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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2017 (2017), No. 232, pp. 1--8.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu}
\thanks{\copyright 2017 Texas State University.}
\vspace{8mm}}

\begin{document}
\title[\hfilneg EJDE-2017/232\hfil Ericksen-Leslie model]
{Local well-posedness for an Ericksen-Leslie's parabolic-hyperbolic
compressible non-isothermal model for liquid crystals}

\author[J. Fan, T. Ozawa \hfil EJDE-2017/232\hfilneg]
{Jishan Fan, Tohru Ozawa}

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

\address{Tohru Ozawa (corresponding author) \newline
Department of Applied Physics,
Waseda University, Tokyo, 169-8555, Japan}
\email{txozawa@waseda.jp}

\dedicatory{Communicated by Mitsuharo Otani}

\thanks{Submitted February 23, 2017. Published September 25, 2017.}
\subjclass[2010]{35Q30, 35Q35, 76N10}
\keywords{Compressible; liquid crystals; local well-posedness}

\begin{abstract}
 In this article we prove the local well-posedness for an Ericksen-Leslie's
 parabolic-hyperbolic compressible non-isothermal model for nematic liquid
 crystals with positive initial density.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{remark}[theorem]{Remark}
\allowdisplaybreaks

\section{Introduction}

We consider the following Ericksen-Leslie system modeling the hydrodynamic 
flow of compressible nematic liquid crystals \cite{1,2,3,4,5}:
\begin{gather}
\partial_t\rho+\operatorname{div}(\rho u)=0,\label{1.1}\\
\begin{aligned}
&\partial_t(\rho u)+\operatorname{div}(\rho u\otimes u)+\nabla p(\rho,\theta)-\mu\Delta u
-(\lambda+\mu)\nabla\operatorname{div} u \\
&=-\nabla\cdot\Big(\nabla d\odot\nabla d-\frac12|\nabla d|^2\mathbb{I}_3\Big),
\end{aligned}\label{1.2}\\
\partial_t(\rho e)+\operatorname{div}(\rho u e)+p\operatorname{div} u-\Delta\theta
 =\frac\mu2|\nabla u+\nabla u^t|^2+\lambda(\operatorname{div} u)^2+|\dot d|^2,\label{1.3}\\
\ddot{d}-\Delta d=d(|\nabla d|^2-|\dot d|^2),\ |d|=1,\quad
 \text{in } \mathbb{R}^3\times(0,\infty),\label{1.4}\\
(\rho,u,\theta,d,d_t)(\cdot,0)=(\rho_0,u_0,\theta_0,d_0,d_1)\quad\text{in }
 \mathbb{R}^3,\; |d_0|=1,\; d_0\cdot d_1=0.\label{1.5}
\end{gather}
Here $\rho,u,\theta$ is the density, velocity and temperature of the fluid, 
and $d$ represents the macroscopic average of the nematic liquid crystals 
orientation field. $e:=C_V\theta$ is the internal energy and $p:=R\rho\theta$ 
is the pressure with positive constants $C_V$ and $R$. 
The viscosity coefficients $\mu$ and $\lambda$ of the fluid satisfy $\mu>0$ 
and $\lambda+\frac23\mu\geq0$. The symbol $\nabla d\odot\nabla d$ denotes 
a matrix whose $(i,j)$th entry is $\partial_id\partial_jd$, $\mathbb{I}_3$ 
is the identity matrix of order $3$, and it is easy to see that 
\[
\operatorname{div}\Big(\nabla d\odot\nabla d-\frac12|\nabla d|^2\mathbb{I}_3\Big)
=-\sum_k\nabla d_k\Delta d_k,\dot d:=d_t+u\cdot\nabla d.
\]
 $u^t$ is the transpose of vector $u$ and $\partial_tu\equiv u_t$.

System \eqref{1.1}-\eqref{1.3} is the well-known full compressible
 Navier-Stokes-Fourier system. When $u=0$, \eqref{1.4} reduces to the 
wave maps system, which is one of the most beautiful and challenging 
nonlinear hyperbolic system. It has captured the attention of mathematicians 
for more than thirty years now. Moreover, the wave maps system is nothing
 other than the Euler-Lagrange system for the nonlinear sigma model,
 which is one of the fundamental problems in classical field theory.

When $\theta$ is a positive constant and the equation \eqref{1.4} is
 replaced by a harmonic heat flow
\begin{equation}
\dot d-\Delta d=d|\nabla d|^2,\label{1.6}
\end{equation}
this problem has received many studies. 
Huang, Wang and Wen \cite{6,7} (see also \cite{8,9}) show the local 
well-posedness of strong solutions with vacuum and prove some regularity 
criteria. Ding, Huang, Wen and Zi \cite{10} (also see \cite{11,12}) 
studied the low Mach number limit. Jiang, Jiang and Wang \cite{13} 
(see also \cite{14}) proved the global existence of weak solutions in 
$\mathbb{R}^2$.

When the fluid is incompressible, i.e., $\operatorname{div} u=0$, the similar model has
 been studied in \cite{15,16}.

The aim of this article is to prove a local-well posedness result when 
$\inf\rho_0\geq 1/C$, we will prove the following result.

\begin{theorem}\label{thm1.1} 
Let $1/C \leq\rho_0\leq C$, $0\leq\theta_0$, $\nabla\rho_0\in H^2$, 
$u_0,\theta_0,\dot d_0,\nabla d_0\in H^3$, with $|d_0|=1, d_0\cdot d_1=0$. 
Then problem \eqref{1.1}-\eqref{1.5} has a unique strong solution 
$(\rho,u,\theta,d)$ satisfying
\begin{equation}
\begin{gathered}
\frac1C\leq\rho\leq C, \quad 0\leq\theta, \quad |d|=1,\\
\nabla\rho\in L^\infty(0,T;H^2),\quad u,\theta,\dot d,\nabla d\in L^\infty(0,T;H^3),\\
u,\theta\in L^2(0,T;H^4), \quad u_t,\theta_t\in L^2(0,T;H^2)
\end{gathered}\label{1.7}
\end{equation}
for some $T>0$.
\end{theorem}

\begin{remark}\label{rmk1.1} \rm
When $n=2$ and taking $d:=\begin{pmatrix}
\cos\phi\\
\sin\phi
\end{pmatrix}$, System \eqref{1.1}-\eqref{1.4} reduces to
\begin{gather*}
\partial_t\rho+\operatorname{div}(\rho u)=0,\\
\begin{aligned}
&\partial_t(\rho u)+\operatorname{div}(\rho u\otimes u)+\nabla p(\rho,\theta)
 -\mu\Delta u-(\lambda+\mu)\nabla\operatorname{div} u\\
&=-\nabla\cdot\Big(\nabla\phi\otimes\nabla\phi
 -\frac12|\nabla\phi|^2\mathbb{I}_2\Big),
\end{aligned}\\
\partial_t(\rho e)+\operatorname{div}(\rho u e)+p\operatorname{div} u-\Delta\theta=\frac\mu2|\nabla u
+\nabla u^t|^2+\lambda(\operatorname{div} u)^2+|\dot\phi|^2,\\
\ddot\phi-\Delta\phi=0.
\end{gather*}
And hence the well-known wave map 
$$
d_{tt}-\Delta d=d(|\nabla d|^2-|d_t|^2)
$$ 
reduces to the wave equation $\phi_{tt}-\Delta\phi=0$.
\end{remark}

\begin{remark} \label{rmk1.2} \rm
Let $d$ be a smooth solution to the system \eqref{1.4} with the initial data 
$(d,d_t)(\cdot,0)=(d_0,d_1)$, if the initial data $(d_0,d_1)$ obeys the 
conditions 
$$
|d_0|=1,\quad  d_0\cdot d_1=0,
$$ 
then we have
$|d|=1$  and $d\cdot d_t=0$
for all times $t$.
\end{remark}

\begin{proof}[Proof of Remark \ref{rmk1.2}]
 Denote $w:=|d|^2-1$, multiplying \eqref{1.4} by $d$, we see that
$$
\ddot w-\Delta w=2w(|\nabla d|^2-|\dot d|^2).
$$
Testing the above equation by $\dot w$, we find that
\begin{align*}
&\frac12\frac{d}{dt}\int(\dot w^2+|\nabla w|^2)dx\\
&=2\int w\dot w(|\nabla d|^2-|\dot d|^2)\,dx
 +\int\Delta w(u\cdot\nabla w)\,dx-\int(u\cdot\nabla)\dot w\cdot\dot w\,dx\\
&= 2\int w\dot w(|\nabla d|^2-|\dot d|^2)\,dx
 -\sum_i\int\partial_ju_i\partial_iw\partial_jw\,dx
 +\frac12\int|\nabla w|^2\operatorname{div} u\,dx\\
&\quad +\frac12\int\dot w^2\operatorname{div} u\,dx \\
&\leq C\int(w^2+\dot w^2+|\nabla w|^2)\,dx.
\end{align*}

On the other hand, we observe that
$$
\frac12\frac{d}{dt}\int w^2\,dx
=\int w(\dot w-u\cdot\nabla w)\,dx\leq C\int(w^2+\dot w^2+|\nabla w|^2)\,dx.
$$
Combining the above two estimates and using the Gronwall inequality,
 we finish the proof.
\end{proof}

We denote
\begin{equation} \label{1.8}
\begin{aligned}
M(t):=
&1+\sup_{0\leq s\leq t}\Big\{\|\frac1\rho(\cdot,s)\|_{L^\infty}
 +\|\rho(\cdot,s)\|_{L^\infty}+\|\nabla\rho(\cdot,s)\|_{H^2}
 +\|u(\cdot,s)\|_{H^3}  \\
&+\|\theta(\cdot,s)\|_{H^3}+\|\dot d(\cdot,s)\|_{H^3}
 +\|\nabla d(\cdot,s)\|_{H^3}\Big\}\\
&+\|u\|_{L^2(0,t;H^4)}+\|u_t\|_{L^2(0,t;H^2)}
 +\|\theta\|_{L^2(0,t;H^4)}+\|\theta_t\|_{L^2(0,t;H^2)}.
\end{aligned}
\end{equation}

\begin{theorem} \label{thm1.2} 
Let $T^*$ be the maximal time of existence for problem \eqref{1.1}-\eqref{1.5} 
in the sense of Theorem \ref{thm1.1}. Then for any $t\in[0,T^*)$, we have that
\begin{equation}
M(t)\leq C_0M(0)\exp(\sqrt t C(M(t)))\label{1.9}
\end{equation}
for some given nondecreasing continuous functions $C_0(\cdot)$ and $C(\cdot)$.
\end{theorem}

It follows from \eqref{1.9}  \cite{17,18,19} that 
\begin{equation}
\sup_{0\leq t\leq T} M(t)\leq C\label{1.10}
\end{equation}
for some $T\in(0,T^*)$.

In the proofs below, we will use the following bilinear commutator 
and product estimates due to Kato-Ponce \cite{20}:
\begin{gather}
\|D^s(fg)-fD^sg\|_{L^p}\leq C(\|\nabla f\|_{L^{p_1}}\|D^{s-1}g\|_{L^{q_1}}
+\|D^sf\|_{L^{p_2}}\|g\|_{L^{q_2}}),\label{1.11}\\
\|D^s(fg)\|_{L^p}\leq C(\|f\|_{L^{p_1}}\|D^sg\|_{L^{q_1}}
+\|D^sf\|_{L^{p_2}}\|g\|_{L^{q_2}})\label{1.12}
\end{gather}
with $s>0$ and $\frac1p=\frac1{p_1}+\frac1{q_1}=\frac{1}{p_2}+\frac1q_2$ 
and $1<p<\infty$.

The proof of the uniqueness part is standard, we omit it here.

It is easy to prove Theorem \ref{thm1.1} by the Galerkin method if 
we have \eqref{1.9} \cite{6}, thus we only need to show a priori 
estimates \eqref{1.9}.

\section{Proof of Theorem \ref{thm1.2}}

Since the physical constants $C_V$ and $R$ do not bring any essential
 difficulties in our arguments, we shall take $C_V=R=1$.
First, it follows from \eqref{1.1} that
\begin{equation}
\rho(x,t)=\rho_0(y(0;x,t))\exp\Big\{-\int_0^t\operatorname{div} u(y(s;x,t),s)ds\Big\},\label{2.1}
\end{equation}
where $y(s;x,t)$ is the characteristic curve defined by 
$$
\frac{dy}{ds}=u(y,s),\quad  y(t;x,t)=x.
$$
Then \eqref{2.1} gives
\begin{equation}
\rho,\frac1\rho\leq C_0\exp(tC(M)).\label{2.2}
\end{equation}
Applying $\nabla$ to \eqref{1.1}, testing by $\nabla\rho$, we see that 
$$
\frac12\frac{d}{dt}\int|\nabla\rho|^2\,dx
=-\int\nabla\operatorname{div}(\rho u)\nabla\rho\,dx\leq C(M),
$$ 
which yields
\begin{equation}
\|\nabla\rho(\cdot,t)\|_{L^2}\leq C_0+tC(M).\label{2.3}
\end{equation}
Applying $D^3$ to \eqref{1.1}, testing by $D^3\rho$, using \eqref{1.11} 
and \eqref{1.12}, we find that
\begin{align*}
&\frac12\frac{d}{dt}\int(D^3\rho)^2\,dx\\
&=-\int(D^3(u\nabla\rho)-u\cdot\nabla D^3\rho)D^3\rho\,dx
 -\int u\cdot\nabla D^3\rho\cdot D^3\rho\,dx \\
&\quad -\int D^3(\rho\operatorname{div} u)D^3\rho\,dx\\
&\leq C(\|\nabla u\|_{L^\infty}\|D^3\rho\|_{L^2}
 +\|\nabla\rho\|_{L^\infty}\|D^3u\|_{L^2})\|D^3\rho\|_{L^2}\\
&\quad +C(\|\rho\|_{L^\infty}\|D^3\operatorname{div} u\|_{L^2}
 +\|\operatorname{div} u\|_{L^\infty}\|D^3\rho\|_{L^2})\|D^3\rho\|_{L^2}\\
&\leq C(M)+C(M)\|D^3\operatorname{div} u\|_{L^2},
\end{align*}
which leads to
\begin{equation}
\|D^3\rho(\cdot,t)\|_{L^2}\leq C+\sqrt tC(M).\label{2.4}
\end{equation}

It is easy to show that
\begin{gather}
\|u(\cdot,t)\|_{H^2}=\|u_0+\int_0^t u_t ds\|_{H^2}
 \leq C_0+\sqrt t C(M),\label{2.5}\\
\|\theta(\cdot,t)\|_{H^2} \leq C_0+\sqrt t C(M).\label{2.6}
\end{gather}
Testing \eqref{1.4} by $\dot d$ and using $d\cdot\dot d=0$, we infer that
 $$
\frac12\frac{d}{dt}\int(|\dot d|^2+|\nabla d|^2)dx
=\int u\cdot\nabla d\cdot\Delta d\,dx
-\int(u\cdot\nabla)\dot d\cdot\dot d\,dx\leq C(M),
$$
 which implies
\begin{equation}
\|\dot d(\cdot,t)\|_{L^2}+\|\nabla d(\cdot,t)\|_{L^2}\leq C_0+tC(M).\label{2.7}
\end{equation}

Taking $D^3$ to \eqref{1.2}, testing by $D^3u$ and using \eqref{1.1}, we derive
\begin{equation}
\begin{aligned}
&\frac12\frac{d}{dt}\int\rho|D^3u|^2\,dx+\mu\int|\nabla D^3 u|^2\,dx
 +(\lambda+\mu)\int(\operatorname{div} D^3 u)^2\,dx \\
&=\int D^3 p\cdot\operatorname{div} D^3 u\,dx-\int(D^3(\rho u\cdot\nabla u)
 -\rho u\cdot\nabla D^3 u)D^3 u\,dx \\
&\quad -\int(D^3(\rho u_t)-\rho D^3u_t)D^3 u\,dx
 -\int(D^3(\nabla d\cdot\Delta d)-\nabla d\cdot\Delta D^3 d)D^3 u\,dx \\
&\quad -\int(D^3 u\cdot\nabla)d\cdot\Delta D^3 d\,dx \\
&=:I_1+I_2+I_3+I_4-I_5.
\end{aligned}\label{2.8}
\end{equation}
Applying $D^3$ to \eqref{1.4} and testing by $D^3\dot d$, we obtain
\begin{equation}
\begin{aligned}
&\frac12\frac{d}{dt}\int(|D^3\dot d|^2+|\nabla D^3 d|^2)\,dx \\
&=-\int(D^3(u\cdot\nabla\dot d)-u\cdot\nabla D^3\dot d)D^3\dot d\,dx
 -\int(u\cdot\nabla)D^3\dot d\cdot D^3\dot d\,dx \\
&\quad +\int D^3(d(|\nabla d|^2-|\dot d|^2))D^3\dot d\,dx
 +\int\Delta D^3 d\cdot(u\cdot\nabla D^3 d)dx \\
&\quad +\int\Delta D^3 d\cdot(D^3(u\cdot\nabla d)-u\cdot\nabla D^3 d
 -(D^3u\cdot\nabla)d)\,dx+I_5 \\
&=:\ell_1+\ell_2+\ell_3+\ell_4+\ell_5+I_5.
\end{aligned}\label{2.9}
\end{equation}
Summing  \eqref{2.8} and \eqref{2.9}, we have
\begin{equation}
\begin{aligned}
&\frac12\frac{d}{dt}\int(\rho|D^3u|^2+|D^3\dot d|^2+|\nabla D^3d|^2)\,dx \\
&+\mu\int|\nabla D^3u|^2\,dx+(\lambda+\mu)\int(\operatorname{div} D^3u)^2\,dx \\
&=\sum_{i=1}^4(I_i+\ell_i)+\ell_5.
\end{aligned}\label{2.10}
\end{equation}
Using \eqref{1.11} and \eqref{1.12}, we bound $I_i$ $(i=1,\cdots,4)$ and
$\ell_i$ $(i=1,\cdots,5)$ as follows.
\begin{gather*}
I_1\leq C(\|\rho\|_{L^\infty}\|D^3\theta\|_{L^2}
 +\|\theta\|_{L^\infty}\|D^3\rho\|_{L^2})\|\operatorname{div} D^3 u\|_{L^2}
\leq C(M)\|\operatorname{div} D^3u\|_{L^2};\\
I_2\leq C(\|\nabla(\rho u)\|_{L^\infty}\|D^3u\|_{L^2}
 +\|\nabla u\|_{L^\infty}\|D^3(\rho u)\|_{L^2})\|D^3 u\|_{L^2}
\leq C(M);\\
I_3\leq C(\|\nabla\rho\|_{L^\infty}\|D^2u_t\|_{L^2}
 +\|u_t\|_{L^\infty}\|D^3\rho\|_{L^2})\|D^3u\|_{L^2}
\leq C(M)\|u_t\|_{H^2};\\
I_4 \leq C\|\nabla^2d\|_{L^\infty}\|D^4d\|_{L^2}\|D^3u\|_{L^2}\leq C(M); \\
\ell_1 \leq C\|\nabla u\|_{L^\infty}\|D^3\dot d\|_{L^2}^2
 +C\|\nabla\dot d\|_{L^\infty}\|D^3u\|_{L^2}\|D^3\dot d\|_{L^2}\leq C(M);\\
\ell_2=\frac12\int|D^3\dot d|^2\operatorname{div} u\,dx\leq C(M);\\
\begin{aligned}
\ell_3&\leq C[\|d\|_{L^\infty}\|D^3(|\nabla d|^2
 -|\dot d|^2)\|_{L^2}+(\|\nabla d\|_{L^\infty}^2
 +\|\dot d\|_{L^\infty}^2)\|D^3d\|_{L^2}]\|D^3\dot d\|_{L^2}\\
&\leq C(M);
\end{aligned} \\
\begin{aligned}
\ell_4&=\sum_{i,j}\int u_i\partial_iD^3d\partial_j^2D^3d\,dx\\
&=-\sum_{i,j}\int\partial_ju_i\partial_iD^3d\partial_jD^3d\,dx
 +\sum_{i,j}\frac12\int\partial_iu_i(\partial_jD^3d)^2\,dx\\
&\leq C\|\nabla u\|_{L^\infty}\|D^4d\|_{L^2}^2\leq C(M);
\end{aligned}\\
\begin{aligned}
\ell_5&=\int\Delta D^3d(C_1D^2 u\cdot\nabla Dd+C_2Du\cdot\nabla D^2d)dx\\
&=-\sum_i\int\partial_iD^3d\partial_i(C_1D^2u\nabla Dd+C_2Du\cdot\nabla D^2d)dx\\
&\leq C\|D^4d\|_{L^2}(\|D^3u\|_{L^2}\|\nabla^2d\|_{L^\infty}
 +\|D^2u\|_{L^3}\|D^3d|_{L^6}+\|\nabla u\|_{L^\infty}\|D^4d\|_{L^2})\\
&\leq C(M).
\end{aligned}
\end{gather*}

Inserting the above estimates into \eqref{2.10}, we have
\begin{equation}
\begin{aligned}
&\frac12\frac{d}{dt}\int(\rho|D^3u|^2+|D^3\dot d|^2+|\nabla D^3d|^2)dx \\
&+\mu\int|\nabla D^3u|^2\,dx+(\lambda+\mu)\int(\operatorname{div} D^3u)^2\,dx \\
&\leq C(M)\|\operatorname{div} D^3u\|_{L^2}+C(M)+C(M)\|u_t\|_{H^2}.
\end{aligned} \label{2.11}
\end{equation}

Integrating the above estimates in $[0,t]$, we arrive at
\begin{equation}
\begin{aligned}
&\|D^3u(\cdot,t)\|_{L^2}^2+\|D^3\dot d(\cdot,t)\|_{L^2}^2
 +\|\nabla D^3d(\cdot,t)\|_{L^2}^2+\int_0^t\int|D^4u|^2\,dx ds \\
&\leq C_0\exp(\sqrt tC(M)).
\end{aligned}\label{2.12}
\end{equation}
On the other hand, it follows from \eqref{1.2} that
\[
u_t=-u\cdot\nabla u+\frac1\rho\Big[\mu\Delta u+(\lambda+\mu)\nabla\operatorname{div} u
-\nabla p-\nabla\cdot(\nabla d\odot\nabla d-\frac12|\nabla d|^2\mathbb{I}_3)
\Big] %\label{2.13}
\]
which easily implies
\begin{equation}
\|u_t\|_{L^2(0,t;H^2)}\leq C_0\exp(\sqrt t C(M)).\label{2.14}
\end{equation}

Applying $D^3$ to \eqref{1.3}, testing by $D^3\theta$ and using \eqref{1.1}, 
\eqref{1.11}, and \eqref{1.12}, we have
\begin{align*}
&\frac12\frac{d}{dt}\int\rho(D^3\theta)^2dx+\int|\nabla D^3\theta|^2\,dx\\
&=-\int(D^3(\rho u\cdot\nabla\theta)-\rho u\cdot\nabla D^3\theta)D^3\theta\,dx
 -\int D^3(p\operatorname{div} u)\cdot D^3\theta\,dx\\
&\quad -\int(D^3(\rho\theta_t)-\rho D^3\theta_t)D^3\theta\,dx \\
&\quad +\int D^3\big[\frac\mu2|\nabla u+\nabla u^t|^2+\lambda(\operatorname{div} u)^2
 +|\dot d|^2\big]D^3\theta\,dx\\
&\leq C(\|\nabla(\rho u)\|_{L^\infty}\|D^3\theta\|_{L^2}
 +\|\nabla\theta\|_{L^\infty}\|D^3(\rho u)\|_{L^2})\|D^3\theta\|_{L^2}\\
&\quad +C(\|p\|_{L^\infty}\|D^3\operatorname{div} u\|_{L^2}
 +\|\operatorname{div} u\|_{L^\infty}\|D^3p\|_{L^2})\|D^3\theta\|_{L^2}\\
&\quad +C(\|\nabla\rho\|_{L^\infty}\|D^2\theta_t\|_{L^2}
 +\|\theta_t\|_{L^\infty}\|D^3\rho\|_{L^2})\|D^3\theta\|_{L^2}\\
&\quad +C(\|\nabla u\|_{L^\infty}\|D^4u\|_{L^2}
 +\|\dot d\|_{L^\infty}\|D^3\dot d\|_{L^2})\|D^3\theta\|_{L^2}\\
&\leq C(M)+C(M)\|D^3\operatorname{div} u\|_{L^2}+C(M)\|\theta_t\|_{H^2}+C(M)\|D^4u\|_{L^2},
\end{align*}
which gives
\begin{equation}
\|D^3\theta(\cdot,t)\|_{L^2}^2+\int_0^t\int|D^4\theta|^2\,dx\, ds
\leq C_0\exp(\sqrt t C(M)).\label{2.15}
\end{equation}
On the other hand, from \eqref{1.3} if follows that
\begin{equation}
\theta_t=-u\cdot\nabla\theta-\frac{p}{\rho}\operatorname{div} u+\frac1\rho
\left[\frac\mu2|\nabla u+\nabla u^t|^2+\lambda(\operatorname{div} u)^2+|\dot d|^2\right],
\label{2.16}
\end{equation}
which easily leads to
\begin{equation}
\|\theta_t\|_{L^2(0,t;H^2)}\leq C_0\exp(\sqrt t C(M)).\label{2.17}
\end{equation}
This completes the proof.

\subsection*{Acknowledgments}
J. Fan is supported by the NSFC (Grant No. 11171154).

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