In this paper, we investigate the existence of a random attractor for the following stochastic plate equations with linear memory and multiplicative noise on bounded domain:

{ u t t + α u t + Δ 2 u + ∫ 0 ∞     μ ( s ) Δ 2 ( u ( t ) − u ( t − s ) ) d s + f ( u ) = g ( x ) + c u ∘ d W d t , u ( x , t ) = u 0 ( x ) ,   u t ( x , t ) = u 1 ( x ) , x ∈ U , t ≤ 0 , u | ∂ U = ∂ u ∂ n | ∂ U = 0 , t ≥ 0. (1.1)

where α and c > 0 are positive constants and μ ( s ) ≥ 0 for every s ∈ ℝ + , U is an open bounded set of ℝ 5 with smooth boundary ∂ U , u = u ( x , t ) is a real function on U × [ 0, + ∞ ) , g ∈ H 0 1 ( U ) ∩ H 2 ( U ) is a given external force and W ( x , t ) is an independent two-sided real-valued wiener process on probability space ( Ω , F , ℙ ) , where

Ω = { ω = ( ω 1 , ω 2 , ⋯ , ω m ) ∈ C ( ℝ , ℝ m ) : ω ( 0 ) = 0 } ,

is endowed with compact-open topology, ℙ is the corresponding wiener measure, and F is the ℙ -completion of Borel σ-algebra on Ω. We identify ω ( t ) with ( W 1 ( t ) , W 2 ( t ) , ⋯ , W m ( t ) ) , i.e.

ω ( t ) = ( W 1 ( t ) , W 2 ( t ) , ⋯ , W m ( t ) ) , t ∈ ℝ .

Then, define the time shift ( θ t ) t ∈ ℝ on Ω by:

θ t ω ( ⋅ ) = ω ( ⋅ + t ) − ω ( t ) , t ∈ ℝ , ω ∈ Ω .

The following conditions are necessary to obtain our main results.

(h₁) The memory kernel μ is assumed to satisfy the following conditions:

{ μ ∈ ℂ 1 ( ℝ + ) ∩ L 1 ( ℝ + ) , μ ( s ) ≥ 0, μ ′ ( s ) ≤ 0 , ∀ s ∈ ℝ + , μ ′ ( s ) + δ μ ( s ) ≤ 0 , ∀ s ∈ ℝ + ,   and   for   some   δ > 0,

and

∫ 0 ∞     μ ( s ) d s < ∞ .

(h₂) The nonlinear term f ∈ ℂ 1 ( ℝ ) with f ( 0 ) = 0 and satisfies the following conditions:

| f ′ ( u ) | ≤ C 1 ( 1 + | u | 4 ) , ∀ u ∈ ℝ , (1.2)

F ( u ) = ∫ 0 u     f ( s ) d s ≥ C 2 ( | u | 6 − 1 ) , ∀ u ∈ ℝ , (1.3)

and

u f ( u ) ≥ C 3 ( F ( s ) − 1 ) , ∀ u ∈ ℝ , (1.4)

where C 1 , C 2 , C 3 > 0 are constants.

Following Dafermos [1] , we introduce a new variable η defined by:

η ( t , x , s ) = u ( t , x ) − u ( t − s , x ) , (1.5)

and let ℜ μ ,2 = L μ 2 ( ℝ + , H 0 2 ( U ) ) be a Hilbert space of H 0 2 ( U ) -valued function on ℝ + with the inner product:

( η 1 , η 2 ) μ ,2 = ∫ 0 ∞     μ ( s ) ( Δ η 1 ( s ) , Δ η 2 ( s ) ) d s , ∀ η 1 , η 2 ∈ ℜ μ ,2 . (1.6)

Set Z = ( u , u t , η ) T , E = H 0 2 ( U ) × L 2 ( U ) × ℜ μ ,2 . Then, the system (1.1) is equivalent to the following initial value problem in the Hilbert space E:

{ Z t = L ( Z ) + N ( Z , t , W ( t ) ) , x ∈ U , t ≥ 0 , Z 0 = ( u 0 ( x ) , u 1 ( x ) , η 0 ( x , s ) ) , ( x , s ) ∈ U × ℝ + , (1.7)

where

{ u ( t , x ) = η ( t , x , s ) = η ( t , x , 0 ) , x ∈ ∂ U , s ∈ ℝ + , t ≥ 0 , u ( t , x ) = u 0 ( x ) , u t ( t , x ) = u 1 ( x ) , x ∈ U , η ( 0 , x , s ) = η 0 ( x , s ) = u ( 0 , x ) − u ( − s , x ) , ( x , s ) ∈ U × ℝ + , (1.8)

L ( Z ) = ( u t − Δ 2 u − α u t − ∫ 0 ∞     μ ( s ) Δ 2 η ( s ) d s u t − η s ) , (1.9)

N ( Z , t , W ( t ) ) = ( 0 − g ( u ) + f ( x ) + c u ∘ d W ( x , t ) d t 0 ) , (1.10)

D ( L ) = { Z ∈ E | u + ∫ 0 ∞     μ ( s ) η ( s ) d s ∈ H 4 ( U ) ∩ H 0 2 ( U ) u t ∈ H 0 2 ( U ) , η ( s ) ∈ H μ 1 ( ℝ + , H 0 2 ( U ) ) , η ( 0 ) = 0 } . (1.11)

The stochastic plate equation is one of the fundamental stochastic partial differential equations (SPDEs) of hyperbolic type, which have been explored in [2] [3] [4] [5] . The behavior of its solutions is significantly different from those of solutions to other SPDEs.

Problem (1.1) models transversal vibration of the extensible elastic plate in a historical space, which is established based on the framework of elastic vibration by Woinowsky-Krieger [6] and Berger [7] . It can also be regarded as an elastoplastic flow equation with some kind of memory effect [1] . When μ = c = 0 , then (1.1) reduces to determined autonomous damped plate equation.

In recent years, there have many results on the dynamics of a variety of systems related to Equation (1.1). The hyperbolic equations with memory have been studied in [8] - [15] and references therein. For instance, Khanmamedov [16] and Yue and Zhong [2] proved the existence of global attractors for plate equations with critical exponent, [17] - [22] obtained the nonlinear damped, and Ma et al. [23] [24] [25] [26] [27] obtained the strongly damped. The existence of random attractors for such system in a bounded domain has been studied in [28] . Furthermore, long-time dynamics of a plate equation with memory and time delay is considered by Feng in [29] , under suitable assumptions on real numbers μ 1 and μ 2 , the quasi-stability property of the system is established and obtained the existence of global attractor, which has finite fractal dimension, and proved the existence of exponential attractors, defined in bounded domain Ω ⊂ ℝ n ( n ≥ 1 ) with a sufficiently smooth boundary ∂ Ω . Shen and Ma in [30] obtained the existence of random attractors for weakly dissipative plate equations with memory and additive noise by defining the energy functionals and using the compactness translation theorem.

Crauel et al. [31] [32] [33] studied the random attractors for stochastic dynamical system. Recently, many authors have established the existence of random attractors for other equations (see [34] - [45] ). In Equation (1.1), there are fewer results and most previous authors have concentrated on the deterministic case, but there is no result of random attractors for Equation (1.1).

To prove the existence of random dynamical system (RDS) for short, the key step is to establish the compactness of the system. For our system (1.7), there are two essential difficulties in proving the compactness. Firstly, the critical growth condition (1.2) of f can be overcome by using the decomposition of solution and more accurate calculation. Secondly, the memory kernel itself, because there is no compact embedding in the history space, we introduce a new variable and define an extended Hilbert space, as well as combine with the compactness transform theorem.

The rest of the paper is organized as follows. In Section 2, we give the existence and uniqueness of the solutions. In Section 3, we devote to uniform estimates and the existence of bounded absorbing sets for the solutions and pullback compactness. In Section 4, the compactness of the random dynamical system is established by the decomposition of solution of the random differential equation into two parts. In Section 5, we prove the asymptotic compactness of the solutions, existence and uniqueness of a random attractor in E.

As mentioned in the introduction, our main purpose is to prove the dynamics of stochastic partial differential equations with multiplicative noise. For that matter, first, we recall some basic concepts related to random attractors for stochastic dynamical systems (see [9] [31] [32] [46] [47] [48] [49] ), which are important for getting our main results. Let ( Ω , F , ℙ ) be a probability space and ( X , d ) be a polish space with the Borel σ-algebra B ( X ) . The distance between x ∈ X and B ⊆ X is denoted by d ( x , B ) . If B ⊆ X and C ⊆ X , the Hausdorff semi-distance from B to C is denoted by d ( B , C ) = sup x ∈ B d ( x , C ) .

Definition 2.1. ( Ω , F , ℙ , ( θ t ) t ∈ ℝ ) is called a metric dynamical system if θ : ℝ × Ω → Ω is ( B ( ℝ ) × F , F ) -measurable, θ 0 is the identity on Ω, θ s + t = θ t ∘ θ s for all s , t ∈ ℝ and θ 0 P = P for all t ∈ ℝ .

Definition 2.2. A mapping Φ ( t , τ , ω , x ) : ℝ + × ℝ × Ω × X → X is called continuous cocycle on X over ℝ and ( Ω , F , ℙ , ( θ t ) t ∈ ℝ ) , if for all τ ∈ ℝ , ω ∈ Ω and t , s ∈ ℝ + , the following conditions are satisfied:

1) Φ ( t , τ , ω , x ) : ℝ + × ℝ × Ω × X → X is a ( B ( ℝ + ) × F , B ( ℝ ) ) measurable mapping.

2) Φ ( 0, τ , ω , x ) is identity on X.

3) Φ ( t + s , τ , ω , x ) = Φ ( t , τ + s , θ s ω , x ) ∘ Φ ( s , τ , ω , x ) .

4) Φ ( t , τ , ω , x ) : X → X is continuous.

Definition 2.3. Let 2 X be the collection of all subsets of X, a set valued mapping ( τ , ω ) ↦ D ( t ω ) : ℝ × Ω ↦ 2 X is called measurable with respect to F in Ω if D ( t , ω ) is a (usually closed) nonempty subset of X and the mapping ω ∈ Ω ↦ d ( X , B ( τ , ω ) ) is ( F , B ( ℝ ) ) -measurable for every fixed x ∈ X and τ ∈ ℝ . Let B = B ( t , ω ) ∈ D ( t , ω ) : τ ∈ ℝ , ω ∈ Ω is called a random set.

Definition 2.4. A random bounded set B = { B ( τ , ω ) : τ ∈ ℝ , ω ∈ Ω } ∈ D of X is called tempered with respect to { θ ( t ) } t ∈ Ω , if for p-a.e ω ∈ Ω ,

l i m t ↦ ∞ e − β t d ( B ( θ − t ω ) ) = 0, ∀ β > 0 ,

where

d ( B ) = sup x ∈ B ‖ x ‖ X .

Definition 2.5. Let D be a collection of random subset of X and K = { K ( τ , ω ) : τ ∈ ℝ , ω ∈ Ω } ∈ D , then K is called an absorbing set of Φ ∈ D if for all τ ∈ ℝ , ω ∈ Ω and B ∈ D , there exists, T = T ( τ , ω , B ) > 0 such that:

Φ ( t , τ , θ − t ω , B ( τ , θ − t ω ) ) ⊆ K ( τ , ω ) , ∀ t ≥ T .

Definition 2.6. Let D be a collection of random subset of X, the Φ is said to be D -pullback asymptotically compact in X if for p-a.e ω ∈ Ω , { Φ ( t n , θ − t n ω , x n ) } n = 1 ∞ has a convergent subsequence in X when t n ↦ ∞ and x n ∈ B ( θ − t n ω ) with { B ( ω ) } ω ∈ Ω ∈ D .

Definition 2.7. Let D be a collection of random subset of X and A = { A ( τ , ω ) : τ ∈ ℝ , ω ∈ Ω } ∈ D , then A is called a D -random attractor (or D -pullback attractor) for Φ, if the following conditions are satisfied: for all t ∈ ℝ + , τ ∈ ℝ and ω ∈ Ω ,

1) A ( τ , ω ) is compact, and ω ↦ d ( x , A ( ω ) ) is measurable for every x ∈ X .

2) A ( τ , ω ) is invariant, that is:

Φ ( t , τ , ω , A ( τ , ω ) ) = A ( τ + t , θ t ω ) , ∀ t ≥ τ .

3) A ( τ , ω ) attracts every set in D , that is for every B = { B ( τ , ω ) : τ ∈ ℝ , ω ∈ Ω } ∈ D ,

l i m t ↦ ∞ d X ( Φ ( t , τ , θ − t ω , B ( τ , θ − t ω ) ) , A ( τ , ω ) ) = 0.

where d X is the Hausdorff semi-distance given by:

d X ( Y , Z ) = sup y ∈ Y inf z ∈ Z ‖ y − z ‖ X   for   any   Y ∈ X   and   Z ∈ X .

Remark 2.8. Let ( Ω , F , ℙ ) be a probability space with wiener measure ℙ , the wiener shift ( θ t ) t ∈ ℝ is defined by:

θ s ω ( t ) = ω ( t + s ) − ω ( s ) , t , s ∈ ℝ ,

then ( Ω , F , ℙ , ( θ t ) t ∈ ℝ ) is an ergodic metric dynamical system.

Lemma 2.9. [31] [32] Let D be a neighborhood-closed collection of ( τ , ω ) -parameterized families of nonempty subsets of X and Φ be a continuous cocycle on X over ℝ and ( Ω , F , ℙ , ( θ t ) t ∈ ℝ ) . Then, Φ has a pullback D -attractor A in D if and only if Φ is pullback D -asymptotically compact in X and Φ has a closed, F -measurable pullback D -absorbing set K ∈ D , the unique pullback D -attractor A = A ( τ , ω ) is given:

A ( τ , ω ) = ∩ r ≥ 0 ∪ t ≥ r   Φ ( t , τ − t , θ − t ω , K ( τ − t , θ − t ω ) ) ¯   , τ ∈ ℝ , ω ∈ Ω .

In this article, we will take D as the collection of all tempered random subsets.

Lemma 2.10. [50] For any k > 0 and any ϕ ∈ H 0 1 ( U ) ∩ L ∞ ( U ) , the following equality holds:

− ∫ U ( | ϕ | k ϕ ) Δ ϕ d x = ( k + 1 ) ( 2 k + 2 ) 2 ∫ U | ∇ | ϕ | k + 2 2 | 2 d x .

Lemma 2.11. [9] Let X 0 , X , X 1 be three Banach spaces such that X 0 ↪ X ↪ X 1 , the first injection being compact. Let Y ⊂ L μ 2 ( ℝ + , X ) satisfy the following hypotheses:

1) Y is bounded in L μ 2 ( ℝ + , X 0 ) ∩ H μ 1 ( ℝ + , X 1 ) .

2) sup η ∈ Y ‖ η ( s ) ‖ X 2 ≤ K 0 , ∀ s ∈ ℝ + for some K 0 > 0 .

Then, Y relatively compact in L μ 2 ( ℝ + , X ) .

From now on, assume that conditions (h₁) - (h₂) hold, the space E and the probability space ( Ω , F , ℙ , ( θ t ) t ∈ ℝ ) are defined in Section 1. Let A = Δ 2 with Neumann boundary condition on U, D ( A ) = H 4 ( U ) ∩ H 0 2 ( U ) . We can define the powers A ν of A for ν ∈ ℝ . The space V 2 ν = D ( A ν 2 ) is the Hilbert space with the following inner product and norm, respectively:

( u , v ) 2 ν = ( A ν 2 u , A ν 2 v ) , ‖ u ‖ 2 ν 2 = ( A ν 2 u , A ν 2 u ) .

The injection V ν 1 ↪ V ν 2 is compact if ν 1 > ν 2 . Then, by the generalized Poincaré inequality, there holds:

‖ u ‖ ν 1 2 ≥ λ 1 ‖ u ‖ ν 2 2 ,

where λ 1 > 0 is the first eigenvalue of A. In particular, V 0 = L 2 ( U ) , V 1 = H 0 1 ( U ) , V 2 = H 0 2 ( U ) , and ( A 1 4 u , A 1 4 v ) = ( ∇ u , ∇ v ) , ∀ u , v ∈ H 0 1 ( U ) . The inner product and norm in L 2 ( U ) is denoted by ( ⋅ , ⋅ ) , ‖   ⋅   ‖ , and in H 0 2 ( U ) is denoted by ( ( ⋅ , ⋅ ) ) , ‖   ⋅   ‖ 2 , respectively. By (h₁), the space ℜ μ ,2 ν = L μ 2 ( ℝ + , V 2 ν ) is a Hilbert space of V 2 ν -valued function on ℝ + with the inner product and norm, respectively:

( η , η 1 ) μ ,2 ν = ∫ 0 ∞     μ ( s ) ( A ν 2 η ( s ) , A ν 2 η 1 ( s ) ) d s , ∀ η , η 1 ∈ V 2 ν , (3.1)

‖ η ‖ μ ,2 ν 2 = ( η , η ) μ ,2 ν = ∫ 0 ∞     μ ( s ) ‖ A ν 2 η ( s ) ‖ 2 d s , (3.2)

and on ℜ μ ,2 ν , the linear operator − ∂ s has domain:

D ( − ∂ s ) = { η ∈ H μ 1 ( ℝ + , V 2 ν ) : η ( 0 ) = 0 } , H μ 1 ( ℝ + , V 2 ν ) = { η : η ( s ) , ∂ s η ∈ L μ 2 ( ℝ + , V 2 ν ) } ,

which generates a right-translation semigroup (see [1] [9] [13] [15] [51] ).

Then, Equation (1.1) can be transformed into the following system:

{ u t t + α u t + Δ 2 u + ∫ 0 ∞     μ ( s ) Δ η ( s ) d s + f ( u ) = g ( x ) + c u ∘ d W d t , η t = − η s + u t , : (3.3)

with the initial-boundary conditions:

{ u ( x ,0 ) = u 0 ( x ) , u t ( x ,0 ) = u 1 ( x ) , x ∈ U , η 0 ( x , s ) = u 0 ( x ) − u 0 ( x , − s ) , ∀ x ∈ U , s ∈ ℝ + . (3.4)

The symbol C and C i ( i = 1,2, ⋯ ) are positive constants, which may change from line to line.

In this section, we show the existence, uniqueness and continuous dependence of (mild) solution of initial problem (1.7) in E, which generates a continuous RDS on E over ℝ and ( Ω , F , ℙ , ( θ t ) t ∈ ℝ ) . For our purpose, we convert the problem (1.7) into a deterministic system with random parameters but without noise terms.

Due to Ornstein-Uhlenbeck process deducing by the Brownian motion, which holds the Itô differential equation:

d z + α z d t = d W ( t ) , (3.5)

and hence, the solution is given by:

z ( θ t ω ) = − α ∫ ∞ 0     e α s ( θ t ω ) ( s ) d s ,   t ∈ ℝ . (3.6)

It is known from [48] [49] , the random variable | z ( ω ) | is tempered and there is a θ t -invariant set Ω ¯ ⊆ Ω of full P measure such that for every ω ∈ Ω ¯ , t ↦ z ( θ t ω ) is continuous in t and:

l i m t → ∞ e − α t | z ( θ − t ω ) | = 0, ∀ α > 0, ω ∈ Ω ¯ . (3.7)

Equation (3.6) has a random fixed point in the sense of random dynamical systems generating a stationary solution known as the stationary Ornstein-Uhlenbeck process (see [31] [32] [36] [52] for more details).

For convenience, in the following, we write Ω ¯ as Ω. Next, we need to transform the stochastic system into deterministic with a random parameter, then show that it generates a random dynamical system.

Let:

w ( t , ω , x ) = u t ( t , ω , x ) + ε u ( t , ω , x ) − c u ( t , ω , x ) z ( θ t ω ) , t > 0 , (3.8)

φ = ( u w η ) = T ε ( u u t η ) = T ε Z , T ε = ( 0 1 0 ε 1 0 0 0 1 ) , (3.9)

where

ε = 2 α 3 + 2 α k + α 2 λ 1 + ( 3 + 2 α k + α 2 λ 1 ) 2 − 24 α k > 0 , k = ‖ μ ‖ L 1 ( ℝ + ) δ > 0 ,

λ 1 ( > 0 ) is the smallest eigenvalue of operator A with Neumann boundary condition on U.

In this paper, we assume that:

| c | < min { α 2 λ 1 − 1 π α ( 2 C 1 λ 1 + 3 + C 1 C 2 ) 2 + 1 π α ( 2 C 1 λ 1 + 3 + C 1 C 2 ) 2 + σ ( 16 ε 2 + α ) α 2 λ 1 16 ε 2 + α , α 2 λ 1 − 1 + ( 2 R 6 ( ω ) + 2 ) λ 1 λ 1 π α + ( 1 + 2 R 6 ( ω ) λ 1 + 2 λ 1 ) 2 λ 1 π α + ( 16 ε 2 + α ) σ 2 ( ω ) α 2 λ 1 16 ε 2 + α α 2 λ 1 − 1 + ( 2 R 6 ( ω ) + 2 ) λ 1 λ 1 π α + ( 1 + 2 R 6 ( ω ) λ 1 + 2 λ 1 ) 2 λ 1 π α + ( 16 ε 2 + α ) σ 3 α 2 λ 1 16 ε 2 + α } , (3.10)

where σ , σ 3 > 0 , σ 2 ( ω ) , R 6 ( ω ) > 0 .

By (3.8) and (1.1), we can obtain the following random evolution equation:

{ u t + ε u − w = c u z ( θ t ω ) , w t − ε ( α − ε ) u + A u + ( α − ε ) w + ∫ 0 ∞     μ ( s ) A η ( s ) d s + f ( u ) = g ( x ) − c z ( θ t ω ) ( w − 2 ε u + c u z ( θ t ω ) ) , η t + η s + ε u − w = c u z ( θ t ω ) . (3.11)

Then, the problem (3.11) is equivalent to the following determined system with random parameter in E:

{ φ ′ + H ( φ ) = Q ( φ , θ t ω , t ) , φ τ ( ω ) = ( u 0 , u 1 + ε u 0 − c u 0 z ( ω ) , η 0 ) T , t ≥ 0, (3.12)

where

H ( φ ) = ( ε u − w − ε ( α − ε ) u + A u + ( α − ε ) w + ∫ 0 ∞     μ ( s ) A η ( s ) d s ε u − w + η s ) = − T ε H T ε ( ψ ) , (3.13)

Q ( φ , θ t ω , t ) = ( c u z ( θ t ω ) − c z ( θ t ω ) ( w − 2 ε u + c u z ( θ t ω ) ) − f ( u ) + g ( x ) c u z ( θ t ω ) ) . (3.14)

In line with [9] [53] , we know that the operator L in (1.9) is the infinitesimal generator of C₀-semigroup e L t of contractions on E for t > 0 . Since − H = T ε L T − ε , and T ε is an isomorphism of E, the operator − H also generates a C₀-semigroup e − H t of contractions on E. By the assumptions (h₂) and the embedding relation H 0 2 ( U ) ↪ L 10 ( U ) , it is easy to check Q ( φ , t , ω ) : E → E is locally Lipschitz continuous with respect to φ , by the classical semigroup theory concerning the (local) existence and uniqueness solution of evolution differential equation [53] , we have the following theorem.

Theorem 3.1. Assume that (h₁) - (h₃) hold. Then, for each ω ∈ Ω and for any φ 0 ∈ E , there exists T > 0 such that (3.12) has a unique mild function φ ( ⋅ , ω , φ 0 ) ∈ C ( [ 0, T ) ; E ) such that φ ( 0, ω , φ 0 ) = φ ( 0 ) satisfies the integral equation:

φ ( t , ω , φ 0 ) = e − H t φ 0 ( ω ) + ∫ 0 t     e − H ( t − s ) Q ( φ ( s , ω , φ 0 ) , θ s ω , s ) d s . (3.15)

Moreover, φ ( t , ω , φ 0 ) is jointly continuous in φ 0 and measurable in ω .

From Theorem 3.1, we know that for P-a.s. each ω ∈ Ω , then the following results hold for all T > 0 :

1) If φ 0 ( ω ) ∈ E then φ ( ⋅ , ω , φ 0 ) ∈ C ( [ 0, T ) ; E ) .

2) φ ( t , ω , φ 0 ) is jointly continuous into t and measurable in ω .

3) The solution mapping of (3.12) satisfies the properties of Random Dynamical System.

We notice that a unique solution φ ( ⋅ , ω , φ 0 ) of (3.12) can define a continuous random dynamical system over ℝ and ( Ω , F , P , ( θ t ) t ∈ ℝ ) . Hence, the solution mapping:

Φ ¯ ( t , ω ) : ℝ × Ω × E ↦ E ,   t ≥ 0, φ ( 0, ω ) = ( u 0 , v 0 , η 0 ) Τ ↦ ( u ( t , ω ) , v ( t , ω ) , η ( t , ω ) ) Τ = φ ( t , ω ) , (3.16)

generates a random dynamical system. Moreover,

Φ ( t , ω ) : φ ( 0, ω ) + ( 0, ε z ( ω ) ,0 ) Τ ↦ φ ( t , ω ) + ( 0, ε z ( θ t ω ) ,0 ) Τ . (3.17)

We also define the following transformation:

ψ 1 = u , ψ 2 = u t + ε u , (3.18)

similar to (3.12), we get that:

{ ψ ′ + H ψ = Q ( ψ , t , ω ) ψ 0 ( ω ) = ( u 0 , v 0 , η 0 ) Τ = ( u 0 , u 1 + ε u 0 , η 0 ) Τ , (3.19)

where

ψ = ( u v η ) , (3.20)

H ( ψ ) = ( ε u − v − ε ( α − ε ) u + A u + ( α − ε ) v + η ε u − v + η s ) (3.21)

and

Q ( ψ , ω , t ) = ( 0 c v z ( θ t ω ) − f ( u ) + g ( x ) 0 ) .

It is easy to see that:

ϒ ( t , ω , Z 0 ) = R ε , θ t ω − 1 Φ ( t , ω ) R ε , θ t ω : Z 0 → Z ( t , ω , Z 0 ) , (3.22)

and

Ψ ( t , ω , ψ 0 ) = T ε ϒ T − ε : ψ τ → ψ ( t , ω , ψ 0 ) , (3.23)

are continuous RDS over ℝ and ( Ω , F , P , ( θ t ) t ∈ ℝ ) associated with system (3.7) and (3.15) respectively.

We introduce the isomorphism T ε Y = ( u , u t , η ) Τ , Y = ( u , v , η ) Τ ∈ E , which has inverse isomorphism T − ε Y = ( u , v − ε u , η ) Τ , it follows that ( θ , ψ ) with mapping:

Ψ = T ε Φ ( t , ω ) T − ε = Ψ ( t , ω ) (3.24)

is a random dynamical system from above discussion, we show that the two RDS are equivalent.

In this section, we will show the existence of a random absorbing set for the RDS φ ( t , ω , φ 0 ( ω ) ) , t ≥ 0 in the space E.

Lemma 4.1. Suppose that (h₁) - (h₂) hold. Then, there exists a closed tempered absorbing ball B 0 ( ω ) ∈ D ( E ) of E, centered at 0 with random radius M 0 ( ω ) > 0 such that for any bounded non-random set B ∈ D ( E ) , there exists a deterministic t B ( ω ) > 0 , such that the solution φ ( t , θ − t ω , φ 0 ( θ − t ω ) ) of (3.12) with initial value ( u 0 , u 1 + ε u 0 − c u 0 z ( ω ) , η 0 ) T ∈ B satisfies, for P-a.s. ω ∈ Ω ,

‖ φ ( t , θ − t ω , φ 0 ( θ − t ω ) ) ‖ E 2 ≤ M 0 ( ω ) , ∀ t ≥ t B ( ω ) , (4.1)

that is,

φ ( t , θ − t ω , B ( θ − t ω ) ) ⊆ B 0 ( ω ) , ∀ t ≥ t B ( ω ) .

Proof. Taking the inner product ( ⋅ , ⋅ ) E of (3.12) with φ ( r ) = ( u ( r ) , w ( r ) , η r ) T , we have:

1 2 d d t ‖ φ ‖ E 2 + ( H ( φ , φ ) ) E = ( Q ( φ , θ t ω , t ) , φ ) . (4.2)

Similar to the proof of Lemma 2 in [54] , we have:

( H ( φ ) , φ ) E = ε ‖ u ‖ 2 2 − ( ( w , u ) ) + ( α − ε ) ‖ w ‖ 2 − ε ( α − ε ) ( u , w ) + ( A u , w )   + ε ( u , η ) μ ,2 + ( ∫ 0 ∞     μ ( s ) A η ( s ) d s , w ) − ( w , η ) μ ,2 + ( η s , η ) μ ,2 = ε ‖ u ‖ 2 2 + ( α − ε ) ‖ w ‖ 2 − ε ( α − ε ) ( u , w ) + ε ( u , η ) μ ,2 + ( η s , η ) μ ,2 . (4.3)

Then, by using (h₁), we find that:

ε ( u , η ) μ ,2 ≥ − k ε 2 ‖ u ‖ 2 2 − δ 4 ‖ η ‖ μ ,2 2 , (4.4)

( η s , η ) μ ,2 ≥ δ 2 ‖ η ‖ μ ,2 2 . (4.5)

Applying (4.3)-(4.5), Hölder inequality, Young inequality and Poincaré inequality, we obtain that:

( H ( φ ) , φ ) E ≥ ε ( 1 − k ε ) ‖ u ‖ 2 2 + ( α − ε ) ‖ w ‖ 2 + δ 4 ‖ η ‖ μ ,2 2 − ε ( α − ε ) ( u , w ) ≥ ε ( 1 − k ε ) ‖ u ‖ 2 2 + ( α − ε ) ‖ w ‖ 2 + δ 4 ‖ η ‖ μ ,2 2 − ε α λ 1 ‖ u ‖ 2 ⋅ ‖ w ‖ = ε 2 ( ‖ u ‖ 2 2 + ‖ w ‖ 2 ) + δ 4 ‖ η ‖ μ ,2 2 + α 2 ‖ w ‖ 2 + ε ( 1 2 − k ε ) ‖ u ‖ 2 2   + ( α 2 − 3 ε 2 ) ‖ w ‖ 2 − ε α λ 1 ‖ u ‖ 2 ⋅ ‖ w ‖ . (4.6)

It follows from a simple computation that:

ε ( 1 2 − k ε ) ( α 2 − 3 ε 2 ) = ε 2 α 2 4 λ 1 . (4.7)

Hence, combining (4.6) and (4.7), we find that:

( H ( φ , φ ) ) E ≥ ε 2 ( ‖ u ‖ 2 2 + ‖ w ‖ 2 ) + δ 4 ‖ η ‖ μ ,2 2 + α 2 ‖ w ‖ 2 . (4.8)

Let us estimate the right hand side of (4.2):

( Q ( φ , θ t ω , t ) , φ ) = ( ( c u z ( θ t ω ) , u ) ) − ( c w z ( θ t ω ) , w ) + ( 2 c ε u z ( θ t ω ) , w )     − ( c 2 u z 2 ( θ t ω ) , w ) − ( f ( u ) , w ) + ( g ( x ) , w ) + ( c u z ( θ t ω ) , η ) μ ,2 . (4.9)

By the Cauchy-Schwartz inequality, we find that:

( ( c u z ( θ t ω ) , u ) ) ≤ | c | | z ( θ t ω ) | ‖ u ‖ 2 2 , (4.10)

( c w z ( θ t ω ) , w ) ≤ | c | | z ( θ t ω ) | ‖ w ‖ 2 , (4.11)

( 2 c ε u z ( θ t ω ) , w ) ≤ 8 ε 2 | c | 2 | z ( θ t ω ) | 2 α λ 1 ‖ u ‖ 2 2 + α 8 ‖ w ‖ 2 , (4.12)

c 2 z 2 ( θ t ω ) | ( u , w ) | ≤ c 2 2 λ 1 | z ( θ t ω ) | 2 ( ‖ u ‖ 2 2 + ‖ w ‖ 2 ) , (4.13)

( g ( x ) , w ) ≤ 2 α ‖ g ( x ) ‖ 2 + α 8 ‖ w ‖ 2 , (4.14)

( c u z ( θ t ω ) , η ) μ ,2 ≤ | c | | z ( θ t ω ) | 2 ( ‖ u ‖ 2 2 + ‖ η ‖ μ ,2 2 ) . (4.15)

Then, we estimate nonlinear term (4.9), by (h₂) and the Hölder inequality, we get that:

( f ( u ) , w ) = ( f ( u ) , u t + ε u − c u z ( θ t ω ) ) = d d t ∫ U     F ( u ) d x + ε ( f ( u ) , u ) − c z ( θ t ω ) ( f ( u ) , u ) . (4.16)

Applying (1.2)-(1.4), we have:

| c z ( θ t ω ) ( f ( u ) , u ) | ≤ C 1 | c | | z ( θ t ω ) | ∫ U ( | u | 2 + | u | 6 ) d x ≤ C 1 C 2 | c | | z ( θ t ω ) | ∫ U ( F ( u ) + C 2 ) d x + C 1 | c | | z ( θ t ω ) | ‖ u ‖ 2 ≤ C 1 C 2 | c | | z ( θ t ω ) | ∫ U     F ( u ) d x + C 1 | U | | c | | z ( θ t ω ) | + C 1 | c | | z ( θ t ω ) | λ 1 ‖ u ‖ 2 2 ; (4.17)

ε ( f ( u ) , u ) ≥ ε C 3 ∫ U     F ( u ) d x − ε C 3 | U | . (4.18)

Thus, due to (4.16)-(4.18), we obtain that:

( f ( u ) , w ) ≥ d d t F ˜ ( u ) − ε C 3 F ˜ ( u ) + ε C 3 | U | − C 1 C 2 | c | | z ( θ t ω ) | F ˜ ( u )     − C 1 | U | | c | | z ( θ t ω ) | − C 1 | c | | z ( θ t ω ) | λ 1 ‖ u ‖ 2 2 , (4.19)

where F ˜ ( u ) = ∫ U   F ( u ) d x .

Collecting (4.2), (4.8), (4.19) and (4.9)-(4.15), we show that:

1 2 d d t ( ‖ φ ‖ E 2 + 2 F ˜ ( u ) ) + ε 2 ( ‖ u ‖ 2 2 + ‖ w ‖ 2 ) + δ 4 ‖ η ‖ μ ,2 2 + α 2 ‖ w ‖ 2 + ε C 3 F ˜ ( u ) ≤ C 1 C 2 | c | | z ( θ t ω ) | F ˜ ( u ) + | c | | z ( θ t ω ) | 2 ‖ η ‖ μ ,2 2 + ( | c | | z ( θ t ω ) | + c 2 | z ( θ t ω ) | 2 2 λ 1 ) ‖ w ‖ 2       + ( C 1 | c | | z ( θ t ω ) | λ 1 + 3 | c | | z ( θ t ω ) | 2 + 8 ε 2 | c | 2 | z ( θ t ω ) | 2 α λ 1 + c 2 | z ( θ t ω ) | 2 2 λ 1 ) ‖ u ‖ 2 2       + ε C 3 | U | + C 1 | U | | c | | z ( θ t ω ) | + 2 α ‖ g ( x ) ‖ 2 ,

choose σ = min { ε , δ 2 , ε C 3 } . Due to ‖ φ ‖ E 2 = ( ‖ u ‖ 2 2 + ‖ w ‖ 2 + ‖ η ‖ μ ,2 2 ) , then we have the following equivalent system:

d d t ( ‖ φ ‖ E 2 + 2 F ˜ ( u ) ) + σ ( φ E 2 + 2 F ˜ ( u ) ) ≤ 2 C 1 C 2 | c | | z ( θ t ω ) | F ˜ ( u ) + | c | | z ( θ t ω ) | ‖ η ‖ μ ,2 2 + ( 2 | c | | z ( θ t ω ) | + c 2 | z ( θ t ω ) | 2 λ 1 ) ‖ w ‖ 2

        + ( 2 C 1 | c | | z ( θ t ω ) | λ 1 + 3 | c | | z ( θ t ω ) | + 16 ε 2 | c | 2 | z ( θ t ω ) | 2 α λ 1 + c 2 | z ( θ t ω ) | 2 λ 1 ) ‖ u ‖ 2 2         + 2 C 3 | U | + 2 C 1 | U | | c | | z ( θ t ω ) | + 4 α ‖ g ( x ) ‖ 2 ,

where

ρ ( t , θ t ω ) = σ − ( 2 C 1 | c | | z ( θ t ω ) | λ 1 + 3 | c | | z ( θ t ω ) | + 16 ε 2 | c | 2 | z ( θ t ω ) | 2 α λ 1   + c 2 | z ( θ t ω ) | 2 λ 1 + C 1 C 2 | c | | z ( θ t ω ) | ) . (4.20)

That is,

d d t ( ‖ φ ‖ E 2 + 2 F ˜ ( u ) ) + ρ ( t , θ t ω ) ( ‖ φ ‖ E 2 + 2 F ˜ ( u ) ) ≤ 2 C 3 | U | + 2 C 1 | U | | c | | z ( θ t ω ) | + 4 α ‖ g ( x ) ‖ 2 . (4.21)

So, applying Gronwall’s Lemma to (4.21), we have:

‖ φ ( t , ω , φ 0 ( ω ) ) ‖ E 2 + 2 F ˜ ( u ) ≤ e − ∫ 0 t   ρ ( s , θ s ω ) d s ( ‖ φ 0 ( ω ) ‖ E 2 + 2 F ˜ ( u 0 ) ) + 2 C 1 | U | | c | ∫ 0 t     e − ∫ s t   ρ ( τ , θ τ ω ) d τ | z ( θ s ω ) | d s       + ( 2 C 3 | U | + 4 α ‖ g ( x ) ‖ 2 ) ∫ 0 t     e − ∫ s t   ρ ( τ , θ τ ω ) d τ d s . (4.22)

Substitiuting ω by θ − t ω , from (4.22), we have:

‖ φ ( t , θ − t ω , φ 0 ( θ − t ω ) ) ‖ E 2 + 2 F ˜ ( u ) ≤ e − ∫ 0 t   ρ ( s − t , θ s − t ω ) d s ( ‖ φ 0 ( θ − t ω ) ‖ E 2 + 2 F ˜ ( u 0 ) ) + 2 C 1 | U | | c | ∫ 0 t     e − ∫ s t   ρ ( τ − t , θ τ − t ω ) d τ | z ( θ s − t ω ) | d s         + ( 2 C 3 | U | + 4 α ‖ g ( x ) ‖ 2 ) ∫ 0 t   e − ∫ s t   ρ ( τ − t , θ τ − t ω ) d τ d s ≤ e − ∫ − t 0   ρ ( s , θ s ω ) d s ( ‖ φ 0 ( θ − t ω ) ‖ E 2 + 2 F ˜ ( u 0 ) ) + 2 C 1 | U | | c | ∫ − t 0     e − ∫ s 0   ρ ( τ , θ τ ω ) d τ | z ( θ s ω ) | d s         + ( 2 C 3 | U | + 4 α ‖ g ( x ) ‖ 2 ) ∫ − t 0     e − ∫ s 0   ρ ( τ , θ τ ω ) d τ d s . (4.23)

Since | z ( θ t ω ) | is stationary and ergodic, it follows from (3.2) and the ergodic theorem that:

l i m t → ∞ 1 t ∫ − t 0 | z ( θ r ω ) | d r = E ( | z ( θ r ω ) | ) = 1 π α , (4.24)

lim t → ∞ 1 t ∫ − t 0 | z ( θ r ω ) | 2 d r = E ( | z ( θ r ω ) | 2 ) = 1 2 α . (4.25)

From (4.24) and (4.25), we know that there exists T 1 ( ω ) > 0 such that for any t ≥ T 1 ( ω ) ,

∫ − t 0 | z ( θ r ω ) | d r < 1 π α t , ∫ − t 0 | z ( θ r ω ) | 2 d r < 1 2 α t . (4.26)

Next, we need to obtain that for any s ≤ − T 1 ,

e ∫ 0 s   ρ ( τ , θ τ ω ) d τ ≤ e σ 2 s . (4.27)

Indeed, by (4.26), we have:

∫ 0 s [ σ − ( 2 C 1 | c | | z ( θ τ ω ) | λ 1 + 3 | c | | z ( θ τ ω ) | + 16 ε 2 | c | 2 | z ( θ τ ω ) | 2 α λ 1 + c 2 | z ( θ τ ω ) | 2 λ 1 + C 1 C 2 | c | | z ( θ τ ω ) | ) ] d τ > σ s − | c | 4 C 1 λ 1 π α s − | c | 6 π α s − | c | 2 8 ε 2 α 2 λ 1 s − | c | 2 1 2 α λ 1 s − | c | 2 C 1 C 2 π α s = − 16 ε 2 + α 2 α 2 λ 1 | c | 2 s − 2 π α ( 2 C 1 λ 1 + 3 + C 1 C 2 ) | c | s + σ s .

In order to obtain (4.27), for any s ≤ − T 1 , there holds:

16 ε 2 + α 2 α 2 λ 1 | c | 2 + 1 π α ( 2 C 1 λ 1 + 3 + C 1 C 2 ) | c | − σ 2 < 0.

Solving this quadratic inequality, we find that:

| c | < α 2 λ 1 − 1 π α ( 2 C 1 λ 1 + 3 + C 1 C 2 ) 2 + 1 π α ( 2 C 1 λ 1 + 3 + C 1 C 2 ) 2 + σ ( 16 ε 2 + α ) α 2 λ 1 16 ε 2 + α .

Since | z ( θ t ω ) | is tempered, it follows from (4.27) that the following integral is bounded:

R 1 ( ω ) = 2 C 1 | U | | c | ∫ − t 0     e − ∫ s 0   ρ ( τ , θ τ ω ) d τ | z ( θ s ω ) | d s   + ( 2 C 3 | U | + 4 α ‖ g ( x ) ‖ 2 ) ∫ − t 0     e − ∫ s 0   ρ ( τ , θ τ ω ) d τ d s . (4.28)

According to (1.3)-(1.5), we have:

C 2 ‖ u ‖ 1 6 − C 2 | U | ≤ ∫ U     F ( u ) d x ≤ 1 C 3 ∫ U     f ( u ) u d x + | U | ≤ 1 C 3 ∫ U ( 1 + | u | 5 ) | u | d x + | U | ≤ 1 C 3 ( ∫ U | u | d x + ∫ U | u | 6 d x ) + | U | ≤ C ( | U | + ‖ u ‖ 1 6 ) . (4.29)

It follows from theorem 3.1 and φ 0 ( θ − t ω ) ∈ B 0 ( θ − t ω ) that:

l i m t → + ∞ e − ∫ − t 0   ρ ( τ , θ τ ω ) d τ [ ‖ φ 0 ( θ − t ω ) ‖ E 2 + 2 F ˜ ( u 0 ) ] = 0. (4.30)

Combining with (4.27)-(4.30), there exits t B ( ω ) ≥ T 1 such that for all t ≥ t B ( ω ) ,

‖ φ ( t , θ − t ω , φ 0 ( θ − t ω ) ) ‖ E 2 ≤ C ( 1 + R 1 ( ω ) ) = M 0 ( ω ) .

Then, we complete the proof. □

In order to obtain regularity estimates later, as in [52] , we decompose the equations (3.3). At first, we will give the following decomposition on nonlinearity f ( u ) = f 1 ( u ) + f 2 ( u ) and f 1 , f 2 ∈ C 1 ( ℝ ) satisfies the following conditions:

{ 1.   | f 1 ( s ) | ≤ C ( | s | + | s | 5 ) , ∀ s ∈ ℝ , 2.   f 1 ( s ) ⋅ s ≥ 0, ∀ s ∈ ℝ , 3.   ∃ k 1 ≥ 1, k 1 F 1 ( s ) − C ≤ s f 1 ( s ) , ∀ s ∈ ℝ , (5.1)

and

{ 1.   | f ′ 2 ( s ) | ≤ C ( 1 + | s | p ) , f 2 ( 0 ) = 0, ∀ s ∈ ℝ , p < 4, 2.   ∃ k 2 ≥ 1, k 2 F 2 ( s ) − C ≤ s f 2 ( s ) , ∀ s ∈ ℝ , 3.   C ˜ 2 ( | s | 6 − 1 ) ≤ F 2 ( s ) , ∀ s ∈ ℝ , (5.2)

where F i ( s ) = ∫ 0 s   f i ( r ) d r , ( i = 1 , 2 ) , C , C ˜ 2 > 0 are constants.

We decompose the solution φ = ( u , w , η ) T of the system (3.12) into the two parts:

φ = φ L + φ N ,

where φ L = ( u L , w L , η L ) , φ N = ( u N , w N , η N ) solve the following equations, respectively:

{ φ ′ L + H ( φ L ) + Q 1 ( φ L ) = 0, φ L ( 0, ω ) = ( u 0 , u 1 + ε u 0 , η 0 ) T , (5.3)

and

{ φ ′ N + H ( φ N ) + Q 2 ( φ , φ L ) = Q ˜ 2 ( φ , θ t ω ) , φ N ( 0, ω ) = ( 0, − c u 0 z ( ω ) ,0 ) T , (5.4)

where

Q 1 ( φ L ) = ( 0 f 1 ( u L ) 0 ) ,   Q 2 ( φ , φ L ) = ( 0 f ( u ) − f 1 ( u L ) 0 ) , Q ˜ 2 ( ω ) = ( c u N z ( θ t ω ) − c z ( θ t ω ) ( w N − 2 ε u N + c u N z ( θ t ω ) ) + g ( x ) c u N z ( θ t ω ) ) . (5.5)

To prove the existence of a compact random attractor for the random dynamical system Φ, we need to get the solutions of systems (5.3) and (5.4), which one decays exponentially and another is bounded in higher regular space. In order to get the regularity estimate, we will prove some priori estimates for the solutions of systems (5.3) on U × [ 0, ∞ ] as follows.

Lemma 5.1. For any P-a.e. ω ∈ Ω , t ≥ 0 , there exists M 1 ( ω ) > 0 such that the solution φ L = ( u L , w L , η L ) T of (5.3) with initial data φ L ( 0, ω ) = ( u 0 , u 1 + ε u 0 , η 0 ) T = φ 0 ( θ − t ω ) + ( 0, c u 0 z ( θ − t ω ) ,0 ) T ∈ B 0 ( θ − t ω ) satisfies:

‖ φ L ( t , θ − t ω , φ L ( 0, θ − t ω ) ) ‖ E 2 ≤ M 1 ( ω ) . (5.6)

Proof. Taking the inner product ( ⋅ , ⋅ ) E of (5.3) with φ L = ( u L , w L , η L ) T , we show that:

1 2 d d t ‖ φ L ‖ E 2 + ( H ( φ L ) , φ L ) E + ( Q 1 ( φ L ) , φ L ) = 0, (5.7)

Similar to the proof of (4.8), we obtain that:

( H ( φ L ) , φ L ) E ≥ ε 2 ( ‖ u L ‖ 2 2 + ‖ w L ‖ 2 ) + α 2 ‖ w L ‖ 2 + δ 4 ‖ η L ‖ μ ,2 2 . (5.8)

Now, we estimate the third term of (5.7). According to (5.1)₃, we get:

( f 1 ( u L ) , w L ) = d d t ∫ U     F 1 ( u L ) d x + ε ∫ U     f 1 ( u L ) ⋅ u L d x ≥ d d t ∫ U     F 1 ( u L ) d x + k 1 ε ∫ U     F 1 ( s ) d x − ε C | U | . (5.9)

Thus, it follows from (5.7)-(5.10) and (5.3) that:

d d t ( ‖ φ L ‖ E 2 + 2 F ˜ 1 ( u L ) ) + σ L ( ‖ φ L ‖ E 2 + 2 F ˜ 1 ( u L ) ) ≤ 2 ε C | U | , (5.10)

where σ L = min { ε , δ 2 , k 1 ε } . By Gronwall’s Lemma to (5.10), we have:

‖ φ L ‖ E 2 + 2 F ˜ 1 ( u L ) ≤ e − σ L t ( ‖ φ L ( 0, ω ) ‖ E 2 + 2 F ˜ 1 ( u 0 ) ) + 2 ε C | U | σ L . (5.11)

According to (5.1), we have:

0 ≤ ∫ U     F ( u ) d x ≤ 1 k 1 ∫ U     f ( u ) u d x + C k 1 | U | ≤ 1 k 1 ∫ U ( 1 + | u | 5 ) | u | d x + C k 1 | U | ≤ C k 1 ( ∫ U | u | d x + ∫ U | u | 6 d x ) + C k 1 | U | ≤ C ( | U | + ‖ u ‖ 1 6 ) . (5.12)

Combining (5.11)-(5.12) with φ L ( 0, θ − t ω ) ∈ B 0 ( θ − t ω ) , we get:

‖ φ L ( t , θ − t ω , φ L ( 0, θ − t ω ) ) ‖ E 2 ≤ e − σ L t ( ‖ φ L ( 0, θ − t ω ) ‖ E 2 + 2 C ( | U | + ‖ u 0 ‖ 1 6 ) ) + 2 ε C | U | σ L = M 1 ( ω ) . (5.13)

So, the proof is completed. □

Lemma 5.2. For any P-a.e. ω ∈ Ω , t ≥ 0 , there exists M 2 ( ω ) > 0, σ 1 ( ω ) ≥ 0 such that the solution φ L = ( u L , w L , η L ) T of (5.3) with initial data φ L ( 0, ω ) = ( u 0 , u 1 + ε u 0 , η 0 ) T ∈ B 0 ( θ − t ω ) satisfies:

‖ φ L ( t , θ − t ω , φ L ( 0, θ − t ω ) ) ‖ E 2 ≤ M 2 ( ω ) e − σ 1 ( ω ) t , t ≥ 0. (5.14)

Proof. We consider (5.1), (5.7) and similar to Lemma 5.1, we conclude that:

0 ≤ F ˜ 1 ( u L ) ≤ C ( ‖ u L ‖ 2 + ‖ u L ‖ L 6 ( U ) 6 ) . (5.15)

Applying interpolation inequality, we have:

‖ u L ‖ L 6 6 ≤ ‖ u L ‖ ⋅ ‖ u L ‖ L 10 5 ≤ ‖ u L ‖ L 10 4 ( ‖ u L ‖ ⋅ ‖ u L ‖ L 10 ) ≤ ‖ u L ‖ L 10 4 ( 1 2 ‖ u L ‖ 2 + 1 2 ‖ u L ‖ L 10 2 ) ≤ C ‖ u L ‖ 2 4 ( ‖ u L ‖ 2 + ‖ u L ‖ 2 2 ) . (5.16)

Hence, combining (5.15)-(5.16) with Lemma 5.1, we find that there exists M 3 ( ω ) > 0 , such that:

‖ u L ‖ 2 2 ≥ 1 M 3 ( ω ) F ˜ 1 ( u L ) . (5.17)

Due to (5.7)-(5.8), (5.1)₂ and (5.17), we can obtain the following result:

d d t ( ‖ φ L ‖ E 2 + 2 F ˜ 1 ( u L ) ) + ε 2 ( ‖ u L ‖ 2 2 + ‖ w L ‖ 2 ) + δ 2 ‖ η L ‖ μ ,2 2 + ε 2 M 3 ( ω ) F ˜ 1 ( u L ) ≤ 0,

that is,

d d t ( ‖ φ L ‖ E 2 + 2 F ˜ 1 ( u L ) ) + σ 1 ( ω ) ( ‖ φ L ‖ E 2 + 2 F ˜ 1 ( u L ) ) ≤ 0, (5.18)

where σ 1 ( ω ) = min { ε 2 , δ 2 , ε 4 M 3 ( ω ) } .

By applying Gronwall’s inequality to (5.18), it yields:

‖ φ L ( t , θ − t ω , φ L ( 0 , θ − t ω ) ) ‖ E 2 ≤ ( ‖ φ L ( 0 , θ − t ω ) ‖ E 2 + F ˜ 1 ( u 0 ) ) e − σ 1 ( ω ) t ≤ ( ‖ φ L ( 0 , θ − t ω ) ‖ E 2 + C ( | U | + ‖ u ‖ 1 6 ) ) e − σ 1 ( ω ) t = M 2 ( ω ) e − σ 1 ( ω ) t . (5.19)

Then, the proof is completed. □

Next, we estimate the component φ N in (5.4).

Lemma 5.3. For any P-a.e. ω ∈ Ω , t ≥ 0 , there exists σ 2 ( ω ) > 0 such that the solution φ N = ( u N , w N , η N ) T of (5.4) with initial data φ N ( 0, ω ) = ( 0, − c u N z ( θ t ω ,0 ) , η 0 ) T ∈ B 0 ( ω ) satisfies:

‖ A 1 + ν 2 u N ‖ 2 + ‖ A ν 2 u N t ‖ 2 + ‖ A ν 2 η N ‖ μ , 2 2 ≤ e σ 2 ( ω ) 2 t P 1 ( θ − t ω ) , (5.20)

where

ν = min { 1 8 , 4 − p 4 } , ∀ 0 ≤ p < 4 , (5.21)

and P 1 ( θ − t ω ) is increasing function.

Proof. By (4.1), (5.6) and φ N = φ − φ L , there exists a random variable R 3 ( ω ) > 0 such that:

max { ‖ φ ( r , θ − t ω , φ ( 0, θ − t ω ) ) ‖ E , ‖ φ N ( ( r , θ − t ω , φ N ( 0, θ − t ω ) ) ) ‖ E } ≤ R 1 ( ω ) , r ≥ − t . (5.22)

Taking the inner product of ( ⋅ , ⋅ ) E of (5.4) with ( A ν φ N , A ν w N , A ν η N ) T , we find that:

1 2 d d t ‖ A ν 2 φ N ‖ E 2 + ( H ( φ N ) , A ν φ N ) E = ( g ( x ) , A ν w N ) − ( c z ( θ t ω ) w N , A ν w N ) + ( 2 ε c u N z ( θ t ω ) , A ν w N )         − ( f ( u ) − f 1 ( u L ) , A ν w N ) − ( c 2 z 2 ( θ t ω ) u N , A ν w N )         + ( ( c u N z ( θ t ω ) , A ν u N ) ) + ( c u N z ( θ t ω ) , A ν η N ) ν ,2 (5.23)

In later calculations, we will use the following embedding relations:

H 2 ν ↪ L 10 5 − 4 ν , H 2 − 2 ν ↪ L 10 1 + 4 ν , H 2 + 2 ν ↪ L 10 1 − 4 ν , L 10 ↪ L 10 p 4 − 4 ν ( 10 p 4 − 4 ν < 10 ) . (5.24)

Similar to the proof of (4.8), we deduce that:

( H ( φ N ) , A ν φ N ) E ≥ ε 2 ( ‖ A ν 2 u N ‖ 2 2 + ‖ A ν 2 w N ‖ 2 ) + α 2 ‖ A ν 2 w N ‖ 2 + δ 4 ‖ A ν 2 η N ‖ μ ,2 2 . (5.25)

Next, we will deal with the right-hand side of (5.23). Using (4.5)-(4.10) and (5.21), we get:

( ( c u N z ( θ t ω ) , A ν u N ) ) ≤ | c | | z ( θ t ω ) | ‖ A ν 2 u N ‖ 2 2 , (5.26)

( c w N z ( θ t ω ) , A ν w N ) ≤ | c | | z ( θ t ω ) | ‖ A ν 2 w N ‖ 2 , (5.27)

( 2 c ε u N z ( θ t ω ) , A ν w N ) ≤ 8 ε 2 | c | 2 | z 2 ( θ t ω ) | α λ 1 ‖ A ν 2 u N ‖ 2 2 + α 8 ‖ A ν 2 w N ‖ 2 , (5.28)

( c 2 u N z 2 ( θ t ω ) , A ν w N ) ≤ | c | 2 | z ( θ t ω ) | 2 2 λ 1 ( ‖ A ν 2 u N ‖ 2 2 + ‖ A ν 2 w N ‖ 2 ) , (5.29)

( g ( x ) , A ν w N ) ≤ 2 α ‖ A ν 2 g ( x ) ‖ 2 + α 8 ‖ A ν 2 w N ‖ 2 , (5.30)

( c u N z ( θ t ω ) , A ν η N ) μ ,2 ≤ | c | | z ( θ t ω ) | 2 λ 1 ( ‖ A ν 2 u N ‖ 2 2 + ‖ A ν 2 η N ‖ μ ,2 2 ) . (5.31)

For the nonlinear term, we have:

( f ( u ) − f 1 ( u L ) , A ν w N ) = ( f 2 ( u L ) , A ν w N ) + ( f ( u ) − f ( u L ) , A ν w N ) = ( f 2 ( u L ) , A ν ( u N t + ε u N − c u N z ( θ t ω ) ) )     + ( f ( u ) − f ( u L ) , A ν ( u N t + ε u N − c u N z ( θ t ω ) ) ) .

Firstly, we deal with the term:

( f 2 ( u L ) , A ν u N t + ε A ν u N A ν u N ) = d d t ( f 2 ( u L ) , A ν u N ) + ε ( f 2 ( u L ) , A ν u N ) − ( f ′ 2 ( u L ) u L t , A ν u N ) , (5.32)

by (5.2)₁, (5.24) and Lemma 5.1, we have:

| ( f ′ 2 ( u L ) u L t , A ν u N ) | ≤ C ∫ U ( 1 + | u L | p ) | u L t | | A ν u N | d x ≤ C ( ∫ U | A ν u N | 10 1 + 4 ν d x ) 1 + 4 ν 10 ⋅ ( ∫ U ( 1 + | u L | p ) 10 4 − 4 ν d x ) 4 − 4 ν 10 ⋅ ( ∫ U | u L t | 2 d x ) 1 2 ≤ C ‖ A ν 2 u N ‖ 2 ⋅ ( 1 + ‖ u L ‖ 2 p ) ⋅ ‖ u L t ‖ ≤ R 2 ( ω ) + ε 8 ‖ A ν 2 u N ‖ 2 2 , (5.33)

and

| ( f 2 ( u L ) , c A ν u N z ( θ t ω ) ) | ≤ C | c | | z ( θ t ω ) | ∫ U ( 1 + | u L | p ) | u L | | A ν u N | d x ≤ C | c | | z ( θ t ω ) | ( ∫ U ( 1 + | u L | p ) 10 4 − 4 ν d x ) 4 − 4 ν 10 ⋅ ( ∫ U | u L | 2 d x ) 1 2 ⋅ ( ∫ U | A ν u N | 10 1 + 4 ν d x ) 1 + 4 ν 10 ≤ C | c | | z ( θ t ω ) | ( 1 + ‖ u L ‖ 2 p ) ⋅ ‖ u L ‖ ⋅ ‖ A ν 2 u N ‖ 2 ≤ R 3 ( ω ) | c | 2 | z ( θ t ω ) | 2 + ε 8 ‖ A ν 2 u N ‖ 2 2 . (5.34)

Secondly, we consider the following term:

( f ( u ) − f ( u L ) , A ν u N t + ε A ν u N ) = d d t ( f ( u ) − f ( u L ) , A ν u N ) + ε ( f ( u ) − f ( u L ) , A ν u N )         − ( f ′ ( u ) u t − f ′ ( u L ) u L t , A ν u N ) = d d t ( f ( u ) − f ( u L ) , A ν u N ) + ε ( f ( u ) − f ( u L ) , A ν u N )         − ( f ′ ( u ) u L t − f ′ ( u L ) u L t , A ν u N ) − ( f ′ ( u ) u N t , A ν u N ) . (5.35)

According to (1.2), (5.24), Lemma 4.1, 5.1 and w N ( t , ω , x ) = u N t ( t , ω , x ) + ε u N ( t , ω , x ) − c u N z ( θ t ω ) , we obtain that:

| ( f ′ ( u ) u L t − f ′ ( u L ) u L t , A ν u N ) | ≤ C ∫ U | u L t | ⋅ | u N | ⋅ ( 1 + | u | 3 + | u L | 3 ) | A ν u N | d x ≤ C ( ∫ U | u L t | 2 d x ) 1 2 ⋅ ( ∫ U | u N | 10 1 − 4 ν d x ) 1 − 4 ν 10 ⋅ ( ∫ U ( 1 + | u | 3 + | u L | 3 ) 10 3 d x ) 3 10         ⋅ ( ∫ U | A ν u N | 10 1 + 4 ν d x ) 1 + 4 ν 10 ≤ R 4 ( ω ) ‖ A ν 2 u N ‖ 2 2 , (5.36)

| ( f ′ ( u ) u N t , A ν u N ) | ≤ C 1 ∫ U | u N t | ⋅ ( 1 + | u | 4 ) | A ν u N | d x ≤ C 1 ( ∫ U | u N t | 10 5 − 4 ν d x ) 5 − 4 ν 10 ⋅ ( ∫ U ( 1 + | u | 4 ) 10 4 d x ) 4 10 ⋅ ( ∫ U | A ν u N | 10 1 + 4 ν d x ) 1 + 4 ν 10 ≤ C ‖ A ν 2 u N t ‖ ⋅ ( 1 + ‖ u ‖ 2 4 ) ⋅ ‖ A ν 2 u N ‖ 2 ≤ R 5 ( ω ) ‖ A ν 2 u N ‖ 2 ⋅ ( ‖ A ν 2 u N ‖ + ε ) , (5.37)

and

| ( f ( u ) − f ( u L ) , c A ν u N z ( θ t ω ) ) | ≤ C | c | | z ( θ t ω ) | ∫ U | u N | ⋅ ( 1 + | u | 4 + | u L | 4 ) | A ν u N | d x ≤ C | c | | z ( θ t ω ) | ( ∫ U | u N | 10 5 − 4 ν d x ) 5 − 4 ν 10 ⋅ ( ∫ U ( 1 + | u | 4 + | u L | 4 ) 10 4 d x ) 4 10         ⋅ ( ∫ U | A ν u N | 10 1 + 4 ν d x ) 1 + 4 ν 10 ≤ C | c | | z ( θ t ω ) | ‖ A ν 2 u N ‖ ‖ A ν 2 u N ‖ 2 ( 1 + ‖ u ‖ 2 4 + ‖ u L ‖ 2 4 ) ≤ R 6 ( ω ) | c | | z ( θ t ω ) | ‖ A ν 2 u N ‖ 2 2 . (5.38)

In addition, by (1.2), (5.2)₁, (5.24) and Lemma 5.1, we find that:

| ( f 2 ( u L ) , A ν u N ) | ≤ C ∫ U | u L | ⋅ ( 1 + | u L | p ) | A ν u N | d x ≤ C ( ∫ U | u L | 2 d x ) 1 2 ⋅ ( ∫ U ( 1 + | u L | p ) 10 4 − 4 ν d x ) 4 − 4 ν 10 ⋅ ( ∫ U | A ν u N | 10 1 + 4 ν d x ) 1 + 4 ν 10 ≤ C ‖ u L ‖ ⋅ ‖ A ν 2 u N ‖ 2 ⋅ ( 1 + ‖ u L ‖ 2 p ) ≤ R 7 ( ω ) ‖ A ν 2 u N ‖ 2 , (5.39)

and

| ( f ( u ) − f ( u L ) , A ν u N ) | ≤ C ∫ U | u N | ⋅ ( 1 + | u | 4 + | u L | 4 ) | A ν u N | d x ≤ C ( ∫ U | u N | 10 5 − 4 ν d x ) 5 − 4 ν 10 ⋅ ( ∫ U ( 1 + | u | 4 + | u L | 4 ) 10 4 d x ) 4 10 ⋅ ( ∫ U | A ν u N | 10 1 + 4 ν d x ) 1 + 4 ν 10 ≤ C ‖ A ν 2 u N ‖ ( 1 + ‖ u ‖ 2 4 + ‖ u L ‖ 2 4 ) ‖ A ν 2 u N ‖ 2

≤ R 8 ( ω ) ‖ A ν 2 u N ‖ 2 2 . (5.40)

Thus, combining with (5.23)-(5.40), we can show that:

d d t ( ‖ A ν 2 φ N ‖ E 2 + 2 ( f ( u ) − f 1 ( u L ) , A ν u N ) ) ≤ ( σ 2 ( ω ) + 1 + 2 R 6 ( ω ) λ 1 + 2 λ 1 λ 1 | c | | z | + 16 ε 2 + α α 2 λ 1 | c | 2 | z | 2 )         ⋅ ( ‖ A ν 2 φ N ‖ E 2 + 2 ( f ( u ) − f 1 ( u L ) , A ν u N ) ) + R 9 ( ω ) + 2 C | c | 2 | z | 2 + C ‖ g ‖ 1 2 , (5.41)

that is,

d d t ( ‖ A ν 2 φ N ‖ E 2 + 2 ( f ( u ) − f 1 ( u L ) , A ν u N ) ) −   ρ 1 ( t , θ t ω ) ( ‖ A ν 2 φ N ‖ E 2 + 2 ( f ( u ) − f 1 ( u L ) , A ν u N ) ) ≤ R 9 ( ω ) + 2 C | c | | z | + C ‖ g ‖ 1 2 , (5.42)

where

ρ 1 ( t , θ t ω ) = σ 2 ( ω ) + 1 + 2 R 6 ( ω ) λ 1 + 2 λ 1 λ 1 | c | | z | + 16 ε 2 + α α λ 1 | c | 2 | z | 2 . (5.43)

Let

Y ( t ) = ‖ A ν 2 φ N ‖ E 2 + 2 ( f ( u ) − f 1 ( u L ) , A ν u N ) .

By Gronwall’s inequality to (5.42), we have:

Y ( t ) ≤ e ∫ 0 t   ρ 1 ( s , θ s ω ) d s Y ( 0 ) + 2 C | c | 2 ∫ 0 t     e ∫ s t   ρ 1 ( τ , θ τ ω ) d τ | z ( θ s ω ) | 2 d s   + ( R 9 ( ω ) + C ‖ g ( x ) ‖ 1 2 ) ∫ 0 t     e ∫ s t   ρ 1 ( τ , θ τ ω ) d τ d s .

Similar to the proof of Lemma 4.1, we have:

Y ( t , θ − t ω , φ ( 0, θ − t ω ) ) ≤ e ∫ − t 0   ρ 1 ( s , θ s ω ) d s Y ( 0, θ − t ω , φ ( 0, θ − t ω ) ) + 2 C | c | 2 ∫ − t 0     e ∫ s 0   ρ 1 ( τ , θ τ ω ) d τ | z ( θ s ω ) | 2 d s         + ( R 10 ( ω ) + C ‖ g ( x ) ‖ 1 2 ) ∫ − t 0   e ∫ s 0   ρ 1 ( τ , θ τ ω ) d τ d s . (5.44)

Due to (4.26) again, there exists σ 2 ( ω ) > 0 such that:

e ∫ s 0   ρ 1 ( τ , θ τ ω ) d τ ≤ e − σ 2 ( ω ) 2 s , (5.45)

where c satisfies (4.10). Thus, it follows from (5.39)-(5.40) and (5.43)-(5.46) that:

‖ A 1 + ν 2 u N ‖ 2 + ‖ A ν 2 u N t ‖ 2 + ‖ A ν 2 η N ‖ μ ,2 2 ≤ e σ 2 ( ω ) 2 t P 1 ( θ − t ω ) , (5.46)

where P ( θ − t ω ) is increasing function. Then, the proof is finished. □

Lemma 5.4. Assume that (h₂) holds. For any ι > 0 , there exists C ι and M ι ( ω ) , such that ℙ -a.s. u t ( s ) = u 1 ( s ) + u 2 ( s ) ,

‖ u 1 ( s ) ‖ 2 + 2 ν ≤ K ι ( ω ) , ∀ − t ≤ s ≤ 0, (5.47)

∫ r 1 r 2 ‖ u 2 ‖ 2 2 d s ≤ ι ( r 2 − r 1 ) + C ι , ∀ − t ≤ r 1 ≤ r 2 ≤ 0. (5.48)

Proof. Combining (5.14), (5.20) with the technique used in [55] , we can finish the proof. □

Lemma 5.5. Assume that (h₁) - (h₂) and (5.1)-(5.2) hold. There exists a random radius M 3 ( ω ) such that for P-a.e. ω ∈ Ω , the solution φ N ( t , θ − t ω , φ N ( 0, θ − t ω ) ) of (5.4) satisfies:

‖ A 1 + ν 2 u N ‖ 2 + ‖ A ν 2 u N t ‖ 2 + ‖ A ν 2 η N ‖ μ ,2 2 ≤ M 3 ( ω ) , (5.49)

where ν is given in (5.21).

Proof. Taking the inner product of ( ⋅ , ⋅ ) E of (5.4) with ( A ν φ N , A ν w N , A ν η N ) T , we find that:

1 2 d d t [ ‖ A ν 2 φ N ‖ E 2 + 2 ( f ( u ) − f 1 ( u L ) , A ν 2 u N ) ] + ( H ( φ ) , φ ) E + ε ( f ( u ) − f 1 ( u L ) , A ν 2 u N ) − ( [ f ′ 1 ( u ) − f ′ 1 ( u L ) ] u t , A ν 2 u N ) − ( f ′ 1 ( u L ) u N t , A ν 2 u N ) − ( f ′ 2 ( u ) u t , A ν 2 u N ) = ( g ( x ) , A ν w N ) − ( c z ( θ t ω ) w N , A ν w N ) + ( 2 ε c u N z ( θ t ω ) , A ν w N )         − ( c 2 z 2 ( θ t ω ) u N , A ν w N ) + ( ( c u N z ( θ t ω ) , A ν u N ) )         + ( c u N z ( θ t ω ) , A ν η N ) ν ,2 − ( f ( u ) − f 1 ( u L ) , c A ν u N z ( θ t ω ) ) (5.50)

First, we deal with the nonlinearity in (5.50). Applying (5.1), (5.6), (5.22) and Hölder’s inequality, we have:

| ( f ′ 1 ( u ) u t − f ′ 1 ( u L ) u t , A ν u N ) | ≤ C ∫ U | u 1 + u 2 | ⋅ | u N | ⋅ ( 1 + | u | 3 + | u L | 3 ) | A ν u N | d x ≤ C ( ∫ U | u 2 | 2 d x ) 1 2 ⋅ ( ∫ U | u N | 10 1 − 4 ν d x ) 1 − 4 ν 10 ⋅ ( ∫ U ( 1 + | u | 3 + | u L | 3 ) 10 3 d x ) 3 10         ⋅ ( ∫ U | A ν u N | 10 1 + 4 ν d x ) 1 + 4 ν 10 + C ( ∫ U | u 1 | 10 1 − 4 ν d x ) 1 − 4 ν 10 ⋅ ( ∫ U | u N | 2 d x ) 1 2         ⋅ ( ∫ U ( 1 + | u | 3 + | u L | 3 ) 10 3 d x ) 3 10 ⋅ ( ∫ U | A ν u N | 10 1 + 4 ν d x ) 1 + 4 ν 10 ≤ R 10 ( ω ) ( ‖ u 2 ‖ 2 + ‖ A 1 + ν 2 u 1 ‖ ) ‖ A ν 2 u N ‖ 2 2 , (5.51)

| ( f ′ 1 ( u L ) u N t , A ν u N ) | ≤ C 1 ∫ U | u N t | ⋅ ( 1 + | u L | 4 ) | A ν u N | d x ≤ C 1 ( ∫ U | u N t | 10 5 − 4 ν d x ) 5 − 4 ν 10 ⋅ ( ∫ U ( 1 + | u L | 4 ) 10 4 d x ) 4 10 ⋅ ( ∫ U | A ν u N | 10 1 + 4 ν d x ) 1 + 4 ν 10 ≤ C ‖ A ν 2 u N t ‖ ⋅ ( 1 + ‖ u L ‖ 2 4 ) ⋅ ‖ A ν 2 w N ‖ 2 ≤ R 11 ( ω ) ‖ A ν 2 w N ‖ 2 ⋅ ( ‖ A ν 2 w N ‖ + ε ) ≤ R 11 ( ω ) ‖ A ν 2 w N ‖ 2 2 + α 4 ‖ A ν 2 w N ‖ 2 + R 12 ( ω ) ‖ A ν 2 w N ‖ 2 , (5.52)

| ( f ′ 2 ( u ) u t , A ν u N ) | ≤ C ∫ U | u 1 + u 2 | ⋅ ( 1 + | u | p ) | A ν u N | d x ≤ C ( ∫ U | u 1 | 10 d x ) 1 10 ⋅ ( ∫ U ( 1 + | u | p ) 10 4 − 4 ν d x ) 4 − 4 ν 10 ⋅ ( ∫ U | A ν u N | 10 5 + 4 ν d x ) 5 + 4 ν 10         + C ( ∫ U | u 2 | 2 d x ) 1 2 ⋅ ( ∫ U ( 1 + | u | p ) 10 4 − 4 ν d x ) 4 − 4 ν 10 ⋅ ( ∫ U | A ν u N | 10 1 + 4 ν d x ) 1 + 4 ν 10

≤ C ( ‖ u 2 ‖ 2 + ‖ A 1 2 u 1 ‖ ) ⋅ ( 1 + ‖ u ‖ 2 p ) ‖ A ν 2 u N ‖ ≤ C ( ‖ u 2 ‖ 2 + ‖ A 1 + ν 2 u 1 ‖ ) ⋅ ( 1 + ‖ u ‖ 2 p ) ‖ A ν 2 u N ‖ 2 ≤ R 13 ( ω ) ( ‖ u 2 ‖ 2 + ‖ A 1 + ν 2 u 1 ‖ ) ‖ A ν 2 u N ‖ 2 . (5.53)

By (5.26)-(5.31), (5.34), (5.38) and (5.50)-(5.53), we obtain that:

d d t Y ˜ ( t ) + ( ρ 3 ( ω , θ t ω ) − R 14 ( ω ) ‖ u 2 ‖ 2 2 ) Y ˜ ( t ) ≤ 4 α ‖ A ν 2 g ‖ 2 + 2 C | c | 2 | z ( θ t ω ) | 2 + R 15 ( ω ) , (5.54)

where

σ 3 = min { ε 2 , δ 2 } ,

Y ˜ ( t ) = ‖ A ν 2 φ N ‖ E 2 + 2 ( f ( u ) − f 1 ( u L ) , A ν 2 u N )

ρ 3 ( ω , θ t ω ) = σ 3 − 16 ε 2 + α α λ 1 | c | 2 | z ( θ t ω ) | 2 − 2 λ 1 + 1 + 2 C λ 1 λ 1 | c | | z ( θ t ω ) | .

By Gronwall’s inequality to (5.54), we get that:

Y ˜ ( t ) ≤ e − ∫ 0 t ( ρ 3 ( s , θ s ω ) − R 14 ( ω ) ‖ u 2 ( s ) ‖ 2 2 ) d s Y ˜ ( 0 ) + 2 C | c | ∫ 0 t     e − ∫ s t ( ρ 3 ( τ , θ τ ω ) − R 14 ( ω ) ‖ u 2 ( τ ) ‖ 2 2 ) d τ | z ( θ s ω ) | 2 d s   + ( R 15 ( ω ) + C ‖ g ( x ) ‖ 1 2 ) ∫ 0 t     e − ∫ s t ( ρ 3 ( τ , θ τ ω ) − R 14 ( ω ) ‖ u 2 ( τ ) ‖ 2 2 ) d τ d s = e − ∫ 0 t ( ρ 3 ( s , θ s ω ) − R 14 ( ω ) ‖ u 2 ( s ) ‖ 2 2 ) d s Y ˜ ( 0 ) + M 4 ( ω ) , (5.55)

where

M 4 ( ω ) = 2 C | c | ∫ 0 t     e − ∫ s t ( ρ 3 ( τ , θ τ ω ) − R 14 ( ω ) ‖ u 2 ( τ ) ‖ 2 2 ) d τ | z ( θ s ω ) | 2 d s   + ( R 15 ( ω ) + C ‖ g ( x ) ‖ 1 2 ) ∫ 0 t     e − ∫ s t ( ρ 3 ( τ , θ τ ω ) − R 14 ( ω ) ‖ u 2 ( τ ) ‖ 2 2 ) d τ d s .

For any ι > 0 such that ι < σ 3 2 R 13 ( ω ) is so small,

e ∫ 0 s   R 14 ( ω ) ‖ u 2 ( τ ) ‖ 2 2 d τ ≤ e 1 4 σ 3 s e R 16 ( ω ) .

Next, similar to the proof of Lemma 4.1, we know that:

e ∫ 0 s   ρ 3 ( τ , θ τ ω ) d τ ≤ e σ 3 2 s , (5.56)

where c satisfies (4.10).

Hence, combining (5.39)-(5.40) with tempered | z ( θ t ω ) | , we obtain that:

‖ A 1 + ν 2 u N ‖ 2 + ‖ A ν 2 u N t ‖ 2 + ‖ A ν 2 η N ‖ μ ,2 2 ≤ e − σ 3 2 t Y ˜ ( 0 ) + M 4 ( ω ) ≤ M 3 ( ω ) .

Then, the proof is completed. □

In this section, we establish the existence of a D -random attractor for the random dynamical system Φ associated with system (3.12) on ℝ 5 , that is, by Lemma 4.1, Φ has a closed random absorbing set in D , which along with the D -pullback asymptotic compactness and then implies the existence of a unique D -random attractor. Next, due to decomposition of solutions, we shall prove the D -pullback asymptotic compactness of Φ (see [56] [57] ).

Since ω ∈ Ω , t ≥ 0 , we get:

η N ( t , θ − t ω , φ 0 ( θ − t ω ) , s ) = { u N ( t , θ − t ω , φ 0 ( θ − t ω ) ) − u N ( t − s , θ − t ω , φ 0 ( θ − t + s ω ) ) ,   s ≤ t , u N ( t , θ − t ω , φ 0 ( θ − t ω ) ) ,   t ≤ s ; (6.1)

η N s ( t , θ − t ω , φ 0 ( θ − t ω ) , s ) = { u N t ( t − s , θ − t + s ω , φ 0 ( θ − t + s ω ) ) ,   0 ≤ s ≤ t , 0, t ≤ s . (6.2)

Denote B ˜ ( ω ) as:

B ˜ ( ω ) = ∪ φ 0 ( θ − t ω ) ∈ B ( θ − t ω ) ∪ t ≥ 0   η N ( t , θ − t ω , φ 0 ( θ − t ω ) , s ) ¯ , s ∈ ℝ + , ω ∈ Ω , t ≥ 0 , (6.3)

is the solution of (3.12), where φ = ( u , w , η ) Τ . Next, it follows from Lemma 5.5 and (6.1)-(6.2) that:

max { ‖ η N s ( t , θ t ω , φ 0 ( θ − t ω ) , s ) ‖ μ ,2 ν 2 , ‖ η N ( t , θ − t ω , φ 0 ( θ − t ω ) , s ) ‖ μ ,2 ν + 2 2 } ≤ M 3 ( ω ) , ∀ s ≥ 0, (6.4)

which implies B ˜ ( ω ) is bounded in L μ 2 ( ℝ + , V 2 ν + 2 ) ∩ H μ 2 ( ℝ + , V 2 μ ) . Also, by Lemmas 4.1, 5.5 and (6.2), there holds:

sup η ∈ B ˜ ( ω ) , s ≥ 0 ‖ Δ η ( s ) ‖ 2 = sup t ≥ sup φ 0 ( θ − t ω ) ∈ B ( θ − t ω ) ‖ Δ η N ( t , θ − t ω , φ 0 ( θ − t ω ) , s ) ‖ 2 ≤ 2 M 0 ( ω ) . (6.5)

By (h₁) and (6.5), for any η ∈ B ˜ ( ω ) , we find that:

‖ η ( s ) ‖ μ ,2 2 = ∫ 0 + ∞   μ ( s ) ‖ Δ η ( s ) ‖ 2 d s ≤ 2 R 1 ( ω ) ∫ 0 + ∞     e − δ s d s ≤ 2 M 0 ( ω ) δ , (6.6)

which shows that B ˜ ( ω ) ⊂ L μ 2 ( ℝ + , H 0 2 ( U ) ) is bounded. It follows from Lemma 2.11 that the set B ˜ ( ω ) is relatively compact in L μ 2 ( ℝ + , H 0 2 ( U ) ) . Next, we investigate the main result about the existence of a random attractor for Random Dynamical System Φ.

Lemma 6.1. Assume that (h₁) - (h₂) hold. Then, for any t ≥ 0, ω ∈ Ω , the RDS Φ associated with (3.12) possesses a uniformly D ( E ) -attracting set Λ ( ω ) ⊂ E and possesses a D ( E ) -random attractor A ( ω ) ⊆ Λ ( ω ) ∩ B 0 ( ω ) .

Proof For any t ≥ 0, ω ∈ Ω , as Lemma 5.5, let B ν ( ω ) be the closed ball in V 2 + 2 ν × V 2 ν of radius M 5 ( ω ) . Set:

Λ ( ω ) = B ν ( ω ) × B ˜ ( ω ) . (6.7)

Then, Λ ( ω ) ∈ D ( E ) . Because V 2 + 2 ν × V 2 ν ↪ H 0 2 ( U ) × L 2 ( U ) is compact, and B ν ( ω ) is compact in H 0 2 ( U ) × L 2 ( U ) . At the same time, B ˜ ( ω ) is compact in ℜ μ ,2 , then Λ ( ω ) is compact in E. Next, we prove the following attraction property of Λ ( ω ) : for every B ( ω ) ∈ D ( E ) ,

l i m t → ∞ d E ( Φ ( t , θ − t ω , B ( θ − t ω ) ) , Λ ( ω ) ) = 0. (6.8)

Indeed, firstly, according to Lemma 4.1, there exists closed, tempered and measurable absorbing set B 0 ( ω ) such that B ∈ D ( E ) , for any t B ( ω ) ≥ 0 ,

φ ( t , θ − t ω , B ( θ − t ω ) ) ⊆ B 0 ( ω ) , ∀ t ≥ t B ( ω ) . (6.9)

Moreover, let:

B 1 ( ω ) = ∪ t ≥ t ( ω , B 0 )   Φ ( t , θ − t ω ) B 0 ( θ − t ω ) .

Assume that t ≥ t B ( ω ) and t 0 = t − t B ( ω ) > t B 0 ( ω ) > 0 . Making use of the property 3) of Φ and (6.9), we deduce that:

φ ( t , θ − t ω , B ( θ − t ω ) ) = φ ( t 0 + t B ( ω ) , θ − t 0 − t B ( ω ) ω , B ( θ − t 0 − t B ( ω ) ω ) ) = φ ( t 0 , θ − t 0 ω , φ ( t B ( ω ) , θ − t 0 − t B ( ω ) ω , B ( θ − t 0 − t B ( ω ) ω ) ) ) ⊆ φ ( t 0 , θ − t 0 ω , B 0 ( θ − t 0 ω ) ) ⊆ B 1 ( ω ) . (6.10)

For any t ≥ t B ( ω ) + t B 0 ( ω ) , choose φ ( t , θ − t ω , φ 0 ( θ − t ω ) ) ∈ φ ( t , θ − t ω , B ( θ − t ω ) ) , where φ 0 ( θ − t ω ) ∈ B ( θ − t ω ) . Due to (6.10) and Lemma 5.1, we have:

φ N ( t , θ − t ω , φ N ( 0, θ − t ω ) ) = φ ( t , θ − t ω , φ 0 ( θ − t ω ) ) − φ L ( t , θ − t ω , φ l ( 0, θ − t ω ) ) ∈ Λ ( ω ) .

So, according to Lemma 5.2, we find that for t ≥ t B ( ω ) + t B 0 ( ω ) ,

inf ψ ∈ Λ ( ω ) ‖ φ ( t , θ − t ω , φ 0 ( θ − t ω ) ) − ψ ‖ E 2 ≤ ‖ φ L ( t , θ − t ω , φ L ( 0, θ − t ω ) ) ‖ E 2 ≤ M 2 ( ω ) e − σ 1 ( ω ) t ,

that is,

dist ( φ ( t , θ − t ω ) B ( θ − t ω ) , Λ ( ω ) ) → 0, t → + ∞

Thus, (6.8) holds. Therefore, applying Lemma 2.11 and Theorem 4.1, we obtain that the RDS Φ possesses a D ( E ) -pullback random attractors A ( ω ) ⊆ Λ ( ω ) ∩ B 0 ( ω ) .

Then, the proof is completed. □

The authors would like to thank anonymous reviewers for their helpful comments that improved the presentation of this work. Mohamed Y. A. Bakhet acknowledges and thanks the financial support from the University of Juba, South Sudan.

The authors declare no conflicts of interest regarding the publication of this paper.

Bakhet, M.Y.A., Dafallah, A.A., Wang, J., Ma, Q.Z., Mosa, F.M., Mohamed, A.E., Lolika, P.O. and Chinor, M.M. (2024) Dynamics of Plate Equations with Memory Driven by Multiplicative Noise on Bounded Domains. Journal of Applied Mathematics and Physics, 12, 1492-1521. https://doi.org/10.4236/jamp.2024.124092