带有分数阶耗散项的三维不可压液晶方程如下

(1.1) $ \left\{\begin{array}{ll}\partial_t u + u\cdot \nabla u + \mu (-\Delta)^{\alpha}u+\nabla p = -\lambda \nabla\cdot (\nabla d\odot \nabla d), \\\partial_t d + u\cdot \nabla d + \gamma (-\Delta)^{\beta} d=-f(d), \\\nabla \cdot u =0, \\u(x, 0) = u_0(x), d(x, 0) = d_0(x), \end{array}\right.$

这里$ u $表示流体速度场, $ d $表示宏观分子方向场, $ p $表示流体压强.参数$\mu$, $\lambda$, $\gamma$都为非负的常数.此外, $\nabla d\odot \nabla d$表示$3\times3$矩阵,其中第$(i, j)$个元素为$\partial_{i}d_{k}\partial_{j} d_{k}(i, j\leq3)$, $f(d)=(|d|^{2}-1)d$.分数阶拉普拉斯算子$(-\Delta)^{\theta}$是通过傅里叶变换定义的,如下

$\widehat{(-\Delta)^{\theta}}f(\xi)=|\xi|^{2\theta}\widehat{f}(\xi). $

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$d=0$, $\alpha=1$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$\alpha\geq\frac{5}{4}$时,在初值$u_{0}$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$\frac{5}{4}$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$\alpha=0$, $\beta>1$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$\alpha\geq 1+\frac{n}{2}$, $\beta=0$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$\alpha\geq\frac{5}{4}$, $\beta\geq1$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$\alpha\geq\frac{5}{4}$, $\beta\geq0$和$\alpha+\beta\geq\frac{7}{4}$,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$\frac{5}{4}$.因为方程组(1.1)中存在高阶导数项$\nabla\cdot(\nabla d\odot\nabla d)$,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$(u_{0}, d_{0})\in H^{s}({\Bbb R} ^{3})\times H^{s+1}({\Bbb R} ^{3})$, $s>\frac{5}{2}$,并且$\nabla\cdot u_{0}=0$.如果$\alpha$和$\beta$满足

$ \alpha\geq \frac{5}{4} , \qquad \beta\geq\frac{5}{4} , $

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

$u, \nabla d\in L^{\infty }(0, T;H^{s}({\Bbb R} ^{n})).$

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$||(u, \nabla d)||_{H^{s}}$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$\mu$, $\lambda$, $\gamma$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${\Bbb R} ^{3}$上积分,通过分部积分可得

$\frac{1}{2}\frac{{\rm d}}{{\rm d} t}||d||_{L^{2}}^{2}+||\Lambda^{\beta}d||_{L^{2}}^{2} + ||d||_{L^{4}}^{4}\leq ||d||_{L^{2}}^{2}, $

这里用到不可压缩条件$\nabla\cdot u=0$.利用Gronwall不等式可得

$\||d||_{L^{2}}^{2}+2\int_{0}^{T}(||d||_{L^{4}}^{4}+||\Lambda^{\beta}d||_{L^{2}}^{2}){\rm d} t\leq C||d_{0}||_{L^{2}}^{2}.$

接下来对方程$(1.1)_{1}$和$(1.1)_{2}$与$u$, $-\Delta d$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

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

下面我们进行$u$和$\nabla d$的$H^1$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

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

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

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

将(2.2)式和(2.3)式相加可得

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

利用交换子估计和Hölder不等式,则有

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

通过分部积分,利用$I_{2}$和$I_{3}$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

(2.6) $ \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}$

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

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

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

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

接下来,利用(2.8)式,我们可得到$u$和$\nabla d$的$H^{s}$估计.

首先在方程$(1.2)_{1}$和$(1.2)_{2}$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$(\Lambda^{s}u, \Lambda^{s+1}d)$做$L^{2}$内积,可以得到

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

由$\nabla\cdot u=0$和引理2.1,有

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

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

(2.11) $ \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}$

类似的$J_{3}$可以估计为

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

对于$J_{4}$,同$I_{4}$项的估计,我们有

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

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.