Blow-up criteria for Boussinesq system and MHD system and Landau-Lifshitz equations in a bounded domain

Abstract

In this paper, we prove some blow-up criteria for the 3D Boussinesq system with zero heat conductivity and MHD system and Landau-Lifshitz equations in a bounded domain.

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.