Global well-posedness of 3D magneto-micropolar fluid equations with mixed partial viscosity near an equilibrium

Abstract

In this paper, we investigate the initial value problem for the 3D magneto-micropolar fluid equations with mixed partial viscosity. The main purpose of this paper is to establish global well-posedness of classical small solutions. More precisely, we prove that the global stability of perturbations near the steady solution is given by a background magnetic field. The proof is mainly based on the energy estimate and bootstrapping argument.

1 Introduction

In this paper, we investigate three-dimensional (3D) magneto-micropolar fluid equations

$$\begin{aligned} \left\{ \begin{array}{l} {\displaystyle \partial _{t}u-(\mu +\chi )\Delta u+u\cdot \nabla u-B\cdot \nabla B+\nabla \left( p+\frac{1}{2}|B|^2\right) -2\chi \nabla \times v=0,}\\ {\displaystyle \partial _{t}v-\gamma \Delta v-\kappa \nabla \nabla \cdot v+4\chi v+u\cdot \nabla v-2\chi \nabla \times u=0,}\\ {\displaystyle \partial _{t}B-\nu \Delta B+u\cdot \nabla B-B\cdot \nabla u=0,}\\ {\displaystyle \nabla \cdot u=0,\nabla \cdot B=0,}\\ \end{array} \right. \end{aligned}$$

(1.1)

where $u=u(x,t)$, $v=v(x,t)$, $B=B(x,t)\in \mathbb {R}^3$, $p=p(x,t)\in \mathbb {R}$ are the velocity of the fluid, the micro-rotational velocity, magnetic field and hydrostatic pressure field, respectively, and the $\mu$, $\chi$, $\gamma$ and $\kappa$ represent the kinematic viscosity, vortex viscosity and spin viscosity, respectively, and $\frac{1}{\nu }$ is the magnetic Reynold.

The incompressible magneto-micropolar fluid equations have made analytic studies a great challenge but offer new opportunities due to their distinctive mathematical features. The local existence and uniqueness of strong solution were proved by Galerkin method in [21]. Later, the global existence of strong solution with the small initial data was established in [18]. We refer to Rojas-Medar and Boldrini [22] for the existence and the uniqueness of weak solutions for 2D incompressible magneto-micropolar fluid equations. Global existence of smooth solutions and global regularity of weak solutions are important topic in the study field of the magneto-micropolar fluid equations. The blow-up criteria for smooth solutions and regularity criteria of weak solutions were obtained in different function spaces, such as Morrey–Campanato space, Besov space and homogeneous Besov space, we may refer to [8, 32, 38, 39, 41] and [43]. Very recently, based on Serrin’s type non-blow up criterion established by [39], Wang and Gu [34] proved global existence of a class of smooth solutions, which ensure the $L^3$ norm is large. When partial viscosities disappear, the case becomes more complex, we may refer to [19, 23, 24] and [28]. Regmi and Wu [19] singled out three special partial dissipation cases and established the global regularity for each case. By fully exploiting the special structure of the system and using the maximal regularity property of the 1D heat operator, Shang and Gu [23] established the global existence of classical solution for 2D magneto-micropolar equations with only velocity dissipation and partial magnetic diffusion. The blow-up criterion for two-dimensional magneto-micropolar fluid equations with partial viscosity has been proved in [28]. Mixed partial viscosity means the viscosity coefficients are different in different directions, even the viscosity coefficients disappear in some directions. Therefore, some nonlinear terms cannot be controlled by the energy functions and the dissipative parts, which causes difficulty in dealing with these nonlinear terms. The regularity of the 2D anisotropic magneto-micropolar fluid equations with vertical kinematic viscosity, horizontal magnetic diffusion and horizontal vortex viscosity is established in Cheng and Liu [3]. Wang and Wang [33] studied the global existence of smooth solutions for 3D magneto-micropolar fluid equations with mixed partial viscosity by energy method. The results in [33] imply that there are two directional viscosities in every equation.

If $v=0$ and $\chi =0$, then the magneto-micropolar fluid equations (1.1) reduce to MHD equations. In additional, if $\mu =\nu =0$, then the MHD equations reduce to ideal MHD equations. The MHD equations govern the dynamics of the velocity and magnetic fields in electrically conducting fluids such as plasmas, liquid metals and salt water (see [12]). The global well-posedness of MHD equations has attracted the attention of many mathematicians, and lots of interesting results were established. On the one hand, we could refer to [6, 9, 15, 17, 27, 29,30,31] for global well-posedness of MHD equations with full viscosity. On the other hand, the global well-posedness problem for the MHD equations with partial viscosity has been successfully investigated in lots of work(see, e.g., [1, 2, 4, 5, 7, 10, 11, 13, 14, 16, 20, 26, 35,36,37, 40, 42, 44] and [45] ). We only recall the global existence of classical solutions to 3D MHD equations with partial viscosity for our purpose. The globally well-posed is proved by Cai and Lei [1] under the assumption that the initial velocity field and the displacement of the initial magnetic field from a nonzero constant are sufficiently small in certain weighted Sobolev spaces. Lin and Zhang [14] proved the global well-posedness to 3D MHD-type equations by the energy method, which depends crucially on the divergence-free condition of the velocity field. Under the condition of small initial data, Wu and Zhu [37] proved that the MHD equations with mixed partial dissipation and magnetic diffusion have a unique global smooth solution. The global well-posedness of smooth solutions to the 3D MHD equations with mixed partial dissipation and magnetic diffusion was proved by Wang and Wang [26]. We also refer to [16] for global well-posedness of classical solutions for a family of special axisymmetric initial data whose swirl components of the velocity field and magnetic vorticity field are trivial.

In this paper, we are interested in the following stability problem of the 3D magneto-micropolar equations with mixed partial viscosity

$$\begin{aligned} \left\{ \begin{array}{l} {\displaystyle \partial _{t}u-\mu _i\partial _i^2u-\mu _j\partial _j^2u-\chi \Delta u+u\cdot \nabla u-B\cdot \nabla B+\nabla \left( p+\frac{1}{2}|B|^2\right) -2\chi \nabla \times v=0,}\\ {\displaystyle \partial _{t}v-\gamma _m\partial _m^2 v-\gamma _n\partial _n^2 v-\kappa \nabla \nabla \cdot v+4\chi v+u\cdot \nabla v-2\chi \nabla \times u=0,}\\ {\displaystyle \partial _{t}B-\nu _k\partial _k^2 B+u\cdot \nabla B-B\cdot \nabla u=0,}\\ {\displaystyle \nabla \cdot u=0,\nabla \cdot B=0,}\\ \end{array} \right. \end{aligned}$$

(1.2)

where $\mu _i,\mu _j, \gamma _m, \gamma _n, \nu _k>0(i,j,k,m,n=1,2,3, i\ne j\ne k,\ m\ne n)$ and $\chi , \kappa \ge 0$. It is easy to find a special solution of (1.2) which is given by the zero velocity field, zero micro-rotational velocity and the background magnetic field $B^0=e_l(l=1 ,2, 3)$, where $e_1=(1,0,0)$, $e_2=(0,1,0)$ and $e_3=(0,0,1)$. The perturbation (u, v, b) around this equilibrium with $b=B-e_l(l=i, j)$ satisfies

$$\begin{aligned} \left\{ \begin{array}{l} {\displaystyle \partial _{t}u-\mu _i\partial _i^2u-\mu _j\partial _j^2u-\chi \Delta u+u\cdot \nabla u-b\cdot \nabla b+\nabla \left( p+\frac{1}{2}|b|^2\right) -2\chi \nabla \times v-\partial _l b=0,}\\ {\displaystyle \partial _{t}v-\gamma _m\partial _m^2 v-\gamma _n\partial _n^2 v-\kappa \nabla \nabla \cdot v+4\chi v+u\cdot \nabla v-2\chi \nabla \times u=0,}\\ {\displaystyle \partial _{t}b-\nu _k\partial _k^2 b+u\cdot \nabla b-b\cdot \nabla u-\partial _l u=0,}\\ {\displaystyle \nabla \cdot u=0,\nabla \cdot b=0.}\\ \end{array} \right. \end{aligned}$$

(1.3)

Inspired by the recent works [33] for 3D magneto-micropolar equations with mixed partial viscosity and [26, 37] for 3D incompressible MHD equations with mixed partial viscosity, the main aim of this paper is to investigate the stability problem on the perturbation (u, v, b). In other words, we shall prove the global small classical solutions to (1.3) with the initial data

$$\begin{aligned} t=0:\;\; u=u_0(x),\;\; v=v_0(x), \;\; b=b_0(x), \;\; x\in \mathbb {R}^3. \end{aligned}$$

(1.4)

Next, we start our main results as follows:

Theorem 1.1

Let $\mu _i,\mu _j, \gamma _m, \gamma _n, \nu _k>0(i,j,k,m,n=1,2,3, i\ne j\ne k,\ m\ne n, l=i, j)$ and $\chi , \kappa \ge 0$. Assume that $(u_0,v_0,b_0)\in H^3(\mathbb {R}^3)$ and $\nabla \cdot u_0=\nabla \cdot b_0=0$. Put

$$\begin{aligned} \mathscr {E}_0=\Vert u_0\Vert _{H^3}+\Vert v_0\Vert _{H^3}+\Vert b_0\Vert _{H^3}. \end{aligned}$$

There exists a constant $\delta >0$ such that if $\mathscr {E}_0\le \delta$, then the problem (1.3), (1.4) has a unique global classical solution $(u,v,b) \in C([0, \infty ); H^3(\mathbb {R}^3))$. Moreover, it holds that for any $t>0$

$$\begin{aligned}&\displaystyle \Vert u(t)\Vert _{H^3}^2+\Vert v(t)\Vert _{H^3}^2+\Vert b(t)\Vert _{H^3}^2+\int \limits _0^t(\Vert \partial _iu(\tau )\Vert _{H^3}^2+\Vert \partial _ju(\tau )\Vert _{H^3}^2+\Vert \partial _mv(\tau )\Vert _{H^3}^2\\&\quad +\,{\displaystyle \Vert \partial _nv(\tau )\Vert _{H^3}^2+\Vert \partial _kb(\tau )\Vert _{H^3}^2+\Vert \partial _lb(\tau )\Vert _{H^2}^2+\kappa \Vert \nabla \cdot v(\tau )\Vert _{H^3})d\tau \le C\mathscr {E}_0.} \end{aligned}$$

Remark 1.2

The result in Theorem 1.1 implies that it only needs one directional viscosity in the equations for b, which reduces to the requirement for the viscosity in [33].

Theorem 1.1 contains eighteen cases; the proof is similar. We only prove Theorem 1.1 with case: $i=m=2$, $j=n=3$, $k=1$ and $l=3$. To obtain global solutions, we need to bound $\Vert u(t)\Vert _{H^3}+\Vert v(t)\Vert _{H^3}+\Vert b(t)\Vert _{H^3}$ via the energy estimates and bootstrapping argument. But there are only two directional velocities and micro-rotational velocity viscosity (in $x_2$ and $x_3$) and one directional magnetic diffusion (in $x_1$); some nonlinear terms cannot be controlled by $\Vert u(t)\Vert _{H^3}+\Vert v(t)\Vert _{H^3}+\Vert b(t)\Vert _{H^3}$ and the dissipative parts $\Vert \partial _2u\Vert _{H^3}$, $\Vert \partial _3u\Vert _{H^3}$, $\Vert \partial _2v\Vert _{H^3}$, $\Vert \partial _3v\Vert _{H^3}$, $\Vert \partial _1b\Vert _{H^3}$. Consequently, we are not able to build a closed differential inequality for

$$\begin{aligned} {\displaystyle E_0(t)}= & {} {\displaystyle \sup _{0\le \tau \le t}\left( \Vert u(\tau )\Vert _{H^3}^2+\Vert v(\tau )\Vert _{H^3}^2+\Vert b(\tau )\Vert _{H^3}^2\right) +2\int \limits _0^t(\mu _2\Vert \partial _2u(\tau )\Vert _{H^3}^2}\\&+\,{\displaystyle \mu _3\Vert \partial _3u(\tau )\Vert _{H^3}^2+\gamma _2\Vert \partial _2v(\tau )\Vert _{H^3}^2+\gamma _3\Vert \partial _3v(\tau )\Vert _{H^3}^2+\nu _1\Vert \partial _1b(\tau )\Vert _{H^3}^2}\\&+\,{\displaystyle \kappa \Vert \nabla \cdot v(\tau )\Vert _{H^3}^2)d\tau ,} \end{aligned}$$

which forces us to introduce suitable extra terms in the energy estimate. To this end, the following term

$$\begin{aligned} E_1(t)=\int \limits _0^t\Vert \partial _3b(\tau )\Vert _{H^2}^2d\tau \end{aligned}$$

is introduced, which serves and achieves our purpose perfectly. Therefore, we could build a closed inequality for $E_0(t)$ and $E_1(t)$; then, global classical solutions follow from bootstrapping argument.

We introduce some notations which are used in this paper. $L^2=L^2(\mathbb {R}^3)$ denotes the usual Lebesgue space with the norm $\Vert \cdot \Vert _{L^2}$. Sobolev space of order m is defined by $H^{m}=H^{m}(\mathbb {R}^3)=\Big \{u\in L^2(\mathbb {R}^3)|\nabla ^m u \in L^2\Big \}$ with the norm $\Vert u\Vert _{H^m}=\Big (\Vert u\Vert ^2_{L^2}+\Vert \nabla ^mu\Vert ^2_{L^2}\Big )^{\frac{1}{2}}$.

The paper is organized as follows: In Sect. 2, we shall establish two a priori estimates and build the closed inequality for $E_0(t)$ and $E_1(t)$. More precisely, we derive the inequalities for $E_0(t)$ and $E_1(t)$ by the energy estimate and the tricky interpolation techniques in Sects. 2.1 and 2.2, respectively. In the last section (Sect. 3), we shall complete the proof of Theorem 1.1 by the bootstrapping argument.

2 Preliminaries

The main aim of this section is to establish the following two a priori estimates

$$\begin{aligned} E_0(t)\le C\left( \mathscr {E}_0+\mathscr {E}^{\frac{3}{2}}_0+E_0(t)^{\frac{3}{2}}+E_0(t)^2+E_1(t)^2\right) \end{aligned}$$

(2.1)

and

$$\begin{aligned} E_1(t)\le C\left( \mathscr {E}_0+E_0(t)+E_0(t)^{\frac{3}{2}}+E_0(t)^2+E_1(t)^2\right) , \end{aligned}$$

(2.2)

respectively.

The following lemma which will play very important roles in proving (2.1) and (2.2) has been obtained in [37].

Lemma 2.1

Assume that the right-hand sides of the following estimates are all bounded, then

$$\begin{aligned} \displaystyle \int \limits _{\mathbb {R}^3}|fgh|\;dx\le & {} C\Vert f\Vert _{L^2}^{\frac{1}{2}}\Vert \partial _1f\Vert _{L^2}^{\frac{1}{2}}\Vert g\Vert _{L^2}^{\frac{1}{2}}\Vert \partial _2g\Vert _{L^2}^{\frac{1}{2}} \Vert h\Vert _{L^2}^{\frac{1}{2}}\Vert \partial _3h\Vert _{L^2}^{\frac{1}{2}}, \end{aligned}$$

(2.3)

$$\begin{aligned} {\displaystyle \int \limits _{\mathbb {R}^3}|fgh\phi |\;dx}\le & {} {\displaystyle C\Vert f\Vert _{L^2}^{\frac{1}{4}}\Vert \partial _2f\Vert _{L^2}^{\frac{1}{4}}\Vert \partial _3f\Vert _{L^2}^{\frac{1}{4}}\Vert \partial _3\partial _2f\Vert _{L^2}^{\frac{1}{4}} \Vert g\Vert _{L^2}^{\frac{1}{4}}\Vert \partial _2g\Vert _{L^2}^{\frac{1}{4}} }\nonumber \\&{\displaystyle \cdot \Vert \partial _3g\Vert _{L^2}^{\frac{1}{4}}\Vert \partial _2\partial _3g\Vert _{L^2}^{\frac{1}{4}} \Vert h\Vert _{L^2}^{\frac{1}{2}}\Vert \partial _1h\Vert _{L^2}^{\frac{1}{2}}\Vert \phi \Vert _{L^2}^{\frac{1}{2}}\Vert \partial _1\phi \Vert _{L^2}^{\frac{1}{2}}} \end{aligned}$$

(2.4)

and

$$\begin{aligned} \begin{aligned} {\displaystyle \int \limits _{\mathbb {R}^3}|fgh|\;dx\le C\Vert f\Vert _{L^2}^{\frac{1}{4}}\Vert \partial _2f\Vert _{L^2}^{\frac{1}{4}}\Vert \partial _3f\Vert _{L^2}^{\frac{1}{4}}\Vert \partial _2\partial _3f\Vert _{L^2}^{\frac{1}{4}}\Vert g\Vert _{L^2}^{\frac{1}{2}}\Vert \partial _1g\Vert _{L^2}^{\frac{1}{2}} \Vert h\Vert _{L^2}}. \end{aligned} \end{aligned}$$

(2.5)

2.1 Proof of (2.1)

Owing to the equivalence of $\Vert (u,v,b)\Vert _{H^3}$ with $\Vert (u,v,b)\Vert _{L^2}+\Vert (u,v,b)\Vert _{\dot{H}^3}$, we estimate the $L^2$ and the homogeneous $\dot{H}^3$-norm of (u, v, b), respectively. By standard energy estimate, $\nabla \cdot u=\nabla \cdot b=0$ and Cauchy inequality, we obtain

$$\begin{aligned}&\ {\displaystyle (\Vert u(t)\Vert _{L^2}^2+\Vert v(t)\Vert _{L^2}^2+\Vert b(t)\Vert _{L^2}^2)+2\int \limits _0^t(\mu _2\Vert \partial _2u\Vert _{L^2}^2+\mu _3\Vert \partial _3u\Vert _{L^2}^2+\gamma _2\Vert \partial _2v\Vert _{L^2}^2}\nonumber \\&\qquad \ {\displaystyle +\,\gamma _3\Vert \partial _3v\Vert _{L^2}^2+\kappa \Vert \nabla \cdot v\Vert _{L^2}^2+\nu _1\Vert \partial _1b\Vert _{L^2}^2)d\tau }\nonumber \\&\quad {\displaystyle \le \Vert u_0\Vert _{L^2}^2+\Vert v_0\Vert _{L^2}^2+\Vert b_0\Vert _{L^2}^2.} \end{aligned}$$

(2.6)

Next we investigate the $\dot{H}^3$-norm. Applying $\partial _i^3(i=1,2,3)$ to (1.3) and then taking the inner product with $(\partial _i^3u,\partial _i^3v,\partial _i^3b)$, we arrive at

$$\begin{aligned}&{\displaystyle \frac{1}{2}\frac{d}{dt}\sum ^3_{i=1}[(\Vert \partial _i^3u\Vert _{L^2}^2+\Vert \partial _i^3v\Vert _{L^2}^2+\Vert \partial _i^3b\Vert _{L^2}^2) +\mu _2\Vert \partial _i^3\partial _2u\Vert _{L^2}^2+\mu _3\Vert \partial _i^3\partial _3u\Vert _{L^2}^2 + \chi \Vert \partial _i^3 \nabla u\Vert _{L^2}^2}\nonumber \\&\quad +\,{\displaystyle \gamma _2\Vert \partial _i^3\partial _2v\Vert _{L^2}^2+\gamma _3\Vert \partial _i^3\partial _3v\Vert _{L^2}^2+\kappa \Vert \partial _i^3\nabla \cdot v\Vert _{L^2}^2+4\chi \Vert \partial _i^3v\Vert _{L^2}^2+\nu _1\Vert \partial _i^3\partial _1b\Vert _{L^2}^2] = \sum _{j=1}^7\mathcal {I}_j,} \end{aligned}$$

(2.7)

where

$$\begin{aligned}&{\displaystyle \mathcal {I}_1=\sum \limits _{i=1}^3\int \limits _{\mathbb {R}^3}(\partial _i^3\partial _3b\cdot \partial _i^3u+\partial _i^3\partial _3u\cdot \partial _i^3b)\;dx,}\\&{\displaystyle \mathcal {I}_2=-\sum \limits _{i=1}^3\int \limits _{\mathbb {R}^3}\partial _i^3(u\cdot \nabla u)\cdot \partial _i^3u\;dx,}\\&{\displaystyle \mathcal {I}_3=\sum \limits _{i=1}^3\int \limits _{\mathbb {R}^3}[\partial _i^3(b\cdot \nabla b)-b\cdot \nabla \partial _i^3b]\cdot \partial _i^3u\;dx,}\\&{\displaystyle \mathcal {I}_4=-\sum \limits _{i=1}^3\int \limits _{\mathbb {R}^3}\partial _i^3(u\cdot \nabla b)\cdot \partial _i^3b\;dx,}\\&{\displaystyle \mathcal {I}_5=\sum \limits _{i=1}^3\int \limits _{\mathbb {R}^3}[\partial _i^3(b\cdot \nabla u)-b\cdot \nabla \partial _i^3u]\cdot \partial _i^3b\;dx,}\\&{\displaystyle \mathcal {I}_6=-\sum \limits _{i=1}^3\int \limits _{\mathbb {R}^3}\partial _i^3(u\cdot \nabla v)\cdot \partial _i^3v\;dx,}\\&{\displaystyle \mathcal {I}_7=-2\chi \sum \limits _{i=1}^3\int \limits _{\mathbb {R}^3}[\partial ^3_i (\nabla \times v) \partial _i^3 u +\partial ^3_i (\nabla \times u) \partial _i^3 v]\;dx.} \end{aligned}$$

By integration by parts, we have

$$\begin{aligned} \mathcal {I}_1=0. \end{aligned}$$

(2.8)

To bound $\mathcal {I}_2$, we decompose it into three pieces as

$$\begin{aligned} {\displaystyle \mathcal {I}_2=-\sum \limits _{i=1}^3\int \limits _{\mathbb {R}^3}\partial _i^3(u\cdot \nabla u)\cdot \partial _i^3u\;dx=:\mathcal {I}_{2,1}+\mathcal {I}_{2,2}+\mathcal {I}_{2,3}.} \end{aligned}$$

By Lemma 2.1 and Cauchy’s inequality, we obtain

$$\begin{aligned} {\displaystyle \mathcal {I}_{2,1}}= & {} {\displaystyle -\int \limits _{\mathbb {R}^3}\partial _1^3(u\cdot \nabla u)\cdot \partial _1^3u\;dx}\\= & {} {\displaystyle -\int \limits _{\mathbb {R}^3}[\partial _1^3(u_2\cdot \partial _2 u)\cdot \partial _1^3u+\partial _1^3(u_3\cdot \partial _3 u)\cdot \partial _1^3u +\partial _1^3(u_1\cdot \partial _1 u)\cdot \partial _1^3u]\;dx}\\= & {} {\displaystyle -\sum \limits _{k=1}^3\mathcal {C}_3^k\int \limits _{\mathbb {R}^3}(\partial _1^ku_2\cdot \partial _1^{3-k}\partial _2 u\cdot \partial _1^3u +\partial _1^ku_3\cdot \partial _1^{3-k}\partial _3 u\cdot \partial _1^3u)\;dx}\\&{\displaystyle -\,\sum \limits _{k=1}^3\mathcal {C}_3^k\int \limits _{\mathbb {R}^3}\partial _1^ku_1\cdot \partial _1^{3-k}\partial _1 u\cdot \partial _1^3u\;dx}\\\le & {} {\displaystyle C\sum \limits _{k=1}^3\Vert \partial _1^ku_2\Vert _{L^2}^{\frac{1}{2}}\Vert \partial _2\partial _1^ku_2\Vert _{L^2}^{\frac{1}{2}} \Vert \partial _2\partial _1^{3-k}u\Vert _{L^2}^{\frac{1}{2}}\Vert \partial _1\partial _2\partial _1^{3-k}u\Vert _{L^2}^{\frac{1}{2}} \Vert \partial _1^3u\Vert _{L^2}^{\frac{1}{2}}\Vert \partial _3\partial _1^3u\Vert _{L^2}^{\frac{1}{2}}}\\&+\,{\displaystyle \Vert \partial _1^ku_3\Vert _{L^2}^{\frac{1}{2}}\Vert \partial _2\partial _1^ku_3\Vert _{L^2}^{\frac{1}{2}} \Vert \partial _3\partial _1^{3-k}u\Vert _{L^2}^{\frac{1}{2}}\Vert \partial _1\partial _3\partial _1^{3-k}u\Vert _{L^2}^{\frac{1}{2}} \Vert \partial _1^3u\Vert _{L^2}^{\frac{1}{2}}\Vert \partial _3\partial _1^3u\Vert _{L^2}^{\frac{1}{2}}}\\&+\,{\displaystyle \Vert \partial _1^{k-1}(\partial _2u_2+\partial _3u_3)\Vert _{L^2}^{\frac{1}{2}}\Vert \partial _1\partial _1^{k-1}(\partial _2u_2+\partial _3u_3)\Vert _{L^2}^{\frac{1}{2}} \Vert \partial _1^{4-k}u\Vert _{L^2}^{\frac{1}{2}}\Vert \partial _2\partial _1^{4-k}u\Vert _{L^2}^{\frac{1}{2}}}\\&\times \,{\displaystyle \Vert \partial _1^3u\Vert _{L^2}^{\frac{1}{2}}\Vert \partial _3\partial _1^3u\Vert _{L^2}^{\frac{1}{2}}}\\\le & {} {\displaystyle C\Vert \partial _2u\Vert _{H^3}^{\frac{3}{2}}\Vert \partial _3u\Vert _{H^3}^{\frac{1}{2}}\Vert u\Vert _{H^3}+\Vert \partial _3u\Vert _{H^3}^{\frac{3}{2}}\Vert \partial _2u\Vert _{H^3}^{\frac{1}{2}}\Vert u\Vert _{H^3}}\\\le & {} {\displaystyle C\Vert u\Vert _{H^3}(\Vert \partial _2u\Vert ^2_{H^3}+\Vert \partial _3u\Vert ^2_{H^3}). } \end{aligned}$$

Hölder’s inequality gives

$$\begin{aligned} {\displaystyle \mathcal {I}_{2,2}} \le C{\displaystyle \Vert \partial _2(u\cdot \nabla u)\Vert _{\dot{H}^2}\Vert \partial _2u\Vert _{\dot{H}^2}\le C\Vert u\Vert _{H^3}\Vert \partial _2u\Vert _{H^3}^2.} \end{aligned}$$

Similarly, it holds that

$$\begin{aligned} {\displaystyle \mathcal {I}_{2,3}} \le C{\displaystyle \Vert \partial _3(u\cdot \nabla u)\Vert _{\dot{H}^2}\Vert \partial _3u\Vert _{\dot{H}^2}\le C\Vert u\Vert _{H^3}\Vert \partial _3u\Vert _{H^3}^2.} \end{aligned}$$

Combining the above three estimates yields

$$\begin{aligned} \mathcal {I}_2\le C\Vert u\Vert _{H^3}(\Vert \partial _2 u\Vert ^2_{H^3}+\Vert \partial _3 u\Vert ^2_{H^3}). \end{aligned}$$

(2.9)

The next term is $\mathcal {I}_3$, which is estimated as

$$\begin{aligned} {\displaystyle \mathcal {I}_3}= & {} {\displaystyle \sum \limits _{i=1}^3\int \limits _{\mathbb {R}^3}[\partial _i^3(b\cdot \nabla b)-b\cdot \nabla \partial _i^3b]\cdot \partial _i^3u\;dx =\sum \limits _{i=1}^3\sum \limits _{k=1}^3\mathcal {C}_3^k\int \limits _{\mathbb {R}^3}\partial _i^kb\cdot \nabla _i^{3-k}b\cdot \partial _i^3u\;dx}\\=: & {} {\displaystyle \mathcal {I}_{3,1}+\mathcal {I}_{3,2}+\mathcal {I}_{3,3}.} \end{aligned}$$

Thanks to Hölder’s inequality and Lemma 2.1, we obtain

$$\begin{aligned} {\displaystyle \mathcal {I}_{3,1}}= & {} {\displaystyle \sum \limits _{k=1}^2\mathcal {C}_3^k\int \limits _{\mathbb {R}^3}\partial _1^kb\cdot \nabla \partial _1^{3-k}b\cdot \partial _1^3u\;dx +\int \limits _{\mathbb {R}^3}\partial _1^3b\cdot \nabla b\cdot \partial _1^3u\;dx}\\\le & {} {\displaystyle C\sum \limits _{k=1}^2\Vert \partial _1^kb\Vert _{L^2}^{\frac{1}{2}}\Vert \partial _3\partial _1^kb\Vert _{L^2}^{\frac{1}{2}} \Vert \nabla \partial _1^{3-k}b\Vert _{L^2}^{\frac{1}{2}}\Vert \partial _1\nabla \partial _1^{3-k}b\Vert _{L^2}^{\frac{1}{2}}\Vert \partial _1^3u\Vert _{L^2}^{\frac{1}{2}} \Vert \partial _2\partial _1^3u\Vert _{L^2}^{\frac{1}{2}}}\\&+\,{\displaystyle \Vert \partial _1^3b\Vert _{L^2}^{\frac{1}{2}}\Vert \partial _1\partial _1^3b\Vert _{L^2}^{\frac{1}{2}}\Vert \nabla b\Vert _{L^2}^{\frac{1}{2}}\Vert \partial _3\nabla b\Vert _{L^2}^{\frac{1}{2}}\Vert \partial _1^3u\Vert _{L^2}^{\frac{1}{2}}\Vert \partial _2\partial _1^3u\Vert _{L^2}^{\frac{1}{2}}}\\\le & {} {\displaystyle C\Vert u\Vert _{H^3}^{\frac{1}{2}}\Vert \partial _2u\Vert _{H^3}^{\frac{1}{2}}\Vert b\Vert _{H^3}^{\frac{1}{2}}\Vert \partial _3b\Vert _{H^2}^{\frac{1}{2}}\Vert \partial _1b\Vert _{H^3}}\\\le & {} {\displaystyle C\Vert u\Vert _{H^3}^{\frac{1}{2}}\Vert b\Vert _{H^3}^{\frac{1}{2}}(\Vert \partial _2u\Vert ^2_{H^3}+\Vert \partial _1b\Vert ^2_{H^3}+\Vert \partial _3b\Vert ^2_{H^2}).} \end{aligned}$$

$\mathcal {I}_{3,2}$ could be treated along the same line; we arrive at

$$\begin{aligned} {\displaystyle \mathcal {I}_{3,2}} \le {\displaystyle C\Vert b\Vert _{H^3}(\Vert \partial _2u\Vert ^2_{H^3}+\Vert \partial _1b\Vert ^2_{H^3}+\Vert \partial _3b\Vert ^2_{H^2}).} \end{aligned}$$

It follows from Hölder’s inequality and the embedding theorem that

$$\begin{aligned} {\displaystyle \mathcal {I}_{3,3}}= & {} {\displaystyle \sum \limits _{k=1}^2\mathcal {C}_3^k\int \limits _{\mathbb {R}^3}\partial _3^kb\cdot \nabla \partial _3^{3-k}b\cdot \partial _3^3u\;dx +\int \limits _{\mathbb {R}^3}\partial _3^3b\cdot \nabla b\cdot \partial _3^3u\;dx}\\\le & {} {\displaystyle C\sum \limits _{k=1}^2\Vert \partial _3^kb\Vert _{L^3}\Vert \nabla \partial _3^{3-k}b\Vert _{L^2}\Vert \partial _3^3u\Vert _{L^6} +\Vert \partial _3^3b\Vert _{L^2}\Vert \nabla b\Vert _{L^3}\Vert \partial _3^3u\Vert _{L^6}}\\\le & {} {\displaystyle C\Vert \partial _3u\Vert _{H^3}\Vert \partial _3b\Vert _{H^2}\Vert b\Vert _{H^3} }\\\le & {} {\displaystyle C\Vert b\Vert _{H^3}(\Vert \partial _3u\Vert ^2_{H^3}+\Vert \partial _3b\Vert _{H^2}).} \end{aligned}$$

Collecting the estimates for $\mathcal {I}_3$ and using Cauchy’s inequality, we obtain

$$\begin{aligned} I_3\le C\left( \Vert u\Vert ^{\frac{1}{2}}_{H^3}\Vert b\Vert ^{\frac{1}{2}}_{H^3}+\Vert b\Vert _{H^3})(\Vert \partial _2 u\Vert ^2_{H^3}+\Vert \partial _3 u\Vert ^2_{H^3}+ \Vert \partial _1 b\Vert ^2_{H^3}+\Vert \partial _3 b\Vert ^2_{H^2}\right) . \end{aligned}$$

(2.10)

We split the next term into three parts as

$$\begin{aligned} {\displaystyle \mathcal {I}_4=-\sum \limits _{i=1}^3\int \limits _{\mathbb {R}^3}\partial _i^3(u\cdot \nabla b)\cdot \partial _i^3bdx =:\mathcal {I}_{4,1}+\mathcal {I}_{4,2}+\mathcal {I}_{4,3}.} \end{aligned}$$

Because $\mathcal {I}_{4,1}$ and $\mathcal {I}_{4,3}$ have partial derivatives in $x_1$ and $x_3$, respectively, we can handle them not difficultly. However, $\mathcal {I}_{4,2}$ involves the partial derivative in $x_2$ and the control of $\mathcal {I}_{4,2}$ is very complex. Lemma 2.1 and Cauchy’s inequality entail that

$$\begin{aligned} {\displaystyle \mathcal {I}_{4,1}}= & {} {\displaystyle -\int \limits _{\mathbb {R}^3}\partial _1^3(u\cdot \nabla b)\cdot \partial _1^3b\;dx}\\= & {} {\displaystyle -\sum \limits _{k=1}^3\mathcal {C}_3^k\int \limits _{\mathbb {R}^3}\partial _1^ku\cdot \nabla \partial _1^{3-k}b\cdot \partial _1^3b\;dx}\\\le & {} {\displaystyle C\Vert \partial _1^ku\Vert _{L^2}^{\frac{1}{2}}\Vert \partial _2\partial _1^ku\Vert _{L^2}^{\frac{1}{2}} \Vert \nabla \partial _1^{3-k}b\Vert _{L^2}^{\frac{1}{2}}\Vert \partial _1\nabla \partial _1^{3-k}b\Vert _{L^2}^{\frac{1}{2}} \Vert \partial _1^3b\Vert _{L^2}^{\frac{1}{2}}\Vert \partial _3\partial _1^3b\Vert _{L^2}^{\frac{1}{2}}}\\\le & {} {\displaystyle C\Vert u\Vert _{H^3}^{\frac{1}{2}}\Vert \partial _2u\Vert _{H^3}^{\frac{1}{2}}\Vert b\Vert _{H^3}^{\frac{1}{2}}\Vert \partial _1b\Vert _{H^3}^{\frac{3}{2}}}\\\le & {} {\displaystyle C\Vert u\Vert _{H^3}^{\frac{1}{2}}\Vert b\Vert _{H^3}^{\frac{1}{2}}(\Vert \partial _2u\Vert ^2_{H^3}+\Vert \partial _1b\Vert ^2_{H^3}).} \end{aligned}$$

The embedding theorem and Cauchy’s inequality give

$$\begin{aligned} {\displaystyle \mathcal {I}_{4,3}}= & {} {\displaystyle -\int \limits _{\mathbb {R}^3}\partial _3^3(u\cdot \nabla b)\cdot \partial _3^3b\;dx}\\= & {} {\displaystyle -\sum \limits _{k=2}^3\mathcal {C}_3^k\int \limits _{\mathbb {R}^3}\partial _3^ku\cdot \nabla \partial _3^{3-k}b\cdot \partial _3^3b\;dx -3\int \limits _{\mathbb {R}^3}\partial _3u\cdot \nabla \partial _3^2b\cdot \partial _3^3b\;dx}\\\le & {} {\displaystyle C\Vert \partial _3^ku\Vert _{L^3}\Vert \nabla \partial _3^{3-k}b\Vert _{L^6}\Vert \partial _3^3b\Vert _{L^2} +\Vert \partial _3u\Vert _{L^\infty }\Vert \nabla \partial _3^2b\Vert _{L^2}\Vert \partial _3^3b\Vert _{L^2}}\\\le & {} {\displaystyle C\Vert b\Vert _{H^3}(\Vert \partial _3u\Vert ^2_{H^3}+\Vert \partial _3b\Vert ^2_{H^2}).} \end{aligned}$$

The difficult term is $\mathcal {I}_{4,2}$; we further decompose it into three parts as

$$\begin{aligned} {\displaystyle \mathcal {I}_{4,2}}= & {} {\displaystyle -\int \limits _{\mathbb {R}^3}\left[ \partial _2^3(u_1\partial _1b)\cdot \partial _2^3b+\partial _2^3(u_2\partial _2b)\cdot \partial _2^3b +\partial _2^3(u_3\partial _3b)\cdot \partial _2^3b\right] \;dx}\\=: & {} {\displaystyle \mathcal {I}_{4,2,1}+\mathcal {I}_{4,2,2}+\mathcal {I}_{4,2,3}.} \end{aligned}$$

The same produce of treating $\mathcal {I}_{43}$ yields

$$\begin{aligned} {\displaystyle \mathcal {I}_{4,2,1}}= & {} {\displaystyle -\sum \limits _{k=1}^3\mathcal {C}_3^k\int \limits _{\mathbb {R}^3}\partial _2^ku_1\cdot \partial _1\partial _2^{3-k}b\cdot \partial _2^3b\;dx}\\\le & {} {\displaystyle C\sum \limits _{k=1}^3\Vert \partial _2^ku_1\Vert _{L^3}\Vert \partial _1\partial _2^{3-k}b\Vert _{L^6}\Vert \partial _2^3b\Vert _{L^2}}\\\le & {} {\displaystyle C\Vert \partial _2u\Vert _{H^3}\Vert b\Vert _{H^3}\Vert \partial _1b\Vert _{H^3}}\\\le & {} {\displaystyle C\Vert b\Vert _{H^3}(\Vert \partial _2u\Vert ^2_{H^3}+\Vert \partial _1b\Vert ^2_{H^3})} \end{aligned}$$

and

$$\begin{aligned} {\displaystyle \mathcal {I}_{4,2,3}}= & {} {\displaystyle -\sum \limits _{k=2}^3\mathcal {C}_3^k\int \limits _{\mathbb {R}^3}\partial _2^ku_3\cdot \partial _3\partial _2^{3-k}b\cdot \partial _2^3b\;dx -3\int \limits _{\mathbb {R}^3}\partial _2u_3\cdot \partial _3\partial _2^2b\cdot \partial _2^3b\;dx}\\\le & {} {\displaystyle C\sum \limits _{k=2}^3\Vert \partial _2^ku_3\Vert _{L^3}\Vert \partial _3\partial _2^{3-k}b\Vert _{L^6}\Vert \partial _2^3b\Vert _{L^2} +\Vert \partial _2u_3\Vert _{L^\infty }\Vert \partial _3\partial _2^2b\Vert _{L^2}\Vert \partial _2^3b\Vert _{L^2}}\\\le & {} {\displaystyle C\Vert b\Vert ^2_{H^3}(\Vert \partial _2u\Vert ^2_{H^3}+\Vert \partial _3b\Vert ^2_{H^2}).} \end{aligned}$$

We now turn to $\mathcal {I}_{4,2,2}$; we further break it down

$$\begin{aligned} {\displaystyle \mathcal {I}_{4,2,2}}= & {} {\displaystyle -\sum \limits _{k=2}^3\mathcal {C}_3^k\int \limits _{\mathbb {R}^3}\partial _2^ku_2\cdot \partial _2^{4-k}b\cdot \partial _2^3b\;dx -3\int \limits _{\mathbb {R}^3}\partial _2u_2\cdot \partial _2^3b\cdot \partial _2^3b\;dx}\\= & {} {\displaystyle -\sum \limits _{k=2}^3\mathcal {C}_3^k\int \limits _{\mathbb {R}^3}\partial _2^ku_2\cdot \partial _2^{4-k}b\cdot \partial _2^3b\;dx +3\int \limits _{\mathbb {R}^3}(\partial _1u_1+\partial _3u_3)\cdot \partial _2^3b\cdot \partial _2^3b\;dx}\\=: & {} {\displaystyle \mathcal {I}_{4,2,2,1}+\mathcal {I}_{4,2,2,2}+\mathcal {I}_{4,2,2,3}.} \end{aligned}$$

We derive from Lemma 2.1 and Cauchy’s inequality

$$\begin{aligned} {\displaystyle \mathcal {I}_{4,2,2,1}}\le & {} {\displaystyle C\sum \limits _{k=2}^3\Vert \partial _2^ku_2\Vert _{L^2}^{\frac{1}{2}}\Vert \partial _2\partial _2^ku_2\Vert _{L^2}^{\frac{1}{2}} \Vert \partial _2^{4-k}b\Vert _{L^2}^{\frac{1}{2}}\Vert \partial _3\partial _2^{4-k}b\Vert _{L^2}^{\frac{1}{2}} \Vert \partial _2^3b\Vert _{L^2}^{\frac{1}{2}}\Vert \partial _1\partial _2^3b\Vert _{L^2}^{\frac{1}{2}}}\\\le & {} {\displaystyle C\Vert \partial _2u\Vert _{H^3}\Vert b\Vert _{H^3}\Vert \partial _1b\Vert _{H^3}^{\frac{1}{2}}\Vert \partial _3b\Vert _{H^2}^{\frac{1}{2}}}\\\le & {} {\displaystyle C\Vert b\Vert _{H^3}(\Vert \partial _2u\Vert ^2_{H^3}+\Vert \partial _1b\Vert ^2_{H^3}+\Vert \partial _3b\Vert ^2_{H^2}).} \end{aligned}$$

From integration by parts and Lemma 2.1, we obtain

$$\begin{aligned} {\displaystyle \mathcal {I}_{4,2,2,2}}= & {} {\displaystyle -6\int \limits _{\mathbb {R}^3}u_1\cdot \partial _2^3b\cdot \partial _1\partial _2^3b\;dx}\\\le & {} { C\displaystyle \Vert u_1\Vert _{L^2}^{\frac{1}{4}}\Vert \partial _2u_1\Vert _{L^2}^{\frac{1}{4}}\Vert \partial _3u_1\Vert _{L^2}^{\frac{1}{4}}\Vert \partial _2\partial _3\partial _2u_1\Vert _{L^2}^{\frac{1}{4}} \Vert \partial _2^3b\Vert _{L^2}^{\frac{1}{2}}\Vert \partial _1\partial _2^3b\Vert _{L^2}^{\frac{1}{2}}\Vert \partial _1\partial _2^3b\Vert _{L^2}}\\\le & {} {\displaystyle C\Vert u\Vert _{H^3}^{\frac{1}{2}}\Vert b\Vert _{H^3}^{\frac{1}{2}}(\Vert \partial _2u\Vert ^2_{H^3}+\Vert \partial _3u\Vert ^2_{H^3}+ \Vert \partial _1b\Vert ^2_{H^3}).} \end{aligned}$$

We cannot estimate $\mathcal {I}_{4,2,2,3}$ to yield a suitable bound directly. If we use Lemma 2.1 to estimate $\mathcal {I}_{4,2,2,3}$ directly,

$$\begin{aligned} {\displaystyle \mathcal {I}_{4,2,2,3}}= & {} {\displaystyle 3\int \limits _{\mathbb {R}^3}\partial _3u_3\cdot \partial _2^3b\cdot \partial _2^3b\;dx}\\\le & {} {\displaystyle C\Vert \partial _3u_3\Vert _{L^2}^{\frac{1}{2}}\Vert \partial _2\partial _3u_3\Vert _{L^2}^{\frac{1}{2}} \Vert \partial _2^3b\Vert _{L^2}^{\frac{1}{2}}\Vert \partial _1\partial _2^3b\Vert _{L^2}^{\frac{1}{2}} \Vert \partial _2^3b\Vert _{L^2}^{\frac{1}{2}}\Vert \partial _3\partial _2^3b\Vert _{L^2}^{\frac{1}{2}},}\\ \end{aligned}$$

there will appear $\Vert \partial _3b\Vert _{H^3}$ and the differential inequality would not be closed. We use the special structure of the equation for b in (1.3) to replace $\partial _3u_3$ as follows

$$\begin{aligned} {\displaystyle \partial _3u_3=\partial _tb_3-\nu _1\partial _1^2b_3+u\cdot \nabla b_3-b\cdot \nabla u_3.} \end{aligned}$$

We write $\mathcal {I}_{4,2,2,3}$ as

$$\begin{aligned} {\displaystyle \mathcal {I}_{4,2,2,3}}= & {} {\displaystyle 3\int \limits _{\mathbb {R}^3}\partial _3u_3\cdot \partial _2^3b\cdot \partial _2^3b\;dx}\\= & {} {\displaystyle 3\int \limits _{\mathbb {R}^3}[\partial _tb_3-\nu _1\partial _1^2b_3+u\cdot \nabla b_3-b\cdot \nabla u_3]\cdot |\partial _2^3b|^2\;dx}\\=: & {} {\displaystyle \mathcal {J}_1+\mathcal {J}_2+\mathcal {J}_3+\mathcal {J}_4.} \end{aligned}$$

In what follows, we can deal with $\mathcal {J}_2$, $\mathcal {J}_3$, $\mathcal {J}_4$. Hölder’s inequality and Cauchy’s inequality give

$$\begin{aligned} {\displaystyle \mathcal {J}_2}= & {} {\displaystyle 6\nu _1\int \limits _{\mathbb {R}^3}\partial _1b_3\cdot \partial _2^3b\cdot \partial _1\partial _2^3b\;dx}\\\le & {} {\displaystyle C\Vert \partial _1b_3\Vert _{L^\infty }\Vert \partial _2^3b\Vert _{L^2}\Vert \partial _1\partial _2^3b\Vert _{L^2}}\\\le & {} {\displaystyle C\Vert b\Vert _{H^3}\Vert \partial _1b\Vert _{H^3}^2.} \end{aligned}$$

From Lemma 2.1 and Cauchy inequality, we obtain

$$\begin{aligned} {\displaystyle \mathcal {J}_3}\le & {} {\displaystyle C\Vert u\Vert _{L^2}^{\frac{1}{4}}\Vert \partial _2u\Vert _{L^2}^{\frac{1}{4}}\Vert \partial _3u\Vert _{L^2}^{\frac{1}{4}}\Vert \partial _2\partial _3u\Vert _{L^2}^{\frac{1}{4}} \Vert \nabla b_3\Vert _{L^2}^{\frac{1}{4}}\Vert \partial _2\nabla b_3\Vert _{L^2}^{\frac{1}{4}}\Vert \partial _3\nabla b_3\Vert _{L^2}^{\frac{1}{4}}}\\&\times \,{\displaystyle \Vert \partial _2\partial _3\nabla b_3\Vert _{L^2}^{\frac{1}{4}}\Vert \partial _2^3b\Vert _{L^2}\Vert \partial _1\partial _2^3b\Vert _{L^2}}\\\le & {} {\displaystyle C\Vert u\Vert _{H^3}^{\frac{1}{2}}\Vert \partial _2u\Vert _{H^3}^{\frac{1}{4}}\Vert \partial _3u\Vert _{H^3}^{\frac{1}{4}} \Vert \partial _1b\Vert _{H^3}\Vert b\Vert _{H^3}^{\frac{3}{2}}\Vert \partial _3b\Vert _{H^2}^{\frac{1}{2}}}\\\le & {} {\displaystyle C\Vert u\Vert _{H^3}^{\frac{1}{2}}\Vert b\Vert _{H^3}^{\frac{3}{2}}(\Vert \partial _2u\Vert ^2_{H^3}+\Vert \partial _3u\Vert ^2_{H^3}+ \Vert \partial _1b\Vert ^2_{H^3}+\Vert \partial _3b\Vert ^2_{H^2})} \end{aligned}$$

and

$$\begin{aligned} {\displaystyle \mathcal {J}_4}\le & {} { \displaystyle C\Vert b\Vert _{L^2}^{\frac{1}{4}}\Vert \partial _2b\Vert _{L^2}^{\frac{1}{4}}\Vert \partial _3b\Vert _{L^2}^{\frac{1}{4}}\Vert \partial _2\partial _3b\Vert _{L^2}^{\frac{1}{4}} \Vert \nabla u_3\Vert _{L^2}^{\frac{1}{4}}\Vert \partial _2\nabla u_3\Vert _{L^2}^{\frac{1}{4}}\Vert \partial _3\nabla u_3\Vert _{L^2}^{\frac{1}{4}}}\\&\times \,{\displaystyle \Vert \partial _2\partial _3\nabla u_3\Vert _{L^2}^{\frac{1}{4}}\Vert \partial _2^3b\Vert _{L^2}\Vert \partial _1\partial _2^3b\Vert _{L^2}}\\\le & {} {\displaystyle C\Vert u\Vert _{H^3}^{\frac{1}{2}}\Vert \partial _2u\Vert _{H^3}^{\frac{1}{4}}\Vert \partial _3u\Vert _{H^3}^{\frac{1}{4}} \Vert \partial _1b\Vert _{H^3}\Vert b\Vert _{H^3}^{\frac{3}{2}}\Vert \partial _3b\Vert _{H^2}^{\frac{1}{2}}.}\\\le & {} {\displaystyle C\Vert u\Vert _{H^3}^{\frac{1}{2}}\Vert b\Vert _{H^3}^{\frac{3}{2}}(\Vert \partial _2u\Vert ^2_{H^3}+\Vert \partial _3u\Vert ^2_{H^3}+ \Vert \partial _1b\Vert ^2_{H^3}+\Vert \partial _3b\Vert ^2_{H^2}).} \end{aligned}$$

We write $\mathcal {J}_1$ as

$$\begin{aligned} {\displaystyle \mathcal {J}_1}= & {} {\displaystyle 3\frac{d}{dt}\int \limits _{\mathbb {R}^3}b_3\cdot |\partial _2^3b|^2dx-3\int \limits _{\mathbb {R}^3}b_3\cdot \partial _t|\partial _2^3b|^2\;dx}\\=: & {} {\displaystyle \mathcal {J}_{1,1}+\mathcal {J}_{1,2}.} \end{aligned}$$

We use the equation of b in (1.3) to estimate $\mathcal {J}_{1,2}$

$$\begin{aligned} {\displaystyle \mathcal {J}_{1,2}}= & {} {\displaystyle -6\int \limits _{\mathbb {R}^3}b_3\cdot \partial _t\partial _2^3b\cdot \partial _2^3b\;dx}\\= & {} {\displaystyle -6\int \limits _{\mathbb {R}^3}b_3\cdot [\nu _1\partial _2^3\partial _1^2b-\partial _2^3(u\cdot \nabla b)+\partial _2^3(b\cdot \nabla u) +\partial _2^3\partial _3u)]\cdot \partial _2^3b\;dx}\\=: & {} {\displaystyle \mathcal {J}_{1,2,1}+\mathcal {J}_{1,2,2}+\mathcal {J}_{1,2,3}+\mathcal {J}_{1,2,4}.} \end{aligned}$$

From integration by parts and Hölder’s inequality, it follows that

$$\begin{aligned} {\displaystyle \mathcal {J}_{1,2,1}}= & {} {\displaystyle 6\nu _1\int \limits _{\mathbb {R}^3}\partial _1(b_3\cdot \partial _2^3b)\cdot \partial _2^3\partial _1b\;dx}\\= & {} {\displaystyle 6\nu _1\int \limits _{\mathbb {R}^3}\partial _1b_3\cdot \partial _2^3b\cdot \partial _2^3\partial _1b\;dx +6\nu _1\int \limits _{\mathbb {R}^3}b_3\cdot |\partial _1\partial _2^3b|^2\;dx}\\\le & {} {\displaystyle C\Vert \partial _1b_3\Vert _{L^\infty }\Vert \partial _2^3b\Vert _{L^2}\Vert \partial _1\partial _2^3b\Vert _{L^2} +\Vert b_3\Vert _{L^\infty }\Vert \partial _1\partial _2^3b\Vert _{L^2}^2}\\\le & {} {\displaystyle C\Vert b\Vert _{H^3}\Vert \partial _1b\Vert _{H^3}^2.} \end{aligned}$$

By using integration by parts and applying Lemma 2.1 and Cauchy’s inequality, we arrive at

$$\begin{aligned} {\displaystyle \mathcal {J}_{1,2,2}}= & {} {\displaystyle 6\sum \limits _{k=1}^3\mathcal {C}_3^k\int \limits _{\mathbb {R}^3}b_3\cdot \partial _2^3b\cdot \partial _2^ku\cdot \nabla \partial _2^{3-k}b\;dx +6\int \limits _{\mathbb {R}^3}b_3\cdot \partial _2^3b\cdot u\cdot \nabla \partial _2^3b\;dx}\\= & {} {\displaystyle 6\sum \limits _{k=1}^3\mathcal {C}_3^k\int \limits _{\mathbb {R}^3}b_3\cdot \partial _2^3b\cdot \partial _2^ku\cdot \nabla \partial _2^{3-k}b\;dx -3\int \limits _{\mathbb {R}^3}\nabla b_3\cdot u\cdot |\partial _2^3b|^2\;dx}\\\le & {} {\displaystyle C\sum \limits _{k=2}^3\Vert b_3\Vert _{L^\infty }\Vert \partial _2^3b\Vert _{L^2}^{\frac{1}{2}}\Vert \partial _1\partial _2^3b\Vert _{L^2}^{\frac{1}{2}} \Vert \partial _2^ku\Vert _{L^2}^{\frac{1}{2}}\Vert \partial _2\partial _2^ku\Vert _{L^2}^{\frac{1}{2}} \Vert \nabla \partial _2^{3-k}b\Vert _{L^2}^{\frac{1}{2}}\Vert \partial _3\nabla \partial _2^{3-k}b\Vert _{L^2}^{\frac{1}{2}}}\\&+\,{\displaystyle \Vert b_3\Vert _{L^2}^{\frac{1}{4}}\Vert \partial _2b_3\Vert _{L^2}^{\frac{1}{4}}\Vert \partial _3b_3\Vert _{L^2}^{\frac{1}{4}}\Vert \partial _2\partial _3b_3\Vert _{L^2}^{\frac{1}{4}} \Vert \partial _2u\Vert _{L^2}^{\frac{1}{4}}\Vert \partial _2\partial _2u\Vert _{L^2}^{\frac{1}{4}}\Vert \partial _3\partial _2u\Vert _{L^2}^{\frac{1}{4}}}\\&\times \,{\displaystyle \Vert \partial _2\partial _3\partial _2u\Vert _{L^2}^{\frac{1}{4}}\Vert \partial _2^3b\Vert _{L^2}^{\frac{1}{2}}\Vert \partial _1\partial _2^3b\Vert _{L^2}^{\frac{1}{2}} \Vert \nabla \partial _2^2b\Vert _{L^2}^{\frac{1}{2}}\Vert \partial _1\nabla \partial _2^2b\Vert _{L^2}^{\frac{1}{2}}+|J_3|}\\\le & {} {\displaystyle C(\Vert b\Vert _{H^3}^2\Vert \partial _1b\Vert _{H^3}^{\frac{1}{2}}\Vert \partial _3b\Vert _{H^2}^{\frac{1}{2}}\Vert \partial _2u\Vert _{H^3}}\\&+\,{\displaystyle \Vert b\Vert _{H^3}^{\frac{3}{2}}\Vert \partial _1b\Vert _{H^3}\Vert \partial _3b\Vert _{H^2}^{\frac{1}{2}}\Vert u\Vert _{H^3}^{\frac{1}{2}}\Vert \partial _2u\Vert _{H^3}^{\frac{1}{4}} \Vert \partial _3u\Vert _{H^3}^{\frac{1}{4}})} \\\le & {} {\displaystyle C(\Vert b\Vert ^2_{H^3}+\Vert u\Vert _{H^3}^{\frac{1}{2}}\Vert b\Vert _{H^3}^{\frac{3}{2}})(\Vert \partial _2u\Vert ^2_{H^3}+\Vert \partial _3u\Vert ^2_{H^3}+\Vert \partial _1b\Vert ^2_{H^3}+ \Vert \partial _3b\Vert ^2_{H^2}).} \end{aligned}$$

Similarly, it holds that

$$\begin{aligned} {\displaystyle \mathcal {J}_{1,2,3}} \le {\displaystyle C\left( \Vert b\Vert ^2_{H^3}+\Vert u\Vert _{H^3}^{\frac{1}{2}}\Vert b\Vert _{H^3}^{\frac{3}{2}}\right) \left( \Vert \partial _2u\Vert ^2_{H^3}+\Vert \partial _3u\Vert ^2_{H^3}+\Vert \partial _1b\Vert ^2_{H^3}+ \Vert \partial _3b\Vert ^2_{H^2}\right) .} \end{aligned}$$

Thanks to Lemma 2.1 and Cauchy’s inequality, we deduce that

$$\begin{aligned} {\displaystyle \mathcal {J}_{1,2,4}}\le & {} {\displaystyle C\Vert b_3\Vert _{L^2}^{\frac{1}{4}}\Vert \partial _2b_3\Vert _{L^2}^{\frac{1}{4}}\Vert \partial _3b_3\Vert _{L^2}^{\frac{1}{4}}\Vert \partial _2\partial _3b_3\Vert _{L^2}^{\frac{1}{4}} \Vert \partial _2^3b\Vert _{L^2}^{\frac{1}{2}}\Vert \partial _1\partial _2^3b\Vert _{L^2}^{\frac{1}{2}}\Vert \partial _3\partial _2^3u\Vert _{L^2}}\\\le & {} {\displaystyle C\Vert b\Vert _{H^3}\Vert \partial _1b\Vert _{H^3}^{\frac{1}{2}}\Vert \partial _3b\Vert _{H^2}^{\frac{1}{2}}\Vert \partial _2u\Vert _{H^3} }\\\le & {} {\displaystyle C\Vert b\Vert _{H^3}(\Vert \partial _2u\Vert ^2_{H^3}+\Vert \partial _1b\Vert ^2_{H^3}+\Vert \partial _3b\Vert ^2_{H^2}). } \end{aligned}$$

It follows from the above estimates for $\mathcal {I}_4$ and Cauchy’s inequality that

$$\begin{aligned} {\displaystyle \mathcal {I}_4 }\le & {} {\displaystyle C(\Vert u\Vert ^{\frac{1}{2}}_{H^3}\Vert b\Vert ^{\frac{1}{2}}_{H^3}+\Vert u\Vert ^{\frac{1}{2}}_{H^3}\Vert b\Vert ^{\frac{3}{2}}_{H^3}+\Vert b\Vert _{H^3}+\Vert b\Vert ^2_{H^3})(\Vert \partial _2 u\Vert ^2_{H^3}+\Vert \partial _3 u\Vert ^2_{H^3}}\nonumber \\&+\,{\displaystyle \Vert \partial _1 b\Vert ^2_{H^3}+\Vert \partial _3 b\Vert ^2_{H^2})+\mathcal {J}_{1, 1}.} \end{aligned}$$

(2.11)

Next we estimate $\mathcal {I}_5$. $\mathcal {I}_5$ is rewritten as

$$\begin{aligned} {\displaystyle \mathcal {I}_5}={\displaystyle \sum \limits _{i=1}^3\int \limits _{\mathbb {R}^3}(\partial _i^3(b\cdot \nabla u)-b\cdot \partial _i^3\nabla u)\cdot \partial _i^3b\;dx =: \mathcal {I}_{5,1}+\mathcal {I}_{5,2}+\mathcal {I}_{5,3}.} \end{aligned}$$

Hölder’s inequality and embedding theorem yield

$$\begin{aligned} {\displaystyle \mathcal {I}_{5,1}}= & {} {\displaystyle \sum \limits _{k=1}^3\mathcal {C}_3^k\int \limits _{\mathbb {R}^3}\partial _1^kb\cdot \nabla \partial _1^{3-k}u\cdot \partial _1^3b\;dx}\\\le & {} {\displaystyle C\Vert \partial _1^kb\Vert _{L^3}\Vert \nabla \partial _1^{3-k}u\Vert _{L^2}\Vert \partial _1^3b\Vert _{L^6}}\\\le & {} {\displaystyle C\Vert u\Vert _{H^3}\Vert \partial _1b\Vert _{H^3}^2} \end{aligned}$$

and

$$\begin{aligned} {\displaystyle \mathcal {I}_{5,3}}= & {} {\displaystyle \sum \limits _{k=1}^3\mathcal {C}_3^k\int \limits _{\mathbb {R}^3}\partial _3^kb\cdot \nabla \partial _3^{3-k}u\cdot \partial _3^3b\;dx}\\\le & {} {\displaystyle C\sum \limits _{k=1}^2\Vert \partial _3^kb\Vert _{L^3}\Vert \nabla \partial _3^{3-k}u\Vert _{L^6}\Vert \partial _3^3b\Vert _{L^2}+\Vert \nabla u\Vert _{L^\infty }\Vert \partial _3^3b\Vert _{L^2}^2}\\\le & {} {\displaystyle C(\Vert b\Vert _{H^3}\Vert \partial _3b\Vert _{H^2}\Vert \partial _3u\Vert _{H^3}+\Vert \partial _3b\Vert _{H^2}^2\Vert u\Vert _{H^3})}\\\le & {} {\displaystyle C(\Vert u\Vert _{H^3}+\Vert b\Vert _{H^3})(\Vert \partial _3u\Vert ^2_{H^3}+\Vert \partial _3b\Vert _{H^2}^2).} \end{aligned}$$

To deal with $I_{5,2}$, we further decomposed it into

$$\begin{aligned} {\displaystyle \mathcal {I}_{5,2}}= & {} {\displaystyle \int \limits _{\mathbb {R}^3}\left\{ [\partial _2^3(b_1\cdot \partial _1u)+\partial _2^3(b_2\cdot \partial _2u) +\partial _2^3(b_3\cdot \partial _3u)] -b\cdot \partial ^3_2\nabla u\right\} \cdot \partial _2^3b\;dx}\\= & {} {\displaystyle \sum \limits _{k=1}^3\mathcal {C}_3^k\int \limits _{\mathbb {R}^3}\partial _2^kb_1\cdot \partial _1\partial _2^{3-k}u\cdot \partial _2^3b\;dx +\sum \limits _{k=1}^3\mathcal {C}_3^k\int \limits _{\mathbb {R}^3}\partial _2^kb_2\cdot \partial _2^{4-k}u\cdot \partial _2^3b\;dx}\\&+\,{\displaystyle \sum \limits _{k=1}^3\mathcal {C}_3^k\int \limits _{\mathbb {R}^3}\partial _2^kb_3\cdot \partial _3\partial _2^{3-k}u\cdot \partial _2^3b\;dx}\\=: & {} {\displaystyle \mathcal {I}_{5,2,1}+\mathcal {I}_{5,2,2}+\mathcal {I}_{5,2,3}.} \end{aligned}$$

We can estimate $\mathcal {I}_{5,2,1}$ and $\mathcal {I}_{5,2,2}$ easily; using integration by parts and Lemma 2.1, we have

$$\begin{aligned} {\displaystyle \mathcal {I}_{5,2,1}}= & {} {\displaystyle -\sum \limits _{k=1}^3\mathcal {C}_3^k\int \limits _{\mathbb {R}^3}\partial _1\partial _2^kb_1\cdot \partial _2^{3-k}u\cdot \partial _2^3b\;dx -\sum \limits _{k=1}^3\mathcal {C}_3^k\int \limits _{\mathbb {R}^3}\partial _2^kb_1\cdot \partial _2^{3-k}u\cdot \partial _1\partial _2^3b\;dx}\\\le & {} {\displaystyle C \sum \limits _{k=1}^3\Vert \partial _1\partial _2^kb_1\Vert _{L^2}\Vert \partial _2^{3-k}u\Vert _{L^2}^{\frac{1}{4}}\Vert \partial _2\partial _2^{3-k}u\Vert _{L^2}^{\frac{1}{4}}\Vert \partial _3\partial _2^{3-k}u\Vert _{L^2}^{\frac{1}{4}} \Vert \partial _2\partial _3\partial _2^{3-k}u\Vert _{L^2}^{\frac{1}{4}}\Vert \partial _2^3b\Vert _{L^2}^{\frac{1}{2}}\Vert \partial _1\partial _2^3b\Vert _{L^2}^{\frac{1}{2}} }\\&+\,{\displaystyle C \sum \limits _{k=1}^3\Vert \partial _1\partial _2^3b\Vert _{L^2}\Vert \partial _2^kb_1\Vert _{L^2}^{\frac{1}{2}}\Vert \partial _1\partial _2^kb_1\Vert _{L^2}^{\frac{1}{2}} \Vert \partial _2^{3-k}u\Vert _{L^2}^{\frac{1}{4}}\Vert \partial _2\partial _2^{3-k}u\Vert _{L^2}^{\frac{1}{4}}\Vert \partial _3\partial _2^{3-k}u\Vert _{L^2}^{\frac{1}{4}} \Vert \partial _2\partial _3\partial _2^{3-k}u\Vert _{L^2}^{\frac{1}{4}}}\\\le & {} {\displaystyle C\Vert b\Vert _{H^3}^{\frac{1}{2}}\Vert \partial _1b\Vert _{H^3}^{\frac{3}{2}}\Vert u\Vert _{H^3}^{\frac{1}{2}}\Vert \partial _2u\Vert _{H^3}^{\frac{1}{2}}}\\\le & {} {\displaystyle C\Vert b\Vert _{H^3}^{\frac{1}{2}}\Vert u\Vert _{H^3}^{\frac{1}{2}}(\Vert \partial _2u\Vert ^2_{H^3}+\Vert \partial _1b\Vert ^2_{H^3}).} \end{aligned}$$

Thanks to $\nabla \cdot b=0$ and Hölder’s inequality, it yields

$$\begin{aligned} {\displaystyle \mathcal {I}_{5,2,2}}= & {} {\displaystyle -\sum \limits _{k=1}^2\mathcal {C}_3^k\int \limits _{\mathbb {R}^3}\partial _2^{k-1}(\partial _1b_1+\partial _3b_3)\cdot \partial _2^{3-k}\partial _2u\cdot \partial _2^3b\;dx +\int \limits _{\mathbb {R}^3}\partial _2^3b_2\cdot \partial _2u\cdot \partial _2^3b\;dx}\\\le & {} {\displaystyle C\sum \limits _{k=1}^2\Vert \partial _2^{k-1}(\partial _1b_1+\partial _3b_3)\Vert _{L^3}\Vert \partial _2^{3-k}\partial _2u\Vert _{L^6}\Vert \partial _2^3b\Vert _{L^2}}\\&{\displaystyle +\,\Vert \partial _2^2(\partial _1b_1+\partial _3b_3)\Vert _{L^2}\Vert \partial _2u\Vert _{L^\infty }\Vert \partial _2^3b\Vert _{L^2}}\\\le & {} {\displaystyle C\Vert b\Vert _{H^3}(\Vert \partial _2u\Vert ^2_{H^3}+\Vert \partial _1b\Vert ^2_{H^3}+\Vert \partial _3b\Vert ^2_{H^2}).} \end{aligned}$$

$\mathcal {I}_{5,2,3}$ cannot be estimate directly; we decomposed it into

$$\begin{aligned} {\displaystyle \mathcal {I}_{5,2,3}}= & {} {\displaystyle \sum \limits _{k=1}^3\mathcal {C}_3^k\int \limits _{\mathbb {R}^3}(\partial _2^kb_3\cdot \partial _3\partial _2^{3-k}u_1\cdot \partial _2^3b_1 +\partial _2^kb_3\cdot \partial _3\partial _2^{3-k}u_2\cdot \partial _2^3b_2+}\\&{\displaystyle \partial _2^kb_3\cdot \partial _3\partial _2^{3-k}u_3\cdot \partial _2^3b_3)\;dx}\\=: & {} {\displaystyle \mathcal {I}_{5,2,3,1}+\mathcal {I}_{5,2,3,2}+\mathcal {I}_{5,2,3,3}.} \end{aligned}$$

We could bound $\mathcal {I}_{5,2,3,2}$ by Hölder’s inequality and the embedding theorem

$$\begin{aligned} {\displaystyle \mathcal {I}_{5,2,3,2}}= & {} {\displaystyle \sum \limits _{k=1}^2\mathcal {C}_3^k\int \limits _{\mathbb {R}^3}\partial _2^kb_3\cdot \partial _3\partial _2^{3-k}u_2\cdot \partial _2^3b_2\;dx +\int \limits _{\mathbb {R}^3}\partial _2^3b_3\cdot \partial _3u_2\cdot \partial _2^3b_2\;dx}\\\le & {} {\displaystyle C\sum \limits _{k=1}^2\Vert \partial _2^kb_3\Vert _{L^3}\Vert \partial _3\partial _2^{3-k}u_2\Vert _{L^6}\Vert \partial _2^3b_2\Vert _{L^2} +\Vert \partial _2^3b_3\Vert _{L^2}\Vert \partial _3u_2\Vert _{L^\infty }\Vert \partial _2^3b_2\Vert _{L^2}}\\\le & {} {\displaystyle C(\Vert \partial _1b\Vert _{H^3}+\Vert \partial _3b\Vert _{H^2})\Vert \partial _3u\Vert _{H^3}\Vert b\Vert _{H^3}}\\\le & {} {\displaystyle C\Vert b\Vert ^2_{H^3}(\Vert \partial _3u\Vert ^2_{H^3}+\Vert \partial _1b\Vert ^2_{H^3}+\Vert \partial _3b\Vert ^2_{H^2}).} \end{aligned}$$

$\mathcal {I}_{5,2,3,3}$ is estimated as by Lemma 2.1

$$\begin{aligned} {\displaystyle \mathcal {I}_{5,2,3,3}}= & {} {\displaystyle \sum \limits _{k=1}^2\mathcal {C}_3^k\int \limits _{\mathbb {R}^3}\partial _2^kb_3\cdot \partial _3\partial _2^{3-k}u_3\cdot \partial _2^3b_3\;dx +\int \limits _{\mathbb {R}^3}\partial _2^3b_3\cdot \partial _3u_3\cdot \partial _2^3b_3\;dx}\\\le & {} {\displaystyle C\sum \limits _{k=1}^2\Vert \partial _2^kb_3\Vert _{L^2}^{\frac{1}{2}}\Vert \partial _3\partial _2^kb_3\Vert _{L^2}^{\frac{1}{2}} \Vert \partial _3\partial _2^{3-k}u_3\Vert _{L^2}^{\frac{1}{2}}\Vert \partial _2\partial _3\partial _2^{3-k}u_3\Vert _{L^2}^{\frac{1}{2}}}\\&\times \,{\displaystyle \Vert \partial _2^3b_3\Vert _{L^2}^{\frac{1}{2}}\Vert \partial _1\partial _2^3b_3\Vert _{L^2}^{\frac{1}{2}} +|\mathcal {I}_{4,2,2,2}|}\\\le & {} {\displaystyle C\Vert \partial _1b\Vert _{H^3}^{\frac{1}{2}}\Vert \partial _3b\Vert _{H^2}^{\frac{1}{2}}\Vert \partial _3u\Vert _{H^3}\Vert b\Vert _{H^3}+|\mathcal {I}_{4,2,2,2}|}\\\le & {} {\displaystyle C\Vert b\Vert _{H^3}(\Vert \partial _3u\Vert ^2_{H^3}+\Vert \partial _1b\Vert ^2_{H^3}+\Vert \partial _3b\Vert ^2_{H^2})+|\mathcal {I}_{4,2,2,2}|.} \end{aligned}$$

The last term $\mathcal {I}_{5,2,3,1}$ contains a part, which cannot be directly handled

$$\begin{aligned} {\displaystyle \mathcal {I}_{5,2,3,1}}= & {} {\displaystyle \sum \limits _{k=1}^2\mathcal {C}_3^k\int \limits _{\mathbb {R}^3}\partial _2^kb_3\cdot \partial _3\partial _2^{3-k}u_1\cdot \partial _2^3b_1\;dx +\int \limits _{\mathbb {R}^3}\partial _2^3b_3\cdot \partial _3u_1\cdot \partial _2^3b_1\;dx}\\\le & {} {\displaystyle C\sum \limits _{k=1}^2\Vert \partial _2^kb_3\Vert _{L^2}^{\frac{1}{2}}\Vert \partial _3\partial _2^kb_3\Vert _{L^2}^{\frac{1}{2}} \Vert \partial _3\partial _2^{3-k}u_1\Vert _{L^2}^{\frac{1}{2}}\Vert \partial _2\partial _3\partial _2^{3-k}u_1\Vert _{L^2}^{\frac{1}{2}}}\\&\times \,{\displaystyle \Vert \partial _2^3b_1\Vert _{L^2}^{\frac{1}{2}}\Vert \partial _1\partial _2^3b_1\Vert _{L^2}^{\frac{1}{2}} +K_1}\\\le & {} {\displaystyle C\Vert \partial _1b\Vert _{H^3}^{\frac{1}{2}}\Vert \partial _3b\Vert _{H^2}^{\frac{1}{2}}\Vert \partial _3u\Vert _{H^3}\Vert b\Vert _{H^3}+K_1 }\\\le & {} {\displaystyle C\Vert b\Vert _{H^3}(\Vert \partial _3u\Vert ^2_{H^3}+\Vert \partial _1b\Vert ^2_{H^3}+\Vert \partial _3b\Vert ^2_{H^2})+K_1. } \end{aligned}$$

To estimate $\mathcal {K}_1$, we shall use the special structure of the equation for b in (1.3) again

$$\begin{aligned} {\displaystyle \partial _3u_1=\partial _tb_1-\nu _1\partial _1^2b_1+u\cdot \nabla b_1-b\cdot \nabla u_1.} \end{aligned}$$

We can write $\mathcal {K}_1$ as

$$\begin{aligned} {\displaystyle \mathcal {K}_1}= & {} {\displaystyle \int \limits _{\mathbb {R}^3}\partial _2^3b_3\cdot (\partial _tb_1-\nu _1\partial _1^2b_1+u\cdot \nabla b_1-b\cdot \nabla u_1)\cdot \partial _2^3b_1\;dx}\\=: & {} {\displaystyle \mathcal {K}_{1,1}+\mathcal {K}_{1,2}+\mathcal {K}_{1,3}+\mathcal {K}_{1,4}.} \end{aligned}$$

We can handle $\mathcal {K}_{1,2}\,\mathcal {K}_{1,3}$ and $\mathcal {K}_{1,4}$ as $\mathcal {J}_2$, $\mathcal {J}_3$ and $\mathcal {J}_4$

$$\begin{aligned} \displaystyle \mathcal {K}_{1,2}\le & {} C\Vert b\Vert _{H^3}\Vert \partial _1b\Vert _{H^3}^2,\\ {\displaystyle \mathcal {K}_{1,3}}\le & {} {\displaystyle C\Vert u\Vert _{H^3}^{\frac{1}{2}}\Vert \partial _2u\Vert _{H^3}^{\frac{1}{4}}\Vert \partial _3u\Vert _{H^3}^{\frac{1}{4}} \Vert b\Vert _{H^3}^{\frac{3}{2}}\Vert \partial _1b\Vert _{H^3}\Vert \partial _3b\Vert _{H^2}^{\frac{1}{2}} }\\\le & {} {\displaystyle C\Vert u\Vert _{H^3}^{\frac{1}{2}}\Vert b\Vert _{H^3}^{\frac{3}{2}}(\Vert \partial _2u\Vert ^2_{H^3}+\Vert \partial _3u\Vert ^2_{H^3} +\Vert \partial _1b\Vert ^2_{H^3}+\Vert \partial _3b\Vert ^2_{H^2}),}\\ {\displaystyle \mathcal {K}_{1,4}}\le & {} {\displaystyle C\Vert u\Vert _{H^3}^{\frac{1}{2}}\Vert \partial _2u\Vert _{H^3}^{\frac{1}{4}}\Vert \partial _3u\Vert _{H^3}^{\frac{1}{4}} \Vert b\Vert _{H^3}^{\frac{3}{2}}\Vert \partial _1b\Vert _{H^3}\Vert \partial _3b\Vert _{H^2}^{\frac{1}{2}} }\\\le & {} {\displaystyle C\Vert u\Vert _{H^3}^{\frac{1}{2}}\Vert b\Vert _{H^3}^{\frac{3}{2}}(\Vert \partial _2u\Vert ^2_{H^3}+\Vert \partial _3u\Vert ^2_{H^3}+ \Vert \partial _1b\Vert ^2_{H^3}+\Vert \partial _3b\Vert ^2_{H^2}). } \end{aligned}$$

$\mathcal {K}_{1, 1}$ could be dealt by integration by parts

$$\begin{aligned} {\displaystyle \mathcal {K}_{1,1}}= & {} {\displaystyle \frac{d}{dt}\int \limits _{\mathbb {R}^3}b_1\cdot \partial _2^3b_3\cdot \partial _2^3b_1\;dx -\int \limits _{\mathbb {R}^3}b_1\cdot \partial _t(\partial _2^3b_3\cdot \partial _2^3b_1)\;dx =: \mathcal {K}_{1,1,1}+\mathcal {K}_{1,1,2}.} \end{aligned}$$

According to the following equations

$$\begin{aligned} \left\{ \begin{array}{lll} {\displaystyle \partial _tb_1=\nu _1\partial _1^2b_1-u\cdot \nabla b_1+b\cdot \nabla u_1+\partial _3u_1,}\\ {\displaystyle \partial _tb_3=\nu _1\partial _1^2b_3-u\cdot \nabla b_3+b\cdot \nabla u_3+\partial _3u_3, } \end{array} \right. \end{aligned}$$

we rewrite $\mathcal {K}_{1,1,2}$ as

$$\begin{aligned} {\displaystyle \mathcal {K}_{1,1,2}}= & {} {\displaystyle -\int \limits _{\mathbb {R}^3}b_1\cdot \partial _2^3b_3\cdot \partial _2^3(\nu _1\partial _1^2b_1-u\cdot \nabla b_1+b\cdot \nabla u_1+\partial _3u_1)\;dx}\\&{\displaystyle -\,\int \limits _{\mathbb {R}^3}b_1\cdot \partial _2^3b_1\cdot \partial _2^3(\nu _1\partial _1^2b_3-u\cdot \nabla b_3+b\cdot \nabla u_3+\partial _3u_3)\;dx}\\= & {} {\displaystyle \int \limits _{\mathbb {R}^3}\{ b_1[\partial _2^3(u\cdot \nabla b_3)\partial _2^3b_1+\partial _2^3(u\cdot \nabla b_1)\partial _2^3b_3] -\nu _1b_1[\partial _2^3\partial _1^2b_3\partial _2^3b_1+\partial _2^3\partial _1^2b_1\partial _2^3b_3]}\\&{\displaystyle -\,b_1[\partial _2^3(b\cdot \nabla u_3)\partial _2^3b_1+\partial _2^3(b\cdot \nabla u_1)\partial _2^3b_3] -[\partial _2^3\partial _3u_3\partial _2^3b_1+\partial _2^3\partial _3u_1\partial _2^3b_3] \}\;dx}\\=: & {} {\displaystyle \mathcal {K}_{1,1,2,1}+\mathcal {K}_{1,1,2,2}+\mathcal {K}_{1,1,2,3}+\mathcal {K}_{1,1,2,4}.} \end{aligned}$$

Similar to the terms $\mathcal {J}_{1,2,1}$, $\mathcal {J}_{1,2,2}$, $\mathcal {J}_{1,2,3}$, $\mathcal {J}_{1,2,4}$, we have

$$\begin{aligned} {\displaystyle \mathcal {K}_{1,1,2,1}}\le & {} {\displaystyle C\Vert b\Vert _{H^3}^2\Vert \partial _1b\Vert _{H^3}^{\frac{1}{2}}\Vert \partial _3b\Vert _{H^2}^{\frac{1}{2}}\Vert \partial _2u\Vert _{H^3} +\Vert b\Vert _{H^3}^{\frac{3}{2}}\Vert \partial _1b\Vert _{H^3}\Vert \partial _3b\Vert _{H^2}^{\frac{1}{2}}\Vert u\Vert ^{\frac{1}{2}}_{H^3} \Vert \partial _2u\Vert ^{\frac{1}{4}}_{H^3}\Vert \partial _3u\Vert ^{\frac{1}{4}}_{H^3} }\\\le & {} {\displaystyle C(\Vert b\Vert ^2_{H^3}+\Vert u\Vert ^{\frac{1}{2}}_{H^3}\Vert b\Vert _{H^3}^{\frac{3}{2}})(\Vert \partial _2u\Vert ^2_{H^3})+\Vert \partial _3u\Vert ^{2}_{H^3}+\Vert \partial _1b\Vert ^2_{H^3}+ \Vert \partial _3b\Vert ^2_{H^2}),}\\ {\displaystyle K_{1,1,2,2}}\le & {} {\displaystyle C\Vert b\Vert _{H^3}\Vert \partial _1b\Vert _{H^3}^2,}\\ {\displaystyle K_{1,1,2,3}}\le & {} {\displaystyle C\Vert b\Vert _{H^3}^2\Vert \partial _1b\Vert _{H^3}^{\frac{1}{2}}\Vert \partial _3b\Vert _{H^2}^{\frac{1}{2}}\Vert \partial _2u\Vert _{H^3} +\Vert b\Vert _{H^3}^{\frac{3}{2}}\Vert \partial _1b\Vert _{H^3}\Vert \partial _3b\Vert _{H^2}^{\frac{1}{2}}\Vert u\Vert _{H^3}^\frac{1}{2}\Vert \partial _2u\Vert _{H^3}^{\frac{1}{4}} \Vert \partial _3u\Vert _{H^3}^{\frac{1}{4}} }\\\le & {} {\displaystyle C(\Vert b\Vert ^2_{H^3}+\Vert u\Vert ^{\frac{1}{2}}_{H^3}\Vert b\Vert _{H^3}^{\frac{3}{2}})(\Vert \partial _2u\Vert ^2_{H^3})+\Vert \partial _3u\Vert ^{2}_{H^3}+\Vert \partial _1b\Vert ^2_{H^3}+ \Vert \partial _3b\Vert ^2_{H^2}),}\\ {\displaystyle K_{1,1,2,4}}\le & {} {\displaystyle C\Vert b\Vert _{H^3}\Vert \partial _1b\Vert _{H^3}^{\frac{1}{2}}\Vert \partial _3b\Vert _{H^2}^{\frac{1}{2}}\Vert \partial _3u\Vert _{H^3},}\\\le & {} {\displaystyle C\Vert b\Vert _{H^3}(\Vert \partial _3u\Vert ^2_{H^3}+\Vert \partial _1b\Vert ^2_{H^3}+\Vert \partial _3b\Vert ^2_{H^2}).} \end{aligned}$$

Thanks to the above estimates for $\mathcal {I}_5$ and Cauchy’s inequality, we deduce that

$$\begin{aligned} \mathcal {I}_5 \le C\left( \Vert u\Vert ^{\frac{1}{2}}_{H^3}\Vert b\Vert ^{\frac{1}{2}}_{H^3}+\Vert u\Vert ^{\frac{1}{2}}_{H^3}\Vert b\Vert ^{\frac{3}{2}}_{H^3}+\Vert b\Vert _{H^3}+\Vert b\Vert ^2_{H^3})(\Vert \partial _2 u\Vert ^2_{H^3}+\Vert \partial _3 u\Vert ^2_{H^3}+ \Vert \partial _1 b\Vert ^2_{H^3}+\Vert \partial _3 b\Vert ^2_{H^2}\right) . \end{aligned}$$

(2.12)

It remains to estimate $\mathcal {I}_6$. We rewrite $\mathcal {I}_6$ as

$$\begin{aligned} {\displaystyle \mathcal {I}_6=-\sum \limits _{i=1}^3\int \limits _{\mathbb {R}^3}\partial _i^3(u\cdot \nabla v)\cdot \partial _i^3v\;dx=:\mathcal {I}_{6,1}+\mathcal {I}_{6,2}+\mathcal {I}_{6,3}.} \end{aligned}$$

Lemma 2.1 and Cauchy’s inequality entail that

$$\begin{aligned} {\displaystyle \mathcal {I}_{6,1}}= & {} {\displaystyle -\int \limits _{\mathbb {R}^3}\partial _1^3(u_1\cdot \partial _1v+u_2\cdot \partial _2v+u_3\cdot \partial _3v)\cdot \partial _1^3v\;dx}\\= & {} {\displaystyle -\sum \limits _{k=1}^3\int \limits _{\mathbb {R}^3}\partial _1^ku_1\cdot \partial _1^{4-k}v\cdot \partial _1^3v\;dx}\\&{\displaystyle -\,\int \limits _{\mathbb {R}^3}(\partial _1^ku_2\cdot \partial _1^{3-k}\partial _2v+\partial _1^ku_3\cdot \partial _1^{3-k}\partial _3v)\cdot \partial _1^3v\;dx}\\\le & {} C{\displaystyle \Vert \partial _1^{k-1}(\partial _2u_2+\partial _3u_3)\Vert _{L^2}^{\frac{1}{2}}\Vert \partial _1\partial _1^{k-1} (\partial _2u_2+\partial _3u_3)\Vert _{L^2}^{\frac{1}{2}}}\\&\times \,{\displaystyle \Vert \partial _1^{4-k}v\Vert _{L^2}^{\frac{1}{2}}\Vert \partial _2\partial _1^{4-k}v\Vert _{L^2}^{\frac{1}{2}} \Vert \partial _1^3v\Vert _{L^2}^{\frac{1}{2}}\Vert \partial _3\partial _1^3v\Vert _{L^2}^{\frac{1}{2}}}\\&+\,{\displaystyle \Vert \partial _1^ku_2\Vert _{L^2}^\frac{1}{2}\Vert \partial _2\partial _1^ku_2\Vert _{L^2}^\frac{1}{2} \Vert \partial _2\partial _1^{3-k}v_2\Vert _{L^2}^\frac{1}{2}\Vert \partial _1\partial _2\partial _1^{3-k}v_2\Vert _{L^2}^\frac{1}{2} \Vert \partial _1^3v\Vert _{L^2}^{\frac{1}{2}}\Vert \partial _3\partial _1^3v\Vert _{L^2}^{\frac{1}{2}}}\\&+\,{\displaystyle \Vert \partial _1^ku_3\Vert _{L^2}^\frac{1}{2}\Vert \partial _2\partial _1^ku_3\Vert _{L^2}^\frac{1}{2} \Vert \partial _3\partial _1^{3-k}v_2\Vert _{L^2}^\frac{1}{2}\Vert \partial _1\partial _3\partial _1^{3-k}v_2\Vert _{L^2}^\frac{1}{2} \Vert \partial _1^3v\Vert _{L^2}^{\frac{1}{2}}\Vert \partial _3\partial _1^3v\Vert _{L^2}^{\frac{1}{2}}}\\\le & {} {\displaystyle C(\Vert \partial _2u\Vert _{H^3}+\Vert \partial _3u\Vert _{H^3})\Vert v\Vert _{H^3}\Vert \partial _2v\Vert _{H^3}^{\frac{1}{2}}\Vert \partial _3v\Vert _{H^3}^{\frac{1}{2}}}\\&+\,{\displaystyle \Vert u\Vert _{H^3}^{\frac{1}{2}}\Vert \partial _2u\Vert _{H^3}^{\frac{1}{2}}\Vert v\Vert _{H^3}^{\frac{1}{2}}\Vert \partial _2v\Vert _{H^3}\Vert \partial _3v\Vert _{H^3}^{\frac{1}{2}} +\Vert u\Vert _{H^3}^{\frac{1}{2}}\Vert \partial _2u\Vert _{H^3}^{\frac{1}{2}}\Vert v\Vert _{H^3}^{\frac{1}{2}}\Vert \partial _3v\Vert _{H^3}^{\frac{3}{2}}}\\\le & {} {\displaystyle C\left( \Vert v\Vert _{H^3}+\Vert u\Vert _{H^3}^{\frac{1}{2}}\Vert v\Vert _{H^3}^{\frac{1}{2}}\right) \left( \Vert \partial _2u\Vert ^2_{H^3}+\Vert \partial _3u\Vert ^2_{H^3}+\Vert \partial _2v\Vert ^2_{H^3} +\Vert \partial _3v\Vert ^2_{H^3}\right) .} \end{aligned}$$

By Hölder inequality and the embedding theorem, we have

$$\begin{aligned} {\displaystyle \mathcal {I}_{6,2}}= & {} {\displaystyle -\sum \limits _{k=1}^3\mathcal {C}_3^k\int \limits _{\mathbb {R}^3}\partial _2^ku\cdot \nabla \partial _2^{3-k}v\cdot \partial _2^3v\;dx}\\\le & {} {\displaystyle C\Vert \partial _2^ku\Vert _{L^3}\Vert \nabla \partial _2^{3-k}v\Vert _{L^2}\Vert \partial _2^3v\Vert _{L^6}}\\\le & {} {\displaystyle C\Vert \partial _2u\Vert _{H^3}\Vert v\Vert _{H^3}\Vert \partial _2v\Vert _{H^3}}\\\le & {} {\displaystyle C\Vert v\Vert _{H^3}(\Vert \partial _2u\Vert ^2_{H^3}+\Vert \partial _2v\Vert ^2_{H^3})} \end{aligned}$$

and

$$\begin{aligned} {\displaystyle \mathcal {I}_{6,3}}= & {} {\displaystyle -\sum \limits _{k=1}^3\mathcal {C}_3^k\int \limits _{\mathbb {R}^3}\partial _3^ku\cdot \nabla \partial _3^{3-k}v\cdot \partial _3^3v\;dx}\\\le & {} {\displaystyle C\Vert \partial _3^ku\Vert _{L^3}\Vert \nabla \partial _3^{3-k}v\Vert _{L^2} \Vert \partial _3^3v\Vert _{L^6}}\\\le & {} {\displaystyle C\Vert \partial _3u\Vert _{H^3}\Vert v\Vert _{H^3}\Vert \partial _3v\Vert _{H^3}}\\\le & {} {\displaystyle C\Vert v\Vert _{H^3}(\Vert \partial _3u\Vert ^2_{H^3}+\Vert \partial _3v\Vert ^2_{H^3}).} \end{aligned}$$

The above estimates for $\mathcal {I}_6$ and Cauchy’s inequality give

$$\begin{aligned} \mathcal {I}_6 \le C\left( \Vert u\Vert ^{\frac{1}{2}}_{H^3}\Vert v\Vert ^{\frac{1}{2}}_{H^3}+\Vert v\Vert _{H^3}\right) \left( \Vert \partial _2 u\Vert ^2_{H^3}+\Vert \partial _3 u\Vert ^2_{H^3}+ \Vert \partial _2 v\Vert ^2_{H^3}+\Vert \partial _3 v\Vert ^2_{H^3}\right) . \end{aligned}$$

(2.13)

By integration by part and Cauchy’s inequality, we arrive at

$$\begin{aligned} {\displaystyle \mathcal {I}_7}= & {} {\displaystyle 4\chi \sum \limits _{i=1}^3\int \limits _{\mathbb {R}^3} \partial ^3_i (\nabla \times u) \partial _i^3 v\;dx}\nonumber \\\le & {} {\displaystyle \chi (\Vert \nabla \times \partial ^3_iu\Vert ^2_{L^2}+4\Vert \partial ^3_iv\Vert ^2_{L^2})}\nonumber \\= & {} {\displaystyle \chi (\Vert \nabla \partial ^3_iu\Vert ^2_{L^2}+4\Vert \partial ^3_iv\Vert ^2_{L^2}).} \end{aligned}$$

(2.14)

Integrating (2.7) with respect to time and combining (2.6), (2.8)–(2.14) yield

$$\begin{aligned} E_0(t)\le & {} {\displaystyle C \mathscr {E}_0+\int \limits _0^t(\mathcal {I}_1+\mathcal {I}_2+\mathcal {I}_3+\mathcal {I}_4+\mathcal {I}_5+\mathcal {I}_6+\mathcal {I}_7)d\tau } \nonumber \\\le & {} C \mathscr {E}_0+C \int \limits _0^t\left( \Vert u\Vert _{H^3}+\Vert v\Vert _{H^3}+\Vert b\Vert _{H^3}+\Vert b\Vert ^2_{H^3}+\Vert u\Vert ^{\frac{1}{2}}_{H^3}\Vert b\Vert ^{\frac{1}{2}}_{H^3} +\Vert u\Vert ^{\frac{1}{2}}_{H^3}\Vert b\Vert ^{\frac{3}{2}}_{H^3}\right. \nonumber \\&\left. + \,\Vert u\Vert ^{\frac{1}{2}}_{H^3}\Vert v\Vert ^{\frac{1}{2}}_{H^3}\right) \cdot \left( \Vert \partial _2 u\Vert ^2_{H^3}+\Vert \partial _3 u\Vert ^2_{H^3}+\Vert \partial _2 v\Vert ^2_{H^3}+\Vert \partial _3 v\Vert ^2_{H^3}+ \Vert \partial _1 b\Vert ^2_{H^3}+\Vert \partial _3 b\Vert ^2_{H^2}\right) d\tau \nonumber \\&+\,\int \limits _0^t\mathcal {J}_{1, 1}d\tau \nonumber \\\le & {} C \mathscr {E}_0+C\left( E^{\frac{3}{2}}_0(t)+E^{2}_0(t)+E^{\frac{1}{2}}_0(t)E_1(t)+E_0(t)E_1(t)\right) \nonumber \\&+\, 3\int \limits _{R^3}b_3\cdot |\partial _2^3b|^2\;dx -3 \int \limits _{\mathbb {R}^3}b_3(x,0)\cdot |\partial _2^3b|^2(x,0)\;dx \nonumber \\\le & {} C \mathscr {E}_0+ C\left( E^{\frac{3}{2}}_0(t)+E^{2}_0(t)+E^2_1(t)\right) \nonumber \\&+\, C\left( \Vert b_3(0)\Vert _{L^\infty }\Vert \partial _2^3b(0)\Vert _{L^2}^2+\Vert b_3(t)\Vert _{L^\infty }\Vert \partial _2^3b(t)\Vert _{L^2}^2\right) \nonumber \\\le & {} C \left( \mathscr {E}_0+\mathscr {E}^{\frac{3}{2}}_0+ E^{\frac{3}{2}}_0(t)+E^{2}_0(t)+E^2_1(t)\right) . \end{aligned}$$

(2.15)

This completes the proof of (2.1).

2.2 Proof of (2.2)

The main purpose of this section is to prove (2.2), namely

$$\begin{aligned} {\displaystyle E_1(t)\le C \left( \mathscr {E}_0+E_0(t)+E_0(t)^{\frac{3}{2}}+E_0(t)^{2}+E_1(t)^{2}\right) .} \end{aligned}$$

To bound the norm $\Vert \partial _3b\Vert _{H^2}$, we need to bound the norm $\Vert \partial _3b\Vert _{L^2}+\Vert \partial _3b\Vert _{\dot{H}^2}$. We first estimate the $L^2$-norm of $\partial _3b$. To this end, we write the equation of u in (1.3) as

$$\begin{aligned} {\displaystyle \partial _3b=\partial _tu-\mu _2\partial _2^2u-\mu _3\partial _3^2u-\chi \Delta u+u\cdot \nabla u-b\cdot \nabla b+\nabla p-2\chi \nabla \times v.} \end{aligned}$$

Taking the inner product of the above equation and $\partial _3b$, we obtain

$$\begin{aligned} {\displaystyle \Vert \partial _3b\Vert _{L^2}^2}= & {} {\displaystyle \int \limits _{\mathbb {R}^3}\partial _tu\cdot \partial _3b\;dx -\mu _2\int \limits _{\mathbb {R}^3}\partial _2^2u\cdot \partial _3b\;dx}\nonumber \\&-\,{\displaystyle \mu _3\int \limits _{\mathbb {R}^3}\partial _3^2u\cdot \partial _3b\;dx+\int \limits _{\mathbb {R}^3}u\cdot \nabla u\cdot \partial _3b\;dx-\int \limits _{\mathbb {R}^3}b\cdot \nabla b\cdot \partial _3b\;dx}\nonumber \\&-\,{\displaystyle \chi \int \limits _{\mathbb {R}^3}\Delta u\cdot \partial _3b\;dx-2\chi \int \limits _{\mathbb {R}^3}(\nabla \times v)\cdot \partial _3b\;dx}\nonumber \\=: & {} {\displaystyle \mathcal {L}_1+\mathcal {L}_2+\mathcal {L}_3+\mathcal {L}_4+\mathcal {L}_5+\mathcal {L}_6+\mathcal {L}_7,} \end{aligned}$$

(2.16)

where we have eliminated the pressure term by using $\nabla \cdot b=0$. We handle $\mathcal {L}_1$ by integrating by parts and the equation of b in (1.3)

$$\begin{aligned} {\displaystyle \mathcal {L}_1}= & {} {\displaystyle \frac{d}{dt}\int \limits _{\mathbb {R}^3}u\cdot \partial _3b\;dx -\int \limits _{\mathbb {R}^3}u\cdot \partial _3(\nu _1\partial _1^2b-u\cdot \nabla b+b\cdot \nabla u+\partial _3u)\;dx}\\=: & {} {\displaystyle \mathcal {L}_{1, 0}+\mathcal {L}_{1,1}+\mathcal {L}_{1,2}+\mathcal {L}_{1,3}+\mathcal {L}_{1,4}.} \end{aligned}$$

It follows from Hölder’s inequality and Cauchy’s inequality that

$$\begin{aligned} {\displaystyle \mathcal {L}_{1,1}}= & {} {\displaystyle -\nu _1\int \limits _{\mathbb {R}^3}u\cdot \partial _3\partial _1^2bdx\le C\Vert \partial _3u\Vert _{H^3}\Vert \partial _1b\Vert _{H^3} \le C(\Vert \partial _3u\Vert ^2_{H^3}+\Vert \partial _1b\Vert ^2_{H^3}).} \end{aligned}$$

Thanks to Lemma 2.1 and Cauchy’s inequality, we arrive at

$$\begin{aligned} {\displaystyle \mathcal {L}_{1,2}}= & {} {\displaystyle \int \limits _{\mathbb {R}^3} u\cdot \partial _3u\cdot \nabla b+u\cdot u\cdot \nabla \partial _3b\;dx}\\\le & {} {\displaystyle C\Vert u\Vert _{L^2}^{\frac{1}{4}}\Vert \partial _2u\Vert _{L^2}^{\frac{1}{4}}\Vert \partial _3u\Vert _{L^2}^{\frac{1}{4}}\Vert \partial _2\partial _3u\Vert _{L^2}^{\frac{1}{4}} \Vert \nabla b\Vert _{L^2}^{\frac{1}{2}}\Vert \nabla \partial _1 b\Vert _{L^2}^{\frac{1}{2}}\Vert \partial _3u\Vert _{L^2}}\\&+\,{\displaystyle C\Vert u\Vert _{L^2}^{\frac{1}{2}}\Vert \partial _2u\Vert _{L^2}^{\frac{1}{2}}\Vert u\Vert _{L^2}^{\frac{1}{2}}\Vert \partial _3u\Vert _{L^2}^{\frac{1}{2}} \Vert \nabla \partial _3 b\Vert _{L^2}^{\frac{1}{2}}\Vert \partial _1\nabla \partial _3 b\Vert _{L^2}^{\frac{1}{2}}}\\\le & {} {\displaystyle C\Vert u\Vert _{H^3}^{\frac{1}{2}}\Vert \partial _2u\Vert _{H^3}^{\frac{1}{4}}\Vert \partial _3u\Vert _{H^3}^{\frac{5}{4}}\Vert b\Vert _{H^3}^{\frac{1}{2}}\Vert \partial _1b\Vert _{H^3}^{\frac{1}{2}}}\\&+\,{\displaystyle C\Vert u\Vert _{H^3}\Vert \partial _2u\Vert _{H^3}^{\frac{1}{2}}\Vert \partial _3u\Vert _{H^3}^{\frac{1}{2}}\Vert \partial _3b\Vert _{H^2}}\\\le & {} {\displaystyle C\Vert u\Vert ^{\frac{1}{2}}_{H^3}\Vert b\Vert ^{\frac{1}{2}}_{H^3}(\Vert \partial _2u\Vert ^2_{H^3}+\Vert \partial _3u\Vert ^2_{H^3}+\Vert \partial _1b\Vert ^2_{H^3})}\\&+\,{\displaystyle \Vert u\Vert ^2_{H^3}(\Vert \partial _2u\Vert ^2_{H^3}+\Vert \partial _3u\Vert ^2_{H^3}) +\epsilon \Vert \partial _3b\Vert ^2_{H^2}.} \end{aligned}$$

and

$$\begin{aligned} {\displaystyle \mathcal {L}_{1,3}}= & {} {\displaystyle -\int \limits _{\mathbb {R}^3} u\cdot \partial _3b\cdot \nabla u+u\cdot b\cdot \nabla \partial _3u\;dx}\\\le & {} {\displaystyle C\Vert u\Vert _{L^2}^{\frac{1}{2}}\Vert \partial _2u\Vert _{L^2}^{\frac{1}{2}}\Vert \partial _3b\Vert _{L^2}^{\frac{1}{2}}\Vert \partial _1\partial _3b\Vert _{L^2}^{\frac{1}{2}} \Vert \nabla u\Vert _{L^2}^{\frac{1}{2}}\Vert \nabla \partial _3 u\Vert _{L^2}^{\frac{1}{2}}}\\&+\,{\displaystyle C\Vert u\Vert _{L^2}^{\frac{1}{2}}\Vert \partial _3u\Vert _{L^2}^{\frac{1}{2}}\Vert b\Vert _{L^2}^{\frac{1}{2}}\Vert \partial _1b\Vert _{L^2}^{\frac{1}{2}} \Vert \nabla \partial _3 u\Vert _{L^2}^{\frac{1}{2}}\Vert \partial _1\nabla \partial _2 u\Vert _{L^2}^{\frac{1}{2}}}\\\le & {} {\displaystyle C\Vert u\Vert _{H^3}\Vert \partial _2u\Vert _{H^3}^{\frac{1}{2}}\Vert \partial _3u\Vert _{H^3}^{\frac{1}{2}}\Vert \partial _3b\Vert _{H^2}}\\&+\,{\displaystyle C\Vert u\Vert ^{\frac{1}{2}}_{H^3}\Vert \partial _2u\Vert _{H^3}^{\frac{1}{2}}\Vert \partial _3u\Vert _{H^3}\Vert b\Vert _{H^3}^{\frac{1}{2}}\Vert \partial _1b\Vert _{H^3}^{\frac{1}{2}} }\\\le & {} {\displaystyle C\Vert u\Vert ^{\frac{1}{2}}_{H^3}\Vert b\Vert ^{\frac{1}{2}}_{H^3}(\Vert \partial _2u\Vert ^2_{H^3}+\Vert \partial _3u\Vert ^2_{H^3}+\Vert \partial _1b\Vert ^2_{H^3})}\\&+\,{\displaystyle \Vert u\Vert ^2_{H^3}(\Vert \partial _2u\Vert ^2_{H^3}+\Vert \partial _3u\Vert ^2_{H^3}) +\epsilon \Vert \partial _3b\Vert ^2_{H^2}.} \end{aligned}$$

Integration by parts yields

$$\begin{aligned} {\displaystyle \mathcal {L}_{1,4}}={\displaystyle -\int \limits _{\mathbb {R}^3} u\cdot \partial _3^2u\;dx =\int \limits _{\mathbb {R}^3} \partial _3u\cdot \partial _3u\;dx\le C\Vert \partial _3u\Vert _{H^3}^2.} \end{aligned}$$

Hölder’s inequality and Cauchy’s inequality entail that

$$\begin{aligned} {\displaystyle \mathcal {L}_2=-\mu _2\int \limits _{\mathbb {R}^3}\partial _2^2u\cdot \partial _3b\;dx\le C\Vert \partial _2u\Vert _{H^3}\Vert \partial _3b\Vert _{H^2} \le C\Vert \partial _2u\Vert ^2_{H^3}+\epsilon \Vert \partial _3b\Vert ^2_{H^2}} \end{aligned}$$

and

$$\begin{aligned} {\displaystyle \mathcal {L}_3=-\mu _3\int \limits _{\mathbb {R}^3}\partial _3^2u\cdot \partial _3b\;dx\le C\Vert \partial _3u\Vert _{H^3}\Vert \partial _3b\Vert _{H^2} \le C\Vert \partial _3u\Vert ^2_{H^3}+\epsilon \Vert \partial _3b\Vert ^2_{H^2}.} \end{aligned}$$

We could deal with $\mathcal {L}_4$ and $\mathcal {L}_5$ by using Lemma 2.1 and Cauchy’s inequality

$$\begin{aligned} {\displaystyle \mathcal {L}_4}= & {} {\displaystyle \int \limits _{\mathbb {R}^3}u\cdot \nabla u\cdot \partial _3b\;dx}\\\le & {} {\displaystyle C\Vert u\Vert _{L^2}^{\frac{1}{2}}\Vert \partial _2u\Vert _{L^2}^{\frac{1}{2}}\Vert \nabla u\Vert _{L^2}^{\frac{1}{2}}\Vert \partial _3\nabla u\Vert _{L^2}^{\frac{1}{2}} \Vert \partial _3b\Vert _{L^2}^{\frac{1}{2}}\Vert \partial _1\partial _3b\Vert _{L^2}^{\frac{1}{2}}}\\\le & {} {\displaystyle C\Vert u\Vert _{H^3}\Vert \partial _2u\Vert _{H^3}^{\frac{1}{2}}\Vert \partial _3u\Vert _{H^3}^{\frac{1}{2}}\Vert \partial _3b\Vert _{H^2}}\\\le & {} {\displaystyle C\Vert u\Vert ^2_{H^3}(\Vert \partial _2u\Vert ^2_{H^3}+\Vert \partial _3u\Vert ^2_{H^3})+\epsilon \Vert \partial _3b\Vert ^2_{H^2}} \end{aligned}$$

and

$$\begin{aligned} {\displaystyle \mathcal {L}_5}= & {} {\displaystyle \int \limits _{\mathbb {R}^3}b\cdot \nabla b\cdot \partial _3b\;dx}\\\le & {} {\displaystyle C\Vert b\Vert _{L^2}^{\frac{1}{2}}\Vert \partial _1b\Vert _{L^2}^{\frac{1}{2}}\Vert \nabla b\Vert _{L^2}^{\frac{1}{2}}\Vert \partial _3\nabla b\Vert _{L^2}^{\frac{1}{2}} \Vert \partial _3b\Vert _{L^2}^{\frac{1}{2}}\Vert \partial _2\partial _3b\Vert _{L^2}^{\frac{1}{2}}}\\\le & {} {\displaystyle C\Vert b\Vert _{H^3}\Vert \partial _1b\Vert _{H^3}^{\frac{1}{2}}\Vert \partial _3b\Vert _{H^2}^{\frac{3}{2}}}\\\le & {} {\displaystyle C\Vert b\Vert _{H^3}(\Vert \partial _1b\Vert ^2_{H^3}+\Vert \partial _3b\Vert ^2_{H^2}).} \end{aligned}$$

By integration by parts, Hölder’s inequality and Cauchy’s inequality, it holds that

$$\begin{aligned} {\displaystyle \mathcal {L}_6}= & {} {\displaystyle -\chi \int \limits _{\mathbb {R}^3}(\partial _1^2u\cdot \partial _3b+\partial _2^2u\cdot \partial _3b+\partial _3^2u\cdot \partial _3b)\;dx}\\= & {} {\displaystyle -\chi \int \limits _{\mathbb {R}^3}(\partial _1\partial _3u\cdot \partial _1b+\partial _2^2u\cdot \partial _3b+\partial _3^2u\cdot \partial _3b)\;dx}\\\le & {} {\displaystyle C(\Vert \partial _3u\Vert _{H^3}\Vert \partial _1b\Vert _{H^3}+\Vert \partial _2u\Vert _{H^3}\Vert \partial _3b\Vert _{H^2}+\Vert \partial _3u\Vert _{H^3}\Vert \partial _3b\Vert _{H^2})}\\\le & {} {\displaystyle C(\Vert \partial _2u\Vert ^2_{H^3}+\Vert \partial _3u\Vert ^2_{H^3}+\Vert \partial _1b\Vert ^2_{H^3})+\epsilon \Vert \partial _3b\Vert ^2_{H^2}} \end{aligned}$$

and

$$\begin{aligned} {\displaystyle \mathcal {L}_7}= & {} {\displaystyle -2\chi \int \limits _{\mathbb {R}^3}[(\partial _2v_3-\partial _3v_2)\cdot \partial _3b_1+(\partial _1v_3-\partial _3v_1)\cdot \partial _3b_2 +(\partial _1v_2-\partial _2v_1)\cdot \partial _3b_3]\;dx}\\= & {} {\displaystyle -2\chi \int \limits _{\mathbb {R}^3}[(\partial _3v_2\partial _1b_3-\partial _3v_3\partial _1b_2)+(\partial _2v_3\partial _3b_1 -\partial _2v_1\partial _2b_3) +(\partial _3v_1\partial _3b_2-\partial _3v_2\partial _3b_1)]\;dx}\\\le & {} {\displaystyle C(\Vert \partial _3v\Vert _{H^3}\Vert \partial _1b\Vert _{H^3}+\Vert \partial _2v\Vert _{H^3}\Vert \partial _3b\Vert _{H^2}+\Vert \partial _3v\Vert _{H^3}\Vert \partial _3v\Vert _{H^2})}\\\le & {} {\displaystyle C(\Vert \partial _2v\Vert ^2_{H^3}+\Vert \partial _3v\Vert ^2_{H^3}+\Vert \partial _1b\Vert ^2_{H^3})+\epsilon \Vert \partial _3b\Vert ^2_{H^2}.} \end{aligned}$$

Collecting the estimates for $\mathcal {L}_i(i=1, \ldots , 7)$ and inserting them into (2.16) yield

$$\begin{aligned} {\displaystyle \Vert \partial _3b\Vert _{L^2}^2 }\le & {} {\displaystyle C(\Vert \partial _2u\Vert ^2_{H^3}+\Vert \partial _3u\Vert ^2_{H^3}+\Vert \partial _2v\Vert ^2_{H^3}+\Vert \partial _3v\Vert ^2_{H^3}+\Vert \partial _1b\Vert ^2_{H^3})} \nonumber \\&+\,{\displaystyle C\left( \Vert u\Vert ^{\frac{1}{2}}_{H^3}\Vert b\Vert ^{\frac{1}{2}}_{H^3}+\Vert u\Vert ^2_{H^3}+\Vert b\Vert _{H^3}\right) \left( \Vert \partial _2u\Vert ^2_{H^3}+\Vert \partial _3u\Vert ^2_{H^3}+\Vert \partial _1b\Vert ^2_{H^3} +\Vert \partial _3b\Vert ^2_{H^2}\right) } \nonumber \\&+\,{\displaystyle 7\epsilon \Vert \partial _3b\Vert ^2_{H^2}+ \mathcal {L}_{1, 0}. } \end{aligned}$$

(2.17)

We have finished the $L^2$ estimate of $\partial _3b$. Now we turn to the $\dot{H}^2$ estimate of $\partial _3b$

$$\begin{aligned} {\displaystyle \sum \limits _{i=1}^3\Vert \partial _i^2\partial _3b\Vert _{L^2}^2}= & {} {\displaystyle \sum \limits _{i=1}^3\int \limits _{\mathbb {R}^3}\partial _i^2\partial _tu\cdot \partial _i^2\partial _3b\;dx -\mu _2\sum \limits _{i=1}^3\int \limits _{\mathbb {R}^3}\partial _i^2\partial _2^2u\cdot \partial _i^2\partial _3bdx}\nonumber \\&-\,{\displaystyle \mu _3\sum \limits _{i=1}^3\int \limits _{\mathbb {R}^3}\partial _i^2\partial _3^2u\cdot \partial _i^2\partial _3b\;dx -\chi \sum \limits _{i=1}^3\int \limits _{\mathbb {R}^3}\partial _i^2\Delta u\cdot \partial _i^2\partial _3b\;dx}\nonumber \\&+\,{\displaystyle \sum \limits _{i=1}^3\int \limits _{\mathbb {R}^3}\partial _i^2(u\cdot \nabla u)\cdot \partial _i^2\partial _3b\;dx -\sum \limits _{i=1}^3\int \limits _{\mathbb {R}^3}\partial _i^2(b\cdot \nabla b)\cdot \partial _i^2\partial _3b\;dx}\nonumber \\&-\,{\displaystyle 2\chi \sum \limits _{i=1}^3\int \limits _{\mathbb {R}^3}\partial _i^2(\nabla \times v)\cdot \partial _i^2\partial _3b\;dx}\nonumber \\=: & {} {\displaystyle \mathcal {M}_1+\mathcal {M}_2+\mathcal {M}_3+\mathcal {M}_4+\mathcal {M}_5+\mathcal {M}_6+\mathcal {M}_7.} \end{aligned}$$

(2.18)

Using integration by parts and the equation of b in (1.3) yields

$$\begin{aligned} {\displaystyle \mathcal {M}_1}= & {} {\displaystyle \frac{d}{dt}\sum \limits _{i=1}^3\int \limits _{\mathbb {R}^3}\partial _i^2u\cdot \partial _3\partial _i^2b\;dx}\\&+\,{\displaystyle \sum \limits _{i=1}^3\int \limits _{\mathbb {R}^3}\partial _i^2u\cdot \partial _3\partial _i^2(u\cdot \nabla b-b\cdot \nabla u-\nu _1\partial _1^2b-\partial _3u)\;dx}\\=: & {} {\displaystyle \mathcal {M}_{1,0}+\mathcal {M}_{1,1}+\mathcal {M}_{1,2}+\mathcal {M}_{1,3}+\mathcal {M}_{1,4}.} \end{aligned}$$

Making use of Lemma 2.1, Hölder’s inequality and Cauchy’s inequality, we derive that

$$\begin{aligned} {\displaystyle \mathcal {M}_{1,1}}= & {} {\displaystyle \int \limits _{\mathbb {R}^3}\partial _3\partial _1^3u\cdot (\partial _1u\cdot \nabla b+u\cdot \partial _1\nabla b) +\partial _2^2u\cdot \partial _2^2\partial _3(u\cdot \nabla b)+\partial _3^2u\cdot \partial _3^2\partial _3(u\cdot \nabla b)\;dx}\\\le & {} {\displaystyle C\Vert \partial _3\partial _1^3u\Vert _{L^2} (\Vert \partial _1u\Vert _{L^2}^{\frac{1}{4}}\Vert \partial _2\partial _1u\Vert _{L^2}^{\frac{1}{4}}\Vert \partial _3\partial _1u\Vert _{L^2}^{\frac{1}{4}} \Vert \partial _2\partial _3\partial _1u\Vert _{L^2}^{\frac{1}{4}}\Vert \nabla b\Vert _{L^2}^{\frac{1}{2}}\Vert \partial _1\nabla b\Vert _{L^2}^{\frac{1}{2}}}\\&+\,{\displaystyle \Vert u\Vert _{L^2}^{\frac{1}{4}}\Vert \partial _2u\Vert _{L^2}^{\frac{1}{4}}\Vert \partial _3u\Vert _{L^2}^{\frac{1}{4}} \Vert \partial _2\partial _3u\Vert _{L^2}^{\frac{1}{4}}\Vert \partial _1\nabla b\Vert _{L^2}^{\frac{1}{2}}\Vert \partial _1\partial _1\nabla b\Vert _{L^2}^{\frac{1}{2}})}\\&+\,{\displaystyle C\Vert \partial _2u\Vert _{H^3}(\Vert u\Vert _{H^3}\Vert \partial _3b\Vert _{H^2}+\Vert \partial _3u\Vert _{H^3}\Vert b\Vert _{H^3})}\\&+\,{\displaystyle C\Vert \partial _3u\Vert _{H^3}(\Vert u\Vert _{H^3}\Vert \partial _3b\Vert _{H^2}+\Vert \partial _3u\Vert _{H^3}\Vert b\Vert _{H^3})}\\\le & {} {\displaystyle C\Vert \partial _3u\Vert _{H^3}^{\frac{5}{4}}\Vert \partial _2u\Vert _{H^3}^{\frac{1}{4}}\Vert u\Vert _{H^3}^{\frac{1}{2}}\Vert b\Vert _{H^3}^{\frac{1}{2}}\Vert \partial _1b\Vert _{H^3}^{\frac{1}{2}} }\\&+\,{\displaystyle C(\Vert \partial _2u\Vert _{H^3}+\Vert \partial _3u\Vert _{H^3})(\Vert u\Vert _{H^3}\Vert \partial _3b\Vert _{H^2}+\Vert \partial _3u\Vert _{H^3}\Vert b\Vert _{H^3})}\\\le & {} {\displaystyle C(\Vert u\Vert ^{\frac{1}{2}}_{H^3}\Vert b\Vert ^{\frac{1}{2}}_{H^3}+\Vert u\Vert ^2_{H^3}+\Vert b\Vert _{H^3})(\Vert \partial _2u\Vert ^2_{H^3}+\Vert \partial _3u\Vert ^2_{H^3} +\Vert \partial _1b\Vert ^2_{H^3})+\epsilon \Vert \partial _3b\Vert ^2_{H^2}} \end{aligned}$$

and

$$\begin{aligned} {\displaystyle \mathcal {M}_{1,2}}= & {} {\displaystyle -\int \limits _{\mathbb {R}^3}\partial _3\partial _1^3u\cdot (\partial _1b\cdot \nabla u+b\cdot \partial _1\nabla u) +\partial _2^2u\cdot \partial _2^2\partial _3(b\cdot \nabla u)+\partial _3^2u\cdot \partial _3^2\partial _3(b\cdot \nabla u)\;dx}\\\le & {} {\displaystyle C\Vert \partial _3\partial _1^3u\Vert _{L^2} (\Vert \partial _1b\Vert _{L^2}^{\frac{1}{4}}\Vert \partial _2\partial _1b\Vert _{L^2}^{\frac{1}{4}}\Vert \partial _3\partial _1b\Vert _{L^2}^{\frac{1}{4}} \Vert \partial _2\partial _3\partial _1b\Vert _{L^2}^{\frac{1}{4}}\Vert \nabla u\Vert _{L^2}^{\frac{1}{2}}\Vert \partial _1\nabla u\Vert _{L^2}^{\frac{1}{2}}}\\&+\,{\displaystyle \Vert \partial _1\nabla u\Vert _{L^2}^{\frac{1}{4}}\Vert \partial _2\partial _1\nabla u\Vert _{L^2}^{\frac{1}{4}}\Vert \partial _3\partial _1\nabla u\Vert _{L^2}^{\frac{1}{4}} \Vert \partial _2\partial _3\partial _1\nabla u\Vert _{L^2}^{\frac{1}{4}}\Vert b\Vert _{L^2}^{\frac{1}{2}}\Vert \partial _1 b\Vert _{L^2}^{\frac{1}{2}})}\\&+\,{\displaystyle C\Vert \partial _2u\Vert _{H^3}(\Vert u\Vert _{H^3}\Vert \partial _3b\Vert _{H^2}+\Vert \partial _3u\Vert _{H^3}\Vert b\Vert _{H^3}) }\\&+\,{\displaystyle C\Vert \partial _3u\Vert _{H^3}(\Vert u\Vert _{H^3}\Vert \partial _3b\Vert _{H^2}+\Vert \partial _3u\Vert _{H^3}\Vert b\Vert _{H^3})}\\\le & {} {\displaystyle C\Vert \partial _3u\Vert _{H^3}^{\frac{5}{4}}\Vert \partial _2u\Vert _{H^3}^{\frac{1}{4}}\Vert u\Vert _{H^3}^{\frac{1}{2}}\Vert b\Vert _{H^3}^{\frac{1}{2}}\Vert \partial _1b\Vert _{H^3}^{\frac{1}{2}} }\\&+\,{\displaystyle C(\Vert \partial _2u\Vert _{H^3}+\Vert \partial _3u\Vert _{H^3})(\Vert u\Vert _{H^3}\Vert \partial _3b\Vert _{H^2}+\Vert \partial _3u\Vert _{H^3}\Vert b\Vert _{H^3})}\\\le & {} {\displaystyle C(\Vert u\Vert ^{\frac{1}{2}}_{H^3}\Vert b\Vert ^{\frac{1}{2}}_{H^3}+\Vert u\Vert ^2_{H^3}+\Vert b\Vert _{H^3})(\Vert \partial _2u\Vert ^2_{H^3}+\Vert \partial _3u\Vert ^2_{H^3} +\Vert \partial _1b\Vert ^2_{H^3})+\epsilon \Vert \partial _3b\Vert ^2_{H^2}.} \end{aligned}$$

It is not difficult to prove that

$$\begin{aligned} {\displaystyle \mathcal {M}_{1,3}}= & {} {\displaystyle -\nu _1\sum \limits _{i=1}^3\int \limits _{\mathbb {R}^3}\partial _3\partial _i^3u\cdot \partial _i\partial _1^2b\;dx}\\\le & {} {\displaystyle C\Vert \partial _3u\Vert _{H^3}\Vert \partial _1b\Vert _{H^3}}\\\le & {} {\displaystyle C\left( \Vert \partial _3u\Vert ^2_{H^3}+\Vert \partial _1b\Vert ^2_{H^3}\right) ,}\\ \displaystyle \mathcal {M}_{1,4}= & {} \sum \limits _{i=1}^3\int \limits _{\mathbb {R}^3}|\partial _3\partial _i^2u|^2\;dx \le C\Vert \partial _3u\Vert _{H^3}^2,\\ {\displaystyle \mathcal {M}_2 }\le & {} {\displaystyle C\Vert \partial _2u\Vert _{H^3}\Vert \partial _3b\Vert _{H^2}}\\\le & {} {\displaystyle C\Vert \partial _2u\Vert ^2_{H^3}+\epsilon \Vert \partial _3b\Vert ^2_{H^2}} \end{aligned}$$

and

$$\begin{aligned} {\displaystyle \mathcal {M}_3 }\le & {} {\displaystyle C\Vert \partial _3u\Vert _{H^3}\Vert \partial _3b\Vert _{H^2}}\\\le & {} {\displaystyle C\Vert \partial _3u\Vert ^2_{H^3}+\epsilon \Vert \partial _3b\Vert ^2_{H^2}.} \end{aligned}$$

Next, we estimate $\mathcal {M}_4$ and $\mathcal {M}_7$ by using integration by parts, Hölder’s inequality and Cauchy’s inequality

$$\begin{aligned} {\displaystyle \mathcal {M}_4}= & {} {\displaystyle -\chi \sum \limits _{i=1}^3\int \limits _{\mathbb {R}^3}\partial _i^2\partial _1\partial _3u\cdot \partial _i^2\partial _1b \;dx- \chi \sum \limits _{i=1}^3\int \limits _{\mathbb {R}^3} \partial _i^2\partial _2^2u\cdot \partial _i^2\partial _3b\;dx -\chi \sum \limits _{i=1}^3\int \limits _{\mathbb {R}^3}\partial _i^2\partial _3^2u\cdot \partial _i^2\partial _3b\;dx}\\\le & {} {\displaystyle C\left( \Vert \partial _3u\Vert _{H^3}\Vert \partial _1b\Vert _{H^3} +\Vert \partial _2u\Vert _{H^3}\Vert \partial _3b\Vert _{H^2}+\Vert \partial _3u\Vert _{H^3}\Vert \partial _3b\Vert _{H^2}\right) }\\\le & {} {\displaystyle C\left( \Vert \partial _2u\Vert ^2_{H^3}+\Vert \partial _3u\Vert ^2_{H^3}+\Vert \partial _1b\Vert ^2_{H^3}\right) +\epsilon \Vert \partial _3b\Vert ^2_{H^2}} \end{aligned}$$

and

$$\begin{aligned} {\displaystyle \mathcal {M}_7}= & {} {\displaystyle -2\chi \sum \limits _{i=1}^3\int \limits _{\mathbb {R}^3}\partial _i^2(\partial _2v_3-\partial _3v_2)\cdot \partial _i^2\partial _3b_1 +\partial _i^2(\partial _3v_1-\partial _1v_3)\cdot \partial _i^2\partial _3b_2}\\&{\displaystyle +\,\partial _i^2(\partial _1v_2-\partial _2v_1)\cdot \partial _i^2\partial _3b_3\;dx}\\\le & {} {\displaystyle C\Vert \partial _2v\Vert _{H^3}\Vert \partial _3b\Vert _{H^2}+C\Vert \partial _3v\Vert _{H^3}\Vert \partial _1b\Vert _{H^3}+C\Vert \partial _3v\Vert _{H^3}\Vert \partial _3b\Vert _{H^2}}\\\le & {} {\displaystyle C(\Vert \partial _2v\Vert ^2_{H^3}+\Vert \partial _3v\Vert ^2_{H^3}+\Vert \partial _1b\Vert ^2_{H^3})+ \epsilon \Vert \partial _3b\Vert ^2_{H^2}.} \end{aligned}$$

To bound $\mathcal {M}_5$, we use Lemma 2.1 and Cauchy’s inequality and obtain

$$\begin{aligned} {\displaystyle \mathcal {M}_5}= & {} {\displaystyle \int \limits _{\mathbb {R}^3}-\partial _3(u\cdot \nabla u)\cdot \partial _1^4b +\partial _2^2(u\cdot \nabla u)\cdot \partial _2^2\partial _3b+\partial _3^2(u\cdot \nabla u)\cdot \partial _3^2\partial _3b\;dx}\\= & {} {\displaystyle \int \limits _{\mathbb {R}^3}-(\partial _3u\cdot \nabla u+u\cdot \partial _3\nabla u))\cdot \partial _1^4b +\partial _2^2(u\cdot \nabla u)\cdot \partial _2^2\partial _3b+\partial _3^2(u\cdot \nabla u)\cdot \partial _3^2\partial _3b\;dx}\\\le & {} \displaystyle C\Vert \partial _1^4b\Vert _{L^2}\left( \Vert \nabla u\Vert _{L^2}^{\frac{1}{4}}\Vert \partial _2\nabla u\Vert _{L^2}^{\frac{1}{4}} \Vert \partial _3\nabla u\Vert _{L^2}^{\frac{1}{4}}\Vert \partial _2\partial _3\nabla u\Vert _{L^2}^{\frac{1}{4}} \Vert \partial _3u\Vert _{L^2}^{\frac{1}{2}}\Vert \partial _1\partial _3u\Vert _{L^2}^{\frac{1}{2}}\right. \\&\left. +\,\displaystyle \Vert u\Vert _{L^2}^{\frac{1}{4}}\Vert \partial _2u\Vert _{L^2}^{\frac{1}{4}}\Vert \partial _3u\Vert _{L^2}^{\frac{1}{4}}\Vert \partial _2\partial _3u\Vert _{L^2}^{\frac{1}{4}} \Vert \partial _3\nabla u\Vert _{L^2}^{\frac{1}{2}}\Vert \partial _1\partial _3\nabla u\Vert _{L^2}^{\frac{1}{2}}\right) \\&+\,{\displaystyle C\Vert u\Vert _{H^3}\Vert \partial _2u\Vert _{H^3}\Vert \partial _3b\Vert _{H^2}+C\Vert u\Vert _{H^3}\Vert \partial _3u\Vert _{H^3}\Vert \partial _3b\Vert _{H^2}}\\\le & {} {\displaystyle C\Vert \partial _1b\Vert _{H^3}\Vert u\Vert _{H^3}\Vert \partial _2u\Vert _{H^3}^{\frac{1}{2}}\Vert \partial _3u\Vert _{H^3}^{\frac{1}{2}} +C\Vert u\Vert _{H^3}\Vert \partial _2u\Vert _{H^3}\Vert \partial _3b\Vert _{H^2}}\\&+\,{\displaystyle C\Vert u\Vert _{H^3}\Vert \partial _3u\Vert _{H^3}\Vert \partial _3b\Vert _{H^2}}\\\le & {} {\displaystyle C\Vert u\Vert _{H^3}\left( \Vert \partial _2u\Vert ^2_{H^3}+\Vert \partial _3u\Vert ^2_{H^3}+\Vert \partial _1b\Vert ^2_{H^3}\right) }\\&+\,{\displaystyle C\Vert u\Vert ^2_{H^3}\left( \Vert \partial _2u\Vert ^2_{H^3}+\Vert \partial _3u\Vert ^2_{H^3}\right) +\epsilon \Vert \partial _3b\Vert ^2_{H^3}.} \end{aligned}$$

Similarly, it holds that

$$\begin{aligned} {\displaystyle \mathcal {M}_6}= & {} {\displaystyle -\int \limits _{\mathbb {R}^3}[\partial _1^2(b\cdot \nabla b)\cdot \partial _1^2\partial _3b+\partial _3^2(b\cdot \nabla b)\cdot \partial _3^2\partial _3b}\\&{\displaystyle +\,\partial _2^2b\cdot \nabla b\cdot \partial _3\partial _2^2b+\partial _2b\cdot \partial _2\nabla b\cdot \partial _3\partial _2^2b+b\cdot \partial _2^2\nabla b\cdot \partial _3\partial _2^2b]\;dx}\\\le & {} {\displaystyle C\Vert \partial _1b\Vert _{H^3}\Vert b\Vert _{H^3}\Vert \partial _3b\Vert _{H^2}+C\Vert \partial _3b\Vert _{H^2}^2\Vert b\Vert _{H^3}}\\&+\,{\displaystyle C\Vert \partial _3\partial _2^2b\Vert _{L^2}(\Vert \nabla b\Vert _{L^2}^{\frac{1}{4}}\Vert \partial _2\nabla b\Vert _{L^2}^{\frac{1}{4}} \Vert \partial _3\nabla b\Vert _{L^2}^{\frac{1}{4}}\Vert \partial _2\partial _3\nabla b\Vert _{L^2}^{\frac{1}{4}} \Vert \partial _2^2b\Vert _{L^2}^{\frac{1}{2}}\Vert \partial _1\partial _2^2b\Vert _{L^2}^{\frac{1}{2}}}\\&+\,{\displaystyle \Vert \partial _2b\Vert _{L^2}^{\frac{1}{4}}\Vert \partial _2\partial _2b\Vert _{L^2}^{\frac{1}{4}} \Vert \partial _3\partial _2b\Vert _{L^2}^{\frac{1}{4}}\Vert \partial _2\partial _3\partial _2b\Vert _{L^2}^{\frac{1}{4}} \Vert \partial _2\nabla b\Vert _{L^2}^{\frac{1}{2}}\Vert \partial _1\partial _2\nabla b\Vert _{L^2}^{\frac{1}{2}}}\\&+\,{\displaystyle \Vert b\Vert _{L^2}^{\frac{1}{4}}\Vert \partial _2b\Vert _{L^2}^{\frac{1}{4}}\Vert \partial _3b\Vert _{L^2}^{\frac{1}{4}}\Vert \partial _2\partial _3b\Vert _{L^2}^{\frac{1}{4}} \Vert \partial _2^2\nabla b\Vert _{L^2}^{\frac{1}{2}}\Vert \partial _1\partial _2^2\nabla b\Vert _{L^2}^{\frac{1}{2}})}\\\le & {} {\displaystyle C(\Vert \partial _1b\Vert _{H^3}\Vert b\Vert _{H^3}\Vert \partial _3b\Vert _{H^2}+\Vert \partial _3b\Vert _{H^2}^2\Vert b\Vert _{H^3} +C\Vert \partial _1b\Vert _{H^3}^{\frac{1}{2}}\Vert b\Vert _{H^3}\Vert \partial _3b\Vert ^{\frac{3}{2}}_{H^2})}\\\le & {} {\displaystyle C\Vert b\Vert _{H^3}(\Vert \partial _1b\Vert ^2_{H^3}+\Vert \partial _3b\Vert ^2_{H^2}).} \end{aligned}$$

We institute the estimates for $\mathcal {M}_i(i=1, \ldots , 7)$ into (2.18) and arrive at

$$\begin{aligned} {\displaystyle \Vert \partial ^2_i\partial _3b\Vert _{L^2}^2 }\le & {} {\displaystyle C(\Vert \partial _2u\Vert ^2_{H^3}+\Vert \partial _3u\Vert ^2_{H^3}+\Vert \partial _2v\Vert ^2_{H^3}+\Vert \partial _3v\Vert ^2_{H^3}+\Vert \partial _1b\Vert ^2_{H^3})} \nonumber \\&+\,{\displaystyle C(\Vert u\Vert ^{\frac{1}{2}}_{H^3}\Vert b\Vert ^{\frac{1}{2}}_{H^3}+\Vert u\Vert ^2_{H^3}+\Vert b\Vert ^2_{H^3}+\Vert b\Vert _{H^3})(\Vert \partial _2u\Vert ^2_{H^3}+\Vert \partial _3u\Vert ^2_{H^3} } \nonumber \\&+\,{\displaystyle \Vert \partial _1b\Vert ^2_{H^3}+\Vert \partial _3b\Vert ^2_{H^2})+7\epsilon \Vert \partial _3b\Vert ^2_{H^2}+ \mathcal {M}_{1, 0}. } \end{aligned}$$

(2.19)

Now, adding (2.17) and (2.19), then integrating in time and using the definition of $E_0(t)$ and $E_1(t)$ and choosing $\epsilon \le \frac{1}{28}$, we obtain

$$\begin{aligned} {\displaystyle E_1(t)}\le & {} {\displaystyle C (E_0(t)+E^{\frac{3}{2}}_0(t)+E^{2}_0(t) +E^{\frac{1}{2}}_0(t)E_1(t)+ E_0(t)E_1(t))}\nonumber \\&+\,{\displaystyle \frac{1}{2}E_1(t)+\int \limits _0^t \mathcal {L}_{1, 0}d\tau +\int \limits _0^t \mathcal {M}_{1, 0}d\tau .} \end{aligned}$$

(2.20)

Noting that

$$\begin{aligned} {\displaystyle \int \limits _0^t\mathcal {L}_{1,0}d\tau }= & {} {\displaystyle \int \limits _{\mathbb {R}^3}u(x,t)\cdot \partial _3b(x,t)dx-\int \limits _{\mathbb {R}^3}u(x,0)\cdot \partial _3b(x,0)\;dx}\\\le & {} {\displaystyle C(E_0(t)+\mathscr {E}_0)} \end{aligned}$$

and

$$\begin{aligned} {\displaystyle \int \limits _0^t\mathcal {M}_{1,0}d\tau }= & {} {\displaystyle \int \limits _{\mathbb {R}^3}\partial _i^2u(x,t)\cdot \partial _3\partial _i^2b(x,t)dx-\int \limits _{\mathbb {R}^3}u(x,0)\cdot \partial _3b(x,0)\;dx}\\\le & {} {\displaystyle C(E_0(t)+\mathscr {E}_0).} \end{aligned}$$

Therefore, we have by Cauchy’s inequality

$$\begin{aligned} {\displaystyle E_1(t)\le C (\mathscr {E}_0 +E_0(t)+E_0(t)^{\frac{3}{2}}+E^2_0(t)+E^2_1(t)).} \end{aligned}$$

This complete the proof of (2.2).

3 Proof of Theorem 1.1

This section is devoted to the proof of Theorem 1.1. We employ the bootstrapping argument(see, e.g.,[25]). It follows from (2.1) and (2.2) that

$$\begin{aligned} E_0(t)+E_1(t)\le C (\mathscr {E}_0+\mathscr {E}^{\frac{3}{2}}_0+E_0(t)^{\frac{3}{2}}+E_0(t)^2+E_1(t)^2), \end{aligned}$$

or, for pure constants $C_0,C_1,C_2$,

$$\begin{aligned} {\displaystyle E_0(t)+E_1(t)}\le & {} {\displaystyle C_0\left( \mathscr {E}_0+\mathscr {E}^{\frac{3}{2}}_0)+C_1(E_0(t)^{\frac{3}{2}}+E_1(t)^{\frac{3}{2}}\right) }\nonumber \\&{\displaystyle +\,C_2(E_0(t)^2+E_1(t)^2).} \end{aligned}$$

(3.1)

To initiate the bootstrapping argument, we make the ansatz

$$\begin{aligned} E_0(t)+E_1(t)\le \min \left\{ \frac{1}{16C_1^2},\frac{1}{4C_2}.\right\} \end{aligned}$$

(3.2)

We then show that (3.1) allows us to conclude that $E_0(t)+E_1(t)$ actually admits an even smaller bound by taking the initial $H^3$-norm $\mathscr {E}_0$ sufficiently small. In fact, when (3.2) holds, (3.1) implies

$$\begin{aligned} {\displaystyle E_0(t)+E_1(t)}\le & {} C{\displaystyle C_0(\mathscr {E}_0+\mathscr {E}^{\frac{3}{2}}_0)+C_1\sqrt{E_0(t)+E_1(t)}(E_0(t)+E_1(t))}\nonumber \\&{\displaystyle +\,C_2(E_0(t)+E_1(t))(E_0(t)+E_1(t))}\nonumber \\\le & {} {\displaystyle C_0(\mathscr {E}_0+\mathscr {E}^{\frac{3}{2}}_0)+\frac{1}{2}(E_0(t)+E_1(t)),} \end{aligned}$$

(3.3)

or

$$\begin{aligned} {\displaystyle E_0(t)+E_1(t)}\le {\displaystyle 2C_0\left( \mathscr {E}_0+\mathscr {E}^{\frac{3}{2}}_0\right) .} \end{aligned}$$

(3.4)

Therefore, if we take $\mathscr {E}_0$ sufficiently small such that

$$\begin{aligned} 2C_0(\mathscr {E}_0+\mathscr {E}^{\frac{3}{2}}_0)\le \min \left\{ \frac{1}{16C_1^2},\frac{1}{4C_2},\right\} \end{aligned}$$

(3.5)

then $E_0(t)+E_1(t)$ actually admits an smaller bound in (3.4) than the one in the ansatz (3.2). Then, the bootstrapping argument assesses that (3.4) holds for all time, provided that (3.5) holds. This completes the proof of Theorem 1.1.