Stabilizing effect of magnetic field on the 2D ideal magnetohydrodynamic flow with mixed partial damping

Abstract

This paper examines the stability of a 2D inviscid MHD system with anisotropic damping near a background magnetic field. It is well known that solutions of the incompressible Euler equations can grow rapidly in time and are thus unstable while solutions of the Euler equations with full damping are stable. Then naturally arises the question of whether solutions of the Euler equations with partial damping are stable. The main purpose of this paper is to give an affirmative answer to this question in the case when the fluid is coupled with the magnetic field through the MHD system with one-component damping. The result presented in this paper especially confirms the stabilizing effects of the magnetic field on the electrically conducting fluids, a phenomenon that has been observed in physical experiments and numerical simulations.

1 Introduction

The MHD system is composed of the Navier–Stokes equations of fluid dynamics and Maxwell’s equations of electromagnetism. It describes the motion of electrically conducting fluids such as plasmas, liquid metals and electrolytes in an electromagnetic field and has a wide range of applications in astrophysics, geophysics, cosmology and engineering (see, e.g., [5, 13, 38]). The MHD equations not only share some mathematically important features with the Euler/Navier–Stokes equations, but also exhibit many more fascinating properties than the fluid equations without the magnetic field. Inspired by the phenomenon observed in physical experiments and numerical simulations that the magnetic field can stabilize electrically conducting fluids (see, e.g., [2, 3, 22, 23]), we aim to explore the smoothing and stabilizing effects of magnetic field on the fluid motion. For this purpose, we consider the following 2D MHD equations with only partial damping in the velocity and the magnetic field,

$$\begin{aligned} \left\{ \begin{aligned}&\partial _t U+U\cdot \nabla U+\nabla P+\nu (U_1, 0)^{\top }= B\cdot \nabla B,\quad x\in \mathbb {R}^2,\,t>0, \\&\partial _tB+U\cdot \nabla B+\eta (0, {B}_2)^{\top } =B\cdot \nabla U, \\&{\nabla \cdot }\,U={\nabla \cdot }\,B=0, \end{aligned} \right. \end{aligned}$$

(1.1)

where $U=(U_1, U_2)^{\top }$, $B=(B_1, B_2)^{\top }$ and P are the velocity field, the magnetic field, and the pressure, respectively. The positive constants $\nu >0$ and $\eta >0$ are the damping coefficients.

There have been substantial developments on two fundamental problems concerning the MHD equations, the global (in time) regularity and stability. In particular, the stability problem near a background magnetic field have recently attracted considerable interests. For the ideal MHD equations, Bardos et al. [4] took advantage of the Elsässer variables to establish the global regularity (in Hölder setting) of perturbations near a strong background magnetic field. Cai and Lei [7] and He et al. [25], via different approaches, successfully solved the stability problem on both the ideal MHD system and its fully dissipative counterpart (with identical viscosity and resistivity) near a background magnetic field. Wei and Zhang [47] allowed the viscosity and resistivity coefficients to be slightly different. The paper of Lin et al. [33] pioneered the study of the stability problem on the incompressible non-resistive MHD equation near a background magnetic field. The 3D problem together with the large-time behavior was solved by Abidi and Zhang [1] and Deng and Zhang [14] in the whole spaces case. Pan et al. [37] dealt with this problem when the spatial domain is a 3D periodic box $\mathbb {T}^3$. Tan and Wang [42] examined the case with the horizontally infinite flat layer $\mathbb {R}^2\times (0,1)$. The approach of Lin et al. [33] on the 2D non-resistive MHD problem is Lagrangian. Ren et al. [39] revisited the stability problem by resorting to the Eulerian energy estimates in anisotropic Sobolev space and obtained explicit time decay rates. Ren et al. [40] proved the global stability in a strip domain, and Chen and Ren [12] considered two types of periodic domains $\mathbb {T}\times \mathbb {R}$ and $\mathbb {T} \times (0, 1)$. Zhang [56] proved the global existence of strong solutions to the Cauchy problem with large initial perturbations, provided that the background magnetic field is sufficiently large. Recently, Jiang and Jiang [28] extended the results [56] to the 2D periodic domains $\mathbb {T}^2$ by using the Lagrangian approach and the odevity conditions proposed in [37], and obtained the asymptotic behaviors of global strong solutions with large initial perturbations. For the 2D inviscid and resistive MHD equations, Zhou and Zhu [57] investigated the stability of perturbations near a background magnetic field on the periodic domain. For the ideal MHD system with velocity damping, Wu et al. [52] studied the stability via the approach of wave equations, and Du et al. [18] proved the exponential stability of a stratified flow in the strip-type doamin $\mathbb {R} \times [0, 1]$. We also refer to [51] for the stability and large-time behavior of the 2D compressible MHD system without magnetic diffusion.

Due to its physical relevance and remarkable enhanced smoothing properties, the stability problem for the incompressible MHD equations with partial dissipation has recently generated a rich array of results. Lin et al. [34] obtained the stability of the 2D MHD equations with vertical velocity dissipation and horizontal magnetic diffusion (see also [31]). A new stability result on 3D MHD equations with horizontal dissipation and vertical magnetic diffusion was achieved by Wu and Zhu [53]. Boardman et al. [6] studied the stability of 2D inviscid and resistive MHD equations with only vertical velocity damping. The stability and large-time behavior of the 2D MHD equations with only vertical velocity dissipation and a damping magnetic field was investigated in [21]. The paper [30] dealt with the anisotropic equations with only (partially) vertical damping magnetic field. In comparison with [21] and [30], the MHD system considered in this current paper contains the least dissipation and damping. It appears that the anisotropic damping required in this paper can not be further reduced.

Many more results on the well-posedness and related issues concerning the incompressible MHD equations are available in the literature. For example, various partial dissipation cases are dealt with in [8, 9, 16, 17, 36], the non-resistive case in [11, 20, 32, 45, 55], the only magnetic diffusion case in [10, 29] and the fractional dissipation case in [15, 44, 48,49,50, 54].

This paper aims to understand the stability of the 2D ideal MHD system (1.1) near the equilibrium state $(U^{(0)},B^{(0)})$,

$$\begin{aligned} U^{(0)} \equiv 0, \quad B^{(0)} \equiv e_1 := (1,0). \end{aligned}$$

Let (u, b) be the perturbation of (U, B) near the steady state $(U^{(0)},B^{(0)})$,

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

The system governing the perturbation is taken to be the following system

$$\begin{aligned} \left\{ \begin{aligned}&\partial _t u+u\cdot \nabla u+\nabla P+\nu {(u_1, 0)^\top }= b\cdot \nabla b+\partial _{1}b,\quad x\in \mathbb {R}^2,\,t>0, \\&\partial _t b+u\cdot \nabla b+\eta {(0, b_2)^\top } =b\cdot \nabla u+\partial _{1}u, \\&{\nabla \cdot }\,u={\nabla \cdot }\,b=0. \end{aligned} \right. \end{aligned}$$

(1.2)

We shall focus on an initial value problem of (1.2) with the Cauchy data:

$$\begin{aligned} u(x, 0) = u_0(x),\,\,\, b(x, 0) = b_0(x). \end{aligned}$$

(1.3)

The motivation for studying the stability problem of (1.2)–(1.3) is twofold. The first is to reveal the phenomenon that the coupling and interaction between the velocity and the magnetic field actually stabilize the fluid motion. Indeed, when $B=0$, (1.1) becomes the 2D incompressible Euler equation with only horizontally damping velocity,

$$\begin{aligned} \left\{ \begin{aligned}&\partial _t U_1+U\cdot \nabla U_1+\partial _1 P+\nu U_1=0, \\&\partial _t U_2+U\cdot \nabla U_2+\partial _2 P=0, \\&{\nabla \cdot }\,U=0. \end{aligned} \right. \end{aligned}$$

(1.4)

The stability problem of (1.4) remains unsolved. To understand the difficulty, we reformulate (1.4) in terms of the following vorticity equation

$$\begin{aligned} \left\{ \begin{aligned}&\partial _t \omega +U\cdot \nabla \omega =\nu \mathcal R_2^2 \omega , \\&U=\nabla ^{\perp }\Delta ^{-1}\omega , \end{aligned} \right. \end{aligned}$$

(1.5)

where $\mathcal R_k = \partial _k (-\Delta )^{-\frac{1}{2}}$ with $k=1,2$ denotes the standard Riesz transform (see, e.g., [24, 41]) and the fractional Laplacian operator is defined via the Fourier transform,

$$\begin{aligned} \widehat{(-\Delta )^\beta f} (\xi ) = |\xi |^{2\beta } \widehat{f}(\xi ). \end{aligned}$$

and $\nabla ^{\perp }=(-\partial _2,\partial _1)$. Unfortunately, the classical Yudovich’s approach used to study the 2D incompressible Euler-like equations do not appear to work for (1.5), since the Riesz transform $\mathcal R_2$ is not known to be bounded in $L^\infty$. In fact, as pointed out by Elgindi [19], the $L^{q}$-norms of $\omega$ are bounded for any $1<q<\infty$, but these $L^{q}$-norms may grow exponentially in q. Therefore, the question of whether the solutions of (1.5) will develop singularity in finite time is an interesting and challenging problem. The first and main purpose of this paper is to show that the magnetic field is able to stabilize the velocity field through the MHD system (1.1). For the recent works on the magnetic inhibition phenomenon (or stability result), we refer to [26, 27, 46] and the references cited therein.

The second motivation is to explore the hidden wave structure and to understand the stability mechanism. To explain this clearly, we apply the Leray projection operator $\mathbb P= I -\nabla \Delta ^{-1}\nabla \cdot$ to the equation (1.2) and separate it into the linear part and the nonlinear part. Due to $\nabla \cdot u = \nabla \cdot b = 0$,

$$\begin{aligned} {\mathbb P}{(u_1, 0)^\top }={(u_1, 0)^\top }-\nabla \Delta ^{-1}\nabla \cdot {(u_1, 0)^\top }=\partial _2^2\Delta ^{-1}u=-{\mathcal R_2^2}u, \end{aligned}$$

and

$$\begin{aligned} {\mathbb P}{(0, b_2)^\top }={(0, b_2)^\top }-\nabla \Delta ^{-1}\nabla \cdot {(0, b_2)^\top }=\partial _1^2\Delta ^{-1}b=-{\mathcal R_1^2}b. \end{aligned}$$

Thus the system (1.2) can be written as

$$\begin{aligned} {\left\{ \begin{array}{ll} \partial _t u=\nu {\mathcal R_2^2}u +\partial _1 b+{\mathbb P} (b\cdot \nabla b-u\cdot \nabla u),\\ \partial _t b=\eta {\mathcal R_1^2}b + \partial _1 u+ {\mathbb P} (b\cdot \nabla u-u\cdot \nabla b),\\ {\nabla \cdot }\,u={\nabla \cdot }\,b=0. \end{array}\right. } \end{aligned}$$

(1.6)

Differentiating (1.6) in t and making several substitutions, we find

$$\begin{aligned} {\left\{ \begin{array}{ll} \partial _{tt} u- (\nu \mathcal R_2^2+\eta \mathcal R_1^2 ) \partial _t u -\partial _1^2 u+ \nu \eta \mathcal R_1^2 \mathcal R_2^2u = N_1, \\ \partial _{tt} b - (\nu \mathcal R_2^2+\eta \mathcal R_1^2 ) \partial _t b -\partial _1^2 b+ \nu \eta \mathcal R_1^2 \mathcal R_2^2b = N_2, \\ {\nabla \cdot }\,u={\nabla \cdot }\,b=0, \end{array}\right. } \end{aligned}$$

(1.7)

where $N_1$ and $N_2$ are the nonlinear terms,

$$\begin{aligned}&N_1 = (\partial _t - \eta \mathcal R_1^2){\mathbb P} (b\cdot \nabla b-u\cdot \nabla u)+\partial _1 {\mathbb P} (b\cdot \nabla u-u\cdot \nabla b), \\&N_2 = (\partial _t - \nu \mathcal R_2^2){\mathbb P}(b\cdot \nabla u-u\cdot \nabla b) +\partial _1 {\mathbb P}(b\cdot \nabla b-u\cdot \nabla u) . \end{aligned}$$

It is surprising that $u ,\ b$ satisfy the same degenerate damped wave equation. The wave structure of (1.7) for (u, b) provides much more stabilization and regularization properties than the original system (1.1). In fact, the wave equation (1.7) indicates that there is a horizontal regularization via the coupling and interaction, and hence, the stability result of the solutions becomes possible.

The main result of this paper is the following stability theorem of global solutions to the Cauchy problem (1.2)–(1.3).

Theorem 1.1

Assume the initial data $(u_0,b_0)\in H^3$ with ${\nabla \cdot }\,u_0={\nabla \cdot }\,b_0=0$. Then there exists a positive constant $\varepsilon >0$, depending only on $\nu$ and $\eta$, such that if

$$\begin{aligned}\Vert (u_0,b_0)\Vert _{H^3}\le \varepsilon ,\end{aligned}$$

then the problem (1.2)–(1.3) has a unique global solution (u, b) on $\mathbb {R}^2\times [0,\infty )$, satisfying

$$\begin{aligned} \Vert (u,b)(t)\Vert _{H^3}^2 + \int _0^t \big (\Vert (u_1,b_2)(\tau )\Vert _{H^3}^2 +\Vert \partial _1u(\tau )\Vert _{H^2}^2\big ) d \tau \le C\varepsilon ^2,\quad \forall \ t\ge 0, \end{aligned}$$

where $C>0$ is a generic positive constant independent of $\varepsilon$ and t.

Since the local-in-time existence result can be shown by the standard method (see, e.g., [35]), our main task is to derive the global-in-time a prior estimates of the solutions. The framework is the bootstrapping argument [43]. Due to the lack of full damping, some serious difficulties arise. To overcome these difficulties, we have to construct a suitable energy functional. It consists of two parts. The first part is the natural $H^3$-energy functional $\mathcal {E}_1(t)$,

$$\begin{aligned} \mathcal {E}_1(t):=\sup _{0\le \tau \le t}\Vert (u,b)(\tau )\Vert _{H^3}^2+2\int _0^t \Big (\nu \Vert u_1(\tau )\Vert _{H^3}^2 +\eta \Vert b_2(\tau )\Vert _{H^3}^2 \Big )d \tau , \end{aligned}$$

(1.8)

The second part $\mathcal {E}_2(t)$ includes the horizontal dissipation piece generated from $\partial _1u$ and indicated by the wave structure of (1.7),

$$\begin{aligned} \mathcal {E}_2(t):= \int _0^t \Vert \partial _1u(\tau )\Vert _{H^2}^2 d \tau . \end{aligned}$$

(1.9)

When applying the standard $L^2$-method to estimate $\mathcal {E}_1(t)$ and $\mathcal {E}_2(t)$, we encounter four of the most difficult terms:

$$\begin{aligned}&\mathrm{Diff}_1 :=\int \partial _1 u_1 |\partial _2^3 b_1|^2 \ d x,\quad \mathrm{Diff}_2 :=\int b_1\partial _2^3b_1\partial _2^3\partial _1u_1 \ d x, \\&\mathrm{Diff}_3:=\int b_1\partial _1u_1|\partial _2^3b_1|^2 \ d x, \quad \mathrm{Diff}_4:=\int b_1^2\partial _2^3b_1\partial _2^3\partial _1u_1 \ d x, \end{aligned}$$

which cannot be well controlled by $\mathcal {E}_1(t)$ and $\mathcal {E}_2(t)$ directly. The strategy here is to use (1.2)$_2$ and (1.2)$_1$ to replace $\partial _1 u_1$ and $\partial _1 b_1$ by

$$\begin{aligned}&\partial _{1}u_1=\partial _t b_1+u\cdot \nabla b_1- b\cdot \nabla u_1, \end{aligned}$$

(1.10)

$$\begin{aligned}&\partial _{1}b_1=\partial _t u_1+u\cdot \nabla u_1+\partial _1P+\nu u_1- b\cdot \nabla b_1. \end{aligned}$$

(1.11)

For example, with the help of (1.10) and (1.11), we find

$$\begin{aligned} \mathrm{Diff}_1&=\int \left( \partial _t b_1+u\cdot \nabla b_1- b\cdot \nabla u_1\right) |\partial _2^3 b_1|^2 \ d x \\&=\frac{d}{d t}\int b_1|\partial _2^3 b_1|^2\ dx-2\int b_1\partial _2^3 b_1\partial _2^3\partial _t b_1 \ dx \\&\quad +\int u\cdot \nabla b_1|\partial _2^3 b_1|^2\ d x- \int b\cdot \nabla u_1|\partial _2^3 b_1|^2 \ d x, \end{aligned}$$

and

$$\begin{aligned} \mathrm{Diff}_2&=-\int \partial _1b_1\partial _2^3 b_1\partial _2^3u_1\ d x-\int b_1\partial _2^3u_1\partial _2^3\partial _1 b_1\ d x\\&=-\int \partial _1b_1\partial _2^3 b_1\partial _2^3u_1\ d x\\&\quad -\int b_1\partial _2^3u_1\partial _2^3\left( \partial _t u_1+u\cdot \nabla u_1+\partial _1P+\nu u_1- b\cdot \nabla b_1\right) \ d x \\&=-\int \partial _1b_1\partial _2^3 b_1\partial _2^3u_1\ d x-\frac{1}{2}\frac{d}{d t}\int b_1|\partial _2^3 u_1|^2\ dx+\frac{1}{2}\int |\partial _2^3 u_1|^2\partial _t b_1\ dx\\&\quad -\int b_1\partial _2^3u_1\partial _2^3\left( u\cdot \nabla u_1+\partial _1P+\nu u_1- b\cdot \nabla b_1\right) \ d x. \end{aligned}$$

The items associated with $\partial _t b_1$ will be handled by using (1.10) again. This process generates many terms. Based upon integration by parts and the anisotropic Sobolev inequalities, it is incredible that all the terms can be bounded by $\mathcal {E}_1(t)$ and $\mathcal {E}_2(t)$, although the process is complicated and lengthy. For the details, we refer to the treatments of $D_i$ with $i=1,\ldots ,4$ in Sect. 2. Collecting these estimates, we are able to establish the energy inequalities stated in Proposition 2.1.

We also make efforts to exploit the full regularization and stabilization effects from the wave structure to understand the large-time behavior of the linearized system. The linearized system of (1.6) reads

$$\begin{aligned} {\left\{ \begin{array}{ll} \partial _t u-\nu {\mathcal R_2^2}u-\partial _1 b=0,\\ \partial _t b-\eta {\mathcal R_1^2}b-\partial _1 u=0,\\ {\nabla \cdot }\,u={\nabla \cdot }\,b=0,\\ u(x,0)=u_0(x), \ b(x,0)=b_0(x), \end{array}\right. } \end{aligned}$$

(1.12)

which can be converted to the linearized system of wave equations (1.7):

$$\begin{aligned} {\left\{ \begin{array}{ll} \partial _{tt} u- (\nu \mathcal R_2^2+\eta \mathcal R_1^2 ) \partial _t u -\partial _1^2 u+ \nu \eta \mathcal R_1^2 \mathcal R_2^2u =0, \\ \partial _{tt} b - (\nu \mathcal R_2^2+\eta \mathcal R_1^2 ) \partial _t b -\partial _1^2 b+ \nu \eta \mathcal R_1^2 \mathcal R_2^2b =0, \\ {\nabla \cdot }\,u={\nabla \cdot }\,b=0,\\ u(x,0)=u_0(x), \ b(x,0)=b_0(x). \end{array}\right. } \end{aligned}$$

(1.13)

We first aim to establish the decay rate of solution for the linearized system (1.12) in negative Sobolev space by careful energy estimates. To state our result precisely, we first define the fractional partial derivative operator $\Lambda _i^\gamma$ with $i=1,\ 2$ and $\gamma \in \mathbb {R}$ by

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

Theorem 1.2

For $\sigma >0$, assume that $(u_0, b_0)$ satisfies

$$\begin{aligned} (\Lambda _1^{-\sigma }, \Lambda _2^{-\sigma })u_0\in H^{1+\sigma }, \ (\Lambda _1^{-\sigma }, \Lambda _2^{-\sigma })b_0\in H^{1+\sigma }, \ \nabla \cdot u_0=\nabla \cdot b_0=0. \end{aligned}$$

Then the corresponding solution (u, b) of (1.12) satisfies

$$\begin{aligned} (u, b)\in L^{\infty }(0, \infty ; H^1), \ (\mathcal R_2u, \mathcal R_1b)\in L^{2}(0, \infty ; H^1). \end{aligned}$$

Moreover,

$$\begin{aligned} \Vert (u, b)(t)\Vert _{H^1}\le C(1+t)^{-\frac{\sigma }{2}},\quad \forall \ t>0, \end{aligned}$$

where C is a generic positive constant depending only on $\nu ,\eta ,\sigma$ and the initial norms.

When the initial data is not in any Sobolev space of negative indices, we can still manage to show the precise decay rates for several quantities.

Theorem 1.3

Assume that

$$\begin{aligned}&(u_0, b_0)\in L^2,\quad (\partial _1u_0, \partial _1b_0)\in L^2,\quad \nabla \cdot u_0=\nabla \cdot b_0=0,\\&(\mathcal R_1\mathcal R_2u_0, \mathcal R_1\mathcal R_2b_0)\in L^2,\quad (\mathcal R_2^2u_0, \mathcal R_1^2b_0)\in L^2. \end{aligned}$$

Then for any $t\ge 0$, the solution (u, b) of (1.12) satisfies,

$$\begin{aligned}&\Vert \partial _t u(t)\Vert _{L^2}+\Vert \partial _1 u(t)\Vert _{L^2}+\Vert \mathcal R_1\mathcal R_2 u(t)\Vert _{L^2}\le C\left( 1+t\right) ^{-\frac{1}{2}}, \\&\Vert \partial _t b(t)\Vert _{L^2}+\Vert \partial _1 b(t)\Vert _{L^2}+\Vert \mathcal R_1\mathcal R_2 b(t)\Vert _{L^2}\le C\left( 1+t\right) ^{-\frac{1}{2}}, \end{aligned}$$

where C is a generic positive constant depending only on $\nu ,\eta$ and the initial norms.

Finally we show that any frequency away from a given area D decays exponentially in time. To do this, we define D by

$$\begin{aligned} D:=\left\{ \xi \in \mathbb {R}^2:\ |\xi _1|<\alpha ~~\text{ and }~~|\xi |^2> \beta |\xi _1| |\xi _2|~\right\} , \end{aligned}$$

(1.14)

where $\alpha >0$ and $\beta >2$ are fixed positive constants. In addition, we set $\widehat{\psi }(\xi )$ to be the following cutoff function in the frequency space,

$$\begin{aligned} \widehat{\psi }(\xi )=\left\{ \begin{array}{lll} &{}0,\quad \;\; &{}\mathrm{if }\quad \xi \in D, \\ &{}1,\quad \;\; &{}\mathrm{if }\quad \xi \in D^c. \end{array} \right. \end{aligned}$$

Obviously,

$$\begin{aligned} \widehat{\psi *f}(\xi ) = \widehat{\psi }(\xi )\, \widehat{f}(\xi ). \end{aligned}$$

(1.15)

Theorem 1.4

Assume the initial data $(u_0, b_0)$ with $\nabla \cdot u_0 =\nabla \cdot b_0=0$ satisfies

$$\begin{aligned}&(\psi *u_0,\,\,\psi *b_0, \,\,\psi *\partial _1u_0,\,\, \psi *\partial _1b_0)\in L^2,\\&(\psi *\mathcal R_1\mathcal R_2u_0,\,\, \psi *\mathcal R_1\mathcal R_2b_0,\,\,\psi *\mathcal R_2^2u_0, \,\,\psi *\mathcal R_1^2b_0)\in L^2 . \end{aligned}$$

Then the corresponding solution (u, b) of (1.12) obeys the following exponential decay estimates,

$$\begin{aligned}&\Vert (\psi *u, \psi *b)\Vert _{L^2}+\Vert (\psi *\partial _1 u, \psi *\partial _1 b)\Vert _{L^2}\\&\qquad + \Vert (\psi *\mathcal R_1\mathcal R_2 u, \psi *\mathcal R_1\mathcal R_2 b) \Vert _{L^2}+\Vert (\psi *\partial _t u, \psi *\partial _t b)\Vert _{L^2}\\&\quad \le C\,e^{-c(\eta ,\nu , \alpha , \beta )\; t}, \end{aligned}$$

where $c=c(\nu , \eta , \alpha , \beta )>0$ depends on $\nu ,\eta ,\alpha$ and $\beta$, and $C=C(u_0, b_0, \nu ,\eta ,\alpha , \beta )>0$ depends additionally on the initial norms.

Remark 1.1

It is an interesting problem to study the decay rates of the solutions to the nonlinear system (1.2). Unfortunately, this seems not easy and is left for the future. In fact, the large-time behavior of the solution depends crucially on the eigenvalues of the wave equation (1.13). Indeed, the characteristic polynomial associated with (1.13) reads

$$\begin{aligned}\lambda ^2+\left( \frac{\nu \xi _2^2}{|\xi |^2}+\frac{\eta \xi _1^2}{|\xi |^2}\right) \lambda +\nu \eta \frac{\xi ^2_1\xi _2^2}{|\xi |^4}+\xi ^2_1=0,\end{aligned}$$

and the roots $\lambda _\mp$ are given by

$$\begin{aligned} \lambda _\mp :=\frac{-\frac{\nu \xi _2^2+\eta \xi _1^2}{|\xi |^2} \mp \sqrt{\Gamma }}{2}\quad {with}\quad \Gamma :=\left( \frac{\nu \xi _2^2+\eta \xi _1^2}{|\xi |^2}\right) ^2-4\left( \nu \eta \frac{\xi ^2_1\xi _2^2}{|\xi |^4}+\xi ^2_1\right) . \end{aligned}$$

By direct calculations, we find

$$\begin{aligned} \lambda _+=-\frac{2\xi ^2_1\left( \nu \eta \frac{\xi _2^2}{|\xi |^4}+1\right) }{\sqrt{\Gamma }+\left( \frac{\nu \xi _2^2}{|\xi |^2}+\frac{\eta \xi _1^2}{|\xi |^2}\right) }\lesssim -\xi _1^2, \end{aligned}$$

provided $\Gamma \ge 0$ and $|\xi _1|$ is sufficiently small. As a result, the heat kernel only admits “one-component" decay. This is the inherent difficulty in the decay analysis of the solutions. Actually, it is also the reason that why we can only obtain the exponential decay away from the domain D.

The rest of this paper is organized as follows. Theorem 1.1 is proven in Sect. 2. The proof of Theorem 1.2 will be carried out in Sect. 3. Section 4 is devoted to the proofs of Theorems 1.3 and 1.4, based on the wave structure (1.13).

2 Proof of Theorem 1.1

This section aims to prove Theorem 1.1. As aforementioned, to establish the stability result in Theorem 1.1, it suffices to prove Proposition 2.1 below.

Proposition 2.1

Let $\mathcal {E}_1(t)$ and $\mathcal {E}_2(t)$ be the same ones as defined in (1.8) and (1.9), respectively. Then there exists a generic positive constant C, depending only on $\nu$ and $\eta$, such that

$$\begin{aligned} \displaystyle \mathcal {E}_{1}(t)&\le C\left( \mathcal {E}_1(0)+\mathcal {E}_1(0)^{\frac{3}{2}}+\mathcal {E}_1(0)^2\right) \nonumber \\&\quad +C\left( \mathcal {E}_1(t)^{\frac{3}{2}}+\mathcal {E}_2(t)^{\frac{3}{2}}\right) +C\left( \mathcal {E}_1(t)^3+\mathcal {E}_2(t)^3\right) \end{aligned}$$

(2.1)

and

$$\begin{aligned} \displaystyle \mathcal {E}_2(t) \le C\mathcal {E}_1(0)+C\mathcal {E}_1(t)+C\mathcal {E}_1(t)^{\frac{3}{2}}+C\mathcal {E}_2(t)^{\frac{3}{2}}. \end{aligned}$$

(2.2)

With Proposition 2.1 at our disposal, Theorem 1.1 can be easily achieved by the bootstrapping argument. For simplicity, we denote by C and $C_i$ ($i=1,2,3$) various generic positive constants, which may depend only on $\nu$ and $\eta$, and may change from line to line.

Proof of Theorem 1.1

It follows from (2.1) and (2.2) that

$$\begin{aligned} \mathcal {E}_1(t)+\mathcal {E}_2(t)&\le C_1\left( \mathcal {E}_1(0)+\mathcal {E}_1(0)^{\frac{3}{2}}+\mathcal {E}_1(0)^2\right) \nonumber \\&\quad +C_2\left( \mathcal {E}_1(t)^{\frac{3}{2}}+\mathcal {E}_2(t)^{\frac{3}{2}} \right) +C_3\left( \mathcal {E}_1(t)^3+\mathcal {E}_2(t)^3\right) . \end{aligned}$$

(2.3)

The bootstrapping argument then allows us to establish the stability result of Theorem 1.1, provided the initial data $\mathcal {E}_1(0)$ is chosen to be sufficiently small such that

$$\begin{aligned} \begin{aligned} C_1\left( \mathcal {E}_1(0)+\mathcal {E}_1(0)^{\frac{3}{2}}+\mathcal {E}_1(0)^2\right) \le \frac{1}{4}\min \left\{ \frac{1}{16 C^2_2}, \left( \frac{1}{4C_3}\right) ^{\frac{1}{2}} \right\} . \end{aligned} \end{aligned}$$

(2.4)

In fact, if we make the ansatz that for $0<T\le \infty$,

$$\begin{aligned} \mathcal {E}_1(t)+\mathcal {E}_2(t) \le \min \left\{ \frac{1}{16 C^2_2},\ \left( \frac{1}{4C_3}\right) ^{\frac{1}{2}} \right\} , \end{aligned}$$

then (2.3) implies

$$\begin{aligned} \mathcal {E}_1(t)+\mathcal {E}_2(t)&\le C_1\left( \mathcal {E}_1(0)+\mathcal {E}_1(0)^{\frac{3}{2}}+\mathcal {E}_1(0)^2\right) \\&\quad +C_2\left( \mathcal {E}_1(t)+\mathcal {E}_2(t)\right) ^{\frac{3}{2}} +C_3\left( \mathcal {E}_1(t) +\mathcal {E}_2(t)\right) ^{3} \\&\le C_1\left( \mathcal {E}_1(0)+\mathcal {E}_1(0)^{\frac{3}{2}}+\mathcal {E}_1(0)^2\right) +\frac{1}{2}\left( \mathcal {E}_1(t) +\mathcal {E}_2(t)\right) , \end{aligned}$$

or

$$\begin{aligned} \mathcal {E}_1(t)+\mathcal {E}_2(t)&\le 2C_1\left( \mathcal {E}_1(0)+\mathcal {E}_1(0)^{\frac{3}{2}}+\mathcal {E}_1(0)^2\right) , \end{aligned}$$

(2.5)

which, combined with the smallness assumption (2.4) on the initial data, leads to

$$\begin{aligned} \mathcal {E}_1(t)+\mathcal {E}_2(t)\le \frac{1}{2}\min \left\{ \frac{1}{16 C^2_2}, \left( \frac{1}{4C_3}\right) ^{\frac{1}{2}} \right\} . \end{aligned}$$

Thus, the bootstrapping argument then asserts that (2.5) holds for all time, provided $\mathcal {E}_1(0)$ fulfills (2.4). The proof of Theorem 1.1 is therefore complete. $\square$

It remains to prove Proposition 2.1. To deal with the nonlinear terms, we need to make use of the anisotropic inequalities (cf. Lemmas 2.1 and 2.2), whose proofs rely on the basic one-dimensional Sobolev inequality

$$\begin{aligned} \Vert g\Vert _{L^{\infty }(\mathbb {R})}\le \sqrt{2}\Vert g\Vert _{L^2(\mathbb {R})}^{\frac{1}{2}}\Vert g'\Vert _{L^2(\mathbb {R})}^{\frac{1}{2}},\end{aligned}$$

and the Minkowski inequality

$$\begin{aligned} \Vert \Vert f\Vert _{L^{q}_y(\mathbb {R}^n)}\Vert _{L^{p}_x(\mathbb {R}^m)}\le \Vert \Vert f\Vert _{L^{p}_x(\mathbb {R}^m)}\Vert _{L^{q}_y(\mathbb {R}^n)},\quad \forall \ 1\le q\le p \le \infty ,\end{aligned}$$

where $f=f(x,y)$ with $x\in \mathbb {R}^m$ and $y\in \mathbb {R}^n$ is a measurable function on $\mathbb {R}^m\times \mathbb {R}^n$.

Lemma 2.1

Assume that f, $\partial _1f$, g and $\partial _2g$ are all in $L^2(\mathbb R^2)$. Then,

$$\begin{aligned} \Vert fg\Vert _{L^2(\mathbb R^2)}\le C\Vert f\Vert _{L^2(\mathbb R^2)}^{\frac{1}{2}}\Vert \partial _1f\Vert _{L^2(\mathbb R^2)}^{\frac{1}{2}}\Vert g\Vert _{L^2(\mathbb R^2)}^{\frac{1}{2}}\Vert \partial _{2}g\Vert _{L^2(\mathbb R^2)}^{\frac{1}{2}}. \end{aligned}$$

Lemma 2.2

The following estimates hold when the right-hand sides are all bounded,

$$\begin{aligned} \Vert f\Vert _{L^\infty (\mathbb R^2)}\le C\Vert f\Vert _{L^2(\mathbb R^2)}^{\frac{1}{4}}\Vert \partial _1f\Vert _{L^2(\mathbb R^2)}^{\frac{1}{4}}\Vert \partial _2f\Vert _{L^2(\mathbb R^2)}^{\frac{1}{4}}\Vert \partial _{12}f\Vert _{L^2(\mathbb R^2)}^{\frac{1}{4}}. \end{aligned}$$

In particular,

$$\begin{aligned}&\Vert f\Vert _{L^{\infty }}\le C\Vert f\Vert _{H^1}^{\frac{1}{2}}\Vert \partial _1f\Vert _{H^1}^{\frac{1}{2}}, \\&\Vert f\Vert _{L^{\infty }}\le C\Vert f\Vert _{H^1}^{\frac{1}{2}}\Vert \partial _2f\Vert _{H^1}^{\frac{1}{2}}. \end{aligned}$$

We are now ready to prove Proposition 2.1. The proofs are split into two steps, which are concerned with the derivations of (2.1) and (2.2), respectively.

2.1 Proof of (2.1)

Due to the equivalence of $\Vert (u, b)\Vert _{H^3}$ with $\Vert (u, b)\Vert _{L^2}+\Vert (u, b)\Vert _{\dot{H}^3}$, it suffices to bound the $L^2$-norm and the homogeneous $\dot{H}^3$-norm of (u, b). First, based on the divergence-free conditions $\nabla \cdot u=\nabla \cdot b=0$, it is easy to check that

$$\begin{aligned} \Vert (u,b)\Vert _{L^2}^2+2\int _0^t\left( \nu \Vert u_1\Vert _{L^2}^2+\eta \Vert b_2\Vert _{L^2}^2\right) d\tau =\Vert (u_0,b_0)\Vert _{L^2}^2. \end{aligned}$$

(2.6)

Next, to estimate the $\dot{H^3}$-norm, applying $\partial _i^3(i=1,2)$ to (1.2) and dotting them with $(\partial _i^3u, \partial _i^3b)$ in $L^2$, we have

$$\begin{aligned} \frac{1}{2}&\frac{d}{d t}\sum _{i=1}^{2}\Vert (\partial _i^3u, \partial _i^3b)\Vert _{L^2}^2+\nu \sum _{i=1}^{2}\Vert \partial _i^3u_1\Vert _{L^2}^2+\eta \sum _{i=1}^{2}\Vert \partial _i^3 b_2\Vert _{L^2}^2\nonumber \\&:=K_1+K_2+K_3+K_4+K_5, \end{aligned}$$

(2.7)

where

$$\begin{aligned}&K_1:=\sum _{i=1}^{2}\int \left( \partial _i^3\partial _1 b \cdot \partial _i^3 u +\partial _i^3\partial _1 u \cdot \partial _i^3 b\right) d x, \\&K_2:=-\sum _{i=1}^{2}\int \partial _i^3(u\cdot \nabla u)\cdot \partial _i^3 u \ d x, \\&K_3:=\sum _{i=1}^{2}\int \left( \partial _i^3(b\cdot \nabla b)-b\cdot \nabla \partial _i^3 b \right) \cdot \partial _i^3 u \ dx, \\&K_4:=-\sum _{i=1}^{2}\int \partial _i^3(u\cdot \nabla b)\cdot \partial _i^3 b \ d x, \\&K_5:=\sum _{i=1}^{2}\int \left( \partial _i^3(b\cdot \nabla u)-b\cdot \nabla \partial _i^3 u \right) \cdot \partial _i^3 b \ dx. \end{aligned}$$

We are now in a position of estimating $K_1,\ldots ,K_5$ term by term. First, integration by parts directly gives

$$\begin{aligned} K_1=0. \end{aligned}$$

(2.8)

To bound $K_2$, we divide it into two parts,

$$\begin{aligned} K_2&=-\int \partial _1^3(u\cdot \nabla u)\cdot \partial _1^3 u \ d x-\int \partial _2^3(u\cdot \nabla u)\cdot \partial _2^3 u \ d x:=K_{21}+K_{22}. \end{aligned}$$

Due to $\nabla \cdot u=0$, by Hölder’s and Sobolev’s inequalities, we obtain

$$\begin{aligned} K_{21}&=-\int (\partial _1^3u\cdot \nabla u+3\partial _1^2u\cdot \nabla \partial _1 u+3\partial _1u\cdot \nabla \partial _1^2 u)\cdot \partial _1^3 u \ d x\nonumber \\&\le C\Vert \partial _1^3 u\Vert _{L^2}\left( \Vert \nabla u\Vert _{L^\infty }\Vert \nabla \partial _1^2 u\Vert _{L^2}+\Vert \partial _1^2 u\Vert _{L^4}\Vert \nabla \partial _1 u\Vert _{L^4}\right) \nonumber \\&\le C\Vert u\Vert _{H^3} \Vert \partial _1 u\Vert ^{2}_{H^2}, \end{aligned}$$

(2.9)

and similarly,

$$\begin{aligned} K_{22}\le C\Vert u\Vert _{H^3} \Vert \partial _2 u\Vert ^{2}_{H^2}, \end{aligned}$$

which, together with (2.9), yields

$$\begin{aligned} K_2\le C\Vert u\Vert _{H^3}\left( \Vert \partial _1 u\Vert ^{2}_{H^2}+\Vert \partial _2u\Vert ^{2}_{H^2} \right) . \end{aligned}$$

(2.10)

To estimate $K_3$, we rewrite it into three items,

$$\begin{aligned} K_3&=\sum _{i=1}^{2} 3\int \partial _i b \cdot \nabla \partial _i^2b \cdot \partial _i^3 u \ d x+\sum _{i=1}^{2} 3\int \partial _i^2b\cdot \nabla \partial _i b\cdot \partial _i^3 u \ d x\\&\quad +\sum _{i=1}^{2}\int \partial _i^3b\cdot \nabla b\cdot \partial _i^3 u \ d x :=K_{31}+K_{32}+K_{33}, \end{aligned}$$

where the first term $K_{31}$ on the right-hand side can be bounded as follows,

$$\begin{aligned} K_{31}&=3\int \left( \partial _1 b \cdot \nabla \partial _1^2 b \cdot \partial _1^3 u+\partial _2 b_1 \partial _1\partial _2^2 b \cdot \partial _2^3 u-\partial _1 b_1 \partial _2^3 b \cdot \partial _2^3 u\right) d x \\&\le C\Vert \partial _1 b\Vert _{L^\infty }\Vert \nabla \partial _1^2 b\Vert _{L^2}\Vert \partial _1^3 u\Vert _{L^2}\\&\quad \quad +C\left( \Vert \partial _2 b_1\Vert _{L^\infty }\Vert \partial _1\partial _2^2 b\Vert _{L^2}+\Vert \partial _1 b_1\Vert _{L^\infty }\Vert \partial _2^3 b\Vert _{L^2}\right) \Vert \partial _2^3 u\Vert _{L^2} \\&\le C\Vert b\Vert _{H^3}\left( \Vert \partial _1 b\Vert ^{2}_{H^2}+\Vert \partial _1u\Vert ^{2}_{H^2}+\Vert \partial _2u\Vert ^{2}_{H^2} \right) . \end{aligned}$$

In a similar manner,

$$\begin{aligned} K_{32}&=3\int \left( \partial _1^2b\cdot \nabla \partial _1 b\cdot \partial _1^3 u+\partial _2^2b_1 \partial _1\partial _2 b\cdot \partial _2^3 u-\partial _1\partial _2 b_1 \partial _2^2 b\cdot \partial _2^3 u\right) d x \\&\le C\Vert \partial _1^2b\Vert _{L^4}\Vert \nabla \partial _1 b\Vert _{L^4}\Vert \partial _1^3 u\Vert _{L^2}+ C\Vert \partial _1\partial _2 b\Vert _{L^4}\Vert \partial _2^2b\Vert _{L^4}\Vert \partial _2^3 u\Vert _{L^2}\\&\le C\Vert b\Vert _{H^3}\left( \Vert \partial _1 b\Vert ^{2}_{H^2}+\Vert \partial _1u\Vert ^{2}_{H^2}+\Vert \partial _2u\Vert ^{2}_{H^2} \right) , \end{aligned}$$

and

$$\begin{aligned} K_{33}&=\int \left( \partial _1^3b\cdot \nabla b\cdot \partial _1^3 u+\partial _2^3 b_1 \partial _1 b\cdot \partial _2^3 u-\partial _1\partial _2^2 b_1 \partial _2 b\cdot \partial _2^3 u\right) d x \\&\le C\Vert \nabla b\Vert _{L^\infty }\Vert \partial _1^3 b\Vert _{L^2}\Vert \partial _1^3 u\Vert _{L^2}\\&\quad \quad +C\left( \Vert \partial _1 b\Vert _{L^\infty }\Vert \partial _2^3 b_1\Vert _{L^2}+\Vert \partial _2 b\Vert _{L^\infty }\Vert \partial _1\partial _2^2 b_1\Vert _{L^2}\right) \Vert \partial _2^3 u\Vert _{L^2}\\&\le C\Vert b\Vert _{H^3}\left( \Vert \partial _1 b\Vert ^{2}_{H^2}+\Vert \partial _1u\Vert ^{2}_{H^2}+\Vert \partial _2u\Vert ^{2}_{H^2} \right) . \end{aligned}$$

Therefore,

$$\begin{aligned} K_3\le C\Vert b\Vert _{H^3}\left( \Vert \partial _1 b\Vert ^{2}_{H^2}+\Vert \partial _1u\Vert ^{2}_{H^2}+\Vert \partial _2u\Vert ^{2}_{H^2} \right) . \end{aligned}$$

(2.11)

In order to estimate $K_4$, we write it in the form:

$$\begin{aligned} K_4&=-\int \partial _1^3(u\cdot \nabla b)\cdot \partial _1^3 b \ dx-\int \partial _2^3(u\cdot \nabla b)\cdot \partial _2^3 b \ d x :=K_{41}+K_{42}, \end{aligned}$$

where the first term $K_{41}$ can be easily bounded by

$$\begin{aligned} K_{41}&=-\int \partial _1^3u\cdot \nabla b\cdot \partial _1^3 b \ dx-3\int \left( \partial _1^2 u \cdot \nabla \partial _1 b+\partial _1 u \cdot \nabla \partial _1^2 b\right) \cdot \partial _1^3 b \ d x\nonumber \\&\le C\Vert \nabla b\Vert _{L^\infty }\Vert \partial _1^3 b\Vert _{L^2}\Vert \partial _1^3 u\Vert _{L^2}\nonumber \\&\quad \quad +C\left( \Vert \partial _1^2 u\Vert _{L^4}\Vert \nabla \partial _1 b\Vert _{L^4}+\Vert \partial _1 u\Vert _{L^\infty }\Vert \nabla \partial _1^2 b\Vert _{L^2}\right) \Vert \partial _1^3 b\Vert _{L^2}\nonumber \\&\le C\Vert b\Vert _{H^3}\left( \Vert \partial _1 b\Vert ^{2}_{H^2}+\Vert \partial _1u\Vert ^{2}_{H^2}\right) . \end{aligned}$$

(2.12)

The second term $K_{42}$ needs more work. First, by virtue of the divergence-free condition $\nabla \cdot u=0$, we split it into three parts:

$$\begin{aligned} K_{42}&= -\int \partial _2^3 u \cdot \nabla b \cdot \partial _2^3 b\ d x -3\int \partial _2^2 u \cdot \nabla \partial _2 b \cdot \partial _2^3 b\ d x\\&\quad -3\int \partial _2 u \cdot \nabla \partial _2^2 b \cdot \partial _2^3 b\ d x :=K_{421}+K_{422}+K_{423}. \end{aligned}$$

For $K_{421}$, we have

$$\begin{aligned} K_{421}&= -\int \partial _2^3 u_1 \partial _1 b \cdot \partial _2^3 b\ d x -\int \partial _2^3 u_2\partial _2 b \cdot \partial _2^3 b\ d x \\&= -\int \partial _2^3 u_1 \partial _1 b \cdot \partial _2^3 b\ d x +\int \partial _2^3 u_2 \partial _1 b_1 \partial _2^3 b_2\ d x\\&\quad +\int \partial _1\partial _2^2 u_1 \partial _2 b_1 \partial _2^3 b_1\ d x:=K_{4211}+K_{4212}+K_{4213}. \end{aligned}$$

where the first two terms $K_{4211}$ and $K_{4212}$ are bounded by

$$\begin{aligned} K_{4211}+K_{4212}&\le C\Vert \partial _1 b\Vert _{L^\infty }\Vert \partial _2^3 u \Vert _{L^2}\Vert \partial _2^3 b\Vert _{L^2}\\&\le C\Vert b\Vert _{H^3}\left( \Vert \partial _1 b\Vert _{H^2}^2+ \Vert \partial _2 u\Vert _{H^2}^2\right) . \end{aligned}$$

For $K_{4213}$, integration by parts twice gives

$$\begin{aligned} K_{4213}&=-\int \partial _2^2 u_1 \partial _1\partial _2 b_1 \partial _2^3 b_1\ d x+\int \partial _2^3 u_1 \partial _2 b_1 \partial _1\partial _2^2 b_1\ d x\\&\quad +\int \partial _2^2 u_1\partial _2^2 b_1\partial _1\partial _2^2 b_1\ d x \\&\le C\Vert \partial _2^2 u_1\Vert _{L^4}\Vert \partial _1\partial _2 b_1\Vert _{L^4}\Vert \partial _2^3 b_1\Vert _{L^2}\\&\quad \quad +C\left( \Vert \partial _2^3 u_1\Vert _{L^2}\Vert \partial _2 b_1\Vert _{L^\infty }+\Vert \partial _2^2 u_1\Vert _{L^4}\Vert \partial _2^2 b_1\Vert _{L^4}\right) \Vert \partial _1\partial _2^2 b\Vert _{L^2}\\&\le C\Vert b\Vert _{H^3}\left( \Vert \partial _1 b\Vert _{H^2}^2+ \Vert \partial _2 u\Vert _{H^2}^2\right) , \end{aligned}$$

which, together with the estimates of $K_{4211}$ and $K_{4212}$, shows that

$$\begin{aligned} K_{421}\le C\Vert b\Vert _{H^3}\left( \Vert \partial _1 b\Vert _{H^2}^2+ \Vert \partial _2 u\Vert _{H^2}^2\right) . \end{aligned}$$

(2.13)

Analogously,

$$\begin{aligned} K_{422}&= -3\int \partial _2^2 u_1\partial _1\partial _2 b \cdot \partial _2^3 b\ d x+3\int \partial _2^2 u_1\partial _1\partial _2^2 b \cdot \partial _2^2 b \ d x\nonumber \\&\le C\Vert b\Vert _{H^3}\left( \Vert \partial _1 b\Vert _{H^2}^2+ \Vert \partial _2 u\Vert _{H^2}^2\right) . \end{aligned}$$

(2.14)

For $K_{423}$, due to $\nabla \cdot u=\nabla \cdot b =0$, we have

$$\begin{aligned} K_{423}&=-3\int \partial _2 u_1\partial _1\partial _2^2 b \cdot \partial _2^3 b\ d x-3\int \partial _2 u_2\partial _1\partial _2^2 b_1 \partial _1\partial _2^2 b_1\ d x\\&\quad +3\int \partial _1 u_1\partial _2^3 b_1 \partial _2^3 b_1\ d x:=K_{4231}+K_{4232}+D_1. \end{aligned}$$

Based upon the Hölder’s and Sobolev’s inequalities, it is easily deduced that

$$\begin{aligned} K_{4231}+K_{4232}&\le C\Vert \partial _2 u\Vert _{L^\infty }\Vert \partial _1\partial _2^2 b\Vert _{L^2}\Vert \nabla ^3 b\Vert _{L^2}\nonumber \\&\le C\Vert b\Vert _{H^3}\left( \Vert \partial _1 b\Vert ^{2}_{H^2}+\Vert \partial _2u\Vert ^{2}_{H^2} \right) . \end{aligned}$$

(2.15)

We now turn to deal with $D_1$, which is one of the most difficult terms. The strategy here is to replace $\partial _1 u_1$ by using the equation of magnetic field,

$$\begin{aligned} \partial _{1}u_1=\partial _t b_1+u\cdot \nabla b_1- b\cdot \nabla u_1. \end{aligned}$$

(2.16)

In terms of (2.16), we can rewrite $D_1$ in the form:

$$\begin{aligned} D_1&=3\int \left( \partial _t b_1+u\cdot \nabla b_1- b\cdot \nabla u_1\right) |\partial _2^3 b_1|^2 \ d x \nonumber \\&=3\frac{d}{d t}\int b_1|\partial _2^3 b_1|^2\ dx-6 \int b_1\partial _2^3 b_1\partial _2^3\partial _t b_1 \ dx\nonumber \\&\quad +3\int \left( u\cdot \nabla b_1\right) |\partial _2^3 b_1|^2\ d x-3\int \left( b\cdot \nabla u_1\right) |\partial _2^3 b_1|^2 \ d x, \end{aligned}$$

(2.17)

where the second term associated with $\partial _tb_1$ on the right side can be written as

$$\begin{aligned}&-6 \int b_1\partial _2^3 b_1\partial _2^3\partial _t b_1 \ dx \nonumber \\&\quad =-6\int b_1\partial _2^3 b_1\partial _2^3(\partial _{1}u_1-u\cdot \nabla b_1+b\cdot \nabla u_1) \ d x\nonumber \\&\quad =-6\int b_1\partial _2^3 b_1\partial _2^3\partial _{1}u_1\ d x +6\int b_1\partial _2^3 b_1\partial _2^3u\cdot \nabla b_1 \ d x \nonumber \\&\qquad +18\int b_1\partial _2^3 b_1\partial _2^2u\cdot \nabla \partial _2 b_1\ dx+18\int b_1\partial _2^3 b_1\partial _2u\cdot \nabla \partial _2^2 b_1 dx \nonumber \\&\qquad +3\int b_1 u\cdot \nabla |\partial _2^3 b_1|^2 \ d x-6\int b_1\partial _2^3 b_1\partial _2^3(b\cdot \nabla u_1)\, d x. \end{aligned}$$

(2.18)

Noting that

$$\begin{aligned} \int b_1 u\cdot \nabla |\partial _2^3 b_1|^2 \ d x+\int u\cdot \nabla b_1|\partial _2^3 b_1|^2\ d x=0, \end{aligned}$$

we obtain after plugging (2.18) into (2.17) that

$$\begin{aligned} D_1&=3\frac{d}{d t}\int b_1|\partial _2^3 b_1|^2\ dx-6\int b_1\partial _2^3 b_1\partial _2^3\partial _{1}u_1\ d x \nonumber \\&\quad +6\int b_1\partial _2^3 b_1\partial _2^3u\cdot \nabla b_1 \ d x +18\int b_1\partial _2^3 b_1\partial _2^2u\cdot \nabla \partial _2 b_1\ dx \nonumber \\&\quad +18\int b_1\partial _2^3 b_1\partial _2u_1 \partial _1\partial _2^2 b_1\ dx+27\int b_1\partial _2u_2 |\partial _2^3 b_1|^2\ dx \nonumber \\&\quad -3\int b_2 \partial _2 u_1|\partial _2^3 b_1|^2 \ d x -6\int b_1\partial _2^3 b_1\partial _2^3b_2\partial _2 u_1\ d x \nonumber \\&\quad -18\int b_1\partial _2^3 b_1\partial _2^2b\cdot \nabla \partial _2 u_1\ d x-18\int b_1\partial _2^3 b_1\partial _2 b\cdot \nabla \partial _2^2 u_1\ d x\nonumber \\&\quad -6\int b_1\partial _2^3 b_1b\cdot \nabla \partial _2^3 u_1\ d x. \end{aligned}$$

(2.19)

Two of the most difficult terms on the right-hand side of (2.19) are the second and sixth terms,

$$\begin{aligned} D_2:=-6\int b_1\partial _2^3b_1\partial _2^3\partial _1u_1 \ d x, \quad D_3:=27\int b_1\partial _2u_2|\partial _2^3b_1|^2 \ d x, \end{aligned}$$

which will be handled by using (2.16) and the equation of velocity,

$$\begin{aligned} \partial _{1}b_1=\partial _t u_1+u\cdot \nabla u_1+\partial _1P+\nu u_1- b\cdot \nabla b_1. \end{aligned}$$

(2.20)

For $D_2$, using (2.16), (2.20) and integrating by parts, we have

$$\begin{aligned} D_2 =&\, 6\int \partial _1b_1\partial _2^3 b_1\partial _2^3u_1\ d x+6\int b_1\partial _2^3u_1\partial _2^3\partial _1 b_1\ d x \nonumber \\ :=&\,J_{1}+6\int b_1\partial _2^3u_1\partial _2^3\left( \partial _t u_1+u\cdot \nabla u_1+\partial _1P+\nu u_1- b\cdot \nabla b_1\right) \ d x \nonumber \\ =&\,J_{1}+3\frac{d}{d t}\int b_1|\partial _2^3 u_1|^2\ dx-3\int |\partial _2^3 u_1|^2(\partial _1 u_1+b\cdot \nabla u_1)\ dx\nonumber \\&+\,6\sum _{k=1}^{3}\mathcal {C}_3^k\int b_1 \partial _2^3 u_1\partial _2^ku\cdot \nabla \partial _2^{3-k} u_1\ dx\nonumber \\&+\,6\int b_1\partial _2^3 u_1\partial _2^3\partial _1 P \ d x+6\nu \int b_1|\partial _2^3 u_1|^2\ d x\nonumber \\&-\,6\sum _{k=1}^{3}\mathcal {C}_3^k \int b_1\partial _2^3 u_1\partial _2^kb\cdot \nabla \partial _2^{3-k} b_1\ dx-6\int b_1\partial _2^3 u_1b\cdot \nabla \partial _2^3 b_1\ d x, \end{aligned}$$

(2.21)

where the symbol $\mathcal {C}_n^k$ denotes the standard combination number, and

$$\begin{aligned}J_{1}:=6\int \partial _1b_1\partial _2^3 b_1\partial _2^3u_1\ d x.\end{aligned}$$

Here, we have also used the divergence-free condition $\nabla \cdot u=0$ to get that

$$\begin{aligned} \int b_1 u\cdot \nabla |\partial _2^3 u_1|^2 \ d x+\int u\cdot \nabla b_1|\partial _2^3 u_1|^2\ d x=0. \end{aligned}$$

To deal with $D_3$, we first infer from (2.16) that

$$\begin{aligned} D_3:=&\, 27\int b_1\partial _2u_2|\partial _2^3b_1|^2 \ d x=-27\int b_1\partial _1u_1|\partial _2^3b_1|^2 \ d x \nonumber \\ =&\,-27\int b_1|\partial _2^3b_1|^2(\partial _t b_1+u\cdot \nabla b_1- b\cdot \nabla u_1) \ d x \nonumber \\ =&\,-\frac{27}{2}\frac{d}{d t}\int b_1^2|\partial _2^3 b_1|^2\ dx+27\int b_1^2\partial _2^3 b_1\partial _2^3\partial _t b_1\ dx\nonumber \\&-\,\frac{27}{2}\int |\partial _2^3b_1|^2u\cdot \nabla b_1^2 \ d x+27\int b_1|\partial _2^3b_1|^2 b\cdot \nabla u_1\ d x, \end{aligned}$$

(2.22)

where, similarly to the derivation of (2.21), the second term on the right-hand side can be written as

$$\begin{aligned}&27\int b_1^2\partial _2^3 b_1\partial _2^3\partial _t b_1\ dx=27\int b_1^2\partial _2^3 b_1\partial _2^3(\partial _{1}u_1-u\cdot \nabla b_1+b\cdot \nabla u_1)\ dx\nonumber \\&\quad =27\int b_1^2\partial _2^3 b_1\partial _2^3\partial _{1}u_1 \ dx-\frac{27}{2}\int b_1^2u\cdot \nabla |\partial _2^3b_1|^2\ dx\nonumber \\&\qquad -27\sum _{k=1}^{3}\mathcal {C}_3^k\int b_1^2\partial _2^3 b_1\partial _2^ku\cdot \nabla \partial _2^{3-k} b_1\ dx\nonumber \\&\qquad +27\sum _{k=1}^{3}\mathcal {C}_3^k\int b_1^2\partial _2^3 b_1\partial _2^kb\cdot \nabla \partial _2^{3-k} u_1\ dx+27\int b_1^2\partial _2^3 b_1b\cdot \nabla \partial _2^3 u_1\ dx. \end{aligned}$$

(2.23)

Thus, inserting (2.23) into (2.22) and noting that

$$\begin{aligned} \int b_1^2 u\cdot \nabla |\partial _2^3 b_1|^2 \ d x+\int u\cdot \nabla b_1^2|\partial _2^3 b_1|^2\ d x=0, \end{aligned}$$

we find

$$\begin{aligned} D_3&=-\frac{27}{2}\frac{d}{d t}\int b_1^2|\partial _2^3 b_1|^2\ dx+27\int b_1^2\partial _2^3 b_1\partial _2^3\partial _{1}u_1 \ dx\nonumber \\&\quad +27\int b_1|\partial _2^3b_1|^2 b\cdot \nabla u_1\ d x-27\sum _{k=1}^{3}\mathcal {C}_3^k\int b_1^2\partial _2^3 b_1\partial _2^ku\cdot \nabla \partial _2^{3-k} b_1\ dx\nonumber \\&\quad +27\sum _{k=1}^{3}\mathcal {C}_3^k\int b_1^2\partial _2^3 b_1\partial _2^kb\cdot \nabla \partial _2^{3-k} u_1\ dx+27\int b_1^2\partial _2^3 b_1b\cdot \nabla \partial _2^3 u_1\ dx. \end{aligned}$$

(2.24)

Clearly, we still need to deal with the second term on the right-hand side of (2.24). In fact, using (2.16) and (2.20) again, we have from integration by parts that

$$\begin{aligned} D_4:=&\,27\int b_1^2\partial _2^3 b_1\partial _2^3\partial _1 u_1\ dx\nonumber \\ =&\,-54\int b_1\partial _1b_1\partial _2^3 b_1\partial _2^3u_1\ dx-27\int b_1^2\partial _2^3 u_1\partial _2^3\partial _1 b_1\ dx\nonumber \\ :=&\,J_2-27\int b_1^2\partial _2^3 u_1\partial _2^3\left( \partial _t u_1+u\cdot \nabla u_1+\partial _1P+\nu u_1- b\cdot \nabla b_1\right) \ dx\nonumber \\ =&\,J_2-\frac{27}{2}\frac{d}{d t}\int b_1^2|\partial _2^3 u_1|^2\ dx+27\int |\partial _2^3 u_1|^2 b_1(\partial _1u_1-u\cdot \nabla b_1+b\cdot \nabla u_1)\ dx\nonumber \\&-\,27\int b_1^2\partial _2^3 u_1 u\cdot \nabla \partial _2^3 u_1\ dx-27\sum _{k=1}^{3}\mathcal {C}_3^k \int b_1^2\partial _2^3 u_1\partial _2^ku\cdot \nabla \partial _2^{3-k} u_1\ dx\nonumber \\&-\,27\int b_1^2\partial _2^3 u_1 \partial _2^3\partial _1P\ dx-27\nu \int b_1^2|\partial _2^3 u_1|^2 \ dx\nonumber \\&+\,27\int b_1^2\partial _2^3 u_1 b\cdot \nabla \partial _2^3 b_1\ dx+27\sum _{k=1}^{3}\mathcal {C}_3^k\int b_1^2\partial _2^3 u_1\partial _2^k b\cdot \nabla \partial _2^{3-k} b_1\ dx, \end{aligned}$$

(2.25)

where $J_2$ is given by

$$\begin{aligned} J_2:=-54\int b_1\partial _1b_1\partial _2^3 b_1\partial _2^3u_1\ dx. \end{aligned}$$

Now, plugging (2.21), (2.24) and (2.25) into (2.19), we obtain after careful rearrangement that

$$\begin{aligned} D_1=&\,3\frac{d}{d t}\int b_1\left( |\partial _2^3 b_1|^2+|\partial _2^3 u_1|^2\right) dx-\frac{27}{2}\frac{d}{d t}\int b_1^2\left( |\partial _2^3 b_1|^2\ dx+ |\partial _2^3 u_1|^2\right) dx\nonumber \\&+\,J_1+J_2+6\int b_1\partial _2^3 b_1\partial _2^3u\cdot \nabla b_1 \ d x+18\int b_1\partial _2^3 b_1\partial _2^2u\cdot \nabla \partial _2 b_1\ dx\nonumber \\&-\,24\int b_1\partial _2^3 b_1\partial _2u_1 \partial _2^3 b_2\ dx -3\int b_2 \partial _2 u_1|\partial _2^3 b_1|^2 \ d x\nonumber \\&-\,18\int b_1\partial _2^3 b_1\partial _2^2b\cdot \nabla \partial _2 u_1\ d x-18\int b_1\partial _2^3 b_1\partial _2 b\cdot \nabla \partial _2^2 u_1\ d x\nonumber \\&+\,6\int b\cdot \nabla b_1(\partial _2^3 u_1\partial _2^3 b_1)\ d x-3\int |\partial _2^3 u_1|^2(\partial _1 u_1+b\cdot \nabla u_1)\ dx\nonumber \\&+\,6\sum _{k=1}^{3}\mathcal {C}_3^k\int b_1 \partial _2^3 u_1\partial _2^ku\cdot \nabla \partial _2^{3-k} u_1\ dx+6\int b_1\partial _2^3 u_1\partial _2^3\partial _1 P \ d x\nonumber \\&+\,6\nu \int b_1|\partial _2^3 u_1|^2\ d x-6\sum _{k=1}^{3}\mathcal {C}_3^k\int b_1\partial _2^3 u_1\partial _2^kb\cdot \nabla \partial _2^{3-k} b_1\ dx \nonumber \\&+\,27\int |\partial _2^3 u_1|^2 b_1(\partial _1u_1-u\cdot \nabla b_1+b\cdot \nabla u_1)\ dx-27\int b_1^2\partial _2^3 u_1 u\cdot \nabla \partial _2^3 u_1\ dx\nonumber \\&-\,27\sum _{k=1}^{3}\mathcal {C}_3^k \int b_1^2\partial _2^3 u_1\partial _2^ku\cdot \nabla \partial _2^{3-k} u_1\ dx-27\int b_1^2\partial _2^3 u_1 \partial _2^3\partial _1P\ dx\nonumber \\&-\,27\nu \int b_1^2|\partial _2^3 u_1|^2 \ dx+27\sum _{k=1}^{3}\mathcal {C}_3^k \int b_1^2\partial _2^3 u_1\partial _2^k b\cdot \nabla \partial _2^{3-k} b_1\ dx \nonumber \\&+\,27\int b_1|\partial _2^3b_1|^2 b\cdot \nabla u_1\ d x-27\sum _{k=1}^{3}\mathcal {C}_3^k \int b_1^2\partial _2^3 b_1\partial _2^ku\cdot \nabla \partial _2^{3-k} b_1\ dx\nonumber \\&+\,27\sum _{k=1}^{3} \mathcal {C}_3^k \int b_1^2\partial _2^3 b_1\partial _2^kb\cdot \nabla \partial _2^{3-k} u_1\ dx-54\int b\cdot \nabla b_1(b_1\partial _2^3 u_1\partial _2^3 b_1)\ d x\nonumber \\ :=&\,I'(t)+J_1+J_2+\cdots +J_{24}, \end{aligned}$$

(2.26)

where we have also used $\nabla \cdot b=0$ and the following simple facts that

$$\begin{aligned} 27&\int b_1^2\partial _2^3 b_1b\cdot \nabla \partial _2^3 u_1\ d x+27\int b_1^2\partial _2^3 u_1b\cdot \nabla \partial _2^3 b_1\ d x\\&=27\int b_1^2b\cdot \nabla (\partial _2^3 u_1\partial _2^3 b_1)\ d x=-54\int b\cdot \nabla b_1(b_1\partial _2^3 u_1\partial _2^3 b_1)\ d x, \end{aligned}$$

and

$$\begin{aligned} -6&\int b_1\partial _2^3 b_1b\cdot \nabla \partial _2^3 u_1\ d x-6\int b_1\partial _2^3 u_1b\cdot \nabla \partial _2^3 b_1\ d x\\&=-6\int b_1b\cdot \nabla (\partial _2^3 u_1\partial _2^3 b_1)\ d x=6\int b\cdot \nabla b_1(\partial _2^3 u_1\partial _2^3 b_1)\ d x. \end{aligned}$$

Next, we need to bound $J_1, J_2,\ldots$ and $J_{24}$ one by one. First, it follows from the Sobolev’s embedding inequality that

$$\begin{aligned} J_{1}+J_{2}&\le C\Vert \partial _1 b\Vert _{L^\infty }\Vert \partial _2^3 u \Vert _{L^2}\Vert \partial _2^3 b\Vert _{L^2}(1+\Vert b_1\Vert _{L^\infty })\\&\le C(\Vert b\Vert _{H^3}+\Vert b\Vert _{H^3}^2)\left( \Vert \partial _1 b\Vert _{H^2}^2+ \Vert \partial _2 u\Vert _{H^2}^2\right) . \end{aligned}$$

For $J_{3}$, $J_{8}$ and $J_{9}$, by Lemma 2.2, we have

$$\begin{aligned} J_3+J_{8}+J_{9}&\le C\Vert b\Vert _{L^\infty }\Vert \nabla b \Vert _{L^\infty }\Vert \partial _2^3 b_1\Vert _{L^2}\Vert \nabla \partial _2^2 u \Vert _{L^2}\\&\le C\Vert b \Vert _{H^1}^{\frac{1}{2}}\Vert \partial _1b \Vert _{H^1}^{\frac{1}{2}}\Vert \nabla b\Vert _{H^1}^{\frac{1}{2}}\Vert \partial _1\nabla b\Vert _{H^1}^{\frac{1}{2}}\Vert b\Vert _{H^3}\Vert \partial _2 u \Vert _{H^2}\\&\le C\Vert b\Vert _{H^3}^2\left( \Vert \partial _1 b\Vert _{H^2}^2+ \Vert \partial _2 u\Vert _{H^2}^2\right) . \end{aligned}$$

For $J_{4}$ and $J_{7}$, we use Lemmas 2.1 and 2.2 to deduce

$$\begin{aligned} J_4+J_{7}&\le C\Vert b\Vert _{L^\infty }\Vert \partial _2^3 b\Vert _{L^2}(\Vert \partial _2^2 u\cdot \nabla \partial _2 b_1\Vert _{L^2}+\Vert \partial _2^2 b\cdot \nabla \partial _2 u_1\Vert _{L^2})\\&\le C\Vert b\Vert _{H^1}^{\frac{1}{2}}\Vert \partial _1b\Vert _{H^1}^{\frac{1}{2}}\Vert b\Vert _{H^3}\Vert \nabla \partial _2 u \Vert _{L^2}^{\frac{1}{2}}\Vert \nabla \partial _2^2 u \Vert _{L^2}^{\frac{1}{2}}\Vert \nabla \partial _2 b\Vert _{L^2}^{\frac{1}{2}}\Vert \nabla \partial _1\partial _2b\Vert _{L^2}^{\frac{1}{2}}\\&\le C\Vert b\Vert _{H^3}^2\left( \Vert \partial _1 b\Vert _{H^2}^2+ \Vert \partial _2 u\Vert _{H^2}^2\right) . \end{aligned}$$

Using $\nabla \cdot b=0$ and the Sobolev’s embedding inequality, we obtain

$$\begin{aligned} J_{5}+J_{6}&\le C\Vert b_1\Vert _{L^\infty }\Vert \partial _2^3 b_1 \Vert _{L^2}\Vert \partial _2 u_1\Vert _{L^\infty }\Vert \partial _2^3 b_2 \Vert _{L^2}\\&\quad +C\Vert b_2\Vert _{L^\infty }\Vert \partial _2 u_1\Vert _{L^\infty }\Vert \partial _2^3 b_1 \Vert _{L^2}^2\\&\le C\Vert b\Vert _{H^3}^2\left( \Vert b_2\Vert _{H^3}^2+\Vert \partial _2 u\Vert _{H^2}^2\right) . \end{aligned}$$

For $J_{10}$, $J_{13}$, $J_{15}$ and $J_{19}$, the Sobolev’s embedding inequality yields

$$\begin{aligned}&J_{10}+J_{13}+J_{15}+J_{19}\\&\quad \le C\Vert \partial _2^3 u_1 \Vert _{L^2}^2\left( \Vert \partial _1 u_1\Vert _{L^\infty }+\Vert b\Vert _{L^\infty }\Vert \nabla u_1\Vert _{L^\infty }+\Vert b_1\Vert _{L^\infty }+\Vert b_1\Vert _{L^\infty }^2\right) \\&\qquad +C\Vert \partial _2^3 u_1 \Vert _{L^2}^2\Vert b_1\Vert _{L^\infty }(\Vert u\Vert _{L^\infty }\Vert \nabla b_1\Vert _{L^\infty }+\Vert b\Vert _{L^\infty }\Vert \nabla u_1\Vert _{L^\infty })\\&\quad \le C\left( \Vert (u,b)\Vert _{H^3}+\Vert (u,b)\Vert _{H^3}^2+\Vert b\Vert _{H^3}^4\right) \Vert \partial _2 u\Vert _{H^2}^2, \end{aligned}$$

and similarly,

$$\begin{aligned} J_{11}&\le C\Vert \partial _2^3 u_1 \Vert _{L^2}\Vert b_1\Vert _{L^\infty }(\Vert \nabla u\Vert _{L^\infty }\Vert \nabla \partial _2^2 u \Vert _{L^2}+\Vert \partial _2^2u\Vert _{L^4}\Vert \nabla \partial _2 u \Vert _{L^4})\\&\le C\Vert (u,b)\Vert _{H^3}^2\Vert \partial _2 u\Vert _{H^2}^2. \end{aligned}$$

To estimate $J_{12}$ and $J_{18}$, we first need to deal with $\Vert \partial _1\partial ^3_2 P\Vert _{L^2}$. In fact, operating ${\nabla \cdot }$ to (1.2)$_1$ yields

$$\begin{aligned}\Delta P=\nabla \cdot (b\cdot \nabla b)-\nabla \cdot (u\cdot \nabla u)-\nu \partial _1u_1,\end{aligned}$$

from which it follows that

$$\begin{aligned} \partial _1\partial ^3_2 P=\partial _1\partial ^3_2\Delta ^{-1}\nabla \cdot (b\cdot \nabla b)-\partial _1\partial ^3_2\Delta ^{-1}\nabla \cdot (u\cdot \nabla u)-\nu \partial _1\partial ^3_2\Delta ^{-1} \partial _1u_1. \end{aligned}$$

(2.27)

Due to $\nabla \cdot b=0$, one has

$$\begin{aligned}\nabla \cdot (b\cdot \nabla b)=\partial _j(b_i\partial _i b_j)=\partial _jb_i\partial _i b_j.\end{aligned}$$

So, using the well known fact that the Riesz operator $\partial _i (-\Delta )^{-\frac{1}{2}}$ with $i=1,2$ is bounded in $L^r$ for any $1<r<\infty$, we deduce

$$\begin{aligned} \Vert \partial _1\partial ^3_2 \Delta ^{-1} \nabla \cdot (b\cdot \nabla b)\Vert _{L^2}=\Vert \partial _1\partial ^3_2 \Delta ^{-1}(\partial _jb_i\partial _i b_j)\Vert _{L^2}\le \Vert \partial _1\partial _2(\partial _jb_i\partial _i b_j)\Vert _{L^2}. \end{aligned}$$

Noting that

$$\begin{aligned} \partial _1\partial _2(\partial _jb_i\partial _i b_j)=\partial _1\partial _2\partial _jb_i\partial _i b_j+\partial _2\partial _jb_i\partial _1\partial _i b_j+\partial _1\partial _jb_i\partial _2\partial _i b_j+\partial _jb_i\partial _1\partial _2\partial _i b_j, \end{aligned}$$

and hence,

$$\begin{aligned} \Vert \partial _1\partial ^3_2 \Delta ^{-1} \nabla \cdot (b\cdot \nabla b)\Vert _{L^2}&\le \Vert \partial _1\partial _2(\partial _jb_i\partial _i b_j)\Vert _{L^2}\nonumber \\&\le C (\Vert \nabla b\Vert _{L^\infty }\Vert \partial _1\partial _2\nabla b\Vert _{L^2}+\Vert \partial _2\nabla b\Vert _{L^4} \Vert \partial _1\nabla b\Vert _{L^4})\nonumber \\&\le C \Vert \nabla b\Vert _{H^2}\Vert \partial _1 b\Vert _{H^2}. \end{aligned}$$

(2.28)

The analogous estimate also holds for $\Vert \partial _1\partial ^3_2 \Delta ^{-1} \nabla \cdot (u\cdot \nabla u)\Vert _{L^2}$, that is,

$$\begin{aligned} \Vert \partial _1\partial ^3_2 \Delta ^{-1} \nabla \cdot (u\cdot \nabla u)\Vert _{L^2}&\le C (\Vert \nabla u\Vert _{L^\infty }\Vert \partial _1\partial _2\nabla u\Vert _{L^2}+\Vert \partial _2\nabla u\Vert _{L^4} \Vert \partial _1\nabla u\Vert _{L^4})\nonumber \\&\le C \Vert \nabla u\Vert _{H^2}\Vert \partial _2 u\Vert _{H^2}. \end{aligned}$$

(2.29)

Thus, inserting (2.28) and (2.29) into (2.27), we arrive at

$$\begin{aligned} \Vert \partial _1\partial ^3_2 P\Vert _{L^2}&\le \Vert \partial _1\partial _2(\partial _jb_i\partial _i b_j)\Vert _{L^2}+\Vert \partial _1\partial _2(\partial _ju_i\partial _i u_j)\Vert _{L^2}+\nu \Vert \partial ^2_1\partial _2 u_1\Vert _{L^2}\nonumber \\&\le C(\Vert \nabla b\Vert _{H^2}\Vert \partial _1 b\Vert _{H^2}+\Vert \nabla u\Vert _{H^2}\Vert \partial _2 u\Vert _{H^2}+\Vert \partial _2 u\Vert _{H^2}). \end{aligned}$$

(2.30)

With (2.30) at our disposal, we can now bound $J_{12}$ and $J_{18}$ by

$$\begin{aligned} J_{12}+J_{18}&\le C\Vert \partial _2^3 u_1 \Vert _{L^2}\Vert \partial _2^3 \partial _1 P\Vert _{L^2}\left( \Vert b_1\Vert _{L^\infty }+\Vert b_1\Vert _{L^\infty }^2\right) \\&\le C\left( \Vert b\Vert _{H^3}+\Vert (u, b)\Vert _{H^3}^2+\Vert b\Vert _{H^3}^3+\Vert b\Vert _{H^3}^4\right) \left( \Vert \partial _1 b\Vert _{H^2}^2+\Vert \partial _2 u\Vert _{H^2}^2\right) . \end{aligned}$$

For $J_{14}$, using Lemma 2.1 and Lemma 2.2, we find

$$\begin{aligned} J_{14}&\le C\Vert \partial _2^3 u_1\Vert _{L^2}\Vert b_1\Vert _{L^\infty }\left( \Vert \nabla b\Vert _{L^\infty }\Vert \nabla \partial _2^2 b\Vert _{L^2}+\Vert \partial _2^2 b\cdot \nabla \partial _2b_1 \Vert _{L^2}\right) \\&\le C\Vert \partial _2^3 u_1\Vert _{L^2}\Vert b_1\Vert _{H^1}^{\frac{1}{2}}\Vert \partial _1b_1\Vert _{H^1}^{\frac{1}{2}}\Vert \nabla b\Vert _{H^1}^{\frac{1}{2}}\Vert \partial _1\nabla b\Vert _{H^1}^{\frac{1}{2}}\Vert \nabla \partial _2^2 b_1\Vert _{L^2}\\&\quad + C\Vert \partial _2^3 u_1\Vert _{L^2}\Vert b_1\Vert _{H^1}^{\frac{1}{2}}\Vert \partial _1b_1\Vert _{H^1}^{\frac{1}{2}}\Vert \partial _2^2b\Vert _{L^2}^{\frac{1}{2}}\Vert \partial _1\partial _2^2b\Vert _{L^2}^{\frac{1}{2}}\Vert \nabla \partial _2b_1\Vert _{L^2}^{\frac{1}{2}}\Vert \nabla \partial _2^2b_1\Vert _{L^2}^{\frac{1}{2}}\\&\le C\Vert b\Vert _{H^3}^2\left( \Vert \partial _1 b\Vert _{H^2}^2+\Vert \partial _2 u\Vert _{H^2}^2\right) . \end{aligned}$$

For $J_{16}$, it is easily seen that

$$\begin{aligned} J_{16}&=-\frac{27}{2}\int b_1^2 u\cdot \nabla |\partial _2^3 u_1|^2\ dx=27\int |\partial _2^3 u_1|^2 b_1 u\cdot \nabla b_1\ dx\\&\le C\Vert \partial _2^3 u_1\Vert ^2_{L^2}\Vert b_1\Vert _{L^\infty }\Vert u\Vert _{L^\infty }\Vert \nabla b_1\Vert _{L^\infty }\\&\le C\left( \Vert u\Vert _{H^3}^2+\Vert b\Vert _{H^3}^4\right) \Vert \partial _2 u\Vert _{H^2}^2. \end{aligned}$$

As the treatment of $J_{14}$, we have

$$\begin{aligned} J_{17}&\le C\Vert b_1\Vert ^2_{L^\infty }\Vert \partial _2^3 u_1 \Vert _{L^2}(\Vert \nabla u\Vert _{L^\infty }\Vert \nabla \partial _2^2 u \Vert _{L^2}+\Vert \partial _2^2u\Vert _{L^4}\Vert \nabla \partial _2 u_1 \Vert _{L^4})\\&\le C\left( \Vert u\Vert _{H^3}^2+\Vert b\Vert _{H^3}^4\right) \Vert \partial _2 u\Vert _{H^2}^2, \end{aligned}$$

and

$$\begin{aligned} J_{20}&\le C\Vert b_1\Vert ^2_{L^\infty }\Vert \partial _2^3 u_1 \Vert _{L^2}(\Vert \nabla b \Vert _{L^\infty }\Vert \nabla \partial _2^2 b \Vert _{L^2}+\Vert \partial _2^2b \Vert _{L^4}\Vert \nabla \partial _2 b_1 \Vert _{L^4})\\&\le C\Vert b\Vert _{H^3}^2\Vert b\Vert _{H^1}\Vert \partial _1b\Vert _{H^1}\Vert \partial _2^3 u_1 \Vert _{L^2}\\&\le C\Vert b\Vert _{H^3}^3\left( \Vert \partial _1b\Vert _{H^2}^2+\Vert \partial _2 u\Vert _{H^2}^2\right) . \end{aligned}$$

Due to $\nabla \cdot u=0$, it holds that $\Vert \nabla u_1\Vert _{L^\infty }=\Vert \partial _2u\Vert _{L^\infty }$. Thus,

$$\begin{aligned} J_{21}+J_{24}&\le C\Vert b\Vert ^2_{L^\infty }\Vert \partial _2^3 b_1\Vert ^2_{L^2}\Vert \nabla u_1 \Vert _{L^\infty }\\&\quad +C\Vert b\Vert ^2_{L^\infty }\Vert \partial _2^3u_1\Vert _{L^2}\Vert \partial _2^3b_1\Vert _{L^2}\Vert \nabla b_1 \Vert _{L^\infty }\\&\le C\Vert b\Vert _{H^1}\Vert \partial _1b\Vert _{H^1}\Vert b\Vert ^2_{H^3}\Vert \partial _2 u \Vert _{H^2}\\&\le C\Vert b\Vert _{H^3}^3\left( \Vert \partial _1b\Vert _{H^2}^2+\Vert \partial _2 u\Vert _{H^2}^2\right) , \end{aligned}$$

and

$$\begin{aligned} J_{22}+J_{23}&\le C\Vert b_1\Vert ^2_{L^\infty }\Vert \partial _2^3 b_1 \Vert _{L^2}\Vert \nabla u_1 \Vert _{L^\infty }\Vert \nabla \partial _2^2 b \Vert _{L^2} \\&\quad +C\Vert b_1\Vert ^2_{L^\infty }\Vert \partial _2^3 b_1 \Vert _{L^2}\Vert \nabla \partial _2 u \Vert _{L^4}\Vert \nabla \partial _2 b \Vert _{L^4}\\&\quad +C\Vert b_1\Vert ^2_{L^\infty }\Vert \partial _2^3 b_1 \Vert _{L^2}\Vert \nabla \partial _2^2u \Vert _{L^2}\Vert \nabla b \Vert _{L^\infty }\\&\le C\Vert b\Vert _{H^3}^2\Vert b\Vert _{H^1}\Vert \partial _1b\Vert _{H^1}\Vert \partial _2 u \Vert _{H^2}\\&\le C\Vert b\Vert _{H^3}^3\left( \Vert \partial _1b\Vert _{H^2}^2+\Vert \partial _2 u\Vert _{H^2}^2\right) . \end{aligned}$$

Thus, noting that $\Vert \partial _1 b\Vert _{H^2}=\Vert \nabla b_2\Vert _{H^2}$, we conclude after inserting the above estimates of $J_1,\ldots ,J_{24}$ in (2.26) and using the Cauchy–Schwarz’s inequality that

$$\begin{aligned} D_1&\le 3\frac{d}{d t}\int b_1\left( |\partial _2^3 b_1|^2+|\partial _2^3 u_1|^2\right) dx-\frac{27}{2}\frac{d}{dt}\int b_1^2\left( |\partial _2^3 b_1|^2+ |\partial _2^3 u_1|^2\right) dx \nonumber \\&\quad + C\left( \Vert (u, b)\Vert _{H^3}+\Vert (u,b)\Vert _{H^3}^4\right) \left( \Vert b_2\Vert _{H^3}^2+\Vert \partial _2 u\Vert _{H^2}^2 \right) . \end{aligned}$$

(2.31)

In view of (2.12), (2.13), (2.14), (2.15) and (2.31), we obtain

$$\begin{aligned} K_{4}&\le 3\frac{d}{d t}\int b_1\left( |\partial _2^3 b_1|^2+|\partial _2^3 u_1|^2\right) dx-\frac{27}{2}\frac{d}{dt}\int b_1^2\left( |\partial _2^3 b_1|^2+ |\partial _2^3 u_1|^2\right) dx \nonumber \\&\quad + C\left( \Vert (u,b)\Vert _{H^3}+\Vert (u,b)\Vert _{H^3}^4\right) \left( \Vert b_2\Vert _{H^3}^2+\Vert \partial _1 u\Vert _{H^2}^2+\Vert \partial _2 u\Vert _{H^2}^2\right) . \end{aligned}$$

(2.32)

It remains to estimate $K_5$. To do this, noting that

$$\begin{aligned} K_5&=\int \left( \partial _1^3(b\cdot \nabla u)-b\cdot \nabla \partial _1^3 u \right) \cdot \partial _1^3 b \ dx\\&\quad +\int \left( \partial _2^3(b\cdot \nabla u)-b\cdot \nabla \partial _2^3 u \right) \cdot \partial _2^3 b \ dx:=K_{51}+K_{52}, \end{aligned}$$

where the first term on the right-hand side can be easily bounded by

$$\begin{aligned} K_{51}&=\int (3\partial _1 b\cdot \nabla \partial _1^2 u+3\partial _1^2 b\cdot \nabla \partial _1 u+\partial _1^3 b\cdot \nabla u) \cdot \partial _1^3b \ dx\nonumber \\&\le C\left( \Vert \partial _1 b\Vert _{L^\infty }\Vert \nabla \partial _1^2 u\Vert _{L^2} +\Vert \partial _1^2 b\Vert _{L^4}\Vert \nabla \partial _1 u\Vert _{L^4}+\Vert \nabla u\Vert _{L^\infty }\Vert \partial _1^3 b\Vert _{L^2}\right) \Vert \partial _1^3 b\Vert _{L^2}\nonumber \\&\le C\Vert u\Vert _{H^3} \Vert \partial _1 b\Vert ^{2}_{H^2}. \end{aligned}$$

(2.33)

To deal with $K_{52}$, we rewrite it as

$$\begin{aligned} K_{52}&= \int \left( 3\partial _2 b\cdot \nabla \partial _2^2 u \cdot \partial _2^3b +3\partial _2^2 b\cdot \nabla \partial _2 u \cdot \partial _2^3b+ \partial _2^3 b\cdot \nabla u \cdot \partial _2^3b \right) dx\\&= 3\int \partial _2 b\cdot \nabla \partial _2^2 u \cdot \partial _2^3b\ dx + 3\int \partial _2^2 b\cdot \nabla \partial _2 u \cdot \partial _2^3b \ dx\\&\quad -\int \partial _1\partial _2^2 b_1\partial _2 u \cdot \partial _2^3b\ dx-\int \partial _2^3 b_1\partial _1 u_2 \partial _1\partial _2^2b_1\ dx\\&\quad +\int \partial _1 u_1|\partial _2^3b_1|^2 \ dx:=K_{521}+K_{522}+K_{523}+K_{524}+\frac{1}{3} D_{1}. \end{aligned}$$

Based upon integration by parts and the divergence-free condition $\nabla \cdot b=0$, we deduce from the Sobolev’s inequalities that

$$\begin{aligned} K_{521}&=3\int \partial _2 b_1\partial _1\partial _2^2 u \cdot \partial _2^3b \ dx+3\int \partial _2 b_2\partial _2^3 u \cdot \partial _2^3b \ dx\nonumber \\&\ =-3\int \partial _1\partial _2 b_1\partial _2^2 u \cdot \partial _2^3b \ dx+3\int \partial _2^2 b_1\partial _2^2 u \cdot \partial _1\partial _2^2b \ dx\nonumber \\&\quad \quad +3\int \partial _2 b_1\partial _2^3 u \cdot \partial _1\partial _2^2b \ dx-3\int \partial _1 b_1\partial _2^3 u \cdot \partial _2^3b \ dx\nonumber \\&\le C\Vert \partial _1\partial _2 b_1\Vert _{L^4}\Vert \partial _2^2 u\Vert _{L^4}\Vert \partial _2^3 b\Vert _{L^2}+C\Vert \partial _2^2 b_1\Vert _{L^4}\Vert \partial _2^2 u\Vert _{L^4}\Vert \partial _1\partial _2^2 b\Vert _{L^2} \nonumber \\&\quad +C\Vert \partial _2b_1\Vert _{L^\infty }\Vert \partial _2^3 u\Vert _{L^2}\Vert \partial _1\partial _2^2 b\Vert _{L^2}+C\Vert \partial _1b_1\Vert _{L^\infty }\Vert \partial _2^3 u\Vert _{L^2}\Vert \partial _2^3 b\Vert _{L^2} \nonumber \\&\le C\Vert b\Vert _{H^3}\left( \Vert \partial _1 b\Vert _{H^2}^2+ \Vert \partial _2 u\Vert _{H^2}^2\right) , \end{aligned}$$

(2.34)

and similarly,

$$\begin{aligned} K_{522}&=3\int \partial _2^2 b_1\partial _1\partial _2 u \cdot \partial _2^3b \ dx+3\int \partial _2^2 b_2\partial _2^2 u \cdot \partial _2^3b \ dx\nonumber \\&\ =-3\int \partial _1\partial _2^2 b_1\partial _2 u \cdot \partial _2^3b \ dx+3\int \partial _2^3 b_1\partial _2 u \cdot \partial _1\partial _2^2b \ dx\nonumber \\&\quad \quad +3\int \partial _2^2 b_1\partial _2^2 u \cdot \partial _1\partial _2^2b \ dx-3\int \partial _1\partial _2 b_1\partial _2^2 u \cdot \partial _2^3b \ dx\nonumber \\&\ \le C(\Vert \partial _1\partial _2^2 b\Vert _{L^2}\Vert \partial _2 u\Vert _{L^\infty }+\Vert \partial _1\partial _2 b_1\Vert _{L^4}\Vert \partial _2^2 u\Vert _{L^4})\Vert \partial _2^3 b\Vert _{L^2}\nonumber \\&\ \quad +C\Vert \partial _2^2 b_1\Vert _{L^4}\Vert \partial _2^2 u\Vert _{L^4}\Vert \partial _1\partial _2^2 b\Vert _{L^2} \nonumber \\&\ \le C\Vert b\Vert _{H^3}\left( \Vert \partial _1 b\Vert _{H^2}^2+ \Vert \partial _2 u\Vert _{H^2}^2\right) . \end{aligned}$$

(2.35)

For $K_{523}$ and $K_{524}$, we have

$$\begin{aligned} K_{523}+K_{524}&\le C\Vert \partial _1\partial _2^2b_1\Vert _{L^2}\Vert \partial _2^3 b\Vert _{L^2}\left( \Vert \partial _2 u\Vert _{L^\infty }+\Vert \partial _1 u_2\Vert _{L^\infty }\right) \nonumber \\&\le C\Vert b\Vert _{H^3}\left( \Vert \partial _1 b\Vert _{H^2}^2+ \Vert \partial _2 u\Vert _{H^2}^2+ \Vert \partial _1 u\Vert _{H^2}^2\right) . \end{aligned}$$

(2.36)

Thus, combining (2.33), (2.34), (2.35), (2.36) with (2.31) gives

$$\begin{aligned} K_{5}&\le \frac{d}{d t}\int b_1\left( |\partial _2^3 b_1|^2+|\partial _2^3 u_1|^2\right) dx-\frac{9}{2}\frac{d}{dt}\int b_1^2\left( |\partial _2^3 b_1|^2+ |\partial _2^3 u_1|^2\right) dx \nonumber \\&\quad + C\left( \Vert (u,b)\Vert _{H^3}+\Vert (u,b)\Vert _{H^3}^4\right) \left( \Vert b_2\Vert _{H^3}^2+\Vert \partial _1 u\Vert _{H^2}^2+\Vert \partial _2 u\Vert _{H^2}^2\right) . \end{aligned}$$

(2.37)

Now, substituting (2.8), (2.10), (2.11), (2.32) and (2.37) into (2.7), we find

$$\begin{aligned} \frac{1}{2}&\frac{d}{d t}\Vert (\nabla ^3 u, \nabla ^3 b)\Vert ^2_{L^2}+\nu \Vert \nabla ^3 u_1\Vert ^2_{L^2}+\eta \Vert \nabla ^3 b_2\Vert ^2_{L^2}\\&\le 4\frac{d}{d t}\int b_1\left( |\partial _2^3 b_1|^2+|\partial _2^3 u_1|^2\right) dx-18\frac{d}{d t}\int b_1^2\left( |\partial _2^3 b_1|^2+|\partial _2^3 u_1|^2\right) dx \\&\quad + C\left( \Vert (u, b)\Vert _{H^3}+\Vert (u, b)\Vert _{H^3}^4\right) \left( \Vert b_2\Vert _{H^3}^2+\Vert \partial _2 u\Vert _{H^2}^2+ \Vert \partial _1 u\Vert _{H^2}^2\right) , \end{aligned}$$

which, integrated over [0, t] and combined with the Sobolev’s inequalities, yields

$$\begin{aligned}&\Vert \nabla ^3(u, b)(t)\Vert ^2_{L^2}+2\int _0^t\left( \nu \Vert \nabla ^3 u_1\Vert ^2_{L^2}+\eta \Vert \nabla ^3 b_2\Vert ^2_{L^2}\right) d\tau \nonumber \\&\quad \le C\left( \Vert (u_0, b_0)\Vert ^2_{H^3}+\Vert (u_0, b_0)\Vert ^3_{H^3}+\Vert (u_0, b_0)\Vert ^4_{H^3}\right) \nonumber \\&\qquad +8\int b_1\left( |\partial _2^3 b_1|^2+|\partial _2^3 u_1|^2\right) dx-36\int b_1^2\left( |\partial _2^3 b_1|^2+|\partial _2^3 u_1|^2\right) dx\nonumber \\&\qquad + C\int _0^t\left( \Vert (u, b)\Vert _{H^3}+\Vert (u, b)\Vert _{H^3}^4\right) \left( \Vert b_2\Vert _{H^3}^2+\Vert \partial _2 u\Vert _{H^2}^2+ \Vert \partial _1 u\Vert _{H^2}^2\right) d \tau \nonumber \\&\quad \le C\left( \Vert (u_0, b_0)\Vert ^2_{H^3}+\Vert (u_0, b_0)\Vert ^3_{H^3}+\Vert (u_0, b_0)\Vert ^4_{H^3}\right) \nonumber \\&\qquad +C\left( \Vert b_1(t)\Vert _{L^\infty } +\Vert b_1(t)\Vert ^2_{L^\infty }\right) \Vert (u, b)(t)\Vert ^2_{H^3}\nonumber \\&\qquad +C\sup _{0\le \tau \le t}\left( \Vert (u, b)\Vert _{H^3}+\Vert (u, b)\Vert _{H^3}^4\right) \int _0^t\left( \Vert (u_1,b_2)\Vert _{H^3}^2+ \Vert \partial _1 u\Vert _{H^2}^2\right) d\tau , \end{aligned}$$

(2.38)

due to the fact that $\Vert \partial _2u\Vert _{H^2}=\Vert \nabla u_1\Vert _{H^2}$. Thus, it readily follows from (2.6) and (2.38) that

$$\begin{aligned} \mathcal {E}_1(t)&\le C\mathcal {E}_1(0)+C\mathcal {E}_1(0)^{\frac{3}{2}}+C\mathcal {E}_1(0)^2\\&\quad +C\mathcal {E}_1(t)^{\frac{3}{2}}+C\mathcal {E}_2(t)^{\frac{3}{2}} +C\mathcal {E}_1(t)^3+C\mathcal {E}_2(t)^3. \end{aligned}$$

The proof of the first assertion (2.1) in Proposition 2.1 is therefore complete.

2.2 Proof of (2.2)

Since $\Vert \partial _1u\Vert _{H^2}\sim \Vert \partial _1u\Vert _{L^2}+\Vert \nabla ^2\partial _1u\Vert _{L^2}$, it suffices to establish the estimates of the following two items:

$$\begin{aligned} \int _0^t \Vert \partial _1u(\tau )\Vert _{L^2}^2 d \tau \quad \mathrm {and} \quad \int _0^t\Vert \nabla ^2\partial _1u(\tau )\Vert _{L^2}^2 d \tau ,\end{aligned}$$

whose proofs are based on the special struture of equation (1.2)$_2$,

$$\begin{aligned} \partial _{1}u=\partial _t b+u\cdot \nabla b+\eta {(0, b_2)^\top }-b\cdot \nabla u. \end{aligned}$$

(2.39)

First, to bound $\Vert \partial _1u(\tau )\Vert _{L^2}$, we multiply (2.39) by $\partial _1 u$ in $L^2$ and integrate by parts over $\mathbb {R}^2$ to get

$$\begin{aligned} \Vert \partial _{1}u\Vert ^2_{L^2}=&\, \int \partial _1 u\cdot \partial _tb \ dx+\int u\cdot \nabla b \cdot \partial _1 u\ dx\nonumber \\&\quad +\eta \int b_2 \partial _1u_2\ dx-\int b\cdot \nabla u \cdot \partial _1 u\ dx\nonumber \\ :=&\,L_1+L_2+L_3+L_4. \end{aligned}$$

(2.40)

Using the velocity equation in (1.2)$_1$ and the fact that $\nabla \cdot b=0$, we have

$$\begin{aligned} L_1=&\,\frac{d}{d t}\int \partial _1 u\cdot b \ dx-\int b\cdot \partial _1\left( \partial _{1}b-\nu {(u_1, 0)^\top }+b\cdot \nabla b-u\cdot \nabla u\right) dx \\ :=&\,L_{11}+L_{12}+L_{13}+L_{14}+L_{15}. \end{aligned}$$

It is easily seen that

$$\begin{aligned} L_{12}+L_{13}&=\int \partial _1 b\cdot \partial _1 b \ dx-\nu \int \partial _1 b_1 u_1\ dx\\&\le C \Vert \partial _1 b\Vert ^2_{H^1}+C\Vert \partial _1b\Vert _{L^2}\Vert u_1\Vert _{L^2}. \end{aligned}$$

Integrating by parts and using Sobolev’s embedding inequality, we find

$$\begin{aligned} L_{14}&=-\int b\cdot \partial _1(b\cdot \nabla b)\ dx=\int \partial _1 b\cdot (b\cdot \nabla b)\ dx\\&=\int b_1 \partial _1 b \cdot \partial _1 b\ dx+\int b_2 \partial _2 b \cdot \partial _1 b\ dx\\&\le C\Vert b_1\Vert _{L^\infty }\Vert \partial _1 b\Vert ^2_{L^2}+C\Vert b_2\Vert _{L^\infty }\Vert \partial _2 b\Vert _{L^2}\Vert \partial _1 b\Vert _{L^2}\\&\le C\Vert b \Vert _{H^2} \Vert b_2\Vert ^2_{H^2}, \end{aligned}$$

where we have used the fact that $\Vert \partial _1b\Vert _{L^2}= \Vert \nabla b_2\Vert _{L^2}$ due to $\nabla \cdot b=0$. By virtue of Lemma 2.1, we have

$$\begin{aligned} L_{15}&=\int b\cdot \partial _1(u\cdot \nabla u) \ dx =-\int \partial _1 b\cdot (u\cdot \nabla u)\ dx \\&\le C \Vert \partial _1 b\Vert _{L^2}\Vert u\Vert ^{\frac{1}{2}}_{L^2}\Vert \partial _2u\Vert ^{\frac{1}{2}}_{L^2}\Vert \nabla u\Vert ^{\frac{1}{2}}_{L^2}\Vert \partial _1 \nabla u\Vert ^{\frac{1}{2}}_{L^2}\\&\le C\Vert u\Vert _{H^2}\left( \Vert \partial _1 b\Vert ^2_{H^1}+\Vert \partial _2u\Vert ^2_{H^2}+\Vert \partial _1 u\Vert ^2_{H^1}\right) . \end{aligned}$$

Thus, collecting the estimates of $L_{12},\ldots ,L_{15}$ together, we obtain

$$\begin{aligned} L_1&\le \frac{d}{d t}\int \partial _1 u\cdot b \ dx+ C \left( \Vert b_2\Vert ^2_{H^2}+\Vert u_1\Vert _{L^2}^2\right) \\&\quad + C\Vert (u,b)\Vert _{H^2}\left( \Vert b_2\Vert ^2_{H^2}+\Vert \partial _2u\Vert ^2_{H^2}+\Vert \partial _1 u\Vert ^2_{H^1}\right) , \end{aligned}$$

since $\Vert \partial _1b\Vert _{H^1}\le \Vert b_2\Vert _{H^2}$. In a similar manner,

$$\begin{aligned} L_2&\le C \Vert \partial _1u\Vert _{L^2}\Vert u\Vert ^{\frac{1}{2}}_{L^2}\Vert \partial _2u\Vert ^{\frac{1}{2}}_{L^2}\Vert \nabla b\Vert ^{\frac{1}{2}}_{L^2}\Vert \partial _1 \nabla b\Vert ^{\frac{1}{2}}_{L^2}\\&\le C\Vert (u,b)\Vert _{H^2}\left( \Vert \partial _1 b\Vert ^2_{H^1}+\Vert \partial _2u\Vert ^2_{H^2}+\Vert \partial _1 u\Vert ^2_{H^1}\right) , \\ L_3&\le C\Vert b_2\Vert _{L^2}\Vert \partial _{1} u_2\Vert _{L^2}\le \frac{1}{2}\Vert \partial _{1} u \Vert ^2_{L^2}+C\Vert b_2\Vert _{L^2}^2, \end{aligned}$$

and

$$\begin{aligned} L_4&\le C \Vert \partial _1u\Vert _{L^2}\Vert b\Vert ^{\frac{1}{2}}_{L^2}\Vert \partial _1b\Vert ^{\frac{1}{2}}_{L^2}\Vert \nabla u\Vert ^{\frac{1}{2}}_{L^2}\Vert \partial _2 \nabla u\Vert ^{\frac{1}{2}}_{L^2}\\&\le C\Vert (u,b)\Vert _{H^2}\left( \Vert \partial _1 b\Vert ^2_{H^1}+\Vert \partial _2u\Vert ^2_{H^2}+\Vert \partial _1 u\Vert ^2_{H^1}\right) , \end{aligned}$$

which, combined with the estimate of $L_1$ and (2.40), shows that

$$\begin{aligned} \Vert \partial _1u\Vert _{L^2}^2&\le 2 \frac{d}{d t}\int \partial _1 u\cdot b \ dx+ C \left( \Vert b_2\Vert ^2_{H^2}+\Vert u_1\Vert _{L^2}^2\right) \nonumber \\&\quad + C\Vert (u,b)\Vert _{H^2}\left( \Vert b_2\Vert ^2_{H^2}+\Vert \partial _2u\Vert ^2_{H^2}+\Vert \partial _1 u\Vert ^2_{H^1}\right) . \end{aligned}$$

(2.41)

This leads to the desired estimate of $\Vert \partial _1 u\Vert _{L^2}$.

Next, we proceed to estimate $\Vert \nabla ^2\partial _1u\Vert _{L^2}$. To do this, applying $\nabla ^2$ to (2.39), and dotting it with $\nabla ^2 \partial _1u$ in $L^2$, we deduce

$$\begin{aligned} \Vert \nabla ^2 \partial _1 u\Vert ^2_{L^2}=&\, \int \nabla ^2\partial _1 u\cdot \partial _t \nabla ^2 b \ dx+\int \nabla ^2(u\cdot \nabla b ) \cdot \nabla ^2 \partial _1 u\ dx\nonumber \\&\quad +\eta \int \nabla ^2\partial _1u_2 \cdot \nabla ^2 b_2 \ dx-\int \nabla ^2 (b\cdot \nabla u) \cdot \nabla ^2\partial _1 u \ dx\nonumber \\ :=&\,M_1+M_2+M_3+M_4. \end{aligned}$$

(2.42)

Owing to (1.2)$_1$ and $\nabla \cdot b=0$, we see that

$$\begin{aligned} M_1&=\frac{d}{d t}\int \nabla ^2\partial _1 u\cdot \nabla ^2 b \ dx\\&\quad -\int \nabla ^2 b\cdot \nabla ^2\partial _1\left( \partial _{1}b-\nu {(u_1, 0)^\top }+b\cdot \nabla b-u\cdot \nabla u\right) dx\\&:=M_{11}+M_{12}+M_{13}+M_{14}+M_{15}. \end{aligned}$$

Integrating by parts gives

$$\begin{aligned} M_{12}+M_{13}&=\int \nabla ^2\partial _1 b\cdot \nabla ^2 \partial _1 b \ dx-\nu \int \partial _1 \nabla ^2 b_1\cdot \nabla ^2 u_1\ dx\\&\le C \Vert \partial _1 b\Vert ^2_{H^2}+C\Vert \partial _1 b\Vert _{H^2}\Vert u_1\Vert _{H^2}. \end{aligned}$$

Due to $\Vert \nabla b_2\Vert _{H^k}=\Vert \partial _1 b\Vert _{H^k}$ for $k=1,2$, we have

$$\begin{aligned} M_{14}&=-\int \nabla ^2 b\cdot \nabla ^2\partial _1(b\cdot \nabla b)\ dx=\int \partial _1 \nabla ^2 b\cdot \nabla ^2(b\cdot \nabla b)\ dx\\&=\int \partial _1 \nabla ^2 b \cdot \left( \nabla ^2 b_1\partial _1 b+ \nabla ^2 b_2\partial _2 b\right) dx \\&\quad +2\int \partial _1 \nabla ^2 b \cdot \left( \nabla b_1\partial _1\nabla b+\nabla b_2\partial _2\nabla b\right) dx\\&\quad +\int \left( b_1|\partial _1\nabla ^2 b|^2 +b_2\partial _2\nabla ^2 b\cdot \partial _1\nabla ^2 b\right) dx\\&\le C\Vert \partial _1\nabla ^2 b\Vert _{L^2} \left( \Vert \partial _1 b\Vert _{L^4}\Vert \nabla ^2 b_1\Vert _{L^4}+ \Vert \partial _2 b\Vert _{L^4}\Vert \nabla ^2 b_2\Vert _{L^4}\right) \\&\quad + C\Vert \partial _1\nabla ^2 b\Vert _{L^2} \left( \Vert \nabla b_1\Vert _{L^4}\Vert \partial _1\nabla b\Vert _{L^4}+ \Vert \nabla b_2\Vert _{L^4}\Vert \partial _2\nabla b\Vert _{L^4}\right) \\&\quad +C\left( \Vert b_1\Vert _{L^\infty }\Vert \partial _1\nabla ^2 b\Vert ^2_{L^2} + \Vert b_2\Vert _{L^\infty }\Vert \partial _2 \nabla ^2 b\Vert _{L^2}\Vert \partial _1 \nabla ^2 b\Vert _{L^2}\right) \\&\le C\Vert b \Vert _{H^3} \Vert b_2\Vert ^2_{H^3}. \end{aligned}$$

Analogously, noting that $\Vert \nabla u_2\Vert _{H^k}=\Vert \partial _1 u\Vert _{H^k}$ and $\Vert \nabla u_1\Vert _{H^k}=\Vert \partial _2 u\Vert _{H^k}$ for $k=1,2$, we obtain

$$\begin{aligned} M_{15}&=\int \nabla ^2 b\cdot \nabla ^2\partial _1(u\cdot \nabla u) \ dx =-\int \partial _1 \nabla ^2 b\cdot \nabla ^2(u\cdot \nabla u)\ dx \\&\le C\Vert \partial _1\nabla ^2 b\Vert _{L^2} \left( \Vert \partial _1 u\Vert _{L^4}\Vert \nabla ^2 u_1\Vert _{L^4}+ \Vert \partial _2 u\Vert _{L^4}\Vert \nabla ^2 u_2\Vert _{L^4}\right) \\&\quad + C\Vert \partial _1\nabla ^2b\Vert _{L^2} \left( \Vert \nabla u_1\Vert _{L^4}\Vert \partial _1\nabla u\Vert _{L^4}+ \Vert \nabla u_2\Vert _{L^4}\Vert \partial _2\nabla u\Vert _{L^4}\right) \\&\quad +C\Vert \partial _1 \nabla ^2 b\Vert _{L^2}\left( \Vert u_1\Vert _{L^\infty }\Vert \partial _1\nabla ^2 u\Vert _{L^2} + \Vert u_2\Vert _{L^\infty }\Vert \partial _2 \nabla ^2 u\Vert _{L^2}\right) \\&\le C\Vert u\Vert _{H^3}\left( \Vert b_2\Vert ^2_{H^3}+\Vert \partial _1 u\Vert ^2_{H^2}+\Vert \partial _2u\Vert ^2_{H^2}\right) . \end{aligned}$$

Hence, in terms of the estimates of $M_{1i}$ with $i=2,\ldots ,5$, we can bound $M_1$ by

$$\begin{aligned} M_1&\le \frac{d}{d t}\int \nabla ^2\partial _1 u\cdot \nabla ^2 b \ dx+C\left( \Vert b_2\Vert _{H^3}^2+\Vert u_1\Vert _{H^3}^2\right) \\&\quad +C\Vert (u,b)\Vert _{H^3}\left( \Vert b_2\Vert ^2_{H^3}+\Vert \partial _1 u\Vert ^2_{H^2}+\Vert \partial _2u\Vert ^2_{H^2}\right) . \end{aligned}$$

For $M_2$, by Lemma 2.1 we infer from integration by parts that

$$\begin{aligned} M_2&= \int \nabla ^2(u\cdot \nabla b ) \cdot \nabla ^2 \partial _1 u\ dx\\&= \int \nabla ^2 u\cdot \nabla b \cdot \nabla ^2 \partial _1 u\ dx+2\int \nabla u_i\cdot \partial _i\nabla b\cdot \nabla ^2 \partial _1 u\ dx\\&\quad +\int u_1\partial _1\nabla ^2 b\cdot \partial _1\nabla ^2 u\ dx-\int \partial _1 u_2\partial _2\nabla ^2 b\cdot \nabla ^2 u\ dx\\&\quad +\int \partial _2 u_2\partial _1\nabla ^2 b\cdot \nabla ^2 u\ dx+\int u_2\partial _1\nabla ^2 b\cdot \partial _2\nabla ^2 u\ dx\\&\le C\Vert \partial _1 \nabla ^2 u\Vert _{L^2} \Vert \nabla ^2 u\Vert ^{\frac{1}{2}}_{L^2}\Vert \partial _2 \nabla ^2 u\Vert ^{\frac{1}{2}}_{L^2}\Vert \nabla b\Vert ^{\frac{1}{2}}_{L^2}\Vert \partial _1\nabla b\Vert ^{\frac{1}{2}}_{L^2}\\&\quad +C\Vert \partial _1 \nabla ^2 u\Vert _{L^2} \Vert \nabla u\Vert ^{\frac{1}{2}}_{L^2}\Vert \partial _2 \nabla u\Vert ^{\frac{1}{2}}_{L^2}\Vert \nabla ^2 b\Vert ^{\frac{1}{2}}_{L^2}\Vert \partial _1\nabla ^2 b\Vert ^{\frac{1}{2}}_{L^2}\\&\quad +C\Vert u_1\Vert _{L^\infty }\Vert \partial _1 \nabla ^2 u\Vert _{L^2}\Vert \partial _1 \nabla ^2 b\Vert _{L^2} +C\Vert \partial _1 u_2\Vert _{L^4}\Vert \partial _2 \nabla ^2 b\Vert _{L^2}\Vert \nabla ^2 u\Vert _{L^4}\\&\quad +C\Vert \partial _2 u_2\Vert _{L^4}\Vert \partial _1 \nabla ^2 b\Vert _{L^2}\Vert \nabla ^2 u\Vert _{L^4}+C\Vert u_2\Vert _{L^\infty }\Vert \partial _1 \nabla ^2 b\Vert _{L^2}\Vert \partial _2\nabla ^2 u\Vert _{L^2}\\&\le C\Vert (u,b)\Vert _{H^3}\left( \Vert \partial _1 u\Vert ^2_{H^2}+ \Vert \partial _2 u\Vert ^2_{H^2}+ \Vert \partial _1 b\Vert ^2_{H^2}\right) . \end{aligned}$$

Obviously, $M_3,\, M_4$ can be bounded as follows.

$$\begin{aligned} M_3\le C \Vert \nabla ^2 b_2\Vert _{L^2}\Vert \nabla ^2 \partial _{1} u_2\Vert _{L^2}\le \frac{1}{2}\Vert \partial _{1} \nabla ^2 u \Vert ^2_{L^2}+C\Vert \nabla ^2 b_2\Vert _{L^2}^2, \end{aligned}$$

and

$$\begin{aligned} M_4&=-\int \nabla ^2\partial _1 u\cdot \left( \nabla ^2 b \cdot \nabla u+2\nabla b_i \cdot \partial _i\nabla u+b_i\partial _i\nabla ^2 u\right) \ dx \\&\le C \Vert \partial _1 \nabla ^2 u\Vert _{L^2} \Vert \nabla ^2 b\Vert _{L^2}^{\frac{1}{2}}\Vert \partial _1 \nabla ^2 b\Vert ^{\frac{1}{2}}_{L^2}\Vert \nabla u\Vert _{L^2}^{\frac{1}{2}}\Vert \partial _2\nabla u\Vert ^{\frac{1}{2}}_{L^2} \\&\quad +C \Vert \partial _1 \nabla ^2 u\Vert _{L^2}\Vert \nabla b\Vert _{L^2}^{\frac{1}{2}}\Vert \partial _1 \nabla b\Vert _{L^2}^{\frac{1}{2}}\Vert \nabla ^2 u\Vert ^{\frac{1}{2}}_{L^2}\Vert \partial _2\nabla ^2u\Vert ^{\frac{1}{2}}_{L^2}\\&\quad +C \Vert b\Vert _{L^\infty }\Vert \nabla ^2 \partial _1 u\Vert _{L^2}\Vert \nabla ^3u\Vert _{L^2} \\&\le C\Vert (u,b)\Vert _{H^3}\left( \Vert \partial _1 b\Vert ^2_{H^2}+\Vert \partial _2u\Vert ^2_{H^2}+\Vert \partial _1 u\Vert ^2_{H^2}\right) . \end{aligned}$$

Thus, it follows from (2.42) and the estimates of $M_i$ ($i=1,\ldots ,4$) that

$$\begin{aligned} \Vert \partial _1\nabla ^2u\Vert ^{2}_{L^2}&\le 2\frac{d}{d t}\int \nabla ^2\partial _1 u\cdot \nabla ^2 b \ dx+C\left( \Vert b_2\Vert _{H^3}^2+\Vert u_1\Vert _{H^3}^2\right) \nonumber \\&\quad +C\Vert (u,b)\Vert _{H^3}\left( \Vert b_2\Vert ^2_{H^3}+\Vert \partial _1 u\Vert ^2_{H^2}+\Vert \partial _2u\Vert ^2_{H^2}\right) . \end{aligned}$$

(2.43)

Now, adding up (2.41) and (2.43), we deduce

$$\begin{aligned} \Vert \partial _1 u\Vert ^2_{H^2}&\le 2\frac{d}{d t}\int \left( \partial _1 u\cdot b + \nabla ^2\partial _1 u\cdot \nabla ^2 b \right) dx +C\left( \Vert u_1\Vert ^2_{H^3}+\Vert b_2\Vert _{H^3}^2\right) \\&\quad +C\Vert (u,b)\Vert _{H^3}\left( \Vert b_2\Vert ^2_{H^3}+\Vert u_1\Vert ^2_{H^3}+\Vert \partial _1 u\Vert ^2_{H^2}\right) , \end{aligned}$$

where we have also used $\Vert \nabla b_2\Vert _{H^k}=\Vert \partial _1 b\Vert _{H^k}$ and $\Vert \partial _2u\Vert _{H^k}=\Vert \nabla u_1\Vert _{H^k}$ for $k=1,2$. As an immediate result,

$$\begin{aligned} \int _0^t\Vert \partial _1 u\Vert ^2_{H^2}d\tau&\le C \Vert (u_0,b_0)\Vert ^2_{H^3} + C\Vert (u, b)\Vert ^2_{H^3}+C\int _0^t \left( \Vert u_1\Vert ^2_{H^3}+\Vert b_2\Vert _{H^3}^2\right) d\tau \\&\quad +C\sup _{0\le \tau \le t}\Vert (u,b)\Vert _{H^3}\int _0^t\left( \Vert b_2\Vert ^2_{H^3}+\Vert u_1\Vert ^2_{H^3}+\Vert \partial _1 u\Vert ^2_{H^2}\right) d\tau , \end{aligned}$$

from which it readily follows that

$$\begin{aligned} \mathcal {E}_2(t)\le C\mathcal {E}_1(0)+C\mathcal {E}_1(t)+C\mathcal {E}_1(t)^{\frac{3}{2}}+C\mathcal {E}_2(t)^{\frac{3}{2}}. \end{aligned}$$

The proof of (2.2) is therefore complete.

3 Proof of Theorem 1.2

This section is devoted to the proof of Theorem 1.2 by making full use of the symmetric structure of linearized system (1.12).

Proof of Theorem 1.2

Taking the inner product of (1.12) with (u, b) in $H^1$, we have

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

(3.1)

where

$$\begin{aligned} A(t)&=\Vert (u,b)(t)\Vert ^2_{L^2}+\Vert (\nabla u, \nabla b)(t)\Vert ^2_{L^2}, \\ B(t)&=2\nu \Vert \mathcal R_2 u(t)\Vert ^2_{L^2}+2\eta \Vert \mathcal R_1 b(t)\Vert ^2_{L^2}+2\nu \Vert \nabla \mathcal R_2 u(t)\Vert ^2_{L^2}+2\eta \Vert \nabla \mathcal R_1 b(t)\Vert ^2_{L^2}. \end{aligned}$$

Next, we compute the norm of (u, b) in anisotropic Sobolev space with negative indices. Applying $\Lambda _1^{-\sigma }$ and $\Lambda _2^{-\sigma }$ to (1.12) and dotting them with $(\Lambda _1^{-\sigma }u, \Lambda _1^{-\sigma } b)$ and $(\Lambda _2^{-\sigma }u, \Lambda _2^{-\sigma } b)$ in $H^{1+\sigma }$, respectively, we find

$$\begin{aligned} \begin{aligned}&\frac{d}{d t}H(t)+ 2\nu \Vert \mathcal R_2(\Lambda _1^{-\sigma }, \Lambda _2^{-\sigma }) u(t)\Vert ^2_{L^2}+2\eta \Vert \mathcal R_1 (\Lambda _1^{-\sigma }, \Lambda _2^{-\sigma })b(t)\Vert ^2_{L^2}\\&\quad +2\nu \Vert \mathcal R_2\Lambda ^{1+\sigma }(\Lambda _1^{-\sigma }, \Lambda _2^{-\sigma }) u(t)\Vert ^2_{L^2}+2\eta \Vert \mathcal R_1 \Lambda ^{1+\sigma }(\Lambda _1^{-\sigma }, \Lambda _2^{-\sigma })b(t)\Vert ^2_{L^2}=0, \end{aligned} \end{aligned}$$

(3.2)

where

$$\begin{aligned} H(t)&=\Vert (\Lambda _1^{-\sigma }, \Lambda _2^{-\sigma }) u(t)\Vert ^2_{L^2}+\Vert (\Lambda _1^{-\sigma }, \Lambda _2^{-\sigma })b(t)\Vert ^2_{L^2}\\&\quad +\Vert \Lambda ^{1+\sigma }(\Lambda _1^{-\sigma }, \Lambda _2^{-\sigma }) u(t)\Vert ^2_{L^2}+\Vert \Lambda ^{1+\sigma }(\Lambda _1^{-\sigma }, \Lambda _2^{-\sigma })b(t)\Vert ^2_{L^2}. \end{aligned}$$

We claim that there exists a generic positive constant $C>0$, depending only on $\nu$ and $\eta$, such that

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

(3.3)

In fact, using Plancherel theorem and Hölder’s inequality, we have from direct calculations that

$$\begin{aligned}&\Vert u(t)\Vert _{L^2}^2\le C\Vert \nabla \mathcal R_2 u(t)\Vert _{L^2}^{\frac{2\sigma }{1+\sigma }}\Vert \Lambda _2^{-\sigma }u(t)\Vert _{L^2}^{\frac{2}{1+\sigma }}\le CB(t)^{\frac{\sigma }{1+\sigma }}H(t)^{\frac{1}{1+\sigma }},\\&\Vert \nabla u(t)\Vert _{L^2}^2\le C\Vert \nabla \mathcal R_2 u(t)\Vert _{L^2}^{\frac{2\sigma }{1+\sigma }}\Vert \Lambda ^{1+\sigma }\Lambda _2^{-\sigma }u(t)\Vert _{L^2}^{\frac{2}{1+\sigma }}\le CB(t)^{\frac{\sigma }{1+\sigma }}H(t)^{\frac{1}{1+\sigma }}, \\&\Vert b(t)\Vert _{L^2}^2\le C\Vert \nabla \mathcal R_1 b(t)\Vert _{L^2}^{\frac{2\sigma }{1+\sigma }}\Vert \Lambda _1^{-\sigma }b(t)\Vert _{L^2}^{\frac{2}{1+\sigma }}\le CB(t)^{\frac{\sigma }{1+\sigma }}H(t)^{\frac{1}{1+\sigma }},\\&\Vert \nabla b(t)\Vert _{L^2}^2\le C\Vert \nabla \mathcal R_1 b(t)\Vert _{L^2}^{\frac{2\sigma }{1+\sigma }}\Vert \Lambda ^{1+\sigma }\Lambda _1^{-\sigma }b(t)\Vert _{L^2}^{\frac{2}{1+\sigma }}\le CB(t)^{\frac{\sigma }{1+\sigma }}H(t)^{\frac{1}{1+\sigma }}, \end{aligned}$$

from which the assertion (3.3) follows.

It is easily seen from (3.2) that H(t) is non-increasing, and $H(t)\le H(0)$. Hence, by (3.3) we have

$$\begin{aligned} A(t)\le CB(t)^{\frac{\sigma }{1+\sigma }}H(0)^{\frac{1}{1+\sigma }} \quad \text{ or } \quad B(t)\ge CH(0)^{-\frac{1}{\sigma }}A(t)^{1+\frac{1}{\sigma }}, \end{aligned}$$

which, inserted in(3.1), yields

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

so that

$$\begin{aligned} A(t)\le \left( A(0)^{-\frac{1}{\sigma }}+\frac{C}{\sigma }H(0)^{-\frac{1}{\sigma }}t\right) ^{-\sigma }. \end{aligned}$$

This finishes the proof of Theorem 1.2. $\square$

4 Proofs of Theorems 1.3 and 1.4

This section aims to prove Theorems 1.3 and 1.4, based on the special wave structure of the linearized system (1.13). To begin, we first recall the following elementary lemma, which provides a precise decay rate for a nonnegative integrable function when it decreases in a generalized sense.

Lemma 4.1

For given positive constants $C_0>0$ and $C_1>0$, assume that $f=f(t)$ is a nonnegative function defined on $[0,\infty )$ and satisfies,

$$\begin{aligned} \int _0^\infty f(\tau ) \,d \tau \le C_0<\infty , \quad \text{ and }\quad f(t) \le C_1 \, f(s), \quad \forall \ 0\le s<t. \end{aligned}$$

Then there exists a positive constant $C_2:=\max \{2C_1 f(0), 4C_0 C_1\}$ such that

$$\begin{aligned} f(t) \le C_2 (1+t)^{-1},\quad \forall \ t\ge 0. \end{aligned}$$

Proof

On the one hand, when $0\le t\le 1$, it holds that

$$\begin{aligned} f(t)\le C_1f(0). \end{aligned}$$

On the other hand, when $t\ge 1$, one has

$$\begin{aligned} C_0\ge \int _{\frac{t}{2}}^tf(\tau )\ d\tau \ge C_1^{-1}\int _{\frac{t}{2}}^tf(t)\ d\tau =\frac{t}{2C_1}f(t), \end{aligned}$$

which implies that

$$\begin{aligned} f(t)\le 2C_0C_1t^{-1},\quad \forall \ t\ge 1. \end{aligned}$$

Combining the above two cases leads to the desired decay estimate. $\square$

We are now ready to prove Theorem 1.3.

Proof of Theorem 1.3

Dotting (1.13)$_1$ with $\partial _{t} u$ in $L^2$, we obtain

$$\begin{aligned}&\frac{1}{2}\frac{d}{d t}\left( \Vert \partial _t u\Vert _{L^2}^2+\nu \eta \Vert \mathcal R_1\mathcal R_2u\Vert ^2_{L^2}+\Vert \partial _1 u\Vert _{L^2}^2\right) \nonumber \\&\quad +\nu \Vert \partial _{t}\mathcal R_2u\Vert _{L^2}^2 +\eta \Vert \partial _t \mathcal R_1 u\Vert _{L^2}^2=0, \end{aligned}$$

(4.1)

and hence,

$$\begin{aligned} \frac{d}{d t}\left( \Vert \partial _t u\Vert _{L^2}^2+\nu \eta \Vert \mathcal R_1\mathcal R_2u\Vert ^2_{L^2}+\Vert \partial _1 u\Vert _{L^2}^2\right) \le 0. \end{aligned}$$

(4.2)

Multiplying (1.13)$_1$ by u in $L^2$ and integrating by parts, we have

$$\begin{aligned}&\frac{1}{2}\frac{d}{d t}\left( \eta \Vert \mathcal R_1 u\Vert _{L^2}^2+\nu \Vert \mathcal R_2 u\Vert _{L^2}^2+2\langle \partial _t u, u\rangle \right) \nonumber \\&\quad +\Vert \partial _1 u\Vert _{L^2}^2+\nu \eta \Vert \mathcal R_{1}\mathcal R_{2} u\Vert _{L^2}^2-\Vert \partial _t u\Vert ^2_{L^2} =0. \end{aligned}$$

(4.3)

where $\langle \cdot , \cdot \rangle$ denotes the standard $L^2$-inner product.

Let $\delta :=\min \left\{ \nu , \eta \right\}$. For a constant $\mu >0$ to be specified later, we obtain after adding (4.1) and $\mu \times$ (4.3) together that

$$\begin{aligned}&\frac{1}{2}\frac{d}{d t}\Big (\Vert \partial _tu\Vert _{L^2}^2+\mu \eta \Vert \mathcal R_1 u\Vert _{L^2}^2+\mu \nu \Vert \mathcal R_2 u\Vert ^2_{L^2}+\nu \eta \Vert \mathcal R_1\mathcal R_2 u\Vert _{L^2}^2+\Vert \partial _1 u\Vert _{L^2}^2+2\mu \langle \partial _t u, u\rangle \Big )\nonumber \\&\qquad \quad +(\delta -\mu )\Vert \partial _t u\Vert _{L^2}^2+\mu \nu \eta \Vert \mathcal R_1\mathcal R_2 u\Vert _{L^2}^2+\mu \Vert \partial _1 u\Vert _{L^2}^2\le 0, \end{aligned}$$

(4.4)

since $\Vert \partial _t \mathcal R_2 u\Vert _{L^2}^2+\Vert \partial _t\mathcal R_1 u\Vert _{L^2}^2=\Vert \partial _t u\Vert _{L^2}^2$. By choosing $\mu =\frac{\delta }{4}$, we see that

$$\begin{aligned} \frac{1}{2}\Vert \partial _t u\Vert _{L^2}^2+\frac{1}{8}\delta ^2\Vert u\Vert _{L^2}^2&\le \Vert \partial _tu\Vert _{L^2}^2+\mu \delta \Vert u\Vert _{L^2}^2+2\mu \langle \partial _t u, u\rangle \nonumber \\&\le \Vert \partial _tu\Vert _{L^2}^2+\mu \eta \Vert \mathcal R_1 u\Vert _{L^2}^2+\mu \nu \Vert \mathcal R_2 u\Vert ^2_{L^2}+2\mu \langle \partial _t u, u\rangle , \end{aligned}$$

(4.5)

Thus, by virtue of (4.5), we deduce after integrating (4.4) over (0, t) that

$$\begin{aligned}&\frac{1}{2}\Vert \partial _tu\Vert _{L^2}^2+\frac{1}{8}\delta ^2\Vert u\Vert _{L^2}^2+\nu \eta \Vert \mathcal R_1\mathcal R_2 u\Vert _{L^2}^2+\Vert \partial _1 u\Vert _{L^2}^2 \\&\qquad + 2\int _0^t \left( \frac{3\delta }{4}\Vert \partial _t u\Vert _{L^2}^2+\mu \nu \eta \Vert \mathcal R_1\mathcal R_2 u\Vert _{L^2}^2+\mu \Vert \partial _1 u\Vert _{L^2}^2\right) d \tau \\&\quad \le C\left( \Vert \partial _tu_0\Vert _{L^2}, \Vert u_0\Vert _{L^2}, \Vert \mathcal R_1\mathcal R_2 u_0\Vert _{L^2}, \Vert \partial _1 u_0\Vert _{L^2}\right) , \end{aligned}$$

and consequently,

$$\begin{aligned} \int _0^\infty \left( \Vert \partial _t u\Vert _{L^2}^2+\Vert \mathcal R_1\mathcal R_2 u\Vert _{L^2}^2+\Vert \partial _1 u\Vert _{L^2}^2\right) d t <\infty . \end{aligned}$$

(4.6)

In view of (4.2) and (4.6), it readily follows from Lemma 4.1 that

$$\begin{aligned} \Vert \partial _t u\Vert _{L^2}^2+\Vert \mathcal R_1\mathcal R_2 u\Vert _{L^2}^2+\Vert \partial _1 u\Vert _{L^2}^2\le C(1+t)^{-1}. \end{aligned}$$

Based upon (1.13)$_2$, one can obtain the same result for b. The proof of Theorm 1.3 is thus complete. $\square$

We proceed to prove Theorem 1.4.

Proof of Theorem 1.4

Let $\psi$ be the Fourier cutoff operator defined in (1.15). Taking the convolution of $\psi$ with (1.13)$_1$ leads to

$$\begin{aligned} \partial _{tt}(\psi *u)-\left( \nu \mathcal R_2^2+\eta \mathcal R_1^2\right) \partial _{t}(\psi *u)-\partial _1^2(\psi *u)+\nu \eta \mathcal R_1^2\mathcal R_2^2(\psi *u) =0. \end{aligned}$$

(4.7)

Dotting (4.7) by $\partial _t(\psi *u)$ in $L^2$ and integrating it by parts, we obtain

$$\begin{aligned}&\frac{d}{d t}\left( \Vert \partial _t(\psi *u)\Vert _{L^2}^2+\Vert \partial _1(\psi *u)\Vert _{L^2}^2+\nu \eta \Vert \mathcal R_1\mathcal R_2(\psi *u)\Vert _{L^2}^2\right) \nonumber \\&\quad +2\nu \Vert \partial _t\mathcal R_2(\psi *u)\Vert _{L^2}^2+2\eta \Vert \partial _t\mathcal R_1(\psi *u)\Vert _{L^2}^2=0. \end{aligned}$$

(4.8)

Similarly, multiplying (4.7) by $\psi *u$ in $L^2$, we have

$$\begin{aligned}&\frac{d}{d t}\left( \nu \Vert \mathcal R_2(\psi *u)\Vert _{L^2}^2+\eta \Vert \mathcal R_1(\psi *u)\Vert _{L^2}^2+2\left\langle \partial _t(\psi *u),\psi *u\right\rangle \right) \nonumber \\&\quad +2\Vert \partial _1(\psi *u)\Vert _{L^2}^2+2\nu \eta \Vert \mathcal R_1\mathcal R_2(\psi *u)\Vert _{L^2}^2-2\Vert \partial _t(\psi *u)\Vert _{L^2}^2=0. \end{aligned}$$

(4.9)

Let $\delta :=\min \left\{ \nu , \eta \right\}$, and $\lambda >0$ be a positive constant to be determined later. Then, operating (4.8)+$\lambda \times$ (4.9) yields

$$\begin{aligned}&\frac{d}{d t} F(t)+2(\delta -\lambda )\Vert \partial _t(\psi *u)\Vert _{L^2}^2\nonumber \\&\quad +2\lambda \Vert \partial _1(\psi *u)\Vert _{L^2}^2+2\lambda \nu \eta \Vert \mathcal R_1\mathcal R_2(\psi *u)\Vert _{L^2}^2\le 0 , \end{aligned}$$

(4.10)

where

$$\begin{aligned} F(t)&:=\Vert \partial _t(\psi *u)\Vert _{L^2}^2+\Vert \partial _1(\psi *u)\Vert _{L^2}^2+\nu \eta \Vert \mathcal R_1\mathcal R_2(\psi *u)\Vert _{L^2}^2\\&\quad +\lambda \nu \Vert \mathcal R_2(\psi *u)\Vert _{L^2}^2+\lambda \eta \Vert \mathcal R_1(\psi *u)\Vert _{L^2}^2+2\lambda \left\langle \partial _t(\psi *u),\psi *u\right\rangle . \end{aligned}$$

Let D be the frequency domain defined in (1.14) and $D^c$ be its complement. Moreover, we divide $D^c$ into two regions:

$$\begin{aligned} A_1=\left\{ \xi \in \mathbb {R}^2:\ |\xi _1|\ge \alpha \right\} , \quad A_2=\left\{ \xi \in \mathbb {R}^2:\ |\xi _1|<\alpha ~~\text{ and }~~|\xi |^2\le \beta |\xi _1| |\xi _2|~\right\} . \end{aligned}$$

We can now bound $\Vert \psi *u\Vert _{L^2}^2$ by $\Vert \partial _1(\psi *u)\Vert _{L^2}^2$ and $\Vert \mathcal R_1\mathcal R_2(\psi *u)\Vert _{L^2}^2$. Indeed,

$$\begin{aligned} \Vert \psi *u\Vert _{L^2}^2&= \left\| \widehat{\psi }\widehat{u}\right\| _{L^2}^2 = \int _{A_1}|\widehat{\psi }\widehat{u}|^2 d\xi +\int _{A_2}|\widehat{\psi }\widehat{u}|^2 d \xi \nonumber \\ {}&\le \alpha ^{-2}\int _{A_1}\xi _1^2|\widehat{\psi }\widehat{u}|^2 d \xi +\beta ^2\int _{A_2}\frac{\xi _1^2\xi _2^2}{|\xi |^4}|\widehat{\psi }\widehat{u}|^2 d \xi \nonumber \\ {}&\le \alpha ^{-2} \Vert \partial _1(\psi *u)\Vert _{L^2}^2+\beta ^2\Vert \mathcal R_1\mathcal R_2(\psi *u)\Vert _{L^2}^2 . \end{aligned}$$

(4.11)

Then, multiplying (4.11) by $\lambda ^2$ and then adding with (4.10), we obtain

$$\begin{aligned}&\frac{d}{d t} F(t) +2(\delta -\lambda )\Vert \partial _t(\psi *u)\Vert _{L^2}^2 +(2\lambda -\lambda ^2\alpha ^{-2})\Vert \partial _1(\psi *u)\Vert _{L^2}^2 \nonumber \\&\qquad +(2\lambda \nu \eta -\lambda ^2\beta ^2)\Vert \mathcal R_1\mathcal R_2(\psi *u)\Vert _{L^2}^2 +\lambda ^2\Vert \psi *u\Vert _{L^2}^2\le 0. \end{aligned}$$

(4.12)

Thus, if $\lambda >0$ is chosen to be such that

$$\begin{aligned} \lambda \le \min \left\{ \frac{1}{2}\delta , \alpha ^2, \frac{\nu \eta }{\beta ^2} \right\} , \end{aligned}$$

then we infer from (4.12) that

$$\begin{aligned}&\frac{d}{d t} F(t) + \delta \Vert \partial _t(\psi *u)\Vert _{L^2}^2 +\lambda \Vert \partial _1(\psi *u)\Vert _{L^2}^2 \nonumber \\&\quad +\lambda \nu \eta \Vert \mathcal R_1\mathcal R_2(\psi *u)\Vert _{L^2}^2 +\lambda ^2\Vert \psi *u\Vert _{L^2}^2\le 0. \end{aligned}$$

(4.13)

Recalling the definition of F and noting that

$$\begin{aligned} 2\lambda \left( \partial _t(\psi *u),\psi *u\right) \le \lambda \Vert \partial _t(\psi *u)\Vert _{L^2}^2+\lambda \Vert \psi *u\Vert _{L^2}^2, \end{aligned}$$

(4.14)

we obtain after operating (4.13)+$\lambda ^2\times$ (4.14) that ($\vartheta :=\max \left\{ \nu , \eta \right\}$)

$$\begin{aligned}&\frac{d}{d t}F+\lambda ^2F+(\delta -\lambda ^2-\lambda ^3)\Vert \partial _t(\psi *u)\Vert _{L^2}^2 +\left( \lambda -\lambda ^2\right) \Vert \partial _1(\psi *u)\Vert _{L^2}^2 \nonumber \\&\quad +\left( \lambda \nu \eta -\lambda ^2\nu \eta \right) \Vert \mathcal R_1\mathcal R_2(\psi *u)\Vert _{L^2}^2 +\lambda ^2(1-\vartheta \lambda -\lambda )\Vert \psi *u\Vert _{L^2}^2\le 0, \end{aligned}$$

(4.15)

If $\lambda >0$ is taken to be sufficiently small such that

$$\begin{aligned} \lambda = \min \left\{ \frac{1}{4}\delta , \alpha ^2, 1,\frac{\nu \eta }{\beta ^2},\frac{1}{\vartheta +1}\right\} , \end{aligned}$$

(4.16)

then it follows from (4.15) that

$$\begin{aligned} \frac{d}{d t}F+\lambda ^2F\le 0 \quad \text{ or }\quad F(t)\le F(0)e^{-\lambda ^2t}. \end{aligned}$$

(4.17)

In view of the simple inequality,

$$\begin{aligned} 2\lambda \left( \partial _t(\psi *u),\psi *u\right) \le \frac{1}{2}\Vert \partial _t(\psi *u)\Vert _{L^2}^2+2\lambda ^2\Vert \psi *u\Vert _{L^2}^2, \end{aligned}$$

one easily has

$$\begin{aligned} \frac{1}{2}\Vert \partial _t(\psi *u)\Vert _{L^2}^2+\frac{1}{2}\lambda \delta \Vert \psi *u\Vert _{L^2}^2+\nu \eta \Vert \mathcal R_1\mathcal R_2(\psi *u)\Vert _{L^2}^2+\Vert \partial _1(\psi *u)\Vert _{L^2}^2\le F(t). \end{aligned}$$

As an immediate consequence of (4.17), we conclude that for $\lambda$ satisfying (4.16),

$$\begin{aligned} \Vert \psi *u\Vert _{L^2}^2+\Vert \partial _t(\psi *u)\Vert _{L^2}^2+\Vert \partial _1(\psi *u)\Vert _{L^2}^2+\Vert \mathcal R_1\mathcal R_2(\psi *u)\Vert _{L^2}^2 \le Ce^{-c(\eta ,\nu ,\alpha ,\beta ) t} \end{aligned}$$

where

$$\begin{aligned} c(\eta ,\nu ,\alpha ,\beta ):=\left( \min \left\{ \frac{1}{4}\delta , \alpha ^2, 1,\frac{\nu \eta }{\beta ^2},\frac{1}{\vartheta +1}\right\} \right) ^2. \end{aligned}$$

The same result also holds for b. The proof of Theorem 1.4 is therefore finished. $\square$