气液两相流是流体动力学的研究范畴之一,被广泛应用于化工材料、大气海洋、航天航空等诸多领域,随着生物、能源与环境保护工程的发展,其理论研究日益得到重视。
Matsumura和Nishihara[1]首先证明了初始小扰动条件下单一流体流动的可压缩Navier-Stokes(NS)方程组三维整体解的存在唯一性及渐进稳定性,其后这一问题得到了进一步的研究[2-3]。在气液两相流的流动中,由于气液物质属性的差别,它们的分布对流场有重大影响,因此需要深刻认识两种流体间界面的扩散运动。由于流相界面的复杂性,对于互不相溶的两相流,其数学模型的建立和分析要远远落后于单一流体。Van der Waals[4]第一次把互不相溶两相流之间的界面看成是一个边界层;沿着这一方向,Cahn和Allen[5]对气液两相流引入界面自由能,提出了刻画气液两相流界面运动的Allen-Cahn方程;Blesgen[6]进一步将描述流体流动的NS方程组与Allen- Cahn方程相结合,建立了气液两相流流动的Navier-Stokes-Allen-Cahn(NSAC)方程组;Feireisl等[7]运用Lions[8]的证明框架得到了NSAC方程组重整化弱解的整体存在性;Kotschote等[9]讨论了NSAC方程组解的局部存在性。此后Ding等[10]证明了NSAC方程组对光滑自由能密度一维初边值问题解的整体存在性,而对于在化学反应工程和石油、天然气等低沸点液体传输中出现的气液混合物的流动,则需要用非光滑自由能密度模型[11-12]来描述。本文针对这一类非光滑自由能密度,通过构造近似方程,采用光滑逼近结合能量估计的方法,证明了对于初始密度不为零的任意初值,一维NSAC方程组周期边值问题整体解是存在唯一的,且其组分浓度差落在有物理意义的区间内,即χ∈[-1, 1]。该研究成果将为气液两相流界面的模拟计算和实验分析提供必要的理论支撑。
1 问题的提出气液两相流的流动通常由NSAC非线性偏微分方程组刻画
$ \left\{ {\begin{array}{*{20}{l}} {{\partial _t}\mathit{\boldsymbol{\rho }} + {\mathop{\rm div}\nolimits} (\rho u) = 0}\\ {{\partial _t}(\rho u) + {\mathop{\rm div}\nolimits} (\rho u \otimes u) = {\mathop{\rm div}\nolimits} \mathit{\boldsymbol{\tilde T}}}\\ {{\partial _t}(\rho \chi ) + {\mathop{\rm div}\nolimits} (\rho u\chi ) = \frac{\delta }{\rho }\Delta \chi - \frac{1}{\delta }\frac{{\partial f}}{{\partial \chi }}} \end{array}} \right. $ | (1) |
式(1)中,ρ、u、χ分别表示气液混合物密度、速度及组分的浓度差,δ表示气液扩散界面的厚度,
$ \left\{ \begin{array}{l} \mathit{\boldsymbol{\tilde T}} = \mathit{\boldsymbol{S}} - \delta \left( {\nabla \chi \otimes \nabla \chi - \frac{{|\nabla \chi {|^2}}}{2}\mathit{\boldsymbol{I}}} \right) - p(\rho , \chi )\mathit{\boldsymbol{I}}\\ f\left( \chi \right) = \left\{ \begin{array}{l} \frac{1}{4}{\left( {{\chi ^2} - 1} \right)^2}, \;\;\;\;\; - 1 \le \chi \le 1\\ + \infty , \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;其他 \end{array} \right. \end{array} \right. $ | (2) |
式(2)中,I为单位矩阵,S为Newton黏性应力,p为压力项,且满足
$ \left\{ {\begin{array}{*{20}{l}} {\mathit{\boldsymbol{S}} = v\left( {\nabla u + {\nabla ^{\rm{T}}}u - \frac{2}{3}{\mathop{\rm div}\nolimits} u\mathit{\boldsymbol{I}}} \right) + \eta {\mathop{\rm div}\nolimits} u\mathit{\boldsymbol{I}}}\\ {p = a{\rho ^r}} \end{array}} \right. $ |
式中,υ>0,η>0,常数项a>0,r>0。
组分浓度差χ满足
$ \chi = {\chi _1} - {\chi _2} $ | (3) |
式(3)中,χi=Mi/M(i=1, 2)为第i种组分的质量分数,Mi为具有体积V的两相流中第i种组分的质量,M为具有体积V的两相流质量。当χ∉[-1, 1]时,关于f的导数在次微分意义下成立。
本文考虑如下一维可压缩NSAC方程组周期边值问题
$ \left\{ {\begin{array}{*{20}{l}} {{\rho _t} + {{(\rho u)}_x} = 0}\\ {\rho {u_t} + \rho u{u_x} + {p_x} = v{u_{xx}} - \frac{\delta }{2}{{\left( {\chi _x^2} \right)}_x}}\\ {\rho {\chi _t} + \rho u{x_x} = \frac{\delta }{\rho }{\chi _{xx}} - \frac{1}{\delta }\frac{{\partial f}}{{\partial \chi }}} \end{array}} \right. $ | (4) |
式(4)满足初值条件
$ {\left. {(\rho , u, \chi )} \right|_{t = 0}} = \left( {{\rho _0}(x), {u_0}(x), {\chi _0}(x)} \right) $ | (5) |
同时满足周期边界条件
$ (\rho , u, \chi )(x + 2L, t) = (\rho , u, \chi )(x, t) $ | (6) |
式中
$ \int_0^{2L} \rho (x, t){\rm{d}}x = \int_0^{2L} {{\rho _0}} (x){\rm{d}}x $ |
定义L2(R)的周期函数空间为
$ \begin{gathered} L_{{\text{per}}}^2 = \{ g|g(x) = g(x + 2L), x \in \mathbb{R}, g(x) \in \hfill \\ {L^2}(0, 2L)\} \hfill \\ \end{gathered} $ |
其范数可表示为
$ \left\| g \right\| = {\left( {\int_0^{2L} | g(x){|^2}{\rm{d}}x} \right)^{\frac{1}{2}}} $ |
令Hperl(l≥0)为Lper2可积且其导数∂xjg(j=1, 2,…,l)也Lper2可积的函数全体,其范数记为
$ {\left\| g \right\|_l} = {\left( {\sum\limits_{j = 0}^l {{{\left\| {\partial _x^jg} \right\|}^2}} } \right)^{\frac{1}{2}}} $ |
式(5)中初值满足如下假设条件
$ \left\{\begin{array}{l}{\left(\rho_{0}, u_{0}\right) \in H_{\mathit{\boldsymbol{per}}}^{1}, \chi_{0} \in H_{\mathit{\boldsymbol{per}}}^{2}} \\ {\rho_{0}>0, -1 \leqslant \chi_{0} \leqslant 1} \\ {\chi_{t}(x, 0)=-u_{0} \mathcal{X}_{0 x}+\frac{\delta}{\rho_{0}^{2}} \mathcal{X}_{0 x x}-\frac{1}{\delta \rho_{0}}\left(\chi_{0}^{3}-\chi_{0}\right)}\end{array}\right. $ | (7) |
定理1 设初值(ρ0, u0, χ0)满足式(7), 则方程组(4)存在唯一的全局解(ρ, u, χ), 对任意的时间T, 都存在正常数C和B,且满足
$ \left\{ \begin{array}{l} \rho \in {L^\infty }\left( {0, T;H_{{\rm{per}}}^1} \right), 0 < \rho \le B\\ {\rho _t} \in {L^\infty }\left( {0, T;L_{{\rm{per}}}^2} \right)\\ u \in {L^\infty }\left( {0, T;H_{{\rm{per}}}^1} \right) \cap {L^2}\left( {0, T;H_{{\rm{per}}}^2} \right)\\ {u_t} \in {L^2}\left( {0, T;L_{{\rm{per}}}^2} \right)\\ \chi \in {L^\infty }\left( {0, T;H_{{\rm{per}}}^2} \right) \cap {L^2}\left( {0, T;H_{{\rm{per}}}^3} \right)\\ {\chi _t} \in {L^\infty }\left( {0, T;L_{{\rm{per}}}^2} \right) \cap {L^2}\left( {0, T;H_{{\rm{per}}}^1} \right), \quad - 1 \le \chi \le 1 \end{array} \right. $ |
式(4)同时满足能量不等式
$ \begin{array}{*{20}{c}} {\mathop {\sup }\limits_{t \in [0, T]} \left( {{{\left\| {\left( {{\rho _t} + {\chi _t}} \right)} \right\|}^2} + \left\| {(\rho , u)} \right\|_1^2 + \left\| \chi \right\|_2^2} \right) + }\\ {\int_0^T {\left( {{{\left\| {{u_t}} \right\|}^2} + \left\| {{\chi _t}} \right\|_1^2 + \left\| u \right\|_2^2 + \left\| \chi \right\|_3^2} \right)} {\rm{d}}t \le C} \end{array} $ |
定理1中的组分浓度差χ可以理解为混合流体的特征函数,假设流体1的特征函数χ=1,流体2特征函数χ=-1,那么流体1、2交界区域为两种流体的混合部分,其特征函数-1 < χ < 1所描述的恰好是两种流体的扩散界面。
定理1的证明需要克服两个困难,首先是自由能密度的非光滑性,本文采用构造近似方程并取光滑逼近的方法克服;其次是整体解的证明,需要得到密度和组分浓度差的上下界估计,本文首先采用路径积分结合能量估计得到密度的上下界估计,然后利用能量积分结合f(χ)的构造以及弱解的存在性得到组分浓度差的上下界估计。特别地,通过光滑逼近的方法对近似解取极限,得出χ一定落在有物理意义的区间,即-1≤χ≤1。
2 定理的证明 2.1 近似解的构造首先,给出式(4)的弱解定义。如果有
$ \left\{ {\begin{array}{*{20}{c}} {\rho (x, t) \in {L^\infty }((0, 2L) \times (0, T)), u(x, t) \in }\\ {{L^2}\left( {0, T;{H^1}(0, 2L)} \right)}\\ {\chi (x, t) \in {L^2}\left( {0, T, {H^2}(0, 2L)} \right), - 1 \le \chi \le 1} \end{array}} \right. $ |
对任意的φ(x, t)∈Cc∞((0, 2L)×(0, T)),都满足
$ \left\{ \begin{array}{l} \int_0^T {\int_0^{2L} {\left( {\rho {\varphi _t} + \rho u{\varphi _x}} \right)} } {\rm{d}}x{\rm{d}}t = 0\\ \begin{array}{*{20}{c}} {\int_0^T {\int_0^{2L} \rho } u{\varphi _t}{\rm{d}}x{\rm{d}}t + \int_0^T {\int_0^{2L} {\left( {\rho {u^2} + P} \right)} } {\varphi _x}{\rm{d}}x{\rm{d}}t = }\\ {\int_0^T {\int_0^{2L} {\left( {{u_x} - \frac{{\chi _x^2}}{2}} \right)} } {\varphi _x}{\rm{d}}x{\rm{d}}t} \end{array}\\ \int_0^T {\int_0^{2L} {\frac{\delta }{\rho }} } {\chi _{xx}}\varphi {\rm{d}}x{\rm{d}}t - \frac{1}{\delta }\int_0^T {\int_0^{2L} {\frac{{\partial f}}{{\partial \chi }}} } (\chi )\varphi {\rm{d}}x{\rm{d}}t = \\ \;\;\;\;\;\;\;\; - \int_0^T {\int_0^{2L} \rho } \chi {\varphi _t}{\rm{d}}x{\rm{d}}t \end{array} \right. $ |
则称(ρ(x, t), u(x, t), χ(x, t))是方程组(4)的弱解。
其次,考虑方程组(4)的近似形式
$ \left\{ \begin{array}{l} \begin{array}{*{20}{l}} {{\rho _t} + {{(\rho u)}_x} = 0}\\ {\rho {u_t} + \rho u{u_x} + {p_x} = v{u_{xx}} - \frac{\delta }{2}{{\left( {\chi _x^2} \right)}_x}} \end{array}\\ \begin{array}{*{20}{l}} {\rho {\chi _t} + \rho u{\chi _x} = \frac{\delta }{\rho }{\chi _{xx}} - \frac{1}{\delta }\frac{{\partial {f_\lambda }}}{{\partial \chi }}}\\ {(\rho , u, \chi )(x + 2L, t) = (\rho , u, \chi )(x, t)}\\ {{{\left. {(\rho , u, \chi )} \right|}_{t = 0}} = \left( {{\rho _0}(x), {u_0}(x), {\chi _0}(x)} \right)} \end{array} \end{array} \right. $ | (8) |
式(8)中fλ满足
$ \begin{array}{l} {f_\lambda }\left( \chi \right) = \\ \left\{ \begin{array}{l} \frac{1}{{2\lambda }}{\left[ {\chi - \left( {1 + \frac{\lambda }{2}} \right)} \right]^2} + \frac{1}{4}{\left( {{\chi ^2} - 1} \right)^2} + \frac{\lambda }{{24}}, \quad \chi \ge 1 + \lambda \\ \frac{1}{4}{\left( {{\chi ^2} - 1} \right)^2} + \frac{{{{(\chi - 1)}^3}}}{{6{\lambda ^2}}}, \quad \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;1 < \chi \le 1 + \lambda \\ \frac{1}{4}{\left( {{\chi ^2} - 1} \right)^2}, \quad \quad \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; - 1 \le \chi \le 1\\ \frac{1}{4}{\left( {{\chi ^2} - 1} \right)^2} - \frac{{{{(\chi + 1)}^3}}}{{6{\lambda ^2}}}, \quad \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; - 1 - \lambda < \chi \le - 1\\ \frac{1}{{2\lambda }}{\left[ {\chi + \left( {1 + \frac{\lambda }{2}} \right)} \right]^2} + \frac{1}{4}{\left( {{\chi ^2} - 1} \right)^2} + \frac{\lambda }{{24}}, \quad \;\chi \le - 1 - \lambda \end{array} \right. \end{array} $ |
且有
$ \left\{ \begin{array}{l} {f_\lambda }(\chi ) \ge \frac{1}{4}{\left( {{\chi ^2} - 1} \right)^2}\\ \frac{{{\partial ^2}{f_\lambda }}}{{\partial {\chi ^2}}}\left( \chi \right)\left\{ {\begin{array}{*{20}{c}} \begin{array}{l} = 3{\chi ^2} - 1, \\ \ge 0, \end{array}&\begin{array}{l} - 1 \le \chi \le 1\\ 其他 \end{array} \end{array}} \right. \end{array} \right. $ | (9) |
构造函数集合
$ \begin{gathered} {X_{{\text{per}}, (\tilde m, \tilde M)}}([0, T]) = \left\{ {(\rho , u, \chi )|(\rho , u) \in {C^0}(0, } \right. \hfill \\ T;H_{{\text{per}}}^1)\chi \in {C^0}\left( {0, T;H_{{\text{per}}}^2} \right), \mathop {\inf }\limits_{x \in \mathbb{R}} \rho \geqslant \tilde m > 0, u \in {L^\infty }(0, T; \hfill \\ H_{{\text{per}}}^1) \cap {L^2}\left( {0, T;H_{{\text{per}}}^2} \right), \chi \in {L^\infty }\left( {0, T;H_{{\text{per}}}^2} \right) \cap {L^2}(0, T; \hfill \\ H_{{\text{per}}}^3), \mathop {\sup }\limits_{x \in \mathbb{R}} \left\{ {{{\left\| {(\rho , u)} \right\|}_1}, {{\left\| \chi \right\|}_2}} \right\} \leqslant \tilde M\} \hfill \\ \end{gathered} $ |
命题1 (局部近似解存在性)
对固定的λ>0, 任意
命题2 (全局弱解存在性)
如果初值(ρ0, u0, χ0)满足假设条件式(7),则方程组(4)存在全局弱解(ρ, u, χ)。
引理1 在命题1的假设下,对于任意给定的T>0, 由近似方程组(8)得到能量估计
$ \begin{array}{l} \int_0^{2L} {\left( {\frac{{\rho {u^2}}}{2} + \mathit{\Phi }(\rho ) + \frac{1}{{4\delta }}\rho {{\left( {{\chi ^2} - 1} \right)}^2} + \frac{\delta }{2}\chi _x^2} \right)} {\rm{d}}x + \\ \int_0^T {\int_0^{2L} {\left( {vu_x^2 + {\mu ^2}} \right)} } {\rm{d}}x{\rm{d}}t \le C \end{array} $ | (10) |
式中
$ \left\{ {\begin{array}{*{20}{l}} {\mathit{\Phi }(\rho ) = \rho \int_{\bar \rho }^\rho {\frac{{p(s) - p(\bar \rho )}}{{{s^2}}}} {\rm{d}}s}\\ {\mu = \frac{1}{\delta }\frac{{\partial {f_\lambda }}}{{\partial \chi }} - \frac{\delta }{\rho }{\chi _{xx}}} \end{array}} \right. $ |
证明 对式(8)的第二个式子乘以u, 并在(0, 2L)上积分,结合式(8)的第一个式子可得
$ \begin{array}{l} \frac{{\rm{d}}}{{{\rm{d}}t}}\int_0^{2L} {\frac{{\rho {u^2}}}{2}} {\rm{d}}x + \int_0^{2L} {{p_x}} u{\rm{d}}x + \int_0^{2L} {\left( {vu_x^2 - \frac{\delta }{2}\chi _x^2{u_x}} \right)} \\ {\rm{d}}x = 0 \end{array} $ | (11) |
对式(8)的第三个式子乘以μ, 并在(0, 2L)上积分得到
$ \begin{array}{*{20}{l}} {\frac{{\rm{d}}}{{{\rm{d}}t}}\int_0^{2L} {\left( {\frac{\delta }{2}\chi _x^2 + \frac{1}{\delta }\rho {f_\lambda }} \right)} {\rm{d}}x + \int_0^{2L} {{\mu ^2}} {\rm{d}}x = - \frac{\delta }{2}}\\ {\int_0^{2L} {{u_x}} \chi _x^2{\rm{d}}x} \end{array} $ | (12) |
构造
$ \mathit{\Phi }{(\rho )_t} + {(\mathit{\Phi }(\rho )u)_x} + (p(\rho ) - p(\bar \rho )){u_x} = 0 $ | (13) |
对式(13)积分有
$ \frac{{\rm{d}}}{{{\rm{d}}t}}\int_0^{2L} \mathit{\Phi } (\rho ){\rm{d}}x + \int_0^{2L} p (\rho ){u_x}{\rm{d}}x = 0 $ | (14) |
将式(12)、式(14)代入式(11),然后在(0, T)上积分,得到式(10),引理1得证。
引理2 在命题1的假设条件下, 对于任意给定的T>0,都有
证明 因
$ \begin{array}{*{20}{c}} {\left| {\chi (x, t)} \right| \le C|\int_0^{2L} \rho (y, t)(\chi (x, t) - \chi (y, t))}\\ {{\rm{d}}y| + C\left| {\int_0^{2L} \rho (y, t)\chi (y, t){\rm{d}}y} \right| \le C\int_0^{2L} {\chi _x^2} {\rm{d}}x + C \le C} \end{array} $ | (15) |
由引理1有
$ \begin{array}{l} \int_0^{2L} \mathit{\Phi } (\rho ){\rm{d}}x = \int_0^{2L} \rho \int_{\bar \rho }^\rho {\frac{{p(s) - p(\bar \rho )}}{{{s^2}}}} {\rm{d}}s{\rm{d}}x = C\\ \int_0^{2L} {{\rho ^r}} {\rm{d}}x + C\int_0^{2L} \rho \ln \rho {\rm{d}}x + C \le C \end{array} $ |
由
$ \int_{0}^{2 L} \rho^{r} \mathit{\boldsymbol{d}} x \leqslant C $ | (16) |
令
$ \left\{ {\begin{array}{*{20}{l}} {{w_t} = {u_x} - \frac{{\chi _x^2}}{2} - \rho {u^2} - p}\\ {{w_x} = \rho u} \end{array}} \right. $ | (17) |
结合引理1、式(16)和式(17)得到
$ \int_0^{2L} {\left( {|w| + \left| {{w_x}} \right|} \right)} {\rm{d}}x \le C + C\int_0^t {\int_0^{2L} {{\rho ^r}} } {\rm{d}}x{\rm{d}}\tau \le C $ | (18) |
对任意(y, s)∈(0, 2L)×(0, T), 使得x(y, t)满足
$ \left\{\begin{array}{ll}{\frac{\mathit{\boldsymbol{d}} x(y, t)}{\mathit{\boldsymbol{d}} t}=u(x(y, t), t), } & {0 \leqslant t<s} \\ {x(y, s)=y, } & {0 \leqslant y \leqslant 1}\end{array}\right. $ |
再令f=exp {w}, 则有
$ \frac{\mathit{\boldsymbol{d}}}{\mathit{\boldsymbol{d}} t}(\rho f)(x(y, t), t)=\left(-\frac{\rho \chi_{x}^{2}}{2}-\rho^{\gamma+1}\right) f \leqslant 0 $ | (19) |
由式(19)可得
$ \rho(y, s) f(y, s) \leqslant \rho(x(y, 0)) f(x(y, 0), 0) \leqslant C $ |
从而由式(18)得到
$ \rho(y, s) \leqslant C \exp \left\{\|w\|_{L^{\infty}((0, 2 L) \times(0, T))}\right\} \leqslant C $ | (20) |
对任意T>0, 将式(8)的第三个式子化为
$ \frac{\delta}{\rho} \chi_{x x}=\frac{1}{\delta} \frac{\partial f_{\lambda}}{\partial \chi}-\mu $ | (21) |
式(21)两边乘以χxx并在[0, 2L]×[0, T]上积分,得到
$ \begin{array}{l} \int_0^T {\int_0^{2L} {\frac{\delta }{\rho }} } \chi _{xx}^2{\rm{d}}x{\rm{d}}t + \int_0^T {\int_0^{2L} {\frac{1}{\delta }} } \frac{{{\partial ^2}{f_\lambda }}}{{\partial {\lambda ^2}}}\chi _x^2{\rm{d}}x{\rm{d}}t \le - \mathop \smallint \nolimits_0^{2T} \\ \int_0^{2L} \mu {{\cal X}_{xx}}{\rm{d}}x{\rm{d}}t \le \frac{1}{2}\int_0^T {\int_0^{2L} {\frac{\delta }{\rho }} } \chi _{xx}^2{\rm{d}}x{\rm{d}}t + \frac{1}{2}\int_0^T {\int_0^{2L} {\frac{\rho }{\delta }} } {\mu ^2}{\rm{d}}x{\rm{d}}t \le \\ \frac{1}{2}\int_0^T {\int_0^{2L} {\frac{\delta }{\rho }} } \chi _{xx}^2{\rm{d}}x{\rm{d}}t + C \end{array} $ | (22) |
结合式(9)、(10)、(15)、(20)、(22)有
$ \int_0^T {\int_0^{2L} {\frac{\delta }{\rho }} } \chi _{xx}^2{\rm{d}}x{\rm{d}}t \le C\int_0^T {\int_0^{2L} {\left| {3{\chi ^2} - 1} \right|} } \chi _x^2{\rm{d}}x{\rm{d}}t + C \le C $ |
进一步可得
$ \int_{0}^{T} \int_{0}^{2 L} \chi_{x x}^{2} \mathit{\boldsymbol{d}} x \mathit{\boldsymbol{d}} t \leqslant C $ | (23) |
引理2得证。
引理3 在命题1的假设条件下, 对于任意给定的T>0, 都有
$ \int_{0}^{2 L} \rho_{x}^{2} \mathit{\boldsymbol{d}} x \leqslant C, \left\|\frac{1}{\rho}\right\|_{L^{\infty}((0, 2 L) \times(0, T))} \leqslant C $ | (24) |
证明 式(8)的第二个式子乘以
$ \begin{array}{l} \frac{1}{2}\frac{{\rm{d}}}{{{\rm{d}}t}}\int_0^{2L} \rho {\left| {{{\left( {\frac{1}{\rho }} \right)}_x}} \right|^2}{\rm{d}}x + \gamma \int_0^{2L} {{\rho ^{\gamma - 3}}} {\rho _x}{\rm{d}}x - \frac{{\rm{d}}}{{{\rm{d}}t}}\\ \int_0^{2L} \rho u{\left( {\frac{1}{\rho }} \right)_x}{\rm{d}}x = \int_0^{2L} {u_x^2} {\rm{d}}x - \int_0^{2L} {\frac{{{\rho _x}}}{{{\rho ^2}}}{\chi _x}} {\chi _{xx}}{\rm{d}}x \le \int_0^{2L} {u_x^2} {\rm{d}}x + \\ C\left( {{{\left\| {\frac{1}{\rho }} \right\|}_{{L^\infty }}} + \int_0^{2L} \rho {{\left| {{{\left( {\frac{1}{\rho }} \right)}_x}} \right|}^2}{\rm{d}}x} \right)\int_0^{2L} {\chi _{xx}^2} {\rm{d}}x + C\\ \int_0^{2L} \rho {\left| {{{\left( {\frac{1}{\rho }} \right)}_x}} \right|^2}{\rm{d}}x + C \end{array} $ | (25) |
其中式(25)第一个等式用到
$ \int_0^{2L} {{{\left( {\frac{1}{\rho }} \right)}_x}} {u_{xx}}{\rm{d}}x = \frac{1}{2}\frac{{\rm{d}}}{{{\rm{d}}t}}\int_0^{2L} \rho {\left| {{{\left( {\frac{1}{\rho }} \right)}_x}} \right|^2}{\rm{d}}x $ |
经计算有
$ \left\|\frac{1}{\rho(x, t)}\right\|_{L^{\infty}} \leqslant C+C \int_{0}^{2 L} \rho\left|\left(\frac{1}{\rho}\right)_{x}\right|^{2} \mathit{\boldsymbol{d}} x $ | (26) |
将式(26)代入式(25),然后对式(25)关于时间t在[0, T]上积分,再结合引理1、式(23)和Gronwalls不等式,得到
$ \int_{0}^{2 L} \rho\left|\left(\frac{1}{\rho}\right)_{x}\right|^{2} \mathit{\boldsymbol{d}} x+\gamma \int_{0}^{T} \int_{0}^{2 L} \rho^{\gamma-3} \rho_{x}^{2} \mathit{\boldsymbol{d}} x \mathit{\boldsymbol{d}} t \leqslant C $ | (27) |
由式(26)、(27)得到
$ \left\|\frac{1}{\rho}\right\|_{L^{\infty}((0, T) \times(0, 2 L))} \leqslant C $ |
进一步由ρ的上下界得到
$ \int_0^{2L} {\rho _x^2} {\rm{d}}x \le C $ | (28) |
引理3得证。
命题2的证明 (ρ, u, χ)是原方程式(4)的弱解,并且-1≤χ≤1。
证明 首先,假设(ρλ, uλ, χλ)是方程组式(4)的逼近解,则其一定在弱意义下满足近似方程式(8)。因为{ρλ}在L∞(0, T; H1)中一致有界,且{ ∂ tρλ}在L2((0, 2L)×(0, T))中一致有界,由Aubin-Lions定理可知,{ρλ}在L2((0, 2L)×(0, T))中强收敛到ρ。
其次,因为{ρλ}在Lγ((0, 2L)×(0, T))中强收敛到ρ, { ∂ xuλ}在L2((0, 2L)×(0, T))中弱收敛到∂xu, {χλ}在L2(0, T; H2(0, 2L))中一致有界, { ∂t χλ}在L2((0, 2L)×(0, T))中一致有界,由Aubin-Lions定理可知,{χλ}在L2(0, T; H1(0, 2L))中强收敛到χ。
最后,{ ∂ xxχλ}在L2((0, 2L)×(0, T))中弱收敛到∂ xxχ, {1/ρλ}在L2((0, 2L)×(0, T))中强收敛到1/ρ。综合上述得到的结果,令λ→0, 可知(ρ, u, χ)满足方程组式(4)定义的弱解,再由f和fλ的构造以及弱解的存在性可得-1≤χ≤1。命题2得证。
2.3 解的能量估计首先给出方程组(4)解的高阶能量估计,在此基础上完成定理1的证明。
命题3 (解的能量估计)如果初值(ρ0, u0, χ0)满足假设条件式(7), 则对于任意给定的T>0, 式(4)的解(ρ, u, χ)满足如下能量不等式
$ \begin{gathered} \mathop {\sup }\limits_{x \in \mathbb{R}} \left\{ {\left\| {(\rho , u)} \right\|_1^2, \left\| \chi \right\|_2^2} \right\} + \int_0^T {\left( {\left\| \rho \right\|_1^2 + } \right.} \hfill \\ \left\| u \right\|_2^2 + \left\| \chi \right\|_3^2){\text{d}}t \leqslant C \hfill \\ \end{gathered} $ |
命题3的证明由命题2结合引理4~5直接给出,其中出现的常数C均只依赖于初值和T。
引理4 在命题2成立的情况下,对于任意给定的T>0, 都有
$ \mathop {\sup }\limits_{t \in [0, T]} \int_0^{2L} {\left( {\chi _t^2 + \chi _{xx}^2} \right)} {\rm{d}}x + \int_0^T {\int_0^{2L} {\left( {\chi _{xt}^2 + \chi _{xxx}^2} \right)} } {\rm{d}}x{\rm{d}}t \le C $ |
证明 将式(4)的第三个式子乘以ρ后对x求导,然后乘以χxt并在[0, 2L]上积分,得到
$ \begin{array}{l} \frac{1}{2}\frac{{\rm{d}}}{{{\rm{d}}t}}\int_0^{2L} {\chi _{xx}^2} {\rm{d}}x + \int_0^{2L} {{\rho ^2}} \chi _{xt}^2{\rm{d}}x \le \frac{1}{2}\int_0^{2L} {{\rho ^2}} \chi _{xt}^2{\rm{d}}x + C\\ \int_0^{2L} {\left( {\rho _x^2\chi _t^2 + {u^2}\rho _x^2\chi _x^2 + u_x^2\chi _x^2 + {u^2}\chi _{xx}^2 + \rho _x^2 + \chi _x^2} \right)} {\rm{d}}x \end{array} $ | (29) |
利用Sobolev嵌入不等式得到
$ \left\| {{\chi _x}} \right\|_{{L^\infty }(0, 2L)}^2 \le C\int_0^{2L} {\left( {\chi _x^2 + \chi _{xx}^2} \right)} {\rm{d}}x $ | (30) |
又因为
$ \begin{array}{l} \left\| {{\chi _t}} \right\|_\infty ^2 \le C(\varepsilon )\int_0^{2L} {\chi _t^2} {\rm{d}}x + \varepsilon \int_0^{2L} {\chi _{xt}^2} {\rm{d}}x \le C(\varepsilon )\\ \int_0^{2L} {\left( {{u^2}\chi _x^2 + \chi _{xx}^2 + 1} \right)} {\rm{d}}x + \varepsilon \int_0^{2L} {\chi _{xt}^2} {\rm{d}}x \le C(\varepsilon )\left( {\begin{array}{*{20}{l}} {\left\| u \right\|_{{L^\infty }}^2} \end{array}} \right.\\ \int_0^{2L} {\chi _x^2} {\rm{d}}x + \int_0^{2L} {\chi _{xx}^2} {\rm{d}}x + 1) + \varepsilon \int_0^{2L} {\chi _{xt}^2} {\rm{d}}x \end{array} $ | (31) |
将式(30)、(31)代入式(29)中,并结合式(28)、引理1及Gronwalls不等式,取ε足够小,得到
$ \int_{0}^{2 L} \chi_{x x}^{2} \mathit{\boldsymbol{d}} x+\int_{0}^{T} \int_{0}^{2 L} \chi_{x t}^{2} \mathit{\boldsymbol{d}} x \mathit{\boldsymbol{d}} t \leqslant C $ | (32) |
将式(4)的第三个式子乘以χt并在[0, T]×[0, 2L]上积分,结合式(32)及引理1得到
$ \int_0^{2L} {\chi _t^2} {\rm{d}}x \le C\int_0^{2L} {\left( {\chi _{xx}^2 + \chi _x^2{u^2}} \right)} {\rm{d}}x + C \le C $ |
对式(4)的第三个式子关于x求导,再乘以χxxx并在[0, T]×[0, 2L]上积分,结合引理1和式(28)、(32)得到
$ \begin{array}{l} \int_0^T {\int_0^{2L} {\chi _{xxx}^2} } {\rm{d}}x{\rm{d}}t \le C\int_0^T {\int_0^{2L} {\left( {\rho _\chi ^2\chi _t^2 + \chi _{xt}^2 + \rho _x^2{u^2}\chi _x^2 + } \right.} } \\ u_x^2\chi _x^2 + {u^2}\chi _{xx}^2 + \rho _x^2 + \chi _x^2){\rm{d}}x{\rm{d}}t \le C\int_0^T {\mathop {\max }\limits_{x \in [0, 2L]} } {u^2}\chi _x^2{\rm{d}}t + C\\ \int_0^T {\mathop {\max }\limits_{x \in [0, 2L]} } \chi _{xx}^2{\rm{d}}t + C \le C\int_0^T {\int_0^{2L} {\left( {{u^2} + u_x^2} \right)} } {\rm{d}}x{\rm{d}}t + \varepsilon \int_0^T {\mathop \smallint \nolimits_0^{2L} } \\ \chi _{xxx}^2{\rm{d}}x{\rm{d}}t + C(\varepsilon ) \end{array} $ | (33) |
式(33)中取ε足够小得到
$ \int_0^T {\int_0^{2L} {\chi _{xxx}^2} } {\rm{d}}x{\rm{d}}t \le C $ |
引理4得证。
引理5 在命题2成立的情况下,对于任意给定的T>0, 都有
$ \mathop {\sup }\limits_{t \in [0, T]} \int_0^{2L} {\left( {\rho _t^2 + u_x^2} \right)} {\rm{d}}x + \int_0^T {\int_0^{2L} {\left( {u_t^2 + u_{xx}^2} \right)} } {\rm{d}}x{\rm{d}}t \le C $ |
证明 第一步,将式(4)的第二个式子乘以ut后在(0, 2L)上积分,利用引理3、4得到
$ \begin{array}{l} \frac{1}{2}\frac{{\rm{d}}}{{{\rm{d}}t}}\int_0^{2L} {u_x^2} {\rm{d}}x + \int_0^{2L} \rho u_t^2{\rm{d}}x \le \frac{1}{2}\int_0^{2L} \rho u_t^2{\rm{d}}x + {C_2}\\ \int_0^{2L} {\left( {{u^2}u_x^2 + \rho _x^2 + \chi _\chi ^2\chi _{xx}^2} \right)} {\rm{d}}x \le \frac{1}{2}\int_0^{2L} \rho u_t^2{\rm{d}}x + C\left\| u \right\|_{{L^\infty }}^2\\ \int_0^{2L} {u_x^2} {\rm{d}}x + C \end{array} $ | (34) |
由式(34)、式(10)和Gronwall不等式可得
$ \int_0^{2L} {u_x^2} {\rm{d}}x + \int_0^T {\int_0^{2L} \rho } u_t^2{\rm{d}}x{\rm{d}}t \le C $ | (35) |
结合式(4)的第二个式子、引理2和式(35),有
$ \begin{array}{*{20}{l}} {\int_0^T {\int_0^{2L} {u_{xx}^2} } {\rm{d}}x{\rm{d}}t \le C + C\int_0^T {\int_0^{2L} {\left( {u_t^2 + {u^2}u_x^2 + \rho _x^2 + \chi _x^2 + } \right.} } }\\ {\chi _{\cal X}^2{\cal X}_{xx}^2){\rm{d}}x{\rm{d}}t \le C} \end{array} $ |
再由式(35)得到
$ \int_0^{2L} {\rho _t^2} {\rm{d}}x \le C\int_0^{2L} {\left( {\rho _x^2{u^2} + u_x^2} \right)} {\rm{d}}x \le C $ |
引理5得证,故定理1得证。
证毕。
3 结束语本文研究了一类具有非光滑自由能密度的一维黏性可压缩NSAC方程组的周期边值问题,克服了自由能密度的非光滑性及密度差与组分浓度差的上下界估计等困难,证明了该方程组整体解的存在唯一性,并证得χ一定落在有物理意义的区间,即-1≤χ≤1。本文结果将为气液两相流界面的模拟计算和实验分析提供必要的理论支撑。
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