一维非等熵欧拉方程组可描述为

(1.1) $ \begin{equation} \left\{ \begin{array}{ll} &\rho_t+(\rho u)_x = 0, \\ &(\rho u)_t+(\rho u^{2}+p)_x = 0, \\ &(\rho E)_t+(\rho uE+pu)_x = 0, \end{array} \right. \end{equation} $

其中$ \rho $, $ u $, $ p $, $ E = \frac{1}{2}u^{2}+e $,分别表示密度,速度,压力,能量.当状态方程满足

(1.2) $ \begin{equation} p(\rho, s) = g(s)-\frac{1}{\rho}f(s), \end{equation} $

$ f(s)>0 $, $ g(s) $是熵$ s $的函数,且$ f'(s)>0 $时, (1.1)–(1.2)式被称为Chaplygin气体方程.由热力学方程

(1.3) $ \begin{equation} T{\rm d}s = {\rm d}e+p{\rm d}\frac{1}{\rho}, \end{equation} $

计算可得

(1.4) $ \begin{equation} (T+\frac{1}{\rho}g'(s)-\frac{1}{2\rho^{2}}f'(s)){\rm d}s = {\rm d}(e+\frac{1}{\rho}g(s)-\frac{1}{2\rho^{2}}f(s)). \end{equation} $

即,存在熵$ s $的函数$ h(s) $使得

(1.5) $ \begin{equation} T = \frac{1}{2\rho^{2}}f'(s)-\frac{1}{\rho}g'(s)+h'(s), \end{equation} $

(1.6) $ \begin{equation} e = \frac{1}{2\rho^{2}}f(s)-\frac{1}{\rho}g(s)+h(s), \end{equation} $

其中$ T $表示温度, $ f'(s)>0, h'(s)>0, g'^2(s)<2f'(s)h'(s) $ (参见文献[1]).

这类气体方程由Chaplygin和Tsien分别在文献[2]和[3]中引入,用于计算在空气动力学中飞机机翼的升力的数学近似.关于Chaplygin气体更多的物理背景,可以参考文献[1, 4-5].在数学上, Chaplygin气体是一类典型的线性退化模型,在经典的弱解意义下非全局可解. 2005年, Brenier首先分析给出等熵Chaplygin气体黎曼问题的$ \delta $ -波解(见文献[6]).随后,文献[7]中利用广义特征分析的方法,详细讨论了初始含有狄拉克测度的广义黎曼问题,并构造了该问题的解.屈和王在文献[8]中证明了该模型等熵时黎曼问题解的稳定性.对于非等熵情形,文献[9]中分析构造压力及内能分别为$ p = \frac{1}{\rho} $, $ e = \frac{1}{2\rho^2}+f(s) $时,密度和内能均含有狄拉克函数的$ \delta $激波,并给出了数值分析.通过对经典黎曼初始扰动,文献[10]中得到初始含有$ \delta $初值问题与经典黎曼问题的本质区别.有关更多的$ \delta $激波理论,可以参看文献[11-14]及其引用文献.

本文考虑方程组(1.1)的黎曼问题,黎曼初值如下

(1.7) $ \begin{equation} (\rho, u, s)(x, t)|_{t = 0} = \left\{ \begin{array}{ll} &(\rho_l, u_l, s_l), x<0, \\ &(\rho_r, u_r, s_r), x>0. \\ \end{array} \right. \end{equation} $

利用特征分析的方法,构造出经典黎曼问题的解.但是,在一般弱解意义下,初始的取值范围限制了弱解的存在性.该文通过Radon测度重新定义弱解^[15],推导了广义Rankine-Hugoniot条件,并构造了经典熵条件下的$ \delta $波解.

本节我们利用特征分析的方法,计算非等熵Chaplygin气体的基本波,并构造经典黎曼问题的弱解.在光滑解意义下,当$ e_s\neq0 $时,方程(1.1)可以等价为

(2.1) $ \begin{equation} \left(\begin{array}{ccc} \rho\\ u\\ s \end{array}\right)_t + \left(\begin{array}{ccc} u&\rho&0\\ { } \frac{f(s)}{\rho^3}&{\quad} u{\quad} &{ } \frac{g'(s)}{\rho}-\frac{f'(s)}{\rho^2}\\ 0&0&u \end{array}\right) \left(\begin{array}{ccc} \rho\\ u\\ s \end{array}\right)_x = 0, \end{equation} $

则系数矩阵的三个特征值分别

(2.2) $ \begin{equation} \lambda_1 = u-\frac{\sqrt{f(s)}}{\rho}, \lambda_2 = u, \lambda_3 = u+\frac{\sqrt{f(s)}}{\rho}, \end{equation} $

相应的右特征向量为

(2.3) $ \begin{equation} \overrightarrow r_1 = (1, -\frac{\sqrt{f(s)}}{\rho^2}, 0)^T, \overrightarrow r_2 = (1, 0, 0)^T, \overrightarrow r_3 = (1, \frac{\sqrt{f(s)}}{\rho^2}, 0)^T, \end{equation} $

由$ \nabla\lambda_i\cdot\overrightarrow r_i = 0\ (i = 1, 2, 3) $,该模型的特征均是线性退化的^[16].

由于该模型具有尺度不变性,我们首先考虑自相似解$ (\rho, u, s)(\xi) (\xi = \frac{x}{t}) $,代入方程(1.1)得

(2.4) $ \begin{equation} \left\{ \begin{array}{ll} &-\xi\rho_{\xi}+(\rho u)_{\xi} = 0, \\ &-\xi(\rho u)_\xi+(\rho u^{2}+p) = 0, \\ &-\xi(\rho E)_\xi+(\rho uE+pu)_\xi = 0. \end{array} \right. \end{equation} $

对于光滑解,方程(2.4)可以表示为

(2.5) $ \begin{equation} \left(\begin{array}{ccc} u-\xi & \rho & 0\\ { } \frac{f(s)}{\rho^3}&{\quad} u-\xi{\quad} &{ } \frac{g'(s)}{\rho}-\frac{f'(s)}{\rho^2}\\ 0 & 0 & u-\xi \end{array}\right) \left(\begin{array}{ccc} \rho\\ u\\ s \end{array}\right)_{\xi} = 0. \end{equation} $

于是,后向稀疏波满足

(2.6) $ \begin{equation} \left\{ \begin{array}{ll} s = s_l, \\ { } u = u_l+\sqrt{f(s)}(\frac{1}{\rho}-\frac{1}{\rho_l}), \rho < \rho_l, \end{array} \right. \end{equation} $

前向稀疏波满足

(2.7) $ \begin{equation} \left\{ \begin{array}{ll} s = s_l, \\ { } u = u_l-\sqrt{f(s)}(\frac{1}{\rho}-\frac{1}{\rho_l}), \rho > \rho_l. \end{array} \right. \end{equation} $

接下来我们将考虑该模型的有界间断解.设间断线为$ \sigma = \frac{{\rm d}x}{{\rm d}t} $,根据Rankine-Hugoniot条件^[16],该间断解满足

(2.8) $ \begin{equation} \left\{ \begin{array}{ll} -\sigma[\rho]+[\rho u] = 0, \\ -\sigma[\rho u]+[\rho u^{2}+p] = 0, \\ -\sigma[\rho E]+[\rho u E+pu] = 0. \end{array} \right. \end{equation} $

这里“$ [\ ] $”表示跨过间断的跳跃.例如, $ [\rho] = \rho_r-\rho_l $,其中$ \rho_l $与$ \rho_r $分别表示$ \rho $在间断处的左右状态.方程(2.8)也可写成

(2.9) $ \begin{equation} \left\{ \begin{array}{ll} (u_r-\sigma)[\rho]+\rho_l[u] = 0, \\ { } \frac{f(s_r)}{\rho_l\rho_r}[\rho]+\rho_r(u_r-\sigma)[u]-\frac{1}{\rho_l}[f(s)]+[g(s)] = 0, \\ { } (u_r-\sigma)(-\frac{\rho_l+\rho_r}{2\rho_l}[g(s)]+\frac{1}{2\rho_l}[f(s)]+\rho_r[h(s)]) = 0, \end{array} \right. \end{equation} $

其中,交换下标$ r $和$ l $的位置,方程组(2.9)仍然成立.为了得到非平凡解,假设方程组(2.9)系数矩阵的秩小于$ 3 $,则可求得接触间断满足

(2.10) $ \begin{equation} \sigma = u_l = u_r, [u] = [p] = 0, [\rho]\neq0, \end{equation} $

后向激波满足

(2.11) $ \begin{equation} \left\{ \begin{array}{ll} s_r = s_l, \\ { } u_r = u_l+\sqrt{f(s)}(\frac{1}{\rho_r}-\frac{1}{\rho_l}), \rho_r>\rho_l. \end{array} \right. \end{equation} $

前向激波满足

(2.12) $ \begin{equation} \left\{ \begin{array}{ll} s_r = s_l, \\ { } u_r = u_l-\sqrt{f(s)}(\frac{1}{\rho_r}-\frac{1}{\rho_l}), \rho_r<\rho_l. \end{array} \right. \end{equation} $

综合上述分析, Riemann问题(1.7)的解可构造如下

(2.13) $ \begin{equation} (\rho, u, s)(x, t) = \left\{ \begin{array}{ll} { } (\rho_l, u_l, s_l), \quad x<(u_l-\frac{\sqrt{f(s_l)}}{\rho_l})t, \\ { } (\rho_1, u, s_l), \quad (u_l-\frac{\sqrt{f(s_l)}}{\rho_l})t<x<ut, \\ { } (\rho_2, u, s_r), \quad ut<x<(u_r+\frac{\sqrt{f(s_r)}}{\rho_r})t, \\ { } (\rho_r, u_r, s_r), \quad (u_r+\frac{\sqrt{f(s_r)}}{\rho_r})t<x, \end{array} \right. \end{equation} $

其中$ \rho_1, \rho_2, u $满足

(2.14) $ \begin{equation} \left\{\begin{array}{ll} { } \rho_1 = \frac{\sqrt{f(s_l)}(\sqrt{f(s_l)}+\sqrt{f(s_r)})}{\sqrt{f(s_r)}(u_r+\frac{\sqrt{f(s_r)}}{\rho_r}-u_l+\frac{\sqrt{f(s_l)}}{\rho_l})+g(s_l)-g(s_r)}, \\ [7mm] { } \rho_2 = \frac{\sqrt{f(s_r)}(\sqrt{f(s_l)}+\sqrt{f(s_r)})}{\sqrt{f(s_l)}(u_r+\frac{\sqrt{f(s_r)}}{\rho_r}-u_l+\frac{\sqrt{f(s_l)}}{\rho_l})-g(s_l)+g(s_r)}, \\ [7mm] { } u = u_l-\frac{\sqrt{f(s_l)}}{\rho_l}+\frac{\sqrt{f(s_r)}(u_r+\frac{\sqrt{f(s_r)}}{\rho_r}-u_l+\frac{\sqrt{f(s_l)}}{\rho_l})+g(s_l)-g(s_r)}{\sqrt{f(s_l)}+\sqrt{f(s_r)}}. \end{array}\right. \end{equation} $

易见,若使得密度$ \rho_1, \rho_2>0 $,初始条件应满足

(2.15) $ \begin{equation} u_r+\frac{\sqrt{f(s_r)}}{\rho_r}-u_l+\frac{\sqrt{f(s_l)}}{\rho_l}>\max \bigg\{\frac{g(s_r)-g(s_l)}{\sqrt{f(s_r)}}, \frac{g(s_l)-g(s_r)}{\sqrt{f(s_l)}}\bigg\}. \end{equation} $

因此,我们有如下结论.

引理2.1 方程组(1.1), (1.2), (1.6)的经典黎曼问题弱解满足(2.13)–(2.14)式的充要条件为初始条件(1.7)满足条件(2.15).

注2.1 上述引理表明,当初值不满足条件(2.15)时,该方程的经典黎曼问题无解.特别地,当初始条件满足

$ u_r+\frac{\sqrt{f(s_r)}}{\rho_r}-u_l+\frac{\sqrt{f(s_l)}}{\rho_l} \rightarrow \max\bigg\{\frac{g(s_r)-g(s_l)}{\sqrt{f(s_r)}}, \frac{g(s_l)-g(s_r)}{\sqrt{f(s_l)}}\bigg\}, $

则有$ \rho_1\rightarrow \infty $或$ \rho_2\rightarrow \infty $,对应有$ u\rightarrow u_l-\frac{\sqrt{f(s_l)}}{\rho_l} $或$ u\rightarrow u_r+\frac{\sqrt{f(s_r)}}{\rho_r} $.

本节将利用测度值解的定义,推导广义Rankine-Hugoniot条件,结合熵条件,证明一般Riemann问题的弱解存在性.为了证明的需要,我们首先引入测度论的一些基本概念.假设$ {\cal B} $是$ {{\Bbb R}} ^{2} $上的博雷尔$ \sigma $ -代数, $ m $是$ {\cal B} $上的Radon测度,则$ m $可以看成是$ C_0({{\Bbb R}} ^{2}) $空间上的有界线性泛函,满足

(3.1) $ \begin{equation} \langle m, \phi\rangle = \int_{{{\Bbb R}} ^{2}}\phi(x, y)m({\rm d}x{\rm d}t), \end{equation} $

其中$ \phi\in C_0({{\Bbb R}} ^{2}) $是任意检验函数.

定义3.1 对于Lipschitz曲线$ L = (x(s), t(s):\alpha\leq s \leq \beta) $上的带权$ \delta $函数$ w(s)\delta_L $定义为

(3.2) $ \begin{equation} \langle w_L\delta_L, \phi\rangle = \int_{\alpha}^{\beta}w_L(s)\phi(x(s), t(s)) {\rm d}s, \forall\phi\in C_0({{\Bbb R}} ^{2}). \end{equation} $

下面我们将给出一般Riemann问题的Radon测度值解的定义.

定义3.2 令$ m^{i}, n^{i}\;(i = 0, 1, 2), \wp^{i}\;(i = 1, 2) $为$ \bar \Omega $上的$ \rm{Radon} $测度,称$ (\rho, u, s) $为问题$ (1.1) $的测度值解,当如下条件成立

i)对任意的$ \phi\in C_0({{\Bbb R}} ^{2}) $,下式成立

(3.3) $ \begin{equation} \left\{ \begin{array}{ll} { }\langle m^{0}, \phi_t\rangle +\langle n^{0}, \phi_x\rangle+\int_{{{\Bbb R}} }\rho_0\phi(0, x){\rm d}x = 0, \\ { }\langle m^{1}, \phi_t\rangle +\langle n^{1}, \phi_x\rangle+\langle \wp^{1}, \phi_{x}\rangle+\int_{{{\Bbb R}} }\rho_0u_0\phi(0, x){\rm d}x = 0, \\ { } \langle m^{2}, \phi_t\rangle +\langle n^{2}, \phi_x\rangle+\langle \wp^{2}, \phi_{x}\rangle+\int_{{{\Bbb R}} }\rho_0E_0\phi(0, x){\rm d}x = 0. \end{array} \right. \end{equation} $

ii) $ m^{0} $是非负的$ \rm{Radon} $测度,满足$ n^{0}\ll m^{0}, (m^k, n^k)\ll (m^{0}, n^{0})\;(k = 1, 2) $,且关于$ m^{0} $的$ \rm{Radon-Nikodym} $导数满足

(3.4) $ \begin{equation} \begin{array}{ll} { } u = \frac{n^{0}({\rm d}x{\rm d}t)}{m^{0}({\rm d}x{\rm d}t)} = \frac{m^{1}({\rm d}x{\rm d}t)}{m^{0}({\rm d}x{\rm d}t)}, \\ { } E = \frac{m^{2}({\rm d}x{\rm d}t)}{m^{0}({\rm d}x{\rm d}t)} \quad {\rm a.e., } \end{array} \end{equation} $

其中$ \lambda\ll \mu $表示测度$ \lambda $关于非负测度$ \mu $是绝对连续的.

iii)间断处熵条件成立.

基于上述定义,我们可以得到如下结论.

定理3.1 方程组(1.1), (1.2), (1.6)的一般$ \rm{Riemann} $问题(1.7)存在测度值解.

证 令$ I_\Omega $为区域$ \Omega\subset{{\Bbb R}} ^{2} $上的特征函数. $ {\cal L}^{2} $是$ {{\Bbb R}} ^{2} $平面上的勒贝格测度.现假设有如下的测度值解

(3.5) $ \begin{equation} \begin{array}{ll} &m^{0} = \rho_{0}I_{\Omega}{\cal L}^{2}+w_{m}^{0}(t)\delta_w, n^{0} = \rho_0u_0I_\Omega{\cal L}^{2}+w_{n}^{0}(t)\delta_w, \\ &m^{1} = \rho_{0}u_0I_{\Omega}{\cal L}^{2}+w_{m}^{1}(t)\delta_w, n^{1} = \rho_0u_0^{2}I_\Omega{\cal L}^{2}+w_{n}^{1}(t)\delta_w, \\ &m^{2} = \rho_{0}E_0I_{\Omega}{\cal L}^{2}+w_{m}^{2}(t)\delta_w, n^{2} = \rho_0u_0E_0I_\Omega{\cal L}^{2}+w_{n}^{2}(t)\delta_w, \\ &\wp^{1} = p_0I_\Omega{\cal L}^{2}, \wp^{2} = u_0p_0I_\Omega{\cal L}^{2}. \end{array} \end{equation} $

根据(3.3)式中的第一个式子,有

(3.6) $ \begin{eqnarray} 0& = &\langle m^{0}, \phi_t\rangle +\langle n^{0}, \phi_x\rangle+\int_{{{\Bbb R}} }\rho_0\phi(0, x){\rm d}x{}\\ & = &{\int\!\!\!\int}_\Omega\rho_0\phi_t{\rm d}x{\rm d}t+\int_{0}^{+\infty}w_{m}^{0}(t)\phi_t{\rm d}t +{\int\!\!\!\int}_\Omega\rho_0u_0\phi_x{\rm d}x{\rm d}t{}\\ & & +\int_{0}^{+\infty}w_{n}^{0}(t)\phi_{x}{\rm d}t+\int_{t = 0}\rho_0\phi(0, x){\rm d}x{}\\ & = &\int_{0}^{+\infty}\phi([\rho]\frac{{\rm d}x}{{\rm d}t}-[\rho u]){\rm d}t+\int_{0}^{+\infty}w_{m}^{0}{\rm d}\phi{}\\ & & -\int_{0}^{+\infty}w_{m}^{0}\phi_x{\rm d}x+\int_{0}^{+\infty}w_{n}^{0}\phi_x{\rm d}t{}\\ & = &\int_{0}^{+\infty}\phi([\rho]\frac{{\rm d}x}{{\rm d}t}-[\rho u]){\rm d}t-w_{m}^{0}(0)\phi(0, 0){}\\ & & -\int_{0}^{+\infty}\phi{\rm d}w_{m}^{0}(t)+\int_{0}^{+\infty}(w_{n}^{0}(t)-w_{m}^{0}(t) \frac{{\rm d}x}{{\rm d}t})\phi_{x}{\rm d}t{}\\ & = &-\int_{0}^{+\infty}\phi(\frac{{\rm d}w_{m}^{0}(t)}{{\rm d}t}-[\rho]\frac{{\rm d}x}{{\rm d}t} +[\rho u]){\rm d}t-w_{m}^{0}(0)\phi(0, 0){}\\ & & +\int_{0}^{+\infty}(w_{n}^{0}(t)-w_{m}^{0}(t)\frac{{\rm d}x}{{\rm d}t})\phi_{x}{\rm d}t. \end{eqnarray} $

由$ \phi $的任意性,可得

(3.7) $ \begin{equation} \begin{array}{ll} { } \frac{{\rm d}w_{m}^{0}(t)}{{\rm d}t} = [\rho]x'(t)-[\rho u], \\ w_{m}^{0}(0) = 0, w_{n}^{0}(t) = w_{m}^{0}(t)x'(t). \end{array} \end{equation} $

同理,根据(3.3)式中的第二式和第三式,分别得到

(3.8) $ \begin{equation} \begin{array}{ll} { }\frac{{\rm d}w_{m}^{1}(t)}{{\rm d}t} = [\rho u]x'(t)-[\rho u^{2}+P], \\ w_{m}^{1}(0) = 0, w_{n}^{1}(t) = w_{m}^{1}(t)x'(t), \end{array} \end{equation} $

(3.9) $ \begin{equation} \begin{array}{ll} { }\frac{{\rm d}w_{m}^{2}(t)}{{\rm d}t} = [\rho E]x'(t)-[\rho uE+up], \\ w_{m}^{2}(0) = 0, w_{n}^{2}(t) = w_{m}^{1}(t)x'(t). \end{array} \end{equation} $

由于初始间断处满足$ x(0) = 0, w_m^i(0) = 0\; (i = 1, 2, 3) $,则对(3.7)–(3.9)式积分可得

(3.10) $ \begin{equation} \left\{ \begin{array}{ll} &w_{m}^{0}(t) = [\rho]x(t)-[\rho u]t, \\ &w_{m}^{1}(t) = [\rho u]x(t)-[\rho u^{2}+p]t, \\ &w_{m}^{2}(t) = [\rho E]x(t)-[\rho uE+up]t. \end{array} \right. \end{equation} $

又由(3.5)式知$ n^{0} = m^{1} $,即, $ w_{n}^{0}(t) = w_{m}^{1}(t) = w_{m}^{0}(t)x'(t) $,于是结合(3.10)式的第一式和第二式,我们有

(3.11) $ \begin{equation} [\rho]x'(t)x(t)-[\rho u]x'(t)t-[\rho u]x(t)+[\rho u^{2}+p]t = 0. \end{equation} $

继续对上式积分,可得

(3.12) $ \begin{equation} [\rho]x^{2}(t)-2[\rho u]x(t)t+[\rho u^{2}+p]t^{2} = 0, \end{equation} $

即,有

i)若$ [\rho] = 0 $,

(3.13) $ \begin{equation} x(t) = \frac{[\rho u^{2}+p]}{2[\rho u]}t, \end{equation} $

ii)若$ [\rho]\neq0 $,

(3.14) $ \begin{equation} x_1(t) = \frac{[\rho u]-\sqrt{[{\rho u}]^{2}-[\rho][\rho u^{2}+p]}}{[\rho]}t, \end{equation} $

(3.15) $ \begin{equation} x_2(t) = \frac{[\rho u]+\sqrt{[{\rho u}]^{2}-[\rho][\rho u^{2}+p]}}{[\rho]}t. \end{equation} $

另外,由熵条件

(3.16) $ \begin{equation} u_r+\frac{\sqrt{f(s_r)}}{\rho_r}\leq x'(t)\leq u_l-\frac{\sqrt{f(s_l)}}{\rho_l}, \end{equation} $

知$ [u]\leq0 $.若间断线$ x(t) = x_1(t) $,

i)当$ [\rho]>0 $,

(3.17) $ \begin{eqnarray} I_1& = &u_r+\frac{\sqrt{f(s_r)}}{\rho_r}-\frac{[\rho u]-\sqrt{[{\rho u}]^{2}-[\rho][\rho u^{2}+p]}}{[\rho]}{}\\ & = &\frac{[\rho]u_r+[\rho]\frac{\sqrt{f(s_r)}}{\rho_r}-[\rho u]+\sqrt{[{\rho u}]^{2}-[\rho][\rho u^{2}+p]}}{[\rho]}{}\\ & = &\frac{u_r\rho_r-u_r\rho_l-\rho_ru_r+\rho_lu_l+[\rho]\frac{\sqrt{f(s_r)}}{\rho_r}+\sqrt{[{\rho u}]^{2}-[\rho][\rho u^{2}+p]}}{[\rho]}{}\\ & = &\frac{-\rho_l[u]+[\rho]\frac{\sqrt{f(s_r)}}{\rho_r}+\sqrt{[{\rho u}]^{2}-[\rho][\rho u^{2}+p]}}{[\rho]}, \end{eqnarray} $

显然$ I_1>0 $,与(3.16)式矛盾.

ii)当$ [\rho]<0 $,

(3.18) $ \begin{eqnarray} I_2& = &\frac{[\rho u]-\sqrt{[{\rho u}]^{2}-[\rho][\rho u^{2}+p]}-[\rho]u_l +[\rho]\frac{\sqrt{f(s_l)}}{\rho_l}}{[\rho]} {}\\ & = &\frac{\rho_r[u]-\sqrt{[{\rho u}]^{2}-[\rho][\rho u^{2}+p]}+[\rho]\frac{\sqrt{f(s_l)}}{\rho_l}}{[\rho]}, \end{eqnarray} $

可知$ I_2>0 $,亦与(3.16)式矛盾.综上,间断线$ x_1(t) $不满足熵条件应舍去.于是,当$ [\rho] = 0 $时,间断线满足(3.13)式,且

(3.19) $ \begin{equation} u_r+\frac{\sqrt{f(s_r)}}{\rho_r}\leq \frac{[\rho u^{2}+p]}{2[\rho u]} \leq u_l-\frac{\sqrt{f(s_l)}}{\rho_l}. \end{equation} $

若$ [\rho] \neq 0 $,间断线满足$ x(t) = x_2(t) $,且

(3.20) $ \begin{equation} u_r+\frac{\sqrt{f(s_r)}}{\rho_r}\leq \frac{[\rho u]+\sqrt{[{\rho u}]^{2}-[\rho][\rho u^{2}+p]}}{[\rho]} \leq u_l-\frac{\sqrt{f(s_l)}}{\rho_l}. \end{equation} $

由(3.10)式,计算可得$ w_m^i(i = 0, 1, 2) $,继而根据(3.7)–(3.9)式可以得到$ w_n^i(i = 0, 1, 2) $.

定理3.1证毕.

注3.1 经典弱解亦可由(3.5)式构造.

注3.2 满足(3.20)式的初始集合非空,例如

(3.21) $ \begin{equation} \left\{ \begin{array}{ll} (\rho_l, u_l, g(s_l), f(s_l)) = (1, 4, 1, 1), \\ (\rho_r, u_r, g(s_r), f(s_r)) = (5, -1, 1, 4). \end{array} \right. \end{equation} $

于是

(3.22) $ \begin{eqnarray} u_r+\frac{\sqrt{f(s_r)}}{\rho_r}& = &-\frac{3}{5}<\frac{[\rho u]+\sqrt{[{\rho u}]^{2}-[\rho][\rho u^{2}+p]}}{[\rho]}{}\\ & = &\frac{\sqrt{3605}-45}{20}<u_l-\frac{\sqrt{f(s_l)}}{\rho_l} = 3. \end{eqnarray} $

注3.3 当状态方程(1.2)和(1.6)满足$ g(s) = 0 $, $ h(s) = 0 $时, $ \delta $ -波解的初始只有在(3.20)式两边取等号时成立.本文在一般形式的状态方程下推广了初始的可解域.

注3.4 如果第一特征与第二特征重叠或者第二与第三特征重叠的情形,在经典的熵条件下,其$ \delta $波速介于两特征速度之间,其分析方法与上述方法相同.