# Asymptotic Behavior of the Kirchhoff Type Stochastic Plate Equation on Unbounded Domains

Authors: Xiaobin Yao; Zhang Zhang
Journal: Complexity
 (2022)

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## Abstract

In this paper, we study the asymptotic behavior of solutions to the Kirchhoff type stochastic plate equation driven by additive noise defined on unbounded domains. We first prove the uniform estimates of solutions and then establish the existence and upper semicontinuity of random attractors.


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## Body

## 1. Introduction

Plate equations can be found in many fields such as certain physical areas as to vibration and elasticity theories of solid mechanics. In this paper, we consider the following Kirchhoff type stochastic plate equation with additive noise defined on(1)utt+αut+Δ2ut+Δ2u+λu−M∇u2Δu+fx,u=gx,t+βhxdWdt,ux,τ=u0x,utx,τ=u1x,where x∈ℝn, t>τ with τ∈ℝ, α,λ>0 and β are positive constants, f is a nonlinearity satisfying certain growth and dissipative conditions, gx,⋅ and h are given functions in Lloc2ℝ,L2ℝn and H2ℝn, respectively, and Wt is a two-sided real-valued Wiener process on a probability space.The functionM⋅ satisfies the following conditions:(1)
M∈C1ℝ, such that(2)M1≤Ms≤M2,whereM1 and M2 are some positive real constants.(2)
LetM^z=∫0zMrdr; for ∀z≥0,(3)Mzz≥M^z≥0.For the nonlinear functionfx,u, we presume fx,⋅∈C2ℝ and let Fx,u=∫0ufx,sds; for x∈ℝn and u∈ℝ, there exist positive constants cii=1,2,3, such that(4)fx,u≤c1uk+η1x,η1∈L2ℝn,(5)fx,uu−c2Fx,u≥η2x,η2∈L1ℝn,(6)Fx,u≥c3uk+1−η3x,η3∈L1ℝn,(7)∂f∂ux,u≤ϖ,where ϖ>0,1≤k≤n+4/n−4. Note that (4) and (5) imply(8)Fx,u≤cu2+uk+1+η12+η2.As for deterministic plate equations, many authors have showed the existence of global attractors (see [1–9]). For the stochastic case, the existence of random attractors for plate equations has been investigated in [10, 11, 12] on bounded domains. In addition, there are results about the existence of random attractors and asymptotic compactness for plate equations on unbounded domains in [13–18].WhenMs≡0 in (1), we have investigated the existence of a random attractor for plate equations with additive noise and nonlinear damping defined on ℝn (see [14]). However, when equation (1) is Kirchhoff type, the problem is not yet considered by any predecessors.To overcome the noncompactness of Sobolev embeddings onℝn, we will apply the idea of uniform estimates on the tails of solutions as in [19, 20] as well as the compactness methods introduced in [21]. More precisely, we first show that the tails of the solutions of (1) are uniformly small outside a bounded domain for large time, and then we derive the asymptotic compactness of solutions in bounded domains by splitting the solutions.This paper is organized as follows. In Section2, we present some notations and proposition about random dynamical systems. In Section 3, we define a continuous cocycle for equation (1) in H2ℝn×L2ℝn. In Section 4, we obtain all necessary uniform estimates of solutions. In Section 5, we show the existence and uniqueness of a random attractor for (1) defined on ℝn.


## 2. Notations

In this section, we present some basic notations and known results on nonautonomous random dynamical systems which can be found in [22, 20].LetX,⋅X be a complete separable metric space and be Ω,F,P,θtt∈ℝ an ergodic metric dynamical system (see [23]).Proposition 1.
LetD be an inclusion closed collection of some families of nonempty subsets of X and Φ be a continuous cocycle on X over Ω,F,P,θtt∈ℝ. Then, Φ has a unique D-pullback random attractor A in D if Φ is D-pullback asymptotically compact in X and Φ has a closed measurable D-pullback absorbing set K in D.
Next, we present criteria concerning upper semicontinuity of nonautonomous random attractors with respect to a parameter.Theorem 1.
LetX,⋅X be a separable Banach space and Φ0 be an autonomous dynamical system with the global attractor A0 in X. Given β>0, suppose that Φβ is the perturbed random dynamical system with a random attractor Aβ∈D and a random absorbing set Eβ∈D. Then, for P−a.e.τ∈ℝ,ω∈Ω,(9)dHAβτ,ω,A0⟶0,asβ⟶0,if the following conditions are satisfied:(i)
There exists some deterministic constantc such that, for P−a.e.τ∈ℝ,ω∈Ω,(10)limsupβ⟶0Eβτ,ωX≤c.(ii)
There existsβ0>0, such that for P−a.e.τ∈ℝ,ω∈Ω,(11)∪0<β≤β0Aβτ,ωis precompact inX.(iii)
ForP−a.e.τ∈ℝ,ω∈Ω, t≥0,βn⟶0, and xn,x∈X with xn⟶x, it holds that(12)limn⟶∞Φβnt,τ,ωxn=Φ0tx,whereEβτ,ωX=supx∈Eβτ,ωxX.


## 3. Cocycles for Stochastic Plate Equation

In this section, we present some basic settings about (1) and prove that it generates a continuous cocycle in H2ℝn×L2ℝn.Let−Δ denote the Laplace operator in ℝn, A=Δ2, with the domain DA=H4ℝn. We can define the powers Aν of A for ν∈ℝ. The space Vν=DAν/4 is a Hilbert space with the following inner product and norm.(13)u,vν=Av/4u,Av/4v,⋅ν=Av/4.Setℋ=H2ℝn×L2ℝn with norm(14)Yℋ=v2+δ2+λ−δαu2+1−δΔu21/2,for Y=u,v⊤∈ℋ, where ⊤ stands for the transposition.Letξ=ut+δu, where δ is a small positive constant whose value will be determined later; then, (1) is equivalent to(15)dudt=ξ−δu,dξdt=δα+A−δ−Au−α+A−δξ−λu+M∇u2Δu−fx,u+gx,t+βhxdWdt,ux,τ=u0x,ξx,τ=ξ0x,where ξ0x=u1x+δu0x,x∈ℝn.Forg: we assume that there exists a positive constant σ such that(16)∫−∞τeσsg⋅,s2ds<∞,∀τ∈ℝ,and(17)limr⟶∞∫−∞τ∫|x|≥reσsg⋅,s2dxds=0,∀τ∈ℝ,where ⋅ denotes the absolute value of real number in ℝ.Denoteωt=Wt=Wt,x,t∈ℝ; then, we consider the Ornstein–Uhlenbeck equation dy+ydt=dWt and the Ornstein–Uhlenbeck process(18)yθtω=−∫−∞0esθtωsds,t∈ℝ.From [24], it is known that the random variable yω is tempered, and there is a θt-invariant set Ω˜⊂Ω of full P measure such that yθtω is continuous in t for every ω∈Ω˜. Put(19)zθtω=zx,θtω=hxyθtω,which solves(20)dz+zdt=hdW.Lemma 1 (see [25]).
For anyε>0, there exists a tempered random variable γ:Ω⟶ℝ+, such that for all t∈ℝ,ω∈Ω,(21)zθtω≤eεtγωh,∇zθtω≤eεtγω∇h,Δzθtω≤eεtγωΔh,where γω satisfies(22)e−ε|t|γω≤γθtω≤eεtγω.
Now, letvt,τ,ω=ξt,τ,ω−βzθtω, and we have(23)dudt=v−δu+βzθtω,dvdt=δ−α−Av+δ−δ+α+A−λ−Au+β1−α+A−δzθtω+M∇u2Δu−fx,u+gx,t,ux,τ,τ=u0x,vx,τ,τ=v0x,where v0x=ξ0x−zθτω,x∈ℝn. We will consider (23) for ω∈Ω˜ and write Ω˜ as Ω from now on.
The well-posedness of the deterministic problem (23) in H2ℝn×L2ℝn can be established by standard methods as in [13, 26–28]. If (2)–(7) are fulfilled, let φβt+τ,τ,θ−τω,φ0β=ut+τ,τ,θ−τω,u0,vt+τ,τ,θ−τω,v0⊤, where φ0β=u0,v0⊤. Then, for every ω∈Ω,τ∈ℝ and φ0β∈ℋℝn, problem (23) has a unique ℱ,ℬH2ℝn×ℬL2ℝn-measurable solution φβ⋅,τ,ω,φ0β∈Cτ,∞,ℋℝn with φβτ,τ,ω,φ0β=φ0β,φβt,τ,ω,φ0β∈ℋ0ℝn being continuous in φ0β for each t>τ. Moreover, for every t,τ,ω,φ0∈ℝ+×ℝ×Ω×ℋℝn, the mapping(24)Φβt,τ,ω,φ0β=φβt+τ,τ,θ−τω,φ0β,generates a continuous cocycle from ℝ+×ℝ×Ω×ℋℝn to ℋℝn over ℝ and Ω,F,P,θtt∈ℝ.


## 4. Uniform Estimates of Solutions

In this section, we derive uniform estimates on the solutions of problem (23) and construct a tempered pullback absorbing set.Letδ∈0,1 be small enough such that(25)δ2+λ−δα>0,1−δ>0,and define σ appearing in (17) by(26)σ=minα−δ,δ,c2δ2.Lemma 2.
Assume that (2)–(7) and (16) hold. Then, for every τ∈ℝ,ω∈Ω, and D=Dτ,ω:τ∈ℝ,ω∈Ω∈D, there exists T=Tτ,ω,D>0 such that for all t≥T,(27)φβτ,τ−t,θ−τω,φ0ℋ2+∫τ−tτeσs−τvs,τ−t,θ−τω,v02ds+∫τ−tτeσs−τus,τ−t,θ−τω,u02ds+∫τ−tτeσs−τΔus,τ−t,θ−τω,u02ds+∫τ−tτeσs−τΔvs,τ−t,θ−τω,u02ds<c+cβ2∫−∞0eσs1+Δzθsω2+∇zθsω2+zθsω2+zθsωH2k+1ds,where φ0β=u0,v0⊤∈Dτ−t,θ−tω and c is a positive constant depending on λ,σ,α, and δ but independent of τ,ω, and D.Proof.
Taking the inner product of232 with v in L2ℝn, we find that(28)12ddtv2=−α−δv,v−λ+δ2−δαu,v−1−δAu,v−Av,v+β1−α+δzθtω,v−βAzθtω,v+gx,t,v+M∇u2Δu−fx,u,v.
By231, we get(29)v=dudt+δu−βzθtω.
Next, we estimate some terms of (28).(30)u,v=u,dudt+δu−βzθtω=12ddtu2+δu2−βu,zθtω≥12ddtu2+δu2−βzθtωu≥12ddtu2+3δ4u2−β23δzθtω2,(31)−Au,v=−Δ2u,dudt+δu−βzθtω=−12ddtΔu2+−δΔu2+βΔu,Δzθtω≤−12ddtΔu2+−δΔu2++βΔzθtωΔu≤−12ddtΔu2+−3δ4Δu2++β23δΔzθtω2,(32)fx,u,v=fx,u,dudt+δu−βzθtω=ddt∫ℝnFx,udx+δfx,u,u−βfx,u,zθtω.
By (5), we get(33)fx,u,u≥c2∫ℝnFx,udx+∫ℝnη2xdx.
By conditions (4) and (6), we obtain(34)βfx,u,zθtω≤β∫ℝnc1uk+η1xzθtωdx≤βη1xzθtω+c1β∫ℝnuk+1dxk/k+1zθtωk+1≤βη1xzθtω+c1β∫ℝnFx,u+η3xdxk/k+1zθtωk+1≤12η1x2+β22zθtω2+δc22∫ℝnFx,udx+δc22∫ℝnη3xdx+cβ2zθtωH2k+1.
Using the Cauchy–Schwarz inequality and the Young inequality, there holds(35)β1−α+δzθtω,v≤21−α+δ2β2α−δzθtω2+α−δ8v2,(36)−βAzθtω,v=−βΔzθtω,Δv≤β22Δzθtω2+12Δv2,(37)gx,t,v≤gx,tv≤2α−δgx,t2+α−δ8v2.
By (2) and (3) we have(38)M∇u2Δu,v=M∇u2Δu,ut+δu−βzθtω=−12ddtM^∇u2+δM∇u2∇u2−βM∇u2∇u,∇zθtω≤−12ddtM^∇u2−δM∇u2∇u2+δ4M∇u2∇u2+cβ2M∇u2∇zθtω2≤−12ddtM^∇u2−3δ4M∇u2∇u2+cβ2∇zθtω2≤−12ddtM^∇u2−3δ4M^∇u2+cβ2∇zθtω2.
By (30)–(38), it follows from (28) that(39)12ddtv2+δ2+λ−δαu2+1−δΔu2+M^∇u2+2∫ℝnFx,udx≤−34α−δv2−34δδ2+λ−δαu2−34δ1−δ∇u2−3δ4M^∇u2−δc22∫ℝnFx,udx−12∇v2+cβ21+Δzθtω2+Δzθtω2+Δzθtω2+Δzθtω2H2k+1+cgx,t2.
By (14) and (26), we get from (39) that(40)ddtφℋ2+2∫ℝnFx,udx+σφℋ2+2∫ℝnFx,udx+12α−δv2+12δδ2+λ−δαu2+12δ1−δΔu2+Δv2+δ2M^∇u2≤cβ21+Δzθtω2+∇zθtω2+zθtω2+zθtωH2k+1+cgx,t2.
Multiplying (40) by eσt and integrating over τ−t,τ and then replacing ω by θ−τω, we get(41)φτ,τ−t,θ−τω,φ0ℋ2+2∫ℝnFx,uτ,τ−t,θ−τω,u0dx+12α−δ∫τ−tτeσs−τvs,τ−t,θ−τω,v02ds+12δδ2+λ−δα∫τ−tτeσs−τus,τ−t,θ−τω,u02ds+12δ1−δ∫τ−tτeσs−τΔus,τ−t,θ−τω,u02ds+∫τ−tτeσs−τΔvs,τ−t,θ−τω,v02ds+δ2∫τ−tτeσsM^Δus,τ−t,θ−τω,u02ds≤e−σtφ0ℋ2+2∫ℝnFx,u0dx+c∫τ−tτeσs−τ1+Δzθs−τω2+Δzθs−τω2+Δzθs−τω2+zθs−τωH2k+1ds+∫τ−tτeσs−τgx,s2≤e−σtφ0ℋ2+2∫ℝnFx,u0dx+cβ2∫−t0eσs1+Δzθsω2+zθsω2+∇zθsω2+zθsωH2k+1ds+∫−∞τeσs−τgx,s2.
It follows from Lemma1 that(42)∫−t0eσsΔzθsω2+Δzθsω2+Δzθsω2+ΔzθsωH2k+1ds≤∫−∞0eσsΔzθsω2+Δzθsω2+Δzθsω2+ΔzθsωH2k+1ds≤∫−∞0eσs/2γ2ωΔh2+∇h2+h2+γk+1ωΔhk+1+∇hk+1+hk+1ds<+∞.
From (8),(43)∫ℝnFx,u0dx≤c1+u02+u0H2k+1.
Becauseφ0=u0,v0⊤∈Dτ−t,θ−tω and D∈D, we get from (43) that(44)limt⟶+∞e−σtφ0ℋ2+2∫ℝnFx,u0dx=0.
Therefore, (17), (42), and (43) deduce the desired result (27).Lemma 3.
Assume that (2)–(7) and (16) hold. Then, there exists a random ball Eβτ,ω:τ∈ℝ,ω∈Ω∈D centered at 0 with random radius(45)ϱτ,ω=c+cβ2∫−∞0eσs1+zθsω2+zθsω2+zθsω2+zθsωH2k+1ds,such that Eβτ,ω:τ∈ℝ,ω∈Ω∈D is a closed measurable D-pullback absorbing set for the continuous cocycle associated with problem (23) in D, that is, for every τ∈ℝ,ω∈Ω, and D=Dτ,ω:τ∈ℝ,ω∈Ω∈D, there exists T=Tτ,ω,D>0, such that for all t≥T,(46)Φβt,τ−t,θ−tω,Dτ−t,θ−tω⊆Aτ,ω.Proof.
This is an immediate consequence of (24) and Lemma 2.
Letρ:ℝn⟶ℝ be a smooth function such that 0≤ρx≤1 for all x∈ℝn, and(47)ρx=0for0≤x≤1;andρx=1forx≥2.
For everyr∈ℕ, let(48)ρrx=ρxr,x∈ℝn.
Then, there exist positive constantsc4,c5,c6, and c7 independent of k such that Δρrx≤1/rc4, Δρrx≤1/rc5, Δρrx≤1/rc6, Δ2ρrx≤1/rc7 for all x∈ℝn and r∈ℕ.
Givenr≥1, denoteℍr=x∈ℝn:x<r and ℝn∖ℍrbe the complement ofℍr. To prove asymptotic compactness of solution on ℝn, we prove the following lemma.Lemma 4.
Assume that (2)–(7) and (16) hold. Then, for every τ∈ℝ,ω∈Ω, and D=Dτ,ω:τ∈ℝ,ω∈Ω∈D, there exist T=Tτ,ω,D,ε>0 and R˜=R˜τ,ω,ε≥1, such that for all t≥T,r≥R˜,(49)φβτ,τ−t,θ−τω,φ0ℋℝn∖ℍr2≤ε,where φ0β=u0,v0⊤∈Dτ−t,θ−tω.Proof.
Taking the inner product of3.72 with ρrxv in L2ℝn, we obtain(50)12ddt∫ℝnρrxv2dx=−α−δ∫ℝnρrxv2dx−λ+δ2−δα∫ℝnρrxuvdx−1−δ∫ℝnρrxAuvdx−∫ℝnρrxAvvdx+β1−α+δ∫ℝnρrxzθtωvdx−β∫ℝnρrxAzθtωvdx+∫ℝnρrxgx,tvdx+∫ℝnρrxM∇u2Δuvdx−∫ℝnρrxfx,uvdx.
Using the Young inequality and the Sobolev interpolation inequality,(51)∇v≤ςv+CςΔv,∀ς>0.
We have(52)∫ℝnρrxuvdx=∫ℝnρrxududt+δu−βzθtωdx=∫ℝnρrx12ddtu2+δu2−βzθtωudx≥12ddt∫ℝnρrxu2dx+δ2∫ℝnρrxu2dx−β22δ∫ℝnρrxzθtω2dx,(53)−∫ℝnρrx⋅Au⋅vdx=−∫ℝnΔ2u⋅ρrx⋅dudt+δu−ϵzθtωdx=−∫ℝnΔu⋅Δρrxdudt+δu−ϵzθtωdx=−∫ℝnΔu⋅Δρrx⋅v+2∇ρrx⋅∇v+ρrx⋅Δdudt+δu−ϵzθtωdx≤c5r∫ℝnΔu⋅vdx+2c4r∫ℝnΔu⋅∇vdx−12ddt∫ℝnρrxΔu2dx−δ∫ℝnρrxΔu2dx+ϵ∫ℝnρrxΔuΔzθtωdx≤c52rΔu2d+v2d+2c4rΔuςv+CςΔv−12ddt∫ℝnρrxΔu2dx−δ∫ℝnρrxΔu2dx+δ2∫ℝnρrxΔu2dx+ϵ22δ∫ℝnρrxΔzθtω2dx≤c5rΔu2+v2+c4rΔu2+2ς2v2+2Cς2Δv2−12ddt∫ℝnρrxΔu2dx−δ2∫ℝnρrxΔu2dx+ϵ22δ∫ℝnρrxΔzθtω2dx,(54)∫ℝnρrxfx,uvdx=∫ℝnρrxfx,ududt+δu−βzθtωdx=ddt∫ℝnρrxFx,udx+δ∫ℝnρrxfx,uudx−β∫ℝnρrxfx,uzθtωdx.
By (6), we get(55)δ∫ℝnρrxfx,uudx≥c2δ∫ℝnρrxFx,udx+δ∫ℝnρrxη2xdx.
By (4) and (6), we get(56)β∫ℝnρrxfx,uzθtωdx≤β∫ℝnρrxc1uk+η1xzθtωdx≤12∫ℝnρrxη1x2dx+β22∫ℝnρrxzθtω2dx+cβ2∫ℝnρrxzθtωk+1dx+c2δ2∫ℝnρrxFx,u+η3xdx.
For the remainder terms, using the Cauchy–Schwarz inequality and the Young inequality, we have(57)−∫ℝnρrx⋅Av⋅vdx=−∫ℝnΔ2v⋅ρrx⋅vdx=−∫ℝnΔv⋅Δρrx⋅vdx=−∫ℝnΔv⋅Δρrx⋅v+2∇ρrx⋅∇v+ρrx⋅Δvdx≤c52rΔv2+v2+2c4rΔvΔv−∫ℝnρkxΔv2dx≤c52rΔv2+v2+2c4rΔvςv+CςΔv−∫ℝnρrxΔv2dx≤c52rΔv2+v2+c4rΔv2+2ς2v2+2Cς2Δv2−∫ℝnρrxΔv2dx,(58)−β∫ℝnρrxAzθtω⋅vdx=−β∫ℝnΔ2zθtω⋅ρrx⋅vdx=−β∫ℝnΔzθtω⋅Δρrx⋅vdx=−β∫ℝnΔzθtω⋅Δρrx⋅v+2∇ρrx⋅∇v+ρrx⋅Δvdx≤c5βr∫ℝnΔzθtω⋅vdx+2c4βr∫ℝnΔzθtω⋅∇vdx+β∫ℝnρrxΔzθtω⋅Δvdx≤c5β2rΔzθtω2+v2+2c4βrΔzθtωςv+CςΔv+β∫ℝnρrxΔzθtω⋅Δvdx≤c5β2rΔzθtω+v2+c4βrΔzθtω2+2ς2v2+2Cς2Δv2+∫ℝnρrxΔv2dx+β24∫ℝnρrxΔzθtω2dx,(59)β1−α+δ∫ℝnρrxzθtωvdx≤1−α+δ2β2α−δ∫ℝnρrxzθtω2dx+α−δ4∫ℝnρrxv2dx,(60)∫ℝnρrxgx,tvdx≤1α−δ∫ℝnρrxgx,t2dx+α−δ4∫ℝnρrxv2dx.(61)∫ℝnρrxM∇u2Δuvdx=∫ℝnρrxM∇u2Δuut+δu−βzθtωdx=−12ddt∫ℝnρrxM^∇u2dx−δ∫ℝnρrxM∇u2∇u2dx+β∫ℝnρrxM∇u2∇u∇zθtωdx≤−12ddt∫ℝnρrxM^∇u2dx−δ∫ℝnρrxM∇u2∇u2dx+δ2∫ℝnρrxM∇u2∇u2dx+cβ2∫ℝnρrxM∇u2∇zθtω2dx≤−12ddt∫ℝnρrxM∇u2dx−δ2∫ℝnρrxM∇u2∇u2dx+cβ2∫ℝnρrx∇zθtω2dx≤−12ddt∫ℝnρrxM^∇u2dx−δ2∫ℝnρrxM^∇u2dx+cβ2∫ℝnρrx∇zθtω2dx.
It follows from (52)–(61) that(62)12ddt∫ℝnρrxv2+δ2+λ−δαu2+1−δΔu2+M^∇u2+2Fx,udx≤crΔv2+v2+Δu2+Δzθtω2−α−δ2∫ℝnρrxv2dx−δδ2+λ−δα2×∫ℝnρrxu2dx−δ1−δ2∫ℝnρrxΔu2dx−c2δ2∫ℝnρrxFx,udx−δ2∫ℝnρrxM^∇u2dx+cβ2∫ℝnρrx1+Δzθtω2+∇zθtω2+∇zθtω2+∇zθtωk+1+gx,t2dx.
Denote(63)X=v2+δ2+λ−δαu2+1−δΔu2+M^∇u2.
By (26), we get(64)12ddt∫ℝnρrxX+2Fx,udx+σ∫ℝnρrxX+2Fx,udx≤crΔv2+v2+Δu2+Δzθtω2+cβ2∫ℝnρrx1+Δzθtω2+∇zθtω2+∇zθtω2+∇zθtωk+1+gx,t2dx.
Multiplying (64) by eσt and integrating over τ−t,τ and then replacing ω by θ−τω, we get(65)∫ℝnρrxXτ,τ−t,θ−τω,X0+2Fx,uτ,τ−t,θ−τω,u0dx≤e−σt∫ℝnρrxX0+2Fx,u0dx+cr∫τ−tτeσs−τΔvs,τ−t,θ−τω,v02+Δvs,τ−t,θ−τω,v02+Δus,τ−t,θ−τω,u02+Δzθs−τω2ds+cβ2∫τ−tτeσs−τ∫ℝnρrx1+Δzθs−τω2+∇zθs−τω2+∇zθs−τω2+∇zθs−τωk+1+gx,s2dxds≤e−σt∫ℝnρrxX0+2Fx,u0dx+cr∫τ−tτeσs−τΔvs,τ−t,θ−τω,v02+vs,τ−t,θ−τω,v02+Δus,τ−t,θ−τω,u02ds+cr∫−t0eσs‖Δzθsω‖2ds+c∫τ−tτeσs∫|x|≥rgx,s2dxds+cβ2∫−t0eσs∫|x|≥r1+Δzθsω2+∇zθsω2+zθsω2+zθsωk+1dxds≤e−σt∫ℝnρrxX0+2Fx,u0dx+cr∫τ−tτeσs−τΔvs,τ−t,θ−τω,v02+vs,τ−t,θ−τω,v02+Δus,τ−t,θ−τω,u02ds+cr∫−∞0eσs∇zθsω2ds+c∫−∞τeσs∫|x|≥rgx,s2dxds+cβ2∫−∞0eσs∫|x|≥r1+Δzθsω2+∇zθsω2+zθsω2+zθsωk+1dxds.
Due toφ0β∈Dτ−t,θ−tω∈D and (43), it is easy to obtain that there exists T1˜=T1˜τ,ε,ω,D>0, such that for all t>T1˜,(66)e−σt∫ℝnρrxX0+2Fx,u0dx≤ε.
By Lemma2, there are T2˜=T2˜τ,ε,ω,D>0 and R1˜=R1˜ε,ω,D>1, such that for all t>T2˜ and r>R1˜,(67)cr2∫τ−tτeσs−τΔvs,τ−t,θ−τω,v02+vs,τ−t,θ−τω,v02+Δus,τ−t,θ−τω,u02ds≤ε.
By Lemma1, there are T3˜=T3˜ε,ω>0 and R2˜=R2˜ε,ω>1, such that for all t>T3˜ and r>R2˜,(68)cβ2∫−∞0eσs∫x≥r1+Δzθsω2+∇zθsω2+zθsω2+zθsωk+1dxds+cr2∫−∞0eσsΔzθsω2ds≤ε.
By equation (17), there is R3˜=R3˜τ,ε>1, such that for all r>R3˜,(69)c∫−∞τeσs∫|x|≥rgx,s2dxds≤ε.
DenoteT˜=maxT1˜,T2˜,T3˜, R˜=maxR1˜,R2˜,R3˜; by (65)–(69), for all t>T˜ and r>R˜, we have(70)∫ℝnρrxXτ,τ−t,θ−τω,X0+2Fx,uτ,τ−t,θ−τω,u0dx≤4ε,which implies(71)φβτ,τ−t,θ−τω,φ0βℋℝn∖ℍr2≤4ε.
Letρ^=1−ρ with ρ given by (48). Fix r≥1 and set(72)u^t,τ,ω,u0^=ρr^xut,τ,ω,u0,v^t,τ,ω,v0^=ρr^xvt,τ,ω,v0,and then φ^βt,τ,ω,φ0^β=u^t,τ,ω,u0^,v^t,τ,ω,v0^⊤ is the solution of problem equation (23) on the bounded domain ℍ2r, where φ0^β=ρr^xφ0β∈ℋℍ2r.
Multiplying (23) by ρr^x and using (72), we find that(73)du^dt=v^−δu^+βρr^xzθtω,dv^dt=−α−δv^−λ+δ2−δαu^−1−δAu^−Av^+β1−α+δρr^xzθtω−βρr^xAzθtω+ρr^xgx,t+M∇u2Δu^−M∇u2Δρr^xu−2M∇u2∇ρr^x∇u−ρr^xfx,u+41−δΔ∇ρr^x∇u+61−δΔρr^xΔu+41−δ∇ρr^xΔ∇u+1−δuAρr^x+4Δ∇ρr^x∇v+6Δρr^xΔv+4∇ρr^xΔ∇v+vAρr^x.
Consider the eigenvalue problem.(74)Au^=λu^inℍ2r,withu^=∂u^∂n=0on∂ℍ2r.
Problem (74) has a family of eigenfunctions eii∈ℕ with the eigenvalues λii∈ℕ:(75)λ1≤λ2≤⋯≤λi≤…,λi⟶+∞i⟶+∞,such that eii∈ℕ is an orthonormal basis of L2ℍ2r. Given n, let Xn=spane1,…,en and Pn:L2ℍ2r⟶Xn be the projection operator.Lemma 5.
Assume that (2)–(7) and (16) hold. Then, for every τ∈ℝ,ω∈Ω, and D=Dτ,ω:τ∈ℝ,ω∈Ω∈D, there exist T^=T^τ,ω,D,ε>0 and R^=R^τ,ω,ε≥1 and N=Nτ,ω,ε>0, such that for all t≥T^,r≥R^, and n≥N,(76)I−Pnφ^βτ,τ−t,θ−τω,φ^0βℋℍ2r2≤ε,where φ^0β=ρr^xφ0β,φ0β=u0,v0⊤∈Dτ−t,θ−tω.Proof.
Letu^n,1=Pnu^,u^n,2=I−Pnu^,v^n,1=Pnv^,v^n,2=I−Pnv^. Applying I−Pn to 731, we obtain(77)v^n,2=du^n,2dt+δu^n,2−βI−Pnρr^xzθtω.
Then, applyingI−Pn to 732 and taking the inner product of the resulting equation with v^n,2 in L2ℍ2r, we have(78)12ddtv^n,22=−α−δv^n,22−λ+δ2−δαu^n,2,v^n,2−1−δAu^n,2,v^n,2−Av^n,2,v^n,2+β1−α+δρr^xzθtω,v^n,2−βρr^xAzθtω,v^n,2+ρr^xgx,t,v^n,2+M‖∇u‖2Δu^n,2,v^n,2−M∇u2Δρr^xu+2M∇u2∇ρr^x∇u,v^n,2−ρr^xfx,u,v^n,2+41−δΔ∇ρr^x∇u+61−δΔρr^xΔu+41−δ∇ρr^xΔ∇u+1−δuAρr^x,v^n,2+4Δ∇ρr^x∇v+6Δρr^xΔv+4∇ρr^xΔ∇v+vAρr^x,v^n,2.
Next, we estimate some terms of (78).(79)u^n,2,v^n,2=u^n,2,du^n,2dt+δu^n,2−βI−Pnρr^xzθtω≥12ddtu^n,22+δu^n,22−βu^n,2⋅I−Pnρr^xzθtω≥12ddtu^n,22+δ2u^n,22−cβ2I−Pnρr^xzθtω2,(80)−Au^n,2,v^n,2=−Δu^n,2,Δdu^n,2dt+δu^n,2−βI−Pnρr^xzθtω≤−12ddtΔu^n,22−δΔu^n,22+βΔu^n,2⋅I−PnΔρr^xzθtω≤−12ddtΔu^n,22−δ2+7α−δΔu^n,22+cβ2I−PnΔρr^xzθtω2.
Denoteθ=nk−1/4k+1; by (4), we get(81)ρr^xfx,u,v^n,2≤c1∫ℝnρr^xukv^n,2dx+∫ℝnρr^xη1xv^n,2dx≤c1uk+1kv^n,2k+1+η1v^n,2≤c1uk+1kΔv^n,2θv^n,21−θ+λn+1−1/2η1Δv^n,2≤c1λn+1θ−1/2uH2kΔv^n,2+λn+1−1/2η1Δv^n,2≤λn+1−1/2Δv^n,2c1λn+1θ2uH2k+η1≤16Δv^n,22+32λn+1−1c1λn+1θ/2uH2k+η12.
For the remainder terms on the right-hand side of (74), we have(82)−βρr^xAzθtω,v^n,2≤βI−Pnρr^xΔzθtω⋅Δv^n,2≤3β22I−Pnρr^xΔzθtω2+16Δv^n,22,(83)β1−α+δρr^xzθtω,v^n,2≤β1−α+δI−Pnρr^xzθtω⋅v^n,2≤cβ2I−Pnρr^xzθtω2+α−δ28v^n,22,(84)ρr^xgx,t,v^n,2≤I−Pnρr^xgx,t⋅v^n,2≤72α−δI−Pnρr^xgx,t2+α−δ14v^n,22,(85)1−δ4Δ∇ρr^x⋅∇u+6Δρr^x⋅Δu+4∇ρr^x⋅Δ∇u+uAρr^x,v^n,2≤4c61−δrλn+1−1/4Δu⋅v^n,2+6c51−δrΔu⋅v^n,2+4c41−δrλn+1−1/4Δu⋅Δv^n,2+c71−δru⋅v^n,2≤cλn+1−−1/2Δu2+13Δv^n,22+cru2+Δu2+5α−δ28v^n,22,(86)4Δ∇ρr^x⋅∇v+6Δρr^x⋅Δv+4∇ρr^x⋅Δ∇v+vAρr^x,v^n,2≤4c6rλn+1−1/4Δv⋅v^n,2+6c5rΔv⋅v^n,2+4c4rλn+1−1/4Δv⋅v^n,2+c7rΔv⋅v^n,2≤cλn+1−1/2Δv2+13Δv^n,22+crv2+Δv2+5α−δ28v^n,22,(87)M∇u2Δu^n,2,v^n,2≤7α−δΔu^n,22+α−δ56v^n,22,(88)M∇u2Δρr^xu+2M∇u2∇ρr^x∇u,v^n,2≤crΔu2+α−δ56v^n,22.
By (30)–(39), we get(89)12ddtv^n,22+δ2+λ−δαu^n,22+1−δΔu^n,22≤−α−δ2v^n,22−δ2δ2+λ−δαu^n,22−δ21−δΔu^n,22+cβ2I−Pnρr^xzθtω2+I−PnΔρr^xzθtω2+I−Pnρ^x2r2Δzθtω2+cI−Pnρr^xgx,t2+cλn+1−1/2Δu2+Δv2+cru2+v2+Δu2+Δv2+32λn+1−1c1λn+1θ/2uH2k+η12.
By (13), (25), and (89), we get(90)ddtφ^n,2βℋℍ2r2≤−σφ^n,2βℋℍ2r2+cβ2I−Pnρr^xzθtω2+I−PnΔρr^xzθtω2+I−Pnρr^xΔzθtω2+cI−Pnρr^xgx,t2+cλn+1−1/2Δu2+Δv2+cru2+v2+Δu2+Δv2+32λn+1−1c1λn+1θ/2uH2k+η12.
Asη1∈L2ℝn, λn⟶∞, there exist N^1=N^1ε>0 and R^1=R^1ε>0 such that for all n>N^1 and r>R^1,(91)ddtφ^n,2βℋℍ2r2≤−σφ^n,2βℋℍ2r2+cI−Pnρr^xgx,t2+cr+εu2+v2+Δu2+Δv2+ε1+uH22k+yθtω2.
Multiplying (91) by eσt and integrating over τ−t,τ and then substituting ω by θ−τω, for all n>N^1 and r>R^1, we have(92)φ^n,2βτ,τ−t,θ−τω,φ^n,2,0βℋℍ2r2≤e−σtφ^n,2,0βℋℍ2r2+c∫−∞τeσs−τI−Pnρr^xgx,s2ds+cr+ε∫τ−tτeσs−τus,τ−t,θ−τω,u02+vs,τ−t,θ−τω,v02+Δus,τ−t,θ−τω,u02+Δvs,τ−t,θ−τω,v02ds+ε∫τ−tτeσs−τ1+us,τ−t,θ−τω,u0H22k+yθs−τω2ds.
By (4), φ0β∈Dτ−t,θ−tω, and Dτ−t,θ−tω∈D, there exist T^1=T^1τ,ε,D,ω>0 and R^1=R^1τ,ε,ω>1, such that if t>T^1 and r>R^1, then(93)e−σtφ^n,2,0βℋℍ2r2≤ε.
For the second term on the right-hand side of (92), by (16), we know that there is N^=N^τ,ε,ω>0, such that for all n>N^,(94)c∫−∞τeσs−τI−Pnρr^xgx,s2ds≤ε.
For the third and fourth terms on the right-hand side of (93), by Lemma 2, there exist T^2=T^2τ,ε,D,ω>0 and R^2τ,ε,ω>1, such that for all t>T^2 and r>R^2, there holds(95)cr+ε∫τ−tτeσs−τus,τ−t,θ−τω,u02+vs,τ−t,θ−τω,v02+Δus,τ−t,θ−τω,u02+Δvs,τ−t,θ−τω,v02ds≤ε.
For the last term on the right-hand side of (93), by Lemma 2, there is T^3=T^3τ,ε,D,ω>0, such that for all t>T^3, it follows that(96)∫τ−tτeσs−τ1+us,τ−t,θ−τω,u0H22k+yθs−τω2ds<∞.
DenoteT^=maxT^1,T^2,T^3 and R^=maxR^1,R^2. Then, by (92)–(96), for all t>T^, r>R^, and n>N^, we get(97)φ^n,2βτ,τ−t,θ−τω,φ^n,2,0βℋℍ2r2≤cε,which completes the proof.


## 5. Random Attractors

In this section, we prove the existence ofD-pullback attractors for stochastic problem (23).Lemma 6.
Assume that (2)–(7) and (16) hold. Then, the solution of problem (23) is asymptotic compactness in ℋℝn; that is, for every τ∈ℝ,ω∈Ω, and B=Bτ,ω:τ∈ℝ,ω∈Ω∈D, the sequence φβτ,τ−tm,θ−τω,φ0,mβ has a convergent subsequence in ℋℝn provided tm⟶∞ and φ0,mβ∈Bτ−tm,θ−tmω.Proof.
We first lettm⟶∞, B∈D, and φ0,mβ∈Bτ−tm,θ−tmω. By Lemma 2, φβτ,τ−tm,θ−τω,φ0,mβ is bounded in ℋℝn; that is, for every τ∈ℝ,ω∈Ω, there exists M1=M1τ,ω,B>0 such for all m>M1,(98)φβτ,τ−tm,θ−τω,φ0,mβℋℝn2≤ϱ2τ,ω.
In addition, it follows from Lemma4 that there exist r1=r1τ,ε,ω>0 and M^2=M^2τ,B,ε,ω>0, such that for every m≥M^2,(99)φβτ,τ−tm,θ−τω,φ0,mβℋℝn∖ℍr12≤ε.
Next, by using Lemma5, there are N=Nτ,ε,ω>0, r2=r2τ,ε,ω≥r1, and M^3=M^3τ,B,ε,ω>0, such that for every m≥M^3,(100)I−PNφ^βτ,τ−tm,θ−τω,φ^0,mβℋℍ2r22≤ε.
Using (73) and (98), we find that PNφ^βτ,τ−tm,θ−τω,φ^0,mβ is bounded in the finite-dimensional space PNℋℍ2r2, which together with (100) implies that φ^βτ,τ−tm,θ−τω,φ^0,mβ is precompact in H2ℍ2r2×L2ℍ2r2.
Note thatρr^x=1 for x≤r2. Recalling (73), we find that φβτ,τ−tm,θ−τω,φ0,mβ is precompact in ℋℍr2, which along with (99) shows the precompactness of this sequence in ℋℝn. This completes the proof.
The main result of this section is given below.Theorem 2.
Assume that (2)–(7) and (16) hold. Then, the continuous cocycle Φβ associated with problem (23) has a unique D-pullback attractor Aβ=Aβτ,ω:τ∈ℝ,ω∈Ω in ℋℝn.Proof.
This is an immediate consequence of Proposition1 and Lemmas 2 and 6.


## 6. Upper Semicontinuity of Pullback Attractors

In this section, we will use Theorem1 to consider an upper semicontinuity of random attractors Aβω when β⟶0. To indicate the dependence of solutions on β, we, respectively, write the solutions of problem (23) as uβ and vβ, that is, uβ,vβ satisfies(101)duβdt=vβ−δuβ+βzθtω,dvβdt=δ−α−Avβ+δ−δ+α+A−λ−Auβ+β1−α+A−δzθtω+M∇uβ2Δuβ−fx,uβ+gx,t,uβτ,τ,x=u0βx,vβτ,τ,x=v0βx.Whenβ=0, the random problem (23) reduces to a deterministic one:(102)du0dt=v0−δu0,dv0dt=δ−α−Av0+δ−δ+α+A−λ−Au0+M∇u02Δu0−fx,u0+gx,t,u0τ,τ,x=u00x,v0τ,τ,x=v00x.By Theorem2, the deterministic and autonomous system Φ0 generated by (102) is readily verified to admit a global attractor A0 in ℋℝn.Theorem 3.
Assume that (2)–(7) and (16) hold. Then, the random dynamical system Φβ generated by (23) has a unique D-pullback attractor Aβτ,ωτ∈ℝ,ω∈Ω in ℋℝn. Moreover, the family Aββ>0 of random attractors is upper semicontinuous.Proof.
By Lemma3 and Theorem 2, Φβ has a closed measurable random absorbing set Eβτ,ω and a unique random attractor Aβ.(i)
In Lemma2, we have proved that the system Φβ possesses a closed random absorbing set Eβ=Eβτ,ωτ∈ℝ,ω∈Ω in D, which is given by(103)Eβτ,ω=u,v∈H2ℝn×L2ℝn:uH2ℝn2+vL2ℝn2≤R1τ,ω.Then, we get(104)limsupβ⟶0Eβτ,ωℋℝn≤c,which deduces condition (i) in Theorem1 immediately.(ii)
Givenβ∈0,1, let E1τ,ω=u,v∈H2ℝn×L2ℝn:uH2ℝn2+vL2ℝn2≤R∗τ,ω, where(105)R∗τ,ω=c+c∫−∞0eσs1+Δzθsω2+∇zθsω2+zθsω2+zθsωH2k+1ds.Then,(106)∪0<β≤1Aβτ,ω⊆∪0<β≤1Eβω⊆E1τ,ω.First, by (106), Lemma 4, and the invariance of Aβτ,ω, we obtain that for every β>0 and P−a.e.τ∈ℝ,ω∈Ω, there exists r0=r0ω,ε≥1 such that(107)∫|x|≥r0ux2+Δux2+vx2dx≤ε,for allu,v∈∪0<β≤1Aβτ,ω.Second, by (106), the proof of Lemma 5, Lemma 6, and the invariance of Aβτ,ω, we know that there exists r1=r1ω,ε≥r0 such that for all r≥r1, the set ∪0<β≤1Aβτ,ω is precompact in ℋℍr, which together with (107) implies that ∪0<β≤1Aβτ,ω is precompact in ℋℝn.(iii)
Letφ0=u0,v0 be a mild solution of (102) with initial data φ0=u0,v0, and U=uβ−u0, V=vβ−v0. By (101) and (102), we get(108)dUdt=V−δU+βzθtω,dVdt=δ−α−AV+δ−δ+α+A−λ−AU+M∇uβ2−M∇u02ΔU−fx,uβ+fx,u0+β1−α+A−δzθtω,Uτ,τ,x=U0x,Vτ,τ,x=V0x.
Taking the inner product of1082 with V in L2ℝn, we get(109)12ddtV2+δ2+λ−δαU2+1−δΔU2≤−34α−δV2−34δδ2+λ−δαU2−34δ1−δΔU2+M∇uβ2−M∇u02ΔU,V+−fx,uβ+fx,u0,V+cβ21+Δzθtω2+zθtω2.
Taking advantage ofMa2−Mb2≤M′supa2,b2a+b⋅a−b and Lemma 2, we get(110)M∇uβ2−Mu02ΔU,V≤M′sup∇uβ2,∇u02∇uβ+∇u0⋅∇uβ−∇u0|ΔU,V≤cΔU2+cV2.
By (7), we get(111)fx,u0−fx,uβ,V≤cU2+cV2,which along with (107)-(108) implies(112)ddtV2+δ2+λ−δαU2+1−δΔU2≤cV2+δ2+λ−δαU2+1−δΔU2+cβ21+Δzθtω2+zθtω2.
Applying Gronwall inequality to (112) over τ,t, we have(113)uβt,τ,ω,u0β−u0t,τ,u00H2ℝn2+vβt,τ,ω,v0β−v0t,τ,v00L2ℝn2≤cect−τu0β−u00H2ℝn2+v0β−v00L2ℝn2+cβ2∫τtect−s1+Δzθsω2+zθsω2ds,which along with (i), (ii), and Theorem 1 completes the proof [29].



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