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Open Mathematics

formerly Central European Journal of Mathematics

Editor-in-Chief: Gianazza, Ugo / Vespri, Vincenzo


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Volume 16, Issue 1

Issues

Volume 13 (2015)

Random attractors for stochastic retarded reaction-diffusion equations with multiplicative white noise on unbounded domains

Xiaoyao Jia
  • Corresponding author
  • Mathematics and Statistics School, Henan University of Science and Technology, No.263 Kai-Yuan Road, Luo-Long District, Luoyang, Henan Province, 471023, China
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/ Xiaoquan Ding
  • Mathematics and Statistics School, Henan University of Science and Technology, No.263 Kai-Yuan Road, Luo-Long District, Luoyang, Henan Province, 471023, China
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/ Juanjuan Gao
  • Mathematics and Statistics School, Henan University of Science and Technology, No.263 Kai-Yuan Road, Luo-Long District, Luoyang, Henan Province, 471023, China
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Published Online: 2018-08-03 | DOI: https://doi.org/10.1515/math-2018-0076

Abstract

In this paper we investigate the stochastic retarded reaction-diffusion equations with multiplicative white noise on unbounded domain ℝn (n ≥ 2). We first transform the retarded reaction-diffusion equations into the deterministic reaction-diffusion equations with random parameter by Ornstein-Uhlenbeck process. Next, we show the original equations generate the random dynamical systems, and prove the existence of random attractors by conjugation relation between two random dynamical systems. In this process, we use the cut-off technique to obtain the pullback asymptotic compactness.

Keywords: Stochastic reaction-diffusion system; Random attractor; Time-delay; Multiplicative white noise

MSC 2010: 35B40; 35B41; 35K45; 35K57

1 Introduction

In this paper we investigate a class of stochastic partial differential equations, which are widely used in quantum field theory, statistical mechanics and financial mathematics. It is known that there are many dynamical systems, depending on both current and historical states. They are referred to as time-decay dynamical systems and can be described by retarded partial differential equations. Hence, it is very important to study the properties of retarded partial differential equations.

In this paper, we consider the asymptotic behavior of the solutions to the following stochastic retarded reaction-diffusion equation with multiplicative noise in the whole space ℝn (n ≥ 2):

du+(λuΔu)dt=(F(x,u(t,x))+G(x,u(th,x))+g(x))dt+ϵudw.(1)

Here λ is a positive constant; h > 0 is the delay time of the system; F and G are given functions satisfying certain conditions which will be given in Section 2; g is a given function defined on ℝn; w is a two-sided real-valued Wiener process on a probability space (Ω, 𝓕, ℙ), with

Ω={ωC(R,R):ω(0)=0},

the Borel σ-algebra 𝓕 on Ω is generated by the compact open topology [2] and ℙ is the corresponding Wiener measure on 𝓕; ϵ is a positive parameter; ∘ denotes the Stratonovich sense in the stochastic term. We identify ω(t) with wt(ω), i.e. wt(ω) = w(t, ω) = ω (t), t ∈ ℝ.

One of the most important problems for stochastic differential equations is to study the asymptotic behavior of the solutions. The asymptotic behavior of random dynamical systems can be described by its random attractors. The existence of attractors for partial differential equations has been studied by many authors, see [3, 12, 16, 17, 20, 21, 22, 23, 24, 25, 26, 27, 29, 31, 32, 34] and the references therein. However, to the best of our knowledge, the studies of attractors for retarded partial differential equations are very few, especially for the stochastic retarded partial differential equations. The attractors for the deterministic retarded partial differential equations were investigated in [6, 7, 9, 33, 35], and random attractors for stochastic retarded partial differential equations were studied in [13, 14, 30, 36]. In [13], the authors studied the existence of random attractor for stochastic retarded reaction-diffusion equations with additive noise, while in [14, 30, 36], the authors studied the random attractors for stochastic retarded lattice dynamical systems. In this paper, we will study the existence of the random attractors for stochastic retarded reaction-diffusion equations with multiplicative white noise on unbounded domain ℝn. In the bounded domain case, the compactness of random absorbing set can be obtained by Sobolev embeddings theory. However, in the unbounded domain case, we can not get compactness in this way, because Sobolev embeddings are not compact. Hence, the cut-off technique will be used to obtain the compactness. This technique has been used by many authors (see [13, 21, 26, 27]).

The remainder of the paper is organized as follows. In section 2, we recall an important theorem which will be used to obtain the existence of random attractor, and transform the stochastic retarded partial differential equations to random ones by Ornstein-Uhlenbeck process. We will show that for each ωΩ, the random equation has a unique solution. In Section 3, we get some useful estimates for the solution of the random equation. At last, we get the existence of the random attractor.

2 Preliminaries and Random Dynamical Systems

In this section we introduce some notations and recall some basic knowledge about random attractors for random dynamical systems (RDS). The reader can refer to [2, 3, 10, 11, 17] for more details.

First, we introduce some notations which will be used in the following. Let ∥⋅∥and (⋅, ⋅) be the norm and inner product in L2(ℝn). For fixed h > 0, let 𝔖 = C ([−h, 0], L2(ℝn)), and let the norm of 𝔖 be ∥f𝔖 = supt∈[−h,0]f(t)∥. Then, it is easy to see that 𝔖 is a Banach space. Let (X, ∥⋅∥X) be a Banach space with Borel σ−algebra 𝓑(X), and let (Ω, 𝓕,ℙ) be a probability space. For ab, t ∈ [a, b]and continuous function uC([ah, b], L2(ℝd)), define ut(s) = u(t + s) for s ∈ [−h, 0]. From the definition, one can see that, for t ∈ [a, b], ut ∈ 𝔖.

Let (X, ∥⋅∥X) be a Banach space with Borel σ−algebra 𝓑(X), and let (Ω, 𝓕,ℙ) be a probability space.

Definition 2.1

(Ω, 𝓕, ℙ, (θt)t∈ℝ) is called a metric dynamical system, if θ:ℝ × ΩΩ is (𝓑(ℝ) × 𝓕, 𝓕)− measurable, θ0 is the identity on Ω, θs+t = θtθs for all s, t ∈ ℝ and θt ℙ = ℙ for all t ∈ ℝ.

Definition 2.2

A stochastic process ϕ is called a continuous random dynamical system (RDS) over (Ω, 𝓕, ℙ, (θt)t∈ℝ), if ϕ is (𝓑([0, +∞)) × 𝓕 × 𝓑(X), 𝓑(X))−neasurable, and for all ωΩ,

  1. the mapping ϕ(t, ω, ⋅) : XX, xϕ(t, ω, x) is continuous for every t ≥ 0;

  2. ϕ(0, ω, ⋅) is the identity on X;

  3. ϕ(s + t, ω, ⋅) = ϕ(t, θs ω, ⋅) ∘ ϕ(s, ω, ⋅) for all s, t > 0.

Definition 2.3

  1. A random set {B(ω)}ωΩ of X is called bounded if there exists x0X and a random variable r(ω) > 0, such that

    B(ω){xX:||xx0||Xr(ω),x0X},forallωΩ.

  2. A random set {B(ω)}ωΩ is called a compact random set if B(ω) is compact for all ωΩ.

  3. A random set {B(ω)}ωΩX is called tempered with respect to (θt)t∈ℝ, if for ℙ − a.e.ωΩ,

    limt+eytsupxB(θtω)||x||X=0forally>0.(2)

  4. A random variable r(ω) is said to be tempered with respect to (θt)t∈ℝ, if for ℙ − a.e.ωΩ,

    limt+eyt|r(θtω)|=0forally>0.(3)

In the following, we assume that ϕ is a continuous RDS on X over (Ω, 𝓕, ℙ, (θt)t∈ℝ), and 𝒟 is a collection of random subsets of X.

Definition 2.4

Let {K(ω)}ωΩ ∈ 𝒟. Then {K(ω)}ωΩ is called a random absorbing set for ϕ in 𝒟 if for every {B(ω)} ∈ 𝒟 and ℙ − a.e. ωΩ, there exists T(B, ω) > 0, such that

ϕ(t,θtω,B(θtω))K(ω)foralltT(B,ω).(4)

Definition 2.5

A random set {A(ω)}ωΩ of X is called a 𝒟-random attractor (or 𝒟-pullback attractor) for ϕ, if the following conditions are satisfied:

  1. {A(ω)}ωΩ is compact, and ωd(x, A(ω)) is measurable for every xX;

  2. {A(ω)}ωΩ is invariant, that is,

    ϕ(t,ω,A(ω))=A(θtω)t0.(5)

  3. {A(ω)}ωΩ attracts every set in 𝒟, that is, for every {B(ω)}ωΩ ∈ 𝒟, ωΩ,

    limt+dist(ϕ(t,θtω,B(θtω)),A(ω))=0,(6)

    where dist(Y,Z)=supyYinfzZ||yz||X is the Hausdorff semi-metrix (YX, ZX).

Definition 2.6

ϕ is said to be 𝒟-pullback asymptotically compact in X, if for all B ∈ 𝒟 and ℙ − a.e. ωΩ, {ϕ(tn,θtnω,xn)}i=1 has a convergent subsequence in X, as tn → ∞, and xnB(θtn ω).

At the end of this section, we refer to [3, 17] for the existence of random attractor for continuous RDS.

Proposition 2.7

Let {K(ω)}ωΩ ∈ 𝒟 be a random absorbing set for the continuous RDS ϕ in 𝒟 and ϕ is 𝒟-pullback asymptotically compact in X. Then ϕ has a unique 𝒟-random attractor {A(ω)}ωΩ which is given by

A(ω)=τ0tτϕ(t,θtω,K(θtω))¯.(7)

In the rest of this paper we will assume that 𝒟 is the collection of all tempered random subsets of 𝔖 and we will prove that the stochastic retarded reaction-diffusion equation has a 𝒟-pullback random attractor.

In the remaining part of this section we show that there is a continuous random dynamical system generated by the following stochastic retarded reaction-diffusion equation on ℝn with the multiplicative noise:

du+(λuΔu)dt=(F(x,u(t,x))+G(x,u(th,x))+g(x))dt+ϵudw,xRn,t>0,(8)

with the initial condition

u(t,x)=u0(t,x),xRn,t[h,0].(9)

Here g is a given function in L2 (ℝn), and F, G are continuous functions satisfying the following conditions:

  • (A1)

    F:ℝn × ℝ → ℝ is a continuous function such that for all x ∈ ℝn and s ∈ ℝ,

    F(x,s)sα1|s|p+β1(x);(10)

    |F(x,s)|α2|s|p1+β2(x);(11)

    sF(x,s)α3;(12)

    |xF(x,s)|β3(x).(13)

    Here α1, α2, α3 are positive constants, p > 2, β1(x), β2(x), β3(x) are nonnegative functions on ℝn, such that β1(x) ∈ L1(ℝn), and β2(x), β3(x) ∈ L2(ℝn).

  • (A2)

    G : ℝn × ℝ → ℝ is a continuous function such that for all x ∈ ℝn and s1, s2 ∈ ℝ,

    |G(x,s1)G(x,s2)|α4|s1s2|;(14)

    |G(x,s)|β4(x)|s|+β5(x).(15)

    Here α4 is a positive constant, β4(x), β5(x) are nonnegative functions on ℝn such that β4(x) ∈ L2pp2(ℝn) ∩ L(ℝn), β5(x) ∈ L2(ℝn).

Example 2.8

For x ∈ ℝn, s, s1, s2 ∈ ℝ and p > 2, let

F(x,s)=α1|s|p1sgn(s);G(x,s)=α4ex2s.

It is easy to check that the functions F and G in Example 2.8 satisfy Condition (A1) and (A2).

In what follows we consider the probability space (Ω, 𝔉, ℙ) which is defined in Section 1. Let

θtω()=ω(+t)ω(t),tR.(16)

Then (Ω, 𝔉, ℙ, (θt)t∈ℝ) is an ergodic metric dynamical system. Since the probability space (Ω,𝔉,ℙ) is canonical, one has

w(t,ω)=ω(t),w(t,θsω)=w(t+s,ω)w(s,ω).(17)

To study the random attractor for problem (8)-(9), we first transform that system into a deterministic system with random parameter. Let

z(θtω)μ0eμs(θtω)(s)ds,tR.

Then, one has that z(θt ω) is the Ornstein-Uhlenbeck process and solves the following equation (see [15] for details):

dz+μzdt=dw(t).(18)

Moreover, the random variable z(θt ω) is tempered, and z(θt ω) is ℙ − a.e. continuous. It follows from Proposition 4.3.3 [2] that there exists a tempered function r(ω) > 0 such that

|z(ω)|+|z(ω)|2r(ω),(19)

where r(ω) satisfies, for α > 0 and for ℙ − a.e. ωΩ,

r(θtω)eα2|t|r(ω),tR.(20)

Then it follows that for α > 0 and for ℙ − a.e. ωΩ,

|z(θtω)|+|z(θtω)|2eα2|t|r(ω),tR,(21)

|z(θthω)|+|z(θthω)|2eα2|th|r(ω)eα2heα2|t|r(ω),tR.(22)

By [9], z(θt ω) has the following properties:

limt±z(θtω)t=0,(23)

limt±1|t|t0|z(θsω)|ds=E|z|,(24)

E|z|E|z|2=12μ.(25)

It follows from (23) and (25) that there exists a positive constant μ > 0 small enough, such that for t > 0 large enough,

2ϵμt0|z(θsω)|dsλ4t;2ϵ|z(θtω)|λ8t.(26)

In this paper we consider the weak solutions of (8)-(9) and (30)-(31).

Definition 2.9

For any u0 ∈ 𝔖, let u : [0,∞) × ΩL2 (ℝn). Suppose that u(⋅, ω, u0) : [0, ∞) → L2 (ℝn) is continuous and ut is measurable. Then we say that u is a weak solution of (8)-(9), if u(t, ω, u0) = u0(t, ω), t ∈ [−h, 0], ωΩ, and uL2 ([0, T], H1(ℝn)) ∩ Lp([0, T], Lp(ℝn)), for any T > 0, and for any ϕC0 (ℝn)

(u(T),ϕ)(u0(0),ϕ)+0T(u(t),λϕΔϕ)dt=0T(F(,u(t))+G(,u(th))+g,ϕ)dt+ϵ0T(ϕ,udw),(27)

ℙ − a.e. ωΩ.

Suppose that u is a weak solution of (8)-(9). Let v(t) = eϵz(θtω) u(t), v0(t) = eϵz(θtω) u0(t) and ψ = eϵz(θtω) ϕ. Then one has that

(u(T),ϕ)(u0(0),ϕ)=0T(ut,ϕ)dt=0T(t(eϵz(θtω)v(t)),ϕ)dt=0T(tv(t),ψ)dt+ϵ0T(ψ,vdz)=(v(T),ψ)(v0(0),ψ)+ϵ0T(ψ,vdw)μϵ0T(ψ,z(θtω)v)dt.(28)

In the last step of (28) we use (18). Hence, it follows from (28) and the definition of weak solution that

(v(T),ψ)(v0(0),ψ)+0T(v(t),λψΔψ)dt=0Teϵz(θtω)(F(,eϵz(θtω)v(t))+G(,eϵz(θthω)v(th))+g,ψ)dt+μϵ0T(ψ,z(θtω)v)dt.(29)

Then function v satisfying (29) is said to be the weak solution of the following equation:

vt+λvΔv=eϵz(θtω)F(x,eϵz(θtω)v(t,x))+eϵz(θtω)G(x,eϵz(θthω)v(th,x))+eϵz(θtω)g(x)+ϵμz(θtω)v,(30)

with the initial condition

v0(t,x)=eϵz(θtω)u0(t,x),xRn,t[h,0].(31)

Equation (30)-(31) can be seen as a deterministic partial differential equation with random coefficients.

We use a similar method as in [4] to obtain the existence of weak solution to equation (30)-(31). First, using the Galerkin method as in [28], we can get that, under the condition (A1) and (A2), for any bounded domain O ⊂ ℝn, (30)-(31) has a unique weak solution v(⋅, ω, v0) ∈ C([0, ∞), L2(O)) ∩ Lloc2 ([0, ∞), H01 (O)) ∩ Llocp ([0, ∞), Lp(O)), for ℙ − a.e. ωΩ. We can take the domain to be a family of balls, and the radius of balls tend to ∞. Then, one can get that vL2(ℝn) is the unique weak solution of (30)-(31) on ℝn. Then, u(t) = eϵz(θtω) v(t) is a unique weak solution to (8)-(9). Similar as Theorem 12 [13], we can show that (30)-(31) generates a continuous random dynamical system (Φ(t))t≥0 over (Ω, 𝓕, ℙ, (θt)t∈ℝ), with

Φ(t,θtω,B(θtω))=vt(,ω,v0),ωΩ,v0S,(32)

and (8)-(9) generates a continuous random dynamical system (Ψ(t))t≥0 over (Ω, 𝓕, ℙ, (θt)t∈ℝ), with

Ψ(t,θtω,B(θtω))=ut(,ω,u0),ωΩ,u0S.(33)

Notice that two dynamical systems are conjugate to each other. Therefore, in the following sections, we only consider the existence of random attractor of (Φ(t))t≥0.

3 Uniform estimates of solutions

In this section we prove the existence of the random attractor for random dynamical system (Φ(t))t≥0. We first give some useful estimates on the mild solutions of equation (30)-(31).

Lemma 3.1

Random dynamical system Φ has a random absorbing set {K(ω)}ωΩ in 𝒟, that is, for any {B(ω)}ωΩ ∈ 𝒟 and ℙ − a.e.ωΩ, there is TB(ω) > 0, such that Φ(t, θtω, B(θt ω)) ⊂ K(ω) for all tTB(ω).

Proof

Taking the inner product of (30) with v, we get that

12ddt||v||2+λ||v||2+||v||2=eϵz(θtω)RnF(x,eϵz(θtω)v(t,x))vdx+eϵz(θtω)RnG(x,eϵz(θthω)v(th,x))vdx+eϵz(θtω)Rng(x)vdx+ϵμz(θtω)||v||2.(34)

We now estimate each term on the right hand side of (34). For the first term, by condition (A1) and Young inequality we have that

eϵz(θtω)RnF(x,eϵz(θtω)v(t,x))vdxe2ϵz(θtω)Rnα1|eϵz(θtω)v|p+β1(x)dx=α1e(p2)ϵz(θtω)||v||Lpp+e2ϵz(θtω)||β1||L1.(35)

Note that by Young inequality, we have that for all a, b, c > 0, α, β, y > 1, and 1α+1β+1y=1,

abcaαα+bββ+cyy.(36)

For the second term on the right hand side of (34), by condition (A2) and (36), we obtain that

eϵz(θtω)RnG(x,eϵz(θthω)v(th,x))vdxeϵz(θtω)Rnβ4(x)|eϵz(θthω)v(th,x)v|dx+eϵz(θtω)Rnβ5(x)|v|dx=Rn(ep2pϵz(θtω)|v|)|v(th,x)|(e22ppϵz(θtω)+ϵz(θthω)β4(x))dx+eϵz(θtω)Rnβ5(x)|v|dxα12e(p2)ϵz(θtω)||v||Lpp+λ4eλ2h||v(th,x)||2+c1e44pp2ϵz(θtω)+2pp2ϵz(θthω)||β4||2pp2+λ4||v||2+1λe2ϵz(θtω)||β5||2,(37)

where c1 is a positive constant depending on α, λ and h. For the third term on the right hand side of (34), by Young inequality,

eϵz(θtω)Rng(x)vdx1λe2ϵz(θtω)||g||2+λ4||v||2.(38)

Then it follows from (34) - (38) that, for all t ≥ 0

ddt||v||2+2||v||2+α1e(p2)ϵz(θtω)||v||Lpp(2ϵμ|z(θtω)|λ)||v||2+λ2eλ2h||v(th,x)||2+c2e2ϵz(θtω)+c3e44pp2ϵz(θtω)+2pp2ϵz(θthω)=(2ϵμ|z(θtω)|λ2)||v||2+λ2(eλ2h||v(th,x)||2||v||2)+c2e2ϵz(θtω)+c3e44pp2ϵz(θtω)+2pp2ϵz(θthω),(39)

with c2=2||β1||L1+2λ||β5||2+2λ||g||2,c3=2c1||β4||2pp2. Applying Gronwall inequality, we obtain that, for all t ≥ 0,

||v(t)||2+20teλ2(st)+2ϵμst|z(θτω)|dτ||v||2ds+α10teλ2(st)+2ϵμst|z(θτω)|dτ+(p2)ϵz(θsω)||v||Lppdseλ2t+2ϵμ0t|z(θτω)|dτ||v0||2+λ20teλ2(st)+2ϵμst|z(θτω)|dτ(eλ2h||v(sh)||2||v||2)ds+c20teλ2(st)+2ϵμst|z(θτω)|dτ2ϵz(θtω)ds+c30teλ2(st)+2ϵμst|z(θτω)|dτ+44pp2ϵz(θsω)+2pp2ϵz(θshω)ds.(40)

Notice that

0teλ2s2ϵμ0s|z(θτω)|dτeλ2h||v(sh)||2ds=htheλ2(s+h)2ϵμ0s+h|z(θτω)|dτeλ2h||v(s)||2ds=h0eλ2s2ϵμ0s+h|z(θτω)|dτ||v(s)||2ds+0theλ2s2ϵμ0s+h|z(θτω)|dτ||v(s)||2dsh||v0||S+0theλ2s2ϵμ0s|z(θτω)|dτ||v(s)||2dsh||v0||S+0teλ2s2ϵμ0s|z(θτω)|dτ||v(s)||2ds.(41)

Hence, it follows from (41) that

λ20teλ2(st)+2ϵμst|z(θtω)|dτ(eλ2h||v(sh)||2||v||2)ds=λ2eλ2t+2ϵμ0t|z(θtω)|dτ0teλ2s2ϵμ0s|z(θtω)|dτeλ2h||v(sh)||2dsλ2eλ2t+2ϵμ0t|z(θtω)|dτ0teλ2s2ϵμ0s|z(θtω)|dτ||v||2dsλh2eλ2t+2ϵμ0t|z(θtω)|dτ||v0||S.(42)

Then, (40) and (42) imply that

||v(t)||2+20teλ2(st)+2ϵμst|z(θtω)|dτ||v||2ds+α10teλ2(st)+2ϵμst|z(θtω)|dτ+(p2)ϵz(θsω)||v||Lppds(1+λh2)eλ2t+2ϵμ0t|z(θtω)|dτ||v0||S2+c20teλ2(st)+2ϵμst|z(θtω)|dτ2ϵz(θtω)ds+c30teλ2(st)+2ϵμst|z(θtω)|dτ+44pp2ϵz(θsω)+2pp2ϵz(θshω)ds.(43)

For fixed σ ∈ [−h, 0], we have that, for t ≥ −σ,

||v(t+σ)||2(1+λh2)eλ2(t+σ)+2ϵμ0t+σ|z(θτω)|dτ||v0||S2+c20t+σeλ2(stσ)+2ϵμst+σ|z(θτω)|dτ2ϵz(θsω)ds+c30t+σeλ2(stσ)+2ϵμst+σ|z(θτω)|dτ+44pp2ϵz(θsω)+2pp2ϵz(θshω)ds(1+λh2)eλ2(ht)+2ϵμ0t|z(θτω)|dτ||v0||S2+c2eλh20teλ2(st)+2ϵμst|z(θτω)|dτ2ϵz(θsω)ds+c3eλh20teλ2(st)+2ϵμst|z(θτω)|dτ+44pp2ϵz(θsω)+2pp2ϵz(θshω)ds.(44)

and for t ∈ [0, −σ],

||v(t+σ)||2||v0||S2(1+λh2)eλ2(ht)+2ϵμ0t|z(θτω)|dτ||v0||S2+c2eλh20teλ2(st)+2ϵμst|z(θτω)|dτ2ϵz(θsω)ds+c3eλh20teλ2(st)+2ϵμst|z(θτω)|dτ+44pp2ϵz(θsω)+2pp2ϵz(θshω)ds.(45)

Due to (44) and (45), we find that for all t ≥ 0

||vt||S2(1+λh2)eλ2(ht)+2ϵμ0t|z(θτω)|dτ||v0||S2+c2eλh20teλ2(st)+2ϵμst|z(θτω)|dτ2ϵz(θsω)ds+c3eλh20teλ2(st)+2ϵμst|z(θτω)|dτ+44pp2ϵz(θsω)+2pp2ϵz(θshω)ds.(46)

Replacing ω by θtω in (46), we get that for all t ≥ 0

||vt(θtω,v0(θtω))||S2(1+λh2)eλ2(ht)+2ϵμ0t|z(θτtω)|dτ||v0(θtω)||S2+c2eλh20teλ2(st)+2ϵμst|z(θτtω)|dτ2ϵz(θstω)ds+c3eλh20teλ2(st)+2ϵμst|z(θτtω)|dτ+44pp2ϵz(θstω)+2pp2ϵz(θsthω)ds.(47)

By (26), one has that for any v0(θt ω) ∈ B(θt ω),

limt+(1+λh2)eλ2(ht)+2ϵμ0t|z(θτtω)|dτ||v0||S2=limt+(1+λh2)eλ2(ht)+2ϵμt0|z(θτω)|dτ||v0||S2limt+(1+λh2)eλ2(ht)+λ4t||v0||S2=0.(48)

It follows from (26) that for any v0(θt ω) ∈ B(θt ω),

c2eλh20teλ2(st)+2ϵμst|z(θτtω)|dτ2ϵz(θstω)ds=c2eλh2t0eλ2s+2ϵμs0|z(θτω)|dτ2ϵz(θsω)dsc2eλh20eλ4s2ϵz(θtω)dsc2eλh20eλ2sλ4sλ8sds<+.(49)

Similarly, we can get that

c3eλh20teλ2(st)+2ϵμst|z(θτtω)|dτ+44pp2ϵz(θstω)+2pp2ϵz(θsthω)ds=c3eλh2t0eλ2s+2ϵμs0|z(θτω)|dτ+44pp2ϵz(θsω)+2pp2ϵz(θshω)dsc3eλh20eλ4s+44pp2ϵz(θsω)+2pp2ϵz(θshω)ds<+.(50)

Set

ρ12(ω)=1+c2eλh20eλ4s2ϵz(θsω)ds+c3eλh20eλ4s+44pp2ϵz(θsω)+2pp2ϵz(θshω)ds.(51)

Then, there exists a TB(ω) > 0, such that for all t > TB(ω),

||vt(θtω,v0(θtω))||S2ρ12(ω).(52)

To show ρ12 (ω) is tempered, we need only to prove that for any y > 0 small enough, the following holds

limt+eytρ12(θtω)=0.(53)

Using (26) again, one has that for y < λ small enough,

limt+eyt0eλ4s2ϵz(θstω)dslimt+eyt0ey4s2ϵz(θstω)ds=limt+eyttey4(s+t)2ϵz(θsω)ds=limt+e34yt(0ey4s2ϵz(θsω)ds+0tey4s2ϵz(θsω)ds)limt+e34yt(0ey4sy8sds+0tey4s+y8sds)=0.(54)

Similarly, we can get that

limt+eyt0eλ4s+44pp2ϵz(θstω)+2pp2ϵz(θsthω)ds=0.(55)

In view of (54) and (55), we find that ρ12 (ω) is tempered. This ends the proof.

In Lemma 3.1, we show that {K(ω)}ωΩ is a random absorbing set for Φ. To prove Φ has a random attractor, we need to show that Φ is 𝒟-pullback asymptotically compact. Therefore, we should obtain some estimates for ∇ v.

Lemma 3.2

There exists a tempered random variable ρ2(ω) > 0 such that for any {B(ω)}ωΩ ∈ 𝒟 and v0(ω) ∈ B(ω), there exists a TB(ω) > 0 such that the solution v of (30)-(31) satisfies, for ℙ-a.e. ωΩ, for all tTB(Ω),

tt+1||v(s,θtω,v0(θtω))||2dsρ2(ω).(56)

Proof

By (43), we have that for all t ≥ 0

20teλ2(st)+2ϵμst|z(θtω)|dτ||v||2ds(1+λh2)eλ2t+2ϵμ0t|z(θtω)|dτ||v0||S2+c20teλ2(st)+2ϵμst|z(θtω)|dτ2ϵz(θtω)ds+c30teλ2(st)+2ϵμst|z(θtω)|dτ+44pp2ϵz(θsω)+2pp2ϵz(θshω)ds.(57)

Let TB(ω) be the positive constant defined in Lemma 4.1. Replacing ω by θt ω in (57), by (48) - (50), one has that for all tTB(ω),

20teλ2(st)+2ϵμst|z(θτtω)|dτ||v(s,θtω,v0(θtω))||2ds(1+λh2)eλ2t+2ϵμ0t|z(θτtω)|dτ||v0(θtω)||S2+c20teλ2(st)+2ϵμst|z(θτtω)|dτ2ϵz(θstω)ds+c30teλ2(st)+2ϵμst|z(θτtω)|dτ+44pp2ϵz(θstω)+2pp2ϵz(θsthω)dsρ12(ω).(58)

In another aspect, for s ∈ [t, t + 1],

0t+1eλ2(st1)+2ϵμst+1|z(θτt1ω)|dτ||v(s,θt1ω,v0(θt1ω))||2dstt+1eλ2(st1)+2ϵμst+1|z(θτt1ω)|dτ||v(s,θt1ω,v0(θt1ω))||2dseλ22ϵμmax1τ0|z(θτω)|tt+1||v(s,θt1ω,v0(θt1ω))||2ds.(59)

It follows from (58) and (59) that

tt+1||v(s,θt1ω,v0(θt1ω))||2dseλ2+2ϵμmax1τ0|z(θτω)|ρ12(ω).(60)

Replacing ω by θ1 ω in the last inequality we obtain that

tt+1||v(s,θtω,v0(θtω))||2dseλ2+2ϵμmax1τ0|z(θτω)|ρ12(ω)ρ2(ω).(61)

This ends the proof. □

Lemma 3.3

There exists a tempered random variable ρ3(ω) > 0 such that for any {B(ω)}ωΩ ∈ 𝒟 and v0(ω) ∈ B(ω), there exists a TB(ω) > 0 such that the solution v of (30)-(31) satisfies, for ℙ-a.e. ωΩ, for all tTB(ω) + h + 1, and σ1, σ2 ∈ [−h, 0],

v(t,θtω,v0(θtω))2ρ3(ω),(62)

t+σ1t+σ2||Δv(s,θtω,v0(θtω))||2dsρ4(ω).(63)

Proof

Taking the inner product of (30) with −Δ v, we get that

12ddt||v||2+λ||v||2+||Δv||2=eϵz(θtω)RnF(x,eϵz(θtω)v(t,x))Δvdxeϵz(θtω)RnG(x,eϵz(θthω)v(th,x))Δvdxeϵz(θtω)Rng(x)Δvdx+ϵμz(θtω)||v||2.(64)

We now estimate each term on the right-hand side of (64). For the first term, by Young inequality and condition (A1), we have

eϵz(θtω)RnF(x,eϵz(θtω)v(t,x))Δvdx=eϵz(θtω)RnFx(x,eϵz(θtω)v)vdx+RnFs(x,eϵz(θtω)v)|v|2dxeϵz(θtω)||β3||||v||+α3||v||2(1+α3)||v||2+14e2ϵz(θtω)||β3||2.(65)

For the second term, it follows from condition (A2) and Young inequality that

eϵz(θtω)RnG(x,eϵz(θthω)v(th,x))Δvdxeϵz(θtω)Rneϵz(θthω)β4(x)|v(th,x)|+β5(x)Δvdx18||Δv||2+2e2ϵz(θtω)+2ϵz(θthω)Rnβ42(x)|v(th,x)|2dx+18||Δv||2+2e2ϵz(θtω)Rnβ52(x)dx14||Δv||2+2e2ϵz(θtω)+2ϵz(θthω)||β4||L2||v(th)||2+2e2ϵz(θtω)||β5||2.(66)

The third term is bounded by

eϵz(θtω)Rng(x)Δvdx14||Δv||2+e2ϵz(θtω)||g||2.(67)

By (64) - (67), we find that for all t ≥ 0,

ddt||v||2+||Δv||22(1+α3)||v||2+c5e2ϵz(θtω)+4e2ϵz(θtω)+2ϵz(θthω)||β4||L2||v(th)||2,(68)

with c5=4||β5||2+2||g||2+12||β3||2. Let TB(ω) be the positive constant defined in Lemma 3.1, take tTB(ω) and s ∈ [t, t + 1]. Integrating (68) over [s, t + 1], we get that

||v(t+1,ω,v0(ω))||2||v(s,ω,v0(ω))||2+2(1+α3)st+1||v(τ,ω,v0(ω))||2dτ+c5st+1e2ϵz(θτω)dτ+4||β4||L2st+1e2ϵz(θτω)+2ϵz(θτhω)||v(τh,ω,v0(ω))||2dτ.(69)

Integrating the above inequality with respect to s over [t, t + 1], we get that

||v(t+1,ω,v0(ω))||22(2+α3)tt+1||v(τ,ω,v0(ω))||2dτ+c5tt+1e2ϵz(θτω)dτ+4||β4||L2tt+1e2ϵz(θτω)+2ϵz(θτhω)||v(τh,ω,v0(ω))||2dτ.(70)

Replacing ω by θt−1 ω in (70), we obtain that

||v(t+1,θt1ω,v0(θt1ω))||22(2+α3)tt+1||v(τ,θt1ω,v0(θt1ω))||2dτ+c5tt+1e2ϵz(θτt1ω)dτ+4||β4||L2tt+1e2ϵz(θτt1ω)+2ϵz(θτth1ω)||v(τh,θt1ω,v0(θt1ω))||2dτ.(71)

It follows from (56) that for all tTB(ω) + h,

2(2+α3)tt+1||v(τ,θt1ω,v0(θt1ω))||2dτ2(2+α3)ρ2(θ1ω),(72)

tt+1e2ϵz(θτt1ω)dτe2ϵmax1τ0|z(θτω)|,(73)

and

tt+1e2ϵz(θτt1ω)+2ϵz(θτth1ω)||v(τh,θt1ω,v0(θt1ω))||2dτe2ϵmax1τ0|z(θτω)|+|z(θτhω)|thth+1||v(τ,θt1ω,v0(θt1ω))||2dτ=e2ϵmax1τ0|z(θτω)|+|z(θτhω)|thth+1||v(τ,θt+h1(θhω),v0(θt+h1(θhω))||2dτe2ϵmax1τ0|z(θτω)|+|z(θτhω)|ρ2(θhω).(74)

It follows from (71) - (74) that, for all tTB(ω) + h

||v(t+1,θt1ω,v0(θt1ω))||22(2+α3)ρ2(θ1ω)+c5e2ϵmax1τ0|z(θτω)|+4||β4||L2e2ϵmax1τ0|z(θτω)|+|z(θτhω)|ρ2(θhω)ρ3(ω).(75)

Then we have that for all tTB(ω) + h +1,

||v(t,θtω,v0(θtω))||2ρ3(ω).(76)

Let th, −hσ1σ2 ≤ 0. Integrating (68) over [t + σ1, t + σ2], we obtain that

||v(t+σ2,ω,v0(ω))||2+t+σ1t+σ2||Δv(τ,ω,v0(ω))||2dτ||v(t+σ1,ω,v0(ω))||2+2(1+α3)t+σ1t+σ2||v(τ,ω,v0(ω))||2dτ+c5t+σ1t+σ2e2ϵz(θτω)dτ+4||β4||L2t+σ1t+σ2e2ϵz(θτω)+2ϵz(θτhω)||v(τh,ω,v0(ω))||2dτ.(77)

Replacing ω by θt ω in the last inequality, we have that

||v(t+σ2,θtω,v0(θtω))||2+t+σ1t+σ2||Δv(τ,θtω,v0(θtω))||2dτ||v(t+σ1,θtω,v0(θtω))||2+2(1+α3)t+σ1t+σ2||v(τ,θtω,v0(θtω))||2dτ+c5t+σ1t+σ2e2ϵz(θτtω)dτ+4||β4||L2t+σ1t+σ2e2ϵz(θτtω)+2ϵz(θτthω)||v(τh,θtω,v0(θtω))||2dτ.(78)

Using (76), we get that

t+σ1t+σ2||v(τ,θtω,v0(θtω))||2dτ=t+σ1t+σ2||v(τ,θτ(θτtω),v0(θτ(θτtω)))||2dτhmaxhτ0ρ3(θτω)).(79)

By (52), we can get that

4||β4||L2t+σ1t+σ2e2ϵz(θτtω)+2ϵz(θτthω)||v(τh,θtω,v0(θtω))||2dτ4||β4||L2max2hτhρ12(θτω)t+σ1t+σ2e2ϵz(θτtω)+2ϵz(θτthω)dτ=4h||β4||L2max2hτhρ12(θτω)e2ϵmaxhτ0|z(θτω)|+|z(θτhω)|.(80)

Notice that

t+σ1t+σ2e2ϵz(θτtω)dτ=σ1σ2e2ϵz(θτω)dτhe2ϵmaxhτ0|z(θτω)|.(81)

It follows from (76), (78)-(81) that

t+σ1t+σ2||Δv(τ,θtω,v0(θtω))||2dτ(1+2h(1+α3))maxhτ0ρ3(θτω)+c5he2ϵmaxhτ0|z(θτω)|+4h||β4||L2max2hτhρ12(θτω)e2ϵmaxhτ0|z(θτω)|+|z(θτhω)|ρ4(ω).(82)

This ends the proof. □

Lemma 3.4

Let B(ω) ∈ 𝒟 and v0(ω) ∈ B(ω). Then for any ϵ1 > 0 and ℙ-a.e. ωΩ, there exists a T = T(B, ω, ϵ1) > 0, and R = R (ω, ϵ1) > 0 such that the solution (30)-(31), satisfies for all tT,

sups[h,0]|x|R|vt(s,θtω,v0(θtω))|dxϵ1.(83)

Proof

Let ρ be a smooth function defined on ℝ+ such that 0 ≤ ρ(s) ≤ 1 for all s ∈ ℝ+, and

ρ(s)=0,0s1,1,s2.(84)

Then there exists a positive constant c0 such that |ρ′(s)| ≤ c0 for all s ∈ ℝ+. Taking the inner product of (30) with ρ|x|2k2v, we get that

12ddtRnρ|x|2k2|v|2dx+λRnρ|x|2k2|v|2dxRnρ|x|2k2vΔvdx=eϵz(θtω)RnF(x,eϵz(θtω)v)ρ|x|2k2vdx+eϵz(θtω)RnG(x,eϵz(θthω)v(th))ρ|x|2k2vdx+eϵz(θtω)Rng(x)ρ|x|2k2vdx+ϵμz(θtω)Rnρ|x|2k2|v|2dx.(85)

We now estimate each term in (85). For the third term on the left hand-side of (85), we have that

Rnρ|x|2k2vΔvdx=Rnρ|x|2k2|v|2dx+Rnρ|x|2k2v2xk2vdx=Rnρ|x|2k2|v|2dx+k|x|2kρ|x|2k2v2xk2vdx,(86)

and

k|x|2kρ|x|2k2v2xk2vdx22kk|x|2k|ρ|x|2k2||v||v|dx4c0kRn|v||v|dx2c0k||v||2+||v||2.(87)

By (86) and (87), one has that

Rnρ|x|2k2vΔvdxRnρ|x|2k2|v|2dx2c0k||v||2+||v||2.(88)

For the first term on the right-hand side of (85) by condition (A1), we have

eϵz(θtω)RnF(x,eϵz(θtω)v)ρ|x|2k2vdx=e2ϵz(θtω)RnF(x,eϵz(θtω)v)ρ|x|2k2eϵz(θtω)vdxα1e(p2)ϵz(θtω)Rnρ|x|2k2|v|pdx+e2ϵz(θtω)Rnρ|x|2k2β1(x)dx.(89)

For the second term on the right-hand side of (85) we have that

eϵz(θtω)RnG(x,eϵz(θthω)v(th))ρ|x|2k2vdxeϵz(θtω)Rnβ4(x)eϵz(θthω)v(th)ρ|x|2k2|v|dx+eϵz(θtω)Rnβ5(x)ρ|x|2k2|v|dx.(90)

By (36), the first term on the right-hand side of (90) is bounded by

eϵz(θtω)Rnβ4(x)eϵz(θthω)v(th)ρ|x|2k2|v|dx=Rnep2pϵz(θtω)|v|ρ1p|x|2k2ρ12|x|2k2|v(th)|ρp22p|x|2k2e22ppϵz(θtω)+ϵz(θthω)β4(x)dxα1Rne(p2)ϵz(θtω)ρ|x|2k2|v|pdx+λ4eλ2hRnρ|x|2k2|v(th)|2dx+c6Rnρ|x|2k2e44pp2ϵz(θtω)+2pp2ϵz(θthω)β4(x)2pp2dx,(91)

and the second term on the right-hand side of (90) is bounded by

eϵz(θtω)Rnβ5(x)ρ|x|2k2|v|dxλ4Rnρ|x|2k2|v|2dx+1λe2ϵz(θtω)Rnρ|x|2k2β52(x)dx.(92)

For the third term on the right-hand side of (85) we have that

eϵz(θtω)Rng(x)ρ|x|2k2vdxλ4Rnρ|x|2k2|v|2dx+1λe2ϵz(θtω)Rnρ|x|2k2g2(x)dx.(93)

It follows from (85) - (93) that

ddtRnρ|x|2k2|v|2dx2ϵμ|z(θtω)|λ2Rnρ|x|2k2|v|2dx+λ2Rnρ|x|2k2eλ2h|v(th)|2|v|2dx+4c0k||v||2+||v||2+2e2ϵz(θtω)Rnρ|x|2k2β1(x)+1λβ52(x)+1λg2(x)dx+2c6Rnρ|x|2k2e44pp2ϵz(θtω)+2pp2ϵz(θthω)β42pp2(x)dx.(94)

Let TB(ω) be the positive constant defined in Lemma 3.1. Set T1(B, ω) ≥ TB(ω) + h +1. Applying Gronwall inequality to (94), we obtain that for tT1,

Rnρ|x|2k2|v|2dxeλ2(T1t)+2ϵμT1t|z(θτω)|dτRnρ|x|2k2|v(T1)|2dx+λ2T1teλ2(st)+2ϵμst|z(θτω)|dτRnρ|x|2k2eλ2h|v(th)|2|v|2dxds+4c0kT1teλ2(st)+2ϵμst|z(θτω)|dτ||v||2+||v||2ds+2T1teλ2(st)+2ϵμst|z(θτω)|dτ2ϵz(θsω)Rnρ|x|2k2β1(x)+1λβ52(x)+1λg2(x)dxds+2c6T1teλ2(st)+2ϵμst|z(θτω)|dτ+44pp2ϵz(θsω)+2pp2ϵz(θshω)Rnρ|x|2k2β42pp2(x)dxds.(95)

If we take tT1 + h, then, by (95), we can get that for all σ ∈ [−h, 0],

Rnρ|x|2k2|v(t+σ)|2dxeλ2(T1tσ)+2ϵμT1t+σ|z(θτω)|dτRnρ|x|2k2|v(T1)|2dx+λ2T1t+σeλ2(stσ)+2ϵμst+σ|z(θτω)|dτRnρ|x|2k2eλ2h|v(sh)|2|v|2dxds+4c0kT1t+σeλ2(stσ)+2ϵμst+σ|z(θτω)|dτ||v||2+||v||2ds+2T1t+σeλ2(stσ)+2ϵμst+σ|z(θτω)|dτ2ϵz(θsω)Rnρ|x|2k2β1(x)+1λβ52(x)+1λg2(x)dxds+2c6T1t+σeλ2(stσ)+2ϵμst+σ|z(θτω)|dτ+44pp2ϵz(θsω)+2pp2ϵz(θshω)Rnρ|x|2k2β42pp2(x)dxds.(96)

It follows that for all tT1 + h,

suphσ0Rnρ|x|2k2|v(t+σ)|2dxeλ2(T1+ht)+2ϵμT1t|z(θτω)|dτ||v(T1)||S2+λ2eλh2T1teλ2(st)+2ϵμst|z(θτω)|dτRnρ|x|2k2eλ2h|v(sh)|2|v|2dxds+4c0keλh2T1teλ2(st)+2ϵμst|z(θτω)|dτ||v||2+||v||2ds+2eλh2T1teλ2(st)+2ϵμst|z(θτω)|dτ2ϵz(θsω)Rnρ|x|2k2β1(x)+1λβ52(x)+1λg2(x)dxds+2c1eλh2T1teλ2(st)+2ϵμst|z(θτω)|dτ+44pp2ϵz(θsω)+2pp2ϵz(θshω)Rnρ|x|2k2β42pp2(x)dxds.(97)

Replacing ω by θtω, we obtain from (97) that for all tT1(B, ω),

suphσ0Rnρ|x|2k2|vt(σ,θtω,v0(θtω))|2dxeλh2eλ2(T1t)+2ϵμT1t|z(θτtω)|dτ||vT1(θt(ω),v0(θt(ω)))||S2+λ2eλh2T1teλ2(st)+2ϵμst|z(θτtω)|dτRnρ|x|2k2eλ2h|v(sh,θtω,v0(θtω))|2|v(s,θtω,v0(θtω))|2dxds+4c0keλh2T1teλ2(st)+2ϵμst|z(θτtω)|dτ||v(s,θtω,v0(θtω))||2+||v(s,θtω,v0(θtω))||2ds+2eλh2T1teλ2(st)+2ϵμst|z(θτtω)|dτ2ϵz(θstω)Rnρ|x|2k2β1(x)+1λβ52(x)+1λg2(x)dxds+2c1eλh2T1teλ2(st)+2ϵμst|z(θτtω)|dτ+44pp2ϵz(θstω)+2pp2ϵz(θsthω)Rnρ|x|2k2β42pp2(x)dxds.(98)

Now we estimate each term on the right hand side of (98). We first replace t by T1 and replace ω by θtω in (46), we have that

||v(T1,θtω,θtv0(ω))||S2(1+λh2)eλ2(hT1)+2ϵμ0T1|z(θτtω)|dτ||v0(θtω)||S2+c2eλ2h0T1eλ2(sT1)+2ϵμsT1|z(θτtω)|dτ2ϵz(θstω)ds+c3eλ2h0T1eλ2(sT1)+2ϵμsT1|z(θτtω)|dτ+44pp2ϵz(θstω)+2pp2ϵz(θsthω)ds.(99)

Therefore, for the first term on the right hand side of (98),

eλh2eλ2(T1t)+2ϵμT1t|z(θτtω)|dτ||vT1(θt(ω),v0(θt(ω)))||S2(1+λh2)eλheλ2t+2ϵμ0t|z(θτtω)|dτ||v0(θtω)||S2+c2eλh0T1eλ2(st)+2ϵμst|z(θτtω)|dτ2ϵz(θstω)ds+c3eλh0T1eλ2(st)+2ϵμst|z(θτtω)|dτ+44pp2ϵz(θstω)+2pp2ϵz(θsthω)ds.(100)

We now estimate the terms in (100) as follows. For the first term, by (26), we find that for tT1 and v0(θtω) ∈ B(θtω),

limt+(1+λh2)eλheλ2t+2ϵμ0t|z(θτtω)|dτ||v0(θtω)||S2=limt+(1+λh2)eλheλ2t+2ϵμt0|z(θτω)|dτ||v0(θtω)||S2limt+(1+λh2)eλheλ2t+λ4t||v0(θtω)||S2=0.(101)

For the second term, by (26)

limt+c2eλh0T1eλ2(st)+2ϵμst|z(θτtω)|dτ2ϵz(θstω)ds=limt+c2eλh0T1eλ2(st)+2ϵμst0|z(θτω)|dτ2ϵz(θstω)dslimt+c2eλh0T1eλ2(st)λ4(st)λ8(st)ds=0.(102)

Similarly, we can get the following estimate for the third term on the right-hand side of (100):

limt+c3eλh0T1eλ2(st)+2ϵμst|z(θτtω)|dτ+44pp2ϵz(θstω)+2pp2ϵz(θsthω)ds=0.(103)

It follows from (100) - (103) that

limt+eλh2eλ2(T1t)+2ϵμT1t|z(θτtω)|dτ||vT1(θt(ω),v0(θt(ω)))||S2=0,(104)

which implies that for any ϵ1 > 0, there is a T2 = T2(B, ω, ϵ) > T1 such that

eλh2eλ2(T1t)+2ϵμT1t|z(θτtω)|dτ||vT1(θt(ω),v0(θt(ω)))||S2ϵ1.(105)

Next, we estimate the second term on the right hand side of (98). Similarly as (42), we can obtain that

λ2eλh2T1teλ2(st)+2ϵμst|z(θτtω)|dτRnρ|x|2k2eλ2h|v(sh,θtω,v0(θtω))|2|v(s,θtω,v0(θtω))|2dxdsλ2eλh2eλ2(T1t)+2ϵμ0t|z(θτtω)|dτRnρ|x|2k2suphσ0|vT1(σ,θtω,v0(θtω))|2dxdsλ2eλh2eλ2(T1t)+2ϵμ0t|z(θτtω)|dτ||vT1(θtω,v0(θtω))||S.(106)

By (104) and (106), we know that for any ϵ1 > 0, there exists a T3 = T3(B, ω, ϵ) > T1 such that

λ2eλh2T1teλ2(st)+2ϵμst|z(θτtω)|dτ××Rnρ|x|2k2eλ2h|v(sh,θtω,v0(θtω))|2|v(s,θtω,v0(θtω))|2dxdsϵ1.(107)

For the third term on the right hand side of (98), by Lemma 3.1 and Lemma 3.3, we obtain that

T1teλ2(st)+2ϵμst|z(θτtω)|dτ||v(s,θtω,v0(θtω))||2+||v(s,θtω,v0(θtω))||2dsT1teλ2(st)+2ϵμst|z(θτtω)|dτρ12(θstω)+ρ3(θstω)ds=T1t0eλ2s+2ϵμs0|z(θτω)|dτρ12(θsω)+ρ3(θsω)ds0eλ2sλ4sρ12(θsω)+ρ3(θsω)ds<.(108)

This implies that there exists a T4 = T4(B, ω, ϵ) > T1 and R2 = R2(ω, ϵ1) such that for all tT4 and kR2,

4c0keλh2T1teλ2(st)+2ϵμst|z(θτtω)|dτ||v(s,θtω,v0(θtω))||2+||v(s,θtω,v0(θtω))||2dsϵ1.(109)

Note that β1L1(ℝn), β5L2(ℝn), gL2(ℝn). Thus, there is R3 = R3(ω, ϵ1) such that for all kR3

Rnρ|x|2k2β1(x)+1λβ52(x)+1λg2(x)dx|x|kβ1(x)+1λβ52(x)+1λg2(x)dxϵ1.(110)

Then, for the fourth term on the right hand side of (98) we have that

2eλh2T1teλ2(st)+2ϵμst|z(θτtω)|dτ2ϵz(θstω)Rnρ|x|2k2β1(x)+1λβ52(x)+1λg2(x)dxds2ϵ1eλh2T1teλ2(st)+2ϵμst|z(θτtω)|dτ2ϵz(θstω)ds=2ϵ1eλh2T1t0eλ2s+2ϵμs0|z(θτω)|dτ2ϵz(θsω)ds2ϵ1eλh20eλ2sλ4sλ8sds=16λeλh2ϵ1.(111)

Notice that β4L2pp2(Rn). Thus, there is R4 = R4(ϵ1, ω) such that for all kR4,

Rnρ|x|2k2β42pp2(x)dx|x|kβ42pp2(x)dxϵ1.(112)

Then, similarly as (111), there exists a constant c > 0, such that

2c1eλh2T1teλ2(st)+2ϵμst|z(θτtω)|dτ+44pp2ϵz(θstω)+2pp2ϵz(θsthω)Rnρ|x|2k2β42pp2(x)dxdscϵ1.(113)

Let T = max {T1, T2, T3. T4} and R = max {R1, R2, R3}. Then, it follows from (98), (105), (107), (109), (111) and (113) that

suphσ0Rnρ|x|2k2|vt(σ,θtω,v0(θtω))|2dx(c+16λeλh2+3)ϵ1,(114)

This ends the proof.□

Lemma 3.5

Let B(ω) ∈ 𝒟 and v0(ω) ∈ B(ω). Then there are tempered random variables ρ5(ω), ρ6(ω), such that the solution of (30)-(31) satisfies for all tTB(ω) + 2h + 1, and σ, σ1, σ2 ∈ [–h, 0],

||vt(σ,θtω,v0(θtω))||H1(Rn)2ρ5(ω);(115)

||vt(σ1,θtω,v0(θtω))vt(σ2,θtω,v0(θtω))||ρ6(ω)|σ1σ2|1/2.(116)

Proof

By Lemma 3.1 and Lemma 3.3, we get that for all tTB(ω) + h + 1, and σ ∈ [–h,0 ],

||vt(σ,θtω,v0(θtω))||H1(Rn)2=||v(t+σ,θtω,v0(θtω))||2+||v(t+σ,θtω,v0(θtω))||2maxhσ0ρ12(θσω)+ρ3(θσω)ρ5(ω).(117)

For σ1, σ2 ∈ [–h, 0] (assuming σ1σ2 for simplicity), one has that,

||vt(σ1,θtω,v0(θtω))vt(σ2,θtω,v0(θtω))||2=||v(t+σ1,θtω,v0(θtω))v(t+σ2,θtω,v0(θtω))||2=||t+σ1t+σ2ddtv(s,θtω,v0(θtω))||2dsλt+σ1t+σ2||v(s,θtω,v0(θtω))||ds+t+σ1t+σ2||Δv(s,θtω,v0(θtω))||ds+t+σ1t+σ2||eϵz(θstω)F(x,eϵz(θstω)v(s,θtω,v0(θtω)))||ds+t+σ1t+σ2||eϵz(θstω)G(x,eϵz(θsthω)v(sh,θtω,v0(θtω)))||ds+t+σ1t+σ2||eϵz(θstω)g||ds+ϵμt+σ1t+σ2||z(θstω)v(s,θtω,v0(θtω))||ds.(118)

Now, we estimate each term on the right hand side of (118). By Lemma 3.1, and Lemma 3.3 we find that for all tTB(ω) + 2h + 1,

t+σ1t+σ2||v(s,θtω,v0(θtω))||ds=t+σ1t+σ2||v(s,θs(θstω),v0(θs(θstω)))||dsmaxhτ0ρ1(θτω)|σ1σ2|;(119)

t+σ1t+σ2||z(θstω)v(s,θtω,v0(θtω))||dsmaxhτ0z(θτω)ρ1(θτω)|σ1σ2|;(120)

t+σ1t+σ2||Δv(s,θtω,v0(θtω))||dst+σ1t+σ2||Δv(s,θtω,v0(θtω))||2ds1/2|σ1σ2|1/2ρ4(ω)|σ1σ2|1/2;(121)

and

t+σ1t+σ2||eϵz(θstω)g||dsmaxhτ0eϵ|z(θτω)|||g|||σ1σ2|.(122)

Using conditions (A1) and (A2), we obtain

t+σ1t+σ2||eϵz(θstω)F(x,eϵz(θstω)v(s,θtω,v0(θtω)))||dst+σ1t+σ2α2e(p2)ϵz(θstω)||v(s,θtω,v0(θtω))||p1+eϵz(θstω)||β2||dsα2maxhτ0e(p+2)ϵ|z(θτω)|ρ1p1(θτω)+maxhτ0eϵ|z(θτω)|||β2|||σ1σ2|,(123)

and

t+σ1t+σ2||eϵz(θstω)G(x,eϵz(θsthω)v(sh,θtω,v0(θtω)))||dsmaxhτ0eϵ|z(θτω)|+ϵ|z(θτhω)|ρ1(θτhω)+||β5|||σ1σ2|.(124)

The lemma follows from (119) to (124). This ends the proof.□

Next, we use Ascoli theorem to show that Φ is 𝒟 pullback asymptotically compact.

Lemma 3.6

The random dynamical system Φ is 𝒟 pullback asymptotically compact in 𝔖; that is, for ℙ-a.e. ωΩ, the sequence Φ(tn,θtn,v0(θtn))}n=1 has a convergent subsequence in 𝔖 provided tn → ∞ and v0(θtn) ∈ B(θtnω).

Proof

Let Qk be the set of {x ∈ ℝn : |x| < k}. Since tn → ∞, there exists N = N(B, ω) such that tn > TB(ω) + 2h + 1, for all n > N. Using Lemma 3.5, we obtain that for n > N and σ ∈ [–h, 0],

||vtn(σ,θtnω,v0(θtnω))||H1(Qk)ρ5(ω).(125)

This implies that {Φ(tn,θtnω,v0(θtnω))}n=N is relatively compact in L2(Qk), since the embedding from H1(Qk) to L2(Qk) is compact. From Lemma 3.5, we can also obtain that for all n > N, and σ1, σ2 ∈ [–h, 0],

||vtn(σ1,θtnω,v0(θtnω))vtn(σ1,θtnω,v0(θtnω))||L2(Qk)ρ6(ω)|σ1σ2|1/2.(126)

This means that {Φ(tn,θtnω,v0(θtnω))}n=N is equicontinuous. By Ascoli Theorem we have that for fixed kN,{vtn(σ,θtnω,v0(θtnω))}n=1 is relatively compact in C([–h, 0];L2(Qk)). Therefore, for each k ∈ ℕ, there exists a subsequence {Φ(tni, θtni ω, v0(θtni ω))} converges to ηk(⋅, ω) in C([–h, 0]; L2(Qk)). Notice that, for fixed σ ∈ [–h, 0] and ωΩ, ηk+1(σ, ω) coincides with ηk(σ, ω) on Qk. Hence, for any ϵ > 0, there exists N1 = N1(ω, ϵ) ∈ ℕ, such that for all i > N1,

supσ[h,0]||vtni(σ,θtniω,v0(θtniω))η(σ,ω)||L2(Qk)ϵ.(127)

By Lemma 3.1, we have that for all σ ∈ [–h, 0], and i > N,

||vtni(σ,θtniω,v0(θtniω))||L2(Qk)supσ[h,0]||vtni(σ,θtniω,v0(θtniω))||L2(Rn)ρ1(ω).(128)

Hence, for all k ∈ ℕ,

||η(σ,ω)||L2(Qk)limi||vtni(σ,θtniω,v0(θtniω))||L2(Qk)ρ1(ω).(129)

It follows that,

||η(σ,ω)||L2(Rn)supkN||η(σ,ω)||L2(Qk)ρ1(ω).(130)

Therefore, η(σ, ω) ∈ L2(ℝn). By Lemma 3.4, for every B ∈ 𝒟 and ϵ1 > 0, there exist K = K(ω, ϵ1) and N2 = N2(ω, ϵ1), such that for all i > N2 and k > K,

supσ[h,0]||vtni(σ,θtni,v0(θtni))||L2(|x|k)ϵ1.(131)

Notice that for all l > k > K and σ ∈ [–h, 0],

||ξ(σ,ω)||L2(k|x|<l)=limi||vtni(σ,θtniω,v0(θtniω))||L2(k|x|<l)supσ[h,0],i>N2||vtni(σ,θtniω,v0(θtniω))||L2(|x|k)ϵ1.(132)

It follows that

supσ[h,0]||ξ(σ,ω)||L2(|x|k)supσ[h,0],l>k||ξ(σ,ω)||L2(k|x|<l)ϵ1.(133)

Then, (113), (127) and (131) imply that for all i > max N1, N2,

supσ[h,0]||vtni(σ,θtniω,v0(θtniω))ξ(σ,ω)||supσ[h,0]||vtni(σ,θtniω,v0(θtniω))ξ(σ,ω)||L2(Qk)+2supσ[h,0]||vtni(σ,θtniω,v0(θtniω))||L2(|x|k)+2supσ[h,0]||ξ(σ,ω)||L2(|x|k)5ϵ1.(134)

This ends the proof.□

Theorem 3.7

The random dynamical system Φ has a unique 𝒟-random attractor in 𝔖.

Proof

By Lemma 3.1, we know that Φ has a random absorbing set, and by Lemma 3.6, we obtain that Φ is 𝒟-pullback asymptotically compact in 𝔖. Therefore, by Proposition 2.1, Φ has an unique 𝒟-pullback attractor. This ends the proof.□

Acknowledgement

This work is partially supported by the National Natural Science Foundation of China under Grants 11301153 and 11271110, the Key Programs for Science and Technology of the Education Department of Henan Province under Grand 12A110007, and the Scientific Research Funds of Henan University of Science and Technology.

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About the article

Received: 2017-12-10

Accepted: 2018-06-12

Published Online: 2018-08-03


Competing interests: The authors declare that there is no conflict of interest regarding the publication of this paper.


Citation Information: Open Mathematics, Volume 16, Issue 1, Pages 862–884, ISSN (Online) 2391-5455, DOI: https://doi.org/10.1515/math-2018-0076.

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© 2018 Jia et al., published by De Gruyter. This work is licensed under the Creative Commons Attribution-NonCommercial-NoDerivatives 4.0 License. BY-NC-ND 4.0

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