\begin{equation}gin{document}

\mathcal{A} ketitle

\begin{equation}gin{abstract} For the  initial boundary value problem of   compressible barotropic Navier-Stokes equations in one-dimensional bounded domains with   general density-dependent viscosity and large  external force,  we prove that there exists   a unique  global  classical  solution with large initial data containing vacuum.  Furthermore, we show that the density is bounded from above independently of time which in particular yields the large time behavior of the solution as time tends to infinity: the density and the velocity converge  to the steady states in $L^p$ and  in $W^{1,p}$  ($1\le p<+\infty$) respectively. Moreover, the decay rate in time of the solution is shown to be exponential.
Finally, we also prove  that the spatial gradient of the density
 will blow up as time tends to infinity when vacuum states appear initially even at one point.

\end{eqnarray}d{abstract}

\thetaxtbf{Keywords}:  compressible  Navier-Stokes equations; density-dependent  viscosity; vacuum;  global classical solution; large-time behavior.

\section{Introduction and main results}
  We consider the one-dimensional
compressible  Navier-Stokes equations which read as follows:
\begin{equation}\label{1d}
\begin{equation}gin{cases} \rho_t + (\rho u)_x = 0,\\
 (\rho u)_t + (\rho u^2)_x + [P(\rho)]_x -[\mu(\rho)u_x]_x=\rho f.
\end{eqnarray}d{cases}
\end{equation}
Here $t\gammae 0$ is time, $x\in \Omega=(0,1)$ is the spatial coordinate,
 $\rho$ and $ u $  represent respectively the fluid density and velocity.
The pressure $P$ is given by
\begin{equation}\label{n2}
P(\rho) = A\rho^{\gammaammamma}\, ( A>0,\, \gammaamma>1).
\end{equation}
 In the sequel, without loss of generality, we set $A= 1.$ The viscosity   $\mu(\rho)$   satisfies
\begin{equation}\label{1d1}
~~0<\begin{equation}gin{aligned}r{\mu}\le \mu(\rho),~~\forall~\rho\gammae 0,
\end{equation} where $\begin{equation}gin{aligned}r{\mu}$ is  a given positive constant. The function $f=f(x)$ is the external force.
 We look for the solutions, $(\rho(x,t),u(x,t)),$ to
 the initial-boundary-value problem with  the    boundary conditions:
\begin{equation}\label{1d2}
u(0,t)=u(1,t)=0,~~~t\gammae 0,
\end{equation}  and the initial ones:
\begin{equation} \label{n3}
(\rho,\rho u)(x,0)=(\rho_0 ,\rho_0u_0)(x ).
\end{equation}

There is huge literature on the studies of the global existence and large time behavior of solutions to the compressible Navier-Stokes equations.
For the initial density away from vacuum, there are  many results concerning the global existence and large-time dynamics of solutions to the
 one-dimensional(1D) problem, see \cite{Kaz,serre1,serre2,Hof,v1989,is2002,is2003,lij1d}  and the reference therein.

For the case of density-dependent viscosity,       Liu-Xin-Yang \cite{lxy} obtains that  the viscosity of a gas depends on the temperature for the non-isentropic case, and thus on
the density for the isentropic case. For $\mu(\rho)=\rho^\theta$, under different restrictions
on the index $\theta$ and the regularities  of initial data,  the global existence  of solutions to 1D compressible Navier-Stokes
equations is investigated  in \cite{ok,jxz,yyz,qyz,fz,yz,ljcmp2008} and the references therein.

 When $\mu(\rho)$  admits a positive constant lower bound,
 the global well-posedness of solutions without initial vacuum to 1D problem is discussed extensively (see \cite{kawohl, zl89,v1989,is2002,is2003}  and  the references therein).

 Recently, in addition to \eqref{1d1},  under some additional stringent condition on $\mu\in C^2[0,\infty)$:
\begin{equation}\label{wen}
  \mu(\rho)\le C(1+P(\rho)) ~~\forall~\rho\gammae 0,
\end{equation} Ding-Wen-Zhu \cite{wen2011} proves the  global existence of classical large solutions to   \eqref{1d}-\eqref{n2} with vacuum. However, since the   upper bound of the density obtained in   \cite{wen2011}  depends crucially on   time,   nothing is known concerning the large-time behavior of the solutions in \cite{wen2011}. For the initial density away from vacuum,  Stra\~skraba-Zlotnik \cite{is2002,is2003}     proves the large time behavior of the solution $(\rho,u)$ to \eqref{1d}.  More precisely, they show that the solution $(\rho,u)$ tends to the stationary case $(\rho_s,0)$ as time tends to infinity. Here,  the  stationary density $\rho_s$  is a solution to the  stationary problem:
  \begin{equation}\label{1s}
\begin{equation}gin{cases}  [P(\rho_s)]_x=\rho_s f,~~~x\in (0,1),\\
\int_0^1\rho_sdx=\int_0^1\rho_0dx.\end{eqnarray}d{cases}
\end{equation}

In this paper, we will study  the global existence and large-time behavior of  strong and  classical solutions to \eqref{1d}-\eqref{n3} not only for   general density-dependent  viscosity \eqref{n3} which is independent of $P(\rho)$ but also for the density containing vacuum initially.   

Before stating the main results, we first explain the notations and
conventions used throughout this paper.
For   $1\le r\le \infty, k\gammae 1$,
$$L^r=L^r(0,1 ),\quad W^{k,r}  = W^{k,r}(0,1) , \quad H^k = W^{k,2}(0,1).$$
Moreover,  without loss of generality, assume that the initial density $\rho_0$ satisfies
 \begin{equation}\label{oy3.7} \int_0^1  \rho_0dx=1.\end{equation}

Our first result concerns the global existence of strong solutions with large initial data.

 \begin{equation}gin{theorem}\label{th11}   Suppose that $f\in H^1$, $\mu \in C^1[0,\infty),$ and the initial data $(\rho_0,u_0)$ satisfy
\begin{equation}\label{1d3a}
  0\le\rho_0\in H^1,  ~u_0\in   H^1_0. \end{equation}  Then, there exists a unique strong  solution $(\rho,u)$ to the  problem  \eqref{1d}-\eqref{n3} satisfying for any $0< T<\infty,$
   \begin{equation}
   \label{1d5}\begin{equation}gin{cases}
   \rho  \in C([0,T]; H^1)\cap H^1(0,T;L^2), \\
    u\in L^\infty(0,T; H^1_0) \cap L^2(0,T;H^2), \\t^{1/2}u \in L^\infty(0,T; H^2),
~~ t^{1/2}u_t\in L^2(0,T; H^1_0)  .\end{eqnarray}d{cases} \end{equation}
 Moreover,  the density remains uniformly  bounded for all time, that is,
 \begin{equation}\label{1d6}
  \sup_{0\le t<\infty}\|\rho(\cdot,t)\|_{L^\infty} <\infty,
\end{equation}  and the following large-time behavior holds:
\begin{equation} \label{1.12'}\lim_{t\rightarrow \infty} \| u(\cdot,t)\|_{W^{1,p}} =0,~~~~\forall~p\in [1,\infty).\end{equation}
\end{eqnarray}d{theorem}

Then the following result shows that the strong solutions obtain by Theorem \ref{th11} become classical provided initial data $(\rho_0,u_0)$ satisfy some additional conditions.
  \begin{equation}gin{theorem}\label{th1}  In addition to \eqref{1d3a}, suppose that $f\in H^1$, $\mu \in C^2[0,\infty),$  and the initial data $(\rho_0,u_0)$ satisfy
\begin{equation}\label{1d3}
 \rho_0\in H^2, ~P(\rho_0)\in H^2,~u_0\in H^2\cap H^1_0, \end{equation}
and the following compatibility condition:
 \begin{equation}\label{1d4}
[\mu(\rho_0)u_{0x}]_x-[P(\rho_0)]_x=\rho_0^{1/2}g ,  \end{equation}
for some $g\in L^2$. Then, the strong solution $(\rho,u)$ obtained by Theorem \ref{th11} becomes  classical and satisfies for any $0< T<\infty,$
   \begin{equation}
   \label{1d5'}\begin{equation}gin{cases}
   \rho,   P(\rho) \in C([0,T]; H^2), \\
    u\in C([0,T]; H^2) \cap L^2(0,T;H^3), \\
 u_t\in L^2(0,T; H_0^1),~~t^{1/2}u \in L^\infty(0,T; H^3),\\
t^{1/2}u_t \in L^\infty(0,T; H^1)\cap L^2(0,T; H^2),~~  t^{1/2}\rho^{1/2}u_{tt}\in L^2(0,T; L^2).\end{eqnarray}d{cases} \end{equation}

\end{eqnarray}d{theorem}

For further  studying the large-time behavior of the strong solutions,   we  first state some known results about the existence and uniqueness of  positive solutions to the stationary problem \eqref{1s}, which has been  discussed extensively under different conditions (see for example \cite{zl1,zaa95,is2002,is2003}).
 \begin{equation}gin{lem}[\cite{zaa95}]\label{le2}   If $f\in L^\infty$ satisfies
 \begin{equation}\label{niu}\int_0^1 \left(1- \gammaammamma^{-1}\right)^{\frac{1}{\gammaammamma-1}} \left(\int_0^x f(y)dy-\min_{[0,1]}\int_0^x f(y)dy\right)^{\frac{1}{\gammaammamma-1}} dx<\int_0^1\rho_0dx,\end{equation} there exists a unique positive solution $\rho_s$ to \eqref{1s} which satisfies
 \begin{equation}\label{pre1}
\rho_s\in W^{1,\infty},~~  0<K_1\le \rho_s\le K_2,
\end{equation}
where $K_1$ and $K_2$  are positive constants depending on $\|f\|_{L^\infty}$.
\end{eqnarray}d{lem}

Then, we have the following result concerning  the  large time asymptotic behavior  of  strong solutions.

\begin{equation}gin{theorem}\label{th2}   Under the same conditions as in Theorem \ref{th11}, for  $f$ satisfying \eqref{niu},   there are    positive constants $\alphalpha $ and $C$ depending only on the initial data and $\|f\|_{H^1}$  such that for any $p\in [1,\infty)$ and any $t>1,$
\begin{equation}\labelbel{ed1}
 \|\rho(\cdot,t)-\rho_s(\cdot)\|_{L^p}+ \|u(\cdot,t)\|_{W^{1,p}}\le C  e^{-\frac{\alphalpha }{ p} t}.
\end{equation}
 Moreover, if  there exists some point $x_0\in [0,1]$ such that $\rho_0(x_0)=0,$  then the spatial gradient of the density the unique strong solution $(\rho, u)$ to  the  problem  \eqref{1d}-\eqref{n3}  has to blow up as $t\rightarrow \infty$ in the following sense,
\begin{equation}\labelbel{bp}
\lim_{t\rightarrow \infty}\|\rho_x(\cdot,t)\|_{L^r}=\infty,~~~\forall~r\in (1,\infty).
\end{equation}
\end{eqnarray}d{theorem}

A few remarks are in order:

\begin{equation}gin{remark}\label{re1}
It should be noted here that the solution  $(\rho,u)$  obtained in Theorem \ref{th1} is actually a classical one to  \eqref{1d}-\eqref{n3}.
Indeed, by the Sobolev embedding theorems, we have
$$H^k(0,1)\hookrightarrow C^{k-\frac{1}{2}}[0,1],~~~\mbox{for}~k=1,\cdots,3, $$
which together with   \eqref{1d5'}  gives
\begin{equation} \labelbel{ccc}(\rho,\,P,\, u) \in C([0 ,T];C^{1+\frac{1}{2}}[0,1]),~~ \rho_t   \in C([0 ,T];C^{\frac{1}{2}}[0,1]).\end{equation}
Furthermore, one can deduce from \eqref{1d5'} that  for any $0<\tau<T$,
$$u \in L^\infty(\tau,T; H^3),~~~u_t \in L^\infty(\tau,T; H^1)\cap L^2(\tau,T; H^2),$$
which yields that for $0<\alphalpha <1/2$,
\begin{equation} \labelbel{ccc1}u \in   C([\tau,T]; C^{2,\alphalpha pha}[0,1]),~~~u_t \in   C([\tau,T]; C^{\alphalpha pha}[0,1]).\end{equation}
 Hence, it follows from \eqref{ccc} and \eqref{ccc1}   that  $(\rho,u)$   is  a classical solution to  \eqref{1d}-\eqref{n3}.
\end{eqnarray}d{remark}

\begin{equation}gin{remark} \label{re4} To obtain the global  existence  of strong solutions in Theorem \ref{th11}, we do not need the additional  compatibility condition \eqref{1d4}. Indeed, we only need the initial data satisfying  the compatibility condition \eqref{1d4} for some $g\in L^2$     in proving the global well-posedness of classical solution in our Theorem \ref{th1}, which is in sharp contrast to \cite{wen2011} where they need   $g\in H^1.$ Therefore, our theorems  essentially weaken those assumptions on the compatibility condition in \cite{wen2011}.
\end{eqnarray}d{remark}

\begin{equation}gin{remark} \label{re2}
In our result, the only restriction  on $\mu\in C^2[0,\infty)$ is  \eqref{1d1}, which is much more general than those in \cite{wen2011} where   in addition to \eqref{1d1},   they need an additional stringent condition   \eqref{wen} which  plays a crucial role in the analysis of \cite{wen2011}.
\end{eqnarray}d{remark}

\begin{equation}gin{remark}\label{re3} In Theorem \ref{th11},
we obtain the time-independent upper bound of the density  in \eqref{1d6} and  the large-time behavior of  the velocity \eqref{1.12'}, which are in sharp contrast to \cite{wen2011} where  the corresponding a priori ones  depend on time.
Moreover, these results also generalized the similar ones in \cite{is2002,is2003} where they need   initial density strictly away from vacuum to the case that the density allows vacuum intially.
\end{eqnarray}d{remark}

\begin{equation}gin{remark}\label{re5}
In contrast to  the results in \cite{is2002,is2003}, where     the spatial $L^2$-norm  $\rho_x$ is  proved to be bounded independently of time provided the density strictly away from vacuum, our Theorem \ref{th2}  shows that  the spatial $L^r$-norm ($r>1$) of $\rho_x$  will blow up when time goes to infinity provided  vacuum appears initially. 
\end{eqnarray}d{remark}

We now make some comments on the analysis of this paper. We begin with the local existence theorem (see  \cite{coi1} or Lemma \ref{le3} below) of classical solutions
to problem \eqref{1d}-\eqref{n3} with  the initial density  strictly away from vacuum.
Then, we prove that the local strong (classical) solution with vacuum also exists and thus extend the local existence time to be global. Hence, we need some
global a priori estimates which are independent of the lower bound of density. It turns out that
the key issue is to derive both the time-independent lower order estimates  and
the time-dependent higher order ones (see Section 2).

It should be noted  that the methods used in  Ding-Wen-Zhu \cite{wen2011} can not be adapted here.
Indeed, on the one hand, the analysis in \cite{wen2011} relies heavily on the special assumption on  viscosity $\mu(\rho)$ (see \eqref{wen}), that is $\mu(\rho)$ should be bounded by   $P(\rho)$ pointwisely.
On the other hand, it seems difficult to study  the large-time behavior of solutions since the a priori  estimates obtained in \cite{wen2011} are all   time-depending. To overcome these difficulties,
motivated by   Li-Xin \cite{lijde,lx1},
we succeed in obtaining  the  key uniform upper bound of the density by making full use
of Zlotnik  inequality (see Lemma \ref{la3.2}),  and bounding the $L^2$-norm of $u_x$ according to the  material derivative  $\dot u$ (see Lemma \ref{la3.5}).
The time-dependent higher order estimates of $(\rho, u)$ are derived by some standard arguments and
the time-weighted estimates due to \cite{H3} (see also \cite{hlx1,lx1,lisiam}).
Next, with both the uniform upper bound of the density  and the time-independent lower order estimates at hand,
 we use the methods owing to Stra\v{s}kraba-Zlotnik \cite{is2002,is2003} and
 thus prove  the following large-time behavior
  $$  \lim_{t\rightarrow \infty}\left(\| u(\cdot,t)\|_{W^{1,p}}+\| \rho-\rho_s\|_{L^{p}}\right)=0,~~~~\forall~p\in [1,\infty). $$
Finally, using a key testing function $\rho_s^{-1}\int_0^x(\rho-\rho_s)dy$
 motivated by Huang-Li-Xin \cite{jmpa06} and Li-Zhang-Zhao \cite{lisiam},
  we derive the desired  exponential decay rate estimate \eqref{ed1}
  in Theorem \ref{th2}(see Section 4).

The rest of the paper is organized as follows: In Section 2, we will  derive the necessary
a priori estimates on smooth solutions. The main results,
Theorems \ref{th11}--\ref{th1}    and \ref{th2},  are proved in Sections 3   and   4 respectively. 

 \section{A priori estimates}

In this section, we will establish some necessary a priori bounds
for smooth solutions to the problem \eqref{1d}-\eqref{n3} to extend the local classical solution guaranteed by following
Lemma \ref{le3}, whose proof can be completed by similar arguments as in  \cite{coi1,HLM}.
 \begin{equation}gin{lemma}\label{le3}   Assume that $f\in H^2$ and  the initial data $(\rho_0, u_0)$ satisfies
   \begin{equation}gin{eqnarray}n\label{pre4}
0<\delta\le\rho_0,~~ \rho_0\in H^3, ~P(\rho_0)\in H^3,~u_0\in H^3\cap H^1_0.
\end{eqnarray}n
Then, there exists a small time $T_0>0$ and a unique classical solution $(\rho,u)$ to the problem \eqref{1d}-\eqref{n3}  on $(0,1)\tildemes(0, T_0]$  such that
   \begin{equation}gin{eqnarray}nn
   \label{pre3}\begin{equation}gin{cases}
   \rho,   P(\rho) \in C([0,T_0]; H^3), \\
 u\in C([0,T_0]; H^3\cap H_0^1) \cap L^2(0,T_0;H^4),\\
 \rho_t \in C([0,T_0]; H^2),~~
   u_t\in  C([0,T_0]; H^1_0)\cap L^2(0,T_0;H^2).\end{eqnarray}d{cases} \end{eqnarray}n
\end{eqnarray}d{lemma}

Let $T>0$ be a fixed time and $(\rho,u)$ be
the smooth solution to  problem \eqref{1d}-\eqref{n3} with the initial data  $(\rho_0,u_0)$ satisfying the  assumptions in Lemma \ref{le3}.

 \subsection{\label{se3} A priori estimates(I): lower order estimates}

This subsection is concerned with the time-independent lower order estimates of  the solution $(\rho,u)$.  In this subsection, we will use the convention that $C$ denotes a generic positive constant depending  on
 $\gammaammamma, \begin{equation}gin{aligned}r\mu, \|\rho_0\|_{H^1}, \|u_0\|_{H^1}$, and $ \|f\|_{H^1}$

but independent of $T$, and  use $C(a)$ to emphasize that $C$ depends on $a.$

First, we state the following Zlotnik  inequality, whose proof can be found in \cite{zl1}, will be used to get the
uniform (in time) upper bound of the density.
\begin{equation}gin{lemma}[\cite{zl1}]\label{le1}   Let the function $y$ satisfy
\begin{equation}gin{eqnarray}n y'(t)\le  g(y)+b'(t) \mbox{  on  } [0,T] ,\quad y(0)=y^0, \end{eqnarray}n
with $ g\in C(R)$ and $y,b\in W^{1,1}(0,T).$ If $g(\infty)=-\infty$
and \begin{equation}gin{eqnarray}n\label{a100} b(t_2) -b(t_1) \le N_0 +N_1(t_2-t_1)\end{eqnarray}n for all
$0\le t_1<t_2\le T$
  with some $N_0\gammae 0$ and $N_1\gammae 0,$ then
\begin{equation}gin{eqnarray}n y(t)\le \mathcal{A} x\left\{y^0,\omegaverline{\zeta} \right\}+N_0<\infty
\mbox{ on
 } [0,T],
\end{eqnarray}n
  where $\omegaverline{\zeta} $ is a constant such
that \begin{equation}gin{eqnarray}n\label{a101} g(\zeta)\le -N_1 \quad\mbox{ for }\quad \zeta\gammae \omegaverline{\zeta}.\end{eqnarray}n
\end{eqnarray}d{lemma}

Then,  we will give the key time-independent upper bound for the density as follows.
 \begin{equation}gin{lemma} \label{la3.2}
There is a positive constant $C$   such that for any $(x,t)\in [0,1]\tildemes[0,T],$
  \begin{equation}\label{gj3.2}
0\le \rho(x,t)\le C .
\end{equation}
\end{eqnarray}d{lemma}
{\it Proof.}    First, standard energy estimate leads to
 \begin{equation}\label{gj3.1} \begin{equation}gin{aligned}
&\sup_{0\le t\le T} \int_0^1 \left(\frac{1}{2}\rho u^2+\frac{1}{\gammaammamma-1}P\right)dx+\int_0^T\int_0^1\mu(\rho)u_x^2dxds\\
&\le C+\sup_{0\le t\le T} \int_0^1 \rho \int_0^xfdydx\\
&\le C+\|f\|_{L^\infty}\sup_{0\le  t\le T} \int_0^1 \rho  dx\\
&\le C,
\end{aligned}\end{equation}
where in the last inequality one has used the following fact:
 \begin{equation}\begin{equation}gin{aligned}\label{03.2}
& \int_0^1 \rho  dx=\int_0^1\rho_0dx=1
\end{aligned}\end{equation}
owing to  \eqref{oy3.7} and $ \eqref{1d}_1$.

Next, integrating $\eqref{1d}_2$ over $(0,x)$ gives
  \begin{equation}\label{l31}\begin{equation}gin{aligned}
 -\mu(\rho)u_x=&-P(\rho)+\int_0^x\rho fdy -\frac{\partialartial}{\partialartial t}\int_0^x\rho udy-\rho u^2\\&+\left(P(\rho)+\rho u^2-\mu(\rho)u_x\right)(0,t),
\end{aligned}\end{equation}
which in particular implies  \begin{equation}\label{l32}\begin{equation}gin{aligned}
&\left(P(\rho)+\rho u^2-\mu(\rho)u_x\right)(0,t)\\&=-\int_0^1\mu(\rho)u_xdx+\int_0^1 P(\rho)dx-\int_0^1 \int_0^x\rho fdydx \\
&\quad+ \frac{d}{dt}\int_0^1\int_0^x\rho udydx+\int_0^1 \rho u^2dx.
\end{aligned}\end{equation}
Combining this with   \eqref{l31} shows that for $D_t\triangleq \frac{\partialartial}{\partialartial t} +u  \frac{\partialartial}{\partialartial x}$
  \begin{equation}\label{l3}\begin{equation}gin{aligned}
&-\mu(\rho)u_x+P(\rho) \\
&= \int_0^x\rho fdy-\int_0^1\int_0^x\rho fdydx+\int_0^1\rho u^2dx+\int_0^1P(\rho)dx\\
&\quad+D_t\left(\int_0^1\rho\int_2^\rho \mu(s)s^{-2}dsdx+\int_0^1\int_0^x\rho udydx-\int_0^x\rho udy\right)\\
&\le  C+D_t\left(b_1(t)+b_2(t)+b_3(t)\right),
\end{aligned}\end{equation}
where one has used \eqref{gj3.1}  and the following fact
  \begin{equation}gin{eqnarray}n\begin{equation}gin{aligned}
-\int_0^1\mu(\rho) u_xdx=D_t\int_0^1\rho\int_2^\rho \mu(s)s^{-2}dsdx
\end{aligned}\end{eqnarray}n
due to $\eqref{1d}_1$.

Next,  it follows from $\eqref{1d}_1$ that
  \begin{equation}gin{eqnarray}n-\mu(\rho)u_x=
D_t\int_1^\rho \mu(s)s^{-1}ds ,
\end{eqnarray}n which together with
  \eqref{l3}   gives
  \begin{equation}\label{l5}\begin{equation}gin{aligned}
D_t\int_1^\rho \mu(s)s^{-1}ds\le -P(\rho)+C+D_t\left(b_1(t)+b_2(t)+b_3(t)\right).
\end{aligned}\end{equation}

 Then, on the one hand, we have
\begin{equation}\label{l8} \begin{equation}gin{aligned}
b_1(t) \le 2^{-1}\int_1^{\mathcal{A} x\{\sup\limits_{\Omegamega_T}\rho,2\}}\mu(s)s^{-1}ds,
\end{aligned}\end{equation}
where $\Omegamega_T=[0,1]\tildemes[0, T]$.
On the other hand, one deduces from \eqref{gj3.1} that
\begin{equation}\label{l9} \begin{equation}gin{aligned}
|b_2(t)|+|b_3(t)|&\le C\int_0^1\int_0^x\rho dy\int_0^x\rho u^2dydx+\int_0^1\rho u^2dx\int_0^1\rho dx \le C.
\end{aligned}\end{equation}
Finally, applying the Zlotnik  inequality (Lemma \ref{le1}) to \eqref{l5}, we get after using  \eqref{l8} and \eqref{l9} that
  \begin{equation}gin{eqnarray}n\begin{equation}gin{aligned}
\int_1^\rho \mu(s)s^{-1}ds\le C+2^{-1}\int_1^{\mathcal{A} x\{\sup\limits_{\Omegamega_T}\rho,2\}}\mu(s)s^{-1}ds,
\end{aligned}\end{eqnarray}n
which together with \eqref{1d1}     implies \eqref{gj3.2} and completes the proof of Lemma \ref{la3.2}. 
 $\Box$

\begin{equation}gin{lemma} \label{la3.5}
Let $(\rho,u)$ be a smooth solution of
 \eqref{1d}-\eqref{n3}  on $[0,1] \tildemes [0,T].$ Then
 \begin{equation}
\begin{equation}gin{aligned}\label{gj3.5}
   \sup_{0\le t\le T}  \| u \|_{H^1}^2+\int_0^T \|\rho^{1/2}\dot u\|_{L^2}^2 dt\le C,
\end{aligned}
\end{equation} where and in what follows, we denote
$ \dot u\triangleq u_t+uu_x. $
\end{eqnarray}d{lemma}

{\it Proof.}   First, multiplying $\eqref{1d1}_{2}$  by  $ \dot u$ and   integrating the resulting equation by parts yield
\begin{equation}
\begin{equation}gin{aligned}\label{3.2}
  &\frac{1}{2}\frac{d}{dt}\int_0^1 \mu(\rho)u_x^2dx+\int_0^1\rho\dot u^2dx\\
  &=\frac{d}{dt}\left(\int_0^1  P(\rho) u_xdx+\int_0^1  \rho fudx\right)\\
  &\quad-\frac{1}{2}\int_0^1 [\mu(\rho)-\mu'(\rho)\rho]u_x^3dx+\gammaamma\int_0^1 P(\rho)u_x^2dx-\int_0^1 \rho u^2 f_xdx\\
  &\le\frac{d}{dt}\left(\int_0^1 P(\rho) u_xdx+\int_0^1 \rho  fudx\right)\\
  &\quad+C \|u_x\|_{L^3}^3+C \|u_x\|_{L^2}^2+C  \|u\|_{L^\infty}^2\|f_x\|_{L^2}\\
    &\le\frac{d}{dt}\left(\int_0^1  P(\rho) u_xdx+\int_0^1 \rho fudx\right) +C \|u_x\|_{L^\infty}\|u_x\|_{L^2}^2+C\|u_x\|_{L^2}^2,
\end{aligned}
\end{equation}
where one has used \eqref{gj3.2} and the following fact:
\begin{equation}
\begin{equation}gin{aligned}\label{3.5}
  \|u\|_{L^\infty}\le C\|u\|_{L^2}^{1/2}\|u_x\|_{L^2}^{1/2}\le C\|u_x\|_{L^2}.
\end{aligned}
\end{equation}

 Then, using \eqref{1d}$_2$, \eqref{1d1}, \eqref{gj3.1}, and \eqref{gj3.2}, we have
 \begin{equation}
\begin{equation}gin{aligned}\label{3.7}
  \|u_x\|_{L^\infty} \le  &C  \|\mu(\rho)u_x-P(\rho)\|_{L^\infty}+C\|P(\rho)\|_{L^\infty}\\
    \le &C  \|\mu(\rho)u_x-P(\rho)\|_{L^1}+C  \|(\mu(\rho)u_x-P(\rho))_x\|_{L^1}+C \\
    \le & C\|\mu(\rho)u_x\|_{L^1}+C\|P(\rho)\|_{L^1}+C\|\rho\dot u\|_{L^1}+C\|\rho f\|_{L^1}+C \\
    \le & C \int_0^1\mu(\rho)u_x^2dx+C \|\rho^{1/2}\dot u\|_{L^2}+C ,
\end{aligned}
\end{equation}
which together with \eqref{3.2} and Young's inequality gives
 \begin{equation}
\begin{equation}gin{aligned}\label{3.8}
  & \frac{d}{dt}B(t)+\frac{1}{2}\int_0^1\rho\dot u^2dx \le  C\left(\int_0^1\mu(\rho)u_x^2dx\right)^2+C\int_0^1\mu(\rho)u_x^2dx,
\end{aligned}
\end{equation}
where
$$B(t)\triangleq   \frac{1}{2} \int_0^1 \mu(\rho)u_x^2dx-\int_0^1 P(\rho)u_xdx-\int_0^1 \rho fudx
$$
satisfies
 \begin{equation}
\begin{equation}gin{aligned}\label{3.6}
\frac{1}{4}\|\sqrt{\mu(\rho)}u_x\|_{L^2}^2-C\le  B(t)\le  C\|\sqrt{\mu(\rho)}u_x\|_{L^2}^2+C
\end{aligned}
\end{equation}
owing to the following estimate:
 \begin{equation}gin{eqnarray}n\label{3.10}
   \int_0^1 P(\rho) u_xdx+ \int_0^1 \rho fudx

      \le \frac{1}{4}\int_0^1 \mu(\rho)u_x^2dx+C,
\end{eqnarray}n
due to \eqref{1d1}, \eqref{gj3.1}, and \eqref{gj3.2}.

Hence,  Gronwall's inequality together with \eqref{3.8}, \eqref{3.6}, and  \eqref{gj3.1} implies \eqref{gj3.5} and thus finishes the proof of Lemma \ref{la3.5}. 
 $\Box$

\begin{equation}gin{lemma} \label{la3.6}
Let $(\rho,u)$ be a smooth solution of
 \eqref{1d}-\eqref{n3}  on $(0,1) \tildemes [0,T].$ Then
\begin{equation}
\begin{equation}gin{aligned}\label{gj3.6}
 \sup_{0\le t\le T}\sigma \left(\|u\|^2_{W^{1,\infty} }+\|\rho^{1/2}\dot u\|_{L^2}^2\right) +\int_0^T \sigma\|  \dot u_x\|_{L^2}^2 dt\le C,
 \end{aligned}
\end{equation} where and in what follows, $\sigma(t)\triangleq\min \{1,t\}.$
\end{eqnarray}d{lemma}

{\it Proof.}   First, operating $\partialartial_t+(u\cdot~)_x$ to $ (\ref{1d})_2 $ yields that
  \begin{equation}
\begin{equation}gin{aligned}\label{3.12}
  \rho \dot u_t+\rho u\dot u_x-[\mu(\rho) \dot u_x]_x=-\gammaammamma [P(\rho)u_x]_x-[(\mu(\rho)+\mu'(\rho)\rho)u_x^2]_x+\rho uf_x,
\end{aligned}
\end{equation} which
multiplied by $\dot{u} $ gives
 \begin{equation}
\begin{equation}gin{aligned}\label{3.13}
  &\frac{1}{2}\frac{d}{dt}\int_0^1 \rho|\dot u|^2dx+\int_0^1 \mu(\rho) |\dot u_x|^2dx\\
  &=\gammaammamma \int_0^1 P(\rho)u_x\dot u_xdx+\int_0^1 (\mu(\rho)+\mu'(\rho)\rho)u_x^2 \dot u_xdx +\int_0^1 \rho uf_x \dot udx\\
  &\le C \|u_x\|_{L^2} \|\dot u_x\|_{L^2}(1+\|u_x\|_{L^\infty} )+C \|u\|_{L^\infty}\|f_x\|_{L^2}\|\rho^{1/2}\dot u\|_{L^2}\\
  &\le C \|u_x\|_{L^2} \|\dot u_x\|_{L^2}(1+ \|\rho^{1/2}\dot u\|_{L^2} )+C \|u_x\|_{L^2}\|\rho^{1/2}\dot u\|_{L^2}\\
  &\le \varepsilon\|\dot u_x\|_{L^2}^2+C(\varepsilon) \|u_x\|_{L^2}^2  +C  \|\rho^{1/2}\dot u\|_{L^2}^2
\end{aligned}
\end{equation}
due to \eqref{3.7} and  \eqref{gj3.5}.

Then, multiplying \eqref{3.13} by $\sigma$ and integrating the resultant inequality  over $(0,T)$, we obtain after  choosing $\varepsilon$ suitably small and using \eqref{gj3.5}  and \eqref{gj3.1} that
 \begin{equation}
\begin{equation}gin{aligned}\label{3.16'}
   &\sup_{0\le t\le T} \sigma \int_0^1 \rho|\dot u|^2dx+\int_0^T \sigma\int_0^1 \mu(\rho) |\dot u_x|^2dxdt \le C,
\end{aligned}
\end{equation}
which together with \eqref{3.5}, \eqref{3.7},   and \eqref{gj3.5} leads to
 \begin{equation}gin{eqnarray}n\label{gj3}
   &\sup_{0\le t\le T} \sigma \|u_x\|^2_{L^\infty} \le C.
\end{eqnarray}n
Combining this  with \eqref{3.16'} gives \eqref{gj3.6}. The proof of Lemma \ref{la3.6} is finished. 
 $\Box$

 \subsection{\label{se3'} A priori estimates(II): higher order estimates}
 In this subsection, we prove the higher-order estimates of the smooth solution  $(\rho,u)$ to the problem \eqref{1d}-\eqref{n3}.

 \begin{equation}gin{lemma} \label{la4.1} For any given $T>0$, there exists a   positive constant $C$ depending  on $T$,
 $\gammaammamma, \begin{equation}gin{aligned}r\mu, \|\rho_0\|_{H^1}, \|u_0\|_{H^1}$, and $ \|f\|_{H^1}$ such that
\begin{equation}
\begin{equation}gin{aligned}\label{gj4.1}
   \sup_{0\le t\le T} \left(\|\rho_x\|_{L^2}^2+\|\rho_t\|_{L^2}^2\right) \le C,
\end{aligned}
\end{equation}
and
\begin{equation}
\begin{equation}gin{aligned}\label{ngj4.1}
   \sup_{0\le t\le T} \left(\sigma\|\rho^{1/2}u_t\|_{L^2}^2 +\sigma\| u_{xx}\|_{L^2}^2\right) +\int_0^T \left( \| u_{xx}\|_{L^2}^2+\sigma\| u_{xt}\|_{L^2}^2\right)dt\le C.
\end{aligned}
\end{equation}
\end{eqnarray}d{lemma}

{\it Proof.}   First, differentiating $\eqref{1d}_1$ with respect to $x$ gives
\begin{equation}
\begin{equation}gin{aligned}\label{a1}
   \rho_{xt}+\rho_{xx}u+2\rho_xu_x+\rho u_{xx}=0.
\end{aligned}
\end{equation}
Multiplying \eqref{a1} by $\rho_x$ and integrating the resulting equation by parts yield that
\begin{equation}\begin{equation}gin{aligned}\label{a2}
   \frac{d}{dt}\|\rho_x\|_{L^2} \le C\|u_x\|_{L^\infty}\|\rho_x\|_{L^2} +C \|u_{xx}\|_{L^2}
\end{aligned}
\end{equation}
due to \eqref{gj3.2}.

Next, it is easy to deduce from $\eqref{1d}_2$ that
\begin{equation}
\begin{equation}gin{aligned}\label{a7}
\mu(\rho)u_{xx}= \rho \dot u+P_x-\rho f-\mu'(\rho)\rho_xu_x,
\end{aligned}
\end{equation}
which together with  \eqref{n3}, \eqref{gj3.2}, \eqref{gj3.5}, and \eqref{gj3.6} yields
\begin{equation}
\begin{equation}gin{aligned}\label{a9}
\|u_{xx}\|_{L^2}&\le  C\| \rho^{1/2} \dot u\|_{L^2}+C\| \rho_x\|_{L^2}(\| u_x\|_{L^\infty}+1)+C\\&\le C\sigma^{-1/2}(1+ \| \rho_x\|_{L^2}).
\end{aligned}
\end{equation}

Submitting \eqref{a9} into \eqref{a2} and using Gronwall's inequality, one gets
\begin{equation}
\begin{equation}gin{aligned}\label{a5}
   \sup_{0\le t\le T}  \|\rho_x\|_{L^2}  \le C,
\end{aligned}
\end{equation}
which along with $\eqref{1d}_1$ and \eqref{gj3.5} leads to
\begin{equation}
\begin{equation}gin{aligned}\label{a6}
 \|\rho_t\|_{L^2}  & \le \|u \|_{L^\infty} \|\rho_x\|_{L^2} +C\| u_x\|_{L^2} \le C.
\end{aligned}
\end{equation}
The combination of \eqref{a5} with  \eqref{a6} leads to \eqref{gj4.1}.

Now, it follows from \eqref{gj3.2}, \eqref{3.7}, \eqref{gj3.5}, \eqref{a9}, and \eqref{gj4.1} that
\begin{equation}\begin{equation}gin{aligned}\label{b5}
 \|\rho^{1/2}u_t\|_{L^2}^2&\le \|\rho^{1/2}\dot u\|_{L^2}^2+ \|\rho^{1/2}u u_x\|_{L^2}^2\\
 &\le \|\rho^{1/2}\dot u\|_{L^2}^2+  \begin{equation}gin{aligned}r \rho \|u\|_{L^\infty}^2\| u_x\|_{L^2}^2\le \|\rho^{1/2}\dot u\|_{L^2}^2+ C,\\
 \|u_{xt}\|_{L^2}^2&= \|(\dot u-uu_x)_x\|_{L^2}^2\le  \|\dot u_x\|_{L^2}^2+ \|u_x^2\|_{L^2}^2+ \|uu_{xx}\|_{L^2}^2\\
 &\le \|\dot u_x\|_{L^2}^2+ \|u_x\|_{L^\infty}^2\|u_x\|_{L^2}^2+ \|u\|_{L^\infty}^2\|u_{xx}\|_{L^2}^2\\
 &\le \|\dot u_x\|_{L^2}^2+\|\rho^{1/2}\dot u\|_{L^2}^2+ C.
\end{aligned}\end{equation}

Hence, \eqref{ngj4.1} is a direct consequence of \eqref{b5}, \eqref{gj3.6},  \eqref{a9}, and \eqref{gj3.5}. We complete the proof of Lemma \ref{la4.1}. 
 $\Box$

From now on, assume that $(\rho,u)$ is a smooth solution of the problem \eqref{1d}-\eqref{n3} with the smooth initial data satisfying the condition in Theorem \ref{th1} and $\mu(\cdot)\in C^2[0,\infty)$. In the following of this subsection, the general constant $C$  may depend on $T$, $\gammaammamma, \begin{equation}gin{aligned}r\mu, \|\rho_0\|_{H^2}, \|P(\rho_0)\|_{H^2}, \|u_0\|_{H^2}$,   $ \|f\|_{H^1}$, and $\|g\|_{L^2}$ with $g$ as in \eqref{1d4}.

\begin{equation}gin{lemma} \label{nla4.3}  For any given $T>0$, there exists a   positive constant $C$ such that
\begin{equation}
\begin{equation}gin{aligned}\label{gj4.2}
   \sup_{0\le t\le T} \left(\| u_{xx}\|_{L^2} + \| u \|_{W^{1,\infty}} + \|\rho^{1/2}u_t\|_{L^2}^2 \right)+\int_0^T \| u_{xt}\|_{L^2}^2 dt\le C.
\end{aligned}
\end{equation}
\end{eqnarray}d{lemma}

{\it Proof.}   It follows from  the compatibility condition \eqref{1d4} that we can set 
\begin{equation}gin{eqnarray}n\labelbel{gy1}\rho^{1/2}\dot u(x,t=0) =g+\rho_0^{1/2}f\in L^2.\end{eqnarray}n
Integrating \eqref{3.13} over $(0,T)$, we obtain after  using \eqref{gj3.5}  and \eqref{gj3.1} that
 \begin{equation}gin{eqnarray}n
   &\sup_{0\le t\le T}  \int_0^1 \rho|\dot u|^2dx+\int_0^T  \int_0^1 \mu(\rho) |\dot u_x|^2dxdt \le C,
\end{eqnarray}n
which together with \eqref{b5}, \eqref{3.5}, \eqref{3.7},   and \eqref{gj3.5}  gives \eqref{gj4.2} and completes the proof of Lemma \ref{nla4.3}.  
 $\Box$

The following higher order estimates of the solutions  are used to guarantee
the extension of local classical solution to be a global one, whose proof  are similar to those in \cite{wen2011}, see also \cite{hlx1,lx1,lisiam} considering the high dimensional case. And, we also sketch them here for completeness.

\begin{equation}gin{lemma} \label{la4.3}  For any given $T>0$, there exists a   positive constant $C$ such that
\begin{equation}
\begin{equation}gin{aligned}\label{gj4.3}
   &\sup_{0\le t\le T} \left( \|\rho_{xx}\|_{L^2}^2 + \|P_{xx}\|_{L^2}^2 +\|\rho_{xt}\|_{L^2}^2 + \|P_{xt}\|_{L^2}^2 \right) \\
   &\quad\qquad+\int_0^T \left(\|u_{xxx}\|_{L^2}^2 + \|\rho_{tt}\|_{L^2}^2 + \|P_{tt}\|_{L^2}^2 \right) dt\le C.
\end{aligned}
\end{equation}
\end{eqnarray}d{lemma}

{\it Proof.}  
Differentiating $\eqref{a1}$ with respect to $x$ gives
\begin{equation}
\begin{equation}gin{aligned}\label{c1}
   (\rho_{xx})_t+\rho_{xxx}u+3\rho_{xx}u_x+3\rho_x u_{xx}+\rho u_{xxx}=0.
\end{aligned}
\end{equation}
Multiplying \eqref{c1} by $\rho_{xx}$ and integrating the resulting equation  by parts, it holds that
\begin{equation}
\begin{equation}gin{aligned}\label{c2}
     \frac{1}{2}\frac{d}{dt}\|\rho_{xx}\|_{L^2}^2&=-\frac{5}{2}\int_0^1 \rho_{xx}^2u_xdx-3\int_0^1 \rho_x\rho_{xx}u_{xx}dx-\int_0^1 \rho \rho_{xx}u_{xxx}dx\\
     &\le C\|u_x\|_{L^\infty} \|\rho_{xx}\|_{L^2}^2+C\|\rho_x\|_{L^\infty}\| u_{xx}\|_{L^2}\|\rho_{xx}\|_{L^2}+C\| u_{xxx}\|_{L^2}\|\rho_{xx}\|_{L^2}\\
     &\le C\|\rho_{xx}\|_{L^2}^2+C\| u_{xxx}\|_{L^2}^2+C,
\end{aligned}
\end{equation}
where in the last inequality one has  used  \eqref{gj4.2}, \eqref{gj4.1}, and the following estimate:
\begin{equation}
\begin{equation}gin{aligned}\label{c3}
     \|\rho_{x}\|_{L^\infty}\le  \|\rho_{x}\|_{L^1}+ \| \rho_{xx}\|_{L^1} \le  C + C\| \rho_{xx}\|_{L^2}.
\end{aligned}
\end{equation}

Since  $P(\rho)$ satisfies
\begin{equation}
\begin{equation}gin{aligned}\label{c4}
P_t+P_xu+\gammaammamma Pu_x=0,
\end{aligned}
\end{equation}
following  the same arguments as \eqref{c2}, one has
\begin{equation}
\begin{equation}gin{aligned}\label{c5}
     \frac{1}{2}\frac{d}{dt}\|P_{xx}\|_{L^2}^2\le C  \|P_{xx}\|_{L^2}^2+C \| u_{xxx}\|_{L^2}^2+C.
\end{aligned}
\end{equation}
Hence, the combination of \eqref{c5} with  \eqref{c2} implies that
\begin{equation}
\begin{equation}gin{aligned}\label{c6}
     \frac{1}{2}\frac{d}{dt}\left(\|\rho_{xx}\|_{L^2}^2+\|P_{xx}\|_{L^2}^2\right)\le C\left(\|\rho_{xx}\|_{L^2}^2+\|P_{xx}\|_{L^2}^2\right)+C+C\| u_{xxx}\|_{L^2}^2.
\end{aligned}
\end{equation}

  In order to estimate $\| u_{xxx}\|_{L^2}^2$, differentiating $\eqref{a7}$ with respect to $x$ gives
\begin{equation}
\begin{equation}gin{aligned}\label{c7}
 \mu(\rho)u_{xxx}= &\rho_x u_t+\rho_x uu_x+ \rho u_{xt}+\rho u_x^2+\rho uu_{xx}+P_{xx}\\
 &-\rho_x f-\rho f_x-\mu''(\rho)\rho_x^2u_x-\mu'(\rho)\rho_{xx}u_x-2\mu'(\rho)\rho_xu_{xx}.
\end{aligned}
\end{equation}
 Similar to \eqref{3.5}, we also have
\begin{equation}
\begin{equation}gin{aligned}\label{c15}
 \| u_t\|_{L^\infty}\le C \|u_{xt}\|_{L^2},
\end{aligned} 
\end{equation} which  along with \eqref{c7}, \eqref{n3}, \eqref{gj3.2}, \eqref{gj4.1}, \eqref{c3}, and \eqref{gj4.2} implies that
\begin{equation}
\begin{equation}gin{aligned}\label{c8}
 \| u_{xxx}\|_{L^2}^2
 &\le C\|\rho_x\|_{L^2}^2\|u_t\|_{L^\infty}^2+C\|\rho_x\|_{L^2}^2\|u\|_{L^\infty}^2\|u_x\|_{L^\infty}^2
 +C\|u_{xt}\|_{L^2}^2+C\|u_x\|_{L^4}^4\\
 &\quad +C\|u\|_{L^\infty}^2\|u_{xx}\|_{L^2}^2+C\|P_{xx}\|_{L^2}^2+C\|\rho_x\|_{L^2}^2\|f\|_{L^\infty}^2+C\|f_x\|_{L^2}^2 \\
 &\quad +C\|\rho_x\|_{L^\infty}^2\|\rho_x\|_{L^2}^2\|u_x\|_{L^\infty}^2+C\|u_x\|_{L^\infty}^2\|\rho_{xx}\|_{L^2}^2+C\|\rho_x\|_{L^\infty}^2\|u_{xx}\|_{L^2}^2\\
 &\le C \left(1+\|u_{xt}\|_{L^2}^2+\|\rho_{xx}\|_{L^2}^2+\|P_{xx}\|_{L^2}^2\right).
\end{aligned}
\end{equation}

Submitting \eqref{c8} into \eqref{c6}, one obtains after using   Gronwall's inequality and \eqref{gj4.2} that
\begin{equation}
\begin{equation}gin{aligned}\label{c11}
 \sup_{0\le t\le T} \left( \|\rho_{xx}\|_{L^2}^2 + \|P_{xx}\|_{L^2}^2
\right)\le C.\end{aligned}
\end{equation}
This together with \eqref{a1}, \eqref{gj4.2},  and \eqref{gj4.1} implies that
\begin{equation}
\begin{equation}gin{aligned}\label{c12}
 \|\rho_{xt}\|_{L^2}^2\le \|\rho_{xx}\|_{L^2}^2\| u\|_{L^\infty}^2+ C\|\rho_{x}\|_{L^2}^2\| u_x\|_{L^\infty}^2+ C\|u_{xx}\|_{L^2}^2\le C.
\end{aligned}
\end{equation}

Differentiating $\eqref{1d}_1$ with respect to $t$ gives
\begin{equation}gin{eqnarray}n
   \rho_{tt}=-\rho_{xt}u-\rho_xu_t-\rho_tu_x-\rho u_{xt}
\end{eqnarray}n
which combined with  \eqref{c12}, \eqref{gj4.1},  \eqref{c15}, and \eqref{gj4.2}  implies that
\begin{equation}
\begin{equation}gin{aligned}\label{c14}
\int_0^T \|\rho_{tt}\|_{L^2}^2dt&\le \int_0^T \left(\|\rho_{xt}\|_{L^2}^2\| u\|_{L^\infty}^2+ \|\rho_{x}\|_{L^2}^2\| u_t\|_{L^\infty}^2 \right)dt\\& \quad+C\int_0^T \left(    \|\rho_{t}\|_{L^2}^2\| u_x\|_{L^\infty}^2+ C\|u_{xt}\|_{L^2}^2\right)dt\\
&\le C\int_0^T \left(1+  \|u_{xt}\|_{L^2}^2\right)dt\le C.
\end{aligned}
\end{equation}
Similarly, one can get
 \begin{equation}gin{eqnarray}n
 \|P_{xt}\|_{L^2}^2+\int_0^T \|P_{tt}\|_{L^2}^2dt\le C,
\end{eqnarray}n
which together with  \eqref{c8}, \eqref{gj4.2},  \eqref{c11}, \eqref{c12},  and \eqref{c14} implies  \eqref{gj4.3}. The proof of Lemma \ref{la4.3} is completed. 
 $\Box$

\begin{equation}gin{lemma} \label{la4.4}  For any given $T>0$, there exists a   positive constant $C $ such that
\begin{equation}
\begin{equation}gin{aligned}\label{gj4.4}
   &\sup_{0\le t\le T} \left( t\|u_{xt}\|_{L^2}^2 + t\| u_{xxx}\|_{L^2}^2 \right) +\int_0^T  \left(t\|\rho^{1/2}u_{tt}\|_{L^2}^2+t\|u_{xxt}\|_{L^2}^2\right) dt\le C.
\end{aligned}
\end{equation}
\end{eqnarray}d{lemma}

{\it Proof.}   Differentiating $\eqref{1d}_2$ with respect to $t$ gives
\begin{equation}
\begin{equation}gin{aligned}\label{b1}
\rho u_{tt}+\rho uu_{xt}-[\mu(\rho)u_x]_{xt}=-\rho_t(u_t+uu_x)-\rho u_tu_x-P_{xt}+\rho_tf.
\end{aligned}
\end{equation}
Multiplying \eqref{b1}  by $u_{tt}$ and integrating the resulting equation   by parts lead to
\begin{equation}
\begin{equation}gin{aligned}\label{e1}
 &\frac{1}{2}\frac{d}{dt}\int_0^1 \mu u_{xt}^2dx+\|\rho^{1/2}u_{tt}\|_{L^2}^2\\
 &=\frac{1}{2}\int_0^1 \mu_t u_{xt}^2dx-\int_0^1 \mu_tu_xu_{xtt}dx+\int_0^1\rho_t fu_{tt}dx-\int_0^1\rho_t\dot uu_{tt}dx\\
 &\quad-\int_0^1 \rho u_tu_xu_{tt}dx-\int_0^1 \rho uu_{xt}u_{tt}dx-\int_0^1 P_{xt}u_{tt}dx\\
 &\triangleq \sum_{i=1}^7 J_i.
\end{aligned}
\end{equation}

The term $J_i$ on the right hand of \eqref{e1} can be estimated as follows.

Using   \eqref{gj4.1} and \eqref{gj4.3}, it holds
\begin{equation}
\begin{equation}gin{aligned}\label{e2}
\|\rho_t\|_{L^\infty}
\le C\| \rho_t\|_{L^2}+C\|\rho_{xt}\|_{L^2}\le C,
\end{aligned}
\end{equation}
which implies that
\begin{equation}
\begin{equation}gin{aligned}\label{e3}
J_1\le C\|\mu_t\|_{L^\infty}\|u_{xt}\|_{L^2}^2\le C\|\rho_t\|_{L^\infty}\|u_{xt}\|_{L^2}^2 \le C\|u_{xt}\|_{L^2}^2.
\end{aligned}
\end{equation}

It deduces from   \eqref{gj3.2}, \eqref{gj4.1}, and \eqref{e2} that
\begin{equation}gin{eqnarray}n
 \|\mu_{tt}\|_{L^2}^2 &\le  \|\mu''(\rho)\rho_t^2\|_{L^2}^2+ \|\mu'(\rho)\rho_{tt}\|_{L^2}^2\\
 &\le C\|\rho_t\|_{L^\infty}^2\|\rho_t\|_{L^2}^2+ C\|\rho_{tt}\|_{L^2}^2\\
 &\le C + C\|\rho_{tt}\|_{L^2}^2,
\end{eqnarray}n
which along with  \eqref{gj4.2} and  \eqref{e2} gives
\begin{equation}
\begin{equation}gin{aligned}\label{e4}
J_2&=-\frac{d}{dt}\int_0^1 \mu_tu_x u_{xt} dx+\int_0^1 \mu_{tt}u_x u_{xt} dx+\int_0^1 \mu_t u_{xt}^2 dx\\
&\le-\frac{d}{dt}\int_0^1 \mu_tu_x u_{xt} dx+ C\|u_x\|_{L^\infty}\|\mu_{tt}\|_{L^2}\|u_{xt}\|_{L^2} + C\|\mu_t\|_{L^\infty}\|u_{xt}\|_{L^2}^2\\
&\le  -\frac{d}{dt}\int_0^1 \mu_tu_x u_{xt} dx+C\|u_{xt}\|_{L^2}^2+C+C\|\rho_{tt}\|_{L^2}^2.
\end{aligned}
\end{equation}

Next, it is easy to derive  from \eqref{c15} that
\begin{equation}
\begin{equation}gin{aligned}\label{e6}
J_3&=\frac{d}{dt}\int_0^1 \rho_t f u_{t} dx-\int_0^1 \rho_{tt}fu_t dx\\
&\le \frac{d}{dt}\int_0^1 \rho_t f u_{t} dx+ C\|u_t\|_{L^\infty}\|\rho_{tt}\|_{L^2}\|f\|_{L^2} \\
&\le \frac{d}{dt}\int_0^1 \rho_t f u_{t} dx+ C\|u_{xt}\|_{L^2}^2+C\|\rho_{tt}\|_{L^2}^2.
\end{aligned}
\end{equation}

For the term  $J_4$, we have
\begin{equation}
\begin{equation}gin{aligned}\label{e7}
J_4&=\int_0^1 (\rho u)_x\dot uu_{tt}dx=-\int_0^1 \rho u\dot uu_{xtt}dx-\int_0^1 \rho u\dot u_xu_{tt}dx\\
&=-\frac{d}{dt}\int_0^1 \rho u \dot uu_{xt}dx+\int_0^1 (\rho u)_t\dot uu_{xt}dx+\int_0^1 \rho u\dot u_tu_{xt}dx-\int_0^1 \rho u\dot u_xu_{tt}dx\\
&\triangleq -\frac{d}{dt}\int_0^1 \rho u \dot uu_{xt}dx+J_{4,1}+J_{4,2}+J_{4,3}.
\end{aligned}
\end{equation}
One deduces from \eqref{gj4.2},  \eqref{gj4.1},   and \eqref{c15} that
\begin{equation}
\begin{equation}gin{aligned}\label{e8}\rhoonumber\\tag
 J_{4,1}&=\int_0^1 \rho_t u \dot uu_{xt}dx+\int_0^1 \rho u_t\dot uu_{xt}dx\\
 &\le C\left(\|u_t\|_{L^\infty}+\|uu_x\|_{L^\infty}\right) \|u_{xt}\|_{L^2}\|u\|_{L^\infty}\|\rho_t\|_{L^2}+ C\|\rho^{1/2}\dot u\|_{L^2}\|u_t\|_{L^\infty}\|u_{xt}\|_{L^2}\\
  &\le C +C\|u_{xt}\|_{L^2}^2,\\
 J_{4,2}&=\int_0^1 \rho u u_{tt}u_{xt}dx+\int_0^1 \rho u u_tu_xu_{xt}dx+\int_0^1 \rho u^2u_{xt}^2dx\\
 &\le C\|\rho^{1/2}u\|_{L^\infty} \|u_{xt}\|_{L^2} \|\rho^{1/2} u_{tt}\|_{L^2}\\
 &\quad+ C\|\rho u\|_{L^\infty}\|u_t\|_{L^\infty}\|u_{xt}\|_{L^2}\|u_x\|_{L^2}+C\|\rho u^2\|_{L^\infty} \|u_{xt}\|_{L^2}^2\\
 &\le \varepsilon \|\rho^{1/2} u_{tt}\|_{L^2}^2+C\|u_{xt}\|_{L^2}^2,\\
 J_{4,3}&\le C\|\rho^{1/2}u\|_{L^\infty} \|\dot u_{x}\|_{L^2} \|\rho^{1/2} u_{tt}\|_{L^2}\le \varepsilon \|\rho^{1/2} u_{tt}\|_{L^2}^2+C\|u_{xt}\|_{L^2}^2+C.
\end{aligned}
\end{equation}
This combined with \eqref{e7} yields
\begin{equation}
\begin{equation}gin{aligned}\label{e9}
J_4 \le -\frac{d}{dt}\int_0^1 \rho u \dot uu_{xt}dx+\varepsilon \|\rho^{1/2} u_{tt}\|_{L^2}^2+C\|u_{xt}\|_{L^2}^2+C .
\end{aligned}
\end{equation}

Moreover, by virtue of  \eqref{gj4.2}, it holds that
\begin{equation}
\begin{equation}gin{aligned}\label{e10}
J_5+J_6&\le  C\|u_x\|_{L^\infty} \|\rho^{1/2}u_{tt}\|_{L^2}\|\rho^{1/2}u_t\|_{L^2}+C\|\rho^{1/2}u\|_{L^\infty} \|\rho^{1/2}u_{tt}\|_{L^2}\|u_{xt}\|_{L^2}\\
&\le \varepsilon \|\rho^{1/2} u_{tt}\|_{L^2}^2+C\|u_{xt}\|_{L^2}^2+C.
\end{aligned}
\end{equation}

Finally, we can estimate $J_7$ as follows,
\begin{equation}
\begin{equation}gin{aligned}\label{e11}
J_7 &=\int_0^1 P_tu_{xtt}dx= \frac{d}{dt}\int_0^1 P_tu_{xt}dx-\int_0^1 P_{tt}u_{xt}dx\\
&\le \frac{d}{dt}\int_0^1 P_tu_{xt}dx +C\|P_{tt}\|_{L^2}^2+C\|u_{xt}\|_{L^2}^2.
\end{aligned}
\end{equation}

Submitting  \eqref{e3}-\eqref{e11} into \eqref{e1} and choosing $\varepsilon$ suitably small, we have
\begin{equation}
\begin{equation}gin{aligned}\label{e13}
\frac{d}{dt}\Pi(t)+\|\rho^{1/2}u_{tt}\|_{L^2}^2\le C\|u_{xt}\|_{L^2}^2+C +C\|\rho_{tt}\|_{L^2}^2+C\|P_{tt}\|_{L^2}^2.
\end{aligned}
\end{equation}
where
\begin{equation}gin{eqnarray}n
\begin{equation}gin{aligned}\label{e12}
\Pi(t)&\triangleq  \int_0^1 \mu u_{xt}^2dx+\int_0^1 \mu_tu_x u_{xt} dx+\int_0^1 \rho u \dot uu_{xt}dx\\
&\quad-\int_0^1 P_tu_{xt}dx-\int_0^1 \rho_t f u_{t} dx
\end{aligned}
\end{eqnarray}n
satisfies
\begin{equation}
\begin{equation}gin{aligned}\label{e15}
 \frac{\begin{equation}gin{aligned}r\mu}{2}\|u_{xt}\|_{L^2}^2-C \le \Pi(t) \le C\|u_{xt}\|_{L^2}^2+C
\end{aligned}
\end{equation}
owing to  the following estimates:
\begin{equation}gin{eqnarray}n
\begin{equation}gin{aligned}\label{e14}
 &\left|\int_0^1 \mu_tu_x u_{xt} dx+\int_0^1 \rho u \dot uu_{xt}dx-\int_0^1 P_tu_{xt}dx-\int_0^1 \rho_t f u_{t} dx\right|\\
&\le C\|\rho_t\|_{L^2} \| u_{x}\|_{L^\infty} \|  u_{xt}\|_{L^2}+C\|\rho^{1/2}u\|_{L^\infty} \|\rho^{1/2}\dot u\|_{L^2} \| u_{xt}\|_{L^2}\\
&\quad+C \|P_t\|_{L^2} \| u_{xt}\|_{L^2}+C\|u_t\|_{L^\infty} \|\rho_t\|_{L^2} \|f\|_{L^2}\\
&\le  \frac{\begin{equation}gin{aligned}r\mu}{2}\|u_{xt}\|_{L^2}^2+C,
\end{aligned}
\end{eqnarray}n
where one has used \eqref{gj4.2},  \eqref{gj4.1}, and \eqref{c15}.

Hence, the Gronwall's inequality   together with \eqref{e13}, \eqref{e15}, \eqref{gj4.2}, and \eqref{gj4.3}  gives
\begin{equation}
\begin{equation}gin{aligned}\label{e16}
   \sup_{0\le t\le T}  t\|u_{xt}\|_{L^2}^2 +\int_0^T t\|\rho^{1/2}u_{tt}\|_{L^2}^2 dt\le C,
\end{aligned}
\end{equation}
which along with  \eqref{c8} and  \eqref{gj4.3} implies
 \begin{equation}
\begin{equation}gin{aligned}\label{e18}
    \sup_{0\le t\le T}  t\|u_{xxx}\|_{L^2}^2 \le C.
\end{aligned}
\end{equation}

Next, it follows from \eqref{b1} that
 \begin{equation}gin{eqnarray}n
\mu(\rho)u_{xxt}&=\rho u_{tt}+\rho_t\dot u+\rho u_tu_x+\rho uu_{xt}+P_{xt}-\rho_tf\\
&\quad-\mu''(\rho)\rho_t\rho_xu_{x}-\mu'(\rho)\rho_{xt}u_{x}-\mu'(\rho)\rho_{x}u_{xt}-\mu'(\rho)\rho_{t}u_{xx}.
\end{eqnarray}n
Combining this   with  \eqref{c15},  \eqref{gj4.1}, \eqref{gj4.2}, \eqref{gj4.3}, and \eqref{e2} yields that
 \begin{equation}
\begin{equation}gin{aligned}\label{Re20}\rhoonumber\\tag
\|u_{xxt}\|_{L^2}^2 
&\le C\|\rho^{1/2} u_{tt}\|_{L^2}^2+ C\|\dot u\|_{L^\infty}^2\|\rho_t\|_{L^2}^2+C\|u_x\|_{L^\infty}^2\|\rho^{1/2}u_t\|_{L^2}^2\\
&\quad+C\|\rho u\|_{L^\infty}^2\|u_{xt}\|_{L^2}^2+C\|P_{xt}\|_{L^2}^2+C\|f\|_{L^\infty}^2\|\rho_t\|_{L^2}^2+C\|\rho_t\|_{L^\infty}^2\|u_{xx}\|_{L^2}^2\\
&\quad+C\|u_x\|_{L^\infty}^2\|\rho_t\|_{L^\infty}^2\|\rho_x\|_{L^2}^2+C\|u_x\|_{L^\infty}^2\|\rho_{xt}\|_{L^2}^2+C\|\rho_x\|_{L^\infty}^2\|u_{xt}\|_{L^2}^2\\
&\le C\|\rho^{1/2} u_{tt}\|_{L^2}^2+ C\|u_{xt}\|_{L^2}^2+ C,
\end{aligned}
\end{equation}
which along with  \eqref{e16} implies that
 \begin{equation}
\begin{equation}gin{aligned}\label{e21}
\int_0^T t\|u_{xxt}\|_{L^2}^2dt\le  C.
\end{aligned}
\end{equation}
The combination of  \eqref{e16}--\eqref{e18}  with  \eqref{e21} yields \eqref{gj4.4} and   completes   the proof of Lemma \ref{la4.4}. 
 $\Box$

\section{Proof of Theorems \ref{th11} and \ref{th1}}

{\it Proof of Theorem \ref{th11}.}  With all the a priori estimates obtained in section 2 at hand, we will divide the proof into three steps.

\thetaxtbf{Step 1}. We prove the local existence and uniqueness of the strong solution when the initial density contains vacuum. That is, Theorem \ref{th11} holds for some $T_0>0$.

Let  $(\rho_{0},u_{0},f)$  be as  in Theorem \ref{th11},  we construct
\begin{equation}\labelbel{gy2}\rho_0^{\deltalta}=\hat\rho_0^\delta+\deltalta,~~u_0^{\deltalta}=u_0\alphast j_\deltalta,~~f^{\deltalta}=f\alphast j_\deltalta,\end{equation}
where $j_\deltalta$ is the standard mollifying kernel of width $\deltalta$ and  $0\le \hat\rho_0^\delta\in  C_0^\infty (0,1)$ satisfies
   \begin{equation}\labelbel{ma5.2} \begin{equation}gin{aligned} \hat\rho_0^\delta\rightarrow  \rho_{0} ~\quad {\rm in}\,\, H^1,~~~  {\rm as}~\delta\rightarrow 0.\end{aligned}\end{equation}
   Thus, we have
      \begin{equation}\labelbel{ma6.1} \begin{equation}gin{aligned}  &\rho_0^{\deltalta}\rightarrow  \rho_{0},~~u_0^{\deltalta}\rightarrow  u_{0},~~f^\deltalta\rightarrow  f, ~~&\quad {\rm in}\,\, H^1, ~~~  {\rm as}~\delta\rightarrow 0,\end{aligned}\end{equation}
      and
\begin{equation}\labelbel{gyma6.1} \begin{equation}gin{aligned} \|\rho_0^{\deltalta}\|_{H^1} \le  C+C\|\rho_0 \|_{H^1},  \,\,  \|u_0^{\deltalta}\|_{H^1}\le C\|u_0\|_{H^1}, \,\,  \|f^{\deltalta}\|_{H^1}\le C\|f\|_{H^1}.\end{aligned}\end{equation}

 By virtue of  Lemma \ref{le3}, the initial boundary problem \eqref{1d}-\eqref{n3}  with the initial data $(\rho_0^{\deltalta},  u_0^\deltalta)$ has a classical solution  $(\rho^{\deltalta},  u^\deltalta)$ on $(0,1)\tildemes [0,T_0]$. Furthermore, the estimates obtained  in  Lemmas \ref{la3.2}--\ref{la4.1}   show that the solution $(\rho^{\deltalta},  u^\deltalta)$ satisfies for any $0<T<+\infty$,
\begin{equation}\begin{equation}gin{aligned}
   \label{8.6}
&  \sup_{0\le t\le T} \left(\|(\rho^\deltalta,\mu(\rho^\deltalta),P(\rho^\deltalta))\|_{H^1}+\| \rho^\deltalta_t\|_{L^2}+\| \rho^\deltalta u^\deltalta \|_{L^2}+ \|u^\deltalta\|_{H^1}\right.\\
&\quad\quad\quad\left.+\sqrt{t}\|\sqrt{\rho^\deltalta}u^\deltalta_t\|_{L^2}+\sqrt{t}\|u^\deltalta_{xx}\|_{L^2}\right) +\int_0^T\left( \|u^\deltalta\|_{H^2}^2+ t\|u^\deltalta_{xt}\|_{L^2}^2\right)dt\le \begin{equation}gin{aligned}r C, \end{aligned}\end{equation}
where $\begin{equation}gin{aligned}r C$ is independent of $\deltalta$.   With all the estimate \eqref{8.6}  at hand,  we  find  that the sequence
$(\rho^{\deltalta},u^{\deltalta})$ converges, up to the extraction of subsequences, to some limit $(\rho,u)$   in the
obvious weak sense.  Then letting $\deltalta\rightarrow 0$, we deduce from \eqref{8.6} that  $(\rho,u)$ is a strong solution of \eqref{1d}-\eqref{n3}   on $(0,1)\tildemes (0,T_0]$ satisfying
\begin{equation}
   \label{all}\begin{equation}gin{cases}
 \rho \in L^{\infty}(0,T_0;\,H^1),~~\rho_t \in L^{\infty}(0,T_0;\,L^2),\\
    u\in L^{\infty}(0,T_0; \, H^1_0)\cap L^2(0,T_0; \,H^2) \\
 t^{1/2}u\in L^{\infty}(0,T_0; \,H^2),~~ t^{1/2}u_t\in L^2(0,T_0; \,H^1),\\
  t^{1/2}\rho^{1/2}u_{t}\in L^{2}(0,T_0;\,L^2).\end{eqnarray}d{cases} \end{equation}

 Then, the uniqueness of the strong solution $(\rho,u)$ is guaranteed by the regularities \eqref{all}. For the detailed proof,  please see \cite{coi1,liliang}.

  \thetaxtbf{Step 2}.  We will extend the local existence time $T_0$ of  strong solution to be infinity and thus prove the global existence result.

Let $T^*$ be the  maximal time of existence for the strong solution. Then, $T^*\gammae T_0$. For any $0<\tau< T\le T^*$ with $T$ finite, one deduce from
$$ u\in L^{\infty}(0,T; \, H^1_0)\cap L^2(0,T; \,H^2),~~u_t\in L^2(\tau,T;\,H^1),$$
that
 \begin{equation} \label{x3} u \in C([\tau ,T]; \,  H^1_0).\end{equation}
Furthermore, it follows from
$$\rho \in L^{\infty}(0,T;\,H^1),~~\rho_t \in L^{\infty}(0,T;\,L^2)$$ that
  \begin{equation} \label{nx2} \rho \in C([0 ,T];H^1).\end{equation}

 Defining
  \begin{equation}gin{eqnarray}n (\rho^*, u^*)\triangleq(\rho,   u)(x,T^*)=\lim_{t\rightarrow T^*}(\rho, u)(x,t),\end{eqnarray}n
we   derive from \eqref{x3} and  \eqref{nx2} that $ (\rho^*, u^*)$ satisfies the initial condition \eqref{1d3a} at $t=T^*$.

Hence, we take $ (\rho^*, u^*)$ as the initial data at $t=T^*$ and then use the local existence theory to extend  the strong solution beyond the maximum existence time $T^*$. This contradicts the assumption on $T^*$. We finally show that $T^*$ could be  infinity and prove the global existence of the strong solution.

  \thetaxtbf{Step 3}. It remains   to prove \eqref{1.12'}. Direct calculations lead to
\begin{equation}gin{eqnarray}n\begin{equation}gin{aligned}\labelbel{ts7}
\left| \frac{d}{dt}\|u_x\|_{L^2}^2\right|& = \left|2\int_0^1 u_xu_{xt}dx\right|\\&=\left|2\int_0^1 u_x\left(\dot u_x-(uu_x)_x\right)dx \right|\\
 & =\left|2\int_0^1 u_x \dot u_xdx- \int_0^1 u_x^3 dx\right|\\
 &\le  C\|\dot u_x\|_{L^2}^2+C(1+\|u_x\|_{L^\infty})\|u_x\|_{L^2}^2,

\end{aligned}\end{eqnarray}n

which together with \eqref{gj3.1} and \eqref{gj3.6} yields
\begin{equation}gin{eqnarray}n\begin{equation}gin{aligned}\labelbel{ts8}
\int_1^\infty\left(\|u_x\|_{L^2}^2+\left|\frac{d}{dt}\|u_x\|_{L^2}^2\right|\right)dt\le C.
\end{aligned}\end{eqnarray}n
Thus,
\begin{equation}gin{eqnarray}n\labelbel{ts3}
\lim_{t\rightarrow \infty}\|u_x\|_{L^2}^2(t)=0,
\end{eqnarray}n
which combined with \eqref{gj3.6}  gives
\begin{equation}\label{gj3.4'}
\lim_{t\rightarrow \infty} \|u\|_{W^{1,p}} =0,~~~~\forall~p\in[1,\infty).
\end{equation}

The proof of Theorem \ref{th11} is finished.

$\Box$

{\it Proof of Theorem \ref{th1}.}  With the higher-order estimates in Lemmas \ref{nla4.3}--\ref{la4.4} at hand, the proof of Theorem \ref{th1} is similar to those of Theorem \ref{th11} and is omitted here for simplicity. 
$\Box$

\section{Proof of Theorem  \ref{th2}}

 The proof of Theorem \ref{th2} is divided into two steps as follows.

\thetaxtbf{Step 1}. We will prove
\begin{equation}\begin{equation}gin{aligned}\label{gj3.4}   \lim_{t\rightarrow0} \|\rho(\cdot,t)-\rho_s(\cdot)\|_{L^p}\rightarrow 0,~~\forall~p\in[1,\infty).\end{aligned}\end{equation}

Considering the function
\begin{equation}gin{eqnarray}n\labelbel{st3.3}
\mathcal{A} thcal{P}(t)=\int_0^1[P(\rho)-\tildelde{P}(\rho)]^2dx
\end{eqnarray}n
with
\begin{equation}gin{eqnarray}n\labelbel{st3.2}
 \tildelde{P}(\rho)= \tildelde{P}(\rho(x,t))\triangleq\int_0^1 P(\rho)dx+\int_0^x\rho fdy-\int_0^1\int_0^x\rho fdydx,
\end{eqnarray}n
we claim that
\begin{equation}gin{equation}\labelbel{st3.4}
    \lim_{t\rightarrow\infty}\int_{t-1}^t\left(\mathcal{A} thcal{P}(\tau)+\left|\frac{d}{d\tau}\mathcal{A} thcal{P}(\tau)\right|\right)d\tau=0.
\end{eqnarray}d{equation}

With \eqref{st3.4} at hand, one can derive the desired \eqref{gj3.4} with the same arguments as those in \cite{is2002,is2003}. For reader's convenience, we sketch them here for completeness. Indeed,
for any $t>1$ and $s\in(t-1,t)$, it holds
\begin{equation}gin{eqnarray}n \labelbel{st15}
    \mathcal{A} thcal{P}(t) \leq\int_{t-1}^t\left(\mathcal{A} thcal{P}(s)+\left|\frac{d}{d\tau}\mathcal{A} thcal{P}\right|\right)d\tau,
\end{eqnarray}n
which together with \eqref{st3.4} implies that
\begin{equation}\labelbel{st3.8}
\lim_{t\rightarrow\infty}\mathcal{A} thcal{P}(t)=\lim_{t\rightarrow\infty}\|P(\rho(\cdot, t))-\tildelde{P}(\rho(\cdot, t))\|_{L^2}^2=0.
\end{equation}
Let $t_n\rightarrow\infty$ be an arbitrary sequence, the uniform upper bound of the density \eqref{gj3.2} imlpies that there exist  a function $\Omegamegaega  idetilde{\rho}\in L^\infty$ and a positive constant $P_c$ such that for some subsequence $\{t_n'\}\subset\{t_n\}$,
\begin{equation}gin{equation}\labelbel{st3.9}
    \rho(\cdot,t_n')\rightarrow\Omegamegaega  idetilde{\rho}(\cdot)\thetaxt{  weakly * in $L^\infty$},~~~~
    \int_0^1 P(\rho(\cdot,t_n'))dx\rightarrow P_c.
\end{eqnarray}d{equation}
Clearly, $0\leq\Omegamegaega  idetilde{\rho}\leq C$. The standard compactness argument together with \eqref{st3.9} yields
\begin{equation}gin{equation}\labelbel{st3.10}
    \tildelde{P}\left(\rho(\cdot,t_n')\right)\rightarrow P_s(\cdot)\triangleq P_c
    +\int_0^x\Omegamegaega  idetilde{\rho}fdx-\int_0^1\int_0^x\Omegamegaega  idetilde{\rho}fdydx~~~\thetaxt{  in~~$C([0,1])$},
\end{eqnarray}d{equation}
which along with \eqref{st3.8} leads to
\begin{equation}gin{eqnarray}n\labelbel{st16}
P(\rho(\cdot,t_n'))\rightarrow P_s(\cdot) ~~~\thetaxt{  in~~$L^2$}.
\end{eqnarray}n
Consequently, it holds that for some subsequence  $\{t_n''\}\subset\{t_n'\}$,
 $$P(\rho(x,t_n''))\rightarrow P_s(x) ~~~~\thetaxt{a.e.~ in}~(0,1).$$
  The continuity and monotonicity of $P(\cdot)$ deduce that the inverse function of $P$, denoted by $P^{-1}$,  is continuous,  and then
$$\rho(x,t_n'')\rightarrow \rho_s(x)\triangleq P^{-1}(P_s(x)) ~~~~\thetaxt{a.e.~ in}~(0,1),$$
where $\rho_s\in C([0,1])$ and $\rho_s >0$.
 This together with Lebesgue dominated theorem and \eqref{gj3.2} implies that
\begin{equation}gin{equation}\labelbel{st3.11}
    \rho(\cdot,t_n'')\rightarrow\rho_s(\cdot),~~\thetaxt{ in $L^p$},~~~
    \forall~p \in[1,\infty),
\end{eqnarray}d{equation}
 which along with \eqref{st3.9} yields $$\rho_s=\Omegamegaega  idetilde{\rho}.$$
 According to the definition of $P_s$ in \eqref{st3.10}, it holds
$$P(\rho_s)=P_s=P_c
    +\int_0^x\rho_sfdx-\int_0^1\int_0^x\rho_sfdydx,$$
which together  with \eqref{st3.11} and \eqref{03.2} yields that
\begin{equation}\rhoonumber\\tag [P(\rho_s)]_x= \rho_sf,~~~ \int_0^1 \rho_s dx=1.\end{equation}
Hence, we show that   $\rho_s$ is indeed the solution to the stationary problem \eqref{1s} due to Lemma \ref{le2}. And, \eqref{gj3.4} is a direct consequence of  \eqref{st3.11}.

Now, it remains to prove \eqref{st3.4}.
Denoting
\begin{equation}gin{eqnarray}n\labelbel{st1}
\Psi=\int_0^x (P(\rho)-\tildelde{P}(\rho))dx,
\end{eqnarray}n which satisfies  $\Psi(0,t)=\Psi(1,t)=0,$  after using integration by parts and \eqref{1d}$_2$, we rewrite $\mathcal{A} thcal{P}(t)$    as
\begin{equation}\begin{equation}gin{aligned}\labelbel{st4}
\mathcal{A} thcal{P}(t)&=\int_0^1(P(\rho)-\tildelde{P}(\rho))d\Psi  =-\int_0^1(P(\rho)-\tildelde{P}(\rho))_x\Psi dx\\

    &=\int_0^1\left((\rho u)_t+(\rho u^2)_x-[\mu(\rho)u_x]_x\right)\Psi dx\\
    &=\frac{d}{dt}\int_0^1\rho u\Psi dx-\int_0^1\rho u\Psi_t dx-\int_0^1\left(\rho u^2-\mu(\rho)u_x\right)\Psi_x dx.
\end{aligned}\end{equation}

 It follows from \eqref{gj3.1} and \eqref{gj3.2} that
\begin{equation}\begin{equation}gin{aligned}\labelbel{st2}
|\Psi|+|\Psi_x|&=\left|\int_0^x\left(P(\rho)-\int_0^1P(\rho)dy-\int_0^x\rho fdy+\int_0^1\int_0^\xi\rho fdyd\xi\right)dx\right|\\
   &\quad+\left|P(\rho)-\int_0^1P(\rho)dx-\int_0^x\rho fdy+\int_0^1\int_0^x\rho fdydx\right|\leq C,
\end{aligned}\end{equation}

and thus\begin{equation}\begin{equation}gin{aligned}\labelbel{st5}
 \left|-\int_0^1\left(\rho u^2-\mu(\rho)u_x\right)\Psi_x dx \right|\le C\|u_x\|_{L^2}^2+C\|u_x\|_{L^2}.
\end{aligned}\end{equation}

Since
$$P_t+(Pu)_x+(\gammaammamma-1)Pu_x=0,$$
one deduces from integration by parts and \eqref{1d}$_1$ that
\begin{equation}gin{eqnarray}n\begin{equation}gin{aligned}\labelbel{st6}
   &(P(\rho)-\tildelde{P}(\rho))_t\\
   &=-(Pu)_x-(\gammaammamma-1)Pu_x-\int_0^1 P_tdx-\int_0^x\rho_t fdy+\int_0^1\int_0^x\rho_t fdydx\\
   &=-(Pu)_x-(\gammaammamma-1)Pu_x+(\gammaammamma-1)\int_0^1Pu_xdx+\int_0^x(\rho u)_y fdy-\int_0^1\int_0^x(\rho u)_y fdydx\\
   &=-(Pu)_x-(\gammaammamma-1)Pu_x+(\gammaammamma-1)\int_0^1Pu_xdx\\
   &\quad+\rho uf-\int_0^x \rho u  f_ydy-\int_0^1 \rho u  f dx+\int_0^1\int_0^x \rho u  f_ydydx.
\end{aligned}\end{eqnarray}n
Combining this, \eqref{gj3.2}, and \eqref{3.5}  gives
\begin{equation}\begin{equation}gin{aligned}\labelbel{st7}
    |\Psi_t| &=\left|\int_0^x(P(\rho)-\tildelde{P}(\rho))_tdx\right|\\
    &=\left|-\rho^\gammaammamma u-(\gammaammamma-1)\int_0^x\rho^\gammaammamma u_xdx+(\gammaammamma-1) \int_0^1\rho^\gammaammamma u_xdx\right.\\
   &\quad +\left.\int_0^x\left(\rho uf-\int_0^x \rho u  f_ydy-\int_0^1 \rho u  f dx+\int_0^1\int_0^x \rho u  f_ydydx\right)dx \right|\\
   &\le C(\begin{equation}gin{aligned}r\rho,\|f\|_{H^1})\|u\|_{L^2}+C(\begin{equation}gin{aligned}r\rho)\|u\|_{L^\infty}+C(\begin{equation}gin{aligned}r\rho)\|u_x\|_{L^2}\le C\|u_x\|_{L^2}.
\end{aligned}\end{equation}
One thus gets
\begin{equation}\begin{equation}gin{aligned}\labelbel{st8}
 \int_0^1 \rho u\Psi_tdx\leq C\|u_x\|_{L^2}\|u\|_{L^\infty}\int_0^1\rho dx\leq C\|u_x\|_{L^2}^2.
\end{aligned}\end{equation}

Hence, on the one hand, it follows from \eqref{st2},  \eqref{st4}, \eqref{st5}, and \eqref{st8} that
\begin{equation}\begin{equation}gin{aligned}\labelbel{st9}
 \int_{t-1}^t \mathcal{A} thcal{P}(\tau)d\tau&=\int_0^1\rho u\Psi dx(t)-\int_0^1\rho u\Psi dx(t-1)\\
 &\quad- \int_{t-1}^t\int_0^1\rho u\Psi_t dxd\tau- \int_{t-1}^t\int_0^1\left(\rho u^2-\mu(\rho)u_x\right)\Psi_x d\tau\\
 &\le C \|u\|_{L^2}+C\int_{t-1}^t\left(\|u_x\|_{L^2}^2+\|u_x\|_{L^2}\right)d\tau\\
 &\le C \|u_x\|_{L^2}+C\int_{t-1}^t\|u_x\|_{L^2}^2d\tau+C\left(\int_{t-1}^t\|u_x\|_{L^2}^2d\tau\right)^{1/2}.
\end{aligned}\end{equation}
On the other hand, it deduces from \eqref{st2} and \eqref{st7} that
\begin{equation}\begin{equation}gin{aligned}\labelbel{st11}
\int_{t-1}^t \left|\frac{d}{d\tau}\mathcal{A} thcal{P}(\tau)\right|d\tau
&=2\int_{t-1}^t \left|\int_0^1 \Psi_x(P(\rho)-\tildelde{P}(\rho))_\tau dx\right|d\tau\\
& \le C\int_{t-1}^t \|u_x\|_{L^2} d\tau\leq C\left(\int_{t-1}^t\|u_x\|_{L^2}^2d\tau\right)^{1/2}.
\end{aligned}\end{equation}

 Then, the desired \eqref{st3.4} is deduced directly from \eqref{st9}, \eqref{st11}, \eqref{gj3.4'}, and \eqref{gj3.1}.

\thetaxtbf{Step 2}.  Now, we are  in a position to prove \eqref{ed1}. The method  used here is  motivated by Huang-Li-Xin \cite{jmpa06} and Li-Zhang-Zhao \cite{lisiam}.

Thanks to \eqref{1s}, the momentum equation \eqref{1d}$_2$ can be rewritten as
\begin{equation}gin{eqnarray}n\begin{equation}gin{aligned}\labelbel{ed1.1}
 \rho u_t + \rho u u_x + [P(\rho)]_x-\rho\rho_s^{-1}[P(\rho_s)]_x=[\mu(\rho)u_x]_x,
\end{aligned}\end{eqnarray}n
which multiplied by $u$, we obtain after using integration by parts and  \eqref{1d}$_1$ that
\begin{equation}\begin{equation}gin{aligned}\labelbel{ed1.2}
 \frac{d}{dt}\left(\frac{1}{2}\int_0^1 \rho u^2dx+\int_0^1 G(\rho)dx\right)+\int_0^1 \mu(\rho) u_x^2dx=0,
\end{aligned}\end{equation}
where
\begin{equation}\begin{equation}gin{aligned}\labelbel{ed2.0}
G(\rho)&\triangleq \int_{\rho_s}^\rho  \int_{\rho_s}^r \frac{P'(\xi)}{\xi}d\xi dr =\rho  \int_{\rho_s}^\rho \frac{P(\xi)-P(\rho_s)}{\xi^2} d\xi\\
&=\frac{1}{\gammaammamma-1}\left(P(\rho)-P(\rho_s)-P'(\rho_s)(\rho-\rho_s)\right).
\end{aligned}\end{equation}
Clearly,  it follows from \eqref{gj3.2} and \eqref{pre1} that there are positive constants $M_1$ and $M_2$ depending only on $\gammaammamma$, $\begin{equation}gin{aligned}r\rho$, $K_1$, and $K_2$ such that
\begin{equation}\begin{equation}gin{aligned}\labelbel{ed2.1}
 M_1(\rho-\rho_s)^2\le G(\rho)\le M_2(\rho-\rho_s)^2. \end{aligned}\end{equation}

Next, it follows from  \eqref{1d}$_2$  and \eqref{1s} that
\begin{equation}\begin{equation}gin{aligned}\labelbel{ed2.3}
-[P(\rho)-P(\rho_s)]_x+\rho_s^{-1}(\rho-\rho_s)P'(\rho_s) (\rho_s)_x=(\rho u)_t+(\rho u^2)_x-[\mu(\rho)u_x]_x.
\end{aligned}\end{equation}
For  $\Phi (x,t)\triangleq\int_0^x(\rho-\rho_s)dy,$

it holds that \begin{equation}\begin{equation}gin{aligned}\labelbel{ed2.7}\Phi (0,t)= \Phi (1,t)=0\end{aligned}\end{equation} owing to \eqref{1s} and \eqref{03.2}.
Multiplying \eqref{ed2.3} by   $\rho_s^{-1}\Phi (x,t)$ gives
\begin{equation}\begin{equation}gin{aligned}\labelbel{ed2.5}
&-\int_0^1 \left[\rho_s^{-1}(P(\rho)-P(\rho_s))\right]_x\Phi  dx\\
&=\int_0^1 \rho_s^{-2}(\rho_s)_x \left[P(\rho)-P(\rho_s)-P'(\rho_s)(\rho-\rho_s) \right]\Phi  dx\\
&\quad+\int_0^1 \rho_s^{-1}(\rho u)_t\Phi  dx+\int_0^1 \rho_s^{-1}(\rho u^2)_x\Phi  dx-\int_0^1 \rho_s^{-1} [\mu(\rho)u_x]_x\Phi  dx\\
&\triangleq \sum_{i=1}^4\begin{equation}gin{aligned}r J_i.
\end{aligned}\end{equation}

First, integration by parts combined with \eqref{ed2.7} and \eqref{pre1} gives
\begin{equation}\begin{equation}gin{aligned}\labelbel{ed2.6}
-\int_0^1 \left[\rho_s^{-1}(P(\rho)-P(\rho_s))\right]_x\Phi  dx 
&=\int_0^1  \rho_s^{-1}P'(\hat\rho)(\rho-\rho_s)^2dx \\&\gammae C_1 \|\rho-\rho_s\|_{L^2}^2,
\end{aligned}\end{equation}
where $0<\hat\rho \in (\min\{\rho, \rho_s\}, \mathcal{A} x\{\rho, \rho_s\})$ and $C_1$ is a positive constant dependent of $\gammaammamma$, $\begin{equation}gin{aligned}r \rho$, $K_1$, and $K_2$.

Next, the terms on the right hand of \eqref{ed2.5} can be estimated as follows. On the one hand, it follows from \eqref{gj3.4} that there is some $T^*>1$ such that for $t>T^*$,
\begin{equation}\begin{equation}gin{aligned}\labelbel{ed1.8}\rhoonumber\\tag
 \|\rho-\rho_s\|_{L^2}  \le \frac{C_1}{4},
\end{aligned}\end{equation}
which along with  \eqref{pre1} and \eqref{ed2.0} yields that
\begin{equation}\begin{equation}gin{aligned}\labelbel{ed2.8}
\begin{equation}gin{aligned}r J_1 
 \le C\|\rho-\rho_s\|_{L^2}\|\rho-\rho_s\|_{L^2}^2
&\le \frac{C_1}{4}\|\rho-\rho_s\|_{L^2}^2.
\end{aligned}\end{equation}
On the other hand, the integration by parts  together with  \eqref{pre1} and \eqref{gj3.2} implies that
\begin{equation}\begin{equation}gin{aligned}\labelbel{ed2.9}
\begin{equation}gin{aligned}r J_2 &= \frac{d}{dt}\int_0^1  \rho_s^{-1}(\rho u) \int_0^x(\rho-\rho_s)dy dx -\int_0^1 \rho_s^{-1} \rho u  \int_0^x \rho_t dy dx \\
&=\frac{d}{dt}\int_0^1  \rho_s^{-1}(\rho u) \int_0^x(\rho-\rho_s)dy dx+\int_0^1 \rho_s^{-1} \rho^2 u^2 dx\\
& \le \frac{d}{dt}\int_0^1  \rho_s^{-1}(\rho u) \int_0^x(\rho-\rho_s)dy dx+ C\|u_x\|_{L^2}^2,
\end{aligned}\end{equation}
and
\begin{equation}\begin{equation}gin{aligned}\labelbel{ed2.10}
\begin{equation}gin{aligned}r J_3 +\begin{equation}gin{aligned}r J_4 &=  \int_0^1 \rho_s^{-2} (\rho_s)_x \rho u^2 \int_0^x(\rho-\rho_s)dy dx- \int_0^1 \rho_s^{-1}   \rho u^2 (\rho-\rho_s) dx\\
&\quad+\int_0^1 \rho_s^{-1}  \mu(\rho)u_x (\rho-\rho_s)dx-\int_0^1 \rho_s^{-2} (\rho_s)_x  \mu(\rho)u_x \int_0^x(\rho-\rho_s)dy dx\\
&  \le C \|u_x\|_{L^2}^2+C \|u_x\|_{L^2} \|\rho-\rho_s\|_{L^2}\\
&\le  \frac{C_1}{4} \|\rho-\rho_s\|_{L^2}^2+C\|u_x\|_{L^2}^2.
\end{aligned}\end{equation}

 Substituting \eqref{ed2.6}--\eqref{ed2.10} into \eqref{ed2.5} derives
 \begin{equation}\begin{equation}gin{aligned}\labelbel{ed3.0}
C_1\|\rho-\rho_s\|_{L^2}^2 \le 2\frac{d}{dt}\int_0^1  \rho_s^{-1}(\rho u) \int_0^x(\rho-\rho_s)dy dx+ C\|u_x\|_{L^2}^2,
\end{aligned}\end{equation}

Since
 \begin{equation}\begin{equation}gin{aligned}\labelbel{ed3.4}\rhoonumber\\tag
\left|2 \int_0^1  \rho_s^{-1}(\rho u) \int_0^x(\rho-\rho_s)dy dx\right|
&\le   C\|\sqrt{\rho} u\|_{L^2}^2+  C\|\rho-\rho_s\|_{L^2}^2,
\end{aligned}\end{equation}
  adding \eqref{ed3.0} multiplied  by some suitably small $\end{eqnarray}d{theorem}a$ to \eqref{ed1.2} gives
 \begin{equation}\begin{equation}gin{aligned}\labelbel{ed3.1}
&\frac{d}{dt}W(t)+\end{eqnarray}d{theorem}a C_1\|\rho-\rho_s\|_{L^2}^2 +\frac{1}{2}\int_0^1 \mu(\rho) u_x^2dx\le 0,
\end{aligned}\end{equation}
where
 \begin{equation}gin{eqnarray}n
 W(t)\triangleq \frac{1}{2}\int_0^1 \rho u^2dx+\int_0^1 G(\rho)dx-2\end{eqnarray}d{theorem}a \int_0^1  \rho_s^{-1}(\rho u) \int_0^x(\rho-\rho_s)dy dx
\end{eqnarray}n
satisfies
 \begin{equation}\begin{equation}gin{aligned}\labelbel{ed3.5}
\frac{1}{4}\int_0^1 \rho u^2dx+\frac{M_1}{2}\|\rho-\rho_s\|_{L^2}^2\le W(t)\le C\|\sqrt{\rho} u\|_{L^2}^2+ C\|\rho-\rho_s\|_{L^2}^2.
\end{aligned}\end{equation}

Furthermore, one has
 \begin{equation}\begin{equation}gin{aligned}\labelbel{ed3.2}
 \int_0^1 \rho u^2dx \le C\|u \|_{L^2}^2\le \frac{C_2^{-1}}{2}\|\sqrt{\mu(\rho)}u_x \|_{L^2}^2,

\end{aligned}\end{equation}
where $C_2>0$ is a positive constant depending on $\begin{equation}gin{aligned}r \mu$.

The combination of \eqref{ed3.1} with \eqref{ed3.2} gives
\begin{equation}\begin{equation}gin{aligned}\labelbel{ed3.6}
&\frac{d}{dt}W(t)+\alphalpha  \|\rho-\rho_s\|_{L^2}^2 +\alphalpha   \int_0^1 \rho u^2dx \le 0,
\end{aligned}\end{equation}
where $\alphalpha $ is a positive constant depending on $\end{eqnarray}d{theorem}a$, $C_1$, and $C_2$. Hence, Gronwall's inequality combined with \eqref{ed3.5} and \eqref{ed3.6} shows
\begin{equation}\begin{equation}gin{aligned}\labelbel{ed3.7}
\|\sqrt{\rho} u\|_{L^2}^2+  \|\rho-\rho_s\|_{L^2}^2 \le Ce^{-\alphalpha  t},~~~\mbox{for}~t>T^*.
\end{aligned}\end{equation}

In what follows, we will prove the exponential decay rate for the  $L^2$-norm of $u_x$.
 First, multiplying \eqref{ed1.2} by $e^{\frac{\alphalpha }{2}t}$, we get after using \eqref{ed2.1} and \eqref{ed3.7} 
\begin{equation}\begin{equation}gin{aligned}\labelbel{ed4.0}
 &\frac{d}{dt}\left(e^{\frac{\alphalpha }{2}t}\|\sqrt{\rho} u\|_{L^2}^2+2e^{\frac{\alphalpha }{2}t}\int_0^1 G(\rho)dx\right)+e^{\frac{\alphalpha }{2}t}\|u_x\|_{L^2}^2\\
 &\le Ce^{\frac{\alphalpha }{2}t}\|\sqrt{\rho} u\|_{L^2}^2+Ce^{\frac{\alphalpha }{2}t}\|\rho-\rho_s\|_{L^2}^2\\
 &\le Ce^{-\frac{\alphalpha }{2}t}.
\end{aligned}\end{equation}
Integrating \eqref{ed4.0} over $[T^*,t]$ gives
\begin{equation}\begin{equation}gin{aligned}\labelbel{ed4.1}
e^{\alphalpha  t}\|\sqrt{\rho} u\|_{L^2}^2+e^{\alphalpha  t}\|\rho-\rho_s\|_{L^2}^2 +\int_{T^*}^te^{\frac{\alphalpha }{2}t}\|u_x\|_{L^2}^2dt\le C,~~~\mbox{for}~t>T^*,
\end{aligned}\end{equation}  due to \eqref{ed3.7}.

Next, multiplying \eqref{ed2.3} by $\dot u$ and integrating the resulting equation by parts, it holds for $t>T^*$,
\begin{equation}\begin{equation}gin{aligned}\labelbel{ed4.2}
 &\frac{d}{dt}\left(\|\sqrt{\mu(\rho)} u_x\|_{L^2}^2\right)+\|\sqrt{\rho} \dot u\|_{L^2}^2\\
 &=-\frac{1}{2}\int_0^1 [\mu(\rho)-\mu'(\rho)\rho]u_x^3dx+\int_0^1 [P(\rho)-P(\rho_s)] \dot u_xdx\\
 &\quad +\int_0^1 \rho_s^{-1}(\rho-\rho_s)P'(\rho_s) (\rho_s)_x\dot udx \\
 &\le C\|u_x\|_{L^\infty}\|u_x\|_{L^2}^2+C\|\rho-\rho_s\|_{L^2}\left(\|\dot u\|_{L^2}+\|\dot u_x\|_{L^2}\right)\\
  &\le C\|u_x\|_{L^2}^2+C\|\rho-\rho_s\|_{L^2}\|\dot u_x\|_{L^2},
\end{aligned}\end{equation}
where in the last inequality one has used \eqref{gj3.6} and Poincar\'e inequality. Integrating \eqref{ed4.2} multiplied  by $e^{\frac{\alphalpha }{4}t}$ over $[T^*,t]$, we thus obtain after using \eqref{ed4.1} and \eqref{gj3.6} that \begin{equation}gin{eqnarray}n
 & e^{\frac{\alphalpha }{4}t}\|u_x\|_{L^2}^2 +\int_{T^*}^te^{\frac{\alphalpha }{4}t}\|\sqrt{\rho} \dot u\|_{L^2}^2dt  \le C,~~~\mbox{for}~t>T^*,
\end{eqnarray}n which together with    \eqref{ed4.1},   \eqref{gj3.4'},  and \eqref{gj3.4} gives  \eqref{ed1}.

Finally, the proof of \eqref{bp} is similar as that of \cite[Theorem 1.2]{lijde}(see also \cite{hlx1,jmpa06}).
The proof of  Theorem \ref{th2} is completed.  
$\Box$

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