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

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

(1.2) $ ((\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, $

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

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

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

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

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

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

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

初值为

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

在无穷远处,当$|x|\rightarrow\infty$时

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

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

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

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

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

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

初值为

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

边界条件为

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

且满足以下假设

(1.16) $ \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.$

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

(1.17) $ (\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}, $

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

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

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

(1.19) $ \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. $

为了证明定理1.1,我们考虑以下线性方程组

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

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

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

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

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

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

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

(2.6) $ 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}). $

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

(2.7) $ \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. $

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

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

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

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

(2.10) $ \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}, $

(2.11) $ \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}, $

(2.12) ${\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}, $

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

(2.13) $ \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.$

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

(2.14) $\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}), $

(2.15) $ -\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}, $

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

(2.17) $-\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}, $

(2.18) $ -\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}, $

(2.13)式关于$t$积分,由(1.4)式和(2.14)-(2.18)式

(2.19) $ \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}).$

由Gronwall不等式和(2.19)式,我们得到

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

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

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

由$(2.21)$式的正则性估计,我们有

(2.22) $\|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}), $

则有

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

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

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

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

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

(2.26) $ 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}. $

且由(2.1)式, Hölder不等式和Cauchy不等式,有

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

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

这里$\epsilon\in(0, 1)$足够小.同样的我们有

(2.29) $ 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}. $

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

(2.30) $ \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}$

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

另一方面,由

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

有

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

(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

(2.33) $ \|\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}) , $

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

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

由(2.20)式的正则性,我们有

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

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

(2.36) $ \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}). $

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

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

利用文献[18]的同样方法,我们有

(2.38) $ \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}$

由(2.1)式,我们有

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

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

(2.41) $ 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}}, $

(2.42) $ \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}$

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

(2.43) $ \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}$

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

(2.44) $ \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 $

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

(2.45) $ \|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}. $

由(2.35)式和(2.45)式,有

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

由下面正则性

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

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

(2.48) $ \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}, $

(2.49) $ \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}, $

(2.50) $ \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}, $

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

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

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

(3.2) $ \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}, $

(3.3) $ \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}, $

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

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

(3.5) $ \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.$

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), $

(3.6) $ \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}, $

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

(3.7) $ \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. $

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

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

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

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

我们解决(3.9)式的等式右边如下

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

这里

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

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

则有

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

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

(3.13) $ \| 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), $

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

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

利用(3.13)和(3.9)式,可以得到以下弱收敛

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

由(3.14)和(3.15)式有以下强收敛

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

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

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

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

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

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

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

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

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

(3.23) $ \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.$

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$R>R_{1}$,

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

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

(3.25) $ (\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.$

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.

为了证明定理1.1,我们首先考虑下面热方程

(4.1) $ \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.$

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

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

且

(4.3) $ \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).$

定义$c_{0}$:

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

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

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

由(4.1)式知

(4.6) $ \|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}).$

结合(2.48)-(2.51)式和(4.3)-(4.6)式,我们有

(4.7) $ \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}.$

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

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

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

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

(4.10) $ \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, $

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

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

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

(4.13) $ \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}}, $

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

(4.14) $ \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}}, $

(4.15) $ \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}}, $

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

(4.16) $ \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}}, $

对于$0\leq t\leq T_{\ast}$.

同样地,我们有

(4.17) $ \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}}, $

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

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

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

(4.19) $ \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}}, $

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

(4.20) $ \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).$

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

(4.21) $ 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, $

由Granwall不等式有

(4.22) $ 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.$

选择足够小的$\varepsilon>0$和$T^{\ast}>0$得到

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

由(4.22)-(4.23)式推出

(4.24) $ 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, $

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

(4.25) $ \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, $

这样意味着

(4.26) $ \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.$

因此,我们得到

(4.27) $ \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.$

因此,当$k\rightarrow\infty$时

(4.28) $ \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.$

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

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

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) $ \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.$

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

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

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

即

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

对于$\gamma>1$.