引言

具有扩散界面的非混相两相流模型描述了两种不同流体的流动及流体间扩散界面的运动，其相关研究成果广泛应用于航空航天、水利工程、化学工程等领域。因此，针对两相流的扩散界面模型进行研究具有重要的理论意义和应用价值。

关于描述单一流体运动的Navier-Stokes方程的研究已有不少。相比于单相流，两相流之间存在相互作用和扩散界面，其模型的建立与分析更为复杂。Van der Waals^[1]最先将互不相溶的两相流之间的界面视为一个有厚度的界面层。之后，Blesgen^[2]将描述单一流体流动的Navier-Stokes方程和描述两种流体在其分界面相互作用的Allen-Cahn方程耦合在一起，提出了Navier-Stokes-Allen-Cahn(NSAC)方程组。关于一维NSAC模型的数学研究结果已有很多，Chen等^[3]证明了初始真空状态下强解和经典解的存在唯一性；Ding等^[4]考虑了NSAC方程组的自由边界问题，证明了强解的存在唯一性；孙颖等^[5]证明了NSAC方程组周期边值问题整体解的存在性；Chen等^[6]证明了非等熵NSAC方程组初边值问题存在唯一的全局强解；Feireisl等^[7]、Chen等^[8]证明了三维可压缩NSAC模型弱解的存在性。

本文研究三维可压缩Navier-Stokes-Allen-Cahn方程组的Cauchy问题。与前人考虑的初始条件不同，我们假设相场(即组分浓度差ϕ)在无穷远处趋于1或-1两种状态。因此，除了克服NSAC方程组的强非线性和耦合性，还需要估计ϕ²-1所带来的困难项。针对此类Cauchy问题，本文在初始小扰动的假设条件下通过能量方法证明了全局强解的存在唯一性。

1 问题的提出及主要定理

可压缩非混相两相流的流动通常由以下的NSAC非线性偏微分方程组描述

$ \left\{\begin{array}{l} \rho_t+\operatorname{div}(\rho \boldsymbol{u})=0 \\ (\rho \boldsymbol{u})_t+\operatorname{div}(\rho \boldsymbol{u} \otimes \boldsymbol{u})=\operatorname{div} \boldsymbol{T} \\ (\rho \phi)_t+\operatorname{div}(\rho \phi \boldsymbol{u})=-\mu \\ \rho \mu=\rho \frac{\partial f}{\partial \phi}-\operatorname{div}\left(\rho \frac{\partial f}{\partial \nabla \phi}\right) \end{array}\right. $

(1)

式中，ρ=ρ₁+ρ₂表示混合流体的总密度；u表示流速，且ρu=ρ₁u₁+ρ₂u₂, ρ_i、u_i(i=1, 2)分别为第i种组分的密度和流速；ϕ=ϕ₁-ϕ₂表示组分间的浓度差，其中ϕ_i=ρ_i/ρ；μ表示化学势；f表示界面自由能密度。Cauchy应力张量T表示为

$ \boldsymbol{T}=2 \nu D(\boldsymbol{u})+\lambda(\operatorname{div} \boldsymbol{u}) \boldsymbol{I}-P \boldsymbol{I}-\rho \nabla \phi \otimes \frac{\partial f}{\partial \nabla \phi} $

(2)

式中，ν>0, λ>0为黏性系数且满足$\lambda+\frac{2}{N} \nu \geqslant 0 $；D(u)=($\nabla $u+$\nabla $^Tu)/2为应变张量；I为单位矩阵；P=p(ρ)-$ \frac{\varepsilon}{2}|\nabla \phi|^2$为总压力，其中压力p是关于密度ρ的光滑函数，且满足p′(ρ)>0，ε为扩散界面厚度；f有以下形式

$ f=\frac{1}{4 \varepsilon}\left(1-\phi^2\right)^2+\frac{\varepsilon}{2 \rho}|\nabla \phi|^2 $

(3)

本文研究的问题的初始条件为

$ \left\{\begin{array}{l} (\rho, \boldsymbol{u}, \phi)(x, 0)=\left(\rho_0, \boldsymbol{u}_0, \phi_0\right)(x) \\ \left(\rho_0, \boldsymbol{u}_0, \left|\phi_0\right|\right) \stackrel{x \rightarrow \pm \infty}{\longrightarrow}(\bar{\rho}, 0, 1) \end{array}\right. $

(4)

式中，ρ为给定的正实数，$\left|\phi_0\right| \stackrel{x \rightarrow \pm \infty}{\longrightarrow} 1 $1表示两相流初始浓度差ϕ₀在无穷远处为1或-1。

本文的主要结论如下。

定理1   假设初始值(ρ₀, u₀, ϕ₀)满足条件

$ \begin{array}{l} \;\;\;\;\;\;\;\left(\rho_0-\bar{\rho}, \boldsymbol{u}_0\right) \in H^3\left(\boldsymbol{R}^3\right), \inf\limits _{x \in R^3} \rho_0(x)>0, \\ \nabla \phi_0 \in H^2\left(\boldsymbol{R}^3\right), \phi_0^2-1 \in L^2\left(\boldsymbol{R}^3\right) \end{array} $

(5)

存在δ>0，使得如果

$ \;\;\;\;\;\;\;\left\|\left(\rho_0-\bar{\rho}, \boldsymbol{u}_0\right)\right\|_{H^3}+\left\|\nabla \phi_0\right\|_{H^2}+\left\|\phi_0^2-1\right\| \leqslant\\\delta $

(6)

则方程(1)~(4)存在唯一的全局解(ρ, u, ϕ)满足

$ \begin{array}{l} \;\;\;\;\;(\rho-\bar{\rho}, \boldsymbol{u}) \in C\left([0, \infty] ; H^3\left(\boldsymbol{R}^3\right)\right), \phi^2-1 \in \\ C\left([0, \infty] ; L^2\left(\boldsymbol{R}^3\right)\right), \nabla \phi \in C\left([0, \infty] ; H^2\left(\boldsymbol{R}^3\right)\right), \\ \nabla \phi \in L^2\left([0, \infty] ; H^2\left(\boldsymbol{R}^3\right)\right), \nabla \rho \in L^2([0, \infty] ; \\ \left.H^2\left(\boldsymbol{R}^3\right)\right), \nabla \boldsymbol{u} \in L^2\left([0, \infty] ; H^3\left(\boldsymbol{R}^3\right)\right) \end{array} $

(7)

且

$ \begin{array}{l} \;\;\;\;\;\;\|(\rho-\bar{\rho}, \boldsymbol{u})\|_{H^3}^2+\|\nabla \phi\|_{H^2}^2+\left\|\phi^2-1\right\|^2+ \\ \int_0^t\left(\|\nabla \rho\|_{H^2}^2+\|(\nabla \boldsymbol{u}, \nabla \phi)\|_{H^3}^2\right) \mathrm{d} \tau \leqslant C\left(\|\left(\rho_0-\right.\right. \\ \left.\left.\bar{\rho}, \boldsymbol{u}_0\right)\left\|_{H^3}^2+\right\| \nabla \phi_0\left\|_{H^2}^2+\right\| \phi_0^2-1 \|^2\right) \end{array} $

(8)

有以下两点需要注意：①式(6)结合Sobolev嵌入定理，δ足够小时，有$\inf\limits _{x \in \boldsymbol{R}^3} \phi_0^2>\frac{1}{3} $，其物理意义为在初始时刻，两相流出现分层且分层区域即$\phi_0^2 <\frac{1}{3} $的测度为零；②记‖ ·‖=‖ ·‖_L²。

2 主要定理的证明

定理1中证明全局解的存在性的方法：首先证明局部解的存在性，再利用能量估计的方法得到解的一致估计，最后延拓到全局解的存在性。在证明过程中，不仅要估计‖(ρ-ρ, u)‖_H³+‖Δϕ‖_H²，还要估计‖ϕ²-1‖。为克服这些难点，在先验估计中需要假设‖(ρ₀-ρ, u₀)‖_H³+‖Δ ϕ₀‖_H²和‖ϕ₀²-1‖均具有小性。

2.1 局部解的存在性

首先，证明局部解的存在性。

令σ=ρ-ρ，定义解空间，对$\forall $m>0, M>0，有

$ \begin{array}{l} X_{m, M}([0, T])=\{(\sigma, \boldsymbol{u}, \phi) \mid(\sigma, \boldsymbol{u}) \in C([0, \\ \left.T] ; H^3\left(\boldsymbol{R}^3\right)\right), \phi^2-1 \in C\left([0, T] ; L^2\left(\boldsymbol{R}^3\right)\right), \nabla \phi \in \\ C\left([0, T] ; H^3\left(\boldsymbol{R}^3\right)\right), \nabla \sigma \in L^2\left([0, T] ; H^2\left(\boldsymbol{R}^3\right)\right), \\ \nabla \boldsymbol{u} \in L^2\left([0, T] ; H^3\left(\boldsymbol{R}^3\right)\right), \nabla \phi \in L^2([0, T] ; \\ \left.H^3\left(\boldsymbol{R}^3\right)\right), \sup _{t \in[0, T]}\left(\|(\sigma, \boldsymbol{u})(t)\|_{H^3}+\|\nabla \phi\|_{H^2}+\right. \\ \left.\left\|\phi^2-1\right\|\right) \leqslant M, \inf\limits _{\substack{x \in \boldsymbol{R}^3 \\ t \in[0, T]}} \phi^2(x, t)-\frac{1}{3}>m ; \inf\limits_{\substack{x \in \boldsymbol{R}^3 \\ t \in[0, T]}} \rho(x, \\ t) \geqslant m\} \end{array} $

(9)

由经典的Schauder不动点方法得到如下命题。

命题1   对$ \forall$M>0, m>0，若初值(ρ0, u0, ϕ0) 满足条件‖(ρ0-ρ, u0)‖H32+‖Δϕ0‖H22+‖ϕ02-1‖2≤M, $\inf\limits _{x \in \boldsymbol{R}^3} \phi_0^2-\frac{1}{3}>m>0 $和$\inf\limits _{x \in \boldsymbol{R}^3} \rho_0(x)> $m, 则存在T*>0，使得方程(1)~(4)存在唯一局部解(ρ, u, ϕ)∈ $X_{\frac{m}{2}, 2 M}\left(\left[0, T^*\right]\right) $。

2.2 全局解的存在性

接着，将方程组(1)写成如下的线性化形式。

$ \left\{\begin{array}{l} \sigma_t+\bar{\rho} \operatorname{div} \boldsymbol{u}=g_1 \\ u_t-\frac{\nu}{\bar{\rho}} \Delta \boldsymbol{u}-\frac{(\nu+\lambda)}{\bar{\rho}} \nabla \operatorname{div} \boldsymbol{u}+\frac{p^{\prime}(\bar{\rho})}{\bar{\rho}} \nabla \sigma+ \\ \;\;\;\;\;\;\;\frac{\varepsilon}{\bar{\rho}} \nabla \phi \Delta \phi=g_2 \\ \rho \phi_t+\rho \boldsymbol{u} \cdot \nabla \phi=-\mu \\ \rho \mu=\frac{\rho}{\varepsilon}\left(\phi^3-\phi\right)-\varepsilon \Delta \phi \end{array}\right. $

(10)

其中，非齐次项g₁和g₂的定义为

$ \begin{array}{l} \left\{ {\begin{array}{*{20}{l}} {{g_1} = - {\mathop{\rm div}\nolimits} (\sigma \mathit{\boldsymbol{u}})}\\ {{g_2} = - (\mathit{\boldsymbol{u}} \cdot \nabla )\mathit{\boldsymbol{u}} + {h_1}(\sigma )\nabla \sigma - {h_2}(\sigma )(\nu \Delta \mathit{\boldsymbol{u}} + } \end{array}} \right.\\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;(\nu + \lambda )\nabla {\mathop{\rm div}\nolimits} \mathit{\boldsymbol{u}} - \varepsilon \nabla \phi \Delta \phi ) \end{array} $

这里$h_1(\sigma)=\frac{p^{\prime}(\bar{\rho})}{\bar{\rho}}-\frac{p^{\prime}(\rho)}{\rho} $，$h_2(\sigma)=\frac{1}{\bar{\rho}}-\frac{1}{\rho} $。

式(10)中第二个式子推导过程中用到

$ \operatorname{div}(\nabla \phi \otimes \nabla \phi)=\nabla\left(\frac{|\nabla \phi|^2}{2}\right)+\nabla \phi \Delta \phi $

由定义的解空间(9)，结合Sobolev嵌入定理可知, 存在M₀>0，使得$ \forall$0 < M < M₀，有

$ \frac{\bar{\rho}}{2} \leqslant \rho(x, t) \leqslant 2 \bar{\rho}, 3 \phi^2-1>m_0=\inf\limits_{x \in \boldsymbol{R}^3}\left(3 \phi_0^2-1\right) $

(11)

命题2   假设(ρ₀, u₀, ϕ₀)满足(σ₀, u₀)∈H³(R³), ϕ₀²-1∈L²(R³), 对$\forall $T>0, 设(σ, u, ϕ)∈X_{m, M}([0, T])为方程组(10)的局部解，则存在一个只依赖于初值和T的常数C，使得

$ \begin{array}{l} \;\;\;\;\;\;\;\;\|(\sigma, \boldsymbol{u})\|_{H^3}^2+\|\nabla \phi\|_{H^2}^2+\left\|\phi^2-1\right\|^2+ \\ \int_0^t\left(\|\nabla \sigma\|_{H^2}^2+\|(\nabla \boldsymbol{u}, \nabla \phi)\|_{H^3}^2\right) \mathrm{d} \tau \leqslant C\left(\|\left(\sigma_0, \right.\right. \\ \left.\left.\boldsymbol{u}_0\right)\left\|_{H^3}^2+\right\| \nabla \phi_0\left\|_{H^2}^2+\right\| \phi_0^2-1 \|^2\right) \end{array} $

(12)

命题2可由以下4个引理得到。

引理1   设(σ, u, ϕ)∈X_{m, M}([0, T])为方程组(10)的局部解，有

$ \;\;\;\;\;\;\;\left\|\left(\sigma, \boldsymbol{u}, \phi^2-1, \nabla \phi\right)\right\|^2+\int_0^t \|(\mu, \nabla \boldsymbol{u}, \Delta \phi,\\ \nabla \phi)\left\|^2 \mathrm{~d} \tau \leqslant C\right\|\left(\sigma_0, u_0, \phi_0^2-1, \nabla \phi_0\right) \|^2 $

(13)

和

$ \|\phi\|_{L^{\infty}} \leqslant 1 $

(14)

证明：

定义

$ G(\rho)=\rho \int_{\bar{\rho}}^\rho \frac{p(\xi)-p(\bar{\rho})}{\xi^2} \mathrm{~d} \xi, \rho>0 $

(15)

由式(15)和质量守恒方程得到

$ \begin{array}{l} \;\;\;\;\;\;\;\rho G^{\prime}(\rho)=G(\rho)+(p(\rho)-p(\bar{\rho})), \rho G^{\prime \prime}(\rho)= \\ p^{\prime}(\rho), [G(\rho)]_t+\operatorname{div}(G(\rho) \boldsymbol{u})+(p(\rho)-p(\bar{\rho})) ·\\ \operatorname{div} \boldsymbol{u}=0 \end{array} $

用式(1)中第二个式子乘以u，关于空间变量x积分，有

$ \;\;\;\;\;\frac{\mathrm{d}}{\mathrm{d} t} \int\left(\frac{1}{2} \rho \boldsymbol{u}^2+G(\rho)\right) \mathrm{d} x+\nu \int|\nabla \boldsymbol{u}|^2 \mathrm{~d} x+(\nu+\\ \lambda \int|\operatorname{div} \boldsymbol{u}|^2 \mathrm{~d} x-\varepsilon \int \boldsymbol{u} \cdot \nabla \phi \Delta \phi \mathrm{d} x=0 $

(16)

式(10)中第三个式子乘以μ, 结合式(10)中第四个式子，关于x积分，利用分部积分得到

$ \begin{aligned} \frac{\varepsilon}{2} & \frac{\mathrm{d}}{\mathrm{d} t} \int|\nabla \phi|^2 \mathrm{~d} x+\frac{1}{4 \varepsilon} \frac{\mathrm{d}}{\mathrm{d} t} \int \rho\left(\phi^2-1\right)^2 \mathrm{~d} x+\\ \int \mu^2 \mathrm{~d} x &=-\varepsilon \int \boldsymbol{u} \cdot \nabla \phi \Delta \phi \mathrm{d} x \end{aligned} $

(17)

将式(16)和式(17)相加，得到

$ \begin{array}{l} \;\;\;\;\;\;\;\frac{\mathrm{d}}{\mathrm{d} t} \int\left(\frac{1}{2} \rho \boldsymbol{u}^2+G(\rho)+\frac{\varepsilon}{2}|\nabla \phi|^2+\frac{\rho}{4 \varepsilon} .\right. \\ \left.\left(\phi^2-1\right)^2\right) \mathrm{d} x+\nu\|\nabla \boldsymbol{u}\|^2+\|\mu\|^2 \leqslant 0 \end{array} $

(18)

由式(9)、式(11)、式(15)得到

$ c_{\bar{\rho}}(\rho-\bar{\rho})^2 \leqslant G(\rho) \leqslant C_{\bar{\rho}}(\rho-\bar{\rho})^2 $

(19)

将式(18)在[0, T]上积分，结合式(19)可得

$ \;\;\;\;\;\;\;\;\;\left\|\left(\sigma, \boldsymbol{u}, \phi^2-1, \nabla \phi\right)\right\|^2+\int_0^t\|(\mu, \nabla \boldsymbol{u})\|^2 \mathrm{~d} \tau \leqslant\\C E_0 $

(20)

其中，E₀=‖(σ₀, u₀, ϕ₀²-1, Δϕ₀)‖²。

设Ω_n=[n, n+1)³, n=0, ±1, ±2, …，由式(20)可得

$ \;\;\;\;\;\;\;\int_{\mathit{\Omega}_n} \phi^4 \mathrm{~d} x \leqslant 2 \int_{\mathit{\Omega}_n} \phi^2 \mathrm{~d} x+E_0 \leqslant \frac{1}{2} \int_{\mathit{\Omega}_n} \phi^4 \mathrm{~d} x+(2+\\ \left.E_0\right) $

于是有

$ \int_{\mathit{\Omega}_n} \phi^4 \mathrm{~d} x \leqslant 2\left(2+E_0\right) $

因此可得到

$ \int_{\mathit{\Omega}_n} \phi(x, t) \mathrm{d} x \leqslant\left(\int_{\mathit{\Omega}_n} \phi^4 \mathrm{~d} x\right)^{\frac{1}{4}} \leqslant\left(4+2 E_0\right)^{\frac{1}{4}} $

(21)

利用式(21)得到

$ \begin{array}{c} |\phi(x, t)| \leqslant\left|\int_{\mathit{\Omega}_n}(\phi(x, t)-\phi(y, t)) \mathrm{d} y\right|+ \\ \left|\int_{\mathit{\Omega}_n} \phi(y, t) \mathrm{d} y\right| \leqslant C\left(\int_{\mathit{\Omega}_n}|\nabla \phi|^2\right)^{\frac{1}{2}}+\left(4+2 E_0\right)^{\frac{1}{4}} \end{array} $

(22)

由式(22)和式(20)，得到式(14)。

接着，用式(10)中第四个式子乘以-Δϕ, 再关于x积分，有

$ \begin{array}{l} \;\;\;\;\;\;\varepsilon {\left\| {\Delta \phi } \right\|^2} + {\underline {\frac{1}{\varepsilon }\int \rho \left( {3{\phi ^2} - 1} \right)|\nabla \phi {|^2}{\rm{d}}x} _{{I_1}}} = \\ - \int \rho \mu \Delta \phi {\rm{d}}x - \frac{1}{\varepsilon }\int {\left( {{\phi ^2} - 1} \right)} \phi \nabla \phi \cdot \nabla \sigma {\rm{d}}x = :{I_2} + {I_3} \end{array} $

(23)

利用式(9)、式(11)和Hölder不等式得

$ \begin{array}{l} \;\;\;\;\;\;\;I_1 \geqslant \frac{\bar{\rho} m_0}{2 \varepsilon}\|\nabla \phi\|^2, I_2 \leqslant 2 \bar{\rho}\|\mu\|\|\Delta \phi\| \leqslant \frac{\varepsilon}{4} \cdot \\ \|\Delta \phi\|^2+\frac{4 \bar{\rho}^2}{\varepsilon}\|\mu\|^2, I_3 \leqslant\left\|\left(\phi^2-1\right)\right\|_{L^6}\|\phi \nabla \phi\| \\ \|\nabla \sigma\|_{L^3} \leqslant M\|\nabla \phi\|^2 \end{array} $

将I_i(i=1, 2, 3)代入式(23)，当M适当小时，有

$ \varepsilon^2\|\Delta \phi\|^2+\|\nabla \phi\|^2 \leqslant\|\mu\|^2 $

(24)

联合式(24)与式(20)得到式(13), 引理1得证。

引理2   设(σ, u, ϕ)∈X_{m, M}([0, T])为方程组(10)的局部解，有

$ \;\;\;\;\;\;\frac{\mathrm{d}}{\mathrm{d} t}\left\|\nabla^{k+1} \phi\right\|^2+\left\|\nabla^{k+2} \phi\right\|^2 \leqslant M\left(\left\|\nabla^{k+1} \sigma\right\|^2+\right.\\ \left.\left\|\nabla^{k+1} \phi\right\|^2+\left\|\nabla^{k+2} \boldsymbol{u}\right\|^2\right), k=1, 2 $

(25)

证明：结合式(10)中第三式和第四式，可以得到

$ \phi_t+\boldsymbol{u} \cdot \nabla \phi-\frac{\varepsilon}{\rho^2} \Delta \phi+\frac{\phi^2-1}{\varepsilon \rho} \phi=0 $

(26)

对式(26)求$\nabla^k $，关于空间变量x积分有

$ \;\;\;\;\;\frac{1}{2} \frac{\mathrm{d}}{\mathrm{d} t}\left\|\nabla^{k+1} \phi\right\|^2+\int \frac{\varepsilon}{\rho}\left|\nabla^k \Delta \phi\right|^2 \mathrm{~d} x=\int \nabla^k(\boldsymbol{u}·\\ \nabla \phi) \Delta \nabla^k \phi \mathrm{d} x-\varepsilon \int \nabla^k\left(\frac{1}{\rho^2} \Delta \phi\right) \Delta \nabla^k \phi \mathrm{d} x+\\ \frac{1}{\varepsilon} \int \nabla^k\left(\frac{\phi^2-1}{\rho} \phi\right) \Delta \nabla^k \phi \mathrm{d} x=: I_4+I_5+I_6 $

(27)

利用莱布尼茨公式、Hölder不等式和Gagliardo-Nirenberg不等式, 可估计I₄为

$ \begin{array}{l} \;\;\;\;\;\;\;I_4=\sum\limits_{0 \leqslant l \leqslant k} C_k^l \int \nabla^l \boldsymbol{u} \cdot \nabla^{k-l+1} \phi \Delta \nabla^k \phi \mathrm{d} x \leqslant \sum\limits_{0 \leqslant l \leqslant k} \| \nabla^l \cdot \\ \boldsymbol{u} \cdot \nabla^{k-l+1} \phi\|\| \nabla^{k+2} \phi \| \end{array} $

当l≤ $\left[\frac{k+1}{2}\right] $时，有

$ \begin{array}{l} \;\;\;\;\;\;\left\|\nabla^l \boldsymbol{u} \cdot \nabla^{k-l+1} \phi\right\| \leqslant\left\|\nabla^l \boldsymbol{u}\right\|_{L^3}\left\|\nabla^{k-l+1} \phi\right\|_{L^6} \leqslant \\ \left\|\nabla^\alpha \boldsymbol{u}\right\|^{1-\frac{l}{k+1}} \end{array} $

$ \;\;\;\;\;\;\left\|\nabla^{k+2} \boldsymbol{u}\right\|^{\frac{l}{k+1}}\|\nabla \phi\|^{\frac{l}{k+1}}\left\|\nabla^{k+2} \phi\right\|^{1-\frac{l}{k+1}} \leqslant\\ \boldsymbol{M}\left\|\nabla^{k+2} \boldsymbol{u}\right\|^{1-\frac{l}{k+1}} $

$ \left\|\nabla^{k+2} \phi\right\|^{1-\frac{1}{k+1}} \leqslant M\left(\left\|\nabla^{k+2} \boldsymbol{u}\right\|+\left\|\nabla^{k+2} \phi\right\|\right) $

由$\frac{l-1}{3} $= $\left(\frac{\alpha}{3}-\frac{1}{2}\right)\left(1-\frac{l}{k+1}\right) $+ $ \left(\frac{k+2}{3}-\frac{1}{2}\right)\frac{l}{k+1}$，可确定α= $\frac{1}{2}-\frac{l}{2(k+1-l)} \in\left[0, \frac{1}{2}\right] $。

当$\left[\frac{k+1}{2}\right]+1 \leqslant l \leqslant k $时，有

$ \;\;\;\;\;\;\;\left\|\nabla^l \boldsymbol{u} \cdot \nabla^{k-l+1} \phi\right\| \leqslant\left\|\nabla^l \boldsymbol{u}\right\|_{L^6}\left\|\nabla^{k-l+1} \phi\right\|_{L^3} \leqslant\\ \|\boldsymbol{u}\|^{1-\frac{l+1}{k+2}} $

$ \;\;\;\;\;\;\;\left\|\nabla^{k+2} \boldsymbol{u}\right\|^{\frac{l+1}{k+2}}\left\|\nabla^\alpha \phi\right\|^{\frac{l+1}{k+2}}\left\|\nabla^{k+2} \phi\right\|^{1-\frac{l+1}{k+2}} \leqslant\\ M\left\|\nabla^{k+2} \boldsymbol{u}\right\|^{\frac{l+1}{k+2}} $

$ \left\|\nabla^{k+2} \phi\right\|^{1-\frac{l+1}{k+2}} \leqslant M\left(\left\|\nabla^{k+2} \boldsymbol{u}\right\|+\left\|\nabla^{k+2} \phi\right\|\right) $

由$\frac{k-l}{3} $= $ \left(\frac{\alpha}{3}-\frac{1}{2}\right) \frac{l+1}{k+2}$+$ (1-\left.\frac{l+1}{k+2}\right)$，可确定α= $\frac{k+2}{2(l+1)} \in\left[\frac{1}{2}, 1\right] $。

因此，可得到

$ \left|I_4\right| \leqslant M\left(\left\|\nabla^{k+2} \boldsymbol{u}\right\|^2+\left\|\nabla^{k+2} \phi\right\|^2\right) $

(28)

接下来，对I₅进行估计。

$ \;\;\;\;\;\;I_5=\sum\limits_{1 \leqslant l \leqslant k} C_k^l \int \nabla^l\left(\frac{1}{\rho^2}\right) \nabla^{k-l} \Delta \phi \Delta \nabla^k \phi \mathrm{d} x \leqslant\\ \sum\limits_{1 \leqslant l \leqslant k} C_k^l\left\|\int \nabla^l\left(\frac{1}{\rho^2}\right) \nabla^{k-l} \Delta \phi\right\|\left\|\nabla^{k+2} \phi\right\| $

当$1 \leqslant l \leqslant\left[\frac{k+1}{2}\right] $时，有

$ \;\;\;\;\;\left\|\int \nabla^l\left(\frac{1}{\rho^2}\right) \nabla^{k-l} \Delta \phi\right\| \leqslant\left\|\nabla^l \sigma\right\|_{L^6}·\\ \left\|\nabla^{k-l+2} \phi\right\|_{L^3} \leqslant\left\|\nabla^\alpha \sigma\right\|^{1-\frac{l-1}{k+1}}\left\|\nabla^{k+1} \sigma\right\|^{\frac{l-1}{k+1}}·\\ \|\nabla \phi\|^{\frac{l-1}{k+1}}\left\|\nabla^{k+2} \phi\right\|^{1-\frac{l-1}{k+1}} \leqslant M\left\|\nabla^{k+1} \sigma\right\|^{\frac{l-1}{k+1}}·\\ \left\|\nabla^{k+2} \phi\right\|^{1-\frac{l-1}{k+1}} \leqslant M\left(\left\|\nabla^{k+1} \sigma\right\|+\left\|\nabla^{k+2} \phi\right\|\right) $

由$\frac{l-1}{3} $=$\left(\frac{\alpha}{3}-\frac{1}{2}\right)\left(1-\frac{l-1}{k+1}\right) $+$\left(\frac{k+1}{3}-\frac{1}{2}\right) $$\frac{l-1}{k+1} $, 可确定α=$\frac{3}{2}+\frac{3(l-1)}{2(k+2-l)} \in\left[\frac{3}{2}, 3\right) $。

当$1+\left[\frac{k+1}{2}\right] \leqslant l \leqslant k $时，有

$ \;\;\;\;\;\;\;\left\|\int \nabla^l\left(\frac{1}{\rho^2}\right) \nabla^{k-l} \Delta \phi\right\| \leqslant\left\|\nabla^l \sigma\right\|_{L^3}·\\ \left\|\nabla^{k-l+2} \phi\right\|_{L^3} \leqslant\|\sigma\|^{1-\frac{l+1}{k+1}}\left\|\nabla^{k+1} \sigma\right\|^{\frac{l+1}{k+1}}\left\|\nabla^\alpha \phi\right\|^{\frac{l+1}{k+1}}·\\ \left\|\nabla^{k+2} \phi\right\|^{1-\frac{l+1}{k+1}} \leqslant M\left\|\nabla^{k+1} \sigma\right\|^{\frac{l+1}{k+1}}\left\|\nabla^{k+2} \phi\right\|^{1-\frac{l+1}{k+1}} \leqslant\\ M\left(\left\|\nabla^{k+1} \sigma\right\|+\left\|\nabla^{k+2} \phi\right\|\right) $

由$\frac{k-l+1}{3} $= $ \left(\frac{\alpha}{3}-\frac{1}{2}\right) \frac{l+1}{k+1}$+$ \left(\frac{k+2}{3}-\frac{1}{2}\right)(1-\frac{l+1}{k+1})$，可确定α=$1+\frac{3(k+1)}{2(l+1)} \in\left[\frac{5}{2}, 4\right) $。

于是得到

$ \left|I_5\right| \leqslant M\left(\left\|\nabla^{k+1} \sigma\right\|^2+\left\|\nabla^{k+2} \phi\right\|^2\right) $

(29)

最后，估计I₆如下。

$ \;\;\;\;\;I_6=\int \frac{\phi^2-1}{\rho} \nabla^k \phi \Delta \nabla^k \phi \mathrm{d} x+\sum\limits_{1 \leqslant l \leqslant k} C_k^l \int \nabla^l·\\ \left(\frac{\phi^2-1}{\rho}\right) \nabla^{k-1} \phi \Delta \nabla^k \phi \mathrm{d} x=-\int \frac{\phi^2-1}{\rho}\left|\nabla^{k+1} \phi\right|^2 \mathrm{~d} x-\\ \int \nabla\left(\frac{\phi^2-1}{\rho}\right) \nabla^k \phi \nabla^{k+1} \phi \mathrm{d} x+\sum\limits_{1 \leqslant l \leqslant k} C_k^l \int \nabla^l\left(\frac{\phi^2-1}{\rho}\right)·\\ \nabla^{k-l} \phi \Delta \nabla^k \phi \mathrm{d} x=: I_6^1+I_6^2+I_6^3 $

逐项估计可得

$ \;\;\;\;I_6^1 \leqslant\left\|\phi^2-1\right\|_{L^3}\left\|\nabla^{k+1} \phi\right\|\left\|\nabla^{k+1} \phi\right\|_{L^6} \leqslant\\ C\left\|\phi^2-1\right\|^{\frac{1}{2}}\left\|\nabla\left(\phi^2-1\right)\right\|^{\frac{1}{2}}\left\|\nabla^{k+1} \phi\right\|\left\|\nabla^{k+2} \phi\right\| \leqslant\\ M\left(\left\|\nabla^{k+1} \phi\right\|^2+\left\|\nabla^{k+2} \phi\right\|^2\right) $

$ \;\;\;\;\;\;\;I_6^2 \leqslant\left\|\nabla\left(\frac{\phi^2-1}{\rho}\right)\right\|\left\|\nabla^k \phi\right\|_{L^6}\left\|\nabla^{k+1} \phi\right\|_{L^3} \leqslant\\ C(\|\nabla \sigma\|+\|\nabla \phi\|)\left\|\nabla^{k+1} \phi\right\|^{\frac{3}{2}}\left\|\nabla^{k+2} \phi\right\|^{\frac{1}{2}} \leqslant\\ M\left(\left\|\nabla^{k+1} \phi\right\|^2+\left\|\nabla^{k+2} \phi\right\|^2\right) $

$ I_6^3 \leqslant \sum\limits_{1 \leqslant l \leqslant k} C_k^l\left\|\nabla^l\left(\frac{\phi^2-1}{\rho}\right) \nabla^{k-l} \phi\right\|\left\|\nabla^{k+2} \phi\right\| $

当1≤l≤$\left[\frac{k}{2}\right] $时，有

$ \;\;\;\;\;\left\|\nabla^l\left(\frac{\phi^2-1}{\rho}\right) \nabla^{k-l} \phi\right\| \leqslant\left(\left\|\nabla^l \sigma\right\|_{L^3}+\right.\\ \left.\nabla^l \phi \|_{L^3}\right)\left\|\nabla^{k-l} \phi\right\|_{L^6} \leqslant\left(\left\|\nabla^\alpha \sigma\right\|^{1-\frac{l}{k}}\left\|\nabla^{k+1} \sigma\right\|^{\frac{l}{k}}+\right.\\ \left.\left\|\nabla^\alpha \phi\right\|^{1-\frac{l}{k}}\left\|\nabla^{k+1} \phi\right\|^{\frac{l}{k}}\right)\|\nabla \phi\|^{\frac{l}{k}}\left\|\nabla^{k+1} \phi\right\|^{1-\frac{l}{k}} \leqslant\\ M\left(\left\|\nabla^{k+1} \sigma\right\|+\left\|\nabla^{k+1} \phi\right\|\right) $

由$\frac{l-1}{3} $=$\left(\frac{\alpha}{3}-\frac{1}{2}\right)\left(1-\frac{l}{k}\right) $+$\left(\frac{k+1}{3}-\frac{1}{2}\right)\frac{l}{k} $, 可确定α=$1-\frac{k}{2(k-l)} \in\left[0, \frac{1}{2}\right) $。

当$1+\left[\frac{k}{2}\right] \leqslant l \leqslant k $时，有

$ \;\;\;\;\;\;\left\|\nabla^l\left(\frac{\phi^2-1}{\rho}\right) \nabla^{k-l} \phi\right\| \leqslant\left(\left\|\nabla^l \sigma\right\|_{L^6}+\right.\\ \left.\left\|\nabla^l \phi\right\|_{L^6}\right)\left\|\nabla^{k-l} \phi\right\|_{L^3} \leqslant\left(\|\sigma\|^{1-\frac{l+1}{k+1}}\left\|\nabla^{k+1} \sigma\right\|^{\frac{l+1}{k+1}}+\right.\\ \left.\|\phi\|^{1-\frac{l+1}{k+1}}\left\|\nabla^{k+1} \phi\right\|^{\frac{l+1}{k+1}}\right)\left\|\nabla^\alpha \phi\right\|^{\frac{l+1}{k+1}}\left\|\nabla^{k+1} \phi\right\|^{1-\frac{l+1}{k+1}} \leqslant\\ M\left(\left\|\nabla^{k+1} \sigma\right\|+\left\|\nabla^{k+1} \phi\right\|\right) $

由$\frac{k-l-1}{3} $=$ \left(\frac{\alpha}{3}-\frac{1}{2}\right) \frac{l+1}{k+1}$+$ \left(\frac{k+1}{3}-\frac{1}{2}\right)(1-\frac{l+1}{k+1})$，可确定α=$\frac{k+1}{2(l+1)} \in\left[\frac{1}{2}, 1\right) $。于是得到

$ \begin{array}{l} \;\;\;\;\;\;\;I_6^3 \leqslant M\left(\left\|\nabla^{k+1} \phi\right\|^2+\left\|\nabla^{k+2} \phi\right\|^2+\right. \\ \left.\left\|\nabla^{k+1} \sigma\right\|^2\right) \end{array} $

由I₆ⁱ(i=1, 2, 3)的估计可得

$ \begin{array}{l} \;\;\;\;\;\;\left|I_6\right| \leqslant M\left(\left\|\nabla^{k+1} \phi\right\|^2+\left\|\nabla^{k+2} \phi\right\|^2+\right. \\ \left.\left\|\nabla^{k+1} \sigma\right\|^2\right) \end{array} $

(30)

将式(28)~(30)代入式(27)，得到式(25)，引理2得证。

引理3   设(σ, u, ϕ)∈X_{m, M}([0, T])为方程组(10)的局部解，有

$ \begin{array}{l} \;\;\;\;\;\;\;\frac{\mathrm{d}}{\mathrm{d} t}\left(\left\|\nabla^k \boldsymbol{u}\right\|^2+\left\|\nabla^k \sigma\right\|^2\right)+\left\|\nabla^{k+1} \boldsymbol{u}\right\|^2 \leqslant \\ M\left(\left\|\nabla^k \sigma\right\|^2+\left\|\nabla^{k+1} \phi\right\|^2\right), k=0, 1, 2, 3 \end{array} $

(31)

证明：对式(10)中第一个式子和第二个式子求$\nabla^k $，再乘以$\nabla^k \boldsymbol{u} $，两式相加后，关于空间变量x积分，得

$ \begin{array}{l} \;\;\;\;\;\;\;\frac{1}{2} \frac{\mathrm{d}}{\mathrm{d} t}\left(\left\|\nabla^k \boldsymbol{u}\right\|^2+\frac{p^{\prime}(\bar{\rho})}{\bar{\rho}^2}\left\|\nabla^k \sigma\right\|^2\right)+\int\left(\frac{\nu}{\bar{\rho}}\right. \\ \left.\left|\nabla^{k+1} \boldsymbol{u}\right|^2+\frac{(\nu+\lambda)}{\bar{\rho}}\left|\operatorname{div} \nabla^k \boldsymbol{u}\right|^2\right) \mathrm{d} x=I_7 \end{array} $

(32)

其中，

$ \;\;\;\;\;\;\;I_7=\frac{p^{\prime}(\bar{\rho})}{\bar{\rho}^2} \int \nabla^k \sigma \nabla^k \operatorname{div}(\sigma \boldsymbol{u}) \mathrm{d} x-\int \nabla^k[(\boldsymbol{u} \cdot \nabla) \boldsymbol{u}-\\ \left.h_1(\sigma) \nabla \sigma\right] \cdot \nabla^k \boldsymbol{u} \mathrm{~d} x-\int \nabla^k\left[h_2(\sigma)\right.(\nu \Delta \boldsymbol{u}+(\nu+\lambda)·\\ \nabla \operatorname{div} \boldsymbol{u})] \cdot \nabla^k \boldsymbol{u} \mathrm{d} x-\frac{\varepsilon}{\bar{\rho}} \int \nabla^k(\nabla \phi \Delta \phi) \cdot \nabla^k \boldsymbol{u} \mathrm{d} x+\varepsilon \int \nabla^k \cdot\\ \left(h_2(\sigma) \nabla \phi \Delta \phi\right) \cdot \nabla^k \boldsymbol{u} \mathrm{d} x=: I_7^1+I_7^2+I_7^3+I_7^4+I_7^5 $

(33)

根据文献[9]中的引理2.1，可以估计I₇ⁱ(i=1, 2, 3)为

$ I_7^i \leqslant M\left(\left\|\nabla^k \sigma\right\|^2+\left\|\nabla^{k+1} \boldsymbol{u}\right\|^2\right), i=1, 2, 3 $

下面对I₇⁴进行估计。

如果k=0，有

$ \begin{array}{l} \;\;\;\quad I_7^4 \leqslant\|\nabla \phi\|\|\nabla \phi\|_{L^{\infty}}\|\nabla \boldsymbol{u}\| \leqslant M\left(\|\nabla \phi\|^2+\right. \\ \left.\|\nabla \boldsymbol{u}\|^2\right) \end{array} $

如果k=1，有

$ \begin{array}{l} \;\;\;\;\;I_7^4 \leqslant\|\nabla \phi\|_{L^{\infty}}\left\|\nabla^2 \phi\right\|\left\|\nabla^2 \boldsymbol{u}\right\| \leqslant M\left(\left\|\nabla^2 \phi\right\|^2+\right. \\ \left.\left\|\nabla^2 \boldsymbol{u}\right\|^2\right) \end{array} $

如果k≥2，利用莱布尼茨公式、Hölder不等式和Gagliardo-Nirenberg不等式，有

$ \;\;\;\;\;\;I_7^4=-\int \nabla^{k-1}(\nabla \phi \Delta \phi) \cdot \nabla^{k+1} \boldsymbol{u} \mathrm{d} x=\sum\limits_{0 \leqslant l \leqslant k-1} C_{k-1}^l \cdot\\ \int \nabla^l \nabla \phi \nabla^{k-l-1} \Delta \phi \cdot \nabla^{k+1} \boldsymbol{u} \mathrm{~d} x \leqslant \sum\limits_{0 \leqslant l \leqslant k-1} C_{k-1}^l \| \nabla^{l+1}\cdot\\ \phi \nabla^{k-l+1} \phi\|\| \nabla^{k+1} \boldsymbol{u} \| $

当$l \leqslant\left[\frac{k+1}{2}\right] $时，有

$ \;\;\;\;\;\left\|\nabla^{l+1} \phi \nabla^{k-l+1} \phi\right\| \leqslant\left\|\nabla^{l+1} \phi\right\|_{L^3}\left\|\nabla^{k-l+1} \phi\right\|_{L^6} \leqslant\\ \left\|\nabla^\alpha \phi\right\|^{1-\frac{l-1}{k-1}}\left\|\nabla^{k+1} \phi\right\|^{\frac{l-1}{k-1}}\left\|\nabla^2 \phi\right\|^{\frac{l-1}{k-1}} \| \nabla^{k+1} \cdot\\ \phi\left\|^{1-\frac{l-1}{k-1}} \leqslant M\right\| \nabla^{k+1} \phi \| $

由$\frac{l}{3} $=$\left(\frac{\alpha}{3}-\frac{1}{2}\right)\left(1-\frac{l-1}{k-1}\right) $+$\left(\frac{k+1}{3}-\frac{1}{2}\right)\frac{l-1}{k-1} $，可确定α=$2+\frac{k-1}{2(k-l)} \in(2, 3] $。

当$\left[\frac{k+1}{2}\right]+1 \leqslant l \leqslant k-1 $时，有

$ \begin{array}{l} \;\;\;\;\;\;\left\|\nabla^{l+1} \phi \nabla^{k-l+1} \phi\right\| \leqslant\left\|\nabla^{l+1} \phi\right\|_{L^6}\left\|\nabla^{k-l+1} \phi\right\|_{L^3} \leqslant \\ \|\nabla \phi\|^{1-\frac{l+1}{k}} \end{array} $

$ \begin{array}{l} \;\;\;\;\;\;\left\|\nabla^{k+1} \phi\right\|^{\frac{l+1}{k}}\left\|\nabla^\alpha \phi\right\|^{\frac{l+1}{k}}\left\|\nabla^{k+1} \phi\right\|^{1-\frac{l+1}{k}} \leqslant \\ M\left\|\nabla^{k+1} \phi\right\| \end{array} $

由$\frac{k-l}{3} $ =$\left(\frac{\alpha}{3}-\frac{1}{2}\right) \frac{l+1}{k} $+$ \left(\frac{k+1}{3}-\frac{1}{2}\right)(1-\frac{l+1}{k})$，可确定α=$1+\frac{3 k}{2(l+1)} \in[2, 4] $。

因此，可得到

$ I_7^4 \leqslant M\left(\left\|\nabla^{k+1} \phi\right\|^2+\left\|\nabla^{k+1} \boldsymbol{u}\right\|^2\right) $

同样可以得到

$ I_7^5 \leqslant M\left(\left\|\nabla^{k+1} \phi\right\|^2+\left\|\nabla^{k+1} \boldsymbol{u}\right\|^2\right) $

将以上估计代入式(33)有

$ I_7 \leqslant M\left(\left\|\nabla^k \sigma\right\|^2+\left\|\nabla^{k+1} \phi\right\|^2+\left\|\nabla^{k+1} \boldsymbol{u}\right\|^2\right) $

(34)

将式(34)代入式(32)，就可以得到式(31), 引理3得证。

引理4   设(σ, u, ϕ)∈X_{m, M}([0, T])为方程组(10)的局部解，有

$ \;\;\;\;\;\frac{\mathrm{d}}{\mathrm{d} t} \int \nabla^k \boldsymbol{u} \cdot \nabla^{k+1} \sigma \mathrm{d} x+\left\|\nabla^{k+1} \sigma\right\|^2 \leqslant M\\ \left(\left\|\nabla^{k+2} \boldsymbol{u}\right\|^2+\left\|\nabla^{k+2} \phi\right\|^2\right)+\left\|\nabla^{k+1} \boldsymbol{u}\right\|^2, k=0, 1\\ 2 $

(35)

证明：式(10)中第二个式子乘以$\nabla^{k+1} \sigma $，关于空间变量x积分有

$ \;\;\;\;\;\;\;\frac{\mathrm{d}}{\mathrm{d} t} \int \nabla^k \boldsymbol{u} \cdot \nabla^{k+1} \sigma \mathrm{d} x+\frac{p^{\prime}(\bar{\rho})}{\bar{\rho}}\left\|\nabla^{k+1} \sigma\right\|^2-\\ \bar{\rho} \int\left(\operatorname{div} \nabla^k \boldsymbol{u}\right)^2 \mathrm{~d} x=I_8 $

(36)

其中，

$ \;\;\;\;\;I_8=\int \nabla^k \operatorname{div}(\sigma \boldsymbol{u}) \operatorname{div} \nabla^k \boldsymbol{u} \mathrm{d} x-\int \nabla^k[(\boldsymbol{u} \cdot \nabla) \boldsymbol{u}-\\ \left.h_1(\sigma) \nabla \sigma\right] \cdot \nabla^{k+1} \sigma \mathrm{d} x-\int \nabla^k\left[h_2(\sigma)\right.(\nu \Delta \boldsymbol{u}+(\nu+\lambda)\\ \nabla \operatorname{div} u)] \cdot \nabla^{k+1} \sigma \mathrm{d} x-\frac{\varepsilon}{\bar{\rho}} \int \nabla^k(\nabla \phi \Delta \phi) \cdot \nabla^{k+1} \sigma \mathrm{d} x+\\ \varepsilon \int \nabla^k\left(h_2(\sigma) \nabla \phi \Delta \phi\right) \cdot \nabla^{k+1} \sigma \mathrm{d} x=: I_8^1+I_8^2+I_8^3+I_8^4+\\ I_8^5 $

(37)

根据文献[9]中的引理2.2，可以估计I₈ⁱ(i=1, 2, 3)为

$ I_8^1 \leqslant M\left(\left\|\nabla^{k+1} \sigma\right\|^2+\left\|\nabla^{k+1} \boldsymbol{u}\right\|^2\right) $

$ I_8^2+I_8^3 \leqslant M\left(\left\|\nabla^{k+1} \sigma\right\|^2+\left\|\nabla^{k+2} \boldsymbol{u}\right\|^2\right) $

关于I₈⁴的估计如下。

如果k=0，有

$ \begin{array}{l} \;\;\;\;I_8^4 \leqslant\|\nabla \phi\|_{L^{\infty}}\left\|\nabla^2 \phi\right\|\|\nabla \sigma\| \leqslant M\left(\left\|\nabla^2 \phi\right\|^2+\right. \\ \left.\|\nabla \sigma\|^2\right) \end{array} $

如果k≥1，有

$ \begin{array}{l} \;\;\;\;\;\;\;I_8^4=\sum\limits_{0 \leqslant l \leqslant k} C_k^l \int \nabla^l \nabla \phi \nabla^{k-l} \Delta \phi \cdot \nabla^{k+1} \sigma \mathrm{d} x \leqslant \sum\limits_{0 \leqslant l \leqslant k} C_k^l \cdot \\ \left\|\nabla^{l+1} \phi \nabla^{k-l+2} \phi\right\|\left\|\nabla^{k+1} \sigma\right\| \end{array} $

当$ l \leqslant\left[\frac{k+1}{2}\right]$时，有

$ \begin{array}{l} \;\;\;\;\;\left\|\nabla^{l+1} \phi \nabla^{k-l+2} \phi\right\| \leqslant\left\|\nabla^{l+1} \phi\right\|_{L^3}\left\|\nabla^{k-l+2} \phi\right\|_{L^6} \leqslant \\ \left\|\nabla^\alpha \phi\right\|^{1-\frac{l-1}{k-1}} \end{array} $

$ \begin{array}{l} \;\;\;\;\;\left\|\nabla^{k+2} \phi\right\|^{\frac{l-1}{k-1}}\left\|\nabla^3 \phi\right\|^{\frac{l-1}{k-1}}\left\|\nabla^{k+2} \phi\right\|^{1-\frac{l-1}{k-1}} \leqslant \\ M\left\|\nabla^{k+2} \phi\right\| \end{array} $

由$\frac{l}{3} $=$\left(\frac{\alpha}{3}-\frac{1}{2}\right)\left(1-\frac{l-1}{k-1}\right) $+$\left(\frac{k+2}{3}-\frac{1}{2}\right)\frac{l-1}{k-1} $，可确定α=$ 3-\frac{k-1}{2(k-l)} \in[2, 3]$。

当$\left[\frac{k+1}{2}\right]+1 \leqslant l \leqslant k $时，有

$ \begin{array}{l} \;\;\;\;\;\left\|\nabla^{l+1} \phi \nabla^{k-l+2} \phi\right\| \leqslant\left\|\nabla^{l+1} \phi\right\|_{L^6}\left\|\nabla^{k-l+2} \phi\right\|_{L^3} \leqslant \\ \left\|\nabla^2 \phi\right\|^{1-\frac{l}{k}} \end{array} $

$ \;\;\;\;\;\left\|\nabla^{l+1} \phi \nabla^{k-l+2} \phi\right\| \leqslant\left\|\nabla^{l+1} \phi\right\|_{L^6}\left\|\nabla^{k-l+2} \phi\right\|_{L^3} \leqslant\\ M\left\|\nabla^{k+2} \phi\right\| $

由$\frac{k-l+1}{3} $=$\left(\frac{\alpha}{3}-\frac{1}{2}\right) \frac{l}{k} $+$\left(\frac{k+2}{3}-\frac{1}{2}\right)(1-\frac{l}{k}) $，可确定α=$2+\frac{k}{2 l} \in\left[\frac{5}{2}, 3\right) $。

因此，

$ I_8^4 \leqslant M\left(\left\|\nabla^{k+2} \phi\right\|^2+\left\|\nabla^{k+1} \sigma\right\|^2\right) $

同样可以得到

$ I_8^5 \leqslant M\left(\left\|\nabla^{k+2} \phi\right\|^2+\left\|\nabla^{k+1} \sigma\right\|^2\right) $

将以上估计代入式(37)，有

$ \begin{array}{l} \;\;\;\;\;\;\;I_8 \leqslant M\left(\left\|\nabla^{k+2} \phi\right\|^2+\left\|\nabla^{k+1} \sigma\right\|^2+\left\|\nabla^{k+2} \boldsymbol{u}\right\|^2+\right. \\ \left.\left\|\nabla^{k+1} \boldsymbol{u}\right\|^2\right) \end{array} $

(38)

将式(38)代入式(36)，得到式(35)，引理4得证。结合引理1~4，命题2得证，进一步定理1得证。

3 结束语

本文研究了三维Navier-Stokes-Allen-Cahn方程组Cauchy问题的适定性。在初值小扰动的假设条件下，克服了由相场ϕ产生的强非线性项(div($\nabla \phi \otimes \nabla \phi-\frac{1}{2}|\nabla \phi|^2 \boldsymbol{I }$)和(1-ϕ²)ϕ)给能量估计带来的困难，证明了该方程组全局解的存在唯一性。所得结果可为非混相两相流的模拟计算和实验研究提供必要的理论支撑。