本文考虑Chaplygin气体的二维定常等熵可压Euler方程组, 由如下质量和动量守恒方程组成

(1.1) $ \begin{equation} \left\{\begin{array}{ll}(\rho u)_{x}+(\rho v)_{y} = 0, &\\(\rho u^{2}+p)_{x}+(\rho uv)_{y} = 0, &\\(\rho uv)_{x}+(\rho v^{2}+p)_{y} = 0, \end{array}\right. \end{equation} $

其中, $ \rho $和$ (u, v) $分别表示质量密度和流体的速度, $ p $为压强, 状态方程为$ p = g(s)-\frac{f(s)}{\rho} $(参看文献[1-2]), 在等熵系统中$ s $为常数, 为方便起见, 取$ p = -\frac{1}{\rho} $.

Chaplygin气体在空气动力学中起着至关重要的作用, 其引入一方面作为用于近似计算空气动力学中机翼的升力问题^[1], 另一方面是近年来研究较热的有关暗物质的模型^[3]. 由此可见, 无论从理论还是从应用角度, Chaplygin气体在双曲守恒律方程的研究中占据着重要地位. 文献[4]中Qu和Yuan给出了Chaplygin气体一维活塞问题的Radon测度解的定义, 证明了当活塞运动的Mach数适当大时在活塞表面形成质量集中, 无法用Lebesgue积分解刻画, 而用Radon测度解来刻画能很好地解决这一问题, 得到解的存在性. 文献[5]中Qu和Yuan等证明了来流Mach数趋于无穷时, 会在楔体表面形成质量集中. 本文主要研究二维定常Chaplygin气体绕直楔流动的相关问题, 在Radon测度解的意义下, 对Mach数大于1和楔角在$ [0, \frac{\pi}{2}] $的情形进行完整分析.

方程组(1.1) 可写成一般的守恒律形式

(1.2) $ \begin{equation} \partial_{x}F(U)+\partial_{y}G(U) = 0, \ U = (\rho, u, v)^{T}, \end{equation} $

其中$ F(U) = (\rho u, \rho u^{2}+p, \rho uv)^{T}, G(U) = (\rho v, \rho uv, \rho v^{2}+p)^{T} $.

直楔为$ \{(x, y)\in {{\Bbb R}} ^{2}:x\geq 0, 0\leq y\leq ax \} $, 其中$ a = \tan\theta $, $ \theta\in(0, \frac{\pi}{2}) $表示楔面的倾角.因此, 我们考虑的区域为$ \Omega = \{(x, y)\in {{\Bbb R}} ^{2}:x> 0, y >ax \} $. 直楔的表面为$ W = \{(x, y)\in {{\Bbb R}} ^{2}:x\geq 0, y = ax \} $.

假定气体在直楔表面满足滑移条件

(1.3) $ \begin{equation} v = au . \end{equation} $

在直线$ I = \{(x, y)\in {{\Bbb R}} ^{2}:x = 0, y >0 \} $上, 气体的状态为

(1.4) $ \begin{equation} U = U_{\infty}, \quad 在I上, \end{equation} $

其中$ U_{\infty} = (\rho_{\infty}, u_{\infty}, 0)^{T} $为常数.

经过量纲分析(参看文献[4]), 方程组(1.1) 的初值可化为

(1.5) $ \begin{equation} U = U_{0} = (1, 1, 0)^{T}, \end{equation} $

且压强与马赫数的关系为

(1.6) $ \begin{equation} p_{0} = -\frac{1}{M^{2}_{0}}, \end{equation} $

(1.7) $ \begin{equation} p_{1} = -\frac{1}{\rho}\frac{1}{M^{2}_{0}}, \end{equation} $

其中$ M_{0} $为来流的马赫数, 由$ M_{0} = \frac{q}{c} $给出, $ q $为气体的速度大小, $ c $为音速, 满足$ c = \sqrt{\frac{\partial p}{\partial \rho}} $.

定义2.1^[5]   称$ U\in L^{\infty}(\Omega) $为方程(1.2)(1.3)(1.5) 的积分弱解, 如果对于任意$ \phi \in C^{1}_{0}({{\Bbb R}} ^{2}) $, 成立

(2.1) $ \begin{equation} \int_{\Omega}(F(U)\partial_{x}\phi+G(U)\partial_{y}\phi){\rm d}x{\rm d}y = \int^{\infty}_{0}(aF(U|_{w})-G(U|_{w}))\phi(x, ax){\rm d}x -\int^{\infty}_{0}F(U_{0})\phi(0, y){\rm d}y. \end{equation} $

方程(1.2)(1.3)(1.5) 的求解构成了一个Riemann问题(参看文献[6-8]), 下文中我们将利用自相似方法构造特殊类型的解来解该Riemann问题. 假设Riemann问题的形式解为$ U(x, y) = V(\frac{x}{y}) $, 令$ \eta = \frac{x}{y} $, 且解具有分片常数的形式

$ V(\eta) = \left\{\begin{array}{ll}V_{0} = (1, 1, 0)^{T}, &{ } 0\leq \eta<\frac{1}{\sigma}, \\V_{1}, &{ } \frac{1}{\sigma}<\eta\leq\frac{1}{a}.\end{array}\right. $

在间断$ \eta = \frac{1}{\sigma} $处, $ V_{1} $和$ \sigma $满足如下的Rankine-Hugoniot条件

(2.2) $ \begin{equation} \left\{\begin{array}{ll} { } \sigma(\rho_{1} u_{1}-\rho_{0} u_{0}) = \rho_{1} v_{1}-\rho_{0} v_{0}, \\ { } \sigma(\rho_{1} u_{1}^{2}+p_{1}-\rho_{0} u_{0}^{2}-p_{0}) = \rho_{1} u_{1}v_{1}-\rho_{0} u_{0}v_{0}, \\ { } \sigma(\rho_{1} u_{1}v_{1}-\rho_{0} u_{0}v_{0}) = \rho_{1} v_{1}^{2}+p_{1}-\rho_{0}v_{0}^{2}-p_{0}, \end{array}\right. \end{equation} $

将式(1.6)和(1.7) 代入式(2.2), 得到波后Chaplygin气体的物理量为

(2.3) $ \begin{equation} \rho_{1} = \frac{\sqrt{M^{2}_{0}-1}+\tan\theta}{\sqrt{M^{2}_{0}-1}+(1-M^{2}_{0})\tan\theta}, \end{equation} $

(2.4) $ \begin{equation} p_{1} = -\frac{1}{M^{2}_{0}}\frac{\sqrt{M^{2}_{0}-1}+(1-M^{2}_{0})\tan\theta}{\sqrt{M^{2}_{0}-1}+\tan\theta}, \end{equation} $

(2.5) $ \begin{equation} v_{1} = -\sqrt{M^{2}_{0}-1}(u_{1}-1), \end{equation} $

(2.6) $ \begin{equation} \sigma = \frac{1}{\sqrt{M^{2}_{0}-1}}. \end{equation} $

当激波间断与直楔表面重合, 此时满足$ \sigma = \frac{1}{\sqrt{M^{2}_{0}-1}} = \tan\theta $, 解得$ M_{0} = \frac{\sqrt{1+\tan^{2}\theta}}{\tan\theta} = \frac{\sqrt{1+a^{2}}}{a} $, 称$ \frac{\sqrt{1+a^{2}}}{a} $为来流Mach数的临界值, 记为$ M^{\ast}_{0} $.

由式(2.3) 知, 当$ M_{0}\rightarrow M^{\ast}_{0} $时, 成立

(2.7) $ \begin{equation} \lim\limits_{M_{0}\rightarrow M^{\ast}_{0}}\rho_{1} = \lim\limits_{M_{0}\rightarrow M^{\ast}_{0}}\frac{\sqrt{M^{2}_{0}-1}+\tan\theta}{\sqrt{M^{2}_{0}-1}+(1-M^{2}_{0})\tan\theta} = \infty, \end{equation} $

由此可见, 当来流Mach数$ M_{0} $趋于$ M^{\ast}_{0} $时, 密度趋于无穷, 此时不能用积分弱解来刻画, 所以在下文第3节我们引入Radon测度解的定义, 把积分弱解进行了推广.

在第4节我们将详细地讨论来流Mach数$ M_{0} $趋于三种极限情况:

$M_{0}\rightarrow M^{\ast-}_{0}$ (引理4.1); $M_{0}\rightarrow \infty$ (引理4.2); $M_{0}\rightarrow 1^{+}$ (引理4.3).

为方便后文讨论, 我们先给出如下引理.

引理2.1  方程(1.2)(1.3)(1.5) 存在含激波的分片光滑解的条件为来流Mach数$ M_{0} $满足$ 1<M_{0}<\frac{\sqrt{1+a^{2}}}{a} $, 且波后的Mach数$ M_{1}>1 $, 为超音速流.

证   若方程(1.2)(1.3)(1.5) 存在积分弱解, 则在间断处相应的Rankine-Hugoniot条件成立, 由密度的非负性结合表达式(2.3) 可知, 来流Mach数$ M_{0} $的范围为$ 1<M_{0}<M^{\ast}_{0} $.

波后的Mach数:

(2.8) $ \begin{eqnarray} M_{1} = \frac{q_{1}}{c_{1}} = \frac{\sqrt{u^{2}_{1}+v^{2}_{1}}}{\sqrt{\frac{\partial P_{1}}{\partial\rho_{1}}}} = \sqrt{u^{2}_{1}+(M^{2}_{0}-1)(u_{1}-1)^{2}}\rho_{1}M_{0}, \end{eqnarray} $

将式(1.7)和(2.3)代入式(2.8), 得

$ M_{1} = \frac{M_{0}\sqrt{1+a^{2}}}{1-\sqrt{M^{2}_{0}-1}a}. $

因为来流Mach数$ M_{0} $的范围为$ 1<M_{0}<\frac{\sqrt{1+a^{2}}}{a} $, 所以$ \frac{1}{M_{0}}-\sqrt{1-\frac{1}{M^{2}_{0}}}a\in(0, 1) $, 于是

$ M_{1} = \frac{M_{0}\sqrt{1+a^{2}}}{1-\sqrt{M^{2}_{0}-1}a} = \frac{\sqrt{1+a^{2}}}{\frac{1}{M_{0}}-\sqrt{1-\frac{1}{M^{2}_{0}}}a}>\sqrt{1+a^{2}}\geq1, $

得出波后的Mach数$ M_{1}>1 $, 为超音速流.

引理2.2  当来流Mach数趋于$ M_0^{*-} $时, 对于波后Chaplygin气体物理量, 我们有下面的结果

(2.9) $ \begin{equation} \lim\limits_{M_{0}\rightarrow M^{\ast-}_{0}}u_{1} = \frac{1}{1+a^{2}}, \end{equation} $

(2.10) $ \begin{equation} \lim\limits_{M_{0}\rightarrow M^{\ast-}_{0}}v_{1} = \frac{a}{1+a^{2}}, \end{equation} $

(2.11) $ \begin{equation} \lim\limits_{M_{0}\rightarrow M^{\ast-}_{0}}p_{1} = 0, \end{equation} $

(2.12) $ \begin{equation} \lim\limits_{M_{0}\rightarrow M^{\ast-}_{0}}(1-a{\sqrt{M^{2}_{0}-1}})\rho_{1} = a^{2}+1, \end{equation} $

(2.13) $ \begin{equation} \lim\limits_{M_{0}\rightarrow M^{\ast-}_{0}}\rho_{1}(\sigma-a) = a^{3}+a, \end{equation} $

(2.14) $ \begin{equation} \lim\limits_{M_{0}\rightarrow M^{\ast-}_{0}}\frac{p_{1}}{\rho_{1}} = 0 . \end{equation} $

证   因为$ M_{0}\rightarrow M^{\ast-}_{0} $时, 极限过程没有奇性产生, 所以根据文献[5]中的结论, 可得式(2.9)和(2.10).

对表达式(2.3)(2.4), 直接取极限$ M_{0}\rightarrow M^{\ast-}_{0} $, 可得式(2.11)(2.14).

结合式(2.3)和(2.6), 得

$ \lim\limits_{M_{0}\rightarrow M^{\ast-}_{0}}(1-a{\sqrt{M^{2}_{0}-1}})\rho_{1} = \lim\limits_{M_{0}\rightarrow M^{\ast-}_{0}}\frac{(1-a\times\frac{1}{a})(\sqrt{M^{2}_{0}-1}+a)}{\sqrt{M^{2}_{0}-1}+(1-M^{2}_{0})a} = 0, $

$ \lim\limits_{M_{0}\rightarrow M^{\ast-}_{0}}\rho_{1}(\sigma-a) = \lim\limits_{M_{0}\rightarrow M^{\ast-}_{0}}( \frac{1}{\sqrt{M^{2}_{0}-1}}-a)\times\frac{\sqrt{M^{2}_{0}-1}+\tan\theta}{\sqrt{M^{2}_{0}-1}+(1-M^{2}_{0})\tan\theta} = a^{3}+a. $

引理$ 2.2 $得证.

注2.1  通过引理2.2, 可知来流Mach数趋于$ M_0^{*-} $时, 激波斜率趋于楔面的斜率$ a $, 密度会趋于无穷.

综上, 描述Chaplygin气体绕流问题的方程(1.2)(1.3)(1.5) 存在分片常数$ V(\eta) $形式解的条件为来流Mach数$ M_{0} $的范围为$ 1<M_{0}<M^{\ast}_{0} $, 当Mach数$ M_{0} $大于等于$ M^{\ast}_{0} $时, 该问题不存在分片常数的积分弱解, 并且当$ M_{0} $趋于$ M^{\ast}_{0} $时, 波后的密度趋于无穷, 不仅如此, 质量还会在楔面上形成集中现象. 由此, 我们考虑在Radon测度解的框架下来研究该问题解的存在性, 并且证明上述积分弱解也是Radon测度解. 在这个意义上, Radon测度解是积分弱解的一个推广, 并且确实可以解决利用积分弱解无法刻画的问题.

令$ B $为Euclid平面$ {{\Bbb R}} ^{2} $上的Borel $ \sigma $代数, 考虑$ ( {{\Bbb R}} ^{2}, B) $上的Radon测度, 记做

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

其中配对关系为Radon测度$ m $和试验函数$ \phi \in C_{0}({{\Bbb R}} ^{2}) $, $ \mu\ll\nu $表示$ \mu $关于$ \nu $绝对连续(参看文献[9-10]).

定义3.1^[5]   令$ L $为Lipschitz曲线, 其参数方程为$ \bigg\{\begin{array}{cc}x = x(t), &\\y = y(t), &\end{array} t\in[0, T) $, $ \omega_{L}(t)\in L^{1}_{loc}(0, T) $.在$ L\subset R^{2} $上带有权重$ w_{L} $的Dirac测度^[11]由下述等式给出

(3.2) $ \begin{equation} \langle \omega_{L}\delta_{L}, \phi\rangle = \int^{T}_{0}\omega_{L}(t)\phi(x(t), y(t))\sqrt{x'^{2}(t)+y'^{2}(t)}{\rm d}t, \quad \forall \phi \in C_{0}({{\Bbb R}} ^{2}) . \end{equation} $

定义3.2^[5]   对固定Mach数$ M_{0} $, $ 0<M_{0}\leq \infty $, 令$ \varrho, m^{0}, m^{1}, m^{2}, n^{0}, n^{1}, n^{2}, \wp $为$ \overline{\Omega} $上的Radon测度, $ \omega_{p}\in L^{1}_{loc}({{\Bbb R}} ^{+}\bigcup \{0\}) $, 且$ \omega_{p}\geq0 $, 称$ (\varrho, u, v, \omega_{p}) $为方程(1.2)(1.3)(1.5) 的Radon测度解, 如果满足

i) $ \varrho $为非负Radon测度, 使得$ \wp\ll\varrho, (m^{0}, n^{0})\ll\varrho $, $ (m^{1}, n^{1})\ll(m^{0}, n^{0}) $, $ (m^{2}, n^{2})\ll(m^{1}, n^{1}) $, 且相应的Radon-Nikodym导数$ \varrho $-a.e. 意义下满足

(3.3) $ \begin{equation} u = \frac{m^{0}({\rm d}x{\rm d}y)}{\varrho({\rm d}x{\rm d}x)} = \frac{\frac{m^{1}({\rm d}x{\rm d}y)}{\varrho({\rm d}x{\rm d}x)}}{\frac{m^{0}({\rm d}x{\rm d}y)}{\varrho({\rm d}x{\rm d}x)}} = \frac{\frac{n^{1}({\rm d}x{\rm d}y)}{\varrho({\rm d}x{\rm d}x)}}{\frac{n^{0}({\rm d}x{\rm d}y)}{\varrho({\rm d}x{\rm d}x)}}, \end{equation} $

(3.4) $ \begin{equation} v = \frac{n^{0}({\rm d}x{\rm d}y)}{\varrho({\rm d}x{\rm d}y)} = \frac{\frac{m^{2}({\rm d}x{\rm d}y)}{\varrho({\rm d}x{\rm d}y)}}{\frac{m^{0}({\rm d}x{\rm d}y)}{\varrho({\rm d}x{\rm d}y)}} = \frac{\frac{n^{2}({\rm d}x{\rm d}y)}{\varrho({\rm d}x{\rm d}y)}}{\frac{n^{0}({\rm d}x{\rm d}y)}{\varrho({\rm d}x{\rm d}y)}}, \end{equation} $

ii) $ \forall \phi \in C^{1}_{0}({{\Bbb R}} ^{2}) $, 成立

(3.5) $ \begin{equation} \langle m^{0}, \partial_{x}\phi\rangle+\langle n^{0}, \partial_{y}\phi\rangle+\int^{\infty}_{0}(\rho_{0}u_{0})\phi(0, y){\rm d}y = 0, \end{equation} $

(3.6) $ \begin{equation} \langle m^{1}, \partial_{x}\phi\rangle+\langle n^{1}, \partial_{y}\phi\rangle+\langle \wp, \partial_{x}\phi\rangle+\langle \omega_{p} n_{1}\delta_{w}, \phi\rangle+\int^{\infty}_{0}(\rho_{0}u^{2}_{0}+p_{0})\phi(0, y){\rm d}y = 0, \end{equation} $

(3.7) $ \begin{equation} \langle m^{2}, \partial_{x}\phi\rangle+\langle n^{2}, \partial_{y}\phi\rangle+\langle \wp, \partial_{y}\phi\rangle+\langle \omega_{p} n_{2}\delta_{w}, \phi\rangle+\int^{\infty}_{0}(\rho_{0}u_{0}v_{0})\phi(0, y){\rm d}y = 0. \end{equation} $

注3.1  $ \overrightarrow{n} = (n_{1}, n_{2}) = \frac{(-a, 1)}{\sqrt{1+a^{2}}} $为直楔的单位内法向量.

引理3.1  定义3.2中Radon测度解是定义2.1中积分弱解的推广.

证   由积分弱解的定义(2.1) 式, 变形得

$ \begin{eqnarray*} && \int_{\Omega}(F(U)\partial_{x}\phi+G(U)\partial_{y}\phi){\rm d}x{\rm d}y+\int^{\infty}_{0}F(U_{0})\phi(0, y){\rm d}y\\&&-\int^{\infty}_{0}(aF(U|_{w})-G(U|_{w}))\phi(x, ax){\rm d}x = 0, \end{eqnarray*} $

结合散度定理^[12], 式(2.1)可改写为

$ \langle F(U), \partial_{x}\phi\rangle+\langle G(U), \partial_{y}\phi\rangle+\int^{\infty}_{0}F(U_{0})\phi(0, y){\rm d}y = 0, $

这与定义3.2中Radon测度解的定义一致, 由此可见, 积分弱解也是Radon测度解, 即定义3.2中Radon测度解是定义2.1中积分弱解的推广.

由引理2.1知, 方程(1.2)(1.3)(1.5) 存在含激波的分片光滑解的条件为来流Mach数$ M_{0} $满足$ 1<M_{0}<\frac{\sqrt{1+a^{2}}}{a} $. 当来流Mach数大于等于$ \frac{\sqrt{1+a^{2}}}{a} $时, 不能用积分弱解表示, 此时考虑Radon测度解.

下面我们给出当来流Mach数$ M_{0} $大于等于临界Mach数$ M^{\ast}_{0} $时, 方程(1.2)(1.3)(1.5) 的Radon测度解的存在性.

引理3.2  来流Mach数$ M_{0} $满足$ M_{0}\geq M^{\ast}_{0} = \frac{\sqrt{1+a^{2}}}{a} $时, 方程(1.2)(1.3)(1.5) 存在Radon测度解.

证   要证明方程(1.2)(1.3)(1.5) 存在Radon测度解, 只需将其Radon测度解求解出来即完成证明.

设$ I_{\Omega} $为区域$ \Omega $上的特征函数, 即$ I_{\Omega}(x, y) = \bigg\{\begin{array}{cc}1, &(x, y)\in\Omega\\0, & \rm{其它} \end{array} $.令

(3.8) $ \begin{equation} m^{0} = \rho_{0}u_{0}I_{\Omega}L^{2}+w^{0}_{m}(x)\delta_{w} = I_{\Omega}L^{2}+w^{0}_{m}(x)\delta_{w}, \end{equation} $

(3.9) $ \begin{equation} n^{0} = \rho_{0}v_{0}I_{\Omega}L^{2}+w^{0}_{n}(x)\delta_{w} = w^{0}_{n}(x)\delta_{w}, \end{equation} $

(3.10) $ \begin{equation} \wp = -\frac{1}{M^{2}_{0}}I_{\Omega}L^{2}, \end{equation} $

(3.11) $ \begin{equation} m^{1} = I_{\Omega}L^{2}+w^{1}_{m}(x)\delta_{w}, \end{equation} $

(3.12) $ \begin{equation} n^{1} = w^{1}_{n}(x)\delta_{w}, \end{equation} $

(3.13) $ \begin{equation} m^{2} = w^{2}_{m}(x)\delta_{w}, \end{equation} $

(3.14) $ \begin{equation} n^{2} = w^{2}_{n}(x)\delta_{w}, \end{equation} $

其中, $ w^{0}_{m}(x), w^{1}_{m}(x), w^{2}_{m}(x), w^{0}_{n}(x), w^{1}_{n}(x), w^{2}_{n}(x) $为待定函数.

将式(3.8)(3.9) 代入式(3.5), 得

(3.15) $ \begin{eqnarray} && -\sqrt{1+a^{2}}w^{0}_{m}(0)\phi(0, 0)+a\int^{\infty}_{0}\phi(x, ax){\rm d}x +\sqrt{1+a^{2}}\int^{\infty}_{0}(-aw^{0}_{m}(x)+w^{0}_{n}(x))\partial_{y}\phi(x, ax){\rm d}x{}\\ && - \sqrt{1+a^{2}}\int^{\infty}_{0}\frac{\rm d}{{\rm d}x}w^{0}_{m}(x)\phi(x, ax){\rm d}x = 0 . \end{eqnarray} $

由$ \phi $的任意性, 得

$ \begin{eqnarray*} \left\{\begin{array}{ll}w^{0}_{n}(x) = aw^{0}_{m}(x), \\ [2mm]{ } a = \sqrt{1+a^{2}}\frac{\rm d}{{\rm d}x}w^{0}_{m}(x), \\w^{0}_{m}(0) = 0.\end{array}\right. \end{eqnarray*} $

解得

(3.16) $ \begin{equation} \left\{\begin{array}{ll}{ } w^{0}_{m}(x) = \frac{a}{\sqrt{1+a^{2}}}x, \\ [3mm]{ } w^{0}_{n}(x) = \frac{a^{2}}{\sqrt{1+a^{2}}}x.\end{array}\right. \end{equation} $

类似地, 将式(3.10)(3.11)(3.12) 代入式(3.6), 得

(3.17) $ \begin{equation} \left\{\begin{array}{ll} w^{1}_{n}(x) = aw^{1}_{m}(x), \\{ } a(1-\frac{1}{M^{2}_{0}})+\sqrt{1+a^{2}}w_{p}n_{1} = \sqrt{1+a^{2}}\frac{\rm d}{{\rm d}x}w^{1}_{m}(x), \\ [2mm]w^{1}_{m}(0) = 0.\end{array}\right. \end{equation} $

将式(3.10)(3.13)(3.14) 代入式(3.7), 得

(3.18) $ \begin{equation} \left\{\begin{array}{ll}w^{2}_{m}(0) = 0, \\{ } \frac{1}{M^{2}_{0}}+\sqrt{1+a^{2}}[w_{p}n_{2}-\frac{\rm d}{{\rm d}x}w^{2}_{m}(x)] = 0, \\{ } w^{2}_{n}(x) = aw^{2}_{m}(x), \end{array}\right. \end{equation} $

利用直楔上的滑移条件(1.3), 结合式(3.17)(3.18), 得到

(3.19) $ \begin{equation} \left\{\begin{array}{ll}{ } w^{1}_{m}(x) = \frac{ax}{(1+a^{2})\sqrt{1+a^{2}}}, \\{ } w^{2}_{m}(x) = \frac{a^{2}x}{(1+a^{2})\sqrt{1+a^{2}}}, \\{ } w_{p}(x) = \frac{a^{2}}{1+a^{2}}-\frac{1}{M^{2}_{0}}, \\{ } w^{1}_{n}(x) = \frac{a^{2}x}{(1+a^{2})\sqrt{1+a^{2}}}, \\{ } w^{2}_{n}(x) = \frac{a^{3}x}{(1+a^{2})\sqrt{1+a^{2}}} .\end{array}\right. \end{equation} $

综上, 描述Chaplygin气体绕流问题的方程(1.2)(1.3)(1.5) 的Radon测度解的密度测度为

(3.20) $ \begin{equation} \varrho = I_{\Omega}L^{2}+a\sqrt{1+a^{2}}x\delta_{w}, \end{equation} $

速度为

(3.21) $ \begin{equation} u = I_{\Omega}+\frac{1}{1+a^{2}}I_{w}, \end{equation} $

(3.22) $ \begin{equation} v = \frac{a}{1+a^{2}}I_{w}. \end{equation} $

于是方程(1.2)(1.3)(1.5) 的Radon测度解的存在性得证.

在第3节中, 我们讨论了描述Chaplygin气体绕流问题方程的Radon测度解及一般Mach数$ M_{0} $的可解性, 本节将详细地讨论来流Mach数$ M_{0} $趋于三种极限状态情况：$ M_{0}\rightarrow M^{\ast-}_{0} $ (引理4.1), $ M_{0}\rightarrow \infty $ (引理4.2), $ M_{0}\rightarrow 1^{+} $ (引理4.3).

引理4.1  来流Mach数$ M_{0}\rightarrow M^{\ast-}_{0} $时, 方程(1.2)(1.3)(1.5) 通过积分弱解的定义取极限得到的结果与构造方法得到的Radon测度解结果相同, Radon测度解弱收敛到密度包含Dirac测度的奇异测度解, 其中Dirac测度的支撑在直楔表面.

证   由第2节可知, 对任意试验函数$ \phi \in C_{0}({{\Bbb R}} ^{2}) $, 由于$ \eta = \frac{x}{y} $, 有

(4.1) $ \begin{eqnarray} \int^{\frac{1}{a}}_{0}\rho v\phi(\eta){\rm d}\eta& = &\int^{\frac{1}{\sigma}}_{0}\rho_{0} v_{0}\phi(\eta){\rm d}\eta + \int^{\frac{1}{a}}_{\frac{1}{\sigma}}\rho_{1} v_{1}\phi(\eta){\rm d}\eta\\& = &\int^{\frac{1}{\sigma}}_{0}\phi(\eta){\rm d}\eta+\rho_{1} v_{1}\int^{\frac{1}{a}}_{\frac{1}{\sigma}}\phi(\eta){\rm d}\eta. \end{eqnarray} $

由于$ { } \lim\limits_{M_{0}\rightarrow M^{\ast-}_{0}} \sigma = a $, 所以

(4.2) $ \begin{equation} \lim\limits_{M_{0}\rightarrow M^{\ast-}_{0}}\int^{\frac{1}{\sigma}}_{0}\phi(\eta){\rm d}\eta = \int^{\frac{1}{a}}_{0}\phi(\eta){\rm d}\eta, \end{equation} $

(4.3) $ \begin{eqnarray} \lim\limits_{M_{0}\rightarrow M^{\ast-}_{0}}\rho_{1} v_{1}\int^{\frac{1}{a}}_{\frac{1}{\sigma}}\phi(\eta){\rm d}\eta & = &\lim\limits_{M_{0}\rightarrow M^{\ast-}_{0}}\rho_{1} v_{1}(\frac{1}{a}-\frac{1}{\sigma})\frac{1}{\frac{1}{a}-\frac{1}{\sigma}}\int^{\frac{1}{a}}_{\frac{1}{\sigma}}\phi(\eta){\rm d}\eta\\& = &\frac{(a^{3}+a)\frac{a}{1+a^{2}}}{a^{2}}\phi(\frac{1}{a}) = \phi(\frac{1}{a}). \end{eqnarray} $

令$ \psi(x, y) \in C_{0}({{\Bbb R}} ^{2}) $, 满足$ \phi(\eta, y) = \psi(\eta y, y) $, 则

$ \begin{eqnarray} \lim\limits_{M_{0}\rightarrow M^{\ast-}_{0}}\int^{\infty}_{0}\int^{\frac{1}{a}}_{0}\rho v\psi(x, y){\rm d}x{\rm d}y & = &\int^{\infty}_{0}\lim\limits_{M_{0}\rightarrow M^{\ast-}_{0}}\int^{\frac{1}{a}}_{0}\rho v\phi(\eta, y){\rm d}\eta {\rm d}y\\& = &\int^{\infty}_{0}\int^{\frac{1}{a}}_{0}\phi(\eta, y)y{\rm d}\eta {\rm d}y+\int^{\infty}_{0}\phi(\frac{1}{a}, y)y{\rm d}y\\& = &\int_{\Omega}\rho_{0}v_{0}\psi(x, y){\rm d}x{\rm d}y+\int^{\infty}_{0}a^{2}x\psi(x, ax){\rm d}x\\& = &\int_{\Omega}\rho_{0}v_{0}\psi(x, y){\rm d}x{\rm d}y+\int^{\infty}_{0}w^{0}_{n}(x)\sqrt{1+a^{2}}\psi(x, ax){\rm d}x, \end{eqnarray} $

得到

(4.4) $ \begin{equation} w^{0}_{n}(x) = \frac{a^{2}x}{\sqrt{1+a^{2}}}, \end{equation} $

同理, 可得

(4.5) $ \begin{equation} w^{1}_{n}(x) = \frac{a^{2}x}{(1+a^{2})\sqrt{1+a^{2}}}, \end{equation} $

(4.6) $ \begin{equation} w^{2}_{n}(x) = \frac{a^{3}x}{(1+a^{2})\sqrt{1+a^{2}}}, \end{equation} $

这与第3节中求得的式(3.16)和(3.19)一致.

其次, 由压强的关系(1.6)–(1.7), 知

(4.7) $ \begin{equation} p(x, y) = \left\{\begin{array}{ll}{ } p_{0} = -\frac{1}{M^{2}_{0}, }&{ } 0\leq \frac{x}{y} <\frac{1}{\sigma}, \\p_{1}, &{ } \frac{1}{\sigma}<\frac{x}{y}\leq\frac{1}{a}.\end{array}\right. \end{equation} $

当$ M_{0}\rightarrow M^{\ast-}_{0} $时, 有$ p_{0}\rightarrow -\frac{a^{2}}{a^{2}+1}, p_{1}\rightarrow 0, p\rightarrow -\frac{a^{2}}{a^{2}+1} $, 在上述测度收敛意义下成立, 于是得到

$ \lim\limits_{M_{0}\rightarrow M^{\ast-}_{0}}\rho_{0}v_{0} = 0, $

$ \lim\limits_{M_{0}\rightarrow M^{\ast-}_{0}}\rho_{0}u_{0}v_{0} = 0, $

$ \lim\limits_{M_{0}\rightarrow M^{\ast-}_{0}}\rho_{0}v^{2}_{0}+p_{0} = -\frac{a^{2}}{a^{2}+1}. $

对式(4.4)–(4.6)关于$ M_{0}\rightarrow M^{\ast-}_{0} $取极限, 得

(4.8) $ \begin{equation} (n^{0}(M_{0}), n^{1}(M_{0}), n^{2}(M_{0}))\rightarrow (n^{0}, n^{1}, n^{2}) . \end{equation} $

同理, 可得

(4.9) $ \begin{equation} (m^{0}(M_{0}), m^{1}(M_{0}), m^{2}(M_{0}))\rightarrow (m^{0}, m^{1}, m^{2}) . \end{equation} $

令

$ \begin{eqnarray*} \left\{\begin{array}{ll}m^{0} = \rho_{1} u_{1}L^{2}, n^{0} = \rho_{1} v_{1}L^{2}, \\m^{1} = \rho_{1} u_{1}^{2}L^{2}, n^{1} = \rho_{1} u_{1}v_{1}L^{2}, \\m^{2} = \rho_{1} u_{1}v_{1}L^{2}, n^{2} = \rho_{1} v_{1}^{2}L^{2}, \\\wp = p_{1}L^{2}, \\{ } w_{p}n_{1} = \frac{-a}{\sqrt{1+a^{2}}}p_{1}, \\{ } w_{p}n_{2} = \frac{1}{\sqrt{1+a^{2}}}p_{1}.\end{array}\right. \end{eqnarray*} $

则式(2.1) 可以改写为

(4.10) $ \begin{equation} \langle m^{0}, \partial_{x}\phi\rangle+\langle n^{0}, \partial_{y}\phi\rangle+\int^{\infty}_{0}\phi(0, y){\rm d}y = 0, \end{equation} $

(4.11) $ \begin{eqnarray} && \langle m^{1}, \partial_{x}\phi\rangle+\langle n^{1}, \partial_{y}\phi\rangle+\langle \wp, \partial_{x}\phi\rangle-\int^{\infty}_{0}\frac{a}{\sqrt{a^{2}+1}}p_{1}\phi(x, ax)\sqrt{1+a^{2}}{\rm d}x {}\\&& +\int^{\infty}_{0}(1+p_{0})\phi(0, y){\rm d}y = 0, \end{eqnarray} $

(4.12) $ \begin{equation} \langle m^{2}, \partial_{x}\phi\rangle+\langle n^{2}, \partial_{y}\phi\rangle+\langle \wp, \partial_{x}\phi\rangle+\int^{\infty} _{0}\frac{1}{\sqrt{a^{2}+1}}p_{1}\phi(x, ax)\sqrt{1+a^{2}}{\rm d}x = 0. \end{equation} $

由于$ \langle\wp, \phi\rangle = \int_{\Omega}p\phi {\rm d}x{\rm d}y $, 所以$ \langle\wp, \phi\rangle\rightarrow -\frac{a^{2}}{a^{2}+1}\int_{\Omega} \phi {\rm d}x{\rm d}y = 0, $得到

(4.13) $ \begin{equation} -\int^{\infty}_{0}\frac{a}{\sqrt{a^{2}+1}}p_{1}\phi(x, ax)\sqrt{1+a^{2}}{\rm d}x\rightarrow \langle w_{p}n_{1}\delta_{w}, \phi \rangle , \end{equation} $

(4.14) $ \begin{equation} \int^{\infty}_{0}\frac{1}{\sqrt{a^{2}+1}}p_{1}\phi(x, ax)\sqrt{1+a^{2}}{\rm d}x\rightarrow \langle w_{p}n_{2}\delta_{w}, \phi \rangle . \end{equation} $

根据式(4.8)–(4.9), 当$ M_{0}\rightarrow M^{\ast-}_{0} $时, 式(3.5)–(3.7) 成立.

注4.1  由第1节可知, 楔面的斜率$ a = \tan\theta $, $ \theta\in(0, \frac{\pi}{2}) $表示楔面的倾角, 当$ \theta = 0 $时, $ a = 0 $, 此时$ \frac{\sqrt{1+a^{2}}}{a} = \infty $, 该情形归结为下文引理4.2;当$ \theta = \frac{\pi}{2} $时, $ a = \infty $, 此时$ \frac{\sqrt{1+a^{2}}}{a} = 1 $, 该情形归结为下文引理4.3.

引理4.2  来流Mach数$ M_{0}\rightarrow \infty $时, 得到极限流

$ \begin{eqnarray*} \left\{\begin{array}{ll}{ } \varrho = I_{\Omega}L^{2}+a\sqrt{1+a^{2}}x\delta_{w}, \\{ } u = I_{\Omega}+\frac{1}{1+a^{2}}I_{w}, \\{ } v = \frac{a}{1+a^{2}}I_{w}, \\{ } p = \frac{a^{2}}{1+a^{2}}.\end{array}\right. \end{eqnarray*} $

证   当$ M_{0}\geq M^{\ast}_{0} $时, 不存在积分弱解意义下形如$ V(\eta) $的分片常数解, 此时考虑Radon测度解. 按照第3节中定义3.2构造Radon测度解, 经过计算, 得到如下结果

(4.15) $ \begin{equation} \left\{\begin{array}{ll}{ } w^{0}_{m}(x) = \frac{a}{\sqrt{1+a^{2}}}x, \\{ } w^{0}_{n}(x) = \frac{a^{2}}{\sqrt{1+a^{2}}}x, \\{ } w^{1}_{m}(x) = \frac{ax}{(1+a^{2})\sqrt{1+a^{2}}}, \\{ } w^{1}_{n}(x) = \frac{a^{2}x}{(1+a^{2})\sqrt{1+a^{2}}}, \\{ } w^{2}_{m}(x) = \frac{a^{2}x}{(1+a^{2})\sqrt{1+a^{2}}}, \\{ } w^{2}_{n}(x) = \frac{a^{3}x}{(1+a^{2})\sqrt{1+a^{2}}}, \\{ } w_{p}(x) = \frac{a^{2}}{1+a^{2}}-\frac{1}{M^{2}_{0}} .\end{array}\right. \end{equation} $

当$ M_{0}\rightarrow \infty $时, 有

$ w_{p}(x)\rightarrow \frac{a^{2}}{1+a^{2}}, $

(4.16) $ \begin{equation} \left\{\begin{array}{ll}{ } w^{0}_{n}(x)\rightarrow \frac{a^{2}}{\sqrt{1+a^{2}}}x, \\ { } w^{1}_{n}(x)\rightarrow \frac{a^{2}}{(1+a^{2})\sqrt{1+a^{2}}}x, \\ { } w^{2}_{n}(x)\rightarrow \frac{a^{3}}{(1+a^{2})\sqrt{1+a^{2}}}x . \end{array}\right. \end{equation} $

(4.17) $ \begin{equation} \left\{\begin{array}{ll} { } w^{0}_{m}(x)\rightarrow \frac{a}{\sqrt{1+a^{2}}}x, \\ { } w^{1}_{m}(x)\rightarrow \frac{a}{(1+a^{2})\sqrt{1+a^{2}}}x, \\ { } w^{2}_{m}(x)\rightarrow \frac{a^{2}}{(1+a^{2})\sqrt{1+a^{2}}}x . \end{array}\right. \end{equation} $

该结果与文献[5]中多方气体的结果一致.

注4.2  通过引理4.2, 可知来流Mach数$ M_{0} $趋于无穷时, 在定义3.2 Radon测度解意义下得到极限流, 结果与文献[5]中多方气体结果相同.

引理4.3  来流Mach数$ M_{0}\rightarrow 1^{+} $时, 激波斜率趋于无穷, 此时激波极限位置垂直于来流, 得到密度趋于无穷的静止气体.

证   当来流Mach数$ M_{0} $趋于$ 1^{+} $时, 我们有如下的结果

(4.18) $ \begin{equation} \lim\limits_{M_{0}\rightarrow 1^{+}} u_{1} = 0, \end{equation} $

(4.19) $ \begin{equation} \lim\limits_{M_{0}\rightarrow 1^{+}} v_{1} = 0, \end{equation} $

(4.20) $ \begin{equation} \lim\limits_{M_{0}\rightarrow 1^{+}} p_{1} = 0, \end{equation} $

(4.21) $ \begin{equation} \lim\limits_{M_{0}\rightarrow 1^{+}} \sqrt{M^{2}_{0}-1}\rho_{1} = a, \end{equation} $

(4.22) $ \begin{equation} \lim\limits_{M_{0}\rightarrow 1^{+}}\frac{ p_{1}}{\rho_{1}} = 0, \end{equation} $

(4.23) $ \begin{equation} \lim\limits_{M_{0}\rightarrow 1^{+}} \frac{1}{\sigma} = 0 . \end{equation} $

由式(2.5) 得

$ \lim\limits_{M_{0}\rightarrow 1^{+}} v_{1} = \lim\limits_{M_{0}\rightarrow 1^{+}}-\sqrt{M^{2}_{0}-1}(u_{1}-1) = 0, $

又由滑移条件(1.3) 知, $ v $和$ u $为同阶量, 所以式(4.18)(4.19) 成立.

对表达式(2.4), 直接取极限$ M_{0}\rightarrow 1^{+} $, 可得式(4.20).

$ \begin{eqnarray*} \lim\limits_{M_{0}\rightarrow 1^{+}} \sqrt{M^{2}_{0}-1}\rho_{1} = \lim\limits_{M_{0}\rightarrow 1^{+}}\sqrt{M^{2}_{0}-1}\times\frac{\sqrt{M^{2}_{0}-1}+a}{\sqrt{M^{2}_{0}-1}(1-\sqrt{M^{2}_{0}-1}a)} = a, \end{eqnarray*} $

由式(2.3)–(2.4) 知

$ \begin{eqnarray*} \lim\limits_{M_{0}\rightarrow 1^{+}}\frac{ p_{1}}{\rho_{1}} = \lim\limits_{M_{0}\rightarrow 1^{+}}\frac{-(a\sqrt{M^{2}_{0}-1}-1)^{2}(M^{2}_{0}-1)}{M^{2}_{0}(\sqrt{M^{2}_{0}-1}+a)^{2}} = 0, \end{eqnarray*} $

由式(2.6) 知

$ \lim\limits_{M_{0}\rightarrow 1^{+}} \frac{1}{\sigma} = \lim\limits_{M_{0}\rightarrow 1^{+}}\frac{v_{1}}{1-u_{1}} = \lim\limits_{M_{0}\rightarrow 1^{+}}\sqrt{M^{2}_{0}-1} = 0, $

此时激波斜率趋于无穷, 激波极限位置垂直于来流.由式(4.1) 知

(4.24) $ \begin{eqnarray} \int^{\frac{1}{a}}_{0}\rho v\phi(\eta){\rm d}\eta = \int^{\frac{1}{\sigma}}_{0}\phi(\eta){\rm d}\eta+\rho_{1}v_{1}\int^{\frac{1}{a}}_{\frac{1}{\sigma}}\phi(\eta){\rm d}\eta, \end{eqnarray} $

而$ \lim\limits_{M_{0}\rightarrow 1^{+}} \frac{1}{\sigma} = 0 $, 所以得到

$ \lim\limits_{M_{0}\rightarrow 1^{+}}\int^{\frac{1}{a}}_{0}\rho v\phi(\eta){\rm d}\eta = \lim\limits_{M_{0}\rightarrow 1^{+}}\rho_{1}v_{1}\int^{\frac{1}{a}}_{0}\phi(\eta){\rm d}\eta. $

由式(2.3)和(2.5)得

$ \lim\limits_{M_{0}\rightarrow 1^{+}}\rho_{1}v_{1} = \lim\limits_{M_{0}\rightarrow 1^{+}}\frac{\sqrt{M^{2}_{0}-1}+a}{\sqrt{M^{2}_{0}-1}(1-\sqrt{M^{2}_{0}-1}a)}[-\sqrt{M^{2}_{0}-1}(u_{1}-1)] = a, $

所以

$ \lim\limits_{M_{0}\rightarrow 1^{+}}\int^{\frac{1}{a}}_{0}\rho_{1}v_{1}\phi(\eta){\rm d}\eta = a\int^{\frac{1}{a}}_{0}\phi(\eta){\rm d}\eta . $

进而

$ \begin{eqnarray*} \lim\limits_{M_{0}\rightarrow 1^{+} }\int^{\infty}_{0}\int^{\frac{1}{a}}_{0}\rho_{1}v_{1}\psi(x, y){\rm d}x{\rm d}y& = &\int^{\infty}_{0}\int^{\frac{1}{a}}_{0}\rho_{1}v_{1}\phi(\eta, y){\rm d}\eta {\rm d}y\nonumber\\& = &a\int^{\infty}_{0}\int^{\frac{1}{a}}_{0}y\phi(\eta y, y){\rm d}\eta {\rm d}y\nonumber\\& = &a\int^{\infty}_{0}\int^{\frac{1}{a}}_{0}a^{2}x\psi(a^{2}x^{2}\eta, ax){\rm d}\eta {\rm d}x\nonumber\\& = &\int^{\infty}_{0}\int^{\frac{1}{a}}_{0}w^{0}_{n}(x)\sqrt{1+a^{2}}\psi(a^{2}x^{2}\eta, ax){\rm d}\eta {\rm d}x, \end{eqnarray*} $

得到

$ w^{0}_{n}(x) = \frac{a^{3}x}{\sqrt{1+a^{2}}}, $

又因为

$ \lim\limits_{M_{0}\rightarrow 1^{+}}\rho_{1}u_{1}v_{1} = 0, \; \; \; \lim\limits_{M_{0}\rightarrow 1^{+}}\rho_{1}v^{2}_{1}+p_{1} = 0, $

同理, 将上述过程中$ \rho_{1}v_{1} $换为$ \rho_{1}u_{1}v_{1} $, $ \rho_{1}v^{2}_{1}+p_{1} $, 得到

$ w^{1}_{n}(x) = w^{2}_{n}(x) = 0. $

由式(2.3) 知

$ \lim\limits_{M_{0}\rightarrow 1^{+}}\rho_{1} = \lim\limits_{M_{0}\rightarrow 1^{+}}\frac{\sqrt{M^{2}_{0}-1}+\tan\theta}{\sqrt{M^{2}_{0}-1}+(1-M^{2}_{0})\tan\theta} = \infty, $

此时密度趋于无穷, 又因为

$ \lim\limits_{M_{0}\rightarrow 1^{+}}\rho_{1}u_{1} = \lim\limits_{M_{0}\rightarrow 1^{+}}\rho_{1}\frac{v_{1}}{a} = 1, \; \; \; \lim\limits_{M_{0}\rightarrow 1^{+}}\rho_{1}u^{2}_{1}+p_{1} = 0, $

同上述过程类似, 可得

$ w^{0}_{m}(x) = \frac{a^{2}x}{\sqrt{1+a^{2}}}, \; \; \; w^{1}_{m}(x) = w^{2}_{m}(x) = 0. $

所以, 当来流Mach数$ M_{0}\rightarrow 1^{+} $时, 得到密度趋于无穷的静止气体.

注4.3  通过引理4.3, 可知来流Mach数$ M_{0} $趋于$ 1^{+} $时, 激波斜率趋于无穷, 激波极限位置垂直于来流, 波后Chaplygin气体静止.

根据前面的讨论, 可以总结得到如下定理:

定理5.1  当来流Mach数$ M_{0}>1 $时, 描述Chaplygin气体绕流问题方程(1.2)(1.3)(1.5) 存在Radon测度解. 特别地

来流Mach数$ M_{0} $范围为$ 1<M_{0}<M^{\ast}_{0} $时, 此时$ \sigma <a = \tan\theta $, 方程(1.2)(1.3)(1.5) 存在定义2.1积分弱解意义下间断解, 且波后的Mach数恒大于1, 为超音速流. 进一步, 当$ M_{0}\rightarrow 1^{+} $时, 得到密度趋于无穷的静止气体.

来流Mach数$ M_{0}\geq M^{\ast}_{0} $时, 方程(1.2)(1.3)(1.5) 存在Radon测度解, 且当$ M_{0}\rightarrow M^{\ast-}_{0} $时, 通过定义2.1积分弱解取极限得到的结果与构造方法得到的Radon测度解结果相同, Radon测度解弱收敛到密度包含Dirac测度的奇异测度解, 其中Dirac测度的支撑在直楔表面. 进一步, 当$ M_{0}\rightarrow \infty $时, 得到极限流, 与文献[5]中多方气体结果相同.