On the Kneser property for reaction-diffusion equations in some unbounded domains with an H^{-1}-valued non-autonomous forcing term

On the Kneser property for reaction-diffusion equations in some unbounded domains with an H−1-valued non-autonomous forcing term M. Anguiano1, F. Morillas2, J. Valero3 1Dpto. Ecuaciones Diferenciales y An´alisis Num´erico Universidad de Sevilla, Apdo. de Correos 1160, 41080-Sevilla, Spain e-mail: anguian[email protected] 2Universitat de Val`encia, Departament d’Economia Aplicada, Facultat d’Economia, Campus dels Tarongers s/n, 46022 Val`encia, Spain. e-mail: Fran[email protected] 3Universidad Miguel Hern´andez, Centro de Investigaci´on Operativa Avda. Universidad s/n, Elche (Alicante), 03202, Spain e-mail: [email protected] Abstract In this paper we prove the Kneser property for a reaction-diffusion equation on an unbounded domain satisfying the Poincar´e inequality with an externalforcetakingvaluesinthespaceH−1. Using this property of solutions we check also the connectedness of the associated global pullback attractor. We study also similar properties for systems of reaction-diffusion equations in which the domain is the whole RN. Finally, the results are applied to a generalized logistic equation. AMS Subject Classification: 35B40, 35B41, 35K55, 35K57, 37B25, 58C06 Keywords: reaction-diffusion system, setvalued dynamical system, global attractor, Kneser property, unbounded domain 1Introductionandsettingoftheproblem When we consider a partial differential equation with non-uniqueness of the Cauchy problem it is interesting to consider the Kneser property, that is, the connectedness and compactness of the set of values attained by the solutions at any moment of time. In particular, this problem has been studied for reaction-diffusion equations by several authors so far. In this direction some results are known for scalar reaction-diffusion equations in bounded domains in the case where the nonlinearity has at most linear growth [12]. Such results were extended later on in [14, 15] for systems of reaction-diffusion equations with nonlinearities having more than linear growth (see also [13]), with applications to the complex Ginzburg-Landau equation and the Lotka-Volterra system with diffusion, among others. Also, the Kneser property for degenerate reaction-diffusion equations was considered in [3, 4]. When the domain is unbounded the problem has additional technical difficulties. A first result in this direction was given [16], in which it is studied a scalar reaction-diffusion on unbounded domains in which the nonlinear term is equal to |u|1 2.In [22] the results of [15] were extended to unbounded domains. However, due to technical difficulties it was necessary to assume an additional condition concerning the derivatives of the nonlinear terms. In this paper we improve the method of the proof given in [22] in order to avoid such condition. 1 Manuscript We consider first the following problem. Let Ω⊂RNbe a nonempty unbounded open set and suppose that Ωsatisfies the Poincar´e inequality, i.e., there exists a constant λ1>0suchthat !Ω|u(x)|2dx ≤λ−1 1!Ω|∇u(x)|2dx, ∀u∈H1 0(Ω) . (1) This condition is satisfied when the set Ωis bounded in one direction. For example, in the two-dimensional case we can take the strip Ω={(x, y):a≤y≤b, −∞ <x<∞}. Let us consider the following problem for a non-autonomous reaction-diffusion equation with zero Dirichlet boundary condition in Ω,      ∂u ∂t−(u=f(x, u)+h(t) in Ω×(τ,+∞), u=0 on ∂Ω×(τ,+∞), u(x, τ)=uτ(x),x∈Ω, (2) where τ∈R,uτ∈L2(Ω), h∈L2 loc(R;H−1(Ω)) and f:Ω×R→Ris such that f(·,u) is a measurable function for any u∈R,f(x, ·)∈C(R) for almost every x∈Ω, and satisfies that there exist constants α1>0, α2>0, and p≥2 and positive functions C1(x)∈L1(Ω)∩L∞(Ω), C2(x)∈L1(Ω) such that |f(x, s)|p p−1≤α1|s|p+C1(x)∀s∈R,x∈Ω, (3) f(x, s)s≤−α2|s|p+C2(x)∀s∈R,x∈Ω. (4) We observe that |f(x, s)|≤α p−1 p 1|s|p−1+C1(x)p−1 p∀s∈R,x∈Ω. (5) By -·-L2(Ω)we denote the norm in L2(Ω), by -·-H1 0(Ω)=-∇·-L2(Ω)the norm in H1 0(Ω) and by -·-∗ the norm in H−1(Ω). We will use (·,·) to denote the scalar product in L2(Ω) or [L2(Ω)]N,and .·,·/to denote either the duality product between H−1(Ω) and H1 0(Ω) or between Lp"(Ω) and Lp(Ω), where 1 p+1 p"= 1. Note that the function fis just continuous on u, and that no condition on the derivative is imposed. In the second section we prove that the Kneser property holds for this problem. In the previous paper [2] concerning equation (2) we proved the existence of a global compact pullback attractor. It is interesting to prove that the global attractor is connected. Using the Kneser property we are able to obtain such result, which is given in the third section. We observe that in [2] the function C1(x) belongs just to L1(Ω), but for the Kneser property we need the stronger condition C1(x)∈ L1(Ω)∩L∞(Ω). In the fourth section we consider a system of reaction-diffusion equations with Ω=RN, which was studied before in [22]. Using a similar technique as for problem (2) we improve the results of that paper, proving the Kneser property and the connectedness of the global attractor without using an extra condition on the derivative of the nonlinear term of the equation. Instead, as in problem 2, we have to assume that the functions appearing in the growth condition of the nonlinear term belong to L∞(Ω). In the last section we apply these results to a generalized logistic equation. 2TheKneserproperty In this section we shall prove that the set of values attained by the solutions of equation (2) at any moment of time is connected. For this aim for each τ∈Rand uτ∈L2(Ω) let us denote by S(τ,uτ)the set of all weak solutions of (2) defined for all t≥τ. Such a set is non-empty as in [2] it is proved that at least one weak global solution exists for any τ∈Rand uτ∈L2(Ω). We define a multi-valued map U:R2 d×L2(Ω)→P(L2(Ω)) by U(t, τ, uτ)={u(t): u∈S(τ,uτ)},τ≤t, uτ∈L2(Ω),(6) 2 where P(L2(Ω)) is the set of all non-empty subsets of L2(Ω). In [2] it is shown that the multi-valued mapping Udefined by (6) is a strict multi-valued nonautonomous dynamical system on L2(Ω) (see Definition 9). Our aim is to prove the connectedness of the set U(t, τ, uτ)⊂L2(Ω) for any t≥τ.WenotethatthecompactnessofU(t, τ, uτ) in L2(Ω) is a consequence of Proposition 16 in [2], as in that paper it is shown that U(t, τ, uτ) is precompact and closed for any uτ. We shall obtain now that U(t, τ, uτ) is conected in L2(Ω) and for this aim we need some preliminary lemmas. We take a sequence 0 <% k<1 converging to 0 as k→∞and define a sequence of smooth functions ψk:R+−→ [0,1] satisfying ψk(s) :=            1, if 0 ≤s≤√%k, 0≤ψk≤1, if √%k≤s≤2√%k, 0, if 2√%k≤s≤1/%k, 0≤ψk≤1, if 1/%k≤s≤1/%k+1, 1, if s≥1/%k+1. Let ρ"k:R−→ R+be a mollifier, that is, ρ"k∈C∞ 0(R;R), suppρ"k⊂B"k,&Rρ"k(s)ds =1and ρ"k(s)≥0 for all s∈R,whereB"k={u∈R:|u|≤%k}. We define the following approximating function fk(x, u) := ψk(|u|)'C1 0|u|p−2u+f(x, 0)(+(1−ψk(|u|)) !R ρ"k(s)f(x, u −s)ds, where k≥1, p≥2, and C1 0is a negative constant. Then it is easy to check that for a.a. x∈Ω, sup |u|≤A))fk(x, u)−f(x, u)))−→ 0, as k−→ ∞,foranyA>0. Lemma 1 Assume (3)-(4). Then the function fksatisfy conditions (3)-(4), i.e., there exist constants *α1,*α2>0,andpositivefunctions * C1(x)∈L1(Ω)∩L∞(Ω)and * C2(x)∈L1(Ω),notdependingonk, such that ))fk(x, u)))p p−1≤*α1|u|p+* C1(x)∀u∈R,x∈Ω,(7) fk(x, u)u≤−*α2|u|p+* C2(x)∀u∈R,x∈Ω,(8) for kgreat enough. Proof. Indeed, for the first property we have the following cases. 1) If 0 ≤|u|≤√%kor |u|≥1/%k+1,thenwehave |fk(x, u)|≤))C1 0))|u|p−1+|f(x, 0)|, so (3) yields |fk(x, u)|p p−1≤21 p−1))C1 0))p p−1|u|p+2 1 p−1|f(x, 0)|p p−1 ≤21 p−1))C1 0))p p−1|u|p+2 1 p−1C1(x). 2) If √%k≤|u|≤2√%kor 1/%k≤|u|≤1/%k+ 1, then using (5) we have |fk(x, u)|≤))C1 0))|u|p−1+|f(x, 0)| +!R ρ"k(s)(α p−1 p 1|u−s|p−1+C1(x)p−1 p)ds ≤))C1 0))|u|p−1+2C1(x)p−1 p +α p−1 p 12p−2!R ρ"k(s)'|u|p−1+|s|p−1(ds ≤2C1(x)p−1 p++))C1 0))+α p−1 p 12p−1,|u|p−1, 3 and then |fk(x, u)|p p−1≤2p+1 p−1C1(x)+2 1 p−1+))C1 0))+α p−1 p 12p−1,p p−1 |u|p. 3) If 2√%k≤|u|≤1/%k, then arguing as in the previous case we have |fk(x, u)|≤C1(x)p−1 p++α p−1 p 12p−1,|u|p−1, and then |fk(x, u)|p p−1≤21 p−1C1(x)+2 1 p−12pα1|u|p. Finally, we obtain |fk(x, u)|p p−1≤* C1(x)+*α1|u|p, where * C1(x) := 2p+1 p−1C1(x)∈L1(Ω)∩L∞(Ω), and *α1:= 2 1 p−1+))C1 0))+α p−1 p 12p−1,p p−1 >0. On the other hand, note that fk(x, u)u=ψk(|u|)C1 0|u|p+ψk(|u|)f(x, 0)u+(1−ψk(|u|)) !R ρ"k(s)f(x, u −s)uds.(9) Hence, for the second property we have the following cases. 1) If 2√%k≤|u|≤1/%k, then using (4), the Young inequality ab ≤ap εp−1p+εbp" p"and (3) we obtain !R ρ"k(s)f(x, u −s)uds =!R ρ"k(s)f(x, u −s)(u−s)ds +!R ρ"k(s)f(x, u −s)sds ≤!R ρ"k(s)(−α2|u−s|p+C2(x)) ds +α2 p%α1!R ρ"k(s)|f(x, u −s)|p"ds +1 p+α1 α2,p−1!R ρ"k(s)|s|pds ≤!R ρ"k(s)(−α2|u−s|p+C2(x)) ds +α2 p%α1!R ρ"k(s)(α1|u−s|p+C1(x)) ds +1 p+α1 α2,p−1!R ρ"k(s)|s|pds ≤+−α2+α2 p%,!R ρ"k(s)+|u|p 2p−1−|s|p,ds +1 p+α1 α2,p−1!R ρ"k(s)|s|pds +C2(x), (10) where p%=p p−1is the conjugate exponent of p,andC2(x) := C2(x)+ α2 p"α1C1(x), and where in the last inequality we have used |u|p=|u−s+s|p≤2p−1(|u−s|p+|s|p). We observe that |u|≥√%k,sothatforklarge enough, |s|p≤%p k≤1 2p% p 2 k≤1 2p|u|p. Then from (10) we obtain !R ρ"k(s)f(x, u −s)uds ≤+−α2+α2 p%,1 2p|u|p+1 p+α1 α2,p−1!R ρ"k(s)|s|pds +C2(x). (11) 4 Since for klarge enough, we have |s|p≤%p k≤−p+α2 α1,p−1+−α2+α2 p%,1 2p+1 % p 2 k≤−p+α2 α1,p−1+−α2+α2 p%,1 2p+1 |u|p, we obtain !R ρ"k(s)f(x, u −s)uds ≤β|u|p+C2(x), (12) where β='−α2+α2 p"(1 2p+1 <0andC2(x)∈L1(Ω) is a positive function. 2) If 0 ≤|u|≤√%kor |u|≥1/%k+ 1, then using the Young inequality ab ≤εap p+bp" ε 1 p−1p"and (3) we obtain fk(x, u)u=C1 0|u|p+f(x, 0)u≤C1 0+1−1 p,|u|p+1 p%(−C1 0)1 p−1 C1(x) ≤C1 0 2|u|p+1 p%(−C1 0)1 p−1 C1(x), where C1 0 2is a negative constant and 1 p"(−C1 0)1 p−1C1(x)∈L1(Ω) is a positive function. 3) If √%k≤|u|≤2√%kor 1/%k≤|u|≤1/%k+ 1, we argue as in the first case to obtain (12). From (9), using the Young inequality and (3), we have fk(x, u)u≤ψk(|u|)C1 0|u|p−ψk(|u|)C1 0 p|u|p+ψk(|u|)1 p%(−C1 0)1 p−1 C1(x) +(1−ψk(|u|)) -β|u|p+C2(x). ≤ψk(|u|)C1 0 2|u|p+(1−ψk(|u|)) β|u|p+/C2(x)+ 1 p%(−C1 0)1 p−1 C1(x)0 ≤1 β|u|p+* C2(x), where 1 β:= max 2β,C1 0 23<0and * C2∈L1(Ω) is a positive function. Then, we have fk(x, u)u≤1 β|u|p+* C2(x), where 1 β<0and * C2(x)∈L1(Ω). Lemma 2 Assume (3)-(4). Then, there exist D"ksuch that ∂fk ∂u(x, u)≤D"k,∀u∈R,fora.a.x∈Ω.(13) Proof. We note that ∂ ∂uψk(|u|) is uniformly bounded on R.Wehave ∂fk ∂u(x, u)=C1 0ψk(|u|)(p−1) |u|p−2+C1 0|u|p−2u∂ ∂uψk(|u|) +f(x, 0) ∂ ∂uψk(|u|)+(1−ψk(|u|)) !R ρ% "k(u−s)f(x, s)ds −∂ ∂uψk(|u|)!R ρ"k(s)f(x, u −s)ds.(14) We consider each term in (14). 5 •As C1 0is a negative constant, for the first term, we have C1 0ψk(|u|)(p−1) |u|p−2≤0. •For the second term, we get ))))C1 0|u|p−2u∂ ∂uψk(|u|)))))≤))C1 0))+1 %k +1 ,p−1 Cψk. •For the third term using (5), we obtain ))))f(x, 0) ∂ ∂uψk(|u|)))))≤|f(x, 0)|Cψk≤C1(x)p−1 pCψk≤-C1-p−1 p ∞Cψk. •For the fourth term, we have to consider several cases. If 0 ≤|u|≤√%kor |u|≥1/%k+ 1, we obtain (1 −ψk(|u|)) !R ρ% "k(u−s)f(x, s)ds =0. If √%k<|u|<1/%k+ 1, then using (5) we have ))))(1 −ψk(|u|)) !R ρ% "k(u−s)f(x, s)ds)))) ≤!B!k))ρ% "k(s)))+α p−1 p 1|u−s|p−1+C1(x)p−1 p,ds ≤-C1-p−1 p ∞!R))ρ% "k(s)))ds +2 p−1α p−1 p 1(1/%k+1) p−1!R))ρ% "k(s)))ds ≤D"k, as ρ"k∈C∞ 0(R;R). •For the last term, if √%k<|u|<1/%k+ 1, using (5) we have ))))−∂ ∂uψk(|u|)!R ρ"k(s)f(x, u −s)ds)))) ≤)))) ∂ ∂uψk(|u|)))))!R ρ"k(s)α p−1 p 1|u−s|p−1ds +)))) ∂ ∂uψk(|u|)))))!R ρ"k(s)C1(x)p−1 pds ≤)))) ∂ ∂uψk(|u|)))))-C1-p−1 p ∞+)))) ∂ ∂uψk(|u|)))))2p−2α p−1 p 1!B!k ρ"k(s)'|u|p−1+|s|p−1(ds ≤Cψk-C1-p−1 p ∞+Cψk2p−1α p−1 p 1(1/%k+1) p−1=D"k. In other case ψku (|u|) = 0. Then (13) holds. Let T>τbe arbitrary. In order to prove the Kneser property let us consider the following auxiliary problem      ∂u ∂t−(u=fk(x, u)+h(t) in Ω×(γ,+∞), u=0 on ∂Ω×(γ,+∞), u(x, γ)=uγ(x),x∈Ω, (15) 6 where γ∈[τ,T]. In view of Lemma 1 for all k≥1 the function fksatisfies (3) and (4), so that by [2, Theorem 2] for any uγ∈L2(Ω) problem (15) has at least one weak solution uk γ(·)definedon[γ,T]. Using Lemma 2 it is standard to check that for the difference w(t) of two solutions we have 1 2 d dt -w(t)-2 L2(Ω)+-∇w(t)-2 L2(Ω)≤D"k-w(t)-2 L2(Ω). Hence, from Gronwall’s lemma we obtain -w(t)-L2(Ω)≤-w(γ)-L2(Ω)eD!k(t−γ),(16) so that the solution is unique. We need some preliminary estimates. Lemma 3 Suppose that Ω⊂RNis a non-empty open unbounded set which satisfies (1) and suppose that f:Ω×R1→ Ris such that f(·,u)is a measurable function for any u∈R,f(x, ·)∈C(R)for almost every x∈Ω,andsatisfies(3)and(4). Leth=4N i=1 ∂hi ∂xi,withhi∈L2 loc(R;L2(Ω)) for all 1≤i≤N, such that N 5 i=1 !t −∞ eλ1s-hi(s)-2 L2(Ω)ds < +∞∀t∈R.(17) Then there exists R=R(B,T)(not depending neither on γnor k), where Bis a bounded set of L2(Ω), such that 66uk γ(t)66L2(Ω)≤R,∀t∈[γ,T],(18) and 66uk γ(·)66Lp(γ,T;Lp(Ω)) ≤R,(19) for any uγ∈B,whereuk γ(·)is the unique solution to (15) with uk γ(γ)=uγ. Proof. We note that using (7) and (8) for fkone can easily obtain that the functions uk γsatisfy the estimate 1 2 d dt 66uk γ(t)662 L2(Ω)+1 266∇uk γ(t)662 L2(Ω)+*α266uk γ(t)66p Lp(Ω)≤1 2-h(t)-2 ∗+666* C2666L1(Ω). Integrating between γto t,wehave 66uk γ(t)662 L2(Ω)+!t γ66∇uk γ(s)662 L2(Ω)ds +2*α2!t γ66uk γ(s)66p Lp(Ω)ds (20) ≤-uγ-2 L2(Ω)+!T γ-h(s)-2 ∗ds +2666* C2666L1(Ω)(T−γ). Hence, (18) and (19) follow. Lemma 4 We suppose the same assumptions for Ω,fand has in Lemma 3. Let Kbe a relatively compact set in L2(Ω).Then,forallτ≤Tand ε>0there exists M=M(γ,T,ε,K)such that !Ω∩{|x|RN≥2m}))uk γ(x, t)))2dx ≤ε,∀t∈[γ,T],∀γ∈[τ,T],∀m≥M,∀k, for any uγ∈K,whereuk γ(·)is the unique solution to (15) with uk γ(γ)=uγ. Proof. Thanks to Lemma 1 we have that fksatisfies (7) and (8). If we argue as in [2, Lemma 15] we obtain a similar estimate for uk γ. Remark 5 Condition (17) is necessary in order to prove Lemma 4, but not for Lemma 3. However, for the sake of clarity we use the same conditions in all lemmas. 7 Theorem 6 Under the assumptions for Ω,fand has in Lemma 3, the set U(t, τ, uτ)is connected in L2(Ω)for any t∈[τ,T]. Proof. The case t=τis obvious. Suppose that for some t∗∈(τ,T]thesetU(t∗,τ,u τ) is not connected. Then there exist two compact sets A1,A2⊂L2(Ω) such that A1∪A2=U(t∗,τ,u τ), A1∩A2=∅. Let u1(·), u2(·)∈S(τ,uτ)besuchthatu1(t∗)∈E1,u2(t∗)∈E2,whereE1,E2are disjoint open neighborhoods of A1,A2, respectively. Let uk i(t, γ), i=1,2,beequaltoui(t), if t∈[τ,γ], and let uk i(t, γ) be the unique solution of problem (15), if t∈[γ,T]. The proof of the following lemma is the same as in [14, Theorem 5] or [22, Lemma 8], so that we omit it. Lemma 7 The maps γ−→ uk(t, γ)are continuous for each fixed k≥1and t∈[τ,T]. Now we put γ(λ)=7τ−(T−τ)λ, if λ∈[−1,0], τ+(T−τ)λ, if λ∈[0,1], and define the function ϕk(λ)(t)=7uk 1(t, γ(λ)) if λ∈[−1,0], uk 2(t, γ(λ)) if λ∈[0,1]. We have ϕk(−1)(t)=uk 1(t, T )=u1(t), ϕk(1)(t)=uk 2(t, T )=u2(t). The map λ1→ ϕk(λ)(t)∈L2(Ω) is continuous for any fixed k≥1, t∈[τ,T](notethatuk 1(t, τ)=uk 2(t, τ)) and ϕk(−1)(t∗)∈E1,ϕk(1)(t∗)∈ E2, so that there exists λk∈[−1,1] such that ϕk(λk)(t∗)/∈E1∪E2.Denoteuk(t)=ϕk(λk)(t). Note that for each k≥1 either uk(t)=uk 1(t, γ(λk)) or uk(t)=uk 2(t, γ(λk)). For some subsequence it is equal to one of them, say uk 1(t, γ(λk)). Now we shall consider the function uk 1(t, γ(λk)), t∈[τ,T]. We have uk(t)=7u1(t), if t∈[τ,γ(λk)], uk 1(t, γ(λk)), if t∈[γ(λk),T], where γ(λk)−→ γ0∈[τ,T]. We define the functions 1 fk(t, x, v)=7f(x, v), if t∈[τ,γ(λk)], fk(x, v), if t∈(γ(λk),T]. By continuity u1(γ(λk)) −→ u1(γ0), as k−→ ∞. Further, by (20), 8uk(·)9is bounded in L∞-τ,T;L2(Ω).∩L2-τ,T;H1 0(Ω).∩Lp(τ,T;Lp(Ω)) . (21) It follows also that duk dt is bounded in Lp"'τ,T;Lp"(Ω)(+L2-τ,T;H−1(Ω).. Then for some function u=u(x, t)wehave uk,uweakly in L2-τ,T;H1 0(Ω).,(22) uk,uweakly in Lp(τ,T;Lp(Ω)) , duk dt ,du dt weakly in Lp"'τ,T;Lp"(Ω)(+L2-τ,T;H−1(Ω)., uk∗ ,uweakly star in L∞-τ,T;L2(Ω).. Arguing in a similar way as in [24, p.75] we first deduce lim a→0sup k!T−a τ66uk(t+a)−uk(t)662 L2(Ω)dt =0, (23) for all T>τ. 8 Now, for all m∈Z,m≥1, we denote Ωm=Ω∩8x∈RN:|x|RN<m 9, where |·|RNdenotes the Euclidean norm in RN.Letφ∈C1([0,+∞)) be a function such that 0≤φ(s)≤1, φ(s)=1 ∀s∈[0,1], φ(s)=0 ∀s≥2. For each kand m≥1, we define uk,m(x, t)=φ/|x|2 RN m20uk(x, t)∀x∈Ω2m,∀k,∀m≥1. (24) We obtain from (21) that, for all m≥1, the sequence 8uk,m9k≥1is bounded in L∞-τ,T;L2(Ω2m).∩ Lp(τ,T;Lp(Ω2m)) ∩L2-τ,T;H1 0(Ω2m)., for all T>τ. In particular, it follows that lim a→0sup k/!τ+a τ66uk,m(t)662 L2(Ω2m)dt +!T T−a66uk,m(t)662 L2(Ω2m)dt0=0. On the other hand, from (23) we deduce that for m≥1, lim a→0sup k/!T−a τ66uk,m(t+a)−uk,m(t)662 L2(Ω2m)dt0=0. Moreover, as Ω2mis a bounded set, H1 0(Ω2m) is included in L2(Ω2m) with compact injection. Then, by the Compactness Theorem 13.3 and Remark 13.1 of [25] with X=L2(Ω2m), Y=H1 0(Ω2m), p=2andG=8uk,m9k≥1, we obtain that 8uk,m9k≥1is relatively compact in L2-τ,T;L2(Ω2m)., and thus, taking into account that uk,m(x, t)=uk(x, t) for all x∈Ωm, we deduce that, in particular, for all m≥12uk |Ωm3k≥1is precompact in L2-τ,T;L2(Ωm)..(25) It is not difficult to conclude from (25), (22), via a diagonal procedure, the existence of a subsequence of 8uk9k≥1(which we denote as the sequence) such that uk |Ωm(x, t)−→ u|Ωm(x, t) for a.a. (x, t)∈Ωm×(τ,T) , for all m≥1. (26) It is easy to obtain (see [20, Chapter 3]) that duk |Ωm dt is bounded in the space Lp"'τ,T;Lp"(Ωm)(+ L2-τ,T;H−1(Ωm)., which is continuously embedded in Lq(τ,T;H−s(Ωm)) for s=max21,N'1 p"−1 2(3. From (26), 1 fk(x, uk |Ωm(x, t)) −→ f(x, u|Ωm(x, t)) for a.a. (x, t)∈Ωm×(τ,T), and then the boundedness of 1 fk(x, uk |Ωm) in Lp"'τ,T;Lp"(Ωm)(implies that 1 fk(x, uk |Ωm)convergestof(x, u|Ωm) weakly in Lp"'τ,T;Lp"(Ωm)(for any m≥1 (see [17]). Also, we note that (25), Lemma 4 and [2, Lemma 15] imply (up to a subsequence) that uk−→ ustrongly in L2-τ,T;L2(Ω).,(27) uk(t)−→ u(t) in L2(Ω) for a.a. t∈(τ,T). 9 Remark 23 The results given in [22] are not applicable to problem (41), as the condition on the derivative used in that paper is not satisfied. On the other hand, the results of the previous section can be applied also to the complex Ginzburg-Landau equation (as done in [22]). 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