一维空间中可压缩变指数增长非牛顿流体模型的强解

The Strong Solutions to the Compressible Non-Newtonian Fluids with Variable Exponential Growth in One Dimension

摘要: 本文在一维有界区间上研究了一类可压缩变指数增长的流体模型。此类模型是带有p(x)-Lapalace粘性项的可压缩非牛顿流体模型。我们通过构造逼近解，应用能量估计，克服变指数带来的强非线性性质，得到了非牛顿粘性参数1 < p(x) < 2，且初始密度存在真空的情况下，此类可压缩非牛顿流体模型初边值问题强解的存在唯一性。

Abstract: In this paper, a class of compressible non-Newtonian fluid with variable exponential is studied on one-dimensional bounded interval. This model is a compressible non-Newtonian fluid model with a p(x)-Laplace viscosity term. By constructing an approximate solution and applying energy estimation to overcome the nonlinear property of strong viscous term, we obtain the existence and uniqueness of the strong solution to electroviscous fluid under the condition of the non-Newtonian viscous parameter 1 < p(x) < 2 and vacuum at the initial density.

文章引用：苏敏, 佟丽宁. 一维空间中可压缩变指数增长非牛顿流体模型的强解[J]. 理论数学, 2020, 10(4): 362-373. https://doi.org/10.12677/PM.2020.104046

1. 引言

本文在一维有界区间上研究了如下可压缩变指数增长非牛顿流体模型的初边值问题，

$\{\begin{array}{l}{\rho }_t+{\left(\rho u\right)}_x=0,\\ {\left(\rho u\right)}_t+{\left(\rho u^2\right)}_x-{\mu }_1{u}_{xx}-{\mu }_0{\left[{\left(1+{\left|u_x\right|}^2\right)}^{\frac{p\left(x\right)-2}{2}}{u}_x\right]}_x+{\pi }_x=0,\\ \pi =a{\rho }^{\gamma }\text{,}a>0,\gamma >0\end{array}$ (1)

带有初边值条件：

$\{\begin{array}{l}{\left(\rho ,u\right)|}_{t=0}=\left({\rho }_0,u_0\right),x\in I\equiv \left[0,1\right],\\ {u|}_{x=0}={u|}_{x=1}=0,t\in \left[0,T\right],\end{array}$ (2)

其中 $\rho ,u,\pi$ 分别表示密度，速度和压力。 $1<p\left(x\right)<2$ 为已知函数， ${\mu }_0>0,{\mu }_1>0$ 为定常数。

此类变指数增长流体模型通常用来描述电粘性流体的流动。1949年Winslow [1] 发现电粘性流体的特性是能够根据提供的电场强度来改变流体的粘性，作为一种新功能材料，广泛运用于可变减震器、喷墨印刷机、消振装置、离合器、阀、密封等，特别倾向于振动控制方面的应用。

当p(x)是常数，模型为带有Carreau’s Law结构的可压缩非牛顿流模型，关于强解的存在唯一性有很多结果见 [2] [3] [4]。当p(x)为函数时，Rajagopal，Růzǐčka [5] 首先得到了不可压缩流体模型，该模型由质量、线动量、角动量、能量的一般平衡定律、克劳修斯–杜赫不等式形式的热力学第二定律和Minkowskian形式的麦克斯韦方程导出。Růzǐčka在 [6] 中对其强解进行了进一步的研究。Diening [7] 对于不可压电粘性流体的理论和数值结果做了研究。还有很多关于变指数增长抛物型方程的结果可以在 [8] [9] [10] 中找到。由于模型强的非线性性和奇异性，对于变指数增长模型的研究，特别是，可压缩变指数非牛顿流体的研究还比较单薄，需要我们进一步探索。

本文研究的主要结论如下：

定理1：假设 $p\left(x\right)$ 满足 $1<p_{\infty }\le p\left(x\right)\le p_0<2$ ( $p_{\infty },p_0$ 为常数)，且具有一阶连续导数，初值条件 $\left({\rho }_0,u_0\right)$ 满足

$\{\begin{array}{l}0\le {\rho }_0\in H^1\left(I\right),\\ u_0\in H_0^1\left(I\right)\cap H^2\left(I\right)\end{array}$ (3)

和相容性条件：

$-{\mu }_1{u}_{0xx}-{\mu }_0{\left[{\left(1+{\left|u_{0x}\right|}^2\right)}^{\frac{p\left(x\right)-2}{2}}{u}_{0x}\right]}_x+{\pi }_x\left({\rho }_0\right)={\rho }^{\frac{1}{2}}g,g\in L^2\left(I\right)$ (4)

则存在 $T_{*}\in \left(0,\infty \right)$，使得问题(1)~(2)存在唯一强解 $\left(\rho ,u\right)$ 满足

$\begin{array}{l}0\le \rho \in C\left(\left[0,T_{*}\right];H^1\left(I\right)\right);\sqrt{\rho }{u}_t\in L^{\infty }\left(0,T_{*};L^2\left(I\right)\right);\sqrt{\rho }{u}_x\in L^{\infty }\left(0,T_{*};L^2\left(I\right)\right);\\ u\in C\left(\left[0,T_{*}\right];H_0^1\left(I\right)\right)\cap L^2\left(0,T^{*};H^2\left(I\right)\right);u_t\in L^2\left(0,T_{*};H_0^1\left(I\right)\right);\\ {\left(1+{\left|u_x^k\right|}^2\right)}^{\frac{p\left(x\right)}{2}}\in L^{\infty }\left(\left[0,T_{*}\right];L^1\left(I\right)\right);\\ {\left({\left(1+{\left|u_x\right|}^2\right)}^{\frac{p\left(x\right)-2}{2}}{u}_x\right)}_x\in C\left(\left[0,T_{*}\right];L^2\left(I\right)\right).\end{array}$ (5)

本文共有三节，在第二节中，我们将通过迭代的方法构造逼近方程，得到逼近解的先验估计，得到初值远离真空时，模型强解的存在性。应用第二节得到的与密度下界无关的一致先验估计，定理一的证明在第三节完成。

2. 先验估计

在这一部分，我们将考虑初值远离真空的情形。假设初始密度具有正的下界，通过迭代的方法构造了初边值问题(1)~(2)的逼近问题，并得到逼近解与密度下界无关的一致能量估计。

2.1. 逼近解的构造

首先，我们假设 ${\rho }_0$ 足够光滑具有正的下界，即存在一个正数 $\delta \left(0<\delta \ll 1\right)$ 使得 ${\rho }_0\ge \delta$。接下来，我们运用文献 [11] 中迭代方法构造逼近方程并得到逼近解，

• 定义。

• 对 $k\ge 1$，给定，则根据一阶双曲守恒律(见 [12])和p(x)-Laplace抛物型方程(见 [8])的存在性理论，得到如下初边值问题具有唯一的强解 $\left({\rho }^k,u^k\right)$ ：

${\rho }_t^k+u^{k-1}{\rho }_x^k+u_x^{k-1}{\rho }^k=0,$ (6)

${\rho }^k{u}_t^k+{\rho }^k{u}^{k-1}{u}_x^k-{\mu }_1{u}_{xx}^k-{\mu }_0{\left[{\left(1+{\left|u_x^k\right|}^2\right)}^{\frac{p\left(x\right)-2}{2}}{u}_x^k\right]}_x+{\pi }_x^k=0$ (7)

满足如下初边值条件

${\left({\rho }^k,u^k\right)|}_{t=0}=\left({\rho }_0,u_0\right),x\in \left[0,1\right],{u^k|}_{x=0}={u^k|}_{x=1}=0,t\in \left[0,T\right].$ (8)

其中 $0<{\rho }_0\in H^2\left(I\right),u\in H_0^1\cap H^2$ 且满足相容性条件(4)。

2.2. 一致能量估计

本小节中，我们将对问题(6)~(8)的解 $\left({\rho }^k,u^k\right)$ 进行一致先验估计。

为了得到 $\left({\rho }^k,u^k\right)$ 的估计，我们定义函数 ${\Phi }_K\left(t\right)$ ：

$\begin{array}{l}{\Phi }_K\left(t\right)=\underset{1\le k\le K}{\max }\underset{0\le s\le t}{\sup }(1+{\Vert \sqrt{{\rho }^k}{u}_t^k\left(s\right)\Vert }_{L^2\left(I\right)}^2+{\Vert \sqrt{{\rho }^k}{u}_x^k\left(s\right)\Vert }_{L^2\left(I\right)}^2\begin{array}{c}\\ \\ \end{array}\\ +{\Vert u_x^k\left(s\right)\Vert }_{L^2\left(I\right)}^2+{\Vert {\left(1+{\left|u_x^k\right|}^2\right)}^{\frac{p\left(x\right)}{4}}\Vert }_{L^2\left(I\right)}^2+{\Vert {\rho }^k\left(s\right)\Vert }_{H^1\left(I\right)}^2),\end{array}$ (9)

其中 $K\ge 1$ 是一个足够大的正整数。为了表述方便，定义下文中C为与 ${\mu }_1,{\mu }_0,p_{\infty },p_0$ 以及 ${\left|{\rho }_0\right|}_{H^1\left(I\right)},{\left|g\right|}_{L^2}$ 有关常数。

下面，我们将估计中的每一项。

引理1 假设 $\left({\rho }^k,u^k\right)$ 是方程(6)~(7)的光滑解，存在 $T\in \left(0,\infty \right)$ 使得

${\displaystyle {\int }_0^t{\Vert \sqrt{{\rho }^k}{u}_t^k\Vert }_{L^2\left(I\right)}^2\text{d}s+\underset{0\le s\le t}{\sup }\left({\Vert u_x^k\Vert }_{L^2\left(I\right)}^2+{\displaystyle {\int }_0^1{\left(1+{\left|u_x^k\right|}^2\right)}^{\frac{p\left(x\right)}{2}}\text{d}x}\right)}\le C\left(1+{\displaystyle {\int }_0^t{\Phi }_K^{\gamma +\frac{3}{2}}}\left(t\right)\text{d}s\right),$

对任意 $t\in \left[0,T\right]$ 成立。

证明：用 $u_t^k$ 乘以(7)式两端，然后关于变量x在[0,1]上积分有

$\begin{array}{l}{\displaystyle {\int }_0^1{\rho }^k{\left|u_t^k\right|}^2\text{d}x}+\frac{{\mu }_1}{2}\frac{\text{d}}{\text{d}t}{\displaystyle {\int }_0^1{\left|u_x^k\left(t\right)\right|}^2\text{d}x}+{\mu }_0\frac{\text{d}}{\text{d}t}{\displaystyle {\int }_0^t\frac{\text{1}}{p\left(x\right)}}\left[{\left(1+{\left|u_x^k\right|}^2\right)}^{\frac{p\left(x\right)}{2}}\right]\text{d}x\\ =\frac{\text{d}}{\text{d}t}{\displaystyle {\int }_0^1{\pi }^k{u}_x^k}\text{d}x+{\displaystyle {\int }_0^t{\displaystyle {\int }_0^1\left(-{\rho }^k{u}^{k-1}{u}_x^k{u}_t^k-{\pi }_t^k{u}_x^k\right)\text{d}x}}.\end{array}$ (10)

对于(10)关于变量t在 $\left[0,t\right]$ 上积分，我们可以得到，

$\begin{array}{l}{\displaystyle {\int }_0^t{\Vert \sqrt{{\rho }^k}{u}_t^k\Vert }_{L^2\left(I\right)}^2\text{d}s}+\frac{{\mu }_1}{2}{\displaystyle {\int }_0^1{\left|u_x^k\right|}^2}\text{d}x+\frac{{\mu }_0}{p_{\infty }}{\displaystyle {\int }_0^t{\left(1+{\left|u_x^k\right|}^2\right)}^{\frac{p\left(x\right)}{2}}\left(t\right)\text{d}x}\\ \le \frac{{\mu }_1}{2}{\displaystyle {\int }_0^1{\left|u_x^k\right|}^2\left(0\right)\text{d}x}+\frac{{\mu }_0}{p_{\infty }}{\displaystyle {\int }_0^t{\left(1+{\left|u_x^k\right|}^2\right)}^{\frac{p\left(x\right)}{2}}\left(0\right)\text{d}x}+{\displaystyle {\int }_0^1{\pi }^k{u}_x^k\left(t\right)\text{d}x}\\ -{\displaystyle {\int }_0^1{\pi }^k{u}_x^k\left(0\right)\text{d}x}+{\displaystyle {\int }_0^t{\displaystyle {\int }_0^1\left(-{\rho }^k{u}^{k-1}{u}_x^k{u}_t^k-{\pi }_t^k{u}_x^k\right)\text{d}x\text{d}s}}\\ \le C+{\displaystyle {\int }_0^1\left|{\pi }^k{u}_x^k\left(t\right)\right|\text{d}x}+{\displaystyle {\int }_0^t{\displaystyle {\int }_0^1\left|-{\rho }^k{u}^{k-1}{u}_x^k{u}_t^k-{\pi }_t^k{u}_x^k\right|\text{d}x\text{d}s}}\end{array}$ (11)

应使用Sobolev不等式和Young’s不等式，有

$\begin{array}{l}{\displaystyle {\int }_{\text{0}}^t{\Vert \sqrt{{\rho }^k}{u}_t^k\Vert }_{L^2\left(I\right)}^{\text{2}}\text{d}s+\frac{{\mu }_1}{4}{\Vert u_x^k\Vert }_{L^2\left(I\right)}^2+\frac{{\mu }_0}{p_{\infty }}{\displaystyle {\int }_0^t{\left(1+{\left|u_x^k\right|}^2\right)}^{\frac{p\left(x\right)}{2}}}\text{d}x}\\ \le C+C{\Vert {\pi }^k\left(t\right)\Vert }_{L^2\left(I\right)}^2+C{\displaystyle {\int }_0^t{\Vert u^{k-1}\Vert }_{L^{\infty }\left(I\right)}{\Vert \sqrt{{\rho }^k}{u}_t^k\Vert }_{L^2\left(I\right)}{\Vert \sqrt{{\rho }^k}{u}_x^k\Vert }_{L^2\left(I\right)}\text{d}s}+{\displaystyle {\int }_0^{\text{t}}{\displaystyle {\int }_0^1\left|{\pi }_t^k{u}_x^k\right|\text{d}x\text{d}s}}\\ ={\displaystyle \underset{i=1}{\overset{3}{\sum }}{K}_i}.\end{array}$ (12)

接下来依次估计 $K_i,i=1,2,3$。应用Sobolev不等式和Young’s不等式，我们可以计算

$\begin{array}{c}{K}_1={\displaystyle {\int }_0^1{\left|{\pi }^k\left(t\right)\right|}^2\text{d}x}={\displaystyle {\int }_0^1{\left|{\pi }^k\left(0\right)\right|}^2}\text{d}x+{\displaystyle {\int }_0^t\left(\frac{\partial }{\partial s}{\displaystyle {\int }_0^t\left|{\pi }^k\left(s\right)\right|\text{d}x}\right)\text{d}s}\\ ={\displaystyle {\int }_0^1{\left|{\pi }^k\left(0\right)\right|}^2}\text{d}x+{\displaystyle {\int }_0^t{\displaystyle {\int }_0^1{\pi }^k\left(\rho \right){\left({\pi }^k\left(\rho \right)\right)}^{\prime }\left(-{\rho }_x^k{u}^{k-1}-\rho {}^{k}u{}_x^{k-1}\right)\text{d}x}\text{d}s}\\ \le C+C{\displaystyle {\int }_0^t{\Vert {\pi }^k\left(\rho \right)\Vert }_{L^{\infty }\left(I\right)}{\Vert {\rho }^k\Vert }_{L^{\infty }\left(I\right)}^{\gamma -1}{\Vert {\rho }^k\left(s\right)\Vert }_{H^1\left(I\right)}{\Vert u_x^{k-1}\Vert }_{L^2\left(I\right)}\text{d}s}\\ \le C\left(1+{\displaystyle {\int }_0^t{\Phi }_K^{\gamma +\frac{1}{2}}\left(t\right)\text{d}s}\right);\end{array}$

$K_2=C{\displaystyle {\int }_0^t{\Vert u^{k-1}\Vert }_{L^{\infty }\left(I\right)}{\Vert \sqrt{{\rho }^k}{u}_t^k\Vert }_{L^2\left(I\right)}{\Vert \sqrt{{\rho }^k}{u}_x^k\Vert }_{L^2\left(I\right)}\text{d}s}\le C{\displaystyle {\int }_0^t{\Phi }_K^{\frac{3}{2}}\left(t\right)\text{d}s}$

$\begin{array}{c}{K}_3={\displaystyle {\int }_0^t{\displaystyle {\int }_0^1\left|{\pi }_t^k{u}_x^k\right|\text{d}x\text{d}s}}\le {\displaystyle {\int }_0^t{\displaystyle {\int }_0^1\left|{\left({\rho }^k\right)}^{\gamma -1}{\rho }^k{u}_x^{k-1}{u}_x^k+{\left({\rho }^k\right)}^{\gamma -1}{\rho }_x^k{u}^{k-1}{u}_x^k\right|\text{d}x\text{d}s}}\\ \le {\displaystyle {\int }_0^t({\Vert {\rho }^k\Vert }_{L^{\infty }\left(I\right)}^{\gamma -1}{\Vert {\rho }^k\Vert }_{L^{\infty }\left(I\right)}^{\frac{1}{2}}{\Vert \sqrt{{\rho }^k}{u}_x^k\Vert }_{L^2\left(I\right)}{\Vert u_x^{k-1}\Vert }_{L^2\left(I\right)}}\\ +{\Vert {\rho }^k\Vert }_{L^{\infty }\left(I\right)}^{\gamma -\frac{\text{3}}{2}}{\Vert {\rho }_x^k\Vert }_{L^2\left(I\right)}{\Vert \sqrt{{\rho }^k}{u}_x^k\Vert }_{L^2\left(I\right)}{\Vert u^{k-1}\Vert }_{L^{\infty }\left(I\right)})\text{d}s\\ \le C{\displaystyle {\int }_0^t{\Phi }^{\gamma +\frac{3}{2}}}\text{d}s.\end{array}$

将的估计代入(12)，有

${\displaystyle {\int }_{\text{0}}^t{\Vert \sqrt{{\rho }^k}{u}_t^k\Vert }_{L^2\left(I\right)}^2\text{d}s}+\frac{{\mu }_1}{\text{4}}{\Vert u_x^k\Vert }_{L^2\left(I\right)}^2+\frac{{\mu }_0}{p_{\infty }}{\displaystyle {\int }_0^t{\left(1+{\left|u_x^k\right|}^2\right)}^{\frac{p\left(x\right)}{2}}}\text{d}x\le C\left(1+{\displaystyle {\int }_0^t{\Phi }_K^{\gamma +\frac{3}{2}}\left(t\right)\text{d}s}\right).$ (13)

引理1的证明完成。

引理2：若是(6)~(7)的解，那么存在 $T\in \left(0,\infty \right)$，使得

$\begin{array}{l}\underset{1\le k\le K}{\max }(\underset{0\le s\le t}{\sup }{\Vert \sqrt{{\rho }^k}{u}_t^k\left(t\right)\Vert }_{L^2\left(I\right)}^2+{\Vert \sqrt{{\rho }^k}{u}_x^k\left(t\right)\Vert }_{L^2\left(I\right)}^2+{\displaystyle {\int }_0^t{\Vert u_{xt}^k\Vert }_{L^2\left(I\right)}^2\text{d}s}+{\displaystyle {\int }_0^t{\Vert u_{xx}^k\Vert }_{L^2\left(I\right)}^2\text{d}s}\\ +{{\displaystyle {\int }_0^t{\displaystyle {\int }_0^1\left(1+{\left|u_x^k\right|}^2\right)}}}^{\frac{p\left(x\right)-2}{2}}{\left|u_{xx}^k\right|}^2\text{d}x\text{d}s+{\displaystyle {\int }_0^t{\displaystyle {\int }_0^1\left({\left(1+{\left|u_x^k\right|}^{\text{2}}\right)}^{\frac{p\left(x\right)-2}{2}}\right){\left|u_{xt}^k\right|}^2}}\text{d}x\text{d}s)\\ \le C{\displaystyle {\int }_0^t{\Phi }_K^{\gamma +\frac{7}{2}}\left(t\right)\text{d}s}\end{array}$

证明：对(7)关于t求导有

${\left({\rho }^k{u}_t^k\right)}_t+{\left({\rho }^k{u}^{k-1}{u}_x^k\right)}_t-{\mu }_1{\left(u_{xx}^k\right)}_t-{\mu }_0{\left[{\left(1+{\left|u_x^k\right|}^2\right)}^{\frac{p\left(x\right)-2}{2}}{u}_x^k\right]}_{xt}+{\pi }_{xt}^k=0$ (14)

将方程(14)两端同乘以 $u_t^k$ 然后在(0,1)上关于x积分，可以得到

$\begin{array}{l}\frac{1}{2}\frac{\text{d}}{\text{d}t}{\displaystyle {\int }_0^1{\rho }^k{\left(u_t^k\right)}^2}\text{d}x+{\mu }_1{\displaystyle {\int }_0^1{\left(u_{xt}^k\right)}^2}\text{d}x+{\mu }_0{\displaystyle {\int }_0^1{\left[{\left(1+{\left|u_x^k\right|}^2\right)}^{\frac{p\left(x\right)-2}{2}}{u}_x^k\right]}_t{u}_{xt}^k}\text{d}x\\ ={\displaystyle {\int }_0^1{\pi }_t^k{u}_{xt}^k}\text{d}x+{\displaystyle {\int }_0^1{\left({\rho }^k{u}^{k-1}\right)}_x\left(\frac{1}{2}{\left(u_t^k\right)}^2+u^{k-1}{u}_x^k{u}_t^k\right)\text{d}x}\\ -{\displaystyle {\int }_0^1{\rho }^k}{u}_t^{k-1}{u}_x^k{u}_t^k\text{d}x-{\displaystyle {\int }_0^1\left({\rho }^k{u}^{k-1}\right)u_{xt}^k{u}_t^k}\text{d}x\end{array}$ (15)

因为 $1<p_{\infty }\le p\left(x\right)\le p_0<2$，我们可以计算

(16)

另一方面，应用质量守恒方程(6)有

${\pi }_t^k=-\gamma {\pi }^k{u}_x^{k-1}-{\pi }_x^k{u}^{k-1}$ (17)

由(15)、(16)和(17)，得到如下不等式，

$\begin{array}{l}\frac{1}{2}\frac{\text{d}}{\text{d}t}{\Vert \sqrt{{\rho }^k}{u}_t^k\Vert }_2^2+{\mu }_1{\displaystyle {\int }_0^1{\left|u_{xt}^k\right|}^2\text{d}x}+{\mu }_0{\displaystyle {\int }_0^1\left(p\left(x\right)-1\right){\left(1+{\left|u_x^k\right|}^2\right)}^{\frac{p\left(x\right)-2}{2}}}{\left(u_{xt}^k\right)}^2\text{d}x\\ \le {\displaystyle {\int }_0^1\left|{\pi }_x^k\right|\left|u^{k-1}\right|\left|u_{xt}^k\right|\text{d}x}+\gamma {\displaystyle {\int }_0^1\left|{\pi }^k\right|\left|u_x^{k-1}\right|\left|u_{xt}^k\right|\text{d}x}+2{\displaystyle {\int }_0^1{\rho }^k\left|u^{k-1}\right|\left|u_t^k\right|\left|u_{xt}^k\right|\text{d}x}\\ +{\displaystyle {\int }_0^1{\rho }^k\left|u^{k-1}\right|\left|u_x^{k-1}\right|\left|u_x^k\right|\left|u_t^k\right|\text{d}x}+{\displaystyle {\int }_0^1{\rho }^k{\left|u^{k-1}\right|}^2\left|u_{xx}^k\right|\left|u_t^k\right|\text{d}x}\\ +{\displaystyle {\int }_0^1{\rho }^k{\left|u^{k-1}\right|}^2\left|u_x^k\right|\left|u_{xt}^k\right|\text{d}x}+{\displaystyle {\int }_{\text{0}}^{\text{1}}{\rho }^k\left|u_t^{k-1}\right|\left|u_x^k\right|\left|u_t^k\right|\text{d}x}\\ ={\displaystyle \underset{j=1}{\overset{7}{\sum }}{I}_j.}\end{array}$ (18)

应用Sobolev不等式，Hȍlder不等式和Young’s不等式，我们可以估计 $I_j,j=1,\cdots ,7$。

$I_1={\displaystyle {\int }_0^1\left|{\pi }_x^k\right|\left|u^{k-1}\right|\left|u_{xt}^k\right|\text{d}x}\le {\Vert {\pi }_x^k\Vert }_{L^2\left(I\right)}{\Vert u^{k-1}\Vert }_{L^{\infty }\left(I\right)}{\Vert u_{xt}\Vert }_{L^2\left(I\right)}\le C{\Phi }_K^{\gamma +1}\left(t\right)+\frac{{\mu }_1}{6}{\Vert u_{xt}^k\Vert }_{L^2\left(I\right)}^2;$

$I_2={\displaystyle {\int }_0^1\gamma \left|{\pi }^k\right|\left|u_x^{k-1}\right|\left|u_{xt}^k\right|\text{d}x}\le C{\Vert {\pi }^k\Vert }_{L^{\infty }\left(I\right)}{\Vert u_x^{k-1}\Vert }_{L^2\left(I\right)}{\Vert u_{xt}^k\Vert }_{L^2\left(I\right)}\le C{\Phi }_K^{\gamma +1}\left(t\right)+\frac{{\mu }_1}{6}{\Vert u_{xt}^k\Vert }_{L^2\left(I\right)}^2;$

$\begin{array}{c}{I}_3=2{\displaystyle {\int }_0^1{\rho }^k\left|u^{k-1}\right|\left|u_t^k\right|\left|u_{xt}^k\right|\text{d}x}\le 2{\Vert {\rho }^k\Vert }_{L^{\infty }\left(I\right)}^{\frac{1}{2}}{\Vert u^{k-1}\Vert }_{L^{\infty }\left(I\right)}{\Vert \sqrt{{\rho }^k}{u}_t^k\Vert }_{L^2\left(I\right)}{\Vert u_{xt}^k\Vert }_{L^2\left(I\right)}\\ \le C{\Phi }_K^{\frac{5}{2}}\left(t\right)+\frac{{\mu }_1}{6}{\Vert u_{xt}^k\Vert }_{L^2\left(I\right)}^2;\end{array}$

$\begin{array}{c}{I}_4={\displaystyle {\int }_0^1{\rho }^k\left|u^{k-1}\right|\left|u_x^{k-1}\right|\left|u_x^k\right|\left|u_t^k\right|\text{d}x}\le {\Vert u^{k-1}\Vert }_{L^{\infty }\left(I\right)}{\Vert u_x^{k-1}\Vert }_{L^{\infty }\left(I\right)}{\Vert \sqrt{{\rho }^k}{u}_x^k\Vert }_{L^2\left(I\right)}{\Vert \sqrt{{\rho }^k}{u}_t^k\Vert }_{L^2\left(I\right)}\\ \le C{\Phi }_K^3\left(t\right)+{\eta }_1{\Vert u_{xx}^{k-1}\Vert }_{L^2\left(I\right)}^2\end{array}$

$\begin{array}{c}{I}_5={\displaystyle {\int }_0^1{\rho }^k{\left|u^{k-1}\right|}^2\left|u_{xx}^k\right|\left|u_t^k\right|\text{d}x}\le {\Vert {\rho }^k\Vert }_{L^{\infty }\left(I\right)}^{\frac{1}{2}}{\Vert u^{k-1}\Vert }_{L^{\infty }\left(I\right)}^2{\Vert \sqrt{{\rho }^k}{u}_t^k\Vert }_{L^2\left(I\right)}{\Vert u_{xx}^k\Vert }_{L^2\left(I\right)}\\ \le C{\Phi }_K^{\frac{7}{2}}\left(x\right)+\frac{{\mu }_1}{5}{\Vert u_{xx}^k\Vert }_{L^2}^2;\end{array}$

$\begin{array}{c}{I}_6={\displaystyle {\int }_0^1\left|{\rho }^k\right|{\left|u^{k-1}\right|}^2\left|u_x^k\right|\left|u_{xt}^k\right|\text{d}x}\le {\Vert {\rho }^k\Vert }_{L^{\infty }\left(I\right)}^{\frac{1}{2}}{\Vert u^{k-1}\Vert }_{L^{\infty }\left(I\right)}^2{\Vert \sqrt{{\rho }^k}{u}_x^k\Vert }_{L^2\left(I\right)}{\Vert u_{xt}^k\Vert }_{L^2\left(I\right)}\\ \le C{\Phi }_K^{\frac{7}{2}}\left(t\right)+\frac{{\mu }_1}{6}{\Vert u_{xt}^k\Vert }_{L^2\left(I\right)}^2;\end{array}$

$\begin{array}{c}{I}_7={\displaystyle {\int }_0^1\left|{\rho }^k\right|\left|u_t^{k-1}\right|\left|u_x^k\right|\left|u_t^k\right|\text{d}x}\le {\Vert u_t^{k-1}\Vert }_{L^{\infty }\left(I\right)}{\Vert \sqrt{{\rho }^k}{u}_x^k\Vert }_{L^2\left(I\right)}{\Vert \sqrt{{\rho }^k}{u}_t^k\Vert }_{L^2\left(I\right)}\\ \le C{\Phi }_K^2\left(t\right)+{\eta }_2{\Vert u_{xt}^{k-1}\Vert }_{L^2\left(I\right)}^2.\end{array}$

其中 ${\eta }_1,{\eta }_2$ 为与有关的足够小的常数。将上述估计代入(18)中有：

$\begin{array}{l}\frac{1}{2}\frac{\text{d}}{\text{d}t}{\Vert \sqrt{{\rho }^k}{u}_t^k\Vert }_2^2+\frac{{\mu }_1}{3}{\displaystyle {\int }_0^1{\left|u_{xt}^k\right|}^2}\text{d}x+{\mu }_0{\displaystyle {\int }_0^1\left(p\left(x\right)-1\right){\left(1+{\left|u_x^k\right|}^2\right)}^{\frac{p\left(x\right)-2}{2}}}{\left(u_{xt}^k\right)}^2\text{d}x\\ \le C{\Phi }_K^{\gamma +\frac{7}{2}}\left(t\right)+\frac{{\mu }_1}{5}{\Vert u_{xx}^k\Vert }_{L^2\left(I\right)}^2+{\eta }_1{\Vert u_{xx}^{k-1}\Vert }_{L^2\left(I\right)}^2+{\eta }_2{\Vert u_{xt}^{k-1}\Vert }_{L^2\left(I\right)}^2.\end{array}$ (19)

接下来，我们估计 ${\Vert \sqrt{{\rho }^k}{u}_x^k\left(t\right)\Vert }_{L^2\left(I\right)}$。将方程(7)两端同乘以 $-u_{xx}^k$，然后在 $\left(0,\text{1}\right)$ 上关于x积分，由方程

(6)和分部积分，我们得到

$\begin{array}{l}\frac{1}{2}\frac{\text{d}}{\text{d}t}{\displaystyle {\int }_0^1{\rho }^k{\left(u_x^k\right)}^2}\text{d}x+{\mu }_1{\displaystyle {\int }_0^1{\left(u_{xx}^k\right)}^2}\text{d}x+{\mu }_0{\displaystyle {\int }_0^1{\left[{\left(1+{\left|u_x^k\right|}^2\right)}^{\frac{p\left(x\right)-2}{2}}{u}_x^k\right]}_x{u}_{xx}^k}\text{d}x\\ ={\displaystyle {\int }_0^1{\pi }_x^k{u}_{xx}^k}\text{d}x+{\displaystyle {\int }_0^1{\rho }_t^k{\left(u_x^k\right)}^2}\text{d}x-{\displaystyle {\int }_0^1{\rho }_x^k{u}_t^k{u}_x^k}\text{d}x.\end{array}$ (20)

我们首先计算方程(20)左端的第三项，

$\begin{array}{l}{\left[{\left(1+{\left|u_x^k\right|}^2\right)}^{\frac{p\left(x\right)-2}{2}}{u}_x^k\right]}_x{u}_{xx}^k={\left(e^{\frac{p\left(x\right)-2}{2}\ln \left(1+{\left|u_x^k\right|}^2\right)}{u}_x^k\right)}_x{u}_{xx}^k\\ ={\left(1+{\left|u_x^k\right|}^2\right)}^{\frac{p\left(x\right)-2}{2}}{\left(u_{xx}^k\right)}^2+{\left(1+{\left|u_x^k\right|}^2\right)}^{\frac{p\left(x\right)-2}{2}}\left(\frac{p_x\left(x\right)}{2}\ln \left(1+{\left|u_x^k\right|}^2\right)+\frac{p\left(x\right)-2}{2}\frac{2\left|u_x^k\right|u_{xx}^k}{1+{\left|u_x^k\right|}^2}\right)u_x^k{u}_{xx}^k\\ =\frac{p_x\left(x\right)}{2}{\left(1+{\left|u_x^k\right|}^2\right)}^{\frac{p\left(x\right)-2}{2}}{u}_x^k{u}_{xx}^k\ln \left(1+{\left|u_x^k\right|}^2\right)+{\left(1+{\left|u_x^k\right|}^2\right)}^{\frac{p\left(x\right)-2}{2}}{\left(u_{xx}^k\right)}^2\left[\frac{\left(p\left(x\right)-2\right){\left(u_x^k\right)}^2}{1+{\left|u_x^k\right|}^2}+1\right]\\ \ge \frac{p_x\left(x\right)}{2}{\left(1+{\left|u_x^k\right|}^2\right)}^{\frac{p\left(x\right)-2}{2}}{u}_x^k{u}_{xx}^k\ln \left(1+{\left|u_x^k\right|}^2\right)+\left(p\left(x\right)-1\right){\left(1+{\left|u_x^k\right|}^2\right)}^{\frac{p\left(x\right)-2}{2}}{\left(u_{xx}^k\right)}^2.\end{array}$

将上式代入(20)，有

(21)

为了估计(21)，我们需要仔细计算上式右端第四项。应用Hȍlder不等式和Young’s不等式，我们可以得到

(22)

其中 ${\Vert p_x\left(x\right)\Vert }_{\infty }\le C$ 和 $1<p_{\infty }\le p\left(x\right)\le p_0<2$。把(22)代入(21)，应用方程(6)有

$\begin{array}{l}\frac{1}{2}\frac{\text{d}}{\text{d}t}{\Vert \sqrt{{\rho }^k}{u}_x^k\Vert }_{L^2\left(I\right)}^2+{\mu }_1{\displaystyle {\int }_0^1{\left(u_{xx}^k\right)}^2\text{d}x}+\frac{{\mu }_0}{2}{\displaystyle {\int }_0^1\left(p\left(x\right)-1\right){\left(1+{\left|u_x^k\right|}^2\right)}^{\frac{p\left(x\right)-2}{2}}}{\left(u_{xx}^k\right)}^2\text{d}x\\ \le {\displaystyle {\int }_0^1\left|{\pi }_x^k\right|\left|u_x^k\right|\text{d}x}+{\displaystyle {\int }_0^1\left|{\rho }_x^k\right|\left|u^k\right|{\left|u_x^k\right|}^2\text{d}x}+{\displaystyle {\int }_0^1\left|{\rho }^k\right|{\left|u_x^k\right|}^3\text{d}x}+{\displaystyle {\int }_0^1\left|{\rho }^k\right|\left|u_{xx}^k\right|\left|u_t^k\right|\text{d}x}\\ +{\displaystyle {\int }_0^1\left|{\rho }^k\right|\left|u_x^k\right|\left|u_{xt}^k\right|\text{d}x}+{\displaystyle {\int }_0^1{\left(u_x^k\right)}^2\text{d}x}+C{\displaystyle {\int }_0^1\left|{\left(1+{\left|u_x^k\right|}^2\right)}^{\frac{p\left(x\right)}{2}}\right|\left|{\ln }^2\left(1+{\left|u_x^k\right|}^2\right)\right|\text{d}x}\\ ={\displaystyle \underset{i=1}{\overset{7}{\sum }}{J}_i.}\end{array}$ (23)

应用Sobolev不等式和Young’s不等式，有

$J_1={\displaystyle {\int }_0^1\left|{\pi }_x^k\right|\left|u_x^k\right|\text{d}x}\le {\Vert {\rho }_x^k\Vert }_{L^2\left(I\right)}^{\gamma }{\Vert u_x^k\Vert }_{L^2\left(I\right)}\le {\Phi }_K^{\frac{\gamma }{2}+\frac{1}{2}}\left(t\right);$

$J_2={\displaystyle {\int }_0^1\left|{\rho }_x^k\right|\left|u^k\right|{\left|u_x^k\right|}^2\text{d}x}\le {\Vert u^k\Vert }_{L^{\infty }\left(I\right)}{\Vert {\rho }_x^k\Vert }_{L^2\left(I\right)}{\Vert u_x^k\Vert }_{L^2\left(I\right)}{\Vert u_x^k\Vert }_{L^{\infty }\left(I\right)}\le C{\Phi }_K^3\left(t\right)+\frac{{\mu }_1}{5}{\Vert u_{xx}^k\Vert }_{L^2\left(I\right)}^2;$

$J_3={\displaystyle {\int }_0^1\left|{\rho }^k\right|{\left|u_x^k\right|}^3\text{d}x}\le {\Vert {\rho }^k\Vert }_{L^{\infty }\left(I\right)}^{\frac{1}{2}}{\Vert \sqrt{{\rho }^k}{u}_x^k\Vert }_{L^2\left(I\right)}{\Vert u_x^k\Vert }_{L^2\left(I\right)}{\Vert u_x^k\Vert }_{L^{\infty }\left(I\right)}\le C{\Phi }_K^{\frac{5}{2}}\left(t\right)+\frac{{\mu }_1}{5}{\Vert u_{xx}^k\Vert }_{L^2\left(I\right)}^2;$

$J_4={\displaystyle {\int }_0^1\left|{\rho }^k\right|\left|u_{xx}^k\right|\left|u_t^k\right|\text{d}x}\le {\Vert {\rho }^k\Vert }_{L^{\infty }\left(I\right)}^{\frac{1}{2}}{\Vert \sqrt{{\rho }^k}{u}_t^k\Vert }_{L^2\left(I\right)}{\Vert u_{xx}^k\Vert }_{L^2\left(I\right)}\le C{\Phi }_K^{\frac{3}{2}}\left(t\right)+\frac{{\mu }_1}{5}{\Vert u_{xx}^k\Vert }_{L^2\left(I\right)}^2$

$J_5={\displaystyle {\int }_0^1\left|{\rho }^k\right|\left|u_x^k\right|\left|u_{xt}^k\right|\text{d}x}\le {\Vert {\rho }^k\Vert }_{L^{\infty }\left(I\right)}^{\frac{1}{2}}{\Vert \sqrt{{\rho }^k}{u}_x^k\Vert }_{L^2\left(I\right)}{\Vert u_{xt}^k\Vert }_{L^2\left(I\right)}\le {\Phi }_K^{\frac{3}{2}}\left(t\right)+\frac{{\mu }_1}{6}{\Vert u_{xt}^k\Vert }_{L^2\left(I\right)}^2$

$J_6={\displaystyle {\int }_0^1{\left(u_x^k\right)}^2\text{d}x}\le {\Phi }_K(t)$

对于 $1<p_{\infty }\le p\left(x\right)\le p_0<2$ 和 $2<s<3$，在Sobolev不等式的帮助下，我们可以推出

$\begin{array}{c}{J}_7=C{\displaystyle {\int }_0^1\left|{\left(1+{\left|u_x^k\right|}^2\right)}^{\frac{p\left(x\right)}{2}}\right|\left|{\ln }^2\left(1+{\left|u_x^k\right|}^2\right)\right|\text{d}x}\\ \le C{\left[{\displaystyle {\int }_0^1{\left({\left|{\left(1+{\left|u_x^k\right|}^2\right)}^{\frac{p\left(x\right)}{2}}\right|}^2\right)}^{\frac{s}{2}}\text{d}x}\right]}^{\frac{2}{s}}{\left[{\displaystyle {\int }_0^1{\left|{\ln }^2\left(1+{\left|u_x^k\right|}^2\right)\right|}^{\frac{s}{s-2}}\text{d}x}\right]}^{\frac{s-2}{s}}\\ \le C{\Vert {\left(1+{\left|u_x^k\right|}^2\right)}^{\frac{p\left(x\right)}{4}}\Vert }_{L^s}^2{\left[{\displaystyle {\int }_0^1{\left|\ln \left(1+{\left|u_x^k\right|}^2\right)\right|}^{\frac{2s}{s-2}}\text{d}x}\right]}^{\frac{s-2}{s}}\end{array}$

$\begin{array}{l}\le C{\Vert {\left(1+{\left|u_x^k\right|}^2\right)}^{\frac{p\left(x\right)}{4}}\Vert }_{L^s}^2{\left[{\displaystyle {\int }_0^1{\left(\frac{8}{\left(s-2\right)p\left(x\right)}{\left|\ln \left(1+{\left|u_x^k\right|}^2\right)\right|}^{\frac{\left(s-2\right)p\left(x\right)}{8}}\right)}^{\frac{2s}{s-2}}\text{d}x}\right]}^{\frac{s-2}{s}}\\ \le C{\Vert {\left(1+{\left|u_x^k\right|}^2\right)}^{\frac{p\left(x\right)}{4}}\Vert }_{L^s}^2{\left[{\displaystyle {\int }_0^1{\left(\frac{8}{\left(s-2\right)p\left(x\right)}{\left|\left(1+{\left|u_x^k\right|}^2\right)\right|}^{\frac{\left(s-2\right)p\left(x\right)}{8}}\right)}^{\frac{2s}{s-2}}\text{d}x}\right]}^{\frac{s-2}{s}}\\ \le C{\Vert {\left(1+{\left|u_x^k\right|}^2\right)}^{\frac{p\left(x\right)}{4}}\Vert }_{L^s}^2{\left|{\left(1+{\left|u_x^k\right|}^2\right)}^{\frac{p\left(x\right)}{4}}\right|}_{L^s}^{s-2}=C{\left|{\left(1+{\left|u_x^k\right|}^2\right)}^{\frac{p\left(x\right)}{4}}\right|}_{L^s}^s\end{array}$

应用Sobolev嵌入定理和 $2<s<3$，下式成立，

(24)

类似(22)中的估计，我们可以得到

${\Vert {\left({\left(1+{\left|u_x^k\right|}^2\right)}^{\frac{p\left(x\right)}{4}}\right)}_x\Vert }_{L^2}^2\le C\left({\displaystyle {\int }_0^1{\left(1+{\left|u_x^k\right|}^2\right)}^{\frac{p\left(x\right)}{2}}{\ln }^2\left(1+{\left|u_x^k\right|}^2\right)\text{d}x}+{\displaystyle {\int }_0^1\left|{\left(1+{\left|u_x^k\right|}^2\right)}^{\frac{p\left(x\right)-2}{2}}\right|}{\left(u_{xx}^k\right)}^2\text{d}x\right).$ (25)

因此，联立(24)和(25)，因为 $1<p\left(x\right)<2$， $2<s<3$，选取足够小的 $\epsilon$，我们可以得到 $J_7$ 的估计，

$J_7\le C{\Vert {\left(1+{\left|u_x^k\right|}^2\right)}^{\frac{p\left(x\right)}{4}}\Vert }_{L^2}^4+\epsilon C{\displaystyle {\int }_0^1\left|{\left(1+{\left|u_x^k\right|}^2\right)}^{\frac{p\left(x\right)-2}{2}}\right|{\left(u_{xx}^k\right)}^2}\text{d}x\le C{\Phi }_K^2\left(t\right)+\frac{{\mu }_0\left(p_{\infty }-1\right)}{2}{\displaystyle {\int }_0^1\left|{\left(1+{\left|u_x^k\right|}^2\right)}^{\frac{p\left(x\right)-2}{2}}\right|{\left(u_{xx}^k\right)}^2\text{d}x}.$

收集的估计，带入(23)，可以得到

(26)

将(19)和(26)相加，可以得到

$\begin{array}{l}\frac{1}{2}\frac{\text{d}}{\text{d}t}\left({\Vert \sqrt{{\rho }^k}{u}_t^k\Vert }_{L^2\left(I\right)}^2+{\Vert \sqrt{{\rho }^k}{u}_x^k\Vert }_{L^2\left(I\right)}^2\right)+\frac{{\mu }_0\left(p_{\infty }-1\right)}{2}{\displaystyle {\int }_0^1{\left(1+{\left|u_x^k\right|}^2\right)}^{\frac{p\left(x\right)-2}{2}}{\left|u_{xx}^k\right|}^2}\text{d}x\\ +{\mu }_0\left(p_{\infty }-1\right){\displaystyle {\int }_0^1{\left(1+{\left|u_x^k\right|}^2\right)}^{\frac{p\left(x\right)-2}{2}}{\left|u_{xt}^k\right|}^2\text{d}x}+\frac{{\mu }_1}{6}{\displaystyle {\int }_0^1{\left|u_{xt}^k\right|}^2}\text{d}x+\frac{{\mu }_1}{5}{\displaystyle {\int }_0^1{\left|u_{xx}^k\right|}^2}\text{d}x\\ \le C{\Phi }_K^{\gamma +\frac{7}{2}}\left(t\right)+{\eta }_1{\Vert u_{xx}^{k-1}\Vert }_{L^2\left(I\right)}^2+{\eta }_2{\Vert u_{xt}^{k-1}\Vert }_{L^2\left(I\right)}^2.\end{array}$ (27)

将上述不等式在 $\left(\tau ,t\right)\subset \left(0,t\right)$ 上积分，同时选取足够小的，对正整数 $K\ge 1$，我们有

$\begin{array}{l}\underset{1\le k\le K}{\max }(\underset{\tau \le s\le t}{\sup }{\Vert \sqrt{{\rho }^k}{u}_t^k\left(s\right)\Vert }_{L^2\left(I\right)}^2+{\Vert \sqrt{{\rho }^k}{u}_x^k\left(s\right)\Vert }_{L^2\left(I\right)}^2+{\displaystyle {\int }_{\tau }^t{\Vert u_{xt}^k\Vert }_{L^2\left(I\right)}^2\text{d}s}+{\displaystyle {\int }_{\tau }^t{\Vert u_{xx}^k\Vert }_{L^2\left(I\right)}^2}\text{d}s\begin{array}{c}\underset{}{\overset{}{}}\\ \end{array}\\ +{\displaystyle {\int }_{\tau }^t{\displaystyle {\int }_0^1{\left(1+{\left|u_x^k\right|}^2\right)}^{\frac{p\left(x\right)-2}{2}}{\left|u_{xx}^k\right|}^2\text{d}x}\text{d}s}+{\displaystyle {\int }_{\tau }^t{\displaystyle {\int }_0^1{\left(1+{\left|u_x^k\right|}^2\right)}^{\frac{p\left(x\right)-2}{2}}}}{\left|u_{xt}^k\right|}^2\text{d}x\text{d}s)\\ \le {\Vert \sqrt{{\rho }^k}{u}_t^k\left(\tau \right)\Vert }_{L^2\left(I\right)}^2+{\Vert \sqrt{{\rho }^k}{u}_x^k\left(\tau \right)\Vert }_{L^2\left(I\right)}^2+C{\displaystyle {\int }_{\tau }^t{\Phi }_K^{\gamma +\frac{7}{2}}}\left(s\right)\text{d}s.\end{array}$

由相容性条件(4)、方程(7)和 $\left({\rho }_0,u_0\right)$ 的正则性，可以得到

由此，令 $\tau \to 0$，我们可以得到

$\begin{array}{l}\underset{1\le k\le K}{\max }(\underset{0\le s\le t}{\sup }{\Vert \sqrt{{\rho }^k}{u}_t^k\left(t\right)\Vert }_{L^2\left(I\right)}^2+{\Vert \sqrt{{\rho }^k}{u}_x^k\left(t\right)\Vert }_{L^2\left(I\right)}^2+{\displaystyle {\int }_0^t{\Vert u_{xt}^k\Vert }_{L^2\left(I\right)}^2\text{d}s}+{\displaystyle {\int }_0^t{\Vert u_{xx}^k\Vert }_{L^2\left(I\right)}^2}\text{d}s\begin{array}{c}\underset{}{\overset{}{}}\\ \end{array}\\ +{\displaystyle {\int }_0^t{\displaystyle {\int }_0^1{\left(1+{\left|u_x^k\right|}^2\right)}^{\frac{p\left(x\right)-2}{2}}{\left|u_{xx}^k\right|}^2\text{d}x}\text{d}s}+{\displaystyle {\int }_0^t{\displaystyle {\int }_0^1{\left(1+{\left|u_x^k\right|}^2\right)}^{\frac{p\left(x\right)-2}{2}}}}{\left|u_{xt}^k\right|}^2\text{d}x\text{d}s)\\ \le C\left(1+{\displaystyle {\int }_0^t{\Phi }_K^{\gamma +\frac{7}{2}}}\left(t\right)\text{d}s\right).\end{array}$

引理2证明完成。

引理3：假设 $\left({\rho }^k,u^k\right)$ 是方程(6)~(7)的解，存在 $T\in \left(0,\infty \right)$，对任意的 $t\in \left[0,T\right]$，

$\underset{0\le s\le t}{\sup }{\Vert {\rho }^k\left(t\right)\Vert }_{H^1\left(I\right)}^2\le C\exp \left({\displaystyle {\int }_0^{t}C{\Phi }_K^{\gamma +\frac{7}{2}}\text{d}s}\right)$

证明：用 ${\rho }^k$ 乘以方程(6)两端，并关于x在 $\left[0,1\right]$ 上积分，可以得到如下不等式

$\frac{\text{d}}{\text{d}t}{\displaystyle {\int }_0^1{\left|{\rho }^k\left(t\right)\right|}^2\text{d}x}\le {\displaystyle {\int }_0^1{\left|{\rho }^k\right|}^2\left|u_x^{k-1}\right|\left(t\right)\text{d}x\le C{\Vert u_{xx}^{k-1}\Vert }_{L^2\left(I\right)}}{\Vert {\rho }^k\Vert }_{L^2\left(I\right)}^2.$ (28)

另一方面，(6)关于x求导后两边乘以 ${\rho }_x^k$ 并关于x在 $\left[0,1\right]$ 上积分，得

$\begin{array}{c}\frac{1}{2}\frac{\text{d}}{\text{d}t}{\displaystyle {\int }_0^1{\left|{\rho }_x^k\left(t\right)\right|}^2}\text{d}x=-{\displaystyle {\int }_0^1\left[\frac{3}{2}{u}_x^{k-1}{\left({\rho }_x^k\right)}^2+{\rho }^k{\rho }_x^k{u}_{xx}^{k-1}\right]\text{d}x}\\ \le C\left({\Vert u_{xx}^{k-1}\Vert }_{L^2\left(I\right)}{\Vert {\rho }_x^k\Vert }_{L^2\left(I\right)}^2+{\Vert {\rho }^k\Vert }_{L^{\infty }\left(I\right)}{\Vert {\rho }_x^k\Vert }_{L^2\left(I\right)}{\Vert u_{xx}^{k-1}\Vert }_{L^2\left(I\right)}\right).\end{array}$ (29)

综合(28)和(29)式，有

$\frac{\text{d}}{\text{d}t}{\Vert {\rho }^k\left(t\right)\Vert }_{H^1\left(I\right)}^2\le C{\Vert {\rho }^k\Vert }_{H^1\left(I\right)}^2{\Vert u_{xx}^{k-1}\Vert }_{L^2\left(I\right)},$ (30)

由Gronwall’s不等式和引理1，上式可推得

${\Vert {\rho }^k\left(t\right)\Vert }_{H^1\left(I\right)}\le {\Vert {\rho }_0\Vert }_{H^1\left(I\right)}\exp \left(C{\displaystyle {\int }_0^t{\Vert u_{xx}^{k-1}\Vert }_{L^2\left(I\right)}\text{d}s}\right)\le C\exp \left({\displaystyle {\int }_0^{t}C{\Phi }_K^{\gamma +\frac{7}{2}}\text{d}s}\right).$ (31)

引理3得证。

应用引理1~3和的定义可得

${\Phi }_K\left(t\right)\le C\exp \left(C{\displaystyle {\int }_0^t{\Phi }_K^{\gamma +\frac{7}{2}}\left(s\right)\text{d}s}\right).$ (32)

容易证明存在 $T_{*}=C^{-\gamma +\frac{7}{2}}{e}_2^{-C\left(\gamma +\frac{7}{2}\right)}$，使得

${\displaystyle {\int }_0^{T_{*}}{\Phi }_K^{\gamma +\frac{7}{2}}\left(t\right)\text{d}s}<C$

因此，我们得到逼近解 $\left({\rho }^k,u^k\right)$ 的能量估计：对 $0\le t\le T_{*}$，下式成立

$\begin{array}{l}\left(\underset{0\le t\le T_{*}}{\sup }{\Vert \sqrt{{\rho }^k}{u}_t^k\left(t\right)\Vert }_{L^2\left(I\right)}^2+{\Vert \sqrt{{\rho }^k}{u}_x^k\left(t\right)\Vert }_{L^2\left(I\right)}^2+{\Vert u_x^k\Vert }_{L^2\left(I\right)}^2+{\displaystyle {\int }_0^1{\left(1+{\left|u_x^k\right|}^2\right)}^{\frac{p\left(x\right)}{2}}\text{d}x}+{\Vert {\rho }^k\left(t\right)\Vert }_{H^1\left(I\right)}^2\right)\\ +{\displaystyle {\int }_0^{T_{*}}{\Vert u_{xt}^k\Vert }_{L^2\left(I\right)}^2\text{d}s}+{\displaystyle {\int }_0^{T_{*}}{\Vert u_{xx}^k\Vert }_{L^2\left(I\right)}^2\text{d}s}+{\displaystyle {\int }_0^{T_{*}}{\displaystyle {\int }_0^1{\left(1+{\left|u_x^k\right|}^2\right)}^{\frac{p\left(x\right)-2}{2}}{\left|u_{xx}^k\right|}^2}\text{d}x}\text{d}s\\ +{\displaystyle {\int }_0^{T_{*}}{\displaystyle {\int }_0^1{\left(1+{\left|u_x^k\right|}^2\right)}^{\frac{p\left(x\right)-2}{2}}{\left|u_{xt}^k\right|}^2\text{d}x\text{d}s}}\le C\end{array}$

3. 定理1的证明

在第二节中得到了逼近解的与密度下界无关的能量估计，应用 [2] 的方法，我们可以证明当 $k\to \infty ,\delta \to 0$ 时，逼近解 $\left({\rho }^k,u^k\right)$ 强收敛，进一步可以证明极限 $\left(\rho ,u\right)$ 为问题(1)~(2)的唯一解，至此，我们完成了定理1的证明。

基金项目

由国家自然科学基金项目资助(No.11901379)。

参考文献

[1]	Winslow, W.M. (1949) Induced Fibration of Suspensions. Journal of Applied Physics, 20, 1137-1140. https://doi.org/10.1063/1.1698285
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[4]	Yuan, H.J. and Li, H.P. (2012) Existence and Uniqueness of Solutions for a Class of Non-Newtonian Fluids with Vacuum and Damping. Journal of Mathematical Analysis and Applications, 391, 223-239. https://doi.org/10.1016/j.jmaa.2012.02.015
[5]	Rajagopal, K.R. and Růzǐčka, M, (1996) On the Modeling of Eletrorheological Materials. Mechanics Research Communications, 23, 401-407. https://doi.org/10.1016/0093-6413(96)00038-9
[6]	Růzǐčka, M. (2000) Electrorheological Fluids: Modeling and Mathematical Theory. Springer-Verlag, Berlin. https://doi.org/10.1007/BFb0104029
[7]	Diening, L. (2002) Theoretical and Numerical Results for Electrorheological Fluid. University of Freiburg, Germany. https://www.researchgate.net/publication/29756045
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[12]	Melikyan. A. (1998) Generalized Characteristics of First Order PDEs. Hamilton Printing, New York. https://doi.org/10.1007/978-1-4612-1758-9

1. 引言

本文在一维有界区间上研究了如下可压缩变指数增长非牛顿流体模型的初边值问题，

$\{\begin{array}{l}{\rho }_t+{\left(\rho u\right)}_x=0,\\ {\left(\rho u\right)}_t+{\left(\rho u^2\right)}_x-{\mu }_1{u}_{xx}-{\mu }_0{\left[{\left(1+{\left|u_x\right|}^2\right)}^{\frac{p\left(x\right)-2}{2}}{u}_x\right]}_x+{\pi }_x=0,\\ \pi =a{\rho }^{\gamma }\text{,}a>0,\gamma >0\end{array}$ (1)

带有初边值条件：

$\{\begin{array}{l}{\left(\rho ,u\right)|}_{t=0}=\left({\rho }_0,u_0\right),x\in I\equiv \left[0,1\right],\\ {u|}_{x=0}={u|}_{x=1}=0,t\in \left[0,T\right],\end{array}$ (2)

其中 $\rho ,u,\pi$ 分别表示密度，速度和压力。 $1<p\left(x\right)<2$ 为已知函数， ${\mu }_0>0,{\mu }_1>0$ 为定常数。

此类变指数增长流体模型通常用来描述电粘性流体的流动。1949年Winslow [1] 发现电粘性流体的特性是能够根据提供的电场强度来改变流体的粘性，作为一种新功能材料，广泛运用于可变减震器、喷墨印刷机、消振装置、离合器、阀、密封等，特别倾向于振动控制方面的应用。

当p(x)是常数，模型为带有Carreau’s Law结构的可压缩非牛顿流模型，关于强解的存在唯一性有很多结果见 [2] [3] [4]。当p(x)为函数时，Rajagopal，Růzǐčka [5] 首先得到了不可压缩流体模型，该模型由质量、线动量、角动量、能量的一般平衡定律、克劳修斯–杜赫不等式形式的热力学第二定律和Minkowskian形式的麦克斯韦方程导出。Růzǐčka在 [6] 中对其强解进行了进一步的研究。Diening [7] 对于不可压电粘性流体的理论和数值结果做了研究。还有很多关于变指数增长抛物型方程的结果可以在 [8] [9] [10] 中找到。由于模型强的非线性性和奇异性，对于变指数增长模型的研究，特别是，可压缩变指数非牛顿流体的研究还比较单薄，需要我们进一步探索。

本文研究的主要结论如下：

定理1：假设 $p\left(x\right)$ 满足 $1<p_{\infty }\le p\left(x\right)\le p_0<2$ ( $p_{\infty },p_0$ 为常数)，且具有一阶连续导数，初值条件 $\left({\rho }_0,u_0\right)$ 满足

$\{\begin{array}{l}0\le {\rho }_0\in H^1\left(I\right),\\ u_0\in H_0^1\left(I\right)\cap H^2\left(I\right)\end{array}$ (3)

和相容性条件：

$-{\mu }_1{u}_{0xx}-{\mu }_0{\left[{\left(1+{\left|u_{0x}\right|}^2\right)}^{\frac{p\left(x\right)-2}{2}}{u}_{0x}\right]}_x+{\pi }_x\left({\rho }_0\right)={\rho }^{\frac{1}{2}}g,g\in L^2\left(I\right)$ (4)

则存在 $T_{*}\in \left(0,\infty \right)$，使得问题(1)~(2)存在唯一强解 $\left(\rho ,u\right)$ 满足

$\begin{array}{l}0\le \rho \in C\left(\left[0,T_{*}\right];H^1\left(I\right)\right);\sqrt{\rho }{u}_t\in L^{\infty }\left(0,T_{*};L^2\left(I\right)\right);\sqrt{\rho }{u}_x\in L^{\infty }\left(0,T_{*};L^2\left(I\right)\right);\\ u\in C\left(\left[0,T_{*}\right];H_0^1\left(I\right)\right)\cap L^2\left(0,T^{*};H^2\left(I\right)\right);u_t\in L^2\left(0,T_{*};H_0^1\left(I\right)\right);\\ {\left(1+{\left|u_x^k\right|}^2\right)}^{\frac{p\left(x\right)}{2}}\in L^{\infty }\left(\left[0,T_{*}\right];L^1\left(I\right)\right);\\ {\left({\left(1+{\left|u_x\right|}^2\right)}^{\frac{p\left(x\right)-2}{2}}{u}_x\right)}_x\in C\left(\left[0,T_{*}\right];L^2\left(I\right)\right).\end{array}$ (5)

本文共有三节，在第二节中，我们将通过迭代的方法构造逼近方程，得到逼近解的先验估计，得到初值远离真空时，模型强解的存在性。应用第二节得到的与密度下界无关的一致先验估计，定理一的证明在第三节完成。

2. 先验估计

在这一部分，我们将考虑初值远离真空的情形。假设初始密度具有正的下界，通过迭代的方法构造了初边值问题(1)~(2)的逼近问题，并得到逼近解与密度下界无关的一致能量估计。

2.1. 逼近解的构造

首先，我们假设 ${\rho }_0$ 足够光滑具有正的下界，即存在一个正数 $\delta \left(0<\delta \ll 1\right)$ 使得 ${\rho }_0\ge \delta$。接下来，我们运用文献 [11] 中迭代方法构造逼近方程并得到逼近解，

• 定义。

• 对 $k\ge 1$，给定，则根据一阶双曲守恒律(见 [12])和p(x)-Laplace抛物型方程(见 [8])的存在性理论，得到如下初边值问题具有唯一的强解 $\left({\rho }^k,u^k\right)$ ：

${\rho }_t^k+u^{k-1}{\rho }_x^k+u_x^{k-1}{\rho }^k=0,$ (6)

${\rho }^k{u}_t^k+{\rho }^k{u}^{k-1}{u}_x^k-{\mu }_1{u}_{xx}^k-{\mu }_0{\left[{\left(1+{\left|u_x^k\right|}^2\right)}^{\frac{p\left(x\right)-2}{2}}{u}_x^k\right]}_x+{\pi }_x^k=0$ (7)

满足如下初边值条件

${\left({\rho }^k,u^k\right)|}_{t=0}=\left({\rho }_0,u_0\right),x\in \left[0,1\right],{u^k|}_{x=0}={u^k|}_{x=1}=0,t\in \left[0,T\right].$ (8)

其中 $0<{\rho }_0\in H^2\left(I\right),u\in H_0^1\cap H^2$ 且满足相容性条件(4)。

2.2. 一致能量估计

本小节中，我们将对问题(6)~(8)的解 $\left({\rho }^k,u^k\right)$ 进行一致先验估计。

为了得到 $\left({\rho }^k,u^k\right)$ 的估计，我们定义函数 ${\Phi }_K\left(t\right)$ ：

$\begin{array}{l}{\Phi }_K\left(t\right)=\underset{1\le k\le K}{\max }\underset{0\le s\le t}{\sup }(1+{\Vert \sqrt{{\rho }^k}{u}_t^k\left(s\right)\Vert }_{L^2\left(I\right)}^2+{\Vert \sqrt{{\rho }^k}{u}_x^k\left(s\right)\Vert }_{L^2\left(I\right)}^2\begin{array}{c}\\ \\ \end{array}\\ +{\Vert u_x^k\left(s\right)\Vert }_{L^2\left(I\right)}^2+{\Vert {\left(1+{\left|u_x^k\right|}^2\right)}^{\frac{p\left(x\right)}{4}}\Vert }_{L^2\left(I\right)}^2+{\Vert {\rho }^k\left(s\right)\Vert }_{H^1\left(I\right)}^2),\end{array}$ (9)

其中 $K\ge 1$ 是一个足够大的正整数。为了表述方便，定义下文中C为与 ${\mu }_1,{\mu }_0,p_{\infty },p_0$ 以及 ${\left|{\rho }_0\right|}_{H^1\left(I\right)},{\left|g\right|}_{L^2}$ 有关常数。

下面，我们将估计中的每一项。

引理1 假设 $\left({\rho }^k,u^k\right)$ 是方程(6)~(7)的光滑解，存在 $T\in \left(0,\infty \right)$ 使得

${\displaystyle {\int }_0^t{\Vert \sqrt{{\rho }^k}{u}_t^k\Vert }_{L^2\left(I\right)}^2\text{d}s+\underset{0\le s\le t}{\sup }\left({\Vert u_x^k\Vert }_{L^2\left(I\right)}^2+{\displaystyle {\int }_0^1{\left(1+{\left|u_x^k\right|}^2\right)}^{\frac{p\left(x\right)}{2}}\text{d}x}\right)}\le C\left(1+{\displaystyle {\int }_0^t{\Phi }_K^{\gamma +\frac{3}{2}}}\left(t\right)\text{d}s\right),$

对任意 $t\in \left[0,T\right]$ 成立。

证明：用 $u_t^k$ 乘以(7)式两端，然后关于变量x在[0,1]上积分有

$\begin{array}{l}{\displaystyle {\int }_0^1{\rho }^k{\left|u_t^k\right|}^2\text{d}x}+\frac{{\mu }_1}{2}\frac{\text{d}}{\text{d}t}{\displaystyle {\int }_0^1{\left|u_x^k\left(t\right)\right|}^2\text{d}x}+{\mu }_0\frac{\text{d}}{\text{d}t}{\displaystyle {\int }_0^t\frac{\text{1}}{p\left(x\right)}}\left[{\left(1+{\left|u_x^k\right|}^2\right)}^{\frac{p\left(x\right)}{2}}\right]\text{d}x\\ =\frac{\text{d}}{\text{d}t}{\displaystyle {\int }_0^1{\pi }^k{u}_x^k}\text{d}x+{\displaystyle {\int }_0^t{\displaystyle {\int }_0^1\left(-{\rho }^k{u}^{k-1}{u}_x^k{u}_t^k-{\pi }_t^k{u}_x^k\right)\text{d}x}}.\end{array}$ (10)

对于(10)关于变量t在 $\left[0,t\right]$ 上积分，我们可以得到，

$\begin{array}{l}{\displaystyle {\int }_0^t{\Vert \sqrt{{\rho }^k}{u}_t^k\Vert }_{L^2\left(I\right)}^2\text{d}s}+\frac{{\mu }_1}{2}{\displaystyle {\int }_0^1{\left|u_x^k\right|}^2}\text{d}x+\frac{{\mu }_0}{p_{\infty }}{\displaystyle {\int }_0^t{\left(1+{\left|u_x^k\right|}^2\right)}^{\frac{p\left(x\right)}{2}}\left(t\right)\text{d}x}\\ \le \frac{{\mu }_1}{2}{\displaystyle {\int }_0^1{\left|u_x^k\right|}^2\left(0\right)\text{d}x}+\frac{{\mu }_0}{p_{\infty }}{\displaystyle {\int }_0^t{\left(1+{\left|u_x^k\right|}^2\right)}^{\frac{p\left(x\right)}{2}}\left(0\right)\text{d}x}+{\displaystyle {\int }_0^1{\pi }^k{u}_x^k\left(t\right)\text{d}x}\\ -{\displaystyle {\int }_0^1{\pi }^k{u}_x^k\left(0\right)\text{d}x}+{\displaystyle {\int }_0^t{\displaystyle {\int }_0^1\left(-{\rho }^k{u}^{k-1}{u}_x^k{u}_t^k-{\pi }_t^k{u}_x^k\right)\text{d}x\text{d}s}}\\ \le C+{\displaystyle {\int }_0^1\left|{\pi }^k{u}_x^k\left(t\right)\right|\text{d}x}+{\displaystyle {\int }_0^t{\displaystyle {\int }_0^1\left|-{\rho }^k{u}^{k-1}{u}_x^k{u}_t^k-{\pi }_t^k{u}_x^k\right|\text{d}x\text{d}s}}\end{array}$ (11)

应使用Sobolev不等式和Young’s不等式，有

$\begin{array}{l}{\displaystyle {\int }_{\text{0}}^t{\Vert \sqrt{{\rho }^k}{u}_t^k\Vert }_{L^2\left(I\right)}^{\text{2}}\text{d}s+\frac{{\mu }_1}{4}{\Vert u_x^k\Vert }_{L^2\left(I\right)}^2+\frac{{\mu }_0}{p_{\infty }}{\displaystyle {\int }_0^t{\left(1+{\left|u_x^k\right|}^2\right)}^{\frac{p\left(x\right)}{2}}}\text{d}x}\\ \le C+C{\Vert {\pi }^k\left(t\right)\Vert }_{L^2\left(I\right)}^2+C{\displaystyle {\int }_0^t{\Vert u^{k-1}\Vert }_{L^{\infty }\left(I\right)}{\Vert \sqrt{{\rho }^k}{u}_t^k\Vert }_{L^2\left(I\right)}{\Vert \sqrt{{\rho }^k}{u}_x^k\Vert }_{L^2\left(I\right)}\text{d}s}+{\displaystyle {\int }_0^{\text{t}}{\displaystyle {\int }_0^1\left|{\pi }_t^k{u}_x^k\right|\text{d}x\text{d}s}}\\ ={\displaystyle \underset{i=1}{\overset{3}{\sum }}{K}_i}.\end{array}$ (12)

接下来依次估计 $K_i,i=1,2,3$。应用Sobolev不等式和Young’s不等式，我们可以计算

$\begin{array}{c}{K}_1={\displaystyle {\int }_0^1{\left|{\pi }^k\left(t\right)\right|}^2\text{d}x}={\displaystyle {\int }_0^1{\left|{\pi }^k\left(0\right)\right|}^2}\text{d}x+{\displaystyle {\int }_0^t\left(\frac{\partial }{\partial s}{\displaystyle {\int }_0^t\left|{\pi }^k\left(s\right)\right|\text{d}x}\right)\text{d}s}\\ ={\displaystyle {\int }_0^1{\left|{\pi }^k\left(0\right)\right|}^2}\text{d}x+{\displaystyle {\int }_0^t{\displaystyle {\int }_0^1{\pi }^k\left(\rho \right){\left({\pi }^k\left(\rho \right)\right)}^{\prime }\left(-{\rho }_x^k{u}^{k-1}-\rho {}^{k}u{}_x^{k-1}\right)\text{d}x}\text{d}s}\\ \le C+C{\displaystyle {\int }_0^t{\Vert {\pi }^k\left(\rho \right)\Vert }_{L^{\infty }\left(I\right)}{\Vert {\rho }^k\Vert }_{L^{\infty }\left(I\right)}^{\gamma -1}{\Vert {\rho }^k\left(s\right)\Vert }_{H^1\left(I\right)}{\Vert u_x^{k-1}\Vert }_{L^2\left(I\right)}\text{d}s}\\ \le C\left(1+{\displaystyle {\int }_0^t{\Phi }_K^{\gamma +\frac{1}{2}}\left(t\right)\text{d}s}\right);\end{array}$

$K_2=C{\displaystyle {\int }_0^t{\Vert u^{k-1}\Vert }_{L^{\infty }\left(I\right)}{\Vert \sqrt{{\rho }^k}{u}_t^k\Vert }_{L^2\left(I\right)}{\Vert \sqrt{{\rho }^k}{u}_x^k\Vert }_{L^2\left(I\right)}\text{d}s}\le C{\displaystyle {\int }_0^t{\Phi }_K^{\frac{3}{2}}\left(t\right)\text{d}s}$

$\begin{array}{c}{K}_3={\displaystyle {\int }_0^t{\displaystyle {\int }_0^1\left|{\pi }_t^k{u}_x^k\right|\text{d}x\text{d}s}}\le {\displaystyle {\int }_0^t{\displaystyle {\int }_0^1\left|{\left({\rho }^k\right)}^{\gamma -1}{\rho }^k{u}_x^{k-1}{u}_x^k+{\left({\rho }^k\right)}^{\gamma -1}{\rho }_x^k{u}^{k-1}{u}_x^k\right|\text{d}x\text{d}s}}\\ \le {\displaystyle {\int }_0^t({\Vert {\rho }^k\Vert }_{L^{\infty }\left(I\right)}^{\gamma -1}{\Vert {\rho }^k\Vert }_{L^{\infty }\left(I\right)}^{\frac{1}{2}}{\Vert \sqrt{{\rho }^k}{u}_x^k\Vert }_{L^2\left(I\right)}{\Vert u_x^{k-1}\Vert }_{L^2\left(I\right)}}\\ +{\Vert {\rho }^k\Vert }_{L^{\infty }\left(I\right)}^{\gamma -\frac{\text{3}}{2}}{\Vert {\rho }_x^k\Vert }_{L^2\left(I\right)}{\Vert \sqrt{{\rho }^k}{u}_x^k\Vert }_{L^2\left(I\right)}{\Vert u^{k-1}\Vert }_{L^{\infty }\left(I\right)})\text{d}s\\ \le C{\displaystyle {\int }_0^t{\Phi }^{\gamma +\frac{3}{2}}}\text{d}s.\end{array}$

将的估计代入(12)，有

${\displaystyle {\int }_{\text{0}}^t{\Vert \sqrt{{\rho }^k}{u}_t^k\Vert }_{L^2\left(I\right)}^2\text{d}s}+\frac{{\mu }_1}{\text{4}}{\Vert u_x^k\Vert }_{L^2\left(I\right)}^2+\frac{{\mu }_0}{p_{\infty }}{\displaystyle {\int }_0^t{\left(1+{\left|u_x^k\right|}^2\right)}^{\frac{p\left(x\right)}{2}}}\text{d}x\le C\left(1+{\displaystyle {\int }_0^t{\Phi }_K^{\gamma +\frac{3}{2}}\left(t\right)\text{d}s}\right).$ (13)

引理1的证明完成。

引理2：若是(6)~(7)的解，那么存在 $T\in \left(0,\infty \right)$，使得

$\begin{array}{l}\underset{1\le k\le K}{\max }(\underset{0\le s\le t}{\sup }{\Vert \sqrt{{\rho }^k}{u}_t^k\left(t\right)\Vert }_{L^2\left(I\right)}^2+{\Vert \sqrt{{\rho }^k}{u}_x^k\left(t\right)\Vert }_{L^2\left(I\right)}^2+{\displaystyle {\int }_0^t{\Vert u_{xt}^k\Vert }_{L^2\left(I\right)}^2\text{d}s}+{\displaystyle {\int }_0^t{\Vert u_{xx}^k\Vert }_{L^2\left(I\right)}^2\text{d}s}\\ +{{\displaystyle {\int }_0^t{\displaystyle {\int }_0^1\left(1+{\left|u_x^k\right|}^2\right)}}}^{\frac{p\left(x\right)-2}{2}}{\left|u_{xx}^k\right|}^2\text{d}x\text{d}s+{\displaystyle {\int }_0^t{\displaystyle {\int }_0^1\left({\left(1+{\left|u_x^k\right|}^{\text{2}}\right)}^{\frac{p\left(x\right)-2}{2}}\right){\left|u_{xt}^k\right|}^2}}\text{d}x\text{d}s)\\ \le C{\displaystyle {\int }_0^t{\Phi }_K^{\gamma +\frac{7}{2}}\left(t\right)\text{d}s}\end{array}$

证明：对(7)关于t求导有

${\left({\rho }^k{u}_t^k\right)}_t+{\left({\rho }^k{u}^{k-1}{u}_x^k\right)}_t-{\mu }_1{\left(u_{xx}^k\right)}_t-{\mu }_0{\left[{\left(1+{\left|u_x^k\right|}^2\right)}^{\frac{p\left(x\right)-2}{2}}{u}_x^k\right]}_{xt}+{\pi }_{xt}^k=0$ (14)

将方程(14)两端同乘以 $u_t^k$ 然后在(0,1)上关于x积分，可以得到

$\begin{array}{l}\frac{1}{2}\frac{\text{d}}{\text{d}t}{\displaystyle {\int }_0^1{\rho }^k{\left(u_t^k\right)}^2}\text{d}x+{\mu }_1{\displaystyle {\int }_0^1{\left(u_{xt}^k\right)}^2}\text{d}x+{\mu }_0{\displaystyle {\int }_0^1{\left[{\left(1+{\left|u_x^k\right|}^2\right)}^{\frac{p\left(x\right)-2}{2}}{u}_x^k\right]}_t{u}_{xt}^k}\text{d}x\\ ={\displaystyle {\int }_0^1{\pi }_t^k{u}_{xt}^k}\text{d}x+{\displaystyle {\int }_0^1{\left({\rho }^k{u}^{k-1}\right)}_x\left(\frac{1}{2}{\left(u_t^k\right)}^2+u^{k-1}{u}_x^k{u}_t^k\right)\text{d}x}\\ -{\displaystyle {\int }_0^1{\rho }^k}{u}_t^{k-1}{u}_x^k{u}_t^k\text{d}x-{\displaystyle {\int }_0^1\left({\rho }^k{u}^{k-1}\right)u_{xt}^k{u}_t^k}\text{d}x\end{array}$ (15)

因为 $1<p_{\infty }\le p\left(x\right)\le p_0<2$，我们可以计算

(16)

另一方面，应用质量守恒方程(6)有

${\pi }_t^k=-\gamma {\pi }^k{u}_x^{k-1}-{\pi }_x^k{u}^{k-1}$ (17)

由(15)、(16)和(17)，得到如下不等式，

$\begin{array}{l}\frac{1}{2}\frac{\text{d}}{\text{d}t}{\Vert \sqrt{{\rho }^k}{u}_t^k\Vert }_2^2+{\mu }_1{\displaystyle {\int }_0^1{\left|u_{xt}^k\right|}^2\text{d}x}+{\mu }_0{\displaystyle {\int }_0^1\left(p\left(x\right)-1\right){\left(1+{\left|u_x^k\right|}^2\right)}^{\frac{p\left(x\right)-2}{2}}}{\left(u_{xt}^k\right)}^2\text{d}x\\ \le {\displaystyle {\int }_0^1\left|{\pi }_x^k\right|\left|u^{k-1}\right|\left|u_{xt}^k\right|\text{d}x}+\gamma {\displaystyle {\int }_0^1\left|{\pi }^k\right|\left|u_x^{k-1}\right|\left|u_{xt}^k\right|\text{d}x}+2{\displaystyle {\int }_0^1{\rho }^k\left|u^{k-1}\right|\left|u_t^k\right|\left|u_{xt}^k\right|\text{d}x}\\ +{\displaystyle {\int }_0^1{\rho }^k\left|u^{k-1}\right|\left|u_x^{k-1}\right|\left|u_x^k\right|\left|u_t^k\right|\text{d}x}+{\displaystyle {\int }_0^1{\rho }^k{\left|u^{k-1}\right|}^2\left|u_{xx}^k\right|\left|u_t^k\right|\text{d}x}\\ +{\displaystyle {\int }_0^1{\rho }^k{\left|u^{k-1}\right|}^2\left|u_x^k\right|\left|u_{xt}^k\right|\text{d}x}+{\displaystyle {\int }_{\text{0}}^{\text{1}}{\rho }^k\left|u_t^{k-1}\right|\left|u_x^k\right|\left|u_t^k\right|\text{d}x}\\ ={\displaystyle \underset{j=1}{\overset{7}{\sum }}{I}_j.}\end{array}$ (18)

应用Sobolev不等式，Hȍlder不等式和Young’s不等式，我们可以估计 $I_j,j=1,\cdots ,7$。

$I_1={\displaystyle {\int }_0^1\left|{\pi }_x^k\right|\left|u^{k-1}\right|\left|u_{xt}^k\right|\text{d}x}\le {\Vert {\pi }_x^k\Vert }_{L^2\left(I\right)}{\Vert u^{k-1}\Vert }_{L^{\infty }\left(I\right)}{\Vert u_{xt}\Vert }_{L^2\left(I\right)}\le C{\Phi }_K^{\gamma +1}\left(t\right)+\frac{{\mu }_1}{6}{\Vert u_{xt}^k\Vert }_{L^2\left(I\right)}^2;$

$I_2={\displaystyle {\int }_0^1\gamma \left|{\pi }^k\right|\left|u_x^{k-1}\right|\left|u_{xt}^k\right|\text{d}x}\le C{\Vert {\pi }^k\Vert }_{L^{\infty }\left(I\right)}{\Vert u_x^{k-1}\Vert }_{L^2\left(I\right)}{\Vert u_{xt}^k\Vert }_{L^2\left(I\right)}\le C{\Phi }_K^{\gamma +1}\left(t\right)+\frac{{\mu }_1}{6}{\Vert u_{xt}^k\Vert }_{L^2\left(I\right)}^2;$

$\begin{array}{c}{I}_3=2{\displaystyle {\int }_0^1{\rho }^k\left|u^{k-1}\right|\left|u_t^k\right|\left|u_{xt}^k\right|\text{d}x}\le 2{\Vert {\rho }^k\Vert }_{L^{\infty }\left(I\right)}^{\frac{1}{2}}{\Vert u^{k-1}\Vert }_{L^{\infty }\left(I\right)}{\Vert \sqrt{{\rho }^k}{u}_t^k\Vert }_{L^2\left(I\right)}{\Vert u_{xt}^k\Vert }_{L^2\left(I\right)}\\ \le C{\Phi }_K^{\frac{5}{2}}\left(t\right)+\frac{{\mu }_1}{6}{\Vert u_{xt}^k\Vert }_{L^2\left(I\right)}^2;\end{array}$

$\begin{array}{c}{I}_4={\displaystyle {\int }_0^1{\rho }^k\left|u^{k-1}\right|\left|u_x^{k-1}\right|\left|u_x^k\right|\left|u_t^k\right|\text{d}x}\le {\Vert u^{k-1}\Vert }_{L^{\infty }\left(I\right)}{\Vert u_x^{k-1}\Vert }_{L^{\infty }\left(I\right)}{\Vert \sqrt{{\rho }^k}{u}_x^k\Vert }_{L^2\left(I\right)}{\Vert \sqrt{{\rho }^k}{u}_t^k\Vert }_{L^2\left(I\right)}\\ \le C{\Phi }_K^3\left(t\right)+{\eta }_1{\Vert u_{xx}^{k-1}\Vert }_{L^2\left(I\right)}^2\end{array}$

$\begin{array}{c}{I}_5={\displaystyle {\int }_0^1{\rho }^k{\left|u^{k-1}\right|}^2\left|u_{xx}^k\right|\left|u_t^k\right|\text{d}x}\le {\Vert {\rho }^k\Vert }_{L^{\infty }\left(I\right)}^{\frac{1}{2}}{\Vert u^{k-1}\Vert }_{L^{\infty }\left(I\right)}^2{\Vert \sqrt{{\rho }^k}{u}_t^k\Vert }_{L^2\left(I\right)}{\Vert u_{xx}^k\Vert }_{L^2\left(I\right)}\\ \le C{\Phi }_K^{\frac{7}{2}}\left(x\right)+\frac{{\mu }_1}{5}{\Vert u_{xx}^k\Vert }_{L^2}^2;\end{array}$

$\begin{array}{c}{I}_6={\displaystyle {\int }_0^1\left|{\rho }^k\right|{\left|u^{k-1}\right|}^2\left|u_x^k\right|\left|u_{xt}^k\right|\text{d}x}\le {\Vert {\rho }^k\Vert }_{L^{\infty }\left(I\right)}^{\frac{1}{2}}{\Vert u^{k-1}\Vert }_{L^{\infty }\left(I\right)}^2{\Vert \sqrt{{\rho }^k}{u}_x^k\Vert }_{L^2\left(I\right)}{\Vert u_{xt}^k\Vert }_{L^2\left(I\right)}\\ \le C{\Phi }_K^{\frac{7}{2}}\left(t\right)+\frac{{\mu }_1}{6}{\Vert u_{xt}^k\Vert }_{L^2\left(I\right)}^2;\end{array}$

$\begin{array}{c}{I}_7={\displaystyle {\int }_0^1\left|{\rho }^k\right|\left|u_t^{k-1}\right|\left|u_x^k\right|\left|u_t^k\right|\text{d}x}\le {\Vert u_t^{k-1}\Vert }_{L^{\infty }\left(I\right)}{\Vert \sqrt{{\rho }^k}{u}_x^k\Vert }_{L^2\left(I\right)}{\Vert \sqrt{{\rho }^k}{u}_t^k\Vert }_{L^2\left(I\right)}\\ \le C{\Phi }_K^2\left(t\right)+{\eta }_2{\Vert u_{xt}^{k-1}\Vert }_{L^2\left(I\right)}^2.\end{array}$

其中 ${\eta }_1,{\eta }_2$ 为与有关的足够小的常数。将上述估计代入(18)中有：

$\begin{array}{l}\frac{1}{2}\frac{\text{d}}{\text{d}t}{\Vert \sqrt{{\rho }^k}{u}_t^k\Vert }_2^2+\frac{{\mu }_1}{3}{\displaystyle {\int }_0^1{\left|u_{xt}^k\right|}^2}\text{d}x+{\mu }_0{\displaystyle {\int }_0^1\left(p\left(x\right)-1\right){\left(1+{\left|u_x^k\right|}^2\right)}^{\frac{p\left(x\right)-2}{2}}}{\left(u_{xt}^k\right)}^2\text{d}x\\ \le C{\Phi }_K^{\gamma +\frac{7}{2}}\left(t\right)+\frac{{\mu }_1}{5}{\Vert u_{xx}^k\Vert }_{L^2\left(I\right)}^2+{\eta }_1{\Vert u_{xx}^{k-1}\Vert }_{L^2\left(I\right)}^2+{\eta }_2{\Vert u_{xt}^{k-1}\Vert }_{L^2\left(I\right)}^2.\end{array}$ (19)

接下来，我们估计 ${\Vert \sqrt{{\rho }^k}{u}_x^k\left(t\right)\Vert }_{L^2\left(I\right)}$。将方程(7)两端同乘以 $-u_{xx}^k$，然后在 $\left(0,\text{1}\right)$ 上关于x积分，由方程

(6)和分部积分，我们得到

$\begin{array}{l}\frac{1}{2}\frac{\text{d}}{\text{d}t}{\displaystyle {\int }_0^1{\rho }^k{\left(u_x^k\right)}^2}\text{d}x+{\mu }_1{\displaystyle {\int }_0^1{\left(u_{xx}^k\right)}^2}\text{d}x+{\mu }_0{\displaystyle {\int }_0^1{\left[{\left(1+{\left|u_x^k\right|}^2\right)}^{\frac{p\left(x\right)-2}{2}}{u}_x^k\right]}_x{u}_{xx}^k}\text{d}x\\ ={\displaystyle {\int }_0^1{\pi }_x^k{u}_{xx}^k}\text{d}x+{\displaystyle {\int }_0^1{\rho }_t^k{\left(u_x^k\right)}^2}\text{d}x-{\displaystyle {\int }_0^1{\rho }_x^k{u}_t^k{u}_x^k}\text{d}x.\end{array}$ (20)

我们首先计算方程(20)左端的第三项，

$\begin{array}{l}{\left[{\left(1+{\left|u_x^k\right|}^2\right)}^{\frac{p\left(x\right)-2}{2}}{u}_x^k\right]}_x{u}_{xx}^k={\left(e^{\frac{p\left(x\right)-2}{2}\ln \left(1+{\left|u_x^k\right|}^2\right)}{u}_x^k\right)}_x{u}_{xx}^k\\ ={\left(1+{\left|u_x^k\right|}^2\right)}^{\frac{p\left(x\right)-2}{2}}{\left(u_{xx}^k\right)}^2+{\left(1+{\left|u_x^k\right|}^2\right)}^{\frac{p\left(x\right)-2}{2}}\left(\frac{p_x\left(x\right)}{2}\ln \left(1+{\left|u_x^k\right|}^2\right)+\frac{p\left(x\right)-2}{2}\frac{2\left|u_x^k\right|u_{xx}^k}{1+{\left|u_x^k\right|}^2}\right)u_x^k{u}_{xx}^k\\ =\frac{p_x\left(x\right)}{2}{\left(1+{\left|u_x^k\right|}^2\right)}^{\frac{p\left(x\right)-2}{2}}{u}_x^k{u}_{xx}^k\ln \left(1+{\left|u_x^k\right|}^2\right)+{\left(1+{\left|u_x^k\right|}^2\right)}^{\frac{p\left(x\right)-2}{2}}{\left(u_{xx}^k\right)}^2\left[\frac{\left(p\left(x\right)-2\right){\left(u_x^k\right)}^2}{1+{\left|u_x^k\right|}^2}+1\right]\\ \ge \frac{p_x\left(x\right)}{2}{\left(1+{\left|u_x^k\right|}^2\right)}^{\frac{p\left(x\right)-2}{2}}{u}_x^k{u}_{xx}^k\ln \left(1+{\left|u_x^k\right|}^2\right)+\left(p\left(x\right)-1\right){\left(1+{\left|u_x^k\right|}^2\right)}^{\frac{p\left(x\right)-2}{2}}{\left(u_{xx}^k\right)}^2.\end{array}$

将上式代入(20)，有

(21)

为了估计(21)，我们需要仔细计算上式右端第四项。应用Hȍlder不等式和Young’s不等式，我们可以得到

(22)

其中 ${\Vert p_x\left(x\right)\Vert }_{\infty }\le C$ 和 $1<p_{\infty }\le p\left(x\right)\le p_0<2$。把(22)代入(21)，应用方程(6)有

$\begin{array}{l}\frac{1}{2}\frac{\text{d}}{\text{d}t}{\Vert \sqrt{{\rho }^k}{u}_x^k\Vert }_{L^2\left(I\right)}^2+{\mu }_1{\displaystyle {\int }_0^1{\left(u_{xx}^k\right)}^2\text{d}x}+\frac{{\mu }_0}{2}{\displaystyle {\int }_0^1\left(p\left(x\right)-1\right){\left(1+{\left|u_x^k\right|}^2\right)}^{\frac{p\left(x\right)-2}{2}}}{\left(u_{xx}^k\right)}^2\text{d}x\\ \le {\displaystyle {\int }_0^1\left|{\pi }_x^k\right|\left|u_x^k\right|\text{d}x}+{\displaystyle {\int }_0^1\left|{\rho }_x^k\right|\left|u^k\right|{\left|u_x^k\right|}^2\text{d}x}+{\displaystyle {\int }_0^1\left|{\rho }^k\right|{\left|u_x^k\right|}^3\text{d}x}+{\displaystyle {\int }_0^1\left|{\rho }^k\right|\left|u_{xx}^k\right|\left|u_t^k\right|\text{d}x}\\ +{\displaystyle {\int }_0^1\left|{\rho }^k\right|\left|u_x^k\right|\left|u_{xt}^k\right|\text{d}x}+{\displaystyle {\int }_0^1{\left(u_x^k\right)}^2\text{d}x}+C{\displaystyle {\int }_0^1\left|{\left(1+{\left|u_x^k\right|}^2\right)}^{\frac{p\left(x\right)}{2}}\right|\left|{\ln }^2\left(1+{\left|u_x^k\right|}^2\right)\right|\text{d}x}\\ ={\displaystyle \underset{i=1}{\overset{7}{\sum }}{J}_i.}\end{array}$ (23)

应用Sobolev不等式和Young’s不等式，有

$J_1={\displaystyle {\int }_0^1\left|{\pi }_x^k\right|\left|u_x^k\right|\text{d}x}\le {\Vert {\rho }_x^k\Vert }_{L^2\left(I\right)}^{\gamma }{\Vert u_x^k\Vert }_{L^2\left(I\right)}\le {\Phi }_K^{\frac{\gamma }{2}+\frac{1}{2}}\left(t\right);$

$J_2={\displaystyle {\int }_0^1\left|{\rho }_x^k\right|\left|u^k\right|{\left|u_x^k\right|}^2\text{d}x}\le {\Vert u^k\Vert }_{L^{\infty }\left(I\right)}{\Vert {\rho }_x^k\Vert }_{L^2\left(I\right)}{\Vert u_x^k\Vert }_{L^2\left(I\right)}{\Vert u_x^k\Vert }_{L^{\infty }\left(I\right)}\le C{\Phi }_K^3\left(t\right)+\frac{{\mu }_1}{5}{\Vert u_{xx}^k\Vert }_{L^2\left(I\right)}^2;$

$J_3={\displaystyle {\int }_0^1\left|{\rho }^k\right|{\left|u_x^k\right|}^3\text{d}x}\le {\Vert {\rho }^k\Vert }_{L^{\infty }\left(I\right)}^{\frac{1}{2}}{\Vert \sqrt{{\rho }^k}{u}_x^k\Vert }_{L^2\left(I\right)}{\Vert u_x^k\Vert }_{L^2\left(I\right)}{\Vert u_x^k\Vert }_{L^{\infty }\left(I\right)}\le C{\Phi }_K^{\frac{5}{2}}\left(t\right)+\frac{{\mu }_1}{5}{\Vert u_{xx}^k\Vert }_{L^2\left(I\right)}^2;$

$J_4={\displaystyle {\int }_0^1\left|{\rho }^k\right|\left|u_{xx}^k\right|\left|u_t^k\right|\text{d}x}\le {\Vert {\rho }^k\Vert }_{L^{\infty }\left(I\right)}^{\frac{1}{2}}{\Vert \sqrt{{\rho }^k}{u}_t^k\Vert }_{L^2\left(I\right)}{\Vert u_{xx}^k\Vert }_{L^2\left(I\right)}\le C{\Phi }_K^{\frac{3}{2}}\left(t\right)+\frac{{\mu }_1}{5}{\Vert u_{xx}^k\Vert }_{L^2\left(I\right)}^2$

$J_5={\displaystyle {\int }_0^1\left|{\rho }^k\right|\left|u_x^k\right|\left|u_{xt}^k\right|\text{d}x}\le {\Vert {\rho }^k\Vert }_{L^{\infty }\left(I\right)}^{\frac{1}{2}}{\Vert \sqrt{{\rho }^k}{u}_x^k\Vert }_{L^2\left(I\right)}{\Vert u_{xt}^k\Vert }_{L^2\left(I\right)}\le {\Phi }_K^{\frac{3}{2}}\left(t\right)+\frac{{\mu }_1}{6}{\Vert u_{xt}^k\Vert }_{L^2\left(I\right)}^2$

$J_6={\displaystyle {\int }_0^1{\left(u_x^k\right)}^2\text{d}x}\le {\Phi }_K(t)$

对于 $1<p_{\infty }\le p\left(x\right)\le p_0<2$ 和 $2<s<3$，在Sobolev不等式的帮助下，我们可以推出

$\begin{array}{c}{J}_7=C{\displaystyle {\int }_0^1\left|{\left(1+{\left|u_x^k\right|}^2\right)}^{\frac{p\left(x\right)}{2}}\right|\left|{\ln }^2\left(1+{\left|u_x^k\right|}^2\right)\right|\text{d}x}\\ \le C{\left[{\displaystyle {\int }_0^1{\left({\left|{\left(1+{\left|u_x^k\right|}^2\right)}^{\frac{p\left(x\right)}{2}}\right|}^2\right)}^{\frac{s}{2}}\text{d}x}\right]}^{\frac{2}{s}}{\left[{\displaystyle {\int }_0^1{\left|{\ln }^2\left(1+{\left|u_x^k\right|}^2\right)\right|}^{\frac{s}{s-2}}\text{d}x}\right]}^{\frac{s-2}{s}}\\ \le C{\Vert {\left(1+{\left|u_x^k\right|}^2\right)}^{\frac{p\left(x\right)}{4}}\Vert }_{L^s}^2{\left[{\displaystyle {\int }_0^1{\left|\ln \left(1+{\left|u_x^k\right|}^2\right)\right|}^{\frac{2s}{s-2}}\text{d}x}\right]}^{\frac{s-2}{s}}\end{array}$

$\begin{array}{l}\le C{\Vert {\left(1+{\left|u_x^k\right|}^2\right)}^{\frac{p\left(x\right)}{4}}\Vert }_{L^s}^2{\left[{\displaystyle {\int }_0^1{\left(\frac{8}{\left(s-2\right)p\left(x\right)}{\left|\ln \left(1+{\left|u_x^k\right|}^2\right)\right|}^{\frac{\left(s-2\right)p\left(x\right)}{8}}\right)}^{\frac{2s}{s-2}}\text{d}x}\right]}^{\frac{s-2}{s}}\\ \le C{\Vert {\left(1+{\left|u_x^k\right|}^2\right)}^{\frac{p\left(x\right)}{4}}\Vert }_{L^s}^2{\left[{\displaystyle {\int }_0^1{\left(\frac{8}{\left(s-2\right)p\left(x\right)}{\left|\left(1+{\left|u_x^k\right|}^2\right)\right|}^{\frac{\left(s-2\right)p\left(x\right)}{8}}\right)}^{\frac{2s}{s-2}}\text{d}x}\right]}^{\frac{s-2}{s}}\\ \le C{\Vert {\left(1+{\left|u_x^k\right|}^2\right)}^{\frac{p\left(x\right)}{4}}\Vert }_{L^s}^2{\left|{\left(1+{\left|u_x^k\right|}^2\right)}^{\frac{p\left(x\right)}{4}}\right|}_{L^s}^{s-2}=C{\left|{\left(1+{\left|u_x^k\right|}^2\right)}^{\frac{p\left(x\right)}{4}}\right|}_{L^s}^s\end{array}$

应用Sobolev嵌入定理和 $2<s<3$，下式成立，

(24)

类似(22)中的估计，我们可以得到

${\Vert {\left({\left(1+{\left|u_x^k\right|}^2\right)}^{\frac{p\left(x\right)}{4}}\right)}_x\Vert }_{L^2}^2\le C\left({\displaystyle {\int }_0^1{\left(1+{\left|u_x^k\right|}^2\right)}^{\frac{p\left(x\right)}{2}}{\ln }^2\left(1+{\left|u_x^k\right|}^2\right)\text{d}x}+{\displaystyle {\int }_0^1\left|{\left(1+{\left|u_x^k\right|}^2\right)}^{\frac{p\left(x\right)-2}{2}}\right|}{\left(u_{xx}^k\right)}^2\text{d}x\right).$ (25)

因此，联立(24)和(25)，因为 $1<p\left(x\right)<2$， $2<s<3$，选取足够小的 $\epsilon$，我们可以得到 $J_7$ 的估计，

$J_7\le C{\Vert {\left(1+{\left|u_x^k\right|}^2\right)}^{\frac{p\left(x\right)}{4}}\Vert }_{L^2}^4+\epsilon C{\displaystyle {\int }_0^1\left|{\left(1+{\left|u_x^k\right|}^2\right)}^{\frac{p\left(x\right)-2}{2}}\right|{\left(u_{xx}^k\right)}^2}\text{d}x\le C{\Phi }_K^2\left(t\right)+\frac{{\mu }_0\left(p_{\infty }-1\right)}{2}{\displaystyle {\int }_0^1\left|{\left(1+{\left|u_x^k\right|}^2\right)}^{\frac{p\left(x\right)-2}{2}}\right|{\left(u_{xx}^k\right)}^2\text{d}x}.$

收集的估计，带入(23)，可以得到

(26)

将(19)和(26)相加，可以得到

$\begin{array}{l}\frac{1}{2}\frac{\text{d}}{\text{d}t}\left({\Vert \sqrt{{\rho }^k}{u}_t^k\Vert }_{L^2\left(I\right)}^2+{\Vert \sqrt{{\rho }^k}{u}_x^k\Vert }_{L^2\left(I\right)}^2\right)+\frac{{\mu }_0\left(p_{\infty }-1\right)}{2}{\displaystyle {\int }_0^1{\left(1+{\left|u_x^k\right|}^2\right)}^{\frac{p\left(x\right)-2}{2}}{\left|u_{xx}^k\right|}^2}\text{d}x\\ +{\mu }_0\left(p_{\infty }-1\right){\displaystyle {\int }_0^1{\left(1+{\left|u_x^k\right|}^2\right)}^{\frac{p\left(x\right)-2}{2}}{\left|u_{xt}^k\right|}^2\text{d}x}+\frac{{\mu }_1}{6}{\displaystyle {\int }_0^1{\left|u_{xt}^k\right|}^2}\text{d}x+\frac{{\mu }_1}{5}{\displaystyle {\int }_0^1{\left|u_{xx}^k\right|}^2}\text{d}x\\ \le C{\Phi }_K^{\gamma +\frac{7}{2}}\left(t\right)+{\eta }_1{\Vert u_{xx}^{k-1}\Vert }_{L^2\left(I\right)}^2+{\eta }_2{\Vert u_{xt}^{k-1}\Vert }_{L^2\left(I\right)}^2.\end{array}$ (27)

将上述不等式在 $\left(\tau ,t\right)\subset \left(0,t\right)$ 上积分，同时选取足够小的，对正整数 $K\ge 1$，我们有

$\begin{array}{l}\underset{1\le k\le K}{\max }(\underset{\tau \le s\le t}{\sup }{\Vert \sqrt{{\rho }^k}{u}_t^k\left(s\right)\Vert }_{L^2\left(I\right)}^2+{\Vert \sqrt{{\rho }^k}{u}_x^k\left(s\right)\Vert }_{L^2\left(I\right)}^2+{\displaystyle {\int }_{\tau }^t{\Vert u_{xt}^k\Vert }_{L^2\left(I\right)}^2\text{d}s}+{\displaystyle {\int }_{\tau }^t{\Vert u_{xx}^k\Vert }_{L^2\left(I\right)}^2}\text{d}s\begin{array}{c}\underset{}{\overset{}{}}\\ \end{array}\\ +{\displaystyle {\int }_{\tau }^t{\displaystyle {\int }_0^1{\left(1+{\left|u_x^k\right|}^2\right)}^{\frac{p\left(x\right)-2}{2}}{\left|u_{xx}^k\right|}^2\text{d}x}\text{d}s}+{\displaystyle {\int }_{\tau }^t{\displaystyle {\int }_0^1{\left(1+{\left|u_x^k\right|}^2\right)}^{\frac{p\left(x\right)-2}{2}}}}{\left|u_{xt}^k\right|}^2\text{d}x\text{d}s)\\ \le {\Vert \sqrt{{\rho }^k}{u}_t^k\left(\tau \right)\Vert }_{L^2\left(I\right)}^2+{\Vert \sqrt{{\rho }^k}{u}_x^k\left(\tau \right)\Vert }_{L^2\left(I\right)}^2+C{\displaystyle {\int }_{\tau }^t{\Phi }_K^{\gamma +\frac{7}{2}}}\left(s\right)\text{d}s.\end{array}$

由相容性条件(4)、方程(7)和 $\left({\rho }_0,u_0\right)$ 的正则性，可以得到

由此，令 $\tau \to 0$，我们可以得到

$\begin{array}{l}\underset{1\le k\le K}{\max }(\underset{0\le s\le t}{\sup }{\Vert \sqrt{{\rho }^k}{u}_t^k\left(t\right)\Vert }_{L^2\left(I\right)}^2+{\Vert \sqrt{{\rho }^k}{u}_x^k\left(t\right)\Vert }_{L^2\left(I\right)}^2+{\displaystyle {\int }_0^t{\Vert u_{xt}^k\Vert }_{L^2\left(I\right)}^2\text{d}s}+{\displaystyle {\int }_0^t{\Vert u_{xx}^k\Vert }_{L^2\left(I\right)}^2}\text{d}s\begin{array}{c}\underset{}{\overset{}{}}\\ \end{array}\\ +{\displaystyle {\int }_0^t{\displaystyle {\int }_0^1{\left(1+{\left|u_x^k\right|}^2\right)}^{\frac{p\left(x\right)-2}{2}}{\left|u_{xx}^k\right|}^2\text{d}x}\text{d}s}+{\displaystyle {\int }_0^t{\displaystyle {\int }_0^1{\left(1+{\left|u_x^k\right|}^2\right)}^{\frac{p\left(x\right)-2}{2}}}}{\left|u_{xt}^k\right|}^2\text{d}x\text{d}s)\\ \le C\left(1+{\displaystyle {\int }_0^t{\Phi }_K^{\gamma +\frac{7}{2}}}\left(t\right)\text{d}s\right).\end{array}$

引理2证明完成。

引理3：假设 $\left({\rho }^k,u^k\right)$ 是方程(6)~(7)的解，存在 $T\in \left(0,\infty \right)$，对任意的 $t\in \left[0,T\right]$，

$\underset{0\le s\le t}{\sup }{\Vert {\rho }^k\left(t\right)\Vert }_{H^1\left(I\right)}^2\le C\exp \left({\displaystyle {\int }_0^{t}C{\Phi }_K^{\gamma +\frac{7}{2}}\text{d}s}\right)$

证明：用 ${\rho }^k$ 乘以方程(6)两端，并关于x在 $\left[0,1\right]$ 上积分，可以得到如下不等式

$\frac{\text{d}}{\text{d}t}{\displaystyle {\int }_0^1{\left|{\rho }^k\left(t\right)\right|}^2\text{d}x}\le {\displaystyle {\int }_0^1{\left|{\rho }^k\right|}^2\left|u_x^{k-1}\right|\left(t\right)\text{d}x\le C{\Vert u_{xx}^{k-1}\Vert }_{L^2\left(I\right)}}{\Vert {\rho }^k\Vert }_{L^2\left(I\right)}^2.$ (28)

另一方面，(6)关于x求导后两边乘以 ${\rho }_x^k$ 并关于x在 $\left[0,1\right]$ 上积分，得

$\begin{array}{c}\frac{1}{2}\frac{\text{d}}{\text{d}t}{\displaystyle {\int }_0^1{\left|{\rho }_x^k\left(t\right)\right|}^2}\text{d}x=-{\displaystyle {\int }_0^1\left[\frac{3}{2}{u}_x^{k-1}{\left({\rho }_x^k\right)}^2+{\rho }^k{\rho }_x^k{u}_{xx}^{k-1}\right]\text{d}x}\\ \le C\left({\Vert u_{xx}^{k-1}\Vert }_{L^2\left(I\right)}{\Vert {\rho }_x^k\Vert }_{L^2\left(I\right)}^2+{\Vert {\rho }^k\Vert }_{L^{\infty }\left(I\right)}{\Vert {\rho }_x^k\Vert }_{L^2\left(I\right)}{\Vert u_{xx}^{k-1}\Vert }_{L^2\left(I\right)}\right).\end{array}$ (29)

综合(28)和(29)式，有

$\frac{\text{d}}{\text{d}t}{\Vert {\rho }^k\left(t\right)\Vert }_{H^1\left(I\right)}^2\le C{\Vert {\rho }^k\Vert }_{H^1\left(I\right)}^2{\Vert u_{xx}^{k-1}\Vert }_{L^2\left(I\right)},$ (30)

由Gronwall’s不等式和引理1，上式可推得

${\Vert {\rho }^k\left(t\right)\Vert }_{H^1\left(I\right)}\le {\Vert {\rho }_0\Vert }_{H^1\left(I\right)}\exp \left(C{\displaystyle {\int }_0^t{\Vert u_{xx}^{k-1}\Vert }_{L^2\left(I\right)}\text{d}s}\right)\le C\exp \left({\displaystyle {\int }_0^{t}C{\Phi }_K^{\gamma +\frac{7}{2}}\text{d}s}\right).$ (31)

引理3得证。

应用引理1~3和的定义可得

${\Phi }_K\left(t\right)\le C\exp \left(C{\displaystyle {\int }_0^t{\Phi }_K^{\gamma +\frac{7}{2}}\left(s\right)\text{d}s}\right).$ (32)

容易证明存在 $T_{*}=C^{-\gamma +\frac{7}{2}}{e}_2^{-C\left(\gamma +\frac{7}{2}\right)}$，使得

${\displaystyle {\int }_0^{T_{*}}{\Phi }_K^{\gamma +\frac{7}{2}}\left(t\right)\text{d}s}<C$

因此，我们得到逼近解 $\left({\rho }^k,u^k\right)$ 的能量估计：对 $0\le t\le T_{*}$，下式成立

$\begin{array}{l}\left(\underset{0\le t\le T_{*}}{\sup }{\Vert \sqrt{{\rho }^k}{u}_t^k\left(t\right)\Vert }_{L^2\left(I\right)}^2+{\Vert \sqrt{{\rho }^k}{u}_x^k\left(t\right)\Vert }_{L^2\left(I\right)}^2+{\Vert u_x^k\Vert }_{L^2\left(I\right)}^2+{\displaystyle {\int }_0^1{\left(1+{\left|u_x^k\right|}^2\right)}^{\frac{p\left(x\right)}{2}}\text{d}x}+{\Vert {\rho }^k\left(t\right)\Vert }_{H^1\left(I\right)}^2\right)\\ +{\displaystyle {\int }_0^{T_{*}}{\Vert u_{xt}^k\Vert }_{L^2\left(I\right)}^2\text{d}s}+{\displaystyle {\int }_0^{T_{*}}{\Vert u_{xx}^k\Vert }_{L^2\left(I\right)}^2\text{d}s}+{\displaystyle {\int }_0^{T_{*}}{\displaystyle {\int }_0^1{\left(1+{\left|u_x^k\right|}^2\right)}^{\frac{p\left(x\right)-2}{2}}{\left|u_{xx}^k\right|}^2}\text{d}x}\text{d}s\\ +{\displaystyle {\int }_0^{T_{*}}{\displaystyle {\int }_0^1{\left(1+{\left|u_x^k\right|}^2\right)}^{\frac{p\left(x\right)-2}{2}}{\left|u_{xt}^k\right|}^2\text{d}x\text{d}s}}\le C\end{array}$

3. 定理1的证明

在第二节中得到了逼近解的与密度下界无关的能量估计，应用 [2] 的方法，我们可以证明当 $k\to \infty ,\delta \to 0$ 时，逼近解 $\left({\rho }^k,u^k\right)$ 强收敛，进一步可以证明极限 $\left(\rho ,u\right)$ 为问题(1)~(2)的唯一解，至此，我们完成了定理1的证明。

基金项目

由国家自然科学基金项目资助(No.11901379)。