\documentstyle[twoside,leqno]{article}
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\markboth{\hfil Weak Solutions \hfil EJDE--1996/04}%
{EJDE--1996/04\hfil Kazufumi Ito\hfil}
\begin{document}
\title{\vspace{-1in}\parbox{\linewidth}{\footnotesize\noindent
{\sc  Electronic Journal of Differential Equations},
Vol.\ {\bf 1996}(1996), No.\ 4, pp. 1--17. \newline
ISSN 1072-6691. URL: http://ejde.math.swt.edu or http://ejde.math.unt.edu
\newline ftp (login: ftp) 147.26.103.110 or 129.120.3.113 }
\vspace{\bigskipamount} \\
Weak Solutions to the One-dimensional Non-Isentropic Gas Dynamics
by the Vanishing Viscosity Method
\thanks{ {\em 1991 Mathematics Subject Classifications:}
35L65.\newline\indent
{\em Key words and phrases:} Compensated compactness, Conservation laws, 
Entropy.
\newline\indent
\copyright 1996 Southwest Texas State University  and University of
North Texas.\newline\indent
Published June 18, 1996.} }
\date{}
\author{Kazufumi Ito}
\renewcommand{\theequation}{\arabic{section}.\arabic{equation}}
\maketitle
\begin{abstract} 
 In this paper we consider the non-isentropic
equations of gas dynamics with the entropy preserved.
Equations are formulated so that the problem
is reduced into the $2 \times 2$ system of conservation laws
with a forcing term in momentum equation.
The method of compensated compactness is then applied to prove the existence
of weak solution in the vanishing viscosity method.
\end{abstract}

\newcommand{\ds}{\displaystyle}

\section{Introduction}
Consider the one-dimensional gas dynamics equation in
the Eulerian coordinate
$$\begin{array}{l}
\ds \rho_t+(\rho u)_x=0 \\
\ds (\rho u)_t+(\rho u^2+p)_x=0 \\
\ds s_t+us_x=0.
\end{array} \leqno (1.1) $$
where $\rho,\;u,\;p$ and $s$ denote the density, velocity, pressure
and entropy. Other relevant quantities are the internal energy $e$
and the temperature $T$. We assume that the gas is ideal, so that
the equation of state is given by
$$
p=R\rho T
$$
and that it is polytropic, so that $e=c_v\, T$ and
$$
p=(\gamma-1)e^{s/c_v}\rho^{\gamma}
\leqno (1.2) $$
where $\gamma=c_p/c_v >1$ and $R=c_p-c_v$. Define $\phi$ by
$$
\phi^{1-\gamma}=\gamma(\gamma-1)\,e^{s/c_v}.
\leqno (1.3) $$
Then, $\phi$ satisfies
$$
\phi_t+u\,\phi_x=0.
$$
Thus, we consider the Cauchy problem (equivalent to (1.1))
$$\begin{array}{l}
\ds \rho_t+m_x=0 \\ 
\ds m_t+(m^2/\rho+p)_x=0 \\ 
\ds \phi_t+u\,\phi_x=0\, .
\end{array} \leqno (1.4) $$
where
$$
m=\rho u \quad\mbox{and}\quad p=\frac{1}{\gamma}\,\phi^{1-\gamma}
\rho^{\gamma},
\leqno (1.5) $$
with smooth initial data $(\rho_0,m_0)$ in  $L^{\infty}(R^2)$ that approaches
a constant state $(\bar{\rho},\bar{m})$ at infinity and satisfies
$\rho_0(x) \ge \delta_1>0$, and $\phi_0$ in $W^{1,\infty}(R)$ that satisfies
$(\phi_0)_x$ converges to 0 at infinity and
$$
\phi_0(x) \ge \delta_2>0\quad\mbox{and}\quad
(\phi_0)_x(x)\ge 0\;\;(\mbox{or}\;\;(\phi_0)_x(x) \le 0).
\leqno (1.6) $$

Consider the conservation form of the gas dynamics
$$\begin{array}{l}
\ds \rho_t+(\rho u)_x=0 \\ 
\ds (\rho u)_t+(\rho u^2+p)_x=0 \\ 
\ds [\rho\,(\frac{1}{2}\,u^2+e)]_t+(\rho u\,[\frac{1}{2}\,u^2+e]+pu)_x=0.
\end{array} \leqno (1.7) $$
System (1.7) can be written as the hyperbolic system of conservation laws
$$
y_t+f(y)_x=0
$$
where $y=y(t,x)=(\rho,\rho u,\rho\,(\frac{1}{2}\,u^2+e))\in R^3$ and $f$
is a smooth nonlinear mapping from $R^3$ to $R^3$.
System (1.4) is equivalent to system (1.7) when solutions are smooth but not
necessarily when solutions are weak (e.g.,[Sm, Chapters 16-17]).
It is proved in Corollary 3.6 that the viscosity limit of solutions
to (1.10) satisfies $\eta_t+q_x \le 0$ in the sense of distributions, i.e.,
the third equation of (1.7), the conservation of energy $\eta_t+q_x=0$
is replaced by the non-energy production. We also note that the isentropic
solution ($\phi=\mbox{const}$) [Di1] is a weak solution of (1.4)
but not necessarily of (1.7).

In this paper we show the existence of weak solutions to (1.4)-(1.5)
using the vanishing viscosity method. The function 
$(\rho,m,\phi) \in L^{\infty}(\Omega)\times L^{\infty}(\Omega)\times
W^{1,\infty}(\Omega)$ with $\Omega=[0,\tau] \times R$
is a weak solution of (1.4)-(1.5) if $\phi$ satisfies the third equation
of (1.4) a.e in $\Omega$ and
$$
\int^{\tau}_0\int^{\infty}_{-\infty}
(v\cdot\psi_t+F(\rho,m,\phi)\cdot\psi_x)\,dx\,dt=0
\leqno (1.8) $$
for all $\psi\in C^{\infty}_c(\Omega;R^2)$ where $v=(\rho,m)$ and
$$
F(\rho,m,\phi)=(m,m^2/\rho+p)
\leqno (1.9) $$
We consider the viscous equation of (1.4) with equal diffusion rates
$$\begin{array}{l}
\ds \rho_t+m_x=\epsilon\,\rho_{xx} \\
\ds m_t+(m^2/\rho+p)_x=\epsilon\,m_{xx} \\
\ds \phi_t+u\,\phi_x=\epsilon\phi_{xx}.
\end{array} \leqno (1.10) $$
It will be shown in Theorem 3.5 that the solutions
$(\rho^{\epsilon},m^{\epsilon},\phi^{\epsilon})$ to (1.10)
converge to a locally defined (in time) weak solution $(\rho,m,\phi)$
of (1.4).

Our approach is based on the following observation.
Suppose $\phi$ is a constant. Then equation (1.4) reduces to the isentropic
gas dynamics.  For the isentropic equation it is shown in DiPerna [Di1] that
(1.4) has a weak solution by the vanishing viscosity method and using
the theory of compensated compactness. The key steps in [Di1] are given
as follows.
First, if $\phi$ is a constant and $\theta=(\gamma-1)/2$ then
$$
w=G_1(\rho,m,\phi)=\frac{m}{\rho}+\frac{1}{\theta}\,\phi^{-\theta}\rho^{\theta}
\quad\mbox{and} \quad -z=G_2(\rho,m,\phi)=-\frac{m}{\rho}
+\frac{1}{\theta}\,\phi^{-\theta}\rho^{\theta}
\leqno (1.11) $$
are the Riemann invariants so that $\nabla_v G_1$ and $\nabla_v G_2$ are
the two left eigenvectors of the $2\times 2$  matrix
$$
\nabla_v F=\left(\begin{array}{cc} 0 & 1 \\
\ds -\frac{m^2}{\rho^2}+\rho^{2\theta}\phi^{-2\theta} & \ds \frac{2m}{\rho}
\end{array}\right).
$$
where $\phi$ is assumed to be a positive constant.
The method of invariant regions ([CCS],[Sm]) is applied  to $G_1,\;G_2$
to obtain that $0\le \rho^{\epsilon}\le\mbox{const}$, $|m^{\epsilon}/\rho^{\epsilon}|
\le\mbox{const}$. Then, there exist a subsequence of $v^{\epsilon}=(\rho^{\epsilon},
m^{\epsilon})$, still denoted by $v^{\epsilon}$ and
a Young measure $\nu_{t,x}$ such that for each $\Phi\in C(R^2)$ we have
$\Phi(v^{\epsilon})$ converges weak star to $\bar{\Phi}$ in
$L^{\infty}(\Omega)$ where
$$
\bar{\Phi}(t,x)=\langle \nu,\Phi\rangle=\int_{\Omega} \Phi(y)\,d\nu_{t,x}(y),
\;\mbox{a.e.}\; (t,x)\in \Omega.
$$
Using the entropy fields [La] and the div-curl theorem of Murat [Mu] and
Tatar [Ta] for bilinear maps in the weak topology,
$$
\langle \nu ,\eta_1q_2-\eta_2 q_1\rangle=\langle \nu,\eta_1\rangle
\langle \nu,q_2 \rangle-\langle \nu,\eta_2\rangle \langle \nu, q_1\rangle
\leqno (1.12) $$
for all entropy/entropy flux pairs $(\eta_i,q_i)$ so that
$$
\nabla_v\eta\nabla_v F=\nabla_v q.
$$
Then, using the weak entropy pairs (i.e., $\eta(0,\cdot)=0$) it is shown that
$\nu$ reduces to a point mass,
i.e., $v_{\epsilon}$ converges to $v$, a.e. in $\Omega$.

We will apply the method described above for the non-isentropic equation
(1.10).  We need to overcome the two
major difficulties. First, $G_1,\;G_2$ are no longer the Riemann invariants
of the $3 \times 3$ matrix $M$:
$$
M=\nabla F=\left(\begin{array}{ccc} 0 & 1 & 0 \\
\ds -\frac{m^2}{\rho^2}+\rho^{2\theta}\phi^{-2\theta} & \ds \frac{2m}{\rho}
& \ds -\frac{2\theta}{\gamma}\rho^{\gamma}\phi^{-2\theta-1} \\
0 & 0 &\ds \frac{m}{\rho}\end{array}\right).
$$
Next, equation (1.12) should be extended to the $3 \times 3$ system.
We resolve these difficulties by the following steps. Note (see Lemma 2.4)
that $G_i=G_i(\rho,m,\phi),\;i=1,2$ satisfies
$$
(G_i)_t+\lambda_i\nabla G_i\cdot y_x+\frac{1}{\gamma}\left\{
\begin{array}{ll} \ds \rho^{2\theta}\phi^{-2\theta-1}\phi_x, & i=1 \\
\ds -\rho^{2\theta}\phi^{-2\theta-1}\phi_x, & i=2 \end{array} \right.=
\epsilon\,((G_i)_{xx}-\nabla^2 G_i(y_x,y_x))
\leqno (1.13) $$
where $y=(\rho,m,\phi)$ is a solution to (1.10) and $\lambda_1=u+\rho^{\theta}
\phi^{-\theta},\;\lambda_2=u-\rho^{\theta}\phi^{-\theta}$ are
the eigenvalues of
the $2 \times 2$ matrix $\nabla_v F$. Here, $G_i,\;i=1,2$  are quasi-convex
functions of $(\rho,m,\phi)$ (see Lemma 2.5), i.e.,
$$
r\cdot\nabla G_i=0\;\;\mbox{implies}\;\;\nabla^2 G_i(r,r)\ge 0.
$$
Note that from the third equation of (1.10) that
$\phi^{\epsilon} \ge \delta_2$ and
$|\phi^{\epsilon}|_{\infty} \le |\phi_0|_{\infty}$ (see Lemma 2.1) and
moreover $\phi^{\epsilon}_x$ satisfies
$$
(\phi^{\epsilon}_x)_t+(u^{\epsilon}\,\phi^{\epsilon}_x)_x=
\epsilon\,(\phi^{\epsilon}_x)_{xx}
$$
Observing that if $(\rho,m,\phi)$ is a solution to (1.10) then
$\xi=\mbox{log}(\frac{\phi_x}{\rho})$ satisfies
$$
\xi_t+u\,\xi_x=\epsilon\,(\xi_{xx}+|(\mbox{log}\phi_x)_x|^2-
|(\mbox{log}\rho)_x|^2),
\leqno (1.14) $$
we show that if $(\phi_0)_x \ge 0$ (resp. $\le 0$)
then $\phi^{\epsilon}_x \ge 0$
(resp. $\le 0$) and $|\phi^{\epsilon}_x|\le c\,\rho^{\epsilon}$ in $\Omega$
provided that $|(\phi_0)_x|\le c\,\rho_0$ in $R$ (see Lemma 2.3).
It thus follows from (1.13) and the quasi-convexity of $G_i,\;i=1,2$ that
$\max_{x} G_2(t,x) \le \max_{x} G_2(0,x)$ (resp.
$\max_{x} G_1(t,x) \le  \max_{x} G_1(0,x)$) and
$$
|\rho^{2\theta}\phi^{-2\theta-1}\phi_x|\le c\,\left(\frac{\rho}{\phi}\right)
^{2\theta+1}.
$$
By the maximum principle (see Theorem 2.6), there exists a $\tau=\tau_c>0$
with $c \to \tau_c$ monotonically decreasing and $\tau_0=\infty$ such that
$\max_{t\in [0,\tau],\,x\in R}\; G_1(t,x)$ is
less than a constant independent of
$\epsilon>0$. Hence, we obtain $0\le \rho^{\epsilon}\le\mbox{const}$,
$|m^{\epsilon}/\rho^{\epsilon}|\le\mbox{const}$ and $|\phi_x| \le\mbox{const}$
in $\Omega$.

Second, in contrast to the isentropic case, the system (1.10) is not
endowed with a rich family of entropy-entropy flux pairs. Thus,
in order to prove that the Young measures $\nu_{t,x}$ of a weakly star
convergent subsequence of $(\rho^{\epsilon},m^{\epsilon},\phi^{\epsilon})$
reduce to a point mass, we first
note that $\{\phi^{\epsilon}(t,x)\}$ is precompact in $L^2_{loc}(\Omega)$ and
thus $\phi^{\epsilon}$ converges $\phi$ a.e. in $\Omega$ (see Lemma 3.4).
Also, we note that dividing the first two equations (1.10) by $\phi$, we obtain
$$\begin{array}{l}
\ds \hat{\rho}_t+\hat{m}_x=\epsilon\,(\hat{\rho}_{xx}+2\frac{\rho_x\phi_x}
{\phi^2}) \\
\ds \hat{m}_t+(\hat{m}^2/\hat{\rho}+\frac{1}{\gamma}\,\hat{\rho}^{\gamma})_x
+\frac{p\phi_x}{\phi^2}=\epsilon\,(\hat{m}_{xx}+2\,\frac{m_x\phi_x}
{\phi^2}).
\end{array} \leqno (1.15) $$
where $\ds \hat{\rho}=\frac{\rho}{\phi}$ and $\ds \hat{m}=\frac{m}{\phi}$
(see Lemma 3.2). This implies that $\hat{v}=(\hat{\rho},\hat{m})$ satisfies
the (viscous) isentropic gas-dynamics with the forcing term
$-p\phi_x/\phi^2$ in the momentum equation.
Since $\ds \frac{p^{\epsilon}\phi^{\epsilon}_x}{(\phi^{\epsilon})^2}
\in L^{\infty}(\Omega)$ uniformly in
$\epsilon>0$, thus $\ds \{ \frac{p^{\epsilon}\phi^{\epsilon}_x}
{(\phi^{\epsilon})^2}\}_{\epsilon>0}$ is precompact
in $H^{-1,q}_{loc}(\Omega),\; 1 \le q <2$. Hence, the method of compensated
compactness in [Di1],[Ch] can be
applied to the functions $(\hat{\rho}^{\epsilon},\hat{m}^{\epsilon})$ to show
that $\nu_{t,x}$ is a point mass provided that $1<\gamma\le 5/3$.

Regarding work on existence of weak solutions for conservation laws,
we refer the reader to an excellent treatise by DiPerna [Di3] and references
therein.
Concerning basic framework on conservation laws, we refer the reader to
[La],[Sm] and for the functional analytic framework of compensated compactness
we refer [Mu],[Ta1],[Ev] and [Di2].
For scalar conservation laws the vanishing viscosity method is employed
(e.g., in [Ol],[Kr] and references in [Sm]) to define the unique entropy
solution. Also, the vanishing viscosity method is used to develop
the viscosity solution to the Hamilton-Jacobi equation in [CL].
The finite-difference methods (e.g., Lax-Friedrichs and Gudunov schemes)
are also used to construct weak solutions to a scalar and
$2 \times 2$ system of
conservation laws (e.g., see [Di2],[Ch] and [Sm]).

In the case where the initial
data have small total variation, Glimm [Gl] proved the global existence
of BV-solutions for a general class of hyperbolic systems as the strong
limit of random choice approximations.  However, the problem of existence of
solutions to (1.7) with large initial data is still unsolved.
In [CD] the vanishing viscosity method is applied to
the system (1.7) under a special class of constitutive relations
in Lagrangian coordinates.

\section{The Viscosity Method}

In this section we establish the uniform $L^{\infty}$ bound of
$y^{\epsilon}=(\rho^{\epsilon},m^{\epsilon},\phi^{\epsilon})$.

\noindent{\bf Lemma 2.1} {\it If $\phi \in C^{1,2}([0,\tau]\times R)$
satisfies $\phi_t+u\,\phi_x=\epsilon\,\phi_{xx}$ then
$$
\min_x\, \phi_0(x) \le \phi(t,x) \le \max_x\, \phi_0(x).
$$}
\noindent{\bf Proof:} Using the same arguments as in the proof of
Theorem 2.6, we can show that $\max_x\,\phi(t,x) \le \max_x\,\phi_0(x)$ and
$\min_x\,\phi(t,x) \ge \min_x\,\phi_0(x)$.  $\Box$
\vspace{2mm}

It will be shown in Section 3 (see (3.4) and Lemma 3.1) that
the normalized mechanical energy
$$
E(\rho,u,\phi)=\frac{1}{2}\,\rho(u-\bar{u})^2
+\frac{1}{\gamma}\,(\rho^{\gamma}-\gamma\,\bar{\rho}^{\gamma-1}
(\gamma-\bar{\gamma})-\bar{\rho}^{\gamma})\phi^{1-\gamma}
\leqno (2.1) $$
satisfies
$$
\int^{\infty}_{-\infty}E(\rho(t,x),u(t,x),\phi(t,x))\,dx
\le \int^{\infty}_{-\infty}E(\rho_0(x),u_0(x),\phi_0(x))\,dx.
\leqno (2.2) $$
The following lemma shows the lower bound of $\rho_{\epsilon}$.
\vspace{2mm}

\noindent{\bf Lemma 2.2} {\it If $\rho \in C^{1,2}([0,\tau] \times R)$
satisfies
$$
\rho_t+(u\,\rho)_x=\epsilon\,\rho_{xx}
\leqno (2.3) $$
with $\rho(0,\cdot) \ge 0$ and $u \in C^1(\Omega)$, then $\rho(t,\cdot)
\ge 0$. Moreover, if $\rho(0,\cdot) \ge \delta>0$ and
$$
\int^{\tau}_0\int^{\infty}_{-\infty}\rho\,|u-u_0|^2\,dx\,dt \le \mbox{const},
\leqno (2.4) $$
then $\rho(t,\cdot)\ge \delta(\epsilon,\tau)>0$ on $(0,\tau)$.}
\vspace{2mm}

\noindent{\bf Proof:} Choose $\psi=\min\,(\rho(t,x),0)$. Then we have
$$
\int^{\infty}_{-\infty} \frac{1}{2}\,|\psi(t,x)|^2
+\int^t_0\int^{\infty}_{-\infty}
(\epsilon\,|\psi_x|^2-\psi u\,\psi_x)\,dx\,ds=0.
$$
By the H\"older inequality, we obtain
$$
\int^{\infty}_{-\infty}|\psi(t,0)|^2 \le \frac{|u|_{\infty}}{2\epsilon}
\int^t_0\int^{\infty}_{-\infty} |\psi|^2\,dx\,ds
$$
where $|u|_{\infty}=\sup_{(t,x) \in (0,\tau)\times R}\,|u(t,x)|$, and the
Gronwall's inequality implies $\psi=0$. Thus, $\rho \ge 0$.

Next, we prove $\rho(t,\cdot) \ge \delta=\delta(\epsilon,\tau)>0$
if $\varphi(0,\cdot )\ge\delta>0$ by using the Stampacchia's lemma
(e.g., see [FI],[Tr]), i.e., suppose
$\chi(c)$ is a nonnegative, non-increasing function on $[c_0,\infty)$,
and there exist positive constants $K,\;s$ and $t$ such that
$$
\chi(\hat{c})\le Kc^s(\hat{c}-c)^{-s}\,\chi(c)^{1+t}
\quad\mbox{for all}\;\;\hat{c}>c \ge c_0,
$$
then
$$
\chi(c^*)=0 \quad \mbox{for}\;\;c^*=2c_0\,(1+2^{\frac{1+2t}{t^2}}
K^{\frac{1+t}{st}}\,\chi(c_0)^{\frac{1+t}{s}}).
$$
First, we establish {\em a priori} bound.
We consider the class $K$ [Di1] of strictly convex $C^2$ functions $h$
with following properties:
$$
h(\bar{\rho})=h^{\prime}(\bar{\rho})=0,\quad
h(\rho)=\rho^{-\alpha}\;\;\mbox{on}\;\;(0,\bar{\rho}/2)\;
\mbox{for some} \;\;0 < \alpha < 1.
$$
Premultiplying the first equation of (1.10) by $h^{\prime}(\rho)$
we obtain
$$
h(\rho)_t+(h^{\prime}(\rho)\rho u)_x-h^{\prime\prime}(\rho)\rho_x\rho u=
\epsilon\,(h(\rho)_{xx}-h^{\prime\prime}(\rho)\rho_x^2)
$$
Integration of this over $(0,t) \times R$ yields
\begin{eqnarray*} 
\lefteqn{\int^{\infty}_{-\infty} h(\rho(t,x))-h(\rho_0(x))\,dx
+\epsilon\int^t_0\int^{\infty}_{-\infty} h^{\prime\prime}
(\rho)\rho_x^2\,dx\,dt}  \\
\hfill&=&\int^t_0\int^{\infty}_{-\infty} h^{\prime\prime}(\rho)\rho_{x}
\rho(u-\bar{u})\,dx\,dt\,. \end{eqnarray*}
Note that
$$
h^{\prime\prime}(\rho)\rho_{x}\rho(u-\bar{u})\le\frac{\epsilon}{2}\,
h^{\prime\prime}(\rho)\rho_x^2+\frac{1}{2\epsilon}\,h^{\prime\prime}(\rho)
\rho^2(u-\bar{u})^2.
$$
Since there exists some constant $\beta>0$ such that
$$\begin{array}{l}
\rho^2\,h^{\prime\prime}(\rho)\le \beta\,\rho\quad \mbox{for}\;\;
\bar{\rho}/2  \le \rho \le M
\\
\rho^2\,h^{\prime\prime}(\rho) \le \beta\,h(\rho)\quad \mbox{for}\;\;
0 < \rho <\bar{\rho}/2
\end{array} $$
it follows that $\rho^2\,h^{\prime\prime}(\rho)(u-\bar{u})^2\le \beta\,
(\rho\,(u-\bar{u})^2+h(\rho))$.  Hence,
\begin{eqnarray*} 
\lefteqn{\int^{\infty}_{-\infty} h(\rho(t,x))-h(\rho_0(x))\,dx
+\frac{\epsilon}{2}\int^t_0\int^{\infty}_{-\infty} h^{\prime\prime}(\rho)
\rho_x^2\,dx\,dt} \\
&&\le \frac{\beta}{2\epsilon}\int^t_0\int^{\infty}_{-\infty}\rho(u-\bar{u})^2
+h(\rho)\,dx\,dt\,. 
\end{eqnarray*}
and it follows from (2.4) and Gronwall's inequality that
$$
\int^{\infty}_{-\infty}h(\rho(t,x))\,dx\le \mbox{const on}\;\; [0,\tau].
\leqno (2.5) $$
Set $\eta=1/\rho$. Then from (2.3) $\eta$ satisfies
$$
\eta_t+u\,\eta_x-u_x\eta=\epsilon\,(\eta_{xx}-\frac{2|\eta_x|^2}{\eta}).
$$
We further introduce $\hat{\eta}=e^{\omega\,t}\eta$ with $\omega>0$
to be determined later. The equation for $\hat{\eta}$ becomes
$$
\hat{\eta}_t+\omega\,\hat{\eta}+u\,\hat{\eta}_x-u_x\hat{\eta}
=\epsilon\,(\hat{\eta}_{xx}-\frac{2|\hat{\eta}_x|^2}{\hat{\eta}}).
\leqno (2.6) $$
Define $\xi=\xi_c=\max\,(0,\hat{\eta}-c)$ with $c \ge c_0=\delta^{-1}$.
Pre-multiplication of (2.6) by $\xi^3$ and integration over $R$  yields
$$
\int^{\infty}_{-\infty}(\frac{1}{4}\,(\xi^4)_t+\omega\,(\xi^4+c\,\xi^3)+
\frac{3\epsilon}{4}|(\xi^2)_x|^2) \le -\int^{\infty}_{-\infty}
u\,(5\,\xi^3+3c\,\xi^2)\xi_x\,dx.
\leqno (2.7) $$
Note that
$$
-\int^{\infty}_{-\infty} 5u\,\xi^3\xi_x \le
-\int^{\infty}_{-\infty}\frac{5}{2}u\xi^2(\xi^2)_x\,dx\le \frac{\epsilon}{4}
\int^{\infty}_{-\infty}|(\xi^2)_x|^2\,dx+\frac{25}{4\epsilon}|u|^2_{\infty}
\int^{\infty}_{-\infty}|\xi|^4\,dx
$$
To estimate the second term on the right hand side of (2.7), we define
$$
I_c(t)=\{x \in R:\xi(t,x) > 0 \}=\{x \in R:\eta(t,x) > c\,e^{\omega t}\}.
$$
Then, using the H\"older inequality, we have
$$\begin{array}{l}
\ds -\int^{\infty}_{-\infty} 3c\,\xi^2\xi_x\,dx=
-\int^{\infty}_{-\infty}\frac{3c}{2}\,u\xi(\xi^2)_x\,dx
\\
\ds \quad \le\frac{3c}{2}|u|_{\infty}\left(
\int^{\infty}_{-\infty}|(\xi^2)_x|^2\,dx\right)^{1/2}
\left(\int^{\infty}_{-\infty}|\xi|^6\,dx\right)^{1/6}|I_c(t)|^{1/3}
\\
\ds \quad \le \frac{\epsilon}{4}\int^{\infty}_{-\infty}|(\xi^2)_x|^2\,dx
+\frac{9c^2}{4\epsilon}\,|u|_{\infty}^2
\left(\int^{\infty}_{-\infty}|\xi|^6\,dx\right)^{1/3}|I_c(t)|^{2/3}.
\end{array} $$
Substituting these estimates into (2.7), choosing 
$\ds \omega=\frac{\epsilon}{4}+\frac{25}{4\epsilon}\,|u|_{\infty}^2$, 
and integrating on the interval $[0,t]$, we obtain
\setcounter{equation}{7}
\begin{eqnarray}
\lefteqn{\int^{\infty}_{-\infty} |\xi|^4\,dx+\epsilon\int^t_0
\int^{\infty}_{-\infty} (|\xi^2|^2+|(\xi^2)_x|^2)\,dx\,ds} \nonumber\\
 &&\le \frac{9c^2}{\epsilon} \,|u|_{\infty}^2\int^t_0
\left(\int^{\infty}_{-\infty}|\xi|^6\,dx\right)^{1/3}|I_c(s)|^{2/3}\,ds.
\end{eqnarray} 
Since
$$
|\xi^2|_{\infty} \le \sqrt{2}\left(\int^{\infty}_{-\infty}
(|\xi^2|^2+|(\xi^2)_x|^2)\,dx\right)^{1/2}\,,
$$
we have
$$
\left(\int^{\infty}_{-\infty}|\xi|^6\,dx\right)^{1/3}\le 2^{1/6}
\left(\int^{\infty}_{-\infty}(|\xi^2|^2+|(\xi^2)_x|^2)\,dx\right)^{1/2}.
$$
Thus, from (2.8),
\begin{eqnarray*}
\lefteqn{\int^{\infty}_{-\infty} |\xi|^4\,dx+\epsilon
\int^t_0\int^{\infty}_{-\infty} (|\xi^2|^2+|(\xi^2)_x|^2)\,dx\,ds} \\
&&\le 2^{1/3}\frac{81c^4}{4\epsilon}\,|u|_{\infty}^4\int^t_0|I_c(s)|^{4/3}\,ds
\le Kc^4\,\chi(c)^{4/3}
\end{eqnarray*}
where we define
$$
\chi(c)=\sup_{t\in [0,\tau]}\;|I_c(t)|\quad\mbox{and} \quad
K=2^{1/3}\frac{81\tau}{4\epsilon}.
$$
Clearly $\chi(c)$ is nonnegative, non-increasing on $[c_0,\infty)$.
Moreover, for $\hat{c}>c$,
$$
\int^{\infty}_{-\infty}|\xi|^4\,dx \ge \int_{I_{\hat{c}}}|\xi|^4\,dx
\ge(\hat{c}-c)^4\,I_{\hat{c}}(t).
$$
Hence,
$$
I_{\hat{c}}(t)\le K\,c^4(\hat{c}-c)^{-4}\chi(c)^{4/3}
$$
and by taking the sup over $t$ we obtain
$$
\chi(\hat{c})\le Kc^4(\hat{c}-c)^{-4}\,\chi(c)^{4/3}.
$$
It follows from (2.5) that $\chi(2/\bar{\rho}) < \infty$ and thus
from  Stampacchia's lemma that $\chi(c^*)=0$ for some
$c^* \ge c_0$ and hence
$$
\eta(t,x) \le c^*e^{\omega t} \quad\mbox{and}\quad  \rho(t,x)\ge
(c^*)^{-1}\,e^{-\omega t}\,,
$$
where $c^*$ depends on $\epsilon$ and $\tau$.   $\Box$
\vspace{2mm}

The following lemma shows that $|\phi^{\epsilon}_x(t,\cdot)|$ is uniformly
bounded by  $\rho^{\epsilon}(t,\cdot)$ for every $t \in [0,\tau]$ under
assumption (1.7).
\vspace{2mm}

\noindent{\bf Lemma 2.3}   {\it Assume that $\phi_0$ satisfies (1.7) and
$|(\phi_0)_x| \le c\,\rho_0$ in $R$. Then $|\phi_x(t,\cdot)|
\le c\, \rho(t,x)$ in $R$ for every $t \in [0,\tau]$.}
\vspace{2mm}

\noindent{\bf Proof:} If the initial condition $(\rho_0,m_0,\phi_0)$
is sufficiently smooth and $(\phi_0)_x \ge \delta_3$
then the solution to (1.10)
satisfies $\rho,\;u,\;\phi \in C^3(\Omega)$ and $\phi_x(t,\cdot)>0$.
Note that $\phi_x$ satisfies
$$
(\phi_x)_t+(u\,\phi_x)_x=\epsilon\,(\phi_x)_{xx}.
\leqno (2.9) $$
Then, it is not difficult to show that
if we define $\xi=\mbox{log}(\frac{\phi_x}{\rho})$ then $\xi$ satisfies
$$
\xi_t+u\,\xi_x=\epsilon\,(\xi_{xx}+|(\mbox{log}\phi_x)_x|^2-
|(\mbox{log}\rho)_x|^2)
$$
Suppose $\xi(t,x_0)=\max_x\;\xi(t,x)$. Then 
$$\xi_x(t,x_0)=(\log\phi_x)_x(t,x_0)-(\log\rho)_x(t,x_0)=0$$ and
$\xi_{xx}(t,x_0) \le 0$. Thus, $\partial_t(\max_x\;\xi(t,x)) \le 0$,
which implies the lemma.  Since the solution to (2.9)  continuously
depends on the initial data $(\phi_0)_x$ the estimate holds for
when $(\phi_0)_x \ge 0$.    $\Box$
\vspace{2mm}

The following lemmas provide the technical properties of the functions
$G_i(t)$, $i=1,2$ defined by (1.11).
\vspace{2mm}

\noindent{\bf Lemma 2.4} {\it If $y=(\rho,m,\phi) \in
C^{1,2}((0,\tau)\times R)^3$ is a solution to (1.10), then (1.13) holds.}
\vspace{2mm}

\noindent{\bf Proof:} First, note that the $3 \times 3$ matrix $M=\nabla F$
has the eigenvalues $\ds \lambda_1=\frac{m}{\rho}+\rho^{\theta}\phi^{-\theta},
\;\lambda_2=\frac{m}{\rho}-\rho^{\theta}\phi^{-\theta}$ and $\ds\frac{m}{\rho}$
and that $\nabla_v G_i,\;i=1,2$ are the left-eigenvectors of the sub-matrix
$\nabla_v F$ corresponding to $\lambda_i$. Thus,
$$
(G_1)_t+\lambda_1\,(\nabla G_i\cdot y_x+\rho^{\theta}\phi^{-\theta-1}
\phi_x)
-(\frac{2\theta}{\gamma}\rho^{2\theta-1}\phi^{-2\theta-1}
+u\rho^{\theta}
\phi^{-\theta-1})\phi_x=\epsilon\nabla G_1\cdot y_{xx}\,.
 $$
Since $\nabla G_1\cdot y_{xx}=(G_1)_{xx}-\nabla^2G_1(y_x,y_x)$ we obtain
(1.13) for $G_1$. The same calculation applies to $G_2$.  $\Box$
\vspace{2mm}

\noindent{\bf Lemma 2.5}   {\it  If $\rho>0,\;\phi>0$ then $G_i,\;i=1,2$, are
quasi-convex.}
\vspace{2mm}

\noindent{\bf Proof:} We prove $\ds G_1=\frac{m}{\rho}+\frac{1}{\theta}\,
\rho^{\theta}\phi^{-\theta}$ is quai-convex. The same proof applies to $G_2$.
Note that
$$
\nabla G_1=\left(\begin{array}{c} \ds -\frac{m}{\rho^2}+\rho^{\theta-1}
\phi^{-\theta}
\\ \ds \frac{1}{\rho} \\ \ds -\rho^{\theta}\phi^{-\theta-1} \end{array}
\right)$$
$$
\nabla^2 G_1=\left(\begin{array}{ccc} \ds-\frac{2m}{\rho^3}+
(\theta-1)\,\rho^{\theta-2}\phi^{-\theta} & \ds -\frac{1}{\rho^2} & \ds
-\theta\,\rho^{\theta-1}\phi^{-\theta-1} \\
\ds -\frac{1}{\rho^2}  & 0 & 0
\\ \ds -\theta\,\rho^{\theta-1}\phi^{-\theta-1} & 0 & \ds
(\theta+1)\,\rho^{\theta}\phi^{-\theta-2}\end{array}\right).
$$
If $r=(X,Y,Z)$ satisfies $r\cdot\nabla G_1$ then $\ds Y=-\frac{m}{\rho}+
\rho^{\theta}\phi^{-\theta}\,X +\rho^{\theta+1}\phi^{-\theta-1}\,Z$. Thus,
$$\begin{array}{l}
\ds \nabla^2 G_i(r,r)=(\theta+1)\,(\rho^{\theta-2}\phi^{-\theta}\,X^2-2\,
\rho^{\theta-1}\phi^{-\theta-1}\,XZ+\rho^{\theta}\phi^{-\theta-2}\,Z^2)
\\
\ds\qquad =(\theta+1)\rho^{\theta-2}\phi^{-\theta-2}\,(\phi\,X-\rho\,Z)^2
\ge 0. \quad \Box
\end{array} $$
\vspace{2mm}

We now state the main result of this section that establishes the uniform
$L^{\infty}$-bound of $(\rho^{\epsilon},m^{\epsilon},\phi^{\epsilon}_x)$
in $\epsilon>0$.
\vspace{2mm}

\noindent{\bf Theorem 2.6}  {\it  Suppose $\phi_0$ satisfies (1.7) and
$|(\phi_0)_x| \le c \rho_0$ in $R$. Then, there exists a $\tau=\tau_c>0$
with $c \to \tau_c$ monotonically decreasing and $\tau_0=\infty$ such that
$0 \le \rho^{\epsilon} \le\mbox{const}$, $|\frac{m^{\epsilon}}{\rho^{\epsilon}}|
\le\mbox{const}$ and $|\phi_x|\le\mbox{const}$ in $\Omega=[0,\tau] \times R$.}
\vspace{2mm}

\noindent{\bf Proof:} Suppose that $(\phi_0)_x \le 0$. Then, it follows
from Lemmas 2.3-2.6 that $\max_x G_2(t,x) \le \max_x G_2(0,x)=A$.
Hence, $\ds \frac{m}{\rho}+A \ge \frac{1}{\theta}\,\rho^{\theta}\phi^{-\theta}
\ge 0$. Set $G=A+G_1$. It then follows from Lemma 2.4 that
$$
G_t+\lambda_1\nabla G\cdot y_x+\frac{1}{\gamma}\,\rho^{2\theta}
\phi^{-2\theta-1}\phi_x=\epsilon\,(G_{xx}-\nabla^2 G(y_x,y_x))
$$
Let $G^k=G(kh),\;h>0$. Then,
$$
G^k-G^{k-1}+\lambda_1\nabla G^k\cdot y^k_x+\frac{1}{\gamma}\,
(\rho^k)^{2\theta}(\phi^k)^{-2\theta-1}(\phi^k)_x
=\epsilon\,(G^k_{xx}-\nabla^2 G^k(y^k_x,y^k_x))+\varepsilon(h)
$$
where $\varepsilon(h)/h \to 0$ as $h \to 0^+$. Suppose $G^k(x_0)=\max_x\;
G^k(x)$. Then, $G^k_x(x_0)=(\nabla G^k\cdot y^k_x)(x_0)=0$
and $G^k_{xx}(x_0) \le 0$.
It follows from Lemmas 2.3 and 2.5 that if $\psi(t)=\max_x\;G(t,x)$ then
$$
\psi(kh)-\psi((k-1)h)-\varepsilon(h) \le \frac{c}{\gamma}\,(\frac{\rho}
{\phi}(x_0))^{2\theta+1} \le \frac{c}{\gamma}\theta^{(2\theta+1)/\theta}
\psi(kh)^{(2\theta+1)/\theta}.
$$
Taking the limit $h \to 0^+$, we obtain
$$
\psi(t)-\psi(0) \le \int^t_0 \frac{c}{\gamma}\,\theta^{(2\theta+1)/\theta}
\,\psi(\tau)^{(2\theta+1)/\theta}\,d\tau
$$
and thus
$$
\psi(t) \le \left(\frac{\psi(0)^{(\theta+1)/\theta}}{1-\frac{c}{\gamma}
(\frac{\theta+1}{\theta})\theta^{(2\theta+1)/\theta}
\psi(0)^{(\theta+1)/\theta}\,t}\right)^{\theta/(\theta+1)}.
\leqno (2.10) $$
In fact, if
$$
s(t)=\psi(0)+\int^t_0 \frac{c}{\gamma}\,\theta^{(2\theta+1)/\theta}
\,\psi(\tau)^{(2\theta+1)/\theta}\,d\tau
$$
then $\psi(t) \le s(t)$ and $\dot{s}\le \frac{c}{\gamma} \,
\theta^{(2\theta+1)/\theta}
s^{(2\theta+1)/\theta}$, which implies (2.10). Since
$$
0 \le \frac{1}{\theta}\,\rho^{\theta}\phi^{-\theta} \le \frac{1}{2}\,(G_1+G_2)
\quad\mbox{and}\quad -G_2 \le \frac{m}{\rho} \le G_1\,,
$$
the lemma follows from (2.9).  $\Box$

\section{Compensated Compactness}

In this section we show that the sequence $\{(\rho^{\epsilon},m^{\epsilon},
\phi^{\epsilon})\}_{\epsilon>0}$ has a subsequence that converges to
a weak solution of (1.10) a.e in $\Omega$ using the method of compensated
compactness.  First note that the mechanical energy
$$
\eta=\frac{1}{2}\,\frac{m^2}{\rho}+\frac{1}{\gamma(\gamma-1)}\,
\rho^{\gamma}\phi^{-\gamma+1}
\leqno (3.1) $$
and the corresponding entropy-flux
$$
\quad q=\frac{\rho}{2}\,(\frac{m}{\rho})^3+\frac{1}{\gamma-1}\,
\frac{m}{\rho}\rho^{\gamma}\phi^{-\gamma+1}
\leqno (3.2) $$
form an entropy pair, i.e.,
$$
\nabla\eta\,M=\nabla q
\leqno (3.3) $$
In order to treat solutions approaching a nonzero state at infinity,
we consider a normalized entropy pair
$$\begin{array}{l}
\tilde{\eta}=\eta(y)-\eta((\bar{v},\phi))-\nabla_v\eta((\bar{v},\phi))
(v-\bar{v}),
\\
\tilde{q}=q(y)-q((\bar{v},\phi))-\nabla_v\eta((\bar{v},\phi))F(y)
\end{array} $$
where $v=(\rho,m),\;\bar{v}=(\bar{\rho},\bar{m})$ and $y=(v,\phi)$.
Premultiplying (1.10) by $\nabla\tilde{\eta}$, we obtain
$$
\tilde{\eta}_t+\tilde{q}_x=\epsilon\,(\tilde{\eta}_{xx}-\nabla^2\eta(y_x,y_x)).
$$
Integration over $\Omega$ yields an energy estimate
$$
\int^{\infty}_{-\infty}\tilde{\eta}(t,x)\,dx+\epsilon
\int^t_0\int^{\infty}_{-\infty}
\nabla^2\eta(y_x,y_x)\,dx\,dt=\int^{\infty}_{-\infty}\tilde{\eta}(0,x)\,dx.
\leqno (3.4) $$
The following lemma implies the energy estimate (2.2) where
$\tilde{\eta}(y)=E(\rho,u,\phi)$.
\vspace{2mm}

\noindent{\bf Lemma 3.1}  {\it For $\rho>0,\;\phi>0$,
$\nabla^2\eta$ is non-negative.}
\vspace{2mm}

\noindent{\bf Proof:} Note that
$$
\nabla^2\eta=\left(\begin{array}{ccc} \ds \frac{m^2}{\rho^3}+\rho^{\gamma-2}
\phi^{-\gamma+1} & \ds -\frac{m}{\rho^2} & \ds -\rho^{\gamma-1}\phi^{-\gamma}
\\ \ds -\frac{m}{\rho^2} & \ds \frac{1}{\rho} & 0 \\
\ds -\rho^{\gamma-1}\phi^{-\gamma} & 0 & \ds \rho^{\gamma}\phi^{-\gamma-1}
\end{array}\right).
$$
Thus,
$$
\nabla^2\eta\,(y_x,y_x)=\frac{1}{\rho}\,(\frac{m}{\rho}\,\rho_x-m_x)^2
+\rho^{\gamma-2}\phi^{-\gamma-1}\,(\phi\,\rho_x-\rho\,\phi_x)^2 \ge 0
\leqno (3.5) $$
for $y_x=(\rho_x,m_x,\phi_x)$.  $\Box$
\vspace{2mm}

The following lemma establishes the viscosity estimate which is essential for
the method of compensated compactness.
\vspace{2mm}

\noindent{\bf Lemma 3.2}  {\it Assume that $1< \gamma \le 2$
and $\int^{\infty}_{-\infty} \tilde{\eta}(0,x)\,dx<\infty$.
Then, if  $(\rho,m,\phi)$ is a solution of (1.10)
$$
\epsilon\int^{\tau}_0\int^{\infty}_{-\infty}(|\rho_x(t,x)|^2+|m_x(t,x)|^2)
\,dx\,dt\le \mbox{const}
$$
where $\tau>0$ is defined in Theorem 2.6}
\vspace{2mm}

\noindent{\bf Proof:}  From (2.9) and Lemma 2.2 we have
$$
\int^{\infty}_{-\infty} |\phi_x(t,x)|\,dx=
\int^{\infty}_{-\infty}|\phi_x(0,x)|\,dx\,, \quad t \in [0,\tau].
$$
It thus follows from Theorem 2.6 that
$$
\int^{\tau}_0\int^{\infty}_{-\infty} |\phi_x(t,x)|^2\,dx\le \mbox{const}.
$$
Since $0< \rho(t,x),\;\phi(t,x) \le\mbox{const}$ in $\Omega$ it follows from
(3.5) that
$$
\nabla^2\,\eta(y_x(t,x),y_x(t,x))+|\phi_x(t,x)|^2 \ge c_1\,|y_x(t,x)|^2
$$
for some $c_1>0$.  Hence, the lemma follows from (3.4). $\Box$
\vspace{2mm}

We apply the method of compensated compactness for the function
$\hat{v}^{\epsilon}$ defined by
$$
\hat{v}^{\epsilon}=(\hat{\rho}^{\epsilon},\hat{m}^{\epsilon})
=(\frac{\rho^{\epsilon}}{\phi^{\epsilon}},\frac{m^{\epsilon}}{\phi^{\epsilon}})
$$
The function $\hat{v}^{\epsilon}$ satisfies the $2 \times 2$ viscous
conservation law (1.15) with the forcing term which is in $L^{\infty}(\Omega)$.
Based on this observation we have
\vspace{2mm}

\noindent{\bf Lemma 3.3}  {\it Assume that the conditions in Theorem 2.6 
are satisfied and that $\int^{\infty}_{-\infty} \tilde{\eta}(0,x)\,dx < 
\infty$. Then, for
$1<\gamma\le 2$, the measure set
$$
\eta(\hat{v}^{\epsilon})_t+q(\hat{v}^{\epsilon})_x
$$
lies in a compact subset of $H^{-1}_{loc}(\Omega)$ for all weak
entropy/entropy flux pair $(\eta,q)$ of $\nabla_v F$, where
$\ds \hat{v}^{\epsilon}=(\frac{\rho^{\epsilon}}{\phi^{\epsilon}},
\frac{m^{\epsilon}}{\phi^{\epsilon}})$.}
\vspace{2mm}

\noindent{\bf Proof:} Suppose $(\rho,m,\phi)$ is a solution to (1.10).
Then, dividing the first two equations of (1.10) by $\phi$, we obtain
(1.15) for $\ds \hat{\rho}=\frac{\rho}{\phi}$ and $\ds \hat{m}
=\frac{m}{\phi}$.
Let $(\eta,q)$ be a weak entropy/entropy flux pair, i.e.,
$$
\nabla\eta\nabla_v F=\nabla q \quad \mbox{and} \quad \eta(0,\cdot)=0.
\leqno (3.6) $$
It can be shown that for $0<\rho \le\mbox{const}$, $|\frac{m}{\rho}|\le\mbox{const}$
$$
|\nabla\eta|\le\mbox{const}\quad\mbox{and}\quad
|\nabla^2\eta(r,r)|\le\mbox{const}\,\nabla^2\eta^*(r,r)
\leqno (3.7) $$
where
$$
\eta^*=\frac{1}{2}\rho\,(\frac{m}{\rho})^2
+\frac{1}{\gamma(\gamma-1)}\,\rho^{\gamma}
$$
is the mechanical energy, $r$ is any vector in $R^2$ and constant is
independent of $r$. Premultiplying (1.15) by $\nabla\eta$, we obtain
$$
\eta(\hat{v})_t+q(\hat{v})_x=\epsilon\,(\eta(\hat{v})_{xx}-\nabla^2\eta
(\hat{v}_x,\hat{v}_x))+\nabla\eta(\hat{v})A
$$
where
$$
A=2\epsilon\left(\frac{\rho_x\phi_x}{\phi^2},\frac{m_x\phi_x}{\phi^2}\right)-
\left(0,\frac{p\phi_x}{\phi^2}\right)
$$
It follows from Theorem 2.6 that
$\ds \frac{p^{\epsilon}\phi^{\epsilon}_x}{(\phi^{\epsilon})^2} \in
L^{\infty}(\Omega)$ uniformly in $\epsilon>0$.  It follows from
Lemma 3.2 and Theorem 2.6 that
$$
\epsilon^{1/2}\left(\frac{\rho^{\epsilon}_x\phi^{\epsilon}_x}{(\phi^{\epsilon})
^2},\frac{m^{\epsilon}_x\phi^{\epsilon}_x}{(\phi^{\epsilon})^2}\right)
\in L^2(\Omega)
$$
uniformly in $\epsilon>0$. Thus, $\{\nabla\eta(v^{\epsilon})A^{\epsilon}\}
_{\epsilon>0}$ is precompact in $W^{-1,q}_{loc}(\Omega)$, $1 \le q <2$.
Since
$$
\int^{\tau}_0 \int^{\infty}_{-\infty} \epsilon\,|\hat{v}^{\epsilon}_x(t,x)|^2
\,dx\,dt \le \mbox{const}
$$
The set $\{\epsilon\nabla\eta\hat{v}^{\epsilon}_x\}_{\epsilon>0}$ is
precompact in $L^2(\Omega)$ and so
is $\{\epsilon\eta(\hat{v}^{\epsilon})_{xx}\}_{\epsilon>0}$ in 
$H^{-1}(\Omega)$.  Hence, the lemma follows from
the fact  that  if set $S$ is compact in $W^{-1,q}(U)$ and bounded in
$W^{-1,r}(U)$ then $S$ is compact in $H^{-1}(U)$ for $1 \le q < 2 < r$ and
any bounded and open set $U$ in $R^2$. [Ev] $\Box$
\vspace{2mm}

In the next lemma we prove that the sequence $\{\phi^{\epsilon}\}_{\epsilon>0}$
is precompact in $L^2_{loc}(\Omega)$.
\vspace{2mm}

\noindent{\bf Lemma 3.4} {\it For $\epsilon>0$ and $\tau>0$ defined in
Theorem 2.6
$$
\int^{\tau}_0\int^{\infty}_{-\infty}(|\phi^{\epsilon}_t|^2
+|\phi^{\epsilon}_x|^2)\,dx\,dt\le \mbox{const}.
$$
Thus, the family $\{\phi^{\epsilon}(t,x)\}_{\epsilon>0}$
is compact in $L^2(U)$ for any bounded rectangle $U=(0,\tau) \times (-L,L)$.}
\vspace{2mm}

\noindent{\bf Proof:} Premultiplying (1.10) by $\phi_{xx}$ and integrating
in $(0,\tau) \times R$, we obtain
\begin{eqnarray*} \lefteqn{
\frac{1}{2}\int^{\infty}_{-\infty}|\phi_x(\tau,x)|^2\,dx+\frac{\epsilon}{2}
\int^{\tau}_0\int^{\infty}_{-\infty}|\phi_{xx}|^2\,dx\,dt} \\
&&\le\frac{1}{2}\int^{\infty}_{-\infty}|\phi_x(0,x)|^2\,dx
+\frac{1}{2\epsilon}\,|u|^2_{\infty}\int^{\tau}_0
\int^{\infty}_{-\infty}|\phi_x|^2\,dx\,dt.
\end{eqnarray*}
where $|u|_{\infty}=\sup_{(t,x)\in (0,\tau) \times R}\,|u(t,x)|$. Thus,
$$
\int^{\tau}_0\int^{\infty}_{-\infty}|\epsilon\,\phi_{xx}|^2\,dx\,dt
\le |u|^2_{\infty} \int^t_0\int^{\infty}_{-\infty}|\phi_x|^2\,dx\,dt
+\epsilon\int^{\infty}_{-\infty}|\phi_x(0,x)|^2\,dx
$$
and
$$
\int^{\tau}_0\int^{\infty}_{-\infty}|\phi_t|^2\,dx\,dt \le 4|u|^2_{\infty}
\int^{\tau}_0\int^{\infty}_{-\infty}|\phi_x|^2\,dx\,dt+
2\epsilon\int^{\infty}_{-\infty}|\phi_x(0,x)|^2\,dx
$$
which proves the lemma.
\vspace{2mm}

Now, we state the main result of the paper.
\vspace{2mm}

\noindent{\bf Theorem 3.5}  {\it Assume that the conditions in Theorem 2.6
are satisfied and $\int \tilde{\eta}(0,x)\,dx < \infty$. Then, for
$1<\gamma\le 5/3$, there exists a subsequence of $(\rho^{\epsilon},
m^{\epsilon},\phi^{\epsilon})$ such that
$$
(\rho^{\epsilon}(t,x),m^{\epsilon}(t,x),\phi^{\epsilon}(t,x)) \to (\rho(t,x),
m(t,x),\phi(t,x))\quad\mbox{a.e. in}\;\; \Omega= [0,\tau] \times R.
\leqno (3.8) $$
where  the triple $(\rho,m,\phi) \in L^{\infty}_{+}(\Omega) \times
L^{\infty}(\Omega) \times W^{1,\infty}(\Omega)$ is a weak solution to (1.4).}
\vspace{2mm}

\noindent{\bf Proof:} It follows from Lemma 3.3 that there exists
a subsequence of $(\hat{\rho}^{\epsilon},\hat{m}^{\epsilon})$ such that
$$
(\hat{\rho}^{\epsilon}(t,x),\hat{m}^{\epsilon}(t,x)) \to (\hat{\rho}(t,x),
\hat{m}(t,x)) \quad \mbox{a.e. in}\;\; \Omega.
$$
by applying the results of [Di1] and [Ch].  It follows from Lemma 3.4 that
using a standard diagonal process, there is a subsequence of
$\phi^{\epsilon}(t,x)$ that converges a.e. in $\Omega$, weakly in
$H^1(\Omega)$ and weakly-star in $W^{1,\infty}(\Omega)$ to $\phi$.
Define $\rho(t,x)=\hat{\rho}(t,x)\phi(t,x),\; m(t,x)=\hat{m}(t,x)\phi(t,x)$
a.e. $(t,x) \in \Omega$.  Then, the statement (3.8) holds.
It follows from the first two equations of (1.10) that
$$
\int^{\tau}_0\int^{\infty}_{-\infty} \left((\rho^{\epsilon},m^{\epsilon})
\cdot(\psi_t-\epsilon\,\psi_{xx})+F(\rho^{\epsilon},m^{\epsilon},
\phi^{\epsilon})\cdot\psi_x\right)\,dx\,dt=0
$$
for all $\psi \in C^{\infty}_c(\Omega;R^2)$. It thus follows from (3.8) and
the dominated convergence theorem that (1.8) is satisfied.
It follows from the third equation of (1.10) that
$$
\int^{\tau}_0\int^{\infty}_{-\infty}\left( (\phi^{\epsilon}_t+u^{\epsilon}
\phi^{\epsilon}_x)
\,\xi+\epsilon\,\phi_x\xi_x\right)\,dx\,dt=0
$$
for all $\xi \in C^{\infty}_c(\Omega;R)$. Since $u^{\epsilon} \to u$
in $L^2(U)$ for any bounded rectangle $U= [0,\tau] \times [-L,L]$ and
$\phi^{\epsilon} \to \phi$ weakly in $H^1(\Omega)$ it follows that
$$
\int^{\tau}_0\int^{\infty}_{-\infty} (\phi_t+u\phi_x)\,\xi\,dx\,dt=0
$$
for all $\xi \in C^{\infty}_c(\Omega;R)$. Hence $\phi$ satisfies (1.4)
a.e. in $\Omega$.  $\Box$
\vspace{2mm}

\noindent{\bf Corollary 3.6} {\it Suppose the entropy pair $(\eta,q)$
is defined by (3.1)-(3.2). Then
$$
\int^{\tau}_0\int^{\infty}_{-\infty}\left(\eta\,\xi_t+q\,\xi_x\right)\,dx\,dt
\ge 0
\leqno (3.9) $$
for all $\xi \in C^{\infty}_c(\Omega;R)$ satisfying $\xi \ge 0$.
That is, the third equation of (1.1)
is replaced by the inequality $\eta_t+q_x \le 0$
in the sense of distributions.}
\vspace{2mm}

\noindent{\bf Proof:} It follows from (3.3) that
$$
\int^{\tau}_0\int^{\infty}_{-\infty} \left(\eta^{\epsilon}\,
(\xi_t-\epsilon\,\xi_{xx})
+q^{\epsilon}\,\xi_x\right)\,dx\,dt=\epsilon\int^{\tau}_0
\int^{\infty}_{-\infty}
\nabla^2\eta(y^{\epsilon},y^{\epsilon})\,\xi \,dx\,dt
$$
for all $\psi \in C^{\infty}_c(\Omega;R^2)$ satisfying $\xi \ge 0$.
It follows from Lemma 3.1 that
the right hand side of this equality is nonnegative. Thus, by taking the
limit as $\epsilon \to 0^+$ we obtain (3.9) $\Box$
\vspace{5mm}

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\vspace{2mm}

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\end{enumerate}

{\sc Kazufumi Ito \\
Center for Research in Scientific Computation\\
North Carolina State University\\
Raleigh, North Carolina 27695-8205}\\ 
E-mail: kito@eos.ncsu.edu

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