2. Formulation of the problem and main results

In this section, we present the two-dimensional transonic shock problem with external force under suitable perturbation of exit pressure and the main results. The 2-D steady compressible isentropic Euler system with external force is of the form

(2.1)\begin{equation} \begin{cases} \partial_{x_1} (\rho u_1)+ \partial_{x_2}(\rho u_2)=0,\\ \partial_{x_1} (\rho u_1^2+P(\rho))+ \partial_{x_2}(\rho u_1 u_2)=\rho \partial_{x_1} \Phi,\\ \partial_{x_1} (\rho u_1u_2)+ \partial_{x_2}(\rho u_2^2+P(\rho))=\rho \partial_{x_2} \Phi, \end{cases} \end{equation}

where $(u_1,\,u_2)={\bf u}:\mathbb {R}^2\rightarrow \mathbb {R}^2$ is the unknown velocity filed and $\rho :\mathbb {R}^2\rightarrow \mathbb {R}$ is the density, and $\Phi (x_1,\,x_2)$ is a given potential function of external force. For the ideal polytropic gas, the equation of state is given by $P(\rho )=A \rho ^{\gamma }$, here $A$ and $\gamma$ ($1<\gamma <\infty$) are positive constants. We take $A=1$ throughout this paper for the convenience.

To this end, let's first focus on the 1-D steady compressible flow with an external force on an interval $I=[L_0,\, L_1]$, which is governed by

(2.2)\begin{equation} \begin{cases} (\bar{\rho} \bar{u})'(x_1)=0,\\ \bar{\rho} \bar{u} \bar{u}'+ \frac{{\rm d}}{{\rm d}x_1} P(\bar{\rho})= \bar{\rho} \bar{f}(x_1),\\ \bar{\rho}(L_0)=\rho_0>0,\quad \bar{u}(L_0)= u_0>0, \end{cases} \end{equation}

where we assume that the flow state at the entrance $x_1=L_0$ is supersonic, meaning that $u_0^2>c^2(\rho _0)=\gamma \rho _0^{\gamma -1}$.

Denote $J= \bar {\rho } \bar {u}=\rho _0 u_0>0$, then it follows from (2.2) that

(2.3)\begin{equation} \begin{cases} \bar{\rho}(x_1)=\frac{J}{\bar{u}(x_1)},\\ ((\bar{u})^{\gamma+1}-\gamma J^{\gamma-1}) \bar{u}'= \bar{u}^{\gamma} \bar{f}. \end{cases} \end{equation}

Also one has

(2.4)\begin{align} & \bar{u}'= \frac{\bar{u}\bar{f}}{\bar{u}^2-c^2(\bar{\rho})},\quad\bar{\rho}'={-}\frac{\bar{\rho}\bar{f}}{\bar{u}^2-c^2(\bar{\rho})}, \end{align}

(2.5)\begin{align} & \frac{{\rm d}}{{\rm d}x_1} \bar{M}^2(x_1)=\frac{(\gamma+1) \bar{M}^2}{\bar{M}^2-1} \frac{\bar{f}}{c^2(\bar{\rho})}, \end{align}

where $\bar {M}(x_1)={\bar {u}(x_1)}/{c(\bar {\rho })}$ is the Mach number.

Since $\bar {M}^2(L_0)>1$, it follows from (2.5) that if the external force satisfies

(2.6)\begin{equation} \bar{f}(x_1)>0, \quad \forall L_0< x_1< L_1, \end{equation}

then problem (2.2) has a global supersonic solution $(\bar {\rho }^-,\, \bar {u}^-)$ on $[L_0,\,L_1]$. If one prescribes a large enough end pressure at $x_1=L_1$, a shock will form at some point $x_1=L_s\in (L_0,\,L_1)$ and the gas is compressed and slowed down to subsonic speed, the gas pressure will increase to match the given end pressure. Mathematically, one looks for a shock $x_1=L_s$ and smooth functions $(\bar {\rho }^{\pm },\, \bar {u}^{\pm },\, \bar {P}^{\pm })$ defined on $I^+=[L_{s},\, L_1]$ and $I^-=[L_0,\,L_s]$ respectively, which solves (2.3) on $I^{\pm }$ with the jump at the shock $x_1=L_{s}\in (L_0,\, L_1)$ satisfying the physical entropy condition $[\bar {P}(L_{s})]=\bar {P}^+(L_s)-\bar {P}^-(L_s)>0$ and the Rankine–Hugoniot conditions

(2.7)\begin{equation} \begin{cases} [{\bar \rho} {\bar u}](L_s)=0,\\ [{\bar\rho} {\bar u}^2+P({\bar\rho})](L_s)=0. \end{cases} \end{equation}

and also the boundary conditions

(2.8)\begin{align} & \rho(L_0)=\rho_{0},\ u(L_0)=u_{0}>0, \end{align}

(2.9)\begin{align} & \bar{P}(L_1)= P_{e}. \end{align}

We will show that there is a unique transonic shock solution to the 1-D Euler system when the end pressure $P_e$ lies in a suitable interval. Such a problem will be solved by a shooting method employing the monotonicity relation between the shock position and the end pressure.

Proof. The existence and uniqueness of smooth supersonic flow $(\bar {u}^-,\, \bar {\rho }^-)$ starting from $(\rho _0,\,u_0)$ on $[L_0,\,L_1]$ is trivial. Suppose the shock occurs at $x_1=L_s\in (L_0,\, L_1)$, then it is well-known that there exists a unique subsonic state $(\bar {u}^+(L_s),\, \bar {\rho }^+(L_s))$ satisfying the Rankine–Hugoniot conditions (2.7) and the entropy condition. With $(\bar {u}^+(L_s),\, \bar {\rho }^+(L_s))$ as the initial data, equation (2.2) has a unique smooth solution $(\bar {u}^+,\, \bar {\rho }^+)$ on $[L_s,\,L_1]$. Denote $P_e= (\bar {\rho }^+(L_1))^{\gamma }$. In the following, we show that the monotonicity between the shock position $x_1=L_s$ and the exit pressure $P_e=(\bar {\rho }^+(L_1))^{\gamma }$. $\bar {\rho }^+(L_1)$ is regarded as a function of $L_s$. Since $(\bar {\rho }^+ \bar {u}^+)(L_s)=(\bar {\rho }^- \bar {u}^-)(L_s)=J=\rho _0 u_0>0$, then

(2.10)\begin{equation} \bar{u}^-(L_s)+ \frac{ J^{\gamma-1}}{(\bar{u}^-(L_s))^{\gamma}}=\bar{u}^+(L_s)+ \frac{ J^{\gamma-1}}{(\bar{u}^+(L_s))^{\gamma}}. \end{equation}

It follows from the second equation in (2.2) that

\begin{align*} & \frac12 (\bar{u}^+(L_1))^2+\frac{\gamma}{\gamma-1}({\bar\rho}^+(L_1))^{\gamma-1}- \bar\Phi(L_1)= \frac12 (\bar{u}^+(L_s))^2\\ & \quad +\frac{\gamma}{\gamma-1}(\bar{\rho}^+(L_s))^{\gamma-1}- \bar\Phi(L_s). \end{align*}

Differentiating with respect to $L_s$, one deduces that

(2.11)\begin{align} & \left(\gamma(\bar{\rho}^+(L_1))^{\gamma-2}-\frac{J^2}{(\bar{\rho}^+(L_1))^3}\right)\frac{{\rm d} \bar{\rho}^+(L_1)}{{\rm d}L_s}\nonumber\\ & \quad=\left(\gamma(\bar{\rho}^+(L_s))^{\gamma-2}-\frac{J^2}{(\bar{\rho}^+(L_s))^3}\right)\frac{{\rm d} \bar{\rho}^+(L_s)}{{\rm d}L_s}-\bar{f}(L_s)=:I. \end{align}

Also (2.10) yields that

\[ \left\{1-\frac{\gamma J^{\gamma-1}}{(\bar{u}^+(L_s))^{\gamma+1}}\right\}\frac{{\rm d} \bar{u}^+(L_s)}{{\rm d}L_s}=\left\{1-\frac{\gamma J^{\gamma-1}}{(\bar{u}^-(L_s))^{\gamma+1}}\right\}\frac{{\rm d} \bar{u}^-(L_s)}{{\rm d}L_s}=\frac{\bar{f}(L_s)}{\bar{u}^-(L_s)}. \]

Finally, we conclude that

\begin{align*} I& ={-}\left\{\gamma (\rho^+(L_s))^{\gamma-1}-\frac{J^2}{(\rho^+(L_s))^2}\right\}\frac{1}{\bar{u}^+(L_s)}\frac{{\rm d}\bar{u}^+(L_s)}{{\rm d}L_s}-\bar{f}(L_s)\\ & =\frac{\bar{f}(L_s)(\bar{u}^+(L_s)-\bar{u}^-(L_s))}{\bar{u}^-(L_s)}<0. \end{align*}

Since the coefficients

\[ \gamma(\bar{\rho}^+(L_1))^{\gamma-2}-\frac{J^2}{(\bar{\rho}^+(L_1))^3}>0, \]

then (2.11) implies that the end density $\bar \rho ^+(L_1)$ is a strictly decreasing function of the shock position $x_1=L_s$. It follows that the end pressure $P_e=(\bar \rho ^+(L_1))^{\gamma }$ is a strictly decreasing and continuous differentiable function on the shock position $x_1=L_s$. In particular, when $L_s=L_0$ and $L_s=L_1$, there are two different end pressure $P_1,\,P_2$ with $P_0>P_1$. Hence, by the monotonicity, one can obtain a transonic shock for the end pressure $P_e\in (P_1,\,P_0)$.

The one-dimensional transonic shock solution $(\bar {u}^{\pm },\, \bar {\rho }^{\pm })$ with a shock occurring at $x_1= L_s$ constructed in lemma 2.1 will be called the background solution in this paper. The extension of the subsonic flow $(\bar u^+(x_1),\,\bar \rho ^+(x_1))$ of the background solution to $L_s-\delta _0< x_1< L_1$ for a small positive number $\delta _0$ will be denoted by $(\hat u^+(x_1),\,\hat \rho ^+(x_1))$.

It is natural to focus on the structural stability of these transonic shock flows. For simplicity, we only investigate the structural stability under suitable small perturbations of the end pressure. Therefore, the supersonic incoming flow is unchanged and remains to be $(\bar {u}^-(x_1),\, 0,\, \bar {\rho }^-(x_1))$.

Assume that the possible shock curve $\Sigma$ and the flow behind the shock are denoted by $x_1=\xi (x_2)$ and $(u_1^+,\,u_2^+,\,P^+)(x)$ respectively (see figure 1). Let $\Omega ^+=\{(x_1,\,x_2): \xi (x_2)< x_1< L_1,\, -1< x_2<1\}$ denotes the subsonic region of the flow. Then the Rankine–Hugoniot conditions on $\Sigma$ gives

(2.12)\begin{equation} \begin{cases} [\rho u_1]- \xi'(x_2) [\rho u_2]=0,\\ [\rho u_1^2+P]- \xi'(x_2) [\rho u_1 u_2]=0,\\ [\rho u_1 u_2]- \xi'(x_2) [\rho u_2^2+P]=0. \end{cases} \end{equation}

In addition, the pressure $P$ satisfies the physical entropy conditions

(2.13)\begin{equation} P^+(x)>P^-(x) \quad\text{on}\,\Sigma. \end{equation}

Since the flow is tangent to the nozzle walls $x_2=\pm 1$, then

(2.14)\begin{equation} u_2^+(x_1,\pm 1)=0. \end{equation}

The end pressure is perturbed by

(2.15)\begin{equation} P^+(L_1,x_2)=P_e+\epsilon P_{ex}(x_2), \end{equation}

due to some technical reasons, we may readily suppose that $P_{ex}(x_2)=P_e^{{1}/{\gamma }}\hat P_{ex}(x_2)\in C^{2,\alpha }([-1,\,1]) (\alpha \in (0,\,1))$ satisfies the compatibility conditions

(2.16)\begin{equation} \hat P_{ex}'({\pm} 1)=0. \end{equation}

Figure 1. Nozzle.

The following theorem gives the main results of this paper.

Theorem 2.3 Suppose that (2.6), (2.8) and (2.9) hold. Then there exist positive constant $\epsilon _0>0$ such that, for all $\epsilon \in (0,\,\epsilon _0]$, system (2.1) with boundary conditions (2.12)–(2.15) has a unique transonic shock solution $(u_1^+(x),\,u_2^+(x),\, P^+(x);\xi (x_2))$ which admits the following properties:

(i). The shock $x_1=\xi (x_2)\in C^{3,\alpha }([-1,\,1])$, and satisfies

(2.17)\begin{equation} \|\xi(x_2)-L_s\|_{C^{3,\alpha}([{-}1,1])}\leq C\epsilon, \end{equation}

where the positive constant $C$ only depends on the background solution, the exit pressure and $\alpha$.

(ii). The velocity and pressure in subsonic region $(u_1^+,\,u_2^+,\,P^+)(x)\in (C^{2,\alpha }(\bar \Omega ^+))^3$, and there holds

(2.18)\begin{equation} \|(u_1^+,u_2^+,P^+)(x)-(\hat u,0,\hat P)\|_{C^{2,\alpha}(\bar\Omega^+)}\leq C\epsilon,\end{equation}

where $\Omega ^+=\{(x_1,\,x_2): \xi (x_2)< x_1< L_1,\, -1< x_2<1\}$ is the subsonic region and $(\hat u,\,0,\,\hat P)=(\hat u(x_1),\,0,\,P(\hat \rho (x_1)))$ is the extended background solution.

Our proof is influenced by the approach developed in [Reference Li, Xin and Yin16, Reference Li, Xin and Yin17, Reference Li, Xin and Yin20], yet the reformulation of the problem is different from there. It is well-known that steady Euler equations are hyperbolic-elliptic coupled in the subsonic region. The entropy and Bernoulli's function are conserved along the particle path, while the pressure and the flow angle satisfy a first-order elliptic system in the subsonic region. These facts are widely used in the structural stability analysis for the transonic shock problems in flat or divergent nozzles, one may refer to [Reference Chen3, Reference Chen, Chen and Song7, Reference Chen, Chen and Feldman8, Reference Li, Xin and Yin16, Reference Li, Xin and Yin17, Reference Li, Xin and Yin20, Reference Serre22, Reference Xin and Yin32, Reference Yuan33] and the references therein. Here we resort to a different decomposition based on the deformation and curl of the velocity developed in [Reference Weng and Xin27, Reference Weng28] for three-dimensional steady Euler and Euler–Poisson systems. The idea in that decomposition is to rewrite the density equation as a Frobenius inner product of a symmetric matrix and the deformation matrix by using Bernoulli's law. The vorticity is resolved by an algebraic equation of Bernoulli's function and the entropy. We should mention that there are several different decompositions to the three-dimensional steady Euler system [Reference Chen and Xie2, Reference Chen4, Reference Chen5, Reference Liu, Xu and Yuan21, Reference Weng23, Reference Xin and Yin32] developed by many researchers for different purposes. An interesting issue that deserves further discussion is when using the deformation-curl decomposition to deal with the transonic shock problem, the end pressure boundary condition becomes non-local since it involves the information from the shock front. However, this non-local boundary condition reduces to be local after introducing the potential function.

3. Reformulation of the problem

Different from previous works on transonic shock problems [Reference Chen3, Reference Chen, Chen and Song7, Reference Chen, Chen and Feldman8, Reference Li, Xin and Yin16, Reference Li, Xin and Yin17, Reference Li, Xin and Yin20], we will use the deformation-curl decomposition developed in [Reference Weng and Xin27, Reference Weng28] for steady Euler system to decompose the original system (2.1) into an equivalent system (3.3), where the hyperbolic quantity $B$ and elliptic quantities $u_1,\,u_2$ are effectively decoupled in subsonic regions. To this end, define the Bernoulli's function

(3.1)\begin{equation} B=\frac{1}{2} |{\bf u}|^2 + h(\rho)-\Phi, \end{equation}

where $h(s)=\frac {\gamma }{\gamma -1}s^{\gamma -1}$ is the enthalpy function. Hence, the density can be expressed by the Bernoulli function and velocity field as

(3.2)\begin{equation} \rho=H(B,\Phi,|{\bf u}|^2)=\left[\frac{\gamma-1}{\gamma}\left(B+\Phi-\frac{1}{2}|{\bf u}|^2\right)\right]^{{1}/{\gamma-1}}. \end{equation}

Consequently, the 2-D Euler system (2.1) with unknown function $(u_1,\,u_2,\,P)$ is equivalent to the following system

(3.3)\begin{equation} \begin{cases} \displaystyle \sum_{i,j=1}^2(c^2(H)\delta_{ij}- u_i u_j) \partial_i u_j + u_1\partial_1 \Phi + u_2 \partial_2 \Phi=0,\\ \partial_1 u_2 -\partial_2 u_1 ={-}\frac{\partial_2 B}{u_1},\\ u_1 \partial_1 B + u_2 \partial_2 B=0, \end{cases} \end{equation}

with unknown function $(u_1,\,u_2,\,B)$.

The shock curve is determined by

(3.4)\begin{equation} \xi'(x_2)= \frac{[\rho u_1 u_2]}{[\rho u_2^2+P]}(\xi(x_2),x_2),\quad x_2\in({-}1,1). \end{equation}

Furthermore, it follows from the R-H conditions (2.12) that

(3.5)\begin{equation} \begin{cases} [\rho u_1]= \frac{[\rho u_2] [\rho u_1 u_2]}{[\rho u_2^2 +P]},\\ [\rho u_1^2+ P(\rho)]= \frac{([\rho u_1 u_2])^2}{[\rho u_2^2 +P]}. \end{cases} \end{equation}

A direct computation by using (3.5) shows that on $x_1=\xi (x_2)$

(3.6)\begin{align} & (\rho^+(\xi(x_2),x_2)- \bar{\rho}^+(L_s),u_1^+(\xi(x_2),x_2)- \bar{u}^+(L_s))\nonumber\\ & \quad=(h_1,h_2)(\rho^-(\xi(x_2))-\bar\rho^-(L_s), u^-(\xi(x_2))-\bar u^-(L_s),(u_2^+(\xi(x_2),x_2))^2) \end{align}

here $h_i(0,\,0,\,0)=0$ for $i=1,\,2$. In addition, we have

(3.7)\begin{equation} \begin{cases} \begin{aligned} \frac{\partial h_1}{\partial(\rho^-{-}\bar\rho^-)}|_{(0,0,0)} & =2\bar u^-(L_s)\frac{\bar u^+{-}\bar u^-}{(\bar u^+(L_s))^2-c^2(\bar\rho^+(L_s))}+1,\\ \frac{\partial h_1}{\partial(u^-{-}\bar u^-)}|_{(0,0,0)} & =2 \bar\rho^-(L_s)\frac{\bar u^+{-}\bar u^-}{(\bar u^+(L_s))^2-c^2(\bar\rho^+(L_s))}, \\ \frac{\partial h_1}{\partial(u_2^+)^2}|_{(0,0,0)} & =\frac{(\bar\rho^+(L_s)\bar u^+(L_s))^2}{\bar P^+(L_s)-\bar P^-(L_s)}\frac{1}{(\bar u^+(L_s))^2-c^2(\bar\rho^+(L_s))}, \end{aligned} \end{cases} \end{equation}

and

(3.8)\begin{equation} \begin{cases} \begin{aligned} \frac{\partial h_2}{\partial(\rho^-{-}\bar\rho^-)}|_{(0,0,0)} & ={-}(\gamma-1)\frac{\bar P^+(L_s)-\bar P^-(L_s)}{(\bar\rho^+(L_s))^2\bar u^+(L_s)}\frac{\bar u^+\bar u^-}{(\bar u^+(L_s))^2-c^2(\bar\rho^+(L_s))},\\ \frac{\partial h_2}{\partial(u^-{-}\bar u^-)}|_{(0,0,0)} & =\frac{2 \bar\rho^-(L_s)\bar u^+(L_s)}{\bar\rho^+(L_s)}\frac{\bar u^-\bar u^+}{(\bar u^+(L_s))^2-c^2(\bar\rho^+(L_s))}+\frac{\bar\rho^-(L_s)}{\bar\rho^+(L_s)}, \\ \frac{\partial h_2}{\partial(u_2^+)^2}|_{(0,0,0)} & =\frac{\bar\rho^+(L_s)\bar u^+(L_s)}{\bar P^+(L_s)-\bar P^-(L_s)}\frac{c^2(\bar\rho^+(L_s))}{(\bar u^+(L_s))^2-c^2(\bar\rho^+(L_s))}. \end{aligned} \end{cases} \end{equation}

By substituting (3.6) into (3.1), we conclude that there is a function $h_3$ such that

(3.9)\begin{align} & B^+(\xi(x_2),x_2)- \bar{B}^+(L_s)= h_3(\rho^-(\xi(x_2))\nonumber\\ & \quad -\bar\rho^-(L_s), u^-(\xi(x_2))-\bar u^-(L_s),(u_2^+(\xi(x_2),x_2))^2). \end{align}

Thus, theorem 2.3 is established as long as we solve problem (3.3)–(3.4) with boundary conditions (3.5), (2.14)–(2.15). In order to deal with the free boundary value problem (3.3)–(3.4), we introduce the following transformation to reduce it into a fixed boundary value problem. Setting

(3.10)\begin{equation} y_1=\frac{x_1-\xi(x_2)}{L_1-\xi(x_2)}(L_1-L_s) +L_s,\quad y_2=x_2, \end{equation}

then, the domain $\Omega ^+=\{(x_1,\,x_2): \xi (x_2)< x_1< L_1,\, -1< x_2<1\}$ is changed into

(3.11)\begin{equation} Q=\{(y_1,y_2): L_s< y_1< L_1, - 1< y_2<1\}. \end{equation}

The inverse change variable gives

\[ x_1=\xi(y_2)+\frac{L_1-\xi(y_2)}{L_1-L_s}(y_1-L_s)= y_1+\frac{L_1-y_1}{L_1-L_s}(\xi(y_2)-L_s),\quad x_2=y_2.\nonumber \]

We now set for $y\in Q$

\[ (\tilde u_j,\tilde\rho,\tilde B,\tilde\Phi)(y_1,y_2)=( u_j,\rho, B,\Phi)\left(\frac{L_1-\xi(y_2)}{L_1-L_s}(y_1-L_s)+\xi(y_2), y_2\right),\quad j=1,2. \]

Shock equation (3.4) becomes

(3.12)\begin{equation} \begin{aligned} \xi'(y_2) & = \frac{(\rho u_1 u_2)(\xi(y_2),y_2)}{P^+(\rho)(\xi(y_2),y_2)-P^-(\xi(y_2))+\rho (u_2)^2(\xi(y_2),y_2)},\\ & =\frac{(\tilde\rho \tilde u_1 \tilde u_2)(L_s,y_2)}{P^+(\tilde\rho)(L_s,y_2)-P^-(\xi(y_2))+\tilde\rho (\tilde u_2)^2(L_s,y_2)},\, y_2\in({-}1,1), \end{aligned} \end{equation}

and system (3.3) is changed into

(3.13)\begin{align} & (c^2(\tilde\rho)-\tilde u_1^2)\frac{L_1-L_s}{L_1-\xi(y_2)} \partial_{y_1}\tilde u_1+c^2(\tilde\rho)\partial_{y_2}\tilde u_2+\tilde u_1\frac{L_1-L_s}{L_1-\xi(y_2)} \partial_{y_1}\tilde \Phi =F_1(\tilde{{\bf u}}, \tilde{B}), \end{align}

(3.14)\begin{align} & \frac{L_1-L_s}{L_1-\xi(y_2)} \partial_{y_1}\tilde u_2-\partial_{y_2}\tilde u_1-\frac{y_1-L_1}{L_1-\xi(y_2)}\xi'(y_2)\partial_{y_1}\tilde u_1+ \frac{\partial_{y_2}\tilde B}{\tilde u_1}=F_2(\tilde {\textbf u},\tilde B),\nonumber\\ & \tilde u_1\frac{L_1-L_s}{L_1-\xi(y_2)} \partial_{y_1}\tilde B+\tilde u_2\partial_{y_2}\tilde B+\frac{y_1-L_1}{L_1-\xi(y_2)}\xi'(y_2)\tilde u_2\partial_{y_1}\tilde B=0, \end{align}

where

\begin{align*} & F_1(\tilde{{\bf u}}, \tilde{B})= \tilde u_2^2\partial_{y_2}\tilde u_2-(c^2(\tilde\rho)-\tilde u_2^2)\frac{y_1-L_1}{L_1-\xi(y_2)}\xi'(y_2)\partial_{y_1}\tilde u_2 +\tilde u_1\tilde u_2\frac{L_1-L_s}{L_1-\xi(y_2)} \partial_{y_1}\tilde u_2\\ & \quad+\tilde u_2\tilde u_1(\partial_{y_2}\tilde u_1+\frac{y_1-L_1}{L_1-\xi(y_2)}\xi'(y_2)\partial_{y_1}\tilde u_1) -\tilde u_2(\partial_{y_2}\tilde \Phi+\frac{y_1-L_1}{L_1-\xi(y_2)}\xi'(y_2)\partial_{y_1}\tilde \Phi),\\ & F_2(\tilde{{\bf u}}, \tilde{B})={-}{\frac{y_1-L_1}{L_1-\xi(y_2)}\frac{\xi'(y_2)}{\tilde u_1} \partial_{y_1}\tilde B}.\nonumber \end{align*}

Consider the perturbed functions $v_i(y_1,\,y_2)$, $i=1,\,2,\,3,\,4,$ as

\begin{align*} & v_1(y_1,y_2)=\tilde{u}_1(y_1,y_2)-\bar{u}^+(y_1),\quad v_2(y_1,y_2)= \tilde{u}_2(y_1,y_2),\\ & v_3(y_1,y_2)=\tilde{B}(y_1,y_2)-\bar{B}^+,\quad v_4(y_2)=\xi(y_2)-L_s, \end{align*}

and define the vector functions

(3.15)\begin{equation} V(y_1,y_2)=(v_1(y_1,y_2),v_2(y_1,y_2),v_3(y_1,y_2),v_4(y_2)).\end{equation}

It follows from (3.12) that the shock satisfies

(3.16)\begin{equation} v_4'(y_2)=\frac{\tilde\rho (\bar{u} +v_1) v_2(L_s, y_2)}{P^+(\tilde\rho)(L_s,y_2)-P^-(\xi(y_2))+\tilde\rho (\tilde u_2)^2(L_s,y_2)}. \end{equation}

Through a direct computation, one can derive from (3.9) and (3.14) that the Bernoulli function satisfies a transport-type equation

(3.17)\begin{equation} \begin{cases} [(\bar u^+{+} v_1)(L_1-L_s)+v_2(y_1-L_1)v_4'(y_2)]\partial_{y_1} v_3+v_2(L_1- v_4-L_s)\partial_{y_2} v_3=0,\\ v_3(L_s,y_2)= b_3 v_4(y_2) + R_3(y_2), \end{cases} \end{equation}

where

(3.18)\begin{equation} b_3=\frac{\bar\rho^-(L_s)-\bar\rho^+(L_s)}{\bar\rho^+(L_s)}\bar f(L_s), \end{equation}

and $R_3(y_2)=R_3(V(L_s,\,y_2))=O(|V(L_s,\,y_2)|)^2$ is an error term of second order. We may readily drop superscribe $+$ on the background solutions if there is no risk of confusion. And the first-order system for $v_1,\, v_2$ is given by,

(3.19)\begin{equation} \begin{cases} (c^2(\bar\rho^+)-(\bar u^+)^2)\partial_{y_1} v_1+c^2(\bar\rho^+)\partial_{y_2}v_2+B_1(y_1)v_1\\ \quad+\,B_3(y_1)v_3+B_4(y_1)v_4(y_2)=F_3(V,\nabla V),\\ \partial_{y_1}v_2-\partial_{y_2}v_1+\frac{L_1-y_1}{L_1-L_s}\bar u'v_4'+\frac{\partial_{y_2}v_3}{\bar u}=F_4(V,\nabla V), \end{cases} \end{equation}

where $F_3,\, F_4$ represent the remainder term of the second order with respect to $V$ and $\nabla V$, and

\begin{align*} B_1(y_1)& =\bar f(y_1)-(\gamma+1)\bar u\bar u'=\frac{\gamma \bar u^2+c^2(\bar\rho^+)}{c^2(\bar\rho^+)-\bar u^2}\bar f>0,\\ B_3(y_1)& =(\gamma-1)\bar u',\\ B_4(y_1)& =\frac{1}{L_1-L_s}[(\gamma-1)\bar f(y_1)(L_1-y_1)\bar u'-\bar u \bar f+\bar u\bar f'(y_1)(L_1-y_1)]. \end{align*}

It's obvious that the second formula in (3.6) gives the boundary condition of $v_1$ on the entrance $x_1=L_s$. Meanwhile, formula (3.2) after changing the variable becomes

(3.20)\begin{equation} \begin{aligned} c^2(\tilde\rho)(y) & =\gamma\tilde\rho^{\gamma-1}=(\gamma-1)\left(\tilde B-\frac{1}{2}|\tilde{\bf u}|^2+\bar \Phi\right)\\ & =(\gamma-1)\left(\bar B^+{+}v_3-\frac{1}{2}(\bar u^+{+}v_1)^2-\frac{1}{2}v_2^2+\bar \Phi\right)\\ & =c^2(\bar\rho^+)+(\gamma-1)\left(v_3-\bar u^+v_1-\frac{1}{2}v_1^2-\frac{1}{2}v_2^2\right), \end{aligned} \end{equation}

which together with (2.15) gives the boundary condition of $v_1$ on the exit $x_1=L_1$. Hence, the boundary conditions to system (3.19) read as follow

(3.21)\begin{equation} \begin{cases} v_1(L_s,y_2)=b_2v_4(y_2)+R_2(y_2),\\ v_1(L_1,y_2)=\frac{1}{\bar u(L_1)}(v_3(L_1,y_2)-\epsilon \hat P_{ex}(y_2))+R_4(y_2),\\ v_2(y_1,\pm 1)=0, \end{cases} \end{equation}

here

(3.22)\begin{equation} b_2=\frac{\bar{\rho}^-(L_s)\bar{u}^+(L_s)}{\bar{\rho}^+(L_s) [(\bar{u}^+(L_s))^2-c^2(\bar{\rho}^+(L_s))]}\bar{f}(L_s)<0, \end{equation}

and $R_2(y_2)=R_2(V(L_s,\,y_2))=O(|V(L_s,\,y_2)|)^2$, $R_4(y_2)=R_4(V(L_1,\,y_2))=O(|V (L_1,\,y_2)|)^2$ are error terms of second order. Based on our reformulation, theorem 2.3 follows from the following results.

Theorem 3.1 Under the same assumptions as in theorem 2.3, there exists a positive constant $\epsilon _0>0$ such that for each $\epsilon \in (0,\,\epsilon _0]$, system (3.16)–(3.21) has a unique solution $V\in (C^{2,\alpha }(\bar {Q}))^3\times C^{3,\alpha }([-1,\,1])$ satisfying the following estimate

(3.23)\begin{equation} \sum_{i=1}^{3}\| v_i\|_{C^{2,\alpha}(\bar Q)}+\| v_4\|_{C^{3,\alpha}([{-}1,1])}\leq C\epsilon, \end{equation}

where the constant $C$ depends only on the background solution, the exit pressure and $\alpha \in (0,\,1)$.

4. Iteration scheme and the linear system

In the first part of this section, we construct an iteration scheme for the non-linear system, and the problem is reduced to the solvability of corresponding linear systems. Indeed, it turns out that the linear system is a non-local elliptic equation of second order with a free parameter denoting the relative location of the shock position on the wall $x_2=-1$. Then we study the existence, uniqueness and regularity for this linear system in the remainder part of this section.

4.1. Iteration scheme

Inspired by [Reference Li, Xin and Yin17], we will develop an iteration scheme to prove theroem 3.1. Consider the Banach space

(4.1)\begin{align} {\mathcal{V}}_{\delta}& :=\left\{V:\displaystyle\sum_{i=1}^{3}\|v_i\|_{C^{2,\alpha}(\bar Q)}+|v_4|_{C^{3,\alpha}[{-}1,1]}\leq \delta;\partial_{y_2}v_{j}(y_1,\pm 1)=0,\right.\nonumber\\ & \quad \left.\vphantom{\frac{1}{2}}j=1,3;v_2(y_1,\pm 1)=\partial_{y_2}^2v_2(y_1,\pm 1)=0; v_4'({\pm} 1)=v_4^{(3)}({\pm} 1)=0\right\}, \end{align}

here $\delta >0$ is a small constant to be determined later. For a fix $\hat V\in {\mathcal {V}}_{\delta }$, equivalently, we have the following quantity

\[ (\hat v_1,\hat v_2,\hat B, \hat\rho, \hat P,\hat\xi)(y). \]

We now define the linearized scheme to problem (3.16)–(3.21) as follows.

Firstly, thanks to (3.16), $v_4$ is determined by

(4.2)\begin{equation} v_4'(y_2)=b_0 v_2(L_s,y_2) +F_5(\hat V)(L_s,y_2), \end{equation}

where

\begin{align*} & b_0=\frac{(\bar{\rho} \bar{u})(L_s)}{P^+(\bar\rho)(L_s)-P^-(L_s)}>0,\\ & F_5(y_2)=\left\{\frac{\hat\rho \hat u_1}{P^+(\hat\rho)(L_s,y_2)-P^-(\hat\xi(y_2))+\hat\rho (\hat u_2)^2(L_s,y_2)} -b_0\right\}v_2(L_s,y_2), \end{align*}

hence, one can express the shock as

(4.3)\begin{equation} v_4(y_2)=v_4({-}1)+\int_{{-}1}^{y_2}b_0v_2(L_s,\tau) \,{\rm d}\tau + R_5(y_2), \end{equation}

where $R_5(y_2)=\int _{-1}^{y_2}F_5(\tau ) \,{\rm d}\tau$ is a error term of second order. Due to $\hat V\in {\mathcal {V}}_{\delta }$, we have

(4.4)\begin{equation} F_5({\pm} 1)=F_5^{''}({\pm} 1)=0,\,\|F_5\|_{C^{k,\alpha}[{-}1,1]}\leq C\delta\|\hat v_2\|_{C^{k,\alpha}(\bar Q)},\quad k=0,1,2. \end{equation}

Secondly, using (3.17), we get the linearized transport equation for $v_3$:

\[ [(\bar u + \hat v_1)(L_1-L_s)+\hat v_2(y_1-L_1)\hat v_4'(y_2)]\partial_{y_1} v_3+\hat v_2(L_1- \hat v_4-L_s)\partial_{y_2} v_3=0 \text{ in}\,Q, \]

with initial data

\[ v_3(L_s,y_2)= b_3 v_4(y_2) + R_3(y_2). \]

Thus, it can be solved by characteristic methods. Let $y_2(s;\beta )$ be the characteristics going through $(y_1,\, y_2)$ with $y_2(L_s)=\beta$, i.e.

(4.5)\begin{equation} \begin{cases} \dfrac{{\rm d} y_2}{{\rm d}s}(s;\beta)=\dfrac{\hat v_2(L_1- \hat v_4-L_s)}{(\bar u + \hat v_1)(L_1-L_s)+\hat v_2(y_1-L_1)\hat v_4'(y_2)},\quad L_s< s< L_1,\\ y_2(L_s)=\beta. \end{cases} \end{equation}

It is noted that $\beta$ can be also regarded as a function of $y=(y_1,\,y_2)$, this is denoted by $\beta =\beta (y)$, which leads to

(4.6)\begin{equation} v_3(y_1,y_2)=v_3(L_s,\beta(y))=b_3 v_4(y_2)+ F_6(y), \end{equation}

where

\[ F_6(y)=b_3\int_{y_2}^{\beta(y)} v_4'(\tau) \,{\rm d}\tau +R_3(\hat V(L_s,\beta(y))) \]

is an error term of second order. Furthermore, we have

(4.7)\begin{align} & \partial_{y_2}F_6(y_1,\pm 1)=0,\nonumber\\ & \|F_6\|_{C^{k,\alpha}(\bar Q)}\leq C\delta \left(\sum_{i=1}^{3}\|\hat v_i\|_{C^{k,\alpha}(\bar Q)}+\|\hat v_4\|_{C^{k+1,\alpha}(\bar Q)}\right),\quad k=0,1,2. \end{align}

It remains to determine the velocity $v_1,\,v_2$ and the shock position difference $v_4(-1)$ on the wall $x_2=-1$. Substituting (4.3) and (4.6) into (3.19) and (3.21), we get the following linearized system for $v_1,\,v_2$ with an unknown parameter $v_4(-1)$:

(4.8)\begin{equation} \begin{cases} \partial_{y_1} v_1+\dfrac{1}{1-\bar M^2}\partial_{y_2}v_2+\dfrac{B_1(y_1)}{c^2(\bar\rho^+)-(\bar u^+)^2}v_1\\ \quad+\,\dfrac{B_3(y_1)b_3+B_4(y_1)}{c^2(\bar\rho^+)-(\bar u^+)^2} (v_4({-}1)+\int_{{-}1}^{y_2}b_0v_2(L_s,\tau) \,{\rm d}\tau) =G_1(y),\\ \partial_{y_1}v_2-\partial_{y_2}[v_1-\lambda(y_1) (v_4({-}1)+\int_{{-}1}^{y_2}b_0v_2(L_s,\tau) \,{\rm d}\tau)] =G_2(y), \end{cases} \end{equation}

and the boundary conditions

(4.9)\begin{equation} \begin{cases} v_1(L_s,y_2)=b_2(v_4({-}1)+\int_{{-}1}^{y_2}b_0v_2(L_s,\tau) \,{\rm d}\tau)+R_6(y_2),\\ v_1(L_1,y_2)=\dfrac{b_3}{\bar u(L_1)}(v_4({-}1)+\int_{{-}1}^{y_2}b_0v_2(L_s,\tau) \,{\rm d}\tau)- \dfrac{\epsilon \hat P_{ex}(y_2)}{\bar u(L_1)}+R_7(y_2),\\ v_2(y_1,\pm 1)=0, \end{cases} \end{equation}

where

\begin{align*} & \lambda(y_1)=\frac{L_1-y_1}{L_1-L_s}\bar u'+\frac{b_3}{\bar u},\\ & R_6(y_2)=b_2R_5(y_2)+R_2(y_2),\\ & R_7(y_2)=\frac{b_3R_5(y_2)+F_6(L_1,y_2)}{\bar u(L_1)}+R_4(y_2),\\ & G_1(y)=\frac{F_3(\hat V,\nabla \hat V)-B_3(y_1)R_5(y_2)-B_3(y_1)F_6(y)-B_4(y_1)R_5(y_2)}{(c^2(\bar\rho^+)-(\bar u^+)^2)},\\ & G_2(y)=F_4(\hat V,\nabla \hat V)-\frac{\partial_{y_2}F_6(y)}{\bar u}+\lambda(y_1)\partial_{y_2}R_5(y_2). \end{align*}

It follows from (4.4), (4.7) and a simple calculation that

\begin{align*} & \partial_{y_2}G_1(y_1,\pm 1)=0,\,G_2(y_1,\pm 1)=0, \\ & \|G_i\|_{C^{k-1,\alpha}(\bar Q)}\leq C\delta \|\hat V\|_{C^{k,\alpha}(\bar Q)},\quad i,k=1,2, \end{align*}

and

(4.10)\begin{align} & R_6'({\pm} 1)=0,\,R_7'({\pm} 1)=0,\nonumber\\ & \|(R_6,R_7)\|_{C^{k,\alpha}[{-}1,1]}\leq C\delta\|\hat V\|_{C^{k,\alpha}(\bar Q)},\,k=0,1,2. \end{align}

The second equation in (4.8) implies that there is a potential function $\phi (y)$ that satisfies

(4.11)\begin{equation} \begin{cases} \partial_{y_2}\phi=v_2,\,\,\phi(L_s,-1)=0,\\ \partial_{y_1}\phi=v_1-\lambda(y_1) (v_4({-}1)+\int_{{-}1}^{y_2}b_0v_2(L_s,\tau) \,{\rm d}\tau)+\int_{{-}1}^{y_2}G_2(y_1,\tau)\,{\rm d}\tau, \end{cases} \end{equation}

it follows that $v_1,\,v_2$ can be represented by

(4.12)\begin{equation} \begin{cases} v_2=\partial_{y_2}\phi,\\ v_1=\partial_{y_1}\phi+\lambda(y_1)[v_4({-}1)+b_0\phi(L_s,y_2)]-\int_{{-}1}^{y_2}G_2(y_1,\tau)\,{\rm d}\tau. \end{cases} \end{equation}

Substituting (4.12) into the first equation in (4.8), we conclude that $\phi$ satisfies the following non-local elliptic equation of the second with an unknown constant $v_4(-1)$

(4.13)\begin{align} & \partial_{y_1}^{2}\phi+\frac{1}{1-\bar M^2}\partial_{y_2}^{2}\phi+\lambda_{1}(y_1)\partial_{y_1}\phi+\lambda_{0}(y_1)b_0\left(\frac{v_4({-}1)}{b_0}+\phi(L_s,y_2)\right)\nonumber\\ & \quad=G_1(y)+\lambda_1(y_1)\int_{{-}1}^{y_2}G_2(y_1,\tau)\,{\rm d}\tau+\partial_{y_1}\int_{{-}1}^{y_2}G_2(y_1,\tau)\,{\rm d}\tau, \end{align}

where

\begin{align*} & \lambda_{1}(y_1)=\frac{B_1(y_1)}{c^2(\bar\rho)-\bar u^2}>0,\\ & \lambda_{0}(y_1)=\frac{1}{c^2(\bar\rho)-\bar u^2}[(c^2(\bar\rho)-\bar u^2)\lambda'+B_1\lambda+B_3b_3+B_4]. \end{align*}

Similarly, substituting (4.12) into boundary conditions (4.9), combined with the boundary condition of $\phi$ in (4.11), we have

(4.14)\begin{equation} \begin{cases} \partial_{y_1}\phi(L_s,y_2)=b_0(b_2-\lambda(L_s))\left(\dfrac{v_4({-}1)}{b_0}\!+\!\phi(L_s,y_2)\right)+R_6(y_2)+\int_{{-}1}^{y_2}G_2(L_s,\tau)\,{\rm d}\tau,\\ \partial_{y_1}\phi(L_1,y_2)={-} \dfrac{\epsilon \hat P_{ex}(y_2)}{\bar u(L_1)}+R_7(y_2)+\int_{{-}1}^{y_2}G_2(L_1,\tau)\,{\rm d}\tau,\\ \partial_{y_2}\phi(y_1,\pm 1)=\phi(L_s,-1)=0. \end{cases} \end{equation}

So far, we have reduced problem (4.8)–(4.9) into a non-local elliptic equation of $\phi$ with an unknown constant $v_4(-1)$. Hence, it is sufficient to establish the solvability and regularity of problem (4.13)–(4.14) to study the original problem. We are going to do it in the next subsection.

4.2. A non-local elliptic equation with a free constant

In this section, we prove the existence, uniqueness and regularity of problem (4.13). To this end, we consider the following more concise form of second-order elliptic system with an unknown constant $\kappa$

(4.15)\begin{equation} \begin{cases} \partial_{y_1}^2\phi+a_2(y_1)\partial_{y_2}^2\phi+a_1(y_1)\partial_{y_1}\phi-a_0(y_1)(\kappa+\phi(L_s,y_2))=\partial_{y_1}f,\text{ in}\,Q,\\ \partial_{y_1}\phi(L_s,y_2)-a_3(\kappa+\phi(L_s,y_2))=g_1(y_2),\\ \partial_{y_1}\phi(L_1,y_2)=g_2(y_2),\\ \partial_{y_2}\phi(y_1,\pm 1)=\phi(L_s,-1)=0,\end{cases} \end{equation}

where the smooth coefficients $a_i(y_1)$, i=0, 1, 2 and the constant $a_3$ satisfy

(4.16)\begin{equation} a_1(y_1)< C_1,\,0< C_0< a_i(y)< C_1,\quad i=0,2,3, \end{equation}

and the parameter $\kappa$ is a constant to be determined with the solution itself.

The first lemma implies that the inhomogeneous problem corresponding to system (4.15) without the unknown constant has a unique weak solution.

Lemma 4.1 Suppose that $f\in L^2(Q)$ and $g_i\in L^2(-1,\,1),\, i=1,\,2$, then there exists a suitable large positive constant $K$, such that the following inhomogeneous second-order elliptic equation

(4.17)\begin{equation} \begin{cases} -\partial_{y_1}^2\phi-a_2(y_1)\partial_{y_2}^2\phi-a_1(y_1)\partial_{y_1}\phi+K\phi+a_0(y_1)\phi(L_s,y_2)=\partial_{y_1}f, \text{ in}\,Q,\\ \partial_{y_1}\phi(L_s,y_2)-a_3\phi(L_s,y_2)=g_1(y_2),\\ \partial_{y_1}\phi(L_1,y_2)=g_2(y_2),\\ \partial_{y_2}\phi(y_1,\pm 1)=0,\end{cases} \end{equation}

admits a unique weak solution $\phi \in H^1(Q)$ satisfying

(4.18)\begin{equation} \|\phi\|_{H^1(Q)}\leq C(\|(g_1,g_2)\|_{L^2({-}1,1)}+\|f\|_{L^2(Q)}), \end{equation}

for some positive constant $C>0$.

Proof. For $\phi,\,\psi \in H^1(Q)$, define the bilinear form

\begin{align*} {\mathcal{B}}[\phi,\psi]& =\int_{Q}\partial_{y_1}\phi\partial_{y_2}\psi {\rm d}y+\int_{Q} a_2(y_1)\partial_{y_2}\phi\partial_{y_2}\psi {\rm d}y-\int_{Q}a_1(y_1)\psi\partial_{y_1}\phi {\rm d}y\nonumber\\ & \quad+K\int_{Q}\phi\psi {\rm d}y+\int_{Q}a_0(y_1)\phi(L_s,y_2)\psi {\rm d}y+a_3\int_{{-}1}^{1}\phi(L_s,y_2)\psi(L_s,y_2) {\rm d}y_2, \end{align*}

and the linear functional on $H^1(Q)$

\[ l(\psi)=\int_{{-}1}^{1} g_2(y_2) \psi(L_1,y_2){\rm d} y_2-\int_{{-}1}^{1} g_1(y_2)\psi(L_s,y_2){\rm d}y_2-\int_{Q}\partial_{y_1}f\psi {\rm d}y. \]

It's obviously that the linear functional $l(\psi )$ on $H^1(Q)$ is continuous, i.e.

(4.19)\begin{equation} |l(\psi)|\leq C(\|(g_1,g_2)\|_{L^2({-}1,1)}+\|f\|_{L^2(Q)})\|\psi\|_{H^1(Q)}, \end{equation}

where we have used the trace theorem. So, what we need to do is just verify that the conditions of the Lax–Milgram Theorem are satisfied for the bilinear form ${\mathcal {B}}$. The boundedness of ${\mathcal {B}}_{K}$ is trivial, we will show that ${\mathcal {B}}_{K}$ is also coercive. Denote $\Lambda =\min \{1,\,C_0\}>0$, then

\begin{align*} & \Lambda\int_{Q}|\nabla\phi|^2 {\rm d}y+K\int_{Q}|\phi|^2{\rm d}y\leq {\mathcal{B}}[\phi,\phi]+\int_{Q}a_1(y_1)\phi\partial_{y_1}\phi {\rm d}y\\ & \quad- \int_{Q}a_0(y_1)\phi(L_s,y_2)\phi(y_1,y_2) {\rm d}y -a_3\int_{{-}1}^{1} |\phi(L_s,y_2)|^2 {\rm d}y_2, \end{align*}

Cauchy's inequality gives

\[ \int_{Q}a_1(y_1)\phi\partial_{y_1}\phi {\rm d}y\leq C_1\epsilon \int_{Q}|\nabla\phi|^2 {\rm d}y +\frac{C_1}{4\epsilon}\int_{Q}|\phi|^2{\rm d}y, \]

and

\[ \int_{Q}a_0(y_1)\phi(L_s,y_2)\phi(y_1,y_2) {\rm d}y\leq C_{tr}(L_1-L_s)C_1\epsilon \int_{Q}|\nabla\phi|^2 {\rm d}y +\frac{C_1}{4\epsilon}\int_{Q}|\phi|^2{\rm d}y. \]

Then, fix $\epsilon _0$ such that $C_1\epsilon _0(1+(L_1-L_s)C_{tr})<{\Lambda }/{2}$, and choosing $K=\max \{\Lambda,\,{C_1}/{\epsilon _0}\}$, thanks to the positivity of $a_3$, we obtain

\[ {\mathcal{B}}[\phi,\phi]\geq\frac{\Lambda}{2}\|\phi\|_{H^1(Q)}, \]

the unique existence follows from the Lax–Milgram Theorem and (4.19) gives the estimates (4.18). Thus, the proof is complete.

The unique existence of regular solution to non-local system (4.15) is established in the following proposition.

Proposition 4.2 For any $f\in C^{1,\alpha }(\bar Q)$, $g_i\in C^{\alpha }(\bar Q)$, there is a unique weak solution $(\phi,\,\kappa )$, such that $\phi \in H^1(Q)$ and the following estimate holds

(4.20)\begin{equation} \|\phi\|_{H^1(Q)}+|\kappa|\leq C(\|f\|_{C^{\alpha}(\bar Q)}+|(g_1,g_2)|_{C^{1,\alpha}[{-}1,1]}). \end{equation}

Moreover, if the compatibility conditions

(4.21)\begin{equation} \partial_{y_2}f(y_1,-1)=\partial_{y_2}f(y_1,1)=0,\,g_i'({-}1)=g_i'(1)=0,\,i=1,2, \end{equation}

are fulfilled, then $\phi \in C^{2,\alpha }(\bar Q)$

(4.22)\begin{equation} \|\phi\|_{C^{1,\alpha}(\bar Q)}\leq C(\|f\|_{C^{\alpha}(\bar Q)}+|(g_1,g_2)|_{C^{\alpha}[{-}1,1]}+\|\phi\|_{H^1(Q)}+|\kappa|), \end{equation}

and

(4.23)\begin{equation} \|\phi\|_{C^{2,\alpha}(\bar Q)}\leq C(\|f\|_{C^{1,\alpha}(\bar Q)}+|(g_1,g_2)|_{C^{1,\alpha}[{-}1,1]}+\|\phi\|_{H^1(Q)}+|\kappa|). \end{equation}

for some positive constant $C>0$.

Proof. The proof is divided into two steps.

Step 1: Regularity of weak solutions. We will use the symmetric extension methods to exclude the possible singularities that may appear at the corner, which implies that the weak solution $\phi \in H^1(Q)$ to system (4.15) is essentially more regular. To this end, introduce the notation

\[ Q^{*}:=\{(y_1,y_2): L_s< y_1< L_1, - 2< y_2<2\}. \]

and define the extended function $\phi ^*(y)$ on $Q^*$ as

(4.24)\begin{equation} \phi^*(y)= \begin{cases} \phi(y_1,2-y_2),\, 1< y_2<2,\\ \phi(y_1,y_2), - 1< y_2<1,\\ \phi(y_1,-2-y_2),-2< y_2<{-}1. \end{cases} \end{equation}

Then the extended function $\phi ^*$ satisfies

(4.25)\begin{equation} \begin{cases} \partial_{y_1}^2\phi^*+a_2(y_1)\partial_{y_2}^2\phi^*+a_1(y_1)\partial_{y_1}\phi^*-a_0(y_1)(\kappa+\phi^*(L_s,y_2))=\partial_{y_1}f^*, \text{in}\,Q^*,\\ \partial_{y_1}\phi^*(L_s,y_2)-a_3(\kappa+\phi^*(L_s,y_2))=g^*_1(y_2),\\ \partial_{y_1}\phi^*(L_1,y_2)=g^*_2(y_2),\\ \partial_{y_2}\phi^*(y_1,\pm 1)=\phi^*(L_s,-1)=0, \end{cases} \end{equation}

Using the standard interior and the boundary estimates for the second-order linear elliptic equations in [Reference Gilbarg and Tudinger13], we obtain that $\phi ^*(y)\in C^{1,\alpha }(Q^*)$ and

(4.26)\begin{equation} \|\phi^*\|_{C^{1,\alpha}(Q^*)}\leq C(\|f^*\|_{C^{1,\alpha}(Q^*)}+\|(g_1,g_2)\|_{C^{1,\alpha}[{-}2,2]}+\|\phi^*(L_s,y_2)\|_{L^2[{-}2,2]}+|\kappa|), \end{equation}

which implies that $\phi ^*(L_s,\,y_2)\in C^{1,\alpha }[-2,\,2]$. Use estimate (4.26) again to conclude that $\phi ^*(y)\in C^{2,\alpha }(Q^*)$ and

(4.27)\begin{equation} \|\phi^*\|_{C^{1,\alpha}(Q^*)}\leq C(\|f^*\|_{C^{1,\alpha}(Q^*)}+\|(g_1,g_2)\|_{C^{1,\alpha}[{-}2,2]}+\|\phi^*(0,y_2)\|_{L^2[{-}2,2]}+|\kappa|). \end{equation}

Then (4.22) and (4.23) follow immediately.

Step 2: Existence and uniqueness of weak solutions. Due to the linearity, any solution $\phi$ to problem (4.15) can be decomposed as $\phi =\phi _1+\phi _2$, where $\phi _i$, i=1,2, satisfy the following equation respectively

(4.28)\begin{equation} \begin{cases} \partial_{y_1}^2\phi_1+a_2(y_1)\partial_{y_2}^2\phi_1+a_1(y_1)\partial_{y_1}\phi_1-a_0(y_1)\phi_1(L_s,y_2)=\partial_{y_1}f,\,\text{in}\,Q,\\ \partial_{y_1}\phi_1(L_s,y_2)-a_3\phi_1(L_s,y_2)=g_1(y_2),\\ \partial_{y_1}\phi_1(L_1,y_2)=g_2(y_2),\\ \partial_{y_2}\phi_1(y_1,\pm 1)=0, \end{cases} \end{equation}

and

(4.29)\begin{equation} \begin{cases} \partial_{y_1}^2\phi_2+a_2(y_1)\partial_{y_2}^2\phi_2+a_1(y_1)\partial_{y_1}\phi_2-a_0(y_1)(\kappa+\phi_2(L_s,y_2))=0, \text{ in}\,Q,\\ \partial_{y_1}\phi_2(L_s,y_2)-a_3(\kappa+\phi_2(L_s,y_2))=0,\\ \partial_{y_1}\phi_2(L_1,y_2)=\partial_{y_2}\phi_2(y_1,\pm 1)=0,\\ \phi_2(L_s,-1)={-}\phi_1(L_s,-1). \end{cases} \end{equation}

Combing lemma 4.1 with the Fredholm alternative theorem, one can easily derive that (4.28) has a unique $H^1(Q)$ solution $\phi _1$ which satisfies (4.20). On the other hand, one can prove that the only weak solution to (4.29) must be $(\phi _2,\,\kappa )=(-\phi _1(L_s,\,-1),\,\phi _1(L_s,\,-1))$ by applying the maximum principle. For the detailed proof, one can refer to lemma 4.1 in [Reference Li, Xin and Yin17]. Thus, the proof is complete.

At this point, we can easily illustrate the well-posedness to reformulated problem (4.13)–(4.14).

Lemma 4.3 Problem (4.13)–(4.14) has a unique solution $(\phi,\,\kappa )\in C^{2,\alpha }(\bar Q)\times \mathbb {R}$ satisfying

(4.30)\begin{equation} \begin{aligned} & \|\phi\|_{C^{k,\alpha}(\bar Q)}+|\kappa|\leq C(\|(G_1,G_2)\|_{C^{k-1,\alpha}(\bar Q)}\\ & \quad+\|(R_6(y_2),R_7(y_2))\|_{C^{1,\alpha}[{-}1,1]}+\|\epsilon\hat P_{ex}(y_2)\|_{C^{1,\alpha}[{-}1,1]}),\quad k=1,2, \end{aligned} \end{equation}

for some positive constant C.

Proof. It suffices to verify solvability condition (4.16) for problem (4.13)–(4.14). A direct but tedious computation shows that

\begin{align*} & a_0(y_1)={-}b_0\lambda_{0}(y_1)={-}\frac{2 c^2(\bar\rho^+)\bar f b_0b_3}{\bar u(c^2(\bar\rho^+)-\bar u^2)^2}>0,\\ & a_1(y_1)=\lambda_{1}(y_1)=\frac{\gamma \bar u^2+c^2(\bar\rho^+)}{(c^2(\bar\rho^+)-\bar u^2)^2}\bar f>0,\\ & a_2(y_1)=\frac{1}{1-\bar M^2}>0,\\ & a_3=b_0(b_2-\lambda(L_s))=b_0\frac{c^2(\bar\rho^+)(\bar\rho^+{-}\bar\rho^-)\bar f(L_s)}{\bar\rho^+\bar u(c^2(\bar\rho^+)-\bar u^2)}>0, \end{align*}

since the background solution is subsonic and smooth, the upper bound is trivial. Hence, proposition 4.2 implies that there exists a unique weak solution $(\phi,\,\kappa )$, and estimate (4.30) follows from (4.22) and (4.23).

In view of the analysis of problem (4.13)–(4.14), the well-posedness of equation (4.8)–(4.9) follows.

Lemma 4.4 Problem (4.8)–(4.9) admits a unique solution $(v_1,\,v_2,\,v_4(-1))\in (C^{2,\alpha }(\bar Q))^2\times \mathbb {R}$ satisfying

(4.31)\begin{equation} \begin{aligned} & \|(v_1,v_2)\|_{C^{k,\alpha}(\bar Q)}+|v_4({-}1)|\leq C(\|(G_1,G_2)\|_{C^{k-1,\alpha}(\bar Q)}\\ & \quad+\|(R_6(y_2),R_7(y_2))\|_{C^{k,\alpha}[{-}1,1]}+\|\epsilon\hat P_{ex}(y_2)\|_{C^{k,\alpha}[{-}1,1]}),\quad k=1,2, \end{aligned} \end{equation}

and the compatibility conditions

(4.32)\begin{equation} \partial_{y_2}v_1(y_1,\pm 1)=0,\quad\partial_{y_2}v_2(y_1,\pm 1)=\partial^2_{y_2}v_2(y_1,\pm 1)=0. \end{equation}

Proof. Combine (4.12) and (4.30), we conclude that there is a unique solution $(v_1,\,v_2,\,v_4(-1))\in (C^{1,\alpha }(\bar Q))^2\times \mathbb {R}$ such that

(4.33)\begin{align} & \|(v_1,v_2)\|_{C^{\alpha}(\bar Q)}+|{\textbf v}_4({-}1)|\leq C(\|(G_1,G_2)\|_{C^{\alpha}(\bar Q)} \nonumber\\ & \quad+\|(R_6(y_2),R_7(y_2))\|_{C^{1,\alpha}[{-}1,1]}+\|\epsilon\hat P_{ex}(y_2)\|_{C^{1,\alpha}[{-}1,1]}). \end{align}

The similar estimates also hold true for $\|(v_1,\,v_2)\|_{C^{1,\alpha }(\bar Q)}$, but we can derive an even better estimate by rewriting (4.8) into an elliptic equation of $v_1$. To this end, we first rewrite system (4.8) as

(4.34)\begin{equation} \begin{cases} \partial_{y_1} v_1+\frac{1}{1-\bar M^2}\partial_{y_2}v_2+\lambda_1(y_1)v_1=\mathcal{G}_1(y),\\ \partial_{y_1}v_2-\partial_{y_2}v_1 =\mathcal{G}_2(y),\\ v_1(L_s,y_2)=\mathcal{R}_6(y_2),\\ v_1(L_1,y_2)=\mathcal{R}_7(y_2),\\ v_2(y_1,\pm 1)=0, \end{cases} \end{equation}

where

\begin{align*} & \mathcal{G}_1(y)=G_1(y)-\frac{B_3(y_1)b_3+B_4(y_1)}{(c^2(\bar\rho^+) -(\bar u^+)^2)}(v_4({-}1)+\int_{{-}1}^{y_2}b_0 v_2(L_s,\tau) \,{\rm d}\tau),\\ & \mathcal{G}_2(y)=G_2(y)+b_0 v_2(L_s,y_2),\\ & \mathcal{R}_6(y_2)=b_2(v_4({-}1)+\int_{{-}1}^{y_2}b_0v_2(L_s,\tau) \,{\rm d}\tau)+R_6(y_2),\\ & \mathcal{R}_7(y_2)=\frac{b_3}{\bar u(L_1)}(v_4({-}1)+\int_{{-}1}^{y_2}b_0v_2(L_s,\tau) \,{\rm d}\tau)- \frac{\epsilon \hat P_{ex}(y_2)}{\bar u(L_1)}+R_7(y_2),\nonumber \end{align*}

and

(4.35)\begin{equation} \partial_{y_2}\mathcal{G}_1(y_1,\pm 1)=0,\,\mathcal{G}_2(y_1,\pm 1)=0,\,\mathcal{R}'_6({\pm} 1)=0,\,\mathcal{R}'_7({\pm} 1)=0.\end{equation}

Note that (4.12) and the boundary conditions of $\phi$ imply that

(4.36)\begin{equation} v_2(y_1,\pm 1)=0,\,\partial_{y_2}v_1(y_1,\pm 1)=0.\end{equation}

A simple cancellation yields to

(4.37)\begin{equation} \begin{cases} \partial_{y_1}((1-\bar M^2)\partial_{y_1} v_1)+\partial^2_{y_2}v_1+\partial_{y_1}(\lambda_1(1-\bar M^2)v_1)=\partial_{y_1}((1-\bar M^2)\mathcal{G}_1)-\partial_{y_2}\mathcal{G}_2,\\ v_1(L_s,y_2)=\mathcal{R}_6(y_2),\,v_1(L_1,y_2)=\mathcal{R}_7(y_2),\\ \partial_{y_2}v_1(y_1,\pm 1)=0. \end{cases} \end{equation}

Since the boundary conditions in (4.37) are compatible at the corners, we obtain the following estimates for $v_1$ by using the symmetric extension defined by (4.24)

(4.38)\begin{equation} \begin{aligned} \|v_1\|_{C^{k,\alpha}(\bar Q)} & \leq \,C(\|v_2\|_{C^{k-1,\alpha}(\bar Q)}+|v_4({-}1)|+\|(G_1,G_2)\|_{C^{k-1,\alpha}(\bar Q)}\\ & \quad+\|\hat P_{ex}\|_{C^{k,\alpha}[{-}1,1]}+\|(R_6,R_7)\|_{C^{k,\alpha}[{-}1,1]}),\quad k=1,2. \end{aligned} \end{equation}

These estimates together with (4.34) also imply that

(4.39)\begin{equation} \begin{aligned} \|v_2\|_{C^{k,\alpha}(\bar Q)} & \leq \,C(\|v_2\|_{C^{k-1,\alpha}(\bar Q)}+|v_4({-}1)|+\|(G_1,G_2)\|_{C^{k-1,\alpha}(\bar Q)}\\ & \quad+\|\hat P_{ex}\|_{C^{k,\alpha}[{-}1,1]}+\|(R_6,R_7)\|_{C^{k,\alpha}[{-}1,1]}),\quad k=1,2. \end{aligned} \end{equation}

Hence, (4.31) follows from (4.33) and (4.38)–(4.39). Finally, differentiating the first equation in (4.34) with respect to $x_2$ and combining with $\partial _{y_2}\mathcal {G}_1(y_1,\,\pm 1)=0$ and $\partial _{y_2}v_1(y_1,\,\pm 1)=0$, we obtain

\[ \partial_{y_2}^2v_2(y_1,\pm 1)=0, \]

which gives compatibility condition (4.32). Thus, the proof is complete.

5. A priori estimates and proofs of main results

In this section, we will use the Banach contraction mapping theorem to prove theorem 3.1. Given any $\hat {V}\in \mathcal {V}_{\delta }$, we could establish some a priori estimates to the linearized problems defined in subsection 4.1, and construct a contractible mapping from ${\mathcal {V}}_{\delta }$ into itself so that there exists a unique fixed point, which is the solutions obtained in theorem 3.1 and the proof of theorem 3.1 will be finished.

Lemma 4.4 implies that there is a unique solution $(v_1,\,v_2,\,v_4(-1))\in (C^{2,\alpha }(\bar Q))^2\times \mathbb {R}$ to system (4.8)–(4.9) satisfying

(5.1)\begin{equation} \begin{aligned} & \|(v_1,v_2)\|_{C^{2,\alpha}(\bar Q)}+|v_4({-}1)|\leq C(\|(G_1,G_2)\|_{C^{1,\alpha}(\bar Q)}\\ & \quad+\|(R_6(y_2),R_7(y_2))\|_{C^{2,\alpha}[{-}1,1]}+\|\epsilon\hat P_{ex}(y_2)\|_{C^{2,\alpha}[{-}1,1]})\\ & \leq C(\epsilon+\delta^2), \end{aligned} \end{equation}

and compatibility condition (4.32) holds true.

The shock curve $v_4$ is given by (4.3), which satisfies

(5.2)\begin{equation} \|v_4\|_{C^{3,\alpha}[{-}1,1]}\leq C(|v_4({-}1)|+\|v_2\|_{C^{2,\alpha}(\bar Q)}+\|F_5\|_{C^{2,\alpha}(\bar Q)})\leq C(\epsilon+\delta^2). \end{equation}

Moreover, it follows from (4.32) and (4.4) that

(5.3)\begin{equation} v_4'({\pm} 1)=0=v_4^{(3)}({\pm} 1). \end{equation}

It remains to solve $v_3$. Due to (4.6), combining with (4.7) and (5.3), we obtain the estimate

(5.4)\begin{equation} \|v_3\|_{C^{2,\alpha}(\bar Q)}\leq C(\|v_4\|_{C^{2,\alpha}[{-}1,1]}+\|F_6\|_{C^{2,\alpha}(\bar Q)})\leq C(\epsilon+\delta^2), \end{equation}

and the compatibility condition

(5.5)\begin{equation} \partial_{y_2}v_3(y_1,\pm 1)=0. \end{equation}

Taking $\delta =O(1)\epsilon$, then for any given $\hat V\in {\mathcal {V}}_{\delta }$ we can define a continuous mapping $T:{\mathcal {V}}_{\delta } \rightarrow {\mathcal {V}}_{\delta }$ as

(5.6)\begin{equation} T\hat V=V, \end{equation}

due to the iteration scheme introduced in the previous section and estimates (5.1)–(5.5). Finally, we show that the mapping is also contractible in the space $(C^{1,\alpha }(\bar Q))^3\times C^{2,\alpha }[-1,\,1]$.

For arbitrarily given two states $\hat V_i=(\hat v_1^i,\,\hat v_2^i,\,\hat v_3^i,\,\hat v_4^i)\in {\mathcal {V}}_{\delta },\,i=1,\,2$ with the corresponding fluid variable $(\hat u_1^i,\,\hat u_2^i,\,\hat B^i,\,\hat \xi ^i)$, set

\[ V_i=T\hat V_i,\quad i=1,2, \]

where $V_i=(v_1^i,\,v_2^i,\,v_3^i,\,v_4^i)$. For the convenience, we denote $\hat W=\hat V_1-\hat V_2$ and $W=V_1-V_2$, or equivalently,

\[ \hat w_k=\hat v_k^1-\hat v_k^2,\quad w_k=v_k^1-v_k^2,\quad 1\leq k \leq 4. \]

Equation (4.2) implies that

(5.7)\begin{equation} w_4'=b_0w_2+O(\epsilon)\sum_{i=1}^{4}\hat w_i, \end{equation}

which yields that

(5.8)\begin{equation} \|w_4'\|_{C^{1,\alpha}[{-}1,1]}\leq C\|w_2\|_{C^{1,\alpha}(\bar Q)}+C\epsilon \left(\sum_{i=1}^{3}\|\hat w_i\|_{C^{1,\alpha}(\bar Q)}+\|\hat w_4\|_{C^{1,\alpha}[{-}1,1]}\right). \end{equation}

It follows from (4.6) that

(5.9)\begin{equation} w_3=b_3 w_4-b_3\int_{\beta_1(y)}^{\beta_2(y)}(\hat v_4^1(\tau))'\,{\rm d}\tau+b_3\int_{y_2}^{\beta_2(y)}\hat w_4'(\tau)\,{\rm d}\tau+O(\epsilon)\sum_{i=1}^{3}\hat w_i, \end{equation}

where $\beta _i,\,i=1,\,2$ is the initial position such that the corresponding characteristic $y_2^i(s,\,\beta _i)$ going through $(y_1,\,y_2)$ with $y_2^i(L_s)=\beta _i$. It is easy to verify that

(5.10)\begin{equation} \|\beta_1(y)-\beta_2(y)\|_{C^{1,\alpha}(\bar Q)}\leq C(\|\hat w_1\|_{C^{1,\alpha}(\bar Q)}+\|\hat w_2\|_{C^{1,\alpha}(\bar Q)}+\|\hat w_4\|_{C^{2,\alpha}[{-}1,1]}), \end{equation}

thus,

(5.11)\begin{equation} \|w_3\|_{C^{1,\alpha}(\bar Q)}\leq C\| w_4\|_{C^{1,\alpha}[{-}1,1]}+C\epsilon\left(\sum_{i=1}^{3}\|\hat w_i\|_{C^{1,\alpha}(\bar Q)}+\|\hat w_4\|_{C^{2,\alpha}[{-}1,1]}\right). \end{equation}

It is straightforward to show that $w_1,\,w_2$ satisfies

(5.12)\begin{equation} \begin{cases} \partial_{y_1} w_1+\frac{1}{1-\bar M^2}\partial_{y_2}w_2+\lambda_1(y_1)w_1+\lambda_2(y_1)(w_4({-}1)+\int_{{-}1}^{y_2}b_0w_2(L_s,\tau)\,{\rm d}\tau)\\ \quad = \sum_{i=1}^{4}(O(\epsilon)\hat w_i+O(\epsilon)\nabla \hat w_i)+O(\epsilon)(\beta_1-\beta_2)+O(1)\int_{y_2}^{\beta_2(y)}\hat w_4'(\tau)\,{\rm d}\tau,\\ \partial_{y_1}w_2-\partial_{y_2}[w_1-\lambda(y_1) (w_4({-}1)+\int_{{-}1}^{y_2}b_0w_2(L_s,\tau) \,{\rm d}\tau)] \\ = \sum_{i=1}^{4}(O(\epsilon)\hat w_i+O(\epsilon)\nabla \hat w_i)+O(\epsilon)\partial_{y_2}(\beta_1-\beta_2)+O(1)\partial_{y_2}\int_{y_2}^{\beta_2(y)}\hat w_4'(\tau)\,{\rm d}\tau, \end{cases} \end{equation}

and the boundary conditions

(5.13)\begin{equation} \begin{cases} w_1(L_s,y_2)=b_2(\hat w_4({-}1)+\int_{{-}1}^{y_2}b_0\hat w_2(L_s,\tau) \,{\rm d}\tau)+\sum_{i=1}^{4}O(\epsilon)\hat w_i,\\ w_1(L_1,y_2)=\frac{b_3}{\bar u(L_1)}(\hat w_4({-}1)+\int_{{-}1}^{y_2}b_0\hat w_2(L_s,\tau) \,{\rm d}\tau)+\sum_{i=1}^{4} O(\epsilon)\hat w_i\\ \quad +\, O(\epsilon)(\beta_1-\beta_2)(L_1,y_2) +O(1)\int_{y_2}^{\beta_2(L_1,y_2)}\hat w_4'(\tau)\,{\rm d}\tau,\\ w_2(y_1,\pm 1)=0, \end{cases} \end{equation}

where

\[ \lambda_2(y_1)=\frac{B_3(y_1)b_3+B_4(y_1)}{c^2(\bar\rho^+)-(\bar u^+)^2}. \]

Then, applying estimate (4.31) to system (5.12)–(5.13) with $k=1$ and together with (5.10), we obtain

(5.14)\begin{equation} \|(w_1,w_2)\|_{C^{1,\alpha}(\bar Q)}+|w_4({-}1)|\leq C\epsilon\left(\sum_{i=1}^{3}\|\hat w_i\|_{C^{1,\alpha}(\bar Q)}+\|\hat w_4\|_{C^{2,\alpha}[{-}1,1]}\right). \end{equation}

Finally, collecting all these estimates above leads to

(5.15)\begin{equation} \sum_{i=1}^{3}\| w_i\|_{C^{1,\alpha}(\bar Q)}+\| w_4\|_{C^{2,\alpha}[{-}1,1]}\leq C\epsilon\left(\sum_{i=1}^{3}\|\hat w_i\|_{C^{1,\alpha}(\bar Q)}+\|\hat w_4\|_{C^{2,\alpha}[{-}1,1]}\right). \end{equation}

By (5.15), there is a small constant $\epsilon _0$ such that for all $\epsilon \in (0,\,\epsilon _0]$, the mapping $T$ defined by (5.6) is contractible in the Banach space $(C^{1,\alpha }(\bar Q))^3\times C^{2,\alpha }[-1,\,1]$. Therefore, there exists a unique solution $V$ in $(C^{1,\alpha }(\bar Q))^3\times C^{2,\alpha }[-1,\,1]$. Due to lemma 4.4 and a priori estimates (5.2), (5.4), we know that $V$ also belongs to ${\mathcal {V}}_{\delta }$. It follows that $V$ satisfies estimates (3.23). Thus, the proof of theorem 3.1 is complete. Theorem 2.3 is a direct inference of theorem 3.1. We omit the details.