Optimal Decay Estimates for the 2D Micropolar Equations with Mixed Velocity Dissipation

Abstract

This paper investigates the large time behavior of solutions to the 2D micropolar equations with partial dissipation. When there is mixed velocity dissipation and positive angular viscosity, we obtain the optimal decay estimates to the global solutions and their higher order derivatives without any small assumptions on the initial data. Moreover, when there is mixed velocity dissipation only, we prove the existence and upper bounds of decay estimates of the global strong solutions provided that the initial data is sufficiently small.

1. Introduction

The micropolar fluid motion model, first proposed by Eringen [1], exhibits micro-rotational effects and micro-rotational inertia in fluid motion systems and covers many more phenomena such as fluids consisting of particles suspended in a viscous medium (see, e.g., [2,3]). The two-dimensional (2D) incompressible micropolar equations are expressed as

$$\left\{\begin{array}{c}{\partial }_t\mathit{u}-(\nu +{\displaystyle \frac{\kappa }{2}})\Delta \mathit{u}+\mathit{u}·\nabla \mathit{u}+\nabla p=\kappa \nabla \times \omega ,\hfill \\ {\partial }_t\omega -\gamma \Delta \omega +\mathit{u}·\nabla \omega +2\kappa \omega =\kappa \nabla \times \mathit{u},\hfill \\ \nabla ·\mathit{u}=0,\hfill \\ {\mathit{u}|}_{t=0}={\mathit{u}}_0{(x,y),\omega |}_{t=0}={\omega }_0(x,y),\hfill \end{array}\right.$$

(1)

where $(x,y)\in {\mathbb{R}}^2$ and $t>0$. The unknown quantities $\mathit{u}=(u_1(x,y,t),u_2(x,y,t))$ is the velocity vector field, $p=p(x,y,t)$ is the scalar pressure and $\omega =\omega (x,y,t)$ is the scalar micro-rotation angular velocity of the rotation of the particles of the fluid. The non-dimensional parameters $\nu \ge 0$ is the Newtonian kinematic viscosity, $\kappa >0$ is the dynamic micro-rotation viscosity and $\gamma \ge 0$ is the angular viscosity. Here we denote $\nabla \times \omega =({\partial }_y\omega ,-{\partial }_x\omega )$ and $\nabla \times \mathit{u}={\partial }_x{u}_2-{\partial }_y{u}_1$.

The micropolar fluid model is particularly suitable for describing flows with suspended particles or micro-rotational effects, such as polymeric suspensions, liquid crystals, and certain biological fluids. However, its applicability is primarily limited to laminar and mildly fluctuating regimes where micro-rotational inertia plays a significant role. In fully developed turbulent flows, the closure of angular momentum equations and the scale separation between micro-rotation and macro-velocity may restrict the direct use of the model without additional turbulence modeling. In this work, we focus on the mathematical analysis of the 2D system under smoothness assumptions typical for well-posedness theory, which aligns with laminar or weakly turbulent settings.

The 2D micropolar system is a class of coupled nonlinear parabolic-elliptic partial differential systems, whose well-posedness and large time behavior have been extensively studied. For example, when there is full dissipation, Galdi-Rionero [4] and Lukaszewicz [2] proved the global existence of weak solutions and the global well-posedness of smooth solution, respectively. Dong and Chen [5] obtained the sharp time $L^2$ decay estimates by using the spectral decomposition of linearized equations. When there is partial dissipation, Dong and Zhang [6] derived the global regularity of smooth solution of system (1) with only velocity dissipation $\Delta \mathit{u}$ by introducing an elegant combined quantity. Later, Guo, Jia and Dong [7] obtained the upper bound of decay estimates for the global weak solutions to the same micropolar system as in [6]. Chen [8] proved the existence and uniqueness of smooth solution of system (1) with partial dissipation $({\partial }_{yy}\mathit{u},{\partial }_{xx}\omega )$. Dong, Li and Wu [9] examined the global well-posedness of system (1) with only angular viscosity dissipation $\Delta \omega$. For more results on the global well-posedness and large time behavior of micropolar equations and related fluid models, such as the magneto-micropolar equations and the Oldroyd-B model, one may refer to [10,11,12,13,14,15,16,17] and references therein.

When micro-rotation effects are neglected, system (1) reduces to the 2D Navier–Stokes equations with full dissipation, the optimal decay rates of the solutions and their higher derivatives were established by Schonbek [18] and Schonbek-Wiegner [19]. Certain components of the dissipation can become small and be ignored in certain physical regimes and under suitable scaling, one important case is the Prandtl’s boundary layer equations (see, e.g., [20,21]). For the 2D anisotropic Navier–Stokes equations with horizontal dissipation ${\partial }_{xx}\mathit{u}$, Dong, Wu, Xu and Zhu [22] studied the stability and large-time behavior of the solutions in the domain $\mathbb{T}\times \mathbb{R}$ with $\mathbb{T}$ being a $1D$ periodic box. Recently, for the 2D anisotropic Navier–Stokes equations with mixed partial dissipation $({\partial }_{yy}{u}_1,{\partial }_{xx}{u}_2)$ in the whole space ${\mathbb{R}}^2$, Shang and Zhou [23] obtained the optimal decay properties to global solutions and their higher order derivatives without any small assumptions on the initial data. In a related context, recent studies on rotating flows and non-standard boundary conditions have also contributed to the understanding of long-time dynamics in fluid systems; see, e.g., Egashira and Takada [24] for the 3D rotating Navier–Stokes equations, and Baranovskii [25] for the Navier–Stokes–Voigt model with slip conditions.

Motivated by the previous works [7,13,16,23], the main purpose of this paper is to investigate the large time decay behavior and the upper bounds of decay estimates for solutions to the following 2D anisotropic micropolar equations with mixed partial velocity dissipation:

$$\left\{\begin{array}{c}{\partial }_t{u}_1-(\nu +{\displaystyle \frac{\kappa }{2}}){\partial }_{yy}{u}_1+\mathit{u}·\nabla u_1+{\partial }_{x}p=\kappa {\partial }_y\omega ,\hfill \\ {\partial }_t{u}_2-(\nu +{\displaystyle \frac{\kappa }{2}}){\partial }_{xx}{u}_2+\mathit{u}·\nabla u_2+{\partial }_{y}p=-\kappa {\partial }_x\omega ,\hfill \\ {\partial }_t\omega -\gamma \Delta \omega +\mathit{u}·\nabla \omega +2\kappa \omega =\kappa {\partial }_x{u}_2-\kappa {\partial }_y{u}_1,\hfill \\ {\partial }_x{u}_1+{\partial }_y{u}_2=0,\hfill \\ {(\mathit{u},\omega )|}_{t=0}=({\mathit{u}}_0,{\omega }_0).\hfill \end{array}\right.$$

(2)

Considering the system (2) with positive and zero angular viscosities, respectively, our main results are stated as follows.

Theorem 1.

Let $\nu >{\displaystyle \frac{\kappa }{2}}$ and $\gamma >0$. Suppose that the initial data $({\mathit{u}}_0,{\omega }_0)\in L^2\left({\mathbb{R}}^2\right)$ with $\nabla ·{\mathit{u}}_0=0$, then the global weak solution of system (2) obeys the following decay properties:

(i) For any $s\ge 0$,

$$\begin{array}{c}\hfill {(1+t)}^{{\displaystyle \frac{s}{2}}}{\parallel ({\mathsf{\Lambda }}^s\mathit{u},{\mathsf{\Lambda }}^s\omega )\left(t\right)\parallel }_{L^2}\to 0,ast\to \infty .\end{array}$$

(3)

(ii) If $({\mathit{u}}_0,{\omega }_0)\in {\dot{H}}^{-\sigma }\left({\mathbb{R}}^2\right)\cap {\dot{H}}^s\left({\mathbb{R}}^2\right)$ with $0<\sigma <1$ and $s\in {\mathbb{Z}}_{\ge 0}$, then for all $t>0$ and $0\le m\le s$,

$$\begin{array}{c}\hfill {\parallel (\mathit{u},\omega )\left(t\right)\parallel }_{H^m}^2+{\displaystyle \frac{\delta }{2}}{\int }_0^t{\parallel (\mathsf{\Lambda }\mathit{u},\mathsf{\Lambda }\omega )\left(\tau \right)\parallel }_{H^m}^{2}d\tau \le C{\parallel ({\mathit{u}}_0,{\omega }_0)\parallel }_{H^m}^2{e}^{C\parallel ({\mathit{u}}_0,{\omega }_0){\parallel }_{L^2}^2},\end{array}$$

(4)

and

$$\begin{array}{c}\hfill {\parallel ({\mathsf{\Lambda }}^m\mathit{u},{\mathsf{\Lambda }}^m\omega )\left(t\right)\parallel }_{L^2}\le C{(1+t)}^{-{\displaystyle \frac{m}{2}}-{\displaystyle \frac{\sigma }{2}}}.\end{array}$$

(5)

Here $\mathsf{\Lambda }:={(-\Delta )}^{{\displaystyle \frac{1}{2}}}$ is the Zygmund operator and the positive constant δ depends on $\nu ,\kappa$ and γ only.

Theorem 2.

Let $\nu >\kappa$ and $\gamma =0$. Suppose that the initial data $({\mathit{u}}_0,{\omega }_0)\in H^{-\sigma }\left({\mathbb{R}}^2\right)\cap H^s\left({\mathbb{R}}^2\right)$ with $\nabla ·{\mathit{u}}_0=0$, $0<\sigma <1$ and $s>2$. Then there exists a positive constant ${\epsilon }_0$ such that for all $0<\epsilon <{\epsilon }_0$, if

$$\begin{array}{c}\hfill \parallel {\mathit{u}}_0{\parallel }_{H^s}+{\parallel {\omega }_0\parallel }_{H^s}<\epsilon ,\end{array}$$

(6)

then system (2) admits a unique global solution $(\mathit{u},\omega )$ satisfies, for all $t>0$ and $0\le m\le s$,

$$\begin{array}{c}\hfill {\parallel (\mathit{u},\omega )\left(t\right)\parallel }_{H^m}^2+{\displaystyle \frac{{\delta }_0}{2}}{\int }_0^t{\parallel (\mathsf{\Lambda }\mathit{u},\omega )\left(\tau \right)\parallel }_{H^m}^{2}d\tau \le C{\epsilon }^2,\end{array}$$

(7)

and

$$\begin{array}{c}\hfill {\parallel ({\mathsf{\Lambda }}^m\mathit{u},{\mathsf{\Lambda }}^m\omega )\left(t\right)\parallel }_{L^2}\le C{(1+t)}^{-{\displaystyle \frac{m}{2}}-{\displaystyle \frac{\sigma }{2}}}.\end{array}$$

(8)

Moreover, for all $t>0$ and $0\le m\le s-1$,

$$\begin{array}{c}\hfill {\parallel {\mathsf{\Lambda }}^m\omega \left(t\right)\parallel }_{L^2}\le C{(1+t)}^{-{\displaystyle \frac{m+1}{2}}-{\displaystyle \frac{\sigma }{2}}}.\end{array}$$

(9)

Here the positive constant ${\delta }_0$ only depends on ν and κ.

Remark 1.

The decay rates of $\mathit{u}$ derived in Theorem 1 (i.e., (5)) and Theorem 2 (i.e., (8)) are respectively optimal, as they coincide with those of heat equations. Although angular viscosity is ignored in Theorem 2, the decay rate of ω in (9) is higher than that of $\mathit{u}$ in (8) owing to the effect of the damping mechanism.

Remark 2.

The present study distinguishes itself from and extends the aforementioned works in several key aspects. Compared to [7,13], which established upper bounds for decay in related micropolar systems, here we prove optimal decay estimates for the system with mixed velocity dissipation and positive angular viscosity (Theorem 1), without requiring the initial data to be small. When angular viscosity is absent, we not only establish the existence of global solutions for small data but also provide explicit decay rates (Theorem 2), refining the qualitative results in some prior works. Furthermore, while [23] focused on the anisotropic Navier–Stokes equations, our analysis targets the coupled and physically richer micropolar system (2). The decay property (9) for the micro-rotation field ω under zero angular viscosity, which is faster than that for the velocity field due to the damping term $2\kappa \omega$, represents a new observation not present in the Navier–Stokes counterpart [23] or the Oldroyd-B model studies like [16]. These results provide a more complete picture of the long-time dynamics for micropolar fluids with anisotropic dissipation.

The micropolar fluid model is particularly effective for describing complex fluids whose behavior is significantly influenced by internal microstructure and rotational degrees of freedom. Concrete examples include blood flow, where red blood cells spin relative to the plasma; liquid crystals, whose orientational order governs flow dynamics; microstructured suspensions such as polymeric or colloidal fluids; drilling muds in petroleum engineering, where particle rotation alters rheological properties; and certain superfluids or turbulent flows with intrinsic angular momentum. In these systems, understanding the long-time decay behavior of both the translational velocity $\mathit{u}$ and the micro-rotation $\omega$ is essential for predicting flow stability, energy dissipation, mixing efficiency, and the relaxation of disturbances. The mixed dissipation structure examined here—where different velocity components experience distinct dissipative mechanisms—naturally arises in anisotropic media, near boundaries, or within engineered microchannels. Consequently, the optimal and sharp decay estimates established in this work offer quantitative guidelines that can support the modeling, simulation, and control of such physical and industrial processes. Moreover, recent simulation-oriented studies further demonstrate the importance of coupled effects in micropolar flows under complex conditions. For instance, thermoelastic responses in fiber-reinforced porous media under a three-phase-lag model [26] and magneto-thermo-convective micropolar flow analysis on rotating non-isothermal cones [27] illustrate how viscosity, magnetic fields, and thermal dissipation interact with micro-rotation in realistic geometries and multi-physical settings. These works underline the relevance of accurate decay estimates for velocity and micro-rotation in simulating and controlling such advanced engineering and material processes.

The rest of this paper is structured as follows. Section 2 provides two fundamental lemmas, while the proofs of Theorems 1 and 2 are presented in Section 3 and Section 4, respectively. Throughout this paper, C denotes a generic positive constant, whose value may vary from line to line. To simplify the notation, we write ${\parallel f\parallel }_{L^2}={\parallel f\parallel }_{L^2\left({\mathbb{R}}^2\right)}$, ${\parallel f\parallel }_{H^m}={\parallel f\parallel }_{H^m\left({\mathbb{R}}^2\right)}$ and ${\parallel (\mathit{u},\omega )\parallel }_X^2={\parallel \mathit{u}\parallel }_X^2+{\parallel \omega \parallel }_X^2$.

2. Preliminaries

We first recall the following property of the mixed partial dissipation in the energy framework (see [28] (Lemma 2.1)).

Lemma 1.

Let $\mathit{u}=(u_1,u_2)$ be a divergence free vector. Then

$$\begin{array}{c}\hfill {\int }_{{\mathbb{R}}^2}\left(-{\partial }_{yy}{u}_1{\mathsf{\Lambda }}^{2\alpha }{u}_1-{\partial }_{xx}{u}_2{\mathsf{\Lambda }}^{2\alpha }{u}_2\right)dxdy\ge {\displaystyle \frac{1}{2}}{\parallel {\mathsf{\Lambda }}^{1+\alpha }\mathit{u}\parallel }_{L^2}^2\end{array}$$

(10)

holds for any $\alpha \in \mathbb{R}$. Here ${\mathsf{\Lambda }}^{2\alpha }={(-\Delta )}^{\alpha }$ is defined via Fourier transform as $\mathcal{F}\left({\mathsf{\Lambda }}^{2\alpha }f\right)\left(\xi \right)={\left|\xi \right|}^{2\alpha }\mathcal{F}f\left(\xi \right)$ with

$$\begin{array}{c}\hfill \mathcal{F}f\left(\xi \right)=\widehat{f}\left(\xi \right)={\left(2\pi \right)}^{-1}{\int }_{{\mathbb{R}}^2}{e}^{-\mathrm{i}(x,y)·\xi }f(x,y)dxdy.\end{array}$$

Proof. 

Due to the divergence-free condition ${\partial }_x{u}_1+{\partial }_y{u}_2=0$, (10) can be easily derived by integration by parts as follows.

$$\begin{array}{ccc}& & {\int }_{{\mathbb{R}}^2}\left(-{\partial }_{yy}{u}_1{\mathsf{\Lambda }}^{2\alpha }{u}_1-{\partial }_{xx}{u}_2{\mathsf{\Lambda }}^{2\alpha }{u}_2\right)dxdy\hfill \\ & =& {\int }_{{\mathbb{R}}^2}\left({\partial }_{yy}{u}_1\Delta {\mathsf{\Lambda }}^{2\alpha -2}{u}_1+{\partial }_{xx}{u}_2\Delta {\mathsf{\Lambda }}^{2\alpha -2}{u}_2\right)dxdy\hfill \\ & =& {\int }_{{\mathbb{R}}^2}\left(|\nabla {\partial }_y{\mathsf{\Lambda }}^{\alpha -1}{u}_1{|}^2+{|\nabla {\partial }_x{\mathsf{\Lambda }}^{\alpha -1}{u}_2|}^2\right)dxdy\hfill \\ & =& {\int }_{{\mathbb{R}}^2}\left(|{\partial }_{xy}{\mathsf{\Lambda }}^{\alpha -1}{u}_1{|}^2+|{\partial }_{yy}{\mathsf{\Lambda }}^{\alpha -1}{u}_1{|}^2+|{\partial }_{xx}{\mathsf{\Lambda }}^{\alpha -1}{u}_2{|}^2+{\left|{\partial }_{yx}{\mathsf{\Lambda }}^{\alpha -1}{u}_2\right|}^2\right)dxdy\hfill \\ & =& {\int }_{{\mathbb{R}}^2}\left(|{\partial }_{yy}{\mathsf{\Lambda }}^{\alpha -1}{u}_2{|}^2+|{\partial }_{yy}{\mathsf{\Lambda }}^{\alpha -1}{u}_1{|}^2+|{\partial }_{xx}{\mathsf{\Lambda }}^{\alpha -1}{u}_2{|}^2+{\left|{\partial }_{xx}{\mathsf{\Lambda }}^{\alpha -1}{u}_1\right|}^2\right)dxdy\hfill \\ & \ge & {\displaystyle \frac{1}{2}}{\int }_{{\mathbb{R}}^2}|\Delta {\mathsf{\Lambda }}^{\alpha -1}{\mathit{u}|}^{2}dxdy={\displaystyle \frac{1}{2}}{\parallel {\mathsf{\Lambda }}^{1+\alpha }\mathit{u}\parallel }_{L^2}^2.\hfill \end{array}$$

□

Next, we present the commutator estimates (see, e.g., [29]), which is a significant tool to derive the a priori estimates of solutions.

Lemma 2.

Let $r>0$, $0<p<\infty$ and ${\displaystyle \frac{1}{p}}={\displaystyle \frac{1}{p_1}}+{\displaystyle \frac{1}{q_1}}={\displaystyle \frac{1}{p_2}}+{\displaystyle \frac{1}{q_2}}$ with $q_1,p_2\in (0,\infty )$ and $p_1,q_2\in [0,\infty ]$. Then there exists a constant C depending on the indices $r,p,p_1,q_1,p_2$ and $q_2$ such that

$$\begin{array}{c}\hfill \parallel {\mathsf{\Lambda }}^r{\left(uv\right)\parallel }_{L^p}\le C\left({\parallel u\parallel }_{L^{p_1}}\parallel {\mathsf{\Lambda }}^r{v\parallel }_{L^{q_1}}+\parallel {\mathsf{\Lambda }}^r{u\parallel }_{L^{p_2}}{\parallel v\parallel }_{L^{q_2}}\right),\end{array}$$

(11)

$$\begin{array}{c}\hfill \parallel [{\mathsf{\Lambda }}^r,u]{v\parallel }_{L^p}\le {C(\parallel \nabla u\parallel }_{L^{p_1}}\parallel {\mathsf{\Lambda }}^{r-1}{v\parallel }_{L^{q_1}}+\parallel {\mathsf{\Lambda }}^r{u\parallel }_{L^{p_2}}{\parallel v\parallel }_{L^{q_2}}),\end{array}$$

(12)

where $[{\mathsf{\Lambda }}^r,u]v={\mathsf{\Lambda }}^r\left(uv\right)-u{\mathsf{\Lambda }}^{r}v$.

3. Proof of Theorem 1

As the local well-posedness and eventual regularity of system (2) in $H^m$ with any $m\ge 0$ can be obtained via standard approaches (see, e.g., [30]), this section aims to establish the global a priori estimates of the solutions to prove Theorem 1. We remark that the proof presented below is formal and can be rigorously applied to approximate solutions.

As preparation, we give the following proposition, which contains the global uniform a priori $L^2$-estimates of the solutions and an elementary estimate for the solutions under the Fourier transform.

Proposition 1.

Let the assumptions of Theorem 1 hold. Then the weak solution $(\mathit{u},\omega )$ of system (2) satisfies, for any $t>0$,

$$\begin{array}{c}\hfill {\displaystyle \frac{d}{dt}}{\parallel (\mathit{u},\omega )\left(t\right)\parallel }_{L^2}^2+\delta {\parallel (\mathsf{\Lambda }\mathit{u},\mathsf{\Lambda }\omega )\left(t\right)\parallel }_{L^2}^2\le 0,\end{array}$$

(13)

and

$$\begin{array}{c}\hfill {\parallel (\mathit{u},\omega )\left(t\right)\parallel }_{L^2}^2+\delta {\int }_0^t{\parallel (\mathsf{\Lambda }\mathit{u},\mathsf{\Lambda }\omega )\left(\tau \right)\parallel }_{L^2}^{2}d\tau \le {\parallel ({\mathit{u}}_0,{\omega }_0)\parallel }_{L^2}^2.\end{array}$$

(14)

Moreover, for any $\xi =({\xi }_x,{\xi }_y)\in {\mathbb{R}}^2$ and $t>0$,

$$\begin{array}{cc}\hfill |\widehat{\mathit{u}}(\xi ,t){|}^2+|\widehat{\omega }(\xi ,t)& {|}^2\le e^{-4\delta min\left\{\right|{\xi }_x{|}^2,\left|{\xi }_y{|}^2\right\}t}\left(\right|{\widehat{\mathit{u}}}_0{\left(\xi \right)|}^2+|{\widehat{\omega }}_0\left(\xi \right){|}^2)\hfill \\ \hfill & +{C\left|\xi \right|}^2{\left({\int }_0^t{e}^{-2\delta min\left\{\right|{\xi }_x{|}^2,\left|{\xi }_y{|}^2\right\}(t-\tau )}{\parallel (\mathit{u},\omega )\left(\tau \right)\parallel }_{L^2}^{2}d\tau \right)}^2.\hfill \end{array}$$

(15)

Here the positive constant $\delta :=min\{\nu -{\displaystyle \frac{\kappa }{2}},2\gamma \}$.

Proof. 

Taking the $L^2$-inner product of Equations (2)₁–(2)₃ with $u_1$, $u_2$ and $\omega$, respectively, then summing up the resulting equations, we apply the H$\ddot{o}$lder and Young inequalities to obtain

$$\begin{array}{cc}\hfill {\displaystyle \frac{1}{2}}{\displaystyle \frac{d}{dt}}{\parallel (\mathit{u},\omega )\left(t\right)\parallel }_{L^2}^2& +(\nu +{\displaystyle \frac{\kappa }{2}})\left(\parallel {\partial }_y{u}_1{\parallel }_{L^2}^2+{\parallel {\partial }_x{u}_2\parallel }_{L^2}^2\right)+{\gamma \parallel \mathsf{\Lambda }\omega \parallel }_{L^2}^2+2\kappa {\parallel \omega \parallel }_{L^2}^2\hfill \\ \hfill & =\kappa {\int }_{{\mathbb{R}}^2}({\partial }_y\omega u_1-{\partial }_x\omega u_2+{\partial }_x{u}_2\omega -{\partial }_y{u}_1\omega )dxdy\hfill \\ \hfill & =2\kappa {\int }_{{\mathbb{R}}^2}({\partial }_x{u}_2-{\partial }_y{u}_1)\omega dxdy\hfill \\ \hfill & \le 2\kappa \left(\parallel {\partial }_x{u}_2{\parallel }_{L^2}+{\parallel {\partial }_y{u}_1\parallel }_{L^2}\right){\parallel \omega \parallel }_{L^2}\hfill \\ \hfill & \le \kappa \left(\parallel {\partial }_y{u}_1{\parallel }_{L^2}^2+{\parallel {\partial }_x{u}_2\parallel }_{L^2}^2\right)+2\kappa {\parallel \omega \parallel }_{L^2}^2,\hfill \end{array}$$

(16)

where we have used the fact that

$$\begin{array}{c}\hfill {\int }_{{\mathbb{R}}^2}\mathit{u}·\nabla \mathit{u}·\mathit{u}dxdy={\int }_{{\mathbb{R}}^2}\mathit{u}·\nabla \omega ·\omega dxdy=0,\end{array}$$

due to the divergence free condition of $\mathit{u}$. Moreover, (10) for $\alpha =0$ gives that

$$\begin{array}{c}\hfill {\parallel {\partial }_y{u}_1\parallel }_{L^2}^2+\parallel {\partial }_x{u}_2{\parallel }_{L^2}^2\ge {\displaystyle \frac{1}{2}}{\parallel \mathsf{\Lambda }\mathit{u}\parallel }_{L^2}^2.\end{array}$$

Plugging this into (16) yields (13). Integrating (13) over the time interval $(0,t)$ leads to (14).

Applying the Fourier transform $\mathcal{F}$ to system (2), we have

$$\left\{\begin{array}{c}{\partial }_t{\widehat{u}}_1+(\nu +{\displaystyle \frac{\kappa }{2}}){\left|{\xi }_y\right|}^2{\widehat{u}}_1=-\mathrm{i}{\xi }_x\widehat{p}+\mathrm{i}\kappa {\xi }_y\widehat{\omega }+\mathcal{F}[-\mathit{u}·\nabla u_1],\hfill \\ {\partial }_t{\widehat{u}}_2+(\nu +{\displaystyle \frac{\kappa }{2}}){\left|{\xi }_x\right|}^2{\widehat{u}}_2=-\mathrm{i}{\xi }_y\widehat{p}-\mathrm{i}\kappa {\xi }_x\widehat{\omega }+\mathcal{F}[-\mathit{u}·\nabla u_2],\hfill \\ {\partial }_t\widehat{\omega }+\gamma {\left|\xi \right|}^2\widehat{\omega }+2\kappa \widehat{\omega }=\mathrm{i}\kappa {\xi }_x{\widehat{u}}_2-\mathrm{i}\kappa {\xi }_y{\widehat{u}}_1+\mathcal{F}[-\mathit{u}·\nabla \omega ],\hfill \\ \mathrm{i}{\xi }_x{\widehat{u}}_1+\mathrm{i}{\xi }_y{\widehat{u}}_2=0,\hfill \\ (\widehat{\mathit{u}},\widehat{\omega }){|}_{t=0}=({\widehat{\mathit{u}}}_0,{\widehat{\omega }}_0).\hfill \end{array}\right.$$

(17)

Taking the dot product of Equations ${\left(17\right)}_1–{\left(17\right)}_3$ with conjugated functions ${\overline{\widehat{u}}}_1$, ${\overline{\widehat{u}}}_2$ and $\overline{\widehat{\omega }}$, respectively, multiplying ${\left(17\right)}_1–{\left(17\right)}_3$ conjugated by ${\widehat{u}}_1$, ${\widehat{u}}_2$ and $\widehat{\omega }$, respectively, and summing them up, noting that ${|\widehat{\mathit{u}}|}^2=\widehat{\mathit{u}}\overline{\widehat{\mathit{u}}}$ and ${|\widehat{\mathit{u}}|}^2={|{\widehat{u}}_1|}^2+{\left|{\widehat{u}}_2\right|}^2$, we obtain

$$\begin{array}{cc}\hfill {\partial }_t\left(|\widehat{\mathit{u}}{|}^2+{\left|\widehat{\omega }\right|}^2\right)& +(2\nu +\kappa )\left(|{\xi }_y{|}^2|{\widehat{u}}_1{|}^2+{|{\xi }_x|}^2{\left|{\widehat{u}}_2\right|}^2\right)+{2\gamma \left|\xi \right|}^2|\widehat{\omega }{|}^2+4\kappa {\left|\widehat{\omega }\right|}^2\hfill \\ \hfill & \le -\mathrm{i}{\xi }_x\widehat{p}·{\overline{\widehat{u}}}_1+\mathrm{i}{\xi }_x\overline{\widehat{p}}·{\widehat{u}}_1-\mathrm{i}{\xi }_y\widehat{p}·{\overline{\widehat{u}}}_2+\mathrm{i}{\xi }_y\overline{\widehat{p}}·{\widehat{u}}_2\hfill \\ \hfill & +\mathrm{i}\kappa {\xi }_y\widehat{\omega }·{\overline{\widehat{u}}}_1-\mathrm{i}\kappa {\xi }_y\overline{\widehat{\omega }}·{\widehat{u}}_1-\mathrm{i}\kappa {\xi }_x\widehat{\omega }·{\overline{\widehat{u}}}_2+\mathrm{i}\kappa {\xi }_x\overline{\widehat{\omega }}·{\widehat{u}}_2\hfill \\ \hfill & +\mathrm{i}\kappa {\xi }_x{\widehat{u}}_2·\overline{\widehat{\omega }}-\mathrm{i}\kappa {\xi }_x{\overline{\widehat{u}}}_2·\widehat{\omega }-\mathrm{i}\kappa {\xi }_y{\widehat{u}}_1·\overline{\widehat{\omega }}+\mathrm{i}\kappa {\xi }_y{\overline{\widehat{u}}}_1·\widehat{\omega }\hfill \\ \hfill & +\mathcal{F}[-\mathit{u}·\nabla u_1]·{\overline{\widehat{u}}}_1+\overline{\mathcal{F}[-\mathit{u}·\nabla u_1]}·{\widehat{u}}_1\hfill \\ \hfill & +\mathcal{F}[-\mathit{u}·\nabla u_2]·{\overline{\widehat{u}}}_2+\overline{\mathcal{F}[-\mathit{u}·\nabla u_2]}·{\widehat{u}}_2\hfill \\ \hfill & +\mathcal{F}[-\mathit{u}·\nabla \omega ]·\overline{\widehat{\omega }}+\overline{\mathcal{F}[-\mathit{u}·\nabla \omega ]}·\widehat{\omega }\hfill \\ \hfill & :=\sum _{j=1}^{18}{J}_j.\hfill \end{array}$$

(18)

Thanks to ${\left(17\right)}_4$, we have

$$J_1+J_3=J_2+J_4=0.$$

By virtue of $\nabla ·\mathit{u}=0$ and the H$\ddot{o}$lder inequality, we yield

$$J_{13}+J_{14}+J_{15}+J_{16}\le 2\left|\xi \right|\left|\mathcal{F}(\mathit{u}\otimes \mathit{u})\right||\widehat{\mathit{u}}{|\le 2|\xi |\parallel \mathit{u}\otimes \mathit{u}\parallel }_{L^1}|\widehat{\mathit{u}}{|\le 2|\xi |\parallel \mathit{u}\parallel }_{L^2}^2\left|\widehat{\mathit{u}}\right|.$$

Similarly,

$$J_{17}+J_{18}\le 2\left|\xi \right|\left|\mathcal{F}(\mathit{u}\otimes \omega )\right||\widehat{\omega }{|\le 2|\xi |\parallel \mathit{u}\otimes \omega \parallel }_{L^1}|\widehat{\omega }{|\le |\xi |\parallel (\mathit{u},\omega )\parallel }_{L^2}^2\left|\widehat{\omega }\right|.$$

Using Young inequality, we get

$$\sum _{J=5}^{12}{J}_j\le 4\kappa \left(|{\xi }_y\left|\right|{\widehat{u}}_1|+|{\xi }_x\left|\right|{\widehat{u}}_2|\right)\left|\widehat{\omega }\right|\le 2\kappa \left(|{\xi }_y{|}^2|{\widehat{u}}_1{|}^2+|{\xi }_x{|}^2{\left|{\widehat{u}}_2\right|}^2\right)+4\kappa {\left|\widehat{\omega }\right|}^2.$$

Denoting $M\left(t\right)=\sqrt{|\widehat{\mathit{u}}{\left(t\right)|}^2+{\left|\widehat{\omega }\left(t\right)\right|}^2}$, after inserting all the above estimates into (18), we obtain

$$\begin{array}{c}\hfill {\partial }_t{\left[M\left(t\right)\right]}^2+2\delta min\left\{\right|{\xi }_x{|}^2,|{\xi }_y{|}^2{\}\left[M\left(t\right)\right]}^2\le {C\left|\xi \right|\parallel (\mathit{u},\omega )\parallel }_{L^2}^{2}M\left(t\right),\end{array}$$

(19)

or

$$\begin{array}{c}\hfill {\displaystyle \frac{d}{dt}}\left(e^{2\delta min\left\{\right|{\xi }_x{|}^2,\left|{\xi }_y{|}^2\right\}t}M\left(t\right)\right)\le {C\left|\xi \right|\parallel (\mathit{u},\omega )\parallel }_{L^2}^2{e}^{2\delta min\left\{\right|{\xi }_x{|}^2,\left|{\xi }_y{|}^2\right\}t}.\end{array}$$

Integrating this with respect to time gives the desired result (15). □

Now we are ready to present the proof of Theorem 1, which comprises four parts.

Proof of Theorem 1.

Part 1: (3) is valid for $s=0$.

For any $t\ge 0$, setting

$$B\left(t\right)=\{\xi \in {\mathbb{R}}^2|\left|\xi \right|\le \rho \left(t\right)\}$$

with $\rho \left(t\right)>0$ to be chosen later. By Plancherel theorem, we have

$$\begin{array}{cc}\hfill {\parallel (\mathsf{\Lambda }\mathit{u},\mathsf{\Lambda }\omega )\parallel }_{L^2}^2& ={\int }_{{\mathbb{R}}^2}{\left|\xi \right|}^2\left(|\widehat{\mathit{u}}{\left(\xi \right)|}^2+{\left|\widehat{\omega }\left(\xi \right)\right|}^2\right)d\xi \hfill \\ \hfill & \ge {\int }_{{\mathbb{R}}^2\setminus B\left(t\right)}{\left|\xi \right|}^2\left(|\widehat{\mathit{u}}{\left(\xi \right)|}^2+{\left|\widehat{\omega }\left(\xi \right)\right|}^2\right)d\xi \hfill \\ \hfill & \ge {\rho }^2\left(t\right){\int }_{{\mathbb{R}}^2\setminus B\left(t\right)}\left(|\widehat{\mathit{u}}{\left(\xi \right)|}^2+{\left|\widehat{\omega }\left(\xi \right)\right|}^2\right)d\xi \hfill \\ \hfill & ={\rho }^2\left(t\right){\parallel (\mathit{u},\omega )\parallel }_{L^2}^2-{\rho }^2\left(t\right){\int }_{B\left(t\right)}\left(|\widehat{\mathit{u}}{\left(\xi \right)|}^2+{\left|\widehat{\omega }\left(\xi \right)\right|}^2\right)d\xi .\hfill \end{array}$$

(20)

Inserting (20) into (13), combining with (15), we obtain

$$\begin{array}{cc}\hfill & {\displaystyle \frac{d}{dt}}{\parallel (\mathit{u},\omega )\parallel }_{L^2}^2+\delta {\rho }^2\left(t\right){\parallel (\mathit{u},\omega )\parallel }_{L^2}^2\le \delta {\rho }^2\left(t\right){\int }_{B\left(t\right)}\left(|\widehat{\mathit{u}}{\left(\xi \right)|}^2+{\left|\widehat{\omega }\left(\xi \right)\right|}^2\right)d\xi \hfill \\ \hfill & \le \delta {\rho }^2\left(t\right){\int }_{\left|\xi \right|\le \rho \left(t\right)}{e}^{-4\delta min\left\{\right|{\xi }_x{|}^2,\left|{\xi }_y{|}^2\right\}t}\left(|{\widehat{\mathit{u}}}_0{\left(\xi \right)|}^2+{\left|{\widehat{\omega }}_0\left(\xi \right)\right|}^2\right)d\xi \hfill \\ \hfill & +C\delta {\rho }^2\left(t\right){\int }_{\left|\xi \right|\le \rho \left(t\right)}{\left|\xi \right|}^2{\left({\int }_0^t{\parallel (\mathit{u},\omega )\left(\tau \right)\parallel }_{L^2}^{2}d\tau \right)}^{2}d\xi .\hfill \end{array}$$

(21)

Choosing

$$\rho \left(t\right)={\left({\displaystyle \frac{k}{\delta (e+t)log(e+t)}}\right)}^{\frac{1}{2}}$$

with $k>3$ in (21). Multiplying the resulting inequality by ${log}^k(e+t)$, we apply the energy inequality (14) to derive

$$\begin{array}{cc}\hfill {\displaystyle \frac{d}{dt}}& \left({log}^k(e+t){\parallel (\mathit{u},\omega )\left(t\right)\parallel }_{L^2}^2\right)\hfill \\ \hfill & \le C{\displaystyle \frac{{log}^{k-1}(e+t)}{e+t}}{\int }_{\left|\xi \right|\le \rho \left(t\right)}{e}^{-4\delta min\left\{\right|{\xi }_x{|}^2,\left|{\xi }_y{|}^2\right\}t}\left(|{\widehat{\mathit{u}}}_0{\left(\xi \right)|}^2+{\left|{\widehat{\omega }}_0\left(\xi \right)\right|}^2\right)d\xi \hfill \\ \hfill & +C{\displaystyle \frac{{log}^{k-3}(e+t)}{{(e+t)}^3}}{\left({\int }_0^t{\parallel (\mathit{u},\omega )\left(\tau \right)\parallel }_{L^2}^{2}d\tau \right)}^2\hfill \\ \hfill & \le C{\displaystyle \frac{{log}^{k-1}(e+t)}{e+t}}{\int }_{\left|\xi \right|\le \rho \left(t\right)}{e}^{-4\delta min\left\{\right|{\xi }_x{|}^2,\left|{\xi }_y{|}^2\right\}t}\left(|{\widehat{\mathit{u}}}_0{\left(\xi \right)|}^2+{\left|{\widehat{\omega }}_0\left(\xi \right)\right|}^2\right)d\xi \hfill \\ \hfill & +C\parallel ({\mathit{u}}_0,{\omega }_0){\parallel }_{L^2}^4{\displaystyle \frac{{log}^{k-3}(e+t)}{e+t}}.\hfill \end{array}$$

Integrating this with respect to time, we arrive at

$$\begin{array}{cc}\hfill & {\parallel (\mathit{u},\omega )\left(t\right)\parallel }_{L^2}^2\le {log}^{-k}(e+t){\parallel ({\mathit{u}}_0,{\omega }_0)\parallel }_{L^2}^2\hfill \\ \hfill & +C{log}^{-k}(e+t){\int }_0^t{\displaystyle \frac{{log}^{k-1}(e+\tau )}{e+\tau }}{\int }_{\left|\xi \right|\le \rho \left(\tau \right)}{e}^{-4\delta min\left\{\right|{\xi }_x{|}^2,\left|{\xi }_y{|}^2\right\}\tau }\left(|{\widehat{\mathit{u}}}_0{\left(\xi \right)|}^2+{\left|{\widehat{\omega }}_0\left(\xi \right)\right|}^2\right)d\xi d\tau \hfill \\ \hfill & +C\parallel ({\mathit{u}}_0,{\omega }_0){\parallel }_{L^2}^4{log}^{-2}(e+t).\hfill \end{array}$$

(22)

Noting the fact that

$${\int }_{\left|\xi \right|\le \rho \left(t\right)}{e}^{-4\delta min\left\{\right|{\xi }_x{|}^2,\left|{\xi }_y{|}^2\right\}t}\left(|{\widehat{\mathit{u}}}_0{\left(\xi \right)|}^2+{\left|{\widehat{\omega }}_0\left(\xi \right)\right|}^2\right)d\xi \to 0,ast\to \infty .$$

Thus, it follows from (22) that

$$\begin{array}{c}\hfill {\parallel (\mathit{u},\omega )\left(t\right)\parallel }_{L^2}\to 0,\mathrm{as}t\to \infty ,\end{array}$$

(23)

which shows (3) holds for $s=0$.

Part 2: (3) is valid for any $s>0$.

By taking the $L^2$ inner product of the first three equations in system (2) with ${\mathsf{\Lambda }}^{2s}{u}_1$, ${\mathsf{\Lambda }}^{2s}{u}_2$ and ${\mathsf{\Lambda }}^{2s}\omega$, respectively, summing the results and integrating by parts, we yield

$$\begin{array}{cc}\hfill {\displaystyle \frac{1}{2}}{\displaystyle \frac{d}{dt}}& {\parallel ({\mathsf{\Lambda }}^s\mathit{u},{\mathsf{\Lambda }}^s\omega )\left(t\right)\parallel }_{L^2}^2+(\nu +{\displaystyle \frac{\kappa }{2}})\left(\parallel {\mathsf{\Lambda }}^s{\partial }_y{u}_1{\parallel }_{L^2}^2+{\parallel {\mathsf{\Lambda }}^s{\partial }_x{u}_2\parallel }_{L^2}^2\right)+\gamma {\parallel {\mathsf{\Lambda }}^{s+1}\omega \parallel }_{L^2}^2\hfill \\ \hfill & +2\kappa \parallel {\mathsf{\Lambda }}^s{\omega \parallel }_{L^2}^2=\kappa {\int }_{{\mathbb{R}}^2}\left({\partial }_y\omega ·{\mathsf{\Lambda }}^{2s}{u}_1-{\partial }_x\omega ·{\mathsf{\Lambda }}^{2s}{u}_2\right)dxdy\hfill \\ \hfill & +\kappa {\int }_{{\mathbb{R}}^2}({\partial }_x{u}_2-{\partial }_y{u}_1)·{\mathsf{\Lambda }}^{2s}\omega dxdy\hfill \\ \hfill & -{\int }_{{\mathbb{R}}^2}(\mathit{u}·\nabla \mathit{u})·{\mathsf{\Lambda }}^{2s}\mathit{u}dxdy-{\int }_{{\mathbb{R}}^2}(\mathit{u}·\nabla \omega )·{\mathsf{\Lambda }}^{2s}\omega dxdy\hfill \\ \hfill & =2\kappa {\int }_{{\mathbb{R}}^2}({\mathsf{\Lambda }}^s{\partial }_x{u}_2-{\mathsf{\Lambda }}^s{\partial }_y{u}_1)·{\mathsf{\Lambda }}^s\omega dxdy\hfill \\ \hfill & -{\int }_{{\mathbb{R}}^2}{\mathsf{\Lambda }}^s(\mathit{u}·\nabla \mathit{u})·{\mathsf{\Lambda }}^s\mathit{u}dxdy-{\int }_{{\mathbb{R}}^2}{\mathsf{\Lambda }}^s(\mathit{u}·\nabla \omega )·{\mathsf{\Lambda }}^s\omega dxdy\hfill \\ \hfill & :=L_1+L_2+L_3.\hfill \end{array}$$

(24)

For $L_1$, applying H$\ddot{o}$lder and Young inequalities, we have

$$\begin{array}{c}\hfill L_1\le \kappa \left(\parallel {\mathsf{\Lambda }}^s{\partial }_y{u}_1{\parallel }_{L^2}^2+{\parallel {\mathsf{\Lambda }}^s{\partial }_x{u}_2\parallel }_{L^2}^2\right)+2\kappa {\parallel {\mathsf{\Lambda }}^s\omega \parallel }_{L^2}^2.\end{array}$$

(25)

Due to $\nabla ·\mathit{u}=0$, by virtue of H$\ddot{o}$lder inequality combined with commutator estimate (11), $L_2$ can be bounded as

$$\begin{array}{cc}\hfill L_2& =-{\int }_{{\mathbb{R}}^2}{\mathsf{\Lambda }}^s\nabla ·(\mathit{u}\otimes \mathit{u})·{\mathsf{\Lambda }}^s\mathit{u}dxdy\hfill \\ \hfill & ={\int }_{{\mathbb{R}}^2}{\mathsf{\Lambda }}^s(\mathit{u}\otimes \mathit{u})·{\mathsf{\Lambda }}^{s+1}\mathit{u}dxdy\hfill \\ \hfill & \le {C\parallel \mathit{u}\parallel }_{L^{\infty }}\parallel {\mathsf{\Lambda }}^s{\mathit{u}\parallel }_{L^2}{\parallel {\mathsf{\Lambda }}^{s+1}\mathit{u}\parallel }_{L^2}\hfill \\ \hfill & \le {C\parallel \mathit{u}\parallel }_{L^2}{\parallel {\mathsf{\Lambda }}^{s+1}\mathit{u}\parallel }_{L^2}^2,\hfill \end{array}$$

where the last inequality has used the following Gagliardo–Nirenberg inequalities:

$$\begin{array}{c}\hfill {\parallel \mathit{u}\parallel }_{L^{\infty }}\le {C\parallel \mathit{u}\parallel }_{L^2}^{{\displaystyle \frac{s}{s+1}}}{\parallel {\mathsf{\Lambda }}^{s+1}\mathit{u}\parallel }_{L^2}^{{\displaystyle \frac{1}{s+1}}},\end{array}$$

and

$$\begin{array}{c}\hfill {\parallel {\mathsf{\Lambda }}^s\mathit{u}\parallel }_{L^2}\le {C\parallel \mathit{u}\parallel }_{L^2}^{{\displaystyle \frac{1}{s+1}}}{\parallel {\mathsf{\Lambda }}^{s+1}\mathit{u}\parallel }_{L^2}^{{\displaystyle \frac{s}{s+1}}}.\end{array}$$

(26)

Similarly,

$$\begin{array}{cc}\hfill L_3& \le C\left({\parallel \mathit{u}\parallel }_{L^{\infty }}\parallel {\mathsf{\Lambda }}^s{\omega \parallel }_{L^2}+{\parallel \omega \parallel }_{L^{\infty }}{\parallel {\mathsf{\Lambda }}^s\mathit{u}\parallel }_{L^2}\right){\parallel {\mathsf{\Lambda }}^{s+1}\omega \parallel }_{L^2}\hfill \\ \hfill & \le C\left({\parallel \mathit{u}\parallel }_{L^2}+{\parallel \omega \parallel }_{L^2}\right)\left(\parallel {\mathsf{\Lambda }}^{s+1}{\mathit{u}\parallel }_{L^2}^2+{\parallel {\mathsf{\Lambda }}^{s+1}\omega \parallel }_{L^2}^2\right).\hfill \end{array}$$

In addition, by taking advantage of (10) for any $\alpha =s>0$, we obtain

$$\begin{array}{c}\hfill {\parallel {\mathsf{\Lambda }}^s{\partial }_y{u}_1\parallel }_{L^2}^2+\parallel {\mathsf{\Lambda }}^s{\partial }_x{u}_2{\parallel }_{L^2}^2\ge {\displaystyle \frac{1}{2}}{\parallel {\mathsf{\Lambda }}^{s+1}\mathit{u}\parallel }_{L^2}^2.\end{array}$$

(27)

Therefore, by inserting the above upper bounds of $L_1-L_3$ and (27) into (24), we have

$$\begin{array}{c}\hfill {\displaystyle \frac{d}{dt}}\parallel ({\mathsf{\Lambda }}^s\mathit{u},{\mathsf{\Lambda }}^s\omega ){\parallel }_{L^2}^2+\delta \parallel ({\mathsf{\Lambda }}^{s+1}\mathit{u},{\mathsf{\Lambda }}^{s+1}\omega ){\parallel }_{L^2}^2\le C\parallel (\mathit{u},\omega ){\parallel }_{L^2}{\parallel ({\mathsf{\Lambda }}^{s+1}\mathit{u},{\mathsf{\Lambda }}^{s+1}\omega )\parallel }_{L^2}^2.\end{array}$$

(28)

It follows from (23) that there exists a time $T_0>0$ such that ${\parallel (\mathit{u},\omega )\left(t\right)\parallel }_{L^2}\le {\displaystyle \frac{\delta }{2C}}$ for $t\ge T_0$. Thus, for any $t\ge T_0$, (28) becomes

$${\displaystyle \frac{d}{dt}}\parallel ({\mathsf{\Lambda }}^s\mathit{u},{\mathsf{\Lambda }}^s\omega ){\parallel }_{L^2}^2+{\displaystyle \frac{\delta }{2}}{\parallel ({\mathsf{\Lambda }}^{s+1}\mathit{u},{\mathsf{\Lambda }}^{s+1}\omega )\parallel }_{L^2}^2\le 0.$$

Taking (26) into account, we have

$${\displaystyle \frac{d}{dt}}\parallel ({\mathsf{\Lambda }}^s\mathit{u},{\mathsf{\Lambda }}^s\omega )\left(t\right){\parallel }_{L^2}^2+C\parallel (\mathit{u},\omega )\left(t\right){\parallel }_{L^2}^{-\frac{2}{s}}{\parallel ({\mathsf{\Lambda }}^s\mathit{u},{\mathsf{\Lambda }}^s\omega )\left(t\right)\parallel }_{L^2}^{\frac{2(s+1)}{s}}\le 0.$$

By integrating this over time interval $[T_0,t]$, we find that

$$\begin{array}{c}\hfill {\parallel ({\mathsf{\Lambda }}^s\mathit{u},{\mathsf{\Lambda }}^s\omega )\left(t\right)\parallel }_{L^2}^2\le {\left({\displaystyle \frac{1}{{\parallel ({\mathsf{\Lambda }}^s\mathit{u},{\mathsf{\Lambda }}^s\omega )\left(T_0\right)\parallel }_{L^2}^{-\frac{2}{s}}+C{\int }_{T_0}^t{\parallel (\mathit{u},\omega )\left(\tau \right)\parallel }_{L^2}^{-\frac{2}{s}}d\tau }}\right)}^s.\end{array}$$

Multiplying this by ${(1+t)}^s$, together with the Robita law and (23), we obtain

$$\begin{array}{cc}\hfill {(1+t)}^s{\parallel ({\mathsf{\Lambda }}^s\mathit{u},{\mathsf{\Lambda }}^s\omega )\left(t\right)\parallel }_{L^2}^2& \le {\left({\displaystyle \frac{1+t}{{\parallel ({\mathsf{\Lambda }}^s\mathit{u},{\mathsf{\Lambda }}^s\omega )\left(T_0\right)\parallel }_{L^2}^{-\frac{2}{s}}+C{\int }_{T_0}^t{\parallel (\mathit{u},\omega )\left(\tau \right)\parallel }_{L^2}^{-\frac{2}{s}}d\tau }}\right)}^s\hfill \\ \hfill & \to 0,ast\to \infty ,\hfill \end{array}$$

which shows (3) holds for any $s>0$.

Part 3: Proof of (5) for $m=0$.

Since $({\mathit{u}}_0,{\omega }_0)\in {\dot{H}}^{-\sigma }$, namely,

$$\begin{array}{c}\hfill {\parallel ({\mathsf{\Lambda }}^{-\sigma }{\mathit{u}}_0,{\mathsf{\Lambda }}^{-\sigma }{\omega }_0)\parallel }_{L^2}^2={\int }_{{\mathbb{R}}^2}{\left|\xi \right|}^{-2\sigma }\left(|{\widehat{\mathit{u}}}_0{|}^2+{\left|{\widehat{\omega }}_0\right|}^2\right)d\xi <\infty .\end{array}$$

(29)

Then we deduce from (22) that, for $k>3$,

$$\begin{array}{cc}\hfill {\parallel (\mathit{u},\omega )\left(t\right)\parallel }_{L^2}^2& \le C{log}^{-2}(e+t)+C{log}^{-k}(e+t)\hfill \\ \hfill & +C\parallel ({\mathsf{\Lambda }}^{-\sigma }{\mathit{u}}_0,{\mathsf{\Lambda }}^{-\sigma }{\omega }_0){\parallel }_{L^2}^2{log}^{-k}(e+t){\int }_0^t{\displaystyle \frac{{log}^{k-1-2\sigma }(e+\tau )}{e+\tau }}d\tau \hfill \\ \hfill & \le C{log}^{-2\sigma }(e+t).\hfill \end{array}$$

(30)

This decay estimate can be further improved. In fact, based on (29) and H$\ddot{o}$lder inequality, we further derive from (21) that

$$\begin{array}{cc}\hfill & {\displaystyle \frac{d}{dt}}{\parallel (\mathit{u},\omega )\parallel }_{L^2}^2+\delta {\rho }^2\left(t\right){\parallel (\mathit{u},\omega )\parallel }_{L^2}^2\hfill \\ \hfill & \le \delta {\left[\rho \left(t\right)\right]}^{2+2\sigma }\parallel ({\mathsf{\Lambda }}^{-\sigma }{\mathit{u}}_0,{\mathsf{\Lambda }}^{-\sigma }{\omega }_0){\parallel }_{L^2}^2+C\delta {\rho }^4\left(t\right)·t{\int }_0^t{\parallel (\mathit{u},\omega )\left(\tau \right)\parallel }_{L^2}^{4}d\tau .\hfill \end{array}$$

(31)

Choosing

$$\rho \left(t\right)={\left({\displaystyle \frac{k}{\delta (1+t)}}\right)}^{{\displaystyle \frac{1}{2}}}$$

in (31), and multiplying the result by ${(1+t)}^k$, we obtain

$$\begin{array}{cc}\hfill {\displaystyle \frac{d}{dt}}\left({(1+t)}^k{\parallel (\mathit{u},\omega )\left(t\right)\parallel }_{L^2}^2\right)& \le C{(1+t)}^{k-1-\sigma }{\parallel ({\mathsf{\Lambda }}^{-\sigma }{\mathit{u}}_0,{\mathsf{\Lambda }}^{-\sigma }{\omega }_0)\parallel }_{L^2}^2\hfill \\ \hfill & +C{(1+t)}^{k-2}{\int }_0^t{\parallel (\mathit{u},\omega )\left(\tau \right)\parallel }_{L^2}^{4}d\tau .\hfill \end{array}$$

Integrating this over $(0,t)$ and combining with (30), we derive

$$\begin{array}{cc}\hfill & {(1+t)}^{\sigma }{\parallel (\mathit{u},\omega )\left(t\right)\parallel }_{L^2}^2\hfill \\ \hfill & \le C+C{(1+t)}^{\sigma -k}+C{(1+t)}^{\sigma -1}{\int }_0^t{log}^{-2\sigma }(e+\tau ){\parallel (\mathit{u},\omega )\left(\tau \right)\parallel }_{L^2}^{2}d\tau \hfill \\ \hfill & \le C+C{(1+t)}^{\sigma -1}{\int }_0^t{(1+\tau )}^{-\sigma }{log}^{-2\sigma }(e+\tau ){(1+\tau )}^{\sigma }{\parallel (\mathit{u},\omega )\left(\tau \right)\parallel }_{L^2}^{2}d\tau .\hfill \end{array}$$

Denoting

$$f\left(t\right)={(1+t)}^{\sigma }{\parallel (\mathit{u},\omega )\left(t\right)\parallel }_{L^2}^2,g\left(\tau \right)={(1+t)}^{\sigma -1}{(1+\tau )}^{-\sigma }{log}^{-2\sigma }(e+\tau ),$$

which satisfy

$$\begin{array}{c}\hfill f\left(t\right)\le C+C{\int }_0^{t}g\left(\tau \right)f\left(\tau \right)d\tau .\end{array}$$

(32)

By H$\ddot{o}$lder inequality, it is easy to check that

$$\begin{array}{cc}\hfill {\int }_0^{t}g\left(\tau \right)d\tau & ={(1+t)}^{\sigma -1}{\int }_0^t{(1+\tau )}^{-\sigma }{log}^{-2\sigma }(e+\tau )d\tau \hfill \\ \hfill & \le {(1+t)}^{\sigma -1}{t}^{1-\sigma }{\left({\int }_0^{\infty }{(1+\tau )}^{-1}{log}^{-2}(e+\tau )d\tau \right)}^{\sigma }\hfill \\ \hfill & <\infty .\hfill \end{array}$$

Applying Gronwall inequality to (32), we get

$$f\left(t\right)\le C[1+e^{C{\int }_0^{t}g\left(\tau \right)d\tau }{\int }_0^{t}g\left(\tau \right)d\tau ]\le C,$$

which means

$$\begin{array}{c}\hfill {\parallel (\mathit{u},\omega )\left(t\right)\parallel }_{L^2}\le C{(1+t)}^{-{\displaystyle \frac{\sigma }{2}}}.\end{array}$$

(33)

Therefore, the desired decay estimate of (5) for $m=0$ is obtained.

Part 4: Proof of (4) and (5) for $0<m\le s$.

Once the case $m=s$ is established, the general case $0<m<s$ immediately follows by (33), (36) and the following interpolation inequality

$$\begin{array}{c}\hfill {\parallel ({\mathsf{\Lambda }}^m\mathit{u},{\mathsf{\Lambda }}^m\omega )\parallel }_{L^2}\le C{\parallel (\mathit{u},\omega )\parallel }_{L^2}^{1-{\displaystyle \frac{m}{s}}}{\parallel ({\mathsf{\Lambda }}^s\mathit{u},{\mathsf{\Lambda }}^s\omega )\parallel }_{L^2}^{\frac{m}{s}}.\end{array}$$

(34)

To this end, we reevaluate $L_2-L_3$ shown in the ${\dot{H}}^s$-estimates (24) as follows.

By utilizing the commutator estimate (12) and the H$\ddot{o}$lder and Young inequalities, we have

$$\begin{array}{cc}\hfill L_2& =-{\int }_{{\mathbb{R}}^2}[{\mathsf{\Lambda }}^s,\mathit{u}·\nabla ]\mathit{u}·{\mathsf{\Lambda }}^s\mathit{u}dxdy\hfill \\ \hfill & \le {C\parallel \nabla \mathit{u}\parallel }_{L^2}{\parallel {\mathsf{\Lambda }}^s\mathit{u}\parallel }_{L^4}^2\hfill \\ \hfill & \le {C\parallel \mathsf{\Lambda }\mathit{u}\parallel }_{L^2}\parallel {\mathsf{\Lambda }}^s{\mathit{u}\parallel }_{L^2}{\parallel {\mathsf{\Lambda }}^{s+1}\mathit{u}\parallel }_{L^2}\hfill \\ \hfill & \le {\displaystyle \frac{\delta }{4}}\parallel {\mathsf{\Lambda }}^{s+1}{\mathit{u}\parallel }_{L^2}^2+{C\parallel \mathsf{\Lambda }\mathit{u}\parallel }_{L^2}^2{\parallel {\mathsf{\Lambda }}^s\mathit{u}\parallel }_{L^2}^2,\hfill \end{array}$$

where we have used the Gagliardo–Nirenberg inequality

$${\parallel {\mathsf{\Lambda }}^s\mathit{u}\parallel }_{L^4}\le C\parallel {\mathsf{\Lambda }}^s{\mathit{u}\parallel }_{L^2}^{\frac{1}{2}}{\parallel {\mathsf{\Lambda }}^{s+1}\mathit{u}\parallel }_{L^2}^{\frac{1}{2}}.$$

Similarly,

$$\begin{array}{cc}\hfill L_3& =-{\int }_{{\mathbb{R}}^2}[{\mathsf{\Lambda }}^s,\mathit{u}·\nabla ]\omega ·{\mathsf{\Lambda }}^s\omega dxdy\hfill \\ \hfill & \le {C\parallel \nabla \mathit{u}\parallel }_{L^2}\parallel {\mathsf{\Lambda }}^s{\omega \parallel }_{L^4}^2+{C\parallel \nabla \omega \parallel }_{L^2}\parallel {\mathsf{\Lambda }}^s{\mathit{u}\parallel }_{L^4}{\parallel {\mathsf{\Lambda }}^s\omega \parallel }_{L^4}\hfill \\ \hfill & \le {C\parallel (\mathsf{\Lambda }\mathit{u},\mathsf{\Lambda }\omega )\parallel }_{L^2}(\parallel {\mathsf{\Lambda }}^s{\mathit{u}\parallel }_{L^2}\parallel {\mathsf{\Lambda }}^{s+1}{\mathit{u}\parallel }_{L^2}+\parallel {\mathsf{\Lambda }}^s{\omega \parallel }_{L^2}\parallel {\mathsf{\Lambda }}^{s+1}\omega {\parallel }_{L^2})\hfill \\ \hfill & \le {\displaystyle \frac{\delta }{4}}\parallel {\mathsf{\Lambda }}^{s+1}{\mathit{u}\parallel }_{L^2}^2+{\displaystyle \frac{\delta }{2}}\parallel {\mathsf{\Lambda }}^{s+1}{\omega \parallel }_{L^2}^2+{C\parallel (\mathsf{\Lambda }\mathit{u},\mathsf{\Lambda }\omega )\parallel }_{L^2}^2{\parallel ({\mathsf{\Lambda }}^s\mathit{u},{\mathsf{\Lambda }}^s\omega )\parallel }_{L^2}^2.\hfill \end{array}$$

Plugging above new bounds into (24), together with (25) and (27), one has

$$\begin{array}{c}\hfill {\displaystyle \frac{d}{dt}}\parallel ({\mathsf{\Lambda }}^s\mathit{u},{\mathsf{\Lambda }}^s\omega ){\parallel }_{L^2}^2+{\displaystyle \frac{\delta }{2}}\parallel ({\mathsf{\Lambda }}^{s+1}\mathit{u},{\mathsf{\Lambda }}^{s+1}\omega ){\parallel }_{L^2}^2\le C\parallel (\mathsf{\Lambda }\mathit{u},\mathsf{\Lambda }\omega ){\parallel }_{L^2}^2{\parallel ({\mathsf{\Lambda }}^s\mathit{u},{\mathsf{\Lambda }}^s\omega )\parallel }_{L^2}^2.\end{array}$$

(35)

Combining this with (13), by Gronwall inequality and (14), we thus obtain (4).

To complete this part, it remains to show

$$\begin{array}{c}\hfill {\parallel ({\mathsf{\Lambda }}^s\mathit{u},{\mathsf{\Lambda }}^s\omega )\left(t\right)\parallel }_{L^2}^2\le C{(1+t)}^{-(s+\sigma )}\end{array}$$

(36)

for any nonnegative integer s, which can be achieved by the induction method. Obviously, (33) implies that (36) holds for $s=0$. Assume that for any integer $s\ge 1$, we have

$$\begin{array}{c}\hfill {\parallel ({\mathsf{\Lambda }}^{s-1}\mathit{u},{\mathsf{\Lambda }}^{s-1}\omega )\left(t\right)\parallel }_{L^2}^2\le C{(1+t)}^{-(s-1+\sigma )}\end{array}$$

(37)

For any $0\le t_0<t$, multiplying (35) by $t-t_0$, we get

$$\begin{array}{cc}\hfill {\displaystyle \frac{d}{dt}}& \left((t-t_0){\parallel ({\mathsf{\Lambda }}^s\mathit{u},{\mathsf{\Lambda }}^s\omega )\parallel }_{L^2}^2\right)+{\displaystyle \frac{\delta }{2}}(t-t_0){\parallel ({\mathsf{\Lambda }}^{s+1}\mathit{u},{\mathsf{\Lambda }}^{s+1}\omega )\parallel }_{L^2}^2\hfill \\ \hfill & \le \parallel ({\mathsf{\Lambda }}^s\mathit{u},{\mathsf{\Lambda }}^s\omega ){\parallel }_{L^2}^2+C(t-t_0)\parallel (\mathsf{\Lambda }\mathit{u},\mathsf{\Lambda }\omega ){\parallel }_{L^2}^2{\parallel ({\mathsf{\Lambda }}^s\mathit{u},{\mathsf{\Lambda }}^s\omega )\parallel }_{L^2}^2.\hfill \end{array}$$

Applying Gronwall inequality, together with (14), we obtain

$$\begin{array}{cc}\hfill (t-t_0){\parallel ({\mathsf{\Lambda }}^s\mathit{u},{\mathsf{\Lambda }}^s\omega )\left(t\right)\parallel }_{L^2}^2& +{\displaystyle \frac{\delta }{2}}{\int }_{t_0}^t(\tau -t_0){\parallel ({\mathsf{\Lambda }}^{s+1}\mathit{u},{\mathsf{\Lambda }}^{s+1}\omega )\left(\tau \right)\parallel }_{L^2}^{2}d\tau \hfill \\ \hfill & \le e^{C{\int }_{t_0}^t{\parallel (\mathsf{\Lambda }\mathit{u},\mathsf{\Lambda }\omega )\left(\tau \right)\parallel }_{L^2}^{2}d\tau }{\int }_{t_0}^t{\parallel ({\mathsf{\Lambda }}^s\mathit{u},{\mathsf{\Lambda }}^s\omega )\left(\tau \right)\parallel }_{L^2}^{2}d\tau \hfill \\ \hfill & \le C{\int }_{t_0}^t{\parallel ({\mathsf{\Lambda }}^s\mathit{u},{\mathsf{\Lambda }}^s\omega )\left(\tau \right)\parallel }_{L^2}^{2}d\tau .\hfill \end{array}$$

(38)

On the other hand, we know from (35) that

$$\begin{array}{c}\hfill {\displaystyle \frac{d}{dt}}\parallel ({\mathsf{\Lambda }}^{s-1}\mathit{u},{\mathsf{\Lambda }}^{s-1}\omega ){\parallel }_{L^2}^2+{\displaystyle \frac{\delta }{2}}\parallel ({\mathsf{\Lambda }}^s\mathit{u},{\mathsf{\Lambda }}^s\omega ){\parallel }_{L^2}^2\le C\parallel (\mathsf{\Lambda }\mathit{u},\mathsf{\Lambda }\omega ){\parallel }_{L^2}^2{\parallel ({\mathsf{\Lambda }}^{s-1}\mathit{u},{\mathsf{\Lambda }}^{s-1}\omega )\parallel }_{L^2}^2.\end{array}$$

Applying Gronwall inequality again gives

$$\begin{array}{cc}\hfill \parallel ({\mathsf{\Lambda }}^{s-1}\mathit{u},{\mathsf{\Lambda }}^{s-1}\omega )\left(t\right){\parallel }_{L^2}^2& +{\displaystyle \frac{\delta }{2}}{\int }_{t_0}^t{\parallel ({\mathsf{\Lambda }}^s\mathit{u},{\mathsf{\Lambda }}^s\omega )\left(\tau \right)\parallel }_{L^2}^{2}d\tau \hfill \\ \hfill & \le e^{C{\int }_{t_0}^t{\parallel (\mathsf{\Lambda }\mathit{u},\mathsf{\Lambda }\omega )\left(\tau \right)\parallel }_{L^2}^{2}d\tau }{\parallel ({\mathsf{\Lambda }}^{s-1}\mathit{u},{\mathsf{\Lambda }}^{s-1}\omega )\left(t_0\right)\parallel }_{L^2}^2.\hfill \end{array}$$

(39)

Choosing $t_0={\displaystyle \frac{t}{2}}$, we obtain from (14) and (37)–(39) that

$$\begin{array}{cc}\hfill \frac{t}{2}{\parallel ({\mathsf{\Lambda }}^s\mathit{u},{\mathsf{\Lambda }}^s\omega )\left(t\right)\parallel }_{L^2}^2& \le C{\int }_{{\displaystyle \frac{t}{2}}}^t{\parallel ({\mathsf{\Lambda }}^s\mathit{u},{\mathsf{\Lambda }}^s\omega )\left(\tau \right)\parallel }_{L^2}^{2}d\tau \hfill \\ \hfill & \le C\parallel ({\mathsf{\Lambda }}^{s-1}\mathit{u},{\mathsf{\Lambda }}^{s-1}\omega )\left(\frac{t}{2}\right){\parallel }_{L^2}^2\hfill \\ \hfill & \le C{(1+\frac{t}{2})}^{-(s-1+\sigma )},\hfill \end{array}$$

which implies (36). Therefore we complete the proof of Theorem 1. □

4. Proof of Theorem 2

Before proving Theorem 2, we first establish the existence of a unique global solution to system (2) with $\gamma =0$, provided that the initial data $({\mathit{u}}_0,{\omega }_0)$ is sufficiently small in $H^s$. To this end, it suffices to derive the global a priori $H^s$-estimates under the condition (6). Therefore, the following proposition on the global a priori ${\dot{H}}^m$-estimates is needed.

Proposition 2.

Let the assumptions of Theorem 2 hold. Then for all $t>0$ and $0\le m\le s$,

$$\begin{array}{c}\hfill {\displaystyle \frac{d}{dt}}\parallel ({\mathsf{\Lambda }}^m\mathit{u},{\mathsf{\Lambda }}^m\omega )\left(t\right){\parallel }_{L^2}^2+{\displaystyle \frac{{\delta }_0}{2}}(\parallel {\mathsf{\Lambda }}^{m+1}{\mathit{u}\parallel }_{L^2}^2+\parallel {\mathsf{\Lambda }}^m\omega {\parallel }_{L^2}^2)\le 0,\end{array}$$

(40)

and

$$\begin{array}{cc}\hfill {\parallel ({\mathsf{\Lambda }}^m\mathit{u},{\mathsf{\Lambda }}^m\omega )\left(t\right)\parallel }_{L^2}^2+{\displaystyle \frac{{\delta }_0}{2}}{\int }_0^t& (\parallel {\mathsf{\Lambda }}^{m+1}\mathit{u}\left(\tau \right){\parallel }_{L^2}^2+\parallel {\mathsf{\Lambda }}^m\omega \left(\tau \right){\parallel }_{L^2}^2)d\tau \hfill \\ \hfill & \le \parallel ({\mathsf{\Lambda }}^m{\mathit{u}}_0,{\mathsf{\Lambda }}^m{\omega }_0){\parallel }_{L^2}^2,\hfill \end{array}$$

(41)

where the positive constant ${\delta }_0:=min\{{\displaystyle \frac{\nu -\kappa }{2}},{\displaystyle \frac{4\nu \kappa }{\nu +2\kappa }}\}$.

Proof. 

Taking the $L^2$ inner product to the first three equations of system (2) with ${\mathsf{\Lambda }}^{2m}{u}_1$, ${\mathsf{\Lambda }}^{2m}{u}_2$ and ${\mathsf{\Lambda }}^{2m}\omega$, respectively, summing them up and after integration by parts, we yield

$$\begin{array}{cc}\hfill {\displaystyle \frac{1}{2}}{\displaystyle \frac{d}{dt}}\parallel ({\mathsf{\Lambda }}^m\mathit{u},& {\mathsf{\Lambda }}^m{\omega )\left(t\right)\parallel }_{L^2}^2+(\nu +{\displaystyle \frac{\kappa }{2}})\left(\parallel {\mathsf{\Lambda }}^m{\partial }_y{u}_1{\parallel }_{L^2}^2+{\parallel {\mathsf{\Lambda }}^m{\partial }_x{u}_2\parallel }_{L^2}^2\right)+2\kappa {\parallel {\mathsf{\Lambda }}^m\omega \parallel }_{L^2}^2\hfill \\ \hfill & =\kappa {\int }_{{\mathbb{R}}^2}\left({\partial }_y\omega ·{\mathsf{\Lambda }}^{2m}{u}_1-{\partial }_x\omega ·{\mathsf{\Lambda }}^{2m}{u}_2\right)dxdy\hfill \\ \hfill & +\kappa {\int }_{{\mathbb{R}}^2}({\partial }_x{u}_2-{\partial }_y{u}_1)·{\mathsf{\Lambda }}^{2m}\omega dxdy\hfill \\ \hfill & -{\int }_{{\mathbb{R}}^2}(\mathit{u}·\nabla \mathit{u})·{\mathsf{\Lambda }}^{2m}\mathit{u}dxdy-{\int }_{{\mathbb{R}}^2}(\mathit{u}·\nabla \omega )·{\mathsf{\Lambda }}^{2m}\omega dxdy\hfill \\ \hfill & =2\kappa {\int }_{{\mathbb{R}}^2}({\mathsf{\Lambda }}^m{\partial }_x{u}_2-{\mathsf{\Lambda }}^m{\partial }_y{u}_1)·{\mathsf{\Lambda }}^m\omega dxdy\hfill \\ \hfill & -{\int }_{{\mathbb{R}}^2}{\mathsf{\Lambda }}^m(\mathit{u}·\nabla \mathit{u})·{\mathsf{\Lambda }}^m\mathit{u}dxdy-{\int }_{{\mathbb{R}}^2}{\mathsf{\Lambda }}^m(\mathit{u}·\nabla \omega )·{\mathsf{\Lambda }}^m\omega dxdy\hfill \\ \hfill & :=M_1+M_2+M_3.\hfill \end{array}$$

(42)

For $M_1$, by H$\ddot{o}$lder and Young inequalities, we have

$$\begin{array}{c}\hfill M_1\le {\displaystyle \frac{\nu +2\kappa }{2}}\left(\parallel {\mathsf{\Lambda }}^m{\partial }_y{u}_1{\parallel }_{L^2}^2+{\parallel {\mathsf{\Lambda }}^m{\partial }_x{u}_2\parallel }_{L^2}^2\right)+{\displaystyle \frac{4{\kappa }^2}{\nu +2\kappa }}{\parallel {\mathsf{\Lambda }}^m\omega \parallel }_{L^2}^2.\end{array}$$

(43)

Thanks to $\nabla ·\mathit{u}=0$, using H$\ddot{o}$lder inequality and commutator estimate (11), $M_2$ can be bounded as

$$\begin{array}{cc}\hfill M_2& =-{\int }_{{\mathbb{R}}^2}{\mathsf{\Lambda }}^m\nabla ·(\mathit{u}\otimes \mathit{u})·{\mathsf{\Lambda }}^m\mathit{u}dxdy\hfill \\ \hfill & ={\int }_{{\mathbb{R}}^2}{\mathsf{\Lambda }}^m(\mathit{u}\otimes \mathit{u})·{\mathsf{\Lambda }}^{m+1}\mathit{u}dxdy\hfill \\ \hfill & \le {C\parallel \mathit{u}\parallel }_{L^6}\parallel {\mathsf{\Lambda }}^m{\mathit{u}\parallel }_{L^3}{\parallel {\mathsf{\Lambda }}^{m+1}\mathit{u}\parallel }_{L^2}\hfill \\ \hfill & \le {C\parallel \mathit{u}\parallel }_{L^2}{\parallel {\mathsf{\Lambda }}^{m+1}\mathit{u}\parallel }_{L^2}^2,\hfill \end{array}$$

where we have used the following two Gagliardo–Nirenberg inequalities:

$${\parallel \mathit{u}\parallel }_{L^6}\le {C\parallel \mathit{u}\parallel }_{L^2}^{1-{\displaystyle \frac{2}{3(m+1)}}}{\parallel {\mathsf{\Lambda }}^{m+1}\mathit{u}\parallel }_{L^2}^{{\displaystyle \frac{2}{3(m+1)}}},$$

$${\parallel {\mathsf{\Lambda }}^m\mathit{u}\parallel }_{L^3}\le {C\parallel \mathit{u}\parallel }_{L^2}^{1-{\displaystyle \frac{3m+1}{3(m+1)}}}{\parallel {\mathsf{\Lambda }}^{m+1}\mathit{u}\parallel }_{L^2}^{{\displaystyle \frac{3m+1}{3(m+1)}}}.$$

Taking advantage of commutator estimate (12) and Sobolev embedding theorem, we estimate $M_3$ as

$$\begin{array}{cc}\hfill M_3& =-{\int }_{{\mathbb{R}}^2}[{\mathsf{\Lambda }}^m,\mathit{u}·\nabla ]\omega ·{\mathsf{\Lambda }}^m\omega dxdy\hfill \\ \hfill & \le C\left({\parallel \nabla \mathit{u}\parallel }_{L^{\infty }}\parallel {\mathsf{\Lambda }}^m{\omega \parallel }_{L^2}+{\parallel \omega \parallel }_{L^{\infty }}{\parallel {\mathsf{\Lambda }}^{m+1}\mathit{u}\parallel }_{L^2}\right){\parallel {\mathsf{\Lambda }}^m\omega \parallel }_{L^2}\hfill \\ \hfill & \le C\left({\parallel \mathit{u}\parallel }_{H^s}\parallel {\mathsf{\Lambda }}^m{\omega \parallel }_{L^2}+{\parallel \omega \parallel }_{H^s}{\parallel {\mathsf{\Lambda }}^{m+1}\mathit{u}\parallel }_{L^2}\right){\parallel {\mathsf{\Lambda }}^m\omega \parallel }_{L^2}.\hfill \end{array}$$

Making use of (10) for any $\alpha =m>0$ and collecting the above bounds on $M_1-M_3$ into (42), we obtain

$$\begin{array}{cc}\hfill {\displaystyle \frac{d}{dt}}\parallel ({\mathsf{\Lambda }}^m\mathit{u},& {\mathsf{\Lambda }}^m{\omega )\left(t\right)\parallel }_{L^2}^2+{\displaystyle \frac{\nu -\kappa }{2}}\parallel {\mathsf{\Lambda }}^{m+1}{\mathit{u}\parallel }_{L^2}^2+{\displaystyle \frac{4\nu \kappa }{\nu +2\kappa }}{\parallel {\mathsf{\Lambda }}^m\omega \parallel }_{L^2}^2\hfill \\ \hfill & \le {C(\parallel \mathit{u}\parallel }_{H^s}+{\parallel \omega \parallel }_{H^s})\left(\parallel {\mathsf{\Lambda }}^{m+1}{\mathit{u}\parallel }_{L^2}^2+{\parallel {\mathsf{\Lambda }}^m\omega \parallel }_{L^2}^2\right)\hfill \end{array}$$

(44)

It follows from (42) and (43) with $m=0$ that

$$\begin{array}{c}\hfill {\displaystyle \frac{d}{dt}}{\parallel (\mathit{u},\omega )\left(t\right)\parallel }_{L^2}^2+{\displaystyle \frac{\nu -\kappa }{2}}{\parallel \mathsf{\Lambda }\mathit{u}\parallel }_{L^2}^2+{\displaystyle \frac{4\nu \kappa }{\nu +2\kappa }}{\parallel \omega \parallel }_{L^2}^2\le 0.\end{array}$$

(45)

Adding (44) and (45) leads to

$$\begin{array}{cc}\hfill {\displaystyle \frac{d}{dt}}{\parallel (\mathit{u},\omega )\left(t\right)\parallel }_{H^m}^2& +{\delta }_0{(\parallel \mathsf{\Lambda }\mathit{u}\parallel }_{H^m}^2+{\parallel \omega \parallel }_{H^m}^2)\hfill \\ \hfill & \le {C(\parallel \mathit{u}\parallel }_{H^s}+{\parallel \omega \parallel }_{H^s})\left({\parallel \mathsf{\Lambda }\mathit{u}\parallel }_{H^m}^2+{\parallel \omega \parallel }_{H^m}^2\right).\hfill \end{array}$$

(46)

Taking the small condition (6) into account, that is, when the initial data $({\mathit{u}}_0,{\omega }_0)$ obey, for $0<\epsilon <{\epsilon }_0={\displaystyle \frac{\sqrt{2}{\delta }_0}{4C}}$,

$$\begin{array}{c}\hfill \parallel {\mathit{u}}_0{\parallel }_{H^s}+{\parallel {\omega }_0\parallel }_{H^s}<\epsilon ,\end{array}$$

then (46) with $m=s$ indicates that the corresponding local solution remains small for all time $t>0$, i.e.,

$$\begin{array}{c}\hfill {\parallel \mathit{u}\left(t\right)\parallel }_{H^s}+{\parallel \omega \left(t\right)\parallel }_{H^s}<\sqrt{2}\epsilon .\end{array}$$

(47)

We illustrate this by contradiction. Indeed, assume that (47) does not hold. Then there exists a time $T^{*}<T$ (the first time) such that

$$\begin{array}{c}\hfill \parallel \mathit{u}\left(T^{*}\right){\parallel }_{H^s}+{\parallel \omega \left(T^{*}\right)\parallel }_{H^s}=\sqrt{2}\epsilon .\end{array}$$

Thus, we have

$$\begin{array}{c}\hfill {\parallel \mathit{u}\left(t\right)\parallel }_{H^s}+{\parallel \omega \left(t\right)\parallel }_{H^s}\le \sqrt{2}\epsilon ,0\le t\le T^{*}.\end{array}$$

This, together with (46) with $m=s$, gives that

$$\begin{array}{c}\hfill {\displaystyle \frac{d}{dt}}{\parallel (\mathit{u},\omega )\left(t\right)\parallel }_{H^s}^2+({\delta }_0-\sqrt{2}C\epsilon ){(\parallel \mathsf{\Lambda }\mathit{u}\parallel }_{H^s}^2+{\parallel \omega \parallel }_{H^s}^2)\le 0,0\le t\le T^{*}.\end{array}$$

Whence we obtain

$$\begin{array}{c}\hfill \parallel \mathit{u}\left(T^{*}\right){\parallel }_{H^s}^2+\parallel \omega \left(T^{*}\right){\parallel }_{H^s}^2\le \parallel {\mathit{u}}_0{\parallel }_{H^s}^2+{\parallel {\omega }_0\parallel }_{H^s}^2<{\epsilon }^2,\end{array}$$

namely,

$$\begin{array}{c}\hfill \parallel \mathit{u}\left(T^{*}\right){\parallel }_{H^s}+{\parallel \omega \left(T^{*}\right)\parallel }_{H^s}<\sqrt{2}\epsilon ,\end{array}$$

which leads to a contradiction. Therefore, we obtain the uniform bound (47). By further applying the bootstrapping argument, the global well-posedness of the small solutions for system (2) with $\gamma =0$ is established. Combining (47) with (44) and (45) yields (40). We then derive (41) by integrating (40) over the time interval $(0,t)$. In addition, substituting (47) into (46) and integrating the resulting inequality with respect to time allows us to obtain (7). We thus complete the proof of Proposition 2. □

Another preparation is the following proposition on the ${\dot{H}}^{-\sigma }$-estimates for the global solution.

Proposition 3.

Let the assumptions of Theorem 2 hold. Then the solution $(\mathit{u},\omega )$ satisfies, for $0<\sigma <1$,

$$\begin{array}{cc}\hfill & {\displaystyle \frac{d}{dt}}{\parallel ({\mathsf{\Lambda }}^{-\sigma }\mathit{u},{\mathsf{\Lambda }}^{-\sigma }\omega )\left(t\right)\parallel }_{L^2}^2+{\delta }_0(\parallel {\mathsf{\Lambda }}^{1-\sigma }{\mathit{u}\parallel }_{L^2}^2+\parallel {\mathsf{\Lambda }}^{-\sigma }\omega {\parallel }_{L^2}^2)\hfill \\ \hfill & \le {C\parallel \mathit{u}\parallel }_{L^2}^{\sigma }{\parallel \mathsf{\Lambda }\mathit{u}\parallel }_{L^2}^{1-\sigma }{\parallel (\mathsf{\Lambda }\mathit{u},\mathsf{\Lambda }\omega )\parallel }_{L^2}{\parallel ({\mathsf{\Lambda }}^{-\sigma }\mathit{u},{\mathsf{\Lambda }}^{-\sigma }\omega )\parallel }_{L^2}.\hfill \end{array}$$

(48)

Proof. 

Applying ${\mathsf{\Lambda }}^{-\sigma }$ to the first three equations of system (2), respectively, taking the $L^2$-inner product with $({\mathsf{\Lambda }}^{-\sigma }{u}_1,{\mathsf{\Lambda }}^{-\sigma }{u}_2,{\mathsf{\Lambda }}^{-\sigma }\omega )$, and after integration by parts, we get

$$\begin{array}{cc}\hfill {\displaystyle \frac{1}{2}}{\displaystyle \frac{d}{dt}}& {\parallel ({\mathsf{\Lambda }}^{-\sigma }\mathit{u},{\mathsf{\Lambda }}^{-\sigma }\omega )\left(t\right)\parallel }_{L^2}^2+(\nu +{\displaystyle \frac{\kappa }{2}})\left(\parallel {\mathsf{\Lambda }}^{-\sigma }{\partial }_y{u}_1{\parallel }_{L^2}^2+{\parallel {\mathsf{\Lambda }}^{-\sigma }{\partial }_x{u}_2\parallel }_{L^2}^2\right)\hfill \\ \hfill & +2\kappa \parallel {\mathsf{\Lambda }}^{-\sigma }{\omega \parallel }_{L^2}^2=\kappa {\int }_{{\mathbb{R}}^2}\left({\mathsf{\Lambda }}^{-\sigma }{\partial }_y\omega ·{\mathsf{\Lambda }}^{-\sigma }{u}_1-{\mathsf{\Lambda }}^{-\sigma }{\partial }_x\omega ·{\mathsf{\Lambda }}^{-\sigma }{u}_2\right)dxdy\hfill \\ \hfill & +\kappa {\int }_{{\mathbb{R}}^2}{\mathsf{\Lambda }}^{-\sigma }({\partial }_x{u}_2-{\partial }_y{u}_1)·{\mathsf{\Lambda }}^{-\sigma }\omega dxdy\hfill \\ \hfill & -{\int }_{{\mathbb{R}}^2}{\mathsf{\Lambda }}^{-\sigma }(\mathit{u}·\nabla \mathit{u})·{\mathsf{\Lambda }}^{-\sigma }\mathit{u}dxdy-{\int }_{{\mathbb{R}}^2}{\mathsf{\Lambda }}^{-\sigma }(\mathit{u}·\nabla \omega )·{\mathsf{\Lambda }}^{-\sigma }\omega dxdy\hfill \\ \hfill & =2\kappa {\int }_{{\mathbb{R}}^2}({\mathsf{\Lambda }}^{-\sigma }{\partial }_x{u}_2-{\mathsf{\Lambda }}^{-\sigma }{\partial }_y{u}_1)·{\mathsf{\Lambda }}^{-\sigma }\omega dxdy\hfill \\ \hfill & -{\int }_{{\mathbb{R}}^2}{\mathsf{\Lambda }}^{-\sigma }(\mathit{u}·\nabla \mathit{u})·{\mathsf{\Lambda }}^{-\sigma }\mathit{u}dxdy-{\int }_{{\mathbb{R}}^2}{\mathsf{\Lambda }}^{-\sigma }(\mathit{u}·\nabla \omega )·{\mathsf{\Lambda }}^{-\sigma }\omega dxdy\hfill \\ \hfill & =:N_1+N_2+N_3.\hfill \end{array}$$

(49)

Similar to the estimate of $M_1$, we have

$$\begin{array}{c}\hfill N_1\le {\displaystyle \frac{\nu +2\kappa }{2}}\left(\parallel {\mathsf{\Lambda }}^{-\sigma }{\partial }_y{u}_1{\parallel }_{L^2}^2+{\parallel {\mathsf{\Lambda }}^{-\sigma }{\partial }_x{u}_2\parallel }_{L^2}^2\right)+{\displaystyle \frac{4{\kappa }^2}{\nu +2\kappa }}{\parallel {\mathsf{\Lambda }}^{-\sigma }\omega \parallel }_{L^2}^2.\end{array}$$

(50)

By employing H$\ddot{o}$lder inequality to the nonlinear term $N_2$, we obtain

$$\begin{array}{cc}\hfill N_2& \le \parallel {\mathsf{\Lambda }}^{-\sigma }{(\mathit{u}·\nabla \mathit{u})\parallel }_{L^2}{\parallel {\mathsf{\Lambda }}^{-\sigma }\mathit{u}\parallel }_{L^2}\hfill \\ \hfill & \le {C\parallel \mathit{u}·\nabla \mathit{u}\parallel }_{L^{{\displaystyle \frac{2}{1+\sigma }}}}{\parallel {\mathsf{\Lambda }}^{-\sigma }\mathit{u}\parallel }_{L^2}\hfill \\ \hfill & \le {C\parallel \mathit{u}\parallel }_{L^{{\displaystyle \frac{2}{\sigma }}}}{\parallel \nabla \mathit{u}\parallel }_{L^2}{\parallel {\mathsf{\Lambda }}^{-\sigma }\mathit{u}\parallel }_{L^2}\hfill \\ \hfill & \le {C\parallel \mathit{u}\parallel }_{L^2}^{\sigma }{\parallel \mathsf{\Lambda }\mathit{u}\parallel }_{L^2}^{2-\sigma }{\parallel {\mathsf{\Lambda }}^{-\sigma }\mathit{u}\parallel }_{L^2},\hfill \end{array}$$

(51)

where we have used the following Sobolev embedding theorem

$$\begin{array}{c}\hfill L^p\left({\mathbb{R}}^2\right)\hookrightarrow {\dot{W}}^{-\sigma ,q}\left({\mathbb{R}}^2\right),\mathrm{if}{\displaystyle \frac{1}{q}}={\displaystyle \frac{1}{p}}-{\displaystyle \frac{\sigma }{2}},\end{array}$$

and the following Gagliardo–Nirenberg inequality

$$\begin{array}{c}\hfill {\parallel \mathit{u}\parallel }_{L^{{\displaystyle \frac{2}{\sigma }}}}\le {C\parallel \mathit{u}\parallel }_{L^2}^{\sigma }{\parallel \mathsf{\Lambda }\mathit{u}\parallel }_{L^2}^{1-\sigma }.\end{array}$$

In a similar manner, we have

$$\begin{array}{cc}\hfill N_3& \le \parallel {\mathsf{\Lambda }}^{-\sigma }{(\mathit{u}·\nabla \omega )\parallel }_{L^2}{\parallel {\mathsf{\Lambda }}^{-\sigma }\omega \parallel }_{L^2}\hfill \\ \hfill & \le {C\parallel \mathit{u}·\nabla \omega \parallel }_{L^{{\displaystyle \frac{2}{1+\sigma }}}}{\parallel {\mathsf{\Lambda }}^{-\sigma }\omega \parallel }_{L^2}\hfill \\ \hfill & \le {C\parallel \mathit{u}\parallel }_{L^{{\displaystyle \frac{2}{\sigma }}}}{\parallel \nabla \omega \parallel }_{L^2}{\parallel {\mathsf{\Lambda }}^{-\sigma }\omega \parallel }_{L^2}\hfill \\ \hfill & \le {C\parallel \mathit{u}\parallel }_{L^2}^{\sigma }{\parallel \mathsf{\Lambda }\mathit{u}\parallel }_{L^2}^{1-\sigma }{\parallel \mathsf{\Lambda }\omega \parallel }_{L^2}{\parallel {\mathsf{\Lambda }}^{-\sigma }\omega \parallel }_{L^2}.\hfill \end{array}$$

(52)

Inserting (50)–(52) into (49), together with (10) for $\alpha =-\sigma$, one yields (48) immediately. □

Now, we give the proof of Theorem 2.

Proof of Theorem 2.

Let

$$\begin{array}{c}\hfill C_0:=\parallel {\mathsf{\Lambda }}^{-\sigma }{\mathit{u}}_0{\parallel }_{L^2}^2+{\parallel {\mathsf{\Lambda }}^{-\sigma }{\omega }_0\parallel }_{L^2}^2.\end{array}$$

(53)

We make the ansatz that

$$\begin{array}{c}\hfill {\parallel {\mathsf{\Lambda }}^{-\sigma }\mathit{u}\left(t\right)\parallel }_{L^2}^2+{\parallel {\mathsf{\Lambda }}^{-\sigma }\omega \left(t\right)\parallel }_{L^2}^2\le 2C_0,\mathrm{for}\mathrm{all}t\in [0,T].\end{array}$$

(54)

If, for all $t\in [0,T]$,

$$\begin{array}{c}\hfill {\parallel {\mathsf{\Lambda }}^{-\sigma }\mathit{u}\left(t\right)\parallel }_{L^2}^2+{\parallel {\mathsf{\Lambda }}^{-\sigma }\omega \left(t\right)\parallel }_{L^2}^2\le {\displaystyle \frac{3C_0}{2}}\end{array}$$

(55)

holds true. Then, by employing the bootstapping argument, we obtain $T=\infty$, which indicates that the solution exists for all positive time, i.e., the system is globally well-posed. Thus, (55) holds for all $0\le t<\infty$.

Therefore, we shall show the improved inequality (55) in the rest of the proof. On the one hand, thanks to the ansatz (54) and the following interpolation inequality

$$\begin{array}{c}\hfill {\parallel {\mathsf{\Lambda }}^s\mathit{u}\parallel }_{L^2}\le C\parallel {\mathsf{\Lambda }}^{-\sigma }{\mathit{u}\parallel }_{L^2}^{{\displaystyle \frac{1}{s+\sigma +1}}}{\parallel {\mathsf{\Lambda }}^{s+1}\mathit{u}\parallel }_{L^2}^{{\displaystyle \frac{s+\sigma }{s+\sigma +1}}},\end{array}$$

(56)

we get

$$\begin{array}{c}\hfill {\parallel {\mathsf{\Lambda }}^{s+1}\mathit{u}\parallel }_{L^2}\ge C{\parallel {\mathsf{\Lambda }}^s\mathit{u}\parallel }_{L^2}^{{\displaystyle \frac{s+\sigma +1}{s+\sigma }}}.\end{array}$$

Due to the assumption (6), without loss of generality, we may assume that $\parallel {\mathsf{\Lambda }}^s{\omega \parallel }_{L^2}^2\le 1$. Inserting the above inequality into (40) with $m=s$, we have

$$\begin{array}{c}\hfill {\displaystyle \frac{d}{dt}}\parallel ({\mathsf{\Lambda }}^s\mathit{u},{\mathsf{\Lambda }}^s\omega )\left(t\right){\parallel }_{L^2}^2+C(\parallel {\mathsf{\Lambda }}^s{\mathit{u}\parallel }_{L^2}^2+\parallel {\mathsf{\Lambda }}^s\omega {{\parallel }_{L^2}^2)}^{{\displaystyle \frac{s+\sigma +1}{s+\sigma }}}\le 0,\end{array}$$

(57)

which gives

$$\begin{array}{c}\hfill {\parallel ({\mathsf{\Lambda }}^s\mathit{u},{\mathsf{\Lambda }}^s\omega )\left(t\right)\parallel }_{L^2}\le C{(1+t)}^{-{\displaystyle \frac{s}{2}}-{\displaystyle \frac{\sigma }{2}}}.\end{array}$$

(58)

Similarly, it follows from (40) with $m=0$, (54) and (56) with $s=0$ that

$$\begin{array}{c}\hfill {\parallel (\mathit{u},\omega )\left(t\right)\parallel }_{L^2}\le C{(1+t)}^{-{\displaystyle \frac{\sigma }{2}}}.\end{array}$$

(59)

Therefore, combining (58) and (59) with the interpolation inequality (32), we obtain that, for any $0\le m\le s$ and all $0\le t\le T$,

$$\begin{array}{c}\hfill {\parallel ({\mathsf{\Lambda }}^m\mathit{u},{\mathsf{\Lambda }}^m\omega )\left(t\right)\parallel }_{L^2}\le C{(1+t)}^{-{\displaystyle \frac{m}{2}}-{\displaystyle \frac{\sigma }{2}}}.\end{array}$$

(60)

On the other hand, thanks to the constant ${\delta }_0$ is a positive one, it follows from (48) that

$$\begin{array}{cc}\hfill & {\displaystyle \frac{d}{dt}}{\parallel ({\mathsf{\Lambda }}^{-\sigma }\mathit{u},{\mathsf{\Lambda }}^{-\sigma }\omega )\left(t\right)\parallel }_{L^2}^2\hfill \\ \hfill & \le {C\parallel \mathit{u}\parallel }_{L^2}^{\sigma }{\parallel \mathsf{\Lambda }\mathit{u}\parallel }_{L^2}^{1-\sigma }{\parallel (\mathsf{\Lambda }\mathit{u},\mathsf{\Lambda }\omega )\parallel }_{L^2}{\parallel ({\mathsf{\Lambda }}^{-\sigma }\mathit{u},{\mathsf{\Lambda }}^{-\sigma }\omega )\parallel }_{L^2}.\hfill \end{array}$$

Integrating this over $[0,t]$, we derive that, for $0<\lambda <1$,

$$\begin{array}{cc}\hfill & \parallel ({\mathsf{\Lambda }}^{-\sigma }\mathit{u},{\mathsf{\Lambda }}^{-\sigma }\omega )\left(t\right){\parallel }_{L^2}^2\hfill \\ \hfill & \le \parallel ({\mathsf{\Lambda }}^{-\sigma }{\mathit{u}}_0,{\mathsf{\Lambda }}^{-\sigma }{\omega }_0)\left(t\right){\parallel }_{L^2}^2\hfill \\ \hfill & +C\underset{0\le \tau \le t}{sup}\parallel ({\mathsf{\Lambda }}^{-\sigma }\mathit{u},{\mathsf{\Lambda }}^{-\sigma }\omega )\left(\tau \right){\parallel }_{L^2}{\int }_0^t{\parallel \mathit{u}\parallel }_{L^2}^{\sigma }{\parallel \mathsf{\Lambda }\mathit{u}\parallel }_{L^2}^{1-\sigma }{\parallel (\mathsf{\Lambda }\mathit{u},\mathsf{\Lambda }\omega )\parallel }_{L^2}\left(\tau \right)d\tau \hfill \\ \hfill & \le \parallel ({\mathsf{\Lambda }}^{-\sigma }{\mathit{u}}_0,{\mathsf{\Lambda }}^{-\sigma }{\omega }_0)\left(t\right){\parallel }_{L^2}^2\hfill \\ \hfill & +C\underset{0\le \tau \le t}{sup}\parallel ({\mathsf{\Lambda }}^{-\sigma }\mathit{u},{\mathsf{\Lambda }}^{-\sigma }\omega )\left(\tau \right){\parallel }_{L^2}{\int }_0^t{\parallel \mathit{u}\parallel }_{L^2}^{\lambda }{\parallel \mathit{u}\parallel }_{L^2}^{\sigma -\lambda }{\parallel (\mathsf{\Lambda }\mathit{u},\mathsf{\Lambda }\omega )\parallel }_{L^2}^{2-\sigma }\left(\tau \right)d\tau \hfill \\ \hfill & \le C_0+C\sqrt{C_0}{\int }_0^t{\parallel \mathit{u}\parallel }_{L^2}^{\lambda }{(1+\tau )}^{-{\displaystyle \frac{\sigma }{2}}(\sigma -\lambda )}{(1+\tau )}^{(-{\displaystyle \frac{1}{2}}-{\displaystyle \frac{\sigma }{2}})(2-\sigma )}d\tau \hfill \\ \hfill & \le C_0+C\sqrt{C_0}{\int }_0^t{\parallel \mathit{u}\parallel }_{L^2}^{\lambda }{(1+\tau )}^{-(1+{\displaystyle \frac{\sigma }{2}}-{\displaystyle \frac{\sigma }{2}}\lambda )}d\tau ,\hfill \end{array}$$

(61)

where we have used (60) with $m=0$ and $m=1$. Noting the fact that

$${\parallel \mathit{u}\parallel }_{L^2}\le C\epsilon ,$$

due to the smallness condition (6). Therefore, (61) can be further bounded as

$$\begin{array}{cc}\hfill \parallel ({\mathsf{\Lambda }}^{-\sigma }\mathit{u},{\mathsf{\Lambda }}^{-\sigma }\omega )\left(t\right){\parallel }_{L^2}^2& \le C_0+C{\epsilon }^{\lambda }\sqrt{C_0}{\int }_0^t{(1+\tau )}^{-(1+{\displaystyle \frac{\sigma }{2}}-{\displaystyle \frac{\sigma }{2}}\lambda )}d\tau \hfill \\ \hfill & \le C_0+C{\epsilon }^{\lambda }\sqrt{C_0}.\hfill \end{array}$$

(62)

By choosing $\epsilon$ sufficiently small such that

$$C{\epsilon }^{\lambda }\sqrt{C_0}\le {\displaystyle \frac{C_0}{2}},$$

then (62) gives (55) for all $t\in [0,T]$, which closes the proof. Thus, the bootstrapping argument yields the global $T=\infty$, and it further ensures that (55) and (60) hold for all $0\le t<\infty$.

To complete the proof of Theorem 2, it remains to prove (9). To this end, taking the $L^2$-inner product to the equation of $\omega$ of system (2) with ${\mathsf{\Lambda }}^{2m}\omega$, integrating by parts and using H$\ddot{o}$lder inequality, commutator estimate (11) and Young inequality, we obtain

$$\begin{array}{cc}\hfill & {\displaystyle \frac{1}{2}}{\displaystyle \frac{d}{dt}}\parallel {\mathsf{\Lambda }}^m{\omega \parallel }_{L^2}^2+2\kappa {\parallel {\mathsf{\Lambda }}^m\omega \parallel }_{L^2}^2\hfill \\ \hfill & =\kappa {\int }_{{\mathbb{R}}^2}{\mathsf{\Lambda }}^m({\partial }_x{u}_2-{\partial }_y{u}_1)·{\mathsf{\Lambda }}^m\omega dxdy-{\int }_{{\mathbb{R}}^2}{\mathsf{\Lambda }}^m(\mathit{u}·\nabla \omega )·{\mathsf{\Lambda }}^m\omega dxdy\hfill \\ \hfill & \le C\left(\parallel {\mathsf{\Lambda }}^{m+1}{\mathit{u}\parallel }_{L^2}+\parallel {\mathsf{\Lambda }}^{m+1}{\mathit{u}\parallel }_{L^2}{\parallel \omega \parallel }_{L^{\infty }}+\parallel {\mathsf{\Lambda }}^{m+1}{\omega \parallel }_{L^2}{\parallel \mathit{u}\parallel }_{L^{\infty }}\right){\parallel {\mathsf{\Lambda }}^m\omega \parallel }_{L^2}\hfill \\ \hfill & \le {\displaystyle \frac{3\kappa }{2}}{\parallel {\mathsf{\Lambda }}^m\omega \parallel }_{L^2}^2+C\left(\parallel {\mathsf{\Lambda }}^{m+1}{\mathit{u}\parallel }_{L^2}^2+{\parallel {\mathsf{\Lambda }}^{m+1}\omega \parallel }_{L^2}^2\right),\hfill \end{array}$$

(63)

where we have used the Sobolev embedding theorem and (41) with $m=s$.

(63) implies that

$$\begin{array}{c}\hfill {\displaystyle \frac{d}{dt}}\parallel {\mathsf{\Lambda }}^m{\omega \parallel }_{L^2}^2+\kappa {\parallel {\mathsf{\Lambda }}^m\omega \parallel }_{L^2}^2\le C\left(\parallel {\mathsf{\Lambda }}^{m+1}{\mathit{u}\parallel }_{L^2}^2+{\parallel {\mathsf{\Lambda }}^{m+1}\omega \parallel }_{L^2}^2\right),\end{array}$$

or

$$\begin{array}{c}\hfill {\displaystyle \frac{d}{dt}}\left(e^{\kappa t}{\parallel {\mathsf{\Lambda }}^m\omega \left(t\right)\parallel }_{L^2}^2\right)\le C{e}^{\kappa t}\left(\parallel {\mathsf{\Lambda }}^{m+1}{\mathit{u}\parallel }_{L^2}^2+{\parallel {\mathsf{\Lambda }}^{m+1}\omega \parallel }_{L^2}^2\right).\end{array}$$

Integrating the above inequality over $[0,t]$, together with (8), we have for all $0\le m+1\le s$,

$$\begin{array}{cc}\hfill & {\parallel {\mathsf{\Lambda }}^m\omega \left(t\right)\parallel }_{L^2}^2\hfill \\ \hfill & \le e^{-\kappa t}{\parallel {\mathsf{\Lambda }}^m{\omega }_0\parallel }_{L^2}^2+C{\int }_0^t{e}^{-\kappa (t-\tau )}\left(\parallel {\mathsf{\Lambda }}^{m+1}{\mathit{u}\parallel }_{L^2}^2+{\parallel {\mathsf{\Lambda }}^{m+1}\omega \parallel }_{L^2}^2\right)d\tau \hfill \\ \hfill & \le C{e}^{-\kappa t}+C{\int }_0^t{e}^{-\kappa (t-\tau )}{(1+\tau )}^{-(m+1+\sigma )}d\tau \hfill \\ \hfill & \le C{e}^{-\kappa t}+C{\int }_0^{{\displaystyle \frac{t}{2}}}{e}^{-\kappa (t-\tau )}{(1+\tau )}^{-(m+1+\sigma )}d\tau +C{\int }_{{\displaystyle \frac{t}{2}}}^t{e}^{-\kappa (t-\tau )}{(1+\tau )}^{-(m+1+\sigma )}d\tau \hfill \\ \hfill & \le C{e}^{-\kappa t}+C{e}^{-{\displaystyle \frac{\kappa }{2}}t}+C{(1+t)}^{-(m+1+\sigma )}\hfill \\ \hfill & \le C{(1+t)}^{-(m+1+\sigma )},\hfill \end{array}$$

which shows (9). We thus complete the proof of Theorem 2. □

5. Conclusions

This paper studies the large-time behavior of the 2D micropolar equations with mixed velocity dissipation, focusing on two physical relevant regimes: with and without the angular viscosity. When the angular viscosity is present ($\gamma >0$), we establish optimal decay estimates for global solutions and their higher-order derivatives without any smallness assumptions on the initial data. The derived decay rates coincide with those of the classical heat equation, indicating that the dissipation structure of the system governs the long-time dynamics. When the angular viscosity is absent ($\gamma =0$), we prove the existence and uniqueness of global strong solutions for small initial data, and derive sharp decay bounds: the velocity field decays as ${(1+t)}^{-({\displaystyle \frac{m}{2}}+{\displaystyle \frac{\sigma }{2}})}$, while the micro-rotation field decays faster due to damping effects (see (9)).

These results extend earlier work on micropolar fluids with partial dissipation, such as those by Dong et al. [6,9] and Shang et al. [23], and contribute to the mathematical theory of anisotropic fluid models with mixed dissipation—a setting relevant to physical contexts such as boundary layers and anisotropic media. Methodologically, the analysis illustrates how mixed dissipation and damping mechanisms can be utilized to establish optimal decay even in the absence of full viscosity.

A natural extension of this work is to consider more complex settings beyond the whole space ${\mathbb{R}}^2$, such as flows in bounded or semi-bounded domains (e.g., channels or near walls) under physically relevant boundary conditions. Future studies could also incorporate magnetic effects, fractional dissipation, or temperature-dependent viscosity, thereby bridging theoretical analysis with applications in magneto-micropolar fluids, porous media, and anisotropic flows. Additionally, numerical simulations—employing finite-difference or spectral methods—could complement the analytical results by visualizing anisotropic decay patterns, verifying the sharpness of the derived rates, and exploring scenarios outside the current theoretical assumptions.