Random attractors for stochastic two-compartment Gray-Scott equations with a multiplicative noise

Xiaoyao Jia; Juanjuan Gao; Xiaoquan Ding

doi:10.1515/math-2016-0052

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Random attractors for stochastic two-compartment Gray-Scott equations with a multiplicative noise

Xiaoyao Jia , Juanjuan Gao and Xiaoquan Ding

Published/Copyright: September 8, 2016

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$Open Mathematics$

From the journal Open Mathematics Volume 14 Issue 1

Abstract

In this paper, we consider the existence of a pullback attractor for the random dynamical system generated by stochastic two-compartment Gray-Scott equation for a multiplicative noise with the homogeneous Neumann boundary condition on a bounded domain of space dimension n ≤ 3. We first show that the stochastic Gray-Scott equation generates a random dynamical system by transforming this stochastic equation into a random one. We also show that the existence of a random attractor for the stochastic equation follows from the conjugation relation between systems. Then, we prove pullback asymptotical compactness of solutions through the uniform estimate on the solutions. Finally, we obtain the existence of a pullback attractor.

Keywords: Gray-Scott equation; Random attractor; Random dynamical system; Multiplicative noise

MSC 2010: 35K45; 35B40

1 Introduction

In this paper, we consider the following coupled stochastic two-compartment Gray-Scott equation, which is a reaction-diffusion system with multiplicative noise:

∂u~∂t=d1Δu~−F+ku~+u~2v~+D1w~−u~+σu~∘dWtdt,(1)

∂v~∂t=d2Δv~+F1−v~−u~2v~+D2y~−v~+σv~∘dWtdt,(2)

∂w~∂t=d1Δw~−F+kw~+w~2y~+D1u~−w~+σw~∘dWtdt,(3)

∂y~∂t=d2Δy~+F1−y~−w~2y~+D2v~−y~+σy~∘dWtdt,(4)

for t > 0, on a bounded domain D. Here D is an open bounded set of ℝⁿ (n ≤ 3), and it has a locally Lipschitz coutinuous boundary ∂D. Suppose that the equations have the following homogeneous Neumann boundary condition:

∂u~∂vt,x=∂v~∂vt,x=∂w~∂vt,x=∂y~∂vt,x=0,t>0,x∈∂D,(5)

where ∂∂v is the outward normal derivative, and have the following initial condition

u~0,x=u~0x,v~0,x=v~0x,w~0,x=w~0x,y~0,x=y~x,x∈D.(6)

Here d₁, d₂, F, k, D₁ and D₂ are positive constants; σ is a positive parameter; Δ is the Laplacian operator with respect to x∈D;u~x,t, u~x,t, u~x,t and y~x,t are real functions of x ∈ D; and W_t is a two-sided real-valued Wiener process on a probability space (Ω, F , ℙ). Here

Ω=ω∈CR,R:ω0=0

the Borel sigma-algebra F on Ω is generated by the compact open topology (see [1]), ℙ is the corresponding Wiener measure on F o denotes the Stratonovich sense in the stochastic term. We identify ω(t) with W_t (ω), i.e. W_t (ω) = W(t, ω) = ω(t), t ∈ ℝ.

The Gray-Scott equation is a kind of very important reaction-diffusion system, which arises from many chemical or biological systems [2-5]. This equation has been researched by many authors (see [2-10]). One of the most important problems in mathematical physics is the asymptotic behavior of dynamical system, which has been developed greatly in recent years. For the deterministic system, the global attractor is a very important tool to study the asymptotic behavior of dynamical system (see [9-16]). If σ = 0, system (1)-(4) reduces to the two-compartment Gray-Scott equation without random terms, which has been investigated by You [10], where we proved the existence of the global attractor for the coupled two-compartment Gray-Scott equations with homogeneous Neumann boundary condition on a bounded domain.

Stochastic differential equations of this type arise from many chemical or biological systems when random spatiotemporal force is taken into consideration. These random perturbations play important roles in macroscopic phenomena. To study the properties of stochastic dynamical systems, the concept of pullback random attractor is introduced [1, 17, 18]. The existence of random attractors for stochastic dynamical systems has been studied [6, 7, 19-21]. In this paper, we study the existence of random attractor for stochastic two-compartment Gray-Scott equation on bounded domain D of space dimension n ≤ 3.

The paper is organized as follows. In Section 2, we recall a theorem about the existence of random pullback attractor for random dynamical system, and transform the stochastic system (1)-(6) into a continuous random dynamical system (18)-(19) Ornstein-Uhlenbeck process. Moreover, we show that, for each ω, the random dynamical system has a unique global solution. In Section 3, we obtain some uniform estimates of solutions for system (18)-(19) as t → ∞. These estimates are used to prove the existence of bounded absorbing sets and the asymptotic compactness of the solutions. In the last section, we obtain the existence of a pullback random attractor.

The following notations will be used throughout this paper. ‖ · ‖ and (·, ·) denote the norm and the inner product in L²(D) or [L²(D)]⁴ respectively. ‖ · ‖L^p and ‖ · ‖H¹ are used to denote the norm in L^p(D) and H¹(D).

By the Poincaré’s inequality, there is a constant γ > 0 such that

∇ϕ2≥γϕ2,forϕ∈H01Dor[H01D]4.(7)

Note that H01D→L6Dforn≤3. There exists a constant η > 0 such that the following embedding inequality holds:

fH12≥ηfL62,forf∈H01Dor[H01(D)]4.(8)

2 RDS Generated by Stochastic Gray-Scott equation

In this section, we first recall a theorem for the existence of random attractors. Please note that here we omit the basic knowledge about random dynamical systems (RDS) and random attractor. Reader can refer to [1, 11, 17-19] for these knowledge.

Suppose that (X, ‖ · ‖_X) is a separable Banach space with Borel σ—algebra Β(X), and (Ω, F , ℙ) is a probability space. Let (Ω, F , ℙ, (θ_t) _t∈ℝ) be a metric dynamical system, and assume that ϕ is a continuous RDS on X over (Ω, F , ℙ, (θ_t) _t∈ℝ). We recall a Proposition which will be used to prove the existence of random attractor ([11], [19]) for RDS.

Proposition 2.1

Suppose D is the collection of random subsets of X, and {K(ω)}_ω∈Ω, ∈ D is a random absorbing set for RDS ϕ in D and ϕ is D-pullback asymptotically compact in X. Then ϕ has a unique D-random attractor {A(ω)}_ω∈Ωwhich has the following form

Aω=∩τ≥0∪t≥τϕt,θ−tω,Kθ−tω.¯(9)

Next, we shall show that system (1)-(6) generates a random dynamical system. For our purpose, we first transform this stochastic system into a deterministic dynamical with random attractor. Assume that (Ω, ℱ, ℙ) is the probability space defined in Section 1. Define (θ_t) _t∈ℝ on Ω by

θtω⋅=ω⋅+t−ωt,t∈R,

then (Ω, ℱ, ℙ, (θ_t) _t∈ℝ) is a metric dynamical system.

Set g~=(u~,v~,w~,y~)T, then the system (1)-(6) can be rewritten as follows:

dg~dt=Ag~+F~g~+σg~∘dWtdt,t>0(10)

g~0,x=g~x=u~x,v~x,w~x,y~xT,x∈D,(11)

where

A=d1Δ0000d2Δ0000d1Δ0000d2Δ,F~g~=−F+ku~+u~2v~+D1w~−u~F1−v~−u~2v~+D2y~−v~−F+kw~+w~2y~+D1u~−w~F1−y~−w~2y~+D2v~−y~.

To transform the stochastic system into a deterministic system with random parameter, we introduce the following one-dimensional Ornstein-Uhlenbeck process:

dz+zdt=dWt.(12)

From [13], we know that the stationary solution of Ornstein-Uhlenbeck process has the following form:

zθtω≡−∫−∞0esθtωsds,t∈R

Moreover, the random variable z(θ_tω) is tempered, and ℙ—a.e. ω ∈ Ω, t ↦ z(θ_tω) is continuous in t, and satisfies the properties (see [1, 11, 13]):

limt→±∞⁡zθtωt=0;limt→±∞⁡1t∫0tzθsωds=0(13)

Set (ut,vt,wt,yt)T=e−σzθtωu~t,v~t,w~t,y~tT Then, we obtain the equivalent system of (10) and (11) as:

∂u∂t=d1Δu−F+k−σzθtωu+e2σzθtωu2v+D1w−u,(14)

∂v∂t=d2Δv+Fe−σzθtω+σzθtω−Fv−e2σzθtωu2v+D2y−v,(15)

∂w∂t=d1Δw−F+k−σzθtωw+e2σzθtωw2y+D1u−w,(16)

∂y∂t=d2Δy+Fe−σzθtω+σzθtω−Fy−e2σzθtωw2y+D2v−y,(17)

that is, g = (u, v, w, y)^T satisfies

dgdt=Ag+Fg,ω+σzθtωg,t>0,(18)

g0,x=g0x=e−σzωg~0x=(u0x,v0x,w0x,y0x)T,x∈D,(19)

with

Fg,w=−F+ku+e2σzθtωu2v+D1w−uFe−σzθtω−Fv−e2σzθtωu2v+D2y−v−F+kw+e2σzθtωw2y+D1u−wFe−σzθtω−Fy−e2σzθtωw2y+D2v−y.

Notice that for ℙ—a.e. ω ∈ Ω, F(g, ω) is locally Lipschtiz continuous with respect to g. In [10], You proved that the deterministic system has a unique solution by the Galekin method. Similar to deterministic system, by the Galekin method, for ℙ—a.e. ω ∈ Ω, we can prove that for g₀ ∈ [L²(D)]⁴, (18)-(19) has a unique solution g(·, ω, g₀) ∈ C([0, ∞), [L²(D)]⁴) ∩ L²((0, ∞), [H¹(D)]⁴) with g(0, ω, g₀) = g₀. Moreover, similarly to Lemma 3 of [10], we can prove that g(t, ω, g₀) is a unique, global, weak solution with respect to g₀ ∈ [L²(D)]⁴, for t ∈ [0, ∞). This shows that (18) and (19) generate a continuous random dynamical system (φ(t))_{t ≥ 0} over (Ω, ℱ, ℙ, (θ_t) _t∈ℝ) with

φt,ω,g0=gt,ω,g0,∀t,ω,g0∈R+×Ω×[L2D]4.(20)

Now assume that ϕ : ℝ⁺ × Ω × [L²(D)]⁴ → [L²(D)]⁴ is given by

ϕt,ω,g~0=g~t,ω,g~0=gt,ω,e−σzωg0eσzθtω.(21)

Then ϕ is a continuous dynamical system associated to (10)-(11). Notice that two dynamical systems are conjugate to each other. Thus, in the following sections, we consider only the existence of a random attractor of φ.

3 Uniform estimates of solutions

To find the existence of the random attractor, we first need to obtain some uniform estimates of the solutions. Therefore in this section we first prove the uniform estimates about the solution of the two-compartment stochastic Gray-Scott equation on D, as t → + ∞. We assume that D is a collection of all tempered random subsets of [L²(D)]₄. First, we define some functions which will be used in this section. Set

Y1t,x=ut,x+vt,x+wt,x+yt,x,Y1,0=u0+v0+w0+y0;Y2t,x=ut,x+wt,x,Y2,0=u0+w0;Y3t,x=ut,x+vt,x−wt,x−yt,x,Y3,0=u0+v0−w0−y0;Y4t,x=vt,x−yt,x;Y5t,x=ut,x−wt,x.

The next lemma shows that φ has a random absorbing set in D.

Lemma 3.1

Random dynamical system φ has a random absorbing set {K(ω)}_ω∈Ω, in D, that is, for any {K(ω)}_ω∈Ω ∈ D, and for ℙ — a.e. ω ∈ Ω, there is T_B (ω) > 0 such that φ(t, θ_—tω, B(θ_—tω)) ⊂ K(ω), for any t > T_B (ω).

Proof

Taking the inner products of (15) and (17) with v and y respectively, and adding them up, we get

12ddtv2+y2+d2∇v2∇y2=F∫De−σzθtωv+ydx+∫D(σzθtω−F)v2+y2dx−∫De2σzθtωu2v2+w2y2dx−∫DD2(y−v)2dx(22)

=σzθtω−F2v2+y2−F2∫D(v−e−σzθtω)2+(y−e−σzθtω)2dx+FDe−2σzθtω−∫DD2(y−v)2dx(23)

≤σzθtω−F2v2+y2+FDe−2σzθtω.(24)

Gronwall’s inequality yields that

v2+y2≤e∫0t2σzθτωdτ−Ftv02+y02+2FDe∫0t2σzθτωdτ−Ft∫0te∫0s−2σzθτωdτ+Fs−2σzθsωds−2d2e∫0t2σzθτωdτ−Ft∫0te∫0s−2σzθτωdτ+Fs∇v2+∇y2ds.(25)

Replacing ω by θ_—tω in the above inequality, we obtain that, for all t ≥ 0,

vt,θ−tω,g0θ−tω2+yt,θ−tω,g0θ−tω2≤e∫0t2σzθτ−tωdτ−Ftv0θ−tω2+y0θ−tω2+2FDe∫0t2σzθτ−tωdτ−Ft∫0te∫0s−2σzθτ−tωdτ+Fs−2σzθs−tωds−2d2e∫0t2σzθτ−tωdτ−Ft∫0te∫0s−2σzθτ−tωdτ+Fs∇vt,θ−tω,g0θ−tω2(26)

+∇yt,θ−tω,g0θ−tω2ds.(27)

Using the properties of the Ornstein-Uhlenbeck process (13), for any g₀ (θ_—tω) ∈ B(θ_—tω), we obtain that

limt→+∞e∫0t2σzθτ−tωdτ−Fty0θ−tω2+v0θ−tω2=limt→+∞e∫−t02σzθτωdτ−Fty0θ−tω2v0θ−tω2=0,(28)

and

2FDe∫0t2σzθτ−tωdτ−Ft∫0te∫0s−2σzθτ−tωdτ+Fs−2σzθs−tωds=2FD∫0te∫st2σzθτ−tωdτ+Fs−t−2σzθs−tωds=2FD∫−t0e∫s02σzθτωdτ+Fs−2σzθsωds≤2FD∫−∞0e∫s02σzθτωdτ+Fs−2σzθsωds<+∞.(29)

Thus, there exists a T_B (ω) > 0, such that, for all t > T_B (ω),

vt,θ−tω,g0θ−tω2+yt,θ−tω,g0θ−tω2≤ρ02ω,(30)

ρ02ω=1+2FD∫−∞0e∫s02σzθτωdτ+Fs−2σzθsωds.(31)

It is easy to check that ρ02ω is a tempered random variable. To estimate u and w, we use (14), (15), (16) and (17) to get the equation for Y₁ (t, x). Now

∂Y1∂t=d1ΔY1−F+k−σzθtωY1+d2−d1Δv+y+kv+y+2Fe−σzθtω.(32)

Taking the inner product of (32) and Y₁, then apply the Holder’s inequality and Poincaré’s inequality (7), we get:

12ddtY12+d1∇Y12+F+k−σzθtωY12=∫D(d2−d1)Δv+yY1dx+∫Dkv+yY1dx+∫D2Fe−σzθtωY1dx≤d1−d2∇v+y∇Y1+kv+yY1+2FD12e−σzθtωY1≤d12∇Y12+(d1−d2)22d1∇v+y2(33)

+k2Y12+k2γ∇v+y2+F2Y12+2FDe−2σzθtω.(34)

Therefore,

ddtY12+d1∇Y12+F−2σzθtωY12≤(d1−d2)2d1+kγ∇v+y2+4FDe−2σzθtω.

Applying Gronwall’s inequality, we get that

Y12≤e∫0t2σzθτωdτ−FtY1,02+(d1−d2)2d1+kγe∫0t2σzθτωdτ−Ft∫0te∫0s−2σzθτωdτ+Fs∇v+y2ds+4FDe∫0t2σzθτωdτ−Ft∫0te∫0s−2σzθτωdτ+Fs−2σzθsωds−d1e∫0t2σzθτωdτ−Ft∫0te∫0s−2σzθτωdτ+Fs∇Y12ds.(35)

By replacing ω by θ_—tω in the above inequality, it follows that

Y1t,θ−tω,g0θ−tω2≤e∫0t2σzθτ−tωdτ−FtY1,0θ−tω2+(d1−d2)2d1+kγe∫0t2σzθτ−tωdτ−Ft×∫0te∫0s−2σzθτ−tωdτ+Fs∇v+ys,θ−tω,g0θ−tω2ds+4FDe∫0t2σzθτ−tωdτ−Ft∫0te∫0s−2σzθτ−tωdτ+Fs−2σzθs−tωds−d1e∫0t2σzθτ−tωdτ−Ft∫0te∫0s−2σzθτ−tωdτ+Fs∇Y1s,θ−tω,g0θ−tω2ds.(36)

Similar to (28), we have, for any g₀ (θ_—tω) ∈ B(θ_—tω),

limt→+∞e∫0t2σzθτ−tωdτ−FtY1,0θ−tω2=limt→+∞e∫−t02σzθτωdτ−FtY1,0θ−tω2=0.(37)

By (27)-(30), we obtain that, for all t ≥ T_B (ω)

2d2e∫0t2σzθτ−tωdτ−Ft∫0te∫0s−2σzθτ−tωdτ+Fs∇vs,θ−tω,g0θ−tω2+∇ys,θ−tω,g0θ−tω2ds≤ρ02ω.(38)

Let c1=12(d1−d2)2d1d2+kγd2 then the above inequality yields that for all t ≥ T_B (ω),

(d1−d2)2d1+kγe∫0t2σzθτ−tωdτ−Ft∫0te∫0s−2σzθτ−tωdτ+Fs∇vs,θ−tω,g0θ−tω2+∇ys,θ−tω,g0θ−tω2ds≤c1ρ02ω.(39)

Therefore (29), (36) and (39) imply that

Y1t,θ−tω,g0θ−tω2≤c1+2ρ02ω.(40)

Hence, we obtain from (30) and (40) that

Y2t,θ−tω,g0θ−tω2=Y1t,θ−tω,g0θ−tω−v+yt,θ−tω,g0θ−tω2≤Y1t,θ−tω,g0θ−tω2+v+yt,θ−tω,g0θ−tω22Y1t,θ−tω,g0θ−tω2+4vt,θ−tω,g0θ−tω2+4yt,θ−tω,g0θ−tω2≤2C1+8ρ02ω.(41)

It is easy to check that Y₃ (t, x) satisfies the following equation

∂Y3∂t=d1ΔY3−F+k−σzθtωY3+2D1w−u+2D2y−v+kv−y+d2−d1Δv−y=d1ΔY3−F+k+2D1−σzθtωY3+k+2D1−2D2v−y+d2−d1Δv−y.(42)

Taking the inner product of the last equation with Y₃, we obtain that

12ddtY32+d1∇YL32+F+k+2D1−σzθtωY32=∫D(k+2D1−2D2)v−yY3dx−∫D(d2−d1)∇v−y∇Y3dx≤k+2D1−2D2v−yY3+d1−d2∇v−y∇Y3≤(d1−d2)22d1+(k+2D1−2D2)22γF+k+2D1∇v−y2+d12∇Y32+F+k+2D12Y32.(43)

In the last step we used Poincaré’s inequality (7). Thus,

ddtY32+d1∇Y32+F−2σzθtωY32≤c2∇v2+∇y2

with c2=2(d1−d2)2d1+2(k+2D1−2D2)2γF+k+2D1. It follows from Gronwall’s inequality that

Y32≤e∫0t2σzθτωdτ−FtY3,02−d1e∫0t2σzθτωdτ−Ft∫0te∫0s−2σzθτωdτ+Fs∇Y32ds

+c2e∫0t2σzθτωdτ−Ft∫0te∫0s−2σzθτωdτ+Fs∇v2+∇y2ds.(44)

Replacing ω by θ_—tω, we find that, for all t ≥ 0,

Y3t,θ−tω,g0θ−tω2≤e∫0t2σzθτ−tωdτ−FtY3,0θ−tω2−d1e∫0t2σzθτ−tωdτ−Ft∫0te∫0s−2σzθτ−tωdτ+Fs∇Y3t,θ−tω,g0θ−tω2ds+c2e∫0t2σzθτ−tωdτ−Ft∫0te∫0s−2σzθτ−tωdτ+Fs∇vt,θ−tω,g0θ−tω2+∇yt,θ−tω,g0θ−tω2ds.(45)

Similar to (28) and (39), we obtain that

limt→+∞e∫0t2σzθτ−tωdτ−FtY3,0θ−tω2=0(46)

and

c2e∫0t2σzθτ−tωdτ−Ft∫0te∫0s−2σzθτ−tωdτ+Fs∇vt,θ−tω,goθ−tω2+∇yt,θ−tω,g0θ−tω+∇yt,θ−tω,g0θ−tω2ds≤C22d2ρ02ω.(47)

Set c3=c22d2+1, then by (45)-(47), one has that, for all t ≥ T_B (ω),

Y3t,θ−tω,g0θ−tω2≤c3ρ02ω.(48)

It follows from (30) and (48) that, for all t ≥ T_B (ω),

Y5t,θ−tω,g0θ−tω2=Y3t,θ−tω,g0θ−tω−Y4t,θ−tω,g0θ−tω2≤2Y3t,θ−tω,g0θ−tω2+Y4t,θ−tω,g0θ−tω2≤2Y3t,θ−tω,g0θ−tω2+4vt,θ−tω,g0θ−tω2+yt,θ−tω,g0θ−tω2=c4ρ02ω(49)

with c₄ = 2c₃ + 4. Consequently, we obtain that, for all t ≥ T_B (ω),

ut,θ−tω,g0θ−tω2+wt,θ−tω,g0θ−tω2=12Y2t,θ−tω,g0θ−tω+Y5t,θ−tω,g0θ−tω2+12Y2t,θ−tω,g0θ−tω−Y5t,θ−tω,g0θ−tω2≤Y2t,θ−tω,g0θ−tω2+Y5t,θ−tω,g0θ−tω2≤c5ρ02ω(50)

with c₅ = 2c₁ + 8 + c₄. We finally obtain that, for all t ≥ T_B (ω),

gt,θ−tω,g0θ−tω≤c5+1ρ02ω.(51)

It is easy to check that ρ02ω is tempered. This ends the proof.

□

Lemma 3.2

There exists a random variableρ₁ (ω), such that, for anyB(ω) ∈ D, and g₀ (ω) ∈ B(ω), for ℙ—a.e. ω∈ Ω, there is aT_B (ω) > 0, such that, for any t ≥ T_B (ω), the following estimate holds

vt,θ−tω,g0θ−tωL66+yt,θ−tω,g0θ−tωL66≤ρ1ω.(52)

Proof

Because the unique global weak solution of system (18)-(19) satisfies g(t, ω, g₀ (ω)) ∈ C((0, ∞), [L²(D)]⁴) ∩ L²((0, ∞), [H¹(D)]⁴), then, for any initial value g₀ ∈ [L²(D)]⁴, there exists a small time t₀ ∈ (0, 1) such that,

gt0,ω,g0ω∈H1D]4→∗L6D]4.(53)

This means that the weak solution g(t, ω, g₀ (ω)) becomes a strong solution on [t₀, +∞) and satisfies g(t, ω, g₀ (ω)) ∈ C([t₀, ∞), [H¹(D)]⁴) ∩ L²([t₀, ∞), [H²(D)]⁴) ⊂ C([t₀, ∞), [L⁶(D)]⁴). Thus, without loss of generality, we can assume that g₀ ∈ [L⁶(D)]⁴. Taking the inner products of (15) and (17) with v⁵ and y⁵ respectively, and adding them up, we obtain

16ddtvL66+yL66+5d2v2∇v2+y2∇y2=F∫De−σzθtωv5+y5−y6+v6dx+∫Dσzθtωv6+y6dx−∫De2σzθtωu2v6+w2y6dx+D2∫D(y−v)v5−y5dx.(54)

We now estimate all terms on the right hand side of (54). For the fourth term, by Young’s inequality, we get that

∫D(y−v)v5−y5dx=∫D−v6−y6+yv5+vy5dx≤∫D−v6−y6+16y6+56v6+16v6+56y6dx=0.(55)

For the first term, by Young’s inequality,

F∫De−σzθtωv5+y5−y6+v6dx≤F∫D56v6+y6+13e−6σzθtω−y6+v6dx≤−F6vL66+yL66+F3De−6σzθtω.(56)

By (54)-(56), we arrive at the following estimate, for all t ≥ 0

ddtvL66+yL66≤6σzθtω−FvL66+yL66+2FDe−6σzθtω.(57)

By Gronwall’s inequality, we obtain that, for all t ≥ 0,

vL66+yL66≤e∫0t6σzθτωdτ−Ftv0L66+y0L66+2FDe∫0t6σzθτωdτ−Ft∫0te∫0s−6σzθτωdτ+Fs−6σzθsωds.(58)

Replacing ω by θ_—tω in the above inequality, we have that, for all t ≥ 0,

vt,θ−tω,g0θ−tωL66+yt,θ−tω,g0θ−tωL66

≤e∫0t6σz(θτ−tω)dτ−Ft(∥v0(θ−tω)∥L66+∥y0(θ−tω)∥L66)+2F|D|e∫0t6σz(θτ−tω)dτ−Ft∫0te∫0s−6σz(θτ−tω)dτ+Fs−6σz(θs−tω)ds=e∫0t6σz(θτω)dτ−Ft(∥v0(θ−tω)∥L66+∥y0(θ−tω)∥L66)+2F|D|∫−t0e(∫s0−6?z(θ−tω)dτ+Fs−6σz(θsω))ds.(59)

For g0(θ−tω)∈B(θ−tω), the properties of the Ornstein-Uhlenbeck process implies that, for all t≥TB(ω)

limt→+∞e∫−t06σz(θτω)dτ−Ft(||v0(θ−tω)||L66+||y0(θ−tω)||L66)=0,(60)

and

2F|D|∫−t0e∫s0−6σz(θτω)dτ+Fs−6σz(θsω)ds≤2F|D|∫−∞0e∫s0−6σz(θτω)dτ+Fs−6σz(θsω)ds<∞.(61)

Set

ρ1(ω)≡1+2F|D|∫−∞0e∫s0−6σz(θτω)dτ+Fs−6σz(θsω)ds.(62)

Then there exists a T_B(ω) > 0, independent of σ, such that for all t > T_B(ω),

||υ(t, θ−tω,g0(θ−tω))||L66+||y(t,θ−tω,g0(θ−tω))||L66≤ρ1(ω).(63)

□

Lemma 3.3

There exists a random variable ρ₂(ω) > 0 such that, for anyB(ω)∈D,G0(ω)∈B(ω), for ℙ – a.e. ω ∈ Ω there exists a T_B(ω) > 0 such that, for all t ≥ T_B(ω), the following estimate holds,

∫tt+1||∇g(s,θ−t−1ω,g0(θ−t−1ω))||2ds≤ρ2(ω).(64)

Proof

Using (27), (28) and (29), we obtain:

2d2e∫0t2σz(θτ−tω)dτ−Ft∫0te∫0t2σz(θτ−tω)dτ−Fs(||∇υ(s,θ−tω,g0(θ−tω))+||∇y(s,θ−tω,g0(θ−tω)||2)ds≤ρ02(ω).(65)

Setting t = t + 1, we have:

2d2e∫0t+12σz(θτ−tω)dτ−F(t+1)∫0t+1e∫0s2σz(θτ−tω)dτ−Fs(||∇υ(s,θ−t−1ω,g0(θ−t−1ω))||2+||∇y(s,θ−t−1ω,g0(θ−t−1ω)||2)ds≥2d2∫tt+1e∫st+12σz(θτ−t−1ω)dτ+F(s−t−1)(||∇υ(s,θ−t−1ω,g0(θ−t−1ω))||2+||∇y(s,θ−t−1ω,g0(θ−t−1ω))||2ds≥2d2e−2σmax−1≤τ≤0|z(θτω)|−F∫tt+1||∇v(s,θ−t−1ω,g0(θ−t−1ω))||2+∇y(s,θ−t−1ω,g0(θ−t−1ω))||2ds.

Let T_B(ω) be the constant defined in Lemma 3.1. It follows that, for all t ≥ T_B(ω),

∫tt+1||∇v(s,θ−t−1ω,g0(θ−t−1ω))||2+||∇y(s,θ−t−1ω,g0(θ−t−1ω)||2ds≤c6ρ02(ω),(66)

with c6=12d2e2σmax−1≤τ≤0 |z(θτω)|+F. Similarly, by (36)-(40) and (45)-(48), we get:

∫tt+1||∇Y1(s,θ−t−1ω,g0(θ−t−1ω))||2ds≤c7ρ02(ω)∫tt+1||∇Y3(s,θ−t−1ω,g0(θ−t−1ω))||2ds≤c8ρ02(ω)

where c7=c1+2d1e2σmax−1≤τ≤0 |z(θτω)|+F,c8=c3d1e2σmax−1≤τ≤0 |z(θτω)|+F. Set c₉ = 2c₇ + 4c₆, and c₁₀ = 2c₈ + 4c₆. It follows that, for all t ≥ T_B(ω),

∫tt+1||∇Y1(s,θ−t−1ω,g0(θ−t−1ω))||2ds∫tt+1||∇Y1(s,θ−t−1ω,g0(θ−t−1ω))||2−∇(v+y)(s,θ−t−1ω,g0(θ−t−1ω))||2ds≤∫tt+12||∇Y1(s,θ−t−1ω,g0(θ−t−1ω))||2+4(||∇v(s,θ−t−1ω,g0(θ−t−1))||2 +||∇y(s,θ−t−1ω,g0(θ−t−1ω))||2)ds≤c9ρ02(ω),(67)

and

∫tt+1||∇Y5(s,θ−t−1ω,g0(θ−t−1ω))||2ds∫tt+1||∇Y3(s,θ−t−1ω,g0(θ−t−1ω))−∇Y4(s,θ−t−1ω,g0(θ−t−1ω))||2ds≤2∫tt+1||∇Y3(s,θ−t−1ω,g0(θ−t−1ω))||2+||∇Y4(s,θ−t−1ω,g0(θ−t−1ω))||2ds≤2∫tt+1||∇Y3(s,θ−t−1ω,g0(θ−t−1ω))||2+2(∥∇v(s,θ−t−1ω,g0(θ−t−1ω))||2ds +||∇y(s,θ−t−1ω,g0(θ−t−1ω))||2)ds≤c9ρ02(ω).(68)

Therefore, by (67) and (68), we get

∫tt+1∥∇u(s,θ−t−1ω,g0(θ−t−1ω))∥2+∥∇w(s,θ−t−1ω,g0(θ−t−1ω))∥2ds=14∫tt+1∥∇Y2(s,θ−t−1ω,g0(θ−t−1ω))+∇Y5(s,θ−t−1ω,g0(θ−t−1ω))∥2 +∥∇Y2(s,θ−t−1ω,g0(θ−t−1ω))−∇Y5(s,θ−t−1ω,g0(θ−t−1ω))∥2ds≤∫tt+1∥∇Y2(s,θ−t−1ω,g0(θ−t−1ω))∥2+∥∇Y5(s,θ−t−1ω,g0(θ−t−1ω))∥2ds≤(c9+c10)ρ02(ω).(69)

Finally, by (66) and (69), we obtain that, for all t ≥ T_B(ω),

∫tt+1∥∇g(s,θ−t−1ω,g0(θ−t−1ω))∥2≤(c6+c9+c10)ρ02(ω)≡ρ2(ω).(70)

□

Lemma 3.4

There exists a random variable ρ₃(ω), such that, for anyB(ω)∈D, andg₀(ω), ∈ B(ω), for ω – a.e. ω ∈ Ω there is a T_B(ω) > 0, such that for all t ≥ T_B(ω), the following estimate holds,

∥∇u(t,θ−tω,g0(θ−tω))∥2+∥∇w(t,θ−1ω,g0(θ−1ω)∥2≤ρ3(ω).

Proof

Taking the inner products of (14) and (16) with –Δu and –Δw respectively, and then summing them up, we obtain that,

12ddt(∥∇u∥2+∥∇w∥2)+d1(∥Δu∥2+∥Δw∥2)+(F+k−σz(θtω))(∥∇u∥2+∥∇w∥2)=−∫De2σz(θtω)(u2υΔu+w2yΔw)dx+D1∫D(w−u)(Δw−Δu)dx.(72)

Now, we estimate each term on the right hand side of (72). For the second term,

∫D(w−u)(Δw−Δu)dx=−∫D(∇w−∇u)2dx≤0.(73)

For the first term, by Hölder’s inequality and (8), we have that

− ∫De2σz(θtω)(u2υΔu+w2yΔw)dx≤d1(∥Δu∥2+∥Δw∥2)+14d1e4σz(θtω)∫Du4υ2+w2y2dx≤d1(∥Δu∥2+∥Δw∥2)+14d1e4σz(θtω)(∥u∥L64∥υ∥L62+∥w∥L64∥y∥L62)≤d1(∥Δu∥2+∥Δw∥2)+12d1η2e4σz(θtω)((∥u∥4+∥∇u∥4)∥υ∥L62+(∥w∥4+∥∇w∥4)∥y∥L62).(74)

It follows from (72)-(74) that

ddt(∥∇u∥2+∥∇w∥2)+2(F+k−σz(θtω))(∥∇u∥2+∥∇w∥2)≤1d1η2e4σz(θtω)[(∥u∥4+∥∇u∥4)∥υ∥L62+(∥ω∥4+∥∇ω∥4)∥y∥L62]≤1d1η2e4σz(θtω)(∥u∥2+∥ω∥2)2(∥υ∥L62+∥y∥L62)+1d1η2e4σz(θtω)(∥υ∥L62+∥y∥L62)(∥∇u∥2+∥∇ω∥2)2.(75)

Hence, for all t ≥ 0

ddt(∥∇u∥2+∥∇ω∥2)≤[1d1η2e4σz(θtω)(∥υ∥L62+∥y∥L62)(∥∇u∥2+∥∇ω∥2)−2(F+k−σz(θtω))](∥∇u∥2+∥∇ω∥2)+1d1η2e4σz(θtω)(∥u∥2+∥ω∥2)2(∥υ∥L62+∥y∥L62).(76)

Replacing t by s, and replacing ω by θ-t-1 ω in the last inequality, for every s < 0, we obtain that

dds(∥∇u(s,θ−t−1ω,g0(θ−t−1ω))∥2+∥∇ω(s,θ−t−1ω,g0(θ−t−1ω))∥2)≤[1d1η2e4σz(θs−t−1ω)(∥υ(s,θ−t−1ω,g0(θ−t−1ω))∥L62+∥y(s,θ−t−1ω,g0(θ−t−1ω))∥L62) ×(∥∇u(s,θ−t−1ω,g0(θ−t−1ω))∥2+∥∇ω(s,θ−t−1ω,g0(θ−t−1ω))∥2)−2(F+k−σz(θs−t−1ω))] ×(∥∇u(s,θ−t−1ω,g0(θ−t−1ω))∥2+∥∇ω(s,θ−t−1ω,g0(θ−t−1ω))∥2) +1d1η2e4σz(θs−t−1ω)(∥u(s,θ−t−1ω,g0(θ−t−1ω))∥2+∥ω(s,θ−t−1ω,g0(θ−t−1ω))∥2)2 ×(∥υ(s,θ−t−1ω,g0(θ−t−1ω))∥L62+∥y(s,θ−t−1ω,g0(θ−t−1ω))∥L62),(77)

which can be rewritten as

dpds≤β(s)ρ(s)+α(s),(78)

with

ρ(s)=∥∇u(s,θ−t−1ω,g0(θ−t−1ω))∥2+∥∇ω(s,θ−t−1ω,g0(θ−t−1ω))∥2;β(s)=1d1η2e4σz(θs−t−1ω)(∥υ(s,θ−t−1ω,g0(θ−t−1ω))∥L62+∥y(s,θ−t−1ω,g0(θ−t−1ω))∥L62) ×(∥∇u(s,θ−t−1ω,g0(θ−t−1ω))∥2+∥∇ω(s,θ−t−1ω,g0(θ−t−1ω))∥2)−2(F+k−σz(θs−t−1ω));α(s)=1d1η2e4σz(θs−t−1ω)(∥u(s,θ−t−1ω,g0(θ−t−1ω))∥2+∥ω(s,θ−t−1ω,g0(θ−t−1ω))∥2)2 ×(∥∇υ(s,θ−t−1ω,g0(θ−t−1ω))∥L62+∥y(s,θ−t−1ω,g0(θ−t−1ω))∥L62).(79)

We use uniform Gronwall inequality to estimate

∥∇u(s,θ−t−1ω,g0(θ−t−1ω))∥2+∥∇ω(s,θ−t−1ω,g0(θ−t−1ω))∥2.

We need to compute ∫^t+1_tβ(s)ds and ∫^t+1_tα(s)ds first. By (50) and (52), we have that, for t < T_B(ω)

∥u(s,θ−t−1ω,g0(θ−t−1ω))∥2+∥ω(s,θ−t−1ω,g0(θ−t−1ω))∥2∥u(s,θ−s(θs−t−1ω),g0(θ−s(θs−t−1ω)))∥2+∥ω(s,θ−s(θs−t−1ω),g0(θ−s(θs−t−1ω)))∥2≤c5ρ02(θs−t−1ω).(80)

and

∥υ(s,θ−t−1ω,g0(θ−t−1ω))∥L62+∥y(s,θ−t−1ω,g0(θ−t−1ω))∥L62=∥u(s,θ−s(θs−t−1ω),g0(θ−s(θs−t−1ω)))∥L62+∥ω(s,θ−s(θs−t−1ω),g0(θ−s(θs−t−1ω)))∥L62≤c11ρ11/3(θs−t−1ω).(81)

It follows from (80) and (81) that

∫tt+1α(s)ds≤c52c11d1η2∫tt+1e4σ|z(θτω)|ρ04σ(θs−t−1ω)ρ11/3(θs−t−1ω)ds ≤c52c11d1η2max−1≤τ≤0[e4σ|z(θτω)|ρ04σ(θτω)ρ11/3(θτω)]≡M1(ω).(82)

Next, we use Lemma 4.3 to estimate ∫tt+1β(s)ds.

Therefore, applying uniform Gronwall inequality, we can get that

∥∇u(t+1,θ−t−1ω,g0(t+1,θ−t−1ω))∥2+∥∇w(t+1,θ−t−1ω,g0(t+1,θ−t−1ω))∥2≤(M1(ω)+ρ2(ω))eM2(ω)≡ρ3(ω).(84)

□

Lemma 3.5

There exists a random variableρ₄(ω), such that, for anyB(ω)∈Dandg₀(ω), for ω – a.e. ω ∈ Ω there is a T_B(ω) > 0, such that for all t ≥ T_B(ω), the following estimate holds,

∥∇υ(t,θ−tω,g0(θ−tω))∥2+∥∇y(t,θ−tω,g0(θ−tω))∥2≤ρ4(ω)(85)

Proof

Taking the inner products of (15) and (17) with –Δu and –Δw respectively, and then summing them up, we obtain that,

12ddt(∥∇υ∥2+∥∇y∥2)+d2(∥∇υ∥2+∥∇y∥2)=−∫DFe−σz(θtω)(Δυ+Δy)dx+(σz(θtω)−F)(∥∇υ∥2+∥∇y∥2) +e2σz(θtω)∫Du2υΔυ+w2yΔydx−D2∫D(y−υ)Δ(υ−y)dx.(86)

Now we estimate each term on the right hand side of (86). For the first and fourth term, by Green’s formula,

∫DFe−σz(θtω)(Δυ+Δy)dx=0, −D2∫D(y−υ)Δ(υ−y)dx=−∫D∥∇(υ−y)∥2dx≤0.

As for the third term,

e2σz(θtω)∫Du2υΔυ+w2yΔydx≤d2(∥Δυ∥2+∥Δy∥2)+e4σz(θtω)4d2∫Du4υ2+w4y2dx.(87)

It follows from Young’s inequality that

ddt(∥∇υ∥2+∥∇y∥2)+2(F−σz(θtω))(∥∇υ∥2+∥∇y∥2)≤e4σz(θtω)2d2(∥u∥L64∥υ∥L62+∥ω∥L64∥y∥L62)≤e4σz(θtω)2d2η2(∥u∥H14∥υ∥L62+∥ω∥H14∥y∥L62)≤e4σz(θtω)d2η2[(∥u∥4+∥∇u∥4)∥υ∥L62+(∥ω∥4+∥∇ω∥4)∥y∥L62]≤e4σz(θtω)d2η2[(∥u∥2+∥ω∥2)2+(∥∇u∥2+∥∇ω∥2)2](∥υ∥L62+∥y∥L62).(88)

Replacing t by s and replacing ω by θ_-t-1ω in the last inequality, we obtain:

ddt(∥∇υ(s,θ−t−1ω,g0(θ−t−1ω))∥2+∥∇y(s,θ−t−1ω,g0(θ−t−1ω))∥2) +2(F−σz(θs−t−1ω))(∥∇υ(s,θ−t−1ω,g0(θ−t−1ω))∥2+∥∇y(s,θ−t−1ω,g0(θ−t−1ω))∥2)≤e4σz(θs−t−1ω)d2η2[(∥u(s,θ−t−1ω,g0(θ−t−1ω))∥2+∥ω(s,θ−t−1ω,g0(θ−t−1ω))∥2)2 +(∥∇u(s,θ−t−1ω,g0(θ−t−1ω))∥2+∥∇ω(s,θ−t−1ω,g0(θ−t−1ω))∥2)2] ×(∥υ(s,θ−t−1ω,g0(θ−t−1ω))∥L62+∥y(s,θ−t−1ω,g0(θ−t−1ω))∥L62).

It can be rewritten as

dρds≤β(s)ρ(s)+α(s)(89)

with

ρ(s)=∥∇υ(s,θ−t−1ω,g0(θ−t−1ω))∥2+∥∇y(s,θ−t−1ω,g0(θ−t−1ω))∥2;β(s)=−2(F−σz(θs−t−1ω));α(s)=e4σz(θs−t−1ω)d2η2[(∥u(s,θ−t−1ω,g0(θ−t−1ω))∥2+∥ω(s,θ−t−1ω,g0(θ−t−1ω))∥2)2 +(∥∇u(s,θ−t−1ω,g0(θ−t−1ω))∥2+∥∇ω(s,θ−t−1ω,g0(θ−t−1ω))∥2)2] ×(∥υ(s,θ−t−1ω,g0(θ−t−1ω))∥L62+∥y(s,θ−t−1ω,g0(θ−t−1ω))∥L62).(90)

Similarly to Lemma 4.4, we need to estimate ∫tt+1α(s)ds,∫tt+1β(s)ds and ∫tt+1ρ(s)ds.

∫tt+1β(s)ds≤2(F+αmax−1≤τ≤0|z(θτω)|)≡M3(ω)(91)

By (80), (81) and Lemma 4.4, we have that, for t>TB(ω),

α(s)≤c11d2η2e4σ|z(θs−t−1ω)|[c52ρ04(θs−t−1ω)+ρ32(θs−t−1ω)]ρ11/3(θs−t−1ω)(92)

It follows that

∫tt+1α(s)ds≤c11d2η2max−1≤τ≤0[e4σ|z(θτω|(c52ρ04(θτω)+ρ32(θτω))ρ11/3(θτω)]≡M4(ω)(93)

By using uniform Gronwall inequality and (66), (93) and (91), we get that

∥∇υ(t+1,θ−t−1ω,g0(θ−t−1ω))∥2+∥∇y(t+1,θ−t−1ω,g0(θ−t−1ω))∥2≤(M4(ω)+ρ2(ω))eM3(ω)≡ρ4(ω)(94)

□

4 Existence of random attractors

In this section we use Proposition 2.1 to prove the existence of a pullback attractor.

Theorem 4.1

The random dynamical system ϕ has a unique D–pullback random attractor in [L²(D)]⁴.

Proof

In Lemma 3.1, we find that the system has a bounded absorbing set. By Lemma 3.4 and Lemma 3.5, we know that for any g0(θ−tω)∈B0(θ−tω), the weak solution g0(t,θ−tω,g0(θ−tω)) is bounded in [H¹(D)]⁴. Since the embedding [H1(D)]4→[L2(D)]4 is a compact mapping, This shows that the random dynamical system φ is asymptotically compact. Hence, by Proposition 2.1, we obtain the existence of a unique D-pullback random attractor {A1(ω)}ω∈Ω for φ in [L²(D)]⁴.

Since ϕ and φ are conjugated by the random homeomorphism T(ω,ξ)=eσz(θtω)ξ(ω), then by Proposition 1.8.3 in [17],ϕ has a unique D-random attractor {A2(ω)}ω∈Ω in [L²(D)]⁴ which is given by

A2(ω)={eσz(θtω)ξ(ω):ξ(ω)∈A1(ω)}(95)

□

Competing interests

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

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Received: 2016-2-4

Accepted: 2016-6-25

Published Online: 2016-9-8

Published in Print: 2016-1-1

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