Random Attractors for the Stochastic FitzHugh-Nagumo System on Unbounded Domains
aa r X i v : . [ m a t h . A P ] M a y Random Attractors for the Stochastic FitzHugh-Nagumo Systemon Unbounded Domains
Bixiang Wang ∗ Department of Mathematics, New Mexico Institute of Mining and TechnologySocorro, NM 87801, USAEmail: [email protected]
Abstract
The existence of a random attractor for the stochastic FitzHugh-Nagumo system defined onan unbounded domain is established. The pullback asymptotic compactness of the stochasticsystem is proved by uniform estimates on solutions for large space and time variables. Theseestimates are obtained by a cut-off technique.
Key words.
Stochastic FitzHugh-Nagumo system, random attractor, pullback attractor, asymp-totic compactness.
MSC 2000.
Primary 37L55. Secondary 60H15, 35B40, 35B41.
In this paper, we investigate the asymptotic behavior of solutions of the following stochasticFitzHugh-Nagumo system defined on R n : du + ( λu − ∆ u + αv ) dt = ( f ( x, u ) + g ) dt + φ dw , x ∈ R n , t > ,dv + ( δv − βu ) dt = hdt + φ dw , x ∈ R n , t > , where λ, α, δ and β are positive constants, g ∈ L ( R n ) and h ∈ H ( R n ) are given, φ ∈ H ( R n ) ∩ W ,p ( R n ) for some p ≥ φ ∈ H ( R n ), f is a nonlinear function satisfying certain dissipativeconditions, w and w are independent two-sided real-valued Wiener processes on a probabilityspace which will be specified later. ∗ Supported in part by NSF grant DMS-0703521
1t is known that the asymptotic behavior of a random system is determined by a pullback randomattractor. The concept of random attractors was introduced in [12, 13], which is an analogue ofglobal attractors for deterministic dynamical systems as studied in [5, 16, 23, 26, 29]. When PDEsare defined in bounded domains, the existence of random attractors has been investigated by manyauthors, see, e.g., [3, 10, 11, 12, 13] and the references therein. However, in the case of unbounded domains, the existence of random attractors is not well understood. It seems that the work [9]by Bates, Lu and Wang is the only existence result of random attractors for PDEs defined onunbounded domains, where a random attractor for the Reaction-Diffusion equation on R n wasestablished. In this paper, we will study such attractors for the stochastic FitzHugh-Nagumosystem defined on unbounded domains.Notice that Sobolev embeddings are not compact when domains are unbounded. This introducesa major obstacle for proving the existence of attractors for PDEs on unbounded domains. Forsome deterministic PDEs, such difficulty can be overcome by the energy equation approach thatwas developed by Ball in [6, 7] and used in [14, 15, 17, 20, 21, 25, 31]). Under certain circumstances,the tail-estimates method must be used to deal with the problem caused by the unboundednessof domains. This approach was developed in [30] for deterministic parabolic equations and usedin [1, 2, 4, 19, 22, 24, 27, 28]. Recently, the authors of [9] extended the tail-estimates method tothe stochastic parabolic equations. In this paper, we will use the idea of uniform estimates on thetails of solutions to investigate the asymptotic behavior of the stochastic FitzHugh-Nagumo systemdefined on R n .This paper is organized as follows. In the next section, we review the pullback random attractorstheory for random dynamical systems. In Section 3, we define a continuous random dynamicalsystem for the stochastic FitzHugh-Nagumo system on R n . Then we derive the uniform estimatesof solutions in Section 4, which include the uniform estimates on the tails of solutions. Finally, weestablish the asymptotic compactness of the random dynamical system and prove the existence ofa pullback random attractor.In the sequel, we adopt the following notations. We denote by k · k and ( · , · ) the norm and theinner product of L ( R n ), respectively. The norm of a given Banach space X is written as k · k X .We also use k · k p to denote the norm of L p ( R n ). The letters c and c i ( i = 1 , , . . . ) are generic2ositive constants which may change their values from line to line or even in the same line. In this section, we recall some basic concepts related to random attractors for stochastic dynamicalsystems. The reader is referred to [3, 8, 11, 13] for more details.Let ( X, k · k X ) be a separable Hilbert space with Borel σ -algebra B ( X ), and (Ω , F , P ) be aprobability space. Definition 2.1. (Ω , F , P, ( θ t ) t ∈ R ) is called a metric dynamical system if θ : R × Ω → Ω is( B ( R ) × F , F )-measurable, θ is the identity on Ω, θ s + t = θ t ◦ θ s for all s, t ∈ R and θ t P = P forall t ∈ R . Definition 2.2.
A continuous random dynamical system (RDS) on X over a metric dynamicalsystem (Ω , F , P, ( θ t ) t ∈ R ) is a mappingΦ : R + × Ω × X → X ( t, ω, x ) Φ( t, ω, x ) , which is ( B ( R + ) × F × B ( X ) , B ( X ))-measurable and satisfies, for P -a.e. ω ∈ Ω,(i) Φ(0 , ω, · ) is the identity on X ;(ii) Φ( t + s, ω, · ) = Φ( t, θ s ω, · ) ◦ Φ( s, ω, · ) for all t, s ∈ R + ;(iii) Φ( t, ω, · ) : X → X is continuous for all t ∈ R + .Hereafter, we always assume that Φ is a continuous RDS on X over (Ω , F , P, ( θ t ) t ∈ R ). Definition 2.3.
A random bounded set { B ( ω ) } ω ∈ Ω of X is called tempered with respect to ( θ t ) t ∈ R if for P -a.e. ω ∈ Ω, lim t →∞ e − βt d ( B ( θ − t ω )) = 0 for all β > , where d ( B ) = sup x ∈ B k x k X . Definition 2.4.
Let D be a collection of random subsets of X . Then D is called inclusion-closedif D = { D ( ω ) } ω ∈ Ω ∈ D and ˜ D = { ˜ D ( ω ) ⊆ X : ω ∈ Ω } with ˜ D ( ω ) ⊆ D ( ω ) for all ω ∈ Ω imply that˜ D ∈ D . 3 efinition 2.5. Let D be a collection of random subsets of X and { K ( ω ) } ω ∈ Ω ∈ D . Then { K ( ω ) } ω ∈ Ω is called an absorbing set of Φ in D if for every B ∈ D and P -a.e. ω ∈ Ω, there exists t B ( ω ) > t, θ − t ω, B ( θ − t ω )) ⊆ K ( ω ) for all t ≥ t B ( ω ) . Definition 2.6.
Let D be a collection of random subsets of X . Then Φ is said to be D -pullbackasymptotically compact in X if for P -a.e. ω ∈ Ω, { Φ( t n , θ − t n ω, x n ) } ∞ n =1 has a convergent subse-quence in X whenever t n → ∞ , and x n ∈ B ( θ − t n ω ) with { B ( ω ) } ω ∈ Ω ∈ D . Definition 2.7.
Let D be a collection of random subsets of X and {A ( ω ) } ω ∈ Ω ∈ D . Then {A ( ω ) } ω ∈ Ω is called a D -random attractor (or D -pullback attractor) for Φ if the following con-ditions are satisfied, for P -a.e. ω ∈ Ω,(i) A ( ω ) is compact, and ω d ( x, A ( ω )) is measurable for every x ∈ X ;(ii) {A ( ω ) } ω ∈ Ω is invariant, that is,Φ( t, ω, A ( ω )) = A ( θ t ω ) , ∀ t ≥ {A ( ω ) } ω ∈ Ω attracts every set in D , that is, for every B = { B ( ω ) } ω ∈ Ω ∈ D ,lim t →∞ d (Φ( t, θ − t ω, B ( θ − t ω )) , A ( ω )) = 0 , where d is the Hausdorff semi-metric given by d ( Y, Z ) = sup y ∈ Y inf z ∈ Z k y − z k X for any Y ⊆ X and Z ⊆ X .The following existence result on a random attractor for a continuous RDS can be found in [8, 13]. Proposition 2.8.
Let D be an inclusion-closed collection of random subsets of X and Φ a con-tinuous RDS on X over (Ω , F , P, ( θ t ) t ∈ R ) . Suppose that { K ( ω ) } ω ∈ Ω is a closed absorbing set of Φ in D and Φ is D -pullback asymptotically compact in X . Then Φ has a unique D -random attractor {A ( ω ) } ω ∈ Ω which is given by A ( ω ) = \ τ ≥ [ t ≥ τ Φ( t, θ − t ω, K ( θ − t ω )) . In this paper, we will take D as the collection of all tempered random subsets of L ( R n ) × L ( R n )and prove the stochastic FitzHugh-Nagumo system on R n has a D -random attractor.4 Stochastic FitzHugh-Nagumo system on R n In this section, we discuss the existence of a continuous random dynamical system for the stochasticFitzHugh-Nagumo system defined on R n : du + ( λu − ∆ u + αv ) dt = ( f ( x, u ) + g ) dt + φ dw , x ∈ R n , t > , (3.1) dv + ( δv − βu ) dt = hdt + φ dw , x ∈ R n , t > , (3.2)with the initial conditions: u ( x,
0) = u ( x ) , v ( x,
0) = v ( x ) , x ∈ R n , (3.3)where λ, α, δ and β are positive constants, g ∈ L ( R n ) and h ∈ H ( R n ) are given, φ ∈ H ( R n ) ∩ W ,p ( R n ) for some p ≥ φ ∈ H ( R n ), w and w are independent two-sided real-valued Wienerprocesses on a probability space which will be specified below, and f is a nonlinear function satis-fying the conditions, ∀ x ∈ R n and s ∈ R , f ( x, s ) s ≤ − α | s | p + ψ ( x ) , (3.4) | f ( x, s ) | ≤ α | s | p − + ψ ( x ) , (3.5) ∂f∂s ( x, s ) ≤ β, (3.6) | ∂f∂x ( x, s ) | ≤ ψ ( x ) , (3.7)where α , α and β are positive constants, ψ ∈ L ( R n ) ∩ L ∞ ( R n ), and ψ ∈ L ( R n ) ∩ L q ( R n ) with q + p = 1, and ψ ∈ L ( R n ).In the sequel, we consider the probability space (Ω , F , P ) whereΩ = { ω = ( ω , ω ) ∈ C ( R , R ) : ω (0) = 0 } , F is the Borel σ -algebra induced by the compact-open topology of Ω, and P the correspondingWiener measure on (Ω , F ). Then we have( w ( t, ω ) , w ( t, ω )) = ω ( t ) , t ∈ R . θ t ω ( · ) = ω ( · + t ) − ω ( t ) , ω ∈ Ω , t ∈ R . Then (Ω , F , P, ( θ t ) t ∈ R ) is a measurable dynamical system.Consider the stationary solutions of the one-dimensional equations dy + λy dt = dw ( t ) , (3.8)and dy + δy dt = dw ( t ) . (3.9)The solutions to (3.8) and (3.9) are given by y ( θ t ω ) = − λ Z −∞ e λτ ( θ t ω )( τ ) dτ, t ∈ R , and y ( θ t ω ) = − δ Z −∞ e δτ ( θ t ω )( τ ) dτ, t ∈ R . Note that there exists a θ t -invariant set ˜Ω ⊆ Ω of full P measure such that y j ( θ t ω j ) ( j = 1 , t for every ω ∈ ˜Ω, and the random variable | y j ( ω j ) | is tempered (see, e.g., [8]).Therefore, it follows from Proposition 4.3.3 in [3] that there exists a tempered function r ( ω ) > X j =1 (cid:0) | y j ( ω j ) | + | y j ( ω j ) | p (cid:1) ≤ r ( ω ) . (3.10)where r ( ω ) satisfies, for P -a.e. ω ∈ Ω, r ( θ t ω ) ≤ e η | t | r ( ω ) , t ∈ R , (3.11)where η = min { λ, δ } . Then it follows from (3.10)-(3.11) that, for P -a.e. ω ∈ Ω, X j =1 (cid:0) | y j ( θ t ω j ) | + | y j ( θ t ω j ) | p (cid:1) ≤ e η | t | r ( ω ) , t ∈ R . (3.12)Putting z j ( θ t ω ) = φ j y j ( θ t ω j ), by (3.8) and (3.9) we have dz + λz dt = φ dw , dz + δz dt = φ dw . Let ˜ u = u ( t ) − z ( θ t ω ) and ˜ v = v ( t ) − z ( θ t ω ), where ( u, v ) satisfies (3.1)-(3.3). Then for (˜ u, ˜ v ), wehave d ˜ udt + λ ˜ u − ∆˜ u + α ˜ v = f ( x, ˜ u + z ( θ t ω )) + g + ∆ z ( θ t ω ) − αz ( θ t ω ) , (3.13)and d ˜ vdt + δ ˜ v − β ˜ u = h + βz ( θ t ω ) . (3.14)By a Galerkin method as in [18], it can be proved that if f satisfies (3.4)-(3.7), then for P -a.e. ω ∈ Ω and for all (˜ u , ˜ v ) ∈ L ( R n ) × L ( R n ), system (3.13)-(3.14) has a unique solution(˜ u ( · , ω, ˜ u ) , ˜ v ( · , ω, ˜ v )) ∈ C ([0 , ∞ ) , L ( R n ) × L ( R n )) with ˜ u (0 , ω, ˜ u ) = ˜ u and ˜ v (0 , ω, ˜ v ) = ˜ v .Further, the solution (˜ u ( t, ω, ˜ u ) , ˜ v ( t, ω, ˜ v )) is continuous with respect to (˜ u , ˜ v ) in L ( R n ) × L ( R n )for all t ≥
0. Throughout this paper, we always write u ( t, ω, u ) = ˜ u ( t, ω, u − z ( ω )) + z ( θ t ω ) , v ( t, ω, v ) = ˜ v ( t, ω, v − z ( ω )) + z ( θ t ω ) . (3.15)Then ( u, v ) is a solution of problem (3.1)-(3.3) in some sense. We now define a mapping Φ : R + × Ω × ( L ( R n ) × L ( R n )) → L ( R n ) × L ( R n ) byΦ( t, ω, ( u , v )) = ( u ( t, ω, u ) , v ( t, ω, v )) , ∀ ( t, ω, ( u , v )) ∈ R + × Ω × ( L ( R n ) × L ( R n )) . (3.16)Note that Φ satisfies conditions (i), (ii) and (iii) in Definition 2.2. Therefore, Φ is a continuousrandom dynamical system associated with the stochastic FitzHugh-Nagumo system on R n . Inthe next two sections, we establish uniform estimates for Φ and prove the existence of a randomattractor. In this section, we derive uniform estimates on the solutions of the stochastic FitzHugh-Nagumosystem defined on R n when t → ∞ . These estimates are necessary for proving the existence ofbounded absorbing sets and the asymptotic compactness of the random dynamical system asso-ciated with the system. Particularly, we will show that the tails of the solutions for large spacevariables are uniformly small when time is sufficiently large.7rom now on, we always assume that D is the collection of all tempered subsets of L ( R n ) × L ( R n )with respect to (Ω , F , P, ( θ t ) t ∈ R ). The next lemma shows that Φ has an absorbing set in D . Lemma 4.1.
Assume that g, h ∈ L ( R n ) and (3.4) - (3.7) hold. Then there exists { K ( ω ) } ω ∈ Ω ∈ D such that { K ( ω ) } ω ∈ Ω is a closed absorbing set for Φ in D , that is, for any B = { B ( ω ) } ω ∈ Ω ∈ D and P -a.e. ω ∈ Ω , there is T B ( ω ) > such that Φ( t, θ − t ω, B ( θ − t ω )) ⊆ K ( ω ) for all t ≥ T B ( ω ) . Proof.
Taking the inner product of (3.13) with β ˜ u in L ( R n ) we find that12 β ddt k ˜ u k + λβ k ˜ u k + β k∇ ˜ u k + αβ (˜ v, ˜ u )= β ( f ( x, ˜ u + z ( θ t ω )) , ˜ u ) + β ( g, ˜ u ) + β (∆ z ( θ t ω ) , ˜ u ) − αβ ( z ( θ t ω ) , ˜ u ) . (4.1)Taking the inner product of (3.14) with α ˜ v in L ( R n ) we find that12 α ddt k ˜ v k + αδ k ˜ v k − αβ (˜ v, ˜ u ) = α ( h, ˜ v ) + αβ ( z ( θ t ω ) , ˜ v ) . (4.2)Adding (4.1) and (4.2), we obtain that12 ddt (cid:0) β k ˜ u k + α k ˜ v k (cid:1) + λβ k ˜ u k + αδ k ˜ v k + β k∇ ˜ u k = β ( f ( x, ˜ u + z ( θ t ω )) , ˜ u ) + β ( g, ˜ u ) + β (∆ z ( θ t ω ) , ˜ u ) − αβ ( z ( θ t ω ) , ˜ u ) + α ( h, ˜ v ) + αβ ( z ( θ t ω ) , ˜ v ) . (4.3)We now estimate every term on the right-hand side of (4.3). For the nonlinear term, by (3.4)-(3.5)we obtain that( f ( x, ˜ u + z ( θ t ω )) , ˜ u ) = ( f ( x, ˜ u + z ( θ t ω )) , ˜ u + z ( θ t ω )) − ( f ( x, ˜ u + z ( θ t ω )) , z ( θ t ω )) ≤ − α Z R n | u | p dx + Z R n ψ ( x ) dx − Z R n f ( x, u ) z ( θ t ω ) dx ≤ − α Z R n | u | p dx + Z R n ψ ( x ) dx + α Z R n | u | p − | | z ( θ t ω ) | dx + Z R n | ψ | | z ( θ t ω ) | dx ≤ − α k u k pp + k ψ k + 12 α k u k pp + c k z ( θ t ω ) k pp + 12 k ψ k + 12 k z ( θ t ω ) k ≤ − α k u k pp + c ( k z ( θ t ω ) k pp + k z ( θ t ω ) k ) + c . (4.4)8he second term on the right-hand side of (4.3) is bounded by β | ( g, ˜ u ) | ≤ β k g kk ˜ u k ≤ λβ k ˜ u k + βλ k g k . (4.5)For the third term on the right-hand side of (4.3), we have β | ( ∇ z ( θ t ω ) , ∇ ˜ u ) | ≤ β k∇ z ( θ t ω ) k + 12 β k∇ ˜ u k (4.6)Similarly, the remaining terms on the right-hand side of (4.3) is bounded by αβ | ( z ( θ t ω ) , ˜ u ) | + α | ( h, ˜ v ) | + αβ | ( z ( θ t ω ) , ˜ v ) |≤ λβ k ˜ u k + 1 λ α β k z ( θ t ω ) k + 14 αδ k ˜ v k + αδ k h k + 14 αδ k ˜ v k + αβ δ k z ( θ t ω ) k . ≤ λβ k ˜ u k + 1 λ α β k z ( θ t ω ) k + 12 αδ k ˜ v k + αδ k h k + αβ δ k z ( θ t ω ) k . (4.7)Then it follows from (4.3)-(4.7) that ddt (cid:0) β k ˜ u k + α k ˜ v k (cid:1) + λβ k ˜ u k + αδ k ˜ v k + β k∇ ˜ u k + α k u k pp ≤ c (cid:0) k z ( θ t ω ) k pp + k z ( θ t ω ) k + k∇ z ( θ t ω ) k + k z ( θ t ω ) k (cid:1) + c . (4.8)Note that z ( θ t ω ) = φ y ( θ t ω ), z ( θ t ω ) = φ y ( θ t ω ), φ ∈ H ( R n ) and φ ∈ H ( R n ) ∩ W ,p ( R n ).Therefore, the right-hand of (4.8) is bounded by c X j =1 ( | y j ( θ t ω j ) | p + | y j ( θ t ω j ) | ) + c = p ( θ t ω ) + c . (4.9)By (3.12), we find that for P -a.e. ω ∈ Ω, p ( θ τ ω ) ≤ c e η | τ | r ( ω ) , ∀ τ ∈ R . (4.10)It follows from (4.8)-(4.9) that, for all t ≥ ddt (cid:0) β k ˜ u k + α k ˜ v k (cid:1) + λβ k ˜ u k + αδ k ˜ v k + β k∇ ˜ u k + α k u k pp ≤ p ( θ t ω ) + c , (4.11)which implies that, for all t ≥ ddt (cid:0) β k ˜ u k + α k ˜ v k (cid:1) + λβ k ˜ u k + αδ k ˜ v k ≤ p ( θ t ω ) + c . η = min { λ, δ } . We get that, for all t ≥ ddt (cid:0) β k ˜ u k + α k ˜ v k (cid:1) + η (cid:0) β k ˜ u k + α k ˜ v k (cid:1) ≤ p ( θ t ω ) + c . (4.12)By Gronwall’s lemma, we find that, for all t ≥ β k ˜ u ( t, ω, ˜ u ( ω )) k + α k ˜ v ( t, ω, ˜ v ( ω )) k ≤ e − ηt (cid:0) β k ˜ u ( ω ) k + α k ˜ v ( ω ) k (cid:1) + Z t e η ( τ − t ) p ( θ τ ω ) dτ + c η . (4.13)By replacing ω by θ − t ω , we get from (4.13) and (4.10) that, for all t ≥ β k ˜ u ( t, θ − t ω, ˜ u ( θ − t ω )) k + α k ˜ v ( t, θ − t ω, ˜ v ( θ − t ω )) k ≤ e − ηt (cid:0) β k ˜ u ( θ − t ω ) k + α k ˜ v ( θ − t ω ) k (cid:1) + Z t e η ( τ − t ) p ( θ τ − t ω ) dτ + c η . ≤ e − ηt (cid:0) β k ˜ u ( θ − t ω ) k + α k ˜ v ( θ − t ω ) k (cid:1) + Z − t e ητ p ( θ τ ω ) dτ + c η . ≤ e − ηt (cid:0) β k ˜ u ( θ − t ω ) k + α k ˜ v ( θ − t ω ) k (cid:1) + c Z − t e ητ r ( ω ) dτ + c η . ≤ e − ηt (cid:0) β k ˜ u ( θ − t ω ) k + α k ˜ v ( θ − t ω ) k (cid:1) + 2 c η r ( ω ) + c η . (4.14)By (3.15) and (3.16) we haveΦ( t, ω, ( u , v )( ω )) = (˜ u ( t, ω, u ( ω ) − z ( ω )) + z ( θ t ω ) , ˜ v ( t, ω, v ( ω ) − z ( ω )) + z ( θ t ω )) . So by (4.14) we get that, for all t ≥ k Φ( t, θ − t ω, ( u , v )( θ − t ω )) k = k ˜ u ( t, θ − t ω, u ( θ − t ω ) − z ( θ − t ω )) + z ( ω ) k + k ˜ v ( t, θ − t ω, v ( θ − t ω ) − z ( θ − t ω )) + z ( ω ) k ≤ k ˜ u ( t, θ − t ω, u ( θ − t ω ) − z ( θ − t ω )) k + 2 k z ( ω ) k +2 k ˜ v ( t, θ − t ω, v ( θ − t ω ) − z ( θ − t ω )) k + 2 k z ( ω ) k ≤ (cid:18) α + 1 β (cid:19) e − ηt (cid:0) β k u ( θ − t ω ) − z ( θ − t ω ) k + α k v ( θ − t ω ) − z ( θ − t ω ) k (cid:1) k z ( ω ) k + 2 k z ( ω ) k + c r ( ω ) + c ≤ c e − ηt (cid:0) k u ( θ − t ω ) k + k v ( θ − t ω ) k + k z ( θ − t ω ) k + k z ( θ − t ω ) k (cid:1) + 2 k z ( ω ) k + 2 k z ( ω ) k + c r ( ω ) + c (4.15)Note that k z ( ω ) k , k z ( ω ) k and { B ( ω ) } ω ∈ Ω ∈ D are tempered. Therefore, for ( u , v )( θ − t ω ) ∈ B ( θ − t ω ), then there is T B ( ω ) > t ≥ T B ( ω ), c e − ηt (cid:0) k u ( θ − t ω ) k + k v ( θ − t ω ) k + k z ( θ − t ω ) k + k z ( θ − t ω ) k (cid:1) ≤ c r ( ω ) + c , which along with (4.15) shows that, for all t ≥ T B ( ω ), k Φ( t, θ − t ω, ( u , v )( θ − t ω )) k ≤ (cid:0) k z ( ω ) k + k z ( ω ) k + c r ( ω ) + c (cid:1) . (4.16)Given ω ∈ Ω, denote by K ( ω ) = { ( u, v ) ∈ L ( R n ) × L ( R n ) : k u k + k v k ≤ (cid:0) k z ( ω ) k + k z ( ω ) k + c r ( ω ) + c (cid:1) } . Then { K ( ω ) } ω ∈ Ω ∈ D . Further, (4.16) shows that { K ( ω ) } ω ∈ Ω is an absorbing set for Φ in D , whichcompletes the proof. Lemma 4.2.
Assume that g, h ∈ L ( R n ) and (3.4) - (3.7) hold. Let B = { B ( ω ) } ω ∈ Ω ∈ D . Then forevery T ≥ and P -a.e. ω ∈ Ω , u ( t, ω, u ( ω )) and ˜ u ( t, ω, ˜ u ( ω )) satisfy, for all t ≥ T , Z tT e η ( s − t ) k u ( s, θ − t ω, u ( θ − t ω )) k pp ds ≤ ce − ηt (cid:0) k ˜ u ( θ − t ω ) k + k ˜ v ( θ − t ω ) k (cid:1) + c (1 + r ( ω )) , (4.17) Z tT e η ( s − t ) k∇ ˜ u ( s, θ − t ω, ˜ u ( θ − t ω )) k ds ≤ ce − ηt (cid:0) k ˜ u ( θ − t ω ) k + k ˜ v ( θ − t ω ) k (cid:1) + c (1 + r ( ω )) , (4.18) where c is a positive deterministic constant independent of T , and r ( ω ) is the tempered functionin (3.10) .Proof. First replacing t by T and then replacing ω by θ − t ω in (4.13), we find that β k ˜ u ( T , θ − t ω, ˜ u ( θ − t ω )) k + α k ˜ v ( T , θ − t ω, ˜ v ( θ − t ω )) k ≤ e − ηT (cid:0) β k ˜ u ( θ − t ω ) k + α k ˜ v ( θ − t ω ) k (cid:1) + Z T e η ( τ − T ) p ( θ τ − t ω ) dτ + c e η ( T − t ) and then simplify to get that e η ( T − t ) (cid:0) β k ˜ u ( T , θ − t ω, ˜ u ( θ − t ω )) k + α k ˜ v ( T , θ − t ω, ˜ v ( θ − t ω )) k (cid:1) ≤ e − ηt (cid:0) β k ˜ u ( θ − t ω ) k + α k ˜ v ( θ − t ω ) k (cid:1) + Z T e η ( τ − t ) p ( θ τ − t ω ) dτ + ce η ( T − t ) . (4.19)By (4.10), the second term on the right-hand side of (4.19) satisfies Z T e η ( τ − t ) p ( θ τ − t ω ) dτ = Z T − t − t e ητ p ( θ τ ω ) dτ ≤ c r ( ω ) Z T − t − t e ητ dτ ≤ η c r ( ω ) e η ( T − t ) . (4.20)From (4.19)-(4.20) it follows that e η ( T − t ) (cid:0) β k ˜ u ( T , θ − t ω, ˜ u ( θ − t ω )) k + α k ˜ v ( T , θ − t ω, ˜ v ( θ − t ω )) k (cid:1) ≤ e − ηt (cid:0) β k ˜ u ( θ − t ω ) k + α k ˜ v ( θ − t ω ) k (cid:1) + 2 η c r ( ω ) e η ( T − t ) + ce η ( T − t ) . (4.21)Note that (4.11) implies, for all t ≥ ddt (cid:0) β k ˜ u k + α k ˜ v k (cid:1) + η (cid:0) β k ˜ u k + α k ˜ v k (cid:1) + β k∇ ˜ u k + α k u k pp ≤ p ( θ t ω ) + c. Multiplying the above by e ηt and then integrating over ( T , t ), we get that, for all t ≥ T , β k ˜ u ( t, ω, ˜ u ( ω )) k + α k ˜ v ( t, ω, ˜ v ( ω )) k + Z tT e η ( s − t ) k∇ ˜ u ( s, ω, ˜ u ( ω )) k ds + α Z tT e η ( s − t ) k u ( s, ω, u ( ω )) k pp ds ≤ e η ( T − t ) (cid:0) β k ˜ u ( T , ω, ˜ u ( ω )) k + α k ˜ v ( T , ω, ˜ v ( ω )) k (cid:1) + Z tT e η ( s − t ) p ( θ s ω ) ds + c Z tT e η ( s − t ) ds. (4.22)Dropping the first two terms on the left-hand side of (4.22) and replacing ω by θ − t ω , we obtainthat, for all t ≥ T , Z tT e η ( s − t ) k∇ ˜ u ( s, θ − t ω, ˜ u ( θ − t ω )) k ds + α Z tT e η ( s − t ) k u ( s, θ − t ω, u ( θ − t ω )) k pp ds ≤ e η ( T − t ) (cid:0) β k ˜ u ( T , θ − t ω, ˜ u ( θ − t ω )) k + α k ˜ v ( T , θ − t ω, ˜ v ( θ − t ω )) k (cid:1) + Z tT e η ( s − t ) p ( θ s − t ω ) ds + c Z tT e η ( s − t ) ds. e η ( T − t ) (cid:0) β k ˜ u ( T , θ − t ω, ˜ u ( θ − t ω )) k + α k ˜ v ( T , θ − t ω, ˜ v ( θ − t ω )) k (cid:1) + Z T − t e ητ p ( θ τ ω ) dτ + cη (4.23)By (4.10), the second term on the right-hand side of (4.23) satisfies, for t ≥ T , Z T − t e ητ p ( θ τ ω ) dτ ≤ c r ( ω ) Z T − t e ητ dτ ≤ η c r ( ω ) . (4.24)Then, using (4.21) and (4.24), it follows from (4.23) that Z tT e η ( s − t ) k∇ ˜ u ( s, θ − t ω, ˜ u ( θ − t ω )) k ds + α Z tT e η ( s − t ) k u ( s, θ − t ω, u ( θ − t ω )) k pp ds ≤ e − ηt (cid:0) β k ˜ u ( θ − t ω ) k + α k ˜ v ( θ − t ω ) k (cid:1) + c (1 + r ( ω )) . This completes the proof.As a special case of Lemma 4.2, we have the following uniform estimates.
Lemma 4.3.
Assume that g, h ∈ L ( R n ) and (3.4) - (3.7) hold. Let B = { B ( ω ) } ω ∈ Ω ∈ D . Then for P -a.e. ω ∈ Ω , there exists T B ( ω ) > such that for all t ≥ T B ( ω ) , Z t +1 t k u ( s, θ − t − ω, u ( θ − t − ω )) k pp ds ≤ c (1 + r ( ω )) , Z t +1 t k∇ ˜ u ( s, θ − t − ω, ˜ u ( θ − t − ω )) k ds ≤ c (1 + r ( ω )) , where c is a positive deterministic constant and r ( ω ) is the tempered function in (3.10) .Proof. First replacing t by t + 1 and then replacing T by t in (4.18), we find that Z t +1 t e η ( s − t − k∇ ˜ u ( s, θ − t − ω, ˜ u ( θ − t − ω )) k ds ≤ ce − η ( t +1) (cid:0) k ˜ u ( θ − t − ω ) k + k ˜ v ( θ − t − ω ) k (cid:1) + c (1 + r ( ω )) . (4.25)Note that e η ( s − t − ≥ e − η for s ∈ [ t, t + 1]. Hence from (4.25) we get that e − η Z t +1 t k∇ ˜ u ( s, θ − t − ω, ˜ u ( θ − t − ω )) k ds ≤ ce − η ( t +1) (cid:0) k ˜ u ( θ − t − ω ) k + k ˜ v ( θ − t − ω ) k (cid:1) + c (1 + r ( ω )) . ce − η ( t +1) (cid:0) k u ( θ − t − ω ) k + k v ( θ − t − ω ) k (cid:1) + ce − η ( t +1) (cid:0) k z ( θ − t − ω ) k + k z ( θ − t − ω ) k (cid:1) + c (1 + r ( ω )) . (4.26)Since k u ( ω ) k , k v ( ω ) k , k z ( ω ) k and k z ( ω ) k are tempered, there is T B ( ω ) > t ≥ T B ( ω ), ce − η ( t +1) (cid:0) k u ( θ − t − ω ) k + k v ( θ − t − ω ) k + k z ( θ − t − ω ) k + k z ( θ − t − ω ) k (cid:1) ≤ c (1 + r ( ω )) , which along with (4.26) shows that, for all t ≥ T B ( ω ), Z t +1 t k∇ ˜ u ( s, θ − t − ω, ˜ u ( θ − t − ω )) k ds ≤ e η (1 + r ( ω )) . (4.27)Using (4.17) and repeating the above process, we can also find that, for t ≥ T B ( ω ), Z t +1 t k u ( s, θ − t − ω, u ( θ − t − ω )) k pp ds ≤ e η (1 + r ( ω )) . (4.28)Then the lemma follows from (4.27)-(4.28). Lemma 4.4.
Assume that g, h ∈ L ( R n ) and (3.4) - (3.7) hold. Let B = { B ( ω ) } ω ∈ Ω ∈ D . Then for P -a.e. ω ∈ Ω , there exists T B ( ω ) > such that for all t ≥ T B ( ω ) , Z t +1 t k∇ u ( s, θ − t − ω, u ( θ − t − ω ) k ds ≤ c (1 + r ( ω )) , where c is a positive deterministic constant and r ( ω ) is the tempered function in (3.10) .Proof. Let T B ( ω ) be the positive constant in Lemma 4.3, take t ≥ T B ( ω ) and s ∈ ( t, t + 1). By(3.15) we find that k∇ u ( s, θ − t − ω, u ( θ − t − ω ) k = k∇ ˜ u ( s, θ − t − ω, ˜ u ( θ − t − ω )) + ∇ z ( θ s − t − ω ) k ≤ k∇ ˜ u ( s, θ − t − ω, ˜ u ( θ − t − ω )) k + 2 k∇ z ( θ s − t − ω ) k . (4.29)By (3.12) we have2 k∇ z ( θ s − t − ω ) k = 2 k∇ φ k | y ( θ s − t − ω ) | ≤ ce η ( t +1 − s ) r ( ω ) ≤ ce η r ( ω ) . (4.30)Now integrating (4.29) with respect to s over ( t, t + 1), by Lemma 4.3 and inequality (4.30), we getthat Z t +1 t k∇ u ( s, θ − t − ω, u ( θ − t − ω ) k ds ≤ c + c r ( ω ) . (4.31)Then the lemma follows from (4.31). 14ext, we derive uniform estimates on u in H ( R n ). Lemma 4.5.
Assume that g, h ∈ L ( R n ) and (3.4) - (3.7) hold. Let B = { B ( ω ) } ω ∈ Ω ∈ D . Then for P -a.e. ω ∈ Ω , there exists T B ( ω ) > such that for all t ≥ T B ( ω ) , k∇ u ( t, θ − t ω, u ( θ − t ω )) k ≤ c (1 + r ( ω )) , where c is a positive deterministic constant and r ( ω ) is the tempered function in (3.10) .Proof. Taking the inner product of (3.13) with ∆˜ u in L ( R n ), we get that12 ddt k∇ ˜ u k + λ k∇ ˜ u k + k ∆˜ u k = α (˜ v, ∆˜ u ) − Z R n f ( x, u )∆˜ udx − ( g + ∆ z ( θ t ω ) − αz ( θ t ω ) , ∆˜ u ) . (4.32)Note that the first term on the right-hand side of (4.32) is bounded by α | (˜ v, ∆˜ u ) | ≤ α k ˜ v kk ∆˜ u k ≤ k ∆˜ u k + α k ˜ v k . (4.33)For the nonlinear term in (4.32), by (3.5)-(3.7), we have − Z R n f ( x, u ) ∆˜ udx = − Z R n f ( x, u ) ∆ udx + Z R n f ( x, u ) ∆ z ( θ t ω ) dx = Z R n ∂f∂x ( x, u ) ∇ udx + Z R n ∂f∂u ( x, u ) |∇ u | dx + Z R n f ( x, u )∆ z ( θ t ω ) dx ≤ k ψ kk∇ u k + β k∇ u k + Z R n | f ( x, u ) | | ∆ z ( θ t ω ) | dx ≤ k ψ kk∇ u k + β k∇ u k + α Z R n | u | p − | ∆ z ( θ t ω ) | dx + Z R n | ψ ( x ) | | ∆ z ( θ t ω ) | dx ≤ c k∇ u k + α q Z R n | u | p dx + α p Z R n | ∆ z ( θ t ω ) | p dx + c ( k ψ k + k ψ k ) + c k ∆ z ( θ t ω ) k . ≤ c (cid:0) k∇ u k + k u k pp (cid:1) + c (cid:0) k ∆ z ( θ t ω ) k + k ∆ z ( θ t ω ) k pp + 1 (cid:1) . (4.34)On the other hand, the last term on the right-hand side of (4.32) is bounded by | ( g, ∆˜ u ) | + | (∆ z ( θ t ω ) , ∆˜ u ) | + α | ( z ( θ t ω ) , ∆˜ u ) |≤ k ∆˜ u k + c (cid:0) k g k + k ∆ z ( θ t ω ) k + k z ( θ t ω ) k (cid:1) . (4.35)15y (4.32)-(4.35) we get that ddt k∇ ˜ u k + 2 λ k∇ ˜ u k + k ∆˜ u k ≤ α k ˜ v k + c (cid:0) k∇ u k + k u k pp (cid:1) + c (cid:0) k ∆ z ( θ t ω ) k + k ∆ z ( θ t ω ) k pp + k z ( θ t ω ) k + 1 (cid:1) . (4.36)Let p ( θ t ω ) = c (cid:0) k ∆ z ( θ t ω ) k + k ∆ z ( θ t ω ) k pp + k z ( θ t ω ) k + 1 (cid:1) . (4.37)Since z ( θ t ω ) = φ y ( θ t ω ) and z ( θ t ω ) = φ y ( θ t ω ) with φ ∈ H ( R n ) ∩ W ,p ( R n ) and φ ∈ H ( R n ), we find that there are positive constants c and c such that p ( θ t ω ) ≤ c X j =1 (cid:0) | y j ( θ t ω j ) | + | y j ( θ t ω j ) | pp (cid:1) + c , which along with (3.12) shows that p ( θ t ω ) ≤ c e η | t | r ( ω ) + c , ∀ t ∈ R . (4.38)By (4.36)-(4.37), we find that ddt k∇ ˜ u k ≤ c (cid:0) k∇ u k + k u k pp + k ˜ v k (cid:1) + p ( θ t ω ) . (4.39)Let T B ( ω ) be the positive constant in Lemma 4.3, take t ≥ T B ( ω ) and s ∈ ( t, t + 1). Then integrate(4.39) over ( s, t + 1) to get k∇ ˜ u ( t + 1 , ω, ˜ u ( ω )) k ≤ k∇ ˜ u ( s, ω, ˜ u ( ω )) k + Z t +1 s p ( θ τ ω ) dτ + c Z t +1 s (cid:0) k∇ u ( τ, ω, u ( ω )) k + k u ( τ, ω, u ( ω )) k pp + k ˜ v ( τ, ω, ˜ v ( ω )) k (cid:1) dτ ≤ k∇ ˜ u ( s, ω, ˜ u ( ω )) k + Z t +1 t p ( θ τ ω ) dτ + c Z t +1 t (cid:0) k∇ u ( τ, ω, u ( ω )) k + k u ( τ, ω, u ( ω )) k pp + k ˜ v ( τ, ω, ˜ v ( ω )) k (cid:1) dτ. Now integrating the above with respect to s over ( t, t + 1), we find that k∇ ˜ u ( t + 1 , ω, ˜ u ( ω )) k ≤ Z t +1 t k∇ ˜ u ( s, ω, ˜ u ( ω )) k ds + Z t +1 t p ( θ τ ω ) dτ + c Z t +1 t (cid:0) k∇ u ( τ, ω, u ( ω )) k + k u ( τ, ω, u ( ω )) k pp + k ˜ v ( τ, ω, ˜ v ( ω )) k (cid:1) dτ. ω by θ − t − ω , we obtain that k∇ ˜ u ( t + 1 , θ − t − ω, ˜ u ( θ − t − ω )) k ≤ Z t +1 t k∇ ˜ u ( s, θ − t − ω, ˜ u ( θ − t − ω )) k ds + Z t +1 t p ( θ τ − t − ω ) dτ + c Z t +1 t (cid:0) k∇ u ( τ, θ − t − ω, u ( θ − t − ω )) k + k u ( τ, θ − t − ω, u ( θ − t − ω )) k pp (cid:1) dτ + c Z t +1 t k ˜ v ( τ, θ − t − ω, ˜ v ( θ − t − ω )) k dτ. (4.40)By Lemmas 4.3 and 4.4, it follows from (4.40) and (4.38) that, for all t ≥ T B ( ω ), k∇ ˜ u ( t + 1 , θ − t − ω, ˜ u ( θ − t − ω )) k ≤ c + c r ( ω ) + Z − p ( θ s ω ) ds ≤ c + c r ( ω ) + Z − (cid:16) c e − η s r ( ω ) + c (cid:17) ds ≤ c + c r ( ω ) . (4.41)Then by (4.41) and (3.15), we have, for all t ≥ T B ( ω ), k∇ u ( t + 1 , θ − t − ω, u ( θ − t − ω ) k = k∇ ˜ u ( t + 1 , θ − t − ω, ˜ u ( θ − t − ω )) + ∇ z ( ω ) k ≤ k∇ ˜ u ( t + 1 , θ − t − ω, ˜ u ( θ − t − ω )) k + 2 k∇ z ( ω ) k ≤ c + c r ( ω ) , which completes the proof. Lemma 4.6.
Assume that g, h ∈ L ( R n ) and (3.4) - (3.7) hold. Let B = { B ( ω ) } ω ∈ Ω ∈ D . Then forevery ǫ > and P -a.e. ω ∈ Ω , there exist T = T B ( ω, ǫ ) > and R = R ( ω, ǫ ) > such that for all t ≥ T , Z | x |≥ R (cid:0) | ˜ u ( t, θ − t ω, ˜ u ( θ − t ω )) | + | ˜ v ( t, θ − t ω, ˜ v ( θ − t ω )) | (cid:1) dx ≤ ǫ. Proof.
Let ρ be a smooth function defined on R + such that 0 ≤ ρ ( s ) ≤ s ∈ R + , and ρ ( s ) = (cid:26) ≤ s ≤ s ≥ . Then there exists a positive constant c such that | ρ ′ ( s ) | ≤ c for all s ∈ R + .Taking the inner product of (3.13) with βρ ( | x | k )˜ u in L ( R n ) we find that12 β ddt Z R n ρ ( | x | k ) | ˜ u | dx + λβ Z R n ρ ( | x | k ) | ˜ u | dx − β Z R n ρ ( | x | k )˜ u ∆˜ udx + αβ Z R n ρ ( | x | k )˜ u ˜ vdx β Z R n f ( x, u ) ρ ( | x | k )˜ udx + β Z R n ( g + ∆ z ( θ t ω ) − αz ( θ t ω )) ρ ( | x | k )˜ udx. (4.42)Taking the inner product of (3.14) with αρ ( | x | k )˜ v in L ( R n ) we find that12 α ddt Z R n ρ ( | x | k ) | ˜ v | dx + αδ Z R n ρ ( | x | k ) | ˜ v | dx − αβ Z R n ρ ( | x | k )˜ u ˜ vdx = α Z R n ρ ( | x | k ) h ˜ vdx + αβ Z R n ρ ( | x | k ) z ( θ t ω )˜ vdx. (4.43)Adding (4.2) and (4.3), we obtain that12 ddt (cid:18) β Z R n ρ ( | x | k ) | ˜ u | dx + α Z R n ρ ( | x | k ) | ˜ v | dx (cid:19) + λβ Z R n ρ ( | x | k ) | ˜ u | dx + αδ Z R n ρ ( | x | k ) | ˜ v | dx − β Z R n ρ ( | x | k )˜ u ∆˜ udx = β Z R n f ( x, u ) ρ ( | x | k )˜ udx + β Z R n ( g + ∆ z ( θ t ω ) − αz ( θ t ω )) ρ ( | x | k )˜ udx + α Z R n ρ ( | x | k ) h ˜ vdx + αβ Z R n ρ ( | x | k ) z ( θ t ω )˜ vdx. (4.44)We now estimate the terms in (4.44) as follows. First we have − β Z R n ρ ( | x | k )˜ u ∆˜ udx = β Z R n |∇ ˜ u | ρ ( | x | k ) dx + β Z R n ˜ uρ ′ ( | x | k ) 2 xk · ∇ ˜ udx = β Z R n |∇ ˜ u | ρ ( | x | k ) dx + β Z k ≤| x |≤√ k ˜ uρ ′ ( | x | k ) 2 xk · ∇ ˜ udx. (4.45)Note that the second term on the right-hand side of (4.45) is bounded by β | Z k ≤| x |≤√ k ˜ uρ ′ ( | x | k ) 2 xk · ∇ ˜ udx | ≤ √ k β Z k ≤| x |≤√ k | ˜ u | | ρ ′ ( | x | k ) | |∇ ˜ u | dx ≤ ck Z R n | ˜ u | |∇ ˜ u | dx ≤ ck ( k ˜ u k + k∇ ˜ u k ) . (4.46)By (4.45)-(4.46), we find that − β Z R n ρ ( | x | k )˜ u ∆˜ udx ≥ β Z R n |∇ ˜ u | ρ ( | x | k ) dx − ck ( k ˜ u k + k∇ ˜ u k ) . (4.47)For the nonlinear term in (4.44), we have β Z R n f ( x, u ) ρ ( | x | k )˜ udx = β Z R n f ( x, u ) ρ ( | x | k ) udx − β Z R n f ( x, u ) ρ ( | x | k ) z ( θ t ω ) dx. (4.48)18y (3.4), the first term on the right-hand side of (4.48) is bounded by β Z R n f ( x, u ) ρ ( | x | k ) udx ≤ − α β Z R n | u | p ρ ( | x | k ) dx + β Z R n ψ ρ ( | x | k ) dx. (4.49)By (3.5), the second term on the right-hand side of (4.48) is bounded by β | Z R n f ( x, u ) ρ ( | x | k ) z ( θ t ω ) dx |≤ α β Z R n | u | p − ρ ( | x | k ) | z ( θ t ω ) | dx + β Z R n | ψ | ρ ( | x | k ) | z ( θ t ω ) | dx ≤ α β Z R n | u | p ρ ( | x | k ) dx + c Z R n | z ( θ t ω ) | p ρ ( | x | k ) dx + 12 β Z R n | z ( θ t ω ) | ρ ( | x | k ) dx + 12 β Z R n ψ ρ ( | x | k ) dx. (4.50)Then it follows from (4.48)-(4.50) that β Z R n f ( x, u ) ρ ( | x | k )˜ udx ≤ − α β Z R n | u | p ρ ( | x | k ) dx + β Z R n ψ ρ ( | x | k ) dx + 12 β Z R n ψ ρ ( | x | k ) dx + c Z R n (cid:0) | z ( θ t ω ) | p + | z ( θ t ω ) | (cid:1) ρ ( | x | k ) dx. (4.51)For the second term on the right-hand side of (4.44), we have that β | Z R n ( g + ∆ z ( θ t ω ) − αz ( θ t ω )) ρ ( | x | k )˜ udx |≤ λβ Z R n ρ ( | x | k ) | ˜ u | dx + c Z R n ( g + | ∆ z ( θ t ω ) | + | z ( θ t ω ) | ) ρ ( | x | k ) dx. (4.52)For the last two terms on the right-hand side of (4.44), we find that α Z R n ρ ( | x | k ) h ˜ vdx + αβ Z R n ρ ( | x | k ) z ( θ t ω )˜ vdx ≤ αδ Z R n ρ ( | x | k ) | ˜ v | dx + c Z R n ρ ( | x | k ) (cid:0) | z ( θ t ω ) | + | h | (cid:1) dx. (4.53)Finally, by (4.44), (4.47) and (4.51)-(4.53), we obtain that12 ddt (cid:18) β Z R n ρ ( | x | k ) | ˜ u | dx + α Z R n ρ ( | x | k ) | ˜ v | dx (cid:19) + 12 λβ Z R n ρ ( | x | k ) | ˜ u | dx + 12 αδ Z R n ρ ( | x | k ) | ˜ v | dx
19 12 α β Z R n ρ ( | x | k ) | u | p dx + β Z R n ρ ( | x | k ) |∇ ˜ u | dx ≤ ck ( k∇ ˜ u k + k ˜ u k ) + c Z R n (cid:0) | ψ | + | ψ | + g + h (cid:1) ρ ( | x | k ) dx + c Z R n (cid:0) | ∆ z ( θ t ω ) | + | z ( θ t ω ) | + | z ( θ t ω ) | p + | z ( θ t ω ) | (cid:1) ρ ( | x | k ) dx. (4.54)Note that (4.54) implies that ddt (cid:18) β Z R n ρ ( | x | k ) | ˜ u | dx + α Z R n ρ ( | x | k ) | ˜ v | dx (cid:19) + η (cid:18) β Z R n ρ ( | x | k ) | ˜ u | dx + α Z R n ρ ( | x | k ) | ˜ v | dx (cid:19) ≤ ck ( k∇ ˜ u k + k ˜ u k ) + c Z R n (cid:0) | ψ | + | ψ | + g + h (cid:1) ρ ( | x | k ) dx + c Z R n (cid:0) | ∆ z ( θ t ω ) | + | z ( θ t ω ) | + | z ( θ t ω ) | p + | z ( θ t ω ) | (cid:1) ρ ( | x | k ) dx. (4.55)By Lemmas 4.1 and 4.5, there is T = T ( B, ω ) > t ≥ T , k ˜ u ( t, θ − t ω, ˜ u ( θ − t ω )) k H ( R n ) ≤ c (1 + r ( ω )) . (4.56)Now integrating (4.55) over ( T , t ), we get that, for all t ≥ T , β Z R n ρ ( | x | k ) | ˜ u ( t, ω, ˜ u ( ω )) | dx + α Z R n ρ ( | x | k ) | ˜ v ( t, ω, ˜ v ( ω )) | dx ≤ e η ( T − t ) (cid:18) β Z R n ρ ( | x | k ) | ˜ u ( T , ω, ˜ u ( ω )) | dx + α Z R n ρ ( | x | k ) | ˜ v ( T , ω, ˜ v ( ω )) | dx (cid:19) + ck Z tT e η ( s − t ) (cid:0) k∇ ˜ u ( s, ω, ˜ u ( ω )) k + k ˜ u ( s, ω, ˜ u ( ω )) k (cid:1) ds + c Z tT e η ( s − t ) Z R n (cid:0) | ψ | + | ψ | + g + h (cid:1) ρ ( | x | k ) dxds + c Z tT e η ( s − t ) Z R n (cid:0) | ∆ z ( θ s ω ) | + | z ( θ s ω ) | + | z ( θ s ω ) | p + | z ( θ s ω ) | (cid:1) ρ ( | x | k ) dxds. (4.57)Replacing ω by θ − t ω , we obtain from (4.57) that, for all t ≥ T , β Z R n ρ ( | x | k ) | ˜ u ( t, θ − t ω, ˜ u ( θ − t ω )) | dx + α Z R n ρ ( | x | k ) | ˜ v ( t, θ − t ω, ˜ v ( θ − t ω )) | dx ≤ βe η ( T − t ) Z R n ρ ( | x | k ) | ˜ u ( T , θ − t ω, ˜ u ( θ − t ω )) | dx αe η ( T − t ) Z R n ρ ( | x | k ) | ˜ v ( T , θ − t ω, ˜ v ( θ − t ω )) | dx + ck Z tT e η ( s − t ) k∇ ˜ u ( s, θ − t ω, ˜ u ( θ − t ω )) k ds + ck Z tT e η ( s − t ) k ˜ u ( s, θ − t ω, ˜ u ( θ − t ω )) k ds + c Z tT e η ( s − t ) Z R n (cid:0) | ψ | + | ψ | + g + h (cid:1) ρ ( | x | k ) dxds + c Z tT e η ( s − t ) Z R n (cid:0) | ∆ z ( θ s − t ω ) | + | z ( θ s − t ω ) | + | z ( θ s − t ω ) | p + | z ( θ s − t ω ) | (cid:1) ρ ( | x | k ) dxds. (4.58)In what follows, we estimate the terms in (4.58). First replacing t by T and then replacing ω by θ − t ω in (4.13), we have the following bounds for the first two terms on the right-hand side of (4.58). βe η ( T − t ) Z R n ρ ( | x | k ) | ˜ u ( T , θ − t ω, ˜ u ( θ − t ω )) | dx + αe η ( T − t ) Z R n ρ ( | x | k ) | ˜ v ( T , θ − t ω, ˜ v ( θ − t ω )) | dx ≤ e η ( T − t ) (cid:18) e − ηT (cid:0) β k ˜ u ( θ − t ω ) k + α k ˜ v ( θ − t ω ) k (cid:1) + Z T e η ( τ − T ) p ( θ τ − t ω ) dτ + c (cid:19) ≤ e − ηt (cid:0) β k ˜ u ( θ − t ω ) k + α k ˜ v ( θ − t ω ) k (cid:1) + Z T e η ( τ − t ) p ( θ τ − t ω ) dτ + ce η ( T − t ) ≤ e − ηt (cid:0) β k ˜ u ( θ − t ω ) k + α k ˜ v ( θ − t ω ) k (cid:1) + Z T − t − t e ητ p ( θ τ ω ) dτ + ce η ( T − t ) ≤ e − ηt (cid:0) β k ˜ u ( θ − t ω ) k + α k ˜ v ( θ − t ω ) k (cid:1) + Z T − t − t e ητ c r ( ω ) dτ + ce η ( T − t ) ≤ e − ηt (cid:0) β k ˜ u ( θ − t ω ) k + α k ˜ v ( θ − t ω ) k (cid:1) + 2 η c r ( ω ) e η ( T − t ) + ce η ( T − t ) (4.59)where we have used (4.10). By (4.59), we find that, given ǫ >
0, there is T = ( B, ω, ǫ ) > T suchthat for all t ≥ T , e η ( T − t ) (cid:18)Z R n ρ ( | x | k ) (cid:0) β | ˜ u ( T , θ − t ω, ˜ u ( θ − t ω )) | + α | ˜ v ( T , θ − t ω, ˜ v ( θ − t ω )) | (cid:1) dx (cid:19) ≤ ǫ. (4.60)By Lemma 4.2, there is T = T ( B, ω ) > T such that the third term on the right-hand side of(4.58) satisfies, for all t ≥ T , ck Z tT e η ( s − t ) k∇ ˜ u ( s, θ − t ω, ˜ u ( θ − t ω )) k ds ≤ ck (1 + r ( ω )) . R = R ( ω, ǫ ) > t ≥ T and k ≥ R , ck Z tT e λ ( s − t ) k∇ v ( s, θ − t ω, v ( θ − t ω )) k ds ≤ ǫ. (4.61)First replacing t by s and then replacing ω by θ − t ω in (4.13), we find that the fourth term on theright-hand side of (4.58) satisfies ck Z tT e η ( s − t ) k ˜ u ( s, θ − t ω, ˜ u ( θ − t ω )) k ds ≤ ck Z tT e − ηt (cid:0) β k ˜ u ( θ − t ω ) k + α k ˜ v ( θ − t ω ) k (cid:1) ds + ck Z tT e η ( s − t ) Z s e η ( τ − s ) p ( θ τ − t ω ) dτ ds + ck Z tT e λ ( s − t ) ds ≤ ck e − ηt ( t − T ) (cid:0) β k ˜ u ( θ − t ω ) k + α k ˜ v ( θ − t ω ) k (cid:1) + ck + ck Z tT Z s e η ( τ − t ) p ( θ τ − t ω ) dτ ds ≤ ck e − ηt ( t − T ) (cid:0) β k ˜ u ( θ − t ω ) k + α k ˜ v ( θ − t ω ) k (cid:1) + ck + ck Z tT Z s − t − t e ητ p ( θ τ ω ) dτ ds ≤ ck e − ηt ( t − T ) (cid:0) β k ˜ u ( θ − t ω ) k + α k ˜ v ( θ − t ω ) k (cid:1) + ck + ck r ( ω ) Z tT Z s − t − t e ητ dτ ds ≤ ck e − ηt ( t − T ) (cid:0) β k ˜ u ( θ − t ω ) k + α k ˜ v ( θ − t ω ) k (cid:1) + ck + 4 cη k r ( ω ) , which shows that, there is T = T ( B, ω, ǫ ) > T and R = R ( ω, ǫ ) such that for all t ≥ T and k ≥ R , ck Z tT e η ( s − t ) k ˜ u ( s, θ − t ω, ˜ u ( θ − t ω )) k ds ≤ ǫ. (4.62)Note that ψ ∈ L ( R n ) and ψ , g, h ∈ L ( R n ). Therefore, there is R = R ( ǫ ) such that for all k ≥ R , Z | x |≥ k (cid:0) | ψ | + | ψ | + | g | + | h | (cid:1) dx ≤ ǫ. Then for the fifth term on the right-hand side of (4.58), we have c Z tT e η ( s − t ) Z R n (cid:0) | ψ | + | ψ | + | g | + | h | (cid:1) ρ ( | x | k ) dxds c Z tT e η ( s − t ) Z | x |≥ k (cid:0) | ψ | + | ψ | + | g | + | h | (cid:1) dxds ≤ cǫ Z tT e η ( s − t ) ds ≤ cǫ, (4.63)where c is independent of ǫ . Note that and φ ∈ H ( R n ) ∩ W ,p ( R n ) and φ ∈ H ( R n ). Hencethere is R = R ( ω, ǫ ) such that for all k ≥ R , Z | x |≥ k (cid:0) | φ ( x ) | + | φ ( x ) | p + | ∆ φ ( x ) | + | φ ( x ) | (cid:1) dx ≤ ǫr ( ω ) , (4.64)where r ( ω ) is the tempered function in (3.10). By (4.64) and (3.10)-(3.11), we have the followingbounds for the last term on the right-hand side of (4.58): c Z tT e η ( s − t ) Z R n (cid:0) | ∆ z ( θ s − t ω ) | + | z ( θ s − t ω ) | + | z ( θ s − t ω ) | p + | z ( θ s − t ω ) | (cid:1) ρ ( | x | k ) dxds ≤ c Z tT e η ( s − t ) Z | x |≥ k (cid:0) | ∆ z ( θ s − t ω ) | + | z ( θ s − t ω ) | + | z ( θ s − t ω ) | p + | z ( θ s − t ω ) | (cid:1) dxds ≤ c Z tT e η ( s − t ) Z | x |≥ k (cid:0) | ∆ φ | | y ( θ s − t ω ) | + | φ | | y ( θ s − t ω ) | + | φ | p | y ( θ s − t ω ) | p + | φ | | y ( θ s − t ω ) | (cid:1) dxds ≤ cǫr ( ω ) Z tT e η ( s − t ) 2 X j =1 (cid:0) | y j ( θ s − t ω j ) | + | y j ( θ s − t ω j ) | p (cid:1) ds ≤ cǫr ( ω ) Z tT e η ( s − t ) r ( θ s − t ω ) ds ≤ cǫr ( ω ) Z T − t e ητ r ( θ τ ω ) dτ ≤ cǫr ( ω ) Z T − t e ητ r ( ω ) dτ ≤ cǫ. (4.65)Let T = T ( B, ω, ǫ ) = max { T , T , T , T } and R = R ( ω, ǫ ) = max { R , R , R , R } . Then itfollows from (4.58), (4.60)-(4.65) that, for all t ≥ T and k ≥ R , β Z R n ρ ( | x | k ) | ˜ u ( t, θ − t ω, ˜ u ( θ − t ω )) | dx + α Z R n ρ ( | x | k ) | ˜ v ( t, θ − t ω, ˜ v ( θ − t ω )) | dx ≤ cǫ, where c is independent of ǫ , and hence, for all t ≥ T and k ≥ R , Z | x |≥√ k (cid:0) β | ˜ u ( t, θ − t ω, ˜ u ( θ − t ω )) | dx + α | ˜ v ( t, θ − t ω, ˜ v ( θ − t ω )) | (cid:1) dx ≤ cǫ. This completes the proof.We now derive uniform estimates on the tails of u and v in L ( R n ). Lemma 4.7.
Assume that g, h ∈ L ( R n ) and (3.4) - (3.7) hold. Let B = { B ( ω ) } ω ∈ Ω ∈ D . Then forevery ǫ > and P -a.e. ω ∈ Ω , there exist T = T B ( ω, ǫ ) > and R = R ( ω, ǫ ) > such that, for all t ≥ T , Z | x |≥ R (cid:0) | u ( t, θ − t ω, u ( θ − t ω )) | + | v ( t, θ − t ω, v ( θ − t ω )) | (cid:1) dx ≤ ǫ. roof. Let T and R be the constants in Lemma 4.6. By (4.64) and (3.10) we have, for all t ≥ T and k ≥ R , Z | x |≥ R (cid:0) | z ( ω ) | + | z ( ω ) | (cid:1) dx = Z | x |≥ R (cid:0) | φ ( x ) | | y ( ω ) | + | φ ( x ) | | y ( ω ) | (cid:1) dx ≤ ǫr ( ω ) (cid:0) | y ( ω ) | + | y ( ω ) | (cid:1) ≤ ǫ. (4.66)Then by (4.66) and Lemma 4.6, we get that, for all t ≥ T and k ≥ R , Z | x |≥ R (cid:0) | u ( t, θ − t ω, u ( θ − t ω )) | + | v ( t, θ − t ω, v ( θ − t ω )) | (cid:1) dx = Z | x |≥ R (cid:0) | ˜ u ( t, θ − t ω, ˜ u ( θ − t ω )) + z ( ω ) | + | ˜ v ( t, θ − t ω, ˜ v ( θ − t ω )) + z ( ω ) | (cid:1) dx ≤ Z | x |≥ R (cid:0) | ˜ u ( t, θ − t ω, ˜ u ( θ − t ω )) | + | ˜ v ( t, θ − t ω, ˜ v ( θ − t ω )) | + | z ( ω ) | + z ( ω ) | (cid:1) dx ≤ ǫ, which completes the proof.Note that equation (3.14) has no any smoothing effect on the solutions, that is, if the initialcondition ˜ v ( ω ) is in L ( R n ) only, then for any t ≥
0, ˜ v ( t, ω, ˜ v ( ω )) only belongs to L ( R n ), but not H ( R n ). Therefore, the compactness of Sobolev embeddings cannot be used directly to derive theasymptotic compactness of the solution operator. To overcome the difficulty, we need to decomposethe solution operator as in the deterministic case. Let ˜ v and ˜ v be the solutions of the followingproblems, respectively, ( d ˜ v dt + δ ˜ v = 0 , ˜ v (0) = ˜ v , (4.67)and ( d ˜ v dt + δ ˜ v = β ˜ u + h + βz ( θ t ω ) , ˜ v (0) = 0 . (4.68)Then we find that ˜ v = ˜ v + ˜ v . Let ( u ( ω ) , v ( ω )) ∈ B ( ω ) with { B ( ω ) } ω ∈ Ω ∈ D , and ˜ v ( ω ) = v ( ω ) − z ( ω ). Then the solution ˜ v ( t, ω, ˜ v ( ω )) of (4.67) satisfies that k ˜ v ( t, ω, ˜ v ( ω )) k = k ˜ v ( t, ω, v ( ω ) − z ( ω )) k = e − δt k v ( ω ) − z ( ω ) k , and k ˜ v ( t, θ − t ω, v ( θ − t ω ) − z ( θ − t ω )) k = e − δt k v ( θ − t ω ) − z ( θ − t ω ) k → , as t → ∞ . (4.69)On the other hand, for the solutions of problem (4.68), we have the following estimates.24 emma 4.8. Assume that g ∈ L ( R n ) , h ∈ H ( R n ) , and (3.4) - (3.7) hold. Let B = { B ( ω ) } ω ∈ Ω ∈D . Then for P -a.e. ω ∈ Ω , there exists T = T B ( ω ) > such that for all t ≥ T , k∇ ˜ v ( t, θ − t ω, k ≤ c (1 + r ( ω )) , where c is a positive deterministic constant and r ( ω ) is the tempered function in (3.10) .Proof. Taking the inner product of (4.68) with ∆˜ v in L ( R n ), we get that12 ddt k∇ ˜ v k + δ k∇ ˜ v k = − β (˜ u, ∆˜ v ) − ( h, ∆˜ v ) − β ( z ( θ t ω ) , ∆˜ v ) . (4.70)Note that the first term on the right-hand side of (4.70) is bounded by β | (˜ u, ∆˜ v ) | ≤ β k∇ ˜ u kk∇ ˜ v k ≤ δ k∇ ˜ v k + 2 β δ k∇ ˜ u k . (4.71)For the second term on the right-hand side of (4.70), we have | ( h, ∆˜ v ) | ≤ k∇ h kk∇ ˜ v k ≤ δ k∇ ˜ v k + 2 δ k h k H . (4.72)The last term on the right-hand side of (4.70) is bounded by β k∇ z ( θ t ω ) kk∇ ˜ v k ≤ δ k∇ ˜ v k + 2 β δ k∇ z ( θ t ω ) k ≤ δ k∇ ˜ v k + c | y ( θ t ω ) | . (4.73)Then it follows from (4.70)-(4.73) that ddt k∇ ˜ v k + δ k∇ ˜ v k ≤ c + c k∇ ˜ u k + c | y ( θ t ω ) | , which implies that, for all t ≥ ddt k∇ ˜ v k + η k∇ ˜ v k ≤ c + c k∇ ˜ u k + c | y ( θ t ω ) | . (4.74)Integrating (4.74) on (0 , t ), we obtain that, for all t ≥ k∇ ˜ v ( t, ω, k ≤ c Z t e η ( s − t ) ds + c Z t e η ( s − t ) k∇ ˜ u ( s, ω, ˜ u ( ω )) k ds + c Z t e η ( s − t ) | y ( θ s ω ) | ds ≤ cη + c Z t e η ( s − t ) k∇ ˜ u ( s, ω, ˜ u ( ω )) k ds + c Z t e η ( s − t ) | y ( θ s ω ) | ds. (4.75)25eplacing ω by θ − t ω in (4.75), by (3.12) and Lemma 4.2 we get that, for all t ≥ k∇ ˜ v ( t, θ − t ω, k ≤ cη + c Z t e η ( s − t ) k∇ ˜ u ( s, θ − t ω, ˜ u ( θ − t ω )) k ds + c Z t e η ( s − t ) | y ( θ s − t ω ) | ds ≤ ce − ηt (cid:0) k ˜ u ( θ − t ω ) k + k ˜ v ( θ − t ω ) k (cid:1) + c (1 + r ( ω )) + c Z − t e ητ | y ( θ τ ω ) | dτ. ≤ ce − ηt (cid:0) k ˜ u ( θ − t ω ) k + k ˜ v ( θ − t ω ) k (cid:1) + c (1 + r ( ω )) + cr ( ω ) Z − t e ητ dτ. ≤ ce − ηt (cid:0) k ˜ u ( θ − t ω ) k + k ˜ v ( θ − t ω ) k (cid:1) + c (1 + r ( ω )) , which shows that there is T > t ≥ T , k∇ ˜ v ( t, θ − t ω, k ≤ r ( ω )) . The proof is complete.
In this section, we prove the existence of a D -random attractor for the random dynamical systemΦ associated with the stochastic FitzHugh-Nagumo system on R n . To this end, we first establishthe D -pullback asymptotic compactness of Φ.In the sequel, for every t ∈ R + , ω ∈ Ω and ( u , v ) ∈ L ( R n ) × L ( R n ), we denote byΦ ( t, ω, ( u , v )) = (0 , ˜ v ( t, ω, v − z ( ω ))) ∈ L ( R n ) × L ( R n ) , and Φ ( t, ω, ( u , v )) = ( u ( t, ω, u ) , ˜ v ( t, ω,
0) + z ( θ t ω )) ∈ L ( R n ) × L ( R n ) , where u ( t, ω, u ) is given by (3.15), ˜ v ( t, ω, v − z ( ω )) is the solution of (4.67) with ˜ v (0) = v − z ( ω ), and ˜ v ( t, ω,
0) is the solution of (4.68) with ˜ v (0) = 0. Then we find thatΦ( t, ω, ( u , v )) = Φ ( t, ω, ( u , v )) + Φ ( t, ω, ( u , v )) . The D -pullback asymptotic compactness of Φ is given by the following lemma.26 emma 5.1. Assume that g ∈ L ( R n ) , h ∈ H ( R n ) and (3.4) - (3.7) hold. Then the randomdynamical system Φ is D -pullback asymptotically compact in L ( R n ) × L ( R n ) ; that is, for P -a.e. ω ∈ Ω , the sequence { Φ( t n , θ − t n ω, ( u ,n ( θ − t n ω ) , v ,n ( θ − t n ω )) } has a convergent subsequencein L ( R n ) × L ( R n ) provided t n → ∞ , B = { B ( ω ) } ω ∈ Ω ∈ D and ( u ,n ( θ − t n ω ) , v ,n ( θ − t n ω )) ∈ B ( θ − t n ω ) .Proof. Suppose that t n → ∞ , B = { B ( ω ) } ω ∈ Ω ∈ D and ( u ,n ( θ − t n ω ) , v ,n ( θ − t n ω )) ∈ B ( θ − t n ω ).Then by Lemma 4.1 and (4.69), we find that, for P -a.e. ω ∈ Ω, { Φ ( t n , θ − t n ω, ( u ,n ( θ − t n ω ) , v ,n ( θ − t n ω ))) } ∞ n =1 is bounded in L ( R n ) × L ( R n ) . Hence, there is ( ξ , ξ ) ∈ L ( R n ) × L ( R n ) such that, up to a subsequence,Φ ( t n , θ − t n ω, ( u ,n ( θ − t n ω ) , v ,n ( θ − t n ω ))) → ( ξ , ξ ) weakly in L ( R n ) × L ( R n ) . (5.1)In what follows, we prove the weak convergence of (5.1) is actually strong convergence. Given ǫ > T = T ( B, ω, ǫ ) and R = R ( ω, ǫ ) such thatfor all t ≥ T , Z | x |≥ R | Φ ( t, θ − t ω, ( u ( θ − t ω ) , v ( θ − t ω ))) | dx ≤ ǫ. (5.2)Since t n → ∞ , there is N = N ( B, ω, ǫ ) such that t n ≥ T for every n ≥ N . Hence it follows from(5.2) that for all n ≥ N , Z | x |≥ R | Φ ( t n , θ − t n ω, ( u ,n ( θ − t n ω ) , v ,n ( θ − t n ω ))) | dx ≤ ǫ. (5.3)On the other hand, By Lemmas 4.1, 4.5 and 4.8, there is T = T ( B, ω ) such that for all t ≥ T , k Φ ( t, θ − t ω, ( u ( θ − t ω ) , v ( θ − t ω ))) k H ( R n ) × H ( R n ) ≤ c (1 + r ( ω )) . (5.4)Let N = N ( B, ω ) be large enough such that t n ≥ T for n ≥ N . Then by (5.4) we find that, forall n ≥ N , k Φ ( t n , θ − t n ω, ( u ,n ( θ − t n ω ) , v ,n ( θ − t n ω ))) k H ( R n ) × H ( R n ) ≤ c (1 + r ( ω )) . (5.5)Denote by Q R = { x ∈ R n : | x | ≤ R } . By the compactness of embedding H ( Q R ) × H ( Q R ) ֒ → L ( Q R ) × L ( Q R ), it follows from (5.5) that, up to a subsequence,Φ ( t n , θ − t n ω, ( u ,n ( θ − t n ω ) , v ,n ( θ − t n ω ))) → ( ξ , ξ ) strongly in L ( Q R ) × L ( Q R ) , ǫ >
0, there exists N = N ( B, ω, ǫ ) such that for all n ≥ N , k Φ ( t n , θ − t n ω, ( u ,n ( θ − t n ω ) , v ,n ( θ − t n ω ))) − ( ξ , ξ ) k L ( Q R ) × L ( Q R ) ≤ ǫ. (5.6)Note that ( ξ , ξ ) ∈ L ( R n ) × L ( R n ). Therefore there exists R = R ( ǫ ) such that Z | x |≥ R (cid:0) | ξ ( x ) | + | ξ ( x ) | (cid:1) dx ≤ ǫ. (5.7)Let R = max { R , R } and N = max { N , N } . We find that k Φ ( t n , θ − t n ω, ( u ,n ( θ − t n ω ) , v ,n ( θ − t n ω ))) − ( ξ , ξ ) k L ( R n ) × L ( R n ) ≤ Z | x |≤ R | Φ ( t n , θ − t n ω, ( u ,n ( θ − t n ω ) , v ,n ( θ − t n ω ))) − ( ξ , ξ ) | dx + Z | x |≥ R | Φ ( t n , θ − t n ω, ( u ,n ( θ − t n ω ) , v ,n ( θ − t n ω ))) − ( ξ , ξ ) | dx ≤ Z | x |≤ R | Φ ( t n , θ − t n ω, ( u ,n ( θ − t n ω ) , v ,n ( θ − t n ω ))) − ( ξ , ξ ) | dx +2 Z | x |≥ R | Φ ( t n , θ − t n ω, ( u ,n ( θ − t n ω ) , v ,n ( θ − t n ω ))) | dx + 2 Z | x |≥ R ( | ξ ( x ) | + | ξ ( x ) | dx By (5.3), (5.6)-(5.7), it follows from the above that, for all n ≥ N , k Φ ( t n , θ − t n ω, ( u ,n ( θ − t n ω ) , v ,n ( θ − t n ω ))) − ( ξ , ξ ) k L ( R n ) × L ( R n ) ≤ ǫ, which shows thatΦ ( t n , θ − t n ω, ( u ,n ( θ − t n ω ) , v ,n ( θ − t n ω ))) → ( ξ , ξ ) strongly in L ( R n ) × L ( R n ) . (5.8)On the other hand, we have k Φ( t n , θ − t n ω, ( u ,n ( θ − t n ω ) , v ,n ( θ − t n ω ))) − ( ξ , ξ ) k L ( R n ) × L ( R n ) ≤ k Φ ( t n , θ − t n ω, ( u ,n ( θ − t n ω ) , v ,n ( θ − t n ω ))) − ( ξ , ξ ) k L ( R n ) × L ( R n ) + k Φ ( t n , θ − t n ω, ( u ,n ( θ − t n ω ) , v ,n ( θ − t n ω ))) k L ( R n ) × L ( R n ) ≤ k Φ ( t n , θ − t n ω, ( u ,n ( θ − t n ω ) , v ,n ( θ − t n ω ))) − ( ξ , ξ ) k L ( R n ) × L ( R n ) + k ˜ v ( t n , θ − t n ω, v ,n ( θ − t n ω ) − z ( θ − t n ω )) k L ( R n ) × L ( R n ) k Φ ( t n , θ − t n ω, ( u ,n ( θ − t n ω ) , v ,n ( θ − t n ω ))) − ( ξ , ξ ) k L ( R n ) × L ( R n ) + e − δt n k v ,n ( θ − t n ω ) − z ( θ − t n ω )) k L ( R n ) (5.9)Then, by (4.69), (5.8) and (5.9) we find that k Φ( t n , θ − t n ω, ( u ,n ( θ − t n ω ) , v ,n ( θ − t n ω ))) − ( ξ , ξ ) k L ( R n ) × L ( R n ) → , which completes the proof.We are now in a position to present our main result: the existence of a D -random attractor forΦ in L ( R n ) × L ( R n ). Theorem 5.2.
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