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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2019 (2019), No. 14, pp. 1--13.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu}
\thanks{\copyright 2019 Texas State University.}
\vspace{8mm}}

\begin{document}
\title[\hfilneg EJDE-2019/14\hfil  Compressible nematic liquid crystal flows]
{Low mach number limit of compressible nematic liquid crystal
flows with well-prepared initial data in a 3D bounded domain}

\author[B. Guo, L. Zeng, G. Ni \hfil EJDE-2019/14\hfilneg]
{Boling Guo, Lan Zeng, Guoxi Ni}

\address{Boling Guo \newline
Institute of Applied Physics and Computational Mathematics,
Beijing, 100088, China}
\email{gbl@iapcm.ac.cn}

\address{Lan Zeng (corresponding author) \newline
Graduate School of China Academy of Engineering Physics,
Beijing, 100088,  China}
\email{zenglan1206@126.com}

\address{Guoxi Ni \newline
Institute of Applied Physics and Computational Mathematics,
Beijing, 100088, China}
\email{gxni@iapcm.ac.cn}

\dedicatory{Communicated by Hongjie Dong}

\thanks{Submitted March 15, 2018. Published January 28, 2019.}
\subjclass[2010]{35Q35, 35M33, 75A15, 76N99}
\keywords{Low Mach number limit; compressible nematic liquid crystal flows;
\hfill\break\indent  bounded domain}

\begin{abstract}
 In this article, we consider the low Mach number limit of the compressible
 nematic liquid crystal flows in a 3D bounded domain.
 We establish the uniform estimates with respect to the Mach number for
 the strong solutions with large initial data in a short time interval.
 Consequently, we obtain the convergence of the compressible nematic
 liquid crystal system to the incompressible nematic liquid crystals
 system as the Mach number tends to zero.
\end{abstract}

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

\section{Introduction}

In this article, we establish the uniform estimates of strong solutions
with respect to the Mach number in a bounded domain $\Omega\subset \mathbb{R}^3$
to the compressible nematic liquid crystal flows \cite{Huang}.
\begin{gather}\label{1.1}
 {\rho}_t+\operatorname{div}({\rho}{u})=0, \\
\label{1.2}
( {\rho}{u})_t+\operatorname{div}( {\rho}{u}\otimes{u})
+\frac{1}{\epsilon^2}\nabla P( \rho)-\mu\Delta u
- (\lambda+\mu)\nabla\operatorname{div} u=-\nabla d\cdot\Delta d , \\
\label{1.3}
 {d}_t+ u\cdot\nabla d =\Delta d+|\nabla d|^2d,\quad |d|=1,
\end{gather}
where the unknowns $\rho,u$ and $d$ stand for the density, velocity,
 and the macroscopic of the nematic liquid crystal orientation field,
respectively. The pressure $P(\rho)$ is a $C^1$ function satisfying $P'(\cdot)>0$
 and $P'(0)=0$, such as the well-known $\gamma-$law $P(\rho)=a\rho^\gamma(\gamma>1)$
 which satisfies the assumptions. The parameter $\epsilon>0$ is the scaled Mach number. The physical constants $\mu$ and $\lambda$ denote the shear viscosity
and bulk viscosity of the flow and satisfy
\begin{equation*}
\mu>0, \quad 2\mu+3\lambda\geq0.
\end{equation*}

In fluid mechanics, the Mach number is an important physical quantity to
determine whether the fluid is compressible or incompressible.
If the Mach number is small, the fluid should behave asymptotically like an
incompressible one, provided velocity and viscosity are small.
As a result, the low Mach number limit problem has attracted much attention
in recent years. When $d$ is a constant vector field, the system
\eqref{1.1}-\eqref{1.3} becomes the compressible Navier-Stokes system,
of which the low Mach number limit problem has obtained a great number
of results in the past decades. The readers may refer
to \cite{Dou,Lionss,Masmoudi,Ou, Xiao}, for instance, and the references
therein for details.

Furthermore, a lot of progress on the low Mach number limit for the compressible
nematic liquid crystal equations have been made. In \cite{Ding}, the authors
concerned the low Mach number limit of system \eqref{1.1}-\eqref{1.3} with
periodic boundary conditions. In \cite{Bie}, Bie, Bo, Wang and Yao obtained
global existence and the low Mach number limit for compressible flow of
liquid crystals in critical spaces. Particularly, for the bounded domain case,
the low Mach number limit of weak solutions to the compressible flow of liquid
crystals was proved in \cite{Wang}, and Yang \cite{Yang} firstly studied
the low Mach number limit of the strong solution to system \eqref{1.1}-\eqref{1.3}
provided the initial data small enough. Motivated by the articles mentioned above,
in this paper, we intend to establish the low Mach number limit of the strong
solution for the system \eqref{1.1}-\eqref{1.3} with the lager initial data in
a short time interval. The main difficulty comparing to the periodic case \cite{Ding}
and the whole space case \cite{Bie} is the uniform
high-norm estimates with respect to the Mach number and a time interval
independent of the Mach number.
In a bounded domain, after integrating by parts for the high-order derivatives,
 we have to estimate the boundary term which we will skillfully apply the slip
conditions to control.

The low Mach number fluid can be regarded as a perturbation near the background
isentropic fluid, where the density is usually set to be constant.
Hence, we introduce the density variation by $\sigma^{\epsilon}$ as follows,
\begin{equation*}
\rho^{\epsilon}=1+\epsilon\sigma^{\epsilon},
\end{equation*}
and we will take $P'(1)=1$. Then the non-dimensional system \eqref{1.1}-\eqref{1.3}
can be rewritten as the form
\begin{gather}\label{1.4}
 \sigma^{\epsilon}_t+\operatorname{div}(\sigma^{\epsilon} u^{\epsilon})
+\frac{1}{\epsilon}\operatorname{div} u^{\epsilon}=0, \\
\label{1.5}
\rho^{\epsilon}(u^{\epsilon}_t+u^{\epsilon}\cdot \nabla u^{\epsilon})
+\frac{1}{\epsilon} P'(1+\epsilon\sigma^{\epsilon})\nabla\sigma^{\epsilon}-\mu
\Delta u^{\epsilon}-(\lambda+\mu)\nabla\operatorname{div}u^{\epsilon}
=-\nabla d^{\epsilon}\cdot\Delta d^{\epsilon}, \\
\label{1.6}
 {d}^{\epsilon}_t+ u^{\epsilon}\cdot\nabla d^{\epsilon}
=(\Delta d^{\epsilon}+|\nabla d^{\epsilon}|^2 d^{\epsilon}),\quad |d^{\epsilon}|=1.
\end{gather}
System \eqref{1.4}-\eqref{1.6} is supplemented with the  initial
and boundary value conditions,
\begin{gather}\label{1.7}
(\sigma^{\epsilon},u^{\epsilon},d^{\epsilon})(\cdot,0)
=(\sigma^{\epsilon}_0,u^{\epsilon}_0,d^{\epsilon}_0)(\cdot) \quad\text{in }\Omega,\\
\label{1.8}
u^{\epsilon}\cdot n=0,\quad\operatorname{curl} u^{\epsilon} \times n=0,
\quad \frac{\partial d^{\epsilon}}{\partial n}=0,  \quad\text{on }\partial\Omega,
\end{gather}
where $n$ is the unit outer normal vector to the smooth boundary $\partial\Omega$.

Firstly, the local existence results for problem \eqref{1.4}-\eqref{1.8}
can be established in a similar way as in \cite{Huang}.

\begin{proposition}[(Local solution)]\label{prop1}
 Let $\Omega\subset \mathbb{R}^3$ be a bounded, simply connected domain with
smooth boundary $\partial\Omega$. Assume the initial data
$(\sigma^{\epsilon}_0,u^{\epsilon}_0,d^{\epsilon}_0)$ satisfy the condition
\begin{equation}\label{1.9}
\begin{gathered}
(\partial^k_t\sigma^{\epsilon}(0),\quad
\partial^k_t u^{\epsilon}(0))\in H^{2-k}(\Omega),\quad
\partial^k_t d^{\epsilon}(0)\in H^{3-k}(\Omega),\quad k=0,1,2,\\
\int_{\Omega}\sigma_0dx=0,\quad 1+\epsilon\sigma_0^{\epsilon}\geq m,
\end{gathered}
\end{equation}
for some constant $m>0$.
Moreover, under the compatibility conditions
\begin{equation}\label{1.10}
\begin{gathered}
\partial^k_t u^{\epsilon}(0)\cdot n=0,\quad
n\times\operatorname{curl}u^{\epsilon}_0=n\times\operatorname{curl}u^{\epsilon}_t(0)=0,\quad
\text{on }\partial \Omega,\; k=0,1,\\
\partial^k_t \frac{ \partial d^{\epsilon}(0)}{\partial n}=0\quad
\text{on }\partial \Omega,\; k=0,1.
\end{gathered}
\end{equation}
There exists a constant $T^{\epsilon}>0$ such that the initial
boundary value problem \eqref{1.4}-\eqref{1.8} has a unique solution
 $(\sigma^{\epsilon},u^{\epsilon},d^{\epsilon})$ satisfying
\begin{gather*}
1+\epsilon\sigma^{\epsilon}>0\quad\text{in }\Omega\times(0,T^{\epsilon}),\\
\partial^k_t\sigma^{\epsilon}\in C([0,T^{\epsilon}],H^{2-k}),\\
\partial^k_t u^{\epsilon}\in C([0,T^{\epsilon}],H^{2-k})
 \cap L^2(0,T^{\epsilon};H^{3-k}),\\
\partial ^k_t d^{\epsilon}\in C([0,T^{\epsilon}],H^{3-k})
 \cap  L^2(0,T^{\epsilon};H^{4-k}),\quad k=0,1,2.
\end{gather*}
\end{proposition}

To simplify the statement, we used $\sigma^{\epsilon}_t(0)$ to denote
the quantity $\sigma^{\epsilon}_t|_{t=0}$ which can be obtained from \eqref{1.4}.
The other quantities are defined in a similar way.
For simplicity, we denote
\begin{align*}
M^{\epsilon}(t)&=\sup_{0\leq s\leq t}
\Big\{\|(\sigma^{\epsilon},u^{\epsilon},\nabla d^{\epsilon})(\cdot,s)\|_{H^2}
+\|(\sigma^{\epsilon}_s,u_s^{\epsilon},\nabla d_s^{\epsilon})(\cdot,s)\|_{H^1}
+\big\|\frac{1}{1+\epsilon\sigma(\cdot,s)}\big\|_{L^{\infty}}\\
&\quad +\epsilon\|(\sigma^{\epsilon}_{ss},u_{ss}^{\epsilon},\nabla d_{ss}^{\epsilon})(\cdot,s)\|_{L^2}\Big\}
+\Big\{\int^t_0\Big(\|u^{\epsilon}\|^2_{H^3}+\|u_s^{\epsilon}\|_{H^2} \\
&\quad +\|\epsilon(\sigma^{\epsilon}_{ss},u_{ss}^{\epsilon},\nabla d_{ss}^{\epsilon})
 \|_{H^1}\Big)ds\Big\}^{1/2}.
\end{align*}
Then, we state the main results in this article as follows.

\begin{theorem}\label{thm1}
Assume that $(\sigma^{\epsilon},u^{\epsilon},d^{\epsilon})$ is the solution
 obtained in Proposition \ref{prop1}, and the initial datum
$(\sigma^{\epsilon}_0, u^{\epsilon}_0, d^{\epsilon}_0)$ further satisfies
\begin{equation*}
\|(\sigma^{\epsilon}_0,u^{\epsilon}_0,\nabla d^{\epsilon}_0)\|_{H^2}
+\|(\sigma^{\epsilon}_t,u^{\epsilon}_t,
\nabla d^{\epsilon}_t)(0)\|_{H^1}+\epsilon\|
 (\sigma^{\epsilon}_{tt},u^{\epsilon}_{tt},
\nabla d^{\epsilon}_{tt})(0)\|_{L^2}\leq D_0.
\end{equation*}
 Then there exist two positive constants $T_0$ and $D$ such that
$(\sigma^{\epsilon},u^{\epsilon},d^{\epsilon})$ satisfies the uniform estimates
\begin{equation}\label{1.11}
M^{\epsilon}(T_0 )\leq  D,
\end{equation}
where $D_0,T_0$ and $D$ are constants independent of $\epsilon\in(0,1)$.
\end{theorem}

Based on the above uniform estimates, by applying the Arzel\`a-Ascolis theorem,
we can prove the following convergence result in a standard way.

\begin{theorem}\label{thm2}
Let $(\sigma^{\epsilon},u^{\epsilon},d^{\epsilon})$ be the solution obtained
 in Theorem \ref{thm1}, and the initial data
$(\sigma^{\epsilon}_0,u^{\epsilon}_0,d^{\epsilon}_0)$ further satisfies that
\begin{equation}
\begin{gathered}
(u^{\epsilon}_0,\nabla d^{\epsilon}_0) \to (u_0, \nabla d_0)\quad
\text{strongly in $H^s$ for all $0\leq s<2$ as }\epsilon \to 0,\\
\epsilon\sigma^{\epsilon}_0 \to 0 \quad\text{strongly in $H^s$
for all $0\leq s<1$  as } \epsilon \to 0,
\end{gathered}
\end{equation}
Then $(\rho^{\epsilon},u^{\epsilon},\nabla d^{\epsilon})\to (1,u,\nabla d)$
strongly in $C([0,T_0];H^1)$ as the Mach number $\epsilon\to0$, and
there exists a function $\pi(x,t)$ such that $(u,\pi,d)$ satisfies the
following classical incompressible nematic crystal equations
\begin{equation}\label{1.12}
\begin{gathered}
 u_t+u\cdot \nabla u+\nabla \pi-\mu
\Delta u=-\nabla d\cdot\Delta d,\\
\operatorname{div}u=0,\\
{d}_t+ u\cdot\nabla d=\Delta d+|\nabla d|^2 d,\quad |d|=1,
 \end{gathered}
\end{equation}
with the  initial and boundary conditions
\begin{equation}
\begin{gathered}
(u,d)|_{t=0}=(u_0,d_0), \quad\text{in } \Omega\\
u\cdot n=0,\quad \operatorname{curl}u\times n=0, \quad
\frac{\partial d}{\partial n}=0, \quad\text{on }\partial\Omega.
\end{gathered}
\end{equation}
\end{theorem}

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

We use the methods applied in \cite{Fan,Dou,Fann}.
According to  similar arguments to those in \cite{Dou,Fan}, we know that to
prove Theorem \ref{thm1} it is suffices to prove that
\begin{equation}\label{2.22}
M^{\epsilon}(T_0)\leq C_0(M^{\epsilon}_0)\exp(t^{1/4}C(M^{\epsilon}(t))),
\end{equation}
for all $t\in [0,T^{\epsilon}]$ and for some given nondecreasing continuous
functions $C_0(\cdot)$ and $C(\cdot)$.

For the sake of simplicity, we will drop the superscript $\epsilon$ of
$\sigma^{\epsilon},u^{\epsilon}, d^{\epsilon}$ and so on.
 Moreover, in the following, we will write $M^{\epsilon}(t)$ and
$M^{\epsilon}_0$ as $M$ and $M_0$, respectively. The symbol $C$ denotes
a generic constant and its value may change from line to line.

Firstly, we list some lemmas which will be  used throughout this paper.

\begin{lemma}[\cite{Lions}]\label{lemma2.3}
Let $\Omega$ be a bounded domain in $\mathbb{R}^{N}$ with smooth
 boundary $\partial\Omega$ and outward normal $n$. For any $u\in H^1(\Omega)$
with $u\cdot n=0$ or $u\times n=0$ on $\partial\Omega$, there exists a positive
constant $C$ independent of $u$ such that
\begin{equation}\label{2.3}
\|u\|_{L^2(\Omega)}\leq C(\|\operatorname{div}u\|_{L^2(\Omega)}
+\|\operatorname{curl}u\|_{L^2(\Omega)}),
\end{equation}
where the vorticity $\operatorname{curl}u=(\partial_2u_3-\partial_3u_2,
\partial_3u_1-\partial_1u_3,\partial_1u_2-\partial_2u_1)^{T}$.
\end{lemma}

\begin{lemma}[\cite{Xiao}]\label{lemma2.1}
Let $\Omega$ be a bounded domain in $\mathbb{R}^{N}$ with smooth boundary
$\partial\Omega$ and outward normal $n$. Then, for any $u\in H^1(\Omega)$,
$s\geq1$, there exists a constant $C>0$ independent of $u$, such that
\begin{equation}\label{2.1}
\|u\|_{H^s(\Omega)}
\leq C(\|\operatorname{div} u\|_{H^{s-1}(\Omega)}
+\|\operatorname{curl}u\|_{H^{s-1}(\Omega)}
+\|u\times  n\|_{H^{s-\frac{1}{2}}(\partial\Omega)}+\|u\|_{H^{s-1}(\Omega)}).
\end{equation}
\end{lemma}

\begin{lemma}[\cite{Bourguignon}]\label{lemma2.2}
Let $\Omega$ be a bounded domain in $\mathbb{R}^{N}$ with smooth boundary
 $\partial\Omega$ and outward normal $n$. Then, for any $u\in H^1(\Omega)$,
 $s\geq1$, there exists a constant $C>0$ independent of $u$, such that
\begin{equation}\label{2.2}
\begin{aligned}
\|u\|_{H^s(\Omega)}
&\leq C(\|\operatorname{div}u\|_{H^{s-1}(\Omega)}
+\|\operatorname{curl}u\|_{H^{s-1}(\Omega)}
+\|u\cdot n\|_{H^{s-\frac{1}{2}}(\partial\Omega)} \\
&\quad +\|u\|_{H^{s-1}(\Omega)}).
\end{aligned}
\end{equation}
\end{lemma}

From Lemmas \ref{lemma2.3}, \ref{lemma2.1} and \ref{lemma2.2},  we have
\begin{equation}\label{2.4}
\|\operatorname{curl}u\|_{H^2}
\leq C(\|\Delta\operatorname{curl}u\|_{L^2}+\|u\|_{H^2}),
\end{equation}
for $u\cdot n=0$ and $\operatorname{curl}u\times n=0$ on $\partial \Omega$.
In fact, the latter one gives (see \cite{Bendali,Fann})
\begin{equation*}
\operatorname{curl}\operatorname{curl}u\cdot n=0 \quad \text{on }\partial\Omega.
\end{equation*}

Firstly, we know that $\rho$ and its derivatives always appear as a coefficient
of $u$ and its derivatives. Thus, for simplicity, we use the standard
energy method in \cite{Dou,Ou} to obtain
\begin{equation}\label{4.10}
\|\rho(\cdot,t)\|_{H^2}+\|\rho_t(\cdot,t)\|_{H^1}
 +\|\rho_{tt}(\cdot,t)\|_{L^2}+\|\frac{1}{\rho}(\cdot,t)\|_{L^{\infty}}
\leq C_0(M_0)(\sqrt tC(M)).
\end{equation}

Now we use the method in \cite{Dou,Fan,Yang} to prove a priori estimates on
$\sigma, u$ and $d$. Multiplying \eqref{1.4}-\eqref{1.5} by $\sigma$ and $u$,
respectively, and integrating over $\Omega\times(0,t)$, we obtain
\begin{equation}\label{2.5}
\begin{aligned}
&\frac{1}{2}\|(\sigma,\sqrt{\rho}u)\|^2_{L^2}+
\int^t_0\|(\sqrt{\mu}\operatorname{curl} u,\sqrt{\lambda+2\mu}\operatorname{div}u)\|^2_{L^2}ds\\
&=-\frac{1}{2}\int^t_0\int_{\Omega}\sigma^2\operatorname{div}u\,dx\,ds
+\int^t_0\int_{\Omega}\frac{P'(1)-P'(1+\epsilon\sigma)}{\epsilon}u\nabla\sigma
  \,dx\,ds\\
&\quad -\int^t_0\int_{\Omega}(u\cdot\nabla) d\cdot \Delta d \,dx\,ds
 +\frac{1}{2}\|(\sigma_0,\sqrt{\rho_0}u_0)\|^2_{L^2}\\
&\leq C_0(M_0)+C\int^t_0\|\nabla\sigma\|_{L^2}^2\|\nabla u\|^2ds
 +C\int^t_0\|u\|_{L^6}\|\sigma\|_{L^3}\|\nabla\sigma\|_{L^2}ds\\
&\quad +C\int^t_0\|u\|_{L^6}\|\nabla d\|_{L^3}\|\Delta d\|_{L^2}ds\\
&\leq C_0(M_0)\exp (tC(M)),
\end{aligned}
\end{equation}
where we have used
\begin{equation}\label{3.21}
-\Delta u=-\nabla\operatorname{div}u+\operatorname{curl}\operatorname{curl}u.
\end{equation}
Multiplying \eqref{1.5} by $\nabla\operatorname{div}u$ and integrating the
result over $\Omega\times(0,t)$, we find
\begin{align*}
&(\lambda+2\mu)\int^t_0\|\nabla\operatorname{div}u\|_{L^2}^2ds
 -\frac{1}{\epsilon}\int^t_0\int_{\Omega}\nabla\operatorname{div}
 u\cdot\nabla\sigma \,dx\,ds\\
&=\int^t_0\int_{\Omega}(\rho u_t+\rho u\cdot\nabla u
 +\nabla d\cdot\Delta d)\nabla\operatorname{div}u \,dx\,ds\\
&+\int^t_0\int_{\Omega}
\frac{P'(1+\epsilon\sigma)-P'(1)}{\epsilon}\nabla\sigma\cdot\nabla
 \operatorname{div}u\,dx\,ds\\
&=-\frac{1}{2}\int_{\Omega}\rho(\operatorname{div}u)^2dx
 +\frac{1}{2}\int_{\Omega}\rho(\operatorname{div}u_0)^2dx
 +\frac{1}{2}\int^t_0\int_{\Omega}\rho_t (\operatorname{div}u)^2\,dx\,ds\\
&\quad -\int^t_0\int_{\Omega}\nabla\rho\cdot u_t\operatorname{div}u \,dx\,ds
+\int^t_0\int_{\Omega}(\rho u\cdot\nabla u+\nabla\cdot\Delta d)
 \nabla\operatorname{div}u\,dx\,ds\\
&\quad +\int^t_0\int_{\Omega}
\frac{P'(1+\epsilon\sigma)-P'(1)}{\epsilon}\nabla\sigma\cdot\nabla
 \operatorname{div}u\,dx\,ds.
\end{align*}
Then we obtain
\begin{equation}\label{4.6}
\begin{aligned}
&\int_{\Omega}\rho(\operatorname{div}u)^2dx
 +\int^t_0\|\nabla\operatorname{div}u\|_{L^2}^2ds
 -\frac{1}{\epsilon}\int^t_0\int_{\Omega}\nabla\operatorname{div}u\cdot
 \nabla\sigma \,dx\,ds\\
&\leq C_0(M_0)+ \int^t_0(\|\nabla u\|_{L^2}\|u\|_{H^2}+\|\nabla d\|_{H^2}
 \|\Delta d\|_{L^2} \\
&\quad +\|\sigma\|_{H^2}\|\nabla\sigma\|_{L^2})
 \|\nabla\operatorname{div}u\|_{L^2}ds\\
&\leq C_0(M_0)\exp(tC(M)).
\end{aligned}
\end{equation}
To eliminate the singular term in \eqref{4.6}, we take $\nabla$ to \eqref{1.4}
 and multiply the result by $\nabla\sigma$ to find
\begin{equation}\label{4.7}
\begin{aligned}
&\frac{1}{2}\int_{\Omega}|\nabla\sigma|^2dx
 +\frac{1}{\epsilon}\int^t_0\int_{\Omega}\nabla\sigma\cdot\nabla\operatorname{div}
 u \,dx\,ds\\
&=\frac{1}{2}\int_{\Omega}|\nabla\sigma_0|^2dx-\int^t_0\int_{\Omega}
 \nabla\operatorname{div}(\sigma u)\cdot\nabla\sigma dx\\
&\leq C_0(M_0)\exp(tC(M)).
\end{aligned}
\end{equation}
Summing  \eqref{4.6} and \eqref{4.7}, we obtain
\begin{equation}\label{4.8}
\|(\operatorname{div}u,\nabla\sigma)\|^2_{L^2}
+\int^t_0\|\nabla\operatorname{div}u\|^2_{L^2}ds\leq C_0(M_0)\exp(tC(M)).
\end{equation}
Denote $\omega=\operatorname{curl}u$. Taking $\operatorname{curl}$ to \eqref{1.4},
 we have
\begin{equation}\label{2.8}
 \rho\partial_t\omega+\rho u\cdot\nabla\omega-\mu\Delta\omega= f,
\end{equation}
where $f=\nabla\rho\times\partial_t u+\nabla(\rho u_i)\times\partial_i u
-\nabla\Delta d_j\times \nabla d_j$.
Multiplying \eqref{2.8} by $\omega$, we obtain
\begin{equation}\label{2.9}
\|\operatorname{curl}u\|_{L^2}^2+\int^t_0\int_{\Omega}|\operatorname{curl}\operatorname{curl}u|^2\,dx\,ds
\leq C_0(M_0)\exp(\sqrt tC(M)).
\end{equation}

From Lemma \ref{lemma2.2} and the boundary condition
$\frac{\partial d}{\partial n}=0$ on $\partial\Omega$, we know that
\begin{equation}\label{3.4}
\|\nabla d\|_{H^1}\leq C(\|\operatorname{div}\nabla d\|_{L^2}+\|\operatorname{curl}\nabla d\|_{L^2})=C\|\Delta d\|_{L^2}
\end{equation}
Applying  $\nabla$ to \eqref{1.6}, we have
\begin{equation}\label{3.2}
\nabla d_t-\nabla\Delta d=\nabla(|\nabla d|^2 d)-\nabla(u\cdot\nabla d).
\end{equation}
Multiplying \eqref{3.2} by $\nabla d_t$  and integrating over $\Omega\times(0,t)$,
we obtain
\begin{equation}\label{4.2}
\begin{aligned}
&\frac{1}{2}\int_{\Omega}|\Delta d|^2dx+\int^t_0\int_{\Omega}|\nabla d_t|^2\,dx\,ds\\
&=\frac{1}{2}\int_{\Omega}|\Delta d_0 |^2dx
+\int^t_0\int_{\Omega}(\nabla(|\nabla d|^2d)-\nabla(u\cdot\nabla d ))
 \cdot \nabla d_t \,dx\,ds\\
&\leq \int^t_0\int_{\Omega}(|\nabla d|^3+|\nabla d||\nabla^2d|
 +|\nabla u||\nabla d|+|u||\nabla^2d|)\nabla d_t \,dx\,ds\\
&\leq \int^t_0\| d\|^2_{H^3}(\|\nabla d\|_{L^2}+\|\nabla^2 d\|_{L^2}
 +\|\nabla u\|_{L^2})\|\nabla d_t\|_{L^2}ds\\
&\quad +\int^t_0\|u\|_{H^2}\|\nabla^2 d\|_{L^2}\|\nabla d_t\|_{L^2}ds\\
&\leq C_0(M_0)\exp(tC(M)).
\end{aligned}
\end{equation}
Combining \eqref{3.4} with \eqref{4.2}, we obtain
\begin{equation}\label{3.5}
\|\nabla d\|_{H^1}^2+\int^t_0\int_{\Omega}|\nabla d_t|^2\,dx\,ds
\leq C_0(M_0)\exp(tC(M)).
\end{equation}
Multiplying \eqref{2.8} by $\partial_t\omega-\Delta \omega$, we obtain
\begin{equation}\label{2.10}
\begin{aligned}
&\frac{\mu}{2}\frac{d}{dt}\int_{\Omega}|\operatorname{curl}\operatorname{curl}u|^2dx
 +\int_{\Omega}(\mu|\Delta\omega|^2+\rho|\omega_t|^2)dx\\
&=\int_{\Omega}\rho\omega_t\Delta\omega dx
 -\int_{\Omega}\rho(u\cdot\nabla)\omega(\omega_t-\Delta \omega)dx
 +\int_{\Omega}f(\omega_t-\Delta \omega)dx\\
&=I_1+I_2+I_3,
\end{aligned}
\end{equation}
where by using \eqref{3.21}, we have
\begin{align*}
-\mu\int_{\Omega}\Delta\omega\cdot\omega_t dx
=&\mu\int_{\Omega}\operatorname{curl}\operatorname{curl}\omega\cdot\omega_t dx\\
=&\mu\int_{\Omega}\operatorname{curl}\omega\cdot\operatorname{curl}\omega_t dx
 +\int_{\partial\Omega}(\omega_t\times n)\operatorname{curl}\omega dS\\
=&\frac{\mu}{2}\frac{d}{dt}\int_{\Omega}|\operatorname{curl}\omega|^2dx.
\end{align*}
Then, we estimate $I_1$, $I_2$ and $I_3$ as follows.
\begin{align*}
I_1=&-\int_{\Omega}\rho\omega_t\operatorname{curl}\operatorname{curl}\omega dx\\
=&-\int_{\Omega}\rho\operatorname{curl}\omega\operatorname{curl}\omega_t dx
 -\int_{\Omega}\operatorname{curl}\omega\cdot(\nabla\rho\times \omega_t)dx\\
=&-\frac{1}{2}\frac{d}{dt}\int_{\Omega}\rho|\operatorname{curl}\omega|^2dx
 +C\|\rho\|_{L^{\infty}}\|\operatorname{curl}\omega\|_{L^2}
 \|\nabla\operatorname{curl}\omega\|_{L^2}\\
&+\|\nabla\rho\|_{L^6}\|\operatorname{curl}\omega\|_{L^3}\|\omega_t\|_{L^2}\\
\leq&-\frac{1}{2}\frac{d}{dt}\int_{\Omega}\rho|\operatorname{curl}\omega|^2dx
 +C\|\rho\|_{H^2}\|u\|^2_{H^2}\|u\|_{H^3}\\
&+\|\rho\|_{H^2}\|u_t\|_{H^1}\|u\|_{H^2}^{1/2}\|u\|_{H^3}^{1/2},
\end{align*}
\begin{align*}
|I_2|
\leq& C\|\rho\|_{L^{\infty}}\|\nabla\omega\|_{L^2}(\|\omega_t\|_{L^2}
 +\|\Delta\omega\|_{L^2})\\
\leq& C\|\rho\|_{H^2}\|u\|_{H^2}(\|u\|_t\|_{H^1}+\|u\|_{H^3})
\end{align*}
and
\[
|I_3|\leq C\|f\|_{L^2}(\|u_t\|_{H^1}+\|u\|_{H^3}),
\]
where
\begin{align*}
\|f\|
\leq &C\|(|\nabla\rho||\partial_tu|,|\nabla(\rho u)||\nabla u|,
|\nabla^3d||\nabla d|)\|_{L^2}\\
\leq&C\|\rho\|_{H^2}\|\partial_tu\|^{1/2}_{L^2}\|\partial_t u\|_{H^1}^{1/2}
+C\|\rho\|_{H^2}\|u\|^{1/2}_{H^1}\|u\|^{3/2}_{H^2}+C\|d\|^2_{H^3}\\
\leq& C(M).
\end{align*}
Substituting the above estimates into \eqref{2.10} and integrating over
$(0,t)$, we obtain
\begin{equation}\label{2.11}
\begin{aligned}
&\|\operatorname{curl}\operatorname{curl}u\|^2_{L^2}
 +\int^t_0\int_{\Omega}(|\Delta\operatorname{curl}u|^2
 +|\operatorname{curl}u_t|^2)\,dx\,ds \\
&\leq C_0(M_0)\exp(\sqrt tC(M)).
\end{aligned}
\end{equation}
Applying $\partial_t$ to \eqref{1.4} and \eqref{1.5}, respectively, we obtain
\begin{gather}\label{2.14}
\sigma_{tt}+\frac{1}{\epsilon}\operatorname{div} u_t
=-\operatorname{div}(\sigma u)_t, \\
\label{2.15}
\begin{aligned}
&\rho u_{tt}+\rho u\cdot\nabla u_t-\mu\Delta u_t
-(\lambda+\mu)\nabla\operatorname{div}u_t \\
&=-\rho_tu_t-(\rho u)_t\cdot\nabla u
 -\frac{1}{\epsilon}(P'(1+\epsilon\sigma)\nabla\sigma)_t
 -\nabla d_t\cdot \Delta d-\nabla d\cdot\Delta d_t.
\end{aligned}
\end{gather}
Multiplying \eqref{2.15} by $-\nabla\operatorname{div}u$, we have
\begin{equation}\label{2.12}
\begin{aligned}
&\frac{\lambda+2\mu}{2}\int_{\Omega}|\nabla\operatorname{div}u|^2dx
 -\frac{P'(1)}{\epsilon}\int^t_0\int_{\Omega}\nabla\sigma_t\nabla
 \operatorname{div}u \,dx\,ds\\
&=\frac{\lambda+2\mu}{2}\int_{\Omega}|\nabla\operatorname{div}u_0|^2dx \\
&\quad  + \int^t_0\int_{\Omega}\left(\frac{P'(1+\epsilon\sigma)
 -P'(1)}{\epsilon}\nabla\sigma\right)_t\nabla\operatorname{div}u\,dx\,ds\\
&\quad +\int^t_0\int_{\Omega}(\rho u_{tt}+\rho u\cdot\nabla u_t
 +\rho_t u_t-(\rho u)_t\cdot\nabla u)\nabla\operatorname{div}u\,dx\,ds\\
&\quad +\int^t_0\int_{\Omega}(\nabla d_t\cdot \Delta d+\nabla d\cdot\Delta d_t)
 \nabla\operatorname{div}u\,dx\,ds\\
&=\frac{\lambda+2\mu}{2}\int_{\Omega}|\nabla\operatorname{div}u_0|^2dx+I_4+I_5+I_6.
\end{aligned}
\end{equation}
We estimate $I_4, I_5$ and $I_6$ as follows.
\begin{gather*}
|I_4|\leq C \int^t_0\|\sigma\|_{H^2}\|\nabla\operatorname{div}u\|_{L^2}
(\|\sigma_t\|_{L^3}+\|\nabla\sigma_t\|_{L^2})ds\leq tC(M), \\
|I_5|\leq C\int^t_0\|\rho\|_{H^2}\|u_{tt}\|_{L^2}\|u\|_{H^2}ds+tC(M)\leq  t C(M),\\
|I_6|\leq \int^t_0\|d\|_{H^2}\|\nabla\operatorname{div}u\|_{L^2}(\|\nabla d_t\|_{L^3}
+\|\Delta d_t\|_{L^2})ds\leq tC(M).
\end{gather*}

To eliminate the singular term, we apply $\nabla$ to \eqref{1.4} and multiply
the result by $\nabla\sigma_t$ to obtain
\begin{equation}\label{2.13}
\begin{aligned}
&\int^t_0\|\nabla\sigma_t\|^2_{L^2}ds+\frac{1}{\epsilon}
 \int^t_0\int_{\Omega}\nabla\sigma_t\nabla\operatorname{div}u \,dx\,ds\\
&=-\int^t_0\int_{\Omega}\nabla\operatorname{div}(\sigma u)\cdot
 \nabla\sigma_t \,dx\,ds\leq  tC(M).
\end{aligned}
\end{equation}
Summing  \eqref{2.12} and \eqref{2.13}, we have
\begin{equation}\label{4.1}
\int_{\Omega}|\nabla\operatorname{div}u|^2dx+\int^t_0\|\nabla\sigma_t\|^2_{L^2}ds
\leq C_0(M_0)\exp( tC(M)).
\end{equation}
Applying  $\partial_i$ to \eqref{1.5} and multiplying the result by
$\partial_i\nabla\operatorname{div}u$, we have
\begin{equation}\label{2.18}
\begin{aligned}
&\int^t_0\int_{\Omega}|\partial_i\nabla\operatorname{div}u|^2\,dx\,ds
 -\frac{1}{\epsilon}\int^t_0\int_{\Omega}\partial_i\nabla\sigma
 \cdot\partial_i\nabla\operatorname{div}u \,dx\,ds\\
&\leq \int^t_0\int_{\Omega}(|\nabla(\rho u_t+\rho u\cdot \nabla u)|^2
 +|\nabla(\sigma\nabla\sigma)|^2
 +|\nabla(\nabla d\cdot\Delta d)|^2)\,dx\,ds\\
&\leq tC(M).
\end{aligned}
\end{equation}
To eliminate the singular term, taking $\partial_i\nabla$ to \eqref{1.4}
and multiplying the result by $\partial_i\nabla\sigma$, we obtain
\begin{equation}\label{2.19}
\begin{aligned}
&\frac{1}{2}\int_{\Omega}|\partial_i\nabla\sigma|^2dx
 +\frac{1}{\epsilon}\int^t_0\int_{\Omega}\partial_i\nabla\operatorname{div}u
 \cdot\partial_i\nabla\sigma \,dx\,ds\\
&=\frac{1}{2}\int_{\Omega}|\partial_i\nabla\sigma_0|^2dx
 +\int^t_0\int_{\Omega}\partial_i\nabla(\sigma\operatorname{div}u
 +\nabla\sigma\cdot u)\partial_i\nabla\sigma \,dx\,ds\\
&\leq C_0(M_0)+\int^t_0\|u\|_{H^3}\|\sigma\|^2_{H^2}ds\\
&\leq C_0(M_0)\exp(\sqrt tC(M)).
\end{aligned}
\end{equation}
Summing \eqref{2.18} with \eqref{2.19}, we obtain
\begin{equation}\label{2.20}
\|\nabla^2\sigma\|^2_{L^2}+\int^t_0\int_{\Omega}|\nabla^2
\operatorname{div}u|^2\,dx\,ds\leq C_0(M_0)\exp(\sqrt tC(M)).
\end{equation}

To obtain a priori estimate on $\|d\|_{L^{\infty}_t(H^3)}$, we use
 elliptic regularity theory, \eqref{3.5}, \eqref{2.11} and \eqref{4.1}.
\begin{align*}
&\|\nabla d\|_{H^2} \\
&\leq C\|\nabla d_t\|_{L^2}+C\|\nabla u \cdot\nabla d\|_{L^2}
 +C\|u\cdot\nabla^2 d\|_{L^2}+C\||\nabla d|^3\|_{L^2}
 +\||\nabla d||\nabla^2 d|\|_{L^2}\\
&\leq C \|\nabla d_t\|_{L^2}+C\|\nabla u\|_{L^3}\|\nabla d\|_{L^6}
 +C\|u\|_{L^6}\|\nabla^2 d\|_{L^3}+C\|\nabla d\|_{L^6}\|\nabla^2d\|_{L^3}\\
&\leq C\|\nabla d_t\|_{L^2}+\frac{1}{2}\|\nabla d\|_{H^2}+C_0(M_0)\exp( t C(M)),
\end{align*}
where we used Nirenberg's interpolation inequality and Young inequality.
Then, we conclude that
\begin{equation}\label{3.6}
\|\nabla d\|_{H^2}\leq C\|\nabla d_t\|_{L^2}+C_0(M_0)\exp(tC(M)).
\end{equation}
Hence, to obtain the estimate on $\|\nabla d\|_{L^{\infty}_t(H^2)}$,
it is sufficient to estimate $\|\nabla d_t\|_{L^{\infty}_t(L^2)}$.
Taking $\partial_t$ to \eqref{1.6}, we obtain
\begin{equation}\label{3.7}
d_{tt}+(u\cdot\nabla d)_t=\Delta d_t+(|\nabla d|^2 d)_t.\\
\end{equation}
Multiplying \eqref{3.7} by $-\Delta d_t$ and integrating over $\Omega\times(0,t)$,
we have
\begin{equation}\label{3.8}
\begin{aligned}
&\frac{1}{2}\int_{\Omega}|\nabla d_t|^2dx+\int^t_0\int_{\Omega}|\Delta d_t|^2\,dx\,dt\\
&=\frac{1}{2}\int_{\Omega}|\nabla d_t(0)|^2dx 
 +\int^t_0\int_{\Omega}\Big(u_t\cdot\nabla d+u\cdot \nabla d_t \\
&\quad -|\nabla d|^2d_t  -d\partial_t|\nabla d|^2\Big)\Delta d_t \,dx\,ds\\
&\leq C_0(M_0)+C\int^t_0(\|\nabla d \|_{L^{\infty}}\|u_t\|_{L^2}
 +\|u\|_{L^{\infty}}\|\nabla d_t\|_{L^2})\|\Delta d_t\|_{L^2}ds\\
&\quad +C\int^t_0(\|\nabla d\|^2_{L^{\infty}}\|d_t\|_{L^2}
 +\|\nabla d\|_{L^{\infty}}\|\nabla d_t\|_{L^2})\|\Delta d_t\|_{L^2}ds\\
&\leq C_0(M_0)\exp(t C(M)).
\end{aligned}
\end{equation}
Substituting \eqref{3.8} into \eqref{3.6}, we obtain
\begin{equation}\label{3.9}
\|\nabla d\|_{H^2}\leq C_0(M_0)\exp(t C(M)).
\end{equation}

Then, by using calculations similar to those in \cite{Dou}, we can obtain
 the basic a priori estimates for $\sigma_t,u_t$.
Multiplying \eqref{2.14}, \eqref{2.15} by $\sigma_t$ and $u_t$, respectively
and integrating over $\Omega\times (0,t)$, we obtain
\begin{equation}\label{2.21}
(\|\sigma_t\|^2_{L^2}+\|u_t\|_{L^2}^2)
+\int^t_0\|(\operatorname{curl}u_t,\operatorname{div}u_t)\|^2_{L^2}ds
\leq C_0(M_0)\exp(t C(M)).
\end{equation}

Multiplying \eqref{2.14}, \eqref{2.15} by $-\Delta\sigma_t$ and
$-\nabla\operatorname{div}u_t$, respectively, we obtain
\begin{equation}\label{2.23}
\begin{aligned}
&\frac{1}{2}\int_{\Omega}|\nabla\sigma_t|^2dx
 +\frac{1}{\epsilon}\int^t_0\int_{\Omega}
 \nabla\operatorname{div}u_t\cdot\nabla\sigma_t\,dx\,ds\\
&=\frac{1}{2}\int_{\Omega}|\nabla\sigma_t(0)|^2dx
 +\int^t_0\int_{\Omega}\operatorname{div}(\sigma_tu
 +\sigma u_t)\Delta\sigma_t \,dx\,ds\\
&=\frac{1}{2}\int_{\Omega}|\nabla\sigma_t(0)|^2dx+I_7,
\end{aligned}
\end{equation}
where
\begin{equation}\label{2.24}
\begin{aligned}
I_7=&\int^t_0\int_{\Omega}u\cdot\nabla\sigma_t\Delta\sigma_t \,dx\,ds
 -\int^t_0\int_{\Omega}\nabla(\sigma_t\operatorname{div}u
 +u_t\nabla\sigma+\sigma\operatorname{div}u_t)\,dx\,ds\\
=&-\int^t_0\int_{\Omega}\partial_ju_i\partial_i\sigma_t\partial_j\sigma_t\,dx\,ds
+\frac{1}{2}\int^t_0\int_{\Omega}\operatorname{div}u|\nabla\sigma_t|^2\,dx\,ds\\
&-\int^t_0\int_{\Omega}\nabla(\sigma_t\operatorname{div}u
 +u_t\nabla\sigma+\sigma\operatorname{div}u_t)\,dx\,ds\\
\leq& tC(M)+C(M)\int^t_0\|u\|_{H^3}ds+C(M)\int^t_0\|u_t\|_{H^2}ds\\
\leq& \sqrt t C(M),
\end{aligned}
\end{equation}
and
\begin{equation}\label{2.25}
\begin{aligned}
&\frac{1}{2}\int_{\Omega}\rho(\operatorname{div}u_t)^2dx
 +(\lambda+2\mu)\int^t_0\int_{\Omega}|\nabla\operatorname{div}u_t|^2dx \\
&-\frac{P'(1)}{\epsilon}\int^t_0\int_{\Omega}\nabla\sigma_t\cdot\nabla
 \operatorname{div}u_t\,dx\,ds\\
&=\frac{1}{2}\int_{\Omega}\rho_0(\operatorname{div}u_t(0))^2dx
 +\int^t_0\int_{\Omega}\Big(\frac{P'(1+\epsilon\sigma)-P'(1)}
 {\epsilon}\nabla\sigma\Big)_t\nabla\operatorname{div}u_t\,dx\,ds\\
&\quad +\int^t_0\int_{\Omega}(\frac{\epsilon}{2}\sigma_t(\operatorname{div}u_t)^2
 -\epsilon u_{tt}\cdot\nabla\sigma\operatorname{div}u_t)\,dx\,ds \\
&\quad -\int^t_0\int_{\Omega}(\rho_t u_t+(\rho u\cdot\nabla u)_t)
 \nabla\operatorname{div}u_t\,dx\,ds\\
&\quad +\int^t_0\int_{\Omega}(\nabla d_t\cdot \Delta d
 +\nabla d\cdot\Delta d_t)\nabla\operatorname{div}u_t \,dx\,ds\\
&\leq C_0(M_0)+\sqrt t C(M).
\end{aligned}
\end{equation}
Summing \eqref{2.23}, \eqref{2.24} and \eqref{2.25}, we obtain
\begin{equation}\label{2.26}
\begin{aligned}
&\int_{\Omega}(|\nabla\sigma_t|^2+(\operatorname{div}u_t)^2)dx
+\int^t_0\int_{\Omega}|\nabla\operatorname{div}u_t|^2\,dx\,ds\\
&\leq C_0(M_0)\exp(\sqrt tC(M)).
\end{aligned}
\end{equation}
To complete the estimate of $\|u_t\|_{L_t^{\infty}(H^1)}$, we apply
$\partial_t$ to \eqref{2.8} to obtain
\begin{equation}\label{2.27}
\begin{aligned}
&\rho_t\omega_t+\rho\omega_{tt}+(\rho u)_t\cdot\nabla\omega
 +\rho u\cdot \nabla\omega_t-\mu\Delta\omega_t \\
&=\nabla\rho_t\times u_t+\nabla\rho\times u_{tt}
 +\nabla\Delta(d_j)_t\times\nabla d_j \\
&\quad +\nabla\Delta d_j\times\nabla (d_j)_t+\nabla(\rho u_i)_t\times\partial_i u
 +\nabla(\rho u_i)\times\partial_i u_t.
\end{aligned}
\end{equation}
Multiplying \eqref{2.27} by $\omega_t$ in $L^2(\Omega\times(0,t))$, we deduce that
\begin{equation}\label{2.28}
\begin{aligned}
&\frac{1}{2}\int_{\Omega}\rho|\omega_t|^2dx+\mu\int^t_0
 \int_{\Omega}|{\rm curl }\omega_t|^2\,dx\,ds\\
&=\frac{1}{2}\int_{\Omega}\rho|\omega_t(0)|^2dx
 +\int^t_0\int_{\Omega}\Big(\frac{\epsilon}{2}\sigma_t|\omega_t|^2
 -\epsilon\sigma_t\omega_t-(\rho u)_t\cdot\nabla\omega-\rho u\cdot
 \nabla\omega_t\Big)\omega_t \,dx\,ds\\
&\quad +\epsilon\int^t_0\int_{\Omega}(\nabla\sigma_t\times u_t
 +\nabla\sigma\times u_{tt})\omega_t \,dx\,ds
 +\int^t_0\int_{\Omega}\nabla\Delta(d_j)_t\times \nabla d_k\omega_t \,dx\,ds\\
&\quad +\int^t_0\int_{\Omega}(\nabla\Delta d_j\times\nabla (d_j)_t
 +\nabla(\rho u_i)_t\times\partial_i u
 +\nabla(\rho u_i)\times\partial_i u_t)\omega_t \,dx\,ds\\
&=C_0(M_0)+I_8+I_9+I_{10}+I_{11},
\end{aligned}
\end{equation}
where, by using \eqref{2.21} and integrating by parts, we have
\begin{align*}
-\mu\int^t_0\int_{\Omega}\Delta\omega\cdot \omega_t \,dx\,ds
=&\mu\int^t_0\int_{\Omega}\operatorname{curl}\operatorname{curl}\omega_t\cdot\omega_t
 \,dx\,ds\\
=&\mu\int^t_0\int_{\Omega}|\operatorname{curl} \omega_t|^2\,dx\,ds
 +\mu\int^t_0\int_{\partial\Omega}\operatorname{curl}\omega_t\cdot
 (\omega_t\times n)dS\\
=&\mu\int^t_0\int_{\Omega}|{\rm curl }\omega_t|^2\,dx\,ds.
\end{align*}
We estimate $I_{i}$ $(i=8,9,10,11)$ as follows.
\begin{align*}
I_{10}=&\int^t_0\int_{\Omega}\nabla\Delta(d_j)_t\cdot
 (\nabla d_j\times\omega_t)\,dx\,ds\\
=&-\int^t_0\int_{\Omega}\Delta(d_j)_t\operatorname{div}(\nabla d_j\times\omega_t)
 \,dx\,ds+\int^t_0\int_{\partial\Omega}\Delta(d_j)_t
 \nabla (d_j)_t\cdot(\omega_t\times n)dS\\
=&\int^t_0\int_{\Omega}\Delta(d_j)_t\nabla d_j\cdot\operatorname{curl}
 \omega_t \,dx\,ds\leq \sqrt t C(M).
\end{align*}
With  calculations similar to those in \cite{Dou}, we have
\begin{equation*}
|J_8|+|J_9|+|J_{11}|\leq \sqrt tC(M).
\end{equation*}
Substituting the above estimates into \eqref{2.28}, we obtain
 \begin{equation}\label{2.29}
\int_{\Omega}\rho|\operatorname{curl}u_t|^2dx
+\int^t_0\int_{\Omega}|{\rm curlcurl }u_t|^2\,dx\,ds
\leq C_0(M_0)\exp C(\sqrt tC(M)).
\end{equation}

Now, we have a priori estimate on $\|\nabla d_{t}\|_{L^{\infty}_t(H^1)}$.
Multiplying \eqref{3.7} by $-\Delta d_{tt}$ and integrating over
$\Omega\times(0,t)$, we obtain
\begin{equation}\label{3.10}
\begin{aligned}
&\frac{1}{2}\int_{\Omega}|\Delta d_t|^2dx
 +\int^t_0\int_{\Omega}|\nabla d_{tt}|^2\,dx\,ds\\
&=\frac{1}{2}\int_{\Omega}|\Delta d_t(0)|^2dx
 +\int^t_0\int_{\Omega}[(u\cdot\nabla d)_t-(|\nabla d|^2d)_t]\Delta d_{tt}\,dx\,ds\\
&\leq C_0(M_0)+ C\int^t_0(\|\nabla d\|_{H^2}\|u_t\|_{L^2}
 +\|u\|_{H^2}\|\nabla d_{t}\|_{L^2}\\
&\quad +\|\nabla d\|_{H^2}\|\nabla d_{t}\|_{L^2})\|\Delta d_{tt}\|_{L^2}ds\\
&\leq C_0(M_0)\exp(\sqrt tC(M)).
\end{aligned}
\end{equation}
By the same reasoning  as for \eqref{3.4}, we conclude that
\begin{equation}\label{3.11}
\|\nabla d_t\|^2_{H^1}+\int^t_0\int_{\Omega}|\nabla d_{tt}|^2\,dx\,ds
\leq C_0(M_0)\exp(\sqrt tC(M)).
\end{equation}

Finally, we only need to estimate
$\epsilon \sigma_{tt},\epsilon u_{tt},\epsilon\nabla d_{tt}$
to close the energy estimates.
 Multiplying $\partial_{tt}$ \eqref{1.4}, $\partial_{tt}$\eqref{1.5},
$\partial_{tt}$\eqref{1.6} by
$\epsilon^2\sigma_{tt}$, $\epsilon^2 u_{tt}$ and $\epsilon^2 \Delta d_{tt}$,
respectively, and integrating over $\Omega\times(0,t)$, we  derive that
\begin{equation}\label{3.12}
\epsilon\|(\sigma_{tt},u_{tt},\nabla d_{tt})\|^2_{L^2}
+\epsilon\int^t_0\|(u_{tt},\nabla d_{tt})\|^2_{H^1}ds
\leq C_0(M_0)\exp(t^{1/4}C(M)).
\end{equation}
Collecting the estimates obtained in \eqref{2.5}, \eqref{4.8},
\eqref{2.9}, \eqref{3.5}, \eqref{2.11}, \eqref{4.1}, \eqref{2.20}, \eqref{3.9},
\eqref{2.21}, \eqref{2.26}, \eqref{2.29}, \eqref{3.11}, and \eqref{3.12}, we have
\begin{equation}\label{3.13}
\begin{aligned}
&\|(\sigma,u)\|_{L^2}+\|(\operatorname{div}u,\operatorname{curl}u,
 \operatorname{curl}\operatorname{curl}u,\nabla\operatorname{div}u)\|_{L^2}
 +\|(\nabla\sigma,\nabla d)\|_{H^1}+\|\nabla d\|_{H^2}\\
&+\|(\sigma_t,u_t)\|_{L^2}+\|(\nabla\sigma_t,\operatorname{div}u_t,
 \operatorname{curl}u_t)\|_{L^2}+\|\nabla d_t\|_{H^1}
 +\epsilon\|(\sigma_{tt},u_{tt},\nabla d_{tt})\|_{L^2}\\
&+\|(\operatorname{div}u,\operatorname{curl}u,\operatorname{curl}
 \operatorname{curl}u)\|_{L^2_t(L^2)}+\|(\nabla^2\operatorname{div}u,
 \Delta\operatorname{curl}u)\|_{L^2_t(L^2)}\\
&+\|(\operatorname{div}u_t,\operatorname{curl}u_t,\operatorname{curl}
 \operatorname{curl}u_t,\nabla\operatorname{div}u_t)\|_{L^2_t(L^2)}
 +\epsilon\|(\sigma_{tt},u_{tt},\nabla d_{tt})\|_{L^2_t(H^1)}\\
&\leq C_0(M_0)\exp(t^{1/4}C(M)).
\end{aligned}
\end{equation}
Thus, \eqref{2.22} holds. this completes the proof of Theorem \ref{thm1}.

\subsection*{Acknowledgements}
B. Guo wass supported by NSFC under grant numbers 11731014, 11571254.
G. Ni is supported by NSFC under grant number 11771055 and by the  Science
Challenge Project under grand number TZ2016002.

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\end{document}