PULLBACK D-ATTRACTORS FOR THE NON-AUTONOMOUS NEWTON-BOUSSINESQ EQUATION IN TWO-DIMENSIONAL BOUNDED DOMAIN. Xue-Li Song.

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1 DISCRETE AND CONTINUOUS doi: /dcds DYNAMICAL SYSTEMS Volume 32, Number 3, March 2012 pp PULLBACK D-ATTRACTORS FOR THE NON-AUTONOMOUS NEWTON-BOUSSINESQ EQUATION IN TWO-DIMENSIONAL BOUNDED DOMAIN Xue-Li Song College of science, Xi an Jiaotong University Xi an, , China and College of science, Xi an University of Science and Technology Xi an, , China Yan-Ren Hou College of science, Xi an Jiaotong University Xi an , China Abstract. We investigate the asymptotic behavior of solutions of a class of non-autonomous Newton-Boussinesq equation in two-dimensional bounded domain. The existence of pullback global attractors is proved in L 2 () L 2 () and H 1 () H 1 (), respectively. 1. Introduction. In this paper, we investigate the asymptotic behavior of solutions of the non-autonomous Newton-Boussinesq equation. Let R 2 be a bounded domain. Consider the system of equations defined in R + : t ζ + u x ζ + v y ζ = ζ R a x η + f(x, y, t), (1.1) Ψ = ζ, u = Ψ y, v = Ψ x, (1.2) t η + u x η + v y η = 1 η + g(x, y, t), (1.3) where u = (u, v) is the velocity field of the fluid, η is the flow temperature, Ψ is the flow function and ζ is the vortex. The positive constants and R a are the Prandtl number and the Rayleigh number, respectively. f and g are external terms. Notice that system (1.1)-(1.3) can be rewritten as follows: for every (x, y) and t > 0, ζ t ζ + J(Ψ, ζ) + R a η = f(x, y, t), x (1.4) Ψ = ζ, (1.5) η t 1 η + J(Ψ, η) = g(x, y, t), (1.6) 2000 Mathematics Subject Classification. Primary: 35B40; Secondary: 35B41, 37L30. Key words and phrases. Pullback attractor, Asymptotic compactness, Newton-Boussinesq equation. The authors are supported by National Natural Science Foundation of China under the grants and

2 992 XUE-LI SONG AND YAN-REN HOU with the boundary conditions and the initial conditions where the function J is given by ζ = 0, η = 0, Ψ = 0, (1.7) ζ(x, y, τ) = ζ τ (x, y), η(x, y, τ) = η τ (x, y), (1.8) J(u, v) = u y v x u x v y. (1.9) The Newton-Boussinesq equation describes many physical phenomena such as Benard flow. If the domain is bounded, the existence, uniqueness and the asymptotic behavior of solutions of system (1.1)-(1.3) have been studied by several authors(see [1]-[4]). And in [5], the authors have proved the existence of a global attractor for the autonomous Newton-Boussinesq equation defined in a two-dimensional channel. As far as we know, there is no results on uniform or pullback attractors for non-autonomous Newton-Boussinesq equation. In this paper, we will investigate the asymptotic behavior of solutions of nonautonomous Newton-Boussinesq equation when the system is defined in a twodimensional bounded domain. In particular, we will prove that the system has pullback attractors in L 2 () L 2 () and H 1 () H 1 (), respectively. And the pullback asymptotic compactness of the system will be proved by using uniform a priori estimates of solutions. The existence and uniqueness of solutions for problem (1.4)-(1.8) can be proved by a standard method as in [1] for the autonomous case. More precisely, for every (ζ τ, η τ ) L 2 () L 2 (), problem (1.4)-(1.8) has a unique solution (ζ, η) such that for every T > 0, ζ C 0 ([τ, ), L 2 ()) L 2 (τ, T ; H 1 0 ()), η C 0 ([τ, ), L 2 ()) L 2 (τ, T ; H 1 0 ()). Furthermore, the solution is continuous with respect to initial data (ζ τ, η τ ) in L 2 () L 2 (). It is easy to verify that J satisfies: J(u, v)vdxdy = 0, for all u H 1 (), v H 2 () H0 1 (), (1.10) J(u, v) C u H 2 v H 2, for all u H 2 (), v H 2 (), (1.11) J(u, v) C u H 3 v, for all u H 3 (), v H 1 (). (1.12) Throughout this paper, we frequently use the Poincaré inequality u λ u, u H 1 0 (), where λ is a positive constant. The following notations will be used throughout this paper. We denote by and (, ) the norm and inner product in L 2 () and use p to denote the norm in L p () (p 2). The norm of any Banach space X is written as X. The letter C is a generic positive constant which may change its values from line to line or even in the same line.

3 PULLBACK D-ATTRACTORS FOR NEWTON-BOUSSINESQ EQUATION Preliminaries and abstract results. In this section, we recall some definitions and results concerning the pullback attractor. The reader is referred to [6]-[9] for details. Let Θ be a non-empty set, X be a metric space. And we denote by D the collection of families of subsets of X: D = {D = {D(ω)} ω Θ : D(ω) X for every ω Θ}. Definition 2.1. A family of mapping {θ t } t R from Θ to itself is called a family of shift operators on Θ if {θ t } t R satisfies the group properties: (1) θ 0 ω = ω, ω Θ; (2) θ t (θ τ ω) = θ t+τ ω, ω Θ and t, τ R. Definition 2.2. Let {θ t } t R be a family of shift operators on Θ. Then a continuous θ-cocycle φ on X is a mapping φ : R + Θ X X, (t, ω, x) φ(t, ω, x), which satisfies, for all ω Θ and t, τ R +, (1) φ(0, ω, ) is the identity on X; (2) φ(t + τ, ω, ) = φ(t, θ τ ω, ) φ(τ, ω, ); (3) φ(t, ω, ) : X X is continuous. Definition 2.3. Let D be a collection of families of subsets of X. Then D is called inclusion-closed if D = {D(ω)} ω Θ D and D = { D(ω) X : ω Θ} with D(ω) D(ω) for all ω Θ imply that D D. Definition 2.4. Let D be a collection of families of subsets of X and {K(ω)} ω Θ D. Then {K(ω)} ω Θ is called a pullback absorbing set for φ in D if, for every B D and ω Θ, there exists T (ω, B) > 0 such that φ(t, θ t ω, B(θ t ω)) K(ω) for all t T (ω, B). Definition 2.5. Let D be a collection of families of subsets of X. Then φ is said to be D-pullback asymptotically compact in X if, for every ω Θ, {φ(t n, θ tn ω, x n )} n=1 has a convergent subsequence in X whenever t n, and x n B(θ tn ω) with {B(ω)} ω Θ D. Definition 2.6. Let D be a collection of families of subsets of X and {A(ω)} ω Θ D. Then {A(ω)} ω Θ is called a D-pullback global attractor for φ if the following conditions are satisfied, for every ω Θ, (1) A(ω) is compact; (2) {A(ω)} ω Θ is invariant, that is, φ(t, ω, A(ω)) = A(θ t ω), t 0; (3) {A(ω)} ω Θ attracts every set in D, that is, for every B = {B(ω)} ω Θ D, lim dist(φ(t, θ tω, B(θ t ω)), A(ω)) = 0, t where dist is the Hausdorff semi-metric given by dist(a, B) = sup a A inf b B a b X for any A X and B X. Proposition 1. Let D be an inclusion-closed collection of families of subsets of X and φ a continuous θ-cocycle on X. Suppose that {K(ω)} ω Θ D is a closed absorbing set for φ in D and φ is D-pullback asymptotically compact in X. Then φ has a unique D-pullback global attractor {A(ω)} ω Θ D which is given by

4 994 XUE-LI SONG AND YAN-REN HOU A(ω) = τ 0 φ(t, θ t ω, K(θ t ω)). t τ To construct a cocycle φ for problem (1.4)-(1.8), we denote by Θ = R, and define a shift operator θ t on Θ for every t R by: θ t (τ) = t + τ, τ R. Let φ be a mapping from R + Θ (L 2 () L 2 ()) to L 2 () L 2 () given by φ(t, τ, (ζ τ, η τ )) = (ζ(t + τ, τ, ζ τ ), η(t + τ, τ, η τ )), where t 0, τ R, (ζ τ, η τ ) L 2 () L 2 (), and (ζ, η) is the solution of problem (1.4)-(1.8). By the uniqueness of solutions, we find that for every t, s 0, τ R and (ζ τ, η τ ) L 2 () L 2 (), φ(t + s, τ, (ζ τ, η τ )) = φ(t, s + τ, (φ(s, τ, (ζ τ, η τ )))). Then we see that φ is a continuous θ-cocycle on L 2 () L 2 (). For convenience, if E L 2 () L 2 (), we denote E = sup x L2 () L 2 (). x E Let D = {D(t)} t R be a family of subsets of L 2 () L 2 (), i.e., D(t) L 2 () L 2 () for every t R. In this paper, we are interested in a family D = {D(t)} t R satisfying lim t eσt D(t) 2 = 0, (2.1) 1 1 where σ is a positive constant satisfying 0 < σ < min{ 2λ 2, 2λ }. We write the 2 collection of all families satisfying (2.1) as D σ, that is, D σ = {D = {D(t)} t R : D satisfies (2.1)}. (2.2) As we will see later, for deriving the uniform estimates of solutions, we need the following conditions for the external force terms where K is a positive constant. e σξ [ f(ξ) 2 + g(ξ) 2 ]dξ < K <, τ R, (2.3) 3. Uniform estimates of solutions. In this section, we derive uniform estimates of solutions of problem (1.4)-(1.8) when t. These estimates are necessary for proving the existence of a bounded pullback absorbing set and the pullback asymptotic compactness of the θ-cocycle φ associated with the system. Lemma 3.1. For every τ R and D = {D(t)} t R D σ, there exists T = T (τ, D) > 0 such that for all t T, and ζ(τ, τ t, ζ 0 (τ t)) 2 + η(τ, τ t, η 0 (τ t)) 2 Ce στ (3.1) e σξ( ) ζ(ξ, τ t, ζ 0 (τ t)) 2 + η(ξ, τ t, η 0 (τ t)) 2 dξ C (3.2)

5 PULLBACK D-ATTRACTORS FOR NEWTON-BOUSSINESQ EQUATION 995 e σξ( ) ζ(ξ, τ t, ζ 0 (τ t)) 2 + η(ξ, τ t, η 0 (τ t)) 2 dξ C (3.3) where (ζ 0 (τ t), η 0 (τ t)) D(τ t), and C is a positive constant independent of τ or D. Proof. Taking the inner product of (1.6) with η in L 2 () and using (1.10) we obtain 1 d 2 dt η η 2 = gηdxdy g η 1 (3.4) 4λ 2 η 2 +C g(t) 2. So d dt η η λ 2 η 2 C g(t) 2. (3.5) Multiplying (3.5) by e σt and then integrating it between τ t and τ with t 0, we get, e στ η(τ, τ t, η 0 (τ t)) e σ() η 0 (τ t) 2 +(σ 1 2λ 2 ) e σξ g(ξ) 2 dξ. e σξ η(ξ, τ t, η 0 (τ t)) 2 dξ e σξ η(ξ, τ t, η 0 (τ t)) 2 dξ (3.6) Noticing that (ζ 0 (τ t), η 0 (τ t)) D(τ t) and D = {D(t)} t R D σ, and we find that for every τ R, there exists T 1 = T 1 (τ, D) > 0 such that for all t T 1, e σ() η 0 (τ t) 2 which along with (3.6) shows that, for all t T 1, η(τ, τ t, η 0 (τ t)) e στ 1 + ( 2λ 2 σ)e στ Ce στ From (3.7) we have e σξ η(ξ, τ t, η 0 (τ t)) 2 dξ e σξ η(ξ, τ t, η 0 (τ t)) 2 dξ e σξ [ f(ξ) 2 + g(ξ) 2 ]dξ. η(τ, τ t, η 0 (τ t)) 2 Ce στ e σξ η(ξ, τ t, η 0 (τ t)) 2 dξ C e σξ η(ξ, τ t, η 0 (τ t)) 2 dξ C (3.7) e σξ [ f(ξ) 2 + g(ξ) 2 ]dξ; (3.8) e σξ [ f(ξ) 2 + g(ξ) 2 ]dξ; (3.9) e σξ [ f(ξ) 2 + g(ξ) 2 ]dξ. (3.10)

6 996 XUE-LI SONG AND YAN-REN HOU Taking the inner product of (1.4) with ζ in L 2 () and using (1.10) we obtain: 1 d 2 dt ζ 2 + ζ 2 + R a η x ζdxdy = f(t)ζdxdy, (3.11) because R a η x ζdxdy = R a f(t)ζdxdy f(t) ζ λ f(t) ζ so taking (3.11)-(3.13) into account, we get ζ x ηdxdy R a ζ η ζ 2 η 2, 4 ζ 2 4 (3.12) f(t) 2, (3.13) d dt ζ ζ λ 2 ζ 2 C η 2 +C f(t) 2. (3.14) Multiplying (3.14) by e σt and then integrating it between τ t and τ, we obtain e στ ζ(τ, τ t, ζ 0 (τ t)) e σ() ζ 0 (τ t) 2 +(σ 1 2λ 2 ) e σξ η(ξ, τ t, η 0 (τ t)) 2 dξ e σξ ζ(ξ, τ t, ζ 0 (τ t)) 2 dξ e σξ ζ(ξ, τ t, ζ 0 (τ t)) 2 dξ e σξ f(ξ) 2 dξ. (3.15) Noticing that (ζ 0 (τ t), η 0 (τ t)) D(τ t) and D = {D(t)} t R D σ, we find that for every τ R, there exists T 2 = T 2 (τ, D) > 0 such that for all t T 2, e σ() ζ 0 (τ t) 2 which along with (3.15) and (3.9) shows that, for all t T = max{t 1, T 2 } > 0, ζ(τ, τ t, ζ 0 (τ t)) e στ e σξ ζ(ξ, τ t, ζ 0 (τ t)) 2 dξ + ( 1 2λ 2 σ)e στ e σξ ζ(ξ, τ t, ζ 0 (τ t)) 2 dξ Ce στ e σξ [ f(ξ) 2 + g(ξ) 2 ]dξ. From (3.16) we have ζ(τ, τ t, ζ 0 (τ t)) 2 Ce στ e σξ ζ(ξ, τ t, ζ 0 (τ t)) 2 dξ C (3.16) e σξ [ f(ξ) 2 + g(ξ) 2 ]dξ; (3.17) e σξ [ f(ξ) 2 + g(ξ) 2 ]dξ. (3.18) e σξ ζ(ξ,, ζ 0 ()) 2 dξ C e σξ [ f(ξ) 2 + g(ξ) 2 ]dξ. (3.19) Thus, the proof is completed.

7 PULLBACK D-ATTRACTORS FOR NEWTON-BOUSSINESQ EQUATION 997 Lemma 3.2. For every τ R and D = {D(t)} t R D σ, there exists T = T (τ, D) > 2 such that for all t T, e σξ ( ζ(ξ, τ t, ζ 0 (τ t)) 2 + η(ξ, τ t, η 0 (τ t)) 2 )dξ C e σξ [ f(ξ) 2 + g(ξ) 2 ]dξ; (3.20) e σξ ( ζ(ξ, τ t, ζ 0 (τ t)) 2 + η(ξ, τ t, η 0 (τ t)) 2 )dξ (3.21) C where (ζ 0 (τ t), η 0 (τ t)) D(τ t), and C is a positive constant independent of τ or D. Proof. Note that (3.5) implies d dt η λ 2 η 2 C g(t) 2. (3.22) Let s [τ 2, τ] and t 2. Multiplying (3.22) by e σt, then relabeling t as ξ and integrating it with respect to ξ over (τ t, s), we have e σs η(s, τ t, η 0 (τ t)) 2 1 +( 2λ 2 σ) e σ() η 0 (τ t) 2 +C s e σξ η(ξ, τ t, η 0 (τ t)) 2 dξ e σξ g(ξ) 2 dξ. (3.23) Noticing that (ζ 0 (τ t), η 0 (τ t)) D(τ t) and D = {D(t)} t R D σ, and we find that for every τ R, there exists T 3 = T 3 (τ, D) > 2 such that for all t T 3, So e σ() η 0 (τ t) 2 s e σs η(s, τ t, η 0 (τ t)) 2 + C e σξ [ f(ξ) 2 + g(ξ) 2 ]dξ. e σξ η(ξ, τ t, η 0 (τ t)) 2 dξ (3.24) for any s [τ 2, τ]. Multiplying (3.5) by e σt and integrating it over (τ 2, τ), by (3.24) we get that, for all t T 3, e στ η(τ, τ t, η 0 (τ t)) ( 2λ 2 σ) e σξ η(ξ, τ t, η 0 (τ t)) 2 dξ e σξ η(ξ, τ t, η 0 (τ t)) 2 dξ e σ() η(τ 2, τ t, η 0 (τ t)) 2 +C C e σξ [ f(ξ) 2 + g(ξ) 2 ]dξ. e σξ g(ξ) 2 dξ (3.25)

8 998 XUE-LI SONG AND YAN-REN HOU From (3.25) we have e σξ η(ξ, τ t, η 0 (τ t)) 2 dξ C e σξ η(ξ, τ t, η 0 (τ t)) 2 dξ C e σξ [ f(ξ) 2 + g(ξ) 2 ]dξ. Note that (3.14) implies e σξ [ f(ξ) 2 + g(ξ) 2 ]dξ; (3.26) (3.27) d dt ζ λ 2 ζ 2 C η 2 +C f(t) 2. (3.28) Let s [τ 2, τ] and t 2, multiplying (3.28) by e σt, then relabeling t as ξ and integrating it with respect to ξ over (τ t, s), we have e σs ζ(s, τ t, ζ 0 (τ t)) 2 +( 1 2λ 2 σ) e σ() ζ 0 (τ t) 2 +C s e σξ f(ξ) 2 dξ. s s e σξ ζ(ξ, τ t, ζ 0 (τ t)) 2 dξ e σξ η(ξ, τ t, η 0 (τ t)) 2 dξ (3.29) Noticing that (ζ 0 (τ t), η 0 (τ t)) D(τ t) and D = {D(t)} t R D σ, and we find that for every τ R, there exists T 4 = T 4 (τ, D) > 2 such that for all t T 4, e σ() ζ 0 (τ t) 2 which along with (3.29) and (3.24) implies that: there exists T 5 = max{t 3, T 4 } > 2, such that for all t T 5 and for any s [τ 2, τ], e σs ζ(s, τ t, ζ 0 (τ t)) 2 C e σξ [ f(ξ) 2 + g(ξ) 2 ]dξ. (3.30) Multiplying (3.14) by e σt and integrating it over (τ 2, τ), using (3.26) and (3.30) we get e στ ζ(τ, τ t, ζ 0 (τ t)) e σξ ζ(ξ, τ t, ζ 0 (τ t)) 2 dξ 2 + ( 1 2λ 2 σ) e σξ ζ(ξ, τ t, ζ 0 (τ t)) 2 dξ e σ() ζ(τ 2, τ t, ζ 0 (τ t)) 2 e σξ η(ξ, τ t, η 0 (τ t)) 2 dξ C e σξ [ f(ξ) 2 + g(ξ) 2 ]dξ. Thanks to (3.31), we have e σξ ζ(ξ, τ t, ζ 0 (τ t)) 2 dξ C e σξ f(ξ) 2 dξ. (3.31) e σξ [ f(ξ) 2 + g(ξ) 2 ]dξ; (3.32)

9 PULLBACK D-ATTRACTORS FOR NEWTON-BOUSSINESQ EQUATION 999 which completes the proof. e σξ ζ(ξ, τ t, ζ 0 (τ t)) 2 dξ C (3.33) Corollary 1. For every τ R and D = {D(t)} t R D σ, there exists T = T (τ, D) > 2 such that for all t T, [ ζ(ξ, τ t, ζ 0 (τ t)) 2 + η(ξ, τ t, η 0 (τ t)) 2 ]dξ Ce στ [ ζ(ξ, τ t,ζ 0 (τ t)) 2 + η(ξ, τ t, η 0 (τ t)) 2 ]dξ Ce στ (3.34) (3.35) where (ζ 0 (τ t), η 0 (τ t)) D(τ t), and C is a positive constant independent of τ or D. Proof. It is straightforward from Lemma 3.2. Lemma 3.3. For every τ R and D = {D(t)} t R D σ, there exists T = T (τ, D) > 2 such that for all t T, ζ(τ, τ t, ζ 0 (τ t)) 2 + η(τ, τ t, η 0 (τ t)) 2 C[e στ + e 3στ ] (3.36) where (ζ 0 (τ t), η 0 (τ t)) D(τ t), and C is a positive constant independent of τ or D. Proof. Taking the inner product of (1.4) with ζ in L 2 () we get 1 d 2 dt ζ 2 + ζ 2 = J(Ψ, ζ) ζdxdy + R a η x ζdxdy f ζdxdy. (3.37) Notice that the first term on the right-hand side of (3.37) is given by J(Ψ, ζ) ζdxdy = Ψ y ζ x ζdxdy Ψ x ζ y ζdxdy. (3.38) We now estimate the first term on the right-hand side of (3.38). By Hölder inequality and Nirenberg-Gagliardo inequality, we have Ψ y ζ x ζdxdy Ψ y 4 ζ x 4 ζ C Ψ y 1/2 Ψ y 1/2 H 1 ζ x 1/2 ζ x 1/2 H 1 ζ C Ψ H 2 ζ 1/2 ζ 3/2 C ζ ζ 1/2 ζ 3/2 1 8 ζ 2 +C ζ 4 ζ 2. (3.39)

10 1000 XUE-LI SONG AND YAN-REN HOU Similarly, we have Ψ x ζ y ζdxdy 1 8 ζ 2 +C ζ 4 ζ 2. (3.40) It is obvious that the last two terms on the right-hand side of (3.37) are bounded by R a η x ζdxdy + f ζdxdy 1 4 ζ 2 +C η 2 +C f 2. (3.41) Now taking (3.37)-(3.41) into account, the usage of Lemma 3.1 leads to d dt ζ 2 + ζ 2 C ζ 4 ζ 2 +C η 2 +C f 2 C[r 0 (t)] 2 ζ 2 +C η 2 +C f 2 Ce 2σt ζ 2 +C η 2 +C f 2, (3.42) where r 0 (t) = Ce σt t eσξ [ f(ξ) 2 + g(ξ) 2 ]dξ < CKe σt Ce σt. Let s τ and t 2. Multiplying (3.42) by e σt, then relabeling t as ξ and integrating it with respect to ξ over (s, τ) with τ 1 s τ, we find that e στ ζ(τ, τ t, ζ 0 (τ t)) 2 e σs ζ(s, τ t, ζ 0 (τ t)) 2 +C + σ s s s e σξ ζ(ξ, τ t, ζ 0 (τ t)) 2 dξ e σξ η(ξ, τ t, η 0 (τ t)) 2 dξ e σξ ζ(ξ, τ t, ζ 0 (τ t)) 2 dξ e σs ζ(s, τ t, ζ 0 (τ t)) 2 +C s e σξ f(ξ) 2 dξ e σξ f(ξ) 2 dξ e 2σ() e σξ ζ(ξ, τ t, ζ 0 (τ t)) 2 dξ e σξ η(ξ, τ t, η 0 (τ t)) 2 dξ + σ e σξ ζ(ξ, τ t, ζ 0 (τ t)) 2 dξ. (3.43)

11 PULLBACK D-ATTRACTORS FOR NEWTON-BOUSSINESQ EQUATION 1001 We now integrate (3.43) with respect to s over (τ 1, τ), and by (3.21) we have e στ ζ(τ, τ t, ζ 0 (τ t)) 2 e σs ζ(s, τ t, ζ 0 (τ t)) 2 ds e 2σ() e σξ ζ(ξ, τ t, ζ 0 (τ t)) 2 dξ e σξ η(ξ, τ t, η 0 (τ t)) 2 dξ + σ e σξ ζ(ξ, τ t, ζ 0 (τ t)) 2 dξ C[1 + e 2σ() ] e σξ [ f(ξ) 2 + g(ξ) 2 ]dξ C[1 + e 2στ ] e σξ [ f(ξ) 2 + g(ξ) 2 ]dξ. e σξ f(ξ) 2 dξ (3.44) So ζ(τ, τ t, ζ 0 (τ t)) 2 C[e στ + e 3στ ] e σξ [ f(ξ) 2 + g(ξ) 2 ]dξ. (3.45) Taking the inner product of (1.6) with η in L 2 () we get 1 d 2 dt η η 2 = J(Ψ, η) ηdxdy g ηdxdy. (3.46) By arguments similar to getting (3.39) and (3.40), we obtain that J(Ψ, η) ηdxdy = Ψ y η x ηdxdy 1 η 2 +C ζ 4 η 2, 4 Ψ x η y ηdxdy (3.47) and g ηdxdy 1 η 2 +C g(t) 2. (3.48) 4 Taking (3.46)-(3.48) into account, the usage of Lemma 3.1 leads to d dt η η 2 C ζ 4 η 2 +C g(t) 2 Ce 2σt η 2 +C g(t) 2. (3.49)

12 1002 XUE-LI SONG AND YAN-REN HOU Let s τ and t 2. Multiplying (3.49) by e σt, then relabeling t as ξ and integrating it with respect to ξ over (s, τ) with τ 1 s τ, we find that e στ η(τ, τ t, η 0 (τ t)) 2 e σs η(s, τ t, η 0 (τ t)) 2 +C + σ s s e σξ η(ξ, τ t, η 0 (τ t)) 2 dξ e σξ η(ξ, τ t, η 0 (τ t)) 2 dξ e σs η(s, τ t, η 0 (τ t)) 2 +C s e σξ g(ξ) 2 dξ e σξ g(ξ) 2 dξ e 2σ() e σξ η(ξ, τ t, η 0 (τ t)) 2 dξ + σ e σξ η(ξ, τ t, η 0 (τ t)) 2 dξ. (3.50) We now integrate (3.50) with respect to s over (τ 1, τ), and according to (3.21), we have So e στ η(τ, τ t, η 0 (τ t)) 2 e σs η(s, τ t, η 0 (τ t)) 2 ds e 2σ() e σξ η(ξ, τ t, η 0 (τ t)) 2 dξ + σ e σξ η(ξ, τ t, η 0 (τ t)) 2 dξ C[1 + e 2σ() ] e σξ [ f(ξ) 2 + g(ξ) 2 ]dξ C[1 + e 2στ ] e σξ [ f(ξ) 2 + g(ξ) 2 ]dξ. η(τ, τ t, η 0 (τ t)) 2 C[e στ + e 3στ ] e σξ g(ξ) 2 dξ (3.51) (3.52) which completes the proof. Lemma 3.4. For every τ R and D = {D(t)} t R D σ, there exists T = T (τ, D) > 2 such that for all t T, e σξ [ ζ(ξ, τ t, ζ 0 (τ t)) 2 + η(ξ, τ t, η 0 (τ t)) 2 ]dξ C[1 + e 2στ ] (3.53)

13 PULLBACK D-ATTRACTORS FOR NEWTON-BOUSSINESQ EQUATION 1003 [ ζ(ξ, τ t, ζ 0 (τ t)) 2 + η(ξ, τ t, η 0 (τ t)) 2 ]dξ C[e στ + e 3στ ] (3.54) where (ζ 0 (τ t), η 0 (τ t)) D(τ t), and C is a positive constant independent of τ or D. Proof. Like the proof of Lemma 3.3, we can also prove that: for every τ R and D = {D(t)} t R D σ, there exists T = T (τ, D) > 2 such that for all t T, ζ(τ 1, τ t, ζ 0 (τ t)) 2 + η(τ 1, τ t, η 0 (τ t)) 2 C[e στ + e 3στ ] (3.55) where (ζ 0 (τ t), η 0 (τ t)) D(τ t), and C is a positive constant independent of τ or D. Now, multiplying (3.42) by e σt and then integrating it over (τ 1, τ), according to (3.55) and (3.21), we have e σξ ζ(ξ, τ t, ζ 0 (τ t)) 2 dξ e σ() ζ(τ 1, τ t, ζ 0 (τ t)) 2 +C + σ e σξ ζ(ξ, τ t, ζ 0 (τ t)) 2 dξ C[1 + e 2στ ] e σξ ζ(ξ, τ t, ζ 0 (τ t)) 2 dξ e σξ η(ξ, τ t, η 0 (τ t)) 2 dξ e σξ [ f(ξ) 2 + g(ξ) 2 ]dξ. e σξ f(ξ) 2 dξ (3.56) Similarly, multiplying (3.49) by e σt and then integrating it over (τ 1, τ), according to (3.55) and (3.21), we can get e σξ η(ξ, τ t, η 0 (τ t)) 2 dξ C[1 + e 2στ ] e σξ [ f(ξ) 2 + g(ξ) 2 ]dξ. (3.57) From (3.56) and (3.57) we can get (3.53) immediately. And (3.54) is straightforward from (3.53). Lemma 3.5. For every τ R and D = {D(t)} t R D σ, there exists T = T (τ, D) > 2 such that for all t T, [ ζ ξ (ξ, τ t, ζ 0 (τ t)) 2 + η ξ (ξ, τ t, η 0 (τ t)) 2 ]dξ C[e στ + e 2στ + e 3στ + e 4στ ] (3.58)

14 1004 XUE-LI SONG AND YAN-REN HOU where (ζ 0 (τ t), η 0 (τ t)) D(τ t), and C is a positive constant independent of τ or D. Proof. From (1.6) and (1.11), we have η t 1 η + J(Ψ, η) + g(t) 1 η +C ζ η + g(t). So η t 2 C[ η 2 + ζ 2 η 2 + g(t) 2 ]. (3.59) Relabeling t as ξ and integrating (3.59) with respect to ξ over (τ 1, τ), by Lemma 3.1 and Lemma 3.4, we have η ξ (ξ, τ t, η 0 (τ t)) 2 dξ C η(ξ, τ t, η 0 (τ t)) 2 dξ C[e στ + e 3στ ] e σξ η(ξ, τ t, η 0 (τ t)) 2 dξ e σ() [e στ + e 3στ ] e στ e σξ [ f(ξ) 2 + g(ξ) 2 ]dξ e σξ g(ξ) 2 dξ C[e στ + e 2στ + e 3στ + e 4στ ] g(ξ) 2 dξ e σξ [ f(ξ) 2 + g(ξ) 2 ]dξ e σξ [ f(ξ) 2 + g(ξ) 2 ]dξ. From (1.4) and (1.11), we have ζ t ζ + J(Ψ, ζ) + R a η + f(t) ζ +C ζ ζ + R a η + f(t). So ζ t 2 C[ ζ 2 + ζ 2 ζ 2 + η 2 + f(t) 2 ]. (3.60)

15 PULLBACK D-ATTRACTORS FOR NEWTON-BOUSSINESQ EQUATION 1005 Relabeling t as ξ and integrating (3.60) with respect to ξ over (τ 1, τ), by Lemma 3.1, Corollary 1 and Lemma 3.4, we have ζ ξ (ξ, τ t, ζ 0 (τ t)) 2 dξ C ζ(ξ, τ t, ζ 0 (τ t)) 2 dξ η(ξ, τ t, η 0 (τ t)) 2 dξ f(ξ) 2 dξ e σξ ζ(ξ, τ t, ζ 0 (τ t)) 2 dξ C[e στ + e 2στ + e 3στ + e 4στ ] e σξ [ f(ξ) 2 + g(ξ) 2 ]dξ. Lemma 3.6. Let df dt, dg dt L 2 loc (R; L2 ()). Then for every τ R and D = {D(t)} t R D σ, there exists T = T (τ, D) > 2 such that for all t T, ζ τ (τ, τ t, ζ 0 (τ t)) 2 + η τ (τ, τ t, η 0 (τ t)) 2 ( C[e στ + e 2στ + e 3στ + e 4στ ] e σξ [ f(ξ) 2 + g(ξ) 2 ]dξ ) { τ [ f ξ (ξ) 2 + g ξ (ξ) 2 ]dξ exp C[e στ + e 3στ ] } e σξ [ f(ξ) 2 + g(ξ) 2 ]dξ, (3.61) where (ζ 0 (τ t), η 0 (τ t)) D(τ t), and C is a positive constant independent of τ or D. Proof. Let ζ = dζ dη dt and η = dt. Differentiating (1.4) and (1.6) with respect to t admits and d ζ dt ζ + J(Ψ t, ζ) + J(Ψ, ζ) + R a η x = f t, (3.62) d η dt 1 η + J(Ψ t, η) + J(Ψ, η) = g t. (3.63) By taking the inner product of (3.63) with η in L 2 () we get 1 d 2 dt η η 2 = (J(Ψ t, η), η) + (g t, η). (3.64) Let us estimate the first term on the right-hand side of (3.64). Due to (J(Ψ t, η), η) = Ψ ty η x η Ψ tx η y η,

16 1006 XUE-LI SONG AND YAN-REN HOU and Ψ ty η x η Ψ ty 4 η x 4 η C Ψ ty 1/2 Ψ ty 1/2 H 1 η x 1/2 η x 1/2 H 1 η C ζ η η C η 2 +C ζ 2 η 2. Similarly, we have So Thanks to Ψ tx η y η C η 2 +C ζ 2 η 2. (J(Ψ t, η), η) C η 2 +C ζ 2 η 2. (3.65) by taking (3.64)-(3.66) into account, we have (g t, η) C g t 2 +C η 2, (3.66) d dt η η 2 C ζ 2 η 2 +C η 2 +C g t 2. (3.67) Taking the inner product of (3.62) with ζ in L 2 (), we get 1 d 2 dt ζ 2 + ζ 2 = (J(Ψ t, ζ), ζ) R a ( η x, ζ) + (f t, ζ). (3.68) By arguments similar to deriving (3.65), we obtain that (J(Ψ t, ζ), ζ) C ζ 2 +C ζ 2 ζ 2. (3.69) And bacause R a ( η x, ζ) R a η P ζ 1 η 2 +C r P ζ 2, (3.70) r so taking (3.68)-(3.71) into account, we have (f t, ζ) C f t 2 +C ζ 2, (3.71) d dt ζ 2 2 η 2 +C ζ 2 ζ 2 +C ζ 2 +C f t 2. (3.72) Adding (3.67) to (3.72), we have d dt ( ζ 2 + η 2 ) C ζ 2 ( ζ 2 + η 2 ) ( ζ 2 + η 2 ) ( f t 2 + g t 2 ) C( ζ 2 + η 2 )( ζ 2 + η 2 +1) ( f t 2 + g t 2 ). (3.73)

17 PULLBACK D-ATTRACTORS FOR NEWTON-BOUSSINESQ EQUATION 1007 By the uniform Gronwall lemma, Lemma 3.4 and Lemma 3.5, we have ζ(τ, τ t, ζ 0 (τ t)) 2 + η(τ, τ t, η 0 (τ t)) 2 ( [ ζ(ξ, τ t, ζ 0 (τ t)) 2 + η(ξ, τ t, η 0 (τ t)) 2 ]dξ ) { [ f ξ (ξ) 2 + g ξ (ξ) 2 ]dξ exp ζ 0 (τ t)) 2 + η(ξ, τ t, η 0 (τ t)) 2 ]dξ ( C[e στ + e 2στ + e 3στ + e 4στ ] C } [ ζ(ξ, τ t, e σξ [ f(ξ) 2 + g(ξ) 2 ]dξ ) { τ [ f ξ (ξ) 2 + g ξ (ξ) 2 ]dξ exp C[e στ + e 3στ ] } e σξ [ f(ξ) 2 + g(ξ) 2 ]dξ (3.74) 4. Existence of pullback attractors. In this section, we will prove the existence of D σ -pullback attractors for problem (1.4)-(1.8) in L 2 () L 2 () and H 1 () H 1 (), respectively. Theorem 4.1. Problem (1.4)-(1.8) has a unique D σ -pullback global attractor {A(τ)} τ R in L 2 () L 2 (). Proof. For τ R, denote by { B 1 (τ) = (ζ, η) L 2 () L 2 () : ζ 2 + η 2 } Ce στ e σξ [ f(ξ) 2 + g(ξ) 2 ]dξ. Thanks to Lemma 3.1, we know that B 1 = {B 1 (τ)} τ R D σ is a D σ -pullback absorbing set for φ in L 2 () L 2 (). Now, we will prove that for every τ R, D = {D(t)} t R D σ, and t n, (ζ 0,n, η 0,n ) D(τ t n ), the sequence φ(t n, τ t n, (ζ 0,n, η 0,n )) has a convergent subsequence in L 2 () L 2 (), i.e., φ is D σ -pullback asymptotically compact in L 2 () L 2 (). By Lemma 3.3, there exist C = C(τ) > 0 and N = N(τ, D) > 0 such that, for all n N, φ(t n, τ t n, (ζ 0,n, η 0,n )) H 1 () H 1 () C. (4.1) By (4.1) and the compactness of embedding H 1 () H 1 () L 2 () L 2 (), the sequence φ(t n, τ t n, (ζ 0,n, η 0,n )) is precompact in L 2 () L 2 (), which completes the proof. Lemma 4.2. Let df dt, dg dt L2 loc (R; L2 ()). Then φ is D σ -pullback asymptotically compact in H 1 () H 1 (), that is, for every τ R, D = {D(t)} t R D σ, and t n, (ζ 0,n, η 0,n ) D(τ t n ), the sequence φ(t n, τ t n, (ζ 0,n, η 0,n )) has a convergent subsequence in H 1 () H 1 ().

18 1008 XUE-LI SONG AND YAN-REN HOU Proof. We will show that the sequence φ(t n, τ t n, (ζ 0,n, η 0,n )) has a Cauchy subsequence in H 1 () H 1 (). By Theorem 4.1, φ(t n, τ t n, (ζ 0,n, η 0,n )) is precompact in L 2 () L 2 (). This means that, up to a subsequence, (ζ(τ, τ t n, ζ 0,n ), η(τ, τ t n, η 0,n )) is a Cauchy sequence in L 2 () L 2 (). (4.2) From (1.4) we find that, for any n, m 1, ( ζ ( τ, τ t n, ζ 0,n ) ζ(τ, τ t m, ζ 0,m ) ) = ( ζ τ (τ, τ t n, ζ 0,n ) ζ τ (τ, τ t m, ζ 0,m ) ) ( J(Ψ, ζ(τ, τ t n, ζ 0,n )) J(Ψ, ζ(τ, τ t m, ζ 0,m )) ) R a ( ηx (τ, τ t n, η 0,n ) η x (τ, τ t m, η 0,m ) ). (4.3) Multiplying (4.3) by ζ(τ, τ t n, ζ 0,n ) ζ(τ, τ t m, ζ 0,m ) and then integrating it on, by (1.10), we get ζ(τ, τ t n, ζ 0,n ) ζ(τ, τ t m, ζ 0,m ) 2 ζ τ (τ, τ t n, ζ 0,n ) ζ τ (τ, τ t m, ζ 0,m ) ζ(τ, τ t n, ζ 0,n ) ζ(τ, τ t m, ζ 0,m ) + R a η(τ, τ t n, η 0,n ) η(τ, τ t m, η 0,m ) ζ(τ, τ t n, ζ 0,n ) ζ(τ, τ t m, ζ 0,m ). (4.4) By Lemma 3.3 and Lemma 3.6, there is N = N(τ, D) > 0 such that, for all n N, and η(τ, τ t n, η 0,n ) η(τ, τ t m, η 0,m ) C ζ τ (τ, τ t n, ζ 0,n ) ζ τ (τ, τ t m, ζ 0,m ) C, which along with (4.2) and (4.4) implies ζ(τ, τ t n, ζ 0,n ) is a Cauchy sequence in L 2 (). (4.5) From (1.6) we find that, for any n, m 1, 1 ( η(τ, τ tn, η 0,n ) η(τ, τ t m, η 0,m ) ) = ( η τ (τ, τ t n, η 0,n ) η τ (τ, τ t m, η 0,m ) ) ( J(Ψ, η(τ, τ t n, η 0,n )) J(Ψ, η(τ, τ t m, η 0,m )) ). (4.6) Multiplying (4.6) by η(τ, τ t n, η 0,n ) η(τ, τ t m, η 0,m ) and then integrating it on, by (1.10), we get 1 η(τ, τ t n, η 0,n ) η(τ, τ t m, η 0,m ) 2 η τ (τ, τ t n, η 0,n ) η τ (τ, τ t m, η 0,m ) η(τ, τ t n, η 0,n ) η(τ, τ t m, η 0,m ) By Lemma 3.6, there is N = N(τ, D) > 0 such that, for all n N, η τ (τ, τ t n, η 0,n ) η τ (τ, τ t m, η 0,m ) C, which along with (4.2) and (4.7) implies (4.7) η(τ, τ t n, η 0,n ) is a Cauchy sequence in L 2 (). (4.8)

19 PULLBACK D-ATTRACTORS FOR NEWTON-BOUSSINESQ EQUATION 1009 Then (4.5) and (4.8) imply that ( ζ(τ, τ t n, ζ 0,n ), η(τ, τ t n, η 0,n ) ) is a Cauchy sequence in H 1 () H 1 (). Theorem 4.3. Let df dt, dg dt L2 loc (R; L2 ()). Then problem (1.4)-(1.8) has a D σ - pullback global attractor {A(τ)} τ R in H 1 () H 1 (), that is, for all τ R, (1) A(τ) is compact in H 1 () H 1 (); (2) {A(τ)} τ R is invariant, that is, φ(t, τ, A(τ)) = A(t + τ), t 0; (3) {A(τ)} τ R attracts every set in D σ with respect to the H 1 () H 1 () norm; that is, for all B = {B(τ)} τ R D σ, lim dist H t 1 () H 1 ()(φ(t, τ t, B(τ t)), A(τ)) = 0. Proof. We will prove that the attractor {A(τ)} τ R in L 2 () L 2 () obtained in Theorem 4.1 is actually a D σ -pullback global attractor for φ in H 1 () H 1 (). And the proof is similar with the proof of Theorem 6.4 in [9], so we omit it. Acknowledgments. The authors express their gratitude to the anonymous referee for his/her valuable comments and suggestions, which lead to the present improved version of our paper. REFERENCES [1] B. Guo, Spectral method for solving the two-dimensional Newton-Boussinesq equations, Acta. Math. Appl. Sinica (English Ser.), 5 (1989), [2] B. Guo, Nonlinear Galerkin methods for solving two-dimensional Newton-Boussinesq equations, Chinese Ann. Math. Ser. B, 16 (1995), [3] B. Guo and B. Wang, Gevrey class regularity and approximate inertial manifolds for the Newton-Boussinesq equations, Chinese Ann. Math. Ser. B, 19 (1998), [4] B. Guo and B. Wang, Approximate inertial manifolds to the Newton-Boussinesq equations, J. Partial Differential Equations, 9 (1996), [5] G. Fucci, B. Wang and P. Singh, Asymptotic behavior of the Newton-Boussinesq equation in a two-dimensional channel, Nonlinear Anal., 70 (2009), [6] T. Caraballo, G. Lukaszewicz and J. Real, Pullback attractors for asymptotically compact non-autonomous dynamical systems, Nonlinear Anal., 64 (2006), [7] T. Caraballo, G. Lukaszewicz and J. Real, Pullback attractors for non-autonomous 2D- Navier-Stokes equations in some unbounded domains, C. R. Acad. Sci. Paris, 342 (2006), [8] B. Wang, Pullback attractors for the non-autonomous FitzHugh-Nagumo system on unbounded domains, Nonlinear Anal., 70 (2009), [9] B. Wang and R. Jones, Asymptotic behavior of a class of non-autonomous degenerate parabolic equations, Nonlinear Anal., 72 (2010), Received September 2010; revised May address: shirley 212@126.com address: yrhou@mail.xjtu.edu.cn