## Asymptotic Behavior for the Damped Boussinesq

Cai Xiaojing,, Zhou Yanjie

School of Mathematics and Statistics, Beijing Technology and Business University, Beijing 100048

 基金资助: 国家重点研发计划课题.  2017YFC1500301国家自然基金联合基金重点项目合作项目.  U1839206北京工商大学科研启动基金.  19008020161

 Fund supported: the National Key Research and Development Project.  2017YFC1500301the Joint Key Project of the NSFC and the China Earthquake Administration.  U1839206the Research Foundation for Advanced Talents of Beijing Technology and Business University.  19008020161

Abstract

This paper mainly focus on the asymptotic behavior of the solutions for the 3D Boussinesq equations with damping term. We obtain the explicit time decay rate of the solutions by means of the Fourier splitting method. In addition, we get the upper-bound of the time decay results compared with the heat equation.

Keywords： Boussinesq Equations ; Damping term ; Fourier Splitting Method ; Decay

Cai Xiaojing, Zhou Yanjie. Asymptotic Behavior for the Damped Boussinesq. Acta Mathematica Scientia[J], 2021, 41(5): 1415-1427 doi:

## 1 引言

$$$\left\{\begin{array}{llll} \partial_{t}u-\mu\Delta u+u\cdot\nabla u+\gamma_1|u|^{\alpha-1}u+\nabla p = \theta f, {\quad} & t\geq 0, x\in {{\Bbb R}} ^3, \\ \partial_{t}\theta-\nu\Delta\theta+u\cdot\nabla \theta+\gamma_2|\theta|^{\beta-1}\theta = 0, & t\geq 0, x\in {{\Bbb R}} ^3, \\ {\rm {div}}\; u = 0, & t\geq 0, x\in {{\Bbb R}} ^3, \end{array}\right.$$$

$$$u\mid_{t = 0} = u_0(x), \; \; \theta\mid_{t = 0} = \theta_0(x), \; \; \; \; x\in {{\Bbb R}} ^3,$$$

## 2 温度$\theta$的衰减性

$$$\|\theta(t)\|_2^2+2\int_0^t\|\nabla\theta(s)\|_2^2{\rm d}s+2\int_0^t\|\theta(s)\|_{\beta+1}^{\beta+1}{\rm d}s = \|\theta_0\|_2^2,$$$

$$$\|u(t)\|_2^2+2\int_0^t\|\nabla u(s)\|_2^2{\rm d}s+2\int_0^t\|u(s)\|_{\alpha+1}^{\alpha+1}{\rm d}s\leq \|u_0\|_2^2+2\int_0^t\int_{{{\Bbb R}} ^3}|\theta fu|{\rm d}x{\rm d}s.$$$

在方程组$(1.1)$的第二式两端乘以$\theta$并在${{\Bbb R}} ^3$上积分, 应用Plancherel定理, 有

$$$\frac{\rm d}{{\rm d}t}\int_{{{\Bbb R}} ^3}|\hat{\theta}|^2{\rm d}\xi+2\int_{{{\Bbb R}} ^3}|\theta|^{\beta+1}{\rm d}x = -2\int_{{{\Bbb R}} ^3}|\xi|^2|\hat{\theta}|^2{\rm d}\xi.$$$

$Q(t)$是球心在原点且半径为$r(t)$的球, 即

$$$Q(t) = \bigg\{\xi: |\xi|\leq r(t) = \bigg[\frac{3}{2(t+1)}\bigg]^{\frac{1}{2}}\bigg\},$$$

$\begin{eqnarray} && \frac{\rm d}{{\rm d}t}\int_{{{\Bbb R}} ^3}|\hat{\theta}|^2{\rm d}\xi+2\int_{{{\Bbb R}} ^3}|\theta|^{\beta+1}{\rm d}x{}\\ & = &-2 \int_{{{\Bbb R}} ^3}|\xi|^2|\hat{\theta}|^2{\rm d}\xi \leq -\frac{3}{t+1}\int_{{{\Bbb R}} ^3}|\xi|^2|\hat{\theta}|^2{\rm d}\xi+\int_{Q(t)} \bigg[\frac{3}{t+1}-2|\xi|^2\bigg]|\hat{\theta}|^2{\rm d}\xi, \end{eqnarray}$

$$$\frac{\rm d}{{\rm d}t}[(t+1)^3\int_{{{\Bbb R}} ^3}|\hat{\theta}|^2{\rm d}\xi]\leq 3(t+1)^2\int_{Q(t)}|\hat{\theta}|^2{\rm d}\xi.$$$

$$$\hat{\theta}(\xi, t) = \hat{\theta}(\xi, 0)e^{-|\xi|^{2}t}-\int_{0}^{t}e^{-|\xi|^{2}(t-s)}A(\xi, s)\; {\rm d}s.$$$

$$$|\widehat{u\cdot \nabla \theta }| = |\int_{{{\Bbb R}}^{3}}{\rm {div}}(u\theta)\exp(-{\rm i}\xi x)\; {\rm d}x|\leq\sum\limits_i\int_{{{\Bbb R}}^{3}}|u_{i}\theta||\xi_{i}|\; {\rm d}x \leq C|\xi|\|\theta\|_2\|u\|_2,$$$

$$$|A(\xi, t)|\leq C|\xi|\|\theta\|_2\|u\|_2+C\|\theta\|_{\beta}^{\beta}.$$$

$$$\|\theta\|_{\beta}\leq C\|\theta\|_{2}^\eta\|\nabla\theta\|_{2}^{1-\eta},$$$

$$$|B(\xi, t)|\leq C|\xi|+C\|u\|_{\alpha+1}^{\frac{(\alpha-2)(\alpha+1)}{\alpha-1}}+C(t+1)^{-\mu}.$$$

$\alpha\geq3$时, 有

$$$\|u\|_{2}^{2}\leq C(t+1)^{-\frac{1}{2}}.$$$

(3.1) 式可以变形为

$$$\frac{\rm d}{{\rm d}t}\bigg[(t+1)^3\int_{{{\Bbb R}} ^3}|\hat{u}|^2{\rm d}\xi\bigg] \leq 3(t+1)^2\int_{Q(t)}|\hat{u}|^2{\rm d}\xi+C(t+1)^{\frac{14-3\beta}{4}}.$$$

i) $2\leq\alpha\leq6.$

$$$|B(\xi, t)|\leq C|\xi|+C\|\nabla u\|_{2}^{\frac{3}{2}\alpha-3}+C(t+1)^{-\mu}.$$$

$$$|\hat{u}(\xi, t)| \leq |\hat{u}_0|+C|\xi|^{-1}+C(t+1)^{1-\mu}+C|\xi|^{-\frac{10-3\alpha}{2}},$$$

$$$\|u\|_2^2\leq C(t+1)^{-\frac{1}{2}}+(t+1)^{-\frac{3\alpha-7}{2}}+(t+1)^{-\frac{3\beta-6}{4}}.$$$

ii) $\alpha\geq2. $$(3.5) 式有 $$|B(\xi, t)|\leq C|\xi|+C\|u\|_{\alpha+1}^{\frac{(\alpha-2)(\alpha+1)}{\alpha-1}}+C(t+1)^{-\mu}.$$ 重复上面的过程, 有如下结论成立 这里 \alpha>\frac{7}{3}. 如果 \frac{3\alpha-7}{2(\alpha-1)}\geq\frac{1}{2}, \; \frac{3\beta-6}{4}\leq\frac{1}{2},$$ \alpha\geq3, \; \frac{7}{3}<\beta\leq\frac{8}{3},$

## 4 解的上界估计

$$$\left\{\begin{array}{llll} \partial_{t}v-\Delta v = f, \; v\mid_{t = 0} = u_0(x), \\ \partial_{t}w-\Delta w = 0, \; w\mid_{t = 0} = \theta_0(x). \end{array}\right.$$$

(1) $\|\nabla v\|_\infty\leq C(t + 1)^{-\frac{n}{4}-\frac{\delta}{2}-\frac{1}{2}};$

(2) $\|v\|_\infty\leq C(t + 1)^{-\frac{n}{4}-\frac{\delta}{2}}.$

$$$\left\{\begin{array}{llll} \partial_{t}U-\Delta U+u\cdot\nabla u+|u|^{\alpha-1}u+\nabla p = \theta f-f, \\ \partial_{t}V-\Delta V+u\cdot\nabla \theta+|\theta|^{\beta-1}\theta = 0, \\ {\rm {div}}\; U = 0, \\ U\mid_{t = 0} = 0, \; \; V\mid_{t = 0} = 0. \end{array}\right.$$$

$\begin{eqnarray} &&\frac{1}{2}\frac{\rm d}{{\rm d}t}\|U\|_2^2+\|\nabla U\|_2^2{}\\ & = &-\int_{{{\Bbb R}} ^3}u\cdot\nabla u U{\rm d}x-\int_{{{\Bbb R}} ^3}|u|^{\alpha-1}uU{\rm d}x+\int_{{{\Bbb R}} ^3}(\theta-1) fU{\rm d}x{}\\ & = &\int_{{{\Bbb R}} ^3}u\cdot\nabla u v{\rm d}x-\int_{{{\Bbb R}} ^3}|u|^{\alpha+1}{\rm d}x+\int_{{{\Bbb R}} ^3}|u|^{\alpha-1}uv{\rm d}x+\int_{{{\Bbb R}} ^3}(\theta-1) fU{\rm d}x{}\\ &\leq&\|u\|_2^2\|\nabla v\|_\infty-\|u\|^{\alpha+1}_{\alpha+1}+\frac{1}{2}\|u\|^{\alpha+1}_{\alpha+1}+C\|v\|^{\alpha+1}_{\alpha+1}+\|U\|_2\|\theta-1\|_\infty\|f\|_2. \end{eqnarray}$

$(4.3)$式的第二个式子两端乘以$V$并在${{\Bbb R}} ^3$上积分, 有

$\begin{eqnarray} \frac{1}{2}\frac{\rm d}{{\rm d}t}\|V\|_2^2+\|\nabla V\|_2^2& = &-\int_{{{\Bbb R}} ^3}u\cdot\nabla \theta V{\rm d}x-\int_{{{\Bbb R}} ^3}|\theta|^{\beta-1}\theta V{\rm d}x{}\\ & = &\int_{{{\Bbb R}} ^3}u\cdot\nabla \theta w{\rm d}x-\int_{{{\Bbb R}} ^3}|\theta|^{\beta+1}{\rm d}x+\int_{{{\Bbb R}} ^3}|\theta|^{\beta-1}\theta w{\rm d}x{}\\ &\leq&\|u\|_2\|\theta\|_2\|\nabla w\|_\infty-\|\theta\|^{\beta+1}_{\beta+1}+\frac{1}{2}\|\theta\|^{\beta+1}_{\beta+1}+C\|w\|^{\beta+1}_{\beta+1}, \end{eqnarray}$

$(4.4)$式和$(4.5)$式相加有

$\begin{eqnarray} &&\frac{\rm d}{{\rm d}t}(\|U\|_2^2+\|V\|_2^2)+2(\|\nabla U\|_2^2+\|\nabla V\|_2^2)+\|u\|^{\alpha+1}_{\alpha+1}+\|\theta\|^{\beta+1}_{\beta+1}{}\\ &\leq&\|u\|_2^2\|\nabla v\|_\infty+\|u\|_2\|\theta\|_2\|\nabla w\|_\infty+C\|v\|^{\alpha+1}_{\alpha+1} +\|\theta-1\|_\infty\|f\|_2 \|U\|_2+C\|w\|^{\beta+1}_{\beta+1}{}\\ &\leq&(\|u\|_2^2+\|\theta\|_2^2)(\|\nabla v\|_\infty+\|\nabla w\|_\infty) +C(\|v\|^{\alpha+1}_{\alpha+1}+C\|w\|^{\beta+1}_{\beta+1})+\|\theta-1\|_\infty\|f\|_2 \|U\|_2.{\qquad} \end{eqnarray}$

$\begin{eqnarray} &&\frac{\rm d}{{\rm d}t}\{(\|U\|_2^2+\|V\|_2^2)G(t)\}{}\\ &\leq& 2G(t)[g^2(\|U\|_2^2+\|V\|_2^2)-(\|\nabla U\|_2^2+\|\nabla V\|_2^2)]{}\\ &&+G(t)(\|u\|_2^2+\|\theta\|_2^2)(\|\nabla v\|_\infty+\|\nabla w\|_\infty){}\\ &&+2 G(t)(\|v\|^{\alpha+1}_{\alpha+1}+G(t)\|w\|^{\beta+1}_{\beta+1})+G(t)\|\theta-1\|_\infty\|f\|_2 \|U\|_2, {\quad} \end{eqnarray}$

$\begin{eqnarray} &&g^2(t)(\|U\|_2^2+\|V\|_2^2)-(\|\nabla U\|_2^2+\|\nabla V\|_2^2){}\\ & = & \int_{{{\Bbb R}} ^3}(g^2(t)-|\xi|^2)(|\hat{U}|^2+|\hat{V}|^2){\rm d}\xi \leq \int_{|\xi|\leq g(t)}(g^2(t)-|\xi|^2)(|\hat{U}|^2+|\hat{V}|^2){\rm d}\xi{}\\ &\leq& g^2(t) \int_{|\xi|\leq g(t)}(|\hat{U}|^2+|\hat{V}|^2){\rm d}\xi, {\quad} \end{eqnarray}$

$\begin{eqnarray} &&\frac{\rm d}{{\rm d}t}\{G(t)\|U\|_2^2+\|V\|_2^2)\}{}\\ &\leq & 2g^2(t)G(t) \int_{|\xi|\leq g(t)}(|\hat{U}|^2+|\hat{V}|^2){\rm d}\xi +G(t)(\|u\|_2^2+\|\theta\|_2^2)(\|\nabla v\|_\infty+\|\nabla w\|_\infty){}\\ &&+2 G(t)(\|v\|^{\alpha+1}_{\alpha+1}+\|w\|^{\beta+1}_{\beta+1})+G(t)\|\theta-1\|_\infty\|f\|_2 \|U\|_2. \end{eqnarray}$

$\begin{eqnarray} |\hat{U}(\xi, t)|+|\hat{V}(\xi, t)|&\leq & C|\xi|\int_{0}^{t}(\|u\|_2^2+\|\theta\|_2^2){\rm d}s+\int_{0}^{t}(\|u\|_\alpha^\alpha+\|\theta\|_\beta^\beta){\rm d}s{}\\ &&+ C \|\theta_0\|_{\infty}\int_{0}^{t}(s+1)^{-\mu}{\rm d}s. \end{eqnarray}$

$$$|\hat{U}(\xi, t)|+|\hat{V}(\xi, t)|\leq C|\xi|\int_{0}^{t}(\|u\|_2^2+\|\theta\|_2^2){\rm d}s+C.$$$

$$$\varphi(t)\leq \varphi(0)+C\int_{0}^{t}\|f\|_2{\rm d}s = C.$$$

$\begin{eqnarray} \int_{|\xi|\leq g(t)}(|\hat{U}(\xi, t)|^2+|\hat{V}(\xi, t)|^2){\rm d}\xi &\leq & C\Phi^2(t)\int_{|\xi|\leq g(t)}|\xi|^2{\rm d}\xi+C\int_{|\xi|\leq g(t)}{\rm d}\xi{}\\ &\leq& C\Phi^2(t)g^5(t)+Cg^3(t), \end{eqnarray}$

$\begin{eqnarray} &&\frac{\rm d}{{\rm d}t}\{G(t)(\|U\|_2^2+\|V\|_2^2)\}{}\\ &\leq& G(t)[\Phi^2(t)g^7(t)+g^5(t)]+G(t)\varphi(t)(\|\nabla v\|_\infty+\|\nabla w\|_\infty){}\\ &&+2 G(t)(\|v\|^{\alpha+1}_{\alpha+1}+\|w\|^{\beta+1}_{\beta+1})+G(t)\|\theta-1\|_\infty\|f\|_2 \|U\|_2. \end{eqnarray}$

$\begin{eqnarray} &&\frac{\rm d}{{\rm d}t}\{G(t)(\|U\|_2^2+\|V\|_2^2)\}{}\\ &\leq& G(t)[\Phi^2(t)g^7(t)+g^5(t)+\varphi(t)(t+1)^{-\frac{3}{4}-\frac{\delta}{2}-\frac{1}{2}}{}\\ &&+(\|v\|_\infty^{\alpha-1}\|v\|^2_{2}+\|w\|_\infty^{\beta-1}\|w\|^2_{2})+(t+1)^{-\mu}]{}\\ & \leq & G(t)[\Phi^2(t)g^7(t)+g^5(t)]+\varphi(t)(t+1)^{-\frac{3}{4}-\frac{\delta}{2}-\frac{1}{2}}{}\\ &&+[(t+1)^{-(\frac{3}{4}-\frac{\delta}{2})(\alpha-1)-\frac{\delta}{2}}+(t+1)^{-(\frac{3}{4}-\frac{\delta}{2})(\beta-1)-\frac{\delta}{2}} +(t+1)^{-\mu}]. \end{eqnarray}$

$$$G(t) = e^{2\int_{0}^{t}g^2(s){\rm d}s} = (t+1)^{\gamma}.$$$

$\begin{eqnarray} &&(t+1)^{\gamma}(\|U\|_2^2+\|V\|_2^2){}\\ &\leq & C(\|U(1)\|_2^2+\|V(1)\|_2^2)+(t+1)^{-\frac{3}{4}-\frac{\delta}{2}-\frac{1}{2}+\gamma}\int_1^t\varphi(t){\rm d}s{}\\ &&+C\Phi^2(t)\int_1^t(s+1)^{-\frac{7}{2}}(s+1)^{\gamma}{\rm d}s{}\\ &&+\int_1^t(s+1)^{-\frac{5}{2}}(s+1)^{\gamma}{\rm d}s+\int_1^t(s+1)^{\gamma}(s+1)^{-(\frac{3}{4}-\frac{\delta}{2})(\alpha-1)-\frac{\delta}{2}}{\rm d}s{}\\ &&+\int_1^t(s+1)^{\gamma}(s+1)^{-(\frac{3}{4}-\frac{\delta}{2})(\beta-1)-\frac{\delta}{2}}{\rm d}s +\int_1^t(s+1)^{\gamma}(s+1)^{-\mu}{\rm d}s. \end{eqnarray}$

$\begin{eqnarray} \|U\|_2^2+\|V\|_2^2&\leq& C(t+1)^{-\gamma}+C\Phi^2(t)(t+1)^{-\frac{5}{2}}+(t+1)^{-\frac{3}{4}-\frac{\delta}{2}-\frac{1}{2}}\Phi(t){}\\ &&+(t+1)^{-\frac{3}{2}}+(t+1)^{-(\frac{3}{4}-\frac{\delta}{2})(\alpha-1)-\frac{\delta}{2}+1}{}\\ &&+(t+1)^{-(\frac{3}{4}-\frac{\delta}{2})(\beta-1)-\frac{\delta}{2}+1} +(t+1)^{-\mu+1}. \end{eqnarray}$

$\begin{eqnarray} \varphi(t)&\leq &C(t + 1)^{-\delta}+C\Phi^2(t)(t+1)^{-\frac{5}{2}}+(t+1)^{-\frac{3}{2}}+(t+1)^{-\mu+1}{}\\ &&+(t+1)^{-(\frac{3}{4}-\frac{\delta}{2})(\alpha-1)-\frac{\delta}{2}+1}+(t+1)^{-(\frac{3}{4}-\frac{\delta}{2})(\beta-1)-\frac{\delta}{2}+1}, \end{eqnarray}$

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