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Existence and upper semicontinuity of random attractors for non-autonomous stochastic strongly damped sine–Gordon equation on unbounded domains
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Existence and upper semicontinuity of random attractors for non-autonomous stochastic strongly damped sine–Gordon equation on unbounded domains✩ Zhaojuan Wang ∗ , Yanan Liu School of Mathematical Science, Huaiyin Normal University, Huaian 223300, PR China
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Article history: Received 12 October 2016 Received in revised form 14 January 2017 Accepted 26 January 2017 Available online xxxx Keywords: Stochastic strongly damped sine–Gordon equation Unbounded domains Random attractor
In this paper we study the asymptotic behavior of solutions of the non-autonomous stochastic strongly damped sine–Gordon equation driven by multiplicative noise defined on an unbounded domain. First we introduce a continuous cocycle for the equation and establish the pullback asymptotic compactness of solutions. Second we consider the existence and upper semicontinuity of random attractors for the equation. © 2017 Elsevier Ltd. All rights reserved.
1. Introduction A random attractor, introduced by Crauel et al. [1] and Crauel and Flandoli [2] to capture the essential dynamics with possibly extremely wide fluctuations, is a generalization of the global attractor of autonomous differential equations for random dynamical systems. In recent years, random attractors for non-autonomous stochastic partial differential equations have been investigated in [3,4] in bounded domains and in [5–9] on unbounded domains. In this paper, we consider the existence of a random attractor for the following non-autonomous stochastic strongly damped sine–Gordon equation with multiplicative noise defined in the entire space Rn (n ∈ N): utt − α 1ut − 1u + ut + λu + µ sin u = g (x, t ) + cu ◦
dW dt
,
(1.1)
with the initial value conditions u(τ , x) = uτ (x),
ut (τ , x) = u1τ (x),
x ∈ Rn ,
(1.2)
where ∆ is the Laplacian with respect to the variable x ∈ R ; u = u(t , x) is a real function of x ∈ R and t > τ , τ ∈ R; α > 0 is the strong damping coefficient; λ > 0, µ and c are constants; g (x, ·) is a given function in L2loc (R, L2 (Rn )); ◦ denotes the Stratonovich sense in the stochastic term; W (t ) is a two-sided real-valued Wiener process on a probability space. If g does not depend on time, then we call Eq. (1.1) an autonomous equation. n
n
✩ The authors are supported by National Natural Science Foundation of China (Nos. 11326114, 11401244); Natural Science Research Project of Ordinary Universities in Jiangsu Province (No. 14KJB110003); Undergraduate Innovation and Entrepreneurship Training Program of Jiangsu Province (No. 201616003xj). ∗ Corresponding author. E-mail address: [email protected] (Z. Wang).
http://dx.doi.org/10.1016/j.camwa.2017.01.015 0898-1221/© 2017 Elsevier Ltd. All rights reserved.
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Sine–Gordon equation describes the dynamics of continuous Josephson junctions (see [10]) and has a wide range of applications in physics. Recently, several authors [11–13] studied the attractors for a stochastic damped sine–Gordon equation on a bounded domain, but not on unbounded domains. So far as we know, there were no results on random attractors for non-autonomous stochastic strongly damped sine–Gordon equation (1.1) on unbounded domains. In general, the existence of global random attractor depends on some kind compactness (see, e.g., [14,1,2]). The main difficulty of this paper is to prove the asymptotic compactness of solutions, because Sobolev compact embedding is lost for unbounded domain. In order to overcome the difficulty, we use uniform estimates on the tails of solutions outside a bounded ball in Rn , then decompose the solutions in a bounded domain in terms of eigenfunctions of negative Laplacian as in [15,16]. On are of great physical interest. It is the other hand, the cases of g (x, t ) depending on time and multiplicative noise cu ◦ dW dt therefore important to investigate the existence of attractors for Eq. (1.1). In these cases, we need two separate parametric spaces to deal with the deterministic perturbations g (x, t ) as well as the stochastic perturbations cu ◦ dW . dt This paper is organized as follows. In the next section, we recall some basic settings for Eq. (1.1) and define a continuous cocycle in H 1 (Rn )× L2 (Rn ). In Section 3, we first derive all necessary uniform estimates of solutions, then prove the existence and uniqueness of tempered pullback random attractors for Eq. (1.1). In Section 4, we prove the upper semicontinuity of random attractors as the coefficient of the white noise term tends to zero. Throughout this paper, we use ∥ · ∥ and ⟨·, ·⟩ to denote the norm and the inner product of L2 (Rn ), respectively. The norm of a Banach space X is generally written as ∥ · ∥X . The letters c and ci (i = 1, 2, . . .) are used to denote positive constants whose values are not significant in the context.
2. Mathematical settings Let (Ω , F , P ) be the standard probability space where Ω = {ω ∈ C (R, R) : ω(0) = 0}, F is the Borel σ -algebra induced by the compact open topology of Ω , and P is the Wiener measure on (Ω , F ) (see [17]). There is a classical group {θt }t ∈R acting on (Ω , F , P ) which is defined by θt ω(·) = ω(· + t ) − ω(t ), for ω ∈ Ω , t ∈ R. Then (Ω , F , P , {θt }t ∈R ) is a metric dynamical system. Denote the inner and norm of L2 (U ) as ⟨·, ·⟩ and ∥ · ∥, respectively, and E (U ) = H 1 (U ) × L2 (U ), U ⊆ Rn , endowed with the usual norm
1 ∥Y ∥H 1 (U )×L2 (U ) = ∥∇ u∥2 + ∥u∥2 + ∥v∥2 2
for Y = (u, v)⊤ ∈ E (U ),
(2.1)
where ⊤ stands for the transposition. We define a new norm ∥ · ∥E (U ) by
1 ∥Y ∥E (U ) = ∥v∥2 + δ 2 + λ − δ ∥u∥2 + 1 − αδ ∥∇ u∥2 2 ,
(2.2)
where δ satisfied (3.6), for Y = (u, v)⊤ ∈ E (U ). It is easy to check that ∥ · ∥E (U ) is equivalent to the usual norm ∥ · ∥H 1 (U )×L2 (U ) in (2.1). For our purpose, it is convenient to convert the problem (1.1)–(1.2) into a deterministic system with a random parameter, and then show that it has a cocycle on E (Rn ) over R and (Ω , F , P , {θt }t ∈R ). We identify ω(t ) with W (t ), i.e., ω(t ) = W (t ), t ∈ R. Consider Ornstein–Uhlenbeck equation dz + α zdt = dW (t ), and Ornstein–Uhlenbeck process z (θt ω) = 0 −α −∞ eαs (θt ω)(s)ds, t ∈ R. From [18,19], it is known that the random variable |z (θt ω)| is tempered, and there is a
⊂ Ω of full P measure such that z (θt ω) is continuous in t for every ω ∈ Ω . θt -invariant set Ω Lemma 2.1 (See [12]). For the Ornstein–Uhlenbeck process z (θt ω), we have the following results
|z (θt ω)| = 0, |t | 1 t lim z (θs ω)ds = 0,
(2.3)
lim
t →∞
t →∞
lim
t →∞
lim
t →∞
t
1
t
t
t 1
(2.4)
0
1
|z (θs ω)|ds = √ 0 t
|z (θs ω)|2 ds = 0
πα
1 2α
.
,
(2.5)
(2.6)
Let v(t , τ , ω) = ut (t , τ , ω) + δ u(t , τ , ω) − cu(t , τ , ω)z (θt ω), where δ is as in (2.2), then (1.1)–(1.2) can be rewritten as the equivalent system with random coefficients but without white noise
Z. Wang, Y. Liu / Computers and Mathematics with Applications (
)
du = v − δ u + cuz (θt ω), dt dv = α 1v + (1 − αδ)1u + (δ − 1)v + (δ − λ − δ 2 )u + α cz (θt ω)1u dt + 2c δ z (θt ω) − c 2 z 2 (θt ω) u − cz (θt ω)v − µ sin u + g (x, t ), u(τ , x) = uτ (x), v(τ , x) = vτ (x),
–
3
(2.7)
where vτ (x) = u1τ (x) + δ uτ (x) − cuτ (x)z (ω), x ∈ Rn . and write Ω as Ω from now on. When deriving uniform estimates on solutions, we also We will consider (2.7) for ω ∈ Ω need the following condition on g:
0
δs
e 2 ∥g (·, s + τ )∥2 ds < ∞,
∀ τ ∈ R,
(2.8)
−∞
which implies that
0
δs
e 2 |g (x, s + τ )|2 dxds = 0,
lim
k→∞
−∞
∀ τ ∈ R,
(2.9)
|x|>k
where | · | denotes the absolute value of real number in R. Put ϕ (c ) (t + τ , τ , θ−τ ω, ϕτ(c ) ) = (u(t + τ , τ , θ−τ ω, uτ , vτ ), v(t + τ , τ , θ−τ ω, uτ , vτ ))⊤ , where ϕτ(c ) = (uτ , vτ )⊤ . By the classical theory concerning the existence and uniqueness of the solutions [20,21], one may show that for every ω ∈ Ω , τ ∈ R and ϕτ(c ) ∈ E (Rn ), problem (2.7) has a unique solution ϕ (c ) (·, τ , ω, ϕτ(c ) ) ∈ C ([τ , ∞), E (Rn )) with ϕ (c ) (τ , τ , ω, ϕτ(c ) ) = ϕτ(c ) . In addition, for each t > τ , ϕ (c ) (t , τ , ω, ϕτ(c ) ) is (F , B (H 1 (Rn )) × B (L2 (Rn )))–measurable and continuous in ϕτ with respect to the usual norm of E (Rn ). Based on this fact, one can define a continuous cocycle for (2.7). Let Φc be a mapping, Φc : R+ × R × Ω × E (Rn ) → E (Rn ) given by
Φc (t , τ , ω, ϕτ(c ) ) = ϕ (c ) (t + τ , τ , θ−τ ω, ϕτ(c ) )
(2.10)
for every (t , τ , ω, ϕτ(c ) ) ∈ R+ × R × Ω × E (Rn ). Then it is easy to prove that Φc is a continuous cocycle over R and (Ω , F , P , {θt }t ∈R ) on E (Rn ).
⊤ ⊤ ⊤ = a, b − δ a + caz (θt ω) , a, b ∈ E (Rn ) with an inverse ⊤ ⊤ −1 homeomorphism P (c , δ, θt ω) a, b = a, +δ a − caz (θt ω) . Then, the transformation ¯ c (t , τ , ω, (uτ , u1τ )) = P (c , δ, θt ω)Φc (t , τ , ω, ϕτ(c ) )P −1 (c , δ, θτ ω) Φ (2.11) Introducing the homeomorphism P (c , δ, θt ω) a, b
generates a continuous cocycle with (1.1)–(1.2) over R and (Ω , F , P , {θt }t ∈R ) on E (Rn ). c has a random attractor Note that these two continuous cocycles are equivalent. By (2.11), it is easy to check that Φ provided Φc possesses a random attractor. Then, we only need to consider the continuous cocycle Φc . 3. Existence of random attractors We first provide some sufficient conditions for the existence of random attractors for non-autonomous RDSs in [7]. Similar results can be found in [17,22,14,1,2] for autonomous random dynamical systems. In the following, let (X , ∥ · ∥X ) be a separable Banach space, D (X ) be the collection of all tempered families of nonempty bounded subsets of X . Remember that a family D = {D(τ , ω) : τ ∈ R, ω ∈ Ω } of nonempty bounded subsets of X is said to be tempered if for every σ > 0, τ ∈ R and ω ∈ Ω , the following holds: lim eσ t ∥D(τ + t , θt ω)∥X = 0,
t →−∞
(3.1)
where ∥D∥X = supx∈D ∥x∥X . The cocycle Ψ is said to be D (X )-pullback asymptotically compact in X if for all τ ∈ R and ω ∈ Ω , the sequence
{Ψ (tn , τ − tn , θ−tn ω, xn )}∞ n=1 has a convergent subsequence in X ,
(3.2)
whenever tn → ∞, and xn ∈ D(τ − tn , θ−tn ω) with {D(τ , ω) : τ ∈ R, ω ∈ Ω } ∈ D (X ). Definition 3.1. Let Ψ be a continuous cocycle on X over R and (Ω , F , P , {θt }t ∈R ). (1) A family B = {B(τ , ω) : τ ∈ R, ω ∈ Ω } ∈ D (X ) is called a D (X )-pullback absorbing set for Ψ if for all τ ∈ R and ω ∈ Ω and for every D ∈ D (X ), there exists T = T (D, τ , ω) > 0 such that
Ψ (t , τ − t , θ−t ω, D(τ − t , θ−t ω)) ⊆ B(τ , ω) for all t ≥ T .
(3.3)
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(2) A family A = {A(τ , ω) : τ ∈ R, ω ∈ Ω } ∈ D (X ) is called a D (X )-pullback attractor for Ψ if for all t ∈ R+ , τ ∈ R and ω ∈ Ω , (i) A(τ , ω) is compact in X and is measurable in ω with respect to F ; (ii) A is invariant, that is, Ψ (t , τ , ω, A(τ , ω)) = A(τ + t , θt ω); (iii) For every D = {D(τ , ω) : τ ∈ R, ω ∈ Ω } ∈ D (X ), lim dH (Ψ (t , τ − t , θ−t ω, D(τ − t , θ−t ω)), A(τ , ω)) = 0,
t →∞
(3.4)
where dH is the Hausdorff semi-distance given by dH (F , G) = supu∈F infv∈G ∥u − v∥X , for any F , G ⊂ X . Proposition 3.2. Let Ψ be a continuous cocycle on X over R and (Ω , F , P , {θt }t ∈R ). Suppose Ψ is D (X )-pullback asymptotically compact in X and has a closed measurable D (X )-pullback absorbing set K in D (X ). Then Ψ has a unique D (X )-pullback attractor A in D (X ) which is given by, for each τ ∈ R and ω ∈ Ω ,
A(τ , ω) =
Ψ (t , τ − t , θ−t ω, K (τ − t , θ−t ω)).
(3.5)
τ >0 t >τ
Next, we will use Proposition 3.2 to prove the existence of a random attractor for the continuous cocycle Φc in E (Rn ). Let δ ∈ (0, 12 ) be small enough such that
δ 2 + λ − δ > 0,
1 − αδ > 0.
(3.6)
Throughout this section we assume that
2|c |δ 2|c | c2 c2 γ =δ− √ − max , (δ 2 + λ − δ + 1) √ + + |µ| . 2(1 − αδ) 2α πα πα
(3.7)
Lemma 3.3. Assume that g (x, ·) ∈ L2loc (R, L2 (Rn )) and (2.8) hold. If γ > 0, then for every τ ∈ R, ω ∈ Ω , and D = {D(τ , ω) : τ ∈ R, ω ∈ Ω } ∈ D (E (Rn )), there exists T = T (τ , ω, D) > 0 such that for all t > T , δ τ τs 32δ −2|cz (θr −τ ω)|−R(θr −τ ω) dr (c ) ∥ϕ (c ) (τ , τ − t , θ−τ ω, ϕτ(c−) t )∥2E (Rn ) + ∥ϕ (s, τ − t , θ−τ ω, ϕτ(c−) t )∥2E (Rn ) ds e 2 τ −t τ s 3δ −2|cz (θr −τ ω)|−R(θr −τ ω) dr ∥∇v(s, τ − t , θ−τ ω, ϕτ(c−) t )∥2 ds +α eτ 2 τ −t
< ϱ(c , τ , ω),
(3.8)
where
ϱ(c , τ , ω) = c + c
0
s 3δ
e
0
2
−2|cz (θr ω)|−R(θr ω) dr
∥g (x, s + τ )∥2 ds,
(3.9)
−∞
ϕτ(c−) t = (uτ −t , vτ −t )⊤ ∈ D(τ − t , θ−t ω), c is a positive constant depending on λ, µ, α and δ , but independent of τ , ω and D, R(θt ω) is given by α R(θt ω) = max |cz (θt ω)|2 , (δ 2 + λ − δ + 1) 2|c δ z (θt ω)| + |cz (θt ω)|2 + |µ| . (3.10) 1 − αδ Proof. Taking the inner product of the second equation of (2.7) with v in L2 (Rn ), we find that 1 d 2 dt
∥v∥2 = −α∥∇v∥2 + 1 − αδ 1u, v − (1 − δ)∥v∥2 − (δ 2 + λ − δ) u, v + α cz (θt ω) 1u, v + 2c δ z (θt ω) − c 2 z 2 (θt ω) u, v − cz (θt ω)∥v∥2 − µ sin u, v + g (x, t ), v .
(3.11)
By the first equation of (2.7), we have
v=
du dt
+ δ u − cuz (θt ω).
(3.12)
Then by (3.12), we have that
1 d 1u, v = − ∥∇ u∥2 + δ∥∇ u∥2 − cz (θt ω)∥∇ u∥2 2 dt
6 −
1 d 2 dt
∥∇ u∥2 + δ − |cz (θt ω)| ∥∇ u∥2 ,
(3.13)
Z. Wang, Y. Liu / Computers and Mathematics with Applications (
u, v
= >
1 d 2 dt 1 d 2 dt
∥u∥2 + δ∥u∥2 − cz (θt ω)∥u∥2
)
–
5
∥u∥2 + δ − |cz (θt ω)| ∥u∥2 .
(3.14)
By the Young inequality and the Cauchy–Schwarz inequality, we have
α cz (θt ω) 1u, v 6 |α cz (θt ω)| · ∥∇ u∥ · ∥∇v∥ α α 6 |cz (θt ω)|2 · ∥∇ u∥2 + ∥∇v∥2 , 2
2c δ z (θt ω) − c 2 z 2 (θt ω) u, v 6 |c δ z (θt ω)| +
g (x, t ), v 6
1 4(1 − 2δ)
(3.15)
2
1 2
|cz (θt ω)|2
∥u∥2 + ∥v∥2 ,
(3.16)
∥g (x, t )∥2 + (1 − 2δ)∥v∥2 ,
(3.17)
|µ| 2 −µ sin u, v 6 ∥u∥ + ∥v∥2 .
(3.18)
2
By (3.13)–(3.18), it then follows from (3.11) that d
∥v∥2 + δ 2 + λ − δ ∥u∥2 + 1 − αδ ∥∇ u∥2 dt 6 −α∥∇v∥2 − 2 δ − |cz (θt ω)| ∥v∥2 + δ 2 + λ − δ ∥u∥2 + 1 − αδ ∥∇ u∥2 + α|cz (θt ω)|2 · ∥∇ u∥2 + 2|c δ z (θt ω)| + |cz (θt ω)|2 + |µ| ∥u∥2 + ∥v∥2 +
1 2(1 − 2δ)
∥g (x, t )∥2 .
(3.19)
Recalling the new norm ∥ · ∥E (U ) in (2.2), then we obtain from (3.10) and (3.19) that d dt
∥ϕ (c ) ∥2E (Rn ) + α∥∇v∥2 6 −2δ + 2|cz (θt ω)| + R(θt ω) ∥ϕ (c ) ∥2E (Rn ) +
1 2(1 − 2δ)
∥g (x, t )∥2 .
(3.20)
t 3δ Multiplying (3.20) by e 0 ( 2 −2|cz (θr ω)|−R(θr ω))dr and then integrating over (τ − t , τ ) with t > 0, we have
∥ϕ (c ) (τ , τ − t , ω, ϕτ(c−) t )∥2E (Rn ) δ τ τs 32δ −2|cz (θr ω)|−R(θr ω) dr (c ) e + ∥ϕ (s, τ − t , ω, ϕτ(c−) t )∥2E (Rn ) ds 2 τ −t τ s 3δ −2|cz (θr ω)|−R(θr ω) dr ∥∇v(s, τ − t , ω, ϕτ(c−) t )∥2 ds +α eτ 2 τ −t τ −t 3δ −2|cz (θr ω)|−R(θr ω) dr τ 2
6e
∥ϕτ(c−) t ∥2E (Rn ) +
1
τ
2(1 − 2δ) τ −t
e
s 3δ τ
2
−2|cz (θr ω)|−R(θr ω) dr
∥g (x, s)∥2 ds.
(3.21)
Substituting ω by θ−τ ω, then we have from (3.21) that
∥ϕ (c ) (τ , τ − t , θ−τ ω, ϕτ(c−) t )∥2E (Rn ) δ τ τs 32δ −2|cz (θr −τ ω)|−R(θr −τ ω) dr (c ) + e ∥ϕ (s, τ − t , θ−τ ω, ϕτ(c−) t )∥2E (Rn ) ds 2 τ −t τ s 3δ −2|cz (θr −τ ω)|−R(θr −τ ω) dr +α ∥∇v(s, τ − t , θ−τ ω, ϕτ(c−) t )∥2 ds eτ 2 τ −t τ s τ −t 3δ 3δ 1 −2|cz (θr −τ ω)|−R(θr −τ ω) dr −2|cz (θr −τ ω)|−R(θr −τ ω) dr 2 6eτ ∥ϕτ(c−) t ∥2E (Rn ) + eτ 2 ∥g (x, s)∥2 ds 2(1 − 2δ) τ −t 0 s −t 3δ 3δ 1 −2|cz (θr ω)|−R(θr ω) dr −2|cz (θr ω)|−R(θr ω) dr 6e0 2 ∥ϕτ(c−) t ∥2E (Rn ) + e0 2 ∥g (x, s + τ )∥2 ds. 2(1 − 2δ) −∞
(3.22)
By Lemma 2.1, there exists T1 (ω) such that for all t > T1 (ω),
0
2|cz (θr ω)| + R(θr ω) dr <
−t
2|c | 2|c |δ c2 c2 + max , (δ 2 + λ − δ + 1) √ + + |µ| t. √ 2(1 − αδ) 2α πα πα
(3.23)
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Then, it follows from (3.6), (3.7) and (3.23) that for all s 6 −T1 (ω), e
s 3δ 0
2
−2|cz (θr ω)|−R(θr ω) dr
δ
6 e 2 s.
(3.24)
By (2.8) and (3.24), we have 0
e
s 3δ 0
−2|cz (θr ω)|−R(θr ω) dr
2
∥g (x, s + τ )∥2 ds < +∞.
(3.25)
−∞ (c )
Due to ϕτ −t ∈ D(τ − t , θ−t ω) and D ∈ D (E (Rn )), we have from (3.24), −t 3δ
lim e
0
2
−2|cz (θr ω)|−R(θr ω) dr
t →+∞
∥ϕτ(c−) t ∥2E (Rn ) = 0.
(3.26)
Therefore, there exists T = T (τ , ω, D) > T1 (ω) such that for all t > T , −t 3δ 0
e
−2|cz (θr ω)|−R(θr ω) dr
2
∥ϕτ(c−) t ∥2E (Rn ) 6 1.
Thus the lemma follows from (3.22) and (3.25).
In order to estimate the tails of solutions outside a bounded ball in Rn , we choose a smooth function ρ , such that 0 6 ρ(s) 6 1 for s ∈ R, and
ρ(s) =
0, 1,
0 6 |s| 6 1, |s| > 2,
(3.27)
and there exist constants µ1 , µ2 , such that |ρ ′ (s)| 6 µ1 , |ρ ′′ (s)| 6 µ2 for s ∈ R. Given k > 1, denote by Hk = {x ∈ Rn : |x| < k} and Rn \ Hk the complement of Hk . To prove asymptotic compactness of solution on Rn , we prove the following lemmas. Lemma 3.4. Assume that g (x, ·) ∈ L2loc (R, L2 (Rn )) and (2.8) hold. If γ > 0, then for every τ ∈ R, ω ∈ Ω , and D = {D(τ , ω) : τ ∈ R, ω ∈ Ω } ∈ D (E (Rn )), there exist T = T (τ , ω, D, ε) > 0 and K = K (τ , ω, ε) > 1, such that for all t > T , k > K ,
∥ϕ (c ) (τ , τ − t , θ−τ ω, ϕτ(c−) t )∥2E (Rn \Hk ) 6 ε,
(3.28)
(c )
where ϕτ −t = (uτ −t , vτ −t )⊤ ∈ D(τ − t , θ−t ω). Proof. Taking the inner product of the second equation of (2.7) with ρ
1 d 2 dt
ρ
Rn
|x|2 v in L2 (Rn ), we obtain k2
|x|2 |v|2 dx 2 k
|x|2 v dx + ( 1 − αδ) ( 1 u )ρ v dx k2 k2 Rn Rn |x|2 |x|2 ρ 2 uv dx − (1 − δ + cz (θt ω)) ρ 2 |v|2 dx − (δ 2 + λ − δ) k k Rn Rn |x|2 |x|2 + α cz (θt ω) (1u)ρ 2 v dx + 2c δ z (θt ω) − c 2 z 2 (θt ω) ρ 2 uv dx
=α
(1v)ρ
|x|2
k
Rn
+ Rn
ρ
|x|2 k2
g (x, t )v dx − µ
Rn
Rn
ρ
|x|2 k2
k
v sin udx.
(3.29)
By using (3.12), we get that
Rn
(1u)ρ
|x|2 2x |x|2 du ′ v dx = − (∇ u ) + δ u − cuz (θ ω) ρ t 2 2 2 k
Rn
k
k
dt
|x|2 du +ρ 2 ∇ + δ u − cuz (θt ω) dx k dt |x|2 2µ1 |x| 1 d 6 |(∇ u )v| dx − ρ 2 |∇ u|2 dx √ 2 k<|x|< 2k
k
+ cz (θt ω) − δ
2 dt
Rn
|x|2 ρ 2 |∇ u|2 dx k
Rn
k
Z. Wang, Y. Liu / Computers and Mathematics with Applications (
√
2µ1
6
k
1 d ∥∇ u∥2 + ∥v∥2 −
2 dt
Rn
ρ
|x|2 k2
Rn
ρ
)
–
7
|x|2 |∇ u|2 dx 2 k
|x|2 ρ 2 |∇ u|2 dx,
+ |cz (θt ω)| − δ k Rn |x|2 du = ρ 2 u· + δ u − cuz (θt ω) dx
uv dx
k
Rn
1 d
>
2 dt
dt
ρ
Rn
(3.30)
|x|2 k2
|u|2 dx + δ − |cz (θt ω)|
Rn
ρ
|x|2 |u|2 dx. 2 k
(3.31)
By the Cauchy–Schwarz inequality and the Young inequality, we obtain
α
Rn
(1v)ρ
|x|2 k2
2x |x|2 |x|2 ′ v + ρ ∇v dx ρ k2 k2 k2 Rn |x|2 2αµ1 |x| 6 |(∇v)v| dx − α ρ |∇v|2 dx √ k2 k2 k<|x|< 2k Rn √ |x|2 2αµ1 2 2 6 ∥∇v∥ + ∥v∥ − α ρ 2 |∇v|2 dx,
v dx = −α
(∇v)
k
α cz (θt ω)
Rn
(1u)ρ
|x|2 k2
Rn
|x|2 2x ′ |x|2 v dx = −α cz (θt ω) ∇u ρ v + ρ ∇v dx k2 k2 k2 Rn 2µ1 |x| | ∇ u v|dx 6 |α cz (θt ω)| √ k2 k<|x|< 2k |x|2 + |α cz (θt ω)| ρ 2 |∇v| · |∇ u|dx k Rn √ |x|2 α 2µ1 |α cz (θt ω)| 6 ∥∇ u∥2 + ∥v∥2 + ρ 2 |∇v|2 dx k 2 Rn k |x|2 α|cz (θt ω)| 2 + ρ 2 |∇ u|2 dx, 2
2c δ z (θt ω) − c 2 z 2 (θt ω)
Rn
ρ
|x|2 k2
−µ Rn
ρ
g (x, t )v dx 6
|x|2 k2
(3.32)
k
Rn
ρ
|x|2
4(1 − 2δ)
v sin udx 6
|µ|
uv dx 6 |c δ z (θt ω)| +
k2
1
2
Rn
k
Rn
Rn
ρ
ρ
|cz (θt ω)|2
2
Rn
|x|2 2 | g ( x , t )| dx + ( 1 − 2 δ) 2 k
|x|2 k2
Rn
ρ ρ
|x|2 |u|2 + |v|2 dx, 2 k
|x|2 k2
|v|2 dx,
|u|2 + |v|2 dx.
(3.33)
(3.34)
(3.35)
(3.36)
Then it follows from (3.30)–(3.36) that 1 d
|x|2
2 2 2 2 |v| + δ + λ − δ | u | + ( 1 − αδ)|∇ u | dx 2 dt Rn k2 |x|2 2 2 6 (|cz (θt ω)| − δ) ρ 2 |v| dx + (δ + λ − δ)(|cz (θt ω)| − δ) ρ
k
Rn
|x|2
α|cz (θt ω)| 2
Rn
ρ
|x|2 k2
|u|2 dx
|x|2 + (1 − αδ)(|cz (θt ω)| − δ) ρ 2 |∇ u| dx + ρ 2 |∇ u|2 dx k 2 k Rn Rn √ √ 2αµ1 2µ1 (1 − αδ + |α cz (θt ω)|) + ∥∇v∥2 + ∥v∥2 + ∥∇ u∥2 + ∥v∥2 k k |x|2 |cz (θt ω)|2 |µ| + |c δ z (θt ω)| + + ρ 2 |u|2 + |v|2 dx
2
−
α 2
Rn
|x|2 ρ 2 |∇v|2 dx + k
2
2
k
Rn
1 4(1 − 2δ)
Rn
ρ
|x|2 k2
|g (x, t )|2 dx.
(3.37)
Let
φ (c ) = |v|2 + δ 2 + λ − δ |u|2 + (1 − αδ)|∇ u|2 .
(3.38)
8
Z. Wang, Y. Liu / Computers and Mathematics with Applications (
)
–
It then follows from (3.37) and (3.38) that d
dt
Rn
ρ
|x|2 (c ) φ dx 6 −2δ + 2|cz (θt ω)| + R(θt ω) 2 k
+
c1
Rn
ρ
|x|2 k2
φ (c ) dx
c1 |z (θt ω)| ∥∇ u∥2 + ∥v∥2 + ∥∇v∥2 + ∥∇ u∥2 + ∥v∥2
k
k
+ c1 Rn
|x|2 ρ 2 |g (x, t )|2 dx
(3.39)
k
t where R(θt ω) is as in (3.10). Multiplying (3.39) by e 0 (2δ−2|cz (θr ω)|−R(θr ω))dr and then integrating over (τ − t , τ ) with t > 0, we have
Rn
ρ
|x|2 |x|2 τ −t (c ) (c ) (2δ−2|cz (θr ω)|−R(θr ω))dr τ φ (τ , τ − t , ω, φ ) dx 6 e ρ 2 φτ(c−) t dx τ −t 2 k k Rn c1 τ s (2δ−2|cz (θr ω)|−R(θr ω))dr eτ |z (θs ω)| ∥∇ u(s, τ − t , ω, ϕτ(c−) t )∥2 + k τ −t + ∥v(s, τ − t , ω, ϕτ(c−) t )∥2 ds c1 τ s (2δ−2|cz (θr ω)|−R(θr ω))dr eτ + ∥∇v(s, τ − t , ω, ϕτ(c−) t )∥2 k τ −t + ∥v(s, τ − t , ω, ϕτ(c−) t )∥2 + ∥∇ u(s, τ − t , ω, ϕτ(c−) t )∥2 ds τ |x|2 s + c1 e τ (2δ−2|cz (θr ω)|−R(θr ω))dr ρ 2 |g (x, s)|2 dxds. (3.40) τ −t
Rn
k
Replacing ω by θ−τ ω in (3.40), then we have that
|x|2 φ (c ) (τ , τ − t , θ−τ ω, φτ(c−) t )dx 2 n k R |x|2 −t ρ 2 φτ(c−) t dx 6 e 0 (2δ−2|cz (θr ω)|−R(θr ω))dr k Rn c1 τ s (2δ−2|cz (θr −τ ω)|−R(θr −τ ω))dr |z (θs−τ ω)| ∥∇ u(s, τ − t , θ−τ ω, ϕτ(c−) t )∥2 + eτ k τ −t + ∥v(s, τ − t , θ−τ ω, ϕτ(c−) t )∥2 ds c1 τ s (2δ−2|cz (θr −τ ω)|−R(θr −τ ω))dr + eτ ∥∇v(s, τ − t , θ−τ ω, ϕτ(c−) t )∥2 k τ −t + ∥v(s, τ − t , θ−τ ω, ϕτ(c−) t )∥2 + ∥∇ u(s, τ − t , θ−τ ω, ϕτ(c−) t )∥2 ds 0 |x|2 s (2δ−2|cz (θr ω)|−R(θr ω))dr 0 + c1 e ρ 2 |g (x, s + τ )|2 dxds. ρ
−∞
(3.41)
k
|x|>k
We next estimate each term on the right-hand side of (3.41). By Lemma 2.1, (3.6) and (3.7), there exists T1 = T1 (ω) such that for all s 6 −T1 , s e 0 (2δ−2|cz (θr ω)|−R(θr ω))dr 6 e
Due to t > T2 ,
ϕτ(c−) t
e
−t 0
s 3δ 0
2
−2|cz (θr ω)|−R(θr ω) dr
δ
6 e 2 s.
(3.42)
∈ D(τ − t , θ−t ω) ∈ D (E (Rn )), we get from (3.42) that, there exists T2 = T2 (τ , ω, D, ε) > T1 , such that for all
(2δ−2|cz (θr ω)|−R(θr ω))dr
Rn
ρ
|x|2 δ φτ(c−) t dx 6 e− 2 t ∥ϕτ(c−) t ∥2E (Rn ) < ε. 2
(3.43)
k
By (3.42), Lemmas 2.1 and 3.3, there are T3 = T3 (τ , ω, D, ε) > 0 and K1 = K1 (τ , ω, ε) > 1, such that for all t > T3 and k > K1 ,
c1 k
τ τ −t
s (c ) (c ) e τ (2δ−2|cz (θr −τ ω)|−R(θr −τ ω))dr |z (θs−τ ω)| ∥∇ u(s, τ − t , θ−τ ω, ϕτ −t )∥2 + ∥v(s, τ − t , θ−τ ω, ϕτ −t )∥2 ds
Z. Wang, Y. Liu / Computers and Mathematics with Applications (
c1
6
τ
e
k
τ −t
δ(s−τ ) 2
|z (θs−τ ω)| · e
s 3δ τ
2
)
–
9
−2|cz (θr −τ ω)|−R(θr −τ ω) dr
∥∇ u(s, τ − t , θ−τ ω, ϕτ(c−) t )∥2
+ ∥v(s, τ − t , θ−τ ω, ϕτ(c−) t )∥2 ds < ε.
(3.44)
By Lemma 3.3, there are T4 = T4 (τ , ω, D, ε) > 0 and K2 = K2 (τ , ω, ε) > 1, such that for all t > T4 and k > K2 , τ
c1 k
s (c ) e τ (2δ−2|cz (θr −τ ω)|−R(θr −τ ω))dr ∥∇v(s, τ − t , θ−τ ω, ϕτ −t )∥2
τ −t
+ ∥v(s, τ − t , θ−τ ω, ϕτ(c−) t )∥2 + ∥∇ u(s, τ − t , θ−τ ω, ϕτ(c−) t )∥2 ds < ε.
(3.45)
By using condition (2.9) and (3.42), there exists K3 = K3 (τ , ω, ε) > 1, such that for all k > K3 ,
0
c1
s e 0 (2δ−2|cz (θr ω)|−R(θr ω))dr
−∞
ρ |x|>k
|x|2 |g (x, s + τ )|2 dxds < ε. 2
(3.46)
k
Letting T = max T1 , T2 , T3 , T4 , K = max K1 , K2 , K3 , then combining (3.43)–(3.46), we have for all t > T and k > K ,
Rn
ρ
|x|2 k2
φ (c ) (τ , τ − t , θ−τ ω, φτ(c−) t )dx < 4ε,
which implies
∥ϕ (c ) (τ , τ − t , θ−τ ω, ϕτ(c−) t )∥2E (Rn \Hk ) < 4ε. Then we complete the proof.
Let ρ = 1 − ρ with ρ given by (3.27). Fix k > 1 and set
|x|2 u(t , τ , ω, uτ , vτ ) = ρ 2 u(t , τ , ω, ϕτ(c ) ), k |x|2 v(t , τ , ω, uτ , vτ ) = ρ 2 v(t , τ , ω, ϕτ(c ) ). k |x|2 Multiplying (2.7) by ρ k2 and using (3.47), we find that d u v − δ u + c uz (θt ω) dt = |x|2 |x|2 d v = α 1 v − αv 1 ρ 2 − 2α∇v∇ ρ 2 + (δ − 1) v + (1 − αδ)1 u k dt |x|2 k|x|2 ρ 2 + (δ − λ − δ 2 ) u − (1 − αδ)u1 ρ 2 − 2(1 − αδ)∇ u∇ k k 2 2 | x| |x| + α cz (θt ω)1 u − α cz (θt ω)u1 ρ 2 − α cz (θt ω)∇ u∇ ρ 2 k k |x|2 |x|2 2 2 + (2c δ z (θt ω) − c z (θt ω)) u − cz (θt ω) v − ρ 2 f (u) + ρ 2 g (x, t ). k
(3.47)
(3.48)
k
Considering the eigenvalue problem
− 1 u= u in H2k , with u = 0 on ∂ H2k .
(3.49)
The problem (3.49) has a family of eigenfunctions {ei }i∈N with the eigenvalues {λi }i∈N :
λ1 6 λ2 6 · · · 6 λi 6 · · · , λi → +∞(i → +∞), such that {ei }i∈N is an orthonormal basis of L2 (H2k ). Given n ∈ N, let Xn = span{e1 , . . . , en } and Pn : L2 (H2k ) → Xn be the projection operator. Lemma 3.5. Assume that g (x, ·) ∈ L2loc (R, L2 (Rn )) and (2.8) hold. If γ > 0, then for every τ ∈ R, ω ∈ Ω , and D = {D(τ , ω) : τ ∈ R, ω ∈ Ω } ∈ D (E (Rn )), there exist T = T (τ , ω, D, ε) > 0, K = K (τ , ω, ε) > 1 and N = N (τ , ω, ε) > 0, such that for all t > T , k > K and n > N,
∥(I − Pn ) ϕ (c ) (τ , τ − t , θ−τ ω, ϕτ(c−) t )∥2E (H2k ) < ε, 2 where ϕτ(c−) t = ρ |kx|2 ϕτ(c−) t , ϕτ(c−) t = (uτ −t , vτ −t )⊤ ∈ D(τ − t , θ−t ω).
(3.50)
10
Z. Wang, Y. Liu / Computers and Mathematics with Applications (
)
–
Proof. Let un,1 = Pn u, un,2 = (I − Pn ) u, vn,1 = Pn v and vn,2 = (I − Pn ) v . Applying I − Pn to the first equation of (3.48), we obtain
vn,2 =
d un , 2 dt
+ δ un,2 − c un,2 z (θt ω).
(3.51)
Then applying I − Pn to the second equation of (3.48) and taking the inner product of the resulting equation with vn,2 in L2 (H2k ), we have 1 d 2 dt
|x|2 |x|2 ρ 2 , vn,2 ∥ vn,2 ∥2 = −α∥∇ vn,2 ∥2 − α v 1 ρ 2 + 2∇v∇ k
k
|x|2 |x|2 + (1 − αδ) 1 un,2 , vn,2 − (1 − αδ + α cz (θt ω)) u1 ρ 2 + 2∇ u∇ ρ 2 , vn,2 k k + (δ − λ − δ 2 ) un,2 , vn,2 + α cz (θt ω) 1 un,2 , vn,2 + 2c δ z (θt ω) − c 2 z 2 (θt ω) un,2 , vn,2 + δ − 1 − cz (θt ω) ∥ vn,2 ∥2 |x|2 |x|2 −µ ρ 2 sin u, vn,2 + vn,2 . ρ 2 g (x, t ), k
k
(3.52)
Substituting vn,2 in (3.51) into the third and fifth terms on the left-hand side of (3.52), we have
d un , 2 + δ un,2 − c un,2 z (θt ω) 1 un,2 , vn,2 = − ∇ un,2 , ∇ dt
1 d
∥∇ un,2 ∥2 − δ − |cz (θt ω)| ∥∇ un,2 ∥2 , 2 dt d un,2 + δ un,2 − c un,2 z (θt ω) un,2 , vn,2 = un,2 , 6 −
(3.53)
dt
>
1 d 2 dt
∥ un,2 ∥2 + δ − |cz (θt ω)| ∥ un,2 ∥2 .
(3.54)
Using the Cauchy–Schwarz inequality and the Young inequality, we obtain
|x|2 |x|2 4x2 |x|2 2 2 ′ |x|2 4x ′ |x| ρ 2 , vn,2 = −α v 4 ρ ′′ 2 + 2 ρ + ∇v · ρ , v −α v 1 ρ 2 + 2∇v∇ n,2 k k k k k k2 k2 k2 2 1 − 2δ 32 µ 2α 2 (8µ2 + 2µ1 )2 6 ∥v∥2 + 2 1 ∥∇v∥2 + ∥ vn,2 ∥2 , (3.55) 1 − 2δ k4 k 4 |x|2 |x|2 −(1 − αδ + α cz (θt ω)) u1 ρ 2 + 2∇ u∇ ρ 2 , vn,2 k
4x2
= −(1 − αδ + α cz (θt ω)) u
k4
ρ ′′
|x|2 k2
+
k 2
ρ ′ 2
k
|x|2 k2
+
4x
∇u · ρ ′ 2
k
|x|2 , v n , 2 2 k
1 − 2δ 32µ21 2(1 − αδ + α c |z (θt ω)|) (4µ2 + µ1 )2 2 2 6 ∥ u ∥ + ∥∇ u ∥ + ∥ vn,2 ∥2 , 1 − 2δ k4 k2 2 α cz (θt ω) 1 un,2 , vn,2 = −α cz (θt ω) ∇ un,2 , ∇ vn,2 α|cz (θt ω)|2 α 6 ∥∇ un,2 ∥ + ∥∇ vn,2 ∥, 2
2
2
2c δ z (θt ω) − c 2 z 2 (θt ω) un,2 , vn,2 6 |c δ z (θt ω)| +
|x|2 ρ 2 g (x, t ), vn,2 6 k
1 1 − 2δ
|cz (θt ω)|2 2
∥ un,2 ∥2 + ∥ vn,2 ∥2 ,
2 1 − 2 δ |x|2 ρ 2 g (x, t ) + ∥ vn,2 ∥2 , (I − Pn ) k
4
|x|2 |x|2 vn,2 6 |µ| · (I − Pn ) −µ ρ 2 sin u, ρ 2 sin u · ∥ vn,2 ∥ k k |x|2 6 |µ| · (I − Pn ) ρ 2 u · ∥ vn,2 ∥ k |µ| 6 ∥ un,2 ∥2 + ∥ vn,2 ∥2 . 2
(3.56)
(3.57) (3.58) (3.59)
(3.60)
Z. Wang, Y. Liu / Computers and Mathematics with Applications (
)
–
11
By recalling the new norm ∥ · ∥E (U ) in (2.2), and R(θt ω) in (3.10), then, we can get from (3.53)–(3.60) that d dt
(c ) 2 vn,2 ∥2 + −2δ + 2|cz (θt ω)| + R(θt ω) ∥ ϕn,2 ∥E (H2k ) ∥ ϕn(,c2) ∥2E (H2k ) 6 −α∥∇ +
c3 k4
∥ u∥ 2 c3 ∥∇ u∥2 ∥u∥2 + ∥v∥2 + 2 ∥∇ u∥2 + ∥∇v∥2 + c3 |z (θt ω)|2 + 4 2 k
k
k
2 |x|2 + c3 0(I − Pn ) ρ 2 g (x, t ) . k t
Multiplying (3.61) by e
0 2δ−2|cz (θr ω)|−R(θr ω) dr
ϕn(c,2) ,τ −t )∥2E (H2k ) 6 e ∥ ϕn(,c2) (τ , τ − t , ω,
(3.61)
and then integrating over (τ − t , τ ) with t > 0, we have
τ −t
∥ ϕn(c,2) ,τ −t ∥2E (H2k ) c3 τ τ 2δ−2|cz (θr ω)|−R(θr ω) dr + 4 e ∥u(s, τ − t , ω, ϕτ(c−) t )∥2 k τ −t + ∥v(s, τ − t , ω, ϕτ(c−) t )∥2 ds c3 τ τs 2δ−2|cz (θr ω)|−R(θr ω) dr + 2 e ∥∇ u(s, τ − t , ω, ϕτ(c−) t )∥2 k τ −t + ∥∇v(s, τ − t , ω, ϕτ(c−) t )∥2 ds τ ∥u(s, τ − t , ω, ϕ (c ) )∥2 s τ −t 2δ−2|cz (θr ω)|−R(θr ω) dr + c3 eτ |z (θs ω)|2 4 τ
2δ−2|cz (θr ω)|−R(θr ω) dr
s
k
τ −t
+
∥∇ u(s, τ −
k2
+ c3
(c ) t , ω, ϕτ −t )∥2
τ
2 |x|2 ρ 2 g (x, s) ds. (I − Pn )
s
e
ds
τ 2δ−2|cz (θr ω)|−R(θr ω) dr
k
τ −t
(3.62)
Substituting ω by θ−τ ω in (3.62), we have
∥ ϕn(,c2) (τ , τ − t , θ−τ ω, ϕn(c,2) ,τ −t )∥2E (H2k ) −t 2δ−2|cz (θr ω)|−R(θr ω) dr 6e0 ∥ ϕn(c,2) ,τ −t ∥2E (H2k ) τ s c3 2δ−2|cz (θr −τ ω)|−R(θr −τ ω) dr eτ ∥u(s, τ − t , θ−τ ω, ϕτ(c−) t )∥2 + ∥v(s, τ − t , θ−τ ω, ϕτ(c−) t )∥2 ds + 4 k τ −t c3 τ τs 2δ−2|cz (θr −τ ω)|−R(θr −τ ω) dr e ∥∇ u(s, τ − t , θ−τ ω, ϕτ(c−) t )∥2 + 2 k τ −t + ∥∇v(s, τ − t , θ−τ ω, ϕτ(c−) t )∥2 ds τ ∥u(s, τ − t , θ ω, ϕ (c ) )∥2 s −τ τ −t 2δ−2|cz (θr −τ ω)|−R(θr −τ ω) dr + c3 eτ |z (θs−τ ω)|2 4 k
τ −t
+
∥∇ u(s, τ −
(c ) t , θ−τ ω, ϕτ −t )∥2
k2
0
+ c3
e
ds
2 |x|2 ρ 2 g (x, s + τ ) ds. (I − Pn )
s
0 2δ−2|cz (θr ω)|−R(θr ω) dr
(3.63)
k
−t
(c )
In what follows, we estimate the terms on the right-hand side of (3.63). Due to ϕτ −t ∈ D(τ − t , θ−t ω) ∈ D (E (Rn )), we get from (3.42) that, there exists T1 = T1 (τ , ω, D, ε) > 0, such that if t > T1 , then e
−t
2δ−2|cz (θr ω)|−R(θr ω) dr
0
∥ ϕn(c,2) ,τ −t ∥2E (H2k ) < ε.
(3.64)
For the second and third terms on the right-hand side of (3.63), by Lemma 3.3, we obtain that there exist T2 T2 (τ , ω, D, ε) > 0 and K1 = K1 (τ , ω, ε) > 1, such that for all t > T2 and k > K1 ,
c3 k4
τ
e τ −t
s
τ 2δ−2|cz (θr −τ ω)|−R(θr −τ ω) dr
∥u(s, τ − t , θ−τ ω, ϕτ(c−) t )∥2 + ∥v(s, τ − t , θ−τ ω, ϕτ(c−) t )∥2 ds
=
12
Z. Wang, Y. Liu / Computers and Mathematics with Applications (
c3
+
τ
e
k2
–
2δ−2|cz (θr −τ ω)|−R(θr −τ ω) dr ∥∇ u(s, τ − t , θ−τ ω, ϕτ(c−) t )∥2 + ∥∇v(s, τ − t , θ−τ ω, ϕτ(c−) t )∥2 ds
τ
τ −t
< ε.
)
s
(3.65)
Since z (ω) is tempered, by Lemmas 2.1 and 3.3, there exist T3 = T3 (τ , ω, D, ε) > 0 and K2 = K2 (τ , ω, ε) > 1, such that for all t > T3 and k > K2 , we have
c3
τ
e
s
τ 2δ−2|cz (θr −τ ω)|−R(θr −τ ω) dr
τ −t τ
6 c3
e τ −t
δ(s−τ ) 2
|z (θs−τ ω)| · e
∥∇ u(s, τ −
+
s
2
|z (θs−τ ω)|2
k4
3δ −2|cz (θr −τ ω)|−R(θr −τ ω) 2
τ
dr
+
∥∇ u(s, τ − t , θ−τ ω, ϕτ(c−) t )∥2 k2
ds
∥u(s, τ − t , θ ω, ϕ (c ) )∥2 −τ τ −t k4
(c ) t , θ−τ ω, ϕτ −t )∥2
ds
k2
< ε.
∥u(s, τ − t , θ ω, ϕ (c ) )∥2 −τ τ −t
(3.66)
Next, we estimate the last term on the right-hand side of (3.63), by (2.8) and (3.42), there are N = N (τ , ω, ε) > 0, T4 = T4 (τ , ω, ε) > 0 and K3 = K3 (τ , ω, ε) > 1, such that for all n > N, t > T4 and k > K3 ,
0
c3
2 |x|2 ρ 2 g (x, s + τ ) ds < ε. (I − Pn )
s
e
0 2δ−2|cz (θr ω)|−R(θr ω) dr
k
−t
(3.67)
Let T = max T1 , T2 , T3 , T4 and K = max K1 , K2 , K3 . Then, it follows from (3.64), (3.66) and (3.67) that, for all t > T , k > K and n > N,
∥ ϕn(,c2) (τ , τ − t , θ−τ ω, ϕn(,c2),τ −t )∥2E (H2k ) < 4ε. This completes the proof.
(3.68)
We now prove the existence of D (E (Rn ))-pullback attractors for the random problem (2.7) in D (E (Rn )). Theorem 3.6. Assume that g (x, ·) ∈ L2loc (R, L2 (Rn )) and (2.8) hold. If γ > 0, then the continuous cocycle Φc associated with problem (2.7) has a unique D (E (Rn ))-pullback attractor Ac = {Ac (τ , ω) : τ ∈ R, ω ∈ Ω } in D (E (Rn )). Proof. By (2.10) and Lemma 3.3, we notice that the continuous cocycle Φc has a closed measurable D (E (Rn ))-pullback absorbing set in E (Rn ). Next, we will prove that the continuous cocycle Φc is asymptotically compact in E (Rn ). Let tm → ∞, B ∈ D (E (Rn )), and (c ) ϕτ −tm ∈ B(τ − tm , θ−tm ω). By using Lemma 3.3, we have that {ϕ (c ) τ , τ − tm , θ−τ ω, ϕτ(c−) tm } is a bounded in E (Rn ); that is, for every τ ∈ R and ω ∈ Ω , there exists M1 = M1 (τ , ω, B) > 0 such that for all m > M1 ,
(c ) ϕ τ , τ − tm , θ−τ ω, ϕτ(c−) t 2 n < ϱ(c , τ , ω), m E (R )
(3.69)
where ϱ(c , τ , ω) is as in (3.9). By Lemma 3.4, there are k1 = k1 (τ , ω, ε) and M2 = M2 (τ , ω, B, ε) > 0, such that for every m > M2 ,
(c ) ϕ τ , τ − tm , θ−τ ω, ϕτ(c−) t 2 n m E (R \H
k1 )
< ε.
(3.70)
In addition, by Lemma 3.5, there exist N = N (τ , ω, ε) > 0, k2 = k2 (τ , ω, ε) > k1 and M3 = M3 (τ , ω, B, ε) > 0, such that for every m > M3 ,
2 (I − PN ) ϕ (c ) τ , τ − tm , θ−τ ω, ϕτ(c−) tm E (H
2k2 )
< ε.
(3.71) (c )
ϕ (c ) τ , τ − tm , θ−τ ω, ϕτ −tm } is a bounded in PN E (H2k2 ), which associates with (3.71) By (3.47) and (3.69), we get that {PN
(c )
implies that { ϕ (c ) τ , τ − tm , θ−τ ω, ϕτ −tm } is precompact in H01 (H2k2 ) × L2 (H2k2 ).
|x|2
Recalling (3.47) and the fact that ρ
k22
= 1 for |x| < k2 , we have that {ϕ (c ) τ , τ − tm , θ−τ ω, ϕτ(c−) tm } is precompact in
E (Hk2 ), which along with (2.10) and (3.70) shows that the continuous cocycle Φc is asymptotically compact in D (E (Rn )). Then, by Proposition 3.2, the continuous cocycle Φc associated with (2.7) has a unique D (E (Rn ))-pullback attractor in D (E (Rn )).
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3.1. Remark In this section we studied the existence of the D (E (Rn ))-pullback attractor for the non-autonomous stochastic strongly damped sine–Gordon equation driven by multiplicative noise. The coefficient c of the multiplicative noise term cu ◦ dW dt needs to be suitably small, this is because that the multiplicative noise cu ◦
dW dt
depends on the state variable u.
4. Upper semicontinuity of D (E (Rn ))-pullback attractors First, we present a criteria concerning upper semicontinuity of non-autonomous random attractors with respect to a parameter in [23]. Theorem 4.1. Let (X , ∥ · ∥X ) be a separable Banach space, Ψc be a continuous cocycle on X over R and (Ω , F , P , {θt }t ∈R ). Suppose that (i) Ψc has a closed measurable random absorbing set Kc = {Kc (τ , ω) : τ ∈ R, ω ∈ Ω } in D (X ) and a unique random attractor Ac = {Ac (τ , ω) : τ ∈ R, ω ∈ Ω } in D (X ). (ii) There exists a map ς : R → R such that for each τ ∈ R, ω ∈ Ω , K0 (τ ) = {u ∈ X : ∥u∥X 6 ς (τ )} and lim sup ∥Kc (τ , ω)∥X = lim sup c →0
c →0
sup
x∈Kc (τ ,ω)
∥x∥X 6 ς (τ ).
(4.1)
(iii) There exists ε > 0, such that for every τ ∈ R and ω ∈ Ω , |c |6ε Ac (τ , ω) is precompact in X . (iv) For t > 0, τ ∈ R, ω ∈ Ω , cn → 0 when n → ∞, and xn , x0 ∈ X with xn → x0 when n → ∞, it holds:
lim Ψcn (t , τ , ω, xn ) = Ψ0 (t , τ , x0 ).
(4.2)
n→∞
Then for τ ∈ R and ω ∈ Ω , dH (Ac (τ , ω), A0 (τ )) =
sup
inf
u∈Ac (τ ,ω) v∈A0 (τ )
∥u − v∥X → 0 as c → 0.
(4.3)
Next, we will use Theorem 4.1 to consider an upper semicontinuity of random attractors Ac (τ , ω) when c → 0. To indicate the dependence of solutions on c, we will write the solution of (2.7) as ϕ (c ) = (u(c ) , v (c ) ), that is, (u(c ) , v (c ) ) satisfies
(c ) du = v (c ) − δ u(c ) + cu(c ) z (θt ω), dt (c ) dv = α 1v (c ) + (1 − αδ)1u(c ) + (δ − 1)v (c ) + (δ − λ − δ 2 )u(c ) + α cz (θt ω)1u(c ) dt + 2c δ z (θt ω) − c 2 z 2 (θt ω) u(c ) − cz (θt ω)v (c ) − µ sin u(c ) + g (x, t ), (c ) u (τ , x) = u(τc ) (x), v (c ) (τ , x) = vτ(c ) (x).
(4.4)
When c = 0, the random problem (2.7) reduces to a deterministic one:
(0) du (0) (0) dt = v − δ u , (0) dv = α 1v (0) + (1 − αδ)1u(0) + (δ − 1)v (0) + (δ − λ − δ 2 )u(0) − µ sin u(0) + g (x, t ), (dt (0) u 0) (τ , x) = u(τ0) (x), v (0) (τ , x) = vτ(0) (x) = u1τ (x) + δ u(τ0) (x).
(4.5)
Let Φ0 be the continuous deterministic cocycle associated with problem (4.5) on E (Rn ) over R and {θt }t ∈R , and D0 (E (Rn )) be the collection of tempered families of deterministic nonempty bounded subsets of E (Rn ). Assume that g (x, ·) ∈ L2loc (R, L2 (Rn )), (2.8) and γ > 0 hold. Then by Theorem 3.6, we have that the continuous cocycle Φ0 has a unique D0 (E (Rn ))-pullback attractor A0 (τ ). Theorem 4.2. Assume that g (x, ·) ∈ L2loc (R, L2 (Rn )) and (2.8) hold. If γ > 0, then for every τ ∈ R and ω ∈ Ω , we have dH (Ac (τ , ω), A0 (τ )) =
sup
inf
1
(∥∇(u(c ) − u(0) )∥2 + ∥u(c ) − u(0) ∥2 + ∥v (c ) − v (0) ∥2 ) 2
(0) (0) (u(c ) ,v (c ) )∈Ac (τ ,ω) (u ,v )∈A0 (τ )
→ 0 as c → 0.
(4.6)
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Proof. (i) By Lemma 3.3 and Theorem 3.6, Φc has a closed measurable random absorbing set Bc = {Bc (τ , ω) : τ ∈ R, ω ∈ Ω } ∈ D (E (Rn )), where Bc (τ , ω) = {ϕ (c ) ∈ E (Rn ) : ∥ϕ (c ) ∥2E (Rn ) 6 ϱ(c , τ , ω)}, and a unique random attractor Ac = {Ac (τ , ω) : τ ∈ R, ω ∈ Ω } in D (E (Rn )), for each τ ∈ R and ω ∈ Ω , Ac (τ , ω) ⊆ Bc (τ , ω). (ii) Given |c | 6 1. By (3.9), we have
ϱ(c , τ , ω) 6 ϱ(1, τ , ω) < ∞, and lim sup ϱ(c , τ , ω) 6 ϱ(1, τ , ω). c →0
So, for every τ ∈ R and ω ∈ Ω , lim sup ∥Bc (τ , ω)∥ρ = lim sup c →0
c →0
sup
x∈Bc (τ ,ω)
1
∥x∥E (Rn ) 6 ϱ 2 (1, τ , ω).
(4.7)
Let B1 (τ , ω) = {ϕ (c ) ∈ E (Rn ) : ∥ϕ (c ) ∥2E (Rn ) 6 ϱ(1, τ , ω)}, then
Ac (τ , ω) ⊆
|c |61
Bc (τ , ω) ⊆ B1 (τ , ω).
(4.8)
|c |61
(iii) Given |c | 6 1. Let us prove the precompactness of |c |61 Ac (τ , ω) for every τ ∈ R and ω ∈ Ω . For one thing, by (4.8), Lemma 3.4 and the invariance of Ac (τ , ω), for every ε > 0, c > 0, τ ∈ R, ω ∈ Ω , there exist T = T (τ , ω, B1 , c , ε) > 0 (c ) and R = R(τ , ω, c , ε) > 1, such that for all t > T , r > R and ϕτ −t ∈ Ac (τ − t , θ−t ω), the solution ϕ (c ) of (4.4) satisfies
sup ϕ (c ) ∈
Ac (τ ,ω)
∥ϕ (c ) (τ , τ − t , θ−τ ω, ϕτ(c−) t )∥2E (Rn \Hr ) < ε.
|c |61
For another, by (4.8) we find that the set |c |61 Ac (τ , ω) is precompact in E (Hr ) and hence |c |61 Ac (τ , ω) is precompact in E (Rn ). (iv) Let ϕ (0) = (u(0) , v (0) ) be a mild solution of (4.5) with initial data ϕτ(0) = (u(τ0) , vτ(0) ), and U = u(c ) − u(0) , V = v (c ) −v (0) . Let
ϕm(c ) (t , τ , ω, ϕτ(c ) ) = (u(mc ) (t , τ , ω, ϕτ(c ) ), vm(c ) (t , τ , ω, ϕτ(c ) )) and
ϕm(0) (t , τ , ϕτ(0) ) = (u(m0) (t , τ , ϕτ(0) ), vm(0) (t , τ , ϕτ(0) )) be the solutions of the following random differential equations with initial data
(c ) dum = vm(c ) − δ u(mc ) + cu(mc ) z (θt ω), dt (c ) dvm = α 1vm(c ) + (1 − αδ)1u(mc ) + (δ − 1)vm(c ) + (δ − λ − δ 2 )u(mc ) + α cz (θt ω)1u(mc ) dt + 2c δ z (θt ω) − c 2 z 2 (θt ω) u(mc ) − cz (θt ω)vm(c ) − µ sin u(mc ) + g (x, t ), (c ) um (τ , x) = ucτ (x), vm(c ) (τ , x) = vτc (x),
(4.9)
and
(0) dum = vm(0) − δ u(m0) , dt (0) dvm = α 1vm(0) + (1 − αδ)1u(m0) + (δ − 1)vm(0) + (δ − λ − δ 2 )u(m0) − µ sin u(m0) + g (x, t ), dt (0) um (τ , x) = u(τ0) (x), vm(0) (τ , x) = vτ(0) (x), (c )
(4.10)
(0)
respectively. Then ϕm (·, τ , ω, ϕτ(c ) ), ϕm (·, τ , ϕτ(0) ) ∈ C ([τ , +∞), E (Rn )) and satisfy the differential equations (4.9) and (4.10) respectively. Moreover, ϕ (c ) (t , τ , ω, ϕτ(c ) ) and ϕ (0) (t , τ , ϕτ(0) ) are limit functions of subsequences of
{ϕm(c ) (t , τ , ω, ϕτ(c ) )} and {ϕm(0) (t , τ , ϕτ(0) )} ∈ E (Rn ). So ϕ (c ) (t , τ , ω, ϕτ(c ) ) − ϕ (0) (t , τ , ϕτ(0) ) is a limit function of a subsequence
Z. Wang, Y. Liu / Computers and Mathematics with Applications (
(c )
)
–
15
(0)
of {ϕm (t , τ , ω, ϕτ(c ) ) − ϕm (t , τ , ϕτ(0) )} in E (Rn ), and (Um (t , τ , ω, ϕτ(c ) , ϕτ(0) ), Vm (t , τ , ω, ϕτ(c ) , ϕτ(0) )) satisfies
dU m (c ) dt = Vm − δ Um + cum z (θt ω), dV m = α 1Vm + (1 − αδ)1Um + (δ − 1)Vm + (δ − λ − δ 2 )Um + α cz (θt ω)1u(mc ) dt (c ) + 2c δ z (θt ω) − c 2 z 2 (θt ω) u(mc ) − cz (θt ω)vm − µ sin u(mc ) + µ sin u(m0) , (c ) (0) (c ) (0) Um (τ ) = uτ − uτ , Vm (τ ) = vτ − vτ .
(4.11)
Taking the inner product of the second equation of (4.11) with V m (t ) in L2 (Rn ), we find that 1 d
∥Vm ∥2 = −α∥∇ Vm ∥2 + 1 − αδ 1Um , Vm − (1 − δ)∥Vm ∥2 2 dt − (δ 2 + λ − δ) Um , Vm + α cz (θt ω) 1u(mc ) , Vm + 2c δ z (θt ω) − c 2 z 2 (θt ω) u(mc ) , Vm − cz (θt ω) vm(c ) , Vm − µ sin u(mc ) − sin u(m0) , Vm .
(4.12)
We now estimate the terms in (4.12):
|cz (θt ω)|2 (c ) 2 2 − 3δ −cz (θt ω) vm(c ) , Vm 6 ∥vm ∥ + ∥ Vm ∥ 2 , 2 − 3δ 4 |2c δ z (θt ω) − c 2 z 2 (θt ω)|2 (c ) 2 2 − 3δ ∥ um ∥ + ∥ Vm ∥ 2 , 2c δ z (θt ω) − c 2 z 2 (θt ω) u(mc ) , Vm 6 2 − 3δ 4 α cz (θt ω) 1u(mc ) , Vm = −α cz (θt ω) ∇ u(mc ) , ∇ Vm α |cz (θt ω)|2 · ∥∇ u(mc ) ∥2 + α∥∇ Vm ∥2 , 6
(4.13) (4.14)
(4.15)
4
1 − αδ 1Um , Vm = − 1 − αδ ∇ Um , ∇
1 d
dU
m
dt
+ δ Um − cu(mc ) z (θt ω)
∥∇ Um ∥2 + δ∥∇ Um ∥2 − ∥∇ Um ∥ · ∥cz (θt ω)∇ u(mc ) ∥ 2 dt 1 d δ |cz (θt ω)|2 6 − 1 − αδ ∥∇ Um ∥2 + ∥∇ Um ∥2 − ∥∇ u(mc ) ∥2 , 2 dt 2 2δ dU m (δ 2 + λ − δ) Um , Vm = δ 2 + λ − δ Um , + δ Um − cu(mc ) z (θt ω) 1 d dt > (δ 2 + λ − δ) ∥Um ∥2 + δ∥Um ∥2 − ∥Um ∥ · ∥cu(mc ) z (θt ω)∥ 2 dt 1 d δ |cz (θt ω)|2 (c ) 2 2 ∥Um ∥2 + ∥Um ∥2 − ∥um ∥ , > (δ + λ − δ) 2 dt 2 2δ − |µ| sin u(mc ) − sin u(m0) , Vm 6 |µ| · ∥ sin u(mc ) − sin u(m0) ∥ · ∥Vm ∥ |µ| 6 (∥Vm ∥2 + ∥Um ∥2 ). 6 − 1 − αδ
2
(4.16)
(4.17)
(4.18)
Then it follows from (4.17)–(4.18) that 1 d
∥Vm ∥2 + 1 − αδ ∥∇ Um ∥2 + δ 2 + λ − δ ∥Um ∥2 2 dt |µ| δ 6− ∥Vm ∥2 + 1 − αδ ∥∇ Um ∥2 + δ 2 + λ − δ ∥Um ∥2 + (∥Vm ∥2 + ∥Um ∥2 ) 2 |cz (θt ω)|2
2
|2c δ z (θt ω) − c 2 z 2 (θt ω)|2 (c ) 2 + ∥vm(c ) ∥2 + ∥ um ∥ 2 − 3δ 2 − 3δ α δ 2 + λ − δ |cz (θt ω)|2 (c ) 2 1 − αδ + + |cz (θt ω)|2 · ∥∇ u(mc ) ∥2 + ∥ um ∥ 4 2δ 2δ 6− ∥Vm ∥2 + 1 − αδ ∥∇ Um ∥2 + δ 2 + λ − δ ∥Um ∥2 2 + H2 (c , t , ω) ∥vm(c ) ∥2 + ∥u(mc ) ∥2 + ∥∇ u(mc ) ∥2 ,
(4.19)
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where = δ − |µ|(1 + δ 2 + λ − δ) > 0, H2 (c , t , ω) = |cz (θt ω)| inequality to (4.19) from τ to t + τ , we have
2 2 1+|2δ−cz (θt ω)| 2−3δ
)
–
2 + α4 + (1−αδ+δ2δ +λ−δ) . Applying Gronwall’s
∥Vm (t + τ , τ , ω, ϕτ(c ) , ϕτ(0) )∥2 + 1 − αδ ∥∇ Um (t + τ , τ , ω, ϕτ(c ) , ϕτ(0) )∥2 + δ 2 + λ − δ ∥Um (t + τ , τ , ω, ϕτ(c ) , ϕτ(0) )∥2 6 e− t ∥vτ(c ) − vτ(0) ∥2 + 1 − αδ ∥∇(u(τc ) − u(τ0) )∥2 + δ 2 + λ − δ ∥u(τc ) − u(τ0) ∥2 t +τ (c ) e− (t +τ −s) H2 (c , s, ω) ∥vm +2 (s, τ , ω, ϕτ(c ) )∥2 τ
+ ∥um (s, τ , ω, ϕτ(c ) )∥2 + ∥∇ u(mc ) (s, τ , ω, ϕτ(c ) )∥2 ds. (c )
(4.20)
We now replace ω in the above by θ−τ ω to yield
∥Vm (t + τ , τ , θ−τ ω, ϕτ(c ) , ϕτ(0) )∥2 + (1 − αδ)∥∇ Um (t + τ , τ , θ−τ ω, ϕτ(c ) , ϕτ(0) )∥2 + δ 2 + λ − δ ∥Um (t + τ , τ , θ−τ ω, ϕτ(c ) , ϕτ(0) )∥2 6 e− t ∥vτ(c ) − vτ(0) ∥2 + 1 − αδ ∥∇(u(τc ) − u(τ0) )∥2 + δ 2 + λ − δ ∥u(τc ) − u(τ0) ∥2 t (c ) (s, τ , θ−τ ω, ϕτ(c ) )∥2 +2 e− (t −s) H2 (c , s, θ−τ ω) ∥vm 0
+ ∥u(mc ) (s, τ , θ−τ ω, ϕτ(c ) )∥2 + ∥∇ u(mc ) (s, τ , θ−τ ω, ϕτ(c ) )∥2 ds.
(4.21)
From (4.21), we see that for τ ∈ R, t ∈ R+ , ω ∈ Ω , c → 0 and ϕτ(c ) , ϕτ(0) ∈ E (Rn ) with ϕτ(c ) → ϕτ(0) , (c ) lim ϕm (t + τ , τ , θ−τ ω, ϕτ(c ) ) = ϕm(0) (t + τ , τ , ϕτ(0) ) in E (Rn ).
c →0
(4.22)
Let {cn } ⊂ [−1, 1] be a sequence of numbers with cn → 0 when n → +∞. Then ϕ (c ) (t + τ , τ , θ−τ ω, ϕτ(c ) ) and
ϕ (0) (t + τ , τ , ϕτ(0) ) being limit functions of subsequences of {ϕm(c ) (t + τ , τ , θ−τ ω, ϕτ(c ) )} and {ϕm(0) (t + τ , τ , ϕτ(0) )} in E (Rn ) imply that for τ ∈ R, t ∈ R+ , ω ∈ Ω , cn → 0 and ϕτ(cn ) , ϕτ(0) ∈ E (Rn ) with ϕτ(cn ) → ϕτ(0) , the following holds: lim ϕ (cn ) (t + τ , τ , θ−τ ω, ϕτ(cn ) ) = ϕ (0) (t + τ , τ , ϕτ(0) ) in E (Rn ).
n→∞
The proof is completed.
(4.23)
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Computers and Mathematics with Applications journal homepage: www.elsevier.com/locate/camwa
Existence and upper semicontinuity of random attractors for non-autonomous stochastic strongly damped sine–Gordon equation on unbounded domains✩ Zhaojuan Wang ∗ , Yanan Liu School of Mathematical Science, Huaiyin Normal University, Huaian 223300, PR China
article
abstract
info
Article history: Received 12 October 2016 Received in revised form 14 January 2017 Accepted 26 January 2017 Available online xxxx Keywords: Stochastic strongly damped sine–Gordon equation Unbounded domains Random attractor
In this paper we study the asymptotic behavior of solutions of the non-autonomous stochastic strongly damped sine–Gordon equation driven by multiplicative noise defined on an unbounded domain. First we introduce a continuous cocycle for the equation and establish the pullback asymptotic compactness of solutions. Second we consider the existence and upper semicontinuity of random attractors for the equation. © 2017 Elsevier Ltd. All rights reserved.
1. Introduction A random attractor, introduced by Crauel et al. [1] and Crauel and Flandoli [2] to capture the essential dynamics with possibly extremely wide fluctuations, is a generalization of the global attractor of autonomous differential equations for random dynamical systems. In recent years, random attractors for non-autonomous stochastic partial differential equations have been investigated in [3,4] in bounded domains and in [5–9] on unbounded domains. In this paper, we consider the existence of a random attractor for the following non-autonomous stochastic strongly damped sine–Gordon equation with multiplicative noise defined in the entire space Rn (n ∈ N): utt − α 1ut − 1u + ut + λu + µ sin u = g (x, t ) + cu ◦
dW dt
,
(1.1)
with the initial value conditions u(τ , x) = uτ (x),
ut (τ , x) = u1τ (x),
x ∈ Rn ,
(1.2)
where ∆ is the Laplacian with respect to the variable x ∈ R ; u = u(t , x) is a real function of x ∈ R and t > τ , τ ∈ R; α > 0 is the strong damping coefficient; λ > 0, µ and c are constants; g (x, ·) is a given function in L2loc (R, L2 (Rn )); ◦ denotes the Stratonovich sense in the stochastic term; W (t ) is a two-sided real-valued Wiener process on a probability space. If g does not depend on time, then we call Eq. (1.1) an autonomous equation. n
n
✩ The authors are supported by National Natural Science Foundation of China (Nos. 11326114, 11401244); Natural Science Research Project of Ordinary Universities in Jiangsu Province (No. 14KJB110003); Undergraduate Innovation and Entrepreneurship Training Program of Jiangsu Province (No. 201616003xj). ∗ Corresponding author. E-mail address: [email protected] (Z. Wang).
http://dx.doi.org/10.1016/j.camwa.2017.01.015 0898-1221/© 2017 Elsevier Ltd. All rights reserved.
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)
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Sine–Gordon equation describes the dynamics of continuous Josephson junctions (see [10]) and has a wide range of applications in physics. Recently, several authors [11–13] studied the attractors for a stochastic damped sine–Gordon equation on a bounded domain, but not on unbounded domains. So far as we know, there were no results on random attractors for non-autonomous stochastic strongly damped sine–Gordon equation (1.1) on unbounded domains. In general, the existence of global random attractor depends on some kind compactness (see, e.g., [14,1,2]). The main difficulty of this paper is to prove the asymptotic compactness of solutions, because Sobolev compact embedding is lost for unbounded domain. In order to overcome the difficulty, we use uniform estimates on the tails of solutions outside a bounded ball in Rn , then decompose the solutions in a bounded domain in terms of eigenfunctions of negative Laplacian as in [15,16]. On are of great physical interest. It is the other hand, the cases of g (x, t ) depending on time and multiplicative noise cu ◦ dW dt therefore important to investigate the existence of attractors for Eq. (1.1). In these cases, we need two separate parametric spaces to deal with the deterministic perturbations g (x, t ) as well as the stochastic perturbations cu ◦ dW . dt This paper is organized as follows. In the next section, we recall some basic settings for Eq. (1.1) and define a continuous cocycle in H 1 (Rn )× L2 (Rn ). In Section 3, we first derive all necessary uniform estimates of solutions, then prove the existence and uniqueness of tempered pullback random attractors for Eq. (1.1). In Section 4, we prove the upper semicontinuity of random attractors as the coefficient of the white noise term tends to zero. Throughout this paper, we use ∥ · ∥ and ⟨·, ·⟩ to denote the norm and the inner product of L2 (Rn ), respectively. The norm of a Banach space X is generally written as ∥ · ∥X . The letters c and ci (i = 1, 2, . . .) are used to denote positive constants whose values are not significant in the context.
2. Mathematical settings Let (Ω , F , P ) be the standard probability space where Ω = {ω ∈ C (R, R) : ω(0) = 0}, F is the Borel σ -algebra induced by the compact open topology of Ω , and P is the Wiener measure on (Ω , F ) (see [17]). There is a classical group {θt }t ∈R acting on (Ω , F , P ) which is defined by θt ω(·) = ω(· + t ) − ω(t ), for ω ∈ Ω , t ∈ R. Then (Ω , F , P , {θt }t ∈R ) is a metric dynamical system. Denote the inner and norm of L2 (U ) as ⟨·, ·⟩ and ∥ · ∥, respectively, and E (U ) = H 1 (U ) × L2 (U ), U ⊆ Rn , endowed with the usual norm
1 ∥Y ∥H 1 (U )×L2 (U ) = ∥∇ u∥2 + ∥u∥2 + ∥v∥2 2
for Y = (u, v)⊤ ∈ E (U ),
(2.1)
where ⊤ stands for the transposition. We define a new norm ∥ · ∥E (U ) by
1 ∥Y ∥E (U ) = ∥v∥2 + δ 2 + λ − δ ∥u∥2 + 1 − αδ ∥∇ u∥2 2 ,
(2.2)
where δ satisfied (3.6), for Y = (u, v)⊤ ∈ E (U ). It is easy to check that ∥ · ∥E (U ) is equivalent to the usual norm ∥ · ∥H 1 (U )×L2 (U ) in (2.1). For our purpose, it is convenient to convert the problem (1.1)–(1.2) into a deterministic system with a random parameter, and then show that it has a cocycle on E (Rn ) over R and (Ω , F , P , {θt }t ∈R ). We identify ω(t ) with W (t ), i.e., ω(t ) = W (t ), t ∈ R. Consider Ornstein–Uhlenbeck equation dz + α zdt = dW (t ), and Ornstein–Uhlenbeck process z (θt ω) = 0 −α −∞ eαs (θt ω)(s)ds, t ∈ R. From [18,19], it is known that the random variable |z (θt ω)| is tempered, and there is a
⊂ Ω of full P measure such that z (θt ω) is continuous in t for every ω ∈ Ω . θt -invariant set Ω Lemma 2.1 (See [12]). For the Ornstein–Uhlenbeck process z (θt ω), we have the following results
|z (θt ω)| = 0, |t | 1 t lim z (θs ω)ds = 0,
(2.3)
lim
t →∞
t →∞
lim
t →∞
lim
t →∞
t
1
t
t
t 1
(2.4)
0
1
|z (θs ω)|ds = √ 0 t
|z (θs ω)|2 ds = 0
πα
1 2α
.
,
(2.5)
(2.6)
Let v(t , τ , ω) = ut (t , τ , ω) + δ u(t , τ , ω) − cu(t , τ , ω)z (θt ω), where δ is as in (2.2), then (1.1)–(1.2) can be rewritten as the equivalent system with random coefficients but without white noise
Z. Wang, Y. Liu / Computers and Mathematics with Applications (
)
du = v − δ u + cuz (θt ω), dt dv = α 1v + (1 − αδ)1u + (δ − 1)v + (δ − λ − δ 2 )u + α cz (θt ω)1u dt + 2c δ z (θt ω) − c 2 z 2 (θt ω) u − cz (θt ω)v − µ sin u + g (x, t ), u(τ , x) = uτ (x), v(τ , x) = vτ (x),
–
3
(2.7)
where vτ (x) = u1τ (x) + δ uτ (x) − cuτ (x)z (ω), x ∈ Rn . and write Ω as Ω from now on. When deriving uniform estimates on solutions, we also We will consider (2.7) for ω ∈ Ω need the following condition on g:
0
δs
e 2 ∥g (·, s + τ )∥2 ds < ∞,
∀ τ ∈ R,
(2.8)
−∞
which implies that
0
δs
e 2 |g (x, s + τ )|2 dxds = 0,
lim
k→∞
−∞
∀ τ ∈ R,
(2.9)
|x|>k
where | · | denotes the absolute value of real number in R. Put ϕ (c ) (t + τ , τ , θ−τ ω, ϕτ(c ) ) = (u(t + τ , τ , θ−τ ω, uτ , vτ ), v(t + τ , τ , θ−τ ω, uτ , vτ ))⊤ , where ϕτ(c ) = (uτ , vτ )⊤ . By the classical theory concerning the existence and uniqueness of the solutions [20,21], one may show that for every ω ∈ Ω , τ ∈ R and ϕτ(c ) ∈ E (Rn ), problem (2.7) has a unique solution ϕ (c ) (·, τ , ω, ϕτ(c ) ) ∈ C ([τ , ∞), E (Rn )) with ϕ (c ) (τ , τ , ω, ϕτ(c ) ) = ϕτ(c ) . In addition, for each t > τ , ϕ (c ) (t , τ , ω, ϕτ(c ) ) is (F , B (H 1 (Rn )) × B (L2 (Rn )))–measurable and continuous in ϕτ with respect to the usual norm of E (Rn ). Based on this fact, one can define a continuous cocycle for (2.7). Let Φc be a mapping, Φc : R+ × R × Ω × E (Rn ) → E (Rn ) given by
Φc (t , τ , ω, ϕτ(c ) ) = ϕ (c ) (t + τ , τ , θ−τ ω, ϕτ(c ) )
(2.10)
for every (t , τ , ω, ϕτ(c ) ) ∈ R+ × R × Ω × E (Rn ). Then it is easy to prove that Φc is a continuous cocycle over R and (Ω , F , P , {θt }t ∈R ) on E (Rn ).
⊤ ⊤ ⊤ = a, b − δ a + caz (θt ω) , a, b ∈ E (Rn ) with an inverse ⊤ ⊤ −1 homeomorphism P (c , δ, θt ω) a, b = a, +δ a − caz (θt ω) . Then, the transformation ¯ c (t , τ , ω, (uτ , u1τ )) = P (c , δ, θt ω)Φc (t , τ , ω, ϕτ(c ) )P −1 (c , δ, θτ ω) Φ (2.11) Introducing the homeomorphism P (c , δ, θt ω) a, b
generates a continuous cocycle with (1.1)–(1.2) over R and (Ω , F , P , {θt }t ∈R ) on E (Rn ). c has a random attractor Note that these two continuous cocycles are equivalent. By (2.11), it is easy to check that Φ provided Φc possesses a random attractor. Then, we only need to consider the continuous cocycle Φc . 3. Existence of random attractors We first provide some sufficient conditions for the existence of random attractors for non-autonomous RDSs in [7]. Similar results can be found in [17,22,14,1,2] for autonomous random dynamical systems. In the following, let (X , ∥ · ∥X ) be a separable Banach space, D (X ) be the collection of all tempered families of nonempty bounded subsets of X . Remember that a family D = {D(τ , ω) : τ ∈ R, ω ∈ Ω } of nonempty bounded subsets of X is said to be tempered if for every σ > 0, τ ∈ R and ω ∈ Ω , the following holds: lim eσ t ∥D(τ + t , θt ω)∥X = 0,
t →−∞
(3.1)
where ∥D∥X = supx∈D ∥x∥X . The cocycle Ψ is said to be D (X )-pullback asymptotically compact in X if for all τ ∈ R and ω ∈ Ω , the sequence
{Ψ (tn , τ − tn , θ−tn ω, xn )}∞ n=1 has a convergent subsequence in X ,
(3.2)
whenever tn → ∞, and xn ∈ D(τ − tn , θ−tn ω) with {D(τ , ω) : τ ∈ R, ω ∈ Ω } ∈ D (X ). Definition 3.1. Let Ψ be a continuous cocycle on X over R and (Ω , F , P , {θt }t ∈R ). (1) A family B = {B(τ , ω) : τ ∈ R, ω ∈ Ω } ∈ D (X ) is called a D (X )-pullback absorbing set for Ψ if for all τ ∈ R and ω ∈ Ω and for every D ∈ D (X ), there exists T = T (D, τ , ω) > 0 such that
Ψ (t , τ − t , θ−t ω, D(τ − t , θ−t ω)) ⊆ B(τ , ω) for all t ≥ T .
(3.3)
4
Z. Wang, Y. Liu / Computers and Mathematics with Applications (
)
–
(2) A family A = {A(τ , ω) : τ ∈ R, ω ∈ Ω } ∈ D (X ) is called a D (X )-pullback attractor for Ψ if for all t ∈ R+ , τ ∈ R and ω ∈ Ω , (i) A(τ , ω) is compact in X and is measurable in ω with respect to F ; (ii) A is invariant, that is, Ψ (t , τ , ω, A(τ , ω)) = A(τ + t , θt ω); (iii) For every D = {D(τ , ω) : τ ∈ R, ω ∈ Ω } ∈ D (X ), lim dH (Ψ (t , τ − t , θ−t ω, D(τ − t , θ−t ω)), A(τ , ω)) = 0,
t →∞
(3.4)
where dH is the Hausdorff semi-distance given by dH (F , G) = supu∈F infv∈G ∥u − v∥X , for any F , G ⊂ X . Proposition 3.2. Let Ψ be a continuous cocycle on X over R and (Ω , F , P , {θt }t ∈R ). Suppose Ψ is D (X )-pullback asymptotically compact in X and has a closed measurable D (X )-pullback absorbing set K in D (X ). Then Ψ has a unique D (X )-pullback attractor A in D (X ) which is given by, for each τ ∈ R and ω ∈ Ω ,
A(τ , ω) =
Ψ (t , τ − t , θ−t ω, K (τ − t , θ−t ω)).
(3.5)
τ >0 t >τ
Next, we will use Proposition 3.2 to prove the existence of a random attractor for the continuous cocycle Φc in E (Rn ). Let δ ∈ (0, 12 ) be small enough such that
δ 2 + λ − δ > 0,
1 − αδ > 0.
(3.6)
Throughout this section we assume that
2|c |δ 2|c | c2 c2 γ =δ− √ − max , (δ 2 + λ − δ + 1) √ + + |µ| . 2(1 − αδ) 2α πα πα
(3.7)
Lemma 3.3. Assume that g (x, ·) ∈ L2loc (R, L2 (Rn )) and (2.8) hold. If γ > 0, then for every τ ∈ R, ω ∈ Ω , and D = {D(τ , ω) : τ ∈ R, ω ∈ Ω } ∈ D (E (Rn )), there exists T = T (τ , ω, D) > 0 such that for all t > T , δ τ τs 32δ −2|cz (θr −τ ω)|−R(θr −τ ω) dr (c ) ∥ϕ (c ) (τ , τ − t , θ−τ ω, ϕτ(c−) t )∥2E (Rn ) + ∥ϕ (s, τ − t , θ−τ ω, ϕτ(c−) t )∥2E (Rn ) ds e 2 τ −t τ s 3δ −2|cz (θr −τ ω)|−R(θr −τ ω) dr ∥∇v(s, τ − t , θ−τ ω, ϕτ(c−) t )∥2 ds +α eτ 2 τ −t
< ϱ(c , τ , ω),
(3.8)
where
ϱ(c , τ , ω) = c + c
0
s 3δ
e
0
2
−2|cz (θr ω)|−R(θr ω) dr
∥g (x, s + τ )∥2 ds,
(3.9)
−∞
ϕτ(c−) t = (uτ −t , vτ −t )⊤ ∈ D(τ − t , θ−t ω), c is a positive constant depending on λ, µ, α and δ , but independent of τ , ω and D, R(θt ω) is given by α R(θt ω) = max |cz (θt ω)|2 , (δ 2 + λ − δ + 1) 2|c δ z (θt ω)| + |cz (θt ω)|2 + |µ| . (3.10) 1 − αδ Proof. Taking the inner product of the second equation of (2.7) with v in L2 (Rn ), we find that 1 d 2 dt
∥v∥2 = −α∥∇v∥2 + 1 − αδ 1u, v − (1 − δ)∥v∥2 − (δ 2 + λ − δ) u, v + α cz (θt ω) 1u, v + 2c δ z (θt ω) − c 2 z 2 (θt ω) u, v − cz (θt ω)∥v∥2 − µ sin u, v + g (x, t ), v .
(3.11)
By the first equation of (2.7), we have
v=
du dt
+ δ u − cuz (θt ω).
(3.12)
Then by (3.12), we have that
1 d 1u, v = − ∥∇ u∥2 + δ∥∇ u∥2 − cz (θt ω)∥∇ u∥2 2 dt
6 −
1 d 2 dt
∥∇ u∥2 + δ − |cz (θt ω)| ∥∇ u∥2 ,
(3.13)
Z. Wang, Y. Liu / Computers and Mathematics with Applications (
u, v
= >
1 d 2 dt 1 d 2 dt
∥u∥2 + δ∥u∥2 − cz (θt ω)∥u∥2
)
–
5
∥u∥2 + δ − |cz (θt ω)| ∥u∥2 .
(3.14)
By the Young inequality and the Cauchy–Schwarz inequality, we have
α cz (θt ω) 1u, v 6 |α cz (θt ω)| · ∥∇ u∥ · ∥∇v∥ α α 6 |cz (θt ω)|2 · ∥∇ u∥2 + ∥∇v∥2 , 2
2c δ z (θt ω) − c 2 z 2 (θt ω) u, v 6 |c δ z (θt ω)| +
g (x, t ), v 6
1 4(1 − 2δ)
(3.15)
2
1 2
|cz (θt ω)|2
∥u∥2 + ∥v∥2 ,
(3.16)
∥g (x, t )∥2 + (1 − 2δ)∥v∥2 ,
(3.17)
|µ| 2 −µ sin u, v 6 ∥u∥ + ∥v∥2 .
(3.18)
2
By (3.13)–(3.18), it then follows from (3.11) that d
∥v∥2 + δ 2 + λ − δ ∥u∥2 + 1 − αδ ∥∇ u∥2 dt 6 −α∥∇v∥2 − 2 δ − |cz (θt ω)| ∥v∥2 + δ 2 + λ − δ ∥u∥2 + 1 − αδ ∥∇ u∥2 + α|cz (θt ω)|2 · ∥∇ u∥2 + 2|c δ z (θt ω)| + |cz (θt ω)|2 + |µ| ∥u∥2 + ∥v∥2 +
1 2(1 − 2δ)
∥g (x, t )∥2 .
(3.19)
Recalling the new norm ∥ · ∥E (U ) in (2.2), then we obtain from (3.10) and (3.19) that d dt
∥ϕ (c ) ∥2E (Rn ) + α∥∇v∥2 6 −2δ + 2|cz (θt ω)| + R(θt ω) ∥ϕ (c ) ∥2E (Rn ) +
1 2(1 − 2δ)
∥g (x, t )∥2 .
(3.20)
t 3δ Multiplying (3.20) by e 0 ( 2 −2|cz (θr ω)|−R(θr ω))dr and then integrating over (τ − t , τ ) with t > 0, we have
∥ϕ (c ) (τ , τ − t , ω, ϕτ(c−) t )∥2E (Rn ) δ τ τs 32δ −2|cz (θr ω)|−R(θr ω) dr (c ) e + ∥ϕ (s, τ − t , ω, ϕτ(c−) t )∥2E (Rn ) ds 2 τ −t τ s 3δ −2|cz (θr ω)|−R(θr ω) dr ∥∇v(s, τ − t , ω, ϕτ(c−) t )∥2 ds +α eτ 2 τ −t τ −t 3δ −2|cz (θr ω)|−R(θr ω) dr τ 2
6e
∥ϕτ(c−) t ∥2E (Rn ) +
1
τ
2(1 − 2δ) τ −t
e
s 3δ τ
2
−2|cz (θr ω)|−R(θr ω) dr
∥g (x, s)∥2 ds.
(3.21)
Substituting ω by θ−τ ω, then we have from (3.21) that
∥ϕ (c ) (τ , τ − t , θ−τ ω, ϕτ(c−) t )∥2E (Rn ) δ τ τs 32δ −2|cz (θr −τ ω)|−R(θr −τ ω) dr (c ) + e ∥ϕ (s, τ − t , θ−τ ω, ϕτ(c−) t )∥2E (Rn ) ds 2 τ −t τ s 3δ −2|cz (θr −τ ω)|−R(θr −τ ω) dr +α ∥∇v(s, τ − t , θ−τ ω, ϕτ(c−) t )∥2 ds eτ 2 τ −t τ s τ −t 3δ 3δ 1 −2|cz (θr −τ ω)|−R(θr −τ ω) dr −2|cz (θr −τ ω)|−R(θr −τ ω) dr 2 6eτ ∥ϕτ(c−) t ∥2E (Rn ) + eτ 2 ∥g (x, s)∥2 ds 2(1 − 2δ) τ −t 0 s −t 3δ 3δ 1 −2|cz (θr ω)|−R(θr ω) dr −2|cz (θr ω)|−R(θr ω) dr 6e0 2 ∥ϕτ(c−) t ∥2E (Rn ) + e0 2 ∥g (x, s + τ )∥2 ds. 2(1 − 2δ) −∞
(3.22)
By Lemma 2.1, there exists T1 (ω) such that for all t > T1 (ω),
0
2|cz (θr ω)| + R(θr ω) dr <
−t
2|c | 2|c |δ c2 c2 + max , (δ 2 + λ − δ + 1) √ + + |µ| t. √ 2(1 − αδ) 2α πα πα
(3.23)
6
Z. Wang, Y. Liu / Computers and Mathematics with Applications (
)
–
Then, it follows from (3.6), (3.7) and (3.23) that for all s 6 −T1 (ω), e
s 3δ 0
2
−2|cz (θr ω)|−R(θr ω) dr
δ
6 e 2 s.
(3.24)
By (2.8) and (3.24), we have 0
e
s 3δ 0
−2|cz (θr ω)|−R(θr ω) dr
2
∥g (x, s + τ )∥2 ds < +∞.
(3.25)
−∞ (c )
Due to ϕτ −t ∈ D(τ − t , θ−t ω) and D ∈ D (E (Rn )), we have from (3.24), −t 3δ
lim e
0
2
−2|cz (θr ω)|−R(θr ω) dr
t →+∞
∥ϕτ(c−) t ∥2E (Rn ) = 0.
(3.26)
Therefore, there exists T = T (τ , ω, D) > T1 (ω) such that for all t > T , −t 3δ 0
e
−2|cz (θr ω)|−R(θr ω) dr
2
∥ϕτ(c−) t ∥2E (Rn ) 6 1.
Thus the lemma follows from (3.22) and (3.25).
In order to estimate the tails of solutions outside a bounded ball in Rn , we choose a smooth function ρ , such that 0 6 ρ(s) 6 1 for s ∈ R, and
ρ(s) =
0, 1,
0 6 |s| 6 1, |s| > 2,
(3.27)
and there exist constants µ1 , µ2 , such that |ρ ′ (s)| 6 µ1 , |ρ ′′ (s)| 6 µ2 for s ∈ R. Given k > 1, denote by Hk = {x ∈ Rn : |x| < k} and Rn \ Hk the complement of Hk . To prove asymptotic compactness of solution on Rn , we prove the following lemmas. Lemma 3.4. Assume that g (x, ·) ∈ L2loc (R, L2 (Rn )) and (2.8) hold. If γ > 0, then for every τ ∈ R, ω ∈ Ω , and D = {D(τ , ω) : τ ∈ R, ω ∈ Ω } ∈ D (E (Rn )), there exist T = T (τ , ω, D, ε) > 0 and K = K (τ , ω, ε) > 1, such that for all t > T , k > K ,
∥ϕ (c ) (τ , τ − t , θ−τ ω, ϕτ(c−) t )∥2E (Rn \Hk ) 6 ε,
(3.28)
(c )
where ϕτ −t = (uτ −t , vτ −t )⊤ ∈ D(τ − t , θ−t ω). Proof. Taking the inner product of the second equation of (2.7) with ρ
1 d 2 dt
ρ
Rn
|x|2 v in L2 (Rn ), we obtain k2
|x|2 |v|2 dx 2 k
|x|2 v dx + ( 1 − αδ) ( 1 u )ρ v dx k2 k2 Rn Rn |x|2 |x|2 ρ 2 uv dx − (1 − δ + cz (θt ω)) ρ 2 |v|2 dx − (δ 2 + λ − δ) k k Rn Rn |x|2 |x|2 + α cz (θt ω) (1u)ρ 2 v dx + 2c δ z (θt ω) − c 2 z 2 (θt ω) ρ 2 uv dx
=α
(1v)ρ
|x|2
k
Rn
+ Rn
ρ
|x|2 k2
g (x, t )v dx − µ
Rn
Rn
ρ
|x|2 k2
k
v sin udx.
(3.29)
By using (3.12), we get that
Rn
(1u)ρ
|x|2 2x |x|2 du ′ v dx = − (∇ u ) + δ u − cuz (θ ω) ρ t 2 2 2 k
Rn
k
k
dt
|x|2 du +ρ 2 ∇ + δ u − cuz (θt ω) dx k dt |x|2 2µ1 |x| 1 d 6 |(∇ u )v| dx − ρ 2 |∇ u|2 dx √ 2 k<|x|< 2k
k
+ cz (θt ω) − δ
2 dt
Rn
|x|2 ρ 2 |∇ u|2 dx k
Rn
k
Z. Wang, Y. Liu / Computers and Mathematics with Applications (
√
2µ1
6
k
1 d ∥∇ u∥2 + ∥v∥2 −
2 dt
Rn
ρ
|x|2 k2
Rn
ρ
)
–
7
|x|2 |∇ u|2 dx 2 k
|x|2 ρ 2 |∇ u|2 dx,
+ |cz (θt ω)| − δ k Rn |x|2 du = ρ 2 u· + δ u − cuz (θt ω) dx
uv dx
k
Rn
1 d
>
2 dt
dt
ρ
Rn
(3.30)
|x|2 k2
|u|2 dx + δ − |cz (θt ω)|
Rn
ρ
|x|2 |u|2 dx. 2 k
(3.31)
By the Cauchy–Schwarz inequality and the Young inequality, we obtain
α
Rn
(1v)ρ
|x|2 k2
2x |x|2 |x|2 ′ v + ρ ∇v dx ρ k2 k2 k2 Rn |x|2 2αµ1 |x| 6 |(∇v)v| dx − α ρ |∇v|2 dx √ k2 k2 k<|x|< 2k Rn √ |x|2 2αµ1 2 2 6 ∥∇v∥ + ∥v∥ − α ρ 2 |∇v|2 dx,
v dx = −α
(∇v)
k
α cz (θt ω)
Rn
(1u)ρ
|x|2 k2
Rn
|x|2 2x ′ |x|2 v dx = −α cz (θt ω) ∇u ρ v + ρ ∇v dx k2 k2 k2 Rn 2µ1 |x| | ∇ u v|dx 6 |α cz (θt ω)| √ k2 k<|x|< 2k |x|2 + |α cz (θt ω)| ρ 2 |∇v| · |∇ u|dx k Rn √ |x|2 α 2µ1 |α cz (θt ω)| 6 ∥∇ u∥2 + ∥v∥2 + ρ 2 |∇v|2 dx k 2 Rn k |x|2 α|cz (θt ω)| 2 + ρ 2 |∇ u|2 dx, 2
2c δ z (θt ω) − c 2 z 2 (θt ω)
Rn
ρ
|x|2 k2
−µ Rn
ρ
g (x, t )v dx 6
|x|2 k2
(3.32)
k
Rn
ρ
|x|2
4(1 − 2δ)
v sin udx 6
|µ|
uv dx 6 |c δ z (θt ω)| +
k2
1
2
Rn
k
Rn
Rn
ρ
ρ
|cz (θt ω)|2
2
Rn
|x|2 2 | g ( x , t )| dx + ( 1 − 2 δ) 2 k
|x|2 k2
Rn
ρ ρ
|x|2 |u|2 + |v|2 dx, 2 k
|x|2 k2
|v|2 dx,
|u|2 + |v|2 dx.
(3.33)
(3.34)
(3.35)
(3.36)
Then it follows from (3.30)–(3.36) that 1 d
|x|2
2 2 2 2 |v| + δ + λ − δ | u | + ( 1 − αδ)|∇ u | dx 2 dt Rn k2 |x|2 2 2 6 (|cz (θt ω)| − δ) ρ 2 |v| dx + (δ + λ − δ)(|cz (θt ω)| − δ) ρ
k
Rn
|x|2
α|cz (θt ω)| 2
Rn
ρ
|x|2 k2
|u|2 dx
|x|2 + (1 − αδ)(|cz (θt ω)| − δ) ρ 2 |∇ u| dx + ρ 2 |∇ u|2 dx k 2 k Rn Rn √ √ 2αµ1 2µ1 (1 − αδ + |α cz (θt ω)|) + ∥∇v∥2 + ∥v∥2 + ∥∇ u∥2 + ∥v∥2 k k |x|2 |cz (θt ω)|2 |µ| + |c δ z (θt ω)| + + ρ 2 |u|2 + |v|2 dx
2
−
α 2
Rn
|x|2 ρ 2 |∇v|2 dx + k
2
2
k
Rn
1 4(1 − 2δ)
Rn
ρ
|x|2 k2
|g (x, t )|2 dx.
(3.37)
Let
φ (c ) = |v|2 + δ 2 + λ − δ |u|2 + (1 − αδ)|∇ u|2 .
(3.38)
8
Z. Wang, Y. Liu / Computers and Mathematics with Applications (
)
–
It then follows from (3.37) and (3.38) that d
dt
Rn
ρ
|x|2 (c ) φ dx 6 −2δ + 2|cz (θt ω)| + R(θt ω) 2 k
+
c1
Rn
ρ
|x|2 k2
φ (c ) dx
c1 |z (θt ω)| ∥∇ u∥2 + ∥v∥2 + ∥∇v∥2 + ∥∇ u∥2 + ∥v∥2
k
k
+ c1 Rn
|x|2 ρ 2 |g (x, t )|2 dx
(3.39)
k
t where R(θt ω) is as in (3.10). Multiplying (3.39) by e 0 (2δ−2|cz (θr ω)|−R(θr ω))dr and then integrating over (τ − t , τ ) with t > 0, we have
Rn
ρ
|x|2 |x|2 τ −t (c ) (c ) (2δ−2|cz (θr ω)|−R(θr ω))dr τ φ (τ , τ − t , ω, φ ) dx 6 e ρ 2 φτ(c−) t dx τ −t 2 k k Rn c1 τ s (2δ−2|cz (θr ω)|−R(θr ω))dr eτ |z (θs ω)| ∥∇ u(s, τ − t , ω, ϕτ(c−) t )∥2 + k τ −t + ∥v(s, τ − t , ω, ϕτ(c−) t )∥2 ds c1 τ s (2δ−2|cz (θr ω)|−R(θr ω))dr eτ + ∥∇v(s, τ − t , ω, ϕτ(c−) t )∥2 k τ −t + ∥v(s, τ − t , ω, ϕτ(c−) t )∥2 + ∥∇ u(s, τ − t , ω, ϕτ(c−) t )∥2 ds τ |x|2 s + c1 e τ (2δ−2|cz (θr ω)|−R(θr ω))dr ρ 2 |g (x, s)|2 dxds. (3.40) τ −t
Rn
k
Replacing ω by θ−τ ω in (3.40), then we have that
|x|2 φ (c ) (τ , τ − t , θ−τ ω, φτ(c−) t )dx 2 n k R |x|2 −t ρ 2 φτ(c−) t dx 6 e 0 (2δ−2|cz (θr ω)|−R(θr ω))dr k Rn c1 τ s (2δ−2|cz (θr −τ ω)|−R(θr −τ ω))dr |z (θs−τ ω)| ∥∇ u(s, τ − t , θ−τ ω, ϕτ(c−) t )∥2 + eτ k τ −t + ∥v(s, τ − t , θ−τ ω, ϕτ(c−) t )∥2 ds c1 τ s (2δ−2|cz (θr −τ ω)|−R(θr −τ ω))dr + eτ ∥∇v(s, τ − t , θ−τ ω, ϕτ(c−) t )∥2 k τ −t + ∥v(s, τ − t , θ−τ ω, ϕτ(c−) t )∥2 + ∥∇ u(s, τ − t , θ−τ ω, ϕτ(c−) t )∥2 ds 0 |x|2 s (2δ−2|cz (θr ω)|−R(θr ω))dr 0 + c1 e ρ 2 |g (x, s + τ )|2 dxds. ρ
−∞
(3.41)
k
|x|>k
We next estimate each term on the right-hand side of (3.41). By Lemma 2.1, (3.6) and (3.7), there exists T1 = T1 (ω) such that for all s 6 −T1 , s e 0 (2δ−2|cz (θr ω)|−R(θr ω))dr 6 e
Due to t > T2 ,
ϕτ(c−) t
e
−t 0
s 3δ 0
2
−2|cz (θr ω)|−R(θr ω) dr
δ
6 e 2 s.
(3.42)
∈ D(τ − t , θ−t ω) ∈ D (E (Rn )), we get from (3.42) that, there exists T2 = T2 (τ , ω, D, ε) > T1 , such that for all
(2δ−2|cz (θr ω)|−R(θr ω))dr
Rn
ρ
|x|2 δ φτ(c−) t dx 6 e− 2 t ∥ϕτ(c−) t ∥2E (Rn ) < ε. 2
(3.43)
k
By (3.42), Lemmas 2.1 and 3.3, there are T3 = T3 (τ , ω, D, ε) > 0 and K1 = K1 (τ , ω, ε) > 1, such that for all t > T3 and k > K1 ,
c1 k
τ τ −t
s (c ) (c ) e τ (2δ−2|cz (θr −τ ω)|−R(θr −τ ω))dr |z (θs−τ ω)| ∥∇ u(s, τ − t , θ−τ ω, ϕτ −t )∥2 + ∥v(s, τ − t , θ−τ ω, ϕτ −t )∥2 ds
Z. Wang, Y. Liu / Computers and Mathematics with Applications (
c1
6
τ
e
k
τ −t
δ(s−τ ) 2
|z (θs−τ ω)| · e
s 3δ τ
2
)
–
9
−2|cz (θr −τ ω)|−R(θr −τ ω) dr
∥∇ u(s, τ − t , θ−τ ω, ϕτ(c−) t )∥2
+ ∥v(s, τ − t , θ−τ ω, ϕτ(c−) t )∥2 ds < ε.
(3.44)
By Lemma 3.3, there are T4 = T4 (τ , ω, D, ε) > 0 and K2 = K2 (τ , ω, ε) > 1, such that for all t > T4 and k > K2 , τ
c1 k
s (c ) e τ (2δ−2|cz (θr −τ ω)|−R(θr −τ ω))dr ∥∇v(s, τ − t , θ−τ ω, ϕτ −t )∥2
τ −t
+ ∥v(s, τ − t , θ−τ ω, ϕτ(c−) t )∥2 + ∥∇ u(s, τ − t , θ−τ ω, ϕτ(c−) t )∥2 ds < ε.
(3.45)
By using condition (2.9) and (3.42), there exists K3 = K3 (τ , ω, ε) > 1, such that for all k > K3 ,
0
c1
s e 0 (2δ−2|cz (θr ω)|−R(θr ω))dr
−∞
ρ |x|>k
|x|2 |g (x, s + τ )|2 dxds < ε. 2
(3.46)
k
Letting T = max T1 , T2 , T3 , T4 , K = max K1 , K2 , K3 , then combining (3.43)–(3.46), we have for all t > T and k > K ,
Rn
ρ
|x|2 k2
φ (c ) (τ , τ − t , θ−τ ω, φτ(c−) t )dx < 4ε,
which implies
∥ϕ (c ) (τ , τ − t , θ−τ ω, ϕτ(c−) t )∥2E (Rn \Hk ) < 4ε. Then we complete the proof.
Let ρ = 1 − ρ with ρ given by (3.27). Fix k > 1 and set
|x|2 u(t , τ , ω, uτ , vτ ) = ρ 2 u(t , τ , ω, ϕτ(c ) ), k |x|2 v(t , τ , ω, uτ , vτ ) = ρ 2 v(t , τ , ω, ϕτ(c ) ). k |x|2 Multiplying (2.7) by ρ k2 and using (3.47), we find that d u v − δ u + c uz (θt ω) dt = |x|2 |x|2 d v = α 1 v − αv 1 ρ 2 − 2α∇v∇ ρ 2 + (δ − 1) v + (1 − αδ)1 u k dt |x|2 k|x|2 ρ 2 + (δ − λ − δ 2 ) u − (1 − αδ)u1 ρ 2 − 2(1 − αδ)∇ u∇ k k 2 2 | x| |x| + α cz (θt ω)1 u − α cz (θt ω)u1 ρ 2 − α cz (θt ω)∇ u∇ ρ 2 k k |x|2 |x|2 2 2 + (2c δ z (θt ω) − c z (θt ω)) u − cz (θt ω) v − ρ 2 f (u) + ρ 2 g (x, t ). k
(3.47)
(3.48)
k
Considering the eigenvalue problem
− 1 u= u in H2k , with u = 0 on ∂ H2k .
(3.49)
The problem (3.49) has a family of eigenfunctions {ei }i∈N with the eigenvalues {λi }i∈N :
λ1 6 λ2 6 · · · 6 λi 6 · · · , λi → +∞(i → +∞), such that {ei }i∈N is an orthonormal basis of L2 (H2k ). Given n ∈ N, let Xn = span{e1 , . . . , en } and Pn : L2 (H2k ) → Xn be the projection operator. Lemma 3.5. Assume that g (x, ·) ∈ L2loc (R, L2 (Rn )) and (2.8) hold. If γ > 0, then for every τ ∈ R, ω ∈ Ω , and D = {D(τ , ω) : τ ∈ R, ω ∈ Ω } ∈ D (E (Rn )), there exist T = T (τ , ω, D, ε) > 0, K = K (τ , ω, ε) > 1 and N = N (τ , ω, ε) > 0, such that for all t > T , k > K and n > N,
∥(I − Pn ) ϕ (c ) (τ , τ − t , θ−τ ω, ϕτ(c−) t )∥2E (H2k ) < ε, 2 where ϕτ(c−) t = ρ |kx|2 ϕτ(c−) t , ϕτ(c−) t = (uτ −t , vτ −t )⊤ ∈ D(τ − t , θ−t ω).
(3.50)
10
Z. Wang, Y. Liu / Computers and Mathematics with Applications (
)
–
Proof. Let un,1 = Pn u, un,2 = (I − Pn ) u, vn,1 = Pn v and vn,2 = (I − Pn ) v . Applying I − Pn to the first equation of (3.48), we obtain
vn,2 =
d un , 2 dt
+ δ un,2 − c un,2 z (θt ω).
(3.51)
Then applying I − Pn to the second equation of (3.48) and taking the inner product of the resulting equation with vn,2 in L2 (H2k ), we have 1 d 2 dt
|x|2 |x|2 ρ 2 , vn,2 ∥ vn,2 ∥2 = −α∥∇ vn,2 ∥2 − α v 1 ρ 2 + 2∇v∇ k
k
|x|2 |x|2 + (1 − αδ) 1 un,2 , vn,2 − (1 − αδ + α cz (θt ω)) u1 ρ 2 + 2∇ u∇ ρ 2 , vn,2 k k + (δ − λ − δ 2 ) un,2 , vn,2 + α cz (θt ω) 1 un,2 , vn,2 + 2c δ z (θt ω) − c 2 z 2 (θt ω) un,2 , vn,2 + δ − 1 − cz (θt ω) ∥ vn,2 ∥2 |x|2 |x|2 −µ ρ 2 sin u, vn,2 + vn,2 . ρ 2 g (x, t ), k
k
(3.52)
Substituting vn,2 in (3.51) into the third and fifth terms on the left-hand side of (3.52), we have
d un , 2 + δ un,2 − c un,2 z (θt ω) 1 un,2 , vn,2 = − ∇ un,2 , ∇ dt
1 d
∥∇ un,2 ∥2 − δ − |cz (θt ω)| ∥∇ un,2 ∥2 , 2 dt d un,2 + δ un,2 − c un,2 z (θt ω) un,2 , vn,2 = un,2 , 6 −
(3.53)
dt
>
1 d 2 dt
∥ un,2 ∥2 + δ − |cz (θt ω)| ∥ un,2 ∥2 .
(3.54)
Using the Cauchy–Schwarz inequality and the Young inequality, we obtain
|x|2 |x|2 4x2 |x|2 2 2 ′ |x|2 4x ′ |x| ρ 2 , vn,2 = −α v 4 ρ ′′ 2 + 2 ρ + ∇v · ρ , v −α v 1 ρ 2 + 2∇v∇ n,2 k k k k k k2 k2 k2 2 1 − 2δ 32 µ 2α 2 (8µ2 + 2µ1 )2 6 ∥v∥2 + 2 1 ∥∇v∥2 + ∥ vn,2 ∥2 , (3.55) 1 − 2δ k4 k 4 |x|2 |x|2 −(1 − αδ + α cz (θt ω)) u1 ρ 2 + 2∇ u∇ ρ 2 , vn,2 k
4x2
= −(1 − αδ + α cz (θt ω)) u
k4
ρ ′′
|x|2 k2
+
k 2
ρ ′ 2
k
|x|2 k2
+
4x
∇u · ρ ′ 2
k
|x|2 , v n , 2 2 k
1 − 2δ 32µ21 2(1 − αδ + α c |z (θt ω)|) (4µ2 + µ1 )2 2 2 6 ∥ u ∥ + ∥∇ u ∥ + ∥ vn,2 ∥2 , 1 − 2δ k4 k2 2 α cz (θt ω) 1 un,2 , vn,2 = −α cz (θt ω) ∇ un,2 , ∇ vn,2 α|cz (θt ω)|2 α 6 ∥∇ un,2 ∥ + ∥∇ vn,2 ∥, 2
2
2
2c δ z (θt ω) − c 2 z 2 (θt ω) un,2 , vn,2 6 |c δ z (θt ω)| +
|x|2 ρ 2 g (x, t ), vn,2 6 k
1 1 − 2δ
|cz (θt ω)|2 2
∥ un,2 ∥2 + ∥ vn,2 ∥2 ,
2 1 − 2 δ |x|2 ρ 2 g (x, t ) + ∥ vn,2 ∥2 , (I − Pn ) k
4
|x|2 |x|2 vn,2 6 |µ| · (I − Pn ) −µ ρ 2 sin u, ρ 2 sin u · ∥ vn,2 ∥ k k |x|2 6 |µ| · (I − Pn ) ρ 2 u · ∥ vn,2 ∥ k |µ| 6 ∥ un,2 ∥2 + ∥ vn,2 ∥2 . 2
(3.56)
(3.57) (3.58) (3.59)
(3.60)
Z. Wang, Y. Liu / Computers and Mathematics with Applications (
)
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11
By recalling the new norm ∥ · ∥E (U ) in (2.2), and R(θt ω) in (3.10), then, we can get from (3.53)–(3.60) that d dt
(c ) 2 vn,2 ∥2 + −2δ + 2|cz (θt ω)| + R(θt ω) ∥ ϕn,2 ∥E (H2k ) ∥ ϕn(,c2) ∥2E (H2k ) 6 −α∥∇ +
c3 k4
∥ u∥ 2 c3 ∥∇ u∥2 ∥u∥2 + ∥v∥2 + 2 ∥∇ u∥2 + ∥∇v∥2 + c3 |z (θt ω)|2 + 4 2 k
k
k
2 |x|2 + c3 0(I − Pn ) ρ 2 g (x, t ) . k t
Multiplying (3.61) by e
0 2δ−2|cz (θr ω)|−R(θr ω) dr
ϕn(c,2) ,τ −t )∥2E (H2k ) 6 e ∥ ϕn(,c2) (τ , τ − t , ω,
(3.61)
and then integrating over (τ − t , τ ) with t > 0, we have
τ −t
∥ ϕn(c,2) ,τ −t ∥2E (H2k ) c3 τ τ 2δ−2|cz (θr ω)|−R(θr ω) dr + 4 e ∥u(s, τ − t , ω, ϕτ(c−) t )∥2 k τ −t + ∥v(s, τ − t , ω, ϕτ(c−) t )∥2 ds c3 τ τs 2δ−2|cz (θr ω)|−R(θr ω) dr + 2 e ∥∇ u(s, τ − t , ω, ϕτ(c−) t )∥2 k τ −t + ∥∇v(s, τ − t , ω, ϕτ(c−) t )∥2 ds τ ∥u(s, τ − t , ω, ϕ (c ) )∥2 s τ −t 2δ−2|cz (θr ω)|−R(θr ω) dr + c3 eτ |z (θs ω)|2 4 τ
2δ−2|cz (θr ω)|−R(θr ω) dr
s
k
τ −t
+
∥∇ u(s, τ −
k2
+ c3
(c ) t , ω, ϕτ −t )∥2
τ
2 |x|2 ρ 2 g (x, s) ds. (I − Pn )
s
e
ds
τ 2δ−2|cz (θr ω)|−R(θr ω) dr
k
τ −t
(3.62)
Substituting ω by θ−τ ω in (3.62), we have
∥ ϕn(,c2) (τ , τ − t , θ−τ ω, ϕn(c,2) ,τ −t )∥2E (H2k ) −t 2δ−2|cz (θr ω)|−R(θr ω) dr 6e0 ∥ ϕn(c,2) ,τ −t ∥2E (H2k ) τ s c3 2δ−2|cz (θr −τ ω)|−R(θr −τ ω) dr eτ ∥u(s, τ − t , θ−τ ω, ϕτ(c−) t )∥2 + ∥v(s, τ − t , θ−τ ω, ϕτ(c−) t )∥2 ds + 4 k τ −t c3 τ τs 2δ−2|cz (θr −τ ω)|−R(θr −τ ω) dr e ∥∇ u(s, τ − t , θ−τ ω, ϕτ(c−) t )∥2 + 2 k τ −t + ∥∇v(s, τ − t , θ−τ ω, ϕτ(c−) t )∥2 ds τ ∥u(s, τ − t , θ ω, ϕ (c ) )∥2 s −τ τ −t 2δ−2|cz (θr −τ ω)|−R(θr −τ ω) dr + c3 eτ |z (θs−τ ω)|2 4 k
τ −t
+
∥∇ u(s, τ −
(c ) t , θ−τ ω, ϕτ −t )∥2
k2
0
+ c3
e
ds
2 |x|2 ρ 2 g (x, s + τ ) ds. (I − Pn )
s
0 2δ−2|cz (θr ω)|−R(θr ω) dr
(3.63)
k
−t
(c )
In what follows, we estimate the terms on the right-hand side of (3.63). Due to ϕτ −t ∈ D(τ − t , θ−t ω) ∈ D (E (Rn )), we get from (3.42) that, there exists T1 = T1 (τ , ω, D, ε) > 0, such that if t > T1 , then e
−t
2δ−2|cz (θr ω)|−R(θr ω) dr
0
∥ ϕn(c,2) ,τ −t ∥2E (H2k ) < ε.
(3.64)
For the second and third terms on the right-hand side of (3.63), by Lemma 3.3, we obtain that there exist T2 T2 (τ , ω, D, ε) > 0 and K1 = K1 (τ , ω, ε) > 1, such that for all t > T2 and k > K1 ,
c3 k4
τ
e τ −t
s
τ 2δ−2|cz (θr −τ ω)|−R(θr −τ ω) dr
∥u(s, τ − t , θ−τ ω, ϕτ(c−) t )∥2 + ∥v(s, τ − t , θ−τ ω, ϕτ(c−) t )∥2 ds
=
12
Z. Wang, Y. Liu / Computers and Mathematics with Applications (
c3
+
τ
e
k2
–
2δ−2|cz (θr −τ ω)|−R(θr −τ ω) dr ∥∇ u(s, τ − t , θ−τ ω, ϕτ(c−) t )∥2 + ∥∇v(s, τ − t , θ−τ ω, ϕτ(c−) t )∥2 ds
τ
τ −t
< ε.
)
s
(3.65)
Since z (ω) is tempered, by Lemmas 2.1 and 3.3, there exist T3 = T3 (τ , ω, D, ε) > 0 and K2 = K2 (τ , ω, ε) > 1, such that for all t > T3 and k > K2 , we have
c3
τ
e
s
τ 2δ−2|cz (θr −τ ω)|−R(θr −τ ω) dr
τ −t τ
6 c3
e τ −t
δ(s−τ ) 2
|z (θs−τ ω)| · e
∥∇ u(s, τ −
+
s
2
|z (θs−τ ω)|2
k4
3δ −2|cz (θr −τ ω)|−R(θr −τ ω) 2
τ
dr
+
∥∇ u(s, τ − t , θ−τ ω, ϕτ(c−) t )∥2 k2
ds
∥u(s, τ − t , θ ω, ϕ (c ) )∥2 −τ τ −t k4
(c ) t , θ−τ ω, ϕτ −t )∥2
ds
k2
< ε.
∥u(s, τ − t , θ ω, ϕ (c ) )∥2 −τ τ −t
(3.66)
Next, we estimate the last term on the right-hand side of (3.63), by (2.8) and (3.42), there are N = N (τ , ω, ε) > 0, T4 = T4 (τ , ω, ε) > 0 and K3 = K3 (τ , ω, ε) > 1, such that for all n > N, t > T4 and k > K3 ,
0
c3
2 |x|2 ρ 2 g (x, s + τ ) ds < ε. (I − Pn )
s
e
0 2δ−2|cz (θr ω)|−R(θr ω) dr
k
−t
(3.67)
Let T = max T1 , T2 , T3 , T4 and K = max K1 , K2 , K3 . Then, it follows from (3.64), (3.66) and (3.67) that, for all t > T , k > K and n > N,
∥ ϕn(,c2) (τ , τ − t , θ−τ ω, ϕn(,c2),τ −t )∥2E (H2k ) < 4ε. This completes the proof.
(3.68)
We now prove the existence of D (E (Rn ))-pullback attractors for the random problem (2.7) in D (E (Rn )). Theorem 3.6. Assume that g (x, ·) ∈ L2loc (R, L2 (Rn )) and (2.8) hold. If γ > 0, then the continuous cocycle Φc associated with problem (2.7) has a unique D (E (Rn ))-pullback attractor Ac = {Ac (τ , ω) : τ ∈ R, ω ∈ Ω } in D (E (Rn )). Proof. By (2.10) and Lemma 3.3, we notice that the continuous cocycle Φc has a closed measurable D (E (Rn ))-pullback absorbing set in E (Rn ). Next, we will prove that the continuous cocycle Φc is asymptotically compact in E (Rn ). Let tm → ∞, B ∈ D (E (Rn )), and (c ) ϕτ −tm ∈ B(τ − tm , θ−tm ω). By using Lemma 3.3, we have that {ϕ (c ) τ , τ − tm , θ−τ ω, ϕτ(c−) tm } is a bounded in E (Rn ); that is, for every τ ∈ R and ω ∈ Ω , there exists M1 = M1 (τ , ω, B) > 0 such that for all m > M1 ,
(c ) ϕ τ , τ − tm , θ−τ ω, ϕτ(c−) t 2 n < ϱ(c , τ , ω), m E (R )
(3.69)
where ϱ(c , τ , ω) is as in (3.9). By Lemma 3.4, there are k1 = k1 (τ , ω, ε) and M2 = M2 (τ , ω, B, ε) > 0, such that for every m > M2 ,
(c ) ϕ τ , τ − tm , θ−τ ω, ϕτ(c−) t 2 n m E (R \H
k1 )
< ε.
(3.70)
In addition, by Lemma 3.5, there exist N = N (τ , ω, ε) > 0, k2 = k2 (τ , ω, ε) > k1 and M3 = M3 (τ , ω, B, ε) > 0, such that for every m > M3 ,
2 (I − PN ) ϕ (c ) τ , τ − tm , θ−τ ω, ϕτ(c−) tm E (H
2k2 )
< ε.
(3.71) (c )
ϕ (c ) τ , τ − tm , θ−τ ω, ϕτ −tm } is a bounded in PN E (H2k2 ), which associates with (3.71) By (3.47) and (3.69), we get that {PN
(c )
implies that { ϕ (c ) τ , τ − tm , θ−τ ω, ϕτ −tm } is precompact in H01 (H2k2 ) × L2 (H2k2 ).
|x|2
Recalling (3.47) and the fact that ρ
k22
= 1 for |x| < k2 , we have that {ϕ (c ) τ , τ − tm , θ−τ ω, ϕτ(c−) tm } is precompact in
E (Hk2 ), which along with (2.10) and (3.70) shows that the continuous cocycle Φc is asymptotically compact in D (E (Rn )). Then, by Proposition 3.2, the continuous cocycle Φc associated with (2.7) has a unique D (E (Rn ))-pullback attractor in D (E (Rn )).
Z. Wang, Y. Liu / Computers and Mathematics with Applications (
)
–
13
3.1. Remark In this section we studied the existence of the D (E (Rn ))-pullback attractor for the non-autonomous stochastic strongly damped sine–Gordon equation driven by multiplicative noise. The coefficient c of the multiplicative noise term cu ◦ dW dt needs to be suitably small, this is because that the multiplicative noise cu ◦
dW dt
depends on the state variable u.
4. Upper semicontinuity of D (E (Rn ))-pullback attractors First, we present a criteria concerning upper semicontinuity of non-autonomous random attractors with respect to a parameter in [23]. Theorem 4.1. Let (X , ∥ · ∥X ) be a separable Banach space, Ψc be a continuous cocycle on X over R and (Ω , F , P , {θt }t ∈R ). Suppose that (i) Ψc has a closed measurable random absorbing set Kc = {Kc (τ , ω) : τ ∈ R, ω ∈ Ω } in D (X ) and a unique random attractor Ac = {Ac (τ , ω) : τ ∈ R, ω ∈ Ω } in D (X ). (ii) There exists a map ς : R → R such that for each τ ∈ R, ω ∈ Ω , K0 (τ ) = {u ∈ X : ∥u∥X 6 ς (τ )} and lim sup ∥Kc (τ , ω)∥X = lim sup c →0
c →0
sup
x∈Kc (τ ,ω)
∥x∥X 6 ς (τ ).
(4.1)
(iii) There exists ε > 0, such that for every τ ∈ R and ω ∈ Ω , |c |6ε Ac (τ , ω) is precompact in X . (iv) For t > 0, τ ∈ R, ω ∈ Ω , cn → 0 when n → ∞, and xn , x0 ∈ X with xn → x0 when n → ∞, it holds:
lim Ψcn (t , τ , ω, xn ) = Ψ0 (t , τ , x0 ).
(4.2)
n→∞
Then for τ ∈ R and ω ∈ Ω , dH (Ac (τ , ω), A0 (τ )) =
sup
inf
u∈Ac (τ ,ω) v∈A0 (τ )
∥u − v∥X → 0 as c → 0.
(4.3)
Next, we will use Theorem 4.1 to consider an upper semicontinuity of random attractors Ac (τ , ω) when c → 0. To indicate the dependence of solutions on c, we will write the solution of (2.7) as ϕ (c ) = (u(c ) , v (c ) ), that is, (u(c ) , v (c ) ) satisfies
(c ) du = v (c ) − δ u(c ) + cu(c ) z (θt ω), dt (c ) dv = α 1v (c ) + (1 − αδ)1u(c ) + (δ − 1)v (c ) + (δ − λ − δ 2 )u(c ) + α cz (θt ω)1u(c ) dt + 2c δ z (θt ω) − c 2 z 2 (θt ω) u(c ) − cz (θt ω)v (c ) − µ sin u(c ) + g (x, t ), (c ) u (τ , x) = u(τc ) (x), v (c ) (τ , x) = vτ(c ) (x).
(4.4)
When c = 0, the random problem (2.7) reduces to a deterministic one:
(0) du (0) (0) dt = v − δ u , (0) dv = α 1v (0) + (1 − αδ)1u(0) + (δ − 1)v (0) + (δ − λ − δ 2 )u(0) − µ sin u(0) + g (x, t ), (dt (0) u 0) (τ , x) = u(τ0) (x), v (0) (τ , x) = vτ(0) (x) = u1τ (x) + δ u(τ0) (x).
(4.5)
Let Φ0 be the continuous deterministic cocycle associated with problem (4.5) on E (Rn ) over R and {θt }t ∈R , and D0 (E (Rn )) be the collection of tempered families of deterministic nonempty bounded subsets of E (Rn ). Assume that g (x, ·) ∈ L2loc (R, L2 (Rn )), (2.8) and γ > 0 hold. Then by Theorem 3.6, we have that the continuous cocycle Φ0 has a unique D0 (E (Rn ))-pullback attractor A0 (τ ). Theorem 4.2. Assume that g (x, ·) ∈ L2loc (R, L2 (Rn )) and (2.8) hold. If γ > 0, then for every τ ∈ R and ω ∈ Ω , we have dH (Ac (τ , ω), A0 (τ )) =
sup
inf
1
(∥∇(u(c ) − u(0) )∥2 + ∥u(c ) − u(0) ∥2 + ∥v (c ) − v (0) ∥2 ) 2
(0) (0) (u(c ) ,v (c ) )∈Ac (τ ,ω) (u ,v )∈A0 (τ )
→ 0 as c → 0.
(4.6)
14
Z. Wang, Y. Liu / Computers and Mathematics with Applications (
)
–
Proof. (i) By Lemma 3.3 and Theorem 3.6, Φc has a closed measurable random absorbing set Bc = {Bc (τ , ω) : τ ∈ R, ω ∈ Ω } ∈ D (E (Rn )), where Bc (τ , ω) = {ϕ (c ) ∈ E (Rn ) : ∥ϕ (c ) ∥2E (Rn ) 6 ϱ(c , τ , ω)}, and a unique random attractor Ac = {Ac (τ , ω) : τ ∈ R, ω ∈ Ω } in D (E (Rn )), for each τ ∈ R and ω ∈ Ω , Ac (τ , ω) ⊆ Bc (τ , ω). (ii) Given |c | 6 1. By (3.9), we have
ϱ(c , τ , ω) 6 ϱ(1, τ , ω) < ∞, and lim sup ϱ(c , τ , ω) 6 ϱ(1, τ , ω). c →0
So, for every τ ∈ R and ω ∈ Ω , lim sup ∥Bc (τ , ω)∥ρ = lim sup c →0
c →0
sup
x∈Bc (τ ,ω)
1
∥x∥E (Rn ) 6 ϱ 2 (1, τ , ω).
(4.7)
Let B1 (τ , ω) = {ϕ (c ) ∈ E (Rn ) : ∥ϕ (c ) ∥2E (Rn ) 6 ϱ(1, τ , ω)}, then
Ac (τ , ω) ⊆
|c |61
Bc (τ , ω) ⊆ B1 (τ , ω).
(4.8)
|c |61
(iii) Given |c | 6 1. Let us prove the precompactness of |c |61 Ac (τ , ω) for every τ ∈ R and ω ∈ Ω . For one thing, by (4.8), Lemma 3.4 and the invariance of Ac (τ , ω), for every ε > 0, c > 0, τ ∈ R, ω ∈ Ω , there exist T = T (τ , ω, B1 , c , ε) > 0 (c ) and R = R(τ , ω, c , ε) > 1, such that for all t > T , r > R and ϕτ −t ∈ Ac (τ − t , θ−t ω), the solution ϕ (c ) of (4.4) satisfies
sup ϕ (c ) ∈
Ac (τ ,ω)
∥ϕ (c ) (τ , τ − t , θ−τ ω, ϕτ(c−) t )∥2E (Rn \Hr ) < ε.
|c |61
For another, by (4.8) we find that the set |c |61 Ac (τ , ω) is precompact in E (Hr ) and hence |c |61 Ac (τ , ω) is precompact in E (Rn ). (iv) Let ϕ (0) = (u(0) , v (0) ) be a mild solution of (4.5) with initial data ϕτ(0) = (u(τ0) , vτ(0) ), and U = u(c ) − u(0) , V = v (c ) −v (0) . Let
ϕm(c ) (t , τ , ω, ϕτ(c ) ) = (u(mc ) (t , τ , ω, ϕτ(c ) ), vm(c ) (t , τ , ω, ϕτ(c ) )) and
ϕm(0) (t , τ , ϕτ(0) ) = (u(m0) (t , τ , ϕτ(0) ), vm(0) (t , τ , ϕτ(0) )) be the solutions of the following random differential equations with initial data
(c ) dum = vm(c ) − δ u(mc ) + cu(mc ) z (θt ω), dt (c ) dvm = α 1vm(c ) + (1 − αδ)1u(mc ) + (δ − 1)vm(c ) + (δ − λ − δ 2 )u(mc ) + α cz (θt ω)1u(mc ) dt + 2c δ z (θt ω) − c 2 z 2 (θt ω) u(mc ) − cz (θt ω)vm(c ) − µ sin u(mc ) + g (x, t ), (c ) um (τ , x) = ucτ (x), vm(c ) (τ , x) = vτc (x),
(4.9)
and
(0) dum = vm(0) − δ u(m0) , dt (0) dvm = α 1vm(0) + (1 − αδ)1u(m0) + (δ − 1)vm(0) + (δ − λ − δ 2 )u(m0) − µ sin u(m0) + g (x, t ), dt (0) um (τ , x) = u(τ0) (x), vm(0) (τ , x) = vτ(0) (x), (c )
(4.10)
(0)
respectively. Then ϕm (·, τ , ω, ϕτ(c ) ), ϕm (·, τ , ϕτ(0) ) ∈ C ([τ , +∞), E (Rn )) and satisfy the differential equations (4.9) and (4.10) respectively. Moreover, ϕ (c ) (t , τ , ω, ϕτ(c ) ) and ϕ (0) (t , τ , ϕτ(0) ) are limit functions of subsequences of
{ϕm(c ) (t , τ , ω, ϕτ(c ) )} and {ϕm(0) (t , τ , ϕτ(0) )} ∈ E (Rn ). So ϕ (c ) (t , τ , ω, ϕτ(c ) ) − ϕ (0) (t , τ , ϕτ(0) ) is a limit function of a subsequence
Z. Wang, Y. Liu / Computers and Mathematics with Applications (
(c )
)
–
15
(0)
of {ϕm (t , τ , ω, ϕτ(c ) ) − ϕm (t , τ , ϕτ(0) )} in E (Rn ), and (Um (t , τ , ω, ϕτ(c ) , ϕτ(0) ), Vm (t , τ , ω, ϕτ(c ) , ϕτ(0) )) satisfies
dU m (c ) dt = Vm − δ Um + cum z (θt ω), dV m = α 1Vm + (1 − αδ)1Um + (δ − 1)Vm + (δ − λ − δ 2 )Um + α cz (θt ω)1u(mc ) dt (c ) + 2c δ z (θt ω) − c 2 z 2 (θt ω) u(mc ) − cz (θt ω)vm − µ sin u(mc ) + µ sin u(m0) , (c ) (0) (c ) (0) Um (τ ) = uτ − uτ , Vm (τ ) = vτ − vτ .
(4.11)
Taking the inner product of the second equation of (4.11) with V m (t ) in L2 (Rn ), we find that 1 d
∥Vm ∥2 = −α∥∇ Vm ∥2 + 1 − αδ 1Um , Vm − (1 − δ)∥Vm ∥2 2 dt − (δ 2 + λ − δ) Um , Vm + α cz (θt ω) 1u(mc ) , Vm + 2c δ z (θt ω) − c 2 z 2 (θt ω) u(mc ) , Vm − cz (θt ω) vm(c ) , Vm − µ sin u(mc ) − sin u(m0) , Vm .
(4.12)
We now estimate the terms in (4.12):
|cz (θt ω)|2 (c ) 2 2 − 3δ −cz (θt ω) vm(c ) , Vm 6 ∥vm ∥ + ∥ Vm ∥ 2 , 2 − 3δ 4 |2c δ z (θt ω) − c 2 z 2 (θt ω)|2 (c ) 2 2 − 3δ ∥ um ∥ + ∥ Vm ∥ 2 , 2c δ z (θt ω) − c 2 z 2 (θt ω) u(mc ) , Vm 6 2 − 3δ 4 α cz (θt ω) 1u(mc ) , Vm = −α cz (θt ω) ∇ u(mc ) , ∇ Vm α |cz (θt ω)|2 · ∥∇ u(mc ) ∥2 + α∥∇ Vm ∥2 , 6
(4.13) (4.14)
(4.15)
4
1 − αδ 1Um , Vm = − 1 − αδ ∇ Um , ∇
1 d
dU
m
dt
+ δ Um − cu(mc ) z (θt ω)
∥∇ Um ∥2 + δ∥∇ Um ∥2 − ∥∇ Um ∥ · ∥cz (θt ω)∇ u(mc ) ∥ 2 dt 1 d δ |cz (θt ω)|2 6 − 1 − αδ ∥∇ Um ∥2 + ∥∇ Um ∥2 − ∥∇ u(mc ) ∥2 , 2 dt 2 2δ dU m (δ 2 + λ − δ) Um , Vm = δ 2 + λ − δ Um , + δ Um − cu(mc ) z (θt ω) 1 d dt > (δ 2 + λ − δ) ∥Um ∥2 + δ∥Um ∥2 − ∥Um ∥ · ∥cu(mc ) z (θt ω)∥ 2 dt 1 d δ |cz (θt ω)|2 (c ) 2 2 ∥Um ∥2 + ∥Um ∥2 − ∥um ∥ , > (δ + λ − δ) 2 dt 2 2δ − |µ| sin u(mc ) − sin u(m0) , Vm 6 |µ| · ∥ sin u(mc ) − sin u(m0) ∥ · ∥Vm ∥ |µ| 6 (∥Vm ∥2 + ∥Um ∥2 ). 6 − 1 − αδ
2
(4.16)
(4.17)
(4.18)
Then it follows from (4.17)–(4.18) that 1 d
∥Vm ∥2 + 1 − αδ ∥∇ Um ∥2 + δ 2 + λ − δ ∥Um ∥2 2 dt |µ| δ 6− ∥Vm ∥2 + 1 − αδ ∥∇ Um ∥2 + δ 2 + λ − δ ∥Um ∥2 + (∥Vm ∥2 + ∥Um ∥2 ) 2 |cz (θt ω)|2
2
|2c δ z (θt ω) − c 2 z 2 (θt ω)|2 (c ) 2 + ∥vm(c ) ∥2 + ∥ um ∥ 2 − 3δ 2 − 3δ α δ 2 + λ − δ |cz (θt ω)|2 (c ) 2 1 − αδ + + |cz (θt ω)|2 · ∥∇ u(mc ) ∥2 + ∥ um ∥ 4 2δ 2δ 6− ∥Vm ∥2 + 1 − αδ ∥∇ Um ∥2 + δ 2 + λ − δ ∥Um ∥2 2 + H2 (c , t , ω) ∥vm(c ) ∥2 + ∥u(mc ) ∥2 + ∥∇ u(mc ) ∥2 ,
(4.19)
16
Z. Wang, Y. Liu / Computers and Mathematics with Applications (
where = δ − |µ|(1 + δ 2 + λ − δ) > 0, H2 (c , t , ω) = |cz (θt ω)| inequality to (4.19) from τ to t + τ , we have
2 2 1+|2δ−cz (θt ω)| 2−3δ
)
–
2 + α4 + (1−αδ+δ2δ +λ−δ) . Applying Gronwall’s
∥Vm (t + τ , τ , ω, ϕτ(c ) , ϕτ(0) )∥2 + 1 − αδ ∥∇ Um (t + τ , τ , ω, ϕτ(c ) , ϕτ(0) )∥2 + δ 2 + λ − δ ∥Um (t + τ , τ , ω, ϕτ(c ) , ϕτ(0) )∥2 6 e− t ∥vτ(c ) − vτ(0) ∥2 + 1 − αδ ∥∇(u(τc ) − u(τ0) )∥2 + δ 2 + λ − δ ∥u(τc ) − u(τ0) ∥2 t +τ (c ) e− (t +τ −s) H2 (c , s, ω) ∥vm +2 (s, τ , ω, ϕτ(c ) )∥2 τ
+ ∥um (s, τ , ω, ϕτ(c ) )∥2 + ∥∇ u(mc ) (s, τ , ω, ϕτ(c ) )∥2 ds. (c )
(4.20)
We now replace ω in the above by θ−τ ω to yield
∥Vm (t + τ , τ , θ−τ ω, ϕτ(c ) , ϕτ(0) )∥2 + (1 − αδ)∥∇ Um (t + τ , τ , θ−τ ω, ϕτ(c ) , ϕτ(0) )∥2 + δ 2 + λ − δ ∥Um (t + τ , τ , θ−τ ω, ϕτ(c ) , ϕτ(0) )∥2 6 e− t ∥vτ(c ) − vτ(0) ∥2 + 1 − αδ ∥∇(u(τc ) − u(τ0) )∥2 + δ 2 + λ − δ ∥u(τc ) − u(τ0) ∥2 t (c ) (s, τ , θ−τ ω, ϕτ(c ) )∥2 +2 e− (t −s) H2 (c , s, θ−τ ω) ∥vm 0
+ ∥u(mc ) (s, τ , θ−τ ω, ϕτ(c ) )∥2 + ∥∇ u(mc ) (s, τ , θ−τ ω, ϕτ(c ) )∥2 ds.
(4.21)
From (4.21), we see that for τ ∈ R, t ∈ R+ , ω ∈ Ω , c → 0 and ϕτ(c ) , ϕτ(0) ∈ E (Rn ) with ϕτ(c ) → ϕτ(0) , (c ) lim ϕm (t + τ , τ , θ−τ ω, ϕτ(c ) ) = ϕm(0) (t + τ , τ , ϕτ(0) ) in E (Rn ).
c →0
(4.22)
Let {cn } ⊂ [−1, 1] be a sequence of numbers with cn → 0 when n → +∞. Then ϕ (c ) (t + τ , τ , θ−τ ω, ϕτ(c ) ) and
ϕ (0) (t + τ , τ , ϕτ(0) ) being limit functions of subsequences of {ϕm(c ) (t + τ , τ , θ−τ ω, ϕτ(c ) )} and {ϕm(0) (t + τ , τ , ϕτ(0) )} in E (Rn ) imply that for τ ∈ R, t ∈ R+ , ω ∈ Ω , cn → 0 and ϕτ(cn ) , ϕτ(0) ∈ E (Rn ) with ϕτ(cn ) → ϕτ(0) , the following holds: lim ϕ (cn ) (t + τ , τ , θ−τ ω, ϕτ(cn ) ) = ϕ (0) (t + τ , τ , ϕτ(0) ) in E (Rn ).
n→∞
The proof is completed.
(4.23)
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