# Remarks on Liouville type theorems for the steady MHD and Hall-MHD equations

Xiaomeng Chen  School of Mathematical Sciences, Dalian University of
Technology, Dalian, 116024, China<EMAIL_ADDRESS>,  Shuai Li  School of
Mathematical Sciences, Dalian University of Technology, Dalian, 116024, China
<EMAIL_ADDRESS>and  Wendong Wang  School of Mathematical Sciences, Dalian
University of Technology, Dalian, 116024, China<EMAIL_ADDRESS>
(Date: June 18, 2021)

###### Abstract.

In this note we investigate Liouville type theorems for the steady three
dimensional MHD and Hall-MHD equations, and show that the velocity field $u$
and the magnetic field $B$ are vanishing provided that $B\in
L^{6,\infty}(\mathbb{R}^{3})$ and $u\in BMO^{-1}(\mathbb{R}^{3})$, which state
that the velocity field plays an important role. Moreover, the similar result
holds in the case of partial viscosity or diffusivity for the three
dimensional MHD equations.

###### Key words and phrases:

Liouville type theorems; MHD equations; Hall-MHD equations; stationary Stokes
system

###### 2010 Mathematics Subject Classification:

35Q30, 76D03, 76D07

##  1\. Introduction

Consider the following general system including the so called
magnetohydrodynamics (MHD) and Hall-MHD equations:

$\displaystyle\left\\{\begin{array}[]{llll}\displaystyle-\Delta u+u\cdot\nabla
u+\nabla p=B\cdot\nabla B,~{}~{}\rm{in}~{}~{}\mathbb{R}^{3},\\\
\displaystyle-\Delta B+u\cdot\nabla B-B\cdot\nabla
u=\alpha\nabla\times((\nabla\times B)\times
B),~{}~{}\rm{in}~{}~{}\mathbb{R}^{3},\\\
\displaystyle{\rm{div}}~{}u=0,~{}~{}~{}{\rm{div}}~{}B=0,~{}~{}\rm{in}~{}~{}\mathbb{R}^{3},\end{array}\right.$
(1.4)

where $u=u(x)=(u_{1},u_{2},u_{3})$ is the velocity field of the fluid flows,
$B=B(x)=(B_{1},B_{2},B_{3})$ is the magnetic field, $p=p(x)$ is the pressure
of the flows, and $\alpha\in\mathbb{R}$ is an absolute constant. When
$\alpha=0$, the system (1.4) is the MHD system, which describes the steady
state of the magnetic properties of electrically conducting fluids, including
plasmas, liquid metals, etc; for the physical background and mathematical
theory we refer to Schnack [26] and the references therein. When $\alpha=1$,
this system governs the dynamics plasma flows of strong shear magnetic fields
as in the solar flares, and has many important applications in the
astrophysics (for example, see Chae-Degond-Liu [5]).

In this note, we focus on the Liouville type properties of the above system,
which is motivated by the development of Navier-Stokes equations. When $B=0$,
the system (1.4) reduces to the standard Navier-Stokes equations, and a very
challenging open question is if there exists a nontrivial solution when the
Dirichlet integral $\int_{\mathbb{R}^{3}}|\nabla u|^{2}dx$ is finite, which
dates back to Leray’s celebrated paper [20] and is explicitly written in
Galdi’s book ([12], Remark X. 9.4, pp.729; see also Tsai’s book [31], pp.23).
This uniqueness problem, or equivalently the Liouville type problem is widely
open. Galdi proved the above Liouville type theorem by assuming $u\in
L^{\frac{9}{2}}(\mathbb{R}^{3})$ in [12]. Chae in [3] showed the condition
$\triangle u\in L^{\frac{6}{5}}(\mathbb{R}^{3})$ is sufficient for the
vanishing property of $u$ by exploring the maximum principle of the head
pressure. Also, Chae-Wolf gave an improvement of logarithmic form for Galdi’s
result in [7] by assuming that
$\int_{\mathbb{R}^{3}}|u|^{\frac{9}{2}}\\{\ln(2+\frac{1}{|u|})\\}^{-1}dx<\infty$.
Seregin obtained the conditional criterion $u\in BMO^{-1}(\mathbb{R}^{3})$ in
[28]. More references, we refer to [18, 19, 29, 30, 2, 8, 4] and the
references therein. Relatively speaking, the two-dimensional case is more
easier due to the vorticity of the 2D Navier-Stokes equations satisfies a nice
elliptic equation, to which the maximum principle applies. For example,
Gilbarg-Weinberger [14] obtained Liouville type theorem provided the Dirichlet
energy is finite. As a different type of Liouville property for the 2D Navier-
Stokes equations, Koch-Nadirashvili-Seregin-Sverak [17] showed that any
bounded solution is trivial solution, say $u\equiv C$, as a byproduct of their
results on the non-stationary case (see also [1] for the unbounded velocity).

However, for the MHD tpye system like (1.4), the situation is quite different.
Due to the lack of maximum principle, there is not much progress in the study
of MHD equation. For the two-dimensional MHD equations, Liouville type
theorems were proved by assuming the smallness of the norm of the magnetic
field in [33] and [32], and we refer to the recent paper in [10] by removing
the smallness assumption. For the three-dimensional MHD equations, Chae-Weng
[6] proved that if a smooth solution to (1.4) in $\mathbb{R}^{3}$ with finite
Dirichlet integral

$\displaystyle\int_{\mathbb{R}^{3}}|\nabla u|^{2}+|\nabla B|^{2}dx<+\infty,$

and $u\in L^{3}(\mathbb{R}^{3})$, then the solutions $(u,B)$ is identically
zero. Schulz [27] proved that if a smooth solution of the stationary MHD
equations is in $L^{6}(\mathbb{R}^{3})$ and $u,B\in BMO^{-1}(\mathbb{R}^{3})$,
then it is identically zero. Recently, Chae-Wolf [9] show that $L^{6}$ mean
oscillations of the potential function of the velocity and magnetic field have
certain linear growth by using the technique of [8]. More references, we refer
to [37, 21, 22, 36, 11] and the references therein.

Motivated by some numerical experiments in [25], which seem to indicate that
the velocity field should play a more important role than the magnetic field
in the regularity theory of solutions to the MHD equations. Our goal is to get
rid of the extra magnetic conditions in [6] or [27]. Recall the definition of
the space $BMO^{-1}(\mathbb{R}^{3})$ as follows. For a measurable set
$E\subset\mathbb{R}^{3}$, we denote by $|E|$ the 3-dimensional Lebesgue
measure of $E$, and for $f\in L^{1}(E)$ we use the notation

$\displaystyle f_{E}=\frac{1}{|E|}\int_{E}fdx,$ (1.5)

Combining (1.5) with Hölder’s inequality, we know for all $f\in L^{p}(E)$

$\displaystyle||f-f_{E}||_{L^{p}(E)}\leq C||f||_{L^{p}(E)}.$

Then we say that $u\in BMO^{-1}(\mathbb{R}^{3})$, if there exists
$\mathbf{\Phi}\in C^{\infty}(\mathbb{R}^{3},\mathbb{R}^{3\times 3})$ such that
$\nabla\cdot\mathbf{\Phi}=u$, and for all $r>0$, there holds

$\displaystyle\sup_{r>0}\frac{1}{|B(r)|}\int_{B(r)}|\mathbf{\Phi}-\mathbf{\Phi}_{B(r)}|dx<\infty,$

i.e. $\mathbf{\Phi}\in BMO(\mathbb{R}^{3})$. And a remarkable property of BMO
space is

$\displaystyle\sup_{r>0}\frac{1}{|B(r)|}\int_{B(r)}|\mathbf{\Phi}-\mathbf{\Phi}_{B(r)}|^{p}dx<\infty,\quad\forall~{}~{}1\leq
p<+\infty.$ (1.6)

Next, the definition of the space $L^{p,q}(\mathbb{R}^{n})$ is as follow: Let
$\Omega\subset\mathbb{R}^{n}$ and $1\leq p,l\leq\infty$, we say that a
measurable function $f$ is in $L^{p,l}(\Omega)$ if
$\|f\|_{L^{p,l}(\Omega)}<+\infty$, where

$\|f\|_{L^{p,l}}(\Omega)=\left\\{\begin{array}[]{llll}\left(\int^{\infty}_{0}\lambda^{l-1}|\\{x\in\Omega;|f|>\lambda\\}|^{\frac{l}{p}}d\lambda\right)^{\frac{1}{l}},\quad~{}{\rm
for}~{}r<\infty;\\\
sup_{\lambda>0}\lambda|\\{x\in\Omega;|f|>\lambda\\}|^{\frac{1}{p}},\quad~{}{\rm
for}~{}r=\infty.\end{array}\right.$

We use Hölder inequality in Lorentz spaceas follows ([24]). Assume $1\leq
p_{1},p_{2}\leq\infty$, $1\leq q_{1},q_{2}\leq\infty$ and $u\in
L^{p_{1},q_{1}}(\Omega)$, $v\in L^{p_{2},q_{2}}(\Omega)$. Then $uv\in
L^{p,q}(\Omega)$ and $\frac{1}{q}\leq\frac{1}{q_{1}}+\frac{1}{q_{2}}$ and the
inequality

$\displaystyle||uv||_{L^{p,q}(\Omega)}\leq
C||u||_{L^{p_{1},q_{1}}(\Omega)}||v||_{L^{p_{2},q_{2}}(\Omega)}$ (1.7)

is valid,where $\frac{1}{p}=\frac{1}{p_{1}}+\frac{1}{p_{2}}$.

Our main result is as follows:

######  Theorem 1.1.

Let $(u,B,p)\in C^{\infty}(\mathbb{R}^{3})\times
C^{\infty}(\mathbb{R}^{3})\times C^{\infty}(\mathbb{R}^{3})$ be a solution of
(1.4) with $B\in L^{6,\infty}(\mathbb{R}^{3})$. Assume that $u\in
BMO^{-1}(\mathbb{R}^{3})$. Then $u=B=0$.

######  Remark 1.2.

The above theorem shows that the velocity field should play a more important
role than the magnetic field in the uniqueness theory, which is similar as in
the regularity theory. For example, [16] and [34] have presented some
regularity criterions to the MHD equations in terms of the velocity field
only.

######  Remark 1.3.

The above result generalized Seregin’s result in [28] or [29] to the MHD case
or Hall-MHD case when $B\equiv 0$, which also improved Chae-Weng’s result [6],
where they assumed $\nabla u,\nabla B\in L^{2}(\mathbb{R}^{3})$ and $u\in
L^{3}(\mathbb{R}^{3})$. It also improved Schulz’s theorem [27] by removing the
additional conditions $B\in BMO^{-1}(\mathbb{R}^{3})$ and $u\in
L^{6}(\mathbb{R}^{3})$. Moreover, we relaxed the condition of $B$ to the
Lorentz space of $L^{6,\infty}(\mathbb{R}^{3})$. A new observation is the use
of delicate $L^{q}$ estimates of stationary Stokes system, which is
independent of interest.

For the MHD equations with partial viscosity or diffusivity, we have the
following similar conclusions. Consider the MHD system (1.4) as follows:

$\displaystyle\left\\{\begin{array}[]{llll}\displaystyle-\lambda_{1}\partial_{11}u-\lambda_{2}\partial_{22}u-\lambda_{3}\partial_{33}u+u\cdot\nabla
u+\nabla p=B\cdot\nabla B,~{}~{}\rm{in}~{}~{}\mathbb{R}^{3},\\\
\displaystyle-\mu_{1}\partial_{11}B-\mu_{2}\partial_{22}B-\mu_{3}\partial_{33}B+u\cdot\nabla
B=B\cdot\nabla u,\\\
\displaystyle{\rm{div}}~{}u=0,~{}~{}~{}{\rm{div}}~{}B=0,\\\
\displaystyle\lambda_{i}\geq 0,~{}~{}~{}\mu_{i}\geq 0,i=1,2,3.\\\
\end{array}\right.$ (1.12)

For the system (1.12), we have

######  Theorem 1.4.

Let $u,b$ be a smooth solutions of the system (1.12). Let’s further assume
that $B\in L^{6}(\mathbb{R}^{3})$ and $u\in L^{3}(\mathbb{R}^{3})$, then
$u,B\equiv 0$ if the following condition holds

$\displaystyle\lambda_{1}+\lambda_{2}+\lambda_{3}>0,~{}~{}~{}\mu_{1}+\mu_{2}+\mu_{3}>0.$
(1.13)

######  Remark 1.5.

When $\lambda_{1}+\lambda_{2}+\lambda_{3}=0$, but $\mu_{1}+\mu_{2}+\mu_{3}>0$,
one can derive that $B\equiv 0$ and $u$ satisfies the three dimensional Euler
equations. It is impossible to deduce that $u$ is vanishing from the bounded-
ness of $L^{q}$ norm. For example, we can refer to the counterexample
belonging to $C_{0}^{\infty}(\mathbb{R}^{3})$ in [13]. The same is true for
another situation of $\mu_{1}+\mu_{2}+\mu_{3}=0$.

In the proof, we need the following lemma (see, for example, Lemma A.5 [35]).

######  Lemma 1.6.

Let $f,g\in\dot{W}^{1,2}(\mathbb{R}^{3})\cap BMO^{-1}(\mathbb{R}^{3})$. Then
there holds

$\displaystyle||fg||_{L^{2}}\leq
C\left(||f||_{\dot{W}^{1,2}}||g||_{BMO^{-1}}+||f||_{BMO^{-1}}||g||_{\dot{W}^{1,2}}\right).$

We also need an interpolation inequality in Lorentz space (see, for example,
Theorem 2.1 [23]).

######  Lemma 1.7.

Let $\Omega$ be a domain in $\mathbb{R}^{n}$. Let $0<q<p<r\leq+\infty$ and
$\alpha>0$. If $f\in L^{q,\infty}(\Omega)\cap L^{r,\infty}(B(r))$, then $f\in
L^{p,\alpha}(\Omega)$ and

$\displaystyle||f||_{L^{p,\alpha}(\Omega)}\leq
C||f||_{L^{q,\infty}(\Omega)}^{\theta}||f||_{L^{r,\infty}(\Omega)}^{1-\theta},$

where $C=C(q,r,p,\alpha)>0$ and

$\displaystyle\frac{1}{p}=\frac{\theta}{q}+\frac{1-\theta}{r}.$

######  Remark 1.8.

The lemma 1.7 follows from Theorem 2.1 in [23], where the domain is the whole
space $\mathbb{R}^{n}$. Note that
$||f\chi_{\Omega}||_{L^{q,\infty}(\mathbb{R}^{n})}=||f||_{L^{q,\infty}(\Omega)}$,
which show that the lemma holds for the general domain $\Omega$ in
$\mathbb{R}^{n}$.

Throughout this article, $C(\alpha_{1},\cdots,\alpha_{n})$ denotes a constant
depending on $\alpha_{1},\cdots,\alpha_{n}$, which may be different from line
to line.

##  2\. Proof of Theorem 1.1

Before starting the proof of Theorem 1.1, we state the following interior
$L^{q}$ estimates for the steady Stokes system of an alternative version and
sketch its proof.

######  Lemma 2.9.

Let $B(r)\subset B(R)\subset\mathbb{R}^{n}$ with $n\geq 2$ be concentric balls
with $0<r<R$. Assume that $v$ is a very weak solution of the following Stokes
system

$\displaystyle-\Delta v_{i}+\partial_{i}p=g_{i},\quad{\rm div}\,v=0\quad{\rm
in}~{}B(R),$ (2.1)

where $g_{i}\in L^{q}(B(R))$, $1<q<\infty$. Then $v\in W_{loc}^{2,q}(B(R))$
and there exists a function $p\in W_{loc}^{1,q}(B(R))$. Moreover, there holds

$\displaystyle||\nabla^{2}v||_{L^{q}(B(r))}+||\nabla p||_{L^{q}(B(r))}\leq
C(n,q)\left(||g||_{L^{q}(B(R))}+(R-r)^{-2}||v||_{L^{q}(B(R))}\right).$ (2.2)

######  Remark 2.10.

A very weak solution $v$ of the system of (2.1) in
$\Omega\subset\mathbb{R}^{n}$ is defined as follows. The divergence-free
vector field $v\in L_{loc}^{1}(\Omega)$ satisfies

$\displaystyle\int v\cdot\triangle\zeta=-\int
g\cdot\zeta,\quad\forall~{}\zeta\in C_{c,\sigma}^{\infty}(\Omega),$

where $C_{c,\sigma}^{\infty}(\Omega)$ denotes any times differentiable
divergence-free vector fields with compact support in $\Omega.$

###### Proof.

At first, recall Lemma 2.12 in [31] and we have

$\displaystyle||\nabla^{2}v||_{L^{q}(B(r))}\leq
C(n,q,r,R)\left(||g||_{L^{q}(B(R))}+||v||_{L^{1}(B(R))}\right).$

Specially,

$\displaystyle||\nabla^{2}v||_{L^{q}(B(1))}\leq
C(n,q)\left(||g||_{L^{q}(B(2))}+||v||_{L^{1}(B(2))}\right).$

Hence, for any $0<r<R$ and $2r\leq R$, by scaling we get

$\displaystyle||\nabla^{2}v||_{L^{q}(B(r))}\leq
C(n,q)\left(||g||_{L^{q}(B(2r))}+r^{-n-2+\frac{n}{q}}||v||_{L^{1}(B(2r))}\right).$
(2.3)

Secondly, for any fixed $r>0$ with $r<R$, and $x_{0}\in B(r)$, let
$10d=R-r>0$, then

$\displaystyle\bigcup_{x_{0}\in B(r)}B(x_{0},d)\supset B(r),$

where $B(x_{0},d)$ denotes the ball of radius $d$ centered at $x_{0}.$ Due to
the lemma of Vitali, there exists finite points, $x_{1},\cdots,x_{k}\in B(r)$
such that $B(x_{1},d),\cdots,B(x_{k},d)$ are disjoint balls satisfying

$\displaystyle B(r)\subset\bigcup_{j=1}^{k}B(x_{j},5d),$

and at most $K_{0}$ finite balls with $K_{0}\leq 6^{n}$ have a joint point in
$B(R)$. Thence, by (2.3) and Hölder inequality we have

$\displaystyle||\nabla^{2}v||_{L^{q}(B(r))}^{q}$ $\displaystyle\leq$
$\displaystyle\sum_{j=1}^{k}||\nabla^{2}v||_{L^{q}(B(x_{j},5d))}^{q}$
$\displaystyle\leq$ $\displaystyle
C(n,q)\sum_{j=1}^{k}\left(||g||_{L^{q}(B(x_{j},10d))}^{q}+d^{-(n+2)q+n}||v||_{L^{1}(B(x_{j},10d))}^{q}\right)$
$\displaystyle\leq$ $\displaystyle
C(n,q)\sum_{j=1}^{k}\left(||g||_{L^{q}(B(x_{j},10d))}^{q}+d^{-2q}||v||_{L^{q}(B(x_{j},10d))}^{q}\right)$

Note that at most $K_{1}$ finite balls of $B(x_{j},10d)$ with $K_{1}\leq
11^{n}$ have a joint point in $B(R)$. Then

$\displaystyle||\nabla^{2}v||_{L^{q}(B(r))}^{q}$ $\displaystyle\leq$
$\displaystyle
C(n,q)11^{n}\left(||g||_{L^{q}(B(R))}^{q}+d^{-2q}||v||_{L^{q}(B(R))}^{q}\right)$
$\displaystyle\leq$ $\displaystyle
C(n,q)\left(||g||_{L^{q}(B(R))}^{q}+(R-r)^{-2q}||v||_{L^{q}(B(R))}^{q}\right)$

by using $10d=R-r$ and

$\displaystyle B(R)\supset\bigcup_{j=1}^{k}B(x_{j},10d).$

At last, using the equations of (2.1), we get

$\displaystyle||\nabla p||_{L^{q}(B(r))}$ $\displaystyle\leq$ $\displaystyle
C(n,q)\left(||g||_{L^{q}(B(R))}+(R-r)^{-2}||v||_{L^{q}(B(R))}\right)$

The proof is complete. ∎

Next we prove Theorem 1.1.

###### Proof.

Step I. Cacciopolli type inequality.

First, we introduce some cut-off functions of $\phi_{i}$ with $i=1,\cdots,3$
for $r>1$ and $\rho,\tau>0$, which will be used later. Assume that
$\phi_{i}(x)\in C_{0}^{\infty}(\mathbb{R}^{3})$ with $0\leq\phi_{i}\leq 1$ for
$i=1,\cdots,3$ satisfying the following conditions:
i) $r\leq\rho<\tau\leq 2r$;
ii)

$\phi_{1}(x)=\phi_{1}(|x|)=\left\\{\begin{array}[]{llll}1,\quad~{}{\rm
in}~{}B_{\rho};\\\ 0,\quad~{}{\rm in}~{}\mathbb{R}^{3}\backslash
B(\frac{\tau+2\rho}{3});\\\ \end{array}\right.$ (2.4)

iii)

$\phi_{2}(x)=\phi_{2}(|x|)=\left\\{\begin{array}[]{llll}1,\quad~{}{\rm
in}~{}B(\frac{\tau+2\rho}{3});\\\ 0,\quad~{}{\rm
in}~{}\mathbb{R}^{3}\backslash B(\frac{2\tau+\rho}{3});\\\ \end{array}\right.$
(2.5)

iv)

$\phi_{3}(x)=\phi_{3}(|x|)=\left\\{\begin{array}[]{llll}1,\quad~{}{\rm
in}~{}B(\frac{2\tau+\rho}{3});\\\ 0,\quad~{}{\rm
in}~{}\mathbb{R}^{3}\backslash B(\tau);\\\ \end{array}\right.$ (2.6)

v)

$\displaystyle|\nabla\phi_{i}|\leq\frac{C}{\tau-\rho},\quad|\nabla^{2}\phi_{i}|\leq\frac{C}{(\tau-\rho)^{2}},\quad
i=1,\cdots,3.$

Secondly, multiply $(\ref{eq:MHD})_{1}$ by $u\phi_{1}$, $(\ref{eq:MHD})_{2}$
by $(B-b)\phi_{1}$, and integrate them over $B(2r)$. Then by integration by
parts we obtain

$\displaystyle\int_{B(2r)}|\nabla u|^{2}\phi_{1}+|\nabla B|^{2}\phi_{1}dx$
$\displaystyle=$
$\displaystyle\frac{1}{2}\int_{B(2r)}|u|^{2}\Delta\phi_{1}dx+\frac{1}{2}\int_{B(2r)}|u|^{2}u\cdot\nabla\phi_{1}dx$
$\displaystyle+\int_{B(2r)}(p-a)u\cdot\nabla\phi_{1}dx+\frac{1}{2}\int_{B(2r)}|B-b|^{2}\Delta\phi_{1}dx$
$\displaystyle+\frac{1}{2}\int_{B(2r)}|B-b|^{2}u\cdot\nabla\phi_{1}dx-\int_{B(2r)}u\cdot(B-b)B\cdot\nabla\phi_{1}dx$
$\displaystyle-\alpha\int_{B(2r)}((\nabla\times(B-b))\times
B)\cdot((B-b)\times\nabla\phi_{1})dx$ $\displaystyle\doteq$ $\displaystyle
I_{1}+\cdots+I_{7},$

where $a$ and $b$ are constant vectors in $B(2r)$, to be decided, and we used
the following formula for the last term of the right hand

$\displaystyle\int\nabla\times F\cdot G\phi dx$ $\displaystyle=$
$\displaystyle\int\nabla\times G\cdot F\phi dx+\int G\times F\cdot\nabla\phi
dx$ $\displaystyle=$ $\displaystyle\int\nabla\times G\cdot F\phi dx-\int
F\cdot(G\times\nabla\phi)dx.$

For the terms of $I_{1}$ and $I_{2}$, it’s easy to deduce that

$\displaystyle|I_{1}|\leq
C(\tau-\rho)^{-2}\int_{B(\frac{\tau+2\rho}{3})\setminus B(\rho)}|u|^{2}dx.$
(2.7)

and

$\displaystyle|I_{2}|\leq
C(\tau-\rho)^{-1}\int_{B(\frac{\tau+2\rho}{3})\setminus B(\rho)}|u|^{3}dx.$
(2.8)

For the term of $I_{4}$, by (1.7) we have

$\displaystyle|I_{4}|$ $\displaystyle\leq$ $\displaystyle
C(\tau-\rho)^{-2}\int_{B(\frac{\tau+2\rho}{3})\setminus B(\rho)}|B-b|^{2}dx$
(2.9) $\displaystyle\leq$ $\displaystyle
C(\tau-\rho)^{-2}||B-b||_{L^{6,\infty}(B(\frac{2\tau+\rho}{3})\backslash
B(\rho))}^{2}||1||_{L^{\frac{3}{2},1}(B(2r))}$ $\displaystyle\leq$
$\displaystyle
Cr^{2}(\tau-\rho)^{-2}||B-b||_{L^{6,\infty}(B(\frac{2\tau+\rho}{3})\backslash
B(\rho))}^{2}.$

For the term $I_{5}$, using $\nabla\cdot\mathbf{\Phi}=u$, integration by parts
and Hölder inequality, we have

$\displaystyle|I_{5}|$ $\displaystyle=$
$\displaystyle\left|\int_{B(\frac{\tau+2\rho}{3})\backslash
B(\rho)}|B-b|^{2}\partial_{i}(\Phi_{ij}-\bar{\Phi}_{ij})\partial_{j}\phi_{1}dx\right|$
$\displaystyle\leq$ $\displaystyle
C(\tau-\rho)^{-1}\int_{B(\frac{\tau+2\rho}{3})\backslash B(\rho)}|B-b||\nabla
B||\mathbf{\Phi}-\bar{\mathbf{\Phi}}|dx$
$\displaystyle+C(\tau-\rho)^{-2}\int_{B(\frac{\tau+2\rho}{3})\backslash
B(\rho)}|B-b|^{2}|\mathbf{\Phi}-\bar{\mathbf{\Phi}}|dx$ $\displaystyle\leq$
$\displaystyle
C(\tau-\rho)^{-1}||B-b||_{L^{6,\infty}(B(\frac{\tau+2\rho}{3})\backslash
B(\rho))}||\nabla B||_{L^{2}(B(\frac{\tau+2\rho}{3})\backslash
B(\rho))}||\mathbf{\Phi}-\bar{\mathbf{\Phi}}||_{L^{3,2}(B(\frac{\tau+2\rho}{3}))}$
$\displaystyle+C(\tau-\rho)^{-2}||B-b||_{L^{6,\infty}(B(\frac{\tau+2\rho}{3})\backslash
B(\rho))}^{2}||\mathbf{\Phi}-\bar{\mathbf{\Phi}}||_{L^{\frac{3}{2},1}(B(\frac{\tau+2\rho}{3}))}.$

Choosing $\bar{\mathbf{\Phi}}=\mathbf{\Phi}_{B(2r)}$ and
$b=B_{B(\frac{4r}{3})\backslash B(r)}$, by Lemma 1.7 and (1.6), we have

$\displaystyle||\mathbf{\Phi}-\bar{\mathbf{\Phi}}||_{L^{3,2}(B(2r))}$
$\displaystyle\leq$ $\displaystyle
C||\mathbf{\Phi}-\bar{\mathbf{\Phi}}||_{L^{2}(B(2r))}^{\frac{1}{3}}||\mathbf{\Phi}-\bar{\mathbf{\Phi}}||_{L^{4}(B(2r))}^{\frac{2}{3}}\leq
Cr,$
$\displaystyle||\mathbf{\Phi}-\bar{\mathbf{\Phi}}||_{L^{\frac{3}{2},1}(B(2r))}$
$\displaystyle\leq$ $\displaystyle
C||\mathbf{\Phi}-\bar{\mathbf{\Phi}}||_{L^{\frac{5}{4}}(B(2r))}^{\frac{5}{12}}||\mathbf{\Phi}-\bar{\mathbf{\Phi}}||_{L^{\frac{7}{4}}(B(2r))}^{\frac{7}{12}}\leq
Cr^{2},$

and by $B\in L^{6,\infty}(\mathbb{R}^{3})$

$\displaystyle||b||_{L^{6,\infty}(B(\frac{\tau+2\rho}{3})\backslash B(\rho)}$
$\displaystyle\leq$ $\displaystyle||b||_{L^{6,\infty}(B(2r))}\leq
Cr^{\frac{1}{2}}|b|$ (2.10) $\displaystyle\leq$ $\displaystyle
Cr^{\frac{1}{2}-3}||B||_{L^{6,\infty}(B(2r))}||1||_{L^{\frac{6}{5},1}(B(2r))}\leq
C.$

Then we get

$\displaystyle|I_{5}|$ $\displaystyle\leq$ $\displaystyle
Cr(\tau-\rho)^{-1}||\nabla B||_{L^{2}(B(\frac{2\tau+\rho}{3})\backslash
B(\rho))}$ (2.11)
$\displaystyle+Cr^{2}(\tau-\rho)^{-2}||B-b||_{L^{6,\infty}(B(\frac{2\tau+\rho}{3})\backslash
B(\rho))}^{2}.$

The term of $I_{6}$ can be controlled by

$\displaystyle|I_{6}|$ $\displaystyle=$
$\displaystyle\left|\int_{B(\frac{\tau+2\rho}{3})\backslash
B(\rho)}\partial_{i}(\Phi_{ij}-\bar{\Phi}_{ij})(B-b)_{j}B_{k}\partial_{k}\phi_{1}dx\right|$
$\displaystyle\leq$ $\displaystyle
C(\tau-\rho)^{-1}\int_{B(\frac{\tau+2\rho}{3})\backslash
B(\rho)}|\mathbf{\Phi}-\bar{\mathbf{\Phi}}||B-b||\nabla B|dx$
$\displaystyle+C(\tau-\rho)^{-1}\int_{B(\frac{\tau+2\rho}{3})\backslash
B(\rho)}|\mathbf{\Phi}-\bar{\mathbf{\Phi}}||B||\nabla B|dx$
$\displaystyle+C(\tau-\rho)^{-2}\int_{B(\frac{\tau+2\rho}{3})\backslash
B(\rho)}|\mathbf{\Phi}-\bar{\mathbf{\Phi}}||B-b||B|dx.$

Similarly as $I_{5}$, we have

$\displaystyle|I_{6}|$ $\displaystyle\leq$ $\displaystyle
Cr(\tau-\rho)^{-1}||\nabla B||_{L^{2}(B(\frac{\tau+2\rho}{3})\backslash
B(\rho))}$ (2.12)
$\displaystyle+Cr^{2}(\tau-\rho)^{-2}||B-b||_{L^{6,\infty}(B(\frac{\tau+2\rho}{3})\backslash
B(\rho))}.$

Using (1.7) and (2.10), the term of $I_{7}$ can be controlled by

$\displaystyle|I_{7}|$ $\displaystyle\leq$ $\displaystyle
C(\tau-\rho)^{-1}\int_{B(\frac{\tau+2\rho}{3})\backslash B(\rho)}|\nabla
B||B||B-b|dx$ (2.13) $\displaystyle\leq$ $\displaystyle
C(\tau-\rho)^{-1}||\nabla B||_{L^{2}(B(\frac{\tau+2\rho}{3})\backslash
B(\rho))}||B||_{L^{6,\infty}(B(\frac{\tau+2\rho}{3}))}||B-b||_{L^{6,\infty}(B(\frac{\tau+2\rho}{3}))}||1||_{L^{6,2}(B(\frac{\tau+2\rho}{3}))}$
$\displaystyle\leq$ $\displaystyle Cr^{\frac{1}{2}}(\tau-\rho)^{-1}||\nabla
B||_{L^{2}(B(\frac{\tau+2\rho}{3})\backslash B(\rho))}.$

At last, we estimate the term of $I_{3}$, and by $\nabla\cdot\mathbf{\Phi}=u$
we get

$\displaystyle|I_{3}|$ $\displaystyle=$
$\displaystyle\left|\int_{B(\frac{\tau+2\rho}{3})\backslash
B(\rho)}(p-a)\partial_{i}(\Phi_{ij}-\bar{\Phi}_{ij})\partial_{j}\phi_{1}dx\right|$
$\displaystyle=$ $\displaystyle\left|\int_{B(\frac{\tau+2\rho}{3})\backslash
B(\rho)}(p-a)(\Phi_{ij}-\bar{\Phi}_{ij})\partial_{ij}\phi_{1}dx+\int_{B(\frac{\tau+2\rho}{3})\backslash
B(\rho)}\partial_{i}p(\Phi_{ij}-\bar{\Phi}_{ij})\partial_{j}\phi_{1}dx\right|$
$\displaystyle\leq$ $\displaystyle
C(\tau-\rho)^{-2}||p-a||_{L^{\frac{3s}{3-s}}(B(\frac{\tau+2\rho}{3})\backslash
B(\rho))}||\mathbf{\Phi}-\mathbf{\bar{\Phi}}||_{L^{\frac{3s}{4s-3}}(B(\frac{\tau+2\rho}{3})\backslash
B(\rho))}$ $\displaystyle+C(\tau-\rho)^{-1}||\nabla
p||_{L^{s}(B(\frac{\tau+2\rho}{3})\backslash
B(\rho))}||\mathbf{\Phi}-\mathbf{\bar{\Phi}}||_{L^{\frac{s}{s-1}}(B(\frac{\tau+2\rho}{3})\backslash
B(\rho))}$ $\displaystyle\leq$ $\displaystyle
Cr^{\frac{4s-3}{s}}(\tau-\rho)^{-2}||p-a||_{L^{\frac{3s}{3-s}}(B(\frac{\tau+2\rho}{3})\backslash
B(\rho))}+Cr^{\frac{3s-3}{s}}(\tau-\rho)^{-1}||\nabla
p||_{L^{s}(B(\frac{\tau+2\rho}{3})\backslash B(\rho))},$

where $s\in(1,3)$.

At last, collecting (2.7)-(2.9), (2.11)-(2.13) and the estimates of $I_{3}$,
noting $r>1$ and $||B-b||_{L^{6,\infty}(B(2r))}\leq C$ from (2.10) we arrive
at the following Cacciopolli type inequality

$\displaystyle\int_{B(\rho)}|\nabla u|^{2}+|\nabla B|^{2}dx$
$\displaystyle\leq C(\tau-\rho)^{-2}\int_{B(\frac{\tau+2\rho}{3})\setminus
B(\rho)}|u|^{2}dx+C(\tau-\rho)^{-1}\int_{B(\frac{\tau+2\rho}{3})\setminus
B(\rho)}|u|^{3}dx$ $\displaystyle+Cr(\tau-\rho)^{-1}||\nabla
B||_{L^{2}(B(\frac{\tau+2\rho}{3})\backslash
B(\rho))}+Cr^{2}(\tau-\rho)^{-2}||B-b||_{L^{6,\infty}(B(\frac{\tau+2\rho}{3})\backslash
B(\rho))}$
$\displaystyle+Cr^{\frac{4s-3}{s}}(\tau-\rho)^{-2}||p-a||_{L^{\frac{3s}{3-s}}(B(\frac{\tau+2\rho}{3})\backslash
B(\rho))}+Cr^{\frac{3s-3}{s}}(\tau-\rho)^{-1}||\nabla
p||_{L^{s}(B(\frac{\tau+2\rho}{3})\backslash B(\rho))}$ (2.14)

Step II. The bounded-ness of energy integral.

Firstly, we estimate the first two terms of (2) by following the estimate of
[8]. Using $\nabla\cdot\mathbf{\Phi}=u$, integration by parts and Hölder
inequality, we have

$\displaystyle\int_{B(\frac{2\tau+\rho}{3})}|u|^{3}\phi_{2}^{3}dx$
$\displaystyle=$
$\displaystyle\int_{B(\frac{2\tau+\rho}{3})}\partial_{i}(\Phi_{ij}-(\Phi_{ij})_{B(2r)})u_{j}|u|\phi_{2}^{3}dx$
$\displaystyle=$
$\displaystyle-\int_{B(\frac{2\tau+\rho}{3})}(\Phi_{ij}-(\Phi_{ij})_{B(2r)})\partial_{i}(u_{j}|u|)\phi_{2}^{3}dx-\int_{B(\frac{2\tau+\rho}{3})}(\Phi_{ij}-(\Phi_{ij})_{B(2r)})u_{j}|u|\partial_{i}(\phi_{2}^{3})dx$
$\displaystyle\leq$ $\displaystyle
C\left(\int_{B(\frac{2\tau+\rho}{3})}|\nabla
u|^{2}dx\right)^{\frac{1}{2}}\left(\int_{B(\frac{2\tau+\rho}{3})}|\mathbf{\Phi}-(\mathbf{\Phi})_{B(2r)}|^{6}dx\right)^{\frac{1}{6}}\left(\int_{B(\frac{2\tau+\rho}{3})}|u\phi_{2}|^{3}dx\right)^{\frac{1}{3}}$
$\displaystyle+$ $\displaystyle
C(\tau-\rho)^{-1}\left(\int_{B(\frac{2\tau+\rho}{3})}|\mathbf{\Phi}-(\mathbf{\Phi})_{B(2r)}|^{3}dx\right)^{\frac{1}{3}}\left(\int_{B(\frac{2\tau+\rho}{3})}|u\phi_{2}|^{3}dx\right)^{\frac{2}{3}},$

which implies from Young inequality that

$\displaystyle\int_{B(\frac{2\tau+\rho}{3})}|u|^{3}\phi_{2}^{3}dx\leq
Cr^{\frac{3}{4}}\left(\int_{B(\frac{2\tau+\rho}{3})}|\nabla
u|^{2}dx\right)^{\frac{3}{4}}+Cr^{3}(\tau-\rho)^{-3}.$

Then we have

$\displaystyle(\tau-\rho)^{-1}||u||_{L^{3}(B(\frac{\tau+2\rho}{3})\setminus
B(\rho))}^{3}$ $\displaystyle\leq$ $\displaystyle
Cr^{\frac{3}{4}}(\tau-\rho)^{-1}\left(\int_{B(\frac{2\tau+\rho}{3})}|\nabla
u|^{2}dx\right)^{\frac{3}{4}}$ (2.15) $\displaystyle+Cr^{3}(\tau-\rho)^{-4}.$

By Hölder inequality, we get

$\displaystyle(\tau-\rho)^{-2}||u||_{L^{2}(B(\frac{\tau+2\rho}{3})\setminus
B(\rho))}^{2}\leq
C(\tau-\rho)^{-2}r||u||_{L^{3}(B(\frac{\tau+2\rho}{3})\setminus B(\rho))}^{2}$
(2.16) $\displaystyle\leq$ $\displaystyle
Cr^{\frac{3}{2}}(\tau-\rho)^{-2}\left(\int_{B(\frac{2\tau+\rho}{3})}|\nabla
u|^{2}dx\right)^{\frac{1}{2}}+Cr^{3}(\tau-\rho)^{-4}.$

Let’s substitute (2.15) and (2.16) into (2). Note that
$||B-b||_{L^{6,\infty}(B(2r))}\leq C$ from (2.10) and by Young inequality we
arrive at

$\displaystyle\int_{B(\rho)}|\nabla u|^{2}+|\nabla B|^{2}dx$
$\displaystyle\leq$ $\displaystyle\frac{1}{4}||\nabla
u||_{L^{2}(B(\frac{\tau+2\rho}{3}))}^{2}+\frac{1}{4}||\nabla
B||_{L^{2}(B(\frac{\tau+2\rho}{3}))}^{2}$
$\displaystyle+Cr^{\frac{3s-3}{s}}(\tau-\rho)^{-1}||\nabla
p||_{L^{s}(B(\frac{\tau+2\rho}{3}))}+Cr^{4}(\tau-\rho)^{-4}$
$\displaystyle+Cr^{\frac{4s-3}{s}}(\tau-\rho)^{-2}||p-a||_{L^{\frac{3s}{3-s}}(B(\frac{\tau+2\rho}{3}))},\quad
1<s<3.$

Next we deal with the terms involving the pressure. By Lemma 2.9, the estimate
of (2.2) ($n=3,q=s$ and $g=-u\cdot\nabla u+B\cdot\nabla B$) implies

$\displaystyle||\nabla p||_{L^{s}(B(\frac{\tau+2\rho}{3}))}$
$\displaystyle\leq$ $\displaystyle C(n,s)\left(||u\cdot\nabla u-B\cdot\nabla
B||_{L^{s}(B(\frac{2\tau+\rho}{3}))}+(\tau-\rho)^{-2}||u||_{L^{s}(B(\frac{2\tau+\rho}{3}))}\right)$
(2.18) $\displaystyle\doteq$ $\displaystyle C(n,s)\left(J_{1}+J_{2}\right).$

On one hand, for $J_{1}$, using Hölder inequality we have

$\displaystyle(J_{1})^{s}$ $\displaystyle=$
$\displaystyle\int_{B(\frac{2\tau+\rho}{3})}|u\cdot\nabla u-B\cdot\nabla
B|^{s}dx$ $\displaystyle\leq$ $\displaystyle
C\left(\int_{B(\frac{2\tau+\rho}{3})}|\nabla
u|^{2}dx\right)^{\frac{s}{2}}\left(\int_{B(\frac{2\tau+\rho}{3})}|u|^{\frac{2s}{2-s}}dx\right)^{\frac{2-s}{2}}$
$\displaystyle+C\left(\int_{B(\frac{2\tau+\rho}{3})}|\nabla
B|^{2}dx\right)^{\frac{s}{2}}||B||_{L^{6,\infty}(B(\frac{2\tau+\rho}{3}))}^{s}||1||_{L^{\frac{3}{3-2s},\frac{2}{2-s}}(B(\frac{2\tau+\rho}{3}))}$
$\displaystyle\leq$ $\displaystyle
C\left(\int_{B(\frac{2\tau+\rho}{3})}|\nabla
u|^{2}dx\right)^{\frac{s}{2}}\left(\int_{B(\frac{2\tau+\rho}{3})}|u|^{\frac{2s}{2-s}}dx\right)^{\frac{2-s}{2}}$
$\displaystyle+Cr^{3-2s}\left(\int_{B(\frac{2\tau+\rho}{3})}|\nabla
B|^{2}dx\right)^{\frac{s}{2}},\quad 1<s<\frac{3}{2}.$

Due to the cut-off function (2.6), we have

$\displaystyle\int_{B(\frac{2\tau+\rho}{3})}|u|^{\frac{2s}{2-s}}dx\leq\int_{B(\tau)}|u\phi_{3}|^{\frac{2s}{2-s}}dx,$

hence,

$\displaystyle\int_{B(\tau)}|u\phi_{3}|^{\frac{2s}{2-s}}dx$ $\displaystyle=$
$\displaystyle\int_{B(\tau)}\partial_{i}(\Phi_{ij}-(\Phi_{ij})_{B(\tau)})u_{j}|u|^{\frac{4s-4}{2-s}}\phi_{3}^{\frac{2s}{2-s}}dx$
$\displaystyle=$
$\displaystyle-\int_{B(\tau)}(\Phi_{ij}-(\Phi_{ij})_{B(\tau)})\partial_{i}(u_{j}|u|^{\frac{4s-4}{2-s}})\phi_{3}^{\frac{2s}{2-s}}dx-\int_{B(\tau)}(\Phi_{ij}-(\Phi_{ij})_{B(\tau)})u_{j}|u|^{\frac{4s-4}{2-s}}\partial_{i}(\phi_{3}^{\frac{2s}{2-s}})dx$
$\displaystyle\leq$ $\displaystyle C\left(\int_{B(\tau)}|\nabla
u|^{2}dx\right)^{\frac{1}{2}}\left(\int_{B(\tau)}|u\phi_{3}|^{\frac{2s}{2-s}}dx\right)^{\frac{2s-2}{s}}\left(\int_{B(\tau)}|\mathbf{\Phi}-\mathbf{\Phi}_{B(\tau)}|^{\frac{2s}{4-3s}}dx\right)^{\frac{4-3s}{2s}}$
$\displaystyle+(\tau-\rho)^{-1}\left(\int_{B(\tau)}|u\phi_{3}|^{\frac{2s}{2-s}}dx\right)^{\frac{3s-2}{2s}}\left(\int_{B(\tau)}|\mathbf{\Phi}-\mathbf{\Phi}_{B(\tau)}|^{\frac{2s}{2-s}}dx\right)^{\frac{2-s}{2s}}$

holds for all $s\in(1,\frac{4}{3})$. Using (1.6) and Young inequality, we have

$\displaystyle\int_{B(\tau)}|u\phi_{3}|^{\frac{2s}{2-s}}dx\leq
Cr^{\frac{12-9s}{2(2-s)}}||\nabla
u||_{L^{2}(B(\tau))}^{\frac{s}{2-s}}+Cr^{3}(\tau-\rho)^{-\frac{2s}{2-s}}.$
(2.19)

Then we have

$\displaystyle(J_{1})^{s}$ $\displaystyle\leq$ $\displaystyle
Cr^{\frac{12-9s}{4}}||\nabla
u||_{L^{2}(B(\tau))}^{\frac{3s}{2}}+Cr^{\frac{6-3s}{2}}(\tau-\rho)^{-s}||\nabla
u||_{L^{2}(B(\tau))}^{s}$ $\displaystyle+Cr^{3-2s}||\nabla
B||_{L^{2}(B(\tau))}^{s}.$

Using Hölder inequality and (2.19), the term $J_{2}$ can be controlled by

$\displaystyle(J_{2})^{s}$ $\displaystyle=$
$\displaystyle(\tau-\rho)^{-2s}\int_{B(\frac{2\tau+\rho}{3})}|u|^{s}dx$
$\displaystyle\leq$ $\displaystyle
Cr^{\frac{3s}{2}}(\tau-\rho)^{-2s}\left(\int_{B(\tau)}|u\phi_{3}|^{\frac{2s}{2-s}}dx\right)^{\frac{2-s}{2}}$
$\displaystyle\leq$ $\displaystyle
Cr^{\frac{12-3s}{4}}(\tau-\rho)^{-2s}||\nabla
u||_{L^{2}(B(\tau))}^{\frac{s}{2}}+Cr^{3}(\tau-\rho)^{-3s},$

Combing the estimates of $J_{1}$ and $J_{2}$, by (2.18) we have

$\displaystyle||\nabla p||_{L^{s}(B(\frac{\tau+2\rho}{3}))}$ (2.20)
$\displaystyle\leq$ $\displaystyle Cr^{\frac{12-9s}{4s}}||\nabla
u||_{L^{2}(B(\tau))}^{\frac{3}{2}}+Cr^{\frac{6-3s}{2s}}(\tau-\rho)^{-1}||\nabla
u||_{L^{2}(B(\tau))}$ $\displaystyle+Cr^{\frac{3-2s}{s}}||\nabla
B||_{L^{2}(B(\tau))}+Cr^{\frac{12-3s}{4s}}(\tau-\rho)^{-2}||\nabla
u||_{L^{2}(B(\tau))}^{\frac{1}{2}}$
$\displaystyle+Cr^{\frac{3}{s}}(\tau-\rho)^{-3},\quad s\in(1,\frac{4}{3}).$

Furthermore, Poincaré inequality implies

$\displaystyle||p-p_{B(\frac{\tau+2\rho}{3})}||_{L^{\frac{3s}{3-s}}(B(\frac{\tau+2\rho}{3}))}\leq
C||\nabla p||_{L^{s}(B(\frac{\tau+2\rho}{3}))}.$ (2.21)

Then choosing $a=p_{B(\frac{\tau+2\rho}{3})}$ in (2), (2.20) and (2.21) imply

$\displaystyle\int_{B(\rho)}|\nabla u|^{2}+|\nabla B|^{2}dx$
$\displaystyle\leq$ $\displaystyle\frac{1}{2}\int_{B(\tau)}|\nabla
u|^{2}+|\nabla B|^{2}dx+Cr^{4}(\tau-\rho)^{-4}$
$\displaystyle+Cr^{\frac{7}{4}}(\tau-\rho)^{-2}||\nabla
u||_{L^{2}(B(\tau))}^{\frac{3}{2}}+Cr^{\frac{5}{2}}(\tau-\rho)^{-3}||\nabla
u||_{L^{2}(B(\tau))}$ $\displaystyle+Cr^{2}(\tau-\rho)^{-2}||\nabla
B||_{L^{2}(B(\tau))}+Cr^{\frac{13}{4}}(\tau-\rho)^{-4}||\nabla
u||_{L^{2}(B(\tau))}^{\frac{1}{2}}+Cr^{4}(\tau-\rho)^{-5}$

Apply the iteration lemma (see, for example, Lemma 3.1 in [15]), we have

$\displaystyle\int_{B(\rho)}|\nabla u|^{2}+|\nabla B|^{2}dx$
$\displaystyle\leq$ $\displaystyle Cr^{8}(\tau-\rho)^{-8}.$

Choosing $\rho=r$ and $\tau=2r$, letting $r\rightarrow+\infty$ we have

$\displaystyle\int_{\mathbb{R}^{3}}|\nabla u|^{2}+|\nabla B|^{2}dx\leq C.$
(2.22)

Step III. Vanishing property of Dirichlet integral.

Next we prove the vanishing property of Dirichlet integral. It follows from
(2.22) that $u,B\in\dot{W}^{1,2}(\mathbb{R}^{3})$. Using Lemma 1.6 and the
condition $u\in BMO^{-1}(\mathbb{R}^{3})$, we have

$\displaystyle|||u|^{2}||_{L^{2}(\mathbb{R}^{3})}\leq
C||u||_{\dot{W}^{1,2}(\mathbb{R}^{3})}||u||_{BMO^{-1}(\mathbb{R}^{3})},$

which means $u\in L^{4}(\mathbb{R}^{3})$. We claim that

$\displaystyle u\in L^{p}(\mathbb{R}^{3})\quad{\rm for~{}~{}all}\quad
p\in[4,6].$ (2.23)

Indeed, for a new cut-off function $\zeta$ satisfying $\zeta=1$ in
$B(\frac{4}{3}r)$ and $\zeta=0$ on $B(2r)^{c}$, we get

$\displaystyle\left(\int_{\mathbb{R}^{3}}|u\zeta|^{6}dx\right)^{\frac{1}{3}}\leq\int_{\mathbb{R}^{3}}|\nabla(u\zeta)|^{2}dx\leq
C\int_{B(2r)}|\nabla
u|^{2}dx+Cr^{-\frac{1}{2}}\left(\int_{B(2r)}|u|^{4}dx\right)^{\frac{1}{2}},$

which implies $u\in L^{6}(\mathbb{R}^{3})$. Similarly,

$\displaystyle B\in L^{6}(\mathbb{R}^{3}),$ (2.24)

since

$\displaystyle\left(\int_{\mathbb{R}^{3}}|B\zeta|^{6}dx\right)^{\frac{1}{3}}\leq
C\int_{B(2r)}|\nabla
B|^{2}dx+Cr^{-2}||B||_{L^{6,\infty}(B(2r))}^{2}||1||_{L^{\frac{3}{2},1}(B(2r))}<\infty.$

Recall the Cacciopolli inequality (2), and choose $\rho=r$, $\tau=2r$,
$s=\frac{3}{2}$ and $b=B_{B(\frac{4}{3}r)\setminus B(r)}$. Then

$\displaystyle\int_{B(r)}|\nabla u|^{2}+|\nabla B|^{2}dx$ $\displaystyle\leq$
$\displaystyle Cr^{-2}||u||_{L^{2}(B(\frac{4}{3}r)\setminus
B(r))}^{2}+Cr^{-1}||u||_{L^{3}(B(\frac{4}{3}r)\setminus B(r))}^{3}$ (2.25)
$\displaystyle+C||B-b||_{L^{6,\infty}(B(\frac{4}{3}r)\backslash
B(r))}+C||\nabla B||_{L^{2}(B(\frac{4}{3}r)\backslash B(r))}$
$\displaystyle+C||\nabla p||_{L^{\frac{3}{2}}(B(\frac{4}{3}r)\backslash
B(r))}+C||p-a||_{L^{3}(B(\frac{4}{3}r)\backslash B(r))}.$

Using Hölder inequality, it follows that

$\displaystyle r^{-2}||u||_{L^{2}(B(\frac{4}{3}r)\setminus B(r))}^{2}\leq
Cr^{-\frac{1}{2}}||u||_{L^{4}(B(\frac{4}{3}r)\setminus B(r))}^{2}\rightarrow
0,$

and

$\displaystyle r^{-1}||u||_{L^{3}(B(\frac{4}{3}r)\setminus B(r))}^{3}\leq
Cr^{-\frac{1}{4}}||u||_{L^{4}(B(\frac{4}{3}r)\setminus B(r))}^{3}\rightarrow
0,$

as $r\rightarrow\infty$ since (2.23). Moreover, note that

$\displaystyle||B-b||_{L^{6}(B(\frac{4}{3}r)\setminus B(r))}\leq C||\nabla
B||_{L^{2}(B(\frac{4}{3}r)\setminus B(r))}\rightarrow 0\quad{\rm as}\quad
r\rightarrow\infty,$

which implies the third and the fourth terms are vanishing as
$r\rightarrow\infty$.

For the pressure, by (2.24) and (2.2) in Lemma 2.9, we have

$\displaystyle||\nabla p||_{L^{\frac{3}{2}}(B(\frac{4}{3}))}$
$\displaystyle\leq$ $\displaystyle C||u||_{L^{6}(B(2r))}||\nabla
u||_{L^{2}(B(2r))}+C||B||_{L^{6}(B(2r))}||\nabla B||_{L^{2}(B(2r))}$
$\displaystyle+Cr^{-1/2}||u||_{L^{6}(B(2r))}.$

Letting $r\rightarrow\infty$, we obtain $\nabla p\in
L^{\frac{3}{2}}(\mathbb{R}^{3})$, which yields $||\nabla
p||_{L^{\frac{3}{2}}(B(\frac{4}{3}r)\backslash B(r))}\rightarrow 0$. Choosing
$a=p_{B(\frac{4}{3}r)\backslash B(r)}$, it immediately implies

$\displaystyle||p-a||_{L^{3}(B(\frac{4}{3}r)\backslash B(r))}\leq||\nabla
p||_{L^{\frac{3}{2}}(B(\frac{4}{3}r)\backslash B(r))}\rightarrow 0.$

Hence, (2.25) implies

$\displaystyle\int_{B(r)}|\nabla u|^{2}+|\nabla B|^{2}dx\rightarrow 0\quad{\rm
as}\quad r\rightarrow\infty,$

which means $u\equiv 0$ and $B\equiv 0$, since $u\in L^{6}(\mathbb{R}^{3})$
and $B\in L^{6,\infty}(\mathbb{R}^{3})$. The proof is complete. ∎

##  3\. Proof of Theorem 1.4

###### Proof.

Assume that $\eta(x)\in C_{0}^{\infty}(\mathbb{R}^{3})$ with $0\leq\eta\leq
1$, satisfying that
i) for $r>1$, $r\leq\rho<\tau\leq 2r$,

$\eta(x)=\eta(|x|)=\left\\{\begin{array}[]{llll}1,\quad~{}{\rm
in}~{}B(\rho);\\\ 0,\quad~{}{\rm in}~{}\mathbb{R}^{3}\backslash
B(\tau)\end{array}\right.$

ii)

$\displaystyle|\nabla\eta|\leq\frac{C}{\tau-\rho},\quad|\nabla^{2}\eta|\leq\frac{C}{(\tau-\rho)^{2}}.$

Since $B(\tau)$ is a star-like domain(see P38, [12]), due to Theorem III 3.1
in [12], there exists a constant $C(s)$ and a vector-valued function
$w:B(\tau)\rightarrow\mathbb{R}^{3}$ such that $w\in W^{1,s}_{0}(B(\tau))$,
and $\nabla\cdot w=u\cdot\nabla\eta$ satisfying

$\displaystyle\int_{B(\tau)}|\nabla w(x)|^{s}\,dxdy\leq
C(s)\int_{B(\tau)}|u\cdot\nabla\eta|^{s}\,dx,\quad 1<s<\infty.$ (3.1)

Then multiplying $(\ref{eq:VMHD})_{1}$ by $u\eta-w$, integrating it over
$B(2r)$ and integration by parts yield that

$\displaystyle\lambda_{1}\int_{B(2r)}|\partial_{1}u|^{2}\eta
dx+\lambda_{2}\int_{B(2r)}|\partial_{2}u|^{2}\eta
dx+\lambda_{3}\int_{B(2r)}|\partial_{3}u|^{2}\eta dx$ (3.2) $\displaystyle=$
$\displaystyle\lambda_{1}\int_{B(2r)}\partial_{1}u\cdot\partial_{1}wdx+\lambda_{2}\int_{B(2r)}\partial_{2}u\cdot\partial_{2}wdx+\lambda_{3}\int_{B(2r)}\partial_{3}u\cdot\partial_{3}wdx$
$\displaystyle+$
$\displaystyle\frac{1}{2}\lambda_{1}\int_{B(2r)}|u|^{2}\partial_{11}\eta
dx+\frac{1}{2}\lambda_{2}\int_{B(2r)}|u|^{2}\partial_{22}\eta
dx+\frac{1}{2}\lambda_{3}\int_{B(2r)}|u|^{2}\partial_{33}\eta dx$
$\displaystyle-$ $\displaystyle\int_{B(2r)}u\cdot\nabla w\cdot
udx+\frac{1}{2}\int_{B(2r)}|u|^{2}u\cdot\nabla\eta dx+\int_{B(2r)}B\cdot\nabla
w\cdot Bdx$ $\displaystyle+$ $\displaystyle\int_{B(2r)}B\cdot\nabla B\cdot
u\eta dx.$

Similarly, by multiplying $(\ref{eq:VMHD})_{2}$ by $B\eta$ we derive

$\displaystyle\mu_{1}\int_{B(2r)}|\partial_{1}B|^{2}\eta
dx+\mu_{2}\int_{B(2r)}|\partial_{2}B|^{2}\eta
dx+\mu_{3}\int_{B(2r)}|\partial_{3}B|^{2}\eta dx$ $\displaystyle=$
$\displaystyle\frac{1}{2}\mu_{1}\int_{B(2r)}|B|^{2}\partial_{11}\eta
dx+\frac{1}{2}\mu_{2}\int_{B(2r)}|B|^{2}\partial_{22}\eta
dx+\frac{1}{2}\mu_{3}\int_{B(2r)}|B|^{2}\partial_{33}\eta dx$ $\displaystyle+$
$\displaystyle\frac{1}{2}\int_{B(2r)}|B|^{2}u\cdot\nabla\eta
dx-\int_{B(2r)}B\cdot\nabla B\cdot u\eta dx-\int_{B(2r)}B\cdot
u(B\cdot\nabla)\eta dx.$

Letting

$\displaystyle
G(r)=\sum_{i=1}^{3}\lambda_{i}\int_{B(r)}|\partial_{i}u|^{2}dx+\sum_{i=1}^{3}\mu_{i}\int_{B(r)}|\partial_{i}B|^{2}dx,$

and taking the sum of (3.2) and (3), we have

$\displaystyle G(\rho)$ $\displaystyle\leq$
$\displaystyle\lambda_{1}\int_{B(2r)}\partial_{1}u\cdot\partial_{1}wdx+\lambda_{2}\int_{B(2r)}\partial_{2}u\cdot\partial_{2}wdx+\lambda_{3}\int_{B(2r)}\partial_{3}u\cdot\partial_{3}wdx$
$\displaystyle+$
$\displaystyle\frac{1}{2}\lambda_{1}\int_{B(2r)}|u|^{2}\partial_{11}\eta
dx+\frac{1}{2}\lambda_{2}\int_{B(2r)}|u|^{2}\partial_{22}\eta
dx+\frac{1}{2}\lambda_{3}\int_{B(2r)}|u|^{2}\partial_{33}\eta dx$
$\displaystyle-$ $\displaystyle\int_{B(2r)}u\cdot\nabla w\cdot
udx+\frac{1}{2}\int_{B(2r)}|u|^{2}u\cdot\nabla\eta dx+\int_{B(2r)}B\cdot\nabla
w\cdot Bdx$ $\displaystyle+$
$\displaystyle\frac{1}{2}\mu_{1}\int_{B(2r)}|B|^{2}\partial_{11}\eta
dx+\frac{1}{2}\mu_{2}\int_{B(2r)}|B|^{2}\partial_{22}\eta
dx+\frac{1}{2}\mu_{3}\int_{B(2r)}|B|^{2}\partial_{33}\eta dx$ $\displaystyle+$
$\displaystyle\frac{1}{2}\int_{B(2r)}|B|^{2}u\cdot\nabla\eta
dx-\int_{B(2r)}B\cdot u(B\cdot\nabla)\eta dx\doteq\sum_{i=1}^{14}I_{i}.$

Since the terms $I_{1}$, $I_{2}$ and $I_{3}$ are similar, we compute the first
term. Letting $T(\rho,\tau)=B(\tau)\setminus B(\rho)$, using Hölder and Young
inequality, (3.1) implies

$\displaystyle|I_{1}|$ $\displaystyle\leq$
$\displaystyle\lambda_{1}\int_{B(\tau)}|\partial_{1}u||\partial_{1}w|dx$
$\displaystyle\leq$
$\displaystyle\varepsilon\lambda_{1}\int_{B(\tau)}|\partial_{1}u|^{2}dx+C\lambda_{1}\int_{B(\tau)}|\partial_{1}w|^{2}dx$
$\displaystyle\leq$ $\displaystyle\varepsilon
G(\tau)+C\lambda_{1}(\tau-\rho)^{-2}\int_{T(\rho,\tau)}|u|^{2}dx$
$\displaystyle\leq$ $\displaystyle\varepsilon
G(\tau)+C\lambda_{1}r(\tau-\rho)^{-2}||u||_{L^{3}(T(\rho,\tau))}^{2},$

where $\varepsilon>0$, to be decided. Similarly,

$\displaystyle|I_{2}|+|I_{3}|\leq\varepsilon
G(\tau)+C(\lambda_{2}+\lambda_{3})r(\tau-\rho)^{-2}||u||_{L^{3}(T(\rho,\tau))}^{2}.$

The estimates of $I_{4}$, $I_{5}$ and $I_{6}$ can be obtained directly from
Hölder’s inequality:

$\displaystyle|I_{4}|+|I_{5}|+|I_{6}|\leq
C(\lambda_{1}+\lambda_{2}+\lambda_{3})r(\tau-\rho)^{-2}||u||_{L^{3}(T(\rho,\tau))}^{2}.$

Similarly, the terms of $I_{10}$, $I_{11}$ and $I_{12}$ are also obtained:

$\displaystyle|I_{10}|+|I_{11}|+|I_{12}|\leq
C(\mu_{1}+\mu_{2}+\mu_{3})r^{2}(\tau-\rho)^{-2}||B||_{L^{6}(T(\rho,\tau))}^{2}.$

For the term $I_{7}$, using (3.1) again, we have

$\displaystyle|I_{7}|$ $\displaystyle\leq$
$\displaystyle\int_{B(2r)}|u|^{2}|\nabla
w|dx\leq\left(\int_{B(\tau)}|u|^{3}dx\right)^{\frac{2}{3}}\left(\int_{B(\tau)}|\nabla
w|^{3}dx\right)^{\frac{1}{3}}$ $\displaystyle\leq$ $\displaystyle
C(\tau-\rho)^{-1}||u||_{L^{3}(T(\rho,\tau))}||u||_{L^{3}(B(\tau))}^{2},$

and

$\displaystyle|I_{8}|\leq
C(\tau-\rho)^{-1}||u||_{L^{3}(T(\rho,\tau))}||u||_{L^{3}(B(\tau))}^{2}.$

Similarly,

$\displaystyle|I_{9}|$ $\displaystyle\leq$
$\displaystyle\int_{B(2r)}|B|^{2}|\nabla w|dx\leq
Cr||B||_{L^{6}(B(\tau))}^{2}\left(\int_{B(\tau)}|\nabla
w|^{3}dx\right)^{\frac{1}{3}}$ $\displaystyle\leq$ $\displaystyle
Cr(\tau-\rho)^{-1}||B||_{L^{6}(B(\tau))}^{2}||u||_{L^{3}(T(\rho,\tau))}.$

The terms $I_{13}$ and $I_{14}$ can be treated as the term $I_{9}$:

$\displaystyle|I_{13}|+|I_{14}|\leq
Cr(\tau-\rho)^{-1}||B||_{L^{6}(T(\rho,\tau))}^{2}||u||_{L^{3}(T(\rho,\tau))}.$

Collecting all the terms of $I_{i}$, we arrive at

$\displaystyle G(\rho)$ $\displaystyle\leq$ $\displaystyle 2\varepsilon
G(\tau)+C(\lambda_{1}+\lambda_{2}+\lambda_{3})r(\tau-\rho)^{-2}||u||_{L^{3}(T(\rho,\tau))}^{2}$
$\displaystyle+C(\mu_{1}+\mu_{2}+\mu_{3})r^{2}(\tau-\rho)^{-2}||B||_{L^{6}(T(\rho,\tau))}^{2}$
$\displaystyle+C(\tau-\rho)^{-1}||u||_{L^{3}(T(\rho,\tau))}||u||_{L^{3}(B(\tau))}^{2}+Cr(\tau-\rho)^{-1}||B||_{L^{6}(B(\tau))}^{2}||u||_{L^{3}(T(\rho,\tau))}$
$\displaystyle\leq$ $\displaystyle 2\varepsilon
G(\tau)+C(\lambda_{1}+\lambda_{2}+\lambda_{3})r(\tau-\rho)^{-2}||u||_{L^{3}(T(r,2r))}^{2}$
$\displaystyle+C(\mu_{1}+\mu_{2}+\mu_{3})r^{2}(\tau-\rho)^{-2}||B||_{L^{6}(T(r,2r))}^{2}$
$\displaystyle+C(\tau-\rho)^{-1}||u||_{L^{3}(T(r,2r))}||u||_{L^{3}(B(\tau))}^{2}+Cr(\tau-\rho)^{-1}||B||_{L^{6}(B(\tau))}^{2}||u||_{L^{3}(T(r,2r))}.$

Choosing $\varepsilon=\frac{1}{4}$ and using the iteration in [15], we have

$\displaystyle G(\rho)$ $\displaystyle\leq$ $\displaystyle
C(\lambda_{1}+\lambda_{2}+\lambda_{3})r(\tau-\rho)^{-2}||u||_{L^{3}(T(r,2r))}^{2}$
$\displaystyle+C(\mu_{1}+\mu_{2}+\mu_{3})r^{2}(\tau-\rho)^{-2}||B||_{L^{6}(T(r,2r))}^{2}$
$\displaystyle+C(\tau-\rho)^{-1}||u||_{L^{3}(T(r,2r))}||u||_{L^{3}(B(\tau))}^{2}+Cr(\tau-\rho)^{-1}||B||_{L^{6}(B(\tau))}^{2}||u||_{L^{3}(T(r,2r))}.$

Taking $\rho=r$ and $\tau=2r$, letting $r\rightarrow\infty$, we have

$\displaystyle\sum_{i=1}^{3}\lambda_{i}\int_{\mathbb{R}^{3}}|\partial_{i}u|^{2}dx+\sum_{i=1}^{3}\mu_{i}\int_{\mathbb{R}^{3}}|\partial_{i}B|^{2}dx=0,$

since $u\in L^{3}(\mathbb{R}^{3})$ and $B\in L^{6}(\mathbb{R}^{3})$. Then
there exist $i\in\\{1,2,3\\}$ and $j\in\\{1,2,3\\}$, such that
$\partial_{i}u,\partial_{j}B\equiv 0$ due to the known condition (1.13).
Without loss of generality, we assume that $\partial_{1}u=\partial_{1}B=0$,
which means $u(x_{1},x_{2},x_{3})=u(x_{2},x_{3})$ and
$B(x_{1},x_{2},x_{3})=B(x_{2},x_{3})$. Then

$\displaystyle\int_{\mathbb{R}^{2}}\int_{-\infty}^{+\infty}|u(x_{2},x_{3})|^{3}dx_{1}dx_{2}dx_{3}=\infty,$

if $u$ is a nontrivial solution, which is a contradiction with $u\in
L^{3}(\mathbb{R}^{3})$. Thus we have $u\equiv 0$ in $\mathbb{R}^{3}$. Using
the same idea, we have $B\equiv 0$ in $\mathbb{R}^{3}$. The proof is complete.
∎

Acknowledgments. W. Wang was supported by NSFC under grant 12071054 and
11671067.

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