Regularity criteria for the 3D generalized MHD and Hall-MHD systems

Abstract

In this paper, we consider regularity criteria for the 3D generalized MHD and Hall-MHD systems with fractional dissipative terms. Some scaling invariant regularity criteria are established for the two systems. Global regularity for the Hall-MHD equation is also proved for the case $\alpha \ge \frac{5}{4}, \beta \ge \frac{7}{4}$.

1 Introduction

We consider the following generalized MHD system:

$$\begin{aligned}&u_t+u\cdot \nabla u+\nabla p+\Lambda ^{2\alpha }u-b\cdot \nabla b=0,\quad (x,t)\in \mathbb {R}^3\times \mathbb {R}^+ , \end{aligned}$$

(1.1)

$$\begin{aligned}&b_t+u\cdot \nabla b+\Lambda ^{2\beta } b -b\cdot \nabla u=0, \quad (x,t)\in \mathbb {R}^3\times \mathbb {R}^+ ,\end{aligned}$$

(1.2)

$$\begin{aligned}&\mathrm {div} u=\mathrm {div} b=0,\quad (x,t)\in \mathbb {R}^3\times \mathbb {R}^+,\end{aligned}$$

(1.3)

$$\begin{aligned}&(u,b)(x,0)=(u_0,b_0),\quad x\in \mathbb {R}^3, \end{aligned}$$

(1.4)

here $u = u(x,t)\in \mathbb {R}^3,~ b = b(x,t)\in \mathbb {R}^3,~ p = p(x, t)\in \mathbb {R}$ represent the unknown velocity field, the magnetic field, and the pressure, respectively. $\alpha \ge 0, \beta \ge 0$ are real parameters. We identify the case $\alpha =\beta =0$ as the GMHD system with zero magnetic and zero velocity diffusivity. $\Lambda =(-\Delta )^\frac{1}{2}$ is defined in terms of Fourier transform by $\widehat{\Lambda f}(\xi )=|\xi |\hat{f}(\xi )$.

The existence of weak solutions for (1.1)–(1.4) is given in [29] for any $u_0, b_0\in L^2(\mathbb {R}^3)$ with $\mathrm {div} u_0 = \mathrm {div} b_0 = 0$ in $\mathbb {R}^3$. It is showed that if $\alpha \ge \frac{1}{2}+\frac{N}{4}, \alpha +\beta \ge 1+\frac{N}{2}$, then the solution (u, b)(x, t) remains smooth for all time (refer [29, 30] for details). The special case $\alpha =\beta =\frac{5}{4}$ for 3D can also be found in [33] via a different approach. For the 2D case, global existence results were established in [14, 19, 24]. In [33], Zhou also proved that if $1\le \alpha =\beta <\frac{5}{4}$

$$\begin{aligned} u(x,t)\in L^p(0,T;L^q), \quad with \quad \frac{2\alpha }{p}+\frac{3}{q}\le 2\alpha -1, \quad \frac{3}{2\alpha -1}<q\le \infty \end{aligned}$$

or

$$\begin{aligned} \Lambda ^\alpha u(x,t)\in L^p(0,T;L^q), \quad with \quad \frac{2\alpha }{p}+\frac{3}{q}\le 3\alpha -1, \quad \frac{3}{3\alpha -1}<q\le \frac{3}{\alpha -1}, \end{aligned}$$

then the solution remains smooth on (0, T]. Other regularity criteria were shown in [9, 11, 18]

When $\alpha =\beta =1$, the system (1.1)–(1.4) is reduced to the classical MHD system

$$\begin{aligned}&u_t+u\cdot \nabla u+\nabla p-\Delta u-b\cdot \nabla b=0,\quad (x,t)\in \mathbb {R}^3\times \mathbb {R}^+ , \\&b_t+u\cdot \nabla b-\Delta b -b\cdot \nabla u=0, \quad (x,t)\in \mathbb {R}^3\times \mathbb {R}^+ , \\&\mathrm {div} u=\mathrm {div} b=0, \quad (x,t)\in \mathbb {R}^3\times \mathbb {R}^+, \\&(u,b)(x,0)=(u_0,b_0),\quad x\in \mathbb {R}^3. \end{aligned}$$

In [22], it was proved that the classical MHD system is locally well posed for any given initial datum $u_0, b_0\in H^s, s\ge 3$. Recently, some regularity criteria for Navier-Stokes equations [1, 23] have been extended to the MHD system in [16, 21, 31]. Later, Zhou established some Serrin-type regularity criteria on the pressure in [32]. The Beale-Kato-Majda-type regularity criterion($curl\, u, curl\, b\in L^1(0,T;L^\infty )$) was given in [3] for the ideal MHD ($\alpha =\beta =0$) system. From [33] (or [2] for the Navier-Stokes equations), we know that if $\alpha =\beta$ and (u, b) is a solution to the system (1.1)–(1.4), then $(u_\lambda , b_\lambda )$ with any $\lambda >0$ is also a solution, where $u_\lambda (x,t)=\lambda ^{2\alpha -1}u(\lambda x,\lambda ^{2\alpha }t)$ and $b_\lambda (x,t)=\lambda ^{2\alpha -1}b(\lambda x,\lambda ^{2\alpha }t)$. By direct calculation, we obtain that the norms $\Vert u\Vert _{L^{p,q}}$ and $\Vert \Lambda ^\gamma u\Vert _{L^{p,q}}$ are scaling dimension zero for $\frac{2\alpha }{p}+\frac{3}{q}=2\alpha -1$ and $\frac{2\alpha }{p}+\frac{3}{q}=2\alpha +\gamma -1$, respectively.

Very recently, Jiang and Zhou proved the local existence and uniqueness of strong solutions for the generalized MHD system as follows

Theorem 1.1

[17] For $s>\max \{\frac{n}{2}+1-\alpha ,1\}$, and initial data $(u_0,b_0)\in H^s(\mathbb {R}^n)$ with $div u_0=div b_0=0$, there exists a time $T_*$ such that (1.1)–(1.2) have a unique solution $(u,b)\in C(0,T_*;H^s(\mathbb {R}^n))$.

The generalized 3D Hall-MHD system reads

$$\begin{aligned}&u_t+u\cdot \nabla u+\Lambda ^{2\alpha } u+\nabla P=(\nabla \times B)\times B, \end{aligned}$$

(1.5)

$$\begin{aligned}&B_t -\nabla \times (u\times B)+\nabla \times ((\nabla \times B)\times B)+\Lambda ^{2\beta } B=0,\end{aligned}$$

(1.6)

$$\begin{aligned}&\mathrm {div}\,u=\mathrm {div}\,B=0. \end{aligned}$$

(1.7)

One can rewrite (1.5)–(1.7) as

$$\begin{aligned}&u_t+u\cdot \nabla u+\Lambda ^{2\alpha } u+\nabla \left( P+\frac{|B|^2}{2}\right) = B\cdot \nabla B,\end{aligned}$$

(1.8)

$$\begin{aligned}&B_t+ u\cdot \nabla B+\nabla \times ((\nabla \times B)\times B)= B\cdot \nabla u-\Lambda ^{2\beta } B,\end{aligned}$$

(1.9)

$$\begin{aligned}&\mathrm {div}\,u=\mathrm {div}\,B=0. \end{aligned}$$

(1.10)

From (1.8)–(1.10), we know that the generalized Hall-MHD system is reduced to the GMHD system (1.1)–(1.4) when the Hall term $\nabla \times ((\nabla \times B)\times B)$ is neglected. Chae and his collaborators got the local existence and uniqueness of smooth solutions in [4, 8]. A blow-up criterion as

$$\begin{aligned} \int _0^T \Vert \omega \Vert _{\dot{B}_{\infty ,\infty }^0}+\frac{1+\Vert u\Vert _{L^\infty }+\Vert B\Vert _{L^\infty }+\Vert \nabla B\Vert _{L^\infty }}{1+\log (1+\Vert u\Vert _{L^\infty }+\Vert B\Vert _{L^\infty }+\Vert \nabla B\Vert _{L^\infty })}d\tau \end{aligned}$$

was also established. Later, Chae and Lee [5] proved the Serrin-type criterion

$$\begin{aligned} u\in L^p(0,T;L^q),~ \nabla B\in L^t(0,T;L^s)~~ with ~~ 2/p+3/q\le 1 ~~and ~~ 2/t+3/s\le 1 \end{aligned}$$

and the criterion in the BMO space

$$\begin{aligned} (u,~ \nabla B)\in L^2(0,T; BMO). \end{aligned}$$

Other regularity criterions can be found in [10, 12, 13, 15, 25–28]. The temporal decay and singularity formation are investigated in [6, 7].

Now, we introduce some notations which will be used in this paper. We use $\Vert \cdot \Vert _{L^p}$ to denote the $L^p(\mathbb {R}^3)$ norm. Throughout this paper, C denotes a generic positive constant (generally large), it may be different from line to line. We use $\hat{f}$ to denote the Fourier transform of f. We introduce the norm $L^{p,q}$

$$\begin{aligned}\Vert f\Vert _{L^{p,q}}={\left\{ \begin{array}{ll} \left( \int _0^t\Vert f(\cdot ,\tau )\Vert _{L^q}^pd\tau \right) ^{\frac{1}{p}}, \,\,\,\, if\,\, 1\le p<\infty ,\\ {ess sup}_{~0<\tau <t}~\Vert f\Vert _{L^q},\,\,\,\, if\,\, p=\infty .\end{array}\right. } \end{aligned}$$

The rest of the paper is organized as follows. In Sect. 2, regularity criteria for the generalized MHD equation will be established. In Sect. 3, some regularity criteria and a global regularity are established for the generalized Hall-MHD system.

2 Regularity criteria for the generalized MHD equation

This section devotes to obtain some scaling invariant regularity criteria for the generalized system (1.1)–(1.4) when $0\le \alpha ,~\beta <1$. Our main results are the following Theorems. The first one is for large $\alpha$ and $\beta$.

Theorem 2.1

For $1>\alpha ,~\beta \ge \frac{3}{4}$, assume that the initial velocity and magnetic field $u_0,b_0\in H^s(\mathbb {R}^3), s>\frac{5}{2}-\alpha$ and (u, b)(x, t) is a local strong solution of the system (1.1)–(1.4). If (u, b)(x, t) satisfies

$$\begin{aligned}&u(x,t)\in L^{p,q}, \quad with \quad \frac{2\alpha }{p}+\frac{3}{q}\le 2\alpha -1, \quad \frac{3}{2\alpha -1}<q\le \frac{6\alpha }{2\alpha -1}, \end{aligned}$$

(2.1)

$$\begin{aligned}&b(x,t)\in L^{p_1,q_1}, \quad with \quad \frac{2\beta }{p_1}+\frac{3}{q_1}\le 2\beta -1, \quad \frac{3}{2\beta -1}<q_1\le \frac{6\beta }{2\beta -1}\qquad \end{aligned}$$

(2.2)

or

$$\begin{aligned}&\Lambda ^{\alpha }u(x,t)\in L^{p,q}, \quad with \quad \frac{2\alpha }{p}+\frac{3}{q}\le 3\alpha -1, \quad \frac{3}{2\alpha -1}<q\le \frac{6\alpha }{3\alpha -1}, \end{aligned}$$

(2.3)

$$\begin{aligned}&\Lambda ^{\beta }b(x,t)\in L^{p_1,q_1}, \quad with \quad \frac{2\beta }{p_1}+\frac{3}{q_1}\le 3\beta -1, \quad \frac{3}{2\beta -1}<q_1\le \frac{6\beta }{3\beta -1},\qquad \qquad \end{aligned}$$

(2.4)

then, the solution remains smooth on (0, T].

The following theorems are established for the cases $\alpha$ or $\beta$ small.

Theorem 2.2

For $0<\alpha =\beta <1$, assume that the initial velocity and magnetic field $u_0,b_0\in H^s(\mathbb {R}^3), s>\frac{5}{2}-\alpha$ and (u, b)(x, t) is a local smooth solution of the system (1.1)–(1.4). If (u, b)(x, t) satisfies

$$\begin{aligned} \int _0^T\Vert \nabla u\Vert _{L^p}^\frac{1}{1-\nu }+\Vert \nabla b\Vert _{L^p}^\frac{1}{1-\nu }d\tau \le C(T), \end{aligned}$$

(2.5)

here $\nu =\max \{\nu _1,\nu _2\}, \nu _1=\frac{3}{2p\alpha }, \nu _2=\frac{3}{2p\beta }, \max \{\frac{3}{2\alpha },\frac{3}{2\beta }\}<p\le \infty$. Then the solution remains smooth on (0, T].

Theorem 2.3

For $\alpha =0,\beta >0$, assume that the initial velocity and magnetic field $u_0,b_0\in H^s(\mathbb {R}^3), s>\frac{5}{2}-\alpha$ and (u, b)(x, t) is a local smooth solution of the system (1.1)–(1.4). If (u, b)(x, t) satisfies

$$\begin{aligned}&\int _0^T\Vert \nabla b\Vert _{L^\frac{3}{\beta }}^2+\Vert \nabla u\Vert _{L^\infty }^2d\tau \le C(T)\quad if \quad \beta <1, \end{aligned}$$

(2.6)

$$\begin{aligned}&\int _0^T\Vert \nabla u\Vert _{L^\infty }d\tau \le C(T)\quad if \quad \beta \ge 1. \end{aligned}$$

(2.7)

Then the solution remains smooth on (0, T].

Theorem 2.4

For $\alpha >0,\beta =0$, assume that the initial velocity and magnetic field $u_0,b_0\in H^s(\mathbb {R}^3), s>\frac{5}{2}-\alpha$ and (u, b)(x, t) is a local smooth solution of the system (1.1)–(1.4). If (u, b)(x, t) satisfies

$$\begin{aligned} \int _0^T\Vert \nabla u\Vert _{L^\infty }+\Vert \nabla b\Vert _{L^p}^\frac{2}{2-\theta }d\tau \le C(T), \end{aligned}$$

(2.8)

or

$$\begin{aligned} \int _0^T\Vert \nabla u\Vert _{L^\infty }+\Vert \nabla b\Vert _{L^{\frac{3}{\alpha }}}^2d\tau \le C(T). \end{aligned}$$

(2.9)

Here, $\theta =\frac{3}{2p\alpha }\le 1$. Then the solution remains smooth on (0, T].

Remark 2.1

If $\alpha =\beta$, the regularity criteria in Theorem 2.1 and 2.2 are all scaling invariant.

2.1 Proof of Theorem 2.1

In this section, we consider the case $1> \alpha ,~~\beta \ge \frac{3}{4}$. Multiplying (1.1) and (1.2) by u and b, integrating over $\mathbb {R}^3$ and adding the resulting equations together we obtain

$$\begin{aligned} \Vert u\Vert _{L^2}^2(t)+\Vert b\Vert _{L^2}^2(t)+2\int _0^t\Vert \Lambda ^{\alpha } u\Vert _{L^2}^2+\Vert \Lambda ^{\beta } b\Vert _{L^2}^2ds=\Vert u_0\Vert _{L^2}^2+\Vert b_0\Vert _{L^2}^2.\qquad \end{aligned}$$

(2.10)

Multiplying (1.1) and (1.2) by $\Delta u$ and $\Delta b$, after integration by parts and taking the divergence free property into account, we have

$$\begin{aligned}&\frac{1}{2}\frac{d}{dt}\left( \Vert \nabla u\Vert _{L^2}^2+\Vert \nabla b\Vert _{L^2}^2\right) +\Vert \Lambda ^{\alpha +1}u\Vert _{L^2}^2+\Vert \Lambda ^{\beta +1}b\Vert _{L^2}^2\nonumber \\&\le \int _{\mathbb {R}^3} u\cdot \nabla u \cdot \Delta u-b\cdot \nabla b\cdot \Delta u+u\cdot \nabla b \cdot \Delta b-b\cdot \nabla u \cdot \Delta bdx\nonumber \\&=I_1+I_2+I_3+I_4. \end{aligned}$$

(2.11)

Actually, for the $H^1-$estimates, we only need $\alpha ,~\beta \ge \frac{1}{2}$. Due to $\mathrm {div}~ u=\mathrm {div}~ b=0$, we can estimate the four terms as follows:

$$\begin{aligned} |I_1|\le & {} \Vert \nabla u\Vert _{L^3}^3\le C\Vert u\Vert _{L^q}^{(1-\theta )\delta }\Vert \Lambda ^{1+\alpha }u\Vert _{L^2}^{\theta \delta } \Vert \nabla u\Vert _{L^2}^{(1-\rho )(3-\delta )}\Vert \Lambda ^{1+\alpha }u\Vert _{L^2}^{\rho (3-\delta )}\\\le & {} C\Vert u\Vert _{L^q}^{\frac{2(1-\theta )\delta }{2-\theta \delta -\rho (3-\delta )}} \Vert \nabla u\Vert _{L^2}^{\frac{2(1-\rho )(3-\delta )}{2-\theta \delta -\rho (3-\delta )}}+ \frac{1}{4}\Vert \Lambda ^{1+\alpha }u\Vert _{L^2}^2. \end{aligned}$$

where we have used the Gagliardo–Nirenberg inequality:

$$\begin{aligned} \frac{1}{3}= & {} \frac{1}{3}+\left( \frac{1}{2}-\frac{1+\alpha }{3}\right) \theta +\frac{1-\theta }{q},\quad \frac{1}{1+\alpha }\le \theta \le 1,\\ \frac{1}{3}= & {} \left( \frac{1}{2}-\frac{\alpha }{3}\right) \rho +\frac{1-\rho }{2},\quad 0\le \rho \le 1. \end{aligned}$$

Let $\frac{2(1-\rho )(3-\delta )}{2-\theta \delta -\rho (3-\delta )}=2$, then it yields

$$\begin{aligned} \delta =\frac{1}{1-\theta }. \end{aligned}$$

By direct calculation, we have $\frac{2(1-\theta )\delta }{2-\theta \delta -\rho (3-\delta )}=\frac{2}{(1-\rho )(3-\delta )}=\frac{2\alpha }{2\alpha -1-\frac{3}{q}}$. Then

$$\begin{aligned} |I_1|\le C\Vert u\Vert _{L^q}^{\frac{2\alpha }{2\alpha -1-\frac{3}{q}}}\Vert \nabla u\Vert _{L^2}^2+\frac{1}{4}\Vert \Lambda ^{1+\alpha }u\Vert _{L^2}^2. \end{aligned}$$

(2.12)

Similarly, we can estimate the other three terms:

$$\begin{aligned} |I_2|+|I_3|+|I_4|\le & {} \Vert \nabla u\Vert _{L^3}\Vert \nabla b\Vert _{L^3}^2\le \Vert \nabla u\Vert _{L^3}^3+\Vert \nabla b\Vert _{L^3}^3\\\le & {} C\Vert u\Vert _{L^q}^{\frac{2\alpha }{2\alpha -1-\frac{3}{q}}}\Vert \nabla u\Vert _{L^2}^2 + C\Vert b\Vert _{L^{q_1}}^{\frac{2\beta }{2\beta -1-\frac{3}{q_1}}}\Vert \nabla b\Vert _{L^2}^2\\&+\, \frac{1}{4}\Vert \Lambda ^{1+\alpha }u\Vert _{L^2}^2+\frac{1}{2}\Vert \Lambda ^{1+\beta }b\Vert _{L^2}^2. \end{aligned}$$

Combining the above estimates to (2.11), we have

$$\begin{aligned}&\frac{d}{dt}\big (\Vert \nabla u\Vert _{L^2}^2+\Vert \nabla b\Vert _{L^2}^2\big )+\Vert \Lambda ^{\alpha +1}u\Vert _{L^2}^2+\Vert \Lambda ^{\beta +1}b\Vert _{L^2}^2\\&\le C\left( \Vert u\Vert _{L^q}^{\frac{2\alpha }{2\alpha -1-\frac{3}{q}}}+ \Vert b\Vert _{L^{q_1}}^{\frac{2\beta }{2\beta -1-\frac{3}{q_1}}}\right) \left( \Vert \nabla u\Vert _{L^2}^2+\Vert \nabla b\Vert _{L^2}^2\right) . \end{aligned}$$

Then (2.1), (2.2) and (2.10) guarantee the $H^1-$estimation

$$\begin{aligned}&u\in L^\infty (0,T;H^1(\mathbb {R}^3))\cap L^2(0,T;H^{1+\alpha }(\mathbb {R}^3)), \end{aligned}$$

(2.13)

$$\begin{aligned}&b\in L^\infty (0,T;H^1(\mathbb {R}^3))\cap L^2(0,T;H^{1+\beta }(\mathbb {R}^3)). \end{aligned}$$

(2.14)

On the other hand, we can give the $H^1-$estimate as follows:

$$\begin{aligned} |I_1|\le & {} \Vert \nabla u\Vert _{L^3}^3\le C\Vert \Lambda ^\alpha u\Vert _{L^q}^{(1-\theta )\delta }\Vert \Lambda ^{1+\alpha }u\Vert _{L^2}^{\theta \delta } \Vert \nabla u\Vert _{L^2}^{(1-\rho )(3-\delta )}\Vert \Lambda ^{1+\alpha }u\Vert _{L^2}^{\rho (3-\delta )}\\\le & {} C\Vert \Lambda ^\alpha u\Vert _{L^q}^{\frac{2(1-\theta )\delta }{2-\theta \delta -\rho (3-\delta )}} \Vert \nabla u\Vert _{L^2}^{\frac{2(1-\rho )(3-\delta )}{2-\theta \delta -\rho (3-\delta )}}+ \frac{1}{4}\Vert \Lambda ^{1+\alpha }u\Vert _{L^2}^2, \end{aligned}$$

where we have used the Gagliardo–Nirenberg inequality:

$$\begin{aligned}&\frac{1}{3}=\frac{1-\alpha }{3}+\left( \frac{1}{2}-\frac{1}{3}\right) \theta +\frac{1-\theta }{q},\quad 1-\alpha \le \theta \le 1,\\&\frac{1}{3}=\left( \frac{1}{2}-\frac{\alpha }{3}\right) \rho +\frac{1-\rho }{2},\quad 0\le \rho \le 1. \end{aligned}$$

Here we need $\theta =\frac{1/q-\alpha /3}{1/q-1/6}$, that means $q\le \frac{6\alpha }{3\alpha -1}$. Let $\frac{2(1-\rho )(3-\delta )}{2-\theta \delta -\rho (3-\delta )}=2$, then it yields

$$\begin{aligned} \delta =\frac{1}{1-\theta }. \end{aligned}$$

By direct calculation, we have $\frac{2(1-\theta )\delta }{2-\theta \delta -\rho (3-\delta )}=\frac{2}{(1-\rho )(3-\delta )}=\frac{2\alpha }{3\alpha -1-\frac{3}{q}}$. Then

$$\begin{aligned} |I_1|\le & {} \Vert \nabla u\Vert _{L^3}^3\le C\Vert u\Vert _{L^q}^{(1-\theta )\delta }\Vert \Lambda ^{1+\alpha }u\Vert _{L^2}^{\theta \delta } \Vert \nabla u\Vert _{L^2}^{(1-\rho )(3-\delta )}\Vert \Lambda ^{1+\alpha }u\Vert _{L^2}^{\rho (3-\delta )}\nonumber \\\le & {} C\Vert \Lambda ^\alpha u\Vert _{L^q}^{\frac{2\alpha }{3\alpha -1-\frac{3}{q}}}\Vert \nabla u\Vert _{L^2}^2+ \frac{1}{4}\Vert \Lambda ^{1+\alpha }u\Vert _{L^2}^2. \end{aligned}$$

(2.15)

Similarly, we can estimate the other three terms:

$$\begin{aligned} |I_2|+|I_3|+|I_4|\le & {} \Vert \nabla u\Vert _{L^3}\Vert \nabla b\Vert _{L^3}^2\le \Vert \nabla u\Vert _{L^3}^3+\Vert \nabla b\Vert _{L^3}^3\\\le & {} C\Vert \Lambda ^\alpha u\Vert _{L^q}^{\frac{2\alpha }{3\alpha -1-\frac{3}{q}}}\Vert \nabla u\Vert _{L^2}^2 + C\Vert \Lambda ^\beta b\Vert _{L^{q_1}}^{\frac{2\beta }{3\beta -1-\frac{3}{q_1}}}\Vert \nabla b\Vert _{L^2}^2\\&+\, \frac{1}{4}\Vert \Lambda ^{1+\alpha }u\Vert _{L^2}^2+ \frac{1}{2}\Vert \Lambda ^{1+\beta }b\Vert _{L^2}^2. \end{aligned}$$

Combining the above estimates together, we have

$$\begin{aligned}&\frac{d}{dt}\big (\Vert \nabla u\Vert _{L^2}^2+\Vert \nabla b\Vert _{L^2}^2\big )+\Vert \Lambda ^{\alpha +1}u\Vert _{L^2}^2+\Vert \Lambda ^{\beta +1}b\Vert _{L^2}^2\\&\le C\left( \Vert \Lambda ^\alpha u\Vert _{L^q}^{\frac{2\alpha }{3\alpha -1-\frac{3}{q}}}+\Vert \Lambda ^\beta b\Vert _{L^{q_1}}^{\frac{2\beta }{3\beta -1-\frac{3}{q_1}}}\right) \left( \Vert \nabla u\Vert _{L^2}^2+\Vert \nabla b\Vert _{L^2}^2\right) . \end{aligned}$$

Then (2.3), (2.4) and (2.10) guarantee the $H^1-$estimation (2.13) and (2.14).

Now, we give the $H^2-$estimates

$$\begin{aligned}&\frac{1}{2}\frac{d}{dt}\big (\Vert \Delta u\Vert _{L^2}^2+\Vert \Delta b\Vert _{L^2}^2\big )+\Vert \Lambda ^{\alpha +2}u\Vert _{L^2}^2+\Vert \Lambda ^{\beta +2}b\Vert _{L^2}^2\nonumber \\&=\int _{\mathbb {R}^3}-\Delta (u\cdot \nabla u)\Delta u+\Delta (b\cdot \nabla b)\Delta u-\Delta (u\cdot \nabla b)\Delta b+\Delta (b\cdot \nabla u)\Delta bdx\nonumber \\&\le C\int _{\mathbb {R}^3} |\nabla u||\Delta u||\Delta u|+|\nabla u||\Delta b||\Delta b|+|\nabla b||\Delta u||\Delta b|dx\nonumber \\&=II_1+II_2+II_3. \end{aligned}$$

(2.16)

By using the Hölder, Hardy–Littlewood–Sobolev, Gagliardo–Nirenberg and Young inequalities, we estimate the three terms as follows.

$$\begin{aligned} II_1\le & {} \Vert \nabla u\Vert _{L^4}\Vert \Delta u\Vert _{L^\frac{8}{3}}^2\nonumber \\\le & {} C\Vert \nabla u\Vert _{\dot{H}^\frac{3}{4}}\Vert \Delta u\Vert _{L^2}\Vert \Lambda ^{\frac{11}{4}}u\Vert _{L^2}\nonumber \\\le & {} C\Vert \Delta u\Vert _{L^2}^2\Vert \nabla u\Vert _{\dot{H}^\frac{3}{4}}^2+\frac{1}{4}\Vert \Lambda ^\frac{11}{4}u\Vert _{L^2}^2. \end{aligned}$$

(2.17)

$$\begin{aligned} II_2\le & {} \Vert \nabla u\Vert _{L^4}\Vert \Delta b\Vert _{L^\frac{8}{3}}^2\nonumber \\\le & {} C\Vert \nabla u\Vert _{\dot{H}^\frac{3}{4}}\Vert \Delta b\Vert _{L^2}\Vert \Lambda ^{\frac{11}{4}}b\Vert _{L^2}\nonumber \\\le & {} C\Vert \Delta u\Vert _{L^2}^2\Vert \nabla b\Vert _{\dot{H}^\frac{3}{4}}^2+\frac{1}{4}\Vert \Lambda ^\frac{11}{4}b\Vert _{L^2}^2. \end{aligned}$$

(2.18)

$$\begin{aligned} II_3\le & {} \Vert \nabla b\Vert _{L^4}\Vert \Delta u\Vert _{L^\frac{8}{3}}\Vert \Delta b\Vert _{L^\frac{8}{3}}\nonumber \\\le & {} C\Vert \nabla b\Vert _{L^4}\Vert \Delta u\Vert _{L^\frac{8}{3}}^2+\Vert \nabla b\Vert _{L^4}\Vert \Delta b\Vert _{L^\frac{8}{3}}^2\nonumber \\\le & {} C\Vert \nabla b\Vert _{\dot{H}^\frac{3}{4}}\Vert \Delta u\Vert _{L^2}\Vert \Lambda ^{\frac{11}{4}}u\Vert _{L^2} +C\Vert \nabla b\Vert _{\dot{H}^\frac{3}{4}}\Vert \Delta b\Vert _{L^2}\Vert \Lambda ^{\frac{11}{4}}b\Vert _{L^2}\nonumber \\\le & {} C\Vert \nabla b\Vert _{\dot{H}^\frac{3}{4}}^2\Vert \Delta u\Vert _{L^2}^2 +C\Vert \nabla b\Vert _{\dot{H}^\frac{3}{4}}^2\Vert \Delta b\Vert _{L^2}^2 \nonumber \\&+\,\frac{1}{4}\Vert \Lambda ^{\frac{11}{4}}u\Vert _{L^2}+\frac{1}{4}\Vert \Lambda ^{\frac{11}{4}}b\Vert _{L^2} \end{aligned}$$

(2.19)

Combining the above estimates to (2.16), we get

$$\begin{aligned}&\frac{d}{dt}\big (\Vert \Delta u\Vert _{L^2}^2+\Vert \Delta b\Vert _{L^2}^2\big )+\Vert \Lambda ^{\alpha +2}u\Vert _{L^2}^2+\Vert \Lambda ^{\beta +2}b\Vert _{L^2}^2\nonumber \\&\le C\big (\Vert \Delta u\Vert _{L^2}^2+\Vert \Delta b\Vert _{L^2}^2\big )\big ( \Vert \nabla u\Vert _{\dot{H}^\frac{3}{4}}^2+ \Vert \nabla b\Vert _{\dot{H}^\frac{3}{4}}^2\big ). \end{aligned}$$

(2.20)

Then, we complete the proof of Theorem 2.1 by (2.13), (2.14) and the Gronwall’s inequality.

2.2 Proof of Theorem 2.2

Recall (2.11) and (2.16), we will estimate $I_i,~ (i=1,2,3,4)$, and $II_j,~( j=1,2,3)$. Firstly, we give the $H^1-$estimates as,

$$\begin{aligned}&\frac{1}{2}\frac{d}{dt}\bigg (\Vert \nabla u\Vert _{L^2}^2+\Vert \nabla b\Vert _{L^2}^2\bigg )+\Vert \Lambda ^{\alpha +1} u\Vert _{L^2}^2+\Vert \Lambda ^{\beta +1}b\Vert _{L^2}^2 \nonumber \\&\quad \le C\Vert \nabla u\Vert _{L^p}\bigg (\Vert \nabla u\Vert _{L^{2q}}^2+\Vert \nabla b\Vert _{L^{2q}}^2\bigg )\nonumber \\&\quad \le C\bigg (\Vert \nabla u\Vert _{L^2}^{2-2\nu _1}\Vert \Lambda ^{\alpha +1} u\Vert _{L^2}^{2\nu _1} +\Vert \nabla b\Vert _{L^2}^{2-2\nu _2}\Vert \Lambda ^{\beta +1} b\Vert _{L^2}^{2\nu _2}\bigg )\Vert \nabla u\Vert _{L^p} \nonumber \\&\quad \le C\Vert \nabla u\Vert _{L^2}^2\Vert \nabla u\Vert _{L^p}^{\frac{1}{1-\nu _1}}+C\Vert \nabla b\Vert _{L^2}^2\Vert \nabla u\Vert _{L^p}^{\frac{1}{1-\nu _2}}+ \frac{1}{2}\Vert \Lambda ^{\alpha +1} u\Vert _{L^2}^2\nonumber \\&\qquad +\frac{1}{2}\Vert \Lambda ^{\beta +1} b\Vert _{L^2}^2, \end{aligned}$$

(2.21)

Here we have used the Gagliardo–Nirenberg inequality. The constants satisfy:

$$\begin{aligned}&\frac{1}{p}+\frac{1}{q}=1,\\&\frac{1}{2q}=\left( \frac{1}{2}-\frac{\alpha }{3}\right) \nu _1+\frac{1-\nu _1}{2},\,\,\,\,0\le \nu _1\le 1,\\&\frac{1}{2q}=\left( \frac{1}{2}-\frac{\beta }{3}\right) \nu _2+\frac{1-\nu _2}{2},\,\,\,\,0\le \nu _2\le 1. \end{aligned}$$

By direct calculation, we have

$$\begin{aligned} \nu _1=\frac{3}{2p\alpha }\quad and \quad \nu _2=\frac{3}{2p\beta }. \end{aligned}$$

By (2.5), (2.21) and the Gronwall’s inequality, we get

$$\begin{aligned} \Vert \nabla u\Vert _{L^2}^2+\Vert \nabla b\Vert _{L^2}^2+\int _0^t\Vert \Lambda ^{\alpha +1} u\Vert _{L^2}^2+\Vert \Lambda ^{\beta +1} b\Vert _{L^2}^2dt\le C(T). \end{aligned}$$

In order to give the $H^2-$estimation for (u, b), we should estimate $II_j~( j=1,2,3)$ which are defined in (2.16).

$$\begin{aligned} |II_1|\le & {} C\Vert \nabla u\Vert _{L^p}\Vert \Delta u\Vert _{L^{2q}}^2 \nonumber \\\le & {} C\Vert \nabla u\Vert _{L^p}\Vert \Delta u\Vert _{L^2}^{2(1-\nu _1)}\Vert \Lambda ^{2+\alpha } u\Vert _{L^2}^{2\nu _1} \nonumber \\\le & {} C\Vert \nabla u\Vert _{L^p}^{\frac{1}{1-\nu _1}}\Vert \Delta u\Vert _{L^2}^2+\frac{1}{4}\Vert \Lambda ^{2+\alpha } u\Vert _{L^2}^{2}. \end{aligned}$$

(2.22)

Similarly,

$$\begin{aligned}&|II_2|+|II_3|\nonumber \\&\quad \le C(\Vert \nabla u\Vert _{L^p}+\Vert \nabla b\Vert _{L^p})\bigg (\Vert \Delta u\Vert _{L^{2q}}^2+\Vert \Delta b\Vert _{L^{2q}}^2\bigg ) \nonumber \\&\quad \le C\bigg (\Vert \nabla u\Vert _{L^p}^{\frac{1}{1-\nu _1}}+\Vert \nabla b\Vert _{L^p}^{\frac{1}{1-\nu _1}}+\Vert \nabla u\Vert _{L^p}^{\frac{1}{1-\nu _2}}+\Vert \nabla b\Vert _{L^p}^{\frac{1}{1-\nu _2}}\bigg )\bigg (\Vert \Delta u\Vert _{L^2}^2+\Vert \Delta b\Vert _{L^2}^2\bigg )\nonumber \\&\quad +\frac{1}{4}\Vert \Lambda ^{2+\alpha } u\Vert _{L^2}^{2}+\frac{1}{2}\Vert \Lambda ^{2+\beta } b\Vert _{L^2}^{2}. \end{aligned}$$

(2.23)

Combining (2.22) and (2.23) to (2.16), we get

$$\begin{aligned}&\frac{d}{dt}\bigg (\Vert \Delta u\Vert _{L^2}^2+\Vert \Delta b\Vert _{L^2}^2\bigg )+\Vert \Lambda ^{\alpha +2}u\Vert _{L^2}^2+\Vert \Lambda ^{\beta +2}b\Vert _{L^2}^2\\&\le C\left( \Vert \nabla u\Vert _{L^p}^{\frac{1}{1-\nu _1}}+\Vert \nabla b\Vert _{L^p}^{\frac{1}{1-\nu _1}}+\Vert \nabla u\Vert _{L^p}^{\frac{1}{1-\nu _2}}+\Vert \nabla b\Vert _{L^p}^{\frac{1}{1-\nu _2}}\right) \left( \Vert \Delta u\Vert _{L^2}^2+\Vert \Delta b\Vert _{L^2}^2\right) . \end{aligned}$$

By the Gronwall’s inequality and (2.5), we have

$$\begin{aligned} \Vert \Delta u\Vert _{L^2}^2+\Vert \Delta b\Vert _{L^2}^2+\int _0^T\Vert \Lambda ^{\alpha +2} u\Vert _{L^2}^2+\Vert \Lambda ^{\beta +2}b\Vert _{L^2}^2d\tau \le C(T). \end{aligned}$$

Actually, when $\alpha >\frac{1}{2}$, we complete the proof by the local solution in [17]. Then, we need to show the $H^3-$estimation for $0<\alpha \le \frac{1}{2}$. Taking $\Lambda ^3$ to (1.1) and (1.2), multiplying (1.1) and (1.2) by $\Lambda ^3 u$ and $\Lambda ^3u$, after integration by parts and taking the divergence free property into account, we have the following energy estimate

$$\begin{aligned}&{\frac{d}{dt}\bigg (\Vert \Lambda ^3u\Vert _{L^2}^2+\Vert \Lambda ^3 b\Vert _{L^2}^2\bigg )+\Vert \Lambda ^{\alpha +3}u\Vert _{L^2}^2+\Vert \Lambda ^{\beta +3}b\Vert _{L^2}^2}\\&\quad \le \int _{\mathbb {R}^3}-\Lambda ^3(u\cdot \nabla u-b\cdot \nabla b)\Lambda ^3 u-\Lambda ^3(u\cdot \nabla b-b\cdot \nabla u)\Lambda ^3bdx\\&\quad \le \int _{\mathbb {R}^3}-\Lambda ^3((u\cdot \nabla u)-u\cdot \nabla \Lambda ^3 u)\Lambda ^3 u+(\Lambda ^3(b\cdot \nabla b)-b\cdot \nabla \Lambda ^3 b)\Lambda ^3 u\\&\qquad \int _{\mathbb {R}^3}-\Lambda ^3((u\cdot \nabla b)-u\cdot \nabla \Lambda ^3 b)\Lambda ^3 b+(\Lambda ^3(b\cdot \nabla u)-b\cdot \nabla \Lambda ^3 u)\Lambda ^3 u\\&\quad \le C\bigg (\Vert \nabla u\Vert _{L^p}+\Vert \nabla b\Vert _{L^p}\bigg )\bigg (\Vert \Lambda ^3u\Vert _{L^{2q}}^2+\Vert \Lambda ^3b\Vert _{L^{2q}}^2 \bigg )\\&\quad \le C\bigg (\Vert \nabla u\Vert _{L^p}^{\frac{1}{1-\nu _1}}+\Vert \nabla b\Vert _{L^p}^{\frac{1}{1-\nu _1}}+\Vert \nabla u\Vert _{L^p}^{\frac{1}{1-\nu _2}}+\Vert \nabla b\Vert _{L^p}^{\frac{1}{1-\nu _2}}\bigg )\bigg (\Vert \Lambda ^3 u\Vert _{L^2}^2+\Vert \Lambda ^3 b\Vert _{L^2}^2\bigg ) \\&\qquad +\,\frac{1}{2}\Vert \Lambda ^{\alpha +3}u\Vert _{L^2}^2+\frac{1}{2}\Vert \Lambda ^{\beta +3}b\Vert _{L^2}^2, \end{aligned}$$

where we have used the Gagliardo–Nirenberg inequality and the bilinear commutator estimates [20]

(2.24)

with $s>0, \frac{1}{p}=\frac{1}{p_1}+\frac{1}{q_1}=\frac{1}{p_2}+\frac{1}{q_2}$.

This complete the proof of Theorem 2.2.

2.3 Proof of Theorem 2.3 and 2.4

For the $H^1-$estimates, we have

$$\begin{aligned} |I_1|+|I_2|+|I_3|+|I_4|\le C\Vert \nabla u\Vert _{L^\infty }\bigg (\Vert \nabla u\Vert _{L^2}^2+\Vert \nabla b\Vert _{L^2}^2\bigg ). \end{aligned}$$

For the $H^2-$estimates s

$$\begin{aligned} |II_1|+|II_2|\le C\Vert \nabla u\Vert _{L^\infty }\bigg (\Vert \Delta u\Vert _{L^2}^2+\Vert \Delta b\Vert _{L^2}^2\bigg ). \end{aligned}$$

(2.25)

The difference between the proof of Theorem 2.3 and 2.4 lies in the estimation for $II_3$.

For the case $\alpha =0, \beta >0$, if $\beta <1$,

$$\begin{aligned} |II_3|\le & {} C\Vert \Delta u\Vert _{L^2}\Vert \nabla b\Vert _{L^\frac{3}{\beta }}\Vert \Lambda ^2b\Vert _{L^\frac{6}{3-2\beta }} \le C\Vert \Delta u\Vert _{L^2}\Vert \nabla b\Vert _{L^\frac{3}{\beta }}\Vert \Lambda ^{2+\beta }b\Vert _{L^2}\nonumber \\\le & {} C\Vert \Delta u\Vert _{L^2}^2\Vert \nabla b\Vert _{L^\frac{3}{\beta }}^2+\frac{1}{2}\Vert \Lambda ^{2+\beta }b\Vert _{L^2}^2. \end{aligned}$$

(2.26)

If $\beta \ge 1$,

$$\begin{aligned} |II_3|\le & {} \int _0^T|II_2|+|\nabla b||\nabla u||\Lambda ^3b|d\tau \nonumber \\\le & {} C\Vert \nabla u\Vert _{L^\infty }\Vert \Lambda ^2 b\Vert _{L^2}^2+\Vert \nabla u\Vert _{L^\infty }\Vert \nabla b\Vert _{L^2}\Vert \Lambda ^3 b\Vert _{L^2}\nonumber \\\le & {} C\Vert \nabla u\Vert _{L^\infty }\bigg (\Vert \Lambda ^2 b\Vert _{L^2}^2+1\bigg )+\frac{1}{2}\Vert \Lambda ^3 b\Vert _{L^2}^2. \end{aligned}$$

(2.27)

Combining (2.25), (2.26) or (2.27) together, we have

$$\begin{aligned} (u, b)\in L^\infty (0,T;H^2(\mathbb {R}^3))~~ and ~~b\in L^2(0,T;H^{2+\beta }(\mathbb {R}^3)). \end{aligned}$$

(2.28)

Now, we need to show $H^3-$estimation for $\alpha =0,0<\beta \le \frac{1}{2}$. Taking $\Lambda ^3$ to (1.1) and (1.2), multiplying (1.1) and (1.2) by $\Lambda ^3 u$ and $\Lambda ^3b$, after integration by parts and taking the divergence free property into account, we have the following energy estimate

$$\begin{aligned}&{\frac{1}{2}\frac{d}{dt}\bigg (\Vert \Lambda ^3u\Vert _{L^2}^2+\Vert \Lambda ^3 b\Vert _{L^2}^2\bigg )+\Vert \Lambda ^{\beta +3}b\Vert _{L^2}^2}\nonumber \\&\quad \le \int _{\mathbb {R}^3}-\Lambda ^3(u\cdot \nabla u-b\cdot \nabla b)\Lambda ^3 u-\Lambda ^3(u\cdot \nabla b-b\cdot \nabla u)\Lambda ^3bdx\nonumber \\&\quad \le \int _{\mathbb {R}^3}-\Lambda ^3((u\cdot \nabla u)-u\cdot \nabla \Lambda ^3 u)\Lambda ^3 u +(\Lambda ^3(b\cdot \nabla b)-b\cdot \nabla \Lambda ^3 b)\Lambda ^3 u\nonumber \\&\quad \int _{\mathbb {R}^3}-\Lambda ^3((u\cdot \nabla b)-u\cdot \nabla \Lambda ^3 b)\Lambda ^3 b +(\Lambda ^3(b\cdot \nabla u)-b\cdot \nabla \Lambda ^3 u)\Lambda ^3 b\nonumber \\&\quad =III_1+III_2+III_3+III_4. \end{aligned}$$

(2.29)

The three terms can be estimated as follows: By using (2.24), $III_1$ can be estimated as

$$\begin{aligned} |III_1|\le \Vert \nabla u\Vert _{L^\infty }\Vert \Lambda ^3u\Vert _{L^2}^2. \end{aligned}$$

(2.30)

For $\beta <1$, we can estimate $III_i,~(i=2,3,4)$

$$\begin{aligned} |III_2|\le & {} C\Vert \Lambda ^3u\Vert _{L^2}\Vert \nabla b\Vert _{L^\frac{3}{\beta }}\Vert \Lambda ^3 b\Vert _{L^\frac{6}{3-2\beta }}\nonumber \\\le & {} C\Vert \Lambda ^3 u\Vert _{L^2}\Vert \nabla b\Vert _{L^\frac{3}{\beta }}\Vert \Lambda ^{3+\beta }b\Vert _{L^2}\nonumber \\\le & {} C\Vert \Lambda ^3 u\Vert _{L^2}^2\Vert \nabla b\Vert _{L^\frac{3}{\beta }}^2+\frac{1}{4}\Vert \Lambda ^{3+\beta }b\Vert _{L^2}^2. \end{aligned}$$

(2.31)

$$\begin{aligned} |III_3|+|III_4|\le & {} C\Vert \Lambda ^3u\Vert _{L^2}\Vert \nabla b\Vert _{L^\frac{3}{\beta }}\Vert \Lambda ^3 b\Vert _{L^\frac{6}{3-2\beta }}\nonumber \\&+\,\Vert \Lambda ^3b\Vert _{L^\frac{6}{3+2\beta }}\Vert \nabla u\Vert _{L^\infty }\Vert \Lambda ^3 b\Vert _{L^\frac{6}{3-2\beta }}\nonumber \\\le & {} C\Vert \Lambda ^3 u\Vert _{L^2}^2\Vert \nabla b\Vert _{L^\frac{3}{\beta }}^2+C\Vert \nabla u\Vert _{L^\infty }^2\Vert \Lambda ^3 u\Vert _{L^2}^2+\frac{1}{4}\Vert \Lambda ^{3+\beta }b\Vert _{L^2}^2.\nonumber \\ \end{aligned}$$

(2.32)

For $\beta \ge 1$, by using (2.24), $III_i,~(i=2,3,4)$ can be estimated as

$$\begin{aligned} |III_2|+|III_3|+|III_4|\le C\Vert \nabla u\Vert _{L^\infty }(\Vert \Lambda ^3u\Vert _{L^2}^2+\Vert \Lambda ^3b\Vert _{L^2}^2). \end{aligned}$$

(2.33)

Combining (2.30), (2.31), (2.32) or (2.33) to (2.29), we complete the proof of Theorem 2.3 by (2.6), (2.7) and the Gronwall’s inequality.

For the case $\alpha >0, \beta =0$,

$$\begin{aligned} II_3\le & {} C\Vert \Delta b\Vert _{L^2}\Vert \nabla b\Vert _{L^p}\Vert \Delta u\Vert _{L^q}\\\le & {} C\Vert \Delta b\Vert _{L^2}\Vert \nabla b\Vert _{L^p}\Vert \Delta u\Vert _{L^2}^{1-\theta }\Vert \Lambda ^{2+\alpha } u\Vert _{L^2}^\theta \\\le & {} C\Vert \nabla b\Vert _{L^p}^{\frac{2}{2-\theta }}\bigg (\Vert \Delta u\Vert _{L^2}^2+\Vert \Delta b\Vert _{L^2}^2+1\bigg )+\Vert \Lambda ^{2+\alpha } u\Vert _{L^2}^2, \end{aligned}$$

here $\theta =\frac{3}{2p\alpha }, \frac{1}{p}+\frac{1}{q}=\frac{1}{2}$. Or

$$\begin{aligned} |II_3|\le & {} C\Vert \Delta b\Vert _{L^2}\Vert \nabla b\Vert _{L^\frac{3}{\alpha }}\Vert \Delta u\Vert _{L^\frac{6}{3-2\alpha }} \le C\Vert \Delta b\Vert _{L^2}\Vert \nabla b\Vert _{L^\frac{3}{\alpha }}\Vert \Lambda ^{2+\alpha }u\Vert _{L^2}\\\le & {} C\Vert \Delta b\Vert _{L^2}^2\Vert \nabla b\Vert _{L^\frac{3}{\alpha }}^2+\frac{1}{2}\Vert \Lambda ^{2+\alpha }u\Vert _{L^2}^2. \end{aligned}$$

Now, we need to show the $H^3-$estimation

$$\begin{aligned} |III_2|\le & {} C\Vert \Lambda ^3b\Vert _{L^2}\Vert \nabla b\Vert _{L^\frac{3}{\alpha }}\Vert \Lambda ^3 u\Vert _{L^\frac{6}{3-2\alpha }}\nonumber \\\le & {} C\Vert \Lambda ^3 b\Vert _{L^2}\Vert \nabla b\Vert _{L^\frac{3}{\alpha }}\Vert \Lambda ^{3+\alpha }u\Vert _{L^2}\\\le & {} C\Vert \Lambda ^3 b\Vert _{L^2}^2\Vert \nabla b\Vert _{L^\frac{3}{\alpha }}^2+\frac{1}{4}\Vert \Lambda ^{3+\beta }u\Vert _{L^2}^2.\\ |III_3|+|III_4|\le & {} C\Vert \Lambda ^3u\Vert _{L^\frac{6}{3-2\alpha }}\Vert \nabla b\Vert _{L^\frac{3}{\alpha }}\Vert \Lambda ^3 b\Vert _{L^2} +C\Vert \nabla u\Vert _{L^\infty }\Vert \Lambda ^3b\Vert _{L^2}^2\\\le & {} C\left( \Vert \nabla u\Vert _{L^\infty }+\Vert \nabla b\Vert _{L^\frac{3}{\alpha }}^2\right) \Vert \Lambda ^3 b\Vert _{L^2}^2+ \frac{1}{2}\Vert \Lambda ^3u\Vert _{L^2}^2. \end{aligned}$$

Or

$$\begin{aligned}&|III_2|+|III_3|+|III_4|\le C\bigg (\Vert \nabla b\Vert _{L^p}^{\frac{2}{2-\theta }}+\Vert \nabla u\Vert _{L^\infty }\bigg ) \bigg (\Vert \Lambda ^3 u\Vert _{L^2}^2\nonumber \\&+\,\Vert \Lambda ^3 b\Vert _{L^2}^2\bigg )+\Vert \Lambda ^{3+\alpha } u\Vert _{L^2}^2. \end{aligned}$$

This complete the proof of Theorem 2.1 by (2.8), (2.9) and the Gronwall’s inequality.

3 Regularity criteria for the generalized Hall-MHD equation

In this section, the generalized incompressible Hall-MHD equations (1.5)–(1.7) are investigated in three dimension. Now, we establish the global regularity for the Hall-MHD equation.

Theorem 3.1

Assume $\alpha \ge \frac{5}{4}, \beta \ge \frac{7}{4}$, and $(u_0,B_0)\in H^s(\mathbb {R}^3), s>2\beta +\frac{3}{2}$ with $div u_0=div B_0=0$, then the 3D generalized incompressible Hall-MHD equation (1.5)–(1.7) has a global classical solution.

Proof

Actually, we only need to show the case $\alpha =\frac{5}{4}$ and $\beta =\frac{7}{4}$.

Note that if $div B_0=0$, then the divergence free condition is propagated by (1.6) (see[5] in detail).

Multiplying (1.8) and (1.9) by u and B, integrating over $\mathbb {R}^3$ and adding the resulting equations together we obtain

$$\begin{aligned} \Vert u\Vert _{L^2}^2+\Vert B\Vert _{L^2}^2+\int _0^T\Vert \Lambda ^\alpha u\Vert _{L^2}^2+\Vert \Lambda ^\beta B\Vert _{L^2}^2d\tau \le C. \end{aligned}$$

(3.1)

Multiplying (1.8) and (1.9) by $\Delta u$ and $\Delta B$, after integration by parts and taking the divergence free property into account, we have

$$\begin{aligned}&{\frac{1}{2}\frac{d}{dt}\bigg (\Vert \nabla u\Vert _{L^2}^2+\Vert \nabla B\Vert _{L^2}^2\bigg )+\Vert \Lambda ^{\alpha +1} u\Vert _{L^2}^2+\Vert \Lambda ^{\beta +1} B\Vert _{L^2}^2}\nonumber \\&=\int _{\mathbb {R}^3}u\cdot \nabla u\Delta udx-\int _{\mathbb {R}^3}B\cdot \nabla B\Delta udx+\int _{\mathbb {R}^3}u\cdot \nabla B\Delta Bdx\nonumber \\&\quad -\int _{\mathbb {R}^3}B\cdot \nabla u\Delta Bdx+\int _{\mathbb {R}^3}\nabla \times ((\nabla \times B)\times B)\Delta Bdx\nonumber \\&=\tilde{I}_1+\tilde{I}_2+\tilde{I}_3+\tilde{I}_4+\tilde{I}_5. \end{aligned}$$

(3.2)

We will estimate the five terms in the right hand side.

$$\begin{aligned} |\tilde{I}_1|\le & {} C\Vert \nabla u\Vert _{L^p}\Vert \nabla u\Vert _{L^{2q}}^2\quad \left( \frac{1}{p}+\frac{1}{q}=1\right) \\\le & {} C\Vert u\Vert _{L^2}^{1-\delta }\Vert \Lambda ^\alpha u\Vert _{L^2}^\delta \Vert \nabla u\Vert _{L^2}^{2(1-\theta )}\Vert \Lambda ^{\alpha +1}u\Vert _{L^2}^{2\theta }\\\le & {} \frac{1}{2}\Vert \Lambda ^{\alpha +1}u\Vert _{L^2}^2+C\Vert \Lambda ^\alpha u\Vert _{L^2}^{\delta /(1-\theta )}\Vert \nabla u\Vert _{L^2}^2, \end{aligned}$$

here, we have used

$$\begin{aligned}&\frac{1}{p}-\frac{1}{3}=\delta \left( \frac{1}{2}-\frac{\alpha }{3}\right) +\frac{1-\delta }{2},\\&\frac{1}{2q}=\theta \left( \frac{1}{2}-\frac{\alpha }{3}\right) +\frac{1-\theta }{2}, \end{aligned}$$

and we need $\frac{\delta }{1-\theta }\le 2$, that means $\alpha \ge \frac{5}{4}$.

$$\begin{aligned} |\tilde{I}_5|= & {} \left| -\int _{\mathbb {R}^3}\partial _i((\nabla \times B)\times B)\partial _i\nabla \times Bdx\right| \\= & {} \left| -\int _{\mathbb {R}^3}((\nabla \times B)\times \partial _i B)\partial _i\nabla \times Bdx\right| \\\le & {} \Vert \nabla B\Vert _{L^{\frac{8}{3}}}^2\Vert \Delta B\Vert _{L^4}\\\le & {} \Vert \nabla B\Vert _{L^2}\Vert \Lambda ^\frac{7}{4} B\Vert _{L^2}\Vert B\Vert _{\dot{H}^\frac{11}{4}}\\\le & {} \Vert \nabla B\Vert _{L^2}^2\Vert \Lambda ^\frac{7}{4} B\Vert _{L^2}^2+\Vert B\Vert _{\dot{H}^\frac{11}{4}}^2, \end{aligned}$$

here, we have used the Gagliardo–Nirenberg, Sobolev inequalities.

$$\begin{aligned} |\tilde{I}_2,\tilde{I}_3,\tilde{I}_4|\le & {} \Vert \nabla u\Vert _{L^4}\Vert \nabla B\Vert _{L^\frac{8}{3}}^2\\\le & {} \Vert u\Vert _{\dot{H}^\frac{7}{4}}\Vert \nabla B\Vert _{L^2}\Vert \Lambda ^\frac{7}{4} B\Vert _{L^2}\\\le & {} \Vert \nabla B\Vert _{L^2}^2\Vert \Lambda ^\frac{7}{4} B\Vert _{L^2}^2+\Vert u\Vert _{\dot{H}^\frac{7}{4}}^2. \end{aligned}$$

Combining the above estimates to (3.2), by the Gronwall’s inequality, we get

$$\begin{aligned} u\in L^\infty (0,T;H^1)\cap L^2(0,T;H^{\alpha +1}),\\ b\in L^\infty (0,T;H^1)\cap L^2(0,T;H^{\beta +1}). \end{aligned}$$

For the $H^2-$estimates

$$\begin{aligned}&\frac{1}{2}\frac{d}{dt}\bigg (\Vert \Delta u\Vert _{L^2}^2+\Vert \Delta B\Vert _{L^2}^2\bigg )+\Vert \Lambda ^{\alpha +2}u\Vert _{L^2}^2+\Vert \Lambda ^{\beta +2}B\Vert _{L^2}^2\nonumber \\&\quad =\int _{\mathbb {R}^3}-\Delta (u\cdot \nabla u)\Delta u+\Delta (B\cdot \nabla B)\Delta u-\Delta (u\cdot \nabla B)\Delta B \nonumber \\&\qquad +\,\Delta (B\cdot \nabla u)\Delta B+\Delta (\nabla \times ((\nabla \times B)\times B))\Delta Bdx\nonumber \\&\quad \le C\int _{\mathbb {R}^3} |\nabla u||\Delta u||\Delta u|+|\nabla u||\Delta B||\Delta B| +|\nabla B||\Delta u||\Delta B|\nonumber \\&\qquad +\,|\nabla B||\Delta B||\Lambda ^3 B|dx\nonumber \\&\quad =\tilde{II}_1+\tilde{II}_2+\tilde{II}_3+\tilde{II}_4. \end{aligned}$$

(3.3)

The estimate of $\tilde{II}_1, \tilde{II}_2, \tilde{II}_3$ is the same to (2.17), (2.18) and (2.19).

$$\begin{aligned} \tilde{II}_4\le & {} C\Vert \nabla B\Vert _{L^4}\Vert \Delta B\Vert _{L^{\frac{8}{3}}}\Vert \Lambda ^3 B\Vert _{L^{\frac{8}{3}}}\\\le & {} C\Vert \nabla B\Vert _{\dot{H}^\frac{3}{4}}\Vert \Delta B\Vert _{L^2}^{1-\theta }\Vert \Lambda ^{\beta +2} B\Vert _{L^2}^\theta \Vert \Delta B\Vert _{L^2}^{1-\delta }\Vert \Lambda ^{2+\beta } B\Vert _{L^2}^\delta \\\le & {} \frac{1}{4}\Vert \Lambda ^{\beta +2} B\Vert _{L^2}+\Vert \nabla B\Vert _{\dot{H}^\frac{3}{4}}^{2}\Vert \Delta B\Vert _{L^2}^2. \end{aligned}$$

Here we have used

$$\begin{aligned} \frac{3}{8}= & {} \frac{1}{3}+\theta \left( \frac{1}{2}-\frac{\frac{7}{4}}{3}\right) +\frac{1-\theta }{2},\,\,\,\,then\,\,\,\,\theta =\frac{11}{14},\\ \frac{3}{8}= & {} \delta \left( \frac{1}{2}-\frac{\frac{7}{4}}{3}\right) +\frac{1-\delta }{2},\,\,\,\,then\,\,\,\,\delta =\frac{3}{14}. \end{aligned}$$

Putting above estimates together, we complete the proof by the Gronwall’s inequality. $\square$

Then, we will establish some regularity criteria for the case with $\alpha <\frac{5}{4}$ and $\beta <\frac{7}{4}$.

Theorem 3.2

For $\frac{5}{4}>\alpha \ge \frac{3}{4}, \frac{7}{4}>\beta \ge 1$, assume that the initial value $(u_0,B_0)\in H^s(\mathbb {R}^3)\times H^s(\mathbb {R}^3)$ for $s\ge 3$ with $div u_0=div B_0=0$ satisfying

$$\begin{aligned} \nabla B\in L^{t}(0,T,L^{s}(\mathbb {R}^3)~~with ~~\frac{2\beta }{t}+\frac{3}{s}\le 2\beta -1, \quad \frac{3}{2\beta -1}<q\le \infty , \end{aligned}$$

(3.4)

and

$$\begin{aligned} u\in L^{p}(0,T,L^{q}(\mathbb {R}^3))~~ with ~~\frac{2\alpha }{p}+\frac{3}{q}\le 2\alpha -1, \quad \frac{3}{2\alpha -1}<q\le \frac{6\alpha }{2\alpha -1},\qquad \end{aligned}$$

(3.5)

or

$$\begin{aligned} \Lambda ^{\alpha }u\in L^{p}(0,T,L^{q}(\mathbb {R}^3))~~ with ~~\frac{2\alpha }{p}+\frac{3}{q}\le 3\alpha -1, \quad \frac{3}{2\alpha -1}<q\le \frac{6\alpha }{3\alpha -1}.\qquad \end{aligned}$$

(3.6)

Then the corresponding strong solution (u, B)(x, t) remains smooth on [0, T].

Theorem 3.3

For $\alpha >0, \frac{7}{4}>\beta \ge 1$, assume that the initial value $(u_0,B_0)\in H^s(\mathbb {R}^3)\times H^s(\mathbb {R}^3)$ for $s\ge 3$ with $div u_0=div B_0=0$ satisfying (3.4) and

$$\begin{aligned} \nabla u\in L^{p}(0,T,L^{q}(\mathbb {R}^3))~~ with ~~\frac{2\alpha }{p}+\frac{3}{q}\le 2\alpha , \quad \frac{3}{2\alpha }<q\le \infty . \end{aligned}$$

(3.7)

Then the corresponding strong solution (u, B)(x, t) remains smoothness on [0, T].

Theorem 3.4

For $\alpha =0, \frac{7}{4}>\beta \ge 1$, assume that the initial value $(u_0,B_0)\in H^s(\mathbb {R}^3)\times H^s(\mathbb {R}^3)$ for $s\ge 3$ with $div u_0=div B_0=0$ satisfying (3.4) and

$$\begin{aligned} \nabla u\in L^1(0,T;L^\infty ). \end{aligned}$$

(3.8)

Then the corresponding strong solution (u, B) remains smoothness on [0, T].

Remark 3.1

If $\beta \ge \frac{7}{4}$, we can neglect the condition (3.4) that is added on $\nabla B$. It is interesting and difficult to establish some regularity criteria for $\beta <1$.

Proof of Theorem 3.2

The estimate $\tilde{I}_1$ is given by (2.12) or (2.15). $|\tilde{I}_2|+|\tilde{I}_3|+|\tilde{I}_4|\le C\Vert \nabla u\Vert _{L^3}^3+C\Vert \nabla B\Vert _{L^3}^3=J_1+J_2$. The estimate for $J_1$ is the same as $\tilde{I}_1$.

$$\begin{aligned} J_2\le & {} C\Vert \nabla B\Vert _{L^2}^{2-2\nu _2}\Vert \Lambda ^{\beta +1} B\Vert _{L^2}^{2\nu _2}\Vert \nabla B\Vert _{L^p}\\\le & {} C\Vert \nabla B\Vert _{L^p}^{\frac{1}{1-\nu _2}} \Vert \nabla B\Vert _{L^2}^2+\frac{1}{4}\Vert \Lambda ^{\beta +1} B\Vert _{L^2}^{2}. \end{aligned}$$

Here we need the criterion $\nabla B\in L^{p,q}$ with $\frac{2\beta }{p}+\frac{3}{q}\le 2\beta$.

$$\begin{aligned} |\tilde{I}_5|= & {} \left| -\int _{\mathbb {R}^3}\partial _i((\nabla \times B)\times B)\partial _i\nabla \times Bdx\right| \nonumber \\= & {} \left| -\int _{\mathbb {R}^3}((\nabla \times B)\times \partial _i B)\partial _i\nabla \times Bdx\right| \nonumber \\\le & {} C\Vert \nabla B\Vert _{L^q}\Vert \nabla B\Vert _{L^\mu }\Vert \Delta B\Vert _{L^\nu }\nonumber \\\le & {} C\Vert \nabla B\Vert _{L^q}\Vert \nabla B\Vert _{L^2}^\theta \Vert \Lambda ^{\beta +1}B\Vert _{L^2}^{1-\theta } \Vert \nabla B\Vert _{L^2}^\delta \Vert \Lambda ^{\beta +1} B\Vert _{L^2}^{1-\delta }\nonumber \\\le & {} \frac{1}{4}\Vert \Lambda ^{\beta +1}B\Vert _{L^2}^2+C\Vert \nabla B\Vert _{L^q}^{2/(\theta +\delta )}\Vert \nabla B\Vert _{L^2}^2, \end{aligned}$$

(3.9)

here

$$\begin{aligned} \frac{1}{\mu }-\frac{1}{3}= & {} \theta \left( \frac{1}{2}-\frac{1}{3}\right) +(1-\theta )\left( \frac{1}{2}-\frac{\beta +1}{3}\right) ,\\ \frac{1}{\nu }-\frac{2}{3}= & {} \delta \left( \frac{1}{2}-\frac{1}{3}\right) +(1-\delta ) \left( \frac{1}{2}-\frac{\beta +1}{3}\right) . \end{aligned}$$

That means $\frac{2}{\theta +\delta }=\frac{2\beta }{2\beta -1-3/q}\le p$.

By (3.4)–(3.6), we get $u\in L^\infty (0,T;H^1)\cap L^2(0,T;H^{1+\alpha })$ and $B\in L^\infty (0,T;H^1)\cap L^2(0,T;H^{1+\beta })$.

In order to give the $H^2-$estimates, we should give the estimates for $\tilde{II}_i( i=1,2,3,4)$. The estimates for $\tilde{II}_1,\tilde{II}_2,\tilde{II}_3$ is the same as that in (2.17)–(2.19).

$$\begin{aligned} \tilde{II}_4\le & {} \Vert \nabla B\Vert _{L^q}\Vert \Delta B\Vert _{L^\mu }\Vert \Lambda ^3 B\Vert _{L^\nu }\nonumber \\\le & {} \Vert \nabla B\Vert _{L^q}\Vert \Delta B\Vert _{L^2}^\theta \Vert \Lambda ^{\beta +2} B\Vert _{L^2}^{1-\theta } \Vert \Delta B\Vert _{L^2}^\delta \Vert \Lambda ^{2+\beta } B\Vert _{L^2}^{1-\delta }\nonumber \\\le & {} \frac{1}{4}\Vert \Lambda ^{\beta +2} B\Vert _{L^2}+\Vert \nabla B\Vert _{L^q}^{2/(\theta +\delta )}\Vert \Delta B\Vert _{L^2}^2. \end{aligned}$$

(3.10)

Combing (2.17)–(2.10) and (3.10) to (3.3), we complete the proof of Theorem 3.2 by the condition (3.4)–(3.6) and the Gronwall’s inequality.

Proof of Theorem 3.3

For the $H^1-$estimation:

where $\nu _1=\frac{3}{2p\alpha }$ and $\nu _2=\frac{3}{2p\beta }$. For the $H^2-$estimation, $\tilde{II}_1$ is same to (2.22),

$$\begin{aligned} \tilde{II}_2\le & {} C\Vert \nabla u\Vert _{L^3}\Vert \Delta B\Vert _{L^3}^2\nonumber \\\le & {} C\Vert \nabla u\Vert _{L^2}^\frac{1}{2}\Vert \Delta u\Vert _{L^2}^\frac{1}{2}\Vert \Delta B\Vert _{L^2}\Vert \Lambda ^3 B\Vert _{L^2}\nonumber \\\le & {} C\bigg (1+\Vert \Delta u\Vert _{L^2}^2\bigg )\Vert \Delta B\Vert _{L^2}^2+\frac{1}{4}\Vert \Lambda ^3 B\Vert _{L^2}^2. \end{aligned}$$

(3.11)

and

$$\begin{aligned} \tilde{II}_3&\le C\Vert \nabla B\Vert _{L^4}\Vert \Delta u\Vert _{L^2}\Vert \Delta B\Vert _{L^4}\nonumber \\&\le C\Vert \nabla B\Vert _{L^2}^\frac{1}{4}\Vert \Delta B\Vert _{L^2}\Vert \Delta u\Vert _{L^2}\Vert \Lambda ^3 B\Vert _{L^2}^\frac{3}{4}\nonumber \\&\le C\Vert \Delta B\Vert _{L^2}^2\Vert \Delta u\Vert _{L^2}^2+\frac{1}{4}\Vert \Lambda ^3 B\Vert _{L^2}^\frac{3}{2}. \end{aligned}$$

(3.12)

Combing (2.22), (3.10)–(3.12) to (3.3), we complete the proof of Theorem 3.3 by (3.7) and the Gronwall’s inequality.

Proof of Theorem 3.4

For the $H^1-$estimation:

$$\begin{aligned} |\tilde{I}_1|+|\tilde{I}_2|+|\tilde{I}_3|+|\tilde{I}_4| \le C\Vert \nabla u\Vert _{L^\infty }(\Vert \nabla u\Vert _{L^2}^2+\Vert \nabla B\Vert _{L^2}^2). \end{aligned}$$

For the $H^2$ estimation, $\tilde{II}_3$ is same as (2.27),

$$\begin{aligned} \tilde{II}_1, \tilde{II}_2\le C\Vert \nabla u\Vert _{L^\infty }\Vert \Delta B\Vert _{L^2}^2, \end{aligned}$$

Combining the above estimates and (3.10)–(3.3), we complete the proof of Theorem 3.4 by (3.8) and the Gronwall’s inequality.