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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2015 (2015), No. 179, pp. 1--18.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2015 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2015/179\hfil Regularity for Hall-mhd systems]
{Global regularity for generalized Hall magneto-hydrodynamics systems}

\author[R. Wan \hfil EJDE-2015/179\hfilneg]
{Renhui Wan}

\address{Renhui Wan \newline
Department of mahtematics \\
Zhejiang University \\
Hangzhou, Zhejiang 310027, China}
\email{rhwanmath@zju.edu.cn, rhwanmath@163.com}

\thanks{Submitted November 11, 2014. Published June 29, 2015.}
\subjclass[2010]{35Q35, 35B65, 76W05}
\keywords{Generalized Hall-mhd system; global regularity; classical solutions}

\begin{abstract}
 In this article, we consider the tridimensional generalized
 Hall magneto-hydrodynamics (Hall-MHD) system,
 with $(-\Delta)^\alpha u$ and $(-\Delta)^\beta b$. For $\alpha\ge 5/4$,
 $\beta\ge 7/4$, we obtain the global regularity of classical solutions.
 For $0<\alpha<5/4$ and $1/2<\beta<7/4$, with small data,
 the system also possesses  global classical solutions. In addition,
 for the standard Hall-MHD system, $\alpha=\beta=1$, by adding a suitable condition,
 we give a positive answer to the open question in \cite{[3]}.
 At last, we study the  regularity criterions of generalized Hall-MHD system.
 In particular, we prove the regularity criterion in terms of horizontal
 gradient $\nabla_{h}u,\nabla_{h}b$ for
 $1<\alpha<5/4$, $5/4\le\beta<7/4$.
\end{abstract}

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

\section{Introduction}

The tridimensional incompressible generalized Hall-MHD system is governed by
\begin{equation} \label{a1}
\begin{gathered}
 \partial_{t}u+u\cdot\nabla u+(-\Delta)^\alpha u+\nabla p=b\cdot\nabla b,\\
 \partial_{t}b+u\cdot\nabla b-b\cdot\nabla u+(-\Delta)^\beta b
  =-\nabla\times(J\times b),\\
 \operatorname{div}u=\operatorname{div}b=0,\\
 u(0,x)=u_0(x),\quad b(0,x)=b_0(x),
\end{gathered}
\end{equation}
where $t\ge 0$, $x\in  \mathbb{R}^3$,  $p,u,b$ stand for scalar pressure,
velocity vector and magnetic field vector, respectively, $J=\nabla\times b$
is the current density, $u_0(x)$, $b_0(x)$ are the initial velocity
and magnetic field, $\alpha,\beta\ge0$ are  constants and $(-\Delta)^\alpha$
is defined by
$$
\widehat{(-\Delta)^\alpha f}(\xi)=|\xi|^{2\alpha}\widehat{f}(\xi),
$$
and we denote $(-\Delta)^{1/2}$ by $\Lambda$.
For $\alpha=\beta=1$, the system reduces to the standard Hall-MHD system,
which can be obtained from kinetic models (cf. \cite{[1]}).
Hall-MHD is required in many physics problem, such as magnetic
reconnection \cite{[6]}, star formation \cite{[7]}. In \cite{[2]},
for $\alpha=0$, $\beta=1$, local classical solutions were obtained and
 Beale-Kato-Majda type blow-up criterion was  also established.
In \cite{[3]}, for $\alpha=\beta=1$, some blow up criterions and small data
results on global existence  were established.

 Recently,  Chae, Wan and Wu \cite{[4]} considered  the generalized Hall-MHD
\eqref{a1}, and obtained the local well-posedness for $\alpha=0$
(without velocity diffusion), $\beta>1/2$.
So it is natural to ask a question:
\begin{quote}
Can the global classical solutions of \eqref{a1} be obtained with some
conditions on $(\alpha,\beta)$?
\end{quote}
One of the  main goals  is to give a positive answer to the question.
Some related work about generalized Navier-Stokes equations and generalized MHD
equations can be seen \cite{[12],[13],[14],[15],[16],[17]}.
Our first result is stated as follows.

\begin{theorem} \label{thm1.1}
Let $T>0,\alpha\ge5/4$, $\beta\ge 7/4$.
Let $(u_0,b_0)\in H^s(\mathbb{R}^3)$, $s>5/2$ satisfying
$\operatorname{div}u_0=\operatorname{div}b_0=0$.
Then \eqref{a1} has a unique global classical solution $(u,b)$ such that
$$
(u,b)\in L^\infty([0,T]; H^s(\mathbb{R}^3)).
$$
\end{theorem}

\begin{remark} \label{rmk10} \rm
The space $H^s(\mathbb{R}^3)$ can be equipped with the norm
$$
\|f\|_{H^s}=\|f\|_{L^2}+\|f\|_{\dot{H}^s},
$$
where
$$
\|f\|_{\dot{H}^s}^2=\sum_{j\in \mathbb{Z}}2^{2js}\|\Delta_j f\|_{L^2}^2.
$$
More details can be seen in section 2.
\end{remark}

\begin{remark} \label{rmk1} \rm
If $b=0$, \eqref{a1} reduces to the generlized Navier-Stokes equations.
Under the scaling transformation $u_{l}=l^{2\alpha-1}u(l^{2\alpha}{t},lx)$,
$p_{l}=l^{4\alpha-2}p(l^{2\alpha}t,lx)$, for $\alpha\ge\frac{5}{4}$,
it is well-known that the generalized Navier-Stokes equations is critical
and subcritical, which can lead the global regular solutions.
We refer to \cite{[12]} and \cite{[13]}.
If $u=0$, \eqref{a1} reduces to the following simple Hall problem
\begin{equation}{\label{a2}}
\partial_{t}b+(-\Delta)^\beta b=\nabla\times ((\nabla \times b)\times b),
\end{equation}
which is scaling invariance under $b_{l}=l^{2\beta-2}b(l^{2\beta}t,lx)$.
The corresponding energy is
\begin{align*}
E(b_{l})&=\operatorname{ess\,sup}_{l^{2\beta}t}\|b_{l}\|_{L^2}^2
 +\int_0^{t}\|\Lambda^{\beta}b_{l}\|_{L^2}^2d\tau\\
&=l^{4\beta-7}\big\{\operatorname{ess\,sup}_{t}\|b\|_{L^2}^2
 +\int_0^{t}\|\Lambda^\beta b\|_{L^2}^2d\tau\big\}
=l^{4\beta-7}E(b).
\end{align*}
This implies that the simple Hall problem \eqref{a2} is critical system
for $\beta=7/4$ and subcritical system with $\beta>7/4$.
As generalized Navier-Stokes equations, the condition on $(\alpha,\beta)$
seems optimal.
\end{remark}

In addition, with small initial data, we can obtain the global small
classical solution for $\alpha\in(0,5/4)$ and
$\beta \in(1/2,7/4)$.

\begin{theorem} \label{thm1.3}
Let $u_0,b_0,s$ be as Theorem \ref{thm1.1}.
Let $\alpha\in(0,5/4)$, $\beta \in(1/2,7/4)$, and if
\begin{equation}{\label{a3}}
\|u_0\|_{H^s}+\|b_0\|_{H^s}<\epsilon,
\end{equation}
where $\epsilon$ is sufficient small, then there exists a unique global
 classical solution $(u,b)$ of \eqref{a1}.
\end{theorem}

\begin{remark}\label{rmk2} \rm
For $\alpha=\beta=1$, this work was proved in \cite{[2]}, which can be
considered  a special case in Theorem \ref{thm1.3}.
\end{remark}

For $\alpha=\beta=1$, the authors in \cite{[3]}  improved the condition
\eqref{a3} by only assuming that
\begin{equation} \label{a4}
\|u_0\|_{\dot{H}^{3/2}}+\|b_0\|_{\dot{H}^{3/2}}<\epsilon\quad
\text{or} \quad
\|u_0\|_{\dot{B}_{2,1}^{1/2}} + \|b_0\|_{{\dot{B}}_{2,1}^{3/2}}<\epsilon,
\end{equation}
where
\[
\|f\|_{\dot{B}_{2,1}^s}=\sum_{j\in\mathbb{Z}}2^{js}\|\Delta_j f\|_{L^2}
\]
 (For more details see section 2.),  $\epsilon$ is sufficient small.
And they also gave an open question, whether the condition
\begin{equation}{\label{a5}}
\|u_0\|_{\dot{H}^{1/2}}+\|b_0\|_{\dot{H}^{3/2}}<\epsilon
\end{equation}
can lead the the desired result?
 Motivated by these, we give a theorem as follows.

\begin{theorem} \label{thm1.5}
Let $u_0,b_0,s$ be as Theorem \ref{thm1.1}.
Let $\alpha=\beta=1$. If $(u_0,b_0)$ satisfies \eqref{a5},
 with an additional condition, for all $t\ge0$,
\begin{equation}{\label{a6}}
\|b(t)\|_{L^\infty}\le C_0<2,
\end{equation}
then there  exists a unique global classical solution $(u,b)$ of \eqref{a1}.
\end{theorem}

\begin{remark}\label{rmk3} \rm
It seems that the condition \eqref{a6} is too strong due to global
assumption. However, we find that \eqref{a4} can be propagated to any $t$,
which lead to the global small of $ \|b(t)\|_{{\dot{{H}}}^{3/2}}$ or
 $ \|b(t)\|_{{\dot{{B}}}_{2,1}^{3/2}}$, while \eqref{a6} fails.
In addition, ${\dot{B}}_{2,1}^{3/2}\hookrightarrow L^{\infty}$,
and we do not need sufficient small of $\|b(t)\|_{L^{\infty}}$,
which seems to make \eqref{a5} and \eqref{a6} weaker than the second
condition of \eqref{a4} in some sense.
\end{remark}

For $\alpha=\beta=1$, the authors in \cite{[3]} also establish some
regularity criterions involving $u$ and $\nabla b$. Since the presence of
Hall-term $\nabla\times((\nabla\times b)\times b)$, we find that
$$
(u,b)\in L^\infty(0,T;H^1(\mathbb{R}^3))\cap L^2(0,T;H^2(\mathbb{R}^3))
$$
can not ensure the  regularity criterion established in \cite{[3]}.
Therefore, unlike MHD system, establishing the regularity criterion in terms
of horizontal gradient $\nabla_{h} u$ and $\nabla_{h} b$ seems formidable
and interesting. But for system \eqref{a1} with
$1<\alpha<5/4 $, $5/4 \le \beta<7/4 $, we can achieve this goal.
 We have the following theorems, where Theorem \ref{thm1.6} can be
considered as a component of the proof of Theorem \ref{thm1.7}.

\begin{theorem} \label{thm1.6}
Let $T>0$, $\alpha\in(1,5/4)$, $\beta\in(1,7/4)$.
Let $u_0,b_0,s$ be as Theorem \ref{thm1.1} and $(u,b)$ be the local
classical solution to \eqref{a1}. If
\begin{equation}{\label{a30}}
\int_0^{T}\|u\|_{L^{p_{1}}}^{q_{1}}+\|\nabla b\|_{L^{p_2}}^{q_2}dt<\infty
\end{equation}
for
\begin{gather*}
\frac{3}{p_{1}}+\frac{2\alpha}{q_{1}}
\le\min\big\{2\alpha-1,(1-\frac{\alpha}{\beta})\frac{3}{p_{1}}
+(2-\frac{1}{\beta})\alpha\big\}, \\
 p_{1}\in (\frac{3}{2\beta-1},\frac{3}{\beta-1}]
\cap(\frac{3}{2\alpha-1},\frac{3}{\alpha-1}], \\
\frac{3}{p_2}+\frac{2\beta}{q_2}\le2\beta-1, \quad
 p_2\in(\frac{3}{2\beta-1},\frac{3}{\beta-1}],
\end{gather*}
then $(u,b)$ remains regular in $[0,T]$.
\end{theorem}

\begin{remark}\label{rmk4} \rm
For $1<\alpha=\beta<5/4 $, the condition of index reduce to
$$
\frac{3}{p_{1}}+\frac{2\alpha}{q_{1}}\le2\alpha-1\quad\text{and}\quad
\frac{3}{p_2}+\frac{2\alpha}{q_2}\le2\beta-1,
$$
which is scaling invariance under the transformation in Remark \ref{rmk1}.
We can also establish the regularity criterion in terms of $\nabla u$ and
$\nabla b$, that is
$$
\int_0^{T}\|\nabla u\|_{L^{p_{1}}}^{q_{1}}
+\|\nabla b\|_{L^{p_2}}^{q_2}d\tau<\infty
$$
for
\begin{gather*}
\frac{3}{p_{1}}+\frac{2\alpha}{q_{1}}
\le\min\big\{2\alpha,(1-\frac{\alpha}{\beta})\frac{3}{p_{1}}+2\alpha\big\},\quad
\max\big\{\frac{3}{2\alpha},\frac{3}{2\beta}\big\}<p_{1}\le\infty
\\
\frac{3}{p_2}+\frac{2\beta}{q_2}\le2\beta-1,\quad
 p_2\in\big(\frac{3}{2\beta-1},\frac{3}{\beta-1}\big].
\end{gather*}
The proof is similar to the previous proofs.
\end{remark}

\begin{theorem}\label{thm1.7}
Let $T>0$, $\alpha\in(1,5/4),\beta\in[5/4,\frac{7}{4})$.
Let $u_0,b_0,s$ be as Theorem \ref{thm1.1} and $(u,b)$ be the local
classical solution to \eqref{a1}. If
\begin{equation}{\label{a31}}
\int_0^{T}\|\nabla_{h}u\|_{L^{p_{1}}}^{q_{1}}
+\|\nabla_{h} b\|_{L^{p_2}}^{q_2}dt<\infty
\end{equation}
for
\begin{gather*}
\frac{3}{p_{1}}+\frac{2\alpha}{q_{1}}\le2\alpha,\quad
\frac{3}{2\alpha}<p_{1}\le\infty, \\
\frac{3}{p_2}+\frac{2\beta}{q_2}\le 2\beta-1,\quad
\frac{3}{2\beta-1}<p_2\le\frac{3}{\alpha},
\end{gather*}
then $(u,b)$ remains regular in $[0,T]$.
\end{theorem}

\begin{remark}\label{rmk5} \rm
We do not know whether the regularity criterion \eqref{a31} can be established
for $\alpha\in(1,\frac{5}{4}),\beta\in(1,\frac{5}{4})$, since we only observe that
\begin{gather*}
u\in L^\infty(0,T;H^1(\mathbb{R}^3))\cap L^2(0,T;H^{\alpha+1}(\mathbb{R}^3)),
\\
b\in L^\infty(0,T;H^1(\mathbb{R}^3))\cap L^2(0,T;H^{\beta+1}(\mathbb{R}^3))
\end{gather*}
can lead to \eqref{a30} for
$\alpha\in(1,5/4),\beta\in[5/4,7/4)$.
\end{remark}

This article is organized as follows.
In section 2, we give some notation and preliminaries.
In the third section, we prove Theorem \ref{thm1.1}.
In the fourth section, we give the proof of Theorem \ref{thm1.3}.
In the fifth section, we give the proof of Theorem \ref{thm1.5}.
We prove Theorem \ref{thm1.6} in the following section.
At the last section, we prove  and Theorem \ref{thm1.7}.

\section{Preliminaries}

Let $\mathfrak{B}=\{\xi\in\mathbb{R}^d,\ |\xi|\le\frac{4}{3}\}$ and
$\mathfrak{C}=\{\xi\in\mathbb{R}^d: 3/4\le|\xi|\le8/3\}$.
Choose two nonnegative smooth radial function $\chi, \varphi$ supported,
respectively, in $\mathfrak{B}$ and $\mathfrak{C}$ such that
\begin{gather*}
\chi(\xi)+\sum_{j\ge0}\varphi(2^{-j}\xi)=1, \quad   \xi\in\mathbb{R}^d,
\\
\sum_{j\in\mathbb{Z}}\varphi(2^{-j}\xi)=1,\quad \xi\in\mathbb{R}^d\setminus\{0\}.
\end{gather*}
We denote $\varphi_j=\varphi(2^{-j}\xi)$,
$h=\mathfrak{F}^{-1}\varphi$ and $\tilde{h}=\mathfrak{F}^{-1}\chi$,
where $\mathfrak{F}^{-1}$ stands for the inverse Fourier transform.
Then the dyadic blocks
$\Delta_j$ and $S_j$ can be defined as follows
\begin{gather*}
\Delta_jf=\varphi(2^{-j}D)f=2^{jd}\int_{\mathbb{R}^d}h(2^jy)f(x-y)dy,\\
S_jf=\sum_{k\le j-1}\Delta_{k}f=\chi(2^{-j}D)f
=2^{jd}\int_{\mathbb{R}^d}\tilde{h}(2^jy)f(x-y)dy.
\end{gather*}
Formally, $\Delta_j=S_j-S_{j-1}$ is a frequency projection to annulus
$\{C_{1}2^j\le|\xi|\le C_22^j\}$, and $S_j$ is a frequency projection
to the ball $\{|\xi|\le C2^j\}$. One can easily verify that with
our choice of $\varphi$,
$$
\Delta_j\Delta_{k}f=0\text{ if } |j-k|\ge2\quad \text{and}\quad
\Delta_j(S_{k-1}f\Delta_{k}f)=0\text{ if }|j-k|\ge5.
$$
With the introduction of $\Delta_j$ and $S_j$, let us recall the
definition of the  Besov space.

 Let $s\in \mathbb{R}$, $(p,q)\in[1,\infty]^2$, the homogeneous space
 $\dot{B}_{p,q}^s$ is defined by
$$
\dot{B}_{p,q}^{s}=\{f\in \mathfrak{S}': \|f\|_{\dot{B}_{p,q}^{s}}<\infty\},
$$
where
\[
\|f\|_{\dot{B}_{p,q}^s}=\begin{cases}
\big(\sum_{j\in \mathbb{Z}}2^{sjq}\|\Delta_jf\|_{L^p}^{q}\big)^{1/q},
&\text{for }  1\le q<\infty,\\
\sup_{j\in\mathbb{Z}}2^{sj}\|\Delta_jf\|_{L^p}, &\text{for } q=\infty,
\end{cases}
\]
and $\mathfrak{S}'$ denotes the dual space of
\[
\mathfrak{S}=\{f\in\mathcal{S}(\mathbb{R}^d):
 \partial^{\alpha}\hat{f}(0)=0: \forall \alpha\in  \mathbb{N}^d
\text{multi-index}\}
\]
and can be identified by the quotient space of $\mathcal{S'}/\mathcal{P}$
with the polynomials space $\mathcal{P}$.

 We also provide the  definition for the  inhomogeneous Besov space.
For $s>0$, and $(p,q)\in [1,\infty]^2$,  the inhomogeneous Besov space
$B_{p,q}^s$ can be defined as follows
$$
{B}_{p,q}^{s}=\{f\in \mathcal{S'}(\mathbb {R}^d): \|f\|_{{B}_{p,q}^{s}}<\infty\},
$$
where
$$
\|f\|_{B_{p,q}^s}=\|f\|_{L^p}+\|f\|_{\dot{B}_{p,q}^s}.
$$
Additionally, when $p=q=2$, the Besov space and Sobolev space are equivalence;
that is
$$
\dot{H}^s\approx\dot{B}_{2,2}^s,\quad  H^s\approx B_{2,2}^s.
$$
Bernstein's inequalities are useful in this paper, so that we give it in
the following proposition.

\begin{proposition}\label{prop2.1}
Let $\alpha\ge0$. Let $1\le p\le q\le \infty$.
\begin{itemize}
\item[(1)] If $f$ satisfies
$$
\operatorname{supp} \widehat{f} \subset \{\xi\in \mathbb{R}^d:  |\xi| \le K 2^j \},
$$
for some integer $j$ and a constant $K>0$, then
$$
\|(-\Delta)^\alpha f\|_{L^q(\mathbb{R}^d)} \le C_1\, 2^{2\alpha j +
j d(\frac{1}{p}-\frac{1}{q})} \|f\|_{L^p(\mathbb{R}^d)}.
$$

\item[(2)] If $f$ satisfies
\begin{equation*}%\label{spp}
\operatorname{supp} \widehat{f} \subset \{\xi\in \mathbb{R}^d: K_12^j
\le |\xi| \le K_2 2^j \}
\end{equation*}
for some integer $j$ and constants $0<K_1\le K_2$, then
$$
C_1 2^{2\alpha j} \|f\|_{L^q(\mathbb{R}^d)}
\le \|(-\Delta)^\alpha f\|_{L^q(\mathbb{R}^d)}
\le C_2\, 2^{2\alpha j + j d(\frac{1}{p}-\frac{1}{q})} \|f\|_{L^p(\mathbb{R}^d)},
$$
where $C_1$ and $C_2$ are constants depending on $\alpha,p$ and $q$
only.
\end{itemize}
\end{proposition}

For more details about Besov space such as  some useful embedding relations,
 we refer to \cite{[8],[9]}. Thanks to the Proposition \ref{prop2.1},
 we can see that $\forall\ s>0$,
\begin{equation}\label{A1}
\|f\|_{H^s}\approx \|f\|_{L^2}+(\sum_{j\ge 0}2^{2js}\|\Delta_j f\|_{L^2}^2)^{1/2},
\end{equation}
which will be frequently used in our proof.

\begin{proposition}[\cite{[11]}] \label{prop2.2}
Let $1\le p_{1}$, $p_2\le\infty$, $\sigma>0$, $\frac{d}{p_i}-\sigma_i>0(i=1,2)$
and assume that $\sigma-\sigma_2+\frac{d}{p_2}>0$. Then the following
inequality holds
\begin{equation}\label{a7}
\begin{aligned}
&\Big(\sum_{j\in \mathbb{Z}}2^{2j\sigma}\|[f,\Delta_j]\cdot\nabla
g\|_2^2\Big)^{1/2}\\
&\le C\Big(\|\nabla f\|_{\dot{B}_{p_{1},\infty}^{\sigma_{1}}}
\|g\|_{\dot{B}_{2,2}^{\sigma-\sigma_{1}
 +\frac{d}{p_{1}}}}+\|\nabla g\|_{\dot{B}_{p_2,\infty}^{\sigma_2}}
\|f\|_{\dot{B}_{2,2}^{\sigma-\sigma_2+\frac{d}{p_2}}}\Big).
\end{aligned}
\end{equation}
\end{proposition}

\section{Proof of Theorem \ref{thm1.1}}

From \cite{[4]}, one can see that the local well-posedness of \eqref{a1} holds
for $\alpha\ge0, \beta>1/2$. So we only need establish the global regularity,
i.e. for all $0\le t\le T$,
$$
\|u(t)\|_{H^s}+\|b(t)\|_{H^s}\le C(s,T,\|u_0\|_{H^s},\|u_0\|_{H^s}).
$$
At first, we give the energy estimate. Taking inner product with $(u,b)$,
 integrating by parts and integrating in time $[0,t]$,
\begin{equation}{\label{a8}}
\|(u(t),b(t)\|_{L^2}^2+2\int_0^{t}\|\Lambda^\alpha u\|_{L^2}^2d\tau
+2\int_0^{t}\|\Lambda^\beta b\|_{L^2}^2d\tau\le\|(u_0,b_0)\|_{L^2}^2.
\end{equation}
Then we establish the $H^s$ estimate. Applying the operator $\Delta_{q}$
to \eqref{a1}, taking the inner product with $(\Delta_{q}u,\Delta_{q}b)$, by
cancelation property, integrating by parts, we obtain
\begin{equation}\label{a9}
\begin{aligned}
&\frac{1}{2}\frac{d}{dt}(\|\Delta_{q} u\|_{L^2}^2
 +\|\Delta_{q} b\|_{L^2}^2)+ \|\Lambda^\alpha\Delta_{q} u\|_{L^2}^2
 + \|\Lambda^\beta\Delta_{q} b\|_{L^2}^2\\
&=-\int_{\mathbb{R}^3}[\Delta_{q},u\cdot\nabla] u\cdot\Delta_{q}u
 +\int_{\mathbb{R}^3}[\Delta_{q},b\cdot\nabla] b\cdot\Delta_{q}u
 -\int_{\mathbb{R}^3}[\Delta_{q},u\cdot\nabla] b\cdot\Delta_{q}b\\
&\quad +\int_{\mathbb{R}^3}[\Delta_{q},b\cdot\nabla] u\cdot\Delta_{q}b
 +\int_{\mathbb{R}^3}[\Delta_{q},b\times]J\cdot\Delta_{q}J\\
&\le \|[\Delta_{q},u\cdot\nabla] u\|_{L^2}\|\Delta_{q}u\|_{L^2}
 +\|[\Delta_{q},b\cdot\nabla ] b\|_{L^2}\|\Delta_{q}u\|_{L^2}\\
&\quad +\|[\Delta_{q},u\cdot\nabla] b\|_{L^2}\|\Delta_{q}b\|_{L^2}
 +\|[\Delta_{q},b\cdot\nabla] u\|_{L^2}\|\Delta_{q}b\|_{L^2} \\
&\quad +\|[\Delta_{q},b\times]J\|_{L^2}\|\Delta_{q}J\|_{L^2}\\
&=L_{1}(t)+L_2(t)+L_3(t)+L_{4}(t)+L_{5}(t).
\end{aligned}
\end{equation}
By the paraproduct decomposition, H\"{o}lder's inequality, commutator estimate
\cite[page. 110]{[8]}, and Bernstein's inequality,
\begin{equation}{\label{a10}}
\begin{aligned}
|L_{1}(t)|
&\le\sum_{|k-q|\le4}\|\Delta_{q}(S_{k-1}u\cdot\nabla\Delta_{k}u)
 -S_{k-1}u\cdot\nabla\Delta_{q}\Delta_{k}u\|_{L^2}\|\Delta_{q}u\|_{L^2}\\
&\quad +\sum_{|k-q|\le4}\|\Delta_{q}(\Delta_{k}u\cdot\nabla S_{k-1}u)
 -\Delta_{k}u\cdot\nabla\Delta_{q}S_{k-1}u\|_{L^2}\|\Delta_{q}u\|_{L^2}\\
&\quad +\sum_{k\ge q-3}\|\Delta_{q}(\tilde{\Delta}_{k}u\cdot\nabla\Delta_{k}u)
 -\tilde{\Delta}_{k}u\cdot\nabla\Delta_{q}\Delta_{k}u\|_{L^2}
 \|\Delta_{q}u\|_{L^2}\\
&\le C\|\nabla S_{q-1}u\|_{L^\infty}\|\Delta_{q}u\|_{L^2}^2
 +C\|\Delta_{q}u\|_{L^2}\sum_{k\ge q-3}
 \|\nabla\Delta_{k}u\|_{L^\infty}\|\tilde{\Delta}_{k}u\|_{L^2},
\end{aligned}
\end{equation}
where $\tilde{\Delta}_{k}= \Delta_{k-1}+\Delta_{k}+\Delta_{k+1}$. Similarly,
\begin{gather*}
|L_2(t)|\le C\|\nabla S_{q-1}b\|_{L^\infty}\|\Delta_{q}b\|_{L^2}
 \|\Delta_{q}u\|_{L^2}+C\|\Delta_{q}u\|_{L^2}\sum_{k\ge q-3}
 \|\nabla\Delta_{k}b\|_{L^\infty}\|\tilde{\Delta}_{k}b\|_{L^2}, \\
\begin{aligned}
|L_3(t)|&\le C\|\nabla S_{q-1}u\|_{L^\infty}\|\Delta_{q}b\|_{L^2}^2
 +\|\nabla S_{q-1}b\|_{L^\infty}\|\Delta_{q}u\|_{L^2}\|\Delta_{q}b\|_{L^2}\\
&\quad +C\|\Delta_{q}b\|_{L^2}\sum_{k\ge q-3}\|\nabla\Delta_{k}b\|_{L^\infty}
 \|\tilde{\Delta}_{k}u\|_{L^2},
\end{aligned} \\
\begin{aligned}
|L_{4}(t)|&\le C\|\nabla S_{q-1}b\|_{L^\infty}\|\Delta_{q}u\|_{L^2}
 \|\Delta_{q}b\|_{L^2}+\|\nabla S_{q-1}u\|_{L^\infty}\|\Delta_{q}b\|_{L^2}^2\\
&\quad +C\|\Delta_{q}b\|_{L^2}\sum_{k\ge q-3}\|\nabla\Delta_{k}u\|_{L^\infty}
\|\tilde{\Delta}_{k}b\|_{L^2},
\end{aligned} \\
|L_{5}(t)|\le C2^q\|\nabla S_{q-1}b\|_{L^\infty}\|\Delta_{q}b\|_{L^2}^2
+ C2^q\|\Delta_{q}b\|_{L^2}\sum_{k\ge q-3}\|\nabla\Delta_{k}b\|_{L^\infty}
\|\tilde{\Delta}_{k}b\|_{L^2}.
\end{gather*}
Multiplying $2^{2sq}$ and taking the summation over $q\ge0$ in \eqref{a9},
\begin{equation}{\label{a11}}
\begin{aligned}
&\frac{1}{2}\frac{d}{dt}
\Big\{\sum_{q\ge0}2^{2sq}(\|\Delta_{q} u\|_{L^2}^2+\|\Delta_{q} b\|_{L^2}^2)\Big\}\\
&+ \sum_{q\ge0}2^{2(s+\alpha) q}\|\Delta_{q} u\|_{L^2}^2
 +\sum_{q\ge0}2^{2(s+\beta) q}\|\Delta_{q} b\|_{L^2}^2\\
&\le\sum_{q\ge0}2^{2sq}(L_{1}(t)+\dots+L_{5}(t)).
\end{aligned}
\end{equation}
By \eqref{a10}, we have
\begin{align*}
\sum_{q\ge0}2^{2sq}|L_{1}(t)|
&\le C\sum_{q\ge0}2^{2sq}\|\nabla S_{q-1}u\|_{L^\infty}\|\Delta_{q}u\|_{L^2}^2\\
&\quad +C\sum_{q\ge0}2^{2sq}\|\Delta_{q}u\|_{L^2}
 \sum_{k\ge q-3}\|\nabla\Delta_{k}u\|_{L^\infty}\|\tilde{\Delta}_{k}u\|_{L^2}\\
&=L_{11}(t)+L_{12}(t).
\end{align*}
For the estimate of $L_{11}(t)$. Using Bernstein's inequality, H\"{o}lder's
inequality and Young's inequality,
\begin{align*}
|L_{11}(t)|
&\le C\sum_{q\ge0}2^{2sq}\|\Delta_{q}u\|_{L^2}^2
 \sum_{m\le q-2}2^{5m/2}\|\Delta_{m}u\|_{L^2}\\
&\le C\sum_{q\ge0}2^{(s+\alpha)q}\|\Delta_{q}u\|_{L^2}2^{(s-\alpha)q}
 \|\Delta_{q}u\|_{L^2}\sum_{m\le q-2}2^{5m/2}\|\Delta_{m}u\|_{L^2}\\
&\le \frac{1}{8}\|\Lambda^\alpha u\|_{\dot{H}^s}^2
+\underbrace{C\sum_{q\ge0}2^{2(s-\alpha)q}\|\Delta_{q}u\|_{L^2}^2
\Big(\sum_{m\le q-2}2^{5m/2}\|\Delta_{m}u\|_{L^2}\Big)^2}_{L_{111}},
\end{align*}
where
\begin{align*}
|L_{111}|
&= C\sum_{q\ge0}2^{2sq}\|\Delta_{q}u\|_{L^2}^2
\Big(\sum_{m\le q-2}2^{\frac{5}{2}m-\alpha q}\|\Delta_{m}u\|_{L^2}\Big)^2\\
&=C\sum_{q\ge0}2^{2sq}\|\Delta_{q}u\|_{L^2}^2
\Big(\sum_{m\le -2}2^{\frac{5}{2}m-\alpha q}\|\Delta_{m}u\|_{L^2}\\
&\quad +\sum_{-1\le m\le q-2}2^{(\frac{5}{2}-\alpha)m-\alpha q}\|\Lambda^\alpha
 \Delta_m u\|_{L^2}\Big)^2\\
&\le C \sum_{q\ge0}2^{2sq}\|\Delta_{q}u\|_{L^2}^2
\Big(\sum_{m\le -2}2^{5m/2}\|u\|_{L^2}\\
&\quad +\sum_{-1\le m\le q-2}
2^{\alpha(m-q)}2^{(\frac{5}{2}-2\alpha)m}
\|\Lambda^\alpha  u\|_{L^2}\Big)^2\\
&\le C(\|u\|_{L^2}^2+\|\Lambda^\alpha u\|_{L^2}^2)\|u\|_{H^s}^2.
\end{align*}
Thus, we obtain
$$
|L_{11}(t)|\le C(\|u\|_{L^2}^2+\|\Lambda^\alpha u\|_{L^2}^2)\|u\|_{H^s}^2
+\frac{1}{8}\|\Lambda^\alpha u\|_{\dot{H}^s}^2.
$$
Similarly,
\begin{align*}
&|L_{12}(t)|\\
&\le C\sum_{q\ge0}2^{2sq}\|\Delta_{q}u\|_{L^2}
 \sum_{k\ge q-3}\|\Delta_{k}u\|_{L^2}2^{5k/2}\|\tilde{\Delta}_{k}u\|_{L^2}\\
&\le C \Big(\sum_{p\ge 0}2^{2sq}\|\Delta_q u\|_{L^2}^2\Big)^{1/2}
\Big\{\sum_{q\ge 0}2^{2sq}(\sum_{k\ge q-3}\|\tilde{\Delta}_k u\|_{L^2}2^{k(\frac{5}{2}-\alpha)}
\|\Lambda^\alpha \Delta_k u\|_{L^2})^2\Big\}^{1/2}\\
&\le C\|\Lambda^\alpha u\|_{L^2}\|u\|_{H^s}
\Big\{\sum_{q\ge 0}2^{2sq}(\sum_{k\ge q-3}2^{k(\frac{5}{2}-2\alpha)}
 \|\tilde{\Delta}_k\Lambda^\alpha u\|_{L^2})^2\Big\}^{1/2}\\
&\le C\|\Lambda^\alpha u\|_{L^2}\|u\|_{H^s}
\Big\{\sum_{q\ge 0}2^{2sq}(\sum_{k\ge q-3}\|\tilde{\Delta}_k\Lambda^\alpha
 u\|_{L^2})^2\Big\}^{1/2}\\
&\le C\|\Lambda^\alpha u\|_{L^2}\|u\|_{H^s}\|u\|_{\dot{H}^{s+\alpha}}\\
&\le C\|\Lambda^\alpha u\|_{L^2}^2\|u\|_{H^s}^2
 +\frac{1}{8}\|u\|_{\dot{H}^{s+\alpha}}^2,
\end{align*}
here we have used Young's inequality for series for the fifth inequality; that is,
\begin{align*}
\Big\{\sum_{q\ge 0}2^{2sq}(\sum_{k\ge q-3}
 \|\tilde{\Delta}_k\Lambda^\alpha u\|_{L^2})^2\Big\}^{1/2}
&\le \Big\{\sum_{q\in\mathbb{Z}}(\sum_{k\ge q-3}2^{s(q-k)}2^{sk}
 \|\tilde{\Delta}_k\Lambda^\alpha u\|_{L^2})^2\Big\}^{1/2}\\
&\le C\|2^{-sk}\mathbf{1}_{k\ge -3}\|_{l^1(\mathbb{Z})}
 \|2^{sk}\|\tilde{\Delta}_k\Lambda^\alpha u\|_{L^2}\|_{l^2(\mathbb{Z})}\\
&\le C\|u\|_{\dot{H}^{s+\alpha}}.
\end{align*}
Collecting the estimates above, we have
$$
\sum_{q\ge0}2^{2sq}|L_{1}(t)|\le C(\|u\|_{L^2}^2
+\|\Lambda^\alpha u\|_{L^2}^2)\|u\|_{H^s}^2
+\frac{1}{4}\|\Lambda^\alpha u\|_{H^s}^2.
$$
Similarly,
\begin{gather*}
\sum_{q\ge0}2^{2sq}|L_2(t)|\le C(\|b\|_{L^2}^2
 +\|\Lambda^\beta b\|_{L^2}^2)\|b\|_{H^s}^2
 +\frac{1}{4}\|\Lambda^\alpha u\|_{H^s}^2,
\\
\sum_{q\ge0}2^{2sq}|L_3(t)|\le C(\|(u,b)\|_{L^2}^2
 +\|(\Lambda^\alpha u,\Lambda^\beta b)\|_{L^2}^2)\|b\|_{H^s}^2
 +\frac{1}{4}\|(\Lambda^\alpha u,\Lambda^\beta b)\|_{H^s}^2,
\\
\sum_{q\ge0}2^{2sq}|L_{4}(t)|
 \le C(\|(u,b)\|_{L^2}^2+\|(\Lambda^\alpha u,\Lambda^\beta b)\|_{L^2}^2)
\|b\|_{H^s}^2+\frac{1}{4}\|(\Lambda^\alpha u,\Lambda^\beta b)\|_{H^s}^2.
\end{gather*}
Now, we  estimate the last term.
\begin{align*}
\sum_{q\ge0}2^{2sq}|L_{5}(t)|
&\le C\sum_{q\ge0}2^{(2s+1)q}\|\Delta_{q}b\|_{L^2}^2\sum_{m\le q-2}
\|\nabla\Delta_{m}b\|_{L^\infty}\\
&\quad +C\sum_{q\ge0}2^{(2s+1)q}\|\Delta_{q}b\|_{L^2}\sum_{k\ge q-3}
 \|\nabla\Delta_{k}b\|_{L^\infty}\|\tilde{\Delta}_{k}b\|_{L^2}\\
&=L_{51}(t)+L_{52}(t).
\end{align*}
Similar to the estimate of $L_{11}(t)$, we have
\begin{align*}
&|L_{51}(t)|\\
&\le C\sum_{q\ge0}2^{(s+\beta)q}\|\Delta_{q}b\|_{L^2}2^{(s+1-\beta)q}
 \|\Delta_{q}b\|_{L^2}\sum_{m\le q-2}2^{5m/2}\|\Delta_{m}b\|_{L^2}\\
&\le \frac{1}{8}\sum_{q\ge0}2^{2(s+\beta)q}\|\Delta_{q}b\|_{L^2}^2
+\underbrace{C\sum_{q\ge0}2^{2(s+1-\beta)q}\|\Delta_{q}b\|_{L^2}^2
\Big(\sum_{m\le q-2}2^{5m/2}\|\Delta_{m}b\|_{L^2}\Big)^2}_{L_{511}},
\end{align*}
where
\begin{align*}
|L_{511}|
&\le C\sum_{q\ge0}2^{2sq}\|\Delta_{q}b\|_{L^2}^2
\Big(\sum_{m\le -2}2^{(1-\beta)q}2^{5m/2}\|\Delta_{m}b\|_{L^2}\\
&\quad +\sum_{-1\le m\le q-2}2^{(1-\beta)q} 2^{(\frac{5}{2}-\beta)m}
 \|\Lambda^\beta\Delta_m b\|_{L^2}\Big)^2\\
&\le C \sum_{q\ge0}2^{2sq}\|\Delta_{q}b\|_{L^2}^2
\sum(\sum_{m\le -2}2^{5m/2}\|b\|_{L^2}\\
&\quad +\sum_{-1\le m\le q-2}2^{(1-\beta)(q-m)}2^{(\frac{7}{2}-2\beta)m}
 \|\Lambda^\beta \Delta_m b\|_{L^2}\Big)^2\\
&\le C (\|b\|_{L^2}^2+\|\Lambda^\beta b\|_{L^2}^2)
\sum_{q\ge0}2^{2sq}\|\Delta_{q}b\|_{L^2}^2\\
&\le C(\|b\|_{L^2}^2+\|\Lambda^\beta b\|_{L^2}^2)\|b\|_{H^s}^2.
\end{align*}
So we have
$$
|L_{51}(t)|\le \frac{1}{8}\|b\|_{\dot{H}^{s+\beta}}^2
+ C(\|b\|_{L^2}^2+\|\Lambda^\beta b\|_{L^2}^2)\|b\|_{H^s}^2.
$$
Similar to the estimate of $L_{12}(t)$, we have
\begin{align*}
|L_{52}(t)|
&\le C\sum_{q\ge 0}2^{sq}\|\Delta_q b\|_{L^2}2^{(s+1)q}
 \sum_{k\ge q-3}\|\nabla \Delta_q b\|_{L^\infty}\|\tilde{\Delta}_k b\|_{L^2}\\
&\le C \Big(\sum_{q\ge 0}2^{2sq}\|\Delta_q b\|_{L^2}^2\Big)^{1/2}
\Big\{\sum_{q\ge 0}2^{2(s+1)q}(\sum_{k\ge q-3}\|\nabla\Delta_k b\|_{L^\infty}
\|\tilde{\Delta}_k b\|_{L^2})^2\Big\}^{1/2} \\
&\le C\|b\|_{H^s}\Big\{\sum_{q\ge 0}2^{2(s+1)q}
\Big(\sum_{k\ge q-3}2^{k(\frac{5}{2}-\beta)}
\|\Lambda^\beta\Delta_k b\|_{L^2}2^{-k\beta}\|\Lambda^\beta
\tilde{\Delta}_k b\|_{L^2} \Big)^2\Big\}^{1/2}\\
&\le C\|b\|_{H^s}\Big\{\sum_{q\ge 0}2^{2(s+1)q}(\sum_{k\ge q-3}
 2^{-k}2^{k(\frac{7}{2}-2\beta)}
\|\Lambda^\beta\Delta_k b\|_{L^2}\|\Lambda^\beta b\|_{L^2} )^2\Big\}^{1/2}\\
&\le C\|\Lambda^\beta b\|_{L^2} \|b\|_{H^s}
\Big\{\sum_{q\ge 0}(\sum_{k\ge q-3}2^{(s+1)(q-k)}2^{ks}
\|\Lambda^\beta\Delta_k b\|_{L^2})^2\Big\}^{1/2}\\
&\le C\|\Lambda^\beta b\|_{L^2} \|b\|_{H^s}\|b\|_{\dot{H}^{s+\beta}}
\le C\|\Lambda^\beta b\|_{L^2}^2 \|b\|_{H^s}^2
+\frac{1}{8}\|b\|_{\dot{H}^{s+\beta}}^2,
\end{align*}
here we have used Young's inequality for the fifth inequality. Therefore,
$$
\sum_{q\ge0}2^{2sq}|L_{5}(t)|\le C(\|\Lambda^\beta b\|_{L^2}^2
+\|b\|_{L^2}^2)\|b\|_{H^s}^2+\frac{1}{4}\|\Lambda^\beta b\|_{\dot{H}^s}^2.
$$
Collecting the estimate above in \eqref{a9}, integrating in $[0,t]$,
 together with \eqref{a8} and \eqref{A1} yields
\begin{align*}
&\|u(t)\|_{H^s}^2+\|b(t)\|_{H^s}^2
 +\int_0^{t}\|\Lambda^\alpha u(\tau)\|_{H^s}^2d\tau
 +\int_0^{t}\|\Lambda^\beta b(\tau)\|_{H^s}^2d\tau\\
&\le C(\|(\Lambda^\alpha u,\Lambda^\beta b)\|_{L^2}^2
 +\|(u,b)\|_{L^2}^2)(\|u\|_{H^s}^2+\|b\|_{H^s}^2)
 +\|u_0\|_{H^s}^2+\|b_0\|_{H^s}^2.
\end{align*}
Applying Gronwall's lemma, with \eqref{a8}, for $0\le t\le T$,
\begin{align*}
&\|u(t)\|_{H^s}^2+\|b(t)\|_{H^s}^2+\int_0^{t}\|\Lambda^\alpha u(\tau)\|_{H^s}^2d\tau+\int_0^{t}\|\Lambda^\beta b(\tau)\|_{H^s}^2d\tau\\
&\le (\|u_0\|_{H^s}^2+\|b_0\|_{H^s}^2)\exp
\Big\{C\int_0^{t}(\|(\Lambda^\alpha u,\Lambda^\beta b)\|_{L^2}^2
 +\|(u,b)\|_{L^2}^2)d\tau\Big\}\\
&\le (\|u_0\|_{H^s}^2+\|b_0\|_{H^s}^2)
\exp\left(C\|(u_0,b_0)\|_{L^2}^2(1+T)\right).
\end{align*}
This completes the proof of Theorem \ref{thm1.1}.

\section{Proof of Theorem \ref{thm1.3}}

As in the proof of Theorem \ref{thm1.1}, we only need to show the global regularity.
Since $\|u\|_{H^s}=\|u\|_{L^2}+\|u\|_{\dot{H}^s}$, we obtain
\begin{align*}
&\|u(t)\|_{H^s}^2+\|b(t)\|_{H^s}^2+2\int_0^{t}\|\Lambda^\alpha
 u(\tau)\|_{H^s}^2d\tau+2\int_0^{t}\|\Lambda^\beta b(\tau)\|_{H^s}^2d\tau\\
&\le \|(u_0,b_0)\|_{H^s}^2+C\int_0^{t}\|(\nabla u,\nabla b)
 \|_{L^\infty}\|(u,b)\|_{\dot{H}^s}^2d\tau+2\int_0^{t}
 \|\nabla b\|_{L^\infty}\|b\|_{\dot{H}^{s+\frac{1}{2}}}^2d\tau\\
&\le \|(u_0,b_0)\|_{H^s}^2+C\int_0^{t}\|(u,b)\|_{H^s}
 (\|\Lambda^\alpha u\|_{H^s}^2+\|\Lambda^\beta b\|_{H^s}^2)d\tau.
\end{align*}
A similar estimate  can be found in \cite{[4]}.
Choosing $\epsilon$ so small that $\|(u_0,b_0)\|_{H^s}^2<\frac{1}{2C^2}$,
 which implies that $C\|(u_0,b_0)\|_{H^s}< 1$.
Suppose there exists a first time $T^\star$, such that
\begin{equation}\label{A2}
\|u(T^\star)\|_{H^s}^2+\|b(T^\star)\|_{H^s}^2\ge\frac{1}{2C^2}.
\end{equation}
This leads to
$$
\|u(T^\star)\|_{H^s}^2+\|u(T^\star)\|_{H^s}^2+\int_0^{T^\star}
\|\Lambda^\alpha u(t)\|_{H^s}^2
+\|\Lambda^\beta b(t)\|_{H^s}^2dt\le \|(u_0,b_0)\|_{H^s}^2
<\frac{1}{2C^2},
$$
which  contradicts  \eqref{A2}.  Hence, for all  $t\ge0$, we get global small
solution satisfying
$$
\|u(t)\|_{H^s}^2+\|b(t)\|_{H^s}^2
+\int_0^{t}\|\Lambda^\alpha u(\tau)\|_{H^s}^2d\tau
+\int_0^{t}\|\Lambda^\beta b(\tau)\|_{H^s}^2d\tau\le\|(u_0,b_0)\|_{H^s}^2.
$$
This concludes the proof of Theorem \ref{thm1.3}.

\section{Proof of Theorem \ref{thm1.5}}

It  suffices to establish the following, for all  $t\ge0$,
$$
\int_0^{t}\|u\|_{\dot{H}^{3/2}}^2+\|\nabla b\|_{\dot{H}^{3/2}}^2d\tau<\infty,
$$
since \cite{[3]} provided the blow up criterion in the $BMO$ space  and
$\dot{H}^{3/2}\hookrightarrow BMO$  \cite{[8]}. As the operation in section 3,
we obtain
\begin{equation}\label{a12}
\begin{aligned}
&\frac{1}{2}\frac{d}{dt}(\|\Delta_{q} u\|_{L^2}^2+\|\Delta_{q} b\|_{L^2}^2)
+ \|\nabla\Delta_{q} u\|_{L^2}^2+ \|\nabla\Delta_{q} b\|_{L^2}^2\\
&=-\int_{\mathbb{R}^3}[\Delta_{q},u\cdot\nabla] u\cdot\Delta_{q}u
 +\int_{\mathbb{R}^3}[\Delta_{q},b\cdot\nabla] b\cdot\Delta_{q}u
 -\int_{\mathbb{R}^3}[\Delta_{q},u\cdot\nabla] b\cdot\Delta_{q}b\\
&\quad -\int_{\mathbb{R}^3}[\Delta_{q},b\cdot\nabla] u\cdot\Delta_{q}b
 +\int_{\mathbb{R}^3}[\Delta_{q},b\times]J\cdot\Delta_{q}J\\
&\le\|[\Delta_{q},u\cdot\nabla] u\|_{L^2}\|\Delta_{q}u\|_{L^2}
 +\|[\Delta_{q},b\cdot\nabla ] b\|_{L^2}\|\Delta_{q}u\|_{L^2}\\
&\quad +\|[\Delta_{q},u\cdot\nabla] b\|_{L^2}\|\Delta_{q}b\|_{L^2}
  +\|[\Delta_{q},b\cdot\nabla] u\|_{L^2}\|\Delta_{q}b\|_{L^2}\\
&\quad +\|[\Delta_{q},b\times]J\|_{L^2}\|\Delta_{q}J\|_{L^2}\\
&=IL_{1}(t)+\dots+IL_{5}(t).
\end{aligned}
\end{equation}
Multiplying \eqref{a12}  by $2^q$ and summing over $q\in\mathbb{Z}$,
\begin{equation}\label{a13}
\frac{1}{2}\frac{d}{dt}(\| u\|_{\dot{H}^{1/2}}^2
+\|b\|_{\dot{H}^{1/2}}^2)+ \|\nabla u\|_{\dot{H}^{1/2}}^2
+ \|\nabla b\|_{\dot{H}^{1/2}}^2\le \sum_{q\in\mathbb{Z}}2^{q}|IL_{1}(t)
+\dots +IL_{5}(t)|.
\end{equation}
By  H\"{o}lder's inequality and choosing $\sigma=\sigma_i=\frac{1}{2}$,
$p_i=2$, $i=1,2$ in \eqref{a7},
$$
\sum_{q\in\mathbb{Z}}2^q|IL_{1}(t)|
\le\Big(\sum_{q\in \mathbb{Z}}2^q\|[\Delta_{q},u]\cdot\nabla u\|_{L^2}^2\Big)^{1/2}
 \|u\|_{\dot{H}^{1/2}}\le C\|u\|_{\dot{H}^{1/2}}\|u\|_{\dot{H}^{3/2}}^2.
$$
Similarly,
\begin{gather*}
\sum_{q\in\mathbb{Z}}2^q|IL_2(t)|
 \le C\|u\|_{\dot{H}^{1/2}}\|b\|_{\dot{H}^{3/2}}^2,\\
\sum_{q\in\mathbb{Z}}2^q|IL_3(t)|
 \le C\|u\|_{\dot{H}^{1/2}}(\|u\|_{\dot{H}^{3/2}}^2+\|b\|_{\dot{H}^{3/2}}^2),
\\
\sum_{q\in\mathbb{Z}}2^q|IL_{4}(t)|\le C\|u\|_{\dot{H}^{1/2}}
 (\|u\|_{\dot{H}^{3/2}}^2+\|b\|_{\dot{H}^{3/2}}^2),\quad
\sum_{q\in\mathbb{Z}}2^q|IL_{5}(t)|\le C\|b\|_{\dot{H}^{3/2}}^3.
\end{gather*}
Plugging the inequalities above in \eqref{a12}, we have
\begin{equation}\label{a14}
\begin{aligned}
&\frac{1}{2}\frac{d}{dt}\|(u,b)\|_{\dot{H}^{1/2}}^2
 +\|\nabla u\|_{\dot{H}^{1/2}}^2+\|\nabla b\|_{\dot{H}^{1/2}}^2\\
&\le C(\|(u,b)\|_{\dot{H}^{1/2}}+\|b\|_{\dot{H}^{3/2}})(\|u\|_{\dot{H}^{3/2}}^2
 +\|b\|_{\dot{H}^{3/2}}^2).
\end{aligned}
\end{equation}
Next, we give the $\dot{H}^{3/2}$ estimate of $b$.  With a similar process,
we obtain
\begin{align*}
&\frac{1}{2}\frac{d}{dt}\|\Delta_{q} b\|_{L^2}^2
 +\|\nabla\Delta_{q} b\|_{L^2}^2\\
&=\int_{\mathbb{R}^3}[\Delta_{q},b\cdot\nabla] u\cdot\Delta_{q}b
 +\int_{\mathbb{R}^3}b\cdot\nabla\Delta_{q}u\cdot \Delta_{q}b\\
&\quad -\int_{\mathbb{R}^3}[\Delta_{q},u\cdot\nabla] b\cdot\Delta_{q}b
 +\int_{\mathbb{R}^3}[\Delta_{q},b\times]J\cdot\Delta_{q}J\\
&\le \|[\Delta_{q},b\cdot\nabla] u\|_{L^2}\|\Delta_{q}b\|_{L^2}
 +\|b\cdot\nabla\Delta_{q}u\|_{L^2}\|\Delta_{q}b\|_{L^2}\\
&\quad +\|[\Delta_{q},u\cdot\nabla] b\|_{L^2}\|\Delta_{q}b\|_{L^2}
 +\|[\Delta_{q},b\times]J\|_{L^2}\|\Delta_{q}J\|_{L^2}\\
&=K_{1}(t)+\dots+K_{5}(t).
\end{align*}
Multiplying $2^{3q}$ and summing over $q\in\mathbb{Z}$,
\begin{equation}\label{a15}
\frac{1}{2}\frac{d}{dt}\|b\|_{\dot{H}^{3/2}}^2
+\|b\|_{\dot{H}^{5/2}}^2
\le \sum_{q\in\mathbb{Z}}2^{3q}|K_{1}(t)+\dots+K_{5}(t)|.
\end{equation}
By H\"{o}lder's inequality and choosing $\sigma=\sigma_i=\frac{1}{2}$,
$p_i=2$, $i=1,2$ in \eqref{a7},
\begin{align*}
\sum_{q\in\mathbb{Z}}2^{3q}|K_{1}(t)|
&=\sum_{q\in\mathbb{Z}}2^{3q}\|[\Delta_{q},b\cdot\nabla] u\|_{L^2}
 \|\Delta_{q}b\|_{L^2}\\
&\le \Big(\sum_{q\in \mathbb{Z}}2^q\|[\Delta_{q},b\cdot\nabla] u \|_{L^2}^2
 \Big)^{1/2}\|b\|_{\dot{H}^{5/2}}\\
&\le C\|b\|_{\dot{H}^{3/2}}(\|u\|_{\dot{H}^{3/2}}^2+\|b\|_{\dot{H}^{5/2}}^2).
\end{align*}
Similarly,
$$
\sum_{q\in\mathbb{Z}}2^{3q}|K_3(t)|
\le C\|b\|_{\dot{H}^{3/2}}(\|b\|_{\dot{H}^{5/2}}^2+\|u\|_{\dot{H}^{3/2}}^2).
$$
By H\"{o}lder's inequality and choosing $\sigma=3/2$,
$\sigma_i=\frac{1}{2}$, $p_i=2$, $i=1,2$ in \eqref{a7}, then
$$
\sum_{q\in\mathbb{Z}}2^{3q}|K_{4}(t)|
\le \Big(\sum_{q\in\mathbb{Z}}2^{3q}\|[\Delta_{q},b\times]J\|_{L^2}^2\Big)^{1/2}
\|J\|_{\dot{H}^{3/2}}\le C\|b\|_{\dot{H}^{3/2}}\|b\|_{\dot{H}^{5/2}}^2.
$$
Using H\"{o}lder's inequality and condition \eqref{a6},
\begin{align*}
\sum_{q\in\mathbb{Z}}2^{3q}|K_2(t)|
&\le\Big(\sum_{q\in\mathbb{Z}}2^q\|b\cdot\nabla\Delta_{q}u\|_{L^2}^2\Big)^{1/2}
 \|b\|_{\dot{H}^{5/2}}\\
&\le \|b\|_{L^\infty}\|u\|_{\dot{H}^{3/2}}\|b\|_{\dot{H}^{5/2}}\\
&\le \frac{C_0}{2}(\|u\|_{\dot{H}^{3/2}}^2+\|b\|_{\dot{H}^{5/2}}^2).
\end{align*}
Plugging the inequalities above in \eqref{a15}, combining with \eqref{a14}
and integrating the resulting inequality in $[0,t]$ gives
\begin{align*}
&\|(u(t),b(t))\|_{\dot{H}^{1/2}}^2
 +\|b(t)\|_{\dot{H}^{3/2}}^2+(2-C_0)\int_0^{t}
\|(u(\tau),b(\tau))\|_{\dot{H}^{3/2}}^2+\|b(\tau)\|_{\dot{H}^{5/2}}^2d\tau\\
&\le C\int_0^{t}(\|(u(\tau),b(\tau))\|_{\dot{H}^{1/2}}
 +\|b(\tau)\|_{\dot{H}^{3/2}})(\|(u(\tau),b(\tau))\|_{\dot{H}^{3/2}}^2
+\|b(\tau)\|_{\dot{H}^{5/2}}^2)d\tau\\
&\quad +\|(u_0,b_0)\|_{\dot{H}^{1/2}}^2+\|b_0\|_{\dot{H}^{3/2}}^2.
\end{align*}
Choose $\epsilon$ so small that
$3(\|u_0\|_{\dot{H}^{1/2}}^2+\|b_0\|_{\dot{H}^{1/2}}^2
 +\|b_0\|_{\dot{H}^{3/2}}^2)<(\frac{2-C_0}{4C})^2$,
 which implies
$$
C\Big(\|u_0\|_{\dot{H}^{1/2}}+\|b_0\|_{\dot{H}^{1/2}}
+\|b_0\|_{\dot{H}^{3/2}}\Big)<\frac{2-C_0}{4}.
$$
Suppose there exists a first time $T^\star$ such that
$$
3\Big(\|u(T^\star)\|_{\dot{H}^{1/2}}^2
+\|b(T^\star)\|_{\dot{H}^{1/2}}^2+\|b(T^\star)\|_{\dot{H}^{3/2}}^2\Big)
\ge\big(\frac{2-C_0}{4C}\big)^2,
$$
which can easily get a contradiction. Hence, for all $t\ge0$, we obtain
\begin{align*}
&\|(u(t),b(t))\|_{\dot{H}^{1/2}}^2+\|b(t)\|_{\dot{H}^{3/2}}^2+
{(2-C_0)}\int_0^{t}\|(u(\tau),b(\tau))\|_{\dot{H}^{3/2}}^2
 +\|b(\tau)\|_{\dot{H}^{5/2}}^2d\tau\\
&\le \|(u_0,b_0)\|_{\dot{H}^{1/2}}^2+\|b_0\|_{\dot{H}^{3/2}}^2.
\end{align*}
This completes the proof of Theorem \ref{thm1.5}.

\section{Proof of Theorem \ref{thm1.6}}

Our proof contains two steps, $H^1$ estimates and $H^s$ estimates.
\smallskip

\noindent\textbf{Step 1: $H^1$ Estimates.}
Using a similar procedure as in \cite{[3]}, we have
\begin{equation}\label{a32}
\begin{aligned}
&\frac{1}{2}\frac{d}{dt}(\|\nabla u\|_{L^2}^2
 +\|\nabla b\|_{L^2}^2)+\|\nabla\Lambda^\alpha u\|_{L^2}^2
 +\|\nabla\Lambda^\beta b\|_{L^2}^2\\
&=-\int_{\mathbb{R}^3}\nabla(u\cdot\nabla u)\cdot \nabla u\,dx
+\int_{\mathbb{R}^3}\nabla(b\cdot\nabla b)\cdot \nabla u\,dx
+\int_{\mathbb{R}^3}\nabla(b\cdot\nabla u)\cdot \nabla b\,dx\\
&\quad -\int_{\mathbb{R}^3}\nabla(u\cdot\nabla b)\cdot \nabla b\,dx
-\int_{\mathbb{R}^3}\nabla(J \times b)\cdot\nabla J\, dx\\
&=I_{1}(t)+\dots+I_{5}(t).
\end{aligned}
\end{equation}
By integrate by parts, H\"{o}lder's inequality, interpolation inequality
and Young's inequality, let $\theta_{1}=\frac{3-(\alpha-1)p_{1}}{\alpha p_{1}}$,
we obtain
\begin{align*}
|I_{1}(t)|
&\le |\int_{\mathbb{R}^3}(u\cdot \nabla)u \cdot\Delta u\,dx|\\
&\le\|u\|_{L^{p_{1}}}\|\nabla u\|_{L^{\frac{6p_{1}}{(1+2\alpha)p_{1}-6}}}
 \|\Delta u\|_{L^{\frac{6}{5-2\alpha}}}\\
&\le\|u\|_{L^{p_{1}}}\|\nabla u\|_{L^2}^{1-\theta_{1}}
 \|\nabla\Lambda^{\alpha}u\|_{L^2}^{1+\theta_{1}}\\
&\le C\|u\|_{L^{p_{1}}}^\frac{2\alpha p_{1}}{(2\alpha-1)p_{1}-3}
 \|\nabla u\|_{L^2}^2+\frac{1}{8}\|\nabla \Lambda^\alpha u\|_{L^2}^2.
\end{align*}
By cancelation property,
$$
\int_{\mathbb{R}^3}(b\cdot\nabla)\partial_ib\cdot\partial_iu\,dx
+\int_{\mathbb{R}^3}(b\cdot\nabla)\partial_iu\cdot\partial_ib\,dx=0,
$$
integrate by parts, H\"{o}lder inequality and Young's inequality,
\begin{align*}
|I_2(t)+I_3(t)|
&=\Big|\int_{\mathbb{R}^3}(b\cdot\nabla)\partial_ib\cdot\partial_iu\,dx
 +\int_{\mathbb{R}^3}(\partial_ib\cdot\nabla)b\cdot\partial_iu\,dx\\
&\quad +\int_{\mathbb{R}^3}(b\cdot\nabla)\partial_iu\cdot\partial_ib\,dx
 +\int_{\mathbb{R}^3}(\partial_ib\cdot\nabla)u\cdot\partial_ib\,dx\Big|\\
&=\Big| \int_{\mathbb{R}^3}(\partial_ib\cdot\nabla)b\cdot\partial_iu\,dx
 +\int_{\mathbb{R}^3}(\partial_ib\cdot\nabla)u\cdot\partial_ib\,dx\Big|\\
&\le \int_{\mathbb{R}^3}{|u||\nabla b||\nabla^2 b|}\,dx\\
&\le C\|u\|_{L^{p_{1}}}^\frac{2\beta p_{1}}{(2\beta-1)p_{1}-3}\|\nabla b|_{L^2}^2
 +\frac{1}{8}\|\nabla \Lambda^\beta b|_{L^2}^2.
\end{align*}
Similarly,
\begin{gather*}
|I_{4}(t)|\le C\|u\|_{L^{p_{1}}}^\frac{2\beta p_{1}}{(2\beta-1)p_{1}-3}
\|\nabla b\|_{L^2}^2+\frac{1}{8}\|\nabla \Lambda^\beta b|_{L^2}^2,
\\
|I_{5}(t)|=\int_{\mathbb{R}^3}|\nabla b||\nabla b||\nabla^2 b|
 \le C\|\nabla b\|_{L^{p_2}}^\frac{2\beta p_2}{(2\beta-1)p_2-3}
 \|\nabla b\|_{L^2}^2+\frac{1}{8}\|\nabla \Lambda^\beta b|_{L^2}^2,
\end{gather*}
where it would be used
$$
\int_{\mathbb{R}^3}\ [(\nabla J)\times b]\cdot (\nabla J)\,dx=0.
$$
Collecting the inequalities above in \eqref{a32}, together with energy estimate,
integrating in time and by Gronwall's lemma yields that
\begin{equation}\label{a41}
\begin{aligned}
&\|u\|_{H^1}^2+\|b\|_{H^1}^2+\int_0^{t}\|\Lambda^\alpha u\|_{H^1}^2d\tau
 +\int_0^{t}\|\Lambda^\beta b\|_{H^1}^2d\tau\\
&\le \|(u_0,b_0)\|_{H^1}^2
\exp\big\{C\int_0^{t}(\|u\|_{L^{p_{1}}}^\frac{2\alpha p_{1}}{(2\alpha-1)p_{1}-3}
+\|u\|_{L^{p_{1}}}^\frac{2\beta p_{1}}{(2\beta-1)p_{1}-3}
 +\|\nabla b\|_{L^{p_2}}^\frac{2\beta p_2}{(2\beta-1)p_2-3})d\tau\big\}.
\end{aligned}
\end{equation}

\noindent\textbf{Step 2: $H^s$ Estimates.}
 Applying the operator $\Lambda^s$ to \eqref{a1}, taking the inner product
with $(\Lambda^s u,\Lambda^s b)$, with energy estimate, we have
\begin{equation}\label{a33}
\begin{aligned}
&\frac{1}{2}\frac{d}{dt}(\|u\|_{H^s}^2+\|b\|_{H^s}^2)
 +\|\Lambda^\alpha u\|_{H^s}^2+\|\Lambda^\beta b\|_{H^s}^2\\
&=-\int_{\mathbb{R}^3}\Lambda^s(u\cdot\nabla)u\cdot\Lambda^s u\,dx
 +\int_{\mathbb{R}^3}\Lambda^s(b\cdot\nabla)b\cdot\Lambda^s u\,dx
 +\int_{\mathbb{R}^3}\Lambda^s(b\cdot\nabla)u\cdot\Lambda^s b\,dx\\
&\quad -\int_{\mathbb{R}^3}\Lambda^s(u\cdot\nabla)b\cdot\Lambda^s b\,dx
 -\int_{\mathbb{R}^3}\Lambda^s((\nabla\times b)\times b)\cdot
 \Lambda^s(\nabla\times b)dx\\
&=II_{1}(t)+\dots+II_{5}(t).
\end{aligned}
\end{equation}
Using H\"{o}lder's inequality and Young's inequality,
\begin{align*}
|II_{1}(t)|
&\le\|\Lambda^{s+1-\alpha}(u\otimes u)\|_{L^2}\|\Lambda^{s+\alpha}u\|_{L^2}\\
&\le C\|u\|_{L^\infty}\|u\|_{H^{s+1-\alpha}}\|\Lambda^\alpha u\|_{H^s}\\
&\le C\|u\|_{L^\infty}^2\|u\|_{H^s}^2+\frac{1}{8}\|\Lambda^\alpha u\|_{H^s}^2.
\end{align*}
Similarly,
\begin{gather*}
|II_2(t)|\le C\|b\|_{L^\infty}^2\|b\|_{H^s}^2+\frac{1}{8}
 \|\Lambda^\alpha u\|_{H^s}^2,\\
|II_3(t),II_{4}(t)|\le C(\|u\|_{L^\infty}^2
 +\|b\|_{L^\infty}^2)(\|u\|_{H^s}^2+\|b\|_{H^s}^2)
 +\frac{1}{8}\|\Lambda^\beta b\|_{H^s}^2.
\end{gather*}
For the estimate of $II_{5}(t)$, let
$\theta_2=\frac{3-(\beta-1)p_2}{\beta p_2}$, then
\begin{align*}
|II_{5}(t)|
&\le\|\Lambda^s[(\nabla\times b)\times b]-[\Lambda^s(\nabla\times b)]
 \times b\|_{L^\frac{6}{2\beta+1}}\|\nabla\Lambda^s b\|_{L^\frac{6}{5-2\beta}}\\
&\le C\|\nabla b\|_{L^{p_2}}\|\Lambda^s b\|_{L^\frac{6p_2}{(1+2\beta)p_2-6}}
 \|\Lambda^\beta b\|_{H^s}\\
&\le C\|\nabla b\|_{L^{p_2}}\|\Lambda^s b\|_{L^2}^{1-\theta_2}
 \|\Lambda^\beta b\|_{H^s}^{1+\theta_2}\\
&\le C\|\nabla b\|_{L^{p_2}}^{\frac{2\beta p_2}{(2\beta-1)p_2-3}}
 \|\Lambda^s b\|_{L^2}^2+\frac{1}{8}\|\Lambda^\beta b\|_{L^2}^2.
\end{align*}
Plugging the estimate above in  \eqref{a33}, applying Gronwall's lemma, we have
\begin{align*}
&\|u\|_{H^s}^2+\|b\|_{H^s}^2+\int_0^{t}\|\Lambda^\alpha u\|_{H^s}^2d\tau
 +\int_0^{t}\|\Lambda^\beta b\|_{H^s}^2d\tau\\
&\le \|(u_0,b_0)\|_{H^s}^2\exp\{C\exp\int_0^{t}(\|(u,b)\|_{L^\infty}^2
 +\|\nabla b\|_{L^{p_2}}^\frac{2\beta p_2}{(2\beta-1)p_2-3})d\tau\}.
\end{align*}
Combining the interpolation
$$
\|u\|_{L^\infty}\le C\|u\|_{H^1}^{1-\frac{1}{2\alpha}}
\|\nabla\Lambda^\alpha u\|_{L^2}^\frac{1}{2\alpha},\quad
\|b\|_{L^\infty}\le C\| b\|_{H^1}^{1-\frac{1}{2\beta}}
\|\nabla\Lambda^\beta b\|_{L^2}^\frac{1}{2\beta}
$$
with \eqref{a30} and \eqref{a41} yields the desired results.
This completes the proof of Theorem \ref{thm1.6}.

\section{Proof of Theorem \ref{thm1.7}}
Using similar operation as the proof of Theorem \ref{thm1.6}, we obtain
\begin{equation}\label{a34}
\begin{aligned}
&\frac{1}{2}\frac{d}{dt}(\|\nabla u\|_{L^2}^2+\|\nabla b\|_{L^2}^2)
 + \|\Delta u\|_{L^2}^2+ \|\Delta b\|_{L^2}^2\\
&=\sum_{i,j,k=1}^3\int_{\mathbb{R}^3}\partial_{k}(u_i\partial_iu_j)
 \partial_{k}u_j
 + \sum_{i,j,k=1}^3\int_{\mathbb{R}^3}\partial_{k}(b_i\partial_ib_j)
 \partial_{k}u_j\\
&\quad + \sum_{i,j,k=1}^3\int_{\mathbb{R}^3}
 \partial_{k}(b_i\partial_iu_j)\partial_{k}b_j
 -\sum_{i,j,k=1}^3\int_{\mathbb{R}^3}\partial_{k}(u_i\partial_i
 b_j)\partial_{k}b_j \\
&\quad -\sum_{i=1}^3\int_{\mathbb{R}^3}\partial_i(J\times b)
 \cdot\partial_iJ \\
&=III_{1}(t)+\dots+III_{5}(t).
\end{aligned}
\end{equation}
Integrating by parts, and using
$$
\int_{\mathbb{R}^3}\ u_i\partial_i \partial_k u_j \partial_k u_j\,dx=0,
$$
we can rewrite $III_{1}(t)$ as
\begin{align*}
III_{1}(t)
&=-\sum_{i,j,k=1}^3\int_{\mathbb{R}^3}\partial_{k}u_i
 \partial_iu_j\partial_{k}u_j\\
&=-\sum_{i=1}^2\sum_{j,k=1}^3\int_{\mathbb{R}^3}\partial_{k}u_i
 \partial_iu_j\partial_{k}u_j
- \sum_{k=1}^2\sum_{j=1}^3\int_{\mathbb{R}^3}\partial_{k}u_3
 \partial_3u_j\partial_{k}u_j\\
&\quad +\sum_{j=1}^3\int_{\mathbb{R}^3}(\partial_{1}u_{1}
 +\partial_2u_2)\partial_3u_j\partial_3u_j.
\end{align*}
Similarly,
\begin{align*}
&III_2(t)+III_3(t)\\
&=\sum_{i,j,k=1}^3\int_{\mathbb{R}^3}\partial_{k}b_i\partial_i
 b_j\partial_{k}u_j
+\sum_{i,j,k=1}^3\int_{\mathbb{R}^3}\partial_{k}b_i\partial_i
 u_j\partial_{k}b_j\\
&=\sum_{k=1}^2\sum_{i,j=1}^3\int_{\mathbb{R}^3}\partial_{k}b_i
 \partial_ib_j\partial_{k}u_j
+\sum_{i=1}^2\sum_{j=1}^3\int_{\mathbb{R}^3}\partial_3b_i
 \partial_ib_j\partial_3u_j\\
&\quad +\sum_{i=1}^2\sum_{j=1}^3\int_{\mathbb{R}^3}\partial_3b_i
 \partial_iu_j\partial_3b_j
-\sum_{j=1}^3\int_{\mathbb{R}^3}(\partial_{1}b_{1}+\partial_2b_2)
 (\partial_3b_j\partial_3u_j+\partial_3u_j\partial_3b_j),
\end{align*}

\begin{align*}
III_{4}(t)&=-\sum_{i,j,k=1}^3\int_{\mathbb{R}^3}\partial_{k}u_i
 \partial_ib_j\partial_{k}b_j\\
&=\sum_{k=1}^2\sum_{i,j=1}^3\int_{\mathbb{R}^3}\partial_{k}u_i
 \partial_ib_j\partial_{k}b_j
 +\sum_{i=1}^2\sum_{j=1}^3\int_{\mathbb{R}^3}\partial_3u_i
 \partial_ib_j\partial_3b_j\\
&\quad +\sum_{j=1}^3\int_{\mathbb{R}^3}(\partial_{1}u_{1}
 +\partial_2u_2)\partial_3b_j\partial_3b_j.
\end{align*}
Using the cancelation property as before,
\begin{align*}
III_{5}(t)
&=-\sum_{i=1}^3\int_{\mathbb{R}^3}\{\partial_i((\nabla\times b)\times b)
 \cdot\partial_i(\nabla\times b)-(\partial_i(\nabla\times b)\times b)
 \cdot\partial_i(\nabla\times b)\}\\
&=-\sum_{i=1}^2\int_{\mathbb{R}^3}(\nabla\times b)\times
 \partial_ib\cdot\partial_i(\nabla\times b)
\underbrace{-\int_{\mathbb{R}^3}(\nabla\times b)\times
\partial_3b\cdot\partial_3(\nabla\times b)}_{III_{51}},
\end{align*}
where
\begin{align*}
III_{51}
&=-\int_{\mathbb{R}^3}\{\partial_3b_3(\partial_3b_{1}-\partial_{1}b_3)
-\partial_3b_2(\partial_{1}b_2-\partial_2b_{1})\}\partial_3
 (\partial_2b_3-\partial_3b_2)\\
&\quad -\int_{\mathbb{R}^3}\{\partial_3b_{1}(\partial_{1}b_2
 -\partial_2b_{1}) -\partial_3b_3(\partial_2b_3-\partial_3b_2)\}
 \partial_3(\partial_3b_{1}-\partial_{1}b_3)\\
&\quad -\int_{\mathbb{R}^3}\{\partial_3b_2(\partial_2b_3-\partial_3b_2)
-\partial_3b_{1}(\partial_3b_{1}-\partial_{1}b_3)\}
\partial_3(\partial_{1}b_2-\partial_2b_2).
\end{align*}
Therefore, we can easily obtain the  estimates
\begin{equation}\label{a35}
\begin{gathered}
|III_{1}(t)|\le C\int_{\mathbb{R}^3}|\nabla_{h}u||\nabla u|^2dx,\\
|III_2(t)+III_3(t)|\le C\int_{\mathbb{R}^3}|\nabla_{h}u||\nabla b|^2dx
 +\underbrace{\int_{\mathbb{R}^3}|\nabla_{h}b||\nabla b||\nabla u|dx}_{\Xi},\\
|III_{4}(t)|\le C\int_{\mathbb{R}^3}|\nabla_{h}u||\nabla b|^2dx
 +\int_{\mathbb{R}^3}|\nabla_{h}b||\nabla b|^2dx,\\
|III_{5}(t)|\le C\int_{\mathbb{R}^3}|\nabla_{h}b||\nabla b||\Delta b|dx.
\end{gathered}
\end{equation}
Using H\"{o}lder's inequality, interpolation inequality and Young's inequality,
let $\theta_3=\frac{3}{2p_{1}\alpha}$,
\begin{align*}
|III_{1}(t)|
&\le C\|\nabla_{h}u\|_{L^{p_{1}}}\|\nabla u\|_{L^2}^{2(1-\theta_3)}
 \|\nabla\Lambda^\alpha u\|_{L^2}^{2\theta_3}\\
&\le C\|\nabla_{h}u\|_{L^{p_{1}}}^\frac{1}{1-\theta_3}\|\nabla u\|_{L^2}^2
 +\frac{1}{8}\|\nabla\Lambda^\alpha u\|_{L^2}^2\\
&=C\|\nabla_{h}u\|_{L^{p_{1}}}^\frac{2p_{1}\alpha}{2p_{1}\alpha-3}
 \|\nabla u\|_{L^2}^2+\frac{1}{8}\|\nabla\Lambda^\alpha u\|_{L^2}^2.
\end{align*}
For the estimate of the second term, let
$\theta_{4}=\frac{6-2p_2\alpha}{2\beta p_2}$, we have
\begin{align*}
\Xi
&\le\|\nabla_{h}b\|_{L^{p_2}}\|\nabla b\|_{L^\frac{6p_2}{(3+2\alpha)p_2-6}}\|
 \nabla u\|_{L^\frac{6}{3-2\alpha}}\\
&\le \|\nabla_{h}b\|_{L^{p_2}}\|\nabla b\|_{L^2}^{1-\theta_{4}}
 \|\nabla\Lambda^\beta b\|_{L^2}^{\theta_{4}}\|\nabla\Lambda^\alpha u\|_{L^2}\\
&\le C\|\nabla_{h}b\|_{L^{p_2}}^\frac{2}{1-\theta_{4}}\|\nabla b\|_{L^2}^2
 +\frac{1}{8}\|(\nabla\Lambda^\alpha u,\nabla\Lambda^\alpha b\|_{L^2}^2\\
&\le C\|\nabla_{h}b\|_{L^{p_2}}^\frac{2p_2\beta}{(\alpha+\beta)p_2-3}
 \|\nabla b\|_{L^2}^2+\frac{1}{8}\|(\nabla\Lambda^\alpha u,
 \nabla\Lambda^\alpha b\|_{L^2}^2.
\end{align*}
Thus,
\begin{align*}
|III_2(t)+III_3(t)|
&\le C(\|\nabla_{h}u\|_{L^{p_{1}}}^\frac{2p_{1}\beta}{2p_{1}\beta-3}
 +\|\nabla_{h}b\|_{L^{p_2}}^\frac{2p_2\beta}{(\alpha+\beta)p_2-3})
 \|\nabla b\|_{L^2}^2\\
&\quad+\frac{1}{8}(\|\nabla\Lambda^\alpha u\|_{L^2}^2
 +\|\nabla \Lambda^\beta b\|_{L^2}^2).
\end{align*}
Similarly,
\begin{gather*}
|III_{4}(t)|\le C(\|\nabla_{h}u\|_{L^{p_{1}}}^\frac{2p_{1}\beta}{2p_{1}\beta-3}
 +\|\nabla_{h}b\|_{L^{p_2}}^\frac{2p_2\beta}{2p_2\beta-3})\|\nabla b\|_{L^2}^2
+\frac{1}{4}\|\nabla\Lambda^\beta b\|_{L^2}^2,\\
|III_{5}(t)|\le C\|\nabla_{h}b\|_{L^{p_2}}^\frac{2\beta p_2}{(2\beta-1)p_2-3}
\|\nabla b\|_{L^2}^2+\frac{1}{8}\|\nabla \Lambda^\beta b\|_{L^2}^2.
\end{gather*}
Plugging the inequalities above in \eqref{a34}, integrating in $[0,t]$,
with energy estimate,
\begin{align*}
&\|(u,b)\|_{H^1}^2+\int_0^{t}\|\Lambda^\alpha u\|_{H^1}^2d\tau
 +\int_0^{t}\|\Lambda^\beta b\|_{H^1}^2d\tau\\
&\le C\int_0^{t}(\|\nabla_{h}b\|_{L^{p_2}}^\frac{2\beta p_2}{(2\beta-1)p_2-3}
 +\|\nabla_{h}u\|_{L^{p_{1}}}^\frac{2p_{1}\alpha}{2p_{1}\alpha-3}
 +\|\nabla_{h}u\|_{L^{p_{1}}}^\frac{2p_{1}\beta}{2p_{1}\beta-3}\\
&\quad +\|\nabla_{h}b\|_{L^{p_2}}^\frac{2p_2\beta}{(\alpha+\beta)p_2-3}
  +\|\nabla_{h}b\|_{L^{p_2}}^\frac{2p_2\beta}{2p_2\beta-3})\|(u,b)\|_{H^1}^2
  +\|(u_0,b_0)\|_{H^1}^2.
\end{align*}
Thanks to  Gronwall's lemma, with assumption \eqref{a31}, we obtain
$$
\|(u,b)\|_{H^1}^2+\int_0^{T}\|\Lambda^\alpha u\|_{H^1}^2dt
+\int_0^{T}\|\Lambda^\beta b\|_{H^1}^2dt<\infty,
$$
which implies that
$$
\sup_{0\le t\le T}\|u\|_{L^3}<\infty, \quad
\int_0^{T}\|\nabla b\|_{L^3}^\frac{\beta}{\beta-1}dt<\infty.
$$
Combining with regularity criterion \eqref{a31} can lead the desired result.
So we complete the proof of Theorem \ref{thm1.7}.

\subsection*{Acknowledgments}
The author would like to thank  Prof. Dongho Chae and Prof. Zhifei Zhang
for their interest and comments.


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\end{document}