\documentclass[reqno]{amsart}
\usepackage{hyperref}
\usepackage{graphicx}

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2017 (2017), No. 156, pp. 1--18.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu}
\thanks{\copyright 2017 Texas State University.}
\vspace{8mm}}

\begin{document}
\title[\hfilneg EJDE-2017/156\hfil Axisymmetric solutions of the 2D NWS]
{Axisymmetric solutions of a two-dimensional
nonlinear wave system with a two-constant equation of state}

\author[G. Wang, Y. Hu, H. Liu \hfil EJDE-2017/156\hfilneg]
{Guodong Wang, Yanbo Hu, Huayong Liu}

\address{Guodong Wang (corresponding author)\newline
School of Mathematics and Physics,
Anhui Jianzhu University of China,
Hefei 230601, China}
\email{gdwang@163.com}

\address{Yanbo Hu \newline
Department of Mathematics,
 Hangzhou Normal University of China,
Hangzhou 310036, China}
\email{yanbo.hu@hotmail.com}

\address{Huayong Liu \newline
School of Mathematics and Physics,
Anhui Jianzhu University of China,
Hefei 230601, China}
\email{liuhy@ahjzu.edu.cn}

\thanks{Submitted February 17, 2017. Published June 28, 2017.}
\subjclass[2010]{35L65, 35J70, 35R35}
\keywords{Nonlinear wave system; generalized Chaplygin gas; 
\hfill\break\indent axisymmetry;
decoupled system}

\begin{abstract}
 We study a special class of Riemann problem with axisymmetry for
 two-dimensional nonlinear wave equations with the equation of state
 $p=A_1\rho^{\gamma_1}+A_2\rho^{\gamma_2}$, $A_i<0$, $-3<\gamma_i<-1$
 ($i=1,2)$. The main difficulty lies in that the equations can not
 be directly reduced to an autonomous system of ordinary differential
 equations. To solve it, we use the axisymmetry and self-similarity
 assumptions to reduce the equations to a decoupled system which includes
 three components of solution. By solving the decoupled system, we obtain the
 structures of the corresponding solutions and their existence.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\allowdisplaybreaks

\section{Introduction}\label{S1}

This article concerns a special class of Riemann
problem to the two-dimensional nonlinear wave system
\begin{equation} \label{NLWeqn}
\begin{gathered}
 \rho_t+(\rho u)_x+(\rho v)_y=0, \\
 (\rho u)_t+p_x=0, \\
 (\rho v)_t+p_y=0,
 \end{gathered}
\end{equation}
with the equation of state
\begin{align}\label{SOS}
p(\rho)=A_1\rho^{\gamma_1}+A_2\rho^{\gamma_2},
\end{align}
where the variables $(u,v)$, $\rho$ represent the velocity and the
density respectively, and $A_i<0$, $-3<\gamma_i<-1$ ($i=1, 2$).
 System \eqref{NLWeqn} can be obtained
either by starting with the isentropic gas dynamics equations and
neglecting the quadratic terms in the velocity or by writing the
nonlinear wave equation as a first-order system
\cite{SBE2002,SBE2006}. If the equation of state is taken as the
following form

\begin{equation}\label{GGCG}
p=A\rho^{\gamma}
\end{equation}
with $-1\leq\gamma<0$ and $A<0$, \eqref{GGCG} is called as the
generalized Chaplygin gas. For the case $\gamma=-1$ and $A=-1$,
\eqref{GGCG} was introduced by Chaplygin\cite{Chap1904}, Tsien and
von Karman \cite{Tsien1939,Karman1941} as a suitable mathematical
approximation for calculating the lifting force on a wing of an
airplane in aerodynamics. Sen and Scherrer generalized this model to
allow for the case $\gamma<-1$. The transient Chaplygin gas model
provides a possible mechanism to allow for a currently accelerating
universe without a future horizon and can be taken to be a model for
dark energy alone \cite{Sen2005}.


The initial data for a general Riemann problem are constant along
radial direction from the origin and piecewise constant as a
function of angle. There are a set of conjectures offered by Zhang
and Zheng in \cite{ZZ90} for the solutions to the two-dimensional
Riemann problem. Following Zhang and Zheng's work, many efforts have
been made to prove these conjectures for the compressible Euler
equations or some reduced systems in the past twenty years
\cite{LZY,Z}. Unfortunately, until now, none of them has been
completely proved due to the complicated structure of solutions. For
related results, we can consult the survey \cite{LSZZ} and
references cited therein.

The study of two-dimensional Riemann problem (or shock reflection)
have been devoted for system \eqref{NLWeqn} with general polytropic
gas
\cite{SBE2006, ChDX2014,HW2014,HW2015,EHK2010,EHK2012,EL2013a,EL2013b,Kim2016,
 TSK2006}.
\v{C}ani\'e, Keyfitz, Kim studied the Mach stem of
shock solution by solving a free boundary problem \cite{SBE2006}.
Tesdall, Sanders and Keyfitz presented numerical solutions for the
nonlinear wave system which is used to describe the Mach reflection
of weak shock waves \cite{TSK2006}. In particular, Kim and Lee
studied the configuration that the transonic shock interacted with
the sonic circle and obtained a global transonic solution to this
configuration. Kim established a global transonic solution to the
interaction of a transonic shock with a rarefaction
wave\cite{EHK2010,EHK2012,EL2013a,EL2013b,Kim2016}. Furthermore,
Chen et al. \cite{ChDX2014} established a global theory of existence
and optimal regularity for a shock diffraction problem. Hu ang Wang
\cite{HW2014,HW2015} studied the semi-hyperbolic patches of
solutions to the two-dimensional nonlinear wave system for Chaplygin
gases and constructed a global classical solution to the interaction
of two arbitrary planar rarefaction waves.

For a special class of Riemann problem with axisymmetry for the
two-dimensional isentropic Euler equations with the equation of
state $p=A\rho^{\gamma}$, $\gamma>1$, Zhang and Zheng constructed
rigorously a three-parameter family of self-similar, globally
bounded, and continuous weak solutions for all positive time to the
Euler equations \cite{ZZ97,ZZ98,Z09}. Under the assumption that
$v=0$, Hu \cite{Hu2014} constructed a family of self-similar and
global bounded weak solutions to the two-dimensional isentropic
Euler equations with the equation of state \eqref{SOS} and $A_i>0$,
$\gamma_i>1$ ($i=1, 2$) for axisymmetry initial data. As for the
related works, we refer the reader to
\cite{ChG1996,Guo1,Guo2,Hu2012}.


As a continuation of the paper \cite{Hu2014}, we will consider the
system
\begin{align}\label{phieqn}
\begin{gathered}
 \rho_t+(\rho u)_x+(\rho v)_y=0,\\
 (\rho u)_t+p_x=0,\\
 \varphi_t+(\varphi u)_x+(\varphi v)_y=0,\\
 (\rho v)_t+p_y=0,
 \end{gathered}
\end{align}
with the equation of state
\begin{align}\label{WSOS}
p=p_1+p_2=A_1\rho^{\gamma_1}+A_2\varphi^{\gamma_2},
\end{align}
for axisymmetry initial data,
where $p_1=A_1\rho^{\gamma_1},
p_2=A_2\varphi^{\gamma_2}$, $A_i<0, -3<\gamma_i<-1$ ($i=1, 2$),
$\rho\geq0$ and $\varphi\geq0$. It is easy to see that, if
$\varphi=\rho$, $\gamma_1=\gamma_2$ and $A_1=A_2$, then system
\eqref{phieqn} reduces to the system in \cite{WH2017}. The purpose
of the present paper is to investigate the two-dimensional Riemann
problem with axisymmetry to the nonlinear wave system \eqref{NLWeqn}
and \eqref{SOS}. We will construct rigorously a family of
self-similar and global solutions for all positive time to
\eqref{phieqn} with \eqref{WSOS}. The difference with this paper
\cite{Hu2014} is that we do not assume $v=0$ here, since we may
decouple the first three equations from the fourth equation in
\eqref{phieqn} by the axisymmetry and self-similarity assumptions.
Moreover, we find that the shock solutions appear in the case
$u_0>0$, instead of $u_0<0$. And then, we obtain the detailed
structures of solutions as well as their existence for system
\eqref{phieqn} with \eqref{WSOS}. If $\varphi=\rho$ at time zero in
the solutions obtained from system \eqref{phieqn}, we may obtain the
existence of solutions and their structures for the special class of
Riemann problem to system \eqref{NLWeqn} with \eqref{SOS}.

The rest of this paper is organized as follows. In section 2, we
present some preliminaries, give axisymmetry induction, intermediate
field equations and the Rankine-Hugoniot relations. The global
solutions for two cases $u_0\leq0$ and $u_0>0$ are constructed in
sections 3 and 4, respectively. The conclusion is delivered in
Section 5.

\section{Preliminaries}

We impose axisymmetry to the system \eqref{phieqn}. That is that the
solutions $(\rho,u,v,\varphi)$ have the property
\begin{equation}\label{Axisy}
\begin{gathered}
\rho(t,r,\theta)=\rho(t,r,0),\\
\varphi(t,r,\theta)=\varphi(t,r,0),\\
\begin{pmatrix}
u(t,r,\theta) \\ v(t,r,\theta)
\end{pmatrix}
=\begin{pmatrix}
\cos\theta & -\sin\theta \\ \sin\theta & \cos\theta
\end{pmatrix}
\begin{pmatrix}
u(t,r,0) \\ v(t,r,0)
\end{pmatrix},
\end{gathered}
\end{equation}
for all $t\geq0$, $\theta\in\mathbb{R}$ and $r>0$, where
$(r,\theta)$ are the polar coordinates of the $(x,y)$-plane.
Substituting \eqref{Axisy} into \eqref{phieqn} for the smooth
solutions reduces
\begin{equation}\label{Utr}
\begin{gathered}
\rho_{t}+(\rho u)_{r}+\frac{\rho u}{r}=0,\\
(\rho u)_t+(p_1+p_2)_r=0,\\
\varphi_{t}+(\varphi u)_{r}+\frac{\varphi u}{r}=0,\\
(\rho v)_t=0,
\end{gathered}
\end{equation}
where $u=u(t,r,0)$ and $v=v(t,r,0)$ are the radial and pure
rotational velocities, respectively. We require the Riemann initial
data as follows
\begin{equation*}%\label{AxisyRP}
\big(\rho(0,r,\theta),u(0,r,\theta),v(0,r,\theta),\varphi(0,r,\theta)\big)=
\big(\rho_0(\theta),u_0(\theta),v_0(\theta),\varphi_0(\theta)\big),
\end{equation*}
which implies, by the axisymmetry condition \eqref{Axisy}, that our
data are limited to
\begin{equation}\label{ThetaTr}
\begin{gathered}
\rho(0,r,\theta)=\rho_0,\\
\varphi(0,r,\theta)=\varphi_0,\\
u(0,r,\theta)=u_0\cos\theta-v_0\sin\theta,\\
v(0,r,\theta)=u_0\sin\theta+v_0\cos\theta,
\end{gathered}
\end{equation}
where $\rho_0>0$, $\varphi_0>0$, $u_0$ and $v_0$ are arbitrary
constants.

Since the problem \eqref{Utr} is invariant under self-similar
transformations, we look for self-similar solutions
$(\rho,u,v,\varphi)(\xi)(\xi=r/t)$. Thus, we obtain from \eqref{Utr}
that
\begin{equation}\label{DUVPXi}
\begin{gathered}
\rho_{\xi}=-\frac{\rho u}{p_1'+\overline{\varphi}-\xi^2},\\
u_{\xi}=-\frac{u(p_1'+\overline{\varphi}-\xi u)}{\xi(p_1'+\overline{\varphi}-\xi^2)},
\\
\overline{\varphi}_{\xi}=-\frac{(\gamma_2-1)
\overline{\varphi}u}{p_1'+\overline{\varphi}-\xi^2},\\
v_{\xi}=\frac{uv}{p_1'+\overline{\varphi}-\xi^2},
\end{gathered}
\end{equation}
and
\begin{equation}\label{infty0}
\lim_{\xi\to+\infty}(\rho,u,\overline{\varphi},v)
=(\rho_0,u_0,\gamma_2p_2(\varphi_0)/\rho_0,v_0),
\end{equation}
where $\overline{\varphi}=\gamma_2p_2(\varphi)/\rho$. It is easy to
see that the first three equations of \eqref{DUVPXi} do not involve the
component $v$. Hence, we consider a subsystem for the
initial pure radial flow
\begin{equation}\label{DUPXi}
\begin{gathered}
\rho_{\xi}=-\frac{\rho u}{p_1'+\overline{\varphi}-\xi^2},\\
u_{\xi}=-\frac{u(p_1'+\overline{\varphi}-\xi u)}{\xi(p_1'
 +\overline{\varphi}-\xi^2)},\\
\overline{\varphi}_{\xi}=-\frac{(\gamma_2-1)\overline{\varphi}u}{p_1'
 +\overline{\varphi}-\xi^2},
\end{gathered}
\end{equation}
and
\begin{equation}\label{infty}
\lim_{\xi\to+\infty}(\rho,u,\overline{\varphi})
=(\rho_0,u_0,\gamma_2p_2(\varphi_0)/\rho_0).
\end{equation}

In this paper, by \eqref{DUPXi} and \eqref{infty}, we will
construct global solutions to the problem \eqref{DUVPXi} and
\eqref{infty0} for any $\rho_0>0$, $\varphi_0>0$ and
$u_0\in\mathbb{R},v_0\in\mathbb{R}$.

\subsection{Far-field solutions and intermediate field equations}

 Let $s=1/\xi$, then \eqref{DUPXi} and \eqref{infty}
become
\begin{equation}\label{DUPs}
\begin{gathered}
\frac{d\rho}{ds}=\frac{\rho u}{s^2(p_1'+\overline{\varphi})-1},\\
\frac{du}{ds}=\frac{u(sp_1'+s\overline{\varphi}- u)}
{s^2(p_1'+\overline{\varphi})-1},\\
\frac{d\overline{\varphi}}{ds}=\frac{(\gamma_2-1)
\overline{\varphi}u}{s^2(p_1'+\overline{\varphi})-1},
\end{gathered}
\end{equation}
and
\begin{equation}\label{Us}
(\rho,u,\overline{\varphi})|_{s=0}=(\rho_0,u_0,\gamma_2p_2(\varphi_0)/\rho_0).
\end{equation}
We can see that \eqref{DUPs} is well-posed and has a unique local
solution for any initial datum with $\rho_0>0$ and $\varphi_0>0$.

By introducing the variables
\begin{equation}\label{HIK}
I=su,\quad K=s\sqrt{p_1'(\rho)},\quad
H=s\sqrt{\overline{\varphi}},
\end{equation}
system \eqref{DUPs} can be put into the form
\begin{equation}\label{HIKs}
\begin{gathered}
 s\frac{dI}{ds}=\frac{I(1+I-2K^2-2H^2)}{1-K^2-H^2},\\
 s\frac{dK}{ds}=
\frac{K(1-\frac{\gamma_1-1}{2}I-K^2-H^2)}{1-K^2-H^2},\\
s\frac{dH}{ds}=\frac{H(1-\frac{\gamma_2-1}{2}I-K^2-H^2)}{1-K^2-H^2}.
\end{gathered}
\end{equation}
Introducing a new parameter $\tau$, it is easy to get from
\eqref{HIKs} that
\begin{equation}\label{HIKtau}
\begin{gathered}
 \frac{dI}{d\tau}=I(1+I-2K^2-2H^2),\\
 \frac{dK}{d\tau}=K(1-\frac{\gamma_1-1}{2}I-K^2-H^2),\\
 \frac{dH}{d\tau}=H(1-\frac{\gamma_2-1}{2}I-K^2-H^2).
\end{gathered}
\end{equation}
and
\begin{equation}\label{stau}
 \frac{ds}{d\tau}=s(1-K^2-H^2).
\end{equation}
Corresponding to the initial data \eqref{HIK}, we will look for
solutions of \eqref{HIKtau} and \eqref{stau} with the initial
condition
\begin{equation}\label{IKs0}
(I,K,H)\sim
s(u_0,\sqrt{p_1'(\rho_0)},\sqrt{\gamma_2p_2(\varphi_0)/\rho_0}),\quad
\text{as } s\to 0^+.
\end{equation}

\subsection{Rankine-Hugoniot relation}

In self-similar coordinates $(\xi,\eta)=(x/t,y/t)$, system
\eqref{phieqn} can be rewritten as
\begin{equation*}%\label{SelfEqn}
\begin{gathered}
 -\xi\rho_{\xi}-\eta\rho_{\eta}+(\rho u)_{\xi}+(\rho v)_{\eta}=0,
 \\
 -\xi(\rho u)_{\xi}-\eta(\rho u)_{\eta}+(p_1+p_2)_{\xi}=0, \\
 -\xi\varphi_{\xi}-\eta\varphi_{\eta}+(\varphi u)_{\xi}+(\varphi v)_{\eta}=0,\\
 -\xi(\rho v)_{\xi}-\eta(\rho v)_{\eta}+(p_1+p_2)_{\eta}=0.
 \end{gathered}
\end{equation*}
Let a discontinuous curve be given $\eta=\eta(\xi)$ and
the slope of the curve be denoted by $\sigma=\eta'(\xi)$, then the
Rankine-Hugoniot relation is
\begin{equation}\label{RH_EQN}
\begin{gathered}
 [(u-\xi)\rho]\sigma=[(\eta-v)\rho],\\
 [\xi\rho u-p_1-p_2]\sigma=[\eta\rho u],\\
 [(u-\xi)\varphi]\sigma=[(\eta-v)\varphi],\\
 [\xi\rho v]\sigma=[\eta\rho v-p_1-p_2],
\end{gathered}
\end{equation}
where $\sigma=\eta'(\xi)$ and $[h]=h_r-h_{\ell}$, $h_{\ell}, h_r$
are the limit states on the left- and right-hand sides of the
discontinuity curve $\eta=\eta(\xi)$, respectively. When axisymmetry
is given, the discontinuity curve has an infinite slope at the
$\xi$-axis, that is $\sigma=\infty$ in \eqref{RH_EQN}. So we can
obtain
\begin{equation*} %\label{RH_EQN1}
\begin{gathered}
 \xi[\rho]=[\rho u],\\
 \xi[\rho u]=[p_1+p_2],\\
 \xi[\varphi]=[\varphi u],\\
 [\rho v]=0,
\end{gathered}
\end{equation*}
which yields the slip line
\begin{equation*}\label{JEqn}
\xi=0,\quad p_{1\ell}+p_{2\ell}=p_{1r}+p_{2r},
\end{equation*}
and the forward or backward discontinuity
\begin{equation}\label{RH_EQN2}
\begin{gathered}
 \xi=\pm\Big(\frac{[p_1+p_2]}{[\rho]}\Big)^{1/2},\\
 (u_r-\xi)\rho_r=(u_{\ell}-\xi)\rho_{\ell},\\
 (u_r-\xi)\varphi_r=(u_{\ell}-\xi)\varphi_{\ell},\\
 \rho_rv_r=\rho_1v_{\ell}.
\end{gathered}
\end{equation}
Recalling $\xi=1/s$ and \eqref{HIK}, the system \eqref{RH_EQN2} can
be transformed to
\begin{equation}\label{RH_IK}
\begin{gathered}
 (1-I_r)K_r^{\frac{2}{\gamma_1-1}}
 =(1-I_{\ell})K_{\ell}^{\frac{2}{\gamma_1-1}},\\
 (1-I_r)H_r^{\frac{2}{\gamma_2}}K_r^{\frac{2}{\gamma_2(\gamma_1-1)}}
 =(1-I_{\ell})H_{\ell}^{\frac{2}{\gamma_2}}K_{\ell}^{\frac{2}{\gamma_2(\gamma_1-1)}},\\
 K_r^{\frac{2}{\gamma_1-1}}-K_{\ell}^{\frac{2}{\gamma_1-1}}
=\frac{1}{\gamma_1}\big(K_r^{\frac{2\gamma_1}{\gamma_1-1}}-K_{\ell}^{\frac{2\gamma_1}{\gamma_1-1}}\big)
+\frac{1}{\gamma_2}\big(H_r^2K_r^{\frac{2}{\gamma_1-1}}-H_{\ell}^2K_{\ell}^{\frac{2}{\gamma_1-1}}\big),\\
\rho_r v_r=\rho_{\ell}v_{\ell}.
\end{gathered}
\end{equation}


\section{Solutions for $u_0\leq 0$}

In this section, we construct global bounded continuous solutions to
problem \eqref{DUPXi} and \eqref{infty} with initial negative
radial velocity. Here we only consider the case
$\gamma_1\neq\gamma_2$, the result for case $\gamma_1=\gamma_2$,
which corresponds to the generalized Chaplygin gas, can be found in
\cite{WH2017}. If $u_0=0$, we get from \eqref{DUPs} that
$\rho=\rho_0$, $\overline{\varphi}=\gamma_2p_2(\varphi_0)/\rho_0$
and $u=0$ is a trivial solution. So we only consider the case
$u_0<0$ in the present section and divide this into two cases:
$-3<\gamma_1<\gamma_2<-1$ and $-3<\gamma_2<\gamma_1<-1$.

We introduce
\begin{gather*}
A:=1-\frac{\gamma_2-1}{2}I-K^2-H^2,\\ 
B:=1+I-2K^2-2H^2,\\
C:=1-\frac{\gamma_1-1}{2}I-K^2-H^2,\\
D:=1-K^2-H^2.
\end{gather*}
Then \eqref{HIKtau} and \eqref{stau} become
\begin{gather*}
\frac{dI}{d\tau}=IB,\quad\frac{dK}{d\tau}=KC,\quad\frac{dH}{d\tau}=HA,\label{ABC}\\
\frac{ds}{d\tau}=D.\label{D}
\end{gather*}

\subsection{Case one: $-3<\gamma_1<\gamma_2<-1$}

We denote the set
\begin{equation}\label{Omg1}
\begin{aligned}
\Omega_1:\Big\{&
I<0,\; K>0,\; H>0;\quad 
 B>0 \text{ for } -1<I\leq 1/\gamma_2,\\
& A>0 \text{ for } 1/\gamma_2 \leq I<0,
\Big\}
\end{aligned}
\end{equation}
see Figure \ref{fig1}.

\begin{figure}[htbp] %\label{CB1}
\begin{center}
 \includegraphics[width=0.9\textwidth]{fig1} % CB1.eps
\end{center}
\caption{Integral curves for $u_0<0$ in $\Omega_1$.}
 \label{fig1}
\end{figure}

It is not difficult to show that the far-fields solutions starting
at $s=0^+$ enter the region $\Omega_1$ and do not leave $\Omega_1$
as $s$ increases. Notice that $s>0$ is increasing function of $\tau$
in $\Omega_1$ by \eqref{D} and one can find that the stationary
points of \eqref{ABC} in the closure $\overline{\Omega}_1$ are
$(H,I,K)=(1,0,0)$, $(H,I,K)=(0,0,1)$, $(H,I,K)=(0,-1,0)$,
$Q_1=(0,\frac{1}{\gamma_1},\sqrt{\frac{1+\gamma_1}{2\gamma_1}})$,
$Q_2=(\sqrt{\frac{1+\gamma_2}{2\gamma_2}},\frac{1}{\gamma_2},0)$,
and the points on the edge
\begin{equation*}
E:\quad K^2+H^2=1,\quad I=0,\quad K>0,\quad H>0.
\end{equation*}
Therefore, there is no stationary point in the open region
$\Omega_1$ and all the stationary points are on the boundary of
$\Omega_1$.

\begin{lemma}\label{lem3.1}
Solutions inside $\Omega_1$ do not leave $\Omega_1$ from its sides
(excluding possibly edge or corners) as $s$ increases.
\end{lemma}

\begin{proof}
It is easy to see from \eqref{ABC} that the sides of $\Omega_1$ in
the surfaces $H=0$, $I=0$ or $K=0$ are invariant regions. We need
only to verify that no solution leaves $\Omega_1$ from the following
surfaces $\widetilde{A}$ and $\widetilde{B}$
\begin{gather*}
\widetilde{A}:=\big\{(H,I,K): A=0,\; H>0,\; K>0,\;
 1/\gamma_2 <I<0\big\},\\
\widetilde{B}:=\big\{(H,I,K): B=0,\; H>0,\; K>0,\;
 -1<I<1/\gamma_2 \big\}.
\end{gather*}
We easily obtain that in the coordinate order $(H,I,K)$ the outward
normal of surface $\widetilde{B}$ is given by
\begin{equation*}
\vec{n}_{_{\widetilde{B}}}=\big(4H, -1, 4K\big).
\end{equation*}
Noting that $A<0$ and $C<0$ and calculating the inner product of the
normal $\vec{n}_{_{\widetilde{B}}}$ with the tangent vector of an
integral curve of \eqref{ABC} on the surface $\widetilde{B}$ yields
\begin{equation*}
\vec{n}_{_{\widetilde{B}}}\cdot\Big(\frac{dH}{d\tau},\frac{dI}{d\tau},
\frac{dK}{d\tau}\Big)
=4H^2A-IB+4K^2C<0,
\end{equation*}
which implies that no solution leaves $\Omega_1$ from the surface
$\widetilde{B}$ as $s$ increases.

An outward normal of the surface $\widetilde{A}$ is given by
\begin{equation*}
\vec{n}_{_{\widetilde{A}}}=\big(2H,\ \frac{\gamma_2-1}{2},\ 2K\big)
\end{equation*}
in the order $(H,I,K)$. We similarly compute the inner product of
the normal $\vec{n}_{_{\widetilde{A}}}$ with the tangent vector of
the integral curve on the surface $\vec{n}_{_{\widetilde{A}}}$ to
obtain
\begin{equation*}
\vec{n}_{_{\widetilde{A}}}\cdot\Big(\frac{dH}{d\tau},
\frac{dI}{d\tau},\frac{dK}{d\tau}\Big)
=2H^2A+\frac{\gamma_2-1}{2}IB+2K^2C<0,
\end{equation*}
because $B<0$, $C<0$ and $\frac{\gamma_2-1}{2}<0$ on the surface
$\widetilde{A}$. Hence, no solution leaves $\Omega_1$ from the
surface $\widetilde{A}$. Thus, we complete the proof.
\end{proof}

Now, we study the local structure of integral curves at the
stationary points of \eqref{ABC}. We first notice that no integral
curve from inside $\Omega_1$ goes to $(H,I,K)=(0,0,1)$ or $Q_1$,
since the component $H$, by \eqref{ABC}, is an increasing function
of $\tau$ in $\Omega_1$ for $I\in\big(\frac{1}{\gamma_2},0\big)$.

Setting
\begin{equation*}
\hat{H}=H-\sqrt{\frac{1+\gamma_2}{2\gamma_2}},\quad\hat{I}
=I-\frac{1}{\gamma_2},\quad \hat{K}=K,
\end{equation*}
and linearizing system \eqref{ABC} at point $Q_2$ yields
\begin{equation}\label{LQ1}
\begin{gathered}
\frac{d\hat{I}}{d\tau}=\frac{1}{\gamma_2}\hat{I}
 -\frac{4}{\gamma_2}\sqrt{\frac{1+\gamma_2}{2\gamma_2}}\hat{H},\\
\frac{d\hat{K}}{d\tau}=\frac{\gamma_2-\gamma_1}{2\gamma_2}\hat{K},\\
\frac{d\hat{H}}{d\tau}=-\frac{\gamma_2-1}{2}\sqrt{\frac{1+\gamma_2}{2\gamma_2}}\hat{I}
 -\frac{1+\gamma_2}{\gamma_2}\hat{H}.
\end{gathered}
\end{equation}
We compute the eigenvalues to get
\begin{equation*}
\lambda_{Q_2}^0=\frac{\gamma_2-\gamma_1}{2\gamma_2}<0,\quad
\lambda_{Q_2}^-=-\frac{1+\gamma_2}{\gamma_2}<0,\quad
 \lambda_{Q_2}^+=1,
\end{equation*}
which indicate that the stationary point $Q_2$ is hyperbolic.
Similarly, we obtain that the stationary point $Q_1$ is also
hyperbolic. For each integral curve ending at $Q_2$, we find that
$s\to+\infty$ and thus $\xi\to0^+$ as
$\tau\to+\infty$ due to $D=\frac{\gamma_2-1}{2\gamma_2}>0$.
Hence, this family of integral curves (called the transitional
solutions) yields smooth solutions of problem \eqref{DUPXi} and
\eqref{infty} in the entire region $\xi\in(0,+\infty)$.

Now, we linearize system \eqref{ABC} at stationary points on curve
$E\cup\{(H,I,K)=(1,0,0)\}$. Setting
\begin{equation*}
\hat{H}=H-\alpha,\quad\hat{I}=I,\quad\hat{K}=K-\sqrt{1-\alpha^2}
\end{equation*}
for $0<\alpha\leq1$, one can obtain
\begin{equation}\label{LE}
\begin{gathered}
\frac{d\hat{I}}{d\tau}=-\hat{I},\\
\hspace{-0.1cm}\frac{d\hat{K}}{d\tau}
=-\frac{\gamma_1-1}{2}\sqrt{1-\alpha^2}\,\hat{I}
 -2(1-\alpha^2)\hat{K}-2\alpha\sqrt{1-\alpha^2}\,\hat{H},\\
\hspace{-0.1cm}\frac{d\hat{H}}{d\tau}=-\frac{\gamma_2-1}{2}\alpha\hat{I}
 -2\alpha\sqrt{1-\alpha^2}\,\hat{K}-2\alpha^2\hat{H}.
\end{gathered}
\end{equation}
which has three eigenvalues $\lambda_1=-2$, $\lambda_2=-1$ and
$\lambda_3=0$. So, solutions of \eqref{ABC} near
$E\cup\{(H,I,K)=(1,0,0)\}$ approach $E\cup\{(H,I,K)=(1,0,0)\}$
exponentially as $\tau\to+\infty$. From equation \eqref{D},
we find
\begin{equation*}
\ln\Big(\frac{s}{s_0}\Big)=\int_{\tau_0}^{\tau}\big(1-K^2-H^2\big)d\tau,
\end{equation*}
where $s_0=s(\tau_0)>0$. Hence, $s$ approaches a finite number as
$\tau\to+\infty$ since $1-K^2-H^2$ approaches zero
exponentially. We linearize system \eqref{ABC} at point
$(H,I,K)=(0,-1,0)$ and get
\begin{equation*}%\label{L-1}
\begin{gathered}
\frac{d\hat{I}}{d\tau}=-\hat{I},\\
\frac{d\hat{K}}{d\tau}=\frac{1+\gamma_1}{2}\hat{K},\\
\frac{d\hat{H}}{d\tau}=\frac{1+\gamma_2}{2}\hat{I},
\end{gathered}
\end{equation*}
which has three eigenvalues $\lambda_1=-1$,
$\lambda_2=\frac{1+\gamma_1}{2}<0$ and
$\lambda_3=\frac{1+\gamma_2}{2}<0$. Similarly, we conclude that the
parameter $s$ approaches a finite number as
$\tau\to+\infty$.

We now depict the integral curves inside $\Omega_1$ in Figure \ref{fig1}. It
is useful to observe that there exists a stable manifold of system
\eqref{ABC} at the point $Q_2$ (see \cite{Kelley,Carr} for details)
which contains the transitional curve in $H=0$, the transitional
integral curve in $K=0$ and the heteroclinic orbit from the point
$Q_1$. There are three kinds of integral curves relative to this
manifold. The first kind consists of integral curves that are below
the manifold and go to the stationary point $(0,-1,0)$. Each of the
second king is right on the manifold and goes to the stationary
point $Q_2$. The third kind consists of integral curves that are
above the manifold and go to $E\cup\{(1,0,0)\}$. None of the
integral curves from inside $\Omega_1$ goes to point
$(H,I,K)=(0,0,1)$ or $Q_1$.

We construct global solutions for \eqref{HIKs} when $u_0<0$ and
$-3<\gamma_1<\gamma_2<-1$. For each integral curve which ends at
point $(H,I,K)=(0,-1,0)$, there exists a solution
$(1/\rho,u,\overline{\varphi})=\Big(\big(\frac{K^2}{-A_1\gamma
s^2}\big)^{\frac{1}{1-\gamma_1}}, I/s,H^2/s^2\Big)$ defined for
$s\in(0,s_1^{*}]$ for some positive number $s_1^{*}<+\infty$, and
then continue the solution by $1/\rho=0$, $\overline{\varphi}=0$ in
$s\in[s_1^{*},+\infty)$. We here do not need to specify the function
$u$ in the above state since system \eqref{ABC} has $1/\rho$ or
$\overline{\varphi}$ as a factor in every term. For each integral
curve ending at a point on $E\cup\{(1,0,0)\}$, there exists a
solution defined for $s\in(0,s_2^{*}]$ for some positive
$s_2^{*}<+\infty$. We next continue the solution by the constant
state
$(1/\rho,u,\overline{\varphi})=(1/\rho(s_2^{*}),0,\overline{\varphi}(s_2^{*}))$
in $s\in[s_2^{*},+\infty)$.


\subsection{Case two: $-3<\gamma_2<\gamma_1<-1$}

It can be verified that all far-field solutions of system
\eqref{HIKs} with $u_0<0$, $\rho_0>0$ and $\overline{\varphi}_0>0$
enter the domain $\Omega_2$ in $s>0$ where $\Omega_2\in\mathbb{R}^3$
be the set of points $(H,I,K)$ satisfying
\begin{equation}\label{Omg2}
\Omega_2:
\begin{aligned}\Big\{&
I<0,\; K>0,\; H>0;\quad
 B>0 \text{ for } -1<I\leq 1/\gamma_1,\\
& A>0\text{ for } 1/\gamma_1\leq I<0\Big\},
\end{aligned}
\end{equation}
see Figure \ref{fig2}.
 In this case, system \eqref{ABC} has similar stationary points with
 the case $\gamma_1<\gamma_2$. Analogously, there is no stationary
 point in the interior of $\Omega_2$. So, we can get the following
 lemma.


 \begin{figure}[htbp]
\begin{center}
 \includegraphics[width=0.9\textwidth]{fig2} % CB2.eps
\end{center}
\caption{Integral curves for $u_0<0$ in $\Omega_2$.}
 \label{fig2}
\end{figure}


\begin{lemma}\label{lem3.2}
Solutions inside $\Omega_2$ do not leave $\Omega_2$ from its sides
(excluding possibly edge or corners) as $s$ increases.
\end{lemma}

\begin{proof}
It is easy to see from \eqref{ABC} that the sides of $\Omega_2$ in
the surfaces $H=0$, $I=0$ or $K=0$ are invariant regions. We need
only to verify that no solution leaves $\Omega_2$ from the following
surfaces $\widetilde{B}$ and $\widetilde{C}$
\begin{gather*}
\widetilde{B}:=\Big\{(H,I,K): B=0,\; H>0,\; K>0,\;
 -1<I<\frac{1}{\gamma_1}\Big\},\\
\widetilde{C}:=\Big\{(H,I,K): C=0,\; H>0,\; K>0,\;
 \frac{1}{\gamma_1}<I<0\Big\}.
\end{gather*}

Analogous to the proof of Lemma \ref{lem3.1}, we can obtain that no
solution leaves $\Omega_2$ from the surface $\widetilde{B}$ as $s$
increases. An outward normal of the surface $\widetilde{C}$ is
\begin{equation*}
\vec{n}_{_{\widetilde{C}}}=\big(2H,\ \frac{\gamma_1-1}{2},\ 2K\big)
\end{equation*}
in the order $(H,I,K)$. We directly compute on the surface
$\widetilde{C}$
\begin{equation*}
\vec{n}_{_{\widetilde{C}}}\cdot\Big(\frac{dH}{d\tau},\frac{dI}{d\tau},
\frac{dK}{d\tau}\Big)
=2H^2A+\frac{\gamma_1-1}{2}IB+2K^2C<0,
\end{equation*}
since $A<0$, $B<0$ and $\frac{\gamma_1-1}{2}<0$ on the surface
$\widetilde{C}$. Hence, no solution leaves $\Omega_2$ from the
surface $\widetilde{C}$. These above indicate that the Lemma is
true.
\end{proof}

Similar to the case $\gamma_1<\gamma_2$, no integral curve inside
$\Omega_2$ goes to points $(H,I,K)=(1,0,0)$ or $Q_2$, since the
function $K$ is an increasing function of $\tau$ in $\Omega_2$ for
$I\in(0,1/\gamma_1)$ by \eqref{ABC}. For the local structure of
solutions at $E\cup\{(0,0,1)\}$ and the point $(H,I,K)=(0,-1,0)$, we
may see the related results for the case $\gamma_1<\gamma_2$ and
here omit the details. We next consider the local structure of
solutions at the stationary point $Q_1$. Setting
\begin{equation*}
\hat{H}=H,\quad\hat{I}=I-\frac{1}{\gamma_1},\quad
\hat{K}=K-\sqrt{\frac{1+\gamma_1}{2\gamma_1}},
\end{equation*}
and linearizing system \eqref{ABC} gives
\begin{equation*}%\label{LQ12}
\begin{gathered}
\frac{d\hat{I}}{d\tau}=\frac{1+\gamma_1}{2\gamma_1}\hat{I}
 -\frac{4}{\gamma_1}
\sqrt{\frac{1+\gamma_1}{2\gamma_1}}\hat{K},\\
\frac{d\hat{K}}{d\tau}=
 -\frac{\gamma_1-1}{2}\sqrt{\frac{1+\gamma_1}{2\gamma_1}}\hat{I}
 -\frac{1+\gamma_1}{\gamma_1}\hat{K},\\
\frac{d\hat{H}}{d\tau}=\frac{\gamma_1-\gamma_2}{2\gamma_1}\hat{H},
\end{gathered}
\end{equation*}
which three eigenvectors are
\begin{equation*}
\lambda_{Q_1}=\frac{\gamma_1-\gamma_2}{2\gamma_1},\quad
\lambda^{\pm}_{Q_1}=\frac{(1+\gamma_1)\pm\sqrt{(1+\gamma_1)
(25\gamma_1-7)}}{-4\gamma_1}.
\end{equation*}
In this case, we easily see that $\lambda_{Q_1}<0$,
$\lambda^{-}_{Q_1}<0$ and $\lambda^{+}_{Q_1}>0$, which indicates
that the stationary point $Q_1$ is hyperbolic. The hyperbolicity of
the stationary point $Q_2$ can be established in a similar way.

The integral curves in $\Omega_2$ in Figure \ref{fig2} can be depicted as
follows. There also exists a stable manifold of system \eqref{ABC}
for $-3<\gamma_2<\gamma_1<-1$ at the point $Q_1$ which contains the
transitional integral curve in $H=0$, the transitional integral
curve in $K=0$ and the heteroclinic orbit from the point $Q_2$.
There are three kinds of integral curves relative to this stable
manifold. The first kind consists of integral curves that are below
the manifold and go to the stationary point $(0,-1,0)$. Each of the
second kind is right on the manifold and goes to the stationary
point $Q_1$. The third kind consists of integral curves that are
above the manifold and go to $E\cup\{(0,0,1)\}$. None of the
integral curves from inside $\Omega_2$ goes to $(1,0,0)$ or $Q_2$.

Now the construction of the global solutions of system \eqref{ABC}
for $u_0<0$ and $-3<\gamma_2<\gamma_1<-1$. For each integral curve
ending at $(H,I,K)=(0,-1,0)$, we continue the solution by vacuum
state $1/\rho=0, \overline{\varphi}=0$ in $s\in[s^{*}_3,+\infty)$
for some positive $s^{*}_3<+\infty$. We also do not specify the
function $u$ in the vacuum. For each solution curve ending on
$E\cup\{(0,0,1)\}$, we continue the solution by the constat state
$(1/\rho(s_{4}^*), 0, \overline{\varphi}(s_{4}^*))$ in
$s\in[s_{4}^*, +\infty)$ for some positive finite number $s_{4}^*$.
The integral curves ending at point $Q_1$ (that is the second kind)
are already defined for all $s>0$ since $\tau\to+\infty$ and
the right hand side of equation \eqref{stau}) does not vanish at
$Q_1$.

Furthermore, from the last equation of \eqref{DUVPXi}, we have
\begin{align}\label{vR}
v(s)=v_0\exp\Big(\int_0^{s}\frac{u}{1-K^2-H^2}ds\Big),
\end{align}
which is a finite number for $u<0$ and $1-K^2-H^2>0$.

In summary, we have constructed global bounded continuous solutions to
system \eqref{DUVPXi} and \eqref{infty0} in the case $u_0\leq0$.

\section{Solutions for $u_0>0$}

In this section, we construct global solutions for problem
\eqref{DUPXi} and \eqref{infty} with initial positive pure radial
velocity for the case $\gamma_1<\gamma_2$. For the other case
$\gamma_1>\gamma_2$, we can obtain the same results, similarly.

It is not difficult to prove that the integral curves of system
\eqref{ABC} starting at the origin with $u_0>0, \rho_0>0$ and
$\overline{\varphi}_0>0$ enter the region
$\Omega_3:=\Omega_{31}\cup\Omega_{32}$, where
\begin{gather*}
\Omega_{31}:=\{(H,I,K): H>0,\; I>0,\ K>0,\; B>0,\; 0<I<1\},\\
\Omega_{32}:=\{(H,I,K): H>0,\; I>0,\ K>0,\; D>0,\; I>1\},
\end{gather*}
see Figure \ref{fig3}. Denote $\Omega_3':=\{H>0,I>0,K>0,D<0,B<0\}$.
Obviously, according to \eqref{D}, the variable $s$ is increasing in
region $\Omega_3$. We conclude, however, that the integral curves in
$\Omega_3$ do not always lie in it.

\begin{figure}[htbp]
\begin{center}
 \includegraphics[width=0.9\textwidth]{fig3} % S.eps
\end{center}
\caption{Integral curves for $u_0>0$ in the case
$\gamma_1<\gamma_2$.}
 \label{fig3}
\end{figure}


\begin{lemma}\label{lem4.1}
Each integral curve of system \eqref{ABC} from the origin passes
through the surface $D=0$ or $B=0$ at a finite point $(\tau, H, I,
K)$ and enters the domain $\Omega_3'$, and then crosses the surface
$C=0$ at a finite $\tau$ from $C>0$ to $C<0$.
\end{lemma}

\begin{proof}
To demonstrate the first part of the Lemma, we divide it
into two parts.

(1) If the integral curves from the origin do not enter
$\Omega_{32}$.
It is easy to obtain from $B>0$ that $2(K^2+H^2)<1+I$, from which,
we find that
\[
C=1-\frac{\gamma_1-1}{2}I-(K^2+H^2)\geq(K^2+H^2)-\frac{\gamma_1+1}{2}I\geq
K^2.
\]
From the equation for $K$, we find that
\begin{eqnarray*}
\frac{{\rm d}K}{{\rm d}\tau}=KC\geq K^3
\end{eqnarray*}
which implies that $K$ blows up at a finite $\tau=\hat{\tau}$.
Combining this and the fact $0<I<1$, we conclude that the integral
curve must pass through the surface $B=0$ and $D=0$, and then enters
the domain $\Omega_3'$.

(2) If the integral curves from the origin leave the region
$\Omega_{31}$ and then enter the region $\Omega_{32}$.
 Similar to the above procedure, the solution curves must pass through
the surface $D=0$ at finite $\overline{\tau}$. Now, we prove that
the above curves must pass through the surface $B=0$ at a finite
$\tau$, and then enter the domain $\Omega_3'$.

Suppose, on the contrary, such an integral curve is always under the
surface $B=0$. It follows from the $I$ equation that $I$ is
increasing for all $\tau$. Using the $K$ equation, $B>0$ and $I>1$,
we have
\begin{equation*}
\frac{d\ln
K}{d\tau}=1-\frac{\gamma_1-1}{2}I-(K^2+H^2)\leq1-\frac{\gamma_1-1}{2}I,
\end{equation*}
which means that
\begin{equation*}
\int_{\tau_0}^{\tau_m}\big(1-\frac{\gamma_1-1}{2}I\big)d\tau=+\infty,
\end{equation*}
for some positive number $\tau_m$, from which one can achieve
\begin{equation}\label{4.1eqn}
\int_{\tau_0}^{\tau_m}\big(-\frac{1+\gamma_1}{2}I+K^2+H^2\big)d\tau=+\infty.
\end{equation}
We arrange the $K$ equation in the following form
\begin{equation*}
\frac{dK}{d\tau}=K\big(B-\frac{1+\gamma_1}{2}I+K^2+H^2\big).
\end{equation*}
Integrating the above equality with respect to $\tau$ from $\tau_0$
to $\tau$ yields
\begin{equation}\label{Kvalue}
K=K_0\exp\big(\int_{\tau_0}^{\tau}Bd\tau\big)\cdot
\exp\Big(\int_{\tau_0}^{\tau}\big(-\frac{1+\gamma_1}{2}I+K^2+H^2\big)d\tau\Big),
\end{equation}
where $K_0=K(\tau_0)$. One can also obtain from the $I$ equation
that
\begin{equation}\label{Ivalue}
I=I_0\exp\Big(\int_{\tau_0}^{\tau}Bd\tau\Big),
\end{equation}
where $I_0=I(\tau_0)$. Combining \eqref{4.1eqn} with \eqref{Kvalue}
and \eqref{Ivalue} gives
\begin{eqnarray*}
\frac{K}{I}=\frac{K_0}{I_0}
\exp\Big(\int_{\tau_0}^{\tau}\big(-\frac{1+\gamma_1}{2}I+K^2+H^2\big)d\tau\Big)
\to +\infty,\quad\text{as } \tau\to\tau_m.
\end{eqnarray*}
On the other hand, one can obtain from the assumption $B>0$ and the
fact $I>1$ that
\begin{equation*}
2K^2<2(K^2+H^2)<1+I<(1+I)^2,
\end{equation*}
which gives
\begin{equation*}
\frac{\sqrt{2}K}{I}<1+\frac{1}{I}.
\end{equation*}
The right-hand side of the above inequality is bounded. So we get a
contradiction.

Next, we prove the second part of the Lemma, that is that all
integral curves from $\Omega_3'$ cross the surface $C=0$ at a finite
$\tau$. One can obtain that all integral curves from $\Omega_3'$ in
$I>0$ must end at $\overline{E}:=E\cup\{(1,0,0),(0,0,1)\}$. By the
linearization equation \eqref{LE} of \eqref{ABC}, we get that the
eigenvalues of \eqref{LE} are $\lambda_1=-2$, $\lambda_2=-1$ and
$\lambda_3=0$, and the corresponding eigenvectors are in turn
\begin{gather*}
\vec{r}_1=(\alpha,0,\sqrt{1-\alpha^2}),\\
\vec{r}_2=\Big(\alpha\big[(1+\gamma_2-2\gamma_1)-2(\gamma_2-\gamma_1)\alpha^2\big],\;
2,\; [(1-\gamma_1)-2(\gamma_2-\gamma_1)\alpha^2]\sqrt{1-\alpha^2}\Big),\\
\vec{r}_3=\big(\sqrt{1-\alpha^2},0,-\alpha\big)
\end{gather*}
for $0\leq\alpha\leq1$. Since $2\gamma_2<1+\gamma_1$ for
$-3<\gamma_1<\gamma_2<-1$, we compute the inner product of the
normal $\vec{n}_{_{\widetilde{C}}}$ with the vector $\vec{r}_2$ on
$\overline{E}$ to obtain
\begin{align*}
\vec{n}_{_{\widetilde{C}}}\cdot\vec{r}_2
&= \alpha^2\big[2(\gamma_2-\gamma_1)(1-\alpha^2)+1-\gamma_2\big]
 +(1-\alpha^2)\big[(1-\gamma_1)-2(\gamma_2-\gamma_1)\alpha^2\big] \\
&\quad +(\gamma_1-1)/2\\
&= (\gamma_1-\gamma_2)\alpha^2+(1-\gamma_1)/2>0,
\end{align*}
which indicates that each integral curve from $\Omega_3'$ will pass
through the surface $C=0$ at a finite $\tau$ and then go to a
stationary point on $\overline{E}$ along the direction $\vec{r}_2$.
\end{proof}

Using \eqref{D}, we can see that the variable $s(\tau)$
starts to decrease as $\tau$ increases in the domain $\{D<0\}$,
which means that no continuous solutions exist in the case $u_0>0$.
Hence we need to use discontinuous solutions to construct global
solutions.

Denote the subscripts $r$ and $\ell$ the front (the side close to
$\xi=+\infty$) and behind (the side close to the origin) states,
respectively. Here we have ignored the backward discontinuous
solutions since we will not need it. The entropy condition requires
that $\rho_r<\rho_\ell$, or equivalently, by \eqref{RH_IK},
\begin{equation}\label{EntropyCnd}
K_r>K_{\ell},\quad I_r<I_{\ell}\quad {\rm or}\quad H_r>H_{\ell}.
\end{equation}
Using the R-H relation \eqref{RH_IK}, we get the following lemma.

\begin{lemma}\label{lem4.2}
For any state $(H_{\ell}, I_{\ell}, K_{\ell})$ in the domain
$\{D>0\}\cap\{I_{\ell}>0\}$, there exists a state $(H_r, I_r, K_r)$
in the domain $\{D<0\}\cap\{I_r>0\}$ satisfying the R-H relation
\eqref{RH_IK} and the entropy condition \eqref{EntropyCnd}.
\end{lemma}

\begin{proof}
It is easy to see from the third equation of \eqref{RH_IK} that
\begin{equation}\label{fgKrl}
\begin{aligned}
1 &= \underbrace{\frac{1}{\gamma_1}\big(K_r^{\frac{2\gamma_1}{\gamma_1-1}}
 -K_{\ell}^{\frac{2\gamma_1}{\gamma_1-1}}\big)
\Big(K_r^{\frac{2}{\gamma_1-1}}-K_{\ell}^{\frac{2}{\gamma_1-1}}
 \Big)^{-1}}_{f(K_{\ell},K_r)}\nonumber\\
&\quad +\underbrace{\frac{1}{\gamma_2}\big(H_r^2K_r^{\frac{2}{\gamma_1-1}}
 -H_{\ell}^2K_{\ell}^{\frac{2}{\gamma_1-1}}\big)
\Big(K_r^{\frac{2}{\gamma_1-1}}-K_{\ell}^{\frac{2}{\gamma_1-1}}
 \Big)^{-1}}_{g(K_{\ell},K_r)}.
\end{aligned}
\end{equation}
Setting
\[
x=K_r^{\frac{2}{\gamma_1-1}} \quad\text{and}\quad
x_0=K_{\ell}^{\frac{2}{\gamma_1-1}},
\]
we have
\[
f(K_{\ell},K_r):=f(x_0,x)=\frac{1}{\gamma_1}
\frac{x^{\gamma_1}-x_0^{\gamma_1}}{x-x_0}.
\]
Since $\big(x^{\gamma_1}/\gamma_1\big)^{\prime}=x^{\gamma_1-1}>0$,
$\big(x^{\gamma_1}/\gamma_1\big)^{\prime\prime}=(\gamma_1-1)x^{\gamma_1-2}<0$
for $0<x<x_0$ and $-3<\gamma_1<-1$, we obtain from the entropy
condition \eqref{EntropyCnd} that
\begin{equation}\label{fKrl}
K_{\ell}^2<f(K_{\ell},K_r)<K_r^2.
\end{equation}
Now, we claim that
\begin{equation} \label{gKrl}
H_{\ell}^2<g(K_{\ell},K_r)<H_r^2.
\end{equation}
In fact, from \eqref{RH_IK}, we have
\[
H_{\ell}^2=\Big(\frac{K_{\ell}}{K_r}\Big)^{\frac{2(\gamma_2-1)}{\gamma_1-1}}H_r^2
:=y^{-(\gamma_2-1)}H_r^2,
\]
where $0<y=\big(K_r/K_{\ell}\big)^{2/(\gamma_1-1)}<1$. Thus
\[
g(K_{\ell},K_r)=
y^{-(\gamma_2-1)}H_r^2\cdot\frac{1}{\gamma_2}\frac{y^{\gamma_2}-1}{y-1},
\]
which gives
\begin{align*}
K_{\ell}^2= y^{-(\gamma_2-1)}H_r^2<g(K_{\ell},K_r)<H_r^2.
\end{align*}
Combining \eqref{fgKrl} and \eqref{fKrl} and \eqref{gKrl} gives
\begin{align}\label{KHrl1}
K_{\ell}^2+H_{\ell}^2<1<K_r^2+H_r^2,
\end{align}
from which it follows that $(H_r, I_r, K_r)\in\{D<0\}\cap\{I_r>0\}$.
\end{proof}

From the R-H relation and the entropy condition, we only have three
equations for four variables $(H(I_{\ell}), I_{\ell}, K(I_{\ell}))$
and $(H_r, I_r, K_r)$, so we need one more condition to find a
unique shock wave transition. Here $H_{\ell}=H(I_{\ell}),
K_{\ell}=K(I_{\ell})$ are determined by the integral curve of system
\eqref{ABC}.

According to \eqref{D}, the direction of $\tau$ is opposite to that
of $s$ in the domain $\{D<0\}$. Therefore, shock transition from a
state $(H_{\ell}, I_{\ell}, K_{\ell})$ in $D>0$ to a state $(H_r,
I_r, K_r)$ on the plane $I=0$ is the only one leading to a global
solution, since the integral curve on the plane $I=0$ yields
stationary constant $\rho$ and $\varphi$ solutions to system
\eqref{phieqn}. Thus, we require that
\begin{equation}\label{Ir0}
I_r=0.
\end{equation}

\begin{lemma}\label{lem4.3}
For any integral curve in the quadrant $\{H>0, I>0, K>0\}$ of the
autonomous system \eqref{ABC}, there exists a solution $(H_r, I_r,
K_r, H_{\ell}, I_{\ell}, K_{\ell})$ with $I_{\ell}>0$ satisfying the
R-H relation \eqref{RH_IK}, the entropy condition \eqref{EntropyCnd}
and the condition \eqref{Ir0}.
\end{lemma}

\begin{proof}
We establish this lemma by the method of continuity. Since each
integral curve of system \eqref{ABC} in the quadrant $\{H>0, I>0,
K>0\}$ crosses the surface $D=0$ at a finite $\tau$ by Lemma
\ref{lem4.1}, then if we take the cross point as $(H_{\ell},
I_{\ell}, K_{\ell})$, the solution to the R-H relation \eqref{RH_IK}
would be the same point, i.e., $(H_{\ell}, I_{\ell}, K_{\ell})=(H_r,
I_r, K_r)$. Now moving from this point down the integral curve, we
notice by the R-H relation \eqref{RH_IK} that $I_r$ will become
minus when $(H_{\ell}, I_{\ell}, K_{\ell})$ is near the origin. In
fact, if not, we then have $K^2_r+H^2_r>1$ by \eqref{KHrl1}. Thus,
one has by the R-H relation \eqref{RH_IK}
\begin{align*}
(1-I_{\ell})(K^{\frac{2}{\gamma_1-1}}_{\ell}
+H^{\frac{2}{\gamma_2}}_{\ell}K^{\frac{2}{\gamma_2(\gamma_1-1)}}_{\ell})
&=(1-I_r)(K^{\frac{2}{\gamma_1-1}}_r
+H^{\frac{2}{\gamma_2}}_r K^{\frac{2}{\gamma_2(\gamma_1-1)}}_r)\\
&\leq (K^{\frac{2}{\gamma_1-1}}_r +H^{\frac{2}{\gamma_2}}_r
K^{\frac{2}{\gamma_2(\gamma_1-1)}}_r)<c_*\hspace{0.43cm}
\end{align*}
for some positive constant $c_*$, which contradicts to the fact
$(H_{\ell}, I_{\ell}, K_{\ell})\to(0,0,0)$. Hence, we have
$I_r>0$ when the point $(H_{\ell}, I_{\ell}, K_{\ell})$ near the
origin. By continuity, we must have a solution
$(H_r, I_r, K_r, H_{\ell}, I_{\ell}, K_{\ell})$ with $I_{\ell}>0$ for equations
\eqref{RH_IK} and \eqref{Ir0} and the entropy condition
\eqref{EntropyCnd}.
\end{proof}

From \eqref{ABC} and \eqref{D}, we get a subsystem on the plane
$I=0$, that reads
\begin{gather*}
\frac{{\rm d}K}{{\rm d}\tau}=K(1-H^2-K^2), \\
\frac{{\rm d}H}{{\rm d}\tau}=H(1-H^2-K^2), \\
\frac{{\rm d}s}{{\rm d}\tau}=s(1-H^2-K^2),
\end{gather*}
which means that
\begin{align*}
\frac{{\rm d}K}{{\rm d}s}=\frac{K}{s}, \quad
\frac{{\rm d}H}{{\rm d}s}=\frac{H}{s},
\end{align*}
which imply that $\frac{K}{s}$ and $\frac{H}{s}$ are constant.
Then a point $(H, I, K)=( H_r, 0, K_r)$ on the plane $I=0$ gives a
constant solution to system \eqref{DUPs}
\begin{align*}
u=0, \quad \frac{1}{\rho}=\big(-A_1\gamma_1\big)^{\frac{1}{\gamma_1-1
}}(\xi^{**}K_{r})^{\frac{2}{1-\gamma_1}}, \quad
\overline{\varphi}=(\xi^{**}H_{r})^2, \quad(0<\xi<\xi^{**}),
\end{align*}
where $\xi^{**}=1/s^{**}$ is the radial coordinate $\xi$ of the
shock location.

We are now ready to construct global solutions for \eqref{DUPs} in
the case $u_0>0$. For any integral curve of \eqref{ABC}, there
exists a solution $(H_{\ell}, I_{\ell}, K_{\ell}, H_r, I_r, K_r)$
satisfying Lemma \ref{lem4.3} such that $(H_{\ell}, I_{\ell},
K_{\ell})$ on the integral curve and $(H_r, I_r, K_r)$ on the plane
$I=0$. For this integral curve from $(H, I, K)=(0, 0, 0)$ to
$(H, I,K)=(H_{\ell}, I_{\ell}, K_{\ell})$, we have a solution
$\Big(1/\rho,u, \overline{\varphi})=\big((K^2/(-A_1\gamma_1s^2))^{\frac{1}{1-\gamma_1
}}, I/s, H^2/s^2\big)$ defined for $s\in(0, s^{**}]$ for some
positive number $s^{**}<+\infty$ such that
$(H_{\ell}, I_{\ell},K_{\ell})=(H(s^{**}), I(s^{**}), K(s^{**}))$.
The solution for
$s\in[s^{**}, +\infty)$ is specified as $(1/\rho, u,
\overline{\varphi})=((K_{r}^2/(-A_1\gamma_1
s^{**2}))^{\frac{1}{1-\gamma_1}}, 0, H_{r}^2/s^{**2})$. We use a shock
wave to connect the two states $(1/\rho, u,
\overline{\varphi})=((K_{r}^2/(-A_1\gamma_1
s^{**2}))^{\frac{1}{1-\gamma_1}}, 0, H_{r}^2/s^{**2})$ and 
\[
(1/\rho, u, \overline{\varphi})=((K_{\ell}^2/(-A_1\gamma_1
s^{**2}))^{\frac{1}{1-\gamma_1}}, I_{\ell}/s^{**},
H_{\ell}^2/s^{**2})
\]
 at $s=s^{**}$. Here $s^{**}$ (i.e., the shock
location $\xi^{**}$) is determined by the R-H relation
\eqref{RH_IK}, the entropy condition \eqref{EntropyCnd}, the
compatibility condition \eqref{Ir0} and the autonomous system
\eqref{ABC}.

From the last equation in \eqref{RH_IK}, we have the $v$ component
of the solution is
\begin{align*}
v=\frac{\rho_{\ell}v_{\ell}}{\rho}.
\end{align*}


\section{Conclusions}
We first summarize the results for system \eqref{phieqn}.

\begin{theorem}
For any datum $(\rho_0, u_0, v_0, \varphi_0)$ with $\rho_0>0$ and
$\varphi_0>0$, there exists a global solution $(\rho, u, v, \varphi)$
to the initial-value problem \eqref{phieqn} and \eqref{ThetaTr}.
\end{theorem}

Similar to \cite{Hu2014}, we obtain from the system \eqref{DUPs}
for $\rho>0, \varphi>0$ that
$$
\frac{{\rm d}\overline{\varphi}}{\overline{\varphi}}
=(\gamma_2-1)\frac{{\rm d}\rho}{\rho} \quad {\rm or} \quad
\frac{{\rm d}\varphi}{\varphi}=\frac{{\rm d}\rho}{\rho},
$$
which means that our solutions satisfy
$\varphi/\rho=\varphi_0/\rho_0$ in the smooth case. If we restrict
$\varphi_0=\rho_0$ at initial time, then we have $\varphi=\rho$ in
the smooth case. Moreover, the constant states continued the smooth
parts of solutions for \eqref{DUVPXi} are also the constant
solutions to system \eqref{NLWeqn}. One can easily check that the
shock solution of system \eqref{phieqn} is also a shock solution to
system \eqref{NLWeqn} with \eqref{SOS} if the initial data satisfies
$\varphi_0=\rho_0$. Therefore, we have the following theorem.

\begin{theorem}
For any datum $(\rho_0, u_0, v_0)$ with $\rho_0>0$, there exists a
global solution $(\rho, u, v)$ to the initial-value problem
\eqref{NLWeqn} and \eqref{ThetaTr} when
$p(\rho)=A_1\rho^{\gamma_1}+A_2\rho^{\gamma_2}$ for any constants
$A_i<0, -3<\gamma_i<-1\ (i=1,2)$.
\end{theorem}


\subsection*{Acknowledgements}

G. D. Wang was supported by the Anhui Provincial Natural Science Foundation
(1508085M A08).
Y. B. Hu was supported by the National Science Foundation of China (11301128
and 11571088), Zhejiang Provincial Natural
Science Foundation (LY17A010019) and China Postdoctoral Science Foundation
(2015M580286).

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\end{document}