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

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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.

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 , Kim , and Kang and Kim  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 . But whether this local solution can exist globally is an outstanding problem. Kang and Kim  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 .

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  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  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  showed the existence and uniqueness of local smooth solutions. When $$\Omega:=\mathbb{R}^{n}$$ ($$n=2,3,4$$), Fan and Ozawa  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.

Preliminary lemmas

In the following proofs, we will use the logarithmic Sobolev inequality 

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

and the following three lemmas.

Lemma 2.1

()

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 . 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 , 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.

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}

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

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

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.

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 

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

and the generalized Hölder inequality 

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

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

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

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

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

This completes the proof of Theorem 1.4.

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Acknowledgements

J. Fan is partially supported by NSFC (No. 11171154), Junpin Yin is supported by the NSFC (Grant No. U1430103) and Beijing Center for Mathematics and Information Interdisciplinary Sciences (BCMIIS). The authors would like to thank the referee for reading the paper carefully and for the valuable comments, which improved the presentation of the paper.

Author information

Correspondence to Junping Yin.

Competing interests

The authors declare that they have no competing interests.

Authors’ contributions

The authors declare that the study was realized in collaboration with the same responsibility. All authors read and approved the final manuscript.

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