本文主要研究以下$ {{\Bbb R}} ^{n}, n\geq 2 $空间中的不可压液晶方程组模型

(1.1) $ \begin{equation} \left \{\begin{array}{l} u_t-\Delta u+u\cdot\nabla u +\nabla P = -\rm{div}(\nabla d \odot \nabla d), \\ \rm{div}\ u = 0, \\ d_t+u\cdot \nabla d-\Delta d = f(d), \\ \end{array} \right. \end{equation} $

这里$ u = (u_1, u_2, u_3, \cdots , u_{n}) $是流体的速度场, $ P $为流体的压强, $ d = (d_1, d_2, d_3, \cdots , d_{n}) $为液晶的方向矢, $ f(d) = \frac{1}{\sigma^2}(|d|^2-1)d $. 这里$ \nabla d \odot \nabla d $为一个对称张量, $ (\nabla d \odot \nabla d)_{ij} $的表达式为$ \nabla_i d \cdot \nabla_j d $. 本文考虑该方程组成立的时空区域为$ Q_{T} = \Omega\times [0, T] $, 其中$ \Omega $为$ {{\Bbb R}} ^n $中的一个光滑区域.方程组的初边值条件为

$ u(x, 0) = u_0(x), \quad {\rm{div}}(u_0) = 0, \quad d(x, 0) = d_0(x), \quad x\in \Omega, $

$ u(x, t) = 0, \quad d(x, t) = d_0(x), \quad (x, t)\in \partial \Omega \times [0, T]. $

该方程组(1.1) 是刻画液晶流体的Ericksen-Leslie模型的简化版本, 也是目前为止能够保持原模型的非线性性质的最简化模型了(参见文献[1]). 在文章[2-3]中, 刘宪高等证明了判别不可压液晶方程组解的正则性的Serrin条件, 也可参见文献[4-5], 这些论文中揭示了以下函数类

$ L^{s, r}(Q_{T})\equiv L^{r}(0, T;L^{s}(\Omega)), \frac{2}{r}+\frac{n}{s} = 1. $

在研究液晶方程组的弱解的正则性中起着非常重要的作用. 本文中称这一类的弱解为Serrin解. 林芳华和柳春在文献[1]中首次证明了当初值$ u_{0}\in L^2, d_{0} \in H^{1} $方程组具有整体弱解, 并且当$ u_{0}\in H^{1}, d_{0} \in H^{2} $具有局部光滑解. 他们同时在文献[6]中证明了当弱解满足C-K-N条件(参见文献[7])时, 弱解具有局部光滑性. 王德华等在文章[8]中证明了不可压液晶方程组满足一定可导条件的小初值局部强解和整体弱解的存在性. 有关液晶方程组的更多结果可以参见文献[9-10] 及其中引用的参考文献. 本文我们主要关注液晶方程组(2.1)在更一般初值条件下的Serrin解的存在唯一性问题. 和文章[1, 8]一样, 为了计算的简便, 我们文中假设函数$ f(d) = 0 $并且$ \Omega = {{\Bbb R}} ^{n} $, 具体来说我们的主要结果如下:

定理1.1  设$ n \leq \sigma < s <\infty $,

$ \frac{2}{r}+\frac{n}{s} = \frac{n}{\sigma}, $

对于任意的

$ u_{0}\in H({{\Bbb R}} ^{n})\cap L^{\sigma}({{\Bbb R}} ^{n}), \ d_{0}\in H^{1}({{\Bbb R}} ^{n})\cap W^{1, \sigma}({{\Bbb R}} ^{n}), $

存在$ T >0 $使得不可压液晶方程组(0.1)在区域$ Q_{T} = {{\Bbb R}} ^{n}\times(0, T) $存在唯一弱解$ u, d $, 满足$ u\in L^{s, r}({{\Bbb R}} ^{n}\times (0, T)), \nabla d\in L^{s, r}({{\Bbb R}} ^{n}\times (0, T)) $. 进一步, 解的生命区间$ T $具有如下估计:

$ \rm(1) $如果$ \sigma > n $

$ T \geq \frac{C}{({\parallel u_{0} \parallel}_{\sigma}+{\parallel \nabla d_{0} \parallel}_{\sigma})^{1/\beta_{1}}}, \ \beta_{1} = \frac{1}{2}(1-\frac{n}{\sigma}), $

$ \rm(2) $如果$ \sigma = n $

$ T \geq \Big(\frac{C-\parallel u_{0}-u_{0\eta}\parallel_{n}-\parallel \nabla d _{0}-\nabla d_{0\eta}\parallel_{n}}{\parallel u_{0\eta}\parallel_{n}+\parallel \nabla d_{0\eta}\parallel_{n}}\Big)^{1/\beta_{2}}, \ \beta_{2} = \frac{1}{2}(1-\frac{n}{q}), $

这里$ C = C(n, s, \sigma) $为正常数, $ q \in (n, s) $, $ u_{0\eta}, \nabla d_{0\eta} $为函数$ u_0, \nabla d_{0} $关于空间变量的磨光函数, 并取$ \eta $足够小以满足条件$ \parallel u_{0}-u_{0\eta}\parallel_{n} +\parallel \nabla d_{0}-\nabla d_{0\eta}\parallel_{n}< C $.

注1.1  对于任意给定的初值条件$ u_{0}\in L^{p}\cap H, d_{0}\in W^{1, p}, p\geq n $, 我们得到了局部Serrin解的存在性. 在$ n = 2 $的情形, 定理的假设是自然的假设, 由文章[2, 11] 的Serrin正则性判定准则, 所得的弱解就是经典解, 而且是在全时空存在. 也就是说我们得到了在二维情形不可压液晶方程的光滑解的整体存在性. 但是在空间三维情形$ {{\Bbb R}} ^{3} $, 我们定理要求的初始条件是$ u_{0}\in L^3\cap H $, $ d_{0}\in W^{1, 3} $, 这和自然的条件$ u_{0}\in L^2\cap H, d_{0}\in W^{1, 2} $还是有一定的差距, 所以我们只能得到局部光滑解的存在性.

为了证明本定理, 我们先给出方程组(1.1)的弱解定义. 设$ Q $为时空区域$ {{\Bbb R}} ^n\times {{\Bbb R}} $, 空间$ H $中的测试函数$ D $为

$ H: = \left\{ X: X\in L^{2}({{\Bbb R}} ^n), \ {\rm{div}}\ X = 0 \right\}. $

$ D: = \left\{ X: X\in C_0^{\infty}(Q: {{\Bbb R}} ^n), \ {\rm{div}}\ X = 0 \right\}. $

定义2.1  如果$ (u, d) $在分布的意义下满足方程组(1.1), 我们就称其为方程组(1.1)的弱解$ Q $:

$ \rm1. $对任意的函数$ \varphi \in C_0^{\infty}(Q) $,

(2.1) $ \begin{equation} \int_Q \langle u, \nabla \varphi \rangle\ {\rm d}x{\rm d}t = 0; \end{equation} $

$ \rm2. $对任意的$ \Phi \in D $,

(2.2) $ \begin{equation} \int_Q \langle u, \Phi_t \rangle+\langle u, \Delta \Phi \rangle+\langle u, u \cdot \nabla \Phi \rangle\ {\rm d}x{\rm d}t = -\int_Q \langle \nabla d \odot \nabla d, \nabla \Phi \rangle \ {\rm d}x{\rm d}t; \end{equation} $

$ \rm3. $对任意的$ \Psi \in C^{\infty}_{0}(Q;{{\Bbb R}}^n) $,

(2.3) $ \begin{equation} \int_Q \langle d, \Psi_t \rangle+\langle d, \Delta \Psi \rangle+\langle d, u \cdot \nabla \Psi \rangle\ {\rm d}x{\rm d}t = -\int_Q \langle F(d), \Psi \rangle \ {\rm d}x{\rm d}t, \end{equation} $

这里$ \langle u, v \rangle $表示$ {{\Bbb R}} ^n $中的内积.

注2.1  文献[1]已经证明, 当$ u_{0}\in L^{2}, d_{0}\in H^{1} $, 方程组(1.1)存在弱解$ (u, d) $, 并且

$ u\in L^{2, \infty}(Q), |\nabla u|\in L^{2, 2}(Q), |\nabla d|\in L^{2, \infty}(Q), |\nabla^2 d|\in L^{2, 2}(Q). \label{rE2.1} $

所以我们的定理的初始条件自然保证了整体弱解的存在性.

本文的主要定理可以看成经典的热方程的能量估计和线性Stokes方程的连续逼近的一个直接应用. 在正式证明之前我们先给出线性Stokes方程组的存在性的结果:

(2.4) $ \begin{equation} \left \{\begin{array}{l} u_t-\Delta u +\nabla P = \rm{div}(F), \\ \rm{div}\ u = 0, \\ u(x, 0) = u_0(x), \\ \end{array} \right. \end{equation} $

这里$ F $为给定的二阶张量场, $ {\rm{div}} {F}_{j} = \partial_{i}F_{ij} $. 和Navier-Stokes方程组以及液晶方程组类似.我们称满足下式的$ u $为方程组(2.4)的弱解

(2.5) $ \begin{equation} \int_Q {\langle u, \Phi_t \rangle-\langle \nabla u, \nabla \Phi \rangle }{\rm d}x{\rm d}t = \int_Q \langle F, \nabla \Phi \rangle \ {\rm d}x{\rm d}t-\int_{{{\Bbb R}} ^n}\langle u_{0}, \Phi(0) \rangle {\rm d}x, \ \ \rm{对任意} \ \Phi \in D. \end{equation} $

根据以下的Helmholtz-Weyl型的分解引理(参见文献[12]), 只要设$ \partial_{i}F_{ij}-\partial_{j}p = \partial_{i}G_{ij} $就可以把Stokes方程看成非齐次热方程.

引理2.1  设$ {\bf F} = \{F_{ij}\} $为二阶张量场, 且满足

$ F_{ij}\in L^{r}({{\Bbb R}} ^{n}), \ \partial_{i}F_{ij}\in L^{s}({{\Bbb R}} ^{n}), \ j = 1, \cdots , n, \ 1<r, s <\infty, $

则存在一个张量场$ {\bf G} = \{G_{ij}\} $, 以及一个数量场$ p $, 使得以下各式成立$ i, j = 1, \cdots , n $

$ \begin{eqnarray*} &&\partial_{j}\partial_{i}G_{ij} = 0, \\ &&G_{ij}\in L^{r}({{\Bbb R}} ^{n}), \ \partial_{i}G_{ij}\in L^{s}({{\Bbb R}} ^{n}), p\in L^{r}({{\Bbb R}} ^{n}), \ \partial_{j}p\in L^{s}({{\Bbb R}} ^{n}), \\ &&\partial_{i}F_{ij} = \partial_{i}G_{ij}+\partial_{j}p, \\ && \parallel G_{ij}\parallel_{r}\leq c(n, r)(\parallel F_{ij}\parallel_{r}), \\ && \parallel \partial_{i}G_{ij}\parallel_{s}\leq c(n, s)(\parallel \partial_{i}F_{ij} \parallel_{s}). \end{eqnarray*} $

也就是说我们把Stokes方程(2.4) 化为以下的非齐次热方程

(2.6) $ \begin{equation} \left \{\begin{array}{l} u_t-\Delta u = \rm{div}(G), \ (x, t)\in {{\Bbb R}} ^{n}\times (0, \infty), \\ u(x, 0) = u_0(x), \\ \end{array} \right. \end{equation} $

以下关于热方程(2.6)的解的估计的引理2.2是本文的关键引理, 相关的证明参见文献[4], 文献[13-15] 也有类似结果.

引理2.2  对于问题(2.6), 设$ n < s < \infty, 1 < q_{1} < \infty, \frac{1}{s_{1}} = \frac{1}{s}+\frac{1}{2}, \frac{1}{q_{2}} = 1-\frac{n}{2s}, \alpha = \frac{1}{2}(1-\frac{n}{s}), $且

(2.7) $ {\bf G}\in L^{\frac{s}{2}, q_{1}}({{\Bbb R}} ^{n}\times (0, T)),\\ \begin{equation} {\rm{div}}{\bf G}\in L^{s_{1}, q_{2}}({{\Bbb R}} ^{n}\times (0, T)), \nonumber \end{equation} \\ t^{2\alpha}{\bf G}\in BC([0, T);L^{\frac{s}{2}}({{\Bbb R}} ^{n})),\\t^{\alpha}{\bf G}\in BC([0, T);L^{s_{1}}({{\Bbb R}} ^{n})), $

同时假设

$ u_{0}\in H({{\Bbb R}} ^{n}), $

$ W\ast u_{0}\in L^{s, r}({{\Bbb R}} ^{n}\times (0, T)), $

$ t^{\alpha} W\ast u_{0}\in BC([0, T);L^{s}({{\Bbb R}} ^{n})), $

这里$ W(x, t) = \frac{1}{(4\pi t)^{\frac{n}{2}}}exp(-\frac{x^{2}}{4t}) $为Gauss-Weierstrass热核函数, 且指标满足

$ \frac{1}{r} = \frac{1}{q_{1}}-\frac{1}{2}(1-\frac{n}{s}), $

则问题(2.4)存在唯一弱解$ u $

$ u\in L^{s, r}({{\Bbb R}} ^{n}\times (0, T)), $

$ t^{\alpha} u\in BC([0, T);L^{s}({{\Bbb R}} ^{n})), $

并且此弱解$ u $满足以下估计

(2.8) $ \begin{equation} \parallel u \parallel_{s, r, T}\ \leq\ \parallel W \ast u_{0} \parallel_{s, r, T}+C\parallel G \parallel_{s/2, q_{1}, T}, \end{equation} $

(2.9) $ \begin{equation} \ll u\gg_{s, \alpha, T}\ \leq\ \ll W \ast u_{0} \gg_{s, \alpha, T}+C\ll G \gg_{s/2, 2\alpha, T}, \end{equation} $

(2.10) $ \begin{equation} \parallel u \parallel_{2, \infty, T} \leq\ \parallel u_{0}\parallel_2+C\ll G \gg_{s_{1}, \alpha, T}, \end{equation} $

(2.11) $ \begin{equation} \parallel \nabla u \parallel_{2, 2, T} \ \leq \ C\parallel u_{0}\parallel_2+\parallel \rm{div} G \parallel_{s_{1}, q_{2}, T}. \end{equation} $

这里$ t^{\alpha}u\in BC([0, T);L^{q}({{\Bbb R}} ^{n})), \ u\in L^{s, r}({{\Bbb R}} ^{n}\times (0, T)) $, 相应的范数定义为

$ \parallel u\parallel_{s, r, T}\triangleq \parallel u\parallel_{L^{s, r}({{\Bbb R}} ^{n}\times (0, T))}, \ \ll u\gg_{s, \alpha, T}\triangleq \sup \limits_{t\in [0, T)}t^{\alpha}\parallel u\parallel_{s}. $

在证明本文的主要定理之前, 为方便读者, 我们还需要列出以下的引理的结果. 引理的第一个结论利用Young不等式直接可得, 第二部分只要使用Marcinkiewicz插值定理^[13].

引理2.3  设$ u_{0}\in L^{\sigma}({{\Bbb R}} ^{n}), \ 1 < \sigma, s \leq \infty $, 且

$ \frac{1}{r} = \frac{n}{2}(\frac{1}{\sigma}-\frac{1}{s}), $

则存在正常数$ C = C(n, s, \sigma), T $, 使得对任意的$ t\in (0, T) $以下性质成立

(2.12) $ \begin{equation} \parallel W\ast u_{0}\parallel_{s} \leq Ct^{-\frac{1}{r}}\parallel u_{0} \parallel_{\sigma}, \ \sigma \leq s. \end{equation} $

(2.13) $ \begin{equation} \parallel W\ast u_{0}\parallel_{s, r, T} \leq C\parallel u_{0} \parallel_{\sigma}, \ \sigma < s. \end{equation} $

本节我们证明主要定理1.1, 即以下不可压液晶方程初值问题的解的存在性唯一性

(3.1) $ \begin{equation} \left \{\begin{array}{l} u_t-\Delta u+u\cdot\nabla u +\nabla P = -{\rm{div}}(\nabla d \odot \nabla d), \\ {\rm{div}}\ u = 0, \\ d_t+u\cdot \nabla d-\Delta d = 0, \\ u(x, 0) = u_{0}, d(x, 0) = d_{0}, \\ \end{array} \right. \end{equation} $

证  我们主要采用的连续逼近的方法, 证明过程中用$ C $代表至多依赖于$ n, s, \sigma $且与函数变量无关的常数. 首先关于$ d $的方程两边进行梯度求导得到

$ \nabla d_{t}-\Delta (\nabla d) = -\nabla (u\cdot \nabla d), $

令

$ u_{1}(x, t) = W(t)*u_0, \nabla d = e, \nabla d(x, 0) = \nabla d_{0} = e_{0}, e_{1}(x, t) = W(t)\ast e_{0}, $

设$ u_{k+1}, e_{k+1} $是以下Stokes型问题的解

(3.2) $ \begin{equation} \left \{\begin{array}{l} u_{k+1, t}-\Delta u_{k+1} = -{\rm{div}}(u_{k}\otimes u_{k})-{\rm{div}}(e_{k} \odot e_{k}), \\ e_{k+1, t}-\Delta (e_{k+1}) = -\nabla (u_{k}\cdot e_{k}), \\ u_{k+1}(x, 0) = u_{0}, e_{k+1}(x, 0) = e_{0}, \\ \end{array} \right. \end{equation} $

需要指出的是, 以上方程的解都是在时空区域$ {{\Bbb R}} ^{n}\times(0, T) $上以弱解的意义成立. 由引理2.2结合Hölder不等式, 我们得到

(3.3) $ \begin{eqnarray} \parallel u_{k+1}\parallel_{s, r, T} &\leq & \parallel u_{1}\parallel_{s, r, T}+C(\parallel u_{k}^{2}\parallel_{s/2, q_{1}, T}+\parallel e_{k}^{2}\parallel_{s/2, q_{1}, T}){}\\ &\leq & \parallel u_{1}\parallel_{s, r, T}+C(\parallel u_{k}\parallel^{2}_{s, 2q_{1}, T}+\parallel e_{k}\parallel^{2}_{s, 2q_{1}, T}){}\\ &\leq& \parallel u_{1}\parallel_{s, r, T}+CT^{\beta_1}(\parallel u_{k}\parallel^{2}_{s, r, T}+\parallel e_{k}\parallel^{2}_{s, r, T}), \end{eqnarray} $

以及

(3.4) $ \begin{eqnarray} \parallel e_{k+1}\parallel_{s, r, T} &\leq& \parallel e_{1}\parallel_{s, r, T}+C(\parallel u_{k}e_{k}\parallel_{s/2, q_{1}, T} ) {}\\ &\leq& \parallel e_{1}\parallel_{s, r, T}+C(\parallel u_{k}\parallel_{s, 2q_{1}, T}\parallel e_k\parallel_{s, 2q_{1}, T}){}\\ &\leq& \parallel e_{1}\parallel_{s, r, T}+CT^{\beta_1}(\parallel u_{k}\parallel_{s, r, T}\cdot\parallel e_{k}\parallel_{s, r, T}), \end{eqnarray} $

这里, 我们用到不等式

$ |u|^{2}_{2q_{1}} \leq |u|^{2}_{r}|1|^{2}_{\beta} = T^{\beta_{1}}|u|_{r}^{2}, \ \beta_{1} = \frac{2}{\beta} = \frac{1}{q_1}-\frac{2}{r} = \frac{1}{2}(1-\frac{n}{\sigma}). $

由式(2.9), 直接计算可以导出

$ \begin{eqnarray*} \ll u_{k+1}\gg_{s, \alpha, T} &\leq& \ll u_{1} \gg_{s, \alpha, T}+C(\ll u_{k}^{2}\gg_{s/2, 2\alpha, T}+\ll e_{k}^{2}\gg_{s/2, 2\alpha, T})\\ &\leq& \ll u_{1} \gg_{s, \alpha, T}+C(\ll u_{k}\gg^{2}_{s, \alpha, T}+\ll e_{k}\gg^{2}_{s, \alpha, T}), \\ \ll e_{k+1}\gg_{s, \alpha, T} &\leq& \ll e_{1} \gg_{s, \alpha, T}+C(\ll u_{k} e_{k}\gg_{s/2, 2\alpha, T})\\ &\leq&\ll e_{1} \gg_{s, \alpha, T}+C(\ll u_{k}\gg_{s, \alpha, T}\cdot\ll e_{k}\gg_{s, \alpha, T}), \end{eqnarray*} $

又由式(2.10), 得到

(3.5) $ \begin{eqnarray} \parallel u_{k+1}\parallel_{2, \infty, T} &\leq& \parallel u_{0}\parallel_2+C(\ll u_{k}^{2}\gg_{s_{1}, \alpha, T}+\ll e_{k}^{2}\gg_{s_{1}, \alpha, T})\\ &\leq &\parallel u_{0}\parallel_2+C(\parallel u_{k}\parallel_{2, \infty, T}\ll u_{k}\gg_{s, \alpha, T}+ \parallel e_{k}\parallel_{2, \infty, T}\ll e_{k}\gg_{s, \alpha, T}, \end{eqnarray} $

(3.6) $ \begin{eqnarray} \parallel e_{k+1}\parallel_{2, \infty, T} &\leq&\parallel e_{0}\parallel_2+C\ll u_{k} e_{k}\gg_{s_{1}, \alpha, T}{}\\ &\leq &\parallel e_{0}\parallel_2+C\parallel u_{k}\parallel_{2, \infty, T}\ll e_{k}\gg_{s, \alpha, T}, \end{eqnarray} $

这里我们用到了不等式

$ \ll u_{k}e_{k}\gg_{s_{1}, \alpha, T} = \sup\limits_{t\in[0, T]}t^{\alpha}|u_{k} e_{k}|_{s_{1}} \leq \sup\limits_{t\in[0, T]}t^{\alpha}|u_{k}|_{2, \infty}|e_{k}|_{s, T}, $

$ \ll u_{k}^{2}\gg_{s_{1}, \alpha, T} = \sup\limits_{t\in[0, T]}t^{\alpha}|u_{k}^{2}|_{s_{1}} \leq \sup\limits_{t\in[0, T]}t^{\alpha}|u_{k}|_{2, \infty}|u_{k}|_{s, T}. $

由式(2.11), 得到

$ \begin{eqnarray*} \parallel \nabla u_{k+1}\parallel_{2, 2, T} &\leq&C(\parallel u_{0}\parallel_2+\parallel u_{k}\nabla u_{k}\parallel_{s_{1}, q_{2}, T}+\parallel e_{k}\nabla e_{k}\parallel_{s_{1}, q_{2}, T})\\ &\leq&C(\parallel u_{0}\parallel_2+\parallel u_{k}\parallel_{s, \frac{2s}{s-n}, T}\parallel \nabla u_{k}\parallel_{2, 2, T}+\parallel e_{k}\parallel_{s, \frac{2s}{s-n}, T}\parallel \nabla e_{k}\parallel_{2, 2, T})\\ &\leq&C[\parallel u_{0}\parallel_2+T^{\beta_{1}}(\parallel u_{k}\parallel_{s, r, T}\parallel \nabla u_{k}\parallel_{2, 2, T}+\parallel e_{k}\parallel_{s, r, T}\parallel \nabla e_{k}\parallel_{2, 2, T})], \\ \parallel \nabla e_{k+1}\parallel_{2, 2, T} &\leq&C(\parallel e_{0}\parallel_2+\parallel u_{k}\nabla e_{k} \parallel_{s_{1}, q_{2}, T}+\parallel e_{k}\nabla u_{k} \parallel_{s_{1}, q_{2}, T})\\ &\leq &C(\parallel e_{0}\parallel_2+\parallel u_{k}\parallel_{s, \frac{2s}{s-n}, T}\parallel \nabla e_{k}\parallel_{2, 2, T}+\parallel e_{k}\parallel_{s, \frac{2s}{s-n}, T}\parallel \nabla u_{k}\parallel_{2, 2, T})\\ &\leq&C(\parallel e_{0}\parallel_2+T^{\beta_{1}}\parallel u_{k}\parallel_{s, r, T}\parallel \nabla e_{k}\parallel_{2, 2, T}+T^{\beta_{1}}\parallel e_{k}\parallel_{s, r, T}\parallel \nabla u_{k}\parallel_{2, 2, T}), \end{eqnarray*} $

以上最后两步, 用到了以下的Hölder不等式

$ \parallel u\cdot v\parallel_{s_{1}, q_{2}} \leq \parallel u\parallel_{s, 2s/(s-n)}\parallel v\parallel_{2, 2}, $

$ \parallel u\cdot v\parallel_{2s/(s-n)} \leq \parallel u\parallel_{r}\parallel v\parallel_{\beta}. $

这里的指数满足

$ \frac{1}{s_{1}} = \frac{1}{s}+\frac{1}{2}, \ \frac{1}{q_{2}} = 1-\frac{n}{2s} = \frac{1}{2}+\frac{s-n}{2s}, \ \beta_{1} = \frac{1}{\beta} = \frac{s-n}{2s}-\frac{1}{r} = \frac{1}{2}(1-\frac{n}{\sigma}). $

我们断言以下的解序列在$ L^{s, r}(Q_{T}) $中一致有界

$ \parallel u_{k+1} \parallel_{s, r, T} , \ \parallel e_{k+1} \parallel_{s, r, T}, \ \ll u_{k+1} \gg_{s, \alpha, T} , \ \ll e_{k+1} \gg_{s, \alpha, T}, k = 0, 1, 2, \cdots . $

设

(3.7) $ \begin{equation} \parallel u_{1} \parallel_{s, r, T}\leq K_{1}, \ \ll u_{1} \gg_{s, \alpha, T}\leq K_{2}, \end{equation} $

(3.8) $ \begin{equation} \parallel e_{1} \parallel_{s, r, T}\leq K_{3}, \ \ll e_{1} \gg_{s, \alpha, T}\leq K_{4}, \end{equation} $

这里$ K_{1}, K_{2}, K_{3}, K_{4} $为待定常数.

我们接下来证明以下一致估计

(3.9) $ \begin{equation} \parallel u_{k+1} \parallel_{s, r, T}\leq 2(K_{1}+K_{3}), \ \parallel e_{k+1} \parallel_{s, r, T}\leq 2(K_{1}+K_{3}), k = 0, 1, 2, \cdots , \end{equation} $

(3.10) $ \begin{equation} \ll u_{k+1} \gg_{s, \alpha, T}\leq 2(K_{2}+K_{4}), \ \ll e_{k+1} \gg_{s, \alpha, T}\leq 2(K_{2}+K_{4}), k = 0, 1, 2, \cdots . \end{equation} $

由式(3.3), (3.4), 可得

(3.11) $ \begin{eqnarray} \parallel u_{k+1} \parallel_{s, r, T} &\leq& K_{1}+CT^{\beta_{1}}(K_{1}+K_{3})^{2} {}\\ &\leq& (K_{1}+K_{3})(1+CT^{\beta_{1}}(K_{1}+K_{3})), \end{eqnarray} $

(3.12) $ \begin{eqnarray} \ll u_{k+1} \gg_{s, \alpha, T} & \leq &K_{2}+C(K_{2}+K_{4})^{2}{}\\ & \leq &(K_{2}+K_{4})(1+C(K_{2}+K_{4})), \end{eqnarray} $

(3.13) $ \begin{eqnarray} \parallel e_{k+1} \parallel_{s, r, T} &\leq&K_{3}+CT^{\beta_{1}}(K_{1}+K_{3})^{2}{}\\ &\leq& (K_{1}+K_{3})(1+CT^{\beta_{1}}(K_{1}+K_{3})), \end{eqnarray} $

(3.14) $ \begin{eqnarray} \ll e_{k+1} \gg_{s, \alpha, T} &\leq& K_{4}+C(K_{2}+K_{4})^{2}{}\\ &\leq& (K_{2}+K_{4})(1+C(K_{2}+K_{4})). \end{eqnarray} $

于是, 要式(3.9), (3.10)成立, 只要下列条件满足即可

(3.15) $ \begin{equation} CT^{\beta_{1}}(K_{1}+K_{3}) < 1, \end{equation} $

(3.16) $ \begin{equation} C(K_{2}+K_{4}) < 1, \end{equation} $

接下来, 我们显示式(3.7), (3.8) 在以下两种情形下成立.

情形1  如果$ \sigma > n $, 由引理2.3知

$ \parallel u_{1}\parallel_{s, r, T} \leq C\parallel u_{0}\parallel_{\sigma}, \ \parallel e_{1}\parallel_{s, r, T} \leq C\parallel e_{0}\parallel_{\sigma}, $

于是令

$ K_{1} = C\parallel u_{0}\parallel_{\sigma}, \ K_{3} = C\parallel \nabla d_{0}\parallel_{\sigma}, $

要式(3.9)成立, 只要选取适当的$ T $满足

$ CT^{\beta_{1}}(\parallel u_{0}\parallel_{\sigma}+\parallel e_{0}\parallel_{\sigma}) < 1. $

类似地, 根据

$ t^{\alpha}\parallel u_{1}\parallel_{s} = t^{(1-n/s)/2}\parallel u_{1}\parallel_{s} \leq Ct^{-\frac{1}{r}}t^{(1-n/s)/2}\parallel u_{0}\parallel_{\sigma}, $

$ t^{\alpha}\parallel \nabla d_{1}\parallel_{s} = t^{(1-n/s)/2}\parallel \nabla d_{1}\parallel_{s} \leq Ct^{-\frac{1}{r}}t^{(1-n/s)/2}\parallel e_{0}\parallel_{\sigma}, $

令

$ t^{-\frac{1}{r}}t^{(1-n/s)/2} = t^{\frac{1}{2}(1-\frac{n}{\sigma})} = t^{\beta_{1}}, \sigma > n, \beta_{1}> 0, $

同时取

$ K_{2} = CT^{\beta_{1}}\parallel u_{0}\parallel_{\sigma}, \ K_{4} = CT^{\beta_{1}}\parallel e_{0}\parallel_{\sigma}, $

只要选取适当的$ T $满足

$ CT^{\beta_{1}}(\parallel u_{0}\parallel_{\sigma}+\parallel e_{0}\parallel_{\sigma}) < 1. $

容易得出式(3.10)成立.

情形2  如果$ \sigma = n $, 令$ u_{0\eta} = u_{0}\ast J_{\eta} \in L^{q}({{\Bbb R}} ^{n}), \ q \in (1, \infty], $这里$ J_{\eta} $为支集在$ (-\eta, \eta) $中的光滑紧支函数, $ \int_{-\infty}^{+\infty}J_{\eta}{\rm d}s = 1 $, $ u_{1\eta} = w\ast u_{0\eta} $对任意的$ q\in (n, s) $, 有

$ \begin{eqnarray*} \parallel u_{1}\parallel_{s, r, T} &\leq & \parallel u_{1}-u_{1\eta}+u_{1\eta}\parallel_{s, r, T}\\ &\leq & \parallel u_{1}-u_{1\eta}\parallel_{s, r, T}+\parallel u_{1\eta}\parallel_{s, r, T}\\ &\leq & \parallel u_{0}-u_{0\eta}\parallel_{n}+CT^{\beta_{2}}\parallel u_{0\eta}\parallel_{q}, \\ \ll u_{1}\gg_{\alpha, s, T}& = &\sup\limits_{t\in[0, T]}t^{\alpha}|u_{1}|_{s}\\ &\leq & \sup\limits_{t\in[0, T]}t^{\alpha}(\parallel u_{1}-u_{1\eta}+u_{1\eta}\parallel_{s})\\ &\leq & \sup\limits_{t\in[0, T]}t^{\alpha}(\parallel u_{1}-u_{1\eta}\parallel_{s}+\parallel u_{1\eta}\parallel_{s})\\ &\leq &(t^{-\alpha}\parallel u_{0}-u_{0\eta}\parallel_{n}+CT^{-\frac{1}{r'}}\parallel u_{0\eta}\parallel_{q})\\ & = & \parallel u_{0}-u_{0\eta}\parallel_{n}+CT^{\beta_{2}}\parallel u_{0\eta}\parallel_{q}), \end{eqnarray*} $

这里指数满足

$ \frac{1}{r'} = \frac{n}{2}(\frac{1}{q}-\frac{1}{s}), \beta_{2} = \frac{1}{r}-\frac{1}{r'} = \frac{1}{2}(1-\frac{n}{q}). $

同理可得

$ \begin{eqnarray*} \parallel e_{1}\parallel_{s, r, T} &\leq & \parallel e_{1}-e_{1\eta}+e_{1\eta}\parallel_{s, r, T}\\ &\leq & \parallel e_{1}-e_{1\eta}\parallel_{s, r, T}+\parallel e_{1\eta}\parallel_{s, r, T}\\ &\leq & \parallel e_{0}-e_{0\eta}\parallel_{n}+CT^{\beta_{2}}\parallel e_{0\eta}\parallel_{q}, \\ \ll e_{1}\gg_{\alpha, s, T}& = &\sup\limits_{t\in[0, T]}t^{\alpha}|e_{1}|_{s}\\ &\leq & \sup\limits_{t\in[0, T]}t^{\alpha}(\parallel e_{1}-e_{1\eta}+e_{1\eta}\parallel_{s})\\ &\leq & \sup\limits_{t\in[0, T]}t^{\alpha}(\parallel e_{1}-e_{1\eta}\parallel_{s}+\parallel e_{1\eta}\parallel_{s})\\ &\leq & t^{\alpha}(t^{-\alpha}\parallel e_{0}-e_{0\eta}\parallel_{n}+CT^{-\frac{1}{r'}}\parallel e_{0\eta}\parallel_{q})\\ & = & \parallel e_{0}-e_{0\eta}\parallel_{n}+CT^{\beta_{2}}\parallel e_{0\eta}\parallel_{q}). \end{eqnarray*} $

令

$ K_{1} = K_{2} = \parallel u_{0}-u_{0\eta}\parallel_{n}+CT^{\beta_{2}}\parallel u_{0\eta}\parallel_{q}, $

$ K_{3} = K_{4} = \parallel e_{0}-e_{0\eta}\parallel_{n}+CT^{\beta_{2}}\parallel e_{0\eta}\parallel_{q}, $

让$ \eta $和$ T $取足够小, 满足

(3.17) $ \begin{equation} \parallel u_{0}-u_{0\eta}\parallel_{n}+\parallel e_{0}-e_{0\eta}\parallel_{n}+CT^{\beta_{2}}(\parallel u_{0\eta}\parallel_{q}+\parallel e_{0\eta}\parallel_{q})< 1. \end{equation} $

即有式(3.7), (3.8)成立. 于是我们得到

$ \parallel u_{k+1} \parallel_{s, r, T}\leq 2(K_{1}+K_{3}), \ \ll u_{k+1} \gg_{s, \alpha, T}\leq 2(K_{2}+K_{4}), k = 1, 2, 3, \cdots, $

$ \parallel e_{k+1} \parallel_{s, r, T}\leq 2(K_{1}+K_{3}), \ \ll e_{k+1} \gg_{s, \alpha, T}\leq 2(K_{2}+K_{4}), k = 1, 2, 3, \cdots. $

由式(3.5), (3.6), 可推出

$ \parallel u_{k+1} \parallel_{2, \infty, T}\leq \parallel u_{0} \parallel_{2}+C(\parallel u_{k} \parallel_{2, \infty, T}\ll u_{k}\gg_{s, \alpha, T}+\parallel e_{k} \parallel_{2, \infty, T}<<e_{k}>>_{s, \alpha, T}), $

$ \parallel e_{k+1} \parallel_{2, \infty, T}\leq \parallel e_{0} \parallel_{2}+C(\parallel e_{k} \parallel_{2, \infty, T} \ll u_{k}\gg_{s, \alpha, T}). $

于是在$ \sigma > n $的情形, 我们有

(3.18) $ \begin{eqnarray} \parallel u_{k+1} \parallel_{2, \infty, T} & \leq &\parallel u_{0} \parallel_{2}+CT^{\beta_{1}}(\parallel u_{0} \parallel_{2}\parallel u_{0} \parallel_{\sigma}+\parallel e_{0} \parallel_{2}\parallel e_{0} \parallel_{\sigma} ){}\\ & \leq &(\parallel u_{0} \parallel_{2}+\parallel e_{0} \parallel_{2})(1+CT^{\beta_{1}}(\parallel u_{0} \parallel_{\sigma}+\parallel e_{0} \parallel_{\sigma})), \end{eqnarray} $

(3.19) $ \begin{eqnarray} \parallel e_{k+1} \parallel_{2, \infty, T} &\leq &\parallel e_{0} \parallel_{2}+CT^{\beta_{1}}\parallel u_{0} \parallel_{2}\parallel e_{0} \parallel_{\sigma} {}\\ &\leq &\parallel e_{0} \parallel_{2}(1+CT^{\beta_{1}}\parallel e_{0} \parallel_{\sigma}), \end{eqnarray} $

选取适当的$ T $使得

$ CT^{\beta_{1}}(\parallel u_{0} \parallel_{\sigma}+\parallel e_{0}\parallel_{\sigma}) <1, $

则有

$ \parallel u_{k+1} \parallel_{2, \infty, T} \leq 2(\parallel u_{0} \parallel_{2}+\parallel e_{0} \parallel_{2}), $

$ \parallel e_{k+1} \parallel_{2, \infty, T} \leq 2\parallel e_{0} \parallel_{2}. $

当$ \sigma = n $时, 在式(3.9), (3.10)中, 可以取适当的$ \eta, T $使得$ K_{2}, K_{4} $充分小, 使得

$ \parallel \nabla u_{k+1} \parallel_{2, 2, T} \leq C\parallel u_{0} \parallel_{2}, $

$ \parallel \nabla e_{k+1} \parallel_{2, 2, T} \leq C\parallel e_{0} \parallel_{2}. $

接下来我们证明序列$ u_{k}, e_{k}\ k = 1, 2, \cdots $收敛到空间$ L^{s, r} $中的一个弱解$ u(x, t), e(x, t) $. 为此, 我们把$ (u_{k+1}, d_{k+1}) $和$ (u_{k}, d_{k}), \ k = 1, 2, \cdots $分别满足的方程组相减, 得到一个新的初值为零的方程组. 对于新的方程组的解($ u_{k+1}-u_{k} $, $ d_{k+1}-d_{k} $)利用引理2.2, 由式(3.9), (3.10), 可得

$ \begin{eqnarray*} \parallel u_{k+1}-u_{k}\parallel_{s, r, T}&\leq& CT^{\beta_{1}}(\parallel u_{k}\parallel_{s, r, T}+\parallel u_{k-1}\parallel_{s, r, T})\parallel u_{k}-u_{k-1}\parallel_{s, r, T}\\ &&+CT^{\beta_{1}}(\parallel e_{k}\parallel_{s, r, T}+\parallel e_{k-1}\parallel_{s, r, T})\parallel e_{k}-e_{k-1}\parallel_{s, r, T}\\ & \leq& CT^{\beta_{1}}(\parallel u_{k}\parallel_{s, r, T}+\parallel u_{k-1}\parallel_{s, r, T}+\parallel e_{k}\parallel_{s, r, T}+\parallel e_{k-1}\parallel_{s, r, T}) \\ &&\cdot (\parallel u_{k}-u_{k-1}\parallel_{s, r, T}+ \parallel e_{k}-e_{k-1}\parallel_{s, r, T}), \\ \parallel e_{k+1}-e_{k}\parallel_{s, r, T} &\leq&CT^{\beta_{1}}(\parallel e_{k}\parallel_{s, r, T}\parallel u_{k}-u_{k-1}\parallel_{s, r, T}+\parallel u_{k-1}\parallel_{s, r, T}\parallel e_{k}-e_{k-1}\parallel_{s, r, T})\\ &\leq& CT^{\beta_{1}}(\parallel e_{k}\parallel_{s, r, T}+\parallel u_{k-1}\parallel_{s, r, T})\\ &&\cdot (\parallel u_{k}-u_{k-1}\parallel_{s, r, T}+\parallel e_{k}-e_{k-1}\parallel_{s, r, T}), \end{eqnarray*} $

由于$ \parallel u_{k+1}\parallel_{s, r, T}, \ \parallel e_{k+1}\parallel_{s, r, T}, $的一致有界性我们得到

$ \begin{eqnarray*} &&\parallel u_{k+1}-u_{k}\parallel_{s, r, T}+\parallel e_{k+1}-e_{k}\parallel_{s, r, T}\\ &\leq& CT^{\beta_{1}}(\parallel u_{0}\parallel_{\sigma}+\parallel e_{0}\parallel_{\sigma})(\parallel u_{k}-u_{k-1}\parallel_{s, r, T} +\parallel e_{k}-e_{k-1}\parallel_{s, r, T}), \end{eqnarray*} $

于是选取适当的正常数$ T $满足

$ CT^{\beta_{1}}(\parallel u_{0}\parallel_{\sigma}+\parallel e_{0}\parallel_{\sigma})< \alpha < 1, $

有

$ \parallel u_{k+1}-u_{k}\parallel_{s, r, T}+\parallel e_{k+1}-e_{k}\parallel_{s, r, T} \leq \alpha(\parallel u_{k}-u_{k-1}\parallel_{s, r, T} +\parallel e_{k}-e_{k-1}\parallel_{s, r, T}). $

由压缩映像原理知$ u_{k}, e_{k}, k = 1, 2, \cdots $为空间$ L^{s, r} $中的的柯西列, 于是存在$ u\in L^{s, r}, \ e \in L^{s, r} $, 满足

$ u_{k}\rightarrow u, \ k\rightarrow \infty, $

$ \ e_{k}\rightarrow e, \ k\rightarrow \infty, $

同时, 根据估计式

$ \parallel u_{k+1} \parallel_{2, \infty, T} \leq C\parallel u_{0}\parallel_{2}, \ \parallel e_{k+1} \parallel_{2, \infty, T} \leq C\parallel e_{0} \parallel_{2}, $

$ \parallel \nabla u_{k+1} \parallel_{2, 2, T} \leq C\parallel u_{0}\parallel_{2}, \ \parallel \nabla e_{k+1} \parallel_{2, 2, T} \leq C\parallel e_{0} \parallel_{2}, $

得到

$ u\in L^{\infty}((0, T);L^{2}({{\Bbb R}} ^{n}))\cap L^{2}((0, T);H^{1}({{\Bbb R}} ^{n})), $

$ d\in L^{\infty}((0, T);H^{1}({{\Bbb R}} ^{n}))\cap L^{2}((0, T);H^{2}({{\Bbb R}} ^{n})), $

于是$ u, d $即为我们所求的弱解.

证  设$ (u_{1}, d_{1}, P_{1}) $和$ (u_{2}, d_{2}, P_{2}) $为方程组(3.1)的在$ L^{s, r}(Q_{T}) $空间中的两个解. 即$ (u_{1}, e_{1}, P_{1}) $和$ (u_{2}, e_{2}, P_{2}) $分别满足以下方程

(3.20) $ \begin{equation} \left \{\begin{array}{l} u_t-\Delta u+u\cdot\nabla u +\nabla P = -{\rm{div}}(e_{i} \odot e_{i}), \\ {\rm{div}}\ u = 0, \\ e_t+\nabla (u\cdot e)-\Delta e = 0, \\ u(x, 0) = u_{0}, e_{i}(x, 0) = e_{0}, \\ \end{array} \right. \end{equation} $

把$ (u_{1}, e_{1}, P_{1}) $和$ (u_{2}, e_{2}, P_{2}) $满足的方程相减, 并令

$ w = u_{1}-u_{2}, h = e_{1}-e_{2}, e_{1} = \nabla d_{1}, e_{2} = \nabla d_{2}. $

可知$ (w, h, P_{1}-P_{2}) $在弱解的意义下满足

(3.21) $ \begin{equation} \left \{\begin{array}{l} w_t-\Delta w+\nabla (P_{1}-P_{2}) = -w\cdot\nabla u_{1} -u_{2}\cdot\nabla w-{\rm{div}}(h^{T} e_{1} +e^{T}_{2}\cdot h ), \\ {\rm{div}}\ w = 0, \\ h_t-\Delta h = - e_{1}\nabla w-h\nabla u_{2}-w\cdot \nabla e_{1}-u_{2}\cdot \nabla h, \\ w(x, 0) = 0, e(x, 0) = 0, \\ \end{array} \right. \end{equation} $

以上方程组(3.21)中第一个方程两边乘以$ w $并在区间$ Q_{T} $积分得到

(3.22) $ \begin{equation} \parallel w(t)\parallel^{2}_{2}+2\int^{t}_{0}{\parallel\nabla w\parallel^{2}_{2} = \int^{t}_{0}{-(w\cdot\nabla u_{1}, w) -(u_{2}\cdot\nabla w, w)-({\rm{div}}(h^{T} e_{1} +e^{T}_{2}\cdot h), w)}}, \end{equation} $

同理在方程组(3.22)中第三个方程两边同乘$ h $并在区间$ Q_{T} $积分得到

(3.23) $ \begin{equation} \parallel h(t)\parallel^{2}_{2}+2\int^{t}_{0}{\parallel h\parallel^{2}_{2} = \int^{t}_{0}{-(e_{1}\nabla w, h)-(h\nabla u_{2}, h)-(w\cdot \nabla e_{1}, h)-(u_{2}\cdot \nabla h, h)}}, \end{equation} $

对以上两个等式的右端项分别应用以下的估计(参见文献[16]的引理2.4或者文献[12]的引理4.1)

(3.24) $ \begin{equation} \int^{t}_{0}(v\nabla w, u)\leq C\left(\int^{t}_{0}\parallel \nabla w\parallel^{2}_{2}\right)^{\frac{1}{2}}\left(\int^{t}_{0}\parallel \nabla v\parallel^{2}_{2}\right)^{\frac{n}{2s}}\left(\int^{t}_{0}\parallel \nabla u\parallel^{r}_{s}\parallel v\parallel^{2}_{2}\right)^{\frac{1}{r}}, \end{equation} $

注意到

$ (u_{2}\nabla h, h) = 0, (u_{2}\nabla w, w) = 0, (u\nabla w, v) = -(u\nabla v, w), (w\cdot\nabla u_{1}, w) = -(w\cdot\nabla w, u_{1}). $

$ \max\left(\int^{t}_{0}\parallel \nabla u_{i}\parallel^{2}_{2}, \int^{t}_{0}\parallel \nabla e_{i}\parallel^{2}_{2}, \int^{t}_{0}\parallel u_{i}\parallel^{r}_{s}, \int^{t}_{0}\parallel e_{i}\parallel^{r}_{s}\right) = M< \infty, i = 1, 2. $

$ \max\left(\int^{t}_{0}\parallel \nabla w \parallel^{2}_{2}, \int^{t}_{0}\parallel \nabla h \parallel^{2}_{2} , \int^{t}_{0}\parallel w \parallel^{r}_{s}, \int^{t}_{0}\parallel h \parallel^{r}_{s}\right) = N< \infty, $

结合Young不等式

$ abc\leq \frac{\varepsilon}{p} a^{p}+\frac{\varepsilon}{q} b^{q}+C(\varepsilon)\frac{1}{r'} c^{r'}, \frac{1}{p}+\frac{1}{q}+\frac{1}{r'} = 1. $

这里我们取$ p = 2, q = \frac{2s}{n}, r' = r $, 和$ \varepsilon $足够小得到

(3.25) $ \begin{eqnarray} \int^{t}_{0}-(w\nabla u_{1}, w)& = &\int^{t}_{0}(w\nabla w, u_{1}){}\\ &\leq& C\left(\int^{t}_{0}\parallel \nabla w\parallel^{2}_{2}\right)^{\frac{1}{2}}\left(\int^{t}_{0}\parallel \nabla w\parallel^{2}_{2}\right)^{\frac{n}{2s}}\left(\int^{t}_{0}\parallel u_{1}\parallel^{r}_{s}\parallel w\parallel^{2}_{2}\right)^{\frac{1}{r}}\\ &\leq& \varepsilon\left(\int^{t}_{0}\parallel \nabla w\parallel^{2}_{2}\right)+C(\varepsilon)\left(\int^{t}_{0}\parallel u_{1}\parallel^{r}_{s}\parallel w\parallel^{2}_{2}\right)^{\frac{2}{r}}, \end{eqnarray} $

(3.26) $ \begin{eqnarray} &&\int^{t}_{0}{-({\rm{div}}(h^{T} e_{1} +e^{T}_{2}\cdot h), w)}{}\\ & = &\int^{t}_{0}{((h^{T} e_{1} +e^{T}_{2}\cdot h), \nabla w)}\\ &\leq & C\left(\int^{t}_{0}\parallel \nabla w\parallel^{2}_{2}\right)^{\frac{1}{2}}\left(\int^{t}_{0}\parallel \nabla h\parallel^{2}_{2}\right)^{\frac{n}{2s}}\left(\int^{t}_{0}(\parallel e_{1} \parallel^{r}_{s}+\parallel e_{2} \parallel^{r}_{s})\parallel h\parallel^{2}_{2}\right)^{\frac{1}{r}} \\ &\leq& \varepsilon\left(\int^{t}_{0}\parallel \nabla w\parallel^{2}_{2}+\int^{t}_{0}\parallel \nabla h\parallel^{2}_{2}\right)+C(\varepsilon)\left(\int^{t}_{0}(\parallel e_{1}\parallel^{r}_{s}+\parallel e_{2}\parallel^{r}_{s})\parallel h\parallel^{2}_{2}\right), \end{eqnarray} $

(3.27) $ \begin{eqnarray} &&\int^{t}_{0}{-( e_{1}\nabla w +h\nabla u_{2}+w\nabla e_{1}, h)}{}\\ & = &\int^{t}_{0}{-( e_{1}\nabla w, h)+(u_{2}\nabla h, h)+(w\nabla h, e_{1})}\\ &\leq & C\left(\int^{t}_{0}\parallel \nabla w\parallel^{2}_{2}\right)^{\frac{1}{2}}\left(\int^{t}_{0}\parallel \nabla h\parallel^{2}_{2}\right)^{\frac{n}{2s}}\left(\int^{t}_{0}(\parallel e_{1} \parallel^{r}_{s})\parallel h\parallel^{2}_{2}\right)^{\frac{1}{r}} \nonumber\\ &&+C\left(\int^{t}_{0}\parallel \nabla h\parallel^{2}_{2}\right)^{\frac{1}{2}}\left(\int^{t}_{0}\parallel \nabla h\parallel^{2}_{2}\right)^{\frac{n}{2s}}\left(\int^{t}_{0}(\parallel u_{2} \parallel^{r}_{s})\parallel h\parallel^{2}_{2}\right)^{\frac{1}{r}}\\ &&+C\left(\int^{t}_{0}\parallel \nabla h\parallel^{2}_{2}\right)^{\frac{1}{2}}\left(\int^{t}_{0}\parallel \nabla w\parallel^{2}_{2}\right)^{\frac{n}{2s}}\left(\int^{t}_{0}(\parallel e_{1} \parallel^{r}_{s})\parallel w\parallel^{2}_{2}\right)^{\frac{1}{r}}\\ &\leq& \varepsilon\left(\int^{t}_{0}\parallel \nabla w\parallel^{2}_{2}+\int^{t}_{0}\parallel \nabla h\parallel^{2}_{2}\right)\\ &&+C(\varepsilon)\left(\int^{t}_{0}(\parallel e_{1}\parallel^{r}_{s}+\parallel u_{2}\parallel^{r}_{s})\parallel h\parallel^{2}_{2}+\parallel e_{1} \parallel^{r}_{s}\parallel w\parallel^{2}_{2}\right). \end{eqnarray} $

记

$ Y(t) = \parallel w(t)\parallel^{2}_{2}+\parallel h(t)\parallel^{2}_{2}, \ Y(0) = 0. $

把式(3.25)–(3.27) 代入式(3.22)和(3.23)得到

(3.28) $ \begin{equation} Y(t)\leq C(\varepsilon)\left(\int^{t}_{0}a(t)Y(t)\right), \ Y(0) = 0, \end{equation} $

这里$ a(t) = \sum\limits^{2}_{i = 1}(\parallel e_{i}\parallel^{r}_{s}+\parallel u_{i}\parallel^{r}_{s}) $. 由Gronwall不等式可得

$ Y(t) = 0, $

即

$ u_{1} = u_{1}, e_{1} = {e}_{2}, \ a.e. . $

于是$ P_{1} = P_{2} $, 得到解在Serrin类中的唯一性. 证毕.