Stabilization Effects of Magnetic Field on a 2D Anisotropic MHD System with Partial Dissipation

Abstract

To uncover that the magnetic field mechanism can stabilize electrically conducting turbulent fluids, we investigate the stability of a special two dimensional anisotropic MHD system with vertical dissipation in the horizontal velocity component and partial magnetic damping near a background magnetic field. Since the MHD system has only vertical dissipation in the horizontal velocity and vertical magnetic damping, the stability issue and large time behavior problem of the linearized magneto-hydrodynamic system is not trivial. By performing refined energy estimates on the linear system coupled with a careful analysis of the nonlinearities, the stability of a MHD-type system near a background magnetic field is justified for the initial data belonging to $H^{3}(\mathbf{R}^{2})$ space. The authors also build the explicit decay rates of the linearized system.

1 Introduction

The standard two dimensional incompressible magnetohydrodynamic(MHD) equations obey

$$ \left \{ \begin{aligned} &\partial _{t}U+U\cdot \nabla U-\nu \Delta U+\nabla \Pi =B\cdot \nabla B, \\ &\partial _{t}B+U\cdot \nabla B-\mu \Delta B=B\cdot \nabla U, \\ &\nabla \cdot U=\nabla \cdot B=0, \\ &U(x,0)=U_{0}(x),\quad B(x,0)=B_{0}(x), \end{aligned} \right . $$

(1.1)

where $U(t,x)=(U_{1}(t,x),U_{2}(t,x)), B(t,x)=(B_{1}(t,x),B_{2}(t,x))$ are the velocity field and the magnetic field, respectively, $\Pi$ is scalar pressure. $\nu \ge 0,\mu \ge 0$ denote the kinematic viscosity and the magnetic diffusivity. The system (1.1) comes from the geophysics, astrophysics, cosmology and has been applied in engineering (see [10, 19]). Many mathematicians and physicians have devoted efforts to addressing some fundamental issues on the MHD equations such as well-posedness, stability and large time behavior problems. In 1972, Duvaut and Lions [12] built the local well-posedness in Sobolev spaces and established the global existence with small initial data. Later on, Sermange and Temam [23] established the global well-posedness of the MHD equations (1.1). Since then, there is a large amount of literature on the global well-posedness and regularity issue of the MHD equations with various partial or fractional dissipation (see [6–9]).

In this paper, we will investigate the incompressible MHD fluid system equations:

$$\{\begin{array}{rl}& {\partial }_{t}U+U\cdot \mathrm{\nabla }U-\nu \left(\begin{array}{c}{\partial }_2^2{U}_1\\ 0\end{array}\right)+\mathrm{\nabla }\mathrm{\Pi }=B\cdot \mathrm{\nabla }B,\\ & {\partial }_{t}B+U\cdot \mathrm{\nabla }B+\eta \left(\begin{array}{c}0\\ B_2\end{array}\right)=B\cdot \mathrm{\nabla }U,\\ & \mathrm{\nabla }\cdot U=\mathrm{\nabla }\cdot B=0,\\ & U{|}_{t=0}=U_0,B{|}_{t=0}=B_0.\end{array}$$

(1.2)

Let

$$\begin{aligned} U^{(0)}=0,\ \ \ \ B^{(0)}=e_{1}=(1,0), \end{aligned}$$

which is a steady solution of (1.2). The perturbation $(u,b)$ with

$$\begin{aligned} u=U-U^{(0)},\ \ \ \ b=B-B^{(0)}. \end{aligned}$$

satisfying

$$\{\begin{array}{rl}& {\partial }_{t}u+u\cdot \mathrm{\nabla }u-\nu \left(\begin{array}{c}{\partial }_2^2{u}_1\\ 0\end{array}\right)+\mathrm{\nabla }\pi =b\cdot \mathrm{\nabla }b+{\partial }_{1}b,\\ & {\partial }_{t}b+u\cdot \mathrm{\nabla }b+\eta \left(\begin{array}{c}0\\ b_2\end{array}\right)=b\cdot \mathrm{\nabla }u+{\partial }_{1}u,\\ & \mathrm{\nabla }\cdot u=\mathrm{\nabla }\cdot b=0,\\ & u{|}_{t=0}=u_0,b{|}_{t=0}=b_0.\end{array}$$

(1.3)

During the past thirty years, more and more attention has been paid to the stability and large time behavior problems to the MHD equations with partial dissipations near a background magnetic field. Some studies which stated above are concerned with the fundamental nonlinear phenomena associated with electrically conducting fluids. Many physicians found that the magnetic field can stabilize electrically conducting fluids (see [1, 2, 13, 14]). It is very interesting to understand the stability problem concerning the partially dissipative MHD equations near the background magnetic field. There are substantial developments on the stability problem of the MHD equations with various dissipation near a background magnetic field (see [3, 5, 11, 15, 17, 20–22, 24–26]). The purpose of this paper is to investigate the smoothing and stabilizing effect of the magnetic field on the fluid motion.

Recently, Lai, Wu and Zhang [16] consider the following MHD system

$$\{\begin{array}{rl}& {\partial }_{t}u+u\cdot \mathrm{\nabla }u-\nu {\partial }_2^{2}u+\mathrm{\nabla }\pi =b\cdot \mathrm{\nabla }b+{\partial }_{1}b,\\ & {\partial }_{t}b+u\cdot \mathrm{\nabla }b+\eta \left(\begin{array}{c}0\\ b_2\end{array}\right)=b\cdot \mathrm{\nabla }u+{\partial }_{1}u,\\ & \mathrm{\nabla }\cdot u=\mathrm{\nabla }\cdot b=0,\\ & u{|}_{t=0}=u_0,b{|}_{t=0}=b_0.\end{array}$$

(1.4)

They established the stability of the system (1.4) near a magnetic background. Later on, Lai informed us that they have obtained the stability issue of the system (1.3) with $\partial _{2}^{2}u_{1}$ replaced by the damping term $u_{1}$ in a coming paper.

For convenience, we assume that $\nu =\eta =1$ throughout this paper. Applying the Helmholtz-Leray projection $\mathbb{P}=I-\nabla \triangle ^{-1}\nabla \cdot$ to the equation (1.3) and using the fact that $\nabla \cdot u=\nabla \cdot b=0$, one has

$$\begin{array}{rl}\mathbb{P}& \left(\begin{array}{c}{\partial }_2^2{u}_1\\ 0\end{array}\right)={\mathrm{△}}^{-1}{\partial }_2^{4}u\triangleq {\mathcal{T}}_2^{2}u,\\ \mathbb{P}& \left(\begin{array}{c}0\\ b_2\end{array}\right)={\mathrm{△}}^{-1}{\partial }_1^{2}b\triangleq -{\mathcal{R}}_1^{2}b.\end{array}$$

Then the system (1.3) converts into

$$ \left \{ \begin{aligned} &\partial _{t} u=\mathcal{T}_{2}^{2}u+\partial _{1}b+\mathbb{P}(b \cdot \nabla b-u\cdot \nabla u), \\ &\partial _{t} b=\mathcal{R}_{1}^{2}b+\partial _{1}u+\mathbb{P}(b \cdot \nabla u-u\cdot \nabla b), \\ &\nabla \cdot u=\nabla \cdot b=0, \\ &u|_{t=0}= u_{0},\ b|_{t=0}=b_{0}. \end{aligned} \right . $$

(1.5)

Differentiating (1.5) in $t$ and making several substitutions, we can convert (1.3) into the following new system

$$ \left \{ \begin{aligned} &\partial _{tt} u-(\mathcal{R}_{1}^{2}+\mathcal{T}_{2}^{2})\partial _{t}u- \partial _{1}^{2}u+\mathcal{R}_{1}^{2}\mathcal{T}_{2}^{2}u=N_{1}, \\ &\partial _{tt} b-(\mathcal{R}_{1}^{2}+\mathcal{T}_{2}^{2})\partial _{t}b- \partial _{1}^{2}b+\mathcal{R}_{1}^{2}\mathcal{T}_{2}^{2}b=N_{2}, \\ &\nabla \cdot u=\nabla \cdot b=0, \\ &u|_{t=0}= u_{0},\ b|_{t=0}=b_{0}, \end{aligned} \right . $$

(1.6)

where $N_{1}$ and $N_{2}$ are the nonlinear terms,

$$\begin{aligned} N_{1}=&(\partial _{t}-\mathcal{R}_{1}^{2})\mathbb{P}(b\cdot \nabla b-u \cdot \nabla u)+\partial _{1}\mathbb{P}(b\cdot \nabla u-u\cdot \nabla b), \\ N_{2}=&(\partial _{t}-\mathcal{T}_{2}^{2})\mathbb{P}(b\cdot \nabla u-u \cdot \nabla b)+\partial _{1}\mathbb{P}(b\cdot \nabla b-u\cdot \nabla u). \end{aligned}$$

Our stability result can be stated as follows.

Theorem 1.1

Assume $(u_{0}, b_{0})\in H^{3}(\mathbf {R}^{2})$ with $\nabla \cdot u_{0}=\nabla \cdot b_{0}=0$. Then there exists a sufficient small $\varepsilon >0$ such that, if

$$\begin{aligned} \|u_{0}\|_{H^{3}}+\|b_{0}\|_{H^{3}}\le \varepsilon . \end{aligned}$$

(1.7)

Then (1.3) has a unique global solution that remains uniformly bounded for all time,

$$\begin{aligned} \|u\|_{H^{3}}^{2}+\|b\|_{H^{3}}^{2}+\int _{0}^{t}\|\partial _{2}u_{1}( \tau )\|_{H^{3}}^{2}d\tau +\int _{0}^{t}\| b_{2}(\tau )\|_{H^{3}}^{2}d \tau +\int _{0}^{t}\|\partial _{1}u(\tau )\|_{H^{2}}^{2}d\tau \le C \varepsilon ^{2}, \end{aligned}$$

where $C>0$ is a pure constant.

Remark 1.2

Due to the lack of the vertical dissipation $\partial _{2}^{2}u_{2}$, the system (1.3) has less dissipation than that of the system (1.4). We need to establish the $H^{3}$-stability and build more subtle energy estimates than that in [16], which obtained the stability in $H^{2}$ space.

Remark 1.3

Now we explain why we need the $H^{3}$ regularity imposed on the initial data. The most difficult term encountering in the proof is $\int _{\mathbf {R}^{2}}\partial _{1}^{2} u_{2}\partial _{2}b_{1} \partial _{1}^{2}b_{2}dx$. In order to establish the stability, we are obliged to prove the bad term is uniformly integrable over $[0,t]$. Thanks to the Hölder inequality, Sobolev inequality, one has

$$\begin{aligned} \int _{0}^{t}\int _{\mathbf {R}^{2}}\partial _{1}^{2} u_{2}\partial _{2}b_{1} \partial _{1}^{2}b_{2}dxd\tau &\le \int _{0}^{t}\|\partial _{1}^{2}u_{2} \|_{L^{2}}\|\partial _{2}b_{1}\|_{L^{\infty}}\|\partial _{1}^{2}b_{2} \|_{L^{2}}d\tau \\ &\le C\int _{0}^{t}\|\partial _{1}u_{2}\|_{H^{2}}\|\partial _{2}b_{1} \|_{H^{2}}\|b_{2}\|_{H^{3}}d\tau \\ &\le C\bigg(\int _{0}^{t}\|\partial _{1}u_{2}\|_{H^{2}}^{2}d\tau \bigg)^{\frac{1}{2}}\bigg(\int _{0}^{t}\|b_{2}\|_{H^{3}}^{2}d\tau \bigg)^{ \frac{1}{2}}\|b_{1}\|_{L_{t}^{\infty}(H^{3})}. \end{aligned}$$

We would like to remark that it is natural to consider the stability problem in anisotropic Sobolev or Besov spaces. To examine this issue, one need to build a subtle energy structure to build an priori estimate. For instance, in the references (see [4, 17]), the authors deal with similar operators (Riesz) in an anisotropic functional setting.

Our second main result explores the large-time behavior of solutions to the linearized system

$$ \left \{ \begin{aligned} &\partial _{t} u-\mathcal{T}_{2}^{2}u-\partial _{1}b=0, \\ &\partial _{t} b-\mathcal{R}_{1}^{2}b-\partial _{1}u=0, \\ &\nabla \cdot u=\nabla \cdot b=0, \\ &u|_{t=0}= u_{0},\ b|_{t=0}=b_{0}, \end{aligned} \right . $$

(1.8)

which can be converted to the following system of wave equations

$$ \left \{ \begin{aligned} &\partial _{tt} u-(\mathcal{R}_{1}^{2}+\mathcal{T}_{2}^{2})\partial _{t}u- \partial _{1}^{2}u+\mathcal{R}_{1}^{2}\mathcal{T}_{2}^{2}u=0, \\ &\partial _{tt} b-(\mathcal{R}_{1}^{2}+\mathcal{T}_{2}^{2})\partial _{t}b- \partial _{1}^{2}b+\mathcal{R}_{1}^{2}\mathcal{T}_{2}^{2}b=0, \\ &\nabla \cdot u=\nabla \cdot b=0, \\ &u|_{t=0}= u_{0},\ b|_{t=0}=b_{0}. \end{aligned} \right . $$

(1.9)

According to Lemma 2.3, we need to show that $\int _{0}^{t}(\|u(\tau )\|_{H^{1}}^{2}+\|b(\tau )\|_{H^{1}}^{2}d \tau$ is finite which requires $(\Lambda _{1}^{-\sigma},\Lambda _{2}^{-2\sigma})u_{0}\in H^{1+\sigma}, \ \ (\Lambda _{1}^{-\sigma},\Lambda _{2}^{-2\sigma})b_{0}\in H^{1+ \sigma}$. We refer the readers to Sect. 4 for details.

Now we give the definition of the partial fractional derivative $\Lambda _{i}^{\gamma}$ with $i=1,\ 2$ and $\gamma \in \mathbf {R}$ as follows

$$\begin{aligned} \widehat{\Lambda _{i}^{\gamma }f}(\xi )=|\xi _{i}|^{\gamma }\hat{f}( \xi ). \end{aligned}$$

Theorem 1.4

Let $\sigma >0$. Assume $(u_{0}, b_{0})$ satisfies

$$\begin{aligned} (\Lambda _{1}^{-\sigma},\Lambda _{2}^{-2\sigma})u_{0}\in H^{1+\sigma}, \ \ (\Lambda _{1}^{-\sigma},\Lambda _{2}^{-2\sigma})b_{0}\in H^{1+ \sigma},\ \ \nabla \cdot u_{0}=\nabla \cdot b_{0}=0. \end{aligned}$$

Then the corresponding solution $(u,b)$ of (1.8) satisfies

$$\begin{aligned} (u,b)\in L^{\infty}(0,\infty ;H^{1}),\ \ \ (\mathcal{T}_{2}u, \mathcal{R}_{1}b)\in L^{2}(0,\infty ;H^{1}). \end{aligned}$$

Moreover,

$$\begin{aligned} \|u(t)\|_{H^{1}}+\|b(t)\|_{H^{1}}\le C(1+t)^{-\frac{\sigma}{2}}, \end{aligned}$$

for any $t>0$, where $C>0$ is a positive constant depending on $u_{0}$ and $b_{0}$,

The next theorem is to evaluate the decay estimates of $\partial _{1}u$ and $\partial _{1}b$, the extra terms generalized by the perturbation near the background magnetic field $B_{0}=(1,0,0)$.

Theorem 1.5

Assume $(u_{0}, b_{0})$ satisfies

$$\begin{aligned} &\nabla \cdot u_{0}=\nabla \cdot b_{0}=0,\ \ (u_{0},b_{0})\in H^{1}, \ \ (\mathcal{T}_{2}^{2}u_{0},\mathcal{R}_{1}^{2}b_{0})\in L^{2}, \\ &(\Lambda \Lambda _{1}^{-1}\mathcal{T}_{2}^{2}u_{0},\Lambda \Lambda _{1}^{-1} \mathcal{T}_{2}^{2}b_{0})\in L^{2},\ \ (\Lambda _{1}^{-1}\mathcal{R}_{1} \mathcal{T}_{2}u_{0},\Lambda _{1}^{-1}\mathcal{R}_{1}\mathcal{T}_{2}b_{0}) \in L^{2}. \end{aligned}$$

Then the corresponding solution $(u,b)$ of (1.9) satisfies

$$\begin{aligned} \|\partial _{t}u(t)\|_{L^{2}}&+\|\partial _{1}u(t)\|_{L^{2}}+\| \mathcal{R}_{1}\mathcal{T}_{2}u(t)\|_{L^{2}} \le C(1+t)^{-\frac{1}{2}}, \\ \|\partial _{t}b(t)\|_{L^{2}}&+\|\partial _{1}b(t)\|_{L^{2}}+\| \mathcal{R}_{1}\mathcal{T}_{2}b(t)\|_{L^{2}} \le C(1+t)^{-\frac{1}{2}}, \end{aligned}$$

for any $t>0$, where $C>0$ is a constant depending on $u_{0}$ and $b_{0}$.

A brief outline of the proof. We first present the main ideas in the proof of Theorem 1.1. The method we use to prove Theorem 1.1 is based on a bootstrapping argument. A natural part of the energy functional is

$$\begin{aligned} E_{1}(t)=\|u\|_{H^{3}}^{2}+\|b\|_{H^{3}}^{2}+\int _{0}^{t}\|\partial _{2}u_{1}( \tau )\|_{H^{3}}^{2}d\tau +\int _{0}^{t}\| b_{2}(\tau )\|_{H^{3}}^{2}d \tau . \end{aligned}$$

(1.10)

This part is not sufficient to bound the bad term $\int \partial _{1}u_{1}(\partial _{2}^{3}b_{1})^{2}dx$ emerged in the $H^{3}$ estimate. We have to seek another part of the energy functional

$$\begin{aligned} E_{2}(t)=\int _{0}^{t}\|\partial _{1}u(\tau )\|_{H^{2}}^{2}d\tau , \end{aligned}$$

(1.11)

which can help us to control those bad terms. We need also verify that $E_{2}(t)$ can be bounded by a combination of $E_{1}(t)$ and $E_{2}(t)$.

Let $E(t)=E_{1}(t)+\delta E_{2}(t)$. From Proposition 3.1 and Proposition 3.2, one has

$$\begin{aligned} E_{1}(t)\le & C(E_{1}(0)+E_{1}^{\frac{3}{2}}(0)+E_{1}^{2}(0))+C(E_{1}^{ \frac{3}{2}}(t)+E_{2}^{\frac{3}{2}}(t)) \\ &+C(E_{1}^{2}(t)+E_{2}^{2}(t))+C(E_{1}^{\frac{5}{2}}(t)+E_{2}^{ \frac{5}{2}}(t)), \end{aligned}$$

and

$$\begin{aligned} E_{2}(t)\le CE_{1}(0)+CE_{1}(t)+CE_{1}^{\frac{3}{2}}(t)+CE_{2}^{ \frac{3}{2}}(t). \end{aligned}$$

These two inequalities which together with the definition of $E(t)$ give rise to

$$\begin{aligned} E(t)\le E(0)+CE^{\frac{3}{2}}(t)+CE^{2}(t)+CE^{\frac{5}{2}}(t). \end{aligned}$$

Thus a bootstrap argument guarantees us the global existence and stability.

We would like to remark that it is very difficult to obtain the large time behavior of the nonlinear problem. The characteristic polynomial associated with (1.9) is given by

$$\begin{aligned} \lambda ^{2}+\frac{\xi _{1}^{2}+\xi _{2}^{4}}{|\xi |^{2}}\lambda + \xi _{1}^{2}+\frac{\xi _{1}^{2}\xi _{2}^{4}}{|\xi |^{4}}=0, \end{aligned}$$

(1.12)

whose eigenvalues are

$$\begin{aligned} \lambda _{1}= \frac{-(\xi _{1}^{2}+\xi _{2}^{4})+\sqrt{(\xi _{1}^{2}-\xi _{2}^{4})^{2}-4\xi _{1}^{2}|\xi |^{4}}}{|\xi |^{2}}, \quad \lambda _{2}= \frac{-(\xi _{1}^{2}+\xi _{2}^{4})-\sqrt{(\xi _{1}^{2}-\xi _{2}^{4})^{2}-4\xi _{1}^{2}|\xi |^{4}}}{|\xi |^{2}}. \end{aligned}$$

It is very easy to verify that $\lambda _{1}\sim 0$ as $\xi \to 0$. When the solution is represented by the integral form, the degeneracy of $\lambda _{1}$ makes the decay evaluation of nonlinear terms extremely difficult. It is not easy to estabilish the explicit decay rate of linearized system due to the degenerate of the eigenvalue $\lambda _{1}$.

The rest of this paper is organized as follows. Section 2 presents some crucial Lemmas. The proof of Theorem 1.1 is performed in Sect. 3. Section 4 is devoted to the proof of Theorem 1.4 and that of Theorem 1.5.

2 Preliminaries

In this section, we state some important lemmas which can be found in [18].

Lemma 2.1

Assume that $f,\ g,\ h,\ \partial _{1}f$ and $\partial _{2}g$ are all in $L^{2}(\mathbf {R}^{2})$. Then there exists a generic constant $C>0$ such that

$$\begin{aligned} \int _{\mathbf {R}^{2}}|fgh|dx\le C\|f\|_{L^{2}(\mathbf {R}^{2})}^{ \frac{1}{2}}\|\partial _{1}f\|_{L^{2}(\mathbf {R}^{2})}^{\frac{1}{2}}\|g \|_{L^{2}(\mathbf {R}^{2})}^{\frac{1}{2}}\|\partial _{2}g\|_{L^{2}( \mathbf {R}^{2})}^{\frac{1}{2}}\|h\|_{L^{2}(\mathbf {R}^{2})}. \end{aligned}$$

Lemma 2.2

Assume that $f, \partial _{1}f,\partial _{2}f$ are bounded in $H^{1}(\mathbf {R}^{2})$, it holds that

$$\begin{aligned} \|f\|_{L^{\infty}(\mathbf {R}^{2})}\le C\|f\|_{L^{2}(\mathbf {R}^{2})}^{ \frac{1}{4}}\|\partial _{1}f\|_{L^{2}(\mathbf {R}^{2})}^{\frac{1}{4}}\| \partial _{2}f\|_{L^{2}(\mathbf {R}^{2})}^{\frac{1}{4}}\|\partial _{1} \partial _{2}f\|_{L^{2}(\mathbf {R}^{2})}^{\frac{1}{4}}, \end{aligned}$$

and

$$\begin{aligned} \|f\|_{L^{\infty}(\mathbf {R}^{2})}\le &C\|f\|_{H^{1}(\mathbf {R}^{2})}^{ \frac{1}{2}}\|\partial _{1}f\|_{H^{1}(\mathbf {R}^{2})}^{\frac{1}{2}}, \\ \|f\|_{L^{\infty}(\mathbf {R}^{2})}\le &C\|f\|_{H^{1}(\mathbf {R}^{2})}^{ \frac{1}{2}}\|\partial _{2}f\|_{H^{1}(\mathbf {R}^{2})}^{\frac{1}{2}}. \end{aligned}$$

Lemma 2.3

For given positive constants $C_{0}, C_{1}, C_{2}>0$, assume that $f=f(t)$ is a nonnegative function defined on $[0,\infty )$ and satisfies,

$$\begin{aligned} \int _{0}^{\infty }f(\tau )\le C_{0}< \infty ,\ \ f(t)\le C_{1}f(s),\ \ 0\le s< t. \end{aligned}$$

Then there exists a positive constant $C_{2}=\max \{2C_{1}f(0),4C_{0}C_{1}\}$ such that

$$\begin{aligned} f(t)\le C_{2}(1+t)^{-1},\ \ \forall t\geq 0. \end{aligned}$$

3 Proof of Theorem 1.1

In this section, we will prove Theorem 1.1. We first introduce the following energy functional,

$$\begin{aligned} E(t)=E_{1}(t)+E_{2}(t), \end{aligned}$$

where

$$ \begin{aligned} E_{1}(t)=\mathop{sup}_{0\le \tau \le t}(\|u(\tau )\|_{H^{3}}^{2}+\|b( \tau )\|_{H^{3}}^{2})+\int _{0}^{t}\|\partial _{2} u_{1}(\tau )\|_{H^{3}}^{2}+ \int _{0}^{t}\| b_{2}(\tau )\|_{H^{3}}^{2}, \end{aligned} $$

(3.1)

$$ \begin{aligned} E_{2}(t)=\int _{0}^{t}\|\partial _{1}u(\tau )\|_{H^{2}}^{2}. \end{aligned} $$

(3.2)

3.1 A Priori Estimates

Proposition 3.1

It holds that

$$\begin{aligned} E_{1}(t)\le & C(E_{1}(0)+E_{1}^{\frac{3}{2}}(0)+E_{1}^{2}(0))+C(E_{1}^{ \frac{3}{2}}(t)+E_{2}^{\frac{3}{2}}(t)) \\ &+C(E_{1}^{2}(t)+E_{2}^{2}(t))+C(E_{1}^{\frac{5}{2}}(t)+E_{2}^{ \frac{5}{2}}(t)), \end{aligned}$$

where $C$ is a pure positive constant.

Proof

Step 1. $\boldsymbol{L^{2}}$ -estimates

Taking the $L^{2}$- inner product to the equations (1.3) with $(u,b)$, one obtains

$$\begin{aligned} \frac{1}{2}\frac{d}{dt}&\|(u, b)\|_{L^{2}}^{2}+\|\partial _{2} u_{1} \|_{L^{2}}^{2}+\|b_{2}\|_{L^{2}}^{2}=0. \end{aligned}$$

(3.3)

Step 2. $\boldsymbol{\dot{H}^{3}}$-estimates Applying $\partial _{i}^{3}\ (i=1,2)$ to (1.3), and then taking the $L^{2}$- inner product with $(\partial _{i}^{3} u,\partial _{i}^{3}b)$ to obtain

$$\begin{aligned} \frac{1}{2}\frac{d}{dt}&\|(\partial _{i}^{3} u,\partial _{i}^{3} b)\|_{L^{2}}^{2}+ \|\partial _{2}\partial _{i}^{3} u_{1}\|_{L^{2}}^{2}+\|\partial _{i}^{3} b_{2}\|_{L^{2}}^{2}=-\int \partial _{i}^{3}(u\cdot \nabla u)\cdot \partial _{i}^{3} udx \\ &+\int \partial _{i}^{3}(b\cdot \nabla b)\cdot \partial _{i}^{3} udx- \int \partial _{i}^{3}(u\cdot \nabla b)\cdot \partial _{i}^{3}bdx+ \int \partial _{i}^{3}(b\cdot \nabla u)\cdot \partial _{i}^{3} bdx \\ &=-\sum _{1\leq \alpha \leq 3} C_{3}^{\alpha}\int \partial _{i}^{ \alpha }u\cdot \nabla \partial _{i}^{3-\alpha} u\cdot \partial _{i}^{3} udx+\sum _{1\leq \alpha \leq 3} C_{3}^{\alpha}\int \partial _{i}^{ \alpha }b\cdot \nabla \partial _{i}^{3-\alpha}b\cdot \partial _{i}^{3} udx \\ &-\sum _{1\leq \alpha \leq 3} C_{3}^{\alpha}\int \partial _{i}^{ \alpha }u\cdot \nabla \partial _{i}^{3-\alpha} b\cdot \partial _{i}^{3} bdx+\sum _{1\leq \alpha \leq 3} C_{3}^{\alpha}\int \partial _{i}^{ \alpha }b\cdot \nabla \partial _{i}^{3-\alpha} u\cdot \partial _{i}^{3} bdx \\ &=I_{1}+I_{2}+I_{3}+I_{4}, \end{aligned}$$

where we use the fact that $\nabla \cdot u=\nabla \cdot b=0$. We write

$$\begin{aligned} I_{1}=&-\sum _{1\leq \alpha \leq 3} C_{3}^{\alpha}\int \partial _{i}^{ \alpha }u\cdot \nabla \partial _{i}^{3-\alpha} u\cdot \partial _{i}^{3} udx \\ =&-\sum _{1\leq \alpha \leq 3} C_{3}^{\alpha}\int \partial _{1}^{ \alpha }u\cdot \nabla \partial _{1}^{3-\alpha} u\cdot \partial _{1}^{3} udx-\sum _{1\leq \alpha \leq 3} C_{3}^{\alpha}\int \partial _{2}^{ \alpha }u\cdot \nabla \partial _{2}^{3-\alpha} u\cdot \partial _{2}^{3} udx \\ =&I_{11}+I_{12}. \end{aligned}$$

According to the Hölder inequality, one gets

$$\begin{aligned} I_{11}=&-\sum _{1\leq \alpha \leq 3} C_{3}^{\alpha}\int \partial _{1}^{ \alpha }u\cdot \nabla \partial _{1}^{3-\alpha} u\cdot \partial _{1}^{3} udx \\ =&-3\int \partial _{1}u\cdot \nabla \partial _{1}^{2}u\cdot \partial _{1}^{3}udx-3 \int \partial _{1}^{2}u\cdot \nabla \partial _{1}u\cdot \partial _{1}^{3}udx -\int \partial _{1}^{3}u\cdot \nabla u\cdot \partial _{1}^{3}udx \\ \le &C\|\partial _{1}^{3}u\|_{L^{2}}(\|\partial _{1}u\|_{L^{\infty}} \|\nabla \partial _{1}^{2}u\|_{L^{2}}+\|\partial _{1}^{2}u\|_{L^{4}} \|\nabla \partial _{1}u\|_{L^{4}}+\|\partial _{1}^{3}u\|_{L^{2}}\| \nabla u\|_{L^{\infty}}) \\ \le &C\|u\|_{H^{3}}\|\partial _{1}u\|_{H^{2}}^{2}. \end{aligned}$$

To deal with $I_{12}$, we decompose it into the following three parts

$$\begin{aligned} I_{12}=&-\sum _{1\leq \alpha \leq 3} C_{3}^{\alpha}\int \partial _{2}^{ \alpha }u\cdot \nabla \partial _{2}^{3-\alpha} u\cdot \partial _{2}^{3} udx \\ =&-\sum _{1\leq \alpha \leq 3} C_{3}^{\alpha}\int \partial _{2}^{ \alpha }u_{1}\partial _{1}\partial _{2}^{3-\alpha} u\cdot \partial _{2}^{3} udx-\sum _{1\leq \alpha \leq 3} C_{3}^{\alpha}\int \partial _{2}^{ \alpha }u_{2}\partial _{2}^{4-\alpha} u_{1}\partial _{2}^{3} u_{1}dx \\ &-\sum _{1\leq \alpha \leq 3} C_{3}^{\alpha}\int \partial _{2}^{ \alpha }u_{2}\partial _{2}^{4-\alpha} u_{2}\partial _{2}^{3} u_{2}dx \\ =&I_{121}+I_{122}+I_{123}. \end{aligned}$$

Similar to estimate $I_{11}$, we find

$$\begin{aligned} I_{121}=&-\sum _{1\leq \alpha \leq 3} C_{3}^{\alpha}\int \partial _{2}^{ \alpha }u_{1}\partial _{1}\partial _{2}^{3-\alpha} u\cdot \partial _{2}^{3} udx \\ =&-3\int \partial _{2} u_{1}\partial _{1}\partial _{2}^{2} u\cdot \partial _{2}^{3} udx -3\int \partial _{2}^{2} u_{1}\partial _{1} \partial _{2} u\cdot \partial _{2}^{3} udx -\int \partial _{2}^{3} u_{1} \partial _{1} u\cdot \partial _{2}^{3} udx \\ \le & C\|\partial _{2}^{3} u\|_{L^{2}}(\| \partial _{2} u_{1}\|_{L^{ \infty}}\|\partial _{1}\partial _{2}^{2} u\|_{L^{2}}+\|\partial _{2}^{2} u_{1}\|_{L^{4}}\|\partial _{1}\partial _{2} u\|_{L^{4}}+\|\partial _{2}^{3} u_{1}\|_{L^{2}}\|\partial _{1} u\|_{L^{\infty}}) \\ \le &C\|u\|_{H^{3}}(\|\partial _{2} u_{1}\|_{H^{3}}^{2}+\|\partial _{1} u\|_{H^{2}}^{2}). \end{aligned}$$

Thanks to the Hölder inequality, we have

$$\begin{aligned} I_{122}=&-\sum _{1\leq \alpha \leq 3} C_{3}^{\alpha}\int \partial _{2}^{ \alpha }u_{2}\partial _{2}^{4-\alpha} u_{1}\partial _{2}^{3} u_{1}dx \\ =&-3\int \partial _{2} u_{2}\partial _{2}^{3} u_{1}\partial _{2}^{3} u_{1}dx -3\int \partial _{2}^{2} u_{2}\partial _{2}^{2} u_{1}\partial _{2}^{3} u_{1}dx -\int \partial _{2}^{3} u_{2}\partial _{2} u_{1}\partial _{2}^{3} u_{1}dx \\ \le &C\|\partial _{2}^{3} u_{1}\|_{L^{2}}(\|\partial _{2} u_{2}\|_{L^{ \infty}}\|\partial _{2}^{3} u_{1}\|_{L^{2}}+\|\partial _{2}^{2} u_{2} \|_{L^{2}} \|\partial _{2}^{2} u_{1}\|_{L^{\infty}}+\|\partial _{2}^{3} u_{2}\|_{L^{2}}\|\partial _{2} u_{1}\|_{L^{\infty}}) \\ \le &C\|u\|_{H^{3}}\|\partial _{2} u_{1}\|_{H^{3}}^{2}. \end{aligned}$$

According to the Hölder inequality and the fact that $\partial _{2}u_{2}=-\partial _{1}u_{1}$, we get

$$\begin{aligned} I_{123}=&-\sum _{1\leq \alpha \leq 3} C_{3}^{\alpha}\int \partial _{2}^{ \alpha }u_{2}\partial _{2}^{4-\alpha} u_{2}\partial _{2}^{3} u_{2}dx \\ =&-4\int \partial _{2} u_{2}\partial _{2}^{3} u_{2}\partial _{2}^{3} u_{2}dx -3\int \partial _{2}^{2} u_{2}\partial _{2}^{2} u_{2}\partial _{2}^{3} u_{2}dx \\ =&-4\int \partial _{2} u_{2}\partial _{2}^{2} \partial _{1}u_{1} \partial _{2}^{2}\partial _{1} u_{1}dx -3\int \partial _{2}^{2} u_{2} \partial _{2}\partial _{1} u_{1}\partial _{2}^{2}\partial _{1} u_{1}dx \\ \le &C\|\partial _{2}^{2}\partial _{1} u_{1}\|_{L^{2}}(\|\partial _{2}^{2} \partial _{1} u_{1}\|_{L^{2}}\|\partial _{2} u_{2}\|_{L^{\infty}}+\| \partial _{2}^{2} u_{2}\|_{L^{4}}\|\partial _{2}\partial _{1} u_{1}\|_{L^{4}}) \\ \le &C\|u\|_{H^{3}}\|\partial _{2} u_{1}\|_{H^{3}}^{2}. \end{aligned}$$

Hence, we get

$$\begin{aligned} I_{1}\le C\|u\|_{H^{3}}(\|\partial _{2} u_{1}\|_{H^{3}}^{2}+\| \partial _{1} u\|_{H^{2}}^{2}). \end{aligned}$$

To estimate $I_{2}$, we split it into five parts

$$\begin{aligned} I_{2}=&\sum _{1\leq \alpha \leq 3} C_{3}^{\alpha}\int \partial _{i}^{ \alpha }b\cdot \nabla \partial _{i}^{3-\alpha}b\cdot \partial _{i}^{3} udx \\ =&\sum _{1\leq \alpha \leq 3} C_{3}^{\alpha}\int \partial _{1}^{ \alpha }b\cdot \nabla \partial _{1}^{3-\alpha}b\cdot \partial _{1}^{3} udx+\sum _{1\leq \alpha \leq 3} C_{3}^{\alpha}\int \partial _{2}^{ \alpha }b\cdot \nabla \partial _{2}^{3-\alpha}b\cdot \partial _{2}^{3} udx \\ =&\sum _{1\leq \alpha \leq 3} C_{3}^{\alpha}\int \partial _{1}^{ \alpha }b\cdot \nabla \partial _{1}^{3-\alpha}b\cdot \partial _{1}^{3} udx+\sum _{1\leq \alpha \leq 3} C_{3}^{\alpha}\int \partial _{2}^{ \alpha }b_{1}\partial _{1} \partial _{2}^{3-\alpha}b_{1}\partial _{2}^{3} u_{1}dx \\ &+\sum _{1\leq \alpha \leq 3} C_{3}^{\alpha}\int \partial _{2}^{ \alpha }b_{1}\partial _{1} \partial _{2}^{3-\alpha}b_{2}\partial _{2}^{3} u_{2}dx +\sum _{1\leq \alpha \leq 3} C_{3}^{\alpha}\int \partial _{2}^{ \alpha }b_{2} \partial _{2}^{4-\alpha}b_{1}\partial _{2}^{3} u_{1}dx \\ &+\sum _{1\leq \alpha \leq 3} C_{3}^{\alpha}\int \partial _{2}^{ \alpha }b_{2} \partial _{2}^{4-\alpha}b_{2}\partial _{2}^{3} u_{2}dx \\ =&I_{21}+I_{22}+\cdots +I_{25}. \end{aligned}$$

According to the Hölder inequality and the Sobolev inequality, we conclude

$$\begin{aligned} I_{21}=&\sum _{1\leq \alpha \leq 3} C_{3}^{\alpha}\int \partial _{1}^{ \alpha }b\cdot \nabla \partial _{1}^{3-\alpha}b\cdot \partial _{1}^{3} udx \\ =&3\int \partial _{1} b\cdot \nabla \partial _{1}^{2}b\cdot \partial _{1}^{3} udx+3\int \partial _{1}^{2} b\cdot \nabla \partial _{1}b\cdot \partial _{1}^{3} udx+\int \partial _{1}^{3} b\cdot \nabla b\cdot \partial _{1}^{3} udx \\ \le &C\|\partial _{1}^{3} u\|_{L^{2}}(\|\partial _{1} b\|_{L^{\infty}} \|\nabla \partial _{1}^{2}b\|_{L^{2}}+\|\partial _{1}^{2} b\|_{L^{4}} \|\nabla \partial _{1}b\|_{L^{4}}+\|\partial _{1}^{3} b\|_{L^{2}}\| \nabla b\|_{L^{\infty}}) \\ \le &C\|b\|_{H^{3}}(\|b_{2}\|_{H^{3}}^{2}+\|\partial _{1} u\|_{H^{2}}^{2}), \end{aligned}$$

where we have used the fact that $\|\partial _{1}b\|_{H^{2}}\le C\|b_{2}\|_{H^{3}}$. Similarly to the derivation of $I_{21}$, we have

$$\begin{aligned} I_{22}=&\sum _{1\leq \alpha \leq 3} C_{3}^{\alpha}\int \partial _{2}^{ \alpha }b_{1}\partial _{1} \partial _{2}^{3-\alpha}b_{1}\partial _{2}^{3} u_{1}dx \\ =&3\int \partial _{2} b_{1}\partial _{1} \partial _{2}^{2}b_{1} \partial _{2}^{3} u_{1}dx +3\int \partial _{2}^{2} b_{1}\partial _{1} \partial _{2}b_{1}\partial _{2}^{3} u_{1}dx +\int \partial _{2}^{3} b_{1} \partial _{1} b_{1}\partial _{2}^{3} u_{1}dx \\ \le &C\|\partial _{2}^{3} u_{1}\|_{L^{2}}(\|\partial _{2} b_{1}\|_{L^{ \infty}}\|\partial _{1} \partial _{2}^{2}b_{1}\|_{L^{2}}+\|\partial _{2}^{2} b_{1}\|_{L^{4}}\|\partial _{1} \partial _{2}b_{1}\|_{L^{4}}+ \| \partial _{2}^{3}b_{1}\|_{L^{2}}\|\partial _{1} b_{1}\|_{L^{\infty}}) \\ \le &C\|b\|_{H^{3}}(\|b_{2}\|_{H^{3}}^{2}+\|\partial _{2} u_{1}\|_{H^{3}}^{2}). \end{aligned}$$

Thanks to the Hölder inequality again, we have

$$\begin{aligned} I_{23}=&\sum _{1\leq \alpha \leq 3} C_{3}^{\alpha}\int \partial _{2}^{ \alpha }b_{1}\partial _{1} \partial _{2}^{3-\alpha}b_{2}\partial _{2}^{3} u_{2}dx \\ =&3\int \partial _{2} b_{1}\partial _{1} \partial _{2}^{2}b_{2} \partial _{2}^{3} u_{2}dx +3\int \partial _{2}^{2} b_{1}\partial _{1} \partial _{2}b_{2}\partial _{2}^{3} u_{2}dx +\int \partial _{2}^{3} b_{1} \partial _{1}b_{2}\partial _{2}^{3} u_{2}dx \\ \le &C\|\partial _{2}^{3} u_{2}\|_{L^{2}}(\|\partial _{2} b_{1}\|_{L^{ \infty}}\|\partial _{1} \partial _{2}^{2}b_{2}\|_{L^{2}}+\|\partial _{2}^{2} b_{1}\|_{L^{4}}\|\partial _{1} \partial _{2}b_{2}\|_{L^{4}}+ \| \partial _{2}^{3} b_{1}\|_{L^{2}}\|\partial _{1} b_{2}\|_{L^{\infty}}) \\ \le &C\|b\|_{H^{3}}(\|b_{2}\|_{H^{3}}^{2}+\|\partial _{2} u_{1}\|_{H^{3}}^{2}), \end{aligned}$$

$$\begin{aligned} I_{24}=&\sum _{1\leq \alpha \leq 3} C_{3}^{\alpha}\int \partial _{2}^{ \alpha }b_{2} \partial _{2}^{4-\alpha}b_{1}\partial _{2}^{3} u_{1}dx \\ =&3\int \partial _{2} b_{2} \partial _{2}^{3}b_{1}\partial _{2}^{3} u_{1}dx +3\int \partial _{2}^{2} b_{2} \partial _{2}^{2}b_{1}\partial _{2}^{3} u_{1}dx +\int \partial _{2}^{3}b_{2} \partial _{2}b_{1}\partial _{2}^{3} u_{1}dx \\ \le &C\|\partial _{2}^{3} u_{1}\|_{L^{2}}(\|\partial _{2} b_{2}\|_{L^{ \infty}}\| \partial _{2}^{3}b_{1}\|_{L^{2}}+\|\partial _{2}^{2} b_{2} \|_{L^{4}}\|\partial _{2}^{2}b_{1}\|_{L^{4}}+ \|\partial _{2}^{3} b_{2} \|_{L^{2}}\|\partial _{2} b_{1}\|_{L^{\infty}}) \\ \le &C\|b\|_{H^{3}}(\|b_{2}\|_{H^{3}}^{2}+\|\partial _{2} u_{1}\|_{H^{3}}^{2}), \end{aligned}$$

and

$$\begin{aligned} I_{25}=&\sum _{1\leq \alpha \leq 3} C_{3}^{\alpha}\int \partial _{2}^{ \alpha }b_{2} \partial _{2}^{4-\alpha}b_{2}\partial _{2}^{3} u_{2}dx \\ =&4\int \partial _{2} b_{2} \partial _{2}^{3}b_{2}\partial _{2}^{3} u_{2}dx +3\int \partial _{2}^{2} b_{2} \partial _{2}^{2}b_{2}\partial _{2}^{3} u_{2}dx \\ \le &C\|\partial _{2}^{3} u_{2}\|_{L^{2}}(\|\partial _{2} b_{2}\|_{L^{ \infty}}\| \partial _{2}^{3}b_{2}\|_{L^{2}}+\|\partial _{2}^{2} b_{2} \|_{L^{4}}^{2}) \\ \le &C\|u\|_{H^{3}}\|b_{2}\|_{H^{3}}^{2}. \end{aligned}$$

Thus we get

$$\begin{aligned} I_{2}\le C(\|u\|_{H^{3}}+\|b\|_{H^{3}})(\|b_{2}\|_{H^{3}}^{2}+\| \partial _{2} u_{1}\|_{H^{3}}^{2}+\|\partial _{1} u\|_{H^{2}}^{2}). \end{aligned}$$

Now we write

$$\begin{aligned} I_{3}=&-\sum _{1\leq \alpha \leq 3} C_{3}^{\alpha}\int \partial _{i}^{ \alpha }u\cdot \nabla \partial _{i}^{3-\alpha} b\cdot \partial _{i}^{3} bdx \\ =&-\sum _{1\leq \alpha \leq 3} C_{3}^{\alpha}\int \partial _{1}^{ \alpha }u\cdot \nabla \partial _{1}^{3-\alpha} b\cdot \partial _{1}^{3} bdx-\sum _{1\leq \alpha \leq 3} C_{3}^{\alpha}\int \partial _{2}^{ \alpha }u\cdot \nabla \partial _{2}^{3-\alpha} b\cdot \partial _{2}^{3} bdx \\ =&-\sum _{1\leq \alpha \leq 3} C_{3}^{\alpha}\int \partial _{1}^{ \alpha }u\cdot \nabla \partial _{1}^{3-\alpha} b\cdot \partial _{1}^{3} bdx-\sum _{1\leq \alpha \leq 3} C_{3}^{\alpha}\int \partial _{2}^{ \alpha }u_{1}\partial _{1}\partial _{2}^{3-\alpha} b\cdot \partial _{2}^{3} bdx \\ &-\sum _{1\leq \alpha \leq 3} C_{3}^{\alpha}\int \partial _{2}^{ \alpha }u_{2}\partial _{2}^{4-\alpha} b\cdot \partial _{2}^{3} bdx \\ =&I_{31}+I_{32}+I_{33}. \end{aligned}$$

Thanks to the Hölder inequality and Sobolev embedding, we get

$$\begin{aligned} I_{31}=&-\sum _{1\leq \alpha \leq 3} C_{3}^{\alpha}\int \partial _{1}^{ \alpha }u\cdot \nabla \partial _{1}^{3-\alpha} b\cdot \partial _{1}^{3} bdx \\ =&-3\int \partial _{1} u\cdot \nabla \partial _{1}^{2} b\cdot \partial _{1}^{3} bdx -3\int \partial _{1}^{2} u\cdot \nabla \partial _{1} b\cdot \partial _{1}^{3} bdx -\int \partial _{1}^{3} u \cdot \nabla b\cdot \partial _{1}^{3} bdx \\ \le &C\|\partial _{1}^{3} b\|_{L^{2}}(\|\partial _{1} u\|_{L^{\infty}} \|\nabla \partial _{1}^{2} b\|_{L^{2}}+\|\partial _{1}^{2} u\|_{L^{4}} \|\nabla \partial _{1} b\|_{L^{4}}+\|\partial _{1}^{3} u\|_{L^{2}}\| \nabla b\|_{L^{\infty}}) \\ \le &C\|b\|_{H^{3}}(\|b_{2}\|_{H^{3}}^{2}+\|\partial _{1} u\|_{H^{2}}^{2}). \end{aligned}$$

Similarly,

$$\begin{aligned} I_{32}=&-\sum _{1\leq \alpha \leq 3} C_{3}^{\alpha}\int \partial _{2}^{ \alpha }u_{1}\partial _{1}\partial _{2}^{3-\alpha} b\cdot \partial _{2}^{3} bdx \\ =&-3\int \partial _{2} u_{1}\partial _{1}\partial _{2}^{2} b\cdot \partial _{2}^{3} bdx -3\int \partial _{2}^{2} u_{1}\partial _{1} \partial _{2} b\cdot \partial _{2}^{3} bdx -\int \partial _{2}^{3} u_{1} \partial _{1} b\cdot \partial _{2}^{3} bdx \\ \le &C\|\partial _{2}^{3} b\|_{L^{2}}(\|\partial _{2} u_{1}\|_{L^{ \infty}}\|\partial _{1}\partial _{2}^{2} b\|_{L^{2}}+\|\partial _{2}^{2} u_{1}\|_{L^{4}}\|\partial _{1}\partial _{2} b\|_{L^{4}}+\|\partial _{2}^{3} u_{1}\|_{L^{2}}\|\partial _{1} b\|_{L^{\infty}}) \\ \le &C\|b\|_{H^{3}}(\|b_{2}\|_{H^{3}}^{2}+\|\partial _{2} u_{1}\|_{H^{3}}^{2}). \end{aligned}$$

To bound $I_{33}$, we further decompose it into three terms

$$\begin{aligned} I_{33}=&-\sum _{1\leq \alpha \leq 3} C_{3}^{\alpha}\int \partial _{2}^{ \alpha }u_{2}\partial _{2}^{4-\alpha} b\cdot \partial _{2}^{3} bdx \\ =&-3\int \partial _{2} u_{2}\partial _{2}^{3} b\cdot \partial _{2}^{3} bdx -3\int \partial _{2}^{2} u_{2}\partial _{2}^{2} b\cdot \partial _{2}^{3} bdx -\int \partial _{2}^{3} u_{2}\partial _{2} b\cdot \partial _{2}^{3} bdx \\ =&I_{331}+I_{332}+I_{333}. \end{aligned}$$

By the Hölder inequality and the Sobolev inequality, we obtain

$$\begin{aligned} I_{331}=&-3\int \partial _{2} u_{2}\partial _{2}^{3} b\cdot \partial _{2}^{3} bdx \\ =&-3\int \partial _{2} u_{2}\partial _{2}^{3} b_{1}\partial _{2}^{3} b_{1}dx-3 \int \partial _{2} u_{2}\partial _{2}^{3} b_{2}\partial _{2}^{3} b_{2}dx \\ \le &C\|\partial _{2} u_{2}\|_{L^{\infty}}\|\partial _{2}^{3} b_{2}\|_{L^{2}}^{2}+3 \int \partial _{1} u_{1}(\partial _{2}^{3} b_{1})^{2}dx \\ \le &C\|u\|_{H^{3}}\|b_{2}\|_{H^{3}}^{2}+3\int \partial _{1} u_{1}( \partial _{2}^{3} b_{1})^{2}dx. \end{aligned}$$

Integrating by parts and using the fact that $\partial _{2}u_{2}=-\partial _{1}u_{1}$, we get

$$\begin{aligned} I_{332}=&-\frac{3}{2}\int \partial _{2}^{2} u_{2}\partial _{2}( \partial _{2}^{2} b)^{2}dx=\frac{3}{2}\int \partial _{2}^{3} u_{2}( \partial _{2}^{2} b)^{2}dx \\ =&-\frac{3}{2}\int \partial _{2}^{2} \partial _{1}u_{1}(\partial _{2}^{2} b)^{2}dx=3\int \partial _{2}^{2} u_{1}\partial _{2}^{2} b\cdot \partial _{1}\partial _{2}^{2} bdx \\ \le &C\|\partial _{2}^{2} u_{1}\|_{L^{4}}\|\partial _{2}^{2} b\|_{L^{4}} \|\partial _{1}\partial _{2}^{2} b\|_{L^{2}} \\ \le &C\|b\|_{H^{3}}(\|b_{2}\|_{H^{3}}^{2}+\|\partial _{2} u_{1}\|_{H^{3}}^{2}). \end{aligned}$$

In view of the divergence-free conditions, integration by parts and the Hölder inequality, one can obtain

$$\begin{aligned} I_{333}=&\int \partial _{2}^{2}\partial _{1} u_{1}\partial _{2} b \cdot \partial _{2}^{3} bdx \\ =&-\int \partial _{2}^{2} u_{1}\partial _{2}\partial _{1} b\cdot \partial _{2}^{3} bdx-\int \partial _{2}^{2} u_{1}\partial _{2} b \cdot \partial _{2}^{3}\partial _{1} bdx \\ =&-\int \partial _{2}^{2} u_{1}\partial _{2}\partial _{1} b\cdot \partial _{2}^{3} bdx+\int \partial _{2}^{3} u_{1}\partial _{2} b \cdot \partial _{2}^{2}\partial _{1} bdx+\int \partial _{2}^{2} u_{1} \partial _{2}^{2} b\cdot \partial _{2}^{2}\partial _{1} bdx \\ \le &C\|\partial _{2}^{2} u_{1}\|_{L^{4}}\|\partial _{2}\partial _{1} b \|_{L^{4}}\|\partial _{2}^{3} b\|_{L^{2}}+ (\|\partial _{2}^{3} u_{1} \|_{L^{2}}\|\partial _{2} b\|_{L^{\infty}}+ \|\partial _{2}^{2} u_{1} \|_{L^{4}}\|\partial _{2}^{2} b\|_{L^{4}})\|\partial _{2}^{2} \partial _{1} b\|_{L^{2}} \\ \ \le &C\|b\|_{H^{3}}(\|b_{2}\|_{H^{3}}^{2}+\|\partial _{2} u_{1}\|_{H^{3}}^{2}). \end{aligned}$$

Then, it leads to

$$\begin{aligned} I_{3}\le C(\|u\|_{H^{3}}+\|b\|_{H^{3}})(\|b_{2}\|_{H^{3}}^{2}+\| \partial _{2} u_{1}\|_{H^{3}}^{2}+\|\partial _{1} u\|_{H^{2}}^{2})+3 \int \partial _{1} u_{1}(\partial _{2}^{3} b_{1})^{2}dx. \end{aligned}$$

In order to bound $I_{4}$, we decompose it into the following terms

$$\begin{aligned} I_{4}=&\sum _{1\leq \alpha \leq 3} C_{3}^{\alpha}\int \partial _{i}^{ \alpha }b\cdot \nabla \partial _{i}^{3-\alpha} u\cdot \partial _{i}^{3} bdx \\ =&\sum _{1\leq \alpha \leq 3} C_{3}^{\alpha}\int \partial _{1}^{ \alpha }b\cdot \nabla \partial _{1}^{3-\alpha} u\cdot \partial _{1}^{3} bdx+\sum _{1\leq \alpha \leq 3} C_{3}^{\alpha}\int \partial _{2}^{ \alpha }b\cdot \nabla \partial _{2}^{3-\alpha} u\cdot \partial _{2}^{3} bdx \\ =&\sum _{1\leq \alpha \leq 3} C_{3}^{\alpha}\int \partial _{1}^{ \alpha }b\cdot \nabla \partial _{1}^{3-\alpha} u\cdot \partial _{1}^{3} bdx +\sum _{1\leq \alpha \leq 3} C_{3}^{\alpha}\int \partial _{2}^{ \alpha }b_{1}\partial _{1}\partial _{2}^{3-\alpha} u_{1}\partial _{2}^{3} b_{1}dx \\ &+\sum _{1\leq \alpha \leq 3} C_{3}^{\alpha}\int \partial _{2}^{ \alpha }b_{1}\partial _{1}\partial _{2}^{3-\alpha} u_{2}\partial _{2}^{3} b_{2}dx +\sum _{1\leq \alpha \leq 3} C_{3}^{\alpha}\int \partial _{2}^{ \alpha }b_{2}\partial _{2}^{4-\alpha} u_{1}\partial _{2}^{3} b_{1}dx \\ &+\sum _{1\leq \alpha \leq 3} C_{3}^{\alpha}\int \partial _{2}^{ \alpha }b_{2}\partial _{2}^{4-\alpha} u_{2}\partial _{2}^{3} b_{2}dx \\ =&I_{41}+I_{42}+\cdots +I_{45}. \end{aligned}$$

Thanks to the Hölder inequality, one has

$$\begin{aligned} I_{41}=&\sum _{1\leq \alpha \leq 3} C_{3}^{\alpha}\int \partial _{1}^{ \alpha }b\cdot \nabla \partial _{1}^{3-\alpha} u\cdot \partial _{1}^{3} bdx \\ =&3\int \partial _{1} b\cdot \nabla \partial _{1}^{2} u\cdot \partial _{1}^{3} bdx +3\int \partial _{1}^{2} b\cdot \nabla \partial _{1} u\cdot \partial _{1}^{3} bdx +\int \partial _{1}^{3} b \cdot \nabla u\cdot \partial _{1}^{3} bdx \\ \le &C\|\partial _{1}^{3} b\|_{L^{2}}(\|\partial _{1} b\|_{L^{\infty}} \|\nabla \partial _{1}^{2} u\|_{L^{2}}+\|\partial _{1}^{2} b\|_{L^{4}} \|\nabla \partial _{1} u\|_{L^{4}}+\|\partial _{1}^{3} b\|_{L^{2}}\| \nabla u\|_{L^{\infty}}) \\ \le &C\|u\|_{H^{3}}\|b_{2}\|_{H^{3}}^{2}. \end{aligned}$$

By using integration by parts several times and the Hölder inequality, we deduce

$$\begin{aligned} I_{42}=&\sum _{1\leq \alpha \leq 3} C_{3}^{\alpha}\int \partial _{2}^{ \alpha }b_{1}\partial _{1}\partial _{2}^{3-\alpha} u_{1}\partial _{2}^{3} b_{1}dx \\ =&3\int \partial _{2} b_{1}\partial _{1}\partial _{2}^{2} u_{1} \partial _{2}^{3} b_{1}dx +3\int \partial _{2}^{2} b_{1}\partial _{1} \partial _{2} u_{1}\partial _{2}^{3} b_{1}dx +\int \partial _{1} u_{1}( \partial _{2}^{3} b_{1})^{2}dx \\ =&-3\int \partial _{2}\partial _{1} b_{1}\partial _{2}^{2} u_{1} \partial _{2}^{3} b_{1}dx -3\int \partial _{2} b_{1}\partial _{2}^{2} u_{1} \partial _{2}^{3}\partial _{1} b_{1}dx+\frac{3}{2}\int \partial _{1} \partial _{2} u_{1}\partial _{2}(\partial _{2}^{2} b_{1})^{2}dx\\ & + \int \partial _{1} u_{1}(\partial _{2}^{3} b_{1})^{2}dx \\ =&-3\int \partial _{2}\partial _{1} b_{1}\partial _{2}^{2} u_{1} \partial _{2}^{3} b_{1}dx +3\int \partial _{2}^{2} b_{1}\partial _{2}^{2} u_{1}\partial _{2}^{2}\partial _{1} b_{1}dx \\ &+3\int \partial _{2} b_{1}\partial _{2}^{3} u_{1}\partial _{2}^{2} \partial _{1} b_{1}dx -\frac{3}{2}\int \partial _{1}\partial _{2}^{2} u_{1}( \partial _{2}^{2} b_{1})^{2}dx +\int \partial _{1} u_{1}(\partial _{2}^{3} b_{1})^{2}dx \\ =&-3\int \partial _{2}\partial _{1} b_{1}\partial _{2}^{2} u_{1} \partial _{2}^{3} b_{1}dx +3\int \partial _{2}^{2} b_{1}\partial _{2}^{2} u_{1}\partial _{2}^{2}\partial _{1} b_{1}dx \\ &+3\int \partial _{2} b_{1}\partial _{2}^{3} u_{1}\partial _{2}^{2} \partial _{1} b_{1}dx +3\int \partial _{2}^{2} u_{1}\partial _{2}^{2} b_{1} \partial _{1}\partial _{2}^{2} b_{1}dx +\int \partial _{1} u_{1}( \partial _{2}^{3} b_{1})^{2}dx \\ \le &C\|\partial _{2}\partial _{1} b_{1}\|_{L^{4}}\|\partial _{2}^{2} u_{1} \|_{L^{4}}\|\partial _{2}^{3} b_{1}\|_{L^{2}}+C\|\partial _{2}^{2} b_{1} \|_{L^{4}}\|\partial _{2}^{2} u_{1}\|_{L^{4}}\|\partial _{2}^{2} \partial _{1} b_{1}\|_{L^{2}} \\ &+C\|\partial _{2} b_{1}\|_{L^{\infty}}\|\partial _{2}^{3} u_{1}\|_{L^{2}} \|\partial _{2}^{2}\partial _{1} b_{1}\|_{L^{2}}+C\|\partial _{2}^{2} u_{1} \|_{L^{4}}\|\partial _{2}^{2} b_{1}\|_{L^{4}}\|\partial _{1}\partial _{2}^{2} b_{1}\|_{L^{2}}\\ &+\int \partial _{1} u_{1}(\partial _{2}^{3} b_{1})^{2}dx \\ \le &C\|b\|_{H^{3}}(\|b_{2}\|_{H^{3}}^{2}+\|\partial _{2} u_{1}\|_{H^{3}}^{2})+ \int \partial _{1} u_{1}(\partial _{2}^{3} b_{1})^{2}dx. \end{aligned}$$

According to the Hölder inequality and Sobolev inequality again, one arrives at

$$\begin{aligned} I_{43}=&\sum _{1\leq \alpha \leq 3} C_{3}^{\alpha}\int \partial _{2}^{ \alpha }b_{1}\partial _{1}\partial _{2}^{3-\alpha} u_{2}\partial _{2}^{3} b_{2}dx \\ =&3\int \partial _{2} b_{1}\partial _{1}\partial _{2}^{2} u_{2} \partial _{2}^{3} b_{2}dx +3\int \partial _{2}^{2} b_{1}\partial _{1} \partial _{2} u_{2}\partial _{2}^{3} b_{2}dx +\int \partial _{2}^{3} b_{1} \partial _{1} u_{2}\partial _{2}^{3} b_{2}dx \\ \le &C\|\partial _{2}^{3} b_{2}\|_{L^{2}}(\|\partial _{2} b_{1}\|_{L^{ \infty}}\|\partial _{1}\partial _{2}^{2} u_{2}\|_{L^{2}}+\|\partial _{2}^{2} b_{1}\|_{L^{4}}\|\partial _{1}\partial _{2} u_{2}\|_{L^{4}}+\| \partial _{2}^{3} b_{1}\|_{L^{2}}\|\partial _{1} u_{2}\|_{L^{\infty}}) \\ \le &C\|b\|_{H^{3}}(\|b_{2}\|_{H^{3}}^{2}+\|\partial _{1} u\|_{H^{2}}^{2}). \end{aligned}$$

Similarly, it leads to

$$\begin{aligned} I_{44}=&\sum _{1\leq \alpha \leq 3} C_{3}^{\alpha}\int \partial _{2}^{ \alpha }b_{2}\partial _{2}^{4-\alpha} u_{1}\partial _{2}^{3} b_{1}dx \\ =&3\int \partial _{2} b_{2}\partial _{2}^{3} u_{1}\partial _{2}^{3} b_{1}dx +3\int \partial _{2}^{2} b_{2}\partial _{2}^{2} u_{1}\partial _{2}^{3} b_{1}dx +\int \partial _{2}^{3} b_{2}\partial _{2} u_{1}\partial _{2}^{3} b_{1}dx \\ \le &C\|\partial _{2}^{3} b_{1}\|_{L^{2}}(\|\partial _{2} b_{2}\|_{L^{ \infty}}\|\partial _{2}^{3} u_{1}\|_{L^{2}}+\|\partial _{2}^{2} b_{2} \|_{L^{4}}\|\partial _{2}^{2} u_{1}\|_{L^{4}}+\|\partial _{2}^{3} b_{2} \|_{L^{2}}\|\partial _{2} u_{1}\|_{L^{\infty}}) \\ \le &C\|b\|_{H^{3}}(\|b_{2}\|_{H^{3}}^{2}+\|\partial _{2} u_{1}\|_{H^{3}}^{2}). \end{aligned}$$

and

$$\begin{aligned} I_{45}=&\sum _{1\leq \alpha \leq 3} C_{3}^{\alpha}\int \partial _{2}^{ \alpha }b_{2}\partial _{2}^{4-\alpha} u_{2}\partial _{2}^{3} b_{2}dx \\ =&3\int \partial _{2} b_{2}\partial _{2}^{3} u_{2}\partial _{2}^{3} b_{2}dx +3\int \partial _{2}^{2} b_{2}\partial _{2}^{2} u_{2}\partial _{2}^{3} b_{2}dx +\int \partial _{2}^{3} b_{2}\partial _{2} u_{2}\partial _{2}^{3} b_{2}dx \\ \le &C\|\partial _{2}^{3} b_{2}\|_{L^{2}}(\|\partial _{2} b_{2}\|_{L^{ \infty}}\|\partial _{2}^{3} u_{2}\|_{L^{2}}+\|\partial _{2}^{2} b_{2} \|_{L^{4}}\|\partial _{2}^{2} u_{2}\|_{L^{4}}+\|\partial _{2}^{3} b_{2} \|_{L^{2}}\|\partial _{2} u_{2}\|_{L^{\infty}}) \\ \le &C\|u\|_{H^{3}}\|b_{2}\|_{H^{3}}^{2}. \end{aligned}$$

Collecting the estimate from $I_{41}$ to $I_{45}$ to give

$$\begin{aligned} I_{4}\le &C(\|u\|_{H^{3}}+\|b\|_{H^{3}})(\|b_{2}\|_{H^{3}}^{2}+\| \partial _{2} u_{1}\|_{H^{3}}^{2}+\|\partial _{1} u\|_{H^{2}}^{2})+ \int \partial _{1} u_{1}(\partial _{2}^{3} b_{1})^{2}dx. \end{aligned}$$

Then one has

$$ \begin{aligned} \frac{1}{2}&\frac{d}{dt}\|(u,b)\|_{H^{3}}^{2}+\|\partial _{2}u_{1}\|_{H^{3}}^{2}+ \|b_{2}\|_{H^{3}}^{2} \\ \le &C\|(u,b)\|_{H^{3}}(\|b_{2}\|_{H^{3}}^{2}+\|\partial _{2} u_{1}\|_{H^{3}}^{2}+ \|\partial _{1} u\|_{H^{2}}^{2})+4\int \partial _{1} u_{1}(\partial _{2}^{3} b_{1})^{2}dx. \end{aligned} $$

(3.4)

To handle the worst term $\int \partial _{1} u_{1}(\partial _{2}^{3} b_{1})^{2}dx$, thanks to the following equation

$$\begin{aligned} \partial _{1}u_{1}=\partial _{t}b_{1}+u\cdot \nabla b_{1}-b\cdot \nabla u_{1}, \end{aligned}$$

(3.5)

one can verify

$$\begin{aligned} \int \partial _{1}& u_{1}(\partial _{2}^{3} b_{1})^{2}dx=\int ( \partial _{t}b_{1}+u\cdot \nabla b_{1}-b\cdot \nabla u_{1})(\partial _{2}^{3} b_{1})^{2}dx \\ =&\frac{d}{dt}\int b_{1}(\partial _{2}^{3} b_{1})^{2}dx-2\int b_{1} \partial _{2}^{3} b_{1}\partial _{2}^{3}\partial _{t} b_{1}+\int (u \cdot \nabla b_{1})(\partial _{2}^{3} b_{1})^{2}dx\\ &-\int (b\cdot \nabla u_{1})(\partial _{2}^{3} b_{1})^{2}dx \\ =&\frac{d}{dt}\int b_{1}(\partial _{2}^{3} b_{1})^{2}dx+2\int b_{1} \partial _{2}^{3} b_{1}\partial _{2}^{3}(u\cdot \nabla b_{1})dx-2 \int b_{1}\partial _{2}^{3} b_{1}\partial _{2}^{3}(b\cdot \nabla u_{1})dx \\ &-2\int b_{1}\partial _{2}^{3} b_{1}\partial _{2}^{3}\partial _{1} u_{1}dx +\int (u\cdot \nabla b_{1})(\partial _{2}^{3} b_{1})^{2}dx-\int (b \cdot \nabla u_{1})(\partial _{2}^{3} b_{1})^{2}dx \\ =&\frac{d}{dt}\int b_{1}(\partial _{2}^{3} b_{1})^{2}dx+2\sum _{1\le \alpha \le 3}C_{3}^{\alpha}\int b_{1}\partial _{2}^{3} b_{1}\partial _{2}^{ \alpha }u\cdot \nabla \partial _{2}^{3-\alpha} b_{1}dx+2\int b_{1} \partial _{2}^{3}b_{1}u\cdot \nabla \partial _{2}^{3}b_{1}dx \\ &-2\int b_{1}\partial _{2}^{3} b_{1}\partial _{2}^{3}(b\cdot \nabla u_{1})dx-2 \int b_{1}\partial _{2}^{3} b_{1}\partial _{2}^{3}\partial _{1} u_{1}dx \\ &+\int (u\cdot \nabla b_{1})(\partial _{2}^{3} b_{1})^{2}dx-\int (b \cdot \nabla u_{1})(\partial _{2}^{3} b_{1})^{2}dx \\ =&J_{1}+J_{2}+\cdots +J_{7}. \end{aligned}$$

By $\nabla \cdot u=0$, we can easily verify

$$\begin{aligned} J_{3}+J_{6}=&2\int b_{1}\partial _{2}^{3}b_{1}u\cdot \nabla \partial _{2}^{3}b_{1}dx+ \int (u\cdot \nabla b_{1})(\partial _{2}^{3} b_{1})^{2}dx \\ =&\int b_{1}u\cdot \nabla (\partial _{2}^{3}b_{1})^{2}dx+\int (u \cdot \nabla b_{1})(\partial _{2}^{3} b_{1})^{2}dx \\ =&\int u\cdot \nabla ( b_{1}(\partial _{2}^{3} b_{1})^{2})dx \\ =&0. \end{aligned}$$

To deal with $J_{2}$, we first split it into the following four terms,

$$\begin{aligned} J_{2}=&2\sum _{1\le \alpha \le 3}C_{3}^{\alpha}\int b_{1}\partial _{2}^{3} b_{1}\partial _{2}^{\alpha }u\cdot \nabla \partial _{2}^{3-\alpha} b_{1}dx \\ =&2\sum _{1\le \alpha \le 3}C_{3}^{\alpha}\int b_{1}\partial _{2}^{3} b_{1} \partial _{2}^{\alpha }u_{1}\partial _{1}\partial _{2}^{3-\alpha} b_{1}dx +2\sum _{1\le \alpha \le 3}C_{3}^{\alpha}\int b_{1}\partial _{2}^{3} b_{1} \partial _{2}^{\alpha }u_{2}\partial _{2}^{4-\alpha} b_{1}dx \\ =&2\sum _{1\le \alpha \le 3}C_{3}^{\alpha}\int b_{1}\partial _{2}^{3} b_{1} \partial _{2}^{\alpha }u_{1}\partial _{1}\partial _{2}^{3-\alpha} b_{1}dx+6 \int b_{1}\partial _{2}^{3} b_{1}\partial _{2} u_{2}\partial _{2}^{3} b_{1}dx \\ &+6\int b_{1}\partial _{2}^{3} b_{1}\partial _{2}^{2} u_{2}\partial _{2}^{2} b_{1}dx+2\int b_{1}\partial _{2}^{3} b_{1}\partial _{2}^{3} u_{2} \partial _{2} b_{1}dx \\ =&J_{21}+J_{22}+J_{23}+J_{24}. \end{aligned}$$

For the first term, the Hölder inequality guarantees that

$$\begin{aligned} J_{21}=&2\sum _{1\le \alpha \le 3}C_{3}^{\alpha}\int b_{1}\partial _{2}^{3} b_{1}\partial _{2}^{\alpha }u_{1}\partial _{1}\partial _{2}^{3-\alpha} b_{1}dx \\ =&6\int b_{1}\partial _{2}^{3} b_{1}\partial _{2} u_{1}\partial _{1} \partial _{2}^{2} b_{1}dx+6\int b_{1}\partial _{2}^{3} b_{1}\partial _{2}^{2} u_{1}\partial _{1}\partial _{2} b_{1}dx+2\int b_{1}\partial _{2}^{3} b_{1} \partial _{2}^{3} u_{1}\partial _{1} b_{1}dx \\ \le &C\|b_{1}\|_{L^{\infty}}\|\partial _{2}^{3} b_{1}\|_{L^{2}}(\| \partial _{2} u_{1}\|_{L^{\infty}}\|\partial _{1}\partial _{2}^{2} b_{1} \|_{L^{2}}+\|\partial _{2}^{2} u_{1}\|_{L^{4}}\|\partial _{1} \partial _{2} b_{1}\|_{L^{4}}\\ &+\|\partial _{2}^{3} u_{1}\|_{L^{2}}\| \partial _{1} b_{1}\|_{L^{\infty}}) \\ \le &C\|b\|_{H^{3}}^{2}(\|b_{2}\|_{H^{3}}^{2}+\|\partial _{2} u_{1}\|_{H^{3}}^{2}). \end{aligned}$$

In term of (3.5), we get

$$\begin{aligned} J_{22}=&6\int b_{1}\partial _{2}^{3} b_{1}\partial _{2} u_{2} \partial _{2}^{3} b_{1}dx=-6\int b_{1} \partial _{1} u_{1}(\partial _{2}^{3} b_{1})^{2}dx \\ =&-6\int b_{1} (\partial _{t}b_{1}+u\cdot \nabla b_{1}-b\cdot \nabla u_{1})( \partial _{2}^{3} b_{1})^{2}dx \\ =&-3\int \partial _{t}(b_{1})^{2}(\partial _{2}^{3} b_{1})^{2}dx-6 \int b_{1}(u\cdot \nabla b_{1})(\partial _{2}^{3} b_{1})^{2}dx+6\int b_{1}(b \cdot \nabla u_{1})(\partial _{2}^{3} b_{1})^{2}dx \\ =&-3\frac{d}{dt}\int (b_{1})^{2}(\partial _{2}^{3} b_{1})^{2}dx+6 \int (b_{1})^{2}\partial _{2}^{3} b_{1}\partial _{2}^{3} \partial _{t}b_{1}dx \\ &-6\int b_{1}(u\cdot \nabla b_{1})(\partial _{2}^{3} b_{1})^{2}dx+6 \int b_{1}(b\cdot \nabla u_{1})(\partial _{2}^{3} b_{1})^{2}dx \\ =&-3\frac{d}{dt}\int (b_{1})^{2}(\partial _{2}^{3} b_{1})^{2}dx-6 \int (b_{1})^{2}\partial _{2}^{3} b_{1}\partial _{2}^{3} (u\cdot \nabla b_{1})dx\\ &+6\int (b_{1})^{2}\partial _{2}^{3} b_{1}\partial _{2}^{3} (b\cdot \nabla u_{1})dx \\ &+6\int (b_{1})^{2}\partial _{2}^{3} b_{1}\partial _{2}^{3} \partial _{1} u_{1}dx-6\int b_{1}(u\cdot \nabla b_{1})(\partial _{2}^{3} b_{1})^{2}dx+6 \int b_{1}(b\cdot \nabla u_{1})(\partial _{2}^{3} b_{1})^{2}dx \\ =&-3\frac{d}{dt}\int (b_{1})^{2}(\partial _{2}^{3} b_{1})^{2}dx-6 \sum _{1\le \alpha \le 3}C_{3}^{\alpha}\int (b_{1})^{2}\partial _{2}^{3} b_{1}\partial _{2}^{\alpha }u\cdot \nabla \partial _{2}^{3-\alpha} b_{1}dx \\ &-6\int (b_{1})^{2}\partial _{2}^{3} b_{1}u\cdot \nabla \partial _{2}^{3} b_{1}dx+6\int (b_{1})^{2}\partial _{2}^{3} b_{1}\partial _{2}^{3} (b \cdot \nabla u_{1})dx \\ &+6\int (b_{1})^{2}\partial _{2}^{3} b_{1}\partial _{2}^{3} \partial _{1} u_{1}dx-6\int b_{1}(u\cdot \nabla b_{1})(\partial _{2}^{3} b_{1})^{2}dx+6 \int b_{1}(b\cdot \nabla u_{1})(\partial _{2}^{3} b_{1})^{2}dx \\ =&N_{1}+N_{2}+\cdots +N_{7}. \end{aligned}$$

According to $\nabla \cdot u=0$, we obtain

$$\begin{aligned} N_{3}+N_{6}=&-6\int (b_{1})^{2}\partial _{2}^{3} b_{1}u\cdot \nabla \partial _{2}^{3} b_{1}dx-6\int b_{1}(u\cdot \nabla b_{1})(\partial _{2}^{3} b_{1})^{2}dx \\ =&-3\int u\cdot \nabla (\partial _{2}^{3} b_{1})^{2}(b_{1})^{2}dx-3 \int u\cdot \nabla (b_{1})^{2}(\partial _{2}^{3} b_{1})^{2}dx \\ =&-3\int u\cdot \nabla ((\partial _{2}^{3} b_{1})^{2}(b_{1})^{2})dx \\ =&0. \end{aligned}$$

Thanks to the Hölder inequality and Lemma 2.2, we have

$$\begin{aligned} N_{2}=&-6\sum _{1\le \alpha \le 3}C_{3}^{\alpha}\int (b_{1})^{2} \partial _{2}^{3} b_{1}\partial _{2}^{\alpha }u\cdot \nabla \partial _{2}^{3- \alpha} b_{1}dx \\ =&-6\sum _{1\le \alpha \le 3}C_{3}^{\alpha}\int (b_{1})^{2}\partial _{2}^{3} b_{1}\partial _{2}^{\alpha }u_{1}\partial _{1} \partial _{2}^{3- \alpha} b_{1}dx\\ &-6\sum _{1\le \alpha \le 3}C_{3}^{\alpha}\int (b_{1})^{2} \partial _{2}^{3} b_{1}\partial _{2}^{\alpha }u_{2}\partial _{2}^{4- \alpha} b_{1}dx \\ =&-18\int (b_{1})^{2}\partial _{2}^{3} b_{1}\partial _{2} u_{1} \partial _{1} \partial _{2}^{2} b_{1}dx -18\int (b_{1})^{2}\partial _{2}^{3} b_{1}\partial _{2}^{2} u_{1}\partial _{1} \partial _{2} b_{1}dx\\ & -6 \int (b_{1})^{2}\partial _{2}^{3} b_{1}\partial _{2}^{3} u_{1} \partial _{1} b_{1}dx \\ &-18\int (b_{1})^{2}\partial _{2}^{3} b_{1}\partial _{2} u_{2} \partial _{2}^{3} b_{1}dx -18\int (b_{1})^{2}\partial _{2}^{3} b_{1} \partial _{2}^{2} u_{2}\partial _{2}^{2} b_{1}dx\\ & -6\int (b_{1})^{2} \partial _{2}^{3} b_{1}\partial _{2}^{3} u_{2}\partial _{2} b_{1}dx \\ \le &C\|b_{1}\|_{L^{\infty}}^{2}\|\partial _{2}^{3} b_{1}\|_{L^{2}}( \|\partial _{2} u_{1}\|_{L^{\infty}}\|\partial _{1} \partial _{2}^{2} b_{1} \|_{L^{2}}+\|\partial _{2}^{2} u_{1}\|_{L^{\infty}}\|\partial _{1} \partial _{2} b_{1}\|_{L^{2}} \\ &+\|\partial _{2}^{3} u_{1}\|_{L^{2}}\|\partial _{1} b_{1}\|_{L^{ \infty}}+\|\partial _{2} u_{2}\|_{L^{\infty}}\|\partial _{2}^{3} b_{1} \|_{L^{2}}+\|\partial _{2}^{2} u_{2}\|_{L^{\infty}}\|\partial _{2}^{2} b_{1}\|_{L^{2}}\\ &+\|\partial _{2}^{3} u_{2}\|_{L^{2}}\|\partial _{2} b_{1} \|_{L^{\infty}}) \\ \le &C\|b_{1}\|_{H^{1}}\|\partial _{1}b_{1}\|_{H^{1}}\|b\|_{H^{3}}^{2}( \|\partial _{2}u_{1}\|_{H^{3}}+\|\partial _{1}u\|_{H^{2}}) \\ \le &C\|b\|_{H^{3}}^{3}(\|b_{2}\|_{H^{3}}^{2}+\|\partial _{2}u_{1}\|_{H^{3}}^{2}+ \|\partial _{1}u\|_{H^{2}}^{2}). \end{aligned}$$

By the Leibniz formula and Lemma 2.2, we infer that

$$\begin{aligned} N_{4}=&6\int (b_{1})^{2}\partial _{2}^{3} b_{1}\partial _{2}^{3} (b \cdot \nabla u_{1})dx \\ =&6\sum _{0\le \alpha \le 3}C_{3}^{\alpha}\int (b_{1})^{2}\partial _{2}^{3} b_{1}\partial _{2}^{\alpha }b\cdot \nabla \partial _{2}^{3-\alpha}u_{1}dx \\ =&6\int (b_{1})^{2}\partial _{2}^{3} b_{1} b\cdot \nabla \partial _{2}^{3}u_{1}dx +18\int (b_{1})^{2}\partial _{2}^{3} b_{1}\partial _{2} b\cdot \nabla \partial _{2}^{2}u_{1}dx \\ &+18\int (b_{1})^{2}\partial _{2}^{3} b_{1}\partial _{2}^{2} b\cdot \nabla \partial _{2}u_{1}dx +6\int (b_{1})^{2}\partial _{2}^{3} b_{1} \partial _{2}^{3} b\cdot \nabla u_{1}dx \\ =&6\int (b_{1})^{2}\partial _{2}^{3} b_{1} b\cdot \nabla \partial _{2}^{3}u_{1}dx +18\int (b_{1})^{2}\partial _{2}^{3} b_{1}\partial _{2} b\cdot \nabla \partial _{2}^{2}u_{1}dx \\ &+18\int (b_{1})^{2}\partial _{2}^{3} b_{1}\partial _{2}^{2} b\cdot \nabla \partial _{2}u_{1}dx +6\int (b_{1})^{2}\partial _{2}^{3} b_{1} \partial _{2}^{3} b_{1}\partial _{1}u_{1}dx\\ &+6\int (b_{1})^{2} \partial _{2}^{3} b_{1}\partial _{2}^{3} b_{2}\partial _{2} u_{1}dx \\ \le &C\|b_{1}\|_{L^{\infty}}^{2}\|\partial _{2}^{3} b_{1}\|_{L^{2}}( \|b\|_{L^{\infty}}\|\nabla \partial _{2}^{3}u_{1}\|_{L^{2}}+\| \partial _{2} b\|_{L^{4}}\|\nabla \partial _{2}^{2}u_{1}\|_{L^{4}} \\ &+\|\partial _{2}^{2} b\|_{L^{4}}\|\nabla \partial _{2}u_{1}\|_{L^{4}}+ \|\partial _{2}^{3} b_{1}\|_{L^{2}}\|\partial _{1}u_{1}\|_{L^{\infty}}+ \|\partial _{2}^{3} b_{2}\|_{L^{2}}\|\partial _{2} u_{1}\|_{L^{\infty}}) \\ \le &C\|b_{1}\|_{H^{1}}\|\partial _{1}b_{1}\|_{H^{1}}\|b\|_{H^{3}}^{2}( \|\partial _{2}u_{1}\|_{H^{3}}^{2}+\|\partial _{1}u\|_{H^{2}}^{2}) \\ \le &C\|b\|_{H^{3}}^{3}(\|b_{2}\|_{H^{3}}^{2}+\|\partial _{2}u_{1}\|_{H^{3}}^{2}+ \|\partial _{1}u\|_{H^{2}}^{2}). \end{aligned}$$

We infer from the Hölder inequality, Lemma 2.2 and the Sobolev inequality

$$\begin{aligned} N_{5}=&6\int (b_{1})^{2}\partial _{2}^{3} b_{1}\partial _{2}^{3} \partial _{1} u_{1}dx \\ \le &C\|b_{1}\|_{L^{\infty}}^{2}\|\partial _{2}^{3} b_{1}\|_{L^{2}}\| \partial _{2}^{3} \partial _{1} u_{1}\|_{L^{2}} \\ \le &C\|b_{1}\|_{H^{1}}\|\partial _{1}b_{1}\|_{H^{1}}\|\partial _{2}^{3} b_{1}\|_{L^{2}}\|\partial _{2}^{3} \partial _{1} u_{1}\|_{L^{2}} \\ \le &C\|b\|_{H^{3}}^{2}(\|b_{2}\|_{H^{3}}^{2}+\|\partial _{2}u_{1}\|_{H^{3}}^{2}). \end{aligned}$$

Reasoning with the same method yields

$$\begin{aligned} N_{7}=&6\int b_{1}(b\cdot \nabla u_{1})(\partial _{2}^{3} b_{1})^{2}dx \\ =&6\int b_{1}b_{1}\partial _{1}u_{1}(\partial _{2}^{3} b_{1})^{2}dx+6 \int b_{1}b_{2}\partial _{2} u_{1}(\partial _{2}^{3} b_{1})^{2}dx \\ \le &C\|b_{1}\|_{L^{\infty}}\|\partial _{2}^{3} b_{1}\|_{L^{2}}^{2}( \|b_{1}\|_{L^{\infty}}\|\partial _{1}u_{1}\|_{L^{\infty}}+ \|b_{2}\|_{L^{ \infty}}\|\partial _{2} u_{1}\|_{L^{\infty}}) \\ \le &C\|b_{1}\|_{H^{1}}^{\frac{1}{2}}\|\partial _{1}b_{1}\|_{H^{1}}^{ \frac{1}{2}}\|b\|_{H^{3}}^{2} (\|b_{1}\|_{H^{1}}^{\frac{1}{2}}\| \partial _{1}b_{1}\|_{H^{1}}^{\frac{1}{2}}\|\partial _{1}u\|_{H^{2}}+ \|b_{2}\|_{H^{2}}\|\partial _{2}u_{1}\|_{H^{2}}) \\ \le &C\|b\|_{H^{3}}^{3}(\|b_{2}\|_{H^{3}}^{2}+\|\partial _{2}u_{1}\|_{H^{3}}^{2}+ \|\partial _{1}u\|_{H^{2}}^{2}). \end{aligned}$$

Collecting all the estimate from $N_{1}$ to $N_{7}$ yields

$$\begin{aligned} J_{22}\le -3\frac{d}{dt}\int (b_{1})^{2}(\partial _{2}^{3}b_{1})^{2}+C( \|b\|_{H^{3}}^{2}+\|b\|_{H^{3}}^{3})(\|b_{2}\|_{H^{3}}^{2}+\| \partial _{2}u_{1}\|_{H^{3}}^{2}+\|\partial _{1}u\|_{H^{2}}^{2}). \end{aligned}$$

(3.6)

With the help of $\nabla \cdot u=0$ and integration by parts, we infer that

$$\begin{aligned} J_{23}=&6\int b_{1}\partial _{2}^{3} b_{1}\partial _{2}^{2} u_{2} \partial _{2}^{2} b_{1}dx \\ =&3\int b_{1}\partial _{2}^{2} u_{2}\partial _{2}(\partial _{2}^{2} b_{1})^{2}dx=-3 \int b_{1}\partial _{2}\partial _{1}u_{1}\partial _{2}(\partial _{2}^{2} b_{1})^{2}dx \\ =&3\int \partial _{2}b_{1}\partial _{2}\partial _{1}u_{1}(\partial _{2}^{2} b_{1})^{2}dx+3\int b_{1}\partial _{2}^{2}\partial _{1}u_{1}(\partial _{2}^{2} b_{1})^{2}dx \\ =&-3\int \partial _{2}\partial _{1}b_{1}\partial _{2}u_{1}(\partial _{2}^{2} b_{1})^{2}dx -6\int \partial _{2}b_{1}\partial _{2}u_{1}\partial _{2}^{2} b_{1}\partial _{1}\partial _{2}^{2} b_{1}dx \\ &-3\int \partial _{1}b_{1}\partial _{2}^{2}u_{1}(\partial _{2}^{2} b_{1})^{2}dx -6\int b_{1}\partial _{2}^{2}u_{1}\partial _{2}^{2} b_{1}\partial _{1} \partial _{2}^{2} b_{1}dx \\ \le &C\|\partial _{2}\partial _{1}b_{1}\|_{L^{2}}\|\partial _{2}u_{1} \|_{L^{\infty}}\|\partial _{2}^{2} b_{1}\|_{L^{4}}^{2}+C\|\partial _{2}b_{1} \|_{L^{4}}\|\partial _{2}u_{1}\|_{L^{\infty}}\|\partial _{2}^{2} b_{1} \|_{L^{4}}\|\partial _{1}\partial _{2}^{2} b_{1}\|_{L^{2}} \\ &+C\|\partial _{1}b_{1}\|_{L^{\infty}}\|\partial _{2}^{2}u_{1}\|_{L^{2}} \|\partial _{2}^{2} b_{1}\|_{L^{4}}^{2}+C\|b_{1}\|_{L^{\infty}}\| \partial _{2}^{2}u_{1}\|_{L^{4}}\|\partial _{2}^{2} b_{1}\|_{L^{4}}\| \partial _{1}\partial _{2}^{2} b_{1}\|_{L^{2}} \\ \le &C\|b\|_{H^{3}}^{2}(\|b_{2}\|_{H^{3}}^{2}+\|\partial _{2} u_{1}\|_{H^{3}}^{2}). \end{aligned}$$

Making use of the fact that $\partial _{2}u_{2}=-\partial _{1}u_{1}$ and integrating by parts, one gets

$$\begin{aligned} J_{24}=&2\int b_{1}\partial _{2}^{3} b_{1}\partial _{2}^{3} u_{2} \partial _{2} b_{1}dx=-2\int b_{1}\partial _{2}^{3} b_{1}\partial _{2}^{2} \partial _{1} u_{1}\partial _{2} b_{1}dx \\ =&2\int \partial _{1}b_{1}\partial _{2}^{3} b_{1}\partial _{2}^{2} u_{1} \partial _{2} b_{1}dx+2\int b_{1}\partial _{2}^{3} b_{1}\partial _{2}^{2} u_{1}\partial _{2} \partial _{1} b_{1}dx+2\int b_{1}\partial _{2}^{3} \partial _{1} b_{1}\partial _{2}^{2} u_{1}\partial _{2} b_{1}dx \\ =&2\int \partial _{1}b_{1}\partial _{2}^{3} b_{1}\partial _{2}^{2} u_{1} \partial _{2} b_{1}dx+2\int b_{1}\partial _{2}^{3} b_{1}\partial _{2}^{2} u_{1}\partial _{2} \partial _{1} b_{1}dx \\ &-2\int \partial _{2}b_{1}\partial _{2}^{2}\partial _{1} b_{1} \partial _{2}^{2} u_{1}\partial _{2} b_{1}dx-2\int b_{1}\partial _{2}^{2} \partial _{1} b_{1}\partial _{2}^{3} u_{1}\partial _{2} b_{1}dx\\ &-2 \int b_{1}\partial _{2}^{2}\partial _{1} b_{1}\partial _{2}^{2} u_{1} \partial _{2}^{2} b_{1}dx \\ \le &C\|\partial _{2}^{3} b_{1}\|_{L^{2}}\|\partial _{2}^{2} u_{1}\|_{L^{4}}( \|\partial _{1}b_{1}\|_{L^{\infty}}\|\partial _{2} b_{1}\|_{L^{4}}+\|b_{1} \|_{L^{\infty}}\|\partial _{2} \partial _{1} b_{1}\|_{L^{4}}) \\ +&C\|\partial _{2}^{2}\partial _{1} b_{1}\|_{L^{2}}(\|\partial _{2}b_{1} \|_{L^{4}}^{2}\|\partial _{2}^{2} u_{1}\|_{L^{\infty}}+\|b_{1}\|_{L^{ \infty}}\|\partial _{2}^{3} u_{1}\|_{L^{2}}\|\partial _{2} b_{1}\|_{L^{ \infty}}\\ &+ \|b_{1}\|_{L^{\infty}}\|\partial _{2}^{2} u_{1}\|_{L^{ \infty}}\|\partial _{2}^{2} b_{1}\|_{L^{2}}) \\ \le &C\|b\|_{H^{3}}^{2}(\|b_{2}\|_{H^{3}}^{2}+\|\partial _{2} u_{1}\|_{H^{3}}^{2}). \end{aligned}$$

Thus, it reaches

$$\begin{aligned} J_{2}\le -3\frac{d}{dt}\int (b_{1})^{2}(\partial _{2}^{3}b_{1})^{2}+C( \|b\|_{H^{3}}^{2}+\|b\|_{H^{3}}^{3})(\|b_{2}\|_{H^{3}}^{2}+\| \partial _{2}u_{1}\|_{H^{3}}^{2}+\|\partial _{1}u\|_{H^{2}}^{2}). \end{aligned}$$

To deal with $J_{4}$. We split it into the following terms

$$\begin{aligned} J_{4}=&-2\int b_{1}\partial _{2}^{3} b_{1}\partial _{2}^{3}(b\cdot \nabla u_{1})dx \\ =&-2\sum _{0\le \alpha \le 3}C_{3}^{\alpha}\int b_{1}\partial _{2}^{3} b_{1}\partial _{2}^{\alpha }b\cdot \nabla \partial _{2}^{3-\alpha}u_{1}dx \\ =&-2\int b_{1}\partial _{2}^{3} b_{1} b\cdot \nabla \partial _{2}^{3}u_{1}dx -6\int b_{1}\partial _{2}^{3} b_{1}\partial _{2} b\cdot \nabla \partial _{2}^{2}u_{1}dx \\ &-6\int b_{1}\partial _{2}^{3} b_{1}\partial _{2}^{2} b\cdot \nabla \partial _{2}u_{1}dx -2\int b_{1}\partial _{2}^{3} b_{1}\partial _{2}^{3} b\cdot \nabla u_{1}dx \\ =&J_{41}+J_{42}+J_{43}+J_{44}. \end{aligned}$$

Thanks to the Hölder inequality and Lemma 2.2, we obtain

$$\begin{aligned} J_{41}=&-2\int b_{1}\partial _{2}^{3} b_{1} b\cdot \nabla \partial _{2}^{3}u_{1}dx \\ \le &C\|b_{1}\|_{L^{\infty}}\|\partial _{2}^{3} b_{1}\|_{L^{2}} \|b\|_{L^{ \infty}}\|\nabla \partial _{2}^{3}u_{1}\|_{L^{2}} \\ \le &C\|b_{1}\|_{H^{1}}^{\frac{1}{2}}\|\partial _{1}b_{1}\|_{H^{1}}^{ \frac{1}{2}} \|b\|_{H^{1}}^{\frac{1}{2}}\|\partial _{1}b\|_{H^{1}}^{ \frac{1}{2}}\|\partial _{2}^{3} b_{1}\|_{L^{2}} \|\nabla \partial _{2}^{3}u_{1} \|_{L^{2}} \\ \le &C\|b\|_{H^{3}}^{2}(\|b_{2}\|_{H^{3}}^{2}+\|\partial _{2}u_{1}\|_{H^{3}}^{2}). \end{aligned}$$

For $J_{42}$, we deduce from integration by parts, Hölder’s inequality and Sobolev’s inequality

$$\begin{aligned} J_{42}=&-6\int b_{1}\partial _{2}^{3} b_{1}\partial _{2} b\cdot \nabla \partial _{2}^{2}u_{1}dx \\ =&-6\int b_{1}\partial _{2}^{3} b_{1}\partial _{2} b_{1}\partial _{1} \partial _{2}^{2}u_{1}dx-6\int b_{1}\partial _{2}^{3} b_{1}\partial _{2} b_{2}\partial _{2}^{3}u_{1}dx \\ =&6\int \partial _{1}b_{1}\partial _{2}^{3} b_{1}\partial _{2} b_{1} \partial _{2}^{2}u_{1}dx +6\int b_{1}\partial _{2}^{3}\partial _{1} b_{1} \partial _{2} b_{1}\partial _{2}^{2}u_{1}dx \\ &+6\int b_{1}\partial _{2}^{3} b_{1}\partial _{2} \partial _{1}b_{1} \partial _{2}^{2}u_{1}dx -6\int b_{1}\partial _{2}^{3} b_{1}\partial _{2} b_{2}\partial _{2}^{3}u_{1}dx \\ =&6\int \partial _{1}b_{1}\partial _{2}^{3} b_{1}\partial _{2} b_{1} \partial _{2}^{2}u_{1}dx -6\int \partial _{2}b_{1}\partial _{2}^{2} \partial _{1} b_{1}\partial _{2} b_{1}\partial _{2}^{2}u_{1}dx -6 \int b_{1}\partial _{2}^{2}\partial _{1} b_{1}\partial _{2}^{2} b_{1} \partial _{2}^{2}u_{1}dx \\ &-6\int b_{1}\partial _{2}^{2}\partial _{1} b_{1}\partial _{2} b_{1} \partial _{2}^{3}u_{1}dx +6\int b_{1}\partial _{2}^{3} b_{1}\partial _{2} \partial _{1}b_{1} \partial _{2}^{2}u_{1}dx -6\int b_{1}\partial _{2}^{3} b_{1}\partial _{2} b_{2}\partial _{2}^{3}u_{1}dx \\ \le &C\|\partial _{1}b_{1}\|_{L^{\infty}}\|\partial _{2}^{3} b_{1}\|_{L^{2}} \|\partial _{2} b_{1}\|_{L^{4}} \|\partial _{2}^{2}u_{1}\|_{L^{4}}+C \|\partial _{2}b_{1}\|_{L^{4}}^{2}\|\partial _{2}^{2}\partial _{1} b_{1} \|_{L^{2}}\|\partial _{2}^{2}u_{1}\|_{L^{\infty}} \\ &+C\|b_{1}\|_{L^{\infty}}\|\partial _{2}^{2}\partial _{1} b_{1}\|_{L^{2}}( \|\partial _{2}^{2} b_{1}\|_{L^{4}}\|\partial _{2}^{2}u_{1}\|_{L^{4}}+ \|\partial _{2} b_{1}\|_{L^{\infty}}\|\partial _{2}^{3}u_{1}\|_{L^{2}}) \\ &+C\|b_{1}\|_{L^{\infty}}\|\partial _{2}^{3} b_{1}\|_{L^{2}}(\| \partial _{2} \partial _{1}b_{1}\|_{L^{4}} \|\partial _{2}^{2}u_{1}\|_{L^{4}}+ \|\partial _{2} b_{2}\|_{L^{\infty}}\|\partial _{2}^{3}u_{1}\|_{L^{2}}) \\ \le &C\|b\|_{H^{3}}^{2}(\|b_{2}\|_{H^{3}}^{2}+\|\partial _{2}u_{1}\|_{H^{3}}^{2}). \end{aligned}$$

With the help of integration by parts and Hölder’s inequality, it reaches

$$\begin{aligned} J_{43}=&-6\int b_{1}\partial _{2}^{3} b_{1}\partial _{2}^{2} b\cdot \nabla \partial _{2}u_{1}dx \\ =&-6\int b_{1}\partial _{2}^{3} b_{1}\partial _{2}^{2} b_{1}\partial _{1} \partial _{2}u_{1}dx-6\int b_{1}\partial _{2}^{3} b_{1}\partial _{2}^{2} b_{2} \partial _{2}^{2}u_{1}dx \\ =&-3\int b_{1}\partial _{1} \partial _{2}u_{1}\partial _{2}(\partial _{2}^{2} b_{1})^{2}dx-6\int b_{1}\partial _{2}^{3} b_{1}\partial _{2}^{2} b_{2} \partial _{2}^{2}u_{1}dx \\ =&3\int \partial _{2}b_{1}\partial _{1} \partial _{2}u_{1}(\partial _{2}^{2} b_{1})^{2}dx +3\int b_{1}\partial _{1} \partial _{2}^{2}u_{1}( \partial _{2}^{2} b_{1})^{2}dx -6\int b_{1}\partial _{2}^{3} b_{1} \partial _{2}^{2} b_{2} \partial _{2}^{2}u_{1}dx \\ =&-3\int \partial _{2}\partial _{1}b_{1} \partial _{2}u_{1}(\partial _{2}^{2} b_{1})^{2}dx -6\int \partial _{2}b_{1}\partial _{2}u_{1}\partial _{2}^{2} b_{1}\partial _{1}\partial _{2}^{2} b_{1}dx \\ &-3\int \partial _{1}b_{1} \partial _{2}^{2}u_{1}(\partial _{2}^{2} b_{1})^{2}dx -6\int b_{1}\partial _{2}^{2}u_{1}\partial _{2}^{2} b_{1}\partial _{1} \partial _{2}^{2} b_{1} dx -6\int b_{1}\partial _{2}^{3} b_{1} \partial _{2}^{2} b_{2} \partial _{2}^{2}u_{1}dx \\ \le &C\|\partial _{2}u_{1}\|_{L^{\infty}}\|\partial _{2}^{2} b_{1}\|_{L^{4}}( \|\partial _{2}^{2} b_{1}\|_{L^{4}}\|\partial _{2}\partial _{1}b_{1} \|_{L^{2}}+\|\partial _{2} b_{1}\|_{L^{4}}\|\partial _{2}^{2} \partial _{1}b_{1}\|_{L^{2}}) \\ &+C\|\partial _{1}b_{1}\|_{L^{\infty}}\|\partial _{2}^{2}u_{1}\|_{L^{2}} \|\partial _{2}^{2} b_{1}\|_{L^{4}}^{2} +C \|b_{1}\|_{L^{\infty}}\| \partial _{2}^{2}u_{1}\|_{L^{4}}\|\partial _{2}^{2} b_{1}\|_{L^{4}}\| \partial _{1}\partial _{2}^{2} b_{1}\|_{L^{2}} \\ &+C \|b_{1}\|_{L^{\infty}}\|\partial _{2}^{3} b_{1}\|_{L^{2}}\| \partial _{2}^{2} b_{2}\|_{L^{4}} \|\partial _{2}^{2}u_{1}\|_{L^{4}} \\ \le &C\|b\|_{H^{3}}^{2}(\|b_{2}\|_{H^{3}}^{2}+\|\partial _{2}u_{1}\|_{H^{3}}^{2}). \end{aligned}$$

From the Hölder’s inequality and the estimate of $J_{22}$ in (3.6), it follows that

$$\begin{aligned} J_{44}=&-2\int b_{1}\partial _{2}^{3} b_{1}\partial _{2}^{3} b\cdot \nabla u_{1}dx \\ =&-2\int b_{1}\partial _{2}^{3} b_{1}\partial _{2}^{3} b_{1}\partial _{1}u_{1}dx-2 \int b_{1}\partial _{2}^{3} b_{1}\partial _{2}^{3} b_{2}\partial _{2}u_{1}dx \\ \le &-2\int b_{1}\partial _{2}^{3} b_{1}\partial _{2}^{3} b_{1} \partial _{1}u_{1}dx+C\|b_{1}\|_{L^{\infty}}\|\partial _{2}^{3} b_{1} \|_{L^{2}}\|\partial _{2}^{3} b_{2}\|_{L^{2}}\|\partial _{2}u_{1}\|_{L^{ \infty}} \\ \le &C\|b\|_{H^{3}}^{2}(\|b_{2}\|_{H^{3}}^{2}+\|\partial _{2}u_{1}\|_{H^{3}}^{2})+ \frac{1}{3}J_{22} \\ \le &-\frac{d}{dt}\int (b_{1})^{2}(\partial _{2}^{3}b_{1})^{2}+C(\|b \|_{H^{3}}^{2}+\|b\|_{H^{3}}^{3})(\|b_{2}\|_{H^{3}}^{2}+\|\partial _{2}u_{1} \|_{H^{3}}^{2}+\|\partial _{1}u\|_{H^{2}}^{2}). \end{aligned}$$

So,

$$\begin{aligned} J_{4}\le -\frac{d}{dt}\int (b_{1})^{2}(\partial _{2}^{3}b_{1})^{2}+C( \|b\|_{H^{3}}^{2}+\|b\|_{H^{3}}^{3})(\|b_{2}\|_{H^{3}}^{2}+\| \partial _{2}u_{1}\|_{H^{3}}^{2}+\|\partial _{1}u\|_{H^{2}}^{2}). \end{aligned}$$

Integration by parts and the Hölder inequality lead to

$$\begin{aligned} J_{5}=&-2\int b_{1}\partial _{2}^{3} b_{1}\partial _{2}^{3}\partial _{1} u_{1}dx \\ =&2\int \partial _{1}b_{1}\partial _{2}^{3} b_{1}\partial _{2}^{3} u_{1}dx+2 \int b_{1}\partial _{2}^{3} \partial _{1}b_{1}\partial _{2}^{3}u_{1}dx \\ =&2\int \partial _{1}b_{1}\partial _{2}^{3} b_{1}\partial _{2}^{3}u_{1}dx-2 \int \partial _{2}b_{1}\partial _{2}^{2} \partial _{1}b_{1}\partial _{2}^{3}u_{1}dx-2 \int b_{1}\partial _{2}^{2} \partial _{1}b_{1}\partial _{2}^{4}u_{1}dx \\ \le &C\|\partial _{2}^{3}u_{1}\|_{L^{2}}(\|\partial _{1}b_{1}\|_{L^{ \infty}}\|\partial _{2}^{3} b_{1}\|_{L^{2}}+\|\partial _{2}b_{1}\|_{L^{ \infty}}\|\partial _{2}^{2} \partial _{1}b_{1}\|_{L^{2}})\\ &+C\|b_{1}\|_{L^{ \infty}}\|\partial _{2}^{2} \partial _{1}b_{1}\|_{L^{2}}\|\partial _{2}^{4}u_{1} \|_{L^{2}} \\ \le &C\|b\|_{H^{3}}(\|b_{2}\|_{H^{3}}^{2}+\|\partial _{2}u_{1}\|_{H^{3}}^{2}). \end{aligned}$$

Thanks to the Hölder inequality and (3.6), we find

$$\begin{aligned} J_{7}=&-\int (b\cdot \nabla u_{1})(\partial _{2}^{3} b_{1})^{2}dx \\ =&-\int b_{1}\partial _{1}u_{1}(\partial _{2}^{3} b_{1})^{2}dx-\int b_{2} \partial _{2}u_{1}(\partial _{2}^{3} b_{1})^{2}dx \\ \le &\|b_{2}\|_{L^{\infty}}\|\partial _{2}u_{1}\|_{L^{\infty}}\| \partial _{2}^{3} b_{1}\|_{L^{2}}^{2}-\int b_{1}\partial _{1}u_{1}( \partial _{2}^{3} b_{1})^{2}dx \\ \le &C\|b\|_{H^{3}}^{2}(\|b_{2}\|_{H^{3}}^{2}+\|\partial _{2}u_{1}\|_{H^{3}}^{2})+ \frac{1}{6}J_{22} \\ \le &-\frac{1}{2}\frac{d}{dt}\int (b_{1})^{2}(\partial _{2}^{3}b_{1})^{2}dx+C( \|b\|_{H^{3}}^{2}+\|b\|_{H^{3}}^{3})(\|b_{2}\|_{H^{3}}^{2}+\| \partial _{2}u_{1}\|_{H^{3}}^{2}+\|\partial _{1}u\|_{H^{2}}^{2}). \end{aligned}$$

Collecting all the estimates from $J_{1}$ to $J_{7}$, we get

$$ \begin{aligned} \int \partial _{1}u_{1}(\partial _{2}^{3}b_{1})^{2}dx\le & \frac{d}{dt}\int b_{1}(\partial _{2}^{3}b_{1})^{2}dx-\frac{9}{2} \frac{d}{dt}\int (b_{1})^{2}(\partial _{2}^{3}b_{1})^{2}dx \\ &+C(\|b\|_{H^{3}}+\|b\|_{H^{3}}^{2}+\|b\|_{H^{3}}^{3})\\ &\times(\|b_{2}\|_{H^{3}}^{2}+ \|\partial _{2}u_{1}\|_{H^{3}}^{2}+\|\partial _{1}u\|_{H^{2}}^{2}). \end{aligned} $$

(3.7)

Substituting (3.7) into (3.4), it shows that

$$\begin{aligned} \frac{1}{2}&\frac{d}{dt}\|(u,b)\|_{H^{3}}^{2}+\|\partial _{2}u_{1}\|_{H^{3}}^{2}+ \|b_{2}\|_{H^{3}}^{2} \\ \le &4\frac{d}{dt}\int b_{1}(\partial _{2}^{3}b_{1})^{2}dx-18 \frac{d}{dt}\int (b_{1})^{2}(\partial _{2}^{3}b_{1})^{2}dx \\ &+C(\|b\|_{H^{3}}+\|b\|_{H^{3}}^{2}+\|b\|_{H^{3}}^{3})(\|b_{2}\|_{H^{3}}^{2}+ \|\partial _{2}u_{1}\|_{H^{3}}^{2}+\|\partial _{1}u\|_{H^{2}}^{2}), \end{aligned}$$

which implies

$$\begin{aligned} \|(u,b)\|_{H^{3}}^{2}&+2\int _{0}^{t}(\|\partial _{2}u_{1}\|_{H^{3}}^{2}+ \|b_{2}\|_{H^{3}}^{2}) \\ \le &\|(u_{0},b_{0})\|_{H^{3}}^{2}+8\int b_{1}(\partial _{2}^{3}b_{1})^{2}dx-8 \int b_{1}(x,0)(\partial _{2}^{3}b_{1})^{2}(x,0)dx \\ &-36\int (b_{1})^{2}(\partial _{2}^{3}b_{1})^{2}dx+36\int (b_{1})^{2}(x,0)( \partial _{2}^{3}b_{1})^{2}(x,0)dx \\ &+C\int _{0}^{t}(\|b\|_{H^{3}}+\|b\|_{H^{3}}^{2}+\|b\|_{H^{3}}^{3})( \|b_{2}\|_{H^{3}}^{2}+\|\partial _{2}u_{1}\|_{H^{3}}^{2}+\|\partial _{1}u \|_{H^{2}}^{2}) \\ \le &\|(u_{0},b_{0})\|_{H^{3}}^{2}+C(\|b_{0}\|_{L^{\infty}}+\|b_{0}\|_{L^{ \infty}}^{2})\|\partial _{2}^{3}b_{0}\|_{L^{2}}^{2}\\ &+ C(\|b_{1}\|_{L^{ \infty}}+\|b_{1}\|_{L^{\infty}}^{2})\|\partial _{2}^{3}b_{1}\|_{L^{2}}^{2} \\ &+C\sup _{0\le \tau \le t}(\|b\|_{H^{3}}+\|b\|_{H^{3}}^{2}+\|b\|_{H^{3}}^{3}) \int _{0}^{t}(\|b_{2}\|_{H^{3}}^{2}+\|\partial _{2}u_{1}\|_{H^{3}}^{2}+ \|\partial _{1}u\|_{H^{2}}^{2}). \end{aligned}$$

Therefore,

$$ \begin{aligned} E_{1}(t)\le & C(E_{1}(0)+E_{1}^{\frac{3}{2}}(0)+E_{1}^{2}(0))+C(E_{1}^{ \frac{3}{2}}(t)+E_{2}^{\frac{3}{2}}(t)) \\ &+C(E_{1}^{2}(t)+E_{2}^{2}(t))+C(E_{1}^{\frac{5}{2}}(t)+E_{2}^{ \frac{5}{2}}(t)). \end{aligned} $$

(3.8)

Hence we finish the proof of Proposition 3.1. □

In order to close the energy estimate, we need the following result.

Proposition 3.2

Suppose that $E_{1}(t)$ and $E_{2}(t)$ are defined as in (3.1) and (3.2). Then exists a positive constant $C>0$, such that

$$\begin{aligned} E_{2}(t)\le CE_{1}(0)+CE_{1}(t)+CE_{1}^{\frac{3}{2}}(t)+CE_{2}^{ \frac{3}{2}}(t). \end{aligned}$$

Proof

We first rewrite the second equation of the system (1.3) as

$${\partial }_{1}u={\partial }_{t}b+u\cdot \mathrm{\nabla }b+\left(\begin{array}{c}0\\ b_2\end{array}\right)-b\cdot \mathrm{\nabla }u.$$

(3.9)

Taking the $L^{2}$-inner product to (3.9) with $\partial _{1}u$ yields

$$\begin{array}{rl}{\parallel {\partial }_{1}u\parallel }_{L^2}^2=& 〈{\partial }_{t}b,{\partial }_{1}u〉+〈u\cdot \mathrm{\nabla }b,{\partial }_{1}u〉+〈b_2,{\partial }_1{u}_2〉-〈b\cdot \mathrm{\nabla }u,{\partial }_{1}u〉\\ =& \frac{d}{dt}〈b,{\partial }_{1}u〉-〈b,{\partial }_1(-u\cdot \mathrm{\nabla }u-\mathrm{\nabla }\pi +(\begin{array}{c}{\partial }_2^2{u}_1\\ 0\end{array})+b\cdot \mathrm{\nabla }b+{\partial }_{1}b)〉\\ & +〈u\cdot \mathrm{\nabla }b,{\partial }_{1}u〉+〈b_2,{\partial }_1{u}_2〉-〈b\cdot \mathrm{\nabla }u,{\partial }_{1}u〉\\ =& \frac{d}{dt}〈b,{\partial }_{1}u〉+{\parallel {\partial }_{1}b\parallel }_{L^2}^2+〈b,{\partial }_1(u\cdot \mathrm{\nabla }u)〉-〈b_1,{\partial }_1{\partial }_2^2{u}_1〉\\ & -〈b,{\partial }_1(b\cdot \mathrm{\nabla }b)〉+〈u\cdot \mathrm{\nabla }b,{\partial }_{1}u〉+〈b_2,{\partial }_1{u}_2〉-〈b\cdot \mathrm{\nabla }u,{\partial }_{1}u〉\\ =& \frac{d}{dt}〈b,{\partial }_{1}u〉+{\parallel {\partial }_{1}b\parallel }_{L^2}^2+M_1+M_2+\cdots +M_6,\end{array}$$

where we use the equation

$${\partial }_{t}u=-u\cdot \mathrm{\nabla }u-\mathrm{\nabla }\pi +\left(\begin{array}{c}{\partial }_2^2{u}_1\\ 0\end{array}\right)+b\cdot \mathrm{\nabla }b+{\partial }_{1}b.$$

Taking advantage of integration by parts and the Hölder inequality, we see that

$$\begin{aligned} M_{1}=&\int b\cdot \partial _{1}(u\cdot \nabla u)dx=-\int \partial _{1} b\cdot (u\cdot \nabla u)dx \\ =&-\int \partial _{1}b\cdot u_{1}\partial _{1} udx-\int \partial _{1}b_{1} u_{2}\partial _{2} u_{1}dx-\int \partial _{1}b_{2} u_{2}\partial _{2} u_{2}dx \\ \le &\|\partial _{1}b\|_{L^{2}}\| u_{1}\|_{L^{\infty}}\|\partial _{1} u \|_{L^{2}}+\| \partial _{1}b_{1}\|_{L^{2}} \|u_{2}\|_{L^{\infty}}\| \partial _{2} u_{1}\|_{L^{2}}+\| \partial _{1}b_{2}\|_{L^{2}}\| u_{2} \|_{L^{\infty}}\|\partial _{2} u_{2}\|_{L^{2}} \\ \le &C\|u\|_{H^{2}}(\|b_{2}\|_{H^{1}}^{2}+\|\partial _{2}u_{1}\|_{L^{2}}^{2}+ \|\partial _{1}u\|_{L^{2}}^{2}). \end{aligned}$$

It is easy to see

$$\begin{aligned} M_{2}=&\int \partial _{1} b_{1}\partial _{2}^{2}u_{1}dx \\ \le &\|\partial _{1} b_{1}\|_{L^{2}}\|\partial _{2}^{2}u_{1}\|_{L^{2}} \\ \le &C(\|b_{2}\|_{H^{1}}^{2}+\|\partial _{2}u_{1}\|_{H^{1}}^{2}). \end{aligned}$$

From integration by parts and Hölder’s inequality, it follows

$$\begin{aligned} M_{3}=&-\int b\cdot \partial _{1}(b\cdot \nabla b)dx=\int \partial _{1}b \cdot (b\cdot \nabla b)dx \\ =&\int \partial _{1}b\cdot b_{1}\partial _{1} bdx+\int \partial _{1}b \cdot b_{2}\partial _{2} bdx \\ \le &\| b_{1}\|_{L^{\infty}}\|\partial _{1} b\|_{L^{2}}^{2}+\| \partial _{1}b\|_{L^{2}}\| b_{2}\|_{L^{4}}\|\partial _{2} b\|_{L^{4}} \\ \le &C\|b\|_{H^{2}}\|b_{2}\|_{H^{1}}^{2}. \end{aligned}$$

From Lemma 2.1, we obviously deduce

$$\begin{aligned} M_{4}=&\int u\cdot \nabla b\cdot \partial _{1}udx \\ =&\int u_{1}\partial _{1} b\cdot \partial _{1}udx+\int u_{2}\partial _{2} b\cdot \partial _{1}udx \\ \le &\|u_{1}\|_{L^{\infty}}\|\partial _{1} b\|_{L^{2}}\|\partial _{1}u \|_{L^{2}}+\|u_{2}\|_{L^{2}}^{\frac{1}{2}}\|\partial _{2}u_{2}\|_{L^{2}}^{ \frac{1}{2}}\|\partial _{2} b\|_{L^{2}}^{\frac{1}{2}}\|\partial _{2} \partial _{1} b\|_{L^{2}}^{\frac{1}{2}} \|\partial _{1}u\|_{L^{2}} \\ \le &C(\|u\|_{H^{2}}+\|b\|_{H^{2}})(\|b_{2}\|_{H^{2}}^{2}+\|\partial _{1}u \|_{L^{2}}^{2}). \end{aligned}$$

Young’s inequality ensures

$$\begin{aligned} M_{5}\le &C\|b_{2}\|_{L^{2}}^{2}+\frac{1}{2}\|\partial _{1}u_{2}\|_{L^{2}}^{2}. \end{aligned}$$

Thanks to the Hölder inequality, we find

$$\begin{aligned} M_{6}=&-\int b\cdot \nabla u\cdot \partial _{1}udx \\ =&-\int b_{1}\partial _{1} u\cdot \partial _{1}udx-\int b_{2} \partial _{2} u\cdot \partial _{1}udx \\ \le &\|b_{1}\|_{L^{\infty}}\|\partial _{1} u\|_{L^{2}}^{2}+\|b_{2}\|_{L^{ \infty}}\|\partial _{2} u\|_{L^{2}}\|\partial _{1}u\|_{L^{2}} \\ \le &C(\|u\|_{H^{1}}+\|b\|_{H^{2}})(\|b_{2}\|_{H^{2}}^{2}+\|\partial _{1}u \|_{L^{2}}^{2}). \end{aligned}$$

Collecting the bounds for $M_{1}$ through $M_{6}$, one has

$$ \begin{aligned} \frac{1}{2}\|\partial _{1}u\|_{L^{2}}^{2}\le &\frac{d}{dt}\langle b, \partial _{1}u\rangle +C(\|b_{2}\|_{H^{1}}^{2}+\|\partial _{2}u_{1}\|_{H^{1}}^{2}) \\ &+C(\|u\|_{H^{2}}+\|b\|_{H^{2}})(\|b_{2}\|_{H^{2}}^{2}+\|\partial _{2}u_{1} \|_{H^{1}}^{2}+\|\partial _{1}u\|_{L^{2}}^{2}). \end{aligned} $$

(3.10)

Now we turn to establish $\dot{H}^{2}$-norm of $\partial _{1}u$. Applying $\partial _{i}^{2}$ to (3.9) and then taking $L^{2}$-inner product to (3.9) with $\partial _{i}^{2}\partial _{1}u$ to conclude

$$\begin{array}{rl}{\parallel {\partial }_1{\partial }_i^{2}u\parallel }_{L^2}^2=& 〈{\partial }_t{\partial }_i^{2}b,{\partial }_1{\partial }_i^{2}u〉+〈{\partial }_i^2(u\cdot \mathrm{\nabla }b),{\partial }_1{\partial }_i^{2}u〉+〈{\partial }_i^2{b}_2,{\partial }_1{\partial }_i^2{u}_2〉-〈{\partial }_i^2(b\cdot \mathrm{\nabla }u),{\partial }_1{\partial }_i^{2}u〉\\ =& \frac{d}{dt}〈{\partial }_i^{2}b,{\partial }_1{\partial }_i^{2}u〉-〈{\partial }_i^{2}b,{\partial }_1{\partial }_i^2(-u\cdot \mathrm{\nabla }u-\mathrm{\nabla }\pi +(\begin{array}{c}{\partial }_2^2{u}_1\\ 0\end{array})+b\cdot \mathrm{\nabla }b+{\partial }_{1}b)〉\\ & +〈{\partial }_i^2(u\cdot \mathrm{\nabla }b),{\partial }_1{\partial }_i^{2}u〉+〈{\partial }_i^2{b}_2,{\partial }_1{\partial }_i^2{u}_2〉-〈{\partial }_i^2(b\cdot \mathrm{\nabla }u),{\partial }_1{\partial }_i^{2}u〉\\ =& \frac{d}{dt}〈{\partial }_i^{2}b,{\partial }_1{\partial }_i^{2}u〉+{\parallel {\partial }_1{\partial }_i^{2}b\parallel }_{L^2}^2+〈{\partial }_i^{2}b,{\partial }_1{\partial }_i^2(u\cdot \mathrm{\nabla }u)〉-〈{\partial }_i^2{b}_1,{\partial }_1{\partial }_i^2{\partial }_2^2{u}_1〉\\ & -〈{\partial }_i^{2}b,{\partial }_1{\partial }_i^2(b\cdot \mathrm{\nabla }b)〉+〈{\partial }_i^2(u\cdot \mathrm{\nabla }b),{\partial }_1{\partial }_i^{2}u〉+〈{\partial }_i^2{b}_2,{\partial }_1{\partial }_i^2{u}_2〉\\ & -〈{\partial }_i^2(b\cdot \mathrm{\nabla }u),{\partial }_1{\partial }_i^{2}u〉\\ =& \frac{d}{dt}〈{\partial }_i^{2}b,{\partial }_1{\partial }_i^{2}u〉+{\parallel {\partial }_1{\partial }_i^{2}b\parallel }_{L^2}^2+K_1+K_2+\cdots +K_6.\end{array}$$

According to integration by parts and Leibniz’s formula, $K_{1}$ can be divided into the following terms

$$\begin{aligned} K_{1}=&\int \partial _{i}^{2}b\cdot \partial _{1}\partial _{i}^{2}(u \cdot \nabla u)dx=-\int \partial _{i}^{2}\partial _{1}b\cdot \partial _{i}^{2}(u\cdot \nabla u)dx \\ =&-\sum _{0\le \alpha \le 2}C_{2}^{\alpha}\int \partial _{i}^{2} \partial _{1}b\cdot \partial _{i}^{\alpha }u\cdot \nabla \partial _{i}^{2- \alpha}udx \\ =&-\sum _{0\le \alpha \le 2}C_{2}^{\alpha}\int \partial _{i}^{2} \partial _{1}b\cdot \partial _{i}^{\alpha }u_{1}\partial _{1} \partial _{i}^{2-\alpha}udx -\sum _{0\le \alpha \le 2}C_{2}^{\alpha} \int \partial _{i}^{2}\partial _{1}b_{1}\partial _{i}^{\alpha }u_{2} \partial _{2} \partial _{i}^{2-\alpha}u_{1}dx \\ &-\sum _{0\le \alpha \le 2}C_{2}^{\alpha}\int \partial _{i}^{2} \partial _{1}b_{2}\partial _{i}^{\alpha }u_{2}\partial _{2} \partial _{i}^{2- \alpha}u_{2}dx \\ =&K_{11}+K_{12}+K_{13}. \end{aligned}$$

Thanks to the Hölder inequality and the Sobolev inequality, we can get

$$\begin{aligned} K_{11}=&-\sum _{0\le \alpha \le 2}C_{2}^{\alpha}\int \partial _{i}^{2} \partial _{1}b\cdot \partial _{i}^{\alpha }u_{1}\partial _{1} \partial _{i}^{2-\alpha}udx \\ =&-\int \partial _{i}^{2}\partial _{1}b\cdot u_{1}\partial _{1} \partial _{i}^{2}udx -2\int \partial _{i}^{2}\partial _{1}b\cdot \partial _{i} u_{1}\partial _{1} \partial _{i}udx -\int \partial _{i}^{2} \partial _{1}b\cdot \partial _{i}^{2} u_{1}\partial _{1} udx \\ \le &\|\partial _{i}^{2}\partial _{1}b\|_{L^{2}} (\|u_{1}\|_{L^{ \infty}}\|\partial _{1} \partial _{i}^{2}u\|_{L^{2}} +2\|\partial _{i} u_{1}\|_{L^{4}}\|\partial _{1} \partial _{i}u\|_{L^{4}} +\|\partial _{i}^{2} u_{1}\|_{L^{4}}\|\partial _{1} u\|_{L^{4}}) \\ \le &C\|u\|_{H^{3}}(\|b_{2}\|_{H^{3}}^{2}+\|\partial _{1}u\|_{H^{2}}^{2}). \end{aligned}$$

Reasoning in the same way gives

$$\begin{aligned} K_{12}=&-\sum _{0\le \alpha \le 2}C_{2}^{\alpha}\int \partial _{i}^{2} \partial _{1}b_{1}\partial _{i}^{\alpha }u_{2}\partial _{2} \partial _{i}^{2- \alpha}u_{1}dx \\ \le &C\sum _{0\le \alpha \le 2}\|\partial _{i}^{2}\partial _{1}b_{1} \|_{L^{2}}\|\partial _{i}^{\alpha }u_{2}\|_{L^{4}}\|\partial _{2} \partial _{i}^{2-\alpha}u_{1}\|_{L^{4}} \\ \le &C\|u\|_{H^{3}}(\|b_{2}\|_{H^{3}}^{2}+\|\partial _{2}u_{1}\|_{H^{3}}^{2}), \end{aligned}$$

and

$$\begin{aligned} K_{13}=&-\sum _{0\le \alpha \le 2}C_{2}^{\alpha}\int \partial _{i}^{2} \partial _{1}b_{2}\partial _{i}^{\alpha }u_{2}\partial _{2} \partial _{i}^{2- \alpha}u_{2}dx \\ =&-\int \partial _{i}^{2}\partial _{1}b_{2} u_{2}\partial _{2} \partial _{i}^{2}u_{2}dx -2\int \partial _{i}^{2}\partial _{1}b_{2} \partial _{i} u_{2}\partial _{2} \partial _{i}u_{2}dx -\int \partial _{i}^{2} \partial _{1}b_{2}\partial _{i}^{2} u_{2}\partial _{2} u_{2}dx \\ \le &C\|\partial _{i}^{2}\partial _{1}b_{2}\|_{L^{2}}(\|u_{2}\|_{L^{ \infty}}\|\partial _{2} \partial _{i}^{2}u_{2}\|_{L^{2}} +\|\partial _{i} u_{2}\|_{L^{4}} \|\partial _{2} \partial _{i}u_{2}\|_{L^{4}}+\| \partial _{i}^{2} u_{2}\|_{L^{4}}\|\partial _{2} u_{2}\|_{L^{4}}) \\ \le &C\|u\|_{H^{3}}(\|b_{2}\|_{H^{3}}^{2}+\|\partial _{1}u\|_{H^{2}}^{2}). \end{aligned}$$

So, we have

$$\begin{aligned} K_{1}\le C\|u\|_{H^{3}}(\|b_{2}\|_{H^{3}}^{2}+\|\partial _{2}u_{1}\|_{H^{3}}^{2}+ \|\partial _{1}u\|_{H^{2}}^{2}). \end{aligned}$$

Integrating by parts and using the Young inequality, we get

$$\begin{aligned} K_{2}=&-\int \partial _{i}^{2}b_{1}\partial _{1}\partial _{i}^{2} \partial _{2}^{2}u_{1}dx=\int \partial _{1} \partial _{i}^{2}b_{1} \partial _{i}^{2}\partial _{2}^{2}u_{1}dx \\ \le &C(\|b_{2}\|_{H^{3}}^{2}+\|\partial _{2}u_{1}\|_{H^{3}}^{2}). \end{aligned}$$

Thanks to integration by parts and the Leibniz law, we obtain

$$\begin{aligned} K_{3}=&-\int \partial _{i}^{2}b\cdot \partial _{1}\partial _{i}^{2}(b \cdot \nabla b)dx \\ =&\int \partial _{1}\partial _{i}^{2}b\cdot \partial _{i}^{2}(b\cdot \nabla b)dx=\sum _{0\le \alpha \le 2}C_{2}^{\alpha}\int \partial _{1} \partial _{i}^{2}b\cdot \partial _{i}^{\alpha }b\cdot \nabla \partial _{i}^{2-\alpha}bdx \\ =&\sum _{0\le \alpha \le 2}C_{2}^{\alpha}\int \partial _{1}\partial _{i}^{2}b \cdot \partial _{i}^{\alpha }b_{1}\partial _{1} \partial _{i}^{2- \alpha}bdx +\sum _{0\le \alpha \le 2}C_{2}^{\alpha}\int \partial _{1} \partial _{i}^{2}b\cdot \partial _{i}^{\alpha }b_{2}\partial _{2} \partial _{i}^{2-\alpha}bdx \\ =&\int \partial _{1}\partial _{i}^{2}b\cdot b_{1}\partial _{1} \partial _{i}^{2}bdx +2\int \partial _{1}\partial _{i}^{2}b\cdot \partial _{i} b_{1}\partial _{1} \partial _{i}bdx +\int \partial _{1} \partial _{i}^{2}b\cdot \partial _{i}^{2} b_{1}\partial _{1} bdx \\ &+\int \partial _{1}\partial _{i}^{2}b\cdot b_{2}\partial _{2} \partial _{i}^{2}bdx +2\int \partial _{1}\partial _{i}^{2}b\cdot \partial _{i} b_{2}\partial _{2} \partial _{i}bdx +\int \partial _{1} \partial _{i}^{2}b\cdot \partial _{i}^{2} b_{2}\partial _{2} bdx \\ \le &C\|\partial _{1}\partial _{i}^{2}b\|_{L^{2}}(\|b_{1}\|_{L^{ \infty}}\|\partial _{1} \partial _{i}^{2}b\|_{L^{2}}+\|\partial _{i} b_{1} \|_{L^{4}}\|\partial _{1} \partial _{i}b\|_{L^{4}}+\|\partial _{i}^{2} b_{1}\|_{L^{4}}\|\partial _{1} b\|_{L^{4}} \\ &+\|b_{2}\|_{L^{\infty}}\|\partial _{2} \partial _{i}^{2}b\|_{L^{2}}+ \|\partial _{i} b_{2}\|_{L^{4}}\|\partial _{2} \partial _{i}b\|_{L^{4}} +\|\partial _{i}^{2} b_{2}\|_{L^{4}}\|\partial _{2} b\|_{L^{4}}) \\ \le &C\|b\|_{H^{3}}\|b_{2}\|_{H^{3}}^{2}. \end{aligned}$$

Using the Leibniz law again, we decompose $K_{4}$ into the following form:

$$\begin{aligned} K_{4}=&\int \partial _{i}^{2}(u\cdot \nabla b)\cdot \partial _{1} \partial _{i}^{2}udx \\ =&\sum _{0\le \alpha \le 2}C_{2}^{\alpha}\int \partial _{i}^{\alpha }u \cdot \nabla \partial _{i}^{2-\alpha} b\cdot \partial _{1}\partial _{i}^{2}udx \\ =&\sum _{0\le \alpha \le 2}C_{2}^{\alpha}\int \partial _{i}^{\alpha }u_{1} \partial _{1} \partial _{i}^{2-\alpha} b\cdot \partial _{1}\partial _{i}^{2}udx+ \sum _{0\le \alpha \le 2}C_{2}^{\alpha}\int \partial _{i}^{\alpha }u_{2} \partial _{2} \partial _{i}^{2-\alpha} b\cdot \partial _{1}\partial _{i}^{2}udx \\ =&K_{41}+K_{42}. \end{aligned}$$

By the Hölder inequality and Sobolev inequality, we thus get

$$\begin{aligned} K_{41}=&\sum _{0\le \alpha \le 2}C_{2}^{\alpha}\int \partial _{i}^{ \alpha }u_{1}\partial _{1} \partial _{i}^{2-\alpha} b\cdot \partial _{1} \partial _{i}^{2}udx \\ =&\int u_{1}\partial _{1} \partial _{i}^{2} b\cdot \partial _{1} \partial _{i}^{2}udx +2\int \partial _{i} u_{1}\partial _{1} \partial _{i} b\cdot \partial _{1}\partial _{i}^{2}udx +\int \partial _{i}^{2} u_{1}\partial _{1} b\cdot \partial _{1}\partial _{i}^{2}udx \\ \le &C\|\partial _{1}\partial _{i}^{2}u\|_{L^{2}}(\|u_{1}\|_{L^{ \infty}}\|\partial _{1} \partial _{i}^{2} b\|_{L^{2}}+\|\partial _{i} u_{1} \|_{L^{4}}\|\partial _{1} \partial _{i} b\|_{L^{4}}+\|\partial _{i}^{2} u_{1}\|_{L^{4}}\|\partial _{1} b\|_{L^{4}}) \\ \le &C\|u\|_{H^{3}}(\|b_{2}\|_{H^{3}}^{2}+\|\partial _{1}u\|_{H^{2}}^{2}). \end{aligned}$$

We first split $K_{42}$ into the following terms

$$\begin{aligned} K_{42}=&\sum _{0\le \alpha \le 2}C_{2}^{\alpha}\int \partial _{i}^{ \alpha }u_{2}\partial _{2} \partial _{i}^{2-\alpha} b\cdot \partial _{1} \partial _{i}^{2}udx \\ =&\int u_{2}\partial _{2} \partial _{i}^{2} b\cdot \partial _{1} \partial _{i}^{2}udx +2\int \partial _{i} u_{2}\partial _{2} \partial _{i} b\cdot \partial _{1}\partial _{i}^{2}udx +\int \partial _{i}^{2} u_{2}\partial _{2} b\cdot \partial _{1}\partial _{i}^{2}udx \\ =&K_{421}+K_{422}+K_{423}. \end{aligned}$$

Using integration by parts and the Hölder inequality, we can obtain

$$\begin{aligned} K_{421}=&\int u_{2}\partial _{2} \partial _{i}^{2} b\cdot \partial _{1} \partial _{i}^{2}udx \\ =&\int u_{2}\partial _{2} \partial _{i}^{2} b_{1}\partial _{1} \partial _{i}^{2}u_{1}dx+\int u_{2}\partial _{2} \partial _{i}^{2} b_{2} \partial _{1}\partial _{i}^{2}u_{2}dx \\ =&-\int \partial _{1}u_{2}\partial _{2} \partial _{i}^{2} b_{1} \partial _{i}^{2}u_{1}dx -\int u_{2}\partial _{2} \partial _{i}^{2} \partial _{1}b_{1}\partial _{i}^{2}u_{1}dx +\int u_{2}\partial _{2} \partial _{i}^{2} b_{2}\partial _{1}\partial _{i}^{2}u_{2}dx \\ =&-\int \partial _{1}u_{2}\partial _{2} \partial _{1}^{2} b_{1} \partial _{1}^{2}u_{1}dx -\int \partial _{1}u_{2} \partial _{2}^{3} b_{1} \partial _{2}^{2}u_{1}dx \\ &+\int \partial _{2}u_{2} \partial _{i}^{2} \partial _{1}b_{1} \partial _{i}^{2}u_{1}dx +\int u_{2} \partial _{i}^{2} \partial _{1}b_{1} \partial _{i}^{2}\partial _{2}u_{1}dx -\int u_{2}\partial _{i}^{2} \partial _{1} b_{1}\partial _{1}\partial _{i}^{2}u_{2}dx \\ \le &\|\partial _{1}u_{2}\|_{L^{\infty}}(\|\partial _{2} \partial _{1}^{2} b_{1}\|_{L^{2}}\|\partial _{1}^{2}u_{1}\|_{L^{2}}+\|\partial _{2}^{3} b_{1} \|_{L^{2}}\|\partial _{2}^{2}u_{1}\|_{L^{2}}) \\ &+\|\partial _{i}^{2} \partial _{1} b_{1}\|_{L^{2}}(\|\partial _{2}u_{2} \|_{L^{\infty}} \|\partial _{i}^{2}u_{1}\|_{L^{2}} +\|u_{2}\|_{L^{ \infty}}\| \partial _{i}^{2}\partial _{2}u_{1}\|_{L^{2}} +\|u_{2}\|_{L^{ \infty}}\|\partial _{1}\partial _{i}^{2}u_{2}\|_{L^{2}}) \\ \le &C(\|u\|_{H^{3}}+\|b\|_{H^{3}})(\|b_{2}\|_{H^{3}}^{2}+\|\partial _{2}u_{1} \|_{H^{3}}^{2}+\|\partial _{1}u\|_{H^{2}}^{2}). \end{aligned}$$

Due to Lemma 2.1, the last two terms in $K_{42}$, one can get the following upper bounds

$$\begin{aligned} K_{422}+K_{423}=&2\int \partial _{i} u_{2}\partial _{2} \partial _{i} b \cdot \partial _{1}\partial _{i}^{2}udx +\int \partial _{i}^{2} u_{2} \partial _{2} b\cdot \partial _{1}\partial _{i}^{2}udx \\ \le &C\|\partial _{i} u_{2}\|_{L^{2}}^{\frac{1}{2}}\|\partial _{i} \partial _{2} u_{2}\|_{L^{2}}^{\frac{1}{2}}\|\partial _{2} \partial _{i} b\|_{L^{2}}^{\frac{1}{2}}\|\partial _{2} \partial _{i} \partial _{1}b \|_{L^{2}}^{\frac{1}{2}}\|\partial _{1}\partial _{i}^{2}u\|_{L^{2}} \\ &+ C\|\partial _{i}^{2} u_{2}\|_{L^{2}}^{\frac{1}{2}}\|\partial _{i}^{2} \partial _{2} u_{2}\|_{L^{2}}^{\frac{1}{2}}\|\partial _{2} b\|_{L^{2}}^{ \frac{1}{2}}\|\partial _{2}\partial _{1}b\|_{L^{2}}^{\frac{1}{2}}\| \partial _{1}\partial _{i}^{2}u\|_{L^{2}} \\ \le &C(\|u\|_{H^{3}}+\|b\|_{H^{3}})(\|b_{2}\|_{H^{3}}^{2}+\|\partial _{1}u \|_{H^{2}}^{2}). \end{aligned}$$

So,

$$\begin{aligned} K_{4}\le &C(\|u\|_{H^{3}}+\|b\|_{H^{3}})(\|b_{2}\|_{H^{3}}^{2}+\| \partial _{2}u_{1}\|_{H^{3}}^{2}+\|\partial _{1}u\|_{H^{2}}^{2}). \end{aligned}$$

By the Hölder inequality and Sobolev inequality, we deduce

$$\begin{aligned} K_{5}=&\int \partial _{i}^{2}b_{2}\partial _{1}\partial _{i}^{2}u_{2}dx \\ \le &\|\partial _{i}^{2} b_{2}\|_{L^{2}}\|\partial _{1}\partial _{i}^{2}u_{2} \|_{L^{2}} \\ \le &C\|b_{2}\|_{H^{3}}^{2}+\frac{1}{2}\|\partial _{1}\partial _{i}^{2}u \|_{L^{2}}^{2}. \end{aligned}$$

Thanks to the Leibniz formula and Hölder inequality, we have

$$\begin{aligned} K_{6}=&-\int \partial _{i}^{2}(b\cdot \nabla u)\cdot \partial _{1} \partial _{i}^{2}udx \\ =&-\sum _{0\le \alpha \le 2}C_{2}^{\alpha}\int \partial _{i}^{\alpha }b \cdot \nabla \partial _{i}^{2-\alpha} u \cdot \partial _{1}\partial _{i}^{2}udx \\ =&-\sum _{0\le \alpha \le 2}C_{2}^{\alpha}\int \partial _{i}^{\alpha }b_{1} \partial _{1}\partial _{i}^{2-\alpha} u \cdot \partial _{1}\partial _{i}^{2}udx- \sum _{0\le \alpha \le 2}C_{2}^{\alpha}\int \partial _{i}^{\alpha }b_{2} \partial _{2}\partial _{i}^{2-\alpha} u \cdot \partial _{1}\partial _{i}^{2}udx \\ =&-\int b_{1}\partial _{1}\partial _{i}^{2} u \cdot \partial _{1} \partial _{i}^{2}udx -2\int \partial _{i} b_{1}\partial _{1}\partial _{i} u \cdot \partial _{1}\partial _{i}^{2}udx -\int \partial _{i}^{2} b_{1} \partial _{1} u \cdot \partial _{1}\partial _{i}^{2}udx \\ &-\int b_{2}\partial _{2}\partial _{i}^{2} u \cdot \partial _{1} \partial _{i}^{2}udx -2\int \partial _{i} b_{2}\partial _{2}\partial _{i}u \cdot \partial _{1}\partial _{i}^{2}udx -\int \partial _{i}^{2} b_{2} \partial _{2}u \cdot \partial _{1}\partial _{i}^{2}udx \\ \le &C\|\partial _{1}\partial _{i}^{2}u\|_{L^{2}}(\|b_{1}\|_{L^{ \infty}}\|\partial _{1}\partial _{i}^{2} u\|_{L^{2}}+\|\partial _{i} b_{1} \|_{L^{4}}\|\partial _{1}\partial _{i} u\|_{L^{4}}+\|\partial _{i}^{2} b_{1}\|_{L^{4}}\|\partial _{1} u\|_{L^{4}} \\ &+\|b_{2}\|_{L^{\infty}}\|\partial _{2}\partial _{i}^{2} u\|_{L^{2}}+ \|\partial _{i} b_{2}\|_{L^{4}}\|\partial _{2}\partial _{i}u\|_{L^{4}}+ \|\partial _{i}^{2} b_{2}\|_{L^{4}}\|\partial _{2}u\|_{L^{4}}) \\ \le &C(\|u\|_{H^{3}}+\|b\|_{H^{3}})(\|b_{2}\|_{H^{3}}^{2}+\|\partial _{1}u \|_{H^{2}}^{2}). \end{aligned}$$

Combining all the estimates from $K_{1}$ to $K_{6}$, we find

$$ \begin{aligned} \frac{1}{2}\|\partial _{1}\partial _{i}^{2}u\|_{L^{2}}^{2}\le & \frac{d}{dt}\langle \partial _{i}^{2}b,\partial _{i}^{2} \partial _{1}u \rangle +C(\|b_{2}\|_{H^{3}}^{2}+\|\partial _{2}u_{1}\|_{H^{3}}^{2}) \\ &+C(\|u\|_{H^{3}}+\|b\|_{H^{3}})(\|b_{2}\|_{H^{3}}^{2}+\|\partial _{2}u_{1} \|_{H^{3}}^{2}+\|\partial _{1}u\|_{H^{2}}^{2}). \end{aligned} $$

(3.11)

Putting (3.10) and (3.11) together gives

$$ \begin{aligned} \frac{1}{2}\|\partial _{1}u\|_{H^{2}}^{2}\le &\frac{d}{dt}\langle b, \partial _{1} u\rangle +\frac{d}{dt}\langle \partial _{i}^{2}b, \partial _{i}^{2}\partial _{1} u\rangle +C(\|b_{2}\|_{H^{3}}^{2}+\| \partial _{2}u_{1}\|_{H^{3}}^{2}) \\ &+C(\|u\|_{H^{3}}+\|b\|_{H^{3}})(\|b_{2}\|_{H^{3}}^{2}+\|\partial _{2}u_{1} \|_{H^{3}}^{2}+\|\partial _{1}u\|_{H^{2}}^{2}). \end{aligned} $$

(3.12)

Integrating (3.12) over $[0,t]$ leads to

$$\begin{aligned} \int _{0}^{t}\|\partial _{1}u\|_{H^{2}}^{2}\le &2\int b\cdot \partial _{1}udx-2\int b(x,0)\cdot \partial _{1}u(x,0)dx+2\int \partial _{i}^{2}b\cdot \partial _{i}^{2} \partial _{1}udx \\ &-2\int \partial _{i}^{2}b(x,0)\cdot \partial _{i}^{2} \partial _{1}u(x,0)dx+C \int _{0}^{t}(\|b_{2}\|_{H^{3}}^{2}+\|\partial _{2}u_{1}\|_{H^{3}}^{2}) \\ &+C\int _{0}^{t}(\|u\|_{H^{3}}+\|b\|_{H^{3}})(\|b_{2}\|_{H^{3}}^{2}+ \|\partial _{2}u_{1}\|_{H^{3}}^{2}+\|\partial _{1}u\|_{H^{2}}^{2}) \\ \le &C(\|b_{0}\|_{L^{2}}\|\partial _{1}u_{0}\|_{L^{2}}+\|\partial _{i}^{2}b_{0} \|_{L^{2}}\|\partial _{i}^{2}\partial _{1}u_{0}\|_{L^{2}}) \\ &+C(\|b\|_{L^{2}}\|\partial _{1}u\|_{L^{2}}+\|\partial _{i}^{2}b\|_{L^{2}} \|\partial _{i}^{2}\partial _{1}u\|_{L^{2}}) +C\int _{0}^{t}(\|b_{2} \|_{H^{3}}^{2}+\|\partial _{2}u_{1}\|_{H^{3}}^{2}) \\ &+C\sup _{0\le \tau \le t}(\|u\|_{H^{3}}+\|b\|_{H^{3}})\int _{0}^{t}( \|b_{2}\|_{H^{3}}^{2}+\|\partial _{2}u_{1}\|_{H^{3}}^{2}+\|\partial _{1}u \|_{H^{2}}^{2}), \end{aligned}$$

which implies that

$$\begin{aligned} E_{2}(t)\le CE_{1}(0)+CE_{1}(t)+CE_{1}^{\frac{3}{2}}(t)+CE_{2}^{ \frac{3}{2}}(t). \end{aligned}$$

(3.13)

Hence we finish the proof of Proposition 3.2. □

3.2. Proof of Theorem 1.1

Proof

Multiplying (3.13) by $\frac{1}{2C}$, and then adding the resulting inequality to (3.8), one has

$$\begin{aligned} E_{1}(t)+\frac{1}{2C}E_{2}(t)\le &C(E_{1}(0)+E_{1}^{\frac{3}{2}}(0)+E_{1}^{2}(0))+C(E_{1}^{ \frac{3}{2}}(t)+E_{2}^{\frac{3}{2}}(t)) \\ &+\frac{1}{2}E_{1}(t)+C(E_{1}^{2}(t)+E_{2}^{2}(t))+C(E_{1}^{ \frac{5}{2}}(t)+E_{2}^{\frac{5}{2}}(t)), \end{aligned}$$

which together with the definition of $E(t)$ implies

$$\begin{aligned} E(t)\le C_{1}(E(0)+E^{\frac{3}{2}}(0)+E^{2}(0))+C_{2}E^{\frac{3}{2}}(t)+C_{3}E^{2}(t)+C_{4}E^{ \frac{5}{2}}(t). \end{aligned}$$

We take $\|(u_{0},b_{0})\|_{H^{3}}$ to be sufficiently small,

$$\begin{aligned} C_{1}(E(0)+E^{\frac{3}{2}}(0)+E^{2}(0))\le \frac{1}{4}\min \bigg\{ \frac{1}{36C_{2}^{2}},\frac{1}{6C_{3}},(\frac{1}{6C_{4}})^{ \frac{2}{3}}\bigg\} . \end{aligned}$$

The bootstrapping argument starts with the ansatz that

$$\begin{aligned} E(t)\le \min \bigg\{ \frac{1}{36C_{2}^{2}},\frac{1}{6C_{3}},( \frac{1}{6C_{4}})^{\frac{2}{3}}\bigg\} . \end{aligned}$$

Then we can infer that

$$\begin{aligned} E(t)\le C_{1}(E(0)+E^{\frac{3}{2}}(0)+E^{2}(0))+\frac{1}{2}E(t), \end{aligned}$$

which gives

$$\begin{aligned} E(t)\le & 2C_{1}(E(0)+E^{\frac{3}{2}}(0)+E^{2}(0)) \\ \le &\frac{1}{2}\min \bigg\{ \frac{1}{36C_{2}^{2}},\frac{1}{6C_{3}},( \frac{1}{6C_{4}})^{\frac{2}{3}}\bigg\} , \end{aligned}$$

This inequality implies the desired estimate. Theorem 1.1 is completed. □

4 Proof of Theorem 1.4 and Theorem 1.5

This section proves Theorem 1.4 and Theorem 1.5. We are now ready to prove Theorem 1.4.

Proof

From (1.8), we find

$$\begin{aligned} \frac{1}{2}\frac{d}{dt}\|(u,b)\|_{L^{2}}^{2}+\|\mathcal{T}_{2}u\|_{L^{2}}^{2}+ \|\mathcal{R}_{1}b\|_{L^{2}}^{2}=0, \end{aligned}$$

(4.1)

and

$$\begin{aligned} \frac{1}{2}\frac{d}{dt}\|(\nabla u,\nabla b)\|_{L^{2}}^{2}+\| \mathcal{T}_{2}\nabla u\|_{L^{2}}^{2}+\|\mathcal{R}_{1}\nabla b\|_{L^{2}}^{2}=0. \end{aligned}$$

(4.2)

From (4.1) and (4.2), we find

$$\begin{aligned} \frac{d}{dt}A(t)+B(t)=0, \end{aligned}$$

(4.3)

where

$$\begin{aligned} A(t)=&\|u\|_{L^{2}}^{2}+\|b\|_{L^{2}}^{2}+\|\nabla u\|_{L^{2}}^{2}+\| \nabla b\|_{L^{2}}^{2}, \\ B(t)=&2\|\mathcal{T}_{2}u\|_{L^{2}}^{2}+2\|\mathcal{R}_{1}b\|_{L^{2}}^{2}+2 \|\mathcal{T}_{2}\nabla u\|_{L^{2}}^{2}+2\|\mathcal{R}_{1}\nabla b\|_{L^{2}}^{2}. \end{aligned}$$

Applying $\Lambda _{1}^{-\sigma},\ \Lambda _{2}^{-2\sigma}$ to (1.8) and dotting them with $(\Lambda _{1}^{-\sigma}u,\ \Lambda _{1}^{-\sigma}b)$ and $(\Lambda _{2}^{-2\sigma}u,\ \Lambda _{2}^{-2\sigma}b)$ in $H^{1+\sigma}$, respectively, we obtain

$$ \begin{aligned} \frac{d}{dt}&L(t)+2\|\mathcal{T}_{2}(\Lambda _{1}^{-\sigma},\Lambda _{2}^{-2 \sigma}) u\|_{L^{2}}^{2} +2\|\mathcal{R}_{1}(\Lambda _{1}^{-\sigma}, \Lambda _{2}^{-2\sigma}) b\|_{L^{2}}^{2} \\ &+2\|\mathcal{T}_{2}\Lambda ^{1+\sigma}(\Lambda _{1}^{-\sigma}, \Lambda _{2}^{-2\sigma}) u\|_{L^{2}}^{2} +2\|\mathcal{R}_{1}\Lambda ^{1+ \sigma}(\Lambda _{1}^{-\sigma},\Lambda _{2}^{-2\sigma}) b\|_{L^{2}}^{2}=0, \end{aligned} $$

(4.4)

where

$$\begin{aligned} L(t)=&\|(\Lambda _{1}^{-\sigma},\Lambda _{2}^{-2\sigma}) u\|_{L^{2}}^{2} +\|(\Lambda _{1}^{-\sigma},\Lambda _{2}^{-2\sigma}) b\|_{L^{2}}^{2} \\ &+\|\Lambda ^{1+\sigma}(\Lambda _{1}^{-\sigma},\Lambda _{2}^{-2\sigma}) u\|_{L^{2}}^{2} +\|\Lambda ^{1+\sigma}(\Lambda _{1}^{-\sigma}, \Lambda _{2}^{-2\sigma}) b\|_{L^{2}}^{2}. \end{aligned}$$

It’s easy to get from (4.3)

$$ \begin{aligned} L(t)\le L(0). \end{aligned} $$

(4.5)

The next job is to prove that

$$ \begin{aligned} A(t)\le CB^{\frac{\sigma}{1+\sigma}}(t)L^{\frac{1}{1+\sigma}}(t). \end{aligned} $$

(4.6)

Applying the Plancherel theorem and the Hölder inequality to obtain

$$\begin{aligned} \|u\|_{L^{2}}^{2}=&\int |\hat{u}(\xi ,t)|^{2}d\xi \\ =&\int (|\xi _{2}|^{4}|\hat{u}(\xi ,t)|^{2})^{\frac{\sigma}{1+\sigma}} (|\xi _{2}|^{-4\sigma}|\hat{u}(\xi ,t)|^{2})^{\frac{1}{1+\sigma}}d \xi \\ \le &(\int |\xi |^{-2}|\xi _{2}|^{4}|\xi |^{2}|\hat{u}(\xi ,t)|^{2}d \xi )^{\frac{\sigma}{1+\sigma}} (\int |\xi _{2}|^{-4\sigma}|\hat{u}( \xi ,t)|^{2}d\xi )^{\frac{1}{1+\sigma}} \\ \le &\|\mathcal{T}_{2}\nabla u\|_{L^{2}}^{\frac{2\sigma}{1+\sigma}}\| \Lambda _{2}^{-2\sigma}u\|_{L^{2}}^{\frac{2}{1+\sigma}}. \end{aligned}$$

and

$$\begin{aligned} \|\nabla u\|_{L^{2}}^{2}=&\int |\xi |^{2}|\hat{u}(\xi ,t)|^{2}d\xi \\ =&\int (|\xi _{2}|^{4}|\hat{u}(\xi ,t)|^{2})^{\frac{\sigma}{1+\sigma}} (|\xi _{2}|^{-4\sigma}|\xi |^{2(1+\sigma )}|\hat{u}(\xi ,t)|^{2})^{ \frac{1}{1+\sigma}}d\xi \\ \le &(\int |\xi |^{-2}|\xi _{2}|^{4}|\xi |^{2}|\hat{u}(\xi ,t)|^{2}d \xi )^{\frac{\sigma}{1+\sigma}} (\int |\xi _{2}|^{-4\sigma}|\xi |^{2(1+ \sigma )}|\hat{u}(\xi ,t)|^{2}d\xi )^{\frac{1}{1+\sigma}} \\ \le &\|\mathcal{T}_{2}\nabla u\|_{L^{2}}^{\frac{2\sigma}{1+\sigma}}\| \Lambda ^{1+\sigma}\Lambda _{2}^{-2\sigma}u\|_{L^{2}}^{ \frac{2}{1+\sigma}}. \end{aligned}$$

Repeating the same argument gives

$$\begin{aligned} \|b\|_{L^{2}}^{2}=&\int |\hat{b}(\xi ,t)|^{2}d\xi \\ =&\int (|\xi _{1}|^{2}|\hat{b}(\xi ,t)|^{2})^{\frac{\sigma}{1+\sigma}} (|\xi _{1}|^{-2\sigma}|\hat{b}(\xi ,t)|^{2})^{\frac{1}{1+\sigma}}d \xi \\ \le &(\int |\xi |^{-2}|\xi _{1}|^{2}|\xi |^{2}|\hat{b}(\xi ,t)|^{2}d \xi )^{\frac{\sigma}{1+\sigma}} (\int |\xi _{1}|^{-2\sigma}|\hat{b}( \xi ,t)|^{2}d\xi )^{\frac{1}{1+\sigma}} \\ \le &\|\mathcal{R}_{1}\nabla b\|_{L^{2}}^{\frac{2\sigma}{1+\sigma}}\| \Lambda _{1}^{-\sigma}b\|_{L^{2}}^{\frac{2}{1+\sigma}}. \end{aligned}$$

and

$$\begin{aligned} \|\nabla b\|_{L^{2}}^{2}=&\int |\xi |^{2}|\hat{b}(\xi ,t)|^{2}d\xi \\ =&\int (|\xi _{1}|^{2}|\hat{b}(\xi ,t)|^{2})^{\frac{\sigma}{1+\sigma}} (|\xi _{1}|^{-2\sigma}|\xi |^{2(1+\sigma )}|\hat{b}(\xi ,t)|^{2})^{ \frac{1}{1+\sigma}}d\xi \\ \le &(\int |\xi |^{-2}|\xi _{1}|^{2}|\xi |^{2}|\hat{b}(\xi ,t)|^{2}d \xi )^{\frac{\sigma}{1+\sigma}} (\int |\xi _{1}|^{-2\sigma}|\xi |^{2(1+ \sigma )}|\hat{b}(\xi ,t)|^{2}d\xi )^{\frac{1}{1+\sigma}} \\ \le &\|\mathcal{R}_{1}\nabla b\|_{L^{2}}^{\frac{2\sigma}{1+\sigma}}\| \Lambda ^{1+\sigma}\Lambda _{1}^{-\sigma}b\|_{L^{2}}^{ \frac{2}{1+\sigma}}. \end{aligned}$$

Combining the above four inequalities can obtain (4.6). Combining (4.5) and (4.6), we immediately get

$$\begin{aligned} B(t)\geq CL^{-\frac{1}{\sigma}}(0)A^{1+\frac{1}{\sigma}}(t), \end{aligned}$$

which together with (4.3) yields

$$\begin{aligned} \frac{d}{dt}A(t)+CL^{-\frac{1}{\sigma}}(0)A^{1+\frac{1}{\sigma}}(t) \le 0. \end{aligned}$$

This inequality gives rise to

$$\begin{aligned} A(t)\le \bigg(A^{-\frac{1}{\sigma}}(0)+\frac{C}{\sigma}L^{- \frac{1}{\sigma}}(0)t\bigg)^{-\sigma}, \end{aligned}$$

which completes the proof of Theorem 1.4. □

We proceed to prove Theorem 1.5.

Proof

Taking the $L^{2}$-inner product to the equation (1.9)₁ with $\partial _{t} u$, we have

$$ \begin{aligned} \frac{1}{2}\frac{d}{dt}&(\|\partial _{t}u\|_{L^{2}}^{2}+\|\mathcal{R}_{1} \mathcal{T}_{2}u\|_{L^{2}}^{2}+\|\partial _{1}u\|_{L^{2}}^{2}) \\ &+\|\partial _{t}\mathcal{R}_{1}u\|_{L^{2}}^{2}+\|\partial _{t} \mathcal{T}_{2}u\|_{L^{2}}^{2}=0. \end{aligned} $$

(4.7)

Similarly,

$$ \begin{aligned} \frac{1}{2}\frac{d}{dt}&(\|\partial _{t}\Lambda \Lambda _{1}^{-1}u\|_{L^{2}}^{2}+ \|\mathcal{T}_{2}u\|_{L^{2}}^{2}+\|\Lambda u\|_{L^{2}}^{2}) \\ &+\|\partial _{t}u\|_{L^{2}}^{2}+\|\partial _{t}\mathcal{T}_{2} \Lambda \Lambda _{1}^{-1}u\|_{L^{2}}^{2}=0. \end{aligned} $$

(4.8)

and

$$ \begin{aligned} \frac{1}{2}\frac{d}{dt}&(\|\partial _{t}\Lambda _{1}^{-1}u\|_{L^{2}}^{2}+ \|\mathcal{R}_{1}\mathcal{T}_{2}\Lambda _{1}^{-1}u\|_{L^{2}}^{2}+\|u \|_{L^{2}}^{2}) \\ &+\|\partial _{t}\mathcal{R}_{1}\Lambda _{1}^{-1}u\|_{L^{2}}^{2}+\| \partial _{t}\mathcal{T}_{2}\Lambda _{1}^{-1}u\|_{L^{2}}^{2}=0. \end{aligned} $$

(4.9)

Taking $L^{2}$-inner product to (1.9)₁ with $u$, we obtain

$$ \begin{aligned} \frac{1}{2}\frac{d}{dt}&(\|\mathcal{R}_{1}u\|_{L^{2}}^{2}+\| \mathcal{T}_{2}u\|_{L^{2}}^{2}+2\langle \partial _{t} u,u\rangle ) \\ &+\|\mathcal{R}_{1}\mathcal{T}_{2}u\|_{L^{2}}^{2}+\|\partial _{1}u\|_{L^{2}}^{2}- \|\partial _{t} u\|_{L^{2}}^{2}=0. \end{aligned} $$

(4.10)

Combining (4.7), (4.8), (4.9) and (4.10) yields

$$ \begin{aligned} \frac{1}{2}\frac{d}{dt}&(\|\partial _{t}u\|_{L^{2}}^{2}+\|\mathcal{R}_{1} \mathcal{T}_{2}u\|_{L^{2}}^{2}+\|\partial _{1}u\|_{L^{2}}^{2} +\| \partial _{t}\Lambda \Lambda _{1}^{-1}u\|_{L^{2}}^{2}+\|\mathcal{T}_{2}u \|_{L^{2}}^{2} \\ &+\|\Lambda u\|_{L^{2}}^{2} +\|\partial _{t}\Lambda _{1}^{-1}u\|_{L^{2}}^{2}+ \|\mathcal{R}_{1}\mathcal{T}_{2}\Lambda _{1}^{-1}u\|_{L^{2}}^{2}+\|u \|_{L^{2}}^{2} +\frac{1}{2}\|\mathcal{R}_{1}u\|_{L^{2}}^{2} \\ &+\frac{1}{2}\|\mathcal{T}_{2}u\|_{L^{2}}^{2} +\langle \partial _{t} u,u \rangle )+\|\partial _{t}\mathcal{R}_{1}u\|_{L^{2}}^{2} +\|\partial _{t} \mathcal{T}_{2}u\|_{L^{2}}^{2}+\|\partial _{t}u\|_{L^{2}}^{2} \\ &+\|\partial _{t}\mathcal{T}_{2}\Lambda \Lambda _{1}^{-1}u\|_{L^{2}}^{2} +\|\partial _{t}\mathcal{R}_{1}\Lambda _{1}^{-1}u\|_{L^{2}}^{2}+\| \partial _{t}\mathcal{T}_{2}\Lambda _{1}^{-1}u\|_{L^{2}}^{2} \\ &+\frac{1}{2}\|\mathcal{R}_{1}\mathcal{T}_{2}u\|_{L^{2}}^{2} + \frac{1}{2}\|\partial _{1}u\|_{L^{2}}^{2}-\frac{1}{2}\|\partial _{t} u \|_{L^{2}}^{2}=0. \end{aligned} $$

(4.11)

We can easily verify that

$$\begin{aligned} \frac{d}{dt}\|\partial _{t}u\|_{L^{2}}^{2}+\|u\|_{L^{2}}^{2}+\langle \partial _{t} u,u\rangle \geq \frac{1}{2}\|\partial _{t}u\|_{L^{2}}^{2}+ \|u\|_{L^{2}}^{2}. \end{aligned}$$

Thus, by integrating both sides of (4.11) over $(0,t)$, we derive

$$\begin{aligned} \int _{0}^{t}&(\|\partial _{t}u\|_{L^{2}}^{2}+\|\mathcal{R}_{1} \mathcal{T}_{2}u\|_{L^{2}}^{2}+\|\partial _{1}u\|_{L^{2}}^{2})d\tau \\ &\le C(\|u_{0}\|_{H^{1}}^{2}+\|\partial _{t}u_{0}\|_{L^{2}}^{2}+\| \Lambda \Lambda _{1}^{-1}\partial _{t}u_{0}\|_{L^{2}}^{2} +\|\Lambda _{1}^{-1} \partial _{t}u_{0}\|_{L^{2}}^{2}+\|\mathcal{R}_{1}\mathcal{T}_{2} \Lambda _{1}^{-1}u_{0}\|_{L^{2}}^{2}). \end{aligned}$$

It follows from (4.7) that

$$\begin{aligned} \frac{d}{dt}(\|\partial _{t}u\|_{L^{2}}^{2}+\|\mathcal{R}_{1} \mathcal{T}_{2}u\|_{L^{2}}^{2} +\|\partial _{1}u\|_{L^{2}}^{2})\le 0, \end{aligned}$$

which together with Lemma 2.3 implies

$$\begin{aligned} \|\partial _{t}u\|_{L^{2}}^{2}+\|\mathcal{R}_{1}\mathcal{T}_{2}u\|_{L^{2}}^{2} +\|\partial _{1}u\|_{L^{2}}^{2}\le C(1+t)^{-1}. \end{aligned}$$

We can prove the same result is true for $b$ along the same method. The proof of Theorem 1.5 is thus completed. □