The Well-Posedness of Three-Dimensional Hall-Magnetohydrodynamics Model with Partial Viscosity Terms

Abstract

The paper is concerned with three-dimensional Hall-magnetohydrodynamics for partial viscosity terms. By applying basic inequality and energy estimates, we prove the global existence of classical solution for small initial data. Additionally, the local existence of classical solution is also obtained. The smallness conditions are given by the suitable Sobolev norms. Furthermore, the existence of classical solutions with large initial data can be obtained when the coefficients of viscosity sufficiently large.

1. Introduction

This paper investigates the following three-dimensional Hall-MHD system (Hall-MHD):

$$\begin{array}{c}\hfill \left\{\begin{array}{c}{\partial }_{t}b-\Delta b=(b·\nabla )v-\nabla \times ((\nabla \times b)\times b)-(v·\nabla )b,\hfill \\ {\partial }_{t}v+\nabla (p+\pi )+(v·\nabla )v={\nu }_1{v}_{x_1{x}_1}+{\nu }_2{v}_{x_2{x}_2}+{\nu }_3{v}_{x_3{x}_3}+(b·\nabla )b,\hfill \\ \nabla ·b=\nabla ·v=0,\hfill \\ {v|}_{t=0}=v_0{,b|}_{t=0}=b_0.\hfill \end{array}\right.\end{array}$$

(1)

where $b,v$ represent the magnetic field and velocity vector field, p denotes the scalar pressure, $\pi$ is a scalar, and ${\nu }_1,{\nu }_2,{\nu }_3\ge 0$ are the kinematic viscosity. The Hall-MHD is indeed needed for such problems as magnetic reconnection in space plasmas, star formation, neutron stars and geo-dynamo. Compared with the MHD system, the Hall-MHD equations have the Hall term $\nabla \times ((\nabla \times b)\times b)$, which plays an important role in magnetic reconnection, which is happening in the case of large magnetic shear. Hall-MHD is an essential feature in relation to the problem of magnetic reconnection; it corresponds to changes in the topology of magnetic field lines which are ubiquitously observed in space.

There exist many results on incompressible Hall-magnetohydrodynamic equations: In paper [1]’s given derivation of model (1), Chae, Degond and Liu [2] established well-posedness for Hall-magnetohydrodynamics. Chae and Lee [3] established the blow-up criterion and small data global existence for the Hall-magnetohydrodynamics. The papers [4,5] investigated temporal decay and singularity formation for the Hall-magnetohydrodynamic equations. Paper [6] studied the partial regularity for the steady Hall-magnetohydrodynamics system followed by paper [7]. Ye [8] obtained the well-posedness results for the 3D incompressible Hall-MHD equations with fractional dissipation. Liu [9] established the global existence to Hall-MHD system. Dai [10] obtained almost certain well-posedness for Hall-MHD.

Dumas and Sueur [11] derived weak solutions to the Maxwell–Landau–Lifshitz equations and to Hall-magnetohydrodynamic equations. Paper [12] established well-posedness for the axisymmetric incompressible viscous Hall-magnetohydrodynamic equations. Homann and Grauer [13] investigated the bifurcation analysis of magnetic reconnection in Hall-MHD systems. Li [14] investigated the local well-posedness of a 3D ideal Hall-MHD system with an azimuthal magnetic field. The global well-posedness of 3D incompressible hyper-dissipative Hall-MHD equations in anisotropic Besov spaces was obtained in [15]. Danchin and Tan [16] established the well-posedness of the Hall-magnetohydrodynamics system in critical spaces.

To inspire this research, we recall related important contributions to incompressible Hall-MHD equations. Fei and Xiang [17] researched Hall-magnetohydrodynamics for partial dissipation, and obtained a blow-up criterion of classical solutions. Du [18] investigated 2.5D Hall-magnetohydrodynamics with partial dissipation and proved the classical solutions that followed in [19]. Two blow-up criteria of smooth solutions and two improved classical solutions with small initial data for the Hall-magnetohydrodynamics with partial viscosity terms were established in paper [20,21]. Additionally, 3D magnetohydrodynamics with partial diffusion have been analysed by Ye and Zhu [22], who obtained the existence of the global classical solutions.

2. Preliminaries

This paper concerns three-dimensional Hall-magnetohydrodynamics (1) with partial viscosity—that is, the case of ${\nu }_2=0,{\nu }_1,{\nu }_3>0$. We should point out that this case is mainly a mathematical simplification, but it is theoretically significant in mathematics due to the partial viscosity terms leading to no smoothing effect on the vertical derivative. We show the global well-posedness and, more precisely, we prove the global existence and the local existence of classical solutions for system (1) with partial viscosity terms.

In this paper, ${\parallel ·\parallel }_2\triangleq {\parallel ·\parallel }_{L^2},C$ represent a generic positive constant, and ${\partial }_k$ and $v_k$ indicate the corresponding k-th components of $\nabla ,v$, respectively.

$${\nabla }^{\beta }:={\partial }^{\left|\beta \right|}/{\partial }_1^{{\beta }_1}{\partial }_2^{{\beta }_2}{\partial }_3^{{\beta }_3},$$

where $\beta =({\beta }_1,{\beta }_2,{\beta }_3)\in {(\mathbb{N}\cup \left\{0\right\})}^3$ with $\left|\beta \right|={\beta }_1+{\beta }_2+{\beta }_3\le 3$.

$$\zeta :=min\{{\nu }_1,{\nu }_3\}>0.$$

$${\nabla }_p:=({\partial }_1,0,{\partial }_3).$$

3. Results

The paper researches the global well-posedness for model (1) with ${\nu }_2=0,{\nu }_1,{\nu }_3>0$.

Theorem 1.

Suppose 3D Hall-magnetohydrodynamics (1) with ${\nu }_2=0,$ and ${\nu }_1,{\nu }_3>0$, and assume $\nabla ·v_0=\nabla ·b_0=0$ and $(b_0,v_0)\in H^3({\mathbb{R}}^3)$. Then, there exists $T=T(\parallel v_0{\parallel }_{H^3},\parallel b_0{\parallel }_{H^3})$ such that there exists a unique solution $(b,v)\in L^{\infty }(0,T;H^3({\mathbb{R}}^3))$.

Based on the local existence of classical solutions for system (1) with the partial viscosity terms in Theorem 1, we can establish the following global-in-time existence of the classical solution for the system (1) with ${\nu }_2=0,{\nu }_1,{\nu }_3>0$.

Theorem 2.

Consider model (1) with ${\nu }_2=0,$ and ${\nu }_1,{\nu }_3>0$, and assume $(v_0,b_0)\in H^3({\mathbb{R}}^3)$ with $\nabla ·b_0=\nabla ·v_0=0$. Then, there exists ${ϵ}_0>0$, if

$$\begin{array}{c}\hfill {\displaystyle \frac{{(\parallel b_0\parallel }_2^2+{\parallel v_0\parallel }_2^2)(\parallel v_0{\parallel }_{H^2}^2+\parallel b_0{\parallel }_{H^2}^2)}{{\zeta }^4}}\le {ϵ}_0.\end{array}$$

(2)

System (1) has a unique classical solution $(b,v)$, satisfying

$$\begin{array}{c}\hfill (b,v)\in L^{\infty }(0,T;H^3({\mathbb{R}}^3)),({\nabla }_{p}v,\nabla b)\in L^2(0,T;H^3({\mathbb{R}}^3)),\end{array}$$

(3)

Remark 1.

Compared to previous results, the existence of classical solutions with large initial data can be obtained when the coefficients of viscosity are sufficiently large. In addition, the initial data $(\parallel v_0{\parallel }_2^2+\parallel b_0{\parallel }_2^2)(\parallel v_0{\parallel }_{H^2}^2+\parallel b_0{\parallel }_{H^2}^2)$ instead of ($\parallel v_0{\parallel }_{H^m}^2+{\parallel b_0\parallel }_{H^m}^2,m>{\displaystyle \frac{5}{2}}$) in paper [2] are sufficiently small.

Remark 2.

It will be interesting if one could obtain the global regularity in terms of only one component of the velocity in system (1), as there is an essential difficulty due to the quadratic Hall term.

4. Local Existence

Proof of Theorem 1.

Applying ${\nabla }^{\beta }$ to the first two equations of (1), and taking the inner product of the resulting equations with ${\nabla }^{\beta }v$ and ${\nabla }^{\beta }b$, respectively, and adding them together, we obtain

$$\begin{array}{cc}\hfill & {\displaystyle \frac{1}{2}}{\displaystyle \frac{d}{dt}}{(\parallel v\parallel }_{H^3}^2+{\parallel b\parallel }_{H^3}^2)+{\nu }_1\parallel {\partial }_1{v\parallel }_{H^3}^2+{\nu }_3\parallel {\partial }_3{v\parallel }_{H^3}^2+{\parallel \Delta b\parallel }_{H^3}^2\\ \hfill & =-\sum _{0<\left|\beta \right|\le 3}{\int }_{{\mathbb{R}}^3}{\nabla }^{\beta }(v·\nabla v)·{\nabla }^{\beta }vdx+\sum _{0<\left|\beta \right|\le 3}{\int }_{{\mathbb{R}}^3}{\nabla }^{\beta }(b·\nabla b)·{\nabla }^{\beta }vdx\\ \hfill & -\sum _{0<\left|\beta \right|\le 3}{\int }_{{\mathbb{R}}^3}{\nabla }^{\beta }(v·\nabla b)·{\nabla }^{\beta }bdx+\sum _{0<\left|\beta \right|\le 3}{\int }_{{\mathbb{R}}^3}{\nabla }^{\beta }(b·\nabla v)·{\nabla }^{\beta }vdx\\ \hfill & -\sum _{0<\left|\beta \right|\le 3}{\int }_{{\mathbb{R}}^3}{\nabla }^{\beta }(\nabla \times ((\nabla \times b)\times b))·{\nabla }^{\beta }bdx\\ \hfill & =Q_1+Q_2+Q_3+Q_4+Q_5.\end{array}$$

(4)

Notice that $\left|\beta \right|=0$, and the above equality becomes (11). Using calculus inequality and Sobolev inequality, one obtains

$$\begin{array}{cc}\hfill |Q_1|& =|\sum _{0<\left|\beta \right|\le 3}{\int }_{{\mathbb{R}}^3}[{\nabla }^{\beta }(v·\nabla )v-(v·\nabla ){\nabla }^{\beta }v]·{\nabla }^{\beta }vdx|\\ \hfill & \le \sum _{0<\left|\beta \right|\le 3}\parallel [{\nabla }^{\beta }(v·\nabla )v-(v·\nabla ){\nabla }^{\beta }v]{\parallel }_2{\parallel v\parallel }_{H^3}\\ \hfill & \le {C(\parallel \nabla v\parallel }_{L^{\infty }}{\parallel \nabla v\parallel }_{H^2}+{\parallel v\parallel }_{H^3}{\parallel \nabla v\parallel }_{L^{\infty }}{)\parallel v\parallel }_{H^3}\\ \hfill & \le {C\parallel v\parallel }_{H^3}^3.\end{array}$$

(5)

Applying Leibnitz formula and Sobolev inequality, we obtain

$$\begin{array}{cc}\hfill |Q_2|& \le {C(\parallel b\parallel }_{H^3}{\parallel \nabla b\parallel }_{L^{\infty }}+{\parallel b\parallel }_{L^{\infty }}{\parallel \nabla b\parallel }_{H^3}{)\parallel v\parallel }_{H^3}\\ \hfill & \le {C\parallel b\parallel }_{H^3}^2{\parallel v\parallel }_{H^3}+{C\parallel \nabla b\parallel }_{H^3}{\parallel b\parallel }_{H^3}{\parallel v\parallel }_{H^3}\\ \hfill & \le {\displaystyle \frac{1}{8}}{\parallel \nabla b\parallel }_{H^3}^2+{C\parallel b\parallel }_{H^3}^2{\parallel v\parallel }_{H^3}^2.\end{array}$$

(6)

In a similar manner, we obtain

$$\begin{array}{c}\hfill |Q_3|\le {\displaystyle \frac{1}{8}}{\parallel \nabla b\parallel }_{H^3}^2+{C\parallel b\parallel }_{H^3}^2{\parallel v\parallel }_{H^3}^2.\end{array}$$

(7)

Integrating by parts in $Q_4$, and then similarly to the above calculation, one obtains

$$\begin{array}{c}\hfill |Q_4|\le {\displaystyle \frac{1}{8}}{\parallel \nabla b\parallel }_{H^3}^2+{C\parallel b\parallel }_{H^3}^2{\parallel v\parallel }_{H^3}^2.\end{array}$$

(8)

By cancellation property, calculus inequality and Sobolev inequality, one can estimate $Q_5$ as

$$\begin{array}{cc}\hfill |Q_5|& =|\sum _{0<\left|\beta \right|\le 3}{\int }_{{\mathbb{R}}^3}[{\nabla }^{\beta }((\nabla \times b)\times b)-{\nabla }^{\beta }(\nabla \times b)\times b]·{\nabla }^{\beta }(\nabla \times b)dx|\\ \hfill & \le \sum _{0<\left|\beta \right|\le 3}\parallel {\nabla }^{\beta }((\nabla \times b)\times b)-{\nabla }^{\beta }{(\nabla \times b)\times b\parallel }_2{\parallel \nabla b\parallel }_{H^3}\\ \hfill & \le {C(\parallel \nabla b\parallel }_{L^{\infty }}{\parallel \nabla b\parallel }_{H^2}+{\parallel b\parallel }_{H^3}{\parallel \nabla b\parallel }_{L^{\infty }}{)\parallel \nabla b\parallel }_{H^3}\\ \hfill & \le {C\parallel b\parallel }_{H^3}^2{\parallel \nabla b\parallel }_{H^3}\\ \hfill & \le {\displaystyle \frac{1}{8}}{\parallel \nabla b\parallel }_{H^3}^2+C{\parallel b\parallel }_{H^3}^4.\end{array}$$

(9)

Combining (4)–(9), we obtain

$$\begin{array}{c}\hfill {\displaystyle \frac{d}{dt}}{(\parallel v\parallel }_{H^3}^2+{\parallel b\parallel }_{H^3}^2)+\zeta \parallel {\nabla }_p{v\parallel }_{H^3}^2+{\parallel \nabla b\parallel }_{H^3}^2\le {C(\parallel v\parallel }_{H^3}^2+{\parallel b\parallel }_{H^3}^2{)}^2.\end{array}$$

(10)

Applying nonlinear Gronwall’s inequality to (10), and choosing $T={\displaystyle \frac{1}{2C(\parallel v_0{\parallel }_{H^3}^2+\parallel b_0{\parallel }_{H^3}^2)}}$, we can then obtain

$${\parallel v\parallel }_{H^3}+{\parallel b\parallel }_{H^3}\le 2(\parallel v_0{\parallel }_{H^3}+\parallel b_0{\parallel }_{H^3})\forall t\in (0,T),$$

which completes the proof of Theorem 1. □

5. Priori Estimates

In this section, we require the following useful inequality.

Lemma 1

([23]). Assume that $h,f,g,{\nabla }_{p}f,{\partial }_{3}g,{\nabla }_{p}h\in L^2({\mathbb{R}}^3)$; then, the following inequality holds:

$${\int }_{{\mathbb{R}}^3}\left|hgf\right|dx\le {C\parallel f\parallel }_2^{{\displaystyle \frac{1}{2}}}{\parallel g\parallel }_2^{{\displaystyle \frac{1}{2}}}{\parallel h\parallel }_2^{{\displaystyle \frac{1}{2}}}\parallel {\nabla }_p{f\parallel }_2^{{\displaystyle \frac{1}{2}}}\parallel {\partial }_3{g\parallel }_2^{{\displaystyle \frac{1}{2}}}{\parallel {\nabla }_{p}h\parallel }_2^{{\displaystyle \frac{1}{2}}}.$$

Given a classical solution $(v,b)$ on $(0,T)\times {\mathbb{R}}^3$ to (1) with ${\nu }_2=0,{\nu }_1,{\nu }_3>0$, we define

$$G(t)\triangleq \underset{0\le s\le t}{sup}{(\parallel \nabla v(s)\parallel }_2^2+{\parallel \nabla b(s)\parallel }_2^2+\parallel {\nabla }^2{v(s)\parallel }_2^2+\parallel {\nabla }^{2}b(s){\parallel }_2^2),0\le t\le T.$$

Under ${\nu }_2=0,$ and ${\nu }_1,{\nu }_3>0$, and taking scalar products of the first two equations of model (1) with v and b, respectively, one obtains

$$\begin{array}{c}\hfill {\displaystyle \frac{1}{2}}{\displaystyle \frac{d}{dt}}{(\parallel v\parallel }_2^2+{\parallel b\parallel }_2^2)+\zeta (\parallel {\nabla }_p{v\parallel }_2^2+{\parallel \nabla b\parallel }_2^2)=0.\end{array}$$

(11)

Based on the above equality, one can obtain the energy inequality as following.

Lemma 2.

Suppose $(b,v)$ solves (1) with the ${\nu }_2=0,{\nu }_1,{\nu }_3>0$ on $[0,T]\times {\mathbb{R}}^3$; then,

$$\begin{array}{c}\hfill {\parallel v\parallel }_2^2+{\parallel b\parallel }_2^2+2\zeta {\int }_0^T(\parallel {\nabla }_p{v\parallel }_2^2+{\parallel \nabla b\parallel }_2^2)dt\le \parallel v_0{\parallel }_2^2+{\parallel b_0\parallel }_2^2.\end{array}$$

(12)

Next, we will structure energy estimates of $b,v$.

Lemma 3.

Suppose $(v,b)$ solves (1) with ${\nu }_2=0,$ and ${\nu }_1,{\nu }_3>0$ on $[0,T]\times {\mathbb{R}}^3$, if

$$\begin{array}{c}\hfill G(T)\le 2(\parallel \nabla v_0{\parallel }_2^2+\parallel \nabla b_0{\parallel }_2^2+\parallel {\nabla }^2{v}_0{\parallel }_2^2+\parallel {\nabla }^2{b}_0{\parallel }_2^2),\end{array}$$

(13)

and (2) holds, and ${ϵ}_0$ is sufficiently small. Then,

$$\begin{array}{c}\underset{0\le t\le T}{sup}{(\parallel \nabla b\parallel }_2^2+{\parallel \nabla v\parallel }_2^2)+2\zeta {\int }_0^T(\parallel {\nabla }^2{b(t)\parallel }_2^2+\parallel {\nabla }_p\nabla v{\parallel }_2^2)dt\hfill \\ \le \parallel \nabla b_0{\parallel }_2^2+{\parallel \nabla v_0\parallel }_2^2.\end{array}$$

(14)

Proof of Lemma 3.

Applying ∇ to the first two equations of model (1), taking the inner products with $\nabla b,\nabla v$ and then summing them together results in

$$\begin{array}{cc}\hfill & {\displaystyle \frac{1}{2}}{\displaystyle \frac{d}{dt}}{(\parallel \nabla v\parallel }_2^2+{\parallel \nabla b\parallel }_2^2)+{\nu }_1\parallel \nabla {\partial }_1{v\parallel }_2^2+{\nu }_3\parallel \nabla {\partial }_3{v\parallel }_2^2+{\parallel \Delta b\parallel }_2^2\hfill \\ \hfill & =-{\int }_{{\mathbb{R}}^3}\nabla v·\nabla (v·\nabla )vdx-{\int }_{{\mathbb{R}}^3}\nabla b·\nabla (v·\nabla )bdx+{\int }_{{\mathbb{R}}^3}\nabla v·\nabla (b·\nabla )bdx\\ \hfill & +{\int }_{{\mathbb{R}}^3}\nabla b·\nabla (b·\nabla )vdx-{\int }_{{\mathbb{R}}^3}\nabla b·\nabla [\nabla \times ((\nabla \times b)\times b)]dx\\ \hfill & ={\Sigma }_{i=1}^5{K}_i.\hfill \end{array}$$

(15)

One can split $K_1$ into three parts:

$$\begin{array}{cc}\hfill K_1& =-{\int }_{{\mathbb{R}}^3}({\nabla }_{p}v·\nabla )v·{\nabla }_{p}vdx-{\int }_{{\mathbb{R}}^3}({\partial }_2{v}_p·{\nabla }_p)v·{\partial }_{2}vdx+{\int }_{{\mathbb{R}}^3}({\nabla }_p·v_p){\partial }_{3}v·{\partial }_{2}vdx\\ \hfill & =K_{11}+K_{12}+K_{13}.\end{array}$$

Applying Hölder inequality, Young’s inequality and interpolation result in

$$\begin{array}{cc}\hfill K_{11}& \le {C\parallel \nabla v\parallel }_2\parallel {\nabla }_p{v\parallel }_{L^3}\parallel {\nabla }_p{v\parallel }_{L^6}\\ \hfill & \le {C\parallel \nabla v\parallel }_2\parallel {\nabla }_p{v\parallel }_2^{{\displaystyle \frac{1}{2}}}\parallel \nabla {\nabla }_p{v\parallel }_2^{{\displaystyle \frac{1}{2}}}\parallel \nabla {\nabla }_p{v\parallel }_2\\ \hfill & \le C\parallel \nabla {\nabla }_p{v\parallel }_2^{{\displaystyle \frac{3}{2}}}{\parallel \nabla v\parallel }_2\parallel {\nabla }_p{v\parallel }_2^{{\displaystyle \frac{1}{2}}}\\ \hfill & \le {\displaystyle \frac{C}{{\zeta }^3}}{\parallel \nabla v\parallel }_2^4\parallel {\nabla }_p{v\parallel }_2^2+{\displaystyle \frac{\zeta }{6}}{\parallel \nabla {\nabla }_{p}v\parallel }_2^2.\end{array}$$

By Lemma 1, one has

$$\begin{array}{cc}\hfill K_{12}& \le C\parallel {\partial }_2{v\parallel }_2^{{\displaystyle \frac{1}{2}}}\parallel {\nabla }_p{v\parallel }_2^{{\displaystyle \frac{1}{2}}}\parallel {\partial }_2{v\parallel }_2^{{\displaystyle \frac{1}{2}}}\parallel {\partial }_2{\nabla }_p{v\parallel }_2^{{\displaystyle \frac{1}{2}}}\parallel {\partial }_2{\nabla }_p{v\parallel }_2^{{\displaystyle \frac{1}{2}}}\parallel {\partial }_2{\nabla }_p{v\parallel }_2^{{\displaystyle \frac{1}{2}}}\\ \hfill & \le C\parallel \nabla {\nabla }_p{v\parallel }_2^{{\displaystyle \frac{3}{2}}}{\parallel \nabla v\parallel }_2\parallel {\nabla }_p{v\parallel }_2^{{\displaystyle \frac{1}{2}}}\\ \hfill & \le {\displaystyle \frac{C}{{\zeta }^3}}{\parallel \nabla v\parallel }_2^4\parallel {\nabla }_p{v\parallel }_2^2+{\displaystyle \frac{\zeta }{6}}{\parallel \nabla {\nabla }_{p}v\parallel }_2^2.\end{array}$$

In a similar manner, we can obtain

$$K_{13}\le {\displaystyle \frac{C}{{\zeta }^3}}{\parallel \nabla v\parallel }_2^4\parallel {\nabla }_p{v\parallel }_2^2+{\displaystyle \frac{\zeta }{6}}{\parallel \nabla {\nabla }_{p}v\parallel }_2^2.$$

Therefore,

$$\begin{array}{c}\hfill K_1\le {\displaystyle \frac{C}{{\zeta }^3}}{\parallel \nabla v|}_2^4\parallel {\nabla }_p{v\parallel }_2^2+{\displaystyle \frac{\zeta }{2}}{\parallel \nabla {\nabla }_{p}v\parallel }_2^2.\end{array}$$

(16)

By Hölder inequality, we obtain

$$\begin{array}{cc}\hfill |K_2|& =|{\int }_{{\mathbb{R}}^3}\Delta b·(v·\nabla )bdx|\\ \hfill & \le {C\parallel \nabla b\parallel }_{L^3}{\parallel v\parallel }_{L^6}{\parallel \Delta b\parallel }_2\\ \hfill & \le {C\parallel \Delta b\parallel }_2^{{\displaystyle \frac{1}{2}}}{\parallel \nabla b\parallel }_2^{{\displaystyle \frac{1}{2}}}{\parallel \nabla v\parallel }_2{\parallel \Delta v\parallel }_2\\ \hfill & \le {C\parallel \Delta b\parallel }_2^{{\displaystyle \frac{3}{2}}}{\parallel \nabla b\parallel }_2^{{\displaystyle \frac{1}{2}}}{\parallel \nabla v\parallel }_2\\ \hfill & \le {\displaystyle \frac{C}{{\zeta }^3}}{\parallel \nabla v\parallel }_2^4{\parallel \nabla b\parallel }_2^2+{\displaystyle \frac{\zeta }{8}}{\parallel \Delta b\parallel }_2^2.\end{array}$$

(17)

Using the cancellation property, we rewrite $K_3+K_4$ as

$$K_3+K_4={\int }_{{\mathbb{R}}^3}(\nabla b·\nabla )b·\nabla vdx+{\int }_{{\mathbb{R}}^3}(\nabla b·\nabla )v·\nabla bdx$$

Hence, one can apply the same procedure as $K_2$ to obtain

$$\begin{array}{cc}\hfill |K_3+K_4|& \le {\displaystyle \frac{C}{{\zeta }^3}}{\parallel \nabla v\parallel }_2^4{\parallel \nabla b\parallel }_2^2+{\displaystyle \frac{\zeta }{4}}{\parallel \Delta b\parallel }_2^2.\end{array}$$

(18)

Based on cancellation property, one obtains

$$\begin{array}{cc}\hfill |K_5|& =|-{\int }_{{\mathbb{R}}^3}\nabla (\nabla \times b)·\nabla ((\nabla \times b)\times b)dx|\\ \hfill & \le {C\parallel \nabla b\parallel }_4^2{\parallel \Delta b\parallel }_2\\ \hfill & \le C\parallel {\nabla }^2{b\parallel }_2^{{\displaystyle \frac{3}{2}}}{\parallel \nabla b\parallel }_2^{{\displaystyle \frac{1}{2}}}{\parallel {\nabla }^{2}b\parallel }_2\\ \hfill & \le {\displaystyle \frac{C}{{\zeta }^3}}{\parallel \Delta b\parallel }_2^4{\parallel \nabla b\parallel }_2^2+{\displaystyle \frac{\zeta }{8}}{\parallel \Delta b\parallel }_2^2.\end{array}$$

(19)

Combining (15)–(19) results in

$$\begin{array}{c}{\displaystyle \frac{d}{dt}}{(\parallel \nabla v\parallel }_2^2+{\parallel \nabla b\parallel }_2^2)+\zeta (\parallel \nabla {\nabla }_p{v\parallel }_2^2+{\parallel \Delta b\parallel }_2^2)\\ \le {\displaystyle \frac{C}{{\zeta }^3}}{(\parallel \nabla v\parallel }_2^4+\parallel {\nabla }^2{b\parallel }_2^4{)(\parallel \nabla b\parallel }_2^2+\parallel {\nabla }_{p}v{\parallel }_2^2).\end{array}$$

Therefore,

$$\begin{array}{cc}\hfill & \underset{0\le t\le T}{sup}{(\parallel \nabla v\parallel }_2^2+{\parallel \nabla b\parallel }_2^2)+\zeta {\int }_0^T(\parallel \nabla {\nabla }_p{v\parallel }_2^2+{\parallel \Delta b\parallel }_2^2)dt\\ \hfill & \le {\displaystyle \frac{C}{{\zeta }^3}}\underset{0\le t\le T}{sup}{(\parallel \nabla v(t)\parallel }_2^4+\parallel {\nabla }^2{b(t)\parallel }_2^4){\int }_0^T{(\parallel \nabla b\parallel }_2^2+\parallel {\nabla }_{p}v{\parallel }_2^2)dt\\ \hfill & +(\parallel \nabla b_0{\parallel }_2^2+\parallel \nabla v_0{\parallel }_2^2).\end{array}$$

(20)

Collecting Lemma 2, (13), (20), one obtains

$$\begin{array}{c}{\displaystyle \underset{0\le t\le T}{sup}}{(\parallel \nabla b\parallel }_2^2+{\parallel \nabla v\parallel }_2^2)+\zeta {\int }_0^T(\parallel {\nabla }^2{b\parallel }_2^2+\parallel \nabla {\nabla }_p{v\parallel }_2^2)dt\\ \le C{\displaystyle \frac{(\parallel v_0{\parallel }_2^2+\parallel b_0{\parallel }_2^2)(\parallel \nabla v_0{\parallel }_2^2+\parallel \nabla b_0{\parallel }_2^2+\parallel {\nabla }^2{v}_0{\parallel }_2^2+\parallel {\nabla }^2{b}_0{\parallel }_2^2)}{{\zeta }^4}}\\ \times {\displaystyle \underset{0\le t\le T}{sup}}{(\parallel \nabla v\parallel }_2^2+{\parallel \nabla b\parallel }_2^2)+(\parallel \nabla v_0{\parallel }_2^2+\parallel \nabla b_0{\parallel }_2^2),\end{array}$$

which, together with (2), and upon choosing a sufficiently small ${ϵ}_0$, yields (14). □

Lemma 4.

Under the same conditions of Lemma 3,

$$\begin{array}{cc}\hfill & \underset{0\le t\le T}{sup}(\parallel {\nabla }^2{v\parallel }_2^2+\parallel {\nabla }^2{b\parallel }_2^2)+2\zeta {\int }_0^T(\parallel {\nabla }^2{\nabla }_p{v\parallel }_2^2+\parallel {\nabla }^{3}b{\parallel }_2^2)dt\\ \hfill & \le \parallel {\nabla }^2{v}_0{\parallel }_2^2+{\parallel {\nabla }^2{b}_0\parallel }_2^2.\end{array}$$

(21)

Proof of Lemma 4.

Similarly to the derivation of (15), one has

$$\begin{array}{cc}\hfill & {\displaystyle \frac{1}{2}}{\displaystyle \frac{d}{dt}}(\parallel {\nabla }^2{b\parallel }_2^2+\parallel {\nabla }^2{v\parallel }_2^2)+{\nu }_1\parallel {\nabla }^2{\partial }_1{v\parallel }_2^2+{\nu }_3\parallel {\nabla }^2{\partial }_3{v\parallel }_2^2+{\parallel {\nabla }^{3}b\parallel }_2^2\\ \hfill & =-\int \Delta (v·\nabla )v·\Delta vdx-\int \Delta {\nabla }^{2}b·(v·\nabla )bdx+\int \Delta v·\Delta (b·\nabla )bdx\\ \hfill & +\int \Delta (b·\nabla )v·\Delta bdx-\int \Delta b·\Delta [\nabla \times ((\nabla \times b)\times b)]dx\\ \hfill & =T_1+T_2+T_3+T_4+T_5.\end{array}$$

(22)

$T_1$ can be rewritten as

$$\begin{array}{cc}\hfill T_1& =-{\int }_{{\mathbb{R}}^3}({\nabla }^{2}v·\nabla )v·{\nabla }^{2}vdx-2{\int }_{{\mathbb{R}}^3}(\nabla v·\nabla )\nabla v·{\nabla }^{2}vdx\\ & =T_{11}+T_{12}.\end{array}$$

Using the same procedure as $H^2$ estimates in [19], we can split $T_{11}$ and $T_{12}$ into three parts, respectively.

$$\begin{array}{cc}\hfill T_{11}& =-{\int }_{{\mathbb{R}}^3}(\nabla {\nabla }_{p}v·\nabla )v·\nabla {\nabla }_{p}vdx\\ \hfill & -{\int }_{{\mathbb{R}}^3}({\partial }_2^2{v}_p·{\nabla }_p)v·{\partial }_2^{2}vdx+{\int }_{{\mathbb{R}}^3}({\partial }_2{\nabla }_p·v_p){\partial }_{2}v·{\partial }_2^{2}vdx\\ \hfill & =T_{111}+T_{112}+T_{113},\end{array}$$

and

$$\begin{array}{cc}\hfill T_{12}& =-2{\int }_{{\mathbb{R}}^3}(\nabla v·\nabla ){\nabla }_{p}v·\nabla {\nabla }_{p}vdx\\ \hfill & -2{\int }_{{\mathbb{R}}^3}({\partial }_2{v}_p·{\nabla }_p){\partial }_{2}v·{\partial }_2^{2}vdx+2{\int }_{{\mathbb{R}}^3}({\nabla }_p·v_p){\partial }_2^{2}v·{\partial }_2^{2}vdx\\ \hfill & =T_{121}+T_{122}+T_{123}.\end{array}$$

By Hölder and Young’s inequality, we obtain

$$\begin{array}{cc}\hfill T_{111}& \le {C\parallel \nabla v\parallel }_{L^3}{\parallel \nabla {\nabla }_{p}v\parallel }_{L^3}^2\\ \hfill & \le {C\parallel \nabla v\parallel }_2^{{\displaystyle \frac{1}{2}}}\parallel {\nabla }^2{v\parallel }_2^{{\displaystyle \frac{1}{2}}}\parallel {\nabla }^2{\nabla }_p{v\parallel }_2^{{\displaystyle \frac{3}{2}}}{\parallel {\nabla }_{p}u\parallel }_2^{{\displaystyle \frac{1}{2}}}\\ \hfill & \le C\parallel {\nabla }^2{\nabla }_p{v\parallel }_2^{{\displaystyle \frac{3}{2}}}{(\parallel \nabla v\parallel }_2+\parallel {\nabla }^2{v\parallel }_2)\parallel {\nabla }_p{v\parallel }_2^{{\displaystyle \frac{1}{2}}}\\ \hfill & \le {\displaystyle \frac{C}{{\zeta }^3}}{(\parallel \nabla v\parallel }_2^4+\parallel {\nabla }^2{v\parallel }_2^4)\parallel {\nabla }_p{v\parallel }_2^2+{\displaystyle \frac{\zeta }{12}}{\parallel {\nabla }^2{\nabla }_{p}v\parallel }_2^2.\end{array}$$

Applying Lemma 1, we estimate $T_{112},T_{113}$ as follows.

$$\begin{array}{cc}\hfill T_{112}& \le C\parallel {\partial }_2^2{v\parallel }_2^{{\displaystyle \frac{1}{2}}}\parallel {\nabla }_p{v\parallel }_2^{{\displaystyle \frac{1}{2}}}\parallel {\partial }_2^2{v\parallel }_2^{{\displaystyle \frac{1}{2}}}\parallel {\partial }_2^2{\nabla }_p{v\parallel }_2^{{\displaystyle \frac{1}{2}}}\parallel {\partial }_3{\nabla }_p{v\parallel }_2^{{\displaystyle \frac{1}{2}}}{\parallel {\partial }_2^2{\nabla }_{p}v\parallel }_2^{{\displaystyle \frac{1}{2}}}\\ \hfill & \le C\parallel {\nabla }^2{\nabla }_p{v\parallel }_2\parallel \nabla {\nabla }_p{v\parallel }_2^{{\displaystyle \frac{1}{2}}}\parallel {\nabla }^2{v\parallel }_2{\parallel {\nabla }_{p}v\parallel }_2^{{\displaystyle \frac{1}{2}}}\\ \hfill & \le C\parallel {\nabla }^2{\nabla }_p{v\parallel }_2^{{\displaystyle \frac{3}{2}}}\parallel {\nabla }^2{v\parallel }_2{\parallel {\nabla }_{p}v\parallel }_2^{{\displaystyle \frac{1}{2}}}\\ \hfill & \le {\displaystyle \frac{C}{{\zeta }^3}}\parallel {\nabla }^2{v\parallel }_2^4\parallel {\nabla }_p{v\parallel }_2^2+{\displaystyle \frac{\zeta }{12}}{\parallel {\nabla }^2{\nabla }_{p}v\parallel }_2^2,\end{array}$$

and

$$\begin{array}{cc}\hfill T_{113}& \le C\parallel {\partial }_2{\nabla }_p{v\parallel }_2^{{\displaystyle \frac{1}{2}}}\parallel {\partial }_2{v\parallel }_2^{{\displaystyle \frac{1}{2}}}\parallel {\partial }_2^2{v\parallel }_2^{{\displaystyle \frac{1}{2}}}\parallel {\partial }_2^2{\nabla }_p{v\parallel }_2^{{\displaystyle \frac{1}{2}}}\parallel {\partial }_2{\nabla }_p{v\parallel }_2^{{\displaystyle \frac{1}{2}}}{\parallel {\partial }_2^2{\nabla }_{p}v\parallel }_2^{{\displaystyle \frac{1}{2}}}\\ \hfill & \le C\parallel {\nabla }^2{\nabla }_p{v\parallel }_2\parallel \nabla {\nabla }_p{v\parallel }_2\parallel {\nabla }^2{v\parallel }_2^{{\displaystyle \frac{1}{2}}}{\parallel \nabla v\parallel }_2^{{\displaystyle \frac{1}{2}}}\\ \hfill & \le C\parallel {\nabla }^2{\nabla }_p{v\parallel }_2^{{\displaystyle \frac{3}{2}}}\parallel {\nabla }^2{v\parallel }_2^{{\displaystyle \frac{1}{2}}}{\parallel \nabla v\parallel }_2^{{\displaystyle \frac{1}{2}}}{\parallel {\nabla }_{p}v\parallel }_2^{{\displaystyle \frac{1}{2}}}\\ \hfill & \le C\parallel {\nabla }^2{\nabla }_p{v\parallel }_2^{{\displaystyle \frac{3}{2}}}(\parallel {\nabla }^2{v\parallel }_2+{\parallel \nabla v\parallel }_2)\parallel {\nabla }_p{v\parallel }_2^{{\displaystyle \frac{1}{2}}}\\ \hfill & \le {\displaystyle \frac{C}{{\zeta }^3}}(\parallel {\nabla }^2{v\parallel }_2^4+{\parallel \nabla v\parallel }_2^4)\parallel {\nabla }_p{v\parallel }_2^2+{\displaystyle \frac{\zeta }{12}}{\parallel {\nabla }^2{\nabla }_{p}v\parallel }_2^2.\end{array}$$

Therefore,

$$T_{11}\le {\displaystyle \frac{C}{{\zeta }^3}}(\parallel {\nabla }^2{v\parallel }_2^4+{\parallel \nabla v\parallel }_2^4)\parallel {\nabla }_p{v\parallel }_2^2+{\displaystyle \frac{\zeta }{4}}{\parallel {\nabla }^2{\nabla }_{p}v\parallel }_2^2.$$

Clearly, $T_{121},T_{122},$ and $T_{123}$ can be estimated as $T_{111},T_{113},$ and $T_{112}$; thus, one has

$$T_{12}\le {\displaystyle \frac{C}{{\zeta }^3}}(\parallel {\nabla }^2{v\parallel }_2^4+{\parallel \nabla v\parallel }_2^4)\parallel {\nabla }_p{v\parallel }_2^2+{\displaystyle \frac{\zeta }{4}}{\parallel {\nabla }^2{\nabla }_{p}v\parallel }_2^2.$$

Hence, we can obtain

$$\begin{array}{c}\hfill T_1\le {\displaystyle \frac{C}{{\zeta }^3}}(\parallel {\nabla }^2{v\parallel }_2^4+{\parallel \nabla v\parallel }_2^4)\parallel {\nabla }_p{v\parallel }_2^2+{\displaystyle \frac{\zeta }{2}}{\parallel {\nabla }^2{\nabla }_{p}v\parallel }_2^2.\end{array}$$

(23)

One decomposes $T_2$ into two terms:

$$\begin{array}{cc}\hfill T_2& =-{\int }_{{\mathbb{R}}^3}({\nabla }^{2}v·\nabla )b·{\nabla }^{2}bdx-2{\int }_{{\mathbb{R}}^3}(\nabla v·\nabla )\nabla b·{\nabla }^{2}bdx\\ & =T_{21}+T_{22}.\end{array}$$

Using interpolation, one obtains

$$\begin{array}{cc}\hfill T_{21}& \le C\parallel {\nabla }^2{v\parallel }_2{\parallel \nabla b\parallel }_{L^3}{\parallel {\nabla }^{2}b\parallel }_{L^6}\\ \hfill & \le C\parallel {\nabla }^2{v\parallel }_2{\parallel \nabla b\parallel }_2^{{\displaystyle \frac{1}{2}}}\parallel {\nabla }^2{b\parallel }_2^{{\displaystyle \frac{1}{2}}}{\parallel {\nabla }^{3}b\parallel }_2\\ \hfill & \le C\parallel {\nabla }^3{b\parallel }_2^{{\displaystyle \frac{3}{2}}}\parallel {\nabla }^2{v\parallel }_2{\parallel \nabla b\parallel }_2^{{\displaystyle \frac{1}{2}}}\\ \hfill & \le {\displaystyle \frac{C}{{\zeta }^3}}\parallel {\nabla }^2{v\parallel }_2^4{\parallel \nabla b\parallel }_2^2+{\displaystyle \frac{\zeta }{14}}{\parallel {\nabla }^{3}b\parallel }_2^2,\end{array}$$

and

$$\begin{array}{cc}\hfill T_{22}& \le {C\parallel \nabla v\parallel }_{L^3}{\parallel {\nabla }^{2}b\parallel }_{L^3}^2\\ \hfill & \le C\parallel {\nabla }^2{v\parallel }_2^{{\displaystyle \frac{1}{2}}}{\parallel \nabla v\parallel }_2^{{\displaystyle \frac{1}{2}}}{\parallel \nabla b\parallel }_2^{{\displaystyle \frac{1}{2}}}{\parallel {\nabla }^{3}b\parallel }_2^{{\displaystyle \frac{3}{2}}}\\ \hfill & \le C\parallel {\nabla }^3{b\parallel }_2^{{\displaystyle \frac{3}{2}}}(\parallel {\nabla }^2{v\parallel }_2+{\parallel \nabla v\parallel }_2{)\parallel \nabla b\parallel }_2^{{\displaystyle \frac{1}{2}}}\\ \hfill & \le {\displaystyle \frac{C}{{\zeta }^3}}(\parallel {\nabla }^2{v\parallel }_2^4+{\parallel \nabla v\parallel }_2^4{)\parallel \nabla b\parallel }_2^2+{\displaystyle \frac{\zeta }{14}}{\parallel {\nabla }^{3}b\parallel }_2^2.\end{array}$$

Thus, one obtains

$$\begin{array}{c}\hfill T_2\le {\displaystyle \frac{C}{{\zeta }^3}}(\parallel {\nabla }^2{v\parallel }_2^4+{\parallel \nabla v\parallel }_2^4{)\parallel \nabla b\parallel }_2^2+{\displaystyle \frac{\zeta }{7}}{\parallel {\nabla }^{3}b\parallel }_2^2.\end{array}$$

(24)

Based on cancellation property, one can write $T_3+T_4$ into four terms:

$$\begin{array}{cc}\hfill & {\int }_{{\mathbb{R}}^3}({\nabla }^{2}b·\nabla )b·{\nabla }^{2}vdx+{\int }_{{\mathbb{R}}^3}({\nabla }^{2}b·\nabla )v·{\nabla }^{2}bdx\\ \hfill & +2{\int }_{{\mathbb{R}}^3}(\nabla b·\nabla )\nabla b·{\nabla }^{2}vdx+2{\int }_{{\mathbb{R}}^3}(\nabla b·\nabla )\nabla v·{\nabla }^{2}bdx\\ \hfill & =T_{341}+T_{342}+T_{343}+T_{344}.\end{array}$$

($T_{341},T_{343},T_{344}$), and $T_{342}$ can be estimated as $T_{21}$ and $T_{22}$; hence, one has

$$\begin{array}{c}\hfill |T_3+T_4|\le {\displaystyle \frac{C}{{\zeta }^3}}(\parallel {\nabla }^2{v\parallel }_2^4+{\parallel \nabla v\parallel }_2^4{)\parallel \nabla b\parallel }_2^2+{\displaystyle \frac{2\zeta }{7}}{\parallel {\nabla }^{3}b\parallel }_2^2.\end{array}$$

(25)

Similarly to (19), we obtain

$$\begin{array}{cc}\hfill |T_5|& =|{\int }_{{\mathbb{R}}^3}\Delta ((\nabla \times b)\times b)·\Delta (\nabla \times b)dx|\\ \hfill & \le C\parallel {\nabla }^3{b\parallel }_2{\parallel \nabla b\parallel }_{L^6}{\parallel {\nabla }^{2}b\parallel }_{L^3}\\ \hfill & \le C\parallel {\nabla }^3{b\parallel }_2\parallel {\nabla }^2{b\parallel }_2\parallel {\nabla }^2{b\parallel }_2^{{\displaystyle \frac{1}{2}}}{\parallel {\nabla }^{3}b\parallel }_2^{{\displaystyle \frac{1}{2}}}\\ \hfill & \le C\parallel {\nabla }^3{b\parallel }_2^{{\displaystyle \frac{3}{2}}}{\parallel {\nabla }^{2}b\parallel }_2^{{\displaystyle \frac{3}{2}}}\\ \hfill & \le {\displaystyle \frac{C}{{\zeta }^3}}\parallel {\nabla }^2{b\parallel }_2^4\parallel {\nabla }^2{b\parallel }_2^2+{\displaystyle \frac{\zeta }{14}}{\parallel {\nabla }^{3}b\parallel }_2^2.\end{array}$$

(26)

Putting (22)–(26) together results in

$$\begin{array}{cc}\hfill & {\displaystyle \frac{d}{dt}}(\parallel {\nabla }^2{v(t)\parallel }_2^2+\parallel {\nabla }^2{b(t)\parallel }_2^2)+\zeta (\parallel {\nabla }^2{\nabla }_p{v\parallel }_2^2+\parallel {\nabla }^{3}b{\parallel }_2^2)\\ \hfill & \le {\displaystyle \frac{C}{{\zeta }^3}}(\parallel {\nabla }^2{v\parallel }_2^4+\parallel {\nabla }^2{b\parallel }_2^4+{\parallel \nabla v\parallel }_2^4{)(\parallel \nabla b\parallel }_2^2+\parallel {\nabla }_{p}v{\parallel }_2^2).\end{array}$$

Hence, one obtains

$$\begin{array}{cc}\hfill & \underset{0\le t\le T}{sup}{(\parallel \Delta b\parallel }_2^2+\parallel {\nabla }^2{v\parallel }_2^2)+\zeta {\int }_0^T(\parallel {\nabla }^2{\nabla }_p{v(t)\parallel }_2^2+\parallel {\nabla }^{3}b(t){\parallel }_2^2)dt\\ \hfill & \le {\displaystyle \frac{C}{{\zeta }^3}}\underset{0\le t\le T}{sup}(\parallel {\nabla }^2{v(t)\parallel }_2^4+\parallel {\nabla }^2{b(t)\parallel }_2^4+{\parallel \nabla v\parallel }_2^4)\\ \hfill & \times {\int }_0^T(\parallel {\nabla }_p{v\parallel }_2^2+{\parallel \nabla b\parallel }_2^2)dt+(\parallel {\nabla }^2{b}_0{\parallel }_2^2+\parallel {\nabla }^2{v}_0{\parallel }_2^2).\end{array}$$

(27)

Combining Lemma 2, (13) and (27), we obtain

$$\begin{array}{cc}\hfill & \underset{0\le t\le T}{sup}(\parallel {\nabla }^2{v\parallel }_2^2+\parallel {\nabla }^2{b\parallel }_2^2)+\zeta {\int }_0^T(\parallel \Delta {\nabla }_p{v\parallel }_2^2+\parallel {\nabla }^{3}b{\parallel }_2^2)dt\\ \hfill & \le C{\displaystyle \frac{(\parallel v_0{\parallel }_2^2+\parallel b_0{\parallel }_2^2)(\parallel v_0{\parallel }_{H^2}^2+\parallel b_0{\parallel }_{H^2}^2)}{{\zeta }^4}}\\ \hfill & \times \underset{0\le t\le T}{sup}(\parallel {\nabla }^2{v\parallel }_2^2+\parallel {\nabla }^2{b\parallel }_2^2+{\parallel \nabla v\parallel }_2^2)+(\parallel {\nabla }^2{v}_0{\parallel }_2^2+\parallel {\nabla }^2{b}_0{\parallel }_2^2),\end{array}$$

which, together with (2), and taking ${ϵ}_0$ as sufficiently small, results in (21). □

Proposition 1.

Suppose $(v,b)$ solves the system (1) with case ${\nu }_2=0,$ and ${\nu }_1,{\nu }_3>0$, and there exists ${ϵ}_0>0$; if (2) and (13) hold, then

$$\begin{array}{c}\hfill G(T)\le (\parallel \nabla v_0{\parallel }_2^2+\parallel \nabla b_0{\parallel }_2^2+\parallel {\nabla }^2{v}_0{\parallel }_2^2+\parallel {\nabla }^2{b}_0{\parallel }_2^2).\end{array}$$

(28)

Proof of Proposition 1.

Based on the following Lemmas 3 and 4, adding (14) and (21) results in

$$\begin{array}{cc}\hfill & \underset{0\le t\le T}{sup}{(\parallel \nabla v\parallel }_2^2+{\parallel \nabla b\parallel }_2^2\parallel {\nabla }^2{v\parallel }_2^2+\parallel {\nabla }^{2}b{\parallel }_2^2)\\ \hfill & +2\zeta {\int }_0^T(\parallel \nabla {\nabla }_p{v\parallel }_2^2+\parallel {\nabla }^2{b\parallel }_2^2+\parallel {\nabla }^2{\nabla }_p{v\parallel }_2^2+\parallel {\nabla }^{3}b{\parallel }_2^2)dt\\ \hfill & \le \parallel \nabla v_0{\parallel }_2^2+\parallel \nabla b_0{\parallel }_2^2\parallel {\nabla }^2{v}_0{\parallel }_2^2+{\parallel {\nabla }^2{b}_0\parallel }_2^2.\end{array}$$

Therefore,

$$\begin{array}{cc}\hfill & \underset{0\le t\le T}{sup}{(\parallel \nabla v\parallel }_2^2+{\parallel \nabla b\parallel }_2^2\parallel {\nabla }^2{v\parallel }_2^2+\parallel {\nabla }^{2}b{\parallel }_2^2)\\ \hfill & \le (\parallel \nabla v_0{\parallel }_2^2+\parallel \nabla b_0{\parallel }_2^2\parallel {\nabla }^2{v}_0{\parallel }_2^2+\parallel {\nabla }^2{b}_0{\parallel }_2^2).\end{array}$$

This completes the proof of Proposition 1. □

From Proposition 1, one deduces energy estimates as follows.

Proposition 2.

Under the same conditions as in Proposition 1,

$$\begin{array}{cc}\hfill & \underset{0\le t\le T}{sup}(\parallel {\nabla }^3{v(t)\parallel }_2^2+\parallel {\nabla }^{3}b(t){\parallel }_2^2)\\ \hfill & +2\zeta {\int }_0^T(\parallel {\nabla }^3{\nabla }_p{v\parallel }_2^2+\parallel {\nabla }^{4}b{\parallel }_2^2)dt\\ \hfill & \le C(\parallel {\nabla }^3{v}_0{\parallel }_2^2+\parallel {\nabla }^3{b}_0{\parallel }_2^2).\end{array}$$

(29)

Proof of Proposition  2.

Similarly to (15), one obtains

$$\begin{array}{cc}\hfill & {\displaystyle \frac{1}{2}}{\displaystyle \frac{d}{dt}}(\parallel {\nabla }^3{v(t)\parallel }_2^2+\parallel {\nabla }^{3}b(t){\parallel }_2^2)\\ \hfill & +{\nu }_1\parallel {\nabla }^3{\partial }_1{v\parallel }_2^2+{\nu }_3\parallel {\nabla }^3{\partial }_2{v\parallel }_2^2+{\parallel {\nabla }^{4}b\parallel }_2^2\\ \hfill & =-{\int }_{{\mathbb{R}}^3}{\nabla }^3(v·\nabla )v·{\nabla }^{3}vdx-{\int }_{{\mathbb{R}}^3}{\nabla }^3(v·\nabla )b·{\nabla }^{3}bdx\\ \hfill & +{\int }_{{\mathbb{R}}^3}{\nabla }^3(b·\nabla )b·{\nabla }^{3}vdx+{\int }_{{\mathbb{R}}^3}{\nabla }^3(b·\nabla )v·{\nabla }^{3}bdx\\ \hfill & -{\int }_{{\mathbb{R}}^3}{\nabla }^3[\nabla \times ((\nabla \times b)\times b)]·{\nabla }^{3}bdx\\ \hfill & =X_1+X_2+X_3+X_4+X_5.\end{array}$$

(30)

One can rewrite $X_1$ into three parts:

$$\begin{array}{cc}\hfill & -{\int }_{{\mathbb{R}}^3}({\nabla }^{3}v·\nabla )v·{\nabla }^{3}vdx-3{\int }_{{\mathbb{R}}^3}({\nabla }^{2}v·\nabla )\nabla v·{\nabla }^{3}vdx\\ \hfill & -3{\int }_{{\mathbb{R}}^3}(\nabla v·\nabla ){\nabla }^{2}v·{\nabla }^{3}vdx=X_{11}+X_{12}+X_{13}.\end{array}$$

Similarly to the $H^3$ estimates in paper [19], $X_{11}$ and $X_{12}$ can be decomposed into three parts, respectively.

$$\begin{array}{cc}\hfill & -{\int }_{{\mathbb{R}}^3}({\nabla }^2{\nabla }_{p}v·\nabla )v·{\nabla }^2{\nabla }_{p}vdx-{\int }_{{\mathbb{R}}^3}({\partial }_2^3{v}_p·{\nabla }_p)v·{\partial }_2^{3}vdx\\ \hfill & +{\int }_{{\mathbb{R}}^3}({\partial }_2^2{\nabla }_p·v_p){\partial }_{2}v·{\partial }_2^{3}vdx=X_{111}+X_{112}+X_{113},\end{array}$$

and

$$\begin{array}{cc}\hfill & -3{\int }_{{\mathbb{R}}^3}(\nabla {\nabla }_{p}v·\nabla )\nabla v·{\nabla }^2{\nabla }_{p}vdx-3{\int }_{{\mathbb{R}}^3}({\partial }_2^2{v}_p·{\nabla }_p){\partial }_{2}v·{\partial }_2^{2}vdx\\ \hfill & +3{\int }_{{\mathbb{R}}^3}({\partial }_2{\nabla }_p·v_p){\partial }_2^{2}v·{\partial }_2^{2}vdx=X_{121}+X_{122}+X_{123}.\end{array}$$

Using Hölder inequality results in

$$\begin{array}{cc}\hfill X_{111}& \le C\parallel {\nabla }^2{\nabla }_p{v\parallel }_2{\parallel \nabla v\parallel }_{L^6}{\parallel {\nabla }^2{\nabla }_{p}v\parallel }_{L^3}\\ \hfill & \le C\parallel {\nabla }^2{\nabla }_p{v\parallel }_2\parallel {\nabla }^2{v\parallel }_2\parallel {\nabla }^3{\nabla }_p{v\parallel }_2^{{\displaystyle \frac{1}{2}}}{\parallel {\nabla }^2{\nabla }_{p}v\parallel }_2^{{\displaystyle \frac{1}{2}}}\\ \hfill & \le C\parallel {\nabla }^3{\nabla }_p{v\parallel }_2^{{\displaystyle \frac{1}{2}}}\parallel {\nabla }^2{\nabla }_p{v\parallel }_2^{{\displaystyle \frac{3}{2}}}{\parallel {\nabla }^{2}v\parallel }_2\\ \hfill & \le C\parallel {\nabla }^3{\nabla }_p{v\parallel }_2^{{\displaystyle \frac{1}{2}}}\parallel {\nabla }^3{v\parallel }_2^{{\displaystyle \frac{3}{2}}}{\parallel {\nabla }^{2}v\parallel }_2\\ \hfill & \le C\parallel {\nabla }^2{v\parallel }_2^{{\displaystyle \frac{4}{3}}}\parallel {\nabla }^3{v\parallel }_2^2+{\displaystyle \frac{\zeta }{18}}{\parallel {\nabla }^3{\nabla }_{p}v\parallel }_2^2.\end{array}$$

By Lemma 1, one can estimate $X_{112},X_{113}$ as follows:

$$\begin{array}{cc}\hfill X_{112}& \le C\parallel {\partial }_2^3{v\parallel }_2^{{\displaystyle \frac{1}{2}}}\parallel {\nabla }_p{v\parallel }_2^{{\displaystyle \frac{1}{2}}}\parallel {\partial }_2^3{u\parallel }_2^{{\displaystyle \frac{1}{2}}}\parallel {\partial }_2^3{\nabla }_p{v\parallel }_2^{{\displaystyle \frac{1}{2}}}\parallel {\partial }_2{\nabla }_p{v\parallel }_2^{{\displaystyle \frac{1}{2}}}{\parallel {\partial }_2^3{\nabla }_{p}v\parallel }_2^{{\displaystyle \frac{1}{2}}}\\ \hfill & \le C\parallel {\nabla }^3{\nabla }_p{v\parallel }_2\parallel {\nabla }^3{v\parallel }_2\parallel \nabla {\nabla }_p{v\parallel }_2^{{\displaystyle \frac{1}{2}}}{\parallel \nabla v\parallel }_2^{{\displaystyle \frac{1}{2}}}\\ \hfill & \le C\parallel {\nabla }^3{\nabla }_p{v\parallel }_2\parallel {\nabla }^3{v\parallel }_2(\parallel \nabla {\nabla }_p{v\parallel }_2+{\parallel \nabla v\parallel }_2)\\ \hfill & \le C(\parallel \nabla {\nabla }_p{v\parallel }_2^2+{\parallel \nabla v\parallel }_2^2)\parallel {\nabla }^3{v\parallel }_2^2+{\displaystyle \frac{\zeta }{18}}{\parallel {\nabla }^3{\nabla }_{p}v\parallel }_2^2,\end{array}$$

and

$$\begin{array}{cc}\hfill X_{113}& \le C\parallel {\partial }_2^2{\nabla }_p{v\parallel }_2^{{\displaystyle \frac{1}{2}}}\parallel {\partial }_2{v\parallel }_2^{{\displaystyle \frac{1}{2}}}\parallel {\partial }_2^3{v\parallel }_2^{{\displaystyle \frac{1}{2}}}\parallel {\partial }_2^3{\nabla }_p{v\parallel }_2^{{\displaystyle \frac{1}{2}}}\parallel {\partial }_2{\nabla }_p{v\parallel }_2^{{\displaystyle \frac{1}{2}}}{\parallel {\partial }_2^3{\nabla }_{p}v\parallel }_2^{{\displaystyle \frac{1}{2}}}\\ \hfill & \le C\parallel {\nabla }^3{\nabla }_p{v\parallel }_2\parallel {\nabla }^3{v\parallel }_2^{{\displaystyle \frac{1}{2}}}\parallel {\nabla }^2{\nabla }_p{v\parallel }_2^{{\displaystyle \frac{1}{2}}}\parallel \nabla {\nabla }_p{v\parallel }_2^{{\displaystyle \frac{1}{2}}}{\parallel \nabla v\parallel }_2^{{\displaystyle \frac{1}{2}}}\\ \hfill & \le C\parallel {\nabla }^3{\nabla }_p{v\parallel }_2\parallel {\nabla }^3{v\parallel }_2(\parallel \nabla {\nabla }_p{v\parallel }_2+{\parallel \nabla v\parallel }_2)\\ \hfill & \le C(\parallel \nabla {\nabla }_p{v\parallel }_2^2+{\parallel \nabla v\parallel }_2^2)\parallel {\nabla }^3{v\parallel }_2^2+{\displaystyle \frac{\zeta }{18}}{\parallel {\nabla }^3{\nabla }_{p}v\parallel }_2^2.\end{array}$$

Therefore,

$$X_{11}\le C(\parallel \nabla {\nabla }_p{v\parallel }_2^2+{\parallel \nabla v\parallel }_2^2+\parallel {\nabla }^2{v\parallel }_2^2)\parallel {\nabla }^3{v\parallel }_2^2+{\displaystyle \frac{\zeta }{6}}{\parallel {\nabla }^3{\nabla }_{p}v\parallel }_2^2.$$

Via applying a similar process as with $X_{111}$, one obtains

$$\begin{array}{cc}\hfill X_{121}& \le C\parallel \nabla {\nabla }_p{v\parallel }_{L^6}\parallel {\nabla }^2{v\parallel }_2{\parallel {\nabla }^2{\nabla }_{p}v\parallel }_{L^3}\\ \hfill & \le C\parallel {\nabla }^2{\nabla }_p{v\parallel }_2\parallel {\nabla }^2{v\parallel }_2\parallel {\nabla }^3{\nabla }_p{v\parallel }_2^{{\displaystyle \frac{1}{2}}}{\parallel {\nabla }^2{\nabla }_{p}v\parallel }_2^{{\displaystyle \frac{1}{2}}}\\ \hfill & \le C\parallel {\nabla }^3{\nabla }_p{v\parallel }_2^{{\displaystyle \frac{1}{2}}}\parallel {\nabla }^2{\nabla }_p{v\parallel }_2^{{\displaystyle \frac{3}{2}}}{\parallel {\nabla }^{2}v\parallel }_2\\ \hfill & \le C\parallel {\nabla }^3{\nabla }_p{v\parallel }_2^{{\displaystyle \frac{1}{2}}}\parallel {\nabla }^3{v\parallel }_2^{{\displaystyle \frac{3}{2}}}{\parallel {\nabla }^{2}v\parallel }_2\\ \hfill & \le C\parallel {\nabla }^2{v\parallel }_2^{{\displaystyle \frac{4}{3}}}\parallel {\nabla }^3{v\parallel }_2^2+{\displaystyle \frac{\zeta }{18}}{\parallel {\nabla }^3{\nabla }_{p}v\parallel }_2^2.\end{array}$$

Based on Lemma 1, we estimate $X_{122}$ as follows:

$$\begin{array}{cc}\hfill X_{122}& \le C\parallel {\partial }_2^2{v\parallel }_2^{{\displaystyle \frac{1}{2}}}\parallel {\partial }_2{\nabla }_p{v\parallel }_2^{{\displaystyle \frac{1}{2}}}\parallel {\partial }_2^3{v\parallel }_2^{{\displaystyle \frac{1}{2}}}\parallel {\partial }_2^2{\nabla }_p{v\parallel }_2^{{\displaystyle \frac{1}{2}}}\parallel {\partial }_2^2{\nabla }_p{v\parallel }_2^{{\displaystyle \frac{1}{2}}}{\parallel {\partial }_2^3{\nabla }_{p}v\parallel }_2^{{\displaystyle \frac{1}{2}}}\\ \hfill & \le C\parallel {\nabla }^3{\nabla }_p{v\parallel }_2^{{\displaystyle \frac{1}{2}}}\parallel {\nabla }^2{\nabla }_p{v\parallel }_2\parallel {\nabla }^3{v\parallel }_2^{{\displaystyle \frac{1}{2}}}\parallel \nabla {\nabla }_p{v\parallel }_2^{{\displaystyle \frac{1}{2}}}{\parallel {\nabla }^{2}v\parallel }_2^{{\displaystyle \frac{1}{2}}}\\ \hfill & \le C\parallel {\nabla }^3{\nabla }_p{v\parallel }_2^{{\displaystyle \frac{1}{2}}}\parallel {\nabla }^3{v\parallel }_2^{{\displaystyle \frac{3}{2}}}{\parallel {\nabla }^{2}v\parallel }_2\\ \hfill & \le C\parallel {\nabla }^2{v\parallel }_2^{{\displaystyle \frac{4}{3}}}\parallel {\nabla }^3{v\parallel }_2^2+{\displaystyle \frac{\zeta }{18}}{\parallel {\nabla }^3{\nabla }_{p}v\parallel }_2^2.\end{array}$$

One can apply same procedure as $X_{122}$ to obtain

$$X_{123}\le C\parallel {\nabla }^2{v\parallel }_2^{{\displaystyle \frac{4}{3}}}\parallel {\nabla }^3{v\parallel }_2^2+{\displaystyle \frac{\zeta }{18}}{\parallel {\nabla }^3{\nabla }_{p}v\parallel }_2^2.$$

Thus,

$$X_{12}\le C\parallel {\nabla }^2{v\parallel }_2^{{\displaystyle \frac{4}{3}}}\parallel {\nabla }^3{v\parallel }_2^2+{\displaystyle \frac{\zeta }{6}}{\parallel {\nabla }^3{\nabla }_{p}v\parallel }_2^2.$$

Obviously, $X_{13}$ can be estimated as $X_{11}$; hence,

$$X_{13}\le C(\parallel \nabla {\nabla }_p{v\parallel }_2^2+{\parallel \nabla v\parallel }_2^2+\parallel {\nabla }^2{v\parallel }_2^2)\parallel {\nabla }^3{v\parallel }_2^2+{\displaystyle \frac{\zeta }{6}}{\parallel {\nabla }^3{\nabla }_{p}v\parallel }_2^2.$$

Collecting these above estimates, we obtain

$$\begin{array}{c}\hfill X_1\le C(\parallel \nabla {\nabla }_p{v\parallel }_2^2+\parallel {\nabla }^2{v\parallel }_2^2)\parallel {\nabla }^3{v\parallel }_2^2+{\displaystyle \frac{\zeta }{2}}{\parallel {\nabla }^3{\nabla }_{p}v\parallel }_2^2.\end{array}$$

(31)

One can write $X_2$ into three parts:

$$\begin{array}{cc}\hfill & -3{\int }_{{\mathbb{R}}^3}(\nabla v·\nabla ){\nabla }^{2}b·{\nabla }^{3}bdx-3{\int }_{{\mathbb{R}}^3}({\nabla }^{2}v·\nabla )\nabla b·{\nabla }^{3}bdx\\ \hfill & -{\int }_{{\mathbb{R}}^3}({\nabla }^{3}v·\nabla )v·{\nabla }^{3}vdx=X_{21}+X_{22}+X_{23}.\end{array}$$

Applying Young’s inequality and interpolation results in

$$\begin{array}{cc}\hfill X_{21}& \le {C\parallel \nabla v\parallel }_{L^6}\parallel {\nabla }^3{b\parallel }_2{\parallel {\nabla }^{3}b\parallel }_{L^3}\\ \hfill & \le C\parallel {\nabla }^2{v\parallel }_2\parallel {\nabla }^3{b\parallel }_2\parallel {\nabla }^3{b\parallel }_2^{{\displaystyle \frac{1}{2}}}{\parallel {\nabla }^{4}b\parallel }_2^{{\displaystyle \frac{1}{2}}}\\ \hfill & \le C\parallel {\nabla }^2{v\parallel }_2\parallel {\nabla }^3{b\parallel }_2^{{\displaystyle \frac{3}{2}}}{\parallel {\nabla }^{4}b\parallel }_2^{{\displaystyle \frac{1}{2}}}\\ \hfill & \le C\parallel {\nabla }^2{v\parallel }_2^{{\displaystyle \frac{4}{3}}}\parallel {\nabla }^3{b\parallel }_2^2+{\displaystyle \frac{\zeta }{22}}{\parallel {\nabla }^{4}b\parallel }_2^2.\end{array}$$

In a similar manner, one can obtain

$$\begin{array}{cc}\hfill X_{22}& \le C\parallel {\nabla }^2{v\parallel }_2\parallel {\nabla }^2{b\parallel }_{L^6}{\parallel {\nabla }^{3}b\parallel }_{L^3}\\ \hfill & \le C\parallel {\nabla }^2{v\parallel }_2\parallel {\nabla }^3{b\parallel }_2\parallel {\nabla }^3{B\parallel }_2^{{\displaystyle \frac{1}{2}}}{\parallel {\nabla }^{4}b\parallel }_2^{{\displaystyle \frac{1}{2}}}\\ \hfill & \le C\parallel {\nabla }^2{v\parallel }_2\parallel {\nabla }^3{b\parallel }_2^{{\displaystyle \frac{3}{2}}}{\parallel {\nabla }^{4}b\parallel }_2^{{\displaystyle \frac{1}{2}}}\\ \hfill & \le C\parallel {\nabla }^2{v\parallel }_2^{{\displaystyle \frac{4}{3}}}\parallel {\nabla }^3{b\parallel }_2^2+{\displaystyle \frac{\zeta }{22}}{\parallel {\nabla }^{4}b\parallel }_2^2.\end{array}$$

and

$$\begin{array}{cc}\hfill X_{23}& \le C\parallel {\nabla }^3{v\parallel }_2{\parallel \nabla b\parallel }_{L^3}{\parallel {\nabla }^{3}b\parallel }_6\\ \hfill & \le C\parallel {\nabla }^3{v\parallel }_2{\parallel \nabla b\parallel }_2^{{\displaystyle \frac{1}{2}}}\parallel {\nabla }^2{b\parallel }_2^{{\displaystyle \frac{1}{2}}}{\parallel {\Delta }^{2}b\parallel }_2\\ \hfill & \le C\parallel {\nabla }^3{v\parallel }_2{(\parallel \nabla b\parallel }_2+{\parallel \Delta b\parallel }_2)\parallel {\Delta }^2{b\parallel }_2\\ \hfill & \le {\displaystyle \frac{\zeta }{22}}\parallel {\Delta }^2{b\parallel }_2^2+{C(\parallel \Delta b\parallel }_2^2+{\parallel \nabla b\parallel }_2^2)\parallel {\nabla }^3{v\parallel }_2^2.\end{array}$$

Hence, we obtain

$$\begin{array}{c}\hfill X_2\le C(\parallel {\nabla }^2{v\parallel }_2^{{\displaystyle \frac{4}{3}}}+\parallel {\nabla }^2{b\parallel }_2^2)(\parallel {\nabla }^3{v\parallel }_2^2+\parallel {\nabla }^3{b\parallel }_2^2)+{\displaystyle \frac{3\zeta }{22}}{\parallel {\nabla }^{4}b\parallel }_2^2.\end{array}$$

(32)

Applying cancellation property, one writes $X_3+X_4$ into six parts:

$$\begin{array}{cc}\hfill & {\int }_{{\mathbb{R}}^3}({\nabla }^{3}b·\nabla )b·{\nabla }^{3}vdx+3{\int }_{{\mathbb{R}}^3}({\nabla }^{2}b·\nabla ){\nabla }^{3}v·\nabla bdx\\ \hfill & +3{\int }_{{\mathbb{R}}^3}(\nabla b·\nabla ){\nabla }^{2}b·{\nabla }^{3}vdx+{\int }_{{\mathbb{R}}^3}({\nabla }^{3}b·\nabla )v·{\nabla }^{3}bdx\\ \hfill & +3{\int }_{{\mathbb{R}}^3}(\nabla b·\nabla ){\nabla }^{3}b·{\nabla }^{2}vdx+3{\int }_{{\mathbb{R}}^3}({\nabla }^{2}b·\nabla ){\nabla }^{3}b·\nabla udx\\ \hfill & =X_{341}+X_{342}+X_{343}+X_{344}+X_{345}+X_{346}.\end{array}$$

$X_{342}$ can be further divided into two terms:

$$-3{\int }_{{\mathbb{R}}^3}({\nabla }^{3}b·\nabla )\nabla b·{\nabla }^{2}vdx-3{\int }_{{\mathbb{R}}^3}({\nabla }^{2}b·\nabla ){\nabla }^{2}b·{\nabla }^{2}vdx.$$

$(X_{341},X_{343},X_{346}),(X_{342},X_{345}),$ and $X_{344}$ can be estimated as $X_{21},X_{22},$ and $X_{23}$; therefore,

$$\begin{array}{c}\hfill |X_3+X_4|\le C(\parallel {\nabla }^2{v\parallel }_2^{{\displaystyle \frac{4}{3}}}+\parallel {\nabla }^2{b\parallel }_2^2)(\parallel {\nabla }^3{v\parallel }_2^2+\parallel {\nabla }^3{b\parallel }_2^2)+{\displaystyle \frac{7\zeta }{22}}{\parallel {\nabla }^{4}b\parallel }_2^2.\end{array}$$

(33)

One can apply a similar process as in (13) to deduce that

$$\begin{array}{cc}\hfill |X_5|& =|-{\int }_{{\mathbb{R}}^3}{\nabla }^3(\nabla \times b)·{\nabla }^3[(\nabla \times b)\times b]dx|\\ \hfill & \le C\parallel {\nabla }^4{b\parallel }_2{\parallel \nabla b\parallel }_{L^6}{\parallel {\nabla }^{3}b\parallel }_{L^3}\\ \hfill & \le C\parallel {\nabla }^4{b\parallel }_2\parallel {\nabla }^2{b\parallel }_2\parallel {\nabla }^3{b\parallel }_2^{{\displaystyle \frac{1}{2}}}{\parallel {\nabla }^{4}b\parallel }_2^{{\displaystyle \frac{1}{2}}}\\ \hfill & \le C\parallel {\nabla }^4{b\parallel }_2^{{\displaystyle \frac{3}{2}}}\parallel {\nabla }^2{b\parallel }_2{\parallel {\nabla }^{3}b\parallel }_2^{{\displaystyle \frac{1}{2}}}\\ \hfill & \le C\parallel {\nabla }^2{b\parallel }_2^4\parallel {\nabla }^3{b\parallel }_2^2+{\displaystyle \frac{\zeta }{22}}{\parallel {\nabla }^{4}b\parallel }_2^2.\end{array}$$

(34)

Combining (30)–(34) results in

$$\begin{array}{cc}\hfill & {\displaystyle \frac{d}{dt}}(\parallel {\nabla }^3{b\parallel }_2^2+\parallel {\nabla }^3{v\parallel }_2^2)+\zeta (\parallel {\nabla }^3{\nabla }_p{v\parallel }_2^2+\parallel {\nabla }^{4}b{\parallel }_2^2)\\ \hfill & \le C(\parallel {\nabla }^2{v\parallel }_2^2+\parallel {\nabla }^2{b\parallel }_2^4+{\parallel \nabla b\parallel }_2^2)(\parallel {\nabla }^3{v\parallel }_2^2+\parallel {\nabla }^{3}b{\parallel }_2^2).\end{array}$$

Based on Gronwall’s inequality, we obtain

$$\begin{array}{cc}\hfill & \underset{0\le t\le T}{sup}(\parallel {\nabla }^3{v\parallel }_2^2+\parallel {\nabla }^3{b\parallel }_2^2)+\zeta {\int }_0^T(\parallel {\nabla }^3{\nabla }_p{v\parallel }_2^2+\parallel {\nabla }^{4}b{\parallel }_2^2)dt\\ \hfill & \le (\parallel {\nabla }^3{v}_0{\parallel }_2^2+\parallel {\nabla }^3{b}_0{\parallel }_2^2)[C{\int }_0^T{(\parallel \nabla b\parallel }_2^2+{\parallel \Delta b\parallel }_2^4+{\parallel \Delta v\parallel }_2^2)dt]\\ \hfill & \times exp(C{\int }_0^T(\parallel {\nabla }^2{v\parallel }_2^2+\parallel {\nabla }^2{b\parallel }_2^4+{\parallel \nabla b\parallel }_2^2)dt).\end{array}$$

The above inequality, together with Lemmas 3 and 4, results in (29). □

6. Global Existence

Proof of Theorem 2.

Based on the Theorem 1 and Propositions 1 and 2, one can imply the following:

$$\begin{array}{cc}\hfill & {\displaystyle \frac{d}{dt}}{(\parallel b\parallel }_{H^3}^2+{\parallel v\parallel }_{H^3}^2)+\zeta (\parallel {\nabla }_p{v\parallel }_{H^3}^2+{\parallel \nabla b\parallel }_{H^3}^2)\\ \hfill & \le {C(\parallel v\parallel }_{H^2}^2+{\parallel b\parallel }_{H^2}^4+{\parallel b\parallel }_{H^1}^2{)(\parallel v\parallel }_{H^3}^2+{\parallel b\parallel }_{H^3}^2).\end{array}$$

Applying Gronwall’s inequality results in

$$\begin{array}{cc}\hfill & \underset{0\le t\le T}{sup}{(\parallel v\parallel }_{H^3}^2+{\parallel b\parallel }_{H^3}^2)+\zeta {\int }_0^T(\parallel {\nabla }_p{v\parallel }_{H^3}^2+{\parallel \nabla b\parallel }_{H^3}^2)dt\\ \hfill & \le C{\displaystyle \frac{(\parallel v_0{\parallel }_2^2+\parallel b_0{\parallel }_2^2)(\parallel v_0{\parallel }_{H^2}^2+\parallel b_0{\parallel }_{H^2}^2)}{{\zeta }^4}}\\ \hfill & \times \underset{0\le t\le T}{sup}{(\parallel v\parallel }_{H^3}^2+{\parallel b\parallel }_{H^3}^2+{\parallel v\parallel }_{H^1}^2)+(\parallel v_0{\parallel }_{H^3}^2+\parallel b_0{\parallel }_{H^3}^2).\end{array}$$

We can observe that $(\parallel v_0{\parallel }_2^2+\parallel b_0{\parallel }_2^2)(\parallel v_0{\parallel }_{H^2}^2+\parallel b_0{\parallel }_{H^2}^2)/{\zeta }^4\le {ϵ}_0$; therefore, if ${ϵ}_0<{\displaystyle \frac{1}{2C}}$, then

$$\begin{array}{cc}\hfill & {\parallel v\parallel }_{H^3}^2+{\parallel b\parallel }_{H^3}^2+2\zeta {\int }_0^T(\parallel {\nabla }_p{v\parallel }_{H^3}^2+{\parallel \nabla b\parallel }_{H^3}^2)dt\\ \hfill & \le 2(\parallel v_0{\parallel }_{H^3}^2+\parallel b_0{\parallel }_{H^3}^2),\end{array}$$

which ends the existence of classical solutions that can be obtained; one can use the $L^2$-method to prove the uniqueness of classical solution. □

7. Conclusions

In this paper, we establish the global existence and the local existence of classical solutions to the incompressible resistive Hall-magnetohydrodynamic equations with partial viscosity terms. There exist two main difficulties: the first one is due to the partial viscosity terms that mean there is no smoothing effect on the vertical derivative; the other one arises from the Hall term, which is quadratic in the magnetic field and involves second-order derivatives. To overcome these two difficulties, we apply delicate estimates for the energy method. It is very important to establish the global regularity for three-dimensional Hall-magnetohydrodynamic equations with partial magnetic diffusion, which need new technical means to overcome the difficulties that arise from the Hall term.