The existence of a global attractor for the -dimensional long wave–short wave resonance...

Nonlinear Analysis 73 (2010) 3767–3778

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Nonlinear Analysis

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The existence of a global attractor for the (2+ 1)-dimensional longwave–short wave resonance interaction equationHong Lu, Jie Xin ∗School of Mathematics and Information, Ludong University, Yantai City, Shandong Province, 264025, PR China

a r t i c l e i n f o

Article history:Received 4 January 2010Accepted 7 August 2010

MSC:35B4137L05

Keywords:Long wave–short wave resonanceinteraction equationGlobal attractorDecomposition of the solution semigroup

a b s t r a c t

In this paper, we prove the existence of a global attractor and an asymptotic smoothingeffect of the solution for the (2 + 1)-dimensional long wave–short wave resonanceinteraction equation.

© 2010 Elsevier Ltd. All rights reserved.

1. Introduction

Long wave–short wave resonance equations arise in the study of the interaction of surface waves with both gravity andcapillary modes present, and also in the analysis of internal waves as well as Rossby waves as in [1]. In plasma physicsthey describe the resonance of high-frequency electron plasma oscillation, and the associated low-frequency ion densityperturbation as in [2]. In the Hamiltonian case they are of the inverse scattering type and have solitary waves as in [3,4].Due to their rich physical andmathematical properties the longwave–short wave resonance equations have drawn great

attention frommanyphysicists andmathematicians. For one-dimensional longwave–shortwave (LS)-type equations, BolingGuo obtained the global solution in [5]. The existence of a global attractor was studied in [6–8].The (2+1)-dimensional longwave–short wave equations describing the interaction of a longwavewith a packet of short

waves arise in fluid mechanics. In [9], Oikawa et al. obtained the interaction equations describing the resonance interactionbetween a long interfacial wave and a short surface wave of (2 + 1) dimensions (two spatial and one temporal) in a two-layer fluid for the case in which they propagate obliquely to each other. In [10], a global smooth solution for the initial valueproblem for the (2+ 1)-dimensional long wave–short wave resonance interaction equations was proved.In this paper, we consider the following (2+ 1)-dimensional long wave–short wave resonance interaction equations:

i(ut + uy)−4u+ nu+ iαu = f (x, y), (1.1)

nt + βn+ γ |u|2x = g(x, y) (1.2)

with the initial condition

u(x, y, 0) = u0(x, y), n(x, y, 0) = n0(x, y), −∞ < x <∞, −∞ < y <∞ (1.3)

∗ Corresponding author. Tel.: +86 535 6681439; fax: +86 535 6681264.E-mail address: [email protected] (J. Xin).

0362-546X/$ – see front matter© 2010 Elsevier Ltd. All rights reserved.doi:10.1016/j.na.2010.08.001

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3768 H. Lu, J. Xin / Nonlinear Analysis 73 (2010) 3767–3778

where the unknown complex valued function u(x, y, t) is a short surfacewave packet, and the unknown real valued functionn(x, y, t) is a long interfacial wave. The functions f (x, y) and g(x, y) are given real functions. The constants α, β are positive,and γ ∈ R \ {0} is real.Throughout this paper, we denote the spaces of complex valued and real valued functions by the same symbols. We

denote by ‖ · ‖ the norm of H = L2(R2) with usual inner product (·, ·), denote by ‖ · ‖p the norm of Lp(R2) for all1 6 p 6∞ (‖ · ‖2 = ‖ · ‖) and denote by ‖ · ‖Hk the norm of a usual Sobolev space H

k(R2) for all 1 6 k <∞.In this paper, we use the following periodic boundary value problem (1.1), (1.2) and (1.4) to approximate the initial value

problem (1.1)–(1.3) by using uniform estimates and a priori estimates in Section 2 and the method of [11]:u(x, y, 0) = u0(x, y), n(x, y, 0) = n0(x, y), (x, y) ∈ R2u(x+ 2D, y, t) = u(x, y+ 2D, t) = u(x, y, t),n(x+ 2D, y, t) = n(x, y+ 2D, t) = n(x, y, t), D > 0, (x, y) ∈ R2, t > 0.

(1.4)

We arrange thework as follows. In Section 2, we obtain the a priori estimates and the bounded absorbing set. In Section 3,we obtain the existence of global smooth solutions by using Galerkin’s method, and the continuity of the semigroup. InSection 4, we decompose the solution semigroup into two parts, one being relatively compact in X1(H2 × H1(R2)) and theother uniformly bounded in X2(H3 × H2(R2)). In Section 5, we get the asymptotical smoothing of S(t), so we prove theexistence of a global attractor.

2. A priori estimates

Lemma 2.1. If f ∈ L2, then for the solution of the problem (1.1)–(1.3), we have

u ∈ L∞([0,+∞); L2(R2))

and

‖u‖2 6 ‖u0‖2e−αt +‖f ‖2

α2(1− e−αt). (2.1)

Moreover, there exist positive constants C and t0(R) such that

‖u‖ 6 C, ∀t > t0(R),

whenever ‖u0‖ 6 R.

Proof. Taking the inner product of (1.1) with 2uwe get

(iut + iuy −4u+ nu+ iαu, 2u) = (f , 2u). (2.2)

Taking the imaginary part of (2.2), we get

ddt‖u‖2 + 2α‖u‖2 = 2 Im(f , u) 6 2‖f ‖‖u‖ 6 α‖u‖2 +

‖f ‖2

α.

By the Gronwall inequality, we have

‖u‖2 6 ‖u0‖2e−αt +‖f ‖2

αte−αt 6 ‖u0‖2e−αt +

‖f ‖2

α2(1− e−αt).

Thus, we derive (2.1). �

Here, and throughout, we use T to denote an arbitrary positive constant, and use C to denote different positive constants,which depend only on the initial data and the positive constant T .

Lemma 2.2. Suppose that f , g ∈ L2, u0 ∈ H1 and n0 ∈ L2, u and n solve the problem (1.1)–(1.3). Then we have

(u, n) ∈ L∞([0,+∞);H1 × L2(R2)),

and

‖∇u‖2 + ‖n‖2 6 C . (2.3)

Moreover, there exist positive constants ρ and t1(R) such that

‖∇u‖2 + ‖n‖2 6 ρ2, ∀t > t1(R),

whenever ‖u0‖ + ‖n0‖ 6 R.

H. Lu, J. Xin / Nonlinear Analysis 73 (2010) 3767–3778 3769

Proof. Taking the inner product of (1.1) with 2(ut + αu), we have

(iut + iuy −4u+ nu+ iαu, 2(ut + αu)) = (f , 2(ut + αu)). (2.4)

Taking the real part of (2.4), we get

ddt

(‖∇u‖2 +

∫n|u|2dxdy− 2 Re

∫f udxdy− Im(uy, u)

)+ 2α‖∇u‖2 + 2α

∫n|u|2dxdy− 2α Re

∫f udxdy− 2α Im(uy, u)−

∫|u|2ntdxdy = 0. (2.5)

By (1.2), we have

−

∫|u|2ntdxdy = −

∫|u|2(−βn− γ |u|2x + g)dxdy

= β

∫n|u|2dxdy−

∫g|u|2dxdy. (2.6)

Taking the inner product of (1.2) with 2n, we have

ddt‖n‖2 + 2β‖n‖2 + 2γ

∫n|u|2xdxdy = 2

∫gndxdy. (2.7)

Note that, by (1.1),∫n|u|2xdxdy = 2 Re

∫ux(un)dxdy

= −2 Re∫ux(iut + iuy −4u+ iαu− f )dxdy

= 2 Im∫uxutdxdy+ 2 Im

∫uxuydxdy+ 2α Im

∫uuxdxdy+ 2 Re

∫f uxdxdy

=ddtIm∫uuxdxdy+ 2 Im

∫uxuydxdy+ 2α Im

∫uuxdxdy+ 2 Re

∫f uxdxdy. (2.8)

By adding (2.5) and (2.7), and putting (2.6) and (2.8) into the result, we get

ddtH(ξ(t))+ F(ξ(t)) = 0, (2.9)

where

H(ξ(t)) = ‖∇u‖2 + ‖n‖2 +∫n|u|2dxdy− 2 Re

∫f udxdy− Im(uy, u)+ 2γ Im

∫uuxdxdy, (2.10)

F(ξ(t)) = 2α‖∇u‖2 + β‖n‖2 + (2α + β)∫n|u|2dxdy− 2α Re

∫f udxdy− 2α Im(uy, u)

+ 4αγ Im∫uuxdxdy− 2

∫gndxdy+ 4γ Re

∫f uxdxdy+ 4γ Im

∫uxuydxdy−

∫g|u|2dxdy. (2.11)

Here, we have used the notation ξ(t) = (u(t), n(t)).

Note that

H(ξ(t)) > ‖∇u‖2 + ‖n‖2 +∫n|u|2dxdy− 2‖f ‖‖u‖ − ‖∇u‖‖u‖ − 2|γ |‖u‖‖∇u‖

>

(12− |γ |ε

)‖∇u‖2 + ‖n‖2 +

∫n|u|2dxdy− C(‖f ‖2 + ‖u‖2), (2.12)

H(ξ(t)) 6 ‖∇u‖2 + ‖n‖2 +∫n|u|2dxdy+ 2‖f ‖‖u‖ + ‖∇u‖‖u‖ + 2|γ |‖u‖‖∇u‖

6

(|γ | +

32

)‖∇u‖2 + ‖n‖2 +

∫n|u|2dxdy+ C(‖f ‖2 + ‖u‖2),


F(ξ(t)) > 2α‖∇u‖2 + 2β‖n‖2 + (2α + β)∫n|u|2dxdy− α(‖f ‖2 + ‖u‖2)

− 4α|γ |(C(ε1)‖u‖2 + ε1‖∇u‖2)− C(ε2)‖g‖2 − ε2‖n‖2 − 4|γ |(C(ε3)‖f ‖2

+ ε3‖∇u‖2)− 4|γ |‖∇u‖2 − C(ε4)‖u‖2 − ε4‖∇u‖2 − 2α(ε5‖∇u‖2 + C(ε5)‖u‖2)> (2α − 4α|γ |ε1 − 4|γ |ε3 − ε4 − 4|γ | − 2αε5)‖∇u‖2 + (2β − ε2)‖n‖2

+ (2α + β)∫n|u|2dxdy− C(‖f ‖2 + ‖u‖2 + ‖g‖2).

Without loss of generality, wemay assume that |γ | < 2α. Let ε1, ε2, ε3, ε4, ε5 be sufficiently small; then there exists α1 > 0such that

α1H(ξ(t)) 6 F(ξ(t))+ C(‖f ‖2 + ‖u‖2 + ‖g‖2). (2.13)

By (2.9) and (2.13), we get

ddtH(ξ(t))+ α1H(ξ(t)) 6 C .


H(ξ(t)) 6 H(ξ0)e−α1t +cα 1(1− e−α1t), (2.14)

where ξ0 = (u0, n0). By (2.12) and the proof of Lemma 2.1, again we have

‖∇u‖2 + ‖n‖2 6 CH(ξ0)e−α1t + C . (2.15)

The proof of the lemma is completed. �

Lemma 2.3. Suppose that f , g ∈ H1, u0 ∈ H2 and n0 ∈ H1, u and n solve the problem (1.1)–(1.3). Then we have

(u, n) ∈ L∞([0,+∞);H2 × H1(R2)).


‖u‖H2 + ‖n‖H1 6 C, ∀ t > t2(R),

whenever ‖u0‖H2 + ‖n0‖H1 6 R.

Proof. Taking the inner product of (1.1) with 2(4ut + α4u), and taking the real part, we have

ddt

(‖4u‖2 − 2 Re

∫nu4udxdy+ 2 Re

∫f4udxdy+ Im(uy,4u)

)+ 2α‖4u‖2 + 2 Re

∫ntu4udxdy

+ 2 Re∫nut4udxdy+ 2α Im(uy,4u)− 2α Re

∫nu4udxdy+ 2α Re

∫f4udxdy = 0. (2.16)

From (1.1) we infer that

Re∫nut4udxdy = −Re

∫i(f − iuy +4u− nu− iαu)n4udxdy

= Im∫fn4udxdy− Re

∫nuy4udxdy− Im

∫n2u4udxdy− α Re

∫un4udxdy. (2.17)

From (1.2) we have

Re∫ntu4udxdy = Re

∫u(−βn− γ |u|2x + g)4udxdy

= −β Re∫un4udxdy− γ Re

∫u|u|2x4udxdy+ Re

∫gu4udxdy. (2.18)

Inserting the above two equalities into (2.16), we obtain

ddt

(‖4u‖2 − 2 Re

∫nu4udxdy+ 2 Re

∫f4udxdy+ Im(uy,4u)

)+ 2α‖4u‖2 − (4α + 2β) Re

∫nu4udxdy+ 2α Re

∫f4udxdy+ 2α Im(uy,4u)+ 2 Im

∫fn4udxdy

− 2 Im∫n2u4udxdy− 2γ Re

∫u|u|2x4udxdy+ 2 Re

∫gu4udxdy− 2 Re

∫nuy4udxdy = 0. (2.19)


By (1.2), we have

(∇(nt + βn+ γ |u|2x − g), 2∇n) = 0,

that is,

ddt‖∇n‖2 + 2β‖∇n‖2 + 4γ Re

∫(∇uxu+ ux∇u)∇ndxdy− 2

∫∇g∇ndxdy = 0. (2.20)

Note that

Re∫∇uxu∇ndxdy = Re

∫∇(un)∇uxdxdy− Re

∫∇u∇uxndxdy

and

Re∫∇(un)∇uxdxdy = Re

∫∇(f − iut − iuy +4u− iαu)∇uxdxdy

= Re∫∇f∇uxdxdy+ Im

∫∇ut∇uxdxdy+ Im

∫∇uy∇uxdxdy+ α Im

∫∇u∇uxdxdy,

while integrating by parts yields

Im∫∇ut∇uxdxdy =

ddtIm∫∇u∇uxdxdy− Im

∫∇u∇uxtdxdy

=ddtIm∫∇u∇uxdxdy− Im

∫∇ut∇uxdxdy.

So we have

Im∫∇ut∇uxdxdy =

12ddtIm∫∇u∇uxdxdy.

Thus

Re∫∇(un)∇uxdxdy = Re

∫∇f∇uxdxdy+

12ddtIm∫∇u∇uxdxdy+ Im

∫∇uy∇uxdxdy+ α Im

∫∇u∇uxdxdy.

Therefore,

Re∫∇uxu∇ndxdy =

12ddtIm∫∇u∇uxdxdy+ α Im

∫∇u∇uxdxdy+ Re

∫∇f∇uxdxdy

+ Im∫∇uy∇uxdxdy− Re

∫∇u∇uxndxdy.

Inserting the above inequalities into (2.20), we have

ddt

(‖∇n‖2 + 2γ Im

∫∇u∇uxdxdy

)+ 2β‖∇n‖2 + 4αγ Im

∫∇u∇uxdxdy+ 4γ Re

∫∇f∇uxdxdy

+ 4γ Im∫∇uy∇uxdxdy− 4γ Re

∫∇u∇uxndxdy+ 4γ Re

∫ux∇u∇ndxdy− 2

∫∇g∇ndxdy = 0. (2.21)

Then, by adding (2.19) and (2.21), we get

ddtH1(ξ(t))+ 2αH1(ξ(t)) = K1(ξ(t)), (2.22)

where

H1(ξ(t)) = ‖4u‖2 + ‖∇n‖2 − 2 Re∫nu∇udxdy+ Im(uy,4u)+ 2 Re

∫f4udxdy

+ 2γ Im∫∇u∇uxdxdy, (2.23)


K1(ξ(t)) = 2(α − β)‖∇n‖2 + 2β∫nu∇udxdy+ 2α Re

∫f4udxdy− 2 Im

∫fn4udxdy

+ 2 Im∫n2u4udxdy+ 2γ Re

∫u|u|2x4udxdy− 2 Re

∫ug4udxdy

+ 2 Re∫nuy4udxdy− 4γ Re

∫∇f∇uxdxdy− 4γ

∫∇uy∇uxdxdy

+ 4γ Re∫∇u∇uxndxdy− 4γ Re

∫ux∇u∇ndxdy+ 2

∫∇g∇ndxdy. (2.24)

The indefinite sign terms in K1(ξ(t)) are estimated as∣∣∣∣Re ∫ u|u|2x4udxdy∣∣∣∣ 6 ‖4u‖‖u‖28‖∇u‖4 6 C‖4u‖ 32 ‖∇u‖2‖u‖ 12 6 ε1‖4u‖2 + C(ε1)‖u‖16H1 ,∣∣∣∣Re ∫ ug4udxdy∣∣∣∣ 6 ‖ug‖‖4u‖ 6 ‖4u‖‖g‖4‖u‖4 6 ε1‖4u‖2 + C(ε1)‖u‖2H1 ,∣∣∣∣Re ∫ ∇f∇uxdxdy∣∣∣∣ 6 ‖∇f ‖‖∇ux‖ 6 ε1‖4u‖2 + C(ε1),∣∣∣∣Re ∫ ∇u∇uxndxdy∣∣∣∣ 6 C‖4u‖‖∇u‖4‖n‖4 6 C‖4u‖ 32 ‖∇n‖ 12 6 ε1‖4u‖2 + C(ε1)‖∇n‖2,where ε1 > 0 is a arbitrary number. The rest of the terms can be controlled similarly and more simply. Recall that we haveassumed α 6 β . Thus, by choosing a suitable ε1, we have

H1(ξ(t)) >12(‖4u‖2 + ‖∇n‖2)− C, (2.25)

and there exists β2 > 0 such that

K1(ξ(t)) 612β2(‖4u‖2 + ‖∇n‖2)+ C 6 β2H1(ξ(t))+ C, (2.26)

where C = C(‖f ‖H1 , ‖g‖H1 , ‖u‖H1 , ‖n‖). So we obtain

ddtH1(ξ(t))+ β2H1(ξ(t)) 6 C . (2.27)


H1(ξ(t)) 6 H1(ξ0)e−β2t +Cβ2(1− e−β2t), (2.28)

where ξ0 = (u0, n0). By (2.25), again we have

‖4u‖2 + ‖∇n‖2 6 2H1(ξ0)e−β2t + C . (2.29)


Similarly, we have the following lemma.

Lemma 2.4. Suppose that f , g ∈ H2, u0 ∈ H3 and n0 ∈ H2, u and n solve the problem (1.1)–(1.3). Then we have

(u, n) ∈ L∞([0,+∞);H3 × H2(R2)).


‖u‖H3 + ‖n‖H2 6 C, ∀t > t3(R),

whenever ‖u0‖H3 + ‖n0‖H2 6 R.

3. Continuity of the semigroup

We use the periodic initial value problem (1.1), (1.2) and (1.4) to approximate the initial value problem (1.1)–(1.3) byusing the uniform estimates of Section 2 and the method of [11–13]; we get the following theorem.


Theorem 3.1. Suppose that f , g ∈ H1. For any (u0, n0) ∈ X1, (1.1)–(1.3) admit a unique global solution

(u, n) ∈ L∞([0,+∞); X1).

Moreover, (1.1) and (1.2) define a continuous dynamic system S(t) on X1, which has a bounded absorbing set B1 ⊂ X1.

Proof. Firstly we can get the existence of a local solution, and then by using the a priori estimate, extend the local solutionto a global solution. By the continuity of S(t), we derive the uniqueness. In fact, suppose that there are the two solutions(u1, n1) and (u2, n2) of (1.1)–(1.3). Let

(u, n) = (u1 − u2, n1 − n2), (u0, n0) = (u01 − u02, n01 − n02).

Then (u, n) satisfy

i(ut + uy)−4u+ n1u1 − n2u2 + iαu = 0, (3.1)

nt + βn+ γ (|u1|2 − |u2|2)x = 0, (3.2)

u(x, y, 0) = 0, n(x, y, 0) = 0. (3.3)

Note that n1u1 − n2u2 = nu1 + n2u, |u1|2x − |u2|2x = uxu1 + u2xu+ uu1x + u2ux. It is easy to prove

‖u‖2H2 + ‖n‖2H1 6 C(‖u0‖

2H2 + ‖n0‖

2H1)e

ct .

So S(t) is continuous, and the existence of a bounded absorbing set was proved in Lemma 2.3. �

Similarly, we have:

Theorem 3.2. Suppose that f , g ∈ H1. For any (u0, n0) ∈ X2, (1.1)–(1.3) admit a unique global solution

(u, n) ∈ L∞([0,+∞); X2).

Moreover, (1.1) and (1.2) define a continuous dynamic system S(t) on X2, which has a bounded absorbing set B2 ⊂ X2.

4. Decomposition of the semigroup

S(t) is a semigroup, since f , g ∈ H2 are independent of t . Assume B ⊂ X2(H3×H2(R2)) is a bounded set; then S(t)B ⊂ X2is bounded too. In this section, we shall decompose S(t) into two parts, and prove the asymptotic smoothing of S(t) bydecomposing S(t). In other words, for one part, S1(t), of S(t), α(S1(t)B) → 0, as t → ∞; for the other part, S2(t), it isrelatively compact in X1.For a set A ⊂ X1, its measure of non-compactness is defined as

α(A) = inf{d > 0|A has a finite open cover of sets of diameter < d}.

So

α(S(t)B) 6 α(S1(t)B)+ α(S2(t)B) = α(S1(t)B)→ 0, t →∞.

Let B ⊂ X2, supξ∈B ‖ξ‖X2 6 R, and (u, n) = S(t)(u0, n0) solve the problem (1.1)–(1.3) with initial value (u0, n0) ∈ B. Wenote that (u, n) is uniformly bounded in X2.LetXL(z) ∈ C∞0 (R

2), z = (x, y) ∈ R2, 0 < XL 6 1, and

XL(z) ={1, |z| 6 L;0, |z| > 1+ L.

So for any η ∈ R2, |η| ∈ (0, 1], there exists L(|η|) > 0 (sufficiently large) such that

‖f − fη‖H2 6 |η|, fη = fXL(η);

‖g − gη‖H2 6 |η|, gη = gXL(η);

‖|u|2x − |u|2xXL(η)‖H2 6 |η|.

Assume (uη, nη) are the solutions of the following problem:

i(uηt + uηy)−4uη + nuη + iαuη − i|η|4uη = f − fη − i|η|4u, (4.1)

nηt + βnη + γ |u|2x(1−XL) = g − gη, (4.2)

uη(x, y, 0) = u0(x, y), nη(x, y, 0) = n0(x, y). (4.3)


Assume that S1(t)(u0, n0) = (uη, nη), (ϕη, ψη) = S2(t)(u0, n0) = S(t)(u0, n0) − S1(t)(u0, n0) = (u − uη, n − nη); then(ϕη, ψη) satisfy

i(ϕηt + ϕηy)−4ϕη + nϕη + iαϕη − i|η|4ϕη = fη, (4.4)

ψηt + βψη + γ |u|2xXL = gη, (4.5)

ϕη(x, y, 0) = 0, ψη(x, y, 0) = 0. (4.6)

Lemma 4.1. There exists a positive constant C, such that

‖uη‖2 + ‖∇uη‖2 + ‖4uη‖2 + ‖nη‖2 + ‖∇nη‖2 6 C(|η|), ∀t > t∗(|η|), |η| ∈ (0, 1].

Proof. Taking the inner product of (4.1) with 2uη , and taking the imaginary part, we have

ddt‖uη‖2 + 2α‖uη‖2 + 2|η|‖∇uη‖2 = 2 Im(f − fη, uη)+ 2|η|Im(∇u,∇uη)

6 α‖uη‖2 + |η|‖∇uη‖2 + C‖f − fη‖2 + |η|‖∇u‖2.

Soddt‖uη‖2 + α‖uη‖2 + |η|‖∇uη‖2 6 C(|η|).


‖uη‖2 6 ‖u0‖2e−αt +c(|η|)α

(1− e−αt).

For any t > 0, 0 < |η| 6 1, we have ‖uη‖2 6 C , and there exists t1 = t1(R) > 0 such that

‖uη‖2 6 C(|η|), ∀t > t1, |η| ∈ (0, 1],

while ‖u0‖2e−αt 6 Re−αt 6 |η|.Taking the inner product of (4.1) with −4uη , taking the imaginary part, and using the Gagliardo–Nirenberg inequality,

we haveddt‖∇uη‖ + 2α‖∇uη‖2 + 2|η|‖4uη‖2 = −2 Im(f − fη,4uη)+ 2|η|Im(4u,4uη)+ 2 Im(nuη,4uη)

6 2‖f − fη‖‖4uη‖ + 2‖4uη‖‖n‖4‖uη‖4 + 2|η|‖4u‖‖4uη‖

6 α‖∇uη‖2 + |η|‖4uη‖2 + C(‖f − fη‖2 + ‖n‖2‖∇n‖2 + ‖4u‖2).

Soddt‖∇uη‖ + α‖∇uη‖2 + |η|‖4uη‖2 6 C(|η|).


‖∇uη‖2 6 ‖∇u(0)‖2e−αt +C(|η|)α

(1− e−αt).

For any t > 0, 0 < |η| 6 1, we have ‖∇uη‖2 6 C , and there exists t2 = t2(R) > t1(|η|) such that

‖∇uη‖2 6 C(|η|), ∀t > t2, |η| ∈ (0, 1],

while ‖∇u(0)‖2e−αt 6 Re−αt 6 |η|.Taking the inner product of (4.1) with 242 uη , and taking the imaginary part, we have

ddt‖4uη‖ + 2α‖4uη‖2 + 2|η|‖∇4uη‖2

= 2 Im(4f −4fη,4uη)+ 2|η|Re(∇4u,∇4uη)− 2 Im(4nuη,4uη)− 2 Im(∇n∇uη,4uη)6 2‖4f −4fη‖‖4uη‖ + 2‖4n‖‖uη‖4‖4uη‖4 + 2|η|‖∇4u‖‖∇4uη‖ + 2‖4uη‖‖∇n‖4‖∇uη‖4

6 α‖4uη‖2 + |η|‖∇4uη‖2 + C(‖f − fη‖2H2 + ‖4n‖43 ‖uη‖2 + ‖n‖H2‖∇uη‖

2).

Soddt‖4uη‖ + 2α‖4uη‖2 + 2|η|‖∇4uη‖2 6 C(|η|)+ C(‖4n‖

43 ‖uη‖2 + ‖n‖H2‖∇uη‖

2).



‖4uη‖2 6 ‖4u(0)‖2e−αt +1α{C(‖4n‖

43 ‖uη‖2 + ‖n‖H2‖∇uη‖

2)+ C(|η|)}(1− e−αt).

Note that ‖∇n‖, ‖4n‖ are bounded, and ‖uη‖2 6 C(|η|), ‖∇uη‖2 6 C(|η|) (∀t > t2), so for any t > 0, 0 < |η| 6 1, we have‖4uη‖2 6 C , and there exist t3 = t3(R) > t2(|η|) such that

‖4uη‖2 6 C(|η|), ∀t > t3, |η| ∈ (0, 1],

while ‖4u(0)‖2e−αt 6 Re−αt 6 |η|.Taking the inner product of (4.2) with 2nη , we get

ddt‖nη‖2 + 2β‖nη‖2 6 2|γ |‖|u|2x − |u|

2xXL‖‖nη‖ + 2‖g − gη‖‖nη‖

6 β‖nη‖2 + C(|η|).


‖nη‖2 6 ‖n0‖2e−βt +C(|η|)β

(1− e−βt).

So there exists t4 = t4(R) such that

‖nη‖2 6 C(|η|), ∀ t > t4, |η| ∈ (0, 1],

while ‖n0‖2e−βt 6 Re−βt 6 |η|.Taking the inner product of (4.2) with 24nη , we get

ddt‖∇nη‖2 + 2β‖∇nη‖2 = 2(∇(γ |u|2x(1−XL)),−∇nη)+ 2(∇(g − gη),∇nη)

6 β‖∇nη‖2 + C(‖|u|2x(1−XL)‖H1 + ‖g(1−XL)‖H1)

6 β‖∇nη‖2 + C(|η|).

So

ddt‖∇nη‖2 + β‖∇nη‖2 6 C(|η|).


‖∇nη‖2 6 ‖∇n(0)‖2e−βt +C(|η|)β

(1− e−βt).

For any t > 0, 0 < |η| 6 1, we have ‖∇uη‖2 6 C , and there exists t5 = t5(R) > t4 such that

‖∇nη‖2 6 C(|η|), ∀t > t5, |η| ∈ (0, 1],

while ‖∇n(0)‖2e−βt 6 Re−βt 6 |η|. Choosing t∗(|η|) = max{t1, t2, t3, t4, t5}, we conclude that

‖uη‖2, ‖∇uη‖2, ‖4uη‖2, ‖nη‖2, ‖∇nη‖2 6 C(|η|), ∀t > t∗(|η|), |η| ∈ (0, 1].


Lemma 4.2. There exist positive constants C1(|η|), C2(|η|) such that

‖|z|ϕη‖2 + ‖|z|∇ϕη‖2 + ‖|z|4ϕη‖2 6 C1(|η|),

‖|z|ψη‖2 + ‖|z|∇ψη‖2 6 C2(|η|), ∀|η| ∈ (0, 1].

Proof. Taking the inner product of (4.4) with 2|z|2ϕη , and taking the imaginary part, we have

ddt‖|z|ϕη‖2 + 2α‖|z|ϕη‖2 + 2|η|‖|z|∇ϕη‖2 = Im(fη, |z|2ϕη)+ 4|η|Re(z∇ϕη, ϕη)

− 4 Im(z∇ϕη, ϕη)+∫|z|2y|ϕη|

2dz. (4.7)


The right side of (4.7) 6 (4+ 4|η|)‖|z|∇ϕη‖‖ϕη‖ + 2‖|z|fη‖ + C‖|z|ϕη‖2

6 |η|‖|z|∇ϕη‖2 +(4|η|+ 4

)‖ϕη‖

2+ α‖|z|ϕη‖2 +

2α‖|z|fη‖2

6 |η|‖|z|∇ϕη‖2 + α‖|z|ϕη‖2 +(4|η|+ 4

)‖ϕη‖

2+2α‖|z|fη‖2.

Soddt‖|z|ϕη‖2 + α‖|z|ϕη‖2 + |η|‖|z|∇ϕη‖2 6

(4|η|+ 4

)‖ϕη‖

2+2α‖|z|fη‖2.

Since fη possesses a compact support set, ‖|z|fη‖ is bounded. And ‖ϕη‖ = ‖u − uη‖ 6 ‖u‖ + ‖uη‖ is bounded. By theGronwall inequality, we have

‖|z|ϕη‖2 6 C1(|η|), ∀t > 0.

Taking the inner product of (4.4) with 2|z|24ϕη , and taking the imaginary part, we have

ddt‖|z|∇ϕη‖2 + 2α‖|z|∇ϕη‖2 + 2|η|‖|z|4ϕη‖2

= −4 Im(zϕηt ,∇ϕη)+ 2 Im(|z|ϕηy,4ϕη)+ 2 Im∫|z|2nϕη4ϕηϕηdz − 4α Re

∫zϕη∇ϕηdz − 2 Im

∫fη|z|24ϕηdz.

Note that∣∣∣∣−4 Im(zϕηt ,∇ϕη)− 4α Re ∫ zϕη∇ϕηdz∣∣∣∣ 6 4‖ϕηt‖‖|z|∇ϕη‖ + 4α‖ϕη‖‖|z|∇ϕη‖6α

2‖|z|∇ϕη‖2 + C(‖ϕηt‖2 + ‖ϕη‖2),∣∣∣∣2 Im ∫ |z|2nϕη4ϕηϕηdz − 4α Re ∫ zϕη∇ϕηdz∣∣∣∣ 6 2‖n‖∞‖|z|ϕη‖‖|z|4ϕη‖ + ‖|z|fη‖‖|z|4ϕη‖6|η|

2‖|z|4ϕη‖2 + C(‖n‖H2‖|z|ϕη‖

2+ ‖|z|fη‖2),

|2 Im(|z|ϕηy,4ϕη)| 6 2‖|z|∇ϕη‖‖|z|4ϕη‖

6α

2‖|z|∇ϕη‖2 +

|η|

2‖|z|4ϕη‖2.

Since (u, n) are bounded in H3 × H2(R2), and (uη, nη) are bounded in H2 × H1(R2), (φη, ψη) = (u − uη, n − nη) are alsobounded in H2 × H1(R2). So

‖ϕη‖H2 , ‖|z|ϕη‖, ‖|z|fη‖ 6 C(|η|), ‖n‖, ‖n‖H2 6 C .

By (4.4), we have

ϕηt = −ifη − ϕηy − i4ϕη + inϕη − αϕη + |η|4ϕη.

So

‖ϕηt‖ 6 ‖fη‖ + ‖∇ϕη‖ + ‖4ϕη‖ + ‖n‖∞‖ϕη‖ + α‖ϕη‖ + |η|‖4ϕη‖ 6 C(|η|), ∀t > 0.

Then we haveddt‖|z|∇ϕη‖2 + α‖|z|∇ϕη‖2 + |η|‖|z|4ϕη‖2 6 C(|η|).


‖|z|∇ϕη‖2 6 C1(|η|), ∀t > 0.

By (4.4), we have

i(4ϕηt +4ϕηy)−42 ϕη + n4ϕη + iα4ϕη − i|η| 42 ϕη = 4fη −4nϕη − 2∇n∇ϕη. (4.8)

Taking the inner product of (4.8) with 2|z|24ϕη , and taking the imaginary part, we have

ddt‖|z|4ϕη‖2 + 2α‖|z|4ϕη‖2 + 2|η|‖|z|∇4ϕη‖2

= −2 Im∫|z|24ϕηy44ϕηdz − 4 Im

∫z4ϕη∇4ϕηdz − 4|η|Re

∫z4ϕη∇4ϕηdz


+ 2 Im∫4fη|z|24ϕηdz − 2 Im

∫|z|24nϕη4ϕηdz − 4 Im

∫∇n|z|2∇ϕη4ϕηdz

6 (6+ 4|η|)‖|z|∇4ϕη‖‖4ϕη‖ + 2‖|z|4fη‖‖4ϕη‖‖|z|4ϕη‖

+ 2 Im∫∇n(2zϕη4ϕη + |z|2∇ϕηϕη + |z|2ϕη∇4ϕη)dz + 4 Im

∫∇n|z|24ϕη4ϕηdz

6 α‖|z|4ϕη‖2 + |η|‖|z|∇4ϕη‖2 + C(‖∇n‖‖|z|ϕη‖4‖4ϕη‖4 + ‖∇n‖‖|z|∇ϕη‖4‖|z|4ϕη‖4+‖4n‖‖|z|ϕη‖4‖|z|4ϕη‖4 + ‖∇n‖‖|z|∇ϕη‖4‖|z|4ϕη‖4 + ‖∇n‖‖|z|ϕη‖4‖4ϕη‖4)

6 α‖|z|4ϕη‖2 + |η|‖|z|∇4ϕη‖2 + C(|η|),

that is,ddt‖|z|4ϕη‖2 + α‖|z|4ϕη‖2 + |η|‖|z|∇4ϕη‖2 6 C(|η|).


‖|z|4ϕη‖2 6 C1(|η|).

Taking the inner product of (4.5) with 2|z|2ψη , we get

ddt‖|z|ψη‖2 + 2β‖|z|ψη‖2 = 2

∫gη|z|2ψηdz − 2γ

∫|z|2|u|2xXLψηdz

6 2‖|z|gη‖‖|z|ψη‖ + 2|γ |‖|z||u|2xXL‖‖|z|ψη‖

6 β‖|z|ψη‖2 +2β‖|z|gη‖2 +

2|γ |β‖|z||u|2xXL‖

2.

Since gη, |u|2xXL possess compact support sets, ‖|z|gη‖, ‖|z||u|2xXL‖ are bounded. By the Gronwall inequality, we have

‖|z|ψη‖2 6 C2(|η|).

Similarly, taking the inner product of (4.5) with 2|z|24ψη , we get

ddt‖|z|∇ψη‖2 + 2β‖|z|∇ψη‖2 = −2

∫zψηt∇ψηdz − 4β

∫zψη∇ψηdz − 4γ

∫|z||u|2xXL∇ψηdz

− 2∫|z|2∇gη∇ψηdz + 4

∫zgη∇ψηdz

6 2‖ψηt‖‖|z|∇ψη‖ + 4β‖ψη‖‖|z|∇ψη‖ + 2|γ |‖|z|∇(|u|2xXL)‖‖|z|∇ψη‖

+ 4|γ |‖|z|∇ψη‖‖|u|2xXL‖ + 2‖|z|∇gη‖‖|z|∇ψη‖ + 4‖gη‖‖|z|∇ψη‖.

Since n is bounded in H2(R2), nη is bounded in H1(R2). Then ψη = n − nη is bounded too in H1(R2). gη = gXL, |u|2xXL

possess compact support sets, and ‖|z|gη‖, ‖|z|∇gη‖, ‖|z||u|2xXL‖, ‖|z|∇(|u|2xXL)‖ are bounded. By (4.5), we get

ψηt = gη − βψη − γ |u|2xXL.

Then

‖ψηt‖ 6 ‖gη‖ + β‖ψη‖ + |γ |‖|u|2xXL‖, ∀t > 0.

So we get

ddt‖|z|∇ψη‖2 + 2β‖|z|∇ψη‖2 6 β‖|z|∇ψη‖2 + C(|η|).


‖|z|∇ψη‖2 6 C2(|η|), ∀t > 0.


5. The existence of a global attractor

In this section, by [14] we will get our main result.

Theorem 5.1. Assume that f , g ∈ H1. Then the solution operator S(t) is a semigroup and possesses a global attractor A ⊂

H2 × H1(R2) satisfying


(i) A is compact in H2 × H1(R2);(ii) S(t)A = A,∀t > 0;(iii) ∀B ⊂ H3 × H2(R2), limt→∞ dist(S(t)B,A) = limt→∞ supξ∈B distH2×H1(S(t)ξ ,A) = 0.

By the theory of [15,16], we get:

Lemma 5.1. Assume r > r1 (r, r1 are integral); then Hr(Rn) ∩ Hr1(Rn, (1+ |z|2)dz) is compactly imbedded into Hr1(Rn).

We can prove Theorem 5.1 by the Kuratowski α-measure. From Lemmas 4.2 and 5.1, we obtained that S2(t) from X2 toX1 is compact. Therefore for any bounded set B ⊂ S2, we have

α(S2(t)B) = 0, ∀t > 0.

From Lemma 4.1, we obtained, for ∀ε > 0, that there exist η and t0 > 0 such that

‖S1(t)(u0, n0)‖ < ε, ∀t > t0

and (u0, n0) ∈ B, B ⊂ X2 is bounded, that is, for ∀η > 0,

α(S1(t)B) 6 2ε, t > t0.

Then, we have

α(S(t)B) 6 α(S1(t)B)+ α(S2(t)B) = α(S1(t)B) 6 2ε, t > t0.

So

limt→∞

α(S(t)B) = 0.

Therefore we get asymptotical smoothing of S(t). Then we have proved Theorem 5.1, by the theory of [17].

Acknowledgements

The authors were supported by the NSF of China (No. 10626046), China Postdoctoral Science Foundation (No.20070410487), the Project of Shandong Province Higher Educational Science and Technology Program and the Project ofLudong University Discipline Construction Foundation.

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