diffusion equations with arbitrary polynomial growth

First, we prove by the`-trajectory method that the exponential attractor in L2Ωwith g ∈ H− 1Ω.. Second, by proving the semigroup satisfying discrete squeezing property, we obtain the exp

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Nonlinear Analysis

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Exponential attractors for reaction–diffusion equations with arbitrary polynomial growth

Yansheng Zhong∗, Chengkui Zhong

School of Mathematics and Statistics, Lanzhou University, Lanzhou, 730000, PR China

a r t i c l e i n f o

Article history:

Received 8 August 2008

Accepted 31 October 2008

Keywords:

Exponential attractor

Asymptotic a priori estimate

Discrete squeezing property

Semigroup

Supercritical nonlinearity

Global attractor

a b s t r a c t

In this paper, we study exponential attractors for an equation with arbitrary polynomial

growth nonlinearity f and inhomogeneous term g First, we prove by the`-trajectory

method that the exponential attractor in L2(Ω)with g ∈ H− 1(Ω) Second, by proving the semigroup satisfying discrete squeezing property, we obtain the exponential attractor

in H1(Ω)with g∈L2(Ω) Because the solutions without higher regularity than L 2p− 2(Ω)

for g belong only to L2(Ω)in the equation, the general method by proving the Lipschitz

continuity between L 2p− 2(Ω)and L2(Ω)does not work in our case Therefore, we give

a new method (presented in a theorem) to obtain an exponential attractor in a stronger

topology space i.e., L 2p− 2(Ω)with g∈G (stated in a definition) when it is out of reach for the other known techniques

© 2008 Elsevier Ltd All rights reserved

1 Introduction

In this paper, we consider the existence of an exponential attractor for the following reaction–diffusion equation







∂u

∂t −∆u+f(u) =g inΩ×R+

u=0 on∂Ω×R+,

u(x,0) =u0 inΩ,

(1.1)

whereΩis a bounded smooth domain in Rn(n>3), f is a C2function satisfying

and

and inhomogeneous term g in H− 1(Ω),L2(Ω),G (seeDefinition 2.1), respectively

Exponential attractors or attractors for the reaction–diffusion equation have been extensively considered in many monographs and lectures see eg., [1,5,7–9,17,18] and the references therein Exponential attractors for unbounded domain

or non-autonomous with subcritical nonlinearity have been considered in [1,5,7,8] In [9], the authors obtained finite

dimensions of a global attractor in L∞(Ω)with nonlinearity f satisfying arbitrary growth when g∈L∞(Ω) Also the finite

dimension of a global attractor in L2(Ω)has been proved in [18] when nonlinearity is supercritical and g∈L2(Ω) In [17], the

authors obtained a global attractor in L2(Ω), H1(Ω), L p(Ω)with arbitrary polynomial growth nonlinearity and g∈H− 1(Ω)

∗Corresponding author.

E-mail address:zhongyansheng04@gmail.com (Y Zhong).

0362-546X/$ – see front matter © 2008 Elsevier Ltd All rights reserved.

At the same time, they also proved the existence of a global attractor in L2(Ω), H2(Ω), L 2p−2(Ω)with arbitrary polynomial

growth nonlinearity and g ∈L2(Ω)

Let us come back to our problem

First, to obtain an exponential attractor in L2(Ω)with g ∈ H−1(Ω), we apply the`-trajectory method as in [10] It is

worth noticing that g∈H− 1(Ω)is different from [10] with g∈L2(Ω)

Second, to obtain an exponential attractor in H1(Ω)with g ∈ L2(Ω), we prove the semigroup satisfies a discrete

squeezing property in H1(Ω) Obviously, it is a more regular result than [18] with finite dimensions of a global attractor

in L2(Ω)when g∈L2(Ω)

In particular, we construct an exponential attractor in L 2p−2(Ω)with g ∈ G Since there is no higher regularity than

L 2p− 2(Ω)for solutions of(1.1)when g∈G belonging only to L2(Ω)(seeRemark 2.1.b), then general method by proving the

Lipschitz continuity between L 2p− 2(Ω)and L2(Ω)(see [9]) does not work in our case To overcome this difficulty, we give a new method, namely asymptotic a priori estimate, which concerns the relations of radius and cardinal number of coverings for some set in two different topologies (seeLemma 2.1,Remark 2.4andTheorem 2.1) In the concrete applications, we can first obtain an exponential attractor in a weaker topology space by a general method In some sense, it implies the change

of radius and cardinal number of covering for sets{S n(B)}∞

n= 1, where B is a bounded set And then, the change of radius and

cardinal number of covering for{S n(B)}∞

n= 1in the stronger topology space is easily obtained (see the proof ofTheorem 2.1) Furthermore, we obtain the existence of an exponential attractor in the stronger topology space

This article is organized as follows In Section2, we first recall some basic results, and then, give our important technique tool, i.e., Theorem 2.1, which will guarantee the existence of an exponential attractor in a stronger topology space In Section3, we prove the existence of exponential attractors in L2(Ω)with g ∈ H− 1(Ω), i.e.,Theorem 3.1 In Section4,

we obtain an exponential attractor in H1(Ω)with g ∈L2(Ω), i.e.,Theorem 4.1 In Section5, we prove the existence of an

exponential attractor in L 2p−2(Ω)with g∈G, i.e.,Theorem 5.2

Throughout this paper we use the following notation: Hilbert spaces L2(Ω), H1(Ω)equipped with the usual scalar products and norms(·, ·)and((·, ·)) In particular,| · |X denote the Banach space X norm.Ωis a bounded smooth domain

in Rn C denotes any positive constant which may be different from line to line even in the same line (sometimes for special differentiation, we also denote the different positive constants by C1,C2, )

2 Preliminary results

We start with the definition of the set G

Definition 2.1 We define the set G by a class of functions:

G= {g∈L2(Ω)|g ∈ ∪2<γ <∞∪C∈R X C,γ}

where X C,γ is a set which is denoted by the functions satisfying the following property:∃constant C∈R+andγ ∈ (2, ∞),

∀  >0, such that if E⊂Ωand m(E)6C· γ −2γ2, we have

Z

E

|g|262

where m(E)denote the Lebesgue measure of E.

Remark 2.1 a.∀g ∈Lγ(Ω)(γ ∈ (2, ∞)), then g∈G In fact, by the Hölder inequality, we deduce

Z

E

|g|2dx6

Z

E

(|g|2)γ2dx

2

γ Z

E 1dx

γ − 2 γ

6

Z

E

|g|γdx

2 γ

· (m(E))γ −γ2

6

Z

Ω

|g|γdx

2 γ

· (m(E))γ −γ2.

Choosing the constant C = (|g|Lγ( Ω ))−γ −2γ2 in ‘‘m(E)’’ above, by calculating, we have

Z

E

|g|2dx6|g|2Lγ( Ω )·

 (|g|Lγ( Ω ))−γ −2γ2 · γ −2γ2

γ − 2 γ

62

b there exist functions g ∈G belonging only to L2(Ω)

Now, we review the concept of an exponential attractor(see [2,3,5,6,8,13] for a detailed exposition)

Definition 2.2 Let E be a metric space and let L: E → E be a map We define a discrete semigroup{S n,n ∈ Z+}by

S n x=Ln x, x∈E A setM⊂E is an exponential attractor for the map L if the following properties hold:

1 The setMis compact in E and has finite fractal dimension

2 The setMis semi-invariant under the map L:L(M) ⊂M

3 The setMis an exponentially attracting set for the semigroup S n, i,e there exists a constantα >0 such that, for every

bounded subset B⊂E, there exists a constant C=C(B)such that

distE(S n B,M)6Cd−αn,

where dist denotes the non-symmetric Hausdorff distance between sets

Remark 2.2 We have given the definition of exponential attractors for discrete times(n ∈ Z+) The extension of this definition to the continuous case(t ∈R+)is straightforward (see e.g [3])

Remark 2.3 We note that the existence of an exponential attractorMfor the map L automatically implies the existence of the global attractorAand the embeddingA⊂M We note however that, in contrast to the global attractor, an exponential attractor is not uniquely defined

Definition 2.3 S is said to satisfy the discrete squeezing property on B if there exists an orthogonal projection P N of rank N such that for every u andvin B,

kP N(Su−Sv)k6k (I−P N)(Su−Sv)k ⇒ kSu−Svk6 1

8ku− vk.

The following lemma concerns the covering of sets in two different topologies (see [17] Lemma 5.3)

Lemma 2.1 For any >0, the bounded subset B of L p(Ω)(p>0)has a finite-net in L p(Ω)if there exists a positive constant

M=M()which depends on, such that

(i) B has a finite(3M)(q−p)/q(

2)p/q -net in L q(Ω)for some q, q>0;

(ii)



R

Ω (|u|>M)|u|p

1 /p

<2−(2p+ 2 )/pfor any u∈B, whereΩ(|u|>M) = {x∈Ω||u(x)|>M}.

Remark 2.4 From the proof of this lemma (see [17] Lemma 5.3), we not only obtain a finite-set in L p(Ω)for B but also obtain that both covering sets in L p(Ω)and L q(Ω)have the same cardinal number Furthermore, we can choose the same central points of both covering sets

Now, we give our main theorem which describes our new technique to construct an exponential attractor in a stronger topology space

Theorem 2.1 Assume that p > q > 0 andΩ ⊂ Rn is bounded Let {S n}∞n=1be a discrete Lipschitz continuous semigroup

on L p(Ω)and L q(Ω)and B0is a positively invariant bounded absorbing set in L p(Ω), respectively, and satisfying the following conditions:

(i) {S n}∞n=1has an exponential attractor in L q(Ω)with the radius of covering sets decreasing as R

2i,i = 1,2, , that is

S i(B0) ⊂ ∪Mi

j= 1



B L q( Ω )(a i j,R

2i) ∩S i(B0) ,i=1,2, , where R =supx∈B0kxkL p( Ω )and B L q( Ω )(a i j,R

2i)denote a ball with center a i j of the radius R

2i in L q(Ω)and the cardinal number M i =N0·K i

0,i=1,2, , where N0,K0are positive constants.

(ii) For any >0, there exist positive constants M=M() =C· q−γγ(γ >q), such that

Z

Ω (|u|>M)

|u|p

1 /p

<2−(2p+ 2 )/p for any u∈B0.

Then,{S n}∞

n= 1has an exponential attractor in L p(Ω).

Remark 2.5 In the concrete application ofTheorem 2.1, M=M() =C· q−γγ, whereγdepending on inhomogeneous term

g, for example see ourTheorem 5.1, and general q=2 By general method see [6,3,11], it is easy to construct an exponential

attractor in L2(Ω)such that the radius of the covering set decreases as 2R i,i = 1,2, orθi R,i = 1,2, , θ <1 and

the cardinal number increases as M i=N0·K0i,i=1,2, Here in ourTheorem 2.1, we only prove the case2R i, the other similar cases are also true by the same argument in the following proof ofTheorem 2.1

Now, we proveTheorem 2.1.

Proof of Theorem 2.1 We will proceed by induction.

Step 1 We consider-net in L p(Ω)for the set S m(B0)(m will be fixed later) FromLemma 2.1and the assumption (ii), it

follows that S m(B0)has a finite-net in L p(Ω)if it has a finite(3M1)(q−p)/q(

2)p/q -net in L q(Ω)with M1=C· γ /(q− γ ) From

our assumption (i), we know that there exist i0∈N such that

R

2i0 6(3M1)(q−p)/q 

2

p/q

Letting m=i0and N] (3M1)(q−p)/q(

2)p/q−net

L q( Ω )=N]( R

2i0)L q( Ω ), and combining withRemark 2.4and the assumption

(i), therefore, for the set S i0(B0), we have

N]( −net)L p( Ω )= N]

 (3M1)(q−p)/q 

2

p/q

−net



L q( Ω )

≡ N] R

2i0



L q( Ω )

=N0·K i0

Step 2 Let be replaced by 

2 Similarly, we consider 

2-net in L p(Ω)for the set S i0 +m1(B0)(m1will be fixed later) Using Lemma 2.1and the assumption (ii) again, we know that S i0 +m1(B0)has a finite 

2-net in L p(Ω)if it has a finite (3M2)(q−p)/q(2

2)p/q -net in L q(Ω)with M2 = C · (

2)γ /(q− γ ) From the assumption (i), we know that there exist k

1 ∈ N such that

R

2i0 +k1 6(3M2)(q−p)/q

 2 2

p/q

2i0 +k1 − 1, for given

Letting m1 = k1 and N](3M2)(q−p)/q(2

2)p/q−net



L q( Ω ) = N

2i0+k1)L q( Ω ) and combining withRemark 2.4and the

assumption (i), therefore, for the set S i0 +k1(B0), we have

N] 

2−net



L p( Ω )= N

] (3M2)(q−p)/q

 2 2

p/q

−net

!

L q( Ω )

≡ N] R

2i0 +k1



L q( Ω )

=N0·K i0 +k1

Step 3 Let 

2i− 1 be replaced by 

2i, similarly, we consider 

2i -net in L p(Ω)for the set S i0 +k1 +k2 +···+k i− 1 +m i(B0)(m iwill be fixed later) FromLemma 2.1and the assumption (ii), it follows that S i0 +k1 +k2 +···+k i− 1 +m i(B0)has a finite 

2i -net in L p(Ω)if it has a finite(3M i)(q−p)/q(2i

2)p/q -net in L q(Ω)with M i =C· (

2i)γ /(q− γ ) From the assumption (i), there exist k

i∈N such that

R

2i0 +k1 +···k i 6(3M i)(q−p)/q



2i

2

p/q

2i0 +k1 +···+k i− 1, for given 

Letting m i = k i and N] (3M i)q−p/q(2i

2)p/q−net



L q( Ω )

= N]( R

2i0+k1+···ki)L q( Ω )and combining withRemark 2.4and the

assumption (i), therefore for the set S i0 +k1 +···+k i(B0), we have

N] 

2i −net



L p( Ω ) =N

] (3M i)q−p/q



2i

2

p/q

−net

!

L q( Ω )

≡N] R

2i0 +k1 +···k i



L q( Ω )

=N0·K i0 +k1 +···k i

Now,(3M i− 1)(q−p)/q(2i− 1

2 )p/qon(3M i)(q−p)/q(2i

2)p/q together with M i=C· (

2i)γ /(q− γ )by above procedure, we have

(3M i− 1)(q−p)/q

 

2i− 1 2

p/q

(3M i)(q−p)/q



2i

p/q =



M i− 1

M i

(q−p)/q

·2p/q



 

2i− 1

γ /(q− γ )

 

2i

γ /(q− γ )(q−p)/q

·2p/q

= 2γ /(q− γ )
(q−p)/q

·2p/q

From (2.7), noticing that the ratio of (3M i− 1)(q−p)/q(2i−1

2 )p/q and (3M i)(q−p)/q(2i

2)p/q is a constant for given γ ,p,q,

independent of,i, and combining with inequalities(2.1),(2.3)and(2.5)and the assumption (i), therefore, we can choose

k1 = k2 = · · · = k i = · · · ≡ [ (2)γ (q−p)/(q− γ )q· 2p/q] +1, where [a] is the integer part of a and in what follows

m0,[ (2)γ (q−p)/(q− γ )q·2p/q] +1, such that

S i0 +k1 +···+k i(B0) =S i0 +i·m0(B0)

=S i0[S m0]i(B0) ⊂

Ni

[

j= 1



B L p( Ω )



a i j, 

2i



∩S i0[S m0]i(B0) , i=1,2, (2.8)

where a i j ∈S i0[S m0]i(B0)and the balls(B L p( Ω )(a i j,

2i))in L p(Ω) Again from the assumption (i) and combining with estimates

(2.2),(2.4)and(2.6), it follows that

Ni=N0·K i0 +i·m0

0 =N0K i0

0 · [K m0

Combining(2.8)and(2.9), it is easy to deduce that the existence of an exponential attractorMdfor the discrete semigroup (S m0)n

in L p(Ω)and from the Lipschitz property of discrete semigroup{S n}∞n=1, we can set

M1d=

m0

[

i= 1

S i(Md).

Using the same argument as in [3], it is easy to deduce thatMd is an exponential attractor in L p(Ω)for discrete semigroup {S n}∞

n= 1 

Finally, we formulate the celebrated so-called Aubin–Lions lemma

Lemma 2.2 Let p1∈ (1, ∞], p2∈ [1, ∞) Let X be a Banach space and Y,Z be separable and reflexive Banach spaces such that

Y,→,→X,→Z Then for anyτ ∈ (0, ∞),

{u∈L p1(0, τ;Y);u0∈L p2(0, τ;Z)} ,→,→L p1(0, τ;X). (2.10)

3 Exponential attractor in L2(Ω)with g∈H−1(Ω)

In this section, we will prove an exponential attractor in L2(Ω)for(1.1)with g ∈H− 1(Ω) We start with the following general existence and uniqueness of solutions which can be obtained by the normal Faedo–Galerkin methods, see [14,16] for details

Lemma 3.1 Let the assumptions(1.2)and(1.3)hold and g ∈H− 1(Ω) Then for any initial date u0 ∈ L2(Ω)and any T >0,

there exists a unique solution u for(1.1)which satisfies

u∈L2(0,T;H01(Ω)) ∩L p(0,T;L p(Ω)), ∀T >0,

u∈C(R+;L2(Ω)),

and the mapping u0→u(t)is continuous in L2(Ω).

If, furthermore, g∈L2(Ω)and u0∈H1(Ω), then

u∈C([0,T);H1(Ω)) ∩L2(0,T;H2(Ω)), ∀T >0.

ByLemma 3.1, we can define the operator semigroup{S(t)}t>0 in L2(Ω)for both g∈H−1(Ω)and g∈L2(Ω)as follows:

which is continuous in L2(Ω)

Remark 3.1 From [17], we note that the semigroup S(t)has a global attractor in H01(Ω)for(1.1)with g ∈ H−1(Ω) It

implies that the semigroup S(t)possesses a bounded and positively invariant absorbing setB0in H1(Ω) More precisely,

we will takeB0 = ∪t>t0S(t)B0, where B0is a bounded absorbing set in H1(Ω)and t0is a time such that S(t)B0 ⊂B0for

t>t0and where the closure is taken in the weak topology of H1(Ω) For the setB0, the following facts are true

Lemma 3.2 Let the assumptions(1.2)and(1.3)hold and g∈H−1(Ω)andB0be defined as above Then, for any u0∈B0, the solution u(t) =S(t)u0∈B0, for any t>0 Furthermore, if u0is the limit, for the weak topology of H1(Ω), of a sequence{u 0n}, where u 0n ∈ ∪t>0 S(t)B0, for every n, then there exists at least a subsequence of solutions,{u n(t) = S(t)u 0n}converges to the solution u(t)for the weak topology of H01(Ω), for any t>0.

Proof For convenience, we denote g by D i f i+h(x)(=Σn

i= 0D i f i+h(x)), where f i,h∈L2(Ω)(i=1, n) Now multiplying(1.1)by u, after the standard integration by parts and using the assumption(1.3), we have

1

2

d

dt|u|

2

2+ |∇u|22+

Z

Ω

f(u)u= hD i f i,ui + hh,ui

= − Z

Ω

˜

which implies that

d

dt|u|

2

2+ |∇u|22+

Z

Ω

wheref˜ = (f1, ,f n),|˜f|2=Σn

i= 1|f i|2andh · idenotes the L2-inner product From(3.3), it follows that

|u(t)|2

26e−αt|u(0)|2

whereαis a positive constant

Meanwhile, let F(s) = Rs

0f(τ)dτ; then by(1.3)again, it follows that

˜

Therefore,

˜

C1

Z

Ω

|u|p−k|Ω|6

Z

Ω

F(u)6C˜2

Z

Ω

Also noticing that

by(3.5)–(3.7), we infer from(3.3)that

d

dt|u|

2+C



|∇u+ ˜f|2+2

Z

Ω

F(u)



Integrating the inequality above from t to t+1, we have

Z t+ 1

t



|∇u+ ˜f|2+2

Z

Ω

F(u)



6C(|˜f|2, λ1, |h|2, |Ω| ) + |u(0)|2, (3.9)

On the other hand, multiplying(1.1)by u t, we obtain

|u t|2+1

2

d

dt|∇u|

2+ d

dt

Z

Ω

F(u) = hD i f i,u ti + hh,u ti

= −d

By the Hölder inequality and the Cauchy inequality, it follows from(3.7)and(3.10)that

d

dt



|∇u+ ˜f|22+2

Z

Ω

F(u)



Combining with(3.9)and(3.11), by the uniform Gronwall inequality, we obtain

|∇u(t) + ˜f|22+2

Z

Ω

F(u(t))6|h|22+C(|˜f|22, λ1, |h|22, |Ω| ) + |u0|22, ∀t>0. (3.12) Combining with the definition ofB , therefore S(t)u ∈B ,∀t>0 for u ∈ ∪ B

Now we assume that u0 is the weak limit of H01(Ω), of a sequence{u 0n}, where u 0n ∈ ∪t>t0S(t)B0 Now we set

vn(t) = u n(t) −u(t)(where u n(t), u(t)are the solutions, to the initial value u 0n , u0, respectively), which satisfies the following equation:

(d

dtvn(t) −∆vn(t) + `(t)vn(t) =0,

where l(t) := R1

0 f0(su n(t) + (1−s)u(t))ds Since f0(u) > − κ, then, obviously, l(t) > − κalso Since{u 0n}is bounded

sequence in H01(Ω)and H01(Ω) ,→,→ L2(Ω), It implies that there exists a subsequence of u 0n converging in L2(Ω) And

since u 0n weak converge to u0in H1(Ω), it follows that u 0n converge to u0in L2(Ω)

Now multiplying(3.13)byvn(t)and integrating overΩ, we have

d

Using the Gronwall inequality, we obtain that

|u n(t) −u(t)|2

Taking the limit on both sides of(3.15), we have

lim

Again using the the fact that u n(t) ∈ B0is bounded in H1(Ω), it implies that u n(t)has a weak converging subsequence {u n j(t)},∀t >0 Combining with(3.16), it is easy to deduce that the subsequence{u n j(t)}is weakly converging to u(t)in

H1(Ω)for any t>0 Therefore, u(t) ∈B0andLemma 3.2is proved 

Now we consider difference of two solutions of(1.1)starting fromB0

Proposition 3.1 Let the assumptions of Lemma 3.2 hold and let u1and u2be two solutions of (1.1)starting fromB0 Then, there exists a constant` >0 such that

Z 2 `

`

|∇ (u1−u2)(s)|2

2ds6C

Z `

0

|u1(s) −u2(s)|2

where the positive constant C is independent of u1(0)and u2(0).

Proof We setv(t) =u1(t) −u2(t) This function satisfies the equation

dv

where l(t) := R1

0 f0(su1(t) + (1−s)u2(t))ds Since f0(u)>− κ, then, obviously, l(t)>− κalso

Multiplying Eq.(3.18)byv(t)and integrating overΩ, we obtain the inequality

d

dt| v|2

2+2|∇ v|2

262κ|v|2

Using the Gronwall inequality, we have

| v(τ)|2

26e2κ(τ−s)| v(s)|2

and taking s∈ (0, `)and integrating(3.19)overτ ∈ (s,2`) Thus, we obtain

| v(2`)|2

2+2

Z 2 `

s

|∇ v(τ)|2

2dτ62` Z 2`

s

| v(τ)|2

2dτ + |v(s)|2

Inserting estimate(3.20)into the right-hand side of estimate(3.21), we obtain the following inequality:

Z 2 `

`

|∇ v(τ)|2

2dτ 6C| v(s)|2

where the positive constant depends on` Integrating(3.22)over s∈ (0, `)then yields the inequality(3.17)and completes the proof ofProposition 3.1 

We now have

Proposition 3.2 Let the assumptions of Proposition 3.1 hold and`be defined as above Then,

∂

∂t(u1−u2)

L2 (`, 2 `;H−1 ( Ω )) 6

Proof We have, again settingv =u1−u2,

∂v

∂t L2 (`, 2 `;H−1 ( Ω ))

=sup

ϕ

Z 2 `

`

 ∂v

∂t, ϕ

dt

whereϕ ∈L2(`,2`;H1(Ω)),k ϕkL2 (`, 2 `;H1 ( Ω ))=1, andh· , ·idenotes the duality product between H1(Ω)and H− 1(Ω) We then have, noting that

∂v

∂t =1v −l(t)v,

∂v

∂t L2 (`, 2 `;H−1 ( Ω ))

6sup

ϕ

Z 2 `

`

|∇ vk∇ϕ|dt+

Z 2 `

`

Furthermore,

Z 2 `

`

Z 2 `

`

|l(t)v||ϕ|dt 6 κkvkL2 (`, 2 `;L2 ( Ω ))

We thus deduce from(3.26)and(3.27)that

∂v

∂t L2 (`, 2 `;H−1 ( Ω ))

6Ck vkL2 (`, 2 `;H1( Ω )), which yields, owing to(3.17), estimate(3.23), i.e.,

∂v

∂t L2 (`, 2 `;H−1 ( Ω ))

6Ck vkL2 (`, 2 `;L2 ( Ω ))

and completes the proof ofProposition 3.2 

Thanks toPropositions 3.1and3.2andLemma 2.2, we now prove that the existence of exponential attractor in L2(Ω)for

Eq.(1.1)with g∈H− 1(Ω) To do so, we use the method of`-trajectories (see [11] for more details)

We introduce the space of trajectories

X`= { w : (0, `) →L2(Ω), wis a solution of(1.1)on(0, `)},

where`is as above, which we endow with the topology of L2(`,2`;L2(Ω)) We then set

B`= { w ∈X`, w(0) ∈B0}

For the setB`, we have the following result

Proposition 3.3 Let the assumptions of Proposition 3.1 hold andB`be defined as above Then,B`is closed in the topology of

L2(0, `;L2(Ω)).

Proof Let{ wn}be a sequence of trajectories belonging toB`which converges to somew FromLemma 3.2, we note that

wn(t) ∈B0, for every t ∈ [0, `]and for every n Using the same argument as inLemma 3.2, we can pass to the limit in the equation to prove that, at least for a subsequence,wnconverges to a solutionwof the equation on[0, `]which is weakly continuous from[0, `]onto H1(Ω)and thatw(0) ∈B0 Therefore, this yields thatB`is a complete metric space (we note that this is not necessarily the case forX`, see [11,13]) and completes the proof ofProposition 3.3 

We then define the operators L(t) :X`→X`, t>0, by(L(t)w)(s) =u(t+s), s∈ [0, `], where u is the unique solution of

Eq.(1.1)such that u|[0,`]= w We finally setL=L(`) Then, it follows from estimate(3.17)that, ifw1, w2∈B`,

and it follows from estimate(3.23)that

∂

∂t(Lw1−Lw2)

L2 ( 0 ,`;H−1 ( Ω ))

Furthermore, we have the following result:

Proposition 3.4 Let the assumptions of Proposition 3.1 hold and let the operator L(t)be defined as above Then, L(t)is Lipschitz fromB`ontoX`,∀t>0, and that the mapping t→L(t)wis Lipschitz,∀ w ∈X`.

Proof Let the space X=L2(Ω)in Lemma 2.1 of [11] and combining with estimate(3.20), it is easy to deduce L(t)is Lipschitz fromB`ontoX`,∀t >0

Now we prove the mapping t → L(t)wis Lipschitz,∀ w ∈ X` Similarly, Let X = L2(Ω)in Lemma 2.2 of [11], it is equivalent to prove that{ χ0; χ ∈ C`}be bounded in space L q(0, `;L2(Ω))with some q ∈ (1, ∞], this is easily deduced from the following lemma

Lemma 3.3 Let the assumptions of Proposition 3.1 hold Then, for any 26 p< ∞and any bounded subset B ⊂L2(Ω), the following estimate is valid:

Z

Ω

|u t(s)|p

6C(e−αt|u0|22+1) for any u0∈B, (3.30)

whereαis a positive constant and the constant C depends on p, `,n (space dimensional),λ1,g, |Ω|and u t(s) = d

dt(S(t)u0)|t=s

Proof Using a similar argument as Lemma 5.20 of [17], we can proveLemma 3.3 

From estimate(3.30), we complete the proof ofProposition 3.4 

Remark 3.2 Thanks to estimates(3.28)and(3.29), it follows (see [11] for the details of the proof) that the semigroup

L(t)possesses an exponential attractorM`onB`, that is,M`is compact for the topology ofX`, is positively invariant (i.e., L(t)M` ⊂ M`, ∀t > 0), has a finite fractal dimension and attracts exponentially fast(B`)(again, for the topology

ofX`)

We now introduce the mapping e:X`→L2(Ω)defined by e(w) = w(`)(i.e., e maps an`-trajectory onto its endpoint)

For the mapping e, we have

Proposition 3.5 Let the assumptions of Proposition 3.1 hold and let the mapping e be defined as above Then, e is Lipschitz.

Proof Let the space X=L2(Ω)in Lemma 2.1 of [11] and combining with estimate(3.20), it is easy to deduceProposition 3.5



Now, we have

Theorem 3.1 Let the assumptions(1.2)and(1.3)hold and g ∈H− 1(Ω) Then, for the semigroup S(t)generated by Eq.(1.1)

possesses an exponential attractorMin L2(Ω) Moreover, these exponential attractors can be chosen such that

and there exists a constantα >0 such that for every bounded subset B of L2(Ω), there exists a constant C0=C0(B)such that

Based on the above results and applying the method as in [10,11], we can deduceTheorem 3.1

4 Exponential attractor in H0 1(Ω)with g∈L2(Ω)

In this section, we show that an exponential attractor in H01(Ω)for Eq.(1.1)with g∈L2(Ω) We begin with the following fact

Since semigroup S(t)has a global attractor in H1(Ω)for Eq.(1.1)with g∈L2(Ω), see [17], it implies that the semigroup

S(t)is asymptotically compact in H1(Ω) From [4], to obtain our result, it remains to prove the semigroup S(t)satisfies the

discrete squeezing property in H01(Ω), that is the following lemma:

Lemma 4.1 Let the assumptions of Theorem 4.1 hold Then, there exist T∗ and a small constantδ > 0, such that for any

u0, v0∈S(δ)B, if

kP N0(S(T∗)u0−S(T∗)v0)kH1( Ω )6k (I−P N0)(S(T∗)u0−S(T∗)v0)kH1( Ω ),

then

k (I−P N0)(S(T∗)u0−S(T∗)v0)kH1( Ω )6

1

where B is an arbitrary bounded set in L2(Ω).

Proof Let Q N0= (I−P N0)andw(t) =u(t) − v(t),w2=Q N0(w)

Obviously, the functionw(t)satisfies the equation

dw

where l(t) := R1

0 f0(su1(t) + (1−s)u2(t))ds Since f0(u)>− κ, then, obviously, l(t)>− κalso

Now, we multiply(4.2)by−∆w2, we have

d

2dtk∇ w2k2+ (∆w,∆w2) = (−l(t)w,∆w2)6κ(w,∆w2).

By orthogonal property, we have

(∆w,∆w2)>λN0(∆w, w2) = λN0(∆w2, w2) = λN0(∇w2, ∇w2)

κ(w,∆w2) = κ(w2,∆w2) = κ(∇w2, ∇w2).

So, we have

d

2dtk∇ w2k2+ λN0(∇w2, ∇w2)6κ(∇w2, ∇w2).

SinceλN0→ +∞, obviously, we can choose N0such thatλN0 > κ Therefore

d

dtk∇ w2k2+2(λN0− κ)(∇w2, ∇w2)60.

By the Gronwall inequality, we have

k w2(t)k2

H1 ( Ω )6e

− 2 (λN0− κ)tk w20k2

H1 ( Ω )6e

− 2 (λN0− κ)tk w0k2

H1 ( Ω ).

Therefore, there exists a T∗such thatk (I−P N0)(S(T∗)u0−S(T∗)v0)kH1( Ω )= k w2(T∗)kH1( Ω ) 6 161k w0kH1 ( Ω ) = 161ku0−

v0kH1 ( Ω ), and ifkP N0(S(T∗)u0−S(T∗)v0)kH1 ( Ω )6k (I−P N0)(S(T∗)u0−S(T∗)v0)kH1 ( Ω ), then

kS(T∗)u0−S(T∗)v0kH1 ( Ω )6 k (I−P N0+P N0)S(T∗)u0−S(T∗)v0kH1 ( Ω )

6 2k (I−P N0)(S(T∗)u0−S(T∗)v0)kH1( Ω )

6 1

8ku0− v0kH1 ( Ω ).

It implies that S(t)satisfies the discrete squeezing property and the proof ofLemma 4.1is finished

Corollary 4.1 Let the assumptions of Theorem 4.1 hold Then, the semigroup S(t)is uniformly Hölder continuous on[0,T∗]×B1

in the topology of H1(Ω), i.e.,

|S(t1)u10−S(t2)u20|H1 ( Ω )6C(|u10−u20|H1 ( Ω )+ |t1−t2|1) (4.3)

for u i0 ∈ B1and t1,t2 6 T∗whereB1 = ∪t>t

0S(t)B0, B0is a bounded absorbing set in H1(Ω)and t0is a time such that

S(t)B0⊂B0for t>t0and where the closure is taken in the strong topology of H1(Ω)and the constant C depends onB1,T∗.

Proof First, it is easy to obtain the Lipschitz continuity with respect to the initial condition fromLemma 4.1 It remains

to prove the Lipschitz continuity with respect to the time t From Lemma 3.1, we know that the solution u(t) ∈

C([0,T] ,H1(Ω))for the initial value belongs toB1, therefore

Since semigroup S(t)has... easy to obtain the Lipschitz continuity with respect to the initial condition fromLemma 4.1 It remains

to prove the Lipschitz continuity with respect to the time t From Lemma... X=L2(Ω)in Lemma 2.1 of [11] and combining with estimate(3.20), it is easy to deduce L(t)is Lipschitz fromB`ontoX`,∀t