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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2013 (2013), No. 107, pp. 1--22.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2013 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2013/107\hfil Serrin blow-up criterion]
{Serrin blow-up criterion for strong solutions to the 3-D
compressible nematic liquid crystal flows with vacuum}

\author[Q. Liu \hfil EJDE-2013/107\hfilneg]
{Qiao Liu} 

\address{Qiao Liu \newline
College of Mathematics and Computer Science,
Hunan Normal University, Changsha,  Hunan 410081,  China}
\email{liuqao2005@163.com}

\thanks{Submitted September 12, 2012. Published April 24, 2013.}
\subjclass[2000]{76A15, 76N10, 35B65, 35Q35}
\keywords{Compressible nematic liquid crystal flows; strong solution;
\hfill\break\indent  Serrin's criterion; blow-up criterion; compressible
Navier-Stokes equations}

\begin{abstract}
 In this article, we extend the well-known Serrin's blow-up criterion
 for solutions of the 3-D incompressible Navier-Stokes equations to
 the 3-D compressible nematic liquid crystal flows where the initial
 vacuum is allowed. It is proved that for the initial-boundary value
 problem of the 3-D compressible nematic liquid crystal flows in a
 bounded domain, the strong solution exists globally if the velocity
 satisfies the Serrin's condition and $L^1(0,T;L^{\infty})$-norm of
 the gradient of the velocity is bounded.
\end{abstract}

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

\section{Introduction}\label{Int}

The time evolution of the density, the velocity and the orientation
of a compressible nematic liquid crystal (NLC) flows occupying a
bounded domain $\Omega$ of $\mathbb{R}^3$ can be described by the
system
\begin{gather}  \label{eq1.1}
\partial_t\rho+\operatorname{div} (\rho u)=0,\quad 
(x,t)\in \Omega\times(0,+\infty),\\
\label{eq1.2}
\begin{aligned}
&{\partial_t}(\rho u) +\operatorname{div}(\rho u\otimes  u)
+\nabla{P}\\
&=\mu\Delta u-\lambda\nabla\cdot(\nabla d
 \odot\nabla d  -(\frac{1}{2}|\nabla d|^2+F(d))I), (x,t)\in \Omega\times(0,+\infty),
\end{aligned}\\
\label{eq1.3}
\partial_td+(u\cdot\nabla)d=\nu(\Delta d-f(d)), \quad
(x,t)\in \Omega\times(0,+\infty),
\end{gather}
together with the initial value:
\begin{equation} \label{eq1.4}
\rho(0,x)=\rho_0(x)\geq 0, \quad u(0,x)=u_0(x),\quad
d(0,x)=d_0(x), \quad\forall x\in\Omega,
\end{equation}
and the boundary conditions:
\begin{equation} \label{eq1.5}
u(t,x)=0,\quad d(t,x)=d_0(x),\quad |d_0(x)|=1, \quad\forall
(t,x)\in [0,+\infty)\times \partial\Omega.
\end{equation}
Here we denote by $\rho$ the unknown density,
$u=(u_1,u_2,u_3)$ the unknown velocity,
$d=(d_1,d_2,d_3)$ the unknown orientation parameter of the
nematic liquid crystal material, and $P=P(\rho)$ the pressure
function. $\mu,\lambda$ and $\nu$ are positive viscosity
coefficients. The unusual term $\nabla d\odot \nabla d$ denotes the
$3\times 3$ matrix, whose $(i,j)$-th element is given by
$\sum_{k=1}^3\partial_{i}d_{k}\partial_{j}d_{k}$. $I$ is the
$3\times 3$ unit matrix. $f(d)$ is a polynomial function of $d$
which satisfies $f(d)=\frac{\partial}{\partial_{d}}F(d)$, where
$F(d)$ is the bulk part of the elastic energy, usually we choose
$F(d)$ to be the Ginzburg-Landau penalization; i.e.,
\[
F(d)=\frac{1}{4\sigma^2}(|d|^2-1)^2,\quad
f(d)=\frac{1}{\sigma^2}(|d|^2-1)d,
\]
where $\sigma$ is a positive constant. In what follows, we will
assume $\sigma=1$ since it does not play a special role in our
discussion. Throughout this paper, we adopt the following simplified
notations for standard Sobolev spaces
\[
L^q:=L^q(\Omega),\quad W^{k,p}:=W^{k,p}(\Omega),\quad
H^{k}:=H^{k}(\Omega)=W^{k,2},\quad H_0^1:=H_0^1(\Omega),
\]
where $1\leq p$, $q\leq \infty$ and $k\in \mathbb{N}$.


System \eqref{eq1.1}--\eqref{eq1.5} is a simplified
version of Ericksen-Lesile system modeling the flow of compressible
nematic liquid crystals materials, and the hydrodynamic theory of
liquid crystals was established by Ericksen \cite{ER1,ER2} and
Leslie \cite{LE} in the 1960's. In \cite{WY}, Wang and Yu
established the global existence and large-time behavior of weak
solutions for the initial-boundary value problem
\eqref{eq1.1}--\eqref{eq1.5}. When the direction $d$ does not
appear, the system \eqref{eq1.1}--\eqref{eq1.5} becomes  the
compressible Navier-Stokes (CNS) equations. Matsumura and Nishida
\cite{MN} obtained global existence of smooth solutions for the
initial data is a small perturbation of a non--vacuum equilibrium.
For the existence of solutions for arbitrary initial value, Lions
\cite{PL} and Feireisl \cite{EF1,EF} established the global
existence of weak solution to the CNS equations. Cho et
al \cite{CK1,CCK,CK2} proved that the existence and uniqueness of
local strong solutions of the CNS equations in the case where
initial density need not to be positive and may vanish in an open
set. Xin in \cite{X} showed that there is no global smooth solution
to the Cauchy problem of the CNS equations with a nontrivial
compactly supported initial density. Hence, there are many works
\cite{CCK,FJ,FJO,HX,HLX1,HLX2,SWZ1,SWZ2} trying to establish
blow-up criterion for the strong solution to the CNS equations. In
particular, it is proved in \cite{HLX2} by Huang, Li and Xin that
the Serrin's blow--up criterion (see \cite{JS}) for the
incompressible Navier--Stokes equations still holds for the CNS
equations; i.e., if $T^{*}$ is the maximal time of existence strong
solution, then
\begin{equation} \label{eq1.6}
\lim_{T\to T^{*}}(\|\operatorname{div}
u\|_{L^1(0,T;L^{\infty})}+\|\rho^{1/2}
u\|_{L^s(0,T;L^r)})=\infty
\end{equation}
or
\begin{equation} \label{eq1.7}
\lim_{T\to
T^{*}}(\|\rho\|_{L^1(0,T;L^{\infty})}+\|\rho^{1/2}
u\|_{L^s(0,T;L^r)})=\infty,
\end{equation}
where $r$ and $s$ satisfy $\frac{2}{s}+\frac{3}{r}\leq 1$,
$3< r\leq \infty$. Huang et al \cite{HX,HLX1} established that the
Beale-Kato-Majda criterion (see \cite{BKM}) for the ideal
incompressible flows still holds for the CNS equations; that is,
\[
\lim_{T\to T^{*}}\int_0^T\|\nabla u\|_{L^{\infty}}\,\mathrm{d} t=\infty.
\]
Sun, Wang and Zhang in \cite{SWZ1} (see also \cite{HLX2}) obtained
another Beale--Kato--Majda criterion in terms of the density, i.e.,
\[
\lim\sup_{T\to T^{*}}\|\rho
\|_{L^{\infty}(0,T;L^{\infty})}=\infty.
\]
Recently, Wen and Zhu in \cite{WZ} established a blow-up criterion
of the strong solution for the CNS equations in terms of the
density,
\[
\limsup_{T\to T^{*}}\|\rho
\|_{L^{\infty}(0,T;L^q)}=\infty,
\]
for some $1<q<\infty$ large enough.


When $\rho$ is a positive constant, the system
\eqref{eq1.1}-\eqref{eq1.3} becomes  the incompressible nematic
liquid crystal (INLC) equations, the global-in-time weak solutions
and local-in-time strong solution have been studied by Lin and Liu
\cite{LL1}. Later, in \cite{LL2}, they further proved that the
one-dimensional spacetime Hausdorff measure of the singular set of
the so-called suitable weak solutions is zero. In \cite{HW}, Hu and
Wang established global existence of strong solutions and
weak--strong uniqueness for initial data belonging to the Besov
spaces of positive order under some smallness assumptions. Liu and
Cui in \cite{LC} obtained that the blow--up criterion \eqref{eq1.6}
or \eqref{eq1.7} still holds for the solution of the INLC equations.
We also refer \cite{HKL,L,LLW,LW1,LW,SL} and the references cited
therein for other related work on the INLC equations.

Inspired by the above mentioned works on blow-up criteria of the
strong solutions to the CNS equations and the INLC equations,
particularly the results of Huang et al \cite{HX,HLX1} and Sun et
al \cite{SWZ1,SWZ2}, we want to investigate the similar problem for
the compressible nematic liquid crystal flow
\eqref{eq1.1}--\eqref{eq1.5}.

When the initial vacuum is allowed, the well-posedness and a
blow-up criterion for strong solutions to the compressible nematic
liquid crystal flows \eqref{eq1.1}-\eqref{eq1.5} were established by
Liu et al \cite{LL,LLH}. More precisely, under the
assumption of the  pressure $P$ satisfies
\begin{equation} \label{eq1.8}
P=P(\cdot)\in C^1[0,\infty),\quad P(0)=0.
\end{equation}
They established the following result.

\begin{theorem}[\cite{LL,LLH}] \label{thm1.1}
 Suppose that the initial value
$(\rho_0,u_0,d_0)$ satisfies the regularity
conditions
\[
0\leq \rho_0\in W^{1,6},\quad u_0\in H_0^1\cap H^2, \quad d_0\in H^3,
\]
and the compatibility condition
\begin{equation} \label{eq1.9}
\mu \Delta u_0-\lambda \operatorname{div} (\nabla d_0\odot\nabla
d_0-(\frac{1}{2}|\nabla d_0|^2+F(d_0))I)-\nabla
P(\rho_0)=\sqrt{\rho}g 
\end{equation}
 for some function $g\in L^2$.
Then there exist a small $T\in (0,\infty)$ and a unique strong
solution $(\rho,u,d)$ to the system \eqref{eq1.1}-\eqref{eq1.3}
with initial boundary condition \eqref{eq1.4}-\eqref{eq1.5} such
that
\begin{gather*}
0\leq \rho\in C([0,T);W^{1,6}),\quad \rho_t\in C([0,T);L^{6}),\\
u\in C([0,T);H^1_0\cap H^2)\cap L^2(0,T;W^{2,6}),\quad
u_t\in L^2(0,T;H^1_0),\\
d\in C([0,T);H^3),\quad d_t\in C([0,T);H^1_0)\cap
L^2(0,T;H^2),\\
d_{tt}\in L^2(0,T;L^2),\quad
\sqrt{\rho}u_t\in C([0,T);L^2).
\end{gather*}
Moreover, let $T^{*}$ be the maximal existence time of the solution.
If $T^{*}<\infty$, then there holds
\begin{equation} \label{eq1.10}
\lim_{T\to T^{*}} \int_0^T(\|\nabla
u\|^{\beta}_{\alpha}+\|u\|_{W^{1,\infty}})\,\mathrm{d}t=\infty,
\end{equation}
%-----------------(eq1.10)-------------------
where $\alpha,\beta$ satisfying $\frac{3}{\alpha}+\frac{2}{\beta}<2$
and $\beta\geq 4$.
\end{theorem}

Recently Huang, Wang and Wen \cite{HWW1,HWW2} considered a similar, but not
equivalent,  system of partial differential equations modeling compressible 
nematic liquid crystal flows, they obtained the existence of local in time strong
solution and two blow--up criteria under some suitable assumptions
on $u$ and $d$ or on  $\rho$ and $d$.

The purpose of this article is to obtain a blow-up criterion for the
strong solutions to the compressible liquid crystal flow only in
terms of the velocity. Our main result is stated as follows

\begin{theorem}\label{thm1.2}
Assume that $(\rho, u, d)$ is the strong solution constructed in
Theorem \ref{thm1.1}, and $T^{*}$ be the maximal existence time of
the solution. If $T^{*}<\infty$, then we have
\begin{equation} \label{eq1.11}
\lim_{T\to T^{*}}\{\|\nabla
u\|_{L^1(0,T;L^{\infty})}+\|u\|_{L^s(0,T;L^r)}\}=\infty,
\end{equation}
for all $r,s$ satisfying
\begin{equation} \label{eq1.12}
\frac{2}{s}+\frac{3}{r}\leq 1, \quad 3< r\leq +\infty.
\end{equation}
\end{theorem}


The remaining of this article is devoted to proving Theorem
\ref{thm1.2}. The main idea used here is similar as the papers
\cite{HLX1,HLX2} of Huang et al, who studied the blow-up criterion
of strong solutions to the 3D CNS equations. The key issue in our
proofs is derive the suitable higher norm estimates of the strong
solution $(\rho,u,d)$. Some of the new difficulties appears due to
the fact that  system \eqref{eq1.1}--\eqref{eq1.3} is the coupling
of the CNS equations \eqref{eq1.1}--\eqref{eq1.2} and the liquid
crystal equation \eqref{eq1.3}. To proceed, some new estimates are
needed. In fact, we found that under the assumption of $L^1(0,T;
L^{\infty})$-norm of the divergence of velocity implies that both
the time-independent upper bound for the density and the
$L^{\infty}(0,T;L^q)$-norm with $2\leq q\leq \infty$ of $d$,
and when the opposite to \eqref{eq1.11} holds, the bound of the
$L^{\infty}(0,T;L^q)$-norm with $2\leq q\leq \infty$ of $\nabla d$
can be directly from the liquid crystal equation \eqref{eq1.3}.
These properties are important for us to establish the higher order
norm estimates of $\rho$, $u$ and $d$.


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

 In what follows, we assume that $(\rho, u, d)$ is the unique strong
solution to the system \eqref{eq1.1}--\eqref{eq1.3} with
initial-boundary condition \eqref{eq1.4}--\eqref{eq1.5} constructed
in Theorem \ref{thm1.1}, and $T^{*}$ is the maximal existence time
of the solution. We assume that the opposite to \eqref{eq1.11}
holds; i.e.,
\[
\lim_{T\to T^{*}} \{\|\nabla
u\|_{L^1(0,T;L^{\infty})}+\|u\|_{L^s(0,T;L^r)}\}\leq M<\infty.
\]
Hence, for all $0<T<T^{*}$, there holds
\begin{equation} \label{eq2.1}
\int_0^T\|\nabla u\|_{L^{\infty}}+\|u\|_{L^r}^s\,\mathrm{d}t\leq
M<\infty,
\end{equation}
from which we will prove the same regularity at time $T^{*}$ as the
initial value, a contradiction to the maximality of $T^{*}$.
Throughout the paper, we will use $C$ to denote a generic positive
constant depending only on $\mu,\lambda,\nu, M, T,\Omega$ and the
initial data. Here, we notice that thanks to the assumptions
\eqref{eq2.1} and \eqref{eq1.12}, we have
\begin{equation} \label{eq2.2}
\int_0^T\|u\|_{L^{\infty}}^2\,\mathrm{d}t\leq M<\infty.
\end{equation}

Under  assumption \eqref{eq2.1}, it is easy to obtain the
$L^{\infty}$-norm bounds of the density $\rho$ and the $L^q$-norm
(for all $q\geq 2$) of the orientation parameter $d$ by using the
mass conservation equation \eqref{eq1.1} and the liquid crystal
equation \eqref{eq1.3} respectively. More precisely, we have the
following Lemma.

\begin{lemma}\label{lem2.1}
Assume that
\begin{equation} \label{eq2.3}
\int_0^T\|\operatorname{div}u\|_{L^{\infty}}\,\mathrm{d}t\leq
M,\quad   0\leq T< T^{*},
\end{equation}
then
\begin{gather}\label{eq2.4}
\sup_{0\leq t\leq T}\|\rho\|_{L^{\infty}}\leq C \quad \forall 0<T<T^{*};\\
     \label{eq2.5}
\sup_{0\leq t\leq T}\|d\|_{L^q}\leq C\quad\text{for all }
q\geq 2 \text{ and } 0<T<T^{*},
\end{gather}
where the constant $C$ independent of $q$.
\end{lemma}


\begin{proof}
The proof of estimate \eqref{eq2.4} is essentially due to Huang and
Xin \cite{HX}, for reader's convenience, we sketch it here.
Multiplying the mass conservation equation \eqref{eq1.1} by
$p\rho^{p-1}$ with $p>1$, it follows that
\[
\partial_t(\rho^p)
+\operatorname{div}(\rho^pu)+(p-1)\rho^p\operatorname{div}u=0.
\]
Integrating the above equality over $\Omega$ yields
\[
\partial_t\|\rho\|_{L^p}^p\leq
(p-1)\|\operatorname{div}u\|_{L^{\infty}}\|\rho\|_{L^p}^p;
\]
i.e.,
\begin{equation} \label{eq2.6}
\partial_t\|\rho\|_{L^p}\leq
\frac{(p-1)}{p}\|\operatorname{div}u\|_{L^{\infty}}\|\rho\|_{L^p}.
\end{equation}
Condition \eqref{eq2.3} and  estimate \eqref{eq2.6} imply
\[
\sup_{0\leq t\leq T}\|\rho\|_{L^p}\leq C\quad  \forall p>1,
\]
where $C$ is a positive constant independent of $p$, letting
$p\to\infty$, we obtain \eqref{eq2.4}, and this completes
the first part of the proof.

To prove the estimate \eqref{eq2.5}, we  multiply the liquid crystal
equation \eqref{eq1.3} by $q|d|^{q-2}d$ with $q\geq 2$, integrate
it over $\Omega$, and make use of the boundary condition
\eqref{eq1.5}, then there holds
\begin{equation} \label{eq2.7}
\begin{aligned}
&\frac{d}{dt}\|d\|_{L^q}^q+\int_{\Omega}(q\nu |\nabla
d|^2|d|^{q-2}+q(q-2)\nu|d|^{q-2}|\nabla
|d||^2)\,\mathrm{d}x+q\nu\int_{\Omega }
|d|^{q+2}\,\mathrm{d}x \\
&= -\sum_{i=1}^3\int_{\Omega} u_{i}\partial_{i}
(|d|^q)\,\mathrm{d}x+q\nu\int_{\Omega}|d|^q\,\mathrm{d}x \\
&\leq \int_{\Omega} |\operatorname{div}u|
|d|^q\,\mathrm{d}x
+\Big(\frac{q\nu}{2}\int_{\Omega}|d|^{q+2}\,\mathrm{d}x\Big)^{\frac{q}{q+2}}
 (2|\Omega|)^{\frac{2}{q+2}} \\
&\leq  C\|\operatorname{div}
u\|_{L^{\infty}}\|d\|_{q}^q+\frac{q\nu}{2}\int_{\Omega}|d|^{q+2}\,\mathrm{d}x+C,
\end{aligned}
\end{equation}
where we have used the fact that $d\nabla d=|d|\nabla|d|$ implies
\begin{align*}
-q\int_{\Omega}\Delta d |d|^{q-2}d\,\mathrm{d}x
&= q\int_{\Omega}|\nabla
d|^2|d|^{q-2}\,\mathrm{d}x+q\int_{\Omega}\nabla d \cdot d\nabla
|d|^{q-2}\,\mathrm{d}x\\
&= q\int_{\Omega}|\nabla
d|^2|d|^{q-2}\,\mathrm{d}x+q\int_{\Omega}|\nabla d|
|d|(q-2)|d|^{q-2}\nabla
|d|\,\mathrm{d}x\\
&=q\int_{\Omega}|\nabla
d|^2|d|^{q-2}\,\mathrm{d}x+q(q-2)\int_{\Omega}|d|^{q-1}|\nabla
|d||^2\,\mathrm{d}x
\end{align*}
and the fact that $f(d)=(|d|^2-1)d$. It follows from inequality
\eqref{eq2.7} that
\[
\frac{d}{dt}\|d\|_{L^q}^q\leq C\|\operatorname{div}
u\|_{L^{\infty}}\|d\|_{L^q}^q+C,
\]
which together with the Gronwall's inequality imply
\begin{equation} \label{eq2.8}
\sup_{0\leq t\leq T}\|d\|_{L^q}\leq C\quad\text{ for all }q\geq 2,
\end{equation}
where $C$ is a positive constant independent of $q$. By letting
$q\to\infty$, we notice that the estimate \eqref{eq2.5}
still holds.
\end{proof}

By assumption \eqref{eq1.8} on the pressure $P$ and
the estimate \eqref{eq2.4}, it is easy to obtain
\begin{equation} \label{eq2.9}
\sup_{0\leq t\leq T}\{\|P(\rho)\|_{L^{\infty}},\|P'(\rho)\|_{L^{\infty}}\}\leq
C<\infty.
\end{equation}
Now, let us derive the stand energy inequality.

\begin{lemma}\label{lem2.2}
There holds
\begin{equation} \label{eq2.10}
\sup_{0\leq t\leq T}\int_{\Omega}(\rho|u|^2+|\nabla
d|^2+2F(d))\,\mathrm{d}x
+\int_0^T\int_{\Omega}|\nabla u|^2+ |\Delta d-f(d)|^2\,\mathrm{d}x\,
\mathrm{d}t\leq C.
\end{equation}
\end{lemma}

\begin{proof}
Multiplying the momentum equation \eqref{eq1.2} by $u$,
integrating it over $\Omega$, making use of the mass conversation
equation \eqref{eq1.1} and the boundary condition \eqref{eq1.5},
it follows that
\begin{equation} \label{eq2.11}
\frac{1}{2}\frac{d}{dt}\int_{\Omega}\rho|u|^2\,\mathrm{d}x
+\mu\int_{\Omega}|\nabla u|^2\,\mathrm{d}x
=-\int_{\Omega} u\nabla P\,\mathrm{d}x
-\lambda\int_{\Omega}(u\cdot\nabla)d\cdot(\Delta d -f(d))\,\mathrm{d}x,
\end{equation}
where we have used the equality
$\operatorname{div}(\nabla d\odot \nabla d)=(\nabla d)^T\Delta d
+\nabla\left(\frac{|\nabla d|^2}{2}\right)$.
 Multiplying the liquid crystal equation
\eqref{eq1.3} by $\Delta d-f(d)$ and integrating it over $\Omega$,
we obtain
\begin{equation} \label{eq2.12}
\frac{d}{dt}\int_{\Omega}(\frac{1}{2}|\nabla
d|^2+F(d))\,\mathrm{d}x+\nu \int_{\Omega}|\Delta
d-f(d)|^2\,\mathrm{d}x=\int_{\Omega}(u\cdot \nabla)d\cdot(\Delta
d-f(d))\,\mathrm{d}x.
\end{equation}
%----------------(eq2.12)--------------------
Combining \eqref{eq2.11} and \eqref{eq2.12}
\begin{equation} \label{eq2.13}
\begin{aligned}
&\frac{d}{dt}\int_{\Omega}[\frac{1}{2}(\rho |u|^2+\lambda |\nabla
d|^2)+\lambda F(d)]\,\mathrm{d}x+\mu\int_{\Omega }|\nabla
u|^2\,\mathrm{d}x+\lambda\nu \int_{\Omega }|\Delta
d-f(d)|^2\,\mathrm{d}x \\
&= -\int_{\Omega } u\nabla P\,\mathrm{d}x\\
&=\int_{\Omega}P\operatorname{div}u\,\mathrm{d}x\leq
\varepsilon\int_{\Omega }|\nabla u|^2\,\mathrm{d}x+C \varepsilon^{-1},
\end{aligned}
\end{equation}
where we have used the estimates \eqref{eq2.4}, \eqref{eq2.9} and
the Young inequality. Taking $\varepsilon$ small enough in
\eqref{eq2.13} and applying the Gronwall's inequality, we can
establish the estimate \eqref{eq2.10} immediately.
\end{proof}

In the next lemma, we will derive some estimates of the direction
field $d$.

\begin{lemma}\label{lem2.3}
Under the assumption \eqref{eq2.1}, it holds that for $0\leq T<T^{*}$
\begin{gather} \label{eq2.14}
\sup_{0\leq t\leq T}\|\nabla d\|_{L^q}\leq C\quad
\text{for all } 2\leq q\leq \infty;\\
 \label{eq2.15}
\sup_{0\leq t\leq T}\|\nabla d\|_{L^2}^2+\int_0^T\|
d_t\|_{L^2}^2\,\mathrm{d}t\leq C;\\
  \label{eq2.16}
\sup_{0\leq t\leq T}\| d\|_{H^2}^2+\int_0^T\| \nabla
d_t\|_{L^2}^2\,\mathrm{d}t\leq C;\\
  \label{eq2.17}
\int_0^T\|\nabla d \|_{H^2}^2\,\mathrm{d}t\leq C.
\end{gather}
\end{lemma}

\begin{proof}
Multiply the gradient of the liquid crystal equation
\eqref{eq1.3} by $q|\nabla d|^{q-2}\nabla d$ with $q\geq 2$,
integrating it over $\Omega$, and using the boundary condition
\eqref{eq1.5}, then there holds
\begin{align*}
&\frac{d}{dt}\|\nabla d\|_{L^q}^q+\int_{\Omega}(q\nu |\nabla
(\nabla d)|^2|\nabla d|^{q-2}+q(q-2)\nu |\nabla |\nabla
d||^2|\nabla d|^{q-2})\,\mathrm{d}x
\\
&= -\sum_{i=1}^3\int_{\Omega}u_{i}\partial_{i}(|\nabla
d|^q)\,\mathrm{d}x-\sum_{i=1}^3q\int_{\Omega}\nabla
u_{i}\partial_{i}d |\nabla d|^{q-2}\nabla d \,\mathrm{d}x\\
&\quad -\nu q\int_{\Omega}\nabla(|d|^2d)|\nabla d|^{q-2}\nabla d\,\mathrm{d}x
+\nu q \int_{\Omega}|\nabla d|^q\,\mathrm{d}x 
\\
&= \sum_{i=1}^3\int_{\Omega}\partial_{i}u_{i}|\nabla
d|^q\,\mathrm{d}x-\sum_{i=1}^3q\int_{\Omega }\nabla
u_{i}\partial_{i}d |\nabla d|^{q-2}\nabla d \,\mathrm{d}x\\
&\quad -\nu q\int_{\Omega}\nabla(|d|^2d)|\nabla d|^{q-2}\nabla
d\,\mathrm{d}x
 +\nu q \int_{\Omega}|\nabla d|^q\,\mathrm{d}x 
\\
&=\int_{\Omega}\operatorname{div}u |\nabla d|^q\,\mathrm{d}x+\sum_{i=1}^3q\int_{\Omega}u_{i}
\partial_{i}\nabla d |\nabla d|^{q-2}\nabla d\,\mathrm{d}x\\
&\quad +\sum_{i=1}^3q\int_{\Omega}u_{i}
 \partial_{i} d \nabla(|\nabla d|^{q-2}\nabla d)\,\mathrm{d}x 
 -\nu q\int_{\Omega}\nabla(|d|^2d)|\nabla d|^{q-2}\nabla
 d\,\mathrm{d}x\\
&\quad +\nu q \int_{\Omega}|\nabla d|^q\,\mathrm{d}x 
\\
&\leq  \int_{\Omega}|\operatorname{div}u||\nabla d|^q\,\mathrm{d}x+
2q\int_{\Omega}|u||\nabla^2d||\nabla
d|^{q-1}\,\mathrm{d}x\\
&\quad +q(q-1)\int_{\Omega}|u||\nabla|\nabla d|||\nabla
d|^{q-1}\,\mathrm{d}x 
-\nu q\int_{\Omega}\nabla(|d|^2d)|\nabla d|^{q-2}\nabla
d\,\mathrm{d}x+\nu q\|\nabla d\|_{L^q}^q
\\
&\leq  \|\operatorname{div} u\|_{L^{\infty}}\|\nabla d\|_{L^q}^q
+\varepsilon(\||\nabla^2d||\nabla
d|^{\frac{q-2}{2}}\|_{L^2}^2+\||\nabla|\nabla d|||\nabla
d|^{\frac{q-2}{2}}\|_{L^2}^2)\\
&\quad +C\varepsilon^{-1}\||u||\nabla d|^{\frac{q}{2}}\|_{L^2}^2 
  -\nu q\int_{\Omega}\nabla(|d|^2d)|\nabla d|^{q-2}\nabla
d\,\mathrm{d}x+\nu q\|\nabla d\|_{L^q}^q
 \\
&= \|\operatorname{div} u\|_{L^{\infty}}\|\nabla d\|_{L^q}^q
+\varepsilon(\||\nabla^2d||\nabla
d|^{\frac{q-2}{2}}\|_{L^2}^2+\||\nabla|\nabla d|||\nabla
d|^{\frac{q-2}{2}}\|_{L^2}^2)\\
&\quad +C\varepsilon^{-1}\||u||\nabla d|^{\frac{q}{2}}\|_{L^2}^2 
-\nu q\int_{\Omega}|d|^2\nabla d|\nabla d|^{q-2}\nabla
d\,\mathrm{d}x\\
&\quad -\nu q\int_{\Omega}d\nabla (|d|^2)|\nabla d|^{q-2}\nabla
d\,\mathrm{d}x+\nu q\|\nabla d\|_{L^q}^q 
\\
&=\|\operatorname{div} u\|_{L^{\infty}}\|\nabla d\|_{L^q}^q
+\varepsilon(\||\nabla^2d||\nabla
d|^{\frac{q-2}{2}}\|_{L^2}^2+\||\nabla|\nabla d|||\nabla
d|^{\frac{q-2}{2}}\|_{L^2}^2)
\\
&\quad +C\varepsilon^{-1}\|u\|_{L^{\infty}}^2\|\nabla d\|_{L^q}^q 
-3\nu q\int_{\Omega}|d|^2|\nabla d|^q\,\mathrm{d}x+\nu q\|\nabla d\|_{L^q}^q
\\
&\leq  C(\|\operatorname{div}
u\|_{L^{\infty}}+\varepsilon^{-1}\|u\|_{L^{\infty}}^2+1)\|\nabla
d\|_{L^q}^q+\varepsilon(\||\nabla^2d||\nabla
d|^{\frac{q-2}{2}}\|_{L^2}^2\\
&\quad +\||\nabla|\nabla d|||\nabla d|^{\frac{q-2}{2}}\|_{L^2}^2),
\end{align*}
where we have used the facts that 
$\nabla |d|^2=2|d|\nabla
|d|=2|d|\frac{d\cdot\nabla d}{|d|}=2d\nabla d$, and 
$|\nabla d|\nabla |\nabla d|= \nabla (\nabla d)\cdot \nabla d$ implies
\begin{align*}
&-q\int_{\Omega}\nabla\Delta d |\nabla d|^{q-2}\nabla
d\,\mathrm{d}x\\
&=q\int_{\Omega}|\nabla(\nabla d)|^2|\nabla
d|^{q-2}\,\mathrm{d}x+q\int_{\Omega}\nabla (\nabla d) \cdot \nabla
d\nabla
|\nabla d|^{q-2}\,\mathrm{d}x\\
&= q\int_{\Omega}|\nabla(\nabla d)|^2|\nabla
d|^{q-2}\,\mathrm{d}x+q\int_{\Omega}\nabla |(\nabla d)| |\nabla
d|(q-2)|\nabla d|^{q-2}\nabla|\nabla d|\,\mathrm{d}x\\
&= q\int_{\Omega}|\nabla(\nabla d)|^2|\nabla
d|^{q-2}\,\mathrm{d}x+q(q-2)\int_{\Omega}|\nabla d|^{q-1}|\nabla |\nabla
d||^2\,\mathrm{d}x.
\end{align*}
Taking $\varepsilon$ small enough, it follows form the above
inequality that
\[
\frac{d}{dt}\|\nabla d\|_{L^q}^q\leq C (\|\operatorname{div}
u\|_{L^{\infty}}+\|u\|_{L^{\infty}}^2+1)\|\nabla d\|_{L^q}^q,
\]
which, together with the Gronwall's inequality, \eqref{eq2.1} and
\eqref{eq2.2}, imply
\begin{equation} \label{eq2.18}
\sup_{0\leq t\leq T}\|\nabla d\|_{L^q}\leq C\quad \text{for all }
q\geq 2.
\end{equation}
Letting $q\to\infty$, estimate \eqref{eq2.18} still holds.
Thus the estimate \eqref{eq2.14} holds.

To prove  estimate \eqref{eq2.15}, multiplying the liquid crystal
equation \eqref{eq1.3} by $ d_t$, integrating it over $\Omega$,
and using the boundary condition \eqref{eq1.5}, then
\begin{equation} \label{eq2.19}
\begin{aligned}
&\|d_t\|_{L^2}^2+\frac{\nu}{2}\frac{d}{dt}\|\nabla d\|_{L^2}^2\\
&=-\int_{\Omega}(u\cdot \nabla)d d_t\,\mathrm{d}x
 -\nu\int_{\Omega}f(d)d_t\,\mathrm{d}x \\
&= -\int_{\Omega}(u\cdot \nabla)d
d_t\,\mathrm{d}x-\nu\int_{\Omega}(|d|^2-1)d
d_t\,\mathrm{d}x \\
&\leq  \|u\cdot\nabla
d\|_{L^2}\|d_t\|_{L^2}+(\|d\|_{L^{\infty}}^2+1)\|d\|_{L^2}\|d_t\|_{L^2} \\
&\leq  \frac{1}{4}\|d_t\|_{L^2}^2+C(\|u\cdot\nabla
d\|_{L^2}^2+1)\leq
\frac{1}{4}\|d_t\|_{L^2}^2+C\|u\|_{L^r}^2\|\nabla
d\|_{L^{\frac{2r}{r-2}}}^2+C \\
&\leq \frac{1}{4}\|d_t\|_{L^2}^2+C\|u\|_{L^r}^2\|\nabla
d\|_{L^2}^{\frac{2(r-3)}{r}}\|\nabla^2d\|_{L^2}^{\frac{6}{r}}+C \\
&\leq \frac{1}{4}\|d_t\|_{L^2}^2+\varepsilon\|\nabla^2d\|_{L^2}^2
+C\varepsilon^{-1}\|u\|_{L^r}^{\frac{2r}{r-3}}\|\nabla
d\|_{L^2}^2+C,
\end{aligned}
\end{equation}
where we have used the H\"{o}lder inequality, estimate
\eqref{eq2.5}, the interpolation inequality and the Young
inequality. By applying the standard elliptic regularity result to
the liquid crystal equation \eqref{eq1.3}, then we have
\begin{align*}
\|\nabla^2 d\|_{L^2}^2&\leq C(\|\Delta
d\|_{L^2}^2+\|d_0\|_{H^2}^2) \\
&\leq  C(\|d_t\|_{L^2}^2+\|u\cdot\nabla
d\|_{L^2}^2+\|f(d)\|_{L^2}^2+\|d_0\|_{H^2}^2) \\
&\leq  C(\|d_t\|_{L^2}^2+\|u\|_{L^r}^2\|\nabla
d\|_{L^{\frac{2r}{r-2}}}^2+\|d\|_{L^{\infty}}^{4}
 \|d\|_{L^2}^2+\|d\|_{L^{\infty}}^2\|d\|_{L^2}^2+C)  \\
&\leq C(\|d_t\|_{L^2}^2+\eta\|\nabla^2d\|_{L^2}^2
 +\eta^{-1}\|u\|_{L^r}^{\frac{2r}{r-3}}\|\nabla
d\|_{L^2}^2+C).
\end{align*}
Taking $\eta$ small enough, the above inequality implies that
\begin{equation} \label{eq2.20}
\|\nabla^2d\|_{L^2}^2\leq
C(\|d_t\|_{L^2}^2+\|u\|_{L^r}^{\frac{2r}{r-3}}\|\nabla
d\|_{L^2}^2+C).
\end{equation}
Inserting the inequality \eqref{eq2.20} in \eqref{eq2.19} gives
\[
\|d_t\|_{L^2}^2+\nu\frac{d}{dt}\|\nabla d\|_{L^2}^2
\leq C\|u\|_{L^r}^{\frac{2r}{r-3}}\|\nabla d\|_{L^2}^2+C,
\]
which, together with the Gronwall's inequality,  give estimate
\eqref{eq2.15}.

To prove the estimate \eqref{eq2.16}, we first multiply the liquid
crystal equation \eqref{eq1.3} by $\Delta d_t$,  integrate it
over $\Omega$ and make use of the boundary condition \eqref{eq1.5},
then we have
\begin{equation} \label{eq2.21}
\begin{aligned}
&\frac{d}{dt}\int_{\Omega}\nu|\Delta
d|^2\,\mathrm{d}x+\int_{\Omega}|\nabla d_t|^2\,\mathrm{d}x \\
&= \int_{\Omega} u\cdot \nabla d\Delta
d_t\,\mathrm{d}x+\nu\int_{\Omega}(|d|^2-1)d\Delta
d_t\,\mathrm{d}x \\
&= \sum_{i,j=1}^3\int_{\Omega}u_{i}\partial_{i}d\partial_{j}^2d_t\,\mathrm{d}x-\nu\int_{\Omega}\nabla(|d|^2d)\nabla
d_t\,\mathrm{d}x+\nu\int_{\Omega}\nabla d\nabla d_t\,\mathrm{d}x \\
&= -\sum_{i,j=1}^3\int_{\Omega}\partial_{j}u_{i}\partial_{i}d\partial_{j}
 d_t\,\mathrm{d}x-\sum_{i,j=1}^3\int_{\Omega}
u_{i}\partial_{i}\partial_{j}d\partial_{j}d_t\,\mathrm{d}x\\
&\quad -\nu\int_{\Omega}\nabla(|d|^2d)\nabla d_t\,\mathrm{d}x
 +\nu\int_{\Omega}\nabla d\nabla d_t\,\mathrm{d}x \\
&\leq  C(\|\nabla u\|_{L^2}\|\nabla d\|_{L^{\infty}}\|\nabla
d_t\|_{L^2}+\||u||\nabla^2 d|\|_{L^2}\|\nabla
d_t\|_{L^2}\\
&\quad +\|d\|_{L^{\infty}}^2\|\nabla
d\|_{L^2}\|\nabla d_t\|_{L^2}+\|\nabla
d\|_{L^2}\|\nabla d_t\|_{L^2}) \\
&\leq  \varepsilon \|\nabla
d_t\|_{L^2}^2+C\varepsilon^{-1}(\|\nabla
u\|_{L^2}^2+\||u||\nabla^2 d|\|_{L^2}^2+1) \\
&\leq \varepsilon \|\nabla
d_t\|_{L^2}^2+C\varepsilon^{-1}(\|\nabla
u\|_{L^2}^2+\|u\|_{L^r}^2\|\nabla^2
d\|_{L^{\frac{2r}{r-2}}}^2+1)
 \\
&\leq \varepsilon \|\nabla
d_t\|_{L^2}^2+C\varepsilon^{-1}(\|\nabla
u\|_{L^2}^2+\|u\|_{L^r}^2\|\nabla^2
d\|_{L^2}^{\frac{2(r-3)}{r}}\|\nabla^3d\|_{L^2}^{\frac{6}{r}}+1)
 \\
&\leq \varepsilon \|\nabla
d_t\|_{L^2}^2+C\varepsilon^{-1}(\|\nabla
u\|_{L^2}^2+C\varepsilon^{-2}\|u\|_{L^r}^{\frac{2r}{r-3}}\|\nabla^2
d\|_{L^2}^2+\varepsilon^2\|\nabla^3d\|_{L^2}^2+1),
\end{aligned}
\end{equation}
where we have used the H\"{o}lder inequality, the interpolation
inequality, the Young inequality, estimates \eqref{eq2.5},
\eqref{eq2.10} and \eqref{eq2.14}. Now, differentiating the liquid
crystal equation \eqref{eq1.3} with respect to space variable
gives us
\begin{equation} \label{eq2.22}
\nu\Delta(\nabla d)=\nabla d_t+(\nabla
u\cdot\nabla)d+(u\cdot\nabla)\nabla d+\nu \nabla[(|d|^2-1)d].
\end{equation}
By applying the standard elliptic regularity result to
\eqref{eq2.22}, one can estimate the term $\|\nabla^3 d\|_{L^2}$
as follows
\begin{align*}
\|\nabla^3 d\|_{L^2}&\leq  C(\|\nabla
d_t\|_{L^2}+\|(\nabla
u\cdot\nabla)d\|_{L^2}+\|(u\cdot\nabla)\nabla d\|_{L^2} \\
&\quad +\|\nabla[(|d|^2-1)d]\|_{L^2}+\|d_0\|_{H^3}+\|\nabla
 d\|_{L^2}) \\
&\leq  C\Big(\|\nabla d_t\|_{L^2}+\|\nabla u\|_{L^2}\|\nabla
d\|_{L^{\infty}}+\|u\|_{L^r}\|\nabla^2d\|_{L^{\frac{2r}{r-2}}}
+\|d\|_{L^{\infty}}^2\|\nabla
d\|_{L^2}\\
&\quad +\|\nabla d\|_{L^2}+1\Big) \\
&\leq  C\big(\|\nabla d_t\|_{L^2}+\|\nabla
u\|_{L^2}+\|u\|_{L^r}\|\nabla^2d\|_{L^2}^{\frac{r-3}{r}}\|\nabla^3
d\|_{L^2}^{\frac{3}{r}}+C\big) \\
&\leq C\big( \|\nabla d_t\|_{L^2}+\|\nabla u\|_{L^2}+ \eta
\|\nabla^3
d\|_{L^2}+\eta^{-1}\|u\|_{L^r}^{\frac{r}{r-3}}\|\nabla^2d\|_{L^2}+C\big),
\end{align*}
where  we have used the estimates \eqref{eq2.5}, \eqref{eq2.10},
\eqref{eq2.14}, \eqref{eq2.15} and the Young inequality. Taking
$\eta$ small enough, it follows from the above estimates that
\begin{equation} \label{eq2.23}
\|\nabla^3 d\|_{L^2}\leq C (\|\nabla d_t\|_{L^2}+\|\nabla
u\|_{L^2}+\|u\|_{L^r}^{\frac{r}{r-3}}\|\nabla^2d\|_{L^2}+1).
\end{equation}
Inserting the estimate \eqref{eq2.23} in \eqref{eq2.21}, and
taking $\varepsilon$ small enough, it follows that
\[
\frac{d}{dt}\int_{\Omega}|\Delta
d|^2\,\mathrm{d}x+\int_{\Omega}|\nabla d_t|^2\,\mathrm{d}x\leq
C(\|\nabla
u\|_{L^2}^2+\|u\|_{L^r}^{\frac{2r}{r-3}}\|\nabla^2
d\|_{L^2}^2+1),
\]
which, together with the standard elliptic regularity result
$\|\nabla^2 d\|_{L^2}\leq C(\|\Delta
d\|_{L^2}+\|d_0\|_{H^2})$ and the Gronwall's inequality, give
that the estimate \eqref{eq2.16} holds.

Now, we will prove the estimate \eqref{eq2.17}. By using the
standard elliptic regularity result $\|\nabla d\|_{H^2}\leq
C(\|\nabla^3 d\|_{L^2}+\|\nabla d\|_{L^2}+\|d_0\|_{H^3})$,
it follows that
\begin{align*}
\int_0^T\|\nabla d\|_{H^2}^2\,\mathrm{d}t&\leq
C\int_0^T(\|\nabla^3
d\|_{L^2}^2+\|\nabla d\|_{L^2}^2+\|d_0\|_{H^3}^2)\,\mathrm{d}t \\
&\leq  C\int_0^T(\|\nabla d_t\|_{L^2}^2+\|\nabla
u\|_{L^2}^2+\|u\|_{L^r}^{\frac{2r}{r-3}}\|\nabla^2d\|_{L^2}^2+\|\nabla
d\|_{L^2}^2+C)\,\mathrm{d}t \\
&\leq  C\int_0^T(\|\nabla d_t\|_{L^2}^2+\|\nabla
u\|_{L^2}^2+\|u\|_{L^r}^{\frac{2r}{r-3}}\|d\|_{H^2}^2+C)\,\mathrm{d}t \\
&\leq  C\int_0^T(\|\nabla d_t\|_{L^2}^2+\|\nabla
u\|_{L^2}^2+\|u\|_{L^r}^{\frac{2r}{r-3}}+C)\,\mathrm{d}t\leq C,
\end{align*}
where we have used the estimates \eqref{eq2.23}, \eqref{eq2.10},
\eqref{eq2.15} and \eqref{eq2.16}. This completes the proof of Lemma
\ref{lem2.3}.
\end{proof}


The following two lemmas give some estimates of the velocity $u$.

\begin{lemma}\label{lem2.4}
Under assumption \eqref{eq2.1}, it holds that for $0\leq T<T^{*}$,
\begin{gather} \label{eq2.24}
\sup_{0\leq t\leq T}(\|\nabla u\|_{L^2}^2+\|\nabla
\rho\|_{L^2}^2)+\int_0^T\|\rho^{1/2}u_t\|_{L^2}^2\,\mathrm{d}t\leq
C,\\
   \label{eq2.25}
\int_0^T\|u\|_{H^2}^2\,\mathrm{d}t\leq C.
\end{gather}
\end{lemma}

\begin{proof}
Notice that the momentum equation \eqref{eq1.2} can be rewritten
as
\begin{equation} \label{eq2.26}
\rho u_t+\rho u\cdot \nabla u +\nabla P=\mu\Delta u-\lambda
(\nabla d)^T(\Delta d-f(d)).
\end{equation}
We multiply  \eqref{eq2.26} by $u_t$ and integrate it over
$\Omega$, to obtains the inequality
\begin{equation} \label{eq2.27}
\begin{aligned}
&\frac{\mu}{2}\frac{d}{dt}\|\nabla u\|_{L^2}^2+\int_{\Omega
}\rho|u_t|^2\,\mathrm{d}x \\
&= -\int_{\Omega}\rho u\cdot\nabla u
u_t\,\mathrm{d}x+\int_{\Omega}P\operatorname{div}u_t\,\mathrm{d}x
-\lambda\int_{\Omega}(u_t\cdot\nabla)d(\Delta
d-f(d))\,\mathrm{d}x \\
&\leq \frac{1}{2}\int_{\Omega}\rho|u_t|^2\,\mathrm{d}x+2\int_{\Omega}\rho
|u|^2|\nabla u|^2\,\mathrm{d}x
 +\int_{\Omega}P\operatorname{div}u_t\,\mathrm{d}x\\
&\quad -\lambda\int_{\Omega}(u_t\cdot\nabla)d(\Delta d-f(d))\,\mathrm{d}x.
\end{aligned}
\end{equation}
Combining the mass conservation equation \eqref{eq1.1} and the
assumption \eqref{eq2.1}, it follows that the pressure $P$ satisfies
the  equation
\begin{equation} \label{eq2.28}
P_t+P'(\rho)\nabla\rho\cdot u+P'(\rho)\rho\operatorname{div} u=0,
\end{equation}
then we have
\begin{equation} \label{eq2.29}
\begin{aligned}
\int_{\Omega}P\operatorname{div}u_t\,\mathrm{d}x
&= \frac{d}{dt} \int_{\Omega}P\operatorname{div}u\,\mathrm{d}x
 -\int_{\Omega}P_t\operatorname{div}u\,\mathrm{d}x \\
&= \frac{d}{dt} \int_{\Omega}P\operatorname{div}u\,\mathrm{d}x+
\int_{\Omega}P'(\rho)(\nabla\rho \cdot
u+\rho\operatorname{div}u)\operatorname{div}u\,\mathrm{d}x.
\end{aligned}
\end{equation}
Inserting the above equation \eqref{eq2.29} in  \eqref{eq2.27},
and integrating it over $[0,T]$ give that
\begin{equation} \label{eq2.30}
\begin{aligned}
&\|\nabla u\|_{L^2}^2+\int_0^T\int_{\Omega}\rho|u_t|^2\,\mathrm{d}x\,\mathrm{d}t \\
&\leq  C+C\int_0^T\int_{\Omega}\rho |u|^2|\nabla
u|^2\,\mathrm{d}x\,\mathrm{d}t+C\int_{\Omega}
P(\rho)\operatorname{div}u\,\mathrm{d}x\\
\quad &+C\int_0^T\int_{\Omega} P'(\rho)(\nabla \rho\cdot
u+\rho\operatorname{div}u)\operatorname{div}u\,\mathrm{d}x\,\mathrm{d}t \\
&\quad -\lambda\int_0^T\int_{\Omega}(u_t\cdot\nabla)d(\Delta
d-f(d))\,\mathrm{d}x\,\mathrm{d}t.
\end{aligned}
\end{equation}
Here, we use the facts $\|\nabla u_0\|_{L^2}\leq C<\infty$ and
$\int_{\Omega}P(\rho_0)\operatorname{div}u_0\,\mathrm{d}x\leq
C<\infty$. To estimate the term $\int_0^T\int_{\Omega}\rho
|u|^2|\nabla u|^2\,\mathrm{d}x\,\mathrm{d}t$, we first have
\begin{equation} \label{eq2.31}
\begin{aligned}
\int_{\Omega}\rho |u|^2|\nabla u|^2\,\mathrm{d}x
&\leq C\int_{\Omega}|u|^2|\nabla u|^2\,\mathrm{d}x \\
&\leq C\|u\|^2_{L^r}\|\nabla u\|_{L^{\frac{2r}{r-2}}}^2\leq
C\|u\|_{L^r}^2\|\nabla u\|_{L^2}^{\frac{2r-6}{r}}\|\nabla
u\|_{H^1}^{\frac{6}{r}}  \\
&\leq C \varepsilon\|\nabla
u\|_{H^1}^2+C\varepsilon^{-1}\|u\|_{L^r}^{\frac{2r}{r-3}}\|\nabla
u\|_{L^2}^2,
\end{aligned}
\end{equation}
where we have used the H\"{o}lder's inequality, the interpolation
inequality and the Young inequality. By using the standard elliptic
estimate on \eqref{eq2.26} yields
\begin{align*}
\|u\|_{H^2}
&\leq C(\|\rho u_t\|_{L^2}+\|\rho u\cdot\nabla
u\|_{L^2}+\|\nabla P\|_{L^2}+\|(\nabla d)^T(\Delta d-f(d))\|_{L^2}) \\
&\leq C(\|\rho^{1/2}u_t\|_{L^2}+\|\rho\|_{L^{\infty}}\|
u\cdot\nabla u\|_{L^2} +\|(\nabla
d)^T(d_t+(u\cdot\nabla)d)\|_{L^2} \\
&\leq C(\|\rho^{1/2}u_t\|_{L^2}+\| u\|_{L^r}\|\nabla
u\|_{L^r}^{\frac{2r}{r-2}} +\|(\nabla
d)^T(d_t+(u\cdot\nabla)d)\|_{L^2} \\
&\leq C(\|\rho^{1/2}u_t\|_{L^2}+\|u\|_{L^r}\|\nabla
u\|_{L^2}^{1-\frac{3}{r}}\|\nabla u\|_{H^1}^{\frac{3}{r}}
+\|(\nabla d)^T(d_t+(u\cdot\nabla)d)\|_{L^2} \\
&\leq C\Big(\|\rho^{1/2}u_t\|_{L^2}+\sigma\|\nabla
u\|_{H^1}+\sigma^{-1}\|u\|_{L^r}^{\frac{r}{r-3}}\|\nabla
u\|_{L^2}+\|\nabla \rho\|_{L^2} \\
&\quad +\|\nabla d\|_{L^3}\|d_t\|_{L^{6}}
+\|\nabla d\|_{L^{6}}^2\|u\|_{L^{6}}\Big) \\
&\leq C\Big\|\rho^{1/2}u_t\|_{L^2}+\sigma\|\nabla
u\|_{H^1}+\sigma^{-1}\|u\|_{L^r}^{\frac{r}{r-3}}\|\nabla
u\|_{L^2}+\|\nabla \rho\|_{L^2}+\|d_t\|_{L^{6}}\\
&\quad +\|\nabla u\|_{L^2}+1\Big) \\
&\leq C\Big(\|\rho^{1/2}u_t\|_{L^2}+\sigma\|\nabla
u\|_{H^1}+\sigma^{-1}\|u\|_{L^r}^{\frac{r}{r-3}}\|\nabla
u\|_{L^2}+\|\nabla \rho\|_{L^2}\\
&\quad +\|\nabla d_t\|_{L^2}+\|\nabla u\|_{L^2}+1\Big),
\end{align*}
where we have used the estimates \eqref{eq2.6}, \eqref{eq2.9},
\eqref{eq2.10} and \eqref{eq2.14} in the last inequality. Taking
$\sigma$ small enough yields
\begin{equation} \label{eq2.32}
\|u\|_{H^2}\leq C(\|\rho^{1/2}u_t\|_{L^2}+\|u\|_{L^r}^{\frac{r}{r-3}}\|\nabla
u\|_{L^2}+\|\nabla \rho\|_{L^2}+\|\nabla
d_t\|_{L^2}+\|\nabla u\|_{L^2}+1).
\end{equation}
Inserting estimate \eqref{eq2.32} into \eqref{eq2.31}, it follows
that
\begin{equation} \label{eq2.33}
\begin{aligned}
\int_0^T\int_{\Omega}\rho|u|^2|\nabla
u|^2\,\mathrm{d}x\,\mathrm{d}t 
&\leq \int_0^T(\varepsilon\|\nabla
u\|_{H^1}^2+C\varepsilon^{-1}\|u\|_{L^r}^{\frac{2r}{r-3}}\|\nabla
u\|_{L^2}^2)\,\mathrm{d}t \\
&\leq
C\varepsilon\int_0^T(\|\rho^{1/2}u_t\|_{L^2}^2+\|\nabla
\rho\|_{L^2}^2+\|\nabla
d_t\|_{L^2}^2+1)\,\mathrm{d}t\\
&\quad +C\varepsilon^{-1}\int_0^T(\|u\|_{L^r}^{\frac{2r}{r-3}}+1)\|\nabla
u\|_{L^2}^2\,\mathrm{d}t \\
&\leq
C\varepsilon\int_0^T(\|\rho^{1/2}u_t\|_{L^2}^2+\|\nabla
\rho\|_{L^2}^2)\,\mathrm{d}t\\
&\quad +C\varepsilon^{-1}\int_0^T(\|u\|_{L^r}^{\frac{2r}{r-3}}+1)\|\nabla
u\|_{L^2}^2\,\mathrm{d}t+C,
\end{aligned}
\end{equation}
where we have used the estimate \eqref{eq2.16}. For the term
$-\lambda\int_0^T\int_{\Omega}(u_t\cdot\nabla)d(\Delta
d-f(d))\,\mathrm{d}x\,\mathrm{d}t$, we have
 %\label{eq2.34}
\begin{align}
&-\int_{\Omega}(u_t\cdot\nabla)d(\Delta d-f(d))\,\mathrm{d}x \nonumber\\
&=\sum_{i,j=1}^3(\int_{\Omega}\partial_{j}u_{it}\partial_{i}d\partial_{j}
 d\,\mathrm{d}x
 +\int_{\Omega}u_{it}\partial_{i}\partial_{j}d\partial_{j}d\,\mathrm{d}x)
 +\int_{\Omega}u_t\cdot\nabla d f(d)\,\mathrm{d}x \nonumber\\
&= \sum_{i,j=1}^3(\int_{\Omega}\partial_{j}u_{it}\partial_{i}d\partial_{j}
 d\,\mathrm{d}x
 -\frac{1}{2}\int_{\Omega}\partial_{i}u_{it}|\partial_{j}d|^2\,\mathrm{d}x)
 - \sum_{i=1}^3\int_{\Omega}\partial_{i}u_{it}(\frac{|d|^{4}}{4}-\frac{|d|^2}{2})
 \,\mathrm{d}x \nonumber\\
&= \sum_{i,j=1}^3\Big\{\frac{d}{dt}\int_{\Omega}(\partial_{j}u_{i}\partial_{i}
 d\partial_{j}d-\frac{1}{2}\partial_{i}u_{i}|\partial_{j}d|^2)\,\mathrm{d}x
 -\int_{\Omega}\partial_{j}u_{i}\partial_{i}d_t\partial_{j}d\,\mathrm{d}x \nonumber\\
&\quad  -\int_{\Omega}\partial_{j}u_{i}\partial_{i}d\partial_{j}d_t\,\mathrm{d}x 
 +\int_{\Omega}\partial_{i}u_{i}\partial_{j}d_t\partial_{j}d\,\mathrm{d}x \nonumber\\
&\quad -\frac{d}{dt}\int_{\Omega}(\frac{\partial_{i}u_{i}
 |d|^{4}}{4}-\frac{\partial_{i}u_{i}|d|^2}{2})\,\mathrm{d}x
 +\int_{\Omega}\partial_{i}u_{i}(|d|^3d_t-|d|d_t)\,\mathrm{d}x\Big\} \nonumber\\
&\leq \sum_{i,j=1}^3\frac{d}{dt}\int_{\Omega}(\partial_{j}u_{i}\partial_{i}
 d\partial_{j}d
 -\frac{1}{2}\partial_{i}u_{i}|\partial_{j}d|^2-\frac{1}{4}\partial_{i}u_{i}
 |d|^{4}+\frac{1}{2}\partial_{i}u_{i}|d|^2)\,\mathrm{d}x \nonumber\\
&\quad +C\|\nabla u\|_{L^2}\|\nabla d\|_{L^{\infty}}\|\nabla d_t\|_{L^2}
 +C\|\nabla u\|_{L^2}\|d_t\|_{L^2}(\|d\|_{L^{\infty}}^2+1)\|d\|_{L^{\infty}} \nonumber\\
&\leq \sum_{i,j=1}^3\frac{d}{dt}\int_{\Omega}
 (\partial_{j}u_{i}\partial_{i}d\partial_{j}d
 -\frac{1}{2}\partial_{i}u_{i}|\partial_{j}d|^2-\frac{1}{4}\partial_{i}u_{i}
 |d|^{4}+\frac{1}{2}\partial_{i}u_{i}|d|^2)\,\mathrm{d}x \nonumber\\
&\quad +C\|\nabla u\|_{L^2}^2+C\|\nabla d_t\|_{L^2}^2+C\|d_t\|_{L^2}^2 \nonumber\\
&= \frac{d}{dt}\int_{\Omega}((\nabla u\cdot \nabla d)\nabla d
 -\frac{1}{2}\operatorname{div} u|\nabla d|^2-\frac{1}{4}\operatorname{div}
  u|d|^{4}+\frac{1}{2}\operatorname{div}u|d|^2)\,\mathrm{d}x \nonumber\\
&\quad +C(\|\nabla u\|_{L^2}^2+\|\nabla d_t\|_{L^2}^2+\|d_t\|_{L^2}^2), \label{eq2.34}
\end{align}
where we have used estimate \eqref{eq2.10} in the last inequality.
Integrating \eqref{eq2.34} over $[0,T]$, and by using Lemmas
\ref{lem2.1}--\ref{lem2.3} give
\begin{equation} \label{eq2.35}
\begin{aligned}
&-\lambda\int_0^T\int_{\Omega}(u_t\cdot\nabla)d(\Delta
 d-f(d))\,\mathrm{d}x\,\mathrm{d}t \\
&\leq  C\int_{\Omega}(|\nabla u||\nabla
 d|^2+|\operatorname{div} u|(|\nabla
 d|^2+|d|^{4}+|d|^2))\,\mathrm{d}x\\
&\quad +C\int_0^T(\|\nabla  u\|_{L^2}^2+\|\nabla
 d_t\|_{L^2}^2+\|d_t\|_{L^2}^2)\,\mathrm{d}t \\
&\leq  C\|\nabla u\|_{L^2}\|\nabla d\|_{L^{4}}^2+C\|\nabla
 u\|_{L^2}(\|\nabla
 d\|_{L^{4}}^2+\|d\|_{L^{8}}^{4}+\|d\|_{L^{4}}^2)+C \\
&\leq  \frac{1}{4}\|\nabla u\|_{L^2}^2+C
\end{aligned}
\end{equation}
To estimate the remaining term of the right side of estimate
\eqref{eq2.30}, by using Lemma \ref{lem2.1}, the estimates
\eqref{eq2.9} and \eqref{eq2.10}, we obtain
\begin{equation} \label{eq2.36}
\int_{\Omega}P(\rho)\operatorname{div}u\,\mathrm{d}x\leq
\frac{1}{4}\|\nabla
u\|_{L^2}^2+C\int_{\Omega}|P(\rho)|^2\,\mathrm{d}x\leq\frac{1}{4}\|\nabla
u\|_{L^2}^2+C;
\end{equation}
\begin{equation} \label{eq2.37}
\begin{aligned}
&\int_0^T\int_{\Omega}P'(\rho)(\nabla \rho\cdot
u)\operatorname{div} u\,\mathrm{d}x\,\mathrm{d}t \\
&\leq C\int_0^T\|P'(\rho)\|_{L^3}\|\nabla
 \rho\|_{L^2}\|u\|_{L^{6}}\|\operatorname{div}u\|_{L^{\infty}}\,\mathrm{d}t \\
&\leq C\int_0^T\|P'(\rho)\|_{L^{\infty}}\|\nabla
 \rho\|_{L^2}\|u\|_{L^{6}}\|\operatorname{div}u\|_{L^{\infty}}\,\mathrm{d}t \\
&\leq C\int_0^T\|\nabla \rho\|_{L^2}\|\nabla u\|_{L^2}\|\operatorname{div}u\|_{L^{\infty}}\,\mathrm{d}t \\
&\leq C\int_0^T\|\nabla \rho\|_{L^2}^2\|\operatorname{div}
u\|_{L^{\infty}}\,\mathrm{d}t+C\int_0^T\|\nabla
u\|_{L^2}^2\|\operatorname{div}
u\|_{L^{\infty}}\,\mathrm{d}t \\
&\leq C\int_0^T\|\nabla
\rho\|_{L^2}^2\|\nabla
u\|_{L^{\infty}}\,\mathrm{d}t+\int_0^T\|\nabla
u\|_{L^2}^2\|\nabla
u\|_{L^{\infty}}\,\mathrm{d}t;
\end{aligned}
\end{equation}
\begin{equation}\label{eq2.38}
\int_0^T\int_{\Omega}P'(\rho)\rho |\operatorname{div}
u|^2\,\mathrm{d}x\,\mathrm{d}t\leq
C\int_0^T\|\rho\|_{L^{\infty}}\|\nabla
u\|_{L^2}^2\,\mathrm{d}t\leq C\int_0^T\|\nabla
u\|_{L^2}^2\,\mathrm{d}t\leq C;
\end{equation}
Inserting estimates \eqref{eq2.33}, \eqref{eq2.35}--\eqref{eq2.38}
into \eqref{eq2.30}, we obtain
\begin{equation} \label{eq2.39}
\begin{aligned}
&\|\nabla u\|_{L^2}^2+\int_0^T\int_{\Omega}\rho |u_t|^2\,\mathrm{d}x\,\mathrm{d}t \\
&\leq C\varepsilon\int_0^T(\|\rho^{1/2}u_t\|_{L^2}^2+\|\nabla
\rho\|_{L^2}^2)\,\mathrm{d}t+C\varepsilon^{-1}\int_0^T(\|u\|_{L^r}^{\frac{2r}{r-3}}+1)\|\nabla
u\|_{L^2}^2\,\mathrm{d}t \\
&\quad +C\int_0^T\|\nabla \rho\|_{L^2}^2\|\nabla
u\|_{L^{\infty}}\,\mathrm{d}t+C.
\end{aligned}
\end{equation}
Now, applying the operator $\nabla$ to the mass conservation
equation \eqref{eq1.1}, then multiplying it by $\nabla \rho$ and
integrating it over $\Omega$ yield
\begin{align*}
\frac{d}{dt}\|\nabla\rho\|_{L^2}^2
&= -\int_{\Omega}|\nabla
\rho|^2\operatorname{div}u\,\mathrm{d}x-2\int_{\Omega}\rho\nabla\rho\nabla\operatorname{div}u\,\mathrm{d}x
-2\int_{\Omega}(\nabla \rho\cdot\nabla u)\nabla
\rho\,\mathrm{d}x \\
&\leq  C\|\nabla\rho\|_{L^2}^2\|\nabla u\|_{L^{\infty}}+C\|\nabla
\rho\|_{L^2}\|\nabla\operatorname{div}u\|_{L^2};
\end{align*}
that is,
\begin{equation} \label{eq2.40}
\frac{d}{dt}\|\nabla \rho\|_{L^2}\leq C\|\nabla
\rho\|_{L^2}\|\nabla u\|_{L^{\infty}}+C\|u\|_{H^2}.
\end{equation}
By applying the Gronwall's inequality,  for $ t\in (0,T]$, we obtain
\begin{equation} \label{eq2.41}
\begin{aligned}
\|\nabla \rho\|_{L^2}
&\leq  (\|\rho_0\|_{H^1}+\int_0^{t}\|
u\|_{H^2}\,\mathrm{d}\tau)\text{exp} (C\int_0^{t}\|\nabla
u\|_{L^{\infty}}\,\mathrm{d}\tau) \\
&\leq  C(1+\int_0^{t}\| u\|_{H^2}\,\mathrm{d}\tau).
\end{aligned}
\end{equation}
Inserting \eqref{eq2.32} into \eqref{eq2.41}, for $\xi>0$, gives
\begin{equation} \label{eq2.42}
\begin{aligned}
\xi\|\nabla \rho\|_{L^2}^2
&\leq C\xi(1+\int_0^T(\|\rho^{1/2}u_t\|_{L^2}^2+\|u\|_{L^{\infty}}^2\|\nabla
u\|_{L^2}^2+\|\nabla \rho\|_{L^2}^2+\|\nabla
d_t\|_{L^2}^2)\,\mathrm{d}t) \\
&\leq C\xi(1+\int_0^T(\|\rho^{1/2}u_t\|_{L^2}^2+\|u\|_{L^{\infty}}^2\|\nabla
u\|_{L^2}^2+\|\nabla \rho\|_{L^2}^2)\,\mathrm{d}t),
\end{aligned}
\end{equation}
where we have used the estimate \eqref{eq2.16} again. Combining
\eqref{eq2.39} and \eqref{eq2.42}, and taking $ \varepsilon<\xi$
small enough, it follows that
\begin{equation} \label{eq2.43}
\begin{aligned}
&\|\nabla u\|_{L^2}^2+\frac{\xi}{2}\|\nabla\rho\|_{L^2}^2
 +\frac{1}{2}\int_0^T\int_{\Omega}\rho|u_t|^2\,\mathrm{d}x\,\mathrm{d}t \\
&\leq  C(1+\xi)\int_0^T(\|\nabla u\|_{L^2}^2+\|\nabla \rho\|_{L^2}^2)
(\|\nabla u\|_{L^{\infty}}+\|u\|_{L^r}^{\frac{2r}{r-3}}
 +\|u\|_{L^{\infty}}^2+1)\,\mathrm{d}t+C.
\end{aligned}
\end{equation}
By using the Gronwall's inequality and noticing that the assumption
\eqref{eq2.1}, we can deduce that
\[
\sup_{0\leq t\leq T}(\|\nabla u\|_{L^2}^2+\|\nabla
\rho\|_{L^2}^2)+\int_0^T\int_{\Omega}\rho|u_t|^2\,\mathrm{d}x\,\mathrm{d}t\leq
C.
\]

For \eqref{eq2.25}, it follows from \eqref{eq2.32} that
\[
\int_0^T\|u\|_{H^2}^2\,\mathrm{d}t\leq
C\int_0^T(\|\rho^{1/2}u_t\|_{L^2}^2+\|\nabla
u\|_{L^2}^2+\|\nabla d_t\|_{L^2}^2+1)\,\mathrm{d}t\leq C,
\]
where we have used the estimates \eqref{eq2.10}, \eqref{eq2.16} and
\eqref{eq2.24} in the last inequality above.  This completes the
proof of Lemma \ref{lem2.4}.
\end{proof}



\begin{lemma}\label{lem2.5}
Under the assumption \eqref{eq2.1}, it holds that for $0\leq T<T^{*}$,
\begin{equation} \label{eq2.44}
\begin{aligned}
&\sup_{0\leq t\leq
T}(\|\rho^{1/2}u_t\|_{L^2}^2+\|d_t\|_{L^2}^2+\|\nabla
d_t\|_{L^2}^2)\\
&+\int_0^T(\|\nabla u_t\|_{L^2}^2+\|(\Delta
d-f(d))_t\|_{L^2}^2)\,\mathrm{d}t\leq C.
\end{aligned}
\end{equation}
\end{lemma}

\begin{proof}
Differentiating  equation \eqref{eq2.26} with respect to time,
multiplying the resulting equation by $u_t$,  integrating it over
$\Omega $, and using the equation \eqref{eq1.1} yields
\begin{equation} \label{eq2.45}
\begin{aligned}
&\frac{1}{2}\frac{d}{dt}\int_{\Omega}\rho|u_t|^2\,\mathrm{d}x
 +\mu\int_{\Omega}|\nabla
 u_t|^2\,\mathrm{d}x-\int_{\Omega}P_t\operatorname{div}u_t\,\mathrm{d}x \\
&= -\int_{\Omega}\rho u\cdot\nabla({|u_t|^2}+(u\cdot\nabla )u
 u_t)+\rho (u_t\cdot\nabla)u  u_t\,\mathrm{d}x \\
&\quad -\lambda\int_{\Omega}(u_t\cdot\nabla )d_t(\Delta
 d-f(d))\,\mathrm{d}t 
-\lambda\int_{\Omega}(u_t\cdot\nabla)d(\Delta
 d-f(d))_t\,\mathrm{d}x \\
&= \int_{\Omega}\nabla\rho \cdot
 u{|u_t|^2}+\rho\operatorname{div} u|u_t|^2-\rho u\cdot
 \nabla((u\cdot\nabla )u u_t)-\rho (u_t\cdot\nabla)u
 u_t\,\mathrm{d}x \\
&\quad -\lambda\int_{\Omega}(u_t\cdot\nabla )d_t(\Delta
 d-f(d))\,\mathrm{d}t -\lambda\int_{\Omega}(u_t\cdot\nabla)d(\Delta
 d-f(d))_t\,\mathrm{d}x.
\end{aligned}
\end{equation}

Differentiating the liquid crystal equation \eqref{eq1.3} with
respect to time gives
\begin{equation} \label{eq2.46}
(u_t\cdot\nabla)d=\nu (\Delta d-f(d))_t-d_{tt}-(u\cdot\nabla)d_t.
\end{equation}
Multiplying the above equality with $(\Delta d-f(d))_t$ and
integrating it over $\Omega$, one obtains the equality
\begin{equation} \label{eq2.47}
\begin{aligned}
&\int_{\Omega}(u_t\cdot\nabla)d(\Delta
d-f(d))_t\,\mathrm{d}x \\
&= \int_{\Omega} (\nu|(\Delta d-f(d))_t|^2-d_{tt}\Delta d_t+
d_{tt}f(d)_t-(u\cdot\nabla)d_t(\Delta
d-f(d))_t)\,\mathrm{d}x \\
&= \int_{\Omega}\nu|(\Delta
d-f(d))_t|^2\,\mathrm{d}x+\frac{1}{2}\frac{d}{dt}\|\nabla
d_t\|_{L^2}^2-\int_{\Omega}((u_t\cdot\nabla)d) f(d)_t\,\mathrm{d}x \\
&\quad +\int_{\Omega}\nu f(d)_t(\Delta
d-f(d))_t\,\mathrm{d}x-\int_{\Omega} ((u\cdot \nabla) d_t)(\Delta
d-f(d))_t\,\mathrm{d}x \\
&\quad -\int_{\Omega} ((u\cdot \nabla) d_t)f(d)_t\,\mathrm{d}x.
\end{aligned}
\end{equation}
From  \eqref{eq2.28}, we have
\begin{equation} \label{eq2.48}
\int_{\Omega}P_t\operatorname{div}u_t\,\mathrm{d}x
=-\int_{\Omega}P'(\rho)(\nabla \rho\cdot
u+\rho\operatorname{div}u)\operatorname{div}u_t\,\mathrm{d}x.
\end{equation}
Inserting  \eqref{eq2.47} and \eqref{eq2.48} into \eqref{eq2.45}
gives
\begin{equation} \label{eq2.49}
\begin{aligned}
&\frac{d}{dt}\int_{\Omega}(\frac{1}{2}\rho |u_t|^2+\lambda
|\nabla d_t|^2)\,\mathrm{d}x+\mu\|\nabla
u_t\|_{L^2}^2+\lambda\nu\|(\Delta
d-f(d))_t\|_{L^2}^2 \\
&\leq C\int_{\Omega}\Big(|\nabla\rho||u||u_t|^2+\rho|\operatorname{div}u||u_t|^2+\rho|u||u_t||\nabla
u|^2+\rho |u|^2|u_t||\nabla^2u|\\
&\quad +\rho |u|^2|\nabla u||\nabla u_t|
  +\rho|u_t|^2|\nabla u|\Big)\,\mathrm{d}x 
 +C\int_{\Omega}\Big(|(u_t\cdot\nabla)d f(d)_t|\\
&\quad +|(\Delta d-f(d))_tf(d)_t|+|( u\cdot\nabla )d_t(\Delta
d-f(d))_t|
 +|((u\cdot\nabla )d_t)f(d)_t|\Big)\,\mathrm{d}x \\
&\quad +C\int_{\Omega}|(u_t\cdot\nabla)d_t(\Delta
d-f(d))|\,\mathrm{d}x\\
&\quad +C\int_{\Omega}|P'(\rho)||\nabla
\rho||u||\operatorname{div}u_t|+\rho|P'(\rho)||\operatorname{div}u||\operatorname{div}u_t|\,\mathrm{d}x \\
&=  \sum_{j=1}^{13}J_{j}.
\end{aligned}
\end{equation}
We will estimate $J_{j}$ term by term. In the following
calculations, we will make extensive use of Sobolev embedding,
H\"{o}lder inequality, Lemmas \ref{lem2.1}--\ref{lem2.4}, and
 estimate \eqref{eq2.5}
\begin{gather*}
J_1\leq C\|\nabla
\rho\|_{L^2}\|u_t\|_{L^{6}}^2\|u\|_{L^{6}}\leq C\|\nabla
u_t\|_{L^2}^2\|\nabla u\|_{L^2}\leq \varepsilon\|\nabla
u_t\|_{L^2}^2+C\varepsilon^{-1};
\\
J_2=\int_{\Omega}\rho |\operatorname{div} u||u_t|^2\,\mathrm{d}x
\leq C \|\nabla u\|_{L^{\infty}}\|\rho^{1/2}
u_t\|_{L^2}^2;
\\
\begin{aligned}
J_3&=\int_{\Omega}\rho |u||\nabla u|^2|u_t|\,\mathrm{d}x\leq
C\|\rho\|_{L^{\infty}}\|u\|_{L^{6}}\|u_t\|_{L^{6}}\|\nabla
u\|_{L^2}\|\nabla u\|_{L^{6}}\\
&\leq C\|\nabla u_t\|_{L^2}\|\nabla
u\|_{L^2}^2\|u\|_{H^2}\leq \varepsilon\|\nabla
u_t\|_{L^2}^2+C\varepsilon^{-1}\|u\|_{H^2}^2;
\end{aligned}\\
\begin{aligned}
J_{4}&=\int_{\Omega}\rho|u|^2|\nabla^2 u||u_t|\,\mathrm{d}x\leq
 C \|\rho\|_{L^{\infty}}\|u_t\|_{L^{6}}\|u\|_{L^{6}}^2\|\nabla^2u\|_{L^2}\\
&\leq C\|\nabla u_t\|_{L^2}\|\nabla
u\|_{L^2}^2\|u\|_{H^2}\leq \varepsilon\|\nabla
u_t\|_{L^2}^2+C\varepsilon^{-1}\|u\|_{H^2}^2;
\end{aligned}\\
\begin{aligned}
J_{5}&=\int_{\Omega}\rho|u|^2|\nabla u||\nabla u_t|\,\mathrm{d}x\leq
C\|\rho\|_{L^{\infty}}\|u\|_{L^{6}}^2\|\nabla
u\|_{L^{6}}\|\nabla u_t\|_{L^2}\\
&\leq C\|\nabla u_t\|_{L^2}\|\nabla
u\|_{L^2}^2\|u\|_{H^2}\leq \varepsilon\|\nabla
u_t\|_{L^2}^2+C\varepsilon^{-1}\|u\|_{H^2}^2;
\end{aligned}\\
\begin{aligned}
J_{6}&=\int_{\Omega}\rho |u_t|^2|\nabla u|\,\mathrm{d}x\\
&\leq C\|\rho^{1/2} u_t\|_{L^3}^2\|\nabla u\|_{L^3}\\
&\leq C\|\rho^{1/2}u_t\|_{L^2}\|\rho^{1/2}u_t\|_{L^{6}}\|\nabla
u\|_{L^2}^{1/2}\|\nabla^2u\|_{L^2}^{1/2}\\
&\leq C \|\rho^{1/2}u_t\|_{L^2}\|\rho^{1/2}\|_{L^{\infty}}\|u_t\|_{L^{6}}\|\nabla
u\|_{L^2}^{1/2}\|\nabla^2u\|_{L^2}^{1/2}\\
&\leq C \|\rho^{1/2}u_t\|_{L^2}\|\nabla
u_t\|_{L^2}\|\nabla^2u\|_{L^2}^{1/2}\\
&\leq \varepsilon\|\nabla u_t\|_{L^2}^2+C\varepsilon^{-1}
 \|\rho^{1/2}u_t\|_{L^2}^2\|u\|_{H^2}\\
&\leq \varepsilon\|\nabla u_t\|_{L^2}^2+C\varepsilon^{-1}
 \|\rho^{1/2}u_t\|_{L^2}^2(\|u\|_{H^2}^2+1);
\end{aligned}\\
\begin{aligned}
J_{7}&\leq C\|u_t\|_{L^2}\|\nabla
d\|_{L^3}(\|d\|_{L^{\infty}}^2+1)\|d_t\|_{L^{6}}\\
&\leq C\| \rho^{1/2}u_t\|_{L^2}\|\nabla
d_t\|_{L^2}\leq \varepsilon\|\rho^{1/2}
u_t\|_{L^2}^2+C\varepsilon^{-1} \|\nabla d_t\|_{L^2}^2;
\end{aligned}
\\
J_{8}\leq C\|(\Delta d-f(d))_t\|_{L^2}\|f(d)_t\|_{L^2}\leq
\varepsilon\|(\Delta
d-f(d))_t\|_{L^2}^2+C\varepsilon^{-1}\|d_t\|_{L^2}^2;
\\
\begin{aligned}
J_{9}&=\int_{\Omega}|(u\cdot)d_t(\Delta d- f(d))_t|\,\mathrm{d}x\leq
C\|(\Delta d -f(d))_t\|_{L^2}\||u||\nabla d_t|\|_{L^2}\\
&\leq \varepsilon \|(\Delta
d-f(d))_t\|_{L^2}^2+C\varepsilon^{-1}\|u\|_{L^{\infty}}^2\|\nabla
d_t\|_{L^2}^2;
\end{aligned}\\
\begin{aligned}
J_{10}&=\int_{\Omega}|((u\cdot\nabla)d_t)f(d)_t|\,\mathrm{d}x\\
&=\int_{\Omega}|((u\cdot\nabla)d_t)(|d|^2d-d)_t|\,\mathrm{d}x\\
&\leq C\int_{\Omega}|u||\nabla d_t|(|d|^2|d_t|+|d_t|)\,\mathrm{d}x \\
&\leq C(\|\nabla d_t\|_{L^2}\|d\|_{L^{\infty}}^2\||u||d_t|\|_{L^2}+\|\nabla d_t\|_{L^2}\||u||d_t|\|_{L^2}) \\
&\leq \frac{\varepsilon}{2}\|\nabla
d_t\|_{L^2}^2+C\varepsilon^{-1}\||u||d_t|\|_{L^2}^2
\leq \frac{\varepsilon}{2}\|\nabla d_t\|_{L^2}^2+C\varepsilon^{-1}\|u\|_{L^r}^2\|d_t\|_{L^{\frac{2r}{r-2}}}^2 \\
&\leq \frac{\varepsilon}{2}\|\nabla d_t\|_{L^2}^2
+C\varepsilon^{-1}\|u\|_{L^r}^2\|d_t\|_{L^2}^{\frac{2(r-3)}{r}}\|\nabla d_t\|_{L^2}^{\frac{6}{r}}\\
&\leq \varepsilon\|\nabla d_t\|_{L^2}^2
+C\varepsilon^{-2}\|u\|_{L^r}^{\frac{2r}{r-3}}\|d_t\|_{L^2}^2;
\end{aligned}\\
\begin{aligned}
J_{11}&\leq C\|u_t\|_{L^{6}}\|\nabla d_t\|_{L^2}\|\Delta
d\|_{L^3}+C\|u_t\|_{L^2}\|\nabla
d_t\|_{L^2}(\|d\|_{L^{\infty}}^2+1)\|d\|_{L^{\infty}}\\
&\leq C\|\nabla u_t\|_{L^2}\|\nabla
d_t\|_{L^2}\|d\|_{H^2}^{1/2}\|\nabla
d\|_{H^2}^{1/2}+C\|\rho^{1/2}u_t\|_{L{2}}\|\nabla d_t\|_{L^2}\\
&\leq \varepsilon\|\nabla
u_t\|_{L^2}^2+C\varepsilon^{-1}\|\nabla
d_t\|_{L^2}^2(\|\nabla d\|_{H^2}^2+1)+\varepsilon\|\rho^{1/2}u_t\|_{L^2}^2
+C\varepsilon^{-1}\|\nabla d_t\|_{L^2}^2\\
&\leq \varepsilon\|\nabla u_t\|_{L^2}^2
 +\varepsilon\|\rho^{1/2}u_t\|_{L^2}^2+C\varepsilon^{-1}\|\nabla
d_t\|_{L^2}^2(\|\nabla d\|_{H^2}^2+1);
\end{aligned}\\
\begin{aligned}
J_{12}&\leq C\|\nabla \rho\|_{L^2}\|u\|_{L^{\infty}}\|\nabla
u_t\|_{L^2}\leq C\|\nabla u_t\|_{L^2}\|u\|_{H^2}\\
&\leq \varepsilon\|\nabla
u_t\|_{L^2}^2+C\varepsilon^{-1}\|u\|_{H^2}^2,
\end{aligned}\\
J_{13}\leq C\|\rho\|_{L^{\infty}}\|\nabla u\|_{L^2}\|\nabla
u_t\|_{L^2}\leq \varepsilon\|\nabla
u_t\|_{L^2}^2+C\varepsilon^{-1}.
\end{gather*}
 Substituting all the estimates of $J_{j}$ into
\eqref{eq2.49}, and taking $\varepsilon$ small enough, we obtain
\begin{equation} \label{eq2.50}
\begin{aligned}
&\frac{d}{dt}\int_{\Omega}(\rho |u_t|^2+|\nabla
d_t|^2)\,\mathrm{d}x+\|\nabla u_t\|_{L^2}^2+\|(\Delta
d-f(d))_t\|_{L^2}^2 \\
&\leq  C\|\rho^{1/2}u_t\|_{L^2}^2(\|\operatorname{div}
u\|_{L^{\infty}}+\|u\|_{H^2}^2+C)+C\|\nabla
d_t\|_{L^2}^2(\|u\|_{L^{\infty}}^2\\
&\quad +\|\nabla d\|_{H^2}^2+C) 
 +C\|d_t\|_{L^2}^2(\|u\|_{L^r}^{\frac{2r}{r-3}}+1)+C\|u\|_{H^2}^2+C \\
&\leq C(\|\rho^{1/2}u_t\|_{L^2}^2+\|d_t\|_{L^2}^2+\|\nabla
d_t\|_{L^2}^2)(\|\nabla
u\|_{L^{\infty}}+\|u\|_{L^{\infty}}^2\\
&\quad +\|u\|_{L^r}^{\frac{2r}{r-3}}+\|u\|_{H^2}^2+\|\nabla
d\|_{H^2}^2+C) +C\|u\|_{H^2}^2+C.
\end{aligned}
\end{equation}
To estimate the terms $\|d_t\|_{L^2}$, by multiplying
 \eqref{eq2.46} with $d_t$, it follows that
\begin{equation} \label{eq2.51}
\begin{aligned}
&\frac{1}{2}\frac{d}{dt}\int_{\Omega}|d_t|^2\mathrm{d}x+\nu\|\nabla
d_t\|_{L^2}^2\\
&=-\int_{\Omega} (u_t\cdot\nabla) d d_t\,\mathrm{d}x
 -\int_{\Omega} (u\cdot\nabla) d_t
d_t\,\mathrm{d}x-\nu\int_{\Omega}f(d)_t d_t\,\mathrm{d}x \\
&=-\int_{\Omega} (u_t\cdot\nabla) d d_t\,\mathrm{d}x+\int_{\Omega}
\operatorname{div}u \frac{|d_t|^2}{2}
\,\mathrm{d}x-\nu\int_{\Omega}f(d)_t d_t\,\mathrm{d}x \\
&\leq C(\|u_t\|_{L^{6}}\|\nabla
d\|_{L^3}\|d_t\|_{L^2}+\|\nabla
u\|_{L^{\infty}}\|d_t\|_{L^2}^2+\|d\|_{L^{\infty}}^2\|d_t\|_{L^2}^2+\|d_t\|_{L^2}^2) \\
&\leq \varepsilon\|\nabla
u_t\|_{L^2}^2+C\varepsilon^{-1}\|d_t\|_{L^2}^2+C\|\nabla
u\|_{L^{\infty}}\|d_t\|_{L^2}^2,
\end{aligned}
\end{equation}
where we have used the estimate \eqref{eq2.14} and  the Cauchy
inequality. Combining the estimates \eqref{eq2.50} and
\eqref{eq2.51} together, and let $\varepsilon$  small enough, it
follows that
\begin{align*}
&\frac{d}{dt}\int_{\Omega}(\rho |u_t|^2+|d_t|^2+|\nabla
d_t|^2)\,\mathrm{d}x+\|\nabla u_t\|_{L^2}^2+\|(\Delta
d-f(d))_t\|_{L^2}^2 \\
&\leq C(\|\rho^{1/2}u_t\|_{L^2}^2+\|d_t\|_{L^2}^2+\|\nabla
d_t\|_{L^2}^2)(\|\nabla
u\|_{L^{\infty}}+\|u\|_{L^{\infty}}^2+\|u\|_{L^r}^{\frac{2r}{r-3}}
+\|u\|_{H^2}^2\\
&\quad +\|\nabla d\|_{H^2}^2+C) +C\|u\|_{H^2}^2+C.
\end{align*}
Applying the Gronwall's inequality to the above estimate, we deduce
\begin{equation} \label{eq2.52}
\begin{aligned}
&\sup_{0\leq t\leq T}\int_{\Omega }(\rho
|u_t|^2+|d_t|^2+|\nabla
d_t|^2)\,\mathrm{d}x+\int_0^T\|\nabla
u_t\|_{L^2}^2+\|(\Delta
d-f(d))_t\|_{L^2}^2\,\mathrm{d}t \\
&\leq C\int_0^T(\|u\|_{H^2}^2+1)\,\mathrm{d}t\exp\{C\int_0^T(\|\nabla
u\|_{L^{\infty}}+\|u\|_{L^{\infty}}^2+\|u\|_{L^r}^{\frac{2r}{r-3}}\\
&\quad +\|u\|_{H^2}^2+\|\nabla d\|_{H^2}^2+C)\,\mathrm{d}t\} 
\leq  C,
\end{aligned}
\end{equation}
where we have used estimates \eqref{eq2.17} and \eqref{eq2.25},  and
the assumption \eqref{eq2.1} in the last inequality. This completes
the proof of Lemma \ref{lem2.5}.
\end{proof}

The following lemma gives the higher order norm estimates of $u$,
$d$ and $\rho$.

\begin{lemma}\label{lem2.6}
Under  assumption \eqref{eq2.1}, it holds that for $0\leq T<T^{*}$,
\begin{gather}     \label{eq2.53}
\sup_{0\leq t\leq T}(\|u\|_{H^2}+\|\nabla d\|_{H^2})\leq C;\\
   \label{eq2.54}
\sup_{0\leq t\leq T}\|\nabla \rho\|_{L^{6}}
+\int_0^T\|\nabla^2 u\|_{L^{6}}^2\,\mathrm{d}t\leq C.
\end{gather}
\end{lemma}

\begin{proof}
From estimates \eqref{eq2.6}, \eqref{eq2.24} and \eqref{eq2.44}, we
have
\begin{equation} \label{eq2.55}
\|u\|_{H^2}\leq C (\|\rho^{1/2}u_t\|_{L^2}+\|\nabla
u\|_{L^2}+\|\nabla \rho\|_{L^2}+\|\nabla d_t\|_{L^2})\leq C.
\end{equation}
By the standard elliptic regularity result for the liquid crystal
equation \eqref{eq1.3}, one obtains
\begin{equation} \label{eq2.56}
\begin{aligned}
\|\nabla^3 d\|_{L^2}
&\leq  C(\|\nabla
d_t\|_{L^2}+\|\nabla(u\cdot \nabla d)\|_{L^2}+\|\nabla
f(d)\|_{L^2}+\|d_0\|_{H^3}) \\
&\leq  C(\|\nabla d_t\|_{L^2}+\|\nabla u\|_{L^2}\|\nabla
d\|_{L^{\infty}}+\||u||\nabla^2d|\|_{L^2}\\
&\quad +\|\nabla d\|_{L^2} (\|d\|_{L^{\infty}}^2+1)+\|d_0\|_{H^3}) \\
&\leq  C(\|\nabla d_t\|_{L^2}+\|\nabla
u\|_{L^2}+\||u||\nabla^2d|\|_{L^2}+C).
\end{aligned}
\end{equation}
Notice that
\begin{align*}
\||u||\nabla^2d|\|_{L^2}&\leq  C
\|u\|_{L^{6}}\|\nabla^2d\|_{L^3}\leq C\|\nabla
u\|_{L^2}\|\nabla d\|_{L^{6}}^{1/2}\|\nabla^3
d\|_{L^2}^{1/2} \\
&\leq  C \|\nabla u\|_{L^2}\|\nabla^3 d\|_{L^2}^{1/2}
\leq \eta \|\nabla^3 d\|_{L^2}+C\eta^{-1}\|\nabla
u\|_{L^2}^2 \\
&\leq  \eta \|\nabla^3 d\|_{L^2}+C\eta^{-1},
\end{align*}
where we have used the estimates \eqref{eq2.14} and \eqref{eq2.24}.
Taking $\eta$ small enough, the above estimate and \eqref{eq2.56}
imply that
\[
\||u||\nabla^2d|\|_{L^2}
\leq  C(\|\nabla d_t\|_{L^2}+\|\nabla u\|_{L^2}+C).
\]
Hence
\[
\|\nabla^3 d\|_{L^2}
\leq  C(\|\nabla d_t\|_{L^2}+\|\nabla u\|_{L^2}+C),
\]
and
\begin{equation} \label{eq2.57}
\begin{aligned}
\|\nabla d\|_{H^2}
&\leq  C(\|\nabla^3d\|_{L^2}+\|\nabla d\|_{L^2}) \\
&\leq C(\|\nabla d_t\|_{L^2}+\|\nabla u\|_{L^2}+\|\nabla d\|_{L^2}+C)\leq C,
\end{aligned}
\end{equation}
where we have used the estimates \eqref{eq2.14}, \eqref{eq2.24} and
\eqref{eq2.44} in the last inequality. Combining the estimates
\eqref{eq2.55} and \eqref{eq2.57} above gives the estimate
\eqref{eq2.53}.

To prove \eqref{eq2.54}. Applying the operator $\nabla$ to the mass
conservation equation \eqref{eq1.1}, then multiplying the
resulting equation by $6|\nabla \rho|^{4}\nabla\rho$ and integrating
it over $\Omega$ give
\begin{align*}
&\frac{d}{dt}\|\nabla \rho\|_{L^{6}}^{6}\\
&= -6\int_{\Omega}|\nabla \rho|^{6}\nabla
u\,\mathrm{d}x-\int_{\Omega }\nabla(|\nabla \rho|^{6})\cdot
u\,\mathrm{d}x-6\int_{\Omega}|\nabla\rho|^{6}\operatorname{div}u\,\mathrm{d}x\\
&\quad -6\int_{\Omega}\rho|\nabla\rho|^{4}\nabla\rho\nabla\operatorname{div}u
 \,\mathrm{d}x \\
&=  -6\int_{\Omega}|\nabla \rho|^{6}\nabla
u\,\mathrm{d}x-5\int_{\Omega}|\nabla\rho|^{6}\operatorname{div}u\,\mathrm{d}x
-6\int_{\Omega}\rho|\nabla\rho|^{4}\nabla\rho\nabla\operatorname{div}u\,\mathrm{d}x   \\
&\leq C\|\operatorname{div} u\|_{L^{\infty}}\|\nabla
\rho\|_{L^{6}}^{6}+C\|\rho\|_{L^{\infty}}\|\nabla
\operatorname{div}u\|_{L^{6}}\|\nabla\rho\|_{L^{6}}^{5} \\
&\leq C\|\operatorname{div} u\|_{L^{\infty}}\|\nabla
\rho\|_{L^{6}}^{6}+C\|\nabla
\operatorname{div}u\|_{L^{6}}\|\nabla\rho\|_{L^{6}}^{5};
\end{align*}
that is,
\begin{equation} \label{eq2.58}
\frac{d}{dt}\|\nabla\rho\|_{L^{6}}\leq C\|\nabla
u\|_{L^{\infty}}\|\nabla\rho\|_{L^{6}}+C\|\nabla\operatorname{div}
u\|_{L^{6}}.
\end{equation}
Using the Gronwall's inequality in the above estimate gives
\begin{equation} \label{eq2.59}
\begin{aligned}
\|\nabla \rho\|_{L^{6}}
&\leq (\|\rho_0\|_{W^{1,6}}+C\int_0^T\|\nabla\operatorname{div}u\|_{L^{6}}
 \,\mathrm{d}t)\exp\{C\int_0^T\|\nabla
u\|_{L^{\infty}}\,\mathrm{d}t\} \\
&\leq  C(\int_0^T\|\nabla^2u\|_{L^{6}}\,\mathrm{d}t+1).
\end{aligned}
\end{equation}
Applying the standard elliptic regularity result
$\|\nabla^2u\|_{L^{6}}\leq C\|\Delta u\|_{L^{6}}$, H\"{o}lder
inequality, Sobolev embedding, the estimates \eqref{eq2.10} and
\eqref{eq2.46}, we have
\begin{equation} \label{eq2.60}
\begin{aligned}
\|\nabla^2u\|_{L^{6}}
&\leq  C(\|\rho u_t\|_{L^{6}}+\|\rho
u\cdot\nabla u\|_{L^{6}}+\|\nabla P\|_{L^{6}}+\|(\nabla
d)^T(\Delta d-f(d))\|_{L^{6}}) \\
&\leq C(\|\nabla u_t\|_{L^2}+\|u\|_{L^{\infty}}\|\nabla
u\|_{L^{6}}+\|\nabla \rho\|_{L^{6}}+\|\nabla
d\|_{L^{\infty}}\|\Delta d\|_{L^2}\\
&\quad +\|\nabla d\|_{L^{6}}\|f(d)\|_{L^{\infty}}) \\
&\leq  C(\|\nabla u_t\|_{L^2}+\|u\|_{H^2}^2+\|\nabla
\rho\|_{L^{6}}+\|d\|_{H^3}^2+\|d\|_{H^2}) \\
&\leq  C(\|\nabla u_t\|_{L^2}+\|\nabla \rho\|_{L^{6}}+1),
\end{aligned}
\end{equation}
where we have used estimate \eqref{eq2.53} and the fact that
$\|d\|_{H^3}\leq C(\|\nabla d\|_{H^2}+\|d\|_{L^2})$. Inserting
the estimate \eqref{eq2.60} into \eqref{eq2.59} yields
\begin{align*}
\|\nabla \rho\|_{L^{6}}&\leq  C\int_0^T(\|\nabla
u_t\|_{L^2}+\|\nabla \rho\|_{L^{6}}+1)\,\mathrm{d}t \\
&\leq C\int_0^T(\|\nabla u_t\|_{L^2}^2+\|\nabla
\rho\|_{L^{6}}+1)\,\mathrm{d}t \\
&\leq  C\int_0^T\|\nabla\rho\|_{L^{6}}\,\mathrm{d}t+C,
\end{align*}
where we have used the estimate \eqref{eq2.38}, then applying the
Gronwall's inequality, we have
\begin{equation} \label{eq2.61}
\sup_{0\leq t\leq T}\|\nabla \rho\|_{L^{6}}\leq C.
\end{equation}
From \eqref{eq2.60} and \eqref{eq2.61}, we have
\begin{equation} \label{eq2.62}
\int_0^T\|\nabla^2 u\|_{L^{6}}^2\,\mathrm{d}t\leq
C\Big(\int_0^T\|\nabla u_t\|_{L^2}^2\,\mathrm{d}t+\sup_{0\leq
t\leq T}\|\nabla \rho\|_{L^{6}}^2+C\Big)\leq C.
\end{equation}
It is easy to show that the estimate \eqref{eq2.54} follows from
\eqref{eq2.59} and \eqref{eq2.62} immediately. This completes the
proof of Lemma \ref{lem2.6}.
\end{proof}

\begin{proof}[Proof of Theorem \ref{thm1.2}]
 From the existence result of Theorem \ref{thm1.1}, we know $\|u(t)\|_{H^2}$,
$\|\rho(t)\|_{W^{1,6}}$, $\|d(t)\|_{H^3}$ and
$\|\rho^{1/2}u_t(t)\|_{L^2}$ are all continuous on time
interval $[0,T^{*})$. From the above Lemmas
\ref{lem2.1}--\ref{lem2.6}, we known there holds
\begin{equation} \label{eq2.63}
\begin{aligned}
&(\|u\|_{H^2},\|\rho\|_{W^{1,6}},\|d\|_{H^3},
 \|\rho^{1/2}u_t\|_{L^2})|_{t=T^{*}} \\
&= \lim_{t\to
T^{*}}(\|u(t)\|_{H^2},\|\rho(t)\|_{W^{1,6}},\|d(t)\|_{H^3},
 \|\rho^{1/2}u_t(t)\|_{L^2})\leq C.
\end{aligned}
\end{equation}
Furthermore,
\begin{equation} \label{eq2.64}
\rho^{1/2} u_t+\rho^{1/2} u\cdot\nabla u\in L^{\infty}([0,T^{*}];L^2)
\end{equation}
and for all $T\in (0,T^{*}]$,
\begin{equation} \label{eq2.65}
(\mu \Delta u-\lambda \operatorname{div} (\nabla d\odot\nabla
d-(\frac{1}{2}|\nabla d|^2+F(d))I)-\nabla P)(T)=(\rho
u_t+\rho u\cdot\nabla u)(T)=\sqrt{\rho}g(T)
\end{equation}
where $g(T):= (\rho^{1/2} u_t+\rho^{1/2} u\cdot\nabla u)(\cdot,T)\in L^2 $.
Therefore, from \eqref{eq2.64}
and \eqref{eq2.65}, we can take $(\rho,u,d)|_{t=T^{*}}$ as the
initial data and apply Theorem \ref{thm1.1} to extend the local
strong solution beyond $T^{*}$, this contradicts with the maximality
of $T^{*}$, hence, the assumption \eqref{eq2.1} does not hold.
This completes the proof of Theorem \ref{thm1.2}.
\end{proof}

\subsection*{Acknowledgments}
This research was supported by grant 11171357 from the National Natural
Science Foundation of China.
The author would like to thank  the anonymous referees  for
their careful reading of the manuscript, and for the valuable suggestions.



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