# (

Authors: Gang Wang; Yanbin Tang
Journal: Abstract and Applied Analysis
 (2013)

---

## Abstract

We study the random dynamical system generated by a stochastic reaction-diffusion equation with additive noise on the whole spaceℝn and prove the existence of an (L2,H1)-random attractor for such a random dynamical system. The nonlinearity f is supposed to satisfy the growth of arbitrary order p-1 (p≥2). The (L2,H1)-asymptotic compactness of the random dynamical system is obtained by using an extended version of the tail estimate method introduced by Wang (1999) and the cut-off technique.


---

## Body

## 1. Introduction

In this paper, we consider the asymptotic behavior of solutions to the following stochastic reaction-diffusion equation (SRDE) with additive noise in the entire spaceℝn:
(1)du+(λu-Δu)dt=(f(x,u)+g(x))dt+∑j=1m‍hjdwj,hhhhhhhhhhhhhhhhhhhhhhhhhhhhhx∈ℝn,t>0,
with the initial condition
(2)u(x,0)=u0(x),x∈ℝn,
where λ is a positive constant, g is a given function in L2(ℝn), for each j=1,…,m,hj∈H2(ℝn)⋂‍W2,p(ℝn)⋂‍L∞(ℝn) for some p≥2,  {wj}j=1m are independent two-sided real-valued Wiener processes on a probability space which will be specified below, and f is a nonlinear function satisfying the following conditions (see, e.g., [1, 2]). For all x∈ℝn and s∈ℝ,
(3)f(x,s)s≤-α1|s|p+ψ1(x),(4)|f(x,s)|≤α2|s|p-1+ψ2(x),(5)∂f∂s(x,s)≤β,(6)|∂f∂x(x,s)|≤ψ3(x),
where α1,α2 and β are positive constants, ψ1∈L1(ℝn)⋂‍L∞(ℝn), ψ2∈L2(ℝn)⋂‍Lp(ℝn)⋂‍L4(ℝn), and ψ3∈L2(ℝn).As we know, the asymptotic behavior of a random dynamical system (RDS) is characterized by random attractors, which were first introduced by Crauel and Flandoli [3] and Schmalfuss [4] and then developed in [1, 2, 5–12] and among others. Recently, the existence of random attractors of the RDS associated with problem (1)-(2) was studied by many authors. For example, in [1, 2] the authors proved the existence of (L2(ℝn),L2(ℝn))-random attractor and (L2(ℝn),Lp(ℝn))-random attractor, respectively, in the case of additive noise. Wang and Zhou obtained (L2(ℝn),L2(ℝn))-random attractor in [12] and Li et al. proved the existence of (L2(D),Lp(D))-random attractor in bounded domains in [10] in the case of multiplicative noise. A necessary and sufficient condition for the existence of random attractors for the so-called quasicontinuous RDS was established in [9], and in the most recent papers [13, 14], the author employed this result to prove the existence of random attractors for some reaction-diffusion equations with additive noise and multiplicative noise on H1, respectively, when the domain is bounded. In this paper, we study the existence of (L2(ℝn),H1(ℝn))-random attractor with additive noise for the same problem in the entire space ℝn.For our problem, there are two difficulties when we consider the existence of(L2,H1)-random attractor. The first is the lack of compactness of Sobolev embeddings when the domain is unbounded. It is worth mentioning that in deterministic case differential equations of this type were extensively studied in both autonomous and nonautonomous cases and in both bounded domains and unbounded domains [15–29]. In the case of unbounded domains the difficulty of noncompact embeddings can be overcome by the energy equation approach introduced by Ball in [30, 31] and other methods. We are interested in the method used in [22] for the deterministic version of the initial problem (1)-(2) on ℝn. In [22] the author approached ℝn by a bounded ball and found that the approximation error of the norm of solutions is arbitrary small uniformly for large time, and thus they proved asymptotic compactness by passing the limit of the energy equation. More recently, the idea of the tail estimate was used in [1] to prove the existence of random attractor in L2(ℝn) for the SRDE (1)-(2). In this paper, we use an extended version of the tail estimate described in [22] to overcome the difficulty of noncompact embeddings.Another difficulty is that one can not differentiate the stochastic equation with respect to timet in usual sense. In the case of deterministic equation, by differentiating the reaction-diffusion equation with respect to t, one can prove the existence of (L2(ℝn),H1(ℝn)) or (L2(D),H01(D)) (D⊂ℝn is bounded) attractors; see [24, 27, 29, 32] for autonomous equations and [23, 26, 33] for non-autonomous equations. But in stochastic case this idea breaks down, since, as we know, neither the Winner process nor the Ornstein-Uhlenbeck process is differentiable with respect to t in usual sense. However, this is only a matter of method or estimate. In [25], the author used a result for compactness in H1(ℝn) introduced in [17] to establish the asymptotic compactness in H1(ℝn) without differentiating the equation. Unfortunately, the growth order p is restricted in that case. In this paper, we overcome this drawback by using an appropriate estimate motivated by the works in [19] and the estimate is accurate enough so that we needn't differentiate the equation as usual.This paper is organized as follows. In Section2, we recall some basic notions of bispaces random attractors for RDS. In Section 3, we transform the problem (1)-(2) into a parameterized evolution equation and obtain the corresponding RDS. In Section 4, we give some uniform estimates of the solutions as t→∞. In Section 5, we prove the asymptotic compactness and the existence of an (L2(ℝn),H1(ℝn))-random attractor.Throughout this paper, we denote by∥·∥X the norm of Banach space  X and by (·,·)H the inner product in Hilbert space H. The inner product and norm of L2(ℝn) are written as (·,·) and ∥·∥, respectively. We also use ∥u∥r to denote the norm of u∈Lr(ℝn) (r≥1, r≠2) and |u| to denote the modular of u. The letter c denotes any positive constant which may be different from line to line or even in the same line (sometimes for special case, we also denote the different positive constants by ci(i=1,2,…)).


## 2. Preliminaries and Abstract Results

In this section, we first recall some basic concepts related to random attractors for RDS (see [1, 3, 5–8, 34] for more detail) and then give some abstract results on the existence of (L2(ℝn),H1(ℝn))-random attractors.


### 2.1. Preliminaries

LetX, Y be two Banach spaces with Borel σ-algebra  ℬ(X) and  ℬ(Y), respectively, and let  (Ω,ℱ,ℙ) be a probability space.Definition 1.
(
Ω
,
ℱ
,
ℙ
,
(
θ
t
)
t
∈
ℝ
) is called a metric dynamical system (MDS) if  θ:ℝ×Ω→Ω is (ℬ(ℝ)×ℱ,ℱ)-measurable, and  θ0 is the identity on Ω, θs+t =  θs∘θt for all s, t∈ℝ and  θtℙ=ℙ for all  t∈ℝ.Definition 2.
An RDS onX over an MDS (Ω,ℱ,ℙ,(θt)t∈ℝ) is a mapping ϕ:ℝ+×Ω×X→X, (t,ω,x)→ϕ(t,ω,x) which is  (ℬ(ℝ+)×ℱ×ℬ(X),ℬ(X))-measurable and satisfies that, for  ℙ-a.e. ω∈Ω,(i)
ϕ(0,ω,·) = id on X;(ii)
ϕ(t+s,ω,·)=ϕ(t,θsω,·)∘ϕ(s,ω,·) (cocycle property) on X for all s,t∈ℝ+.
An RDS is said to be continuous onX if ϕ(t,ω):X→X is continuous for all t∈ℝ+ and  ℙ-a.e. ω∈Ω.Definition 3.
A random set is a set-valued mapA:Ω→2X, ω↦A(ω), which satisfies that, for each x∈X, the map ω↦d(x,A(ω)) is measurable. A random set {A(ω)} is called a random closed (compact) set if A(ω) is closed (compact) for all ω∈Ω. A random set {A(ω)} is called a random bounded set if there exist x0∈X and a random variable r(ω)>0 such that, for all ω∈Ω,
(7)A(ω)⊂{x∈X:d(x,x0)≤r(ω)}.Definition 4.
(
1
) A random bounded set {B(ω)}ω∈Ω of X is called tempered with respect to (θt)t∈ℝ if, for  ℙ-a.e. ω∈Ω,
(8)limt→∞e-βtd(B(θ-tω))=0,∀β>0,
where
(9)d(B)=supx∈B∥x∥X.
(
2
) A random variable r(ω)≥0 is called tempered with respect to (θt)t∈ℝ if, for  ℙ-a.e. ω∈Ω,
(10)limt→∞e-βtr(θ-tω)=0,∀β>0.
Next, we introduce some notions about the bi-spaces random attractors which are motivated by the works in [2, 20, 25, 35]. We assume that ϕ is an RDS on X and Y over an MDS (Ω,ℱ,ℙ,(θt)t∈ℝ), respectively. Let  𝒫(X) denote the family of all nonempty subsets of X and 𝒮X the class of all families D^={D(ω)}ω∈Ω⊂𝒫(X). 𝒫(Y), and  𝒮Y can be defined in the same way. We consider the given nonempty subclasses 𝒟X,𝒟Y, where 𝒟X⊂𝒮X, 𝒟Y⊂𝒮Y.Definition 5.
A familyB^={B(ω)}ω∈Ω∈𝒟Y is said to be (X,Y)-random absorbing for ϕ if, for every D^={D(ω)}ω∈Ω∈𝒟X, there exists TD^(ω)>0 such that, for  ℙ-a.e. ω∈Ω,
(11)ϕ(t,θ-tω,D(θ-tω))⊂B(ω),∀t≥TD^(ω).Definition 6.
A familyC^={C(ω)}ω∈Ω∈𝒟Y is said to be (X,Y)-random attracting for ϕ if, for every D^={D(ω)}ω∈Ω∈𝒟X,  we have, for ℙ-a.e. ω∈Ω,
(12)limt→∞dY(ϕ(t,θ-tω,D(θ-tω)),C(ω))=0,
where dY(C1,C2) denotes the Hausdorff semi-distance between C1 and C2 in Y; that is,
(13)dY(C1,C2)=supx∈C1infy∈C2∥x-y∥YforC1,C2⊂Y.Definition 7.
The RDSϕ is said to be (X,Y)-asymptotically compact if, for ℙ-a.e. ω∈Ω, {ϕ(tn,θ-tnω,xn)}n=1∞ has a convergent subsequence in Y whenever tn→∞ and xn∈D(θ-tnω) with D^={D(ω)}ω∈Ω∈𝒟X.Definition 8.
A random set A^={A(ω)}ω∈Ω∈𝒮Y is said to be an (X,Y)-random attractor if the following conditions are satisfied for ℙ-a.e. ω∈Ω, (i)
A(ω) is closed in X and compact in Y;(ii)
A^={A(ω)}ω∈Ω is invariant; that is, ϕ(t,ω,A(ω))=A(θtω) for all t≥0;(iii)
A^={A(ω)}ω∈Ω attracts every random set in 𝒟X in the norm topology of Y in the sense of (12).


### 2.2. Abstract Results

Now, we present the main abstract results. Recall that a collection𝒟 of random subsets is called inclusion closed if whenever {E(ω)}ω∈Ω is an arbitrary random set and {F(ω)}ω∈Ω is in  𝒟 with E(ω)⊂F(ω) for all ω∈Ω, then {E(ω)}ω∈Ω∈𝒟.The following theorem is an adaptation of a result of [25] to the case of RDS. The proof is similar to that of [25], and here we omit it.Theorem 9.
Letϕ be a continuous RDS on X and an RDS on Y over (Ω,ℱ,ℙ,(θt)t∈ℝ), respectively, and  𝒟X and 𝒟Y are inclusion closed.(i)
Case  1  (X=Y)  (see [1]). Assume that the family B^0={B0(ω)}ω∈Ω∈𝒟X is a closed (X,X)-random absorbing set for ϕ and ϕ is (X,X)-asymptotically compact. Then ϕ has a unique (X,X)-random attractor A^={A(ω)}ω∈Ω which is given by
(14)A(ω)=⋂s≥0‍⋃t≥s‍ϕ(t,θ-tω,B0(θ-tω))¯X,whereA¯X denotes the closure of A with respect to the norm topology in X.(ii)
Case  2  (X≠Y). If the assumptions in (i) are satisfied, moreover, we assume that  B^1={B1(ω)}ω∈Ω∈𝒟Y is (X,Y)-random absorbing and ϕ is (X,Y)-asymptotically compact. Then ϕ has an (X,Y)-random attractor A′^={A′(ω)}ω∈Ω which is given by
(15)A′(ω)=⋂s≥0‍⋃t≥s‍ϕ(t,θ-tω,B0(θ-tω)⋂‍B1(θ-tω))¯XA′(ω)=⋂s≥0‍⋃t≥s‍ϕ(t,θ-tω,B0(θ-tω)⋂‍B1(θ-tω))¯Y,whereB^0={B0(ω)}ω∈Ω∈𝒟X is the (X,X)-random absorbing set in (i).In the following of this paper we only considerX=L2(ℝn), Y=H1(ℝn), and 𝒟X=𝒟2, 𝒟Y=𝒟v, where 𝒟2 and 𝒟v denote the collections of all tempered random subsets of L2(ℝn) and H1(ℝn), respectively.Theorem 10.
Assume thatϕ is an RDS on L2(ℝn) and H1(ℝn), respectively, and then ϕ is (L2(ℝn),H1(ℝn))-asymptotically compact if(i)
for everyD^={D(ω)}ω∈Ω∈𝒟2, ℙ-a.e. ω∈Ω and every ɛ>0, there exist k0=k0(ω,ɛ)>0 and TD^(ω,ɛ)>0 such that, for all t≥TD^(ω,ɛ),
(16)∥χQk0c·ϕ(t,θ-tω,x)∥H1(Qk0c)≤ɛ,∀x∈D(θ-tω),(ii)
χQk·ϕ is (L2(ℝn),H1(Qk))-asymptotically compact, ∀k≥1,  where Qk0={x∈ℝn:|x|≤k0}, Qk0c=ℝn∖Qk0, Qk={x∈ℝn:|x|≤k}, and χA is the identical function on A.Proof.
It suffices to check that, for allD^={D(ω)}ω∈Ω∈𝒟2 and  ℙ-a.e. ω∈Ω, we can extract a Cauchy subsequence {ϕ(tn′,θ-tn′ω,xn′)} from {ϕ(tn,θ-tnω,xn)}, whenever tn→∞ and xn∈D(θ-tnω). We assume that there is Ω~⊂Ω of full  ℙ-measure such that assumption (i) holds for every ω∈Ω~. We now fix ω∈Ω~ and ɛ>0, and then by (i) there exist k0=k0(ω,ɛ)>0 and TD^(ω,ɛ)>0 such that for all t≥TD^(ω,ɛ),
(17)∥χQk0c·ϕ(t,θ-tω,x)∥H1(Qk0c)≤ɛ,∀x∈D(θ-tω).
On the other hand, by (ii), χQk·ϕ is (L2(ℝn),H1(Qk))-asymptotically compact, for all k≥1. For the above k0, there is a subsequence {ϕ(tn′,θ-tn′ω,xn′)} such that {χQk0·ϕ(tn′,θ-tn′ω,xn′)} is convergent in H1(Qk0). Therefore, there exists an integer N(D^,ω,ɛ)>0 such that  for all n′,m′≥N(D^,ω,ɛ), we have
(18)∥ϕ(tn′,θ-tn′ω,xn′)-ϕ(tm′,θ-tm′ω,xm′)∥H1(ℝn)2=∥χQk0·(ϕ(tn′,θ-tn′ω,xn′)-ϕ(tm′,θ-tm′ω,xm′))χQk0∥H1(Qk0)2+∥χQk0c·(ϕ(tn′,θ-tn′ω,xn′)-ϕ(tm′,θ-tm′ω,xm′))χQk0∥H1(Qk0c)2≤∥χQk0·ϕ(tn′,θ-tn′ω,xn′)-χQk0·ϕ(tm′,θ-tm′ω,xm′)∥H1(Qk0)2+2∥χQk0c·ϕ(tn′,θ-tn′ω,xn′)∥H1(Qk0c)2+2∥χQk0c·ϕ(tm′,θ-tm′ω,xm′)∥H1(Qk0c)2≤5ɛ.
The proof is complete.Remark 11.
If we replaceH1(ℝn) by other Banach spaces in Theorem 10, such as L2(ℝn), Lp(ℝn) and  H2(ℝn), the corresponding results also hold true. In particular, in the deterministic case, it is the exact method used in [22] when H1(ℝn) is replaced by L2(ℝn).


## 2.1. Preliminaries

LetX, Y be two Banach spaces with Borel σ-algebra  ℬ(X) and  ℬ(Y), respectively, and let  (Ω,ℱ,ℙ) be a probability space.Definition 1.
(
Ω
,
ℱ
,
ℙ
,
(
θ
t
)
t
∈
ℝ
) is called a metric dynamical system (MDS) if  θ:ℝ×Ω→Ω is (ℬ(ℝ)×ℱ,ℱ)-measurable, and  θ0 is the identity on Ω, θs+t =  θs∘θt for all s, t∈ℝ and  θtℙ=ℙ for all  t∈ℝ.Definition 2.
An RDS onX over an MDS (Ω,ℱ,ℙ,(θt)t∈ℝ) is a mapping ϕ:ℝ+×Ω×X→X, (t,ω,x)→ϕ(t,ω,x) which is  (ℬ(ℝ+)×ℱ×ℬ(X),ℬ(X))-measurable and satisfies that, for  ℙ-a.e. ω∈Ω,(i)
ϕ(0,ω,·) = id on X;(ii)
ϕ(t+s,ω,·)=ϕ(t,θsω,·)∘ϕ(s,ω,·) (cocycle property) on X for all s,t∈ℝ+.
An RDS is said to be continuous onX if ϕ(t,ω):X→X is continuous for all t∈ℝ+ and  ℙ-a.e. ω∈Ω.Definition 3.
A random set is a set-valued mapA:Ω→2X, ω↦A(ω), which satisfies that, for each x∈X, the map ω↦d(x,A(ω)) is measurable. A random set {A(ω)} is called a random closed (compact) set if A(ω) is closed (compact) for all ω∈Ω. A random set {A(ω)} is called a random bounded set if there exist x0∈X and a random variable r(ω)>0 such that, for all ω∈Ω,
(7)A(ω)⊂{x∈X:d(x,x0)≤r(ω)}.Definition 4.
(
1
) A random bounded set {B(ω)}ω∈Ω of X is called tempered with respect to (θt)t∈ℝ if, for  ℙ-a.e. ω∈Ω,
(8)limt→∞e-βtd(B(θ-tω))=0,∀β>0,
where
(9)d(B)=supx∈B∥x∥X.
(
2
) A random variable r(ω)≥0 is called tempered with respect to (θt)t∈ℝ if, for  ℙ-a.e. ω∈Ω,
(10)limt→∞e-βtr(θ-tω)=0,∀β>0.
Next, we introduce some notions about the bi-spaces random attractors which are motivated by the works in [2, 20, 25, 35]. We assume that ϕ is an RDS on X and Y over an MDS (Ω,ℱ,ℙ,(θt)t∈ℝ), respectively. Let  𝒫(X) denote the family of all nonempty subsets of X and 𝒮X the class of all families D^={D(ω)}ω∈Ω⊂𝒫(X). 𝒫(Y), and  𝒮Y can be defined in the same way. We consider the given nonempty subclasses 𝒟X,𝒟Y, where 𝒟X⊂𝒮X, 𝒟Y⊂𝒮Y.Definition 5.
A familyB^={B(ω)}ω∈Ω∈𝒟Y is said to be (X,Y)-random absorbing for ϕ if, for every D^={D(ω)}ω∈Ω∈𝒟X, there exists TD^(ω)>0 such that, for  ℙ-a.e. ω∈Ω,
(11)ϕ(t,θ-tω,D(θ-tω))⊂B(ω),∀t≥TD^(ω).Definition 6.
A familyC^={C(ω)}ω∈Ω∈𝒟Y is said to be (X,Y)-random attracting for ϕ if, for every D^={D(ω)}ω∈Ω∈𝒟X,  we have, for ℙ-a.e. ω∈Ω,
(12)limt→∞dY(ϕ(t,θ-tω,D(θ-tω)),C(ω))=0,
where dY(C1,C2) denotes the Hausdorff semi-distance between C1 and C2 in Y; that is,
(13)dY(C1,C2)=supx∈C1infy∈C2∥x-y∥YforC1,C2⊂Y.Definition 7.
The RDSϕ is said to be (X,Y)-asymptotically compact if, for ℙ-a.e. ω∈Ω, {ϕ(tn,θ-tnω,xn)}n=1∞ has a convergent subsequence in Y whenever tn→∞ and xn∈D(θ-tnω) with D^={D(ω)}ω∈Ω∈𝒟X.Definition 8.
A random set A^={A(ω)}ω∈Ω∈𝒮Y is said to be an (X,Y)-random attractor if the following conditions are satisfied for ℙ-a.e. ω∈Ω, (i)
A(ω) is closed in X and compact in Y;(ii)
A^={A(ω)}ω∈Ω is invariant; that is, ϕ(t,ω,A(ω))=A(θtω) for all t≥0;(iii)
A^={A(ω)}ω∈Ω attracts every random set in 𝒟X in the norm topology of Y in the sense of (12).


## 2.2. Abstract Results

Now, we present the main abstract results. Recall that a collection𝒟 of random subsets is called inclusion closed if whenever {E(ω)}ω∈Ω is an arbitrary random set and {F(ω)}ω∈Ω is in  𝒟 with E(ω)⊂F(ω) for all ω∈Ω, then {E(ω)}ω∈Ω∈𝒟.The following theorem is an adaptation of a result of [25] to the case of RDS. The proof is similar to that of [25], and here we omit it.Theorem 9.
Letϕ be a continuous RDS on X and an RDS on Y over (Ω,ℱ,ℙ,(θt)t∈ℝ), respectively, and  𝒟X and 𝒟Y are inclusion closed.(i)
Case  1  (X=Y)  (see [1]). Assume that the family B^0={B0(ω)}ω∈Ω∈𝒟X is a closed (X,X)-random absorbing set for ϕ and ϕ is (X,X)-asymptotically compact. Then ϕ has a unique (X,X)-random attractor A^={A(ω)}ω∈Ω which is given by
(14)A(ω)=⋂s≥0‍⋃t≥s‍ϕ(t,θ-tω,B0(θ-tω))¯X,whereA¯X denotes the closure of A with respect to the norm topology in X.(ii)
Case  2  (X≠Y). If the assumptions in (i) are satisfied, moreover, we assume that  B^1={B1(ω)}ω∈Ω∈𝒟Y is (X,Y)-random absorbing and ϕ is (X,Y)-asymptotically compact. Then ϕ has an (X,Y)-random attractor A′^={A′(ω)}ω∈Ω which is given by
(15)A′(ω)=⋂s≥0‍⋃t≥s‍ϕ(t,θ-tω,B0(θ-tω)⋂‍B1(θ-tω))¯XA′(ω)=⋂s≥0‍⋃t≥s‍ϕ(t,θ-tω,B0(θ-tω)⋂‍B1(θ-tω))¯Y,whereB^0={B0(ω)}ω∈Ω∈𝒟X is the (X,X)-random absorbing set in (i).In the following of this paper we only considerX=L2(ℝn), Y=H1(ℝn), and 𝒟X=𝒟2, 𝒟Y=𝒟v, where 𝒟2 and 𝒟v denote the collections of all tempered random subsets of L2(ℝn) and H1(ℝn), respectively.Theorem 10.
Assume thatϕ is an RDS on L2(ℝn) and H1(ℝn), respectively, and then ϕ is (L2(ℝn),H1(ℝn))-asymptotically compact if(i)
for everyD^={D(ω)}ω∈Ω∈𝒟2, ℙ-a.e. ω∈Ω and every ɛ>0, there exist k0=k0(ω,ɛ)>0 and TD^(ω,ɛ)>0 such that, for all t≥TD^(ω,ɛ),
(16)∥χQk0c·ϕ(t,θ-tω,x)∥H1(Qk0c)≤ɛ,∀x∈D(θ-tω),(ii)
χQk·ϕ is (L2(ℝn),H1(Qk))-asymptotically compact, ∀k≥1,  where Qk0={x∈ℝn:|x|≤k0}, Qk0c=ℝn∖Qk0, Qk={x∈ℝn:|x|≤k}, and χA is the identical function on A.Proof.
It suffices to check that, for allD^={D(ω)}ω∈Ω∈𝒟2 and  ℙ-a.e. ω∈Ω, we can extract a Cauchy subsequence {ϕ(tn′,θ-tn′ω,xn′)} from {ϕ(tn,θ-tnω,xn)}, whenever tn→∞ and xn∈D(θ-tnω). We assume that there is Ω~⊂Ω of full  ℙ-measure such that assumption (i) holds for every ω∈Ω~. We now fix ω∈Ω~ and ɛ>0, and then by (i) there exist k0=k0(ω,ɛ)>0 and TD^(ω,ɛ)>0 such that for all t≥TD^(ω,ɛ),
(17)∥χQk0c·ϕ(t,θ-tω,x)∥H1(Qk0c)≤ɛ,∀x∈D(θ-tω).
On the other hand, by (ii), χQk·ϕ is (L2(ℝn),H1(Qk))-asymptotically compact, for all k≥1. For the above k0, there is a subsequence {ϕ(tn′,θ-tn′ω,xn′)} such that {χQk0·ϕ(tn′,θ-tn′ω,xn′)} is convergent in H1(Qk0). Therefore, there exists an integer N(D^,ω,ɛ)>0 such that  for all n′,m′≥N(D^,ω,ɛ), we have
(18)∥ϕ(tn′,θ-tn′ω,xn′)-ϕ(tm′,θ-tm′ω,xm′)∥H1(ℝn)2=∥χQk0·(ϕ(tn′,θ-tn′ω,xn′)-ϕ(tm′,θ-tm′ω,xm′))χQk0∥H1(Qk0)2+∥χQk0c·(ϕ(tn′,θ-tn′ω,xn′)-ϕ(tm′,θ-tm′ω,xm′))χQk0∥H1(Qk0c)2≤∥χQk0·ϕ(tn′,θ-tn′ω,xn′)-χQk0·ϕ(tm′,θ-tm′ω,xm′)∥H1(Qk0)2+2∥χQk0c·ϕ(tn′,θ-tn′ω,xn′)∥H1(Qk0c)2+2∥χQk0c·ϕ(tm′,θ-tm′ω,xm′)∥H1(Qk0c)2≤5ɛ.
The proof is complete.Remark 11.
If we replaceH1(ℝn) by other Banach spaces in Theorem 10, such as L2(ℝn), Lp(ℝn) and  H2(ℝn), the corresponding results also hold true. In particular, in the deterministic case, it is the exact method used in [22] when H1(ℝn) is replaced by L2(ℝn).


## 3. The Reaction-Diffusion Equation onℝn with Additive Noise

We consider the probability space(Ω,ℱ,ℙ) where
(19)Ω={ω=(ω1,ω2,…,ωm)∈C(ℝ,ℝm):ω(0)=0},ℱ is the Borel σ-algebra induced by the compact-open topology of Ω, and  ℙ the corresponding Wiener measure on (Ω,ℱ). Then we will identify ω with
(20)W(t)≡(w1(t),w2(t),…,wm(t))=ω(t)fort∈ℝ.
Define the time shift by(21)θtω(·)=ω(·+t)-ω(t),ω∈Ω,t∈ℝ,
and then (Ω,ℱ,ℙ,(θt)t∈ℝ) is an MDS.We now translate the stochastic equation (1)-(2) into a deterministic equation with a random parameter.To this end, we consider the one-dimensional Ornstein-Uhlenbeck process given by(22)zj(t)=zj(θtωj)≡-λ∫-∞0‍eλτ(θtωj)(τ)dτ,∀t∈ℝ,
which solves the Itô differential equation
(23)dzj+λzjdt=dωj(t).
Note that the random variable |zj(ωj)| is tempered and zj(θtωj) is ℙ-a.e. continuous in t∈ℝ. Therefore, it follows from the Proposition 4.3.3 in [34] that there exists a tempered function r(ω)>0 such that
(24)p(ω)≔∑j=1m‍(|zj(ωj)|2+|zj(ωj)|4+|zj(ωj)|p+|zj(ωj)|2p+|zj(ωj)|2p-2)≤r(ω),
where r(ω) satisfies that for ℙ-a.e. ω∈Ω,
(25)r(θtω)≤e(λ/2)|t|r(ω),∀t∈ℝ.
Therefore, for  ℙ-a.e. ω∈Ω,
(26)p(θtω)≤e(λ/2)|t|r(ω),∀t∈ℝ.
Putting
(27)z(θtω)=∑j=1m‍hjzj(θtωj),
then by (23) we have
(28)dz+λzdt=∑j=1m‍hjdωj.Remark 12.
From (24) and (27), we can easily show that the sum
(29)∥z(ω)∥2+∥z(ω)∥pp+∥z(ω)∥2p2p+∥z(ω)∥2p-22p-2+∥z(ω)∥44+∥∇z(ω)∥2+∥∇z(ω)∥2+∥Δz(ω)∥pp,
is bounded by p(ω) with a deterministic positive constant c0. In the following of this paper, we use the symbols p(ω) and r(ω) to denote the random variables in (24).In order to show that the initial problem (1)-(2) generates an RDS, we set v(t)=u(t)-z(θtω). Then we can consider the following evolution equation with random parameter but without white noise:
(30)∂v∂t+λv-Δv=f(x,v+z(θtω))+g+Δz(θtω),
with initial value condition
(31)v(x,0)=v0(ω)=u0-z(ω).From [1, 2], we see that for  ℙ-a.e. ω∈Ω and all v0∈L2(ℝn), the parameterized evolution equation (30)-(31) with conditions (3)–(6) has a unique solution
(32)v(·,ω,v0)∈C([0,∞);L2(ℝn))⋂Lp((0,T);Lp(ℝn))⋂L2((0,T);H1(ℝn)),
for every T>0. Furthermore, v(t,ω,v0) is continuous with respect to v0 in L2(ℝn), for all t≥0.As(33)u(t,ω,u0)=v(t,ω,u0-z(ω))+z(θtω),
the process u is the solution to the problem (1)-(2) in a certain sense. We now define a mapping ϕ:ℝ+×Ω×L2(ℝn)→L2(ℝn) by
(34)ϕ(t,ω,u0)=u(t,ω,u0)=v(t,ω,u0-z(ω))+z(θtω),
for all  (t,ω,u0)∈ℝ+×Ω×L2(ℝn). Then ϕ is a continuous RDS on L2(ℝn) and an RDS on H1(ℝn) respectively associated with the initial value problem of SRDE (1)-(2) on ℝn.Theorem 13 (see [1, 2]).
Assume thatg∈L2(ℝn) and (3)–(6) hold. Then the RDS ϕ generated by (1)-(2) has a unique (L2(ℝn),L2(ℝn))-random attractor A^2 and has a unique (L2(ℝn),Lp(ℝn))-random attractor A^p; furthermore, we have A^2=A^p.


## 4. Uniform Estimates of Solutions

### 4.1.(L2(ℝn),H1(ℝn))-Random Absorbing Set and Some Useful Estimates

The next lemma shows thatϕ has a tempered (L2(ℝn),H1(ℝn))-random absorbing set.Lemma 14 (see [1]).
Assume thatg∈L2(ℝn) and (3)–(6) hold. Let D^={D(ω)}ω∈Ω∈𝒟2 and u0(ω)∈D(ω). Then, for ℙ-a.e. ω∈Ω, there exists TD^(ω)>0 such that,  for all t≥TD^(ω),
(35)∥u(t,θ-tω,u0(θ-tω))∥H1(ℝn)2≤c(1+r(ω)),
where c is a constant independent of t, ω, and u0.We now give some new estimates for the solutionv(t,ω,v0) of (30)-(31).Lemma 15.
Assume thatg∈L2(ℝn) and (3)–(6) hold. Let D^={D(ω)}ω∈Ω∈𝒟2 and u0(ω)∈D(ω). Then, for  ℙ-a.e. ω∈Ω, there exists TD^(ω)≥2, such that the solution v(t,ω,v0(ω)) of (30)-(31) satisfies that,  for all t≥TD^(ω) and  for all s∈[t-1,t+1],
(36)∥v(s,θ-t-1ω,v0(θ-t-1ω))∥2≤c(1+r(ω)),(37)∥v(s,θ-t-1ω,v0(θ-t-1ω))∥pp≤c(1+r(ω)),(38)∥∇v(s,θ-t-1ω,v0(θ-t-1ω))∥2≤c(1+r(ω)),
where c is a constant independent of s, t, and ω.Proof.
The first assertion was proved in [2] in the case of s∈[t,t+1], and the case for s∈[t-1,t+1] can be obtained by slightly modifying the proof of Lemma 4.4 in [2] (in fact, for s∈[t-2,t+1], (36) also holds true, and we will use this result in (55)), and here we omit it.
Now, we prove the second assertion. Multiplying (30) with |v|p-2v and integrating over ℝn, we get
(39)1pddt∥v∥pp+λ∥v∥pp+(-Δv,|v|p-2v)=∫ℝn‍f(x,v+z(θtω))|v|p-2vdx+(g+Δz(θtω),|v|p-2v).
Since p≥2, we have
(40)(-Δv,|v|p-2v)=(∇v,∇(|v|p-2v))=(∇v,(p-2)|v|p-4v2∇v)+(∇v,|v|p-2∇v)≥0.
For the nonlinearity, similar to (4.6) and (4.8) in [2], we have
(41)f(x,v+z(θtω))v≤-α12p|v|p+c(|z(θtω)|p+|z(θtω)|2)+ψ1+12ψ22,(42)∫ℝn‍f(x,v+z(θtω))|v|p-2vdx≤-α12p+1∥v∥2p-22p-2+λ∥v∥pp+c(∥z(θtω)∥2p-22p-2+∥z(θtω)∥pp)+c(∥ψ1∥p/2p/2+∥ψ2∥pp),(43)(g+Δz(θtω),|v|p-2v)≤∥g∥∥v∥2p-2p-1+∥Δz(θtω)∥∥v∥2p-2p-1≤α12p+2∥v∥2p-22p-2+c∥Δz(θtω)∥2+c∥g∥2.
From (39)-(40) and (42)-(43), we get
(44)ddt∥v∥pp+c∥v∥2p-22p-2≤c1(∥z(θtω)∥2p-22p-2+∥z(θtω)∥pp+∥Δz(θtω)∥2)+c1≤c2p(θtω)+c2.
On the other hand, multiplying (30) by  v and integrating over ℝn, we get the results in [1]:
(45)ddt∥v∥2+λ∥v∥2+∥∇v∥2+α1∥u∥pp≤cp(θtω)+c,(46)∥v(t,ω,v0(ω))∥2≤e-λt∥v0(ω)∥2+c∫0t‍eλ(τ-t)p(θτω)dτ+c.
By Hölder inequality, ∥u∥pp≥21-p∥v∥pp-∥z(θtω)∥pp, we can convert (45) into
(47)ddt∥v∥2+λ∥v∥2+∥∇v∥2+α121-p∥v∥pp≤cp(θtω)+c.
For any t≥0, integrating (47) over (t,t+1) and using (46), we get
(48)∫tt+1‍∥v(s)∥ppds≤c(∫tt+1‍∥v(t)∥2+∫tt+1‍p(θτω)dτ+1)≤c(e-λt∥v0(ω)∥2+∫0t‍eλ(τ-t)p(θτω)dτ+∫tt+1‍p(θτω)dτ+1).
Next, fix s∈(t,t+1) and integrate (44) over (s,t+1) to get
(49)∥v(t+1,ω,v0(ω))∥pp≤∥v(s,ω,v0(ω))∥pp+c2∫st+1‍p(θτω)dτ+c2(t+1-s)≤∥v(s,ω,v0(ω))∥pp+c2∫tt+1‍p(θτω)dτ+c2.
Integrating the above inequality with respect to s over (t,t+1) and using (48), we obtain, for all t≥0,
(50)∥v(t+1,ω,v0(ω))∥pp≤∫tt+1‍∥v(s,ω,v0(ω))∥ppds+c2∫tt+1‍p(θsω)ds+c2≤c(e-λt∥v0(ω)∥2+∫0t‍eλ(τ-t)p(θτω)dτ+∫tt+1‍p(θτω)dτ+1)≤c(e-λ(t+1)∥v0(ω)∥2+∫0t+1‍eλ(τ-t-1)p(θτω)dτ+∫tt+1‍p(θτω)dτ+1).
Replacing t+1 by s first, then substituting  θ-t-1ω for ω in the aforementioned inequality, and noting that s∈[t-1,t+1](t≥2), we have
(51)∥v(s,θ-t-1ω,v0(θ-t-1ω))∥pp≤c[e-λs∥v0(θ-t-1ω)∥2+∫0s‍eλ(τ-s)p(θτ-t-1ω)dτ+∫s-1s‍p(θτ-t-1ω)dτ+1]≤c[∫t-2t+1‍eλ(t+1-s)e-λ(t+1)∥v0(θ-t-1ω)∥2+∫0t+1‍eλ(t+1-s)eλ(τ-t-1)p(θτ-t-1ω)dτ+∫t-2t+1‍p(θτ-t-1ω)dτ+1]≤c[∫t-2t+1‍e-λ(t+1)∥v0(θ-t-1ω)∥2+∫0t+1‍eλ(τ-t-1)p(θτ-t-1ω)dτ+∫t-2t+1‍p(θτ-t-1ω)dτ+1]=c[e-λ(t+1)∥v0(θ-t-1ω)∥2+∫-t-10‍eλτp(θτω)dτ+∫-30‍p(θτω)dτ+1]≤c[e-λ(t+1)∥v0(θ-t-1ω)∥2+r(ω)∫-t-10‍e(λ/2)τdτ+r(ω)∫-30‍e-(λ/2)τdτ+1],
where we have used (26) in the pervious inequality. Noting that u0(ω)∈D(ω) with D^={D(ω)}ω∈Ω∈𝒟2 and v0(ω)=u0(ω)-z(ω), we get from (51) that there exists TD^(ω)≥2 such that,  for all t≥TD^(ω) and for all s∈[t-1,t+1],
(52)∥v(s,θ-t-1ω,v0(θ-t-1ω))∥pp≤c(1+r(ω)).
That is, (37) holds true.
To prove (38), we take the inner product of (30) with -Δv in L2(Ω), and using  (4.31) in [1], we get
(53)ddt∥∇v∥2≤c(∥∇u∥2+∥u∥pp)+c(∥Δz(θtω)∥2+∥Δz(θtω)∥pp+1).
This implies that
(54)ddt∥∇v∥2≤c(∥∇v∥2+∥v∥pp)+c(p(θtω)+1).
Integrating (47) over (t-2,t+1) and substituting θ-t-1ω for ω, then from (36), we get
(55)∫t-2t+1‍∥∇v(s,θ-t-1ω,v0(θ-t-1ω))∥2ds≤∥v(t-2,θ-t-1ω,v0(θ-t-1ω))∥2+c(∫t-2t+1‍p(θτ-t-1ω)dτ+1)≤c(1+r(ω)),∀t≥TD^(ω).
Obviously, from (51) we can easily see that (37) also holds for s∈[t-2,t+1]; then, by (37), (47), (54)-(55), and a similar procedure as the proof of (37), one can show that,  for all t≥TD^(ω), and for all s∈[t-1,t+1],
(56)∥∇v(s,θ-t-1ω,v0(θ-t-1ω))∥2≤c(1+r(ω)).
The proof is complete.Lemma 16.
Assume thatg∈L2(ℝn) and (3)–(6) hold. Let D^={D(ω)}ω∈Ω∈𝒟2 and u0(ω)∈D(ω). Then, for  ℙ-a.e. ω∈Ω, there exists TD^(ω)>0, such that the solution v(t,ω,v0(ω)) of (30)-(31) satisfies that for all t≥TD^(ω),
(57)∫tt+1‍∥v(s,θ-t-1ω,v0(θ-t-1ω))∥2p-22p-2ds≤c(1+r(ω)),(58)∫tt+1‍∥vs(s,θ-t-1ω,v0(θ-t-1ω))∥2ds≤c(1+r(ω)).Proof.
Integrating (44) over (t,t+1), we get
(59)∫tt+1‍∥v(s)∥2p-22p-2ds≤c(∥v(t)∥pp+∫tt+1‍p(θτω)dτ+1).
Replacing ω by θ-t-1ω in the aforementioned inequality, it yields that
(60)∫tt+1‍∥v(s,θ-t-1ω,v0(θ-t-1ω))∥2p-22p-2ds≤c(∫tt+1‍∥v(t,θ-t-1ω,v0(θ-t-1ω))∥pp+∫tt+1‍p(θτ-t-1ω)dτ+1)=c(∫-10‍∥v(t,θ-t-1ω,v0(θ-t-1ω))∥pp+∫-10‍p(θτω)dτ+1)≤c(∫-10‍∥v(t,θ-t-1ω,v0(θ-t-1ω))∥pp+r(ω)∫-10‍e-(λ/2)τdτ+1).
Lemma 15 and the aforementioned inequality imply that there exists TD^(ω)>0, ∀t≥TD^(ω), such that (57) holds.
Next, taking the inner product of (30) with vt in L2(ℝn), and using (4), we obtain
(61)∥vt∥2+λ2ddt∥v∥2+12ddt∥∇v∥2=∫ℝn‍f(x,u)vtdx+∫ℝn‍gvtdx+∫ℝn‍Δz(θtω)vtdx≤α2∫ℝn‍|u|p-1|vt|dx+∫ℝn‍|ψ2||vt|dx+∫ℝn‍|g||vt|dx+∫ℝn‍|Δz(θtω)||vt|dx≤12∥vt∥2+c(∥u∥2p-22p-2+∥ψ2∥2+∥g∥2+∥Δz(θtω)∥2)≤12∥vt∥2+c(∥z(θtω)∥2p-22p-2∥v∥2p-22p-2+∥ψ2∥2+∥g∥2≤12∥vt∥2+hc+∥Δz(θtω)∥2+∥z(θtω)∥2p-22p-2),
that is,
(62)∥vt∥2+λddt∥v∥2+ddt∥∇v∥2≤c(∥v∥2p-22p-2+∥ψ2∥2+∥g∥2+∥Δz(θtω)∥2+∥z(θtω)∥2p-22p-2).
We now integrate (62) over (t,t+1) to obtain
(63)∫tt+1‍∥vs(s,ω,v0(ω))∥2ds≤c(∥v(t,ω,v0(ω))∥2+∥∇v(t,ω,v0(ω))∥2)+c∫tt+1‍(∥z(θτω)∥2p-22p-2+∥Δz(θτω)∥2)dτ+c∫tt+1‍∥v(τ,ω,v0(ω))∥2p-22p-2dτ+c(∥ψ2∥2+∥g∥2).
Replacing ω by θ-t-1ω, we get
(64)∫tt+1‍∥vs(s,θ-t-1ω,v0(θ-t-1ω))∥2ds≤c(∥v(t,θ-t-1ω,v0(θ-t-1ω))∥2+∥∇v(t,θ-t-1ω,v0(θ-t-1ω))∥2)+c∫tt+1‍(∥z(θτ-t-1ω)∥2p-22p-2+∥Δz(θτ-t-1ω)∥2)dτ+c∫tt+1‍∥v(s,θ-t-1ω,v0(θ-t-1ω))∥2p-22p-2ds+c(∥ψ2∥2+∥g∥2).
Equations (36), (38) and (57) together imply that (58) is also true. The proof is complete.


### 4.2. Tail Estimate inH1(ℝn)

We next estimate “the tail” of the solution to the problem (1)-(2) in H1(ℝn).Lemma 17.
Assume thatg∈L2(ℝn) and (3)–(6) hold. Let D^={D(ω)}ω∈Ω∈𝒟2 and u0(ω)∈D(ω). Then, for  ℙ-a.e. ω∈Ω and for every ɛ>0, there exist T*=TD^*(ω,ɛ)>0 and R*=R*(ω,ɛ), such that the solution v(t,ω,v0(ω)) of (30)-(31) satisfies that  ∀t≥T*,
(65)∫|x|≥R*‍(|v(t,θ-tω,v0(θ-tω))|2+|∇v(t,θ-tω,v0(θ-tω))|2)dx≤ɛ.Proof.
By Lemma 4.6 in [1], it suffices to prove that
(66)∫|x|≥R*‍|∇v(t,θ-tω,v0(θ-tω))|2dx≤ɛ.
Let φ be a smooth function defined on ℝ+ such that 0≤φ(s)≤1, for all s∈ℝ+, and
(67)φ(s)={0,0≤s≤1,1,s≥2.
Then there is a positive constant c such that |φ′(s)|+|φ′′(s)|≤c for all s∈ℝ+.
Multiplying (30) by -φ(|x|2/k2)Δv and integrating with respect to x over ℝn, we get
(68)12ddt∫ℝn‍φ(|x|2k2)|∇v|2dx+∫ℝn‍(∇φ(|x|2k2)·∇v)vtdx+λ∫ℝn‍φ(|x|2k2)|∇v|2dx+λ∫ℝn‍(∇φ(|x|2k2)·∇v)vdx+∫ℝn‍φ(|x|2k2)|Δv|2dx=-∫ℝn‍φ(|x|2k2)f(x,u)Δvdx-∫ℝn‍φ(|x|2k2)(g+Δz(θtω))Δvdx.
The second term of the left-hand side is bounded by
(69)∫ℝn‍|∇φ(|x|2k2)||∇v||vt|dx=∫ℝn‍|φ′(|x|2k2)||2xk2||∇v||vt|dx≤ck∫k≤|x|≤2k‍|∇v||vt|dx≤ck(∥vt∥2+∥∇v∥2).
Similarly, the forth term of the left-hand side of (68) is bounded by
(70)λ∫ℝn‍|∇φ(|x|2k2)||∇v||v|dx≤ck(∥v∥2+∥∇v∥2).
For the last term of the right-hand side of (68), we have
(71)|∫ℝn‍φ(|x|2k2)(g+Δz(θtω))Δvdx|≤12∫ℝn‍φ(|x|2k2)|Δv|2dx+∫ℝn‍φ(|x|2k2)|g|2dx+∫ℝn‍φ(|x|2k2)|Δz(θtω)|2dx.
We next consider the nonlinear term in (68). Since
(72)-∫ℝn‍φ(|x|2k2)f(x,u)Δvdx=-∫ℝn‍φ(|x|2k2)f(x,u)Δudx+∫ℝn‍φ(|x|2k2)f(x,u)Δz(θtω)dx=∫ℝn‍f(x,u)(∇φ(|x|2k2)·∇u)dx+∫ℝn‍φ(|x|2k2)(∂∂xf(x,u)·∇u)dx+∫ℝn‍φ(|x|2k2)∂∂uf(x,u)|∇u|2dx+∫ℝn‍φ(|x|2k2)f(x,u)Δz(θtω)dx.
We now estimate each term in the right-hand side of (72). Using (4), the property of  φ, and Cauchy's inequality, we see that the first term of the right-hand side of (72) is bounded by
(73)∫ℝn‍|∇φ(|x|2k2)||f(x,u)||∇u|dx≤α2∫ℝn‍|∇φ(|x|2k2)||u|p-1|∇u|dx+∫ℝn‍|∇φ(|x|2k2)||ψ2||∇u|dx≤c(∫ℝn‍|∇φ(|x|2k2)||u|2p-2dx+∫ℝn‍|∇φ(|x|2k2)||∇u|2dx)+ck∫k≤|x|≤2k‍|ψ2||∇u|dx≤ck(∥u∥2p-22p-2+∥∇u∥2+1).
By (6), we can estimate the second term of the right-hand side of (72) as follows:
(74)∫ℝn‍|φ(|x|2k2)||∂∂xf(x,u)||∇u|dx≤∫ℝn‍|φ(|x|2k2)||ψ3||∇u|dx≤λ4∫ℝn‍φ(|x|2k2)|∇u|2dx+cλ∫ℝn‍φ(|x|2k2)|ψ3|2dx≤λ2∫ℝn‍φ(|x|2k2)|∇v|2dx+c∫ℝn‍φ(|x|2k2)|∇z(θtω)|2dx+c∫ℝn‍φ(|x|2k2)|ψ3|2dx.
For the third term of the right-hand side of (72), by using (5), we have
(75)∫ℝn‍φ(|x|2k2)∂∂uf(x,u)|∇u|2dx≤β∫ℝn‍φ(|x|2k2)|∇u|2dx≤2β∫ℝn‍φ(|x|2k2)|∇v|2dx+2β∫ℝn‍φ(|x|2k2)|∇z(θtω)|2dx.
For the last term of the right-hand side of (72), by using (4) and Young's inequality, we find
(76)|∫ℝn‍φ(|x|2k2)f(x,u)Δz(θtω)dx|≤α2∫ℝn‍φ(|x|2k2)|u|p-1|Δz(θtω)|dx+∫ℝn‍φ(|x|2k2)|ψ2||Δz(θtω)|dx≤c∫ℝn‍φ(|x|2k2)|u|pdx+c∫ℝn‍φ(|x|2k2)|Δz(θtω)|pdx+12∫ℝn‍φ(|x|2k2)|ψ2|2dx+12∫ℝn‍φ(|x|2k2)|Δz(θtω)|2dx.
Putting (73)–(76) together into (72), it yields that
(77)-∫ℝn‍φ(|x|2k2)f(x,u)Δvdx≤λ+4β2∫ℝn‍φ(|x|2k2)|∇v|2dx+c∫ℝn‍φ(|x|2k2)|u|pdx+ck(∥u∥2p-22p-2+∥∇u∥2+1)+c∫ℝn‍φ(|x|2k2)(|ψ2|2+|ψ3|2)dx+c∫ℝn‍φ(|x|2k2)×(|∇z(θtω)|2+|Δz(θtω)|2+|Δz(θtω)|p)dx.
Then by (68)–(71) and (77), we get
(78)ddt∫ℝn‍φ(|x|2k2)|∇v|2dx+λ∫ℝn‍φ(|x|2k2)|∇v|2dx+∫ℝn‍φ(|x|2k2)|Δv|2dx≤ck(∥vt∥2+∥∇v∥2+∥v∥2+∥u∥2p-22p-2+∥∇u∥2+1)+4β∫ℝn‍φ(|x|2k2)|∇v|2dx+c∫ℝn‍φ(|x|2k2)|u|pdx+c∫ℝn‍φ(|x|2k2)(|∇z(θtω)|2+|Δz(θtω)|2+|Δz(θtω)|p)dx+c∫ℝn‍φ(|x|2k2)(|g|2+|ψ2|2+|ψ3|2)dx.
In particular,
(79)ddt∫ℝn‍φ(|x|2k2)|∇v(t)|2dx≤4β∫ℝn‍φ(|x|2k2)|∇v(t)|2dx+c∫ℝn‍φ(|x|2k2)|u(t)|pdx+ck(∥z(θtω)∥2p-22p-2∥vt(t)∥2+∥∇v(t)∥2+∥v(t)∥2+∥v(t)∥2p-22p-2+∥z(θtω)∥2p-22p-2+∥∇z(θtω)∥2+1∥z(θtω)∥2p-22p-2)+c∫ℝn‍φ(|x|2k2)(|∇z(θtω)|2+|Δz(θtω)|2+|Δz(θtω)|p)dx+c∫ℝn‍φ(|x|2k2)(|g|2+|ψ2|2+|ψ3|2)dx.
Let s∈(t,t+1) and integrate the aforementioned inequality from s to t+1:
(80)∫ℝn‍φ(|x|2k2)|∇v(t+1,ω,v0(ω))|2dx≤∫ℝn‍φ(|x|2k2)|∇v(s,ω,v0(ω))|2dx+4β∫tt+1‍∫ℝn‍φ(|x|2k2)|∇v(τ,ω,v0(ω))|2dxdτ+c∫tt+1‍∫ℝn‍φ(|x|2k2)|u(τ,ω,u0(ω))|pdxdτ+ck(∫tt+1‍∥vτ(τ,ω,v0(ω))∥2dτ+∫tt+1‍∥∇v(τ,ω,v0(ω))∥2dτ+∫tt+1‍∥v(τ,ω,v0(ω))∥2dτ+∫tt+1‍∥v(τ,ω,v0(ω))∥2p-22p-2dτ+∫tt+1‍∥z(θτω)∥2p-22p-2dτ+∫tt+1‍∥∇z(θτω)∥2dτ+1)+c∫tt+1‍∫ℝn‍φ(|x|2k2)×(|∇z(θτω)|2+|Δz(θτω)|2+|Δz(θτω)|p)dxdτ+c∫ℝn‍φ(|x|2k2)(|g|2+|ψ2|2+|ψ3|2)dx.
Integrating the aforementioned inequality with respect to s over (t,t+1), and replacing ω by θ-t-1ω, we obtain
(81)∫ℝn‍φ(|x|2k2)|∇v(t+1,θ-t-1ω,v0(θ-t-1ω))|2dx≤c3∫tt+1‍∫ℝn‍φ(|x|2k2)×|∇v(τ,θ-t-1ω,v0(θ-t-1ω))|2dxdτ+c3∫tt+1‍∫ℝn‍φ(|x|2k2)×|u(τ,θ-t-1ω,u0(θ-t-1ω))|pdxdτ+c3k(∫tt+1‍∥vτ(τ,θ-t-1ω,v0(θ-t-1ω))∥2dτ+∫tt+1‍∥∇v(τ,θ-t-1ω,v0(θ-t-1ω))∥2dτ+∫tt+1‍∥v(τ,θ-t-1ω,v0(θ-t-1ω))∥2dτ+∫tt+1‍∥v(τ,θ-t-1ω,v0(θ-t-1ω))∥2p-22p-2dτ+∫tt+1‍∥z(θτ-t-1ω)∥2p-22p-2dτ+∫tt+1‍∥∇z(θτ-t-1ω)∥2dτ+1)+c3∫tt+1‍∫ℝn‍φ(|x|2k2)×(|∇z(θτ-t-1ω)|2+|Δz(θτ-t-1ω)|2+|Δz(θτ-t-1ω)|p)dxdτ+c3∫ℝn‍φ(|x|2k2)(|g|2+|ψ2|2+|ψ3|2)dx.
In the sequel, we estimate each term in the right-hand side of (81) to show that they are arbitrary small when k and t are large enough.
To estimate the first term in the right-hand side of (81), we cite the result (4.46) in [1]; that is,
(82)12ddt∫ℝn‍φ(|x|2k2)|v|2dx+12α1∫ℝn‍φ(|x|2k2)|u|pdx+∫ℝn‍φ(|x|2k2)|∇v|2dx≤ck(∥∇v∥2+∥v∥2)+∫ℝn‍φ(|x|2k2)(1λ|g|2+|ψ1|+12|ψ2|2)dx+c∫ℝn‍φ(|x|2k2)(|Δz(θtω)|2+|z(θtω)|2+|z(θtω)|p)dx.
Integrating the aforementioned inequality over (t,t+1) and replacing ω by θ-t-1ω we get
(83)∫tt+1‍∫ℝn‍φ(|x|2k2)|∇v(s,θ-t-1ω,v0(θ-t-1ω))|2dxds≤12∫ℝn‍φ(|x|2k2)|v(t,θ-t-1ω,v0(θ-t-1ω))|2dx+ck(∫tt+1‍∥∇v(s,θ-t-1ω,v0(θ-t-1ω))∥2ds+∫tt+1‍∥v(s,θ-t-1ω,v0(θ-t-1ω))∥2ds)+∫ℝn‍φ(|x|2k2)(1λ|g|2+|ψ1|+12|ψ2|2)dx+c∫tt+1‍∫ℝn‍φ(|x|2k2)×(|Δz(θs-t-1ω)|2+|z(θs-t-1ω)|2+|z(θs-t-1ω)|p)dxds.
To estimate the first term of the right-hand side of (83), we need a result in [1]. Substituting θ-t-1ω for ω of (4.49) in [1], we get
(84)∫ℝn‍φ(|x|2k2)|v(t,θ-t-1ω,v0(θ-t-1ω))|2dx≤eλ(T-t)∫ℝn‍φ(|x|2k2)|v(T,θ-t-1ω,v0(θ-t-1ω))|2dx+ck∫Tt‍eλ(s-t)(∥v(s,θ-t-1ω,v0(θ-t-1ω))∥2+∥∇v(s,θ-t-1ω,v0(θ-t-1ω))∥2)ds+∫Tt‍eλ(s-t)∫ℝn‍φ(|x|2k2)×(2λ|g|2+2|ψ1|+|ψ2|2)dxds+c∫Tt‍eλ(s-t)∫ℝn‍φ(|x|2k2)×(|Δz(θs-t-1ω)|2+|z(θs-t-1ω)|2+|z(θs-t-1ω)|p)dxds≤ceλ(T-t-1)∫ℝn‍φ(|x|2k2)×|v(T,θ-t-1ω,v0(θ-t-1ω))|2dx+ck∫Tt+1‍eλ(s-t-1)(∥v(s,θ-t-1ω,v0(θ-t-1ω))∥2+∥∇v(s,θ-t-1ω,v0(θ-t-1ω))∥2)ds+c∫Tt+1‍eλ(s-t-1)∫ℝn‍φ(|x|2k2)×(2λ|g|2+2|ψ1|+|ψ2|2)dxds+c∫Tt+1‍eλ(s-t-1)∫ℝn‍φ(|x|2k2)×(|Δz(θs-t-1ω)|2+|z(θs-t-1ω)|2+|z(θs-t-1ω)|p)dxds,
where T=TD^(ω) is an absorbing time in Lemma 14 and t≥T. It is a direct result of [1] that there exist T1=TD^1(ω,ɛ)≥T, R1=R1(ω,ɛ) such that, for all t≥T1, k≥R1 the right-hand side of the aforementioned inequality is less than or equal to ɛ/20c3, so we get
(85)12∫ℝn‍φ(|x|2k2)|v(t,θ-t-1ω,v0(θ-t-1ω))|2dx≤ɛ40c3.
By Lemma 15 we find that there exists T2=TD^2(ω)>0, for all t≥T2,
(86)ck(∫tt+1‍∥∇v(s,θ-t-1ω,v0(θ-t-1ω))∥2ds+∫tt+1‍∥v(s,θ-t-1ω,v0(θ-t-1ω))∥2ds)≤ck(1+r(ω)).
Choosing R2=R2(ω,ɛ)≥0 such that (c/k)(1+r(ω))≤ɛ/40c3 for k≥R2, we deduce that
(87)ck(∫tt+1‍∥∇v(s,θ-t-1ω,v0(θ-t-1ω))∥2ds+∫tt+1‍∥v(s,θ-t-1ω,v0(θ-t-1ω))∥2ds)≤ɛ40c3,∀t≥T2,∀k≥R2.
For the third term of the right-hand side of (83), we have
(88)∫ℝn‍φ(|x|2k2)(1λ|g|2+|ψ1|+12|ψ2|2)dx≤∫|x|≥k(1λ|g|2+|ψ1|+12|ψ2|2)dx,
since ψ1∈L1(ℝn), ψ2∈L2(ℝn), and g∈L2(ℝn) there is a constant R3=R3(ɛ) such that, for all k≥R3,
(89)∫ℝn‍φ(|x|2k2)(1λ|g|2+|ψ1|+12|ψ2|2)dx≤ɛ40c3.
Note that z(θtω)=∑j=1m‍hjzj(θtωj) and hj∈H2(ℝn)⋂‍W2,p(ℝn)⋂‍L∞(ℝn) for j=1,2,…,m. Hence there exists R4=R4(ω,ɛ) such that for all k≥R4 and j=1,2,…,m,
(90)∫|x|≥k‍(|hj(x)|2+|hj(x)|p+|Δhj(x)|2)dx≤λe-(λ/2)ɛ160cc3mpr(ω),
where c is the constant in (83), and thus we have the following estimate:
(91)c∫tt+1‍∫ℝn‍φ(|x|2k2)×(|Δz(θs-t-1ω)|2+|z(θs-t-1ω)|2+|z(θs-t-1ω)|p)dxds≤c∫tt+1‍∫|x|≥k‍(|Δz(θs-t-1ω)|2+|z(θs-t-1ω)|2≤c∫tt+1‍∫|x|≥k‍h+|z(θs-t-1ω)|p)dxds≤cmp∫tt+1‍∑j=1m‍∫|x|≥k(|Δhj|2|zj(θs-t-1ωj)|2‍≤cmp∫tt+1‍∑j=1m‍hhhhhhhh+|hj|2|zj(θs-t-1ωj)|2≤cmp∫tt+1‍∑j=1m‍hhhhhhhh+|hj|p|zj(θs-t-1ωj)|p)dxds=cmp∫tt+1‍[∑j=1m‍|zj(θs-t-1ωj)|2∫|x|≥k‍|Δhj|2dx+∑j=1m‍|zj(θs-t-1ωj)|2∫|x|≥k‍|hj|2dx+∑j=1m‍|zj(θs-t-1ωj)|p∫|x|≥k‍|hj|pdx]ds≤λe-(λ/2)ɛ160cc3mpr(ω)cmp∫tt+1‍∑j=1m‍(2|zj(θs-t-1ωj)|2+|zj(θs-t-1ωj)|p)ds≤λe-(λ/2)ɛ80c3r(ω)∫tt+1‍∑j=1m‍(|zj(θs-t-1ωj)|2+|zj(θs-t-1ωj)|p)ds≤λe-(λ/2)ɛ80c3r(ω)∫tt+1‍p(θs-t-1ω)ds=λe-(λ/2)ɛ80c3r(ω)∫-10‍p(θsω)ds≤λe-(λ/2)ɛ80c3r(ω)r(ω)∫-10‍e-(λ/2)sds≤ɛ40c3.
Let
(92)T1*=T1*(D^,ω,ɛ)=max{T1,T2},R1*=R1*(ω,ɛ)=max{R1,R2,R3,R4}.
From (83), (85), (87), (89), and (91), ∀t≥T1*, k≥R1*, we can estimate the first term of the right-hand side of (81) as
(93)c3∫tt+1‍∫ℝn‍φ(|x|2k2)×|∇v(s,θ-t-1ω,v0(θ-t-1ω))|2dxds≤ɛ10.
Next, from (82) and a similar process as the proof of (93), one can also get the result for the second term of right-hand side of (81); that is, there exist R2*=R2*(ω,ɛ)>0 and T2*=T2*(D^,ω,ɛ)>0 such that, for all t≥T2*, k≥R2*,
(94)c3∫tt+1‍∫ℝn‍φ(|x|2k2)×|u(s,θ-t-1ω,v0(θ-t-1ω))|pdxds≤ɛ10.
From Lemmas15 and 16 we see that there exist T3*=T3*(D^,ω)>0 and R3*=R3*(ω,ɛ)>0 such that ∀t≥T3*,∀k≥R3*,
(95)c3k∫tt+1‍∥∇v(s,θ-t-1ω,v0θ-t-1ω)∥2ds≤ck(1+r(ω))≤ɛ10,c3k∫tt+1‍∥v(s,θ-t-1ω,v0θ-t-1ω)∥2ds≤ck(1+r(ω))≤ɛ10,c3k∫tt+1‍∥vs(s,θ-t-1ω,v0θ-t-1ω)∥2ds≤ck(1+r(ω))≤ɛ10,c3k∫tt+1‍∥v(s,θ-t-1ω,v0θ-t-1ω)∥2p-22p-2ds≤ck(1+r(ω))≤ɛ10.
For the seventh and eighth terms of the right-hand side of (81), we have
(96)∫tt+1‍(∥z(θs-t-1ω)∥2p-22p-2+∥∇z(θs-t-1ω)∥2)ds≤c0∫tt+1‍p(θs-t-1ω)ds=c0∫-10‍p(θsω)ds≤c0r(ω)∫-10‍e-(λ/2)sds≤cr(ω).
So there exists R4*=R4*(ω,ɛ)>0 such that, for all k≥R4* and all  t≥0, we have
(97)c3k∫tt+1‍(∥z(θs-t-1ω)∥2p-22p-2+∥∇z(θs-t-1ω)∥2)ds≤ckr(ω)≤ɛ10.
Similar to the proof of (91), one can show that there exists R5*=R5*(ω,ɛ)>0 such that, for all k≥R5* and all  t≥0, we have
(98)c3∫tt+1‍∫ℝn‍φ(|x|2k2)(|∇z(θs-t-1ω)|2+|Δz(θs-t-1ω)|2+|Δz(θs-t-1ω)|p)dxds≤ɛ10.
Sinceψ2∈L2(ℝn), ψ3∈L2(ℝn), and g∈L2(ℝn), we can easily show that there exists R6*=R6*(ɛ)>0 such that the last term of the right-hand side of (81) is bounded by
(99)c3∫ℝn‍φ(|x|2k2)(|g|2+|ψ2|2+|ψ3|2)dx≤c3∫|x|≥k‍(|g|2+|ψ2|2+|ψ3|2)dx≤ɛ10,
when k≥R6*.
Finally, let(100)R*=R*(ω,ɛ)=max{R1*,R2*,…,R6*},T*=TD^*(ω,ɛ)=max{T1*,T2*,T3*}.
From (81), (93)–(95), (97)–(99) we get
(101)∫ℝn‍φ(|x|2k2)|∇v(t+1,θ-t-1ω,v0(θ-t-1ω))|2dx≤ɛ,t≥T*,k≥R*.
Therefore, ∀t≥T*,
(102)∫|x|≥2R*‍|∇v(t+1,θ-t-1ω,v0(θ-t-1ω))|2dx≤∫ℝn‍φ(|x|2R*2)|∇v(t+1,θ-t-1ω,v0(θ-t-1ω))|2dx≤ɛ.
The proof is complete.Lemma 18.
Assume thatg∈L2(ℝn) and (3)–(6) hold. Let D^={D(ω)}ω∈Ω∈𝒟2 and u0(ω)∈D(ω). Then, for  ℙ-a.e. ω∈Ω and for every ɛ>0, there exist T*=TD^*(ω,ɛ)>0 and R*=R*(ω,ɛ), such that the solution u(t,ω,u0(ω)) of problem (1)-(2) satisfies that, ∀t≥TD^*(ω,ɛ),
(103)∫|x|≥R*‍(|u(t,θ-tω,u0(θ-tω))|2+|∇u(t,θ-tω,u0(θ-tω))|2)dx≤ɛ.Proof.
Considering∀ɛ>0, we choose R5=R5(ω,ɛ) large enough such that
(104)∫|x|≥R5‍(|hj|2+|∇hj|2)dx≤ɛ2m2r(ω),j=1,2,…,m,
and set R*′=max{R*,R5}. Then by Lemma 17 and (34), one can easily show that
(105)∫|x|≥R*′‍(|u(t,θ-tω,u0(θ-tω))|2+|∇u(t,θ-tω,u0(θ-tω))|2)dx≤3ɛ,
with ∀t≥TD^*(ω,ɛ), where R*, TD^*(ω,ɛ) are the constants in Lemma 17. The proof is complete.


### 4.3. Asymptotic Compactness in Bounded Balls

In what follows, we prove the asymptotic compactness in any bounded ball, which together with Lemma18 and Theorem 10 is a necessary condition for verifying the (L2(ℝn),H1(ℝn))-asymptotic compactness. For this purpose, we set ψ=1-φ, where φ is the function described in Lemma 17.For fixedk≥1, define
(106)v~(t,ω,v0(ω))=ψ(|x|2k2)v(t,ω,v0(ω)).
Then v~(t,ω,v0(ω))∈H01(Q2k) and
(107)∥v~(t,ω,v0(ω))∥H01(Q2k)≤c∥v(t,ω,v0(ω))∥H1(ℝn),
where c is a positive constant, independent of t, ω, and k. Then we have
(108)v~t(t)=ψ(|x|2k2)vt(t),Δv~=(Δψ)v+2∇ψ·∇v+ψΔv.
Multiplying (30) by ψ(|x|2/k2), then we can easily show that
(109)v~t+λv~-Δv~=ψf(x,u)+ψg+ψΔz(θtω)-vΔψ-2∇ψ·∇v.
Consider the following eigenvalue problem:
(110)-Δν~=λν~,inQ2κ,ν~∣ϑQ2κ=0,
and then problem (110) has a family of eigenfunctions {ej}j=1∞ with corresponding eigenvalues {λj}j=1∞ such that {ej}j=1∞ forms an orthogonal basis in both L2(Q2k) and H01(Q2k) and
(111)λ1≤λ2≤⋯≤λj⟶∞asj⟶∞.
Given n, let Xn= span{e1,…,en} and Pn:L2(Q2k)→Xn be the projection operator. For any v~∈H01(Q2k), we write
(112)v~=Pmv~+(I-Pm)v~=:v~1+v~2.
In order to prove the asymptotic compactness we need the following lemma, which can be found in [19].Lemma 19 (see [19]).
Letr>0, λ>0, τ∈ℝ, and for s>τ(113)y′(s)+λy(s)≤h(s),
where the functions  y, y′, h are assumed to be locally integrable and y, h nonnegative on the interval t<s<t+r, for some t≥τ. Then, for any δ∈(0,r),
(114)y(t+r)≤e-λ(r-δ)1δ∫tt+r‍y(s)dsy(t+r)≤+e-λ(t+r)∫tt+r‍eλsh(s)ds.
In particular, let δ=(r/2), then
(115)y(t+r)≤e-λ(r/2)2r∫tt+r‍y(s)dsy(t+r)≤+e-λ(t+r)∫tt+reλsh(s)ds.Lemma 20.
Assume thatg∈L2(ℝn) and (3)–(6) hold. Let D^={D(ω)}ω∈Ω∈𝒟2 and u0(ω)∈D(ω). Then, for  ℙ-a.e. ω∈Ω and for every ɛ>0, there exist M(ω,ɛ)>0, η(ω,ɛ)>0, and TD^(ω)>0 such that the solution v(t,ω,v0(ω)) of (30)-(31) satisfies that, ∀t≥TD^(ω),
(116)∫t-ηt‍∫Q2k(v(s,θ-tω,v0(θ-tω))≥M)‍|v(s,θ-tω,v0(θ-tω))|p-1×|v(s,θ-tω,v0(θ-tω))-M|p-1dxds≤ɛ,∫t-ηt‍∫Q2k(v(s,θ-tω,v0(θ-tω))≤-M)‍|v(s,θ-tω,v0(θ-tω))|p-1×|v(s,θ-tω,v0(θ-tω))+M|p-1dxds≤ɛ.Proof.
Ifp=2, one can easily show the results by (36). So in the following we assume p>2. Let
(117)(v-M)+={v-M,x∈Q2k1,0,elsewhere,(v+M)-={v+M,x∈Q2k2,0,elsewhere,
where Q2k1=Q2k(v≥M)=Q2k⋂‍{x:v(t,ω,v0(ω))≥M} and Q2k2=Q2k(v≤-M)=Q2k⋂‍{x:v(t,ω,v0(ω))≤-M}.
Multiplying (30) with |(v-M)+|p-1 and integrating over ℝn, we get
(118)1pddt∥(v-M)+∥pp+λ∫Q2k1‍v(v-M)p-1dx+(p-1)∫Q2k1‍|∇(v-M)|2(v-M)p-2dx=∫Q2k1‍f(x,u)(v-M)p-1dx+∫Q2k1‍g(v-M)p-1dx+∫Q2k1‍Δz(θtω)(v-M)p-1dx.
From (41), when x∈Q2k1, we have
(119)f(x,v+z(θtω))(v-M)p-1=f(x,v+z(θtω))v(v-M)p-1v≤-α12p|v|p(v-M)p-1v+c(|z(θtω)|p+|z(θtω)|2)(v-M)p-1v+(ψ1+12ψ22)(v-M)p-1v≤-α12p|v|p-1(v-M)p-1+cM(|z(θtω)|p+|z(θtω)|2)(v-M)p-1+1M(ψ1+12ψ22)(v-M)p-1≤-α12p|v|p-1(v-M)p-1+c(|z(θtω)|p+|z(θtω)|2)(v-M)p-1+c(ψ1+12ψ22)(v-M)p-1,
where the constant c in the right hand side of (119) is independent of  M when we assume, without loss of generality, that M≥1.
From (118)-(119), we get
(120)1pddt∥(v-M)+∥pp+c4∫Q2k1‍|v|p-1(v-M)p-1dx≤c5∫Q2k1‍(|ψ1|+12|ψ2|2+|g|)(v-M)p-1dx+c5∫Q2k1‍(|z(θtω)|p+|z(θtω)|2+|Δz(θtω)||z(θtω)|p)(v-M)p-1dx.
Applying Cauchy's inequality, we obtain
(121)1pddt∥(v-M)+∥pp+c42∫Q2k1‍|v|p-1(v-M)p-1dx+c42∫Q2k1‍(v-M)2p-2dx≤c42∫Q2k1‍(v-M)2p-2dx+c∫Q2k1(|ψ1|2+|ψ2|4+g2)dx+c∫Q2k1(|z(θtω)|2p+|z(θtω)|4+|Δz(θtω)|2)dx,
that is,
(122)2pddt∥(v-M)+∥pp+c4∫Q2k1|v|p-1(v-M)p-1dx≤c∫Q2k1(|ψ1|2+|ψ2|4+g2)dx+c∫Q2k1(|z(θtω)|2p+|z(θtω)|4+|Δz(θtω)|2)dx,(123)2pddt∥(v-M)+∥pp+c4Mp-2∫Q2k1‍(v-M)pdx≤c∫Q2k1(|ψ1|2+|ψ2|4+g2)dx+c∫Q2k1(|z(θtω)|2p+|z(θtω)|4+|Δz(θtω)|2)dx.
By (123) and Lemma 19 with r=1, we get
(124)∥(v-M)+(t)∥pp≤2e-μ/2∫t-1t‍∥(v-M)+(τ)∥ppdτ+ce-μt∫t-1t‍eμτ(∥z(θτω)∥2p2p+∥z(θτω)∥44+∥Δz(θτω)∥2∥z(θτω)∥2p2p)dτ+ce-μt∫t-1t‍eμτ(∥ψ1∥2+∥ψ2∥44+∥g∥2)dτ,
where μ=c4Mp-2p/2.
Sets∈[t-1,t]; we first substitute s for t in the aforementioned inequality and then we replace ω by θ-tω to get
(125)∥(v(s,θ-tω,v0(θ-tω))-M)+∥pp≤2e-μ/2∫s-1s‍∥(v(τ,θ-tω,v0(θ-tω))-M)+∥ppdτ+ce-μs∫s-1s‍eμτ(∥z(θτ-tω)∥2p2p+∥z(θτ-tω)∥44+∥Δz(θτ-tω)∥2∥z(θτ-tω)∥2p2p)dτ+ce-μs∫s-1s‍eμτdτ(∥ψ1∥2+∥ψ2∥44+∥g∥2)≤2e-μ/2∫t-2t‍∥(v(τ,θ-tω,v0(θ-tω))-M)+∥ppdτ+ce-μs∫s-1s‍eμτ(∥z(θτ-tω)∥2p2p+∥z(θτ-tω)∥44+∥Δz(θτ-tω)∥2∥z(θτ-tω)∥2p2p)dτ+ce-μs∫s-1s‍eμτdτ(∥ψ1∥2+∥ψ2∥44+∥g∥2).
For the first term of the right-hand side of (125), we use Lemma 15:
(126)∫t-2t‍∥(v(τ,θ-tω,v0(θ-tω))-M)+∥ppdτ≤∫t-2t‍∥v(τ,θ-tω,v0(θ-tω))∥ppdτ≤c(1+r(ω)),∀t≥TD^(ω).
Then there exists M1=M1(ω,ɛ)>0 such that, for all t≥TD^(ω) and all M≥M1,
(127)2e-μ/2∫t-2t‍∥(v(τ,θ-tω,v0(θ-tω))-M)+∥ppdτ≤ce-μ/2(1+r(ω))≤ɛ3.
For the second term of the right-hand side of (125), ∀s∈[t-1,t], by (26) we have
(128)ce-μs∫s-1s‍eμτ(∥z(θτ-tω)∥2p2p+∥z(θτ-tω)∥44+∥Δz(θτ-tω)∥2∥z(θτ-tω)∥2p2p)dτ≤c∫s-1s‍eμ(τ-s)p(θτ-tω)dτ=c∫-10‍eμτp(θτ-t+sω)dτ≤cr(ω)∫-10‍eμτe(λ/2)|τ-t+s|dτ≤cr(ω)∫-10‍eμτeλdτ≤cr(ω)∫-10‍eμτdτ≤cr(ω)μ.
This implies that there exists M2=M2(ω,ɛ)>0 such that  ∀M≥M2 we have
(129)ce-μs∫s-1s‍eμτ(∥z(θτ-tω)∥2p2p+∥z(θτ-tω)∥44+∥Δz(θτ-tω)∥2∥z(θτ-tω)∥2p2p)dτ≤ɛ3.
For the last term of the right-hand side of (125), we can easily see that there exists M3=M3(ɛ)>0 such that, for all M≥M3,
(130)ce-μs∫s-1s‍eμτdτ(∥ψ1∥2+∥ψ2∥44+∥g∥2)=c∫-10‍eμτdτ(∥ψ1∥2+∥ψ2∥44+∥g∥2)≤cμ≤ɛ3.
Letting N*=N*(ω,ɛ)=max{M1,M2,M3}, then combining with (125), (127) and (129)–(130), we can show that, for all t≥TD^(ω) and all M≥N*(ω,ɛ),
(131)∥(v(s,θ-tω,v0(θ-tω))-M)+∥pp≤ɛ,∀s∈[t-1,t].
Next, we set(132)η=min{ɛ3c7(∥ψ1∥2+∥ψ2∥44+∥g∥2),12λln(λɛ6c8r(ω)+1),ɛ3c7(∥ψ1∥2+∥ψ2∥44+∥g∥2),1},
where c7=c/c4, c8=c0c7, and c is the constants in (122). Integrating (122) over (t-η,t), we get
(133)∫t-ηt‍∫Q2k1‍|v(s)|p-1(v(s)-M)p-1dxds≤c6∥(v(t-η)-M)+∥pp+c7∫t-ηt‍∫Q2k1‍(|z(θsω)|2p+|z(θsω)|4+|Δz(θsω)|2)dxds+c7∫t-ηt‍∫Q2k1‍(|ψ1|2+|ψ2|4+g2)dxds.
Replacing ω by θ-tω and setting  Q2k*=Q2k(v(s,θ-tω,v0(θ-tω))≥M), we obtain
(134)∫t-ηt‍∫Q2k*‍|v(s,θ-tω,v0(θ-tω))|p-1×(v(s,θ-tω,v0(θ-tω))-M)p-1dxds≤c7∫t-ηt‍∫Q2k*‍(|z(θs-tω)|2p+|z(θs-tω)|4+|Δz(θs-tω)|2)dxds+c6∥(v(t-η,θ-tω,v0(θ-tω))-M)+∥pp+c7η(∥ψ1∥2+∥ψ2∥44+∥g∥2).
For fixed ω∈Ω, by (131) and (132), ∀t≥TD^(ω), ∀M≥M1*(ω,ɛ)=N*(ω,ɛ/3c6),
(135)c6∥(v(t-η,θ-tω,v0(θ-tω))-M)+∥pp≤ɛ3,
where N*(ω,ɛ) is the constant in (131).
Using (132), the first term of the right-hand side of (134) is bounded by
(136)c7∫t-ηt‍(∥z(θτ-tω)∥2p2p+∥z(θτ-tω)∥44+∥Δz(θτ-tω)∥2∥z(θτ-tω)∥2p2p)dτ≤c8∫t-ηt‍p(θτ-tω)dτ=c8∫-η0‍p(θτω)dτ≤c8r(ω)∫-η0‍e-(λ/2)τdτ=2c8λr(ω)(eλη/2-1)≤ɛ3.
For the last term of right-hand side of (134), by using (132), we have
(137)c7η(∥ψ1∥2+∥ψ2∥44+∥g∥2)≤ɛ3.
By (134)–(137), ∀M≥M1*, ∀t≥TD^(ω), and ∀u0(ω)∈D(ω) with v0(ω)=u0(ω)-z(ω), we have
(138)∫t-ηt‍∫Q2k*‍|v(s,θ-tω,v0(θ-tω))|p-1×|v(s,θ-tω,v0(θ-tω))-M|p-1dxds≤ɛ.
Similarly, multiplying (30) with |(v+M)-|p-1, we can prove that there exists M2*=M2*(ω,ɛ) such that  ∀M≥M2*, ∀t≥TD^(ω), and ∀u0(ω)∈D(ω) with v0(ω)=u0(ω)-z(ω), over the region Q2k**=Q2k(v(s,θ-tω,v0(θ-tω))≤-M); we have
(139)∫t-ηt‍∫Q2k**‍|v(s,θ-tω,v0(θ-tω))|p-1×|v(s,θ-tω,v0(θ-tω))+M|p-1dxds≤ɛ.
Let M*=max{M1*,M2*}; from (138) and (139) we can obtain our results. The proof is complete.Remark 21.
The idea of the proof of the above lemma comes from [19] (this idea can be further traced back to Marion [16] and Robinson [18]). We see from (132) that the constant η in Lemma 20 is independent of t, which is different from the η in the Lemma 3.4 in [19]. This is crucial in the following estimates. As we know, the time t will vary to infinite when we consider the asymptotic behavior of an RDS. It means that if η is not a fixed constant with respect to t, the following estimate will be invalid. In other words, if the function g in (1) is dependent on t, our method will fail.Lemma 22.
Assume thatg∈L2(ℝn) and (3)–(6) hold. Let D^={D(ω)}ω∈Ω∈𝒟2 and u0(ω)∈D(ω). Then, for  ℙ-a.e. ω∈Ω and for every ɛ>0 and all k≥1, there exist TD^(ω)>0 and N(k,ω,ɛ)>0 such that the solution v(t,ω,v0(ω)) of (30)-(31) satisfies that, ∀t≥TD^(ω), ∀m≥N(k,ω,ɛ),
(140)∥v~2(t,θ-tω,v0(θ-tω))∥H01(Q2k)2≤ɛ,
where v~2=(I-Pm)v~.Proof.
Multiplying (109) with -Δv~2 and integrating over Q2k, we get
(141)12ddt∥v~2∥H01(Q2k)2+λ∥v~2∥H01(Q2k)2+∥Δv~2∥L2(Q2k)2=-(ψf(x,u),Δv~2)L2(Q2k)-(ψΔz(θtω),Δv~2)L2(Q2k)-(ψg,Δv~2)L2(Q2k)+(vΔψ,Δv~2)L2(Q2k)+2(∇ψ·∇v,Δv~2)L2(Q2k).
We now estimate each term in the right-hand side of the aforementioned equality. For the first term, by using Cauchy inequality and (4), we have
(142)|(ψf(x,u),Δv~2)L2(Q2k)|≤110∥Δv~2∥L2(Q2k)2+c∫Q2k‍|f(x,u)|2dx≤110∥Δv~2∥L2(Q2k)2+c∫Q2k‍|u|2p-2dx+c∫Q2k‍|ψ2|2dx≤110∥Δv~2∥L2(Q2k)2+c∫Q2k‍|v|2p-2dx+c∫Q2k‍|z(θtω)|2p-2dx+c≤110∥Δv~2∥L2(Q2k)2+c∫Q2k(|v|≤M)‍|v|2p-2dx+c∫Q2k(|v|≥M)‍|v|2p-2dx+c∫Q2k‍|z(θtω)|2p-2dx+c≤110∥Δv~2∥L2(Q2k)2+c∫Q2k(|v|≥M)‍|v|2p-2dx+c∥z(θtω)∥2p-22p-2+c(∥ψ2∥2,|Q2k|,p,M),
where |Q2k| denotes the Lebesgue measure of Q2k.
For the second to fifth term, we can estimate them as follows:(143)|(ψΔz(θtω),Δv~2)L2(Q2k)|≤110∥Δv~2∥L2(Q2k)2+c∥Δz(θtω)∥2,(144)|(ψg,Δv~2)L2(Q2k)|≤110∥Δv~2∥L2(Q2k)2+c∥g∥2,(145)|(vΔψ,Δv~2)L2(Q2k)|≤110∥Δv~2∥L2(Q2k)2+c∫Q2k‍|v|2|Δψ|2dx≤110∥Δv~2∥L2(Q2k)2+c∫ℝn‍|v|2|Δψ|2dx≤110∥Δv~2∥L2(Q2k)2+c∫ℝn‍|v|2|φ′′(|x|2k2)4|x|2k4+φ′(|x|2k2)2nk2|2dx≤110∥Δv~2∥L2(Q2k)2+ck4∫k≤|x|≤2k‍|v|2dx≤110∥Δv~2∥L2(Q2k)2+c∥v∥2,(146)2|(∇ψ·∇v,Δv~2)L2(Q2k)|≤110∥Δv~2∥L2(Q2k)2+c∫Q2k‍|∇ψ|2|∇v|2dx≤110∥Δv~2∥L2(Q2k)2+ck2∥∇v∥2≤110∥Δv~2∥L2(Q2k)2+c∥∇v∥2,
where the constant c in (145) and (146) is independent of k, since k≥1.
Combining (141)–(146) we get
(147)ddt∥v~2∥H01(Q2k)2+2λ∥v~2∥H01(Q2k)2+∥Δv~2∥L2(Q2k)2≤c∫Q2k(|v|≥M)‍|v|2p-2dx+c(∥v∥2+∥∇v∥2)+c(∥z(θtω)∥2p-22p-2+∥Δz(θtω)∥2)+c(∥ψ2∥2,∥g∥2,|Q2k|,p,M),
and this implies that
(148)ddt∥v~2∥H01(Q2k)2+λm+1∥v~2∥H01(Q2k)2≤c∫Q2k(|v|≥M)‍|v|2p-2dx+c(∥v∥2+∥∇v∥2)+c(∥z(θtω)∥2p-22p-2+∥Δz(θtω)∥2)+c(∥ψ2∥2,∥g∥2,|Q2k|,p,M).
By using Lemma 19 with r=1, we get
(149)∥v~2(t+1)∥H01(Q2k)2≤2e-λm+1/2∫tt+1‍∥v~2(s)∥H01(Q2k)2ds+ce-λm+1(t+1)∫tt+1‍eλm+1s∫Q2k(|v(s)|≥M)‍|v(s)|2p-2dxds+ce-λm+1(t+1)∫tt+1‍eλm+1s×(∥z(θsω)∥2p-22p-2+∥Δz(θsω)∥2)ds+ce-λm+1(t+1)∫tt+1‍eλm+1s(∥v(s)∥2+∥∇v(s)∥2)ds+c(∥ψ2∥2,∥g∥2,|Q2k|,p,M)e-λm+1(t+1)∫tt+1‍eλm+1sds.
Substituting  θ-t-1ω for ω, we get
(150)∥v~2(t+1,θ-t-1ω,v0(θ-t-1ω))∥H01(Q2k)2≤2e-λm+1/2∫tt+1‍∥v~2(s,θ-t-1ω,v0(θ-t-1ω))∥H01(Q2k)2ds+ce-λm+1(t+1)∫tt+1‍eλm+1s×∫Q2k(|v(*)|≥M)‍|v(*)|2p-2dxds+ce-λm+1(t+1)∫tt+1‍eλm+1s(∥z(θs-t-1ω)∥2p-22p-2+∥Δz(θs-t-1ω)∥2∥z(θs-t-1ω)∥2p-22p-2)ds+ce-λm+1(t+1)∫tt+1‍eλm+1s(∥v(*)∥2+∥∇v(*)∥2)ds+c(∥ψ2∥2,∥g∥2,|Q2k|,p,M)e-λm+1(t+1)∫tt+1‍eλm+1sds,
for simplicity, hereafter, we write v(*)=v(s,θ-t-1ω,v0(θ-t-1ω)).
In the sequel, we estimate each term in the right-hand side of (150). For the first term, we use Lemma 15 and (107) to obtain that, for t≥TD^(ω),
(151)2e-λm+1/2∫tt+1‍∥v~2(s,θ-t-1ω,v0(θ-t-1ω))∥H01(Q2k)2ds≤2e-λm+1/2∫tt+1‍∥v~(s,θ-t-1ω,v0(θ-t-1ω))∥H01(Q2k)2ds≤ce-λm+1/2∫tt+1‍∥v(*)∥H1(ℝn)2ds≤ce-λm+1/2(1+r(ω)).
Thus, there exists N1=N1(ω,ɛ), for all m≥N1 and all t≥TD^(ω), we get
(152)2e-λm+1/2∫tt+1‍∥v~2(s,θ-t-1ω,v0(θ-t-1ω))∥H01(Q2k)2ds≤ɛ5.
For the second term in the right-hand side of (150), we can estimate it as follows:
(153)∫Q2k(|v(*)|≥M)‍|v(*)|2p-2dx=∫Q2k(v(*)≥M)‍|v(*)|2p-2dx+∫Q2k(v(*)≤-M)‍|v(*)|2p-2dx≤2p-2[∫Q2k(v(*)≥M)‍|v(*)|p-1×(|v(*)-M|p-1+Mp-1)dx+∫Q2k(v(*)≤-M)‍|v(*)|p-1×(|v(*)+M|p-1+Mp-1)dx∫Q2k(v(*)≥M)‍]≤2p-2[∫Q2k(v(*)≥M)‍|v(*)|p-1|v(*)-M|p-1dx+∫Q2k(v(*)≤-M)‍|v(*)|p-1|v(*)+M|p-1dx]+2p-2Mp-2∫Q2k(|v(*)|≥M)‍|v(*)|pdx,(154)ce-λm+1(t+1)∫tt+1‍eλm+1s∫Q2k(|v(*)|≥M)‍|v(*)|2p-2dxds=ce-λm+1(t+1)(∫t+1-ηt+1‍+∫tt+1-η‍)eλm+1s×∫Q2k(|v(*)|≥M)‍|v(*)|2p-2dxds≤ce-λm+1(t+1)∫t+1-ηt+1‍eλm+1s2p-2×[∫Q2k(v(*)≥M)‍|v(*)|p-1|v(*)-M|p-1dx+∫Q2k(v(*)≤-M)‍|v(*)|p-1×|v(*)+M|p-1dx∫Q2k(v(*)≥M)‍]ds+c2p-2Mp-2e-λm+1(t+1)×∫t+1-ηt+1‍eλm+1s∫Q2k(|v(*)|≥M)‍|v(*)|pdxds+ce-λm+1(t+1)×∫tt+1-η‍eλm+1s∫Q2k(|v(*)|≥M)‍|v(*)|2p-2dxds.
The first term in the right-hand side of (154) is bounded by
(155)2p-2c∫t+1-ηt+1[∫Q2k(v(*)≥M)‍|v(*)|p-1‍2p-2c∫t+1-ηt+1‍h×|v(*)-M|p-1dx2p-2hc∫t+1-ηt+1‍+∫Q2k(v(*)≤-M)‍|v(*)|p-1|v(*)+M|p-1dx]ds≤ɛ15,∀t≥TD^(ω),
when we choose appropriate η and M by Lemma 20.
For the second term of the right-hand side of (154), we use Lemma 15, ∀t≥TD^(ω),
(156)c2p-2Mp-2e-λm+1(t+1)×∫t+1-ηt+1‍eλm+1s∫Q2k(|v(*)|≥M)‍|v(*)|pdxds≤c2p-2Mp-2e-λm+1(t+1)∫t+1-ηt+1‍eλm+1s∥v(*)∥ppds≤c2p-2Mp-2e-λm+1(t+1)(1+r(ω))∫t+1-ηt+1‍eλm+1sds≤c2p-2Mp-2(1+r(ω))∫t+1-ηt+1‍eλm+1(s-t-1)ds=c2p-2Mp-2(1+r(ω))∫-η0‍eλm+1sds≤c2p-2Mp-2(1+r(ω))λm+1,
where M is determined in (155), and then there exists N21(ω,ɛ)>0, such that  ∀m≥N21,∀t≥TD^(ω), we get
(157)c2p-2Mp-2e-λm+1(t+1)×∫t+1-ηt+1‍eλm+1s∫Q2k(|v(*)|≥M)‍|v(*)|pdxds≤ɛ15.
By Lemma 16, the last term of the right-hand side of (154) is bounded by
(158)ce-λm+1η∫tt+1-η‍∫Q2k(|v(*)|≥M)‍|v(*)|2p-2dxds≤ce-λm+1η∫tt+1‍∥v(*)∥2p-22p-2ds≤ce-λm+1η(1+r(ω)),
for all t≥TD^(ω). This implies that there exists N22=N22(ω,ɛ)>0, such that, for all m≥N22, t≥TD^(ω), we have
(159)ce-λm+1(t+1)×∫tt+1-η‍eλm+1s∫Q2k(|v(*)|≥M)‍|v(*)|2p-2dxds≤ɛ15.
Combining (154), (155), (157), and (159) and setting N2=N2(ω,ɛ)=max{N21(ω,ɛ),N22(ω,ɛ)}, then we have that, for all m≥N2, t≥TD^(ω),
(160)ce-λm+1(t+1)∫tt+1‍eλm+1s×∫Q2k(|v(*)|≥M)‍|v(*)|2p-2dxds≤ɛ5.
The third term of the right-hand side of (150) is bounded by
(161)ce-λm+1(t+1)∫tt+1‍eλm+1sp(θs-t-1ω)ds=c∫tt+1‍eλm+1(s-t-1)p(θs-t-1ω)ds=c∫-10‍eλm+1sp(θsω)ds≤cr(ω)∫-10‍eλm+1se-(λ/2)sds≤cr(ω)λm+1-λ/2,
so there exists N3=N3(ω,ɛ)>0, ∀m≥N3,∀t≥0; we have
(162)ce-λm+1(t+1)×∫tt+1‍eλm+1s(∥z(θs-t-1ω)∥2p-22p-2+∥Δz(θs-t-1ω)∥2)ds≤ɛ5.
For the forth term of the right-hand side of (150), by using Lemma 15, we have
(163)ce-λm+1(t+1)∫tt+1‍eλm+1s(∥v(*)∥2+∥∇v(*)∥2)ds≤ce-λm+1(t+1)∫tt+1‍eλm+1sds(1+r(ω))≤c∫-10‍eλm+1sds(1+r(ω))≤cλm+1(1+r(ω)),∀t≥TD^(ω),
and then there exists N4=N4(ω,ɛ)>0, for all m≥N4 and all t≥TD^(ω),
(164)ce-λm+1(t+1)×∫tt+1‍eλm+1s(∥v(*)∥2+∥∇v(*)∥2)ds≤ɛ5.
For the last term of the right-hand side of (150), we have
(165)c(∥ψ2∥2,∥g∥2,|Q2k|,p,M)e-λm+1(t+1)×∫tt+1‍eλm+1sds≤c(∥ψ2∥2,∥g∥2,|Q2k|,p,M)λm+1,
this implies that there exists N5=N5(k,ɛ) (M is determined in (155)), for all m≥N5 and all t≥0,
(166)c(∥ψ2∥2,∥g∥2,|Q2k|,p,M)e-λm+1(t+1)×∫tt+1‍eλm+1sds≤ɛ5.
Finally, letN(k,ω,ɛ)=max{N1,N2,N3,N4,N5}, and then from (150), (152), (160), (162), (164), and (166) we have that, for all m≥N(k,ω,ɛ), t≥TD^(ω),
(167)∥v~2(t+1,θ-t-1ω,v0(θ-t-1ω))∥H01(Q2k)2≤ɛ.
The proof is complete.Lemma 23.
Assume thatg∈L2(ℝn) and (3)–(6) hold, and ϕ is the RDS generated by problem (1)-(2); then, χQk·ϕ is (L2(ℝn),H1(Qk))-asymptotically compact for all k≥1.Proof.
We first prove that the sequence{v~(tm,θ-tmω,ym)}m=1∞ is precompact in H01(Q2k), for any tm→∞, and any xm∈D(θ-tmω) with D^={D(ω)}ω∈Ω∈𝒟2, where the relationship between xm and ym is determined by (31), that is, ym=xm-z(θ-tmω). By Lemma 15 we see that, for all t≥TD^(ω),
(168)∥v(t,θ-tω,v0(θ-tω))∥H1(ℝn)2≤c(1+r(ω)),
where v0(θ-tω)=u0(θ-tω)-z(θ-tω) with u0(ω)∈D(ω). Thus there is M1=M1(D^,ω) such that, for all m≥M1, we have tm≥TD^(ω) and
(169)∥v(tm,θ-tmω,ym)∥H1(ℝn)2≤c(1+r(ω)),∀m≥M1.
By (107) and the aforementioned inequality we find that
(170)∥v~(tm,θ-tmω,ym)∥H01(Q2k)2≤c(1+r(ω)),∀m≥M1.
Givenɛ>0, it follows from Lemma 22 that there are TD^(ω) and N(k,ω,ɛ) such that, for all t≥TD^(ω),
(171)∥(I-PN)v~(t,θ-tω,v0(θ-tω))∥H01(Q2k)2≤ɛ.
Taking M2=M2(D^,ω) large enough such that tm≥TD^(ω) for m≥M2, then we get from (171) that
(172)∥(I-PN)v~(tm,θ-tmω,ym)∥H01(Q2k)2≤ɛ,∀m≥M2.
On the other hand, (170) shows that the sequence {PNv~(tm,θ-tmω,ym)} is bounded in the finite-dimensional space PNH01(Q2k) and hence is precompact in PNH01(Q2k), which along with (172) implies the precompactness of {v~(tm,θ-tmω,ym)} in H01(Q2k).
Next we prove the(L2(ℝn),H1(Qk))-asymptotic compactness for χQk·ϕ. From the definition of v~, we see that v=v~ in Qk, so v(tm,θ-tmω,ym)}m=1∞ is precompact in H1(Qk). By the relationship
(173)u(tm,θ-tmω,xm)=v(tm,θ-tmω,ym)+z(ω),
and the aforementioed assert, we obtain that {u(tm,θ-tmω,xm)}m=1∞ is precompact in H1(Qk), for all k≥1. The proof is complete.


## 4.1.(L2(ℝn),H1(ℝn))-Random Absorbing Set and Some Useful Estimates

The next lemma shows thatϕ has a tempered (L2(ℝn),H1(ℝn))-random absorbing set.Lemma 14 (see [1]).
Assume thatg∈L2(ℝn) and (3)–(6) hold. Let D^={D(ω)}ω∈Ω∈𝒟2 and u0(ω)∈D(ω). Then, for ℙ-a.e. ω∈Ω, there exists TD^(ω)>0 such that,  for all t≥TD^(ω),
(35)∥u(t,θ-tω,u0(θ-tω))∥H1(ℝn)2≤c(1+r(ω)),
where c is a constant independent of t, ω, and u0.We now give some new estimates for the solutionv(t,ω,v0) of (30)-(31).Lemma 15.
Assume thatg∈L2(ℝn) and (3)–(6) hold. Let D^={D(ω)}ω∈Ω∈𝒟2 and u0(ω)∈D(ω). Then, for  ℙ-a.e. ω∈Ω, there exists TD^(ω)≥2, such that the solution v(t,ω,v0(ω)) of (30)-(31) satisfies that,  for all t≥TD^(ω) and  for all s∈[t-1,t+1],
(36)∥v(s,θ-t-1ω,v0(θ-t-1ω))∥2≤c(1+r(ω)),(37)∥v(s,θ-t-1ω,v0(θ-t-1ω))∥pp≤c(1+r(ω)),(38)∥∇v(s,θ-t-1ω,v0(θ-t-1ω))∥2≤c(1+r(ω)),
where c is a constant independent of s, t, and ω.Proof.
The first assertion was proved in [2] in the case of s∈[t,t+1], and the case for s∈[t-1,t+1] can be obtained by slightly modifying the proof of Lemma 4.4 in [2] (in fact, for s∈[t-2,t+1], (36) also holds true, and we will use this result in (55)), and here we omit it.
Now, we prove the second assertion. Multiplying (30) with |v|p-2v and integrating over ℝn, we get
(39)1pddt∥v∥pp+λ∥v∥pp+(-Δv,|v|p-2v)=∫ℝn‍f(x,v+z(θtω))|v|p-2vdx+(g+Δz(θtω),|v|p-2v).
Since p≥2, we have
(40)(-Δv,|v|p-2v)=(∇v,∇(|v|p-2v))=(∇v,(p-2)|v|p-4v2∇v)+(∇v,|v|p-2∇v)≥0.
For the nonlinearity, similar to (4.6) and (4.8) in [2], we have
(41)f(x,v+z(θtω))v≤-α12p|v|p+c(|z(θtω)|p+|z(θtω)|2)+ψ1+12ψ22,(42)∫ℝn‍f(x,v+z(θtω))|v|p-2vdx≤-α12p+1∥v∥2p-22p-2+λ∥v∥pp+c(∥z(θtω)∥2p-22p-2+∥z(θtω)∥pp)+c(∥ψ1∥p/2p/2+∥ψ2∥pp),(43)(g+Δz(θtω),|v|p-2v)≤∥g∥∥v∥2p-2p-1+∥Δz(θtω)∥∥v∥2p-2p-1≤α12p+2∥v∥2p-22p-2+c∥Δz(θtω)∥2+c∥g∥2.
From (39)-(40) and (42)-(43), we get
(44)ddt∥v∥pp+c∥v∥2p-22p-2≤c1(∥z(θtω)∥2p-22p-2+∥z(θtω)∥pp+∥Δz(θtω)∥2)+c1≤c2p(θtω)+c2.
On the other hand, multiplying (30) by  v and integrating over ℝn, we get the results in [1]:
(45)ddt∥v∥2+λ∥v∥2+∥∇v∥2+α1∥u∥pp≤cp(θtω)+c,(46)∥v(t,ω,v0(ω))∥2≤e-λt∥v0(ω)∥2+c∫0t‍eλ(τ-t)p(θτω)dτ+c.
By Hölder inequality, ∥u∥pp≥21-p∥v∥pp-∥z(θtω)∥pp, we can convert (45) into
(47)ddt∥v∥2+λ∥v∥2+∥∇v∥2+α121-p∥v∥pp≤cp(θtω)+c.
For any t≥0, integrating (47) over (t,t+1) and using (46), we get
(48)∫tt+1‍∥v(s)∥ppds≤c(∫tt+1‍∥v(t)∥2+∫tt+1‍p(θτω)dτ+1)≤c(e-λt∥v0(ω)∥2+∫0t‍eλ(τ-t)p(θτω)dτ+∫tt+1‍p(θτω)dτ+1).
Next, fix s∈(t,t+1) and integrate (44) over (s,t+1) to get
(49)∥v(t+1,ω,v0(ω))∥pp≤∥v(s,ω,v0(ω))∥pp+c2∫st+1‍p(θτω)dτ+c2(t+1-s)≤∥v(s,ω,v0(ω))∥pp+c2∫tt+1‍p(θτω)dτ+c2.
Integrating the above inequality with respect to s over (t,t+1) and using (48), we obtain, for all t≥0,
(50)∥v(t+1,ω,v0(ω))∥pp≤∫tt+1‍∥v(s,ω,v0(ω))∥ppds+c2∫tt+1‍p(θsω)ds+c2≤c(e-λt∥v0(ω)∥2+∫0t‍eλ(τ-t)p(θτω)dτ+∫tt+1‍p(θτω)dτ+1)≤c(e-λ(t+1)∥v0(ω)∥2+∫0t+1‍eλ(τ-t-1)p(θτω)dτ+∫tt+1‍p(θτω)dτ+1).
Replacing t+1 by s first, then substituting  θ-t-1ω for ω in the aforementioned inequality, and noting that s∈[t-1,t+1](t≥2), we have
(51)∥v(s,θ-t-1ω,v0(θ-t-1ω))∥pp≤c[e-λs∥v0(θ-t-1ω)∥2+∫0s‍eλ(τ-s)p(θτ-t-1ω)dτ+∫s-1s‍p(θτ-t-1ω)dτ+1]≤c[∫t-2t+1‍eλ(t+1-s)e-λ(t+1)∥v0(θ-t-1ω)∥2+∫0t+1‍eλ(t+1-s)eλ(τ-t-1)p(θτ-t-1ω)dτ+∫t-2t+1‍p(θτ-t-1ω)dτ+1]≤c[∫t-2t+1‍e-λ(t+1)∥v0(θ-t-1ω)∥2+∫0t+1‍eλ(τ-t-1)p(θτ-t-1ω)dτ+∫t-2t+1‍p(θτ-t-1ω)dτ+1]=c[e-λ(t+1)∥v0(θ-t-1ω)∥2+∫-t-10‍eλτp(θτω)dτ+∫-30‍p(θτω)dτ+1]≤c[e-λ(t+1)∥v0(θ-t-1ω)∥2+r(ω)∫-t-10‍e(λ/2)τdτ+r(ω)∫-30‍e-(λ/2)τdτ+1],
where we have used (26) in the pervious inequality. Noting that u0(ω)∈D(ω) with D^={D(ω)}ω∈Ω∈𝒟2 and v0(ω)=u0(ω)-z(ω), we get from (51) that there exists TD^(ω)≥2 such that,  for all t≥TD^(ω) and for all s∈[t-1,t+1],
(52)∥v(s,θ-t-1ω,v0(θ-t-1ω))∥pp≤c(1+r(ω)).
That is, (37) holds true.
To prove (38), we take the inner product of (30) with -Δv in L2(Ω), and using  (4.31) in [1], we get
(53)ddt∥∇v∥2≤c(∥∇u∥2+∥u∥pp)+c(∥Δz(θtω)∥2+∥Δz(θtω)∥pp+1).
This implies that
(54)ddt∥∇v∥2≤c(∥∇v∥2+∥v∥pp)+c(p(θtω)+1).
Integrating (47) over (t-2,t+1) and substituting θ-t-1ω for ω, then from (36), we get
(55)∫t-2t+1‍∥∇v(s,θ-t-1ω,v0(θ-t-1ω))∥2ds≤∥v(t-2,θ-t-1ω,v0(θ-t-1ω))∥2+c(∫t-2t+1‍p(θτ-t-1ω)dτ+1)≤c(1+r(ω)),∀t≥TD^(ω).
Obviously, from (51) we can easily see that (37) also holds for s∈[t-2,t+1]; then, by (37), (47), (54)-(55), and a similar procedure as the proof of (37), one can show that,  for all t≥TD^(ω), and for all s∈[t-1,t+1],
(56)∥∇v(s,θ-t-1ω,v0(θ-t-1ω))∥2≤c(1+r(ω)).
The proof is complete.Lemma 16.
Assume thatg∈L2(ℝn) and (3)–(6) hold. Let D^={D(ω)}ω∈Ω∈𝒟2 and u0(ω)∈D(ω). Then, for  ℙ-a.e. ω∈Ω, there exists TD^(ω)>0, such that the solution v(t,ω,v0(ω)) of (30)-(31) satisfies that for all t≥TD^(ω),
(57)∫tt+1‍∥v(s,θ-t-1ω,v0(θ-t-1ω))∥2p-22p-2ds≤c(1+r(ω)),(58)∫tt+1‍∥vs(s,θ-t-1ω,v0(θ-t-1ω))∥2ds≤c(1+r(ω)).Proof.
Integrating (44) over (t,t+1), we get
(59)∫tt+1‍∥v(s)∥2p-22p-2ds≤c(∥v(t)∥pp+∫tt+1‍p(θτω)dτ+1).
Replacing ω by θ-t-1ω in the aforementioned inequality, it yields that
(60)∫tt+1‍∥v(s,θ-t-1ω,v0(θ-t-1ω))∥2p-22p-2ds≤c(∫tt+1‍∥v(t,θ-t-1ω,v0(θ-t-1ω))∥pp+∫tt+1‍p(θτ-t-1ω)dτ+1)=c(∫-10‍∥v(t,θ-t-1ω,v0(θ-t-1ω))∥pp+∫-10‍p(θτω)dτ+1)≤c(∫-10‍∥v(t,θ-t-1ω,v0(θ-t-1ω))∥pp+r(ω)∫-10‍e-(λ/2)τdτ+1).
Lemma 15 and the aforementioned inequality imply that there exists TD^(ω)>0, ∀t≥TD^(ω), such that (57) holds.
Next, taking the inner product of (30) with vt in L2(ℝn), and using (4), we obtain
(61)∥vt∥2+λ2ddt∥v∥2+12ddt∥∇v∥2=∫ℝn‍f(x,u)vtdx+∫ℝn‍gvtdx+∫ℝn‍Δz(θtω)vtdx≤α2∫ℝn‍|u|p-1|vt|dx+∫ℝn‍|ψ2||vt|dx+∫ℝn‍|g||vt|dx+∫ℝn‍|Δz(θtω)||vt|dx≤12∥vt∥2+c(∥u∥2p-22p-2+∥ψ2∥2+∥g∥2+∥Δz(θtω)∥2)≤12∥vt∥2+c(∥z(θtω)∥2p-22p-2∥v∥2p-22p-2+∥ψ2∥2+∥g∥2≤12∥vt∥2+hc+∥Δz(θtω)∥2+∥z(θtω)∥2p-22p-2),
that is,
(62)∥vt∥2+λddt∥v∥2+ddt∥∇v∥2≤c(∥v∥2p-22p-2+∥ψ2∥2+∥g∥2+∥Δz(θtω)∥2+∥z(θtω)∥2p-22p-2).
We now integrate (62) over (t,t+1) to obtain
(63)∫tt+1‍∥vs(s,ω,v0(ω))∥2ds≤c(∥v(t,ω,v0(ω))∥2+∥∇v(t,ω,v0(ω))∥2)+c∫tt+1‍(∥z(θτω)∥2p-22p-2+∥Δz(θτω)∥2)dτ+c∫tt+1‍∥v(τ,ω,v0(ω))∥2p-22p-2dτ+c(∥ψ2∥2+∥g∥2).
Replacing ω by θ-t-1ω, we get
(64)∫tt+1‍∥vs(s,θ-t-1ω,v0(θ-t-1ω))∥2ds≤c(∥v(t,θ-t-1ω,v0(θ-t-1ω))∥2+∥∇v(t,θ-t-1ω,v0(θ-t-1ω))∥2)+c∫tt+1‍(∥z(θτ-t-1ω)∥2p-22p-2+∥Δz(θτ-t-1ω)∥2)dτ+c∫tt+1‍∥v(s,θ-t-1ω,v0(θ-t-1ω))∥2p-22p-2ds+c(∥ψ2∥2+∥g∥2).
Equations (36), (38) and (57) together imply that (58) is also true. The proof is complete.


## 4.2. Tail Estimate inH1(ℝn)

We next estimate “the tail” of the solution to the problem (1)-(2) in H1(ℝn).Lemma 17.
Assume thatg∈L2(ℝn) and (3)–(6) hold. Let D^={D(ω)}ω∈Ω∈𝒟2 and u0(ω)∈D(ω). Then, for  ℙ-a.e. ω∈Ω and for every ɛ>0, there exist T*=TD^*(ω,ɛ)>0 and R*=R*(ω,ɛ), such that the solution v(t,ω,v0(ω)) of (30)-(31) satisfies that  ∀t≥T*,
(65)∫|x|≥R*‍(|v(t,θ-tω,v0(θ-tω))|2+|∇v(t,θ-tω,v0(θ-tω))|2)dx≤ɛ.Proof.
By Lemma 4.6 in [1], it suffices to prove that
(66)∫|x|≥R*‍|∇v(t,θ-tω,v0(θ-tω))|2dx≤ɛ.
Let φ be a smooth function defined on ℝ+ such that 0≤φ(s)≤1, for all s∈ℝ+, and
(67)φ(s)={0,0≤s≤1,1,s≥2.
Then there is a positive constant c such that |φ′(s)|+|φ′′(s)|≤c for all s∈ℝ+.
Multiplying (30) by -φ(|x|2/k2)Δv and integrating with respect to x over ℝn, we get
(68)12ddt∫ℝn‍φ(|x|2k2)|∇v|2dx+∫ℝn‍(∇φ(|x|2k2)·∇v)vtdx+λ∫ℝn‍φ(|x|2k2)|∇v|2dx+λ∫ℝn‍(∇φ(|x|2k2)·∇v)vdx+∫ℝn‍φ(|x|2k2)|Δv|2dx=-∫ℝn‍φ(|x|2k2)f(x,u)Δvdx-∫ℝn‍φ(|x|2k2)(g+Δz(θtω))Δvdx.
The second term of the left-hand side is bounded by
(69)∫ℝn‍|∇φ(|x|2k2)||∇v||vt|dx=∫ℝn‍|φ′(|x|2k2)||2xk2||∇v||vt|dx≤ck∫k≤|x|≤2k‍|∇v||vt|dx≤ck(∥vt∥2+∥∇v∥2).
Similarly, the forth term of the left-hand side of (68) is bounded by
(70)λ∫ℝn‍|∇φ(|x|2k2)||∇v||v|dx≤ck(∥v∥2+∥∇v∥2).
For the last term of the right-hand side of (68), we have
(71)|∫ℝn‍φ(|x|2k2)(g+Δz(θtω))Δvdx|≤12∫ℝn‍φ(|x|2k2)|Δv|2dx+∫ℝn‍φ(|x|2k2)|g|2dx+∫ℝn‍φ(|x|2k2)|Δz(θtω)|2dx.
We next consider the nonlinear term in (68). Since
(72)-∫ℝn‍φ(|x|2k2)f(x,u)Δvdx=-∫ℝn‍φ(|x|2k2)f(x,u)Δudx+∫ℝn‍φ(|x|2k2)f(x,u)Δz(θtω)dx=∫ℝn‍f(x,u)(∇φ(|x|2k2)·∇u)dx+∫ℝn‍φ(|x|2k2)(∂∂xf(x,u)·∇u)dx+∫ℝn‍φ(|x|2k2)∂∂uf(x,u)|∇u|2dx+∫ℝn‍φ(|x|2k2)f(x,u)Δz(θtω)dx.
We now estimate each term in the right-hand side of (72). Using (4), the property of  φ, and Cauchy's inequality, we see that the first term of the right-hand side of (72) is bounded by
(73)∫ℝn‍|∇φ(|x|2k2)||f(x,u)||∇u|dx≤α2∫ℝn‍|∇φ(|x|2k2)||u|p-1|∇u|dx+∫ℝn‍|∇φ(|x|2k2)||ψ2||∇u|dx≤c(∫ℝn‍|∇φ(|x|2k2)||u|2p-2dx+∫ℝn‍|∇φ(|x|2k2)||∇u|2dx)+ck∫k≤|x|≤2k‍|ψ2||∇u|dx≤ck(∥u∥2p-22p-2+∥∇u∥2+1).
By (6), we can estimate the second term of the right-hand side of (72) as follows:
(74)∫ℝn‍|φ(|x|2k2)||∂∂xf(x,u)||∇u|dx≤∫ℝn‍|φ(|x|2k2)||ψ3||∇u|dx≤λ4∫ℝn‍φ(|x|2k2)|∇u|2dx+cλ∫ℝn‍φ(|x|2k2)|ψ3|2dx≤λ2∫ℝn‍φ(|x|2k2)|∇v|2dx+c∫ℝn‍φ(|x|2k2)|∇z(θtω)|2dx+c∫ℝn‍φ(|x|2k2)|ψ3|2dx.
For the third term of the right-hand side of (72), by using (5), we have
(75)∫ℝn‍φ(|x|2k2)∂∂uf(x,u)|∇u|2dx≤β∫ℝn‍φ(|x|2k2)|∇u|2dx≤2β∫ℝn‍φ(|x|2k2)|∇v|2dx+2β∫ℝn‍φ(|x|2k2)|∇z(θtω)|2dx.
For the last term of the right-hand side of (72), by using (4) and Young's inequality, we find
(76)|∫ℝn‍φ(|x|2k2)f(x,u)Δz(θtω)dx|≤α2∫ℝn‍φ(|x|2k2)|u|p-1|Δz(θtω)|dx+∫ℝn‍φ(|x|2k2)|ψ2||Δz(θtω)|dx≤c∫ℝn‍φ(|x|2k2)|u|pdx+c∫ℝn‍φ(|x|2k2)|Δz(θtω)|pdx+12∫ℝn‍φ(|x|2k2)|ψ2|2dx+12∫ℝn‍φ(|x|2k2)|Δz(θtω)|2dx.
Putting (73)–(76) together into (72), it yields that
(77)-∫ℝn‍φ(|x|2k2)f(x,u)Δvdx≤λ+4β2∫ℝn‍φ(|x|2k2)|∇v|2dx+c∫ℝn‍φ(|x|2k2)|u|pdx+ck(∥u∥2p-22p-2+∥∇u∥2+1)+c∫ℝn‍φ(|x|2k2)(|ψ2|2+|ψ3|2)dx+c∫ℝn‍φ(|x|2k2)×(|∇z(θtω)|2+|Δz(θtω)|2+|Δz(θtω)|p)dx.
Then by (68)–(71) and (77), we get
(78)ddt∫ℝn‍φ(|x|2k2)|∇v|2dx+λ∫ℝn‍φ(|x|2k2)|∇v|2dx+∫ℝn‍φ(|x|2k2)|Δv|2dx≤ck(∥vt∥2+∥∇v∥2+∥v∥2+∥u∥2p-22p-2+∥∇u∥2+1)+4β∫ℝn‍φ(|x|2k2)|∇v|2dx+c∫ℝn‍φ(|x|2k2)|u|pdx+c∫ℝn‍φ(|x|2k2)(|∇z(θtω)|2+|Δz(θtω)|2+|Δz(θtω)|p)dx+c∫ℝn‍φ(|x|2k2)(|g|2+|ψ2|2+|ψ3|2)dx.
In particular,
(79)ddt∫ℝn‍φ(|x|2k2)|∇v(t)|2dx≤4β∫ℝn‍φ(|x|2k2)|∇v(t)|2dx+c∫ℝn‍φ(|x|2k2)|u(t)|pdx+ck(∥z(θtω)∥2p-22p-2∥vt(t)∥2+∥∇v(t)∥2+∥v(t)∥2+∥v(t)∥2p-22p-2+∥z(θtω)∥2p-22p-2+∥∇z(θtω)∥2+1∥z(θtω)∥2p-22p-2)+c∫ℝn‍φ(|x|2k2)(|∇z(θtω)|2+|Δz(θtω)|2+|Δz(θtω)|p)dx+c∫ℝn‍φ(|x|2k2)(|g|2+|ψ2|2+|ψ3|2)dx.
Let s∈(t,t+1) and integrate the aforementioned inequality from s to t+1:
(80)∫ℝn‍φ(|x|2k2)|∇v(t+1,ω,v0(ω))|2dx≤∫ℝn‍φ(|x|2k2)|∇v(s,ω,v0(ω))|2dx+4β∫tt+1‍∫ℝn‍φ(|x|2k2)|∇v(τ,ω,v0(ω))|2dxdτ+c∫tt+1‍∫ℝn‍φ(|x|2k2)|u(τ,ω,u0(ω))|pdxdτ+ck(∫tt+1‍∥vτ(τ,ω,v0(ω))∥2dτ+∫tt+1‍∥∇v(τ,ω,v0(ω))∥2dτ+∫tt+1‍∥v(τ,ω,v0(ω))∥2dτ+∫tt+1‍∥v(τ,ω,v0(ω))∥2p-22p-2dτ+∫tt+1‍∥z(θτω)∥2p-22p-2dτ+∫tt+1‍∥∇z(θτω)∥2dτ+1)+c∫tt+1‍∫ℝn‍φ(|x|2k2)×(|∇z(θτω)|2+|Δz(θτω)|2+|Δz(θτω)|p)dxdτ+c∫ℝn‍φ(|x|2k2)(|g|2+|ψ2|2+|ψ3|2)dx.
Integrating the aforementioned inequality with respect to s over (t,t+1), and replacing ω by θ-t-1ω, we obtain
(81)∫ℝn‍φ(|x|2k2)|∇v(t+1,θ-t-1ω,v0(θ-t-1ω))|2dx≤c3∫tt+1‍∫ℝn‍φ(|x|2k2)×|∇v(τ,θ-t-1ω,v0(θ-t-1ω))|2dxdτ+c3∫tt+1‍∫ℝn‍φ(|x|2k2)×|u(τ,θ-t-1ω,u0(θ-t-1ω))|pdxdτ+c3k(∫tt+1‍∥vτ(τ,θ-t-1ω,v0(θ-t-1ω))∥2dτ+∫tt+1‍∥∇v(τ,θ-t-1ω,v0(θ-t-1ω))∥2dτ+∫tt+1‍∥v(τ,θ-t-1ω,v0(θ-t-1ω))∥2dτ+∫tt+1‍∥v(τ,θ-t-1ω,v0(θ-t-1ω))∥2p-22p-2dτ+∫tt+1‍∥z(θτ-t-1ω)∥2p-22p-2dτ+∫tt+1‍∥∇z(θτ-t-1ω)∥2dτ+1)+c3∫tt+1‍∫ℝn‍φ(|x|2k2)×(|∇z(θτ-t-1ω)|2+|Δz(θτ-t-1ω)|2+|Δz(θτ-t-1ω)|p)dxdτ+c3∫ℝn‍φ(|x|2k2)(|g|2+|ψ2|2+|ψ3|2)dx.
In the sequel, we estimate each term in the right-hand side of (81) to show that they are arbitrary small when k and t are large enough.
To estimate the first term in the right-hand side of (81), we cite the result (4.46) in [1]; that is,
(82)12ddt∫ℝn‍φ(|x|2k2)|v|2dx+12α1∫ℝn‍φ(|x|2k2)|u|pdx+∫ℝn‍φ(|x|2k2)|∇v|2dx≤ck(∥∇v∥2+∥v∥2)+∫ℝn‍φ(|x|2k2)(1λ|g|2+|ψ1|+12|ψ2|2)dx+c∫ℝn‍φ(|x|2k2)(|Δz(θtω)|2+|z(θtω)|2+|z(θtω)|p)dx.
Integrating the aforementioned inequality over (t,t+1) and replacing ω by θ-t-1ω we get
(83)∫tt+1‍∫ℝn‍φ(|x|2k2)|∇v(s,θ-t-1ω,v0(θ-t-1ω))|2dxds≤12∫ℝn‍φ(|x|2k2)|v(t,θ-t-1ω,v0(θ-t-1ω))|2dx+ck(∫tt+1‍∥∇v(s,θ-t-1ω,v0(θ-t-1ω))∥2ds+∫tt+1‍∥v(s,θ-t-1ω,v0(θ-t-1ω))∥2ds)+∫ℝn‍φ(|x|2k2)(1λ|g|2+|ψ1|+12|ψ2|2)dx+c∫tt+1‍∫ℝn‍φ(|x|2k2)×(|Δz(θs-t-1ω)|2+|z(θs-t-1ω)|2+|z(θs-t-1ω)|p)dxds.
To estimate the first term of the right-hand side of (83), we need a result in [1]. Substituting θ-t-1ω for ω of (4.49) in [1], we get
(84)∫ℝn‍φ(|x|2k2)|v(t,θ-t-1ω,v0(θ-t-1ω))|2dx≤eλ(T-t)∫ℝn‍φ(|x|2k2)|v(T,θ-t-1ω,v0(θ-t-1ω))|2dx+ck∫Tt‍eλ(s-t)(∥v(s,θ-t-1ω,v0(θ-t-1ω))∥2+∥∇v(s,θ-t-1ω,v0(θ-t-1ω))∥2)ds+∫Tt‍eλ(s-t)∫ℝn‍φ(|x|2k2)×(2λ|g|2+2|ψ1|+|ψ2|2)dxds+c∫Tt‍eλ(s-t)∫ℝn‍φ(|x|2k2)×(|Δz(θs-t-1ω)|2+|z(θs-t-1ω)|2+|z(θs-t-1ω)|p)dxds≤ceλ(T-t-1)∫ℝn‍φ(|x|2k2)×|v(T,θ-t-1ω,v0(θ-t-1ω))|2dx+ck∫Tt+1‍eλ(s-t-1)(∥v(s,θ-t-1ω,v0(θ-t-1ω))∥2+∥∇v(s,θ-t-1ω,v0(θ-t-1ω))∥2)ds+c∫Tt+1‍eλ(s-t-1)∫ℝn‍φ(|x|2k2)×(2λ|g|2+2|ψ1|+|ψ2|2)dxds+c∫Tt+1‍eλ(s-t-1)∫ℝn‍φ(|x|2k2)×(|Δz(θs-t-1ω)|2+|z(θs-t-1ω)|2+|z(θs-t-1ω)|p)dxds,
where T=TD^(ω) is an absorbing time in Lemma 14 and t≥T. It is a direct result of [1] that there exist T1=TD^1(ω,ɛ)≥T, R1=R1(ω,ɛ) such that, for all t≥T1, k≥R1 the right-hand side of the aforementioned inequality is less than or equal to ɛ/20c3, so we get
(85)12∫ℝn‍φ(|x|2k2)|v(t,θ-t-1ω,v0(θ-t-1ω))|2dx≤ɛ40c3.
By Lemma 15 we find that there exists T2=TD^2(ω)>0, for all t≥T2,
(86)ck(∫tt+1‍∥∇v(s,θ-t-1ω,v0(θ-t-1ω))∥2ds+∫tt+1‍∥v(s,θ-t-1ω,v0(θ-t-1ω))∥2ds)≤ck(1+r(ω)).
Choosing R2=R2(ω,ɛ)≥0 such that (c/k)(1+r(ω))≤ɛ/40c3 for k≥R2, we deduce that
(87)ck(∫tt+1‍∥∇v(s,θ-t-1ω,v0(θ-t-1ω))∥2ds+∫tt+1‍∥v(s,θ-t-1ω,v0(θ-t-1ω))∥2ds)≤ɛ40c3,∀t≥T2,∀k≥R2.
For the third term of the right-hand side of (83), we have
(88)∫ℝn‍φ(|x|2k2)(1λ|g|2+|ψ1|+12|ψ2|2)dx≤∫|x|≥k(1λ|g|2+|ψ1|+12|ψ2|2)dx,
since ψ1∈L1(ℝn), ψ2∈L2(ℝn), and g∈L2(ℝn) there is a constant R3=R3(ɛ) such that, for all k≥R3,
(89)∫ℝn‍φ(|x|2k2)(1λ|g|2+|ψ1|+12|ψ2|2)dx≤ɛ40c3.
Note that z(θtω)=∑j=1m‍hjzj(θtωj) and hj∈H2(ℝn)⋂‍W2,p(ℝn)⋂‍L∞(ℝn) for j=1,2,…,m. Hence there exists R4=R4(ω,ɛ) such that for all k≥R4 and j=1,2,…,m,
(90)∫|x|≥k‍(|hj(x)|2+|hj(x)|p+|Δhj(x)|2)dx≤λe-(λ/2)ɛ160cc3mpr(ω),
where c is the constant in (83), and thus we have the following estimate:
(91)c∫tt+1‍∫ℝn‍φ(|x|2k2)×(|Δz(θs-t-1ω)|2+|z(θs-t-1ω)|2+|z(θs-t-1ω)|p)dxds≤c∫tt+1‍∫|x|≥k‍(|Δz(θs-t-1ω)|2+|z(θs-t-1ω)|2≤c∫tt+1‍∫|x|≥k‍h+|z(θs-t-1ω)|p)dxds≤cmp∫tt+1‍∑j=1m‍∫|x|≥k(|Δhj|2|zj(θs-t-1ωj)|2‍≤cmp∫tt+1‍∑j=1m‍hhhhhhhh+|hj|2|zj(θs-t-1ωj)|2≤cmp∫tt+1‍∑j=1m‍hhhhhhhh+|hj|p|zj(θs-t-1ωj)|p)dxds=cmp∫tt+1‍[∑j=1m‍|zj(θs-t-1ωj)|2∫|x|≥k‍|Δhj|2dx+∑j=1m‍|zj(θs-t-1ωj)|2∫|x|≥k‍|hj|2dx+∑j=1m‍|zj(θs-t-1ωj)|p∫|x|≥k‍|hj|pdx]ds≤λe-(λ/2)ɛ160cc3mpr(ω)cmp∫tt+1‍∑j=1m‍(2|zj(θs-t-1ωj)|2+|zj(θs-t-1ωj)|p)ds≤λe-(λ/2)ɛ80c3r(ω)∫tt+1‍∑j=1m‍(|zj(θs-t-1ωj)|2+|zj(θs-t-1ωj)|p)ds≤λe-(λ/2)ɛ80c3r(ω)∫tt+1‍p(θs-t-1ω)ds=λe-(λ/2)ɛ80c3r(ω)∫-10‍p(θsω)ds≤λe-(λ/2)ɛ80c3r(ω)r(ω)∫-10‍e-(λ/2)sds≤ɛ40c3.
Let
(92)T1*=T1*(D^,ω,ɛ)=max{T1,T2},R1*=R1*(ω,ɛ)=max{R1,R2,R3,R4}.
From (83), (85), (87), (89), and (91), ∀t≥T1*, k≥R1*, we can estimate the first term of the right-hand side of (81) as
(93)c3∫tt+1‍∫ℝn‍φ(|x|2k2)×|∇v(s,θ-t-1ω,v0(θ-t-1ω))|2dxds≤ɛ10.
Next, from (82) and a similar process as the proof of (93), one can also get the result for the second term of right-hand side of (81); that is, there exist R2*=R2*(ω,ɛ)>0 and T2*=T2*(D^,ω,ɛ)>0 such that, for all t≥T2*, k≥R2*,
(94)c3∫tt+1‍∫ℝn‍φ(|x|2k2)×|u(s,θ-t-1ω,v0(θ-t-1ω))|pdxds≤ɛ10.
From Lemmas15 and 16 we see that there exist T3*=T3*(D^,ω)>0 and R3*=R3*(ω,ɛ)>0 such that ∀t≥T3*,∀k≥R3*,
(95)c3k∫tt+1‍∥∇v(s,θ-t-1ω,v0θ-t-1ω)∥2ds≤ck(1+r(ω))≤ɛ10,c3k∫tt+1‍∥v(s,θ-t-1ω,v0θ-t-1ω)∥2ds≤ck(1+r(ω))≤ɛ10,c3k∫tt+1‍∥vs(s,θ-t-1ω,v0θ-t-1ω)∥2ds≤ck(1+r(ω))≤ɛ10,c3k∫tt+1‍∥v(s,θ-t-1ω,v0θ-t-1ω)∥2p-22p-2ds≤ck(1+r(ω))≤ɛ10.
For the seventh and eighth terms of the right-hand side of (81), we have
(96)∫tt+1‍(∥z(θs-t-1ω)∥2p-22p-2+∥∇z(θs-t-1ω)∥2)ds≤c0∫tt+1‍p(θs-t-1ω)ds=c0∫-10‍p(θsω)ds≤c0r(ω)∫-10‍e-(λ/2)sds≤cr(ω).
So there exists R4*=R4*(ω,ɛ)>0 such that, for all k≥R4* and all  t≥0, we have
(97)c3k∫tt+1‍(∥z(θs-t-1ω)∥2p-22p-2+∥∇z(θs-t-1ω)∥2)ds≤ckr(ω)≤ɛ10.
Similar to the proof of (91), one can show that there exists R5*=R5*(ω,ɛ)>0 such that, for all k≥R5* and all  t≥0, we have
(98)c3∫tt+1‍∫ℝn‍φ(|x|2k2)(|∇z(θs-t-1ω)|2+|Δz(θs-t-1ω)|2+|Δz(θs-t-1ω)|p)dxds≤ɛ10.
Sinceψ2∈L2(ℝn), ψ3∈L2(ℝn), and g∈L2(ℝn), we can easily show that there exists R6*=R6*(ɛ)>0 such that the last term of the right-hand side of (81) is bounded by
(99)c3∫ℝn‍φ(|x|2k2)(|g|2+|ψ2|2+|ψ3|2)dx≤c3∫|x|≥k‍(|g|2+|ψ2|2+|ψ3|2)dx≤ɛ10,
when k≥R6*.
Finally, let(100)R*=R*(ω,ɛ)=max{R1*,R2*,…,R6*},T*=TD^*(ω,ɛ)=max{T1*,T2*,T3*}.
From (81), (93)–(95), (97)–(99) we get
(101)∫ℝn‍φ(|x|2k2)|∇v(t+1,θ-t-1ω,v0(θ-t-1ω))|2dx≤ɛ,t≥T*,k≥R*.
Therefore, ∀t≥T*,
(102)∫|x|≥2R*‍|∇v(t+1,θ-t-1ω,v0(θ-t-1ω))|2dx≤∫ℝn‍φ(|x|2R*2)|∇v(t+1,θ-t-1ω,v0(θ-t-1ω))|2dx≤ɛ.
The proof is complete.Lemma 18.
Assume thatg∈L2(ℝn) and (3)–(6) hold. Let D^={D(ω)}ω∈Ω∈𝒟2 and u0(ω)∈D(ω). Then, for  ℙ-a.e. ω∈Ω and for every ɛ>0, there exist T*=TD^*(ω,ɛ)>0 and R*=R*(ω,ɛ), such that the solution u(t,ω,u0(ω)) of problem (1)-(2) satisfies that, ∀t≥TD^*(ω,ɛ),
(103)∫|x|≥R*‍(|u(t,θ-tω,u0(θ-tω))|2+|∇u(t,θ-tω,u0(θ-tω))|2)dx≤ɛ.Proof.
Considering∀ɛ>0, we choose R5=R5(ω,ɛ) large enough such that
(104)∫|x|≥R5‍(|hj|2+|∇hj|2)dx≤ɛ2m2r(ω),j=1,2,…,m,
and set R*′=max{R*,R5}. Then by Lemma 17 and (34), one can easily show that
(105)∫|x|≥R*′‍(|u(t,θ-tω,u0(θ-tω))|2+|∇u(t,θ-tω,u0(θ-tω))|2)dx≤3ɛ,
with ∀t≥TD^*(ω,ɛ), where R*, TD^*(ω,ɛ) are the constants in Lemma 17. The proof is complete.


## 4.3. Asymptotic Compactness in Bounded Balls

In what follows, we prove the asymptotic compactness in any bounded ball, which together with Lemma18 and Theorem 10 is a necessary condition for verifying the (L2(ℝn),H1(ℝn))-asymptotic compactness. For this purpose, we set ψ=1-φ, where φ is the function described in Lemma 17.For fixedk≥1, define
(106)v~(t,ω,v0(ω))=ψ(|x|2k2)v(t,ω,v0(ω)).
Then v~(t,ω,v0(ω))∈H01(Q2k) and
(107)∥v~(t,ω,v0(ω))∥H01(Q2k)≤c∥v(t,ω,v0(ω))∥H1(ℝn),
where c is a positive constant, independent of t, ω, and k. Then we have
(108)v~t(t)=ψ(|x|2k2)vt(t),Δv~=(Δψ)v+2∇ψ·∇v+ψΔv.
Multiplying (30) by ψ(|x|2/k2), then we can easily show that
(109)v~t+λv~-Δv~=ψf(x,u)+ψg+ψΔz(θtω)-vΔψ-2∇ψ·∇v.
Consider the following eigenvalue problem:
(110)-Δν~=λν~,inQ2κ,ν~∣ϑQ2κ=0,
and then problem (110) has a family of eigenfunctions {ej}j=1∞ with corresponding eigenvalues {λj}j=1∞ such that {ej}j=1∞ forms an orthogonal basis in both L2(Q2k) and H01(Q2k) and
(111)λ1≤λ2≤⋯≤λj⟶∞asj⟶∞.
Given n, let Xn= span{e1,…,en} and Pn:L2(Q2k)→Xn be the projection operator. For any v~∈H01(Q2k), we write
(112)v~=Pmv~+(I-Pm)v~=:v~1+v~2.
In order to prove the asymptotic compactness we need the following lemma, which can be found in [19].Lemma 19 (see [19]).
Letr>0, λ>0, τ∈ℝ, and for s>τ(113)y′(s)+λy(s)≤h(s),
where the functions  y, y′, h are assumed to be locally integrable and y, h nonnegative on the interval t<s<t+r, for some t≥τ. Then, for any δ∈(0,r),
(114)y(t+r)≤e-λ(r-δ)1δ∫tt+r‍y(s)dsy(t+r)≤+e-λ(t+r)∫tt+r‍eλsh(s)ds.
In particular, let δ=(r/2), then
(115)y(t+r)≤e-λ(r/2)2r∫tt+r‍y(s)dsy(t+r)≤+e-λ(t+r)∫tt+reλsh(s)ds.Lemma 20.
Assume thatg∈L2(ℝn) and (3)–(6) hold. Let D^={D(ω)}ω∈Ω∈𝒟2 and u0(ω)∈D(ω). Then, for  ℙ-a.e. ω∈Ω and for every ɛ>0, there exist M(ω,ɛ)>0, η(ω,ɛ)>0, and TD^(ω)>0 such that the solution v(t,ω,v0(ω)) of (30)-(31) satisfies that, ∀t≥TD^(ω),
(116)∫t-ηt‍∫Q2k(v(s,θ-tω,v0(θ-tω))≥M)‍|v(s,θ-tω,v0(θ-tω))|p-1×|v(s,θ-tω,v0(θ-tω))-M|p-1dxds≤ɛ,∫t-ηt‍∫Q2k(v(s,θ-tω,v0(θ-tω))≤-M)‍|v(s,θ-tω,v0(θ-tω))|p-1×|v(s,θ-tω,v0(θ-tω))+M|p-1dxds≤ɛ.Proof.
Ifp=2, one can easily show the results by (36). So in the following we assume p>2. Let
(117)(v-M)+={v-M,x∈Q2k1,0,elsewhere,(v+M)-={v+M,x∈Q2k2,0,elsewhere,
where Q2k1=Q2k(v≥M)=Q2k⋂‍{x:v(t,ω,v0(ω))≥M} and Q2k2=Q2k(v≤-M)=Q2k⋂‍{x:v(t,ω,v0(ω))≤-M}.
Multiplying (30) with |(v-M)+|p-1 and integrating over ℝn, we get
(118)1pddt∥(v-M)+∥pp+λ∫Q2k1‍v(v-M)p-1dx+(p-1)∫Q2k1‍|∇(v-M)|2(v-M)p-2dx=∫Q2k1‍f(x,u)(v-M)p-1dx+∫Q2k1‍g(v-M)p-1dx+∫Q2k1‍Δz(θtω)(v-M)p-1dx.
From (41), when x∈Q2k1, we have
(119)f(x,v+z(θtω))(v-M)p-1=f(x,v+z(θtω))v(v-M)p-1v≤-α12p|v|p(v-M)p-1v+c(|z(θtω)|p+|z(θtω)|2)(v-M)p-1v+(ψ1+12ψ22)(v-M)p-1v≤-α12p|v|p-1(v-M)p-1+cM(|z(θtω)|p+|z(θtω)|2)(v-M)p-1+1M(ψ1+12ψ22)(v-M)p-1≤-α12p|v|p-1(v-M)p-1+c(|z(θtω)|p+|z(θtω)|2)(v-M)p-1+c(ψ1+12ψ22)(v-M)p-1,
where the constant c in the right hand side of (119) is independent of  M when we assume, without loss of generality, that M≥1.
From (118)-(119), we get
(120)1pddt∥(v-M)+∥pp+c4∫Q2k1‍|v|p-1(v-M)p-1dx≤c5∫Q2k1‍(|ψ1|+12|ψ2|2+|g|)(v-M)p-1dx+c5∫Q2k1‍(|z(θtω)|p+|z(θtω)|2+|Δz(θtω)||z(θtω)|p)(v-M)p-1dx.
Applying Cauchy's inequality, we obtain
(121)1pddt∥(v-M)+∥pp+c42∫Q2k1‍|v|p-1(v-M)p-1dx+c42∫Q2k1‍(v-M)2p-2dx≤c42∫Q2k1‍(v-M)2p-2dx+c∫Q2k1(|ψ1|2+|ψ2|4+g2)dx+c∫Q2k1(|z(θtω)|2p+|z(θtω)|4+|Δz(θtω)|2)dx,
that is,
(122)2pddt∥(v-M)+∥pp+c4∫Q2k1|v|p-1(v-M)p-1dx≤c∫Q2k1(|ψ1|2+|ψ2|4+g2)dx+c∫Q2k1(|z(θtω)|2p+|z(θtω)|4+|Δz(θtω)|2)dx,(123)2pddt∥(v-M)+∥pp+c4Mp-2∫Q2k1‍(v-M)pdx≤c∫Q2k1(|ψ1|2+|ψ2|4+g2)dx+c∫Q2k1(|z(θtω)|2p+|z(θtω)|4+|Δz(θtω)|2)dx.
By (123) and Lemma 19 with r=1, we get
(124)∥(v-M)+(t)∥pp≤2e-μ/2∫t-1t‍∥(v-M)+(τ)∥ppdτ+ce-μt∫t-1t‍eμτ(∥z(θτω)∥2p2p+∥z(θτω)∥44+∥Δz(θτω)∥2∥z(θτω)∥2p2p)dτ+ce-μt∫t-1t‍eμτ(∥ψ1∥2+∥ψ2∥44+∥g∥2)dτ,
where μ=c4Mp-2p/2.
Sets∈[t-1,t]; we first substitute s for t in the aforementioned inequality and then we replace ω by θ-tω to get
(125)∥(v(s,θ-tω,v0(θ-tω))-M)+∥pp≤2e-μ/2∫s-1s‍∥(v(τ,θ-tω,v0(θ-tω))-M)+∥ppdτ+ce-μs∫s-1s‍eμτ(∥z(θτ-tω)∥2p2p+∥z(θτ-tω)∥44+∥Δz(θτ-tω)∥2∥z(θτ-tω)∥2p2p)dτ+ce-μs∫s-1s‍eμτdτ(∥ψ1∥2+∥ψ2∥44+∥g∥2)≤2e-μ/2∫t-2t‍∥(v(τ,θ-tω,v0(θ-tω))-M)+∥ppdτ+ce-μs∫s-1s‍eμτ(∥z(θτ-tω)∥2p2p+∥z(θτ-tω)∥44+∥Δz(θτ-tω)∥2∥z(θτ-tω)∥2p2p)dτ+ce-μs∫s-1s‍eμτdτ(∥ψ1∥2+∥ψ2∥44+∥g∥2).
For the first term of the right-hand side of (125), we use Lemma 15:
(126)∫t-2t‍∥(v(τ,θ-tω,v0(θ-tω))-M)+∥ppdτ≤∫t-2t‍∥v(τ,θ-tω,v0(θ-tω))∥ppdτ≤c(1+r(ω)),∀t≥TD^(ω).
Then there exists M1=M1(ω,ɛ)>0 such that, for all t≥TD^(ω) and all M≥M1,
(127)2e-μ/2∫t-2t‍∥(v(τ,θ-tω,v0(θ-tω))-M)+∥ppdτ≤ce-μ/2(1+r(ω))≤ɛ3.
For the second term of the right-hand side of (125), ∀s∈[t-1,t], by (26) we have
(128)ce-μs∫s-1s‍eμτ(∥z(θτ-tω)∥2p2p+∥z(θτ-tω)∥44+∥Δz(θτ-tω)∥2∥z(θτ-tω)∥2p2p)dτ≤c∫s-1s‍eμ(τ-s)p(θτ-tω)dτ=c∫-10‍eμτp(θτ-t+sω)dτ≤cr(ω)∫-10‍eμτe(λ/2)|τ-t+s|dτ≤cr(ω)∫-10‍eμτeλdτ≤cr(ω)∫-10‍eμτdτ≤cr(ω)μ.
This implies that there exists M2=M2(ω,ɛ)>0 such that  ∀M≥M2 we have
(129)ce-μs∫s-1s‍eμτ(∥z(θτ-tω)∥2p2p+∥z(θτ-tω)∥44+∥Δz(θτ-tω)∥2∥z(θτ-tω)∥2p2p)dτ≤ɛ3.
For the last term of the right-hand side of (125), we can easily see that there exists M3=M3(ɛ)>0 such that, for all M≥M3,
(130)ce-μs∫s-1s‍eμτdτ(∥ψ1∥2+∥ψ2∥44+∥g∥2)=c∫-10‍eμτdτ(∥ψ1∥2+∥ψ2∥44+∥g∥2)≤cμ≤ɛ3.
Letting N*=N*(ω,ɛ)=max{M1,M2,M3}, then combining with (125), (127) and (129)–(130), we can show that, for all t≥TD^(ω) and all M≥N*(ω,ɛ),
(131)∥(v(s,θ-tω,v0(θ-tω))-M)+∥pp≤ɛ,∀s∈[t-1,t].
Next, we set(132)η=min{ɛ3c7(∥ψ1∥2+∥ψ2∥44+∥g∥2),12λln(λɛ6c8r(ω)+1),ɛ3c7(∥ψ1∥2+∥ψ2∥44+∥g∥2),1},
where c7=c/c4, c8=c0c7, and c is the constants in (122). Integrating (122) over (t-η,t), we get
(133)∫t-ηt‍∫Q2k1‍|v(s)|p-1(v(s)-M)p-1dxds≤c6∥(v(t-η)-M)+∥pp+c7∫t-ηt‍∫Q2k1‍(|z(θsω)|2p+|z(θsω)|4+|Δz(θsω)|2)dxds+c7∫t-ηt‍∫Q2k1‍(|ψ1|2+|ψ2|4+g2)dxds.
Replacing ω by θ-tω and setting  Q2k*=Q2k(v(s,θ-tω,v0(θ-tω))≥M), we obtain
(134)∫t-ηt‍∫Q2k*‍|v(s,θ-tω,v0(θ-tω))|p-1×(v(s,θ-tω,v0(θ-tω))-M)p-1dxds≤c7∫t-ηt‍∫Q2k*‍(|z(θs-tω)|2p+|z(θs-tω)|4+|Δz(θs-tω)|2)dxds+c6∥(v(t-η,θ-tω,v0(θ-tω))-M)+∥pp+c7η(∥ψ1∥2+∥ψ2∥44+∥g∥2).
For fixed ω∈Ω, by (131) and (132), ∀t≥TD^(ω), ∀M≥M1*(ω,ɛ)=N*(ω,ɛ/3c6),
(135)c6∥(v(t-η,θ-tω,v0(θ-tω))-M)+∥pp≤ɛ3,
where N*(ω,ɛ) is the constant in (131).
Using (132), the first term of the right-hand side of (134) is bounded by
(136)c7∫t-ηt‍(∥z(θτ-tω)∥2p2p+∥z(θτ-tω)∥44+∥Δz(θτ-tω)∥2∥z(θτ-tω)∥2p2p)dτ≤c8∫t-ηt‍p(θτ-tω)dτ=c8∫-η0‍p(θτω)dτ≤c8r(ω)∫-η0‍e-(λ/2)τdτ=2c8λr(ω)(eλη/2-1)≤ɛ3.
For the last term of right-hand side of (134), by using (132), we have
(137)c7η(∥ψ1∥2+∥ψ2∥44+∥g∥2)≤ɛ3.
By (134)–(137), ∀M≥M1*, ∀t≥TD^(ω), and ∀u0(ω)∈D(ω) with v0(ω)=u0(ω)-z(ω), we have
(138)∫t-ηt‍∫Q2k*‍|v(s,θ-tω,v0(θ-tω))|p-1×|v(s,θ-tω,v0(θ-tω))-M|p-1dxds≤ɛ.
Similarly, multiplying (30) with |(v+M)-|p-1, we can prove that there exists M2*=M2*(ω,ɛ) such that  ∀M≥M2*, ∀t≥TD^(ω), and ∀u0(ω)∈D(ω) with v0(ω)=u0(ω)-z(ω), over the region Q2k**=Q2k(v(s,θ-tω,v0(θ-tω))≤-M); we have
(139)∫t-ηt‍∫Q2k**‍|v(s,θ-tω,v0(θ-tω))|p-1×|v(s,θ-tω,v0(θ-tω))+M|p-1dxds≤ɛ.
Let M*=max{M1*,M2*}; from (138) and (139) we can obtain our results. The proof is complete.Remark 21.
The idea of the proof of the above lemma comes from [19] (this idea can be further traced back to Marion [16] and Robinson [18]). We see from (132) that the constant η in Lemma 20 is independent of t, which is different from the η in the Lemma 3.4 in [19]. This is crucial in the following estimates. As we know, the time t will vary to infinite when we consider the asymptotic behavior of an RDS. It means that if η is not a fixed constant with respect to t, the following estimate will be invalid. In other words, if the function g in (1) is dependent on t, our method will fail.Lemma 22.
Assume thatg∈L2(ℝn) and (3)–(6) hold. Let D^={D(ω)}ω∈Ω∈𝒟2 and u0(ω)∈D(ω). Then, for  ℙ-a.e. ω∈Ω and for every ɛ>0 and all k≥1, there exist TD^(ω)>0 and N(k,ω,ɛ)>0 such that the solution v(t,ω,v0(ω)) of (30)-(31) satisfies that, ∀t≥TD^(ω), ∀m≥N(k,ω,ɛ),
(140)∥v~2(t,θ-tω,v0(θ-tω))∥H01(Q2k)2≤ɛ,
where v~2=(I-Pm)v~.Proof.
Multiplying (109) with -Δv~2 and integrating over Q2k, we get
(141)12ddt∥v~2∥H01(Q2k)2+λ∥v~2∥H01(Q2k)2+∥Δv~2∥L2(Q2k)2=-(ψf(x,u),Δv~2)L2(Q2k)-(ψΔz(θtω),Δv~2)L2(Q2k)-(ψg,Δv~2)L2(Q2k)+(vΔψ,Δv~2)L2(Q2k)+2(∇ψ·∇v,Δv~2)L2(Q2k).
We now estimate each term in the right-hand side of the aforementioned equality. For the first term, by using Cauchy inequality and (4), we have
(142)|(ψf(x,u),Δv~2)L2(Q2k)|≤110∥Δv~2∥L2(Q2k)2+c∫Q2k‍|f(x,u)|2dx≤110∥Δv~2∥L2(Q2k)2+c∫Q2k‍|u|2p-2dx+c∫Q2k‍|ψ2|2dx≤110∥Δv~2∥L2(Q2k)2+c∫Q2k‍|v|2p-2dx+c∫Q2k‍|z(θtω)|2p-2dx+c≤110∥Δv~2∥L2(Q2k)2+c∫Q2k(|v|≤M)‍|v|2p-2dx+c∫Q2k(|v|≥M)‍|v|2p-2dx+c∫Q2k‍|z(θtω)|2p-2dx+c≤110∥Δv~2∥L2(Q2k)2+c∫Q2k(|v|≥M)‍|v|2p-2dx+c∥z(θtω)∥2p-22p-2+c(∥ψ2∥2,|Q2k|,p,M),
where |Q2k| denotes the Lebesgue measure of Q2k.
For the second to fifth term, we can estimate them as follows:(143)|(ψΔz(θtω),Δv~2)L2(Q2k)|≤110∥Δv~2∥L2(Q2k)2+c∥Δz(θtω)∥2,(144)|(ψg,Δv~2)L2(Q2k)|≤110∥Δv~2∥L2(Q2k)2+c∥g∥2,(145)|(vΔψ,Δv~2)L2(Q2k)|≤110∥Δv~2∥L2(Q2k)2+c∫Q2k‍|v|2|Δψ|2dx≤110∥Δv~2∥L2(Q2k)2+c∫ℝn‍|v|2|Δψ|2dx≤110∥Δv~2∥L2(Q2k)2+c∫ℝn‍|v|2|φ′′(|x|2k2)4|x|2k4+φ′(|x|2k2)2nk2|2dx≤110∥Δv~2∥L2(Q2k)2+ck4∫k≤|x|≤2k‍|v|2dx≤110∥Δv~2∥L2(Q2k)2+c∥v∥2,(146)2|(∇ψ·∇v,Δv~2)L2(Q2k)|≤110∥Δv~2∥L2(Q2k)2+c∫Q2k‍|∇ψ|2|∇v|2dx≤110∥Δv~2∥L2(Q2k)2+ck2∥∇v∥2≤110∥Δv~2∥L2(Q2k)2+c∥∇v∥2,
where the constant c in (145) and (146) is independent of k, since k≥1.
Combining (141)–(146) we get
(147)ddt∥v~2∥H01(Q2k)2+2λ∥v~2∥H01(Q2k)2+∥Δv~2∥L2(Q2k)2≤c∫Q2k(|v|≥M)‍|v|2p-2dx+c(∥v∥2+∥∇v∥2)+c(∥z(θtω)∥2p-22p-2+∥Δz(θtω)∥2)+c(∥ψ2∥2,∥g∥2,|Q2k|,p,M),
and this implies that
(148)ddt∥v~2∥H01(Q2k)2+λm+1∥v~2∥H01(Q2k)2≤c∫Q2k(|v|≥M)‍|v|2p-2dx+c(∥v∥2+∥∇v∥2)+c(∥z(θtω)∥2p-22p-2+∥Δz(θtω)∥2)+c(∥ψ2∥2,∥g∥2,|Q2k|,p,M).
By using Lemma 19 with r=1, we get
(149)∥v~2(t+1)∥H01(Q2k)2≤2e-λm+1/2∫tt+1‍∥v~2(s)∥H01(Q2k)2ds+ce-λm+1(t+1)∫tt+1‍eλm+1s∫Q2k(|v(s)|≥M)‍|v(s)|2p-2dxds+ce-λm+1(t+1)∫tt+1‍eλm+1s×(∥z(θsω)∥2p-22p-2+∥Δz(θsω)∥2)ds+ce-λm+1(t+1)∫tt+1‍eλm+1s(∥v(s)∥2+∥∇v(s)∥2)ds+c(∥ψ2∥2,∥g∥2,|Q2k|,p,M)e-λm+1(t+1)∫tt+1‍eλm+1sds.
Substituting  θ-t-1ω for ω, we get
(150)∥v~2(t+1,θ-t-1ω,v0(θ-t-1ω))∥H01(Q2k)2≤2e-λm+1/2∫tt+1‍∥v~2(s,θ-t-1ω,v0(θ-t-1ω))∥H01(Q2k)2ds+ce-λm+1(t+1)∫tt+1‍eλm+1s×∫Q2k(|v(*)|≥M)‍|v(*)|2p-2dxds+ce-λm+1(t+1)∫tt+1‍eλm+1s(∥z(θs-t-1ω)∥2p-22p-2+∥Δz(θs-t-1ω)∥2∥z(θs-t-1ω)∥2p-22p-2)ds+ce-λm+1(t+1)∫tt+1‍eλm+1s(∥v(*)∥2+∥∇v(*)∥2)ds+c(∥ψ2∥2,∥g∥2,|Q2k|,p,M)e-λm+1(t+1)∫tt+1‍eλm+1sds,
for simplicity, hereafter, we write v(*)=v(s,θ-t-1ω,v0(θ-t-1ω)).
In the sequel, we estimate each term in the right-hand side of (150). For the first term, we use Lemma 15 and (107) to obtain that, for t≥TD^(ω),
(151)2e-λm+1/2∫tt+1‍∥v~2(s,θ-t-1ω,v0(θ-t-1ω))∥H01(Q2k)2ds≤2e-λm+1/2∫tt+1‍∥v~(s,θ-t-1ω,v0(θ-t-1ω))∥H01(Q2k)2ds≤ce-λm+1/2∫tt+1‍∥v(*)∥H1(ℝn)2ds≤ce-λm+1/2(1+r(ω)).
Thus, there exists N1=N1(ω,ɛ), for all m≥N1 and all t≥TD^(ω), we get
(152)2e-λm+1/2∫tt+1‍∥v~2(s,θ-t-1ω,v0(θ-t-1ω))∥H01(Q2k)2ds≤ɛ5.
For the second term in the right-hand side of (150), we can estimate it as follows:
(153)∫Q2k(|v(*)|≥M)‍|v(*)|2p-2dx=∫Q2k(v(*)≥M)‍|v(*)|2p-2dx+∫Q2k(v(*)≤-M)‍|v(*)|2p-2dx≤2p-2[∫Q2k(v(*)≥M)‍|v(*)|p-1×(|v(*)-M|p-1+Mp-1)dx+∫Q2k(v(*)≤-M)‍|v(*)|p-1×(|v(*)+M|p-1+Mp-1)dx∫Q2k(v(*)≥M)‍]≤2p-2[∫Q2k(v(*)≥M)‍|v(*)|p-1|v(*)-M|p-1dx+∫Q2k(v(*)≤-M)‍|v(*)|p-1|v(*)+M|p-1dx]+2p-2Mp-2∫Q2k(|v(*)|≥M)‍|v(*)|pdx,(154)ce-λm+1(t+1)∫tt+1‍eλm+1s∫Q2k(|v(*)|≥M)‍|v(*)|2p-2dxds=ce-λm+1(t+1)(∫t+1-ηt+1‍+∫tt+1-η‍)eλm+1s×∫Q2k(|v(*)|≥M)‍|v(*)|2p-2dxds≤ce-λm+1(t+1)∫t+1-ηt+1‍eλm+1s2p-2×[∫Q2k(v(*)≥M)‍|v(*)|p-1|v(*)-M|p-1dx+∫Q2k(v(*)≤-M)‍|v(*)|p-1×|v(*)+M|p-1dx∫Q2k(v(*)≥M)‍]ds+c2p-2Mp-2e-λm+1(t+1)×∫t+1-ηt+1‍eλm+1s∫Q2k(|v(*)|≥M)‍|v(*)|pdxds+ce-λm+1(t+1)×∫tt+1-η‍eλm+1s∫Q2k(|v(*)|≥M)‍|v(*)|2p-2dxds.
The first term in the right-hand side of (154) is bounded by
(155)2p-2c∫t+1-ηt+1[∫Q2k(v(*)≥M)‍|v(*)|p-1‍2p-2c∫t+1-ηt+1‍h×|v(*)-M|p-1dx2p-2hc∫t+1-ηt+1‍+∫Q2k(v(*)≤-M)‍|v(*)|p-1|v(*)+M|p-1dx]ds≤ɛ15,∀t≥TD^(ω),
when we choose appropriate η and M by Lemma 20.
For the second term of the right-hand side of (154), we use Lemma 15, ∀t≥TD^(ω),
(156)c2p-2Mp-2e-λm+1(t+1)×∫t+1-ηt+1‍eλm+1s∫Q2k(|v(*)|≥M)‍|v(*)|pdxds≤c2p-2Mp-2e-λm+1(t+1)∫t+1-ηt+1‍eλm+1s∥v(*)∥ppds≤c2p-2Mp-2e-λm+1(t+1)(1+r(ω))∫t+1-ηt+1‍eλm+1sds≤c2p-2Mp-2(1+r(ω))∫t+1-ηt+1‍eλm+1(s-t-1)ds=c2p-2Mp-2(1+r(ω))∫-η0‍eλm+1sds≤c2p-2Mp-2(1+r(ω))λm+1,
where M is determined in (155), and then there exists N21(ω,ɛ)>0, such that  ∀m≥N21,∀t≥TD^(ω), we get
(157)c2p-2Mp-2e-λm+1(t+1)×∫t+1-ηt+1‍eλm+1s∫Q2k(|v(*)|≥M)‍|v(*)|pdxds≤ɛ15.
By Lemma 16, the last term of the right-hand side of (154) is bounded by
(158)ce-λm+1η∫tt+1-η‍∫Q2k(|v(*)|≥M)‍|v(*)|2p-2dxds≤ce-λm+1η∫tt+1‍∥v(*)∥2p-22p-2ds≤ce-λm+1η(1+r(ω)),
for all t≥TD^(ω). This implies that there exists N22=N22(ω,ɛ)>0, such that, for all m≥N22, t≥TD^(ω), we have
(159)ce-λm+1(t+1)×∫tt+1-η‍eλm+1s∫Q2k(|v(*)|≥M)‍|v(*)|2p-2dxds≤ɛ15.
Combining (154), (155), (157), and (159) and setting N2=N2(ω,ɛ)=max{N21(ω,ɛ),N22(ω,ɛ)}, then we have that, for all m≥N2, t≥TD^(ω),
(160)ce-λm+1(t+1)∫tt+1‍eλm+1s×∫Q2k(|v(*)|≥M)‍|v(*)|2p-2dxds≤ɛ5.
The third term of the right-hand side of (150) is bounded by
(161)ce-λm+1(t+1)∫tt+1‍eλm+1sp(θs-t-1ω)ds=c∫tt+1‍eλm+1(s-t-1)p(θs-t-1ω)ds=c∫-10‍eλm+1sp(θsω)ds≤cr(ω)∫-10‍eλm+1se-(λ/2)sds≤cr(ω)λm+1-λ/2,
so there exists N3=N3(ω,ɛ)>0, ∀m≥N3,∀t≥0; we have
(162)ce-λm+1(t+1)×∫tt+1‍eλm+1s(∥z(θs-t-1ω)∥2p-22p-2+∥Δz(θs-t-1ω)∥2)ds≤ɛ5.
For the forth term of the right-hand side of (150), by using Lemma 15, we have
(163)ce-λm+1(t+1)∫tt+1‍eλm+1s(∥v(*)∥2+∥∇v(*)∥2)ds≤ce-λm+1(t+1)∫tt+1‍eλm+1sds(1+r(ω))≤c∫-10‍eλm+1sds(1+r(ω))≤cλm+1(1+r(ω)),∀t≥TD^(ω),
and then there exists N4=N4(ω,ɛ)>0, for all m≥N4 and all t≥TD^(ω),
(164)ce-λm+1(t+1)×∫tt+1‍eλm+1s(∥v(*)∥2+∥∇v(*)∥2)ds≤ɛ5.
For the last term of the right-hand side of (150), we have
(165)c(∥ψ2∥2,∥g∥2,|Q2k|,p,M)e-λm+1(t+1)×∫tt+1‍eλm+1sds≤c(∥ψ2∥2,∥g∥2,|Q2k|,p,M)λm+1,
this implies that there exists N5=N5(k,ɛ) (M is determined in (155)), for all m≥N5 and all t≥0,
(166)c(∥ψ2∥2,∥g∥2,|Q2k|,p,M)e-λm+1(t+1)×∫tt+1‍eλm+1sds≤ɛ5.
Finally, letN(k,ω,ɛ)=max{N1,N2,N3,N4,N5}, and then from (150), (152), (160), (162), (164), and (166) we have that, for all m≥N(k,ω,ɛ), t≥TD^(ω),
(167)∥v~2(t+1,θ-t-1ω,v0(θ-t-1ω))∥H01(Q2k)2≤ɛ.
The proof is complete.Lemma 23.
Assume thatg∈L2(ℝn) and (3)–(6) hold, and ϕ is the RDS generated by problem (1)-(2); then, χQk·ϕ is (L2(ℝn),H1(Qk))-asymptotically compact for all k≥1.Proof.
We first prove that the sequence{v~(tm,θ-tmω,ym)}m=1∞ is precompact in H01(Q2k), for any tm→∞, and any xm∈D(θ-tmω) with D^={D(ω)}ω∈Ω∈𝒟2, where the relationship between xm and ym is determined by (31), that is, ym=xm-z(θ-tmω). By Lemma 15 we see that, for all t≥TD^(ω),
(168)∥v(t,θ-tω,v0(θ-tω))∥H1(ℝn)2≤c(1+r(ω)),
where v0(θ-tω)=u0(θ-tω)-z(θ-tω) with u0(ω)∈D(ω). Thus there is M1=M1(D^,ω) such that, for all m≥M1, we have tm≥TD^(ω) and
(169)∥v(tm,θ-tmω,ym)∥H1(ℝn)2≤c(1+r(ω)),∀m≥M1.
By (107) and the aforementioned inequality we find that
(170)∥v~(tm,θ-tmω,ym)∥H01(Q2k)2≤c(1+r(ω)),∀m≥M1.
Givenɛ>0, it follows from Lemma 22 that there are TD^(ω) and N(k,ω,ɛ) such that, for all t≥TD^(ω),
(171)∥(I-PN)v~(t,θ-tω,v0(θ-tω))∥H01(Q2k)2≤ɛ.
Taking M2=M2(D^,ω) large enough such that tm≥TD^(ω) for m≥M2, then we get from (171) that
(172)∥(I-PN)v~(tm,θ-tmω,ym)∥H01(Q2k)2≤ɛ,∀m≥M2.
On the other hand, (170) shows that the sequence {PNv~(tm,θ-tmω,ym)} is bounded in the finite-dimensional space PNH01(Q2k) and hence is precompact in PNH01(Q2k), which along with (172) implies the precompactness of {v~(tm,θ-tmω,ym)} in H01(Q2k).
Next we prove the(L2(ℝn),H1(Qk))-asymptotic compactness for χQk·ϕ. From the definition of v~, we see that v=v~ in Qk, so v(tm,θ-tmω,ym)}m=1∞ is precompact in H1(Qk). By the relationship
(173)u(tm,θ-tmω,xm)=v(tm,θ-tmω,ym)+z(ω),
and the aforementioed assert, we obtain that {u(tm,θ-tmω,xm)}m=1∞ is precompact in H1(Qk), for all k≥1. The proof is complete.


## 5. Asymptotic Compactness and Random Attractors

In this section, we prove our main result, that is, the existence of an(L2(ℝn),H1(ℝn))-random attractor for the RDS ϕ associated with the initial value problem of SRDE (1)-(2). To this end, we should show the (L2(ℝn),H1(ℝn))-asymptotic compactness of ϕ. From Theorem 10 and Lemmas 18 and 23, we can immediately obtain the asymptotic compactness of ϕ.Lemma 24.
Assume thatg∈L2(ℝn) and (3)–(6) hold. Then the RDS ϕ generated by (1)-(2) is (L2(ℝn),H1(ℝn)) asymptotically compact.Now, we are in a position to present our main result.Theorem 25.
Assume thatg∈L2(ℝn) and (3)–(6) hold. Then the RDS ϕ generated by (1)-(2) has an (L2(ℝn),H1(ℝn))-random attractor A^v.Proof.
The result can be obtained by Theorems9, and 13, Lemmas 14, and 24 immediately.Remark 26.
Our methods can be used to prove the existence of(L2(ℝn),H1(ℝn))-pullback attractors for the following non-autonomous reaction-diffusion equation on unbounded domains:
(174)∂u∂t+λu-Δu=f(x,u)+g(x,t),x∈ℝn,
with the initial condition
(175)u(x,τ)=uτ(x),x∈ℝn,
for every τ∈ℝ and t>τ where the nonlinear term f satisfies (3)–(6) in this paper. g is a given function in Lloc2(ℝ,ℝn) with
(176)∫-∞t‍eλs∥g(s)∥2ds<∞,∀t∈ℝ,
and the result is new in this case.



---

*Source: 279509-2013-12-21.xml*