一维Navier-Stokes-Smoluchowski系统的自由边界问题

Free Boundary Problem for One-Dimensional Navier-Stokes-Smoluchowski Systems

摘要: 研究了一类可压缩粒子流体耦合模型的一维自由边界问题，借助拉格朗日坐标变换建立解的基本能量不等式。

Abstract: The one-dimensional free boundary problem of a particle-fluid interation model is studied in this paper, and the basic energy law is established by Lagrange coordinate transformation.

1. 模型与预备知识

在当代多相流体力学的研究中，颗粒与流体之间的交互作用已成为热点研究方向之一。其应用广泛涉及能源、化工、生物医药等关键领域，成为关联基础科学前沿与重大工程需求的桥梁。本文考虑一类流体粒子相互作用的Navier-Stokes-Smoluchowski模型[1]-[6]：

$\{\begin{array}{l}{\rho }_t+{\left(\rho u\right)}_x=0,\\ {\left(\rho u\right)}_t+{\left(\rho u^2\right)}_x+{\left(P\left(\rho ,\eta \right)\right)}_x+\left(\rho +\eta \right){\Phi }_x+{\eta }_x-{\left(2\mu u_x\right)}_x-{\left(\lambda u_x\right)}_x=0,\\ {\eta }_t+{\left(\eta \left(u-{\Phi }_x\right)\right)}_x-{\eta }_{xx}=0.\end{array}$ (1.1)

其中$\left(x,t\right)\in \left(a\left(t\right),b\left(t\right)\right)\times [0,T]$ 。$\rho$ 为流体密度，$\Phi$ 为外势力(只与重力、浮力相关)，$u$ 为流体速度，$\eta$ 为粒子密度，$\tilde{P}=\tilde{P}\left(\rho ,\eta \right)=A{\rho }^{\gamma }+\eta$ 为正压力，为简单起见，假定常数$\gamma >1$ 和$A>0$ ，粘性系数$\mu$ 和满$\lambda$ 足$\mu >0,2\mu +\lambda \ge 0$ .考虑系统满足初始条件：

${\left(\rho ,u,\eta \right)|}_{t=0}=\left({\rho }_0\left(x\right),u_0\left(x\right),{\eta }_0\left(x\right)\right).$ (1.2)

其中$x\in \left[a\left(0\right),b\left(0\right)\right]$ 。满系统满足速度场的无滑移边界条件以及粒子密度的无流边界条件：

${u\left(x,t\right)|}_{x=a\left(t\right),b\left(t\right)}=0,{\left({\eta }_x+\eta {\Phi }_x\right)|}_{x=a\left(t\right),b\left(t\right)}=0.$ (1.3)

上述模型数学理论的研究始于Ballew以及其合作者[1] [2]，建立了高维弱解的整体存在性以及低分层极限的收敛性。进而，文献[3]-[5]建立了高维强解的局部适定性以及小条件整体适定性。对于一维固定边界问题，文献[6]建立了经典解的整体适定性。而关于该模型的自由边界问题，目前而言结果较少。

借助拉格朗日坐标变换，可以将自由边界转化为固定边界$\left(y,\tau \right)\in \left[0,1\right]\times \left[0,T\right]$ 。考虑拉格朗日坐标变换如下：

$y={\displaystyle {\int }_0^x\rho \left(\xi ,t\right)\text{d}\xi },t=\tau .$ (1.4)

$y_x=\rho \left(x,t\right).$ (1.5)

根据拉格朗日变换，方程组(1.1)可改写为：

$\{\begin{array}{l}{\rho }_{\tau }+{\rho }^{\text{2}}{u}_y=0,\\ u_{\tau }+{\left(P\left(\rho \right)\right)}_y+{\eta }_y=\left(\text{2}\mu +\lambda \right){\left(\rho u_y\right)}_y-\left(\rho +\eta \right){\Phi }_y,\\ {\eta }_{\tau }+\rho \eta u_y-\rho {\left(\rho {\eta }_y\right)}_y-\rho {\left(\rho \eta {\Phi }_y\right)}_y=0.\end{array}$ (1.6)

考虑流体、粒子质量守恒，不妨设：

${\displaystyle {\int }_0^1\rho \left(\xi ,t\right)\text{d}\xi }=1,{\displaystyle {\int }_0^1\eta \left(\xi ,t\right)\text{d}\xi }=1.$ (1.7)

系统的初始条件变为：

${\left(\rho ,u,\eta \right)|}_{\tau =0}=\left({\rho }_0,u_0,{\eta }_0\right),y\in \left[0,1\right].$ (1.8)

边界条件变为：

$u\left(y,\tau \right)=\left(\rho {\eta }_y+\rho \eta {\Phi }_y\right)\left(y,\tau \right)=0,y=0,1,\tau \in \left[0,T\right].$ (1.9)

考虑粒子密度满足结构性条件：

${\displaystyle {\int }_0^{\tau }\underset{x\in I}{\sup }{\left|\eta -{\eta }_s\right|}^{\text{2}}\text{d}\sigma }\le C,\tau \in \left[0,T\right].$ (1.10)

其中，稳态解$\left({\rho }_s,{\eta }_s\right)$ 满足稳态方程：

$\{\begin{array}{l}{P}_y\left({\rho }_s\right)=-{\rho }_s{\Phi }_y,\\ {\left({\eta }_s\right)}_y=-{\eta }_s{\Phi }_y.\end{array}$ (1.11)

定义依赖于压力项与稳态解两个辅助函数：

$\begin{array}{l}{G}_1\left(\rho \right)={\displaystyle {\int }_{{\rho }_s}^{\rho }{\displaystyle {\int }_{{\rho }_s}^r\frac{P^{\prime }\left(s\right)}{s}\text{d}s\text{d}r}}=\rho {\displaystyle {\int }_{{\rho }_s}^{\rho }\frac{P\left(s\right)-P\left({\rho }_s\right)}{s^2}\text{d}s},\\ G_2\left(\eta \right)=\eta {\displaystyle {\int }_{{\eta }_s}^{\eta }\frac{s-{\eta }_s}{s^2}\text{d}s}.\end{array}$ (1.12)

其中，$G_1\left({\rho }_s\right)=0$ ，且

${G^{\prime }}_1\left(\rho \right)=\frac{G_1\left(\rho \right)+P\left(\rho \right)-P\left({\rho }_s\right)}{\rho },{G^{\prime }}_1\left({\rho }_s\right)=0,{G^{″}}_1\left(\rho \right)=\frac{P^{\prime }\left(\rho \right)}{\rho }.$ (1.13)

改写(1.6)₂为

$u_{\tau }+\rho \left(\frac{P_y\left(\rho \right)}{\rho }-\frac{P_y\left({\rho }_s\right)}{{\rho }_s}\right)+\eta \left(\frac{{\eta }_y}{\eta }-\frac{{\left({\eta }_s\right)}_y}{{\eta }_s}\right)=\left(2\mu +\lambda \right){\left(\rho u_y\right)}_y.$ (1.14)

1989年，Mari Okada [7]探讨了一维可压缩Navier-Stokes方程自由边界问题的全局解的存在性以及长时间行为；1992年，Okada和Makino [8]考虑球对称情形的自由边界问题，得到全局弱解的存在性。2013年，黄金锐等人[9]证明了初始密度真空跳跃连接条件下，向列型液晶Ericksen-Leslie简化系统全局经典解的存在唯一性；2015年，黄金锐和丁时进[10]进一步得到了初始真空连续连接条件下全局弱解的存在性。基于对已有文献的研究，本文考虑一维Navier-Stokes-Smoluchowski系统耦合模型的自由边界问题，分析系统的基本能量不等式。

2. 主要结果

主要问题是建立系统(1.6)，(1.8)，(1.9)，(1.10)的基本能量率。

定理1：假定$\left(\rho ,u,\eta \right)$ 为系统(1.6)，(1.8)，(1.9)，(1.10)的全局强解，则存在常数$C\left(E_0,T\right)$ ，使得：

$\begin{array}{l}{\displaystyle {\int }_I\left[\frac{{\left|u\right|}^2}{2}+\frac{G_1\left(\rho \right)}{\rho }+\frac{G_2\left(\eta \right)}{\rho }+\frac{\eta \ln \eta }{\rho }+\frac{2\eta \Phi }{\rho }\right]}\left(\cdot ,\tau \right)\text{d}y+\frac{2\mu +\lambda }{2}{\displaystyle {\int }_0^T{\displaystyle {\int }_I\rho u_y^2\text{d}y\text{d}\sigma }}\\ +{\displaystyle {\int }_0^T{\displaystyle {\int }_{I}2\rho {\left|\frac{{\eta }_y}{\sqrt{\eta }}+\sqrt{\eta }{\Phi }_y\right|}^2\text{d}y\text{d}\sigma }}\le C\left(E_0,T\right).\end{array}$ (2.1)

其中，$\tau \in \left[0,T\right]$ 以及$E_0={\displaystyle {\int }_I\left[\frac{{\left|u_0\right|}^2}{2}+\frac{G_1\left({\rho }_0\right)}{{\rho }_0}+\frac{G_1\left({\eta }_0\right)}{{\rho }_0}+\frac{{\eta }_0\ln {\eta }_0}{{\rho }_0}+\frac{\text{2}{\eta }_0\Phi }{{\rho }_0}\right]\text{d}y}$ .

证明：证明主要分成三步。

步骤一：用u点乘(1.14)并在空间I上进行积分，分部积分并结合边界条件(1.9)，可得：

$\frac{\text{d}}{\text{d}\tau }{\displaystyle {\int }_I\frac{{\left|u\right|}^2}{2}\text{d}y}+{\displaystyle {\int }_I\rho u\left(\frac{P_y\left(\rho \right)}{\rho }-\frac{P_y\left({\rho }_s\right)}{{\rho }_s}\right)\text{d}y}+{\displaystyle {\int }_I\eta u\left(\frac{{\eta }_y}{\eta }-\frac{{\eta }_{sy}}{{\eta }_s}\right)\text{d}y}=-\left(2\mu +\lambda \right){\displaystyle {\int }_I\rho u_y^2}\text{d}y.$ (2.2)

根据定义(1.12)，有

${G^{\prime }}_1\left(\rho \right)={\displaystyle {\int }_{{\rho }_s}^{\rho }\frac{P^{\prime }\left(s\right)}{s}\text{d}s}={\displaystyle {\int }_{{\rho }_s}^{\rho }A\gamma s^{\gamma -2}\text{d}s}=\frac{A\gamma }{\gamma -1}\left({\rho }^{\gamma -1}-{\rho }_s^{\gamma -1}\right).$ (2.3)

从而利用分部积分以及(1.6)₁，可得

$\begin{array}{l}{\displaystyle {\int }_I\rho u\left(\frac{P_y\left(\rho \right)}{\rho }-\frac{P_y\left({\rho }_s\right)}{{\rho }_s}\right)\text{d}y}\\ =A\gamma {\displaystyle {\int }_I\rho u\left({\rho }^{\gamma -\text{2}}{\rho }_y-{\rho }_s^{\gamma -\text{2}}{\left({\rho }_s\right)}_y\right)\text{d}y}={\displaystyle {\int }_I\rho u{\left({G^{\prime }}_1\left(\rho \right)\right)}_y\text{d}y}\\ =-{\displaystyle {\int }_I{\left(\rho u\right)}_y{G^{\prime }}_1\left(\rho \right)\text{d}y}={\displaystyle {\int }_I\left(\frac{{\rho }_{\tau }}{\rho }-u{\rho }_y\right){G^{\prime }}_1\left(\rho \right)\text{d}y}\\ ={\displaystyle {\int }_I\frac{{\rho }_{\tau }}{\rho }{G^{\prime }}_1\left(\rho \right)\text{d}y}+{\displaystyle {\int }_I{u}_y{G}_{\text{1}}\left(\rho \right)\text{d}y}=\frac{\text{d}}{\text{d}\tau }{\displaystyle {\int }_I\frac{G_{\text{1}}\left(\rho \right)}{\rho }\text{d}y.}\end{array}$ (2.4)

另外，由于

$\begin{array}{l}{G}_2\left(\eta \right)=\eta {\left(\ln s+{\eta }_s\frac{1}{s}\right)|}_{{\eta }_s}^{\eta }=\eta \ln \frac{\eta }{{\eta }_s}+{\eta }_s-\eta ,\\ G_2^{\prime }\left(\eta \right)=\frac{G_2\left(\eta \right)+\eta -{\eta }_s}{\eta }=\ln \frac{\eta }{{\eta }_s},\end{array}$ (2.5)

可得

$\begin{array}{c}{\displaystyle {\int }_I\eta u\left(\frac{{\eta }_y}{\eta }-\frac{{\left({\eta }_s\right)}_y}{{\eta }_s}\right)\text{d}y}={\displaystyle {\int }_I\eta u{\left(\ln \frac{\eta }{{\eta }_s}\right)}_y\text{d}y}\\ ={\displaystyle {\int }_I\eta u{\left({G^{\prime }}_2\left(\eta \right)\right)}_y\text{d}y}\\ =-{\displaystyle {\int }_I{\left(\eta u\right)}_y{G^{\prime }}_2\left(\eta \right)\text{d}y}.\end{array}$ (2.6)

再次应用分部积分、(1.6)₃、边界条件(1.9)以及${G^{″}}_2\left(\eta \right)=\frac{1}{\eta }$ ，可得：

$\begin{array}{l}{\displaystyle {\int }_I\eta u\left(\frac{{\eta }_y}{\eta }-\frac{{\eta }_{sy}}{{\eta }_s}\right)\text{d}y}=-{\displaystyle {\int }_I{\left(\eta u\right)}_y{G^{\prime }}_2\left(\eta \right)\text{d}y}\\ ={\displaystyle {\int }_I\left[\frac{{\eta }_{\tau }}{\rho }-{\left(\rho {\eta }_y\right)}_y-{\left(\rho \eta {\Phi }_y\right)}_y\right]{G^{\prime }}_2\left(\eta \right)\text{d}y}-{\displaystyle {\int }_{I}u{\eta }_y{G^{\prime }}_2\left(\eta \right)\text{d}y}\\ ={\displaystyle {\int }_I\frac{{\left(G_2\left(\eta \right)\right)}_{\tau }}{\rho }\text{d}y}+{\displaystyle {\int }_I\left(\rho {\eta }_y+\rho \eta {\Phi }_y\right)}{\left({G^{\prime }}_2\left(\eta \right)\right)}_y\text{d}y-{\displaystyle {\int }_{I}u{\eta }_y{G^{\prime }}_2\left(\eta \right)\text{d}y}\\ =\frac{\text{d}}{\text{d}\tau }{\displaystyle {\int }_I\frac{G_2\left(\eta \right)}{\rho }\text{d}y}+{\displaystyle {\int }_I\frac{G_2\left(\eta \right)}{{\rho }^2}{\rho }_{\tau }\text{d}y}+{\displaystyle {\int }_I\left(\rho {\eta }_y+\rho \eta {\Phi }_y\right){G^{″}}_2\left(\eta \right){\eta }_y\text{d}y}-{\displaystyle {\int }_{I}u{\left(G_2\left(\eta \right)\right)}_y\text{d}y}\\ =\frac{\text{d}}{\text{d}\tau }{\displaystyle {\int }_I\frac{G_2\left(\eta \right)}{\rho }\text{d}y}-{\displaystyle {\int }_I{G}_2\left(\eta \right)u_y\text{d}y}+{\displaystyle {\int }_I\left(\rho {\eta }_y+\rho \eta {\Phi }_y\right)\frac{{\eta }_y}{\eta }\text{d}y}+{\displaystyle {\int }_I{u}_y{G}_2\left(\eta \right)\text{d}y}\\ =\frac{\text{d}}{\text{d}\tau }{\displaystyle {\int }_I\frac{G_2\left(\eta \right)}{\rho }\text{d}y}+{\displaystyle {\int }_I\frac{\rho {\eta }_y^2}{\eta }\text{d}y}+{\displaystyle {\int }_I\rho {\Phi }_y{\eta }_y\text{d}y.}\end{array}$ (2.7)

综合上述(2.2)~(2.5)，可得：

$\frac{\text{d}}{\text{d}\tau }{\displaystyle {\int }_I\left[\frac{{\left|u\right|}^2}{2}+\frac{G_1\left(\rho \right)}{\rho }+\frac{G_1\left(\eta \right)}{\rho }\right]\text{d}y}+\left(2\mu +\lambda \right){\displaystyle {\int }_I\rho u_y^2\text{d}y}+{\displaystyle {\int }_I\frac{\rho }{\eta }{\eta }_y^2\text{d}y}+{\displaystyle {\int }_I\rho {\Phi }_y{\eta }_y\text{d}y}=0.$ (2.8)

步骤二：将$\frac{1}{\rho }\left(\ln \eta +1\right)$ 乘以(1.6)₃并在空间I上积分可得：

$\begin{array}{l}{\displaystyle {\int }_I\frac{{\eta }_{\tau }}{\rho }\left(\ln \eta +1\right)\text{d}y}+{\displaystyle {\int }_I\eta u_y\left(\ln \eta +1\right)\text{d}y}-{\displaystyle {\int }_I{\left(\rho {\eta }_y\right)}_y\left(\ln \eta +1\right)\text{d}y}\\ -{\displaystyle {\int }_I{\left(\rho \eta {\Phi }_y\right)}_y\left(\ln \eta +1\right)\text{d}y}=0.\end{array}$ (2.9)

由于${\left(\ln \eta +1\right)}_y=\frac{{\eta }_y}{\eta }$ ，对上式分部积分并利用(1.6)₁以及(1.9)，整理后得到：

$\frac{\text{d}}{\text{d}\tau }{\displaystyle {\int }_I\frac{\eta \ln \eta }{\rho }\text{d}y}+{\displaystyle {\int }_I\left(\rho \frac{{\eta }_y^2}{\eta }+\rho {\eta }_y{\Phi }_y\right)\text{d}y}=-{\displaystyle {\int }_I\eta u_y\text{d}y}.$ (2.10)

结合(2.8)与(2.10)式得到：

$\begin{array}{l}\frac{\text{d}}{\text{d}\tau }{\displaystyle {\int }_I\left[\frac{{\left|u\right|}^2}{2}+\frac{G_1\left(\rho \right)}{\rho }+\frac{G_1\left(\eta \right)}{\rho }+\frac{\eta \ln \eta }{\rho }\right]\text{d}y}+\left(2\mu +\lambda \right){\displaystyle {\int }_I\rho u_y^2\text{d}y}\\ +2{\displaystyle {\int }_I\frac{\rho }{\eta }{\eta }_y^2\text{d}y}+2{\displaystyle {\int }_I\rho {\Phi }_y{\eta }_y\text{d}y}=-{\displaystyle {\int }_I\eta u_y\text{d}y}.\end{array}$ (2.11)

另外，将$\frac{\Phi }{\rho }$ 乘以(1.6)的第三条式子并在空间I上积分以及分部积分得到：

$\begin{array}{l}{\displaystyle {\int }_I\left[{\eta }_{\tau }\frac{\Phi }{\rho }+\eta u_y\Phi \right]\text{d}y}-{\displaystyle {\int }_I{\left(\rho {\eta }_y\right)}_y\Phi \text{d}y}-{\displaystyle {\int }_I{\left(\rho \eta {\Phi }_y\right)}_y\Phi \text{d}y}\\ =\frac{\text{d}}{\text{d}\tau }{\displaystyle {\int }_I\frac{\eta \Phi }{\rho }\text{d}y}+{\displaystyle {\int }_I\rho {\eta }_y{\Phi }_y\text{d}y}+{\displaystyle {\int }_I\rho \eta {\Phi }_y^2\text{d}y}=0.\end{array}$ (2.12)

结合上述(2.11)与(2.12)式可得：

$\begin{array}{l}\frac{\text{d}}{\text{d}\tau }{\displaystyle {\int }_I\left[\frac{{\left|u\right|}^2}{2}+\frac{G_1\left(\rho \right)}{\rho }+\frac{G_1\left(\eta \right)}{\rho }+\frac{\eta \ln \eta }{\rho }+\frac{\text{2}\eta \Phi }{\rho }\right]\text{d}y+\left(2\mu +\lambda \right){\displaystyle {\int }_I\rho u_y^2\text{d}y}}\\ +2{\displaystyle {\int }_I\rho {\left|\frac{{\eta }_y}{\sqrt{\eta }}+\sqrt{\eta }{\Phi }_y\right|}^2\text{d}y}=-{\displaystyle {\int }_I\eta u_y\text{d}y}.\end{array}$ (2.13)

步骤三：下文证明(2.11)的右端项满足如下不等式：

$-{\displaystyle {\int }_I\eta }{u}_y\text{d}y\le \frac{2\mu +\lambda }{2}{\displaystyle {\int }_I\rho u_y^2\text{d}y}+C\underset{x\in I}{\sup }{\left(\eta -{\eta }_s\right)}^2+C.$ (2.14)

证明：通过引入稳态解，可得

$-{\displaystyle {\int }_I\eta u_y\text{d}y}=-{\displaystyle {\int }_I\left(\eta -{\eta }_s\right)u_y\text{d}y}-{\displaystyle {\int }_I{\eta }_s{u}_y\text{d}y}=-{\displaystyle {\int }_I\frac{\eta -{\eta }_s}{\sqrt{\rho }}\sqrt{\rho }{u}_y\text{d}y}-{\displaystyle {\int }_I{\eta }_s{u}_y\text{d}y}.$ (2.15)

应用Hölder不等式，Cauchy不等式，可得：

$\begin{array}{c}-{\displaystyle {\int }_I\eta u_y\text{d}y}\le {\left({\displaystyle {\int }_I\frac{{\left|\eta -{\eta }_s\right|}^2}{\rho }\text{d}y}\right)}^{\frac{1}{2}}{\left({\displaystyle {\int }_I\rho u_y^2\text{d}y}\right)}^{\frac{1}{2}}-{\displaystyle {\int }_I\frac{{\eta }_s}{\sqrt{\rho }}}\sqrt{\rho }{u}_y\text{d}y\\ \le {\left({\displaystyle {\int }_I\frac{{\left|\eta -{\eta }_s\right|}^2}{\rho }\text{d}y}\right)}^{\frac{1}{2}}{\left({\displaystyle {\int }_I\rho u_y^2\text{d}y}\right)}^{\frac{1}{2}}+\frac{2\mu +\lambda }{4}\left({\displaystyle {\int }_I\rho u_y^2\text{d}y}\right)+C{\displaystyle {\int }_I\frac{\text{1}}{\rho }\text{d}y}.\end{array}$ (2.16)

根据${\rho }_{\tau }+{\rho }^2{u}_y=0$ ，可得：

$\begin{array}{l}{\displaystyle {\int }_I{\displaystyle {\int }_0^{\tau }{\left(\frac{1}{\rho }\right)}_{\tau }\text{d}\sigma \text{d}y}}={\displaystyle {\int }_I{\displaystyle {\int }_0^{\tau }{u}_y}}\text{d}\sigma \text{d}y,\\ {\displaystyle {\int }_I\left(\frac{1}{\rho }-\frac{1}{{\rho }_0}\right)\text{d}y}={\displaystyle {\int }_0^{\tau }{\displaystyle {\int }_I{u}_y\text{d}y\text{d}\sigma }}={\displaystyle {\int }_{\text{0}}^{\tau }{u|}_0^1\text{d}\sigma }=0.\end{array}$ (2.17)

从而${\displaystyle {\int }_I\frac{1}{\rho }\text{d}y}={\displaystyle {\int }_I\frac{1}{{\rho }_0}\text{d}y}$ 。

通过(2.16)式得到：

$\begin{array}{c}-{\displaystyle {\int }_I\eta u_y\text{d}y}\le \frac{2\mu +\lambda }{\text{2}}{\displaystyle {\int }_I\rho u_y^2\text{d}y}+C\underset{x\in I}{\sup }{\left(\eta -{\eta }_s\right)}^2{\displaystyle {\int }_I\frac{\text{1}}{\rho }\text{d}y}+C\\ \le \frac{2\mu +\lambda }{\text{2}}{\displaystyle {\int }_I\rho u_y^2\text{d}y}+C\underset{x\in I}{\sup }{\left(\eta -{\eta }_s\right)}^2+C.\end{array}$ (2.18)

整理(2.13)与(2.18)式得出：

$\begin{array}{l}\frac{\text{d}}{\text{d}\tau }{\displaystyle {\int }_I\left[\frac{{\left|u\right|}^2}{2}+\frac{G_1\left(\rho \right)}{\rho }+\frac{G_1\left(\eta \right)}{\rho }+\frac{\eta \ln \eta }{\rho }+\frac{\text{2}\eta \Phi }{\rho }\right]\text{d}y+\frac{2\mu +\lambda }{2}{\displaystyle {\int }_I\rho u_y^2\text{d}y}}\\ +2{\displaystyle {\int }_I\rho {\left|\frac{{\eta }_y}{\sqrt{\eta }}+\sqrt{\eta }{\Phi }_y\right|}^2\text{d}y}\le C\underset{x\in I}{\sup }{\left(\eta -{\eta }_s\right)}^2+C.\end{array}$ (2.19)

(2.19)两边关于时间方向在$\left[0,\tau \right]$ 上积分，并应用(1.10)，可得，对任意的$\tau \in \left[0,T\right]$ ，有

$\begin{array}{l}{\displaystyle {\int }_I\left[\frac{{\left|u\right|}^2}{2}+\frac{G_1\left(\rho \right)}{\rho }+\frac{G_1\left(\eta \right)}{\rho }+\frac{\eta \ln \eta }{\rho }+\frac{\text{2}\eta \Phi }{\rho }\right]\left(\cdot ,\tau \right)\text{d}y}\\ +\frac{2\mu +\lambda }{2}{\displaystyle {\int }_0^{\tau }{\displaystyle {\int }_I\rho u_y^2\text{d}y}\text{d}\sigma }+2{\displaystyle {\int }_0^{\tau }{{\displaystyle {\int }_I\rho \left|\frac{{\eta }_y}{\sqrt{\eta }}+\sqrt{\eta }{\Phi }_y\right|}}^2\text{d}y\text{d}\sigma }\\ \le {\displaystyle {\int }_I\left[\frac{{\left|u_0\right|}^2}{2}+\frac{G_1\left({\rho }_0\right)}{{\rho }_0}+\frac{G_1\left({\eta }_0\right)}{{\rho }_0}+\frac{{\eta }_0\ln {\eta }_0}{{\rho }_0}+\frac{\text{2}{\eta }_0\Phi }{{\rho }_0}\right]\text{d}y}+C{\displaystyle {\int }_0^{\tau }\underset{x\in I}{\sup }{\left(\eta -{\eta }_s\right)}^2\text{d}\sigma }+C\\ \le C\left(E_0,T\right).\end{array}$ (2.20)

定理1证毕。 □

基金项目

国家自然科学基金面上项目(11971357)；广东省自然科学基金面上项目(2023A1515010213)。

NOTES

^*通讯作者。