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数学物理学报, 2020, 40(4): 918-924 doi:

论文

分数阶扩散的三维液晶方程的整体正则性

李强,

Global Regularity for the 3D Liquid Crystal Equations with Fractional Diffusion

Li Qiang,

收稿日期: 2019-06-20  

Received: 2019-06-20  

作者简介 About authors

李强,E-mail:1220494943@qq.com , E-mail:1220494943@qq.com

摘要

该文考虑的是带有分数阶耗散项(Δ)αu(Δ)βd的广义不可压缩液晶模型.目标是在需要最小的耗散情况下建立整体正则性.在初值充分光滑的情况下,若耗散指标α54, β54,方程组有唯一的整体光滑解.

关键词: 液晶方程 ; 分数阶耗散 ; 整体正则性

Abstract

In this paper, the focus is the global regularity of three-dimensional liquid crystal equations with fractional dissipations (Δ)αu and (Δ)βd. The objective is to establish the global regularity of the fractional liquid crystal equations with the minimal amount of dissipations. And it is obtained that the equations have a global classical solution with sufficiently smooth data if α54 and β54.

Keywords: Liquid crystal equations ; Fractional dissipation ; Global regularity

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本文引用格式

李强. 分数阶扩散的三维液晶方程的整体正则性. 数学物理学报[J], 2020, 40(4): 918-924 doi:

Li Qiang. Global Regularity for the 3D Liquid Crystal Equations with Fractional Diffusion. Acta Mathematica Scientia[J], 2020, 40(4): 918-924 doi:

1 引言

带有分数阶耗散项的三维不可压液晶方程如下

{tu+uu+μ(Δ)αu+p=λ(dd),td+ud+γ(Δ)βd=f(d),u=0,u(x,0)=u0(x),d(x,0)=d0(x),
(1.1)

这里u表示流体速度场, d表示宏观分子方向场, p表示流体压强.参数μ, λ, γ都为非负的常数.此外, dd表示3×3矩阵,其中第(i,j)个元素为idkjdk(i,j3), f(d)=(|d|21)d.分数阶拉普拉斯算子(Δ)θ是通过傅里叶变换定义的,如下

^(Δ)θf(ξ)=|ξ|2θˆf(ξ).

液晶动力学模型是由Ericksen和Leslie首先建立的(参见文献[1-2]),方程组(1.1)作为Ericksen-Leslie系统一个简化的版本,是由Lin在文献[3]中引入来描述向列型液晶流.并且, Lin和Liu在文献[4]中已经证明出液晶方程在二维情况下有唯一的整体光滑解,三维情况下有唯一的局部光滑解.他们也证得了整体弱解的存在性.但是,解的整体正则性依然是一个公开性的问题.

当方向场d=0, α=1时,方程组(1.1)退化为经典的不可压缩纳维斯托克斯方程.众所周知,三维情况下当速度场的耗散指标α54时,在初值u0充分光滑的情况下,纳维斯托克斯方程有唯一的整体光滑解.对于广义的磁流体方程来说, Wu在文献[5]中已经证明得到,在三维时初值充分光滑的情况下,速度场和磁场上的耗散指标不小于54时能够保证解的整体正则性.基于上述结果,在广义的纳维斯托克斯方程、广义的磁流体方程和其他一些类似的方程,例如布辛涅斯克方程、液晶方程、微极流体方程等等上,有大量的工作来研究如何降低系统中的耗散.对于液晶方程,在二维情况下, Wang和Zhou在文献[6]中研究了当α=0, β>1时的这种情况,并且得到了解的整体正则性.在更一般的n维情况下,利用一个对数作用到速度场的耗散项上, Yuan和Wei证明了当α1+n2, β=0时,液晶方程有唯一的整体光滑解[7].最近,对于微极流体方程, Li证得了当α54, β1时的解的整体正则性[8].紧接着在文献[9]中, Wang和Wu等证明了只需要α54, β0α+β74,解的整体正则性就可以得到保证.

受上述文献的启发,我们考虑三维时速度场u和方向场d上都存在耗散的这种情况,并且考虑是否能得到相应的类似结果.答案是显然的,但是方向场d上的耗散指标不能小于54.因为方程组(1.1)中存在高阶导数项(dd),所以方向场d上的耗散不能弱.下面我们给出本文的研究结果.

定理1.1  假设初值(u0,d0)Hs(R3)×Hs+1(R3), s>52,并且u0=0.如果αβ满足

α54,β54,

那么方程组(1.1)有唯一的整体光滑解,并且对于任意给定的T>0,有

u,dL(0,T;Hs(Rn)).

2 定理1.1的证明

本节我们专注于证明定理1.1.由引言已知方程组(1.1)的局部光滑解的存在唯一性已经得到证明,具体的过程读者可以参考文献[10].这里我们只需要得到局部解的先验估计来确保||(u,d)||Hs一致有界.通过基本的能量方法,充分利用交换子估计和Hölder不等式、插值不等式等,一步一步的提高解的正则性,最后得到局部解的高阶导数估计.本文一般的常数简记为C,且每个地方的值一般都不相同.由于参数μ, λ, γ的具体数值在本文的讨论中没有特别的作用,为了简单化,我们通常都把它们假定为常数1来处理.具体过程如下.

首先,在方程(1.1)2两端点乘d,并且在R3上积分,通过分部积分可得

12ddt||d||2L2+||Λβd||2L2+||d||4L4||d||2L2,

这里用到不可压缩条件u=0.利用Gronwall不等式可得

接下来对方程(1.1)_{1}(1.1)_{2}u, -\Delta d对应地做L^{2}内积,再将两式相加,我们有

\frac{1}{2}\frac{{\rm d}}{{\rm d} t}(||u||_{L^{2}}^{2}+||\nabla d||_{L^{2}}^{2})+||\Lambda^{\alpha} u||_{L^{2}}^{2}+||\Lambda^{\beta+1}d||_{L^{2}}^{2}+||d|\nabla d|||_{L^{2}}^{2}+\frac{1}{2}||\nabla|d|^{2}||_{L^{2}}^{2}\leq||\nabla d||_{L^{2}}^{2},

这里由\nabla\cdot u=0,利用了下面的等式

\int_{{\Bbb R}^{3}}u\cdot\nabla u\cdot u{\rm d} x=0, \ \ \int_{{\Bbb R}^{3}}u\cdot\nabla(\frac{|\nabla d|^{2}}{2})=0, \ \ \int_{{\Bbb R}^{3}}\nabla p\cdot u{\rm d} x=0.

因此,通过Gronwall不等式,则有

(||u||_{L^{2}}^{2}+||\nabla d||_{L^{2}}^{2})+2\int_{0}^{T}(||\Lambda^{\alpha} u||_{L^{2}}^{2}+||\Lambda^{\beta+1} d||_{L^{2}}^{2}+||d|\nabla d|||_{L^{2}}^{2}+\frac{1}{2}||\nabla|d|^{2}||_{L^{2}}^{2}){\rm d} t\nonumber \leqC(||u_{0}||_{L^{2}}^{2}+||\nabla d_{0}||_{L^{2}}^{2}).
(2.1)

下面我们进行u\nabla dH^1估计,首先给出一个交换子估计.

引理2.1  令s>0, 1< p<\infty, \frac{1}{p}=\frac{1}{p_1}+\frac{1}{p_2}=\frac{1}{p_3}+\frac{1}{p_4}其中p_2, p_3\in(1, +\infty), p_1, p_4\in[1, +\infty].那么

||\Lambda^{s}(fg)||_{L^{p}}\leq C(||g||_{L^{p_{1}}}||\Lambda^{s}f||_{L^{p_{2}}}+||\Lambda^{s}g||_{L^{p_{3}}} ||f||_{L^{p_{4}}}),

||[\Lambda^{s}, f\cdot \nabla]g||_{L^{p}}\leq C(||\nabla f||_{L^{p_{1}}}||\Lambda^{s}g||_{L^{p_{2}}}+||\Lambda^{s}f||_{L^{p_{3}}}||\nabla g||_{L^{p_{4}}}),

这里[\Lambda^{s}, f\cdot \nabla]g=\Lambda^{s}(f\cdot \nabla g)-f\cdot \nabla\Lambda^{s}g.

在方程(1.1)_{2}两边作用\Delta,然后和\Delta d作内积,我们有

\frac{1}{2}\frac{{\rm d}}{{\rm d} t}||\Delta d||_{L^{2}}^{2}+||\Lambda^{\beta+2} d||_{L^{2}}^{2}=-\int_{{\Bbb R}^{3}}\Delta(u\cdot\nabla d)\cdot\Delta d{\rm d} x-\int_{{\Bbb R}^{3}}\Delta f(d)\cdot\Delta d {\rm d} x.
(2.2)

然后在方程(1.1)_{1}两边点乘-\Delta u并在空间变量上积分,通过分部积分有

\frac{1}{2}\frac{{\rm d}}{{\rm d} t}||\nabla u||_{L^{2}}^{2}+||\Lambda^{\alpha+1}u||_{L^{2}}^{2}=\int_{{\Bbb R}^{3}}(u\cdot\nabla)u\cdot\Delta u{\rm d} x+\int_{{\Bbb R}^{3}}\nabla\cdot(\nabla d \odot\nabla d )\cdot\Delta u{\rm d} x.
(2.3)

将(2.2)式和(2.3)式相加可得

\begin{eqnarray}\label{2.4}&& \frac{1}{2}\frac{{\rm d}}{{\rm d} t}(||\nabla u||_{L^{2}}^{2}+||\Delta d||_{L^{2}}^{2})+||\Lambda^{\alpha+1} u||_{L^{2}}^{2}+||\Lambda^{\beta+2} d||_{L^{2}}^{2}\nonumber\\&=&\int_{{\Bbb R}^{3}}(u\cdot\nabla)u\cdot\Delta u{\rm d} x+\int_{{\Bbb R}^{3}} \nabla\cdot(\nabla d \odot\nabla d )\cdot\Delta u{\rm d} x\nonumber\\&-&\int_{{\Bbb R}^{3}}\Delta(u\cdot\nabla d)\cdot\Delta d{\rm d} x-\int_{{\Bbb R}^{3}}\Delta f(d)\cdot\Delta d {\rm d} x\nonumber\\ &:=& I_{1}+I_{2}+I_{3}+I_{4}.\end{eqnarray}
(2.4)

利用交换子估计和Hölder不等式,则有

\begin{eqnarray}\label{2.5}I_{1}&=&\int_{{\Bbb R}^{3}}[\nabla, u\cdot\nabla]u\cdot\nabla u{\rm d} x\nonumber\\ &\leq&||[\nabla, u\cdot\nabla]u||_{L^{2}}||\nabla u||_{L^{2}}\nonumber\\ &\leq& C||\nabla u||_{L^{\frac{12}{5}}}||\nabla u||_{L^{12}}||\nabla u||_{L^{2}}\nonumber\\ &\leq&C||u||_{H^{\alpha}}||u||_{H^{\alpha+1}}||\nabla u||_{L^{2}}\nonumber\\ &\leq& \frac{1}{2}||u||_{H^{\alpha+1}}^{2}+C||u||_{H^{\alpha}}^{2}||\nabla u||_{L^{2}}^{2}\nonumber\\ &\leq& \frac{1}{2}||\Lambda^{\alpha+1} u||_{L^{2}}^{2}+C||u||_{L^{2}}^{2}+C||u||_{H^{\alpha}}^{2}||\nabla u||_{L^{2}}^{2}.\end{eqnarray}
(2.5)

通过分部积分,利用I_{2}I_{3}中一些项相消的性质,我们有

\begin{eqnarray*}I_{2}+I_{3}&=&\int_{{\Bbb R}^{3}}\partial_{j}(\partial_{i}d_{k}\partial_{j}d_{k})\partial_{ll}u_{i} {\rm d} x -\int_{{\Bbb R}^{3}}\partial_{ll}(u_{i}\partial_{i}d_{k})\partial_{jj} d_{k}{\rm d} x\\ &=&\int_{{\Bbb R}^{3}}\partial_{i}\partial_{j}d_{k}\partial_{j}d_{k}\partial_{ll}u_{i} {\rm d} x +\int_{{\Bbb R}^{3}}\partial_{i}d_{k}\partial_{jj}d_{k}\partial_{ll} u_{i}{\rm d} x +\int_{{\Bbb R}^{3}}\partial_{l}(u_{i}\partial_{i}d_{k})\partial_{l}\partial_{jj} d_{k}{\rm d} x\\ &=&-\int_{{\Bbb R}^{3}}\partial_{il}d_{k}\partial_{jj} d_{k}\partial_{l}u_{i}{\rm d} x-\int_{{\Bbb R}^{3}}\partial_{i}d_{k}\partial_{l}\partial_{jj} d_{k}\partial_{l}u_{i}{\rm d} x\\&&+\int_{{\Bbb R}^{3}}\partial_{l}u_{i}\partial_{i}d_{k}\partial_{l}\partial_{jj} d_{k}{\rm d} x+\int_{{\Bbb R}^{3}}u_{i}\partial_{i}\partial_{l}d_{k}\partial_{l}\partial_{jj} d_{k}{\rm d} x\\ &=&-\int_{{\Bbb R}^{3}}\partial_{il}d_{k}\partial_{jj} d_{k}\partial_{l}u_{i}{\rm d} x+\int_{{\Bbb R}^{3}}u_{i}\partial_{i}\partial_{l}d_{k}\partial_{l}\partial_{jj} d_{k}{\rm d} x\\ &=&-\int_{{\Bbb R}^{3}}\partial_{il}d_{k}\partial_{jj} d_{k}\partial_{l}u_{i}{\rm d} x-\int_{{\Bbb R}^{3}}\partial_{l}u_{i}\partial_{i}\partial_{l}d_{k}\partial_{jj} d_{k}{\rm d} x .\end{eqnarray*}

现由Gagliardo-Nirenberg不等式可得

\begin{eqnarray}\label{2.6} I_{2}+I_{3}&\leq&||\nabla u||_{L^{\frac{12}{5}}}||\Delta d||_{L^{\frac{24}{7}}}^{2}\nonumber \\ &\leq& C||u||_{H^{\alpha}}||\Delta d||_{L^{2}}^{\frac{8\beta-5}{4\beta}}||\Lambda^{\beta+2} d||_{L^{2}}^{\frac{5}{4\beta}}\nonumber\\ &\leq&C||u||_{H^{\alpha}}^{\frac{8\beta}{8\beta-5}}||\Delta d||_{L^{2}}^{2}+\frac{1}{4}||\Lambda^{\beta+2} d||_{L^{2}}^{2}, \end{eqnarray}
(2.6)

这里\beta\geq\frac{5}{4}保证了\frac{8\beta}{8\beta-5}\leq 2,而且这里也是条件\beta\geq\frac{5}{4}需要的地方.对于I_{4},由引理2.1有

\begin{eqnarray}\label{2.7}I_{4}&\leq& \int_{{\Bbb R}^{3}} |\Delta d|^{2}+\Delta(|d|^{2}d)\cdot\Delta d {\rm d} x\nonumber\\ &\leq&||\Delta d||_{L^{2}}^{2}+||\Delta d||_{L^{4}}||\Delta(|d|^{2}d)||_{L^{\frac{4}{3}}}\nonumber\\ &\leq&||\Delta d||_{L^{2}}^{2}+C(||\Delta |d|^{2} ||_{L^{2}}||d||_{L^{4}}+||\Delta d||_{L^{4}}|||d|^{2}||_{L^{2}})||\Delta d||_{L^{4}}\nonumber\\ &\leq&||\Delta d||_{L^{2}}^{2}+C||\Delta d||_{L^{4}}^{2}||d||_{L^{4}}^{2}\nonumber\\ &\leq&||\Delta d||_{L^{2}}^{2}+C||\Delta d||_{L^{2}}^{\frac{4\beta-3}{2\beta}}||d||_{H^{\beta+2}}^{\frac{3}{2\beta}}||d||_{L^{2}}^{\frac{1}{2}}||\nabla d||_{L^{2}}^{\frac{3}{2}}\nonumber\\ &\leq&||\Delta d||_{L^{2}}^{2}+C(||d||_{L^{2}}^{2}+||\nabla d||_{L^{2}}^{2})||\Delta d||_{L^{2}}^{2}+{\frac{1}{4}}||d||_{H^{\beta+2}}^{2}\nonumber\\ &\leq&C||\Delta d||_{L^{2}}^{2}+{\frac{1}{4}}||\Lambda^{\beta+2} d||_{L^{2}}^{2}.\end{eqnarray}
(2.7)

因此,把(2.5), (2.6)和(2.7)式代入(2.4)式我们就有

\begin{eqnarray*}\frac{{\rm d}}{{\rm d} t}(||\nabla u||_{L^{2}}^{2}+||\Delta d||_{L^{2}}^{2})+||\Lambda^{\alpha+1} u||_{L^{2}}^{2}+||\Lambda^{\beta+2} d||_{L^{2}}^{2}\leq C(1+||u||_{H^{\alpha}}^{2})(||\nabla u||_{L^{2}}^{2}+||\Delta d||_{L^{2}}^{2}).\end{eqnarray*}

由(2.1)式可知

\begin{eqnarray*}\int_{0}^{T}(1+||u||_{H^{\alpha}}^{2}){\rm d} t\leq C(||u_{0}||_{2}^{2}+||\nabla d_{0}||_{2}^{2}), \end{eqnarray*}

那么通过Gronwall不等式,可得

\begin{eqnarray}\label{2.8}||\nabla u||_{L^{2}}^{2}+||\Delta d||_{L^{2}}^{2}+\int_{0}^{T}(||\Lambda^{\alpha+1} u||_{L^{2}}^{2}+||\Lambda^{\beta+2} d||_{L^{2}}^{2}){\rm d} t\leq C(||\nabla u_{0}||_{L^{2}}^{2}+||\Delta d_{0}||_{L^{2}}^{2}).\end{eqnarray}
(2.8)

接下来,利用(2.8)式,我们可得到u\nabla dH^{s}估计.

首先在方程(1.2)_{1}(1.2)_{2}两边分别对应作用\Lambda^{s}, \Lambda^{s+1},其中s>\frac{5}{2},然后对应的和(\Lambda^{s}u, \Lambda^{s+1}d)L^{2}内积,可以得到

\begin{eqnarray}\label{2.9}&& \frac{1}{2}\frac{{\rm d}}{{\rm d} t}(||\Lambda^{s}u||_{L^{2}}^{2}+||\Lambda^{s+1} d||_{L^{2}}^{2})+||\Lambda^{s+\alpha}u||_{L^{2}}^{2}+||\Lambda^{s+\beta+1} d||_{L^{2}}^{2}\nonumber\\&=&-\int_{{\Bbb R}^{3}}\Lambda^{s}(u\cdot\nabla u)\cdot\Lambda^{s} u{\rm d} x-\int_{{\Bbb R}^{3}}\Lambda^{s}(\nabla d_{j}\cdot\Delta d_{j})\cdot\Lambda^{s}u{\rm d} x\nonumber\\&-&\int_{{\Bbb R}^{3}}\Lambda^{s+1}(u\cdot\nabla d)\cdot\Lambda^{s+1} d{\rm d} x-\int_{{\Bbb R}^{3}}\Lambda^{s+1}f(d)\cdot\Lambda^{s+1} d {\rm d} x\nonumber\\&:=&J_{1}+J_{2}+J_{3}+J_{4}.\end{eqnarray}
(2.9)

\nabla\cdot u=0和引理2.1,有

\begin{eqnarray}\label{2.10}J_{1}&=&-\int_{{\Bbb R}^{3}}[\Lambda^{s}, u\cdot\nabla]u\cdot\Lambda^{s} u{\rm d} x\nonumber\\&\leq&||[\Lambda^{s}, u\cdot\nabla]u||_{L^{\frac{4}{3}}}||\Lambda^{s}u||_{L^{4}}\nonumber\\ &\leq&C(||\nabla u||_{L^{2}}||\Lambda^{s-1}\nabla u||_{L^{4}}+||\Lambda^{s}u||_{L^{4}}||\nabla u||_{L^{2}})||\Lambda^{s}u||_{L^{4}}\nonumber\\ &\leq&C||\nabla u||_{L^{2}}^{2}||\Lambda^{s}u||_{L^{2}}^{\frac{4\alpha-3}{2\alpha}}||\Lambda^{s+\alpha}u||_{L^{2}}^{\frac{3}{2\alpha}}\nonumber\\ &\leq&C||\Lambda^{s}u||_{L^{2}}^{2}+\frac{1}{4}||\Lambda^{s+\alpha}u||_{L^{2}}^{2}.\end{eqnarray}
(2.10)

然后,利用Sobolev嵌入不等式以及引理2.1, J_{2}可以估计为

\begin{eqnarray*}J_{2}&=&-\int_{{\Bbb R}^{3}}\Lambda^{s}(\partial_{i} d_{j}\cdot\Delta d_{j})\Lambda^{s}u_{i} {\rm d} x\\&\leq&||\Lambda^{s-\alpha}(\nabla d_{j}\cdot\Delta d_{j})||_{L^{2}}||\Lambda^{s+\alpha}u||_{L^{2}}\nonumber\\ &\leq&(||\nabla d||_{L^{6}}||\Lambda^{s-\alpha}\Delta d||_{L^{3}}+ ||\Delta d||_{L^{12}}||\Lambda^{s-\alpha}\nabla d||_{L^\frac{12}{5}})||\Lambda^{s+\alpha}u||_{L^{2}}\\ &\leq&C(||\Delta d||_{L^{2}}||d||_{H^{s+\beta}}+||d||_{H^{\beta+2}}||d||_{H^{s}})||\Lambda^{s+\alpha}u||_{L^{2}}\\ &\leq&C||\Delta d||_{L^{2}}^{2}||d||_{H^{s+\beta}}^{2}+\frac{1}{4}||\Lambda^{s+\alpha}u||_{L^{2}}^{2}+\frac{1}{4}||\Lambda^{s+\alpha}u||_{L^{2}}^{2}+C||d||_{H^{\beta+2}}^{2}||d||_{H^{s}}^{2}\\ &\leq&C||d||_{H^{s+\beta}}^{2}+\frac{1}{2}||\Lambda^{s+\alpha}u||_{L^{2}}^{2}+C||d||_{H^{\beta+2}}^{2}||d||_{H^{s+1}}^{2}.\end{eqnarray*}

由插值不等式和Young不等式,则有

\begin{eqnarray*}||d||_{H^{s+\beta}}^{2}&\leq&C||\Delta d||_{L^{2}}^{\frac{2}{s+\beta-1}}||d||_{H^{s+\beta+1}}^{\frac{2(s+\beta-2)}{s+\beta-1}}\leqC||\Delta d||_{L^{2}}^{2}+\frac{1}{4}||d||_{H^{s+\beta+1}}^{2}.\end{eqnarray*}

因此

\begin{eqnarray}\label{2.11} J_{2}&\leq& C||\Delta d||_{L^{2}}^{2}+\frac{1}{4}||d||_{H^{s+\beta+1}}^{2}+\frac{1}{2}||\Lambda^{s+\alpha}u||_{L^{2}}^{2}+C||d||_{H^{\beta+2}}^{2}||d||_{H^{s+1}}^{2}\nonumber\\ &\leq&\frac{1}{4}||\Lambda^{s+\beta+1} d||_{L^{2}}^{2}+\frac{1}{2}||\Lambda^{s+\alpha}u||_{L^{2}}^{2}+C||d||_{H^{\beta+2}}^{2}||\Lambda^{s+1} d||_{L^{2}}^{2}+C||d||_{H^{\beta+2}}^{2}.\end{eqnarray}
(2.11)

类似的J_{3}可以估计为

\begin{eqnarray}\label{2.12}J_{3}&=&\int_{{\Bbb R}^{3}}\Lambda^{s}(u\cdot\nabla d)\Lambda^{s+2}d{\rm d} x\nonumber\\&\leq&||\Lambda^{s}(u\cdot\nabla d)||_{L^{2}}||\Lambda^{s+2} d||_{L^{2}}\nonumber\\ &\leq&C(||\Lambda^{s}u||_{L^{12}}||\nabla d||_{L^{\frac{12}{5}}}+||\Lambda^{s+1} d||_{L^{12}}||u||_{L^{\frac{12}{5}}})||\Lambda^{s+2} d||_{L^{2}}\nonumber\\ &\leq&C(||u||_{H^{s+\alpha}}||d||_{H^{\frac{5}{4}}}+||d||_{H^{s+\beta+1}}||u||_{H^{\frac{1}{4}}})||\Lambda^{s+2} d||_{L^{2}}\nonumber\\ &\leq&\frac{1}{4}||u||_{H^{s+\alpha}}^{2}+C||\Lambda^{s+2} d||_{L^{2}}^{2}+\frac{1}{8}||d||_{H^{s+\beta+1}}^{2}\nonumber\\ &\leq&\frac{1}{4}||\Lambda^{s+\alpha} u||_{L^{2}}^{2}+\frac{1}{8}||\Lambda^{s+\beta+1}d||_{L^{2}}^{2}+C(||u||_{L^{2}}^{2}+||d||_{L^{2}}^{2})\nonumber\\&&+C||\Delta d||_{L^{2}}^{\frac{2+2\beta}{s+\beta-1}}||\Lambda^{s+\beta+1}d||_{L^{2}}^{\frac{2s-4}{s+\beta-1}}\nonumber\\ &\leq&\frac{1}{4}||\Lambda^{s+\alpha}u||_{L^{2}}^{2}+\frac{1}{4}||\Lambda^{s+\beta+1} d||_{L^{2}}^{2}+C(||u||_{L^{2}}^{2}+||d||_{L^{2}}^{2}+||\Delta d||_{L^{2}}^{2}).\end{eqnarray}
(2.12)

对于J_{4},同I_{4}项的估计,我们有

\begin{eqnarray}\label{2.14}J_{4}&=&\int_{{\Bbb R}^{3}}\Lambda^{s+1}(d-|d|^{2}d)\Lambda^{s+1}d{\rm d} x\nonumber\\ &\leq&||\Lambda^{s+1} d||_{L^{2}}^{2}+||\Lambda^{s+1} d||_{L^{4}}||\Lambda^{s+1}(|d|^{2}d)||_{L^{\frac{4}{3}}}\nonumber\\ &\leq&||\Lambda^{s+1} d||_{L^{2}}^{2}+C(||\Lambda^{s+1} |d|^{2} ||_{L^{2}}||d||_{L^{4}}+||\Lambda^{s+1} d||_{L^{4}}|||d|^{2}||_{L^{2}})||\Lambda^{s+1} d||_{L^{4}}\nonumber\\ &\leq&||\Lambda^{s+1} d||_{L^{2}}^{2}+C||\Lambda^{s+1} d||_{L^{4}}^{2}||d||_{L^{4}}^{2}\nonumber\\ &\leq&||\Lambda^{s+1} d||_{L^{2}}^{2}+C||\Lambda^{s+1} d||_{L^{2}}^{\frac{4\beta-3}{2\beta}}||d||_{H^{s+\beta+1}}^{\frac{3}{2\beta}}||d||_{L^{2}}^{\frac{1}{2}}||\nabla d||_{L^{2}}^{\frac{3}{2}}\nonumber\\ &\leq&||\Lambda^{s+1} d||_{L^{2}}^{2}+C(||d||_{L^{2}}^{2}+||\nabla d||_{L^{2}}^{2})||\Lambda^{s+1} d||_{L^{2}}^{2}+{\frac{1}{4}}||d||_{H^{s+\beta+1}}^{2}\nonumber\\ &\leq&C||\Lambda^{s+1} d||_{L^{2}}^{2}+{\frac{1}{4}}||d||_{H^{s+\beta+1}}^{2}.\end{eqnarray}
(2.13)

联合上述不等式(2.10)--(2.13),代入(2.9)式,我们推得

\begin{eqnarray*}&&\frac{{\rm d}}{{\rm d} t}(||\Lambda^{s}u||_{L^{2}}^{2}+||\Lambda^{s+1} d||_{L^{2}}^{2})+||\Lambda^{s+\alpha}u||_{L^{2}}^{2}+||\Lambda^{s+\beta+1} d||_{L^{2}}^{2}\\ &\leq& C(||\Lambda^{s}u||_{L^{2}}^{2}+||\Lambda^{s+1}d||_{L^{2}}^{2})+C(||u||_{L^{2}}^{2}+||d||_{H^{\beta+2}}^{2}).\end{eqnarray*}

又由(2.8)式可知

\begin{eqnarray*}\int_{0}^{T}||d||_{H^{\beta+2}}^{2}){\rm d} t\leq C(||\nabla u_{0}||_{2}^{2}+||\Delta d_{0}||_{2}^{2}), \end{eqnarray*}

那么Gronwall不等式表明

\begin{eqnarray*}&& ||\Lambda^{s}u||_{L^{2}}^{2}+||\Lambda^{s+1} d||_{L^{2}}^{2}+\int_{0}^{T}(||\Lambda^{s+\alpha}u||_{L^{2}}^{2}+||\Lambda^{s+\beta+1} d||_{L^{2}}^{2}) {\rm d} t\\&\leq& C(||\Lambda^{s}u_{0}||_{L^{2}}^{2}+||\Lambda^{s+1} d_{0}||_{L^{2}}^{2}). \end{eqnarray*}

至此就完成了对定理1.1的证明.

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