三维流体-粒子相互作用模型:Flowing Regime模型的局部解问题

三维流体-粒子相互作用模型:Flowing Regime模型的局部解问题

郑琳^,¹, 王术¹, 李林锐²

Local Classical Solutions of 3D Fluid-Particle Interaction Model: The Flowing Regime

Zheng Lin^,¹, Wang Shu¹, Li Linrui²

通讯作者: 郑琳, E-mail: zhenglin0103@emails.bjut.edu.cn

收稿日期: 2017-11-10

基金资助:

国家自然科学基金. 11371042

Received: 2017-11-10

Fund supported:

the NSFC. 11371042

摘要

该文主要研究了三维流体-粒子相互作用模型：Flowing Regime模型在全空间中的Cauchy问题.证明了局部强解的存在性和唯一性，通过推导强解的光滑性得到了一个局部经典解.

关键词： Flowing Regime模型 ; 局部解 ; Sobolev嵌入

Abstract

In this paper, we concerned with the Cauchy problem for 3D fluid-particle interaction model in the so-called Flowing regime with vacuum in ${{\mathbb{R}}^{3}}$ . We prove both existence and uniqueness of the local strong solution, and then obtain a local classical solution by deriving the smooth effect of the strong solution for t > 0.

Keywords： Flowing Regime ; Local solution ; Sobolev embedding

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本文引用格式

郑琳, 王术, 李林锐. 三维流体-粒子相互作用模型:Flowing Regime模型的局部解问题. 数学物理学报[J], 2018, 38(6): 1173-1192 doi:

Zheng Lin, Wang Shu, Li Linrui. Local Classical Solutions of 3D Fluid-Particle Interaction Model: The Flowing Regime. Acta Mathematica Scientia[J], 2018, 38(6): 1173-1192 doi:

1 引言

流体-粒子相互作用模型(fluid-particle interaction model)在流体中的分散悬浮液粒子的沉降分析中发挥重要作用,在生物、医学、化工和矿物过程中具有非常重要的应用^[1-4].此外,这种交互系统也用于柴油发动机和火箭推进器燃烧理论模型^[5-6].本文中,我们主要研究所谓的Flowing Regime模型,其在全空间 $\mathbb{R}^{3}$ 的表示为

${{\rho }_{t}}+\nabla \cdot (\rho u)=0,$

(1.1)

$((\rho+\beta^{-2}\eta)u)_{t}+{\rm div}((\rho+\beta^{-2}\eta)u\otimes u)+\nabla(p+\eta)- \mu\triangle u-\lambda\nabla(\nabla\cdot u)=-(\alpha\beta^{2}\rho+\eta)\nabla\Phi,$

(1.2)

$\eta_{t}+\nabla\cdot(\eta u)=0,$

(1.3)

其中: $\rho$ 表示流体密度; $u$ 表示速度场; $\eta$ 表示粒子密度,与分布函数 $f(t, x, \xi)$ 的关系为

$\eta(t, x)=\int_{{\mathbb{R}}^{3}}f(t, x, \xi){\rm d}\xi;$

这里流体压力函数 $P_{F}$ 表示为

$p_{F}(\rho)=a\rho^{\gamma}, a>0, \gamma>1;$

$\Phi(x)$ 是不依赖于时间的外力项; $\alpha, \beta$ 是无量纲常数; $\mu, \lambda$ 是粘性系数且满足

$\mu>0, \lambda+\frac{2}{3}\mu\geq 0.$

(1.4)

该模型系统由Carrillo和Goudon推导出来,他们推导出了耦合动力学和流体力学的两种模型^[4].所谓的Bubbling Regime和Flowing Regime.对于Bubbling Regime模型^{[1, 10]}表示的是动力学方程的扩散近似;而Flowing Regime模型,描述了强阻力流动机制和强布朗运动.对于Bubbling Regime,已经有了很多结果^{[8-10, 14, 16]}.但是对于Flowing Regime模型的相关研究却甚少.当 $\eta=0$ 时,方程(1.1)-(1.3)变为Navier-Stokes方程. Cho和Kim在文献[18]中研究了Navier-Stokes方程的Cauchy问题的局部解问题.本文的主要目的是利用同样的方法来解决较为复杂的Flowing Regime模型的局部解问题.

为简单起见,我们令 $\alpha, \beta=1$ ,模型(1.1)-(1.3)重新写为如下

$\rho_{t}+\nabla\cdot(\rho u)=0,$

(1.5)

$\begin{equation} (\rho+\eta)u_{t}+(\rho+\eta)u\cdot\nabla u+\nabla(p+\eta)- \mu\triangle u-\lambda\nabla(\nabla\cdot u)=-(\rho+\eta)\nabla\Phi,\end{equation}$

(1.6)

$\eta_{t}+\nabla\cdot(\eta u)=0,$

(1.7)

初值为

$(\rho, u, \eta)|_{t=0}=(\rho_{0}, u_{0}, \eta_{0}),$

(1.8)

在无穷远处,当 $|x|\rightarrow\infty$ 时

$(\rho, u, \eta)|_{t=0}=(\rho_{0}, u_{0}, \eta_{0}),$

(1.9)

这里,常向量 $(\rho^{\infty}, 0, \eta^{\infty})$ 满足 $\rho^{\infty}, \eta^{\infty}\geq0$ .接下来,为了简单起见令

$Lu=- \mu\triangle u-\lambda\nabla(\nabla\cdot u).$

在接下来的定理中,为了证明 $p_{F}-p^{\infty}_{F}\in C([0, T_{\ast}];H^{3})$ ,主要难点是解决 $p_{F}$ 项,所以我们把(1.4)-(1.8)式重新写为

$\rho_{t}+\nabla\cdot(\rho u)=0,$

(1.10)

$p_{t}+u\cdot\nabla p+\gamma p\nabla\cdot u=0,$

(1.11)

$(\rho+\eta)u_{t}+(\rho+\eta)u\cdot\nabla u+\nabla(p+\eta)+Lu=-(\rho+\eta)\nabla \Phi,$

(1.12)

$\eta_{t}+\nabla\cdot(\eta u)=0,$

(1.13)

初值为

$(\rho, p, u, \eta)|_{t=0}=(\rho_{0}, p_{0}, u_{0}, \eta_{0}), \quad \mathbb{R}^{3},$

(1.14)

边界条件为

$(\rho, p, u, \eta)\rightarrow(\rho^{\infty}, p^{\infty}, 0, \eta^{\infty}), \quad |x|\rightarrow \infty,$

(1.15)

且满足以下假设

$\left \{\begin{array}{l} (\rho_{0}-\rho^{\infty}, p_{0}-p^{\infty}, \eta_{0}-\eta^{\infty})\in H^{3}, \rho^{\infty}, \eta^{\infty}\in \bar{R}_{+}, \rho_{0}, \eta_{0}\geq0, \\ p_{0}=a\rho^{\gamma}_{0}, p^{\infty}=a(\rho^{\infty})^{\gamma}, u_{0} \in D^{1}_{0}\cap D^{3}, \Phi\in H^{4}. \end{array}\right.$

(1.16)

在全篇文章中,我们做一些符号的注释(文献[18]中有详细的说明).

(1) $B_{R}=B_{R}(0), L^{r}=L^{r}(B_{R}), D^{k, r}=\{v\in L^{1}_{\rm loc}(B_{R}):\|v\|_{D^{k, r}}\equiv\|\nabla^{k}v\|_{L^{r}} < \infty\}$ ;

(2) $D^{1}_{0}={v\in L^{6}(B_{R}):\|\nabla v\|_{L^{2}} <\infty; \quad v=0, \quad \partial B_{R}}$ ;

(3) $W^{k, r}=L^{r}\cap D^{k, r}, H^{k}=W^{k, 2}, D^{k}=D^{k, 2}, H^{1}_{0}=L^{2}\cap D^{1}_{0},$

$\|v\|_{D^{1}_{0}}=\|\nabla v\|_{L^{2}}, \|v\|_{D^{k, r}}=\|\nabla^{k}v\|_{L^{r}};$

(4) $\|\cdot\|_{X\cap Y}=\|\cdot\|_{X}+\|\cdot\|_{ Y}$ ,以及经典的Sololev嵌入(对于 $q> 3$ )

$\|v\|_{L^{6}}\leq C\|v\|_{D^{1}_{0}}, \quad \|v\|_{L^{\infty}}\leq C \|v\|_{W^{1, q}}, \quad \|v\|_{L^{\infty}}\leq C \|v\|_{D^{1}_{0}\cap D^{2}}.$

定理1.1 假设

$(\rho_{0}-\rho^{\infty}, p_{0}-p^{\infty}, \eta_{0}-\eta^{\infty})\in H^{3}, \rho^{\infty}, \eta^{\infty}\in \bar{R}_{+}, \rho_{0}, \eta_{0}\geq0, u_{0} \in D^{1}_{0}\cap D^{3}, \Phi\in H^{4},$

(1.17)

这里 $p_{0}=a\rho^{\gamma}_{0}, p^{\infty}=a(\rho^{\infty})^{\gamma}$ 且满足相容性条件

$Lu_{0}+\nabla(p_{0}+\eta_{0})+\eta_{0}\nabla \Phi=-\rho_{0}\nabla \Phi-(\rho_{0}+\eta_{0})g,$

(1.18)

对于 $g\in D^{1}$ , $\sqrt{\rho_{0}}g\in L^{2}$ .则这里存在一个小时间区间 $T_{\ast}\in (0, T)$ 使得问题(1.5)-(1.9)存在唯一强解 $(\rho, p, u, \eta)$ 满足

$\left\{ \begin{array}{l} (\rho-\rho^{\infty}, p-p^{\infty}, \eta-\eta^{\infty})\in C([0, T_{\ast}];H^{3}), (\rho_{t}, p_{t}, \eta_{t})\in L^{\infty}(0, T_{\ast};H^{2}), \\ (\sqrt{\rho}u_{t}, \sqrt{\eta}u_{t})\in L^{\infty}(0, T_{\ast};L^{2}) , \\ u \in C([0, T_{\ast}];D^{1}\cap D^{3})\cap L^{2}(0, T_{\ast};D^{4}), u_{t}\in L^{\infty}(0, T_{\ast};D^{1})\cap L^{2}(0, T_{\ast};D^{2}). \end{array} \right.$

(1.19)

2 线性方程组的先验估计

为了证明定理1.1,我们考虑以下线性方程组

$\rho_{t}+\nabla\cdot(\rho v)=0,$

(2.1)

$p_{t}+v\cdot\nabla p+\gamma p\nabla\cdot v=0,$

(2.2)

$(\rho+\eta)u_{t}+\nabla(p+\eta)+Lu=-(\rho+\eta)v\cdot\nabla v-(\rho+\eta)\nabla \Phi,$

(2.3)

$\eta_{t}+\nabla\cdot(\eta v)=0,$

(2.4)

在 $(t, x)\in (0, T)\times B_{R}$ ,这里 $T$ 是一个大于零的数,初边界条件是(1.14)-(1.15)式且

$u=0, \quad (0, T)\times \partial B_{R}.$

(2.5)

这里 $v:(0, T)\times B_{R}\rightarrow \mathbb{R}^{3}$ 是一个已知向量,具有以下正则性

$v\in C([0, T];D^{1}_{0}\cap D^{3})\cap L^{2}(0, T;D^{4}), v_{t}\in L^{\infty}(0, T;D^{1}_{0})\cap L^{2}(0, T;D^{2}).$

(2.6)

注2.1 当 $B_{R}\subset \mathbb{R}^{3}$ 是有界区域(或全空间),边界条件(1.15) (或(2.5)不是必要的,甚至是可以忽略的.

首先,我们有一个在初值是正时,解的存在性结果.可以参看文献[18]中的引理3.1.

引理2.1 令 $B_{R}\subset \mathbb{R}^{3}$ 是有界区域.在(1.12)和(2.6)式成立的条件下,进一步假设在 $B_{R}$ 中, $\rho_{0}\geq \delta$ ,其中 $\delta>0$ ,且

$v(0)\cdot \nabla v(0)+\nabla \Phi+(\rho_{0}+\eta_{0})^{-1}(Lu_{0}+\nabla(p_{0}+\eta_{0}))\in H^{1}_{0}(B_{R}).$

则存在时间 $T>0$ 使得初值为(2.10)式的系统(2.1)-(2.4)存在唯一解 $(\rho, p, u, \eta)$ 满足

$\left\{\begin{array}{l} (\rho-\rho^{\infty}, p-p^{\infty}, \eta-\eta^{\infty})\in C([0, T];H^{3}(B_{R})), (\rho_{t}, p_{t}, \eta_{t})\in C([0, T];H^{2}(B_{R})), \\ \eta \in C([0, T];H_{0}^{1}(B_{R})\cap H^{3}(B_{R}))\cap L^{2}(0, T;H^{4}(B_{R})), \\ \eta_{t}\in C([0, T];H_{0}^{1}(B_{R}))\cap L^{2}(0, T;H^{2}(B_{R})), \eta_{tt}\in L^{2}(0, T;H^{2}(B_{R})), \rho\geq\underline{\delta}, \end{array} \right.$

(2.7)

对于常数 $\underline{\delta}>0$ .

接下来,我们推导出 $(\rho, p, u, \eta)$ 的不依赖于 $\rho$ 的下界 $\delta$ 和半径 $R$ 的局部先验估计.

我们假设存在一个常数 $c_{0}>1$ 使得

$\begin{array}{*{20}{l}} 1+\rho^{\infty}+p^{\infty}+\eta^{\infty}+\|\rho_{0}-\rho^{\infty}\|_{H^{3}}+\|p_{0}-p^{\infty}\|_{H^{3}}+\|\eta_{0}-\eta^{\infty}\|_{H^{3}}\\ +\|u_{0}\|_{D^{1}_{0}} +\|\sqrt{\rho_{0}+\eta_{0}}g\|_{L^{2}}+\|g\|_{D^{1}_{0}}\leq c_{0}, \end{array}$

(2.8)

这里 $g=-(\rho_{0}+\eta_{0})^{-1} (Lu_{0}+\nabla(p_{0}+\eta_{0}))-\nabla \Phi=u_{t}(0)+v(0)\cdot \nabla v(0)$ ,且假设

$\|v(0)\|_{D^{1}_{0}\cap D^{3}}\leq c_{1},$

(2.9)

$\sup\limits_{0\leq t \leq T_{\ast}}\|v(t)\|_{D^{1}_{0}}+\int^{T_{\ast}}_{0}(\|v(t)\|^{2}_{D^{2}}+\|v_{t}(t)\|^{2}_{L^{2}}){\rm d}t \leq c_{2},$

(2.10)

$\sup\limits_{0\leq t \leq T_{\ast}}\|v(t)\|_{D^{2}}+\int^{T_{\ast}}_{0}(\|v(t)\|^{2}_{D^{3}}+\|v_{t}(t)\|^{2}_{D_{0}^{1}}){\rm d}t \leq c_{3},$

(2.11)

${\rm ess}\sup\limits_{0\leq t \leq T_{\ast}}(\|v(t)\|_{D^{3}}+\|v_{t}(t)\|_{D_{0}^{1}})+\int^{T_{\ast}}_{0}(\|v(t)\|^{2}_{D^{4}}+\|v_{t}(t)\|^{2}_{D^{2}}){\rm d}t \leq c_{4},$

(2.12)

对于时间 $T_{\ast}\in (0, T)$ ,这里 $c_{i}$ $(i=1, 2, 3, 4)$ 和 $T_{\ast}$ 依赖于 $C$ 和 $c_{0}$ ,且有 $1 <c_{0}\leq c_{1}\leq c_{2}\leq c_{3}\leq c_{4}$ .在本文中, $C$ 是仅依赖于 $\mu, \lambda, \gamma, T$ 和 $\|\Phi(x)\|_{H^{4}}$ , $M=M(\cdot)$ 是 $[1, +\infty)$ 上的连续增函数,仅依赖于 $C$ .

我们给出下面 $(\rho, p, u, \eta)$ 的估计.

引理2.2^{[18, Lemma 3.2]} 对于 $0\leq t \leq\min(T_{\ast}, T_{1})$ ,有

$\|\rho(t)\|_{L^{\infty}}+\|\rho-\rho^{\infty}\|_{H^{3}}\leq Cc_{0}, \ \|\eta(t)\|_{L^{\infty}}+\|\eta-\eta^{\infty}\|_{H^{3}}\leq Cc_{0}, \ \|p-p^{\infty}\|_{H^{3}}\leq M(c_{0}),$

$\|\rho_{t}(t)\|_{H^{1}}\leq Cc^{2}_{3}, \ \|\eta_{t}(t)\|_{H^{1}}\leq Cc^{2}_{3}, \quad \|p_{t}(t)\|_{H^{1}}\leq M(c_{0})c^{2}_{3}, \ \int^{t}_{0}\|p_{tt}(t)\|^{2}_{L^{2}}{\rm d}s\leq M(c_{0})c^{8}_{3},$

$\int^{t}_{0}\|\rho_{tt}(t)\|^{2}_{L^{2}}{\rm d}s\leq Cc^{8}_{3}, \quad\int^{t}_{0}\|\eta_{tt}(t)\|^{2}_{L^{2}}{\rm d}s\leq Cc^{8}_{3}, \quad\|\rho_{t}(t)\|_{H^{2}}\leq Cc^{2}_{4}, \quad \|\eta_{t}(t)\|_{H^{2}}\leq Cc^{2}_{4},$

$\|p_{t}(t)\|_{H^{2}}\leq M(c_{0})c^{2}_{4}, \quad \inf\limits_{B_{R}}\rho(t)\geq C^{-1}\delta, \quad \inf\limits_{B_{R}}\eta(t)\geq C^{-1}\delta,$

这里 $T_{1}=c_{4}^{-1}$ .

引理2.3 对于 $0\leq t \leq\min(T_{\ast}, T_{2})$ ,有

$\|u(t)\|^{2}_{D_{0}^{1}}+\int^{t}_{0}\|u\|^{2}_{D^{2}}{\rm d}s\leq M(c_{0}),$

这里 $T_{2}=c_{4}^{-4} <T_{1}$ .

证把(2.3)式乘 $u_{t}$ 然后在 $B_{R}$ 上积分,我们有

$\frac{1}{2}\frac{\rm d}{{\rm d}t}\int(\mu|\nabla u|^{2}+\lambda|\nabla\cdot u|^{2}){\rm d}x +\int \rho|u_{t}|^{2}{\rm d}x+\int \eta|u_{t}|^{2}{\rm d}x\\ = -\int\nabla(p+\eta)\cdot u_{t}{\rm d}x-\int(\rho+\eta)(v\cdot\nabla v)\cdot u_{t}{\rm d}x-\int(\rho+\eta)\nabla\Phi\cdot u_{t}{\rm d}x\\ = \int(p-p^{\infty})\nabla\cdot u_{t}+(\eta-\eta^{\infty})\nabla\cdot u_{t}{\rm d}x-\int(\rho+\eta)(v\cdot\nabla v)\cdot u_{t}{\rm d}x-\int(\rho+\eta)\nabla\Phi\cdot u_{t}{\rm d}x\\ = \frac{\rm d}{{\rm d}t}\int[(p-p^{\infty})\nabla\cdot u+(\eta-\eta^{\infty})\nabla\cdot u]{\rm d}x-\int p_{t}\nabla\cdot u{\rm d}x-\int \eta_{t}\nabla\cdot u{\rm d}x\\ -\int(\rho+\eta)(v\cdot\nabla v)\cdot u_{t}{\rm d}x-\int\rho\nabla\Phi\cdot u_{t}{\rm d}x-\int\eta\nabla\Phi\cdot u_{t}{\rm d}x.$

(2.13)

同文献[18]类似的方法,由引理2.2, (2.9)-(2.12)式和Sobolev嵌入,我们得到(2.13)式的各项估计

$\int[(p-p^{\infty})\nabla\cdot u+(\eta-\eta^{\infty})\nabla\cdot u]{\rm d}x \\ \leq C\|p-p^{\infty}\|^{2}_{L^{2}}+C\|\eta-\eta^{\infty}\|^{2}_{L^{2}}+\frac{\mu}{12}\|\nabla u\|^{2}_{L^{2}} \leq \frac{\mu}{12}\|\nabla u\|^{2}_{L^{2}}+M(c_{0}),$

(2.14)

$-\int p_{t}\nabla\cdot u{\rm d}x\leq \|\nabla u\|^{2}_{L^{2}}+ C\|p_{t}\|^{2}_{L^{2}}\leq \|\nabla u\|^{2}_{L^{2}}+M(c_{0})c^{4}_{3},$

(2.15)

$-\int \eta_{t}\nabla\cdot u{\rm d}x\leq \frac{1}{4}\|\nabla u\|^{2}_{L^{2}}+Cc^{4}_{3},$

(2.16)

$-\int(\rho+\eta)(v\cdot\nabla v)\cdot u_{t}{\rm d}x \leq \|\rho\|^{\frac{1}{2}}_{L^{\infty}}\|v\|^{2}_{D^{1}_{0}\cap D^{2}}\|\sqrt{\rho}u_{t}\|_{L^{2}}+\|\eta\|^{\frac{1}{2}}_{L^{\infty}}\|v\|^{2}_{D^{1}_{0}\cap D^{2}}\|\sqrt{\eta}u_{t}\|_{L^{2}}\\ \leq \frac{1}{4}\|\sqrt{\rho}u_{t}\|^{2}_{L^{2}}+\frac{1}{4}\|\sqrt{\eta}u_{t}\|^{2}_{L^{2}}+Cc^{4}_{3},$

(2.17)

$-\int\rho\nabla\Phi\cdot u_{t}{\rm d}x-\int\eta\nabla\Phi\cdot u_{t}{\rm d}x\leq \frac{1}{4}\|\sqrt{\rho}u_{t}\|^{2}_{L^{2}}+\frac{1}{4}\|\sqrt{\eta}u_{t}\|^{2}_{L^{2}}+Cc_{0},$

(2.18)

(2.13)式关于 $t$ 积分,由(1.4)式和(2.14)-(2.18)式

$\frac{\mu}{6}\|\nabla u\|^{2}_{L^{2}}+\int^{t}_{0}\int \rho|u_{t}|^{2}{\rm d}x{\rm d}s+\int^{t}_{0}\int \eta|u_{t}|^{2}{\rm d}x{\rm d}s\leq \int^{t}_{0}\|\nabla u\|^{2}_{L^{2}} +M(c_{0})c^{4}_{3}t+M(c_{0}).$

(2.19)

由Gronwall不等式和(2.19)式,我们得到

$\|u(t)\|^{2}_{D_{0}^{1}}+\int^{t}_{0}\|u\|^{2}_{D^{2}}{\rm d}s\leq M(c_{0}),$

(2.20)

对于 $0\leq t \leq\min(T_{\ast}, T_{2})$ ,这里 $T_{2}=c_{4}^{-4}$ .

(2.3)式重新写为

$Lu=-\rho u_{t}-\eta u_{t}-\nabla(p+\eta)-(\rho+\eta)v\cdot\nabla v-(\rho+\eta)\nabla\Phi,$

(2.21)

由 $(2.21)$ 式的正则性估计,我们有

$\|u\|_{D^{2}} \leq C[\|\rho\|^{\frac{1}{2}}_{L^{\infty}}\|\sqrt{\rho}u_{t}\|_{L^{2}}+\|\eta\|^{\frac{1}{2}}_{L^{\infty}}\|\sqrt{\eta}u_{t}\|_{L^{2}}+\|\nabla p\|_{L^{2}}+\|\nabla \eta\|_{L^{2}} \\ +\|\rho\|_{L^{\infty}}\|v\|_{D^{1}_{0}}\|v\|_{D^{1}_{0}\cap D^{2}}+\|\eta\|_{L^{\infty}}\|v\|_{D^{1}_{0}}\|v\|_{D^{1}_{0}\cap D^{2}}+(\|\rho\|_{L^{\infty}}+\|\eta\|_{L^{\infty}})\|\nabla\Phi\|_{L^{2}}]\\ \leq M(c_{0})(\|\sqrt{\rho}u_{t}\|_{L^{2}}+\|\sqrt{\eta}u_{t}\|_{L^{2}}+c^{\frac{3}{2}}_{2}c^{\frac{1}{2}}_{3}),$

(2.22)

则有

$\int^{t}_{0}\|u\|_{D^{2}}{\rm d}s\leq M(c_{0}),$

(2.23)

对于 $0\leq t \leq\min(T_{\ast}, T_{2})$ .

引理2.4 对于 $0\leq t \leq\min(T_{\ast}, T_{3})$ ,有

$\|u(t)\|^{2}_{D^{2}}+\|\sqrt{\rho}u_{t}\|^{2}_{L^{2}}+\|\sqrt{\eta}u_{t}\|^{2}_{L^{2}}+\int^{t}_{0}(\|u\|^{2}_{D^{3}}+\|u_{t}\|^{2}_{D_{0}^{1}}){\rm d}s\leq M(c_{1})c^{\frac{3}{2}}_{2}c^{\frac{1}{2}}_{3}),$

这里 $T_{3}=c_{4}^{-9}\leq T_{2}$ .

证 (2.3)式关于时间 $t$ 求导,可得

$\rho u_{tt}+\eta u_{tt}+L u_{t}+\nabla(p_{t}+\eta_{t})=-\rho_{t}u_{t}-\eta_{t}u_{t}-((\rho+\eta)\cdot v\nabla v)_{t}-(\rho_{t}+\eta_{t})\nabla\Phi.$

(2.24)

(2.24)式乘 $u_{t}$ ,然后在 $B_{R}$ 上积分,由(1.4)式,我们有

$\begin{array}{*{20}{l}}\frac{1}{2}\frac{\rm d}{{\rm d}t}\int\rho|u_{t}|^{2}+\eta|u_{t}|^{2}{\rm d}x+\frac{\mu}{3}\int|\nabla u_{t}|^{2}{\rm d}x \\ \leq -\int\nabla(p_{t}+\eta_{t})\cdot u_{t}{\rm d}x-\frac{1}{2}\int\rho_{t}|u_{t}|^{2}+\eta_{t}|u_{t}|^{2}{\rm d}x\\ -\int[((\rho_{t}+\eta_{t}) v\nabla v)+(\rho+\eta)( v\nabla v)_{t}]\cdot u_{t}{\rm d}x-\int(\rho_{t}+\eta_{t})\nabla\Phi\cdot u_{t}{\rm d}x \\=I_{1}+I_{2}+I_{3}+I_{4}.\end{array}$

(2.25)

接下来,我们对(2.25)式逐项估计.首先,由引理2.2和Cauchy不等式,我们有

$I_{1}=\int(p_{t}+\eta_{t})\nabla\cdot u_{t}{\rm d}x\leq\frac{\mu}{24}\|\nabla u_{t}\|^{2}_{L^{2}}+M(c_{0})c^{4}_{3}.$

(2.26)

且由(2.1)式, Hölder不等式和Cauchy不等式,有

$\begin{array}{*{20}{l}}I_{2} = \int\nabla\cdot(\rho v)(\frac{1}{2}\| u_{t}\|^{2})+\nabla\cdot(\eta v)(\frac{1}{2}\| u_{t}\|^{2}){\rm d}x\\ \leq \int\rho |v|| u_{t}|| \nabla u_{t}|+\eta|v|| u_{t}|| \nabla u_{t}| \\ \leq C\|\rho\|^{\frac{3}{4}}_{L^{\infty}}\|v\|^{2}_{D^{1}_{0}}\|\sqrt{\rho}u_{t}\|^{\frac{1}{2}}_{L^{2}}\|\nabla u_{t}\|^{\frac{13}{2}}_{L^{2}}+C\|\eta\|^{\frac{3}{4}}_{L^{\infty}}\|v\|^{2}_{D^{1}_{0}}\|\sqrt{\eta}u_{t}\|^{\frac{1}{2}}_{L^{2}}\|\nabla u_{t}\|^{\frac{13}{2}}_{L^{2}}\\ \leq \frac{\mu}{24}\|\nabla u_{t}\|^{2}_{L^{2}}+Cc^{7}_{2}\|\sqrt{\rho}u_{t}\|^{2}_{L^{2}}+Cc^{7}_{2}\|\sqrt{\eta}u_{t}\|^{2}_{L^{2}}, \end{array}$

(2.27)

$\begin{array}{*{20}{l}} I_{3} \leq C\|\rho_{t}\|_{H^{1}}\|v\|^{2}_{D^{1}_{0}\cap D^{2}}\|\nabla u_{t}\|_{L^{2}}\|\eta_{t}\|_{H^{1}}\|v\|^{2}_{D^{1}_{0}\cap D^{2}}\|\nabla u_{t}\|_{L^{2}} \\+C\|\rho\|^{\frac{3}{4}}_{L^{\infty}}\|\sqrt{\rho}u_{t}\|^{\frac{1}{2}}_{L^{2}}\|\nabla u_{t}\|^{\frac{1}{2}}_{L^{2}}\|v_{t}\|^{2}_{D^{1}_{0}}\|v\|^{2}_{D^{1}_{0}} \\ \leq \frac{\mu}{24}\|\nabla u_{t}\|^{2}_{L^{2}}+\frac{\epsilon}{2}\|v_{t}\|^{2}_{D^{1}_{0}}+C\epsilon^{-2}c^{7}_{2}(\|\sqrt{\rho}u_{t}\|^{2}_{L^{2}} \|\sqrt{\eta}u_{t}\|^{2}_{L^{2}})+Cc^{8}_{3}, \end{array}$

(2.28)

这里 $\epsilon\in(0, 1)$ 足够小.同样的我们有

$I_{4}\leq C(\|\rho_{t}\|_{L^{2}}\|\Phi\|_{H^{4}}+\|\eta_{t}\|_{L^{2}}\|\Phi\|_{H^{4}})\|\nabla u_{t}\|^{2}_{L^{2}}\leq \frac{\mu}{24}\|\nabla u_{t}\|^{2}_{L^{2}}+M(c_{0})c^{4}_{3}.$

(2.29)

把(2.26)-(2.29)式代入(2.25)式,选择 $\epsilon=c^{-1}_{3}$ ,有

$\begin{array}{*{20}{l}}\frac{\rm d}{{\rm d}t}\int\rho|u_{t}|^{2}+\eta|u_{t}|^{2}{\rm d}x+\frac{\mu}{3}\int|\nabla u_{t}|^{2}{\rm d}x \\\leqCc^{9}_{3}(\|\sqrt{\rho}u_{t}\|^{2}_{L^{2}}\|\sqrt{\eta}u_{t}\|^{2}_{L^{2}})+c^{-1}_{3}\|v_{t}\|^{2}_{D^{1}_{0}}+M(c_{0})c^{8}_{3} , \end{array}$

(2.30)

对于 $0\leq t \leq\min(T_{\ast}, T_{2})$ .

另一方面,由

$u_{t}\in C([0, T];H^{1}_{0}) \quad \mbox{及} \quad u_{t}(0)=-v(0)\cdot\nabla v(0)+g ,$

(2.31)

有

$\|\sqrt{\rho}u_{t}(0)\|_{L^{2}}+\|u_{t}(0)\|_{D^{1}_{0}}\leq Cc^{3}_{1}.$

(2.32)

(2.30)式关于时间 $(0, t)$ 积分,我们有

$\begin{array}{*{20}{l}}\|\sqrt{\rho}u_{t}\|_{L^{2}}^{2}+\|\sqrt{\eta}u_{t}\|_{L^{2}}^{2}+\frac{\mu}{3}\int^{t}_{0}\|\nabla u_{t}\|_{L^{2}}^{2}{\rm d}s \\ \leq Cc^{9}_{3}\int^{t}_{0}(\|\sqrt{\rho}u_{t}\|^{2}_{L^{2}}\|\sqrt{\eta}u_{t}\|^{2}_{L^{2}}){\rm d}s+M(c_{1})(1+c^{8}_{3}t). \end{array}$

由Gronwall不等式,有

$\|\sqrt{\rho}u_{t}\|_{L^{2}}^{2}+\|\sqrt{\eta}u_{t}\|_{L^{2}}^{2}+\int^{t}_{0}\| u\|_{D^{1}_{0}}^{2}{\rm d}s\leq M(c_{1}) ,$

(2.33)

对于 $0\leq t \leq\min(T_{\ast}, T_{3})$ ,这里 $T_{3}=c_{4}^{-9}\leq T_{2}$ .

由(2.22)式有

$\|u\|_{D^{2}}\leq M(c_{1})c^{\frac{3}{2}}_{2}c^{\frac{1}{2}}_{3}.$

(2.34)

由(2.20)式的正则性,我们有

$\|u\|_{D^{3}}\leq M(c_{1})(1+\|v\|^{2}_{D^{1}_{0}\cap D^{2}}+\|u_{t}\|_{D^{1}_{0}}),$

(2.35)

则对于 $0\leq t \leq\min(T_{\ast}, T_{3})$ ,有

$\int^{t}_{0}\|u\|^{2}_{D^{3}}{\rm d}s\leq M(c_{1})\int^{t}_{0}(1+\|v\|^{4}_{D^{1}_{0}\cap D^{2}}+\|u_{t}\|^{2}_{D^{1}_{0}}){\rm d}s\leq M(c_{1}).$

(2.36)

证毕.

引理2.5 对于 $0\leq t \leq\min(T_{\ast}, T_{3})$ ,有

$\|u(t)\|^{2}_{D^{3}}+\|u_{t}\|^{2}_{D_{0}^{1}}+\int^{t}_{0}(\|\sqrt{\rho}u_{tt}\|^{2}_{L^{2}}+\|\sqrt{\eta}u_{tt}\|^{2}_{L^{2}} +\|u\|^{2}_{D^{4}}+\|u_{t}\|^{2}_{D^{2}}){\rm d}s\leq M(c_{1})c^{12}_{3}).$

证 (2.24)式乘 $u_{tt}$ 然后在 $B_{R}$ 上积分,我们得到

$\begin{array}{*{20}{l}}\frac{1}{2}\frac{\rm d}{{\rm d}t}\int(\mu|\nabla u_{t}|^{2}+\lambda|\nabla\cdot u_{t}|){\rm d}x+\int\rho|u_{tt}|^{2}+\eta|u_{tt}|^{2}{\rm d}x \\ = -\int\nabla(p_{t}+\eta_{t})\cdot u_{tt}{\rm d}x-\int\rho_{t}u_{t}u_{tt}-\eta_{t}u_{t}u_{tt}{\rm d}x \\-\int[((\rho_{t}+\eta_{t}) v\cdot\nabla v)+(\rho+\eta) (v\cdot\nabla v)_{t}]\cdot u_{tt}{\rm d}x-\int(\rho_{t}+\eta_{t})\nabla\Phi\cdot u_{tt}{\rm d}x \\ = II_{1}+II_{2}+II_{3}+II_{4}++II_{5}. \end{array}$

(2.37)

利用文献[18]的同样方法,我们有

$\begin{array}{*{20}{l}}II_{1} = \frac{\rm d}{{\rm d}t}\int(p_{t}+\eta_{t})\cdot(\nabla\cdot u_{t}) {\rm d}x+\int(p_{tt}+\eta_{tt})\cdot(\nabla\cdot u_{t}) {\rm d}x\\ \leq \frac{\rm d}{{\rm d}t}\int(p_{t}+\eta_{t})\cdot(\nabla\cdot u_{t}) {\rm d}x+\|p_{tt}\|^{2}_{L^{2}}+\|\eta_{tt}\|^{2}_{L^{2}}+\|\nabla u_{t}\|^{2}_{L^{2}}, \end{array}$

(2.38)

由(2.1)式,我们有

$\begin{array}{*{20}{l}}II_{2} = -\frac{\rm d}{{\rm d}t}\int \rho_{t}(\frac{1}{2}|u_{t}|^{2})+ \eta_{t}(\frac{1}{2}|u_{t}|^{2}){\rm d}x+\int \rho_{tt}(\frac{1}{2}|u_{t}|^{2})+\eta_{tt}(\frac{1}{2}|u_{t}|^{2}){\rm d}x \\ \leq -\frac{\rm d}{{\rm d}t}\int (\rho_{t}+ \eta_{t})(\frac{1}{2}|u_{t}|^{2}){\rm d}x\\+\int( |\rho_{t}||v|+|\rho||v_{t}|)|u_{t}||\nabla u_{t}| +( |\eta_{t}||v|+|\eta||v_{t}|)|u_{t}||\nabla u_{t}|{\rm d}x \\ \leq -\frac{\rm d}{{\rm d}t}\int (\rho_{t}+ \eta_{t})(\frac{1}{2}|u_{t}|^{2}){\rm d}x +\|\rho_{t}\|_{H^{1}}\|v\|_{D^{1}_{0}\cap D^{2}}\|\nabla u_{t}\|^{2}_{L^{2}} \\ +\|\rho\|^{\frac{3}{4}}_{L^{\infty}}\|v_{t}\|_{D^{1}_{0}}\|\sqrt{\rho}u_{t}\|^{\frac{1}{2}}_{L^{2}}\|\nabla u_{t}\|^{\frac{3}{2}}_{L^{2}} +\|\eta_{t}\|_{H^{1}}\|v\|_{D^{1}_{0}\cap D^{2}}\|\nabla u_{t}\|^{2}_{L^{2}}\\+\|\eta\|^{\frac{3}{4}}_{L^{\infty}}\|v_{t}\|_{D^{1}_{0}}\|\sqrt{\eta}u_{t}\|^{\frac{1}{2}}_{L^{2}}\|\nabla u_{t}\|^{\frac{3}{2}}_{L^{2}}\\ \leq -\frac{\rm d}{{\rm d}t}\int (\rho_{t}+ \eta_{t})(\frac{1}{2}|u_{t}|^{2}){\rm d}x+Cc^{3}_{3}\|\nabla u_{t}\|^{2}_{L^{2}}\\ +Cc^{-1}_{3}\|v_{t}\|_{D^{1}_{0}}(\|\sqrt{\rho}u_{t}\|^{2}_{L^{2}}+\|\sqrt{\eta}u_{t}\|^{2}_{L^{2}}+\|\nabla u_{t}\|^{2}_{L^{2}}), \end{array}$

(2.39)

$\begin{array}{*{20}{l}}II_{3} = -\frac{\rm d}{{\rm d}t}\int ((\rho_{t}+\eta_{t}) v\cdot\nabla v)\cdot u_{t}+[(\rho_{tt}+\eta_{tt})(v\cdot\nabla v)]\cdot u_{t}+ (\rho_{t}+\eta_{t})(v\cdot\nabla v)_{t}\cdot u_{t}{\rm d}x \\ \leq -\frac{\rm d}{{\rm d}t}\int( (\rho_{t}+\eta_{t}) v\cdot\nabla v)\cdot u_{t}{\rm d}x+C\|\rho_{tt}\|_{L^{2}}\|v\|^{2}_{D^{1}_{0}\cap D^{2}}\|\nabla u_{t}\|_{L^{2}} \\+C\|\rho_{t}\|_{H^{1}}\|v\|_{D^{1}_{0}\cap D^{2}}\|v_{t}\|_{D_{0}^{1}}\|u_{t}\|_{D_{0}^{1}} +C\|\eta_{tt}\|_{L^{2}}\|v\|^{2}_{D^{1}_{0}\cap D^{2}}\|\nabla u_{t}\|_{L^{2}}\\ +C\|\eta_{t}\|_{H^{1}}\|v\|_{D^{1}_{0}\cap D^{2}}\|v_{t}\|_{D_{0}^{1}}\|u_{t}\|_{D_{0}^{1}} \\ \leq -\frac{\rm d}{{\rm d}t}\int ((\rho_{t}+\eta_{t}) v\cdot\nabla v)\cdot u_{t}{\rm d}x+\|u_{t}\|^{2}_{D_{0}^{1}}+Cc^{4}_{3}(\|\rho_{tt}\|^{2}_{L^{2}}+\|\eta_{tt}\|^{2}_{L^{2}}) \\+Cc^{6}_{3}\|v_{t}\|^{2}_{D_{0}^{1}}, \end{array}$

(2.40)

$II_{4}\leq\frac{1}{2}\|\sqrt{\rho}u_{tt}\|^{2}_{L^{2}}+\frac{1}{2}\|\sqrt{\eta}u_{tt}\|^{2}_{L^{2}}+Cc_{0}c^{2}_{3}\|v_{t}\|^{2}_{D_{0}^{1}},$

(2.41)

$\begin{array}{*{20}{l}} II_{5} = -\frac{\rm d}{{\rm d}t}\int (\rho_{t}+\eta_{t})\nabla\Phi\cdot u_{t}{\rm d}x+\int (\rho_{tt}+\eta_{tt})\nabla\Phi\cdot u_{t}{\rm d}x\\ \leq -\frac{\rm d}{{\rm d}t}\int (\rho_{t}+\eta_{t})\nabla\Phi\cdot u_{t}{\rm d}x+\|u_{t}\|^{2}_{D_{0}^{1}}+C(\|\rho_{tt}\|^{2}_{L^{2}}+\|\eta_{tt}\|^{2}_{L^{2}}). \end{array}$

(2.42)

把估计(2.38)-(2.42)代入(2.37)式,由(2.33)式,我们有

$\begin{array}{*{20}{l}} \frac{1}{2}\frac{\rm d}{{\rm d}t}\int(\mu|\nabla u_{t}|^{2}+\lambda|\nabla\cdot u_{t}|){\rm d}x+\int\rho|u_{tt}|^{2}+\eta|u_{tt}|^{2}{\rm d}x \\ \leq \frac{\rm d}{{\rm d}t}\int[(p_{t}+\eta_{t})\cdot(\nabla\cdot u_{t})-(\rho_{t}+ \eta_{t})(\frac{1}{2}|u_{t}|^{2})-(\rho_{t}+\eta_{t}) v\cdot\nabla v\cdot u_{t}-(\rho_{t}+\eta_{t})\nabla\Phi\cdot u_{t}]{\rm d}x \\+C(\|p_{tt}\|^{2}_{L^{2}}+\|\eta_{tt}\|^{2}_{L^{2}})+Cc^{4}_{3}(\|\rho_{tt}\|^{2}_{L^{2}}+\|\eta_{tt}\|^{2}_{L^{2}})+Cc^{6}_{3}\|v_{t}\|^{2}_{D_{0}^{1}} \\+Cc^{3}_{3}\|u_{t}\|^{2}_{D_{0}^{1}}+C(1+c_{3})^{-1}\|v_{t}\|_{D^{1}_{0}}(\|\sqrt{\rho}u_{t}\|^{2}_{L^{2}}+\|\sqrt{\eta}u_{t}\|^{2}_{L^{2}}+\|\nabla u_{t}\|^{2}_{L^{2}}) \\ \leq \frac{\rm d}{{\rm d}t}\Lambda(t)+M(c_{1})(\|p_{tt}\|^{2}_{L^{2}}+c^{4}_{3}(\|\rho_{tt}\|^{2}_{L^{2}}+\|\eta_{tt}\|^{2}_{L^{2}})+c^{6}_{3}\|v_{t}\|^{2}_{D_{0}^{1}}+ c^{3}_{3}\|u_{t}\|^{2}_{D_{0}^{1}}) \\+Cc^{-1}_{3}\|v_{t}\|^{2}_{D_{0}^{1}}\|u_{t}\|^{2}_{D_{0}^{1}}, \end{array}$

(2.43)

这里

$\Lambda(t)=\int[(p_{t}+\eta_{t})\cdot(\nabla\cdot u_{t})-(\rho_{t}+ \eta_{t})(\frac{1}{2}|u_{t}|^{2})-(\rho_{t}+\eta_{t}) v\cdot\nabla v\cdot u_{t}-(\rho_{t}+\eta_{t})\nabla\Phi\cdot u_{t}]{\rm d}x.$

利用引理2.2,引理2.4和(2.32)式,很容易可以得到

$\begin{array}{*{20}{l}}\Lambda(t) \leq \frac{\mu}{12}\|\nabla u_{t}\|^{2}_{L^{2}}+C(\|p_{t}\|^{2}_{L^{2}}+\|\eta_{t}\|^{2}_{L^{2}}+\|\rho_{t}\|^{2}_{L^{2}}) \\+C(\|\sqrt{\rho}u_{t}\|^{2}_{L^{2}}\|\rho\|_{L^{\infty}}\|v\|^{2}_{D^{1}_{0}\cap D^{2}}+\|\sqrt{\eta}u_{t}\|^{2}_{L^{2}}\|\eta\|_{L^{\infty}}\|v\|^{2}_{D^{1}_{0}\cap D^{2}}+\|\rho_{t}\|^{2}_{L^{2}}\|v\|^{2}_{D^{1}_{0}\cap D^{2}}) \\ \leq \frac{\mu}{12}\|\nabla u_{t}\|^{2}_{L^{2}}+M(c_{1})c^{6}_{3} \end{array}$

和

$\Lambda(0)\leq M(c_{1})c^{6}_{3}.$

(2.43)式关于时间在 $(0, t)$ 上积分且利用(1.4)式和引理2.2,我们得到

$\frac{\mu}{12}\|u_{t}\|^{2}_{D_{0}^{1}}+\int^{t}_{0}\|\sqrt{\rho}u_{tt}\|^{2}_{L^{2}}+\|\sqrt{\eta}u_{tt}\|^{2}_{L^{2}}{\rm d}s\leq M(c_{1})c^{12}_{3} +\int^{t}_{0}Cc^{-1}_{3}\|v_{t}\|^{2}_{D_{0}^{1}}\|u_{t}\|^{2}_{D_{0}^{1}}{\rm d}s$

(2.44)

对于 $0\leq t \leq\min(T_{\ast}, T_{3})$ .由Granwall不等式和(2.43)式有对于 $0\leq t \leq\min(T_{\ast}, T_{3})$

$\|u_{t}\|^{2}_{D_{0}^{1}}+\int^{t}_{0}\|\sqrt{\rho}u_{tt}\|^{2}_{L^{2}}+\|\sqrt{\eta}u_{tt}\|^{2}_{L^{2}}{\rm d}s\leq M(c_{1})c^{12}_{3}.$

(2.45)

由(2.35)式和(2.45)式,有

$\|u\|_{D^{3}}\leq M(c_{1})c^{12}_{3}.$

(2.46)

由下面正则性

$\begin{array}{*{20}{l}}\|u_{t}\|_{D^{2}} \leq C(\|\rho\|^{\frac{1}{2}}_{L^{\infty}}\|\sqrt{\rho}u_{tt}\|_{L^{2}}+\|\eta\|^{\frac{1}{2}}_{L^{\infty}}\|\sqrt{\eta}u_{tt}\|_{L^{2}}+\|\nabla p_{t}\|_{L^{2}}+\|\nabla \eta_{t}\|_{L^{2}} \\+\|\rho_{t}\|_{L^{3}}\|u_{t}\|_{D_{0}^{1}}+\|\eta_{t}\|_{L^{3}}\|u_{t}\|_{D_{0}^{1}}+\|\rho_{t}\|_{L^{2}}\|v\|_{L^{\infty}}\|\nabla v\|_{L^{2}} \\+\|\eta_{t}\|_{L^{2}}\|v\|_{L^{\infty}}\|\nabla v\|_{L^{2}}+\|\rho\|_{L^{3}}\|v_{t}\|_{L^{6}}\|\nabla v\|_{L^{2}}+\|\eta\|_{L^{3}}\|v_{t}\|_{L^{6}}\|\nabla v\|_{L^{2}}\\+\|\rho\|_{L^{3}}\|v\|_{L^{6}}\|\nabla v_{t}\|_{L^{2}}+\|\eta\|_{L^{3}}\|v\|_{L^{6}}\|\nabla v_{t}\|_{L^{2}}+\|\rho_{t}\|_{L^{2}}+\|\eta_{t}\|_{L^{2}}, \\ \leq M(c_{0})(\|\sqrt{\rho}u_{t}\|_{L^{2}}+\|\sqrt{\eta}u_{t}\|_{L^{2}}+c^{\frac{3}{2}}_{2}c^{\frac{1}{2}}_{3}), \end{array}$

(2.47)

则对于 $0\leq t \leq\min(T_{\ast}, T_{3})$ ,有

$\int^{0}_{0}\|u_{t}\|^{2}_{D^{2}}\leq M(c_{1})c^{12}_{3}$

同样地对于 $0\leq t \leq\min(T_{\ast}, T_{3})$ ,我们有

$\int^{0}_{0}\|u_{t}\|^{2}_{D^{4}}\leq M(c_{1})c^{12}_{3}.$

证毕.

由引理2.2-2.5,我们可以得到对于 $0\leq t \leq\min(T_{\ast}, T_{3})$ ,有

$\|u(t)\|^{2}_{D_{0}^{1}}+\int^{T_{\ast}}_{0}\|u\|^{2}_{D^{2}}{\rm d}t\leq M(c_{0}),$

$\|u(t)\|^{2}_{D^{2}}+\int^{T_{\ast}}_{0}(\|u\|^{2}_{D^{3}}+\|u_{t}\|^{2}_{D_{0}^{1}}){\rm d}t\leq M(c_{1})c^{\frac{3}{2}}_{2}c^{\frac{1}{2}}_{3},$

$\|u(t)\|^{2}_{D^{3}}+\|u_{t}\|^{2}_{D_{0}^{1}}+\int^{T_{\ast}}_{0} \|u\|^{2}_{D^{4}}+\|u_{t}\|^{2}_{D^{2}}{\rm d}t\leq M(c_{1})c^{12}_{3},$

$\begin{array}{*{20}{l}}\|\rho-\rho^{\infty}\|_{H^{3}}+\|\eta-\eta^{\infty}\|_{H^{3}}+\|p-p^{\infty}\|_{H^{3}}+\|\rho_{t}(t)\|_{H^{2}}+\|\eta_{t}(t)\|_{H^{2}}+\|\eta_{t}(t)\|_{H^{2}}\\+\|\sqrt{\rho}u_{t}(t)\|^{2}_{L^{2}}+\|\sqrt{\eta}u_{t}(t)\|^{2}_{L^{2}}+\int^{T_{\ast}}_{0}\|\sqrt{\rho}u_{tt}\|^{2}_{L^{2}}+\|\sqrt{\eta}u_{tt}\|^{2}_{L^{2}}{\rm d}t\leq M(c_{1})c^{12}_{3}\end{array}$

这里 $M=M(\cdot)$ 是由前面给出的定义.因此,令

$c_{1}=M(c_{0}), \quad\quad c_{2}=M(c_{1}), \quad \quad c_{3}=c^{5}_{2}, \quad\quad c_{4}=c_{2}c^{12}_{3},$

$T_{\ast}=\min(T, T_{3}), \quad\quad T_{3}=(1+c_{4})^{-9},$

我们有以下估计

$\sup\limits_{0\leq t \leq T_{\ast}}\|u(t)\|^{2}_{D_{0}^{1}}+\int^{T_{\ast}}_{0}\|u\|^{2}_{D^{2}}{\rm d}t\leq c_{2},$

(2.48)

$\sup\limits_{0\leq t \leq T_{\ast}} \|u(t)\|^{2}_{D^{2}}+\int^{T_{\ast}}_{0}(\|u\|^{2}_{D^{3}}+\|u_{t}\|^{2}_{D_{0}^{1}}){\rm d}t\leq c_{3},$

(2.49)

$\sup\limits_{0\leq t \leq T_{\ast}} \|u(t)\|^{2}_{D^{3}}+\|u_{t}\|^{2}_{D_{0}^{1}}+\int^{T_{\ast}}_{0} \|u\|^{2}_{D^{4}}+\|u_{t}\|^{2}_{D^{2}}{\rm d}t\leq c_{4},$

(2.50)

$\begin{array}{*{20}{l}}\sup\limits_{0\leq t \leq T_{\ast}}(\|\rho-\rho^{\infty}\|_{H^{3}}+\|\eta-\eta^{\infty}\|_{H^{3}}+\|p-p^{\infty}\|_{H^{3}}+\|\rho_{t}(t)\|_{H^{2}}+\|\eta_{t}(t)\|_{H^{2}}+\|p_{t}(t)\|_{H^{2}}\\+\|\sqrt{\rho}u_{t}(t)\|^{2}_{L^{2}}+\|\sqrt{\eta}u_{t}(t)\|^{2}_{L^{2}})+\int^{T_{\ast}}_{0}\|\sqrt{\rho}u_{tt}\|^{2}_{L^{2}}+\|\sqrt{\eta}u_{tt}\|^{2}_{L^{2}}{\rm d}t\leq c_{4}.\end{array}$

(2.51)

3 线性方程在 $\mathbb{R}^{3}$ 的解

线性系统(2.1)-(2.4)初边值问题的局部解在区域 $(0, T_{\ast})\times \mathbb{R}^{3}$ 的存在性和唯一性将在本章给出.

引理3.1 令 $v$ 是在全空间 $\mathbb{R}^{3}$ 具有 $(2.6)$ 式的正则性的已知函数,这里 $T$ 由 $T_{\ast}$ 代替,且满足

$\|v(0)\|_{D^{1}_{0}\cap D^{3}}\leq c_{1},$

(3.1)

$\sup\limits_{0\leq t \leq T_{\ast}}\|v(t)\|_{D^{1}_{0}}+\int^{T_{\ast}}_{0}(\|v(t)\|^{2}_{D^{2}}+\|v_{t}(t)\|^{2}_{L^{2}}){\rm d}t \leq c_{2},$

(3.2)

$\sup\limits_{0\leq t \leq T_{\ast}}\|v(t)\|_{D^{2}}+\int^{T_{\ast}}_{0}(\|v(t)\|^{2}_{D^{3}}+\|v_{t}(t)\|^{2}_{D_{0}^{1}}){\rm d}t \leq c_{3},$

(3.3)

${\rm ess}\sup\limits_{0\leq t \leq T_{\ast}}(\|v(t)\|_{D^{3}}+\|v_{t}(t)\|_{D_{0}^{1}})+\int^{T_{\ast}}_{0}(\|v(t)\|^{2}_{D^{4}}+\|v_{t}(t)\|^{2}_{D^{2}}){\rm d}t \leq c_{4}.$

(3.4)

则方程(2.1)-(2.4)的初边值问题有唯一解 $(\rho, p, u, \eta)$ 且满足(2.48)-(2.51)式和

$\left\{ \begin{array}{l} (\rho-\rho^{\infty}, p-p^{\infty}, \eta-\eta^{\infty}) \in C([0, T_{\ast}];H^{3}(\mathbb{R}^{3})), (\sqrt{\rho}u_{t}, \sqrt{\eta}u_{t})\in L^{\infty}(0, T_{\ast};L^{2}(\mathbb{R}^{3})) , \\ u \in C([0, T_{\ast}];D^{1}(\mathbb{R}^{3})\cap D^{3}(\mathbb{R}^{3})) \cap L^{2}(0, T_{\ast};D^{4}), \\ u_{t}\in L^{\infty}(0, T_{\ast};D^{1}(\mathbb{R}^{3}))\cap L^{2}(0, T_{\ast};D^{2}(\mathbb{R}^{3})). \end{array}\right.$

(3.5)

证令 $R$ 充分大, $\varphi\in C^{\infty}_{0}(B_{1})$ 满足

$\varphi=1 , \quad B_{\frac{1}{2}}, \quad\quad\quad 0\leq\varphi(x)\leq1.$

对于 $(t, x)\in [0, T_{\ast}]\times B_{R}$ ,我们定义

$\varphi^{R}(x)=\varphi(\frac{x}{R}), \quad g^{R}(x)=\varphi^{R}(x)g(x), \quad v^{R}(x)=\varphi^{R}(x)v(x), \quad \Phi^{R}(x)=\varphi^{R}(x)\Phi(x),$

$\rho_{0}^{R}(x)=\rho_{0}(x)+R^{-3}, \quad p_{0}^{R}(x)=p_{0}(x)+R^{-3}, \quad \eta_{0}^{R}(x)=\eta_{0}(x)+R^{-3},$

(3.6)

令 $u^{R}_{0}\in H^{1}_{0}(B_{R})\cap H^{3}(B_{R})$ 是下面椭圆方程的解

$\left\{\begin{array}{l} Lu^{R}_{0}=F^{R}_{0}\equiv -\nabla (p^{R}_{0}+\eta^{R}_{0})-(\rho^{R}_{0}+\eta^{R}_{0})\nabla\Phi-(\rho^{R}_{0}+\eta^{R}_{0})g^{R} , \\ u^{R}_{0}|_{\partial B_{R}}=0, \end{array}\right.$

(3.7)

且通过定义 $u^{R}_{0}$ 在 $B_{R}$ 外侧为零,把 $u^{R}_{0}$ 扩展到全空间,利用文献[18]的方法,我们可以证得当 $R\rightarrow\infty$ 有下面强收敛

$u^{R}_{0}\rightarrow u_{0}, \quad D^{1}_{0}.$

(3.8)

由相容性条件,我们可以得到 $L(u^{R}_{0}-u_{0})=F^{R}_{0}-F_{0}$ .所以我们有

$\begin{array}{*{20}{l}} \int_{B_{R}}(\mu |\nabla u^{R}_{0}|^{2}+\lambda |\nabla \cdot u^{R}_{0}|^{2}){\rm d}x\\ = \int_{B_{R}}(\mu\nabla u_{0}\cdot \nabla u^{R}_{0}+\lambda(\nabla \cdot u_{0})(\nabla \cdot u^{R}_{0})){\rm d}x+\int_{B_{R}}(F^{R}_{0}-F_{0})\cdot u^{R}_{0}{\rm d}x.\end{array}$

(3.9)

我们解决(3.9)式的等式右边如下

$\begin{array}{*{20}{l}} \int_{B_{R}}(F^{R}_{0}-F_{0})\cdot u^{R}_{0}{\rm d}x = \int_{B_{R}}(p^{R}_{0}-p_{0}+\eta^{R}_{0}-\eta_{0})(\nabla\cdot u^{R}_{0}){\rm d}x \\-\int_{B_{R}}(\eta^{R}_{0}\nabla\Phi^{R}-\eta_{0}\nabla\Phi)u^{R}_{0}{\rm d}x -\int_{B_{R}}(\rho^{R}_{0}\nabla\Phi^{R}-\rho_{0}\nabla\Phi)u^{R}_{0}{\rm d}x\\ -\int_{B_{R}}(\rho_{0}+\eta_{0})(g^{R}-g)u^{R}_{0}{\rm d}x -2R^{-3}\int_{B_{R}}g^{R}u^{R}_{0}{\rm d}x\\ = \Sigma^{5}_{1}III_{i}, \end{array}$

(3.10)

这里

$\begin{array}{*{20}{l}} III_{1} = \int_{B_{R}}(p^{R}_{0}-p_{0})(\nabla\cdot u^{R}_{0}){\rm d}x+\int_{B_{R}}(\eta^{R}_{0}-\eta_{0})(\nabla\cdot u^{R}_{0}){\rm d}x\\ \leq 2\int_{B_{R}}R^{-3}|\nabla\cdot u^{R}_{0}|{\rm d}x\leq CR^{-\frac{3}{2}}\|\nabla u^{R}_{0}\|_{L^{2}}, \end{array}$

(3.11)

$\begin{array}{*{20}{l}} III_{2} = \int_{B_{R}}[(\eta_{0}+R^{-3})\nabla(\varphi^{R}\Phi)-\eta_{0}\nabla\Phi] u^{R}_{0}{\rm d}x \\ = \int_{B_{R}}(\eta_{0}\varphi^{R}\nabla\Phi+\frac{1}{R}\eta_{0}\Phi\nabla\varphi^{R} +R^{-3}\varphi^{R}\nabla\Phi+\frac{1}{R}R^{-3}\Phi\nabla\varphi^{R}-\eta_{0}\nabla\Phi) u^{R}_{0}{\rm d}x\\ = \int_{B_{R}}(\eta_{0}\nabla\Phi(\varphi^{R}-1)+\frac{1}{R}\eta_{0}\Phi\nabla\varphi^{R} +R^{-3}\varphi^{R}\nabla\Phi+\frac{1}{R}R^{-3}\Phi\nabla\varphi^{R}) u^{R}_{0}{\rm d}x, \end{array}$

则有

$\begin{array}{*{20}{l}} III_{2}+\cdots+III_{5}\\ = \int_{B_{R}}(\rho_{0}+\eta_{0})(\varphi^{R}-1)(\nabla\Phi+g)u^{R}_{0}+R^{-3}\varphi^{R}(\nabla\Phi+g)u^{R}_{0}\\ +R^{-1}(\rho_{0}+\eta_{0})\nabla\varphi^{R}\Phi u^{R}_{0}+R^{-4}\Phi \nabla\varphi^{R}u^{R}_{0}{\rm d}x \\ \leq \int_{B_{R}}|Lu_{0}+\nabla(p_{0}+\eta_{0})||\varphi^{R}-1||u^{R}_{0}|+R^{-3}|\varphi^{R}||\nabla\Phi+g||u^{R}_{0}| \\ +R^{-1}|\rho_{0}+\eta_{0}||\nabla\varphi||\Phi|| u^{R}_{0}|+R^{-4}|\Phi| |\nabla\varphi||u^{R}_{0}|{\rm d}x \\ \leq C(\|\nabla u_{0}\|_{L^{2}(B_{R}\setminus B_{\frac{R}{2}})}+\|p_{0}-p^{\infty}\|_{L^{2}(B_{R}\setminus B_{\frac{R}{2}})}+\|\eta_{0}-\eta^{\infty}\|_{L^{2}(B_{R}\setminus B_{\frac{R}{2}})})\|\nabla u^{R}_{0}\|_{L^{2}} \\ +CR^{-1}(\|\rho_{0}-\rho^{\infty}\|_{L^{2}}+\|\eta_{0}-\eta^{\infty}\|_{L^{2}}+\|\rho^{\infty}\|_{L^{2}}+\|\eta^{\infty}\|_{L^{2}})\|\nabla u^{R}_{0}\|_{L^{2}}\\ +CR^{-4}\|\nabla u^{R}_{0}\|_{L^{2}}+CR^{-3}\|\nabla\Phi+g\|_{L^{6}}\| u^{R}_{0}\|_{L^{6}}|B_{R}|^{\frac{2}{3}}.\end{array}$

(3.12)

把估计式(3.11)-(3.12)代入(3.10)式中并结合(3.9)式,我们有

$\| u^{R}_{0}\|_{D_{0}^{1}({{\mathbb{R}}^{3}})}=o(1), \int_{R^{3}}(F^{R}_{0}-F_{0})\cdot u^{R}_{0}{\rm d}x=o(1),$

(3.13)

这里,当 $R\rightarrow \infty$ , $o(1)\rightarrow 0$ .因此,这里存在子列 $\{R_{j}\}$ , $R_{j}\rightarrow \infty$ ,和 $u^{\infty}_{0}$ 使得当 $j\rightarrow\infty$ 有以下弱收敛

$u^{R_{j}}_{0}\rightarrow u^{\infty}_{0}, \quad D^{1}_{0}(\mathbb{R}^{3}) ,$

$Lu^{R_{j}}_{0}\rightarrow Lu^{\infty}_{0}, \quad D^{-1}_{0}(\mathbb{R}^{3}).$

同样地,我们有

$\int_{{\mathbb{R}}^{3}}(F^{R}_{0}-F_{0})\cdot \Psi {\rm d}x\rightarrow 0, \quad R\rightarrow\infty, \quad \forall\Psi\in D^{1}_{0}(\mathbb{R}^{3}),$

由 $L(u^{R}_{0}-u_{0})=F^{R}_{0}-F_{0}$ ,我们有以下弱收敛

$Lu^{R_{j}}_{0}\rightarrow Lu_{0}, \quad D^{-1}_{0}(\mathbb{R}^{3}).$

由极限的唯一性,有 $Lu^{\infty}_{0}= Lu_{0}$ ,即 $u^{\infty}_{0}=u_{0}$ ,有以下弱收敛

$u^{R_{j}}_{0}\rightarrow u_{0}, \quad D^{1}_{0}(\mathbb{R}^{3}).$

(3.14)

利用(3.13)和(3.9)式,可以得到以下弱收敛

$\|\nabla u^{R_{j}}_{0}\|_{L^{2}(\mathbb{R}^{3})}\rightarrow \|\nabla u_{0}\|_{L^{2}(\mathbb{R}^{3})}, \quad j\rightarrow\infty.$

(3.15)

由(3.14)和(3.15)式有以下强收敛

$u^{R_{j}}_{0}\rightarrow u_{0}, \quad D^{1}_{0}(\mathbb{R}^{3}), \quad j\rightarrow\infty.$

(3.16)

综合以上,我们可知每个序列 $\{u^{R}_{0}\}$ 存在一个子列收敛到同样的在 $D^{1}_{0}(\mathbb{R}^{3})$ 中极限 $u_{0}$ ,这意味着当 $R\rightarrow \infty$ . (3.8)式所有序列 $\{u^{R}_{0}\}$ 都收敛到 $u_{0}$ ,得证.

接下来,为了证明引理3.1,我们构造下面近似问题

$\rho_{t}+\nabla\cdot(\rho v^{R})=0,$

(3.17)

$p_{t}+v^{R}\cdot\nabla p+\gamma p\nabla\cdot v^{R}=0,$

(3.18)

$(\rho+\eta)u_{t}+\nabla(p+\eta)+Lu=-(\rho+\eta)v\cdot\nabla v^{R}-(\rho+\eta)\nabla \Phi^{R},$

(3.19)

$\eta_{t}+\nabla\cdot(\eta v^{R})=0,$

(3.20)

$(\rho, p, u, \eta)|_{t=0}=(\rho^{R}_{0}, p^{R}_{0}, u^{R}_{0}, \eta^{R}_{0}), \quad \mathbb{R}^{3},$

(3.21)

$u=0, \quad (0, T)\times \partial B_{R},$

(3.22)

对于 $(t, x)\in (0, T)\times \partial B_{R}$ .由(3.6)式容易得到 $\rho^{R}_{0}>0$ ,则这里方程(3.17)-(3.22)存在唯一强解 $(\rho^{R}, p^{R}, u^{R}, \eta^{R})$ 满足(2.7)式,这里 $T$ 由 $T_{\ast}$ 代替.

由 $v^{R}$ , $\rho^{R}_{0}$ , $p^{R}_{0}$ , $\eta^{R}_{0}$ 和 $g^{R}$ 的定义,当 $R\rightarrow \infty$ ,我们有

$\left\{\begin{array}{l} \|v^{R}-v\|_{C([0, T\ast];D^{1}_{0}\cap D^{3})}+ \|v_{t}^{R}-v_{t}\|_{L^{\infty}(0, T\ast;D^{1}_{0}\cap L^{2}(0, T\ast;D^{2}}\rightarrow \infty, \\ \|\sqrt{\rho^{R}_{0}}g^{R}-\sqrt{\rho_{0}}g\|_{L^{2}} + \|\sqrt{\eta^{R}_{0}}g^{R}-\sqrt{\eta_{0}}g\|_{L^{2}}+\|g^{R}-g\|_{D^{1}_{0}}+\|p^{R}_{0}-p_{0}\|_{H^{3}}\rightarrow 0. \end{array}\right.$

(3.23)

结合(2.8), (3.1)-(3.4), (3.8)和(3.23)式,我们知这里存在一个足够的数 $R_{1}>1$ ,使得对于所有 $R>R_{1}$ ,

$\begin{array}{*{20}{l}} 1+(\rho^{\infty}+p^{\infty}+\eta^{\infty}+R^{-3})+\|\rho^{R}_{0}-(\rho^{\infty}+R^{-3})\|_{H^{3}(B_{R})}\\+\|p^{R}_{0}-(p^{\infty}+R^{-3})\|_{H^{3}(B_{R})} +\|\eta^{R}_{0}-(\eta^{\infty}+R^{-3})\|_{H^{3}(B_{R})}+ \|u^{R}_{0}\|_{D_{0}^{1}(B_{R})}\\+ \|\sqrt{\rho^{R}_{0}}g^{R}\|_{L^{2}(B_{R})}+\|\sqrt{\eta^{R}_{0}}g^{R}\|_{L^{2}(B_{R})}+\|g^{R}\|_{D_{0}^{1}(B_{R})}, \end{array}$

(3.24)

且 $v^{R}$ 由(2.9)-(2.12)式决定,所以估计式(2.48)-(2.51)决定了在 $(0, T_{\ast})\times B_{R}$ 中的解 $(\rho^{R}, p^{R},$ $u^{R}, \eta^{R})$ .由在 $R$ 上的(2.48)-(2.51)式一致性估计知这里存在一个子列 $\{R_{j}\}$ 和 $(\rho, p, u, \eta)$ 使得

$(\rho^{R_{j}}, p^{R_{j}}, u^{R_{j}}, \eta^{R_{j}})\rightarrow (\rho, p, u, \eta)\弱或弱-\ast当j\rightarrow \infty.$

由下半连续性知 $(\rho, p, u, \eta)$ 对于每个 $R>R_{1}$ 在 $(0, T_{\ast})\times B_{R}$ 中满足估计式(2.48)-(2.51),所以 $(\rho, p, u, \eta)$ 也有引理3.5的结论的正则性.

接下来,我们讲证明 $(\rho, p, u, \eta)$ 是方程组(2.1)-(2.4)的解.首先,由(2.48)-(2.51)式,我们很容易得到当 $|x|\rightarrow\infty$ 有 $(\rho, p, u, \eta)\rightarrow (\rho^{\infty}, p^{\infty}, 0, \eta^{\infty})$ .对于任意足够大的数 $R>R_{1}$ , $(\rho^{R_{j}}, p^{R_{j}}, u^{R_{j}}, \eta^{R_{j}})$ 在 $(0, T_{\ast})\times B_{R}$ 中满足(2.48)-(2.51)式,由紧性原理,我们有

$(\rho^{R_{j}}, p^{R_{j}}, u^{R_{j}}, \eta^{R_{j}})\rightarrow (\rho, p, u, \eta), \quad C([0, T_{\ast}];H^{1}( B_{R})), \quad j\rightarrow\infty.$

(3.25)

由(3.6), (3.8)和(3.25)式,知 $(\rho, p, u, \eta)$ 是方程(2.1)-(2.5)的解.所以 $(\rho, p, u, \eta)$ 是方程(2.1)-(2.4)的解,且当 $R$ 足够大时有以下正则性

$(\rho-\rho^{\infty}, p-p^{\infty}, \eta-\eta^{\infty})\in L^{\infty}(0, T_{\ast};H^{3}), u\in L^{\infty}(0, T_{\ast};D^{1}_{0}\cap D^{3})\cap \in L^{2}(0, T_{\ast};D^{4}),$

$u_{t}\in L^{\infty}(0, T_{\ast};D^{1}_{0})\cap \in L^{2}(0, T_{\ast};D^{2}), (\sqrt{\rho}u_{t}, \sqrt{\eta}u_{t})\in L^{\infty}(0, T_{\ast};L^{2}).$

最后,由Lions-Aubin引理,有

$u\in C([0, T_{\ast}];D^{1}_{0}\cap D^{3}),$

由(2.1)式, (2.2)式和文献[18]中的引理2.1,我们有

$(\rho-\rho^{\infty}, p-p^{\infty}, \eta-\eta^{\infty})\in C([0, T_{\ast}];H^{3}).$

初边值条件为(1.14)和(1.15)式时方程(2.1)-(2.4)的解唯一性很容易得证,这里略去.

4 主要结果的证明

为了证明定理1.1,我们首先考虑下面热方程

$\left\{\begin{array}{l} F_{t}-\triangle F=0, \\ F|_{t=0}=F(0)\equiv-\nabla (p_{0}+\eta_{0})-(\rho_{0}+\eta_{0})\nabla\Phi-(\rho_{0}+\eta_{0})g. \end{array} \right.$

(4.1)

由相容性条件有 $F(0)\in H^{1}(\mathbb{R}^{3})$ ,再由基本的能量方法我们可知,以上热方程有唯一解 $F\in C([0, \infty);H^{1})\cap L^{2}([0, \infty);H^{2})$ .

由 $u_{0}\in D^{1}_{0}\cap D^{3}$ 和 $Lu_{0}-F(0)=0\in D^{1}_{0}$ ,可知 $\omega=u_{0}\in C(0, \infty;D^{1}_{0}\cap D^{3})\cap L^{2}(0, \infty;D^{4})$ 是以下线性抛物方程的唯一解

$\left\{\begin{array}{l} \omega_{t}+L\omega=F , \\ \omega(0)=u_{0}, \end{array} \right.$

(4.2)

且

$\sup\limits_{0\leq t \leq T_{\ast}}(\|u^{0}\|_{D^{1}_{0}\cap D^{3}}+\|u^{0}_{t}\|_{D^{1}_{0}})+\int^{T_{\ast}}_{0}(\|u^{0}\|^{2}_{D^{4}}+\|u^{0}_{t}\|^{2}_{D^{2}}) \leq C(\|F(0)\|^{2}_{H^{1}}+\|u_{0}\|^{2}_{D^{1}_{0}\cap D^{3}}+1).$

(4.3)

定义 $c_{0}$ :

$\begin{array}{*{20}{l}} c_{0} = 2+\rho^{\infty}+p^{\infty}+\eta^{\infty}+\|\rho_{0}-\rho^{\infty}\|_{H^{3}} +\|p_{0}-p^{\infty}\|_{H^{3}}+\|\eta_{0}-\eta^{\infty}\|_{H^{3}}\\ +\|u_{0}\|_{D^{1}_{0}} \|\sqrt{\rho_{0}}g\|_{L^{2}}+ \|\sqrt{\eta_{0}}g\|_{L^{2}}+\|g\|_{D^{1}_{0}}, \end{array}$

(4.4)

这里我们选择仅依赖于 $c_{0}$ 和 $C$ 适合的正常数 $c_{1}$ , $c_{2}$ , $c_{3}$ , $c_{4}$ 和 $T_{\ast}$ .对于 $u_{0}$ ,由椭圆性估计有

$\|u_{0}\|_{D^{1}_{0}\cap D^{3}}\leq C(\|F(0)\|_{ H^{1}}+\|u_{0}\|^{2}_{D^{1}_{0}}).$

(4.5)

由(4.1)式知

$\|F(0)\|_{ H^{1}}\leq C \|-\nabla(p_{0}+\eta_{0})- (\rho_{0}+\eta_{0})\nabla \Phi-(\rho_{0}+\eta_{0})g\|_{H^{1}}\leq M(c_{0}).$

(4.6)

结合(2.48)-(2.51)式和(4.3)-(4.6)式,我们有

$\sup\limits_{0\leq t \leq T_{\ast}}(\|u^{0}(t)\|_{D^{1}_{0}\cap D^{3}}+u_{t}^{0}(t)\|_{D^{1}_{0}})+\int^{T_{\ast}}_{0}(u_{t}^{0}(t)\|_{D^{2}}+\|u^{0}(t)\|_{ D^{4}}){\rm d}t\leq M(c_{0})\leq c_{1}.$

(4.7)

下面我们由 $v$ 的正则性定义出 $u^{0}$ ,这样我们可以通过 $v=u^{0}$ 解出(2.1)-(2.4), (1.14)和(1.15)式来定义 $(\rho^{1}, p^{1}, u^{1}, \eta^{1})$ ,且 $(\rho^{1}, p^{1}, u^{1}, \eta^{1})$ 具有(2.50)-(2.51)式的正则性.利用归纳法,假设给定 $u^{k}$ ,这里 $k\geq 1$ ,我们可以得到一组解 $(\rho^{k+1}, p^{k+1}, u^{k+1}, \eta^{k+1})$ 且满足(2.50)-(2.51)式,所以我们构造近似解 $(\rho^{k}, p^{k}, u^{k}, \eta^{k})$ $(k\geq 1)$ .由引理3.1知存在一个常数 $\bar{C}>1$ 使得

$\begin{array}{*{20}{l}} \sup\limits_{0\leq t \leq T_{\ast}}\Big(\|\rho^{k}-\rho^{\infty}\|_{H^{3}}+\|p^{k}-p^{\infty}\|_{H^{3}}+\|\eta^{k}-\eta^{\infty}\|_{H^{3}}+\|\rho_{t}^{k}\|_{H^{2}}+\|\eta_{t}^{k}\|_{H^{2}} \\+\|p_{t}^{k}\|_{H^{2}}+\|u^{k}\|_{D^{1}_{0}\cap D^{3}}\Big)+\mathop{\rm ess\sup}\limits_{0\leq t \leq T_{\ast}}\Big(\|u_{t}^{k}(t)\|_{D^{1}_{0}}+\|\sqrt{\rho^{k}}u^{k}_{t}\|_{L^{2}}+\|\sqrt{\eta^{k}}u^{k}_{t}\|_{L^{2}}\Big)\\+\int^{T_{\ast}}_{0}\Big(\|u_{t}^{k}(t)\|^{2}_{D^{2}}+\|u^{k}\|_{D^{4}}+\|\sqrt{\rho^{k}}u^{k}_{tt}\|^{2}_{L^{2}}+\|\sqrt{\eta^{k}}u^{k}_{tt}\|^{2}_{L^{2}}\Big){\rm d}t\leq \bar{C}, \end{array}$

(4.8)

这里 $\bar{C}$ 是不依赖于 $k$ ,仅依赖于 $c_{0}$ 和 $C$ 的正常数.

接下里,我们需要证明全序列 $\{(\rho^{k}, p^{k}, u^{k}, \eta^{k})\}$ 收敛到原始问题(1.10)-(1.15)的解.定义

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

$\bar{\eta}^{k+1}=\eta^{k+1}-\eta^{k}, \quad p^{k}=p(\rho^{k+1}),$

这里 $p^{k}=a(\rho^{k})^{\gamma}$ ,且方程变为

$\bar{\rho}^{k+1}_{t}+\nabla\cdot(\bar{\rho}^{k+1}u^{k})+\nabla\cdot(\rho^{k}\bar{u}^{k})=0,$

(4.9)

$\bar{p}^{k+1}_{t}+u^{k}\cdot\nabla \bar{p}^{k+1}+\bar{u}^{k}\cdot\nabla p^{k}+\gamma\bar{p}^{k+1}\nabla\cdot u^{k}+\gamma p^{k+1}\nabla\cdot\bar{u}^{k-1} =0,$

(4.10)

$\begin{array}{*{20}{l}} (\rho^{k+1}+\eta^{k+1})\bar{u}^{k+1}_{t}+ (\rho^{k+1}+\eta^{k+1})u^{k}\nabla \bar{u}^{k+1}+L\bar{u}^{k+1}+\nabla(\bar{p}^{k+1}+\bar{\eta}^{k+1})\\ = -(\bar{\rho}^{k+1}+\bar{\eta}^{k+1})(u^{k}_{t}-u^{k-1}\nabla u^{k-1}+\nabla\Phi)\\+(\rho^{k+1}+\eta^{k+1})(u^{k}\nabla \bar{u}^{k+1}-\bar{u}^{k}\nabla u^{k}-u^{k-1}\nabla \bar{u}^{k})\end{array}$

(4.11)

$\bar{\eta}^{k+1}_{t}+\nabla\cdot(\bar{\eta}^{k+1}u^{k})+\nabla\cdot(\eta^{k}\bar{u}^{k})=0,$

(4.12)

把(4.9)式乘 $\bar{\rho}^{k+1}$ 并在 $\mathbb{R}^{3}$ 上积分,有

$\begin{array}{*{20}{l}} \frac{\rm d}{{\rm d}t}\|\bar{\rho}^{k+1}\|^{2}_{L^{2}} \leq C\int (|\nabla u^{k}||\bar{\rho}^{k+1}|^{2}+|\nabla \bar{\rho}^{k+1}||u^{k}||\bar{\rho}^{k+1}|+|\nabla \bar{u}^{k}||\rho^{k}||\bar{\rho}^{k+1}|+|\nabla \rho^{k}||\bar{u}^{k}||\bar{\rho}^{k+1}|){\rm d}x\\ \leq C\int (|\nabla u^{k}||\bar{\rho}^{k+1}|^{2}+(|\nabla \rho^{k}||\bar{u}^{k}|+|\nabla \bar{u}^{k}||\rho^{k}|)|\bar{\rho}^{k+1}|){\rm d}x\quad\quad\quad\quad\quad\\ \leq C\|\nabla u^{k}\|_{L^{\infty}}\|\bar{\rho}^{k+1}\|^{2}_{L^{2}}+C(\|\nabla \rho^{k}\|_{H^{1}}+\|\rho^{k}\|_{L^{\infty}})\|\nabla \bar{u}^{k}\|_{L^{2}}\|\bar{\rho}^{k+1}\|_{L^{2}}, \end{array}$

利用(4.8)式知

$\frac{\rm d}{{\rm d}t}\|\bar{\rho}^{k+1}\|^{2}_{L^{2}}\leq \bar{C}\varepsilon^{-1}\|\bar{\rho}^{k+1}\|^{2}_{L^{2}}+\varepsilon \|\nabla \bar{u}^{k}\|^{2}_{L^{2}},$

(4.13)

对于 $0\leq t\leq T_{\ast}$ ,这里 $\varepsilon\in (0, 1)$ 是个小参数.

同样地对于 $0\leq t\leq T_{\ast}$ ,我们有

$\frac{\rm d}{{\rm d}t}\|\bar{p}^{k+1}\|^{2}_{L^{2}}\leq \bar{C}\varepsilon^{-1}\|\bar{p}^{k+1}\|^{2}_{L^{2}}+\varepsilon \|\nabla \bar{u}^{k}\|^{2}_{L^{2}},$

(4.14)

$\frac{\rm d}{{\rm d}t}\|\bar{\eta}^{k+1}\|^{2}_{L^{2}}\leq \bar{C}\varepsilon^{-1}\|\bar{\eta}^{k+1}\|^{2}_{L^{2}}+\varepsilon \|\nabla \bar{u}^{k}\|^{2}_{L^{2}},$

(4.15)

这里 $\varepsilon\in (0, 1)$ 是个小参数.

把(4.9)式乘sgn $(\bar{\rho}^{k+1})|\bar{\rho}^{k+1}|^{\frac{1}{2}}$ 并在 $\mathbb{R}^{3}$ 上积分有

$\frac{\rm d}{{\rm d}t}\|\bar{\rho}^{k+1}\|^{\frac{3}{2}}_{L^{\frac{3}{2}}}\leq C\|\nabla u^{k}\|_{L^{\infty}}\|\bar{\rho}^{k+1}\|^{\frac{3}{2}}_{L^{\frac{3}{2}}}+C\| \rho^{k}\|_{H^{1}}\|\nabla \bar{u}^{k}\|_{L^{2}}\|\bar{\rho}^{k+1}\|^{\frac{1}{2}}_{L^{\frac{3}{2}}},$

上式乘 $\|\bar{\rho}^{k+1}\|^{\frac{1}{2}}_{L^{\frac{3}{2}}}$ 并利用(4.8)式,有

$\frac{\rm d}{{\rm d}t}\|\bar{\rho}^{k+1}\|^{2}_{L^{\frac{3}{2}}}\leq \bar{C}\varepsilon^{-1}\|\bar{\rho}^{k+1}\|^{2}_{L^{\frac{3}{2}}}+\varepsilon\|\nabla \bar{u}^{k}\|^{2}_{L^{2}},$

(4.16)

对于 $0\leq t\leq T_{\ast}$ .

同样地,我们有

$\frac{\rm d}{{\rm d}t}\|\bar{\eta}^{k+1}\|^{2}_{L^{\frac{3}{2}}}\leq \bar{C}\varepsilon^{-1}\|\bar{\eta}^{k+1}\|^{2}_{L^{\frac{3}{2}}}+\varepsilon\|\nabla \bar{u}^{k}\|^{2}_{L^{2}},$

(4.17)

对于 $0\leq t\leq T_{\ast}$ .

我们把(4.11)式乘 $\bar{u}^{k+1}$ 并在 $\mathbb{R}^{3}$ 上积分有

$\begin{array}{*{20}{l}} \frac{1}{2}\frac{\rm d}{{\rm d}t}\int(\rho^{k+1}+\eta^{k+1})|\bar{u}^{k+1}|^{2}{\rm d}x+\frac{\mu}{3} \|\nabla \bar{u}^{k+1}\|^{2}_{L^{2}} \\ \leq C\int|\bar{\rho}^{k+1}+\bar{\eta}^{k+1}||u^{k}_{t}||\bar{u}^{k+1}|{\rm d}x+C \int |\bar{\rho}^{k+1}+\bar{\eta}^{k+1}||u^{k-1}\nabla u^{k-1}||\bar{u}^{k+1}|{\rm d}x \\ +C \int |\rho^{k+1}+\eta^{k+1}|(|u^{k}||\nabla \bar{u}^{k+1}|+|\bar{u}^{k}||\nabla u^{k}|+|u^{k-1}||\nabla \bar{u}^{k}|)|\bar{u}^{k+1}|{\rm d}x \\ +C \int |\bar{\rho}^{k+1}+\bar{\eta}^{k+1}||\nabla\Phi||\bar{u}^{k+1}|{\rm d}x+C \int |\bar{p}^{k+1}+\bar{\eta}^{k+1}||\nabla \bar{u}^{k+1}|{\rm d}x\\ = \mathop \sum \limits_{i = 1}^5 IV_{i}.\end{array}$

(4.18)

利用Hölder不等式, Cauchy不等式, Sobolev不等式和(4.8)式,有

$\begin{array}{*{20}{l}} IV_{1} \leq C \int |\bar{\rho}^{k+1}||u^{k}_{t}||\bar{u}^{k+1}|{\rm d}x+C \int |\bar{\eta}^{k+1}||u^{k}_{t}||\bar{u}^{k+1}|{\rm d}x\\ \leq C \|\bar{\rho}^{k+1}\|_{L^{\frac{3}{2}}}\|u^{k}_{t}\|_{L^{6}}\|\bar{u}^{k+1}\|_{L^{6}}+C \|\bar{\eta}^{k+1}\|_{L^{\frac{3}{2}}}\|u^{k}_{t}\|_{L^{6}}\|\bar{u}^{k+1}\|_{L^{6}}\\ \leq C \|\bar{\rho}^{k+1}\|_{L^{\frac{3}{2}}}\|u^{k}_{t}\|_{D^{1}_{0}}\|\nabla\bar{u}^{k+1}\|_{L^{2}}+C \|\bar{\eta}^{k+1}\|_{L^{\frac{3}{2}}}\|u^{k}_{t}\|_{D^{1}_{0}}\|\nabla\bar{u}^{k+1}\|_{L^{2}}\\ \leq \frac{\mu}{20}\|\nabla\bar{u}^{k+1}\|^{2}_{L^{2}}+\bar{C} \|\bar{\rho}^{k+1}\|^{2}_{L^{\frac{3}{2}}}+C \|\bar{\eta}^{k+1}\|_{L^{\frac{3}{2}}}, \quad\end{array}$

$\begin{array}{*{20}{l}} IV_{2} \leq C \|\bar{\rho}^{k+1}\|_{L^{2}}\|u^{k-1}\|_{L^{6}}\|\nabla u^{k-1}\|_{L^{6}}\|\bar{u}^{k+1}\|_{L^{6}}\\ + C \|\bar{\eta}^{k+1}\|_{L^{2}}\|u^{k-1}\|_{L^{6}}\|\nabla u^{k-1}\|_{L^{6}}\|\bar{u}^{k+1}\|_{L^{6}}\\ \leq C \|\bar{\rho}^{k+1}\|_{L^{2}}\|u^{k-1}\|_{D^{1}_{0}}\|\nabla u^{k-1}\|_{D^{1}_{0}}\|\bar{u}^{k+1}\|_{D^{1}_{0}}\\ + C \|\bar{\eta}^{k+1}\|_{L^{2}}\|u^{k-1}\|_{D^{1}_{0}}\|\nabla u^{k-1}\|_{D^{1}_{0}}\|\bar{u}^{k+1}\|_{D^{1}_{0}}\\ \leq \frac{\mu}{20}\|\nabla\bar{u}^{k+1}\|^{2}_{L^{2}}+ \bar{C}\|\bar{\rho}^{k+1}\|^{2}_{L^{2}}+\bar{C}\|\bar{\eta}^{k+1}\|^{2}_{L^{2}}, \end{array}$

$IV_{3}\leq \frac{\mu}{20}\|\nabla\bar{u}^{k+1}\|^{2}_{L^{2}}+\bar{C}\|\bar{\rho}^{k+1}\|^{2}_{L^{2}}+\bar{C}\|\bar{\eta}^{k+1}\|^{2}_{L^{2}},$

$\begin{array}{*{20}{l}} IV_{4} \leq C \|\rho^{k+1}\|^{\frac{1}{2}}_{L^{\infty}}\|u^{k}\|_{L^{\infty}}\|\nabla\bar{u}^{k+1}\|_{L^{2}}\|\sqrt{\rho}^{k+1}\bar{u}^{k+1}\|^{2}_{L^{2}} \\ +C \|\eta^{k+1}\|^{\frac{1}{2}}_{L^{\infty}}\|u^{k}\|_{L^{\infty}}\|\nabla\bar{u}^{k+1}\|_{L^{2}}\|\sqrt{\eta}^{k+1}\bar{u}^{k+1}\|^{2}_{L^{2}}\\ +C\|\rho^{k+1}\|^{\frac{1}{2}}_{L^{\infty}}(\|u^{k}\|_{D^{1}_{0}\cap D^{2}}+\|u^{k-1}\|_{D^{1}_{0}\cap D^{2}})\|\nabla\bar{u}^{k}\|_{L^{2}}\|\sqrt{\rho}^{k+1}\bar{u}^{k+1}\|_{L^{2}}\\ +C\|\eta^{k+1}\|^{\frac{1}{2}}_{L^{\infty}}(\|u^{k}\|_{D^{1}_{0}\cap D^{2}}+\|u^{k-1}\|_{D^{1}_{0}\cap D^{2}})\|\nabla\bar{u}^{k}\|_{L^{2}}\|\sqrt{\eta}^{k+1}\bar{u}^{k+1}\|_{L^{2}}\\ \leq \frac{\mu}{20}\|\nabla\bar{u}^{k+1}\|^{2}_{L^{2}}+\varepsilon^{-1}\bar{C}(\|\sqrt{\rho}^{k+1}\bar{u}^{k+1}\|^{2}_{L^{2}}+\|\sqrt{\eta}^{k+1}\bar{u}^{k+1}\|^{2}_{L^{2}}), \end{array}$

$\begin{array}{*{20}{l}} IV_{5} \leq C(\|\bar{p}^{k+1}\|_{L^{2}}+\|\bar{\eta}^{k+1}\|_{L^{2}})\|\nabla\bar{u}^{k+1}\|_{L^{2}} \\ \leq \frac{\mu}{20}\|\nabla\bar{u}^{k+1}\|^{2}_{L^{2}}+\bar{C}\|\bar{p}^{k+1}\|^{2}_{L^{2}}+\bar{C}\|\bar{\eta}^{k+1}\|^{2}_{L^{2}}, \end{array}$

把估计式 $IV_{i}$ 带入(4.17)式,这里 $i=1, \cdots , 5$ , $t\in [0, T_{\ast}]$ ,我们得到

$\frac{\rm d}{{\rm d}t}\int(\rho^{k+1}+\eta^{k+1})|\bar{u}^{k+1}|^{2}{\rm d}x+\mu\|\nabla\bar{u}^{k+1}\|^{2}_{L^{2}}\leq \varepsilon^{-1}\bar{C}A^{k+1}(t)+\varepsilon \|\nabla\bar{u}^{k}\|^{2}_{L^{2}},$

(4.19)

这里

$A^{k+1}=\|\sqrt{\rho}^{k+1}\bar{u}^{k+1}\|^{2}_{L^{2}}+\|\sqrt{\eta}^{k+1}\bar{u}^{k+1}\|^{2}_{L^{2}} +\|\bar{\rho}^{k+1}\|^{2}_{L^{2}\cap L^{\frac{3}{2}}}+\|\bar{\eta}^{k+1}\|^{2}_{L^{2}\cap L^{\frac{3}{2}}}+\|\bar{p}^{k+1}\|^{2}_{L^{2}}.$

令 $B^{k}=\|\nabla\bar{u}^{k}\|^{2}_{L^{2}}$ ,结合(4.13)-(4.17)和(4.19)式,我们得到

$\frac{\rm d}{{\rm d}t}A^{k+1}(t)+\mu B^{k+1}(t)\leq \varepsilon^{-1}\bar{C}A^{k+1}(t)+3\varepsilon B^{k}(t).$

(4.20)

注意 $A^{k+1}(0)=0$ .因此(4.20)式在 $(0, t)$ 上积分有

$A^{k+1}(t)+\mu\int^{t}_{0} B^{k+1}(s){\rm d}s\leq \int^{t}_{0}\varepsilon^{-1}\bar{C}A^{k+1}(s){\rm d}s+3\varepsilon \int^{t}_{0}B^{k}(s){\rm d}s,$

(4.21)

由Granwall不等式有

$A^{k+1}(t)+\int^{t}_{0} B^{k+1}(s){\rm d}s\leq 3\varepsilon(\varepsilon^{-1}\bar{C}te^{\varepsilon^{-1}\bar{C}t}+1) \int^{t}_{0}B^{k}(s){\rm d}s.$

(4.22)

选择足够小的 $\varepsilon>0$ 和 $T^{\ast}>0$ 得到

$T^{\ast}\leq T_{\ast}, e^{\varepsilon^{-1}\bar{C}T^{\ast}} <2, \quad 3\varepsilon(2\ln2+1) <\frac{1}{2},$

(4.23)

由(4.22)-(4.23)式推出

$A^{k+1}(t)+\int^{t}_{0} B^{k+1}(s){\rm d}s\leq \frac{1}{2}\int^{t}_{0}B^{k}(s){\rm d}s,$

(4.24)

对于 $0\leq t\leq T^{\ast}$ , $k\geq1$ .由迭代法,可以得到

$\sup\limits_{0\leq t\leq T^{\ast}}A^{k+1}(t)+\int^{T^{\ast}}_{0} B^{k+1}(s){\rm d}s\leq \frac{1}{2^{k-1}}\int^{t}_{0}B^{2}(s){\rm d}s,$

(4.25)

这样意味着

$\mathop \sum \limits_{k = 1}^\infty \Big(\sup\limits_{0\leq t\leq T^{\ast}}A^{k+1}(t)+\int^{T^{\ast}}_{0} B^{k+1}(s){\rm d}s\Big)\leq \bar{C}\int^{t}_{0}B^{2}(s){\rm d}s <\infty.$

(4.26)

因此,我们得到

$\mathop \sum \limits_{k = 2}^\infty \|\bar{\rho}^{k}\|_{L^{\infty}(0, T^{\ast};L^{\frac{3}{2}})} <\infty, \quad \mathop \sum \limits_{k = 2}^\infty \|\bar{\eta}^{k}\|_{L^{\infty}(0, T^{\ast};L^{\frac{3}{2}})} <\infty, \quad \mathop \sum \limits_{k = 2}^\infty \|\bar{u}^{k}\|_{L^{2}(0, T^{\ast};H^{1})} <\infty.$

(4.27)

因此,当 $k\rightarrow\infty$ 时

$\left\{\begin{array}{ll} \rho^{k}\rightarrow \rho^{1}+\mathop \sum \limits_{i = 2}^\infty \bar{\rho}^{i}, &\quad L^{\infty}(0, T^{\ast};L^{\frac{3}{2}}), \\[4mm] p^{k}\rightarrow p^{1}+\mathop \sum \limits_{i = 2}^\infty \bar{p}^{i}, &\quad L^{\infty}(0, T^{\ast};L^{2}), \\[4mm] \rho^{u}\rightarrow u^{1}+\sum\limits^{\infty}_{i=2}\bar{u}^{i}, &\quad L^{2}(0, T^{\ast};H^{1}), \\[4mm] \eta^{k}\rightarrow \eta^{1}+\mathop \sum \limits_{i = 2}^\infty \bar{\eta}^{i}, &\quad L^{\infty}(0, T^{\ast};L^{\frac{3}{2}}). \end{array}\right.$

(4.28)

由引理3.1,我们可以取一个 $(\rho^{k_{i}}, p^{k_{i}}, u^{k_{i}}, \eta^{k_{i}})$ 中的使得当 $i\rightarrow \infty$ $(\rho^{k_{i}}, p^{k_{i}}, u^{k_{i}}, \eta^{k_{i}})\rightarrow (\rho, p, u, \eta)$ ,弱或弱- $^{\ast}$ ,且 $(\rho, p, u, \eta)$ 满足(4.8)式的正则性,这里 $T_{\ast}$ 由 $T^{\ast}$ 代替.由引理3.1的证明,我们知 $(\rho, p, u, \eta)$ 是系统(1.10)-(1.15)的解.这里唯一性很容易得证,我们略去.

所以,我们得到以下结果

定理4.1 在(1.16)式的条件下,进一步假设以下相容性条件

$Lu_{0}+\nabla(p_{0}+\eta_{0})=-(\rho_{0}+\eta_{0})\nabla \Phi-(\rho_{0}+\eta_{0})g,$

(4.29)

对于 $g\in D^{1}$ , $\sqrt{\rho_{0}}g\in L^{2}$ .则存在小时间区域 $T_{\ast}\in (0, T)$ 使得系统(1.10)-(1.15)存在唯一强解 $(\rho, p, u, \eta)$ 满足

(4.30)

因此,为了证明定理1.1,我们只需证明 $p=a\rho^{\gamma}, \gamma>1$ .定义 $\bar{p}=p-a\rho^{\gamma}, \gamma>1$ ,由(1.10), (1.11)和(1.14)式,有

$\left\{\begin{array}{ll} \bar{p}_{t}+u\cdot \nabla \bar{p}+\gamma \bar{p}(\nabla\cdot u)=0, &(0, T_{\ast})\times \mathbb{R}^{3}, \\ \bar{p}_{0}=0, &\mathbb{R}^{3}. \end{array}\right.$

(4.31)

(4.31)式乘 $\bar{p}$ ,并在 $\mathbb{R}^{3}$ 上积分,易知

$\bar{p}=0, \quad \quad (0, T_{\ast})\times \mathbb{R}^{3},$

(4.32)

即

$p=a\rho^{\gamma}, \quad \quad (0, T_{\ast})\times \mathbb{R}^{3},$

对于 $\gamma>1$ .

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... 流体-粒子相互作用模型(fluid-particle interaction model)在流体中的分散悬浮液粒子的沉降分析中发挥重要作用,在生物、医学、化工和矿物过程中具有非常重要的应用^[1-4].此外,这种交互系统也用于柴油发动机和火箭推进器燃烧理论模型^[5-6].本文中,我们主要研究所谓的Flowing Regime模型,其在全空间

$\mathbb{R}^{3}$

的表示为 ...

... 该模型系统由Carrillo和Goudon推导出来,他们推导出了耦合动力学和流体力学的两种模型^[4].所谓的Bubbling Regime和Flowing Regime.对于Bubbling Regime模型^{[1, 10]}表示的是动力学方程的扩散近似;而Flowing Regime模型,描述了强阻力流动机制和强布朗运动.对于Bubbling Regime,已经有了很多结果^{[8-10, 14, 16]}.但是对于Flowing Regime模型的相关研究却甚少.当

$\eta=0$

时,方程(1.1)-(1.3)变为Navier-Stokes方程. Cho和Kim在文献[18]中研究了Navier-Stokes方程的Cauchy问题的局部解问题.本文的主要目的是利用同样的方法来解决较为复杂的Flowing Regime模型的局部解问题. ...

A modeling of biospray for the upper airways

2005

Large eddy simulation of droplet dispersion for inhomogeneous turbulent wall flow

2006

Stability and asymptotic analysis of a fluid-particle interaction model

2006

$\mathbb{R}^{3}$

的表示为 ...

$\eta=0$

$\mathbb{R}^{3}$

的表示为 ...

Spray combustion and atomization

1958

$\mathbb{R}^{3}$

的表示为 ...

Global classical large solutions to a 1D fluid-particle interaction model:The bubbling regime

2012

$\eta=0$

Strong solutions to a 1D fluid-particle interaction non-newtonian model:The bubbling regime

2013

On the dynamics of a fluid-particle interaction model:the bubbling regime

2011

$\eta=0$

... -10, 14, 16].但是对于Flowing Regime模型的相关研究却甚少.当

$\eta=0$

On the existence of globally defined weak solutions to the Navier-Stokes equations of compressible isentropic fluids

2001

Large-time behavior of solutions to the Navier-Stokes equations of compressible flow

1999

Weakly dissipative solutions and weak-strong uniqueness for the Navier Stokes Smolunchowski system

2013

$\eta=0$

Viscous and inviscid models in fluid-particle interaction

2013

$\eta=0$

1994

On classical solutions of the compressible Navier-Stokes equations with nonnegative initial densities

2006

$\eta=0$

... 在全篇文章中,我们做一些符号的注释(文献[18]中有详细的说明). ...

... 首先,我们有一个在初值是正时,解的存在性结果.可以参看文献[18]中的引理3.1. ...

... 引理2.2^{[18, Lemma 3.2]} 对于

$0\leq t \leq\min(T_{\ast}, T_{1})$

,有 ...

... 同文献[18]类似的方法,由引理2.2, (2.9)-(2.12)式和Sobolev嵌入,我们得到(2.13)式的各项估计 ...

... 利用文献[18]的同样方法,我们有 ...

... 且通过定义

$u^{R}_{0}$

在

$B_{R}$

外侧为零,把

$u^{R}_{0}$

扩展到全空间,利用文献[18]的方法,我们可以证得当

$R\rightarrow\infty$

有下面强收敛 ...

... 由(2.1)式, (2.2)式和文献[18]中的引理2.1,我们有 ...

Unique solvability of the initial boundary value problems for compressible viscous fluids

2004

$\mathbb{R}^{3}$ 的解

〈

〉