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=aρr(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+ρ2u2,ui(i=1, 2)为第i种组分速度;χ=χ1-χ2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。
$ 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)∈
$ 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(
$ 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
设密度常数参数
$ \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
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p′ρ、p″ρρ具体表达式及ρ≤γ<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
$ (\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,
$ \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
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
$ \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,
$ \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
$ \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
$ \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) |
又
$ \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) |
若
$ \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
$ \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,
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得
$ \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
$ \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(p′ρ)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,
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
$ \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p″ρρ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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