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数学物理学报, 2021, 41(6): 1805-1815 doi:

论文

分数阶扩散的MHD-Boussinesq系统的全局正则性

杨静,1, 邓雪梅,1,2, 周艳平,1,2

1 三峡大学理学院 湖北宜昌 443002

2 三峡大学数学研究中心 湖北宜昌 443002

Global Regularity for the MHD-Boussinesq System with Fractional Diffusion

Yang Jing,1, Deng Xuemei,1,2, Zhou Yanping,1,2

1 College of Science, China Three Gorges University, Hubei Yichang 443002

2 Three Gorges Mathematical Research Center, China Three Gorges University, Hubei Yichang 443002

通讯作者: 周艳平, zhyp5208@163.com

收稿日期: 2021-03-12  

基金资助: 国家自然科学基金.  11901346
国家自然科学基金.  11871305
三峡大学学位论文培优基金.  2021SSPY148

Received: 2021-03-12  

Fund supported: the NSFC.  11901346
the NSFC.  11871305
the Research Fund for Excellent Dissertation of China Three Gorges University.  2021SSPY148

作者简介 About authors

杨静,E-mail:1972639378@qq.com , E-mail:1972639378@qq.com

邓雪梅,E-mail:dxmeisx@126.com , E-mail:dxmeisx@126.com

Abstract

In this paper, we investigate the n-dimensional (n2) Magnetohydrodynamics-Boussinesq system with fractional diffusion. When the nonnegative constants α,β and γ satisfy α12+n4, α+β1+n2 and α+γn2, by using the energy methods, we obtain the global existence and uniqueness of solution for the system, which generalizes the existing result.

Keywords: MHD-Boussinesq system ; Global regularity ; Fractional diffusion

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本文引用格式

杨静, 邓雪梅, 周艳平. 分数阶扩散的MHD-Boussinesq系统的全局正则性. 数学物理学报[J], 2021, 41(6): 1805-1815 doi:

Yang Jing, Deng Xuemei, Zhou Yanping. Global Regularity for the MHD-Boussinesq System with Fractional Diffusion. Acta Mathematica Scientia[J], 2021, 41(6): 1805-1815 doi:

1 引言

本文研究具分数阶扩散Magnetohydrodynamics-Boussinesq(MHD-Boussinesq)系统

{ut+uu+p+Λ2αu=hh+ρen,ht+uhhu+Λ2βh=0,ρt+uρ+Λ2γρ=0,divu=divh=0,(u,h,ρ)(x,0)=(u0,h0,ρ0),
(1.1)

其中(x,t)Rn×R+ (n2), u=u(x,t)h=h(x,t)分别表示速度场和磁场, p=p(x,t)ρ=ρ(x,t)分别表示压力和温度. α0, β0γ0是实参数, 并且en=(0,,1).通过傅里叶变换定义分数阶拉普拉斯算子Λ=(Δ)12, 如下:

^Λαf(ξ)=|ξ|αˆf(ξ),  α0.

当流体不受温度的影响, 即ρ=0时, 系统(1.1)退化为广义的MHD方程.关于广义MHD方程的适定性, 已有许多研究和成果.对于二维的情形, Tran等[1]证明了在三种情况下有全局光滑解:α12, β1; 0α<12, 2α+β>2; α2, β=0.0α<12, β13α+2β>3的条件下, Jiu-Zhao[2]得到了一个全局正则解.对于三维情形, 当α=β=1时, Xu等[3]研究了局部经典解的正则性准则; Wang-Wang[4]证明了在空间χ1中Cauchy问题全局解的存在性.对于n维情形, Wu[5]证明了只要满足条件α12+n4β12+n4, 则解具有全局正则性.

当流体不受洛伦兹力的影响, 即h=0时, 系统(1.1)退化为Boussinesq方程.为了研究Boussinesq方程的适定性, 许多学者做了大量的工作.对于二维情形, Hou-Li[6]证明了在α=1, γ=0的情况下解的全局正则性. Chae[7]α=1, γ=0α=0, γ=1情况下也得到了相同的结果.对于三维情形, 当α12+n4, γ=0时, Ye[8]和Yamazaki-Pullman[9]证明了全局解的存在性.

关于MHD-Boussinesq方程解的全局适定性, 最近也有一些研究.对于二维情形, 可参见文献[10, 11].对于三维情形, 在三种情况下: α=β=γ=1; α=β=1, γ=0; α=β=γ=0, Larios - Pei[12]建立了在Sobolev空间中解的局部适定性. Liu等[13]证明了动量方程中带有非线性阻尼项的强解的全局适定性.对于更多分数阶扩散的流体动力学方程的正则性结果, 可参见文献[14-20].

2 主要结果

我们的主要目标是在α12+n4, α+β1+n2α+γn2情况下, 利用系统(1.1)的特殊结构和能量估计方法, 研究解的全局存在性和唯一性.

该文的主要结果陈述如下.

定理2.1  假设初值满足(u0,h0,ρ0)Hs(Rn)divu0=divh0=0, 其中s1, n2.

α12+n4, α+β1+n2, α+γn2,
(2.1)

则系统(1.1)存在唯一全局解(u,h,ρ), 满足对任意T>0, 有

(u,h,ρ)L(0,T;Hs(Rn)), uL2(0,T;Hs+α(Rn)),hL2(0,T;Hs+β(Rn)), ρL2(0,T;Hs+γ(Rn)).
(2.2)

注2.1  当ρ=0时, 系统(1.1)退化为广义MHD方程, Wu[5]证明了α12+n4,β12+n4时并且初值满足(u0,h0)Hs(smax, 则解具有全局正则性.这里, 假设 (u_0, h_0) \in H^s (s\geq1) , 当 \alpha\geq\frac{1}{2}+\frac{n}{4}, \ \alpha+\beta\geq 1+\frac{n}{2} 时, 建立了广义MHD方程的全局适定性, 推广了Wu[5]的结果.

3 主要结果的证明

本节将证明定理2.1, 并将证明分为两部分:全局存在性和唯一性.为此, 首先建立解的先验估计 \|(u, h, \rho)\|_{H^s} \ (s\geq1) .

命题3.1  假设 (u, h, \rho) 是系统(1.1)的解, 则对 s\geq1 t\in [0, T] , 有

\begin{eqnarray} &&\|\Lambda^s{u(t)}\|^2_{L^2} + \|\Lambda^s{h(t)}\|^2_{L^2} + \|\Lambda^s{\rho(t)}\|^2_{L^2} + \int_{0}^{t}(\|\Lambda^{\alpha+s}u(\tau)\|^2_{L^2} +\|\Lambda^{\beta+s}h(\tau)\|^2_{L^2} {}\\ &&+\|\Lambda^{\gamma+s}\rho(\tau)\|^2_{L^2}){\rm d}\tau \leq C. \end{eqnarray}
(3.1)

  证明将分为三个步骤: H^0 估计, H^1 估计及 H^s \ (s>1) 估计.

步骤1 ( H^0 估计)  分别将(1.1) _1 乘以 u , (1.1) _2 乘以 h , (1.1) _3 乘以 \rho , 然后在 {\Bbb R} ^ n 中积分, 并且考虑到 {\rm div} u = {\rm div} h = 0 , 可以得到

\begin{eqnarray} &&\frac{1}{2}\frac{\rm d}{{\rm d}t}(\|u\|^2_{L^2}+\|h\|^2_{L^2}+\|\rho\|^2_{L^2}) + \|\Lambda^{\alpha}u\|^2_{L^2}+\|\Lambda^{\beta}h\|^2_{L^2}+\|\Lambda^{\gamma}\rho\|^2_{L^2} \\ & = &\int_{{\Bbb R} ^ n}h\cdot\nabla h\cdot u \ {\rm d}x+\int_{{\Bbb R} ^ n}\rho e_n\cdot u \ {\rm d}x+\int_{{\Bbb R} ^ n} h\cdot\nabla u\cdot h \ {\rm d}x. \end{eqnarray}
(3.2)

先将第一项和第三项一起处理, 得

\begin{eqnarray} &&\int_{{\Bbb R} ^ n}h\cdot\nabla h\cdot u\ {\rm d}x+\int_{{\Bbb R} ^ n} h\cdot\nabla u\cdot h\ {\rm d}x\\ & = &\int_{{\Bbb R} ^ n}h\cdot\nabla h\cdot u\ {\rm d}x-\int_{{\Bbb R} ^ n}{\rm div}(h\otimes h)\cdot u\ {\rm d}x \\ & = &\int_{{\Bbb R} ^ n}h\cdot\nabla h\cdot u \ {\rm d}x-\int_{{\Bbb R} ^ n}h\cdot\nabla h\cdot u \ {\rm d}x = 0. \end{eqnarray}
(3.3)

注意到

\begin{equation} \int_{{\Bbb R} ^ n}\rho e_n\cdot u\ {\rm d}x \leq \|\rho\|_{L^2}\|u\|_{L^2}\leq\frac{1}{2}\|\rho\|^2_{L^2}+\frac{1}{2}\|u\|^2_{L^2}. \end{equation}
(3.4)

将(3.3)式和(3.4)式代入(3.2)式, 推出

\begin{eqnarray} && \frac{1}{2}\frac{\rm d}{{\rm d}t}(\|u\|^2_{L^2}+\|h\|^2_{L^2}+\|\rho\|^2_{L^2})+(\|\Lambda^{\alpha}u\|^2_{L^2}+\|\Lambda^{\beta}h\|^2_{L^2} +\|\Lambda^{\gamma}\rho\|^2_{L^2}){}\\ &\leq&\frac{1}{2}(\|\rho\|^2_{L^2}+\|u\|^2_{L^2}). \end{eqnarray}
(3.5)

由(3.5)式和Gronwall不等式, 得

\begin{eqnarray} &&\|u(t)\|^2_{L^2}+\|h(t)\|^2_{L^2}+\|\rho(t)\|^2_{L^2}{}\\ &&+2\int_{0}^{t}(\|\Lambda^{\alpha}u(\tau)\|^2_{L^2} +\|\Lambda^{\beta}h(\tau)\|^2_{L^2}+\|\Lambda^{\gamma}\rho(\tau)\|^2_{L^2}) {\rm d}\tau\leq C. \end{eqnarray}
(3.6)

步骤2 ( H^1 估计)  分别将(1.1) _1 乘以 \Delta u , (1.1) _2 乘以 \Delta h , 然后在 {\Bbb R} ^ n 中积分, 可以得到

\begin{eqnarray} &&\frac{1}{2}\frac{\rm d}{{\rm d}t}(\|\nabla u\|^2_{L^2}+\|\nabla h\|^2_{L^2})+\|\Lambda^{\alpha+1}u\|^2_{L^2}+\|\Lambda^{\beta+1}h\|^2_{L^2}\\ & = &-\int_{{\Bbb R} ^ n}\partial_k u^i\partial_i u^j\partial_k u^j\ {\rm d}x-\int_{{\Bbb R} ^ n}(h\cdot\nabla h)\cdot\Delta u\ {\rm d}x-\int_{{\Bbb R} ^ n}\rho e_n\cdot\Delta u \ {\rm d}x\\ &&-\int_{{\Bbb R} ^ n}\partial_k u^i\partial_i h^j\partial_k h^j\ {\rm d}x-\int_{{\Bbb R} ^ n}(h\cdot\nabla u)\cdot\Delta h\ {\rm d}x\\ &: = &I_1+I_2+I_3+I_4+I_5. \end{eqnarray}
(3.7)

为了估计 I_1 , 需要用到Gagliardo-Nirenberg不等式:

\begin{eqnarray} \|\nabla u\|_{L^3}\leq C\|\nabla u\|^a_{L^2}\|\Lambda^\alpha u\|^b_{L^2}\|\Lambda^{\alpha+1} u\|^c_{L^2}, \end{eqnarray}
(3.8)

其中

a = 1-\frac{1}{3\alpha}\big(1+\frac{n}{2}\big), \ b = \frac{1}{3}, \ c = \frac{1}{3\alpha}\big(1-\alpha+\frac{n}{2}\big).

由(3.8)式, Hölder和Young不等式, 得

\begin{eqnarray} I_1&\leq& \|\nabla u\|^3_{L^3}\leq C\|\nabla u\|^{3a}_{L^2}\|\Lambda^\alpha u\|^{3b}_{L^2}\|\Lambda^{\alpha+1}u\|^{3c}_{L^2}\\ &\leq&\varepsilon\|\Lambda^{\alpha+1}u\|^2_{L^2}+C\|\Lambda^\alpha u\|^{\frac{4\alpha}{6\alpha-2-n}}_{L^2}\|\nabla u\|^2_{L^2}. \end{eqnarray}
(3.9)

下面在 \beta<\frac{n}{2} \beta>\frac{n}{2} 两种情况下估计 I_4 .注意到 \beta = \frac{n}{2} 的情况可以通过用 \beta<\frac{n}{2} \beta>\frac{n}{2} 的结果进行插值得到(见下面(3.15)式).后面类似的情况可以同样处理.

\beta<\frac{n}{2} 时, 注意到 \alpha+\beta\geq 1+\frac{n}{2} , 有

\begin{eqnarray} I_4&\leq& C\|\nabla h\|_{L^2}\|\nabla u\|_{L^{\frac{n}{\beta}}}\|\nabla h\|_{L^{\frac{2n}{n-2\beta}}}\\ &\leq& C\|\nabla h\|_{L^2}\|\Lambda^{\frac{n}{2}-\beta+1}u\|_{L^2}\|\Lambda^{\beta+1} h\|_{L^2}\\ &\leq&\varepsilon\|\Lambda^{\beta+1} h\|^2_{L^2}+C(\|u\|^2_{L^2}+\|\Lambda^{\alpha}u\|^2_{L^2})\|\nabla h\|^2_{L^2}. \end{eqnarray}
(3.10)

\beta>\frac{n}{2} 时, 有

\begin{eqnarray} I_4&\leq &C\|\nabla h\|_{L^2}\|\nabla u\|_{L^2}\|\nabla h\|_{L^{\infty}}\\&\leq & C\|\nabla h\|_{L^2}(\|u\|_{L^2}+\|\Lambda^{\alpha} u\|_{L^2})(\|h\|_{L^2}+\|\Lambda^{\beta+1}h\|_{L^2})\\&\leq &\varepsilon\|\Lambda^{\beta+1}h\|^2_{L^2}+C\|h\|^2_{L^2}+C(\|u\|^2_{L^2}+\|\Lambda^{\alpha}u\|^2_{L^2})\|\nabla h\|^2_{L^2}. \end{eqnarray}
(3.11)

对于 I_2+I_5 的估计和 I_4 类似.事实上, 因为 {\rm div} u = {\rm div} h = 0 , 有

\begin{eqnarray} I_2+I_5& = &-\int_{{\Bbb R} ^ n}(h^i\partial_i h^j)\partial_k\partial_ku^j\ {\rm d}x- \int_{{\Bbb R} ^ n}(h^i\partial_i u^j)\partial_k\partial_k h^j\ {\rm d}x\\ & = &\int_{{\Bbb R} ^ n}\partial_k(h^i\partial_i h^j)\partial_k u^j\ {\rm d}x+\int_{{\Bbb R} ^ n} \partial_k(h^i\partial_i u^j)\partial_k h^j\ {\rm d}x\\ & = &\int_{{\Bbb R} ^ n}\partial_k h^i\partial_i h^j\partial_k u^j\ {\rm d}x+\int_{{\Bbb R} ^ n}\partial_k h^i\partial_i u^j\partial_k h^j\ {\rm d}x\\ &\leq &\varepsilon\|\Lambda^{\beta+1}h\|^2_{L^2}+C\|h\|^2_{L^2}+C(\|u\|^2_{L^2}+\|\Lambda^{\alpha}u\|^2_{L^2})\|\nabla h\|^2_{L^2}. \end{eqnarray}
(3.12)

对于 I_3 , 若 \alpha<1 , 注意到 \alpha+\gamma\geq\frac{n}{2} , 有

\begin{eqnarray} I_3& = &-\int_{{\Bbb R} ^ n}\Lambda^{1-\alpha}(\rho e_n)\cdot\Lambda^{\alpha-1}\Delta u\ {\rm d}x\\ &\leq & C\|\Lambda^{1-\alpha}\rho\|_{L^2}\|\Lambda^{\alpha-1}\Delta u\|_{L^2} \\ {} & \leq& C(\|\rho\|_{L^2} +\|\Lambda^{\gamma}\rho\|_{L^2})\|\Lambda^{\alpha+1} u\|_{L^2}\\ &\leq &\varepsilon\|\Lambda^{\alpha+1} u\|^2_{L^2}+C(\|\rho\|^2_{L^2}+\|\Lambda^{\gamma}\rho\|^2_{L^2}). \end{eqnarray}
(3.13)

另一方面, 若 \alpha\geq1 , 有

\begin{eqnarray} I_3&\leq &C\|\rho\|_{L^2}\|\Delta u\|_{L^2} \leq C\|\rho\|_{L^2}\|u\|^{\frac{\alpha-1}{\alpha+1}} _{L^2}\|\Lambda^{\alpha+1}u\|^{\frac{2}{\alpha+1}}_{L^2}\\ &\leq&\varepsilon\|\Lambda^{\alpha+1}u\|^2_{L^2}+C(\|\rho\|^2_{L^2}+\|u\|^2_{L^2}). \end{eqnarray}
(3.14)

将(3.9)–(3.14)式代入(3.7)式, 得

\begin{eqnarray*} &&\frac{\rm d}{{\rm d}t}(\|\nabla u(t)\|^2_{L^2}+\|\nabla h(t)\|^2_{L^2})+\| \Lambda^{\alpha+1}u\|^2_{L^2}+\|\Lambda^{\beta+1}h\|^2_{L^2}\\ &\leq & C(\|\Lambda^{\alpha}u\|^{\frac{4\alpha}{6\alpha-2-n}}_{L^2}+\|\Lambda^{\alpha}u\|^2_{L^2}+\|u\|^2_{L^2}) (\|\nabla u\|^2_{L^2}+\|\nabla h\|^2_{L^2})\\ & &+C(\|u\|^2_{L^2}+\|h\|^2_{L^2}+\|\rho\|^2_{L^2}+\|\Lambda^{\gamma}\rho\|^2_{L^2}). \end{eqnarray*}

由上式, Gronwall不等式, \alpha\geq\frac{1}{2}+\frac{n}{4} 以及(3.6)式, 推出

\begin{eqnarray} \|\nabla u(t)\|^2_{L^2}+\|\nabla h(t)\|^2_{L^2}+\int_{0}^{t}(\|\Lambda^{\alpha+1}u(\tau)\|^2_{L^2}+\|\Lambda^{\beta+1}h(\tau)\|^2_{L^2}){\rm d}\tau\leq C. \end{eqnarray}
(3.15)

将(1.1) _3 式乘以 \Delta \rho , 在 {\Bbb R} ^ n 中积分, 得

\begin{eqnarray} \frac{1}{2}\frac{\rm d}{{\rm d}t}\|\nabla\rho\|^2_{L^2}+\|\Lambda^{\gamma+1}\rho\|^2_{L^2} = -\int_{{\Bbb R} ^ n}\partial_k u^i\partial_i \rho^j\partial_k\rho^j\ {\rm d}x: = M_1. \end{eqnarray}
(3.16)

由Hölder不等式, Sobolev嵌入定理, Young不等式和 \alpha+\gamma\geq\frac{n}{2} , 可得:若 \gamma<\frac{n}{2} , 则

\begin{eqnarray} M_1&\leq &C\|\nabla\rho\|_{L^2}\|\nabla u\|_{L^{\frac{n}{\gamma}}}\|\nabla\rho\|_{L^{\frac{2n}{n-2\gamma}}}\\ &\leq& C\|\nabla\rho\|_{L^2}\|\Lambda^{\frac{n}{2}-\gamma+1}u\|_{L^2}\|\Lambda^{\gamma+1}\rho\|_{L^2}\\ &\leq&\varepsilon\|\Lambda^{\gamma+1}\rho\|^2_{L^2}+C(\|u\|^2_{L^2}+\|\Lambda^{\alpha+1}u\|^2_{L^2})\|\nabla\rho\|^2_{L^2}. \end{eqnarray}
(3.17)

\gamma>\frac{n}{2} , 则有

\begin{eqnarray} M_1&\leq& C\|\nabla\rho\|_{L^2}\|\nabla u\|_{L^2}\|\nabla\rho\|_{L^{\infty}}\\&\leq &C\|\nabla\rho\|_{L^2}(\|u\|_{L^2}+\|\Lambda^{\alpha+1}u\|_{L^2})(\|\rho\|_{L^2}+\|\Lambda^{\gamma+1} \rho\|_{L^2})\\ &\leq&\varepsilon\|\Lambda^{\gamma+1}\rho\|^2_{L^2}+C\|\rho\|^2_{L^2}+C(\|u\|^2_{L^2} +\|\Lambda^{\alpha+1}u\|^2_{L^2})\|\nabla\rho\|^2_{L^2}. \end{eqnarray}
(3.18)

因此, 推出

\begin{eqnarray} \frac{\rm d}{{\rm d}t}\|\nabla\rho\|^2_{L^2}+\|\Lambda^{\gamma+1}\rho\|^2_{L^2}\leq C(\|u\|^2_{L^2}+\|\Lambda^{\alpha+1}u\|^2_{L^2})\|\nabla\rho\|^2_{L^2}+C\|\rho\|^2_{L^2}. \end{eqnarray}
(3.19)

由Gronwall不等式, (3.6)式和(3.15)式, 有

\begin{eqnarray} \|\nabla\rho(t)\|^2_{L^2}+\int_{0}^{t}\|\Lambda^{\gamma+1}\rho(\tau)\|^2_{L^2}{\rm d}\tau\leq C. \end{eqnarray}
(3.20)

至此完成了当 s = 1 时命题3.1的证明.

步骤3 ( H^s 估计 (s>1) )  将算子 \Lambda^s 作用到(1.1) _1 , (1.1) _2 和(1.1) _3 式, 再分别和 \Lambda^s u , \Lambda^s h , \Lambda^s\rho L^2 内积, 合并后有

\begin{eqnarray} \label{2.21} &&\frac{1}{2}\frac{\rm d}{{\rm d}t}(\|\Lambda^s u\|^2_{L^2}+\|\Lambda^s h\|^2_{L^2}+\|\Lambda^s\rho\|^2_{L^2}) +\|\Lambda^{s+\alpha}u\|^2_{L^2}+\|\Lambda^{s+\beta}h\|^2_{L^2}+\|\Lambda^{s+\gamma}\rho\|^2_{L^2}\\ & = &-\int_{{\Bbb R} ^ n}\Lambda^s(u\cdot\nabla u)\cdot\Lambda^s u\ {\rm d}x+\int_{{\Bbb R} ^ n}\Lambda^s(h\cdot\nabla h)\cdot\Lambda^s u\ {\rm d}x+\int_{{\Bbb R} ^ n}\Lambda^s(\rho e_n)\cdot\Lambda^s u\ {\rm d}x\\ &&-\int_{{\Bbb R} ^ n}\Lambda^s(u\cdot\nabla h)\cdot\Lambda^s h\ {\rm d}x+\int_{{\Bbb R} ^ n} \Lambda^s(h\cdot\nabla u)\cdot\Lambda^s h\ {\rm d}x- \int_{{\Bbb R} ^ n}\Lambda^s(u\cdot\nabla \rho)\Lambda^s \rho\ {\rm d}x\\ &: = &K_1+K_2+K_3+K_4+K_5+K_6. \end{eqnarray}
(3.21)

对于 K_3 , 有

\begin{equation} K_3\leq\|\Lambda^s \rho\|_{L^2}\|\Lambda^s u\|_{L^2}\leq C(\|\Lambda^s \rho\|^2_{L^2}+\|\Lambda^s u\|^2_{L^2}). \end{equation}
(3.22)

为了估计 K_1 , 需要下面的Kato-Ponce交换子估计:

\begin{eqnarray} \|[\Lambda^s, f]g\|_{L^r}\leq C(\|\nabla f\|_{L^{p_1}}\|\Lambda^{s-1} g\|_{L^{q_1}}+\|\Lambda^s f\|_{L^{p_2}}\|g\|_{L^{q_2}}), \end{eqnarray}
(3.23)

其中 \frac{1}{r} = \frac{1}{p_i}+\frac{1}{q_i} , 并且 r, p_i, q_i \in[1, \infty] , i = 1, 2 .

由Hölder不等式, (3.23)式, Young不等式和Gagliardo-Nirenberg不等式, 可以得到

\begin{eqnarray} K_1& = &-\int_{{\Bbb R} ^ n}[\Lambda^s, u\cdot\nabla ] u\cdot\Lambda^s u\ {\rm d}x\\&\leq &C\|\nabla u\|_{L^2}\|\Lambda^s u\|^2_{L^4} \\ & \leq &{} C\|\nabla u\|_{L^2}\|\Lambda^s u\|^{\frac{4\alpha-n}{2\alpha}}_{L^2}\|\Lambda^{s+\alpha} u\|^{\frac{n}{2\alpha}}_{L^2}\\ &\leq&\varepsilon\|\Lambda^{s+\alpha} u\|^2_{L^2}+C\|\nabla u\|^{\frac{4\alpha}{4\alpha-n}}_{L^2}\|\Lambda^s u\|^2_{L^2}. \end{eqnarray}
(3.24)

下面将 K_4 的估计分成两种情况: \alpha<\frac{n}{2} , \beta<\frac{n}{2} \alpha>\frac{n}{2} , \beta>\frac{n}{2} . \alpha<\frac{n}{2} , \beta<\frac{n}{2} 时, 由(3.23)式, {\rm div} u = 0 \alpha+\beta\geq1+\frac{n}{2} , 有

\begin{eqnarray} K_4& = &-\int_{{\Bbb R} ^ n}[\Lambda^s, u\cdot\nabla]h\cdot\Lambda^s h\ {\rm d}x\\&\leq & C\|\Lambda^s u\|_{L^{\frac{2n}{n-2\alpha}}}\|\nabla h\|_{L^2}\|\Lambda^s h\|_{L^{\frac{n}{\alpha}}}+ C\|\Lambda^s h\|_{L^2}\|\nabla u\|_{L^{\frac{n}{\beta}}}\|\Lambda^s h\|_{L^{\frac{2n}{n-2\beta}}}\\ &\leq &C\|\Lambda^{s+\alpha} u\|_{L^2}\|\nabla h\|_{L^2}\|\Lambda^s h\|^{\frac{2\alpha+2\beta-n}{2\beta}}_{L^2}\|\Lambda^{s+\beta} h\|^{\frac{n-2\alpha}{2\beta}}_{L^2} \\&&+C\|\Lambda^s h\|_{L^2}\|\Lambda^{\frac{n}{2}-\beta+1} u\|_{L^2}\|\Lambda^{s+\beta} h\|_{L^2} \\ &\leq&\varepsilon\|\Lambda^{s+\alpha} u\|^2_{L^2}+\varepsilon\|\Lambda^{s+\beta} h\|^2_{L^2} +C\big(\|\nabla u\|^2_{L^2}+\|\Lambda^{\alpha+1}u\|^2_{L^2} +\|\nabla h\|^{\frac{4\beta}{2\alpha+2\beta-n}}_{L^2}\big)\|\Lambda^s h\|^2_{L^2}. \end{eqnarray}

\alpha>\frac{n}{2} , \beta>\frac{n}{2} 时, 有

\begin{eqnarray} K_4&\leq &C\|\Lambda^s u\|_{L^{\infty}}\|\nabla h\|_{L^2}\|\Lambda^s h\|_{L^2}+C\|\Lambda^s h\|_{L^2}\|\nabla u\|_{L^2}\|\Lambda^s h\|_{L^{\infty}}\\ &\leq &C\|\Lambda^s u\|^{\frac{2\alpha-n}{2\alpha}}_{L^2}\| \Lambda^{s+\alpha}u\|^{\frac{n}{2\alpha}}_{L^2}\|\nabla h\|_{L^2}\|\Lambda^s h\|_{L^2}\\ &&+C\|\Lambda^s h\|_{L^2}\|\nabla u\|_{L^2}(\|\nabla h\|_{L^2}+\|\Lambda^{s+\beta} h\|_{L^2})\\ &\leq&\varepsilon\|\Lambda^{s+\alpha} u\|^2_{L^2}+\varepsilon\|\Lambda^{s+\beta} h\|^2_{L^2} +C\|\Lambda^s u\|^2_{L^2}+C\|\nabla h\|^2_{L^2}\\ &&+C(\|\nabla u\|^2_{L^2}+\|\nabla h\|^2_{L^2} )\|\Lambda^s h\|^2_{L^2}. \end{eqnarray}

将两种情况合并后可得

\begin{eqnarray} K_4&\leq&\varepsilon\|\Lambda^{s+\alpha} u\|^2_{L^2}+\varepsilon\|\Lambda^{s+\beta}h\|^2_{L^2} +C\|\Lambda^s u\|^2_{L^2}+C\|\nabla h\|^2_{L^2}\\ &&+C\big(\|\nabla u\|^2_{L^2}+\|\nabla h\|^2_{L^2}+\|\Lambda^{\alpha+1} u\|^2_{L^2}+\|\nabla h\|^{\frac{4\beta}{2\alpha+2\beta-n}}_{L^2}\big)\|\Lambda^s h\|^2_{L^2}. \end{eqnarray}
(3.25)

类似于 K_4 的推导, 有

\begin{eqnarray} K_6&\leq&\varepsilon\|\Lambda^{s+\alpha} u\|^2_{L^2}+\varepsilon\|\Lambda^{s+\gamma}\rho\|^2_{L^2} +C\|\Lambda^s u\|^2_{L^2}+C\|\nabla \rho\|^2_{L^2}\\ &&+C\big(\|\nabla u\|^2_{L^2}+\|\nabla \rho\|^2_{L^2}+\|\Lambda^{\alpha+1} u\|^2_{L^2}+\|\nabla\rho\|^{\frac{4\gamma}{2\alpha+2\gamma-n}} _{L^2}\big)\|\Lambda^s \rho\|^2_{L^2}. \end{eqnarray}
(3.26)

最后, 将 K_2 K_5 一起处理, 得

\begin{eqnarray} K_2+K_5& = &\int_{{\Bbb R} ^ n}\Lambda^s({\rm div}{(h\otimes h)})\cdot\Lambda^s u\ {\rm d}x+ \int_{{\Bbb R} ^ n}\Lambda^s(h\cdot\nabla u)\cdot\Lambda^s h\ {\rm d}x\\ & = &-\int_{{\Bbb R} ^ n}\Lambda^s(h\cdot h)\cdot\Lambda^s \nabla u\ {\rm d}x+ \int_{{\Bbb R} ^ n}\Lambda^s(h\cdot\nabla u)\cdot\Lambda^s h\ {\rm d}x. \end{eqnarray}

注意到 \alpha+\beta\geq1+\frac{n}{2} , 若 \alpha<1+\frac{n}{2} \beta<\frac{n}{2} , 有

\begin{eqnarray*} K_2+K_5&\leq &C\|h\|_{L^{\frac{n}{\alpha-1}}}\|\Lambda^s\nabla u\|_{L^{\frac{2n}{2-2\alpha+n}}} \|\Lambda^s h\|_{L^2}+C\|\Lambda^s h\|_{L^{\frac{2n}{n-2\beta}}}\|\nabla u\|_{L^{\frac{n}{\beta}}} \|\Lambda^s h\|_{L^2}\\ &\leq &C\|h\|_{L^{\frac{n}{\alpha-1}}}\|\Lambda^{s+\alpha}u\|_{L^2}\|\Lambda^s h\|_{L^2}+C\|\Lambda^{s+\beta}h\|_{L^2}\|\Lambda^{\frac{n}{2}-\beta+1} u\|_{L^2}\|\Lambda^s h\|_{L^2}\\ &\leq&\varepsilon\|\Lambda^{s+\alpha}u\|^2_{L^2}+\varepsilon\|\Lambda^{s+\beta}h\|_{L^2}\\ &&+ C\big(\|u\|^2_{L^2}+\|\Lambda^{\alpha+1} u\|^2_{L^2}+\| h\|^2_{L^2} +\|\Lambda^{\beta+1} h\|^2_{L^2}\big)\|\Lambda^s h\|^2_{L^2}. \end{eqnarray*}

\alpha>1+\frac{n}{2} \beta>\frac{n}{2} , 有

\begin{eqnarray*} K_2+K_5&\leq &C\|h\|_{L^2}\|\Lambda^s\nabla u\|_{L^\infty}\|\Lambda^s h\|_{L^2}+C\|\Lambda^s h \|_{L^\infty}\|\nabla u\|_{L^2}\|\Lambda^s h\|_{L^2}\\&\leq &C\|h\|_{L^2}\|\Lambda^s u\|^{\frac{2\alpha-n-2}{2\alpha}}_{L^2}\|\Lambda^{s+\alpha} u\|^{\frac{n+2}{2\alpha}}_{L^2}\|\Lambda^s h\|_{L^2}\\ &&+C(\|\nabla h\|_{L^2}+\|\Lambda^{s+\beta} h\|_{L^2})\|\nabla u\|_{L^2}\|\Lambda^s h\|_{L^2}\\&\leq &\varepsilon\|\Lambda^{s+\alpha} u\|^2_{L^2}+\varepsilon\|\Lambda^{s+\beta} h\|^2_{L^2} +C(\|\nabla u\|^2_{L^2}+\|h\|^2_{L^2})\|\Lambda^s h\|^2_{L^2} \\ &&+C\|\Lambda^s u\|^2_{L^2}+C\|\nabla h\|^2_{L^2}. \end{eqnarray*}

将上面两种情况合并后可得

\begin{eqnarray} K_2+K_5&\leq&\varepsilon\|\Lambda^{s+\alpha} u\|^2_{L^2}+\varepsilon\|\Lambda^{s+\beta} h\|^2_{L^2} +C\|\Lambda^s u\|^2_{L^2}+C\|\nabla h\|^2_{L^2}\\ &&+C\big(\|u\|^2_{L^2}+\| h\|^2_{L^2}+\|\nabla u\|^2_{L^2}+\|\Lambda^{\alpha+1} u\|^2_{L^2}+\|\Lambda^{\beta+1} h\|^2_{L^2}\big)\|\Lambda^s h\|^2_{L^2}. \end{eqnarray}
(3.27)

将(3.22)式, (3.24)–(3.27)式代入(3.21)式, 得

\begin{eqnarray*} &&\frac{\rm d}{{\rm d}t}(\|\Lambda^s u\|^2_{L^2}+\|\Lambda^s h\|^2_{L^2}+\|\Lambda^s\rho\|^2_{L^2}) +\|\Lambda^{s+\alpha}u\|^2_{L^2}+\|\Lambda^{s+\beta}h\|^2_{L^2}+\|\Lambda^{s+\gamma}\rho\|^2_{L^2}\\ &\leq& C\big(\| u\|^2_{L^2}+\|h\|^2_{L^2}+\|\nabla u\|^2_{L^2}+\|\nabla h\|^2_{L^2}+\|\nabla\rho\|^2_{L^2} +\|\nabla u\|^{\frac{4\alpha}{4\alpha-n}}_{L^2}\\ &&+\|\nabla h\|^{\frac{4\beta}{2\alpha+2\beta-n}}_{L^2}+\|\nabla\rho\|^{\frac{4\gamma} {2\alpha+2\gamma-n}}_{L^2}+\|\Lambda^{\alpha+1} u\|^2_{L^2}+\|\Lambda^{\beta+1}h\|^2_{L^2}+1\big)\\ &&\times(\|\Lambda^s u\|^2_{L^2}+\|\Lambda^s h\|^2_{L^2}+\|\Lambda^s\rho\|^2_{L^2})+C(\|\nabla h\|^2_{L^2}+\|\nabla\rho\|^2_{L^2}). \end{eqnarray*}

最后, 由Gronwall不等式, (3.6), (3.15)和(3.20)式可以得到 s>1 时命题3.1的结论.

接下来证明定理2.1中解的唯一性, 即

命题3.2  令 T>0 .假设 (u^{(1)}, h^{(1)}, \rho^{(1)}) (u^{(2)}, h^{(2)}, \rho^{(2)}) 是系统(1.1)的两个解, 满足

\begin{eqnarray*} &&(u^{(i)}, h^{(i)}, \rho^{(i)})\in L^{\infty}(0, T;H^1({\Bbb R} ^ n)), \ i = 1, 2, \\ &&(\Lambda^{\alpha+1} u^{(2)}, \Lambda^{\beta+1} h^{(2)}, \Lambda^{\gamma+1}\rho^{(2)})\in L^2(0, T;L^2({\Bbb R} ^ n)), \end{eqnarray*}

则在 {\Bbb R} ^ n \times(0, T) 上, 有

(u^{(1)}, h^{(1)}, \rho^{(1)}) = (u^{(2)}, h^{(2)}, \rho^{(2)}) .

  设 p^{(1)} p^{(2)} 分别是与 (u^{(1)}, h^{(1)}, \rho^{(1)}) (u^{(2)}, h^{(2)}, \rho^{(2)}) 相关的压力, 两个解的差 (\tilde{u}, \tilde{h}, \tilde{\rho}) 定义为

\tilde{u} = u^{(1)}-u^{(2)}, \ \tilde{h} = h^{(1)}-h^{(2)}, \ \tilde{\rho} = \rho^{(1)}-\rho^{(2)}, \ \tilde{p} = p^{(1)}-p^{(2)},

满足

\begin{eqnarray*} \left\{ \begin{array}{lr} \tilde{u}_t+u^{(1)}\cdot\nabla\tilde{u}+\tilde{u}\cdot\nabla u^{(2)}+\nabla\tilde{p}+\Lambda^{2\alpha}\tilde{u} = h^{(1)}\cdot\nabla h^{(1)}-h^{(2)}\cdot\nabla h^{(2)}+\tilde{\rho}e_n, & \\ \tilde{h}_t+u^{(1)}\cdot\nabla\tilde{h}+\tilde{u}\cdot\nabla h^{(2)}-\tilde h\cdot\nabla u^{(1)}-h^{(2)}\cdot\nabla \tilde u+\Lambda^{2\beta}\tilde{h} = 0, & \\ \tilde{\rho}_t+u^{(1)}\cdot\nabla\tilde{\rho}+\tilde{u}\cdot\nabla\rho^{(2)}+\Lambda^{2\gamma}\tilde{\rho} = 0, & \\ {\rm div} {\tilde u} = {\rm div} {\tilde h} = 0, & \\ (\tilde{u}, \tilde{h}, \tilde{\rho})(x, 0) = 0. & \\ \end{array} \right. \end{eqnarray*}

将上面方程组的前三个方程分别与 \tilde{u} , \tilde{h} \tilde{\rho} L^2 内积, 合并后可得

\begin{eqnarray*} &&\frac{1}{2}\frac{\rm d}{{\rm d}t}(\|\tilde{u}(t)\|^2_{L^2}+\|\tilde{h}(t)\|^2_{L^2} +\|\tilde{\rho}(t)\|^2_{L^2})+\|\Lambda^{\alpha}\tilde{u}\|^2_{L^2}+\|\Lambda^{\beta}\tilde{h}\|^2_{L^2} +\|\Lambda^{\gamma}\tilde{\rho}\|^2_{L^2}\\ & = &-\int_{{\Bbb R} ^ n}\tilde{u}\cdot\nabla u^{(2)}\cdot\tilde{u}\ {\rm d}x+\int_{{\Bbb R} ^ n}(h^{(1)}\cdot\nabla h^{(1)}-h^{(2)}\cdot\nabla h^{(2)})\cdot\tilde{u}\ {\rm d}x\\ &&+\int_{{\Bbb R} ^ n}\tilde{\rho}e_n\cdot\tilde{u}\ {\rm d}x-\int_{{\Bbb R} ^ n} \tilde{u}\cdot\nabla h^{(2)}\cdot\tilde{h}\ {\rm d}x+\int_{{\Bbb R} ^ n}\tilde{h}\cdot\nabla u^{(1)}\cdot\tilde{h}\ {\rm d}x\\ &&+\int_{{\Bbb R} ^ n} h^{(2)}\cdot\nabla\tilde{u}\cdot\tilde{h}\ {\rm d}x-\int_{{\Bbb R} ^ n}\tilde{u}\cdot\nabla\rho^{(2)}\tilde{\rho}\ {\rm d}x\\ &: = &N_1+N_2+N_3+N_4+N_5+N_6+N_7, \end{eqnarray*}

其中用到了

\begin{eqnarray*} &&\int_{{\Bbb R} ^ n} u^{(1)}\cdot\nabla\tilde{u}\cdot\tilde{u}\ {\rm d}x = 0, \ \int_{{\Bbb R} ^ n} u^{(1)}\cdot\nabla\tilde{h}\cdot\tilde{h}\ {\rm d}x = 0, \\ &&\int_{{\Bbb R} ^ n} u^{(1)}\cdot\nabla\tilde{\rho}\cdot\tilde{\rho}\ {\rm d}x = 0, \ \int_{{\Bbb R} ^ n}\nabla\tilde{p}\cdot\tilde{u}\ {\rm d}x = 0. \end{eqnarray*}

注意到

N_3\leq\|\tilde{\rho}\|_{L^2}\|{\tilde{u}}\|_{L^2}\leq C(\|\tilde{\rho}\|^2_{L^2}+\|{\tilde{u}}\|^2_{L^2}).

由Hölder, Young和Gagliardo-Nirenberg不等式以及Sobolev嵌入定理, 有:若 \alpha<\frac{n}{2} , 则

\begin{eqnarray*} N_1&\leq&\|\tilde{u}\|_{L^{\frac{n}{\alpha}}}\|\nabla u^{(2)}\|_{L^{\frac{2n}{n-2\alpha}}} \|\tilde{u}\|_{L^2}\\&\leq &C\|\tilde{u}\|^{\frac{4\alpha-n}{2\alpha}}_{L^2}\|\Lambda^{\alpha} \tilde{u}\|^{\frac{n-2\alpha}{2\alpha}}_{L^2}\|\Lambda^{\alpha+1} u^{(2)}\|_{L^2}\|\tilde{u}\|_{L^2}\\ &\leq&\varepsilon\|\Lambda^{\alpha} \tilde u\|^2_{L^2}+C\|\Lambda^{\alpha+1} u^{(2)} \|^{\frac{4\alpha}{6\alpha-n}}_{L^2}\|\tilde{u}\|^2_{L^2}. \end{eqnarray*}

\alpha>\frac{n}{2} , 则

\begin{eqnarray*} N_1&\leq&\|\tilde{u}\|_{L^2}\|\nabla u^{(2)}\|_{L^2}\|\tilde{u}\|_{L^{\infty}}\\ &\leq& C\|\tilde{u}\|_{L^2}\|\nabla u^{(2)}\|_{L^2}\|\tilde{u}\|^{\frac{2\alpha-n}{2\alpha}}_{L^2} \|\Lambda^{\alpha}\tilde{u}\|^{\frac{n}{2\alpha}}_{L^2}\\ &\leq&\varepsilon\|\Lambda^{\alpha}\tilde{u}\|^2_{L^2}+C\|\nabla u^{(2)}\|^{\frac{4\alpha}{4\alpha-n}}_{L^2}\|\tilde{u}\|^2_{L^2}. \end{eqnarray*}

下面估计 N_4 . \beta<\frac{n}{2} , 注意到 \alpha+\beta\geq1+\frac{n}{2} , 有

\begin{eqnarray*} N_4&\leq& C\|\tilde{h}\|_{L^2}\|\nabla h^{(2)}\|_{L^{\frac{2n}{n-2\beta}}}\|\tilde{u}\|_{L^{\frac{n}{\beta}}}\\ &\leq &C\|\tilde{h}\|_{L^2}\|\Lambda^{\beta+1} h^{(2)}\|_{L^2}\|\tilde{u}\|^{\frac{2\alpha+2\beta-n}{2\alpha}} _{L^2}\|\Lambda^{\alpha}\tilde{u}\|^{\frac{n-2\beta}{2\alpha}}_{L^2}\\ &\leq&\varepsilon\|\Lambda^{\alpha}\tilde{u}\|^2_{L^2}+C\|\Lambda^{\beta+1}h^{(2)}\|^{\frac{4\alpha} {4\alpha+2\beta-n}}_{L^2}(\|\tilde{u}\|^2_{L^2}+\|\tilde{h}\|^2_{L^2}). \end{eqnarray*}

\beta>\frac{n}{2} , 则

\begin{eqnarray*} N_4&\leq&\|\tilde{h}\|_{L^{\infty}}\|\tilde{u}\|_{L^2}\|\nabla h^{(2)}\|_{L^2}\\ &\leq &C\|\tilde{h} \|^{\frac{2\beta-n}{2\beta}}_{L^2}\|\Lambda^{\beta}\tilde{h}\|^{\frac{n}{2\beta}}_{L^2} \|\tilde{u}\|_{L^2}\|\nabla h^{(2)}\|_{L^2}\\ &\leq&\varepsilon\|\Lambda^{\beta}\tilde{h} \|^2_{L^2}+C\|\nabla h^{(2)}\|^{\frac{4\beta}{4\beta-n}}_{L^2}(\|\tilde{u}\|^2_{L^2}+\|\tilde{h}\|^2_{L^2}). \end{eqnarray*}

类似地, 对于 N_7 , 若 \gamma<\frac{n}{2} , 有

N_7\leq\varepsilon\|\Lambda^{\alpha}\tilde{u}\|^2_{L^2}+C\|\Lambda^{\gamma+1}\rho^{(2)} \|^{\frac{4\alpha}{4\alpha+2\gamma-n}}_{L^2}(\|\tilde{u}\|^2_{L^2}+\|\tilde{\rho}\|^2_{L^2}).

\gamma>\frac{n}{2} , 有

N_7\leq\varepsilon\|\Lambda^{\gamma}\tilde{\rho}\|^2_{L^2}+C\|\nabla\rho^{(2)}\|^{\frac{ 4\gamma}{4\gamma-n}}_{L^2}(\|\tilde{u}\|^2_{L^2}+\|\tilde{\rho}\|^2_{L^2}).

接下来处理 N_2 , N_5 N_6 .

\begin{eqnarray*} N_2& = &\int_{{\Bbb R} ^ n}({\rm div} {(h^{(1)}\otimes h^{(1)})}-{\rm div} {(h^{(2)}\otimes h^{(2)})})\cdot\tilde{u}\ {\rm d}x\\ & = &\int_{{\Bbb R} ^ n} (-h^{(1)}\cdot h^{(1)}\cdot\nabla\tilde{u}+h^{(2)}\cdot h^{(2)} \cdot\nabla\tilde{u})\ {\rm d}x\\ & = &\int_{{\Bbb R} ^ n} (-h^{(1)}\cdot h^{(1)}\cdot \nabla\tilde{u}+h^{(2)}\cdot h^{(1)}\cdot\nabla\tilde{u}-h^{(2)}\cdot h^{(1)} \cdot\nabla\tilde{u}+h^{(2)}\cdot h^{(2)}\cdot\nabla\tilde{u})\ {\rm d}x\\ & = &\int_{{\Bbb R} ^ n} (-\tilde{h}\cdot h^{(1)}\cdot\nabla\tilde{u}-h^{(2)} \cdot\tilde{h}\cdot\nabla\tilde{u})\ {\rm d}x. \end{eqnarray*}

\begin{eqnarray*} N_2+N_5+N_6& = &\int_{{\Bbb R} ^ n}(-\tilde{h}\cdot h^{(1)}\cdot\nabla \tilde{u}+\tilde{h}\cdot\nabla {u^{(1)}}\cdot\tilde{h})\ {\rm d}x\\ & = &\int_{{\Bbb R} ^ n}(-\tilde{h}\cdot h^{(1)}\cdot\nabla\tilde{u}+ \tilde{h}\cdot h^{(2)}\cdot\nabla\tilde{u}-\tilde{h}\cdot h^{(2)}\cdot\nabla \tilde{u}+\tilde{h}\cdot\nabla u^{(1)}\cdot\tilde{h})\ {\rm d}x\\ & = &\int_{{\Bbb R} ^ n}(-\tilde{h}\cdot\tilde{h}\cdot\nabla\tilde{u}-\tilde{h} \cdot h^{(2)}\cdot\nabla\tilde{u}+\tilde{h}\cdot\nabla {u^{(1)}}\cdot\tilde{h})\ {\rm d}x\\ & = &-\int_{{\Bbb R} ^ n}\tilde{h}\cdot h^{(2)}\cdot\nabla\tilde{u}\ {\rm d}x+\int_{{\Bbb R} ^ n}\tilde{h}\cdot\nabla u^{(2)}\cdot\tilde{h}\ {\rm d}x\\ &: = &Q_1+Q_2. \end{eqnarray*}

对于 Q_1 , 若 \beta<\frac{n}{2}-1 , 由Hölder, Young和Gagliardo-Nirenberg不等式, 得

\begin{eqnarray*} Q_1&\leq& C\|\tilde{h}\|_{L^2}\|h^{(2)}\|_{L^{\frac{2n}{n-2\beta-2}}}\| \nabla\tilde{u}\|_{L^{\frac{n}{\beta+1}}}\\ &\leq& C\|\tilde{h}\|_{L^2}\| \Lambda^{\beta+1}h^{(2)}\|_{L^2}\|\tilde{u}\|^{\frac{2\alpha+2\beta-n}{2\alpha}} _{L^2}\|\Lambda^{\alpha}\tilde{u}\|^{\frac{n-2\beta}{2\alpha}}_{L^2}\\ &\leq&\varepsilon\|\Lambda^{\alpha}\tilde{u}\|^2_{L^2}+C\|\Lambda^{\beta+1}h^{(2)} \|^{\frac{4\alpha}{4\alpha+2\beta-n}}_{L^2}(\|\tilde{h}\|^2_{L^2}+\|\tilde{u}\|^2_{L^2}), \end{eqnarray*}

这里 \alpha+\beta\geq1+\frac{n}{2} . \beta>\frac{n}{2}-1 ,

\begin{eqnarray*} Q_1&\leq& C\|\tilde{h}\|_{L^2}\|h^{(2)}\|_{L^{\infty}}\|\nabla\tilde{u}\|_{L^2}\\ &\leq &C\|\tilde{h}\|_{L^2}(\|\nabla h^{(2)}\|_{L^2}+\|\Lambda^{\beta+1}v^{(2)}\|_{L^2}) (\|\tilde{u}\|_{L^2}+\|\Lambda^{\alpha}\tilde{u}\|_{L^2})\\ &\leq&\varepsilon\|\Lambda^{\alpha}\tilde{u}\|^2_{L^2}+C\|\tilde{u}\|^2_{L^2} +C(\|\Lambda^{\beta+1}h^{(2)}\|^2_{L^2}+\|\nabla h^{(2)}\|^2_{L^2})\|\tilde{h}\|^2_{L^2}. \end{eqnarray*}

对于 Q_2 , 若 \beta<\frac{n}{2} , 注意到 \alpha+\beta\geq1+\frac{n}{2} , 有

\begin{eqnarray*} Q_2&\leq&\|\tilde{h}\|_{L^2}\|\nabla u^{(2)}\|_{L^{\frac{n}{\beta}}}\|\tilde{h}\|_{L^{\frac{2n}{n-2\beta}}}\\ &\leq &C\|\tilde{h}\|_{L^2}\|\Lambda^{\frac{n}{2}-\beta+1} u^{(2)}\|_{L^2}\|\Lambda^{\beta}\tilde{h}\|_{L^2}\\ &\leq&\varepsilon\|\Lambda^{\beta} \tilde{h}\|^2_{L^2}+C(\|\nabla u^{(2)}\|^2_{L^2} +\|\Lambda^{\alpha+1}u^{(2)}\|^2_{L^2})\|\tilde{h}\|^2_{L^2}. \end{eqnarray*}

\beta>\frac{n}{2} , 有

\begin{eqnarray*} Q_2&\leq&\|\tilde{h}\|_{L^2}\|\nabla u^{(2)}\|_{L^2}\|\tilde{h}\|_{L^{\infty}}\\ &\leq& C\|\tilde{h}\|_{L^2}\|\nabla u^{(2)}\|_{L^2}\|\tilde{h}\|^{\frac{2\beta-n} {2\beta}}_{L^2}\|\Lambda^{\beta}\tilde{h}\|^{\frac{n}{2\beta}}_{L^2}\\ &\leq&\varepsilon\|\Lambda^{\beta}\tilde{h}\|^2_{L^2}+C\|\nabla u^{(2)}\|^{\frac{4\beta}{4\beta-n}}_{L^2}\|\tilde{h}\|^2_{L^2}. \end{eqnarray*}

综合上面所有式子, 得到

\begin{eqnarray*} &&\frac{\rm d}{{\rm d}t}(\|\tilde{u}\|^2_{L^2}+\|\tilde{h}\|^2_{L^2}+\|\tilde{\rho}\|^2_{L^2}) +\|\Lambda^{\alpha}\tilde{u}\|^2_{L^2}+\|\Lambda^{\beta}\tilde{h}\|^2_{L^2} +\|\Lambda^{\gamma}\tilde{\rho}\|^2_{L^2}\\ &\leq& C\big(\|\nabla u^{(2)}\|^2_{L^2}+ \|\nabla u^{(2)}\|^{\frac{4\alpha}{4\alpha-n}}_{L^2}+\|\nabla u^{(2)}\| ^{\frac{4\beta}{4\beta-n}}_{L^2}+\|\nabla h^{(2)}\|^2_{L^2}+\|\nabla h^{(2)} \|^{\frac{4\beta}{4\beta-n}}_{L^2}\\ &&+\|\nabla\rho^{(2)}\|^{\frac{4\gamma}{4\gamma-n}} +\|\Lambda^{\alpha+1} u^{(2)}\|^2_{L^2}+\|\Lambda^{\alpha+1} u^{(2)}\|^{\frac{4\alpha}{6\alpha-n}} _{L^2}+\|\Lambda^{\beta+1} h^{(2)}\|^2_{L^2}\\ &&+\|\Lambda^{\beta+1} h^{(2)}\| ^{\frac{4\alpha}{4\alpha+2\beta-n}}_{L^2}+\|\Lambda^{\gamma+1} \rho^{(2)}\| ^{\frac{4\alpha}{4\alpha+2\gamma-n}}_{L^2}+1\big)(\|\tilde{u}\|^2_{L^2}+\|\tilde{h}\|^2_{L^2}+\|\tilde{\rho}\|^2_{L^2}). \end{eqnarray*}

由Gronwall不等式可以得到唯一性, 至此完成了命题3.2的证明.

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