^{2}\backslash S_{1}$.

Then $\widehat{G}_{1}(\xi ,t)$ and $\widehat{G}_{2}(\xi ,t)$ satisfy the following estimates:

(a) $\forall \xi \in S_{1}$,

$$\begin{aligned} Re\lambda _{1}\leq -\frac{1}{2}\bigg( \frac {\mu \xi _{2}^{4}+\kappa \xi _{1}^{4}}{|\xi |^{2}}+\eta \xi _{1}^{2} \bigg), \quad Re\lambda _{2}\leq -\frac{1}{4}\bigg( \frac {\mu \xi _{2}^{4}+\kappa \xi _{1}^{4}}{|\xi |^{2}}+\eta \xi _{1}^{2} \bigg), \end{aligned}$$

$$\begin{array}{rl}& |{\hat{G}}_1(\xi ,t)|\le {\mathit{te}}^{-\frac{1}{4}(\frac{\mu {\xi }_2^4+\kappa {\xi }_1^4}{{|\xi |}^2}+\eta {\xi }_1^2)t},\\ & |{\hat{G}}_2(\xi ,t)|\le {\mathit{Ce}}^{-\frac{\eta }{4}{\xi }_1^{2}t}.\end{array}$$

(4.15)

(b) $\forall \xi \in S_{2}$,

$$\begin{aligned} \lambda _{1}\leq -\frac{3}{4}\bigg( \frac {\mu \xi _{2}^{4}+\kappa \xi _{1}^{4}}{|\xi |^{2}}+\eta \xi _{1}^{2} \bigg), \quad \lambda _{2}\leq -c_{0}\xi _{1}^{2}, \end{aligned}$$

$$\begin{array}{rl}& |{\hat{G}}_1(\xi ,t)|\le \frac{C}{\frac{\mu {\xi }_2^4+\kappa {\xi }_1^4}{{|\xi |}^2}+\eta {\xi }_1^2}(e^{-\frac{3}{4}(\frac{\mu {\xi }_2^4+\kappa {\xi }_1^4}{{|\xi |}^2}+\eta {\xi }_1^2)t}+e^{-c_0{\xi }_1^{2}t}),\\ & |{\hat{G}}_2(\xi ,t)|\le {\mathit{Ce}}^{-c{\xi }_1^{2}t}.\end{array}$$

(4.16)

Proof

(a) For $\xi \in S_{1}$, we divide $S_{1}$ into the following two regions:

$$\begin{aligned} &S_{11}=\left \{\xi \in S_{1}: 0\leq \Gamma \leq \frac{1}{4}\bigg( \frac {\mu \xi _{2}^{4}+\kappa \xi _{1}^{4}}{|\xi |^{2}}+\eta \xi _{1}^{2} \bigg)^{2}\right \}, \\ &S_{12}=\left \{\xi \in S_{1}: \Gamma < 0\right \}. \end{aligned}$$

For any $\xi \in S_{11}$, according to the definition of $\lambda _{1}$ and $\lambda _{2}$ in (4.4) and (4.5), $\lambda _{1}$ and $\lambda _{2}$ are real roots and satisfy

$$\begin{aligned} &\lambda _{1}\leq -\frac{1}{2}\bigg( \frac {\mu \xi _{2}^{4}+\kappa \xi _{1}^{4}}{|\xi |^{2}}+\eta \xi _{1}^{2} \bigg), \\ &\lambda _{2}\leq -\frac{1}{4}\bigg( \frac {\mu \xi _{2}^{4}+\kappa \xi _{1}^{4}}{|\xi |^{2}}+\eta \xi _{1}^{2} \bigg). \end{aligned}$$

By the mean-value theorem, we know

$$\begin{array}{rl}& |{\hat{G}}_1(\xi ,t)|\le {\mathit{te}}^{{\lambda }_{2}t}\le {\mathit{te}}^{-\frac{1}{4}(\frac{\mu {\xi }_2^4+\kappa {\xi }_1^4}{{|\xi |}^2}+\eta {\xi }_1^2)t},\\ & |{\hat{G}}_2(\xi ,t)|\le {\mathit{Ce}}^{-\frac{1}{4}(\frac{\mu {\xi }_2^4+\kappa {\xi }_1^4}{{|\xi |}^2}+\eta {\xi }_1^2)t}.\end{array}$$

For any $\xi \in S_{12}$, $\lambda _{1}$ and $\lambda _{2}$ are a pair of complex conjugate roots, then one has

$$\begin{aligned} \widehat{G}_{1}(\xi ,t)&=e^{-\frac{1}{2}( \frac {\mu \xi _{2}^{4}+\kappa \xi _{1}^{4}}{|\xi |^{2}}+\eta \xi _{1}^{2})t} \frac {e^{\frac {i\sqrt{-\Gamma}}{2}t}-e^{-\frac {i\sqrt{-\Gamma}}{2}t}}{i\sqrt{-\Gamma}}, \\ &=e^{-\frac{1}{2}( \frac {\mu \xi _{2}^{4}+\kappa \xi _{1}^{4}}{|\xi |^{2}}+\eta \xi _{1}^{2})t} \frac {2\sin (\frac {\sqrt{-\Gamma}}{2}t)}{\sqrt{-\Gamma}}. \end{aligned}$$

We can infer from $|\sin{x}|\leq |x|$ that

$$\begin{array}{rl}& |{\hat{G}}_1(\xi ,t)|\le {\mathit{te}}^{-\frac{1}{2}(\frac{\mu {\xi }_2^4+\kappa {\xi }_1^4}{{|\xi |}^2}+\eta {\xi }_1^2)t},\\ & |{\hat{G}}_2(\xi ,t)|\le \frac{1}{2}(e^{\mathit{tRe}{\lambda }_1}+e^{\mathit{tRe}{\lambda }_2})\le {\mathit{Ce}}^{-\frac{1}{2}(\frac{\mu {\xi }_2^4+\kappa {\xi }_1^4}{{|\xi |}^2}+\eta {\xi }_1^2)t}.\end{array}$$

(b) For $\xi \in S_{2}$, $\lambda _{1}$ and $\lambda _{2}$ are real roots, we have

$$\begin{aligned} \lambda _{1}\leq -\frac{3}{4}\bigg( \frac {\mu \xi _{2}^{4}+\kappa \xi _{1}^{4}}{|\xi |^{2}}+\eta \xi _{1}^{2} \bigg), \end{aligned}$$

and

$$\begin{aligned} \lambda _{2}&=-\frac{1}{2}\bigg( \frac {\mu \xi _{2}^{4}+\kappa \xi _{1}^{4}}{|\xi |^{2}}+\eta \xi _{1}^{2}- \sqrt{\Gamma}\bigg) \\ &=-\frac{1}{2} \frac {|\xi |^{-2}(4\xi _{1}^{2}+4\eta \xi _{1}^{2}(\mu \xi _{2}^{4}+\kappa \xi _{1}^{4}))}{|\xi |^{-2}(\mu \xi _{2}^{4}+\kappa \xi _{1}^{4})+\eta \xi _{1}^{2}+\sqrt{\Gamma}} \\ &\leq - \frac {|\xi |^{-2}(\xi _{1}^{2}+\eta \xi _{1}^{2}(\mu \xi _{2}^{4}+\kappa \xi _{1}^{4}))}{|\xi |^{-2}(\mu \xi _{2}^{4}+\kappa \xi _{1}^{4})+\eta \xi _{1}^{2}}. \end{aligned}$$

In order to control $\lambda _{2}$, we further divide $S_{2}$ into the following two regions:

$$\begin{aligned} &S_{21}=\left \{\xi \in S_{2}: |\xi _{1}|\geq |\xi _{2}|\right \}, \\ &S_{22}=\left \{\xi \in S_{2}:|\xi _{1}|< |\xi _{2}|\right \}. \end{aligned}$$

For $\xi \in S_{21}$, we obtain

$$\begin{aligned} \lambda _{2}&\leq - \frac {1+\eta (\mu \xi _{2}^{4}+\kappa \xi _{1}^{4})}{\mu \xi _{2}^{4}\xi _{1}^{-2}+\kappa \xi _{1}^{2}+\eta |\xi |^{2}} \leq - \frac {1+\eta (\mu \xi _{2}^{4}+\kappa \xi _{1}^{4})}{\mu \xi _{2}^{2}+\kappa \xi _{1}^{2}+\eta |\xi |^{2}} \\ &\leq - \frac {1+\eta (\mu \xi _{2}^{4}+\kappa \xi _{1}^{4})}{(\mu +\kappa +2\eta )\xi _{1}^{2}} \leq -\frac {\eta \kappa}{\mu +\kappa +2\eta}\xi _{1}^{2}. \end{aligned}$$

For $\xi \in S_{22}$, one has

$$\begin{aligned} \lambda _{2}\leq - \frac {\xi _{1}^{2}+\eta \xi _{1}^{2}(\mu \xi _{2}^{4}+\kappa \xi _{1}^{4})}{\mu \xi _{2}^{4}+\kappa \xi _{1}^{4}+\eta \xi _{1}^{2}|\xi |^{2}} \le - \frac {\xi _{1}^{2}+\eta \xi _{1}^{2}(\mu \xi _{2}^{4}+\kappa \xi _{1}^{4})}{(\mu +\kappa +2\eta )\xi _{2}^{4}} \le -\frac {\eta \mu}{\mu +\kappa +2\eta}\xi _{1}^{2}. \end{aligned}$$

Let $c_{0}=\min \{\frac {\eta \kappa}{\mu +\kappa +2\eta}, \frac {\eta \mu}{\mu +\kappa +2\eta}\}$. Then we have

$$\begin{aligned} \lambda _{2}\leq -c_{0}\xi _{1}^{2},\quad \text{when} \quad \xi \in S_{2}. \end{aligned}$$

Thanks to $\xi \in S_{2}$, we have

$$\begin{aligned} \lambda _{2}-\lambda _{1}=\sqrt{\Gamma}>\frac{1}{2}\bigg( \frac {\mu \xi _{2}^{4}+\kappa \xi _{1}^{4}}{|\xi |^{2}}+\eta \xi _{1}^{2} \bigg). \end{aligned}$$

Consequently, we can easily obtain the upper bounds for $\widehat{G}_{1}(\xi ,t)$ and $\widehat{G}_{2}(\xi ,t)$ where $c=\min \{\frac{3}{4}\eta ,c_{0}\}$. This completes the proof of Lemma 4.2. □

Now we are ready to prove Theorem 1.3 according to Lemma 4.1 and Lemma 4.2.

Proof

Applying Lemma 4.1 to (1.14) leads to

$$ \left \{ \begin{aligned} &u(x,t)=G_{1}\bigg(\partial _{t}u(x,0)-\frac{1}{2}(\Delta ^{-1}(\mu \partial _{2}^{4}+\kappa \partial _{1}^{4})+\eta \partial _{11})u_{0} \bigg)+G_{2}u_{0}, \\ &\theta (x,t)=G_{1}\bigg(\partial _{t}\theta (x,0)-\frac{1}{2}(\Delta ^{-1}( \mu \partial _{2}^{4}+\kappa \partial _{1}^{4})+\eta \partial _{11}) \theta _{0}\bigg)+G_{2}\theta _{0}. \end{aligned} \right . $$

(4.17)

Setting $t=0$ in the linearized equations (1.9), we get

$$ \left \{ \begin{aligned} \partial _{t}&u_{1}(x,0)=\Delta ^{-1}(\mu \partial _{2}^{4}+\kappa \partial _{1}^{4})u_{10}-\partial _{1}\partial _{2}\Delta ^{-1}\theta _{0}, \\ \partial _{t}&u_{2}(x,0)=\Delta ^{-1}(\mu \partial _{2}^{4}+\kappa \partial _{1}^{4})u_{20}+\partial _{1}\partial _{1}\Delta ^{-1}\theta _{0}, \\ \partial _{t}&\theta (x,0)=\eta \partial _{11}\theta _{0}-u_{20}. \end{aligned} \right . $$

(4.18)

Then, inserting (4.18) into (4.17) yields

$$ \left \{ \begin{aligned} &u_{1}(x,t)=\frac{1}{2}G_{1}\bigg(\Delta ^{-1}(\mu \partial _{2}^{4}+ \kappa \partial _{1}^{4})-\eta \partial _{11}\bigg)u_{10}-\partial _{1} \partial _{2}\Delta ^{-1}G_{1}\theta _{0}+G_{2}u_{10}, \\ &u_{2}(x,t)=\frac{1}{2}G_{1}\bigg(\Delta ^{-1}(\mu \partial _{2}^{4}+ \kappa \partial _{1}^{4})-\eta \partial _{11}\bigg)u_{20}+\partial _{1} \partial _{1}\Delta ^{-1}G_{1}\theta _{0}+G_{2}u_{20}, \\ &\theta (x,t)=-\frac{1}{2}G_{1}\bigg(\Delta ^{-1}(\mu \partial _{2}^{4}+ \kappa \partial _{1}^{4})-\eta \partial _{11}\bigg)\theta _{0}+G_{2} \theta _{0}-G_{1}u_{20}. \end{aligned} \right . $$

(4.19)

(i) To estimate $\|u_{1}\|_{L^{2}}$, by Plancherel’s Theorem and dividing the spatial domain $\mathbf {R}^{2}$ as in Lemma 4.2, we have

$$\begin{aligned} \|u_{1}\|_{L^{2}}=\|\widehat{u}_{1}\|_{L^{2}}\leq &\frac{1}{2}\|\bigg( \frac {\mu \xi _{2}^{4}+\kappa \xi _{1}^{4}}{|\xi |^{2}}-\eta \xi _{1}^{2} \bigg)\widehat{G}_{1}\widehat{u}_{10}\|_{L^{2}(S_{1})} +\| \frac {\xi _{1}\xi _{2}}{|\xi |^{2}}\widehat{G}_{1}\widehat{\theta}_{0} \|_{L^{2}(S_{1})} \\ &+\|\widehat{G}_{2}\widehat{u}_{10}\|_{L^{2}(S_{1})}+\frac{1}{2}\|\bigg( \frac {\mu \xi _{2}^{4}+\kappa \xi _{1}^{4}}{|\xi |^{2}}-\eta \xi _{1}^{2} \bigg)\widehat{G}_{1}\widehat{u}_{10}\|_{L^{2}(S_{2})} \\ &+\|\frac {\xi _{1}\xi _{2}}{|\xi |^{2}}\widehat{G}_{1} \widehat{\theta}_{0}\|_{L^{2}(S_{2})}+\|\widehat{G}_{2}\widehat{u}_{10} \|_{L^{2}(S_{2})} \\ =&K_{1}+K_{2}+K_{3}+K_{4}+K_{5}+K_{6}. \end{aligned}$$

Thanks to (4.15) and the fact that $x^{n}e^{-x}\leq C(n)$ for any $n\geq 0$ and $x\geq 0$.

$$\begin{array}{rl}{K}_1\le & {\parallel (\frac{\mu {\xi }_2^4+\kappa {\xi }_1^4}{{|\xi |}^2}+\eta {\xi }_1^2){\mathit{te}}^{-\frac{1}{4}(\frac{\mu {\xi }_2^4+\kappa {\xi }_1^4}{{|\xi |}^2}+\eta {\xi }_1^2)t}{\hat{u}}_{10}\parallel }_{L^2(S_1)}\\ \le & C{\parallel e^{-\frac{1}{8}(\frac{\mu {\xi }_2^4+\kappa {\xi }_1^4}{{|\xi |}^2}+\eta {\xi }_1^2)t}{\hat{u}}_{10}\parallel }_{L^2}\\ \le & C{\parallel {|{\xi }_1|}^{\sigma }{e}^{-\frac{\eta }{8}{\xi }_1^{2}t}{|{\xi }_1|}^{-\sigma }{\hat{u}}_{10}\parallel }_{L^2}\\ \le & {\mathit{Ct}}^{-\frac{\sigma }{2}}{\parallel {\mathrm{\Lambda }}_1^{-\sigma }{u}_{10}\parallel }_{L^2},\end{array}$$

where $\sigma >0$. By $L^{p}$-boundedness of the Riesz transform and (4.15), we get

$$\begin{array}{rl}{K}_2\le & {\parallel {\hat{G}}_1{\hat{\theta }}_0\parallel }_{L^2(S_1)}\\ \le & {\parallel {\mathit{te}}^{-\frac{1}{4}(\frac{\mu {\xi }_2^4+\kappa {\xi }_1^4}{{|\xi |}^2}+\eta {\xi }_1^2)t}{\hat{\theta }}_0\parallel }_{L^2}\\ \le & {\parallel {\mathit{te}}^{-\frac{\eta }{4}{\xi }_1^{2}t}{\hat{\theta }}_0\parallel }_{L^2}\\ \le & C{\parallel e^{-\frac{\eta }{8}{\xi }_1^{2}t}{\xi }_1^{-2}{\hat{\theta }}_0\parallel }_{L^2}\\ \le & C{\parallel {|{\xi }_1|}^{\sigma }{e}^{-\frac{\eta }{8}{\xi }_1^{2}t}{\xi }_1^{-(\sigma +2)}{\hat{\theta }}_0\parallel }_{L^2}\\ \le & {\mathit{Ct}}^{-\frac{\sigma }{2}}{\parallel {\mathrm{\Lambda }}_1^{-(\sigma +2)}{\theta }_0\parallel }_{L^2}.\end{array}$$

From (4.15), one can follow that

$$\begin{array}{rl}{K}_3\le & C{\parallel e^{-\frac{\eta }{4}{\xi }_1^{2}t}{\hat{u}}_{10}\parallel }_{L^2}\\ \le & {\mathit{Ct}}^{-\frac{\sigma }{2}}{\parallel {\mathrm{\Lambda }}_1^{-\sigma }{u}_{10}\parallel }_{L^2}.\end{array}$$

Similarly, due to (4.16), one gets

$$\begin{array}{rl}{K}_4\le & C{\parallel (\frac{\mu {\xi }_2^4+\kappa {\xi }_1^4}{{|\xi |}^2}-\eta {\xi }_1^2)\frac{1}{\frac{\mu {\xi }_2^4+\kappa {\xi }_1^4}{{|\xi |}^2}+\eta {\xi }_1^2}(e^{-\frac{3}{4}(\frac{\mu {\xi }_2^4+\kappa {\xi }_1^4}{{|\xi |}^2}+\eta {\xi }_1^2)t}+e^{-c_0{\xi }_1^{2}t}){\hat{u}}_{10}\parallel }_{L^2},\\ \le & C{\parallel e^{-c{\xi }_1^{2}t}{\hat{u}}_{10}\parallel }_{L^2}\\ \le & {\mathit{Ct}}^{-\frac{\sigma }{2}}{\parallel {\mathrm{\Lambda }}_1^{-\sigma }{u}_{10}\parallel }_{L^2},\end{array}$$

and

$$\begin{array}{rl}{K}_5\le & C{\parallel \frac{1}{\frac{\mu {\xi }_2^4+\kappa {\xi }_1^4}{{|\xi |}^2}+\eta {\xi }_1^2}(e^{-\frac{3}{4}(\frac{\mu {\xi }_2^4+\kappa {\xi }_1^4}{{|\xi |}^2}+\eta {\xi }_1^2)t}+e^{-c_0{\xi }_1^{2}t}){\hat{\theta }}_0\parallel }_{L^2},\\ \le & C{\parallel {|{\xi }_1|}^{-2}{e}^{-c{\xi }_1^{2}t}{\hat{\theta }}_0\parallel }_{L^2}\\ \le & {\mathit{Ct}}^{-\frac{\sigma }{2}}{\parallel {\mathrm{\Lambda }}_1^{-(\sigma +2)}{\theta }_0\parallel }_{L^2}.\end{array}$$

The estimates for $K_{6}$ are similar to those for $K_{4}$ and the bound is

$$\begin{array}{rl}{K}_6\le & C{\parallel e^{-c{\xi }_1^{2}t}{\hat{u}}_{10}\parallel }_{L^2}\\ \le & {\mathit{Ct}}^{-\frac{\sigma }{2}}{\parallel {\mathrm{\Lambda }}_1^{-\sigma }{u}_{10}\parallel }_{L^2}.\end{array}$$

Combining the estimates from $K_{1}$ and $K_{6}$, we can obtain

$${\parallel u_1\parallel }_{L^2}\le {\mathit{Ct}}^{-\frac{\sigma }{2}}({\parallel {\mathrm{\Lambda }}_1^{-\sigma }{u}_{10}\parallel }_{L^2}+{\parallel {\mathrm{\Lambda }}_1^{-(\sigma +2)}{\theta }_0\parallel }_{L^2}).$$

Similarly,

$${\parallel u_2\parallel }_{L^2}\le {\mathit{Ct}}^{-\frac{\sigma }{2}}({\parallel {\mathrm{\Lambda }}_1^{-\sigma }{u}_{20}\parallel }_{L^2}+{\parallel {\mathrm{\Lambda }}_1^{-(\sigma +2)}{\theta }_0\parallel }_{L^2}),$$

and

$${\parallel \theta \parallel }_{L^2}\le {\mathit{Ct}}^{-\frac{\sigma }{2}}{\parallel {\mathrm{\Lambda }}_1^{-\sigma }{\theta }_0\parallel }_{L^2}.$$

(ii) The bound for $\|\partial _{1}^{m}u\|_{L^{2}}$ and $\|\partial _{1}^{m}\theta \|_{L^{2}}$ are similar to case (i). This completes the proof of Theorem 1.3. □