1 Introduction

The magneto-micropolar equations were introduced in [1] to describe the motion of an incompressible, electrically conducting micropolar fluids in the presence of an arbitrary magnetic field. It belongs to a class of fluids with nonsymmetric stress tensor and includes, as special cases, the classical fluids modeled by the Navier-Stokes equation (see, e.g., [5, 31, 39]), magnetohydrodynamic (MHD) equations (see, e.g., [26]) and micropolar equations (see., e.g., [15, 16]). The 3D incompressible magneto-micropolar fluid equations can be written as:

$$\begin{aligned} \left\{ \begin{array}{lcl} \partial _t u +u \cdot \nabla u = {(\mu +\chi ) \Delta u -\nabla \pi }+b\cdot \nabla b + 2\chi \nabla \times \omega ,\\ \partial _t \omega +u\cdot \nabla \omega -\gamma \nabla \nabla \cdot \omega + 4\chi \omega =\kappa \Delta \omega + 2\chi \nabla \times u, \\ \partial _t b +u\cdot \nabla b =\nu \Delta b + b\cdot \nabla u, \\ \nabla \cdot u =0, ~ \nabla \cdot b =0,\\ u(x, 0) =u_0(x), \omega (x, 0) =\omega _0(x), b(x, 0) =b_0(x), \end{array} \right. \end{aligned}$$
(1)

where \(x = (x_1, x_2, x_3) \in {\mathbb {R}}^3\) and \(t\ge 0\), \(u(x, t), \omega (x, t), b(x, t)\) and \(\pi (x, t)\) 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, and this is an isotropic system. The 3D magneto-micropolar equations reduce to the 2D micropolar equations when

$$\begin{aligned}&u = (u_1(x_1, x_2, t), u_2(x_1, x_2, t), 0), \qquad \pi = \pi (x_1, x_2,t), \\&b = (b_1(x_1, x_2, t), b_2(x_1, x_2, t), 0), \qquad \omega = (0,0,\omega _3(x_1, x_2, t)). \end{aligned}$$

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

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

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

is a scalar function representing the vorticity, and \(\nabla \times \omega = (\partial _{2}\omega , -\partial _{1}\omega )\).

The magneto-micropolar equations play an important role in engineering and physics and have attracted considerable attention from the community of mathematical fluids (see, e.g., [20, 25, 28, 29]). When (2) has full dissipation (namely, \(\mu\), \(\chi\), \(\kappa\), \(\nu > 0\)), the global existence and uniqueness of solutions could be obtained easily (see, e.g., [20, 28]). However, for the inviscid case (namely, (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 (2) with partial dissipation.

This paper aims at a system of the 2D magneto-micropolar equations that is closely related to (2),

$$\begin{aligned} \left\{ \begin{array}{lcl} \partial _{t}u + u\cdot \nabla u = (\mu + \chi ) \Delta u - \nabla \pi + b \cdot \nabla b + 2\chi \nabla \times \omega ,\\ \partial _{t}\omega + u\cdot \nabla \omega + 4\chi \omega = 2\chi \nabla \times u,\\ \partial _{t}b_{1} + u \cdot \nabla b_{1} = \nu \partial _{22} b_{1} + b \cdot \nabla u_{1},\\ \partial _{t}b_{2} + u \cdot \nabla b_{2} = \nu \partial _{11} b_{2} + b \cdot \nabla u_{2},\\ \nabla \cdot u = \nabla \cdot b = 0,\\ u(x,0) = u_{0}(x), \omega (x,0) = \omega _{0}(x), b(x,0) = b_{0}(x). \end{array} \right. \end{aligned}$$
(3)

Physically, the partial dissipation assumption is natural in the study of geophysical fluids. It turns out that, in certain regimes and under suitable scaling, certain dissipation can become small and be ignored. Anisotropic magnetic diffusion also arises in the modeling of reconnecting plasmas. When the resistivity of electrically conducting fluids such as certain plasmas and liquid metal is anisotropic and only in the mixed directions, the mixed magnetic diffusion may be relevant. In addition, mathematically, (3) allows us to explore the smoothing effect and the effect on large time behavior of the anisotropic magnetic diffusion. When the b with partial dissipation and zero angular viscosity, the global regularity problem for (3) can be quite difficult. However, many important progresses have recently been made on this direction (see, e.g., [6,7,8,9,10,11,12,13,14, 17, 23, 30, 36, 38, 42]). In [17, 23, 30], the global regularity of the 2D magneto-micropolar equations with various partial dissipation cases was obtained. Wang, Xu and Liu in [41] proved the uniqueness of global strong solution for the magneto-micropolar equations with zero angular viscosity in a smooth bounded domain. Yamazaki [43] obtained the global regularity of the Cauchy problem for the magneto-micropolar equations with zero angular viscosity

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) with \(\omega = 0\), \(b=0\) and \(\chi =0\), was proposed by the celebrated work of Leray [19]. By introducing the elegant method of Fourier splitting, the algebraic decay rate for weak solutions was first obtained by Schonbek [33]. Later, the result in [33] is sharpened and extended in [34], see also [35]. Recently, Niu and Shang [24] proved the \(L^2\)-decay estimates of weak solutions, and also proved the optimal decay rates of global solutions in \(\dot{H}^s({\mathbb {R}}^3)\)(\(s > \frac{3}{2}\)) and in \(\dot{B}^m_{2, 1}({\mathbb {R}}^3)\) with \(0 \le m \le \frac{1}{2}\). Shang and Gu [37] also proved the global existence of classical solutions for (3). Li [21] proved the \(L^2\)-decay estimates for global solutions of (8) and their derivate with initial data in \(L^1({\mathbb {R}}^2)\). In addition, Li [21] also shown the global stability of these solutions in \(H^s({\mathbb {R}}^2) (s>1)\) and the decay rates of global solutions and their higher derivates.

Motivated by the results of the magneto-micropolar equations [43] and the related fluid models [11, 18]. In this paper, the first theorem states that system (3) has a unique golbal solution when the initial data \((u_0, \omega _0, b_0)\) is sufficiently small in \(H^s({\mathbb {R}}^2)\), and obtain the upper bounds of time decay rates of the global solution to (3) in \(L^2 ({\mathbb {R}}^2)\), as stated in the following theorem.

Theorem 1

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

$$\begin{aligned} \Vert u_0\Vert _{H^s({\mathbb {R}}^2)}^2 + \Vert \omega _0\Vert _{H^s({\mathbb {R}}^2)}^2 + \Vert b_0\Vert _{H^s({\mathbb {R}}^2)}^2 < \epsilon , \end{aligned}$$
(4)

then system (3) has a unique global solution \((u, \omega , b)\) satisfying, for any \(t>0\),

$$\begin{aligned} \Vert u(t)&\Vert _{H^s({\mathbb {R}}^2)}^2 + \Vert \omega (t)\Vert _{H^s({\mathbb {R}}^2)}^2 + \Vert b(t)\Vert _{H^s({\mathbb {R}}^2)}^2 \nonumber \\&+ \int ^t_0 (\Vert \nabla u(\tau )\Vert _{H^s({\mathbb {R}}^2)}^2 + \Vert \omega (\tau )\Vert _{H^s({\mathbb {R}}^2)}^2 + \Vert \nabla b(\tau )\Vert _{H^s({\mathbb {R}}^2)}^2)d\tau \le C \epsilon , \end{aligned}$$
(5)

where \(C>0\) is a constant independent of t.

(II) suppose that \((u_0, \omega _0,b_0)\in L^1({\mathbb {R}}^2)\), then the global solution \((u, \omega , b)\) has the following upper decay rates:

$$\begin{aligned} \Vert u(t)\Vert _{L^2} + \Vert \omega (t)\Vert _{L^2} + \Vert b(t)\Vert _{L^2} \le C (1+t)^{-\frac{1}{2}}. \end{aligned}$$
(6)

Moreover, when \(\mu < \sqrt{3} \chi\), then the global solution \((u, \omega , b)\) of the system (3) has the following upper decay rates

$$\begin{aligned} \Vert \nabla u(t)\Vert _{L^2} + \Vert \omega (t)\Vert _{L^2} + \Vert \nabla b(t)\Vert _{L^2} \le C (1+t)^{-\frac{1}{2}}. \end{aligned}$$
(7)

Finally, we consider the 2D magneto-micropolar equations with partial dissipation for the magnetic field, which can be written as

$$\begin{aligned} \left\{ \begin{array}{lcl} \partial _{t}u + u\cdot \nabla u = (\mu + \chi ) \Delta u - \nabla \pi + b \cdot \nabla b + 2\chi \nabla \times \omega ,\\ \partial _{t}\omega + u\cdot \nabla \omega + 4\chi \omega = \kappa \Delta \omega + 2\chi \nabla \times u,\\ \partial _{t}b_{1} + u \cdot \nabla b_{1} = \nu \partial _{22} b_{1} + b \cdot \nabla u_{1},\\ \partial _{t}b_{2} + u \cdot \nabla b_{2} = \nu \partial _{11} b_{2} + b \cdot \nabla u_{2},\\ \nabla \cdot u = \nabla \cdot b = 0,\\ u(x,0) = u_{0}(x), \omega (x,0) = \omega _{0}(x), b(x,0) = b_{0}(x). \end{array} \right. \end{aligned}$$
(8)

Motivated by the [21, 24], we establish the global existence results to system (8) in Besov spaces \(B^s_{2,1}({\mathbb {R}}^2)\). Furthermore, we study the large time decay rates of these global solutions in the Besove spaces \(B^s_{2,1}({\mathbb {R}}^2)\), as stated in the following theorem.

Theorem 2

Let \(\mu > 0\), \(\chi >0\), \(\nu >0\) and \(\kappa >0\). Assume that \((u_0, \omega _0, b_0) \in B^s_{2, 1}({\mathbb {R}}^2)\) with \(s>0\) and \(\nabla \cdot u_0 = \nabla \cdot b_0 = 0\). Then the following two statements hold:

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

$$\begin{aligned} \Vert u_0\Vert _{\dot{B}^0_{2, 1}({\mathbb {R}}^2)}^2 + \Vert \omega _0\Vert _{\dot{B}^0_{2, 1}({\mathbb {R}}^2)}^2 + \Vert b_0\Vert _{\dot{B}^0_{2, 1}({\mathbb {R}}^2)}^2 < \epsilon , \end{aligned}$$
(9)

then system (8) has a unique global solution \((u, \omega , b)\) satisfying, for any \(t>0\),

$$\begin{aligned} \Vert u(t)&\Vert _{B^s_{2, 1}({\mathbb {R}}^2)} + \Vert \omega (t)\Vert _{B^s_{2, 1}({\mathbb {R}}^2)} + \Vert b(t)\Vert _{B^s_{2, 1}({\mathbb {R}}^2)} +\Vert \nabla ^2 u(\tau )\Vert _{L^1_t(B^s_{2, 1}({\mathbb {R}}^2))}\nonumber \\&+ \Vert \nabla ^2 \omega (\tau )\Vert _{L^1_t(B^1_{2, 1}({\mathbb {R}}^2))} + \Vert \nabla ^2 b(\tau )\Vert _{L^1_t(B^s_{2, 1}({\mathbb {R}}^2))} \le C, \end{aligned}$$
(10)

where \(C>0\) is a constant independent of t.

(II) Let \(s\ge 1\), suppose that \((u_0, \omega _0, b_0 ) \in \dot{H}^{-l}({\mathbb {R}}^2)\) with \(0 \le l < 1\). Then for all real numbers m with \(0 \le m \le s\), the global solution \((u, \omega , b)\) established in (I) satisfies the following decay estimates:

$$\begin{aligned} \Vert u(t)\Vert _{\dot{B}^m_{2, 1}({\mathbb {R}}^2)} + \Vert \omega (t)\Vert _{\dot{B}^m_{2, 1}({\mathbb {R}}^2)} + \Vert b(t)\Vert _{\dot{B}^m_{2, 1}({\mathbb {R}}^2)} \le C (1+t)^{-\frac{m}{2} - \frac{l}{2}}. \end{aligned}$$
(11)

Remark 1

  1. (i)

    Since \(L^p({\mathbb {R}}^2) \hookrightarrow \dot{H}^{-l}({\mathbb {R}}^2)\) when \(l\in [0,1)\) and \(p \in (1,2]\), and \(L^p({\mathbb {R}}^2) \hookrightarrow \dot{B}^{-l}_{2, \infty }({\mathbb {R}}^2)\) when \(l\in (0,1]\) and \(p \in [1,2)\), thus Theorem 2 also hold for \((u_0, \omega _0, b_0) \in L^p({\mathbb {R}}^2)\) with \(p\in [1,2]\).

  2. (ii)

    Because of the divergence free condition \(\nabla \cdot b=0\), then \(\Vert \nabla b\Vert _{L^2({\mathbb {R}}^2)} = \Vert \nabla \times b\Vert _{L^2({\mathbb {R}}^2)}\), thus for the full dissipation 2D magneto-micropolar equations, we also have the same results as Theorem 1 and Theorem 2.

Remark 2

  1. (i)

    Compared to the classical magneto-micropolar equations (1), the full Laplacian operator is replaced by partial magnetic diffusion in systems (3) and (8). Theorem 1 and Theorem 2 indicate that the mixed partial magnetic diffusion has the same effect as the full Laplacian in deriving the large time behavior, in the sense that the decay rates in Theorem 1 and Theorem 2 coincide with the solutions of system (1).

  2. (ii)

    In Theorem 2, by assuming the initial data small in the critical Besov space \(\dot{B}_{2, 1}^{0}\), we can establish the global well-posedness to (8). However, due to the lack of micro-rotational velocity dissipation and the complex structure of the magneto-micropolar equations, it appears difficult to show the global well-posedness in critical Besov space to the solutions of (3).

To prove Theorem 2, we focus on the uniform bounds of \(\Vert (u, \omega , b)\Vert _{B^s_{2,1}}\). As preparation, we firstly show the global existence of solutions with small data in \(\dot{B}^0_{2,1}({\mathbb {R}}^2)\), then used the \(\Vert (u(t), \omega (t), b(t))\Vert _{\dot{B}^0_{2,1}} \le C \epsilon\), to obtain (10). The rest of this paper is divided into four sections. Sections 2 and 3 state the proofs of Theorem 1 and Theorem 2, respectively. An appendix containing the Littlewood-Paley decomposition, the definition of Besov spaces, and several useful calculus inequalities are also given for the convenience of the readers. To simplify the notation, we will write \(\partial _1\) for \(\partial _{x_1}\), \(\partial _2\) for \(\partial _{x_2}\), \(\int f\) for \(\int _{{\mathbb {R}}^2} f dx\), \(\Vert f\Vert _{L^p}\) 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, \(\Vert f\Vert _{\dot{B}^s_{p,r}}\) and \(\Vert f\Vert _{B^s_{p,r}}\) for \(\Vert f\Vert _{\dot{B}^s_{p,r}({\mathbb {R}}^2)}\) and \(\Vert f\Vert _{B^s_{p,r}({\mathbb {R}}^2)}\) respectively, and \(L_t^{q}(\dot{B}^s_{p,r})\) and \(\tilde{L}_t^{q}(\dot{B}^s_{p,r})\) for \(L_t^{q}(\dot{B}^s_{p,r}({\mathbb {R}}^2))\) and \(\tilde{L}_t^{q}(\dot{B}^s_{p,r}({\mathbb {R}}^2))\) respectively.

2 The Proof of Theorem 1

This section is devoted to the proof of Theorem 1. We first prove the global well-posedness part (I) of Theorem 1. As preparation, we give the following global a priori estimates.

Proposition 3

Let \((u_0, \omega _0, b_0) \in L^2\). Then for any \(t>0\), the solution \((u, \omega , b)\) of (3) satisfies

$$\begin{aligned} \frac{1}{2}&\frac{d}{dt} \left( \Vert u(t)\Vert _{L^2}^2 + \Vert \omega (t)\Vert _{L^2}^2 + \Vert b(t)\Vert _{L^2}^2 \right) + \frac{\mu }{2} \Vert \nabla u\Vert _{L^2}^2 + 4\chi \left( 1 - \frac{2\chi }{2\chi + \mu }\right) \Vert \omega \Vert _{L^2}^2 \nonumber \\&+ \nu (\Vert \partial _{y} b_1\Vert _{L^2}^2 + \Vert \partial _{x} b_2\Vert _{L^2}^2) \le 0, \end{aligned}$$
(12)
$$\begin{aligned} \Vert u(t)&\Vert _{L^2}^2 + \Vert \omega (t)\Vert _{L^2}^2 + \Vert b(t)\Vert _{L^2}^2 + \mu \int ^t_0 \Vert \nabla u(\tau )\Vert _{L^2}^2 d \tau + \frac{4 \chi \mu }{\mu + 2 \chi } \int ^t_0 \Vert \omega (\tau )\Vert _{L^2}^2 d \tau \nonumber \\&+ \nu \int ^t_0 \Vert \nabla b(\tau )\Vert _{L^2}^2 d\tau \le \Vert u_0\Vert _{L^2}^2 + \Vert \omega _0\Vert _{L^2}^2 + \Vert b_0\Vert _{L^2}^2. \end{aligned}$$
(13)

Proof

Taking the \(L^2\)-inner product to (3) with \((u, \omega , b_1, b_2)\) we have

$$\begin{aligned} \frac{1}{2}&\frac{d}{dt} \left( \Vert u(t)\Vert _{L^2}^2 + \Vert \omega (t)\Vert _{L^2}^2 + \Vert b(t)\Vert _{L^2}^2 \right) + (\mu + \chi ) \Vert \nabla u\Vert _{L^2}^2 + 4\chi \Vert \omega \Vert _{L^2}^2 \nonumber \\&+ \ \nu (\Vert \partial _{2} b_1\Vert _{L^2}^2 + \Vert \partial _{1} b_2\Vert _{L^2}^2) = 4 \chi \int \nabla \times u \cdot \omega dx, \end{aligned}$$
(14)

where we used the facts that,

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

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

$$\begin{aligned} 4 \chi \int \nabla \times u \cdot \omega dx&\le 4\chi \Vert \nabla u\Vert _{L^2} \Vert \omega \Vert _{L^2} \nonumber \\&\le \left( \frac{\mu }{2} + \chi \right) \Vert \nabla u\Vert _{L^2}^2 + \frac{4 \chi ^2}{\frac{\mu }{2} + \chi } \Vert \omega \Vert _{L^2}^2. \end{aligned}$$
(15)

Inserting (15) into (14), we obtain

$$\begin{aligned} \frac{1}{2}&\frac{d}{dt} \left( \Vert u(t)\Vert _{L^2}^2 + \Vert \omega (t)\Vert _{L^2}^2 + \Vert b(t)\Vert _{L^2}^2 \right) + \frac{\mu }{2} \Vert \nabla u\Vert _{L^2}^2 + 4\chi \left( 1 - \frac{2\chi }{2\chi + \mu }\right) \Vert \omega \Vert _{L^2}^2 \nonumber \\&+ \nu (\Vert \partial _{2} b_1\Vert _{L^2}^2 + \Vert \partial _{1} b_2\Vert _{L^2}^2) \le 0. \end{aligned}$$
(16)

Because of the divergence free condition \(\nabla \cdot b =0\), we have \(\Vert \nabla b\Vert _{L^2({\mathbb {R}}^2)}^2 = \Vert \nabla \times b\Vert _{L^2({\mathbb {R}}^2)}^2 \le 2 \left( \Vert \partial _2 b_1\Vert _{L^2({\mathbb {R}}^2)}^2 + \Vert \partial _1 b_2\Vert _{L^2({\mathbb {R}}^2)}^2 \right)\). Integrating (16) in [0, t], we can get

$$\begin{aligned}\ \Vert u(t)&\Vert _{L^2}^2 + \Vert \omega (t)\Vert _{L^2}^2 + \Vert b(t)\Vert _{L^2}^2 + \mu \int ^t_0 \Vert \nabla u(\tau )\Vert _{L^2}^2 d \tau + \frac{4 \chi \mu }{\mu + 2 \chi } \int ^t_0 \Vert \omega (\tau )\Vert _{L^2}^2 d \tau \nonumber \\&+ \nu \int ^t_0 \Vert \nabla b(\tau )\Vert _{L^2}^2 d\tau \le \Vert u_0\Vert _{L^2}^2 + \Vert \omega _0\Vert _{L^2}^2 + \Vert b_0\Vert _{L^2}^2. \end{aligned}$$

This completes the proof of Proposition 3\(\square\)

Next, we want to establish the global a priori \(H^s\) estimates. Applying \(\dot{\Delta }_j\) to (3), we have

$$\begin{aligned}{} & {} \partial _t \dot{\Delta }_j u + u \cdot \nabla \dot{\Delta }_j u - (\mu + \chi ) \Delta \dot{\Delta }_j u = - \dot{\Delta }_j \nabla \pi - [\dot{\Delta }_j, u\cdot \nabla ]u \nonumber \\{} & {} \quad + \dot{\Delta }_j(b \cdot \nabla b) + 2\chi \dot{\Delta }_j \nabla \times \omega , \end{aligned}$$
(17)
$$\begin{aligned}{} & {} \partial _t \dot{\Delta }_j \omega + u \cdot \nabla \dot{\Delta }_j \omega + 4 \chi \dot{\Delta }_j \omega = - [\dot{\Delta }_j, u\cdot \nabla ]\omega + 2\chi \dot{\Delta }_j \nabla \times u, \end{aligned}$$
(18)
$$\begin{aligned}{} & {} \partial _t \dot{\Delta }_j b_1 + u \cdot \nabla \dot{\Delta }_j b_1 - \nu \partial _{22} \dot{\Delta }_j b_1 = - [\dot{\Delta }_j, u\cdot \nabla ]b_1 + \dot{\Delta }_j(b \cdot \nabla u_1), \end{aligned}$$
(19)
$$\begin{aligned}{} & {} \partial _t \dot{\Delta }_j b_2 + u \cdot \nabla \dot{\Delta }_j b_2 - \nu \partial _{11} \dot{\Delta }_j b_2 = - [\dot{\Delta }_j, u\cdot \nabla ]b_2 + \dot{\Delta }_j(b \cdot \nabla u_2), \end{aligned}$$
(20)

where \([\dot{\Delta }_j, f \cdot \nabla ]g = \dot{\Delta }_j (f \cdot \nabla g) - f \cdot \dot{\Delta }_j (\nabla g)\) is commutator. Dotting (17) - (20) by \(\dot{\Delta }_j u\), \(\dot{\Delta }_j \omega\), \(\dot{\Delta }_j b_1\) and \(\dot{\Delta }_j b_2\) respectively, integrating the resulting equations in \({\mathbb {R}}^2\), and adding them together, we have

$$\begin{aligned}&\frac{1}{2} \frac{d}{dt} (\Vert \dot{\Delta }_j u\Vert ^2_{L^2} + \Vert \dot{\Delta }_j \omega \Vert ^2_{L^2} + \Vert \dot{\Delta }_j b_1\Vert ^2_{L^2} + \Vert \dot{\Delta }_j b_2\Vert ^2_{L^2}) + (\mu + \chi )\Vert \dot{\Delta }_j \nabla u\Vert ^2_{L^2}\\&+ 4 \chi \Vert \dot{\Delta }_j \omega \Vert ^2_{L^2} + \nu (\Vert \dot{\Delta }_j \partial _2 b_1\Vert ^2_{L^2} + \Vert \dot{\Delta }_j\partial _1 b_2\Vert ^2_{L^2})\\ \le&-\int [\dot{\Delta }_j , u \cdot \nabla ]u \cdot \dot{\Delta }_j u + \int [\dot{\Delta }_j , b \cdot \nabla ]b \cdot \dot{\Delta }_j u - \int [\dot{\Delta }_j , u \cdot \nabla ]\omega \cdot \dot{\Delta }_j \omega \\&+ 4 \chi \int \dot{\Delta }_j \nabla \times u \cdot \dot{\Delta }_j \omega - \int [\dot{\Delta }_j , u \cdot \nabla ]b_1 \cdot \dot{\Delta }_j b_1 - \int [\dot{\Delta }_j , u \cdot \nabla ]b_2 \cdot \dot{\Delta }_j b_2 \\&+ \int [\dot{\Delta }_j , b \cdot \nabla ]u_1 \cdot \dot{\Delta }_j b_1 + \int [\dot{\Delta }_j , b \cdot \nabla ]u_2 \cdot \dot{\Delta }_j b_2, \end{aligned}$$

where we used the facts that

$$\begin{aligned} \int b \cdot \nabla \dot{\Delta }_j u_1 \cdot \dot{\Delta }_j b_1 + \int b \cdot \nabla \dot{\Delta }_j u_2 \cdot \dot{\Delta }_j b_2 = \int b \cdot \nabla \dot{\Delta }_j u \cdot \dot{\Delta }_j b = - \int b \cdot \nabla \dot{\Delta }_j b \cdot \dot{\Delta }_j u, \end{aligned}$$

and

$$\begin{aligned} \int \dot{\Delta }_j \nabla \times u \cdot \dot{\Delta }_j \omega = \int \dot{\Delta }_j \nabla \times \omega \cdot \dot{\Delta }_j u. \end{aligned}$$

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

$$\begin{aligned}{} & {} \Vert \dot{\Delta }_j \nabla b \Vert ^2_{L^2} = \Vert \dot{\Delta }_j \nabla \times b \Vert ^2_{L^2} \le 2 (\Vert \dot{\Delta }_j \partial _1 b_2 \Vert ^2_{L^2} + \Vert \dot{\Delta }_j \partial _2 b_1 \Vert ^2_{L^2}), \\{} & {} \quad \Vert \dot{\Delta }_j b \Vert ^2_{L^2} = \Vert \dot{\Delta }_j b_1 \Vert ^2_{L^2} + \Vert \dot{\Delta }_j b_2 \Vert ^2_{L^2}. \end{aligned}$$

Then, we can derive from the above inequalities

$$\begin{aligned} \frac{1}{2}&\frac{d}{dt} (\Vert \dot{\Delta }_j u\Vert ^2_{L^2} + \Vert \dot{\Delta }_j \omega \Vert ^2_{L^2} + \Vert \dot{\Delta }_j b\Vert ^2_{L^2}) + (\mu + \chi )\Vert \dot{\Delta }_j \nabla u\Vert ^2_{L^2} \nonumber \\&+ 4 \chi \Vert \dot{\Delta }_j \omega \Vert ^2_{L^2} + \frac{\nu }{2} \Vert \dot{\Delta }_j \nabla b\Vert ^2_{L^2}\nonumber \\ \le&-\int [\dot{\Delta }_j , u \cdot \nabla ]u \cdot \dot{\Delta }_j u + \int [\dot{\Delta }_j , b \cdot \nabla ]b \cdot \dot{\Delta }_j u - \int [\dot{\Delta }_j , u \cdot \nabla ]\omega \cdot \dot{\Delta }_j \omega \nonumber \\&+ 4 \chi \int \dot{\Delta }_j \nabla \times u \cdot \dot{\Delta }_j \omega - \int [\dot{\Delta }_j , u \cdot \nabla ]b \cdot \dot{\Delta }_j b + \int [\dot{\Delta }_j , b \cdot \nabla ]u \cdot \dot{\Delta }_j b. \end{aligned}$$
(21)

Due to

$$\begin{aligned} \left| 4\chi \int \dot{\Delta }_j \nabla \times u \cdot \dot{\Delta }_j \omega \right| \le \left( \frac{\mu }{2} + \chi \right) \Vert \nabla u\Vert _{L^2}^2 + \frac{8 \chi ^2}{ \mu + 2 \chi } \Vert \omega \Vert ^2_{L^2}, \end{aligned}$$

then, we have

$$\begin{aligned} \frac{1}{2}&\frac{d}{dt} (\Vert \dot{\Delta }_j u\Vert ^2_{L^2} + \Vert \dot{\Delta }_j \omega \Vert ^2_{L^2} + \Vert \dot{\Delta }_j b\Vert ^2_{L^2}) + \frac{\mu }{2}\Vert \dot{\Delta }_j \nabla u\Vert ^2_{L^2} \nonumber \\&+ \frac{\nu }{2} \Vert \dot{\Delta }_j \nabla b\Vert ^2_{L^2} + (4 \chi - \frac{4\chi ^2}{\frac{\mu }{2} + \chi })\Vert \dot{\Delta }_j \omega \Vert ^2_{L^2}\nonumber \\ \le&-\int [\dot{\Delta }_j , u \cdot \nabla ]u \cdot \dot{\Delta }_j u + \int [\dot{\Delta }_j , b \cdot \nabla ]b \cdot \dot{\Delta }_j u - \int [\dot{\Delta }_j , u \cdot \nabla ]\omega \cdot \dot{\Delta }_j \omega \nonumber \\&- \int [\dot{\Delta }_j , u \cdot \nabla ]b \cdot \dot{\Delta }_j b + \int [\dot{\Delta }_j , b \cdot \nabla ]u \cdot \dot{\Delta }_j b. \end{aligned}$$
(22)

Multiplying (22) by \(2^{2sj}\), taking the \(l^2_j\) over \(j \in {\mathbb {Z}}\), nothing that \(\dot{B}^s_{2,2} = \dot{H}^s\) and using Hölder’s inequality, we yield

$$\begin{aligned} \frac{1}{2}&\frac{d}{dt} (\Vert u\Vert ^2_{\dot{H}^s} + \Vert \omega \Vert ^2_{\dot{H}^s} + \Vert b\Vert ^2_{\dot{H}^s}) + \frac{c_0}{2} ( \Vert \nabla u\Vert ^2_{\dot{H}^s} + \Vert \omega \Vert ^2_{\dot{H}^s} + \Vert \nabla b\Vert ^2_{\dot{H}^s}) \nonumber \\ \le&\Vert 2^{sj} \Vert [\dot{\Delta }_j , u \cdot \nabla ]u\Vert _{L^2} \Vert _{l^2_j} \Vert u\Vert _{\dot{H}^s} + \Vert 2^{sj} \Vert [\dot{\Delta }_j , b \cdot \nabla ]b \Vert _{L^2} \Vert _{l^2_j} \Vert u\Vert _{\dot{H}^s} \nonumber \\&+ \Vert 2^{sj} \Vert [\dot{\Delta }_j , u \cdot \nabla ]\omega \Vert _{L^2} \Vert _{l^2_j} \Vert \omega \Vert _{\dot{H}^s} + \Vert 2^{sj} \Vert [\dot{\Delta }_j , u \cdot \nabla ]b\Vert _{L^2} \Vert _{l^2_j} \Vert b\Vert _{\dot{H}^s} \nonumber \\&+ \Vert 2^{sj} \Vert [\dot{\Delta }_j , b \cdot \nabla ]u\Vert _{L^2} \Vert _{l^2_j} \Vert b\Vert _{\dot{H}^s}, \end{aligned}$$

where \(c_0 = \min \left\{\mu , \nu , \frac{8 \chi \mu }{\mu + 2 \chi } \right\}\). Adding the resulting inequality and (12) together, we have

$$\begin{aligned} \frac{d}{dt}&(\Vert u\Vert ^2_{H^s} + \Vert \omega \Vert ^2_{H^s} + \Vert b\Vert ^2_{H^s}) + c_0 ( \Vert \nabla u\Vert ^2_{H^s} + \Vert \omega \Vert ^2_{H^s} + \Vert \nabla b\Vert ^2_{H^s}) \nonumber \\ \le&2\Vert 2^{sj} \Vert [\dot{\Delta }_j , u \cdot \nabla ]u\Vert _{L^2} \Vert _{l^2_j} \Vert u\Vert _{H^s} + 2\Vert 2^{sj} \Vert [\dot{\Delta }_j , b \cdot \nabla ]b \Vert _{L^2} \Vert _{l^2_j} \Vert u\Vert _{H^s} \nonumber \\&+ 2 \Vert 2^{sj} \Vert [\dot{\Delta }_j , u \cdot \nabla ]\omega \Vert _{L^2} \Vert _{l^2_j} \Vert \omega \Vert _{H^s} + 2 \Vert 2^{sj} \Vert [\dot{\Delta }_j , u \cdot \nabla ]b\Vert _{L^2} \Vert _{l^2_j} \Vert b\Vert _{H^s} \nonumber \\&+ 2 \Vert 2^{sj} \Vert [\dot{\Delta }_j , b \cdot \nabla ]u\Vert _{L^2} \Vert _{l^2_j} \Vert b\Vert _{H^s}. \end{aligned}$$
(23)

Using commutator estimate (A5), and nothing that for \(s>1\),

$$\begin{aligned} \Vert f\Vert _{L^{\infty }} \le C \Vert f\Vert _{H^s}, \quad \Vert f\Vert _{\dot{B}^{s-1}_{2,2}} \le C \Vert f\Vert _{B^s_{2,2}} = C\Vert f\Vert _{H^s}, \end{aligned}$$

one obviously derives

$$\begin{aligned} 2\Vert 2^{sj} \Vert [\dot{\Delta }_j, u \cdot \nabla ]u\Vert _{L^2} \Vert _{l^2_j} \le C \Vert \nabla u\Vert _{L^{\infty }} \Vert \nabla u \Vert _{\dot{B}^{s-1}_{2,2}} \le C \Vert \nabla u\Vert ^2_{H^s}. \end{aligned}$$

Similarly, we have

$$\begin{aligned}&\Vert 2^{sj} \Vert [\dot{\Delta }_j , b \cdot \nabla ]b \Vert _{L^2} \Vert _{l^2_j} \le C \Vert \nabla b\Vert ^2_{H^s}, \\&\quad \Vert 2^{sj} \Vert [\dot{\Delta }_j , u \cdot \nabla ]b\Vert _{L^2} \Vert _{l^2_j} \le C( \Vert \nabla u\Vert _{L^{\infty }} \Vert \nabla b \Vert _{\dot{B}^{s-1}_{2,2}} + \Vert \nabla b\Vert _{L^{\infty }} \Vert \nabla u\Vert _{\dot{B}^{s-1}_{2,2}} )\\&\quad \le C \Vert \nabla u\Vert _{H^s} \Vert \nabla b\Vert _{H^s}, \end{aligned}$$

and

$$\begin{aligned} \Vert 2^{sj} \Vert [\dot{\Delta }_j , b \cdot \nabla ]u\Vert _{L^2} \Vert _{l^2_j} \le C \Vert \nabla u\Vert _{H^s} \Vert \nabla b\Vert _{H^s}. \end{aligned}$$

Taking advantage of the commutator estimate (A6), we imply that

$$\begin{aligned} \Vert 2^{sj} \Vert [\dot{\Delta }_j , u \cdot \nabla ]\omega \Vert _{L^2} \Vert _{l^2_j}&\le C( \Vert \nabla u\Vert _{L^{\infty }} \Vert \omega \Vert _{\dot{B}^{s}_{2,2}} + \Vert \omega \Vert _{L^{\infty }} \Vert \nabla u\Vert _{\dot{B}^{s}_{2,2}} )\\&\le C \Vert \nabla u\Vert _{H^s} \Vert \omega \Vert _{H^s}. \end{aligned}$$

Combining the above estimates together, we get

$$\begin{aligned} \frac{d}{dt}&(\Vert u\Vert ^2_{H^s} + \Vert \omega \Vert ^2_{H^s} + \Vert b\Vert ^2_{H^s}) + c_0 ( \Vert \nabla u\Vert ^2_{H^s} + \Vert \omega \Vert ^2_{H^s} + \Vert \nabla b\Vert ^2_{H^s}) \\ \le&C \Vert \nabla u\Vert ^2_{H^s} \Vert u\Vert _{H^s} + C \Vert \nabla b\Vert ^2_{H^s} \Vert u\Vert _{H^s} + \Vert \nabla u\Vert _{H^s} \Vert \omega \Vert ^2_{H^s} \\&+ C \Vert \nabla u\Vert _{H^s} \Vert \nabla b\Vert _{H^s} \Vert b\Vert _{H^s}. \end{aligned}$$

Then the Young inequality leads to

$$\begin{aligned} \frac{d}{dt}&(\Vert u\Vert ^2_{H^s} + \Vert \omega \Vert ^2_{H^s} + \Vert b\Vert ^2_{H^s}) + \frac{c_0}{2} ( \Vert \nabla u\Vert ^2_{H^s} + \Vert \omega \Vert ^2_{H^s} + \Vert \nabla b\Vert ^2_{H^s}) \nonumber \\ \le&C(\Vert u\Vert ^2_{H^s} + \Vert \omega \Vert ^2_{H^s} + \Vert b\Vert ^2_{H^s})^{\frac{1}{2}} (\Vert \omega \Vert ^2_{H^s} + \Vert \nabla u\Vert ^2_{H^s} + \Vert \nabla b\Vert ^2_{H^s}). \end{aligned}$$
(24)

This inequality indicates that, if the initial data \((u_0, \omega _0, b_0)\) satisfy, for \(0< \epsilon < \epsilon _0 = \left(\frac{c_0}{2C} \right)^2\),

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

then the corresponding solution remains for all time. Namely,

$$\begin{aligned} \Vert u(t)\Vert ^2_{H^s} + \Vert \omega (t)\Vert ^2_{H^s} + \Vert b(t)\Vert ^2_{H^s} < \epsilon . \end{aligned}$$
(25)

In fact, if suppose (25) is not true and \(T_0\) is the first time such that (25) 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 (25) holds for any \(0 \le t < T_0\). We can deduce from (24) that for any \(0 \le t \le T_0\),

$$\begin{aligned} \frac{d}{dt}&(\Vert u\Vert ^2_{H^s} + \Vert \omega \Vert ^2_{H^s} + \Vert b\Vert ^2_{H^s}) \nonumber \\&+ \left(\frac{c_0}{2} - C \sqrt{\epsilon }) ( \Vert \nabla u\Vert ^2_{H^s} + \Vert \omega \Vert ^2_{H^s} + \Vert \nabla b\Vert ^2_{H^s} \right) \le 0. \end{aligned}$$
(26)

Therefore,

$$\begin{aligned} \Vert u(t)\Vert ^2_{H^s} + \Vert \omega (t)\Vert ^2_{H^s} + \Vert b(t)\Vert ^2_{H^s} \le \Vert u_0\Vert ^2_{H^s} + \Vert \omega _0\Vert ^2_{H^s} + \Vert b_0\Vert ^2_{H^s} < \epsilon . \end{aligned}$$
(27)

This is a contradiction. Thus, we get the uniform bound of (25). In addition,

$$\begin{aligned} \int ^t_0 (\Vert \nabla u(\tau )\Vert _{H^s({\mathbb {R}}^2)}^2 + \Vert \omega (\tau )\Vert _{H^s({\mathbb {R}}^2)}^2 + \Vert \nabla b(\tau )\Vert _{H^s({\mathbb {R}}^2)}^2)d\tau \le C \epsilon . \end{aligned}$$
(28)

Therefore, the proof of (I) of Theorem 1 is completed.

Next, we start to prove (II) of Theorem 1.

Proposition 4

Let \((u, \omega , b)\) be the global solutions of the system (3) with the initial data \((u_0, \omega _0, b_0) \in (L^1({\mathbb {R}}^2) \cap L^2({\mathbb {R}}^2) )^3\). Then \((u, \omega , b)\) satisfies the following inequality,

$$\begin{aligned}&\vert \hat{u}(\xi , t) \vert + \vert \hat{\omega }(\xi , t) \vert + \vert \hat{b}_1(\xi , t) \vert + \vert \hat{b}_2(\xi , t) \vert \nonumber \\&\le C+ C\vert \xi \vert \int ^{t}_0 \left( \Vert u(\tau )\Vert _{L^2}^2 + \Vert \omega (\tau )\Vert _{L^2}^2 + \Vert b(\tau )\Vert _{L^2}^2 \right) d\tau . \end{aligned}$$
(29)

Proof of Proposition 4

Applying the Fourier transform to system (3), we obtain:

$$\begin{aligned} \left\{ \begin{array}{lcl} \partial _t \hat{u} + (\mu + \chi ) \vert \xi \vert ^2 \hat{u} = -{\mathcal {F}} (\nabla \pi ) + {\mathcal {F}}(b\cdot \nabla b) + 2 \chi i \xi \times \hat{\omega } -{\mathcal {F}}(u \cdot \nabla u),\\ \partial _t \hat{\omega } + 4 \chi \hat{\omega } = 2 \chi i \xi \times \hat{u} - {\mathcal {F}}(u \cdot \nabla \omega ),\\ \partial _t \hat{b}_1 + \nu \vert \xi _2 \vert ^2 \hat{b}_1 = {\mathcal {F}} [b\cdot \nabla u_1 - u \cdot \nabla b_1],\\ \partial _t \hat{b}_2 + \nu \vert \xi _1\vert ^2 \hat{b}_2 = {\mathcal {F}} [b\cdot \nabla u_2 - u \cdot \nabla b_2]. \end{array} \right. \end{aligned}$$
(30)

Multiplying the (30)\(_1\), (30)\(_2\), (30)\(_3\) and (30)\(_4\) by \(\bar{\hat{u}}\), \(\bar{\hat{\omega }}\), \(\bar{\hat{b}}_1\) and \(\bar{\hat{b}}_2\) respectively, and summing up, we have, noting that \(\vert \hat{u}\vert ^2 = \hat{u} \bar{\hat{u}}\)

$$\begin{aligned} \frac{1}{2} \frac{d}{dt}&(\vert \hat{u} \vert ^2 + \vert \hat{\omega } \vert ^2 + \vert \hat{b}_1\vert ^2 + \vert \hat{b}_2 \vert ^2) + (\mu + \chi ) \vert \xi \vert ^2 \vert \hat{u} \vert ^2 + \nu (\vert \xi _2 \vert ^2 \vert \hat{b}_1\vert ^2 + \vert \xi _1\vert ^2 \vert \hat{b}_2 \vert ^2) + 4 \chi \vert \hat{\omega } \vert ^2 \nonumber \\ & =-{\mathcal {F}} (\nabla \pi ) \bar{\hat{u}} + {\mathcal {F}}(b\cdot \nabla b)\bar{\hat{u}} -{\mathcal {F}}(u \cdot \nabla u)\bar{\hat{u}}- {\mathcal {F}}(u \cdot \nabla \omega )\bar{\hat{\omega }} + {\mathcal {F}} (b\cdot \nabla u_1) \bar{\hat{b}}_1 \nonumber \\&\quad - {\mathcal {F}}(u \cdot \nabla b_1)\bar{\hat{b}}_1 + {\mathcal {F}} (b\cdot \nabla u_2) \bar{\hat{b}}_2 - {\mathcal {F}}(u \cdot \nabla b_2)\bar{\hat{b}}_2 + 2\chi i \xi \times \hat{\omega } \bar{\hat{u}} + 2 \chi i \xi \times \hat{u} \bar{\hat{\omega }} \nonumber \\ &= K_1 + K_2 + \cdot \cdot \cdot + K_{10}. \end{aligned}$$
(31)

For \(K_1\), taking divergence to the first equation of (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 \(\vert \hat{u}\vert = \vert \bar{\hat{u}} \vert\)

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

For \(K_2\),

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

Similarly, we obtain

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

Inserting \(K_1\) - \(K_{10}\) into (31), we derive that

$$\begin{aligned} \frac{d}{dt}&(\vert \hat{u} \vert ^2 + \vert \hat{\omega }\vert ^2 + \vert \hat{b}_1\vert ^2 + \vert \hat{b}_2\vert ^2) + \mu \vert \xi \vert ^2 \vert \hat{u} \vert ^2 \\&+ 2 \nu (\vert \xi _2 \vert ^2 \vert \hat{b}_1 \vert ^2 + \vert \xi _1 \vert ^2 \vert \hat{b}_2 \vert ^2) + \frac{8 \chi \mu }{\mu + 2\chi } \vert \hat{\omega } \vert ^2\\ \le&C \vert \xi \vert (\Vert u\Vert ^2_{L^2} + \Vert b\Vert ^2_{L^2} + \Vert \omega \Vert ^2_{L^2})(\vert \hat{u} \vert +\vert \hat{\omega } \vert +\vert \hat{b}_1\vert + \vert \hat{b}_2 \vert ), \end{aligned}$$

which immediately yields

$$\begin{aligned} \partial _t \sqrt{\vert \hat{u} \vert ^2 + \vert \hat{\omega } \vert ^2 + \vert \hat{b}_1\vert ^2 + \vert \hat{b}_2 \vert ^2} \le C \vert \xi \vert (\Vert u\Vert ^2_{L^2} + \Vert b\Vert ^2_{L^2} + \Vert \omega \Vert ^2_{L^2}). \end{aligned}$$
(32)

Integrating (32) in [0, t], we obtain

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

Thus the proof of Proposition 4 is completed. \(\square\)

Next, we obtain the result of Theorem 1 by using Proposition 4 and the generalized Fourier splitting method.

Let

$$\begin{aligned} B(t) = \left\{ \xi \in {\mathbb {R}}^2: \vert \xi \vert ^2 \le \frac{h'(t)}{c_1 h(t)}\right\} , \qquad B^{c}(t) = {\mathbb {R}}^2 \backslash B(t), \end{aligned}$$

where \(h(t) \in C^{\infty } [0, + \infty )\) is a positive function with respect to t and satisfies

$$\begin{aligned} h(0) = 1, \ \ \ h'(t)> 0, \ \ \ and \ \ \ \frac{h'(t)}{c_1 h(t)} \le 1, \ \ \forall \ t> t_0 > 0, \end{aligned}$$
(33)

where \(c_1 = \min \{ \mu , \nu , \frac{4 \chi \mu }{u + 2 \chi }\}\).

Multiplying both side of (12) by h(t), we have

$$\begin{aligned} \frac{d}{dt}&\left[ h(t) (\Vert u(t)\Vert _{L^2}^2 + \Vert \omega (t)\Vert _{L^2}^2 + \Vert b(t)\Vert _{L^2}^2) \right] \nonumber \\&+ c_1 h(t) ( \Vert \nabla u(t)\Vert _{L^2}^2 + \Vert \omega (t)\Vert _{L^2}^2 + \Vert \nabla b (t)\Vert _{L^2}^2) \nonumber \\ \le&h'(t) (\Vert u(t)\Vert _{L^2}^2 + \Vert \omega (t)\Vert _{L^2}^2 + \Vert b(t)\Vert _{L^2}^2). \end{aligned}$$
(34)

By using the Plancherel Theorem for (34), we get

$$\begin{aligned} \frac{d}{dt}&\left[ h(t)\left( \Vert \hat{u}(t)\Vert ^2_{L^2} + \Vert \hat{b}(t)\Vert ^2_{L^2} + \Vert \hat{\omega }(t)\Vert ^2_{L^2} \right) \right] \nonumber \\&+ c_1 h(t) \int _{{\mathbb {R}}^2} \left( \vert \xi \vert ^2(\vert \hat{u}(\xi ) \vert ^2 + \vert \hat{b}(\xi ) \vert ^2) + \vert \hat{\omega }(\xi ) \vert ^2 \right) d\xi \nonumber \\ \le&h'(t) \int _{{\mathbb {R}}^2} \left( \vert \hat{u}(\xi )\vert ^2 +\vert \hat{\omega }(\xi ) \vert ^2 + \vert \hat{b}(\xi ) \vert ^2 \right) d\xi . \end{aligned}$$
(35)

Applying (33), we can obtain

$$\begin{aligned} c_1&h(t) \int _{{\mathbb {R}}^2} \left( \vert \xi \vert ^2\vert \hat{u}(\xi ) \vert ^2 +\vert \hat{\omega }(\xi ) \vert ^2 + \vert \xi \vert ^2\vert \hat{b}(\xi ) \vert ^2 \right) d\xi \nonumber \\&+ h'(t) \int _{B(t)} \left( \vert \hat{u}(\xi ) \vert ^2 + \vert \hat{\omega }(\xi ) \vert ^2 + \vert \hat{b}(\xi ) \vert ^2 \right) d\xi \nonumber \\ \ge&c_1 h(t) \int _{B^c(t)} \left( \vert \xi \vert ^2 \vert \hat{u}(\xi ) \vert ^2 + \vert \hat{\omega }(\xi ) \vert ^2 + \vert \xi \vert ^2 \vert \hat{b}(\xi ) \vert ^2 \right) d\xi \nonumber \\&+ h'(t) \int _{B(t)} \left( \vert \hat{u}(\xi ) \vert ^2 + \vert \hat{\omega }(\xi ) \vert ^2 + \vert \hat{b}(\xi ) \vert ^2 \right) d\xi \nonumber \\ \ge&c_1 h(t) \left( \frac{h'(t)}{c_1 h(t)}\right) \int _{B^c(t)} \left( \vert \hat{u} (\xi ) \vert ^2 + \vert \hat{\omega }(\xi )\vert ^2 + \vert \hat{b}(\xi ) \vert ^2 \right) d\xi \nonumber \\ &+ h'(t) \int _{B(t)} \left( \vert \hat{u}(\xi ) \vert ^2 + \vert \hat{\omega }(\xi ) \vert ^2 + \vert \hat{b}(\xi ) \vert ^2 \right) d\xi \nonumber \\ =& h'(t) \int _{{\mathbb {R}}^2} \left( \vert \hat{u}(\xi ) \vert ^2 + \vert \hat{\omega }(\xi ) \vert ^2 + \vert \hat{b}(\xi ) \vert ^2 \right) d\xi . \end{aligned}$$
(36)

Combining the result of (35) and (36), we get

$$\begin{aligned} \frac{d}{dt}&\left[ h(t) \left( \Vert \hat{u}(t)\Vert ^2_{L^2} + \Vert \hat{b}(t)\Vert ^2_{L^2} + \Vert \hat{\omega }(t)\Vert ^2_{L^2} \right) \right] \nonumber \\&\le h'(t)\int _{B(t)} \left( \vert \hat{u}(\xi ) \vert ^2 + \vert \hat{\omega }(\xi ) \vert ^2 + \vert \hat{b}(\xi ) \vert ^2 \right) d\xi . \end{aligned}$$
(37)

Employing (29), we have

$$\begin{aligned}&\int _{B(t)} \left( \vert \hat{u}(\xi ) \vert ^2 + \vert \hat{\omega }(\xi ) \vert ^2 + \vert \hat{b}(\xi ) \vert ^2 \right) d\xi \nonumber \\&= C \int _{B(t)} \left\{ \vert \xi \vert ^2 \left( \int ^{t}_0 (\Vert u(\tau )\Vert _{L^2}^2 + \Vert \omega (\tau )\Vert _{L^2}^2 + \Vert b(\tau )\Vert _{L^2}^2) d\tau \right) ^2 + 1 \right\} d \xi \nonumber \\&\le \frac{C h'(t)}{h(t)} + \frac{C(h'(t))^2 }{h^2(t)} \left[ \int ^t_0 \left( \Vert u(\tau )\Vert _{L^2}^2 + \Vert \omega (\tau )\Vert _{L^2}^2 + \Vert b(\tau )\Vert _{L^2}^2 \right) d\tau \right] ^2. \end{aligned}$$
(38)

Substituting (38) to (37), we have

$$\begin{aligned} \frac{d}{dt}&\left[ h(t) \left( \Vert \hat{u}(t)\Vert ^2_{L^2} + \Vert \hat{b}(t)\Vert ^2_{L^2} + \Vert \hat{\omega }(t)\Vert ^2_{L^2} \right) \right] \\&\le \frac{C[ h'(t)]^2}{h(t)} + \frac{C[h'(t)]^3 }{h^2(t)} \left[ \int ^t_0 \left( \Vert u(\tau )\Vert _{L^2}^2 + \Vert \omega (\tau )\Vert _{L^2}^2 + \Vert b(\tau )\Vert _{L^2}^2 \right) d\tau \right] ^2. \nonumber \end{aligned}$$
(39)

Next, taking \(h(t) = [\ln (e + t)]^3\), then we have

$$\begin{aligned} \int ^t_0 \frac{[h'(\tau )]^2}{h(\tau )} d\tau&= \int ^t_0 \frac{3^2 \ln ^4(e+\tau )}{(e + \tau )^2 \ln ^3(e + \tau )} d \tau = \int ^t_0 \frac{9 \ln (e + \tau )}{(e + \tau )^2} d \tau \nonumber \\&\le C \int ^t_0 \frac{1}{e + \tau } d \tau \le C \ln (e + t), \end{aligned}$$
(40)

and

$$\begin{aligned} \int ^t_0&\frac{[h'(\tau )]^3 }{h^2(\tau )} \left[ \int ^{\tau }_0 \left( \Vert u(s)\Vert _{L^2}^2 + \Vert \omega (s)\Vert _{L^2}^2 + \Vert b(s)\Vert _{L^2}^2 \right) ds \right] ^2 d\tau \nonumber \\&\le C \int ^t_0 \frac{\tau ^2}{(e + \tau )^3} [\Vert u_0\Vert _{L^2}^2 + \Vert \omega _0\Vert _{L^2}^2 + \Vert b_0\Vert _{L^2}^2]^2 d \tau \nonumber \\&\le C \int ^t_0 \frac{1}{(e + \tau )} d \tau \le C \ln (e + t). \end{aligned}$$
(41)

Combining (39) - (41), we have

$$\begin{aligned}&\Vert u(t)\Vert ^2_{L^2} + \Vert b(t)\Vert ^2_{L^2} + \Vert \omega (t)\Vert ^2_{L^2} \nonumber \\&\le C[\ln (e + t)]^{-3} + C [ \ln (e + t) ]^{-2} \nonumber \\&\le C [ \ln (e + t) ]^{-2}. \end{aligned}$$
(42)

Now, taking \(h(t) = (1+t)^2\) and inserting it into (39), together with (42) and Hölder’s inequality, we have

$$\begin{aligned} (1 & + t)^2\left( \Vert \hat{u}(t)\Vert ^2_{L^2} + \Vert \hat{b}(t)\Vert ^2_{L^2} + \Vert \hat{\omega }(t)\Vert ^2_{L^2} \right) \nonumber \\ \le&(\Vert u_0\Vert _{L^2}^2 + \Vert \omega _0\Vert _{L^2}^2 + \Vert b_0\Vert _{L^2}^2) + {C \int ^t_0 \frac{[ h'(\tau )]^2}{h(\tau )} d \tau } \nonumber \\&+ \int _0^t \frac{[h'(\tau )]^3 }{h^2(\tau )} \left[ \int ^{\tau }_0 \left( \Vert u(s)\Vert _{L^2}^2 + \Vert \omega (s)\Vert _{L^2}^2 + \Vert b(s)\Vert _{L^2}^2 \right) ds \right] ^2 d\tau , \end{aligned}$$
(43)

where

$$\begin{aligned}&C \int ^t_0 \frac{[ h'(\tau )]^2}{h(\tau )} d \tau \le C \int ^t_0 \frac{[2 (1 + \tau )]^2}{(1 + \tau )^2} d \tau \le C (t+ 1), \\&\quad \int _0^t \frac{[h'(\tau )]^3 }{h^2(\tau )} \left[ \int ^{\tau }_0 \left( \Vert u(s)\Vert _{L^2}^2 + \Vert \omega (s)\Vert _{L^2}^2 + \Vert b(s)\Vert _{L^2}^2 \right) ds \right] ^2 d\tau \\&\quad \le C \int ^t_0 \frac{\tau }{(1 + \tau )} \int ^{\tau }_0 \left( \Vert u(s)\Vert _{L^2}^2 + \Vert \omega (s)\Vert _{L^2}^2 + \Vert b(s)\Vert _{L^2}^2 \right) \ln ^{-2}(e+ s) ds d\tau \\&\quad \le C (1+t) \int ^{\tau }_0 \left( \Vert u(s)\Vert _{L^2}^2 + \Vert \omega (s)\Vert _{L^2}^2 + \Vert b(s)\Vert _{L^2}^2 \right) \ln ^{-2}(e+ s) ds. \end{aligned}$$

From (43), we have

$$\begin{aligned}&(1 + t) \left( \Vert \hat{u}(t)\Vert ^2_{L^2} + \Vert \hat{b}(t)\Vert ^2_{L^2} + \Vert \hat{\omega }(t)\Vert ^2_{L^2} \right) \nonumber \\&\le C + C \int ^t_0 (1+s)^{-1} (1+ s) \left( \Vert u(s)\Vert _{L^2}^2 + \Vert \omega (s)\Vert _{L^2}^2 + \Vert b(s)\Vert _{L^2}^2 \right) \ln ^{-2}(e+ s) ds. \end{aligned}$$
(44)

Taking \({\mathcal {N}}(t) = (1 + t) \left( \Vert u(t)\Vert _{L^2}^2 + \Vert \omega (t)\Vert _{L^2}^2 + \Vert b(t)\Vert _{L^2}^2 \right)\), then we have

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

Applying Gronwall’s inequality, we obtain

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

which implies the following decay

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

Therefore, the proof of (6) is completed.

Next, we will prove (7). The vorticity \(\Omega = \nabla \times u\), \(j = \nabla \times b\) satisfies

$$\begin{aligned}{} & {} \partial _{t} \Omega + u\cdot \nabla \Omega - (\mu + \chi ) \Delta \Omega = b \cdot \nabla j - 2\chi \Delta \omega , \end{aligned}$$
(45)
$$\begin{aligned}{} & {} \partial _{t} j + u \cdot \nabla j - \nu \partial _{111} b_{2} + \nu \partial _{222} b_{1} = b \cdot \nabla \Omega + T(\nabla u , \nabla b), \end{aligned}$$
(46)

where

$$\begin{aligned} T(\nabla u, \nabla b) = 2 \partial _{1} b_1 (\partial _1 u_2 + \partial _2 u_1) - 2 \partial _{1} u_1 (\partial _1 b_2 + \partial _2 b_1). \end{aligned}$$

Due to the lack of angular viscosity for the system (3), it is crucial to deal with \(- 2 \chi \omega\) in (45) by introducing a new function \(Z = \Omega - \frac{2 \chi }{\mu + \chi } \omega\) in [11]. Subtracting \(\frac{2\chi }{\mu + \chi } \times\) (3)\(_2\) from (45), we have

$$\begin{aligned} \partial _t Z - (\mu + \chi ) \Delta Z + (u \cdot \nabla ) Z + \frac{4 \chi ^2}{\mu + \chi } Z = \left( \frac{8 \chi ^2}{\mu + \chi } - \frac{8 \chi ^3}{(\mu + \chi )^2}\right) \omega + b \cdot \nabla j. \end{aligned}$$
(47)

Taking the \(L^2\)-inner products of (47), (3)\(_2\) and (46) with Z, \(\omega\) and j, respectively, we obtain

$$\begin{aligned} \frac{1}{2}&\frac{d}{dt} \Vert Z(t)\Vert _{L^2}^2 + (\mu + \chi ) \Vert \nabla Z\Vert _{L^2}^2 + \frac{4 \chi ^2}{\mu + \chi } \Vert Z\Vert _{L^2}^2 \nonumber \\&\le \left( \frac{8 \chi ^2}{\mu + \chi } - \frac{8 \chi ^2}{(\mu + \chi )^2}\right) \Vert Z\Vert _{L^2}\Vert \omega \Vert _{L^2} + \int (b \cdot \nabla j)(\Omega - \frac{2 \chi }{\mu + \chi } \omega ) dx, \end{aligned}$$
(48)
$$\begin{aligned}&\frac{1}{2} \frac{d}{dt} \Vert \omega (t)\Vert ^2_{L^2} + 4 \chi \Vert \omega \Vert ^2_{L^2} \le 2 \chi \Vert Z\Vert _{L^2}\Vert \omega \Vert _{L^2} + \frac{4 \chi ^2}{\mu + \chi } \Vert \omega \Vert _{L^2}^2, \end{aligned}$$
(49)
$$\begin{aligned}&\frac{1}{2} \frac{d}{dt} \Vert j(t)\Vert _{L^2}^2 + I = \int (b \cdot \nabla \Omega j + Tj ) dx,\end{aligned}$$
(50)

where

$$\begin{aligned} I &= \nu \int (- \partial _{111} b_2 + \partial _{222} b_1) j dx\\ &= \nu \int (- \partial _{111} b_2 + \partial _{222} b_1) (\partial _1 b_2 - \partial _2 b_1) dx \\ & =\nu \int (\partial _{11} b_2)^2 + (\partial _{11} b_1)^2 + (\partial _{22} b_1)^2 + (\partial _{22} b_2)^2 dx \equiv H (b, t), \end{aligned}$$

due to the divergence free condition \(\nabla \cdot b = \partial _1 b_1 + \partial _2 b_2\). By Hölder’s inequality

$$\begin{aligned} \int Tj dx&\le C \Vert \nabla u\Vert _{L^2} \Vert \nabla b\Vert _{L^4} \Vert j\Vert _{L^4} \\&\le C \Vert \Omega \Vert _{L^2} \Vert j\Vert _{L^2} \Vert \nabla j\Vert _{L^2} \\&\le C \Vert \Omega \Vert _{L^2}^2 \Vert j\Vert ^2_{L^2} + \frac{\nu }{8} \Vert \nabla j \Vert _{L^2}^2, \end{aligned}$$

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,

$$\begin{aligned} \nu \Vert \nabla j\Vert _{L^2}^2&= \nu \Vert (\partial _1 j , \partial _2 j )\Vert _{L^2}^2 = \nu \Vert ((\partial _{11}b_2 - \partial _{12}b_1), (\partial _{12}b_2 - \partial _{22}b_1))\Vert _{L^2}^2\\&= \nu \Vert ((\partial _{11}b_2 + \partial _{22}b_2), - (\partial _{11}b_1 + \partial _{22}b_1))\Vert _{L^2}^2 \le 4 H(b,t). \end{aligned}$$

Then it follows from the above bounds and (50) that

$$\begin{aligned} \frac{1}{2} \frac{d}{dt} \Vert j\Vert _{L^2}^2 + \frac{1}{2} H(b,t) \le \int b \cdot \nabla \Omega j dx + C \Vert \Omega \Vert _{L^2}^2 \Vert j\Vert _{L^2}^2. \end{aligned}$$
(51)

Combining (48), (49) and (51), we have

$$\begin{aligned} \frac{1}{2} \frac{d}{dt}&(\Vert Z(t)\Vert _{L^2}^2 + \Vert \nabla Z \Vert _{L^2}^2 + \Vert \nabla j\Vert _{L^2}^2) + (\mu + \chi ) \Vert \nabla Z\Vert _{L^2}^2 + \frac{4 \chi ^2}{\mu + \chi } \Vert Z\Vert _{L^2}^2 \nonumber \\&+ 4 \chi \Vert \omega \Vert _{L^2}^2 +\frac{\nu }{8} \Vert \nabla j\Vert _{L^2}^2 \nonumber \\ \le&\left( \frac{8 \chi ^2}{\mu + \chi } - \frac{8 \chi ^3}{(\mu + \chi )^2}\right) \Vert Z\Vert _{L^2}\Vert \omega \Vert _{L^2} + \int (b \cdot \nabla ) j (\Omega - \frac{2 \chi }{\mu + \chi } \omega ) dx \nonumber \\&+ \int b \cdot \nabla \Omega j dx + C \Vert \Omega \Vert _{L^2}^2 \Vert j\Vert _{L^2}^2 + 2 \chi \Vert Z\Vert _{L^2}\Vert \omega \Vert _{L^2} + \frac{4 \chi ^2}{\mu + \chi } \Vert \omega \Vert _{L^2}^2 \nonumber \\ \triangleq&I_1 + I_2 + I_3 + I_4 + I_5 + I_6. \end{aligned}$$
(52)

Next, we consider \(I_1\) - \(I_6\), respectively. Applying the Young inequality

$$\begin{aligned} I_1&= \left( \frac{8 \chi ^2}{\mu +\chi } - \frac{8 \chi ^3}{(\mu + \chi )^2}\right) \Vert Z\Vert _{L^2}\Vert \omega \Vert _{L^2} \nonumber \\&= \frac{8 \chi ^2 \mu }{(\mu + \chi )^2} \Vert Z\Vert _{L^2} \Vert \omega \Vert _{L^2} \nonumber \\&\le \frac{2 \chi ^2 \mu }{(\mu + \chi )^2} \Vert Z\Vert _{L^2}^2 + \frac{8 \chi \mu }{(\mu + \chi )^2} \Vert \omega \Vert _{L^2}^2. \end{aligned}$$

By using Hölder’s inequality, the Gagliardo-Nirenberg inequality, and the Young inequality,

$$\begin{aligned} I_2 + I_3&= \int (b \cdot \nabla ) j (\Omega - \frac{2 \chi }{\mu + \chi }) \omega dx + \int b \cdot \nabla \Omega j dx \nonumber \\&= - \int b \cdot \nabla j (\frac{2 \chi }{\mu + \chi }) \omega dx \\&\le C \Vert b\Vert _{L^{\infty }} \Vert \nabla j\Vert _{L^2} \Vert \omega \Vert _{L^2} \\&\le C \Vert b\Vert _{L^2}^{\frac{1}{2}} \Vert \nabla j\Vert _{L^2}^{\frac{3}{2}} \Vert \omega \Vert _{L^2} \\&\le C \Vert b_0\Vert _{L^2}^{\frac{1}{2}} \Vert \nabla j\Vert _{L^2}^{\frac{3}{2}} \Vert \omega \Vert _{L^2} \\&\le \frac{\nu }{16} \Vert \nabla j\Vert _{L^2}^{2} + C \Vert \omega \Vert _{L^2}^4, \end{aligned}$$

and

$$\begin{aligned} I_5 =&2 \chi \Vert Z\Vert _{L^2} \Vert \omega \Vert _{L^2} \le \chi \Vert Z\Vert _{L^2}^2 + \chi \Vert \omega \Vert _{L^2}^2. \end{aligned}$$

Inserting \(I_1\) - \(I_6\) into (52), we have

$$\begin{aligned} \frac{1}{2}&\frac{d}{dt} \left( \Vert Z(t)\Vert _{L^2}^2 + \Vert \omega (t)\Vert _{L^2}^2 + \Vert j(t)\Vert _{L^2}^2 \right) + (\mu + \chi )\Vert \nabla Z\Vert ^2_{L^2} \nonumber \\&+ \left( \frac{4 \chi ^2}{\mu + \chi } - \chi - \frac{2 \chi ^2 \mu }{(\mu + \chi )^2} \right) \Vert Z\Vert _{L^2}^2 + \frac{\nu }{8} \Vert \nabla j\Vert _{L^2}^2 + 4 \chi \Vert \omega \Vert _{L^2}^2 \nonumber \\ \le&C \Vert \omega \Vert _{L^2}^4 + C \Vert \Omega \Vert _{L^2}^2\Vert j\Vert _{L^2}^2 + \left( \chi + \frac{4 \chi ^2}{(\mu + \chi )} - 4 \chi \right) \Vert \omega \Vert _{L^2}^2 + C \Vert \omega \Vert _{L^2}^2 \nonumber \\ \le&C \Vert \omega \Vert _{L^2}^4 + C \Vert \Omega \Vert _{L^2}^2\Vert j\Vert _{L^2}^2 + C \Vert \omega \Vert _{L^2}^2 \nonumber \\ \le&C (\Vert Z\Vert _{L^2}^2 + \Vert \Omega \Vert _{L^2}^2 + \Vert j\Vert _{L^2}^2) (\Vert \Omega \Vert _{L^2}^2 + \Vert \omega \Vert _{L^2}^2) + C \Vert \omega \Vert _{L^2}^2. \end{aligned}$$
(53)

Due to \(\mu < \sqrt{3} \chi\), then

$$\begin{aligned}&\frac{4 \chi ^2}{(\mu + \chi )} - \chi - \frac{2 \chi ^2 \mu }{(\mu + \chi )^2} \\&= \frac{3 \chi ^2 \mu + 3 \chi ^3 - \mu ^2 \chi - \mu \chi ^2 -2 \chi ^2 \mu }{(\mu + \chi )^2} = \frac{3 \chi ^3 - \mu ^2 \chi }{(\mu + \chi )^2} >0. \end{aligned}$$

Applying Gronwall’s inequality, we have

$$\begin{aligned} \Vert Z&(t)\Vert _{L^2}^2 + \Vert \omega (t)\Vert _{L^2}^2 + \Vert j(t)\Vert _{L^2}^2\nonumber \\ \le&\exp \left\{ C \int ^t_r (\Vert \Omega (s)\Vert _{L^2}^2 + \Vert \omega (s)\Vert _{L^2}^2)ds \right\} \nonumber \\&\times \left( \Vert Z(r)\Vert _{L^2}^2 + \Vert \omega (r)\Vert _{L^2}^2 + \Vert j(r)\Vert _{L^2}^2 + \int _r^t \Vert \omega (s)\Vert _{L^2}^2 ds\right) \nonumber \\ \le&C (\Vert Z(r)\Vert _{L^2}^2 + \Vert \omega (r)\Vert _{L^2}^2 + \Vert j(r)\Vert _{L^2}^2 + \Vert u(r)\Vert _{L^2}^2 + \Vert b(r)\Vert _{L^2}^2 ), \end{aligned}$$
(54)

where we also used (13).

Multiplying (12) by \((1+ t)^n\), by (6), we have

$$\begin{aligned} \frac{d}{dt}&\left[ (1 + t)^n ( \Vert u(t)\Vert _{L^2}^2 + \Vert \omega (r)\Vert _{L^2}^2 + \Vert b(r)\Vert _{L^2}^2)\right] \nonumber \\&+ c_1 (1 + t)^n ( \Vert \nabla u(t)\Vert _{L^2}^2 + \Vert \omega (t)\Vert _{L^2}^2 + \Vert \nabla b(t)\Vert _{L^2}^2) \nonumber \\ \le&n (1 + t)^{n-1} ( \Vert u(t)\Vert _{L^2}^2 + \Vert \omega (r)\Vert _{L^2}^2 + \Vert b(r)\Vert _{L^2}^2) \nonumber \\ \le&n (1+ t)^{n-2}, \end{aligned}$$
(55)

where \(c_1 = \min \{\mu , \frac{4 \chi \mu }{\mu + 2 \chi }, \nu \}\). Integrating (55) in time, we have

$$\begin{aligned} \int ^T_0&(1 + r)^n ( \Vert \nabla u(r)\Vert _{L^2}^2 + \Vert \omega (r)\Vert _{L^2}^2 + \Vert \nabla b(r)\Vert _{L^2}^2) dr \nonumber \\&\le C (1 +t)^{n-1}, \ \ \quad \text{ for } \ \text{ large } \ \ n\ge 6. \end{aligned}$$
(56)

Multiplying (53) by \((1+ r)^n\) and integrating with respect to r over \((\frac{t}{2}, t)\), we have

$$\begin{aligned} \frac{t}{2}&\left( 1 + \frac{t}{2}\right) ^n \left( \Vert Z(t)\Vert _{L^2}^2 + \Vert \omega (t)\Vert _{L^2}^2 + \Vert j(t)\Vert _{L^2}^2 \right) \nonumber \\&\le \int ^{t}_{\frac{t}{2}} (1 + r)^n (\Vert Z(r)\Vert _{L^2}^2 + \Vert \omega (r)\Vert _{L^2}^2 + \Vert j(r)\Vert _{L^2}^2 + \Vert u(r)\Vert _{L^2}^2 + \Vert b(r)\Vert _{L^2}^2 ) dr \nonumber \\&\le C (1+ t)^{n-1} + \int _{\frac{t}{2}}^t (1 + r)^{n-1} dr \nonumber \\&\le C (1 + t)^n, \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{ for } \ \text{ some } \ \text{ given } \ \text{ large } \ n \ge 6. \end{aligned}$$
(57)

Nothing that

$$\begin{aligned}&\left( \frac{1}{2} + \frac{t}{2}\right) ^{n+1} \left( \Vert Z(t)\Vert _{L^2}^2 + \Vert \omega (t)\Vert _{L^2}^2 + \Vert j(t)\Vert _{L^2}^2\right) \nonumber \\&\le \frac{t}{2} \left( 1 + \frac{t}{2}\right) ^{n} \left( \Vert Z(t)\Vert _{L^2}^2 + \Vert \omega (t)\Vert _{L^2}^2 + \Vert j(t)\Vert _{L^2}^2\right) . \end{aligned}$$
(58)

Combining the results of (57) and (48), for \(t \ge 1\), we get

$$\begin{aligned} \Vert Z(t)\Vert _{L^2}^2 + \Vert \omega (t)\Vert _{L^2}^2 + \Vert j(t)\Vert _{L^2}^2 \le (1 + t)^{-1}, \end{aligned}$$

then we have

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

Therefore, the proof of Theorem 1 is completed.

3 The Proof of Theorem 2

This section is devoted to the proof of Theorem 2. Firstly, we first prove the global stability part (I) of Theorem 2. As we know, the key step is to establish the global \(a \ priori \ B^s_{2, 1}({\mathbb {R}}^2)\) estimates of the solution.

Proof of (I) of Theorem 2

As preparation, in the following proposition, we state that system (8) has unique global solution when the initial data \((u_0, \omega _0, b_0)\) is sufficiently small in \(\dot{B}^0_{2,1}({\mathbb {R}}^2)\).

Proposition 5

Assume \((u_0, \omega _0, b_0)\) satisfies the conditions in (I) of Theorem 2 and (9), then system (8) has a unique global solution \((u, \omega , b)\) satisfying, for any \(t > 0\),

$$\begin{aligned} \Vert&u(t)\Vert _{\dot{B}^0_{2,1}} + \Vert \omega (t)\Vert _{\dot{B}^0_{2,1}} + \Vert b(t)\Vert _{\dot{B}^0_{2,1}} \nonumber \\&+ \Vert \nabla ^2 u(t)\Vert _{L^1_t(\dot{B}^0_{2,1})} +\Vert \nabla ^2 \omega (t)\Vert _{L^1_t(\dot{B}^0_{2,1})} + \Vert \nabla ^2 b(t)\Vert _{L^1_t(\dot{B}^0_{2,1})} \le C \epsilon . \end{aligned}$$
(59)

Proof

Now, we turn to establish the global \(a \ priori \ B^s_{2,1}\) estimates. Applying \(\dot{\Delta }_j\) to (8), we have

$$\begin{aligned}{} & {} \partial _t \dot{\Delta }_j u + u \cdot \nabla \dot{\Delta }_j u - (\mu + \chi ) \Delta \dot{\Delta }_j u = - \dot{\Delta }_j \nabla \pi - [\dot{\Delta }_j, u\cdot \nabla ]u \nonumber \\{} & {} \quad + \dot{\Delta }_j(b \cdot \nabla b) + 2\chi \dot{\Delta }_j \nabla \times \omega , \end{aligned}$$
(60)
$$\begin{aligned}{} & {} \partial _t \dot{\Delta }_j \omega + u \cdot \nabla \dot{\Delta }_j \omega - \kappa \Delta \dot{\Delta }_j \omega + 4 \chi \dot{\Delta }_j \omega = - [\dot{\Delta }_j, u\cdot \nabla ]\omega + 2\chi \dot{\Delta }_j \nabla \times u, \end{aligned}$$
(61)
$$\begin{aligned}{} & {} \partial _t \dot{\Delta }_j b_1 + u \cdot \nabla \dot{\Delta }_j b_1 - \nu \partial _{yy} \dot{\Delta }_j b_1 = - [\dot{\Delta }_j, u\cdot \nabla ]b_1 + \dot{\Delta }_j(b \cdot \nabla u_1), \end{aligned}$$
(62)
$$\begin{aligned}{} & {} \partial _t \dot{\Delta }_j b_2 + u \cdot \nabla \dot{\Delta }_j b_2 - \nu \partial _{xx} \dot{\Delta }_j b_2 = - [\dot{\Delta }_j, u\cdot \nabla ]b_2 + \dot{\Delta }_j(b \cdot \nabla u_2), \end{aligned}$$
(63)

where \([\dot{\Delta }_j, f \cdot \nabla ]g = \dot{\Delta }_j (f \cdot \nabla g) - f \cdot \dot{\Delta }_j (\nabla g)\) is commutator. Dotting (60) - (63) by \(\dot{\Delta }_j u\), \(\dot{\Delta }_j \omega\), \(\dot{\Delta }_j b_1\) and \(\dot{\Delta }_j b_2\) respectively, integrating the resulting equations in \({\mathbb {R}}^2\), and adding them together, we have

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

where we used the facts that

$$\begin{aligned} \int b \cdot \nabla \dot{\Delta }_j u_1 \cdot \dot{\Delta }_j b_1 + \int b \cdot \nabla \dot{\Delta }_j u_2 \cdot \dot{\Delta }_j b_2 = \int b \cdot \nabla \dot{\Delta }_j u \cdot \dot{\Delta }_j b = - \int b \cdot \nabla \dot{\Delta }_j b \cdot \dot{\Delta }_j u, \end{aligned}$$

and

$$\begin{aligned} \int \dot{\Delta }_j \nabla \times u \cdot \dot{\Delta }_j \omega = \int \dot{\Delta }_j \nabla \times \omega \cdot \dot{\Delta }_j u. \end{aligned}$$

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

$$\begin{aligned}{} & {} \Vert \dot{\Delta }_j \nabla b \Vert ^2_{L^2} = \Vert \dot{\Delta }_j \nabla \times b \Vert ^2_{L^2} \le 2 (\Vert \dot{\Delta }_j \partial _x b_2 \Vert ^2_{L^2} + \Vert \dot{\Delta }_j \partial _y b_1 \Vert ^2_{L^2}), \\{} & {} \Vert \dot{\Delta }_j b \Vert ^2_{L^2} = \Vert \dot{\Delta }_j b_1 \Vert ^2_{L^2} + \Vert \dot{\Delta }_j b_2 \Vert ^2_{L^2}. \end{aligned}$$

Then, we can derive from the above inequalities

$$\begin{aligned} \frac{1}{2}&\frac{d}{dt} (\Vert \dot{\Delta }_j u\Vert ^2_{L^2} + \Vert \dot{\Delta }_j \omega \Vert ^2_{L^2} + \Vert \dot{\Delta }_j b\Vert ^2_{L^2}) + (\mu + \chi )\Vert \dot{\Delta }_j \nabla u\Vert ^2_{L^2} + \kappa \Vert \dot{\Delta }_j \nabla \omega \Vert ^2_{L^2} \nonumber \\&+ 4 \chi \Vert \dot{\Delta }_j \omega \Vert ^2_{L^2} + \frac{\nu }{2} \Vert \dot{\Delta }_j \nabla b\Vert ^2_{L^2}\nonumber \\ \le&-\int [\dot{\Delta }_j , u \cdot \nabla ]u \cdot \dot{\Delta }_j u + \int [\dot{\Delta }_j , b \cdot \nabla ]b \cdot \dot{\Delta }_j u - \int [\dot{\Delta }_j , u \cdot \nabla ]\omega \cdot \dot{\Delta }_j \omega \nonumber \\&+ 4 \chi \int \dot{\Delta }_j \nabla \times u \cdot \dot{\Delta }_j \omega - \int [\dot{\Delta }_j , u \cdot \nabla ]b \cdot \dot{\Delta }_j b + \int [\dot{\Delta }_j , b \cdot \nabla ]u \cdot \dot{\Delta }_j b. \end{aligned}$$
(64)

Due to

$$\begin{aligned} \left| 4\chi \int \dot{\Delta }_j \nabla \times u \cdot \dot{\Delta }_j \omega \right| \le \left( \frac{\mu }{2} + \chi \right) \Vert \nabla u\Vert _{L^2}^2 + \frac{8 \chi ^2}{ \mu + 2 \chi } \Vert \omega \Vert ^2_{L^2}, \end{aligned}$$

then, we have

$$\begin{aligned} \frac{1}{2}&\frac{d}{dt} \left(\Vert \dot{\Delta }_j u\Vert ^2_{L^2} + \Vert \dot{\Delta }_j \omega \Vert ^2_{L^2} + \Vert \dot{\Delta }_j b\Vert ^2_{L^2}) + \frac{\mu }{2}\Vert \dot{\Delta }_j \nabla u\Vert ^2_{L^2} + \kappa \Vert \dot{\Delta }_j \nabla \omega \Vert ^2_{L^2} \right. \nonumber \\ &+\left. \frac{\nu }{2} \Vert \dot{\Delta }_j \nabla b\Vert ^2_{L^2} + (4 \chi - \frac{4\chi ^2}{\frac{\mu }{2} + \chi } \right)\Vert \dot{\Delta }_j \omega \Vert ^2_{L^2}\nonumber \\ \le&-\int [\dot{\Delta }_j , u \cdot \nabla ]u \cdot \dot{\Delta }_j u + \int [\dot{\Delta }_j , b \cdot \nabla ]b \cdot \dot{\Delta }_j u - \int [\dot{\Delta }_j , u \cdot \nabla ]\omega \cdot \dot{\Delta }_j \omega \nonumber \\&- \int [\dot{\Delta }_j , u \cdot \nabla ]b \cdot \dot{\Delta }_j b + \int [\dot{\Delta }_j , b \cdot \nabla ]u \cdot \dot{\Delta }_j b{.} \end{aligned}$$
(65)

Applying Hölder’s inequality, the Young inequality, Bernstein’s inequality and the divergence free condition \(\nabla \cdot u = \nabla \cdot b = 0\) to (65), we obtain

$$\begin{aligned} \frac{1}{2}&\frac{d}{dt} \left( \Vert \dot{\Delta }_j u\Vert _{L^2}^2 +\Vert \dot{\Delta }_j \omega \Vert _{L^2}^2 +\Vert \dot{\Delta }_j b\Vert _{L^2}^2\right) + \frac{\mu }{2}\tilde{c}_2 2^{2j} \Vert \dot{\Delta }_j u\Vert _{L^2}^2 \nonumber \\&+ \frac{\nu }{2}\tilde{c}_2 2^{2j} \Vert \dot{\Delta }_j b\Vert _{L^2}^2+ \kappa \tilde{c}_2 2^{2j} \Vert \dot{\Delta }_j \omega \Vert _{L^2}^2 + \frac{4\mu \chi }{\mu + 2 \chi } \Vert \dot{\Delta }_j \omega \Vert ^2_{L^2} \nonumber \\ \le&- \int \dot{\Delta }_j (u \cdot \nabla u) \cdot \dot{\Delta }_j u + \int \dot{\Delta }_j (b \cdot \nabla b) \cdot \dot{\Delta }_j u - \int \dot{\Delta }_j (u \cdot \nabla b) \cdot \dot{\Delta }_j b \nonumber \\&+ \int \dot{\Delta }_j (b \cdot \nabla u) \cdot \dot{\Delta }_j b -\int \dot{\Delta }_j (u \cdot \nabla \omega ) \cdot \dot{\Delta }_j \omega \nonumber \\ \le&C 2^j \Vert \dot{\Delta }_j (u \otimes u)\Vert _{L^2}\Vert \dot{\Delta }_j u\Vert _{L^2} + C 2^j \Vert \dot{\Delta }_j (b \otimes b)\Vert _{L^2}\Vert \dot{\Delta }_j u\Vert _{L^2} \nonumber \\&+ C 2^j \Vert \dot{\Delta }_j (u \otimes b)\Vert _{L^2}\Vert \dot{\Delta }_j b\Vert _{L^2}+ C 2^j \Vert \dot{\Delta }_j (u \otimes \omega )\Vert _{L^2}\Vert \dot{\Delta }_j \omega \Vert _{L^2}. \end{aligned}$$
(66)

Then, (66) implies

$$\begin{aligned} \frac{d}{dt}&\sqrt{\Vert \dot{\Delta }_j u\Vert _{L^2}^2 +\Vert \dot{\Delta }_j \omega \Vert _{L^2}^2 +\Vert \dot{\Delta }_j b\Vert _{L^2}^2} + \tilde{c}_2 c_2 2^{2j} \sqrt{\Vert \dot{\Delta }_j u\Vert _{L^2}^2 +\Vert \dot{\Delta }_j \omega \Vert _{L^2}^2 +\Vert \dot{\Delta }_j b\Vert _{L^2}^2} \nonumber \\ \le&C 2^j ( \Vert \dot{\Delta }_j (u \otimes u) \Vert _{L^2} + \Vert \dot{\Delta }_j (b \otimes b) \Vert _{L^2} + \Vert \dot{\Delta }_j (b \otimes u) \Vert _{L^2} + \Vert \dot{\Delta }_j (u \otimes \omega ) \Vert _{L^2}), \end{aligned}$$
(67)

where \(c_2 = \min \{\mu , \nu , 2\kappa \}\). For (67), applying Bernstein’s inequality and integrating it in [0, t], we obtain

$$\begin{aligned} \Vert \dot{\Delta }_j&u\Vert _{L^2} +\Vert \dot{\Delta }_j \omega \Vert _{L^2} +\Vert \dot{\Delta }_j b\Vert _{L^2} + c_2 \tilde{c}_2 c_2^{\star } (\Vert \nabla ^2 u\Vert _{L^1_t L^2} + \Vert \nabla ^2 \omega \Vert _{L^1_t L^2} + \Vert \nabla ^2 b\Vert _{L^1_t L^2})\\ \le&2\Vert \dot{\Delta }_j u_0\Vert _{L^2} +2 \Vert \dot{\Delta }_j \omega _0\Vert _{L^2} + 2 \Vert \dot{\Delta }_j b_0\Vert _{L^2} + C2^j \Vert \dot{\Delta }_j (u \otimes u)\Vert _{L_t^1L^2} \\&+ C2^j \Vert \dot{\Delta }_j (b \otimes b)\Vert _{L_t^1L^2} + C2^j \Vert \dot{\Delta }_j (u \otimes \omega )\Vert _{L_t^1L^2} + C2^j \Vert \dot{\Delta }_j (u \otimes b)\Vert _{L_t^1L^2}. \end{aligned}$$

Taking the \(l^1_j\) over \(j \in {\mathbb {Z}}\), one yields

$$\begin{aligned} \Vert u&\Vert _{\dot{B}^0_{2,1}} +\Vert b\Vert _{\dot{B}^0_{2,1}} + \Vert \omega \Vert _{\dot{B}^0_{2,1}} + c_2 \left( \Vert \nabla ^2 u\Vert _{\tilde{L}^1_t\dot{B}^0_{2,1}} +\Vert \nabla ^2 b\Vert _{\tilde{L}^1_t\dot{B}^0_{2,1}} + \Vert \nabla ^2 \omega \Vert _{\tilde{L}^1_t\dot{B}^0_{2,1}} \right) \nonumber \\ \le&2\Vert u_0\Vert _{\dot{B}^0_{2,1}} + 2\Vert b_0\Vert _{\dot{B}^0_{2,1}} + 2\Vert \omega _0 \Vert _{\dot{B}^0_{2,1}} \nonumber \\&+ C\Vert 2^j \Vert \dot{\Delta }_j (u \otimes u) \Vert _{L^1_t L^2} \Vert _{l^1_j} + C\Vert 2^j \Vert \dot{\Delta }_j (b \otimes b) \Vert _{L^1_t L^2} \Vert _{l^1_j} \nonumber \\&+ C\Vert 2^j \Vert \dot{\Delta }_j (u \otimes b) \Vert _{L^1_t L^2} \Vert _{l^1_j} + C\Vert 2^j \Vert \dot{\Delta }_j (u \otimes \omega ) \Vert _{L^1_t L^2}\Vert _{l^1_j}. \end{aligned}$$
(68)

Denote \(c_3 = 2 c_1 \tilde{c}_2 c_2^{\star }\). Using Lemma 13, Lemma 14, and noting that \(\Vert f\Vert _{L^{\infty }} \le C \Vert f\Vert _{\dot{B}^1_{2.1}}\) and \(\Vert f\Vert _{\tilde{L}^1_t (\dot{B}^s_{2,1})} \approx \Vert f\Vert _{L^1_t(\dot{B}^s_{2,1})}\), we have

$$\begin{aligned} \Vert u&\Vert _{\dot{B}^0_{2,1}} +\Vert b\Vert _{\dot{B}^0_{2,1}} + \Vert \omega \Vert _{\dot{B}^0_{2,1}} + c_3 \left( \Vert \nabla ^2 u\Vert _{L^1_t\dot{B}^0_{2,1}} +\Vert \nabla ^2 b\Vert _{L^1_t\dot{B}^0_{2,1}} + \Vert \nabla ^2 \omega \Vert _{L^1_t\dot{B}^0_{2,1}} \right) \nonumber \\ \le&2\Vert u_0\Vert _{\dot{B}^0_{2,1}} + 2\Vert b_0\Vert _{\dot{B}^0_{2,1}} + 2\Vert \omega _0 \Vert _{\dot{B}^0_{2,1}} + C \int ^t_0 \left( \Vert u(\tau )\Vert _{\dot{B}^0_{2,1}} + \Vert b(\tau )\Vert _{\dot{B}^0_{2,1}} + \Vert \omega (\tau ) \Vert _{\dot{B}^0_{2,1}} \right) \nonumber \\&\times \left( \Vert \nabla ^2 u(\tau )\Vert _{\dot{B}^0_{2,1}} + \Vert \nabla ^2 b(\tau )\Vert _{\dot{B}^0_{2,1}} + \Vert \nabla ^2 \omega (\tau ) \Vert _{\dot{B}^0_{2,1}}\right) d\tau . \end{aligned}$$
(69)

This inequality indicates that, for any \(0< \epsilon < \epsilon _0\),

$$\begin{aligned} \Vert u_0\Vert _{\dot{B}^0_{2,1}} +\Vert b_0\Vert _{\dot{B}^0_{2,1}} + \Vert \omega _0\Vert _{\dot{B}^0_{2,1}} < \epsilon , \end{aligned}$$

then bootstrap argument yields

$$\begin{aligned} \Vert&u(t)\Vert _{\dot{B}^0_{2,1}} + \Vert \omega (t)\Vert _{\dot{B}^0_{2,1}} + \Vert b(t)\Vert _{\dot{B}^0_{2,1}} \nonumber \\&+ \Vert \nabla ^2 u(t)\Vert _{L^1_t(\dot{B}^0_{2,1})} +\Vert \nabla ^2 \omega (t)\Vert _{L^1_t(\dot{B}^0_{2,1})} + \Vert \nabla ^2 b(t)\Vert _{L^1_t(\dot{B}^0_{2,1})} \le C \epsilon , \end{aligned}$$
(70)

which completed the Proposition 5. \(\square\)

Next, we start to prove the decay estimates assertion of (II). As a tool, we first verify the following Proposition in the negative Sobolev space \(\dot{H}^{-l}\), with \(0< l < 1\).

Proposition 6

Let \(c_2 = \min \{\mu , 2\kappa , \nu \}\). Then for \(0< l <1\), we have

$$\begin{aligned} \frac{d}{dt}&(\Vert u\Vert ^2_{\dot{H}^{-l}} + \Vert b \Vert ^2_{\dot{H}^{-l}} + \Vert \omega \Vert ^2_{\dot{H}^{-l}}) + c_1 (\Vert \nabla u\Vert ^2_{\dot{H}^{-l}} + \Vert \nabla b \Vert ^2_{\dot{H}^{-l}} + \Vert \nabla \omega \Vert ^2_{\dot{H}^{-l}}) \nonumber \\ \le&C(\Vert \nabla u\Vert _{L^2} + \Vert \nabla \omega \Vert _{L^2} + \Vert \nabla b\Vert _{L^2})(\Vert \nabla u\Vert ^{1-l}_{L^2} + \Vert \nabla b \Vert ^{1-l}_{L^2})(\Vert u\Vert _{L^2}^l + \Vert b\Vert ^l_{L^2})\nonumber \\&\times (\Vert u\Vert _{\dot{H}^{-l}} + \Vert b \Vert _{\dot{H}^{-l}} + \Vert \omega \Vert _{\dot{H}^{-l}} ). \end{aligned}$$
(71)

Proof of Proposition 6

Due to (64) and with the divergence free condition \(\nabla \cdot u = \nabla \cdot b = 0\), we derive that

$$\begin{aligned} \frac{1}{2}&\frac{d}{dt} (\Vert \dot{\Delta }_j u\Vert ^2_{L^2} + \Vert \dot{\Delta }_j \omega \Vert ^2_{L^2} + \Vert \dot{\Delta }_j b\Vert ^2_{L^2}) + (\mu + \chi )\Vert \dot{\Delta }_j \nabla u\Vert ^2_{L^2} + \kappa \Vert \dot{\Delta }_j \nabla \omega \Vert ^2_{L^2} \\&+ 4 \chi \Vert \dot{\Delta }_j \omega \Vert ^2_{L^2} + \frac{\nu }{2} \Vert \dot{\Delta }_j \nabla b\Vert ^2_{L^2} \\ \le&-\int \dot{\Delta }_j(u \cdot \nabla u) \cdot \dot{\Delta }_j u + \int \dot{\Delta }_j(b \cdot \nabla b) \cdot \dot{\Delta }_j u - \int \dot{\Delta }_j(u \cdot \nabla b) \cdot \dot{\Delta }_j b \\&+ \int \dot{\Delta }_j(b \cdot \nabla u) \cdot \dot{\Delta }_j b - \int \dot{\Delta }_j(u \cdot \nabla \omega ) \cdot \dot{\Delta }_j \omega + 4 \chi \int \dot{\Delta }_j \nabla \times u \cdot \dot{\Delta }_j \omega . \end{aligned}$$

Multiplying the above inequality by \(2^{-2lj}\) and taking the \(l_j^2\) over \(j \in {\mathbb {Z}}\), and noting that \(\dot{B}^{-l}_{2,2} = \dot{H}^{-l}\), we have

$$\begin{aligned} \frac{d}{dt}&(\Vert u\Vert ^2_{\dot{H}^{-l}} + \Vert b \Vert ^2_{\dot{H}^{-l}} + \Vert \omega \Vert ^2_{\dot{H}^{-l}}) + 2(\mu + \chi )\Vert \nabla u\Vert ^2_{\dot{H}^{-l}} \nonumber \\&+ 2\kappa \Vert \nabla \omega \Vert ^2_{\dot{H}^{-l}} + 8 \chi \Vert \omega \Vert ^2_{\dot{H}^{-l}} + \nu \Vert \nabla b\Vert ^2_{\dot{H}^{-l}} \nonumber \\ \le&2 \Vert 2^{-lj} \Vert \dot{\Delta }_j(u \cdot \nabla u)\Vert _{L^2} \Vert _{l_j^2} \Vert u\Vert _{\dot{H}^{-l}} + 2 \Vert 2^{-lj} \Vert \dot{\Delta }_j(b \cdot \nabla b)\Vert _{L^2} \Vert _{l_j^2} \Vert u\Vert _{\dot{H}^{-l}} \nonumber \\&+ 2 \Vert 2^{-lj} \Vert \dot{\Delta }_j(u \cdot \nabla \omega )\Vert _{L^2} \Vert _{l_j^2} \Vert \omega \Vert _{\dot{H}^{-l}} + 8 \chi \Vert 2^{-lj}\Vert \dot{\Delta }_j (\nabla \times u)\Vert _{L^2} \Vert _{l_j^2} \Vert \omega \Vert _{\dot{H}^{-l}}\nonumber \\&+ 2 \Vert 2^{-lj} \Vert \dot{\Delta }_j(u \cdot \nabla b)\Vert _{L^2} \Vert _{l_j^2} \Vert b\Vert _{\dot{H}^{-l}} + 2 \Vert 2^{-lj} \Vert \dot{\Delta }_j(b \cdot \nabla u)\Vert _{L^2} \Vert _{l_j^2} \Vert b\Vert _{\dot{H}^{-l}} \nonumber \\ \le&2 \Vert u \cdot \nabla u\Vert _{\dot{H}^{-l}}\Vert u\Vert _{\dot{H}^{-l}} + 2 \Vert b \cdot \nabla b\Vert _{\dot{H}^{-l}}\Vert u\Vert _{\dot{H}^{-l}} + 2 \Vert u \cdot \nabla \omega \Vert _{\dot{H}^{-l}}\Vert \omega \Vert _{\dot{H}^{-l}} \nonumber \\&+ (\mu + 2\chi )\Vert \nabla u\Vert _{\dot{H}^{-l}}^2 + \frac{8\chi ^2}{\frac{\mu }{2} + \chi }\Vert \omega \Vert ^2_{\dot{H}^{-l}} + 2( \Vert u \cdot \nabla b\Vert _{\dot{H}^{-l}} + \Vert b \cdot \nabla u\Vert _{\dot{H}^{-l}} )\Vert b\Vert _{\dot{H}^{-l}}. \end{aligned}$$
(72)

Applying Lemma 10, Hölder’s inequality and the Gagliardo-Nirenberg inequality, we obtain

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

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

$$\begin{aligned} \Vert u \cdot \nabla u\Vert _{\dot{H}^{-l}} \le C \Vert \nabla u\Vert _{L^2}^{2-l} \Vert u\Vert _{L^2}^{l}. \end{aligned}$$
(73)

Similarly,

$$\begin{aligned}&\Vert b \cdot \nabla b\Vert _{\dot{H}^{-l}} \le C \Vert \nabla b\Vert _{L^2}^{2-l} \Vert b\Vert _{L^2}^{l}, \end{aligned}$$
(74)
$$\begin{aligned}&\Vert u \cdot \nabla b\Vert _{\dot{H}^{-l}}+ \Vert b \cdot \nabla u\Vert _{\dot{H}^{-l}} \nonumber \\&\le C ( \Vert \nabla u\Vert _{L^2}^{1-l} \Vert u\Vert _{L^2}^{l}\Vert \nabla b\Vert _{L^2} + \Vert \nabla b\Vert _{L^2}^{1-l} \Vert b\Vert _{L^2}^{l}\Vert \nabla u\Vert _{L^2}) , \end{aligned}$$
(75)
$$\begin{aligned}&\Vert u \cdot \nabla \omega \Vert _{\dot{H}^{-l}} \le C \Vert \nabla \omega \Vert _{L^2} \Vert \nabla u\Vert _{L^2}^{1-l} \Vert u\Vert _{L^2}^{l}. \end{aligned}$$
(76)

Combining (72) - (76), we obtain

$$\begin{aligned} \frac{d}{dt}&(\Vert u\Vert ^2_{\dot{H}^{-l}} + \Vert b \Vert ^2_{\dot{H}^{-l}} + \Vert \omega \Vert ^2_{\dot{H}^{-l}}) + c_2 (\Vert \nabla u\Vert ^2_{\dot{H}^{-l}} + \Vert \nabla b \Vert ^2_{\dot{H}^{-l}} + \Vert \nabla \omega \Vert ^2_{\dot{H}^{-l}}) \\ \le&C(\Vert \nabla u\Vert _{L^2} + \Vert \nabla \omega \Vert _{L^2} + \Vert \nabla b\Vert _{L^2})(\Vert \nabla u\Vert ^{1-l}_{L^2} + \Vert \nabla b \Vert ^{1-l}_{L^2})(\Vert u\Vert _{L^2}^l + \Vert b\Vert ^l_{L^2})\\&\times (\Vert u\Vert _{\dot{H}^{-l}} + \Vert b \Vert _{\dot{H}^{-l}} + \Vert \omega \Vert _{\dot{H}^{-l}} ), \end{aligned}$$

where \(c_2 = \min \{\mu , 2\kappa , \nu \}\). Thus the proof of Proposition 6 is completed. \(\square\)

Next, we continue to prove (10). Multiplying (67) by \(2^{sj}\) and utilizing Bernstein’s inequality and integrating it in [0, t], finally, taking the \(l^1_j\) over \(j \in {\mathbb {Z}}\), we have

$$\begin{aligned} \Vert u&\Vert _{\dot{B}^s_{2,1}} +\Vert b\Vert _{\dot{B}^s_{2,1}} + \Vert \omega \Vert _{\dot{B}^s_{2,1}} + c_3 \left( \Vert \nabla ^2 u\Vert _{\tilde{L}^1_t(\dot{B}^s_{2,1})} + \Vert \nabla ^2 b\Vert _{\tilde{L}^1_t(\dot{B}^s_{2,1})} + \Vert \nabla ^2 \omega \Vert _{\tilde{L}^1_t(\dot{B}^s_{2,1})} \right) \nonumber \\ \le&2( \Vert u_0\Vert _{\dot{B}^s_{2,1}} +\Vert b_0\Vert _{\dot{B}^s_{2,1}} + \Vert \omega _0\Vert _{\dot{B}^s_{2,1}}) + C\Vert 2^{(s+1)j} \Vert \dot{\Delta }_j (u \otimes u) \Vert _{L^1_t L^2} \Vert _{l^1_j} \nonumber \\&+ C\Vert 2^{(s+1)j} \Vert \dot{\Delta }_j (b \otimes b) \Vert _{L^1_t L^2} \Vert _{l^1_j} + C\Vert 2^{(s+1)j} \Vert \dot{\Delta }_j (u \otimes b) \Vert _{L^1_t L^2} \Vert _{l^1_j} \nonumber \\&+ C\Vert 2^{(s+1)j} \Vert \dot{\Delta }_j (u \otimes \omega ) \Vert _{L^1_t L^2} \Vert _{l^1_j} \nonumber \\ \triangleq&A_1+ A_2 + A_3 + A_4 + A_5 . \end{aligned}$$
(77)

Since \((u_0, \omega _0, b_0) \in B^s_{2,1}({\mathbb {R}}^2)\), we have

$$\begin{aligned} A_1= 2( \Vert u_0\Vert _{\dot{B}^s_{2,1}} +\Vert b_0\Vert _{\dot{B}^s_{2,1}} + \Vert \omega _0\Vert _{\dot{B}^s_{2,1}}) \le C. \end{aligned}$$

By Lemma 13, Lemma 14 and noting that \(\Vert f\Vert _{L^1_t(\dot{B}^s_{2,1})} \approx \Vert f\Vert _{\tilde{L}^1_t(\dot{B}^s_{2,1})}\) and \(\Vert f\Vert _{L^{\infty }} \le C \Vert f\Vert _{\dot{B}^1_{2,1}}\), yield

$$\begin{aligned} A_2 =&C\Vert 2^{(s+1)j} \Vert \dot{\Delta }_j (u \otimes u) \Vert _{L^1_t L^2} \Vert _{l^1_j} \nonumber \\ \le&C \int ^t_0 \Vert u \otimes u (\tau )\Vert _{\dot{B}^{s+1}_{2,1}} d\tau \\ \le&C \int ^t_0 \Vert u(\tau )\Vert _{\dot{B}_{2,1}^{s+1}} \Vert u(\tau )\Vert _{L^{\infty }} d\tau \\ \le&C \int ^t_0 \left( \Vert u(\tau )\Vert ^{\frac{s+1}{s+2}}_{\dot{B}_{2,1}^{s+2}} \Vert u(\tau )\Vert ^{\frac{1}{s+2}}_{\dot{B}_{2,1}^0} \Vert u(\tau )\Vert ^{\frac{1}{s+2}}_{\dot{B}_{2,1}^{s+2}} \Vert u(\tau )\Vert ^{\frac{s+1}{s+2}}_{\dot{B}_{2,1}^0} \right) d\tau \\ \le&C \int ^t_0 \Vert \nabla ^2 u(\tau )\Vert _{\dot{B}_{2,1}^{s}} \Vert u(\tau )\Vert _{\dot{B}_{2,1}^0} d\tau . \end{aligned}$$

For \(A_3\) - \(A_5\), by the similar method as \(A_2\), together with the Young inequality, we get

$$\begin{aligned} A_3 &= C\Vert 2^{(s+1)j} \Vert \dot{\Delta }_j (b \otimes b) \Vert _{L^1_t L^2} \Vert _{l^1_j} \nonumber \\ & \le C \int ^t_0 \Vert b \otimes b (\tau )\Vert _{\dot{B}^{s+1}_{2,1}} d\tau \\ &\le C \int ^t_0 \Vert \nabla ^2 b(\tau )\Vert _{\dot{B}_{2,1}^{s}} \Vert b(\tau )\Vert _{\dot{B}_{2,1}^0} d\tau . \end{aligned}$$

Similarly,

$$\begin{aligned}&A_4 \le C \int ^t_0 (\Vert u(\tau )\Vert _{\dot{B}_{2,1}^0} + \Vert b(\tau )\Vert _{\dot{B}_{2,1}^0})(\Vert \nabla ^2 u(\tau )\Vert _{\dot{B}_{2,1}^{s}} + \Vert \nabla ^2 b(\tau )\Vert _{\dot{B}_{2,1}^{s}}) d\tau ,\\&A_5 \le C \int ^t_0 (\Vert u(\tau )\Vert _{\dot{B}_{2,1}^0} + \Vert \omega (\tau )\Vert _{\dot{B}_{2,1}^0})(\Vert \nabla ^2 u(\tau )\Vert _{\dot{B}_{2,1}^{s}} + \Vert \nabla ^2 \omega (\tau )\Vert _{\dot{B}_{2,1}^{s}}) d\tau . \end{aligned}$$

Inserting \(A_1\) - \(A_5\) into (77), we derive that

$$\begin{aligned} \Vert u&\Vert _{\dot{B}^s_{2,1}} +\Vert b\Vert _{\dot{B}^s_{2,1}} + \Vert \omega \Vert _{\dot{B}^s_{2,1}} + c_3 \left( \Vert \nabla ^2 u\Vert _{L^1_t(\dot{B}^s_{2,1})} + \Vert \nabla ^2 b\Vert _{L^1_t(\dot{B}^s_{2,1})} + \Vert \nabla ^2 \omega \Vert _{L^1_t(\dot{B}^s_{2,1})} \right) \nonumber \\ \le&C + \int ^t_0 (\Vert u(\tau )\Vert _{\dot{B}_{2,1}^0} + \Vert \omega (\tau )\Vert _{\dot{B}_{2,1}^0} + \Vert b (\tau )\Vert _{\dot{B}_{2,1}^0}) \nonumber (\Vert \nabla ^2 u(\tau )\Vert _{\dot{B}_{2,1}^{s}} + \Vert \nabla ^2 \omega (\tau )\Vert _{\dot{B}_{2,1}^{s}} + \Vert \nabla ^2 b (\tau )\Vert _{\dot{B}_{2,1}^{s}}) d\tau . \end{aligned}$$
(78)

Combining (78) and (59) with \(\epsilon\) sufficiently small, we have

$$\begin{aligned}&\Vert u(t)\Vert _{\dot{B}^s_{2,1}} +\Vert b(t)\Vert _{\dot{B}^s_{2,1}} + \Vert \omega (t)\Vert _{\dot{B}^s_{2,1}} \nonumber \\&+ \frac{c_3}{2} \int ^t_0 \left( \Vert \nabla ^2 u(\tau )\Vert _{\dot{B}^s_{2,1}} +\Vert \nabla ^2 b(\tau )\Vert _{\dot{B}^s_{2,1}} + \Vert \nabla ^2 \omega (\tau )\Vert _{\dot{B}^s_{2,1}} \right) d\tau \le C. \end{aligned}$$
(79)

Because \(\dot{B}^0_{2,1} \hookrightarrow \dot{B}^0_{2,2}\) and \(\dot{B}^0_{2,2} \sim \dot{H}^0 \sim L^2\), thus (59) implies

$$\begin{aligned}&\Vert u(t)\Vert _{L^2} + \Vert b(t)\Vert _{L^2} + \Vert \omega (t)\Vert _{L^2} \nonumber \\&+ \int ^t_0 \left( \Vert \nabla ^2 u(\tau )\Vert _{L^2} +\Vert \nabla ^2 b(\tau )\Vert _{L^2} + \Vert \nabla ^2 \omega (\tau )\Vert _{L^2} \right) d\tau < C\epsilon . \end{aligned}$$
(80)

Combining (79) and (80), we have

$$\begin{aligned}&\Vert u(t)\Vert _{B^s_{2,1}} +\Vert b(t)\Vert _{B^s_{2,1}} + \Vert \omega (t)\Vert _{B^s_{2,1}} \nonumber \\&+ \int ^t_0 \left( \Vert \nabla ^2 u(\tau )\Vert _{B^s_{2,1}} +\Vert \nabla ^2 b(\tau )\Vert _{B^s_{2,1}} + \Vert \nabla ^2 \omega (\tau )\Vert _{B^s_{2,1}} \right) d\tau < C, \end{aligned}$$

which completed the proof of (I) in Theorem 2. \(\square\)

We now turn to prove the decay part (II) of Theorem 2.

Proof of (II) of Theorem 2

Multiplying (67) by \(2^{mj}\), and taking the \(l^1_j\) over \(j\in {\mathbb {Z}}\), we obtain

$$\begin{aligned} \frac{d}{dt}&y(t) + c_3 \left( \Vert u\Vert _{\dot{B}^{m+2}_{2,1}} + \Vert b\Vert _{\dot{B}^{m+2}_{2,1}} + \Vert \omega \Vert _{\dot{B}^{m+2}_{2,1}} \right) \\ \le&C\Vert 2^{(m+1)j} \Vert \dot{\Delta }_j (u \otimes u) \Vert _{L^2} \Vert _{l^1_j} + C\Vert 2^{(m+1)j} \Vert \dot{\Delta }_j (b \otimes b) \Vert _{L^2} \Vert _{l^1_j} \nonumber \\&+ C\Vert 2^{(m+1)j} \Vert \dot{\Delta }_j (u \otimes b) \Vert _{L^2} \Vert _{l^1_j} + C\Vert 2^{(m+1)j} \Vert \dot{\Delta }_j (u \otimes \omega ) \Vert _{L^2} \Vert _{l^1_j}, \end{aligned}$$

where \(y(t) = \left\| 2^{mj} \sqrt{\Vert \dot{\Delta }_j u\Vert ^2 + \Vert \dot{\Delta }_j \omega \Vert ^2 +\Vert \dot{\Delta }_j b\Vert ^2 }\right\| _{l^1_j}\). Using Lemma 13 and Lemma 15, together with \(\Vert f\Vert _{L^{\infty }} \le C \Vert f\Vert _{\dot{B}^1_{2,1}}\), we have

$$\begin{aligned} \frac{d}{dt}&y(t) + c_3 \left( \Vert u\Vert _{\dot{B}^{m+2}_{2,1}} + \Vert b\Vert _{\dot{B}^{m+2}_{2,1}} + \Vert \omega \Vert _{\dot{B}^{m+2}_{2,1}} \right) \nonumber \\ \le&C \left( \Vert u\Vert _{\dot{B}^0_{2,1}} +\Vert b\Vert _{\dot{B}^0_{2,1}} + \Vert \omega \Vert _{\dot{B}^0_{2,1}} \right) \left( \Vert u\Vert _{\dot{B}^{m+2}_{2,1}} +\Vert b\Vert _{\dot{B}^{m+2}_{2,1}} + \Vert \omega \Vert _{\dot{B}^{m+2}_{2,1}} \right) . \end{aligned}$$

Then this inequality, together with (59) with \(\epsilon\) sufficiently small, we have

$$\begin{aligned} \frac{d}{dt}&y(t) + \frac{c_2}{2} \left( \Vert u\Vert _{\dot{B}^{m+2}_{2,1}} + \Vert b\Vert _{\dot{B}^{m+2}_{2,1}} + \Vert \omega \Vert _{\dot{B}^{m+2}_{2,1}} \right) \le 0. \end{aligned}$$
(81)

Applying Lemma 15 and \(\dot{B}^s_{2,2} \hookrightarrow \dot{B}^s_{2, \infty }\), we infer that

$$\begin{aligned}{} & {} \Vert u\Vert _{\dot{B}^m_{2,1}} \le C \Vert u\Vert _{\dot{H}^{-l}} ^{\frac{2}{m+l+2}} \Vert \nabla ^2 u\Vert _{\dot{B}^m_{2,1}}^{\frac{m+l}{m+l+2}}, \end{aligned}$$
(82)
$$\begin{aligned}{} & {} \Vert b\Vert _{\dot{B}^m_{2,1}} \le C \Vert b\Vert _{\dot{H}^{-l}} ^{\frac{2}{m+l+2}} \Vert \nabla ^2 b\Vert _{\dot{B}^m_{2,1}}^{\frac{m+l}{m+l+2}}, \end{aligned}$$
(83)

and

$$\begin{aligned} \Vert \omega \Vert _{\dot{B}^m_{2,1}} \le C \Vert \omega \Vert _{\dot{H}^{-l}} ^{\frac{2}{m+l+2}} \Vert \nabla ^2 \omega \Vert _{\dot{B}^m_{2,1}}^{\frac{m+l}{m+l+2}}. \end{aligned}$$
(84)

Therefore, if

$$\begin{aligned} \Vert u\Vert _{\dot{H}^{-l}} + \Vert b\Vert _{\dot{H}^{-l}} + \Vert \omega \Vert _{\dot{H}^{-l}}\le C, \end{aligned}$$
(85)

then we can obtain from (81) - (83), there exists a constant \(a_0 >0\) such that,

$$\begin{aligned} \frac{d}{dt}&y(t)+ a_0 \left( y(t) \right) ^{\frac{m+l+2}{m+l}} \le 0. \end{aligned}$$
(86)

It follows from this that

$$\begin{aligned} y(t) \le C (1 + t)^{- \frac{m+l}{2}}, \end{aligned}$$

which implies

$$\begin{aligned} \Vert u\Vert _{\dot{B}^m_{2,1}} + \Vert b\Vert _{\dot{B}^m_{2,1}} + \Vert \omega \Vert _{\dot{B}^m_{2,1}} \le C (1 + t)^{- \frac{m+l}{2}}, \end{aligned}$$
(87)

which immediately yields (11).

Finally, to make the process more complete, we need to verify that (85) holds for \(0 \le l <1\). To this end, we divide the proof into two cases.

Case 1. (\(l=0\)) Using the fact that \(\dot{H}^0 = L^2\), by (13) we have

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

Then it yields (85).

Case 2. (\(0<l < 1\)) Assume that

$$\begin{aligned} \Vert u_0\Vert ^2_{\dot{H}^{-l}} + \Vert b_0\Vert ^2_{\dot{H}^{-l}} + \Vert \omega _0\Vert ^2_{\dot{H}^{-l}}\le C_0. \end{aligned}$$
(88)

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

$$\begin{aligned} \Vert u(t)\Vert ^2_{\dot{H}^{-l}} + \Vert b(t)\Vert ^2_{\dot{H}^{-l}} + \Vert \omega (t) \Vert ^2_{\dot{H}^{-l}}\le 2C_0. \end{aligned}$$
(89)

If we can derive that for all \(t \in [0, T ]\),

$$\begin{aligned} \Vert u(t)\Vert ^2_{\dot{H}^{-l}} + \Vert b(t)\Vert ^2_{\dot{H}^{-l}} + \Vert \omega (t) \Vert ^2_{\dot{H}^{-l}}\le \frac{3}{2}C_0, \end{aligned}$$
(90)

then an application of the bootstrapping argument would imply that the solution \((u, \omega , b)\) of system (8) satisfies (90) for all \(t \in [0, T]\), which implies (85). With (86) and (87) at our disposal, we show that (90) holds.

With the help of (87) and Lemma 8, we know that

$$\begin{aligned}{} & {} \Vert u(t)\Vert _{L^2} = \Vert u\Vert _{\dot{B}^0_{2,2}} \le C \Vert u\Vert _{\dot{B}^0_{2,1}}, \end{aligned}$$
(91)
$$\begin{aligned}{} & {} \Vert \nabla u\Vert _{L^2} \le C \Vert u\Vert _{\dot{B}^1_{2,1}} \le C \Vert u\Vert _{\dot{B}^0_{2,1}}^{\frac{s-1}{s}} \Vert u\Vert _{\dot{B}^s_{2,1}}^{\frac{1}{s}} \le C (1+\tau )^{-\frac{l+1}{2}}. \end{aligned}$$
(92)

Similarly,

$$\begin{aligned}&\Vert \omega (t)\Vert _{L^2} + \Vert b(t)\Vert _{L^2} = \Vert \omega \Vert _{\dot{B}^0_{2,2}} + \Vert b \Vert _{\dot{B}^0_{2,2}} \le C (\Vert \omega \Vert _{\dot{B}^0_{2,1}} + \Vert \omega \Vert _{\dot{B}^0_{2,1}}), \end{aligned}$$
(93)
$$\begin{aligned}&\Vert \nabla \omega \Vert _{L^2} + \Vert \nabla b\Vert _{L^2} \le C (1 + t)^{- \frac{l+1}{2}}. \end{aligned}$$
(94)

Integrating (71) in [0, t] with \(0<t\le T\), together with (80), and (91) - (94), one infers that

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

By choosing \(\epsilon\) in (80) sufficiently small, then the above inequality yields (90) for all \(t \in [0, t]\), which closes the proof. Then we have (85) and completed the proof of (11). \(\square\)