非线性发展方程是数学学科领域内重要研究分支之一，它是描述动力学的形态随时间而改变的非线性偏微分方程的总称，在流体力学、热传导、量子力学等方面具有重要应用，对于大部分非线性发展方程而言，得到它的精确解是非常困难的，因而可以研究其Cauchy问题解的若干定性性质，从而对其他学科的研究提供理论支持。本文主要考虑具有如下形式的n分支 $\mu$-Camassa-Holm系统Cauchy问题

$\{\begin{array}{l}{m}_{i,t}+2m_i{u}_{i,x}+m_{i,x}{u}_i+{\left(m_i{\displaystyle \underset{j\ne i}{\overset{n}{\sum }}{u}_j}\right)}_x+{\displaystyle \underset{j\ne i}{\overset{n}{\sum }}{m}_j{u}_{j,x}}=0,\\ u_i\left(0,x\right)=u_{i,0}\left(x\right),x\in S,\end{array}$(1.1)

$E\left[{\left(u_i\right)}_{i=1}^n\right]=\frac{1}{2}{\displaystyle {\int }_S\left({\displaystyle \underset{i=1}{\overset{n}{\sum }}{u}_{i,x}^2}\right)\text{d}x}$。

$m_t+u{m}_x+2u_{x}m=0$， $m=\mu \left(u\right)-u_{xx}$， $x\in S$， (1.2)

方程(1.2)是由Khesin，Lenells和Misiołek在文献 [2] 中提出来的，可作为描述具有自作用和外磁场作用的向列液晶的转子的演化模型。对于该方程Cauchy问题解的适定性、尖峰孤子解的存在性以及解的爆破现象等问题的讨论，具体可见文献 [3] [4] 。

当 $n=2$时，问题(1.1)中的系统可简化为具有如下形式的两分支 $\mu$-Camassa-Holm ( $\mu$-CH)系统

$\{\begin{array}{l}{m}_t+2m{u}_x+m_{x}u+{\left(mv\right)}_x+n{v}_x=0,m=\mu \left(u\right)-u_{xx},x\in S,\\ n_t+2n{v}_x+n_{x}v+{\left(nu\right)}_x+m{u}_x=0,n=\mu \left(v\right)-v_{xx},\end{array}$(1.3)

$\underset{x\in S}{\max }{f}^2\left(x\right)\le \frac{\epsilon +2}{24}{\displaystyle {\int }_S{f}_x^2\left(x\right)\text{d}x}+\frac{\epsilon +2}{4\epsilon }{a}_0^2$。

$\underset{x\in S}{\max }{f}^2\left(x\right)\le \frac{1}{12}{\displaystyle {\int }_S{f}_x^2\left(x\right)\text{d}x},f\in H^1$。

$\omega \left(t\right):=\underset{x\in S}{\inf }{u}_x\left(t,x\right)=u_x\left(t,\xi \left(t\right)\right)$。

${\Vert fg\Vert }_{H^r}\le C\left({\Vert f\Vert }_{L^{\infty }}{\Vert g\Vert }_{H^r}+{\Vert f\Vert }_{H^r}{\Vert g\Vert }_{L^{\infty }}\right)$。

${\Vert \left[{\Lambda }^r,f\right]g\Vert }_{L^2}\le C\left({\Vert {\partial }_{x}f\Vert }_{L^{\infty }}{\Vert {\Lambda }^{r-1}g\Vert }_{L^2}+{\Vert {\Lambda }^{r}f\Vert }_{L^2}{\Vert g\Vert }_{L^{\infty }}\right)$。

${\Vert u\Vert }_{L^r\left(\Omega \right)}\le {\Vert u\Vert }_{L^q\left(\Omega \right)}^{\theta }{\Vert u\Vert }_{L^p\left(\Omega \right)}^{1-\theta }$。

最后，证明n分支 $\mu$-Camassa-Holm 系统的一些先验估计。

$\frac{\text{d}}{\text{d}t}\mu \left(u_i\right)=0$，

${\mu }_{i,0}:=\mu \left(u_{i,0}\right)=\mu \left(u_i\right)$， ${\mu }_{i,1}:={\left({\displaystyle {\int }_S{u}_{i,x}^2}\left(0,x\right)\text{d}x\right)}^{\frac{1}{2}}$，

其中 $i=1,2,\cdots ,n$。

$\underset{x\in S}{\max }{\left[u_i\left(t,x\right)-{\mu }_{i,0}\right]}^2\le \frac{1}{12}{\displaystyle {\int }_S{u}_{i,x}^2}\left(t,x\right)\text{d}x=\frac{1}{12}{\displaystyle {\int }_S{u}_{i,x}^2}\left(0,x\right)\text{d}x=\frac{1}{12}{\mu }_{i,1}^2$，

${\Vert u_i\left(t,x\right)-{\mu }_{i,0}\Vert }_{L^{\infty }}\le \frac{\sqrt{3}}{6}{\mu }_{i,1}$， (2.2)

其中 $i=1,2,\cdots ,n$。

其中T依赖于初值 $z_0$。

注3.1 定理3.1中解的最大存在时间T不依赖于s。

证明 考虑如下形式的方程

$\{\begin{array}{l}\frac{\text{d}z}{\text{d}t}+H\left(z\right)z=f\left(z\right),t>0,\\ z\left(0\right)=z_0,\end{array}$

其中 $z={\left(u_1,u_2,\cdots ,u_n\right)}^{\prime }$， $H\left(z\right)=\left(\left({\displaystyle \underset{i=1}{\overset{n}{\sum }}{u}_i}\right){\partial }_x\right)I_n$( $I_n$表示n阶单位矩阵），且

$f\left(z\right)={\left(f_1\left(z\right),f_2\left(z\right),\cdots ,f_n\left(z\right)\right)}^{\prime }$，

其中

$f_i\left(z\right)={\displaystyle \underset{j\ne i}{\overset{n}{\sum }}\mu \left(u_{i,x}{u}_j\right)-{\partial }_x{{\rm A}}^{-1}\left(2{\mu }_{i,0}{u}_i+{\mu }_{i,0}{\displaystyle \underset{j\ne i}{\overset{n}{\sum }}{u}_j}+{\displaystyle \underset{j\ne i}{\overset{n}{\sum }}{\mu }_{j,0}{u}_j}+u_{i,x}{\displaystyle \underset{j\ne i}{\overset{n}{\sum }}{u}_{j,x}}+\frac{1}{2}{u}_{i,x}^2-\frac{1}{2}{\displaystyle \underset{j\ne i}{\overset{n}{\sum }}{u}_{j,x}^2}\right)}$。

令 $X={\left(H^{s-1}\right)}^n$， $Y={\left(H^s\right)}^n$， $Q={\left(1-{\partial }_x^2\right)}^{1/2}{I}_n$，显然 $Q:{\left(H^s\right)}^n\to {\left(H^{s-1}\right)}^n$为等距映射。为了证明定理3.1，需说明 $H\left(z\right)$和 $f\left(z\right)$满足Kato理论 [13] 中的三个条件，根据文献 [17] 中引理4.1~4.5，可以得到下面类似的引理：

引理3.1 算子 $H\left(z\right)=\left(\left({\displaystyle \underset{i=1}{\overset{n}{\sum }}{u}_i}\right){\partial }_x\right)I_n\in G\left({\left(L^2\right)}^n,1,\beta \right)$，其中 $z\in {\left(H^s\right)}^n$， $s>3/2$。

引理3.2 算子 $H\left(z\right)=\left(\left({\displaystyle \underset{i=1}{\overset{n}{\sum }}{u}_i}\right){\partial }_x\right)I_n\in G\left({\left(H^{s-1}\right)}^n,1,\beta \right)$，其中 $z\in {\left(H^s\right)}^n$， $s>3/2$。

引理3.3 令算子 $H\left(z\right)=\left(\left({\displaystyle \underset{i=1}{\overset{n}{\sum }}{u}_i}\right){\partial }_x\right)I_n$，其中 $z\in {\left(H^s\right)}^n$， $s>3/2$，则 $H\left(z\right)\in L\left({\left(H^s\right)}^n,{\left(H^{s-1}\right)}^n\right)$，且

${\Vert \left(H\left(y\right)-H\left(z\right)\right)\omega \Vert }_{{\left(H^{s-1}\right)}^n}\le C_1{\Vert y-z\Vert }_{{\left(H^{s-1}\right)}^n}{\Vert \omega \Vert }_{{\left(H^s\right)}^n},y,z,\omega \in {\left(H^s\right)}^n$。

引理3.4 令 $B\left(z\right)=QH\left(z\right)Q^{-1}-H\left(z\right)$，其中 $z\in {\left(H^s\right)}^n$， $s>3/2$，则 $B\left(z\right)\in L\left({\left(H^{s-1}\right)}^n\right)$且

${\Vert \left(B\left(y\right)-B\left(z\right)\right)\omega \Vert }_{{\left(H^{s-1}\right)}^n}\le C_2{\Vert y-z\Vert }_{{\left(H^s\right)}^n}{\Vert \omega \Vert }_{{\left(H^{s-1}\right)}^n},y,z\in {\left(H^s\right)}^n,\omega \in {\left(H^{s-1}\right)}^n$。

引理3.5 令 $z={\left(u_1,u_2,\cdots ,u_n\right)}^{\prime }\in {\left(H^s\right)}^n$， $s>3/2$， $f\left(z\right)={\left(f_1\left(z\right),f_2\left(z\right),\cdots ,f_n\left(z\right)\right)}^{\prime }$，则f在空间 ${\left(H^s\right)}^n$的有界子集上一致有界，且满足

${\Vert f\left(y\right)-f\left(z\right)\Vert }_{{\left(H^s\right)}^n}\le C_3{\Vert y-z\Vert }_{{\left(H^s\right)}^n},y,z\in {\left(H^s\right)}^n$，

${\Vert f\left(y\right)-f\left(z\right)\Vert }_{{\left(H^{s-1}\right)}^n}\le C_4{\Vert y-z\Vert }_{{\left(H^{s-1}\right)}^n},y,z\in {\left(H^s\right)}^n$。

注3.2上述引理3.3~3.5中出现的C₁，C₂，C₃，C₄均依赖于 $\max \left({\Vert y\Vert }_{{\left(H^s\right)}^n},{\Vert z\Vert }_{{\left(H^s\right)}^n}\right)$。

最后，根据引理3.1~3.5，再结合Kato理论 [13] ，可以得到问题(2.1)解的局部适定性，即定理3.1。

$\frac{1}{2}\frac{\text{d}}{\text{d}t}{\Vert u_i\Vert }_{H^s}^2+{\left({\displaystyle \underset{j=1}{\overset{n}{\sum }}{u}_j}{u}_{i,x},u_i\right)}_s={\left(u_i,f_i\left(z\right)\right)}_s$， (4.1)

其中

$f_i\left(z\right)={\displaystyle \underset{j\ne i}{\overset{n}{\sum }}\mu \left(u_{i,x}{u}_j\right)}-{\partial }_x{{\rm A}}^{-1}\left(2{\mu }_{i,0}{u}_i+{\mu }_{i,0}{\displaystyle \underset{j\ne i}{\overset{n}{\sum }}{u}_j}+{\displaystyle \underset{j\ne i}{\overset{n}{\sum }}{\mu }_{j,0}{u}_j}+u_{i,x}{\displaystyle \underset{j\ne i}{\overset{n}{\sum }}{u}_{j,x}}+\frac{1}{2}{u}_{i,x}^2-\frac{1}{2}{\displaystyle \underset{j\ne i}{\overset{n}{\sum }}{u}_{j,x}^2}\right)$。

$\begin{array}{l}\left|{\left(u_{i,x}{\displaystyle \underset{j=1}{\overset{n}{\sum }}{u}_j,u_i}\right)}_s\right|=\left|{\left(\left[{\Lambda }^s,{\displaystyle \underset{j=1}{\overset{n}{\sum }}{u}_j}\right]u_{i,x},{\Lambda }^s{u}_i\right)}_0+{\left({\displaystyle \underset{j=1}{\overset{n}{\sum }}{u}_j}{\Lambda }^s{u}_{i,x},{\Lambda }^s{u}_i\right)}_0\right|\\ \le {\Vert \left[{\Lambda }^s,{\displaystyle \underset{j=1}{\overset{n}{\sum }}{u}_j}\right]u_{i,x}\Vert }_{L^2}{\Vert {\Lambda }^s{u}_i\Vert }_{L^2}+\frac{1}{2}\left|{\left({\displaystyle \underset{j=1}{\overset{n}{\sum }}{u}_{j,x}}{\Lambda }^s{u}_i,{\Lambda }^s{u}_i\right)}_0\right|\\ \le C\left({\Vert {\displaystyle \underset{j=1}{\overset{n}{\sum }}{u}_{j,x}}\Vert }_{L^{\infty }}{\Vert u_i\Vert }_{H^s}+{\Vert u_{i,x}\Vert }_{L^{\infty }}{\Vert {\displaystyle \underset{j=1}{\overset{n}{\sum }}{u}_j}\Vert }_{H^s}\right){\Vert u_i\Vert }_{H^s}+\frac{1}{2}{\Vert {\displaystyle \underset{j=1}{\overset{n}{\sum }}{u}_{j,x}}\Vert }_{L^{\infty }}{\Vert u_i\Vert }_{H^s}^2\\ \le C\left({\displaystyle \underset{j=1}{\overset{n}{\sum }}{\Vert u_{j,x}\Vert }_{L^{\infty }}}\right)\left({\displaystyle \underset{j=1}{\overset{n}{\sum }}{\Vert u_j\Vert }_{H^s}^2}\right),\end{array}$(4.2)

$\begin{array}{l}\left|{\left(u_i,f_i\left(z\right)\right)}_s\right|=\left|{\left({\displaystyle \underset{j\ne i}{\overset{n}{\sum }}\mu \left(u_{i,x}{u}_j\right)}-{\partial }_x{{\rm A}}^{-1}\left(2{\mu }_{i,0}{u}_i+{\mu }_{i,0}{\displaystyle \underset{j\ne i}{\overset{n}{\sum }}{u}_j}+{\displaystyle \underset{j\ne i}{\overset{n}{\sum }}{\mu }_{j,0}{u}_j}+u_{i,x}{\displaystyle \underset{j\ne i}{\overset{n}{\sum }}{u}_{j,x}}+\frac{1}{2}{u}_{i,x}^2-\frac{1}{2}{\displaystyle \underset{j\ne i}{\overset{n}{\sum }}{u}_{j,x}^2}\right),u_i\right)}_s\right|\\ \le {\displaystyle \underset{j\ne i}{\overset{n}{\sum }}\left|\mu \left(u_{i,x}{u}_j\right)\right|{\Vert u_i\Vert }_{H^s}}+C{\Vert u_i\Vert }_{H^s}\left({\displaystyle \underset{j=1}{\overset{n}{\sum }}{\Vert u_j^2\Vert }_{H^{s-1}}}+{\displaystyle \underset{j=1}{\overset{n}{\sum }}{\Vert u_{j,x}^2\Vert }_{H^{s-1}}}\right)\\ \le C{\Vert u_i\Vert }_{H^s}\left({\displaystyle \underset{j\ne i}{\overset{n}{\sum }}{\Vert u_{i,x}\Vert }_{L^{\infty }}{\Vert u_j\Vert }_{L^{\infty }}+{\displaystyle \underset{j=1}{\overset{n}{\sum }}{\Vert u_j\Vert }_{L^{\infty }}}{\Vert u_j\Vert }_{H^{s-1}}+{\displaystyle \underset{j=1}{\overset{n}{\sum }}{\Vert u_{j,x}\Vert }_{L^{\infty }}}{\Vert u_{j,x}\Vert }_{H^{s-1}}}\right)\\ \le C\left({\displaystyle \underset{j=1}{\overset{n}{\sum }}\left({\Vert u_j\Vert }_{L^{\infty }}+{\Vert u_{j,x}\Vert }_{L^{\infty }}\right)}\right)\left({\displaystyle \underset{j=1}{\overset{n}{\sum }}{\Vert u_j\Vert }_{H^s}^2}\right),\end{array}$(4.3)

$\frac{1}{2}\frac{\text{d}}{\text{d}t}\left({\displaystyle \underset{i=1}{\overset{n}{\sum }}{\Vert u_i\Vert }_{H^s}^2}\right)\le C\left({\displaystyle \underset{j=1}{\overset{n}{\sum }}\left({\Vert u_j\Vert }_{L^{\infty }}+{\Vert u_{j,x}\Vert }_{L^{\infty }}\right)}\left({\displaystyle \underset{j=1}{\overset{n}{\sum }}{\Vert u_j\Vert }_{H^s}^2}\right)\right)$，

${\displaystyle \underset{i=1}{\overset{n}{\sum }}{\Vert u_i\Vert }_{H^s}}\le {\displaystyle \underset{i=1}{\overset{n}{\sum }}{\Vert u_{i,0}\Vert }_{H^s}\exp \left\{C{\displaystyle {\int }_0^t{\displaystyle \underset{i=1}{\overset{n}{\sum }}{\Vert u_{i,x}\Vert }_{L^{\infty }}}}\text{d}\tau \right\}}$， (4.4)

若 $T<\infty$满足 ${\displaystyle {\int }_0^T{\displaystyle \underset{i=1}{\overset{n}{\sum }}{\Vert u_{i,x}\Vert }_{L^{\infty }}}\text{d}\tau <\infty }$，则根据(4.4)式有

$\underset{t\to T}{\lim }\underset{x\in S}{\sup }{\displaystyle \underset{i=1}{\overset{n}{\sum }}{\Vert u_i\Vert }_{H^s}}<\infty$，

于是 $T=\infty$，与假设 $T<\infty$矛盾，这也就完成了定理4.1的证明。

其次，在给出爆破准则前先介绍如下定义：

设 $q\left(t,x\right)$是随着解 $z\left(t,x\right)$发展的例子轨迹，并且满足方程

$\{\begin{array}{l}{q}_t\left(t,x\right)=z\left(t,q\left(t,x\right)\right),\left(t,x\right)\in \left(0,T\right)\times S,\\ q\left(0,x\right)=x,x\in S,\end{array}$(4.5)

其中 $z={\left(u_1,u_2,\cdots ,u_n\right)}^{\prime }\in C^1\left(\left[0,T\right),{\left(H^{s-1}\right)}^n\right)$是Cauchy问题(2.1)的解，初值 $z_0={\left(u_{1,0},u_{2,0},\cdots ,u_{n,0}\right)}^{\prime }\in {\left(H^s\right)}^n$， $s>3/2$， $i=1,2,\cdots ,n$，且T是Cauchy问题(2.1)解的最大存在时间。通过计算有

$q_{tx}\left(t,x\right)=z_x\left(t,q\left(t,x\right)\right)q_x\left(t,x\right)$，

于是对于 $\left(t,x\right)\in \left[0,T\right)\times S$有

$q_x\left(t,x\right)=\exp \left\{{\displaystyle {\int }_0^t{z}_x\left(\tau ,q\left(\tau ,x\right)\right)\text{d}\tau }\right\}>0$，

因此对于任意的 $t\in \left[0,T\right)$， $q\left(t,\cdot \right):S\to S$是微分同胚映射。

接着利用上述定义、守恒律以及定理4.1给出问题(2.1)解的爆破准则。

证明 利用定理3.1以及稠密性定理，接下来考虑 $s=3$的情况。将系统(2.1)中的前n个方程相加并通过整理可得

$\begin{array}{l}{\left({\displaystyle \underset{i=1}{\overset{n}{\sum }}{u}_i}\right)}_t+\left({\displaystyle \underset{i=1}{\overset{n}{\sum }}{u}_i}\right)\left({\displaystyle \underset{i=1}{\overset{n}{\sum }}{u}_{i,x}}\right)\\ =-{\partial }_x{{\rm A}}^{-1}\left[\left(n+1\right){\displaystyle \underset{i=1}{\overset{n}{\sum }}\mu \left(u_i\right)u_i}+{\displaystyle \underset{i=1}{\overset{n}{\sum }}\left(\mu \left(u_i\right){\displaystyle \underset{j\ne i}{\overset{n}{\sum }}{u}_j}\right)}+2{\displaystyle \underset{1\le i<j\le n}{\sum }{u}_{i,x}{u}_{j,x}}-\frac{n-2}{2}\left({\displaystyle \underset{i=1}{\overset{n}{\sum }}{u}_{i,x}^2}\right)\right],\end{array}$

上述等式左右两边对x求偏导同时利用(2.3)式可以得到

$\begin{array}{l}{\left({\displaystyle \underset{i=1}{\overset{n}{\sum }}{u}_{i,x}}\right)}_t+{\displaystyle \underset{i=1}{\overset{n}{\sum }}{u}_{i,x}^2}+\left({\displaystyle \underset{i=1}{\overset{n}{\sum }}{u}_i}\right)\left({\displaystyle \underset{i=1}{\overset{n}{\sum }}{u}_{i,xx}}\right)\\ ={\displaystyle \underset{i=1}{\overset{n}{\sum }}\left(\left(n+1\right){\mu }_{i,0}+{\displaystyle \underset{j\ne i}{\overset{n}{\sum }}{\mu }_{j,0}}\right)\left(u_i-{\mu }_{i,0}\right)}-2{\displaystyle \underset{1\le i<j\le n}{\sum }\mu \left(u_{i,x}{u}_{j,x}\right)}-\frac{n-2}{2}\left({\displaystyle \underset{i=1}{\overset{n}{\sum }}{u}_{i,x}^2}\right)+\frac{n-2}{2}\left({\displaystyle \underset{i=1}{\overset{n}{\sum }}{\mu }_{i,1}^2}\right),\end{array}$

通过整理有

结合

$2{\displaystyle \underset{1\le i<j\le n}{\sum }\mu \left(u_{i,x}{u}_{j,x}\right)}\le {\displaystyle \underset{1\le i<j\le n}{\sum }\left({\mu }_{i,1}^2+{\mu }_{j,1}^2\right)}=\left(n-1\right)\left({\displaystyle \underset{i=1}{\overset{n}{\sum }}{\mu }_{i,1}^2}\right)$

以及(2.2)式对函数f进行估计有

$\begin{array}{l}{\displaystyle {\int }_S\left({\displaystyle \underset{i=1}{\overset{n}{\sum }}{u^{\prime }}_{i,0}}\right)\text{d}x}<-n\left({\displaystyle \underset{i=1}{\overset{n}{\sum }}{\mu }_{i,1}^2}\right){\left[\frac{2\left(n+1\right)}{n-1}\left(\frac{\sqrt{3}}{6}{\displaystyle \underset{i=1}{\overset{n}{\sum }}{\mu }_{i,1}\left|\left(n+1\right){\mu }_{i,0}+{\displaystyle \underset{j\ne i}{\overset{n}{\sum }}{\mu }_{j,0}}\right|}+\frac{3n-4}{2}{\displaystyle \underset{i=1}{\overset{n}{\sum }}{\mu }_{i,1}^2}\right)\right]}^{\frac{n-1}{2}}\\ :\equiv -K\end{array}$

$0<T\le -\frac{2n^{\frac{1}{n-1}}{\left({\displaystyle \underset{i=1}{\overset{n}{\sum }}{\mu }_{i.1}^2}\right)}^{\frac{1}{n-1}}}{{\left({\displaystyle {\int }_S{\left({\displaystyle \underset{i=1}{\overset{n}{\sum }}{u^{\prime }}_{i,0}}\right)}^{n+1}}\text{d}x\right)}^{\frac{1}{n-1}}+{\left[-K{\left({\displaystyle {\int }_S{\left({\displaystyle \underset{i=1}{\overset{n}{\sum }}{u^{\prime }}_{i,0}}\right)}^{n+1}\text{d}x}\right)}^{\frac{-n^2+2n+1}{{\left(n-1\right)}^2}}\right]}^{\frac{n-1}{2}}}$，

$\underset{t\to T}{\lim \inf }\left\{\underset{x\in S}{\inf }\left[{\displaystyle \underset{i=1}{\overset{n}{\sum }}{u}_{i,x}\left(x,t\right)}\right]\right\}=-\infty$。

证明 根据定理3.1，存在 $T_0>0$使得Cauchy问题(2.1)存在唯一的解 $z=\left(u_1,u_2,\cdots ,u_n\right)\in C\left(\left[0,T_0\right);{\left(H^s\right)}^n\right)$。

首先，将(4.7)式左右两边同乘 $\left(n+1\right){\left({\displaystyle \underset{i=1}{\overset{n}{\sum }}{u}_{i,x}}\right)}^n$并对x在 $S$上积分有

$\begin{array}{l}\frac{\text{d}}{\text{d}t}{\displaystyle {\int }_S{\left({\displaystyle \underset{i=1}{\overset{n}{\sum }}{u}_{i,x}}\right)}^{n+1}\text{d}x+\frac{n\left(n+1\right)}{2}}{\displaystyle {\int }_S\left({\displaystyle \underset{i=1}{\overset{n}{\sum }}{u}_{i,x}^2}\right)}{\left({\displaystyle \underset{i=1}{\overset{n}{\sum }}{u}_{i,x}}\right)}^n\text{d}x-{\displaystyle {\int }_S{\left({\displaystyle \underset{i=1}{\overset{n}{\sum }}{u}_{i,x}}\right)}^{n+2}\text{d}x}\\ =\left(n+1\right){\displaystyle {\int }_S{\left({\displaystyle \underset{i=1}{\overset{n}{\sum }}{u}_{i,x}}\right)}^n\left[{\displaystyle \underset{i=1}{\overset{n}{\sum }}\left(\left(n+1\right){\mu }_{i,0}+{\displaystyle \underset{j\ne i}{\overset{n}{\sum }}{\mu }_{j,0}}\right)\left(u_i-{\mu }_{i,0}\right)}-2{\displaystyle \underset{1\le i<j\le n}{\sum }\mu \left(u_{i,x}{u}_{j,x}\right)}+\frac{n-2}{2}\left({\displaystyle \underset{i=1}{\overset{n}{\sum }}{\mu }_{i,1}^2}\right)\right]\text{d}x},\end{array}$

另外根据

${\displaystyle {\int }_S\left({\displaystyle \underset{i=1}{\overset{n}{\sum }}{u}_{i,x}^2}\right)}{\left({\displaystyle \underset{i=1}{\overset{n}{\sum }}{u}_{i,x}}\right)}^n\text{d}x\ge \frac{1}{n}{\displaystyle {\int }_S{\left({\displaystyle \underset{i=1}{\overset{n}{\sum }}{u}_{i,x}}\right)}^{n+2}\text{d}x}$，

可以得到

$\begin{array}{l}\frac{\text{d}}{\text{d}t}{\displaystyle {\int }_S{\left({\displaystyle \underset{i=1}{\overset{n}{\sum }}{u}_{i,x}}\right)}^{n+1}\text{d}x}+\frac{n-1}{2}{\displaystyle {\int }_S{\left({\displaystyle \underset{i=1}{\overset{n}{\sum }}{u}_{i,x}}\right)}^{n+2}\text{d}x}\\ \le \left(n+1\right){\displaystyle {\int }_S{\left({\displaystyle \underset{i=1}{\overset{n}{\sum }}{u}_{i,x}}\right)}^n\left[{\displaystyle \underset{i=1}{\overset{n}{\sum }}\left(\left(n+1\right){\mu }_{i,0}+{\displaystyle \underset{j\ne i}{\overset{n}{\sum }}{\mu }_{j,0}}\right)\left(u_i-{\mu }_{i,0}\right)}-2{\displaystyle \underset{1\le i<j\le n}{\sum }\mu \left(u_{i,x}{u}_{j,x}\right)}+\frac{n-2}{2}\left({\displaystyle \underset{i=1}{\overset{n}{\sum }}{\mu }_{i,1}^2}\right)\right]\text{d}x},\end{array}$(4.13)

接下来对不等式(4.13)左边第二项进行估计。利用引理2.5有

${\Vert {\displaystyle \underset{i=1}{\overset{n}{\sum }}{u}_{i,x}}\Vert }_{L^{n+1}}\le {\Vert {\displaystyle \underset{i=1}{\overset{n}{\sum }}{u}_{i,x}}\Vert }_{L^2}^{\frac{2}{n\left(n+1\right)}}{\Vert {\displaystyle \underset{i=1}{\overset{n}{\sum }}{u}_{i,x}}\Vert }_{L^{n+2}}^{\frac{\left(n-1\right)\left(n+2\right)}{n\left(n+1\right)}}$，

从而可以得到

${\displaystyle {\int }_S{\left({\displaystyle \underset{i=1}{\overset{n}{\sum }}{u}_{i,x}}\right)}^{n+2}\text{d}x}\ge \frac{1}{{\left(n\left({\displaystyle \underset{i=1}{\overset{n}{\sum }}{\mu }_{i,1}^2}\right)\right)}^{\frac{1}{n-1}}}{\left({\displaystyle {\int }_S{\left({\displaystyle \underset{i=1}{\overset{n}{\sum }}{u}_{i,x}}\right)}^{n+1}\text{d}x}\right)}^{\frac{n}{n-1}}$。 (4.14)

紧接着对(4.13)式右边进行估计。同样的方法，利用引理2.5有

${\Vert {\displaystyle \underset{i=1}{\overset{n}{\sum }}{u}_{i,x}}\Vert }_{L^n}\le {\Vert {\displaystyle \underset{i=1}{\overset{n}{\sum }}{u}_{i,x}}\Vert }_{L^2}^{\frac{2}{n\left(n-1\right)}}{\Vert {\displaystyle \underset{i=1}{\overset{n}{\sum }}{u}_{i,x}}\Vert }_{L^{n+1}}^{\frac{\left(n+1\right)\left(n-2\right)}{n\left(n-1\right)}}$，

即

$\left|{\displaystyle {\int }_S{\left({\displaystyle \underset{i=1}{\overset{n}{\sum }}{u}_{i,x}}\right)}^n\text{d}x}\right|\le {\displaystyle {\int }_S{\left|{\displaystyle \underset{i=1}{\overset{n}{\sum }}{u}_{i,x}}\right|}^n\text{d}x}\le n^{\frac{1}{n-1}}{\left({\displaystyle \underset{i=1}{\overset{n}{\sum }}{\mu }_{i,1}^2}\right)}^{\frac{1}{n-1}}{\left({\displaystyle {\int }_S{\left|{\displaystyle \underset{i=1}{\overset{n}{\sum }}{u}_{i,x}}\right|}^{n+1}\text{d}x}\right)}^{\frac{n-2}{n-1}}$，

根据假设 ${\displaystyle \underset{i=1}{\overset{n}{\sum }}{u}_{i,x}}$在 $S$上不变号，则可以得到

$\left|{\displaystyle {\int }_S{\left({\displaystyle \underset{i=1}{\overset{n}{\sum }}{u}_{i,x}}\right)}^n\text{d}x}\right|\le n^{\frac{1}{n-1}}{\left({\displaystyle \underset{i=1}{\overset{n}{\sum }}{\mu }_{i,1}^2}\right)}^{\frac{1}{n-1}}{\left({\displaystyle {\int }_S{\left({\displaystyle \underset{i=1}{\overset{n}{\sum }}{u}_{i,x}}\right)}^{n+1}\text{d}x}\right)}^{\frac{n-2}{n-1}}$。 (4.15)

再结合(2.2)式以及

$2{\displaystyle \underset{1\le i<j\le n}{\sum }\mu \left(u_{i,x}{u}_{j,x}\right)}\le {\displaystyle \underset{1\le i<j\le n}{\sum }\left({\mu }_{i,1}^2+{\mu }_{j,1}^2\right)}=\left(n-1\right)\left({\displaystyle \underset{i=1}{\overset{n}{\sum }}{\mu }_{i,1}^2}\right)$

可以得到

$\begin{array}{l}\left|\left(n+1\right){\displaystyle {\int }_S{\left({\displaystyle \underset{i=1}{\overset{n}{\sum }}{u}_{i,x}}\right)}^n}\left[{\displaystyle \underset{i=1}{\overset{n}{\sum }}\left(\left(n+1\right){\mu }_{i,0}+{\displaystyle \underset{j\ne i}{\overset{n}{\sum }}{\mu }_{j,0}}\right)\left(u_i-{\mu }_{i,0}\right)-2{\displaystyle \underset{1\le i<j\le n}{\sum }\mu \left(u_{i,x}{u}_{j,x}\right)}}+\frac{n-2}{2}\left({\displaystyle \underset{i=1}{\overset{n}{\sum }}{\mu }_{i,1}^2}\right)\right]\text{d}x\right|\\ \le \left(n+1\right)n^{\frac{1}{n-1}}{\left({\displaystyle \underset{i=1}{\overset{n}{\sum }}{\mu }_{i,1}^2}\right)}^{\frac{1}{n-1}}{\left({\displaystyle {\int }_S{\left({\displaystyle \underset{i=1}{\overset{n}{\sum }}{u}_{i,x}}\right)}^{n+1}\text{d}x}\right)}^{\frac{n-2}{n-1}}\left[\frac{\sqrt{3}}{6}{\displaystyle \underset{i=1}{\overset{n}{\sum }}{\mu }_{i,1}\left|\left(n+1\right){\mu }_{i,0}+{\displaystyle \underset{j\ne i}{\overset{n}{\sum }}{\mu }_{j,0}}\right|+\frac{3n-4}{2}\left({\displaystyle \underset{i=1}{\overset{n}{\sum }}{\mu }_{i,1}^2}\right)}\right],\end{array}$(4.16)

最后，将(4.13)式~(4.16)式相结合可得

$\begin{array}{c}\frac{\text{d}}{\text{d}t}{\displaystyle {\int }_S{\left({\displaystyle \underset{i=1}{\overset{n}{\sum }}{u}_{i,x}}\right)}^{n+1}\text{d}x}\le {\left({\displaystyle {\int }_S{\left({\displaystyle \underset{i=1}{\overset{n}{\sum }}{u}_{i,x}}\right)}^{n+1}\text{d}x}\right)}^{\frac{n-2}{n-1}}[-\frac{n-1}{2}\frac{1}{n^{\frac{1}{n-1}}{\left({\displaystyle \underset{i=1}{\overset{n}{\sum }}{\mu }_{i,1}^2}\right)}^{\frac{1}{n-1}}}{\left({\displaystyle {\int }_S{\left({\displaystyle \underset{i=1}{\overset{n}{\sum }}{u}_{i,x}}\right)}^{n+1}\text{d}x}\right)}^{\frac{2}{n-1}}\\ \begin{array}{c}\\ \\ \\ \end{array}+\left(n+1\right)n^{\frac{1}{n-1}}{\left({\displaystyle \underset{i=1}{\overset{n}{\sum }}{\mu }_{i,1}^2}\right)}^{\frac{1}{n-1}}\left[\frac{\sqrt{3}}{6}{\displaystyle \underset{i=1}{\overset{n}{\sum }}{\mu }_{i,1}\left|\left(n+1\right){\mu }_{i,0}+{\displaystyle \underset{j\ne i}{\overset{n}{\sum }}{\mu }_{j,0}}\right|+\frac{3n-4}{2}\left({\displaystyle \underset{i=1}{\overset{n}{\sum }}{\mu }_{i,1}^2}\right)}\right]]\end{array}$(4.17)

令 $V\left(t\right)={\displaystyle {\int }_S{\left({\displaystyle \underset{i=1}{\overset{n}{\sum }}{u}_{i,x}}\right)}^{n+1}\text{d}x}$， $t\in \left[0,T\right)$，

$K:=n\left({\displaystyle \underset{i=1}{\overset{n}{\sum }}{\mu }_{i,1}^2}\right){\left[\frac{2\left(n+1\right)}{n-1}\left(\frac{\sqrt{3}}{6}{\displaystyle \underset{i=1}{\overset{n}{\sum }}{\mu }_{i,1}\left|\left(n+1\right){\mu }_{i,0}+{\displaystyle \underset{j\ne i}{\overset{n}{\sum }}{\mu }_{j,0}}\right|}+\frac{3n-4}{2}{\displaystyle \underset{i=1}{\overset{n}{\sum }}{\mu }_{i,1}^2}\right)\right]}^{\frac{n-1}{2}}$， $c:=\frac{n^{\frac{1}{n-1}}}{n-1}{\left({\displaystyle \underset{i=1}{\overset{n}{\sum }}{\mu }_{i,1}^2}\right)}^{\frac{1}{n-1}}$，

则(4.17)式可写为

$\frac{\text{d}}{\text{d}t}V\left(t\right)\le {\left[V\left(t\right)\right]}^{\frac{n-2}{n-1}}\left(-\frac{1}{2c}{\left(V\left(t\right)\right)}^{\frac{2}{n-1}}+\frac{1}{2c}{K}^{\frac{2}{n-1}}\right)$。 (4.18)

根据假设条件 $V\left(0\right)<-K$，则 $V^{\prime }\left(0\right)<0$，且 $V\left(t\right)$在 $\left[0,T\right)$严格单调递减。令

$c\delta =\frac{1}{2}-\frac{1}{2}{\left(\frac{K}{-V\left(0\right)}\right)}^{\frac{n-1}{2}}\in \left(0,\frac{1}{2}\right)$，

则

${\left(V\left(0\right)\right)}^{\frac{2}{n-1}}=\frac{K^{\frac{2}{n-1}}}{{\left(1-2c\delta \right)}_{}^{{\left(\frac{2}{n-1}\right)}^2}}<{\left(V\left(t\right)\right)}^{\frac{2}{n-1}}$，

即

$K^{\frac{2}{n-1}}<{\left(1-2c\delta \right)}_{}^{{\left(\frac{2}{n-1}\right)}^2}{\left(V\left(t\right)\right)}^{\frac{2}{n-1}}$， (4.19)

将(4.19)式代入到(4.18)式，可以得到

$\frac{\text{d}}{\text{d}t}V\left(t\right)\le -\frac{1}{2c}{\left[V\left(t\right)\right]}^{\frac{n}{n-1}}\left(1-{\left(1-2c\delta \right)}_{}^{\frac{4}{{\left(n-1\right)}^2}}\right)\le -\delta {\left[V\left(t\right)\right]}^{\frac{n}{n-1}}$， $t\in \left[0,T\right)$，