本文主要研究如下多介质等熵气体动力学方程组

(1.1) $ \begin{equation} \left\{ \begin{array}{l} \rho_{it}+(\rho_i u)_x = 0, \quad i = 1, \cdots , n, \\ (\rho u)_t+(\rho u^2+P(\rho))_x = 0 \end{array}\right. \end{equation} $

带有界可测初值

(1.2) $ \begin{equation} (\rho_i(x, 0), u(x, 0)) = (\rho_{i0}(x), u_0(x)), \quad i = 1, \cdots , n \end{equation} $

的柯西问题,其中$ \rho_i $是各个介质的密度, $ \rho $是混合密度, $ u $是速度, $ m = \rho u $是动量, $ P $是压强.其物理背景可参见文献[1-2].

如果在系统(1.1)中取$ n = 1 $,则上述系统简化为如下单介质等熵动力学系统

(1.3) $ \begin{equation} \left\{ \begin{array}{l} \rho_t+(\rho u)_x = 0, \\ (\rho u)_t+(\rho u^2+P(\rho))_x = 0, \end{array}\right. \end{equation} $

对于多方气体,压强$ P = P(\rho) = \frac{1}{\gamma}\rho^\gamma $, $ \gamma $成为绝热指数. (1.3)式的两个方程分别描述了可压缩理想流体在一维管道中流动时的质量守恒和动量守恒.

Nishida和Smoller^[3-4]利用Glimm格式得到了$ \gamma\in[1, 1+\delta) $时系统(1.3)的弱解,其中要求$ \delta $依赖于初值且充分小;之后Diperna^[5]利用补偿列紧方法证明了$ \gamma = 1+\frac{2}{2n+1} $时弱解的全局存在性;丁夏畦,陈贵强和罗佩珠^[6-8]利用Lax-Friedrichs格式结合补偿列紧方法得到了$ \gamma\in(1, \frac{5}{3}] $时全局弱解的存在性; Lions, Perthame和Tamdor^[9]利用补偿列紧方法结合动力学公式构造了新的熵-熵流对,给出了$ \gamma\geq 3 $时弱解的全局存在性; Lions, Perthame和Souganidis^[10]成功推广了这一方法从而填补了$ \gamma\in(\frac{5}{3}, 3) $的空白,并且对整个$ \gamma>1 $提供了一个新的证明.黄飞敏和王振^[11]利用复分析中的解析开拓定理构造了复熵-熵流,并结合补偿列紧方法在$ L^\infty $框架下得到了等温动力学系统($ \gamma = 1 $)弱解的全局存在性.至此,关于多方动力学系统弱解的存在性问题得到了完全解决.

对于一般的压强$ P(\rho) $,陆云光^[12]引入了$ \delta $扰动技巧,对系统(1.3)的柯西问题弱解的全局存在性给出了一个简单的证明;陈贵强和LeFloch^[13]证明了具有特殊压强$ P(\rho) $的带一般有界可测初值$ L^\infty $熵解的全局存在性.关于多介质二次流系统和LeRoux系统的研究可以参看文献[14-15].

这里我们指出,当$ \gamma>1 $时,系统(1.3)的全局存在性结果都是在$ L^\infty $框架下得到的.目前还无法在BV框架下得到弱解的全局存在性.这是因为即使初值远离真空,在大初值条件下也无法得到密度$ \rho(x, t) $的一致下界;另一方面,对于包含真空的初值,即使其变差非常小,此时初等波相互作用也会产生一个一阶增量,这意味着利用目前主流的Glimm相互作用泛函来控制波的强度是行不通的.在这方面,如何巧妙地利用稀疏波的抵消仍然是一个公开问题.

系统(1.1)比系统(1.3)更加复杂,因为此时需要考虑不同介质的密度的初值的振荡沿着线性退化方向的传播.利用线性退化方向的一致BV估计,我们将补偿列紧理论应用到非熵-熵流函数对,避开了线性退化场对应的粘性方程右边产生的奇性,得到了粘性消失解在线性退化方向上的紧性,并且用这一思想研究了如下多介质河流方程组

(1.4) $ \begin{equation} \left\{ \begin{array}{l} \rho_{it}+(\rho_iu)_x = 0, \quad i = 1, \cdots , n, \\ (\rho u)_t+(\rho u^2+P(\rho))_x+a(x)\rho+c\rho uf(u) = 0 \end{array} \right. \end{equation} $

带有界可测初值

(1.5) $ \begin{equation} (\rho_i(x, 0), u(x, 0)) = (\rho_{i0}(x), u_0(x)), \quad i = 1, \cdots , n \end{equation} $

的柯西问题.其中函数$ a(x) $相当于地形的坡度, $ c\rho f(u) $相当于摩擦力, $ f(u)\geq 0 $, $ c $是非负常数.

本节我们证明柯西问题(1.1), (1.2)弱解的全局存在性.我们假设其混合密度和压强分别为

(2.1) $ \begin{equation} \rho = \sum\limits_{i = 1}^n \rho_i, \quad P(\rho) = \frac{(\gamma-1)^2}{4\gamma}\rho^\gamma, \quad \gamma>1, \end{equation} $

其物理背景可以参见文献[2].此外,我们假设其初始密度的振荡是有界的,

(2.2) $ \begin{equation} \bigg(\frac{\rho_i}{\rho_1}\bigg)(\cdot, 0)\in L^\infty({\Bbb R}), \quad \bigg(\frac{\rho_i}{\rho_1}\bigg)_x(\cdot, 0)\in L^1({\Bbb R}), \quad i = 2, \cdots , n. \end{equation} $

经简单计算,系统(1.1)的$ n+1 $个特征值为

(2.3) $ \begin{equation} \lambda_1 = u-\theta \rho^\theta, \quad \lambda_2 = \cdots = \lambda_n = u, \quad \lambda_{n+1} = u+\theta \rho^\theta, \end{equation} $

其对应的Riemann不变量坐标为

(2.4) $ \begin{equation} w_1 = u-\rho^\theta, \quad w_2 = \frac{\rho_2}{\rho_1}, \cdots , w_n = \frac{\rho_n}{\rho_1}, \quad w_{n+1} = u+\rho^\theta, \end{equation} $

其中$ \theta = \frac{\gamma-1}{2}>0 $.从而系统(1.1)是非严格双曲的;其第一个和最后一个特征场在非真空的时是真正非线性的;中间的$ n-1 $个特征场是线性退化的.

在多介质系统(1.1)的右边添加粘性项,考虑如下抛物型系统

(2.5) $ \begin{equation} \left\{ \begin{array}{l} \rho^\varepsilon_{it}+(\rho^\varepsilon_i u^\varepsilon)_x = \varepsilon\rho^\varepsilon_{ixx}, \quad i = 1, \cdots , n, \\ (\rho^\varepsilon u^\varepsilon)_t+(\rho^\varepsilon (u^\varepsilon)^2+P(\rho^\varepsilon))_x = \varepsilon(\rho^\varepsilon u^\varepsilon)_{xx}, \end{array} \right. \end{equation} $

带有界光滑初值

(2.6) $ \begin{equation} (\rho^\varepsilon_i(x, 0), u^\varepsilon(x, 0)) = (\rho_{i0}(x), u_0(x))\ast j^\varepsilon \cdot\phi^\varepsilon(x)+(\varepsilon, 0), \quad i = 1, \cdots , n, \end{equation} $

其中$ j^\varepsilon $是支集在$ [-\varepsilon, \varepsilon] $上的光滑子, $ \phi^\varepsilon(x) $是支集在$ [-2/\varepsilon, 2/\varepsilon] $上的截断函数,且$ \phi^\varepsilon(x) = 1, $$ x\in[-1/\varepsilon, 1/\varepsilon] $.易知

$ \begin{eqnarray*} \left\{ \begin{array}{l} \lim\limits_{|x|\rightarrow\infty}(\rho^\varepsilon_1(x, 0), \cdots , \rho^\varepsilon_n(x, 0), u^\varepsilon(x, 0)) = (\varepsilon, \cdots , \varepsilon, 0), \\ (\rho^\varepsilon_1(x, 0), \cdots , \rho^\varepsilon_n(x, 0), u^\varepsilon(x, 0))\rightarrow (\rho_{10}(x), \cdots , \rho_{n0}(x), u_0(x))\quad {\rm点点成立, }\\ ||(\rho^\varepsilon_1(x, 0), \cdots , \rho^\varepsilon_n(x, 0), u^\varepsilon(x, 0))||_{L^\infty}\leq M, \end{array} \right. \end{eqnarray*} $

其中$ M $是一个只依赖于初值(1.2)而与$ \varepsilon $无关的常数.我们的主要结果如下.

定理2.1  若初值(1.2)是$ L^\infty $有界的, $ \rho_i(x, 0)\geq 0 $, $ i = 1, \cdots , n $.假设(2.1), (2.2)式成立.则对任给的$ \varepsilon>0 $,柯西问题(2.5), (2.6)存在全局有界粘性解$ (\rho^\varepsilon_1(x, t), \cdots , \rho^\varepsilon_n(x, t), u^\varepsilon(x, t)) $满足

(2.7) $ \begin{equation} 0<c(\varepsilon, t)\leq \rho^\varepsilon_i(x, t)\leq M, \quad |u^\varepsilon(x, t)|\leq M, \quad i = 1, \cdots , n, \end{equation} $

其中$ M $是与$ \varepsilon $无关的常数, $ c(\varepsilon, t) $可能会在时间$ t\rightarrow \infty $或者粘性系数$ \varepsilon\rightarrow 0 $时趋于零.

进一步,存在$ (\rho^\varepsilon_1, \cdots , \rho^\varepsilon_n, u^\varepsilon) $的子列(仍记作$ (\rho^\varepsilon_1, \cdots , \rho^\varepsilon_n, u^\varepsilon)) $,使得$ \rho^\varepsilon_i(x, t) $几乎处处收敛到$ \rho_i(x, t) $,而$ u^\varepsilon(x, t) $在集合$ \{(x, t):\rho(x, t)>0\} $上几乎处处收敛到$ u(x, t) $.特别地, $ \rho^\varepsilon_i(x, t)u^\varepsilon(x, t) $几乎处处收敛到$ \rho_i(x, t), u(x, t) $.其中函数$ (\rho_i(x, t), u(x, t)) $是柯西问题(1.1), (1.2)的一个弱解,并且有如下估计

(2.8) $ \begin{equation} \bigg(\frac{\rho_i}{\rho_1}\bigg)(\cdot, t)\in L^\infty({\Bbb R}), \quad \bigg(\frac{\rho_i}{\rho_1}\bigg)_x(\cdot, t)\in L^1({\Bbb R}), \quad i = 2, \cdots , n. \end{equation} $

我们通过下面三个引理证明定理2.1.

引理2.1  柯西问题(2.5), (2.6)存在全局$ L^\infty $有界光滑解$ (\rho^\varepsilon_1(x, t), \cdots , \rho^\varepsilon_n(x, t), u^\varepsilon(x, t)) $.

证  首先由文献[16]中的定理1.0.2(4)直接可以得到

(2.9) $ \begin{equation} \rho^\varepsilon_i(x, t)\geq c(t, \varepsilon)>0, \quad i = 1, \cdots , n, \end{equation} $

其中$ c(\varepsilon, t) $可能会在时间$ t\rightarrow \infty $或者粘性系数$ \varepsilon\rightarrow 0 $时趋于零.

根据拟线性抛物型方程组局部解的存在性结果,我们只需要得到其粘性解$ (\rho^\varepsilon_1, \cdots , \rho^\varepsilon_n, u^\varepsilon) $的先验$ L^\infty $估计.

对(2.5)式的前$ n $个方程求和得

(2.10) $ \begin{equation} \rho^\varepsilon_t+(\rho^\varepsilon u^\varepsilon)_x = \varepsilon \rho^\varepsilon_{xx}, \end{equation} $

将(2.10)式带入(2.5)式的最后一个方程得

(2.11) $ \begin{equation} u^\varepsilon_t+\bigg( \frac{1}{2}(u^\varepsilon)^2+\frac{\gamma-1}{4}(\rho^\varepsilon)^{\gamma-1} \bigg)_x = \varepsilon u^\varepsilon_{xx}+\frac{2\varepsilon}{\rho^\varepsilon}\rho^\varepsilon_x u^\varepsilon_x, \end{equation} $

分别用$ (w_{1\rho}, w_{1u})(\rho^\varepsilon, u^\varepsilon) $和$ (w_{(n+1)\rho}, w_{(n+1)u})(\rho^\varepsilon, u^\varepsilon) $乘以(2.10)和(2.11)式得

(2.12) $ \begin{eqnarray} w^\varepsilon_{1t}+\lambda^\varepsilon_1 w^\varepsilon_{1x}& = &\varepsilon w^\varepsilon_{1xx}+\frac{2\varepsilon}{\rho^\varepsilon}\rho^\varepsilon_x w^\varepsilon_{1x}+\frac{\varepsilon}{4}(\gamma^2-1)(\rho^\varepsilon_x)^2\\ &\geq& \varepsilon w^\varepsilon_{1xx}+\frac{2\varepsilon}{\rho^\varepsilon}\rho^\varepsilon_x w^\varepsilon_{1x}, \end{eqnarray} $

(2.13) $ \begin{eqnarray} w^\varepsilon_{(n+1)t}+\lambda^\varepsilon_{(n+1)} w^\varepsilon_{(n+1)x} & = &\varepsilon w^\varepsilon_{(n+1)xx}+\frac{2\varepsilon}{\rho^\varepsilon}\rho^\varepsilon_x w^\varepsilon_{(n+1)x} -\frac{\varepsilon}{4}(\gamma^2-1)(\rho^\varepsilon_x)^2\\ &\leq &\varepsilon w^\varepsilon_{(n+1)xx}+\frac{2\varepsilon}{\rho^\varepsilon}\rho^\varepsilon_x w^\varepsilon_{(n+1)x}, \end{eqnarray} $

将极值原理应用于(2.12)和(2.13)式可知区域

(2.14) $ \begin{equation} \Sigma = \{(\rho, u):w_1(\rho^\varepsilon, u^\varepsilon)\geq -N, \quad w_{n+1}(\rho^\varepsilon, u^\varepsilon)\leq N\} \end{equation} $

是柯西问题(2.5), (2.6)的一个正不变区域.结合(2.9)式我们有如下先验估计

(2.15) $ \begin{equation} 0<\rho^\varepsilon_i(x, t)\leq M, \quad ||u^\varepsilon(x, t)||_{L^\infty}\leq M, \quad \forall (x, t)\in {\Bbb R}\times{\Bbb R}^+, \end{equation} $

其中常数$ M, N $只与初值(1.2)的$ L^\infty $范数有关,而与$ \varepsilon $无关.

至此我们得到了粘性解$ (\rho^\varepsilon_1, \cdots , \rho^\varepsilon_n, u^\varepsilon) $的一致先验有界估计,这蕴含着粘性解的全局存在性.证毕.

在文献[9]和[10]中, Lions等人对单介质多方气体动力学系统构造了如下动力学熵-熵流对$ (\eta, q) $,

(2.16) $ \begin{equation} \eta(\rho, u) = \int_{{\Bbb R}}G(\rho, \xi-u)g(\xi){\rm d}\xi, \quad q(\rho, u) = \int_{{\Bbb R}}[\theta\xi+(1-\theta)u]G(\rho, \xi-u){\rm d}\xi, \end{equation} $

其中$ g(\cdot) $为任意光滑函数, $ G(\rho, w) = (\rho^{\gamma-1}-w^2)^\lambda_{+} $, $ \lambda = \frac{3-\gamma}{2(\gamma-1)} $, $ x_+ = \max\{x, 0\} $.并将补偿列紧方法应用到上述熵-熵流对得到了粘性解$ (\rho^\varepsilon(x, t), \rho^\varepsilon(x, t)u^\varepsilon(x, t)) $的几乎处处收敛性.

引理2.2^[10]  存在$ (\rho^\varepsilon, u^\varepsilon) $的子列$ ( $仍记作$ (\rho^\varepsilon, u^\varepsilon) $$ ) $,使得$ \rho^\varepsilon(x, t) $几乎处处收敛到$ \rho(x, t) $,而$ u^\varepsilon(x, t) $在集合$ \{(x, t):\rho(x, t)>0\} $上几乎处处收敛到$ u(x, t) $.特别地, $ \rho^\varepsilon(x, t)u^\varepsilon(x, t) $几乎处处收敛到$ \rho(x, t)u(x, t) $.

下面我们利用文献[17-18]中引入的将Div-Curl定理应用到非熵-熵流对的方法,证明每一个介质的密度$ \rho^\varepsilon_i(x, t) $是几乎处处收敛的.

引理2.3  存在$ \rho^\varepsilon_i $的子列$ ( $仍记作$ \rho^\varepsilon_i $$ ) $,使得$ \rho^\varepsilon_i(x, t) $几乎处处收敛到$ \rho_i(x, t) $.并且估计式(2.8)成立.

证  分别在系统(2.5)的第一个和第$ i $个方程两边同时乘以$ -\frac{\rho^\varepsilon_i}{\rho^\varepsilon_1} $和$ \frac{1}{\rho^\varepsilon_1} $,然后求和得

(2.17) $ \begin{equation} \bigg(\frac{\rho^\varepsilon_i}{\rho^\varepsilon_1}\bigg)_t+u^\varepsilon\bigg(\frac{\rho^\varepsilon_i}{\rho^\varepsilon_1}\bigg)_x = \varepsilon\bigg(\frac{\rho^\varepsilon_i}{\rho^\varepsilon_1}\bigg)_{xx}+\frac{2\varepsilon}{\rho^\varepsilon_1}\rho^\varepsilon_{1x}\bigg(\frac{\rho^\varepsilon_i}{\rho^\varepsilon_1}\bigg)_x, \quad i = 2, \cdots , n, \end{equation} $

由极值原理知

(2.18) $ \begin{equation} \bigg\|\frac{\rho^\varepsilon_i}{\rho^\varepsilon_1}(\cdot, t)\bigg\|_{L^\infty}\leq \bigg\|\frac{\rho^\varepsilon_i}{\rho^\varepsilon_1}(\cdot, 0)\bigg\|_{L^\infty}, \quad \forall t\in{\Bbb R}_+. \end{equation} $

在(2.17)式两边同时对$ x $求导并且乘以$ g'(\theta, \alpha) $得

$ \begin{eqnarray*} & &g(\theta, \alpha)_t+(u^\varepsilon g(\theta, \alpha))_x+[g'(\theta, \alpha)\theta-g(\theta, \alpha)]u^\varepsilon_x\\ & = &\varepsilon g(\theta, \alpha)_{xx}-\varepsilon g''(\theta, \alpha)\theta^2_x+\bigg(\frac{2\varepsilon}{\rho^\varepsilon_1}\rho^\varepsilon_{1x}g(\theta, \alpha)\bigg)_x +\bigg(\frac{2\varepsilon}{\rho^\varepsilon_1}\rho^\varepsilon_{1x}\bigg)_x[g'(\theta, \alpha)\theta-g(\theta, \alpha)], \end{eqnarray*} $

其中$ g(\theta, \alpha) = \sqrt{\theta^2+\alpha^2} $是$ \theta $的光滑函数, $ \alpha $是参数, $ \theta = \Big(\frac{\rho^\varepsilon_i}{\rho^\varepsilon_1}\Big)_x $.易知$ g''(\theta)\geq 0 $,且当$ \alpha\rightarrow 0 $时, $ g'(\theta)\rightarrow {\rm sgn}\theta $, $ g(\theta, \alpha)\rightarrow |\theta| $.令$ \alpha\rightarrow 0 $可得

(2.19) $ \begin{equation} |\theta|_t+(u|\theta|)_x\leq \varepsilon |\theta|_{xx}+\bigg(\frac{2\varepsilon}{\rho^\varepsilon_1}\rho^\varepsilon_{1x}|\theta|\bigg)_x, \end{equation} $

将(2.19)式两边同时在$ {\Bbb R}\times[0, t] $积分得

(2.20) $ \begin{equation} \int_{-\infty}^{\infty}|\theta(x, t)|{\rm d}x\leq \int_{-\infty}^\infty |\theta(x, 0)|{\rm d}x\leq M. \end{equation} $

在(2.18)和(2.20)式两边令$ \varepsilon\rightarrow 0 $即得到估计式(2.8).

记$ v^\varepsilon_i = \frac{\rho^\varepsilon_i}{\rho^\varepsilon_1} $, $ i = 2, \cdots , n $,将Div-Curl定理应用于如下两组非熵-熵流函数对

(2.21) $ \begin{equation} (C, v^\varepsilon_i), \quad (\rho^\varepsilon_1, \rho^\varepsilon_1 u^\varepsilon)\quad {\rm和} \quad (C, v^\varepsilon_i), \quad (\rho^\varepsilon_i, \rho^\varepsilon_i u^\varepsilon) \end{equation} $

得

$ \overline{\rho^\varepsilon_1 w^\varepsilon_i} = \overline{\rho^\varepsilon_1}\cdot\overline{v^\varepsilon_i}, \quad \overline{\rho^\varepsilon_1(v^\varepsilon_i)^2} = \overline{\rho^\varepsilon_1 v^\varepsilon_i}\cdot\overline{v^\varepsilon_i}, $

即

(2.22) $ \begin{equation} \overline{\rho^\varepsilon_1(v^\varepsilon_i-v^\varepsilon)^2} = 0, \end{equation} $

其中$ \overline{f^\varepsilon} $表示$ f^\varepsilon $的弱*极限. (2.22)式表明着在区域$ \{(x, t):\rho_1(x, t)>0\} $上, $ v^\varepsilon_i\rightarrow v_i $几乎处处成立,这和引理2.2中$ (\rho^\varepsilon, u^\varepsilon) $的几乎处处收敛性一起蕴含着$ (\rho^\varepsilon_1, \cdots , \rho^\varepsilon_n, u^\varepsilon) $的几乎处处收敛性.而在区域$ \{(x, t):\rho_1(x, t) = 0\} $上,由估计式(2.8)可得$ \rho_i(x, t) = 0 $, $ i = 2, \cdots , n $.至此,我们得到了每个介质密度$ \rho^\varepsilon_i(x, t) $的几乎处处收敛性.证毕.

至此我们已经得到了粘性解$ (\rho^\varepsilon_1, \cdots , \rho^\varepsilon_n, u^\varepsilon) $的几乎处处收敛性,这蕴含着柯西问题(1.1), (1.2)弱解的全局存在性.这里我们指出,上述得到的弱解是满足相容性条件的.事实上,若$ (\eta, q) $是系统(1.1)的一对凸熵-熵流,在(2.5)式的两边同时乘以$ \nabla \eta(\rho^\varepsilon_1, \cdots , \rho^\varepsilon_n, u^\varepsilon) $得

$ \begin{eqnarray*} \eta^\varepsilon_t+q^\varepsilon_x = \varepsilon \eta^\varepsilon_{xx}-\varepsilon (\rho^\varepsilon_{1x}, \cdots , \rho^\varepsilon_{nx}, u^\varepsilon_x)^{\Bbb T}\cdot \nabla^2 \eta^\varepsilon \cdot (\rho^\varepsilon_{1x}, \cdots , \rho^\varepsilon_{nx}, u^\varepsilon_x), \end{eqnarray*} $

从而对任意给定的试验函数$ \phi\in C^\infty_0({\Bbb R}\times{\Bbb R}^+) $, $ \phi(x, t)\geq 0 $,有

$ \begin{eqnarray*} \int_0^\infty\int_{-\infty}^\infty \eta^\varepsilon \phi_t+q^\varepsilon \phi_x {\rm d}x{\rm d}t\geq -\varepsilon\int_0^\infty\int_{-\infty}^\infty \eta^\varepsilon \phi_{xx}{\rm d}x{\rm d}t, \end{eqnarray*} $

令$ \varepsilon\rightarrow0 $即得

(2.23) $ \begin{equation} \int_0^\infty\int_{-\infty}^\infty \eta \phi_t+q \phi_x {\rm d}x{\rm d}t\geq 0. \end{equation} $

注2.1  考虑如下已经化简过的欧拉方程组

(2.24) $ \begin{equation} \left\{ \begin{array}{l} \rho_{it}+(\rho_i u)_x = 0, \quad i = 1, \cdots , n, \\ u_t+\bigg( \frac{u^2}{2}+\frac{\gamma-1}{4}\rho^{\gamma-1}\bigg)_x = 0, \end{array} \right. \end{equation} $

其中$ \gamma\geq 3 $,则此时其混合密度$ \rho $可以包含更广的一类函数

(2.25) $ \begin{equation} \rho = \bigg(\sum\limits_{i = 1}^n\rho_i^p\bigg)^{\frac{1}{p}}, \quad 1\leq p\leq \infty. \end{equation} $

事实上,考虑相应的粘性系统

(2.26) $ \begin{equation} \left\{ \begin{array}{l} \rho^\varepsilon_{it}+(\rho^\varepsilon_i u^\varepsilon)_x = \varepsilon \rho^\varepsilon_{ixx}, \quad i = 1, \cdots , n, \\ u^\varepsilon_t+\bigg( \frac{(u^\varepsilon)^2}{2}+\frac{\gamma-1}{4}(\rho^\varepsilon)^{\gamma-1}\bigg)_x = \varepsilon u^\varepsilon_{xx}, \end{array} \right. \end{equation} $

若$ 1\leq p<\infty $,在系统(2.26)的第$ i $个方程两边同时乘以$ \Big(\frac{\rho^\varepsilon_i}{\rho^\varepsilon}\Big)^{p-1} $, $ i = 1, \cdots , n $,然后求和得

(2.27) $ \begin{equation} \rho^\varepsilon_t+(\rho^\varepsilon u^\varepsilon)_x = \varepsilon \rho^\varepsilon_{xx}-\varepsilon\frac{p-1}{(\rho^\varepsilon)^{p-1}}\sum\limits_{i = 1}^n \bigg[(1-\frac{\rho^\varepsilon_i}{\rho^\varepsilon})(\rho^\varepsilon_i)^{p-2}(\rho^\varepsilon_{ix})^2\bigg]\leq \varepsilon\rho^\varepsilon_{xx}, \end{equation} $

我们记

(2.28) $ \begin{equation} w^\varepsilon = u^\varepsilon-(\rho^\varepsilon)^\theta, \quad z^\varepsilon = u^\varepsilon+(\rho^\varepsilon)^\theta, \quad \lambda^\varepsilon_1 = u^\varepsilon-\theta(\rho^\varepsilon)^\theta, \quad \lambda^\varepsilon_2 = u^\varepsilon+(\rho^\varepsilon)^\theta, \end{equation} $

其中$ \theta = \frac{\gamma-1}{2} $.则由(2.27)和(2.26)式的最后一个方程得

(2.29) $ \begin{equation} w^\varepsilon_t+\lambda^\varepsilon_1 w^\varepsilon_x\geq\varepsilon w^\varepsilon_{xx}, \quad z^\varepsilon_t+\lambda^\varepsilon_2 z^\varepsilon_x \leq \varepsilon z^\varepsilon_{xx}, \end{equation} $

至此我们得到了粘性解的先验一致$ L^\infty $估计,这蕴含着粘性解的全局存在性.

令$ p\rightarrow\infty $,亦可得到$ \rho = \max\{\rho_1, \cdots , \rho_n\} $时粘性解的先验一致$ L^\infty $估计.之后粘性解的紧性证明同引理2.2,引理2.3,不再赘述.

本节我们研究多介质河流方程组Cauchy问题(1.4), (1.5)弱解的全局存在性.与上一节一样,考虑相应的抛物型系统

(3.1) $ \begin{equation} \left\{ \begin{array}{l} \rho^\varepsilon_{it}+(\rho^\varepsilon_i u^\varepsilon)_x = \varepsilon\rho^\varepsilon_{ixx}, \quad i = 1, \cdots , n, \\ (\rho^\varepsilon u^\varepsilon)_t+(\rho^\varepsilon (u^\varepsilon)^2+P(\rho^\varepsilon))_x+a(x)\rho^\varepsilon+c\rho^\varepsilon u^\varepsilon f(u^\varepsilon) = \varepsilon(\rho^\varepsilon u^\varepsilon)_{xx}, \end{array} \right. \end{equation} $

带有界光滑初值

(3.2) $ \begin{equation} (\rho^\varepsilon_i(x, 0), u^\varepsilon(x, 0)) = (\rho_{i0}(x), u_0(x))\ast j^\varepsilon \cdot\phi(\varepsilon)+(\varepsilon, 0), \quad i = 1, \cdots , n, \end{equation} $

其中$ j^\varepsilon $是支集在$ [-\varepsilon, \varepsilon] $上的光滑子, $ \phi^\varepsilon $是支集在$ [-2/\varepsilon, 2/\varepsilon] $上的截断函数,且$ \phi(x) = 1, $$ x\in[-1/\varepsilon, 1/\varepsilon] $.

本节的主要结果如下.

定理3.1  若$ |a(x)|\leq K $, $ f(u)\geq 0 $.初值(1.5)是$ L^\infty $有界的, $ \rho_i(x, 0)\geq 0 $, $ i = 1, \cdots , n $.假设(2.1)和(2.2)式成立.则对任给的$ \varepsilon>0 $,柯西问题(3.1)和(3.2)式存在全局有界粘性解$ (\rho^\varepsilon_1(x, t) $, $ \cdots , \rho^\varepsilon_n(x, t), u^\varepsilon(x, t)) $满足

(3.3) $ \begin{equation} 0<c(\varepsilon, t)\leq \rho^\varepsilon_i(x, t)\leq M(t), \quad |u^\varepsilon(x, t)|\leq M(t), \quad i = 1, \cdots , n, \end{equation} $

其中$ M(t) $是与$ \varepsilon $无关但可能与$ t $相关的常数, $ c(\varepsilon, t) $可能会在时间$ t\rightarrow \infty $或者粘性系数$ \varepsilon\rightarrow 0 $时趋于零.

进一步,存在$ (\rho^\varepsilon_1, \cdots , \rho^\varepsilon_n, u^\varepsilon) $的子列$ ( $仍记作$ (\rho^\varepsilon_1, \cdots , \rho^\varepsilon_n, u^\varepsilon) $$ ) $,使得$ \rho^\varepsilon_i(x, t) $几乎处处收敛到$ \rho_i(x, t) $,而$ u^\varepsilon(x, t) $在集合$ \{(x, t):\rho(x, t)>0\} $上几乎处处收敛到$ u(x, t) $.特别地, $ \rho^\varepsilon_i(x, t)u^\varepsilon(x, t) $几乎处处收敛到$ \rho_i(x, t), u(x, t) $.其中函数$ (\rho_i(x, t), u(x, t)) $是柯西问题(1.4), (1.5)的一个弱解,并且有如下估计

(3.4) $ \begin{equation} \bigg(\frac{\rho_i}{\rho_1}\bigg)(\cdot, t)\in L^\infty({\Bbb R}), \quad \bigg(\frac{\rho_i}{\rho_1}\bigg)_x(\cdot, t)\in L^1({\Bbb R}), \quad i = 2, \cdots , n. \end{equation} $

证  根据上一节的框架,我们只需要得到粘性解的先验$ L^\infty $估计.将(3.1)式的前$ n $个方程求和得

(3.5) $ \begin{equation} \rho^\varepsilon_t+(\rho^\varepsilon u^\varepsilon)_x = \varepsilon \rho^\varepsilon_{xx}, \end{equation} $

将(3.5)式代入(3.1)式的最后一个方程得

(3.6) $ \begin{equation} u^\varepsilon_t+\bigg( \frac{1}{2}(u^\varepsilon)^2+\frac{\gamma-1}{4}(\rho^\varepsilon)^{\gamma-1} \bigg)_x+a(x)+cu^\varepsilon f(u^\varepsilon) = \varepsilon u^\varepsilon_{xx}+\frac{2\varepsilon}{\rho^\varepsilon}\rho^\varepsilon_x u^\varepsilon_x, \end{equation} $

我们记

(3.7) $ \begin{equation} w^\varepsilon = (\rho^\varepsilon)^\theta-u^\varepsilon, \quad z^\varepsilon = (\rho^\varepsilon)^\theta+u^\varepsilon, \quad \lambda^\varepsilon_1 = u^\varepsilon-\theta(\rho^\varepsilon)^\theta, \quad w^\varepsilon = u^\varepsilon+\theta(\rho^\varepsilon)^\theta, \end{equation} $

分别用$ (w^\varepsilon_\rho, w^\varepsilon_u) $和$ (z^\varepsilon_\rho, z^\varepsilon_u) $乘以(3.5)和(3.6)式得

(3.8) $ \begin{eqnarray} & &w^\varepsilon_t+\lambda^\varepsilon_1 w^\varepsilon_x-a(x)+\frac{cf(u^\varepsilon)}{2}(w^\varepsilon-z^\varepsilon)\\ & = &\varepsilon w^\varepsilon_{xx}-\frac{2^\varepsilon}{\rho^\varepsilon}\rho^\varepsilon_x w^\varepsilon_x-\frac{\varepsilon}{4}(\gamma^2-1)(\rho^\varepsilon_x)^2\leq \varepsilon w^\varepsilon_{xx}-\frac{2\varepsilon}{\rho^\varepsilon}\rho^\varepsilon_x w^\varepsilon_{x}, \end{eqnarray} $

(3.9) $ \begin{eqnarray} & &z^\varepsilon_t+\lambda^\varepsilon_2 z^\varepsilon_x+a(x)+\frac{cf(u^\varepsilon)}{2}(z^\varepsilon-w^\varepsilon)\\ & = &\varepsilon z^\varepsilon_{xx}+\frac{2^\varepsilon}{\rho^\varepsilon}\rho^\varepsilon_x z^\varepsilon_x-\frac{\varepsilon}{4}(\gamma^2-1)(\rho^\varepsilon_x)^2\leq \varepsilon z^\varepsilon_{xx}+\frac{2\varepsilon}{\rho^\varepsilon}\rho^\varepsilon_x z^\varepsilon_{x}, \end{eqnarray} $

做变换

$ w^\varepsilon = X+Kt, \quad z^\varepsilon = Y+Kt, $

其中, $ K $是函数$ a(x) $的$ L^\infty $上界.则由(3.8)和(3.9)式得

(3.10) $ \begin{equation} \left\{ \begin{array}{l} X_t+\lambda^\varepsilon_1 X_x+\frac{cf(u^\varepsilon)}{2}(X-Y)\leq \varepsilon X_{xx}-\frac{2\varepsilon}{\rho^\varepsilon}\rho^\varepsilon_x X_x, \\ Y_t+\lambda^\varepsilon_2 Y_x+\frac{cf(u^\varepsilon)}{2}(Y-X)\leq \varepsilon Y_{xx}+\frac{2\varepsilon}{\rho^\varepsilon}\rho^\varepsilon_x Y_x, \end{array} \right. \end{equation} $

以及初值

(3.11) $ \begin{equation} X(x, 0) = w^\varepsilon(x, 0)\leq N_1, \quad Y(x, 0) = z^\varepsilon(x, 0)\leq N_1, \end{equation} $

其中$ N_1 $是只依赖于初值(3.2)的常数.下面用极值原理证明

(3.12) $ \begin{equation} X(x, t)\leq N_1, \quad Y(x, t)\leq N_1, \quad \forall (x, t)\in {\Bbb R}\times[0, T], \end{equation} $

做变换

$ \begin{eqnarray*} \overline{X} = X-N-\frac{N(x^2+CLe^t)}{L^2}, \quad \overline{Y} = Y-N-\frac{N(x^2+CLe^t)}{L^2}, \end{eqnarray*} $

其中$ C, L $为适当的正常数, $ N = N(T, \varepsilon) $为$ X, Y $在$ {\Bbb R}\times[0, T] $中的上界.带入(3.10)和(3.11)式得

(3.13) $ \begin{equation} X_t+\lambda^\varepsilon_1 X_x+\frac{cf(u^\varepsilon)}{2}(X-Y) +\bigg( CLe^t+2\lambda^\varepsilon_1x-2\varepsilon+\frac{4\varepsilon}{\rho}\rho^\varepsilon_xx \bigg)\frac{N}{L^2} \leq \varepsilon X_{xx}-\frac{2\varepsilon}{\rho^\varepsilon}\rho^\varepsilon_x X_x, \end{equation} $

(3.14) $ \begin{equation} Y_t+\lambda^\varepsilon_2 Y_x+\frac{cf(u^\varepsilon)}{2}(Y-X) +\bigg( CLe^t+2\lambda^\varepsilon_2x-2\varepsilon-\frac{4\varepsilon}{\rho}\rho^\varepsilon_xx \bigg)\frac{N}{L^2} \leq \varepsilon Y_{xx}+\frac{2\varepsilon}{\rho^\varepsilon}\rho^\varepsilon_x Y_x, \end{equation} $

以及初边值条件

(3.15) $ \begin{equation} \overline{X}(x, 0)<0, \quad \overline{Y}(x, 0)< 0, \quad \overline{X}(\pm L, t)< 0, \quad \overline{Y}(\pm L, t)<0. \end{equation} $

我们断言

(3.16) $ \begin{equation} \overline{X}(x, t)<0, \quad \overline{Y}(x, t)<0, \quad \forall (x, t)\in (-L, L)\times (0, T). \end{equation} $

否则假设(3.16)式不成立,令

$ \begin{eqnarray*} \bar{t} = \sup \{ t: \overline{X}(x, \tau)<0, \quad \overline{Y}(x, \tau)<0, \quad \forall (x, \tau)\in (-L, L)\times [0, t) \}, \end{eqnarray*} $

不失一般性,存在$ (\bar{x}, \bar{t})\in (-L, L)\times [0, t) $,使得$ \overline{X}(\bar{x}, \bar{t}) = 0 $, $ \overline{Y}(\bar{x}, \bar{t})\leq 0 $,且$ \overline{X}_t(\bar{x}, \bar{t})\geq 0 $, $ \overline{X}_x(\bar{x}, \bar{t}) = 0 $, $ \overline{X}_{xx}(\bar{x}, \bar{t})\leq 0 $,因此

(3.17) $ \begin{equation} \overline{X}_t+\lambda^\varepsilon_1 \overline{X_x}-\varepsilon \overline{X}_{xx}+\frac{2\varepsilon}{\rho^\varepsilon}\rho^\varepsilon_x \overline{X}_x \geq 0. \end{equation} $

如果选取$ C $充分大使得

$ \begin{eqnarray*} CLe^t+2\lambda^\varepsilon_1x-2\varepsilon+\frac{4\varepsilon}{\rho}\rho^\varepsilon_xx>0, \quad \forall (x, t)\in (-L, L)\times (0, T), \end{eqnarray*} $

则(3.17)与(3.13)式矛盾,这就证明了(3.16)式.从而对任给的$ (x, t)\in(-L, L)\times (0, T) $有

$ \begin{eqnarray*} \overline{X}(x, t)<N_1+\frac{N(x^2+CLe^t)}{L^2}, \quad \overline{Y}(x, t)<N_1+\frac{N(x^2+CLe^t)}{L^2}, \end{eqnarray*} $

在上式中令$ \varepsilon\rightarrow 0 $即得(3.12)式,这蕴含着粘性解的$ L^\infty $先验估计(3.3).之后粘性解的紧性证明和定理2.1类似,不再赘述.证毕.

注3.1  和注2.1一样,考虑如下已经化简过的欧拉方程组

(3.18) $ \begin{equation} \left\{ \begin{array}{l} \rho_{it}+(\rho_i u)_x = 0, \quad i = 1, \cdots , n, \\ u_t+\bigg( \frac{u^2}{2}+\frac{\gamma-1}{4}\rho^{\gamma-1}\bigg)_x+a(x)+cuf(u) = 0, \end{array} \right. \end{equation} $

其中$ \gamma\geq 3 $,则此时其混合密度$ \rho $可以包含更广的一类函数

(3.19) $ \begin{equation} \rho = \bigg(\sum\limits_{i = 1}^n\rho_i^p\bigg)^{\frac{1}{p}}, \quad 1\leq p\leq \infty. \end{equation} $

事实上,考虑相应的粘性系统

(3.20) $ \begin{equation} \left\{ \begin{array}{l} \rho^\varepsilon_{it}+(\rho^\varepsilon_i u^\varepsilon)_x = \varepsilon \rho^\varepsilon_{ixx}, \quad i = 1, \cdots , n, \\ u^\varepsilon_t+\bigg( \frac{(u^\varepsilon)^2}{2}+\frac{\gamma-1}{4}(\rho^\varepsilon)^{\gamma-1}\bigg)_x+a(x)+cu^\varepsilon f(u^\varepsilon) = \varepsilon u^\varepsilon_{xx}, \end{array} \right. \end{equation} $

和注2.1一样

(3.21) $ \begin{equation} \rho^\varepsilon_t+(\rho^\varepsilon u^\varepsilon)_x\leq \varepsilon \rho^\varepsilon_{xx}, \end{equation} $

从而

(3.22) $ \begin{equation} \begin{array}{l} w^\varepsilon_t+\lambda^\varepsilon_1 w^\varepsilon_x-a(x)+\frac{cf(u^\varepsilon)}{2}(w^\varepsilon-z^\varepsilon)\leq\varepsilon w^\varepsilon_{xx}, \\ z^\varepsilon_t+\lambda^\varepsilon_2 z^\varepsilon_x +a(x)+\frac{cf(u^\varepsilon)}{2}(z^\varepsilon-w^\varepsilon)\leq \varepsilon z^\varepsilon_{xx}, \end{array} \end{equation} $

其中$ w^\varepsilon $, $ z^\varepsilon $, $ \lambda^\varepsilon_1 $, $ \lambda^\varepsilon_2 $的表达式如(3.7)式.由(3.22)式可以直接得到(3.10)式,之后的证明过程类似,不再赘述.