考虑带有复合源项的一维可压缩流体欧拉方程组

(1.1) $ \begin{align} \left\{\begin{array}{ll} \rho_{t}+(\rho u)_{x}=0, \\ { } u_{t}+(\frac{u^{2}}{2}+p (\rho))_{x}=-\alpha u+\beta, \end{array}\right . \end{align} $

其中, $ \beta $是常数, 表示Coulomb-like摩擦项, $ \alpha> 0 $表示恒耗散系数, $ \rho $表示密度, $ u $表示速度, 以及$ p(\rho) $表示流体压力, 且非线性方程$ p(\rho)=\frac{\theta}{2}\rho^{\gamma-1}, $ 其中$ \theta=\frac{\gamma-1}{2} $, $ \gamma\in(1, 2) $是常数. 绝热指数$ \gamma $在不同的情况下起着重要作用, 例如当$ \gamma=-1 $, $ p(\rho)=-\frac{1}{2\rho^{2}} $称为Chaplygin气体压力, 这是Chaplygin^[1], Tsien^[2]和von Karman^[3]为了计算飞机机翼上升力的数值近似而引入的.

Shen^[4]研究具有Coulomb-like摩擦项的零压欧拉方程组, 并通过变量代换

(1.2) $ \begin{align} v(t, x)=u(t, x)-\beta t \end{align} $

获得其非自相似解. Faccanoni和Mangeney^[5]首次通过引入变量代换(1.2)研究具有Coulomb-like摩擦项的浅水方程组的Riemann问题.

如果$ \alpha=0 $和$ \beta=0 $, 则系统(1.1)为齐次可压缩流体欧拉方程组^[6−9]

(1.3) $ \begin{align} \left\{\begin{array}{ll} \rho_{t}+(\rho u)_{x}=0, \\ { } u_{t}+(\frac{u^{2}}{2}+p (\rho))_{x}=0.\end{array}\right . \end{align} $

系统(1.3)是Earnshaw^[6]于1858年首次用于研究等熵流体, 具有不同的物理背景. 例如, 它是具有远距离相互作用的牛顿动力学标度极限系统, 用于质量在$ {\Bbb R} $中的连续分布^[10−11], 同时也是Vlasov方程的流体动力的极限^[12].

许多学者已深入研究系统(1.3)的解^{[7−9, 13−20]}. 特别的, DiPerna^[14]首次使用Glimm方法对$ 1 <\gamma<3 $情形的Cauchy问题研究全局弱解的存在性定理. Li^[15]在DiPerna^[14]研究的基础上获得$ -1 < \gamma < 1 $情形的Cauchy问题全局弱解. Lu^[7]和Cheng^[8]基于补偿紧致性理论和动力学方程的基本思想, 分别研究$ \gamma > 3 $, $ 1 <\gamma < 3 $情形的Cauchy问题的全局熵解的存在性定理. 对压力$ p $为在实轴上取实值的一整函数情形, Sarrico^[17]构造性的得到Riemann问题在含间断函数和Dirac测度的广义函数的适当的空间中解. 对压力$ p(\rho)=\int^{\rho}_{0}s e^{s}{\rm d}s $情形, Song and Xiao^[18]研究带有源项的系统(1.3)的Cauchy问题全局弱解的存在性定理. 因此, 研究具有源项的可压缩流体流动欧拉方程组的解, 如摩擦、阻尼和松弛效应是很自然的. 这里, 我们集中讨论(1.1)式中提出的源项. 如果$ \alpha=0 $, 那么这种源项, 有时被称为Coulomb-like摩擦项, 是Savage和Hutter^[21]早期提出来描述颗粒流动行为的. 目前, 对一些具有Coulomb-like摩擦项的双曲型方程组解的数学研究非常活跃, 有兴趣的读者可以参考文献[4−5, 22−33].

当$ \gamma\rightarrow 1 $, 系统(1.1)趋于带有复合源项的一维零压欧拉方程

(1.4) $ \begin{align} \left\{\begin{array}{ll} \rho_{t}+(\rho u)_{x}=0, \\ { } u_{t}+(\frac{u^{2}}{2})_{x}=-\alpha u+\beta.\end{array}\right . \end{align} $

如果取$ u_{a}=-1 $, 则系统(1.4)涵盖了重要的欧拉液滴模型^[34−39]. 此外, 如果$ \alpha=0 $和$ \beta=-1 $, 那么系统(1.4) 在文献[25]中模拟了具有大弗劳德数浅流中的剧烈不连续性.

对一维等熵气体欧拉方程组

(1.5) $ \begin{align} \left\{\begin{array}{ll} \rho_{t}+(\rho u)_{x}=0, \\(\rho u)_{t}+ (\rho u^{2}+p(\rho) )_{x}=0, \end{array}\right . \end{align} $

当压力趋于零或常数时, 欧拉方程组(1.5)收敛到零压气体动力学系统. 在早期的开创性论文中, Chen和Liu^[40]通过在模型$ p(\rho)=\varepsilon\rho^{\gamma}/\gamma $$ ( \gamma >1) $中取极限$ \varepsilon \rightarrow 0+ $, 首次证明了多方气体欧拉方程组(1.5) 的Riemann解的delta激波和真空状态的形成, 它严格地描述了数学中的集中和空化现象. 此外, 他们还得到了文献[41]中非等熵流体欧拉方程相同的结果。Li^[42]研究了欧拉方程(1.5) 对等温情形$ (\gamma = 1) $的相同问题. 最近, 当$ \gamma \rightarrow 0 $, 即压力趋于常数时, Ibrahim等人在文献[43]中证明了集中现象也存在于模型$ p(\rho) =\rho^{\gamma} $$ (0 < \gamma < 1) $. 也就是说, 他们用欧拉方程(1.5)的黎曼解的极限行为严格地证明了delta激波的形成. 对于其他一些物理模型, 也有许多结果^[44−58].

在文献[40−43]的激励下, 本文研究了可压缩流体流动的非齐次欧拉方程组(1.1)的黎曼解当绝热指数趋于1时的极限. 与齐次方程不同, 黎曼解是非自相似的, 当$ \gamma \rightarrow1 $时, 我们发现在$ 1 < \gamma < 2 $的情况下, 也存在相同的质量集中现象和真空现象.

本文的结构如下. 在第2和第3节中, 我们分别介绍了方程(1.4), (1.1)的黎曼解的一些结果. 在第4节中, 我们严格证明了当$ \gamma \rightarrow1 $时, 系统(1.1) 的Riemann解形成delta激波和真空状态. 最后, 结论在第5节进行了论述.

在本节, 我们给出系统(1.4)的Riemann问题的解. 对于系统(1.4)相应的齐次型系统, Riemann问题的解可参考文献[59, 60].

通过变量代换

(2.1) $ \begin{align} v(t, x)=\frac{\beta}{\alpha}+(u(t, x)-\frac{\beta}{\alpha})e^{\alpha t}, \end{align} $

系统(1.4) 可改写为以下齐次守恒形式

(2.2) $ \begin{align} \left\{\begin{array}{ll} { } \rho_{t}+(\rho (\frac{\beta}{\alpha}+(v-\frac{\beta}{\alpha})e^{-\alpha t}))_{x}=0, \\ { } v_{t}+(\frac{1}{2}(\frac{\beta}{\alpha}e^{\alpha t}+v-\frac{\beta}{\alpha})(\frac{\beta}{\alpha}+(v-\frac{\beta}{\alpha})e^{-\alpha t}) )_{x}=0.\end{array}\right . \end{align} $

接下来, 我们将研究具有下列初值的系统(2.2) 的Riemann问题

(2.3) $ \begin{align} (\rho, v)(0, x) =\left\{\begin{array}{ll} (\rho_{-}, u_{-}), \, \, \, \, x< 0, \\(\rho_{+}, u_{+}), \, \, \, \, x> 0, \end{array} \right. \end{align} $

其中$ \rho_{\pm}>0 $和$ u_{\pm} $都是常数.

可以看出, 通过变量代换$ (\rho, u)(t, x) =(\rho(t, x), \frac{\beta}{\alpha}+(v(t, x)-\frac{\beta}{\alpha})e^{-\alpha t})) $, 系统(1.4)的Riemann解可通过直接求解系统(2.2)–(2.3)的Riemann问题而得.

系统(2.2)可以重新表示为一个拟线性形式

(2.4) $ \begin{align} \left(\begin{array}{cccc}\rho\\v \end{array}\right)_{t}+\left( \begin{array}{ccc} { } \frac{\beta}{\alpha}+(v-\frac{\beta}{\alpha})e^{-\alpha t}{\quad} & \rho e^{-\alpha t}\\ 0 {\quad} &{ } \frac{\beta}{\alpha}+(v-\frac{\beta}{\alpha})e^{-\alpha t} \end{array}\right)\left(\begin{array}{cccc}\rho\\v \end{array}\right)_{x}=\left(\begin{array}{cccc}0\\0 \end{array}\right). \end{align} $

通过(2.4)式, 很容易看出系统(2.2)有两个相同的特征值$ \lambda =\frac{\beta}{\alpha}+(v-\frac{\beta}{\alpha})e^{-\alpha t} $, 其对应的右特征向量为$ \overrightarrow{r} = (1, 0)^{T}. $ 由于$ \nabla\lambda\cdot \overrightarrow{r}\equiv 0, $ 所以系统(2.2)是完全线性退化的, 相应的基本波都是接触间断.

假设接触间断的波速为$ \sigma(t)=x'(t), $ 则满足下列Rankine-Hugoniot条件

(2.5) $ \begin{align} \left\{ \begin{array}{ll} { } -\sigma(t)[\rho]+[\rho (\frac{\beta}{\alpha}+(v-\frac{\beta}{\alpha})e^{-\alpha t})]=0, \\ { } -\sigma(t)[v]+[\frac{1}{2}(\frac{\beta}{\alpha}e^{\alpha t}+v-\frac{\beta}{\alpha})(\frac{\beta}{\alpha}+(v-\frac{\beta}{\alpha})e^{-\alpha t}) ]=0, \end{array} \right. \end{align} $

其中$ [\rho]=\rho_{r}-\rho_{l} $是$ \rho $穿过间断线的跳跃, $ [\rho(\frac{\beta}{\alpha}+(v-\frac{\beta}{\alpha})e^{-\alpha t})]=\rho_{r}(\frac{\beta}{\alpha}+(v_{r}-\frac{\beta}{\alpha})e^{-\alpha t}) -\rho_{l}(\frac{\beta}{\alpha}+(v_{l}-\frac{\beta}{\alpha})e^{-\alpha t}) $为$ \rho(\frac{\beta}{\alpha}+(v-\frac{\beta}{\alpha})e^{-\alpha t}) $穿过间断线的跳跃, 其中$ \rho_{l}=\rho(t, x(t)-0) $和$ \rho_{r}=\rho(t, x(t)+0) $.

如果$ \sigma(t)\neq 0, $ 则由(2.5)式可得

(2.6) $ \begin{eqnarray} &&\frac{1}{2}(\rho_{r}-\rho_{l})\bigg((\frac{\beta}{\alpha}e^{\alpha t}+v_{r}-\frac{\beta}{\alpha}) (\frac{\beta}{\alpha}+(v_{r}-\frac{\beta}{\alpha})e^{-\alpha t})-(\frac{\beta}{\alpha}e^{\alpha t} +v_{l}-\frac{\beta}{\alpha})(\frac{\beta}{\alpha}+(v_{l}-\frac{\beta}{\alpha})e^{-\alpha t})\bigg){}\\ &=&(v_{r}-v_{l})(\rho_{r}(\frac{\beta}{\alpha}+(v_{r}-\frac{\beta}{\alpha})e^{-\alpha t}) -\rho_{l} (\frac{\beta}{\alpha}+(v_{l}-\frac{\beta}{\alpha})e^{-\alpha t})). \end{eqnarray} $

简化(2.6)式可得

$ e^{-\alpha t}(\rho_{r}+\rho_{l})(v_{r}-v_{l})^{2}=0. $

因此, 两个非真空常状态$ (\rho_{l}, v_{l}) $和$ (\rho_{r}, v_{r}) $可由间断线$ J $连接当且仅当

$ \sigma(t)=\frac{\beta}{\alpha}+(v_{l}-\frac{\beta}{\alpha})e^{-\alpha t}=\frac{\beta}{\alpha}+(v_{r}-\frac{\beta}{\alpha})e^{-\alpha t}, $

即$ v_{l}=v_{r}. $

下面, 我们构造系统(2.2)–(2.3)的Riemann解, 即分别通过接触间断、真空状态和$ \delta $ -激波连接两个常数状态$ (\rho_{\pm}, u_{\pm}) $.

对于$ u_{-}< u_{+} $情形, Riemann解包含两条间断线和真空状态, 即

(2.7) $ \begin{align} (\rho, v)(t, x)=\left\{\begin{array}{ll} (\rho_{-}, u_{-}), { } &-\infty<x<\frac{\beta}{\alpha}t+\frac{1}{\alpha}(u_{-}-\frac{\beta}{\alpha})(1 -e^{-\alpha t}), \\ { } \Big(0, \frac{\beta}{\alpha}+\frac{\alpha(x-\frac{\beta}{\alpha}t)}{1 -e^{-\alpha t}} \Big), {\quad} & { } \frac{\beta}{\alpha}t+\frac{1}{\alpha}(u_{-}-\frac{\beta}{\alpha})(1 -e^{-\alpha t}) \leq x\\ &{\qquad} { } \leq \frac{\beta}{\alpha}t+\frac{1}{\alpha}(u_{+}-\frac{\beta}{\alpha})(1 -e^{-\alpha t}), \\ (\rho_{+}, u_{+}), & { } \frac{\beta}{\alpha}t+\frac{1}{\alpha}(u_{+}-\frac{\beta}{\alpha})(1 -e^{-\alpha t})<x<+\infty. \end{array}\right . \end{align} $

Riemann解可表示为

(2.8) $ \begin{align} (\rho_{-}, u_{-})+J_{1}+Vac+J_{2}+(\rho_{+}, u_{+}), \end{align} $

其中“+”表示“紧接着”.

对于$ u_{-}= u_{+} $情形, Riemann解包含一条间断线

(2.9) $ \begin{align} (\rho, v)(t, x)=\left\{\begin{array}{ll} (\rho_{-}, u_{-}), &{ } -\infty<x<\frac{\beta}{\alpha}t +\frac{1}{\alpha}(u_{-}-\frac{\beta}{\alpha})(1 -e^{-\alpha t}), \\ (\rho_{+}, u_{+}), {\quad} &{ } \frac{\beta}{\alpha}t+\frac{1}{\alpha}(u_{-}-\frac{\beta}{\alpha})(1 -e^{-\alpha t})<x<+\infty. \end{array}\right . \end{align} $

Riemann解可表示为

(2.10) $ \begin{align} (\rho_{-}, u_{-})+ J + (\rho_{+}, u_{+}). \end{align} $

对于$ u_{-}> u_{+} $情形, Riemann解不再是经典波, 出现了$ \delta $ -激波. 则Riemann解为

(2.11) $ \begin{align} (\rho_{-}, u_{-})+ \delta S + (\rho_{+}, u_{+}), \end{align} $

即

(2.12) $ \begin{align} (\rho, v)(t, x)=\left\{\begin{array}{ll} (\rho_{-}, u_{-}), &-\infty<x<x(t), \\(w(t)\delta(x-x(t)), v_{\delta}), {\quad} &x=x(t), \\(\rho_{+}, u_{+}), & x(t)<x<+\infty, \end{array}\right . \end{align} $

其中$ x(t) $, $ w(t) $和$ u_{\delta}(t) =\frac{\beta}{\alpha}+(v_{\delta}-\frac{\beta}{\alpha})e^{-\alpha t} $分别表示$ \delta $ -激波的位置, 权重和速度, $ v_{\delta} $表示$ v $在这个$ \delta $ -激波曲线上对应的函数值. 这个$ \delta $ -激波$ \delta S $满足如下的广义R-H条件

(2.13) $ \begin{align} \left\{ \begin{array}{ll} { } \frac{{\rm d}x(t)}{{\rm d}t}=u_{\delta}(t), \\ { } \frac{{\rm d}w(t)}{{\rm d}t}=u_{\delta}(t) [\rho]-[\rho (\frac{\beta}{\alpha}+(v-\frac{\beta}{\alpha})e^{-\alpha t})], \\ { } u_{\delta}(t) [v]=[\frac{1}{2}(\frac{\beta}{\alpha}e^{\alpha t}+v-\frac{\beta}{\alpha})(\frac{\beta}{\alpha}+(v-\frac{\beta}{\alpha})e^{-\alpha t}) ], \end{array} \right. \end{align} $

其中$ [\rho] = \rho_{+}- \rho_{-} $, 且$ (x, w)(0) =(0, 0). $

计算(2.13)式可得

(2.14) $ \begin{align} \begin{array}{l}{ } v_{\delta}=\frac{1}{2}(u_{-}+u_{+}), \; \; x(t)=\frac{\beta}{\alpha}t+\frac{1}{\alpha}(v_{\delta}-\frac{\beta}{\alpha})(1 -e^{-\alpha t}), \\ { } w(t)=\frac{1}{2\alpha}(\rho_{-}+\rho_{+})(u_{-}-u_{+})(1 -e^{-\alpha t}). \end{array} \end{align} $

由此可知, 这个$ \delta $ -激波满足广义熵条件

(2.15) $ \begin{align} \frac{\beta}{\alpha}+(u_{+}-\frac{\beta}{\alpha})e^{-\alpha t}< u_{\delta}(t) < \frac{\beta}{\alpha}+(u_{-}-\frac{\beta}{\alpha})e^{-\alpha t}. \end{align} $

综上所述, 通过变量代换(2.1), 我们可获得系统(1.4)的Riemann解, 即

(1) 当$ u_{-} >u_{+} $, 有

(2.16) $ \begin{align} (\rho, u)(t, x)=\left\{ \begin{array}{ll} { } (\rho_{-}, \frac{\beta}{\alpha}+(u_{-}-\frac{\beta}{\alpha})e^{-\alpha t}), & { x<x(t) , }\\ (w(t)\delta(x-x(t)), u_{\delta} (t)), & { x=x(t) , }\\ { } (\rho_{+}, \frac{\beta}{\alpha}+(u_{+}-\frac{\beta}{\alpha})e^{-\alpha t}), & { x>x(t) , } \end{array} \right. \end{align} $

其中

(2.17) $ \begin{align} \begin{array}{l} { } x(t)=\frac{\beta}{\alpha}t+\frac{1}{\alpha}(\frac{1}{2}(u_{-}+u_{+})-\frac{\beta}{\alpha})(1 -e^{-\alpha t}), \\ { } w(t)=\frac{1}{2\alpha}(\rho_{-}+\rho_{+})(u_{-}-u_{+})(1 -e^{-\alpha t}), \\ { } u_{\delta}(t)=\frac{\beta}{\alpha}+(\frac{1}{2}(u_{-}+u_{+})-\frac{\beta}{\alpha})e^{-\alpha t}.\end{array} \end{align} $

(2) 当$ u_{-}<u_{+} $, 有

(2.18) $ \begin{align} (\rho, u)(t, x)=\left\{\begin{array}{ll} { } (\rho_{-}, \frac{\beta}{\alpha}+(u_{-}-\frac{\beta}{\alpha})e^{-\alpha t}), & { } -\infty<x<\frac{\beta}{\alpha}t+\frac{1}{\alpha}(u_{-}-\frac{\beta}{\alpha})(1 -e^{-\alpha t}), \\ { } \Big(0, \frac{\beta}{\alpha}+\frac{\alpha(x-\frac{\beta}{\alpha} t)}{e^{\alpha t}-1}\Big), & { } \frac{\beta}{\alpha}t+\frac{1}{\alpha}(u_{-}-\frac{\beta}{\alpha})(1 -e^{-\alpha t}) \leq x\\ &{\qquad} { } \leq \frac{\beta}{\alpha}t+\frac{1}{\alpha}(u_{+}-\frac{\beta}{\alpha})(1 -e^{-\alpha t}), \\ { } (\rho_{+}, \frac{\beta}{\alpha}+(u_{+}-\frac{\beta}{\alpha})e^{-\alpha t}), {\quad} & { } \frac{\beta}{\alpha}t+\frac{1}{\alpha}(u_{+}-\frac{\beta}{\alpha})(1 -e^{-\alpha t})<x<+\infty, \end{array}\right . \end{align} $

其中两条间断线$ J_{1} $和$ J_{2} $的位置和波速与系统(2.2)–(2.3)相同.

(3) 当$ u_{-}=u_{+} $, 有

(2.19) $ \begin{align} (\rho, u)(t, x)=\left\{\begin{array}{ll} { } (\rho_{-}, \frac{\beta}{\alpha}+(u_{-}-\frac{\beta}{\alpha})e^{-\alpha t}), & { } -\infty<x<\frac{\beta}{\alpha}t+\frac{1}{\alpha}(u_{-}-\frac{\beta}{\alpha})(1 -e^{-\alpha t}), \\ { } (\rho_{+}, \frac{\beta}{\alpha}+(u_{+}-\frac{\beta}{\alpha})e^{-\alpha t}), & { } \frac{\beta}{\alpha}t+\frac{1}{\alpha}(u_{-}-\frac{\beta}{\alpha})(1 -e^{-\alpha t})<x<+\infty, \end{array}\right . \end{align} $

其中间断线$ J_{} $的位置和波速与系统(2.2)–(2.3)相同.

本节, 我们构造带有源项的一维可压缩流体欧拉方程组(1.1)的Riemann解.

通过变量代换(2.1), 系统(1.1)可写为下列形式

(3.1) $ \begin{align} \left\{\begin{array}{ll} { } \rho_{t}+(\rho (\frac{\beta}{\alpha}+(v-\frac{\beta}{\alpha})e^{-\alpha t}))_{x}=0, \\ { } v_{t}+(\frac{1}{2}(\frac{\beta}{\alpha}e^{\alpha t}+v-\frac{\beta}{\alpha})(\frac{\beta}{\alpha}+(v-\frac{\beta}{\alpha})e^{-\alpha t}) +\frac{\gamma-1}{4}\rho^{\gamma-1}e^{\alpha t} )_{x}=0.\end{array}\right . \end{align} $

接下来, 我们将研究具有下列初值的系统(3.1)的Riemann问题

(3.2) $ \begin{align} (\rho, v)(0, x) =\left\{\begin{array}{ll} (\rho_{-}, u_{-}), \, \, \, \, x< 0, \\(\rho_{+}, u_{+}), \, \, \, \, x> 0, \end{array} \right. \end{align} $

其中$ \rho_{\pm}>0 $和$ u_{\pm} $是常数.

系统(3.1)可写为下列拟线性形式

(3.3) $ \begin{align} \left(\begin{array}{cc}\rho\\v \end{array}\right)_t+\left( \begin{array}{ccc} { } \frac{\beta}{\alpha}+(v-\frac{\beta}{\alpha})e^{-\alpha t}{\quad} & \rho e^{-\alpha t} \\ { } \frac{(\gamma-1)^{2}}{4}\rho^{\gamma-2}e^{\alpha t}{\quad} &{ } \frac{\beta}{\alpha}+(v-\frac{\beta}{\alpha})e^{-\alpha t} \end{array}\right)\left(\begin{array}{cc}\rho\\v \end{array}\right)_x=\left(\begin{array}{cc}0\\0 \end{array}\right). \end{align} $

由(3.3)式可知, 系统(3.1)的特征值为

$ \lambda_{1}=\frac{\beta}{\alpha}+(v-\frac{\beta}{\alpha})e^{-\alpha t}-\frac{\gamma-1}{2}\rho^{\frac{\gamma-1}{2}}, \, \, \, \, \, \, \, \lambda_{2}=\frac{\beta}{\alpha}+(v-\frac{\beta}{\alpha})e^{-\alpha t} +\frac{\gamma-1}{2}\rho^{\frac{\gamma-1}{2}}, $

对应的右特征向量为

$ \overrightarrow{r}_{1} =(e^{-\alpha t}, -\frac{\gamma-1}{2}\rho^{\frac{\gamma-3}{2}})^{T}, \, \, \overrightarrow{r}_{2} =(e^{-\alpha t}, \frac{\gamma-1}{2}\rho^{\frac{\gamma-3}{2}})^{T}, $

满足

$ \nabla\lambda_{1}\cdot \overrightarrow{r}_{1}=-\frac{(\gamma-1)(\gamma+1)}{4}\rho^{\frac{\gamma-3}{2}}e^{-\alpha t}\neq 0, $

$ \nabla\lambda_{2}\cdot \overrightarrow{r}_{2}=\frac{(\gamma-1)(\gamma+1)}{4}\rho^{\frac{\gamma-3}{2}}e^{-\alpha t}\neq 0. $

因此, 系统(3.1)关于$ \rho>0 $是严格双曲型的, 且两个特征值均为真正非线性的, 相应的基本波为激波或稀疏波.

系统(3.1)的Riemann不变量可选为

$ w =v+\rho^{\frac{\gamma-1}{2}}e^{\alpha t}, \, \, \, \, z =v-\rho^{\frac{\gamma-1}{2}}e^{\alpha t}, $

分别满足$ \bigtriangledown w \cdot \overrightarrow{r_1}=0 $和$ \bigtriangledown z \cdot \overrightarrow{r_2}=0 $.

下面让我们把注意力放在系统(3.1)的基本波上. 我们首先用稀疏波作为基本解. 我们知道如果$ \alpha=0 $和$ \beta=0 $, 也就是说, 在齐次情况下, 存在自相似解$ (\rho, v)(t, x)=(\rho, v)(\xi) $$ (\xi=x/t) $. 但是现在情况则不存在像齐次情形那样的经典自相似解$ (\rho, v)(x/t) $. 在这种情况下, 解决方法必须另辟蹊径. 由于稀疏波是一种满足系统(3.1)的连续解, 可以通过确定每个特征场的积分曲线得到. 注意到$ k $ - 黎曼不变量$ (k = 1, 2) $在$ k $ -稀疏波中是守恒的。因此, 就像文献[23]一样, 我们仍然可以使用稀疏波作为基本解.

在相平面上, 给定一个状态$ (\rho_{-}, u_{-}) $, 稀疏波曲线是通过1 -稀疏波或2 -稀疏波在右侧连接的状态集合, 即

(3.4) $ \begin{align} R_{1}(\rho_{-}, u_{-}):\, \, \left\{\begin{array}{ll} { } \frac{{\rm d}x}{{\rm d}t}=\lambda_{1}=\frac{\beta}{\alpha}+(v-\frac{\beta}{\alpha})e^{-\alpha t}-\frac{\gamma-1}{2}\rho^{\frac{\gamma-1}{2}}, \\ { } v+\rho^{\frac{\gamma-1}{2}}e^{\alpha t}=u_{-}+\rho_{-}^{\frac{\gamma-1}{2}}e^{\alpha t}, \, \, \, \, \rho<\rho_{-}, v>u_{-}, \\ { } \lambda_{1}(\rho_{-}, u_{-})<\lambda_{1}(\rho, v), \end{array} \right. \end{align} $

(3.5) $ \begin{align} R_{2}(\rho_{-}, u_{-}):\, \, \left\{\begin{array}{ll} { } \frac{{\rm d}x}{{\rm d}t}=\lambda_{2}=\frac{\beta}{\alpha}+(v-\frac{\beta}{\alpha})e^{-\alpha t}+\frac{\gamma-1}{2}\rho^{\frac{\gamma-1}{2}}, \\ { } v-\rho^{\frac{\gamma-1}{2}}e^{\alpha t}=u_{-}-\rho_{-}^{\frac{\gamma-1}{2}}e^{\alpha t}, \, \, \, \, \rho>\rho_{-}, v>u_{-}, \\ \lambda_{2}(\rho_{-}, u_{-})<\lambda_{2}(\rho, v).\end{array} \right. \end{align} $

对方程(3.4)的第二个方程$ v $关于$ \rho $求导, 可得

$ \frac{{\rm d}v}{{\rm d}\rho} = -\frac{\gamma-1}{2}\rho^{\frac{\gamma-3}{2}}e^{\alpha t}<0, $

$ \frac{{\rm d}^{2}v}{{\rm d}\rho^{2}} =-\frac{(\gamma-1)(\gamma-3)}{4}\rho^{\frac{\gamma-5}{2}}e^{\alpha t}>0, $

则1 -稀疏波曲线$ R_{1}(\rho_{-}, u_{-}) $在$ (\rho, v) $相平面上是单调递减的凸函数. 同样的, 对(3.5)式的第二个方程$ v $关于$ \rho $求导得$ \frac{{\rm d}v}{{\rm d}\rho}> 0 $和$ \frac{{\rm d}^{2}v}{{\rm d}\rho^{2}} < 0 $, 则2 -稀疏波曲线$ R_{2}(\rho_{-}, u_{-}) $在$ (\rho, v) $相平面上是单调递增的凹函数. 而且通过(3.4)式, 对于1 -稀疏波曲线

$ R_{1}(\rho_{-}, u_{-}): \lim\limits_{\rho\rightarrow 0^{+}}v= u_{-}+\rho_{-}^{\frac{\gamma-1}{2}}e^{\alpha t}, $

即$ R_{1}(\rho_{-}, u_{-}) $与$ v $ -轴上交于点$ (0, \widetilde{v}_{\ast}) $, 其中$ \widetilde{v}_{\ast} $ is determined by $ \widetilde{v}_{\ast}= u_{-}+\rho_{-}^{\frac{\gamma-1}{2}}e^{\alpha t} $. 对于$ R_{2}(\rho_{-}, u_{-}) $, 从(3.5)式可知$ \lim\limits_{\rho\rightarrow +\infty}v=+\infty. $

我们首先计算在1 -稀疏波内部点$ (t, x) $上的Riemann解$ (\rho, v) $. 由(3.4) 式的第一个方程可得

(3.6) $ \begin{align} \frac{x}{t}-\frac{\beta}{\alpha} =(v-\frac{\beta}{\alpha})\frac{1 -e^{-\alpha t}}{\alpha t}-\frac{\gamma-1}{2}\rho^{\frac{\gamma-1}{2}}. \end{align} $

结合(3.6)式和(3.4)式的第二个方程可得

(3.7) $ \begin{align} \left\{\begin{array}{ll} { } \rho(t, x)=\Big(\frac{\rho^{\frac{\gamma-1}{2}}_{-}\frac{e^{\alpha t}-1}{\alpha t} +u_{-}\frac{1 -e^{-\alpha t}}{\alpha t}-\frac{x}{t}+\frac{\beta}{\alpha} \frac{\alpha t-1 +e^{-\alpha t}}{\alpha t}}{\frac{\gamma-1}{2}+\frac{e^{\alpha t}-1}{\alpha t}}\Big)^{\frac{2}{\gamma-1}}, \\ { } v(t, x)=\frac{\frac{x}{t}+\frac{\gamma-1}{2}(u_{-}e^{-\alpha t}+\rho^{\frac{\gamma-1}{2}}_{-})+ \frac{\beta}{\alpha}\frac{1 -e^{-\alpha t}-\alpha t}{\alpha t}}{\frac{\gamma-1}{2}e^{-\alpha t} +\frac{1 -e^{-\alpha t}}{\alpha t}}.\end{array}\right . \end{align} $

同样地, 我们计算在2 -稀疏波内部点$ (t, x) $上的Riemann解$ (\rho, v) $, 由(3.5) 式的第一个方程得到

(3.8) $ \begin{align} \frac{x}{t}-\frac{\beta}{\alpha} =(v-\frac{\beta}{\alpha})\frac{1 -e^{-\alpha t}} {\alpha t}+\frac{\gamma-1}{2}\rho^{\frac{\gamma-1}{2}}. \end{align} $

结合(3.8)式和(3.5) 式的第二个方程可知

(3.9) $ \begin{align} \left\{\begin{array}{ll} { } \rho(t, x)=\Big(\frac{\rho^{\frac{\gamma-1}{2}}_{-}\frac{e^{\alpha t}-1}{\alpha t}-u_{-}\frac{1 -e^{-\alpha t}}{\alpha t}+\frac{x}{t}+\frac{\beta}{\alpha} \frac{1 -e^{-\alpha t}-\alpha t}{\alpha t}}{\frac{\gamma-1}{2}+\frac{e^{\alpha t}-1}{\alpha t}}\Big)^{\frac{2}{\gamma-1}}, \\ { } v(t, x)=\frac{\frac{x}{t}+\frac{\gamma-1}{2}(u_{-}e^{-\alpha t}-\rho^{\frac{\gamma-1}{2}}_{-}) +\frac{\beta}{\alpha}\frac{1 -e^{-\alpha t}-\alpha t}{\alpha t}}{\frac{\gamma-1}{2}e^{-\alpha t} +\frac{1 -e^{-\alpha t}}{\alpha t}}.\end{array}\right . \end{align} $

设$ \sigma(t)=\frac{{\rm d}x (t)}{{\rm d}t} $是有界间断$ x=x(t) $的波速, 则系统(3.1) 满足以下Rankine-Hugoniot条件

(3.10) $ \begin{align} \left\{ \begin{array}{ll} { } -\sigma(t)[\rho]+[\rho (\frac{\beta}{\alpha}+(v-\frac{\beta}{\alpha})e^{-\alpha t})]=0, \\ { } -\sigma(t)[ v]+[\frac{1}{2}(\frac{\beta}{\alpha}e^{\alpha t}+v-\frac{\beta}{\alpha})(\frac{\beta}{\alpha}+(v-\frac{\beta}{\alpha})e^{-\alpha t}) +\frac{\gamma-1}{4}\rho^{\gamma-1}e^{\alpha t} ]=0, \end{array} \right. \end{align} $

其中$ [\rho]=\rho-\rho_{-} $等. 由(3.10)式获得

(3.11) $ \begin{array}{c} \sigma (t)=\frac{[\rho (\frac{\beta}{\alpha}+(v-\frac{\beta}{\alpha})e^{-\alpha t})]}{[\rho]},\\ \frac{v-u_{-}}{\rho-\rho_{-}} =\pm e^{\alpha t}\sqrt{\frac{\frac{\gamma-1}{2}(\rho^{\gamma-1}-\rho_{-}^{\gamma-1})}{(\rho+\rho_{-})(\rho-\rho_{-})}}, \end{array} $

其中$ (\rho_{-}, u_{-}) $和$ (\rho_{}, v_{}) $分别是左状态和右状态.

1 -激波曲线$ S_{1} (\rho_{-}, u_{-}) $

Lax熵条件说明1 -激波的波速$ \sigma_{1} (t) $满足

(3.12) $ \begin{align} \sigma_{1} (t)<\lambda_{1} (\rho_{-}, u_{-}), \, \, \, \, \lambda_{1} (\rho, v)<\sigma_{1} (t)<\lambda_{2} (\rho, v). \end{align} $

通过(3.10)式的第一个方程可知

(3.13) $ \begin{align} \sigma_{1} (t)&{ } =\frac{\rho (\frac{\beta}{\alpha}+(v-\frac{\beta}{\alpha})e^{-\alpha t}) -\rho_{-} (\frac{\beta}{\alpha}+(u_{-}-\frac{\beta}{\alpha})e^{-\alpha t})}{\rho-\rho_{-}} \\ &{ } = \frac{\beta}{\alpha}+(u_{-}-\frac{\beta}{\alpha})e^{-\alpha t}+\frac{\rho e^{-\alpha t}}{\rho-\rho_{-}}(v-u_{-}). \end{align} $

将(3.13)式代入(3.12)式的第一个不等式, 可得

$ \frac{\rho e^{-\alpha t}}{\rho-\rho_{-}}(v-u_{-})<-\frac{\gamma-1}{2}\rho_{-}^{\frac{\gamma-1}{2}}<0, $

这意味着$ v - u_{-} $与$ \rho- \rho_{-} $有不同的符号. 因此, 通过(3.11) 式可得

$ v=u_{-}- e^{\alpha t}\sqrt{\frac{\frac{\gamma-1}{2}(\rho^{\gamma-1}-\rho_{-}^{\gamma-1})}{(\rho+\rho_{-})(\rho-\rho_{-})}}(\rho-\rho_{-}). $

若$ v>u_{-} $, 则$ \rho<\rho_{-} $及

$ \begin{eqnarray*} \sigma_{1} (t)- (\frac{\beta}{\alpha}+(u_{-}-\frac{\beta}{\alpha})e^{-\alpha t})&=&\frac{\rho e^{-\alpha t}}{\rho-\rho_{-}}(v-u_{-}) =-\rho\sqrt{\frac{\frac{\gamma-1}{2}(\rho^{\gamma-1}-\rho_{-}^{\gamma-1})}{(\rho+\rho_{-})(\rho-\rho_{-})}}\\ &=&-\frac{\gamma-1}{2} \overline{\rho}^{\frac{\gamma-2}{2}}\rho\sqrt{\frac{2}{\rho+\rho_{-}}}, \end{eqnarray*} $

对于$ \bar{\rho}\in(\rho, \rho_{-}). $ 通过直接计算可得

$ \begin{eqnarray*} \frac{\gamma-1}{2}\rho_{-}^{\frac{\gamma-1}{2}}-\frac{\gamma-1}{2} \overline{\rho}^{\frac{\gamma-2}{2}}\rho\sqrt{\frac{2}{\rho+\rho_{-}}}&>&\frac{\gamma-1}{2}\bigg(\rho_{-}^{\frac{\gamma-1}{2}}- \rho^{\frac{\gamma-2}{2}}\rho\sqrt{\frac{2}{\rho+\rho_{-}}}\bigg)\\ &>&\frac{\gamma-1}{2}(\rho_{-}^{\frac{\gamma-1}{2}}-\rho^{\frac{\gamma-1}{2}})>0, \end{eqnarray*} $

即

$ \sigma_{1} (t)- (\frac{\beta}{\alpha}+(u_{-}-\frac{\beta}{\alpha})e^{-\alpha t}) >-\frac{\gamma-1}{2}\rho_{-}^{\frac{\gamma-1}{2}}. $

这与$ \sigma_{1} (t)<\lambda_{1} (\rho_{-}, u_{-}) $矛盾. 因此, 给定一个状态$ (\rho_{-}, u_{-}) $, 1 -激波曲线$ S_{1} (\rho_{-}, u_{-}) $是相平面上所有通过1 -激波右连接$ (\rho_{-}, u_{-}) $的状态集合, 即

(3.14) $ \begin{align} S_{1} (\rho_{-}, u_{-}):\, \, \left\{\begin{array}{ll} { } \sigma_{1} (t)=\frac{\beta}{\alpha} +(u_{-}-\frac{\beta}{\alpha})e^{-\alpha t} -\rho\sqrt{\frac{\frac{\gamma-1}{2}(\rho^{\gamma-1}-\rho_{-}^{\gamma-1})}{(\rho+\rho_{-})(\rho-\rho_{-})}}, \\ { } v=u_{-}- e^{\alpha t}\sqrt{\frac{\frac{\gamma-1}{2}(\rho^{\gamma-1}-\rho_{-}^{\gamma-1})} {(\rho+\rho_{-})(\rho-\rho_{-})}}(\rho-\rho_{-}), \; \; \; \\ \rho>\rho_{-}, v<u_{-}.\end{array} \right. \end{align} $

2 -激波曲线$ S_{2} (\rho_{-}, u_{-}) $

同样地, 2 -激波$ S_{2} $的波速$ \sigma_{2} (t) $满足

$ \lambda_{1} (\rho_{-}, u_{-})<\sigma_{2} (t)<\lambda_{2} (\rho_{-}, u_{-}), \, \, \, \, \lambda_{2} (\rho, v)<\sigma_{2} (t). $

则给定一个状态$ (\rho_{-}, u_{-}) $, 2 -激波曲线$ S_{2} (\rho_{-}, u_{-}) $是相平面上所有通过2 -激波右连接$ (\rho_{-}, u_{-}) $的状态集合, 即

(3.15) $ \begin{align} S_{2} (\rho_{-}, u_{-}):\, \, \left\{\begin{array}{ll} { } \sigma_{2} (t)=\frac{\beta}{\alpha}+(u_{-}-\frac{\beta}{\alpha})e^{-\alpha t} +\rho\sqrt{\frac{\frac{\gamma-1}{2}(\rho^{\gamma-1}-\rho_{-}^{\gamma-1})}{(\rho+\rho_{-})(\rho-\rho_{-})}}, \\ { } v=u_{-} +e^{\alpha t}\sqrt{\frac{\frac{\gamma-1}{2}(\rho^{\gamma-1}-\rho_{-}^{\gamma-1})} {(\rho+\rho_{-})(\rho-\rho_{-})}}(\rho-\rho_{-}), \; \; \; \\ \rho<\rho_{-}, v<u_{-}.\end{array} \right. \end{align} $

对(3.14)式的第二个方程$ v $的两边关于$ \rho $求导, 则当$ \rho>\rho_{-} $时, 有

$ \frac{{\rm d}v}{{\rm d}\rho}=-\frac{e^{\alpha t}}{2}\sqrt{\frac{\frac{\gamma-1}{2}(\rho+\rho_{-})} {(\rho^{\gamma-1}-\rho_{-}^{\gamma-1})(\rho-\rho_{-})}} \frac{(\gamma-1)\rho^{\gamma-2}(\rho-\rho_{-})(\rho+\rho_{-})+2\rho_{-}(\rho^{\gamma-1}-\rho_{-}^{\gamma-1})} {(\rho+\rho_{-})^{2}}<0, $

即在相平面$ (\rho, v) $上, 1 -激波曲线$ S_{1} (\rho_{-}, u_{-}) $单调递减. 同样地, 对于(3.15)式, 当$ \rho<\rho_{-} $时, $ \frac{{\rm d}v}{{\rm d}\rho} > 0, $ 即2 -激波曲线$ S_{2} (\rho_{-}, u_{-}) $在相平面$ (\rho, v) $上是单调递增. 由(3.15)式可得

$ \lim\limits_{\rho\rightarrow 0^{+}}v= u_{-}-e^{\alpha t}\sqrt{\frac{\gamma-1}{2}}\rho_{-}^{\frac{\gamma-1}{2}}, $

即$ S_{2} (\rho_{-}, u_{-}) $与$ v $ -轴交于$ (0, \widetilde{v}_{\ast\ast} ) $, 其中$ \widetilde{v}_{\ast\ast} =u_{-}-e^{\alpha t}\sqrt{\frac{\gamma-1}{2}}\rho_{-}^{\frac{\gamma-1}{2}}. $ 对$ S_{1} (\rho_{-}, u_{-}) $, 由(3.14)式可得$ \lim\limits_{\rho\rightarrow +\infty}v= -\infty $.

在相平面$ (\rho, v) $上, 通过给定状态$ (\rho_{-}, u_{-}) $, 我们画出基本波$ R_{j} (\rho_{-}, u_{-}) $和$ S_{j} (\rho_{-}, u_{-}) $$ (j=1, 2). $ 基本波将相平面$ (\rho, v) $划分为五个区域(见图 1). 根据在不同区域的右状态$ (\rho_{+}, u_{+}) $, 我们构造(3.1)和(3.2)式的唯一全局Riemann解即

(1) $ (\rho_{+}, u_{+})\in I(\rho_{-}, u_{-}): $$ (\rho_{-}, u_{-})+R_{1} +(\rho_{\ast}, v_{\ast})+R_{2} +(\rho_{+}, u_{+}); $

(2) $ (\rho_{+}, u_{+})\in II(\rho_{-}, u_{-}): $$ (\rho_{-}, u_{-})+S_{1} +(\rho_{\ast}, v_{\ast})+R_{2} +(\rho_{+}, u_{+}); $

(3) $ (\rho_{+}, u_{+})\in III(\rho_{-}, u_{-}): $$ (\rho_{-}, u_{-})+R_{1} +(\rho_{\ast}, v_{\ast})+S_{2} +(\rho_{+}, u_{+}); $

(4) $ (\rho_{+}, u_{+})\in IV(\rho_{-}, u_{-}): $$ (\rho_{-}, u_{-})+S_{1} +(\rho_{\ast}, v_{\ast})+S_{2} +(\rho_{+}, u_{+}); $

(5) $ (\rho_{+}, u_{+})\in V(\rho_{-}, u_{-}): $$ (\rho_{-}, u_{-})+R_{1} +{\rm Vac}+R_{2} +(\rho_{+}, u_{+}). $

准确地说, 若$ (\rho_{+}, u_{+})\in I(\rho_{-}, u_{-}) $, 则Riemann解包含两条稀疏波$ R_{1} $和$ R_{2} $以及中间状态$ (\rho_{\ast}, v_{\ast}) $, 其中$ (\rho_{\ast}, v_{\ast}) $只取决于

$ \left\{\begin{array}{ll} v_{\ast}+\rho_{\ast}^{\frac{\gamma-1}{2}}e^{\alpha t}=u_{-}+\rho_{-}^{\frac{\gamma-1}{2}}e^{\alpha t}, \\v_{\ast}-\rho_{\ast}^{\frac{\gamma-1}{2}}e^{\alpha t}=u_{+}-\rho_{+}^{\frac{\gamma-1}{2}}e^{\alpha t} .\end{array} \right. $

通过简单计算, 不难发现

(3.16) $ \begin{align} v_{\ast}=\frac{1}{2}(u_{-}+u_{+}+(\rho_{-}^{\frac{\gamma-1}{2}}-\rho_{+}^{\frac{\gamma-1}{2}})e^{\alpha t}), \, \, \, \, \, \rho_{\ast}=\Bigg(\frac{(u_{-}-u_{+})e^{-\alpha t}+\rho_{-}^{\frac{\gamma-1}{2}} +\rho_{+}^{\frac{\gamma-1}{2}}}{2}\Bigg)^{\frac{2}{\gamma-1}}. \end{align} $

因此, 系统(3.1)和(3.2)的Riemann解可写成

(3.17) $ \begin{align} (\rho, v)(t, x)=\left\{\begin{array}{ll} (\rho_{-}, u_{-}), &{ } -\infty<x<\frac{\beta}{\alpha}t+\frac{1}{\alpha}(u_{-}-\frac{\beta}{\alpha})(1 -e^{-\alpha t}) -\frac{\gamma-1}{2}\rho_{-}^{\frac{\gamma-1}{2}}t, \\ R_{1}, &{ } \frac{\beta}{\alpha}t+\frac{1}{\alpha}(u_{-}-\frac{\beta}{\alpha})(1 -e^{-\alpha t}) -\frac{\gamma-1}{2}\rho_{-}^{\frac{\gamma-1}{2}}t\leq x\\ &{\qquad}{ } \leq \frac{\beta}{\alpha}t+\frac{1}{\alpha}(v_{\ast}-\frac{\beta}{\alpha})(1 -e^{-\alpha t})-\frac{\gamma-1}{2}\rho_{\ast}^{\frac{\gamma-1}{2}}t, \\ (\rho_{\ast}, v_{\ast}), &{ } \frac{\beta}{\alpha}t+\frac{1}{\alpha}(v_{\ast}-\frac{\beta}{\alpha})(1 -e^{-\alpha t})-\frac{\gamma-1}{2}\rho_{\ast}^{\frac{\gamma-1}{2}}t<x\\ &{\qquad} { } < \frac{\beta}{\alpha}t+\frac{1}{\alpha}(v_{\ast}-\frac{\beta}{\alpha})(1 -e^{-\alpha t}) +\frac{\gamma-1}{2}\rho_{\ast}^{\frac{\gamma-1}{2}}t, \\ R_{2}, &{ } \frac{\beta}{\alpha}t+\frac{1}{\alpha} (v_{\ast}-\frac{\beta}{\alpha})(1 -e^{-\alpha t})+\frac{\gamma-1}{2}\rho_{\ast}^{\frac{\gamma-1}{2}}t\leq x\\ &{\qquad}{ } \leq \frac{\beta}{\alpha}t+\frac{1}{\alpha}(u_{+}-\frac{\beta}{\alpha})(1 -e^{-\alpha t}) +\frac{\gamma-1}{2}\rho_{+}^{\frac{\gamma-1}{2}}t, \\ (\rho_{+}, u_{+}), & { } \frac{\beta}{\alpha}t+\frac{1} {\alpha}(u_{+}-\frac{\beta}{\alpha})(1 -e^{-\alpha t})+\frac{\gamma-1}{2}\rho_{+}^{\frac{\gamma-1}{2}}t<x<+\infty, \end{array}\right . \end{align} $

其中$ R_{1} $上状态$ (\rho_{1}, v_{1}) $可以由(3.7)式计算, 而$ R_{2} $上状态$ (\rho_{2}, v_{2}) $可以由(3.9)式分别用$ \rho_{\ast} $和$ v_{\ast} $代替$ \rho_{-} $和$ u_{-} $计算.

若$ (\rho_{+}, u_{+})\in II(\rho_{-}, u_{-}) $, 则Riemann解包含1 -激波$ S_{1} $和2 -稀疏波$ R_{2} $以及中间状态$ (\rho_{\ast}, v_{\ast}) $, 其中$ (\rho_{\ast}, v_{\ast}) $只取决于

(3.18) $ \begin{align} \left\{\begin{array}{ll} { } v_{\ast}=u_{-}-e^{\alpha t}\sqrt{\frac{\frac{\gamma-1}{2}(\rho_{\ast}^{\gamma-1}- \rho_{-}^{\gamma-1})}{(\rho_{\ast}+\rho_{-})(\rho_{\ast}-\rho_{-})}}(\rho_{\ast}-\rho_{-}), \\ { } v_{\ast}-\rho_{\ast}^{\frac{\gamma-1}{2}}e^{\alpha t}=u_{+}-\rho_{+}^{\frac{\gamma-1}{2}}e^{\alpha t}. \end{array} \right. \end{align} $

系统(3.1)和(3.2)的Riemann解可写成

(3.19) $ \begin{align} (\rho, v)(t, x)=\left\{\begin{array}{ll} { } (\rho_{-}, u_{-}), &{ } -\infty<x<\frac{\beta}{\alpha}t+\frac{1}{\alpha}(u_{-}-\frac{\beta}{\alpha}) (1 -e^{-\alpha t})-\rho_{\ast} \sqrt{\frac{\frac{\gamma-1}{2}(\rho_{\ast}^{\gamma-1}-\rho_{-}^{\gamma-1})} {(\rho_{\ast}+\rho_{-})(\rho_{\ast}-\rho_{-})}} \, t, \\ (\rho_{\ast}, v_{\ast}), &{ } \frac{\beta}{\alpha}t+\frac{1}{\alpha}(u_{-}-\frac{\beta}{\alpha})(1 -e^{-\alpha t})-\rho_{\ast} \sqrt{\frac{\frac{\gamma-1}{2}(\rho_{\ast}^{\gamma-1}-\rho_{-}^{\gamma-1})}{(\rho_{\ast}+\rho_{-})(\rho_{\ast}-\rho_{-})}} \, t<x\\ &{\qquad} { } <\frac{\beta}{\alpha}t+\frac{1}{\alpha}(v_{\ast}-\frac{\beta}{\alpha})(1 -e^{-\alpha t}) +\frac{\gamma-1}{2}\rho_{\ast}^{\frac{\gamma-1}{2}}t, \\ R_{2}, &{ } \frac{\beta}{\alpha}t+\frac{1} {\alpha}(v_{\ast}-\frac{\beta}{\alpha})(1 -e^{-\alpha t})+\frac{\gamma-1}{2}\rho_{\ast}^{\frac{\gamma-1}{2}}t\leq x\\ &{\qquad}{ } \leq \frac{\beta}{\alpha}t+\frac{1}{\alpha}(u_{+}-\frac{\beta}{\alpha})(1 -e^{-\alpha t}) +\frac{\gamma-1}{2}\rho_{+}^{\frac{\gamma-1}{2}}t, \\ (\rho_{+}, u_{+}), & { } \frac{\beta}{\alpha}t+\frac{1}{\alpha}(u_{+}-\frac{\beta}{\alpha})(1 -e^{-\alpha t}) +\frac{\gamma-1}{2}\rho_{+}^{\frac{\gamma-1}{2}}t<x<+\infty, \end{array}\right . \end{align} $

其中, $ R_{2} $上状态$ (\rho_{2}, v_{2}) $也可以用前面的方法计算, $ S_{1} $的位置由下式确定

(3.20) $ \begin{align} x=\frac{\beta}{\alpha}t+\frac{1}{\alpha}(u_{-}-\frac{\beta}{\alpha})(1 -e^{-\alpha t})-\rho_{\ast} \sqrt{\frac{\frac{\gamma-1}{2}(\rho_{\ast}^{\gamma-1}-\rho_{-}^{\gamma-1})}{(\rho_{\ast}+\rho_{-})(\rho_{\ast}-\rho_{-})}} \, t. \end{align} $

若$ (\rho_{+}, u_{+})\in III(\rho_{-}, u_{-}) $, 则Riemann解包含1 -稀疏波$ R_{1} $和2 -激波$ S_{2} $以及中间状态$ (\rho_{\ast}, v_{\ast}) $, 其中$ (\rho_{\ast}, v_{\ast}) $只取决于

(3.21) $ \begin{align} \left\{\begin{array}{ll} { } v_{\ast}+\rho_{\ast}^{\frac{\gamma-1}{2}}e^{\alpha t}=u_{-}+\rho_{-}^{\frac{\gamma-1}{2}}e^{\alpha t}, \\ { } u_{+}=v_{\ast} +e^{\alpha t}\sqrt{\frac{\frac{\gamma-1}{2}(\rho_{+}^{\gamma-1}-\rho_{\ast}^{\gamma-1})} {(\rho_{+}+\rho_{\ast})(\rho_{+}-\rho_{\ast})}}(\rho_{+} -\rho_{\ast}). \end{array} \right. \end{align} $

系统(3.1)和(3.2)的Riemann解可写成

(3.22) $ \begin{align} (\rho, v)(t, x)=\left\{\begin{array}{ll} (\rho_{-}, u_{-}), &{ } -\infty<x<\frac{\beta}{\alpha}t+\frac{1}{\alpha}(u_{-}-\frac{\beta}{\alpha})(1 -e^{-\alpha t}) -\frac{\gamma-1}{2}\rho_{-}^{\frac{\gamma-1}{2}}t, \\ R_{1}, &{ } \frac{\beta}{\alpha}t+\frac{1}{\alpha}(u_{-}-\frac{\beta}{\alpha})(1 -e^{-\alpha t}) -\frac{\gamma-1}{2}\rho_{-}^{\frac{\gamma-1}{2}}t\leq x\\ &{\qquad} { } \leq \frac{\beta}{\alpha}t+\frac{1}{\alpha}(v_{\ast}-\frac{\beta}{\alpha})(1 -e^{-\alpha t})- \frac{\gamma-1}{2}\rho_{\ast}^{\frac{\gamma-1}{2}}t, \\ (\rho_{\ast}, v_{\ast}), &{ } \frac{\beta}{\alpha}t +\frac{1}{\alpha}(v_{\ast}-\frac{\beta}{\alpha})(1 -e^{-\alpha t})-\frac{\gamma-1}{2}\rho_{\ast}^{\frac{\gamma-1}{2}}t<x\\ &{\qquad}{ } <\frac{\beta}{\alpha}t+\frac{1}{\alpha}(v_{\ast}-\frac{\beta}{\alpha})(1 -e^{-\alpha t})+\rho_{+} \sqrt{\frac{\frac{\gamma-1}{2}(\rho_{+}^{\gamma-1}-\rho_{\ast}^{\gamma-1})} {(\rho_{+}+\rho_{\ast})(\rho_{+}-\rho_{\ast})}} \, t, \\ (\rho_{+}, u_{+}), & { } \frac{\beta}{\alpha}t+\frac{1}{\alpha}(v_{\ast}-\frac{\beta}{\alpha}) (1 -e^{-\alpha t})+\rho_{+} \sqrt{\frac{\frac{\gamma-1}{2}(\rho_{+}^{\gamma-1}- \rho_{\ast}^{\gamma-1})}{(\rho_{+}+\rho_{\ast})(\rho_{+}-\rho_{\ast})}} \, t<x<+\infty, \end{array}\right . \end{align} $

其中$ R_{1} $上状态$ (\rho_{1}, v_{1}) $可以由(3.7)式计算, $ S_{2} $的位置由下式确定

(3.23) $ \begin{align} x=\frac{\beta}{\alpha}t+\frac{1}{\alpha}(v_{\ast}-\frac{\beta}{\alpha}) (1 -e^{-\alpha t})+\rho_{+} \sqrt{\frac{\frac{\gamma-1}{2}(\rho_{+}^{\gamma-1} -\rho_{\ast}^{\gamma-1})}{(\rho_{+}+\rho_{\ast})(\rho_{+}-\rho_{\ast})}} \, t. \end{align} $

若$ (\rho_{+}, u_{+})\in IV(\rho_{-}, u_{-}) $, 则Riemann解由1 -激波$ S_{1} $和2 -激波$ S_{2} $以及中间状态$ (\rho_{\ast}, v_{\ast}) $组成, 其中

(3.24) $ \begin{align} \left\{\begin{array}{ll} { } v_{\ast}=u_{-}-e^{\alpha t}\sqrt{\frac{\frac{\gamma-1}{2}(\rho_{\ast}^{\gamma-1} -\rho_{-}^{\gamma-1})}{(\rho_{\ast}+\rho_{-})(\rho_{\ast} -\rho_{-})}}(\rho_{\ast}-\rho_{-}), \\ { } u_{+}=v_{\ast} +e^{\alpha t}\sqrt{\frac{\frac{\gamma-1}{2}(\rho_{+}^{\gamma-1}-\rho_{\ast}^{\gamma-1})} {(\rho_{+}+\rho_{\ast})(\rho_{+}-\rho_{\ast})}} (\rho_{+}-\rho_{\ast}).\end{array} \right. \end{align} $

系统(3.1)和(3.2)的Riemann解可写成

(3.25) $ \begin{align} (\rho, v)(t, x)=\left\{\begin{array}{ll} { } (\rho_{-}, u_{-}), &{ } -\infty<x<\frac{\beta}{\alpha}t+\frac{1}{\alpha}(u_{-}-\frac{\beta}{\alpha}) (1 -e^{-\alpha t})-\rho_{\ast} \sqrt{\frac{\frac{\gamma-1}{2}(\rho_{\ast}^{\gamma-1}-\rho_{-}^{\gamma-1})} {(\rho_{\ast}+\rho_{-})(\rho_{\ast}-\rho_{-})}} \, t, \\ (\rho_{\ast}, v_{\ast}), &{ } \frac{\beta}{\alpha}t +\frac{1}{\alpha}(u_{-}-\frac{\beta}{\alpha})(1 -e^{-\alpha t})-\rho_{\ast} \sqrt{\frac{\frac{\gamma-1}{2} (\rho_{\ast}^{\gamma-1}-\rho_{-}^{\gamma-1})}{(\rho_{\ast}+\rho_{-})(\rho_{\ast}-\rho_{-})}} t<x\\ &{\qquad}{ } <\frac{\beta}{\alpha}t+\frac{1}{\alpha}(v_{\ast}-\frac{\beta}{\alpha})(1 -e^{-\alpha t})+\rho_{+} \sqrt{\frac{\frac{\gamma-1}{2}(\rho_{+}^{\gamma-1}-\rho_{\ast}^{\gamma-1})}{(\rho_{+}+\rho_{\ast}) (\rho_{+}-\rho_{\ast})}} \, t, \\ (\rho_{+}, u_{+}), & { } \frac{\beta}{\alpha}t+\frac{1}{\alpha} (v_{\ast}-\frac{\beta}{\alpha})(1 -e^{-\alpha t})+\rho_{+} \sqrt{\frac{\frac{\gamma-1}{2} (\rho_{+}^{\gamma-1}-\rho_{\ast}^{\gamma-1})}{(\rho_{+}+\rho_{\ast})(\rho_{+}-\rho_{\ast})}} \, t<x<+\infty, \end{array}\right . \end{align} $

其中$ S_{1} $和$ S_{2} $的位置分别由(3.20)和(3.23)式确定.

通过变量代换(1.2), 可得到系统(1.1)的Riemann解为

(1) $ (\rho_{+}, u_{+})\in I(\rho_{-}, u_{-}): $$ (\rho_{-}, \frac{\beta}{\alpha}+(u_{-}-\frac{\beta}{\alpha})e^{-\alpha t})+R_{1} +(\rho_{\ast}, \frac{\beta}{\alpha}+(v_{\ast}-\frac{\beta}{\alpha})e^{-\alpha t})+R_{2} +(\rho_{+}, \frac{\beta}{\alpha}+(u_{+}-\frac{\beta}{\alpha})e^{-\alpha t}); $

(2) $ (\rho_{+}, u_{+})\in II(\rho_{-}, u_{-}): $$ (\rho_{-}, \frac{\beta}{\alpha}+(u_{-}-\frac{\beta}{\alpha})e^{-\alpha t})+S_{1} +(\rho_{\ast}, \frac{\beta}{\alpha}+(v_{\ast}-\frac{\beta}{\alpha})e^{-\alpha t}) +R_{2} +(\rho_{+}, \frac{\beta}{\alpha}+(u_{+}-\frac{\beta}{\alpha})e^{-\alpha t}); $

(3) $ (\rho_{+}, u_{+})\in III(\rho_{-}, u_{-}): $$ (\rho_{-}, \frac{\beta}{\alpha}+(u_{-}-\frac{\beta} {\alpha})e^{-\alpha t})+R_{1} +(\rho_{\ast}, \frac{\beta}{\alpha}+(v_{\ast}-\frac{\beta}{\alpha}) e^{-\alpha t})+S_{2} +(\rho_{+}, \frac{\beta}{\alpha}+(u_{+}-\frac{\beta}{\alpha})e^{-\alpha t}); $

(4) $ (\rho_{+}, u_{+})\in IV(\rho_{-}, u_{-}): $$ (\rho_{-}, \frac{\beta}{\alpha}+(u_{-}-\frac{\beta}{\alpha}) e^{-\alpha t})+S_{1} +(\rho_{\ast}, \frac{\beta}{\alpha}+(v_{\ast}-\frac{\beta}{\alpha})e^{-\alpha t})+S_{2} +(\rho_{+}, \frac{\beta}{\alpha}+(u_{+}-\frac{\beta}{\alpha})e^{-\alpha t}); $

(5) $ (\rho_{+}, u_{+})\in V(\rho_{-}, u_{-}): $$ (\rho_{-}, \frac{\beta}{\alpha}+(u_{-}-\frac{\beta}{\alpha}) e^{-\alpha t})+R_{1} +{\rm Vac}+R_{2} +(\rho_{+}, \frac{\beta}{\alpha}+(u_{+}-\frac{\beta}{\alpha})e^{-\alpha t}). $

我们以$ (\rho_{+}, u_{+})\in I(\rho_{-}, u_{-}) $为例来说明. 如果$ (\rho_{+}, u_{+})\in I(\rho_{-}, u_{-}) $, 则系统(1.1)的Riemann解可表示为

(3.26) $ \begin{align} (\rho, u)(t, x)=\left\{\begin{array}{ll} { } (\rho_{-}, \frac{\beta}{\alpha}+(u_{-}-\frac{\beta}{\alpha})e^{-\alpha t}), &-\infty<x\\ &{\qquad} { } <\frac{\beta}{\alpha}t+\frac{1}{\alpha}(u_{-}-\frac{\beta}{\alpha})(1 -e^{-\alpha t}) -\frac{\gamma-1}{2}\rho_{-}^{\frac{\gamma-1}{2}}t, \\ { } (\rho_{1}, \frac{\beta}{\alpha}+(v_{1}-\frac{\beta}{\alpha})e^{-\alpha t}), & { } \frac{\beta}{\alpha}t+\frac{1}{\alpha}(u_{-}-\frac{\beta}{\alpha})(1 -e^{-\alpha t}) -\frac{\gamma-1}{2}\rho_{-}^{\frac{\gamma-1}{2}}t\leq x\\ &{\qquad} { } \leq \frac{\beta}{\alpha}t+\frac{1}{\alpha}(v_{\ast}-\frac{\beta}{\alpha}) (1 -e^{-\alpha t})-\frac{\gamma-1}{2}\rho_{\ast}^{\frac{\gamma-1}{2}}t, \\ { } (\rho_{\ast}, \frac{\beta}{\alpha}+(v_{\ast}-\frac{\beta}{\alpha})e^{-\alpha t}), & { } \frac{\beta}{\alpha}t+\frac{1}{\alpha}(v_{\ast}-\frac{\beta}{\alpha})(1 -e^{-\alpha t})- \frac{\gamma-1}{2}\rho_{\ast}^{\frac{\gamma-1}{2}}t<x\\ &{\qquad}{ } < \frac{\beta}{\alpha}t+\frac{1}{\alpha}(v_{\ast}-\frac{\beta}{\alpha}) (1 -e^{-\alpha t})+\frac{\gamma-1}{2}\rho_{\ast}^{\frac{\gamma-1}{2}}t, \\ { } (\rho_{2}, \frac{\beta}{\alpha}+(v_{2}-\frac{\beta}{\alpha})e^{-\alpha t}), & { } \frac{\beta}{\alpha}t+\frac{1}{\alpha}(v_{\ast}-\frac{\beta}{\alpha})( 1 -e^{-\alpha t})+\frac{\gamma-1}{2}\rho_{\ast}^{\frac{\gamma-1}{2}}t\leq x\\ &{\qquad} { } \leq \frac{\beta}{\alpha}t+\frac{1}{\alpha}(u_{+}-\frac{\beta}{\alpha}) (1 -e^{-\alpha t})+\frac{\gamma-1}{2}\rho_{+}^{\frac{\gamma-1}{2}}t, \\ { } (\rho_{+}, \frac{\beta}{\alpha}+(u_{+}-\frac{\beta}{\alpha})e^{-\alpha t}), & { } \frac{\beta}{\alpha}t+\frac{1}{\alpha}(u_{+}-\frac{\beta}{\alpha})(1 -e^{-\alpha t})+\frac{\gamma-1}{2}\rho_{+}^{\frac{\gamma-1}{2}}t\\ &{\qquad} <x<+\infty, \end{array}\right . \end{align} $

其中状态$ (\rho_{1}, v_{1}) $, $ (\rho_{\ast}, v_{\ast}) $和$ (\rho_{2}, v_{2}) $也可以像以前一样计算.

在本节, 我们分别讨论$ u_{+}<u_{-} $和$ u_{+}>u_{-} $两种情形. 令$ \gamma \rightarrow 1 $, 研究系统(1.1) 的Riemann解的变化, 即分别研究系统(1.1)的delta激波和真空状形成.

引理4.1  若$ u_{+}<u_{-} $, 则存在一个足够小的$ \gamma_{0} > 0 $, 使得当$ 1<\gamma <1 +\gamma_{0} $时, $ (\rho_{+}, u_{+})\in IV( \rho_{-}, u_{-}) $.

证  若$ \rho_{+} = \rho_{-} $, 则对$ \forall\gamma\in (1, 2) $, $ (\rho_{+}, u_{+})\in IV( \rho_{-}, u_{-}) $成立. 因此下面我们只需要考虑$ \rho_{+} \neq\rho_{-} $情况.

通过(3.14)和(3.15)式, 易知所有与左状态$ (\rho_{-}, u_{-}) $通过1 -激波$ S_{1} $或2 -激波$ S_{2} $连接的$ (\rho, v) $满足

(4.1) $ \begin{equation} S_{1}: \, \, \, \, v=u_{-}-e^{\alpha t}\sqrt{\frac{\frac{\gamma-1}{2}(\rho^{\gamma-1}-\rho_{-}^{\gamma-1})}{(\rho+\rho_{-})(\rho-\rho_{-})}}(\rho-\rho_{-}), \; \; \; \rho>\rho_{-}, \end{equation} $

(4.2) $ \begin{equation} S_{2}: \, \, \, \, v=u_{-}+e^{\alpha t}\sqrt{\frac{\frac{\gamma-1}{2}(\rho^{\gamma-1}-\rho_{-}^{\gamma-1})}{(\rho+\rho_{-})(\rho-\rho_{-})}}(\rho-\rho_{-}), \; \; \rho<\rho_{-}. \end{equation} $

若$ \rho_{+} \neq\rho_{-} $及$ (\rho_{+}, u_{+})\in IV( \rho_{-}, u_{-}) $, 则由图 1, (4.1) 和(4.2) 式, 可得

(4.3) $ \begin{equation} u_{+}<u_{-}-e^{\alpha t}\sqrt{\frac{\frac{\gamma-1}{2}(\rho_{+}^{\gamma-1}-\rho_{-}^{\gamma-1})}{(\rho_{+}+\rho_{-})(\rho_{+}-\rho_{-})}}(\rho_{+}-\rho_{-}), \; \; \; \mathrm {} \, \, \rho_{+}>\rho_{-}, \end{equation} $

(4.4) $ \begin{equation} u_{+}<u_{-}+e^{\alpha t}\sqrt{\frac{\frac{\gamma-1}{2}(\rho_{+}^{\gamma-1}-\rho_{-}^{\gamma-1})}{(\rho_{+}+\rho_{-})(\rho_{+}-\rho_{-})}}(\rho_{+}-\rho_{-}), \; \; \; \mathrm {} \, \, \rho_{+}<\rho_{-}. \end{equation} $

由(4.3)和(4.4)式, 有

(4.5) $ \begin{equation} e^{\alpha t}\sqrt{\frac{\frac{\gamma-1}{2}(\rho_{+}^{\gamma-1}-\rho_{-}^{\gamma-1})}{\rho_{+}^{2}-\rho_{-}^{2}}}< \frac{u_{-}-u_{+}}{|\rho_{+}-\rho_{-}|}. \end{equation} $

由于

(4.6) $ \begin{equation} \lim\limits_{{\gamma\rightarrow1}}e^{\alpha t}\sqrt{\frac{\frac{\gamma-1}{2}(\rho_{+}^{\gamma-1}-\rho_{-}^{\gamma-1})}{\rho_{+}^{2}-\rho_{-}^{2}}}= e^{\alpha t}\lim\limits_{{\gamma\rightarrow1}}\sqrt{\frac{\frac{\gamma-1}{2}(\rho_{+}^{\gamma-1}-\rho_{-}^{\gamma-1})}{\rho_{+}^{2}-\rho_{-}^{2}}}=0, \end{equation} $

因此存在$ \gamma_{0}>0 $足够小, 当$ 1<\gamma <1 +\gamma_{0} $时, 我们有不等式(4.5)成立. 所以, 当$ 1<\gamma <1 +\gamma_{0} $时, $ (\rho_{+}, u_{+})\in IV( \rho_{-}, u_{-}) $成立. 证毕.

当$ 1<\gamma <1+\gamma_{0} $, 即$ (\rho_{+}, u_{+})\in IV( \rho_{-}, u_{-}) $, 假设中间状态$ (\rho_{\ast}, v_{\ast}) $通过波速为$ \sigma_{1}(t) $的1 -激波$ S_{1} $连接$ (\rho_{-}, u_{-}) $, 且通过波速为$ \sigma_{2}(t) $的2 - 激波$ S_{2} $连接$ (\rho_{+}, u_{+}) $, 则

(4.7) $ \begin{equation} S_{1} :\, \, \left\{\begin{array}{ll} { } \sigma_{1} (t)=\frac{\beta}{\alpha}+(u_{-}-\frac{\beta}{\alpha})e^{-\alpha t} -\rho_{\ast}\sqrt{\frac{\frac{\gamma-1}{2}(\rho_{\ast}^{\gamma-1}-\rho_{-}^{\gamma-1})} {(\rho_{\ast}+\rho_{-})(\rho_{\ast}-\rho_{-})}}, \\ { } v_{\ast}=u_{-}-e^{\alpha t}\sqrt{\frac{\frac{\gamma-1}{2}(\rho_{\ast}^{\gamma-1} -\rho_{-}^{\gamma-1})}{(\rho_{\ast}+\rho_{-})(\rho_{\ast} -\rho_{-})}}(\rho_{\ast}-\rho_{-}), \; \; \; \rho_{\ast}>\rho_{-}, \end{array} \right. \end{equation} $

(4.8) $ \begin{equation} S_{2} :\, \, \left\{\begin{array}{ll} { } \sigma_{2} (t)=\frac{\beta}{\alpha}+(v_{\ast}-\frac{\beta}{\alpha})e^{-\alpha t} +\rho_{+}\sqrt{\frac{\frac{\gamma-1}{2}(\rho_{+}^{\gamma-1}-\rho_{\ast}^{\gamma-1})} {(\rho_{+}+\rho_{\ast})(\rho_{+}-\rho_{\ast})}}, \\ { } u_{+}=v_{\ast} +e^{\alpha t}\sqrt{\frac{\frac{\gamma-1}{2}(\rho_{+}^{\gamma-1}-\rho_{\ast}^{\gamma-1})} {(\rho_{+}+\rho_{\ast})(\rho_{+}-\rho_{\ast})}} (\rho_{+}-\rho_{\ast}), \; \; \; \rho_{\ast}>\rho_{+}.\end{array} \right. \end{equation} $

由(4.7)和(4.8)式, 有

(4.9) $ \begin{eqnarray} u_{-}-u_{+}&=&e^{\alpha t}\sqrt{\frac{\frac{\gamma-1}{2}(\rho_{\ast}^{\gamma-1}-\rho_{-} ^{\gamma-1})}{(\rho_{\ast}+\rho_{-})(\rho_{\ast}-\rho_{-})}}(\rho_{\ast}-\rho_{-}){}\\ &&+ e^{\alpha t}\sqrt{\frac{\frac{\gamma-1}{2}(\rho_{+}^{\gamma-1}-\rho_{\ast}^{\gamma-1})} {(\rho_{\ast}+\rho_{+})(\rho_{+}-\rho_{\ast})}}(\rho_{\ast}-\rho_{+}), \, \, \, \, \rho_{\ast}>\rho_{\pm}. \end{eqnarray} $

则有下列引理.

引理4.2  $ \lim\limits_{\gamma\rightarrow1}\rho_{\ast}=+\infty, $$ \lim\limits_{\gamma\rightarrow1}\frac{\gamma-1}{2}\rho_{\ast}^{\gamma-1}=:a=\frac{((u_{-}-u_{+})e^{-\alpha t})^{2}}{4}. $

证  令$ \lim\limits_{\gamma\rightarrow1}\inf\rho_{\ast}=\alpha $, $ \lim\limits_{\gamma\rightarrow1}\sup\rho_{\ast}=\beta $.

若$ \alpha<\beta $, 则由$ \rho_{\ast}(\gamma) $的连续性可知, 存在数列$ \{\gamma_{n}\}_{n=1}^{\infty}\subseteq(1, 2) $使得

$ \lim\limits_{n\rightarrow +\infty}\gamma_{n}=1, \, \, \mbox{和}\, \, \lim\limits_{n\rightarrow +\infty}\rho_{\ast}(\gamma_{n})=c, $

其中$ c\in(\alpha, \beta). $ 将数列代入(4.9)式的右边, 并令$ n\rightarrow +\infty $, 则有

(4.10) $ \begin{equation} \lim\limits_{n\rightarrow +\infty}\sqrt{\frac{\frac{\gamma_{n}-1}{2}(\rho_{\ast}(\gamma_{n})^{\gamma_{n}-1}-\rho_{\pm}^{\gamma_{n}-1})} {(\rho_{\ast}(\gamma_{n})+\rho_{\pm})(\rho_{\ast}(\gamma_{n})-\rho_{\pm})}}(\rho_{\ast}(\gamma_{n})-\rho_{\pm})=0. \end{equation} $

因此, 由(4.9)式可知

$ u_{-}-u_{+}=0, $

这与假设$ u_{-}>u_{+} $矛盾. 从而可知$ \alpha=\beta $, 即$ \lim\limits_{\gamma\rightarrow1}\rho_{\ast}(\gamma)=\alpha. $

若$ \alpha\in(0, +\infty), $ 那么在(4.9)式取极限时我们也可以得到一个矛盾. 因此$ \alpha=0 $或$ \alpha=+\infty $. 由条件$ \rho_{\ast}>\max\{\rho_{-}, \rho_{+}\} $, 易知

$ \lim\limits_{\gamma\rightarrow1}\rho_{\ast}(\gamma)=\alpha=+\infty. $

对(4.9)式的右边取极限$ \gamma\rightarrow 1 $, 则有

$ \lim\limits_{\gamma\rightarrow 1}\sqrt{\frac{\frac{\gamma-1}{2}(\rho_{\ast}^{\gamma-1}-\rho_{\pm}^{\gamma-1})} {(\rho_{\ast}+\rho_{\pm})(\rho_{\ast}-\rho_{\pm})}}(\rho_{\ast}-\rho_{\pm}) =\lim\limits_{\gamma\rightarrow1}\sqrt{\frac{(\frac{\gamma-1}{2}\rho_{\ast}^{\gamma-1} -\frac{\gamma-1}{2}\rho_{\pm}^{\gamma-1})(\rho_{\ast}-\rho_{\pm})} {\rho_{\ast}+\rho_{\pm}}} =:\sqrt{a}, $

$ u_{-}-u_{+}=2e^{\alpha t}\sqrt{a}, $

从而可知$ a=\frac{((u_{-}-u_{+})e^{-\alpha t})^{2}}{4}. $ 证毕.

引理4.3  若$ u_{-}>u_{+}, $ 则

(4.11) $ \begin{equation} \lim\limits_{\gamma\rightarrow1}u_{\ast}= \lim\limits_{\gamma\rightarrow1}(\frac{\beta}{\alpha}+(v_{\ast}-\frac{\beta}{\alpha})e^{-\alpha t}) =\lim\limits_{\gamma\rightarrow1}\sigma_{1}(t)=\lim\limits_{\gamma\rightarrow1}\sigma_{2}(t)=u_{\delta}(t), \end{equation} $

(4.12) $ \begin{equation} \lim\limits_{\gamma\rightarrow1}\int^{\sigma_{2}(t)}_{\sigma_{1}(t)}\rho_{\ast} {\rm d}\xi=u_{\delta}(t)[\rho]-[\rho (\frac{\beta}{\alpha}+(v-\frac{\beta}{\alpha})e^{-\alpha t})] =\frac{1}{2}(\rho_{-}+\rho_{+})(u_{-}-u_{+})e^{-\alpha t}, \end{equation} $

其中$ [\rho] = \rho_{+}- \rho_{-} $且$ u_{\delta}(t)=\frac{\beta}{\alpha}+(\frac{1}{2} (u_{-}+u_{+})-\frac{\beta}{\alpha})e^{-\alpha t}. $

证  由(2.1), (4.7)和(4.8)式以及引理4.2, 可得

$ \begin{eqnarray*} \lim\limits_{\gamma\rightarrow1}u_{\ast}&=&\lim\limits_{\gamma\rightarrow1}(\frac{\beta} {\alpha}+(v_{\ast}-\frac{\beta}{\alpha})e^{-\alpha t})\\ &=&\frac{\beta}{\alpha}+(u_{-}-\frac{\beta}{\alpha})e^{-\alpha t}-\lim\limits_{\gamma\rightarrow1} \sqrt{\frac{\frac{\gamma-1}{2}(\rho_{\ast}^{\gamma-1}-\rho_{-}^{\gamma-1})} {(\rho_{\ast}+\rho_{-})(\rho_{\ast}-\rho_{-})}}(\rho_{\ast}-\rho_{-})\\ &=&\frac{\beta}{\alpha}+(u_{-}-\frac{\beta}{\alpha})e^{-\alpha t}-\sqrt{a}=\frac{\beta} {\alpha}+(u_{-}-\frac{\beta}{\alpha})e^{-\alpha t}-\frac{1}{2} (u_{-}-u_{+})e^{-\alpha t}=u_{\delta}(t), \end{eqnarray*} $

$ \begin{eqnarray*} \lim\limits_{\gamma\rightarrow1}\sigma_{1}(t)&=& \frac{\beta}{\alpha}+(u_{-}-\frac{\beta}{\alpha})e^{-\alpha t}-\lim\limits_{\gamma\rightarrow1} \rho_{\ast}\sqrt{\frac{\frac{\gamma-1}{2}(\rho_{\ast}^{\gamma-1}-\rho_{-}^{\gamma-1})}{(\rho_{\ast} +\rho_{-})(\rho_{\ast}-\rho_{-})}}\\ &=&\frac{\beta}{\alpha}+(u_{-}-\frac{\beta}{\alpha})e^{-\alpha t}-\sqrt{a}=u_{\delta}(t), \end{eqnarray*} $

$ \begin{eqnarray*} \lim\limits_{\gamma\rightarrow1}\sigma_{2} (t)&=&\lim\limits_{\gamma\rightarrow1} (\frac{\beta}{\alpha}+(v_{\ast}-\frac{\beta}{\alpha})e^{-\alpha t})+\lim\limits_{\gamma\rightarrow1} \rho_{+}\sqrt{\frac{\frac{\gamma-1}{2}(\rho_{+}^{\gamma-1}-\rho_{\ast}^{\gamma-1})} {(\rho_{+}+\rho_{\ast})(\rho_{+}-\rho_{\ast})}} \\&=&\lim\limits_{\gamma\rightarrow1}(\frac{\beta}{\alpha}+(v_{\ast}-\frac{\beta}{\alpha}) e^{-\alpha t})=u_{\delta}(t), \end{eqnarray*} $

因此$ \lim\limits_{\gamma\rightarrow1}u_{\ast}= \lim\limits_{\gamma\rightarrow1} \sigma_{1}(t)=\lim\limits_{\gamma\rightarrow1}\sigma_{2}(t)=u_{\delta}(t). $

类似地, 由(4.7)式, (4.8)式和引理4.2, 可得

(4.13) $ \begin{eqnarray} \lim\limits_{\gamma\rightarrow1}\int^{\sigma_{2}(t)}_{\sigma_{1}(t)}\rho_{\ast}{\rm d}\xi &=&\lim\limits_{\gamma\rightarrow1}\rho_{\ast}(\sigma_{2}(t)-\sigma_{1}(t)){}\\ &=&\lim\limits_{\gamma\rightarrow1}(\rho_{\ast}(v_{\ast}-u_{-})e^{-\alpha t}+\rho_{\ast}\rho_{+}\sqrt{\frac{\frac{\gamma-1}{2}(\rho_{+}^{\gamma-1}-\rho_{\ast}^{\gamma-1})}{(\rho_{+}+\rho_{\ast})(\rho_{+}-\rho_{\ast})}} {} \\ & &+\rho_{\ast}^{2}\sqrt{\frac{\frac{\gamma-1}{2}(\rho_{\ast}^{\gamma-1}-\rho_{-}^{\gamma-1})}{(\rho_{\ast}+\rho_{-})(\rho_{\ast}-\rho_{-})}}) {}\\ & =&\lim\limits_{\gamma\rightarrow1}(-\rho_{\ast}(\rho_{\ast}-\rho_{-})\sqrt{\frac{\frac{\gamma-1}{2}(\rho_{\ast}^{\gamma-1}-\rho_{-}^{\gamma-1})}{(\rho_{\ast} +\rho_{-})(\rho_{\ast}-\rho_{-})}} {}\\ && +\rho_{\ast}\rho_{+}\sqrt{\frac{\frac{\gamma-1}{2}(\rho_{+}^{\gamma-1}-\rho_{\ast}^{\gamma-1})} {(\rho_{+}+\rho_{\ast})(\rho_{+}-\rho_{\ast})}} +\rho_{\ast}^{2}\sqrt{\frac{\frac{\gamma-1}{2}(\rho_{\ast}^{\gamma-1}-\rho_{-}^{\gamma-1})}{(\rho_{\ast}+\rho_{-})(\rho_{\ast}-\rho_{-})}}) {}\\ &=&\lim\limits_{\gamma\rightarrow1}(\rho_{\ast}\rho_{-}\sqrt{\frac{\frac{\gamma-1}{2}(\rho_{\ast}^{\gamma-1}-\rho_{-}^{\gamma-1})}{(\rho_{\ast} +\rho_{-})(\rho_{\ast}-\rho_{-})}} +\rho_{\ast}\rho_{+}\sqrt{\frac{\frac{\gamma-1}{2}(\rho_{+}^{\gamma-1}-\rho_{\ast}^{\gamma-1})}{(\rho_{+}+\rho_{\ast})(\rho_{+}-\rho_{\ast})}}) {} \\ &=&\rho_{-}\sqrt{a}+\rho_{+}\sqrt{a}=(\rho_{-}+\rho_{+})\frac{1}{2} (u_{-}-u_{+})e^{-\alpha t}. \end{eqnarray} $

直接计算, 有

(4.14) $ \begin{eqnarray} & &u_{\delta}(t)[\rho]-[\rho (\frac{\beta}{\alpha}+(v-\frac{\beta}{\alpha})e^{-\alpha t})]{}\\ &=&\frac{\beta}{\alpha}[\rho]+\frac{1}{2}(u_{-}+u_{+})e^{-\alpha t}[\rho]-\frac{\beta}{\alpha} e^{-\alpha t}[\rho]-[\rho (\frac{\beta}{\alpha}+(v-\frac{\beta}{\alpha})e^{-\alpha t})] {}\\ & =&\frac{1}{2}(u_{-}+u_{+})[\rho]e^{-\alpha t}-[\rho v]e^{-\alpha t}{}\\ & =&\big(\frac{1}{2}(u_{-}+u_{+})(\rho_{+}-\rho_{-})-\rho_{+}u_{+}+\rho_{-}u_{-}\big)e^{-\alpha t}{}\\ &=&\frac{1}{2}(\rho_{-}+\rho_{+})(u_{-}-u_{+})e^{-\alpha t}. \end{eqnarray} $

则由(4.13)和(4.14)式可立即得到(4.12)式. 证毕.

注4.1  从引理4.2 –4.3可以得出这样的结论: 当$ \gamma\rightarrow1 $时, 两条激波曲线$ S_{1} $和$ S_{2} $重合, 中间状态的密度$ \rho_{\ast} $变为奇异, $ \rho_{\ast} $的极限具有一个奇点, 该奇点是一个带有速度$ u_{\delta}(t) $的加权狄拉克$ \delta $函数.

注4.2  从引理4.3可以得出这样的结论: 当$ \gamma\rightarrow1 $时, 两条激波$ S_{1} $和$ S_{2} $的速度和系统(1.1)的中间$ u_{\ast} $收敛于$ u_{\delta}(t) $, 它决定了带阻尼和摩擦的零压气体欧拉方程组的$ \delta $ -激波解, 且两条激波之间的中间密度$ \rho_{\ast} $趋于一个加权的$ \delta $测度, 从而形成$ \delta $ -激波.

根据上面的分析, 我们得到了以下结果.

定理4.1  若$ u_{+}<u_{-} $, 当$ \gamma\rightarrow1 $时, 带有初值$ (\rho_{\pm}, u_{\pm}) $的系统(1.1), 在第3节构造的包含两条激波的黎曼解收敛于具有同样初值$ (\rho_{\pm}, u_{\pm}) $的系统(1.4) 的$ \delta $ -激波解.

引理4.4  如果$ u_{-}<u_{+}<u_{-}+2, $ 则存在$ \gamma_{1}>0 $, 使得$ (\rho_{+}, u_{+})\in I(\rho_{-}, u_{-}) $, 当$ 1<\gamma<1+\gamma_{1}. $

证  由(3.4)和(3.5)式可推导出通过1 -稀疏波$ R_{1} $或2 -稀疏波$ R_{2} $与左状态$ (\rho_{-}, u_{-}) $右连接的所有可能的状态$ (\rho, v) $满足

(4.15) $ \begin{equation} R_{1}(\rho_{-}, u_{-}): v+\rho^{\frac{\gamma-1}{2}}e^{\alpha t}=u_{-}+\rho_{-}^{\frac{\gamma-1}{2}}e^{\alpha t}, \, \, \, \, v>u_{-}, \, \, \, \rho<\rho_{-}, \end{equation} $

(4.16) $ \begin{equation} R_{2}(\rho_{-}, u_{-}): v-\rho^{\frac{\gamma-1}{2}}e^{\alpha t}=u_{-}-\rho_{-}^{\frac{\gamma-1}{2}}e^{\alpha t}, \, \, \, \, \, \, v>u_{-}, \, \, \, \rho>\rho_{-}. \end{equation} $

同样地, 由(3.5)式可知, 所有通过2 -稀疏波$ R_{2} $与左状态$ (0, \widetilde{v}_{\ast}) $右连接的状态$ (\rho, v) $满足

(4.17) $ \begin{equation} R_{2}(0, \widetilde{v}_{\ast}): v-\rho^{\frac{\gamma-1}{2}}e^{\alpha t}=u_{-}+\rho_{-}^{\frac{\gamma-1}{2}}e^{\alpha t}, \, \, \, \, \, \, v>u_{-}+\rho_{-}^{\frac{\gamma-1}{2}}e^{\alpha t}, \, \, \, \rho>0. \end{equation} $

若$ u_{-}<u_{+}<u_{-}+2 $, $ \rho_{+}\neq\rho_{-} $及$ (\rho_{+}, u_{+})\in I(\rho_{-}, u_{-}), $ 则结合(4.15)–(4.17)式, 由图 1可得

(4.18) $ \begin{equation} u_{+}>u_{-}+\rho_{-}^{\frac{\gamma-1}{2}}e^{\alpha t}-\rho_{+}^{\frac{\gamma-1}{2}}e^{\alpha t}, \, \, \rho_{+}<\rho_{-}, \end{equation} $

(4.19) $ \begin{equation} u_{+}>u_{-}-\rho_{-}^{\frac{\gamma-1}{2}}e^{\alpha t}+\rho_{+}^{\frac{\gamma-1}{2}}e^{\alpha t}, \, \, \rho_{+}>\rho_{-}, \end{equation} $

(4.20) $ \begin{equation} u_{+}<u_{-}+\rho_{-}^{\frac{\gamma-1}{2}}e^{\alpha t}+\rho_{+}^{\frac{\gamma-1}{2}}e^{\alpha t}, \, \, \rho_{+}>0. \end{equation} $

由(4.18)–(4.20)式, 可知

(4.21) $ \begin{equation} |(\rho_{-}^{\frac{\gamma-1}{2}}-\rho_{+}^{\frac{\gamma-1}{2}})e^{\alpha t}|<u_{+}-u_{-}<(\rho_{-}^{\frac{\gamma-1}{2}}+\rho_{+}^{\frac{\gamma-1}{2}})e^{\alpha t}, \, \, \, \, \, \, \, \, \, \, \rho_{+}>0, \, \rho_{-}>0. \end{equation} $

由于$ \lim\limits_{{\gamma\rightarrow1}}(\rho_{-}^{\frac{\gamma-1}{2}}-\rho_{+}^{\frac{\gamma-1}{2}})e^{\alpha t}=0<u_{+}-u_{-} $和$ \lim\limits_{{\gamma\rightarrow1}}(\rho_{-}^{\frac{\gamma-1}{2}}+\rho_{+}^{\frac{\gamma-1}{2}})e^{\alpha t}=2e^{\alpha t}>2>u_{+}-u_{-}, $ 因此, 存在足够小$ \gamma_{1}>0 $, 使得当$ 1<\gamma <1 +\gamma_{1} $时, 我们有不等式(4.21). 于是, 当$ 1<\gamma <1 +\gamma_{1} $时, 很明显$ (\rho_{+}, u_{+})\in I( \rho_{-}, u_{-}) $. 证毕.

当$ u_{-}<u_{+}<u_{-}+2, $ 由引理4.5, 对任意给定$ \gamma\in (1, 1+\gamma_{1}), $ 带有Riemann初值的系统(1.1)的Riemann解为

(4.22) $ \begin{equation} (\rho_{-}, \frac{\beta}{\alpha}+(u_{-}-\frac{\beta}{\alpha})e^{-\alpha t})+R_{1}+(\rho_{\ast}, \frac{\beta}{\alpha}+(v_{\ast}-\frac{\beta}{\alpha})e^{-\alpha t})+R_{2}+(\rho_{+}, \frac{\beta}{\alpha}+(u_{+}-\frac{\beta}{\alpha})e^{-\alpha t}), \end{equation} $

其中

(4.23) $ \begin{equation} R_{1}:\, \, \left\{\begin{array}{ll} { } \frac{{\rm d}x}{{\rm d}t}=\lambda_{1}=\frac{\beta}{\alpha}+(v-\frac{\beta}{\alpha})e^{-\alpha t}-\frac{\gamma-1}{2}\rho^{\frac{\gamma-1}{2}}, \\ { } v+\rho^{\frac{\gamma-1}{2}}e^{\alpha t}=u_{-}+\rho_{-}^{\frac{\gamma-1}{2}}e^{\alpha t}, \, \, \, \, \rho_{\ast}\leq\rho\leq\rho_{-}, \end{array} \right. \end{equation} $

(4.24) $ \begin{equation} R_{2}:\, \, \left\{\begin{array}{ll} { } \frac{{\rm d}x}{{\rm d}t}=\lambda_{2}=\frac{\beta}{\alpha}+(v-\frac{\beta}{\alpha})e^{-\alpha t}+\frac{\gamma-1}{2}\rho^{\frac{\gamma-1}{2}}, \\ { } v-\rho^{\frac{\gamma-1}{2}}e^{\alpha t}=u_{+}-\rho_{+}^{\frac{\gamma-1}{2}}e^{\alpha t}, \, \, \, \, \rho_{\ast}\leq\rho\leq\rho_{+}.\end{array} \right. \end{equation} $

因此, 从(4.23)和(4.24)式可得

(4.25) $ \begin{equation} u_{+}-u_{-}=\rho_{+}^{\frac{\gamma-1}{2}}e^{\alpha t}+\rho_{-}^{\frac{\gamma-1}{2}}e^{\alpha t}-2\rho_{\ast}^{\frac{\gamma-1}{2}}e^{\alpha t}, \, \, \, \, \, \, \, \, \, \, \, \, \, \rho_{\ast}\leq\rho_{\pm}. \end{equation} $

当$ \gamma\rightarrow1 $时, 真空状态出现.

定理4.2  如果$ u_{-}<u_{+}< u_{-}+2 $. 对任意$ \gamma \in(1, 2) $, 假设$ (\rho, u)(t, x) $是带有初值$ (\rho_{\pm}, u_{\pm}) $的系统(1.1)在第三节构造的包含两条疏散波的黎曼解. 则当$ \gamma\rightarrow1 $时, 真空状态出现, 且两条疏散波变成连接$ (\rho_{\pm}, \frac{\beta}{\alpha}+(u_{\pm}-\frac{\beta}{\alpha})e^{-\alpha t}) $和真空状态$ (\rho =0) $的两条接触间断, 其中真空状态是具有相同初值$ (\rho_{\pm}, u_{\pm}) $的系统(1.4)的真空解.

证  若$ \lim\limits_{\gamma\rightarrow 1}\rho_{\ast}=K\in(0, \min(\rho_{-}, \rho_{+})), $ 则在(4.25)中令$ \gamma\rightarrow 1 $取极限, 可得$ u_{+}=u_{-} $, 这与$ u_{-}<u_{+} $矛盾. 因此$ \lim\limits_{\gamma\rightarrow 1}\rho_{\ast}=0 $, 即当$ \gamma\rightarrow 1 $时, 出现真空状态. 进一步地, 由(4.23)和(4.24)式可得

(4.26) $ \begin{equation} \lim\limits_{\gamma\rightarrow 1}v=u_{-} \, \, {\rm on}\, \, R_{1}, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \lim\limits_{\gamma\rightarrow 1}v=u_{+} \, \, {\rm on}\, \, R_{2}, \end{equation} $

(4.27) $ \begin{equation} \left\{ \begin{array}{ll} { } \lambda_{1}=\frac{\beta}{\alpha}+(\frac{\gamma+1}{2}v-\frac{\gamma-1}{2} u_{-}-\frac{\beta}{\alpha})e^{-\alpha t}-\frac{\gamma-1}{2}\rho_{-}^{\frac{\gamma-1}{2}}, \\ { } \lambda_{2}=\frac{\beta}{\alpha}+(\frac{\gamma+1}{2}v-\frac{\gamma-1}{2} u_{+}-\frac{\beta}{\alpha})e^{-\alpha t}+\frac{\gamma-1}{2}\rho_{-}^{\frac{\gamma-1}{2}}. \end{array} \right. \end{equation} $

从(4.26)和(4.27)式可得

(4.28) $ \begin{equation} \lim\limits_{\gamma\rightarrow 1}\lambda_{1}= \frac{\beta}{\alpha}+(u_{-}-\frac{\beta}{\alpha})e^{-\alpha t}, \, \, \, \, \, \, \, \, \, \, \, \lim\limits_{\gamma\rightarrow 1}\lambda_{2} =\frac{\beta}{\alpha}+(u_{+}-\frac{\beta}{\alpha})e^{-\alpha t}. \end{equation} $

证毕.

引理4.5  如果$ u_{+}>u_{-}+2. $ 则

$ (1) $当$ t<\frac{1}{\alpha}\ln(\frac{u_{+}-u_{-}}{2}) $时, 存在$ \gamma_{2}>0 $, 使得$ (\rho_{+}, u_{+})\in V(\rho_{-}, u_{-}) $当$ 1<\gamma<1+\gamma_{2}; $

$ (2) $当$ t>\frac{1}{\alpha}\ln(\frac{u_{+}-u_{-}}{2}) $时, 存在$ \gamma_{3}>0 $, 使得$ (\rho_{+}, u_{+})\in I(\rho_{-}, u_{-}) $当$ 1<\gamma<1+\gamma_{3}; $

$ (3) $当$ t=\frac{1}{\alpha}\ln(\frac{u_{+}-u_{-}}{2}) $, $ \rho_{-}<1 $和$ \rho_{+}<1 $时, 对任意给定$ \gamma\in (1, 2), $ 有$ (\rho_{+}, u_{+})\in V(\rho_{-}, u_{-}) $;

$ (4) $当$ t=\frac{1}{\alpha}\ln(\frac{u_{+}-u_{-}}{2}), $$ \rho_{-}>1 $和$ \rho_{+}>1 $时, 存在$ \gamma_{4}>0 $, 使得$ (\rho_{+}, u_{+})\in I(\rho_{-}, u_{-}) $当$ 1<\gamma<1+\gamma_{4}. $

证  若$ u_{+}>u_{-}+2 $和$ (\rho_{+}, u_{+})\in V(\rho_{-}, u_{-}), $ 则结合(4.17)式及图 1知

(4.29) $ \begin{equation} u_{+}>u_{-}+(\rho_{-}^{\frac{\gamma-1}{2}}+\rho_{+}^{\frac{\gamma-1}{2}})e^{\alpha t}, \, \, \rho_{+}>0. \end{equation} $

(1) 当$ t<\frac{1}{\alpha}\ln(\frac{u_{+}-u_{-}}{2}) $时, 由于

$ \lim\limits_{{\gamma\rightarrow1}}(u_{-}+(\rho_{-}^{\frac{\gamma-1}{2}}+\rho_{+}^{\frac{\gamma-1}{2}})e^{\alpha t})=u_{-}+2e^{\alpha t}<u_{-}+2e^{\ln(\frac{u_{+}-u_{-}}{2})}=u_{+}, $

因此, 存在足够小$ \gamma_{2}>0 $, 使得当$ 1<\gamma <1 +\gamma_{2} $时, 我们有不等式(4.29). 于是, 当$ 1<\gamma <1 +\gamma_{2} $时, 显然$ (\rho_{+}, u_{+})\in V( \rho_{-}, u_{-}) $.

(2) 当$ t>\frac{1}{\alpha}\ln(\frac{u_{+}-u_{-}}{2}) $时, 由于

$ \lim\limits_{{\gamma\rightarrow1}}(\rho_{-}^{\frac{\gamma-1}{2}}+\rho_{+}^{\frac{\gamma-1}{2}})e^{\alpha t}=2e^{\alpha t}>2e^{\ln(\frac{u_{+}-u_{-}}{2})}=u_{+}-u_{-}, $

$ \lim\limits_{{\gamma\rightarrow1}}(\rho_{-}^{\frac{\gamma-1}{2}}-\rho_{+}^{\frac{\gamma-1}{2}})e^{\alpha t}=0<u_{+}-u_{-}, $

因此, 存在足够小$ \gamma_{3}>0 $, 使得当$ 1<\gamma <1 +\gamma_{3} $时, 我们有不等式(4.21). 于是, 当$ 1<\gamma <1 +\gamma_{3} $时, 显然$ (\rho_{+}, u_{+})\in I( \rho_{-}, u_{-}) $.

(3) 当$ t=\frac{1}{\alpha}\ln(\frac{u_{+}-u_{-}}{2}), $$ \rho_{-}<1 $和$ \rho_{+}<1 $时, 由于

$ u_{-}+(\rho_{-}^{\frac{\gamma-1}{2}}+\rho_{+}^{\frac{\gamma-1}{2}})e^{\alpha t}=u_{-}+(\rho_{-}^{\frac{\gamma-1}{2}}+\rho_{+}^{\frac{\gamma-1}{2}}) \frac{u_{+}-u_{-}}{2}<u_{-}+u_{+}-u_{-}=u_{+}, $

因此, 当$ 1<\gamma <2 $时, 显然$ (\rho_{+}, u_{+})\in V( \rho_{-}, u_{-}) $.

(4) 当$ t=\frac{1}{\alpha}\ln(\frac{u_{+}-u_{-}}{2}), $$ \rho_{-}>1 $和$ \rho_{+}>1 $时, 由于

$ (\rho_{-}^{\frac{\gamma-1}{2}}+\rho_{+}^{\frac{\gamma-1}{2}})e^{\alpha t}=(\rho_{-}^{\frac{\gamma-1}{2}}+\rho_{+}^{\frac{\gamma-1}{2}}) \frac{u_{+}-u_{-}}{2}>2\times\frac{u_{+}-u_{-}}{2}=u_{+}-u_{-} $

和

$ \lim\limits_{{\gamma\rightarrow1}}(\rho_{-}^{\frac{\gamma-1}{2}}-\rho_{+}^{\frac{\gamma-1}{2}})e^{\alpha t}=\lim\limits_{{\gamma\rightarrow1}}(\rho_{-}^{\frac{\gamma-1}{2}}-\rho_{+}^{\frac{\gamma-1}{2}})\frac{u_{+}-u_{-}}{2}=0<u_{+}-u_{-}, $

因此, 存在足够小$ \gamma_{4}>0 $, 使得当$ 1<\gamma <1 +\gamma_{4} $时, 我们有不等式(4.21). 于是, 当$ 1<\gamma <1 +\gamma_{4} $时, 很明显$ (\rho_{+}, u_{+})\in I( \rho_{-}, u_{-}) $. 证毕.

当$ u_{+}>u_{-}+2 $时, 为了简单和不失一般性, 我们可以假设对于任意给定的$ \gamma\in (1, 1+\gamma_{2}), $$ (\rho_{+}, u_{+})\in V( \rho_{-}, u_{-}) $. 那么, 具有黎曼初值$ (\rho_{\pm}, u_{\pm}) $的系统(1.1) 的黎曼解是两个稀疏波$ R_{1} $, $ R_{2} $和它们之间的中间真空状态(用vac表示), 可以表示为

(4.30) $ \begin{equation} (\rho_{-}, \frac{\beta}{\alpha}+(u_{-}-\frac{\beta}{\alpha})e^{-\alpha t})+R_{1}+\mathrm {Vac}+R_{2}+(\rho_{+}, \frac{\beta}{\alpha}+(u_{+}-\frac{\beta}{\alpha})e^{-\alpha t}), \end{equation} $

其中$ R_{1} $和$ R_{2} $为

(4.31) $ \begin{equation} R_{1}:\, \, \left\{\begin{array}{ll} { } \frac{{\rm d}x}{{\rm d}t}=\lambda_{1}=\frac{\beta}{\alpha}+(v-\frac{\beta}{\alpha})e^{-\alpha t}-\frac{\gamma-1}{2}\rho^{\frac{\gamma-1}{2}}, \\ { } v+\rho^{\frac{\gamma-1}{2}}e^{\alpha t}=u_{-}+\rho_{-}^{\frac{\gamma-1}{2}}e^{\alpha t}, \, \, \, \, 0\leq\rho\leq\rho_{-} \end{array} \right. \end{equation} $

和

(4.32) $ \begin{equation} R_{2}:\, \, \left\{\begin{array}{ll} { } \frac{{\rm d}x}{{\rm d}t}=\lambda_{2}^{\gamma}=\frac{\beta}{\alpha}+(v-\frac{\beta}{\alpha})e^{-\alpha t}+\frac{\gamma-1}{2}\rho^{\frac{\gamma-1}{2}}, \\ { } v-\rho^{\frac{\gamma-1}{2}}e^{\alpha t}=u_{+}-\rho_{+}^{\frac{\gamma-1}{2}}e^{\alpha t}, \, \, \, \, 0\leq\rho\leq\rho_{+};\end{array} \right. \end{equation} $

两个稀疏波之间的真空状态为$ (0, \frac{\beta}{\alpha}+(v_{\ast}-\frac{\beta}{\alpha})e^{-\alpha t}) $且$ v_{1}\leq v_{\ast}\leq v_{2} $, 其中

(4.33) $ \begin{equation} v_{1}=u_{-}+\rho_{-}^{\frac{\gamma-1}{2}}e^{\alpha t}, \, \, \, \, \, \, \, \, \, \, \, \, v_{2}=u_{+}-\rho_{+}^{\frac{\gamma-1}{2}}e^{\alpha t}. \end{equation} $

我们可以看到真空状态出现在黎曼解中, 位于两个稀疏波之间.

从(4.31)式中, 很容易看到

$ \lim\limits_{\gamma\rightarrow 1}\lambda_{1}=\frac{\beta}{\alpha}+(u_{-}-\frac{\beta}{\alpha})e^{-\alpha t}, $

这意味着当$ \gamma\rightarrow 1 $时, 稀疏波$ R_{1} $将退化为接触不连续$ J_{1} $, 其中$ J_{1} $的位置由

$ x_{1}(t)=\frac{\beta}{\alpha}t+\frac{1}{\alpha}(u_{-}-\frac{\beta}{\alpha})(1 -e^{-\alpha t}) $

给出, 且$ J_{1} $的传播速度为

$ \sigma_{1}(t)=\frac{\beta}{\alpha}+(u_{-}-\frac{\beta}{\alpha})e^{-\alpha t}. $

类似地, 从(4.32)式, 很容易看到

$ \lim\limits_{\gamma\rightarrow 1}\lambda_{2}=\frac{\beta}{\alpha}+(u_{+}-\frac{\beta}{\alpha})e^{-\alpha t}, $

这意味着当$ \gamma\rightarrow 1 $时, 稀疏波$ R_{2} $将退化为接触不连续$ J_{2} $, 其中$ J_{2} $的位置由

$ x_{2}(t)=\frac{\beta}{\alpha}t+\frac{1}{\alpha}(u_{+}-\frac{\beta}{\alpha}) (1 -e^{-\alpha t}) $

给出, 而$ J_{2} $的传播速度为

$ \sigma_{2}(t)=\frac{\beta}{\alpha}+ (u_{+}-\frac{\beta}{\alpha})e^{-\alpha t}. $

因此, 我们证明了以下结果.

定理4.3  如果$ u_{+}>u_{-}+2 $和$ (\rho_{+}, u_{+})\in V( \rho_{-}, u_{-}) $. 对任意给定的$ \gamma\in (1, 1+\gamma_{2}), $ 假设$ (\rho_{\gamma}, u_{\gamma})(t, x) $是带有黎曼初值$ (\rho_{\pm}, u_{\pm}) $的系统(1.1)在第3节构造的包含两条稀疏波的黎曼解. 则当$ \gamma\rightarrow1 $时, 黎曼解趋于两条连接状态$ (\rho_{\pm}, \frac{\beta}{\alpha}+(u_{\pm}-\frac{\beta}{\alpha})e^{-\alpha t}) $的接触间断且两条接触间断之间的真空状态, 具体为

(4.34) $ \begin{align} \lim\limits_{ \gamma\rightarrow 1}(\rho_{\gamma}, u_{\gamma})(t, x)=\left\{\begin{array}{ll} { } (\rho_{-}, \frac{\beta}{\alpha}+(u_{-}-\frac{\beta}{\alpha})e^{-\alpha t}), \, \, \, \, x< \frac{\beta}{\alpha}t+\frac{1}{\alpha}(u_{-}-\frac{\beta}{\alpha})(1 -e^{-\alpha t}), \\ { } {\rm Vac}, \, \, \, \, \frac{\beta}{\alpha}t+\frac{1}{\alpha}(u_{-}-\frac{\beta}{\alpha})(1 -e^{-\alpha t}) \leq x \leq \frac{\beta}{\alpha}t+\frac{1}{\alpha}(u_{+}-\frac{\beta}{\alpha})(1 -e^{-\alpha t}), \\ { } (\rho_{+}, \frac{\beta}{\alpha}+(u_{+}-\frac{\beta}{\alpha})e^{-\alpha t}), \, \, \, \, x>\frac{\beta}{\alpha}t+\frac{1}{\alpha}(u_{+}-\frac{\beta}{\alpha})(1 -e^{-\alpha t}), \end{array} \right. \end{align} $

这正是具有相同黎曼初值$ (\rho_{\pm}, u_{\pm}) $的无压气体动力学系统(1.4)的黎曼解.

本文给出并研究了具有复合源项且绝热指数趋于1的可压缩流体流动欧拉方程的黎曼解的极限. 源可以表示类库仑摩擦或阻尼, 或两者兼有. 我们发现系统(1.1)中的复合源项具有使特征曲线为抛物曲线的能力, 使系统(1.1)的黎曼解中的所有稀疏波和激波都是弯曲的. 因此, 系统(1.1)的黎曼解不再是自相似的. 同样, 在复合源项的影响下, 系统(1.4) 的黎曼解中的$ \delta $ - 激波也弯曲成$ (t, x) $平面上的抛物线. 具体来说, 作者遵循一条很好的研究路径: 通过变量代换去除源(以引入非自相似解为代价), 并在包括压力的情况下将解与无压系统的解进行比较. 当前的结果从物理角度揭示了出现的$ \delta $ -激波可以用来解释质量集中现象.