# Random Attractors for Stochastic Retarded 2D-Navier-Stokes Equations with Additive Noise

Authors: Xiaoyao Jia; Xiaoquan Ding
Journal: Journal of Function Spaces
 (2018)

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## Abstract

In this paper, the existence and the upper semicontinuity of a pullback attractor for stochastic retarded 2D-Navier-Stokes equation on a bounded domain are obtained. We first transform the stochastic equation into a random equation and then obtain the existence of a random attractor for random equation. Then conjugation relation between two random dynamical systems implies the existence of a random attractor for the stochastic equation. At last, we get the upper semicontinuity of random attractor.


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## Body

## 1. Introduction

In this paper, we study the asymptotic behavior of the solutions for stochastic retarded Navier-Stokes equations with additive noise on a bounded domain inR2:(1)du-μΔudt+u·∇udt=fx+gx,ut-hdt+ϵ∑j=1mhjdwjin0,+∞×Dwhere D is a bounded set in R2, μ is a positive constant, h is the delay time of the system, f,g,hi(i=1,2,…,m) are given functions, ϵ>0 is the strength of noise, and wi(i=1,2,…,m) are two-sided Wiener processes on a probability space which will be specified later.Navier-Stokes equations are basic equations of describing uncompressible viscous fluids, which are very important problems in fluid mechanics. Hence, Navier-Stokes equations have attracted many authors’ attention. As all of us know, the asymptotic behavior of dynamical systems, which can be studied by attractor, is an important problem. Regarding the asymptotic profile of the solution, using similarity variables, some interesting results can be found in [1, 2]. The random attractor for stochastic system was introduced in [3, 4] as an extension of the attractor for deterministic system in [5]. The attractors for dynamical systems have been studied by many authors (see [3, 4, 6–19]). The attractors are also used to study deterministic Navier-Stokes equations (see [9, 20–22]) and stochastic Navier-Stokes equations (see [4, 23, 24]). For the stochastic Navier-Stokes equations with additive noise and without time delay, the pullback attractors have been investigated in [23, 24]. However, as far as we know, there is no result on the random attractors for the stochastic retarded Navier-Stokes equation on bounded domains. So the main task of this paper is to study the asymptotic behavior of random attractor for stochastic Navier-Stokes equation (1).To study the asymptotic behavior of random attractors, we shall establish a random dynamical system for (1) first. Then, we prove the existence and uniqueness of pullback random attractor. To achieve this, we derive the uniform estimates of solutions for Navier-Stokes equation with delays and additive noise. Finally, we study the convergence of random attractors as the noise intensity goes to zero.The paper is organized as follows. In Section2, we recall some notations and results on pullback attractors for random dynamical systems. And we show that the stochastic retarded Navier-Stokes equations generate a random dynamical system. In Section 3, we obtain some uniform estimates of solutions for system (14). 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.


## 2. RDS Generated by Stochastic 2D-Navier-Stokes Equation

In this section, we introduce some notations and recall some results about random attractors for random dynamical systems (RDS). The readers can refer to [3, 4, 6, 8, 10] for more details.Suppose thatϕ is a continuous RDS of X over (Ω,F,P,(θt)t∈R). And suppose D is a collection of subsets of X. We refer to [4, 7] for the existence of random attractor for continuous RDS.Proposition 1.
Let{K(ω)}ω∈Ω∈D be a random absorbing set for the continuous RDS ϕ in D and ϕ be D-pullback asymptotically compact in X. Then ϕ has a unique D-random attractor {A(ω)}ω∈Ω which is given by(2)Aω=⋂τ≥0⋃t≥τϕt,θ-tω,Kθ-tω¯.Next, we refer to [19] for the upper semicontinuity of random attractors.Proposition 2.
Suppose the following conditions are satisfied.
(1) For givenσ>0, suppose ϕσ is a random dynamical system over a metric system (Ω,F,P,(θt)t∈R). For P-a.e. ω∈Ω, t≥0, σn→0, and xn,x∈X with xn→x, there holds(3)limn→+∞ϕσnt,ω,xn=ϕt,x.
(2) Assume thatϕσ has a random attractor Aσ={Aσ(ω)}ω∈Ω∈D and a random absorbing set Eσ={Eσ(ω)}ω∈Ω∈D such that for some deterministic positive constant c and for P-a.e. ω∈Ω,(4)limsupσ→0EσωX≤c,with Eσ(ω)X=supx∈Eσ(ω)xX.
(3) There existsσ0>0 such that for P-a.e. ω∈Ω, ⋃0<σ≤σ0Aσ(ω) is precompact in X.
Then, forP-a.e. ω∈Ω,(5)distAσω,A0→0,asσ→0.In the rest of this paper, we use the following notation. LetD⊂R2 be an open bounded set with regular boundary Γ. Set(6)V=u∈C0∞D2:divu=0,and let H be the closure of V in (L2(D))2 with norm · and inner product (·,·). Let V be the closure of V in (H01(D))2 with norm ·V, and inner product (·,·)V, where for u,v∈(H01(D))2,(7)u,vV=∑i,j=12∫D∂uj∂xi∂vj∂xidx.Let H′ and V′ be dual space of H and V. Then one has V⊂H≡H′⊂V′, and the injections are dense and compact. Let 〈·,·〉 be the duality pairing between V and V′. For u,v∈(H2(D))2, define A:V→V′, by(8)Au,v=u,vV.Let D(A)={u∈(H2(D))2:Au∈H}. Then Au≔-PΔu, u∈D(A), and P is a orthogonal projection from (L2(D))2 to H. Next, we define a trilinear form b on V×V×V by(9)bu,v,w=∑i,j=12∫Dui∂vj∂xiwjdx∀u,v,w∈V.Then, b is a continuous bilinear form and satisfies the following condition (see [5, 22]):(10)bu,v,v=0,foru∈V,v∈H01D2,(11)bu,v,w=-bu,w,v,foru∈V,v,w∈H01D2,(12)bu,v,w≤Cu1/2∇u1/2∇v1/2Av1/2wu∈V,v∈DA,w∈H;u1/2Au1/2∇vwu∈DA,v∈V,w∈H;u∇vw1/2∇w1/2u∈H,v∈V,w∈DA;u1/2∇u1/2∇vw1/2∇w1/2u,v,w∈V.By the Poincaré inequality and the homogeneous Dirichlet boundary, there is a constant γ>0 such that(13)∇ϕ2≥γϕ2,forϕ∈H01DorH01D2.The norm of a general Banach space X is denoted by ·X. For h>0, we denote by ut the function defined on [-h,0] by the relation ut(s)=u(t+s), s∈[-h,0]. Let S denote the Banach space of all continuous functions ξ:[-h,0]→(L2(D))2 endowed with the supremum norm ξS=sups∈[-h,0]ξ(s).In this section, we consider the following stochastic Navier-Stokes equation onD with additive white noise:(14)du-μΔudt+u·∇udt=fx+gx,ut-hdt+ϵ∑j=1mhjdwjin0,+∞×D;divu=0in0,+∞×D;u=0on0,+∞×Γ;ut,x=u0t,xx∈D,t∈-h,0.Here μ is a positive constant, h>0 is the delay time of the system, {wj}j=1m are independent two-sided Wiener processes defined on a probability space on a probability space (Ω,F,P), with(15)Ω=ω=ω1,ω2,…,ωm∈CR,Rm:ω0=0,the Borel σ-algebra F on Ω is generated by the compact open topology [6] and P is the corresponding Wiener measure on F; and f is a given function in [L2(D)]2 and hj(1≤j≤m) are given functions in [H02(D)]2 with divhj=0, and g is a function that satisfies the following conditions.(A1)
g:D×R2→R2 is continuous with g(·,0)=0, and, for all x∈D, and ξ,η∈R2,(16)gx,ξ2≤cg2ξ2+φx2,φx∈L2D2;(17)gx,ξ-gx,η≤lgξ-η;(A2)
μ>cgγ-1.Let(18)θtω·=ω·+t-ωt,t∈R.Then(Ω,F,P,(θt)t∈R) is an ergodic metric dynamical system. Notice that the above probability space is canonical; one has(19)wt,ω=ωt,wt,θsω=wt+s,ω-ws,ω.To study the random attractor for system (14), we first transform that system into a deterministic system with random parameter. For j=1,2,…,m, we introduce the Ornstein-Uhlenbeck process:(20)zjθtωj≡-λ∫-∞0eλsθtωjsds,t∈R.Then, one has that zj(θtωj) solves the following equation (see [11] for details):(21)dzj+λzjdt=dwjt.Moreover, the random variable zj(θtωj) is tempered, and zj(θtωj) is P-a.e. continuous. It follows from Proposition 4.3.3 [6] that there exists a tempered function r(ω)>0 such that(22)∑j=1mzjωj+zjωj2+zjωj3+zjωj4≤rω,where r(ω) satisfies, for P-a.e. ω∈Ω,(23)rθtω≤eα/2trω,t∈R.Here α is a positive constant which will be fixed later in (43) and (44). Then, it follows that, for α>0 and for P-a.e. ω∈Ω,(24)∑j=1mzjθtωj+zjθtωj2+zjθtωj3+zjθtωj4≤eα/2trω,t∈R,(25)∑j=1mzjθt-hωj+zjθt-hωj2+zjθt-hωj3+zjθt-hωj4≤eα/2t-hrω≤eα/2heα/2trω,t∈R.By [3, 25] and Itô formula, one has that(26)limt→±∞zjθtωjt=0,(27)limt→±∞1t∫-t0zjθsωjds=Ezj,(28)Ezj≤Ezj2=12λ.It follows from (26)-(28) that there exists a positive constant λ>0 large enough, and a constant c>0, such that, for all t>0,(29)4c1∫-t0∑j=1mzjθsωjds≤α4t+c;2ϵ∑j=1mzjθtωj≤α8t+c.Let z(θtω)=ϵ∑j=1m‍hjzj(θtωj); then (21) implies that(30)dz+λzdt=ϵ∑j=1mhjdwj.Let v(t)=u(t)-z(θtω). Then v satisfies(31)∂v∂t-μΔv+v+z·∇v+z=fx+gx,vt-h+zθt-hω+λzθtω+μΔzθtω,divv=0,with the boundary condition(32)v=0,on0,+∞×Γ,and with the initial condition(33)vt,x=v0t,x,t∈-h,0,x∈D.Here v0(t,x)=u0(t,x)-z(θtω).For the deterministic problem (31)-(33) with random parameter, under the assumptions (A1) and (A2), by the Galerkin method (using a similar method to Theorem 2.1 [26]), one has that for all >0, ω∈Ω, and u0(t,x)∈S, it has a unique solution v(·,ω,v0)∈C([-h,+∞),V)∩Lloc2((0,+∞),D(A)). Define a mapping Φϵ:R×Ω×S→S by(34)Φϵt,ω,v0=vt·,ω,v0with vt(s,ω,v0)=v(t+s,ω,v0) for s∈[-h,0]. Then, it is easy to verify that Φϵ is a continuous random dynamical system associated with the stochastic Navier-Stokes equation. Notice that u(t)=v(t)+z(θtω). Define(35)Ψϵt,ω,u0=ut·,ω,u0=Φϵt,ω,v0+zθtω.Then Ψϵ is a continuous random dynamical system. Moreover, Φϵ and Ψϵ are conjugated random dynamical system. Therefore, in the following sections, we mainly consider the existences of random attractors for Φϵ.


## 3. Uniform Estimates of Solutions

In this section, we prove the existence of the random attractor for the random dynamical systemΨϵ associated with the stochastic Navier-Stokes equation (14). We first establish the existence of the random attractor for its conjugated random dynamical system Φϵ; then the existence of a D-random attractor for Ψϵ follows from the conjugation relation between Φϵ and Ψϵ.From now on, we assume thatD is the collection of all tempered subsets of S with respect to (Ω,F,P,(θt)t∈R). In the rest of the paper, we use notation ci(i=1,2,…) to denote positive constant which is independent of ϵ.Lemma 3.
Let0<ϵ≤1 and suppose that conditions (A1) and (A2) hold. Then there exists a random ball {Kϵ(ω)}ω∈Ω∈D centered at 0 with random radius ρ1(ω)>0 such that {Kϵ(ω)}ω∈Ω is a random absorbing set for Φϵ in D; that is, for every {B(ω)}ω∈Ω∈D and P-a.e. ω∈Ω, there is TB(ω)>0, independent of ϵ, such that Φϵ(t,θ-tω,B(θ-tω))∈{Kϵ(ω)}ω∈Ω for all t≥TB(ω).Proof.
Taking the inner product of (31) with v, we obtain that(36)12ddtv2+μ∇v2=-bv+z,v+z,v+f,v+gx,vt-h+zθt-hω,v+λzθtω,v+μΔzθtω,v.Now, we estimate each term on the right-hand side of (36). Notice that hj(j=1,2,…,m) are given functions in (H02(D))2. By the definition of b(·,·,·), the first term can be estimated as follows:(37)bv+z,v+z,v=bv+z,v+z,v+z-bv+z,v+z,z=bv+z,v+z,z=bv+z,z,v+z=ϵ∑j=1mbv+z,hjzj,v+z≤c1ϵ∑j=1mzjv+z2≤2c1ϵ∑j=1mzjv2+2c1ϵ∑j=1mzjz2.Here c1 is a positive constant depending only on hj(1≤j≤m). By condition (A2), we can choose β1>0 small enough such that 2μ>(2cg+β1)γ-1. Then, Young’s inequality implies that(38)f,v+λzθtω,v+μΔzθtω,v≤β14v2+1β1f2+β18v2+2λ2β1zθtω2+β18v2+2μ2β1Δzθtω2=β12v2+1β1f2+2λ2β1zθtω2+2μ2β1Δzθtω2.Choose β2>0 small enough such that 2μ>(2+β2)cg+β1γ-1. By condition (A1) and Young’s inequality, we have that(39)gx,vt-h+zθt-hω,v≤cg2v2+12cggx,vt-h+zθt-hω2≤cg2v2+cg2vt-h+zθt-hω2+12cgφ2≤cg2v2+cg21+β2vt-h2+cg21+1β2zθt-hω2+12cgφ2.Then it follows from (36)–(39) that(40)ddtv2≤-2μ∇v2+cg1+β2vt-h2+cg+β1+4c1ϵ∑j=1mzjv2+cg1+1β2zθt-hω2+2β1f2+1cgφ2+4μ2β1Δzθtω2+4λ2β1zθtω2+4c1ϵ∑j=1mzjz2.Notice that z(θtω)=ϵ∑j=1m‍hjzj(θtωj) and hj∈(H02(D))2. Hence there exists a constant c2>0, such that(41)cg1+1β2zθt-hω2+4μ2β1Δzθtω2+4λ2β1zθtω2+4c1ϵ∑j=1mzjz2≤c2ϵ2∑j=1mzjθt-hωj2+∑j=1mzjθtωj2+∑j=1mzjθtωj3≡ϵ2p1θtω,with p1(θtω) independent of ϵ. By (24) and (25), we have that, for P-a.e. ω∈Ω, for any α>0, there exists a constant c3>0, such that(42)p1θτω≤c3e1/2ατrω.Choose α>0,β3>0 small enough such that(43)2μ>2+β2cg+α+β1+β3γ-1,(44)1+β2eαhcg<1+β2cg+β3.Then by (40), (41), and Poincaré inequality (13), we obtain that(45)ddtv2≤-2μ∇v2+cg1+β2vt-h2+cg+β1+4c1ϵ∑j=1mzjv2+c4+ϵ2p1θtω≤-c5∇v2+4c1ϵ∑j=1mzj-αv2+1+β2cgvt-h2-1+β2cg+β3v2+c4+ϵ2p1θtω,with c4=2/β1f2+1/cgφ2, c5=2μ-(2+β2)cg+β1+α+β3γ-1>0. By Gronwall’s inequality, one has that, for all t≥0,(46)vt2≤e-αt+4c1ϵ∫0t∑j=1mzjdsv02-c5e-αt+4c1ϵ∫0t∑j=1mzjds∫0teαs-4c1ϵ∫0s∑j=1mzjdτ∇v2ds-cg1+β2+β3e-αt+4c1ϵ∫0t∑j=1mzjds∫0teαs-4c1ϵ∫0s∑j=1mzjdτv2ds+cg1+β2e-αt+4c1ϵ∫0t∑j=1mzjds∫0teαs-4c1ϵ∫0s∑j=1mzjdτvs-h2ds+e-αt+4c1ϵ∫0t∑j=1mzjds∫0teαs-4c1ϵ∫0s∑j=1mzjdτc4+ϵ2p1θsωds.Now, we estimate the fourth term on the right-hand side of (46). Note that(47)∫0teαs-4c1ϵ∫0s∑j=1mzjdτvs-h2ds=∫-ht-heαs+h-4c1ϵ∫0s+h∑j=1mzjdτvs2ds≤∫-h0eαs+hvs2ds+∫0t-heαs+h-4c1ϵ∫0s+h∑j=1mzjdτvs2ds≤heαhv0S2+eαh∫0teαs-4c1ϵ∫0s∑j=1mzjdτvs2ds.Then, by (44) and (47), we have(48)-cg1+β2+β3e-αt+4c1ϵ∫0t∑j=1mzjds∫0teαs-4c1ϵ∫0s∑j=1mzjdτv2ds+cg1+β2e-αt+4c1ϵ∫0t∑j=1mzjds∫0teαs-4c1ϵ∫0s∑j=1mzjdτvs-h2ds≤cg1+β2heαhe-αt+4c1ϵ∫0t∑j=1mzjdsv0S2.Therefore, (46) and (48) imply(49)vt2≤c6e-αt+4c1ϵ∫0t∑j=1mzjdsv0S2-c5e-αt+4c1ϵ∫0t∑j=1mzjds∫0teαs-4c1ϵ∫0s∑j=1mzjdτ∇v2ds+e-αt+4c1ϵ∫0t∑j=1mzjds∫0teαs-4c1ϵ∫0s∑j=1mzjdτc4+ϵ2p1θsωds≤c6e-αt+4c1ϵ∫0t∑j=1mzjdsv0S2+∫0teαs-t+4c1ϵ∫st∑j=1mzjdτc4+ϵ2p1θsωds,with c6=1+cg(1+β2)heαh. Therefore, for fixed σ∈[-h,0], we have that, for t≥-σ,(50)vt+σ2≤c6e-αt+σ+4c1ϵ∫0t+σ∑j=1mzjdτdsv0S2+∫0t+σeαs-t-σ+4c1ϵ∫st+σ∑j=1mzjdτc4+ϵ2p1θsωds≤c6eαh-t+4c1ϵ∫0t∑j=1mzjdsv0S2+eαh∫0teαs-t+4c1ϵ∫st∑j=1mzjdτc4+ϵ2p1θsωds,and for t∈[0,-σ],(51)vt+σ2≤v0S2≤c6eαh-t+4c1ϵ∫0t∑j=1mzjdsv0S2+eαh∫0teαs-t+4c1ϵ∫st∑j=1mzjdτc4+ϵ2p1θsωds.Thus, we deduce from (50) and (51) that, for all t≥0(52)vtS2≤c6eαh-t+4c1ϵ∫0t∑j=1mzjdsv0S2+eαh∫0teαs-t+4c1ϵ∫st∑j=1mzjdτc4+ϵ2p1θsωds.Replacing ω by θ-tω in the last inequality, we get that, for all t≥0,(53)vtθ-tω,v0θ-tωS2≤c6eαh-t+4c1ϵ∫0t∑j=1mzjθs-tωjdsv0θ-tωS2+eαh∫0teαs-t+4c1ϵ∫st∑j=1mzjθτ-tωjdτc4+ϵ2p1θs-tωds≤c6eαh-t+4c1ϵ∫-t0∑j=1mzjθsωjdsv0θ-tωS2+eαh∫-∞0eαs+4c1ϵ∫s0∑j=1mzjθτωjdτc4+ϵ2p1θsωds.By (29) and (42), one has that, for v0(θ-tω)∈B(θ-tω),(54)limt→+∞c6eαh-t+4c1ϵ∫-t0∑j=1mzjθsωjdsv0θ-tωS2≤limt→+∞c6eαh-t+α/4t+cv0θ-tωS2=0,and(55)eαh∫-∞0eαs+4c1ϵ∫s0∑j=1mzjθτωjdτc4+ϵ2p1θsωds≤eαh∫-∞0eαs-α/4s+cc4+c3e-α/2srωds≤c7+c7rω.Set(56)ρ12ω=1+c7+c7rω.Then, there exist a TB(ω)>0, independent of ϵ, such that, for all t>TB(ω),(57)vtθ-tω,v0θ-tωS2≤ρ12ω.This ends the proof.Lemma 4.
Let0<ϵ≤1 and suppose that conditions (A1) and (A2) hold. Then there exists a tempered random variable ρ2(ω)>0 such that, for any {B(ω)}ω∈Ω∈D and v0(ω)∈B(ω), the solution v of (31) satisfies, for P-a.e. ω∈Ω, for all t≥TB(ω),(58)∫tt+1∇vs,θ-t-1ω,v0θ-t-1ω2ds≤ρ2ω.Here TB is the random variable defined in Lemma 3.Proof.
By (49) we have that(59)c5∫0teαs-t+4c1ϵ∫st∑j=1mzjθτ-tωjdτ∇vs,θ-tω,v0θ-tω2ds≤c6e-αt+4c1ϵ∫0t∑j=1mzjdsv0S2+∫0teαs-t+4c1ϵ∫st∑j=1mzjdτc4+ϵ2p1θsωds.Replacing ω by θ-tω, and by (54) and (55), one has that, for all t≥TB(ω),(60)c5∫0teαs-t+4c1ϵ∫st∑j=1mzjθτ-tωjdτ∇vs,θ-tω,v0θ-tω2ds≤c6e-αt+4c1ϵ∫0t∑j=1mzjθs-tωjdsv0S2+∫0teαs-t+4c1ϵ∫st∑j=1mzjθτ-tωjdτc4+ϵ2p1θsωds≤1+c7+c7rω.On the other hand, we have that(61)c5∫0t+1eαs-t-1+4c1ϵ∫st+1∑j=1mzjθτ-t-1ωjdτ∇vs,θ-t-1ω,v0θ-t-1ω2ds≥c5∫tt+1eαs-t-1+4c1ϵ∫st+1∑j=1mzjθτ-t-1ωjdτ∇vs,θ-t-1ω,v0θ-t-1ω2ds≥c5e-α∫tt+1∇vs,θ-t-1ω,v0θ-t-1ω2ds.Equations (60) and (61) imply that, for all t≥TB(ω),(62)∫tt+1∇vs,θ-t-1ω,v0θ-t-1ω2ds≤eα1+c7+c7rω≡ρ2ω.This ends the proof.Lemma 5.
Let0<ϵ≤1 and suppose that conditions (A1) and (A2) hold. Then, there exist random variables ρ3(ω)>0, ρ4(ω)>0, for any {B(ω)}ω∈Ω∈D, and v0(ω)∈B(ω); the solution v of (31) satisfies the following estimates: for P-a.e. ω∈Ω, for all t≥TB(ω)+2h+1, and -h≤τ1≤τ2≤0(63)∇vt,θ-tω,v0θ-tω2≤ρ3ω,(64)∫t+τ1t+τ2Δvs,θ-tω,v0θ-tω2ds≤ρ4ω.Here, TB is the random variable defined in Lemma 3.Proof.
Taking the inner product of (31) with -Δv, we obtain that(65)12ddt∇v2+μΔv2=bv+z,v+z,Δv-f,Δv-gx,vt-h+zθt-hω,Δv-λzθtω,Δv-μΔzθtω,Δv.Now, we estimate each term on the right-hand side of (65). By (12) and Young’s inequality(66)bv+z,v+z,Δv≤Cv+z1/2∇v+zΔv+z1/2Δv≤Cv+z1/2∇v+zΔv3/2+Cv+z1/2∇v+zΔz1/2Δv≤μ8Δv2+c8v+z2∇v+z4+μ8Δv2+c8v+z∇v+z2Δz≤μ4Δv2+c9v2+z2∇v2+v+zΔz∇v2+c9v2+z2∇z4+v+z∇z2Δz,and(67)-f,Δv-gx,vt-h+zθt-hω,Δv-λzθtω,Δv-μΔzθtω,Δv≤μ8Δv2+2μf2+μ8Δv2+2μgx,vt-h+zθt-hω2+μ8Δv2+2μΔzθtω2+μ8Δv2+2λ2μzθtω2≤μ2Δv2+c10f2+gx,vt-h+zθt-hω2+Δzθtω2+zθtω2.Consequently, it follows from (65)–(67) and Poincaré inequality (13) that(68)ddt∇v2≤-μ2Δv2+F1v,z∇v2+F2v,z≤-μ2γ+F1v,z∇v2+F2v,z,with(69)F1v,z=2c9v2+z2∇v2+v+zΔz,F2v,z=2c9v2+z2∇z4+v+z∇z2Δz+2c10f2+gx,vt-h+zθt-hω2+Δzθtω2+zθtω2.Replacing t by s and replacing ω by θ-t-1ω in (68), we get that, for all t>0,(70)dds∇vs,θ-t-1ω,v0θ-t-1ω2≤-μ2γ+F1vs,θ-t-1ω,v0θ-t-1ω,zθs-t-1ω∇vs,θ-t-1ω,v0θ-t-1ω2+F2vs,θ-t-1ω,v0θ-t-1ω,zθs-t-1ω.Now, we use uniform Gronwall inequality to estimate ∇v(s,θ-t-1ω,v0(θ-t-1ω)). By (57) and (22), we can get that, for all t≥TB(ω),(71)maxt≤s≤t+1vs,θ-t-1ω,v0θ-t-1ω2=maxt≤s≤t+1vs,θ-sθs-t-1ω,v0θ-sθs-t-1ω2≤max-1≤τ≤0ρ12θτω,and(72)maxt≤s≤t+1zθs-t-1ω+zθs-t-1ω2+Δzθs-t-1ω=max-1≤τ≤0zθτω+zθτω2+Δzθτω≤c11max-1≤τ≤0rθτω.Consequently, it follows from (58), (71), and (72) that, for t≥TB(ω),(73)∫tt+1-μ2γ+F1vs,θ-t-1ω,v0θ-t-1ω,zθs-t-1ωds≤c12max-1≤τ≤0ρ12θτω+rθτω∫tt+1∇vs,θ-t-1ω,v0θ-t-1ω2ds+c12max-1≤τ≤0ρ1θτω+rθτωrθτω≤c12max-1≤τ≤0ρ12θτω+rθτωρ2ω+ρ1θτω+rθτωrθτω≡aω.By condition (A1), we get that, for t≥0,(74)gx,vt-h+zθt-hω2≤2cg2vt-h2+zθt-hω2+φx2.It follows from (22), (57), and (74) that, for t≥TB(ω)+h,(75)maxt≤s≤t+1gx,vs-h,θ-t-1ω,v0θ-t-1ω+zθs-h-t-1ω2≤maxt≤s≤t+12cg2vs-h,θ-t-1ω,v0θ-t-1ω2+zθs-h-t-1ω2+φx2≤c13max-1-h≤τ≤-hρ12θτω+rθτω+1.Similarly as (73), we have that, for t≥TB(ω)+h,(76)∫tt+1F2vs,θ-t-1ω,v0θ-t-1ω,zθs-t-1ωds≤c14max-1≤τ≤0ρ12θτω+ρ1θτω+1rθτω+r2θτω+1+c14max-1-h≤τ≤-hρ12θτω+rθτω≡bω.By (58), (70), (73), (76), and uniform Gronwall inequality, we have that for P-a.e. ω∈Ω, for all t≥TB(ω)+h,(77)∇vt+1,θ-t-1ω,v0θ-t-1ω2≤ρ2ω+bωeaω≡ρ3ω.Consequently, we obtain that, for all t>TB(ω)+h+1,(78)∇vt,θ-tω,v0θ-tω2≤ρ3ω.Integrate (68) over [t+τ1,t+τ2]; then one has that(79)∇vt+τ22+μ2∫t+τ1t+τ2Δvs2ds≤∇vt+τ12+∫t+τ1t+τ2F1v,z∇vs2ds+∫t+τ1t+τ2F2v,zdsReplacing ω by θ-tω in (79), we find that(80)μ2∫t+τ1t+τ2Δvs,θ-tω,v0θ-tω2ds≤∇vt+τ1,θ-tω,v0θ-tω2+∫t+τ1t+τ2F1vs,θ-tω,v0θ-tω,zθs-tω∇vs,θ-tω,v0θ-tω2ds+∫t+τ1t+τ2F2vs,θ-tω,v0θ-tω,zθs-tωds.We need to estimate each term on the right-hand side of (80). By (78), one has that, for t≥TB(ω)+h,(81)∇vt+τ1,θ-tω,v0θ-tω2=∇vt+τ1,θ-t-τ1θτ1ω,v0θ-t-τ1θτ1ω2≤max-h≤τ1≤0ρ3θτ1ω.Similarly, we can deduce that, for t≥TB(ω)+2h, t+τ1≤s≤t+τ2(82)vs,θ-tω,v0θ-tω2+vs-h,θ-tω,v0θ-tω2≤max-2h≤τ≤0ρ12θτω,(83)zθs-tω2+zθs-t-hω2≤max-2h≤τ≤0rθτω.Then, it follows from (81)–(83) that(84)∫t+τ1t+τ2F1vs,θ-tω,v0θ-tω,zθs-tω∇vs,θ-tω,v0θ-tω2ds≤c15max-2h≤τ≤0ρ3θτωρ12θτω+rθτω+ρ1θτω+rθτωrθτωρ3θτω,and(85)∫t+τ1t+τ2F2vs,θ-tω,v0θ-tω,zθs-tωds≤c16max-2h≤τ≤0ρ12θτω+rθτωρ3θτω+ρ1θτω+rθτωrθτω+ρ12θτω+rθτω+1.Then, it follows from (80), (84), and (85) that(86)∫t+τ1t+τ2Δvs,θ-tω,v0θ-tω2ds≤c17max-2h≤τ≤0ρ32θτω+rθτω+1ρ12θτω+rθτω+ρ1θτω+rθτωρ3θτω+1rθτω+1≡ρ4ω.This ends the proof.We use Ascoli theorem to prove the existence of a pullback random attractor. Before that, we need to prove the following result.Lemma 6.
For allt≥TB(ω)+2h+1, and for -h≤τ1≤τ2≤0, there exists a random variable ρ5(ω) such that(87)vtτ1,θ-tω,v0θ-tω-vtτ2,θ-tω,v0θ-tω≤ρ5ωτ1-τ21/2.Proof.
By (31),(88)vtτ1,θ-tω,v0θ-tω-vtτ2,θ-tω,v0θ-tω=vt+τ1,θ-tω,v0θ-tω-vt+τ2,θ-tω,v0θ-tω=∫t+τ1t+τ2ddsvs,θ-tω,v0θ-tωds≤∫t+τ1t+τ2ddsvs,θ-tω,v0θ-tωds≤μ∫t+τ1t+τ2Δvs,θ-tω,v0θ-tωds+∫t+τ1t+τ2fds+∫t+τ1t+τ2vs,θ-tω,v0θ-tω+zθs-tω·∇vs,θ-tω,v0θ-tω+zθs-tωds+∫t+τ1t+τ2gx,vs-h,θ-tω,v0θ-tω+zθs-t-hωds+λ∫t+τ1t+τ2zθs-tωds+μ∫t+τ1t+τ2Δzθs-tωds.Lemma 5 is used to estimate the first term on the right-hand side of (88). One has that, for t≥TB(ω)+2h+1,(89)∫t+τ1t+τ2Δvs,θ-tω,v0θ-tωds≤∫t+τ1t+τ2Δvs,θ-tω,v0θ-tω2ds1/2τ1-τ21/2≤ρ41/2ωτ1-τ21/2.For the second term on the right-hand side of (88), we have that(90)∫t+τ1t+τ2fds=fτ1-τ2≤h1/2fτ1-τ21/2.Next, we use Lemmas 3 and 5 to estimate the third and fourth terms on the right-hand side of (88). For t≥TB(ω)+2h+1, we can obtain that(91)∫t+τ1t+τ2vs,θ-tω,v0θ-tω+zθs-tω·∇vs,θ-tω,v0θ-tω+zθs-tωds≤∫t+τ1t+τ2vs,θ-tω,v0θ-tω+zθs-tω·∇vs,θ-tω,v0θ-tω+zθs-tω2ds1/2τ1-τ21/2≤max-h≤τ≤0c19h1/2ρ12θτω+rθτωρ3θτω+rθτω1/2τ1-τ21/2,and(92)∫t+τ1t+τ2gx,vs-h,θ-tω,v0θ-tω+zθs-t-hωds≤∫t+τ1t+τ2gx,vs-h,θ-tω,v0θ-tω+zθs-t-hω2ds1/2τ1-τ21/2≤max-2h≤τ≤-hc20h1/2ρ12θτω+rθτω+11/2τ1-τ21/2.In (92), we use condition (A1). For the last two terms on the right-hand side of (88), by (22), we have(93)λ∫t+τ1t+τ2zθs-tωds+μ∫t+τ1t+τ2Δzθs-tωds≤max-h≤τ≤0c21h1/2rθτωτ1-τ21/2.This completes the proof.Theorem 7.
Let0<ϵ≤1 and suppose that conditions (A1) and (A2) hold. Then, the random dynamical system Φϵ has a unique D-random attractor in S.Proof.
By Lemma3, we know that, for fixed 0<ϵ≤1, Φϵ has a closed random absorbing set K(ω) in S. Next, we use Ascoli theorem to prove that Φϵ is D-pullback asymptotically compact in S. That is, we need only to show that, for P-a.e. ω∈Ω, the sequence {Φϵ(tn,θ-tnω,v0(θ-tnω))}n=1∞ has a convergent subsequence in S, as tn→∞. Because tn→∞, there is N=N(B,ω), such that tn≥TB(ω)+2h+1, as n≥N. By Lemmas 3 and 5, we obtain that, for all n≥N and τ∈[-h,0],(94)vtnτ,θ-tnω,v0θ-tnωH1D2≤max-h≤τ≤0ρ12θτω+ρ3θτω.It follows from H1(D)↪L2(D) that for each τ∈[-h,0], {vtn(τ,θ-tnω,v0(θ-tnω))}n=1∞ is relatively compact in L2(D). By Lemma 6, one has that {vtn(·,θ-tnω,v0(θ-tnω))}n=1∞ is equicontinuous. Ascoli theorem [27] implies that {vtn(·,θ-tnω,v0(θ-tnω))}n=N∞ is relatively compact in C([-h,0];L2(D)). Thus, Proposition 1 implies the existence of a unique D-random attractor in S.Notice thatΦϵ and Ψϵ are conjugated random dynamical system. Hence, it follows from Theorem 7 that Ψϵ has unique D- random attractors in S.


## 4. Upper Semicontinuity of Random Attractor

In this section, we consider the upper semicontinuity of random attractors of Navier-Stokes equation (31), when the intensity of noise tends to zero. We write the solution and the random dynamical system of (31) as vϵ and Φϵ, respectively. Let(95)Kϵω=v∈S:vS2≤Rϵ2ω,with(96)Rϵ2ω=1+eαh∫-∞0eαs+2c1ϵ∫s0∑j=1mzjθτωdτc4+ϵ2p1θsωds.Set R02(ω)=1+c4eαh/α. In Section 3, we prove that Φϵ has a unique D-pullback random attractor Aϵ(ω). It follows from the invariance of the random attractor {Aϵ(ω)}ω∈Ω that, for 0<ϵ≤1, Aϵ(ω)⊆Kϵ(ω). Set(97)Bω=v∈S:vS2≤R12ω,with(98)R12ω=1+eαh∫-∞0eαs+2c1∫s0∑j=1mzjθτωdτc4+p1θsωds.Then, for all 0<ϵ≤1,(99)Kϵω⊆Bω.This implies that(100)⋃0<ϵ≤1Aϵω⊆⋃0<ϵ≤1Kϵω⊆Bω.We will show that, as ϵ→0, the solution of (31) converges to the limiting deterministic system:(101)∂∂tv-μΔv+v·∇v=fx+gx,vt-h.Lemma 8.
For0<ϵ≤1, let vϵ and v be the solution of (31) and (101), respectively. Then, for ω∈Ω, T>0, there exists a positive constant M which is independent of ϵ, such that for all t∈[0,T](102)vϵ-vS2≤Mv0ϵ-v0S2+Mϵ2.Proof.
Letw=vϵ-v. By (31) and (101), we have that(103)∂∂tw-μΔw+vϵ+z·∇vϵ+z-v·∇v=gx,vϵt-h+zθt-hω-gx,vt-h+λzθtω+μΔzθtω.Taking the inner product of (103) with w, we obtain that(104)12ddtw2+μ∇w2=-bvϵ+z,vϵ+z,w+bv,v,w+gx,vϵt-h+zθt-hω-gx,vt-h,w+λzθtω,w+μΔzθtω,w.Now, we estimate each term on the right-hand side of (104). For the first and second terms(105)-bvϵ+z,vϵ+z,w+bv,v,w=-bvϵ,vϵ,w-bvϵ,z,w-bz,vϵ,w-bz,z,w+bv,v,w=-bw,vϵ,w-bvϵ,z,w-bz,vϵ,w-bz,z,w.By (12), (24), and Young inequality, one has(106)
b
w
,
v
ϵ
,
w
≤
C
w
∇
v
ϵ
∇
w
≤
μ
∇
w
2
+
C
4
μ
w
2
∇
v
ϵ
2
,
b
v
ϵ
,
z
,
w
≤
C
v
ϵ
1
/
2
∇
v
ϵ
1
/
2
∇
z
1
/
2
Δ
z
1
/
2
w
≤
w
2
+
C
2
v
ϵ
∇
v
ϵ
∇
z
Δ
z
≤
w
2
+
c
22
ϵ
2
∑
j
=
1
m
z
j
2
v
ϵ
2
+
∇
v
ϵ
2
≤
w
2
+
c
22
ϵ
2
e
α
/
2
t
r
ω
v
ϵ
2
+
∇
v
ϵ
2
,
(107)
b
z
,
v
ϵ
,
w
≤
C
z
1
/
2
Δ
z
1
/
2
∇
v
ϵ
w
≤
w
2
+
C
2
z
∇
z
∇
v
ϵ
2
≤
w
2
+
c
23
ϵ
2
∑
j
=
1
m
z
j
2
∇
v
ϵ
2
≤
w
2
+
c
23
ϵ
2
e
α
/
2
t
r
ω
∇
v
ϵ
2
,
(108)
b
z
,
z
,
w
≤
C
z
1
/
2
Δ
z
1
/
2
∇
z
w
≤
w
2
+
C
2
z
∇
z
∇
z
2
≤
w
2
+
c
24
ϵ
4
∑
j
=
1
m
z
j
4
≤
w
2
+
c
24
ϵ
4
e
α
/
2
t
r
ω
.For the third term on the right-hand side of (104), by condition (A1) and (25), one has that(109)gx,vϵt-h+zθt-hω-gx,vt-h,w≤w2+gx,vϵt-h+zθt-hω-gx,vt-h2≤w2+2lg2wt-h2+2lg2zθt-hω2,and(110)2lg2zθt-hω2≤2clg2ϵ2∑j=1mzj2≤c25ϵ2eα/2trω.For the fourth and fifth terms on the right-hand side of (104),(111)λzθtω,w+μΔzθtω,w≤w2+λzθtω2+w2+μ2Δzθtω2≤2w2+c26ϵ2∑j=1mzj2≤2w2+c26ϵ2eα/2trω.It follows from the above inequalities that we have (112)ddtw2≤C2μ∇vϵ2+12w2+2lg2wt-h2+c27ϵ2eα/2trωvϵ2+∇vϵ2+1≤C2μ∇vϵ2+2lg2+12w2+2lg2wt-h2-w2+c27ϵ2eα/2trωvϵ2+∇vϵ2+1.Using Gronwall inequality, we get that(113)w2≤e∫0tC/2μ∇vϵ2ds+12+2lg2tw02+2lg2e∫0tC/2μ∇vϵ2ds+12+2lg2t∫0te-∫0sC/2μ∇vϵ2dτ-12+2lg2sws-h2-w2ds+c27ϵ2rωe∫0tC/2μ∇vϵ2ds+12+2lg2t∫0te-∫0sC/2μ∇vϵ2dτ-12+2lg2s+α/2svϵ2+∇vϵ2+1ds.Note that(114)
∫
0
t
e
-
∫
0
s
C
/
2
μ
∇
v
ϵ
2
d
τ
-
12
+
2
l
g
2
s
w
s
-
h
2
d
s
=
∫
-
h
t
-
h
e
-
∫
0
s
+
h
C
/
2
μ
∇
v
ϵ
2
d
τ
-
12
+
2
l
g
2
s
+
h
w
s
2
d
s
≤
∫
-
h
0
w
s
2
d
s
+
∫
0
t
e
-
∫
0
s
C
/
2
μ
∇
v
ϵ
2
d
τ
-
12
+
2
l
g
2
s
w
s
2
d
s
≤
h
w
0
S
2
+
∫
0
t
e
-
∫
0
s
C
/
2
μ
∇
v
ϵ
2
d
τ
-
12
+
2
l
g
2
s
w
s
2
d
s
.Then, it follows that(115)w2≤e∫0tC/2μ∇vϵ2ds+12+2lg2tw02+2lg2he∫0tC/2μ∇vϵ2ds+12+2lg2tw0S2+c27ϵ2rωe∫0tC/2μ∇vϵ2ds+12+2lg2t∫0te-∫0sC/2μ∇vϵ2dτ-12+2lg2s+α/2svϵ2+∇vϵ2+1ds≤1+2lg2he∫0tC/2μ∇vϵ2ds+12+2lg2tw0S2+c27ϵ2rω∫0te∫stC/2μ∇vϵ2dτ+12+2lg2t-svϵ2+∇vϵ2+1ds.By (49), one has that, for all t∈[0,T], (46) and (48) imply(116)vϵt2+c5e-αt+4c1ϵ∫0t∑j=1mzjds∫0teαs-4c1ϵ∫0s∑j=1mzjdτ∇vϵ2ds≤c6e-αt+4c1ϵ∫0t∑j=1mzjdsv0ϵS2+∫0teαs-t+4c1ϵ∫st∑j=1mzjdτc4+p1θsωds.Notice that, for all t∈[0,T],(117)c5e-αt+4c1ϵ∫0t∑j=1mzjds∫0teαs-4c1ϵ∫0s∑j=1mzjdτ∇vϵ2ds=c5∫0teαs-t+4c1ϵ∫st∑j=1mzjdτ∇vϵ2ds≥c5e-αt∫0t∇vϵ2ds≥c5e-αT∫0t∇vϵ2ds.By (29) and (42), one has that, for all t∈[0,T],(118)c6e-αt+4c1ϵ∫0t∑j=1mzjdsv0ϵS2≤c6e-αt+α/4t+cv0ϵS2≤c28v0ϵS2,(119)∫0teαs-t+4c1ϵ∫st∑j=1mzjdτc4+p1θsωds≤∫0Teαs-t+4c1ϵ∫0T∑j=1mzjdτc4+c3e1/2αsrωds≤c291+rω.Here c28 and c29 are constants depending on T. Then by (116)–(119),(120)vϵ2+∫0t∇vϵ2ds≤c301+v0ϵS2+rω.It follows from (115) and (120) that one has that, for all t∈[0,T],(121)w2≤c31ec321+v0ϵS2+rωw0S2+c33ϵ2ec321+v0ϵS2+rω1+v0ϵS2+rω,which implies (102).Theorem 9.
Suppose (A1) and (A2) hold. LetAϵ(ω),A0 be random attractor of Φϵ,Φ, respectively. Then, for P-a.e. ω∈Ω,(122)limϵ→0distSAϵω-A0=0.Proof.
Letϵn→0 and v0,n→v0 in S. Then, by Lemma 8, we can obtain that, for P-a.e. ω∈Ω, and t≥0,(123)Φϵnt,ω,v0,n→Φt,v0.Assume that {Kϵ(ω)}ω∈Ω is defined by (95). From the proof of Lemma 3, we find that {Kϵ(ω)}ω∈Ω is an absorbing set of Φϵ in S. It is easy to see that(124)limsupϵ→0Kϵω=limsupϵ→0Rϵω=R0.By Theorem 7 and the invariance of the random attractor {Aϵ(ω)}ω∈Ω, we have that(125)⋃0<σ≤1Aϵω=⋃0<σ≤1Φϵt,ω,Aϵω⊆Φϵt,ω,Bωis precompact in S. Then, (122) follows from Proposition 2.By the conjugated relation betweenΦϵ and Ψϵ, it follows from Theorem 9 that the random attractor of Ψϵ converges to the random attractor of Ψ, as ϵ→0.



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*Source: 3105239-2018-09-13.xml*