DYNAMICS OF THE 3D FRACTIONAL GINZBURG ...

DYNAMICS OF THE 3D FRACTIONAL GINZBURG-LANDAU EQUATION WITH MULTIPLICATIVE NOISE ON AN UNBOUNDED DOMAIN ∗ HONG LU

†,

PETER W. BATES

‡,

¨ SHUJUAN LU

§ , AND

MINGJI ZHANG

¶

Abstract. We study a stochastic fractional complex Ginzburg-Landau equation with multiplicative noise in three spatial dimensions with particular interest in the asymptotic behavior of its solutions. We first transform our equation into a random equation whose solutions generate a random dynamical system. A priori estimates are derived when the nonlinearity satisfies certain growth conditions. Applying the estimates for far-field values of solutions and a cut-off technique, asymptotic compactness is proved. Furthermore, the existence of a random attractor in H 1 (R3 ) of the random dynamical system is established. Key words. stochastic fractional Ginzburg-Landau equation, asymptotic compactness, random attractor, pullback attractor AMS subject classifications. 37L55, 60H15, 35Q99.

1. Introduction A fractional differential equation is an equation that contains fractional derivatives or fractional integrals. The fractional derivative and the fractional integral have a wide range of applications in physics, biology, chemistry and other fields of science, such as kinetic theories of systems with chaotic dynamics ([34, 41]), pseudochaotic dynamics ([42]), dynamics in a complex or porous medium ([13, 26, 35]), random walks with a memory and flights ([24, 33, 40]), obstacle problems ([6, 31]). Recently, some of the classical equations of mathematical physics have been postulated with fractional derivatives to better describe complex phenomena. Of particular interest are the fractional Schr¨odinger equation ([12, 16, 17]), the fractional Landau-Lifshitz equation ([19]), the fractional Landau-Lifshitz-Maxwell equation ([28]) and the fractional Ginzburg-Landau equation ([37]). Small perturbations (such as molecular collisions in gases and liquids and electric fluctuations in resistors [15]) may be neglected during the derivation of these ideal models. However, the perturbations should be included to obtain a more realistic model and to better understand the dynamical behavior of the model. One may represent the micro effects by random perturbations in the dynamics of the macro observable through additive or multiplicative noise in the governing equation. To study a stochastic partial differential equation, a key step is to examine the asymptotic behavior of the random dynamical systems generated by its solutions. Some nice works along these lines are, for example, by Crauel and Flandoli ([7, 8]) who developed the theory of random attractors which closely parallels the deterministic case ([36]), and by Debussche ([11]) who proved that the Hausdorff dimension of the random attractor could be estimated by using global Lyapunov exponents. ∗ † College of Science, China University of Mining and Technology, Jiangsu, China, 221116, ([email protected]). ‡ Department of Mathematics, Michigan State University, 619 Red Cedar Road, East Lansing, MI 48824, USA, ([email protected]). § School of Mathematics and Systems Science & LMIB, Beihang University, Beijing, China, 100191, ([email protected]). ¶ Department of Mathematics, Michigan State University, 619 Red Cedar Road, East Lansing, MI 48823, USA, ([email protected], [email protected]).

1

2

Dynamics of 3D Fractional GL-Equation with Multiplicative Noise

The well-posedness of solutions of fractional partial differential equations has been studied to some extent (See [17, 19, 21, 28]). However, there are not many results for stochastic fractional partial differential equations. In this paper, we examine the asymptotic behavior of solutions of the fractional Ginzburg-Landau equation with multiplicative noise on an unbounded domain. The fractional Ginzburg-Landau equation arises, for example, from the variational Euler-Lagrange equation for fractal media, which can be used to describe dynamical processes in a medium with fractal dispersion in [37]. In [29], the authors analyzed a one-dimensional fractional complex Ginzburg-Landau equation ut + (1 + iν)(−4)α u + (1 + iµ)|u|2σ u = ρu. The well-posedness of solutions was obtained by applying the semigroup method under the condition 1 1 . ≤σ≤ p 2 1 + µ2 − 1 The existence of a global attractor in L2 was also proved when σ = 1. In [23], the dynamics of a two-dimensional fractional complex Ginzburg-Landau equations is studied. A fractional Ginzburg-Landau equation on the line with special nonlinearity and multiplicative noise was analyzed in [22]. In this paper, we consider a general three-dimensional stochastic fractional Ginzburg-Landau equation with multiplicative noise of Stratonovich form defined in the entire space R3 given by du + ((1 + iν)(−4)α u + ρu)dt = f (x,u)dt + βu ◦ dW (t),

x ∈ R3 ,

t > 0 (1.1)

with the initial condition u(x,0) = u0 (x),

x ∈ R3 ,

(1.2)

where u(x,t) is a complex-valued function on R3 × [0,+∞). In (1.1), i is the imaginary unit, ν is a real constants, ρ > 0, α ∈ (1/2,1), and f (x,u) is a nonlinear function, for instance, f (x,u) = −(1 + iµ)|u|2σ u with µ ∈ R and σ > 0. For convenience, we sometimes write it as f = f (x,u, u ¯) or f = f (u), and in the various lemmas that follow we assume f satisfies some of the following conditions: Ref (x,u)¯ u ≤ −β1 |u|2σ+2 + γ1 (x), ¯ σ (¯ ¯ 2 ≤ −βσ |u|2σ |V|2 + |u|2σ−2 (λσ (uV) ¯ 2 +λ Refu |V|2 + Refu¯ (V) uV)2 ), max{|fu |,|fu¯ |} ≤ β2 , ∂f (x,u) ∂x = |fx | ≤ γ2 (x),

(1.3) (1.4) (1.5) (1.6)

for u ∈ C and V ∈ Cn , where σ, βi (i = 1,2) are positive constants, βσ is a positive constantPdepending on σ, λσ is a complex constant depending on σ, and (V)2 = n V · V = i=1 Vi2 , (which is not an inner product on Cn ), and γ1 (x) ∈ L1 (R3 ), γ2 (x) ∈ 2 3 L (R ). The white noise described by a two-sided Wiener process W (t) on a complete probability space results from the fact that small irregularity has to be taken into account in some circumstances.

H. Lu, P. W. Bates, S. L¨ u and M. Zhang

3

Most of the research with respect to random attractors is restricted to L2 . In this work, we obtain the existence of a pullback attractor in H 1 (actually, one can choose the space to be H α ,α ∈ (0,1], but we prefer the stronger regularity of the random attractor in H 1 ). The concept of pullback random attractor, which is an extension of global attractor in deterministic systems (see [2, 20, 30, 32, 36]) was introduced in [8, 14]. In the case of bounded domains, the existence of random attractors for stochastic partial differential equations has been investigated by many authors (see [1, 7, 8, 10, 14] and the references therein). However, the problem is more challenging in the case of unbounded domains. Recently, the existence of random attractors for systems on unbounded domains was studied in [3, 5, 38, 39], which provides guidance for this work. It is well known that asymptotic compactness and the existence of a bounded absorbing set are sufficient to guarantee the existence of a random attractor for a continuous random dynamical system. However, Sobolev embeddings are not compact on an unbounded domain. In this paper, we employ a tail-estimates approach to prove the existence of a compact random attractor. The paper is organized as follows. In section 2, some preliminaries, notations and random attractor theory for random dynamical systems are introduced. In section 3, we define a continuous random dynamical system for the stochastic fractional complex Ginzburg-Landau equation. In section 4, we derive uniform estimates for solutions, which include uniform estimates on far field values of solutions. In section 5, we establish the asymptotic compactness of the solution operator, and then prove the existence of a pullback random attractor. 2. Preliminaries and Notations We first recall some basic concepts related to random attractors for stochastic dynamical systems (see [4, 8, 10] for more details). Let (X,|| · ||X ) be a separable Hilbert space with Borel σ-algebra B(X), and let (Ω,F,P) be a probability space. Definition 2.1. (Ω,F,P,(θt )t∈R ) is called a measurable dynamical systems, if θ : R × Ω → Ω is (B(R) × F,F)−measurable, θ0 = I, θt+s = θt ◦ θs for all t,s ∈ R, and θt A = A for all t ∈ R and A ∈ F. Definition 2.2. A stochastic process φ(t,ω) is called a continuous random dynamical system (RDS) over (Ω,F,P,(θt )t∈R ) if φ is (B(R+ ) × F × B(X),B(X))−measurable, and for all ω ∈ Ω • the mapping φ : R+ × Ω × X → X is continuous; • φ(0,ω) = I on X; • φ(t + s,ω,χ) = φ(t,θs ω,φ(s,ω,χ)) for all t,s ≥ 0 and χ ∈ X (cocycle property). Definition 2.3. A random bounded set {B(ω)}ω∈Ω ⊆ X is called tempered with respect to (θt )t∈R if for P-a.e. ω ∈ Ω and all > 0 lim e−t d(B(θ−t ω)) = 0.

t→∞

where d(B) = supχ∈B kχkX . Consider a continuous random dynamical system φ(t,w) over (Ω,F,P,(θt )t∈R ) and let D be the collection of all tempered random set of X. ˜ = {D(ω) ˜ Definition 2.4. D is called inclusion-closed if D = {D(ω)}ω∈Ω ∈ D and D ⊆ ˜ ˜ X : ω ∈ Ω} with D(ω) ⊆ D(ω) for all ω ∈ Ω imply that D ∈ D.

4


Definition 2.5. Let D be a collection of random subsets of X and {K(ω)}ω∈Ω ∈ D. Then {K(ω)}ω∈Ω is called an absorbing set of φ in D if for all B ∈ D and P-a.e. ω ∈ Ω there exist tB (ω) > 0 such that φ(t,θ−t ω,B(θ−t ω)) ⊆ K(ω),

t ≥ tB (ω).

Definition 2.6. Let D be a collection of random subsets of X. Then φ is said to be Dpullback asymptotically compact in X if for P-a.e. ω ∈ Ω, {φ(tn ,θ−tn ω,χn )}∞ n=1 has a convergent subsequence in X whenever tn → ∞, and χn ∈ B(θ−tn ω) with {B(ω)}ω∈Ω ∈ D. Definition 2.7. Let D be a collection of random subsets of X and {A(ω)}ω∈Ω ∈ D. Then {A(ω)}ω∈Ω is called a D-random attractor (or D-pullback attractor) for φ if the following conditions are satisfied, for P-a.e. ω ∈ Ω, • A(ω) is compact, and ω → d(χ,A(ω)) is measurable for every χ ∈ X; • {A(ω)}ω∈Ω is strictly invariant, i.e., φ(t,ω,A(ω)) = A(θt ω), ∀t ≥ 0 and for a.e.ω ∈ Ω; • {A(ω)}ω∈Ω attracts all sets in D, i.e., for all B ∈ D and a.e. ω ∈ Ω we have lim d(φ(t,θ−t ω,B(θ−t ω)),A(ω)) = 0,

t→∞

where d is the Hausdorff semi-metric given by d(Y,Z) = supy∈Y inf z∈Z ky − zkX , for any Y,Z ⊆ X. According to [9], we can infer the following result. Proposition 2.8. Let D be an inclusion-closed collection of random subsets of X and φ a continuous RDS on X over (Ω,F,P,(θt )t∈R ). Suppose that {K(ω)}ω∈Ω ∈ D is a closed absorbing set of φ and φ is D-pullback asymptotically compact in X. Then φ has a unique D-random attractor which is given by {A(ω)}ω∈Ω with A(ω) =

\ [

φ(t,θ−t ω,K(θ−t ω)).

κ≥0 t≥κ

For convenience, we recall some notation related to the fractional derivative and fractional Sobolev spaces. Firstly, we present the definition and some properties β of (−4)α through Fourier transforms ([18]). The negative powers (−4) 2 (that is, β (−4)− 2 ), Reβ > 0, can be represented by Riesz potentials Z 1 (I β ϕ)(x) = |x − y|−3+β ϕ(y)dy, γ(β) R3 where γ(β) = π 3/2 2β Γ( β2 )/Γ( 32 − β2 ). We consider the Fourier transform Z Φ(ξ) = φ(x)e−i(x·ξ) dz, R3

so (−4)

β 2

can be defined as β

F{(−4) 2 ϕ} = |k|β Φ, β

(−4) 2 ϕ = F −1 {|k|β Φ} =

1 (2π)3

Z R3

|k|β Φeik·x dk,

5


where 4 = ∂ 2 /∂x21 + ∂ 2 /∂x22 + ∂ 2 /∂x23 . Let H 2α (R3 ) denote the complete Sobolev space of order α under the norm: Z kuk2H 2α (R3 ) = (1 + |k|4α )|ˆ u(k)|2 dk. R3

By virtue of the definition of (−4)α , we have the following formula for integration by parts. Lemma 2.9. If f,g ∈ H 2α (Rn ), then the following equation holds. Z Z α (−4) f · gdx = (−4)α1 f · (−4)α2 gdx, (2.1) Rn

Rn

where α1 ,α2 are nonnegative constant and satisfy α1 + α2 = α. Proof. By the definition of (−4)α and Parseval formula, we have Z Z Z (−4)α f · gdx = F −1 {|k|2α fˆ} · gdx = F −1 {|k|2α fˆ} · F −1 gˆdx Rn Rn Rn Z Z 1 1 2α ˆ |k| f · gˆdk = |k|2α1 fˆ· |k|2α2 gˆdk = (2π)n Rn (2π)n Rn Z Z = F −1 {|k|2α1 fˆ} · F −1 {|k|2α2 gˆ}dx = (−4)α1 f · (−4)α2 gdx. Rn

Rn

In addition, the following Gagliardo-Nirenberg inequality([27]) is also frequently used. Lemma 2.10. Let u belong to Lq (Rn ) and its derivatives of order m, Dm u, belong to Lr (Rn ), 1 ≤ q,r ≤ ∞. For the derivatives Dj u, 0 ≤ j < m, the following inequalities hold kDj ukLp ≤ ckDm ukθLr kuk1−θ Lq ,

(2.2)

where 1 j 1 m 1 = + θ( − ) + (1 − θ) , p n r n q for all θ in the interval j ≤ θ ≤ 1, m (the constant c depending only on n,m,j,q,r,θ), with the following exceptional case 1. If j = 0,rm < n,q = ∞, then we make the additional assumption that either u tends to zero at infinite or u ∈ Lq˜ for some finite q˜ > 0. 2. If 1 < r < ∞, and m − j − n/r is a nonnegative integer, then (2.2) holds only for θ satisfying j/m ≤ θ < 1. In the forthcoming discussions, we denote by k · k and (·,·) the norm and the inner product in L2 (R3 ) and use k · kp to denote the norm in Lp (R3 ). Otherwise, the letters c,cj (j = 1,2,···) are generic positive constants which may change their values from line to line or even in the same line.

6


3. Stochastic fractional complex Ginzburg-Landau equation sequel, we consider the probability space (Ω,F,P) where

In the

Ω = {ω ∈ C(R,R) : ω(0) = 0}, F is the Borel σ-algebra induced by the compact-open topology of Ω, and P the corresponding Wiener measure on (Ω,F). Define a shift on ω by θt ω(·) = ω(· + t) − ω(t),

ω ∈ Ω, t ∈ R.

Then (Ω,F,(θt )t∈R ) is a metric dynamical system. In this section, we discuss the existence of a continuous random dynamical system for the stochastic fractional complex Ginzburg-Landau equation perturbed by a multiplicative white noise in the Stratonovich sense. Thanks to the special linear multiplicative noise, the stochastic fractional Ginzburg-Landau equation can be reduced to an equation with random coefficients by a suitable change of variable. To this end, we consider the stationary process Z 0 z(t) = z(t,ω) = z(θt ω) = − eτ (θt ω)(τ )dτ, t ∈ R, −∞

satisfies the stochastic differential equation: dz + zdt = dW (t). Moreover, for any t,s, z(t,θs ω) = z(t + s,ω),

P-a.s..

Here the exceptional set may be a priori depending on t and s. In fact, we suppose that z has a continuous modification. Once this modification is chosen, the exceptional set is independent of t. It is known that the random variable z(ω) is tempered (see ˜ ⊆ Ω of full P measure such that for every [1, 7, 14]), there exists a θt -invariant set Ω ˜ ω ∈ Ω, z(θt ω) is continuous in t; and lim

t→±∞

|z(θt ω)| = 0, |t|

˜ for all ω ∈ Ω,

(3.1)

and 1 lim t→±∞ t

Z

t

z(θt ω)dt = 0,

˜ for all ω ∈ Ω.

(3.2)

0

We rewrite the unknown v(t) as v(t) = e−βz(θt ω) u(t) to obtain the following random differential equation vt = −(1 + iν)(−4)α v + e−βz(θt ω) f (eβz(θt ω) v) + (βz(θt ω) − ρ)v

(3.3)

with the initial data v(x,0) = v0 (x) = e−βz(ω) u0 (x),

x ∈ R3 .

(3.4)

Next, we construct a random dynamical system modeling the stochastic fractional Ginzburg-Landau equation.

7


By the Galerkin method, one can show that if f satisfies (1.3)-(1.6), then in the case of a bounded domain with Dirichlet boundary conditions, for P-a.e. ω ∈ Ω and for all v0 ∈ H 1 , equation (3.3) has a unique solution v(·,ω,v0 ) ∈ C([0,∞),H 1 ) ∩ L2 ((0,T );H 1+α ) with v(0,ω,v0 ) = v0 for every T > 0. This is similar to [21]. Then, following the approach in [25], we take the domain to be a sequence of balls with radius approaching ∞ to deduce the existence of a weak solution of equation (3.3) on R3 . Furthermore, we obtain that v(t,ω,v0 ) is unique and continuous with respect to v0 in H 1 (R3 ) for all t ≥ 0. Let u(t,ω,u0 ) = eβz(θt ω) v(t,ω,e−βz(ω) u0 )). Then the process u is the solution of problem (1.1)-(1.2). We now define a mapping φ : R+ × Ω × H 1 (R3 ) → H 1 (R3 ) by φ(t,ω,u0 ) = u(t,ω,u0 ) = eβz(θt ω) v(t,ω,e−βz(ω) u0 ), for u0 ∈ H 1 (R3 ), t ≥ 0 and for all ω ∈ Ω. It is easy to check that φ satisfies the three conditions in Definition 2.2. Therefore, φ is a continuous random dynamical system associated with problem (3.3) on H 1 (R3 ). Let ϕ(t,ω,v0 ) = v(t,ω,v0 ) for v0 ∈ H 1 (R3 ), t ≥ 0 and for all ω ∈ Ω. Then ϕ is a continuous random dynamical system associated with problem (1.1) on H 1 (R3 ). It is worth noticing that, the two random dynamical systems are equivalent. It is easy to check that φ has a random attractor provided ϕ possesses a random attractor. Then, we only need to consider the random dynamical system ϕ. 4. Uniform estimates of solutions In this section, we deduce uniform estimates on the solutions of the stochastic fractional complex Ginzburg-Landau equation on R3 when t → ∞. These estimates are necessary for proving the existence of bounded absorbing sets and the asymptotic compactness of the random dynamical system associated with the equation. In particular, we will show that the solutions for large space variables are uniformly small when time is sufficiently large. From now on, we always suppose that D is the collection of all tempered random subsets of H 1 (R3 ). First, we derive the following uniform on v in D. Lemma 4.1. Suppose that (1.3) holds. Let B = {B(ω)} ∈ D and v0 (ω) ∈ B(ω), and let %0 > 0 be fixed and 0 < δ < 2ρ. Then for P-a.e. ω ∈ Ω, there exists T0B (ω) > 0 such that for any t ≥ T0B (ω), one has Z

0

δ

e2β

R0 s

z(θτ ω)dτ +(2ρ−δ)s

kv(s + t,θ−t ω,v0 (θ−t ω))k2 ds

(4.1)

−t 2

+ kv(t,θ−t ω,v0 (θ−t ω))k

≤ %20 .

Proof. Taking the inner product in L2 of (3.3) with v and taking the real part, we obtain Z α 1 d kvk2 + k(−4) 2 vk2 = e−βz(θt ω) Re f (eβz(θt ω) v)¯ v dx + (βz(θt ω) − ρ)kvk2 . (4.2) 2 dt R3 By condition (1.3), we have Z −2βz(θt ω) v dx ≤ −β1 e−2βz(θt ω) keβz(θt ω) vk2σ+2 kγ1 (x)kL1 . e−βz(θt ω) Re f (eβz(θt ω) v)¯ 2σ+2 + e R3

8


Then (4.2) can be rewritten as α d kvk2 + 2k(−4) 2 vk2 + 2β1 e−2βz(θt ω) keβz(θt ω) vk2σ+2 2σ+2 dt 2 −2βz(θt ω) ≤ 2(βz(θt ω) − ρ)kvk + 2e kγ1 (x)k1 .

(4.3)

Therefore, d kvk2 + δkvk2 ≤ (2βz(θt ω) − 2ρ + δ)kvk2 + 2e−2βz(θt ω) kγ1 (x)k1 . dt

(4.4)

Here,R ρ > 0, so there exists δ > 0 such that 2ρ > δ > 0. Multiplying (4.4) by t e−2β 0 z(θs ω)ds+(2ρ−δ)t , and integrating over (0,t), we infer that Z t Rt 2 kv(t,ω,v0 (ω))k + δ e2β s z(θτ ω)dτ +(2ρ−δ)(s−t) kv(s,ω,v0 (ω))k2 ds ≤ e2β

Rt 0

0 z(θs ω)ds+(δ−2ρ)t

Z

t

e2β

+2 ≤e

Rt s

kv0 (ω)k2

z(θτ ω)dτ +(2ρ−δ)(s−t)−2βz(θs ω)

0 Rt 2β 0 z(θs ω)ds+(δ−2ρ)t

kv0 (ω)k2 + 2c1

(4.5) kγ1 (x)k1 ds

t

Z

e2β

Rt s

z(θτ ω)dτ +(2ρ−δ)(s−t)−2βz(θs ω)

ds.

0

Substituting ω by θ−t ω, then we deduce from (4.5), Z t R t δ e2β s z(θτ −t ω)dτ +(2ρ−δ)(s−t) kv(s,θ−t ω,v0 (θ−t ω))k2 ds 0

+ kv(t,θ−t ω,v0 (θ−t ω))k2 ≤ e2β

Rt 0

z(θs−t ω)ds+(δ−2ρ)t t

Z

e2β

+ 2c1

Rt s

kv0 (θ−t ω)k2

z(θτ −t ω)dτ +(2ρ−δ)(s−t)−2βz(θs−t ω)

ds.

0

Applying the transformation of variables, one has Z 0 R0 δ e2β s z(θτ ω)dτ +(2ρ−δ)s kv(s + t,θ−t ω,v0 (θ−t ω))k2 ds −t

+ kv(t,θ−t ω,v0 (θ−t ω))k2 ≤ e2β

R0 −t

z(θs ω)ds+(δ−2ρ)t

(4.6)

kv0 (θ−t ω)k2 + 2c1

Z

0

e2β

R0 s

z(θτ ω)dτ +(2ρ−δ)s−2βz(θs ω)

ds.

−t

{B(ω)} ∈ D is tempered, so for any v0 (θ−t ω) ∈ B(θ−t ω), lim e2β

t→+∞

R0 −t


kv0 (θ−t ω)k2 = lim e2β

R0 −t

z(θs ω)ds+(δ−2ρ)t−2βz(θt ω)

t→+∞

(4.7)

=0. Therefore, there exists T0B (ω) > 0 such that for any t ≥ T0B (ω), Z 0 R0 R0 2β −t z(θs ω)ds+(δ−2ρ)t 2 e kv0 (θ−t ω)k + 2c1 e2β s z(θτ ω)dτ +(2ρ−δ)s−2βz(θs ω) ds −t

≤ %20 ,

(4.8)

9


which along with (4.6) shows that, for any t ≥ T0B (ω), 0

Z

e2β

δ

R0 s


kv(s + t,θ−t ω,v0 (θ−t ω))k2 ds

−t 2

+ kv(t,θ−t ω,v0 (θ−t ω))k

(4.9)

≤ %20 .

The proof is complete. Lemma 4.2. Suppose (1.4) and βσ ≤ 2|λσ |. Let B = {B(ω)} ∈ D and v0 (ω) ∈ B(ω), let %1 > 0 be fixed. Then for P-a.e. ω ∈ Ω, there exists T1B (ω) > 0 such that for any t ≥ T1B (ω), we have Z

0

e2β

R0 s

z(θs ω)dτ +(2ρ−δ)s

k(−4)

α+1 2

v(s + t,θ−t ω,v0 (θ−t ω))k2 ds

−t

+

δ 2

Z

0

e2β

R0 s


k∇v(s + t,θ−t ω,v0 (θ−t ω))k2 ds

(4.10)

−t

+ k∇v(t,θ−t ω,v0 (θ−t ω))k2 ≤ %21 . Proof. Taking the inner product in L2 of (3.3) with −4v and taking the real part, we obtain α+1 d k∇vk2+2k(−4) 2 vk2 dt

=−2e−2βz(θt ω) Re f (eβz(θt ω) v),4(eβz(θt ω) v) +2(βz(θt ω)−ρ)k∇vk2 .

(4.11)

Now, we will estimate the first term on the right-hand side of (4.11). For convenience, we set ψ = eβz(θt ω) v. Integrating by parts and using (1.4) and (1.6), then applying the Young’s inequality, we find −Re f (eβz(θt ω) v),4(eβz(θt ω) v) = −Re(f (ψ),4ψ) Z Z ¯ ψ¯ dx + Re ¯ = Re fψ (ψ)|∇ψ|2 + fψ¯ (ψ)∇ψ∇ fx ∇ψdx 3 R3 Z R 2 ¯ 2 +λ ¯ σ (ψ∇ψ) ¯ ≤ −βσ |ψ|2σ |∇ψ|2 + |ψ|2(σ−1) (λσ (ψ∇ψ) ) dx R3 Z + |γ2 (x)||∇v|eβz(θt ω) dx 3 Z R 2 ¯ 2 +λ ¯ σ (ψ∇ψ) ¯ ≤ |ψ|2(σ−1) −βσ |ψ|2 |∇ψ|2 + λσ (ψ∇ψ) dx R3

δ + k∇vk2 + c2 e2βz(θt ω) Z 4 δ = |ψ|2(σ−1) tr(Y M Y H )dx + k∇vk2 + c2 e2βz(θt ω) , 4 R3 where Y=

¯ ψ∇ψ ψ∇ψ¯

H

, M=

− β2σ λσ ¯ σ − βσ λ 2

,

(4.12)

10


and Y H is the conjugate transpose of the matrix Y . We observe that the condition βσ ≤ 2|λσ | implies that the matrix M is nonpositive definite. One can rewrite (4.11) as α+1 δ d k∇vk2 + 2k(−4) 2 vk2 + k∇vk2 ≤(2βz(θt ω) − 2ρ + δ)k∇vk2 dt 2 + 2c2 e2βz(θt ω) .

Multiplying (4.13) by e−2β t

Z

Rt 0

z(θs ω)ds+(2ρ−δ)t

and integrating over (0,t), we infer that

Rt

k∇v(t,ω,v0 (ω))k + e2β s z(θτ ω)dτ +(2ρ−δ)(s−t) k(−4) 0 Z δ t 2β R t z(θτ ω)dτ +(2ρ−δ)(s−t) k∇v(s,ω,v0 (ω))k2 ds + e s 2 0 2

≤e

2β

Rt 0

z(θs ω)ds+(δ−2ρ)t t

Z

e2β

+ 2c2

Rt s

(4.13)

α+1 2

v(s,ω,v0 (ω))k2 ds

(4.14)

2

k∇v0 (ω)k

z(θτ ω)dτ +(2ρ−δ)(s−t)+2βz(θs ω)

ds.

0

Substituting θ−t ω for ω, then we deduce from (4.13) that, t

Z

Rt

α+1

e2β s z(θτ −t ω)dτ +(2ρ−δ)(s−t) k(−4) 2 v(s,θ−t ω,v0 (θ−t ω))k2 ds 0 Z δ t 2β R t z(θτ −t ω)dτ +(2ρ−δ)(s−t) + e s k∇v(s,θ−t ω,v0 (θ−t ω))k2 ds 2 0 + k∇v(t,θ−t ω,v0 (θ−t ω))k2

≤ e2β

Rt 0

z(θs−t ω)ds+(δ−2ρ)t t

Z

e2β

+ 2c2

Rt s

k∇v0 (θ−t ω)k2

z(θτ −t ω)dτ +(2ρ−δ)(s−t)+2βz(θs−t ω)

ds.

0

Changing the variables in the integrals, one has Z

0

e2β

R0


s

k(−4)

α+1 2

v(s + t,θ−t ω,v0 (θ−t ω))k2 ds

−t

δ + 2

0

Z

e2β

R0 s



−t

(4.15)

+ k∇v(t,θ−t ω,v0 (θ−t ω))k2 ≤ e2β

R0 −t


Z

0

+ 2c2

e2β

R0 s


z(θτ ω)dτ +(2ρ−δ)s+2βz(θs ω)

ds.

−t

{B(ω)} ∈ D is tempered, so for any v0 (θ−t ω) ∈ B(θ−t ω), lim e2β

t→+∞

R0 −t


k∇v0 (θ−t ω)k2 = lim e2β

R0 −t

z(θs ω)ds+(δ−2ρ)t+2βz(θt ω)

t→+∞

=0. (4.16)

11


Therefore, there exists T1B (ω) > 0 such that for any t ≥ T1B (ω), Z 0 R0 R0 e2β s z(θτ ω)dτ +(2ρ−δ)s+2βz(θs ω) ds e2β −t z(θs ω)ds+(δ−2ρ)t k∇v0 (θ−t ω)k2 + 2c2 −t

(4.17)

≤ %21 , which along with (4.15) shows that, for any t ≥ T1B (ω), Z 0 R0 α+1 e2β s z(θτ ω)dτ +(2ρ−δ)s k(−4) 2 v(s + t,θ−t ω,v0 (θ−t ω))k2 ds −t

δ + 2

Z

0

e2β

R0 s



−t

+ k∇v(t,θ−t ω,v0 (θ−t ω))k2 ≤ %21 . We complete the proof. Lemma 4.3. Suppose that (1.5) holds. Let B = {B(ω)} ∈ D and v0 (ω) ∈ B(ω). Then for P-a.e. ω ∈ Ω, there exists T1B (ω) > 0 such that for any t ≥ T1B (ω), one has k(−4)

1+α 2

v(t + 1,θ−t−1 ω,v0 (θ−t−1 ω))k2 ≤ %21 + r02 + r12 + r22 , %22 .

(4.18)

Proof. Taking the inner product of (3.3) with (−4)1+α v and taking the real part, we obtain 1+α 1 d k(−4) 2 vk2 + 2k(−4)α+ 2 vk2 dt

= −2(ρ − βz(θt ω))k(−4)

1+α 2

2

vk + 2e

−βz(θt ω)

Re f (e

βz(θt ω)

1+α

v),(−4)

(4.19) v .

We estimate the second term of the right-hand side of (4.19). For convenience, we set ψ = eβz(θt ω) v. Integrating by parts, applying (1.5) and (1.6), and using the H¨older and Young inequalities, we obtain 2e−βz(θt ω) Re f (eβz(θt ω) v),(−4)1+α v = 2e−2βz(θt ω) Re f (eβz(θt ω) v),(−4)1+α (eβz(θt ω) v) = 2e−2βz(θt ω) Re f (ψ),(−4)1+α ψ 1 ≤ 2e−2βz(θt ω) |(fψ (ψ)∇ψ + fψ¯ (ψ)∇ψ¯ + fx ,(−4) 2 +α ψ)| Z Z (4.20) 1 1 ≤ 4β2 e−2βz(θt ω) |∇ψ||(−4) 2 +α ψ|dx + 2e−βz(θt ω) |fx ||(−4) 2 +α v|dx

R3

≤ 4β2 e

−2βz(θt ω)

k(−4)

R3 1 2 +α

ψkk∇ψk + 2e

−βz(θt ω)

1

1

k(−4) 2 +α vkkγ2 (x)k

1

= 4β2 k(−4) 2 +α vkk∇vk + 2e−βz(θt ω) k(−4) 2 +α vkkγ2 (x)k 1

≤ k(−4) 2 +α vk2 + 8β22 k∇vk2 + c3 e−2βz(θt ω) . Substituting (4.20) into (4.19), we deduce that 1+α 1+α 1 d k(−4) 2 vk2 + 2(ρ − βz(θt ω))k(−4) 2 vk2 + k(−4) 2 +α vk2 dt ≤ 8β22 k∇vk2 + c3 e−2βz(θt ω) .

(4.21)

12


This implies that

1+α 1+α 1 d k(−4) 2 vk2 +(2ρ−δ−2βz(θt ω))k(−4) 2 vk2+k(−4) 2 +α vk2 dt ≤ 8β22 k∇vk2+c3 e−2βz(θt ω) .

Taking t ≥ T1B (ω) and s ∈ (t,t + 1), multiplying (4.22) by e−2β integrating (4.21) over (s,t + 1), we get

Rt 0

(4.22)


, and

1+α

k(−4) 2 v(t + 1,ω,v0 (ω))k2 Z t+1 R t+1 1 e2β τ z(θτ1 ω)dτ1 +(δ−2ρ)(t+1−τ ) k(−4) 2 +α v(τ,ω,v0 (ω))k2 dτ + ≤e

s R t+1

2β

s

z(θτ ω)dτ +(δ−2ρ)(t+1−s) t+1

Z

+ 8β22

e2β

s t+1

Z

e2β

+ c3

R t+1 τ

R t+1 τ

k(−4)

1+α 2

v(s,ω,v0 (ω))k2

z(θτ1 ω)dτ1 +(δ−2ρ)(t+1−τ )

(4.23)

k∇v(τ,ω,v0 (ω))k2 dτ

z(θτ1 ω)dτ1 +(δ−2ρ)(t+1−τ )−2βz(θτ ω)

dτ.

s

Integrating (4.23) with respect to s over (t,t + 1), then applying Gagliardo-Nirenberg inequality, we obtain

1+α

k(−4) 2 v(t + 1,ω,v0 (ω))k2 Z t+1 R t+1 1 + e2β τ z(θτ1 ω)dτ1 +(δ−2ρ)(t+1−τ ) k(−4) 2 +α v(τ,ω,v0 (ω))k2 dτ t t+1

Z

e2β

≤

R t+1 s

z(θτ ω)dτ +(δ−2ρ)(t+1−s)

k(−4)

1+α 2

v(s,ω,v0 (ω))k2 ds

t

Z

+ 8β22

t+1

e2β

t t+1

Z

e2β

+ c3 ≤

1 2

Z

t t+1

e2β

R t+1 τ

R t+1 τ

R t+1 s

z(θτ1 ω)dτ1 +(δ−2ρ)(t+1−τ )




dτ

1

k(−4) 2 +α v(s,ω,v0 (ω))k2 ds

t t+1

Z

e2β

+ c4

R t+1 s


kv(s,ω,v0 (ω))k2 ds

t

Z

+ 8β22 Z + c3

t

t+1

e2β

t t+1

e2β

R t+1 τ

R t+1 τ

z(θτ1 ω)dτ1 +(δ−2ρ)(t+1−τ )



dτ.

(4.24)

13


It follows that 1+α

k(−4) 2 v(t + 1,ω,v0 (ω))k2 Z 1 1 t+1 2β R t+1 z(θτ ω)dτ1 +(δ−2ρ)(t+1−τ ) 1 k(−4) 2 +α v(τ,ω,v0 (ω))k2 dτ + e τ 2 t Z t+1 R t+1 ≤ c4 e2β s z(θτ ω)dτ +(δ−2ρ)(t+1−s) kv(s,ω,v0 (ω))k2 ds t

Z

+ 8β22

t+1

e2β

t t+1

Z

e2β

+ c3

R t+1 τ

R t+1 τ

z(θτ1 ω)dτ1 +(δ−2ρ)(t+1−τ )

(4.25)



dτ.

t

Replacing ω by θ−t−1 ω, we infer 1+α

k(−4) 2 v(t + 1,θ−t−1 ω,v0 (θ−t−1 ω))k2 Z 1 1 t+1 2β R t+1 z(θτ −t−1 ω)dτ1 +(δ−2ρ)(t+1−τ ) 1 + e τ k(−4) 2 +α v(τ,θ−t−1 ω,v0 (θ−t−1 ω))k2 dτ 2 t Z t+1 R t+1 ≤ c4 e2β s z(θτ −t−1 ω)dτ +(δ−2ρ)(t+1−s) kv(s,θ−t−1 ω,v0 (θ−t−1 ω))k2 ds t

Z

+ 8β22

t+1

e2β

t t+1

Z

e2β

+ c3

R t+1 τ

R t+1

z(θτ1 −t−1 ω)dτ1 +(δ−2ρ)(t+1−τ )

k∇v(τ,θ−t−1 ω,v0 (θ−t−1 ω))k2 dτ

z(θτ1 −t−1 ω)dτ1 +(δ−2ρ)(t+1−τ )−2βz(θτ −t−1 ω)

τ

dτ.

t

(4.26) Now, we estimate the three terms on the right-hand side of (4.24). For the first term, by Lemma 4.1, for any t ≥ T1B (ω), one has Z

t+1

c4

e2β

R t+1 s

z(θτ −t−1 ω)dτ +(δ−2ρ)(t+1−s)

kv(s,θ−t−1 ω,v0 (θ−t−1 ω))k2 ds

t

≤ c4 %20

(4.27)

0

Z

2β

e

R0

z(θs ω)ds+(2ρ−δ)τ

τ

dτ

, r02 .

−1

For the second term, by Lemma 4.2, for any t ≥ T1B (ω), one has 8β22

Z

t+1

e2β

R t+1 τ

z(θτ1 −t−1 ω)dτ1 +(δ−2ρ)(t+1−τ )

k∇v(τ,θ−t−1 ω,v0 (θ−t−1 ω))k2 dτ

t

≤ 8β22 %21

Z

(4.28)

0

e2β

R0 τ

z(θs ω)ds+(2ρ−δ)τ

dτ , r12 .

−1

For the third term, we have Z c3

t+1

e2β

R t+1 τ

z(θτ1 −t−1 ω)dτ1 +(δ−2ρ)(t+1−τ )−2βz(θτ −t−1 ω)

t

Z

(4.29)

0

≤ c3

e −1

dτ

2β

R0 τ

z(θs ω)ds+(2ρ−δ)τ −2βz(θτ ω)

dτ

, r22 .

14


Substituting (4.27), (4.28) and (4.29) into (4.24) gives k(−4)

1+α 2

v(t + 1,θ−t−1 ω,v0 (θ−t−1 ω))k2 ≤ %21 + r02 + r12 + r22 , %22 ,

(4.30)

which completes the proof. Lemma 4.4. Let B = {B(ω)} ∈ D and v0 (ω) ∈ B(ω). Then for P-a.e. ω ∈ Ω, there exist T ∗ = TB∗ (ω) > 0 and R∗ = R∗ (ω,ε) such that for any t ≥ TB∗ (ω), one has Z |v(t,θ−t ω,v0 (θ−t ω))|2 dx ≤ ε. (4.31) |x|≥R∗

Proof. Take a smooth function χ such that 0 ≤ χ(s) ≤ 1 for all s ≥ 0 and 0, if 0 ≤ s ≤ 1, χ(s) = 1, if s ≥ 2.

(4.32)

There exists a positive constant c such that |χ0 (s)| ≤ c for all s ≥ 0. Taking the real 2 part of the inner product of (3.3) with χ( xk2 )v, we obtain 2 2 Z Z 1 d x x 2 χ |v| dx + (ρ − βz(θ ω)) χ |v|2 dx t 2 2 dt R3 k2 k 3 R 2 2 Z Z x x α −βz(θt ω) βz(θt ω) = −Re(1 + iν) (−4) vχ v ¯ dx + e Re f (e v)χ v¯dx. 2 k k2 3 3 R R (4.33) We estimate each term on the right-hand side of (4.32). For the first term, integrating by parts and applying the H¨ older, Gagliardo-Nirenberg and Young inequalities, we have 2 Z x − Re(1 + iν) (−4)α vχ v¯dx k2 R3 2 2 Z 1 x 2|x| 0 x |v| dx ≤ |1 + iν| |(−4)α− 2 v| χ |∇v| + χ k2 k2 k2 R3 2 Z 0 x 2|x| 1 α− 12 χ ≤ |1 + iν| k(−4)α− 2 vkk∇vk + |(−4) v| |v|dx √ k2 k2 k≤|x|≤ 2k √ Z 2 0 x 2 2 α− 12 α− 21 ≤ |1 + iν| k(−4) vkk∇vk + |(−4) v| χ |v|dx k k≤|x|≤√2k k2 Z 1 c α− 21 ≤ |1 + iν| k(−4)α− 2 vkk∇vk + |(−4) v||v|dx k k≤|x|≤√2k c ≤ c kvk2 + k∇vk2 + k∇vk2 + kvk2 . k (4.34) For the second term, applying (1.3), one has 2 2 Z Z x x −βz(θt ω) βz(θt ω) −2βz(θt ω) v¯dx ≤ e γ1 (x)χ dx e Re f (e v)χ 2 k k2 3 3 R R 2 Z x dx − β1 e−2βz(θt ω) |eβz(θt ω) v|2σ+2 χ k2 3 R 2 Z x ≤ e−2βz(θt ω) |γ1 (x)|χ dx. k2 3 R

(4.35)


15

Using (4.34) and (4.35), (4.32) can be rewritten as 2 2 Z Z x d x 2 |v| dx + (2ρ − δ − 2βz(θt ω)) χ |v|2 dx χ 2 dt R3 k2 k 3 R (4.36) 2 Z c x 2 2 2 2 −2βz(θt ω) ≤ c kvk + k∇vk + k∇vk + kvk + e dx. |γ1 (x)|χ k k2 R3 Rt

Multiplying (4.36) by e−2β 0 z(θs ω)ds+(2ρ−δ)t and integrating over (T1 ,t), we have 2 Z x χ |v(t,ω,v0 (ω))|2 dx k2 R3 2 Z R x 2β Tt z(θs ω)ds+(δ−2ρ)(t−T1 ) 1 |v(T1 ,ω,v0 (ω))|2 dx ≤e χ k2 R3 2 Z Z t Rt x |γ1 (x)|χ e2β s z(θτ ω)dτ +(δ−2ρ)(t−s)−2βz(θs ω) dxds + (4.37) k2 R3 T1 Z t Rt +c e2β s z(θτ ω)dτ +(δ−2ρ)(t−s) kv(s,ω,v0 (ω))k2 + k∇v(s,ω,v0 (ω))k2 ds c + k

T1 t

Z

e2β

Rt s

z(θτ ω)dτ +(δ−2ρ)(t−s)

k∇v(s,ω,v0 (ω))k2 + kv(s,ω,v0 (ω))k2 ds.

T1

Replacing ω by θ−t ω, in (4.37), we deduce that for all t ≥ T1 , 2 Z x |v(t,θ−t ω,v0 (θ−t ω))|2 dx χ k2 R3 2 Z R x 2β Tt z(θs−t ω)ds+(δ−2ρ)(t−T1 ) 1 ≤e χ |v(T1 ,θ−t ω,v0 (θ−t ω))|2 dx k2 R3 2 Z t Z Rt x + e2β s z(θτ −t ω)dτ +(δ−2ρ)(t−s)−2βz(θs−t ω) |γ1 (x)|χ dxds k2 T1 R3 Z t Rt +c e2β s z(θτ −t ω)dτ +(δ−2ρ)(t−s) W(s)ds c + k

T1 t

Z

e2β

Rt s

z(θτ −t ω)dτ +(δ−2ρ)(t−s)

(4.38)

W(s)ds,

T1

where W(x) = kv(x,θ−t ω,v0 (θ−t ω))k2 + k∇v(x,θ−t ω,v0 (θ−t ω))k2 . In what follows, we estimate each term on the right-hand side of (4.38). For the first term, replacing t by T1 and ω by θ−t ω in (4.5), we have 2 Z R x 2β t z(θ ω)ds+(δ−2ρ)(t−T1 ) e T1 s−t χ |v(T1 ,θ−t ω,v0 (θ−t ω))|2 dx k2 R3 Z R 2β t z(θ ω)ds+(δ−2ρ)(t−T1 ) ≤ e T1 s−t |v(T1 ,θ−t ω,v0 (θ−t ω))|2 dx R3 R (4.39) R 2β t z(θ ω)ds+(δ−2ρ)(t−T1 ) 2β 0T1 z(θs−t ω)ds+(1−2ρ)T1 ≤ e T1 s−t e kv0 (θ−t ω)k2 = e2β

Rt

= e2β

R0

0

z(θs−t ω)ds+(δ−2ρ)t

−t


kv0 (θ−t ω)k2

kv0 (θ−t ω)k2 .

16


We find that, given ε > 0, there exists T2 = T2 (B,ω,ε) > T1 such that for all t ≥ T2 , 2 Z R x ε ω)ds+(δ−2ρ)(t−T1 ) 2β t z(θ χ |v(T1 ,θ−t ω,v0 (θ−t ω))|2 dx ≤ . (4.40) e T1 s−t 2 k 4 R3 For the second term, note that γ1 (x) ∈ L1 (R3 ), so there exists R1 = R1 (ε) such that for all k ≥ R1 , we have 2 Z x dx ≤ cε. (4.41) |γ1 (x)|χ k2 |x|≥k Given ε0 > 0, there exists T3 = T3 (ω) > 0 such that for s < −T3 , we have Z t Rt e2β s z(θτ −t ω)dτ +(δ−2ρ)(t−s)−2βz(θs−t ω) ds T1

Z

0

e2β

= T1 −t Z 0

R0 s

e2β

≤


R0 s

ds


(4.42) Z

T1 −T3

ds +

T1 −T3

es(2ρ−δ+ε0 ) ds

T1 −t

≤ c(ω) + c1 (ω). So there exists R1 = R1 (ε,ω) such that for all t ≥ T3 and k ≥ R1 , 2 Z t Z Rt x ε e2β s z(θτ −t ω)dτ +(δ−2ρ)(t−s)−2βz(θs−t ω) |γ1 (x)|χ dxds ≤ . 2 k 4 T1 R3

(4.43)

For the third term, by (4.6) and (4.15), one has Z t Rt c e2β s z(θτ −t ω)dτ +(δ−2ρ)(t−s) W(s)ds T1

Z

0

e2β

≤c ≤ ce

T1 −t R0

2β

T1 −t

R0 s


W(s + t)ds

z(θs ω)ds+(2ρ−δ)(T1 −t)

(4.44)

(kv0 (θ−t ω)k2 + k∇v0 (θ−t ω)k2 ).

Since {B(ω)} ∈ D is tempered, for any v0 (θ−t ω) ∈ B(θ−t ω), lim e

2β

R0 T1 −t

z(θs ω)ds+(2ρ−δ)(T1 −t)

t→+∞

(kv0 (θ−t ω)k2 + k∇v0 (θ−t ω)k2 ) = 0.

(4.45)

Therefore, there exists T4 = T4 (B,ω,ε) > T1 such that for any t ≥ T4 , Z t Rt c e2β s z(θτ −t ω)dτ +(δ−2ρ)(t−s) W(s)ds T1

≤e

2β

R0 T1 −t

z(θs ω)ds+(δ−2ρ)(T1 −t)

ε (kv0 (θ−t ω)k2 + k∇v0 (θ−t ω)k2 ) ≤ . 4

Similarly, there exists R2 = R2 (ω,ε) such that for all t ≥ T4 and k ≥ R2 , Z c t 2β R t z(θτ −t ω)dτ +(δ−2ρ)(t−s) ε e s W(s)ds ≤ . k T1 4

(4.46)

(4.47)


17

Let T ∗ = T ∗ (B,ω,ε) = max{T1 ,T2 ,T3 ,T4 }. Then by (4.40), (4.46) and (4.47), for all t ≥ T ∗ and k ≥ R∗ = max{R1 ,R2 }, one has 2 Z x |v(t,θ−t ω,v0 (θ−t ω))|2 dx ≤ ε. (4.48) χ k2 R3 This implies that for all t ≥ T ∗ and k ≥ R∗ , we have 2 Z Z x |v(t,θ−t ω,v0 (θ−t ω))|2 dx ≤ |v(t,θ−t ω,v0 (θ−t ω))|2 dx ≤ ε. (4.49) χ 2 k 3 |x|≥k R The proof is complete. Lemma 4.5. Let B = {B(ω)} ∈ D and v0 (ω) ∈ B(ω). Then for P-a.e. ω ∈ Ω, there exists T ∗∗ = TB∗∗ (ω) > 0 such that for any t ≥ TB∗∗ (ω), one has Z |∇v(t,θ−t ω,v0 (θ−t ω))|2 dx ≤ ε. (4.50) |x|≥k

Proof. Differentiating (3.3) with respect to x = (x1 ,x2 ,x3 ), then taking the real 2 part of the inner product with χ( xk2 )∇v gives 2 2 Z Z x 1 d x 2 χ |∇v| dx + (ρ − βz(θt ω)) χ |∇v|2 dx 2 2 dt R3 k k2 R3 2 Z x (4.51) = −Re(1 + iν) ((−4)α (∇v))χ ∇¯ v dx k2 R3 2 Z x ∇¯ v dx. + e−βz(θt ω) Re ∇f (eβz(θt ω) v)χ k2 R3 Now, we estimate the right-hand side of (4.51). For the first term, we have 2 Z 1 x α −Re(1 + iν) ((−4) (∇v))χ ∇¯ v dx ≤ |1 + iν|k(−4)α+ 2 vkk∇vk 2 k R3 1 ≤ c k(−4)α+ 2 vk2 + k∇vk2 . For the second term, one has 2 Z x −βz(θt ω) βz(θt ω) ∇¯ v dx e Re ∇f (e v)χ k2 R3 2 2 Z Z x x 2 −βz(θt ω) ≤ 2β2 |∇v| χ dx + e |γ2 (x)||∇v|χ dx 2 k k2 R3 R3 Z 1 1 x2 ≤ (2β2 + )k∇vk2 + e−2βz(θt ω) |γ2 (x)|2 |χ2 dx. 2 2 k2 R3 Substituting (4.52) and (4.53) into (4.51), we deduce that 2 2 Z Z d x x 2 χ |∇v| dx + (2ρ − δ − 2βz(θt ω)) χ |∇v|2 dx dt R3 k2 k2 3 R 2 Z 1 α+ 12 2 2 −2βz(θt ω) 2 2 x ≤ c k(−4) vk + k∇vk + e |γ2 (x)| |χ dx. 2 k2 R3

(4.52)

(4.53)

(4.54)

18


Multiplying (4.54) by e−2β t ≥ T1 ,

Z χ R3

0


x2 |∇v(t,ω,v0 (ω))|2 dx k2 Z

2β

Rt

Z

t

≤e

Rt

z(θs ω)ds+(δ−2ρ)(t−T1 )

T1

χ R3

e2β

+c 1 + 2

Rt s

T1 t

Z

e

2β

z(θτ ω)dτ +(δ−2ρ)(t−s)

Rt s

and integrating over (T1 ,t) gives for all

x2 |∇v(T1 ,ω,v0 (ω))|2 dx k2

1 k(−4)α+ 2 v(s,ω,v0 (ω))k2 + k∇v(s,ω,v0 (ω))k2 ds

z(θτ ω)dτ +(δ−2ρ)(t−s)−2βz(θs ω)

Z

2

|γ2 (x)| |χ R3

T1

2

x2 dxds. k2 (4.55)

Replacing ω by θ−t ω and applying (4.37), for all t ≥ T1 , one has x2 χ |∇v(t,θ−t ω,v0 (θ−t ω))|2 dx k2 R3 Z R 2β t z(θ ω)ds+(δ−2ρ)(t−T1 ) ≤ e T1 s−t |∇v(T1 ,θ−t ω,v0 (θ−t ω))|2 dx 3 R Z t R 1 2β st z(θτ −t ω)dτ +(δ−2ρ)(t−s) +c e k(−4)α+ 2 v(s,θ−t ω,v0 (θ−t ω))k2 ds

Z

T1 t

Z

e2β

+c +

1 2

T1 Z t

Rt

e2β

s


Rt s

(4.56)

k∇v(s,θ−t ω,v0 (θ−t ω))k2 ds

z(θτ −t ω)dτ +(δ−2ρ)(t−s)−2βz(θs−t ω)

Z

|γ2 (x)|2 χ2

R3

T1

x2 dxds. k2

We estimate each term on the right-hand side of (4.56). For the first term, replacing t by T1 and ω by θ−t ω in (4.14), we have e

2β

Rt T1

z(θs−t ω)ds+(δ−2ρ)(t−T1 )

Z

|∇v(T1 ,θ−t ω,v0 (θ−t ω))|2 dx

R3 2β

Rt

=e

2β

Rt

=e

2β

R0

≤e

T1

0

z(θs−t ω)ds+(δ−2ρ)(t−T1 ) 2β

e

z(θs−t ω)ds+(δ−2ρ)t

z(θs ω)ds+(δ−2ρ)t −t

R T1 0

z(θs−t ω)ds+(δ−2ρ)T1


(4.57)

2

k∇v0 (θ−t ω)k

k∇v0 (θ−t ω)k2 .

Since {B(ω)} ∈ D is tempered, for any v0 (θ−t ω) ∈ B(θ−t ω), lim e2β

R0 −t


t→+∞

k∇v0 (θ−t−1 ω)k2 = 0.

(4.58)

Therefore, given ε > 0, there exists T5 = T5 (B,ω,ε) > T1 such that for all t ≥ T5 , e

2β

Rt T1

z(θs−t ω)ds+(δ−2ρ)(t−T1 )

Z

ε |∇v(T1 ,θ−t ω,v0 (θ−t ω))|2 dx ≤ . 4 R3

(4.59)

19


For the second term, one has t

Z

e2β

c

Rt s


1

k(−4)α+ 2 v(s,θ−t ω,v0 (θ−t ω))k2 ds

T1 0

Z

e2β

≤c ≤c

R0 s


T1 −t Z 0 R 2β s0 z(θτ ω)dτ +(2ρ−δ)s

e

1

k(−4)α+ 2 v(s + t,θ−t ω,v0 (θ−t ω))k2 ds

(4.60)

1

k(−4)α+ 2 v(s + t,θ−t ω,v0 (θ−t ω))k2 ds.

−t

Replacing ω by θ−t−1 ω in (4.23), dropping the first term on the left-hand side, and integrating with respect to s over (T1 ,t + 1), we obtain t+1

Z

e2β

R t+1 τ

z(θτ1 −t−1 ω)dτ1 +(δ−2ρ)(t+1−τ )

1

k(−4) 2 +α v(τ,θ−t−1 ω,v0 (θ−t−1 ω))k2 dτ

T1

Z

t+1

e2β

≤

R t+1 s

z(θτ −t−1 ω)dτ +(δ−2ρ)(t+1−s)

k(−4)

1+α 2

v(s,θ−t−1 ω,v0 (θ−t−1 ω))k2 ds

T1

+ 8β22

t+1

Z

e2β

R t+1 s

z(θτ −t−1 ω)dτ +(δ−2ρ)(t+1−s)

k∇v(s,θ−t−1 ω,v0 (θ−t−1 ω))k2 ds

T1

Z

0

e2β

=

R0 s


k(−4)

1+α 2

v(s + t + 1,θ−t−1 ω,v0 (θ−t−1 ω))k2 ds

T1 −t−1

+ 8β22

Z

0

e2β

R0 s


k∇v(s + t + 1,θ−t−1 ω,v0 (θ−t−1 ω))k2 ds

T1 −t−1

Z

0

e2β

≤

R0 s


k(−4)

1+α 2

v(s + t + 1,θ−t−1 ω,v0 (θ−t−1 ω))k2 ds

−t−1

+ 8β22

Z

0

e2β

R0 s


k∇v(s + t + 1,θ−t−1 ω,v0 (θ−t−1 ω))k2 ds.

−t−1

(4.61) By (4.15), for all t ≥ T1B (ω) − 1, we have Z

0

e2β

R0 s


k(−4)

α+1 2

v(s + t + 1,θ−t−1 ω,v0 (θ−t−1 ω))k2 ds

−t−1

Z +δ ≤e

2β

0

e2β

R0 s


−t−1 R0 z(θs ω)ds+(δ−2ρ)(t+1) −t−1

k∇v(s + t + 1,θ−t−1 ω,v0 (θ−t−1 ω))k2 ds

(4.62)

k∇v0 (θ−t−1 ω)k2 .

Substituting (4.62) into (4.61), one has Z

t+1

T1

R t+1

1

e2β τ z(θτ1 −t−1 ω)dτ1 +(δ−2ρ)(t+1−τ ) k(−4) 2 +α v(τ,θ−t−1 ω,v0 (θ−t−1 ω))k2 dτ 0 8β22 2β R−t−1 z(θs ω)ds+(δ−2ρ)(t+1) ≤ 1+ e k∇v0 (θ−t−1 ω)k2 . δ (4.63)

Again, since {B(ω)} ∈ D is tempered, by a similar argument, there exists T6 =

20


T6 (B,ω,ε) > T1B (ω) such that for any t ≥ T6 , 8β 2 2β R 0 z(θ ω)ds+(δ−2ρ)(T1 −t) ε 1 + 2 e T1 −t s (kv0 (θ−t ω)k2 + k∇v0 (θ−t ω)k2 ) ≤ . δ 4

(4.64)

So, we infer that Z t Rt 1 ε c e2β s z(θτ −t ω)dτ +(δ−2ρ)(t−s) k(−4)α+ 2 v(s,θ−t ω,v0 (θ−t ω))k2 ds ≤ . (4.65) 4 T1 For the third term, by (4.46), there exists T4 = T4 (B,ω,ε) > T1 such that for any t ≥ T4 , Z t Rt ε (4.66) c e2β s z(θτ −t ω)dτ +(δ−2ρ)(t−s) k∇v(s,θ−t ω,v0 (θ−t ω))k2 ds ≤ . 4 T1 For the last term, note that γ2 (x) ∈ L2 (R3 ). In a manner similar to the argument for (4.43), there exists R1∗ = R1∗ (ε) such that for all t ≥ T3 and k ≥ R1∗ , Z

t

T1

e2β

Rt s

z(θτ −t ω)dτ +(δ−2ρ)(t−s)−2βz(θs−t ω)

Z

|γ2 (x)|2 χ2

R3

x2 ε dxds ≤ . (4.67) 2 k 4

Let T ∗∗ = T ∗∗ (B,ω,ε) = max{T3 ,T4 ,T5 ,T6 }. Then by (4.59), (4.65) and (4.66), for all t ≥ T ∗∗ and k ≥ R1∗ , one has 2 Z x χ |∇v(t,θ−t ω,v0 (θ−t ω))|2 dx ≤ ε. (4.68) 2 k 3 R This implies that for all t ≥ T ∗∗ and k ≥ R1∗ , 2 Z Z x |∇v(t,θ−t ω,v0 (θ−t ω))|2 dx ≤ χ |∇v(t,θ−t ω,v0 (θ−t ω))|2 dx ≤ ε. (4.69) 2 k 3 |x|≥k R This completes the proof. By Lemmas 4.4 and 4.5, we have Corollary 4.6. Let B = {B(ω)} ∈ D and v0 (ω) ∈ B(ω). Then for P-a.e. ω ∈ Ω, there exists TB? = max{TB∗ (ω),TB∗∗ (ω)} and R∗ = R∗ (ω,ε) such that for any t ≥ TB? (ω), one has kv(t,θ−t ω,v0 (θ−t ω))k2H 1 (|x|≥R∗ ) ≤ ε.

(4.70)

5. Random attractor In this section, we prove the existence of a random attractor for the random dynamical system generated by (3.3) on R3 . From Lemma 4.2, ϕ has a closed random absorbing set in D. The D-pullback asymptotic compactness of ϕ is demonstrated below using the uniform estimates obtained in the previous sections. Lemma 5.1. Assume (1.3)–(1.5) and βσ ≤ 2|λσ |. Then the random dynamical system ϕ is D-pullback asymptotically compact in H 1 (R3 ); that is, for P-a.e. ω ∈ Ω, the sequence ϕ(tn ,θ−tn ω,v0,n (θ−tn ω)) has a convergent subsequence in H 1 (R3 ) provided tn → ∞, B = {B(ω)} ∈ D and v0,n (θ−tn ω) ∈ B(θ−tn ω).

21


Proof. Let tn → ∞, B = {B(ω)} ∈ D and v0,n (θ−tn ω) ∈ B(θ−tn ω). Lemma 4.1 and 4.2, for P-a.e. ω ∈ Ω, we have {ϕ(tn ,θ−tn ω,v0,n (θ−tn ω))}∞ n=1

Applying

is bounded in H 1 (R3 ).

Therefore, there exists η(ω) ∈ H 1 (R3 ) and a subsequence, for convenience, still denoted by {ϕ(tn ,θ−tn ω,v0,n (θ−tn ω))}, such that ϕ(tn ,θ−tn ω,v0,n (θ−tn ω)) → η

weakly in H 1 (R3 ).

(5.1)

Given ε > 0, by Corollary 4.6, there is TB? = max{TB∗ (ω),TB∗∗ (ω)} and R∗ = R∗ (ω,ε) such that for any t ≥ TB? (ω), kϕ(t,θ−t ω,v0 (θ−t ω))k2H 1 (|x|≥R∗ ) ≤ ε.

(5.2)

Since tn → ∞, there exists N1 = N1 (B,ω,ε) such that tn ≥ TB? for all n ≥ N1 . Then, by (5.2), we have for all n ≥ N1 , kϕ(tn ,θ−tn ω,v0,n (θ−tn ω))k2H 1 (|x|≥R∗ ) ≤ ε,

(5.3)

||η||2H 1 (|x|≥R∗ ) ≤ ε.

(5.4)

and hence,

Applying Lemmas 4.1 and 4.3, there exists T2B = max{T0B (ω),T1B (ω)} such that for all t ≥ T2B , kϕ(t,θ−t ω,v0 (θ−t ω))k2H 1+α (R3 ) ≤ %20 + %22 , %23 .

(5.5)

Let N2 = N2 (B,ω) be large enough such that tn ≥ T2B for n ≥ N2 . It follows from (5.5) that, for all n ≥ N2 , kϕ(tn ,θ−tn ω,v0,n (θ−tn ω))k2H 1+α (R3 ) ≤ %23 .

(5.6)

Let BR∗ = {x ∈ R3 : |x| ≤ R∗ } be a ball. By the compactness of the embedding H 1+α (BR∗ ) ,→ H 1 (BR∗ ), from (5.6), we deduce that, up to a subsequence depending ˆR∗ ), which implies that there on R∗ , ϕ(tn ,θ−tn ω,v0,n (θ−tn ω)) → η strongly in H 1 (B exists N3 = N3 (B,ω,ε) ≥ N2 such that for all n ≥ N3 , kϕ(tn ,θ−tn ω,v0,n (θ−tn ω)) − ηk2H 1 (BR∗ ) ≤ ε. Let N ? = max{N1 ,N3 }. Then, from (5.2), (5.3) and (5.4), we have for all n ≥ N ? , kϕ(tn ,θ−tn ω,v0,n (θ−tn ω)) − ηk2H 1 (R3 ) ≤ kϕ(tn ,θ−tn ω,v0,n (θ−tn ω)) − ηk2|x|≤R∗ + kϕ(tn ,θ−tn ω,v0,n (θ−tn ω))||2|x|≥R∗ + ||η||2|x|≥R∗ ≤ 5ε, which implies that ϕ(tn ,θ−tn ω,v0,n (θ−tn ω)) → η This completes the proof.

strongly in H 1 (R3 ).

22


By Proposition 2.8, we have Theorem 5.2. Assume (1.3)–(1.5) and βσ ≤ 2|λσ |. Then the random dynamical system ϕ associated with the fractional Ginzburg-Landau equation with multiplicative noise (1.1) has a unique D-random attractor in H 1 (R3 ). Acknowledgement. We would like to thank the anonymous referees for their valuable suggestions in improving the paper. Peter W. Bates and Mingji Zhang were supported in part by the NSF DMS-0908348 and DMS-1413060. Hong Lu and Shujuan L¨ u were supported by the NSF of China (No.11272024), and China Scholarship Council (CSC)

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