Global Smooth Axisymmetic Solutions of the Boussinesq Equations for Magnetohydrodynamics Convection

Abstract

In this paper, we consider global axisymmetric smooth solutions for the Boussinesq equation for magnetohydrodynamics convection without magnetic diffusion and heat convection. We obtain that for axially symmetric initial data without any smallness restrictions, such a system admits global smooth axially symmetric solutions without swirl.

1 Introduction

In this paper, we consider the following three dimensional incompressible Boussinesq equations for magnetohydrodynamics (MHD) convection

$$\begin{aligned}&\displaystyle u_t+u\cdot \nabla u+\nabla \pi =\mu \Delta u+B\cdot \nabla B+g\theta e_z, \end{aligned}$$

(1.1a)

$$\begin{aligned}&\displaystyle B_t+u\cdot \nabla B=B\cdot \nabla u,\end{aligned}$$

(1.1b)

$$\begin{aligned}&\displaystyle \theta _t+u\cdot \nabla \theta =0,\end{aligned}$$

(1.1c)

$$\begin{aligned}&\displaystyle \nabla \cdot u=\nabla \cdot B=0. \end{aligned}$$

(1.1d)

Here $u=(u_1,u_2,u_3)$ is the velocity, $\pi$ is the pressure, $\theta$ is the temperature fluctuation about a constant, and $B=(B_1,B_2,B_3)$ is the magnetic field, defined on $x\in {\mathbb {R}}^3$ and $t\in {\mathbb {R}}^+$. This system can be used to model the large scale cosmic magnetic fields that are maintained by hydromagnetic dynamos. Physically, the first equation describes the conservation law of the momentum with the effect of the buoyancy force, and the constant $\mu$ is the viscosity. Here $-ge_z$ denotes the direction of the gravity and the original form of the buoyancy term is $g(\theta -\theta _0)e_z$ with $\theta _0$ denoting the temperature distribution of the reference state which can be absorbed in the pressure term and hence is assumed to be zero in this paper. The second equation shows that the electromagnetic field is governed by the Maxwell equation and the third equation describes the temperature fluctuation about a constant state. Here, we have omitted the magnetic diffusion and heat diffusion. For more physics details and numerical simulations, the interested readers may refer to [5, 28, 31, 32] and the references therein. Hereafter, the system is referred to as the Boussinesq–MHD system or BMHD for short.

Global regularity of such a PDE system for large initial data is widely open even if when $\theta =B\equiv 0$. In this case, the system reduces to the 3D classical incompressible Navier–Stokes equation, whose global well-posedness is widely open for large initial data. But under axially symmetric assumptions, global well-posedness of classical solutions without swirl component of velocity field was solved by Ladyzhenskaya [20] and by Ukhovskii and Yudovich [33] independently. More precisely, the Navier–Stokes system has a unique global axisymmetric solution for initial data $u_0\in H^1$ and vorticity $\omega _0$ and $r^{-1}\omega _0\in L^2\cap L^{\infty }$, which can be guaranteed when $u_0\in H^s$ with $s>7/2$ in 3D. The initial regularity was weakened to $u_0\in H^2$ by Leonardi et al. [24] and to $u_0\in H^{1/2}$ by Abidi [1] later on. The main observation is that under axisymmetric assumptions, the vorticity quantity $r^{-1}\omega$ has maximum principle and hence global regularity can be obtained. See also [17] for the global regularity of the axisymmetric Navier–Stokes equation with swirl for a class of large anisotropic initial data. If furthermore, we let the viscosity $\mu$ to be zero, the Navier–Stokes equation reduces to the standard 3D Euler equation describing the motion of an ideal incompressible fluid, whose global in time regularity is a long standing open problem due to possible vortex stretching [4, 27]. To gain insight into this challenging problem, many authors turn to study the 2D Boussinesq equation, i.e., the system (1.1) without magnetic field B, which retains some key features of the 3D Euler or Navier–Stokes equations.

When the magnetic field $B\equiv 0$, the BMHD system reduces to the Boussinesq system, whose weak solutions in $L^p$ was studied in ${\mathbb {R}}^n$ by Cannon and DiBenedetto [10] for general spatial dimensions and the local well-posedness in Sobolev spaces was obtained By Chae and Nam in [13] in ${\mathbb {R}}^2$. In the 2D case, many global well-posedness results are obtained under various viscous conditions. For example, see [12, 18] for global well-posedness in the partial viscosity case and [21] for the initial boundary value problem. See also [2] for global well-posedness in the 2D case with partial vsicosity in Besov spaces. For the 3D case, when the initial data is axisymmetric, global well-posedness was shown by Abidi et al. [3], under the assumption that the initial density/temperature $\theta _0$ does not intersect the Z-axis and the orthogonal projection of the support of $\theta _0$ to the Z-axis is compact.

When the temperature $\theta$ vanishes, the BMHD system reduces to the well-known MHD system, for which there are lots of important results up to date. Concerning the local well-posedness, one may refer to the paper of Sermange and Temam [30] in the case of fully viscosity, where the authors also proved the global well-posedness in the 2D case. Global existence of classical solutions is obtained by Lin et al. [26] under smallness conditions in Sobolev spaces of the initial velocity field and the displacement of the magnetic field from a non-zero constant. See also [11, 19, 29, 36] and the references therein for global well-posedness under different conditions. For partial regularity and various blowup conditions, one may refer to [9, 14,15,16, 23] and the references therein. For global well-posedness in the 3D case, Lin et al. [25, 35] studied global well-posedness of small solutions for MHD-type solutions. For a class of axisymmetric initial data, Lei [22] established the global well-posedness of classical solutions whose the swirl component of the velocity and magnetic vorticity vanish.

For the full BMHD system, there are some theoretical as well as numerical results up to date. Bian et al. [5,6,7] studied the global existence and uniqueness for the initial boundary value problem to the 2D stratified Boussinesq–MHD system without smallness assumptions on the initial data, with temperature-dependent viscosity, thermal diffusivity and electrical conductivity. But few results are known up to date about global well-posedness in the 3D case. In a recent paper [8], the authors proved a global well-posedness result for large initial data for the BMHD system with a nonlinear damping term, with both fluid viscosity and magnetic diffusion. However, it is not known whether global well-posedness holds without the nonlinear damping term even with full velocity viscosity, magnetic diffusion as well as heat diffusion. Numerically, Schrinner et al. [31, 32] studied the global numerical simulations of rotating magnetoconvection and the geodynamo with mean-field description, where mean fields are defined by azimuthal averaging over all values of the zaimuthal coordinate and are axisymmetric about the polar axis. Both the theoretic difficulties and the numerical simulations motivate us to study the radial solutions or the axisymmetric solutions of such a system.

In this paper, we will show that the BMHD system (1.1) in ${\mathbb {R}}^3$ is globally well-posed for a class of large axially symmetric initial data without swirl, even if there is no magnetic diffusion and heat convection. The case when swirl is present will be pursued in short future. Before we state the main result, we introduce the axisymmetric solutions for the BMHD system (1.1).

Let $x=(x_1,x_2,z)\in {\mathbb {R}}^3$ and $r=\sqrt{x_1^2+x_2^2}$. We define the axially symmetric coordinate system $(e_r,e_{\phi },e_z)$ by

$$\begin{aligned} e_r=(x_1/r,x_2/r,0)^{\top },\ \ e_\phi =(-x_2/r,x_1/r,0)^{\top },\ \ e_z=(0,0,1)^{\top }, \end{aligned}$$

where $\phi$ denotes the angle variable. Considering the BMHD system (1.1) in the axially symmetric coordinate $(e_r,e_{\phi },e_z)$, but letting the unknowns depend only on the variables (t, r, z) and be independent of the angular variable $\phi$, we can write

$$\begin{aligned} {\left\{ \begin{array}{ll} u(t,x)=u^r(t,r,z)e_r+u^{\phi }(t,r,z)e_{\phi }+u^z(t,r,z)e_z,\\ B(t,x)=B^r(t,r,z)e_r+B^{\phi }(t,r,z)e_{\phi }+B^z(t,r,z)e_z,\\ \theta (t,x)=\theta (t,r,z), \quad \pi (t,x)=\pi (t,r,z). \end{array}\right. } \end{aligned}$$

(1.2)

Then the BMHD system (1.1) can be equivalently written in the axially symmetric coordinate $(e_r,e_{\phi },e_z)$,

$$\begin{aligned} {\left\{ \begin{array}{ll} \partial _tu^r+u^r\partial _ru^r+u^z\partial _zu^r-\frac{(u^{\phi })^2}{r}+\partial _r\pi =\left( \Delta -\frac{1}{r^2}\right) u^r+B^r\partial _rB^r+B^z\partial _zB^r-\frac{(B^{\phi })^2}{r},\\ \partial _tu^{\phi }+u^r\partial _ru^{\phi }+u^z\partial _zu^{\phi }+\frac{u^ru^{\phi }}{r}=\left( \Delta -\frac{1}{r^2}\right) u^{\phi }+B^r\partial _rB^{\phi }+B^z\partial _zB^{\phi }+\frac{B^rB^{\phi }}{r},\\ \partial _tu^z+u^r\partial _ru^z+u^z\partial _zu^z+\partial _z\pi =\Delta u^z+B^r\partial _rB^z+B^z\partial _zB^z+\theta ,\\ \partial _tB^r+u^r\partial _rB^r+u^z\partial _zB^r=B^r\partial _ru^{r}+B^z\partial _zu^{r},\\ \partial _tB^{\phi }+u^r\partial _rB^{\phi }+u^z\partial _zB^{\phi }+\frac{B^ru^{\phi }}{r}=B^r\partial _ru^{\phi }+B^z\partial _zu^{\phi }+\frac{u^rB^{\phi }}{r},\\ \partial _tB^{z}+u^r\partial _rB^{z}+u^z\partial _zB^z=B^r\partial _ru^z+B^z\partial _zu^z,\\ \partial _t\theta +u^r\partial _r\theta +u^z\partial _z\theta =0. \end{array}\right. } \end{aligned}$$

(1.3)

For such a system, it is not difficult to have the following local existence and uniqueness result.

Lemma 1.1

Let $(u_0,B_0,\theta _0)\in H^2({\mathbb {R}}^3)$ be axially symmetric and $u_0$ and $B_0$ are divergence free. Then there exists exactly one solution $(u,B,\theta ,\pi )$ such that

$$\begin{aligned} \begin{aligned}&(u,B,\theta )\in L^{\infty }(0,T;H^2({\mathbb {R}}^3)), \quad u\in L^2(0,T;H^3({\mathbb {R}}^3)),\\&\left( \frac{\partial u}{\partial t},\frac{\partial B}{\partial t},\frac{\partial \theta }{\partial t}\right) \in L^2(0,t;H^1({\mathbb {R}}^3)),\ \ \ \nabla \pi \in L^{\infty }(0,T;L^2({\mathbb {R}}^3)), \end{aligned} \end{aligned}$$

for some $T>0$. Moreover, $(u,B,\theta ,\pi )$ is axially symmetric.

The proof can be adapted from a similar local existence and uniqueness result for the incompressible Navier–Stokes equations in ${\mathbb {R}}^3$ in [24]. By uniqueness of local classical solutions, it is clear that if $u^{\phi }=B^{r}=B^z=0$ for all later times if they vanish initially. In this case, we have the following simplified system for $(u^r,u^z,B^{\phi },\theta )$:

$$\begin{aligned} {\left\{ \begin{array}{ll} \partial _tu^r+u^r\partial _ru^r+u^z\partial _zu^r+\partial _r\pi =(\Delta -\frac{1}{r^2})u^r-\frac{(B^{\phi })^2}{r},\\ \partial _tu^z+u^r\partial _ru^z+u^z\partial _zu^z+\partial _z\pi =\Delta u^z+\theta ,\\ \partial _tB^{\phi }+u^r\partial _rB^{\phi }+u^z\partial _zB^{\phi }=\frac{u^rB^{\phi }}{r},\\ \partial _t\theta +u^r\partial _r\theta +u^z\partial _z\theta =0. \end{array}\right. } \end{aligned}$$

(1.4)

In this case, the incompressible condition is equivalent to

$$\begin{aligned} \partial _ru^r+\frac{u^r}{r}+\partial _zu^z=0, \end{aligned}$$

and the divergence free condition is automatically satisfied since $B^r=B^z=0$ for all times $t\ge 0$.

Let $\omega =\nabla \times u$ be the vorticity. Then it is computed that $\omega =\omega ^{\phi }e^{\phi }$, where $\omega ^{\phi }=\partial _zu^r-\partial _ru^z$. From the first two equations of (1.4), we have the following equation for $\omega ^{\phi }$,

$$\begin{aligned} \partial _t\omega ^{\phi }+u\cdot \nabla \omega ^{\phi }-\frac{u^r}{r}\omega ^{\phi }=\left( \partial _{rr}+\frac{1}{r}\partial _r+\partial _{zz}-\frac{1}{r^2}\right) \omega ^{\phi }-\frac{2}{r}B^{\phi }\partial _zB^{\phi }-\partial _r\theta . \end{aligned}$$

(1.5)

Further, let $\Pi =B^{\phi }/r$ and $\Omega =\omega ^{\phi }/r$, the system (1.4) gives the following system

$$\begin{aligned} {\left\{ \begin{array}{ll} \partial _t\Omega +u\cdot \nabla \Omega =(\Delta +\frac{2}{r}\partial _r)\Omega -\partial _z\Pi ^2-\frac{\partial _r\theta }{r},\\ \partial _t\Pi +u\cdot \nabla \Pi =0,\\ \partial _t\theta +u\cdot \nabla \theta =0. \end{array}\right. } \end{aligned}$$

(1.6)

We also note that by definition of $\omega$ and $\Omega$, there exists some $\psi ^{\phi }$ such that

$$\begin{aligned} -\left( \Delta +\frac{2}{r}\partial _r\right) (r^{-1}\psi ^{\phi })=\Omega , \quad u=\nabla \times (\psi ^{\phi }e_{\phi }). \end{aligned}$$

The main result is stated in the following.

Theorem 1.1

Suppose that $u_0,B_0$ and $\theta _0$ are all axially symmetric and $u_0,B_0$ are divergence free vectors with $u_0^{\phi }=B_0^r=B_0^z=0$. Moreover, we assume that $u_0,B_0\in H^s({\mathbb {R}}^3)$ with $s\ge 2$ and $r^{-1}B_0^{\phi }\in L^{\infty }({\mathbb {R}}^3)$. Suppose also that $\theta _0\in H^s({\mathbb {R}}^3)$ with $s\ge 2$ such that ${{\,\mathrm{spt}\,}}\theta _0$, the support of $\theta _0$, does not intersect the Z-axis and the projection of ${{\,\mathrm{spt}\,}}\theta _0$ to the Z-axis is compact. Then there exists a unique global solution to the system (1.1) with initial data $(u_0,B_0,\theta _0)$ that satisfies

$$\begin{aligned} \begin{aligned} \Vert \nabla ^su(t)\Vert ^2_{L^2}+\Vert \nabla ^sB(t)\Vert ^2_{L^2}+\Vert \nabla ^s\theta (t)\Vert ^2_{L^2})+\mu \int _0^t\Vert \nabla ^{s+1}u(\tau )\Vert ^2_{L^2}d\tau \lesssim C(t), \end{aligned} \end{aligned}$$

for some $C(t)>0$.

Remark 1.1

Here, we indeed assumed that $\theta _0\in L^{\infty }({\mathbb {R}}^3)$ thanks to Sobolev embedding. We also remark that as pointed out in [3], the assumption that ${{\,\mathrm{spt}\,}}\theta _0$ is away from the Z-axis can be relaxed to by assuming that $\theta$ is a constant $c_0$ near the Z-axis, by taking a change of variable ${\bar{\theta }}=\theta -c_0$ and ${\bar{\pi }}=\pi -c_0z$. We will not go into the details of this point.

Compared to the MHD system considered in [22], we have an extra transport equation of the temperature and an extra singular term $r^{-1}\partial _r\theta$ in the momentum equation in (1.6). This singular term causes difficulties in estimating the $\Vert \Omega (t)\Vert _{L^2}$ in Lemma 2.3, due to the exponential growth of the quantity $\Vert r^{-1}\theta (t)\Vert _{L^2}$ in terms of $\int _0^t\Vert r^{-1}u^r\Vert _{L^{\infty }}d\tau$. More precisely, from (1.6), one has

$$\begin{aligned} \partial _t(r^{-1}\theta ) +u\cdot \nabla (r^{-1}\theta ) +(r^{-1}\theta )(r^{-1}u^r) =0, \end{aligned}$$

which gives the estimate

$$\begin{aligned} \Vert r^{-1}\theta (t)\Vert _{L^p}\lesssim \Vert r^{-1}\theta _0\Vert _{L^p}e^{\int _0^t\Vert r^{-1}u^r\Vert _{L^{\infty }}d\tau }. \end{aligned}$$

To avoid this difficulty, we assume as in [3] that ${{\,\mathrm{spt}\,}}\theta _0$ is away from Z-axis and its projection to Z-axis is compact, and this property is maintained due to the transport equation satisfied by $\theta$ in (1.6). Therefore, not involved in much technicalities, we assume that ${{\,\mathrm{spt}\,}}\theta _0$ is away from the Z-axis to avoid the singularities of last term $r^{-1}\partial _r\theta$ in (1.6) near $r=0$.

In the next section, we will prove theorem 1.1. Throughout this paper, $A\lesssim B$ means there exists some constant $C>0$ such that $A\le CB$.

2 Proof of Theorem 1.1

2.1 Basic Estimates

From the Biot–Savart law, we have the following Lemma.

Lemma 2.1

Let u be a smooth axisymmetric divergence free vector field and $\omega =\omega ^{\phi }e_{\phi }$ be its curl, then

$$\begin{aligned} \begin{aligned} \Vert u\Vert _{L^{\infty }}\lesssim&\Vert \omega ^{\phi }\Vert ^{1/2}_{L^2}\Vert \nabla \omega ^{\phi }\Vert ^{1/2}_{L^2},\\ \Vert r^{-1}u^r\Vert _{L^{\infty }}\lesssim&\Vert \Omega \Vert ^{1/2}_{L^2}\Vert \nabla \Omega \Vert ^{1/2}_{L^2}. \end{aligned} \end{aligned}$$

This lemma was proved in [3]. See also similar estimates in [22] in integral form.

2.2 The Flow Map

First, we cite the following proposition concerning the transport equation satisfied by the temperature $\theta$, which was proved in [3].

Proposition 2.1

Let u be a smooth axisymmetric vector field and $\theta$ be a solution of the transport equation

$$\begin{aligned} \partial _t\theta +u\cdot \nabla \theta =0, \end{aligned}$$

(2.1)

with initial data $\theta (t=0)=\theta _0$.

• (a) Assume that $d({{\,\mathrm{spt}\,}}\theta _0,\{OZ\})=r_0>0$. Then one has for every $t\ge 0$ that

  $$\begin{aligned} d({{\,\mathrm{spt}\,}}\theta (t),\{OZ\})\ge r_0e^{-\int _0^t\Vert r^{-1}u^r\Vert _{L^{\infty }}d\tau }. \end{aligned}$$

• (b) Denote by $\Pi _z$ the orthogonal projector over the z-axis $\{OZ\}$, and assume that $\Pi _z({{\,\mathrm{spt}\,}}\theta _0)$ is a compact set with diameter $d_0$. Then for every $t\ge 0$, one has $\Pi _z({{\,\mathrm{spt}\,}}\theta (t))$ is a compact set with diameter d(t) such that

  $$\begin{aligned} d(t)\le d_0+2\int _0^t\Vert u(\tau )\Vert _{L^{\infty }}d\tau . \end{aligned}$$

With this proposition, one has the following Corollary, which was also proved in [3].

Corollary 2.1

Let u be a smooth axisymmetric divergence free vector field, and $\theta$ be a solution of the transport equation (2.1) with initial data $\theta _0\in L^2\cap L^{\infty }$. Assume further that

$$\begin{aligned} r_0:=d({{\,\mathrm{spt}\,}}\theta _0,\{OZ\})>0, \quad d_0:=diam(\Pi _z({{\,\mathrm{spt}\,}}\theta _0))<\infty , \end{aligned}$$

then we have

$$\begin{aligned} \int _{{\mathbb {R}}^3}r^{-2}\theta ^2(t,x)dx\le & {} r_0^{-2}\Vert \theta _0\Vert ^2_{L^2} \nonumber \\&+2\pi \Vert \theta _0\Vert ^2_{L^{\infty }}\int _0^t\Vert r^{-1}u^r(\tau )\Vert _{L^{\infty }}d\tau \left( d_0+2\int _0^t\Vert u(\tau )\Vert _{L^{\infty }}d\tau \right) . \end{aligned}$$

(2.2)

2.3 Energy Estimates

Here, we first give some $L^2$-estimates for the solutions of the Boussinesq system (1.1).

Lemma 2.2

Let $u_0,B_0\in L^2$ be divergence free, $\theta _0\in L^2\cap L^{\infty }$. Then for every smooth solution $(u,B,\theta )$, it holds that,

$$\begin{aligned} \begin{aligned}&\Vert \theta (t)\Vert _{L^p}\le \Vert \theta _0\Vert _{L^p},\ \ \ \forall p\in [1,\infty ],\\&\Vert \Pi (t)\Vert _{L^p}\le \Vert \Pi _0\Vert _{L^p},\ \ \ \forall p\in [1,\infty ],\\&\Vert u(t)\Vert ^2_{L^2}+\Vert B(t)\Vert ^2_{L^2}+\int _0^t\Vert \nabla u(\tau )\Vert ^2_{L^2}d\tau \lesssim (1+t)e^t. \end{aligned} \end{aligned}$$

(2.3)

Proof

The first two inequalities are standard. Since u is divergence free, by taking $L^2$-estimates for the first two equations, one has

$$\begin{aligned} \frac{1}{2}\frac{d}{dt}\Vert u(t),B(t)\Vert ^2_{L^2}+\mu \Vert \nabla u(\tau )\Vert ^2_{L^2} \le \Vert u(t)\Vert _{L^2}\Vert \theta (t)\Vert _{L^2}\lesssim 1+\Vert u(t)\Vert ^2_{L^2}, \end{aligned}$$

(2.4)

which implies immediately that

$$\begin{aligned} \Vert u(t),B(t)\Vert ^2_{L^2}+\mu \int _0^t\Vert \nabla u(\tau )\Vert ^2_{L^2}d\tau \le e^t(\Vert u_0,B_0\Vert ^2_{L^2}+t)\lesssim (1+t)e^t, \end{aligned}$$

(2.5)

thanks to the Gronwall inequality. $\square$

Next, we give some estimates for $\omega ^{\phi }$ and $\Omega =\omega ^{\phi }/r$.

Lemma 2.3

Suppose that $(u,B,\theta )$ is a smooth solution of the Boussinesq–MHD system (1.6) with initial data $(u_0,B_0,\theta _0)\in H^2$, which satisfies the conditions of Theorem 1.1. Then there holds,

$$\begin{aligned} \Vert \omega ^{\phi }(t)\Vert ^2_{L^2}+\int _0^t\Vert \nabla \omega ^{\phi }(\tau )\Vert _{L^2}^2+\Vert r^{-1}\omega ^{\phi }\Vert ^2_{L^2}d\tau \lesssim C(t), \end{aligned}$$

and

$$\begin{aligned} \Vert \Omega (t)\Vert ^2_{L^2}+\int _0^t\Vert \nabla \Omega (\tau )\Vert _{L^2}^2d\tau +4\pi \int _0^t\int _{{\mathbb {R}}}|\Omega (\tau ,0,z)|^2dz d\tau \lesssim C(t), \end{aligned}$$

for some constant C(t).

Proof

(i). Recall the equation (1.5) for $\omega ^{\phi }$. Take the $L^2$-inner product with $\omega ^{\phi }$ to obtain

$$\begin{aligned} \begin{aligned} \frac{1}{2}\frac{d}{dt}\Vert \omega ^{\phi }(t)\Vert ^2_{L^2}=&\int \omega ^{\phi }\left( \partial _{rr}+\frac{1}{r}\partial _r+\partial _{zz}-\frac{1}{r^2}\right) \omega ^{\phi }dx -\int \omega ^{\phi } (u\cdot \nabla \omega ^{\phi }) dx\\&+\int \frac{u^r}{r}|\omega ^{\phi }|^2dx -\int 2r\Pi \partial _z\Pi \omega ^{\phi }dx -\int \omega ^{\phi }{\partial _r\theta } dx. \end{aligned} \end{aligned}$$

The first integral on the RHS equals to

$$\begin{aligned} \begin{aligned} -\int \omega ^{\phi }\left( \partial _{rr}+\frac{1}{r}\partial _r+\partial _{zz}-\frac{1}{r^2}\right) \omega ^{\phi }dx&= \Vert \partial _r\omega ^{\phi }\Vert ^2_{L^2}+\Vert \partial _z\omega ^{\phi }\Vert ^2_{L^2}+\Vert r^{-1}\omega ^{\phi }\Vert ^2_{L^2}\\&= \Vert \nabla \omega ^{\phi }\Vert ^2_{L^2}+\Vert r^{-1}\omega ^{\phi }\Vert ^2_{L^2}. \end{aligned} \end{aligned}$$

The second integral vanishes, and the third and fourth terms can be estimated as

$$\begin{aligned} \begin{aligned} \left| \int \frac{u^r}{r}|\omega ^{\phi }|^2dx\right| \le&\Vert u^r\Vert _{L^{6}}\Vert \omega ^{\phi }\Vert _{L^3}\Vert \Omega \Vert _{L^2}\le C\Vert u^r\Vert _{L^{6}}\Vert \omega ^{\phi }\Vert ^{1/2}_{L^2}\Vert \nabla \omega ^{\phi }\Vert ^{1/2}_{L^2}\Vert \Omega \Vert _{L^2}\\ \le&\frac{1}{4}\Vert \nabla \omega ^{\phi }\Vert ^2_{L^2} +C\Vert \omega ^{\phi }\Vert ^2_{L^2} +C\Vert \nabla u\Vert ^2_{L^{2}}\Vert \Omega \Vert ^2_{L^2}, \end{aligned} \end{aligned}$$

and

$$\begin{aligned} \begin{aligned} \left| \int 2r\Pi \partial _z\Pi \omega ^{\phi }dx\right|&= \left| \int B^{\phi }\Pi \partial _z\omega ^{\phi }dx\right| \le \Vert \Pi \Vert _{L^{\infty }}\Vert B^{\phi }\Vert _{L^2}\Vert \partial _z\omega ^{\phi }\Vert _{L^2}\\ \le \,&\Vert \Pi _0\Vert _{L^{\infty }}\Vert B^{\phi }\Vert _{L^2}\Vert \partial _z\omega ^{\phi }\Vert _{L^2}\le C(1+t)e^t+\frac{1}{4}\Vert \partial _z\omega ^{\phi }\Vert ^2_{L^2}, \end{aligned} \end{aligned}$$

where we have used Lemma 2.2. For the last integral, it follows from integration by parts that

$$\begin{aligned} \begin{aligned} \left| \int \omega ^{\phi }{\partial _r\theta } dx\right|&= \left| 2\pi \int \omega ^{\phi }{\partial _r\theta } rdrdz\right| \le \left| 2\pi \int \theta \Omega rdrdz\right| + \left| 2\pi \int \theta \partial _r\omega ^{\phi } rdrdz\right| \\&\le \Vert \theta \Vert _{L^2}\Vert \Omega \Vert _{L^2} +\Vert \theta \Vert ^2_{L^2}+\frac{1}{4}\Vert \partial _r\omega ^{\phi }\Vert ^2_{L^2}\\&\le C\Vert \Omega \Vert ^2_{L^2} +C\Vert \theta _0\Vert ^2_{L^2}+\frac{1}{4}\Vert \partial _r\omega ^{\phi }\Vert ^2_{L^2}. \end{aligned} \end{aligned}$$

Therefore, we have

$$\begin{aligned} \begin{aligned} \frac{d}{dt}\Vert \omega ^{\phi }(t)\Vert ^2_{L^2}&+ \Vert \nabla \omega ^{\phi }\Vert ^2_{L^2}+\Vert r^{-1}\omega ^{\phi }\Vert ^2_{L^2} \lesssim (1+t)e^t + \Vert \omega ^{\phi }\Vert ^2_{L^2} +(1+\Vert \nabla u\Vert ^2_{L^{2}})\Vert \Omega \Vert ^2_{L^2}. \end{aligned} \end{aligned}$$

Integrating over [0, t], one has

$$\begin{aligned} \begin{aligned}&\Vert \omega ^{\phi }(t)\Vert ^2_{L^2} + \int _0^t\Vert \nabla \omega ^{\phi }\Vert ^2_{L^2}+\Vert r^{-1}\omega ^{\phi }\Vert ^2_{L^2}d\tau \\&\quad \lesssim te^t + \int _0^t\Vert \omega ^{\phi }(\tau )\Vert ^2_{L^2}d\tau + \int _0^t(1+\Vert \nabla u(\tau )\Vert ^2_{L^{2}})\Vert \Omega (\tau )\Vert ^2_{L^2}d\tau . \end{aligned} \end{aligned}$$

(2.6)

(ii). On the other hand, from the transport equation for $\Pi$ in (1.6), we have for any $\Pi _0\in L^2\cap L^{\infty }$

$$\begin{aligned} \Vert \Pi (t)\Vert _{L^p}\le \Vert \Pi _0\Vert _{L^p}, \quad \forall \, p\in [2,\infty ]. \end{aligned}$$

(2.7)

Note also that

$$\begin{aligned} |\nabla B|^2=|(e_r\partial _r+\frac{1}{r}e_{\phi }\partial _{\phi }+e_z\partial _z)(B^{\phi }e_{\phi })|^2=|\nabla B^{\phi }|^2+|\Pi |^2. \end{aligned}$$

In particular, one has

$$\begin{aligned} \Vert \Pi (t)\Vert _{L^2}\le \Vert \Pi _0\Vert _{L^2}\lesssim \Vert B_0\Vert _{H^1}, \quad \ \Vert \Pi (t)\Vert _{L^4}\le \Vert \Pi _0\Vert _{L^4} \le \Vert \nabla B_0\Vert _{L^4}\lesssim \Vert B_0\Vert _{H^2}. \end{aligned}$$

(iii). Taking $L^2$-inner product of the equation for $\Omega$ in (1.6), we obtain

$$\begin{aligned} \begin{aligned} \frac{1}{2}\frac{d}{dt}\Vert \Omega (t)\Vert ^2_{L^2}=\int \Omega (\Delta +\frac{2}{r}\partial _r)\Omega dx -\int \Omega (u\cdot \nabla \Omega ) dx -\int \Omega \partial _z\Pi ^2dx -\int \Omega \frac{\partial _r\theta }{r} dx. \end{aligned} \end{aligned}$$

For the terms on the RHS, it follows from integration by parts that

$$\begin{aligned} \begin{aligned} -\int \Omega (\Delta +\frac{2}{r}\partial _r)\Omega dx&=\Vert \nabla \Omega \Vert ^2_{L^2}+2\pi \int _{{\mathbb {R}}}|\Omega (t,0,z)|^2dz\\&= \Vert \partial _r\Omega \Vert ^2_{L^2}+\Vert \partial _z\Omega \Vert ^2_{L^2}+2\pi \int _{{\mathbb {R}}}|\Omega (t,0,z)|^2dz, \\ \int \Omega (u\cdot \nabla \Omega ) dx&=0,\\ \left| \int \Omega \partial _z\Pi ^2dx\right|&\le \Vert \Pi \Vert ^2_{L^4}\Vert \partial _z\Omega \Vert _{L^2}\le \frac{1}{2}\Vert \Pi \Vert ^4_{L^4}+\frac{1}{2}\Vert \partial _z\Omega \Vert ^2_{L^2},\\ \left| \int \Omega \frac{\partial _r\theta }{r} dx\right|&= \left| 2\pi \int \Omega {\partial _r\theta }drdz\right| = \left| 2\pi \int \partial _r\Omega \frac{\theta }{r}rdrdz\right| \le \frac{1}{2}\Vert \theta /r\Vert ^2_{L^2}+\frac{1}{2}\Vert \partial _r\Omega \Vert ^2_{L^2}. \end{aligned} \end{aligned}$$

Therefore, we have

$$\begin{aligned} \begin{aligned} \frac{d}{dt}\Vert \Omega (t)\Vert ^2_{L^2}+\Vert \nabla \Omega \Vert ^2_{L^2} +4\pi \int _{{\mathbb {R}}}|\Omega (t,0,z)|^2dz \le \Vert \theta /r\Vert ^2_{L^2}+ \Vert B_0\Vert ^4_{H^2}. \end{aligned} \end{aligned}$$

Integrating over [0, t], one then has

$$\begin{aligned} \begin{aligned}&\Vert \Omega (t)\Vert ^2_{L^2}+\int _0^t\Vert \nabla \Omega (\tau )\Vert ^2_{L^2}d\tau +4\pi \int _0^t\int _{{\mathbb {R}}}|\Omega (\tau ,0,z)|^2dzd\tau \\&\quad \le \Vert \Omega (0)\Vert ^2+\int _0^t\Vert \theta /r(\tau )\Vert ^2_{L^2}d\tau + t\Vert B_0\Vert ^4_{H^2}\\&\quad \lesssim 1+t+t\int _0^t\Vert r^{-1}u^r(\tau )\Vert _{L^{\infty }}d\tau +t\int _0^t\Vert r^{-1}u^r(\tau )\Vert _{L^{\infty }}d\tau \int _0^t\Vert u(\tau )\Vert _{L^{\infty }}d\tau \\&\quad \lesssim 1+t+t\int _0^t\Vert r^{-1}u^r(\tau )\Vert _{L^{\infty }}d\tau + t^{3/2}\int _0^t\Vert r^{-1}u^r(\tau )\Vert _{L^{\infty }}d\tau \left( \int _0^t\Vert u(\tau )\Vert ^2_{L^{\infty }}d\tau \right) ^{1/2}\\&\quad \lesssim 1+t+t\int _0^t\Vert r^{-1}u^r(\tau )\Vert _{L^{\infty }}d\tau +t^3\left( \int _0^t\Vert r^{-1}u^r(\tau )\Vert _{L^{\infty }}d\tau \right) ^2 +\int _0^t\Vert u(\tau )\Vert ^2_{L^{\infty }}d\tau , \end{aligned} \end{aligned}$$

(2.8)

where in the above inequality, we have used Corollary 2.1. For the first two integrals, we use Hölder and Young’s inequalities and Lemma 2.1 to obtain

$$\begin{aligned} \begin{aligned} t^{\alpha }\int _0^t\Vert r^{-1}u^r(\tau )\Vert _{L^{\infty }}d\tau&\le t^{\alpha }\int _0^t\Vert \Omega (\tau )\Vert _{L^2}^{1/2}\Vert \nabla \Omega (\tau )\Vert ^{1/2}_{L^2}d\tau , \\&\le t^{{\alpha }+\frac{1}{2}}\left( \int _0^t \Vert \Omega (\tau )\Vert ^2_{L^2}d\tau \right) ^{1/4}\left( \int _0^t\Vert \nabla \Omega (\tau )\Vert ^2_{L^2}d\tau \right) ^{1/4}. \end{aligned} \end{aligned}$$

Therefore, when $\alpha =1$, we obtain

$$\begin{aligned} \begin{aligned} t\int _0^t\Vert r^{-1}u^r(\tau )\Vert _{L^{\infty }}d\tau&\le t^{3/2}\left( \int _0^t \Vert \Omega (\tau )\Vert ^2_{L^2}d\tau \right) ^{1/4}\left( \int _0^t\Vert \nabla \Omega (\tau )\Vert ^2_{L^2}d\tau \right) ^{1/4}\\&\lesssim t^{2} \left( \int _0^t \Vert \Omega (\tau )\Vert ^2_{L^2}d\tau \right) ^{1/3} +\frac{1}{4}\int _0^t\Vert \nabla \Omega (\tau )\Vert ^2_{L^2}d\tau \\&\lesssim 1+ t^{6} \int _0^t \Vert \Omega (\tau )\Vert ^2_{L^2}d\tau +\frac{1}{4}\int _0^t\Vert \nabla \Omega (\tau )\Vert ^2_{L^2}d\tau , \end{aligned} \end{aligned}$$

and when $\alpha =3/2$, we obtain

$$\begin{aligned} \begin{aligned} t^3\left( \int _0^t\Vert r^{-1}u^r(\tau )\Vert _{L^{\infty }}d\tau \right) ^2&\le t^{4}\left( \int _0^t \Vert \Omega (\tau )\Vert ^2_{L^2}d\tau \right) ^{1/2}\left( \int _0^t\Vert \nabla \Omega (\tau )\Vert ^2_{L^2}d\tau \right) ^{1/2}\\&\lesssim t^{8} \int _0^t \Vert \Omega (\tau )\Vert ^2_{L^2}d\tau +\frac{1}{4}\int _0^t\Vert \nabla \Omega (\tau )\Vert ^2_{L^2}d\tau . \end{aligned} \end{aligned}$$

Thanks again to Lemma 2.1, the last integral in (2.8) can be estimated as

$$\begin{aligned} \begin{aligned} \int _0^t\Vert u(\tau )\Vert ^2_{L^{\infty }}d\tau \le&\int _0^t\Vert \omega ^{\phi }(\tau )\Vert _{L^2}\Vert \nabla \omega ^{\phi }(\tau )\Vert _{L^2}d\tau \\ \le&\frac{1}{2}\int _0^t\Vert \omega ^{\phi }(\tau )\Vert ^2_{L^2}d\tau +\frac{1}{2}\int _0^t\Vert \nabla \omega ^{\phi }(\tau )\Vert ^2_{L^2}d\tau . \end{aligned} \end{aligned}$$

Therefore, we arrive at the following inequality

$$\begin{aligned} \begin{aligned}&\Vert \Omega (t)\Vert ^2_{L^2} +\int _0^t\Vert \nabla \Omega (\tau )\Vert ^2_{L^2}d\tau +4\pi \int _0^t\int _{{\mathbb {R}}}|\Omega (\tau ,0,z)|^2dz d\tau \\&\quad \le C(1+t^8)\left( 1+\int _0^t\Vert \Omega (\tau )\Vert ^2_{L^2}d\tau \right) +C\int _0^t\Vert \omega ^{\phi }(\tau )\Vert ^2_{L^2}d\tau +\frac{1}{2}\int _0^t\Vert \nabla \omega ^{\phi }(\tau )\Vert ^2_{L^2}d\tau . \end{aligned} \end{aligned}$$

(2.9)

(iv). Combining the inequalities (2.6) and (2.9), one has

$$\begin{aligned} \begin{aligned}&\Vert \omega ^{\phi }(t)\Vert ^2_{L^2} +\Vert \Omega (t)\Vert ^2_{L^2} + \int _0^t\Vert \nabla \omega ^{\phi }\Vert ^2_{L^2}+\Vert r^{-1}\omega ^{\phi }\Vert ^2_{L^2}d\tau \\&\quad \quad + \int _0^t\Vert \nabla \Omega (\tau )\Vert ^2_{L^2}d\tau +4\pi \int _0^t\int _{{\mathbb {R}}}|\Omega (\tau ,0,z)|^2dz d\tau \\&\quad \lesssim (1+t^8)\left( 1+\int _0^t(1+\Vert \nabla u(\tau )\Vert ^2_{L^{2}})(\Vert \omega ^{\phi }(\tau )\Vert ^2_{L^2}+\Vert \Omega (\tau )\Vert ^2_{L^2})d\tau \right) . \end{aligned} \end{aligned}$$

(2.10)

Recalling the third inequality in Lemma 2.2, we have by integral Gronwall inequality that

$$\begin{aligned} \begin{aligned} \Vert \omega ^{\phi }(t)\Vert ^2_{L^2}+\Vert \Omega (t)\Vert ^2_{L^2} \lesssim C(t). \end{aligned} \end{aligned}$$

It follows from (2.10) that

$$\begin{aligned} \begin{aligned}&\int _0^t\Vert \partial _r\omega ^{\phi }\Vert ^2_{L^2}+\Vert \partial _z\omega ^{\phi }\Vert ^2_{L^2}+\Vert r^{-1}\omega ^{\phi }\Vert ^2_{L^2}d\tau \\&\quad + \int _0^t\Vert \nabla \Omega (\tau )\Vert ^2_{L^2}d\tau +4\pi \int _0^t\int _{{\mathbb {R}}}|\Omega (\tau ,0,z)|^2dz d\tau \lesssim C(t). \end{aligned} \end{aligned}$$

The proof is complete. $\square$

By using the Biot–Savart law [24], we have the following two corollaries.

Corollary 2.2

Under the assumption of Lemma 2.3, there exists some constant C(t) such that

$$\begin{aligned} \Vert u(t)\Vert ^2_{H^1}+\int _0^t\Vert u(\tau )\Vert ^2_{H^2}d\tau \lesssim C(t). \end{aligned}$$

Corollary 2.3

Under the assumption of Lemma 2.3, there exists some constant C(t) such that

$$\begin{aligned} \int _0^t\Vert r^{-1}u^r(\tau )\Vert _{L^{\infty }}d\tau \lesssim C(t). \end{aligned}$$

Proof

By combing the estimates in Lemma 2.3 and Lemma 2.1,

$$\begin{aligned} \begin{aligned} \int _0^t\Vert r^{-1}u^r(\tau )\Vert _{L^{\infty }}d\tau \le C\sup _{0\le \tau \le t}\Vert \Omega (\tau ,\cdot )\Vert ^{1/2}_{L^2}\int _0^t\Vert \nabla \Omega (\tau )\Vert ^{1/2}_{L^2}d\tau \lesssim C(t). \end{aligned} \end{aligned}$$

$\square$

Lemma 2.4

Suppose that $(u,B,\theta )$ is a smooth solution of the Boussinesq–MHD system (1.6) with initial data $(u_0,B_0,\theta _0)\in H^2$, which satisfies the conditions of Theorem 1.1. Then there holds

$$\begin{aligned} \Vert B^{\phi }(t)\Vert _{L^{p}}\lesssim C(t), \quad \ \forall \, p\in [1,+\infty ]. \end{aligned}$$

Proof

By multiplying the third equation in (1.4) with $p|B^{\phi }|^{p-2}B^{\phi }$ and integrating over ${\mathbb {R}}^3$, one has

$$\begin{aligned} \begin{aligned} \frac{d}{dt}\Vert B^{\phi }(t)\Vert _{L^{p}}\le \Vert r^{-1}u^{r}\Vert _{L^{\infty }}\Vert B^{\phi }(t)\Vert _{L^{p}},\ \ \ \forall \, p\in [1,\infty ]. \end{aligned} \end{aligned}$$

By Gronwall inequality, one has

$$\begin{aligned} \begin{aligned} \Vert B^{\phi }(t)\Vert _{L^{p}}\le \Vert B^{\phi }_0\Vert _{L^{p}}e^{\int _0^t\Vert r^{-1}u^{r}\Vert _{L^{\infty }}d\tau }\lesssim C(t), \end{aligned} \end{aligned}$$

independent of $p>0$. Letting $p\rightarrow \infty$, then we finishe the proof. $\square$

Lemma 2.5

Suppose that $(u,B,\theta )$ is a smooth solution of the Boussinesq–MHD system (1.6) with initial data $(u_0,B_0,\theta _0)\in H^2$, which satisfies the conditions of Theorem 1.1. Then there exists some constant C(t) such that

$$\begin{aligned} \Vert \omega ^{\phi }(t)\Vert _{L^4}^4+\int _0^t\Vert \omega ^{\phi }(\tau )\Vert _{L^{12}}^4d\tau \lesssim C(t). \end{aligned}$$

Proof

Now, we consider the $L^4$-estimate of the vorticity $\omega ^{\phi }$. For this, we multiply the equation (1.5) with $|\omega ^{\phi }|^2\omega ^{\phi }$ and then integrating over ${\mathbb {R}}^3$ to obtain

$$\begin{aligned} \begin{aligned} \frac{1}{4}\frac{d}{dt}\Vert \omega ^{\phi }\Vert ^4_{L^4}&= \int |\omega ^{\phi }|^2\omega ^{\phi } \left( \partial _{rr}+\frac{1}{r}\partial _r+\partial _{zz}-\frac{1}{r^2}\right) \omega ^{\phi } - \int (u\cdot \nabla \omega ^{\phi })|\omega ^{\phi }|^2\omega ^{\phi }\\&\quad + \int \frac{u^r}{r}\omega ^{\phi }|\omega ^{\phi }|^2\omega ^{\phi }-2\int |\omega ^{\phi }|^2\omega ^{\phi }r\Pi \partial _z\Pi -\int |\omega ^{\phi }|^2\omega ^{\phi } \partial _r\theta . \end{aligned} \end{aligned}$$

(2.11)

For the first integral, we can show by integration by parts that

$$\begin{aligned} \begin{aligned} \int |\omega ^{\phi }|^2\omega ^{\phi } \left( \partial _{rr}+\frac{1}{r}\partial _r+\partial _{zz}-\frac{1}{r^2}\right) \omega ^{\phi }dx=-\int \left( \frac{3}{4}|\nabla |\omega ^{\phi }|^2|^2+\frac{|\omega ^{\phi }|^4}{r^2}\right) dx. \end{aligned} \end{aligned}$$

By Hölder and Young’s inequality, (2.7), Lemma 2.3 and 2.4

$$\begin{aligned} \begin{aligned} \left| 2\int |\omega ^{\phi }|^2\omega ^{\phi }r\Pi \partial _z\Pi dx\right| \le&C\Vert \Pi \Vert _{L^{\infty }}\Vert B^{\phi }\Vert _{L^{\infty }}\Vert \partial _z|\omega ^{\phi }|^2\Vert _{L^2}\Vert \omega ^{\phi }\Vert _{L^2}\\ \le&\frac{1}{4}\Vert \partial _z|\omega ^{\phi }|^2\Vert ^2_{L^2} +C(t). \end{aligned} \end{aligned}$$

From integration by parts, it holds that

$$\begin{aligned} \begin{aligned} -\int |\omega ^{\phi }|^2\omega ^{\phi } \partial _r\theta dx&= 6\pi \int \theta |\omega ^{\phi }|^2\partial _r\omega ^{\phi }rdrdz+2\pi \int \theta |\omega ^{\phi }|^2\frac{\omega ^{\phi }}{r}rdrdz\\&= \frac{3}{2}\int \theta \omega ^{\phi }\partial _r(|\omega ^{\phi }|^2)2\pi rdrdz+\int \theta |\omega ^{\phi }|^2\frac{\omega ^{\phi }}{r}2\pi rdrdz, \end{aligned} \end{aligned}$$

and hence

$$\begin{aligned} \begin{aligned} \left| \int |\omega ^{\phi }|^2\omega ^{\phi } \partial _r\theta dx\right| \le \,&\frac{1}{4}\Vert \partial _r|\omega ^{\phi }|^2\Vert ^2_{L^2}+2\Vert \theta \Vert ^2_{L^{\infty }}\Vert \omega ^{\phi }\Vert ^2_{L^2} +\Vert \omega ^{\phi }\Vert ^4_{L^4}+ \Vert \theta \Vert ^2_{L^{\infty }}\Vert r^{-1}\omega ^{\phi }\Vert ^2_{L^2}\\ \le \,&\frac{1}{4}\Vert \partial _r|\omega ^{\phi }|^2\Vert ^2_{L^2} +\Vert \omega ^{\phi }\Vert ^4_{L^4}+ C(t). \end{aligned} \end{aligned}$$

Noting that

$$\begin{aligned} |\partial _r(|\omega ^{\phi }|^2)|^2+|\partial _z(|\omega ^{\phi }|^2)|^2\le |\nabla (|\omega ^{\phi }|^2)|^2 \end{aligned}$$

by direct computation and that the second integral on the right side of (2.11) vanishes by integration by parts, we obtain

$$\begin{aligned} \begin{aligned} \frac{d}{dt}\Vert \omega ^{\phi }\Vert ^4_{L^4}&+\int \left( |\nabla |\omega ^{\phi }|^2|^2+\frac{|\omega ^{\phi }|^4}{r^2}\right) dx \lesssim (1+\Vert r^{-1}{u^r}\Vert _{L^{\infty }})\Vert \omega ^{\phi }\Vert ^4_{L^4}+ C(t). \end{aligned} \end{aligned}$$

Using Gronwall inequality, one has

$$\begin{aligned} \begin{aligned} \Vert \omega ^{\phi }(t)\Vert ^4_{L^4}&+\int _0^t\int \left( |\nabla |\omega ^{\phi }(\tau )|^2|^2+\frac{|\omega ^{\phi }(\tau )|^4}{r^2}\right) dxd\tau \\ \lesssim&e^{\int _0^t\left( 1+\Vert r^{-1}{u^r}\Vert _{L^{\infty }}\right) d\tau }\left( \Vert \omega ^{\phi }_0\Vert ^4_{L^4}+\int _0^tC(\tau ) d\tau \right) \lesssim C(t), \end{aligned} \end{aligned}$$

thanks to Corollary 2.3. $\square$

Lemma 2.6

Under the same conditions of Lemma 2.3, there exists some constant C(t)

$$\begin{aligned} \int _0^t\Vert \nabla u(\tau )\Vert _{L^{\infty }}d\tau \lesssim C(t), \quad \Vert \nabla B(t)\Vert _{L^{\infty }} \lesssim C(t). \end{aligned}$$

Proof

Recalling Lemma 2.5, we have by interpolation that

$$\begin{aligned} \begin{aligned} \Vert u\Vert _{L^{\infty }([0,t];L^{\infty }({\mathbb {R}}^3))} \lesssim \Vert u\Vert _{L^{\infty }([0,t];L^{2}({\mathbb {R}}^3))}+ \Vert \omega ^{\phi }\Vert _{L^{\infty }([0,t];L^{4}({\mathbb {R}}^3))} \lesssim C(t), \end{aligned} \end{aligned}$$

and hence

$$\begin{aligned} \Vert \omega \times u\Vert _{L^4([0,t];L^{12}({\mathbb {R}}^3))} \lesssim C(t). \end{aligned}$$

Rewriting the equation for $\omega =\nabla \times u$, we have

$$\begin{aligned} \begin{aligned} \partial _t\omega +\nabla \times (\omega \times u) = \mu \Delta \omega -\partial _z(\Pi B^{\phi }e_{\phi }) +\nabla \times (\theta e_z). \end{aligned} \end{aligned}$$

Standard estimates show that [34]

$$\begin{aligned} \begin{aligned} \Vert \nabla \omega \Vert _{L^4([0,t];L^{12}({\mathbb {R}}^3))} \lesssim C(t). \end{aligned} \end{aligned}$$

Sobolev embedding then implies that

$$\begin{aligned} \begin{aligned} \Vert \nabla u\Vert _{L^4([0,t];L^{\infty }({\mathbb {R}}^3))} \lesssim C(t). \end{aligned} \end{aligned}$$

(2.12)

Applying $\nabla$ to the third equation in (1.4), we have

$$\begin{aligned} \begin{aligned} \partial _t\nabla B^{\phi } +u\cdot \nabla \nabla B^{\phi } =-\nabla u\cdot \nabla B^{\phi }+\frac{u^r}{r}\nabla B^{\phi } +\nabla u^r\Pi -\frac{u^r}{r}\Pi e_r. \end{aligned} \end{aligned}$$

Multiplying the equation with $|\nabla B^{\phi }|^{p-2}\nabla B^{\phi }$, and then integrating over ${\mathbb {R}}^3$, one has

$$\begin{aligned} \begin{aligned}&\frac{1}{p}\frac{d}{dt}\Vert \nabla B^{\phi }\Vert ^p_{L^p} + \int u\cdot \nabla \nabla B^{\phi }|\nabla B^{\phi }|^{p-2}\nabla B^{\phi } dx\le \Vert \nabla u\Vert _{L^{\infty }}\Vert \nabla B^{\phi }\Vert ^p_{L^p}\\&\quad +\Vert r^{-1}{u^r}\Vert _{L^{\infty }}\Vert \nabla B^{\phi }\Vert ^p_{L^p} +\left( \Vert \nabla u^r\Vert _{L^{\infty }} +\Vert {r^{-1}}{u^r}\Vert _{L^{\infty }}\right) \Vert \Pi \Vert _{L^{p}}\Vert \nabla B^{\phi }\Vert ^{p-1}_{L^p}. \end{aligned} \end{aligned}$$

Since u is divergence free, from integration by parts, one has

$$\begin{aligned} \begin{aligned} \frac{d}{dt}\Vert \nabla B^{\phi }\Vert _{L^p} \le \left( \Vert \nabla u\Vert _{L^{\infty }} +\Vert r^{-1}{u^r}\Vert _{L^{\infty }}\right) \Vert \nabla B^{\phi }\Vert _{L^p} +\left( \Vert \nabla u^r\Vert _{L^{\infty }} +\Vert {r^{-1}}{u^r}\Vert _{L^{\infty }}\right) \Vert \Pi \Vert _{L^{p}}. \end{aligned} \end{aligned}$$

Using Gronwall inequality then gives that

$$\begin{aligned} \begin{aligned} \Vert \nabla B^{\phi }(t)\Vert _{L^p}&\le e^{\int _0^t\big (\Vert \nabla u\Vert _{L^{\infty }} +\Vert r^{-1}{u^r}\Vert _{L^{\infty }}\big )d\tau }\\&\quad \times \left( \Vert \nabla B^{\phi }_0\Vert _{L^p}+ \int _0^t\big (\Vert \nabla u^r\Vert _{L^{\infty }} +\Vert {r^{-1}}{u^r}\Vert _{L^{\infty }}\big )d\tau \right) \lesssim C(t). \end{aligned} \end{aligned}$$

where we have used (2.7), (2.12) and Corollary 2.3. Letting $p\rightarrow \infty$ then implies the result. $\square$

2.4 Proof of Theorem 1.1

Applying the $H^2$ estimate for the system (1.1), we get

$$\begin{aligned} \begin{aligned}&\frac{1}{2}\frac{d}{dt} (\Vert \nabla ^2u(t)\Vert ^2_{L^2}+\Vert \nabla ^2B(t)\Vert ^2_{L^2}+\Vert \nabla ^2\theta (t)\Vert ^2_{L^2})+\mu \Vert \nabla ^3u(t)\Vert ^2_{L^2}\\&\quad = \int \nabla ^2u\nabla ^2(B\cdot \nabla B-u\cdot \nabla u)dx + \int \nabla ^2\theta \nabla ^2u^zdx \\&\quad \quad + \int \nabla ^2B\nabla ^2(B\cdot \nabla u-u\cdot \nabla B) dx + \int \nabla ^2\theta \nabla ^2(u\cdot \nabla \theta ) dx. \end{aligned} \end{aligned}$$

(2.13)

Note that

$$\begin{aligned} \begin{aligned}&\int \nabla ^2u\nabla ^2(B\cdot \nabla B) dx + \int \nabla ^2B\nabla ^2(B\cdot \nabla u) dx\\&\quad =\int \nabla ^2u\cdot [\nabla ^2,B\cdot ]\nabla B + \nabla ^2B\cdot [\nabla ^2,B\cdot ]\nabla u dx +\int \nabla ^2uB\cdot \nabla \nabla ^2B +\nabla ^2BB\cdot \nabla \nabla ^2udx\\&\quad =\int \nabla ^2u\cdot [\nabla ^2,B\cdot ]\nabla B + \nabla ^2B\cdot [\nabla ^2,B\cdot ]\nabla u dx \\&\quad \lesssim (\Vert \nabla u\Vert _{L^{\infty }}+\Vert \nabla B\Vert _{L^{\infty }}) \Vert \nabla ^2u\Vert _{L^2}\Vert \nabla ^2B\Vert _{L^2}, \end{aligned} \end{aligned}$$

where $[\cdot ,\cdot ]$ denotes the commutation and the last integral in the second line cancels thanks to integration by parts and divergence free condition of B. Other terms in (2.13) can be treated similarly, thanks to integration by parts and the divergence free condition of u and B, leading to the estimates

$$\begin{aligned} \begin{aligned}&\frac{1}{2}\frac{d}{dt} (\Vert \nabla ^2u(t)\Vert ^2_{L^2}+\Vert \nabla ^2B(t)\Vert ^2_{L^2}+\Vert \nabla ^2\theta (t)\Vert ^2_{L^2})+\mu \Vert \nabla ^3u(t)\Vert ^2_{L^2}\\&\quad \lesssim (\Vert \nabla u(t)\Vert _{L^{\infty }}+\Vert \nabla B(t)\Vert _{L^{\infty }})(\Vert \nabla ^2u(t)\Vert ^2_{L^{2}}+\Vert \nabla ^2B(t)\Vert ^2_{L^{2}})\\&\quad \quad + (1+\Vert \nabla u(t)\Vert _{L^{\infty }}+\Vert \nabla \theta (t)\Vert _{L^{\infty }})(\Vert \nabla ^2u(t)\Vert ^2_{L^{2}}+\Vert \nabla ^2\theta (t)\Vert ^2_{L^{2}}). \end{aligned} \end{aligned}$$

Gronwall inequality then implies that

$$\begin{aligned} \begin{aligned}&\Vert \nabla ^2u(t)\Vert ^2_{L^2}+\Vert \nabla ^2B(t)\Vert ^2_{L^2}+\Vert \nabla ^2\theta (t)\Vert ^2_{L^2}+\mu \int _0^t\Vert \nabla ^3u(\tau )\Vert ^2_{L^2}d\tau \lesssim C(t). \end{aligned} \end{aligned}$$

Similarly, one can get $H^s$ estimates as follows

$$\begin{aligned} \begin{aligned} \Vert \nabla ^su(t)\Vert ^2_{L^2}+\Vert \nabla ^sB(t)\Vert ^2_{L^2}+\Vert \nabla ^s\theta (t)\Vert ^2_{L^2})+\mu \int _0^t\Vert \nabla ^{s+1}u(\tau )\Vert ^2_{L^2}d\tau \lesssim C(t). \end{aligned} \end{aligned}$$

This completes the proof of Theorem 1.1.