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Research article

On distance signless Laplacian eigenvalues of zero divisor graph of commutative rings

  • Received: 15 December 2021 Revised: 27 March 2022 Accepted: 05 April 2022 Published: 28 April 2022
  • MSC : 05C50, 05C12, 15A18

  • For a simple connected graph G of order n, the distance signless Laplacian matrix is defined by DQ(G)=D(G)+Tr(G), where D(G) and Tr(G) is the distance matrix and the diagonal matrix of vertex transmission degrees, respectively. The zero divisor graph Γ(R) of a finite commutative ring R is a simple graph, whose vertex set is the set of non-zero zero divisors of R and two vertices v,wΓ(R) are edge connected whenever vw=wv=0. In this article, we find the DQ-eigenvalues of zero divisor graph of the ring Zn for general value n=pl11pl22, where p1<p2 are distinct prime numbers and l1,l2N. Further, we investigate the DQ-eigenvalues of zero divisor graphs of local rings and the rings whose associated zero divisor graphs are Hamiltonian. Also, we obtain the trace norm and the Wiener index of Γ(Zn) for some special values of n.

    Citation: Bilal A. Rather, M. Aijaz, Fawad Ali, Nabil Mlaiki, Asad Ullah. On distance signless Laplacian eigenvalues of zero divisor graph of commutative rings[J]. AIMS Mathematics, 2022, 7(7): 12635-12649. doi: 10.3934/math.2022699

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  • For a simple connected graph G of order n, the distance signless Laplacian matrix is defined by DQ(G)=D(G)+Tr(G), where D(G) and Tr(G) is the distance matrix and the diagonal matrix of vertex transmission degrees, respectively. The zero divisor graph Γ(R) of a finite commutative ring R is a simple graph, whose vertex set is the set of non-zero zero divisors of R and two vertices v,wΓ(R) are edge connected whenever vw=wv=0. In this article, we find the DQ-eigenvalues of zero divisor graph of the ring Zn for general value n=pl11pl22, where p1<p2 are distinct prime numbers and l1,l2N. Further, we investigate the DQ-eigenvalues of zero divisor graphs of local rings and the rings whose associated zero divisor graphs are Hamiltonian. Also, we obtain the trace norm and the Wiener index of Γ(Zn) for some special values of n.



    In this paper, we consider the 3D nonlinear damped micropolar equation

    {ut+(u)u(ν+κ)Δu+σ|u|β1u+p=2κ×ω+f1(x,t),ωt+(u)ω+4κωγΔωμω=2κ×u+f2(x,t),u=0,u(x,t)|t=τ=uτ(x),   ω(x,t)|t=τ=ωτ(x), (1.1)

    where (x,t)Ω×[τ,+), τR, ΩR3 is a bounded domain, u=u(x,t) is the fluid velocity, ω=ω(x,t) is the angular velocity, σ is the damping coefficient, which is a positive constant, f1=f1(x,t) and f2=f2(x,t) represent external forces, ν, κ, γ, μ are all positive constants, γ and μ represent the angular viscosities.

    Micropolar flow can describe a fluid with microstructure, that is, a fluid composed of randomly oriented particles suspended in a viscous medium without considering the deformation of fluid particles. Since Eringen first published his paper on the model equation of micropolar fluids in 1966 [5], the formation of modern theory of micropolar fluid dynamics has experienced more than 40 years of development. For the 2D case, many researchers have discussed the long time behavior of micropolar equations (such as [2,4,10,24]). It should be mentioned that some conclusions in the 2D case no longer hold for the 3D case due to different structures of the system. In the 3D case, the work of micropolar equations (1.1) with σ=0, f1=0, and f2=0 has attracted a lot of attention (see [6,14,19]). Galdi and Rionero [6] proved the existence and uniqueness of solutions of 3D incompressible micropolar equations. In a 3D bounded domain, for small initial data Yamaguchi [19] investigated the existence of a global solution to the initial boundary problem for the micropolar system. In [14], Silva and Cruz et al. studied the L2-decay of weak solutions for 3D micropolar equations in the whole space R3. When f1=f2=0, for the Cauchy problem of the 3D incompressible nonlinear damped micropolar equations, Ye [22] discussed the existence and uniqueness of global strong solutions when β=3 and 4σ(ν+κ)>1 or β>3. In [18], Wang and Long showed that strong solutions exist globally for any 1β3 when initial data satisfies some certain conditions. Based on [22], Yang and Liu [20] obtained uniform estimates of the solutions for 3D incompressible micropolar equations with damping, and then they proved the existence of global attractors for 3<β<5. In [7], Li and Xiao investigated the large time decay of the L2-norm of weak solutions when β>145, and considered the upper bounds of the derivatives of the strong solution when β>3. In [21], for 1β<73, Yang, Liu, and Sun proved the existence of trajectory attractors for 3D nonlinear damped micropolar fluids.

    To the best of our knowledge, there are few results on uniform attractors for the three-dimensional micropolar equation with nonlinear damping term. The purpose of this paper is to consider the existence of uniform attractors of system (1.1). When ω=0,κ=0, system (1.1) is reduced to the Navier-Stokes equations with damping. In recent years, some scholars have studied the three-dimensional nonlinear damped Navier-Stokes equations (see [1,13,15,16,23,25]). In order to obtain the desired conclusion, we will use some proof techniques which have been used in the 3D nonlinear damped Navier Stokes equations. Note that, in [20], for the convenience of discussion the authors choose κ,μ=12,γ=1, and ν=32. In this work, we do not specify these parameters, but only require them to be positive real numbers. More importantly, we obtain the existence of uniform attractors in the case of β>3, which undoubtedly expands the range of β when the global attractor exists in [20], i.e., 3<β<5. For the convenience of discussion, similar to [3,8,9,11,16], we make some translational compactness assumption on the external forces term in this paper.

    The organizational structure of this article is as follows: In Section 2, we give some basic definitions and properties of function spaces and process theory which will be used in this paper. In Section 3, using various Sobolev inequalities and Gronwall inequalities, we make some uniform estimates from the space with low regularity to high regularity on the solution of the equation. Based on these uniform estimates, in Section 4 we prove that the family of processes {U(f1,f2)(t,τ)}tτ corresponding to (1.1) has uniform attractors A1 in V1×V2 and A2 in H2(Ω)×H2(Ω), respectively. Furthermore, we prove A1=A2.

    We define the usual functional spaces as follows:

    V1={u(C0(Ω))3:divu=0,Ωudx=0},V2={ω(C0(Ω))3:Ωωdx=0},H1=the closure of V1 in (L2(Ω))3,H2=the closure of V2 in (L2(Ω))3,V1=the closure of V1 in (H1(Ω))3,V2=the closure of V2 in (H1(Ω))3.

    For H1 and H2 we have the inner product

    (u,υ)=Ωuυdx,   u,vH1,or u,vH2,

    and norm 2=22=(,). In this paper, Lp(Ω)=(Lp(Ω))3, and p represents the norm in Lp(Ω).

    We define operators

    Au=PΔu=Δu,   Aω=Δω,  (u,ω)H2×H2,B(u)=B(u,u)=P((u)u),   B(u,ω)=(u)ω,  (u,ω)V1×V2,b(u,υ,ω)=B(u,υ),ω=3i,j=1Ωui(Diυj)ωjdx,  uV1,υ,ωV2,

    where P is the orthogonal projection from L2(Ω) onto H1. Hs(Ω)=(Hs(Ω))3 is the usual Sobolev space, and its norm is defined by Hs=∥As2; as s=2, H2=∥A.

    Let us rewrite system (1.1) as

    {ut+B(u)+(ν+κ)Au+G(u)=2κ×ω+f1(x,t),ωt+B(u,ω)+4κω+γAωμω=2κ×u+f2(x,t),u=0,u(x,t)|t=τ=uτ(x),  ω(x,t)|t=τ=ωτ(x), (2.1)

    where we let G(u)=P(σ|u|β1u).

    The Poincarˊe inequality [17] gives

    λ1uu,λ2ωω,(u,ω)V1×V2, (2.2)
    λ1u∥≤∥Au, λ2ω∥≤∥Aω,(u,ω)H2(Ω)×H2(Ω), (2.3)

    where λ1 is the first eigenvalue of Au, and λ2 is the first eigenvalue of Aω. Let λ=min. Then, we have

    \begin{align*} \lambda(\|u\|^2+\|\omega\|^2)&\leq\|\nabla u\|^2+\|\nabla\omega\|^2,\ \forall(u,\omega)\in V_{1}\times V_{2},\\ \lambda(\|\nabla u\|^2+\|\nabla \omega\|^2)&\leq \| Au\|^2+\|A\omega\|^2, \forall (u,\omega)\in \mathbf{H}^2(\Omega)\times \mathbf{H}^2(\Omega). \end{align*}

    Agmon's inequality [17] gives

    \begin{equation*} \parallel u\parallel_\infty\leq d_1\parallel\nabla u\parallel^{\frac{1}{2}}\parallel\Delta u\parallel^{\frac{1}{2}},\ \forall u\in \mathbf{H}^2(\Omega). \end{equation*}

    The trilinear inequalities [12] give

    \begin{equation} |b(u,v,w)|\leq \parallel u\parallel_\infty\parallel\nabla v\parallel\parallel w\parallel, \forall u\in L^\infty(\Omega), v\in V_1\ \text{or}\ V_2, w\in H_1\ \text{or}\ H_2, \end{equation} (2.4)
    \begin{equation} |b(u,v,w)|\leq k \parallel u\parallel^{\frac{1}{4}}\parallel\nabla u\parallel^{\frac{3}{4}}\parallel\nabla v\parallel\parallel w\parallel^{\frac{1}{4}}\parallel\nabla w\parallel^{\frac{3}{4}}, \forall u,v,w\in V_1\ \text{or}\ V_2, \end{equation} (2.5)
    \begin{equation} |b(u,v,w)|\leq k\parallel\nabla u\parallel\parallel\nabla v\parallel^{\frac{1}{2}}\parallel Av\parallel^{\frac{1}{2}}\parallel w\parallel, \forall u\in V_1\ \text{or}\ V_2, v\in \mathbf{H}^2, w\in H_1\ \text{or}\ H_2. \end{equation} (2.6)

    Recall that a function f(t) is translation bounded (tr.b.) in L_{\mathrm{loc}}^2(\mathbb{R}; \mathbf{L}^2(\Omega)) if

    \parallel f\parallel_{L_b^2}^2 = \parallel f\parallel_{L_b^2(\mathbb{R};\mathbf{L}^2(\Omega))}^2 = \sup\limits_{t\in\mathbb{R}}\int_t^{t+1}\parallel f(t)\parallel^2dt < \infty,

    where L_b^2(\mathbb{R}; \mathbf{L}^2(\Omega)) represents the collection of functions that are tr.b. in L_{\mathrm{loc}}^2(\mathbb{R}; \mathbf{L}^2(\Omega)) . We say that \mathcal{H}(f_0) = \overline{\{f_0(\cdot+t):t\in\mathbb{R}\}} is the shell of f_0 in L_{\mathrm{loc}}^2(\mathbb{R}; \mathbf{L}^2(\Omega)) . If \mathcal{H}(f_0) is compact in L_{\mathrm{loc}}^2(\mathbb{R}; \mathbf{L}^2(\Omega)) , then we say that f_0(x, t)\in L_{\mathrm{loc}}^2(\mathbb{R}; \mathbf{L}^2(\Omega)) is translation compact (tr.c.). We use L_c^2(\mathbb{R}; \mathbf{L}^2(\Omega)) to express the collection of all translation compact functions in L_{\mathrm{loc}}^2(\mathbb{R}; \mathbf{L}^2(\Omega)) .

    Next, we will provide the existence and uniqueness theorems of the solution of Eq (2.1).

    Definition 2.1. A function pair (u, \omega) is said to be a global strong solution to system (2.1) if it satisfies

    (u,\omega)\in L^\infty(\tau,T;V_1\times V_2)\cap L^2(\tau,T; \mathbf{H}^2(\Omega)\times \mathbf{H}^2(\Omega)),
    |u|^{\frac{\beta-1}{2}}\nabla u\in L^2(\tau,T;\mathbf{L}^2(\Omega)),\ \nabla |u|^{\frac{\beta+1}{2}}\in L^2(\tau,T;\mathbf{L}^2(\Omega)),

    for any given T > \tau .

    Theorem 2.1. Suppose (u_\tau, \omega_\tau)\in V_1\times V_2 with \nabla\cdot u_\tau = 0, f_1, f_2\in L_b^2(\mathbb{R}; \mathbf{L}^2(\Omega)) . If \beta = 3 and 4\sigma(\nu+\kappa) > 1 or \beta > 3 , then there exists a unique global strong solution of (2.1).

    Proof. Since the proof method is similar to that of Theorem 1.2 in [22], we omit it here.

    Let \Sigma be a metric space. X , Y are two Banach spaces, and Y\subset X is continuous. \{U_{\sigma}(t, \tau)\}_{t\geq\tau} , \sigma\in\Sigma is a family of processes in Banach space X , i.e., u(t) = U_\sigma(t, \tau)u_\tau, U_\sigma(t, s)U_\sigma(s, \tau) = U_\sigma(t, \tau), \forall t\geq s\geq\tau, \tau\in\mathbb{R}, U_\sigma(\tau, \tau) = I , where \sigma\in\Sigma is a time symbol space. \mathcal{B}(X) is the set of all bounded subsets of X . \mathbb{R}^{\tau} = [\tau, +\infty) .

    For the basic concepts of bi-space uniform absorbing set, uniform attracting set, uniform attractor, uniform compact, and uniform asymptotically compact of the family of processed \{U_\sigma(t, \tau)\}_{t\geq\tau}, \sigma\in\Sigma , one can refer to [9,16].

    Let T(h) be a family of operators acting on \Sigma , satisfying: T(h)\sigma(s) = \sigma(s+h), \forall s\in\mathbb{R} . In this paper, we assume that \Sigma satisfies

    (C1) T(h)\Sigma = \Sigma , \forall h\in\mathbb{R}^{+} ;

    (C2) translation identity:

    \begin{equation*} U_{\sigma}(t+h,\tau+h) = U_{T(h)\sigma}(t,\tau),\ \ \ \forall\sigma\in\Sigma, t\geq\tau, \tau\in\mathbb{R}, h\geq0. \end{equation*}

    Theorem 2.2. [3] If the family of processes \{U_\sigma(t, \tau)\}_{t\geq\tau}, \sigma\in\Sigma is (X, Y) -uniformly (w.r.t. \sigma\in\Sigma ) asymptotically compact, then it has a (X, Y) -uniform (w.r.t. \sigma\in\Sigma ) attractor \mathcal{A}_\Sigma , \mathcal{A}_\Sigma is compact in Y , and it attracts all bounded subsets of X in the topology of Y .

    In this paper, the letter C represents a positive constant. It may represent different values in different lines, or even in the same line.

    In this paper, we chose \mathcal{H}(f_{1}^{0})\times \mathcal{H}(f_{2}^{0}) as the symbol space. Obviously, T(t)(\mathcal{H}((f_{1}^{0})\times \mathcal{H}(f_{2}^{0})) = \mathcal{H}(f_{1}^{0})\times \mathcal{H}(f_{2}^{0}) , for all t\geq 0 . \{T(t)\}_{t\geq0} is defined by

    \begin{eqnarray*} T(h)(f_1(\cdot),f_2(\cdot)) = (f_1(\cdot+h),f_2(\cdot+h)),\ \ \ \forall h\geq 0,(f_1,f_2)\in\mathcal{H}(f_{1}^{0})\times \mathcal{H}(f_{2}^{0}), \end{eqnarray*}

    which is a translation semigroup and is continuous on \mathcal{H}(f_{1}^{0})\times \mathcal{H}(f_{2}^{0}) .

    Thanks to Theorem 2.1, when (u_\tau, \omega_\tau)\in V_1\times V_2 , f_1, f_2\in L_{\mathrm{loc}}^2(\mathbb{R}; \mathbf{L}^2(\Omega)) , and \beta > 3 , we can define a process \{U_{(f_1, f_2)}(t, \tau)\}_{t\geq\tau} in V_1\times V_2 by

    U_{(f_1,f_2)}(t,\tau)(u_\tau,\omega_\tau) = (u(t),\omega(t)),\ t\geq\tau,

    where (u(t), \omega(t)) is the solution of Eq (1.1) with external forces f_1, f_2 and initial data (u_\tau, \omega_\tau) .

    Next, let us assume that the external forces f_1^0(x, t), f_2^0(x, t) are tr.c. in L_{\mathrm{loc}}^2(\mathbb{R}; \mathbf{L}^2(\Omega)) . Then, f_1^0, f_2^0 are tr.b. in L_{\mathrm{loc}}^2(\mathbb{R}; \mathbf{L}^2(\Omega)) , and

    \parallel f_1\parallel_{L_b^2}^2 = \parallel f_1\parallel_{L_b^2(\mathbb{R};\mathbf{L}^2(\Omega))}^2 = \sup\limits_{t\in\mathbb{R}}\int_t^{t+1}\parallel f_1(s)\parallel^2ds\leq\parallel f_1^0\parallel_{L_b^2}^2 < +\infty, \forall f_1\in \mathcal{H}(f_1^0),
    \parallel f_2\parallel_{L_b^2}^2 = \parallel f_2\parallel_{L_b^2(\mathbb{R};\mathbf{L}^2(\Omega))}^2 = \sup\limits_{t\in\mathbb{R}}\int_t^{t+1}\parallel f_2(s)\parallel^2ds\leq\parallel f_2^0\parallel_{L_b^2}^2 < +\infty, \forall f_2\in \mathcal{H}(f_2^0).

    Furthermore, we assume f_1^0, f_2^0 are uniformly bounded in \mathbf{L}^2(\Omega) , i.e., there exists a positive constant K , which satisfies

    \sup\limits_{t\in\mathbb{R}}\parallel f_1^0(x,t)\parallel\leq K,\ \sup\limits_{t\in\mathbb{R}}\parallel f_2^0(x,t)\parallel\leq K.

    Meanwhile, we suppose the derivatives \frac{\mathrm{d}f_1^0}{\mathrm{d}t}, \ \frac{\mathrm{d}f_2^0}{\mathrm{d}t} , labeled as h_1, h_2 , also belong to L_{c}^2(\mathbb{R}; \mathbf{L}^2(\Omega)) .

    Lemma 3.1. Suppose (u_{\tau}, \omega_{\tau})\in V_{1}\times V_{2} and (f_{1}, f_{2})\in \mathcal{H}(f_{1}^{0})\times \mathcal{H}(f_{2}^{0}) . If \beta > 3 then there exists a time t_{0} and constants \rho_1, I_1 such that, for any t\geq t_{0} ,

    \begin{equation} \|u(t)\|^{2}+\|\omega(t)\|^{2}\leq \rho_{1}, \end{equation} (3.1)
    \begin{equation} \int^{t+1}_{t}[\|\nabla u(s)\|^{2}+\|\nabla \omega(s)\|^{2}+\|u(s)\|^{\beta+1}_{\beta+1}+\|\nabla\cdot\omega(s)\|^{2}]ds\leq I_{1} . \end{equation} (3.2)

    Proof. Multiplying (1.1)_{1} and (1.1)_{2} with external forces f_{1}\in\mathcal{H}(f^{0}_{1}) , f_{2}\in\mathcal{H}(f^{0}_{2}) by u and \omega , respectively, and integrating the results equations on \Omega , using H \ddot{o} lder's inequality, Young's inequality, and Poincar \acute{e} 's inequality, it yields

    \begin{eqnarray} &&\quad\frac{1}{2}\frac{d}{dt}(\|u(t)\|^{2}+\|\omega(t)\|^{2})+(\nu+\kappa)\|\nabla u\|^{2}+\gamma\|\nabla \omega\|^{2}+4\kappa\|\omega(t)\|^{2}+\sigma\|u(t)\|^{\beta+1}_{\beta+1}+ \mu\|\nabla\cdot\omega\|^{2}\\ && = 4\kappa\int_{\Omega}\nabla\times u\cdot\omega dx+(f_{1},u(t))+(f_{2},\omega(t))\\ &&\leq \kappa\parallel\nabla u\parallel^2+4\kappa\parallel \omega\parallel^2+\frac{\nu\lambda}{2}\parallel u\parallel^2+\frac{\gamma\lambda}{2}\parallel\omega\parallel^2+\frac{1}{2\nu\lambda}\parallel f_1\parallel^2+\frac{1}{2\gamma\lambda}\parallel f_2\parallel^2\\ &&\leq(\frac{\nu}{2}+\kappa)\|\nabla u\|^{2}+\frac{\gamma}{2}\|\nabla\omega\|^{2}+4\kappa\|\omega(t)\|^{2}+\frac{1}{2\nu\lambda}\parallel f_1\parallel^2+\frac{1}{2\gamma\lambda}\parallel f_2\parallel^2. \end{eqnarray} (3.3)

    So, we can obtain that

    \begin{eqnarray} \frac{d}{dt}(\|u(t)\|^{2}+\|\omega(t)\|^{2})+\nu\|\nabla u\|^{2}+\gamma\|\nabla \omega\|^{2}+2\sigma\|u(t)\|^{\beta+1}_{\beta+1}+2\mu\|\nabla\cdot\omega\|^{2}\leq\frac{1}{\nu\lambda}\|f_{1}(t)\|^{2}+\frac{1}{\gamma\lambda}\|f_{2}(t)\|^{2}, \end{eqnarray} (3.4)

    and by Poincar \acute{e} 's inequality, it yields

    \begin{equation} \frac{d}{dt}(\|u(t)\|^{2}+\|\omega(t)\|^{2})+\lambda\alpha(\|u(t)\|^{2}+\|\omega(t)\|^{2})\leq\frac{1}{\lambda\alpha}(\|f_{1}(t)\|^{2}+\|f_{2}(t)\|^{2}), \end{equation} (3.5)

    where \alpha = \min\{\nu, \gamma\} . So, by Gronwall's inequality, we get

    \begin{align*} \|u(t)\|^{2}+\|\omega(t)\|^{2}&\leq(\|u_{\tau}\|^{2}+\|\omega_{\tau}\|^{2})e^{-\lambda\alpha(t-\tau)}+\frac{1}{\lambda\alpha}\int^{t}_{\tau}e^{-\lambda\alpha(t-s)}(\|f_{1}(s)\|^{2}+\|f_{2}(s)\|^{2})ds\nonumber\\ &\leq(\|u_{\tau}\|^{2}+\|\omega_{\tau}\|^{2})e^{-\lambda\alpha(t-\tau)}+\frac{1}{\lambda\alpha}[\int^{t}_{t-1}e^{-\lambda\alpha(t-s)}(\|f_{1}(s)\|^{2}+\|f_{2}(s)\|^{2})ds\nonumber\\ &\quad+\int^{t-1}_{t-2}e^{-\lambda\alpha(t-s)}(\|f_{1}(s)\|^{2}+\|f_{2}(s)\|^{2})ds+...]\nonumber\\ &\leq(\|u_{\tau}\|^{2}+\|\omega_{\tau}\|^{2})e^{-\lambda\alpha(t-\tau)}+\frac{1}{\lambda\alpha}[1+e^{-\lambda\alpha}+e^{-2\lambda\alpha}+...](\|f_{1}\|^{2}_{L^{2}_{b}}+\|f_{2}\|^{2}_{L^{2}_{b}})\nonumber\\ &\leq(\|u_{\tau}\|^{2}+\|\omega_{\tau}\|^{2})e^{-\lambda\alpha(t-\tau)}+\frac{1}{\lambda\alpha}(1-e^{-\lambda\alpha})^{-1}(\|f_{1}\|^{2}_{L^{2}_{b}}+\|f_{2}\|^{2}_{L^{2}_{b}})\nonumber\\ &\leq(\|u_{\tau}\|^{2}+\|\omega_{\tau}\|^{2})e^{-\lambda\alpha(t-\tau)}+\frac{1}{\lambda\alpha}(1+\frac{1}{\lambda\alpha})(\|f_{1}\|^{2}_{L^{2}_{b}}+\|f_{2}\|^{2}_{L^{2}_{b}}),\ \ \ \forall t\geq\tau. \end{align*}

    Therefore, there must exists a time t_{0}\geq\tau+\frac{1}{\lambda\alpha}\ln\frac{\lambda^{2}\alpha^{2}(\|u_{\tau}\|^{2}+\|\omega_{\tau}\|^{2})}{(1+\lambda\alpha)(\|f_{1}\|^{2}_{L^{2}_{b}}+\|f_{2}\|^{2}_{L^{2}_{b}})}, such that, \forall t\geq t_{0} ,

    \begin{equation} \|u(t)\|^{2}+\|\omega(t)\|^{2}\leq\frac{2}{\lambda\alpha}(1+\frac{1}{\lambda\alpha})(\|f_{1}\|^{2}_{L^{2}_{b}}+\|f_{2}\|^{2}_{L^{2}_{b}})\equiv\rho_{1}. \end{equation} (3.6)

    Taking t\geq t_{0} , integrating (3.4) from t to t+1 , and noticing (3.6), we get

    \begin{eqnarray} &&\quad\int^{t+1}_{t}[\nu\|\nabla u(s)\|^{2}+\gamma\|\nabla \omega(s)\|^{2}+2\sigma\|u(s)\|^{\beta+1}_{\beta+1}+2\mu\|\nabla\cdot\omega(s)\|^{2}]ds\\ &&\leq(\|u(t)\|^{2}+\|\omega(t)\|^{2})+\frac{1}{\nu\lambda}\int^{t+1}_{t}\|f_{1}(s)\|^{2}ds+\frac{1}{\gamma\lambda}\int_t^{t+1}\|f_{2}(s)\|^{2}ds\\ &&\leq\rho_{1}+\frac{1}{\lambda\alpha}(\|f_{1}\|^{2}_{L^{2}_{b}}+\|f_{2}\|^{2}_{L^{2}_{b}}),\ \ \ \forall t\geq t_{0}. \end{eqnarray} (3.7)

    Letting \delta_{1} = \mathrm{min}\{\nu, \gamma, 2\sigma, 2\mu\} , we have

    \begin{eqnarray*} \delta_{1}\int^{t+1}_{t}[\|\nabla u(s)\|^{2}+\|\nabla \omega(s)\|^{2}+\|u(s)\|^{\beta+1}_{\beta+1}+\|\nabla\cdot\omega(s)\|^{2}]ds\leq\rho_{1}+\frac{1}{\lambda\alpha}(\|f_{1}\|^{2}_{L^{2}_{b}}+\|f_{2}\|^{2}_{L^{2}_{b}}),\ \ \ \forall t\geq t_{0}. \end{eqnarray*}

    Letting I_{1} = \frac{1}{\delta_{1}}(\rho_{1}+\frac{1}{\lambda\alpha}(\|f_{1}\|^{2}_{L^{2}_{b}}+\|f_{2}\|^{2}_{L^{2}_{b}})) , we have

    \begin{eqnarray*} \int^{t+1}_{t}[\|\nabla u(s)\|^{2}+\|\nabla \omega(s)\|^{2}+\|u(s)\|^{\beta+1}_{\beta+1}+\|\nabla\cdot\omega(s)\|^{2}]ds\leq I_{1},\ \ \ \forall t\geq t_{0}. \end{eqnarray*}

    This completes the proof of Lemma 3.1.

    Lemma 3.2. Assume \beta > 3 , (u_{\tau}, \omega_{\tau})\in V_{1}\times V_{2} and (f_{1}, f_{2})\in \mathcal{H}(f_{1}^{0})\times \mathcal{H}(f_{2}^{0}) . Then, there exists a time t_{2} and a constant \rho_2 such that

    \begin{eqnarray} \|\nabla u(t)\|^{2}+\|\nabla \omega(t)\|^{2} +\int^{t+1}_{t}(\|Au(s)\|^{2}+\|A\omega(s)\|^{2}+\||u|^{\frac{\beta-1}{2}}\nabla u\|^{2}+\|\nabla |u|^{\frac{\beta+1}{2}}\|^{2})ds\leq \rho_2, \end{eqnarray} (3.8)

    for any t\geq t_{2} .

    Proof. Taking the inner product of -\Delta u in H_1 with the first equation of (1.1), we obtain

    \begin{eqnarray} &&\quad\frac{1}{2}\frac{d}{dt}\|\nabla u\|^{2}+(\nu+\kappa)\|Au\|^{2}+\sigma\||u|^{\frac{\beta-1}{2}}\nabla u\|^{2}+\frac{4\sigma(\beta-1)}{(\beta+1)^{2}}\|\nabla|u|^{\frac{\beta+1}{2}}\|^{2}\\ && = -b(u,u,Au)+2\kappa\int_{\Omega}\nabla\times\omega\cdot Audx+(f_{1}(t),Au). \end{eqnarray} (3.9)

    In [18], we find that, when \beta > 3 ,

    \begin{equation} \int_\Omega (u\cdot\nabla u)\cdot\Delta udx\leq\frac{\nu+\kappa}{4}\parallel \Delta u\parallel^2+\frac{\sigma}{2}\parallel |u|^{\frac{\beta-1}{2}}\nabla u\parallel^2+C_1\parallel\nabla u\parallel^2, \end{equation} (3.10)

    where C_1 = \frac{N^2}{\nu+\kappa}+\frac{N^2}{(\nu+\kappa)(N^{\beta-1}+1)} , and N is sufficiently large such that

    \begin{equation*} N\geq(\frac{2}{\beta-3})^{\frac{1}{\beta-1}}\ \text{and}\ \ \frac{N^2}{(\nu+\kappa)(N^{\beta-1}+1)}\leq\frac{\sigma}{2}. \end{equation*}

    And, because

    \begin{equation} |2\kappa\int_\Omega \nabla\times \omega\cdot Audx|\leq\frac{\nu+\kappa}{4}\parallel\Delta u\parallel^2+\frac{4\kappa^2}{\nu+\kappa}\parallel\nabla \omega\parallel^2, \end{equation} (3.11)
    \begin{equation} |(f_1(t),Au)|\leq\frac{\nu+\kappa}{4}\parallel\Delta u\parallel^2+\frac{\parallel f_1(t)\parallel^2}{\nu+\kappa}, \end{equation} (3.12)

    so combining (3.10)–(3.12) with (3.9), we have

    \begin{align} &\quad\frac{d}{dt}\|\nabla u\|^{2}+\frac{\nu+\kappa}{2}\|Au\|^{2}+\sigma\||u|^{\frac{\beta-1}{2}}\nabla u\|^{2}+\frac{8\sigma(\beta-1)}{(\beta+1)^{2}}\|\nabla|u|^{\frac{\beta+1}{2}}\|^{2}\\ &\leq 2C_1\|\nabla u\|^{2}+\frac{8\kappa^2}{\nu+\kappa}\parallel\nabla \omega\parallel^2+\frac{2\parallel f_1(t)\parallel^2}{\nu+\kappa}\\ &\leq C_2(\|\nabla u\|^{2}+\|\nabla\omega\|^{2}+\|f_{1}(t)\|^{2}), \end{align} (3.13)

    where C_2 = \max\{2C_1, \frac{8\kappa^2}{\nu+\kappa}, \frac{2}{\nu+\kappa}\} .

    Applying uniform Gronwall's inequality to (3.13), we obtaint, \forall t\geq t_{0}+1\equiv t_{1} ,

    \begin{eqnarray} \|\nabla u(t)\|^{2}+\int^{t+1}_{t}\Big(\frac{\nu+\kappa}{2}\|Au(s)\|^{2}+\sigma\||u(s)|^{\frac{\beta-1}{2}}\nabla u(s)\|^{2}+\frac{8\sigma(\beta-1)}{(\beta+1)^{2}}\|\nabla|u(s)|^{\frac{\beta+1}{2}}\|^{2}\Big)ds\leq C_3, \end{eqnarray} (3.14)

    where C_3 is a positive constant dependent on C_2 , I_1 , and \parallel f_1^0\parallel_{L_b^2}^2 .

    Taking the inner product of -\Delta \omega in H_2 with the second equation of (1.1), we get

    \begin{eqnarray} &&\quad\frac{1}{2}\frac{d}{dt}\|\nabla \omega\|^{2}+4\kappa\|\nabla\omega\|^{2}+\gamma\|A\omega\|^{2}+\mu\|\nabla\nabla\cdot\omega\|^{2}\\ && = -b(u,\omega,A\omega)+2\kappa\int_{\Omega}\nabla\times u\cdot A\omega dx+(f_{2}(t),A\omega)\\ &&\leq\frac{3\gamma}{4}\|A\omega\|^{2}+\frac{d_1^2}{\gamma}\|\nabla u\|\|Au\|\|\nabla \omega\|^{2}+\frac{4\kappa^2}{\gamma}\| \nabla u\|^{2}+\frac{1}{\gamma}\|f_{2}(t)\|^{2}. \end{eqnarray} (3.15)

    In the last inequality of (3.15), we used Agmon's inequality and the trilinear inequality. Then,

    \begin{eqnarray} \frac{d}{dt}\|\nabla\omega\|^{2}+\frac{\gamma}{2}\|A\omega\|^{2}+2\mu\|\nabla\nabla\cdot\omega\|^{2}\leq C_4(\|\nabla u\|\|Au\|\|\nabla \omega\|^{2}+\|\nabla u\|^{2}+\|f_{2}(t)\|^{2}), \end{eqnarray} (3.16)

    where C_4 = \max\{\frac{2d_1^2}{\gamma}, \frac{8\kappa^2}{\gamma}, \frac{2}{\gamma}\} .

    By the uniform Gronwall's inequality, we easily obtain that, for t\geq t_{1}+1\equiv t_{2} ,

    \begin{eqnarray} \|\nabla\omega(t)\|^{2}+\int^{t+1}_{t}(\frac{\gamma}{2}\|A\omega(s)\|^{2}+2\mu\|\nabla\nabla\cdot\omega(s)\|^{2})ds\leq C_5,\ \text{for}\ t\geq t_{1}+1\equiv t_{2}, \end{eqnarray} (3.17)

    where C_5 is a positive constant dependent on C_3, C_4 , and \parallel f_2^0\parallel_{L_b^2}^2 .

    Adding (3.14) with (3.17) yields

    \begin{align*} \label{3.14} \|\nabla u(s)\|^{2}+\|\nabla\omega(s)\|^{2}+\int^{t+1}_{t}(\|Au(s)\|^{2}+\|A\omega(s)\|^{2}+\||u(s)|^{\frac{\beta-1}{2}}\nabla u(s)\|^{2}+\|\nabla |u(s)|^{\frac{\beta+1}{2}}\|^{2})ds\leq C, \end{align*}

    for t\geq t_2 . Hence, Lemma 3.2 is proved.

    Lemma 3.3. Suppose that (u_{\tau}, \omega_{\tau})\in V_{1}\times V_{2} and (f_{1}, f_{2})\in \mathcal{H}(f_{1}^{0})\times \mathcal{H}(f_{2}^{0}) . Then, for \beta > 3 , there exists a time t_{3} and a constant \rho_3 such that

    \begin{eqnarray} \|u(t)\|_{\beta+1}+\|\nabla\cdot\omega(t)\|^{2}\leq \rho_{3}, \end{eqnarray} (3.18)

    for any t\geq t_{3} .

    Proof. Multiplying (1.1)_{1} by u_{t} , then integrating the equation over \Omega , we have

    \begin{eqnarray} &&\quad\|u_{t}\|^{2}+\frac{\nu+\kappa}{2}\frac{d}{dt}\|\nabla u\|^{2}+\frac{\sigma}{\beta+1}\frac{d}{dt}\|u(t)\|^{\beta+1}_{\beta+1}\\ && = -b(u,u,u_{t})+2\kappa\int_{\Omega}\nabla\times\omega\cdot u_{t}dx+(f_{1}(t),u_{t})\\ &&\leq\frac{1}{2}\|u_{t}\|^{2}+\frac{3d_1^2}{2\sqrt{\lambda_1}}\parallel\nabla u\parallel^2\parallel Au\parallel^2+6\kappa^2\|\nabla \omega\|^{2}+\frac{3}{2}\|f_{1}(t)\|^{2}. \end{eqnarray} (3.19)

    The trilinear inequality (2.4), Agmon's inequality, and Poincar \acute{e} 's inequality are used in the last inequality of (3.19).

    Hence,

    \begin{eqnarray} (\nu+\kappa)\frac{d}{dt}\|\nabla u\|^{2}+\frac{2\sigma}{\beta+1}\frac{d}{dt}\|u(t)\|^{\beta+1}_{\beta+1}\leq C_6(\parallel\nabla u\parallel^2\parallel Au\parallel^2+\|\nabla \omega\|^{2}+\|f_{1}(t)\|^{2}), \end{eqnarray} (3.20)

    where C_6 = \max\{\frac{3d_1^2}{\sqrt{\lambda_1}}, 12\kappa^2, 3\} .

    By (3.20), using Lemmas 3.1 and 3.2 and the uniform Gronwall's inequality, we have

    \begin{equation} \|u(t)\|_{\beta+1}\leq C,\ \ \forall t\geq t_{2}+1\equiv t_{3}. \end{equation} (3.21)

    Similar to (3.19), multiplying (1.1)_{2} by \omega_{t} and integrating it over \Omega , we get

    \begin{align} &\quad\|\omega_{t}\|^{2}+2\kappa\frac{d}{dt}\|\omega\|^{2}+\frac{\gamma}{2}\frac{d}{dt}\|\nabla \omega\|^{2}+\frac{\mu}{2}\frac{d}{dt}\|\nabla\cdot\omega\|^{2} = -b(u,\omega,\omega_{t})+2\kappa\int_{\Omega}\nabla\times u\cdot\omega_{t}dx+(f_{2}(t),\omega_{t})\\ &\leq\frac{1}{2}\|\omega_{t}\|^{2}+\frac{3d_1^2}{2\sqrt{\lambda_1}}\parallel Au\parallel^{2}\parallel\nabla\omega\parallel^2+6\kappa^2\|\nabla u\|^{2}+\frac{3}{2}\|f_{2}(t)\|^{2}. \end{align} (3.22)

    Hence,

    \begin{align} 4\kappa\frac{d}{dt}\parallel \omega\parallel^2+\gamma\frac{d}{dt}\parallel \nabla\omega\parallel^2+\mu\frac{d}{dt}\parallel\nabla\cdot\omega\parallel^2\leq C_6(\parallel Au\parallel^2\parallel\nabla\omega\parallel^2+\parallel\nabla u\parallel^2+\parallel f_2(t)\parallel^2). \end{align} (3.23)

    By (3.23), using Lemma 3.2 and the uniform Gronwall's inequality, we infer that

    \begin{eqnarray} \|\nabla\cdot\omega(t)\|^{2}\leq C,\ \ \forall t\geq t_{3}. \end{eqnarray} (3.24)

    The proof of Lemma 3.3 is finished.

    Lemma 3.4. Suppose (u_{\tau}, \omega_{\tau})\in V_{1}\times V_{2} and (f_{1}, f_{2})\in \mathcal{H}(f_{1}^{0})\times \mathcal{H}(f_{2}^{0}) . If \beta > 3 , then there exists a time t_{4} and a constant \rho_5 , such that

    \begin{eqnarray} \|u_{t}(s)\|^{2}+\|\omega_{t}(s)\|^{2}\leq \rho_5, \end{eqnarray} (3.25)

    for any s\geq t_{4} .

    Proof. Taking the inner products of u_{t} and \omega_{t} with the first and second equations of (1.1), respectively, and using (3.19) and (3.22), we find

    \begin{align} &\quad\|u_{t}\|^{2}+\|\omega_{t}\|^{2}+\frac{\nu+\kappa}{2}\frac{d}{dt}\|\nabla u\|^{2}+\frac{\gamma}{2}\frac{d}{dt}\|\nabla\omega\|^{2}+2\kappa\frac{d}{dt}\|\omega(t)\|^{2}+\frac{\sigma}{\beta+1}\frac{d}{dt}\|u(t)\|^{\beta+1}_{\beta+1}+\frac{\mu}{2}\frac{d}{dt}\|\nabla\cdot\omega\|^{2}\\ & = -b(u,u,u_{t})-b(u,\omega,\omega_{t})+2\kappa\int_{\Omega}\nabla\times\omega\cdot u_{t}dx+2\kappa\int_{\Omega}\nabla\times u\cdot\omega_{t}dx+(f_{1}(t),u_{t})+(f_{2}(t),\omega_{t})\\ &\leq \frac{1}{2}(\|u_{t}\|^{2}+\|\omega_{t}\|^{2})+C_7(\|f_{1}(t)\|^{2}+\|f_{2}(t)\|^{2}+\|\nabla u\|^{2}\\&\quad+\|\nabla\omega\|^{2}+\parallel\nabla u\parallel^2\parallel Au\parallel^2+\parallel\nabla\omega\parallel^2\parallel Au\parallel^2), \end{align} (3.26)

    where C_7 = \max\{\frac{3d_1^2}{2\sqrt{\lambda_1}}, 6\kappa^2, \frac{3}{2}\} . The trilinear inequality (2.4), Agmon's inequality, and Poincar \acute{e} 's inequality are used in the last inequality of (3.26).

    Integrating (3.26) over [t, t+1] and using Lemmas 3.1–3.3, we get

    \begin{eqnarray} \int^{t+1}_{t}(\|u_{t}(s)\|^{2}+\|\omega_{t}(s)\|^{2})ds\leq \rho_4,\ \forall t\geq t_3, \end{eqnarray} (3.27)

    where \rho_4 is a positive constant dependent on C_7, \rho_2, \rho_3 , \parallel f_1^0\parallel_{L_b^2}^2 , and \parallel f_2^0\parallel_{L_b^2}^2 .

    We now differentiate (2.1)_1 with respect to t , then take the inner product of u_t with the resulting equation to obtain

    \begin{align} &\quad\frac{1}{2}\frac{d}{dt}\parallel u_t\parallel^2+(\nu+\kappa)\parallel\nabla u_t\parallel^2\\ & = -b(u_t,u,u_t)-\int_\Omega G'(u)u_t\cdot u_tdx+2\kappa\int_\Omega\nabla\times \omega_t\cdot u_tdx+(f_{1t},u_t). \end{align} (3.28)

    Then, we differentiate (2.1)_2 with respect to t and take the inner product with \omega_t to obtain

    \begin{align} &\quad\frac{1}{2}\frac{d}{dt}\parallel\omega_t\parallel^2+4\kappa\parallel \omega_t\parallel^2+\gamma\parallel\nabla\omega_t\parallel^2+\mu\parallel\nabla\cdot\omega_t\parallel^2\\ & = -b(u_t,\omega,\omega_t)+2\kappa\int_\Omega \nabla\times u_t\cdot\omega_tdx+(f_{2t},\omega_t). \end{align} (3.29)

    Adding (3.28) with (3.29), we have

    \begin{eqnarray} &&\quad\frac{1}{2}\frac{d}{dt}(\|u_{t}\|^{2}+\|\omega_{t}\|^{2})+(\nu+\kappa)\|\nabla u_{t}\|^{2}+\gamma\|\nabla\omega_{t}\|^{2}+4\kappa\|\omega_{t}\|^{2}+\mu\|\nabla\cdot\omega_{t}\|^{2}\\ &&\leq|b(u_{t},u,u_{t})|+|b(u_{t},\omega,\omega_{t})|+2\kappa\int_{\Omega}\nabla\times\omega_{t}\cdot u_{t}dx+2\kappa\int_{\Omega}\nabla\times u_{t}\cdot\omega_{t}dx\\ &&\quad+(f_{1t},u_{t})+(f_{2t},\omega_{t})-\int_{\Omega}G^{'}(u)u_{t}\cdot u_{t}dx\\ &&: = \sum\limits_{i = 1}^{7} L_{i}. \end{eqnarray} (3.30)

    From Lemma 2.4 in [15], we know that G'(u) is positive definite, so

    \begin{equation} L_7 = -\int_\Omega G'(u)u_t\cdot u_tdx\leq 0. \end{equation} (3.31)

    For L_{1} , using the trilinear inequality (2.5) and Lemma 3.2, we have

    \begin{align} L_{1}&\leq k\|u_{t}\|^{\frac{1}{2}}\|\nabla u_{t}\|^{\frac{3}{2}}\|\nabla u\|\\ &\leq \frac{\nu+\kappa}{4}\|\nabla u_{t}\|^{2}+C\|u_{t}\|^{2}\|\nabla u\|^{4}\\ &\leq\frac{\nu+\kappa}{4}\|\nabla u_{t}\|^{2}+C\|u_{t}\|^{2},\ \text{for}\ t\geq t_2. \end{align} (3.32)

    For L_2 , by H \ddot{o} lder's inequality, Gagliardo-Nirenberg's inequality, and Young's inequality, we have

    \begin{align} L_{2}&\leq C\|u_{t}\|_{4}\|\omega_t\|_{4}\|\nabla\omega\|\\ &\leq C\|u_{t}\|^{\frac{1}{4}}\|\nabla u_{t}\|^{\frac{3}{4}}\|\omega_{t}\|^{\frac{1}{4}}\|\nabla\omega_{t}\|^{\frac{3}{4}}\|\nabla\omega\|\\ &\leq\frac{\nu+\kappa}{4}\|\nabla u_{t}\|^{2}+\frac{\gamma}{4}\|\nabla\omega_{t}\|^{2}+C(\|u_{t}\|^{2}+\|\omega_{t}\|^{2}),\ \text{for}\ t\geq t_2. \end{align} (3.33)
    \begin{align} L_{3}+L_{4}&\leq\frac{\nu+\kappa}{4}\|\nabla u_{t}\|^{2}+\frac{\gamma}{2}\|\nabla \omega_{t}\|^{2}+C(\|u_{t}\|^{2}+\|\omega_{t}\|^{2}). \end{align} (3.34)

    By (3.30)–(3.34), we get

    \begin{align} \frac{d}{dt}(\|u_{t}\|^{2}+\|\omega_{t}\|^{2})&\leq C(\|u_{t}\|^{2}+\|\omega_{t}\|^{2})+(f_{1t},u_{t})+(f_{2t},\omega_{t})\\ &\leq C(\|u_{t}\|^{2}+\|\omega_{t}\|^{2})+\|f_{1t}\|^{2}+\|f_{2t}\|^{2}. \end{align} (3.35)

    Thanks to

    \int_t^{t+1}\parallel f_{1t}(s)\parallel^2ds\leq\parallel f_{1t}\parallel_{L_b^2}^2\leq\parallel h_1\parallel_{L_b^2}^2, \int_t^{t+1}\parallel f_{2t}(s)\parallel^2ds\leq\parallel f_{2t}\parallel_{L_b^2}^2\leq\parallel h_2\parallel_{L_b^2}^2,

    and applying uniform Gronwall's inequality to (3.35), we have for any s\geq t_{3}+1\equiv t_{4} ,

    \begin{eqnarray} \|u_{t}(s)\|^{2}+\|\omega_{t}(s)\|^{2}\leq C. \end{eqnarray} (3.36)

    Thus, Lemma 3.4 is proved.

    Lemma 3.5. Suppose (u_{\tau}, \omega_{\tau})\in V_{1}\times V_{2} and (f_{1}, f_{2})\in \mathcal{H}(f_{1}^{0})\times \mathcal{H}(f_{2}^{0}) . Then, for \beta > 3 , there exists a constant \rho_6 such that

    \begin{eqnarray} \|Au(t)\|^{2}+\|A\omega(t)\|^{2}\leq \rho_6, \end{eqnarray} (3.37)

    for any t\geq t_{4} .

    Proof. Taking the inner product of -\Delta u in H_1 with the first equation of (1.1), we have

    \begin{align} &\quad(\nu+\kappa)\parallel Au\parallel^2+\sigma\parallel|u|^{\frac{\beta-1}{2}}\nabla u\parallel^2+\frac{4\sigma(\beta-1)}{(\beta+1)^2}\parallel\nabla|u|^{\frac{\beta+1}{2}}\parallel^2\\ & = -(u_t,Au)-(B(u),Au)+2\kappa\int_\Omega \nabla\times\omega\cdot Audx+(f_1(t),Au)\\ &\leq\frac{4(\nu+\kappa)}{6}\parallel Au\parallel^2+\frac{3}{2(\nu+\kappa)}\parallel u_t\parallel^2+\frac{3}{2(\nu+\kappa)}\parallel B(u)\parallel^2\\ &\quad+\frac{6\kappa^2}{\nu+\kappa}\parallel\nabla\omega\parallel^2+\frac{3}{2(\nu+\kappa)}\parallel f_1(t)\parallel^2. \end{align} (3.38)

    Because

    \begin{align} \frac{3}{2(\nu+\kappa)}\parallel B(u)\parallel^2&\leq\frac{3}{2(\nu+\kappa)}\parallel u\parallel_\infty^2\parallel\nabla u\parallel^2\\ &\leq \frac{3d_1^2}{2(\nu+\kappa)}\parallel\nabla u\parallel^3\parallel\Delta u\parallel\\ &\leq\frac{\nu+\kappa}{6}\parallel Au\parallel^2+C\parallel\nabla u\parallel^6, \end{align} (3.39)

    combining (3.39) with (3.38), we obtain

    \begin{equation} \frac{\nu+\kappa}{6}\parallel Au\parallel^2\leq \frac{3}{2(\nu+\kappa)}\parallel u_t\parallel^2+C\parallel\nabla u\parallel^6+\frac{6\kappa^2}{\nu+\kappa}\parallel\nabla\omega\parallel^2+\frac{3}{2(\nu+\kappa)}\parallel f_1(t)\parallel^2. \end{equation} (3.40)

    From the assumption of f_1^0(t) , we can easily get

    \begin{equation} \sup\limits_{t\in\mathbb{R}}\parallel f_1(t)\parallel\leq\sup\limits_{t\in\mathbb{R}}\parallel f_{1}^0(t)\parallel\leq K, \forall f_1\in\mathcal{H}(f_1^0). \end{equation} (3.41)

    By Lemmas 3.2 and 3.4, we obtain

    \begin{eqnarray} \|Au(t)\|\leq C,\ \text{for any}\ t\geq t_4. \end{eqnarray} (3.42)

    Taking the inner product of A\omega with (2.1)_{2} , we get

    \begin{align} &\quad\gamma\|A\omega\|^{2}+4\kappa\parallel\nabla \omega\parallel^2+\mu\parallel\nabla\nabla\cdot\omega\parallel^2\\ & = -(\omega_t,A\omega)-(B(u,\omega),A\omega)+2\kappa(\nabla\times u,A\omega)+(f_2(t),A\omega)\\ &\leq\frac{\gamma}{2}\parallel A\omega\parallel^2+C(\parallel \omega_t\parallel^2+\parallel B(u,\omega)\parallel^2+\parallel\nabla u\parallel^2+\parallel f_2(t)\parallel^2). \end{align} (3.43)

    And, by Agmon's inequality,

    \begin{align} \|B(u,\omega)\|^2&\leq C\|u\|_{\infty}^2\|\nabla \omega\|^2\\ &\leq C\|\nabla u\|\|\Delta u\|\|\nabla\omega\|^2\\ &\leq\|Au\|^2+C\parallel\nabla u\parallel^2\parallel\nabla\omega\parallel^4. \end{align} (3.44)

    From the assumption on f_2^0(t) , we can easily obtain

    \begin{equation} \sup\limits_{t\in\mathbb{R}}\parallel f_2(t)\parallel\leq\sup\limits_{t\in\mathbb{R}}\parallel f_{2}^0(t)\parallel\leq K, \forall f_2\in\mathcal{H}(f_2^0). \end{equation} (3.45)

    By Lemma 3.2, Lemma 3.4, (3.42), (3.43), (3.44), and (3.45), we get

    \begin{eqnarray} \|A\omega(t)\|\leq C,\ \text{for any }t\geq t_{4}. \end{eqnarray} (3.46)

    By (3.42) and (3.46), Lemma 3.5 is proved for all t\geq t_{4} .

    Lemma 3.6. Suppose (u_{\tau}, \omega_{\tau})\in V_{1}\times V_{2} and (f_{1}, f_{2})\in \mathcal{H}(f_{1}^{0})\times \mathcal{H}(f_{2}^{0}) . Then, for \beta > 3 , there exists a time t_5 and a constant \rho_{7} satisfying

    \begin{eqnarray} \|\nabla u_{t}(t)\|^{2}+\|\nabla\omega_{t}(t)\|^{2}\leq \rho_7, \forall t\geq t_{5}. \end{eqnarray} (3.47)

    Proof. In the proof of Lemma 3.4, from (3.30)–(3.34) we can also get

    \begin{align} &\quad\frac{d}{dt}(\parallel u_t\parallel^2+\parallel\omega_t\parallel^2)+\frac{\nu+\kappa}{2}\parallel\nabla u_t\parallel^2+\frac{\gamma}{2}\parallel\nabla\omega_t\parallel^2+2\mu\parallel \nabla\cdot\omega_t\parallel^2\\ &\leq C(\parallel u_t\parallel^2+\parallel\omega_t\parallel^2)+\parallel f_1(t)\parallel^2+\parallel f_2(t)\parallel^2. \end{align} (3.48)

    Integrating (3.48) from t to t+1 , and according to Lemma 3.4, we have

    \begin{align} &\quad\int^{t+1}_{t}(\|\nabla u_{t}(s)\|^{2}+\|\nabla\omega_{t}(s)\|^{2}+\|\nabla\cdot\omega_{t}(s)\|^{2})ds\\ &\leq C(\|u_{t}(t)\|^{2}+\|\omega_{t}(t)\|^{2}+\int^{t+1}_{t}(\|u_{t}(s)\|^{2}+\|\omega_{t}(s)\|^{2})ds+\int^{t+1}_{t}\|f_{1t}(s)\|^{2}ds+\int^{t+1}_{t}\|f_{2t}(s)\|^{2}ds)\\ &\leq C+\|h_{1}\|^{2}_{L^{2}_{b}}+\|h_{2}\|^{2}_{L^{2}_{b}}\\ &\leq C,\ \forall t\geq t_4. \end{align} (3.49)

    By Lemma 3.5, we get

    \begin{eqnarray} \|u(t)\|_{H^{2}}+\|\omega(t)\|_{H^{2}}\leq C, \forall t\geq t_4. \end{eqnarray} (3.50)

    So, by Lemma 3.2, applying Agmon's inequality, we get

    \begin{eqnarray} \|u(t)\|_{\infty}+\|\omega(t)\|_{\infty}\leq C, \forall t\geq t_4. \end{eqnarray} (3.51)

    Taking the derivative of (2.1)_{1} and (2.1)_{2} with respect to t , then multiplying by Au_{t} and A\omega_{t} , respectively, and integrating the resulting equations over \Omega , we then have

    \begin{align} &\quad\frac{1}{2}\frac{d}{dt}(\|\nabla u_{t}\|^{2}+\|\nabla\omega_{t}\|^{2})+(\nu+\kappa)\|Au_{t}\|^{2}+\gamma\|A\omega_{t}\|^{2}+4\kappa\|\nabla\omega_{t}\|^{2}+\mu\parallel\nabla\nabla\cdot\omega_t\parallel^2\\ &\leq|b(u_t,u,Au_t)|+|b(u,u_{t},Au_{t})|+|b(u,\omega_{t},A\omega_{t})|+|b(u_{t},\omega,A\omega_{t})|\\ &\ \ \ \ +2\kappa\int_{\Omega}|\nabla\times\omega_{t}\cdot Au_{t}|dx+2\kappa\int_{\Omega}|\nabla\times u_{t}\cdot A\omega_{t}|dx+|\int_{\Omega}G'(u)u_{t}\cdot Au_{t}dx|\\ &\ \ \ \ +(f_{1t},Au_{t})+(f_{2t},A\omega_{t})\\ &: = \sum^{9}_{i = 1}J_{i}. \end{align} (3.52)

    For J_{1} , J_2 , using (2.6) and Lemmas 3.2 and 3.5, we have

    \begin{align} J_{1}&\leq k\|\nabla u_{t}\|\|\nabla u\|^{\frac{1}{2}}\|Au\|^{\frac{1}{2}}\|Au_{t}\|\\ &\leq\frac{\nu+\kappa}{5}\|Au_{t}\|^{2}+C\|\nabla u_{t}\|^{2},\ \forall t\geq t_4, \end{align} (3.53)

    and

    \begin{align} J_{2}&\leq k\|\nabla u\|\|\nabla u_{t}\|^{\frac{1}{2}}\|A u_{t}\|^{\frac{1}{2}}\|Au_{t}\|\\ &\leq k\|\nabla u\|\|\nabla u_{t}\|^{\frac{1}{2}}\|Au_{t}\|^{\frac{3}{2}}\\ &\leq \frac{\nu+\kappa}{5}\|Au_{t}\|^{2}+C\|\nabla u_{t}\|^{2},\ \forall t\geq t_4. \end{align} (3.54)

    For J_{3} and J_{4} , similar to (3.53) and (3.54), we get

    \begin{align} J_{3}&\leq k\|\nabla u\|\|\nabla\omega_t\|^{\frac{1}{2}}\|A\omega_{t}\|^{\frac{1}{2}}\|A\omega_{t}\|\\ &\leq \frac{\gamma}{4}\|A\omega_{t}\|^{2}+C\|\nabla \omega_{t}\|^{2},\ \forall t\geq t_4, \end{align} (3.55)
    \begin{align} J_{4}&\leq k\|\nabla u_{t}\|\|\nabla\omega\|^{\frac{1}{2}}\|A\omega\|^{\frac{1}{2}}\|A\omega_{t}\|\\ &\leq\frac{\gamma}{4}\|A\omega_{t}\|^{2}+C\|\nabla u_{t}\|^{2},\ \forall t\geq t_4. \end{align} (3.56)

    For J_{5} , J_6 , and J_{7} , applying Hölder's inequality and Young's inequality, we have

    \begin{align} J_{5}+J_{6}\leq\frac{\nu+\kappa}{5}\|Au_{t}\|^{2}+\frac{\gamma}{4}\|A\omega_{t}\|^{2}+C(\|\nabla u_{t}\|^{2}+\|\nabla\omega_{t}\|^{2}), \end{align} (3.57)

    and thanks to (3.51),

    \begin{align} J_{7}&\leq C\|u\| ^{\beta-1}_{\infty}\|u_{t}\|\|Au_{t}\|\\ &\leq\frac{\nu+\kappa}{5}\|Au_{t}\|^{2}+C\|u_{t}\|^{2},\ \forall t\geq t_4. \end{align} (3.58)

    For J_8 and J_9 , we have

    \begin{align} J_{8}&\leq\frac{\nu+\kappa}{5}\|Au_{t}\|^{2}+C\|f_{1t}\|^{2}, \end{align} (3.59)
    \begin{align} J_{9}&\leq\frac{\gamma}{4}\|A\omega_{t}\|^{2}+C\|f_{2t}\|^{2}. \end{align} (3.60)

    By (3.52)–(3.60), we obtain

    \begin{eqnarray} \frac{d}{dt}(\|\nabla u_{t}\|^{2}+\|\nabla\omega_{t}\|^{2})\leq C(\|\nabla u_{t}\|^{2}+\|\nabla\omega_{t}\|^{2})+C\|u_{t}\|^{2}+C(\|f_{1t}\|^{2}+\|f_{2t}\|^{2}). \end{eqnarray} (3.61)

    Then, by (3.27), (3.49), and using the uniform Gronwall's lemma, we get

    \begin{eqnarray} \|\nabla u_{t}(s)\|^{2}+\|\nabla\omega_{t}(s)\|^{2}\leq C,\ \forall s\geq t_{4}+1\equiv t_{5}. \end{eqnarray} (3.62)

    Thus, Lemma 3.6 is proved.

    In this section, we consider the existence of the (V_1\times V_2, V_1\times V_2) -uniform (w.r.t. (f_1, f_2)\in \mathcal{H}(f_1^0)\times\mathcal{H}(f_2^0) ) attractor and the (V_1\times V_2, \mathbf{H}^2(\Omega)\times \mathbf{H}^2(\Omega)) -uniform attractor for \{U_{(f_1, f_2)}(t, \tau)\}_{t\geq\tau}, f_1\times f_2\in \mathcal{H}(f^{0}_{1})\times \mathcal{H}(f^{0}_{2}) .

    Lemma 4.1. Suppose \beta > 3 . The family of processes \{U_{(f_{1}, f_{2})}(t, \tau)\}_{t\geq\tau} , f_{1}\times f_{2}\in\mathcal{H}(f^{0}_{1})\times \mathcal{H}(f^{0}_{2}) , corresponding to (2.1) is ((V_{1}\times V_{2})\times(\mathcal{H}(f^{0}_{1})\times \mathcal{H}(f^{0}_{2})), V_{1}\times V_{2}) -continuous for \tau\geq t_5.

    Proof. Let \tau_n\subset [\tau, +\infty) be a time sequence, U_{(f_{1}^{(n)}, f_{2}^{(n)})}(t, \tau)(u_{\tau_n}, \omega_{\tau_n}) = (u^{(n)}(t), \omega^{(n)}(t)) , U_{(f_{1}, f_{2})}(t, \tau)(u_{\tau}, \omega_{\tau}) = (u(t), \omega(t)) and

    \begin{align*} (\bar{u}^{(n)}(t),\bar{\omega}^{(n)}(t))& = (u(t)-u^{(n)}(t),\omega(t)-\omega^{(n)}(t))\\ & = U_{(f_{1},f_{2})}(t,\tau)(u_{\tau},\omega_{\tau})-U_{(f_{1}^{(n)},f_{2}^{(n)})}(t,\tau)(u_{\tau_n},\omega_{\tau_n}). \end{align*}

    It is evident that \bar{u}^{(n)}(t) is the solution of

    \begin{align} \frac{\partial\bar{u}^{(n)}(t)}{\partial t}+B(u)-B(u^{(n)}(t))+(\nu+\kappa)A\bar{u}^{(n)}+G(u)-G(u^{(n)}) = 2\kappa\nabla\times\bar{\omega}^{(n)}+(f_{1}-f_{1}^{(n)}), \end{align} (4.1)

    and \bar{\omega}^{(n)}(t) is the solution of the following system

    \begin{align} \frac{\partial\bar{\omega}^{(n)}(t)}{\partial t}&+B(u,\omega)-B(u^{(n)},\omega^{(n)})+4\kappa\bar{\omega}^{(n)}+\gamma A\bar{\omega}^{(n)}-\mu\nabla\nabla\cdot\bar{\omega}^{(n)} = 2\kappa\nabla\times\bar{u}^{(n)}+(f_{2}-f_{2}^{(n)}), \end{align} (4.2)

    for each n .

    Taking the inner product of (4.1) with A\bar{u}^{(n)} in H_1 , we get

    \begin{align} &\quad\frac{1}{2}\frac{d}{dt}\|\nabla\bar{u}^{(n)}\|^{2}+b(u,u,A\bar{u}^{(n)})-b(u^{(n)},u^{(n)},A\bar{u}^{(n)}) +(\nu+\kappa)\parallel A\bar{u}^{(n)}\parallel^2+(G(u)-G(u^{(n)}),A\bar{u}^{(n)})\\& = 2\kappa(\nabla\times\bar{\omega}^{(n)},A\bar{u}^{(n)})+(f_{1}-f_{1}^{(n)},A\bar{u}^{(n)}). \end{align} (4.3)

    Taking the inner product of (4.2) with A\bar{\omega}^{(n)} in H_2 , we have

    \begin{align} &\quad\frac{1}{2}\frac{d}{dt}\|\nabla\bar{\omega}^{(n)}\|^{2}+b(u,\omega,A\bar{\omega}^{(n)})-b(u^{(n)},\omega^{(n)},A\bar{\omega}^{(n)}) +4\kappa\|\nabla\bar{\omega}^{(n)}\|^2+\gamma\parallel A\bar{\omega}^{(n)}\parallel^2+\mu\parallel\nabla\nabla\cdot\bar{\omega}^{(n)}\parallel^2\\ & = 2\kappa(\nabla\times\bar{u}^{(n)},A\bar{\omega}^{(n)})+(f_{2}-f_{2}^{(n)},A\bar{\omega}^{(n)}). \end{align} (4.4)

    Combining (4.3) with (4.4), we get

    \begin{align} &\quad\frac{1}{2}\frac{d}{dt}(\|\nabla\bar{u}^{(n)}\|^{2}+\|\nabla\bar{\omega}^{(n)}\|^{2})+b(u,u,A\bar{u}^{(n)})-b(u^{(n)},u^{(n)},A\bar{u}^{(n)})+(\nu+\kappa)\|A\bar{u}^{(n)}\|^{2}\\ &\quad+(G(u)-G(u^{(n)}),A\bar{u}^{(n)})+b(u,\omega,A\bar{\omega}^{(n)})-b(u^{(n)},\omega^{(n)},A\bar{\omega}^{(n)})\\ &\quad+4\kappa\parallel\nabla\bar{\omega}^{(n)}\parallel^2+\gamma\parallel A\bar{\omega}^{(n)}\parallel^2+\mu\parallel\nabla\nabla\cdot\bar{\omega}^{(n)}\parallel^2\\ & = 2\kappa(\nabla\times\bar{\omega}^{(n)},A\bar{u}^{(n)})+2\kappa(\nabla\times\bar{u}^{(n)},A\bar{\omega}^{(n)})+(f_{1}-f_{1}^{(n)},A\bar{u}^{(n)}) +(f_{2}-f_{2}^{(n)},A\bar{\omega}^{(n)}). \end{align} (4.5)

    Due to

    \begin{align} b(u,u,A\bar{u}^{(n)})-b(u^{(n)},u^{(n)},A\bar{u}^{(n)})& = b(\bar{u}^{(n)},u,A\bar{u}^{(n)})+b({u}^{(n)},\bar{u}^{(n)},A\bar{u}^{(n)}), \end{align} (4.6)
    \begin{align} b(u,\omega,A\bar{\omega}^{(n)})-b(u^{(n)},\omega^{(n)},A\bar{\omega}^{(n)})& = b(\bar{u}^{(n)},\omega,A\bar{\omega}^{(n)})+b(u^{(n)},\bar{\omega}^{(n)},A\bar{\omega}^{(n)}), \end{align} (4.7)

    and

    \begin{align} |b(\bar{u}^{(n)},u,A\bar{u}^{(n)})|&\leq k\|\nabla\bar{u}^{(n)}\|\|\nabla u\|^{\frac{1}{2}}\|Au\|^{\frac{1}{2}}\|A\bar{u}^{(n)}\|\\&\leq\frac{\nu+k}{5}\|A\bar{u}^{(n)}\|^{2}+C\|\nabla\bar{u}^{(n)}\|^{2}\|\nabla u\|\|Au\|, \end{align} (4.8)
    \begin{align} |b({u}^{(n)},\bar{u}^{(n)},A\bar{u}^{(n)})|&\leq k\|\nabla{u}^{(n)}\|\|\nabla\bar{u}^{(n)}\|^{\frac{1}{2}}\|A\bar{u}^{(n)}\|^{\frac{1}{2}}\|A\bar{u}^{(n)}\|\\ &\leq\frac{\nu+k}{5}\|A\bar{u}^{(n)}\|^{2}+C\|\nabla{u}^{(n)}\|^{4}\|\nabla\bar{u}^{(n)}\|^{2}, \end{align} (4.9)
    \begin{align} b(\bar{u}^{(n)},\omega,A\bar{\omega}^{(n)})&\leq k\|\nabla\bar{u}^{(n)}\|\|\nabla\omega\|^{\frac{1}{2}}\|A\omega\parallel^{\frac{1}{2}}\|A\bar{\omega}^{(n)}\|\\ &\leq\frac{\gamma}{4}\|A\bar{\omega}^{(n)}\|^{2}+C\|\nabla\bar{u}^{(n)}\|^{2}\|\nabla\omega\|\|A\omega\|, \end{align} (4.10)
    \begin{align} b(u^{(n)},\bar{\omega}^{(n)},A\bar{\omega}^{(n)})&\leq k\|\nabla u^{(n)}\|\|\nabla\bar{\omega}^{(n)}\|^{\frac{1}{2}}\|A\bar{\omega}^{(n)}\|^{\frac{1}{2}}\|A\bar{\omega}^{(n)}\|\\ &\leq\frac{\gamma}{4}\|A\bar{\omega}^{(n)}\|^{2}+C\parallel\nabla u^{(n)}\parallel^4\parallel\nabla\bar{\omega}^{(n)}\parallel^2, \end{align} (4.11)
    \begin{align} 2\kappa|(\nabla\times\bar{\omega}^{(n)},A\bar{u}^{(n)})|&\leq 2\kappa\|A\bar{u}^{(n)}\|\|\nabla\bar{\omega}^{(n)}\|\\ &\leq\frac{\nu+k}{5}\|A\bar{u}^{(n)}\|^{2}+C\|\nabla\bar{\omega}^{(n)}\|^{2}, \end{align} (4.12)
    \begin{align} 2\kappa|(\nabla\times\bar{u}^{(n)},A\bar{\omega}^{(n)})\|&\leq 2\kappa\parallel A\bar{\omega}^{(n)}\|\|\nabla\bar{u}^{(n)}\|\\ &\leq\frac{\gamma}{4}\|A\bar{\omega}^{(n)}\|^{2}+C\|\nabla\bar{u}^{(n)}\|^{2}, \end{align} (4.13)
    \begin{align} |(f_{1}-f_{1}^{(n)},A\bar{u}^{(n)})|&\leq\frac{\nu+k}{5}\|A\bar{u}^{(n)}\|^{2}+\frac{5}{4(\nu+\kappa)}\|f_{1}-f_{1}^{(n)}\|^{2}, \end{align} (4.14)
    \begin{align} |(f_{2}-f_{2}^{(n)},A\bar{\omega}^{(n)})|&\leq\frac{\gamma}{4}\|A\bar{\omega}^{(n)}\|^{2}+\frac{1}{\gamma}\|f_{2}-f_{2}^{(n)}\|^{2}, \end{align} (4.15)
    \begin{align} \|G(u)-G(u^{(n)})\|^{2}& = \int_\Omega \big|\sigma|u|^{\beta-1}u-\sigma|u^{(n)}|^{\beta-1}u^{(n)}\big|^2dx\\ &\leq C\int_\Omega [|u|^{\beta-1}|\bar{u}^{(n)}|+\big||u|^{\beta-1}-|u^{(n)}|^{\beta-1}\big|\cdot|u^{(n)}|]^2dx\\ &\leq C\int_\Omega |u|^{2(\beta-1)}|\bar{u}^{(n)}|^2dx+C\int_\Omega[|u|^{\beta-2}+|u^{(n)}|^{\beta-2}]^2|u^{(n)}|^2|\bar{u}^{(n)}|^2dx\\ &\leq C[\parallel u\parallel_\infty^{2(\beta-1)}+(\parallel u\parallel_\infty^{2(\beta-2)}+\parallel u^{(n)}\parallel_{\infty}^{2(\beta-2)})\parallel u^{(n)}\parallel_\infty^2]\parallel\nabla \bar{u}^{(n)}\parallel^2, \end{align} (4.16)

    where \bar{u}^{(n)}(t) = u(t)-u^{n}(t) . In the above inequality, we used the fact that

    |x^p-y^p|\leq cp(|x|^{p-1}+|y|^{p-1})|x-y|

    for any x, y\geq 0 , where c is an absolute constant.

    Therefore,

    \begin{align} (G(u)-G(u^{(n)}),A\bar{u}^{(n)})&\leq\frac{\nu+\kappa}{5}\|A\bar{u}^{(n)}\|^{2}+\frac{5}{4(\nu+\kappa)}\|G(u)-G(u^{(n)})\|^{2}\\ &\leq C[\parallel u\parallel_\infty^{2(\beta-1)}+(\parallel u\parallel_\infty^{2(\beta-2)}+\parallel u^{(n)}\parallel_{\infty}^{2(\beta-2)})\parallel u^{(n)}\parallel_\infty^2]\parallel\nabla \bar{u}^{(n)}\parallel^2\\ &\quad+\frac{\nu+k}{5}\parallel A\bar{u}^{(n)}\parallel^2. \end{align} (4.17)

    By (4.5)–(4.15) and (4.17), we obtain

    \begin{align} \frac{d}{dt}(\|\nabla\bar{u}^{(n)}\|^{2}+\|\nabla\bar{\omega}^{(n)}\|^{2})&\leq C[\parallel u\parallel_\infty^{2(\beta-1)}+(\parallel u\parallel_\infty^{2(\beta-2)}+\parallel u^{(n)}\parallel_{\infty}^{2(\beta-2)})\parallel u^{(n)}\parallel_\infty^2\\ &\quad+\|\nabla u\|\|Au\|+\|\nabla u^{(n)}\|^{4}+\|\nabla\omega\|\|A\omega\|+1]\\ &\quad \cdot(\|\nabla\bar{u}^{(n)}\|^{2}+\|\nabla\bar{\omega}^{(n)}\|^{2})+\frac{5}{2(\nu+\kappa)}\|f_{1}-f_{1}^{(n)}\|^{2}\\ &\quad +\frac{2}{\gamma}\|f_{2}-f_{2}^{(n)}\|^{2}. \end{align} (4.18)

    Using Gronwall's inequality in (4.18) yields

    \begin{align} \|\nabla\bar{u}^{(n)}\|^{2}+\|\nabla\bar{\omega}^{(n)}\|^{2} &\leq\Big(\|\nabla\bar{u}_{\tau}^{(n)}\|^{2}+\|\nabla\bar{\omega}_{\tau}^{(n)}\|^{2}+\frac{5}{2(\nu+\kappa)}\int^{t}_{\tau}\|f_{1}-f_{1}^{(n)}\|^{2}ds\\ &\ \ \ \ +\frac{2}{\gamma}\int_\tau^t \|f_{2}-f_{2}^{(n)}\|^{2}ds\Big)\\ &\ \ \ \ \cdot\exp\Big\{C\int^{t}_{\tau}[\parallel u\parallel_\infty^{2(\beta-1)}+(\parallel u\parallel_\infty^{2(\beta-2)}+\parallel u^{(n)}\parallel_{\infty}^{2(\beta-2)})\parallel u^{(n)}\parallel_\infty^2\\ &\ \ \ \ +\|\nabla u\|\|Au\|+\|\nabla u^{(n)}\|^{4}+\|\nabla\omega\|\|A\omega\|+1]ds\Big\}. \end{align} (4.19)

    From Lemmas 3.2 and 3.5, and using Agmon's inequality, we know that

    \parallel u\parallel_\infty < C, \parallel u^{(n)}\parallel_\infty < C, \forall t\geq t_5.

    So, from Lemmas 3.2–3.5, we have

    \begin{align*} &\exp\Big\{C\int^{t}_{\tau}[\parallel u\parallel_\infty^{2(\beta-1)}+(\parallel u\parallel_\infty^{2(\beta-2)}+\parallel u^{(n)}\parallel_{\infty}^{2(\beta-2)})\parallel u^{(n)}\parallel_\infty^2+\|\nabla u\|\|Au\|\nonumber\\ &+\|\nabla u^{(n)}\|^{4}+\|\nabla\omega\|\|A\omega\|+1]ds\Big\} < +\infty, \end{align*}

    for any given t and \tau , t\geq\tau , \tau\geq t_5 .

    Thus, from (4.19), we have that \{U_{(f_{1}, f_{2})}(t, \tau)\}_{t\geq\tau} , f_{1}\times f_{2}\in\mathcal{H}(f^{0}_{1})\times \mathcal{H}(f^{0}_{2}) is ((V_{1}\times V_{2})\times(\mathcal{H}(f^{0}_{1})\times \mathcal{H}(f^{0}_{2})), V_{1}\times V_{2}) -continuous, for \tau\geq t_5 .

    By Lemma 3.5, the fact of compact imbedding \mathbf{H}^2\times \mathbf{H}^2\hookrightarrow V_{1}\times V_{2} , and Theorem 3.1 in [16], we have the following theorems.

    Theorem 4.1. Suppose \beta > 3 . The family of processes \{U_{(f_{1}, f_{2})}(t, \tau)\}_{t\geq\tau} , f_{1}\times f_{2}\in\mathcal{H}(f^{0}_{1})\times \mathcal{H}(f^{0}_{2}) with respect to problem (1.1) has a (V_1\times V_2, V_1\times V_2) uniform attractor \mathcal{A}_{1} . Moreover,

    \begin{eqnarray} \mathcal{A}_{1} = \bigcup\limits_{(f_{1},f_{2})\in \mathcal{H}(f_1^0)\times\mathcal{H}(f_2^0)}\mathcal{K}_{(f_1,f_2)}(0), \end{eqnarray} (4.20)

    where \mathcal{K}_{(f_{1}, f_{2})}(0) is the section at t = 0 of kernel \mathcal{K}_{(f_{1}, f_{2})} of the processes \{U_{(f_1, f_2)}(t, \tau)\}_{t\geq\tau} .

    Theorem 4.2. Suppose \beta > 3 . The family of processes \{U_{(f_{1}, f_{2})}(t, \tau)\}_{t\geq\tau} , f_{1}\times f_{2}\in\mathcal{H}(f^{0}_{1})\times \mathcal{H}(f^{0}_{2}) with respect to problem (1.1) has a (V_{1}\times V_{2}, \mathbf{H}^{2}(\Omega)\times \mathbf{H}^{2}(\Omega)) -uniform attractor \mathcal{A}_{2} . \mathcal{A}_{2} is compact in \mathbf{H}^{2}(\Omega)\times \mathbf{H}^{2}(\Omega) , and it attracts every bounded subset of V_{1}\times V_{2} in the topology of \mathbf{H}^{2}(\Omega)\times \mathbf{H}^{2}(\Omega) .

    Proof. By Theorem 2.2, we only need to prove that \{U_{(f_{1}, f_{2})}(t, \tau)\}_{t\geq\tau} , f_{1}\times f_{2}\in\mathcal{H}(f^{0}_{1})\times \mathcal{H}(f^{0}_{2}) acting on V_{1}\times V_{2} is (V_{1}\times V_{2}, \mathbf{H}^{2}(\Omega)\times \mathbf{H}^{2}(\Omega)) -uniform (w.r.t.\ \ f_{1}\times f_{2}\in \mathcal{H}(f^{0}_{1})\times \mathcal{H}(f^{0}_{2})) asymptotically compact.

    Thanks to Lemma 3.5, we know that B = \{(u\times\omega)\in \mathbf{H}^2\times \mathbf{H}^2: \|Au\|^{2}+\|A\omega\|^{2}\leq C\} is a bounded (V_{1}\times V_{2}, \mathbf{H}^{2}(\Omega)\times \mathbf{H}^{2}(\Omega)) -uniform absorbing set of \{U_{(f_{1}, f_{2})}(t, \tau)\}_{t\geq\tau} . Then, we just need to prove that, for any \tau_{n}\in\mathbb{R} , any t\rightarrow +\infty , and (u_{\tau_{n}}, \omega_{\tau_n})\in B , \{(u_{n}(t), \omega_{n}(t))\}_{n = 0}^{\infty} is precompact in \textbf{H}^{2}(\Omega)\times\textbf{H}^{2}(\Omega) , where (u_{n}(t), \omega_{n}(t)) = U_{(f_{1}, f_{2})}(t, \tau_{n})(u_{\tau_{n}}, \omega_{\tau_{n}}) .

    Because V_1\hookrightarrow H_1, V_2\hookrightarrow H_2 are compact, from Lemma 3.6 we obtain that \{\frac{d}{dt}u_{n}(t)\}_{n = 0}^{\infty} , \{\frac{d}{dt}\omega_{n}(t)\}_{n = 0}^{\infty} are precompact in H_1 and H_2 , respectively.

    Next, we will prove \{u_{n}(t)\}_{n = 0}^{\infty} , \{\omega_{n}(t)\}_{n = 0}^{\infty} are Cauchy sequences in \mathbf{H}^{2}(\Omega) . From (2.1), we have

    \begin{align} &(\nu+\kappa)(Au_{n_k}(t)-Au_{n_j}(t)) = -\frac{d}{dt}u_{n_k}(t)+\frac{d}{dt}u_{n_j}(t)-B(u_{n_k}(t))+B(u_{n_j}(t))\\ &\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad-G(u_{n_k}(t))+G(u_{n_j}(t))+2\kappa\nabla\times\omega_{n_k}(t)-2\kappa\nabla\times\omega_{n_j}(t). \end{align} (4.21)
    \begin{align} &\gamma(A\omega_{n_k}(t)-A\omega_{n_j}(t))-\mu\nabla\nabla\cdot\omega_{n_k}(t)+\mu\nabla\nabla\cdot\omega_{n_j}(t) = -\frac{d}{dt}\omega_{n_k}(t)+\frac{d}{dt}\omega_{n_j}(t)-B(u_{n_k}(t),\omega_{n_k}(t))\\&\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad+B(u_{n_j}(t),\omega_{n_j}(t))-4\kappa\omega_{n_k}(t)\\ &\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad+4\kappa\omega_{n_j}(t)+2\kappa\nabla\times u_{n_k}(t)-2\kappa\nabla\times u_{n_j}(t). \end{align} (4.22)

    Multiplying (4.21) by Au_{n_k}(t)-Au_{n_j}(t) , we obtain

    \begin{align*} (\nu+\kappa)\parallel Au_{n_k}(t)-Au_{n_j}(t)\parallel^2&\leq\parallel\frac{d}{dt}u_{n_k}(t)-\frac{d}{dt}u_{n_j}(t)\parallel\cdot\parallel Au_{n_k}(t)-Au_{n_j}(t)\parallel+\parallel B(u_{n_k}(t))-B(u_{n_j}(t))\parallel\nonumber\\ & \cdot\parallel Au_{n_k}(t)-Au_{n_j}(t)\parallel+\parallel G(u_{n_k}(t))-G(u_{n_j}(t))\parallel\cdot\parallel Au_{n_k}(t)-Au_{n_j}(t)\parallel\nonumber\\ &\ \ \ +2\kappa\parallel\nabla\omega_{n_k}(t)-\nabla\omega_{n_j}(t)\parallel\cdot\parallel Au_{n_k}(t)-Au_{n_j}(t)\parallel\nonumber\\ &\leq\frac{4(\nu+\kappa)}{5}\parallel Au_{n_k}(t)-Au_{n_j}(t)\parallel^2+\frac{5}{4(\nu+\kappa)}\parallel\frac{d}{dt}u_{n_k}(t)-\frac{d}{dt}u_{n_j}(t)\parallel^2\nonumber\\ &\ \ \ +\frac{5}{4(\nu+\kappa)}\parallel B(u_{n_k}(t))-B(u_{n_j}(t))\parallel^2+\frac{5}{4(\nu+\kappa)}\parallel G(u_{n_k}(t))-G(u_{n_j}(t))\parallel^2\nonumber\\ &\ \ \ +\frac{5\kappa^2}{\nu+\kappa}\parallel\nabla\omega_{n_k}(t)-\nabla\omega_{n_j}(t)\parallel^2, \end{align*}

    so we have

    \begin{align} \frac{\nu+\kappa}{5}\parallel Au_{n_k}(t)-Au_{n_j}(t)\parallel^2 &\leq\frac{5}{4(\nu+\kappa)}\parallel\frac{d}{dt}u_{n_k}(t)-\frac{d}{dt}u_{n_j}(t)\parallel^2\\&\ \ \ +\frac{5}{4(\nu+\kappa)}\parallel B(u_{n_k}(t))-B(u_{n_j}(t))\parallel^2\\ &\ \ \ +\frac{5}{4(\nu+\kappa)}\parallel G(u_{n_k}(t))-G(u_{n_j}(t))\parallel^2\\&\ \ \ +\frac{5\kappa^2}{\nu+\kappa}\parallel\nabla\omega_{n_k}(t)-\nabla\omega_{n_j}(t)\parallel^2. \end{align} (4.23)

    Multiplying (4.22) by A\omega_{n_k}(t)-A\omega_{n_j}(t) we obtain

    \begin{align*} &\quad\gamma\parallel A\omega_{n_k}(t)-A\omega_{n_j}(t)\parallel^2+\mu\parallel\nabla\nabla\cdot(\omega_{n_k}(t)-\omega_{n_j}(t))\parallel^2\nonumber\\ &\leq \parallel\frac{d}{dt}\omega_{n_k}(t)-\frac{d}{dt}\omega_{n_j}(t)\parallel\cdot\parallel A\omega_{n_k}(t)-A\omega_{n_j}(t)\parallel+\parallel B(u_{n_k}(t),\omega_{n_k}(t))-B(u_{n_j}(t),\omega_{n_j}(t))\parallel\nonumber\\ &\ \ \ \cdot\parallel A\omega_{n_k}(t)-A\omega_{n_j}(t)\parallel+4\kappa\parallel\omega_{n_k}(t)-\omega_{n_j}(t)\parallel\cdot\parallel A\omega_{n_k}(t)-A\omega_{n_j}(t)\parallel\nonumber\\ &\ \ \ +2\kappa\parallel\nabla u_{n_k}(t)-\nabla u_{n_j}(t)\parallel\cdot\parallel A\omega_{n_k}(t)-A\omega_{n_j}(t)\parallel\nonumber\\ &\leq\frac{4\gamma}{5}\parallel A\omega_{n_k}(t)-A\omega_{n_j}(t)\parallel^2+\frac{5}{4\gamma}\parallel\frac{d}{dt}\omega_{n_k}(t)-\frac{d}{dt}\omega_{n_j}(t)\parallel^2\nonumber\\ &\ \ \ +\frac{5}{4\gamma}\parallel B(u_{n_k}(t),\omega_{n_k}(t))-B(u_{n_j}(t),\omega_{n_j}(t))\parallel^2+\frac{20\kappa^2}{\gamma}\parallel\omega_{n_k}(t)-\omega_{n_j}(t))\parallel^2\nonumber\\ &\ \ \ +\frac{5\kappa^2}{\gamma}\parallel\nabla u_{n_k}(t)-\nabla u_{n_j}(t)\parallel^2, \end{align*}

    so we get

    \begin{align} &\ \ \ \frac{\gamma}{5}\parallel A\omega_{n_k}(t)-A\omega_{n_j}(t)\parallel^2+\mu\parallel\nabla\nabla\cdot(\omega_{n_k}(t)-\omega_{n_j}(t))\parallel^2\\ &\leq\frac{5}{4\gamma}\parallel\frac{d}{dt}\omega_{n_k}(t)-\frac{d}{dt}\omega_{n_j}(t)\parallel^2+\frac{5}{4\gamma}\parallel B(u_{n_k}(t),\omega_{n_k}(t))-B(u_{n_j}(t),\omega_{n_j}(t))\parallel^2\\ &\ \ \ +\frac{20\kappa^2}{\gamma}\parallel\omega_{n_k}(t)-\omega_{n_j}(t)\parallel^2+\frac{5\kappa^2}{\gamma}\parallel\nabla u_{n_k}(t)-\nabla u_{n_j}(t)\parallel^2. \end{align} (4.24)

    Combining (4.23) with (4.24), we have

    \begin{align} &\ \ \ \frac{\nu+\kappa}{5}\parallel Au_{n_k}(t)-Au_{n_j}(t)\parallel^2+\frac{\gamma}{5}\parallel A\omega_{n_k}(t)-A\omega_{n_j}(t)\parallel^2\\ &\leq \frac{5}{4(\nu+\kappa)}\parallel\frac{d}{dt}u_{n_k}(t)-\frac{d}{dt}u_{n_j}(t)\parallel^2+\frac{5}{4(\nu+\kappa)}\parallel B(u_{n_k}(t))-B(u_{n_j}(t))\parallel^2\\ &\ \ \ +\frac{5}{4(\nu+\kappa)}\parallel G(u_{n_k}(t))-G(u_{n_j}(t))\parallel^2+\frac{5\kappa^2}{\nu+\kappa}\parallel\nabla\omega_{n_k}(t)-\nabla\omega_{n_j}(t)\parallel^2\\ &\ \ \ +\frac{5}{4\gamma}\parallel\frac{d}{dt}\omega_{n_k}(t)-\frac{d}{dt}\omega_{n_j}(t)\parallel^2+\frac{5}{4\gamma}\parallel B(u_{n_k}(t),\omega_{n_k}(t))-B(u_{n_j}(t),\omega_{n_j}(t))\parallel^2\\ &\ \ \ +\frac{20\kappa^2}{\gamma}\parallel\omega_{n_k}(t)-\omega_{n_j}(t)\parallel^2+\frac{5\kappa^2}{\gamma}\parallel\nabla u_{n_k}(t)-\nabla u_{n_j}(t)\parallel^2. \end{align} (4.25)

    Because V_2\hookrightarrow H_2 is compact, from Lemma 3.2 we know that \{\omega_n(t)\}_{n = 0}^\infty is precompact in H_2 . And, using the compactness of embedding \mathbf{H}^2(\Omega)\hookrightarrow V_1, \mathbf{H}^2(\Omega)\hookrightarrow V_2 and Lemma 3.5, we have that \{u_{n}(t)\}_{n = 0}^\infty, \{\omega_n(t)\}_{n = 0}^\infty are precompact in V_1 and V_2 , respectively. Considering V_1\hookrightarrow H_1, V_2\hookrightarrow H_2 are compact, from Lemma 3.6 we know that \{\frac{d}{dt}u_n(t)\}_{n = 0}^\infty , \{\frac{d}{dt}\omega_n(t)\}_{n = 0}^\infty are precompact in H_1 and H_2 , respectively.

    Using (2.6), we have

    \begin{align} &\ \ \ \parallel B(u_{n_k}(t))-B(u_{n_j}(t))\parallel^2\\ &\leq C(\parallel B(u_{n_k}(t),u_{n_k}(t)-u_{n_j}(t))\parallel^2+\parallel B(u_{n_k}(t)-u_{n_j}(t),u_{n_j}(t))\parallel^2)\\ &\leq C(\parallel\nabla u_{n_k}(t)\parallel^2\parallel\nabla(u_{n_k}(t)-u_{n_j}(t))\parallel\parallel A(u_{n_k}(t)-u_{n_j}(t))\parallel\\ &\ \ \ \ +\parallel\nabla(u_{n_k}(t)-u_{n_j}(t))\parallel^2\parallel\nabla u_{n_j}(t)\parallel\parallel Au_{n_j}(t)\parallel)\rightarrow 0, \text {as}\ k,j\rightarrow +\infty, \end{align} (4.26)

    and

    \begin{align} &\ \ \ \ \parallel B(u_{n_k}(t),\omega_{n_k}(t))-B(u_{n_j}(t),\omega_{n_j}(t))\parallel^2\\ &\leq C(\parallel B(u_{n_k}(t),\omega_{n_k}(t)-\omega_{n_j}(t))\parallel^2+\parallel B(u_{n_k}(t)-u_{n_j}(t),\omega_{n_j}(t))\parallel^2)\\ &\leq C(\parallel\nabla u_{n_k}(t)\parallel^2\parallel\nabla(\omega_{n_k}(t)-\omega_{n_j}(t))\parallel\parallel A(\omega_{n_k(t)}-\omega_{n_j}(t))\parallel\\ &\ \ \ \ +\parallel\nabla(u_{n_k}(t)-u_{n_j}(t))\parallel^2\parallel\nabla\omega_{n_j}(t)\parallel\parallel A\omega_{n_j}(t)\parallel)\rightarrow 0,\ \text{as}\ k,j\rightarrow +\infty. \end{align} (4.27)

    From the proof of Lemma 4.2 in [15], we have

    \begin{equation} \parallel G(u_{n_k}(t))-G(u_{n_j}(t))\parallel^2\leq C\parallel u_{n_k}(t)-u_{n_j}(t)\parallel^2\rightarrow 0,\ \text{as}\ k,j\rightarrow +\infty. \end{equation} (4.28)

    Taking into account (4.25)–(4.28), we have

    \begin{equation} \frac{\nu+\kappa}{5}\parallel Au_{n_k}(t)-Au_{n_j}(t)\parallel^2+\frac{\gamma}{5}\parallel A\omega_{n_k}(t)-A\omega_{n_j}(t)\parallel^2\rightarrow 0,\ \text{as}\ k,j\rightarrow +\infty. \end{equation} (4.29)

    (4.29) indicates that the processes \{U_{(f_1, f_2)}(t, \tau)\}_{t\geq\tau} are uniformly asymptotically compact in \mathbf{H}^2(\Omega)\times\mathbf{H}^2(\Omega) . So, by Theorem 2.2, it has a (V_1\times V_2, \mathbf{H}^2(\Omega)\times \mathbf{H}^2(\Omega)) -uniform attractor \mathcal{A}_2 .

    Theorem 4.3. Suppose \beta > 3 . The (V_{1}\times V_{2}, V_{1}\times V_{2}) -uniform attractor \mathcal{A}_{1} of the family of processes \{U_{(f_{1}, f_{2})}(t, \tau)\}_{t\geq\tau} , f_{1}\times f_{2}\in\mathcal{H}(f^{0}_{1})\times \mathcal{H}(f^{0}_{2}) is actually the (V_{1}\times V_{2}, \mathbf{H}^{2}(\Omega)\times \mathbf{H}^{2}(\Omega)) -uniform attractor \mathcal{A}_{2} , i.e., \mathcal{A}_{1} = \mathcal{A}_{2} .

    Proof. First, we will prove \mathcal{A}_{1}\subset\mathcal{A}_{2} . Because \mathcal{A}_{2} is bounded in \mathbf{H}^2(\Omega)\times \mathbf{\mathbf{H}}^2(\Omega) , and the embedding \mathbf{H}^2(\Omega)\times \mathbf{\mathbf{H}}^2(\Omega)\hookrightarrow V_{1}\times V_{2} is continuous, \mathcal{A}_{2} is bounded in V_{1}\times V_{2} . From Theorem 4.2, we know that \mathcal{A}_{2} attracts uniformly all bounded subsets of V_{1}\times V_{2} , so \mathcal{A}_{2} is a bounded uniform attracting set of \{U_{(f_{1}, f_{2})}(t, \tau)\}_{t\geq\tau} , f_{1}\times f_{2}\in\mathcal{H}(f^{0}_{1})\times \mathcal{H}(f^{0}_{2}) in V_{1}\times V_{2} . By the minimality of \mathcal{A}_{1} , we have \mathcal{A}_{1}\subset\mathcal{A}_{2} .

    Now, we will prove \mathcal{A}_{2}\subset\mathcal{A}_{1} . First, we will prove \mathcal{A}_{1} is a (V_{1}\times V_{2}, \mathbf{H}^{2}(\Omega)\times \mathbf{H}^{2}(\Omega)) -uniformly attracting set of \{U_{(f_{1}, f_{2})}(t, \tau)\}_{t\geq\tau} , f_{1}\times f_{2}\in\mathcal{H}(f^{0}_{1})\times \mathcal{H}(f^{0}_{2}) . That is to say, we will prove

    \begin{equation} \lim\limits_{t\rightarrow +\infty}( \sup\limits_{(f_{1},f_{2})\in\mathcal{H}(f^{0}_{1})\times\mathcal{H}(f^{0}_{2})} \mathrm{dist}_{\mathbf{H}^{2}(\Omega)\times \mathbf{H}^{2}(\Omega)}(U_{(f_{1},f_{2})}(t,\tau)B,\mathcal{A}_{1})) = 0, \end{equation} (4.30)

    for any \tau\in\mathbb{R} and B\in \mathcal{B}(V_{1}\times V_{2}) .

    If we suppose (4.30) is not valid, then there must exist some \tau\in\mathbb{R} , B\in \mathcal{B}(V_{1}\times V_{2}) , \varepsilon_{0} > 0 , (f_{1}^{(n)}, f_{2}^{(n)})\in\mathcal{H}(f^{0}_{1})\times \mathcal{H}(f^{0}_{2}) , and t_{n}\rightarrow +\infty , when n\rightarrow +\infty , such that, for all n\geq 1 ,

    \begin{equation} \mathrm{dis t}_{\mathbf{H}^{2}(\Omega)\times \mathbf{H}^{2}(\Omega)}(U_{(f_{1}^{(n)},f_{2}^{(n)})}(t_{n},\tau)B,\mathcal{A}_{1})\geq 2\varepsilon_{0}. \end{equation} (4.31)

    This shows that there exists (u_{n}, \omega_{n})\in B such that

    \begin{eqnarray} \mathrm{dis t}_{\mathbf{H}^{2}(\Omega)\times \mathbf{H}^{2}(\Omega)}(U_{(f_{1}^{(n)},f_{2}^{(n)})}(t_{n},\tau)(u_{n},\omega_{n}) ,\mathcal{A}_{1})\geq \varepsilon_{0}. \end{eqnarray} (4.32)

    In the light of Theorem 4.2, \{U_{(f_{1}, f_{2})}(t, \tau)\}_{t\geq\tau} , f_{1}\times f_{2}\in\mathcal{H}(f^{0}_{1})\times \mathcal{H}(f^{0}_{2}) has a (V_{1}\times V_{2}, \mathbf{H}^{2}(\Omega)\times \mathbf{H}^{2}(\Omega)) -uniform attractor \mathcal{A}_{2} which attracts any bounded subset of V_{1}\times V_{2} in the topology of \mathbf{H}^{2}(\Omega)\times \mathbf{H}^{2}(\Omega) . Therefore, there exists (\zeta, \eta)\in\mathbf{H}^{2}(\Omega)\times \mathbf{H}^{2}(\Omega) and a subsequence of U_{(f_{1}^{(n)}, f_{2}^{(n)})}(t_{n}, \tau)(u_{n}, \omega_{n}) such that

    \begin{eqnarray} U_{(f_{1}^{(n)},f_{2}^{(n)})}(t_{n},\tau)(u_{n},\omega_{n})\rightarrow (\zeta,\eta )\quad\text{strongly in}\ \mathbf{H}^{2}(\Omega)\times \mathbf{H}^{2}(\Omega). \end{eqnarray} (4.33)

    On the other side, the processes \{U_{(f_{1}, f_{2})}(t, \tau)\}_{t\geq\tau} , f_{1}\times f_{2}\in\mathcal{H}(f^{0}_{1})\times \mathcal{H}(f^{0}_{2}) have a (V_{1}\times V_{2}, V_1\times V_2) -uniform attractor \mathcal{A}_{1} , which attracts uniformly any bounded subsets of V_{1}\times V_{2} in the topology of V_{1}\times V_{2} . So, there exists (u, \omega)\in V_{1}\times V_{2} and a subsequence of U_{(f_{1}^{(n)}, f_{2}^{(n)})}(t_{n}, \tau)(u_n, \omega_n) such that

    \begin{eqnarray} U_{(f_{1}^{(n)},f_{2}^{(n)})}(t_{n},\tau)(u_{n},\omega_{n}) \rightarrow (u,\omega)\ \text{strongly in}\ V_{1}\times V_{2}. \end{eqnarray} (4.34)

    From (4.33) and (4.34), we have (u, \omega) = (\zeta, \eta) , so (4.33) can also be written as

    \begin{eqnarray} U_{(f_{1}^{(n)},f_{2}^{(n)})}(t_{n},\tau)(u_{n},\omega_{n})\rightarrow (u,\omega)\ \text{strongly in}\ \mathbf{H}^{2}(\Omega)\times \mathbf{H}^{2}(\Omega). \end{eqnarray} (4.35)

    And, from Theorem 4.1, we know that \mathcal{A}_{1} attracts B , so

    \begin{eqnarray} \lim\limits_{n\rightarrow +\infty}\mathrm{dist}_{V_{1}\times V_{2}}(U_{(f_{1}^{(n)},f_{2}^{(n)})}(t_{n},\tau)(u_{n},\omega_{n}) ,\mathcal{A}_{1}) = 0. \end{eqnarray} (4.36)

    By (4.34), (4.36), and the compactness of \mathcal{A}_{1} in V_{1}\times V_{2} , we have (u, \omega)\in\mathcal{A}_{1} . Considering (4.35), we have

    \begin{eqnarray*} &\quad \lim\limits_{n\rightarrow +\infty}\mathrm{dist}_{\mathbf{H}^{2}(\Omega)\times \mathbf{H}^{2}(\Omega)}(U_{(f_{1}^{(n)},f_{2}^{(n)})}(t_{n},\tau)(u_{n},\omega_{n}), \mathcal{A}_{1})\nonumber\\ &\leq \lim\limits_{n\rightarrow +\infty}\mathrm{dist}_{\mathbf{H}^{2}(\Omega)\times \mathbf{H}^{2}(\Omega)}(U_{(f_{1}^{(n)},f_{2}^{(n)})}(t_{n},\tau)(u_{n},\omega_{n}), (u,\omega))\nonumber\\ & = 0, \end{eqnarray*}

    which contradicts (4.32). Therefore, \mathcal{A}_{1} is a (V_{1}\times V_{2}, \mathbf{H}^{2}(\Omega)\times \mathbf{H}^{2}(\Omega)) -uniform attractor of \{U_{(f_{1}, f_{2})}(t, \tau)\}_{t\geq\tau} , f_{1}\times f_{2}\in\mathcal{H}(f^{0}_{1})\times \mathcal{H}(f^{0}_{2}) , and by the minimality of \mathcal{A}_{2} , we have \mathcal{A}_{2}\subset\mathcal{A}_{1} .

    In this paper, we discussed the existence of uniform attractors of strong solutions for 3D incompressible micropolar equations with nonlinear damping. Based on some translation-compactness assumption on the external forces, and when \beta > 3 , we made a series of uniform estimates on the solutions in various functional spaces. According to these uniform estimates, we proved the existence of uniform attractors for the process operators corresponding to the solution of the equation in V_1\times V_2 and \mathbf{H}^2\times\mathbf{H}^2 , and verified that the uniform attractors in V_1\times V_2 and \mathbf{H}^2\times\mathbf{H}^2 are actually the same.

    The authors declare they have not used Artificial Intelligence (AI) tools in the creation of this article.

    The authors are thankful to the editors and the anonymous reviewers for their valuable suggestions and comments on the manuscript. This work is supported by National Natural Science Foundation of China (Nos. 11601417, 12001420).

    The authors declare no conflict of interest in this paper.



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