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  北京化工大学学报(自然科学版)  2019, Vol. 46 Issue (6): 101-107   DOI: 10.13543/j.bhxbzr.2019.06.015
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引用本文  

王暐翼, 童天娇, 陈亚洲. 一维Navier-Stokes-Cahn-Hilliard方程组解的适定性分析[J]. 北京化工大学学报(自然科学版), 2019, 46(6): 101-107. DOI: 10.13543/j.bhxbzr.2019.06.015.
WANG WeiYi, TONG TianJiao, CHEN YaZhou. Well-posedness of solutions for navier-stokes-cahn-hilliard system in one dimension[J]. Journal of Beijing University of Chemical Technology (Natural Science), 2019, 46(6): 101-107. DOI: 10.13543/j.bhxbzr.2019.06.015.

第一作者

王暐翼, 男, 1995年生, 硕士生.

通信联系人

陈亚洲, E-mail: chenyz@mail.buct.edu.cn

文章历史

收稿日期:2019-05-19
一维Navier-Stokes-Cahn-Hilliard方程组解的适定性分析
王暐翼 , 童天娇 , 陈亚洲     
北京化工大学 数理学院, 北京 100029
摘要:讨论和描述了具有扩散界面的互不相溶气液两相流动的可压缩Navier-Stokes-Cahn-Hilliard(NSCH)方程组的周期边值问题,NSCH方程组中采用了van der Waals状态方程,该状态方程是关于密度非凸的刻画气液相变的经典模型。通过对压力的单调分解并结合能量估计的方法,克服了状态方程非凸性带来的困难,得到了流体密度的上下界估计;对任意初始值(密度不含真空),证明了该问题的一维流动强解是全局存在且唯一的。结果表明,该气液相变问题不会出现激波和真空现象。
关键词Navier-Stokes-Cahn-Hilliard(NSCH)方程组    van der Waals状态方程    气液两相流    
Well-posedness of solutions for Navier-Stokes-Cahn-Hilliard system in one dimension
WANG WeiYi , TONG TianJiao , CHEN YaZhou     
College of Mathematics and Physics, Beijing University of Chemical Technology, Beijing 100029, China
Abstract: This paper is concerned with a diffusive interface model for immiscible gas-liquid binary fluids. The periodic boundary value problem for a compressible Navier-Stokes-Cahn-Hilliard equation is discussed, and the van der Waals equation of state, which is non-convex for density and is the classical model for gas-liquid phase transition, is employed. In addition, we make use of the monotonic decomposition of pressure combined with the energy estimation method to overcome the difficulty caused by the non-convexity of the state equation, and the upper and lower bounds of the density are obtained. For any initial value (density without vacuum), the global existence and uniqueness of strong solutions is thus proved. The results indicate that there is no shock or vacuum phenomenon in the gas-liquid phase change problem.
Key words: Navier-Stokes-Cahn-Hilliard (NSCH) equations    van der Waals equation of state    gas-liquid flow    
引言

气液两相流广泛存在于化工、海洋、航空、生物等领域的生产过程中,对其进行研究具有重要的理论和应用价值。对于两相流接触界面的刻画可分为两种类型:一类是将接触面当作光滑界面,另一类是将互不相溶的两相流界面看成是有厚度的扩散界面层。本文研究第二种类型。Cahn等[1]首先引入界面自由能,提出了著名的Cahn-Hilliard(CH)方程来描述两种互不相溶流体的界面运动情况;Lowengrub等[2]进一步将刻画单一流体流动的Navier-Stoke(NS)方程与CH方程相结合,得到了Navier-Stokes-Cahn-Hilliard(NSCH)方程组;Abels等[3]证明了三维可压缩NSCH方程组全局弱解的存在性, 但是弱解的唯一性还没有得到解决。此后Matthias等[4]得到了三维NSCH方程组的局部强解的存在唯一性;Ding等[5]给出了一维情形下可压缩NSCH方程组全局强解的存在性;Chen等[6]证明了一维带有非光滑自由能密度的可压缩NSCH方程组周期边值问题强解的适定性。

然而以上文献在研究NSCH方程解的适定性问题时所考虑的压力项p采用了理想正压气体模型,即p=r(ρ为密度,常数项a>0, r>1)。本文在前人的研究工作的基础上,讨论带有van der Waals状态方程的一维黏性可压缩NSCH方程组,利用能量积分和压力p的特殊结构,克服压力的非凸性及密度上下界估计的困难,来证明对于初始密度不含真空的任意初值的NSCH方程组,其强解具有全局存在唯一性。

1 模型构造及主要定理

可压缩气液混合两相流流动过程通常由如下的NSCH非线性偏微分方程组来描述

$ \left\{ {\begin{array}{*{20}{l}} {{\partial _t}\rho + {\rm div}(\rho \mathit{\boldsymbol{u}}) = 0}\\ {{\partial _t}(\rho \mathit{\boldsymbol{u}}) + {\rm div}(\rho \mathit{\boldsymbol{u}} \otimes \mathit{\boldsymbol{u}}) = {\rm div}\mathit{\boldsymbol{T}}}\\ {{\partial _t}(\rho \chi ) + {\rm div}(\rho \chi \mathit{\boldsymbol{u}}) = \Delta \mu }\\ {\rho \mu = \frac{1}{\varepsilon }\rho \frac{{\partial f}}{{\partial \chi }} - \varepsilon \Delta \chi } \end{array}} \right. $ (1)

式中,ρ=ρ1+ρ2为混合流体的总密度;u为流体速度,且有ρu=ρ1u1+ρ2u2ui(i=1, 2)为第i种组分速度;χ=χ1χ2为两种组分质量浓度差, 其中χi=Mi/M(i=1, 2),Mi为第i种组分的质量;μ为混合流体的化学势能;常数ε>0为混合流体的界面厚度;T为应力张量,且满足

$ \left\{ {\begin{array}{*{20}{l}} {\mathit{\boldsymbol{T}} = \mathit{\boldsymbol{S}} - \varepsilon \left( {\nabla \chi \otimes \nabla \chi - \frac{{{{\left| {\nabla \chi } \right|}^2}}}{2}\mathit{\boldsymbol{I}}} \right) - p\mathit{\boldsymbol{I}}}\\ {\mathit{\boldsymbol{S}} = \mathit{\boldsymbol{v}}\left( {\left( {\nabla \mathit{\boldsymbol{u}} + {\nabla ^{\rm{T}}}\mathit{\boldsymbol{u}}} \right) - \frac{2}{3}{\rm div}\mathit{\boldsymbol{uI}}} \right) + \eta {\rm div}\mathit{\boldsymbol{uI}}} \end{array}} \right. $ (2)

式(2)中,I为单位矩阵,S为Newtonian黏性张力,p为压力,常数υ>0、η≥0为黏性系数。

自由能密度函数f=f(ρ, χ)定义如下[7-9]

$ f(\rho ,\chi ) = - 3\rho + \frac{{8\mathit{\Theta }}}{3}\ln \frac{\rho }{{3 - \rho }} + \frac{1}{4}{\left( {{\chi ^2} - 1} \right)^2} $ (3)

式中Θ为温度。本文考虑式(1)中的一维情形

$ \left\{ {\begin{array}{*{20}{l}} {{\rho _t} + {{(\rho u)}_x} = 0}\\ {\rho {u_t} + \rho u{u_x} + {p_x} = v{u_{xx}} - \frac{\varepsilon }{2}{{\left( {\chi _x^2} \right)}_x}}\\ {\rho {X_t} + \rho u{\chi _x} = {\mu _{xx}}}\\ {\mu = \frac{1}{\varepsilon }\left( {{\chi ^3} - \chi } \right) - \frac{\varepsilon }{\rho }{\chi _{xx}}} \end{array}} \right. $ (4)

式(4)中,(x, t)∈$ \mathbb{R} $×(0, +∞),u为混合流体在一维情形下的速度。压力p满足van der Waals状态方程

$ p(\rho ) = \left\{ {\begin{array}{*{20}{l}} {{\rho ^2}\frac{{\partial f}}{{\partial \rho }} = - 3{\rho ^2} + \frac{{8\mathit{\Theta }\rho }}{{3 - \rho }},}&{0 \le \rho < 3}\\ { + \infty ,}&{\rho \ge 3} \end{array}} \right. $ (5)

同时有

$ \left\{ {\begin{array}{*{20}{l}} {{p^\prime } = p/{p_{\rm{c}}}}\\ {{\rho ^\prime } = \rho /{\rho _{\rm{c}}}}\\ {{\mathit{\Theta }^\prime } = \mathit{\Theta }/{\mathit{\Theta }_{\rm{c}}}} \end{array}} \right. $

式中,pc, ρc, Θc分别表示临界点处的压力、密度及温度[9]。在不引起混淆的情况下,p′, ρ′, Θ′仍记作p, ρ, Θ

由式(5)知,压力p关于参变量ρ有以下单调性质:

① 当Θ>1时,p为单调增函数;

② 当0<Θ<1时,存在两点α, β∈(0, 3), 使得ρ∈(0, α)∪(β, 3)时,p为单调增函数,ρ∈(α, β)时,p为单调减函数。

方程组(4)的初值及周期边值条件如式(6)

$ \left\{ {\begin{array}{*{20}{l}} {(\rho ,u,\chi )(x,t) = (\rho ,u,\chi )(x + L,t),}&{x \in \mathbb{R},t > 0} \\ {{{\left. {(\rho ,u,\chi )} \right|}_{t = 0}} = \left( {{\rho _0},{u_0},{\chi _0}} \right),}&{x \in \mathbb{R}} \end{array}} \right. $ (6)

定义L2($ \mathbb{R} $)的周期函数空间,有

$ L_{{\text{per}}}^2 = \left\{ {g|g(x) = g(x + L),x \in \mathbb{R},g(x) \in {L^2}\left( {0,L} \right)} \right\} $

其范数可表示为

$ \left\| g \right\| = {\left( {\int_0^L {{{\left| {g(x)} \right|}^2}} {\rm{d}}x} \right)^{\frac{1}{2}}} $

Hperl(l≥0)可定义为

$ H_{{\text{per}}}^l = \left\{ {g \in \mathbb{R}|g(x) \in L_{{\text{per}}}^2,\partial _x^j \in L_{{\text{per}}}^2,j = 1,2, \cdots ,l} \right\} $

其范数可表示为

$ {\left\| g \right\|_l} = {\left( {\sum\limits_{j = 0}^l {{{\left\| {\partial _x^jg} \right\|}^2}} } \right)^{\frac{1}{2}}} $

假设式(4)的初值满足以下条件

$ \left\{ \begin{array}{l} \left( {{\rho _0},{u_0}} \right) \in H_{{\rm{per}}}^2,{\chi _0} \in H_{{\rm{per}}}^4,{\rho _0} \in (0,3)\\ {\rho _t}(x,0) = - {\rho _{0x}}{u_0} - {\rho _0}{u_{0x}}\\ {u_t}(x,0) = \frac{v}{{{\rho _0}}}{u_{0xx}} - {u_0}{u_{0x}} - \frac{{p_\rho ^\prime \left( {{\rho _0}} \right)}}{{{\rho _0}}}{\rho _{0x}} - \\ \;\;\;\;\;\;\;\;\;\;\;\;\frac{\varepsilon }{{{\rho _{0x}}}}{\chi _{0x}}{\chi _{0xx}}\\ {\chi _t}(x,0) = - {u_0}{\chi _{0x}} - \frac{{\varepsilon {\chi _{0xxxx}}}}{{\rho _0^2}} + \frac{{2\varepsilon {\rho _{0x}}}}{{\rho _0^3}}{\chi _{xxx}} + \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\varepsilon \left( {\frac{{{\rho _{0xx}}}}{{\rho _0^3}} - \frac{{2\rho _{0x}^2}}{{\rho _0^4}}} \right){\chi _{0xx}} + \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\frac{{\left( {3\chi _0^2 - 1} \right){\chi _{0xx}}}}{{\varepsilon {\rho _0}}} + \frac{{6{\chi _0}\chi _{0x}^2}}{{\varepsilon {\rho _0}}} \end{array} \right. $ (7)

由式(5)及压力p关于参变量ρ的单调性质可得:当0<Θ<1时,有$ \exists $ζ∈(0, α), 使得p(ζ)=p(β)。

设密度常数参数$ {\tilde \rho } $满足0<$ {\tilde \rho } $ζ<3,定义

$ \mathit{\Phi }(\rho ) = \rho \int_{\tilde \rho }^\rho {\frac{{p(s) - p(\tilde \rho )}}{{{s^2}}}{\rm{d}}s} $

则由式(4)的第一方程可以得到

$ \mathit{\Phi }{(\rho )_t} + {(\mathit{\Phi }(\rho )u)_x} + (p(\rho ) - p(\tilde \rho )){u_x} = 0 $ (8)

对于周期边值问题,可将整个空间$ \mathbb{R} $看作是[0, L]以L为周期的扩展,因此只需在[0, L]上讨论解的存在性。

定理1   假设(ρ0, u0, χ0)满足条件式(7),则方程组(4)~(6)存在唯一的全局强解(ρ, u, χ),使得对任意的T>0,存在常数0<γ<3,且满足

$ \left\{ \begin{gathered} \rho \in {L^\infty }\left( {0,T,H_{{\text{per}}}^2} \right) \hfill \\ {\rho _t} \in {L^\infty }\left( {0,T;L_{{\text{per}}}^2} \right) \hfill \\ u \in {L^\infty }\left( {0,T;H_{{\text{per}}}^2} \right) \cap {L^2}\left( {0,T;H_{{\text{per}}}^3} \right) \hfill \\ {u_t} \in {L^\infty }\left( {0,T;L_{{\text{per}}}^2} \right) \cap {L^2}\left( {0,T;H_{{\text{per}}}^1} \right) \hfill \\ {\chi _t} \in {L^\infty }\left( {0,T;H_{{\text{per}}}^4} \right) \hfill \\ {\chi _t} \in {L^\infty }\left( {0,T;H_{{\text{per}}}^4} \right) \cap {L^2}\left( {0,T;H_{{\text{per}}}^2} \right) \hfill \\ \mu \in {L^\infty }\left( {0,T;H_{{\text{per}}}^2} \right) \cap {L^2}\left( {0,T;H_{{\text{per}}}^4} \right) \hfill \\ {\mu _t} \in {L^2}\left( {0,T;L_{{\text{per}}}^2} \right) \hfill \\ 0 < \rho \leqslant \gamma < 3,{\text{ for all }}(x,t) \in \mathbb{R} \times [0,T] \hfill \\ \end{gathered} \right. $ (9)

且有

$ \begin{array}{l} \mathop {\sup }\limits_{t \in [0,T]} \left\{ {\left\| {(\rho ,u)(t)} \right\|_2^2 + \left\| {\chi (t)} \right\|_4^2 + \left\| {\mu (t)} \right.} \right.\\ \left. {\left. {} \right\|_2^2} \right\} + \int_0^T {\left( {\left\| \rho \right\|_2^2 + \left\| u \right\|_3^2 + \left\| {(\chi ,\mu )(t)} \right\|_4^2} \right)} {\rm{d}}t \le C \end{array} $ (10)

式中,C为只依赖于初值及T的正常数。

2 主要定理的证明

定理1的证明分为两个部分:先证明局部解的存在性,再通过能量方法得到解的一致估计,结合延拓方法得到全局强解的存在性。

证明过程中存在两个难点:一是密度ρ的上下界确定,二是压力p的非凸性。对于第一个难点,可以利用所得基本能量不等式、压力p结构的特殊性以及测度相关知识,通过反证法得到;对于第二个难点,可以根据pρpρρ具体表达式及ργ<3,求出高阶能量估计。

2.1 局部解的存在性

先定义周期解空间,即对∀m>0, M>0, T>0,有

$ \begin{gathered} {\mathit{\boldsymbol{X}}_{{\text{per,}}m,M}}([0,T]) = \left\{ {(\rho ,u,\chi )|\rho \in {C^0}([0,T];} \right. \hfill \\ \left. {H_{{\text{pres}}}^2} \right) \cap {L^2}\left( {0,T;H_{{\text{pres}}}^2} \right) \hfill \\ u \in {C^0}\left( {[0,T];H_{{\text{per}}}^2} \right) \cap {L^2}\left( {0,T;H_{{\text{per}}}^3} \right) \hfill \\ \chi \in {C^0}\left( {[0,T];H_{{\text{per}}}^4} \right) \cap {L^2}\left( {0,T;H_{{\text{per}}}^5} \right) \hfill \\ \mathop {\inf }\limits_{x \in \mathbb{R},t \in [0,T]} \rho (x,t) \geqslant m,\mathop {\sup }\limits_{t \in [0,T]} \left\{ {\left\| {(\rho ,u)} \right\|_2^2,} \right. \hfill \\ \left. {\left. {\left\| \chi \right\|_4^2} \right\} \leqslant M} \right\} \hfill \\ \end{gathered} $ (11)

命题1   对∀m>0, M>0,若初值满足条件$ \inf \limits_{x \in \mathbb{R}} \rho_{0}(x, t) \geqslant m, \left\|\left(\rho_{0}, u_{0}\right)\right\|_{2}^{2} \leqslant M, \left\|\chi_{0}\right\|_{4}^{2} \leqslant M $,则存在一个小时间T*=T*(ρ0, u0, χ0)>0,使得周期边值问题式(4)~(6)存在唯一解(ρ, u, χ),且满足

$ (\rho ,u,\chi ) \in {\mathit{\boldsymbol{X}}_{{\rm{per}},\frac{m}{2},2M}}\left( {\left[ {0,{T^*}} \right]} \right) $

证明:令0<T<+∞, 对任意的m>0, M>0, 构造一个可迭代的序列(ρ(n), u(n), χ(n)), n=1, 2, …, 且满足初始条件(ρ(0), u(0), χ(0))=(ρ0, u0, χ0),利用此可迭代序列构造以下迭代格式

$ \left\{ \begin{array}{l} \rho _t^{(n)} + {\left( {{\rho ^{(n)}}{u^{(n - 1)}}} \right)_x} = 0\\ {\rho ^{(n)}}u_t^{(n)} + {\rho ^{(n)}}{u^{(n - 1)}}u_x^{(n)} + {\left( {p\left( {{\rho ^{(n)}}} \right)} \right)_x} = \\ \;\;\;\;\;\;\;\;\;\;vu_{xx}^{(n)} - \frac{\varepsilon }{2}{\left( {\left( {{\chi ^{(n)}}} \right)_x^2} \right)_x}\\ p\chi _t^{(n)} + {\rho ^{(n)}}{u^{(n - 1)}}\chi _x^{(n)} = \mu _{xx}^{(n)}\\ {\mu ^{(n)}} = \frac{1}{\varepsilon }\left( {{{\left( {{\chi ^{(n - 1)}}} \right)}^3} - {\chi ^{(n - 1)}}} \right) - \frac{{\cal E}}{{{\rho ^{(n)}}}}\chi _{xx}^{(n)}\\ \left( {{\rho ^{(n)}},{u^{(n)}},{\chi ^{(n)}}} \right)(x,t) = \left( {{\rho ^{(n)}},} \right.\\ \;\;\;\;\;\;\left. {{u^{(n)}},{\chi ^{(n)}}} \right)(x + L,t)\\ \left( {{\rho ^{(n)}},{u^{(n)}},{\chi ^{(n)}}} \right)(x,0) = \left( {{\rho _0},{u_0},{\chi _0}} \right)(x) \end{array} \right. $ (12)

利用类似文献[6]的迭代方法,可以证明周期边值问题式(4)~(6)解的局部存在性。

2.2 全局解的存在性

命题2   假设(ρ0, u0, χ0)满足条件式(7),对于任意的T>0,设(ρ, u, χ)∈Xper, m, M([0, T])为方程组式(4)~(6)的局部解,则存在一个只依赖于初值及T的常数C, 使得

$ \begin{array}{l} \mathop {\sup }\limits_{t \in [0,T]} \left\{ {\left\| {(\rho ,u)(t)} \right\|_2^2 + \left\| {\chi (t)} \right\|_4^2 + \left\| {\mu (t)} \right.} \right.\\ \left. {\left. {} \right\|_2^2} \right\} + \int_0^{\rm{T}} {\left( {\left\| \rho \right\|_2^2 + \left\| u \right\|_3^2 + \left\| {(\chi ,\mu )(t)} \right\|_4^2} \right){\rm{d}}t} \le C \end{array} $ (13)

命题2的证明可由以下引理1~4得到。

引理1   在命题2的假设下,对于任意的T>0,都有

$ \begin{array}{l} \mathop {\sup }\limits_{t \in [0,T]} \int_0^L {\left( {\frac{{\rho {u^2}}}{2} + \frac{{\varepsilon \chi _x^2}}{2} + \frac{{\rho {{\left( {{\chi ^2} - 1} \right)}^2}}}{{4\varepsilon }} + \mathit{\Phi }(\rho )} \right){\rm{d}}x} + \\ \int_0^T {\int_0^L {\left( {\mathit{\boldsymbol{v}}u_x^2 + \mu _x^2} \right){\rm{d}}x{\rm{d}}t} } \le C \end{array} $ (14)

证明:将式(4)第二个方程乘以u, 式(4)的第三个方程乘以μ,相加后关于x在[0, L]上积分,得到

$ \begin{array}{l} \frac{{\rm{d}}}{{{\rm{d}}t}}\int_0^L {\left( {\frac{{\rho {u^2}}}{2} + \frac{{\varepsilon \chi _x^2}}{2} + \frac{{\rho {{\left( {{\chi ^2} - 1} \right)}^2}}}{{4\varepsilon }}} \right){\rm{d}}x} + \int_0^L {\left( {\mu _x^2 + } \right.} \\ \left. {vu_x^2 + u{p_x}} \right){\rm{d}}x = 0 \end{array} $ (15)

对式(8)在[0, L]上积分,得

$ \frac{{\rm{d}}}{{{\rm{d}}t}}\int_0^L \mathit{\Phi } (\rho ){\rm{d}}x = \int_0^L u {p_x}(\rho ){\rm{d}}x $ (16)

将式(16)代入式(15),再关于t在[0, T]上积分,得

$ \begin{array}{l} \mathop {\sup }\limits_{t \in [0,T]} \int_0^L {\left( {\frac{{\rho {u^2}}}{2} + \frac{{\varepsilon \chi _x^2}}{2} + \frac{{\rho {{\left( {{\chi ^2} - 1} \right)}^2}}}{{4\varepsilon }} + \mathit{\Phi }(\rho )} \right)} {\rm{d}}x + \\ \int_0^T {\int_0^L {\left( {\mathit{\boldsymbol{v}}u_x^2 + \mu _x^2} \right){\rm{d}}x{\rm{d}}t} } \le {E_0} \end{array} $

式中

$ {E_0} = \int_0^L {\left( {\frac{{{\rho _0}u_0^2}}{2} + \frac{{\varepsilon \chi _{0x}^2}}{2} + \frac{{{\rho _0}{{\left( {\chi _0^2 - 1} \right)}^2}}}{{4\varepsilon }} + \mathit{\Phi }\left( {{\rho _0}} \right)} \right)} {\rm{d}}x $

引理1得证。

引理2   在命题2的假设下,对任意的T>0,都有

$ \mathop {\sup }\limits_{t \in [0,T]} {\left\| {\chi (x,t)} \right\|_{L_{{\rm{per}}}^\infty }} \le C,\mathop {\sup }\limits_{t \in [0,T]} {\left\| {\rho (x,t)} \right\|_{L_{{\rm{per}}}^\infty }} < 3 $ (17)

以及

$ \mathop {\sup }\limits_{t \in [0,T]} \int_0^L \rho {\chi ^2}{\rm{d}}x + \int_0^T {\int_0^L {\frac{{\chi _{xx}^2}}{\rho }} } {\rm{d}}x{\rm{d}}t + \int_0^T {\int_0^L {{\chi ^2}} } \chi _x^2{\rm{d}}x{\rm{d}}t \le C $ (18)

证明:对式(4)第一个方程在[0, L]×[0, t]上积分,得

$ \int_0^L \rho (x,t){\rm{d}}x = \int_0^L {{\rho _0}} (x){\rm{d}}x $ (19)

由式(19)及式(14)可得,$ \int_0^L \rho {\chi ^4}{\rm{d}}x \le C, \int_0^L \rho \chi {\rm{d}}\mathit{x} \le \mathit{C} $,并且有

$ \begin{array}{l} \left| {\chi (x,t)} \right| = \frac{1}{{\int_0^L {{\rho _0}} {\rm{d}}x}}\left| {\chi (x,t)\int_0^L \rho (y,t){\rm{d}}y} \right| \le \\ \frac{1}{{\int_0^L {{\rho _0}} {\rm{d}}x}}\left( {\left| {\int_0^L \rho (y,t)\left( {\int_y^x {{\chi _s}} (s,t){\rm{d}}s} \right){\rm{d}}y} \right| + \left| {\int_0^L {} } \right.} \right.\\ \chi (y,t)\rho (y,t){\rm{d}}y|) \le C\int_0^L {\chi _x^2} {\rm{d}}x + C \end{array} $

另外,若ρ≥3, 则由式(5)可得p(ρ)=+∞,易得Φ(ρ)=+∞, 这与式(14)中$ \mathop {\sup }\limits_{t \in [0, L]} \int_0^L \mathit{\Phi } (\rho ){\rm{d}}x \le {E_0} < + \infty $相矛盾,所以有ρ<3,式(17)得证。

将式(4)第三个方程乘以χ, 并在[0, L]×[0, T]上积分,得

$ \begin{array}{l} \frac{1}{2}\int_0^L \rho {\chi ^2}{\rm{d}}x + \frac{1}{\varepsilon }\int_0^T {\int_0^L 3 } {\chi ^2}\chi _x^2{\rm{d}}x{\rm{d}}t + \varepsilon \int_0^T {\int_0^L {\frac{{\chi _{xx}^2}}{\rho }} } {\rm{d}}x{\rm{d}}t = \\ \frac{1}{\varepsilon }\int_0^L {\int_0^L {\chi _x^2} } {\rm{d}}x{\rm{d}}t + \frac{1}{2}\int_0^L {{\rho _0}} \chi _0^2{\rm{d}}x \le C \end{array} $

式(18)得证。

引理3   在命题2的假设下,对于任意的T>0,都有

$ \left\{ {\begin{array}{*{20}{l}} {\mathop {\sup }\limits_{t \in [0,T]} {{\left\| {\frac{1}{\rho }} \right\|}_{L_{{\rm{per}}}^\infty }} \le C}\\ {\mathop {\sup }\limits_{t \in [0,T]} \int_0^L {\rho _x^2} {\rm{d}}x \le C}\\ {\rho \le \gamma < 3} \end{array}} \right. $ (20)

证明:对式(4)第一个方程关于x求导,再代入式(4)第二个方程,得

$ \begin{array}{l} {(\rho u)_t} + {\left( {\rho {u^2}} \right)_x} + {p^\prime }(\rho ){\rho _x} = v\left[ {\rho \frac{{\rm{d}}}{{{\rm{d}}t}}{{\left( {\frac{1}{\rho }} \right)}_x} + } \right.\\ \left. {\rho u{{\left( {\frac{1}{\rho }} \right)}_{xx}}} \right] - \frac{\varepsilon }{2}{\left( {\chi _x^2} \right)_x} \end{array} $ (21)

将式(21)乘以$ \frac{v}{2}{\left( {\frac{1}{\rho }} \right)_x} $,并在[0, L]上积分,结合式(18)及式(14),并利用文献[7]中p′(ρ)正负性处理含p′(ρ)项的方法,使用Gronwall's不等式可得

$ \begin{array}{l} \mathop {\sup }\limits_{t \in [0,T]} \int_0^L {\left( {\rho {{\left| {{{\left( {\frac{1}{\rho }} \right)}_x}} \right|}^2} + \rho {\chi ^2} + \rho {u^2} + \chi _x^2 + \mathit{\Phi }(\rho ) + } \right.} \\ \left. {\rho {{\left( {{\chi ^2} - 1} \right)}^2}} \right){\rm{d}}x + \int_0^T {\int_0^L {\left( {{u_x} + \mu _x^2} \right)} } {\rm{d}}x{\rm{d}}t \le C \end{array} $ (22)

由式(22)及积分中值定理可知,$ \exists $a(t)∈[0, T]满足$ \rho(a(t), t)=\frac{1}{L} \int_{0}^{L} \rho_{0} \mathrm{d} x $, 所以有

$ \begin{array}{l} \frac{1}{{\rho \left( {x,t} \right)}} = \frac{1}{{\rho \left( {x,t} \right)}} - \frac{1}{{\rho (a(t),t)}} + \frac{1}{{\rho (a(t),t)}} \le \\ C\int_0^L \rho {\left| {{{\left( {\frac{1}{\rho }} \right)}_x}} \right|^2}{\rm{d}}x + C \le C \end{array} $

再结合式(22)可得

$ \mathop {\sup }\limits_{t \in [0,T]} {\left\| {\frac{1}{\rho }} \right\|_{L_{{\rm{per}}}^\infty }} \le C $ (23)

由式(22)、式(17)及Φ(ρ)的定义,并令$ v = \frac{1}{\rho }, \tau = \frac{1}{s} $,则有

$ \begin{array}{l} C \ge \int_0^L {\mathit{\Phi }(\rho ){\rm{d}}x} = \int_0^L - \rho \int_\rho ^\rho {(p(} s) - p(\tilde \rho ))\\ {\rm{d}}\left( {\frac{1}{s}} \right){\rm{d}}x = \int_0^L - \frac{1}{v}\int_v^v {\left( {p\left( {\frac{1}{\tau }} \right) - p\left( {\frac{1}{{\tilde v}}} \right)} \right)} {\rm{d}}\tau {\rm{d}}x = \\ \int_0^L {\frac{1}{v}} \left( {\frac{{8\mathit{\Theta }}}{{3\tilde v - 1}} - \frac{3}{{{{\tilde v}^2}}}} \right)(v - \tilde v){\rm{d}}x - \int_0^L {\frac{1}{v}} \int_v^v {\left( {\frac{{8\mathit{\Theta }}}{{3\tau - 1}} - } \right.} \\ \left. {\frac{3}{{{\tau ^2}}}} \right){\rm{d}}\tau {\rm{d}}x \end{array} $ (24)

式(24)中$ \tilde v = \frac{1}{{\tilde \rho }} $为常数,则$ \;\frac{{8\mathit{\Theta }}}{{3\tilde v - 1}} - \frac{3}{{{{\tilde v}^2}}} $也为常数,则

$ \begin{array}{l} \int_0^L {\frac{1}{v}} \left( {\frac{{8\mathit{\Theta }}}{{3\tilde v - 1}} - \frac{3}{{{{\hat v}^2}}}} \right)(v - \tilde v){\rm{d}}x = C - C\int_0^L {\frac{1}{v}{\rm{d}}x} = \\ C - C\int_0^L {\rho {\rm{d}}x} = C - C\int_0^L {{\rho _0}{\rm{d}}x} = C \end{array} $ (25)

$ \mathop {\sup }\limits_{t \in [0, L]} {\left\| {\frac{1}{v}} \right\|_{L_{{\rm{per }}}^\infty }} < 3, \mathop {\sup }\limits_{t \in [0, T]} \int_0^L {\frac{1}{{{v^2}}}} {\rm{d}}x < C $, 结合式(23)可得$ v=\frac{1}{\rho} \in\left(\frac{1}{3}, M\right] $,并可得到

$ \begin{array}{l} - \int_0^L {\frac{1}{v}} \int_v^v {\left( {\frac{{8\mathit{\Theta }}}{{3\tau - 1}} - \frac{3}{{{\tau ^2}}}} \right){\rm{d}}\tau {\rm{d}}x} = - \frac{{8\mathit{\Theta }}}{3}\mathop \smallint \nolimits_0^L \\ \frac{{\ln (3v - 1)}}{v}{\rm{d}}x - \int_0^L {\frac{3}{{{v^2}}}} {\rm{d}}x + C \ge C - \frac{{8\mathit{\Theta }}}{3}\int_d^L {\frac{{\ln (3v - 1)}}{v}{\rm{d}}x} = \\ C - \frac{{8\mathit{\Theta }}}{3}\int_{v \in \left( {\frac{1}{3},M} \right]} {\frac{{\ln (3v - 1)}}{v}{\rm{d}}x} \end{array} $ (26)

$ \exists $m0, |m0|>0使得$ v \to \frac{1}{3} $在[0, L]\m0上,则有

$ \begin{array}{l} - \frac{{8\mathit{\Theta }}}{3}\int_{v \in \left( {\frac{1}{3},M} \right]} {\frac{{\ln (3v - 1)}}{v}{\rm{d}}x} = - \frac{{8\mathit{\Theta }}}{3}\int_{\left[ {0,L} \right]\backslash {m_0}} {} \\ \frac{{\ln (3v - 1)}}{v}{\rm{d}}x - \frac{{8\mathit{\Theta }}}{3}\int_{{m_0}} {\frac{{\ln (3v - 1)}}{v}{\rm{d}}x} = + \infty \end{array} $ (27)

由式(24)~(27)可得

$ \mathop {\sup }\limits_{t \in [0,T]} \int_0^L {\mathit{\Phi }(\rho ){\rm{d}}x} = + \infty $ (28)

式(28)与$ \mathop {\sup }\limits_{t \in [0, T]} \int_0^L \mathit{\Phi } (\rho ){\rm{d}}x < + \infty $矛盾,即不存在这样的测度大于零的集合m0使得$ v \to \frac{1}{3} $,所以$ \frac{1}{3} < \frac{1}{\gamma } \le v $, 即ργ<3。由式(22)有

$ \int_0^L {\rho _x^2} {\rm{d}}x \le \left\| \rho \right\|_{L_{{\rm{per}}}^\infty }^3\int_0^L \rho {\left| {{{\left( {\frac{1}{\rho }} \right)}_x}} \right|^2}{\rm{d}}x \le C $

引理3得证。

引理4   在命题2的假设下,对任意的T>0都有

$ \left\{ \begin{array}{l} \mathop {\sup }\limits_{t \in [0,T]} \left( {\left\| \chi \right\|_{H_{{\rm{per}}}^4}^2 + \left\| {{\chi _t}} \right\|_{L_{{\rm{per}}}^2}^2} \right) + \\ \;\;\;\;\;\;\;\;\int_0^T {\left( {\left\| \chi \right\|_{H_{{\rm{per}}}^4}^2 + \left\| {{\chi _t}} \right\|_{H_{{\rm{per}}}^2}^2} \right){\rm{d}}t} \le C\\ \mathop {\sup }\limits_{t \in [0,T]} \left( {\left\| u \right\|_{H_{{\rm{per}}}^2}^2 + \left\| {{u_t}} \right\|_{L_{{\rm{per}}}^2}^2} \right) + \\ \;\;\;\;\;\;\;\;\int_0^T {\left( {\left\| u \right\|_{H_{{\rm{per}}}^3}^2 + \left\| {{u_t}} \right\|_{L_{{\rm{per}}}^1}^2} \right){\rm{d}}t} \le C\\ \mathop {\sup }\limits_{t \in [0,T]} \left( {\left\| \rho \right\|_{H_{{\rm{per}}}^2}^2 + \left\| {{\rho _t}} \right\|_{L_{{\rm{per}}}^2}^2} \right) + \int_0^T {\left\| \rho \right\|_{H_{{\rm{per }}}^2}^2} {\rm{d}}t \le C\\ \mathop {\sup }\limits_{t \in [0,T]} \left\| \mu \right\|_{H_{{\rm{per}}}^2}^2 + \int_0^T {\left( {\left\| \mu \right\|_{H_{{\rm{per}}}^4}^2 + \left\| {{\mu _t}} \right\|_{L_{{\rm{per}}}^2}^2} \right)} {\rm{d}}t \le C \end{array} \right. $ (29)

证明:由式(18)及式(20)可得

$ \int_0^T {\int_0^L {\chi _{xx}^2} } {\rm{d}}x{\rm{d}}t \le {\left\| \rho \right\|_{{L^\infty }}}\int_0^T {\int_0^L {\frac{{\chi _{xx}^2}}{\rho }} } {\rm{d}}x{\rm{d}}t \le C $ (30)

对式(4)第四个方程关于x求导,再乘以χxxx, 然后在[0, L]×[0, T]上积分,并由式(14)、(20)得

$ \begin{array}{l} \int_0^T {\int_0^L {\chi _{xxx}^2} } {\rm{d}}x{\rm{d}}t \le C\int_0^T {\int_0^L {\left( {\frac{\rho }{\varepsilon }\left( {3{\chi ^2} - 1} \right) - \frac{{{\mu _x}\rho }}{\varepsilon } + } \right.} } \\ {\left. {\frac{{{\chi _{xx}}{\rho _x}}}{\rho }} \right)^2}{\rm{d}}x{\rm{d}}t \le C + C(\varepsilon )\int_0^T {\int_0^L {\chi _{xx}^2} } {\rm{d}}x{\rm{d}}t + \varepsilon \int_0^T {\int_0^L {\chi _{xxx}^2} } \\ {\rm{d}}x{\rm{d}}t \le C \end{array} $ (31)

将式(4)的第二个方程乘以ut, 再在[0, L]上积分得

$ \begin{array}{l} \frac{1}{2}\frac{{\rm{d}}}{{{\rm{d}}t}}\int_0^L {u_x^2} {\rm{d}}x + \int_0^L \rho u_t^2{\rm{d}}x = - \int_0^L \rho u{u_x}{u_t}{\rm{d}}x - \\ \int_0^L {p_\rho ^\prime {\rho _x}{u_t}{\rm{d}}x} - \varepsilon \int_0^L {{\chi _x}{\chi _{xx}}{u_t}{\rm{d}}x} \le C\int_0^L {u_x^2} {\rm{d}}x + \frac{1}{4}\int_0^L \rho u_t^2\\ {\rm{d}}x + C \end{array} $ (32)

式中,$ p_\rho ^\prime = \frac{{ - 6{\rho ^3} + 36{\rho ^2} - 54\rho + 24\mathit{\Theta }}}{{{{(\rho - 3)}^2}}} $。再结合式(20),pρ的上下界可以确定。

由式(32)并结合Gronwall's不等式,可得

$ \mathop {\sup }\limits_{t \in [0,T]} \int_0^L {u_x^2} {\rm{d}}x + \int_0^T {\int_0^L p } u_t^2{\rm{d}}x{\rm{d}}t \le C $ (33)

将式(4)第三个方程对t求导,再乘以χt后在[0, L]上积分,可得

$ \begin{array}{l} \frac{1}{2}\frac{{\rm{d}}}{{{\rm{d}}t}}\int_0^L \rho \chi _t^2{\rm{d}}x + \int_0^L {\frac{{\chi _{xxt}^2}}{\rho }{\rm{d}}x} = - 2\int_0^L \rho u{\chi _t}{\chi _{xt}}{\rm{d}}x + \\ \int_0^L {{\rho _x}} {u^2}{\chi _t}{\chi _x}{\rm{d}}x + \int_0^L \rho {u_x}u{\chi _x}{\chi _t}{\rm{d}}x - \int_0^L \rho {u_\lambda }{\chi _x}{\chi _t}{\rm{d}}x + \frac{1}{\varepsilon }\mathop \smallint \nolimits_0^L \\ \left( {3{\chi ^2} - 1} \right){\chi _t}{\chi _{xxt}}{\rm{d}}x - \int_0^L {\frac{{{u_x}}}{\rho }} {\chi _{xx}}{\chi _{xxt}}{\rm{d}}x - \int_0^L {\frac{{{\rho _x}}}{\rho }} u{\chi _{xx}}{\chi _{xx{\rm{t}}}}{\rm{d}}x \le \\ \frac{1}{2}\int_0^L {\frac{{\chi _{xxt}^2}}{\rho }{\rm{d}}x} + C\left( {1 + \int_0^L {\left( {u_x^2 + \chi _{xx}^2} \right){\rm{d}}x} } \right)\int_0^L \rho \chi _t^2{\rm{d}}x + C\mathop \smallint \nolimits_0^L \\ \left( {\chi _{xx}^2 + \chi _{xxx}^2 + u_x^2} \right){\rm{d}}x\int_0^L {u_x^2} {\rm{d}}x + \frac{1}{2}\int_0^L \rho u_t^2{\rm{d}}x \end{array} $ (34)

由式(20)~(22)及式(33)~(34), 利用Gronwall's不等式得到

$ \mathop {\sup }\limits_{t \in [0,T]} \int_0^L \rho \chi _t^2{\rm{d}}x + \int_0^T {\int_0^L {\frac{1}{\rho }} } \chi _{xxt}^2{\rm{d}}x{\rm{d}}t \le C $ (35)

由式(4)第二个及第三个方程可得

$ \begin{array}{l} \int_0^L {\frac{{\chi _{xx}^2}}{\rho }{\rm{d}}x} = \int_0^L {\left( {{\chi ^3} - \chi } \right)} {\chi _{xx}}{\rm{d}}x - \frac{1}{\varepsilon }\int_0^L \chi \left( {\rho {\chi _t} + \rho u{\chi _x}} \right)\\ {\rm{d}}x \le \frac{1}{4}\int_0^L {\frac{{\chi _{xx}^2}}{\rho }{\rm{d}}x} + C\int_0^L \rho \chi _t^2{\rm{d}}x + C \end{array} $ (36)

由式(35)、(36)得

$ \int_0^L {\chi _{xx}^2} {\rm{d}}x = {\left\| \rho \right\|_{L_{{\rm{per}}}^\infty }}\int_0^L {\frac{{\chi _{xx}^2}}{\rho }} {\rm{d}}x \le C $ (37)

将式(4)第三个方程关于x求导后乘以μxxx

$ \mathop {\sup }\limits_{t \in [0,T]} \int_0^L {\chi _{xxx}^2{\rm{d}}x} + \int_0^T {\int_0^L {\mu _{xxx}^L{\rm{d}}x{\rm{d}}t} } \le C $ (38)

将式(4)第二个方程乘以$ - \frac{1}{\rho }{u_{xx}} $后在[0, L]上积分得

$ \begin{array}{l} \frac{{\rm{d}}}{{{\rm{d}}t}}\int_0^L {u_x^2} {\rm{d}}x + \mathit{\boldsymbol{v}}\int_0^L {\frac{{u_{xx}^2}}{\rho }{\rm{d}}x} = \int_0^L {{u_{xx}}} \left( {\varepsilon \frac{1}{\rho }{\chi _x}{\chi _{xx}} + } \right.\\ \left. {\frac{{{\rho _x}}}{\rho }p_\rho ^\prime + u{u_x}} \right){\rm{d}}x \le \frac{v}{4}\int_0^L {\frac{{u_{xx}^2}}{\rho }{\rm{d}}x} + C{\left\| {{\chi _x}} \right\|_{{L^\infty }}}\int_0^L {\frac{{\chi _{xx}^2}}{\rho }{\rm{d}}x} + \\ \int_0^L {{{\left( {p_\rho ^\prime } \right)}^2}} \frac{{\rho _x^2}}{\rho }{\rm{d}}x + \left( {\int_0^L {u_x^2} {\rm{d}}x + \int_0^L {u_{xx}^2} {\rm{d}}x} \right)\int_0^L \rho {u^2}{\rm{d}}x \end{array} $ (39)

结合式(39)和Gronwall's不等式及(pρ)2在式(20)条件下的有界性,可得

$ \mathop {\sup }\limits_{t \in [0,T]} \int_0^L {u_x^2} {\rm{d}}x + \int_0^T {\int_0^L {u_{xx}^2} } {\rm{d}}x{\rm{d}}t \le C $ (40)

则由式(4)第一个方程可得

$ \begin{array}{l} \int_0^L {\rho _t^2} {\rm{d}}x = \int_0^L {{{\left( {{\rho _x}u + \rho {u_x}} \right)}^2}} {\rm{d}}x \le {\left\| u \right\|_{{L^2}}}{\left\| {{u_x}} \right\|_{{L^2}}}\\ \int_0^L {\rho _x^2} {\rm{d}}x + \left\| \rho \right\|_{{L^\infty }}^2\int_0^L {u_x^2} {\rm{d}}x \le C \end{array} $ (41)

将式(4)第二个方程关于t求导,等式两边同乘ut后在[0, L]上积分,得

$ \begin{array}{l} \frac{1}{2}\frac{{\rm{d}}}{{{\rm{d}}t}}\int_0^L \rho u_t^2{\rm{d}}x + v\int_0^L {u_{xt}^2} {\rm{d}}x = \int_0^L {p_\rho ^\prime } {\rho _t}{u_{xt}}{\rm{d}}x + \\ \varepsilon \int_0^L {{\chi _{xt}}} {\chi _x}{u_{xt}}{\rm{d}}x - \int_0^L \rho u_t^2{u_x}{\rm{d}}x - 2\int_0^L \rho u{u_{xt{u_t}dx}} - \\ \int_0^L {{\rho _t}} u{u_x}{u_t}{\rm{d}}x \le \left( {C{{\left\| {{u_x}} \right\|}_{{L^2}}}{{\left\| {{u_{xx}}} \right\|}_{{L^2}}} + 2} \right)\int_0^L \rho u_t^2{\rm{d}}x + \\ \left( {C + 2\left\| \rho \right\|_{{L^2}}^{\frac{1}{2}}\left\| {{\rho _x}} \right\|_{{L^2}}^{\frac{1}{2}}{{\left\| u \right\|}_{{L^2}}}{{\left\| {{u_x}} \right\|}_{{L^2}}} + \varepsilon } \right)\int_0^L {u_{tx}^2} {\rm{d}}x + \\ \left( {C + {{\left\| {\frac{1}{\rho }} \right\|}_{{L^\infty }}}{{\left\| u \right\|}_{{L^2}}}{{\left\| {{u_x}} \right\|}_{{L^2}}}} \right)\int_0^L {\rho _t^2} {\rm{d}}x + \varepsilon {\left\| {{\chi _x}} \right\|_{{L^2}}}\\ {\left\| {{\chi _{xx}}} \right\|_{{L^2}}}\int_0^L {\chi _{xt}^2} {\rm{d}}x \end{array} $

由Gronwall's不等式及式(20)~(41)得

$ \mathop {\sup }\limits_{t \in [0,T]} \int_0^L {{u_t}} {\rm{d}}x + \int_0^T {\int_0^L {u_{xt}^2} } {\rm{d}}x{\rm{d}}t \le C $ (42)

将式(4)第二个方程对x求导,得

$ \begin{array}{l} {\rho _x}{u_t} + \rho {u_{xt}} + {\rho _x}u{u_x} + \rho u{u_x} + \rho u_x^2 + \rho u{u_{xx}} + p_{\rho \rho }^{\prime \prime }{\left( {{\rho _x}} \right)^2} + \\ p_\rho ^\prime {\rho _{xx}} = v{u_{xxx}} - \varepsilon \chi _{xx}^2 - \varepsilon {\chi _x}{\chi _{xxx}} \end{array} $ (43)

式(43)中,$ p_{\rho \rho }^{\prime \prime } = \frac{{ - 6{\rho ^3} + 54{\rho ^2} - 162\rho - 48\mathit{\Theta } + 162}}{{{{(\rho - 3)}^3}}} $

将式(21)对x求导,得

$ \begin{array}{l} {(\rho u)_{tx}} + {\left( {\rho {u^2}} \right)_{xx}} + p_{\rho \rho }^{\prime \prime }{\left( {{\rho _x}} \right)^2} + p_\rho ^\prime {\rho _{xx}} = v{\rho _x}{\left( {\frac{1}{\rho }} \right)_{xt}} + \\ v\rho {\left( {\frac{1}{\rho }} \right)_{xxt}} + v{(\rho u)_x}{\left( {\frac{1}{\rho }} \right)_{xx}} + v(\rho u){\left( {\frac{1}{\rho }} \right)_{xxx}} - \\ \frac{\varepsilon }{2}{\left( {\chi _x^2} \right)_{xx}} \end{array} $ (44)

将式(44)乘以$ \frac{1}{\rho }{\left( {\frac{1}{\rho }} \right)_{xx}} $、式(43)乘以$ - \frac{{{u_{xxx}}}}{\rho } $, 再对两式关于x在[0, L]上积分,可得

$ \left\{ {\begin{array}{*{20}{l}} {\mathop {\sup }\limits_{t \in [0,T]} \int_0^L {\rho _{xx}^2} {\rm{d}}x + \int_0^T {\int_0^L {\rho _{xx}^2} } {\rm{d}}x{\rm{d}}t \le C}\\ {\mathop {\sup }\limits_{t \in [0,T]} \int_0^L {u_{xx}^2} {\rm{d}}x + \int_0^T {\int_0^L {u_{xxx}^2} } {\rm{d}}x{\rm{d}}t \le C} \end{array}} \right. $ (45)

由式(20)可以得到pρρ的上下界,由式(4)第三个方程可得

$ \begin{array}{l} \int_0^L {\mu _{xx}^2} {\rm{d}}x = \int_0^L {{{\left( {\rho {\chi _t} + \rho u{\chi _x}} \right)}^2}} {\rm{d}}x \le \int_0^L {\left( {{\rho ^2}\chi _t^2 + } \right.} \\ \left. {{\rho ^2}{u^2}\chi _x^2} \right){\rm{d}}x \le C \end{array} $ (46)

将式(4)第四个方程关于t求导,利用式(31)、(34)及式(4)第一个方程可得

$ \begin{array}{l} \int_0^T {\int_0^L {\mu _t^2} } {\rm{d}}x{\rm{d}}t = \int_0^T {\int_0^L {\left( {\frac{1}{\varepsilon }{{\left( {{\chi ^3} - \chi } \right)}_t} - \frac{\varepsilon }{\rho }} \right.} } \\ {\left. {{\chi _{txx}}} \right)^2}{\rm{d}}x{\rm{d}}t \le C \end{array} $ (47)

将式(4)第四个方程关于x求二次导,可得

$ \begin{array}{l} {\mu _{xx}} = \frac{6}{\varepsilon }\chi \chi _x^2 + \frac{{3{\chi ^2} - 1}}{\varepsilon }{\chi _{xx}} - \varepsilon {\chi _{xxxx}}\frac{1}{\rho } + 2\varepsilon {\chi _{xxx}}\frac{{{\rho _x}}}{{{\rho ^2}}} + \\ \varepsilon {\chi _{xx}}\frac{{{\rho _{xx}}}}{{{\rho ^2}}} - \frac{{2\varepsilon {\chi _{xx}}\rho _x^2}}{{{\rho ^3}}} \end{array} $ (48)

将式(48)两边平方后关于x在[0, L]上积分,得

$ \mathop {\sup }\limits_{t \in [0,T]} \int_0^L {\chi _{xxxx}^2} {\rm{d}}x \le C $ (49)

将式(3)两边关于x求二次导,两边平方后在[0, L]×[0, T]上积分,可得

$ \int_0^T {\int_0^L {\mu _{xxxx}^2} } {\rm{d}}x{\rm{d}}t \le C $ (50)

由式(31)、(33)、(35)及式(37) ~(50),可得式(29),引理4得证。结合引理1~3,命题2得证,进一步定理1得证。

3 结论

本文研究一类非凸压力状态方程的气-液两相流,即带有van der Waals状态方程的NSCH方程组解的适定性。在初值密度不含真空时,利用测度论及能量方法得出,混合流体总密度有严格小于3的上界及大于0的下界,即不会出现真空,最终证明了该方程组全局强解的存在唯一性。通过嵌入定理可推出该问题的解是连续的,即密度和速度不会出现间断,因此不会出现激波现象。

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