Stability and large time decay for the three-dimensional magneto-micropolar equations with mixed partial viscosity

Abstract

This paper focuses on the stability problem and decay estimates for two classes of three-dimensional (3D) magneto-micropolar equations with mixed partial viscosity. When $\mu _3= \nu _3 = \gamma _1 = \gamma _2=0$ and $\chi \Delta u$ replaced by $\chi u$ in (1.2), by fully exploiting the structure of the system, and the method of bootstrapping argument, we prove the global stability of solutions to this system with initial data small in $H^2({\mathbb {R}}^3)$. Furthermore, for these global solutions with initial data in $H^s({\mathbb {R}}^3)$ $(s \ge 3)$ being large, we obtain their global $H^s({\mathbb {R}}^3)$ $(s \ge 3)$ bound which is independent of time. In addition, we obtain the global existence of solutions for small initial data and the decay estimates of these solutions to 3D magneto-micropolar equations with mixed partial viscosity [namely $\mu _3= \gamma _1 = \gamma _2=0$ and $\chi \Delta u$ replaced by $\chi u$ in (1.2)].

1 Introduction

The three-dimensional (3D) incompressible magneto-micropolar fluid equations can be written as

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

(1.1)

where $(x, t) \in {\mathbb {R}}^3 \times {\mathbb {R}}^+$, $u = (u_1(x,t), u_2(x,t), u_3(x,t))$ denote the velocity of fluid, $b = (b_1(x,t), b_2(x,t),$ $b_3(x,t))$ the magnetic field, $\omega = (\omega _1(x,t), \omega _2(x,t), \omega _3(x,t))$ the microrotational velocity, $\pi = \pi (x,t)$ the pressure, and $\mu , \chi$ and $\frac{1}{\nu }$ are, respectively, kinematic viscosity, vortex viscosity and magnetic Reynolds number. $\gamma$ and $\kappa$ are angular viscosities. The magneto-micropolar fluid equations usually describe the motion of aggregates of small solid ferromagnetic particles relative to viscous magnetic fluids under the action of magnetic fields, such as saltwater, ester and fluorocarbon [1, 2]. It has attracted the attention of many physicists and mathematicians due to its important physical background, rich phenomena, mathematical complexity and challenges.

To the full magneto-micropolar fluid equations (1.1), Galdi and Rionero [3] stated the theorem of existence and uniqueness of strong solutions, but without proof. Ahmadi and Shaninpoor [2] studied the stability of solutions for the system. By using the spectral Galerkin method, Rojas-Medar [4] established the local existence and uniqueness of strong solutions. Ortega-Torres and Rojas-Medar [5] proved the global existence of strong solutions with small initial data. For the weak solution, Rojas-Medar and Boldrini [6] established the local existence in two and three dimensions by using the Galerkin method and also proved the uniqueness in the 2D case. The global existence of smooth solutions and the global regularity of weak solutions are important topics in the study field of magneto-micropolar fluid equations. The blow-up criteria for smooth solutions and regularity criteria of weak solutions were obtained in different function spaces, such as Morrey–Campanato space, Besov space and homogeneous Besov space, which we may refer to [7,8,9,10,11]. Recently, based on Serrin’s type non-blow-up criterion established by [7], Wang and Gu [12] have proved the global existence of a class of smooth solutions, which ensures the $L^3$ norm is large.

In this paper, we aim at the following three-dimensional (3D) magneto-micropolar fluid equations with mixed partial viscosity

$$\begin{aligned} \left\{ \begin{array}{lcl} \partial _t u - \mu _{1}\partial _{11} u - \mu _{2}\partial _{22} u - \mu _{3}\partial _{33}u - \chi \Delta u +u \cdot \nabla u + \nabla \pi = b\cdot \nabla b + 2\chi \nabla \times \omega ,\\ \partial _t b - \nu _{1}\partial _{11} b - \nu _{2}\partial _{22} b -\nu _{3}\partial _{33} b +u\cdot \nabla b = b\cdot \nabla u, \\ \partial _t \omega - \gamma _{1} \partial _{11} \omega - \gamma _{2} \partial _{22} \omega - \gamma _{3} \partial _{33} \omega + u\cdot \nabla \omega -\kappa \nabla \nabla \cdot \omega + 4\chi \omega = 2\chi \nabla \times u, \\ \nabla \cdot u =0, ~ \nabla \cdot b =0,\\ u(x, 0) =u_0(x), \omega (x, 0) =\omega _0(x), b(x, 0) =b_0(x). \end{array} \right. \end{aligned}$$

(1.2)

If

$$\begin{aligned} \mu _1 = \mu _2 = \mu _3 = \mu , \quad \nu _1 =\nu _2 =\nu _3 = \nu , \quad \gamma _1 = \gamma _2 = \gamma _3 =\gamma , \end{aligned}$$

then the (1.2) reduces to the standard 3D magneto-micropolar equations (1.1). For notational convenience, we will write $\partial _1$, $\partial _2$ and $\partial _3$ for $\partial _{x_1}$, $\partial _{x_2}$ and $\partial _{x_3}$, respectively.

The focus on this paper will be on the global stability problem of 3D magneto-micropolar equations (1.2) with mixed partial viscosity. Firstly, we consider the case: $\mu _1 = \mu _2 = \mu$, $\nu _1 =\nu _2 = \nu$, $\gamma _3 =\gamma$, $\mu _3 = \nu _3 = \gamma _1 = \gamma _2 = 0$, and $\chi \Delta u$ replaced by $\chi u$. Therefore, (1.2) reduce to

$$\begin{aligned} \left\{ \begin{array}{lcl} \partial _t u - \mu \Delta _h u + \chi u +u \cdot \nabla u + \nabla \pi = b\cdot \nabla b + 2\chi \nabla \times \omega ,\\ \partial _t b - \nu \Delta _h b +u\cdot \nabla b = b\cdot \nabla u, \\ \partial _t \omega - \gamma \partial _{33} \omega + u\cdot \nabla \omega -\kappa \nabla \nabla \cdot \omega + 4\chi \omega = 2\chi \nabla \times u, \\ \nabla \cdot u =0, ~ \nabla \cdot b =0,\\ u(x, 0) =u_0(x), \omega (x, 0) =\omega _0(x), b(x, 0) =b_0(x). \end{array} \right. \end{aligned}$$

(1.3)

Here $\Delta _h = \partial ^2_{x_1} + \partial ^2_{x_2}$, and we use $\nabla _h:= (\partial _1, \partial _2)$, $u_h:= (u_1, u_2)$ and $b_h:= (b_1, b_2)$ for the horizontal gradient, horizontal components of velocity and magnetic field, respectively.

So far, to our best knowledge, the previous well-posedness results on (1.3) mainly focus on the two-dimensional system. Liu [13] proved the global regularity result for the 2D incompressible magneto-micropolar system with the horizontal kinematic dissipation by $\partial _{x_1 x_1}$, the horizontal magnetic diffusion by $\partial _{x_1 x_1}$ and the spin dissipation by the fractional operator $(- \Delta )^{\gamma }(\gamma >1)$. For more well-posedness results to the 2D magneto-micropolar equations with partial dissipation, one refers to [14,15,16,17,18] and the references therein. In addition, when $b=0$ and $\chi =0$, then system (1.1) becomes magnetohydrodynamic (MHD) equations. The existence of small data global solutions to the 3D MHD equations with mixed partial dissipation and magnetic diffusion was obtained by Wang and Wang [19]. Wu and Zhu in [20] proved the global stability of perturbations near the steady solution to the 3D MHD equations with only horizontal velocity dissipation and vertical magnetic diffusion. Recently, Wang and Wang [21] have investigated the global existence of smooth solutions to 3D magneto-micropolar fluid equations with mixed partial viscosity (namely $\mu _1, \mu _2, \gamma _2, \gamma _3, \nu _1, \nu _2>0$, $\mu _3 = \gamma _1 = \nu _3 = 0$, or $\mu _1, \mu _2, \gamma _2, \gamma _3, \nu _2, \nu _3>0$, $\mu _3 = \gamma _1 = \nu _1 = 0$, or $\mu _1, \mu _2, \gamma _1, \gamma _2, \nu _1, \nu _2>0$, $\mu _3 = \gamma _3 = \nu _3 = 0$, or $\mu _1, \mu _2, \gamma _1, \gamma _2, \nu _2, \nu _3>0$, $\mu _3 = \gamma _3 = \nu _1 = 0$ in (1.2)); Wang and Li [22] proved the global well-posedness of 3D magneto-micropolar equations with mixed partial viscosity near an equilibrium. Shang and Zhai [23] proved the stability problem to the 3D anisotropic magnetohydrodynamic (MHD) equations (namely (1.3) with $\omega =0$ and $\chi = 0$).

Motivated by the ideas in [23], we focus on the global stability of the system (1.3), and the first result is the global stability of solutions the system (1.3) in $H^2({\mathbb {R}}^3)$ and the uniform boundedness for time of these global solutions in $H^s({\mathbb {R}}^3)$, as stated in the following theorem.

Theorem 1.1

Let $\mu >0$, $\chi > 0$, $\nu >0$ and $\kappa >0$. Assume $(u_0, b_0, \omega _0) \in H^2({\mathbb {R}}^3)$ with $\nabla \cdot u_0 =0$ and $\nabla \cdot b_0 = 0$. Then there exists $\epsilon > 0$ such that, if

$$\begin{aligned} \Vert u_0\Vert _{H^2({\mathbb {R}}^3)} + \Vert b_0\Vert _{H^2({\mathbb {R}}^3)} + \Vert \omega _0\Vert _{H^2({\mathbb {R}}^3)} \le \epsilon , \end{aligned}$$

(1.4)

and $\mu > 4 \chi$ and $\gamma > 4 \chi$. Then (1.3) has a unique global solution $(u, b, \omega ) \in L^{\infty } (0, \infty ; H^2({\mathbb {R}}^3))$ satisfying

$$\begin{aligned} \Vert u(t)\Vert _{H^2({\mathbb {R}}^3)} + \Vert b(t)\Vert _{H^2({\mathbb {R}}^3)} + \Vert \omega (t)\Vert _{H^2({\mathbb {R}}^3)} \le C \epsilon , \end{aligned}$$

(1.5)

and

$$\begin{aligned}&\Vert u(t) \Vert _{H^2}^2 + \Vert b(t)\Vert _{H^2}^2 + \Vert \omega (t)\Vert _{H^2}^2 \nonumber \\ {}&\quad + \int \limits ^t_0 \left( \Vert \nabla _h u(\tau )\Vert _{H^2}^2 + \Vert \nabla _h b(\tau )\Vert _{H^2}^2 + \Vert \partial _3 \omega (\tau )\Vert _{H^2}^2 + \Vert \omega (\tau )\Vert _{H^2}^2\right) d\tau \nonumber \\ {}&\quad \le C (\Vert u_0\Vert _{H^2}^2 + \Vert b_0\Vert _{H^2}^2 + \Vert \omega _0\Vert _{H^2}^2), \end{aligned}$$

(1.6)

for some constant $C > 0$ and for all $t >0$.

Furthermore, if $(u_0, b_0, \omega _0) \in H^s({\mathbb {R}}^3)$, $s \ge 3$ and (1.4) hold, then the global solution $(u, b, \omega )$ obeys for all $t > 0$,

$$\begin{aligned}&\Vert u(t)\Vert ^2_{H^s({\mathbb {R}}^3)} + \Vert b(t)\Vert ^2_{H^s({\mathbb {R}}^3)} + \Vert \omega (t)\Vert ^2_{H^s({\mathbb {R}}^3)} \nonumber \\ {}&\quad + c_1 \int \limits ^t_0 \left( \Vert \nabla _h u(\tau )\Vert ^2_{H^s({\mathbb {R}}^3)} + \Vert \nabla _h b(\tau )\Vert ^2_{H^s({\mathbb {R}}^3)} + \Vert \partial _3 \omega (\tau )\Vert ^2_{H^s({\mathbb {R}}^3)} \right) d\tau \nonumber \\ {}&\quad \le C_s \left( \Vert u_0\Vert ^2_{H^s({\mathbb {R}}^3)} + \Vert b_0\Vert ^2_{H^s({\mathbb {R}}^3)} + \Vert \omega _0\Vert ^2_{H^s({\mathbb {R}}^3)} \right) , \end{aligned}$$

(1.7)

where $c_1= \min \{\frac{\mu }{2}, \nu , \gamma - 4 \chi , 4\chi - \frac{16 \chi ^2}{\mu }\}$ and the constant $C_s$ depends on $\mu$, $\chi$, $\nu$, $\kappa$, $\gamma$ and the initial data.

Compared with the magnitude of research conducted on the well-posedness problem for the 3D Magneto-micropolar equations with partial dissipation, the large time behavior for the partial dissipation cases also has attracted considerable attention from the community of mathematical fields (see, e.g., [24,25,26,27,28,29,30,31,32,33]). This is an important issue in the fields of partial differential equations. It is well known that the $L^2$ decay problem of weak solutions to the 3D Navier–Stokes equations, i.e., (1.1) with $\omega = 0$ and $b=0$, was proposed by the celebrated work of Leray [34]. By introducing the elegant method of Fourier splitting, Schonbek [35, 36] successfully established the optimal time decay rate of weak solutions of the Navier–Stokes equations, see also [37, 38]. Recently, by the structure to (1.1) and the Fourier splitting method, Li and Shang [39] have established the decay estimates for weak solutions of (1.1) and obtained the same rates as those of the 3D Navier–Stokes equations.

Recently, Shang and Gu [40, 41] have proved the 2D magneto-micropolar equations with only microrotational dissipation and magneto diffusion. Li [42] proved the $L^2$-decay estimates for global solution and their derivative for 2D magneto-micropolar equation with partial dissipation (namely $\nu \Delta b$ replace by $\nu \partial _{yy}b_1 + \nu \partial _{xx} b_2$). Niu and Shang [43] proved the magneto-micropolar equations only have velocity dissipation and magnetic diffusion (namely (1.1) with $\gamma =0$).

Next, we consider the case: $\mu _1 = \mu _2 = \mu$, $\nu _1 =\nu _2 = \nu _3 = \nu$, $\gamma _3 =\gamma$, $\mu _3 = \gamma _1 = \gamma _2 = 0$ and $\chi \Delta u$ replaced by $\chi u$. Therefore, (1.2) reduce to

$$\begin{aligned} \left\{ \begin{array}{lcl} \partial _t u - \mu \Delta _h u + \chi u +u \cdot \nabla u + \nabla \pi = b\cdot \nabla b + 2\chi \nabla \times \omega ,\\ \partial _t b - \nu \Delta b +u\cdot \nabla b = b\cdot \nabla u, \\ \partial _t \omega - \gamma \partial _{33} \omega + u\cdot \nabla \omega -\kappa \nabla \nabla \cdot \omega + 4\chi \omega = 2\chi \nabla \times u, \\ \nabla \cdot u =0, ~ \nabla \cdot b =0,\\ u(x, 0) =u_0(x), \omega (x, 0) =\omega _0(x), b(x, 0) =b_0(x), \end{array} \right. \end{aligned}$$

(1.8)

Before giving the decay results of the solutions for (1.8), we first consider the global stability of solutions to system (1.8), as the following theorem.

Theorem 1.2

Let $\mu >0$, $\chi > 0$, $\nu >0$ and $\kappa >0$. Assume $(u_0, b_0, \omega _0) \in H^2({\mathbb {R}}^3)$ with $\nabla \cdot u_0 =0$ and $\nabla \cdot b_0 = 0$. Then there exists $\epsilon > 0$ such that, if

$$\begin{aligned} \Vert u_0\Vert _{H^2({\mathbb {R}}^3)} + \Vert b_0\Vert _{H^2({\mathbb {R}}^3)} + \Vert \omega _0\Vert _{H^2({\mathbb {R}}^3)} \le \epsilon , \end{aligned}$$

(1.9)

and $\mu > 4 \chi$ and $\gamma > 4 \chi$. Then (1.8) has a unique global solution $(u, b, \omega ) \in L^{\infty } (0, \infty ; H^2({\mathbb {R}}^3))$ satisfying

$$\begin{aligned} \Vert u(t)\Vert _{H^2({\mathbb {R}}^3)} + \Vert b(t)\Vert _{H^2({\mathbb {R}}^3)} + \Vert \omega (t)\Vert _{H^2({\mathbb {R}}^3)} \le C \epsilon , \end{aligned}$$

(1.10)

and

$$\begin{aligned}&\Vert u(t) \Vert _{H^2}^2 + \Vert b(t)\Vert _{H^2}^2 + \Vert \omega (t)\Vert _{H^2}^2 \nonumber \\ {}&\qquad + \int \limits ^t_0 \left( \Vert \nabla _h u(\tau )\Vert _{H^2}^2 + \Vert u(\tau )\Vert _{H^2}^2 + \Vert \nabla b(\tau )\Vert _{H^2}^2 + \Vert \partial _3 \omega (\tau )\Vert _{H^2}^2 + \Vert \omega (\tau )\Vert _{H^2}^2\right) d\tau \nonumber \\ {}&\quad \le C (\Vert u_0\Vert _{H^2}^2 + \Vert b_0\Vert _{H^2}^2 + \Vert \omega _0\Vert _{H^2}^2), \end{aligned}$$

(1.11)

for some constant $C > 0$ and for all $t >0$.

Furthermore, if $(u_0, b_0, \omega _0) \in H^s({\mathbb {R}}^3)$, $s \ge 3$ and (1.9) hold, then the global solution $(u, b, \omega )$ obeys for all $t > 0$,

$$\begin{aligned}&\Vert u(t)\Vert ^2_{H^s({\mathbb {R}}^3)} + \Vert b(t)\Vert ^2_{H^s({\mathbb {R}}^3)} + \Vert \omega (t)\Vert ^2_{H^s({\mathbb {R}}^3)} \nonumber \\ {}&\qquad + c_2 \int \limits ^t_0 \left( \Vert \nabla _h u(\tau )\Vert ^2_{H^s({\mathbb {R}}^3)} + \Vert u(\tau )\Vert ^2_{H^s({\mathbb {R}}^3)}+ \Vert \nabla b(\tau )\Vert ^2_{H^s({\mathbb {R}}^3)} + \Vert \partial _3 \omega (\tau )\Vert ^2_{H^s({\mathbb {R}}^3)} \right) d\tau \nonumber \\ {}&\quad \le C_s \left( \Vert u_0\Vert ^2_{H^s({\mathbb {R}}^3)} + \Vert b_0\Vert ^2_{H^s({\mathbb {R}}^3)} + \Vert \omega _0\Vert ^2_{H^s({\mathbb {R}}^3)} \right) , \end{aligned}$$

(1.12)

where $c_2= \min \{\frac{\mu }{2}, \frac{\chi }{2}, \nu , \gamma - 8 \chi , 4\chi - \frac{16 \chi ^2}{\mu }\}$ and the constant $C_s$ depends on $\mu$, $\chi$, $\nu$, $\kappa$, $\gamma$ and the initial data.

Remark 1.3

The proof of Theorem 1.2 is similar to the proof of Theorem 1.1; then, we omit the details.

At last, using the delicate a priori estimates and the properties of the heat operator, we can establish the following decay results for the global solutions established in Theorem 1.2. More precisely, we have the following theorem.

Theorem 1.4

Let $(u_0, \omega _0) \in H^2({\mathbb {R}}^3)$, $b_0 \in L^1({\mathbb {R}}^3) \cap H^2({\mathbb {R}}^3)$ with $\nabla \cdot u_0 = \nabla \cdot b_0 = 0$. Let $(u, b, \omega )$ be a global solution to system (1.8), and assume that

$$\begin{aligned} \mu> 4\chi ,\quad \gamma > \frac{16 \chi }{3}, \end{aligned}$$

and

$$\begin{aligned} \Vert u_0\Vert _{H^2({\mathbb {R}}^3)} + \Vert b_0\Vert _{H^2({\mathbb {R}}^3)} + \Vert \omega _0\Vert _{H^2({\mathbb {R}}^3)} \le \epsilon . \end{aligned}$$

(1.13)

Then, the following decay properties hold:

$$\begin{aligned} \Vert b(t)\Vert _{L^2} \le C (1+t)^{-\frac{3}{4}}, \quad \Vert \nabla b(t)\Vert _{L^2} \le C (1+t)^{-\frac{5}{4}}. \end{aligned}$$

(1.14)

Moreover, for any $0< \alpha < \frac{1}{4}$, the following decay properties hold:

$$\begin{aligned}&\Vert \nabla ^2b(t)\Vert _{L^{2}} \le C (1+t)^{-\frac{65}{48} + \frac{17}{12} \alpha }, \end{aligned}$$

(1.15)

$$\begin{aligned}&\quad \Vert u(t)\Vert _{L^2} + \Vert \omega (t)\Vert _{L^2} \le C(1+ t)^{-\frac{157}{64} + \frac{17}{16} \alpha }, \end{aligned}$$

(1.16)

$$\begin{aligned}&\quad \Vert \nabla u(t)\Vert _{L^2} + \Vert \nabla \omega (t)\Vert \le C (1+t)^{-\frac{471}{512} + \frac{51}{128} \alpha }, \end{aligned}$$

(1.17)

where the constant C depends on $\mu$, $\chi$, $\nu$, $\kappa$, $\gamma$ and the initial data.

The rest of this paper is divided into two sections. Sect. 2 proves Theorem 1.1 while Sect. 1.4 is devoted to verifying Theorem 1.4. Throughout this manuscript, to simplify the notations, we will write $\int f \textrm{d}x$ for $\int \limits _{{\mathbb {R}}^3} f \textrm{d}x$, $\Vert f\Vert _{L^p}$ for $\Vert f\Vert _{L^p({\mathbb {R}}^3)}$. We write $\Vert f\Vert _{L^p_{x_j}}$ with $j = 1, 2, 3$ for the $L^p$-norm with respect to $x_j$ on ${\mathbb {R}}$ and $\Vert f\Vert _{L^p_{x_j x_k}}$ with $j, k = 1, 2, 3$ for the $L^p$-norm with respect to $(x_j, x_k)$ on ${\mathbb {R}}^2$. We also write $\Vert f\Vert _{L^p_h}$ for the $\Vert f\Vert _{L^p_{x_1x_2}}$ to shorten the notation. In addition, the anisotropic norm $\Vert f\Vert _{L^p_h L^q_{x_3}}: = \Vert \Vert f\Vert _{L_{x_3}^q}\Vert _{L^p_h}$ is also frequently used.

2 The proof of Theorem 1.1

This section focuses its attention on the proof of Theorem 1.1. As preparations, we state three important lemmas, which are frequently used. The first lemma provides an upper bound for the $L^p$-norm of a one-dimensional (1D) function, which serves as a basic ingredient for an anisotropic upper bound (see, e.g., [44]).

Lemma 2.1

Let $2 \le p \le \infty$ and $\Lambda = (- \Delta )^{-\frac{1}{2}}$ be the Zygmund operator. Then there exists a positive constant C such that, for any 1D functions $f \in H^s({\mathbb {R}})$ with $s > \frac{1}{2} - \frac{1}{p}$,

$$\begin{aligned} \Vert f\Vert _{L^p({\mathbb {R}})} \le C \Vert f\Vert _{L^2({\mathbb {R}})}^{1- \frac{1}{s}(\frac{1}{2} - \frac{1}{p})} \Vert \Lambda ^s f\Vert _{L^2({\mathbb {R}})}^{\frac{1}{s}(\frac{1}{2} - \frac{1}{p})}. \end{aligned}$$

In particular, if $p = \infty$ and $s = 1$, then any $f = f(x_3) \in H^1({\mathbb {R}})$ satisfies

$$\begin{aligned} \Vert f\Vert _{L^\infty ({\mathbb {R}})} \le \Vert f\Vert _{L^2({\mathbb {R}})}^{\frac{1}{2}} \Vert \partial _{x_3} f\Vert _{L^2({\mathbb {R}})}^{\frac{1}{2}}. \end{aligned}$$

The second lemma provides an anisotropic upper bound for the integral of a triple product (see, e.g., [20]). It is a very powerful tool in dealing with anisotropic equations.

Lemma 2.2

Assume that f, $\partial _1 f$, $\partial _2 f$, $\partial _1\partial _2 f$, g, $\partial _2 g$, h, $\partial _3 h \in L^2({\mathbb {R}}^3)$. Then

$$\begin{aligned}{} & {} \int \limits _{{\mathbb {R}}^3} |fgh| \textrm{d}x \le C \Vert f\Vert ^{\frac{1}{2}}_{L^2} \Vert \partial _1 f\Vert ^{\frac{1}{2}}_{L^2} \Vert g\Vert ^{\frac{1}{2}}_{L^2} \Vert \partial _2 g\Vert ^{\frac{1}{2}}_{L^2} \Vert h\Vert ^{\frac{1}{2}}_{L^2} \Vert \partial _3 h\Vert ^{\frac{1}{2}}_{L^2}.\\{} & {} \quad \int \limits _{{\mathbb {R}}^3} |fgh| \textrm{d}x \le C \Vert f\Vert ^{\frac{1}{4}}_{L^2} \Vert \partial _1 f\Vert ^{\frac{1}{4}}_{L^2} \Vert \partial _2 f\Vert ^{\frac{1}{4}}_{L^2} \Vert \partial _1 \partial _2 f\Vert ^{\frac{1}{4}}_{L^2} \Vert g\Vert ^{\frac{1}{2}}_{L^2} \Vert \partial _3 g\Vert ^{\frac{1}{2}}_{L^2} \Vert h\Vert _{L^2}. \end{aligned}$$

The third lemma states Minkowski’s inequality. It is an elementary tool that allows us to estimate the Lebesgue norm with a large index first followed by the Lebesgue norm with a smaller index (see, e.g., [45, 46]).

Lemma 2.3

Let $(X_1, \mu _1)$ and $(X_2, \mu _2)$ be two measure spaces. Let f be a nonnegative measurable function over $X_1 \times X_2$. For all $1 \le p \le q \le \infty$, we have

$$\begin{aligned} \left\| \Vert f(\cdot , x_2)\Vert _{L^p(X_1, \mu _1)} \right\| _{L^q(X_2, \mu _2)} \le \left\| \Vert f(x_1, \cdot )\Vert _{L^q(X_2, \mu _2)} \right\| _{L^p(X_1, \mu _1)}. \end{aligned}$$

In particular, for a nonnegative measurable function f over ${\mathbb {R}}^m \times {\mathbb {R}}^n$ and for $1 \le p \le q \le \infty$,

$$\begin{aligned} \left\| \Vert f\Vert _{L^p({\mathbb {R}}^m)} \right\| _{L^q({\mathbb {R}}^n)} \le \left\| \Vert f\Vert _{L^q({\mathbb {R}}^n)} \right\| _{L^p({\mathbb {R}}^m)}.\end{aligned}$$

Next, we are ready to prove Theorem 1.1. The general approach to establish the global existence and regularity results consists of two main steps. The first step assesses the local (in time) well-posedness while the second extends the local solution into a global one by obtaining global (in time) a priori bounds. For the system (1.3) concerned here, the local well-posedness follows from a standard approach such as the contraction mapping principle or successive approximations (see, e.g., [47]). Therefore, the effort of proving the Theorem 1.1 is mainly devoted to the global a priori bounds for $\Vert (u, b, \omega )\Vert _{H^3}$ and $\Vert (u, b, \omega )\Vert _{H^s}$ with $s \ge 3$.

Proof of Theorem 1.1

For the sake of clarity, we divide the proof into two parts.

(1) This part proves (1.5) of Theorem 1.1. Firstly, we examine the global $a \ priori$ $L^2$- bound as follows.

Proposition 2.4

Let $\mu >0$, $\chi > 0$, $\nu >0$ and $\kappa >0$. Assume $(u_0, b_0, \omega _0) \in L^2({\mathbb {R}}^3)$ with $\nabla \cdot u_0 =0$ and $\nabla \cdot b_0 = 0$. If $\mu > 4 \chi$ and $\gamma > 4 \chi$, then the corresponding solution $(u, b, \omega )$ of (1.3) obeys the following uniform bounds, for any $t > 0$

$$\begin{aligned} \frac{\textrm{d}}{\textrm{d}t}&\left( \Vert u(t)\Vert ^2_{L^2} + \Vert b(t)\Vert ^2_{L^2} + \Vert \omega (t)\Vert ^2_{L^2} \right) + \mu \Vert \nabla _h u\Vert ^2_{L^2} + {2 \nu }\Vert \nabla _h b\Vert ^2_{L^2} + 2(\gamma - 4 \chi ) \Vert \partial _3 \omega \Vert ^2_{L^2} \nonumber \\&+ 2(4\chi - \frac{16 \chi ^2}{\mu }) \Vert \omega \Vert _{L^2}^2 \le 0, \end{aligned}$$

(2.1)

or

$$\begin{aligned}&\Vert u(t)\Vert ^2_{L^2} + \Vert b(t)\Vert ^2_{L^2} + \Vert \omega (t)\Vert ^2_{L^2} \nonumber \\ {}&\qquad + 2 c_1 \int \limits ^t_0 \left( \Vert \nabla _h u (\tau )\Vert ^2_{L^2} + \Vert \nabla _h b (\tau )\Vert ^2_{L^2} + \Vert \partial _3 \omega (\tau )\Vert ^2_{L^2} + \Vert \omega (\tau )\Vert _{L^2}^2 \right) d\tau \nonumber \\ {}&\quad \le \Vert u_0\Vert _{L^2}^2 + \Vert b_0\Vert _{L^2}^2 + \Vert \omega _0\Vert _{L^2}^2, \end{aligned}$$

(2.2)

where $c_1= \min \{\frac{\mu }{2}, \nu , \gamma - 4 \chi , 4\chi - \frac{16 \chi ^2}{\mu }\}$.

Proof of Proposition 2.4

Taking the $L^2$-inner product to (1.3) with $(u, b, \omega )$, by integrating by parts, we have

$$\begin{aligned}&\frac{1}{2} \frac{\textrm{d}}{\textrm{d}t} \left( \Vert u(t)\Vert ^2_{L^2} + \Vert b(t)\Vert ^2_{L^2} + \Vert \omega (t)\Vert ^2_{L^2} \right) + \mu \Vert \nabla _h u\Vert ^2_{L^2} + \chi \Vert u\Vert _{L^2}^2 + \nu \Vert \nabla _h b\Vert ^2_{L^2} \\&\qquad + \gamma \Vert \partial _3 \omega \Vert ^2_{L^2} + 4 \chi \Vert \omega \Vert _{L^2}^2 + \kappa \Vert \nabla \cdot \omega \Vert _{L^2}^2\\&\quad = 4\chi \int \nabla \times u \cdot \omega \textrm{d}x \\&\quad = 4 \chi \int (\partial _2 u_3 - \partial _3 u_2, \partial _3 u_1 - \partial _1 u_3, \partial _1 u_2 - \partial _2 u_1) \cdot \omega \textrm{d}x\\&\quad = 4 \chi \int \left( (\partial _2 u_3 - \partial _3 u_2)\omega _1 + (\partial _3 u_1 - \partial _1 u_3)\omega _2 + (\partial _1 u_2 - \partial _2 u_1)\omega _3 \right) \textrm{d}x \\&\quad = 4 \chi \int (\partial _2 u_3 \omega _1 - \partial _1 u_3 \omega _2 + \partial _1 u_2 \omega _3 - \partial _2 u_1 \omega _3)\textrm{d}x - 4 \chi \int (u_1 \partial _3 \omega _2 - u_2 \partial _3 \omega _1)\textrm{d}x\\&\quad \le \frac{\mu }{2}(\Vert \partial _2 u_3\Vert ^2_{L^2} + \Vert \partial _1 u_3\Vert ^2_{L^2} + \Vert \partial _1 u_2\Vert ^2_{L^2} + \Vert \partial _2 u_1\Vert ^2_{L^2}) + \frac{8 \chi ^2}{\mu }(\Vert \omega _1\Vert _{L^2}^2 + \Vert \omega _2\Vert _{L^2}^2 + 2 \Vert \omega _3\Vert _{L^2}^2)\\&\qquad + \chi (\Vert u_1\Vert _{L^2}^2 + \Vert u_2\Vert _{L^2}^2) + 4\chi (\Vert \partial _3 \omega _2\Vert _{L^2}^2 + \Vert \partial _3 \omega _1\Vert _{L^2}^2)\\&\quad \le \frac{\mu }{2} \Vert \nabla _h u\Vert _{L^2}^2 + \frac{16 \chi ^2 }{\mu } \Vert \omega \Vert _{L^2}^2 + \chi \Vert u\Vert _{L^2}^2 + 4 \chi \Vert \partial _3 \omega \Vert _{L^2}^2, \end{aligned}$$

where we have used the facts that

$$\begin{aligned}&\int \nabla \times u \cdot \omega \textrm{d}x = \int \nabla \times \omega \cdot u \textrm{d}x,\\&\quad \int b \cdot \nabla b \cdot u \textrm{d}x + \int b \cdot \nabla u \cdot b \textrm{d}x = 0. \end{aligned}$$

Then we have

$$\begin{aligned}&\frac{\textrm{d}}{\textrm{d}t} \left( \Vert u(t)\Vert ^2_{L^2} + \Vert b(t)\Vert ^2_{L^2} + \Vert \omega (t)\Vert ^2_{L^2} \right) + \mu \Vert \nabla _h u\Vert ^2_{L^2} + 2 \nu \Vert \nabla _h b\Vert ^2_{L^2} + 2(\gamma - 4 \chi ) \Vert \partial _3 \omega \Vert ^2_{L^2} \nonumber \\&\quad + 2(4\chi - \frac{16 \chi ^2}{\mu }) \Vert \omega \Vert _{L^2}^2 \le 0, \end{aligned}$$

which immediately yields (2.1). Integrating the above inequality in [0, t], we can get (2.2) immediately.

$\square$

Applying $\partial _k^2$ with $k=1, 2, 3$ to (1.3)$_1$, (1.3)$_2$ and (1.3)$_3$, dotting the results by $\partial _k^2 u$, $\partial _k^2 b$ and $\partial _k^2 \omega$, respectively, integrating in space domain and adding them together, we obtain

$$\begin{aligned}&\frac{1}{2} \frac{\textrm{d}}{\textrm{d}t} \sum _{k=1}^3 \left( \Vert \partial _k^2 u(t)\Vert _{L^2}^2 +\Vert \partial _k^2 b(t)\Vert _{L^2}^2 + \Vert \partial _k^2 \omega (t)\Vert _{L^2}^2 \right) + \mu \sum _{k=1}^3 \Vert \nabla _h \partial _k^2 u\Vert _{L^2}^2 + \chi \sum _{k=1}^3 \Vert \partial _k^2 u\Vert _{L^2}^2\nonumber \\&\qquad + \nu \sum _{k=1}^3 \Vert \nabla _h \partial _k^2 b\Vert _{L^2}^2 + \gamma \sum _{k=1}^3 \Vert \partial _3 \partial _k^2 \omega \Vert _{L^2}^2 + 4 \chi \sum _{k=1}^3 \Vert \partial _k^2 \omega \Vert _{L^2}^2 + \kappa \sum _{k=1}^3 \Vert \partial _k^2 \nabla \cdot \omega \Vert _{L^2}^2\nonumber \\&\quad = - \sum _{k=1}^3 \int \partial _k^2 (u \cdot \nabla u) \cdot \partial _k^2 u \textrm{d}x + \sum _{k=1}^3 \int \partial _k^2 (b \cdot \nabla b) \cdot \partial _k^2 u \textrm{d}x + 2 \chi \sum _{k=1}^3 \int \nabla \times \partial _k^2 \omega \cdot \partial _k^2 u \textrm{d}x \nonumber \\&\qquad - \sum _{k=1}^3 \int \partial _k^2 (u \cdot \nabla b) \cdot \partial _k^2 b \textrm{d}x + \sum _{k=1}^3 \int \partial _k^2 (b \cdot \nabla u) \cdot \partial _k^2 b \textrm{d}x - \sum _{k=1}^3 \int \partial _k^2 (u \cdot \nabla \omega ) \cdot \partial _k^2 \omega \textrm{d}x \nonumber \\&\qquad + 2 \chi \sum _{k=1}^3 \int \nabla \times \partial _k^2 u \cdot \partial _k^2 \omega \textrm{d}x\nonumber \\&\quad := M_1 + M_2 + M_3 + M_4 + M_5 + M_6 + M_7. \end{aligned}$$

(2.3)

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

$$\begin{aligned} \sum _{k=1}^3 \int u \cdot \nabla \partial _k^2 u \cdot \partial _k^2 u \textrm{d}x = 0. \end{aligned}$$

Thus, we can rewrite $M_1$ as

$$\begin{aligned} M_1&= - \sum _{k=1}^3 \sum _{j=1}^3 \sum _{\alpha = 1}^2 \left( {\begin{array}{c}2\\ \alpha \end{array}}\right) \int \partial _k^{\alpha } u_j \partial _j \partial _k^{2-\alpha } u \cdot \partial _k^2 u \textrm{d}x \\&= - \sum _{k=1}^2 \sum _{j=1}^3 \sum _{\alpha = 1}^2 \left( {\begin{array}{c}2\\ \alpha \end{array}}\right) \int \partial _k^{\alpha } u_j \partial _j \partial _k^{2-\alpha } u \cdot \partial _k^2 u \textrm{d}x - \sum _{j=1}^2 \sum _{\alpha = 1}^2 \left( {\begin{array}{c}2\\ \alpha \end{array}}\right) \int \partial _3^{\alpha } u_j \partial _j \partial _3^{2-\alpha } u \cdot \partial _3^2 u \textrm{d}x\\&\qquad - \sum _{\alpha = 1}^2 \left( {\begin{array}{c}2\\ \alpha \end{array}}\right) \int \partial _3^{\alpha } u_3 \partial _3 \partial _3^{2-\alpha } u \cdot \partial _3^2 u \textrm{d}x\\&:= M_{11} + M_{12} + M_{13}, \end{aligned}$$

where $\left( {\begin{array}{c}2\\ \alpha \end{array}}\right) = \frac{2!}{\alpha !(2 - \alpha )!}$ is the binomial coefficient.

We first bound $M_{11}$. To do this, applying Lemma 2.2, we drive that

$$\begin{aligned} M_{11}&= - \sum _{k=1}^2 \sum _{j=1}^3 \sum _{\alpha = 1}^2 \left( {\begin{array}{c}2\\ \alpha \end{array}}\right) \int \partial _k^{\alpha } u_j \partial _j \partial _k^{2-\alpha } u \cdot \partial _k^2 u \textrm{d}x \\&\le C \sum _{k=1}^2 \sum _{j=1}^3 \sum _{\alpha = 1}^2 \Vert \partial _k^{\alpha } u_j\Vert _{L^2}^{\frac{1}{2}} \Vert \partial _3 \partial _k^{\alpha } u_j\Vert _{L^2}^{\frac{1}{2}} \Vert \partial _j \partial _k^{2 - \alpha } u \Vert _{L^2}^{\frac{1}{2}} \Vert \partial _2 \partial _j \partial _k^{2 - \alpha } u\Vert _{L^2}^{\frac{1}{2}} \Vert \partial _k^2 u\Vert _{L^2}^{\frac{1}{2}} \Vert \partial _1 \partial _k^{2} u\Vert _{L^2}^{\frac{1}{2}} \\&\le C \Vert u\Vert _{H^2} \Vert \nabla _h u\Vert _{H^2}^2. \end{aligned}$$

Again applying Lemma 2.2, yields

$$\begin{aligned} M_{12}&= - \sum _{j=1}^2 \sum _{\alpha = 1}^2 \left( {\begin{array}{c}2\\ \alpha \end{array}}\right) \int \partial _3^{\alpha } u_j \partial _j \partial _3^{2-\alpha } u \cdot \partial _3^2 u \textrm{d}x\\&\le C \sum _{j=1}^2 \sum _{\alpha = 1}^2 \Vert \partial _3^{\alpha } u_j\Vert _{L^2}^{\frac{1}{2}} \Vert \partial _1 \partial _3^{\alpha } u_j\Vert _{L^2}^{\frac{1}{2}} \Vert \partial _j \partial _3^{2 - \alpha } u \Vert _{L^2}^{\frac{1}{2}} \Vert \partial _3 \partial _j \partial _3^{2 - \alpha } u\Vert _{L^2}^{\frac{1}{2}} \Vert \partial _3^2 u\Vert _{L^2}^{\frac{1}{2}} \Vert \partial _2 \partial _3^{2} u\Vert _{L^2}^{\frac{1}{2}} \\&\le C \Vert u\Vert _{H^2} \Vert \nabla _h u\Vert _{H^2}^2. \end{aligned}$$

We cannot estimate the term $M_{13}$ directly. To bound it, the strategy is to use the divergence free condition $\nabla \cdot u = 0$ to convert $\partial _3 u_3$ to $- \nabla _h \cdot u_h$ and using Lemma 2.2. Then it can be bounded by

$$\begin{aligned} M_{13}&= - \sum _{\alpha = 1}^2 \left( {\begin{array}{c}2\\ \alpha \end{array}}\right) \int \partial _3^{\alpha } u_3 \partial _3 \partial _3^{2-\alpha } u \cdot \partial _3^2 u \textrm{d}x \\&= - \sum _{\alpha = 1}^2 \left( {\begin{array}{c}2\\ \alpha \end{array}}\right) \int \partial _3^{\alpha - 1} \nabla _h \cdot u_h \partial _3 \partial _3^{2-\alpha } u \cdot \partial _3^2 u \textrm{d}x\\&\le C \sum _{\alpha = 1}^2 \Vert \partial _3^{\alpha -1} \nabla _h u_h\Vert _{L^2}^{\frac{1}{2}} \Vert \partial _3^{\alpha } \nabla _h u_h\Vert _{L^2}^{\frac{1}{2}} \Vert \partial _3^{3 - \alpha } u \Vert _{L^2}^{\frac{1}{2}} \Vert \partial _1 \partial _3^{3 - \alpha } u\Vert _{L^2}^{\frac{1}{2}} \Vert \partial _3^2 u\Vert _{L^2}^{\frac{1}{2}} \Vert \partial _2 \partial _3^{2} u\Vert _{L^2}^{\frac{1}{2}} \\&\le C \Vert u\Vert _{H^2} \Vert \nabla _h u\Vert _{H^2}^2. \end{aligned}$$

Therefore, we obtain

$$\begin{aligned} M_1 \le C \Vert u\Vert _{H^2} \Vert \nabla _h u\Vert _{H^2}^2. \end{aligned}$$

By integrating by parts, Hölder’s inequality and the Young inequality, we have

$$\begin{aligned} M_3 + M_7&= 2 \chi \sum _{k=1}^3 \int \nabla \times \partial _k^2 \omega \cdot \partial _k^2 u \textrm{d}x + 2 \chi \sum _{k=1}^3 \int \nabla \times \partial _k^2 u \cdot \partial _k^2 \omega \textrm{d}x\\&= 4 \chi \sum _{k=1}^3 \int \nabla \times \partial _k^2 u \cdot \partial _k^2 \omega \textrm{d}x\\&\le \frac{\mu }{2} \sum _{k = 1}^3 \Vert \nabla _h \partial _k^2 u\Vert _{L^2}^2 + \frac{16 \chi ^2}{\mu } \sum _{k = 1}^3 \Vert \partial _k^2 \omega \Vert _{L^2}^2 + \chi \sum _{k = 1}^3 \Vert \partial _k^2 u\Vert _{L^2}^2 + 4 \chi \sum _{k = 1}^3 \Vert \partial _3 \partial _k^2 \omega \Vert _{L^2}^2. \end{aligned}$$

By using the same method as $M_1$, we have

$$\begin{aligned} M_4&= - \sum _{k=1}^3 \sum _{j=1}^3 \sum _{\alpha = 1}^2 \left( {\begin{array}{c}2\\ \alpha \end{array}}\right) \int \partial _k^{\alpha } u_j \partial _j \partial _k^{2-\alpha } b \cdot \partial _k^2 b \textrm{d}x \\&= - \sum _{k=1}^2 \sum _{j=1}^3 \sum _{\alpha = 1}^2 \left( {\begin{array}{c}2\\ \alpha \end{array}}\right) \int \partial _k^{\alpha } u_j \partial _j \partial _k^{2-\alpha } b \cdot \partial _k^2 b \textrm{d}x - \sum _{j=1}^2 \sum _{\alpha = 1}^2 \left( {\begin{array}{c}2\\ \alpha \end{array}}\right) \int \partial _3^{\alpha } u_j \partial _j \partial _3^{2-\alpha } b \cdot \partial _3^2 b \textrm{d}x\\&\quad - \sum _{\alpha = 1}^2 \left( {\begin{array}{c}2\\ \alpha \end{array}}\right) \int \partial _3^{\alpha } u_3 \partial _3 \partial _3^{2-\alpha } b \cdot \partial _3^2 b \textrm{d}x\\&\le C \sum _{k=1}^2 \sum _{j=1}^3 \sum _{\alpha = 1}^2 \Vert \partial _k^{\alpha } u_j\Vert _{L^2}^{\frac{1}{2}} \Vert \partial _3 \partial _k^{\alpha } u_j\Vert _{L^2}^{\frac{1}{2}} \Vert \partial _j \partial _k^{2 - \alpha } b \Vert _{L^2}^{\frac{1}{2}} \Vert \partial _2 \partial _j \partial _k^{2 - \alpha } b\Vert _{L^2}^{\frac{1}{2}} \Vert \partial _k^2 b\Vert _{L^2}^{\frac{1}{2}} \Vert \partial _1 \partial _k^{2} b\Vert _{L^2}^{\frac{1}{2}} \\&\quad + C \sum _{j=1}^2 \sum _{\alpha = 1}^2 \Vert \partial _3^{\alpha } u_j\Vert _{L^2}^{\frac{1}{2}} \Vert \partial _1 \partial _3^{\alpha } u_j\Vert _{L^2}^{\frac{1}{2}} \Vert \partial _j \partial _3^{2 - \alpha } b \Vert _{L^2}^{\frac{1}{2}} \Vert \partial _3 \partial _j \partial _3^{2 - \alpha } b\Vert _{L^2}^{\frac{1}{2}} \Vert \partial _3^2 b\Vert _{L^2}^{\frac{1}{2}} \Vert \partial _2 \partial _3^{2} b\Vert _{L^2}^{\frac{1}{2}} \\&\quad + C \sum ^{2}_{\alpha = 1} \Vert \partial _3^{\alpha -1} \nabla _h u_h \Vert _{L^2}^{\frac{1}{2}} \Vert \partial _3^{\alpha } \nabla _h u_h \Vert _{L^2}^{\frac{1}{2}} \Vert \partial _3^{3 - \alpha } b\Vert _{L^2}^{\frac{1}{2}} \Vert \partial _1\partial _3^{3 - \alpha } b\Vert _{L^2}^{\frac{1}{2}} \Vert \partial _3^{2} b\Vert _{L^2}^{\frac{1}{2}} \Vert \partial _2 \partial _3^{2} b\Vert _{L^2}^{\frac{1}{2}}\\&\le C \Vert b\Vert _{H^2} \Vert \nabla _h u\Vert _{H^2} \Vert \nabla _h b\Vert _{H^2} + C \Vert u\Vert _{H^2}^{\frac{1}{2}} \Vert b\Vert _{H^2}^{\frac{1}{2}} \Vert \nabla _h u\Vert _{H^2}^{\frac{1}{2}} \Vert \nabla _h b\Vert _{H^2}^{\frac{3}{2}} \\&\le C (\Vert u\Vert _{H^2} + \Vert b\Vert _{H^2}) (\Vert \nabla _h u\Vert _{H^2}^2 + \Vert \nabla _h b\Vert _{H^2}^2). \end{aligned}$$

Note that

$$\begin{aligned} \sum _{k=1}^3 \int b \cdot \nabla \partial _k^2 b \cdot \partial _k^2 u \textrm{d}x + \sum _{k=1}^3 \int b \cdot \nabla \partial _k^2 u \cdot \partial _k^2 b \textrm{d}x = 0, \end{aligned}$$

then the estimates similar as $M_1$, we have

$$\begin{aligned} M_2 + M_5&= \sum _{k=1}^3 \sum _{j =1}^3 \sum _{\alpha = 1}^2 \left( {\begin{array}{c}2\\ \alpha \end{array}}\right) \left( \int \partial _k^{\alpha } b_j \partial _j \partial _{k}^{2 - \alpha } b \cdot \partial _k^2 u \textrm{d}x + \int \partial _k^{\alpha } b_j \partial _j \partial _{k}^{2 - \alpha } u \cdot \partial _k^2 b \textrm{d}x \right) \\&\le C (\Vert u\Vert _{H^2} + \Vert b\Vert _{H^2}) (\Vert \nabla _h u\Vert _{H^2}^2 + \Vert \nabla _h b\Vert _{H^2}^2). \end{aligned}$$

Finally, we consider $M_6$,

$$\begin{aligned} M_6&= - \sum _{k=1}^3 \int \partial _{k}^2 (u \cdot \nabla \omega ) \cdot \partial _{k}^2 \omega \textrm{d}x\\&= - \sum _{k=1}^3 \sum _{j=1}^3 \sum _{\alpha = 1}^2 \left( {\begin{array}{c}2\\ \alpha \end{array}}\right) \int \partial _k^{\alpha } u_j \partial _j \partial _k^{2-\alpha } \omega \cdot \partial _k^2 \omega \textrm{d}x \\&= - \sum _{k=1}^2 \sum _{j=1}^3 \sum _{\alpha = 1}^2 \left( {\begin{array}{c}2\\ \alpha \end{array}}\right) \int \partial _k^{\alpha } u_j \partial _j \partial _k^{2-\alpha } \omega \cdot \partial _k^2 \omega \textrm{d}x - \sum _{j=1}^3 \sum _{\alpha = 1}^2 \left( {\begin{array}{c}2\\ \alpha \end{array}}\right) \int \partial _3^{\alpha } u_j \partial _j \partial _3^{2-\alpha } \omega \cdot \partial _3^2 \omega \textrm{d}x\\&:= M_{61} + M_{62}. \end{aligned}$$

Applying the Young inequality, Lemma 2.2 and the fact that $\Vert f\Vert _{L^{\infty }({\mathbb {R}}^3)} \le C \Vert f\Vert _{H^2({\mathbb {R}}^3)}$, we obtain

$$\begin{aligned} M_{61}&= - \sum _{k=1}^2 \sum _{j=1}^3 \sum _{\alpha = 1}^2 \left( {\begin{array}{c}2\\ \alpha \end{array}}\right) \int \partial _k^{\alpha } u_j \partial _j \partial _k^{2-\alpha } \omega \cdot \partial _k^2 \omega \textrm{d}x\\&= - 2 \sum _{k=1}^2 \sum _{j=1}^3 \int \partial _k u_j \partial _j \partial _k \omega \cdot \partial _k^2 \omega \textrm{d}x - \sum _{k=1}^2 \sum _{j=1}^3 \int \partial _k^{2} u_j \partial _j \omega \cdot \partial _k^2 \omega \textrm{d}x\\&\le C \sum _{k=1}^2 \sum _{j=1}^3 \Vert \partial _k u_j\Vert _{L^{\infty }} \Vert \partial _{j} \partial _{k}\omega \Vert _{L^2} \Vert \partial _k^2 \omega \Vert _{L^2} \\&\quad + C \sum _{k=1}^2 \sum _{j=1}^3 \Vert \partial _k^2 u_j\Vert _{L^2}^{\frac{1}{2}} \Vert \partial _k^2 \partial _1 u_j\Vert _{L^2}^{\frac{1}{2}} \Vert \partial _j \omega \Vert _{L^2}^{\frac{1}{2}} \Vert \partial _j \partial _2 \omega \Vert _{L^2}^{\frac{1}{2}} \Vert \partial _k^2 \omega \Vert _{L^2}^{\frac{1}{2}} \Vert \partial _k^2 \partial _3 \omega \Vert _{L^2}^{\frac{1}{2}}\\&\le C \Vert \nabla _h u\Vert _{H^2} \Vert \omega \Vert _{H^2}^2 + C \Vert \omega \Vert _{H^2} (\Vert \nabla _h u\Vert _{H^2} \Vert \omega \Vert _{H^2}^{\frac{1}{2}} \Vert \partial _3 \omega \Vert _{H^2}^{\frac{1}{2}})\\&\le C \Vert \omega \Vert _{H^2} (\Vert \nabla _h u\Vert _{H^2}^2 + \Vert \omega \Vert _{H^2}^2 + \Vert \partial _3 \omega \Vert _{H^2}^2 ). \end{aligned}$$

Similarly as $M_{61}$, by Lemma 2.2, the Young inequality and Hölder’s inequality yield

$$\begin{aligned} M_{62}&= -\sum _{j=1}^3 \sum _{\alpha = 1}^2 \left( {\begin{array}{c}2\\ \alpha \end{array}}\right) \int \partial _3^{\alpha } u_j \partial _j \partial _3^{2-\alpha } \omega \cdot \partial _3^2 \omega \textrm{d}x\\&= - 2 \int \partial _3 u \cdot \nabla \partial _3 \omega \cdot \partial _3^2 \omega \textrm{d}x - \int \partial _3^{2} u \cdot \nabla \omega \cdot \partial _3^2 \omega \textrm{d}x \\&\le C \Vert \partial _3 u\Vert ^{\frac{1}{2}}_{L^2} \Vert \partial _1 \partial _3 u\Vert ^{\frac{1}{2}}_{L^2} \Vert \nabla \partial _3 \omega \Vert ^{\frac{1}{2}}_{L^{2}} \Vert \nabla \partial _3 \partial _2 \omega \Vert ^{\frac{1}{2}}_{L^2} \Vert \partial _3^2 \omega \Vert ^{\frac{1}{2}} \Vert \partial _3^3 \omega \Vert ^{\frac{1}{2}}\\&\quad + C \Vert \partial _3^2 u\Vert ^{\frac{1}{2}}_{L^2} \Vert \partial _1 \partial _3^2 u\Vert ^{\frac{1}{2}}_{L^2} \Vert \nabla \omega \Vert ^{\frac{1}{2}}_{L^{2}} \Vert \nabla \partial _3 \omega \Vert ^{\frac{1}{2}}_{L^2} \Vert \partial _3^2 \omega \Vert ^{\frac{1}{2}} \Vert \partial _3^2 \partial _2 \omega \Vert ^{\frac{1}{2}}\\&\le C \Vert u\Vert _{H^2}^{\frac{1}{2}} \Vert \partial _1 u\Vert _{H^2}^{\frac{1}{2}}\Vert \omega \Vert _{H^2}^{\frac{1}{2}} \Vert \partial _3 \omega \Vert _{H^2} \Vert \omega \Vert _{H^2}^{\frac{1}{2}} + C \Vert u\Vert _{H^2}^{\frac{1}{2}} \Vert \partial _1 u\Vert _{H^2}^{\frac{1}{2}}\Vert \omega \Vert _{H^2}^{\frac{1}{2}} \Vert \partial _3 \omega \Vert _{H^2}^{\frac{1}{2}} \Vert \omega \Vert _{H^2}\\&\le C( \Vert u\Vert _{H^2} + \Vert \omega \Vert _{H^2} ) (\Vert \partial _3 \omega \Vert _{H^2}^2 + \Vert \omega \Vert _{H^2}^2 + \Vert \nabla _h u\Vert _{H^2}^2). \end{aligned}$$

Then we have

$$\begin{aligned} M_6 \le C (\Vert u\Vert _{H^2} + \Vert \omega \Vert _{H^2}) (\Vert \partial _3 \omega \Vert _{H^2}^2 + \Vert \omega \Vert _{H^2}^2 + \Vert \nabla _h u\Vert _{H^2}^2). \end{aligned}$$

Inserting these above estimates in (2.3), we infer that

$$\begin{aligned}&\frac{1}{2} \frac{\textrm{d}}{\textrm{d}t} \sum _{k=1}^3 \left( \Vert \partial _k^2 u(t)\Vert _{L^2}^2 +\Vert \partial _k^2 b(t)\Vert _{L^2}^2 + \Vert \partial _k^2 \omega (t)\Vert _{L^2}^2 \right) \nonumber \\&\qquad + c_1\sum _{k=1}^3\left( \Vert \nabla _h \partial _k^2 u\Vert _{L^2}^2 +\Vert \nabla _h \partial _k^2 b\Vert _{L^2}^2 + \Vert \partial _3 \partial _k^2 \omega \Vert _{L^2}^2 + \Vert \partial _k^2 \omega \Vert _{L^2}^2 \right) \nonumber \\&\le C (\Vert u\Vert _{H^2} + \Vert b\Vert _{H^2} + \Vert \omega \Vert _{H^2}) (\Vert \partial _3 \omega \Vert _{H^2}^2 + \Vert \omega \Vert _{H^2}^2 + \Vert \nabla _h u\Vert _{H^2}^2 + \Vert \nabla _h b\Vert _{H^2}^2), \end{aligned}$$

(2.4)

where $c_1 = \min \{ \frac{\mu }{2}, \nu , \gamma - 4 \chi , 4 \chi - \frac{16 \chi ^2}{\mu }\}.$

Adding (2.1) and (2.4) up, and integrating the result in [0, t], we find

$$\begin{aligned}&\Vert u (t) \Vert _{H^2}^2 + \Vert b(t)\Vert _{H^2}^2 + \Vert \omega (t)\Vert _{H^2}^2 \nonumber \\ {}&\qquad + 2 c_1 \int \limits ^t_0 \left( \Vert \nabla _h u(\tau )\Vert _{H^2}^2 + \Vert \nabla _h b(\tau )\Vert _{H^2}^2 + \Vert \partial _3 \omega (\tau )\Vert _{H^2}^2 + \Vert \omega (\tau )\Vert _{H^2}^2\right) d \tau \nonumber \\ {}&\quad \le \Vert u_0\Vert _{H^2}^2 + \Vert b_0\Vert _{H^2}^2 + \Vert \omega _0\Vert _{H^2}^2 + C \int \limits ^t_0 \left( \Vert u(\tau )\Vert _{H^2} + \Vert b(\tau )\Vert _{H^2} + \Vert \omega (\tau )\Vert _{H^2} \right) \nonumber \\ {}&\qquad \times \left( \Vert \partial _3 \omega (\tau )\Vert _{H^2}^2 + \Vert \omega (\tau )\Vert _{H^2}^2 + \Vert \nabla _h u(\tau )\Vert _{H^2}^2 + \Vert \nabla _h b(\tau )\Vert _{H^2}^2\right) d\tau . \end{aligned}$$

(2.5)

Let

$$\begin{aligned} E(t)&= \sup _{0 \le \tau \le t} \{\Vert u(\tau ) \Vert _{H^2}^2 + \Vert b(\tau )\Vert _{H^2}^2 + \Vert \omega (\tau )\Vert _{H^2}^2\} \nonumber \\ {}&\quad + 2 c_1 \int \limits ^t_0 \left( \Vert \partial _3 \omega (\tau )\Vert _{H^2}^2 + \Vert \omega (\tau )\Vert _{H^2}^2 + \Vert \nabla _h u(\tau )\Vert _{H^2}^2 + \Vert \nabla _h b(\tau )\Vert _{H^2}^2\right) d\tau . \end{aligned}$$

(2.6)

Then (2.5) implies

$$\begin{aligned} E(t) \le E(0) + C_0 E^{\frac{3}{2}}(t). \end{aligned}$$

(2.7)

Now we can start to show (1.5), by using the bootstrapping argument. By choosing $\epsilon \le \frac{1}{4 \sqrt{2} C_0}$ in (1.4), we have

$$\begin{aligned} E(0) \le \frac{1}{32 C^2_0}. \end{aligned}$$

(2.8)

To initial the bootstrapping argument, we assume that

$$\begin{aligned} E(t) \le \frac{1}{4C^2_0}. \end{aligned}$$

(2.9)

Then (2.7)–(2.9) imply

$$\begin{aligned} E(t) \le \frac{5}{32C_0^2}. \end{aligned}$$

(2.10)

This completes the proof (1.5). Substituting (1.5) with $\epsilon < \frac{2 c_1}{C}$ into (2.5), we have

$$\begin{aligned}&\Vert u (t) \Vert _{H^2}^2 + \Vert b(t)\Vert _{H^2}^2 + \Vert \omega (t)\Vert _{H^2}^2 \nonumber \\ {}&\qquad + \int \limits ^t_0 \left( \Vert \nabla _h u(\tau )\Vert _{H^2}^2 + \Vert \nabla _h b(\tau )\Vert _{H^2}^2 + \Vert \partial _3 \omega (\tau )\Vert _{H^2}^2 + \Vert \omega (\tau )\Vert _{H^2}^2\right) d\tau \nonumber \\ {}&\quad \le C (\Vert u_0\Vert _{H^2}^2 + \Vert b_0\Vert _{H^2}^2 + \Vert \omega _0\Vert _{H^2}^2). \end{aligned}$$

(2.11)

(2) Next we start to show (1.7). At the beginning of the proof, we show the uniform bound for these global solutions in $H^3 ({\mathbb {R}}^3)$ as follows.

Proposition 2.5

Let $\mu >0$, $\chi > 0$, $\nu >0$ and $\kappa >0$. Assume $(u_0, b_0, \omega _0) \in H^3({\mathbb {R}}^3)$ with $\nabla \cdot u_0 =0$, $\nabla \cdot b_0 = 0$ and (1.4) hold. If $\mu > 4 \chi$ and $\gamma > 4 \chi$, then the global solution $(u, b, \omega )$ of (1.3) obeys for all $t > 0$,

$$\begin{aligned}&\Vert u (t) \Vert _{H^3}^2 + \Vert b(t)\Vert _{H^3}^2 + \Vert \omega (t)\Vert _{H^3}^2 \nonumber \\ {}&\qquad + \int \limits ^t_0 \left( \Vert \nabla _h u(\tau )\Vert _{H^3}^2 + \Vert \nabla _h b(\tau )\Vert _{H^3}^2 + {\Vert \partial _3 \omega (\tau )\Vert _{H^3}^2} + \Vert \omega (\tau )\Vert _{H^3}^2\right) d\tau \nonumber \\ {}&\quad \le (\Vert u_0\Vert _{H^3}^2 + \Vert b_0\Vert _{H^3}^2 + \Vert \omega _0\Vert _{H^3}^2) e^{C(\Vert u_0\Vert _{H^2}^2 + \Vert b_0\Vert _{H^2}^2 + \Vert \omega _0\Vert _{H^2}^2)}\nonumber \\ {}&\quad \le C_3 (\Vert u_0\Vert _{H^3}^2 + \Vert b_0\Vert _{H^3}^2 + \Vert \omega _0\Vert _{H^3}^2) \end{aligned}$$

(2.12)

proof of Proposition 2.5

Applying $\partial _k^3$ with $k=1,2,3$ to (1.3)$_1$, (1.3)$_2$ and (1.3)$_3$, taking the $L^2$-inner product with $\partial _k^3 u$, $\partial _k^3 b$ and $\partial _k^3 \omega$, respectively, and adding the results up, we have

$$\begin{aligned}&\frac{1}{2} \frac{\textrm{d}}{\textrm{d}t} \sum _{k=1}^3 \left( \Vert \partial _k^3 u(t)\Vert _{L^2}^2 +\Vert \partial _k^3 b(t)\Vert _{L^2}^2 + \Vert \partial _k^3 \omega (t)\Vert _{L^2}^2 \right) + \mu \sum _{k=1}^3 \Vert \nabla _h \partial _k^3 u\Vert _{L^2}^2 + \chi \sum _{k=1}^3 \Vert \partial _k^3 u\Vert _{L^2}^2\nonumber \\&\qquad + \nu \sum _{k=1}^3 \Vert \nabla _h \partial _k^3 b\Vert _{L^2}^2 + \gamma \sum _{k=1}^3 \Vert \partial _3 \partial _k^3 \omega \Vert _{L^2}^2 + 4 \chi \sum _{k=1}^3 \Vert \partial _k^3 \omega \Vert _{L^2}^2 + \kappa \sum _{k=1}^3 \Vert \partial _k^3 \nabla \cdot \omega \Vert _{L^2}^2\nonumber \\&\quad = - \sum _{k=1}^3 \int \partial _k^3 (u \cdot \nabla u) \cdot \partial _k^3 u \textrm{d}x + \sum _{k=1}^3 \int \partial _k^3 (b \cdot \nabla b) \cdot \partial _k^3 u \textrm{d}x + 2 \chi \sum _{k=1}^3 \int \nabla \times \partial _k^3 \omega \cdot \partial _k^3 u \textrm{d}x \nonumber \\&\qquad - \sum _{k=1}^3 \int \partial _k^3 (u \cdot \nabla b) \cdot \partial _k^3 b \textrm{d}x + \sum _{k=1}^3 \int \partial _k^3 (b \cdot \nabla u) \cdot \partial _k^3 b \textrm{d}x - \sum _{k=1}^3 \int \partial _k^3 (u \cdot \nabla \omega ) \cdot \partial _k^3 \omega \textrm{d}x \nonumber \\&\qquad + 2 \chi \sum _{k=1}^3 \int \nabla \times \partial _k^3 u \cdot \partial _k^3 \omega \textrm{d}x\nonumber \\&\quad := N_1 + N_2 + N_3 + N_4 + N_5 + N_6 + N_7. \end{aligned}$$

(2.13)

Thus, we can rewrite $N_1$ as

$$\begin{aligned} N_1&= - \sum _{k=1}^3 \int \partial _k^3 (u \cdot \nabla u) \cdot \partial _k^3 u \textrm{d}x \\&= - \sum _{k=1}^2 \sum _{j=1}^3 \sum _{\alpha = 1}^3 \left( {\begin{array}{c}3\\ \alpha \end{array}}\right) \int \partial _k^{\alpha } u_j \partial _j \partial _k^{3-\alpha } u \cdot \partial _k^3 u \textrm{d}x - \sum _{j=1}^2 \sum _{\alpha = 1}^3 \left( {\begin{array}{c}3\\ \alpha \end{array}}\right) \int \partial _3^{\alpha } u_j \partial _j \partial _3^{3-\alpha } u \cdot \partial _3^3 u \textrm{d}x\\&\quad - \sum _{\alpha = 1}^3 \left( {\begin{array}{c}3\\ \alpha \end{array}}\right) \int \partial _3^{\alpha } u_3 \partial _3 \partial _3^{3-\alpha } u \cdot \partial _3^3 u \textrm{d}x\\&:= N_{11} + N_{12} + N_{13}. \end{aligned}$$

Applying the Young inequality, Lemma 2.2 and the fact that $\Vert f\Vert _{L^{\infty }({\mathbb {R}}^3)} \le C \Vert f\Vert _{H^2({\mathbb {R}}^3)}$, we obtain

$$\begin{aligned} N_{11}&= - \sum _{k=1}^2 \sum _{j=1}^3 \sum _{\alpha = 1}^3 \left( {\begin{array}{c}3\\ \alpha \end{array}}\right) \int \partial _k^{\alpha } u_j \partial _j \partial _k^{3-\alpha } u \cdot \partial _k^3 u \textrm{d}x\\&= - \sum _{k=1}^2 \sum _{j=1}^3 \left( {\begin{array}{c}3\\ 1\end{array}}\right) \int \partial _k u_j \partial _j \partial _k^{2} u \cdot \partial _k^3 u \textrm{d}x - \sum _{k=1}^2 \sum _{j=1}^3 \sum _{\alpha = 2}^3 \left( {\begin{array}{c}3\\ \alpha \end{array}}\right) \int \partial _k^{\alpha } u_j \partial _j \partial _k^{3-\alpha } u \cdot \partial _k^3 u \textrm{d}x\\&\le C \sum _{k=1}^2 \sum _{j=1}^3 \Vert \partial _k u_j\Vert _{L^{\infty }} \Vert \partial _j \partial _k^2 u\Vert _{L^2} \Vert \partial _k^3 u\Vert _{L^2}\\&\quad + C \sum _{k=1}^2 \sum _{j=1}^3 \sum _{\alpha = 2}^3 \Vert \partial _k^{\alpha } u_j\Vert _{L^2}^{\frac{1}{2}} \Vert \partial _3 \partial _k^{\alpha } u_j\Vert _{L^2}^{\frac{1}{2}} \Vert \partial _j \partial _k^{3 - \alpha } u \Vert _{L^2}^{\frac{1}{2}} \Vert \partial _2 \partial _j \partial _k^{3 - \alpha } u\Vert _{L^2}^{\frac{1}{2}} \Vert \partial _k^3 u\Vert _{L^2}^{\frac{1}{2}} \Vert \partial _1 \partial _k^{3} u\Vert _{L^2}^{\frac{1}{2}} \\&\quad \le C \Vert u\Vert _{H^3} \Vert \nabla _h u\Vert _{H^2} \Vert \nabla _h u\Vert _{H^3}\\&\quad \le \frac{c_1}{128} \Vert \nabla _h u\Vert _{H^3}^2 + C \Vert \nabla _h u\Vert _{H^2}^2 \Vert u\Vert _{H^3}^2, \end{aligned}$$

and

$$\begin{aligned} N_{12}&= \sum _{j=1}^2 \sum _{\alpha = 1}^3 \left( {\begin{array}{c}3\\ \alpha \end{array}}\right) \int \partial _3^{\alpha } u_j \partial _j \partial _3^{3-\alpha } u \cdot \partial _3^3 u \textrm{d}x\\&= -3 \sum _{j=1}^2 \int \partial _3 u_j \partial _j \partial _3^2 u \cdot \partial _3^3 u \textrm{d}x - \sum _{j=1}^2 \sum _{\alpha = 2}^3 \left( {\begin{array}{c}3\\ \alpha \end{array}}\right) \int \partial _3^{\alpha } u_j \partial _j \partial _3^{3-\alpha } u \cdot \partial _3^3 u \textrm{d}x\\&\le C \sum _{j=1}^2 \Vert \partial _3 u_j\Vert _{L^2}^{\frac{1}{2}} \Vert \partial _1 \partial _3 u_j\Vert _{L^2}^{\frac{1}{2}} \Vert \partial _j \partial _3^2 u \Vert _{L^2}^{\frac{1}{2}} \Vert \partial _j \partial _3^3 u\Vert _{L^2}^{\frac{1}{2}} \Vert \partial _3^3 u\Vert _{L^2}^{\frac{1}{2}} \Vert \partial _2 \partial _3^{3} u\Vert _{L^2}^{\frac{1}{2}} \\&\quad + C \sum _{j=1}^2 \sum _{\alpha = 2}^3 \Vert \partial _3^{\alpha } u_j\Vert _{L^2}^{\frac{1}{2}} \Vert \partial _1 \partial _3^{\alpha } u_j\Vert _{L^2}^{\frac{1}{2}} \Vert \partial _j \partial _3^{3 - \alpha } u \Vert _{L^2}^{\frac{1}{2}} \Vert \partial _j \partial _3^{4 - \alpha } u\Vert _{L^2}^{\frac{1}{2}} \Vert \partial _3^3 u\Vert _{L^2}^{\frac{1}{2}} \Vert \partial _2 \partial _3^{3} u\Vert _{L^2}^{\frac{1}{2}} \\&\le \frac{c_1}{128} \Vert \nabla _h u\Vert _{H^3}^2 + C \Vert \nabla _h u\Vert _{H^2}^2 \Vert u\Vert _{H^3}^2. \end{aligned}$$

Using the divergence free condition $\nabla \cdot u =0$, together with Lemma 2.2 and the Young inequality, we have

$$\begin{aligned} N_{13}&= - \sum _{\alpha = 1}^3 \left( {\begin{array}{c}3\\ \alpha \end{array}}\right) \int \partial _3^{\alpha } u_3 \partial _3 \partial _3^{3-\alpha } u \cdot \partial _3^3 u \textrm{d}x\\&= 3 \int \partial _h u_h \partial _3^3 u \cdot \partial _3^3 u \textrm{d}x - \sum _{\alpha = 2}^3 \left( {\begin{array}{c}3\\ \alpha \end{array}}\right) \int \partial _3^{\alpha } u_3 \partial _3 \partial _3^{3-\alpha } u \cdot \partial _3^3 u \textrm{d}x\\&\le C \Vert \partial _h u_h\Vert _{L^2}^{\frac{1}{2}} \Vert \partial _3 \partial _h u_h\Vert _{L^2}^{\frac{1}{2}} \Vert \partial _3^3 u \Vert _{L^2}^{\frac{1}{2}} \Vert \partial _2 \partial _3^3 u\Vert _{L^2}^{\frac{1}{2}} \Vert \partial _3^3 u\Vert _{L^2}^{\frac{1}{2}} \Vert \partial _1 \partial _3^{3} u\Vert _{L^2}^{\frac{1}{2}} \\&\quad + C \sum _{\alpha = 2}^3 \Vert \partial _3^{\alpha -1} \partial _h u_h\Vert _{L^2}^{\frac{1}{2}} \Vert \partial _3 \partial _3^{\alpha -1} \partial _h u_h\Vert _{L^2}^{\frac{1}{2}} \Vert \partial _3^{4 - \alpha } u \Vert _{L^2}^{\frac{1}{2}} \Vert \partial _2 \partial _3^{4 - \alpha } u\Vert _{L^2}^{\frac{1}{2}} \Vert \partial _3^3 u\Vert _{L^2}^{\frac{1}{2}} \Vert \partial _1 \partial _3^{3} u\Vert _{L^2}^{\frac{1}{2}} \\&\le \frac{c_1}{128} \Vert \nabla _h u\Vert _{H^3}^2 + C \Vert \nabla _h u\Vert _{H^2}^2 \Vert u\Vert _{H^3}^2. \end{aligned}$$

Then we have

$$\begin{aligned} N_1 \le \frac{c_1}{32} \Vert \nabla _h u\Vert _{H^3}^2 + C \Vert \nabla _h u\Vert _{H^2}^2 \Vert u\Vert _{H^3}^2. \end{aligned}$$

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

$$\begin{aligned} \sum _{k=1}^3 \int b \cdot \nabla \partial _k^3 b \cdot \partial _k^3 u \textrm{d}x + \sum _{k=1}^3 \int b \cdot \nabla \partial _k^3 u \cdot \partial _k^3 b \textrm{d}x = 0. \end{aligned}$$

Then applying the same strategy as above, we get

$$\begin{aligned} N_2 + N_5&= \sum _{k=1}^3 \int \partial _k^3 (b \cdot \nabla b) \cdot \partial _k^3 u \textrm{d}x + \sum _{k=1}^3 \int \partial _k^3 (b \cdot \nabla u) \cdot \partial _k^3 b \textrm{d}x\\&= \sum _{k=1}^3 \sum _{j =1}^3 \sum _{\alpha = 1}^3 \left( {\begin{array}{c}3\\ \alpha \end{array}}\right) \left( \int \partial _k^{\alpha } b_j \partial _j \partial _{k}^{3 - \alpha } b \cdot \partial _k^3 u \textrm{d}x + \int \partial _k^{\alpha } b_j \partial _j \partial _{k}^{3 - \alpha } u \cdot \partial _k^3 b \textrm{d}x \right) \\&\le \frac{c_1}{16} ( \Vert \nabla _h u\Vert _{H^3}^2 + \Vert \nabla _h b\Vert _{H^3}^2) + C ( \Vert \nabla _h u\Vert _{H^2}^2 + \Vert \nabla _h b\Vert _{H^2}^2)(\Vert u\Vert _{H^3}^2 + \Vert b\Vert _{H^3}^2). \end{aligned}$$

Similarly, we have

$$\begin{aligned} N_4&= - \sum _{k=1}^3 \int \partial _k^3 (u \cdot \nabla b) \cdot \partial _k^3 b \textrm{d}x \\&= - \sum _{k=1}^3 \sum _{j=1}^3 \sum _{\alpha = 1}^3 \left( {\begin{array}{c}3\\ \alpha \end{array}}\right) \int \partial _k^{\alpha } u_j \partial _j \partial _k^{3-\alpha } b \cdot \partial _k^3 b \textrm{d}x \\&\le \frac{c_1}{16} ( \Vert \nabla _h u\Vert _{H^3}^2 + \Vert \nabla _h b\Vert _{H^3}^2) + C ( \Vert \nabla _h u\Vert _{H^2}^2 + \Vert \nabla _h b\Vert _{H^2}^2)(\Vert u\Vert _{H^3}^2 + \Vert b\Vert _{H^3}^2). \end{aligned}$$

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

$$\begin{aligned} N_3 + N_7&= 2 \chi \sum _{k=1}^3 \int \nabla \times \partial _k^3 \omega \cdot \partial _k^3 u \textrm{d}x + 2 \chi \sum _{k=1}^3 \int \nabla \times \partial _k^3 u \cdot \partial _k^3 \omega \textrm{d}x\\&= 4 \chi \sum _{k=1}^3 \int \nabla \times \partial _k^3 u \cdot \partial _k^3 \omega \textrm{d}x\\&\le \frac{\mu }{2} \sum _{k = 1}^3 \Vert \nabla _h \partial _k^3 u\Vert _{L^2}^2 + \frac{16 \chi ^2}{\mu } \sum _{k = 1}^3 \Vert \partial _k^3 \omega \Vert _{L^2}^2 + \chi \sum _{k = 1}^3 \Vert \partial _k^3 u\Vert _{L^2}^2 + 4 \chi \sum _{k = 1}^3 {\Vert \partial _3 \partial _k^3 \omega \Vert _{L^2}^2}. \end{aligned}$$

Next, we consider $N_6$,

$$\begin{aligned} N_6&= - \sum _{k=1}^3 \int \partial _{k}^3 (u \cdot \nabla \omega ) \cdot \partial _{k}^3 \omega \textrm{d}x\\&= - \sum _{k=1}^3 \sum _{j=1}^3 \sum _{\alpha = 1}^3 \left( {\begin{array}{c}3\\ \alpha \end{array}}\right) \int \partial _k^{\alpha } u_j \partial _j \partial _k^{3-\alpha } \omega \cdot \partial _k^3 \omega \textrm{d}x \\&= - \sum _{k=1}^2 \sum _{j=1}^3 \sum _{\alpha = 1}^3 \left( {\begin{array}{c}3\\ \alpha \end{array}}\right) \int \partial _k^{\alpha } u_j \partial _j \partial _k^{3-\alpha } \omega \cdot \partial _k^3 \omega \textrm{d}x - \sum _{j=1}^3 \sum _{\alpha = 1}^3\left( {\begin{array}{c}3\\ \alpha \end{array}}\right) \int \partial _3^{\alpha } u_j \partial _j \partial _3^{3-\alpha } \omega \cdot \partial _3^3 \omega \textrm{d}x\\&:= N_{61} + N_{62}. \end{aligned}$$

We can rewrite $N_{61}$ as:

$$\begin{aligned} N_{61}&= - \sum _{k=1}^2 \sum _{j=1}^3 \sum _{\alpha = 1}^3 \left( {\begin{array}{c}3\\ \alpha \end{array}}\right) \int \partial _k^{\alpha } u_j \partial _j \partial _k^{3-\alpha } \omega \cdot \partial _k^3 \omega \textrm{d}x\\&= - \sum _{k=1}^2 \sum _{j=1}^3 \left( {\begin{array}{c}3\\ 1\end{array}}\right) \int \partial _k u_j \partial _j \partial _k^{2} \omega \cdot \partial _k^3 \omega \textrm{d}x - \sum _{k=1}^2 \sum _{j=1}^3 \sum _{\alpha = 2}^3 \left( {\begin{array}{c}3\\ \alpha \end{array}}\right) \int \partial _k^{\alpha } u_j \partial _j \partial _k^{3-\alpha } \omega \cdot \partial _k^3 \omega \textrm{d}x\\&= N_{611} + N_{612}. \end{aligned}$$

For $N_{611}$, by the Young inequality and Hölder inequality, we have

$$\begin{aligned} N_{611}&= - 3 \sum _{k=1}^2 \sum _{j=1}^3 \int \partial _k u_j \partial _j \partial _k^{2} \omega \cdot \partial _k^3 \omega \textrm{d}x \\&\le C \sum _{k=1}^{2} \sum _{j=1}^3 \Vert \partial _k u_j\Vert _{L^{\infty }} \Vert \partial _j \partial _k^2 \omega \Vert _{L^2} \Vert \partial _{k}^3 \omega \Vert _{L^2} \\&\le C \Vert \nabla _h u\Vert _{H^2} \Vert \omega \Vert _{H^3}^2\\&\le \frac{c_1}{64} \Vert \omega \Vert _{H^3}^2 + C (\Vert \nabla _h u\Vert ^2_{H^2} \Vert \omega \Vert _{H^3}^2), \end{aligned}$$

where we also used the fact that $\Vert f\Vert _{L^{\infty }({\mathbb {R}}^3)} \le C \Vert f\Vert _{H^2({\mathbb {R}}^3)}$. By Lemma 2.2 and the Young inequality, we have

$$\begin{aligned} N_{612}&= - \sum _{k=1}^2 \sum _{j=1}^3 \sum _{\alpha = 2}^3 \left( {\begin{array}{c}3\\ \alpha \end{array}}\right) \int \partial _k^{\alpha } u_j \partial _j \partial _k^{3-\alpha } \omega \cdot \partial _k^3 \omega \textrm{d}x\\&\le C \sum _{k=1}^2 \sum _{\alpha = 2}^3 \Vert \partial _k^{\alpha } u \Vert _{L^2}^{\frac{1}{2}} \Vert \partial _1 \partial _k^{\alpha } u \Vert _{L^2}^{\frac{1}{2}} \Vert \nabla \partial _k^{3 - \alpha } \omega \Vert _{L^2}^{\frac{1}{2}} \Vert \partial _2 \nabla \partial _k^{3 - \alpha } \omega \Vert _{L^2}^{\frac{1}{2}} \Vert \partial _k^{3} \omega \Vert _{L^2}^{\frac{1}{2}} \Vert \partial _3 \partial _k^{3} \omega \Vert _{L^2}^{\frac{1}{2}}\\&\le C \Vert \nabla _h u\Vert _{H^2}^{\frac{1}{2}} \Vert \nabla _h u\Vert _{H^3}^{\frac{1}{2}} \Vert \omega \Vert _{H^2}^{\frac{1}{2}} \Vert \omega \Vert _{H^3}^{\frac{1}{2}} \Vert \omega \Vert _{H^3}^{\frac{1}{2}} \Vert \partial _3 \omega \Vert _{H^3}^{\frac{1}{2}} \\&\le C (\Vert \nabla _h u\Vert _{H^2}^2 + \Vert \omega \Vert _{H^2}^2) \Vert \omega \Vert _{H^3}^2 + \frac{c_1}{64} ( \Vert \partial _3 \omega \Vert _{H^3}^2 + \Vert \nabla _h u\Vert _{H^3}^2). \end{aligned}$$

Then, we have

$$\begin{aligned} N_{61} \le C(\Vert \nabla _h u\Vert _{H^2}^2 + \Vert \omega \Vert _{H^2}^2) \Vert \omega \Vert _{H^3}^2 + \frac{c_1}{32} ( \Vert \omega \Vert _{H^3}^2 + \Vert \partial _3 \omega \Vert _{H^3}^2 + \Vert \nabla _h u\Vert _{H^3}^2). \end{aligned}$$

Similarly,

$$\begin{aligned} N_{62}&= -\sum _{j=1}^3 \sum _{\alpha = 1}^3 \left( {\begin{array}{c}3\\ \alpha \end{array}}\right) \int \partial _3^{\alpha } u_j \partial _j \partial _3^{3-\alpha } \omega \cdot \partial _3^3 \omega \textrm{d}x\\&= - \sum _{j=1}^3 \sum _{\alpha = 2}^3 \left( {\begin{array}{c}3\\ \alpha \end{array}}\right) \int \partial _3^{\alpha } u_j \partial _j \partial _3^{3-\alpha } \omega \cdot \partial _3^3 \omega \textrm{d}x - 3 \sum _{j=1}^3 \int \partial _3 u_j \partial _j \partial _3^{2} \omega \cdot \partial _3^3 \omega \textrm{d}x\\&\le C \sum _{\alpha = 2}^3 \Vert \partial _3^{\alpha } u_j\Vert ^{\frac{1}{2}}_{L^2}\Vert \partial _1 \partial _3^{\alpha } u_j\Vert ^{\frac{1}{2}}_{L^2} \Vert \nabla \partial _3^{3- \alpha } \omega \Vert ^{\frac{1}{2}}_{L^2} \Vert \partial _3 \nabla \partial _3^{3- \alpha } \omega \Vert ^{\frac{1}{2}}_{L^2} \Vert \partial _3^{3} \omega \Vert ^{\frac{1}{2}}_{L^2} \Vert \partial _2 \partial _3^{3} \omega \Vert ^{\frac{1}{2}}_{L^2} \\&\quad + C \Vert u\Vert _{H^3} \Vert \partial _3 \omega \Vert _{H^2}\Vert \partial _3 \omega \Vert _{H^3}\\&\le C (\Vert u\Vert _{H^3}^2 + \Vert \omega \Vert _{H^3}^2)(\Vert \omega \Vert _{H^2}^2 + \Vert \partial _3 \omega \Vert _{H^2}^2 )+ \frac{c_1}{32}{( \Vert \partial _3 \omega \Vert _{H^3}^2 + \Vert \nabla _h u\Vert _{H^3}^2 ).} \end{aligned}$$

Combining the estimates $N_{61}$ and $N_{62}$ yields

$$\begin{aligned} N_6&\le \frac{c_1}{16} (\Vert \omega \Vert _{H^3}^2 + \Vert \partial _3 \omega \Vert _{H^3}^2 + \Vert \nabla _h u\Vert _{H^3}^2) \\&\quad + C (\Vert \omega \Vert _{H^3}^2 + \Vert u\Vert _{H^3}^2 )( \Vert \nabla _h u\Vert _{H^2}^2 + \Vert \omega \Vert _{H^2}^2 + \Vert \partial _3 \omega \Vert _{H^2}^2 ). \end{aligned}$$

Inserting the above bounds into (2.13), and combining with (2.1), one obtains

$$\begin{aligned}&\frac{\textrm{d}}{\textrm{d}t} ( \Vert u(t) \Vert _{H^3}^2 + \Vert b(t)\Vert _{H^3}^2 + \Vert \omega (t)\Vert _{H^3}^2) \nonumber \\&\qquad + c_1 \left( \Vert \nabla _h u\Vert _{H^3}^2 + \Vert \nabla _h b\Vert _{H^3}^2 + \Vert \partial _3 \omega \Vert _{H^3}^2 + \Vert \omega \Vert _{H^3}^2\right) \nonumber \\&\quad \le {C\left( \Vert u\Vert ^2_{H^3} + \Vert b\Vert ^2_{H^3} + \Vert \omega \Vert ^2_{H^3} \right) \left( \Vert \partial _3 \omega \Vert _{H^2}^2 + \Vert \omega \Vert _{H^2}^2 + \Vert \nabla _h u\Vert _{H^2}^2 + \Vert \nabla _h b\Vert _{H^2}^2 \right) .} \end{aligned}$$

(2.14)

This together with Gronwall’s inequality and (1.6) implies

$$\begin{aligned}&\Vert u (t) \Vert _{H^3}^2 + \Vert b(t)\Vert _{H^3}^2 + \Vert \omega (t)\Vert _{H^3}^2 \nonumber \\&\qquad + c_1 \int ^t_0 \left( \Vert \nabla _h u(\tau )\Vert _{H^3}^2 + \Vert \nabla _h b(\tau )\Vert _{H^3}^2 + \Vert \partial _3 \omega (\tau )\Vert _{H^3}^2 + \Vert \omega (\tau )\Vert _{H^3}^2\right) d\tau \nonumber \\&\quad \le (\Vert u_0\Vert _{H^3}^2 + \Vert b_0\Vert _{H^3}^2 + \Vert \omega _0\Vert _{H^3}^2) e^{C (\Vert u_0\Vert _{H^2}^2 + \Vert b_0\Vert _{H^2}^2 + \Vert \omega _0\Vert _{H^2}^2) }\nonumber \\&\quad \le C (\Vert u_0\Vert _{H^3}^2 + \Vert b_0\Vert _{H^3}^2 + \Vert \omega _0\Vert _{H^3}^2) . \end{aligned}$$

(2.15)

We set $C_3 = C$, which is (2.12).

$\square$

To complete the proof of (1.7), we use the induction for s. The case s = 3 has been proved in Proposition 2.5. Assume that for $s \ge 3$,

$$\begin{aligned}&\Vert u (t) \Vert _{H^{s-1}}^2 + \Vert b(t)\Vert _{H^{s-1}}^2 + \Vert \omega (t)\Vert _{H^{s-1}}^2 \nonumber \\ {}&\qquad + c_1 \int \limits ^t_0 \left( \Vert \nabla _h u(\tau )\Vert _{H^{s-1}}^2 + \Vert \nabla _h b(\tau )\Vert _{H^{s-1}}^2 + \Vert \partial _3 \omega (\tau )\Vert _{H^{s-1}}^2 + \Vert \omega (\tau )\Vert _{H^{s-1}}^2\right) d\tau \nonumber \\ {}&\quad \le C_{s-1} (\Vert u_0\Vert _{H^{s-1}}^2 + \Vert b_0\Vert _{H^{s-1}}^2 + \Vert \omega _0\Vert _{H^s-1}^2) . \end{aligned}$$

(2.16)

If we have

$$\begin{aligned}&\Vert u (t) \Vert _{H^{s}}^2 + \Vert b(t)\Vert _{H^{s}}^2 + \Vert \omega (t)\Vert _{H^{s}}^2 \nonumber \\ {}&\qquad + c_1 \int \limits ^t_0 \left( \Vert \nabla _h u(\tau )\Vert _{H^{s}}^2 + \Vert \nabla _h b(\tau )\Vert _{H^{s}}^2 + \Vert \partial _3 \omega (\tau )\Vert _{H^{s}}^2 + \Vert \omega (\tau )\Vert _{H^{s}}^2\right) d\tau \nonumber \\ {}&\quad \le C_{s} (\Vert u_0\Vert _{H^{s}}^2 + \Vert b_0\Vert _{H^{s}}^2 + \Vert \omega _0\Vert _{H^s}^2) , \end{aligned}$$

(2.17)

then we yield the desired estimate (1.7). Next, we need to verify that (2.17) is correct. Applying $\partial _k^s$ with $k=1,2,3$ to (1.3)$_1$, (1.3)$_2$ and (1.3)$_3$, taking the $L^2$-inner product with $\partial _k^s u$, $\partial _k^s b$ and $\partial _k^s \omega$, respectively, and adding the results up, we have

$$\begin{aligned}&\frac{1}{2} \frac{\textrm{d}}{\textrm{d}t} \sum _{k=1}^3 \left( \Vert \partial _k^s u(t)\Vert _{L^2}^2 +\Vert \partial _k^s b(t)\Vert _{L^2}^2 + \Vert \partial _k^s \omega (t)\Vert _{L^2}^2 \right) + \mu \sum _{k=1}^3 \Vert \nabla _h \partial _k^s u\Vert _{L^2}^2 + \chi \sum _{k=1}^3 \Vert \partial _k^s u\Vert _{L^2}^2\nonumber \\&\qquad + \nu \sum _{k=1}^3 \Vert \nabla _h \partial _k^s b\Vert _{L^2}^2 + \gamma \sum _{k=1}^3 \Vert \partial _3 \partial _k^s \omega \Vert _{L^2}^2 + 4 \chi \sum _{k=1}^3 \Vert \partial _k^s \omega \Vert _{L^2}^2 + \kappa \sum _{k=1}^3 \Vert \partial _k^s \nabla \cdot \omega \Vert _{L^2}^2\nonumber \\&\quad = - \sum _{k=1}^3 \int \partial _k^s (u \cdot \nabla u) \cdot \partial _k^s u \textrm{d}x + \sum _{k=1}^3 \int \partial _k^s (b \cdot \nabla b) \cdot \partial _k^s u \textrm{d}x + 2 \chi \sum _{k=1}^3 \int \nabla \times \partial _k^s \omega \cdot \partial _k^s u \textrm{d}x \nonumber \\&\qquad - \sum _{k=1}^3 \int \partial _k^s (u \cdot \nabla b) \cdot \partial _k^s b \textrm{d}x + \sum _{k=1}^3 \int \partial _k^s (b \cdot \nabla u) \cdot \partial _k^s b \textrm{d}x - \sum _{k=1}^3 \int \partial _k^s (u \cdot \nabla \omega ) \cdot \partial _k^s \omega \textrm{d}x \nonumber \\&\qquad + 2 \chi \sum _{k=1}^3 \int \nabla \times \partial _k^s u \cdot \partial _k^s \omega \textrm{d}x\nonumber \\&\quad := J_1 + J_2 + J_3 + J_4 + J_5 + J_6 + J_7. \end{aligned}$$

(2.18)

Since the estimate method is very similar to $N_1$ - $N_7$ in the proof of Proposition 2.5, we omit the details for simplicity and have

$$\begin{aligned}&J_1 + J_4 \le \frac{c_1}{16} (\Vert \nabla _h u\Vert ^2_{H^s} + \Vert \nabla _h b\Vert _{H^s}^2) + C (\Vert \nabla _h u\Vert _{H^{s-1}}^2 + \Vert \nabla _h b\Vert _{H^{s-1}}^2) (\Vert u\Vert ^2_{H^s} +\Vert b\Vert ^2_{H^s} ), \\&J_2 + J_5 \le \frac{c_1}{16} (\Vert \nabla _h u\Vert ^2_{H^s} + \Vert \nabla _h b\Vert _{H^s}^2) + C (\Vert \nabla _h u\Vert _{H^{s-1}}^2 + \Vert \nabla _h b\Vert _{H^{s-1}}^2) (\Vert u\Vert ^2_{H^s} +\Vert b\Vert ^2_{H^s} ), \\&J_3 + J_7 \le \frac{\mu }{2} \sum _{k = 1}^3 \Vert \nabla _h \partial _k^s u\Vert _{L^2}^2 + \frac{16 \chi ^2}{\mu } \sum _{k = 1}^3 \Vert \partial _k^s \omega \Vert _{L^2}^2 + \chi \sum _{k = 1}^3 \Vert \partial _k^s u\Vert _{L^2}^2 + 4 \chi \sum _{k = 1}^3 {\Vert \partial _3 \partial _k^s \omega \Vert _{L^2}^2}, \end{aligned}$$

and

$$\begin{aligned} J_6&\le \frac{c_1}{16} (\Vert \nabla _h u\Vert ^2_{H^s} + \Vert \omega \Vert _{H^s}^2 + \Vert \partial _3 \omega \Vert _{H^s}^2) \\&\quad + C (\Vert \nabla _h u\Vert _{H^{s-1}}^2 + \Vert \partial _3 \omega \Vert _{H^{s-1}}^2 + \Vert \omega \Vert _{H^{s-1}}^2 ) (\Vert u\Vert ^2_{H^s} +\Vert \omega \Vert ^2_{H^s} ). \end{aligned}$$

Inserting the above bounds into (2.18), and combining with (2.1), we have

$$\begin{aligned}&\frac{\textrm{d}}{\textrm{d}t} ( \Vert u(t) \Vert _{H^s}^2 + \Vert b(t)\Vert _{H^s}^2 + \Vert \omega (t)\Vert _{H^s}^2) \nonumber \\&\qquad + c_1 \left( \Vert \nabla _h u\Vert _{H^s}^2 + \Vert \nabla _h b\Vert _{H^s}^2 + \Vert \partial _3 \omega \Vert _{H^s}^2 + \Vert \omega \Vert _{H^s}^2\right) \nonumber \\&\quad \le C \left( \Vert u\Vert ^2_{H^s} + \Vert b\Vert ^2_{H^s} + \Vert \omega \Vert ^2_{H^s} \right) \left( \Vert \partial _3 \omega \Vert _{H^{s-1}}^2 + \Vert \omega \Vert _{H^{s-1}}^2 + \Vert \nabla _h u\Vert _{H^{s-1}}^2 + \Vert \nabla _h b\Vert _{H^{s-1}}^2 \right) . \end{aligned}$$

(2.19)

This together with Gronwall’s inequality and (2.16) implies

$$\begin{aligned}&\Vert u (t) \Vert _{H^s}^2 + \Vert b(t)\Vert _{H^s}^2 + \Vert \omega (t)\Vert _{H^s}^2 \nonumber \\ {}&\qquad + c_1 \int \limits ^t_0 \left( \Vert \nabla _h u(\tau )\Vert _{H^s}^2 + \Vert \nabla _h b(\tau )\Vert _{H^s}^2 + \Vert \partial _3 \omega (\tau )\Vert _{H^s}^2 + \Vert \omega (\tau )\Vert _{H^s}^2\right) d\tau \nonumber \\ {}&\quad \le (\Vert u_0\Vert _{H^s}^2 + \Vert b_0\Vert _{H^s}^2 + \Vert \omega _0\Vert _{H^s}^2) \exp \{ CC_{s-1}(\Vert u_0\Vert _{H^{s-1}}^2 + \Vert b_0\Vert _{H^{s-1}}^2 + \Vert \omega _0\Vert _{H^{s-1}}^2) \}\nonumber \\ {}&\quad \le C_s (\Vert u_0\Vert _{H^s}^2 + \Vert b_0\Vert _{H^s}^2 + \Vert \omega _0\Vert _{H^s}^2), \end{aligned}$$

where $C_s$ is dependent on $\mu$, $\chi$, $\nu$, $\kappa$ and the initial data, which is (2.17).

We now briefly explain the uniqueness at the $H^2$-level, which can be quickly established. Let $(u^{(1)}, b^{(1)}, \omega ^{(1)})$ and $(u^{(2)}, b^{(2)}, \omega ^{(2)})$ be two solutions of (1.3) in the regularity class, for $T >0$,

$$\begin{aligned} (u^{(1)}, b^{(1)}, \omega ^{(1)}), \ \ (u^{(2)}, b^{(2)}, \omega ^{(2)}) \in L^{\infty }(0, T; H^2({\mathbb {R}}^3)). \end{aligned}$$

(2.20)

Their difference $({\tilde{u}}, {\tilde{b}}, {\tilde{\omega }})$ with

$$\begin{aligned} {\tilde{u}} = u^{(1)} - u^{(2)}, \quad {\tilde{b}} = b^{(1)} - b^{(2)}, \quad {\tilde{\omega }} = \omega ^{(1)} - \omega ^{(2)} \end{aligned}$$

satisfies

$$\begin{aligned} \left\{ \begin{array}{lcl} \partial _t {\tilde{u}} + u^{(1)} \cdot \nabla {\tilde{u}} + {\tilde{u}} \cdot \nabla u^{(2)} + \chi {\tilde{u}} = - \nabla {\tilde{\pi }} + \mu \Delta _h {\tilde{u}} + b^{(1)}\cdot \nabla {\tilde{b}} + {\tilde{b}} \cdot \nabla b^{(2)} + 2\chi \nabla \times {\tilde{\omega }},\\ \partial _t {\tilde{b}} + u^{(1)} \cdot \nabla {\tilde{b}} + {\tilde{u}} \cdot \nabla b^{(2)}= \nu \Delta _h {\tilde{b}} + b^{(1)} \cdot \nabla {\tilde{u}} + {\tilde{b}} \cdot \nabla u^{(2)}, \\ \partial _t {\tilde{\omega }} + u^{(1)}\cdot \nabla {\tilde{\omega }} + {\tilde{u}} \cdot \nabla \omega ^{(2)} + 4\chi {\tilde{\omega }} = \gamma \partial _{33} {\tilde{\omega }} + \kappa \nabla \nabla \cdot {\tilde{\omega }} + 2\chi \nabla \times {\tilde{u}}, \\ \nabla \cdot {\tilde{u}} =0, ~ \nabla \cdot {\tilde{b}} =0,\\ {\tilde{u}}(x, 0) = 0, {\tilde{\omega }}(x, 0) = 0, {\tilde{b}}(x, 0) = 0, \end{array} \right. \end{aligned}$$

where ${\tilde{\pi }} = \pi ^{(1)} - \pi ^{(2)}$ with $\pi ^{(1)}$ and $\pi ^{(2)}$ being the pressure corresponding to $u^{(1)}$ and $u^{(2)}$, respectively. Taking the $L^2$-inner product to the above system with ${\tilde{u}}$, ${\tilde{b}}$ and ${\tilde{\omega }}$, respectively, we obtain

$$\begin{aligned}&\frac{1}{2} \frac{\textrm{d}}{\textrm{d}t} (\Vert {\tilde{u}}(t)\Vert _{L^2}^2 +\Vert {\tilde{b}}(t)\Vert _{L^2}^2 + \Vert {\tilde{\omega }}(t)\Vert _{L^2}^2) + \mu \Vert \nabla _h {\tilde{u}}\Vert _{L^2}^2 + \nu \Vert \nabla _h {\tilde{b}}\Vert _{L^2}^2 + \gamma \Vert \partial _3 {\tilde{\omega }}\Vert _{L^2}^2 \nonumber \\&\qquad + \chi \Vert {\tilde{u}}\Vert _{L^2}^2 + \kappa \Vert \nabla \cdot {\tilde{\omega }}\Vert _{L^2}^2 + 4 \chi \Vert {\tilde{\omega }}\Vert _{L^2}^2 \nonumber \\&\quad = \int {\tilde{b}} \cdot \nabla b^{(2)} \cdot {\tilde{u}} \textrm{d}x - \int {\tilde{u}} \cdot \nabla u^{(2)} \cdot {\tilde{u}} \textrm{d}x - \int {\tilde{u}} \cdot \nabla b^{(2)} \cdot {\tilde{b}} \textrm{d}x + \int {\tilde{b}} \cdot \nabla u^{(2)} \cdot {\tilde{b}} \textrm{d}x\nonumber \\&\qquad {- \int {\tilde{u}} \cdot \nabla \omega ^{(2)} \cdot {\tilde{\omega }} \textrm{d}x }+ 2 \chi \int \nabla \times {\tilde{\omega }} \cdot {\tilde{u}} \textrm{d}x + 2 \chi \int \nabla \times {\tilde{u}} \cdot {\tilde{\omega }} \textrm{d}x\nonumber \\&\quad = I_1 + I_2 + I_3 + I_4 + I_5 + I_6 + I_7. \end{aligned}$$

(2.21)

Applying Hölder’s inequality, Lemma 2.1, Lemma 2.2, Gagliardo–Nirenberg inequality and the Young inequality yields

$$\begin{aligned} I_1&= \int {\tilde{b}} \cdot \nabla b^{(2)} \cdot {\tilde{u}} \textrm{d}x \\&\le \Vert {\tilde{b}}\Vert _{L^4_{h}L^2_{x_3}} \Vert \nabla b^{(2)}\Vert _{L^2_{h}L^{\infty }_{x_3}} \Vert {\tilde{u}}\Vert _{L^4_{h}L^2_{x_3}}\\&\le \left\| \Vert {\tilde{b}}\Vert _{L^4_{h}} \right\| _{L^2_{x_3}} \left\| \Vert \nabla b^{(2)}\Vert _{L^{\infty }_{x_3}} \right\| _{L^2_{h}} \left\| \Vert {\tilde{u}}\Vert _{L^4_{h}} \right\| _{L^2_{x_3}}\\&\le \left\| \Vert {\tilde{b}}\Vert _{L^2_{h}}^{\frac{1}{2}} \Vert \nabla _h {\tilde{b}}\Vert _{L^2_{h}}^{\frac{1}{2}} \right\| _{L^2_{x_3}} \left\| \Vert \nabla b^{(2)}\Vert _{L^{2}_{x_3}}^{\frac{1}{2}} \Vert \partial _3 \nabla b^{(2)}\Vert _{L^{2}_{x_3}}^{\frac{1}{2}} \right\| _{L^2_{h}} \left\| \Vert {\tilde{u}}\Vert _{L^2_{h}}^{\frac{1}{2}}\Vert \nabla _h {\tilde{u}}\Vert _{L^2_{h}}^{\frac{1}{2}}\right\| _{L^2_{x_3}}\\&\le C \Vert {\tilde{b}}\Vert _{L^2}^{\frac{1}{2}} \Vert \nabla _h {\tilde{b}}\Vert _{L^2}^{\frac{1}{2}} \Vert b^{(2)}\Vert _{H^2} \Vert {\tilde{u}}\Vert _{L^2}^{\frac{1}{2}} \Vert \nabla _h {\tilde{u}}\Vert _{L^2}^{\frac{1}{2}}\\&\le \frac{c_1}{16} ( \Vert \nabla _h {\tilde{u}}\Vert _{L^2}^2 + \Vert \nabla _h {\tilde{b}}\Vert _{L^2}^2 ) + C \Vert b^{(2)}\Vert _{H^2}^2 ( \Vert {\tilde{b}}\Vert _{L^2}^2 + \Vert {\tilde{u}}\Vert _{L^2}^2). \end{aligned}$$

Similarly, we have

$$\begin{aligned} I_2&= - \int {\tilde{u}} \cdot \nabla u^{(2)} \cdot {\tilde{u}} \textrm{d}x\\&\le \Vert {\tilde{u}}\Vert _{L^4_{h}L^2_{x_3}}^2 \Vert \nabla u^{(2)}\Vert _{L^2_{h}L^{\infty }_{x_3}}\\&\le C \Vert {\tilde{u}}\Vert _{L^2} \Vert \nabla _h {\tilde{u}}\Vert _{L^2} \Vert \nabla u^{(2)}\Vert _{L^2}^{\frac{1}{2}} \Vert \partial _3\nabla u^{(2)}\Vert _{L^2}^{\frac{1}{2}}\\&\le \frac{c_1}{16} \Vert \nabla _h {\tilde{u}}\Vert _{L^2}^2 + C \Vert u^{(2)}\Vert ^2_{H^2} \Vert {\tilde{u}}\Vert _{L^2}^2, \\ I_3&= - \int {\tilde{u}} \cdot \nabla b^{(2)} \cdot {\tilde{b}} \textrm{d}x \\&\le \Vert {\tilde{u}}\Vert _{L^4_{h}L^2_{x_3}} \Vert \nabla b^{(2)}\Vert _{L^2_{h}L^{\infty }_{x_3}} \Vert {\tilde{b}}\Vert _{L^4_{h}L^2_{x_3}}\\&\le C \Vert {\tilde{u}}\Vert _{L^2}^{\frac{1}{2}} \Vert \nabla _h {\tilde{u}}\Vert _{L^2}^{\frac{1}{2}} \Vert b^{(2)}\Vert _{H^2} \Vert {\tilde{b}}\Vert _{L^2}^{\frac{1}{2}} \Vert \nabla _h {\tilde{b}}\Vert _{L^2}^{\frac{1}{2}}\\&\le \frac{c_1}{16} ( \Vert \nabla _h {\tilde{u}}\Vert _{L^2}^2 + \Vert \nabla _h {\tilde{b}}\Vert _{L^2}^2 ) + C \Vert b^{(2)}\Vert _{H^2}^2 ( \Vert {\tilde{b}}\Vert _{L^2}^2 + \Vert {\tilde{u}}\Vert _{L^2}^2), \\ I_4&= \int {\tilde{b}} \cdot \nabla u^{(2)} \cdot {\tilde{b}} \textrm{d}x\\&\le \Vert {\tilde{b}}\Vert _{L^4_{h}L^2_{x_3}} \Vert \nabla u^{(2)}\Vert _{L^2_{h}L^{\infty }_{x_3}} \Vert {\tilde{b}}\Vert _{L^4_{h}L^2_{x_3}}\\&\le C \Vert {\tilde{b}}\Vert _{L^2}^{\frac{1}{2}} \Vert \nabla _h {\tilde{b}}\Vert _{L^2}^{\frac{1}{2}} \Vert u^{(2)}\Vert _{H^2} \Vert {\tilde{b}}\Vert _{L^2}^{\frac{1}{2}} \Vert \nabla _h {\tilde{b}}\Vert _{L^2}^{\frac{1}{2}}\\&\le \frac{c_1}{16} \Vert \nabla _h {\tilde{b}}\Vert _{L^2}^2 + C \Vert u^{(2)}\Vert _{H^2}^2 \Vert {\tilde{b}}\Vert _{L^2}^2, \end{aligned}$$

and

$$\begin{aligned} I_5&= {- \int {\tilde{u}} \cdot \nabla \omega ^{(2)} \cdot {\tilde{\omega }} \textrm{d}x}\\&\le \Vert {\tilde{u}}\Vert _{L^4_{h}L^2_{x_3}} \Vert \nabla \omega ^{(2)}\Vert _{L^4_{h}L^2_{x_3}} \Vert {\tilde{\omega }}\Vert _{L^2_{h}L^{\infty }_{x_3}}\\&\le C \Vert {\tilde{u}}\Vert _{L^2}^{\frac{1}{2}} \Vert \nabla _h {\tilde{u}}\Vert _{L^2}^{\frac{1}{2}} \Vert \nabla \omega ^{(2)}\Vert _{L^2}^{\frac{1}{2}} \Vert \nabla \nabla _h \omega ^{(2)}\Vert _{L^2}^{\frac{1}{2}} \Vert {\tilde{\omega }}\Vert _{L^2}^{\frac{1}{2}} \Vert \partial _3 {\tilde{\omega }}\Vert _{L^2}^{\frac{1}{2}}\\&\le C \Vert {\tilde{u}}\Vert _{L^2}^{\frac{1}{2}} \Vert \omega ^{(2)}\Vert _{H^2} ( \Vert \nabla _h {\tilde{u}}\Vert _{L^2}^{\frac{3}{2}} + \Vert {\tilde{\omega }}\Vert _{L^2}^{\frac{3}{2}} + \Vert \partial _3 {\tilde{\omega }}\Vert _{L^2}^{\frac{3}{2}} )\\&\le \frac{c_1}{16} (\Vert \nabla _h {\tilde{u}}\Vert _{L^2}^2 + \Vert \partial _3 {\tilde{\omega }}\Vert _{L^2}^2 + \Vert {\tilde{\omega }}\Vert ^2_{L^2}) + C \Vert {\tilde{u}}\Vert _{L^2}^2 \Vert \omega ^{(2)}\Vert _{H^2}^4. \end{aligned}$$

Finally, we consider $I_6$ and $I_7$, by Hölder’s inequality and the Young inequality, and yield

$$\begin{aligned} I_6 + I_7&= 2 \chi \int \nabla \times {\tilde{\omega }} \cdot {\tilde{u}} \textrm{d}x + 2 \chi \int \nabla \times {\tilde{u}} \cdot {\tilde{\omega }} \textrm{d}x\\&= 4 \chi \int \nabla \times {\tilde{u}} \cdot {\tilde{\omega }} \textrm{d}x\\&\quad \le \frac{\mu }{2} \Vert \nabla _h {\tilde{u}}\Vert _{L^2}^2 + \frac{16 \chi ^2 }{\mu } \Vert {\tilde{\omega }}\Vert _{L^2}^2 + \chi \Vert {\tilde{u}}\Vert _{L^2}^2 + 4 \chi \Vert \partial _3 {\tilde{\omega }}\Vert _{L^2}^2. \end{aligned}$$

Substituting the above estimates into (2.21), it infers that

$$\begin{aligned}&\frac{\textrm{d}}{\textrm{d}t} (\Vert {\tilde{u}}(t)\Vert _{L^2}^2 +\Vert {\tilde{b}}(t)\Vert _{L^2}^2 + \Vert {\tilde{\omega }}(t)\Vert _{L^2}^2) + c_1 (\Vert \nabla _h {\tilde{u}}\Vert _{L^2}^2 + \Vert \nabla _h {\tilde{b}}\Vert _{L^2}^2 + \Vert \partial _3 {\tilde{\omega }}\Vert _{L^2}^2 + \Vert {\tilde{\omega }}\Vert _{L^2}^2) \nonumber \\&\quad \le (\Vert u^{(2)}\Vert _{H^2}^2 + {\Vert b^{(2)}\Vert _{H^2}^2}+ \Vert \omega ^{(2)}\Vert _{H^2}^4)(\Vert {\tilde{u}}\Vert _{L^2}^2 +\Vert {\tilde{b}}\Vert _{L^2}^2 + \Vert {\tilde{\omega }}\Vert _{L^2}^2), \end{aligned}$$

(2.22)

where $c_1= \min \{\frac{\mu }{2}, \nu , \gamma - 4 \chi , 4\chi - \frac{16 \chi ^2}{\mu }\}$. Since $(u^{(2)}, b^{(2)}, \omega ^{(2)})$ is in the regularity class (2.20), we obtain for any $T>0$,

$$\begin{aligned} \int \limits ^T_0 (\Vert u^{(2)}(t)\Vert _{H^2}^2 + \Vert b^{(2)}(t)\Vert _{H^2}^2 + \Vert \omega ^{(2)}(t)\Vert _{H^2}^4) \text {d}t \le C(T) < + \infty . \end{aligned}$$

(2.23)

Then, (2.22) together with Gronwall’s inequality and (2.23) leads to for any $T>0$,

$$\begin{aligned} \Vert {\tilde{u}}(t)\Vert _{L^2}^2 +\Vert {\tilde{b}}(t)\Vert _{L^2}^2 + \Vert {\tilde{\omega }}(t)\Vert _{L^2}^2 \le 0, \quad {\forall t \in [0, T],} \end{aligned}$$

which implies the uniqueness. Thus, we complete the proof of Theorem 1.1.

$\square$

3 The proof of Theorem 1.4

This section is devoted to the proof of Theorem 1.4. Since the proof is slightly long, we divide it into four propositions for clarity. The strategy is as follows: As preparations, we first establish the $H^1$ estimates for $(u, \omega , b )$ in Proposition 3.1. Secondly, we will show the following preliminary estimate $\Vert \nabla u(t)\Vert _{L^2} + \Vert \nabla b(t)\Vert _{L^2} + \Vert \nabla \omega (t)\Vert _{L^2} \le C (1+t)^{-\frac{1}{2}}$, $\Vert u(t)\Vert _{L^2} + \Vert \omega (t)\Vert _{L^2} \le C (1 + t)^{- \frac{5}{6}}$, $\Vert b(t)\Vert _{L^2} \le C (1+t)^{- \frac{3}{4}}$ and $\Vert \nabla b(t)\Vert _{L^2} \le C (1+t)^{- \frac{5}{4}}$ in Propositions 3.2 and 3.5. We give the improve decay estimates $\Vert \nabla ^2 b(t)\Vert _{L^{2}} \le C (1 +t)^{-\frac{65}{48} + \frac{17}{12}\alpha }$ and $\Vert u(t)\Vert _{L^2} + \Vert \omega (t)\Vert _{L^2} \le C (1 + t)^{- \frac{157}{64} + \frac{17}{16} \alpha }$ in Proposition 3.7. Finally, we obtain the decay estimate $\Vert \nabla u(t)\Vert _{L^2} + \Vert \nabla \omega (t)\Vert _{L^2} \le C(1 +t)^{- \frac{471}{512} + \frac{51}{128}\alpha }$ in Proposition 3.8, and thus, the proof of Theorem 1.4 is completed.

Proposition 3.1

Let the assumptions stated in Theorem 1.4 and (1.13) hold. Then for all $t >0$, $(u, \omega , b)$ obey the following uniform bounds

$$\begin{aligned}&\Vert u (t)\Vert ^2_{L^2} + \Vert b(t)\Vert ^2_{L^2} + \Vert \omega (t)\Vert ^2_{L^2} \nonumber \\ {}&\qquad + 2 c_2 \int \limits ^t_0 \left( \Vert \nabla _h u (\tau )\Vert ^2_{L^2} + \Vert u(t)\Vert _{L^2} + \Vert \nabla b (\tau )\Vert ^2_{L^2} + \Vert \partial _3 \omega (\tau )\Vert ^2_{L^2} + \Vert \omega (\tau )\Vert _{L^2}^2 \right) d\tau \nonumber \\ {}&\quad \le \Vert u_0\Vert _{L^2}^2 + \Vert b_0\Vert _{L^2}^2 + \Vert \omega _0\Vert _{L^2}^2, \end{aligned}$$

(3.1)

and

$$\begin{aligned}&\Vert \nabla u(t)\Vert _{L^2}^2 + \Vert \nabla b (t)\Vert _{L^2}^2 + \Vert \nabla \omega (t)\Vert _{L^2}^2 \nonumber \\ {}&\qquad + c_2 \int \limits ^t_0 (\Vert \nabla \nabla _h u(\tau )\Vert _{L^2}^2 +\Vert \nabla u(\tau )\Vert _{L^2}^2 + \Vert \nabla ^2 b (\tau )\Vert _{L^2}^2 + \Vert \nabla \partial _3 \omega (\tau )\Vert _{L^2}^2 + \Vert \nabla \omega (\tau )\Vert _{L^2}^2) d\tau \nonumber \\ {}&\quad \le \left( \Vert \nabla u_0\Vert _{L^2}^2 +\Vert \nabla b_0\Vert _{L^2}^2 + \Vert \nabla \omega _0\Vert _{L^2}^2 \right) e^{{C(\Vert u_0\Vert _{H^2}^2 + \Vert b_0\Vert _{H^2}^2+ \Vert \omega _0\Vert _{H^2}^2 )}}, \end{aligned}$$

(3.2)

where $c_2= \min \{\frac{\mu }{2}, \frac{\chi }{4}, \nu , \gamma - \frac{16 \chi }{3}, 4\chi - \frac{16 \chi ^2}{\mu }\}.$

Proof

Taking the $L^2$-inner product of (1.8) with u, $\omega$ and b, respectively, and then adding the resulting equations together, we yield

$$\begin{aligned}&\frac{1}{2} \frac{\textrm{d}}{\textrm{d}t} \left( \Vert u(t)\Vert ^2_{L^2} + \Vert b(t)\Vert ^2_{L^2} + \Vert \omega (t)\Vert ^2_{L^2} \right) + \mu \Vert \nabla _h u\Vert ^2_{L^2} + \chi \Vert u\Vert _{L^2}^2 + \nu \Vert \nabla b\Vert ^2_{L^2} \nonumber \\&\qquad + \gamma \Vert \partial _3 \omega \Vert ^2_{L^2} + 4 \chi \Vert \omega \Vert _{L^2}^2 + \kappa \Vert \nabla \cdot \omega \Vert _{L^2}^2 \nonumber \\&\quad = 4\chi \int \nabla \times u \cdot \omega \textrm{d}x. \end{aligned}$$

(3.3)

Applying the Young inequality and Hölder’s inequality, we obtain

$$\begin{aligned} 4\chi \int \nabla \times u \cdot \omega \textrm{d}x \le \frac{\mu }{2} \Vert \nabla _h u\Vert _{L^2}^2 + \frac{16 \chi ^2 }{\mu } \Vert \omega \Vert _{L^2}^2 + \frac{3 \chi }{4} \Vert u\Vert _{L^2}^2 + \frac{16 \chi }{3} \Vert \partial _3 \omega \Vert _{L^2}^2. \end{aligned}$$

Inserting this bound into (3.3) and then integrating in time, we yield the desired bound (3.1).

Now we turn to proof (3.2). Applying $\partial _k$ with $k=1, 2, 3$ to (1.8)$_1$, (1.8)$_2$ and (1.8)$_3$, dotting the results by $\partial _k u$, $\partial _k b$ and $\partial _k \omega$, respectively, integrating in space domain and adding them together, we obtain

$$\begin{aligned}&\frac{1}{2} \frac{\textrm{d}}{\textrm{d}t} \sum _{k=1}^3 \left( \Vert \partial _k u(t)\Vert _{L^2}^2 +\Vert \partial _k b(t)\Vert _{L^2}^2 + \Vert \partial _k \omega (t)\Vert _{L^2}^2 \right) + \mu \sum _{k=1}^3 \Vert \nabla _h \partial _k u\Vert _{L^2}^2 + \chi \sum _{k=1}^3 \Vert \partial _k u\Vert _{L^2}^2\nonumber \\&\qquad + \nu \sum _{k=1}^3 \Vert \nabla \partial _k b\Vert _{L^2}^2 + \gamma \sum _{k=1}^3 \Vert \partial _3 \partial _k \omega \Vert _{L^2}^2 + 4 \chi \sum _{k=1}^3 \Vert \partial _k \omega \Vert _{L^2}^2 + \kappa \sum _{k=1}^3 \Vert \partial _k \nabla \cdot \omega \Vert _{L^2}^2\nonumber \\&\quad = - \sum _{k=1}^3 \int \partial _k (u \cdot \nabla u) \cdot \partial _k u \textrm{d}x + \sum _{k=1}^3 \int \partial _k (b \cdot \nabla b) \cdot \partial _k u \textrm{d}x + 2 \chi \sum _{k=1}^3 \int \nabla \times \partial _k \omega \cdot \partial _k u \textrm{d}x \nonumber \\&\qquad - \sum _{k=1}^3 \int \partial _k (u \cdot \nabla b) \cdot \partial _k b \textrm{d}x + \sum _{k=1}^3 \int \partial _k (b \cdot \nabla u) \cdot \partial _k b \textrm{d}x - \sum _{k=1}^3 \int \partial _k (u \cdot \nabla \omega ) \cdot \partial _k \omega \textrm{d}x \nonumber \\&\qquad + 2 \chi \sum _{k=1}^3 \int \nabla \times \partial _k u \cdot \partial _k \omega \textrm{d}x\nonumber \\&\quad := A_1 + A_2 + A_3 + A_4 + A_5 + A_6 + A_7. \end{aligned}$$

(3.4)

Using the divergence free condition $\nabla \cdot u =0$, together with Lemma 2.2 and the Young inequality, one infers

$$\begin{aligned} A_1&= - \sum ^3_{k=1} \sum ^3_{j=1} \int \partial _k u_j \partial _j u \cdot \partial _k u \textrm{d}x\\&= - \sum ^2_{k=1} \sum ^3_{j=1} \int \partial _k u_j \partial _j u \cdot \partial _k u \textrm{d}x - \sum ^2_{j=1} \int \partial _3 u_j \partial _j u \cdot \partial _3 u \textrm{d}x - \int \partial _3 u_3 \partial _3 u \cdot \partial _3 u \textrm{d}x\\&\le C \sum ^2_{k=1} \sum ^3_{j=1} \Vert \partial _k u_j\Vert _{L^2}^{\frac{1}{2}} \Vert \partial _1 \partial _k u_j\Vert _{L^2}^{\frac{1}{2}} \Vert \partial _j u\Vert _{L^2}^{\frac{1}{2}} \Vert \partial _2 \partial _j u\Vert _{L^2}^{\frac{1}{2}} \Vert \partial _k u\Vert _{L^2}^{\frac{1}{2}} \Vert \partial _3 \partial _k u\Vert _{L^2}^{\frac{1}{2}}\\&\quad + \sum ^2_{j=1} \Vert \partial _3 u_j\Vert _{L^2}^{\frac{1}{2}} \Vert \partial _1 \partial _3 u_j\Vert _{L^2}^{\frac{1}{2}} \Vert \partial _j u\Vert _{L^2}^{\frac{1}{2}} \Vert \partial _3 \partial _j u\Vert _{L^2}^{\frac{1}{2}} \Vert \partial _3 u\Vert _{L^2}^{\frac{1}{2}} \Vert \partial _2 \partial _3 u\Vert _{L^2}^{\frac{1}{2}}\\&\quad + \Vert \partial _h u_h\Vert _{L^2}^{\frac{1}{2}} \Vert \partial _3 \partial _h u_h\Vert _{L^2}^{\frac{1}{2}} \Vert \partial _3 u\Vert _{L^2}^{\frac{1}{2}} \Vert \partial _2 \partial _3 u\Vert _{L^2}^{\frac{1}{2}} \Vert \partial _3 u\Vert _{L^2}^{\frac{1}{2}} \Vert \partial _1 \partial _3 u\Vert _{L^2}^{\frac{1}{2}}\\&\le C \Vert \nabla u\Vert ^2_{L^2} \Vert \nabla _h u\Vert ^2_{H^2} + \frac{c_2}{32} \Vert \nabla \nabla _h u\Vert _{L^2}^2. \end{aligned}$$

Similarly,

$$\begin{aligned} A_6&= - \sum _{k=1}^3 \int \partial _k (u \cdot \nabla \omega ) \cdot \partial _k \omega \textrm{d}x\\&= - \sum ^2_{k=1} \sum ^3_{j=1} \int \partial _k u_j \partial _j \omega \cdot \partial _k \omega \textrm{d}x - \sum ^3_{j=1} \int \partial _3 u_j \partial _j \omega \cdot \partial _3 \omega \textrm{d}x\\&\le C \sum ^2_{k=1} \sum ^3_{j=1} \Vert \partial _k u_j\Vert _{L^2}^{\frac{1}{2}} \Vert \partial _1 \partial _k u_j\Vert _{L^2}^{\frac{1}{2}} \Vert \partial _j \omega \Vert _{L^2}^{\frac{1}{2}} \Vert \partial _2 \partial _j \omega \Vert _{L^2}^{\frac{1}{2}} \Vert \partial _k \omega \Vert _{L^2}^{\frac{1}{2}} \Vert \partial _3 \partial _k \omega \Vert _{L^2}^{\frac{1}{2}}\\&\qquad + \sum ^3_{j=1} \Vert \partial _3 u_j\Vert _{L^2}^{\frac{1}{2}} \Vert \partial _1 \partial _3 u_j\Vert _{L^2}^{\frac{1}{2}} \Vert \partial _j \omega \Vert _{L^2}^{\frac{1}{2}} \Vert \partial _3 \partial _j \omega \Vert _{L^2}^{\frac{1}{2}} \Vert \partial _3 \omega \Vert _{L^2}^{\frac{1}{2}} \Vert \partial _2 \partial _3 \omega \Vert _{L^2}^{\frac{1}{2}}\\&\le C( \Vert \nabla u\Vert ^2_{L^2} + \Vert \nabla \omega \Vert ^2_{L^2} )(\Vert \nabla _h u\Vert ^2_{H^2} + \Vert \partial _3 \omega \Vert ^2_{H^2} + \Vert \omega \Vert ^2_{H^2} )+ \frac{c_2}{32} (\Vert \nabla \nabla _h u\Vert _{L^2}^2 + {\Vert \nabla \partial _3 \omega \Vert _{L^2}^2}), \end{aligned}$$

and

$$\begin{aligned} A_4&= - \sum _{k=1}^3 \int \partial _k (u \cdot \nabla b) \cdot \partial _k b \textrm{d}x \\&\le C( \Vert \nabla u\Vert ^2_{L^2} + \Vert \nabla b\Vert ^2_{L^2} )\Vert \nabla b\Vert ^2_{H^2} + \frac{c_2}{32} (\Vert \nabla \nabla _h u\Vert _{L^2}^2 + \Vert \nabla ^2 b\Vert _{L^2}^2). \end{aligned}$$

Due to

$$\begin{aligned} \sum _{k=1}^3 \int b \cdot \nabla \partial _k b \cdot \partial _k u \textrm{d}x + \sum _{k=1}^3 \int b \cdot \nabla \partial _k u \cdot \partial _k b \textrm{d}x = 0, \end{aligned}$$

then, we have

$$\begin{aligned} A_2 + A_5&= \sum _{k=1}^3 \int \partial _k b \cdot \nabla b \cdot \partial _k u \textrm{d}x + \sum _{k=1}^3 \int \partial _k b \cdot \nabla u \cdot \partial _k b \textrm{d}x\\&\le C( \Vert \nabla u\Vert ^2_{L^2} + \Vert \nabla b\Vert ^2_{L^2} )\Vert \nabla b\Vert ^2_{H^2} + \frac{c_2}{32} (\Vert \nabla \nabla _h u\Vert _{L^2}^2 + \Vert \nabla ^2 b\Vert _{L^2}^2). \end{aligned}$$

By the Young inequality, we have

$$\begin{aligned} A_3 + A_7 \le \frac{\mu }{2} \sum ^3_{k=1} \Vert \nabla _h \partial _k u\Vert _{L^2}^2 + \frac{16 \chi ^2 }{\mu } \sum ^3_{k=1}\Vert \partial _k\omega \Vert _{L^2}^2 + \frac{3 \chi }{4} \sum ^3_{k=1} \Vert \partial _k u\Vert _{L^2}^2 + \frac{16 \chi }{3} \sum ^3_{k=1}\Vert \partial _3\partial _k \omega \Vert _{L^2}^2. \end{aligned}$$

Inserting the above estimates into (3.4), we infer that

$$\begin{aligned}&\frac{1}{2} \frac{\textrm{d}}{\textrm{d}t} \left( \Vert \nabla u(t)\Vert _{L^2}^2 +\Vert \nabla b(t)\Vert _{L^2}^2 + \Vert \nabla \omega (t)\Vert _{L^2}^2 \right) + \frac{\mu }{2} \Vert \nabla _h \nabla u\Vert _{L^2}^2 + \frac{\chi }{4} \Vert \nabla u\Vert _{L^2}^2\nonumber \\&\qquad + \nu \Vert \nabla ^2 b\Vert _{L^2}^2 + (\gamma - \frac{16 \chi }{3}) \Vert \partial _3 \nabla \omega \Vert _{L^2}^2 + \left( 4 \chi - \frac{16 \chi ^2}{\mu }\right) \Vert \nabla \omega \Vert _{L^2}^2 + \kappa \Vert \nabla \nabla \cdot \omega \Vert _{L^2}^2\nonumber \\&\quad \le \frac{c_2}{2} (\Vert \nabla \nabla _h u\Vert _{L^2}^2 + \Vert \nabla ^2 b\Vert _{L^2}^2 + \Vert \nabla \partial _3 \omega \Vert _{L^2}^2) \\&\qquad + C ( \Vert \nabla u(t)\Vert _{L^2}^2 +\Vert \nabla b(t)\Vert _{L^2}^2 + \Vert \nabla \omega (t)\Vert _{L^2}^2 ) ( \Vert \nabla _h u\Vert ^2_{H^2} + \Vert \nabla b\Vert ^2_{H^2} + {\Vert \partial _3 \omega \Vert ^2_{H^2}} + \Vert \omega \Vert ^2_{H^2} ), \end{aligned}$$

where $c_2= \min \{\frac{\mu }{2}, \frac{\chi }{4}, \nu , \gamma - \frac{16 \chi }{3}, 4\chi - \frac{16 \chi ^2}{\mu }\}$. Then applying Gronwall’s inequality, we obtain

$$\begin{aligned}&\Vert \nabla u(t)\Vert _{L^2}^2 +\Vert \nabla b(t)\Vert _{L^2}^2 + \Vert \nabla \omega (t)\Vert _{L^2}^2 + c_2 \int \limits ^t_0 \left( \Vert \nabla _h \nabla u(\tau )\Vert _{L^2}^2 + \Vert \nabla u(\tau )\Vert _{L^2}^2 \right. \nonumber \\ {}&\qquad \left. {+ \Vert \nabla ^2 b(\tau )\Vert _{L^2}^2 + \Vert \partial _3 \nabla \omega (\tau )\Vert _{L^2}^2 + \Vert \nabla \omega (\tau )\Vert _{L^2}^2 }\right) d\tau \nonumber \\ {}&\quad \le \left( \Vert \nabla u_0\Vert _{L^2}^2 +\Vert \nabla b_0\Vert _{L^2}^2 + \Vert \nabla \omega _0\Vert _{L^2}^2 \right) \nonumber \\ {}&\qquad \times \exp \left\{ \int \limits ^t_0 ( \Vert \nabla _h u(\tau )\Vert ^2_{H^2} + \Vert \nabla b(\tau )\Vert ^2_{H^2} + \Vert \nabla _3 \omega (\tau ) \Vert ^2_{H^2} + \Vert \omega (\tau )\Vert ^2_{H^2} ) d \tau \right\} \nonumber \\ {}&\quad \le \left( \Vert \nabla u_0\Vert _{L^2}^2 +\Vert \nabla b_0\Vert _{L^2}^2 + \Vert \nabla \omega _0\Vert _{L^2}^2 \right) e^{{C(\Vert u_0\Vert _{H^2}^2 + \Vert b_0\Vert _{H^2}^2+ \Vert \omega _0\Vert _{H^2}^2 )}}, \end{aligned}$$

(3.5)

where we used (1.11) in the last step. $\square$

With Proposition 3.1 at our disposal, we now start to prove our decay estimates.

Proposition 3.2

Let the assumptions stated in Theorem 1.4 hold. Then

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

(3.6)

Proof

Using (3.5) and (1.11), for $0< s <t$, we have

$$\begin{aligned}&\Vert \nabla u(t)\Vert _{L^2}^2 +\Vert \nabla b(t)\Vert _{L^2}^2 + \Vert \nabla \omega (t)\Vert _{L^2}^2 \nonumber \\&\quad \le \left( \Vert \nabla u(s)\Vert _{L^2}^2 +\Vert \nabla b(s)\Vert _{L^2}^2 + \Vert \nabla \omega (s)\Vert _{L^2}^2 \right) e^{{C(\Vert u_0\Vert _{H^2}^2 + \Vert b_0\Vert _{H^2}^2+ \Vert \omega _0\Vert _{H^2}^2 )}}. \end{aligned}$$

(3.7)

By (3.1) and (3.2), we have

$$\begin{aligned} \int \limits ^{\infty }_0 \Vert \nabla b(\tau )\Vert _{L^2}^2 d\tau \le C (\Vert u_0\Vert _{L^2}^2 + \Vert \omega _0\Vert _{L^2}^2 + \Vert b_0\Vert _{L^2}^2), \end{aligned}$$

(3.8)

$$\begin{aligned} \int \limits ^{\infty }_0 \Vert \nabla \omega (\tau )\Vert _{L^2}^2 + \Vert \nabla u(\tau )\Vert _{L^2}^2 d \tau \le C (\Vert u_0\Vert _{H^2}^2 + \Vert \omega _0\Vert _{H^2}^2 + \Vert b_0\Vert _{H^2}^2). \end{aligned}$$

(3.9)

Integrating (3.7) in $(\frac{t}{2}, t)$ with respect to s, together with (3.8)–(3.9), we obtain

$$\begin{aligned}&t(\Vert \nabla u(t)\Vert _{L^2}^2 +\Vert \nabla b(t)\Vert _{L^2}^2 + \Vert \nabla \omega (t)\Vert _{L^2}^2) \nonumber \\ {}&\quad \le 2 e^{{C(\Vert u_0\Vert _{H^2}^2 + \Vert b_0\Vert _{H^2}^2+ \Vert \omega _0\Vert _{H^2}^2 )}} \int \limits ^t_{\frac{t}{2}}\left( \Vert \nabla u(s)\Vert _{L^2}^2 +\Vert \nabla b(s)\Vert _{L^2}^2 + \Vert \nabla \omega (s)\Vert _{L^2}^2 \right) \text {d}s \\ {}&\le C. \end{aligned}$$

Therefore, for $t \ge 1$, we have

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

(3.10)

For $0<t<1$, it follows from (3.2); we have

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

(3.11)

Then (3.10) and (3.11) yield (3.6).

$\square$

To give the decay estimates for $(u, \omega , b)$ and to improve the decay estimate for $\nabla b$, we recall the following estimate for the heat operator (see, e.g., [48]).

Lemma 3.3

Let $m\ge 0$, $a>0$ and $1 \le p \le q \le +\infty$. Then for any $t>0$,

$$\begin{aligned} \Vert \nabla ^m e^{a \Delta t} f\Vert _{L^q({\mathbb {R}}^3)} \le C t^{-\frac{m}{2} - \frac{3}{2}(\frac{1}{p} - \frac{1}{q})} \Vert f\Vert _{L^p({\mathbb {R}}^3)}, \end{aligned}$$

(3.12)

where

$$\begin{aligned} e^{a \Delta t} f(x) = (4 \pi a t)^{-1} \int \limits _{{\mathbb {R}}^3} e^{- \frac{|x-y|^2}{4at}} f(y) dy. \end{aligned}$$

And the following calculus inequalities (see, e.g., [49, 50]) involving fractional differential operators $\Lambda ^s$ with $s>0$ and

$$\begin{aligned} \widehat{\Lambda ^s f}(\xi ) = |\xi |^s {\hat{f}}(\xi ), \quad {\hat{f}}(\xi ) = \int \limits _{{\mathbb {R}}^3} e^{- ix \cdot \xi } f(x) \textrm{d}x. \end{aligned}$$

Lemma 3.4

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

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

where C is constant depending on the indices s, r, $p_1$, $q_1$, $p_2$ and $q_2$.

Now, we can start to establish the desired decay estimates.

Proposition 3.5

Let the assumptions stated in Theorem 1.4 hold. Then

$$\begin{aligned} \Vert u(t)\Vert _{L^2} + \Vert \omega (t)\Vert _{L^2} \le C (1 + t)^{- \frac{5}{6}}, \end{aligned}$$

(3.13)

$$\begin{aligned} \Vert b(t)\Vert _{L^2} \le C (1+t)^{- \frac{3}{4}}, \quad \Vert \nabla b(t)\Vert _{L^2} \le C (1+t)^{- \frac{5}{4}}. \end{aligned}$$

(3.14)

Proof

Taking the $L^2$-inner product to the first and the third equations of (1.8) with u and $\omega$, respectively, and adding the resulting equations together, we obtain

$$\begin{aligned}&\frac{1}{2} \frac{\textrm{d}}{\textrm{d}t} \left( \Vert u(t)\Vert ^2_{L^2} + \Vert \omega (t)\Vert ^2_{L^2} \right) + \mu \Vert \nabla _h u\Vert ^2_{L^2} + \chi \Vert u\Vert _{L^2}^2 + \gamma \Vert \partial _3 \omega \Vert ^2_{L^2} + 4 \chi \Vert \omega \Vert _{L^2}^2 + \kappa \Vert \nabla \cdot \omega \Vert _{L^2}^2 \nonumber \\&\quad = \int b \cdot \nabla b \cdot u \textrm{d}x + 4\chi \int \nabla \times u \cdot \omega \textrm{d}x \nonumber \\&\quad \le C \Vert \nabla b\Vert _{L^2}^2 \Vert u\Vert _{L^3} + \frac{\mu }{2} \Vert \nabla _h u\Vert _{L^2}^2 + \frac{16 \chi ^2 }{\mu } \Vert \omega \Vert _{L^2}^2 + {\frac{3 \chi }{4}} \Vert u\Vert _{L^2}^2 + \frac{16 \chi }{3} \Vert \partial _3 \omega \Vert _{L^2}^2, \end{aligned}$$

(3.15)

where we have used Sobolev’s inequality. Set $c = \min \{\frac{ \chi }{2}, 8\chi - \frac{32\chi ^2}{\mu }\}$.Then integrating (3.15) in time, we yield

$$\begin{aligned}&\Vert u (t)\Vert ^2_{L^2} + \Vert \omega (t)\Vert ^2_{L^2} \nonumber \\ {}&\quad \le e^{-ct}(\Vert u_0\Vert ^2_{L^2} + \Vert \omega _0\Vert ^2_{L^2}) + C \int \limits ^t_0 e^{-c(t-s)} \Vert \nabla b(s)\Vert _{L^2}^2 \Vert u(s)\Vert ^{\frac{1}{2}}_{L^2} \Vert \nabla u(s)\Vert ^{\frac{1}{2}}_{L^2} \text {d}s \nonumber \\ {}&\quad \le e^{-ct}(\Vert u_0\Vert ^2_{L^2} + \Vert \omega _0\Vert ^2_{L^2}) + C (Q_1 + Q_2), \end{aligned}$$

(3.16)

where

$$\begin{aligned} Q_1 = \int \limits ^{\frac{t}{2}}_0 e^{-c(t-s)} \Vert \nabla b(s)\Vert _{L^2}^2 \Vert u(s)\Vert ^{\frac{1}{2}}_{L^2} \Vert \nabla u(s)\Vert ^{\frac{1}{2}}_{L^2} \text {d}s, \end{aligned}$$

$$\begin{aligned} Q_2 = \int \limits ^t_{\frac{t}{2}} e^{-c(t-s)} \Vert \nabla b(s)\Vert _{L^2}^2 \Vert u(s)\Vert ^{\frac{1}{2}}_{L^2} \Vert \nabla u(s)\Vert ^{\frac{1}{2}}_{L^2} \text {d}s.\end{aligned}$$

By (3.1) and (3.2), we get

$$\begin{aligned} Q_1 \le C e^{- \frac{ct}{2}} \int \limits ^{\frac{t}{2}}_0 \Vert \nabla b(s)\Vert _{L^2}^2 \text {d}s \le C e^{- \frac{ct}{2}}. \end{aligned}$$

(3.17)

Set

$$\begin{aligned} {\mathcal {M}}(t) = \sup _{0 \le s \le t} \{(1 + s)^{\frac{1}{2}} (\Vert \nabla u(s)\Vert _{L^2} + \Vert \nabla \omega (s)\Vert _{L^2} + \Vert \nabla b(s)\Vert _{L^2} \}. \end{aligned}$$

Then

$$\begin{aligned} Q_2 \le C {\mathcal {M}}^{\frac{5}{2}}(t) \int \limits ^t_{\frac{t}{2}} e^{-c(t-s)} (1 + s)^{- \frac{5}{4}} \Vert u(s)\Vert _{L^2}^{\frac{1}{2}}\text {d}s. \end{aligned}$$

(3.18)

Set

$$\begin{aligned} {\mathcal {N}}(t) = \sup _{0 \le s \le t} \{(1 + s)^{\frac{5}{6}} (\Vert u(s)\Vert _{L^2} + \Vert \omega (s)\Vert _{L^2})\}. \end{aligned}$$

Inserting (3.17), (3.18) into (3.16), we obtain

$$\begin{aligned} {\mathcal {N}}^2(t) \le C (1+ t)^{- \frac{5}{3}} e^{- \frac{ct}{2}} + C {\mathcal {M}}^{\frac{5}{2}}(t) {\mathcal {N}}^{\frac{1}{2}}(t). \end{aligned}$$

Then Young’s inequality and ${\mathcal {M}}(t) \le C$ lead to the desired result.

To get the decay estimate of b, we write the second equation of (1.8) into integral form.

$$\begin{aligned} b(t)&= e^{\nu \Delta t} b_0 + \int \limits _0^t e^{\nu \Delta (t-s)} (b \cdot \nabla u - u \cdot \nabla b)(s) \textrm{d}s\nonumber \\&= e^{\nu \Delta t} b_0 + \int \limits _0^{\frac{t}{2}} \nabla e^{\nu \Delta (t-s)} (b \otimes u - u \otimes b)(s)\textrm{d}s + \int \limits _{\frac{t}{2}}^t \nabla e^{\nu \Delta (t-s)} (b \otimes u - u \otimes b)(s)\textrm{d}s, \end{aligned}$$

(3.19)

where $f\otimes g = (f_i g_j)$ defines the tensor product. By Lemma 3.3, for $0<t < 1$, we have

$$\begin{aligned} \Vert e^{\nu \Delta t} b_0 \Vert _{L^2} \le C \Vert b_0\Vert _{L^2}, \end{aligned}$$

and for $t \ge 1$,

$$\begin{aligned} \Vert e^{\nu \Delta t} b_0 \Vert _{L^2} \le C t^{-\frac{3}{4}}\Vert b_0\Vert _{L^1}. \end{aligned}$$

Therefore, for any $t>0$,

$$\begin{aligned} \Vert e^{\nu \Delta t} b_0 \Vert _{L^2} \le C(t + 1)^{-\frac{3}{4}}. \end{aligned}$$

(3.20)

Again applying Lemma 3.3, together with (3.13), we obtain for $t \ge 1$,

$$\begin{aligned}&\left\| \int \limits _0^{\frac{t}{2}} \nabla e^{\nu \Delta (t-s)} (b \otimes u - u \otimes b)(s)\textrm{d}s \right\| _{L^2}\nonumber \\&\quad \le C \int \limits _0^{\frac{t}{2}} (t-s)^{-\frac{1}{2} -\frac{3}{4}} \Vert (b \otimes u - u \otimes b)(s)\Vert _{L^1} \textrm{d}s \nonumber \\&\quad \le C \int \limits _0^{\frac{t}{2}} (t-s)^{ -\frac{5}{4}} \Vert b(s)\Vert _{L^2} \Vert u(s)\Vert _{L^2} \textrm{d}s \nonumber \\&\quad \le C \int \limits _0^{\frac{t}{2}} (t-s)^{ -\frac{5}{4}} (1+ s)^{-\frac{5}{6}} \textrm{d}s . \end{aligned}$$

(3.21)

Using Lemma 3.3 and the Gagliardo–Nirenberg inequality, together with (3.6) and (3.13), for any $t>0$, we yield

$$\begin{aligned}&\left\| \int \limits _{\frac{t}{2}}^t \nabla e^{\nu \Delta (t-s)} (b \otimes u - u \otimes b)(s)\textrm{d}s \right\| _{L^2}\nonumber \\&\quad \le C \int \limits _{\frac{t}{2}}^t (t-s)^{\frac{1}{4} -\frac{3}{2p}} \Vert (b \otimes u - u \otimes b)(s)\Vert _{L^p} \textrm{d}s \nonumber \\&\quad \le C \int \limits _{\frac{t}{2}}^t (t-s)^{\frac{1}{4} -\frac{3}{2p}} \Vert b(s)\Vert _{L^{2p}} \Vert u(s)\Vert _{L^{2p}} \textrm{d}s \nonumber \\&\quad \le C \int \limits _{\frac{t}{2}}^t (t-s)^{\frac{1}{4} -\frac{3}{2p}} \Vert u(s)\Vert ^{\frac{3-p}{2p}}_{L^{2}}\Vert \nabla u(s)\Vert ^{\frac{3p-3}{2p}}_{L^{2}} \Vert b(s)\Vert ^{\frac{3-p}{2p}}_{L^{2}}\Vert \nabla b(s)\Vert ^{\frac{3p-3}{2p}}_{L^{2}}\textrm{d}s \nonumber \\&\quad \le C {\mathcal {M}}^{\frac{3p-3}{p}}(t) \int \limits _{\frac{t}{2}}^t (t-s)^{\frac{1}{4} -\frac{3}{2p}} (1+s)^{-\frac{5}{6} \times \frac{3-p}{2p} - \frac{3p-3}{2p}} \textrm{d}s \nonumber \\&\quad \le C (1+ t)^{-\frac{15-2p}{12p}}, \end{aligned}$$

(3.22)

with $\frac{12}{10} < p \le \frac{15}{11}$. Taking the $L^2$-norm for space to (3.19), together with (3.20)–(3.22), we obtain for any $t \ge 1$,

$$\begin{aligned} \Vert b(t)\Vert _{L^2}&\le C (t + 1)^{-\frac{3}{4}} + C (1+ t)^{-\frac{13}{12}} + C (1+ t)^{-\frac{15-2p}{12p}} \nonumber \\&\le C (t + 1)^{-\frac{3}{4}}. \end{aligned}$$

(3.23)

Note that for $0<t <1$, (3.1) implies

$$\begin{aligned} \Vert b(t)\Vert _{L^2}&\le C , \end{aligned}$$

then we immediately obtain the first decay estimate (3.14).

Now we turn to the decay estimate of $\nabla b$. Applying $\nabla$ to (3.19),

$$\begin{aligned} \nabla b(t)&= \nabla e^{\nu \Delta t} b_0 + \int \limits _0^{\frac{t}{2}} \nabla ^2 e^{\nu \Delta (t-s)} (b \otimes u - u \otimes b)(s)\textrm{d}s \nonumber \\&\quad + \int \limits _{\frac{t}{2}}^t \nabla ^2 e^{\nu \Delta (t-s)} (b \otimes u - u \otimes b)(s)\textrm{d}s. \end{aligned}$$

(3.24)

By Lemma 3.3, for $0< t <1$, we have

$$\begin{aligned} \Vert \nabla e^{\nu \Delta t} b_0 \Vert _{L^2} \le C \Vert \nabla b_0\Vert _{L^2}, \end{aligned}$$

and for $t \ge 1$,

$$\begin{aligned} \Vert \nabla e^{\nu \Delta t} b_0 \Vert _{L^2} \le C t^{-\frac{5}{4}}\Vert b_0\Vert _{L^1}. \end{aligned}$$

Therefore, for any $t>0$,

$$\begin{aligned} \Vert \nabla e^{\nu \Delta t} b_0 \Vert _{L^2} \le C(t + 1)^{-\frac{5}{4}}. \end{aligned}$$

(3.25)

Using Lemma 3.3, together with (3.13) and (3.23), we have for any $t \ge 1$,

$$\begin{aligned}&\left\| \int \limits _0^{\frac{t}{2}} \nabla ^2 e^{\nu \Delta (t-s)} (b \otimes u - u \otimes b)(s)\textrm{d}s \right\| _{L^2}\nonumber \\&\quad \le C \int \limits _0^{\frac{t}{2}} (t-s)^{-\frac{2}{2} -\frac{3}{4}} \Vert (b \otimes u - u \otimes b)(s)\Vert _{L^1} \textrm{d}s \nonumber \\&\quad \le C \int \limits _0^{\frac{t}{2}} (t-s)^{ -\frac{7}{4}} \Vert b(s)\Vert _{L^2} \Vert u(s)\Vert _{L^2} \textrm{d}s \nonumber \\&\quad \le C \int \limits _0^{\frac{t}{2}} (t-s)^{ -\frac{7}{4}} (1+ s)^{-\frac{19}{12}} \textrm{d}s. \nonumber \\&\quad \le C (1+ t)^{-\frac{7}{4}}. \end{aligned}$$

(3.26)

Applying Lemmas 3.3 and 3.4, together with (3.6), (3.13) and (3.23), for any $t>0$, we yield

$$\begin{aligned}&\left\| \int \limits _{\frac{t}{2}}^t \nabla ^2 e^{\nu \Delta (t-s)} (b \otimes u - u \otimes b)(s)\textrm{d}s \right\| _{L^2}\nonumber \\&\quad = \left\| \int \limits _{\frac{t}{2}}^t \Lambda ^{- {\beta }} \nabla ^2 e^{\nu \Delta (t-s)} {\Lambda ^{{\beta }}} (b \otimes u - u \otimes b)(s)\textrm{d}s \right\| _{L^2} \nonumber \\&\quad \le C \int \limits _{\frac{t}{2}}^t (t-s)^{- \frac{2-{\beta }}{2}} \Vert \Lambda ^{{\beta }} (b \otimes u - u \otimes b)(s)\Vert _{L^2} \textrm{d}s \nonumber \\&\quad \le C \int \limits _{\frac{t}{2}}^t (t-s)^{- \frac{2-{\beta }}{2}} (\Vert \Lambda ^{{\beta }} b(s) \Vert _{L^4} \Vert u(s)\Vert _{L^4} + \Vert \Lambda ^{{\beta }} u(s) \Vert _{L^4} \Vert b(s)\Vert _{L^4}) \textrm{d}s \nonumber \\&\quad \le C \int \limits _{\frac{t}{2}}^t (t-s)^{- \frac{2-{\beta }}{2}} \left( \Vert u(s)\Vert ^{\frac{1}{4}}_{L^{2}}\Vert \nabla u(s)\Vert ^{\frac{3}{4}}_{L^{2}} \Vert b(s)\Vert ^{\frac{1}{4} - {\beta } }_{L^{2}}\Vert \nabla b(s)\Vert ^{\frac{3}{4} + {\beta }}_{L^{2}} \right. \nonumber \\&\qquad \left. + \Vert b(s)\Vert ^{\frac{1}{4}}_{L^{2}}\Vert \nabla b(s)\Vert ^{\frac{3}{4}}_{L^{2}} \Vert u(s)\Vert ^{\frac{1}{4} - {\beta } }_{L^{2}} \Vert \nabla u(s)\Vert ^{\frac{3}{4} + {\beta }}_{L^{2}} \right) \textrm{d}s \nonumber \\&\quad \le C {\mathcal {M}}^{\frac{3}{2} + {\beta }}(t) \int \limits _{\frac{t}{2}}^t (t-s)^{- \frac{2-{\beta }}{2}} \left( (1+s)^{- (\frac{55}{48} - \frac{{\beta }}{4})} + (1+s)^{- (\frac{55}{48} - \frac{{\beta }}{3})} \right) \textrm{d}s \nonumber \\&\quad \le C (1+t)^{\frac{5}{6}{\beta } - \frac{55}{48}}, \end{aligned}$$

(3.27)

where we set $0 < {\beta } \le \frac{7}{40}$; then,

$$\begin{aligned} C (1+t)^{\frac{5}{6} {\beta } - \frac{55}{48}} \le C (1+t)^{-1}. \end{aligned}$$

Taking the $L^2$-inner for space to (3.24), together with (3.25) - (3.27), for $t \ge 1$, we obtain

$$\begin{aligned} \Vert \nabla b (t)\Vert _{L^2} \le C(t + 1)^{-\frac{5}{4}} + C (1+ t)^{-\frac{7}{4}}+ C (1+t)^{\frac{5}{6} {\beta } - \frac{55}{48}} \le C (1+t)^{-1}. \end{aligned}$$

(3.28)

To improve the decay rates of $\Vert \nabla b \Vert _{L^2}$, when $t \ge 1$, we insert (3.28) into (3.27). Set

$$\begin{aligned} {\mathcal {M}}_1 (t) = \sup _{0 \le s \le t} \{ (1+s) \Vert \nabla b (t)\Vert _{L^2} \}. \end{aligned}$$

Then, we have

$$\begin{aligned}&\left\| \int \limits _{\frac{t}{2}}^t \nabla ^2 e^{\nu \Delta (t-s)} (b \otimes u - u \otimes b)(s)\textrm{d}s \right\| _{L^2}\nonumber \\&\quad \le C \int \limits _{\frac{t}{2}}^t (t-s)^{- \frac{2-{\beta }}{2}} \left( \Vert u(s)\Vert ^{\frac{1}{4}}_{L^{2}}\Vert \nabla u(s)\Vert ^{\frac{3}{4}}_{L^{2}} \Vert b(s)\Vert ^{\frac{1}{4} - {\beta } }_{L^{2}}\Vert \nabla b(s)\Vert ^{\frac{3}{4} + {\beta }}_{L^{2}} \right. \nonumber \\&\qquad \left. + \Vert b(s)\Vert ^{\frac{1}{4}}_{L^{2}}\Vert \nabla b(s)\Vert ^{\frac{3}{4}}_{L^{2}} \Vert u(s)\Vert ^{\frac{1}{4} - {\beta } }_{L^{2}} \Vert \nabla u(s)\Vert ^{\frac{3}{4} + {\beta }}_{L^{2}} \right) \textrm{d}s \nonumber \\&\quad \le C {\mathcal {M}}^{\frac{3}{4}}(t) {\mathcal {M}}_1^{\frac{3}{4} + {\beta }}(t) \int \limits _{\frac{t}{2}}^t (t-s)^{- \frac{2-{\beta }}{2}} (1+s)^{- (\frac{73}{48} + \frac{{\beta }}{4})} \textrm{d}s \nonumber \\&\qquad + C {\mathcal {M}}_1^{\frac{3}{4}}(t) {\mathcal {M}}^{\frac{3}{4} + {\beta }}(t) \int \limits _{\frac{t}{2}}^t (t-s)^{- \frac{2-{\beta }}{2}} (1+s)^{\frac{{\beta }}{3} - \frac{73}{48}} \textrm{d}s \nonumber \\&\quad \le C (1+t )^{\frac{5}{6} {\beta } - \frac{73}{48}}, \end{aligned}$$

(3.29)

where $0 < {\beta } \le \frac{1}{4}$. Together with (3.25), (3.26) and (3.29), for $t \ge 1$, we have

$$\begin{aligned} \Vert \nabla b (t)\Vert _{L^2} \le C(t + 1)^{-\frac{5}{4}} + C (1+ t)^{-\frac{7}{4}}+ C (1+t )^{\frac{5}{6} {\beta } - \frac{73}{48}} \le C (1+t)^{-\frac{5}{4}}. \end{aligned}$$

Note that for $0< t <1$, (3.2) implies

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

Thus, we obtain the second decay estimate in (3.14).

$\square$

Proposition 3.6

Let the assumption stated in Theorem 1.4 hold. Then

$$\begin{aligned} \Vert u(t)\Vert _{L^2} + \Vert \omega (t)\Vert _{L^2} \le C (1 + t)^{- \frac{11}{6}}. \end{aligned}$$

(3.30)

Proof

By the Young inequality, we have

$$\begin{aligned} C \Vert \nabla b\Vert _{L^2}^2 \Vert u\Vert ^{\frac{1}{2}}_{L^2} \Vert \nabla u\Vert ^{\frac{1}{2}}_{L^2} \le \frac{\chi }{8} \Vert u\Vert ^{2}_{L^2} + C \Vert \nabla b\Vert _{L^2}^{\frac{8}{3}} \Vert \nabla u\Vert ^{\frac{2}{3}}_{L^2}. \end{aligned}$$

Inserting it into (3.15),

$$\begin{aligned}&\frac{1}{2} \frac{\textrm{d}}{\textrm{d}t} \left( \Vert u(t)\Vert ^2_{L^2} + \Vert \omega (t)\Vert ^2_{L^2} \right) + \mu \Vert \nabla _h u\Vert ^2_{L^2} + \frac{\chi }{8} \Vert u\Vert _{L^2}^2 + (4 \chi - \frac{16\chi ^2}{\mu }) \Vert \omega \Vert _{L^2}^{2} \nonumber \\&\quad \le C \Vert \nabla b\Vert _{L^2}^{\frac{8}{3}} \Vert \nabla u\Vert ^{\frac{2}{3}}_{L^2}\nonumber \\&\quad \le C (1+t)^{- \frac{11}{3}}. \end{aligned}$$

(3.31)

Set $c_3 = \min \{\frac{\chi }{4}, 8\chi - \frac{32\chi ^2}{\mu }\}$. Integrating (3.31) in time, we obtain

$$\begin{aligned}&\Vert u (t)\Vert _{L^2}^2 + \Vert \omega (t)\Vert _{L^2}^2 \nonumber \\ {}&\quad \le e^{-c_3 t} (\Vert u_0\Vert _{L^2}^2 + \Vert \omega _0\Vert _{L^2}^2 ) + C \int \limits ^t_0 e^{-c_3 (t-s)} (1+s)^{- \frac{11}{3}} \text {d}s \nonumber \\ {}&\quad = e^{-c_3 t} (\Vert u_0\Vert _{L^2}^2 + \Vert \omega _0\Vert _{L^2}^2 ) + C \int \limits ^{\frac{t}{2}}_0 e^{-c_3 (t-s)} (1+s)^{- \frac{11}{3}} \text {d}s \nonumber \\ {}&\qquad + C \int \limits _{\frac{t}{2}}^t e^{-c_3 (t-s)} (1+s)^{- \frac{11}{3}} \text {d}s \nonumber \\ {}&\quad \le C e^{-c_3 t} + C e^{-\frac{c_3 t}{2}} + C (1+t)^{- \frac{11}{3}} \nonumber \\ {}&\quad \le C (1+t)^{- \frac{11}{3}}, \end{aligned}$$

(3.32)

which immediately implies the desired bound. Thus, the proof of Proposition 3.6 is completed.

$\square$

Finally, with Propositions 3.2, 3.5 and 3.6 at our disposal, we can improve the $L^2$ decay for $(u, \omega )$ and $(\nabla u, \nabla \omega )$, and obtain the decay for $\nabla ^2 b$ in $L^{2}$.

Proposition 3.7

Let the assumptions stated in Theorem 1.4 hold. Then for any $0< \alpha < \frac{1}{4}$, we have

$$\begin{aligned} \Vert \nabla ^2b(t)\Vert _{L^{2}} \le C (1+t)^{-\frac{65}{48} + \frac{17}{12} \alpha }, \end{aligned}$$

(3.33)

$$\begin{aligned} \Vert u(t)\Vert _{L^2} + \Vert \omega (t)\Vert _{L^2} \le C t^{- \frac{157}{64} + \frac{17}{16} \alpha }. \end{aligned}$$

(3.34)

Proof

Firstly, we establish the decay estimate of $\nabla ^2 b$. Applying $\nabla ^2$ to (3.19), we have

$$\begin{aligned} \nabla ^2 b(t)&= \nabla ^2 e^{\nu \Delta t} b_0 + \int \limits _0^{\frac{t}{2}} \nabla ^3 e^{\nu \Delta (t-s)} (b \otimes u - u \otimes b)(s)\textrm{d}s \nonumber \\&\quad + \int \limits _{\frac{t}{2}}^t \nabla ^3 e^{\nu \Delta (t-s)} (b \otimes u - u \otimes b)(s)\textrm{d}s. \end{aligned}$$

(3.35)

By Lemma 3.3, for $0< t <1$, we have

$$\begin{aligned} \Vert \nabla ^2 e^{\nu \Delta t} b_0 \Vert _{L^2} \le C \Vert \nabla ^2 b_0\Vert _{L^2}, \end{aligned}$$

and for $t \ge 1$,

$$\begin{aligned} \Vert \nabla ^2 e^{\nu \Delta t} b_0 \Vert _{L^2} \le C t^{-\frac{7}{4}}\Vert b_0\Vert _{L^1}. \end{aligned}$$

Therefore, for any $t>0$,

$$\begin{aligned} \Vert \nabla ^2 e^{\nu \Delta t} b_0 \Vert _{L^2} \le C(t + 1)^{-\frac{7}{4}}. \end{aligned}$$

(3.36)

Using Lemma 3.3, together with (3.23) and (3.30), we have for any $t \ge 1$,

$$\begin{aligned}&\left\| \int \limits _0^{\frac{t}{2}} \nabla ^3 e^{\nu \Delta (t-s)} (b \otimes u - u \otimes b)(s)\textrm{d}s \right\| _{L^2}\nonumber \\&\quad \le C \int \limits _0^{\frac{t}{2}} (t-s)^{-\frac{3}{2} -\frac{3}{4}} \Vert (b \otimes u - u \otimes b)(s)\Vert _{L^1} \textrm{d}s \nonumber \\&\quad \le C \int \limits _0^{\frac{t}{2}} (t-s)^{ -\frac{9}{4}} \Vert b(s)\Vert _{L^2} \Vert u(s)\Vert _{L^2} \textrm{d}s \nonumber \\&\quad \le C \int \limits _0^{\frac{t}{2}} (t-s)^{ -\frac{9}{4}} (1+ s)^{-\frac{31}{12}} \textrm{d}s. \nonumber \\&\quad \le C (1+ t)^{-\frac{9}{4}}. \end{aligned}$$

(3.37)

Applying Lemmas 3.3 and 3.4, for any $t>0$, we yield

$$\begin{aligned}&\left\| \int \limits _{\frac{t}{2}}^t \nabla ^3 e^{\nu \Delta (t-s)} (b \otimes u - u \otimes b)(s)\textrm{d}s \right\| _{L^2}\nonumber \\&\quad = \left\| \int \limits _{\frac{t}{2}}^t \Lambda ^{- \alpha } \nabla ^2 e^{\nu \Delta (t-s)} \Lambda ^{\alpha } \nabla (b \otimes u - u \otimes b)(s)\textrm{d}s \right\| _{L^2} \nonumber \\&\quad \le C \int \limits _{\frac{t}{2}}^t (t-s)^{- \frac{2- \alpha }{2}} \Vert \Lambda ^{1+ \alpha } (b \otimes u - u \otimes b)(s)\Vert _{L^2} \textrm{d}s \nonumber \\&\quad \le C \int \limits _{\frac{t}{2}}^t (t-s)^{- \frac{2-\alpha }{2}} (\Vert \Lambda ^{1+ \alpha } u(s) \Vert _{L^4} \Vert b(s)\Vert _{L^4} + \Vert \Lambda ^{1+\alpha } b(s) \Vert _{L^4} \Vert u(s)\Vert _{L^4}) \textrm{d}s \nonumber \\&\quad \le F_1 + F_2, \end{aligned}$$

(3.38)

where we set $0< \alpha < \frac{1}{4}$. Then by using Gagliardo–Nirenberg inequality, (1.11), (3.14) and (3.30), we have

$$\begin{aligned} F_1&= C \int \limits _{\frac{t}{2}}^t (t-s)^{- \frac{2-\alpha }{2}} \Vert \Lambda ^{1+ \alpha } u(s) \Vert _{L^4} \Vert b(s)\Vert _{L^4} \textrm{d}s\\&\le C \int \limits _{\frac{t}{2}}^t (t-s)^{- \frac{2-\alpha }{2}} (\Vert u(s)\Vert _{L^2}^{\frac{1}{8} - \frac{\alpha }{2}} \Vert \nabla ^2 u(s)\Vert _{L^2}^{\frac{7}{8} + \frac{\alpha }{2}} \Vert b(s)\Vert _{L^2}^{\frac{1}{4} } \Vert \nabla b(s)\Vert _{L^2}^{\frac{3}{4} }) \textrm{d}s\\&\le C \int \limits _{\frac{t}{2}}^t (t-s)^{- \frac{2-\alpha }{2}} \Vert u(s)\Vert _{H^2} (1+s)^{-\frac{11}{6} (\frac{1}{8} - \frac{\alpha }{2}) - \frac{1}{4} \times \frac{3}{4} - \frac{3}{4} \times \frac{5}{4}} \textrm{d}s\\&\le C \int \limits _{\frac{t}{2}}^t (t-s)^{- \frac{2-\alpha }{2}} (1+s)^{\frac{11}{12} \alpha - \frac{65}{48}}\textrm{d}s\\&\le C (1 + t)^{\frac{17}{12} \alpha - \frac{65}{48}}. \end{aligned}$$

Next, we consider $F_2$, set

$$\begin{aligned} {\mathcal {M}}_2(t) = \sup _{0 \le s \le t} \{(1 + s)^{-\frac{17}{12} \alpha + \frac{65}{48}} \Vert \nabla ^2 b(s)\Vert _{L^2} \}, \end{aligned}$$

then we have

$$\begin{aligned} F_2&= C \int \limits ^{t}_{\frac{t}{2}} (t-s)^{-1 + \frac{\alpha }{2}} \Vert \Lambda ^{1 + \alpha } b (s)\Vert _{L^4} \Vert u(s)\Vert _{L^4} \text {d}s\\ {}&\le C \int \limits _{\frac{t}{2}}^t (t-s)^{- \frac{2-\alpha }{2}} (\Vert b(s)\Vert _{L^2}^{\frac{1}{8} - \frac{\alpha }{2}} \Vert \nabla ^2 b(s)\Vert _{L^2}^{\frac{7}{8} + \frac{\alpha }{2}} \Vert u(s)\Vert _{L^2}^{\frac{1}{4} } \Vert \nabla u(s)\Vert _{L^2}^{\frac{3}{4} }) \text {d}s\\ {}&\le C \int \limits _{\frac{t}{2}}^t (t-s)^{- \frac{2-\alpha }{2}} \Vert \nabla ^2 b(s)\Vert _{L^2}^{\frac{7}{8} + \frac{\alpha }{2}} (1+s)^{-\frac{3}{4} (\frac{1}{8} - \frac{\alpha }{2}) - \frac{1}{4} \times \frac{11}{6} - \frac{3}{4} \times \frac{1}{2}} \text {d}s\\ {}&\le C {\mathcal {M}}^{\frac{7}{8} + \frac{\alpha }{2}}_2(t) \int \limits _{\frac{t}{2}}^t (t-s)^{- \frac{2-\alpha }{2}} (1+s)^{-(-\frac{17}{12} \alpha + \frac{65}{48})(\frac{7}{8} + \frac{\alpha }{2}) - \frac{89}{96} + \frac{3}{8} \alpha } \text {d}s\\ {}&\le C {\mathcal {M}}^{\frac{7}{8} + \frac{\alpha }{2}}_2(t) (1+t)^{-(-\frac{17}{12} \alpha + \frac{65}{48})(\frac{7}{8} + \frac{\alpha }{2}) - \frac{89}{96} + \frac{7}{8} \alpha }. \end{aligned}$$

Inserting $F_1$ and $F_2$ into (3.38), and combining (3.36), (3.37) and (3.38), when $t \ge 1$, we have

$$\begin{aligned} {\mathcal {M}}_2(t)&\le C (1+t)^{-\frac{7}{4} -\frac{17}{12} \alpha + \frac{65}{48}} + C (1+t)^{-\frac{9}{4} -\frac{17}{12} \alpha + \frac{65}{48}} \\&\quad + C + C {\mathcal {M}}_2^{\frac{7}{8} + \frac{\alpha }{2}} (1+t)^{-(-\frac{17}{12} \alpha + \frac{65}{48})(\frac{7}{8} + \frac{\alpha }{2}) - \frac{89}{96} + \frac{7}{8} \alpha -\frac{17}{12} \alpha + \frac{65}{48}} , \end{aligned}$$

where $0< \alpha < \frac{1}{4}$. Through calculation, it can be seen that when $0< \alpha < \frac{1}{4}$,

$$\begin{aligned}&-\left( -\frac{17}{12} \alpha + \frac{65}{48}\right) \left( \frac{7}{8} + \frac{\alpha }{2}\right) - \frac{89}{96} + \frac{7}{8} \alpha -\frac{17}{12} \alpha + \frac{65}{48} \\&\quad = \frac{17}{24} \alpha ^2 + \frac{1}{48} \alpha - \frac{291}{384} < 0, \end{aligned}$$

$$\begin{aligned} -\frac{7}{4} -\frac{17}{12} \alpha + \frac{65}{48}< 0, \ \ \text{ and } \ \ -\frac{9}{4} -\frac{17}{12} \alpha + \frac{65}{48} < 0. \end{aligned}$$

Then for the $t \ge 1$, by the Young inequality we yield

$$\begin{aligned} {\mathcal {M}}_2 (t) \le C. \end{aligned}$$

Note that for $0< t <1$, (1.11) implies

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

Thus, we obtain (3.33).

Finally, with (3.33) at our disposal, we can turn to establish the decay estimate (3.34). Again applying Hölder’s inequality, we obtain

$$\begin{aligned}&\int b \cdot \nabla b \cdot u \le \Vert b\Vert _{L^6} \Vert \nabla b \Vert _{L^2} \Vert u\Vert _{L^3} \nonumber \\&\quad \le C \Vert \nabla b \Vert _{L^2}^2 \Vert u\Vert ^{\frac{1}{2}}_{L^2} \Vert \nabla u\Vert ^{\frac{1}{2}}_{L^2}, \end{aligned}$$

(3.39)

and

$$\begin{aligned}&\int b \cdot \nabla b \cdot u \le \Vert b\Vert _{L^{3}} \Vert \nabla b \Vert _{L^6} \Vert u\Vert _{L^2} \nonumber \\&\quad \le C \Vert b\Vert _{L^2}^{\frac{1}{2}} \Vert \nabla b\Vert _{L^2}^{\frac{1}{2}} \Vert \nabla ^2 b\Vert _{L^2} \Vert u\Vert _{L^2}. \end{aligned}$$

(3.40)

Inserting (3.39) and (3.40) into (3.15) and integrating in time yields

$$\begin{aligned} \Vert u(t)\Vert ^2_{L^2} + \Vert \omega (t)\Vert ^2_{L^2} \le C e^{-ct} + Z_1 + Z_2 , \end{aligned}$$

(3.41)

where

$$\begin{aligned} Z_1 = C \int \limits ^{\frac{t}{2}}_0 e^{-c(t-s)} \Vert \nabla b(s) \Vert _{L^2}^2 \Vert u(s)\Vert ^{\frac{1}{2}}_{L^2} \Vert \nabla u(s)\Vert ^{\frac{1}{2}}_{L^2} \text {d}s, \end{aligned}$$

$$\begin{aligned} Z_2 = C \int \limits _{\frac{t}{2}}^t e^{-c(t-s)} \Vert b\Vert _{L^2}^{\frac{1}{2}} \Vert \nabla b\Vert _{L^2}^{\frac{1}{2}} \Vert \nabla ^2 b\Vert _{L^2} \Vert u\Vert _{L^2} \textrm{d}s. \end{aligned}$$

Applying Hölder’s inequality, together with (3.1) and (3.2), we obtain

$$\begin{aligned} Z_1 \le C e^{- \frac{ct}{2}}. \end{aligned}$$

(3.42)

Set

$$\begin{aligned} {\mathcal {N}}_1(t) = \sup _{0\le s\le t} \{(1+s)^{(\frac{157}{64} - \frac{17}{16} \alpha )} ( \Vert u(s)\Vert _{L^2} + \Vert \omega (s)\Vert _{L^2}) \}. \end{aligned}$$

Then (3.14), (3.30) and (3.33) yield

$$\begin{aligned} Z_2 \le {\mathcal {N}}_1(t) \int \limits _{\frac{t}{2}}^t e^{-c(t-s)} (1+ s)^{- \frac{157}{32} + \frac{17}{8} \alpha } \textrm{d}s. \end{aligned}$$

(3.43)

Inserting (3.42) and (3.43) into (3.41), we get

$$\begin{aligned} {\mathcal {N}}_1^2(t) \le C + C {\mathcal {N}}_1(t), \end{aligned}$$

since for any $t>0$, $(1 +t)^{(\frac{157}{64} - \frac{17}{16} \alpha )}e^{-\frac{ct}{2}} \le C$. Then the Young inequality yields

$$\begin{aligned} {\mathcal {N}}_1(t) \le C, \end{aligned}$$

which implies (3.34). Thus, the proof of Proposition 3.7 is completed. $\square$

Proposition 3.8

Let the assumptions stated in Theorem 1.4 hold. Then for any $0< \alpha < \frac{1}{4}$, we have

$$\begin{aligned} \Vert \nabla u(t)\Vert _{L^2} + \Vert \nabla \omega (t)\Vert _{L^2} \le C(1 +t)^{- \frac{471}{512} + \frac{51}{128}\alpha }. \end{aligned}$$

(3.44)

Proof

Applying $\partial _k$ with $k = 1, 2, 3$ to (1.8)$_1$ and (1.8)$_3$, dotting the results by $\partial _k u$ and $\partial _k \omega$, respectively, integrating in space domain and adding them together, we obtain

$$\begin{aligned}&\frac{1}{2} \frac{\textrm{d}}{\textrm{d}t} \sum _{k=1}^3 \left( \Vert \partial _k u(t)\Vert _{L^2}^2 + \Vert \partial _k \omega (t)\Vert _{L^2}^2 \right) + \mu \sum _{k=1}^3 \Vert \nabla _h \partial _k u\Vert _{L^2}^2 + \chi \sum _{k=1}^3 \Vert \partial _k u\Vert _{L^2}^2\nonumber \\&\qquad + \gamma \sum _{k=1}^3 \Vert \partial _3 \partial _k \omega \Vert _{L^2}^2 + 4 \chi \sum _{k=1}^3 \Vert \partial _k \omega \Vert _{L^2}^2 + \kappa \sum _{k=1}^3 \Vert \partial _k \nabla \cdot \omega \Vert _{L^2}^2\nonumber \\&\quad = - \sum _{k=1}^3 \int \partial _k (u \cdot \nabla u) \cdot \partial _k u \textrm{d}x + \sum _{k=1}^3 \int \partial _k (b \cdot \nabla b) \cdot \partial _k u \textrm{d}x + 2 \chi \sum _{k=1}^3 \int \nabla \times \partial _k \omega \cdot \partial _k u \textrm{d}x \nonumber \\&\qquad - \sum _{k=1}^3 \int \partial _k (u \cdot \nabla \omega ) \cdot \partial _k \omega \textrm{d}x + 2 \chi \sum _{k=1}^3 \int \nabla \times \partial _k u \cdot \partial _k \omega \textrm{d}x\nonumber \\&\quad := D_1 + D_2 + D_3 + D_4 + D_5. \end{aligned}$$

(3.45)

Using the divergence free condition $\nabla \cdot u =0$, together with Lemma 2.2 and the Young inequality, we have

$$\begin{aligned} D_1&= - \sum ^3_{k=1} \sum ^3_{j=1} \int \partial _k u_j \partial _j u \cdot \partial _k u \textrm{d}x\\&= - \sum ^2_{k=1} \sum ^3_{j=1} \int \partial _k u_j \partial _j u \cdot \partial _k u \textrm{d}x - \sum ^2_{j=1} \int \partial _3 u_j \partial _j u \cdot \partial _3 u \textrm{d}x - \int \partial _3 u_3 \partial _3 u \cdot \partial _3 u \textrm{d}x\\&\le C \sum ^2_{k=1} \sum ^3_{j=1} \Vert \partial _k u_j\Vert _{L^2}^{\frac{1}{2}} \Vert \partial _1 \partial _k u_j\Vert _{L^2}^{\frac{1}{2}} \Vert \partial _j u\Vert _{L^2}^{\frac{1}{2}} \Vert \partial _2 \partial _j u\Vert _{L^2}^{\frac{1}{2}} \Vert \partial _k u\Vert _{L^2}^{\frac{1}{2}} \Vert \partial _3 \partial _k u\Vert _{L^2}^{\frac{1}{2}}\\&\quad + \sum ^2_{j=1} \Vert \partial _3 u_j\Vert _{L^2}^{\frac{1}{2}} \Vert \partial _1 \partial _3 u_j\Vert _{L^2}^{\frac{1}{2}} \Vert \partial _j u\Vert _{L^2}^{\frac{1}{2}} \Vert \partial _3 \partial _j u\Vert _{L^2}^{\frac{1}{2}} \Vert \partial _3 u\Vert _{L^2}^{\frac{1}{2}} \Vert \partial _2 \partial _3 u\Vert _{L^2}^{\frac{1}{2}}\\&\quad + \Vert \partial _h u_h\Vert _{L^2}^{\frac{1}{2}} \Vert \partial _3 \partial _h u_h\Vert _{L^2}^{\frac{1}{2}} \Vert \partial _3 u\Vert _{L^2}^{\frac{1}{2}} \Vert \partial _2 \partial _3 u\Vert _{L^2}^{\frac{1}{2}} \Vert \partial _3 u\Vert _{L^2}^{\frac{1}{2}} \Vert \partial _1 \partial _3 u\Vert _{L^2}^{\frac{1}{2}}\\&\le C( \Vert \nabla _h u\Vert _{L^2} \Vert \nabla u\Vert ^{\frac{1}{2}}_{L^2} + \Vert \nabla u\Vert _{L^2} \Vert \nabla _h u\Vert ^{\frac{1}{2}}_{L^2})\Vert \nabla \nabla _h u \Vert _{L^2}^{\frac{3}{2}}. \end{aligned}$$

Similarly,

$$\begin{aligned} D_2&= \sum _{k=1}^3 \int \partial _k (b \cdot \nabla b) \cdot \partial _k u \textrm{d}x\\&\le C \Vert \nabla ^2 b\Vert _{L^2} (\Vert \nabla b\Vert _{L^2} \Vert \nabla \nabla _h u\Vert ^{\frac{1}{2}}_{L^2} \Vert \nabla u\Vert ^{\frac{1}{2}}_{L^2} + \Vert b\Vert _{L^{\infty }} \Vert \nabla u\Vert _{L^2} ), \\ D_4&= - \sum _{k=1}^3 \int \partial _k (u \cdot \nabla \omega ) \cdot \partial _k \omega \textrm{d}x \\&\le C \Vert \nabla _h u\Vert ^{\frac{1}{2}}_{L^2} \Vert \nabla \nabla _h u\Vert ^{\frac{1}{2}}_{L^2} \Vert \nabla \omega \Vert ^{\frac{1}{2}}_{L^2} \Vert \partial _3 \nabla \omega \Vert ^{\frac{1}{2}}_{L^2}( \Vert \nabla ^2 \omega \Vert ^{\frac{1}{2}}_{L^2} \Vert \nabla \omega \Vert ^{\frac{1}{2}}_{L^2} + \Vert \partial _3 \omega \Vert ^{\frac{1}{2}}_{L^2} \Vert \nabla \partial _3 \omega \Vert ^{\frac{1}{2}}_{L^2}). \end{aligned}$$

By the Young inequality, we have

$$\begin{aligned} D_3 + D_5 \le \frac{\mu }{2} \sum ^3_{k=1} \Vert \nabla _h \partial _k u\Vert _{L^2}^2 + \frac{16 \chi ^2 }{\mu } \sum ^3_{k=1}\Vert \partial _k\omega \Vert _{L^2}^2 + \frac{3 \chi }{4} \sum ^3_{k=1} \Vert \partial _k u\Vert _{L^2}^2 + \frac{16 \chi }{3} \sum ^3_{k=1}\Vert \partial _3\partial _k \omega \Vert _{L^2}^2. \end{aligned}$$

Inserting the above estimates into (3.45), we infer that

$$\begin{aligned}&\frac{1}{2} \frac{\textrm{d}}{\textrm{d}t} \left( \Vert \nabla u(t)\Vert _{L^2}^2 + \Vert \nabla \omega (t)\Vert _{L^2}^2 \right) + \frac{\mu }{2} \Vert \nabla _h \nabla u\Vert _{L^2}^2 + \frac{\chi }{4} \Vert \nabla u\Vert _{L^2}^2\nonumber \\&\qquad + (\gamma - \frac{16 \chi }{3}) \Vert \partial _3 \nabla \omega \Vert _{L^2}^2 + \left( 4 \chi - \frac{16 \chi ^2}{\mu }\right) \Vert \nabla \omega \Vert _{L^2}^2 + \kappa \Vert \nabla \nabla \cdot \omega \Vert _{L^2}^2\nonumber \\&\quad \le C\Vert \nabla u\Vert ^{\frac{3}{2}}_{L^2}\Vert \nabla \nabla _h u \Vert _{L^2}^{\frac{3}{2}} \\&\qquad + C \Vert \nabla ^2 b\Vert _{L^2} (\Vert \nabla b\Vert _{L^2} \Vert \nabla \nabla _h u\Vert ^{\frac{1}{2}}_{L^2} \Vert \nabla u\Vert ^{\frac{1}{2}}_{L^2} + \Vert b\Vert _{L^{\infty }} \Vert \nabla u\Vert _{L^2} ) \\&\qquad + C \Vert \nabla _h u\Vert ^{\frac{1}{2}}_{L^2} \Vert \nabla \nabla _h u\Vert ^{\frac{1}{2}}_{L^2} \Vert \nabla \omega \Vert _{L^2} \Vert \partial _3 \nabla \omega \Vert ^{\frac{1}{2}}_{L^2} \Vert \nabla ^2 \omega \Vert ^{\frac{1}{2}}_{L^2}. \end{aligned}$$

Then integrating the above inequality in time, set $c = \min \{\frac{ \chi }{2}, 8\chi - \frac{32\chi ^2}{\mu }\}$, we yield

$$\begin{aligned}&\Vert \nabla u (t)\Vert ^2_{L^2} + \Vert \nabla \omega (t)\Vert ^2_{L^2} \nonumber \\&\quad \le e^{-ct}(\Vert \nabla u_0\Vert ^2_{L^2} + \Vert \nabla \omega _0\Vert ^2_{L^2}) + C (J_1 + J_2 + J_3), \end{aligned}$$

(3.46)

where

$$\begin{aligned} J_1= & {} {} \int \limits ^{t}_0 e^{-c(t-s)} \Vert \nabla u\Vert ^{\frac{3}{2}}_{L^2}\Vert \nabla \nabla _h u \Vert _{L^2}^{\frac{3}{2}} \text {d}s, \\ J_2= & {} {} \int \limits ^t_{0} e^{-c(t-s)} \Vert \nabla ^2 b(s)\Vert _{L^2} (\Vert \nabla b(s)\Vert _{L^2} \Vert \nabla \nabla _h u(s)\Vert ^{\frac{1}{2}}_{L^2} \Vert \nabla u(s)\Vert ^{\frac{1}{2}}_{L^2} + \Vert b(s)\Vert _{L^{\infty }} \Vert \nabla u(s)\Vert _{L^2} ) \text {d}s, \\ J_3= & {} {} \int \limits ^{t}_0 e^{-c(t-s)} \Vert \nabla _h u(s)\Vert ^{\frac{1}{2}}_{L^2} \Vert \nabla \nabla _h u(s)\Vert ^{\frac{1}{2}}_{L^2} \Vert \nabla \omega (s)\Vert _{L^2} \Vert \partial _3 \nabla \omega (s)\Vert ^{\frac{1}{2}}_{L^2} \Vert \nabla ^2 \omega (s)\Vert ^{\frac{1}{2}}_{L^2} \text {d}s. \end{aligned}$$

Finally, we need consider $J_1$ - $J_3$, respectively. Applying the Gagliardo– Nirenberg inequality, (3.2) and (3.34), we have,

$$\begin{aligned} J_{1}&= \int \limits ^{\frac{t}{2}}_0 e^{-c(t-s)} \Vert \nabla u(s)\Vert ^{\frac{3}{2}}_{L^2}\Vert \nabla \nabla _h u(s) \Vert _{L^2}^{\frac{3}{2}} \text {d}s + \int \limits _{\frac{t}{2}}^t e^{-c(t-s)} \Vert \nabla u(s)\Vert ^{\frac{3}{2}}_{L^2}\Vert \nabla \nabla _h u(s) \Vert _{L^2}^{\frac{3}{2}} \text {d}s\\ {}&\le C e^{- \frac{ct}{2}} \int \limits ^{\frac{t}{2}}_0 \Vert u(s)\Vert _{H^2} (\Vert \nabla u (s)\Vert ^2_{L^2} + \Vert \nabla \nabla _h u (s)\Vert _{L^2}^2) \text {d}s\\ {}&\quad + C \int \limits _{\frac{t}{2}}^t e^{-c(t-s)} \Vert u(s)\Vert ^{\frac{3}{4}}_{L^2}\Vert \nabla ^2 u(s) \Vert _{L^2}^{\frac{9}{4}} \text {d}s\\ {}&\le C e^{- \frac{ct}{2}} + C (1+t)^{- \frac{471}{256} + \frac{51}{64}\alpha }, \end{aligned}$$

where C is dependent on $\mu$, $\chi$, $\nu$, $\kappa$ and the initial data $\Vert (u_0, b_0, \omega _0)\Vert _{H^2}$. For $J_2$,

$$\begin{aligned} J_2&= \int \limits ^t_{0} e^{-c(t-s)} \Vert \nabla ^2 b(s)\Vert _{L^2} \Vert \nabla b(s)\Vert _{L^2} \Vert \nabla \nabla _h u(s)\Vert ^{\frac{1}{2}}_{L^2} \Vert \nabla u(s)\Vert ^{\frac{1}{2}}_{L^2} \text {d}s\\ {}&\quad + \int \limits ^t_{0} e^{-c(t-s)} \Vert \nabla ^2 b(s)\Vert _{L^2}\Vert b(s)\Vert _{L^{\infty }} \Vert \nabla u(s)\Vert _{L^2} \text {d}s\\ {}&:= J_{21} + J_{22}.\end{aligned}$$

Using the Young inequality, (3.1), (3.2), (3.14) and (3.34), we obtain

$$\begin{aligned} J_{21}&= \int \limits ^{\frac{t}{2}}_0 e^{-c(t-s)} \Vert \nabla ^2 b(s)\Vert _{L^2} \Vert \nabla b(s)\Vert _{L^2} \Vert \nabla \nabla _h u(s)\Vert ^{\frac{1}{2}}_{L^2} \Vert \nabla u(s)\Vert ^{\frac{1}{2}}_{L^2} \text {d}s \\ {}&\qquad + \int \limits _{\frac{t}{2}}^t e^{-c(t-s)} \Vert \nabla ^2 b(s)\Vert _{L^2} \Vert \nabla b(s)\Vert _{L^2} \Vert \nabla \nabla _h u(s)\Vert ^{\frac{1}{2}}_{L^2} \Vert \nabla u(s)\Vert ^{\frac{1}{2}}_{L^2} \text {d}s\\ {}&\quad \le \int \limits ^{\frac{t}{2}}_0 e^{-c(t-s)} \Vert b(s)\Vert _{H^2}( \Vert \nabla b(s)\Vert ^2_{L^2} + \Vert \nabla \nabla _h u(s)\Vert ^{2}_{L^2} + \Vert \nabla u(s)\Vert ^{2}_{L^2} )\text {d}s\\ {}&\qquad + \int \limits _{\frac{t}{2}}^t e^{-c(t-s)} \Vert \nabla ^2 b(s)\Vert _{L^2} \Vert \nabla b(s)\Vert _{L^2} \Vert \nabla \nabla _h u(s)\Vert ^{\frac{1}{2}}_{L^2} \Vert \nabla u(s)\Vert ^{\frac{1}{2}}_{L^2} \text {d}s\\ {}&\quad \le C e^{- \frac{ct}{2}} + \int \limits _{\frac{t}{2}}^t e^{-c(t-s)} \Vert \nabla ^2 b(s)\Vert _{L^2} \Vert \nabla b(s)\Vert _{L^2} \Vert \nabla ^2 u(s)\Vert ^{\frac{3}{4}}_{L^2} \Vert u(s)\Vert ^{\frac{1}{4}}_{L^2} \text {d}s \\ {}&\quad \le C e^{- \frac{ct}{2}} + \int \limits _{\frac{t}{2}}^t e^{-c(t-s)} (1+s)^{-\frac{5}{4} - \frac{1}{4} (\frac{157}{64} - \frac{17}{16} \alpha )} \text {d}s\\ {}&\quad \le C e^{- \frac{ct}{2}} + C(1+t)^{- \frac{477}{256} + \frac{17}{64} \alpha }, \end{aligned}$$

and

$$\begin{aligned} J_{22}&= \int \limits ^{\frac{t}{2}}_{0} e^{-c(t-s)} \Vert \nabla ^2 b(s)\Vert _{L^2}\Vert b(s)\Vert _{L^{\infty }} \Vert \nabla u(s)\Vert _{L^2} \text {d}s \\ {}&\quad + \int \limits _{\frac{t}{2}}^{t} e^{-c(t-s)} \Vert \nabla ^2 b(s)\Vert _{L^2}\Vert b(s)\Vert _{L^{\infty }} \Vert \nabla u(s)\Vert _{L^2} \text {d}s\\ {}&\le \int \limits ^{\frac{t}{2}}_{0} e^{-c(t-s)} \Vert b(s)\Vert _{H^2}(\Vert \nabla ^2 b(s)\Vert ^2_{L^{2}} + \Vert \nabla u(s)\Vert ^2_{L^2} ) \text {d}s\\ {}&\quad + \int \limits _{\frac{t}{2}}^{t} e^{-c(t-s)} \Vert b(s)\Vert ^{\frac{1}{4}}_{L^{2}} \Vert \nabla ^2 b(s)\Vert ^{\frac{7}{4}}_{L^2} \Vert \nabla u(s)\Vert ^{\frac{1}{2}}_{L^2} \Vert \nabla ^2 u(s)\Vert ^{\frac{1}{2}}_{L^2} \text {d}s\\ {}&\le C e^{- \frac{ct}{2}} + C \int \limits _{\frac{t}{2}}^{t} e^{-c(t-s)} (1+s)^{- \frac{1}{4} \times \frac{3}{4} + \frac{7}{4} (\frac{17}{12} \alpha - \frac{65}{48}) - \frac{1}{2} (\frac{157}{64} - \frac{17}{16} \alpha )} \text {d}s\\ {}&\le C e^{- \frac{ct}{2}} + C (1+t)^{-\frac{23248}{6144} + \frac{289}{96} \alpha }. \end{aligned}$$

Then, we have

$$\begin{aligned} J_2&\le C e^{- \frac{ct}{2}} + C(1+t)^{- \frac{477}{256} + \frac{17}{64} \alpha } + C (1+t)^{-\frac{23248}{6144} + \frac{289}{96} \alpha }\\&\le C e^{- \frac{ct}{2}} + C(1+t)^{- \frac{477}{256} + \frac{17}{64} \alpha }, \end{aligned}$$

where $0< \alpha < \frac{1}{4}$. Similarly,

$$\begin{aligned} J_3&= \int \limits ^{\frac{t}{2}}_{0} e^{-c(t-s)} \Vert \nabla _h u(s)\Vert ^{\frac{1}{2}}_{L^2} \Vert \nabla \nabla _h u(s)\Vert ^{\frac{1}{2}}_{L^2} \Vert \nabla \omega (s)\Vert _{L^2} \Vert \partial _3 \nabla \omega (s)\Vert ^{\frac{1}{2}}_{L^2} \Vert \nabla ^2 \omega (s)\Vert ^{\frac{1}{2}}_{L^2} \text {d}s\\ {}&\quad + \int \limits _{\frac{t}{2}}^{t} e^{-c(t-s)} \Vert \nabla _h u(s)\Vert ^{\frac{1}{2}}_{L^2} \Vert \nabla \nabla _h u(s)\Vert ^{\frac{1}{2}}_{L^2} \Vert \nabla \omega (s)\Vert _{L^2} \Vert \partial _3 \nabla \omega (s)\Vert ^{\frac{1}{2}}_{L^2} \Vert \nabla ^2 \omega (s)\Vert ^{\frac{1}{2}}_{L^2} \text {d}s\\ {}&\le \int \limits ^{\frac{t}{2}}_{0} e^{-c(t-s)} \Vert u(s)\Vert _{H^2}^{\frac{1}{2}} \Vert \omega (s)\Vert _{H^2}^{\frac{1}{2}} (\Vert \nabla _h u(s)\Vert ^{2}_{L^2} + \Vert \partial _3 \nabla \omega (s)\Vert ^{2}_{L^2} + \Vert \nabla ^2 \omega (s)\Vert ^{2}_{L^2} )\text {d}s\\ {}&\quad + C \int \limits _{\frac{t}{2}}^{t} e^{-c(t-s)} \Vert \nabla u(s)\Vert ^{\frac{1}{2}}_{L^2} \Vert \nabla \nabla _h u(s)\Vert ^{\frac{1}{2}}_{L^2} \Vert \nabla \omega (s)\Vert _{L^2} \Vert \nabla ^2 \omega (s)\Vert _{L^2}\text {d}s\\ {}&\le C e^{- \frac{ct}{2}} + C \int \limits _{\frac{t}{2}}^{t} e^{-c(t-s)} \Vert \omega (s)\Vert ^{\frac{1}{2}}_{L^2} \Vert \nabla ^2 \omega (s)\Vert ^{\frac{3}{2}}_{L^2} \Vert u(s)\Vert ^{\frac{1}{4}}_{L^2} \Vert \nabla ^2 u(s)\Vert ^{\frac{3}{4}}_{L^2}\text {d}s\\ {}&\le C e^{- \frac{ct}{2}} + C \int \limits _{\frac{t}{2}}^{t} e^{-c(t-s)} (1+s)^{-\frac{3}{4} (\frac{157}{64} - \frac{17}{16} \alpha )}\text {d}s\\ {}&\le C e^{- \frac{ct}{2}} + C (1+t)^{- (\frac{471}{256} - \frac{51}{64} \alpha )}. \end{aligned}$$

Inserting $J_1$–$J_3$ into (3.46), we obtain

$$\begin{aligned}&\Vert \nabla u (t)\Vert ^2_{L^2} + \Vert \nabla \omega (t)\Vert ^2_{L^2} \nonumber \\&\quad \le C e^{- \frac{ct}{2}} + C(1+t)^{- \frac{477}{256} + \frac{17}{64} \alpha } + C(1+t)^{- \frac{471}{256} + \frac{51}{64} \alpha }\le C(1+t)^{- \frac{471}{256} + \frac{51}{64} \alpha }. \end{aligned}$$

which implies (3.44). Thus, the proof of Proposition 3.8 is completed. $\square$