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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2012 (2012), No. 186, pp. 1--6.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2012 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2012/186\hfil Existence of smooth global solutions]
{Existence of smooth global solutions for a 1-D modified
Navier-Stokes-Fourier model}

\author[J. Sun, J. Fan, G. Nakamura \hfil EJDE-2012/186\hfilneg]
{Jianzhu Sun, Jishan Fan, Gen Nakamura}  % in alphabetical order

\address{Jianzhu Sun \newline
 Department of Applied Mathematics\\
 Nanjing Forestry University, Nanjing 210037, China}
\email{jzsun@njfu.com.cn}

\address{Jishan Fan \newline
 Department of Applied Mathematics \\
 Nanjing Forestry University, Nanjing 210037,  China}
\email{fanjishan@njfu.com.cn}

\address{Gen Nakamura \newline
 Department of Mathematics\\
 Hokkaido University, Sapporo, 060-0810, Japan}
\email{gnaka@math.sci.hokudai.ac.jp}

\thanks{Submitted June 18, 2012. Published October 28, 2012.}
\subjclass[2000]{35Q30, 76D03, 76D05}
\keywords{Mass velocity; volume velocity; Navier-Stokes-Fourier equations}

\begin{abstract}
 We prove the existence of strong global solutions of the 1-D
 modified compressible Navier-Stokes-Fourier equations proposed by
 Howard Brenner \cite{2,3}.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{proposition}[theorem]{Proposition}
\newtheorem{definition}[theorem]{Definition}
\newtheorem{remark}[theorem]{Remark}
\allowdisplaybreaks

\section{Introduction}

We consider the  modified Navier-Stokes-Fourier equations proposed by
by Brenner \cite{2,3}:
\begin{gather}
\partial_t\rho+\operatorname{div}(\rho v_m)=0,\label{1.1}
\\
\partial_t(\rho v)+\operatorname{div}(\rho v\otimes v_m)+\nabla
p=\operatorname{div}\mathbb{S},\label{1.2}
\\
\partial_t(\rho(\frac{1}{2}v^2+e))
+\operatorname{div}(\rho(\frac{1}{2}v^2+e)v_m)
+\operatorname{div}(pv)+\operatorname{div}
q=\operatorname{div}(\mathbb{S}v),\label{1.3}
\\
v|_{\partial\Omega}=0,v_m\cdot n|_{\partial\Omega}=\nabla\rho\cdot
n|_{\partial\Omega}=0,q\cdot n|_{\partial\Omega}=\nabla\theta\cdot
n|_{\partial\Omega}=0,\label{1.4}
\\
(\rho,v,\theta)|_{t=0}=(\rho_0,v_0,\theta_0)\quad\text{in }\Omega:=(0,1).\label{1.5}
\end{gather}
where $\rho$ is the mass density, $v$ is the fluid-based
(Lagrangian) volume velocity, $v_m$ is the mass-based (Eulerian)
mass velocity, $p=R\rho\theta$ is the pressure with positive
constant $R>0$, $e=C_V\theta$ the specific internal energy, $\theta$
the temperature, $\mathbb{S}$ the viscous stress tensor, we will
adopt the Newton's rheological law:
\begin{equation}
\mathbb{S}:=\mu\Big(\nabla v+\nabla v^T-\frac{2}{3}\operatorname{div}
v\mathbb{I}\Big)+\eta\operatorname{div} v \mathbb{I},\label{1.6}
\end{equation}
where $\mu\geq0$ and $\eta\geq0$ stand for the shear and bulk
viscosity coefficients, respectively. The relationship between $v_m$ and $v$ is a cornerstone of Brenner's approach. After
a careful study \cite{2,3}, Brenner proposes a universal
constitutive equation in the form:
\begin{equation}
v-v_m=K\nabla\log\rho, \label{1.7}
\end{equation}
with $K\geq0$ a purely phenomenological coefficient.

Moreover, we suppose the heat flux obeys Fourier's law,
specifically,
\begin{equation}
q=-k\nabla\theta,\label{1.8}
\end{equation}
where $k$ is the heat conductivity coefficient.

We will assume $K=1$, $C_v=1$, $R=1$, $\mu>0$, $\eta=0$, and
\begin{equation}
k(\theta):=k_0(1+4\theta^3),\label{1.9}
\end{equation}
with a positive constant $k_0=1$. \eqref{1.9} is physically relevant
as radiation heat conductivity at least for large values of $\theta$
(see \cite{4}).

Very recently, Feireisl and Vasseur \cite{5} proved the
global-in-time existence of weak solutions to the problem
\eqref{1.1}-\eqref{1.5}. Under the conditions that
$\rho_0,\theta_0,v_0\in L^\infty(\Omega)$ and $\rho_0\geq
C>0,\theta_0\geq C>0$ in $\Omega$. Here it should be noted that
similar result for the classical Navier-Stokes-Fourier system
(\eqref{1.1}-\eqref{1.3} with $v=v_m$) have not yet been proved. In
their proof, they obtained the following global-in-time estimates:
\begin{gather}
\|v\|_{L^2(0,T;H^1(\Omega))}\leq C,\label{1.10}\\
\|\theta^{3/2}\|_{L^2(0,T;H^1(\Omega))}\leq C,\label{1.11}\\
\|\nabla\theta\|_{L^2(0,T;L^2(\Omega))}\leq C,\label{1.12}
\end{gather}
where $C$ is a positive constant depending on $\int_\Omega\rho_0dx$,
$\int_\Omega\rho(\frac{1}{2}v_0^2+C_V\theta_0)dx$, and
$\int_\Omega \rho_0s(\rho_0,\theta_0)dx$, the other norms of $\rho_0$
and $v_0,\ \theta_0$.

Our aim in this article is to show the existence of a smooth
global solution to the  problem \eqref{1.1}-\eqref{1.5}.

\begin{theorem}\label{th1.1}
Let $\rho_0,v_0,\theta_0\in H^1(\Omega)$ with
$\inf\rho_0>0,\inf\theta_0>0$ in $\Omega$. Then there exists a
unique strong solution $(\rho,v,\theta)$ to the problem
\eqref{1.1}-\eqref{1.5} satisfying
$$
(\rho,v,\theta)\in L^\infty(0,T;H^1(\Omega))\cap
L^2(0,T;H^2(\Omega)),(\partial_t\rho,\partial_tv,\partial_t\theta)\in
L^2(0,T;L^2(\Omega))
$$
for any given $T>0$ and
\begin{equation}
\inf\rho(x,t)>0,\quad \inf\theta(x,t)>0\quad\text{in }
\Omega\times(0,T).\label{1.13}
\end{equation}
\end{theorem}

\begin{remark}\label{re1.1} \rm
The methods for the one-dimensional classical Navier-Stokes-Fourier
equations \cite{9,7} do not work here. Because
their clever method for proving $0<\frac{1}{C}\leq\rho\leq C<\infty$
does not work here.
\end{remark}

The continuity equation \eqref{1.1} can be rewritten as
\begin{equation}
\partial_t\rho+\operatorname{div}(\rho v)=\Delta\rho.\label{1.14}
\end{equation}
The energy equation \eqref{1.3} can be rewritten as
\begin{equation}
\partial_t(\rho\theta)+\operatorname{div}(\rho v_m\theta)+\operatorname{div} q
=\mathbb{S}:\nabla v-p\operatorname{div} v.\label{1.15}
\end{equation}

\section{Proof of Theorem \ref{th1.1}}

Since it is easy to prove a local existence result for smooth
solution, which is very similar as that in \cite{1},  we omit
the details here. We  need to prove only the a priori estimates for smooth
solutions and omit the proof of the uniqueness which is standard.

Since we take $x\in\Omega:=(0,1)$ and $\partial\Omega=\{0,1\}$,
it follows that $\operatorname{div}=\nabla=\partial_x$,
$\Delta=\partial_x^2$, $\mathbb{S}:=(\frac{4}{3}\mu+\eta)\partial_xv$ and
\eqref{1.4} becomes
$$
v|_{\partial\Omega}=0, \quad\nabla\rho|_{\partial\Omega}
=\frac{\partial\rho}{\partial x}\big|_{\partial\Omega}=0, \quad
\nabla\theta|_{\partial\Omega}
=\frac{\partial\theta}{\partial x}\big|_{\partial\Omega}=0.
$$
First, we note that in 1-D, we have
\begin{equation}
\|\rho\|_{L^\infty}\leq C\|\rho\|_{H^1},\quad
\|\theta\|_{L^\infty}\leq C\|\theta\|_{H^1},\quad
\|v\|_{L^\infty}\leq C\|\nabla v\|_{L^2}.\label{2.1}
\end{equation}

\begin{lemma}\label{le2.1}
If $(\rho,v,\theta)$ is a strong solution, then
\begin{gather*}
\|\rho\|_{L^\infty(0,T;H^1)}+\|\rho\|_{L^2(0,T;H^2)}\leq
C(T),\label{2.2}\\
\|\partial_t\rho\|_{L^2(0,T;L^2)}\leq C(T),\label{2.3}\\
\frac{1}{C(T)}\leq\rho.\label{2.4}
\end{gather*}
\end{lemma}

\begin{proof}
 Testing \eqref{1.14} with $\rho$, using
\eqref{1.10} and \eqref{2.1}, we have
\begin{align*}
\frac{1}{2}\frac{d}{dt}\int\rho^2dx+\int|\nabla\rho|^2dx
&=\int\rho v\nabla\rho dx\\
&\leq \|\rho\|_{L^2}\|v\|_{L^\infty}\|\nabla\rho\|_{L^2}\leq
C\|\nabla v\|_{L^2}\|\rho\|_{L^2}\|\nabla\rho\|_{L^2}\\
&\leq \frac{1}{2}\|\nabla\rho\|_{L^2}^2+C\|\nabla
v\|_{L^2}^2\|\rho\|_{L^2}^2
\end{align*}
which gives
$$
\|\rho\|_{L^\infty(0,T;L^2)}+\|\rho\|_{L^2(0,T;H^1)}\leq C(T).
$$
Similarly, testing \eqref{1.14} with $-\Delta\rho$, using \eqref{1.10}
and \eqref{2.1}, we see that
\begin{align*}
&\frac{1}{2}\frac{d}{dt}\int|\nabla\rho|^2dx+\int|\Delta\rho|^2dx=\int(\rho\operatorname{div}
v+v\nabla\rho)\Delta\rho dx\\
&\leq (\|\rho\|_{L^\infty}\|\operatorname{div}
v\|_{L^2}+\|v\|_{L^\infty}\|\nabla\rho\|_{L^2})\|\Delta\rho\|_{L^2}\\
&\leq C\|\rho\|_{H^1}\|\nabla v\|_{L^2}\|\Delta\rho\|_{L^2}\\
&\leq \frac{1}{2}\|\Delta\rho\|_{L^2}^2+C\|\nabla
v\|_{L^2}^2\|\rho\|_{H^1}^2
\end{align*}
which yields \eqref{2.2}. Here we have
$\operatorname{div} v=\nabla v=\frac{\partial v}{\partial x}$.
Then \eqref{2.3} follows easily from \eqref{1.14} and \eqref{2.2}.

To prove \eqref{2.4}, we multiply \eqref{1.14} by $\frac{1}{\rho}$
to obtain
\begin{align*}
\partial_t\log\rho-\Delta\log\rho
&= |\nabla\log\rho|^2-v\cdot\nabla\log\rho-\operatorname{div}
v \\
&= \Big(\nabla\log\rho-\frac{1}{2}v\Big)^2-\frac{1}{4}v^2-\operatorname{div}
v \\
&\geq -\frac{1}{4}v^2-\operatorname{div} v.\label{2.5}
\end{align*}
By the classical comparison principle, it is easy to infer that
$\log\rho\geq w$,
 with $w$ a solution to the problem
\begin{equation}
\partial_tw-\Delta w=-\frac{1}{4}v^2-\operatorname{div} v, \quad
\nabla w|_{\partial\Omega}=\frac{\partial w}{\partial x}\big|_{\partial\Omega}=0,\quad
 w|_{t=0}=\log\rho_0, \label{2.6}
\end{equation}
with fixed $v$ satisfying \eqref{1.10}.

Testing \eqref{2.6} with $w$, using \eqref{1.10}, we find that
\begin{align*}
\frac{1}{2}\frac{d}{dt}\int w^2dx+\int|\nabla w|^2dx
&\leq (\frac{1}{4}\|v\|_{L^\infty}\|v\|_{L^2}+\|\operatorname{div}
v\|_{L^2})\|w\|_{L^2}\\
&\leq C(\|\nabla v\|_{L^2}+\|\nabla v\|_{L^2}^2)\|w\|_{L^2}
\end{align*}
which gives
$$
\|w\|_{L^\infty(0,T;L^2)}+\|w\|_{L^2(0,T;H^1)}\leq C(T).
$$
Similarly, testing \eqref{2.6} with $-\Delta w$, using \eqref{1.10},
we infer that
\begin{align*}
\frac{1}{2}\frac{d}{dt}\int|\nabla w|^2dx+\int|\Delta w|^2dx
&\leq \big|\int\frac{1}{4}\nabla v^2\cdot\nabla w dx\big|
+|\int\operatorname{div} v\cdot\Delta w dx|\\
&\leq \frac{1}{2}\|v\|_{L^\infty}\|\operatorname{div} v\|_{L^2}\|\nabla w\|_{L^2}
+\|\operatorname{div} v\|_{L^2}\|\Delta w\|_{L^2}\\
&\leq \frac{1}{2}\|\Delta w\|_{L^2}^2+C\|\operatorname{div} v\|_{L^2}^2+C\|\nabla
v\|_{L^2}^2\|\nabla w\|_{L^2}
\end{align*}
which yields
$$
\|w\|_{L^\infty(0,T;H^1)}\leq C(T).
$$
This yields
 $$
\log\rho\geq w\geq -C(T)>-\infty
$$
and thus \eqref{2.4} holds.
The proof is complete.
\end{proof}

Using \eqref{1.1}, \eqref{1.2}, \eqref{2.2}, \eqref{2.4}, $p:=R\rho\theta$,
\eqref{1.11}, \eqref{1.12} and the method in \cite{5}, it is easy to
verify the following lemma.

\begin{lemma}[\cite{5}] \label{le2.2}
If $(\rho,v,\theta)$ is a weak solution, then
\begin{equation}
\|v\|_{L^\infty(0,T;L^m(\Omega))}\leq C(T)\quad\text{for some $m>2$}.\label{2.7}
\end{equation}
\end{lemma}

It follows from \eqref{1.11} and \eqref{2.1} that
\begin{equation}
\|\theta\|_{L^3(0,T;L^\infty(\Omega))}\leq C(T).\label{2.8}
\end{equation}

\begin{lemma}\label{le2.3}
If $(\rho,v,\theta)$ is a strong solution, then
\begin{gather}
\|v\|_{L^\infty(0,T;H^1)}+\|v_t\|_{L^2(0,T;L^2)}\leq
C(T),\label{2.9}\\
\|v\|_{L^2(0,T;H^2)}\leq C(T).\label{2.10}
\end{gather}
\end{lemma}

\begin{proof}
 We start rewriting the momentum equation \eqref{1.2} in the form
\begin{equation}
\rho(\partial_tv+v_m\cdot\nabla v)+R\nabla(\rho\theta)=\mu\Delta
v+\frac{1}{3}\mu\nabla\operatorname{div} v.\label{2.11}
\end{equation}
Testing \eqref{2.11} with $v_t$, using \eqref{2.1}, \eqref{2.2},
\eqref{1.12}, \eqref{2.7} and \eqref{2.8}, we deduce that
\begin{equation}
\begin{split}
&\frac{1}{2}\frac{d}{dt}\int\mu|\nabla v|^2+\frac{1}{3}\mu(\operatorname{div}
v)^2dx+\int\rho v_t^2dx \\
&= -\int\rho v_m\cdot\nabla v\cdot v_t
dx-R\int\nabla(\rho\theta)\cdot v_t dx \\
&= -\int\rho v\cdot\nabla v\cdot v_t dx+\int\nabla\rho\cdot\nabla
v\cdot v_t dx-R\int(\rho\nabla\theta+\theta\nabla\rho)v_t dx \\
&\leq \|\rho\|_{L^\infty}\|v\|_{L^\infty}\|\nabla
v\|_{L^2}\|v_t\|_{L^2}+\|\nabla\rho\|_{L^\infty}\|\nabla
v\|_{L^2}\|v_t\|_{L^2} \\
&\quad +R(\|\rho\|_{L^\infty}\|\nabla\theta\|_{L^2}
+\|\theta\|_{L^\infty}\|\nabla\rho\|_{L^2})\|v_t\|_{L^2} \\
&\leq C\|\nabla
v\|_{L^2}^2\|v_t\|_{L^2}+C\|\Delta\rho\|_{L^2}\|\nabla
v\|_{L^2}\|v_t\|_{L^2} \\
&\quad +C(\|\nabla\theta\|_{L^2}+\|\theta\|_{L^\infty})\|v_t\|_{L^2}.
\end{split}\label{2.12}
\end{equation}
On the other hand, using \eqref{2.11} and the $H^2$-theory of second
order elliptic equations, we have
\begin{equation}
\begin{split}
\|v\|_{H^2}
&\leq C\|\rho\partial_tv+\rho v_m\cdot\nabla
v+R\nabla(\rho\theta)\|_{L^2} \\
&\leq C(\|v_t\|_{L^2}+\|v\cdot\nabla
v\|_{L^2}+\|\nabla\rho\|_{L^\infty}\|\nabla
v\|_{L^2}+\|\nabla\theta\|_{L^2}+\|\theta\|_{L^\infty}) \\
&\leq C(\|v_t\|_{L^2}+\|\nabla
v\|_{L^2}^2+\|\Delta\rho\|_{L^2}\|\nabla
v\|_{L^2}+\|\nabla\theta\|_{L^2}+\|\theta\|_{L^\infty}).
\end{split}\label{2.13}
\end{equation}
Now using \eqref{2.7}, Young's inequality and the Gagliardo-Nirenberg
inequality \cite{6},
$$
\|\nabla v\|_{L^2}^2\leq C\|v\|_{L^m}^{2\alpha}
\|v\|_{H^2}^{2(1-\alpha)}\leq C\|v\|_{H^2}^{2(1-\alpha)}\leq\frac{1}{2C}\|v\|_{H^2}+C,
$$
 with $1-\alpha=\frac{m+2}{3m+2}<\frac{1}{2}\big)$,
we obtain
\begin{equation}
\|v\|_{H^2}\leq C(\|v_t\|_{L^2}+\|\Delta\rho\|_{L^2}\|\nabla
v\|_{L^2}+\|\nabla\theta\|_{L^2}+\|\theta\|_{L^\infty}+C).\label{2.14}
\end{equation}
Combining \eqref{2.12}, \eqref{2.13} and \eqref{2.14} and using Gronwall's inequality,
 we obtain \eqref{2.9} and \eqref{2.10}.
This completes the proof.
\end{proof}

\begin{lemma}\label{le2.4}
Let $K(\theta):=\theta+\theta^4$. If $(\rho,v,\theta)$ is a strong solution, then
\begin{equation}
\|K(\theta)\|_{L^\infty(0,T;L^2)}+\|K(\theta)\|_{L^2(0,T;H^1)}\leq
C(T).\label{2.15}
\end{equation}
\end{lemma}

\begin{proof}
 We start by rewriting the energy equation \eqref{1.15} in the form:
\begin{equation}
\rho\partial_tK(\theta)+\rho v_m\cdot\nabla K(\theta)-\Delta
K(\theta)=(\mathbb{S}:\nabla v-p\operatorname{div} v)K'(\theta).\label{2.16}
\end{equation}
Testing \eqref{2.16} with $K(\theta)$, using \eqref{1.1}, \eqref{2.9},
\eqref{2.2} and \eqref{2.4}, we find that
\begin{align*}
&\frac{1}{2}\frac{d}{dt}\int\rho K^2(\theta)dx+\int|\nabla K(\theta)|^2dx\\
&= \int(\mathbb{S}:\nabla v-p\operatorname{div} v)K'(\theta)K(\theta)dx\\
&\leq \|\mathbb{S}\|_{L^2}\|\nabla
v\|_{L^2}\|K'(\theta)K(\theta)\|_{L^\infty}
+C\|\rho\|_{L^\infty}\|\operatorname{div} v\|_{L^2}\|K(\theta)\|_{L^4}^2\\
&\leq C\|K(\theta)\|_{L^\infty}^{7/4}+C\|K(\theta)\|_{L^4}^2\\
&\leq C\|K(\theta)\|_{L^2}^{7/8}\|K(\theta)\|_{H^1}^{7/8}+\frac{1}{8}\|\nabla
K(\theta)\|_{L^2}^2+C\|K(\theta)\|_{L^2}^2\\
&\leq \frac{1}{4}\|\nabla K(\theta)\|_{L^2}^2+C\|K(\theta)\|_{L^2}^2+C
\end{align*}
which yields \eqref{2.15}. Here we have used the Gagliardo-Nirenberg
inequalities:
\begin{gather*}
\|K(\theta)\|_{L^\infty}\leq C\|K(\theta)\|_{L^2}^{1/2}\|K(\theta)\|_{H^1}^{1/2},\\
\|K(\theta)\|_{L^4}\leq C\|K(\theta)\|_{L^2}^{3/4}\|K(\theta)\|_{H^1}^{1/4}.
\end{gather*}
This completes the proof.
\end{proof}

\begin{lemma}\label{le2.5}
If $(\rho,v,\theta)$ is a strong solution, then
\begin{gather}
\|\theta\|_{L^\infty(0,T;H^1)}+\|\theta\|_{L^2(0,T;H^2)}\leq
C(T),\label{2.17}\\
\|\theta_t\|_{L^2(0,T;L^2)}\leq C(T).\label{2.18}
\end{gather}
\end{lemma}

\begin{proof}
We start by rewriting the energy equation
\eqref{2.16} in the form:
 $$
\partial_tK(\theta)+v_m\cdot\nabla
K(\theta)-\frac{1}{\rho}\Delta K(\theta)=\frac{\mathbb{S}:\nabla
v-p\operatorname{div} v}{\rho}K'(\theta).
$$
Testing the above equation with $-\Delta K(\theta)$, using
\eqref{2.9}, \eqref{2.10}, \eqref{2.2}, \eqref{2.4} and
\eqref{2.15}, we deduce that
\begin{align*}
&\frac{1}{2}\frac{d}{dt}\int|\nabla
K(\theta)|^2dx+\int\frac{1}{\rho}|\Delta K(\theta)|^2dx
\\
&= \int\big[(v-\nabla\log\rho)\nabla
K(\theta)-\frac{\mathbb{S}:\nabla
v-p\operatorname{div} v}{\rho}K'(\theta)\big]\Delta K(\theta)dx
\\
&\leq \Big(\|v\|_{L^\infty}\|\nabla
K(\theta)\|_{L^2}+\left\|\frac{1}{\rho}\right\|_{L^\infty}\|\nabla
\rho\|_{L^\infty}\|\nabla K(\theta)\|_{L^2}
\\
&\quad +\|\frac{1}{\rho}\|_{L^\infty}\|\mathbb{S}:\nabla
v\|_{L^2}\|K'(\theta)\|_{L^\infty}+C\|\operatorname{div}
v\|_{L^2}\|K(\theta)\|_{L^\infty}\Big)\|\Delta K(\theta)\|_{L^2}
 \\
&\leq C(\|K(\theta)\|_{H^1}+\|\rho\|_{H^2}\|K(\theta)\|_{H^1}+\|\nabla
v\|_{L^4}^2\|K(\theta)\|_{L^\infty}^{3/4})\|\Delta
K(\theta)\|_{L^2}
\\
&\leq C(\|K(\theta)\|_{H^1}+\|\rho\|_{H^2}\|K(\theta)\|_{H^1}
+\|v\|_{H^2}^{1/2}\|K(\theta)\|_{H^1}^{3/8})\|\Delta K(\theta)\|_{L^2}
\\
&\leq \frac{1}{2}\|\Delta
K(\theta)\|_{L^2}^2+C\|K(\theta)\|_{H^1}^2+C\|\rho\|_{H^2}^2
\|K(\theta)\|_{H^1}^2+C\|v\|_{H^2}\|K(\theta)\|_{H^1}^{3/4}
\end{align*}
which yields \eqref{2.17}. Here we have used the Gagliardo-Nirenberg
inequalities:
\begin{gather*}
\|\nabla v\|_{L^4}^2 \leq C\|\nabla v\|_{L^2}^{3/2}\|v\|_{H^2}^{1/2},
\|K(\theta)\|_{L^\infty}\leq C\|K(\theta)\|_{L^2}^{1/2}\|K(\theta)\|_{H^1}^{1/2},\\
\|\theta\|_{L^\infty(0,T;L^\infty)}\leq C\|\theta\|_{L^\infty(0,T;H^1)},\\
\|\nabla\theta\|_{L^\infty(0,T;L^2)}\leq C\|\nabla K(\theta)\|_{L^\infty(0,T;L^2)},\\
\|\Delta\theta\|_{L^2(0,T;L^2)}\leq C\|\Delta K(\theta)\|_{L^2(0,T;L^2)}.
\end{gather*}
Equation \eqref{2.18} follows easily from
\eqref{2.16}, \eqref{2.17}, \eqref{2.9}, \eqref{2.10}, and \eqref{2.2}.
This completes the proof.
\end{proof}

\subsection*{Acknowledgments}
This work is partially supported by grant 11171154 from the NSFC .

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\end{document}