Remark on the global regularity of 2D MHD equations

with almost Laplacian magnetic diffusion

Zhuan Ye

Department of Mathematics and Statistics, Jiangsu Normal University,

101 Shanghai Road, Xuzhou 221116, Jiangsu, PR China

E-mail<EMAIL_ADDRESS>

Abstract: Whether or not the classical solutions of the two-dimensional (2D)
incompressible magnetohydrodynamics (MHD) equations with only Laplacian
magnetic diffusion (without velocity dissipation) are globally well-posed is a
difficult problem and remains completely open. In this paper, we establish the
global regularity of solutions to the 2D incompressible MHD equations with
almost Laplacian magnetic diffusion in the whole space. This result can be
regarded as a further improvement and generalization of the previous works.
Consequently, our result is more closer to the resolution of the global
regularity issue on the 2D MHD equations with standard Laplacian magnetic
diffusion.

AMS Subject Classification 2010: 35Q35; 35B35; 35B65; 76D03.

Keywords: Generalized MHD equations; Classical solutions; Global regularity.

##  1\. Introduction

In this article we focus on the global regularity problem concerning the 2D
generalized incompressible magnetohydrodynamic (GMHD) equations of the form in
the whole space

$\left\\{\begin{aligned} &\partial_{t}u+(u\cdot\nabla)u+\nabla
p=(b\cdot\nabla)b,\qquad x\in\mathbb{R}^{2},\,t>0,\\\
&\partial_{t}b+(u\cdot\nabla)b+\mathcal{L}b=(b\cdot\nabla)u,\\\ &\nabla\cdot
u=0,\ \ \ \nabla\cdot b=0,\\\
&u(x,0)=u_{0}(x),\,\,b(x,0)=b_{0}(x),\end{aligned}\right.$ (1.1)

where $u=(u_{1}(x,t),\,u_{2}(x,t))$ denotes the velocity, $p=p(x,t)$ the
scalar pressure and $b=(b_{1}(x,t),\,b_{2}(x,t))$ the magnetic field of the
fluid. $u_{0}(x)$ and $b_{0}(x)$ are the given initial data satisfying
$\nabla\cdot u_{0}=\nabla\cdot b_{0}=0$. The operator $\mathcal{L}$ is a
Fourier multiplier with symbol $|\xi|^{2}g(\xi)$, namely,

$\widehat{\mathcal{L}b}(\xi)=|\xi|^{2}g(\xi)\widehat{b}(\xi),$

with $g(\xi)=g(|\xi|)$ a radial non-decreasing smooth function satisfying the
following two conditions

(a) $g$ obeys that

$g(\xi)>0\quad\mbox{for all}\ \ \xi\neq 0;$

(b) $g$ is of the Mikhlin-H$\rm\ddot{o}$mander type, namely, a constant
$\widetilde{C}>0$ such that

$|\partial_{\xi}^{k}g(\xi)|\leq\widetilde{C}|\xi|^{-k}|g(\xi)|,\quad
k\in\\{1,\,2\\},\ \forall\,\xi\neq 0.$

The MHD equations model the complex interaction between the fluid dynamic
phenomena, such as the magnetic reconnection in astrophysics and geomagnetic
dynamo in geophysics, plasmas, liquid metals, and salt water, etc (see, e.g.,
[6, 13]). The fundamental concept behind MHD is that magnetic fields can
induce currents in a moving conductive fluid, which in turn creates forces on
the fluid and also changes the magnetic field itself. Besides their important
physical applications, the MHD equations and GMHD equations are also
mathematically significant. Due to the physical background and mathematical
relevance, the MHD equations attracted quite a lot of attention lately from
various authors. One of the fundamental problems concerning the MHD equations
is whether physically relevant regular solutions remain smooth for all time or
they develop finite time singularities. Actually, there is a considerable body
of literature on the global regularity of the MHD equations with different
form of the dissipation (see e.g. [5, 4, 3, 7, 8, 9, 16, 17, 18, 19, 20, 23,
21, 22, 24] and the references cited therein), and here for our purpose we
only recall the notable works about the 2D fractional MHD equations

$\left\\{\begin{aligned}
&\partial_{t}u+(u\cdot\nabla)u+(-\Delta)^{\alpha}u+\nabla
p=(b\cdot\nabla)b,\qquad x\in\mathbb{R}^{2},\,t>0,\\\
&\partial_{t}b+(u\cdot\nabla)b+(-\Delta)^{\beta}b=(b\cdot\nabla)u,\\\
&\nabla\cdot u=0,\ \ \ \nabla\cdot b=0,\\\ &u(x,0)=u_{0}(x),\ \
b(x,0)=b_{0}(x),\end{aligned}\right.$ (1.2)

where $(-\Delta)^{\gamma}$ is defined by the Fourier transform, namely

$\widehat{(-\Delta)^{\gamma}f}(\xi)=|\xi|^{2\gamma}\hat{f}(\xi).$

We remark the convention that by $\alpha=0$ we mean that there is no
dissipation in $(\ref{2dGMHD})_{1}$, and similarly $\beta=0$ represents that
there is no diffusion in $(\ref{2dGMHD})_{2}$. It is well-known that the 2D
MHD equations with both Laplacian dissipation and magnetic diffusion
($\alpha=\beta=1$) have the global smooth solution (e.g. [15]). In the
completely inviscid case ($\alpha=\beta=0$), the question of whether smooth
solution of the 2D MHD equations develops singularity in finite time appears
to be out of reach. Therefore, it is natural to examine the MHD equations with
fractional dissipation (see e.g. [5, 7, 8, 9, 12, 14, 17, 23, 21, 22, 25] and
the references cited therein). Let us review several works about the system
(1.2) which are very closely related to our study. To the best of our
knowledge, the issue of the global regularity for 2D resistive MHD
$(\alpha=0,\beta=1)$ is still a challenging open problem (see [4, 11]) as we
are unable to derive the key a priori estimate of $\|\nabla
j\|_{L_{t}^{1}L^{\infty}}$. But if more dissipation is added, then the
corresponding system do admit a unique global regular solution. On the one
hand, when $\alpha>0,\beta=1$, the global regularity issue was solved by Fan
and al. [7]. Very recently, Yuan and Zhao improved this work as they obtained
the global regularity of solutions requiring the dissipative operators weaker
than any power of the fractional Laplacian. On the other hand, when
$\alpha=0,\beta>1$, the global regularity of smooth solutions for the
corresponding system was established in works [5, 9] with quite different
approach, which was also improved by Agelas [1] with the $(-\Delta)^{\beta}b$
($\beta>1$) replaced by $-\Delta\big{(}\log(e-\Delta)\big{)}^{\kappa}b$
($\kappa>1$).

Inspired by the previous works, the aim of this paper is to weaken the
operator $\mathcal{L}$ as possible as one can, without losing the global
regularity of the system (1.1). More precisely, our main result reads as
follows.

######  Theorem 1.1.

Let $(u_{0},b_{0})\in H^{s}(\mathbb{R}^{2})\times H^{s}(\mathbb{R}^{2})$ for
any $s>2$ and satisfy $\nabla\cdot u_{0}=\nabla\cdot b_{0}=0$. Assume that
$g(\xi)=g(|\xi|)$ is a radial non-decreasing smooth function satisfying the
conditions (a)-(b) and the following growth condition: for any given finite
$T\in(0,\,\infty)$ such that

$\displaystyle\int_{A_{T}}^{\infty}{\frac{dr}{rg(r)}}=C_{T}<\infty,$ (1.3)

where $A_{T}>0$ is the unique solution of the following equation

$x^{2}g(x)=\frac{1}{T},$

then the system (1.1) admits a unique global solution such that

$u,\,\,b\in L^{\infty}\big{(}[0,T];H^{s}(\mathbb{R}^{2})\big{)},\ \ \ b\in
L^{2}\big{(}[0,T];H^{s+1}(\mathbb{R}^{2})\big{)}.$

######  Remark 1.2.

On the one hand, $A_{T}$ is a decreasing function of $T$ which obviously
satisfies

$\lim_{T\rightarrow\infty}A_{T}=0.$

On the other hand, $C_{T}$ is a non-decreasing function of $T$ which can
satisfy the condition

$\lim_{T\rightarrow\infty}C_{T}=\infty.$

######  Remark 1.3.

The typical examples of $g$ satisfying the conditions (a)-(b) and (1.3) are

$\begin{split}&g(r)=r^{\mu_{1}}\ \ \mbox{with}\ \mu_{1}>0;\\\
&g(r)=\big{(}\ln(1+r)\big{)}^{\mu_{2}}\ \ \mbox{with}\ \mu_{2}>1;\\\
&g(r)=r^{\mu_{3}}\big{(}\ln(1+r)\big{)}^{\mu_{4}}\ \ \mbox{with}\
\mu_{3}>0,\,\mu_{4}\geq 0;\\\
&g(r)=\ln(1+r)\Big{(}\ln\big{(}1+\ln(1+r)\big{)}\Big{)}^{\mu_{5}}\ \
\mbox{with}\ \mu_{5}>1.\end{split}$

######  Remark 1.4.

In [1], Agelas proved the the global regularity of the smooth solution of the
system (1.1) with the operator given by
$\mathcal{L}=-\Delta\big{(}\log(e-\Delta)\big{)}^{\kappa}$ for $\kappa>1$.
Obviously, one can easily check that
$\big{(}\log(e+|\xi|^{2})\big{)}^{\kappa}$ for $\kappa>1$ satisfies our
conditions (a)-(b) and (1.3). Clearly, our result generalizes the work [1],
and improves the previous works [5, 9] which require the dissipative
$(-\Delta)^{\beta}b$ with $\beta>1$.

######  Remark 1.5.

For the system (1.1), it remains an open problem whether there exists a global
smooth solution when the function $g(\xi)$ is a positive constant.

The plan of this paper is as follows: we collect some useful properties of the
operator $\mathcal{L}$ in the Section 2, and then we prove Theorem 1.1 in the
Section 3. Throughout this paper, the letter $C$ denotes various positive and
finite constants whose exact values are unimportant and may vary from line to
line.

##  2\.  The properties of the operator $\mathcal{L}$

This section is devoted to establishing some properties of the operator
$\mathcal{L}$, which are the key components in proving our main theorem. To
begin with, we consider the following linear inhomogeneous equation

$\left\\{\begin{aligned} &\partial_{t}W+\mathcal{L}W=f,\\\
&W(x,0)=W_{0}(x).\end{aligned}\right.$ (2.1)

Thanks to the Fourier transform method, the solution of the linear
inhomogeneous equation (2.1) can be explicitly given by

$W(t)=K(t)\ast W_{0}+\int_{0}^{t}{K(t-\tau)\ast f(\tau)\,d\tau},$ (2.2)

where the kernel function $K$ satisfies

$\displaystyle K(x,t)=\mathcal{F}^{-1}\big{(}e^{-t|\xi|^{2}g(\xi)}\big{)}(x).$

Now we will establish the following estimate which plays an essential role in
proving our main theorem.

######  Lemma 2.1.

For any $k\geq 0$ and $s>k-1$, there exists a constant $C$ depending only on
$s$ and $k$ such that for any $t>0$

$\displaystyle\int_{\mathbb{R}^{2}}{|\xi|^{2s}g(\xi)^{k}e^{-2t|\xi|^{2}g(\xi)}\,d\xi}\leq
Ct^{-(s+1)}g(A_{t})^{-(s-k+1)},$ (2.3)

where $A_{t}>0$ is the unique solution of the following equation

$\displaystyle x^{2}g(x)=\frac{1}{t}.$ (2.4)

######  Proof of Lemma 2.1.

We split the integral into the following two parts

$\displaystyle\int_{\mathbb{R}^{2}}{|\xi|^{2s}g(\xi)^{k}e^{-2t|\xi|^{2}g(\xi)}\,d\xi}=\int_{|\xi|\leq
R}{|\xi|^{2s}g(\xi)^{k}e^{-2t|\xi|^{2}g(\xi)}\,d\xi}+\int_{|\xi|\geq
R}{|\xi|^{2s}g(\xi)^{k}e^{-2t|\xi|^{2}g(\xi)}\,d\xi},$

where $R>0$ will be fixed hereafter. Recalling $s>k-1$, direct computations
yield that the first part admits the following bound

$\displaystyle\int_{|\xi|\leq
R}{|\xi|^{2s}g(\xi)^{k}e^{-2t|\xi|^{2}g(\xi)}\,d\xi}=$
$\displaystyle\int_{|\xi|\leq
R}{|\xi|^{2s}\big{(}2t|\xi|^{2}\big{)}^{-k}\big{(}2t|\xi|^{2}g(\xi)\big{)}^{k}e^{-2t|\xi|^{2}g(\xi)}\,d\xi}$
$\displaystyle\leq$ $\displaystyle Ct^{-k}\int_{|\xi|\leq
R}{|\xi|^{2s-2k}\,d\xi}$ $\displaystyle\leq$ $\displaystyle
Ct^{-k}\int_{0}^{R}{r^{2s-2k+1}\,dr}$ $\displaystyle\leq$ $\displaystyle
Ct^{-k}R^{2(s-k+1)},$ (2.5)

where in the second line we have used the simple fact

$\displaystyle\max_{\lambda\geq 0}(\lambda^{k}e^{-\lambda})\leq C(k).$ (2.6)

Thanks to the above fact (2.6), by some simple computations, we can estimate
the remainder term as follows

$\displaystyle\int_{|\xi|\geq
R}{|\xi|^{2s}g(\xi)^{k}e^{-2t|\xi|^{2}g(\xi)}\,d\xi}=$
$\displaystyle\int_{|\xi|\geq
R}{|\xi|^{2s}\big{(}t|\xi|^{2}\big{)}^{-k}\big{(}t|\xi|^{2}g(\xi)\big{)}^{k}e^{-t|\xi|^{2}g(\xi)}e^{-t|\xi|^{2}g(\xi)}\,d\xi}$
$\displaystyle\leq$ $\displaystyle Ct^{-k}\int_{|\xi|\geq
R}{|\xi|^{2s-2k}e^{-t|\xi|^{2}g(\xi)}\,d\xi}$ $\displaystyle\leq$
$\displaystyle Ct^{-k}\int_{|\xi|\geq
R}{|\xi|^{2s-2k}e^{-t|\xi|^{2}g(R)}\,d\xi}$ $\displaystyle\leq$ $\displaystyle
Ct^{-k}\int_{R}^{\infty}{r^{2s-2k+1}e^{-tg(R)r^{2}}\,dr}$ $\displaystyle\leq$
$\displaystyle Ct^{-(s+1)}g(R)^{-(s-k+1)}.$ (2.7)

Combining (2) and (2), we get

$\displaystyle\int_{\mathbb{R}^{2}}{|\xi|^{2s}g(\xi)^{k}e^{-2t|\xi|^{2}g(\xi)}\,d\xi}\leq$
$\displaystyle Ct^{-k}R^{2(s-k+1)}+Ct^{-(s+1)}g(R)^{-(s-k+1)}$
$\displaystyle\leq$ $\displaystyle
Ct^{-(s+1)}\Big{(}(R^{2}t)^{(s-k+1)}+g(R)^{-(s-k+1)}\Big{)}.$

Now taking

$R^{2}t=\frac{1}{g(R)}\quad\mbox{or}\quad
R^{2}g(R)=\frac{1}{t}\quad\mbox{or}\quad R=A_{t},$

we eventually deduce

$\displaystyle\int_{\mathbb{R}^{2}}{|\xi|^{2s}g(\xi)^{k}e^{-2t|\xi|^{2}g(\xi)}\,d\xi}\leq
2Ct^{-(s+1)}g(R)^{-(s-k+1)}=2Ct^{-(s+1)}g(A_{t})^{-(s-k+1)}.$

Thus, we get the desired result. This concludes the proof of Lemma 2.1. ∎

With the help of Lemma 2.1, one can conclude the following results.

######  Lemma 2.2.

For any $s\geq 0$, there exists a constant $C$ depending only on $s$ such that
for any $t>0$

$\displaystyle\|K(t)\|_{\dot{H}^{s}}\leq
Ct^{-\frac{s+1}{2}}g(A_{t})^{-\frac{s+1}{2}}.$ (2.8)

In particular, it holds

$\displaystyle\|K(t)\|_{L^{\infty}}\leq Ct^{-1}g(A_{t})^{-1},$ (2.9)

where $A_{t}$ is given by (2.4).

######  Proof of Lemma 2.2.

According to (2.3), one immediately obtains

$\displaystyle\|K(t)\|_{\dot{H}^{s}}^{2}=$
$\displaystyle\int_{\mathbb{R}^{2}}{|\xi|^{2s}|\widehat{K}(\xi,t)|^{2}}\,d\xi$
$\displaystyle=$
$\displaystyle\int_{\mathbb{R}^{2}}{|\xi|^{2s}e^{-2t|\xi|^{2}g(\xi)}}\,d\xi$
$\displaystyle\leq$ $\displaystyle Ct^{-(s+1)}g(A_{t})^{-(s+1)}.$

By the simple interpolation inequality and (2.8), we have

$\displaystyle\|K(t)\|_{L^{\infty}}\leq$ $\displaystyle
C\|K(t)\|_{L^{2}}^{\frac{1}{2}}\|K(t)\|_{\dot{H}^{2}}^{\frac{1}{2}}$
$\displaystyle\leq$ $\displaystyle
Ct^{-\frac{1}{4}}g(A_{t})^{-\frac{1}{4}}t^{-\frac{3}{4}}g(A_{t})^{-\frac{3}{4}}$
$\displaystyle=$ $\displaystyle Ct^{-1}g(A_{t})^{-1}.$

Therefore, we finally conclude the proof of Lemma 2.2. ∎

Next we would like to show the following lemma.

######  Lemma 2.3.

There exists a constant $C$ such that for any $t>0$

$\displaystyle\|\nabla^{2}K(t)\|_{L^{1}}\leq Ct^{-1}g(A_{t})^{-1},$ (2.10)

where $A_{t}$ is given by (2.4).

###### Proof of Lemma 2.3.

Since for any radial function $g$ in $\mathbb{R}^{n}$, one may check that

$\widehat{g}(x)=\check{g}(x),$

where $\check{g}$ denotes the inverse Fourier transform of $g$. Since
$\widehat{K}(.,t)$ is a radial function, then we infer that ${K}(.,t)$ is also
a radial function. Based on this observation and the following interpolation
inequality (see the end of this lemma)

$\displaystyle\|\widehat{h}\|_{L^{1}}\leq
C\|h\|_{L^{2}}^{\frac{1}{2}}\|h\|_{\dot{H}^{2}}^{\frac{1}{2}}$ (2.11)

it yields that

$\displaystyle\|\nabla^{2}K(t)\|_{L^{1}}$
$\displaystyle=\|\mathcal{F}(\xi^{2}\check{K}(\xi,t))\|_{L^{1}}$
$\displaystyle=\|\mathcal{F}(\xi^{2}\widehat{K}(\xi,t))\|_{L^{1}}$
$\displaystyle\leq
C\|\xi^{2}\widehat{K}(\xi,t)\|_{L^{2}}^{\frac{1}{2}}\|\nabla_{\xi}^{2}(\xi^{2}\widehat{K}(\xi,t))\|_{L^{2}}^{\frac{1}{2}},$
(2.12)

where we have denoted by $\xi^{2}$ the matrix $\xi\otimes\xi$. By (2.8), one
gets

$\displaystyle\|\xi^{2}\widehat{K}(\xi,t)\|_{L^{2}}\leq
Ct^{-\frac{3}{2}}g(A_{t})^{-\frac{3}{2}}.$ (2.13)

It follows from some direct computations that

$|\nabla_{\xi}^{2}(\xi^{2}\widehat{K}(\xi,t))|\leq
C(|\widehat{K}(\xi,t)|+|\xi|\,|\nabla_{\xi}\widehat{K}(\xi,t)|+|\xi|^{2}|\nabla_{\xi}^{2}\widehat{K}(\xi,t)|).$

Recalling $\widehat{K}(\xi,t)=e^{-t|\xi|^{2}g(\xi)}$ and the property
$|\partial_{\xi}^{k}g(\xi)|\leq\widetilde{C}|\xi|^{-k}|g(\xi)|,\,k\in\\{1,\,2\\}$,
it is not hard to check

$\displaystyle|\nabla_{\xi}\widehat{K}(\xi,t)|$ $\displaystyle\leq
C(t|\xi|\,|g(\xi)|+t|\xi|^{2}|\partial_{\xi}g(\xi)|)|\widehat{K}(\xi,t)|$
$\displaystyle\leq Ct|\xi|\,|g(\xi)|\,|\widehat{K}(\xi,t)|$

and

$\displaystyle|\nabla_{\xi}^{2}\widehat{K}(\xi,t)|\leq$ $\displaystyle
C(t^{2}|\xi|^{2}|g(\xi)|^{2}+t^{2}|\xi|^{4}|\partial_{\xi}g(\xi)|^{2}+t|g(\xi)|+t|\xi||\partial_{\xi}g(\xi)|+t|\xi|^{2}|\partial_{\xi}^{2}g(\xi)|)$
$\displaystyle\times|\widehat{K}(\xi,t)|$ $\displaystyle\leq$ $\displaystyle
C(t^{2}|\xi|^{2}|g(\xi)|^{2}+t|g(\xi)|)|\widehat{K}(\xi,t)|.$ (2.14)

Therefore, we obtain

$|\nabla_{\xi}^{2}(\xi^{2}\widehat{K}(\xi,t))|\leq
C(1+t|\xi|^{2}|g(\xi)|+t^{2}|\xi|^{4}|g(\xi)|^{2})|\widehat{K}(\xi,t)|.$

We appeal to (2.3) to get

$\displaystyle\|\nabla_{\xi}^{2}(\xi^{2}\widehat{K}(\xi,t))\|_{L^{2}}^{2}$
$\displaystyle\leq
C\int_{\mathbb{R}^{2}}{e^{-2t|\xi|^{2}g(\xi)}\,d\xi}+Ct^{2}\int_{\mathbb{R}^{2}}{|\xi|^{4}g(\xi)^{2}e^{-2t|\xi|^{2}g(\xi)}\,d\xi}$
$\displaystyle\quad+Ct^{4}\int_{\mathbb{R}^{2}}{|\xi|^{8}g(\xi)^{4}e^{-2t|\xi|^{2}g(\xi)}\,d\xi}$
$\displaystyle\leq
Ct^{-1}g(A_{t})^{-1}+Ct^{2}t^{-3}g(A_{t})^{-1}+Ct^{4}t^{-5}g(A_{t})^{-1}$
$\displaystyle\leq Ct^{-1}g(A_{t})^{-1}.$ (2.15)

Inserting (2.13) and (2) into (2) yields

$\|\nabla^{2}K(t)\|_{L^{1}}\leq Ct^{-1}g(A_{t})^{-1},$

which is nothing but the desired estimate (2.10). Finally, let us show (2.11)

$\displaystyle\|\widehat{h}\|_{L^{1}}$
$\displaystyle=\int_{\mathbb{R}^{2}}|\widehat{h}(\xi)|\,d\xi$
$\displaystyle=\int_{|\xi|\leq N}|\widehat{h}(\xi)|\,d\xi+\int_{|\xi|\geq
N}|\widehat{h}(\xi)|\,d\xi$ $\displaystyle=\int_{|\xi|\leq
N}|\widehat{h}(\xi)|\,d\xi+\int_{|\xi|\geq
N}|\xi|^{-2}|\xi|^{2}|\widehat{h}(\xi)|\,d\xi$ $\displaystyle\leq
CN\Big{(}\int_{|\xi|\leq
N}|\widehat{h}(\xi)|^{2}\,d\xi\Big{)}^{\frac{1}{2}}+CN^{-1}\Big{(}\int_{|\xi|\geq
N}|\xi|^{4}|\widehat{h}(\xi)|^{2}\,d\xi\Big{)}^{\frac{1}{2}}$
$\displaystyle\leq CN\|h\|_{L^{2}}+CN^{-1}\|h\|_{\dot{H}^{2}}$
$\displaystyle\leq
C\|h\|_{L^{2}}^{\frac{1}{2}}\|h\|_{\dot{H}^{2}}^{\frac{1}{2}},$

where in the last line we have chosen $N$ as

$N=\Big{(}\frac{\|h\|_{\dot{H}^{2}}}{\|h\|_{L^{2}}}\Big{)}^{\frac{1}{2}}.$

We thus complete the proof of the lemma. ∎

Now we are ready to show the following key lemma.

######  Lemma 2.4.

Under the assumptions of Theorem 1.1, there holds

$\displaystyle\int_{0}^{T}{\|\nabla
K(t)\|_{L^{2}}\,dt}+\int_{0}^{T}{\|K(t)\|_{L^{2}}^{2}\,dt}+\int_{0}^{T}{\|K(t)\|_{L^{\infty}}\,dt}+\int_{0}^{T}{\|\nabla^{2}K(t)\|_{L^{1}}\,dt}\leq
C(T).$

###### Proof of Lemma 2.4.

Making use of Lemma 2.2 and Lemma 3.3, it follows that

$\|\nabla
K(t)\|_{L^{2}}+\|K(t)\|_{L^{2}}^{2}+\|K(t)\|_{L^{\infty}}+\|\nabla^{2}K(t)\|_{L^{1}}\leq
Ct^{-1}g(R)^{-1},$

where $R$ satisfies

$R^{2}g(R)=\frac{1}{t}.$

Taking $R$-derivative of the both sides of the above equation, it is easy to
check

$2Rg(R)+R^{2}g^{\prime}(R)=-\frac{1}{t^{2}}\frac{dt}{dR}$

or equivalently

$dt=-t^{2}\big{(}2Rg(R)+R^{2}g^{\prime}(R)\big{)}dR=-R^{-4}g(R)^{-2}\big{(}2Rg(R)+R^{2}g^{\prime}(R)\big{)}dR.$

By direct computations, we have that

$\displaystyle\int_{0}^{T}{t^{-1}g(R)^{-1}\,dt}$
$\displaystyle=-\int_{+\infty}^{A_{T}}{R^{-2}g(R)^{-2}\big{(}2Rg(R)+R^{2}g^{\prime}(R)\big{)}\,dR}$
$\displaystyle=\int_{A_{T}}^{+\infty}{R^{-2}g(R)^{-2}\big{(}2Rg(R)+R^{2}g^{\prime}(R)\big{)}\,dR}$
$\displaystyle=2\int_{A_{T}}^{+\infty}{R^{-1}g(R)^{-1}\,dR}+\int_{A_{T}}^{+\infty}{g(R)^{-2}g^{\prime}(R)\,dR}$
$\displaystyle=2\int_{A_{T}}^{+\infty}{R^{-1}g(R)^{-1}\,dR}+g(A_{T})^{-1}-g(+\infty)^{-1}$
$\displaystyle\leq 2\int_{A_{T}}^{+\infty}{R^{-1}g(R)^{-1}\,dR}+TA_{T}^{2}$
$\displaystyle\leq C(T),$

where we have applied (1.3) and the fact $A_{T}^{2}g(A_{T})=\frac{1}{T}$ or
$g(A_{T})^{-1}=TA_{T}^{2}$. Consequently, it is obvious to check that

$\int_{0}^{T}{\|\nabla
K(t)\|_{L^{2}}\,dt}+\int_{0}^{T}{\|K(t)\|_{L^{2}}^{2}\,dt}+\int_{0}^{T}{\|K(t)\|_{L^{\infty}}\,dt}+\int_{0}^{T}{\|\nabla^{2}K(t)\|_{L^{1}}\,dt}\leq
C(T).$

This concludes the proof of the lemma. ∎

##  3\.  The proof of Theorem 1.1

This section is devoted to the proof of Theorem 1.1, the global existence and
unique- ness of smooth solution to the system $(\ref{GMHD})$. Before the
proof, we will state a notation. For a quasi-Banach space $X$ and for any
$0<T\leq\infty$, we use standard notation $L^{p}(0,T;X)$ or $L_{T}^{p}(X)$ for
the quasi-Banach space of Bochner measurable functions $f$ from $(0,T)$ to $X$
endowed with the norm

$\|f\|_{L_{T}^{p}(X)}:=\left\\{\begin{aligned}
&\left(\int_{0}^{T}{\|f(.,t)\|_{X}^{p}\,dt}\right)^{\frac{1}{p}},\,\,\,\,\,1\leq
p<\infty,\\\ &\sup_{0\leq t\leq T}\|f(.,t)\|_{X},\qquad\qquad
p=\infty.\end{aligned}\right.$

By the classical hyperbolic method, there exists a finite time $T_{0}$ such
that the system (1.1) is local well-posedness in the interval $[0,\,T_{0}]$ in
$H^{s}$ with $s>2$. Thus, it is sufficient to establish a priori estimates in
the interval $[0,\,T]$ for the given $T>T_{0}$.

We first state the basic $L^{2}$-estimate of the system (1.1).

######  Lemma 3.1.

Let $(u_{0},b_{0})$ satisfy the conditions stated in Theorem 1.1, then it
holds

$\displaystyle\|u(t)\|_{L^{2}}^{2}+\|b(t)\|_{L^{2}}^{2}+2\int_{0}^{t}{\|\mathcal{L}^{\frac{1}{2}}b(\tau)\|_{L^{2}}^{2}\,d\tau}=\|u_{0}\|_{L^{2}}^{2}+\|b_{0}\|_{L^{2}}^{2},$
(3.1)

where $\mathcal{L}^{\frac{1}{2}}$ is defined by

$\widehat{\mathcal{L}^{\frac{1}{2}}b}(\xi)=|\xi|\sqrt{g(\xi)}\,\widehat{b}(\xi).$

###### Proof of Lemma 3.1.

Taking the inner products of $(\ref{GMHD})_{1}$ with $u$ and
$(\ref{GMHD})_{2}$ with $b$, adding the results and invoking the following
cancelation identity

$\int_{\mathbb{R}^{2}}{(b\cdot\nabla)b\cdot
u\,dx}+\int_{\mathbb{R}^{2}}{(b\cdot\nabla)u\cdot b\,dx}=0,$

we obtain that

$\displaystyle\frac{1}{2}\frac{d}{dt}(\|u(t)\|_{L^{2}}^{2}+\|b(t)\|_{L^{2}}^{2})+\|\mathcal{L}^{\frac{1}{2}}b\|_{L^{2}}^{2}=0,$

where we have used the following estimate due to the Plancherel theorem

$\displaystyle\int_{\mathbb{R}^{2}}{\mathcal{L}b\,b\,dx}=\int_{\mathbb{R}^{2}}{\widehat{\mathcal{L}b}(\xi)\,\widehat{b}(\xi)\,d\xi}=\int_{\mathbb{R}^{2}}{|\xi|^{2}{g(\xi)}|\widehat{b}(\xi)|^{2}\,d\xi}=\int_{\mathbb{R}^{2}}{|\widehat{\mathcal{L}^{\frac{1}{2}}b}(\xi)|^{2}\,d\xi}=\|\mathcal{L}^{\frac{1}{2}}b\|_{L^{2}}^{2}.$

Integrating the above inequality implies (3.1). This ends the proof of Lemma
3.1. ∎

In order to get the $H^{1}$ estimate on $(u,b)$, we apply $\nabla\times$ to
the MHD equations (1.1) to obtain the governing equations for the vorticity
$\omega:=\nabla\times u=\partial_{x_{1}}u_{2}-\partial_{x_{2}}u_{1}$ and the
current $j:=\nabla\times b=\partial_{x_{1}}b_{2}-\partial_{x_{2}}b_{1}$ as
follows

$\left\\{\begin{aligned}
&\partial_{t}\omega+(u\cdot\nabla)\omega=(b\cdot\nabla)j,\\\
&\partial_{t}j+(u\cdot\nabla)j+\mathcal{L}j=(b\cdot\nabla)\omega+T(\nabla
u,\nabla b),\end{aligned}\right.$ (3.2)

where $T(\nabla u,\nabla
b)=2\partial_{x_{1}}b_{1}(\partial_{x_{2}}u_{1}+\partial_{x_{1}}u_{2})-2\partial_{x_{1}}u_{1}(\partial_{x_{2}}b_{1}+\partial_{x_{1}}b_{2}).$

We now prove that any classical solution of (1.1) admits a global
$H^{1}$-bound, as stated in the following lemma.

######  Lemma 3.2.

Let $(u_{0},b_{0})$ satisfy the conditions stated in Theorem 1.1, then it
holds for any $t\in[0,\,T]$

$\displaystyle\|\omega(t)\|_{L^{2}}^{2}+\|j(t)\|_{L^{2}}^{2}+\int_{0}^{t}{\|\mathcal{L}^{\frac{1}{2}}j(\tau)\|_{L^{2}}^{2}\,d\tau}\leq
C(T,u_{0},b_{0}).$ (3.3)

###### Proof of Lemma 3.2.

Taking the inner products of $(\ref{VMHD})_{1}$ with $\omega$,
$(\ref{VMHD})_{2}$ with $j$, adding them up and using the incompressible
condition as well as the following fact

$\int_{\mathbb{R}^{2}}{(b\cdot\nabla
j)\omega\,dx}+\int_{\mathbb{R}^{2}}{(b\cdot\nabla\omega)j\,dx}=0,$

it yields

$\displaystyle\frac{1}{2}\frac{d}{dt}(\|\omega(t)\|_{L^{2}}^{2}+\|j(t)\|_{L^{2}}^{2})+\|\mathcal{L}^{\frac{1}{2}}j\|_{L^{2}}^{2}=\int_{\mathbb{R}^{2}}{T(\nabla
u,\nabla b)j\,dx}.$

We obtain by direct computations that

$\displaystyle\|\nabla\phi\|_{L^{2}}^{2}$
$\displaystyle=C\|\xi\widehat{\phi}(\xi)\|_{L^{2}}^{2}$
$\displaystyle=C\int_{\mathbb{R}^{2}}|\xi|^{2}|\widehat{\phi}(\xi)|^{2}\,d\xi$
$\displaystyle=C\int_{|\xi|\leq
1}|\xi|^{2}|\widehat{\phi}(\xi)|^{2}\,d\xi+C\int_{|\xi|\geq
1}|\xi|^{2}|\widehat{\phi}(\xi)|^{2}\,d\xi$ $\displaystyle=C\int_{|\xi|\leq
1}|\widehat{\phi}(\xi)|^{2}\,d\xi+C\int_{|\xi|\geq
1}\frac{1}{g(\xi)}|\xi|^{2}g(\xi)|\widehat{\phi}(\xi)|^{2}\,d\xi$
$\displaystyle\leq
C\int_{\mathbb{R}^{2}}|\widehat{\phi}(\xi)|^{2}\,d\xi+C\int_{|\xi|\geq
1}\frac{1}{g(1)}|\xi|^{2}g(\xi)|\widehat{\phi}(\xi)|^{2}\,d\xi$
$\displaystyle\leq
C_{1}\|\phi\|_{L^{2}}^{2}+C_{2}\|\mathcal{L}^{\frac{1}{2}}\phi\|_{L^{2}}^{2},$

which implies

$\displaystyle\|\nabla\phi\|_{L^{2}}\leq
C_{1}\|\phi\|_{L^{2}}+C_{2}\|\mathcal{L}^{\frac{1}{2}}\phi\|_{L^{2}}.$ (3.4)

Therefore, we get by virtue of the easy interpolation inequality that

$\displaystyle\int_{\mathbb{R}^{2}}{T(\nabla u,\nabla b)j\,dx}$
$\displaystyle\leq C\|\nabla u\|_{L^{2}}\|\nabla b\|_{L^{4}}\|j\|_{L^{4}}$
$\displaystyle\leq C\|\omega\|_{L^{2}}\|j\|_{L^{4}}^{2}$ $\displaystyle\leq
C\|\omega\|_{L^{2}}\|j\|_{L^{2}}\|\nabla j\|_{L^{2}}$ $\displaystyle\leq
C\|\omega\|_{L^{2}}\|\nabla b\|_{L^{2}}\|\nabla j\|_{L^{2}}$
$\displaystyle\leq
C\|\omega\|_{L^{2}}(\|b\|_{L^{2}}+\|\mathcal{L}^{\frac{1}{2}}b\|_{L^{2}})(\|j\|_{L^{2}}+\|\mathcal{L}^{\frac{1}{2}}j\|_{L^{2}})$
$\displaystyle\leq\frac{1}{2}\|\mathcal{L}^{\frac{1}{2}}j\|_{L^{2}}^{2}+C(1+\|b\|_{L^{2}}^{2}+\|\mathcal{L}^{\frac{1}{2}}b\|_{L^{2}}^{2})(\|\omega\|_{L^{2}}^{2}+\|j\|_{L^{2}}^{2}).$

As a result, one has

$\frac{d}{dt}(\|\omega(t)\|_{L^{2}}^{2}+\|j(t)\|_{L^{2}}^{2})+\|\mathcal{L}^{\frac{1}{2}}j\|_{L^{2}}^{2}\leq
C(1+\|b\|_{L^{2}}^{2}+\|\mathcal{L}^{\frac{1}{2}}b\|_{L^{2}}^{2})(\|\omega\|_{L^{2}}^{2}+\|j\|_{L^{2}}^{2}).$

From the classical Gronwall lemma and (3.1), we easily get (3.3). The proof of
the lemma is now achieved. ∎

Making use of the global bounds obtained in Lemma 3.2, we obtain a global
bound for the $L_{t}^{\infty}L^{\infty}$-norm of $b$ as follows.

######  Lemma 3.3.

Let $(u_{0},b_{0})$ satisfy the conditions stated in Theorem 1.1, then it
holds for any $t\in[0,\,T]$

$\displaystyle\|b\|_{L_{t}^{\infty}L^{\infty}}\leq C(T,u_{0},b_{0}).$ (3.5)

###### Proof of Lemma 3.3.

we rewrite the second equation of (1.1) as

$\partial_{t}b+\mathcal{L}b=\nabla\cdot(b\otimes u)-\nabla\cdot(u\otimes b).$

By (2.2), we can check that

$b(t)=K(t)\ast b_{0}+\int_{0}^{t}{K(t-\tau)\ast[\nabla\cdot(b\otimes
u)-\nabla\cdot(u\otimes b)](\tau)\,d\tau}.$

Taking $L^{\infty}$-norm in terms of space variable and using the Young
inequality, it implies

$\displaystyle\|b(t)\|_{L^{\infty}}$ $\displaystyle\leq\|K(t)\ast
b_{0}\|_{L^{\infty}}+\int_{0}^{t}{\|\nabla K(t-\tau)\ast[(b\otimes
u)-(u\otimes b)](\tau)\|_{L^{\infty}}\,d\tau}$
$\displaystyle\leq\|\widehat{K(t)\ast b_{0}}\|_{L^{1}}+C\int_{0}^{t}{\|\nabla
K(t-\tau)\|_{L^{2}}\|[(b\otimes u)-(u\otimes b)](\tau)\|_{L^{2}}\,d\tau}$
$\displaystyle\leq\|\widehat{K(t)}\widehat{b_{0}}\|_{L^{1}}+C\int_{0}^{t}{\|\nabla
K(t-\tau)\|_{L^{2}}\|u(\tau)\|_{L^{4}}\|b(\tau)\|_{L^{4}}\,d\tau}$
$\displaystyle\leq\|\widehat{b_{0}}\|_{L^{1}}+C\|\nabla
K(\tau)\|_{L_{t}^{1}L^{2}}\|u(\tau)\|_{L_{t}^{\infty}L^{4}}\|b(\tau)\|_{L_{t}^{\infty}L^{4}}$
$\displaystyle\leq
C\|b_{0}\|_{L^{2}}^{\frac{1}{2}}\|b_{0}\|_{\dot{H}^{2}}^{\frac{1}{2}}+C(t,u_{0},b_{0})$
$\displaystyle\leq C(t,u_{0},b_{0}),$

where we have used (2.11) and the following fact

$\displaystyle\int_{0}^{t}{\|\nabla K(\tau)\|_{L^{2}}\,d\tau}\leq C(t).$ (3.6)

This allows us to derive

$\displaystyle\|b\|_{L_{t}^{\infty}L^{\infty}}\leq C(t,u_{0},b_{0}),$

which is the desired bound (3.5). Consequently, the proof of Lemma 3.3 is
completed. ∎

With the help of the global bounds obtained in the above lemmas, we obtain a
global bound for the $L_{t}^{2}L^{\infty}$-norm of $j$.

######  Lemma 3.4.

Let $(u_{0},b_{0})$ satisfy the conditions stated in Theorem 1.1, then it
holds for any $t\in[0,\,T]$

$\displaystyle\int_{0}^{t}{\|j(\tau)\|_{L^{\infty}}^{2}\,d\tau}\leq
C(T,u_{0},b_{0}).$ (3.7)

###### Proof of Lemma 3.4.

We rewrite the second equation of (3.2) as

$\displaystyle\partial_{t}j+\mathcal{L}j=\nabla\cdot(b\otimes\omega)-\nabla\cdot(u\otimes
j)+T(\nabla u,\nabla b).$

We again make use of (2.2) to deduce

$\displaystyle j(t)=K(t)\ast
j_{0}+\int_{0}^{t}{K(t-\tau)\ast[\nabla\cdot(b\otimes\omega)-\nabla\cdot(u\otimes
j)+T(\nabla u,\nabla b)](\tau)\,d\tau}.$ (3.8)

Taking $L^{\infty}$-norm in terms of space variable and using the Young
inequality, we conclude

$\displaystyle\|j(t)\|_{L^{\infty}}$ $\displaystyle\leq\|K(t)\ast
j_{0}\|_{L^{\infty}}+\int_{0}^{t}{\|\nabla
K(t-\tau)\ast[(b\otimes\omega)-(u\otimes j)](\tau)\|_{L^{\infty}}\,d\tau}$
$\displaystyle\quad+\int_{0}^{t}{\|K(t-\tau)\ast T(\nabla u,\nabla
b)(\tau)\|_{L^{\infty}}\,d\tau}$ $\displaystyle\leq
C\|K(t)\|_{L^{2}}\|j_{0}\|_{L^{2}}+C\int_{0}^{t}{\|K(t-\tau)\|_{L^{\infty}}\|T(\nabla
u,\nabla b)(\tau)\|_{L^{1}}\,d\tau}$
$\displaystyle\quad+C\int_{0}^{t}{\|\nabla
K(t-\tau)\|_{L^{2}}(\|(b\otimes\omega)(\tau)\|_{L^{2}}+\|(u\otimes
j)(\tau)\|_{L^{2}})\,d\tau}.$

Take $L^{2}$-norm in terms of time variable and use the convolution Young
inequality as well as the estimates (3.5)-(3.7) to show

$\displaystyle\|j(t)\|_{L_{t}^{2}L^{\infty}}$ $\displaystyle\leq
C\|K(t)\|_{L_{t}^{2}L^{2}}\|j_{0}\|_{L^{2}}+C\|K(t)\|_{L_{t}^{1}L^{\infty}}\|T(\nabla
u,\nabla b)\|_{L_{t}^{2}L^{1}}$ $\displaystyle\quad+C\|\nabla
K(t)\|_{L_{t}^{1}L^{2}}(\|b\otimes\omega\|_{L_{t}^{2}L^{2}}+\|u\otimes
j\|_{L_{t}^{2}L^{2}})$ $\displaystyle\leq
C\|K(t)\|_{L_{t}^{2}L^{2}}\|j_{0}\|_{L^{2}}+C\|K(t)\|_{L_{t}^{1}L^{\infty}}\|\omega\|_{L_{t}^{2}L^{2}}\|j\|_{L_{t}^{2}L^{2}}$
$\displaystyle\quad+C\|\nabla
K(t)\|_{L_{t}^{1}L^{2}}(\|b\|_{L_{t}^{2}L^{\infty}}\|\omega\|_{L_{t}^{\infty}L^{2}}+\|u\|_{L_{t}^{4}L^{4}}\|j\|_{L_{t}^{4}L^{4}})$
$\displaystyle\leq C(T,u_{0},b_{0}),$

where we have applied the following facts:

$\displaystyle\int_{0}^{t}{\|K(\tau)\|_{L^{2}}^{2}\,d\tau}\leq
C(T,u_{0},b_{0}),\qquad\int_{0}^{t}{\|K(\tau)\|_{L^{\infty}}\,d\tau}\leq
C(T,u_{0},b_{0}),$ $\|u\|_{L_{t}^{4}L^{4}}\leq
C\|u\|_{L_{t}^{\infty}L^{2}}^{\frac{1}{2}}\|\omega\|_{L_{t}^{2}L^{2}}^{\frac{1}{2}}\leq
C(T,u_{0},b_{0})$

and by (3.4)

$\displaystyle\|j\|_{L_{t}^{4}L^{4}}$ $\displaystyle\leq
C\|j\|_{L_{t}^{\infty}L^{2}}^{\frac{1}{2}}\|\nabla
j\|_{L_{t}^{2}L^{2}}^{\frac{1}{2}}$ $\displaystyle\leq
C\|j\|_{L_{t}^{\infty}L^{2}}^{\frac{1}{2}}(\|j\|_{L_{t}^{2}L^{2}}^{\frac{1}{2}}+\|\mathcal{L}^{\frac{1}{2}}j\|_{L_{t}^{2}L^{2}}^{\frac{1}{2}})$
$\displaystyle\leq C(T,u_{0},b_{0}).$

This completes the proof of Lemma 3.4. ∎

The next lemma plays a significant role in obtaining the global bound for
$(u,\,b)$ in $H^{s}$ with $s>2$ for the system (1.1).

######  Lemma 3.5.

Let $(u_{0},b_{0})$ satisfy the conditions stated in Theorem 1.1, then it
holds for any $t\in[0,\,T]$

$\displaystyle\int_{0}^{t}{\|\nabla j(\tau)\|_{L^{\infty}}\,d\tau}\leq
C(T,u_{0},b_{0}),$ (3.9) $\displaystyle\|\omega(t)\|_{L^{\infty}}\leq
C(T,u_{0},b_{0}),$ (3.10) $\displaystyle\int_{0}^{t}{\|\nabla
b(\tau)\|_{L^{\infty}}^{2}\,d\tau}\leq C(T,u_{0},b_{0}).$ (3.11)

###### Proof of Lemma 3.5.

Coming back to (3.8), we get

$\displaystyle j(t)=K(t)\ast
j_{0}+\int_{0}^{t}{K(t-\tau)\ast[\nabla\cdot(b\otimes\omega)-\nabla\cdot(u\otimes
j)+T(\nabla u,\nabla b)](\tau)\,d\tau}.$

Applying the gradient $\nabla$ to the both sides of the above equality, taking
$L^{\infty}$-norm in terms of space variable and using the Young inequality,
it follows that

$\displaystyle\|\nabla j(t)\|_{L^{\infty}}$ $\displaystyle\leq\|\nabla
K(t)\ast
j_{0}\|_{L^{\infty}}+\int_{0}^{t}{\|\nabla^{2}K(t-\tau)\ast[(b\otimes\omega)-(u\otimes
j)](\tau)\|_{L^{\infty}}\,d\tau}$ $\displaystyle\quad+\int_{0}^{t}{\|\nabla
K(t-\tau)\ast T(\nabla u,\nabla b)(\tau)\|_{L^{\infty}}\,d\tau}$
$\displaystyle\leq C\|\nabla
K(t)\|_{L^{2}}\|j_{0}\|_{L^{2}}+C\int_{0}^{t}{\|\nabla
K(t-\tau)\|_{L^{2}}\|T(\nabla u,\nabla b)(\tau)\|_{L^{2}}\,d\tau}$
$\displaystyle\quad+C\int_{0}^{t}{\|\nabla^{2}K(t-\tau)\|_{L^{1}}(\|(b\otimes\omega)(\tau)\|_{L^{\infty}}+\|(u\otimes
j)(\tau)\|_{L^{\infty}})\,d\tau}.$

Now we obtain by taking $L^{1}$-norm in terms of time variable and appealing
to the convolution Young inequality

$\displaystyle\|\nabla j(t)\|_{L_{t}^{1}L^{\infty}}\leq$ $\displaystyle
C\|\nabla K(t)\|_{L_{t}^{1}L^{2}}\|j_{0}\|_{L^{2}}+C\|\nabla
K(t)\|_{L_{t}^{1}L^{2}}\|T(\nabla u,\nabla b)\|_{L_{t}^{1}L^{2}}$
$\displaystyle+C\|\nabla^{2}K(t)\|_{L_{t}^{1}L^{1}}(\|b\otimes\omega\|_{L_{t}^{1}L^{\infty}}+\|u\otimes
j\|_{L_{t}^{1}L^{\infty}})$ $\displaystyle\leq$ $\displaystyle C\|\nabla
K(t)\|_{L_{t}^{1}L^{2}}\|j_{0}\|_{L^{2}}+C\|\nabla
K(t)\|_{L_{t}^{1}L^{2}}\|\omega\|_{L_{t}^{2}L^{4}}\|j\|_{L_{t}^{2}L^{4}}$
$\displaystyle+C\|\nabla^{2}K(t)\|_{L_{t}^{1}L^{1}}(\|b\omega\|_{L_{t}^{1}L^{\infty}}+\|u\|_{L_{t}^{2}L^{\infty}}\|j\|_{L_{t}^{2}L^{\infty}})$
$\displaystyle\leq$ $\displaystyle C\|\nabla
K(t)\|_{L_{t}^{1}L^{2}}\|j_{0}\|_{L^{2}}$ $\displaystyle+C\|\nabla
K(t)\|_{L_{t}^{1}L^{2}}\|\omega\|_{L_{t}^{\infty}L^{2}}^{\frac{1}{2}}\|\omega\|_{L_{t}^{1}L^{\infty}}^{\frac{1}{2}}\|j\|_{L_{t}^{\infty}L^{2}}^{\frac{1}{2}}\|j\|_{L_{t}^{1}L^{\infty}}^{\frac{1}{2}}$
$\displaystyle+C\|\nabla^{2}K(t)\|_{L_{t}^{1}L^{1}}(\|b\|_{L_{t}^{\infty}L^{\infty}}\|\omega\|_{L_{t}^{1}L^{\infty}}+\|u\|_{L_{t}^{\infty}L^{2}}^{\frac{1}{2}}\|\omega\|_{L_{t}^{1}L^{\infty}}^{\frac{1}{2}}\|j\|_{L_{t}^{2}L^{\infty}})$
$\displaystyle=$ $\displaystyle
H_{1}(t)+H_{2}(t)\|\omega\|_{L_{t}^{1}L^{\infty}}^{\frac{1}{2}}+H_{3}(t)\|\omega\|_{L_{t}^{1}L^{\infty}},$
(3.12)

where $H_{l}(t)$ ($l=1,2,3$) are given by

$\displaystyle H_{1}(t)=C\|\nabla
K(t)\|_{L_{t}^{1}L^{2}}\|j_{0}\|_{L^{2}};\qquad
H_{3}(t)=C\|b\|_{L_{t}^{\infty}L^{\infty}}\|\nabla^{2}K(t)\|_{L_{t}^{1}L^{1}};$
$\displaystyle H_{2}(t)=C(\|\nabla
K(t)\|_{L_{t}^{1}L^{2}}\|\omega\|_{L_{t}^{\infty}L^{2}}^{\frac{1}{2}}\|j\|_{L_{t}^{\infty}L^{2}}^{\frac{1}{2}}\|j\|_{L_{t}^{1}L^{\infty}}^{\frac{1}{2}}+\|\nabla^{2}K(t)\|_{L_{t}^{1}L^{1}}\|u\|_{L_{t}^{\infty}L^{2}}^{\frac{1}{2}}\|j\|_{L_{t}^{2}L^{\infty}}).$

Thanks to (2.10), (3.3), (3.5), (3.6) and (3.7), it is easy to show that
$H_{l}(t)$ ($l=1,2,3$) are non-decreasing functions satisfying

$H_{l}(t)\leq C(T,u_{0},b_{0}),\quad\forall\,t\in[0,\,T].$

Consequently, it thus follows from (3) that

$\displaystyle\int_{0}^{t}\|\nabla j(\tau)\|_{L^{\infty}}\,d\tau$
$\displaystyle\leq
H_{1}(t)+\frac{1}{4}H_{2}(t)+H_{2}(t)\|\omega\|_{L_{t}^{1}L^{\infty}}+H_{3}(t)\|\omega\|_{L_{t}^{1}L^{\infty}}$
$\displaystyle=H_{1}(t)+\frac{1}{4}H_{2}(t)+(H_{2}(t)+H_{3}(t))\int_{0}^{t}\|\omega(\tau)\|_{L^{\infty}}\,d\tau$
$\displaystyle\leq\zeta_{1}(T)+\zeta_{2}(T)\int_{0}^{t}\|\omega(\tau)\|_{L^{\infty}}\,d\tau,$
(3.13)

where

$\zeta_{1}(T)=H_{1}(T)+\frac{1}{4}H_{2}(T)<\infty,\qquad\zeta_{2}(T)=H_{2}(T)+H_{3}(T)<\infty.$

Multiplying the vorticity $\omega$ equation of (3.2) by $|\omega|^{p-2}\omega$
and integrating over $\mathbb{R}^{2}$ with respect to variable $x$, it holds

$\displaystyle\frac{1}{p}\frac{d}{dt}\|\omega(t)\|_{L^{p}}^{p}=$
$\displaystyle\int_{\mathbb{R}^{2}}(b\cdot\nabla j)\omega|\omega|^{p-2}\,dx$
$\displaystyle\leq$ $\displaystyle\|b\|_{L^{\infty}}\|\nabla
j\|_{L^{p}}\|\omega\|_{L^{p}}^{p-1}.$

We thus have

$\frac{d}{dt}\|\omega(t)\|_{L^{p}}\leq\|b\|_{L^{\infty}}\|\nabla j\|_{L^{p}}.$

Integrating in time, the outcome is

$\|\omega(t)\|_{L^{p}}\leq\|\omega(0)\|_{L^{p}}+\int_{0}^{t}\|b(\tau)\|_{L^{\infty}}\|\nabla
j(\tau)\|_{L^{p}}\,d\tau.$

Letting $p\rightarrow\infty$, it has

$\displaystyle\|\omega(t)\|_{L^{\infty}}\leq\|\omega(0)\|_{L^{\infty}}+\int_{0}^{t}\|b(\tau)\|_{L^{\infty}}\|\nabla
j(\tau)\|_{L^{\infty}}\,d\tau.$

By (3.5), we conclude

$\displaystyle\|\omega(t)\|_{L^{\infty}}\leq\|\omega(0)\|_{L^{\infty}}+C(T)\int_{0}^{t}\|\nabla
j(\tau)\|_{L^{\infty}}\,d\tau.$ (3.14)

Now letting

$G(t):=\zeta_{1}(T)+\zeta_{2}(T)\int_{0}^{t}\|\omega(\tau)\|_{L^{\infty}}\,d\tau,$

it follows from (3) that

$\int_{0}^{t}\|\nabla j(\tau)\|_{L^{\infty}}\,d\tau\leq G(t).$

Thanks to (3.14), we easily get

$\displaystyle\frac{d}{dt}G(t)$
$\displaystyle=\zeta_{2}(T)\|\omega(t)\|_{L^{\infty}}$
$\displaystyle\leq\zeta_{2}(T)\big{(}\|\omega(0)\|_{L^{\infty}}+C(T)\int_{0}^{t}\|\nabla
j(\tau)\|_{L^{\infty}}\,d\tau\big{)}$
$\displaystyle\leq\zeta_{2}(T)\big{(}\|\omega(0)\|_{L^{\infty}}+C(T)G(t)\big{)}.$
(3.15)

The following key estimate is an easy consequence of the Gronwall lemma

$G(t)\leq C(T,u_{0},b_{0}),$

which further implies that

$\int_{0}^{t}\|\nabla j(\tau)\|_{L^{\infty}}\,d\tau\leq C(T,u_{0},b_{0}).$

From the above estimate combined with (3.14), we get

$\|\omega(t)\|_{L^{\infty}}\leq C(T,u_{0},b_{0}).$

The following interpolation inequality

$\|\nabla b\|_{L^{\infty}}\leq C\|\nabla b\|_{L^{2}}^{\frac{1}{2}}\|\nabla
j\|_{L^{\infty}}^{\frac{1}{2}},$

as well as (3.3) and (3.9) yields that

$\int_{0}^{t}{\|\nabla b(\tau)\|_{L^{\infty}}^{2}\,d\tau}\leq
C(T,u_{0},b_{0}).$

This achieves the proof of Lemma 3.5. ∎

The global $H^{s}$ estimate To show the global bound for $(u,\,b)$ in $H^{s}$
with $s>2$, we apply $\Upsilon^{s}$ with $\Upsilon:=(I-\Delta)^{\frac{1}{2}}$
to the equations $u$ and $b$, and take the $L^{2}$ inner product of the
resulting equations with $(\Upsilon^{s}u,\,\Upsilon^{s}b)$ to obtain the
energy inequality

$\displaystyle\frac{1}{2}\frac{d}{dt}(\|u(t)\|_{H^{s}}^{2}+\|b(t)\|_{H^{s}}^{2})+\|\mathcal{L}^{\frac{1}{2}}b\|_{H^{s}}^{2}$
$\displaystyle=-\int_{\mathbb{R}^{2}}{[\Upsilon^{s},u\cdot\nabla]u\cdot\Upsilon^{s}u\,dx}+\int_{\mathbb{R}^{2}}{[\Upsilon^{s},b\cdot\nabla]b\cdot\Upsilon^{s}u\,dx}$
$\displaystyle\quad-\int_{\mathbb{R}^{2}}{[\Upsilon^{s},u\cdot\nabla]b\cdot\Upsilon^{s}b\,dx}+\int_{\mathbb{R}^{2}}{[\Upsilon^{s},b\cdot\nabla]u\cdot\Upsilon^{s}b\,dx}$
$\displaystyle=J_{1}+J_{2}+J_{3}+J_{4},$

where $[a,\,b]$ is the standard commutator notation, namely $[a,\,b]=ab-ba$.
By means of the following Kato-Ponce inequality (see [10])

$\|[\Upsilon^{s},f]g\|_{L^{p}}\leq C(\|\nabla
f\|_{L^{\infty}}\|\Upsilon^{s-1}g\|_{L^{p}}+\|g\|_{L^{\infty}}\|\Upsilon^{s}f\|_{L^{p}}),\quad
1<p<\infty,$

one can deduce that

$J_{1}\leq C\|\nabla u\|_{L^{\infty}}\|u\|_{H^{s}}^{2},\qquad J_{2}\leq
C\|\nabla b\|_{L^{\infty}}(\|u\|_{H^{s}}^{2}+\|b\|_{H^{s}}^{2}),$
$J_{3}+J_{4}\leq C(\|\nabla u\|_{L^{\infty}}+\|\nabla
b\|_{L^{\infty}})(\|u\|_{H^{s}}^{2}+\|b\|_{H^{s}}^{2}).$

Consequently, it implies that

$\displaystyle\frac{d}{dt}(\|u(t)\|_{H^{s}}^{2}+\|b(t)\|_{H^{s}}^{2})+\|\mathcal{L}^{\frac{1}{2}}b\|_{H^{s}}^{2}\leq
C(\|\nabla u\|_{L^{\infty}}+\|\nabla
b\|_{L^{\infty}})(\|{u}\|_{H^{s}}^{2}+\|b\|_{H^{s}}^{2}).$

To bound the term $\|\nabla u\|_{L^{\infty}}$ with $\|\omega\|_{L^{\infty}}$,
we need the following Sobolev extrapolation inequality with logarithmic
correction (see e.g., [2])

$\|\nabla u\|_{L^{\infty}(\mathbb{R}^{2})}\leq
C\Big{(}1+\|u\|_{L^{2}(\mathbb{R}^{2})}+\|\omega\|_{L^{\infty}(\mathbb{R}^{2})}\ln(e+\|u\|_{H^{s}(\mathbb{R}^{2})})\Big{)},\quad
s>2.$

Consequently, it enables us to get

$\displaystyle\frac{d}{dt}(\|u(t)\|_{H^{s}}^{2}+\|b(t)\|_{H^{s}}^{2})+\|\mathcal{L}^{\frac{1}{2}}b\|_{H^{s}}^{2}$
$\displaystyle\leq C(1+\|\omega\|_{L^{\infty}}+\|\nabla
b\|_{L^{\infty}})\ln(e+\|u\|_{H^{s}}^{2}+\|b\|_{H^{s}}^{2})(\|{u}\|_{H^{s}}^{2}+\|b\|_{H^{s}}^{2}).$
(3.16)

Applying the log-Gronwall type inequality as well as the estimates (3.10) and
(3.11), we eventually obtain

$\|u(t)\|_{H^{s}}+\|b(t)\|_{H^{s}}+\int_{0}^{t}\|\mathcal{L}^{\frac{1}{2}}b(\tau)\|_{H^{s}}^{2}\,d\tau\leq
C(t),$

which together with (3.4) implies

$\int_{0}^{t}\|b(\tau)\|_{H^{s+1}}^{2}\,d\tau\leq C(t).$

This is nothing but the desired global $H^{s}$ estimates.

With the global bounds in the previous lemmas at our disposal, we are ready to
prove Theorem 1.1.

###### Proof of Theorem 1.1.

With a priori estimates achieved in the previous lemmas, it is a standard
procedure to complete the proof of Theorem 1.1. The proof is achieved by using
a standard procedure. First we seek the solution of a regularized system. In
order to do this, we recall the mollification of $\varrho_{N}f$ given by

$(\varrho_{N}f)(x)=N^{2}\int_{\mathbb{R}^{2}}{\eta\big{(}N(x-y)\big{)}f(y)\,dy},$

where $0\leq\eta(|x|)\in C_{0}^{\infty}(\mathbb{R}^{2})$ satisfies
$\int_{\mathbb{R}^{2}}{\eta(y)\,dy}=1$. Now we regularize our system (1.1) as
follows

$\left\\{\begin{aligned}
&\partial_{t}u^{N}+\mathbb{P}\varrho_{N}((\varrho_{N}u^{N}\cdot\nabla)\varrho_{N}u^{N})=\mathbb{P}\varrho_{N}((\varrho_{N}b^{N}\cdot\nabla)\varrho_{N}b^{N}),\\\
&\partial_{t}b^{N}+\varrho_{N}((\varrho_{N}u^{N}\cdot\nabla)\varrho_{N}b^{N})+\mathcal{J}\varrho_{N}b^{N}=\varrho_{N}((\varrho_{N}b^{N}\cdot\nabla)\varrho_{N}u^{N}),\\\
&\nabla\cdot u^{N}=\nabla\cdot b^{N}=0,\\\
&u^{N}(x,0)=\varrho_{N}u_{0}(x),\quad
b^{N}(x,0)=\varrho_{N}b_{0}(x),\end{aligned}\right.$

where $\mathbb{P}$ denotes the Leray projection operator (onto divergence-free
vector fields). For any fixed $N>0$, using properties of mollifiers and
following the same argument used in proving the previous lemmas, it is not
difficult to establish the global bound, for any $t\in(0,\infty)$,

$\|u^{N}(t)\|_{H^{s}}+\|b^{N}(t)\|_{H^{s}}+\int_{0}^{t}\|\mathcal{L}^{\frac{1}{2}}b^{N}(\tau)\|_{H^{s}}^{2}\,d\tau\leq
C(t)$ $\int_{0}^{t}\|b^{N}(\tau)\|_{H^{s+1}}^{2}\,d\tau\leq C(t).$

Now the standard Alaoglu’s theorem allows us to obtain the global existence of
the classical solution $(u,b)$ to (1.1). The uniqueness can also be easily
established. This completes the proof of Theorem 1.1. ∎

Acknowledgements. The author would like to thank the anonymous referee and the
corresponding editor for their insightful comments and many valuable
suggestions, which greatly improved the exposition of the manuscript. The
author was supported by the Foundation of Jiangsu Normal University (No.
16XLR029), the Natural Science Foundation of Jiangsu Province (No.
BK20170224), the National Natural Science Foundation of China (No. 11701232).

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