带有非局部效应耦合流体模型的局部经典解

The Local Classical Solution of Coupled Fluid Model with Nonlocal Effect

摘要: 本文研究了带有非局部速度趋同效应的耦合模型。该系统通过阻尼项对粒子速度和流体速度进行耦合，描述了群集粒子在粘性不可压流体中的相互作用。本文首先构造逼近方程和逼近解，并应用能量估计的方法，得到逼近解的一致先验估计，然后利用低阶范数收敛，证明具有非局部速度趋同效应的欧拉系统局部解的存在唯一性。

Abstract: This paper studied a coupling model with non-local velocity alignment. In this system, particle ve-locity and fluid velocity are coupled by damping term, and the interaction of cluster particles in viscous incompressible fluid is described. We first construct the approximation equation and the approximation solution, and the uniform a priori estimate of the approximation solution is obtained by using the method of energy coupling estimation, and then the existence and uniqueness of the local solution of the Euler system with non-local velocity convergence effect is proved by using the lower-order norm convergence.

1. 引言

本文研究如下带有非局部速度趋同效应的耦合流体模型：

$\{\begin{array}{l}{\partial }_t\rho +\nabla \cdot \left(\rho u\right)=0\\ {\partial }_{t}u+\left(u\cdot \nabla \right)u=-K_1{\displaystyle {\int }_{{\mathbb{T}}^d}\Gamma \left(x-y\right)\left(u\left(x\right)-u\left(y\right)\right)\rho \left(y\right)\text{d}y+K_2\left(v-u\right),\left(x,t\right)\in {\mathbb{T}}^d\times \left(0,+\infty \right)}\\ {\partial }_{t}v+v\cdot \nabla v+\nabla p=\mu \Delta v+K_2\rho \left(u-v\right),\nabla \cdot v=0,\end{array}$ (1)

满足初值条件：

$\{\begin{array}{l}\left(\rho ,u,v\right)\left(x,0\right)=\left({\rho }_0,u_0,v_0\right)\\ \nabla \cdot v_0=0,\end{array}x\in {\mathbb{T}}^d,$ (2)

其中d为空间维数， ${\mathbb{T}}^d={ℝ}^d/{\mathbb{Z}}^d\left(d\ge 1\right)$ 是d维空间中的周期环形区域， $\rho ,u,v$ 分别表示粒子的密度、粒子的速度和不可压流体的速度， $K_i\left(i=1,2\right)$ 分别表示粒子与粒子，流体与流体相互作用的非负耦合强度常数，速度趋同效应中的通信权重函数 $\Gamma \left(x-y\right)\in L^1\left({\mathbb{T}}^d\right)$ 是一个非常数且非正定矩阵，它反映了系统内部的非局部相互作用。

带有非局部效应的欧拉模型，是反应复杂系统运动的微观粒子模型的宏观表达。此类模型的研究在社会学，经济学和自然科学等不同领域都有重要意义，例如鸟、鱼等小动物的聚集现象，物质的运输或行人的流动，半导体或气体动力学中的带电粒子的运动等。本文研究的这类模型通过阻尼项对粒子速度和流体速度进行耦合，描述了多粒子复杂系统在不压缩流体中的运动状态。

为介绍模型的非局部效应，首先从微观粒子模型出发：

$\begin{array}{l}\frac{\text{d}}{\text{d}t}{x}_i=v_i\\ \frac{\text{d}}{\text{d}t}{v}_i=\frac{1}{N}{\displaystyle \underset{j=1}{\overset{N}{\sum }}F\left(t,x_i-x_j,v_i-v_j\right)}+G\left(t,x_i,v_i\right)\end{array}$ (3)

其中 $\left(x_i,v_i\right)$ 是第i个粒子在d维空间中的位置和速度， $G\left(t,x_i,v_i\right)$ 是第i个粒子的自加速度或阻尼力。 $\frac{1}{N}F\left(t,x_i-x_j,v_i-v_j\right)$ 是第i和第j个粒子之间的弱相互作用力， $F\left(t,x_i-x_j,v_i-v_j\right)$ 常见有两种类型，一种是导致个体之间的吸引和排斥现象的相互作用力，例如三维库伦力 $F\left(t,x_i-x_j,v_i-v_j\right)=\frac{x_i-x_j}{{\left|x_i-x_j\right|}^3}$ ( [1] )；另一种是导致粒子产生速度趋同效应的速度趋同力，例如在描述小动物聚集现象的Cucker-Smale模型中 $F\left(t,x_i-x_j,v_i-v_j\right)=-\left(v_i-v_j\right)\varphi \left(\left|x_i-x_j\right|\right)$ ( [2] )，在速度趋同力的作用下，系统中个体受其他个体的影响调整自己的速度，系统中个体在长时间过程中速度趋于一致。更多带有非局部效应的微观模型可以在 [3] [4] [5] 中找到。

通过取平均场极限，并对系统的密度分布函数作Mono-Kinetic拟设，即设 $f\left(t,x,v\right)=\rho \left(t,x\right){\delta }_{u\left(t,x\right)}\left(v\right)$，可得到宏观无压欧拉系统：

$\begin{array}{l}{\rho }_t+\nabla \cdot \left(\rho u\right)=0\\ {\left(\rho u\right)}_t+\left(\rho u\cdot \nabla \right)u=\rho {\displaystyle \int F\left(t,x-y,u\left(x\right)-u\left(y\right)\right)\rho \left(t,y\right)\text{d}y}+\rho G\left(t,x,u\right)\end{array}$ (4)

其中 $\rho$ 是密度，u是粒子速度，v是流体速度。

关于带有非局部效应的无压欧拉系统(1)，陈丽教授和Göttlich ( [6] )等人研究了二维空间中复杂材料流动的流体力学模型，在模型中非局部效应满足一定的结构条件，光滑解的局部存在性被得到。Carrillo

和Choi等人对一维简化模型( $G=-u,F=\frac{x-y}{\left|x-y\right|}-2\left(x-y\right)$ )解的长时间渐进行为进行分析，并给出了阻尼

抑制激波形成时，初值需要满足的临界条件( [7] )。当模型(4)中不含阻尼项，但非局部排斥力和速度校准力同时存在时，Kiselev和Tan等人通过一系列的工作( [8] [9] [10] )证明，如果速度校准效应的积分核非负对称且具有一定光滑性，则模型具有全局光滑解，这意味着速度较准项有助于防止激波的形成。

而对含有压力项的欧拉方程组，Feireisl等人采用凸积分的方法，证明了一维空间中的Cauchy问题存在无穷多的弱解，在这里模型带有阻尼和非局部速度趋同效应 [11]。最近，Choi研究了带有线性压力项和非局部速度较准项的欧拉系统，给出了经典解的全局存在性 [12]。进一步的，Carrillo等人针对含有非局部效应的可压粘性流体力学模型，得到了弱解的全局存在性，并对解的长时间渐进行为进行了分析 [13]。

对于本文这类欧拉方程与不可压缩Navier-Stokes方程相耦合的流体模型，Ha和他的合作者们在 [14] 中研究了经典解的适定性。值得一提的是，在他们的模型中，非局部效应的通讯权重函数满足对称性，正则性，且具有恒正的下界，在得到解的能量估计中起到积极地作用。与之相比，我们模型中的 $\Gamma \left(x-y\right)\in L^1\left({\mathbb{T}}^d\right)$ 是一个非常数且非正定矩阵，这给解的正则性估计带来了一定的阻碍。

我们的主要结果如下：

定理1假设 $s>\frac{d}{2}+1$，初值条件满足 $\left({\rho }_0>0,u_0,v_0\right)\in H^s\left({\mathbb{T}}^d\right)\times H^{s+2}\left({\mathbb{T}}^d\right)\times H^{s+1}\left({\mathbb{T}}^d\right)$，则系统(1)~(2)存在唯一的局部经典解 $\left(\rho ,u,v\right)$，满足如下正则性，

$\left(\rho ,u,v\right)\in C\left(\left[0,T_0\right];L^2\left({\mathbb{T}}^d\right)\right)\times C\left(\left[0,T_0\right];H^1\left({\mathbb{T}}^d\right)\right)\times C\left(\left[0,T_0\right];L^2\left({\mathbb{T}}^d\right)\right),$

其中 $T_0$ 与初值有关。

在本文中，我们用C表示与 $\mu ,s,d,K_1,K_2$ 有关的正常数。对于任何非负整数s， $H^s\left({\mathbb{T}}^d\right)$ 表示 ${\mathbb{T}}^d$ 上的s阶 $L^2$ -Sobolev空间，为表示简洁，记 $H^s:=H^s\left({\mathbb{T}}^d\right)$。我们在第2节中构造了逼近解，并在第3节得到了逼近解的一致能量估计，进一步的在第4节证明了逼近解的柯西收敛性，在第5节中完成了定理的证明。

2. 逼近解的构造

在这一节中，我们将应用迭代法构造逼近解：

零阶逼近：

$\left({\rho }^0,u^0,v^0\right)\left(x,t\right)=\left({\rho }_0,u_0,v_0\right).$ (5)

$k+1$ 阶逼近：假设k阶逼近解 $\left({\rho }^k,u^k,v^k\right)\left(x,t\right),k\ge 1$ 已给出，则定义 $k+1$ 阶逼近解 $\left({\rho }^{k+1},u^{k+1},v^{k+1}\right)\left(x,t\right)$ 为下列线性系统的解：

${\partial }_t{\rho }^{k+1}+u^k\cdot \nabla {\rho }^{k+1}+{\rho }^{k+1}\nabla \cdot u^k=0$ (6)

${\partial }_t{v}^{k+1}+v^k\cdot \nabla v^{k+1}+\nabla p^{k+1}=\mu \Delta v^{k+1}+K_2{\rho }^{k+1}\left(u^k-v^k\right),\nabla \cdot v^{k+1}=0$ (7)

${\partial }_t{u}^{k+1}+u^k\cdot \nabla u^{k+1}=-K_1{\displaystyle {\int }_{{\mathbb{T}}^d}\Gamma \left(x-y\right)\left(u^k\left(x\right)-u^k\left(y\right)\right){\rho }^{k+1}\left(y\right)\text{d}y+K_2\left(v^{k+1}-u^k\right)}$ (8)

$\left({\rho }^{k+1},u^{k+1},v^{k+1}\right)\left(x,0\right)=\left({\rho }_0,u_0,v_0\right),$ (9)

其中 $\Gamma \left(x-y\right)\in L^1\left({\mathbb{T}}^d\right)$ 为非常数非正定矩阵。应用文献 [15] 中的多维双曲型方程的线性理论，可以得到逼近问题(6)~(9)存在局部经典解 $\left({\rho }^{k+1},u^{k+1},v^{k+1}\right)$。

3. 一致能量估计

为论文中证明部分的表述方便，定义常量M

$M=\sqrt{{\Vert {\rho }_0\Vert }_{H^s}^2+{\Vert u_0\Vert }_{H^{s+2}}^2+{\Vert v_0\Vert }_{H^{s+1}}^2+\delta }$， (10)

其中 $\delta >0$ 为常数。进一步的，选择常数 $T_0>0$，当 $T_0$ 足够小时，可满足

$\begin{array}{l}\left({\text{e}}^{CM{T}_0}-1\right){\Vert {\rho }_0\Vert }_{H^s}^2+\left({\text{e}}^{C\left(M^2+M\right)T_0}-1\right)\Vert v_0\Vert {|}_{H^{s+1}}^2\\ +\left(e^{C\left(M^2+M\right)T_0}-1\right){\Vert u_0\Vert }_{H^{s+1}}^2+{\text{e}}^{C\left(M^2+M\right)T_0}\left(C{M}^2+CM\right)T_0\le \delta .\end{array}$ (11)

其中 $C>0$ 是引理1证明中出现的常数。

引理1假设 $\left({\rho }^k,u^k,v^k\right)$ 是由线性迭代方程组(6)~(9)构造的一组逼近解序列，则对任意 $k\ge 0$，如下估计成立：

$\underset{0\le t\le T_0}{\sup }\left({\Vert {\rho }^k\Vert }_{H^s}+{\Vert u^k\Vert }_{H^{s+2}}+{\Vert v^k\Vert }_{H^{s+1}}\right)\le M.$ (12)

证明我们应用 [16] 中的方法证明引理1中的结论，证明分为如下两步。

Step 1. (初始步骤)：由(5)和(10)易知：

$\underset{0\le t\le T_0}{\sup }\left({\Vert u^0\Vert }_{H^{s+2}}+{\Vert v^0\Vert }_{H^{s+1}}\right)\le M.$ (13)

Step 2. (归纳步骤)：假设：

$\underset{0\le t\le T_0}{\sup }\left({\Vert u^k\Vert }_{H^{s+2}}+{\Vert v^k\Vert }_{H^{s+1}}\right)\le M,$ (14)

其中： $T_0$ 和M是(10)和(11)确定的常数，我们将会证明：

$\underset{0\le t\le T_0}{\sup }\left({\Vert {\rho }^{k+1}\Vert }_{H^s}+{\Vert u^{k+1}\Vert }_{H^{s+2}}+{\Vert v^{k+1}\Vert }_{H^{s+1}}\right)\le M.$ (15)

Step 2.1. ( ${\rho }^{k+1}$ 的估计)：首先用 ${\rho }^{k+1}$ 乘以方程(6)两端，并在 ${\mathbb{T}}^d$ 上关于变量x积分，可以得到

$\begin{array}{c}\frac{1}{2}\frac{\text{d}}{\text{d}t}{\Vert {\rho }^{k+1}\Vert }_{L^2}^2=\frac{1}{2}{\displaystyle {\int }_{{\mathbb{T}}^d}{u}^k}\cdot \nabla {\left({\rho }^{k+1}\right)}^2\text{d}x-{\displaystyle {\int }_{{\mathbb{T}}^d}\left(\nabla \cdot u^k\right)}{\left|{\rho }^{k+1}\right|}^2\text{d}x\\ =-\frac{1}{2}{\displaystyle {\int }_{{\mathbb{T}}^d}\left(\nabla \cdot u^k\right)}{\left|{\rho }^{k+1}\right|}^2\text{d}x\le C{\Vert \nabla u^k\Vert }_{L^{\infty }}{\Vert {\rho }^{k+1}\Vert }_{L^2}^2\le CM{\Vert {\rho }^{k+1}\Vert }_{L^2}^2.\end{array}$ (16)

接下来，对任意 $1\le r\le s$，关于方程(6)作用算子 ${\nabla }^r$，然后两端乘以 ${\nabla }^r{\rho }^{k+1}$，并在 ${\mathbb{T}}^d$ 上关于变量x积分，有

$\begin{array}{l}\frac{1}{2}\frac{\text{d}}{\text{d}t}{\Vert {\nabla }^r{\rho }^{k+1}\Vert }_{L^2}^2=-{\displaystyle {\int }_{{\mathbb{T}}^d}\left[{\nabla }^r\left(u^k\cdot \nabla {\rho }^{k+1}\right)-u^k\cdot {\nabla }^{r+1}{\rho }^{k+1}\right]}\cdot {\nabla }^r{\rho }^{k+1}\text{d}x\\ -{\displaystyle {\int }_{{\mathbb{T}}^d}{u}^k\cdot {\nabla }^{r+1}{\rho }^{k+1}}\cdot {\nabla }^r{\rho }^{k+1}\text{d}x\\ -{\displaystyle {\int }_{{\mathbb{T}}^d}\left[{\nabla }^r\left({\rho }^{k+1}\nabla \cdot u^k\right)-{\rho }^{k+1}{\nabla }^r\left(\nabla \cdot u^k\right)\right]}\cdot {\nabla }^r{\rho }^{k+1}\text{d}x\\ -{\displaystyle {\int }_{{\mathbb{T}}^d}{\rho }^{k+1}{\nabla }^r\left(\nabla \cdot u^k\right)}\cdot {\nabla }^r{\rho }^{k+1}\text{d}x\\ ={\displaystyle \underset{i=1}{\overset{4}{\sum }}{I}_i.}\end{array}$ (17)

利用Hölder不等式，Moser不等式，我们可以依次估计 $I_i,i=1,2,3,4$，

$\begin{array}{c}{I}_1=-{\displaystyle {\int }_{{\mathbb{T}}^d}\left[{\nabla }^r\left(u^k\cdot \nabla {\rho }^{k+1}\right)-u^k\cdot {\nabla }^{r+1}{\rho }^{k+1}\right]}\cdot {\nabla }^r{\rho }^{k+1}\text{d}x\\ \le {\Vert {\nabla }^r\left(u^k\cdot \nabla {\rho }^{k+1}\right)-u^k\cdot {\nabla }^{r+1}{\rho }^{k+1}\Vert }_{L^2}{\Vert {\nabla }^r{\rho }^{k+1}\Vert }_{L^2}\\ \le C\left({\Vert \nabla u^k\Vert }_{L^{\infty }}{\Vert {\nabla }^r{\rho }^{k+1}\Vert }_{L^2}+{\Vert \nabla {\rho }^{k+1}\Vert }_{L^{\infty }}{\Vert {\nabla }^r{u}^k\Vert }_{L^2}\right){\Vert {\nabla }^r{\rho }^{k+1}\Vert }_{L^2}\\ \le CM\left({\Vert {\nabla }^r{\rho }^{k+1}\Vert }_{L^2}+{\Vert \nabla {\rho }^{k+1}\Vert }_{H^{s-1}}\right){\Vert {\nabla }^r{\rho }^{k+1}\Vert }_{L^2};\end{array}$ (18)

$\begin{array}{c}{I}_2=-{\displaystyle {\int }_{{\mathbb{T}}^d}{u}^k\cdot {\nabla }^{r+1}{\rho }^{k+1}}\cdot {\nabla }^r{\rho }^{k+1}\text{d}x\\ =\frac{1}{2}{\displaystyle {\int }_{{\mathbb{T}}^d}\left(\nabla \cdot u^k\right)}{\left|{\nabla }^r{\rho }^{k+1}\right|}^2\text{d}x\\ \le C{\Vert \nabla \cdot u^k\Vert }_{L^{\infty }}{\Vert {\nabla }^r{\rho }^{k+1}\Vert }_{L^2}^2\le CM{\Vert {\nabla }^r{\rho }^{k+1}\Vert }_{L^2}^2;\end{array}$ (19)

$\begin{array}{c}{I}_3=-{\displaystyle {\int }_{{\mathbb{T}}^d}\left[{\nabla }^r\left({\rho }^{k+1}\nabla \cdot u^k\right)-{\rho }^{k+1}{\nabla }^r\left(\nabla \cdot u^k\right)\right]}\cdot {\nabla }^r{\rho }^{k+1}\text{d}x\\ \le {\Vert {\nabla }^r\left({\rho }^{k+1}\nabla \cdot u^k\right)-{\rho }^{k+1}{\nabla }^r\left(\nabla \cdot u^k\right)\Vert }_{L^2}{\Vert {\nabla }^r{\rho }^{k+1}\Vert }_{L^2}\\ \le C\left({\Vert \nabla {\rho }^{k+1}\Vert }_{L^{\infty }}{\Vert {\nabla }^r{u}^k\Vert }_{L^2}+{\Vert \nabla u^k\Vert }_{L^{\infty }}{\Vert {\nabla }^r{\rho }^{k+1}\Vert }_{L^2}\right){\Vert {\nabla }^r{\rho }^{k+1}\Vert }_{L^2}\\ \le CM\left({\Vert \nabla {\rho }^{k+1}\Vert }_{H^{s-1}}+{\Vert {\nabla }^r{\rho }^{k+1}\Vert }_{L^2}\right){\Vert {\nabla }^r{\rho }^{k+1}\Vert }_{L^2};\end{array}$ (20)

$\begin{array}{c}{I}_4=-{\displaystyle {\int }_{{\mathbb{T}}^d}{\rho }^{k+1}{\nabla }^r\left(\nabla \cdot u^k\right)}\cdot {\nabla }^r{\rho }^{k+1}\text{d}x\\ \le C{\Vert {\rho }^{k+1}\Vert }_{L^{\infty }}{\Vert {\nabla }^{r+1}{u}^k\Vert }_{L^2}{\Vert {\nabla }^r{\rho }^{k+1}\Vert }_{L^2}\\ \le CM{\Vert {\rho }^{k+1}\Vert }_{H^s}{\Vert {\nabla }^r{\rho }^{k+1}\Vert }_{L^2}.\end{array}$ (21)

在这里，我们用到了Sobolev不等式：

${\Vert \nabla u\Vert }_{L^{\infty }}\le {\Vert \nabla u\Vert }_{H^{\left[\frac{d}{2}\right]+1}}\le {\Vert \nabla u\Vert }_{H^{s-1}},对s>\frac{d}{2}+1.$ (22)

联立 $I_i,i=1,2,3,4$，我们可以得到

$\frac{1}{2}\frac{\text{d}}{\text{d}t}{\Vert {\nabla }^r{\rho }^{k+1}\Vert }_{L^2}^2\le CM\left({\Vert \nabla {\rho }^{k+1}\Vert }_{H^{s-1}}+{\Vert {\rho }^{k+1}\Vert }_{H^s}+{\Vert {\nabla }^r{\rho }^{k+1}\Vert }_{L^2}\right){\Vert {\nabla }^r{\rho }^{k+1}\Vert }_{L^2}.$ (23)

对(23)关于r从1到s求和，并与(16)相加，有

$\frac{\text{d}}{\text{d}t}{\Vert {\rho }^{k+1}\Vert }_{H^s}^2\le CM{\Vert {\rho }^{k+1}\Vert }_{H^s}^2.$ (24)

应用Gronwall不等式，

$\underset{0\le t\le T_0}{\sup }{\Vert {\rho }^{k+1}\Vert }_{H^s}^2\le {\text{e}}^{CM{T}_0}{\Vert {\rho }_0\Vert }_{H^s}^2\le M^2.$ (25)

由(10)和(11)中M和 $T_0$ 的选取，可得 ${\rho }^{k+1}$ 的估计：

$\underset{0\le t\le T_0}{\sup }{\Vert {\rho }^{k+1}\Vert }_{H^s}\le M.$ (26)

Step 2.2. ( $v{}^{k+1}$ 的估计)：用 $v{}^{k+1}$ 乘以方程(7)的两端，并在 ${\mathbb{T}}^d$ 上关于变量x积分，利用Hölder不等式，Young’s不等式，Sobolev不等式，以及 $\nabla \cdot v=0$，我们得到

$\begin{array}{l}\frac{1}{2}\frac{\text{d}}{\text{d}t}{\Vert v^{k+1}\Vert }_{L^2}^2+\mu {\displaystyle {\int }_{{\mathbb{T}}^d}{\left|\nabla v^{k+1}\right|}^2\text{d}x}\\ =-{\displaystyle {\int }_{{\mathbb{T}}^d}{v}^k}\cdot \nabla v^{k+1}\cdot v^{k+1}\text{d}x+K_2{\displaystyle {\int }_{{\mathbb{T}}^d}{\rho }^{k+1}{v}^{k+1}\left(u^k-v^k\right)\text{d}x}\\ \le \frac{1}{2}{\displaystyle {\int }_{{\mathbb{T}}^d}{\left|v^{k+1}\right|}^2}\left(\nabla \cdot v^k\right)\text{d}x+K_2\left({\Vert u^k\Vert }_{L^{\infty }}+{\Vert v^k\Vert }_{L^{\infty }}\right){\displaystyle {\int }_{{\mathbb{T}}^d}{\rho }^{k+1}{v}^{k+1}\text{d}x}\\ \le CM{\Vert {\rho }^{k+1}\Vert }_{L^2}{\Vert v^{k+1}\Vert }_{L^2}\le C{M}^2{\Vert v^{k+1}\Vert }_{L^2}\le C{M}^2{\Vert v^{k+1}\Vert }_{L^2}^2+C{M}^2.\end{array}$ (27)

对方程(7)作用算子 ${\nabla }^r$ ( $1\le r\le s+1$ )，然后两端乘以 ${\nabla }^r{v}^{k+1}$，并在 ${\mathbb{T}}^d$ 上关于变量x积分，有

$\begin{array}{l}\frac{1}{2}\frac{\text{d}}{\text{d}t}{\Vert {\nabla }^r{v}^{k+1}\Vert }_{L^2}^2+\mu {\displaystyle {\int }_{{\mathbb{T}}^d}{\left|{\nabla }^{r+1}{v}^{k+1}\right|}^2\text{d}x}\\ =-{\displaystyle {\int }_{{\mathbb{T}}^d}{v}^k\cdot {\nabla }^r\left(\nabla v^{k+1}\right)}\cdot {\nabla }^r{v}^{k+1}\text{d}x-{\displaystyle {\int }_{{\mathbb{T}}^d}\left[{\nabla }^r\left(v^k\cdot \nabla v^{k+1}\right)-v^k\cdot {\nabla }^r\left(\nabla v^{k+1}\right)\right]}\cdot {\nabla }^r{v}^{k+1}\text{d}x\\ +K_2{\displaystyle {\int }_{{\mathbb{T}}^d}{\nabla }^r\left({\rho }^{k+1}{u}^k\right)}\cdot {\nabla }^r{v}^{k+1}\text{d}x-K_2{\displaystyle {\int }_{{\mathbb{T}}^d}{\nabla }^r\left({\rho }^{k+1}{v}^k\right)}\cdot {\nabla }^r{v}^{k+1}\text{d}x\\ ={\displaystyle \underset{i=1}{\overset{4}{\sum }}{J}_i.}\end{array}$ (28)

接下来我们依次估计 $J_i,i=1,2,3,4$

$J_1=-{\displaystyle {\int }_{{\mathbb{T}}^d}{v}^k\cdot {\nabla }^r\left(\nabla v^{k+1}\right)}\cdot {\nabla }^r{v}^{k+1}\text{d}x=\frac{1}{2}{\displaystyle {\int }_{{\mathbb{T}}^d}\left(\nabla \cdot v^k\right)}{\left|{\nabla }^r{v}^{k+1}\right|}^2\text{d}x=0;$ (29)

$\begin{array}{l}{J}_2=-{\displaystyle {\int }_{{\mathbb{T}}^d}\left[{\nabla }^r\left(v^k\cdot \nabla v^{k+1}\right)-v^k\cdot {\nabla }^r\left(\nabla v^{k+1}\right)\right]}\cdot {\nabla }^r{v}^{k+1}\text{d}x\\ \le C{\Vert {\nabla }^r\left(v^k\cdot \nabla v^{k+1}\right)-v^k\cdot {\nabla }^r\left(\nabla v^{k+1}\right)\Vert }_{L^2}{\Vert {\nabla }^r{v}^{k+1}\Vert }_{L^2}\\ \le C\left({\Vert \nabla v^k\Vert }_{L^{\infty }}{\Vert {\nabla }^r{v}^{k+1}\Vert }_{L^2}+{\Vert \nabla v^{k+1}\Vert }_{L^{\infty }}{\Vert {\nabla }^r{v}^k\Vert }_{L^2}\right){\Vert {\nabla }^r{v}^{k+1}\Vert }_{L^2}\\ \le CM\left({\Vert {\nabla }^r{v}^{k+1}\Vert }_{L^2}^2+{\Vert v^{k+1}\Vert }_{H^{s+1}}^2\right);\end{array}$ (30)

$\begin{array}{l}{J}_3=K_2{\displaystyle {\int }_{{\mathbb{T}}^d}{\nabla }^r\left({\rho }^{k+1}{u}^k\right)}\cdot {\nabla }^r{v}^{k+1}\text{d}x\le C{\Vert {\nabla }^r\left({\rho }^{k+1}{u}^k\right)\Vert }_{L^2}{\Vert {\nabla }^r{v}^{k+1}\Vert }_{L^2}\\ \le C\left({\Vert {\rho }^{k+1}\Vert }_{L^{\infty }}{\Vert {\nabla }^r{u}^k\Vert }_{L^2}+{\Vert u^k\Vert }_{L^{\infty }}{\Vert {\nabla }^r{\rho }^{k+1}\Vert }_{L^2}\right){\Vert {\nabla }^r{v}^{k+1}\Vert }_{L^2}\\ \le C{M}^2{\Vert {\nabla }^r{v}^{k+1}\Vert }_{L^2}\le C{M}^2{\Vert {\nabla }^r{v}^{k+1}\Vert }_{L^2}^2+C{M}^2;\end{array}$ (31)

$\begin{array}{l}{J}_4=-K_2{\displaystyle {\int }_{{\mathbb{T}}^d}{\nabla }^r\left({\rho }^{k+1}{v}^k\right)}\cdot {\nabla }^r{v}^{k+1}\text{d}x\le C{\Vert {\nabla }^r\left({\rho }^{k+1}{v}^k\right)\Vert }_{L^2}{\Vert {\nabla }^r{v}^{k+1}\Vert }_{L^2}\\ \le C\left({\Vert {\rho }^{k+1}\Vert }_{L^{\infty }}{\Vert {\nabla }^r{v}^k\Vert }_{L^2}+{\Vert v^k\Vert }_{L^{\infty }}{\Vert {\nabla }^r{\rho }^{k+1}\Vert }_{L^2}\right){\Vert {\nabla }^r{v}^{k+1}\Vert }_{L^2}\\ \le C{M}^2{\Vert {\nabla }^r{v}^{k+1}\Vert }_{L^2}\le C{M}^2{\Vert {\nabla }^r{v}^{k+1}\Vert }_{L^2}^2+C{M}^2.\end{array}$ (32)

在这里用到了Hölder不等式，Young’s不等式，Sobolev不等式，Moser不等式。将 $J_i,i=1,2,3,4,5$,的估计带入(28)式，我们得到

$\frac{\text{d}}{\text{d}t}{\Vert {\nabla }^r{v}^{k+1}\Vert }_{L^2}^2+\mu {\displaystyle {\int }_{{\mathbb{T}}^d}{\left|{\nabla }^{r+1}{v}^{k+1}\right|}^2\text{d}x}\le \left(C{M}^2+CM\right){\Vert {\nabla }^r{v}^{k+1}\Vert }_{L^2}^2+CM{\Vert v^{k+1}\Vert }_{H^{s+1}}^2+C{M}^2.$ (33)

对(33)的r从0到 $s+1$ 求和，并与(27)式联立，

$\frac{\text{d}}{\text{d}t}{\Vert v^{k+1}\Vert }_{H^{s+1}}^2+\mu {\displaystyle \underset{r=0}{\overset{s+1}{\sum }}{{\displaystyle {\int }_{{\mathbb{T}}^d}\left|{\nabla }^{r+1}{v}^{k+1}\right|}}^2\text{d}x}\le \left(C{M}^2+CM\right){\Vert v^{k+1}\Vert }_{H^{s+1}}^2+C{M}^2.$ (34)

对(34)关于时间变量t从0到 $T_0$ 积分，可以得到

${\Vert v^{k+1}\Vert }_{H^{s+1}}^2+{\displaystyle {\int }_0^{T_0}\mu {\displaystyle \underset{r=0}{\overset{s+1}{\sum }}{{\displaystyle {\int }_{{\mathbb{T}}^d}\left|{\nabla }^{r+1}{v}^{k+1}\right|}}^2\text{d}x}\text{d}t}\le {\Vert v_0\Vert }_{H^{s+1}}^2+{\displaystyle {\int }_0^{T_0}\left(\left(C{M}^2+CM\right){\Vert v^{k+1}\Vert }_{H^{s+1}}^2+C{M}^2\right)\text{d}t},$ (35)

应用Gronwall不等式，我们得到

$\underset{0\le t\le T_0}{\sup }{\Vert v^{k+1}\Vert }_{H^{s+1}}^2+{\displaystyle {\int }_0^{T_0}\mu {\displaystyle \underset{r=0}{\overset{s+1}{\sum }}{{\displaystyle {\int }_{{\mathbb{T}}^d}\left|{\nabla }^{r+1}{v}^{k+1}\right|}}^2\text{d}x}\text{d}t}\le {\text{e}}^{C\left(M^2+M\right)T_0}\left({\Vert v_0\Vert }_{H^{s+1}}^2+C{M}^2{T}_0\right).$ (36)

由(10)和(11)中M和 $T_0$ 的选取，可得 $v^{k+1}$ 的估计：

$\underset{0\le t\le T_0}{\sup }{\Vert v^{k+1}\Vert }_{H^{s+1}}\le M.$ (37)

Step 2.3. ( $u^{k+1}$ 的估计)：用 $u^{k+1}$ 乘以方程(8)的两端，并在 ${\mathbb{T}}^d$ 上关于变量x积分，

$\begin{array}{c}\frac{1}{2}\frac{\text{d}}{\text{d}t}{\Vert u^{k+1}\Vert }_{L^2}^2=-{\displaystyle {\int }_{{\mathbb{T}}^d}{u}^k\cdot \nabla u^{k+1}\cdot u^{k+1}\text{d}x}+K_2{\displaystyle {\int }_{{\mathbb{T}}^d}{u}^{k+1}\left(v^{k+1}-u^k\right)\text{d}x}\\ -K_1{\displaystyle {\iint }_{{\mathbb{T}}^{2d}}\Gamma \left(x-y\right)\left(u^k\left(x\right)-u^k\left(y\right)\right)u^{k+1}\left(x\right){\rho }^{k+1}\left(y\right)\text{d}x\text{d}y}\\ ={\displaystyle \underset{i=1}{\overset{3}{\sum }}{I}_i.}\end{array}$ (38)

应用Hölder不等式，Young’s不等式，Sobolev不等式，以及(26)和(37)，我们估计 $I_i,i=1,2,3$,

$I_1=-{\displaystyle {\int }_{{\mathbb{T}}^d}{u}^k\cdot \nabla u^{k+1}\cdot u^{k+1}\text{d}x}=\frac{1}{2}{\displaystyle {\int }_{{\mathbb{T}}^d}\left(\nabla \cdot u^k\right){\left|u^{k+1}\right|}^2\text{d}x}=0;$ (39)

$\begin{array}{l}{I}_2=K_1{\displaystyle {\iint }_{{\mathbb{T}}^{2d}}\Gamma \left(x-y\right)\left(u^k\left(x\right)-u^k\left(y\right)\right)u^{k+1}\left(x\right){\rho }^{k+1}\left(y\right)\text{d}x\text{d}y}\\ \le C\left({\Vert u^k\left(y\right)\Vert }_{L^{\infty }}+{\Vert u^k\left(x\right)\Vert }_{L^{\infty }}\right){\displaystyle {\iint }_{{\mathbb{T}}^{2d}}\Gamma \left(x-y\right)u^{k+1}\left(x\right){\rho }^{k+1}\left(y\right)\text{d}x\text{d}y}\\ \le CM{\Vert \Gamma \Vert }_{L^1}{\Vert {\rho }^{k+1}\Vert }_{L^2}{\Vert u^{k+1}\Vert }_{L^2}\le C{M}^2{\Vert u^{k+1}\Vert }_{L^2}\le C{M}^2{\Vert u^{k+1}\Vert }_{L^2}^2+C{M}^2;\end{array}$ (40)

$\begin{array}{l}{I}_3=K_2{\displaystyle {\int }_{{\mathbb{T}}^d}{u}^{k+1}\left(v^{k+1}-u^k\right)\text{d}x}\le C\left({\Vert u^{k+1}\Vert }_{L^2}{\Vert v^{k+1}\Vert }_{L^2}+{\Vert u^{k+1}\Vert }_{L^2}{\Vert u^k\Vert }_{L^2}\right)\\ \le C\left({\Vert u^{k+1}\Vert }_{L^2}^2+{\Vert v^{k+1}\Vert }_{L^2}^2+{\Vert u^{k+1}\Vert }_{L^2}^2+{\Vert u^k\Vert }_{L^2}^2\right)\le C\left(M^2+{\Vert u^{k+1}\Vert }_{L^2}^2\right).\end{array}$ (41)

联立 $I_i,i=1,2,3$，

$\frac{\text{d}}{\text{d}t}{\Vert u^{k+1}\Vert }_{L^2}^2\le C\left(M^2+1\right){\Vert u^{k+1}\Vert }_{L^2}^2+C{M}^2.$ (42)

对方程(8)作用算子 ${\nabla }^r$ ( $1\le r\le s+2$ )，然后两端乘以 ${\nabla }^{r}u{}^{k+1}$，并在 ${\mathbb{T}}^d$ 上关于变量x积分，我们可以得到

$\begin{array}{l}\frac{1}{2}\frac{\text{d}}{\text{d}t}{\Vert {\nabla }^r{u}^{k+1}\Vert }_{L^2}^2=-{\displaystyle {\int }_{{\mathbb{T}}^d}{u}^k\nabla \left({\nabla }^r{u}^{k+1}\right)\cdot {\nabla }^r{u}^{k+1}\text{d}x}-{\displaystyle {\int }_{{\mathbb{T}}^d}\left[{\nabla }^r\left(u^k\cdot \nabla u^{k+1}\right)-u^k\nabla \left({\nabla }^r{u}^{k+1}\right)\right]}\cdot {\nabla }^r{u}^{k+1}\text{d}x\\ -K_1{\displaystyle {\int }_{{\mathbb{T}}^d}{\nabla }_x^r{\displaystyle {\int }_{{\mathbb{T}}^d}\Gamma \left(x-y\right)\cdot u^k\left(x\right)}\cdot {\rho }^{k+1}\left(y\right)\text{d}y}{\nabla }^r{u}^{k+1}\text{d}x\\ +K_1{\displaystyle {\int }_{{\mathbb{T}}^d}{\nabla }_x^r{\displaystyle {\int }_{{\mathbb{T}}^d}\Gamma \left(x-y\right)\cdot u^k\left(y\right)}\cdot {\rho }^{k+1}\left(y\right)\text{d}y}{\nabla }^r{u}^{k+1}\text{d}x\\ +K_2{\displaystyle {\int }_{{\mathbb{T}}^d}{\nabla }^r\left(v^{k+1}-u^k\right)\cdot {\nabla }^r{u}^{k+1}\text{d}x}={\displaystyle \underset{i=1}{\overset{5}{\sum }}{J}_i}.\end{array}$ (43)

由Hölder不等式，Young’s不等式，Sobolev不等式，Moser不等式，我们可以得到，

$J_1=-{\displaystyle {\int }_{{\mathbb{T}}^d}{u}^k\nabla \left({\nabla }^r{u}^{k+1}\right)\cdot {\nabla }^r{u}^{k+1}\text{d}x}=\frac{1}{2}{\displaystyle {\int }_{{\mathbb{T}}^d}\left(\nabla \cdot u^k\right)}{\left|{\nabla }^r{u}^{k+1}\right|}^2\text{d}x=0;$ (44)

$\begin{array}{l}{J}_2=-{\displaystyle {\int }_{{\mathbb{T}}^d}\left[{\nabla }^r\left(u^k\cdot \nabla u^{k+1}\right)-u^k\nabla \left({\nabla }^r{u}^{k+1}\right)\right]}\cdot {\nabla }^r{u}^{k+1}\text{d}x\\ \le C{\Vert {\nabla }^r\left(u^k\cdot \nabla u^{k+1}\right)-u^k\nabla \left({\nabla }^r{u}^{k+1}\right)\Vert }_{L^2}{\Vert {\nabla }^r{u}^{k+1}\Vert }_{L^2}\\ \le C\left({\Vert \nabla u^k\Vert }_{L^{\infty }}{\Vert {\nabla }^r{u}^{k+1}\Vert }_{L^2}+{\Vert \nabla u^{k+1}\Vert }_{L^{\infty }}{\Vert {\nabla }^r{u}^k\Vert }_{L^2}\right){\Vert {\nabla }^r{u}^{k+1}\Vert }_{L^2}\\ \le CM\left({\Vert {\nabla }^r{u}^{k+1}\Vert }_{L^2}^2+{\Vert u^{k+1}\Vert }_{H^{s+2}}^2\right);\end{array}$ (45)

$\begin{array}{l}{J}_3=-K_1{\displaystyle {\int }_{{\mathbb{T}}^d}{\nabla }_x^r{\displaystyle {\int }_{{\mathbb{T}}^d}\Gamma \left(x-y\right)\cdot u^k\left(x\right)}\cdot {\rho }^{k+1}\left(y\right)\text{d}y}{\nabla }^r{u}^{k+1}\text{d}x\\ \le C{\Vert {\nabla }_x^r\left(u^k\cdot \Gamma \ast {\rho }^{k+1}\right)\Vert }_{L^2}{\Vert {\nabla }^r{u}^{k+1}\Vert }_{L^2}\\ \le C\left({\Vert u^k\Vert }_{L^{\infty }}{\Vert {\nabla }_x^r\left(\Gamma \ast {\rho }^{k+1}\right)\Vert }_{L^2}+{\Vert \Gamma \ast {\rho }^{k+1}\Vert }_{L^{\infty }}{\Vert {\nabla }_x^r{u}^k\Vert }_{L^2}\right){\Vert {\nabla }^r{u}^{k+1}\Vert }_{L^2}\\ \le CM\left({\Vert {\nabla }_x^r\left(\Gamma \ast {\rho }^{k+1}\right)\Vert }_{L^2}+{\Vert \Gamma \ast {\rho }^{k+1}\Vert }_{L^{\infty }}\right){\Vert {\nabla }^r{u}^{k+1}\Vert }_{L^2}\\ \le CM\left({\Vert \Gamma \Vert }_{L^1}{\Vert {\nabla }_y^r{\rho }^{k+1}\Vert }_{L^2}+{\Vert \Gamma \Vert }_{L^1}{\Vert {\rho }^{k+1}\Vert }_{L^{\infty }}\right){\Vert {\nabla }^r{u}^{k+1}\Vert }_{L^2}\\ \le C{M}^2{\Vert {\nabla }^r{u}^{k+1}\Vert }_{L^2}\le C{M}^2{\Vert {\nabla }^r{u}^{k+1}\Vert }_{L^2}^2+C{M}^2;\end{array}$ (46)

$\begin{array}{l}{J}_4=K_1{\displaystyle {\int }_{{\mathbb{T}}^d}{\nabla }_x^r{\displaystyle {\int }_{{\mathbb{T}}^d}\Gamma \left(x-y\right)\cdot u^k\left(y\right)}\cdot {\rho }^{k+1}\left(y\right)\text{d}y}{\nabla }^r{u}^{k+1}\text{d}x\\ \le C{\displaystyle {\int }_{{\mathbb{T}}^d}{\displaystyle {\int }_{{\mathbb{T}}^d}\Gamma \left(x-y\right)\cdot {\nabla }_y^r\left(u^k\left(y\right)\cdot {\rho }^{k+1}\left(y\right)\right)\text{d}y}}{\nabla }^r{u}^{k+1}\text{d}x\\ \le C{\Vert \Gamma \Vert }_{L^1}{\Vert {\nabla }^r\left(u^k\cdot {\rho }^{k+1}\right)\Vert }_{L^2}{\Vert {\nabla }^r{u}^{k+1}\Vert }_{L^2}\\ \le C{\Vert \Gamma \Vert }_{L^1}\left({\Vert u^k\Vert }_{L^{\infty }}{\Vert {\nabla }^r{\rho }^{k+1}\Vert }_{L^2}+{\Vert {\rho }^{k+1}\Vert }_{L^{\infty }}{\Vert {\nabla }^r{u}^k\Vert }_{L^2}\right){\Vert {\nabla }^r{u}^{k+1}\Vert }_{L^2}\\ \le C{M}^2{\Vert {\nabla }^r{u}^{k+1}\Vert }_{L^2}\le C{M}^2{\Vert {\nabla }^r{u}^{k+1}\Vert }_{L^2}^2+C{M}^2;\end{array}$ (47)

$\begin{array}{l}{J}_5=K_2{\displaystyle {\int }_{{\mathbb{T}}^d}{\nabla }^r\left(v^{k+1}-u^k\right)\cdot {\nabla }^r{u}^{k+1}\text{d}x}\\ \le C\left({\Vert {\nabla }^r{v}^{k+1}\Vert }_{L^2}{\Vert {\nabla }^r{u}^{k+1}\Vert }_{L^2}+{\Vert {\nabla }^r{u}^k\Vert }_{L^2}{\Vert {\nabla }^r{u}^{k+1}\Vert }_{L^2}\right)\\ \le C\left({\Vert {\nabla }^r{v}^{k+1}\Vert }_{L^2}^2+{\Vert {\nabla }^r{u}^{k+1}\Vert }_{L^2}^2+{\Vert {\nabla }^r{u}^k\Vert }_{L^2}^2+{\Vert {\nabla }^r{u}^{k+1}\Vert }_{L^2}^2\right)\\ \le C\left(M^2+{\Vert {\nabla }^r{u}^{k+1}\Vert }_{L^2}^2+{\Vert {\nabla }^r{v}^{k+1}\Vert }_{L^2}^2\right).\end{array}$ (48)

将 $J_i,i=1,2,3,4,5$ 的估计带入(43)，如下不等式成立

$\frac{\text{d}}{\text{d}t}{\Vert {\nabla }^r{u}^{k+1}\Vert }_{L^2}^2\le C\left(M^2+M+1\right){\Vert {\nabla }^r{u}^{k+1}\Vert }_{L^2}^2+CM{\Vert u^{k+1}\Vert }_{H^{s+2}}^2+C{\Vert {\nabla }^r{v}^{k+1}\Vert }_{L^2}^2+C{M}^2.$ (49)

进一步的，对(49)的r从1到 $s+2$ 求和，并联立(42)，

$\frac{\text{d}}{\text{d}t}{\Vert u^{k+1}\Vert }_{H^{s+2}}^2\le C\left(M^2+M+1\right){\Vert u^{k+1}\Vert }_{H^{s+2}}^2+C{\displaystyle \underset{r=1}{\overset{s+2}{\sum }}{\displaystyle {\int }_{{\mathbb{T}}^d}{\left|{\nabla }^r{v}^{k+1}\right|}^2\text{d}x}}+C{M}^2.$ (50)

对(50)关于时间变量t从0到 $T_0$ 积分，可以得到

${\Vert u^{k+1}\Vert }_{H^{s+2}}^2\le C{\displaystyle {\int }_0^{T_0}{\displaystyle \underset{r=1}{\overset{s+2}{\sum }}{\displaystyle {\int }_{{\mathbb{T}}^d}{\left|{\nabla }^r{v}^{k+1}\right|}^2\text{d}x\text{d}t}}}+{\displaystyle {\int }_0^{T_0}C\left(M^2+M+1\right){\Vert u^{k+1}\Vert }_{H^{s+2}}^2+C{M}^2\text{d}t},$ (51)

利用Gronwall不等式和(36)可得，

$\underset{0\le t\le T_0}{\text{sup}}{\Vert u^{k+1}\Vert }_{H^{s+2}}^2\le {\text{e}}^{C\left(M^2+M+1\right)T_0}\left({\Vert u_0\Vert }_{H^{s+2}}^2+C{M}^2{T}_0\right).$ (52)

由(10)和(11)中M和 $T_0$ 的选取，我们得到 $u^{k+1}$ 的估计

$\underset{0\le t\le T_0}{\text{sup}}{\Vert u^{k+1}\Vert }_{H^{s+2}}\le M.$ (53)

至此，我们完成了归纳过程，引理1证毕。

4. 低阶范数收敛

在这一节中，我们将证明近似解序列 $\left({\rho }^k,u^k,v^k\right)$ 是 $C\left(\left[0,T_0\right];L^2\right)\times C\left(\left[0,T_0\right];H^1\right)\times C\left(\left[0,T_0\right];L^2\right)$ 中的柯西序列。

${\bar{\rho }}^{k+1}={\rho }^{k+1}-{\rho }^k,{\bar{u}}^{k+1}=u^{k+1}-u^k,{\bar{v}}^{k+1}=v^{k+1}-v^k,$

其中 $\left({\rho }^{k+1},u^{k+1},v^{k+1}\right)$ 和 $\left({\rho }^k,u^k,v^k\right)$ 分别满足：

$\begin{array}{l}{\partial }_t{\rho }^{k+1}+u^k\cdot \nabla {\rho }^{k+1}+{\rho }^{k+1}\nabla \cdot u^k=0\\ {\partial }_t{v}^{k+1}+v^k\cdot \nabla v^{k+1}+\nabla p^{k+1}=\mu \Delta v^{k+1}+K_2{\rho }^{k+1}\left(u^k-v^k\right),\nabla \cdot v^{k+1}=0\\ {\partial }_t{u}^{k+1}+u^k\cdot \nabla u^{k+1}=-K_1{\displaystyle {\int }_{{\text{T}}^d}\Gamma \left(x-y\right)\left(u^k\left(x\right)-u^k\left(y\right)\right){\rho }^{k+1}\left(y\right)\text{d}y}+K_2\left(v^{k+1}-u^k\right),\end{array}$ (54)

$\begin{array}{l}{\partial }_t{\rho }^k+u^{k-1}\cdot \nabla {\rho }^k+{\rho }^k\nabla \cdot u^{k-1}=0\\ {\partial }_t{v}^k+v^{k-1}\cdot \nabla v^k+\nabla p^k=\mu \Delta v^k+K_2{\rho }^k\left(u^{k-1}-v^{k-1}\right),\nabla \cdot v^k=0\\ {\partial }_t{u}^k+u^{k-1}\cdot \nabla u^k=-K_1{\displaystyle {\int }_{{\text{T}}^d}\Gamma \left(x-y\right)\left(u^{k-1}\left(x\right)-u^{k-1}\left(y\right)\right){\rho }^k\left(y\right)\text{d}y}+K_2\left(v^k-u^{k-1}\right),\end{array}$ (55)

且具有相同的初值条件：

$\left({\rho }^{k+1},u^{k+1},v^{k+1}\right)\left(x,0\right)=\left({\rho }^k,u^k,v^k\right)\left(x,0\right)=\left({\rho }_0,u_0,v_0\right).$ (56)

应用类似引理1中的方法，我们可以得到 $\left({\bar{\rho }}^{k+1},{\bar{u}}^{k+1},{\bar{v}}^{k+1}\right)$ 的 $L^2$ -模估计。

Step 1. ( ${\Vert {\rho }^{k+1}-{\rho }^k\Vert }_{L^2}$ 的估计)：由(54)₁与(55)₁相减，

${\partial }_t\left({\rho }^{k+1}-{\rho }^k\right)=-\left(u^k-u^{k-1}\right)\cdot \nabla {\rho }^{k+1}-u^{k-1}\cdot \nabla \left({\rho }^{k+1}-{\rho }^k\right)-\left({\rho }^{k+1}-{\rho }^k\right)\cdot \nabla u^k-{\rho }^k\nabla \cdot \left(u^k-u^{k-1}\right).$ (57)

上述方程两端乘以 ${\rho }^{k+1}-{\rho }^k$，并关于变量x在 ${\mathbb{T}}^d$ 上积分，应用第3节相同的方法，

$\begin{array}{c}\frac{1}{2}\frac{\text{d}}{\text{d}t}{\Vert {\rho }^{k+1}-{\rho }^k\Vert }_{L^2}^2=-{\displaystyle {\int }_{{\mathbb{T}}^d}\left(u^k-u^{k-1}\right)\cdot \nabla {\rho }^{k+1}\left({\rho }^{k+1}-{\rho }^k\right)\text{d}x}+\frac{1}{2}{\displaystyle {\int }_{{\mathbb{T}}^d}\left(\nabla \cdot u^{k-1}\right)}{\left|{\rho }^{k+1}-{\rho }^k\right|}^2\text{d}x\\ -{\displaystyle {\int }_{{\mathbb{T}}^d}\left(\nabla \cdot u^k\right)}{\left|{\rho }^{k+1}-{\rho }^k\right|}^2\text{d}x-{\displaystyle {\int }_{{\mathbb{T}}^d}{\rho }^k\nabla \cdot \left(u^k-u^{k-1}\right)}\left({\rho }^{k+1}-{\rho }^k\right)\text{d}x\\ \le CM\left({\Vert {\rho }^{k+1}-{\rho }^k\Vert }_{L^2}^2+{\Vert u^k-u^{k-1}\Vert }_{H^1}^2\right).\end{array}$ (58)

对上述方程在 $\left[0,T_0\right]$ 上关于变量t进行积分，我们得到 ${\Vert {\rho }^{k+1}-{\rho }^k\Vert }_{L^2}$ 的估计。

${\Vert {\rho }^{k+1}-{\rho }^k\Vert }_{L^2}^2\le CM{\displaystyle {\int }_0^{T_0}\left({\Vert {\rho }^{k+1}-{\rho }^k\Vert }_{L^2}^2+{\Vert u^k-u^{k-1}\Vert }_{H^1}^2\right)\text{d}\tau }.$ (59)

Step 2. ( ${\Vert v^{k+1}-v^k\Vert }_{L^2}$ 的估计)：由(54)₂与(55)₂相减，

$\begin{array}{l}\partial t\left(v^{k+1}-v^k\right)-\mu \Delta \left(v^{k+1}-v^k\right)\\ =-\left(v^k-v^{k-1}\right)\cdot \nabla v^{k+1}-v^{k-1}\cdot \nabla \left(v^{k+1}-v^k\right)+K_2{\rho }^{k+1}\left(u^k-u^{k-1}\right)\\ -K_2{\rho }^{k+1}\left(v^k-v^{k-1}\right)+K_2\left({\rho }^{k+1}-{\rho }^k\right)u^{k-1}-K_2\left({\rho }^{k+1}-{\rho }^k\right)v^{k-1}-\nabla \left(p^{k+1}-p^k\right),\\ \nabla \cdot \left(v^{k+1}-v^k\right)=0.\end{array}$ (60)

对上述方程两端乘以 $v^{k+1}-v^k$，并关于变量x在 ${\mathbb{T}}^d$ 上积分，应用第3节相同的方法，

$\begin{array}{l}\frac{1}{2}\frac{\text{d}}{\text{d}t}{\Vert v^{k+1}-v^k\Vert }_{L^2}^2+\mu {\Vert \nabla \left(v^{k+1}-v^k\right)\Vert }_{L^2}^2\\ =-{\displaystyle {\int }_{{\mathbb{T}}^d}\left(v^k-v^{k-1}\right)\cdot \nabla v^{k+1}\cdot \left(v^{k+1}-v^k\right)\text{d}x}+\frac{1}{2}{\displaystyle {\int }_{{\mathbb{T}}^d}\left(\nabla \cdot v^{k-1}\right){\left|v^{k+1}-v^k\right|}^2}\text{d}x\\ +K_2{\displaystyle {\int }_{{\mathbb{T}}^d}{\rho }^{k+1}\left[\left(u^k-u^{k-1}\right)-\left(v^k-v^{k-1}\right)\right]}\cdot \left(v^{k+1}-v^k\right)\text{d}x\\ +K_2{\displaystyle {\int }_{{\mathbb{T}}^d}\left({\rho }^{k+1}-{\rho }^k\right)\left(u^{k-1}-v^{k-1}\right)}\cdot \left(v^{k+1}-v^k\right)\text{d}x\\ \le CM\left({\Vert v^{k+1}-v^k\Vert }_{L^2}^2+{\Vert v^k-v^{k-1}\Vert }_{L^2}^2+{\Vert u^k-u^{k-1}\Vert }_{L^2}^2+{\Vert {\rho }^{k+1}-{\rho }^k\Vert }_{L^2}^2\right).\end{array}$ (61)

对上述方程在 $\left[0,T_0\right]$ 上关于变量t进行积分，我们得到 ${\Vert v^{k+1}-v^k\Vert }_{L^2}$ 的估计。

$\begin{array}{l}{\Vert v^{k+1}-v^k\Vert }_{L^2}^2+\mu {\displaystyle {\int }_0^{T_0}{\Vert \nabla \left(v^{k+1}-v^k\right)\Vert }_{L^2}^2\text{d}\tau }\\ \le CM{\displaystyle {\int }_0^{T_0}\left({\Vert v^{k+1}-v^k\Vert }_{L^2}^2+{\Vert v^k-v^{k-1}\Vert }_{L^2}^2+{\Vert u^k-u^{k-1}\Vert }_{L^2}^2+{\Vert {\rho }^{k+1}-{\rho }^k\Vert }_{L^2}^2\right)\text{d}\tau }.\end{array}$ (62)

Step 3. ( ${\Vert u^{k+1}-u^k\Vert }_{H^1}$ 的估计)：由(54)₃与(55)₃相减，

$\begin{array}{c}\partial t\left(u^{k+1}-u^k\right)=-\left(u^k-u^{k-1}\right)\cdot \nabla u^{k+1}-u^{k-1}\cdot \nabla \left(u^{k+1}-u^k\right)\\ +K_1{\displaystyle {\int }_{{\mathbb{T}}^d}\Gamma \left({\rho }^{k+1}-{\rho }^k\right)\left(u^k\left(y\right)-u^k\left(x\right)\right)\text{d}y}\\ +K_1{\displaystyle {\int }_{{\mathbb{T}}^d}\Gamma {\rho }^k\left[\left(u^k\left(y\right)-u^{k-1}\left(y\right)\right)-\left(u^k\left(x\right)-u^{k-1}\left(x\right)\right)\right]\text{d}y}\\ +K_2\left(v^{k+1}-v^k\right)-K_2\left(u^k-u^{k-1}\right).\end{array}$ (63)

对上述方程两端乘以 $u^{k+1}-u^k$，并关于变量x在 ${\mathbb{T}}^d$ 上积分，应用第3节相同的方法，

$\begin{array}{c}\frac{1}{2}\frac{\text{d}}{\text{d}t}{\Vert u^{k+1}-u^k\Vert }_{L^2}^2=-{\displaystyle {\int }_{{\mathbb{T}}^d}\left(u^k-u^{k-1}\right)\cdot \nabla u^{k+1}\cdot \left(u^{k+1}-u^k\right)\text{d}x}+\frac{1}{2}{\displaystyle {\int }_{{\mathbb{T}}^d}\left(\nabla \cdot u^{k-1}\right)\cdot {\left|u^{k+1}-u^k\right|}^2\text{d}x}\\ +K_1{\displaystyle {\int }_{{\mathbb{T}}^{2d}}\Gamma \left({\rho }^{k+1}-{\rho }^k\right)}\left(u^k\left(y\right)-u^k\left(x\right)\right)\cdot \left(u^{k+1}-u^k\right)\text{d}x\text{d}y\\ +K_1{\displaystyle {\int }_{{\mathbb{T}}^{2d}}\Gamma {\rho }^k}\left[\left(u^k\left(y\right)-u^{k-1}\left(y\right)\right)-\left(u^k\left(x\right)-u^{k-1}\left(x\right)\right)\right]\cdot \left(u^{k+1}-u^k\right)\text{d}x\text{d}y\\ +K_2{\displaystyle {\int }_{{\mathbb{T}}^d}\left(v^{k+1}-v^k\right)\cdot \left(u^{k+1}-u^k\right)\text{d}x}-K_2{\displaystyle {\int }_{{\mathbb{T}}^d}\left(u^k-u^{k-1}\right)\cdot \left(u^{k+1}-u^k\right)\text{d}x}\\ \le CM\left({\Vert u^{k+1}-u^k\Vert }_{L^2}^2+{\Vert u^k-u^{k-1}\Vert }_{L^2}^2+{\Vert v^{k+1}-v^k\Vert }_{L^2}^2+{\Vert {\rho }^{k+1}-{\rho }^k\Vert }_{L^2}^2\right).\end{array}$ (64)

为了得到 ${\Vert {\rho }^{k+1}-{\rho }^k\Vert }_{L^2}^2$ 的收敛性，还需证明 ${\Vert u^{k+1}-u^k\Vert }_{H^{\text{1}}}^{\text{2}}$ 收敛，因此，我们还需得到 ${\Vert \nabla \left(u^{k+1}-u^k\right)\Vert }_{L^{\text{2}}}^{\text{2}}$ 的估计。对方程(63)作用算子 $\nabla$，然后方程两端同乘以 $\nabla \left(u^{k+1}-u^k\right)$，并在 ${\mathbb{T}}^d$ 上关于变量x积分，应用第3节相同的方法，可以得到

$\begin{array}{l}\frac{1}{2}\frac{\text{d}}{\text{d}t}{\Vert \nabla \left(u^{k+1}-u^k\right)\Vert }_{L^2}^2\\ =-{\displaystyle {\int }_{{\mathbb{T}}^d}\nabla \left(u^k-u^{k-1}\right)\cdot \nabla u^{k+1}\cdot \nabla \left(u^{k+1}-u^k\right)\text{d}x}-{\displaystyle {\int }_{{\mathbb{T}}^d}\left(u^k-u^{k-1}\right)\cdot {\nabla }^2{u}^{k+1}\cdot \nabla \left(u^{k+1}-u^k\right)\text{d}x}\\ -{\displaystyle {\int }_{{\mathbb{T}}^d}\left(\nabla \cdot u^{k-1}\right){\left|\nabla \left(u^{k+1}-u^k\right)\right|}^2\text{d}x}+\frac{1}{2}{\displaystyle {\int }_{{\mathbb{T}}^d}\left(\nabla \cdot u^{k-1}\right){\left|\nabla \left(u^{k+1}-u^k\right)\right|}^2\text{d}x}\\ -K_1{\Vert \Gamma \Vert }_{L^1}{\displaystyle {\int }_{{\mathbb{T}}^{2d}}\left({\rho }^{k+1}\left(y\right)-{\rho }^k\left(y\right)\right)}\cdot \nabla u^k\left(x\right)\cdot \nabla \left(u^{k+1}-u^k\right)\text{d}x\text{d}y\end{array}$ (65)

$\begin{array}{l}-K_1{\Vert \Gamma \Vert }_{L^1}{\displaystyle {\int }_{{\mathbb{T}}^{2d}}{\rho }^k\left(y\right)\nabla }\left(u^k\left(x\right)-u^{k-1}\left(x\right)\right)\cdot \nabla \left(u^{k+1}-u^k\right)\text{d}x\text{d}y\\ +K_2{\displaystyle {\int }_{{\mathbb{T}}^d}\nabla \left(v^{k+1}-v^k\right)\cdot \nabla \left(u^{k+1}-u^k\right)\text{d}x}-K_2{\displaystyle {\int }_{{\mathbb{T}}^d}\nabla \left(u^k-u^{k-1}\right)\cdot \nabla \left(u^{k+1}-u^k\right)\text{d}x}\\ \le CM\left({\Vert \nabla \left(u^{k+1}-u^k\right)\Vert }_{L^2}^2+{\Vert u^k-u^{k-1}\Vert }_{H^1}^2+{\Vert {\rho }^{k+1}-{\rho }^k\Vert }_{L^2}^2+\epsilon {\Vert \nabla \left(v^{k+1}-v^k\right)\Vert }_{L^2}^2\right).\end{array}$

其中上述关系中的最后第二项使用了带有 $\epsilon \left(>0\right)$ 的Young’s不等式，

$\begin{array}{l}{\displaystyle {\int }_{{\mathbb{T}}^d}\nabla \left(v^{k+1}-v^k\right)\cdot \nabla \left(u^{k+1}-u^k\right)\text{d}x}\\ \le {\Vert \nabla \left(v^{k+1}-v^k\right)\Vert }_{L^2}{\Vert \nabla \left(u^{k+1}-u^k\right)\Vert }_{L^2}\\ \le \epsilon {\Vert \nabla \left(v^{k+1}-v^k\right)\Vert }_{L^2}^2+\frac{1}{4\epsilon }{\Vert \nabla \left(u^{k+1}-u^k\right)\Vert }_{L^2}^2.\end{array}$ (66)

结合(64)与(65)，并在 $\left[0,T_0\right]$ 上进行积分，

$\begin{array}{l}{\Vert u^{k+1}-u^k\Vert }_{H^1}^2\le CM{\displaystyle {\int }_0^{T_0}({\Vert u^{k+1}-u^k\Vert }_{H^1}^2+{\Vert u^k-u^{k-1}\Vert }_{H^1}^2+{\Vert {\rho }^{k+1}-{\rho }^k\Vert }_{L^2}^2{}_{{}_{}^{}}^{}}\\ +{\Vert v^{k+1}-v^k\Vert }_{L^2}^2+\epsilon {\Vert \nabla \left(v^{k+1}-v^k\right)\Vert }_{L^2}^2)\text{d}\tau .\end{array}$ (67)

结合(59)，(62)，(67)，并选择 $\epsilon$ 使得 $\epsilon \ll \mu$ 可得，

$\begin{array}{l}{\Vert {\rho }^{k+1}-{\rho }^k\Vert }_{L^2}^2+{\Vert u^{k+1}-u^k\Vert }_{H^1}^2+{\Vert v^{k+1}-v^k\Vert }_{L^2}^2\\ \le CM{\displaystyle {\int }_0^{T_0}({\Vert u^{k+1}-u^k\Vert }_{H^1}^2+{\Vert u^k-u^{k-1}\Vert }_{H^1}^2+{\Vert {\rho }^{k+1}-{\rho }^k\Vert }_{L^2}^2}+{\Vert v^{k+1}-v^k\Vert }_{L^2}^2+{\Vert v^k-v^{k+1}\Vert }_{L^2}^2)\text{d}\tau .\end{array}$ (68)

对上式的 $k=1,2,\cdots$ 求和，并利用Gronwall不等式，

${\displaystyle \underset{k=1}{\overset{\infty }{\sum }}\left({\Vert {\rho }^{k+1}-{\rho }^k\Vert }_{L^2}^2+{\Vert u^{k+1}-u^k\Vert }_{H^1}^2+{\Vert v^{k+1}-v^k\Vert }_{L^2}^2\right)}\le C.$ (69)

这就说明 $\left({\rho }^k,u^k,v^k\right)$ 是 $C\left(\left[0,T_0\right];L^2\right)\times C\left(\left[0,T_0\right];H^1\right)\times C\left(\left[0,T_0\right];L^2\right)$ 中的柯西序列。

5. 局部解的存在唯一性

在这一节中，我们完成定理1的证明。

存在性证明：对于近似解 $\left({\rho }^k,u^k,v^k\right)$，我们在第3节中得到了一致有界性，并在第4节中证明了柯西收敛性，结合Gagliardo-Nirenberg不等式，可得到，

$\begin{array}{l}{\rho }^k\to \rho \in C\left(\left[0,T_0\right];H^{s-1}\right)\\ u^k\to u\in C\left(\left[0,T_0\right];H^{s+1}\right)\\ v^k\to v\in C\left(\left[0,T_0\right];H^s\right).\end{array}$ (70)

由(70)易知极限函数 $\left(\rho ,u,v\right)$ 是(1)~(2)在分布意义上的一个解，利用 [15] 中相似论点，可得到 $\left(\rho ,u,v\right)$ 的正则性：

$\begin{array}{l}{\rho }^k\to \rho \in C\left(\left[0,T_0\right];H^s\right)\\ u^k\to u\in C\left(\left[0,T_0\right];H^{s+2}\right)\\ v^k\to v\in C\left(\left[0,T_0\right];H^{s+1}\right).\end{array}$ (71)

利用Sobolev嵌入定理，可知 $\left(\rho ,u,v\right)\in C^1\left(\left[0,T_0\right];{\mathbb{T}}^d\right)$ 是一个经典解，存在性得证。

唯一性证明：设 $\left(\rho ,u,v\right)$ 和 $\left(\tilde{\rho },\tilde{u},\tilde{v}\right)$ 是(1)~(2)具有相同初值条件 $\left({\rho }_0,u_0,v_0\right)$ 的两个经典解，令：

$U\left(t\right)={\Vert \rho -\tilde{\rho }\Vert }_{L^2}^2+{\Vert u-\tilde{u}\Vert }_{H^1}^2+{\Vert v-\tilde{v}\Vert }_{L^2}^2.$

利用与第4节相同的方法，容易证明

$U\left(t\right)\le C{\displaystyle {\int }_0^{T_0}U\left(\tau \right)\text{d}\tau },U\left(0\right)=0.$

应用Gronwall不等式，可以证明 $U\left(t\right)=0$，即：

$\rho \equiv \tilde{\rho }\in C\left(\left[0,T_0\right];L^2\right),u\equiv \tilde{u}\in C\left(\left[0,T_0\right];H^1\right),v\equiv \tilde{v}\in C\left(\left[0,T_0\right];L^2\right).$

于是唯一性得证，定理1的证明完成。

致谢

感谢佟丽宁导师对本文的指导与建议。