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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2018 (2018), No. 06, pp. 1--10.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu}
\thanks{\copyright 2018 Texas State University.}
\vspace{8mm}}

\begin{document}
\title[\hfilneg EJDE-2018/06\hfil Uniform regularity for superfluids]
{Uniform regularity for a mathematical model in superfluidity}

\author[J. Fan, B. Samet, Y. Zhou \hfil EJDE-2018/06\hfilneg]
{Jishan Fan, Bessem Samet, Yong Zhou}

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

\address{Bessem Samet \newline
Department of Mathematics, College of Science,
King Saud University, P.O. Box 2455,
Riyadh 11451, Saudi Arabia}
\email{bessem.samet@gmail.com}

\address{Yong Zhou (corresponding author)\newline
School of Mathematics,
Sun Yat-Sen University,
Zhuhai 518092, China}
\email{zhouyong3@mail.sysu.edu.cn}

\thanks{Submitted May 19, 2017. Published January 4, 2018.}
\subjclass[2010]{35K55, 74A15, 82D50}
\keywords{Ginzburg-Landau equations; superfluids; uniform regularity}

\begin{abstract}
 We prove uniform-in-$\mu$ estimates for a mathematical model in
 superfluidity. Consequently, the limit as $\mu\to0$ can be established.
\end{abstract}

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

\section{Introduction}

Let $\Omega$ be a bounded domain in $\mathbb{R}^3$ with smooth boundary 
$\partial\Omega$, and $\nu$ is the unit outward normal vector to 
$\partial\Omega$. We consider the following mathematical model in 
superfluidity \cite{1}:
\begin{gather}
\gamma\psi_t=\frac{1}{k^2}\Delta\psi-\frac{2i}{k}A\cdot\nabla\psi
-\psi|A|^2+i\beta\psi\operatorname{div} A-\psi(|\psi|^2-1+u),\label{1.1}\\
A_t=\mu\Delta A-|\psi|^2A+\frac{i}{2k}(\psi\nabla\overline\psi
-\overline\psi\nabla\psi)-\nabla u,\label{1.2}\\
u_t-I(u)(|\psi|^2)_t=\Delta u+I(u)\nabla\cdot
\big[-|\psi|^2A+\frac{i}{2k}(\psi\nabla\overline\psi-\overline\psi\nabla\psi)\big],
\label{1.3}
\end{gather}
in $\Omega\times(0,\infty)$ with boundary conditions
\begin{gather}
A\cdot\nu=0,\quad  \operatorname{curl} A\times\nu=0,\quad  \nabla\psi\cdot\nu=0,\label{1.4}\\
\nabla u\cdot\nu=0,\quad \text{on } \partial\Omega\times(0,\infty)\label{1.5}
\end{gather}
and initial data
\begin{equation}
(\psi,A,u)(\cdot,0)=(\psi_0,A_0,u_0)(\cdot)\quad\text{in }
\Omega\subseteq\mathbb{R}^3.\label{1.6}
\end{equation}
The unknowns $\psi, A$, and $u$ are $\mathbb{C}$-valued, $\mathbb{R}^3$-valued, 
and $\mathbb{R}^+$-valued functions, respectively.
 $\overline\psi$ denotes the complex conjugate of 
$\psi, |\psi|^2:=\psi\overline\psi$ is the density of superconducting carriers, 
and $i:=\sqrt{-1}$. $\gamma, k, \mu$, and $\beta:=\frac{1}{k}(k^2\gamma-1)$ 
are positive constants and for simplicity we will take $k=1, \gamma=2$ 
and thus $\beta=1$. The function $I(u)$ is defined by
\begin{equation}
I(u):=\begin{cases}
0,&u<0,\\
1,&u\geq0.
\end{cases} \label{1.7}
\end{equation}

When $u=0$ in \eqref{1.1} and \eqref{1.2}, then the system \eqref{1.1} 
and \eqref{1.2} is the well-known Ginzburg-Landau equations in 
superconductivity with the choice of the Lorentz gauge, which has received 
many studies \cite{2,3,8,9,7,5,4,6}.

In \cite{1}, Berti and Fabrizio proved the global-in-time existence and 
uniqueness of strong solutions when $\psi_0, A_0\in H^1(\Omega)$ and 
$u_0\in L^2(\Omega)$ when \eqref{1.5} is replaced by the homogeneous 
Dirichlet boundary condition $$u=0.$$ However, their proof also works 
here for \eqref{1.5}. But their estimates depend on $\mu$. 
The long-time behavior of the problem \eqref{1.1}-\eqref{1.6} has been 
studied in \cite{10}.

The aim of this paper is to prove global-in-time estimates for solutions
 of \eqref{1.1}-\eqref{1.6} uniform-in $\mu$.
We will prove the following result.

\begin{theorem}\label{th1.1}
Let $0<\mu<1$. Let $\psi_0, u_0\in H^2(\Omega), A_0\in W^{1,q}(\Omega)\ (3<q\leq 6)$, 
with $|\psi_0|\leq 1$ and $u_0\geq 0$ in $\Omega$. Then for any $T>0$, there 
exists a unique strong solution $(\psi_\mu, A_\mu, u_\mu)$ of \eqref{1.1}-\eqref{1.6} 
such that
\begin{equation} \label{1.8}
\begin{gathered}
\psi_\mu\in L^\infty(0,T;H^2)\cap L^2(0,T;H^3),
 \partial_t\psi_\mu\in L^\infty(0,T;L^2)\cap L^2(0,T;H^1), \\
A_\mu\in L^\infty(0,T;W^{1,q}), \partial_tA_\mu\in L^\infty(0,T;L^2),\\
u_\mu\in L^\infty(0,T;H^2)\cap L^2(0,T;W^{2,q}),
\partial_tu_\mu\in L^\infty(0,T;L^2)\cap L^2(0,T;H^1)
\end{gathered}
\end{equation}
with the corresponding norms that are uniformly bounded with respect to $\mu>0$.
\end{theorem}

\begin{remark}\label{re1.1} \rm
As soon as the uniform-in-$\mu$ a priori estimates are established,
 we can easily show by standard compactness arguments that the limit as 
$\mu\to 0$ for \eqref{1.1}-\eqref{1.6} exists.
\end{remark}

We now collect several vector identities and the Gauss-Green formula 
which will be used in the rest of the paper.

\begin{lemma}[{\cite[Theorem 2.1]{11}}] \label{le1.1}
 Let $\Omega$ be a regular bounded domain in $\mathbb{R}^3, A:\Omega\to\mathbb{R}^3$ 
be a sufficiently smooth vector field, and let $1<p<\infty$. Then, the following 
identity holds.
\begin{equation}
\begin{aligned}
&-\int_\Omega\Delta A\cdot A|A|^{p-2}dx \\
&=\int_\Omega|A|^{p-2}|\nabla A|^2dx
 +4\frac{p-2}{p^2}\int_\Omega\big|\nabla|A|^{p/2}\big|^2dx \\
&\quad -\int_{\partial\Omega}|A|^{p-2}(\nu\cdot\nabla)A\cdot A\,dS.
\end{aligned} \label{1.9}
\end{equation}
Moreover, recalling the vector identity:
\begin{equation}
(\nu\cdot\nabla)A\cdot A=(A\cdot\nabla)A\cdot\nu
+(\operatorname{curl} A\times\nu)\cdot A\label{1.10}
\end{equation}
for a sufficiently smooth vector field $A$, we can also deduce that
\begin{equation}
\begin{aligned}
-\int_\Omega\Delta A\cdot A|A|^{p-2}dx
&=\int_\Omega|A|^{p-2}|\nabla A|^2dx
 +4\frac{p-2}{p^2}\int_\Omega\big|\nabla|A|^{p/2}\big|^2dx \\
&\quad-\int_{\partial\Omega}|A|^{p-2}(A\cdot\nabla)A\cdot\nu \,d S \\
&\quad -\int_{\partial\Omega}|A|^{p-2}(\operatorname{curl} A\times\nu)\cdot A \,d S.
\end{aligned}\label{1.11}
\end{equation}
\end{lemma}

\begin{lemma}[{\cite[Lemma 2.2]{12}}] \label{le1.2}
Assume that $A$ is sufficiently smooth, satisfying the boundary condition 
\eqref{1.4} on $\partial\Omega$. Then, the following identity for $B:=\operatorname{curl} A$ holds.
\begin{equation}
-\frac{\partial B}{\partial\nu}\cdot B
=(\epsilon_{1jk}\epsilon_{1\beta\gamma}+\epsilon_{2jk}
\epsilon_{2\beta\gamma} +\epsilon_{3jk}
\epsilon_{3\beta\gamma})B_jB_\beta\partial_k\nu_\gamma\label{1.12}
\end{equation}
on $\partial\Omega$, where $\epsilon_{ijk}$ denotes the totally 
anti-symmetric tensor such that $(a\times b)_i=\epsilon_{ijk}a_jb_k$.
\end{lemma}

\begin{lemma}[{\cite[ Lemma 7.44]{13}, \cite[Corollary 1.7]{14}}]
Let a smooth and bounded open set $\Omega$ be given and let $1<p<\infty$. 
Then the following inequality holds. There exists a constant $C>0$, such that
\begin{equation}
\|f\|_{L^p(\partial\Omega)}
\leq C\|f\|_{L^p(\Omega)}^{1-\frac{1}{p}}\|f\|_{W^{1,p}(\Omega)}^{1/p} \label{1.13}
\end{equation}
for any $f\in W^{1,p}(\Omega)$.
\end{lemma}

\begin{lemma}[\cite{15}] \label{le1.4}
 There exists a constant $C>0$, such that
\begin{equation}
\|f\|_{W^{1,p}(\Omega)}\leq C(\|f\|_{L^p(\Omega)}
+\|\operatorname{div} f\|_{L^p(\Omega)}+\|\operatorname{curl} f\|_{L^p(\Omega)}) \label{1.14}
\end{equation}
for any $1<p<\infty$ and all $f\in W^{1,p}(\Omega)$.
\end{lemma}

When $A$ satisfies $A\cdot\nu=0$ on $\partial\Omega$, we will also use the identity
\begin{equation}
(A\cdot\nabla)A\cdot\nu=-(A\cdot\nabla)\nu\cdot A\quad
\text{on } \partial\Omega\label{1.15}
\end{equation}
for any sufficiently smooth vector field $A$.

\begin{lemma}[\cite{16}] \label{le1.5}
Let $u$ be a smooth solution of the  problem 
\begin{gather*}
u_t-\Delta u=\operatorname{div} g\quad\text{in }\Omega\times(0,T),\\
\frac{\partial u}{\partial\nu}=0\quad\text{on }\partial\Omega\times(0,T),\\
u(\cdot,0)=0\quad\text{in }\Omega
\end{gather*}
for any given $T>0$. Then there exists a constat $C>0$, such that
\begin{equation}
\|\nabla u\|_{L^q(0,T;L^p(\Omega))}
\leq C\|g\|_{L^q(0,T;L^p(\Omega))}.\label{1.16}
\end{equation}
with $1<p,q<\infty$.
\end{lemma}

\section{Proof of main results}

This section is devoted to the proof of Theorem \ref{th1.1}. 
Since it has been proved that the problem \eqref{1.1}-\eqref{1.6} 
has a unique global-in-time strong solution \cite{1}, we only need 
to prove a priori estimates \eqref{1.8} uniformly in $\mu$.
 From now on, we drop the subscript $\mu$.

It follows from \eqref{1.3}, \eqref{1.5} and \eqref{1.6} that
\begin{equation}
u\geq 0\quad\text{if } u_0\geq 0\label{2.1}
\end{equation}
and thus $I(u)\equiv 1$ in \eqref{1.3}.
Then we have
\begin{gather}
2f_t = \Delta f-f(f^2-1+u+V_s^2)\quad\text{in } \Omega\times(0,\infty),\label{2.2}\\
\nabla f\cdot\nu=0\quad\text{on }\partial\Omega\times(0,\infty),\label{2.3}\\
f=f_0\quad\text{in } \Omega\label{2.4}
\end{gather}
where 
$$
f:=|\psi|, \psi:=fe^{i\phi}, V_s:=-A+\nabla\phi.
$$
It follows from \eqref{2.2}, \eqref{2.3}, and \eqref{2.4} that
\begin{equation}
|\psi|\leq 1\quad\text{in } \Omega\times(0,\infty).\label{2.5}
\end{equation}

Testing \eqref{1.1} by $\psi$, taking the real part and using \eqref{2.1}, 
we see that 
$$
\frac{d}{dt}\int|\psi|^2dx+\int|i\nabla\psi+\psi A|^2dx+\int|\psi|^4dx
+\int u|\psi|^2dx=\int|\psi|^2dx,
$$ 
which gives
\begin{equation}
\int_0^T\int|i\nabla\psi+\psi A|^2\,dx\,dt\leq C.\label{2.6}
\end{equation}
Here and in what follows, $C$ will denote a generic positive constant 
independent of $\mu>0$.

Testing \eqref{2.2} by $f$ and using \eqref{2.1}, we find that 
$$
\frac{d}{dt}\int f^2dx+\int|\nabla f|^2dx+\int f^2(f^2+u+V_s^2)dx=\int f^2dx,
$$ 
which reads 
\begin{equation}
\int_0^T\int\big|\nabla|\psi|^2\big|^2\,dx\,dt\leq C.\label{2.7}
\end{equation}

We denote $w:=u-|\psi|^2$.
Testing \eqref{1.3} by $w$, using \eqref{2.5}, \eqref{2.6} and \eqref{2.7}, we get
\begin{align*}
\frac{1}{2}\frac{d}{dt}\int w^2dx+\int|\nabla w|^2dx
&\leq \int\big|\nabla|\psi|^2\big|\cdot|\nabla w|dx
 +\int|i\nabla\psi+\psi A|\cdot|\nabla w|dx\\
&\leq \int(\big|\nabla|\psi|^2\big|^2+|i\nabla\psi
 +\psi A|^2)dx+\frac{1}{2}\int|\nabla w|^2dx,
\end{align*}
which gives
\begin{gather*}
\|w\|_{L^\infty(0,T;L^2)}+\|w\|_{L^2(0,T;H^1)}\leq C,\label{2.8}\\
\|u\|_{L^\infty(0,T;L^2)}+\|u\|_{L^2(0,T;H^1)}\leq C.\label{2.9}
\end{gather*}
Testing \eqref{1.2} by $A$, using \eqref{2.5}, \eqref{2.6} and \eqref{2.9}, 
we deduce that
\begin{align*}
&\frac{1}{2}\frac{d}{dt}\int A^2dx+\mu\int(|\operatorname{div} A|^2+|\operatorname{curl} A|^2)dx\\
&\leq \int|i\nabla\psi+\psi A||\psi||A|dx+\int|\nabla u||A|dx\\
&\leq \int|A|^2dx+\int|i\nabla\psi+\psi A|^2dx+\int|\nabla u|^2dx,
\end{align*}
which implies
\begin{equation}
\|A\|_{L^\infty(0,T;L^2)}+\sqrt\mu\|A\|_{L^2(0,T;H^1)}\leq C.\label{2.10}
\end{equation}
Obviously,  inequalities \eqref{2.5}, \eqref{2.6} and \eqref{2.10} imply
\begin{equation}
\|\psi\|_{L^2(0,T;H^1)}\leq C.\label{2.11}
\end{equation}

Testing \eqref{1.1} by $-\Delta\overline\psi$, taking the real part, and 
using \eqref{2.5}, we have
\begin{equation}
\begin{aligned}
&\frac{d}{dt}\int|\nabla\psi|^2dx+\int|\Delta\psi|^2dx \\
&\leq2\int|A||\nabla\psi||\Delta\psi|dx+\int|\psi||A|^2|\Delta\psi|dx \\
&\quad+\int|\psi||\operatorname{div} A||\Delta\psi|dx+\int|\psi|(|\psi|^2+1+|u|)|\Delta\psi|dx \\
&\leq C(\|A\|_{L^4}\|\nabla\psi\|_{L^4}+\|A\|_{L^4}^2
 +\|\operatorname{div} A\|_{L^2}+\|u\|_{L^2}+1)\|\Delta\psi\|_{L^2} \\
&\leq C(\|A\|_{L^4}\|\Delta\psi\|_{L^2}^{1/2}+\|A\|_{L^4}^2
 +\|\operatorname{div} A\|_{L^2}+\|u\|_{L^2}+1)\|\Delta\psi\|_{L^2} \\
&\leq \frac{1}{16}\|\Delta\psi\|_{L^2}^2+C\|A\|_{L^4}^4
 +C\|\operatorname{div} A\|_{L^2}^2+C\|u\|_{L^2}^2+C,
\end{aligned}\label{2.12}
\end{equation}
where we have used the Gagliardo-Nirenberg inequality:
\begin{equation}
\|\nabla\psi\|_{L^4}^2\leq C\|\psi\|_{L^\infty}\|\Delta\psi\|_{L^2}.\label{2.13}
\end{equation}

Testing \eqref{1.2} by $|A|^2A$, using \eqref{1.11}, \eqref{1.15}, \eqref{2.5} 
and \eqref{1.13}, we derive
\begin{equation}
\begin{aligned}
&\frac{1}{4}\frac{d}{dt}\int|A|^4dx+\mu\int|A|^2|\nabla A|^2dx
 +\frac{\mu}{2}\int\big|\nabla|A|^2\big|^2dx \\
&=\mu\int_{\partial\Omega}|A|^2(A\cdot\nabla)\nu\cdot A\,dS
 -\int\nabla w\cdot|A|^2Adx-\int\nabla|\psi|^2\cdot|A|^2 A dx \\
&\quad-\int\operatorname{Re}\{(i\nabla\psi+\psi A)\overline\psi\}|A|^2A dx \\
&\leq \|\nabla\nu\|_{L^\infty}\mu\int_{\partial\Omega}|A|^4d S
 +(\|\nabla w\|_{L^4}+3\|\nabla\psi\|_{L^4})\|A\|_{L^4}^3 \\
&\leq C\mu\int|A|^4dx+\frac{1}{16}\mu\int\big|\nabla|A|^2\big|^2dx
 +C\|A\|_{L^4}^4 \\
&\quad+\epsilon\int(|\nabla w|^4+|\nabla\psi|^4)dx
\end{aligned} \label{2.14}
\end{equation}
for any $0<\epsilon<1$.

It follows from \eqref{1.2}, \eqref{1.4} and \eqref{1.5} that \cite{7}:
\begin{equation}
\nabla\operatorname{div} A\cdot\nu=0\quad\text{on } \partial\Omega\times(0,\infty).\label{2.15}
\end{equation}
Taking $\operatorname{div}$ to \eqref{1.2}, testing by $\operatorname{div} A$, using \eqref{2.5} 
and \eqref{2.13}, we obtain
\begin{equation}
\begin{aligned}
&\frac{1}{2}\frac{d}{dt}\int|\operatorname{div} A|^2dx+\mu\int|\nabla\operatorname{div} A|^2dx
 +\int|\psi|^2|\operatorname{div} A|^2dx \\
&\leq \int|A|\big|\nabla|\psi|^2\big|\cdot|\operatorname{div} A|dx
 +2\int|\Delta\psi||\operatorname{div} A|dx+\int|\Delta(w+|\psi|^2)||\operatorname{div} A|dx \\
&\leq C\|A\|_{L^4}\|\nabla\psi\|_{L^4}\|\operatorname{div} A\|_{L^2}
 +C(\|\Delta w\|_{L^2}+\|\Delta\psi\|_{L^2}
 +\|\nabla\psi\|_{L^4}^2)\|\operatorname{div} A\|_{L^2} \\
&\leq C\|\operatorname{div} A\|_{L^2}^2+C\|A\|_{L^4}^4+\epsilon\|\Delta w\|_{L^2}^2
 +\epsilon\|\Delta\psi\|_{L^2}^2
\end{aligned}\label{2.16}
\end{equation}
for any $0<\epsilon<1$.
We rewrite \eqref{1.3} as 
\begin{equation}
w_t-\Delta w=\Delta|\psi|^2+\nabla\cdot
\big[-|\psi|^2A +\frac{i}{2}(\psi\nabla\overline\psi-\overline\psi\nabla\psi)\big].\label{2.17}
\end{equation}
Testing \eqref{2.17} by $-\Delta w$, using \eqref{2.5} and \eqref{2.13}, we have
\begin{equation}
\begin{aligned}
&\frac{1}{2}\frac{d}{dt}\int|\nabla w|^2dx+\int|\Delta w|^2dx \\
&\leq C\int(|\Delta\psi|+|\nabla\psi|^2+|\operatorname{div} A|+|\nabla\psi||A|)|\Delta w|dx \\
&\leq \epsilon\int|\Delta w|^2dx+C_0\int|\Delta\psi|^2dx+C\int|\operatorname{div} A|^2dx
+C\int|A|^4dx.
\end{aligned} \label{2.18}
\end{equation}

By Lemma \ref{le1.5}, from \eqref{2.17} and \eqref{2.5} it follows that
\begin{equation}
\int_0^T\int|\nabla w|^4\,dx\,dt\leq C+C\int_0^T\int|\nabla\psi|^4\,dx\,dt
+C\int_0^T\int|A|^4\,dx\,dt. \label{2.19}
\end{equation}

Integrating $2C_0\times \eqref{2.12}+\eqref{2.14}+\eqref{2.16}+\eqref{2.18}$
 over $(0,T)$, using \eqref{2.19}, \eqref{2.5} and \eqref{2.13}, taking 
$\epsilon$ small enough, we have
\begin{gather}
\|\psi\|_{L^\infty(0,T;H^1)}+\|\psi\|_{L^2(0,T;H^2)}\leq C,\label{2.20}\\
\|A\|_{L^\infty(0,T;L^4)}+\|\operatorname{div} A\|_{L^\infty(0,T;L^2)}
 +\sqrt\mu\|\nabla\operatorname{div} A\|_{L^2(0,T;L^2)}\leq C,\label{2.21}\\
\|w\|_{L^\infty(0,T;H^1)}+\|w\|_{L^2(0,T;H^2)}\leq C.\label{2.22}
\end{gather}
Testing \eqref{1.2} with $\operatorname{curl}^2A$, and utilize the fact $\operatorname{curl}\nabla=0$, \eqref{2.5},
 \eqref{2.13}, \eqref{1.14}, \eqref{2.20}, and \eqref{2.21}, we have
\begin{align*}
&\frac{1}{2}\frac{d}{dt}\int|\operatorname{curl} A|^2dx+\mu\int|\operatorname{curl}^2A|^2dx\\
&=-\operatorname{Re}\int\operatorname{curl}[(i\nabla\psi+\psi A)\overline\psi]\operatorname{curl} A dx\\
&=-\operatorname{Re}\int(i\nabla\overline\psi\times\nabla\psi
 +|\psi|^2\operatorname{curl} A+\nabla|\psi|^2\times A)\operatorname{curl} A dx\\
&\leq C(\|\nabla\psi\|_{L^4}^2+\|\nabla\psi\|_{L^4}\|A\|_{L^4})\|\operatorname{curl} A\|_{L^2}\\
&\leq C\|\Delta\psi\|_{L^2}^2+C\|A\|_{L^4}^4+C\|\operatorname{curl} A\|_{L^2}^2,
\end{align*}
which gives
\begin{equation}
\|A\|_{L^\infty(0,T;H^1)}+\sqrt\mu\|A\|_{L^2(0,T;H^2)}\leq C.\label{2.23}
\end{equation}

On the other hand,  from \eqref{1.1}, \eqref{1.2}, \eqref{1.3}, \eqref{2.20},
 \eqref{2.21}, \eqref{2.22} and \eqref{2.23} it follows that
\begin{equation}
\|\psi_t\|_{L^2(0,T;L^2)}+\|A_t\|_{L^2(0,T;L^2)}+\|w_t\|_{L^2(0,T;L^2)}
+\|u_t\|_{L^2(0,T;L^2)}\leq C.\label{2.24}
\end{equation}

Now, taking $\partial_t$ to \eqref{1.1}, testing then by $\overline\psi_t$, 
taking the real part, and employing \eqref{2.5}, \eqref{2.23}, and \eqref{2.22}, 
we have
\begin{equation}
\begin{aligned}
&\frac{d}{dt}\int|\psi_t|^2dx+\int|\nabla\psi_t|^2dx+\int A^2|\psi_t|^2dx \\
&\leq 2\int|A_t||\nabla\psi||\psi_t|dx+2\int|A||\nabla\psi_t||\psi_t|dx+2\int|A||A_t||\psi_t|dx \\
&\quad +\big|\int\psi\overline\psi_t\operatorname{div} A_t dx\big|+C\int|\psi_t|^2dx+C\int|u||\psi_t|^2dx+C\int|u_t||\psi_t|dx \\
&\leq C\|A_t\|_{L^2}\|\nabla\psi\|_{L^3}\|\psi_t\|_{L^6}+C\|A\|_{L^6}\|\nabla\psi_t\|_{L^2}\|\psi_t\|_{L^3} \\
&\quad+C\|A\|_{L^6}\|A_t\|_{L^2}\|\psi_t\|_{L^3}
 +\big|\int A_t(\psi\nabla\overline\psi_t+\overline\psi_t\nabla\psi)dx\big| \\
&\quad +C\int|\psi_t|^2dx+C\|u\|_{L^3}\|\psi_t\|_{L^3}^2+C\|u_t\|_{L^2}\|\psi_t\|_{L^2} \\
&\leq C\|A_t\|_{L^2}\|\nabla\psi\|_{L^3}(\|\psi_t\|_{L^2}+\|\nabla\psi_t\|_{L^2})
+C\|\nabla\psi_t\|_{L^2}\|\psi_t\|_{L^3} \\
&\quad +C\|A_t\|_{L^2}\|\psi_t\|_{L^3}+C\|A_t\|_{L^2}\|\nabla\psi_t\|_{L^2}\\
&\quad +C\int|\psi_t|^2dx
+C\|\psi_t\|_{L^3}^2+C\|u_t\|_{L^2}^2 \\
&\leq \frac{1}{16}\|\nabla\psi_t\|_{L^2}^2+C\|\psi_t\|_{L^2}^2+C\|A_t\|_{L^2}^2+C\|u_t\|_{L^2}^2
+C\|\nabla\psi\|_{L^3}^2\|A_t\|_{L^2}^2.
\end{aligned} \label{2.25}
\end{equation}

Taking $\partial_t$ to \eqref{1.2}, testing then by $A_t$, and making use of 
\eqref{2.5} and \eqref{2.23}, we have
\begin{equation}
\begin{aligned}
&\frac{1}{2}\frac{d}{dt}\int|A_t|^2dx+\mu\int(|\operatorname{div} A_t|^2
+|\operatorname{curl} A_t|^2)dx+\int|\psi|^2|A_t|^2dx \\
&\leq C\int(|\nabla\psi_t|+|\nabla\psi||\psi_t|+|\psi_t||A|)|A_t|dx+C\int|u_t||\operatorname{div} A_t|dx \\
&\leq C(\|\nabla\psi_t\|_{L^2}+\|\nabla\psi\|_{L^6}\|\psi_t\|_{L^3}+\|\psi_t\|_{L^3}\|A\|_{L^6})\|A_t\|_{L^2}
+C\|u_t\|_{L^2}^2 \\
&\quad +C\|\operatorname{div} A_t\|_{L^2}^2 \\
&\leq \frac{1}{16}\|\nabla\psi_t\|_{L^2}^2+C\|\nabla\psi\|_{L^6}^2\|A_t\|_{L^2}^2+C\|\psi_t\|_{L^2}^2 \\
&\quad +C\|A_t\|_{L^2}^2+C\|u_t\|_{L^2}^2+C\|\operatorname{div} A_t\|_{L^2}^2.
\end{aligned} \label{2.26}
\end{equation}

Taking $\operatorname{div}$ to \eqref{1.2}, testing by $\operatorname{div} A_t$, using \eqref{2.5}, \eqref{2.13}, 
\eqref{2.23}, \eqref{2.22} and \eqref{2.20}, we have
\begin{align*}
&\frac{\mu}{2}\frac{d}{dt}\int|\nabla\operatorname{div} A|^2dx+\int|\operatorname{div} A_t|^2dx\\
&\leq C\int(|\operatorname{div} A|+|A||\nabla\psi|+|\Delta u|+|\Delta\psi|
 +|\nabla\psi|^2)|\operatorname{div} A_t|dx\\
&\leq \frac{1}{2}\int|\operatorname{div} A_t|^2dx+C\int|\operatorname{div} A|^2dx+C\int|\Delta u|^2dx \\
&\quad +C\int|\Delta\psi|^2dx+C\|A\|_{L^6}^2\|\nabla\psi\|_{L^3}^2\\
&\leq \frac{1}{2}\int|\operatorname{div} A_t|^2dx+C+C\int(|\Delta u|^2+|\Delta\psi|^2)dx
\end{align*}
which implies
\begin{equation}
\int_0^T\int|\operatorname{div} A_t|^2\,dx\,dt\leq C.\label{2.27}
\end{equation}

Combining \eqref{2.25} and \eqref{2.26}, using \eqref{2.20}, \eqref{2.24},
 \eqref{2.27} and the Gronwall inequality, we arrive at
\begin{gather}
\|\psi_t\|_{L^\infty(0,T;L^2)}+\|\psi_t\|_{L^2(0,T;H^1)}\leq C,\label{2.28}\\
\|A_t\|_{L^\infty(0,T;L^2)}\leq C.\label{2.29}
\end{gather}
It follows from \eqref{1.1}, \eqref{2.5}, \eqref{2.20}, \eqref{2.23}, \eqref{2.22},
 and \eqref{2.28} that
\begin{equation}
\|\psi\|_{L^\infty(0,T;H^2)}+\|\psi\|_{L^2(0,T;H^3)}\leq C.\label{2.30}
\end{equation}

Taking $\operatorname{curl}$ to \eqref{1.2}, testing by $|\operatorname{curl} A|^{q-2}\operatorname{curl} A\ (3<q\leq 6)$, 
using $\operatorname{curl}\nabla=0$, \eqref{1.9}, \eqref{1.12}, \eqref{2.23} and \eqref{2.30}, 
we have
\begin{align*}
&\frac{1}{q}\frac{d}{dt}\int|\operatorname{curl} A|^q dx+\mu\int|\operatorname{curl} A|^{q-2}|\nabla\operatorname{curl} A|^2dx
 +4\frac{q-2}{q^2}\mu\int\left|\nabla|\operatorname{curl} A|^{q/2}\right|^2dx\\
&\leq C\mu\int_{\partial\Omega}|\nabla\nu||\operatorname{curl} A|^q\,dS \\
&\quad -\operatorname{Re}\int(i\nabla\overline\psi+|\psi|^2\operatorname{curl} A
+\nabla|\psi|^2\times A)|\operatorname{curl} A|^{q-2}\operatorname{curl} A dx\\
&\leq C\mu\int_{\partial\Omega}|\operatorname{curl} A|^q\,dS
+C\int(|\nabla\psi|^2+|\nabla\psi||A|)|\operatorname{curl} A|^{q-1}dx\\
&\leq C\mu\int|\operatorname{curl} A|^q dx+2\frac{q-2}{q^2}\mu
\int\left|\nabla|\operatorname{curl} A|^{q/2}\right|^2dx\\
&\quad +C\|\nabla\psi\|_{L^{2q}}^2\|\operatorname{curl} A\|_{L^q}^{q-1}
+C\|\nabla\psi\|_{L^\infty}\|A\|_{L^q}\|\operatorname{curl} A\|_{L^q}^{q-1},
\end{align*}
whence, 
$$
\frac{d}{dt}\|\operatorname{curl} A\|_{L^q}\leq C\|\operatorname{curl} A\|_{L^q}
+C\|\nabla\psi\|_{L^{2q}}^2+C\|\nabla\psi\|_{L^\infty},
$$ 
which implies
\begin{equation}
\|\operatorname{curl} A\|_{L^\infty(0,T;L^q)}\leq C\quad (3<q\leq 6).\label{2.31}
\end{equation}

Taking $\partial_t$ to \eqref{1.3}, testing then by $w_t$, using \eqref{2.5}, 
\eqref{2.28}, \eqref{2.30}, and \eqref{2.23}, we have
\begin{align*}
&\frac{1}{2}\frac{d}{dt}\int w_t^2dx+\int|\nabla w_t|^2dx\\
&\leq C\int(|\nabla\psi_t|+|\psi_t||\nabla\psi|+|A_t|+|\psi_t||A|)|\nabla w_t|dx\\
&\leq \frac{1}{2}\|\nabla w_t\|_{L^2}^2+C\|\nabla\psi_t\|_{L^2}^2
 +C\|\psi_t\|_{L^3}^2
\|\nabla\psi\|_{L^6}^2+C\|A_t\|_{L^2}^2+C\|\psi_t\|_{L^3}^2\|A\|_{L^6}^2\\
&\leq \frac{1}{2}\|\nabla w_t\|_{L^2}^2+C\|\nabla\psi_t\|_{L^2}^2
 +C\|\psi_t\|_{L^3}^2+C\|A_t\|_{L^2}^2,
\end{align*}
which implies
\begin{equation}
\|w_t\|_{L^\infty(0,T;L^2)}+\|w_t\|_{L^2(0,T;H^1)}\leq C.\label{2.32}
\end{equation}

Taking $\operatorname{div}$ to \eqref{1.2}, testing by $|\operatorname{div} A|^{q-2}\operatorname{div} A\ (3<q\leq 6)$, 
using \eqref{2.5}, \eqref{2.23}, and \eqref{2.30}, we have
\begin{align*}
&\frac{1}{q}\frac{d}{dt}\int|\operatorname{div} A|^q dx+\mu\int|\operatorname{div} A|^{q-2}|\nabla\operatorname{div} A|^2dx
+4\frac{q-2}{q^2}\int\left|\nabla|\operatorname{div} A|^{q/2}\right|^2dx\\
&\quad +\int|\psi|^2|\operatorname{div} A|^q dx\\
&\leq C\int(|A||\nabla\psi|+|\Delta\psi|+|\Delta u|)|\operatorname{div} A|^{q-1}dx\\
&\leq C(\|A\|_{L^q}\|\nabla\psi\|_{L^\infty}+\|\Delta\psi\|_{L^q}
 +\|\Delta u\|_{L^q})\|\operatorname{div} A\|_{L^q}^{q-1}\\
&\leq C(\|\nabla\psi\|_{L^\infty}+\|\Delta\psi\|_{L^q}
 +\|\Delta u\|_{L^q})\|\operatorname{div} A\|_{L^q}^{q-1},
\end{align*}
whence
\begin{align*}
\frac{d}{dt}\|\operatorname{div} A\|_{L^q}^2
&\leq C(\|\nabla\psi\|_{L^\infty}+\|\Delta\psi\|_{L^q}
 +\|\Delta u\|_{L^q})\|\operatorname{div} A\|_{L^q},\\
&\leq C(\|\nabla\psi\|_{L^\infty}^2+\|\Delta\psi\|_{L^q}^2
 +\|\Delta u\|_{L^q}^2)+C\|\operatorname{div} A\|_{L^q}^2,
\end{align*}
which implies
\begin{equation}
\begin{aligned}
\|\operatorname{div} A\|_{L^q}^2
&\leq C+C\int_0^t\big(\|\nabla\psi\|_{L^\infty}^2
 +\|\Delta\psi\|_{L^q}^2+\|\Delta u\|_{L^q}^2\big) ds
 +C\int_0^t\|\operatorname{div} A\|_{L^q}^2 ds \\
&\leq C+C\int_0^t\|\Delta u\|_{L^q}^2ds+C\int_0^t\|\operatorname{div} A\|_{L^q}^2 ds.
\end{aligned}\label{2.33}
\end{equation}

By the $L^2(0,T;W^{2,q})$-theory of heat equation, it follows from 
\eqref{1.3}, \eqref{2.5}, \eqref{2.23} and \eqref{2.30}, we have
\begin{equation}
\begin{aligned}
\|\Delta u\|_{L^2(0,t;L^q)}
&\leq C+C\|\Delta\psi\|_{L^2(0,t;L^q)}+C\|\operatorname{div} A\|_{L^2(0,t;L^q)} \\
&\quad +C\|A\|_{L^\infty(0,t;L^q)}\|\nabla\psi\|_{L^2(0,t;L^\infty)} \\
&\leq C+C\|\operatorname{div} A\|_{L^2(0,t;L^q)}.
\end{aligned}\label{2.34}
\end{equation}

Inserting \eqref{2.34} into \eqref{2.33}, we have
\begin{gather}
\|\operatorname{div} A\|_{L^\infty(0,T;L^q)}\leq C\quad (3<q\leq 6),\label{2.35}\\
\|u\|_{L^2(0,T;W^{2,q})}\leq C.\label{2.36}
\end{gather}
It follows from \eqref{1.14}, \eqref{2.31} and \eqref{2.35} that
\begin{equation}
\|A\|_{L^\infty(0,T;W^{1,q})}\leq C\quad (3<q\leq 6).\label{2.37}
\end{equation}
It follows from \eqref{1.3}, \eqref{2.30}, \eqref{2.28}, \eqref{2.32} 
and \eqref{2.37} that 
$$
\|u\|_{L^\infty(0,T;H^2)}\leq C.
$$
This completes the proof.


\begin{remark}\label{re2.1} \rm
We do not need to assume $u_0\geq 0$ in $\Omega$ and then we take $I(u)=1$ 
in \eqref{1.3}. Now we use the Lyapunov functional \cite{1}: 
$$
G(t):=\|\nabla f\|_{L^2}^2+\frac{1}{2}\|f^2-1\|_{L^2}^2
+\|f V_s\|_{L^2}^2+\mu\|\operatorname{curl} V_s\|_{L^2}^2+\|u\|_{L^2}^2\leq G(0)<\infty,
$$ 
to prove that $u\in L^\infty(0,T;L^2)$. Then by the method of 
Stampacchia \cite{17}, it follows from \eqref{2.2}, \eqref{2.3} and 
\eqref{2.4} that 
$$
\|\psi\|_{L^\infty(0,T;L^\infty(\Omega))}\leq C.
$$
Then by the same calculations above, we can complete the  proof.
\end{remark}

\subsection{Acknowledgments}
This work was supported by NSFC (No. 11171154). The second author 
extends his appreciation to Distinguished Scientist Fellowship Program 
(DSFP) at King Saud University (Saudi Arabia).


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\end{document}