大初值有界区域可压缩液晶方程组强解的爆破准则

Blow-Up Criteria for Strong Solutions of Compressible Liquid Crystal Equations in Large Initial Bounded Region

摘要: 液晶在现实生活中有着非常广泛的应用，它们对光、电、磁和热都具有很高灵敏性，根据这一特点可以设计出各种各样具有热–光、磁–光、电–光等物理效应的仪器。关于它的研究遍及了化学、物理学、生物学、材料科学和电子学，形成了各自专门的学科，例如液晶化学、液晶物理、生物液晶、液晶光学等，也引起了国内外数学家和物理学家们广泛的研究兴趣。液晶工业技术中最受重视的问题就是液晶中观察到的分子分布规律、缺陷、相变现象及其动力学规律，这些问题直接关系到液晶设备的制造，而适当的数学模型就是描述这些现象最有力的工具。针对现实中关注的问题，从数学的角度来看，我们关心的问题是相变的数学机理、动力学方程组解的整体存在性和正则性以及强解的爆破准则。本课题考虑的是一维完全可压缩液晶方程无热传导的Cauchy问题，在大初值有界区域的情况下，利用精细的能量估计方法和非线性泛函分析中的方法，证明这个方程组的强解存在爆破准则。

Abstract: Liquid crystals are widely used in real life. They are highly sensitive to light, electricity, magnetism and heat. According to this feature, various kinds of instruments with physical effects such as heat light, magnetism light, electricity light can be designed. Its research covers chemistry, physics, biology, materials science and electronics, and forms their own specialized disciplines, such as liquid crystal chemistry, liquid crystal physics, biological liquid crystal, liquid crystal optics, etc., which also arouses extensive interest of mathematicians and physicists at home and abroad. The most important problem in the liquid crystal industry is the molecular distribution law, defect, phase change phenomenon and its dynamics law observed in the liquid crystal. These problems are directly related to the manufacture of liquid crystal equipment, and the appropriate mathematical model is the most powerful tool to describe these phenomena. In view of the problems concerned in reality, from the mathematical point of view, we are concerned with the mathematical mechanism of phase transition, the global existence and regularity of solutions of dynamic equations and the blow-up criteria of strong solutions. In this paper, we consider the Cauchy problem of one-dimensional fully compressible liquid crystal equations without heat conduction. In the case of a large initial value bounded region, we use the method of precise energy estimation and nonlinear functional analysis to prove the existence of blow-up criteria for the strong solution of this system of equations.

1. 引言

在我们的日常生活中常见的物质形态是固态、液态和气态，而这些状态是由分子或者原子的集合形式所决定的。在这三种物态中分子或者原子不同的运动状况，让我们观察到了不同的特征。而另外一些固体，尽管在常温下有稳定的体积和形状，但它们内部的结构却更像是液体，我们称之为非晶态。而对于一些有机物质，不但有类似于晶体的光学性质，还有类似于液体的流动性，介于晶态和液态之间，我们称之为液晶态。

早在19世纪液晶现象就已经被人们发现，但直到20世纪50年代，人们才得以进一步地认识它的性质，逐渐建立起液晶理论。液晶态的物质在自然界当中也是广泛存在的，据统计，大约每200种有机化合物中就能发现一种液晶分子，相比于自然界中常见的气态、液态和固态，液晶态也被科学家称为第四态。生命系统中也普遍存在着液晶态结构。液晶就是晶体向液体过渡的中间相，它的内部粒子具有各向异性、取向有序、位置无序的特点。

液晶在现实生活中有着非常广泛应用，它们对光、电、磁和热都具有很高灵敏性，根据这一特点可以设计出各种各样具有热–光、磁–光、电–光等物理效应的仪器。鉴于它如此重要的应用背景，对于它的研究遍及了化学、物理学、生物学、材料科学和电子学，形成了各自专门的学科。例如液晶化学、液晶物理、生物液晶、液晶光学等，也引起了国内外数学家和物理学家们广泛的研究兴趣。液晶工业技术中最受重视的问题就是液晶中观察到的分子分布规律、缺陷、相变现象及其动力学规律，这些问题直接关系到液晶设备的制造。而适当的数学模型就是描述这些现象最有力的工具。上世纪50年代以来，数学家和物理学家先后建立了各种各样的数学模型。而针对现实中关注的问题，从数学的角度来看，我们关心的问题是相变的数学机理、动力学方程组解的整体存在性 [1] 和正则性以及强解的爆破准则 [2]。

2. 研究内容及研究方法

2.1. 研究内容

液晶方程在动力学中的重要性不言而喻，但很少对大初值有界区域液晶方程组进行数学分析，特别是大初值有界区域的液晶方程组强解的爆破准则是未知的。本课题研究的重点是，如果方程组具有大初值，且方程组是在有界区域内，则液晶方程组强解的是否会发生爆破现象 [3]。

2.2. 具体研究方法

首先，我们考虑Euler坐标下的大初值有界区域一维完全可压缩的液晶方程组：

$\{\begin{array}{l}{\tilde{\rho }}_t+{\left(\tilde{\rho }\tilde{u}\right)}_x=0\\ {\left(\tilde{\rho }\tilde{u}\right)}_t+{\left(\tilde{\rho }{\tilde{u}}^2+\tilde{P}\right)}_x={\left(\mu u_x\right)}_x-{\tilde{n}}_x{\tilde{n}}_{xx}\\ {\tilde{n}}_t+\tilde{u}{\tilde{n}}_x={\tilde{n}}_{xx}+{\left|{\tilde{n}}_x\right|}^2\tilde{n}\\ {\tilde{P}}_t+\tilde{u}{\tilde{P}}_x+\gamma \tilde{P}{\tilde{u}}_x=\left(\gamma -1\right)\mu {\left({\tilde{u}}_x\right)}^2+\left(\gamma -1\right){\left|{\tilde{n}}_{xx}+{\left|{\tilde{n}}_x\right|}^2\tilde{n}\right|}^2\end{array}$ (*)

其中 $\tilde{\rho }\ge \text{0}$ 代表密度函数； $\tilde{u}$ 代表速度场； $\tilde{n}$ 代表单位的液晶光轴矢量(其中， ${\left|\tilde{n}\right|}^{\text{2}}=1$ )；压强 $\tilde{P}\left(\tilde{\rho }\right)=R\tilde{\rho }\tilde{\theta }$；且

$\tilde{P}:\left[0,+\infty \right)\to R$ 是局部Lipschitz连续函数 (2.1.1)

考虑初始条件

${\left(\tilde{\rho },\tilde{u},\tilde{n},\tilde{P}\right)|}_{t=0}=\left({\tilde{\rho }}_{\text{0}},{\tilde{u}}_0,{\tilde{n}}_0,{\tilde{P}}_0\right)$，其中 ${\tilde{\rho }}_{\text{0}}$ 恒为常数。 (2.1.2)

Ω是有界光滑区域，且

${\left(\tilde{u},\frac{\partial \tilde{n}}{\partial v}\right)|}_{\partial \Omega }=0$ (2.1.3)

其中v是 $\partial \Omega$ 的单位外法向量。

定理2.1.令是(*)在初始条件(2.1.2)和(2.1.3)下的初值边界问题的一个强解，假设 $\tilde{P}$ 满足条件(2.1.1),如果 $\text{0}<T_{\ast }<+\infty$ 是强解存在的关于时间的最大值，那么

${\displaystyle {\int }_{\text{0}}^{T_{\ast }}\left({\Vert {\tilde{u}}_x\Vert }_{L^{\infty }}+{\Vert {\tilde{n}}_x\Vert }_{L^{\infty }}^2\right)\text{d}t}=+\infty$ (2.1.4)

下面来证明定理2.1， $\text{0}<T_{\ast }<+\infty$ 是(*)在初始条件(2.1.2)和(2.1.3)下的初值边界问题存在强解 $\left(\tilde{\rho },\tilde{u},\tilde{n},\tilde{P}\right)$ 的关于时间的最大值， $\left(\tilde{\rho },\tilde{u},\tilde{n},\tilde{P}\right)$ 是(*)在 $\Omega \times \left(0,T\right]$ 上的一个强解，其中 $\text{0}<T<T_{\ast }$，但不是在 $\Omega \times \left(0,T_{\ast }\right]$ 上的强解，假设(2.1.4)是错误的，那么有

${\displaystyle {\int }_{\text{0}}^{T_{\ast }}\left({\Vert {\tilde{u}}_x\Vert }_{L^{\infty }}+{\Vert {\tilde{n}}_x\Vert }_{L^{\infty }}^2\right)\text{d}t}<+\infty$ (2.1.5)

目标是在(2.1.5)式成立的条件下，这个方程组的强解仍然存在，那么便可以说明 $T_{\ast }$ 不是这个方程组强解存在的关于时间的最大值。

引理2.2. $0<T_{\ast }<+\infty$ 是一个关于(*)在初始条件(2.1.2)和(2.1.3)下取得强解的关于时间的最大值，如果(2.1.5)式成立，那么对于任意的 $\text{2}\le r<+\infty$，存在一个常数C，使得

$\underset{0\le t<T_{*}}{\sup }{\Vert {\tilde{n}}_x\Vert }_{L^r}^r+{\displaystyle {\int }_0^{T_{*}}{\displaystyle {\int }_{\Omega }{\left|{\tilde{n}}_x\right|}^{r-2}{\left|{\tilde{n}}_{xx}\right|}^2\text{d}x\text{d}t}}\le C$ (2.2.1)

证明：由(*)₃式两边对x求导得：

${\tilde{n}}_{tx}-{\tilde{n}}_{xxx}=-{\left(u{\tilde{n}}_x\right)}_x+{\left({\left|{\tilde{n}}_x\right|}^2\tilde{n}\right)}_x$ (2.2.2)

在(2.2.2)式两边乘上 $r{\left|{\tilde{n}}_x\right|}^{r-1}{\tilde{n}}_x$，并对所得方程在上做关于x的积分得：

$\begin{array}{l}\frac{\text{d}}{\text{d}t}{\displaystyle \int {\left|{\tilde{n}}_x\right|}^r\text{d}x}+r\left(r-1\right){\displaystyle \int {\left|{\tilde{n}}_x\right|}^{r-2}}{\left|{\tilde{n}}_{xx}\right|}^2\text{d}x\\ =r{\displaystyle \int {\left({\left|{\tilde{n}}_x\right|}^2\tilde{n}\right)}_x{\left|{\tilde{n}}_x\right|}^{r-2}{\tilde{n}}_x\text{d}x}-r{\displaystyle \int {\left(u{\tilde{n}}_x\right)}_x{\left|{\tilde{n}}_x\right|}^{r-2}{\tilde{n}}_x\text{d}x}=I_1+I_2\end{array}$ (2.2.3)

又由

${\left({\left|{\tilde{n}}_x\right|}^2\tilde{n}\right)}_x={\left|{\tilde{n}}_x\right|}^2{\tilde{n}}_x+{\left({\left|{\tilde{n}}_x\right|}^2\right)}_x\tilde{n}$，且 ${\tilde{n}}_x\tilde{n}=0$。

所以

$I_1=r{\displaystyle \int {\left|{\tilde{n}}_x\right|}^{r+2}\text{d}x}\le {\Vert {\tilde{n}}_x\Vert }_{L^{\infty }}^2{\displaystyle \int {\left|{\tilde{n}}_x\right|}^r\text{d}x}$

$\begin{array}{c}{I}_2=-r{\displaystyle \int {\left(\tilde{u}{\tilde{n}}_x\right)}_x}{\left|{\tilde{n}}_x\right|}^{r-2}{\tilde{n}}_x\text{d}x\\ =-r{\displaystyle \int \left(\tilde{u}{\tilde{n}}_{xx}{\left|{\tilde{n}}_x\right|}^{r-2}{\tilde{n}}_x+{\tilde{u}}_x{\tilde{n}}_x{\left|{\tilde{n}}_x\right|}^{r-2}{\tilde{n}}_x\right)\text{d}x}\\ =-{\displaystyle \int \tilde{u}}{\left({\left|{\tilde{n}}_x\right|}^r\right)}_x\text{d}x-r{\displaystyle \int {\tilde{u}}_x}{\left|{\tilde{n}}_x\right|}^r\text{d}x\\ =\left(1-r\right){\displaystyle \int {\tilde{u}}_x}{\left|{\tilde{n}}_x\right|}^r\text{d}x\\ \le {\Vert {\tilde{u}}_x\Vert }_{L^{\infty }}{\displaystyle \int {\left|{\tilde{n}}_x\right|}^r\text{d}x}\end{array}$

将 $I_1,I_2$ 代入(2.2.3)式整理得：

$\begin{array}{l}\frac{\text{d}}{\text{d}t}{\displaystyle \int {\left|{\tilde{n}}_x\right|}^r}\text{d}x+r\left(r-1\right){\displaystyle \int {\left|{\tilde{n}}_x\right|}^{r-2}}{\left|{\tilde{n}}_{xx}\right|}^2\text{d}x\\ \le \left({\Vert {\tilde{n}}_x\Vert }_{L^{\infty }}^2+{\Vert {\tilde{u}}_x\Vert }_{L^{\infty }}\right){\displaystyle \int {\left|{\tilde{n}}_x\right|}^r}\text{d}x\end{array}$

利用(2.1.5)式，由Gronwall不等式，对于任意的 $0\le t<T_{\ast }$ 有：

$\begin{array}{l}{\displaystyle \int {\left|{\tilde{n}}_x\right|}^r}\text{d}x+r\left(r-1\right){\displaystyle {\int }_0^t{\displaystyle {\int }_{\Omega }{\left|{\tilde{n}}_x\right|}^{r-2}{\left|{\tilde{n}}_{xx}\right|}^2\text{d}}x\text{d}s}\\ \le {\displaystyle \int {\left|{\left({\tilde{n}}_0\right)}_x\right|}^r}\text{d}x\cdot \exp \left({\displaystyle {\int }_0^{T_{*}}\left({\Vert {\tilde{n}}_x\Vert }_{L^{\infty }}^2+{\Vert {\tilde{u}}_x\Vert }_{L^{\infty }}\right)\text{d}t}\right)\\ \le C\end{array}$

证毕。

引理2.3. $0<T_{\ast }<+\infty$ 是一个关于(*)在初始条件(2.1.2)和(2.1.3)下取得强解的关于时间的最大值，如果(2.1.5)式成立，存在一个常数C，使得

$\underset{0\le t<T_{*}}{\sup }{\Vert {\tilde{n}}_{xx}\Vert }_{L^2}^2+{\displaystyle {\int }_0^{T_{*}}{\displaystyle {\int }_{\Omega }{\left|{\tilde{n}}_{tx}\right|}^2\text{d}x\text{d}t}}\le C$ (2.3.1)

证明：在(2.2.2)式两边乘上 ${\tilde{n}}_{tx}$，并对所得方程在 $\Omega$ 上做关于x的积分得：

$\begin{array}{l}\frac{\text{d}}{\text{d}t}{\displaystyle \int {\left|{\tilde{n}}_{xx}\right|}^2\text{d}x}+{\displaystyle \int {\left|{\tilde{n}}_{tx}\right|}^2\text{d}x}={\displaystyle \int \left({\left({\left|{\tilde{n}}_x\right|}^2\tilde{n}\right)}_x-{\left(\tilde{u}{\tilde{n}}_x\right)}_x\right){\tilde{n}}_{tx}\text{d}x}\\ \le C\left(\epsilon \right){\displaystyle \int \left({\left|{\tilde{n}}_x\right|}^6+{\left|{\tilde{u}}_x\right|}^2{\left|{\tilde{n}}_x\right|}^2+{\left|\tilde{u}\right|}^2{\left|{\tilde{n}}_{xx}\right|}^2+{\left|{\tilde{n}}_x\right|}^2{\left|{\tilde{n}}_{xx}\right|}^2\right)\text{d}x}+\epsilon {\Vert {\tilde{n}}_{tx}\Vert }_{L^2}^2\\ \le \epsilon {\Vert {\tilde{n}}_{tx}\Vert }_{L^2}^2+C+C{\Vert {\tilde{n}}_x\Vert }_{L^{\infty }}^2\left({\Vert {\tilde{u}}_x\Vert }_{L^2}^2+{\Vert {\tilde{n}}_{xx}\Vert }_{L^2}^2\right)+C{\displaystyle \int {\left|\tilde{u}\right|}^2{\left|{\tilde{n}}_{xx}\right|}^2\text{d}x}\end{array}$ (2.3.2)

利用(2.2.1)式(r = 6)，由Nirenberg’s interpolation不等式得：

$\begin{array}{c}{{\displaystyle \int \left|\tilde{u}\right|}}^2{\left|{\tilde{n}}_{xx}\right|}^2\text{d}x\le {\Vert \tilde{u}\Vert }_{L^6}^2{\Vert {\tilde{n}}_{xx}\Vert }_{L^3}^2\le {\Vert {\tilde{u}}_x\Vert }_{L^2}^2{\Vert {\tilde{n}}_x\Vert }_{L^6}{\Vert {\tilde{n}}_{xxx}\Vert }_{L^2}\\ \le {\Vert {\tilde{u}}_x\Vert }_{L^2}^2{\Vert {\tilde{n}}_{xxx}\Vert }_{L^2}\le C\left(\epsilon \right){\Vert {\tilde{u}}_x\Vert }_{L^2}^4+\epsilon {\Vert {\tilde{n}}_{xxx}\Vert }_{L^2}^2\end{array}$ (2.3.3)

又由

(2.3.4)

将(2.3.3)代入(2.3.4)得：

${\Vert {\tilde{n}}_{xxx}\Vert }_{L^2}^2\le {\Vert {\tilde{n}}_{tx}\Vert }_{L^2}^2+{\Vert {\tilde{n}}_x\Vert }_{L^{\infty }}^2\left({\Vert {\tilde{u}}_x\Vert }_{L^2}^2+{\Vert {\tilde{n}}_{xx}\Vert }_{L^2}^2\right)+{\Vert {\tilde{u}}_x\Vert }_{L^2}^4+C$ (2.3.5)

将(2.3.5)代入(2.3.3)得：

${\displaystyle \int {\left|u\right|}^2{\left|{\tilde{n}}_{xx}\right|}^2\text{d}x}\le C+\epsilon {\Vert {\tilde{n}}_{tx}\Vert }_{L^2}^2+{\Vert {\tilde{n}}_x\Vert }_{L^{\infty }}^2\left({\Vert {\tilde{u}}_x\Vert }_{L^2}^2+{\Vert {\tilde{n}}_{xx}\Vert }_{L^2}^2\right)+{\Vert u_x\Vert }_{L^2}^4$ (2.3.6)

将(2.3.6)代入(2.3.2)，利用(2.1)式，取 $\epsilon$ 适当小得：

$\begin{array}{c}\frac{\text{d}}{\text{d}t}{\displaystyle \int {\left|{\tilde{n}}_{xx}\right|}^2\text{d}x}+{\displaystyle \int {\left|{\tilde{n}}_{tx}\right|}^2\text{d}x}\le C+{\Vert {\tilde{u}}_x\Vert }_{L^2}^4+{\Vert {\tilde{n}}_x\Vert }_{L^{\infty }}^2{\Vert {\tilde{u}}_x\Vert }_{L^2}^2+{\Vert {\tilde{n}}_x\Vert }_{L^{\infty }}^2{\Vert {\tilde{n}}_{xx}\Vert }_{L^2}^2\\ \le C+{\Vert {\tilde{n}}_x\Vert }_{L^{\infty }}^2{\Vert {\tilde{n}}_{xx}\Vert }_{L^2}^2\end{array}$ (2.3.7)

由Gronwall不等式，对于任意的 $0\le t<T_{\ast }$ 有：

$\begin{array}{l}{\displaystyle \int {\left|{\tilde{n}}_{xx}\right|}^2\text{d}x}+{\displaystyle {\int }_{\text{0}}^t{\displaystyle {\int }_{\Omega }{\left|{\tilde{n}}_{tx}\right|}^2\text{d}}x\text{d}s}\\ \le CT\cdot \exp \left(CT\right)+{\displaystyle \int {\left|{\left({\tilde{n}}_0\right)}_{xx}\right|}^2\text{d}x}\cdot \exp \left({\displaystyle {\int }_0^{T_{*}}{\Vert {\tilde{n}}_x\Vert }_{L^{\infty }}^2\text{d}t}\right)\\ \le C\end{array}$ (2.3.8)

证毕。

接下来我们将(2.3.8)结果回代到(2.3.5)，利用(2.1.5)式我们得到

${\Vert {\tilde{n}}_{xxx}\Vert }_{L^2}^2\le {\Vert {\tilde{n}}_{tx}\Vert }_{L^2}^2+{\Vert {\tilde{n}}_x\Vert }_{L^{\infty }}^2\left({\Vert {\tilde{u}}_x\Vert }_{L^2}^2+{\Vert {\tilde{n}}_{xx}\Vert }_{L^2}^2\right)+{\Vert {\tilde{u}}_x\Vert }_{L^2}^4+C\le C$ (2.3.9)

于是有 ${\tilde{n}}_{xxx}\in L_x^2\left(\Omega \right)$，进而得到

${\tilde{n}}_{xx}\in L_x^{\infty }\left(\Omega \right),{\tilde{n}}_x\in L_x^{\infty }\left(\Omega \right)$ (2.3.10)

推理2.3.从(2.1.5)和引理2.2得到的，进而得 ，另一方面，由(2.3.10)得，因此，由(*)₃我们有：

$\left|{\tilde{n}}_t\right|\le \left(\left|\tilde{u}\right|\left|{\tilde{n}}_x\right|+\left|{\tilde{n}}_{xx}\right|+{\left|{\tilde{n}}_{xx}\right|}^2\right)\in L_t^{\infty }{L}_x^{\infty }\left(\Omega \times \left[0,T_{*}\right]\right)$ (2.3.1)

接下来我们进行坐标变换 [4]，设y为拉格朗日坐标，将拉格朗日坐标y和欧拉坐标x之间的坐标变换定义为 $x=\eta \left(y,t\right)$。其中 $\eta \left(y,t\right)$ 是由 $\tilde{u}$ 所确定的映射，即：

$\{\begin{array}{l}{\partial }_t\eta \left(y,t\right)=\tilde{u}\left(\eta \left(y,t\right),t\right)\\ \eta \left(y,0\right)=y\end{array}$

定义函数；于是有 $J_t=u_y$；又因为 ${\left(J\rho \right)}_t=J_t\rho +J{\rho }_t=u_y\rho -J\frac{u_y}{J}\rho =0$，且 ${\rho |}_{t=0}={\rho }_{\text{0}}$， ${J|}_{t=0}=1$；于是。

通过以上的变换我们就可以得到拉格朗日坐标下的液晶方程组：

$\{\begin{array}{l}{J}_t=u_y\\ {\rho }_0{u}_t-\mu {\left(\frac{u_y}{J}\right)}_y+P_y+{\left(\frac{n_y}{J}\right)}_y\frac{n_y}{J}=0\\ n_t=\frac{1}{J}{\left(\frac{n_y}{J}\right)}_y+{\left|\frac{n_y}{J}\right|}^{2}n\\ P_t+\gamma P\frac{u_y}{J}=\left(\gamma -1\right)\mu {\left(\frac{u_y}{J}\right)}^{\text{2}}+\left(\gamma -1\right){\left|\frac{1}{J}{\left(\frac{n_y}{J}\right)}_y+{\left|\frac{n_y}{J}\right|}^{2}n\right|}^2\end{array}$ (**)

且其初始条件为： ${\left(\rho ,u,P,n\right)|}_{t=0}=\left({\rho }_{\text{0}}\left(x\right),u_0\left(x\right),P_0\left(x\right),n_0\left(x\right)\right)$， $x\in {\Omega }_{\text{1}}$，其中 ${\rho }_{\text{0}}\left(x\right)$ 恒为常数。

P仍是局部Lipschitz连续函数， $\Omega$ 变为 $\partial \Omega$ 也仍是有界光滑区域，且

${\left(u,\frac{\partial n}{\partial v_1}\right)|}_{\partial {\Omega }_1}=0$ (3.1.1)

其中是 $\partial {\Omega }_{\text{1}}$ 的单位外法向量。

(2.1.5)式变为

(3.1.2)

3. 强解存在的定义

定义3.1. 给定一个时间 $T\in \left(0,T_{\ast }\right)$，称为上述拉格朗日坐标下的液晶方程组(**)的一个强解,如果它满足以下条件：

$\underset{y\in {\Omega }_{\text{1}},t\in \left(0,T\right)}{\inf }J\left(y,t\right)>0,P\ge 0\text{on}{\Omega }_{\text{1}}\times \left(0,T\right)$；

；

$\sqrt{{\rho }_{\text{0}}}u\in C\left(\left[0,T\right];L^2\right),u_y\in L^{\infty }\left(\left[0,T\right];L^2\right),\left(\sqrt{{\rho }_{\text{0}}}{u}_t,\frac{u_{yy}}{\sqrt{{\rho }_{\text{0}}}}\right)\in L^2\left(\left[0,T\right];L^2\right)$；

；

$P\in C\left(\left[0,T\right];L^2\right),\frac{P_y}{\sqrt{{\rho }_{\text{0}}}}\in L^{\infty }\left(\left[0,T\right];L^2\right),P_t\in L^4\left(\left[0,T\right];L^2\right)$

命题3.1. 方程组(**)存在系统的初始能量 $E_0$，且J存在上界 $\bar{J}$ 和下界。

证明：将(**)₂两边乘上u，并对所得方程做关于y在 ${\Omega }_{\text{1}}$ 上的积分得：

${\displaystyle \underset{{\Omega }_{\text{1}}}{\int }{\left(\frac{1}{2}{\rho }_0{u}^2\right)}_t\text{d}y}+{\displaystyle \underset{{\Omega }_{\text{1}}}{\int }\mu \frac{{\left(u_y\right)}^2}{J}\text{d}y}-{\displaystyle \underset{{\Omega }_{\text{1}}}{\int }P{u}_y\text{d}y}+{\displaystyle \underset{{\Omega }_{\text{1}}}{\int }{\left(\frac{n_y}{J}\right)}_y\frac{n_y}{J}u\text{d}y}=0$ (3.1.3)

又 ${\left(JP\right)}_t=J_{t}P+J{P}_t=P{u}_y+J{P}_t$；

再由(**)4得：

于是有：

${\displaystyle \underset{R}{\int }{\left(\frac{1}{2}{\rho }_0{u}^2+\frac{JP}{\gamma -1}\right)}_t\text{d}y}-{\displaystyle \underset{R}{\int }J{\left|n_t\right|}^2\text{d}y}+{\displaystyle \underset{R}{\int }{\left(\frac{n_y}{J}\right)}_y\frac{n_y}{J}u\text{d}y}=0$ (3.1.4)

将(**)3两边乘上 $\frac{1}{J}{\left(\frac{n_y}{J}\right)}_y+{\left|\frac{n_y}{J}\right|}^{2}n$，并对所得方程做关于y在 ${\Omega }_{\text{1}}$ 上的积分得：

${\displaystyle \underset{R}{\int }J{\left|n_t\right|}^2\text{d}y}={\displaystyle \underset{R}{\int }{n}_t{\left(\frac{n_y}{J}\right)}_y\text{d}y}+{\displaystyle \underset{R}{\int }J{n}_{t}n{\left|\frac{n_y}{J}\right|}^2}$

由 ${\left|\tilde{n}\right|}^2={\left|n\right|}^2=\left|\tilde{n}\right|=1$ 知， $n_{t}n=n_{y}n=0$

于是

${\displaystyle \underset{R}{\int }J{\left|n_t\right|}^2\text{d}y}=-{\displaystyle \underset{R}{\int }{\left(\frac{\frac{1}{2}{\left(n_y\right)}^2}{J}\right)}_t\text{d}y}+{\displaystyle \underset{R}{\int }{\left(\frac{n_y}{J}\right)}_y\frac{n_y}{J}u\text{d}y}$ (3.1.5)

把(3.1.5)式代入(3.1.4)式得

(3.1.6)

对(3.1.6)式关于t在 $\left[0,t\right]$ 上积分得：

${\displaystyle \underset{R}{\int }\left(\frac{1}{2}{\rho }_0{u}^2+\frac{JP}{\gamma -1}+\frac{\frac{1}{2}{\left(n_y\right)}^2}{J}\right)\text{d}y}-{\displaystyle \underset{R}{\int }\left(\frac{1}{2}{\rho }_0{u}_0^2+\frac{P_0}{\gamma -1}+\frac{1}{2}{\left(n_0{}_y\right)}^2\right)\text{d}y}=0$ (3.1.7)

于是可以定义为系统的初始能量。

然后来证明J存在上界：

由 $J_t=u_y$ 得：，进一步有 ${\left|\ln J\right|}_t=\frac{u_y}{J}$，两边同时在 $\left(0,t\right)$ 上积分得：

$\ln J-\ln J_0={\displaystyle {\int }_0^t\frac{u_y}{J}\text{d}t}$，于是有 $J=J_0{\text{e}}^{{\displaystyle {\int }_0^t\frac{u_y}{J}\text{d}t}}\le J_0{\text{e}}^{{\displaystyle {\int }_0^t{\Vert \frac{u_y}{J}\Vert }_{L^{\infty }}\text{d}t}}\le C$，定义J的上界为 $\bar{J}$。

接下来证明J存在下界：对(**)3有：

${\partial }_t\left({\rho }_{0}u\right)-\mu {\partial }_{ty}\left(\ln J\right)+P_y+{\left(\frac{n_y}{J}\right)}_y\frac{n_y}{J}=0$ (3.1.8)

对(3.1.8)式关于t在 $\left[0,t\right]$ 上积分，如何再对所得方程关于y在 $\left[z_0,z\right]$ 上积分得：

$\begin{array}{l}{\displaystyle {\int }_{z_0}^z\left({\rho }_{0}u-{\rho }_0{u}_0\right)\text{d}y}-\mu \left[\ln J\left(z,t\right)-\ln J\left(z_0,t\right)\right]+{\displaystyle {\int }_0^t\left[P\left(z,\tau \right)-P\left(z_0,\tau \right)\right]\text{d}\tau }\\ +\frac{1}{2}{\displaystyle {\int }_0^t\left[{\left(\frac{n_y\left(z,t\right)}{J\left(z,t\right)}\right)}^2-{\left(\frac{n_y\left(z_0,t\right)}{J\left(z_0,t\right)}\right)}^2\right]\text{d}\tau }=0\end{array}$

令 $z_0\to -\infty$ 得： $J\to 1,P\to 0,n_y\to 0$；于是：

${\displaystyle {\int }_{-\infty }^z\left({\rho }_{0}u-{\rho }_0{u}_0\right)\text{d}y}-\mu \ln J\left(z,t\right)+{\displaystyle {\int }_0^{t}P\left(z,\tau \right)\text{d}\tau }+\frac{1}{2}{\displaystyle {\int }_0^t{\left(\frac{n_y\left(z,t\right)}{J\left(z,t\right)}\right)}^2\text{d}\tau }=0$

$\begin{array}{c}J\left(z,t\right)\ge {\text{e}}^{\frac{1}{\mu }{\displaystyle {\int }_{-\infty }^z\left({\rho }_{0}u-{\rho }_0{u}_0\right)\text{d}y}+{\displaystyle {\int }_0^{t}P\left(z,\tau \right)\text{d}\tau }+\frac{1}{2}{\displaystyle {\int }_0^t{\left(\frac{n_y\left(z,t\right)}{J\left(z,t\right)}\right)}^2\text{d}\tau }}\\ \ge {\text{e}}^{\frac{1}{\mu }{\displaystyle {\int }_{-\infty }^z\left({\rho }_{0}u-{\rho }_0{u}_0\right)\text{d}y}}\end{array}$

$\begin{array}{l}\left|{\displaystyle {\int }_{-\infty }^z\left({\rho }_{0}u-{\rho }_0{u}_0\right)\text{d}y}\right|\le {\displaystyle {\int }_{-\infty }^z{\rho }_0\left(\left|u\right|-\left|u_{\text{0}}\right|\right)\text{d}y}\\ \le {\left({\displaystyle \underset{{\Omega }_{\text{1}}}{\int }{\rho }_0\text{d}y}\right)}^{\frac{1}{2}}{\left({\displaystyle \underset{{\Omega }_{\text{1}}}{\int }\frac{{\rho }_0}{2}\left(2{\left|u\right|}^2-2{\left|u_{\text{0}}\right|}^2\right)\text{d}y}\right)}^{\frac{1}{2}}\\ \le {\Vert {\rho }_0\Vert }_1^{\frac{1}{2}}{\left(4E_0\right)}^{\frac{1}{2}}=2{\Vert {\rho }_0\Vert }_1^{\frac{1}{2}}{E}_0^{\frac{1}{2}}\end{array}$

于是 $J\ge {\text{e}}^{\frac{2}{\mu }{\Vert {\rho }_0\Vert }_1^{\frac{1}{2}}{E}_0^{\frac{1}{2}}}$。

定义J的下界为 $\underset{\_}{J}={\text{e}}^{\frac{2}{\mu }{\Vert {\rho }_0\Vert }_1^{\frac{1}{2}}{E}_0^{\frac{1}{2}}}$。

接下来，定义有效粘性通量G为 $G=\mu \frac{u_y}{J}-P$；

命题3.2. 有效粘性通量G有以下估计：

$\underset{0\le t\le T}{\sup }{\Vert G\Vert }_2^2+{\displaystyle {\int }_0^T{\Vert \frac{G_y}{\sqrt{{\rho }_0}}\Vert }_2^2\text{d}t}+{\left({\displaystyle {\int }_0^T{\Vert G\Vert }_{\infty }^4\text{d}t}\right)}^{\frac{1}{2}}\le C{\text{e}}^{CT}{\Vert G_0\Vert }_2^2$

其中正常数C仅取决于 $\gamma ,\mu ,\bar{\rho },E_0$ 和 ${\Vert {\rho }_0\Vert }_{L^1}$

证明：由(**)₃和(**)₄可知：

$G_t=\mu {\left(\frac{u_y}{J}\right)}_t-P_t=\frac{\mu }{J}{\left(\frac{G_y}{{\rho }_0}\right)}_y-\gamma \frac{u_y}{J}G-\frac{\mu }{J}\left(\frac{1}{{\rho }_0}{\left(\frac{n_y}{J}\right)}_y\left(\frac{n_y}{J}\right)\right)-\left(\gamma -1\right){\left|n_t\right|}^2$ (3.1.9)

方程两边同时乘以JG，并在 ${\Omega }_{\text{1}}$ 上做关于y的积分：

$\begin{array}{l}{\displaystyle \underset{{\Omega }_{\text{1}}}{\int }JG{G}_t\text{d}y}-\mu {\displaystyle \underset{{\Omega }_{\text{1}}}{\int }G}\left(\frac{G_y}{{\rho }_0}\right)\text{d}y\\ =-\gamma {\displaystyle \underset{{\Omega }_{\text{1}}}{\int }{u}_y{G}^2\text{d}y}-\mu {\displaystyle \underset{{\Omega }_{\text{1}}}{\int }G}{\left(\frac{1}{{\rho }_0}{\left(\frac{n_y}{J}\right)}_y\frac{n_y}{J}\right)}_y\text{d}y-\left(\gamma -1\right){\displaystyle \underset{{\Omega }_{\text{1}}}{\int }JG}{\left|n_t\right|}^2\text{d}y\end{array}$ (3.1.10)

$\begin{array}{l}\frac{\text{1}}{\text{2}}\frac{\text{d}}{\text{d}t}{\displaystyle \underset{{\Omega }_{\text{1}}}{\int }J{G}^{\text{2}}\text{d}y}+\mu {\displaystyle \underset{{\Omega }_{\text{1}}}{\int }\frac{{\left(G_y\right)}^2}{{\rho }_0}\text{d}y}\\ =\left(\frac{\text{1}}{\text{2}}-\gamma \right){\displaystyle \underset{{\Omega }_{\text{1}}}{\int }{u}_y{G}^2\text{d}y}-\mu {\displaystyle \underset{{\Omega }_{\text{1}}}{\int }G}{\left(\frac{1}{{\rho }_0}{\left(\frac{n_y}{J}\right)}_y\frac{n_y}{J}\right)}_y\text{d}y-\left(\gamma -1\right){\displaystyle \underset{{\Omega }_{\text{1}}}{\int }JG}{\left|n_t\right|}^2\text{d}y\end{array}$ (3.1.11)

对 $\left(\frac{1}{2}-\gamma \right){\displaystyle \underset{{\Omega }_{\text{1}}}{\int }{u}_y{G}^2\text{d}y}$ 的估计，利用Gagliardo-Nirenberg不等式和Young不等式得：

$\begin{array}{c}\left(\frac{1}{2}-\gamma \right){\displaystyle \underset{{\Omega }_{\text{1}}}{\int }{u}_y{G}^2\text{d}y}=\left(2\gamma -1\right){\displaystyle \underset{{\Omega }_{\text{1}}}{\int }uG{G}_y\text{d}y}\le C{\displaystyle \underset{{\Omega }_{\text{1}}}{\int }\left|u\right|\left|G\right|\left|G_y\right|\text{d}y}\\ \le C{\displaystyle \underset{{\Omega }_{\text{1}}}{\int }\sqrt{{\rho }_0}\left|u\right|\left|G\right|\left|\frac{G_y}{\sqrt{{\rho }_0}}\right|\text{d}y}\le C{\Vert \sqrt{{\rho }_0}u\Vert }_{L^2}{\Vert G\Vert }_{L^{\infty }}{\Vert \frac{G_y}{\sqrt{{\rho }_0}}\Vert }_{L^2}\\ \le C{\Vert \sqrt{{\rho }_0}u\Vert }_{L^2}{\Vert G\Vert }_{L^2}^{\frac{1}{2}}{\Vert \frac{G_y}{\sqrt{{\rho }_0}}\Vert }_{L^2}^1{\Vert \frac{G_y}{\sqrt{{\rho }_0}}\Vert }_{L^2}\\ \le C{\Vert G\Vert }_{L^2}^{\frac{1}{2}}{\Vert \frac{G_y}{\sqrt{{\rho }_0}}\Vert }_{L^2}^{\frac{3}{2}}\le \frac{\mu }{8}{\Vert \frac{G_y}{\sqrt{{\rho }_0}}\Vert }_{L^2}^2+C{\Vert G\Vert }_{L^2}^2\end{array}$ (3.1.12)

对 $-\mu {\displaystyle \underset{{\Omega }_1}{\int }G{\left(\frac{1}{{\rho }_0}{\left(\frac{n_y}{J}\right)}_y\frac{n_y}{J}\right)}_y\text{d}y}$ 的估计：

$\begin{array}{l}-\mu {\displaystyle \underset{R}{\int }G{\left(\frac{1}{{\rho }_0}{\left(\frac{n_y}{J}\right)}_y\frac{n_y}{J}\right)}_y\text{d}y}=\mu {\displaystyle \underset{R}{\int }\frac{G_y}{\sqrt{{\rho }_0}}\frac{1}{\sqrt{{\rho }_0}}{\left(\frac{n_y}{J}\right)}_y\frac{n_y}{J}\text{d}y}\\ \le \frac{\mu }{16}{\Vert \frac{G_y}{\sqrt{{\rho }_0}}\Vert }_{L^2}^2+C{\Vert \frac{\text{1}}{\sqrt{{\rho }_0}}\Vert }_{L^2}^2{\Vert \frac{n_y}{J}\Vert }_{L^{\infty }}^2{\Vert {\left(\frac{n_y}{J}\right)}_y\Vert }_{L^2}^2=\frac{\mu }{16}{\Vert \frac{G_y}{\sqrt{{\rho }_0}}\Vert }_{L^2}^2+C\end{array}$ (3.1.13)

由 $n_t={\tilde{n}}_t+\tilde{u}{\tilde{n}}_x$， ${\tilde{n}}_t,{\tilde{n}}_x,\tilde{u}\in L_t^{\infty }{L}_x^{\infty }\left(\Omega \times \left[0,T_{\ast }\right]\right)$，J有上下界，于是有 ${\Vert n_t\Vert }_{L^2\left({\Omega }_1\right)}^2\le C$。

对 $-\left(\gamma -1\right){\displaystyle \underset{R}{\int }JG{\left|n_t\right|}^2\text{d}y}$ 的估计，利用Gagliardo-Nirenberg不等式和Young不等式得：

$\begin{array}{c}-\left(\gamma -1\right){\displaystyle \underset{{\Omega }_1}{\int }JG{\left|n_t\right|}^2\text{d}y}\le \left(\gamma -1\right){\displaystyle \underset{{\Omega }_1}{\int }\bar{J}\left|G\right|}{\left|n_t\right|}^2\text{d}y\le \left(\gamma -1\right)\bar{J}{\Vert G\Vert }_{L^{\infty }}{\Vert n_t\Vert }_{L^2}^2\le \left(\gamma -1\right)\bar{J}{\Vert n_t\Vert }_{L^2}^2{\Vert G\Vert }_{L^2}^{\frac{1}{2}}{\Vert \frac{G_y}{\sqrt{{\rho }_0}}\Vert }_{L^2}^{\frac{1}{2}}\\ \le C{\left(\left(\gamma -1\right)\bar{J}{\Vert n_t\Vert }_{L^2}^2\right)}^2+C{\left({\Vert G\Vert }_{L^2}^{\frac{1}{2}}{\Vert \frac{G_y}{\sqrt{{\rho }_0}}\Vert }_{L^2}^{\frac{1}{2}}\right)}^2\le C+C{\Vert G\Vert }_{L^2}^2+\frac{\mu }{16}{\Vert \frac{G_y}{\sqrt{{\rho }_0}}\Vert }_{L^2}^2\end{array}$ (3.1.14)

联立(3.1.9)~(3.1.12)得：

$\frac{\text{d}}{\text{d}t}{\displaystyle \underset{R}{\int }J{G}^{\text{2}}}\text{d}y+\frac{3\mu }{2}{\Vert \frac{G_y}{\sqrt{{\rho }_0}}\Vert }_{L^2}^2\le C{\Vert G\Vert }_{L^2}^2+C$ (3.1.15)

对(3.13)式两边关于t在[0, T]上积分，又由J有上下界可知：

${\Vert G\Vert }_{L^2}^2\left(t\right)+{\displaystyle {\int }_{\text{0}}^t{\Vert \frac{G_y}{\sqrt{{\rho }_0}}\Vert }_{L^2}^2\text{d}\tau }\le C\left({\displaystyle {\int }_0^t{\Vert G\Vert }_{L^2}^2\text{d}\tau }+{\Vert G_0\Vert }_{L^2}^2\right)+CT$ (3.1.16)

由Gronwall不等式得：

$\underset{0\le t\le T}{\sup }{\Vert G\Vert }_{L^2}^2+{\displaystyle {\int }_0^T{\Vert \frac{G_y}{\sqrt{{\rho }_0}}\Vert }_{L^2}^2\text{d}t}\le C{\text{e}}^{CT}{\Vert G_0\Vert }_{L^2}^2+C_{2}T$ (3.1.17)

再由Gagliardo-Nirenberg不等式：

${\displaystyle {\int }_0^T{\Vert G\Vert }_{L^{\infty }}^4\text{d}t}\le {\displaystyle {\int }_0^T{\Vert G\Vert }_{L^2}^2{\Vert G_y\Vert }_{L^2}^2\text{d}t}\le C\left(T\right)$ (3.1.18)

其中 $C\left(T\right)$ 取决于 $T,\gamma ,\mu ,\bar{\rho },E_0$ 和 ${\Vert {\rho }_{\text{0}}\Vert }_{\text{1}}$ 等。

命题3.3. 对于命题3.2.中的时间T，有如下估计：

${\displaystyle {\int }_0^T\left({\Vert P_t\Vert }_2^4+{\Vert \left(\sqrt{{\rho }_0}{u}_t,\frac{u_{yy}}{\sqrt{{\rho }_0}}\right)\Vert }_2^2\right)\text{d}t}\le C$

其中C是正常数，且仅取决于 $\gamma ,\mu ,\bar{\rho },\underset{\_}{J},\bar{J},{\Vert \frac{{J^{\prime }}_0}{\sqrt{{\rho }_0}}\Vert }_2,{\Vert G_0\Vert }_2,{\Vert P_0\Vert }_2$ 和 ${\Vert \frac{{P^{\prime }}_0}{\sqrt{{\rho }_0}}\Vert }_2$，但独立于 $\underset{\_}{\rho }$。

证明：

$\begin{array}{c}{P}_t+\mu {\left(\frac{u_y}{J}\right)}^2=\gamma \mu {\left(\frac{u_y}{J}\right)}^2-\gamma \frac{u_y}{J}P+\left(\gamma -1\right){\left|n_t\right|}^2\\ =\gamma \frac{u_y}{J}G+\left(\gamma -1\right){\left|n_t\right|}^2\\ \le \mu {\left(\frac{u_y}{J}\right)}^2+\frac{{\gamma }^2}{4\mu }{G}^2+\left(\gamma -1\right){\left|n_t\right|}^2\end{array}$

$P_t\le \frac{{\gamma }^2}{4\mu }{G}^2+\left(\gamma -1\right){\left|n_t\right|}^2$ (3.1.19)

于是：

$P\le P_0+\frac{{\gamma }^2}{4\mu }{\displaystyle {\int }_0^t{G}^2}\text{d}t+\left(\gamma -1\right){\displaystyle {\int }_0^t{\left|n_t\right|}^2}\text{d}t$

由 $n_t={\tilde{n}}_t+\tilde{u}{\tilde{n}}_x$， ${\tilde{n}}_t,{\tilde{n}}_x,\tilde{u}\in L_t^{\infty }{L}_x^{\infty }\left(\Omega \times \left[0,T_{\ast }\right]\right)$，J有上下界，于是有。

于是有：

${\Vert P\Vert }_q\le {\Vert P_0+C\Vert }_q+\frac{{\gamma }^2}{4\mu }{\displaystyle {\int }_0^t{\Vert G\Vert }_{2q}^2\text{d}t}$

当时：

${\Vert P\Vert }_2\le {\Vert P_0+C\Vert }_2+\frac{{\gamma }^2}{4\mu }{\displaystyle {\int }_0^t{\Vert G\Vert }_4^2\text{d}t}\le C$ (3.1.20)

当时：

${\Vert P\Vert }_2\le {\Vert P_0+C\Vert }_{\infty }+\frac{{\gamma }^2}{4\mu }{\displaystyle {\int }_0^t{\Vert G\Vert }_{\infty }^2\text{d}t}\le C$ (3.1.21)

于是

$\underset{0\le t\le T}{\sup }\left({\Vert P\Vert }_2+{\Vert P\Vert }_{\infty }\right)\le C$ (3.1.22)

又由

$P_t+\frac{\text{1}}{\mu }{\left(P+\frac{2-\gamma }{2}G\right)}^2=\frac{{\gamma }^{\text{2}}}{4\mu }{G}^2+\left(\gamma -1\right){\left|n_t\right|}^2$ (3.1.23)

由(3.1.23)两边对y求导，并对所得方程两边乘以 $\frac{P_y}{{\rho }_0}$，然后在R上做关于y的积分，利用Hölder不等式和柯西不等式得：

$\begin{array}{c}\frac{\text{d}}{\text{d}t}{\Vert \frac{P_y}{\sqrt{{\rho }_0}}\Vert }_2^2\le C\left({\Vert P\Vert }_{L^{\infty }}+{\Vert G\Vert }_{L^{\infty }}\right)\left({\Vert \frac{P_y}{\sqrt{{\rho }_0}}\Vert }_{L^2}+{\Vert \frac{G_y}{\sqrt{{\rho }_0}}\Vert }_{L^2}\right){\Vert \frac{P_y}{\sqrt{{\rho }_0}}\Vert }_{L^2}+{\Vert \frac{{\partial }_y{\left|n_t\right|}^2}{\sqrt{{\rho }_0}}\Vert }_{L^2}{\Vert \frac{P_y}{\sqrt{{\rho }_0}}\Vert }_{L^2}\\ \le C_1+C_2{\Vert \frac{{\partial }_y{\left|n_t\right|}^2}{\sqrt{{\rho }_0}}\Vert }_{L^2}^2+C_3{\Vert \frac{P_y}{\sqrt{{\rho }_0}}\Vert }_{L^2}^2\end{array}$

又由

${\Vert \frac{{\partial }_y{\left|n_t\right|}^2}{\sqrt{{\rho }_0}}\Vert }_{L^2}^2={\displaystyle {\int }_{{\Omega }_1}\frac{4{\left(n_t\right)}^2{\left(n_{ty}\right)}^2}{{\rho }_0}\text{d}y}\le \frac{4}{{\rho }_0}{\Vert n_t\Vert }_{L^{\infty }}^2{\Vert n_{ty}\Vert }_{L^2}^2$

而 $n_t={\tilde{n}}_t+\tilde{u}{\tilde{n}}_x$， ${\tilde{n}}_t,{\tilde{n}}_x,\tilde{u}\in L_t^{\infty }{L}_x^{\infty }\left(\Omega \times \left[0,T_{\ast }\right]\right)$，J有上下界，于是有 ${\Vert n_t\Vert }_{L^{\infty }\left({\Omega }_{\text{1}}\right)}^2\le C$。

由 ${\tilde{n}}_t=n_t-\tilde{u}{\tilde{n}}_x$ 得 ${\tilde{n}}_{tx}=\frac{1}{J}{\left(n_t-\tilde{u}{\tilde{n}}_x\right)}_y=\frac{n_{ty}}{J}-\frac{u_y}{J}\frac{n_y}{J}-\frac{u}{J}{\left(\frac{n_y}{J}\right)}_y=\frac{n_{ty}}{J}-{\tilde{u}}_x{\tilde{n}}_x-\tilde{u}{\tilde{n}}_{xx}$，

于是有 $\frac{n_{ty}}{\bar{J}}\le \frac{n_{ty}}{J}={\tilde{n}}_{tx}+{\tilde{u}}_x{\tilde{n}}_x+\tilde{u}{\tilde{n}}_{xx}$，

又由 ${\tilde{n}}_{ty}\in L_x^{\text{2}}\left(\Omega \right)$， ${\tilde{n}}_x\in L_x^{\infty }\left(\Omega \right)$， ${\tilde{n}}_{xx}\in L_x^{\infty }\left(\Omega \right)$， $\tilde{u}\in L_x^{\infty }\left(\Omega \right)$， ${\tilde{u}}_x\in L_x^{\infty }\left(\Omega \right)$，J有上下界，于是有 ${\Vert n_{ty}\Vert }_{L^2\left({\Omega }_{\text{1}}\right)}^2\le C$。

所以得到

$\frac{\text{d}}{\text{d}t}{\Vert \frac{P_y}{\sqrt{{\rho }_0}}\Vert }_2^2\le C_{\text{4}}+C_{\text{3}}{\Vert \frac{P_y}{\sqrt{{\rho }_0}}\Vert }_{L^2}^2$

再由Gronwall不等式得：

$\underset{0\le t\le T}{\sup }{\Vert \frac{P_y}{\sqrt{{\rho }_0}}\Vert }_2^2\le CT{\text{e}}^{CT}+{\Vert \frac{{P^{\prime }}_0}{\sqrt{{\rho }_0}}\Vert }_2{\text{e}}^{CT}\le C$ (3.1.24)

回头看G的定义，我们可以得到：

$J_t=\frac{J}{\mu }\left(G+P\right)$ (3.1.25)

于是有：

$\begin{array}{l}{\Vert J-J_0\Vert }_2+{\Vert J_t\Vert }_2={\Vert {\displaystyle {\int }_0^t{J}_t\text{d}\tau }\Vert }_2+{\Vert J_t\Vert }_2\\ \le \frac{2\bar{J}}{\mu }\left({\displaystyle {\int }_0^t\left({\Vert G\Vert }_2+{\Vert P\Vert }_2\right)\text{d}\tau }+{\Vert G\Vert }_2+{\Vert P\Vert }_2\right)\end{array}$

进一步有：

$\underset{0\le t\le T}{\sup }\left({\Vert J-J_0\Vert }_2+{\Vert J_t\Vert }_2\right)=C\left(\gamma ,\mu ,\bar{\rho },\underset{\_}{J},\bar{J},{\Vert G_0\Vert }_2,{\Vert P_0\Vert }_2\right)$ (3.1.26)

由(3.23)可得：

$J\left(y,t\right)=\exp \left\{\frac{1}{\mu }{\displaystyle {\int }_0^t\left(G+P\right)\text{d}\tau }\right\}{J}_0(y)$

于是有：

$J_y=\left(\frac{1}{\mu }{\displaystyle {\int }_0^t\left(G_y+P_y\right)\text{d}\tau }{J}_0+{J^{\prime }}_0\right)\exp \left\{\frac{1}{\mu }{\displaystyle {\int }_0^t\left(G+P\right)\text{d}\tau }\right\}$

进一步有

$\underset{0\le t\le T}{\text{sup}}{\Vert \frac{J_y}{\sqrt{{\rho }_0}}\Vert }_{L^2}\le \left(\frac{\bar{J}}{\mu }{\displaystyle {\int }_0^T{\Vert \left(\frac{G_y}{\sqrt{{\rho }_0}}+\frac{P_y}{\sqrt{{\rho }_0}}\right)\Vert }_{L^2}\text{d}}t+{\Vert \frac{{J^{\prime }}_{\text{0}}}{\sqrt{{\rho }_0}}\Vert }_{L^2}\right)\exp \left\{\frac{1}{\mu }{\displaystyle {\int }_0^T{\Vert \left(G+P\right)\Vert }_{L^{\infty }}\text{d}}t\right\}\le C$ (3.1.27)

其中C是正常数，且仅取决于 $\gamma ,\mu ,\bar{\rho },\underset{\_}{J},\bar{J},{\Vert \frac{{J^{\prime }}_0}{\sqrt{{\rho }_0}}\Vert }_2,{\Vert G_0\Vert }_2,{\Vert P_0\Vert }_2$ 和 ${\Vert \frac{{P^{\prime }}_0}{\sqrt{{\rho }_0}}\Vert }_2$，但独立于 $\underset{\_}{\rho }$。

再回过头来去看G的定义，注意到 ${\rho }_{\text{0}}{u}_t=G_y-\frac{1}{2}{\left({\left(\frac{n_y}{J}\right)}^2\right)}_y$，又有：

$\underset{0\le t\le T}{\sup }{\Vert u_y\Vert }_{L^2}^2=\underset{0\le t\le T}{\sup }{\Vert \frac{J}{\mu }\left(G+P\right)\Vert }_{L^2}^2\le C\left(\gamma ,\mu ,\bar{\rho },\underset{\_}{J},\bar{J},{\Vert G_0\Vert }_2,{\Vert P_0\Vert }_2\right)$ (3.1.28)

$\begin{array}{l}{\Vert \frac{\text{1}}{\sqrt{{\rho }_{\text{0}}}}{\left(\frac{n_y}{J}\right)}_y\frac{n_y}{J}\Vert }_{L^2}^2={\displaystyle {\int }_{{\Omega }_{\text{1}}}{\left(\frac{\text{1}}{\sqrt{{\rho }_{\text{0}}}}{\left(\frac{n_y}{J}\right)}_y\frac{n_y}{J}\right)}^2}\text{d}y\\ \le \frac{\text{1}}{{\rho }_0}{\Vert \frac{n_y}{J}\Vert }_{L^{\infty }}^2{\Vert {\left(\frac{n_y}{J}\right)}_y\Vert }_{L^2}=\frac{\text{1}}{{\rho }_0}{\Vert \frac{n_y}{J}\Vert }_{L^{\infty }}^2{\Vert {\left(\frac{n_y}{J}\right)}_y\frac{J}{J}\Vert }_{L^2}\\ \le \frac{\text{1}}{{\rho }_0}{\Vert \frac{n_y}{J}\Vert }_{L^{\infty }}^2{\Vert {\left(\frac{n_y}{J}\right)}_y\frac{\bar{J}}{J}\Vert }_{L^2}=\frac{\bar{J}}{{\rho }_0}{\Vert \frac{n_y}{J}\Vert }_{L^{\infty }}^2{\Vert {\left(\frac{n_y}{J}\right)}_y\frac{1}{J}\Vert }_{L^2}\end{array}$

又由 ${\tilde{n}}_x\in L_x^{\infty }\left(\Omega \right)$， ${\tilde{n}}_{xx}\in L_x^{\infty }\left(\Omega \right)$，J有上下界，可知 ${\Vert \frac{\text{1}}{\sqrt{{\rho }_{\text{0}}}}{\left(\frac{n_y}{J}\right)}_y\frac{n_y}{J}\Vert }_{L^2\left({\Omega }_1\right)}^2\le C$。

于是有

${\displaystyle {\int }_0^T{\Vert \sqrt{{\rho }_0}{u}_t\Vert }_{L^2}^2\text{d}t}\le {\displaystyle {\int }_0^T{\Vert \frac{G_y}{\sqrt{{\rho }_0}}\Vert }_{L^2}^2\text{d}t}+{\displaystyle {\int }_0^T{\Vert \frac{\text{1}}{\sqrt{{\rho }_{\text{0}}}}{\left(\frac{n_y}{J}\right)}_y\frac{n_y}{J}\Vert }_{L^2}^2\text{d}t}\le C\left(\gamma ,\mu ,\bar{\rho },\underset{\_}{J},\bar{J},{\Vert G_0\Vert }_{L^2}\right)$ (3.1.29)

于是有

$\begin{array}{c}\underset{0\le t\le T}{\sup }{\Vert \sqrt{{\rho }_{\text{0}}}u\Vert }_{L^2\left({\Omega }_{\text{1}}\right)}=\underset{0\le t\le T}{\sup }{\Vert \sqrt{{\rho }_{\text{0}}}{u}_0+{\displaystyle {\int }_0^t\sqrt{{\rho }_{\text{0}}}{u}_t\text{d}\tau }\Vert }_{L^2\left({\Omega }_{\text{1}}\right)}^2\\ \le {\Vert \sqrt{{\rho }_{\text{0}}}{u}_0\Vert }_{L^2\left({\Omega }_{\text{1}}\right)}^2+{\displaystyle {\int }_0^T{\Vert \sqrt{{\rho }_{\text{0}}}{u}_t\Vert }_{L^2\left({\Omega }_{\text{1}}\right)}\text{d}\tau }\\ \le {\Vert \sqrt{{\rho }_{\text{0}}}{u}_0\Vert }_{L^2\left({\Omega }_{\text{1}}\right)}^2+C\left(\gamma ,\mu ,\bar{\rho },\underset{\_}{J},\bar{J},{\Vert G_0\Vert }_{L^2}\right)\end{array}$

注意到：

$u_{yy}={\left[\frac{J}{\mu }\left(G+P\right)\right]}_y=\frac{J_y}{\mu }\left(G+P\right)+\frac{J}{\mu }\left(G_y+P_y\right)$

再由Hölder不等式得：

${\displaystyle {\int }_0^{T_{\text{0}}}{\Vert \frac{u_{yy}}{\sqrt{{\rho }_0}}\Vert }_{L^2}^2\text{d}t}\le C{\displaystyle {\int }_0^{T_0}\left({\Vert \left(\frac{G_y}{\sqrt{{\rho }_0}},\frac{P_y}{\sqrt{{\rho }_0}}\right)\Vert }_2^2+{\Vert \left(G,P\right)\Vert }_{\infty }^2{\Vert \frac{J_y}{\sqrt{{\rho }_0}}\Vert }_2^2\right)\text{d}t}\le C$ (3.1.30)

于是可得：

${\displaystyle {\int }_0^T{\Vert \frac{u_{yy}}{\sqrt{{\rho }_0}}\Vert }_{L^2}^2\text{d}t}\le {\displaystyle {\int }_0^T\left({\Vert \left(\frac{G_y}{\sqrt{{\rho }_0}},\frac{P_y}{\sqrt{{\rho }_0}}\right)\Vert }_{L^2}^2+{\Vert \left(G,P\right)\Vert }_{L^{\infty }}^2\right)\text{d}t}\le C$

其中C是正常数，且仅取决于 $\gamma ,\mu ,\bar{\rho },\underset{\_}{J},\bar{J},{\Vert \frac{{J^{\prime }}_0}{\sqrt{{\rho }_0}}\Vert }_2,{\Vert G_0\Vert }_2,{\Vert P_0\Vert }_2$ 和 ${\Vert \frac{{P^{\prime }}_0}{\sqrt{{\rho }_0}}\Vert }_2$，但独立于 $\underset{\_}{\rho }$。

证毕。

上文已经知道，接下来由 ${\tilde{n}}_x\in L^{\infty }\left(\Omega \right)$，J有上下界知 $\frac{n_y}{J}\in L^{\infty }\left({\Omega }_1\right)$，又得 $n_y\in L^{\infty }\left({\Omega }_1\right)$。

${\tilde{n}}_{xx}\in L^{\infty }\left(\Omega \right)$，J有上下界知 ${\left(\frac{n_y}{J}\right)}_y\frac{1}{J}\in L^{\infty }\left({\Omega }_1\right)$，又 ${\left(\frac{n_y}{J}\right)}_y\frac{1}{J}=\frac{n_{yy}}{J^2}-\frac{n_y{J}_y}{J^3}$

于是有

又由 ${\tilde{n}}_{xx}\in L^{\infty }\left(\Omega \right)$， ${\tilde{n}}_x\in L^{\infty }\left(\Omega \right)$， $J_y\in L^2\left({\Omega }_1\right)$ 得

$n_{yy}\in L^2(\Omega 1)$

至此定义3.1的所有要求均已满足，即 $\left(J,u,P,n\right)$ 是拉格朗日坐标下液晶方程组(**)的一个强解，也就是说原液晶方程组(*)在满足(2.1.5)式条件下，仍然存在强解，则 $T_{\ast }$ 不是原液晶方程组(*)强解存在的关于时间的最大值。这与 $T_{\ast }$ 的定义相矛盾，因此(2.1.5 )式是错误的，定理2.1.的证明也就完成了。

4. 结论

大初值有界区域一维完全可压缩液晶方程组的强解存在爆破准则。