1 Introduction

Let Ω be a bounded, simply connected domain in \(\mathbb{R}^{3}\) with smooth boundary ∂Ω, and ν be the unit outward normal vector to ∂Ω. First, we consider the regularity criterion of the Boussinesq system with zero heat conductivity:

$$\begin{aligned}& \operatorname {div}u=0, \end{aligned}$$
(1.1)
$$\begin{aligned}& \partial_{t}u+u\cdot\nabla u+\nabla\pi-\Delta u=\theta e_{3}, \end{aligned}$$
(1.2)
$$\begin{aligned}& \partial_{t}\theta+u\cdot\nabla\theta=0\quad \text{in }\Omega \times(0, \infty), \end{aligned}$$
(1.3)
$$\begin{aligned}& u\cdot\nu=0,\quad\quad \operatorname {curl}u\times\nu=0\quad \text{on }\partial\Omega \times(0, \infty), \end{aligned}$$
(1.4)
$$\begin{aligned}& (u,\theta) (\cdot,0)=(u_{0},\theta_{0})\quad \text{in }\Omega \subseteq\mathbb{R}^{3}, \end{aligned}$$
(1.5)

where u, π, and θ denote the unknown velocity vector field, pressure scalar, and temperature scalar of the fluid, respectively. \(\omega:=\operatorname {curl}u\) is the vorticity, and \(e_{3}:=(0,0,1)^{t}\).

When \(\theta=0\), (1.1) and (1.2) are the well-known Navier-Stokes system. Giga [1], Kim [2], and Kang and Kim [3] have proved some Serrin-type regularity criteria.

The first aim of this paper is to prove a new regularity criterion for problem (1.1)-(1.5).

Theorem 1.1

Let \(u_{0}\in H^{3}\) and \(\theta_{0}\in W^{1,p}\) with \(3< p\leq6\) and \(\operatorname {div}u_{0}=0\) in Ω and \(u_{0}\cdot\nu=0\), \(\operatorname {curl}u_{0}\times\nu=0\) on ∂Ω. Let \((u,\theta)\) be a strong solution of problem (1.1)-(1.5). If u satisfies

$$ \nabla u\in L^{1}\bigl(0,T;\operatorname {BMO}(\Omega)\bigr) $$
(1.6)

with \(0< T<\infty\), then the solution \((u,\theta)\) can be extended beyond \(T>0\). Here BMO denotes the space of bounded mean oscillation.

Secondly, we consider the blow-up criterion for the 3D MHD system

$$\begin{aligned}& \operatorname {div}u=\operatorname {div}b=0, \end{aligned}$$
(1.7)
$$\begin{aligned}& \partial_{t}u+u\cdot\nabla u+\nabla \biggl(\pi+\frac{1}{2} \vert b\vert ^{2} \biggr)-\Delta u=b\cdot\nabla b, \end{aligned}$$
(1.8)
$$\begin{aligned}& \partial_{t}b+u\cdot\nabla b-b\cdot\nabla u=\Delta b\quad \text{in } \Omega \times(0,\infty), \end{aligned}$$
(1.9)
$$\begin{aligned}& u\cdot\nu=0,\quad\quad \operatorname {curl}u\times\nu=0,\quad\quad b\cdot\nu=0,\quad\quad \operatorname {curl}b\times\nu=0 \\& \quad \text{on } \partial \Omega\times(0,\infty), \end{aligned}$$
(1.10)
$$\begin{aligned}& (u,b) (\cdot,0)=(u_{0},b_{0})\quad \text{in }\Omega\subseteq \mathbb {R}^{3}. \end{aligned}$$
(1.11)

Here b is the magnetic field of the fluid.

It is well known that problem (1.7)-(1.11) has a unique local strong solution [4]. But whether this local solution can exist globally is an outstanding problem. Kang and Kim [3] proved some Serrin-type regularity criteria.

The second aim of this paper is to prove a new regularity criterion for problem (1.7)-(1.11).

Theorem 1.2

Let \(u_{0},b_{0}\in H^{3}\) with \(\operatorname {div}u_{0}=\operatorname {div}b_{0}=0\) in Ω and \(u_{0}\cdot\nu=b_{0}\cdot\nu=0\), \(\operatorname {curl}u_{0}\times\nu=\operatorname {curl}b_{0}\times\nu =0\) on ∂Ω. Let \((u,b)\) be a strong solution to problem (1.7)-(1.11). If (1.6) holds, then the solution \((u,b)\) can be extended beyond \(T>0\).

Remark 1.1

When \(\Omega:=\mathbb{R}^{3}\), our result gives the well-known regularity criterion

$$\omega:=\operatorname {curl}u\in L^{1}\bigl(0,T;\dot{B}_{\infty,\infty}^{0} \bigr), $$

but the method of proof we use is different from that in [5, 6]. Here \(\dot{B}_{\infty,\infty}^{0}\) denotes the homogeneous Besov space [7].

Next, we consider the following 3D density-dependent MHD equations:

$$\begin{aligned}& \operatorname {div}u=\operatorname {div}b=0, \end{aligned}$$
(1.12)
$$\begin{aligned}& \partial_{t}\rho+\operatorname {div}(\rho u)=0, \end{aligned}$$
(1.13)
$$\begin{aligned}& \partial_{t}(\rho u)+\operatorname {div}(\rho u\otimes u)+\nabla \biggl(\pi+ \frac{1}{2}\vert b\vert ^{2} \biggr)-\Delta u=b\cdot\nabla b, \end{aligned}$$
(1.14)
$$\begin{aligned}& \partial_{t}b+u\cdot\nabla b-b\cdot\nabla u=\Delta b\quad \text{in } \Omega \times(0,\infty), \end{aligned}$$
(1.15)
$$\begin{aligned}& u=0,\quad\quad b\cdot\nu=0,\quad\quad \operatorname {curl}b\times\nu=0\quad \text{on }\partial\Omega\times (0, \infty), \end{aligned}$$
(1.16)
$$\begin{aligned}& (\rho,\rho u,b) (\cdot,0)=(\rho_{0},\rho_{0}u_{0},b_{0}) \quad \text{in }\Omega \subset\mathbb{R}^{3}. \end{aligned}$$
(1.17)

For this problem, Wu [8] proved that if the initial data \(\rho_{0}\), \(u_{0}\), and \(b_{0}\) satisfy

$$ \begin{aligned} &0\leq\rho_{0}\in H^{2},\quad\quad u_{0}\in H_{0}^{1}\cap H^{2}, \quad\quad b_{0}\in H^{2}, \\ & {-}\Delta u_{0}+\nabla \biggl(\pi_{0}+ \frac{1}{2}\vert b_{0}\vert ^{2} \biggr)=b_{0}\cdot\nabla b_{0}+\sqrt{\rho_{0}}g \end{aligned} $$
(1.18)

for some \((\pi_{0},g)\in H^{1}\times L^{2}\), then there exists a positive time \(T_{*}\) and a unique strong solution \((\rho, u, b)\) to problem (1.12)-(1.17) such that

$$ \begin{aligned} &\rho\in C([0,T_{*}];H^{2}),\quad\quad u\in C([0,T_{*}];H_{0}^{1}\cap H^{2})\cap L^{2}(0,T_{*};H^{2}),\\ &u_{t}\in L^{2}(0,T_{*};H_{0}^{1}), \quad\quad \sqrt{\rho}u_{t}\in L^{\infty}(0,T_{*};L^{2}),\\ &b\in L^{\infty}(0,T_{*};H^{2})\cap L^{2}(0,T_{*};H^{3}),\quad\quad b_{t}\in L^{\infty}(0,T_{*};L^{2})\cap L^{2}(0,T_{*};H^{1}). \end{aligned} $$
(1.19)

When \(b=0\), Kim [2] proved the following regularity criterion:

$$ u\in L^{\frac{2s}{s-3}}\bigl(0,T;L_{w}^{s}(\Omega) \bigr)\quad \text{with }3< s\leq \infty. $$
(1.20)

Here \(L_{w}^{s}\) denotes the weak-\(L^{s}\) space, and \(L_{w}^{\infty}=L^{\infty}\).

The aim of this paper is to refine (1.20) as follows.

Theorem 1.3

Let \(\rho_{0}\), \(u_{0}\), and \(b_{0}\) satisfy (1.18). Let \((\rho, u, b)\) be a strong solution of problem (1.12)-(1.17) in the class (1.19). Suppose that u satisfies one of the following two conditions:

$$\begin{aligned}& \mbox{(i)}\quad \int_{0}^{T}\frac{\Vert u(t)\Vert _{L_{w}^{s}}^{\frac{2s}{s-3}}}{1+\log (e+\Vert u(t)\Vert _{L_{w}^{s}})}\,dt< \infty\quad \textit{with }3< s\leq\infty, \end{aligned}$$
(1.21)
$$\begin{aligned}& \mbox{(ii)}\quad u\in L^{2}\bigl(0,T;\operatorname {BMO}(\Omega)\bigr) \end{aligned}$$
(1.22)

with \(0< T<\infty\). Then the solution \((\rho,u,b)\) can be extended beyond \(T>0\).

Finally, we consider the 3D Landau-Lifshitz system:

$$\begin{aligned}& \partial_{t}d-\Delta d=d\vert \nabla d\vert ^{2}+d \times\Delta d,\quad\quad \vert d\vert =1\quad \text{in } \Omega\times(0, \infty), \end{aligned}$$
(1.23)
$$\begin{aligned}& \partial_{\nu}d=0\quad \text{on }\partial\Omega\times(0, \infty), \end{aligned}$$
(1.24)
$$\begin{aligned}& d(\cdot,0)=d_{0},\quad\quad \vert d_{0}\vert =1\quad \text{in }\Omega \subseteq\mathbb {R}^{3}. \end{aligned}$$
(1.25)

Carbou and Fabrie [9] showed the existence and uniqueness of local smooth solutions. When \(\Omega:=\mathbb{R}^{n}\) (\(n=2,3,4\)), Fan and Ozawa [10] proved some regularity criteria. The aim of this paper is to prove a logarithmic blow-up criterion for problem (1.23)-(1.25) when Ω is a bounded domain. We will prove the following:

Theorem 1.4

Let \(d_{0}\in H^{3}(\Omega)\) with \(\vert d_{0}\vert =1\) in Ω and \(\partial _{\nu}d_{0}=0\) on ∂Ω. Let d be a local smooth solution to problem (1.23)-(1.25). If d satisfies

$$ \int_{0}^{T}\frac{\Vert \nabla d\Vert _{L^{q}}^{\frac{2q}{q-3}}}{1+\log(e+\Vert \nabla d\Vert _{L^{q}})}\,dt< \infty\quad \textit{with }3< q\leq\infty $$
(1.26)

and \(0< T<\infty\), then the solution can be extended beyond \(T>0\).

In Section 2, we give some preliminary lemmas, which will be used in the following sections. The proof of Theorem 1.1 for problem (1.1)-(1.5) will be given in Section 3. The new regularly criterion of Theorem 1.2 for the 3D MHD problem (1.7)-(1.11) will be proved in Section 4. In Section 5, we prove Theorem 1.3, and in Section 6, we give the main proof of final Theorem 1.4.

2 Preliminary lemmas

In the following proofs, we will use the logarithmic Sobolev inequality [11]

$$ \Vert \nabla u\Vert _{L^{\infty}}\leq C\bigl(1+\Vert \nabla u\Vert _{\operatorname {BMO}}\log\bigl(e+\Vert u\Vert _{W^{s,p}}\bigr)\bigr)\quad \text{with } s>1+\frac{3}{p} $$
(2.1)

and the following three lemmas.

Lemma 2.1

([12])

Let \(\Omega\subseteq\mathbb{R}^{3}\) be a smooth bounded domain, let \(b:\Omega\rightarrow\mathbb{R}^{3}\) be a smooth vector field, and let \(1< p<\infty\). Then

$$\begin{aligned} - \int_{\Omega}\Delta b\cdot b\vert b\vert ^{p-2}\,dx =& \frac{1}{2} \int_{\Omega} \vert b\vert ^{p-2}\vert \nabla b \vert ^{2}\,dx+4\frac{p-2}{p^{2}} \int_{\Omega}\bigl\vert \nabla \vert b\vert ^{\frac{p}{2}}\bigr\vert ^{2}\,dx \\ &{}- \int_{\partial\Omega} \vert b\vert ^{p-2}(b\cdot\nabla)b\cdot \nu \,d\sigma- \int_{\partial\Omega} \vert b\vert ^{p-2}(\operatorname {curl}b\times\nu)\cdot b \,d\sigma. \end{aligned}$$
(2.2)

Lemma 2.2

([13, 14])

Let Ω be a smooth and bounded open set, and let \(1< p<\infty\). Then we have the estimate

$$ \Vert b\Vert _{L^{p}(\partial\Omega)}\leq C\Vert b\Vert _{L^{p}(\Omega)}^{1-\frac{1}{p}} \Vert b\Vert _{W^{1,p}(\Omega)}^{\frac{1}{p}} $$
(2.3)

for all \(b\in W^{1,p}(\Omega)\).

Lemma 2.3

We have

$$ \Vert f\Vert _{L^{\infty}(\Omega)}\leq C\bigl(1+\Vert f\Vert _{\operatorname {BMO}(\Omega)} \log^{\frac{1}{2}}\bigl(e+\Vert f\Vert _{W^{1,4}(\Omega )}\bigr) \bigr) $$
(2.4)

for all \(f\in W_{0}^{1,4}(\Omega)\).

Proof

When \(\Omega:=\mathbb{R}^{3}\), (2.4) is proved by Ogawa [15]. For a bounded domain Ω in \(\mathbb{R}^{3}\), we define

$$\tilde{f}:= \textstyle\begin{cases} f&\text{in }\Omega,\\ 0&\text{in }\Omega^{c}:=\mathbb{R}^{3}\setminus\Omega. \end{cases} $$

Then we have [13], p.71,

$$\Vert \tilde{f}\Vert _{W^{1,4}(\mathbb{R}^{3})}=\Vert f\Vert _{W^{1,4}(\Omega)}, $$

and it is obvious that

$$\Vert \tilde{f}\Vert _{L^{\infty}(\mathbb{R}^{3})}= \Vert f\Vert _{L^{\infty}(\Omega)},\Vert \tilde{f}\Vert _{\operatorname {BMO}(\mathbb{R}^{3})}=\Vert f\Vert _{\operatorname {BMO}(\Omega)}. $$

Thus, (2.4) is proved. □

Finally, when b satisfies \(b\cdot\nu=0\) on ∂Ω, we will also use the identity

$$ (b\cdot\nabla)b\cdot\nu=-(b\cdot\nabla)\nu\cdot b \quad \text{on } \partial \Omega $$
(2.5)

for any sufficiently smooth vector field b.

3 Proof of Theorem 1.1

Since it is easy to prove that problem (1.1)-(1.5) has a unique local-in-time strong solution, we omit the details. We only need to establish a priori estimates.

First, thanks to the maximum principle, it follows from (1.1) and (1.3) that

$$ \Vert \theta \Vert _{L^{\infty}(0,T;L^{\infty})}\leq C. $$
(3.1)

Testing (1.2) by u and using (1.1) and (3.1), we see that

$$\frac{1}{2}\frac{d}{dt} \int_{\Omega}u^{2} \,dx+ \int_{\Omega} \vert \operatorname {curl}u\vert ^{2}\,dx\leq \int_{\Omega}\theta e_{3}\cdot u \,dx\leq\frac{1}{2} \int_{\Omega}\theta ^{2}\,dx+\frac{1}{2} \int_{\Omega}u^{2} \,dx, $$

which gives

$$ \Vert u\Vert _{L^{\infty}(0,T;L^{2})}+\Vert u\Vert _{L^{2}(0,T;H^{1})}\leq C. $$
(3.2)

Applying curl to (1.2) and setting \(\omega:=\operatorname {curl}u\), we find that

$$ \partial_{t}\omega+u\cdot\nabla\omega-\Delta\omega=\omega\cdot \nabla u+\operatorname {curl}(\theta e_{3}). $$
(3.3)

Testing (3.3) by ω and using (1.1) and (3.1), we infer that

$$\begin{aligned} \frac{1}{2}\frac{d}{dt} \int_{\Omega} \vert \omega \vert ^{2}\,dx+ \int_{\Omega} \vert \operatorname {curl}\omega \vert ^{2}\,dx =& \int_{\Omega}(\omega\cdot\nabla)u\cdot\omega \,dx+ \int _{\Omega}\theta e_{3}\operatorname {curl}\omega \,dx \\ \leq&\Vert \nabla u\Vert _{L^{\infty}} \int_{\Omega}\omega^{2}\,dx+\frac{1}{2} \int _{\Omega} \vert \operatorname {curl}\omega \vert ^{2}\,dx+C, \end{aligned}$$

which implies

$$\begin{aligned} \frac{d}{dt} \int_{\Omega} \vert \omega \vert ^{2}\,dx+ \int_{\Omega} \vert \operatorname {curl}\omega \vert ^{2}\,dx \leq&C \Vert \nabla u\Vert _{L^{\infty}} \int_{\Omega} \vert \omega \vert ^{2}\,dx+C \\ \leq&C\bigl(1+\Vert \nabla u\Vert _{\operatorname {BMO}}\bigr)\log\bigl(e+\Vert u\Vert _{H^{3}}\bigr) \int_{\Omega} \vert \omega \vert ^{2}\,dx+C, \end{aligned}$$

and therefore

$$ \int_{\Omega} \vert \omega \vert ^{2}\,dx+ \int_{t_{0}}^{t}\Vert \operatorname {curl}\omega \Vert _{L^{2}}^{2}\,d\tau \leq C(e+y)^{C_{0}\epsilon}, $$
(3.4)

provided that

$$ \int_{t_{0}}^{t}\Vert \nabla u\Vert _{\operatorname {BMO}}\,d\tau\leq\epsilon \ll 1, $$
(3.5)

and \(y(t):=\sup_{[t_{0},t]}\Vert u\Vert _{H^{3}}\) for any \(0< t_{0}\leq t\leq T\), and \(C_{0}\) is an absolute constant.

Applying \(\partial_{t}\) to (1.2), we deduce that

$$ \partial_{t}^{2}u+u\cdot\nabla u_{t}+\nabla \pi_{t}-\Delta u_{t}=-u_{t}\cdot \nabla u+ \theta_{t} e_{3}. $$
(3.6)

Testing (3.6) by \(u_{t}\) and using (1.1), (1.3), (3.1), and (3.2), we derive

$$\begin{aligned}& \frac{1}{2}\frac{d}{dt} \int_{\Omega} \vert u_{t}\vert ^{2}\,dx+ \int_{\Omega} \vert \operatorname {curl}u_{t}\vert ^{2}\,dx \\& \quad = - \int_{\Omega}u_{t}\cdot\nabla u\cdot u_{t} \,dx+ \int_{\Omega}\theta _{t}e_{3}u_{t} \,dx \\& \quad = - \int_{\Omega}u_{t}\cdot\nabla u\cdot u_{t} \,dx- \int_{\Omega} \operatorname {div}(u\theta)e_{3}u_{t} \,dx \\& \quad = - \int_{\Omega}u_{t}\cdot\nabla u\cdot u_{t} \,dx+ \int_{\Omega}u\theta \nabla(e_{3}u_{t})\,dx \\& \quad \leq \Vert \nabla u\Vert _{L^{\infty}} \int_{\Omega} \vert u_{t}\vert ^{2}\,dx+ \frac{1}{2} \int _{\Omega} \vert \operatorname {curl}u_{t}\vert ^{2}\,dx+C \\& \quad \leq C\bigl(1+\Vert \nabla u\Vert _{\operatorname {BMO}}\bigr)\log(e+y) \int_{\Omega} \vert u_{t}\vert ^{2}\,dx+ \frac{1}{2} \int_{\Omega} \vert \operatorname {curl}u_{t}\vert ^{2}\,dx+C, \end{aligned}$$

which yields

$$ \int_{\Omega} \vert u_{t}\vert ^{2}\,dx+ \int_{t_{0}}^{t} \int_{\Omega} \vert \operatorname {curl}u_{t}\vert ^{2}\,dx \,d \tau \leq C(e+y)^{C_{0}\epsilon}. $$
(3.7)

On the other hand, thanks to the \(H^{2}\)-theory of the Stokes system, if follows from (1.2), (3.1), (3.4), and (3.7) that

$$\begin{aligned} \Vert u\Vert _{H^{2}} \leq&C\Vert {-}\Delta u+\nabla\pi \Vert _{L^{2}} \\ \leq&C\Vert \partial_{t}u+u\cdot\nabla u-\theta e_{3} \Vert _{L^{2}} \\ \leq&C\Vert u_{t}\Vert _{L^{2}}+C\Vert u\Vert _{L^{6}}\Vert \nabla u\Vert _{L^{3}}+C\Vert \theta \Vert _{L^{2}} \\ \leq&C\Vert u_{t}\Vert _{L^{2}}+C\Vert \nabla u\Vert _{L^{2}}^{\frac{3}{2}}\Vert u\Vert _{H^{2}}^{\frac{1}{2}}+C, \end{aligned}$$

which implies

$$ \Vert u\Vert _{H^{2}}\leq C\Vert u_{t}\Vert _{L^{2}}+C\Vert \nabla u\Vert _{L^{2}}^{3}+C\leq C(e+y)^{C_{0}\epsilon}. $$
(3.8)

Applying ∇ to (1.3), testing by \(\vert \nabla\theta \vert ^{p-2}\nabla\theta\) (\(2\leq p<\infty\)), and using (1.1), we get

$$\begin{aligned} \frac{d}{dt}\Vert \nabla\theta \Vert _{L^{p}} \leq&C\Vert \nabla u\Vert _{L^{\infty}} \Vert \nabla\theta \Vert _{L^{p}} \\ \leq&C\bigl(1+\Vert \nabla u\Vert _{\operatorname {BMO}}\bigr)\log(e+y)\Vert \nabla\theta \Vert _{L^{p}}, \end{aligned}$$

which leads to

$$ \Vert \nabla\theta \Vert _{L^{\infty}(t_{0},t;L^{p})}\leq C(e+y)^{C_{0}\epsilon} \quad \text{with } 2\leq p< \infty. $$
(3.9)

Testing (3.6) by \(-\Delta u_{t}+\nabla\pi_{t}\) and using (1.1), (1.3), (3.7), (3.8), and (3.9), we obtain

$$\begin{aligned}& \frac{1}{2}\frac{d}{dt} \int_{\Omega} \vert \operatorname {curl}u_{t}\vert ^{2}\,dx+ \int_{\Omega} \vert {-}\Delta u_{t}+\nabla \pi_{t}\vert ^{2}\,dx \\& \quad = \int_{\Omega}(-u_{t}\cdot\nabla u+\theta_{t}e_{3}-u \cdot\nabla u_{t}) (-\Delta u_{t}+\nabla\pi_{t})\,dx \\& \quad \leq \bigl(\Vert \nabla u\Vert _{L^{6}}\Vert u_{t} \Vert _{L^{3}}+\Vert u\Vert _{L^{\infty}} \Vert \nabla \theta \Vert _{L^{2}}+\Vert u\Vert _{L^{\infty}} \Vert \nabla u_{t}\Vert _{L^{2}}\bigr)\Vert {-}\Delta u_{t}+ \nabla\pi_{t}\Vert _{L^{2}} \\& \quad \leq \Vert u\Vert _{H^{2}}\bigl(\Vert u_{t}\Vert _{H^{1}}+\Vert \nabla\theta \Vert _{L^{2}}\bigr)\Vert {-}\Delta u_{t}+\nabla\pi_{t}\Vert _{L^{2}} \\& \quad \leq \frac{1}{2}\Vert {-}\Delta u_{t}+\nabla\pi_{t} \Vert _{L^{2}}^{2}+C\Vert u\Vert _{H^{2}}^{2} \bigl(\Vert u_{t}\Vert _{H^{1}}^{2}+\Vert \nabla \theta \Vert _{L^{2}}^{2}\bigr), \end{aligned}$$

which leads to

$$ \int_{\Omega} \vert \operatorname {curl}u_{t}\vert ^{2}\,dx+ \int_{t_{0}}^{t}\Vert u_{t}\Vert _{H^{2}}^{2}\,d\tau\leq C(e+y)^{C_{0}\epsilon}. $$
(3.10)

On the other hand, if follows from (3.3), (3.10), (3.9), and (3.8) that

$$\begin{aligned} \Vert u\Vert _{H^{3}} \leq&C\bigl(1+\Vert \Delta\omega \Vert _{L^{2}}\bigr) \\ \leq&C\bigl(1+\bigl\Vert \partial_{t}\omega+u\cdot\nabla\omega- \omega\cdot \nabla u-\operatorname {curl}(\theta e_{3})\bigr\Vert _{L^{2}}\bigr) \\ \leq&C\bigl(1+\Vert \partial_{t}\omega \Vert _{L^{2}}+ \Vert u\Vert _{L^{\infty}} \Vert \nabla \omega \Vert _{L^{2}}+ \Vert \omega \Vert _{L^{4}}\Vert \nabla u\Vert _{L^{4}}+ \Vert \nabla\theta \Vert _{L^{2}}\bigr) \\ \leq&C(e+y)^{C_{0}\epsilon}, \end{aligned}$$

which gives

$$ \Vert u\Vert _{L^{\infty}(0,T;H^{3})}\leq C $$
(3.11)

and

$$ \Vert \theta \Vert _{L^{\infty}(0,T;W^{1,p})}\leq C\quad \text{with }3\leq p \leq6. $$
(3.12)

This completes the proof of Theorem 1.1.

4 Proof of Theorem 1.2

We only need to prove a priori estimates.

First, testing (1.8) by u and using (1.7), we see that

$$ \frac{1}{2}\frac{d}{dt} \int_{\Omega}u^{2}\,dx+ \int_{\Omega} \vert \operatorname {curl}u\vert ^{2}\,dx= \int _{\Omega}(b\cdot\nabla)b\cdot u \,dx. $$
(4.1)

Testing (1.9) by b and using (1.7), we find that

$$ \frac{1}{2}\frac{d}{dt} \int_{\Omega}b^{2} \,dx+ \int_{\Omega} \vert \operatorname {curl}b\vert ^{2}\,dx= \int _{\Omega}(b\cdot\nabla)u\cdot b \,dx. $$
(4.2)

Summing up (4.1) and (4.2), we get the well-known energy inequality

$$ \frac{1}{2}\frac{d}{dt} \int_{\Omega}\bigl(u^{2}+b^{2}\bigr)\,dx+ \int_{\Omega}\bigl(\vert \operatorname {curl}u\vert ^{2}+\vert \operatorname {curl}b\vert ^{2}\bigr)\,dx\leq0. $$
(4.3)

Testing (1.9) by \(\vert b\vert ^{p-2}b\) (\(2\leq p\leq6\)) and using (1.7), (2.2), (2.3), and (2.5), we derive

$$\begin{aligned}& \frac{1}{p}\frac{d}{dt} \int_{\Omega} \vert b\vert ^{p} \,dx+\frac{1}{2} \int_{\Omega} \vert b\vert ^{p-2}\vert \nabla b \vert ^{2}\,dx+4\frac{p-2}{p^{2}} \int_{\Omega}\bigl\vert \nabla \vert b\vert ^{\frac{p}{2}}\bigr\vert ^{2}\,dx \\& \quad = - \int_{\partial\Omega} \vert b\vert ^{p-2}(b\cdot\nabla)\nu\cdot b \,d\sigma+ \int_{\Omega}b\cdot\nabla u\cdot \vert b\vert ^{p-2}b \,dx \\& \quad \leq C \int_{\partial\Omega} \vert b\vert ^{p} \,dx+\Vert \nabla u \Vert _{L^{\infty}} \int _{\Omega} \vert b\vert ^{p} \,dx \\& \quad \leq 2\frac{p-2}{p^{2}} \int_{\Omega}\bigl\vert \nabla \vert b\vert ^{\frac{p}{2}}\bigr\vert ^{2}\,dx+C\bigl(1+\Vert \nabla u\Vert _{L^{\infty}}\bigr) \int_{\Omega} \vert b\vert ^{p} \,dx \\& \quad \leq 2\frac{p-2}{p^{2}} \int_{\Omega}\bigl\vert \nabla \vert b\vert ^{\frac{p}{2}}\bigr\vert ^{2}\,dx+C\bigl(1+\Vert \nabla u\Vert _{\operatorname {BMO}}\bigr) \int_{\Omega} \vert b\vert ^{p} \,dx\log(e+y), \end{aligned}$$

which implies

$$ \Vert b\Vert _{L^{\infty}(t_{0},t;L^{p})}+ \int_{t_{0}}^{t} \int_{\Omega} \vert b\vert ^{2}\vert \nabla b\vert ^{2}\,dx \,d\tau\leq C(e+y)^{C_{0}\epsilon}\quad \text{with }2\leq p \leq6, $$
(4.4)

with the same y and ϵ as in (3.5).

Taking curl to (1.8) and (1.9), respectively, and setting \(\omega:=\operatorname {curl}u\) and \(j:=\operatorname {curl}b\), we infer that

$$\begin{aligned}& \partial_{t}\omega+u\cdot\nabla\omega-\Delta\omega=\omega\cdot \nabla u+b\cdot\nabla j+\sum_{i}\nabla b_{i}\times\partial _{i}b, \end{aligned}$$
(4.5)
$$\begin{aligned}& \partial_{t}j+u\cdot\nabla j-\Delta j=b\cdot\nabla\omega+\sum _{i}\nabla b_{i}\times \partial_{i}u-\sum_{i}\nabla u_{i}\times \partial_{i}b. \end{aligned}$$
(4.6)

Testing (4.5) and (4.6) by ω and j, respectively, summing up the result, and using (1.7), we have

$$\begin{aligned}& \frac{1}{2}\frac{d}{dt} \int_{\Omega}\bigl(\omega^{2}+j^{2}\bigr)\,dx+ \int_{\Omega}\bigl(\vert \operatorname {curl}\omega \vert ^{2}+\vert \operatorname {curl}j\vert ^{2}\bigr)\,dx \\& \quad = \int_{\Omega}(\omega\cdot\nabla)u\cdot\omega \,dx+\sum _{i} \int_{\Omega}(\nabla b_{i}\times\partial_{i}b) \omega \,dx \\& \quad\quad{} +\sum_{i} \int_{\Omega}(\nabla b_{i}\times\partial_{i}u) \cdot j \,dx-\sum_{i} \int_{\Omega}(\nabla u_{i}\times\partial_{i}b) \cdot j \,dx \\& \quad \leq C\Vert \nabla u\Vert _{L^{\infty}} \int_{\Omega}\bigl(\omega^{2}+j^{2}\bigr)\,dx \\& \quad \leq C\bigl(1+\Vert \nabla u\Vert _{\operatorname {BMO}}\bigr) \int_{\Omega}\bigl(\omega^{2}+j^{2}\bigr)\,dx \log(e+y), \end{aligned}$$

which implies

$$ \int_{\Omega}\bigl(\omega^{2}+j^{2}\bigr)\,dx+ \int_{t_{0}}^{t} \int_{\Omega}\bigl(\vert \operatorname {curl}\omega \vert ^{2}+\vert \operatorname {curl}j\vert ^{2}\bigr)\,dx \,d\tau\leq C(e+y)^{C_{0}\epsilon}. $$
(4.7)

Thus, it follows from (1.8), (1.9), and (4.7) that

$$ \int_{t_{0}}^{t} \int_{\Omega}\bigl(\vert u_{t}\vert ^{2}+ \vert b_{t}\vert ^{2}\bigr)\,dx \,d\tau\leq C(e+y)^{C_{0}\epsilon}. $$
(4.8)

Applying \(\partial_{t}\) to (1.8), we have

$$ \partial_{t}^{2}u+u\cdot\nabla u_{t}+\nabla \pi_{t}-\Delta u_{t}=\operatorname {div}(b\otimes b)_{t}-u_{t} \cdot\nabla u. $$
(4.9)

Testing (4.9) by \(u_{t}\) and using (1.7), we get

$$\begin{aligned}& \frac{1}{2}\frac{d}{dt} \int_{\Omega} \vert u_{t}\vert ^{2}\,dx+ \int_{\Omega} \vert \operatorname {curl}u_{t}\vert ^{2}\,dx \\& \quad = -\sum_{i,j} \int_{\Omega}\bigl(b^{i}b^{j} \bigr)_{t}\partial_{j}u_{t}^{i}\,dx- \int _{\Omega}u_{t}\cdot\nabla u\cdot u_{t} \,dx \\& \quad \leq C\Vert b_{t}\Vert _{L^{3}}\Vert b\Vert _{L^{6}}\Vert \nabla u_{t}\Vert _{L^{2}}+\Vert \nabla u\Vert _{L^{2}}\Vert u_{t}\Vert _{L^{4}}^{2} \\& \quad \leq C\Vert b_{t}\Vert _{L^{2}}^{\frac{1}{2}}\Vert \operatorname {curl}b_{t}\Vert _{L^{2}}^{\frac{1}{2}}\Vert \operatorname {curl}u_{t}\Vert _{L^{2}}\Vert b\Vert _{L^{6}}+C\Vert \nabla u\Vert _{L^{2}}\Vert u_{t}\Vert _{L^{2}}^{\frac{1}{2}} \Vert \operatorname {curl}u_{t}\Vert _{L^{2}}^{\frac{3}{2}} \\& \quad \leq \delta \Vert \operatorname {curl}u_{t}\Vert _{L^{2}}^{2}+ \delta \Vert \operatorname {curl}b_{t}\Vert _{L^{2}}^{2}+C\Vert b_{t}\Vert _{L^{2}}^{2}\Vert b\Vert _{L^{6}}^{4}+C\Vert \nabla u\Vert _{L^{2}}^{4} \Vert u_{t}\Vert _{L^{2}}^{2} \end{aligned}$$
(4.10)

for any \(\delta\in(0,1)\).

Applying \(\partial_{t}\) to (1.9), we have

$$ \partial_{t}^{2}b+u\cdot\nabla b_{t}-\Delta b_{t}=b_{t}\cdot\nabla u+b\cdot \nabla u_{t}-u_{t} \cdot\nabla b. $$
(4.11)

Testing (4.11) by \(b_{t}\) and using (1.7), we deduce that

$$\begin{aligned}& \frac{1}{2}\frac{d}{dt} \int_{\Omega} \vert b_{t}\vert ^{2}\,dx+ \int_{\Omega} \vert \operatorname {curl}b_{t}\vert ^{2}\,dx \\& \quad = \int_{\Omega}(b_{t}\cdot\nabla u+b\cdot\nabla u_{t}-u_{t}\cdot\nabla b)b_{t} \,dx \\& \quad \leq \Vert \nabla u\Vert _{L^{2}}\Vert b_{t}\Vert _{L^{4}}^{2}+\Vert b\Vert _{L^{6}}\Vert \nabla u_{t}\Vert _{L^{2}}\Vert b_{t}\Vert _{L^{3}}+\Vert \nabla b\Vert _{L^{2}}\Vert u_{t} \Vert _{L^{4}}\Vert b_{t}\Vert _{L^{4}} \\& \quad \leq \delta \Vert \operatorname {curl}b_{t}\Vert _{L^{2}}^{2}+ \delta \Vert \operatorname {curl}u_{t}\Vert _{L^{2}}^{2} \\& \quad\quad{} +C\Vert \nabla u\Vert _{L^{2}}^{4}\Vert b_{t}\Vert _{L^{2}}^{2}+C\Vert b\Vert _{L^{6}}^{4}\Vert b_{t}\Vert _{L^{2}}^{2}+C \Vert \nabla b\Vert _{L^{2}}^{4}\bigl(\Vert u_{t}\Vert _{L^{2}}^{2}+\Vert b_{t} \Vert _{L^{2}}^{2}\bigr) \end{aligned}$$
(4.12)

for any \(\delta\in(0,1)\).

Combining (4.10) and (4.12), taking δ small enough, and using (4.7) and (4.8), we have

$$ \int_{\Omega}\bigl(\vert u_{t}\vert ^{2}+ \vert b_{t}\vert ^{2}\bigr)\,dx+ \int_{t_{0}}^{t} \int_{\Omega}\bigl(\vert \operatorname {curl}u_{t}\vert ^{2}+\vert \operatorname {curl}b_{t}\vert ^{2}\bigr)\,dx \,d\tau \leq C(e+y)^{C_{0}\epsilon}. $$
(4.13)

It follows from (1.8), (1.9), (4.7), and (4.13) that

$$ \Vert u\Vert _{L^{\infty}(t_{0},t;H^{2})}+\Vert b\Vert _{L^{\infty}(t_{0},t;H^{2})}\leq C(e+y)^{C_{0}\epsilon}. $$
(4.14)

Testing (4.9) by \(\nabla (\pi+\frac{1}{2}\vert b\vert ^{2} )_{t}-\Delta u_{t}\) and using (1.7), we find that

$$\begin{aligned}& \frac{1}{2}\frac{d}{dt} \int_{\Omega} \vert \operatorname {curl}u_{t}\vert ^{2}\,dx+ \int_{\Omega}\biggl\vert \nabla \biggl(\pi+\frac{1}{2} \vert b\vert ^{2} \biggr)_{t}-\Delta u_{t}\biggr\vert ^{2}\,dx \\& \quad = \int_{\Omega}\bigl((b\cdot\nabla b)_{t}-u_{t} \cdot\nabla u-u\cdot\nabla u_{t}\bigr) \biggl(\nabla \biggl(\pi+ \frac{1}{2}\vert b\vert ^{2} \biggr)_{t}-\Delta u_{t} \biggr)\,dx \\& \quad \leq C\bigl(\Vert b\Vert _{L^{\infty}} \Vert \nabla b_{t} \Vert _{L^{2}}+\Vert b_{t}\Vert _{L^{6}}\Vert \nabla b\Vert _{L^{3}}+\Vert u_{t}\Vert _{L^{6}} \Vert \nabla u\Vert _{L^{3}} \\& \quad\quad{} +\Vert u\Vert _{L^{\infty}} \Vert \nabla u_{t}\Vert _{L^{2}}\bigr)\biggl\Vert \nabla \biggl(\pi +\frac{1}{2}\vert b \vert ^{2} \biggr)_{t}-\Delta u_{t}\biggr\Vert _{L^{2}} \\& \quad \leq \frac{1}{4}\biggl\Vert \nabla \biggl(\pi+\frac{1}{2}\vert b\vert ^{2} \biggr)_{t}-\Delta u_{t}\biggr\Vert _{L^{2}}^{2}+C\bigl(\Vert u\Vert _{L^{\infty}}^{2}+ \Vert \nabla u\Vert _{L^{3}}^{2}\bigr)\Vert \nabla u_{t}\Vert _{L^{2}}^{2} \\& \quad\quad{} +C\bigl(\Vert b\Vert _{L^{\infty}}^{2}+\Vert \nabla b\Vert _{L^{3}}^{2}\bigr)\Vert \nabla b_{t}\Vert _{L^{2}}^{2}. \end{aligned}$$
(4.15)

Similarly, testing (4.11) by \(-\Delta b_{t}\), we infer that

$$\begin{aligned}& \frac{1}{2}\frac{d}{dt} \int_{\Omega} \vert \operatorname {curl}b_{t}\vert ^{2}\,dx+ \int_{\Omega} \vert \Delta b_{t}\vert ^{2}\,dx \\& \quad = \int_{\Omega}(u_{t}\cdot\nabla b+u\cdot\nabla b_{t}-b_{t}\cdot\nabla u-b\cdot\nabla u_{t})\Delta b_{t} \,dx \\& \quad \leq \bigl(\Vert u_{t}\Vert _{L^{6}}\Vert \nabla b \Vert _{L^{3}}+\Vert u\Vert _{L^{\infty}} \Vert \nabla b_{t}\Vert _{L^{2}}+\Vert \nabla u\Vert _{L^{3}} \Vert b_{t}\Vert _{L^{6}}+\Vert b\Vert _{L^{\infty}} \Vert \nabla u_{t}\Vert _{L^{2}}\bigr)\Vert \Delta b_{t}\Vert _{L^{2}} \\& \quad \leq \frac{1}{4}\Vert \Delta b_{t}\Vert _{L^{2}}^{2}+C\bigl(\Vert u\Vert _{L^{\infty}}^{2}+ \Vert \nabla u\Vert _{L^{3}}^{2}\bigr)\Vert \nabla b_{t}\Vert _{L^{2}}^{2}+C\bigl(\Vert b\Vert _{L^{\infty}}^{2}+\Vert \nabla b\Vert _{L^{3}}^{2} \bigr)\Vert \nabla u_{t}\Vert _{L^{2}}^{2}. \end{aligned}$$
(4.16)

Combining (4.15) and (4.16) and using (4.14) and (4.13), we have

$$ \int_{\Omega}\bigl(\vert \operatorname {curl}u_{t}\vert ^{2}+\vert \operatorname {curl}b_{t}\vert ^{2}\bigr)\,dx+ \int_{t_{0}}^{t} \int_{\Omega}\bigl(\vert \Delta u_{t}\vert ^{2}+\vert \Delta b_{t}\vert ^{2}\bigr)\,dx \,d\tau \leq C(e+y)^{C_{0}\epsilon }. $$
(4.17)

On the other hand, it follows from (4.5), (4.6), (4.3), (4.17), and (4.14) that

$$\begin{aligned} \bigl\Vert u(t)\bigr\Vert _{H^{3}}+\bigl\Vert b(t)\bigr\Vert _{H^{3}} \leq& C\bigl(1+\Vert \Delta\omega \Vert _{L^{2}}+\Vert \Delta j\Vert _{L^{2}}\bigr) \\ \leq& C\biggl(1+\biggl\Vert \partial_{t}\omega+u\cdot\nabla\omega- \omega\cdot \nabla u-b\cdot\nabla j-\sum_{i}\nabla b_{i}\times\partial_{i}b\biggr\Vert _{L^{2}} \\ &{}+\biggl\Vert \partial_{t}j+u\cdot\nabla j-b\cdot\nabla\omega+\sum _{i}\nabla u_{i}\times \partial_{i}b-\sum_{i}\nabla b_{i}\times \partial_{i}u\biggr\Vert _{L^{2}} \biggr) \\ \leq&C\bigl(e+y(t)\bigr)^{C_{0}\epsilon}, \end{aligned}$$

which yields

$$\Vert u\Vert _{L^{\infty}(0,T;H^{3})}+\Vert b\Vert _{L^{\infty}(0,T;H^{3})}\leq C, $$

This completes the proof of Theorem 1.2.

5 Proof of Theorem 1.3

We only need to establish a priori estimates.

First, it follows from (1.12) and (1.13) that

$$ \Vert \rho \Vert _{L^{\infty}(0,T;L^{\infty})}\leq C. $$
(5.1)

Testing (1.14) by u and using (1.12) and (1.13), we see that

$$ \frac{1}{2}\frac{d}{dt} \int_{\Omega}\rho u^{2}\,dx+ \int_{\Omega} \vert \nabla u\vert ^{2}\,dx= \int_{\Omega}(b\cdot\nabla)b\cdot u \,dx $$
(5.2)

and testing (1.15) by b and using (1.12) and (1.16), we find that

$$ \frac{1}{2}\frac{d}{dt} \int_{\Omega} \vert b\vert ^{2}\,dx+ \int_{\Omega} \vert \operatorname {curl}b\vert ^{2}\,dx= \int _{\Omega}(b\cdot\nabla)u\cdot b \,dx. $$
(5.3)

Summing up (5.2) and (5.3), we get the well-known energy inequality

$$ \frac{1}{2}\frac{d}{dt} \int_{\Omega}\bigl(\rho \vert u\vert ^{2}+\vert b \vert ^{2}\bigr)\,dx+ \int_{\Omega}\bigl(\vert \nabla u\vert ^{2}+\vert \operatorname {curl}b\vert ^{2}\bigr)\,dx\leq0. $$
(5.4)

(I) Let (1.21) hold.

Testing (1.15) by \(\vert b\vert ^{p-2}b\) (\(2\leq p<\infty\)), using (1.12), (2.2), (2.3), and (2.5), setting \(\phi =\vert b\vert ^{\frac{p}{2}}\), and using the Gagliardo-Nirenberg inequality [3]

$$ \Vert \phi \Vert _{L^{\frac{2s}{s-2},2}}\leq C\Vert \phi \Vert _{L^{2}}^{1-\frac{3}{s}}\Vert \phi \Vert _{H^{1}}^{\frac{3}{s}} \quad \text{with }3< s\leq\infty $$
(5.5)

and the generalized Hölder inequality [7]

$$ \Vert fg\Vert _{L^{p,q}}\leq C\Vert f\Vert _{L^{p_{1},q_{1}}}\Vert g \Vert _{L^{p_{2},q_{2}}} $$
(5.6)

with \(\frac{1}{p}=\frac{1}{p_{1}}+\frac{1}{p_{2}}\) and \(\frac{1}{q}=\frac {1}{q_{1}}+\frac{1}{q_{2}}\), we derive

$$\begin{aligned}& \frac{1}{p}\frac{d}{dt} \int_{\Omega} \vert b\vert ^{p} \,dx+\frac{1}{2} \int_{\Omega} \vert b\vert ^{p-2}\vert \nabla b \vert ^{2}\,dx+4\frac{p-2}{p^{2}} \int_{\Omega}\bigl\vert \nabla \vert b\vert ^{\frac{p}{2}}\bigr\vert ^{2}\,dx \\& \quad = - \int_{\partial\Omega} \vert b\vert ^{p-2}(b\cdot\nabla)\nu\cdot b \,d\sigma+ \int_{\Omega}(b\cdot\nabla)u\cdot \vert b\vert ^{p-2}b \,dx \\& \quad \leq \Vert \nabla\nu \Vert _{L^{\infty}} \int_{\partial\Omega} \vert b\vert ^{p} \,d\sigma -\sum _{i} \int_{\Omega}b_{i}u\partial_{i}\bigl(\vert b\vert ^{p-2}b\bigr)\,dx \\& \quad \leq C \int_{\partial\Omega}\phi^{2}\,d\sigma+C \int_{\Omega} \vert u\phi \nabla\phi \vert \,dx \\& \quad \leq C \int_{\partial\Omega}\phi^{2}\,d\sigma+C\Vert u\Vert _{L_{w}^{s}}\Vert \phi \Vert _{L^{\frac{2s}{s-2},2}}\Vert \nabla\phi \Vert _{L^{2}} \\& \quad \leq C\Vert \phi \Vert _{L^{2}}\Vert \phi \Vert _{H^{1}}+C \Vert u\Vert _{L_{w}^{s}}\Vert \phi \Vert _{L^{2}}^{1-\frac{3}{s}} \Vert \nabla\phi \Vert _{L^{2}}^{1+\frac{3}{s}} \\& \quad \leq 2\frac{p-2}{p^{2}} \int_{\Omega} \vert \nabla\phi \vert ^{2}\,dx+C\Vert \phi \Vert _{L^{2}}^{2}+C\Vert u\Vert _{L_{w}^{s}}^{\frac{2s}{s-3}}\Vert \phi \Vert _{L^{2}}^{2}, \end{aligned}$$

which yields

$$\begin{aligned} \frac{d}{dt} \int_{\Omega}\phi^{2}\,dx+C \int_{\Omega} \vert \nabla\phi \vert ^{2}\,dx \leq& C \bigl(1+\Vert u\Vert _{L_{w}^{s}}^{\frac{2s}{s-3}}\bigr)\Vert \phi \Vert _{L^{2}}^{2} \\ \leq &C \biggl(1+\frac{\Vert u\Vert _{L_{w}^{s}}^{\frac{2s}{s-3}}}{1+\log(e+\Vert u\Vert _{L_{w}^{s}})} \biggr)\Vert \phi \Vert _{L^{2}}^{2}\bigl(1+\log\bigl(e+\Vert u\Vert _{L_{w}^{s}} \bigr)\bigr) \\ \leq& C \biggl(1+\frac{\Vert u\Vert _{L_{w}^{s}}^{\frac{2s}{s-3}}}{1+\log(e+\Vert u\Vert _{L_{w}^{s}})} \biggr) \bigl(1+\log(e+y)\bigr)\Vert \phi \Vert _{L^{2}}^{2}, \end{aligned}$$

from which it follows that

$$ \Vert b\Vert _{L^{\infty}(t_{0},t;L^{p})}+ \int_{t_{0}}^{t} \int_{\Omega} \vert b\vert ^{2}\vert \nabla b\vert ^{2}\,dx \,d\tau\leq C\bigl(e+y(t)\bigr)^{C_{0}\epsilon} $$
(5.7)

with

$$y(t):=\sup_{[t_{0},t]}\Vert u\Vert _{W^{1,4}} $$

for any \(0< t_{0}\leq t\leq T\), where \(C_{0}\) is an absolute constant, provided that

$$ \int_{t_{0}}^{T}\frac{\Vert u\Vert _{L_{w}^{s}}^{\frac{2s}{s-3}}}{1+\log(e+\Vert u\Vert _{L_{w}^{s}})}\,d\tau\leq \epsilon \ll 1. $$
(5.8)

Testing (1.14) by \(u_{t}\) and using (1.12) and (1.13), we infer that

$$\begin{aligned} \frac{1}{2}\frac{d}{dt} \int_{\Omega} \vert \nabla u\vert ^{2}\,dx+ \int_{\Omega}\rho \vert u_{t}\vert ^{2}\,dx =&- \int_{\Omega}\rho u\cdot\nabla u\cdot u_{t} \,dx+ \int_{\Omega}b\cdot\nabla b\cdot u_{t} \,dx \\ =&:I_{1}+I_{2}. \end{aligned}$$
(5.9)

We first compute \(I_{2}\):

$$\begin{aligned} I_{2} =& \int_{\Omega} \operatorname {div}(b\otimes b)\cdot u_{t} \,dx=- \int_{\Omega}b\otimes b:\nabla u_{t} \,dx \\ =&-\frac{d}{dt} \int_{\Omega}b\otimes b:\nabla u \,dx+2 \int_{\Omega}b\otimes b_{t}:\nabla u \,dx \\ \leq&-\frac{d}{dt} \int_{\Omega}b\otimes b:\nabla u \,dx+C\Vert b_{t} \Vert _{L^{2}}\Vert b\Vert _{L^{6}}\Vert \nabla u\Vert _{L^{3}} \\ \leq&-\frac{d}{dt} \int_{\Omega}b\otimes b:\nabla u \,dx+C\Vert b_{t} \Vert _{L^{2}}\Vert b\Vert _{L^{6}}\Vert \nabla u\Vert _{L^{2}}^{\frac{1}{2}}\Vert u\Vert _{H^{2}}^{\frac{1}{2}} \\ \leq&-\frac{d}{dt} \int_{\Omega}b\otimes b:\nabla u \,dx+\delta \Vert b_{t} \Vert _{L^{2}}^{2}+\delta \Vert u\Vert _{H^{2}}^{2}+C \Vert b\Vert _{L^{6}}^{4}\Vert \nabla u\Vert _{L^{2}}^{2} \end{aligned}$$
(5.10)

for any \(0<\delta<1\).

We use (5.1), (5.5), and (5.6) to bound \(I_{1}\) as follows:

$$\begin{aligned} I_{1} \leq&\Vert \sqrt{\rho}u_{t}\Vert _{L^{2}} \Vert \sqrt{\rho} \Vert _{L^{\infty}} \Vert u\Vert _{L_{w}^{s}}\Vert \nabla u\Vert _{L^{\frac{2s}{s-2},2}} \\ \leq&C\Vert \sqrt{\rho}u_{t}\Vert _{L^{2}}\Vert u\Vert _{L_{w}^{s}}\Vert \nabla u\Vert _{L^{2}}^{1-\frac{3}{s}}\Vert u \Vert _{H^{2}}^{\frac{3}{s}} \\ \leq&\delta \Vert \sqrt {u}_{t}\Vert _{L^{2}}^{2}+ \delta \Vert u\Vert _{H^{2}}^{2}+C\Vert u\Vert _{L_{w}^{s}}^{\frac{2s}{s-3}}\Vert \nabla u\Vert _{L^{2}}^{2} \end{aligned}$$
(5.11)

for any \(0<\delta<1\).

On the other hand, by the \(H^{2}\)-theory of the Stokes system, using (5.1), (5.5), and (5.6), we obtain

$$\begin{aligned} \Vert u\Vert _{H^{2}} \leq&C\biggl\Vert {-}\Delta u+\nabla \biggl( \pi+\frac{1}{2}\vert b\vert ^{2} \biggr)\biggr\Vert _{L^{2}} \\ \leq&C\Vert \rho\partial_{t} u+\rho u\cdot\nabla u-b\cdot\nabla b \Vert _{L^{2}} \\ \leq&C\Vert \sqrt{\rho}u_{t}\Vert _{L^{2}}+C\Vert u \Vert _{L_{w}^{s}}\Vert \nabla u\Vert _{L^{\frac {2s}{s-2},2}}+C\Vert b\cdot \nabla b\Vert _{L^{2}} \\ \leq&C\Vert \sqrt{\rho}u_{t}\Vert _{L^{2}}+C\Vert u \Vert _{L_{w}^{s}}\Vert \nabla u\Vert _{L^{2}}^{1-\frac{3}{s}} \Vert u\Vert _{H^{2}}^{\frac{3}{s}}+C\Vert b\cdot\nabla b\Vert _{L^{2}}, \end{aligned}$$

which gives

$$ \Vert u\Vert _{H^{2}}\leq C\Vert \sqrt{\rho}u_{t}\Vert _{L^{2}}+C\Vert b\cdot\nabla b\Vert _{L^{2}}+C\Vert u\Vert _{L_{w}^{s}}^{\frac{s}{s-3}}\Vert \nabla u\Vert _{L^{2}}. $$
(5.12)

Testing (1.15) by \(b_{t}-\Delta b\) and using (5.5) and (5.6), we deduce that

$$\begin{aligned}& \frac{d}{dt} \int_{\Omega} \vert \operatorname {curl}b\vert ^{2}\,dx+ \int_{\Omega}\bigl(\vert b_{t}\vert ^{2}+ \vert \Delta b\vert ^{2}\bigr)\,dx \\& \quad = \int_{\Omega}(b\cdot\nabla u-u\cdot\nabla b) (b_{t}- \Delta b)\,dx \\& \quad \leq \bigl(\Vert u\Vert _{L_{w}^{s}}\Vert \nabla b\Vert _{L^{\frac{2s}{s-2},2}}+\Vert b\Vert _{L^{6}}\Vert \nabla u\Vert _{L^{3}}\bigr) \bigl(\Vert b_{t}\Vert _{L^{2}}+ \Vert \Delta b\Vert _{L^{2}}\bigr) \\& \quad \leq C\bigl(\Vert u\Vert _{L_{w}^{s}}\Vert \nabla b\Vert _{L^{2}}^{1-\frac{3}{s}}\Vert b\Vert _{H^{2}}^{\frac{3}{s}}+C \Vert b\Vert _{L^{6}}\Vert \nabla u\Vert _{L^{2}}^{\frac{1}{2}} \Vert u\Vert _{H^{2}}^{\frac{1}{2}}\bigr) \bigl(\Vert b_{t}\Vert _{L^{2}}+\Vert \Delta b\Vert _{L^{2}} \bigr) \\& \quad \leq \frac{1}{2}\bigl(\Vert b_{t}\Vert _{L^{2}}^{2}+ \Vert \Delta b\Vert _{L^{2}}^{2}\bigr)+\delta \Vert u\Vert _{H^{2}}^{2}+C\Vert b\Vert _{L^{6}}^{4} \Vert \nabla u\Vert _{L^{2}}^{2}+C\Vert u\Vert _{L_{w}^{s}}^{\frac{2s}{s-3}}\Vert \nabla b\Vert _{L^{2}}^{2}+C \end{aligned}$$
(5.13)

for any \(0<\delta<1\).

It is easy to compute that

$$\begin{aligned} \frac{d}{dt} \int_{\Omega} \vert b\vert ^{4}\,dx \leq&C \int_{\Omega} \vert b\vert ^{3}\vert b_{t}\vert \,dx \\ \leq&C\Vert b\Vert _{L^{6}}^{3}\Vert b_{t} \Vert _{L^{2}}\leq\delta \Vert b_{t}\Vert _{L^{2}}^{2}+C\Vert b\Vert _{L^{6}}^{6} \end{aligned}$$
(5.14)

for any \(0<\delta<1\).

Combining (5.9), (5.10), (5.11), (5.12), (5.13) and (5.14), and taking δ small enough, we obtain

$$\begin{aligned}& \frac{d}{dt} \int_{\Omega}\bigl(\vert \nabla u\vert ^{2}+\vert \operatorname {curl}b\vert ^{2}+b\otimes b:\nabla u+C_{0}\vert b\vert ^{4}\bigr)\,dx \\& \quad\quad{} + \int_{\Omega}\bigl(\rho \vert u_{t}\vert ^{2}+\vert b_{t}\vert ^{2}+\vert \Delta b \vert ^{2}\bigr)\,dx+\Vert u\Vert _{H^{2}}^{2} \\& \quad \leq C\Vert b\Vert _{L^{6}}^{4}\Vert \nabla u\Vert _{L^{2}}^{2}+C\Vert u\Vert _{L_{w}^{s}}^{\frac{2s}{s-3}}\bigl(\Vert \nabla u\Vert _{L^{2}}^{2}+\Vert \operatorname {curl}b\Vert _{L^{2}}^{2}\bigr)+C\Vert b\cdot\nabla b\Vert _{L^{2}}^{2}+C. \end{aligned}$$
(5.15)

Using (5.4), (5.7), (5.8), and the Gronwall inequality, we have

$$\begin{aligned}& \int_{\Omega}\bigl(\vert \nabla u\vert ^{2}+\vert \operatorname {curl}b\vert ^{2}+b\otimes b:\nabla u+C_{0}\vert b\vert ^{4}\bigr)\,dx \\& \quad \leq \biggl[ \int_{\Omega}\bigl(\vert \nabla u_{0}\vert ^{2}+\vert \operatorname {curl}b_{0}\vert ^{2}+b_{0} \otimes b_{0}:\nabla u_{0}+C_{0}\vert b_{0}\vert ^{4}\bigr)\,dx \\& \quad\quad{} +C\Vert b\Vert _{L^{\infty}(t_{0},t;L^{6})}^{4} \int_{t_{0}}^{t}\Vert \nabla u\Vert _{L^{2}}^{2}\,d\tau+C(t-t_{0})+C \int_{t_{0}}^{t}\Vert b\cdot\nabla b\Vert _{L^{2}}^{2}\,d\tau \biggr] \\& \quad\quad{}\times \exp \biggl( \int_{t_{0}}^{t}\Vert u\Vert _{L_{w}^{s}}^{\frac{2s}{s-3}}\,d\tau \biggr) \\& \quad \leq C(e+y)^{C_{0}\epsilon}\exp \biggl[ \int_{t_{0}}^{t}\frac{\Vert u\Vert _{L_{w}^{s}}^{\frac{2s}{s-3}}}{1+\log(e+\Vert u\Vert _{L_{w}^{s}})} \,d\tau\bigl(1+\log (e+y)\bigr) \biggr] \\& \quad \leq C(e+y)^{C_{0}\epsilon}. \end{aligned}$$
(5.16)

Plugging (5.16) into (5.15) and integrating over \([t_{0},t]\), we have

$$ \int_{t_{0}}^{t} \int_{\Omega}\bigl(\rho \vert u_{t}\vert ^{2}+\vert b_{t}\vert ^{2}+\vert \Delta b \vert ^{2}\bigr)\,dx \,d\tau + \int_{t_{0}}^{t}\Vert u\Vert _{H^{2}}^{2}\,d\tau\leq C(e+y)^{C_{0}\epsilon}. $$
(5.17)

Applying \(\partial_{t}\) to (1.15), testing by \(u_{t}\), and using (1.12) and (1.13), we obtain

$$\begin{aligned}& \frac{1}{2}\frac{d}{dt} \int_{\Omega}\rho \vert u_{t}\vert ^{2}\,dx+ \int_{\Omega} \vert \nabla u_{t}\vert ^{2}\,dx \\& \quad = - \int_{\Omega}\rho u\cdot\nabla \vert u_{t}\vert ^{2}\,dx- \int_{\Omega}\rho u\cdot \nabla(u\cdot\nabla u\cdot u_{t})\,dx \\& \quad\quad{} - \int_{\Omega}\rho u_{t}\cdot\nabla u\cdot u_{t} \,dx+ \int_{\Omega}b\otimes b_{t}:\nabla u_{t} \,dx+ \int_{\Omega}b_{t}\otimes b:\nabla u_{t} \,dx \\& \quad \leq C\Vert u\Vert _{L^{6}}\Vert \sqrt{\rho}u_{t}\Vert _{L^{3}}\Vert \nabla u_{t}\Vert _{L^{2}}+C\Vert u \Vert _{L^{6}}\Vert \nabla u\Vert _{L^{6}}\Vert u_{t}\Vert _{L^{6}}\Vert \nabla u\Vert _{L^{2}} \\& \quad\quad{} +C\Vert u\Vert _{L^{6}}^{2}\Vert \Delta u\Vert _{L^{2}}\Vert u_{t}\Vert _{L^{6}}+C\Vert u\Vert _{L^{6}}^{2}\Vert \nabla u\Vert _{L^{6}}\Vert \nabla u_{t}\Vert _{L^{2}} \\& \quad\quad{} +C\Vert \sqrt{\rho}u_{t}\Vert _{L^{4}}^{2} \Vert \nabla u\Vert _{L^{2}}+C\Vert b\Vert _{L^{6}}\Vert b_{t}\Vert _{L^{3}}\Vert \nabla u_{t}\Vert _{L^{2}} \\& \quad \leq C\Vert \nabla u\Vert _{L^{2}}\Vert \sqrt{\rho}u_{t} \Vert _{L^{2}}^{\frac{1}{2}}\Vert \sqrt{\rho}u_{t}\Vert _{L^{6}}^{\frac{1}{2}}\Vert \nabla u_{t}\Vert _{L^{2}} \\& \quad\quad{} +C\Vert \nabla u\Vert _{L^{2}}^{2}\Vert u\Vert _{H^{2}}\Vert \nabla u_{t}\Vert _{L^{2}}+C\Vert \nabla u\Vert _{L^{2}}\Vert \sqrt{\rho}u_{t}\Vert _{L^{2}}^{\frac{1}{2}}\Vert \sqrt{\rho}u_{t}\Vert _{L^{6}}^{\frac{3}{2}} \\& \quad\quad{} +C\Vert b\Vert _{L^{6}}\Vert b_{t}\Vert _{L^{3}}\Vert \nabla u_{t}\Vert _{L^{2}} \\& \quad \leq C\Vert \nabla u\Vert _{L^{2}}\Vert \sqrt{\rho}u_{t} \Vert _{L^{2}}^{\frac{1}{2}}\Vert \nabla u_{t}\Vert _{L^{2}}^{\frac{3}{2}}+C\Vert \nabla u\Vert _{L^{2}}^{2} \Vert u\Vert _{H^{2}}\Vert \nabla u_{t}\Vert _{L^{2}} \\& \quad\quad{} +C\Vert \nabla u\Vert _{L^{2}}\Vert \sqrt{\rho}u_{t} \Vert _{L^{2}}^{\frac{1}{2}}\Vert \nabla u_{t}\Vert _{L^{2}}^{\frac{3}{2}}+C\Vert b\Vert _{L^{6}}\Vert b_{t}\Vert _{L^{3}}\Vert \nabla u_{t}\Vert _{L^{2}} \\& \quad \leq \frac{1}{4}\Vert \nabla u_{t}\Vert _{L^{2}}^{2}+C\Vert \nabla u\Vert _{L^{2}}^{4} \bigl(\Vert \sqrt {\rho}u_{t}\Vert _{L^{2}}^{2}+ \Vert u\Vert _{H^{2}}^{2}\bigr)+C\Vert b\Vert _{L^{6}}^{2}\Vert b_{t}\Vert _{L^{3}}^{2} \\& \quad \leq \frac{1}{4}\Vert \nabla u_{t}\Vert _{L^{2}}^{2}+C\Vert \nabla u\Vert _{L^{2}}^{4} \bigl(\Vert \sqrt {\rho}u_{t}\Vert _{L^{2}}^{2}+ \Vert u\Vert _{H^{2}}^{2}\bigr)+\frac{1}{4}\Vert \operatorname {curl}b_{t}\Vert _{L^{2}}^{2} \\& \quad\quad{}+C\Vert b\Vert _{L^{6}}^{4}\Vert b_{t}\Vert _{L^{2}}^{2}. \end{aligned}$$
(5.18)

Applying \(\partial_{t}\) to (1.15), testing by \(b_{t}\), and using (1.12), we get

$$\begin{aligned}& \frac{1}{2}\frac{d}{dt} \int_{\Omega} \vert b_{t}\vert ^{2}\,dx+ \int_{\Omega} \vert \operatorname {curl}b_{t}\vert ^{2}\,dx \\& \quad = - \int_{\Omega}(u_{t}\cdot\nabla b-b_{t}\nabla u-b\cdot\nabla u_{t})b_{t} \,dx \\& \quad \leq \Vert u_{t}\Vert _{L^{6}}\Vert \nabla b\Vert _{L^{2}}\Vert b_{t}\Vert _{L^{3}}+\Vert \nabla u \Vert _{L^{2}}\Vert b_{t}\Vert _{L^{4}}^{2}+ \Vert \nabla u_{t}\Vert _{L^{2}}\Vert b\Vert _{L^{6}}\Vert b_{t}\Vert _{L^{3}} \\& \quad \leq \frac{1}{4}\Vert \nabla u_{t}\Vert _{L^{2}}^{2}+\frac{1}{4}\Vert \operatorname {curl}b_{t}\Vert _{L^{2}}^{2}+C\Vert \nabla b\Vert _{L^{2}}^{4} \Vert b_{t}\Vert _{L^{2}}^{2}+C\Vert \nabla u \Vert _{L^{2}}^{4}\Vert b_{t}\Vert _{L^{2}}^{2}. \end{aligned}$$
(5.19)

Combining (5.18) and (5.19) and integrating over \([t_{0},t]\), we have

$$ \int_{\Omega}\bigl(\vert \rho u_{t}\vert ^{2}+\vert b_{t}\vert ^{2}\bigr)\,dx+ \int_{t_{0}}^{t} \int_{\Omega}\bigl(\vert \nabla u_{t}\vert ^{2}+\vert \operatorname {curl}b_{t}\vert ^{2}\bigr)\,dx \,d\tau \leq C(e+y)^{C_{0}\epsilon}. $$
(5.20)

Similarly to (5.12), we deduce that

$$\begin{aligned} \Vert u\Vert _{H^{2}} \leq&C\Vert \sqrt{\rho}u_{t}\Vert _{L^{2}}+C\Vert u\Vert _{L^{6}}\Vert \nabla u\Vert _{L^{3}}+C\Vert b\Vert _{L^{6}}\Vert \nabla b\Vert _{L^{3}} \\ \leq&C\Vert \sqrt{\rho}u_{t}\Vert _{L^{2}}+C\Vert u \Vert _{L^{6}}\Vert \nabla u\Vert _{L^{2}}^{\frac{1}{2}} \Vert u\Vert _{H^{2}}^{\frac{1}{2}}+C\Vert b\Vert _{L^{6}}\Vert \nabla b\Vert _{L^{2}}^{\frac{1}{2}}\Vert b\Vert _{H^{2}}^{\frac{1}{2}}, \end{aligned}$$

which leads to

$$ \Vert u\Vert _{H^{2}}^{2}\leq C\Vert \sqrt{\rho}u_{t}\Vert _{L^{2}}^{2}+C\Vert \nabla u\Vert _{L^{2}}^{6}+C\Vert \nabla b\Vert _{L^{2}}^{6}+ \frac{1}{2}\Vert b\Vert _{H^{2}}^{2}. $$
(5.21)

Similarly, we have

$$\begin{aligned} \Vert b\Vert _{H^{2}} \leq&C\Vert b_{t}+u\cdot\nabla b-b \cdot\nabla u\Vert _{L^{2}} \\ \leq&C\Vert b_{t}\Vert _{L^{2}}+C\Vert u\Vert _{L^{6}}\Vert \nabla b\Vert _{L^{3}}+C\Vert b\Vert _{L^{6}}\Vert \nabla u\Vert _{L^{3}} \\ \leq&C\Vert b_{t}\Vert _{L^{2}}+C\Vert u\Vert _{L^{6}}\Vert \nabla b\Vert _{L^{2}}^{\frac{1}{2}}\Vert b\Vert _{H^{2}}^{\frac{1}{2}}+C\Vert b\Vert _{L^{6}} \Vert \nabla u\Vert _{L^{2}}^{\frac{1}{2}}\Vert u\Vert _{H^{2}}^{\frac{1}{2}}, \end{aligned}$$

which implies

$$ \Vert b\Vert _{H^{2}}^{2}\leq C\Vert b_{t} \Vert _{L^{2}}^{2}+C\Vert \nabla u\Vert _{L^{2}}^{6}+C\Vert \nabla b\Vert _{L^{2}}^{6}+ \frac{1}{2}\Vert u\Vert _{H^{2}}^{2}. $$
(5.22)

Combining (5.21) and (5.22) and using (5.20) and (5.16), we conclude that

$$ \Vert u\Vert _{H^{2}}^{2}+\Vert b\Vert _{H^{2}}^{2}\leq C(e+y)^{C_{0}\epsilon}, $$
(5.23)

and thus

$$ \Vert u\Vert _{L^{\infty}(0,T;H^{2})}+\Vert b\Vert _{L^{\infty}(0,T;H^{2})}\leq C. $$
(5.24)

Now it is standard to prove that

$$\begin{aligned}& \Vert u\Vert _{L^{2}(0,T;H^{3})}+\Vert b\Vert _{L^{2}(0,T;H^{3})}\leq C, \end{aligned}$$
(5.25)
$$\begin{aligned}& \Vert \rho \Vert _{L^{\infty}(0,T;H^{2})}\leq C. \end{aligned}$$
(5.26)

(II) Let (1.22) hold.

Similarly to (5.7), we take \(s=\infty\) and using (2.4), we still get (5.7), provided that

$$ \int_{t_{0}}^{T}\bigl\Vert u(t)\bigr\Vert _{\operatorname {BMO}}^{2}\,dt\leq\epsilon \ll 1. $$
(5.27)

We still have (5.9), (5.10), (5.11) with \(s=\infty \), (5.12) with \(s=\infty\), (5.13) with \(s=\infty\), and (5.14), (5.15) with \(s=\infty\), and then using (5.27) and (2.4), we arrive at (5.16) and (5.17). Then by the same calculations as those in (5.18)-(5.26), we conclude that (5.18)-(5.26) hold.

This completes the proof of Theorem 1.3.

6 Proof of Theorem 1.4

We only need to establish a priori estimates.

First, using the formula \(a\times(b\times c)=(a\cdot c)b-(a\cdot b)c\) and the fact that \(\vert d\vert =1\) implies \(d\Delta d=-\vert \nabla d\vert ^{2}\), we have the following equivalent equation:

$$ \frac{1}{2} d_{t}-\frac{1}{2} d\times d_{t}= \Delta d+d\vert \nabla d\vert ^{2}. $$
(6.1)

Testing (6.1) by \(d_{t}\) and using \((a\times b)\cdot b=0\) and \(d\cdot d_{t}=0\), we get

$$ \frac{d}{dt} \int_{\Omega} \vert \nabla d\vert ^{2}\,dx+ \int_{\Omega} \vert d_{t}\vert ^{2}\,dx\leq 0. $$
(6.2)

Testing (1.23) by \(-\Delta d_{t}\) and using \(\vert d\vert =1\), we find that

$$\begin{aligned}& \frac{1}{2}\frac{d}{dt} \int_{\Omega} \vert \Delta d\vert ^{2}\,dx+ \int_{\Omega} \vert \nabla d_{t}\vert ^{2}\,dx \\& \quad =- \int_{\Omega}\bigl(d\vert \nabla d\vert ^{2}+d\times \Delta d\bigr)\cdot\Delta d_{t} \,dx \\& \quad = \int_{\Omega}\nabla\bigl(d\vert \nabla d\vert ^{2}+d \times\Delta d\bigr)\cdot\nabla d_{t} \,dx \\& \quad \leq C\bigl(\Vert \nabla d\Vert _{L^{q}}\Vert \nabla d\Vert _{L^{\frac{4q}{q-2}}}^{2}+\Vert \nabla d\Vert _{L^{q}}\Vert \Delta d\Vert _{L^{\frac{2q}{q-2}}}+\Vert \nabla\Delta d\Vert _{L^{2}} \bigr)\Vert \nabla d_{t}\Vert _{L^{2}} \\& \quad \leq C\bigl(\Vert \nabla d\Vert _{L^{q}}\Vert \Delta d\Vert _{L^{\frac{2q}{q-2}}}+\Vert \nabla \Delta d\Vert _{L^{2}}\bigr)\Vert \nabla d_{t}\Vert _{L^{2}} \\& \quad \leq C\bigl(\Vert \nabla d\Vert _{L^{q}}\Vert \Delta d\Vert _{L^{2}}^{1-\frac{3}{q}}\Vert d\Vert _{H^{3}}^{\frac{3}{q}}+ \Vert d\Vert _{H^{3}}\bigr)\Vert \nabla d_{t}\Vert _{L^{2}} \\& \quad \leq \frac{1}{4}\Vert \nabla d_{t}\Vert _{L^{2}}^{2}+\delta \Vert d\Vert _{H^{3}}^{2}+C \Vert \nabla d\Vert _{L^{q}}^{\frac{2q}{q-3}}\Vert \Delta d \Vert _{L^{2}}^{2} \end{aligned}$$
(6.3)

for any \(0<\delta<1\). Here we have used the Gagliardo-Nirenberg inequalities

$$\begin{aligned}& \Vert \nabla d\Vert _{L^{\frac{4q}{q-2}}}^{2}\leq C\Vert d\Vert _{L^{\infty}} \Vert \Delta d\Vert _{L^{\frac{2q}{q-2}}}, \end{aligned}$$
(6.4)
$$\begin{aligned}& \Vert \Delta d\Vert _{L^{\frac{2q}{q-2}}}\leq C\Vert \Delta d\Vert _{L^{2}}^{1-\frac{3}{q}}\Vert d\Vert _{H^{3}}^{\frac{3}{q}}. \end{aligned}$$
(6.5)

Applying \(\partial_{i}\) to (1.23), we get

$$\partial_{i}d_{t}-\Delta\partial_{i}d= \partial_{i}\bigl(d\vert \nabla d\vert ^{2}\bigr)+ \partial _{i}d\times\Delta d+d\times\Delta\partial_{i}d. $$

Testing this equation by \(\Delta\partial_{i}d\), summing over i, and using (6.4) and (6.5) and \(\vert d\vert =1\), we obtain

$$\begin{aligned} \Vert d\Vert _{H^{3}} \leq&C\bigl(\Vert d\Vert _{L^{2}}+ \Vert \nabla\Delta d\Vert _{L^{2}}\bigr) \\ \leq&C+C\Vert \nabla d_{t}\Vert _{L^{2}}+C\bigl\Vert \nabla\bigl(d\vert \nabla d\vert ^{2}\bigr)\bigr\Vert _{L^{2}}+\sum_{i}C\Vert \partial_{i}d\times\Delta d\Vert _{L^{2}} \\ \leq&C+C\Vert \nabla d_{t}\Vert _{L^{2}}+C\Vert \nabla d \Vert _{L^{q}}\Vert \nabla d\Vert _{L^{\frac{4q}{q-2}}}^{2}+C \Vert \nabla d\Vert _{L^{q}}\Vert \Delta d\Vert _{L^{\frac{2q}{q-2}}} \\ \leq&C+C\Vert \nabla d_{t}\Vert _{L^{2}}+C\Vert \nabla d \Vert _{L^{q}}\Vert \Delta d\Vert _{L^{\frac{2q}{q-2}}} \\ \leq&C+C\Vert \nabla d_{t}\Vert _{L^{2}}+C\Vert \nabla d \Vert _{L^{q}}\Vert \Delta d\Vert _{L^{2}}^{1-\frac{3}{q}} \Vert d\Vert _{H^{3}}^{\frac{3}{q}}, \end{aligned}$$

which yields

$$ \Vert d\Vert _{H^{3}}\leq C+C\Vert \nabla d_{t}\Vert _{L^{2}}+C\Vert \nabla d\Vert _{L^{q}}^{\frac{q}{q-3}} \Vert \Delta d\Vert _{L^{2}}. $$
(6.6)

Plugging (6.6) into (6.3) and taking δ small enough, we have

$$\begin{aligned}& \frac{d}{dt} \int_{\Omega} \vert \Delta d\vert ^{2}\,dx+ \int_{\Omega} \vert \nabla d_{t}\vert ^{2}\,dx \\& \quad \leq C+C\Vert \nabla d\Vert _{L^{q}}^{\frac{2q}{q-3}}\Vert \Delta d\Vert _{L^{2}}^{2} \\& \quad \leq C+C\frac{\Vert \nabla d\Vert _{L^{q}}^{\frac{2q}{q-3}}}{1+\log(e+\Vert \nabla d\Vert _{L^{q}})}\Vert \Delta d\Vert _{L^{2}}^{2} \log\bigl(e+\Vert \nabla d\Vert _{L^{q}}\bigr) \\& \quad \leq C+C\frac{\Vert \nabla d\Vert _{L^{q}}^{\frac{2q}{q-3}}}{1+\log(e+\Vert \nabla d\Vert _{L^{q}})}\Vert \Delta d\Vert _{L^{2}}^{2} \log(e+y), \end{aligned}$$

which implies

$$ \int_{\Omega} \vert \Delta d\vert ^{2}\,dx+ \int_{t_{0}}^{t} \int_{\Omega} \vert \nabla d_{t}\vert ^{2}\,dx \,d\tau\leq C(e+y)^{C_{0}\epsilon}, $$
(6.7)

provided that

$$\int_{t_{0}}^{T}\frac{\Vert \nabla d\Vert _{L^{q}}^{\frac{2q}{q-3}}}{1+\log(e+\Vert \nabla d\Vert _{L^{q}})}\,d\tau\leq\epsilon \ll 1, $$

with \(y(t):=\sup_{[t_{0},t]}\Vert d\Vert _{H^{3}}\) for any \(0< t_{0}\leq t\leq T\), where \(C_{0}\) is an absolute constant.

It follows from (1.23), (6.6), and (6.7) that

$$ \int_{\Omega} \vert d_{t}\vert ^{2}\,dx+ \int_{t_{0}}^{t}\Vert d\Vert _{H^{3}}^{2}\,d\tau\leq C(e+y)^{C_{0}\epsilon}. $$
(6.8)

Applying \(\partial_{t}\) to (1.23), testing by \(-\Delta d_{t}\), and using \(\vert d\vert =1\), (6.7), and (6.8), we have

$$\begin{aligned}& \frac{1}{2}\frac{d}{dt} \int_{\Omega} \vert \nabla d_{t}\vert ^{2}\,dx+ \int_{\Omega} \vert \Delta d_{t}\vert ^{2}\,dx \\& \quad =- \int_{\Omega}\bigl[\partial_{t}\bigl(d\vert \nabla d \vert ^{2}\bigr)+d_{t}\times \Delta d\bigr]\Delta d_{t} \,dx \\& \quad \leq C\bigl(\Vert \nabla d\Vert _{L^{6}}^{2}\Vert d_{t}\Vert _{L^{6}}+\Vert \nabla d\Vert _{L^{6}} \Vert \nabla d_{t}\Vert _{L^{3}}+\Vert d_{t} \Vert _{L^{\infty}} \Vert \Delta d\Vert _{L^{2}}\bigr)\Vert \Delta d_{t}\Vert _{L^{2}} \\& \quad \leq C\bigl(\Vert \nabla d\Vert _{L^{6}}^{2}\Vert d_{t}\Vert _{L^{6}}+\Vert \Delta d\Vert _{L^{2}} \Vert \nabla d_{t}\Vert _{L^{2}}^{\frac{1}{2}}\Vert \Delta d_{t}\Vert _{L^{2}}^{\frac{1}{2}}+\Vert \Delta d \Vert _{L^{2}}\Vert d_{t}\Vert _{L^{2}}\bigr)\Vert \Delta d_{t}\Vert _{L^{2}} \\& \quad \leq \frac{1}{2}\Vert \Delta d_{t}\Vert _{L^{2}}^{2}+C\Vert d\Vert _{H^{2}}^{4} \Vert d_{t}\Vert _{H^{1}}^{2}+C\Vert d\Vert _{H^{2}}^{2}\Vert d_{t}\Vert _{L^{2}}^{2}, \end{aligned}$$

which implies

$$ \int_{\Omega} \vert \nabla d_{t}\vert ^{2}\,dx+ \int_{t_{0}}^{t}\Vert \Delta d_{t}\Vert _{L^{2}}^{2}\,d\tau \leq C(e+y)^{C_{0}\epsilon}. $$
(6.9)

It follows from (6.6), (6.7), (6.8), and (6.9) that

$$\Vert d\Vert _{H^{3}}\leq C+C\Vert \nabla d_{t}\Vert _{L^{2}}+C\Vert \nabla d\Vert _{L^{6}}^{2}\Vert \Delta d\Vert _{L^{2}}\leq C(e+y)^{C_{0}\epsilon}, $$

which leads to

$$\Vert d\Vert _{L^{\infty}(0,T;H^{3})}\leq C. $$

This completes the proof of Theorem 1.4.