# Continuous Dependence on a Parameter of Exponential Attractors for Nonclassical Diffusion Equations

Authors: Gang Wang; Chaozhu Hu
Journal: Discrete Dynamics in Nature and Society
 (2020)

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## Abstract

In this paper, a new abstract result is given to verify the continuity of exponential attractors with respect to a parameter for the underlying semigroup. We do not impose any compact embedding on the main assumptions in the abstract result which is different from the corresponding result established by Efendiev et al. in 2004. Consequently, it can be used for equations whose solutions have no higher regularity. As an application, we prove the continuity of exponential attractors inH01 for a class of nonclassical diffusion equations with initial datum in H01.


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## Body

## 1. Introduction

In this paper, we study the existence and robustness of exponential attractors for the following nonclassical diffusion equation:(1)utϵ−ϵΔutϵ−Δuϵ+fuϵ=g,x∈Ω,t>0,with the initial-boundary value conditions(2)uϵx,0=u0x,x∈Ω,uϵ=0,on∂Ω,where ϵ∈0,1 and Ω⊂ℝN N≥3 is a bounded open set with smooth boundary ∂Ω. When ϵ=0, it turns out to be the classical reaction-diffusion equation. We assume that g∈L2Ω and the nonlinearity f∈C1ℝ,ℝ satisfies the following (see, e.g., [1]):(F1): there existsν>0 such that f′s≥−ν, ∀s∈ℝ(F2): there existsκ1>0 such that f′s≤κ11+s2/N−2, ∀s∈ℝ(F3):liminfs⟶∞Fs/s2≥0, where Fs=∫0sfrdr(F4): there existsκ2>0 such that liminfs⟶∞sfs−κ2Fs/s2≥0Nonclassical diffusion equations appear in fluid mechanics, soil mechanics, and heat conduction theory (see, e.g., [2]). The long-time behavior of solutions to nonclassical diffusion equations has been extensively studied by many authors for both autonomous and nonautonomous cases [3–9].The global attractor plays an important role in the study of long-time behavior of infinite dimension systems arising from physics and mechanics. It is a compact invariant set and attracts uniformly the bounded sets of the phase space. However, the rate of attraction may be arbitrary, and it may be sensible to perturbations. These drawbacks can be overcome by creating the notion of the exponential attractor [10], which is a compact, positively invariant set of finite dimension and exponentially attracts each bounded set. The existence of the exponential attractor has been extensively studied since 1994, see e.g., [5, 10–17].As discussed in [12], exponential attractors are to be more robust objects under perturbations than global attractors. In general, global attractors are only upper semicontinuous with respect to perturbations, and the lower semicontinuity property is much more delicate and can be established only for some particular cases. However, one can prove the continuity of exponential attractors under perturbations in many cases [5, 13]. In particular, for problems (1) and (2), the existence of a pullback attractor was shown by Anh and Bao in [3] for the subcritical case in H01Ω. They also proved the upper semicontinuity of the pullback attractors. However, this upper semicontinuity of the pullback attractor was established only in L2Ω, and the upper semicontinuity in H01Ω remains an open problem. In this paper, we not only prove the upper and lower semicontinuity of the exponential attractor but also show these continuities in a stronger space, i.e., H01Ω when the initial value only belongs to H01Ω.In [12], see also [17], Efendiev et al. gave an abstract result about the robustness of exponential attractors (Theorem 4.4 in [12]). A main assumption called “compact Lipschitz condition” was proposed in that theorem. The main difficulty when we apply this result to problems (1) and (2) is that the solution to problems (1) and (2) has no regularity as ϵ>0 [3]. For example, if the initial datum u0 belongs to H01Ω, the solution with initial u0=u0 is always in H01Ω and has no higher regularity. Thus, it is impossible to verify the “compact Lipschitz condition” when we want to prove the continuity of exponential attractors in H01Ω. Motivated by [18], we modify the result in [12] to adapt it to our case. Moreover, some of the coefficients are allowed to be dependent on the parameter ϵ which relax the conditions, see Theorem 1 in the following.The rest of this paper is organized as follows. In Section2, we formulate and prove the main abstract result, i.e., Theorem 1. In Section 3, we apply Theorem 1 to the dynamical system generated by (1) and (2) to prove the continuity of the exponential attractors, and we consider two cases according to the constant ν in this section.Throughout this paper, we denote by⋅X the norm of a Banach space X. The inner product and norm of L2ℝn are written as ⋅,⋅ and ⋅, respectively. We also use ur to denote the norm of u∈Lrℝn r≥1,r≠2 and u to denote the modular of u. Letter c is a generic positive constant independent of ϵ which may change its values from line to line even in the same line (sometimes for the special case, we also denote different positive constants by cii=1,2,…).


## 2. The Abstract Result and Its Proof

In this section, we modify Theorem 4.4 in [12] to adapt to problems (1) and (2). We start with the definition of exponential attractors.Definition 1.
LetE be a metric space, B is a bounded set in E, and let S:B⟶B be a map. We define a discrete semigroup Sn,n∈ℤ+ by Snx:=S∘⋯∘S (n times). A set ℳ⊂B is an exponential attractor for the map if the following properties hold:(1)
The setℳ is compact in E and has finite fractal dimension.(2)
The setℳ is positively invariant with respect to S, i.e., Sℳ⊂ℳ.(3)
There exist positive constantsα0 and β0 such that(3)distESnB,ℳ≤α0e−β0n,where distEC1,C2 denotes the Hausdorff semidistance between C1 and C2 in E given by(4)distEC1,C2=supx∈C1infy∈C2x−yE,forC1,C2⊂E.Definition 2.
LetX be a complete metric space endowed with the metric d and M be a bounded closed set in X. Assume that ϱ is a pseudometric [19] defined on M. Let B⊂M and ε>0.(i)
A subsetU in B is said to be ε,ϱ-distinguishable if ϱx,x′>ε for any x,x′∈U,x≠x′. We denote by mϱB,ε the maximal cardinality of an ε,ϱ-distinguishable subset of B.(ii)
Pseudometricϱ is said to be compact on M iff mϱM,ε is finite for every ε>0.(iii)
For anyr>0, we define a local r,ε,ϱ-capacity of set M by the formula(5)CϱM;r,ε=suplnmϱB,ε:B⊂M,diamB≤2r.
We now state and prove the main abstract theorem. The proof is essentially a combination of that in [18] and that of [12].Theorem 1.
LetX be a Banach space and B be a bounded set in X. We assume that there exists a family of operators Sϵ:B⟶B, ϵ∈0,ϵ0, which satisfy, for any ϵ∈0,ϵ0,(i)
Sϵ is Lipschitz on B, i.e., there exists Lϵ>0 such that(6)Sϵx1−Sϵx2X≤Lϵx1−x2X,∀x1,x2∈B.(ii)
There exist constantsη and Kϵ and compact seminorms n1ϵx and n2ϵx on B such that(7)Sϵx1−Sϵx2X≤ηx1−x2X+Kϵn1ϵx1−x2+n2ϵSϵx1−Sϵx2,for anyx1,x2∈B, where 0<η<1 is independent of ϵ and Kϵ>0 is a constant which may be dependent on ϵ(seminorm nx on B is said to be compact iff for any subset B′⊂B if there exists a sequence xn⊂B′ such that nxm−xn⟶0 as n,m⟶∞).(iii)
∀ϵ∈0,ϵ0, ∀i∈ℕ, ∀x∈B,(8)Sϵix−S0ixX≤κiϵα,where κ>0 and α∈0,1 are constants independent of ϵ and for mapping V, Vi=V∘⋯∘V (i times).
Then,∀ϵ∈0,ϵ0, and the discrete dynamical system generated by the iterations of Sϵ possesses an exponential attractor ℳϵ on B such that(1)
The fractal dimension ofℳϵ is bounded in X:(9)dimf,Xℳϵ≤cϵ:=ln21+η−1lnmϵ4Kϵ1+Lϵ21/21−η,wheredimf,XA denotes the fractal dimension of A in X and mϵR is the maximal number of pairs xi,yi in X×X possessing the properties(10)xiX2+yiX2≤R2,n1εxi−xj+n2εyi−yj>1,i≠j.(2)
ℳϵ attracts B in X, uniformly with respect to ϵ,(11)distXSϵiB,ℳϵ≤c1e−c2i,c2>0,i∈ℕ,wherec1 and c2 are independent of ϵ.(3)
The familyℳϵ,ϵ∈0,ϵ0 is continuous at 0:(12)distsym,Xℳϵ,ℳ0≤c3ϵc4,where c3 and c4∈0,1 are independent of ϵ and distsym,X denotes the symmetric Hausdorff distance in X between sets defined by(13)distsym,XA,B:=maxdistXA,B,distXB,A.Proof.
For any fixedϵ∈0,ϵ0, we set ϱϵx,y=Kϵn1ϵx−y+n2ϵSϵx−Sϵy; then, ϱϵ is compact on B in the sense of Definition 2. From [18], we see that the local r,ρ,ϱϵ-capacity of the set B admits the estimate(14)CϱϵB;r,ρ≤lnmϵ2Kϵ1+Lϵ21/2rρ,where mϵR is the maximal number of pairs xi,yi in X×X possessing the properties(15)xiX2+yiX2≤R2,n1ϵxi−xj+n2ϵyi−yj>1,i≠j.
We assume diamB=2R. Let δ=1−η/2 and xi1:i1=1,…,n1 be a maximal δR,ϱϵ-distinguishable subset of B. Then, from (14), we have(16)n1=mϱϵB,δR≤expCϱϵB;R,δR≤mϵ2Kϵ1+Lϵ21/2δ:=Pϵ,B=∪i1=1n1Bi1,Bi1=υ∈B:ϱϵυ,xi1≤δR.
Therefore,(17)SϵB=∪i1=1n1SϵBi1.
Ify1,y2∈Bi1, then from (7), we have(18)Sϵy1−Sϵy2X≤ηy1−y2X+ϱϵy1,xi1+ϱϵy2,xi1≤2ηR+2δR:=2qR,where q=η+δ=1+η/2<1. We set Eϵ0=xi1i1=1n1 and Eϵ1=SϵEϵ0; then,(19)Eϵ1⊂SϵB,SϵEϵ0⊂Eϵ1,♯Eϵ1≤Pϵ≤Pϵ2,distXSϵB,Eϵ1≤2qR.
Next, for any fixedi1∈1,…,ni1, we assume that xi1,i2:i2=1,…,ni1,2 be a maximal δqR,ϱϵ-distinguishable subset of SϵBi1. Then,(20)ni1,2=mϱϵSϵBi1,δqR≤expCϱϵSϵB;qR,δqR≤expCϱϵB;qR,δqR≤Pϵ,SϵB=∪i1=1n1∪i2=1ni1,2Bi1,i2,Bi1,i2=υ∈SϵBi1:ϱϵυ,xi1,i2≤δqR.
Therefore,(21)Sϵ2B=∪i1=1n1∪i2=1ni1,2SϵBi1,i2.
Ify1,y2∈Bi1,i2, then from (7), we have(22)Sϵy1−Sϵy2X≤ηy1−y2X+ϱϵy1,xi1,i2+ϱϵy2,xi1,i2≤2ηqR+2δqR=2q2R.
We setEϵ2=SϵEϵ1∪Sϵxi1,i2, and then we have(23)Eϵ2⊂Sϵ2B,SϵEϵ1⊂Eϵ2,♯Eϵ2≤Pϵ2+Pϵ2≤Pϵ3,distXSϵ2B,Eϵ2≤2q2R.
By the induction procedure, we can find setsEϵi, i∈ℕ, enjoy the following properties:(24)Eϵi⊂SϵiB,(25)SϵEϵi⊂Eϵi+1,(26)♯Eϵi≤Pϵi+1,(27)distXSϵiB,Eϵi≤2qiR.
We now define the exponential attractor for the mapS0:B⟶B as follows:(28)ℳ0′=∪i=1∞E0i,ℳ0=ℳ0′¯X.
Then, from (24)–(27), we see that ℳ0 is indeed an exponential attractor for the map S0:B⟶B (see [12]). For ϵ∈0,ϵ0, one can also construct exponential attractors for Sϵ:B⟶B as above. However, they are not the ones we needed. Totally similar to [12], one can construct exponential attractors ℳϵ based on E0i. We note that the only difference between our construction procedure and that of in [12] is that the number Pϵ here may be dependent on ϵ. However, (26) only contributes to the fractal dimension of ℳϵ. Therefore, (11) and (12) in Theorem 1 hold true. Finally, assertion (9) is a direct result of [18]. The proof is completed.Remark 1.
Similar to [12], we can give an explicit expression for c4, i.e., c4=−αlnq/lnκ−lnq. If Lϵ, Kϵ, n1ϵx, and n2ϵx are all independent ϵ, we can obtain the uniform bound with respect to ϵ of the fractal dimension, and we will apply this abstract result in the last section.


## 3. Application to the Nonclassical Diffusion Equation

### 3.1. Some Useful Estimates of the Solution

Since−Δ−1 is a continuous compact operator in L2Ω, by the classical spectral theorem, there exist a sequence λjj=1∞, 0<λ1≤λ2≤⋯≤λj⟶∞ as j⟶∞ and a family of elements ejj=1∞ of D−Δ, which forms an orthogonal basis in both L2D and H01D such that −Δej=λjej,∀j∈N. Given m, let Xm=spane1,…,em and Pm:L2D⟶Xm be the projection operator. For any υ∈H01Ω, we write υ=Pmυ+I−Pmυ:=υ1+υ2.Lemma 1.
(see [1]). Let ϵ>0. Assume (F1)–(F4) hold and g∈L2Ω. Then, for each u0∈H01Ω, problems (1) and (2) have a unique solution uϵ=uϵt=uϵt;u0 with uϵ∈C0,T,H01Ω and utϵ∈L20,T;H01Ω for any T>0. Moreover, for any fixed t>0, uϵ is continuous in u0.Lemma 2.
(see [1]). Let ϵ=0. Assume (F1)–(F4) hold and g∈L2Ω. Then, for each u0∈H01Ω, problems (1) and (2) have a unique solution u0=u0t=u0t;u0 which satisfies(29)u0∈C0,T;H01Ω∩L20,T;H2Ω,∀T>0.
Also for any fixedt>0, u0 is continuous in u0.
From Lemmas1 and 2, we can define a semigroup Sϵt:H01Ω⟶H01Ω by the expression(30)Sϵtu0=uϵt,t≥0,where uϵt is the solution of (1) and (2).Lemma 3.
(see [1]). Assume (F1)–(F4) hold and g∈L2Ω. Then, for any bounded set D⊂H01Ω, there exist positive constants ED, c0, and T1D such that, for any solution uϵ of problems (1) and (2),(31)∇uϵt≤ED,t≥0,∇uϵt≤c0,t≥T1D,provided u0∈D, where ED, c0, and T1D are independent of ϵ.
SetB=υ∈H01Ω:υH01Ω≤c0, where c0 is the constant in Lemma 3; then, B is a uniformly (with respect to ϵ) bounded absorbing set for Sϵt in H01Ω. We note that the absorbing time is independent of ϵ, and we can choose T1=T1B large enough such that SϵtB⊂B for any t≥T0 and any ϵ∈0,1.
The following lemma gives several priori estimates for the derivativeutϵt of the solution to (1) and (2).Lemma 4.
Under the assumptions (F1)–(F4), for anyD∈H01Ω, D is bounded, and there exists T2D, which is independent of ϵ such that, for all ϵ∈0,1,(32)utϵt2+ϵ∇utϵt2≤c,(33)∫tt+1∇utϵs2ds≤c,for any u0∈D and any t≥T2D, where c is a constant independent of ϵ.Proof.
We first take the inner product of (1) with uϵt in L2Ω to get(34)12ddt∇uϵ2+ϵ∇uϵ2+∇uϵ2+fuϵ,uϵ=g,uϵ.
By condition (F4), for anyθ>0, there exists cθ>0 satisfying(35)fu,u−κ2∫ΩFu+θu2+cθ≥0,∀u∈H01Ω.
Putting (35) into (34), it yields that(36)ddtuϵ2+ϵ∇uϵ2+2∇uϵ2+2κ2∫ΩFuϵ≤2g,uϵ+2θuϵ2+2cθ.
By condition (F3), for anyγ>0, there exists cγ>0 satisfying(37)∫ΩFu+γu2+cγ≥0,∀u∈H01Ω.
Combining (36) and (37), we get(38)ddtuϵ2+ϵ∇uϵ2+2∇uϵ2≤2g,uϵ+2θuϵ2+2cθ+2κ2γuϵ2+2κ2cγ,that is,(39)ddtuϵ2+ϵ∇uϵ2+uϵ2≤2g,uϵ+2θuϵ2+2cθ+2κ2γuϵ2+2κ2cγ.
We chooseα1>0 small enough such that α1λ1<1 and 1−α1>1/2, and apply Poincare inequality in (39) to get(40)ddtuϵ2+ϵ∇uϵ2+α1λ1uϵ2+1−α1uϵ2≤2g,uϵ+2θuϵ2+2cθ+2κ2γuϵ2+2κ2cγ.
Choosingθ,γ small enough and using Ho¨lder inequality, we can get from (40) that(41)ddtuϵ2+ϵ∇uϵ2+α1λ12uϵ2+12∇uϵ2≤c.
We setσ=α1λ1/2; then, 0<σ<1/2. From (41), we obtain(42)ddtuϵ2+ϵ∇uϵ2+σuϵ2+ϵ∇uϵ2≤c,that is,(43)ddteσt∇uϵt2+ϵ∇uϵt2≤ceσt.
Integrating the above inequality from 0 tot, it yields that(44)eσt∇uϵt2+ϵ∇uϵt2≤u02+ϵ∇u02+ceσt,t≥0.
Now, we consider (36) again.
If0<κ2≤1, we can get from (36) that(45)ddtuϵ2+ϵ∇uϵ2+∇uϵ2+κ2∇uϵ2+2∫ΩFuϵ≤2g,uϵ+2θuϵ2+2cθ.
Using a similar process as (39)–(42), we deduce that, for θ small enough,(46)ddtuϵ2+ϵ∇uϵ2+σuϵ2+ϵ∇uϵ2+κ2∇uϵ2+2∫ΩFuϵ≤c.
Therefore,(47)ddteσtuϵ2+ϵ∇uϵ2+κ2eσt∇uϵ2+2∫ΩFuϵ≤ceσt.
Ifκ2>1, we can get from (36) that(48)ddtuϵ2+ϵ∇uϵ2+2∇uϵ2+2∫ΩFuϵ+2κ2−2∫ΩFuϵ≤2g,uϵ+2θuϵ2+2cθ.
Applying (37) to the above, we obtain(49)ddtuϵ2+ϵ∇uϵ2+∇uϵ2+∇uϵ2+2∫ΩFuϵ≤2g,uϵ+2θuϵ2+2cθ+2κ2−2γuϵ2+2κ2−2cγ.
Following a similar process as (39)–(42), we deduce that, for θ, γ small enough,(50)ddtuϵ2+ϵ∇uϵ2+σuϵ2+ϵ∇uϵ2+∇uϵ2+2∫ΩFuϵ≤c.
Therefore,(51)ddteσtuϵ2+ϵ∇uϵ2+eσt∇uϵ2+2∫ΩFuϵ≤ceσt.
From (47) and (51), we see that, for any κ2>0,(52)ddteσtuϵ2+ϵ∇uϵ2+c5eσt∇uϵ2+2∫ΩFuϵ≤ceσt.where c5=κ2 when 0<κ2≤1 and c5=1 when κ2>1.
For anyt>0, we integrate (52) over [t, t + 1] to get(53)∫tt+1eσs∇uϵs2+2∫ΩFuϵsds≤ceσtuϵt2+ϵ∇uϵt2+ceσt.
Putting (44) into (53), we have, for any t>0,(54)∫tt+1eσs∇uϵs2+2∫ΩFuϵsds≤cu02+ϵ∇u02+ceσt.
Next, we multiply (1) with ut and integrate it in Ω to get(55)utϵ2+ϵ∇utϵ2+12ddt∇uϵ2+2∫ΩFuϵ=g,utϵ.
Applying Ho¨lder inequality to the above we get(56)utϵ2+ϵ∇utϵ2+ddt∇uϵ2+2∫ΩFuϵ≤c,t>0.
Thus,(57)eσtutϵ2+ϵ∇utϵ2+ddteσt∇uϵ2+2∫ΩFuϵ≤ceσt+ceσt∇uϵ2+2∫ΩFuϵ,t>0.
Dropping the first term in the left-hand side of (57), from (54) and using uniform Gronwall inequality, we conclude that(58)eσt∇uϵt2+2∫ΩFuϵt≤cu02+ϵ∇u02+ceσt,t≥1.
We now differentiate equation (1) with respect to t to get(59)uttϵ−ϵΔuttϵ−Δutϵ+f′uϵutϵ=0.
Taking the inner product of (59) with utϵ in L2Ω, we obtain(60)ddtutϵ2+ϵ∇utϵ2+2∇utϵ2≤cutϵ2.
Since0<σ<1/2, we have(61)ddtutϵ2+ϵ∇utϵ2+σutϵ2+ϵ∇utϵ2≤cutϵ2.
Hence,(62)ddteσtutϵ2+ϵ∇utϵ2≤ceσtutϵ2+ϵ∇utϵ2.
Integrating (57) over (t, t + 1) and using (58), we deduce that(63)∫tt+1eσsutϵs2+ϵ∇utϵs2ds≤cu02+ϵ∇u02+ceσt,t≥1.
Combining (62) and (63) and using the uniform Gronwall inequality, we obtain that(64)utϵt2+ϵ∇utϵt2≤ce−σtu02+ϵ∇u02+c,t≥2.
Thus, for any boundedD⊂H01Ω, there exists T2D≥2 which is independent of ϵ such that, for any t≥T2D and any u0∈D,(65)utϵt2+ϵ∇utϵt2≤c,where c is independent of ϵ. This proves assertion (32) in Lemma 4.
Finally, integrating (60) over t,t+1, then assertion (33) can be easily verified by using (32). The proof is completed.Lemma 5.
Under conditions (F1)–(F4), for any two solutions of (1) and (2) uϵ1t, uϵ2t (corresponding to ϵ1, ϵ2∈0,1, respectively), with initials starting from B and for any υ∈L2Ω, we have(66)∫Ωfuϵ1t−fuϵ2tυdx≤c6∇wϵ1,ϵ2tυ,t≥0,where wϵ1,ϵ2t=uϵ1t−uϵ2t and c6 is a constant independent of ϵ.Proof.
By (F2), we have(67)∫Ωfuϵ1t−fuϵ2tυdx≤∫Ωlt,ϵ1,ϵ2wϵ1,ϵ2tυdx≤c∫Ω1+uϵ1t2/N−2+uϵ2t2/N−2wϵ1,ϵ2tυdx≤c∫Ωwϵ1,ϵ2tυdx+c∫Ωuϵ1t2/N−2wϵ1,ϵ2tυdx+c∫Ωuϵ2t2/N−2wϵ1,ϵ2tυdx,where lt,ϵ1,ϵ2=∫01f′suϵ1t+1−suϵ2tds.
From (31) and the Sobolev embedding H01Ω⊂L2N/N−2Ω, we have that, for t≥0,(68)∫Ωfuϵ1t−fuϵ2tυdx≤cwϵ1,ϵ2υ+cuϵ12N/N−22/N−2wϵ1,ϵ22NN−2υ+cuϵ22N/N−22/N−2wϵ1,ϵ22N/N−2υ≤c∇wϵ1,ϵ2υ+c∇uϵ12/N−2∇wϵ1,ϵ2υ+c∇uϵ22/N−2∇wϵ1,ϵ2υ≤c6∇wϵ1,ϵ2υ.
The proof is completed.


### 3.2. The Main Result for a General Case:ν∈ℝ+

Now, we verify the conditions in Theorem1 for Sϵt in this case. We first verify condition (i), i.e., the Lipschitz continuity in H01 (actually uniform Lipschtz continuity in H01 since the coefficient is independent of ϵ).Lemma 6.
Under assumptions (F1)–(F4), we have, for anyϵ∈0,1 and any x1,x2∈B,(69)Sϵtx1−Sϵtx2H01Ω≤ec62tx1−x2H01Ω,t≥0,where c6 is the constant in Lemma 5.Proof.
Assume thatu1ϵt and u2ϵt are two solutions of (1) and (2) starting from x1, x2∈B, respectively. We consider the difference wϵt=u1ϵt−u2ϵt, and then wϵt satisfies(70)wtϵt−ϵΔwtϵt−Δwϵt+lt,ϵwϵt=0.
Multiplying (70) with wtϵt and integrating it in Ω, we get(71)wtϵt2+ϵ∇wtϵt2+12ddt∇wϵt2≤∫Ωlt,ϵwϵtwtϵtdx.
Applying Lemma5, we obtain(72)wtϵt2+ϵ∇wtϵt2+12ddt∇wϵt2≤c6∇wϵtwtϵt,t≥0.
We use Ho¨lder inequality to get(73)ddt∇wϵt2≤c62∇wϵt2.
Then, the result follows immediately from Gronwall inequality. The proof is completed.Lemma 7.
Under assumptions (F1)–(F4), we have, for anyx1,x2∈B and any η∈0,1,(i)
Forϵ∈0,ϵ0: there exist T3=T3η, K>0, and compact seminorm n1ϵ,t on B such that(74)Sϵtx1−Sϵtx2H01Ω≤ηx1−x2H01Ω+Kn1ϵ,tx1−x2,t≥T3.(ii)
Forϵ=0: there exist a positive integer M, T3=T3η, and K>0 such that(75)S0tx1−S0tx2H01Ω≤ηx1−x2H01Ω+Kn2S0tx1−S0tx2,t≥T3,for some ϵ0>0, where n2x=PMxH01Ω, M and K are independent of ϵ, η, and t and n1ϵ,t depend on ϵ and t.Proof.
We first take the inner product of (70) with kwϵ in L2Ω (k is a constant which will be fixed later) to get(76)kwtϵ,wϵ+kϵ∇wtϵ,∇wϵ+k∇wϵ2+k∫Ωlt,ϵwϵ2=0.
Using (F1), we get(77)k∇wϵ2≤kwtϵwϵ+kϵ∇wtϵ∇wϵ+kνwϵ2.
Combining (72) and (77), we get(78)wtϵ2+ϵ∇wtϵ2+12ddt∇wϵ2+k∇wϵ2≤kwtϵwϵ+kϵ∇wtϵ∇wϵ+kνwϵ2+c6∇wϵwtϵ.
For the right-hand side of (78), we use Ho¨lder inequality to get(79)kwtϵwϵ+kϵ∇wtϵ∇wϵ+kνwϵ2+c6∇wϵwtϵ≤kν+k22wϵ2+wtϵ2+ϵ2∇wtϵ2+k2ϵ2+c622∇wϵ2.
Putting the above inequality into (78), we obtain(80)ddt∇wϵ2+2k∇wϵ2≤2kν+k2wϵ2+k2ϵ+c62∇wϵ2.
Fixk=c62, and then choose ϵ0 such that ϵ<1/c62, ∀ϵ≤ϵ0. Then, we get from (80) that(81)ddt∇wϵ2+c7∇wϵ2≤cwϵ2,ϵ∈0,ϵ0,c7>0,that is,(82)ddtec7t∇wϵ2≤cec7twϵ2,ϵ∈0,ϵ0.
We integrate the above inequality over0,t to get(83)∇wϵt2≤e−c7t∇wϵ02+c∫0twϵs2ds,ϵ∈0,ϵ0.
For anyη∈0,1, we can choose T3=T3η>0 such that e−c7T3≤η2. Therefore, from (83),(84)∇wϵt2≤η2∇wϵ02+c∫0twϵs2ds,t≥T3,ϵ∈0,ϵ0.
Thus,(85)∇wϵt≤η∇wϵ0+K∫0twϵs2ds1/2,t≥T3,ϵ∈0,ϵ0.
Now, we setn1ϵ,tx1−x2=∫0twϵs2ds1/2. We need to show that n1ϵ,t is compact on B for any t>0 and any ϵ∈0,1. From condition (F2), we deduce that, for any s∈ℝ,(86)fss≤cs+s2+s2N−2/N−2.
Hence,(87)fu,u≤cu2+u2N−2/N−22N−2/N−2≤cu2+u2N−2/N−22N−2/N−2≤cu2+∇u2N−2/N−2≤c∇u2+∇u2N−2/N−2,∀u∈H01Ω,where we have used the Sobolev embedding H01Ω⊂L2N/N−2Ω in the above inequality. From (35) and (87), we obtain(88)κ2∫ΩFu≤θu2+cθ+c∇u2+∇u2N−2/N−2,∀u∈H01Ω,for any θ>0. For any t≥0, we can see from (32) that(89)∫0t∇uϵs2ds≤ct,ϵ∈0,1,for any u0∈B and c is independent of ϵ. Next, we integrate (56) from 0 to t, and we get(90)ϵ∫0t∇utϵs2ds≤ct+c∇u02+2∫ΩFu0.
We first fixθ small enough, and then put (88) into (90) to get, for all u0∈B,(91)ϵ∫0t∇utϵs2ds≤ct+c,ϵ∈0,1.
From (89) and (91), we can easily see that, for any t>0 and any fixed ϵ∈0,1, ℬε,t=uεs:s∈0,t,uε0=u0∈B is bounded in W1,20,t;H01Ω. Therefore, from the compact embedding W1,20,t;H01Ω⊂L20,t;L2Ω, we deduce that n1ϵ,t is compact in B. Thus, assertion (i) is proved.
Assertion (ii) can be easily proved by multiplying (70) (with ϵ=0) with −Δw20, and this procedure is elementary; here, we omit it (see, e.g., [20] for details). The proof is completed.Lemma 8.
Under assumptions (F1)–(F4), we have, for anyϵ∈0,1, any i∈ℕ, and any x∈B, there exists a positive function ℓt independent of ϵ such that, for all t≥T2+1,(92)Sϵitx−S0itxH01Ω≤ℓitϵ1/4.where T2=T2B (see Lemma 4).Proof.
We assume thatuϵt ϵ∈0,1 and u0t are the solutions for the following equations:(93)utϵ−ϵΔutϵ−Δuϵ+fuϵ=g,t>0,x∈Ω,uϵ∂Ω=0,uϵx,0=x1x∈B,x∈Ω,ut0−Δu0+fu0=g,t>0,x∈Ω,u0∂Ω=0,u0x,0=x2x∈B,x∈Ω,respectively. Set w=uϵ−u0, and then w satisfies(94)wt−ϵΔutϵ−Δw+ltw=0,where lt=∫01f′suϵt+1−su0tds. Multiplying (94) with wt and using Lemma 5, it yields(95)wt2+ϵ∇utϵ,∇wt+12ddt∇w2≤c6∇wwt,t≥0.
We chooseσ1=maxc62,2ν; then, the above inequality implies(96)2ϵ∇utϵ2+ddt∇w2≤σ1∇w2+2ϵ∇utϵ∇ut0.
Therefore,(97)ddte−σ1t∇w2≤2ϵe−σ1t∇utϵ∇ut0.
By Lemma4, we see that, for t≥T2,(98)ddte−σ1t∇w2≤cϵ1/2∇ut0≤cϵ1/2+cϵ1/2∇ut02.
Multiplying (94) with w and integrating it in Ω, we obtain(99)12ddtw2+ϵ∇utϵ,∇w+∇w2≤νw2,that is,(100)ddtw2+2∇w2≤σ1w2+2ϵ∇utϵ∇w.
The above inequality implies(101)ddte−σ1tw2+2e−σ1t∇w2≤2ϵ∇utϵ∇w.
Dropping the second term in the left-hand side of (101) and integrating it over 0,t, we get(102)e−σ1twt2≤w02+cϵ∫0t∇utϵs2ds1/2∫0t∇ws2ds1/2.
From (91), we see that, for any ϵ∈0,1,(103)∫0t∇utϵs2ds≤1ϵct+1.
Combining (31), (102), and (103), we deduce that(104)e−σ1twt2≤w02+ct+1ϵ1/2,t≥0.
Now, we integrate (101) over (t, t + 1) to get(105)∫tt+1e−σ1s∇ws2ds≤ce−σ1twt2+cϵ∫tt+1∇utϵs2ds1/2∫tt+1∇ws2ds1/2.
By using (31) and (33), we get that, for t≥T2,(106)∫tt+1∇utϵs2ds≤c,∫tt+1∇ws2ds≤c.
Therefore, from (104)–(106), we deduce that(107)∫tt+1e−σ1s∇ws2ds≤cw02+ct+1ϵ1/2+cϵ≤cw02+ct+1ϵ1/2,t≥T2.
On the contrary, using (33) again, we obtain(108)∫tt+1cϵ1/2+cϵ1/2∇ut0s2ds≤cϵ1/2,t≥T2.
Combining (98), (107), and (108) and using a similar process as the proof of uniform Gronwall inequality, it yields that, for all t≥T2,(109)e−σ1t+1∇wt+12≤cw02+ct+1ϵ1/2+cϵ1/2≤cw02+ct+1ϵ1/2.
This implies that there exists a positive functionℓ′t independent of ϵ such that, for t≥T2+1,(110)Sϵtx1−S0tx2H01Ω≤ℓ′tϵ1/4+ℓ′tx1−x2H01Ω.
Whenx1=x2=x, we get(111)Sϵtx−S0txH01Ω≤ℓ′tϵ1/4,t≥T2+1.
From (110) and (111), we obtain for t≥T2+1,(112)Sϵitx−S0itxH01Ω≤ℓ′tϵ1/4+ℓ′tSϵi−1tx−S0i−1txH01Ω≤ℓ′tϵ1/4+ℓ′tℓ′tϵ1/4+ℓ′tSϵi−2tx−S0i−2txH01Ω=ℓ′tϵ1/4+ℓ′2tϵ1/4+ℓ′2tSϵi−2tx−S0i−2txH01Ω≤ℓ′t+ℓ′2t+⋯+ℓ′itϵ1/4+ℓ′i−1tSϵtx−S0txH01Ω≤2ℓ′tiϵ1/4+ℓ′itϵ1/4≤ℓitϵ1/4.
The proof is completed.
For any fixedη∈0,1, we set T0η=maxT1,T2+1,T3η and Sϵ=SϵT0, and then Sϵ: B⟶B. From Lemmas 6–8, we see that the discrete system Sϵnn (ϵ∈0,ϵ0) satisfies all the assumptions in Theorem 1. Therefore, we have the following.Theorem 2.
Let assumptions (F1)–(F4) hold, and then∀ϵ∈0,ϵ0, and the discrete dynamical system Sϵnn defined above possesses an exponential attractor ℳϵd on B such that(1)
The fractal dimension ofℳϵd is bounded in H01Ω:(113)dimf,H01Ωℳϵd≤cϵ:=ln21+η−1lnmϵ4K1+L21/21−η,whereL=ec62T0 (see Lemma 6) and mϵR is the maximal number of pairs xi,yi in H01Ω×H01Ω possessing the properties(114)ϵ∈0,ϵ0:xiH01Ω2+yiH01Ω2≤R2,n1ϵxi−xj>1,i≠j,ϵ=0:xiH01Ω2+yiH01Ω2≤R2,n2yi−yj>1,i≠j.(2)
ℳϵd attracts B in H01Ω, uniformly with respect to ϵ,(115)distH01ΩSϵiB,ℳϵd≤c8e−c9i,c9>0,i∈ℕ,wherec8 and c9 are independent of ϵ.(3)
The familyℳϵd,ϵ∈0,ϵ0 is continuous at 0 in H01Ω:(116)distsym,H01Ωℳϵd,ℳ0d≤c10ϵc11,where c11=−1/4ln1+η/2/lnℓT0−ln1+η/2 and c10 are independent of ϵ.
To obtain the corresponding result for continuous systemSϵt defined in (30), we need to show the Ho¨lder continuity with respect to time t and the initial conditions. In general, it is difficult to verify the uniform (with respect to ϵ) Ho¨lder continuity with respect to time t when t is small. However, when t is large enough, we have the following.Lemma 9.
Let assumptions (F1)–(F4) hold. Then, for anyT≥T0, the semigroup Sϵt defined in (30) is uniformly Ho¨lder continuous on T0,T×B, i.e.,(117)Sϵt1x1−Sϵt2x2H01Ω≤cx1−x2H01Ω+t1−t21/2,ϵ∈0,1,for x1, x2∈B and T0≤t1, t2≤T and c is independent of ϵ.Proof.
The Lipschitz continuity with respect to the initial conditions is an immediate corollary of Lemma6. It remains to prove the continuity with respect to time t. From Lemmas 1 and 2, we know that the solution uϵt∈C0,T,H01Ω for the initial value belongs to B. Therefore, for any T0≤t1≤t2≤T,(118)uϵt2−uϵt1=∫t1t2utϵsds,which implies that(119)uϵt1−uϵt2H01Ω=∫t1t2utϵsdsH01Ω≤∫t1t2utϵsH01Ωds≤t1−t21/2∫t1t2utϵsH01Ω2ds1/2.
To estimate (119), we integrate (60) from t1 to t2 to get(120)∫t1t2∇utϵs2ds≤c∫t1t2utϵs2ds+cutϵt12+ϵ∇utϵt12.
Sincet1≥T0, we apply (32) to (120) to get(121)∫t1t2∇utϵs2ds≤c,where c is independent of ϵ. Putting (121) into (119), we obtain the result. The proof is completed.
Our main result in this section reads as follows:Theorem 3.
Let assumptions (F1)–(F4) hold. Then, for everyϵ∈0,ϵ0, the semigroup Sϵt generated by equations (1) and (2) possesses an exponential attractor ℳϵc in H01Ω. Moreover, these exponential attractors can be constructed such that(1)
The fractal dimension ofℳϵc is bounded in H01Ω:(122)dimf,H01Ωℳϵc≤cϵ+2.(2)
ℳϵc attracts B in H01Ω, uniformly with respect to ϵ:(123)distH01ΩSϵtB,ℳϵc≤c12e−c13t,c13>0,where c12 and c13 are independent of ϵ.(3)
The familyℳϵc,ϵ∈0,ϵ0 is continuous at 0 in H01Ω:(124)distsym,H01Ωℳϵc,ℳ0c≤c14ϵc11,where c11=−1/4ln1+η/2/lnℓT0−ln1+η/2 and c14 are independent of ϵ.Proof.
Settingℳϵc:=∪t∈T0,2T0Sϵtℳϵd, then we have(125)ℳϵc:=∪t∈T0,2T0Sϵtℳϵd=∪t∈T0,2T0Sϵt−T0SϵT0ℳϵd=∪t∈0,T0SϵtSϵT0ℳϵd.
It follows from its definition that the setSϵT0ℳϵd is also an exponential attractor for the discrete dynamical system Sϵnn. Thus, ℳϵc is the set we needed (for details, see, e.g., [12]). The proof is completed.


### 3.3. The Main Result for a Special Case:0<ν<λ1

If, in addition,ν<λ1, we can get the uniform bound (with respect to ϵ) for the fractal dimension. To this end, we need show that the constants Lϵ and Kϵ and the compact seminorms n1ϵ and n2ϵ in Theorem 1 are all independent of ϵ for the discrete system generated by problems (1) and (2). From the above, we have proved the constant Lϵ is independent of ϵ; thus, we only need the following lemma:Lemma 10.
Let (F1)–(F4) hold; in addition, we assume thatν<λ1, and then we have for any x1,x2∈B and any η∈0,1, there exist constants ϵ0∈0,1, K>0, and T0′η≥T0η (T0 is defined above) and a compact seminorm nx on B such that(126)SϵT0′x1−SϵT0′x2H01Ω≤ηx1−x2H01Ω+Knx1−x2+nSϵT0′x1−SϵT0′x2,ϵ∈0,ϵ0,where T0′η, K, and the compact seminorm nx are all independent of ϵ.Proof.
We assume thatu1ϵt and u2ϵt are two solutions of (1) starting from x1,x2∈B, respectively. We consider the difference wϵt=u1ϵt−u2ϵt; then, wϵt satisfies(127)wtϵt−ϵΔwtϵt−Δwϵt+lt,ϵwϵt=0.
We first take the inner product of (127) with kw2ϵ in L2Ω (k is a constant which will be fixed later) to get(128)kw2,tϵ,w2ϵ+kϵ∇w2,tϵ,∇w2ϵ+k∇w2ϵ2+k∫Ωlt,ϵwϵw2ϵ=0.
Using Lemma5, we get(129)k∇w2ϵ2≤kw2,tϵw2ϵ+kϵ∇w2,tϵ∇w2ϵ+kc∇wϵw2ϵ.
Next, we take the inner product of (127) with w2,tϵ in L2Ω to get(130)w2,tϵ2+ϵ∇w2,tϵ2+12ddt∇w2ϵ2+∫Ωlt,ϵwϵw2,tϵ=0.
We apply Lemma5 again to get(131)w2,tϵ2+ϵ∇w2,tϵ2+12ddt∇w2ϵ2≤c6∇wϵw2,tϵ.
Combining (129) and (131), we get(132)w2,tϵ2+ϵ∇w2,tϵ2+12ddt∇w2ϵ2+k∇w2ϵ2≤kw2,tϵw2ϵ+kϵ∇w2,tϵ∇w2ϵ+kc∇wϵw2ϵ+c6∇wϵw2,tϵ.
For the right-hand side of (132), we use Ho¨lder inequality to get(133)kw2,tϵw2ϵ+kϵ∇w2,tϵ∇w2ϵ+kc∇wϵw2ϵ+c6∇wϵw2,tϵ≤ck2w2ϵ2+12w2,tϵ2+ϵ2∇w2,tϵ2+k2ϵ2∇w2ϵ2+k2∇wϵ2+ckw2ϵ2+c622∇wϵ2+12w2,tϵ2≤ck2w2ϵ2+12w2,tϵ2+ϵ2∇w2,tϵ2+k2ϵ2∇w2ϵ2+k2∇w2ϵ2+k2∇w1ϵ2+ckw2ϵ2+c622∇w2ϵ2+c622∇w1ϵ2+12w2,tϵ2.
Putting the above inequality into (132), we obtain(134)ddt∇w2ϵ2+k∇w2ϵ2≤ck2+kw2ϵ2+c+k∇w1ϵ2+c62+k2ϵ∇w2ϵ2.
Fixk=2c62, and then choose ϵ0 such that ϵ<1/4c62, ∀ϵ≤ϵ0. In the following, we assume ϵ∈0,ϵ0. Then, we get from (134) that(135)ddt∇w2ϵ2+c15∇w2ϵ2≤cw2ϵ2+c∇w1ϵ2,c15>0.
Using Poincare inequality, we obtain(136)ddt∇w2ϵ2+c15∇w2ϵ2≤cλm+1−1∇w2ϵ2+c∇w1ϵ2≤cλm+1−1∇wϵ2+c∇w1ϵ2.
From Lemma6, it yields that(137)ddtec15t∇w2ϵ2≤cλm+1−1ec15tec62t∇wϵ02+cec15t∇w1ϵ2.
We integrate the above inequality over0,t to get(138)∇w2ϵt2≤e−c15t∇w2ϵ02+cλm+1−1ec62t∇wϵ02+c∫0t∇w1ϵs2ds≤e−c15t+cλm+1−1ec62t∇wϵ02+c∫0t∇w1ϵs2ds.
For anyη∈0,1, we first choose T0′=T0′η>T0η such that e−c15T0′≤η2/2, and then fix a positive integer M=Mη such that cλM+1−1ec62T0′≤η2/2. Therefore, from (138),,(139)∇w2ϵT0′2≤η2∇wϵ02+c∫0T0′∇w1ϵs2ds.
Therefore,(140)∇wϵT0′2=∇w1ϵT0′2+∇w2ϵT0′2≤η2∇wϵ02+c∫0T0′∇w1ϵs2ds+∇w1ϵT0′2,that is,(141)∇wϵT0′≤η∇wϵ0+K∫0T0′∇w1ϵs2ds1/2+∇w1ϵT0′.
We setnx=PMxH01. Then, ∇w1ϵT0′=nSϵT0′x1−SϵT0′x2, and n is obviously compact on B.
To estimate the term∫0T0′∇w1ϵs2ds1/2, we multiply (127) with w1ϵt and integrate it in Ω:(142)ddtw1ϵt2+ϵ∇w1ϵt2+2∇w1ϵt2+2lt,ϵwϵt,w1ϵt=0.
By using (F1) and Poincare inequality, we can get from the above inequality(143)ddtw1ϵt2+ϵ∇w1ϵt2+2∇w1ϵt2≤2νw1ϵt2≤2νλ1−1∇w1ϵt2.
Therefore,(144)ddtw1ϵt2+ϵ∇w1ϵt2+β∇w1ϵt2≤0,where β=21−νλ1−1>0. Integrating (144) from 0 to T0′, we get(145)∫0T0′∇w1ϵs2ds≤cw1ϵ02+ϵ∇w1ϵ02≤c∇w1ϵ02.
Thus, for anyϵ∈0,ϵ0,(146)∫0T0′∇w1εs2ds1/2≤cnx1−x2.
Putting (146) into (141), one can obtain the result. The proof is completed.Theorem 5.
Let assumptions (F1)–(F4) hold andν<λ1. Then, for every ϵ∈0,ϵ0, the semigroup Sϵt generated by equations (1) and (2) possesses an exponential attractor ℳϵc in H01Ω. Moreover, these exponential attractors can be constructed such that (2) and (3) in Theorem 3 hold and the fractal dimension of ℳϵ is uniformly (with respect to ϵ) bounded in H01Ω, i.e.,(147)dimf,H01Ωℳϵc≤c,where c is a constant independent of ϵ.


## 3.1. Some Useful Estimates of the Solution

Since−Δ−1 is a continuous compact operator in L2Ω, by the classical spectral theorem, there exist a sequence λjj=1∞, 0<λ1≤λ2≤⋯≤λj⟶∞ as j⟶∞ and a family of elements ejj=1∞ of D−Δ, which forms an orthogonal basis in both L2D and H01D such that −Δej=λjej,∀j∈N. Given m, let Xm=spane1,…,em and Pm:L2D⟶Xm be the projection operator. For any υ∈H01Ω, we write υ=Pmυ+I−Pmυ:=υ1+υ2.Lemma 1.
(see [1]). Let ϵ>0. Assume (F1)–(F4) hold and g∈L2Ω. Then, for each u0∈H01Ω, problems (1) and (2) have a unique solution uϵ=uϵt=uϵt;u0 with uϵ∈C0,T,H01Ω and utϵ∈L20,T;H01Ω for any T>0. Moreover, for any fixed t>0, uϵ is continuous in u0.Lemma 2.
(see [1]). Let ϵ=0. Assume (F1)–(F4) hold and g∈L2Ω. Then, for each u0∈H01Ω, problems (1) and (2) have a unique solution u0=u0t=u0t;u0 which satisfies(29)u0∈C0,T;H01Ω∩L20,T;H2Ω,∀T>0.
Also for any fixedt>0, u0 is continuous in u0.
From Lemmas1 and 2, we can define a semigroup Sϵt:H01Ω⟶H01Ω by the expression(30)Sϵtu0=uϵt,t≥0,where uϵt is the solution of (1) and (2).Lemma 3.
(see [1]). Assume (F1)–(F4) hold and g∈L2Ω. Then, for any bounded set D⊂H01Ω, there exist positive constants ED, c0, and T1D such that, for any solution uϵ of problems (1) and (2),(31)∇uϵt≤ED,t≥0,∇uϵt≤c0,t≥T1D,provided u0∈D, where ED, c0, and T1D are independent of ϵ.
SetB=υ∈H01Ω:υH01Ω≤c0, where c0 is the constant in Lemma 3; then, B is a uniformly (with respect to ϵ) bounded absorbing set for Sϵt in H01Ω. We note that the absorbing time is independent of ϵ, and we can choose T1=T1B large enough such that SϵtB⊂B for any t≥T0 and any ϵ∈0,1.
The following lemma gives several priori estimates for the derivativeutϵt of the solution to (1) and (2).Lemma 4.
Under the assumptions (F1)–(F4), for anyD∈H01Ω, D is bounded, and there exists T2D, which is independent of ϵ such that, for all ϵ∈0,1,(32)utϵt2+ϵ∇utϵt2≤c,(33)∫tt+1∇utϵs2ds≤c,for any u0∈D and any t≥T2D, where c is a constant independent of ϵ.Proof.
We first take the inner product of (1) with uϵt in L2Ω to get(34)12ddt∇uϵ2+ϵ∇uϵ2+∇uϵ2+fuϵ,uϵ=g,uϵ.
By condition (F4), for anyθ>0, there exists cθ>0 satisfying(35)fu,u−κ2∫ΩFu+θu2+cθ≥0,∀u∈H01Ω.
Putting (35) into (34), it yields that(36)ddtuϵ2+ϵ∇uϵ2+2∇uϵ2+2κ2∫ΩFuϵ≤2g,uϵ+2θuϵ2+2cθ.
By condition (F3), for anyγ>0, there exists cγ>0 satisfying(37)∫ΩFu+γu2+cγ≥0,∀u∈H01Ω.
Combining (36) and (37), we get(38)ddtuϵ2+ϵ∇uϵ2+2∇uϵ2≤2g,uϵ+2θuϵ2+2cθ+2κ2γuϵ2+2κ2cγ,that is,(39)ddtuϵ2+ϵ∇uϵ2+uϵ2≤2g,uϵ+2θuϵ2+2cθ+2κ2γuϵ2+2κ2cγ.
We chooseα1>0 small enough such that α1λ1<1 and 1−α1>1/2, and apply Poincare inequality in (39) to get(40)ddtuϵ2+ϵ∇uϵ2+α1λ1uϵ2+1−α1uϵ2≤2g,uϵ+2θuϵ2+2cθ+2κ2γuϵ2+2κ2cγ.
Choosingθ,γ small enough and using Ho¨lder inequality, we can get from (40) that(41)ddtuϵ2+ϵ∇uϵ2+α1λ12uϵ2+12∇uϵ2≤c.
We setσ=α1λ1/2; then, 0<σ<1/2. From (41), we obtain(42)ddtuϵ2+ϵ∇uϵ2+σuϵ2+ϵ∇uϵ2≤c,that is,(43)ddteσt∇uϵt2+ϵ∇uϵt2≤ceσt.
Integrating the above inequality from 0 tot, it yields that(44)eσt∇uϵt2+ϵ∇uϵt2≤u02+ϵ∇u02+ceσt,t≥0.
Now, we consider (36) again.
If0<κ2≤1, we can get from (36) that(45)ddtuϵ2+ϵ∇uϵ2+∇uϵ2+κ2∇uϵ2+2∫ΩFuϵ≤2g,uϵ+2θuϵ2+2cθ.
Using a similar process as (39)–(42), we deduce that, for θ small enough,(46)ddtuϵ2+ϵ∇uϵ2+σuϵ2+ϵ∇uϵ2+κ2∇uϵ2+2∫ΩFuϵ≤c.
Therefore,(47)ddteσtuϵ2+ϵ∇uϵ2+κ2eσt∇uϵ2+2∫ΩFuϵ≤ceσt.
Ifκ2>1, we can get from (36) that(48)ddtuϵ2+ϵ∇uϵ2+2∇uϵ2+2∫ΩFuϵ+2κ2−2∫ΩFuϵ≤2g,uϵ+2θuϵ2+2cθ.
Applying (37) to the above, we obtain(49)ddtuϵ2+ϵ∇uϵ2+∇uϵ2+∇uϵ2+2∫ΩFuϵ≤2g,uϵ+2θuϵ2+2cθ+2κ2−2γuϵ2+2κ2−2cγ.
Following a similar process as (39)–(42), we deduce that, for θ, γ small enough,(50)ddtuϵ2+ϵ∇uϵ2+σuϵ2+ϵ∇uϵ2+∇uϵ2+2∫ΩFuϵ≤c.
Therefore,(51)ddteσtuϵ2+ϵ∇uϵ2+eσt∇uϵ2+2∫ΩFuϵ≤ceσt.
From (47) and (51), we see that, for any κ2>0,(52)ddteσtuϵ2+ϵ∇uϵ2+c5eσt∇uϵ2+2∫ΩFuϵ≤ceσt.where c5=κ2 when 0<κ2≤1 and c5=1 when κ2>1.
For anyt>0, we integrate (52) over [t, t + 1] to get(53)∫tt+1eσs∇uϵs2+2∫ΩFuϵsds≤ceσtuϵt2+ϵ∇uϵt2+ceσt.
Putting (44) into (53), we have, for any t>0,(54)∫tt+1eσs∇uϵs2+2∫ΩFuϵsds≤cu02+ϵ∇u02+ceσt.
Next, we multiply (1) with ut and integrate it in Ω to get(55)utϵ2+ϵ∇utϵ2+12ddt∇uϵ2+2∫ΩFuϵ=g,utϵ.
Applying Ho¨lder inequality to the above we get(56)utϵ2+ϵ∇utϵ2+ddt∇uϵ2+2∫ΩFuϵ≤c,t>0.
Thus,(57)eσtutϵ2+ϵ∇utϵ2+ddteσt∇uϵ2+2∫ΩFuϵ≤ceσt+ceσt∇uϵ2+2∫ΩFuϵ,t>0.
Dropping the first term in the left-hand side of (57), from (54) and using uniform Gronwall inequality, we conclude that(58)eσt∇uϵt2+2∫ΩFuϵt≤cu02+ϵ∇u02+ceσt,t≥1.
We now differentiate equation (1) with respect to t to get(59)uttϵ−ϵΔuttϵ−Δutϵ+f′uϵutϵ=0.
Taking the inner product of (59) with utϵ in L2Ω, we obtain(60)ddtutϵ2+ϵ∇utϵ2+2∇utϵ2≤cutϵ2.
Since0<σ<1/2, we have(61)ddtutϵ2+ϵ∇utϵ2+σutϵ2+ϵ∇utϵ2≤cutϵ2.
Hence,(62)ddteσtutϵ2+ϵ∇utϵ2≤ceσtutϵ2+ϵ∇utϵ2.
Integrating (57) over (t, t + 1) and using (58), we deduce that(63)∫tt+1eσsutϵs2+ϵ∇utϵs2ds≤cu02+ϵ∇u02+ceσt,t≥1.
Combining (62) and (63) and using the uniform Gronwall inequality, we obtain that(64)utϵt2+ϵ∇utϵt2≤ce−σtu02+ϵ∇u02+c,t≥2.
Thus, for any boundedD⊂H01Ω, there exists T2D≥2 which is independent of ϵ such that, for any t≥T2D and any u0∈D,(65)utϵt2+ϵ∇utϵt2≤c,where c is independent of ϵ. This proves assertion (32) in Lemma 4.
Finally, integrating (60) over t,t+1, then assertion (33) can be easily verified by using (32). The proof is completed.Lemma 5.
Under conditions (F1)–(F4), for any two solutions of (1) and (2) uϵ1t, uϵ2t (corresponding to ϵ1, ϵ2∈0,1, respectively), with initials starting from B and for any υ∈L2Ω, we have(66)∫Ωfuϵ1t−fuϵ2tυdx≤c6∇wϵ1,ϵ2tυ,t≥0,where wϵ1,ϵ2t=uϵ1t−uϵ2t and c6 is a constant independent of ϵ.Proof.
By (F2), we have(67)∫Ωfuϵ1t−fuϵ2tυdx≤∫Ωlt,ϵ1,ϵ2wϵ1,ϵ2tυdx≤c∫Ω1+uϵ1t2/N−2+uϵ2t2/N−2wϵ1,ϵ2tυdx≤c∫Ωwϵ1,ϵ2tυdx+c∫Ωuϵ1t2/N−2wϵ1,ϵ2tυdx+c∫Ωuϵ2t2/N−2wϵ1,ϵ2tυdx,where lt,ϵ1,ϵ2=∫01f′suϵ1t+1−suϵ2tds.
From (31) and the Sobolev embedding H01Ω⊂L2N/N−2Ω, we have that, for t≥0,(68)∫Ωfuϵ1t−fuϵ2tυdx≤cwϵ1,ϵ2υ+cuϵ12N/N−22/N−2wϵ1,ϵ22NN−2υ+cuϵ22N/N−22/N−2wϵ1,ϵ22N/N−2υ≤c∇wϵ1,ϵ2υ+c∇uϵ12/N−2∇wϵ1,ϵ2υ+c∇uϵ22/N−2∇wϵ1,ϵ2υ≤c6∇wϵ1,ϵ2υ.
The proof is completed.


## 3.2. The Main Result for a General Case:ν∈ℝ+

Now, we verify the conditions in Theorem1 for Sϵt in this case. We first verify condition (i), i.e., the Lipschitz continuity in H01 (actually uniform Lipschtz continuity in H01 since the coefficient is independent of ϵ).Lemma 6.
Under assumptions (F1)–(F4), we have, for anyϵ∈0,1 and any x1,x2∈B,(69)Sϵtx1−Sϵtx2H01Ω≤ec62tx1−x2H01Ω,t≥0,where c6 is the constant in Lemma 5.Proof.
Assume thatu1ϵt and u2ϵt are two solutions of (1) and (2) starting from x1, x2∈B, respectively. We consider the difference wϵt=u1ϵt−u2ϵt, and then wϵt satisfies(70)wtϵt−ϵΔwtϵt−Δwϵt+lt,ϵwϵt=0.
Multiplying (70) with wtϵt and integrating it in Ω, we get(71)wtϵt2+ϵ∇wtϵt2+12ddt∇wϵt2≤∫Ωlt,ϵwϵtwtϵtdx.
Applying Lemma5, we obtain(72)wtϵt2+ϵ∇wtϵt2+12ddt∇wϵt2≤c6∇wϵtwtϵt,t≥0.
We use Ho¨lder inequality to get(73)ddt∇wϵt2≤c62∇wϵt2.
Then, the result follows immediately from Gronwall inequality. The proof is completed.Lemma 7.
Under assumptions (F1)–(F4), we have, for anyx1,x2∈B and any η∈0,1,(i)
Forϵ∈0,ϵ0: there exist T3=T3η, K>0, and compact seminorm n1ϵ,t on B such that(74)Sϵtx1−Sϵtx2H01Ω≤ηx1−x2H01Ω+Kn1ϵ,tx1−x2,t≥T3.(ii)
Forϵ=0: there exist a positive integer M, T3=T3η, and K>0 such that(75)S0tx1−S0tx2H01Ω≤ηx1−x2H01Ω+Kn2S0tx1−S0tx2,t≥T3,for some ϵ0>0, where n2x=PMxH01Ω, M and K are independent of ϵ, η, and t and n1ϵ,t depend on ϵ and t.Proof.
We first take the inner product of (70) with kwϵ in L2Ω (k is a constant which will be fixed later) to get(76)kwtϵ,wϵ+kϵ∇wtϵ,∇wϵ+k∇wϵ2+k∫Ωlt,ϵwϵ2=0.
Using (F1), we get(77)k∇wϵ2≤kwtϵwϵ+kϵ∇wtϵ∇wϵ+kνwϵ2.
Combining (72) and (77), we get(78)wtϵ2+ϵ∇wtϵ2+12ddt∇wϵ2+k∇wϵ2≤kwtϵwϵ+kϵ∇wtϵ∇wϵ+kνwϵ2+c6∇wϵwtϵ.
For the right-hand side of (78), we use Ho¨lder inequality to get(79)kwtϵwϵ+kϵ∇wtϵ∇wϵ+kνwϵ2+c6∇wϵwtϵ≤kν+k22wϵ2+wtϵ2+ϵ2∇wtϵ2+k2ϵ2+c622∇wϵ2.
Putting the above inequality into (78), we obtain(80)ddt∇wϵ2+2k∇wϵ2≤2kν+k2wϵ2+k2ϵ+c62∇wϵ2.
Fixk=c62, and then choose ϵ0 such that ϵ<1/c62, ∀ϵ≤ϵ0. Then, we get from (80) that(81)ddt∇wϵ2+c7∇wϵ2≤cwϵ2,ϵ∈0,ϵ0,c7>0,that is,(82)ddtec7t∇wϵ2≤cec7twϵ2,ϵ∈0,ϵ0.
We integrate the above inequality over0,t to get(83)∇wϵt2≤e−c7t∇wϵ02+c∫0twϵs2ds,ϵ∈0,ϵ0.
For anyη∈0,1, we can choose T3=T3η>0 such that e−c7T3≤η2. Therefore, from (83),(84)∇wϵt2≤η2∇wϵ02+c∫0twϵs2ds,t≥T3,ϵ∈0,ϵ0.
Thus,(85)∇wϵt≤η∇wϵ0+K∫0twϵs2ds1/2,t≥T3,ϵ∈0,ϵ0.
Now, we setn1ϵ,tx1−x2=∫0twϵs2ds1/2. We need to show that n1ϵ,t is compact on B for any t>0 and any ϵ∈0,1. From condition (F2), we deduce that, for any s∈ℝ,(86)fss≤cs+s2+s2N−2/N−2.
Hence,(87)fu,u≤cu2+u2N−2/N−22N−2/N−2≤cu2+u2N−2/N−22N−2/N−2≤cu2+∇u2N−2/N−2≤c∇u2+∇u2N−2/N−2,∀u∈H01Ω,where we have used the Sobolev embedding H01Ω⊂L2N/N−2Ω in the above inequality. From (35) and (87), we obtain(88)κ2∫ΩFu≤θu2+cθ+c∇u2+∇u2N−2/N−2,∀u∈H01Ω,for any θ>0. For any t≥0, we can see from (32) that(89)∫0t∇uϵs2ds≤ct,ϵ∈0,1,for any u0∈B and c is independent of ϵ. Next, we integrate (56) from 0 to t, and we get(90)ϵ∫0t∇utϵs2ds≤ct+c∇u02+2∫ΩFu0.
We first fixθ small enough, and then put (88) into (90) to get, for all u0∈B,(91)ϵ∫0t∇utϵs2ds≤ct+c,ϵ∈0,1.
From (89) and (91), we can easily see that, for any t>0 and any fixed ϵ∈0,1, ℬε,t=uεs:s∈0,t,uε0=u0∈B is bounded in W1,20,t;H01Ω. Therefore, from the compact embedding W1,20,t;H01Ω⊂L20,t;L2Ω, we deduce that n1ϵ,t is compact in B. Thus, assertion (i) is proved.
Assertion (ii) can be easily proved by multiplying (70) (with ϵ=0) with −Δw20, and this procedure is elementary; here, we omit it (see, e.g., [20] for details). The proof is completed.Lemma 8.
Under assumptions (F1)–(F4), we have, for anyϵ∈0,1, any i∈ℕ, and any x∈B, there exists a positive function ℓt independent of ϵ such that, for all t≥T2+1,(92)Sϵitx−S0itxH01Ω≤ℓitϵ1/4.where T2=T2B (see Lemma 4).Proof.
We assume thatuϵt ϵ∈0,1 and u0t are the solutions for the following equations:(93)utϵ−ϵΔutϵ−Δuϵ+fuϵ=g,t>0,x∈Ω,uϵ∂Ω=0,uϵx,0=x1x∈B,x∈Ω,ut0−Δu0+fu0=g,t>0,x∈Ω,u0∂Ω=0,u0x,0=x2x∈B,x∈Ω,respectively. Set w=uϵ−u0, and then w satisfies(94)wt−ϵΔutϵ−Δw+ltw=0,where lt=∫01f′suϵt+1−su0tds. Multiplying (94) with wt and using Lemma 5, it yields(95)wt2+ϵ∇utϵ,∇wt+12ddt∇w2≤c6∇wwt,t≥0.
We chooseσ1=maxc62,2ν; then, the above inequality implies(96)2ϵ∇utϵ2+ddt∇w2≤σ1∇w2+2ϵ∇utϵ∇ut0.
Therefore,(97)ddte−σ1t∇w2≤2ϵe−σ1t∇utϵ∇ut0.
By Lemma4, we see that, for t≥T2,(98)ddte−σ1t∇w2≤cϵ1/2∇ut0≤cϵ1/2+cϵ1/2∇ut02.
Multiplying (94) with w and integrating it in Ω, we obtain(99)12ddtw2+ϵ∇utϵ,∇w+∇w2≤νw2,that is,(100)ddtw2+2∇w2≤σ1w2+2ϵ∇utϵ∇w.
The above inequality implies(101)ddte−σ1tw2+2e−σ1t∇w2≤2ϵ∇utϵ∇w.
Dropping the second term in the left-hand side of (101) and integrating it over 0,t, we get(102)e−σ1twt2≤w02+cϵ∫0t∇utϵs2ds1/2∫0t∇ws2ds1/2.
From (91), we see that, for any ϵ∈0,1,(103)∫0t∇utϵs2ds≤1ϵct+1.
Combining (31), (102), and (103), we deduce that(104)e−σ1twt2≤w02+ct+1ϵ1/2,t≥0.
Now, we integrate (101) over (t, t + 1) to get(105)∫tt+1e−σ1s∇ws2ds≤ce−σ1twt2+cϵ∫tt+1∇utϵs2ds1/2∫tt+1∇ws2ds1/2.
By using (31) and (33), we get that, for t≥T2,(106)∫tt+1∇utϵs2ds≤c,∫tt+1∇ws2ds≤c.
Therefore, from (104)–(106), we deduce that(107)∫tt+1e−σ1s∇ws2ds≤cw02+ct+1ϵ1/2+cϵ≤cw02+ct+1ϵ1/2,t≥T2.
On the contrary, using (33) again, we obtain(108)∫tt+1cϵ1/2+cϵ1/2∇ut0s2ds≤cϵ1/2,t≥T2.
Combining (98), (107), and (108) and using a similar process as the proof of uniform Gronwall inequality, it yields that, for all t≥T2,(109)e−σ1t+1∇wt+12≤cw02+ct+1ϵ1/2+cϵ1/2≤cw02+ct+1ϵ1/2.
This implies that there exists a positive functionℓ′t independent of ϵ such that, for t≥T2+1,(110)Sϵtx1−S0tx2H01Ω≤ℓ′tϵ1/4+ℓ′tx1−x2H01Ω.
Whenx1=x2=x, we get(111)Sϵtx−S0txH01Ω≤ℓ′tϵ1/4,t≥T2+1.
From (110) and (111), we obtain for t≥T2+1,(112)Sϵitx−S0itxH01Ω≤ℓ′tϵ1/4+ℓ′tSϵi−1tx−S0i−1txH01Ω≤ℓ′tϵ1/4+ℓ′tℓ′tϵ1/4+ℓ′tSϵi−2tx−S0i−2txH01Ω=ℓ′tϵ1/4+ℓ′2tϵ1/4+ℓ′2tSϵi−2tx−S0i−2txH01Ω≤ℓ′t+ℓ′2t+⋯+ℓ′itϵ1/4+ℓ′i−1tSϵtx−S0txH01Ω≤2ℓ′tiϵ1/4+ℓ′itϵ1/4≤ℓitϵ1/4.
The proof is completed.
For any fixedη∈0,1, we set T0η=maxT1,T2+1,T3η and Sϵ=SϵT0, and then Sϵ: B⟶B. From Lemmas 6–8, we see that the discrete system Sϵnn (ϵ∈0,ϵ0) satisfies all the assumptions in Theorem 1. Therefore, we have the following.Theorem 2.
Let assumptions (F1)–(F4) hold, and then∀ϵ∈0,ϵ0, and the discrete dynamical system Sϵnn defined above possesses an exponential attractor ℳϵd on B such that(1)
The fractal dimension ofℳϵd is bounded in H01Ω:(113)dimf,H01Ωℳϵd≤cϵ:=ln21+η−1lnmϵ4K1+L21/21−η,whereL=ec62T0 (see Lemma 6) and mϵR is the maximal number of pairs xi,yi in H01Ω×H01Ω possessing the properties(114)ϵ∈0,ϵ0:xiH01Ω2+yiH01Ω2≤R2,n1ϵxi−xj>1,i≠j,ϵ=0:xiH01Ω2+yiH01Ω2≤R2,n2yi−yj>1,i≠j.(2)
ℳϵd attracts B in H01Ω, uniformly with respect to ϵ,(115)distH01ΩSϵiB,ℳϵd≤c8e−c9i,c9>0,i∈ℕ,wherec8 and c9 are independent of ϵ.(3)
The familyℳϵd,ϵ∈0,ϵ0 is continuous at 0 in H01Ω:(116)distsym,H01Ωℳϵd,ℳ0d≤c10ϵc11,where c11=−1/4ln1+η/2/lnℓT0−ln1+η/2 and c10 are independent of ϵ.
To obtain the corresponding result for continuous systemSϵt defined in (30), we need to show the Ho¨lder continuity with respect to time t and the initial conditions. In general, it is difficult to verify the uniform (with respect to ϵ) Ho¨lder continuity with respect to time t when t is small. However, when t is large enough, we have the following.Lemma 9.
Let assumptions (F1)–(F4) hold. Then, for anyT≥T0, the semigroup Sϵt defined in (30) is uniformly Ho¨lder continuous on T0,T×B, i.e.,(117)Sϵt1x1−Sϵt2x2H01Ω≤cx1−x2H01Ω+t1−t21/2,ϵ∈0,1,for x1, x2∈B and T0≤t1, t2≤T and c is independent of ϵ.Proof.
The Lipschitz continuity with respect to the initial conditions is an immediate corollary of Lemma6. It remains to prove the continuity with respect to time t. From Lemmas 1 and 2, we know that the solution uϵt∈C0,T,H01Ω for the initial value belongs to B. Therefore, for any T0≤t1≤t2≤T,(118)uϵt2−uϵt1=∫t1t2utϵsds,which implies that(119)uϵt1−uϵt2H01Ω=∫t1t2utϵsdsH01Ω≤∫t1t2utϵsH01Ωds≤t1−t21/2∫t1t2utϵsH01Ω2ds1/2.
To estimate (119), we integrate (60) from t1 to t2 to get(120)∫t1t2∇utϵs2ds≤c∫t1t2utϵs2ds+cutϵt12+ϵ∇utϵt12.
Sincet1≥T0, we apply (32) to (120) to get(121)∫t1t2∇utϵs2ds≤c,where c is independent of ϵ. Putting (121) into (119), we obtain the result. The proof is completed.
Our main result in this section reads as follows:Theorem 3.
Let assumptions (F1)–(F4) hold. Then, for everyϵ∈0,ϵ0, the semigroup Sϵt generated by equations (1) and (2) possesses an exponential attractor ℳϵc in H01Ω. Moreover, these exponential attractors can be constructed such that(1)
The fractal dimension ofℳϵc is bounded in H01Ω:(122)dimf,H01Ωℳϵc≤cϵ+2.(2)
ℳϵc attracts B in H01Ω, uniformly with respect to ϵ:(123)distH01ΩSϵtB,ℳϵc≤c12e−c13t,c13>0,where c12 and c13 are independent of ϵ.(3)
The familyℳϵc,ϵ∈0,ϵ0 is continuous at 0 in H01Ω:(124)distsym,H01Ωℳϵc,ℳ0c≤c14ϵc11,where c11=−1/4ln1+η/2/lnℓT0−ln1+η/2 and c14 are independent of ϵ.Proof.
Settingℳϵc:=∪t∈T0,2T0Sϵtℳϵd, then we have(125)ℳϵc:=∪t∈T0,2T0Sϵtℳϵd=∪t∈T0,2T0Sϵt−T0SϵT0ℳϵd=∪t∈0,T0SϵtSϵT0ℳϵd.
It follows from its definition that the setSϵT0ℳϵd is also an exponential attractor for the discrete dynamical system Sϵnn. Thus, ℳϵc is the set we needed (for details, see, e.g., [12]). The proof is completed.


## 3.3. The Main Result for a Special Case:0<ν<λ1

If, in addition,ν<λ1, we can get the uniform bound (with respect to ϵ) for the fractal dimension. To this end, we need show that the constants Lϵ and Kϵ and the compact seminorms n1ϵ and n2ϵ in Theorem 1 are all independent of ϵ for the discrete system generated by problems (1) and (2). From the above, we have proved the constant Lϵ is independent of ϵ; thus, we only need the following lemma:Lemma 10.
Let (F1)–(F4) hold; in addition, we assume thatν<λ1, and then we have for any x1,x2∈B and any η∈0,1, there exist constants ϵ0∈0,1, K>0, and T0′η≥T0η (T0 is defined above) and a compact seminorm nx on B such that(126)SϵT0′x1−SϵT0′x2H01Ω≤ηx1−x2H01Ω+Knx1−x2+nSϵT0′x1−SϵT0′x2,ϵ∈0,ϵ0,where T0′η, K, and the compact seminorm nx are all independent of ϵ.Proof.
We assume thatu1ϵt and u2ϵt are two solutions of (1) starting from x1,x2∈B, respectively. We consider the difference wϵt=u1ϵt−u2ϵt; then, wϵt satisfies(127)wtϵt−ϵΔwtϵt−Δwϵt+lt,ϵwϵt=0.
We first take the inner product of (127) with kw2ϵ in L2Ω (k is a constant which will be fixed later) to get(128)kw2,tϵ,w2ϵ+kϵ∇w2,tϵ,∇w2ϵ+k∇w2ϵ2+k∫Ωlt,ϵwϵw2ϵ=0.
Using Lemma5, we get(129)k∇w2ϵ2≤kw2,tϵw2ϵ+kϵ∇w2,tϵ∇w2ϵ+kc∇wϵw2ϵ.
Next, we take the inner product of (127) with w2,tϵ in L2Ω to get(130)w2,tϵ2+ϵ∇w2,tϵ2+12ddt∇w2ϵ2+∫Ωlt,ϵwϵw2,tϵ=0.
We apply Lemma5 again to get(131)w2,tϵ2+ϵ∇w2,tϵ2+12ddt∇w2ϵ2≤c6∇wϵw2,tϵ.
Combining (129) and (131), we get(132)w2,tϵ2+ϵ∇w2,tϵ2+12ddt∇w2ϵ2+k∇w2ϵ2≤kw2,tϵw2ϵ+kϵ∇w2,tϵ∇w2ϵ+kc∇wϵw2ϵ+c6∇wϵw2,tϵ.
For the right-hand side of (132), we use Ho¨lder inequality to get(133)kw2,tϵw2ϵ+kϵ∇w2,tϵ∇w2ϵ+kc∇wϵw2ϵ+c6∇wϵw2,tϵ≤ck2w2ϵ2+12w2,tϵ2+ϵ2∇w2,tϵ2+k2ϵ2∇w2ϵ2+k2∇wϵ2+ckw2ϵ2+c622∇wϵ2+12w2,tϵ2≤ck2w2ϵ2+12w2,tϵ2+ϵ2∇w2,tϵ2+k2ϵ2∇w2ϵ2+k2∇w2ϵ2+k2∇w1ϵ2+ckw2ϵ2+c622∇w2ϵ2+c622∇w1ϵ2+12w2,tϵ2.
Putting the above inequality into (132), we obtain(134)ddt∇w2ϵ2+k∇w2ϵ2≤ck2+kw2ϵ2+c+k∇w1ϵ2+c62+k2ϵ∇w2ϵ2.
Fixk=2c62, and then choose ϵ0 such that ϵ<1/4c62, ∀ϵ≤ϵ0. In the following, we assume ϵ∈0,ϵ0. Then, we get from (134) that(135)ddt∇w2ϵ2+c15∇w2ϵ2≤cw2ϵ2+c∇w1ϵ2,c15>0.
Using Poincare inequality, we obtain(136)ddt∇w2ϵ2+c15∇w2ϵ2≤cλm+1−1∇w2ϵ2+c∇w1ϵ2≤cλm+1−1∇wϵ2+c∇w1ϵ2.
From Lemma6, it yields that(137)ddtec15t∇w2ϵ2≤cλm+1−1ec15tec62t∇wϵ02+cec15t∇w1ϵ2.
We integrate the above inequality over0,t to get(138)∇w2ϵt2≤e−c15t∇w2ϵ02+cλm+1−1ec62t∇wϵ02+c∫0t∇w1ϵs2ds≤e−c15t+cλm+1−1ec62t∇wϵ02+c∫0t∇w1ϵs2ds.
For anyη∈0,1, we first choose T0′=T0′η>T0η such that e−c15T0′≤η2/2, and then fix a positive integer M=Mη such that cλM+1−1ec62T0′≤η2/2. Therefore, from (138),,(139)∇w2ϵT0′2≤η2∇wϵ02+c∫0T0′∇w1ϵs2ds.
Therefore,(140)∇wϵT0′2=∇w1ϵT0′2+∇w2ϵT0′2≤η2∇wϵ02+c∫0T0′∇w1ϵs2ds+∇w1ϵT0′2,that is,(141)∇wϵT0′≤η∇wϵ0+K∫0T0′∇w1ϵs2ds1/2+∇w1ϵT0′.
We setnx=PMxH01. Then, ∇w1ϵT0′=nSϵT0′x1−SϵT0′x2, and n is obviously compact on B.
To estimate the term∫0T0′∇w1ϵs2ds1/2, we multiply (127) with w1ϵt and integrate it in Ω:(142)ddtw1ϵt2+ϵ∇w1ϵt2+2∇w1ϵt2+2lt,ϵwϵt,w1ϵt=0.
By using (F1) and Poincare inequality, we can get from the above inequality(143)ddtw1ϵt2+ϵ∇w1ϵt2+2∇w1ϵt2≤2νw1ϵt2≤2νλ1−1∇w1ϵt2.
Therefore,(144)ddtw1ϵt2+ϵ∇w1ϵt2+β∇w1ϵt2≤0,where β=21−νλ1−1>0. Integrating (144) from 0 to T0′, we get(145)∫0T0′∇w1ϵs2ds≤cw1ϵ02+ϵ∇w1ϵ02≤c∇w1ϵ02.
Thus, for anyϵ∈0,ϵ0,(146)∫0T0′∇w1εs2ds1/2≤cnx1−x2.
Putting (146) into (141), one can obtain the result. The proof is completed.Theorem 5.
Let assumptions (F1)–(F4) hold andν<λ1. Then, for every ϵ∈0,ϵ0, the semigroup Sϵt generated by equations (1) and (2) possesses an exponential attractor ℳϵc in H01Ω. Moreover, these exponential attractors can be constructed such that (2) and (3) in Theorem 3 hold and the fractal dimension of ℳϵ is uniformly (with respect to ϵ) bounded in H01Ω, i.e.,(147)dimf,H01Ωℳϵc≤c,where c is a constant independent of ϵ.



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*Source: 1025457-2020-04-07.xml*