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Pullback attractor for a nonlocal discrete nonlinear Schrödinger equation with delays

Jardel Morais Pereira

B

Department of Mathematics, Federal University of Santa Catarina, Florianópolis SC, 88040-900, Brazil Received 15 February 2021, appeared 30 December 2021

Communicated by Christian Pötzsche

Abstract. We consider a nonlocal discrete nonlinear Schrödinger equation with delays.

We prove that the process associated with the non-autonomous model possesses a pull- back attractor. As a consequence of our discussion, the existence of a global attractor for the autonomous model is derived.

Keywords: pullback attractor, discrete nonlinear Schrödinger equation, delay terms, global attractor differential equations, difference equations.

2020 Mathematics Subject Classification: 35Q55, 37L60, 37B55, 37L30.

1 Introduction

Discrete Schrödinger equations are widely used as models in Physics and other branches of science (see, e.g., [3,6,11,12,14,19] and the references therein). These discrete equations belong to a large class of lattice dynamical systems which has been the object of extensive research (see, for example, [4,5,7,9,12,13,19,22] and the references therein). Various properties related to the dynamics of such systems have been studied. Among them, the existence of global attractors is a theme which attracts a great deal of attention. However, most of the contributions in this line of research addressed to discrete Schrödinger models are concerned the discrete nonlinear Schrödinger equation (DNLS). In this paper, our main aim is to prove the existence of a pullback attractor for a nonlocal discrete nonlinear Schrödinger equation when delay terms are considered. The model is written as follows

iu˙n(t) +

+ m=−

J(n−m)um(t) +gn(t,unt) +iγun(t) = fn(t), t>τ,n∈Z, un(s) =ψn(s−τ), ∀s∈ [τ−h,τ],

(1.1)

where τ,h, and γ are real numbers with h > 0 and γ > 0. In (1.1), un(t), fn(t), and ψn are complex functions and unt denotes the translation of un at time t, defined by unt(s) = un(t+s),∀s ∈ [−h, 0]. The dispersive coupling parameters J(m) are assumed to be real numbers, symmetric (i.e., J(−m) = J(m), for all positive integerm) and ∑+m=1|J(m)| < +.

BEmail: [email protected]

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This includes important special cases asJ(m) =J0eβ|m| andJ(m) = J0|m|s, whereJ0, β, and sare positive real constants suitably chosen [8].

We assume that the nonlinear termgn(t,unt)in (1.1) includes delay terms as follows gn(t,unt) =g0,n(un(t)) +g1,n(un(t−ρ(t)) +

Z 0

hbn(s,un(t+s))ds. (1.2) Appropriate hypotheses on the functions ρ : R → [0,h], gi,n : CC,i = 0, 1, bn : [0,h]×CC, and fn(t)are stated in Section 2.

Specific deterministic cases of equation (1.1) have been used in the study of physical phe- nomena in which long-range dispersive interactions cannot be disregarded (see the physical discussions in [8]). An example is the model proposed in [17] for the description of the non- linear dynamics of the DNA molecule.

A class of discrete Schrödinger equations of great importance is

iu˙n(t) +pdun(t) +gn(t,unt) +iγun(t) = fn(t), (1.3) where ∆dp = d◦ · · · ◦d, p times, and ∆d is the one-dimensional discrete Laplace operator defined by ∆dun = un+1+un1−2un. Equation (1.3) can be derived from (1.1) by choosing the coupling parametersJ(m)as

J(m) =

2p j=0

2p j

(−1)jδm,jp,

where pis any positive integer andδm,k is the Kronecker delta.

Many contributions on existence and properties of solutions of the DNLS equation (i.e, (1.3) with p = 1, g1,n = bn = 0) and fn independent of time can be found in the literature (see, e.g., [3,4,11,19] and references therein). For example, the existence and approximation of attractors for the DNLS equation were investigated in [11] while the existence of attractors for the DNLS with retarded terms was studied in [4]. Concerning equation (1.1), in [19], the authors studied the existence of localized solutions for the homogeneous case without delays. Later, also for the autonomous deterministic model, the existence of a global attractor in weighted spaces was established in [20]. For the existence of attractors for some non- autonomous lattice dynamical systems with retarded terms of the type (1.2) and references about related works we refer the reader to the article [2]. Still concerning lattice models with nonlocal terms, we would like to mention the papers [1,10,15,18,21].

In this paper, under suitable conditions on the functions ρ, gi,n,i = 0, 1, bn, and fn, we prove the existence of a pullback attractor for theprocessassociated with problem (1.1). As a consequence of our discussion, the existence of a global attractor for the autonomous model is derived.

The paper is organized as follows. In Section 2, we prove that the initial value problem (1.1) is globally well posed. In Section 3, we establish the existence of a pullback attractor for the process associated with problem (1.1) using the results in [16]. Finally, in Section 4, we briefly show how the same ideas of the previous sections can be adapted to prove the existence of a global attractor for the autonomous model.

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2 Existence of solutions

In this section, we discuss the existence of solutions for the problem (1.1). We denote by `p the usual space of complex sequences u= (un)nZ such thatkuk`p <∞, where

kuk`p =

+ n=−

|un|p

!1p

, if 1≤ p< andkuk` =sup

nZ

|un|, if p=∞.

When p=2,`2 is a Hilbert space with the inner product given by (u,v)`2 =

+ n=−

unvn, u,v∈`2, and, in this case, we denote by k · kthe corresponding norm.

For 1 ≤ p < ∞, Lp(−h, 0)denotes the usual Banach space of (class of ) real functions f defined on [−h, 0]such that |f|p is integrable in sense of Lebesgue and we recall that for the

`pspaces the following embedding relation holds:

`q⊂ `p, kuk`p ≤ kuk`q, 1≤q≤ p≤∞.

Regarding the functionsgi,n : CC,i= 0, 1, bn : [−h, 0]×CC , f = (fn(t))nZ, and ρ(t)in (1.1) and (1.2) we assume that

(A1) zg0,n(z)is real for allz ∈Candn∈Z.

(A2) There exist a functionκ∈ L2(−h, 0)and functionsb0,n:CCsuch that

|bn(s,z1)−bn(s,z2)| ≤κ(s)|b0,n(z1)−b0,n(z2)|,

∀s ∈[−h, 0] and∀z1,z2C. We setκ20 :=R0

h|κ(s)|2ds.

(A3) For every R>0 there exist positive constants Lj(R),j=1, 2, such that

|gi,n(z1)−gi,n(z2)| ≤L1(R)|z1−z2|, i=0, 1,

|b0,n(z1)−b0,n(z2)| ≤L2(R)|z1−z2|,

for anyn∈Zand anyz1,z2Csuch that|zj| ≤R, j=1, 2. Moreover,(g0,n(0))nZ∈ `2. (A4) There exist sequences of real numbers k1 = (k1,n)nZ ∈ `, k2 = (k2,n)nZ ∈ `2 and

non-negative real functionsβ1,n(·)∈ L2(−h, 0)andβ2,n(·)∈ L1(−h, 0)such that

|g1,n(z)| ≤k1,n|z|+k2,n and |bn(s,z)| ≤ β1,n(s)|z|+β2,n(s), for alln ∈Z, s∈[−h, 0], and z∈C. We setK1=kk1k`,K2 =kk2k, and

B1 =sup

nZ

Z 0

hβ21,n(s)ds 1/2

<∞, B2 =

" +

n=−

Z 0

hb2,n(s)ds 2#1/2

<∞.

(A5) f ∈C(R;`2). (A6) ρ ∈C(R;[0,h]). (A7) Rt

kf(s)k2ds<∞, ∀t∈R.

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Example 2.1. Let 0 6= χ = (χn)nZ ∈ `p, for some 1≤ p ≤ ∞, and ϕ1 : RR defined by ϕ1(t) = t2

a+bt2, whereaandbare positive real constants. Also define the functionsg1,n:CC, b0,n :CCandbn :[−h, 0]×CCby

g1,n(z) =b0,n(z) =χnϕ1(|z|)z, bn(s,z) =χnϕ1(|z|)z1

h(s+h), ∀n ∈Z, s∈ [−h, 0]and z∈C.

Then, the hypotheses (A2)–(A4) are satisfied with L1(R) = L2(R) =

1 b+√R

ab

kχk`p,

κ(s) = 1

h(s+h), k1,n= 1

b|χn|, k2,n=0, β1,n(s) = 1

bh|χn|(s+h), and β2,n=0.

Conditions (A1) and (A3) concerningg0,nare satisfied, for example, ifg0,n(z) =χnϕ2(|z|)z, withχnas before and any ϕ2 ∈C1(R+;R), such that ϕ2(0) =0.

Now let us write (1.1) as an evolution equation with a retarded term in `2. For any u= (un)nZwe define(Au)n =+m=− J(n−m)um, ∀n∈Z.

Lemma 2.2. A:`2→`2is a bounded operator andkAuk ≤4kJk`1kuk, ∀u∈`2. Proof. See Lemma 2.1 in [20].

We consider the space Eh = C([−h, 0];`2) with the usual norm given by kukEh = maxs∈[−h,0]ku(s)k and define the map g : R×Eh → `2 by (g(t,v))nZ = gn(t,vn), where v(s) = (vn(s))nZ, for anys ∈[−h, 0], and

gn(t,vn) =g0,n(vn(0)) +g1,n(vn(−ρ(t))) +

Z 0

hbn(s,vn(s))ds.

If we set ut = (unt)nZfor any t ≥ τ, then we can write the initial value problem (1.1) in

`2 as

iu˙(t) +Au(t) +g(t,ut) +iγu(t) = f(t), t >τ,

u(s) =ψ(s−τ), ∀s∈[τ−h,τ], (2.1) whereψ(s) = (ψn(s))nZ, for anys∈[−h, 0].

We now define the mapB:R×Eh→ `2by B(t,v) =−i

Av(0) +g(t,v) +iγv(0)− f(t).

Then, problem (2.1) can be rewritten as the following functional equation in`2 du

dt +B(t,ut) =0, t>τ uτ =ψ.

(2.2)

The following two lemmas are sufficient to ensure the existence of a local solution for (2.1).

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Lemma 2.3. Assume that (A2)–(A6) hold. Then the map B is continuous and satisfies the local Lipschitz condition: For any v,w ∈ Eh, with kvkEh ≤ R and kwkEh ≤ R, there exists a positive constant L= L(R)such that

kB(t,v)− B(t,w)k ≤ Lkv−wkEh, ∀t ∈R.

Proof. Using (A2)–(A6) we see thatB is well defined. Fix(t,v)∈R×Ehand consider tm →t inR andvm →v in Eh. We have that

kB(tm,vm)− B(t,v)k ≤ kA(vm(0)−v(0))k+kg(tm,vm)−g(t,v)k

+γkvm(0)−v(0)k+kf(tm)− f(t)k. (2.3) Since the sequence(vm)mN is bounded inEh, then using the assumptions (A2), (A3), and (A6) we can find a positive constant Ldepending only onkvkEh such that

kg(tm,vm)−g(t,v)k2 ≤4

+ n=−

|g0,n(vmn(0))−g0,n(vn(0))|2 +4

+ n=−

|g1,n(vmn(−ρ(tm)))−g1,n(vn(−ρ(t)))|2 +4

+ n=−

Z 0

h

|bn(s,vmn(s))−bn(s,vn(s))|ds 2

≤8L2kvm−vk2E

h+4L2

+ n=−

Z 0

h

|κ(s)| |vmn(s)−vn(s)|ds 2

. (2.4)

Using the Cauchy–Schwarz inequality and the fact thatkvm−vkEh < we can estimate the last term in (2.4) as follows

+ n=−

Z 0

h

|κ(s)| |vmn(s)−vn(s)|ds 2

ds≤κ20

+ n=−

Z 0

h

|vmn(s)−v(s)|2ds

κ02 Z 0

h + n=−

|vmn(s)−v(s)|2ds≤κ20kvm−vk2E

hh.

(2.5)

From (2.3), (2.4), (2.5), (A5), and Lemma 2.2 we deduce the continuity of B. In a similar manner we prove the Lipschitz condition.

Lemma 2.4. Assume that (A2)–(A6) hold. Then the map Bis bounded, i.e., it takes bounded subsets ofR×Eh onto bounded subsets of`2.

Proof. LetObe a bounded subset ofR×Eh. Then, there exists a positive constantRsuch that

|t|2+kvk2E

h ≤ R2, ∀(t,v) ∈ O. Using Lemma2.3 we find a positive constant L = L(R)such that

kB(t,v)k ≤ kB(t,v)− B(t, 0)k+kB(t, 0)k

≤ LR+max

|t|≤RkB(t, 0)k<∞, ∀(t,v)∈ O.

Using Lemmas2.3, 2.4and applying the Theory of Functional Equations to problem (2.2) we deduce the following result of existence of local solution for (2.1).

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Theorem 2.5. Assume that (A2)–(A6) hold. Then, for each ψ ∈ Eh, the initial value problem (2.1) has a unique solution u= u(t)defined in[τ−h,T)such that u∈ C([τ−h,T);`2)∩C1([τ,T);`2). Moreover, if T<thenlimtTku(t)k=∞.

Next let us show that the local solution obtained in Theorem2.5can be extended globally.

Lemma 2.6. Assume that (A1)–(A6) hold. Then the solution u of (2.1)with uτ = ψ∈ Ehsatisfies 1

2 d

dtku(t)k2+ γ

2ku(t)k21

2γkf(t)k2+ (K1ku(t−ρ(t))k+K2)ku(t)k +

"

B1 Z 0

h

ku(t+s)k2ds 1/2

+B2

#

ku(t)k, τ≤t< T.

(2.6)

Proof. Taking the imaginary part of the inner product of equation (2.1) withuin`2, we obtain 1

2 d

dtku(t)k2+Im(Au(t),u(t))`2+γku(t)k2+Im(g(t,ut),u(t))`2 =Im(f(t),u(t))`2, for allτ≤t< T. Since

Im(f(t),u(t))`21

2γkf(t)k2+ γ

2ku(t)k2, (Au(t),u(t))`2 = J(0)ku(t)k2+2

+ m

=1

+ n=−

J(m)Re(un+m(t)un(t)), then, using (A1), we get the inequality

1 2

d

dtku(t)k2+ γ

2ku(t)k21

2γkf(t)k2−Im

+ n=−

g1,n(un(t−ρ(t)))un

−Im

+ n=−

Z 0

hbn(s,un(t+s))ds un, τ≤t <T.

(2.7)

Let us estimate the last two terms in (2.7) using the assumption (A4) and the fact that kutkEh < ∞,τ≤t <T. We have that

−Im

+ n=−

g1,n(un(t−ρ(t)))un

+ n=−

[k1,n|un(t−ρ(t))|+k2,n]|un|

≤(K1ku(t−ρ(t))k+K2)kuk,

(2.8)

−Im

+ n=−

Z 0

hbn(s,un(t+s))ds un

+ n=−

Z 0

h

[β1,n(s)|un(t+s)|+β2,n(s)]ds|un|

" +

n=−

Z 0

hβ21,n(s)ds Z 0

h

|un(t+s)|2ds #1/2

kuk+B2kuk

"

B1 Z 0

h

ku(t+s)k2ds 1/2

+B2

# kuk.

(2.9)

From (2.7)–(2.9) we obtain (2.6).

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We now make the following assumptions on the constants B1, K1, γ, h, and a suitable positive parameterµ, which will be used in Section 3 to define the universe where the pullback attractor will lie in.

(A8) We assume that there exists a positive real numberµsuch that (i) IfK1 >0 andB1 ≥0 then

4B12h<eµhγ γ

2 −µ

(2.10) and

µ>2K1eµh. (2.11)

(ii) IfK1 =0 andB1 >0 then

µ< γ

2 and µ> 4

γB12e2µhh. (2.12)

(iii) IfK1 =B1=0 then µ= γ2 andhis arbitrary.

Remark 2.7. Conditions in (A8) will be used in the next theorem to prove an estimate for the solution of (2.1) that allows us to extend it globally and that will be used in the proofs of Lemmas3.1,3.2 and3.3 in Section 3. It is clear from (2.10) that µ< γ2. We also observe that (2.11) holds if and only if 0 < 2K1 < he1, where he1 is the maximum value of the real function φ(s) = sehs, s≥0. From this we see that 2K1eh< 1 and µ∈(µ1,µ2), where µj,j=1, 2, are the two positive solutions of the equation µeµh =2K1.

Theorem 2.8. Assume that (A1)–(A8) hold. Then, the solution u= u(t)of (2.1)with uτ = ψ∈ Eh exists globally. Moreover, for eachτ< T<∞, the mapI:Eh →C([τ,T];Eh), defined byI(ψ)(t) = ut, ∀τ≤t ≤T, is continuous.

Proof. Assume that (A8)(i) holds. Multiplying (2.6) by eµt and integrating the resulting in- equality over[τ,t]we have, for any positive real constantsεandε0,

eµtku(t)k2 ≤eµτkψk2E

h+ (µγ+ε+ε0)

Z t

τ

eµsku(s)k2ds+ 1 γ

Z t

τ

eµskf(s)k2ds +

2B22 ε

+ K

22

ε0 eµt

µ

+2K1 Z t

τ

eµskusk2E

hds + 2B

12

ε Z t

τ

Z 0

heµt0ku(t0+s)k2ds dt0.

(2.13)

Let us estimate the last term in (2.13) using the initial condition in (2.1). We have Z t

τ

Z 0

heµt0ku(t0+s)k2ds dt0 =

Z 0

h

Z t

τ

eµseµ(t0+s)ku(t0+s)k2dt0ds

≤eµh Z 0

h

Z t

τh

eµσku(σ)k2dσds

=eµhh Z τ

τheµσku(σ)k2dσ+

Z t

τ

eµσku(σ)k2

e

µ(τ+h)h µ kψk2E

h +eµhh Z t

τ

eµσku(σ)k2dσ.

(2.14)

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Substituting (2.14) into (2.13) we get eµtku(t)k2≤ eµτkψk2E

h+ µγ+ε+ε0+2B

12eµhh ε

! Z t

τ

eµsku(s)k2ds +2B

2 1eµhh

µε eµτkψk2E

h+

2B22 ε + K

22

ε0 eµt

µ + 1

γ Z t

τ

eµskf(s)k2ds+2K1 Z t

τ

eµskusk2E

hds.

(2.15)

Using (2.10) we can chooseε= γ2 and ε0 = γ

2 −µ4B

2 1eµhh

γ (2.16)

in (2.15) to obtain

eµtku(t)k2≤eµτ 1+ 4B

12eµhh µγ

! kψk2E

h+

4B22 γ + K

22

ε0 eµt

µ + 1

γ Z t

τ

eµskf(s)k2ds+2K1 Z t

τ

eµskusk2E

hds.

(2.17)

Since ku(s)k ≤ kψkEh, ∀s ∈ [τ−h,τ], then we can replace t in (2.17) by t+σ, with σ∈[−h, 0], to deduce that

eµtkutk2E

h ≤ M(t) +L

Z t

τ

eµskusk2E

hds, whereL=2K1eµhand

M(t) =eµ(τ+h) 1+ 4B

12eµhh µγ

! kψk2E

h+

4B22 γ +K

22

ε0

eµ(t+h) µ + e

µh

γ Z t

τ

eµskf(s)k2ds.

The above inequality implies that eµtkutk2E

h ≤eL(tτ)M(τ) +eLt Z t

τ

eLsM0(s)ds. (2.18) Performing the calculations in (2.18) using M(t) above and the fact thatµ > L by (2.11), we find the following estimate for the solution of (2.1)

kutk2E

h ≤c1kψk2E

he(Lµ)te(µL)τ+ −L µ−L c2+ e

µh

γ Z t

kf(s)k2ds, (2.19) where

c1= eµh 1+ 4B

12eµhh µγ

!

and c2= 4B22

γ + K

22

ε0 eµh

µ . (2.20)

Now, assume that (A8)(ii) holds. For this case we replace (2.14) by eµtku(t)k2 ≤eµτkψk2E

h+ (µγ+ε+ε0)

Z t

τ

eµsku(s)k2ds +2B

21eµhh

µε eµτkψk2E

h+

2B22 ε +K

22

ε0 eµt

µ + 1

γ Z t

τ

eµskf(s)k2ds+2B

21eµhh ε

Z t

τ

eµskusk2E

hds.

(2.21)

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Since thatµ< γ2, then we can chooseε= γ2 andε0 = γ2µin (2.21) and proceed as before to obtain

eµtku(t)k2≤eµ(τ+h) 1+ 4B

21eµhh µγ

! kψk2E

h +

4B22 γ +K

22

ε0

eµ(t+h) µ + e

µh

γ Z t

τ

eµskf(s)k2ds+L Z t

τ

eµskusk2E

hds,

(2.22)

where L = γ4B21e2µhh. By (2.12) we see that µ > L. Therefore, we can deduce the estimate (2.19) with c1 andc2 as in (2.20), with ε0 = γ2µ. Similarly, we can treat the case (A8)(iii) to obtain the estimate

kutk2E

h ≤c01kψk2E

heµteµτ+2c02+ e

µh

γ Z t

kf(s)k2ds, (2.23) where

c01 =2eµh and c02= e

µh

µ2 B22+K22

. (2.24)

From (2.19) or (2.23) and Theorem2.5we conclude that the solution of (2.1) exists globally.

Next, let us prove that the map I is continuous. Fix τ < T < ∞,ψ ∈ Eh and consider ψ1 ∈ Eh such that kψψ1kEh < 1. Let us denote byv = v(t)the solution of (2.1) with initial condition v(s) = ψ1(sτ), ∀s ∈ [τ−h,τ]. Using the estimate (2.19) or (2.23) we can find a positive constantK0depending onkψkEh andTsuch thatkutkEh ≤K0andkvtkEh ≤K0, for all τ≤ t≤T. Then, using the integral representations ofuandv and Lemma2.3it follows that

ku(t)−v(t)k ≤ kψ(0)−ψ1(0)k+

Z t

τ

kB(s,us)− B(s,vs)kds

≤ kψψ1kEh +L(K0)

Z t

τ

kus−vskEhds.

(2.25)

Replacing t in (2.25) by t+σ, with σ ∈ [−h, 0], taking into account that ku(t+σ)− v(t+σ)kEh ≤ kψψ1kEh ift+στ, we obtain

kut−vtkEh ≤ kψψ1kEh+L(K0)

Z t

τ

kus−vskEhds, ∀τ≤t ≤T.

Then, by Gronwall’s inequality, we conclude that kut−vtkEh ≤ eL(K0)(Tτ)kψψ1kEh, which implies the continuity ofI.

3 Existence of a pullback attractor

By Theorem 2.8 we can associate to the initial value problem (2.1) a process{U(t,τ)}tτ of continuous maps U(t,τ) in Eh defined by U(t,τ)ψ = ut, where τ ≤ t and u = u(t) is the global solution of (2.1). In this section, we establish the existence of a pullback attractor for the process {U(t,τ)}tτ using the results obtained in [16]. We are interested in the existence of a pullback attractor for a family of sets depending on time (see [16, Section 3]). Motivated by the estimate (2.19) we consider the setRµ of all functionsr:R→(0,)such that

t→−lime(µL)tr2(t) =0. (3.1)

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Let us denote byDµthe class of all families ˆD={D(t);t ∈R}of nonempty subsets ofEhsuch that D(t) ⊂ BEh[0;rDˆ(t)] := {ψ ∈ Eh;kψkEh ≤ rDˆ(t)}, for some radius rDˆ ∈ Rµ. For the case (A8)(iii) we consider in (3.1) L = 0. In what follows, we will assume that (A8)(i) or (A8)(ii) holds. Suitable modifications will be indicated for the case (A8)(iii). We will also consider L as in the proof of Theorem2.8and the constantsc1,c2, c10 andc20 given by (2.20) and (2.24).

Lemma 3.1. Assume that (A1)–(A8) hold. Then, the family Bˆµ of closed balls Bµ(t) =BEh[0;Rµ(t)], where for each t∈R, the radius Rµ(t)is defined by

R2µ(t) = −L µ−L c2+e

µh

γ Z t

kf(s)k2ds+1, (3.2) is pullbackDµ-absorbing for the process{U(t,τ)}tτ.

Proof. Sinceµ> L, then using (A7), we have

t→−lime(µL)tR2µ(t) = lim

t→−e(µL)t

2µ−L µ−L c2+ e

µh

γ Z t

kf(s)k2ds+1

=0,

which shows that ˆBµ ∈ Dµ. Now, fixed t ∈ R and ˆD ∈ Dµ, there exists a τ0 = τ0(t, ˆD) ≤ t such that

e(µL)τr2Dˆ(τ)<c11e(µL)t, for anyττ0. Then, for anyψ∈ D(τ), using (2.19) we obtain

kU(t,τ)ψk2E

h ≤c1r2Dˆ(τ)e(µL)τe(Lµ)t+−L µ−L c2+ e

µh

γ Z t

kf(s)k2ds

≤R2µ(t).

Therefore,U(t,τ)D(τ)⊂ Bµ(t), for allττ0, which proves that the family ˆBµ is pullback Dµ-absorbing for the process{U(t,τ)}tτ.

In Lemma3.1, in the case (A8)(iii), we take L = 0 and replace c2 by c02 in (3.2). Next, let us prove an estimate for the tails of the solutionsu=u(t)of (2.1) when the initial conditions uτ = ψbelong to Bµ(τ).

Lemma 3.2. Assume that (A1)–(A8) hold. Let Bˆµ be the pullback Dµ-absorbing family defined in Lemma3.1. Then, for anyε>0and any t0 < T, there existτ0=τ0(ε,t0,T, ˆBµ)and a positive integer k=k(ε,T, ˆBµ), such that

s∈[−maxh,0]

|n|>2k

|un(t+s)|2 <ε,ττ0, t ∈[t0,T], for any solution u=u(t)of (2.1)with initial condition uτ ∈ Bµ(τ).

Proof. Assume that (A8)(i) holds. Similarly, we treat the case (A8)(ii). Letuτ =ψ∈Bµ(τ)and consider the corresponding solution u = u(t) of (2.1) defined in [τ,∞). Let θ ∈ C1(R+;R) be a function such thatθ ≡ 0 on [0, 1],θ ≡ 1 on[2,∞), 0 ≤ θ ≤ 1, and |θ0(t)| ≤ 2,∀t ≥ 0.

Let v = (vn(t))nZ, where vn(t) = θ(|nk|)un(t), with k > 0 fixed in Z. In order to simplify notation, we will writeθn =θ |nk|

,kwkθ =+n=− θn|wn|2andkutk2E

h,θ =maxs∈[−h,0]kut(s)k2θ. Taking the imaginary part of the inner product of equation (2.1) with vin`2 we find

1 2

d

dt(u,v)`2 +γ(u,v)`2 =Im(f,v)`2 −Im(Au,v)`2−Im(g(t,ut),v)`2, ∀t≥ τ. (3.3)

(11)

Let us estimate the terms on the right-hand side of (3.3). Since ψ ∈ Bµ(τ) then, using (2.19), we see that

ku(t)k ≤r0, ∀t∈ [τ,T],

withr0 = (c1+1)Rµ(T). Moreover, by the definition ofθ, we have that|θn+mθn| ≤ 2kmand

|θn+mθn| ≤2. Then,

−Im(Au(t),v(t))`2 =−Im (

J(0)ku(t)k2θ+

+ n=−

+ m

=1

J(m)(θn+mθn)un+m(t)un(t) )

+ n=−

+ m

=1

|J(m)| |θn+mθn| |un+m(t)| |un(t)| ≤ν(T,k,l), whereν(T,k,l) = 2k lm=1m|J(m)|+2 ∑+m=l+1|J(m)|r02, l≥1.

Using the hypotheses (A1) and (A4) and proceeding as in the proof of Lemma 2.6 we obtain the estimate

Im(g(t,ut),v(t))`2

+ n=−

θn|g1,n(t,un(t−ρ(t)))| |un(t)|

+

+ n=−

θn

Z 0

h

|bn(s,un(t+s))|ds|un(t)|

≤(K1ku(t−ρ(t))kθ+K2,θ)kukθ +

"

B1 Z 0

h

ku(t+s)k2θds 1/2

+B2,θ

# kukθ,

where B2,θ =+n=− θn R0

hβ2,n(s)ds21/2

andK2,θ = +n=− θnk22,n1/2

. In addition, we know that

−Im(f(t),v(t))`21

2γkf(t)k2θ+ γ

2ku(t)k2θ. Therefore,

d

dtku(t)k2θ+γku(t)k2θ1

γkfk2θ+2(K1ku(t−ρ(t))kθ+K2,θ)ku(t)kθ +2

"

B1 Z 0

h

ku(t+s)k2θds 1/2

+B2,θ

#

ku(t)kθ+2ν(T,k,l), (3.4)

for all τ≤t≤ T.

Now, we multiply (3.4) byeµtand use the inequalities 2

"

B1 Z 0

h

ku(t+s)k2θds 1/2

+B2,θ

#

kukθ4B

21

γ Z 0

h

ku(t+s)k2θds+ 4B

2,θ2

γ + γ 2 kuk2θ,

2(K1ku(t−ρ(t))kθ+K2,θ)ku(t)kθ ≤2K1kutk2E

h,θ+ K

22,θ

ε0 +ε0kuk2θ,

(12)

whereε0 >0, to find d

dt eµtku(t)k2θµγ 2 +ε0

eµtku(t)k2θ+ 1

γeµtkf(t)k2θ+2K1eµtkutk2E

h,θ

+ 4B

2,θ2

γ + K

22,θ

ε0

!

eµt+2ν(T,k,l)eµt +4B

21eµt γ

Z 0

h

ku(t+s)k2θds, ∀τ≤ t≤T.

(3.5)

Integrating (3.5) over[τ,t]and using the following estimate analogous to (2.14) Z t

τ

Z 0

heµt0ku(t+s)k2θds dt0e

µ(τ+h)h

µ kψk2Eh+eµhh Z t

τ

eµsku(s)k2θds, we obtain

eµtku(t)k2θ ≤eµτ 1+4B

12eµhh µγ

! kψk2E

h+ µγ

2 +ε0+4B

21eµhh γ

! Z t

τ

eµsku(s)k2θds + 4B

22,θ

γ + K

22,θ

ε0 +2ν(T,k,l)

!eµt µ +2K1

Z t

τ

eµskusk2E

h,θds + 1

γ Z t

τ

eµskf(s)k2θds.

By condition (2.10) we can chooseε0 as in (2.16) in the above inequality to obtain eµtku(t)k2θ ≤eµτ 1+4B

21eµhh µγ

! kψk2E

h+ 4B

22,θ

γ + K

2,θ2

ε0 +2ν(T,k,l)

!eµt µ +2K1

Z t

τ

eµskusk2E

h,θds+ 1 γ

Z t

τ

eµskf(s)k2θds.

(3.6)

Replacing t by t+σ, with σ ∈ [−h, 0] in (3.6) and using the inequality ku(t+σ)k = kψ(t+σ)k ≤ kψkEh, valid fort+σ <τ, we deduce that

eµtkutk2E

h,θ ≤ Mθ(t) +L Z t

τ

eµskusk2E

h,θds, (3.7)

whereL=2K1eµhand

Mθ(t) =eµ(τ+h) 1+4B

12eµhh µγ

! kψk2E

h+ 4B

22,θ

γ + K

22,θ

ε0 +2ν(T,k,l)

!eµ(t+h) µ +e

µh

γ Z t

τ

eµskf(s)k2θds.

We know thatµ> L. Then, from (3.7) andψ∈ Bµ(τ), we obtain kutk2E

h,θ ≤c1R2µ(τ)e(Lµ)te(µL)τ+ −L

µ−L c2,θ+2(2µ−L) µ(µ−L)e

µhν(T,k,l) + e

µh

γ Z t

kf(s)k2θds, ∀t ≥τ,

(3.8)

参照

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