Stability and time decay rates of the 2D magneto-micropolar equations with partial dissipation

Abstract

This paper studies the stability and decay estimates of solutions to the two-dimensional (2D) magneto-micropolar fluid equations with partial dissipation. We first establish the $L^2$-decay estimates for global solutions and their derivative with initial data in $L^1({\mathbb {R}}^2)$. Furthermore, we show the global stability of these solutions in $H^s(\mathbb {R}^2)$, and the decay rates of these global solutions and their higher derivatives when the initial data belongs to the negative Sobolev space $\dot{H}^{-l}(\mathbb {R}^2)$ (for each $0 \le l < 1$).

1 Introduction

The 3D incompressible magneto-micropolar fluid equations can be written as

$$\begin{aligned} \left\{ \begin{array}{lcl} \partial _t u +u \cdot \nabla u = -\nabla \pi + (\mu +\chi ) \Delta u +b\cdot \nabla b + 2\chi \nabla \times \omega ,\\ \partial _t \omega +u\cdot \nabla \omega -\gamma \nabla \nabla \cdot \omega + 4\chi \omega =\kappa \Delta \omega + 2\chi \nabla \times u, \\ \partial _t b +u\cdot \nabla b =\nu \Delta b + b\cdot \nabla u, \\ \nabla \cdot u =0, ~ \nabla \cdot b =0,\\ u(x, y, z, 0) =u_0(x, y, z), \omega (x, y, z, 0) =\omega _0(x, y, z), b(x, y, z, 0) =b_0(x, y, z), \end{array} \right. \end{aligned}$$

(1.1)

where $(x, y, z)\in {\mathbb {R}}^3$ and $t\ge 0$, $u(x,y,z,t), \omega (x,y,z,t), b(x,y,z,t)$ and $\pi (x,y,z,t)$ denote the velocity of the fluid, microrotational velocity, the magnetic field and the hydrostatic pressure, respectively. $\mu , \chi$ and $\frac{1}{\nu }$ are, respectively, kinematic viscosity, vortex viscosity and magnetic Reynolds number. $\gamma$ and $\kappa$ are angular viscosities. The 3D magneto-micropolar equations reduce to the 2D micropolar equations when

$$\begin{aligned}&u = (u_1(x,y,t), u_2(x,y,t),0), \qquad \pi = \pi (x,y,t), \\&b = (b_1(x,y,t), b_2(x,y,t),0), \qquad \omega = (0,0,\omega _3(x,y,t)). \end{aligned}$$

More explicitly, the 2D incompressible magneto-micropolar fluid equations can be written as

$$\begin{aligned} \left\{ \begin{array}{lcl} \partial _{t}u + u\cdot \nabla u = (\mu + \chi ) \Delta u - \nabla \pi + b \cdot \nabla b + 2\chi \nabla \times \omega ,\\ \partial _{t}\omega + u\cdot \nabla \omega + 4\chi \omega = \kappa \Delta \omega + 2\chi \nabla \times u,\\ \partial _{t}b + u \cdot \nabla b = \nu \Delta b + b \cdot \nabla u,\\ \nabla \cdot u = \nabla \cdot b = 0,\\ u(x,y,0) = u_{0}(x,y), \omega (x,y,0) = \omega _{0}(x,y), b(x,y,0) = b_{0}(x,y), \end{array} \right. \end{aligned}$$

(1.2)

where we have written $u=(u_1, u_2)$, $b=(b_1, b_2)$ and $\omega$ for $\omega _3$ for notational brevity. It is worth noting that, in the 2D case,

$$\begin{aligned} \Omega \equiv \nabla \times u = \partial _{x}u_{2} - \partial _{y}u_{1} \end{aligned}$$

is a scalar function representing the vorticity, and $\nabla \times \omega = (\partial _{y}\omega , -\partial _{x}\omega )$.

The magneto-micropolar equations were introduced in [1] to describe the motion of incompressible, electrically conducting micropolar fluids in the presence of an arbitrary magnetic field. It has attracted considerable attention from the community of mathematical fluids (see, e.g., [2,3,4,5,6,7,8,9,10,11,12]). The magneto-micropolar equations share similarities with the Navier–Stokes equations, but they contain much richer structures than Navier–Stokes. It is well-known that the $L^2$ decay problem of weak solutions to the 3D Navier–Stokes equations, i.e., (1.1) with $\omega = 0$, $b=0$ and $\chi =0$, was proposed by the celebrated work of Leray [13]. By introducing the elegant method of Fourier splitting, the algebraic decay rate for weak solutions was first obtained by Schonbek [14]. Later, the result in [14] is sharpened and extended in [15], see also [16]. Recently, in [17] Zhao obtained the decay rates of solutions for the three-dimensional incompressible Navier–Stokes equations with damping term $|u|^{\beta -1}u$ ($\beta \ge 3$).

When (1.2) has full dissipation (namely, $\mu$, $\chi$, $\kappa$, $\nu >0$), the global existence and uniqueness of solutions can be obtained easily (see, e.g., [8, 18]). For more results related to the well-posedness of solutions, one refers to [19,20,21,22,23,24,25] and the reference therein. However, for the inviscid case (namely, (1.2) with $\mu > 0$, $\chi > 0$, $\kappa = \nu = 0$ and $\Delta u$ replaced by u), the global regularity problem is still a challenging open problem. Therefore, it is natural to study the intermediate cases, namely (1.2) with partial dissipation.

In certain physical regimes and under suitable scaling, the full Laplacian dissipation is reduced to a partial dissipation. One notable example is the Prandtl boundary layer equation in which only the vertical dissipation is included in the horizontal component (see, e.g., [26]). This paper focuses on a system of the 2D incompressible magneto-micropolar equations that is closely related to (1.2),

$$\begin{aligned} \left\{ \begin{array}{lcl} \partial _{t}u + u\cdot \nabla u = (\mu + \chi ) \Delta u - \nabla \pi + b \cdot \nabla b + 2\chi \nabla \times \omega ,\\ \partial _{t}\omega + u\cdot \nabla \omega + 4\chi \omega = \kappa \Delta \omega + 2\chi \nabla \times u,\\ \partial _{t}b_{1} + u \cdot \nabla b_{1} = \nu \partial _{yy} b_{1} + b \cdot \nabla u_{1},\\ \partial _{t}b_{2} + u \cdot \nabla b_{2} = \nu \partial _{xx} b_{2} + b \cdot \nabla u_{2},\\ \nabla \cdot u = \nabla \cdot b = 0,\\ u(x,y,0) = u_{0}(x,y), \omega (x,y,0) = \omega _{0}(x,y), b(x,y,0) = b_{0}(x,y). \end{array} \right. \end{aligned}$$

(1.3)

When the magnetic fluid $b_1 = b_2 = 0$, the system (1.3) reduces to the 2D incompressible micropolar fluid equations, which describe some physical phenomena such as the motion of animal blood, liquid crystals and dilute aqueous polymer solutions. Micropolar fluid model was first proposed by Eringen [27], while the existences of weak and strong solutions were studied by Galdi and Rionero [28] and Yamaguchi [29]. The global well-posedness of the micropolar equations with full viscosity was obtained by Łukaszewicz [18]. Dong and Chen [19] via using Fourier splitting method proved the $L^2$-decay rates for global solutions of the 2D micropolar equations. Guterres, Melo, Nunes and Perusato [12] improved the decay rates of $\omega$ in $L^2({\mathbb {R}}^2)$ and the decay estimates of the higher-order derivatives of $\omega$.

When $\omega = 0$ and $\chi = 0$, the system (1.3) reduces to the 2D incompressible Magnetohydrodynamic (MHD) equations with partial magnetic diffusion. The global existence and regularity have been obtained in [30] and proved the large time decay rates for smooth solutions of the 2D MHD equations with fractional dissipation partial and magnetic diffusion, that is, $\Delta u$ in (1.3) is replaced by $(-\Delta )^{\alpha } u$, and $\alpha$ is required to belong to $(1, \frac{1}{2})$.

Due to the complex structure of (1.2), when there is only partial dissipation, the global regularity problem can be quite difficult. However, many important progress has recently been made on this direction (see, e.g., [7, 25, 30,31,32,33,34,35]). In [25, 32, 34], the global regularity of the 2D magneto-micropolar equations with various partial dissipation cases was obtained. When the magneto-micropolar equations only have velocity dissipation and magnetic diffusion, namely, (1.1) with $\kappa = 0$, Niu and Shang [36] proved the optimal $L^2$-decay estimates of weak solutions, and also obtained the decay rates of global solutions in $\dot{H}^s$($s > \frac{3}{2}$) and $\dot{B}^m_{2, \infty }$ spaces with $0 \le m \le s$ and in $\dot{B}^m_{2, 1}$ with $0 \le m \le \frac{1}{2}$. When (1.2) only has microrotational dissipation and magnetic diffusion(namely, $\Delta u$ replaced by u), the decay estimates of solutions have been obtained by Shang and Gu [37, 38]. Shang and Gu [37] also proved the global existence of classical solutions for (1.3).

In this paper, we study the global well-posedness and the decay estimates of solutions to (1.3). Motivated by [36], we first establish the $L^2$-decay estimates for global solutions and their derivative with initial data in $L^1({\mathbb {R}}^2)$. Secondly, we establish the global existence result to system (1.3) in $H^s({\mathbb {R}}^2)$. Furthermore, we show the decay rates of these global solutions and their higher derivatives.

We now list the main results of the paper. The first result is the decay estimates of global solutions in $L^2$ space.

Theorem 1.1

Let $\mu > 0$, $\chi > 0$, $\nu > 0$ and $\kappa > 0$. Assume that $(u_0, \omega _0, b_0) \in H^1({\mathbb {R}}^2) \cap L^1({\mathbb {R}}^2)$, with $\nabla \cdot u_0 = 0$ and $\nabla \cdot b_0 = 0$. Then, the global solutions $(u, \omega , b)$ of (1.3) satisfies, for all $t > 0$,

$$\begin{aligned} \Vert u(t)\Vert _{L^2} + \Vert \omega (t)\Vert _{L^2} + \Vert b(t)\Vert _{L^2} \le C(1+t)^{- \frac{1}{2}}, \end{aligned}$$

(1.4)

$$\begin{aligned} \Vert \nabla u(t)\Vert _{L^2} + \Vert \nabla \omega (t)\Vert _{L^2} + \Vert \nabla b(t)\Vert _{L^2} \le C(1+t)^{- \frac{1}{2}}. \end{aligned}$$

(1.5)

We remark that the magnetic field b only has partial dissipation in (1.3), the classic Fourier splitting method which relies on decomposing the whole space into two time-dependent sub-dependent does not apply. To do so, we used the method of [39], to obtain the $L^2$-decay estimates for global solutions of (1.3).

The next theorem is devoted to the global existence and the time decay estimates of solutions to system (1.3) with small initial data in Sobolev space. The process of proving the existence of global solutions is similar to [36]. For the sake of simplicity, we only pay attention to the global a priori bounds of $(u, \omega , b)$, and we can obtain the global existence in $H^s({\mathbb {R}}^2)$. Furthermore, by using the interpolation inequality, energy estimates and the technique of Fourier analysis, we obtain the corresponding time decay rates of these solutions.

More precisely, the following theorem establishes a unique global solution when the initial data $(u_0, \omega _0, b_0)$ is sufficiently small in $H^s({\mathbb {R}}^2)$, and obtain the decay rates of these global solutions and their higher derivatives, as stated blow.

Theorem 1.2

Let $\mu > 0$, $\chi >0$, $\nu >0$ and $\kappa >0$. Assume that $(u_0, \omega _0, b_0) \in H^s({\mathbb {R}}^2)$ with $s>0$ and $\nabla \cdot u_0 = \nabla \cdot b_0 = 0$. Then, the following two statements hold:

• (I) Let $s>1$, then there exists a positive constant $\epsilon _0$, such that for all $0< \epsilon < \epsilon _0$, if

  $$\begin{aligned} \Vert u_0\Vert _{H^s({\mathbb {R}}^2)}^2 + \Vert \omega _0\Vert _{H^s({\mathbb {R}}^2)}^2 + \Vert b_0\Vert _{H^s({\mathbb {R}}^2)}^2 < \epsilon , \end{aligned}$$

  (1.6)

  then system (1.3) has a unique global solution $(u, \omega , b)$ satisfying, for any $t>0$,

  $$\begin{aligned} \Vert u(t)&\Vert _{H^s({\mathbb {R}}^2)}^2 + \Vert \omega (t)\Vert _{H^s({\mathbb {R}}^2)}^2 + \Vert b(t)\Vert _{H^s({\mathbb {R}}^2)}^2 \nonumber \\&+ \int \limits ^t_0 (\Vert \nabla u(\tau )\Vert _{H^s({\mathbb {R}}^2)}^2 + \Vert \nabla \omega (\tau )\Vert _{H^s({\mathbb {R}}^2)}^2 + \Vert \nabla b(\tau )\Vert _{H^s({\mathbb {R}}^2)}^2)\mathrm{d}\tau \le C \epsilon , \end{aligned}$$

  (1.7)

  where $C>0$ is a constant independent of t.

• (II) Let $s>1$, suppose that $(u_0, \omega _0, b_0 ) \in \dot{H}^{-l}({\mathbb {R}}^2)$ with $0 \le l < 1$. Then, the global solution $(u, \omega , b)$ satisfies the following decay estimates:

• i) For all real number m with $0 \le m \le s$, we have

  $$\begin{aligned} \Vert D^m u(t)\Vert _{L^2({\mathbb {R}}^2)} + \Vert D^m \omega (t)\Vert _{L^2({\mathbb {R}}^2)} + \Vert D^m b(t)\Vert _{L^2({\mathbb {R}}^2)} \le C (1+t)^{-\frac{m}{2} - \frac{l}{2}}. \end{aligned}$$

  (1.8)

• ii) For $0 \le m \le s-1$, we have

  $$\begin{aligned} \Vert D^m \omega (t)\Vert _{L^2({\mathbb {R}}^2)} \le C (1+t)^{-\frac{(m+1)}{2} - \frac{l}{2}}. \end{aligned}$$

  (1.9)

  This decay estimate of $\omega (t)$ is improved from (1.8) to (1.9).

Remark 1.3

Since $L^p({\mathbb {R}}^2) \hookrightarrow \dot{H}^{-l} ({\mathbb {R}}^2)$ when $0 \le l <1$ and $p \in (1,2]$, thus Theorem 1.2 also holds for $(u_0, \omega _0, b_0) \in L^p({\mathbb {R}}^2)$ with $p \in (1,2]$.

The proof of the global existence part of Theorem 1.2 relies on the global a priori bound for $\Vert (u, \omega , b)\Vert _{H^s({\mathbb {R}}^2)}$. For the proof of (II) in Theorem 1.2, compared with [36], here we used the negative homogeneous Sobolev space $\dot{H}^{-l}({\mathbb {R}}^2)$ to study the decay estimates of (1.3). Since the space dimension $n=2$, Sobolev’s inequality in $L^2$ is invalid, then we used Sobolev–Nirenberg–Gagliardo inequality to overcome this difficulty.

The rest of this paper is divided into four sections. In Sect. 2, we provide some lemmas to be used later. Sections 3 and 4 state the proofs of Theorems 1.1 and 1.2, respectively. To simplify the notation, we will write $\int f$ for $\int \limits _{\mathbb R^2} f \mathrm{d}x$, $\Vert f\Vert _{L^p}$ for $\Vert f\Vert _{L^p({\mathbb {R}}^2)}$, $\Vert f\Vert _{\dot{H}^s}$ and $\Vert f\Vert _{H^s}$ for $\Vert f\Vert _{\dot{H}^s({\mathbb {R}}^2)}$ and $\Vert f\Vert _{H^s( {\mathbb {R}}^2)}$, respectively, $\Vert f\Vert _{\dot{B}^s_{p,r}}$ for $\Vert f\Vert _{\dot{B}^s_{p,r}({\mathbb {R}}^2)}$.

2 Preliminaries

As preparation, we recall the following lemmas.

Lemma 2.1

(commutator estimates, see, e.g., [40, 41]) Let $s>0$, $1<r<\infty$ and $\frac{1}{r}=\frac{1}{p_1}+\frac{1}{q_1}=\frac{1}{p_2}+\frac{1}{q_2}$ with $q_1, p_2\in (1,\infty )$ and $p_1, q_2\in [1,\infty ]$. Then,

$$\begin{aligned} \Vert [\Lambda ^s, f\cdot \nabla ]g \Vert _{L^r} \le C (\Vert \nabla f \Vert _{L^{p_1}}\Vert \Lambda ^{s} g \Vert _{L^{q_1}} + \Vert \Lambda ^{s} f \Vert _{L^{p_2}} \Vert \nabla g\Vert _{L^{q_2}}), \end{aligned}$$

(2.1)

and

$$\begin{aligned} \Vert \Lambda ^s(f\,g)\Vert _{L^r} \le C\, \left( \Vert g\Vert _{L^{p_1}}\, \Vert \Lambda ^s f\Vert _{L^{q_1}} + \Vert \Lambda ^s g\Vert _{L^{p_2}}\,\Vert f\Vert _{L^{q_2}}\right) , \end{aligned}$$

(2.2)

where $\Lambda =(-\Delta )^{\frac{1}{2}}$ denotes the Zygmund operator, and $[\Lambda ^s, f\cdot \nabla ]g = \Lambda ^s(f\cdot \nabla g) - f\cdot \nabla \Lambda ^s g$ and C’s are positive constants depending on the indices $s, r, p_1, q_1, p_2$ and $q_2$.

Lemma 2.2

(see [42]) If p belongs to (1, 2], then $L^{p}({\mathbb {R}}^d)$ embeds continuously in $\dot{H}^s ({\mathbb {R}}^d)$ with $s = \frac{d}{2} - \frac{d}{p}$.

Next, we state the Sobolev–Nirenberg–Gagliardo inequality. For the sake of simplicity, the proof process will not be described here, and the detailed proof can be found in [43].

Lemma 2.3

Let $u \in L^q$ in ${\mathbb {R}}^n$ and its derivatives of order m, $D^m u \in L^r$, $1 \le q,r \le \infty$. For the derivatives $D^j u$, $0 \le j < m$, the following inequality hold

$$\begin{aligned} \Vert D^j u\Vert _{L^p} \le C \Vert D^m u\Vert _{L^r}^a \Vert u\Vert _{L^q}^{1-a}, \end{aligned}$$

(2.3)

where

$$\begin{aligned} \frac{1}{p} = \frac{j}{n} + a\left( \frac{1}{r} - \frac{m}{n} \right) + (1-a) \frac{1}{q}, \end{aligned}$$

for all a in the interval

$$\begin{aligned} \frac{j}{m} \le a \le 1 \end{aligned}$$

(2.4)

(the constant depending only on n, m, j, q, r, a), with the following exceptional cases

1, If $j=0$, $rm < n$, $q = \infty$ then we make the additional assumption that either u tends to zero at infinity or $u \in L^{\tilde{q}}$ for some finite $\tilde{q} > 0$.

2, If $1< r < \infty$, and $m - j - n/r$ is a nonnegative integer then (2.3) holds only for a satisfying $j/m \le a <1$.

Finally, since $\dot{B}^s_{2,2} \sim \dot{H}^s$, we recall the following Besov space interpolation estimate.

Lemma 2.4

(see [44]) Fixed $m> l > k$, and $1 \le p \le q \le r \le \infty$, we have

$$\begin{aligned} \Vert f\Vert _{\dot{B}^{l}_{q, q'} ({\mathbb {R}}^2)} \le \Vert f\Vert _{\dot{B}^{k}_{r, r'}({\mathbb {R}}^2)}^{\theta } \Vert f\Vert _{\dot{B}^{m}_{p, p'}({\mathbb {R}}^2)}^{1 - \theta }. \end{aligned}$$

(2.5)

These parameters satisfy the following restrictions

$$\begin{aligned} l = k \theta + m(1 - \theta ), \quad \frac{1}{q} = \frac{\theta }{r} + \frac{1- \theta }{p}, \quad \frac{1}{q'} = \frac{\theta }{r'} + \frac{1 - \theta }{p'}. \end{aligned}$$

Also $1 \le p' \le q' \le r' \le \infty$ and solving we have $\theta = \frac{m-l}{m-k} \in (0,1]$.

3 The proof of Theorem 1.1

This section is devoted to the proof of Theorem 1.1. As preparations, we verify the following four propositions. The first proposition is as follows.

Proposition 3.1

Assume $\mu > 0$, $\chi > 0$, $\nu > 0$ and $\kappa > 0$ and $(u_0, \omega _0, b_0) \in H^1({\mathbb {R}}^2)$ with $\nabla \cdot u_0 = \nabla \cdot b_0 = 0$. Then, the corresponding solution $(u, \omega , b)$ of (1.3) obeys the following uniform bounds, for any $t > 0$,

$$\begin{aligned} \Vert u(t)\Vert _{L^2}^2 +&\Vert \omega (t)\Vert _{L^2}^2 + \Vert b(t)\Vert _{L^2}^2 + \mu \int \limits ^{t}_{0} \Vert \nabla u(\tau )\Vert _{L^2}^2 \mathrm{d}\tau + \frac{8\mu \chi }{\mu + 2 \chi } \int \limits ^{t}_{0} \Vert \omega (\tau )\Vert _{L^2}^2 \mathrm{d}\tau \nonumber \\ + 2\kappa \int \limits ^{t}_{0}&\Vert \nabla \omega (\tau )\Vert _{L^2}^2 \mathrm{d}\tau + 2\nu \int \limits ^{t}_{0}\left( \Vert \partial _{y} b_1(\tau )\Vert _{L^2}^2 + \Vert \partial _{x} b_2(\tau )\Vert _{L^2}^2 \right) \mathrm{d}\tau \le C, \end{aligned}$$

(3.1)

and

$$\begin{aligned} \Vert&\Omega (t)\Vert _{L^2}^2 + \Vert j(t)\Vert _{L^2}^2 + \Vert \nabla \omega (t)\Vert _{L^2}^2 + \mu \int \limits ^t_0 \Vert \nabla \Omega (\tau )\Vert _{L^2}^2 \mathrm{d}\tau + \kappa \int \limits ^t_0 \Vert \Delta \omega (\tau )\Vert _{L^2}^2 \mathrm{d}\tau \qquad \qquad \quad \nonumber \\&+ \int \limits ^t_0 H(b, \tau ) \mathrm{d}\tau \le C, \end{aligned}$$

(3.2)

where

$$\begin{aligned} H(b,t) = \nu \int \limits _{{\mathbb {R}}^2} \left( (\partial _{xx} b_1)^2 + (\partial _{xx} b_2)^2 + (\partial _{yy} b_1)^2 + (\partial _{yy} b_2)^2\right) \mathrm{d}x. \end{aligned}$$

and C’s are positive constants depending on $\mu$, $\nu$, $\chi$, $\kappa$ and $\Vert (u_0, \omega _0, b_0)\Vert _{H^1}$ only.

Proof of Proposition 3.1

Taking the $L^2$-inner product to (1.3) with $(u, \omega , b_1, b_2)$ and integrating in time yield

$$\begin{aligned}&\frac{1}{2} \frac{\mathrm{d}}{\mathrm{d}t} \left( \Vert u(t)\Vert _{L^2}^2 + \Vert \omega (t)\Vert _{L^2}^2 + \Vert b(t)\Vert _{L^2}^2 \right) + (\mu + \kappa ) \Vert \nabla u\Vert _{L^2}^2 + \kappa \Vert \nabla \omega \Vert _{L^2}^2 + 4\chi \Vert \omega \Vert _{L^2}^2 \nonumber \\&\quad + \nu (\Vert \partial _{y} b_1\Vert _{L^2}^2 + \Vert \partial _{x} b_2\Vert _{L^2}^2) = 4 \chi \int \nabla \times u \cdot \omega \mathrm{d}x, \end{aligned}$$

(3.3)

where we used the facts that,

$$\begin{aligned}&\int (b \cdot \nabla u_1 \cdot b_1 + b \cdot \nabla u_2 \cdot b_2) \mathrm{d}x = \int b \cdot \nabla u \cdot b \mathrm{d}x = - \int b \cdot \nabla b \cdot u \mathrm{d}x,\\&\quad \int \nabla \times u \cdot \omega \mathrm{d}x = \int \nabla \times \omega \cdot u \mathrm{d}x . \end{aligned}$$

By Hölder’s inequality and the Young inequality, we have

$$\begin{aligned}&4 \chi \int \nabla \times u \cdot \omega \mathrm{d}x\nonumber \\&\quad \le 4\chi \Vert \nabla u\Vert _{L^2} \Vert \omega \Vert _{L^2} \nonumber \\&\quad \le \left( \frac{\mu }{2} + \chi \right) \Vert \nabla u\Vert _{L^2}^2 + \frac{4 \chi ^2}{\frac{\mu }{2} + \chi } \Vert \omega \Vert _{L^2}^2. \end{aligned}$$

(3.4)

Inserting (3.4) into (3.3), we obtain

$$\begin{aligned}&\frac{1}{2} \frac{\mathrm{d}}{\mathrm{d}t} \left( \Vert u(t)\Vert _{L^2}^2 + \Vert \omega (t)\Vert _{L^2}^2 + \Vert b(t)\Vert _{L^2}^2 \right) + \frac{\mu }{2} \Vert \nabla u\Vert _{L^2}^2 + \kappa \Vert \nabla \omega \Vert _{L^2}^2 + 4\chi \left( 1 - \frac{2\chi }{2\chi + \mu }\right) \Vert \omega \Vert _{L^2}^2 \nonumber \\&\quad + \nu (\Vert \partial _{y} b_1\Vert _{L^2}^2 + \Vert \partial _{x} b_2\Vert _{L^2}^2) \le 0. \end{aligned}$$

(3.5)

Integrating (3.5) in [0, t], we can get (3.1) immediately.

Now we turn to prove (3.2). The vorticity $\Omega = \nabla \times u$, $\nabla \omega$ and $j = \nabla \times b$ satisfy

$$\begin{aligned} \left\{ \begin{array}{lcl} \partial _{t} \Omega + u\cdot \nabla \Omega - (\mu + \chi ) \Delta \Omega = b \cdot \nabla j - 2\chi \Delta \omega ,\\ \partial _{t} \nabla \omega + \nabla (u\cdot \nabla \omega ) + 4\chi \nabla \omega = \kappa \nabla \Delta \omega + 2\chi \nabla \Omega ,\\ \partial _{t} j + u \cdot \nabla j - \nu \partial _{xxx} b_{2} + \nu \partial _{yyy} b_{1} = b \cdot \nabla \Omega + T(\nabla u , \nabla b),\\ \end{array} \right. \end{aligned}$$

(3.6)

where

$$\begin{aligned} T(\nabla u , \nabla b) = 2 \partial _{x} b_1 (\partial _x u_2 + \partial _y u_1) - 2 \partial _{x} u_1 (\partial _x b_2 + \partial _y b_1). \end{aligned}$$

Taking the $L^2$-inner products of (3.6) with $(\Omega , \nabla \omega , j)$, respectively, we have

$$\begin{aligned}&\frac{1}{2} \frac{\mathrm{d}}{\mathrm{d}t} \Vert \Omega \Vert _{L^2}^2 + (\mu + \chi ) \Vert \nabla \Omega \Vert _{L^2}^2 = \int b \cdot \nabla j \ \Omega \mathrm{d}x - 2 \chi \int \Delta \omega \ \Omega \mathrm{d}x, \end{aligned}$$

(3.7)

$$\begin{aligned}&\quad \frac{1}{2} \frac{\mathrm{d}}{\mathrm{d}t} \Vert \nabla \omega \Vert _{L^2}^2 + 4 \chi \Vert \nabla \omega \Vert _{L^2}^2 + \kappa \Vert \Delta \omega \Vert _{L^2}^2 = 2 \chi \int \nabla \omega \cdot \nabla \Omega \mathrm{d}x - \int \nabla \omega \cdot \nabla u \cdot \nabla \omega \mathrm{d}x , \end{aligned}$$

(3.8)

$$\begin{aligned}&\quad \frac{1}{2} \frac{\mathrm{d}}{\mathrm{d}t} \Vert j \Vert _{L^2}^2 + I = \int ( b \cdot \nabla \Omega j + T j ) \mathrm{d}x, \end{aligned}$$

(3.9)

where

$$\begin{aligned} I&= \nu \int (- \partial _{xxx} b_2 + \partial _{yyy} b_1) j \mathrm{d}x \\&= \nu \int (- \partial _{xxx} b_2 + \partial _{yyy} b_1) (\partial _x b_2 - \partial _y b_1) \mathrm{d}x \\&= \nu \int [(\partial _{xx} b_2)^2 + (\partial _{yy} b_2)^2 + (\partial _{xx} b_1)^2 + (\partial _{yy} b_1)^2] \mathrm{d}x \equiv H(b, t ), \end{aligned}$$

due to the divergence free condition $\nabla \cdot b = \partial _x b_1 + \partial _y b_2 = 0$. By Hölder’s inequality and Sobolev’s inequality,

$$\begin{aligned} \int T j \mathrm{d}x&\le C \Vert \nabla u \Vert _{L^2} \Vert \nabla b\Vert _{L^4} \Vert j\Vert _{L^4} \\&\le C \Vert \Omega \Vert _{L^2} \Vert j\Vert _{L^2} \Vert \nabla j\Vert _{L^2} \le C(\nu ) \Vert \Omega \Vert _{L^2}^2 \Vert j\Vert _{L^2}^2 + \frac{\nu }{8} \Vert \nabla j\Vert _{L^2}^2, \end{aligned}$$

where we have used the fact that the Calderon–Zygmund operators are bounded on $L^p$ $(1< p < \infty )$. It is easy to verify that

$$\begin{aligned} \frac{\nu }{4} \Vert \nabla j\Vert _{L^2}^2 \le H(b,t). \end{aligned}$$

Indeed,

$$\begin{aligned} \nu \Vert \nabla j\Vert _{L^2}^2&= \nu \Vert (\partial _x j , \partial _y j )\Vert _{L^2}^2 = \nu \Vert ((\partial _{xx}b_2 - \partial _{xy}b_1), (\partial _{xy}b_2 - \partial _{yy}b_1))\Vert _{L^2}^2\\&= \nu \Vert ((\partial _{xx}b_2 + \partial _{yy}b_2), - (\partial _{xx}b_1 + \partial _{yy}b_1))\Vert _{L^2}^2 \le 4 H(b,t). \end{aligned}$$

Then, it follows from the above bounds and (3.9) that

$$\begin{aligned} \frac{1}{2} \frac{\mathrm{d}}{\mathrm{d}t} \Vert j\Vert _{L^2}^2 + \frac{1}{2} H(b,t) \le \int b \cdot \nabla \Omega j \mathrm{d}x + C \Vert \Omega \Vert _{L^2}^2 \Vert j\Vert _{L^2}^2. \end{aligned}$$

(3.10)

Combining (3.7), (3.8) and (3.10), we have

$$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}t}&(\Vert \Omega \Vert _{L^2}^2 + \Vert \nabla \omega \Vert _{L^2}^2 + \Vert j\Vert _{L^2}^2) + 2(\mu + \chi )\Vert \nabla \Omega \Vert _{L^2}^2 + 8\chi \Vert \nabla \omega \Vert _{L^2}^2 + 2 \kappa \Vert \Delta \omega \Vert _{L^2}^2 + H(b, t) \nonumber \\&\le 8 \chi \int \nabla \omega \cdot \nabla \Omega \mathrm{d}x - 2 \int \nabla \omega \cdot \nabla u \cdot \nabla \omega \mathrm{d}x + C \Vert \Omega \Vert _{L^2}^2 \Vert j\Vert _{L^2}^2 \nonumber \\&= I_1 + I_2 + I_3, \end{aligned}$$

(3.11)

where we have used the fact that,

$$\begin{aligned} \int b \cdot \nabla \Omega j \mathrm{d}x = \int b \cdot \nabla j \Omega \mathrm{d}x . \end{aligned}$$

Next, we consider $I_1$ and $I_2$, respectively. Applying the Young inequality,

$$\begin{aligned} I_1&= 8 \chi \int \nabla \omega \cdot \nabla \Omega \mathrm{d}x \\&\le (\mu + 2 \chi ) \Vert \nabla \Omega \Vert ^2_{L^2} + \frac{8\chi ^2}{\frac{\mu }{2} + \chi } \Vert \nabla \omega \Vert _{L^2}^2, \end{aligned}$$

and

$$\begin{aligned} I_2&= -2 \int \nabla \omega \cdot \nabla u \cdot \nabla \omega \mathrm{d}x \\&\le C \Vert \nabla u\Vert _{L^2} \Vert \nabla \omega \Vert _{L^4}^2 \\&\le \kappa \Vert \Delta \omega \Vert _{L^2}^2 + C \Vert \nabla u\Vert _{L^2}^2 \Vert \nabla \omega \Vert _{L^2}^2. \end{aligned}$$

Inserting $I_1$ and $I_2$ into (3.11), we obtain

$$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}t}&(\Vert \Omega \Vert _{L^2}^2 + \Vert \nabla \omega \Vert _{L^2}^2 + \Vert j\Vert _{L^2}^2) + \mu \Vert \nabla \Omega \Vert _{L^2}^2 + \frac{8\chi \mu }{\mu + 2 \chi } \Vert \nabla \omega \Vert _{L^2}^2 + \kappa \Vert \Delta \omega \Vert _{L^2}^2 + H(b, t)\\&\le C \left( \Vert j\Vert ^2_{L^2} + \Vert \nabla \omega \Vert ^2_{L^2} \right) \left( \Vert \Omega \Vert _{L^2}^2 + \Vert \nabla \omega \Vert _{L^2}^2 + \Vert j\Vert _{L^2}^2 \right) . \end{aligned}$$

Gronwall’s inequality, together with the fact that

$$\begin{aligned} \Vert j\Vert ^2_{L^2} \le 2 \Vert \partial _x b_2\Vert ^2_{L^2} + 2 \Vert \partial _y b_1\Vert ^2_{L^2} \end{aligned}$$

allows us to conclude that

$$\begin{aligned} \Vert \Omega (t)&\Vert _{L^2}^2 + \Vert \nabla \omega (t) \Vert _{L^2}^2 + \Vert j(t)\Vert _{L^2}^2 + \mu \int \limits ^t_0 \Vert \nabla \Omega (\tau )\Vert _{L^2}^2 \mathrm {d}\tau + \frac{8\chi \mu }{\mu + 2 \chi } \int \limits ^t_0 \Vert \nabla \omega (\tau )\Vert _{L^2}^2 \mathrm {d}\tau \\&+\kappa \int \limits ^t_0 \Vert \Delta \omega (\tau )\Vert _{L^2}^2 \mathrm {d}\tau + \int \limits ^t_0 H(b, \tau ) \mathrm {d}\tau \\ \le&\left( \Vert \Omega _0\Vert _{L^2}^2 + \Vert \nabla \omega _0 \Vert _{L^2}^2 + \Vert j_0\Vert _{L^2}^2 \right) \exp \left\{ C \int \limits ^t_0 \left( \Vert j(\tau )\Vert ^2_{L^2} + \Vert \nabla \omega (\tau )\Vert ^2_{L^2} \right) \mathrm {d}\tau \ \right\} \\ \le&\left( \Vert \nabla u_0\Vert _{L^2}^2 + \Vert \nabla \omega _0 \Vert _{L^2}^2 + \Vert \nabla b_0\Vert _{L^2}^2 \right) \exp \left\{ C \left( \Vert u_0\Vert _{L^2}^2 + \Vert \omega _0 \Vert _{L^2}^2 + \Vert b_0\Vert _{L^2}^2 \right) \right\} , \end{aligned}$$

which immediately yields (3.2). Thus, the proof of Proposition 3.1 is completed. $\square$

Remark 3.2

Because of $\nabla \cdot b =0$, it is easy to verify that

$$\begin{aligned} \Vert \nabla b\Vert _{L^2 ({\mathbb {R}}^2)} = \Vert \nabla \times b\Vert _{L^2 ({\mathbb {R}}^2)} \le 2 \left( \Vert \partial _{y} b_1(\tau )\Vert _{L^2}^2 + \Vert \partial _{x} b_2(\tau )\Vert _{L^2}^2 \right) . \end{aligned}$$

Therefore, we can rewrite (3.1) as

$$\begin{aligned} \Vert u(t)\Vert _{L^2}^2 +&\Vert \omega (t)\Vert _{L^2}^2 + \Vert b(t)\Vert _{L^2}^2 + \mu \int \limits ^{t}_{0} \Vert \nabla u(\tau )\Vert _{L^2}^2 \mathrm{d}\tau + 8\chi \left( 1- \frac{2\chi }{\mu + 2 \chi }\right) \int \limits ^{t}_{0} \Vert \omega (\tau )\Vert _{L^2}^2 \mathrm{d}\tau \nonumber \\ + 2\kappa \int \limits ^{t}_{0}&\Vert \nabla \omega (\tau )\Vert _{L^2}^2 \mathrm{d}\tau + \nu \int \limits ^{t}_{0}\Vert \nabla b(\tau )\Vert _{L^2}^2 \mathrm{d}\tau \le C. \end{aligned}$$

(3.12)

The following proposition provides the decay rates for $\Vert (\nabla u, \nabla \omega , \nabla b)\Vert _{L^2}$.

Proposition 3.3

Suppose (u, $\omega$, b) is a solution of (1.3) with the corresponding initial data $(u_0, \omega _0, b_0) \in H^1({\mathbb {R}}^2)$. Then,

$$\begin{aligned} \Vert \nabla u(t)\Vert _{L^2} + \Vert \nabla \omega (t)\Vert _{L^2} + \Vert \nabla b(t)\Vert _{L^2} \le C (1 + t)^{-\frac{1}{2}}. \end{aligned}$$

(3.13)

Proof of Proposition 3.3

As in proof of Proposition 3.1, we have, for all $0 \le s < t \le \infty$,

$$\begin{aligned} \Vert&u(t)\Vert _{L^2}^2 + \Vert \omega (t)\Vert _{L^2}^2 + \Vert b(t)\Vert _{L^2}^2 + \mu \int \limits ^{t}_{s} \Vert \nabla u(\tau )\Vert _{L^2}^2 \mathrm{d}\tau + 2\kappa \int \limits ^{t}_{s} \Vert \nabla \omega (\tau )\Vert _{L^2}^2 \mathrm{d}\tau \nonumber \\&+ 2\nu \int \limits ^{t}_{s} \left( \Vert \partial _{y} b_1(\tau )\Vert _{L^2}^2 + \Vert \partial _{x} b_2(\tau )\Vert _{L^2}^2 \right) \mathrm{d}\tau \le \Vert u(s)\Vert _{L^2}^2 + \Vert \omega (s)\Vert _{L^2}^2 + \Vert b(s)\Vert _{L^2}^2, \end{aligned}$$

(3.14)

and

$$\begin{aligned} \Vert \Omega (t)&\Vert _{L^2}^2 + \Vert \nabla \omega (t) \Vert _{L^2}^2 + \Vert j(t)\Vert _{L^2}^2 + \mu \int \limits ^t_s \Vert \nabla \Omega (\tau )\Vert _{L^2}^2 \mathrm{d}\tau \nonumber \\&+ \kappa \int \limits ^t_s \Vert \Delta \omega (\tau )\Vert _{L^2}^2 \mathrm{d}\tau + \int \limits ^t_s H(b, \tau ) \mathrm{d}\tau \nonumber \\ \le&\left( \Vert \Omega (s)\Vert _{L^2}^2 + \Vert \nabla \omega (s) \Vert _{L^2}^2 + \Vert j(s)\Vert _{L^2}^2 \right) \exp \left\{ C \left( \Vert u_0\Vert _{L^2}^2 + \Vert \omega _0 \Vert _{L^2}^2 + \Vert b_0\Vert _{L^2}^2 \right) \right\} . \end{aligned}$$

(3.15)

Therefore,

$$\begin{aligned} \int \limits ^{\infty }_{0} \Vert \nabla b (\tau )\Vert _{L^2}^2 \mathrm{d} \tau&= \int \limits _{0}^{\infty } \Vert j(\tau )\Vert _{L^2}^2 \mathrm{d} \tau \\&\le 2 \int \limits ^{\infty }_{0} \left( \Vert \partial _{x}b_2(\tau )\Vert _{L^2}^2 + \Vert \partial _y b_1(\tau )\Vert _{L^2}^2 \right) \mathrm{d}\tau \le C \left( \Vert u_0\Vert _{L^2}^2 + \Vert \omega _0 \Vert _{L^2}^2 + \Vert b_0\Vert _{L^2}^2 \right) , \end{aligned}$$

and

$$\begin{aligned}&\int \limits ^{\infty }_{0} \Vert \nabla \omega (\tau )\Vert _{L^2}^2 \mathrm{d} \tau \le C \left( \Vert u_0\Vert _{L^2}^2 + \Vert \omega _0 \Vert _{L^2}^2 + \Vert b_0\Vert _{L^2}^2 \right) ,\\&\quad \int \limits ^{\infty }_{0} \Vert \nabla u (\tau )\Vert _{L^2}^2 \mathrm{d} \tau \le C \left( \Vert u_0\Vert _{L^2}^2 + \Vert \omega _0 \Vert _{L^2}^2 + \Vert b_0\Vert _{L^2}^2 \right) . \end{aligned}$$

A special consequence is that

$$\begin{aligned} \int \limits ^{t}_{\frac{t}{2}} \left( \Vert \nabla u (\tau )\Vert _{L^2}^2 + \Vert \nabla \omega (\tau )\Vert _{L^2}^2 + \Vert \nabla b (\tau )\Vert _{L^2}^2 \right) \mathrm{d} \tau \longrightarrow 0, \quad as \quad t \rightarrow \infty . \end{aligned}$$

Integrating (3.15) with respect to s in $(\frac{t}{2}, t)$, we have

$$\begin{aligned} t&\left( \Vert \nabla u (t)\Vert _{L^2}^2 + \Vert \nabla \omega (t)\Vert _{L^2}^2 + \Vert \nabla b (t)\Vert _{L^2}^2 \right) \nonumber \\&\le 2 \exp \left\{ C \left( \Vert u_0\Vert _{L^2}^2 + \Vert \omega _0 \Vert _{L^2}^2 + \Vert b_0\Vert _{L^2}^2 \right) \right\} \int \limits ^{t}_{\frac{t}{2}} \left( \Vert \nabla u (\tau )\Vert _{L^2}^2 + \Vert \nabla \omega (\tau )\Vert _{L^2}^2 + \Vert \nabla b (\tau )\Vert _{L^2}^2 \right) \mathrm{d} \tau \\&\le C. \end{aligned}$$

Therefore, for any $t \ge 1$, we get

$$\begin{aligned} \Vert \nabla u (t)\Vert _{L^2}^2 + \Vert \nabla \omega (t)\Vert _{L^2}^2 + \Vert \nabla b (t)\Vert _{L^2}^2 \le C t^{-1} \le C (1 + t)^{-1}. \end{aligned}$$

(3.16)

Moreover, for any $0<t<1$, it follows from (3.15),

$$\begin{aligned} \Vert \nabla u (t)\Vert _{L^2}^2 + \Vert \nabla \omega (t)\Vert _{L^2}^2 + \Vert \nabla b (t)\Vert _{L^2}^2 \le C \le C (1 + t)^{-1}. \end{aligned}$$

(3.17)

Then, (3.16) and (3.17) yield (3.13).

$\square$

The following proposition which will play an important role to drive the decay estimates of Theorem 1.1.

Proposition 3.4

Let $(u, \omega , b)$ be a global solution of system (1.3), with $\nabla \cdot u_0 = \nabla \cdot b_0 = 0$ and $(u_0, \omega _0, b_0) \in L^1({\mathbb {R}}^2) \cap L^2({\mathbb {R}}^2)$. Then, there exists a constant $C > 0$, such that

$$\begin{aligned}&|\hat{u}(\xi , t)| + |\hat{\omega }(\xi , t)| + |\hat{b}_1(\xi , t)| + |\hat{b}_2(\xi , t)| \nonumber \\&\le C + C |\xi | \int \limits ^{t}_0 \left( \Vert u(\tau )\Vert _{L^2}^2 + \Vert \omega (\tau )\Vert _{L^2}^2 + \Vert b(\tau )\Vert _{L^2}^2 \right) \mathrm{d}\tau . \end{aligned}$$

(3.18)

Proof of Proposition 3.3

Applying the Fourier transform to system (1.3), we obtain:

$$\begin{aligned} \left\{ \begin{array}{lcl} \partial _t \hat{u} + (\mu + \chi ) |\xi |^2 \hat{u} = -{\mathcal {F}} (\nabla \pi ) + {\mathcal {F}}(b\cdot \nabla b) + 2 \chi i \xi \times \hat{\omega } -{\mathcal {F}}(u \cdot \nabla u),\\ \partial _t \hat{\omega } + \kappa |\xi |^2 \hat{\omega } + 4 \chi \hat{\omega } = 2 \chi i \xi \times \hat{u} - {\mathcal {F}}(u \cdot \nabla \omega ),\\ \partial _t \hat{b}_1 + \nu |\xi _2|^2 \hat{b}_1 = {\mathcal {F}} [b\cdot \nabla u_1 - u \cdot \nabla b_1],\\ \partial _t \hat{b}_2 + \nu |\xi _1|^2 \hat{b}_2 = {\mathcal {F}} [b\cdot \nabla u_2 - u \cdot \nabla b_2]. \end{array} \right. \end{aligned}$$

(3.19)

Multiplying the (3.19)${}_1$, (3.19)${}_2$, (3.19${}_{3}$ and (3.19)${}_4$ by $\bar{\hat{u}}), \bar{\hat{\omega }}$, $\bar{\hat{b}}_1$ and $\bar{\hat{b}}_2$, respectively, and summing up, we have, noting that $|\hat{u}|^2 = \hat{u} \bar{\hat{u}}$

$$\begin{aligned} \frac{1}{2} \frac{\mathrm{d}}{\mathrm{d}t}&(|\hat{u}|^2 + |\hat{\omega }|^2 + |\hat{b}_1|^2 + |\hat{b}_2|^2) + (\mu + \chi )|\xi |^2 |\hat{u}|^2 + \kappa |\xi |^2 |\hat{\omega }|^2 \nonumber \\&+ \nu (|\xi _2|^2 |\hat{b}_1|^2 + |\xi _1|^2 |\hat{b}_2|^2) + 4 \chi |\hat{\omega }|^2\nonumber \\ =&-{\mathcal {F}} (\nabla \pi ) \bar{\hat{u}} + {\mathcal {F}}(b\cdot \nabla b)\bar{\hat{u}} -{\mathcal {F}}(u \cdot \nabla u)\bar{\hat{u}}- {\mathcal {F}}(u \cdot \nabla \omega )\bar{\hat{\omega }} + {\mathcal {F}} (b\cdot \nabla u_1) \bar{\hat{b}}_1 \nonumber \\&\quad - {\mathcal {F}}(u \cdot \nabla b_1)\bar{\hat{b}}_1 + {\mathcal {F}} (b\cdot \nabla u_2) \bar{\hat{b}}_2 - {\mathcal {F}}(u \cdot \nabla b_2)\bar{\hat{b}}_2 + 2\chi i \xi \times \hat{\omega } \bar{\hat{u}} + 2 \chi i \xi \times \hat{u} \bar{\hat{\omega }} \nonumber \\ =&K_1 + K_2 + \cdot \cdot \cdot + K_{10}. \end{aligned}$$

(3.20)

For $K_1$, taking divergence to the first equation of (1.3), one yields

$$\begin{aligned} \pi = (- \Delta )^{-1} (\nabla \otimes \nabla ) (b \otimes b - u \otimes u). \end{aligned}$$

And taking Fourier transformation obeys, nothing that $|\hat{u}| = |\bar{\hat{u}}|$

$$\begin{aligned} K_1&\le |\xi | |\hat{\pi }| |\bar{\hat{u}}|\\&\le |\xi | (\Vert b \otimes b\Vert _{L^1}+ \Vert u \otimes u\Vert _{L^1} )|\bar{\hat{u}}| \\&\le |\xi | (\Vert b\Vert ^2_{L^2}+ \Vert u\Vert ^2_{L^2}) |\hat{u}| . \end{aligned}$$

For $K_2$,

$$\begin{aligned}&K_2 \le |\xi ||\widehat{b \otimes b}| |\bar{\hat{u}}| \le |\xi | \Vert b \otimes b\Vert _{L^1} |\bar{\hat{u}}| \le |\xi | \Vert b\Vert ^2_{L^2} |\hat{u}|.\\ \end{aligned}$$

Similarly, we obtain

$$\begin{aligned} |K_3+ K_4|&\le 2 |\xi | (\Vert u\Vert ^2_{L^2} + \Vert \omega \Vert ^2_{L^2})(|\hat{u}| + |\hat{\omega }|),\\ |K_5 + K_6|&\le |\xi | (\Vert b\Vert ^2_{L^2} + \Vert u_1\Vert ^2_{L^2} + \Vert b_1\Vert ^2_{L^2} + \Vert u\Vert ^2_{L^2}) |\hat{b}_1|\\&\le 2 |\xi | (\Vert u\Vert ^2_{L^2} + \Vert b\Vert ^2_{L^2} ) |\hat{b}_1|,\\ |K_7 + K_8|&\le 2 |\xi | (\Vert u\Vert ^2_{L^2} + \Vert b\Vert ^2_{L^2} ) |\hat{b}_2|,\\ |K_9 + K_{10}|&\le 4 \chi |\xi | |\hat{\omega }| |\hat{u}|\\&\le (\frac{\mu }{2} + \chi ) |\xi |^2 |\hat{u}|^2 + \frac{8 \chi ^{2}}{\mu + 2\chi } |\hat{\omega }|^2. \end{aligned}$$

Inserting $K_1$–$K_{10}$ into (3.20), we derive that

$$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}t}&(|\hat{u}|^2 + |\hat{\omega }|^2 + |\hat{b}_1|^2 + |\hat{b}_2|^2) + \mu |\xi |^2 |\hat{u}|^2 + 2 \kappa |\xi |^2 |\hat{\omega }|^2\\&+ 2 \nu (|\xi _2|^2 |\hat{b}_1|^2 + |\xi _1|^2 |\hat{b}_2|^2) + \frac{16 \chi \mu }{\mu + 2\chi } |\hat{\omega }|^2\\ \le&C |\xi | (\Vert u\Vert ^2_{L^2} + \Vert b\Vert ^2_{L^2} + \Vert \omega \Vert ^2_{L^2})(|\hat{u}| + |\hat{\omega }| + |\hat{b}_1| + |\hat{b}_2|), \end{aligned}$$

which immediately yields

$$\begin{aligned} \partial _t \sqrt{|\hat{u}|^2 + |\hat{\omega }|^2 + |\hat{b}_1|^2 + |\hat{b}_2|^2} \le C |\xi | (\Vert u\Vert ^2_{L^2} + \Vert b\Vert ^2_{L^2} + \Vert \omega \Vert ^2_{L^2}). \end{aligned}$$

(3.21)

Integrating (3.21) in [0, t], we obtain

$$\begin{aligned}&\sqrt{|\hat{u}(t)|^2 + |\hat{\omega }(t)|^2 + |\hat{b}_1(t)|^2 + |\hat{b}_2(t)|^2} \\&\le \sqrt{|\hat{u}(0)|^2 + |\hat{\omega }(0)|^2 + |\hat{b}_1(0)|^2 + |\hat{b}_2(0)|^2} + C |\xi | \int \limits ^t_0 (\Vert u(\tau )\Vert ^2_{L^2} + \Vert b(\tau )\Vert ^2_{L^2} + \Vert \omega (\tau )\Vert ^2_{L^2})\mathrm{d} \tau \\&\le C + C |\xi | \int \limits ^t_0 (\Vert u(\tau )\Vert ^2_{L^2} + \Vert b(\tau )\Vert ^2_{L^2} + \Vert \omega (\tau )\Vert ^2_{L^2})\mathrm{d} \tau . \end{aligned}$$

Thus, the proof of Proposition 3.3 is completed. $\square$

The following proposition provides an auxiliary logarithmic decay estimate.

Proposition 3.5

Assume that $(u_0, \omega _0, b_0) \in L^1({\mathbb {R}}^2)\cap L^2({\mathbb {R}}^2)$, then the global solution $(u, \omega , b)$ of (1.3), satisfies

$$\begin{aligned} \Vert u(t)\Vert ^2_{L^2({\mathbb {R}}^2)} + \Vert b(t)\Vert ^2_{L^2({\mathbb {R}}^2)} + \Vert \omega (t)\Vert ^2_{L^2({\mathbb {R}}^2)} \le C \ln ^{-2} (e + t). \end{aligned}$$

(3.22)

Proof of Proposition 3.4

Choosing a positive smooth function $f(t) \in$ $C^{\infty } [0, \infty )$ satisfying $f(0) = 1$, $f'(t) > 0$ and

$$\begin{aligned} \frac{f'(t)}{2f(t)} <1, \quad t> t_0 > 0 \end{aligned}$$

multiplying both sides of inequality (3.5) with f(t), one shows that

$$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}t}&\left[ f(t) \left( \Vert u(t)\Vert ^2_{L^2} + \Vert b(t)\Vert ^2_{L^2} + \Vert \omega (t)\Vert ^2_{L^2} \right) \right] + c_1 f(t) \left( \Vert \nabla u(t)\Vert ^2_{L^2} + \Vert \nabla b(t)\Vert ^2_{L^2} + \Vert \nabla \omega (t)\Vert ^2_{L^2} \right) \\&\le f'(t) \left( \Vert u(t)\Vert ^2_{L^2} + \Vert b(t)\Vert ^2_{L^2} + \Vert \omega (t)\Vert ^2_{L^2} \right) , \end{aligned}$$

where $c_1 = min \{\mu , 2 \kappa , \nu \}$. From this and together with Plancherel Theorem,

$$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}t}&\left[ f(t)\left( \Vert \hat{u}(t)\Vert ^2_{L^2} + \Vert \hat{b}(t)\Vert ^2_{L^2} + \Vert \hat{\omega }(t)\Vert ^2_{L^2} \right) \right] + c_1 f(t) \int \limits _{{\mathbb {R}}^2} |\xi |^2 \left( |\hat{u}(\xi )|^2 + |\hat{\omega }(\xi )|^2 + |\hat{b}(\xi )|^2\right) \mathrm{d} \xi \nonumber \\&\le f'(t) \int \limits _{{\mathbb {R}}^2} \left( |\hat{u}(\xi )|^2 + |\hat{\omega }(\xi )|^2 + |\hat{b}(\xi )|^2 \right) \mathrm{d} \xi . \end{aligned}$$

(3.23)

Letting $B(t) = \left\{ \xi \in {\mathbb {R}}^2 \mid |\xi |^2 \le \frac{f'(t)}{c_1 f(t)}\right\}$, then we obtain

$$\begin{aligned}&c_1 f(t) \int \limits _{{\mathbb {R}}^2} |\xi |^2 \left( |\hat{u}(\xi )|^2 + |\hat{\omega }(\xi )|^2 + |\hat{b}(\xi )|^2 \right) \mathrm{d} \xi \\&\ge c_1 f(t) \int \limits _{B^c(t)} |\xi |^2 \left( |\hat{u}(\xi )|^2 + |\hat{\omega }(\xi )|^2 + |\hat{b}(\xi )|^2 \right) \mathrm{d} \xi \\&\ge c_1 f(t) \left( \frac{f'(t)}{c_1 f(t)}\right) \int \limits _{B^c(t)} \left( |\hat{u} (\xi )|^2 + |\hat{\omega }(\xi )|^2 + |\hat{b}(\xi )|^2 \right) \mathrm{d} \xi \\&= f'(t) \int \limits _{{\mathbb {R}}^2} \left( |\hat{u}(\xi )|^2 + |\hat{\omega }(\xi )|^2 + |\hat{b}(\xi )|^2 \right) \mathrm{d} \xi - f'(t)\int \limits _{B(t)} \left( |\hat{u}(\xi )|^2 + |\hat{\omega }(\xi )|^2 + |\hat{b}(\xi )|^2 \right) \mathrm{d} \xi . \end{aligned}$$

Substituting this result into (3.23), we have

$$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}t}&\left[ f(t) \left( \Vert u(t)\Vert ^2_{L^2} + \Vert b(t)\Vert ^2_{L^2} + \Vert \omega (t)\Vert ^2_{L^2} \right) \right] \le f'(t)\int \limits _{B(t)} \left( |\hat{u}(\xi )|^2 + |\hat{\omega }(\xi )|^2 + |\hat{b}(\xi )|^2 \right) \mathrm{d} \xi . \end{aligned}$$

(3.24)

Employing (3.18), we have

$$\begin{aligned}&\int \limits _{B(t)} \left( |\hat{u}(\xi )|^2 + |\hat{\omega }(\xi )|^2 + |\hat{b}(\xi )|^2 \right) \mathrm{d} \xi \\&= C \int \limits _{B(t)} \left\{ |\xi |^2 \left( \int \limits ^{t}_0 (\Vert u(\tau )\Vert _{L^2}^2 + \Vert \omega (\tau )\Vert _{L^2}^2 + \Vert b(\tau )\Vert _{L^2}^2) \mathrm{d}\tau \right) ^2 + 1 \right\} \mathrm{d} \xi \\&\le \frac{C f'(t)}{f(t)} + \frac{C(f'(t))^2 t}{f^2(t)} \int \limits ^t_0 \left( \Vert u(\tau )\Vert _{L^2}^4 + \Vert \omega (\tau )\Vert _{L^2}^4 + \Vert b(\tau )\Vert _{L^2}^4 \right) \mathrm{d}\tau . \end{aligned}$$

Inserting the above inequality into (3.24) and integrating in time from 0 to t $(t>t_0)$, one shows the following integral inequality which is crucial for the large time decay for global solutions

$$\begin{aligned} f(t)&\left( \Vert u(t)\Vert ^2_{L^2} + \Vert b(t)\Vert ^2_{L^2} + \Vert \omega (t)\Vert ^2_{L^2} \right) \nonumber \\ \le&f(0) \left( \Vert u_0\Vert ^2_{L^2} + \Vert b_0\Vert ^2_{L^2} + \Vert \omega _0\Vert ^2_{L^2} \right) + C \int \limits ^t_0 \frac{ (f'(\tau ))^2}{f(\tau )}\mathrm{d}\tau \nonumber \\&+ C \int \limits ^t_0 \frac{(f'(\tau ))^3 \tau }{f^2(\tau )} \int \limits ^\tau _0 \left( \Vert u(s)\Vert _{L^2}^4 + \Vert \omega (s)\Vert _{L^2}^4 + \Vert b(s)\Vert _{L^2}^4 \right) \mathrm{d}s \mathrm{d}\tau . \end{aligned}$$

(3.25)

Letting $f(t) = (\ln (e +t))^3$ in (3.25) and applying energy inequality (3.1), one yields

$$\begin{aligned}&\ln ^3(e+t) \left( \Vert u(t)\Vert ^2_{L^2} + \Vert b(t)\Vert ^2_{L^2} + \Vert \omega (t)\Vert ^2_{L^2} \right) \\&\le C + C \ln (e + t) + \int \limits ^t_0 \frac{\tau ^2}{(e + \tau )^3} \mathrm{d}\tau \left( \Vert u_0\Vert _{L^2}^4 + \Vert \omega _0\Vert _{L^2}^4 + \Vert b_0\Vert _{L^2}^4 \right) \\&\le C + C \ln (e + t). \end{aligned}$$

Thus, we have,

$$\begin{aligned} \Vert u(t)\Vert ^2_{L^2} + \Vert b(t)\Vert ^2_{L^2} + \Vert \omega (t)\Vert ^2_{L^2} \le C \ln ^{-2} (e + t). \end{aligned}$$

This completes the proof of Proposition 3.4.

$\square$

With the above propositions at our disposal, we are ready to prove Theorem 1.1.

Proof of Theorem 1.1

From Proposition 3.3, we completed the proof of (1.5) of Theorem 1.1, next we start to prove (1.4) of Theorem 1.1.

Letting $f(t) = (1+ t)^2$ and inserting it into (3.25), we obtain

$$\begin{aligned}&(1+t)^2 \left( \Vert u(t)\Vert ^2_{L^2} + \Vert b(t)\Vert ^2_{L^2} + \Vert \omega (t)\Vert ^2_{L^2} \right) \\&\le C + C(1+t) + C \int \limits ^t_0 \frac{\tau }{1 + \tau } \int \limits ^\tau _0 \left( \Vert u(s)\Vert _{L^2}^4 + \Vert \omega (s)\Vert _{L^2}^4 + \Vert b(s)\Vert _{L^2}^4 \right) \mathrm{d}s \mathrm{d}\tau \\&\le C(1+t) + C \int \limits ^t_0 \int \limits ^{\tau }_0 \left( \Vert u(s)\Vert _{L^2}^4 + \Vert \omega (s)\Vert _{L^2}^4 + \Vert b(s)\Vert _{L^2}^4 \right) \mathrm{d}s \mathrm{d}\tau \\&\le C (1+t) + C(1+t) \int \limits ^t_0 \left( \Vert u(s)\Vert ^2_{L^2} + \Vert b(s)\Vert ^2_{L^2} + \Vert \omega (s)\Vert ^2_{L^2} \right) \ln ^{-2} (e+s)\mathrm{d}s. \end{aligned}$$

Denoting

$$\begin{aligned} {\mathcal {N}}(t)= \sup _{0 \le \tau \le t} \left\{ (1+ \tau )(\Vert u(\tau )\Vert ^2_{L^2} + \Vert b(\tau )\Vert ^2_{L^2} + \Vert \omega (\tau )\Vert ^2_{L^2})\right\} . \end{aligned}$$

Then, we have

$$\begin{aligned} {\mathcal {N}}(t) \le C + C \int \limits ^t_0 {\mathcal {N}}(s) (1+s)^{-1} \ln ^{-2} (e + s) \mathrm{d}s. \end{aligned}$$

Applying Gronwall’s inequality, we obtain

$$\begin{aligned} {\mathcal {N}}(t) \le C \exp \left\{ \int \limits ^{\infty }_0 (1+ s)^{-1} \ln ^{-2} (e + s) \mathrm{d}s \right\} < C, \end{aligned}$$

which implies the desired decay

$$\begin{aligned} \Vert u(t)\Vert _{L^2} + \Vert b(t)\Vert _{L^2} + \Vert \omega (t)\Vert _{L^2} \le C (1 + t)^{- \frac{1}{2}}, \end{aligned}$$

That is (1.4), and completed the proof of Theorem 1.1. $\square$

4 The proof of Theorem 1.2

This section is devoted to proving Theorem 1.2.

We first prove the global stability part (I) of Theorem 1.2. As we know, it suffices to establish the global $a \ priori \ H^s$ estimates.

Proof of (I) of Theorem 1.2

Applying $\Lambda ^s$ to (1.3), we have

$$\begin{aligned}&\partial _t \Lambda ^s u + u \cdot \nabla \Lambda ^s u - (\mu + \chi ) \Delta \Lambda ^s u = - \Lambda ^s \nabla \pi - [\Lambda ^s , u\cdot \nabla ]u + \Lambda ^s (b \cdot \nabla b) + 2\chi \Lambda ^s \nabla \times \omega , \end{aligned}$$

(4.1)

$$\begin{aligned}&\quad \partial _t \Lambda ^s \omega + u \cdot \nabla \Lambda ^s \omega - \kappa \Delta \Lambda ^s \omega + 4 \chi \Lambda ^s \omega = - [\Lambda ^s , u\cdot \nabla ]\omega + 2\chi \Lambda ^s \nabla \times u, \end{aligned}$$

(4.2)

$$\begin{aligned}&\quad \partial _t \Lambda ^s b_1 + u \cdot \nabla \Lambda ^s b_1 - \nu \partial _{yy} \Lambda ^s b_1 = - [\Lambda ^s , u\cdot \nabla ]b_1 + \Lambda ^s (b \cdot \nabla u_1), \end{aligned}$$

(4.3)

$$\begin{aligned}&\quad \partial _t \Lambda ^s b_2 + u \cdot \nabla \Lambda ^s b_2 - \nu \partial _{xx} \Lambda ^s b_2 = - [\Lambda ^s , u\cdot \nabla ]b_2 + \Lambda ^s (b \cdot \nabla u_2), \end{aligned}$$

(4.4)

where $[\Lambda ^s , f \cdot \nabla ]g =\Lambda ^s (f \cdot \nabla g) - f \cdot \nabla \Lambda ^s g$ is commutator.

Dotting (4.1)–(4.4) by $\Lambda ^s u$, $\Lambda ^s \omega$, $\Lambda ^s b_1$ and $\Lambda ^s b_2$, respectively, integrating the resulting equations in ${\mathbb {R}}^2$, and adding them together, we easily obtain

$$\begin{aligned} \frac{1}{2}&\frac{\mathrm{d}}{\mathrm{d}t} (\Vert \Lambda ^s u\Vert ^2_{L^2} + \Vert \Lambda ^s \omega \Vert ^2_{L^2} + \Vert \Lambda ^s b_1\Vert ^2_{L^2} + \Vert \Lambda ^s b_2\Vert ^2_{L^2}) + (\mu + \chi )\Vert \Lambda ^s \nabla u\Vert ^2_{L^2} + \kappa \Vert \Lambda ^s \nabla \omega \Vert ^2_{L^2}\\&+ 4 \chi \Vert \Lambda ^s \omega \Vert ^2_{L^2} + \nu (\Vert \Lambda ^s \partial _y b_1\Vert ^2_{L^2} + \Vert \Lambda ^s \partial _x b_2\Vert ^2_{L^2})\\ \le&-\int [\Lambda ^s , u \cdot \nabla ]u \cdot \Lambda ^s u + \int [\Lambda ^s, b \cdot \nabla ]b \cdot \Lambda ^s u - \int [\Lambda ^s, u \cdot \nabla ]\omega \cdot \Lambda ^s \omega \\&+ 4 \chi \int \Lambda ^s \nabla \times u \cdot \Lambda ^s \omega - \int [\Lambda ^s , u \cdot \nabla ]b_1 \cdot \Lambda ^s b_1 - \int [\Lambda ^s , u \cdot \nabla ]b_2 \cdot \Lambda ^s b_2 \\&+ \int [\Lambda ^s , b \cdot \nabla ]u_1 \cdot \Lambda ^s b_1 + \int [\Lambda ^s , b \cdot \nabla ]u_2 \cdot \Lambda ^s b_2, \end{aligned}$$

where we used the facts that,

$$\begin{aligned} \int b \cdot \nabla \Lambda ^s u_1 \cdot \Lambda ^s b_1 + \int b \cdot \nabla \Lambda ^s u_2 \cdot \Lambda ^s b_2 = \int b \cdot \nabla \Lambda ^s u \cdot \Lambda ^s b = - \int b \cdot \nabla \Lambda ^s b \cdot \Lambda ^s u, \end{aligned}$$

and

$$\begin{aligned} \int \Lambda ^s \nabla \times u \cdot \Lambda ^s \omega = \int \Lambda ^s \nabla \times \omega \cdot \Lambda ^s u. \end{aligned}$$

Due to the divergence free condition $\nabla \cdot b = 0$, we have

$$\begin{aligned} \Vert \Lambda ^s \nabla b \Vert ^2_{L^2} = \Vert \Lambda ^s \nabla \times b \Vert ^2_{L^2} \le 2 \left( \Vert \Lambda ^s \partial _x b_2 \Vert ^2_{L^2} + \Vert \Lambda ^s \partial _y b_1 \Vert ^2_{L^2} \right) , \end{aligned}$$

$$\begin{aligned} \Vert \Lambda ^s b \Vert ^2_{L^2} = \Vert \Lambda ^s b_1 \Vert ^2_{L^2} + \Vert \Lambda ^s b_2 \Vert ^2_{L^2}. \end{aligned}$$

Then, it follows from the above bounds and facts that

$$\begin{aligned} \frac{1}{2}&\frac{\mathrm{d}}{\mathrm{d}t} \left( \Vert \Lambda ^s u\Vert ^2_{L^2} + \Vert \Lambda ^s \omega \Vert ^2_{L^2} + \Vert \Lambda ^s b\Vert ^2_{L^2} \right) + (\mu + \chi )\Vert \Lambda ^s \nabla u\Vert ^2_{L^2} + \kappa \Vert \Lambda ^s \nabla \omega \Vert ^2_{L^2} \nonumber \\&+ 4 \chi \Vert \Lambda ^s \omega \Vert ^2_{L^2} + \frac{\nu }{2} \Vert \Lambda ^s \nabla b\Vert ^2_{L^2}\nonumber \\ \le&-\int [\Lambda ^s , u \cdot \nabla ]u \cdot \Lambda ^s u + \int [\Lambda ^s , b \cdot \nabla ]b \cdot \Lambda ^s u - \int [\Lambda ^s , u \cdot \nabla ]\omega \cdot \Lambda ^s \omega \nonumber \\&+ 4 \chi \int \Lambda ^s \nabla \times u \cdot \Lambda ^s \omega - \int [\Lambda ^s , u \cdot \nabla ]b \cdot \Lambda ^s b + \int [\Lambda ^s , b \cdot \nabla ]u \cdot \Lambda ^s b \nonumber \\ =&I_1 + \cdot \cdot \cdot + I_6. \end{aligned}$$

(4.5)

We first bound $I_1$. Resorting to the commutator estimate (2.1), we have

$$\begin{aligned} |I_1|&= \left| - \int [\Lambda ^s , u \cdot \nabla ]u \cdot \Lambda ^s u \right| \\&\le \Vert [\Lambda ^s , u \cdot \nabla ]u\Vert _{L^2} \Vert \Lambda ^{s-1} \nabla u\Vert _{L^2}\\&\le C \Vert \nabla u\Vert _{L^{\infty }}\Vert \Lambda ^s u\Vert _{L^2}\Vert \Lambda ^{s-1} \nabla u\Vert _{L^2}. \end{aligned}$$

Similarly,

$$\begin{aligned} |I_2|&= \left| \int [\Lambda ^s , b \cdot \nabla ]b \cdot \Lambda ^s u \right| \\&\le \Vert [\Lambda ^s , b \cdot \nabla ] b \Vert _{L^2} \Vert \Lambda ^{s-1} \nabla u\Vert _{L^2}\\&\le C \Vert \nabla b \Vert _{L^{\infty }}\Vert \Lambda ^s b\Vert _{L^2}\Vert \Lambda ^{s-1} \nabla u\Vert _{L^2}. \end{aligned}$$

Again using the commutator estimate (2.1), we get

$$\begin{aligned} |I_3|&= \left| - \int [\Lambda ^s , u \cdot \nabla ] \omega \cdot \Lambda ^s \omega \right| \\&\le \Vert [\Lambda ^s , u \cdot \nabla ] \omega \Vert _{L^2} \Vert \Lambda ^{s-1} \nabla \omega \Vert _{L^2}\\&\le C ( \Vert \nabla u \Vert _{L^{\infty }} \Vert \Lambda ^s \omega \Vert _{L^2} + \Vert \nabla \omega \Vert _{L^{\infty }} \Vert \Lambda ^s u \Vert _{L^2} )\Vert \Lambda ^{s-1} \nabla \omega \Vert _{L^2}, \end{aligned}$$

and

$$\begin{aligned} |I_5| + |I_6|&= \left| - \int [\Lambda ^s , u \cdot \nabla ] b \cdot \Lambda ^s b \right| + \left| \int [\Lambda ^s , b \cdot \nabla ] u \cdot \Lambda ^s b \right| \\&\le (\Vert [\Lambda ^s , u \cdot \nabla ] b \Vert _{L^2} + \Vert [\Lambda ^s , b \cdot \nabla ] u \Vert _{L^2})\Vert \Lambda ^{s-1} \nabla b \Vert _{L^2}\\&\le C ( \Vert \nabla b \Vert _{L^{\infty }} \Vert \Lambda ^s u \Vert _{L^2} + \Vert \nabla u \Vert _{L^{\infty }} \Vert \Lambda ^s b \Vert _{L^2} )\Vert \Lambda ^{s-1} \nabla b\Vert _{L^2}. \end{aligned}$$

For $I_4$, by the Young inequality

$$\begin{aligned} |I_4|&\le 4 \chi \Vert \Lambda ^s \nabla u\Vert _{L^2} \Vert \Lambda ^s \omega \Vert _{L^2}\\&\le \left( \frac{\mu }{2} + \chi \right) \Vert \Lambda ^s \nabla u\Vert _{L^2}^2 + \frac{4 \chi ^2}{\frac{\mu }{2} + \chi } \Vert \Lambda ^s \omega \Vert ^2_{L^2}. \end{aligned}$$

Inserting the above estimates into (4.5), and note that $\Vert f\Vert _{L^{\infty }} \le C \Vert f\Vert _{H^s}$ with $s>1$, and using the Young inequality, we have

$$\begin{aligned} \frac{1}{2}&\frac{\mathrm{d}}{\mathrm{d}t} (\Vert \Lambda ^s u\Vert ^2_{L^2} + \Vert \Lambda ^s \omega \Vert ^2_{L^2} + \Vert \Lambda ^s b\Vert ^2_{L^2}) + \frac{\mu }{2}\Vert \Lambda ^s \nabla u\Vert ^2_{L^2} + \kappa \Vert \Lambda ^s \nabla \omega \Vert ^2_{L^2} \nonumber \\&+ \frac{8 \mu \chi }{\mu + 2 \chi }\Vert \Lambda ^s \omega \Vert ^2_{L^2} + \frac{\nu }{2} \Vert \Lambda ^s \nabla b\Vert ^2_{L^2}\nonumber \\ \le&C \Vert \Lambda ^s u\Vert _{L^2} (\Vert \nabla u\Vert _{H^s}^2 + \Vert \nabla \omega \Vert _{H^s}^2 + \Vert \nabla b\Vert _{H^s}^2 ) + C \Vert \Lambda ^s \omega \Vert _{L^2} \Vert \nabla u\Vert _{H^s} \Vert \nabla \omega \Vert _{H^s} \nonumber \\&+ C \Vert \Lambda ^s b\Vert _{L^2} \Vert \nabla u\Vert _{H^s} \Vert \nabla b \Vert _{H^s} \nonumber \\ \le&C (\Vert \nabla u\Vert _{H^s}^2 + \Vert \nabla \omega \Vert _{H^s}^2 + \Vert \nabla b\Vert _{H^s}^2 ) (\Vert u\Vert ^2_{H^s} + \Vert \omega \Vert ^2_{H^s} + \Vert b\Vert ^2_{H^s} )^{\frac{1}{2}}. \end{aligned}$$

(4.6)

Substituting (4.6) and (3.5) together, we get

$$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}t}&(\Vert u\Vert ^2_{H^s} + \Vert \omega \Vert ^2_{H^s} + \Vert b\Vert ^2_{H^s}) + c_1 (\Vert \nabla u\Vert ^2_{H^s} + \Vert \nabla \omega \Vert ^2_{H^s} + \Vert \nabla b\Vert ^2_{H^s}) \nonumber \\&\le C (\Vert \nabla u\Vert ^2_{H^s} + \Vert \nabla \omega \Vert ^2_{H^s} + \Vert \nabla b\Vert ^2_{H^s})(\Vert u\Vert ^2_{H^s} + \Vert \omega \Vert ^2_{H^s} + \Vert b\Vert ^2_{H^s})^{\frac{1}{2}} \end{aligned}$$

(4.7)

where $c_1 = \min \{\mu , 2\kappa , \nu \}$. This inequality indicates that, if the initial $(u_0, \omega _0, b_0)$ satisfy, for $0< \epsilon < \epsilon _0 = (\frac{c_1}{C})^2$

$$\begin{aligned} \Vert u_0\Vert ^2_{H^s} + \Vert \omega _0\Vert ^2_{H^s} + \Vert b_0\Vert ^2_{H^s} < \epsilon , \end{aligned}$$

then the corresponding solution remains for all time. That is

$$\begin{aligned} \Vert u(t)\Vert ^2_{H^s} + \Vert \omega (t)\Vert ^2_{H^s} + \Vert b(t)\Vert ^2_{H^s} < \epsilon . \end{aligned}$$

(4.8)

Next, we need to prove that (4.8) is correct. In fact, if suppose (4.8) is not true and $T_0$ is the first time such that (4.8) is violated, i.e.,

$$\begin{aligned} \Vert u(T_0)\Vert ^2_{H^s} + \Vert \omega (T_0)\Vert ^2_{H^s} + \Vert b(T_0)\Vert ^2_{H^s} = \epsilon , \end{aligned}$$

and (4.8) holds for any $0 \le t < T_0$. We can deduce from (4.7) that for any $0 \le t \le T_0$,

$$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}t} (\Vert u\Vert ^2_{H^s} + \Vert \omega \Vert ^2_{H^s} + \Vert b\Vert ^2_{H^s}) + (c_1 - C \sqrt{\epsilon }) (\Vert \nabla u\Vert ^2_{H^s} + \Vert \nabla \omega \Vert ^2_{H^s} + \Vert \nabla b\Vert ^2_{H^s}) \le 0. \end{aligned}$$

Therefore,

$$\begin{aligned} \Vert u\Vert ^2_{H^s} + \Vert \omega \Vert ^2_{H^s} + \Vert b\Vert ^2_{H^s} \le \Vert u_0\Vert ^2_{H^s} + \Vert \omega _0\Vert ^2_{H^s} + \Vert b_0\Vert ^2_{H^s} < \epsilon . \end{aligned}$$

This is a contradiction. Thus, we get the uniform bound of (4.8). In addition,

$$\begin{aligned} \int \limits ^t_0 (\Vert \nabla u(\tau )\Vert ^2_{H^s} + \Vert \nabla \omega (\tau )\Vert ^2_{H^s} + \Vert \nabla b(\tau )\Vert ^2_{H^s}) \mathrm{d} \tau \le C \epsilon . \end{aligned}$$

Therefore, the proof of (I) of Theorem 1.2 is completed.

$\square$

Next, we start to prove (II) of Theorem 1.2. For the simplicity of the proof, we divide the main proof into two parts. Firstly, we give the proof of (1.8), then prove that (1.9) is true. As a tool, we verify the following proposition in the negative Sobolev space $\dot{H}^{-l}$ with $0 \le l <1$.

Proposition 4.1

Let $c_1 = \min \{\mu , 2\kappa , \nu \}$. Then, for $0 \le l <1$, we have

$$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}t}&(\Vert u\Vert ^2_{\dot{H}^{-l}} + \Vert b \Vert ^2_{\dot{H}^{-l}} + \Vert \omega \Vert ^2_{\dot{H}^{-l}}) + c_1 (\Vert \nabla u\Vert ^2_{\dot{H}^{-l}} + \Vert \nabla b \Vert ^2_{\dot{H}^{-l}} + \Vert \nabla \omega \Vert ^2_{\dot{H}^{-l}}) \nonumber \\ \le&C(\Vert \nabla u\Vert _{L^2} + \Vert \nabla \omega \Vert _{L^2} + \Vert \nabla b\Vert _{L^2})(\Vert \nabla u\Vert ^{1-l}_{L^2} + \Vert \nabla b \Vert ^{1-l}_{L^2})(\Vert u\Vert _{L^2}^l + \Vert b\Vert ^l_{L^2})\nonumber \\&\times (\Vert u\Vert _{\dot{H}^{-l}} + \Vert b \Vert _{\dot{H}^{-l}} + \Vert \omega \Vert _{\dot{H}^{-l}} ). \end{aligned}$$

(4.9)

Proof of Proposition 4.1

For $0 \le l < 1$, similar to the process of (4.5) and the divergence free condition $\nabla \cdot u = \nabla \cdot b = 0$, we drive that

$$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}t}&(\Vert u\Vert ^2_{\dot{H}^{-l}} + \Vert b \Vert ^2_{\dot{H}^{-l}} + \Vert \omega \Vert ^2_{\dot{H}^{-l}}) + 2(\mu + \chi )\Vert \nabla u\Vert ^2_{\dot{H}^{-l}} + 2\kappa \Vert \nabla \omega \Vert ^2_{\dot{H}^{-l}} + 8 \chi \Vert \omega \Vert ^2_{\dot{H}^{-l}} + \nu \Vert \nabla b\Vert ^2_{\dot{H}^{-l}} \nonumber \\ \le&2 \Vert u \cdot \nabla u\Vert _{\dot{H}^{-l}}\Vert u\Vert _{\dot{H}^{-l}} + 2 \Vert b \cdot \nabla b\Vert _{\dot{H}^{-l}}\Vert u\Vert _{\dot{H}^{-l}} + 2 \Vert u \cdot \nabla \omega \Vert _{\dot{H}^{-l}}\Vert \omega \Vert _{\dot{H}^{-l}} + (\mu + 2\chi )\Vert \nabla u\Vert _{\dot{H}^{-l}}^2 \nonumber \\&+ \frac{8\chi ^2}{\frac{\mu }{2} + \chi }\Vert \omega \Vert ^2_{\dot{H}^{-l}} + 2( \Vert u \cdot \nabla b\Vert _{\dot{H}^{-l}} + \Vert b \cdot \nabla u\Vert _{\dot{H}^{-l}} )\Vert b\Vert _{\dot{H}^{-l}}. \end{aligned}$$

(4.10)

By Lemma 2.2, Hölder’s inequality and Gagliardo–Nirenberg inequality, we have

$$\begin{aligned}&\Vert u \cdot \nabla u\Vert _{\dot{H}^{-l}} \le C \Vert u \cdot \nabla u\Vert _{L^{\frac{2}{l+1}}},\\&\quad \Vert u \cdot \nabla u\Vert _{L^{\frac{2}{l+1}}} \le C \Vert u\Vert _{L^{\frac{2}{l}}} \Vert \nabla u\Vert _{L^2}, \end{aligned}$$

and

$$\begin{aligned} \Vert u\Vert _{L^{\frac{2}{l}}} \le C \Vert \nabla u\Vert _{L^2}^{1-l} \Vert u\Vert _{L^2}^{l}. \end{aligned}$$

Then,

$$\begin{aligned} \Vert u \cdot \nabla u\Vert _{\dot{H}^{-l}} \le C \Vert \nabla u\Vert _{L^2}^{2-l} \Vert u\Vert _{L^2}^{l}. \end{aligned}$$

(4.11)

Similarly, we have

$$\begin{aligned}&\Vert b \cdot \nabla b\Vert _{\dot{H}^{-l}} \le C \Vert \nabla b\Vert _{L^2}^{2-l} \Vert b\Vert _{L^2}^{l}, \end{aligned}$$

(4.12)

$$\begin{aligned}&\Vert u \cdot \nabla b\Vert _{\dot{H}^{-l}}+ \Vert b \cdot \nabla u\Vert _{\dot{H}^{-l}}\nonumber \\&\qquad \le C ( \Vert \nabla u\Vert _{L^2}^{1-l} \Vert u\Vert _{L^2}^{l}\Vert \nabla b\Vert _{L^2} + \Vert \nabla b\Vert _{L^2}^{1-l} \Vert b\Vert _{L^2}^{l}\Vert \nabla u\Vert _{L^2}) , \end{aligned}$$

(4.13)

$$\begin{aligned}&\Vert u \cdot \nabla \omega \Vert _{\dot{H}^{-l}} \le C \Vert \nabla \omega \Vert _{L^2}^{1-l} \Vert \nabla u\Vert _{L^2}^{1-l} \Vert u\Vert _{L^2}^{l}. \end{aligned}$$

(4.14)

Combining (4.10) and (4.11)–(4.14), we obtain

$$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}t}&(\Vert u\Vert ^2_{\dot{H}^{-l}} + \Vert b \Vert ^2_{\dot{H}^{-l}} + \Vert \omega \Vert ^2_{\dot{H}^{-l}}) + c_1 (\Vert \nabla u\Vert ^2_{\dot{H}^{-l}} + \Vert \nabla b \Vert ^2_{\dot{H}^{-l}} + \Vert \nabla \omega \Vert ^2_{\dot{H}^{-l}}) \\ \le&C(\Vert \nabla u\Vert _{L^2} + \Vert \nabla \omega \Vert _{L^2} + \Vert \nabla b\Vert _{L^2})(\Vert \nabla u\Vert ^{1-l}_{L^2} + \Vert \nabla b \Vert ^{1-l}_{L^2})(\Vert u\Vert _{L^2}^l + \Vert b\Vert ^l_{L^2})\\&\times (\Vert u\Vert _{\dot{H}^{-l}} + \Vert b \Vert _{\dot{H}^{-l}} + \Vert \omega \Vert _{\dot{H}^{-l}} ), \end{aligned}$$

where $c_1 = \min \{\mu , 2\kappa , \nu \}$. Thus, the proof of Proposition 4.1 is completed. $\square$

Proof of (II) of Theorem 1.2

• (1) We first prove (1.8). For $0 \le m \le s$, similarly to the process of (4.5), together with Hölder’s inequality and the divergence free condition $\nabla \cdot u = \nabla \cdot b = 0$, we get

  $$\begin{aligned}&\frac{1}{2} \frac{\mathrm {d}}{\mathrm {d}t} (\Vert \Lambda ^m u\Vert ^2_{L^2} + \Vert \Lambda ^m \omega \Vert ^2_{L^2} + \Vert \Lambda ^m b\Vert ^2_{L^2}) + (\mu + \chi )\Vert \Lambda ^m \nabla u\Vert ^2_{L^2} + \kappa \Vert \Lambda ^m \nabla \omega \Vert ^2_{L^2} \nonumber \\ {}&\quad + 4 \chi \Vert \Lambda ^m \omega \Vert ^2_{L^2} + \frac{\nu }{2} \Vert \Lambda ^m \nabla b\Vert ^2_{L^2} \nonumber \\&\le -\int \Lambda ^m( u \cdot \nabla u) \cdot \Lambda ^m u + \int \Lambda ^m (b \cdot \nabla b) \cdot \Lambda ^m u - \int \Lambda ^m (u \cdot \nabla b) \cdot \Lambda ^m b \nonumber \\ {}&\quad + \int \Lambda ^m (b \cdot \nabla u) \cdot \Lambda ^m b - \int \Lambda ^m (u \cdot \nabla \omega ) \cdot \Lambda ^m \omega + 4 \chi \int \Lambda ^m \nabla \times u \cdot \Lambda ^m \omega \nonumber \\&\le \Vert \Lambda ^{m+1} (u \otimes u)\Vert _{L^2}\Vert \Lambda ^{m} u\Vert _{L^2} + \Vert \Lambda ^{m+1} (b \otimes b)\Vert _{L^2}\Vert \Lambda ^{m} u\Vert _{L^2} + \Vert \Lambda ^{m+1} (u \otimes b)\Vert _{L^2}\Vert \Lambda ^{m} b\Vert _{L^2} \nonumber \\ {}&\quad + \Vert \Lambda ^{m+1} (b \otimes u)\Vert _{L^2}\Vert \Lambda ^{m} b\Vert _{L^2} + \Vert \Lambda ^{m+1} (u \otimes \omega )\Vert _{L^2}\Vert \Lambda ^{m} \omega \Vert _{L^2} + 4 \chi \Vert \Lambda ^{m} \nabla u\Vert ^2_{L^2} \Vert \Lambda ^{m} \nabla \omega \Vert ^2_{L^2} \nonumber \\&= A_1 + A_2 + A_3 + A_4 + A_5 + A_6. \end{aligned}$$

  (4.15)

  For $A_1$, applying the commutator estimate (2.2) and Lemma 2.3 yield

  $$\begin{aligned} A_1 \le C (\Vert u\Vert _{L^{\infty }} \Vert \nabla u\Vert _{\dot{H}^{m}}) \Vert u\Vert _{\dot{H}^m} \le C \Vert u\Vert _{H^s} \Vert \nabla u\Vert ^2_{\dot{H}^m}, \end{aligned}$$

  where we used the facts that

  $$\begin{aligned} \Vert u\Vert _{\dot{H}^{m}} \le C \Vert \nabla u\Vert _{\dot{H}^{m}}^{\frac{m}{m+1}}\Vert u\Vert _{L^2}^{\frac{1}{m+1}}, \quad \Vert u\Vert _{L^{\infty }} \le C \Vert u \Vert _{L^2}^{\frac{m}{m+1}} \Vert \nabla u\Vert _{\dot{H}^{m}}^{\frac{1}{m+1}} , \quad m>0 . \end{aligned}$$

  Using the commutator estimate (2.2), Lemma 2.3 and the Young inequality lead to

  $$\begin{aligned} A_2&\le C (\Vert b\Vert _{L^{\infty }} \Vert \nabla b\Vert _{\dot{H}^{m}}) \Vert u\Vert _{\dot{H}^m} \\&\le C (\Vert b \Vert _{L^2}^{\frac{m}{m+1}} \Vert \nabla b\Vert _{\dot{H}^m}^{\frac{1}{m+1}} \Vert \nabla u\Vert _{\dot{H}^m}^{\frac{m}{m+1}}\Vert u\Vert _{L^2}^{\frac{1}{m+1}}) \Vert \nabla b\Vert _{\dot{H}^{m}}\\&\le C (\Vert b \Vert _{L^2}^{\frac{m}{m+1}} \Vert u\Vert _{L^2}^{\frac{1}{m+1}} \Vert \nabla b\Vert _{\dot{H}^{m}})(\Vert \nabla b\Vert _{\dot{H}^{m}}+ \Vert \nabla u\Vert _{\dot{H}^{m}}) \\&\le (\Vert u\Vert _{H^s} + \Vert b\Vert _{H^s})(\Vert \nabla b\Vert ^2_{\dot{H}^{m}}+ \Vert \nabla u\Vert ^2_{\dot{H}^{m}}). \end{aligned}$$

  Similarly,

  $$\begin{aligned}&A_3 + A_4 \le C (\Vert u\Vert _{H^s} + \Vert b\Vert _{H^s})(\Vert \nabla b\Vert ^2_{\dot{H}^{m}}+ \Vert \nabla u\Vert ^2_{\dot{H}^{m}}),\\&\quad A_5 \le C (\Vert u\Vert _{H^s} + \Vert \omega \Vert _{H^s})(\Vert \nabla \omega \Vert ^2_{\dot{H}^{m}}+ \Vert \nabla u\Vert ^2_{\dot{H}^{m}}). \end{aligned}$$

  For $A_6$, applying the Young inequality, yields

  $$\begin{aligned} A_6 \le (\frac{\mu }{2} + \chi ) \Vert \Lambda ^m\nabla u\Vert ^2_{L^2} + \frac{8 \chi ^2}{\mu + 2 \chi } \Vert \Lambda ^m \omega \Vert ^2_{L^2}. \end{aligned}$$

  Inserting $A_1$–$A_6$ into (4.15), we obtain,

  $$\begin{aligned}&\frac{\mathrm{d}}{\mathrm{d}t} (\Vert u\Vert ^2_{\dot{H}^m} + \Vert \omega \Vert ^2_{\dot{H}^m} + \Vert b\Vert ^2_{\dot{H}^m}) + c_1 (\Vert \nabla u\Vert ^2_{\dot{H}^m} + \Vert \nabla \omega \Vert ^2_{\dot{H}^m} + \Vert \nabla b\Vert ^2_{\dot{H}^m}) \nonumber \\&\quad \le C ( \Vert u\Vert _{H^s} + \Vert \omega \Vert _{H^s} + \Vert b\Vert _{H^s} )(\Vert \nabla u\Vert ^2_{\dot{H}^m} + \Vert \nabla \omega \Vert ^2_{\dot{H}^m} + \Vert \nabla b\Vert ^2_{\dot{H}^m}), \end{aligned}$$

  (4.16)

  where for $0 \le m\le s$. Choosing the $\epsilon$ in (4.8) sufficiently small (i.e.,$\epsilon < (\frac{c_1}{2C})^2$), we have

  $$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}t} (\Vert u\Vert ^2_{\dot{H}^m} + \Vert \omega \Vert ^2_{\dot{H}^m} + \Vert b\Vert ^2_{\dot{H}^m}) + \frac{c_1}{2} (\Vert \nabla u\Vert ^2_{\dot{H}^m} + \Vert \nabla \omega \Vert ^2_{\dot{H}^m} + \Vert \nabla b\Vert ^2_{\dot{H}^m}) \le 0. \end{aligned}$$

  (4.17)

  Because $\dot{B}^s_{2,2} \sim \dot{H}^s$, applying Lemma 2.4, we get

  $$\begin{aligned}&\Vert u\Vert _{\dot{H}^m} \le C \Vert u \Vert ^{\frac{1}{m+l+1}}_{\dot{H}^{-l}} \Vert \nabla u \Vert ^{\frac{m+l}{m+l+1}}_{\dot{H}^{m}}, \end{aligned}$$

  (4.18)
  $$\begin{aligned}&\quad \Vert b\Vert _{\dot{H}^m} \le C \Vert b \Vert ^{\frac{1}{m+l+1}}_{\dot{H}^{-l}} \Vert \nabla b\Vert ^{\frac{m+l}{m+l+1}}_{\dot{H}^{m}}, \end{aligned}$$

  (4.19)
  $$\begin{aligned}&\quad \Vert \omega \Vert _{\dot{H}^m} \le C \Vert \omega \Vert ^{\frac{1}{m+l+1}}_{\dot{H}^{-l}} \Vert \nabla \omega \Vert ^{\frac{m+l}{m+l+1}}_{\dot{H}^{m}}. \end{aligned}$$

  (4.20)

  Therefore, if

  $$\begin{aligned} \Vert u\Vert _{\dot{H}^{-l}} + \Vert b \Vert _{\dot{H}^{-l}} + \Vert \omega \Vert _{\dot{H}^{-l}} \le C, \end{aligned}$$

  (4.21)

  and inserting (4.18)–(4.20) into (4.17), we obtain

  $$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}t} (\Vert u\Vert ^2_{\dot{H}^m} + \Vert \omega \Vert ^2_{\dot{H}^m} + \Vert b\Vert ^2_{\dot{H}^m}) + C (\Vert \nabla u\Vert ^2_{\dot{H}^m} + \Vert \nabla \omega \Vert ^2_{\dot{H}^m} + \Vert \nabla b\Vert ^2_{\dot{H}^m})^{\frac{m+l+1}{m+l}} \le 0. \end{aligned}$$

  (4.22)

  It implies that

  $$\begin{aligned} \Vert u\Vert ^2_{\dot{H}^m} + \Vert \omega \Vert ^2_{\dot{H}^m} + \Vert b\Vert ^2_{\dot{H}^m} \le C (1 + t)^{-l -m}. \end{aligned}$$

  (4.23)

  To make the proof more completed, therefore, we need to verify that (4.21) holds for $0 \le l < 1$.

  Next, we prove that (4.21). Where we will applying the bootstrapping argument. Assume that

  $$\begin{aligned} \Vert u_0\Vert ^2_{\dot{H}^{-l}} + \Vert \omega _0\Vert ^2_{\dot{H}^{-l}} + \Vert b_0\Vert ^2_{\dot{H}^{-l}} \le C_0. \end{aligned}$$

  (4.24)

  Suppose that for all $t \in [0,T]$,

  $$\begin{aligned} \Vert u(t)\Vert ^2_{\dot{H}^{-l}} + \Vert \omega (t)\Vert ^2_{\dot{H}^{-l}} + \Vert b(t)\Vert ^2_{\dot{H}^{-l}} \le 2 C_0. \end{aligned}$$

  (4.25)

  If we can drive that for all $t \in [0,T]$,

  $$\begin{aligned} \Vert u(t)\Vert ^2_{\dot{H}^{-l}} + \Vert \omega (t)\Vert ^2_{\dot{H}^{-l}} + \Vert b(t)\Vert ^2_{\dot{H}^{-l}} \le \frac{3 C_0}{2}, \end{aligned}$$

  (4.26)

  then an application of the bootstrapping argument would imply that the solution $(u, \omega , b)$ of system (1.3) satisfies (4.26) for all $t \in [0,T]$, which implies (4.21). With (4.24) and (4.25) at our disposal, we shall show that (4.26) holds. Integrating (4.9) in [0, t] with $0< t \le T$, together with (4.23), (4.24) and (4.8), one infers that

  $$\begin{aligned} \Vert u(t)&\Vert ^2_{\dot{H}^{-l}} + \Vert b(t) \Vert ^2_{\dot{H}^{-l}} + \Vert \omega (t) \Vert ^2_{\dot{H}^{-l}} \\ \le&\Vert u_0\Vert ^2_{\dot{H}^{-l}} + \Vert b_0 \Vert ^2_{\dot{H}^{-l}} + \Vert \omega _0 \Vert ^2_{\dot{H}^{-l}} \\&+ C\sup _{0 \le \tau \le t} (\Vert u\Vert _{\dot{H}^{-l}} + \Vert b \Vert _{\dot{H}^{-l}} + \Vert \omega \Vert _{\dot{H}^{-l}} ) \int \limits ^t_0 \left( \Vert \nabla u(\tau )\Vert _{L^2} + \Vert \nabla b(\tau )\Vert _{L^2} + \Vert \nabla \omega (\tau )\Vert _{L^2} \right) \\&\times \left( \Vert \nabla u(\tau )\Vert _{L^2}^{1-l} +\Vert \nabla b(\tau )\Vert ^{1-l}_{L^2}\right) \left( \Vert u(\tau )\Vert _{L^2}^l + \Vert b(\tau )\Vert _{L^2}^l \right) \mathrm{d}\tau \\ \le&C_0 + C\epsilon ^l \sup _{0 \le \tau \le t} \left( \Vert u\Vert _{\dot{H}^{-l}} + \Vert b \Vert _{\dot{H}^{-l}} + \Vert \omega \Vert _{\dot{H}^{-l}} \right) \int \limits ^t_0 (1+\tau )^{-\left( \frac{l(1-l)}{2} + 1\right) } \mathrm{d}\tau \\ \le&C_0 + C \epsilon ^l \sup _{0 \le \tau \le t} \left( \Vert u\Vert _{\dot{H}^{-l}} + \Vert b \Vert _{\dot{H}^{-l}} + \Vert \omega \Vert _{\dot{H}^{-l}} \right) . \end{aligned}$$

  By choosing $\epsilon$ in (4.8) sufficiently small, then the above inequality together with the Young inequality yield (4.26) for all $t \in [0,T]$, which close the proof. Then, we complete the proof of (4.21).

• (2) Now, we start to prove (1.9). Applying $D^m$ to the second equation of (1.3), dotting $D^m \omega$ and integrating in ${\mathbb {R}}^2$, we get

  $$\begin{aligned} \frac{1}{2}&\frac{\mathrm{d}}{\mathrm{d}t} \Vert D^m \omega \Vert _{L^2}^2 + \kappa \Vert D^{m+1} \omega \Vert _{L^2}^2 + 4 \chi \Vert D^m \omega \Vert _{L^2}^2 \\&\le 2\chi \int \limits D^m \nabla \times u \cdot D^m \omega \mathrm{d}x - \int \limits D^m (u \cdot \nabla \omega ) \cdot D^m \omega \mathrm{d}x\\&\le C \left( \Vert D^{m+1} u\Vert _{L^2} + \Vert D^{m+1} u\Vert _{L^2}\Vert \omega \Vert _{L^{\infty }} + \Vert u\Vert _{L^{\infty }}\Vert D^{m+1} \omega \Vert _{L^2} \right) \Vert D^m \omega \Vert _{L^2}\\&\le C (\Vert D^{m+1} u\Vert _{L^2} + \Vert D^{m+1} \omega \Vert _{L^2}), \end{aligned}$$

  where we used the commutator estimate (2.2), Hölder’s inequality and (1.7). Multiplying this equation by $e^{4 \chi t}$, integrating the resulting inequality in [0, t], together with (1.7) and (1.8), we have

  $$\begin{aligned} \Vert D^m \omega \Vert _{L^2} \le&e^{- 4 \chi t} \Vert D^m\omega _0 \Vert _{L^2} + C \int \limits ^t_0 e^{- 4\chi (t-\tau )} (\Vert D^{m+1} u(\tau )\Vert _{L^2} + \Vert D^{m+1} \omega (\tau )\Vert _{L^2}) \mathrm{d}\tau \\ \le&C e^{- 4 \chi t} + C \int \limits ^{\frac{t}{2}}_0 e^{- 4\chi (t-\tau )} (\Vert D^{m+1} u(\tau )\Vert _{L^2} + \Vert D^{m+1} \omega (\tau )\Vert _{L^2}) \mathrm{d}\tau \\&+ C \int \limits ^t_{\frac{t}{2}} e^{- 4\chi (t-\tau )} (\Vert D^{m+1} u(\tau )\Vert _{L^2} + \Vert D^{m+1} \omega (\tau )\Vert _{L^2}) \mathrm{d}\tau \\ \le&C e^{- 4 \chi t} + C \int \limits ^t_{\frac{t}{2}} e^{- 4\chi (t-\tau )} (1+\tau )^{- \frac{m+l}{2} - \frac{l}{2}} \mathrm{d}\tau \\&+ C e^{- 2 \chi t} \left( \int \limits ^{\frac{t}{2}}_0 \mathrm{d}\tau \right) ^{\frac{1}{2}} \left( \int \limits ^{\frac{t}{2}}_0 e^{- 4\chi (t-\tau )} (\Vert D^{m+1} u(\tau )\Vert ^2_{L^2} + \Vert D^{m+1} \omega (\tau )\Vert ^2_{L^2}) \mathrm{d}\tau \right) ^{\frac{1}{2}}\\ \le&C e^{- 4 \chi t} + C e^{- 2 \chi t} t^{\frac{1}{2}} + C(1+t)^{- \frac{m+1}{2} - \frac{l}{2}} \\ \le&C (1+\tau )^{- \frac{m+1}{2} - \frac{l}{2}}. \end{aligned}$$

  Thus, we complete (1.9). Therefore, the proof of Theorem 1.2 is completed.

$\square$