## The Two-Dimensional Steady Chaplygin Gas Flows Passing a Straight Wedge

Jia Jia,

School of Mathematical Sciences, East China Normal University, Shanghai 200241

Received: 2020-07-29

Abstract

The purpose of this paper is to investigate the two-dimensional steady supersonic chaplygin gas flows passing a straight wedge. By the definition of Radon measure solution, the accurate expressions are obtained for all cases where the Mach number is greater than 1. It is quite different from the polytropic gas, for the chaplygin gas flows passing problems, there exists a Mach number $M^{\ast}_{0}$, when the Mach number of incoming flows is greater than or equal to $M^{\ast}_{0}$, the quality will be concentrated on the surface of the straight wedge. At this time, there are not piecewise smooth solutions in the Lebesgue sense. The limit analysis is used to prove that the limit obtained by Lebesgue integral is consistent with the solution obtained in the sence of Radon measure solution.

Keywords： Chaplygin gas ; Radon measure solution ; Riemann problem ; Hypersonic limit

Jia Jia. The Two-Dimensional Steady Chaplygin Gas Flows Passing a Straight Wedge. Acta Mathematica Scientia[J], 2021, 41(5): 1270-1282 doi:

## 1 引言

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

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

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

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

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

$$$\sigma = \frac{1}{\sqrt{M^{2}_{0}-1}}.$$$

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

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$1<M_{0}<M^{\ast}_{0}$.

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

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

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

$$$\lim\limits_{M_{0}\rightarrow M^{\ast-}_{0}}p_{1} = 0,$$$

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

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

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

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).

## 3 Radon测度解及一般$M_{0}$的可解性

$B$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

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

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

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

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

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