Global and exponential attractors for the Penrose-Fife system Giulio Schimperna Dipartimento di Matematica, Universit`a di Pavia, Via Ferrata, 1, I-27100 Pavia, Italy, E-mail: [email protected] 8 0 February 2, 2008 0 2 n Abstract a ThePenrose-Fifesystem forphasetransitions isaddressed. Dirichlet boundaryconditionsfor J the temperature are assumed. Existence of global and exponential attractors is proved. Differ- 7 ently from preceding contributions, here the energy balance equation is both singular at 0 and 1 degenerate at ∞. For this reason, the dissipativity of the associated dynamical process is not ] trivial and hasto be proved rathercarefully. P A AMS (MOS) subject classification: 35B41, 35K55,80A22 . h 1 Introduction t a m We consider here the thermodynamically consistent model for phase transitions proposed by Penrose [ and Fife in [17, 18] and represented by the equations 1 v ϑ +λ(χ) +div m(ϑ)∇1 =g, (1.1) 5 t t (cid:16) ϑ(cid:17) 5 1 1 6 χt−∆χ+W′(χ)=λ′(χ) − + . (1.2) 2 (cid:16) ϑ ϑc(cid:17) 1. The system above is settled in a smooth, bounded domain Ω⊂R3, with boundary Γ. The unknowns 0 are the absolute temperature ϑ > 0 and the order parameter χ. The smooth functions λ′, m and 8 W represent the latent heat, the thermal conductivity, and the potential associated to the local 0 phase configuration, respectively, and ϑ > 0 is a critical temperature. Finally, g is a volumic heat : c v source. On the basis of physical considerations,the kinetic equation (1.2) is complemented, as usual, i X with no-flux (i.e., homogeneous Neumann) boundary conditions; instead, various types of meaningful boundaryconditionscanbeassociatedwiththeenergybalance equation(1.1). Weshallconsiderhere r a the Dirichlet boundary conditions. As far as well-posedness is concerned, system (1.1)–(1.2) has been studied in a number of recent works, among which we quote [4, 5, 7, 11, 14, 15, 29], under various assumptions on the data. The paperslistedabovealsocontainamuchmorecomprehensivebibliography. Justarapidsurveyof theliteraturesuggeststhat,indeed,thechoiceoftheboundaryconditionsforϑcangiverisetoseveral different mathematical situations. In particular, the Dirichlet and Robin conditions seem easier to treat than the Neumann ones (cf., e.g., [4, 11] for further comments), due to correspondingly higher coercivity. Another important factor is the expression of the thermal conductivity m. Meaningful choices are given by (cf. [5] for further comments) m(r)∼m r+m r2, m ,m ∈[0,∞). (1.3) 0 ∞ 0 ∞ In particular, m = 0,m > 0 represents the Fourier heat conduction law, which appears to be the 0 ∞ mostdifficult situation[15]since equation(1.1), whichis now linearinϑ, is coupledwith thesingular relation (1.2). Instead, in the case m >0,m =0, the well-posedness issue is simpler (cf. [14, 29]); 0 ∞ however, there is a lack of coercivity for large ϑ, which creates difficulties in the long-time analysis. 1 Finally, the probably simplest situation is that proposed in [5] (see also [6]), i.e., m ,m > 0, since 0 ∞ (1.1) maintains both the singular character at 0 and the coercivity at ∞. In view of these considerations,it is not surprising that the long time behavior of (1.1)–(1.2) isbetter understoodwhenm ,m >0,andinthiscasethe existenceoftheglobalattractorhasbeen 0 ∞ shownin[22,23]. Indeed,testing(1.1)byϑonereadilygetsadissipativeestimateforthetemperature, which permits to constructa uniformly absorbingset and, consequently,the globalattractor. Similar resultsarealsoobtainedin[12,13],whereitisactuallytakenm =0,butatermµ ϑ,withµ >0, ∞ ∞ ∞ is added on the left hand side of (1.1), so that the system is still coercive in ϑ. Speaking of the non-coercive case m = 0, up to our knowledge the only papers devoted to ∞ the large-times analysis of it are [27] (see also [28] for the conserved case) and [10]. In [27], the case of homogeneous Neumann conditions for both unknowns is addressed in one space dimension, and existence of a global attractor is shown in a proper phase space which takes into account the conser- vation(ordissipation)propertiescomingfromthe no-flux conditions. In[10], the (non-homogeneous) Dirichlet case is considered in three space dimensions and ω-limits of single trajectories are studied. It is worth remarking that in both papers the external source g is taken equal to 0. Inthe presentwork,weprovidea further contributionto theanalysisofthe noncoercivecase. Precisely,weassumem >0,m =0,andtakeDirichletboundaryconditionsforϑexactlyasin[10]. 0 ∞ For the resulting problem, we show existence of both global and exponential attractors. Comparing with [10], where the behavior of a single trajectory is investigated, here the proofs are very different andinseveralpointsmoredifficult. Indeed,determiningattractorsmeanstounderstandthebehavior ofbundlesoftrajectories,sothatweneedtofindestimateswhichareuniformnotonlyintime,butalso withrespecttoinitialdatavaryinginaboundedset. Wethentrytominimizetechnicalitiesbymaking some restrictions on data. Namely, we take a constant latent heat (i.e., set λ(χ) = χ), set m(r) = 1 (i.e., m = 1, m = 0), let the critical temperature ϑ be equal to 1, and correspondingly assume 0 ∞ c the Dirichlet condition ϑ ≡ 1 on the boundary. Actually, all these assumptions could by avoided by paying the price of some additional computations in the proofs. More restrictive is, instead, the assumption g =0, which we take exactly as it was done in [10, 27]. We then end up with the system 1 ϑ +χ −∆ − =0, (1.4) t t (cid:16) ϑ(cid:17) 1 χ −∆χ+W′(χ)=1− . (1.5) t ϑ Being g = 0, (1.4)–(1.5) admits a Liapounov functional (and consequently a dissipation integral), and this information will be crucial to overcome the lack of coercivity in ϑ. Actually, the global attractor will be constructed by proving uniform boundedness and asymptotic compactness of single trajectories and taking advantage of the dissipation property. Although this procedure might seem straighforward,the proof presents a number of difficulties. First of all, we have to settle the problem in a phase space X (cf. (2.13) below) where both ϑ and χ are bounded in sufficiently strong norms. The conditions we require on the initial data are in fact more restrictive than what is necessary, e.g., for the mere well-posedness. In particular, we cannot deal with completely general potentials W. Namely,weareforcedtoassumeW beasmoothfunctiondefinedonthewholerealline(like,e.g.,the double well potential W(r) = (r2 −1)2), and, for instance, we cannot treat the singular potentials, i.e., those being identically +∞ outside a bounded interval, like the so-called logarithmic potential W(r)=(r+1)log(r+1)+(1−r)log(1−r)−λr2,whereλ>0. Moreover,wenotethat,inanalogywith the coercive case m >0 studied in [22], X does not have a Banach structure, due to the nonlinear ∞ terms inthe energy,but it is just a metric space. Inthis setting, the key point of ourargumentis the proofofauniformtimeregularizationpropertyforthesolutions,which,inouropinion,canconstitute an interesting issue by itself. Namely, we can show that both ϑ and u=ϑ−1 are uniformly bounded for sufficiently large times, whereas this need not hold for the initial temperature ϑ . Thus, (1.4) 0 eventually loses both the singular and the degenerate character. A further open problem to which we give a positive answer is the existence of exponential attractors for the system (1.4)–(1.5). This is shown by using the so-called method of ℓ-trajectories (cf. [16, 19,20, 21]). However,we cannotproveexponentialattractionin the metric of X (thatkeeps, insomeway,atraceofthenonlinearterms),butareforcedtoworkwithaweakernorm,corresponding in fact to the only contractive estimate which seems to hold for system (1.4)–(1.5). 2 Therestofthispaperisorganizedasfollows. InthenextSection2,wepresentourhypotheses and state our main results. The proofs are collected in Section 3. Acknowledgment. We express our gratitude to Elisabetta Rocca and Riccarda Rossi for fruitful discussions on the subject of this work. 2 Notation and main results Let Ω ⊂ R3 be a smooth bounded domain with boundary Γ. Let us set H := L2(Ω) and denote by (·,·) both the scalar product in H and that in H ×H, and by k·k the induced norm. The symbol k·k indicates the norm in the generic Banach space X. Next, we set V := H1(Ω), V := H1(Ω), X 0 0 and define A:V →V′, hAv,zi := ∇v·∇z, ∀v,z ∈V , (2.1) 0 0 0 Z 0 Ω B :V →V′, hBv,zi:= vz+∇v·∇z , ∀v,z ∈V, (2.2) Z Ω(cid:0) (cid:1) h·,·i and h·,·i denoting the duality pairings between V and V′ = H−1(Ω) and between V and V′, 0 0 0 respectively. It turns out that A and B are the Riesz operators associated to the standard norms in V and V, respectively. 0 Our hypotheses on the potential W are the following: W ∈C2(R;R), W′(0)=0, lim W′(r)r =+∞, (2.3) r→∞ ∃λ≥0: W′′(r)≥−λ ∀r ∈R. (2.4) Inparticular,bythelatterassumption,β(r):=W′(r)+λrisincreasinglymonotone. Next,considering B, witha smallabuse ofnotation,as astrictly positiveunboundedlinearoperatoronH withdomain D(B) = {v ∈ H2(Ω) : ∂nv = 0 on ∂Ω}, we can take real powers of B and set V2s := D(Bs), endowed with the graph norm kvk := kBsvk. Note that V = V. The variational formulation of s 1 system (1.4)–(1.5) takes then the form 1 ϑ +χ +A 1− =0, in V′, (2.5) t t (cid:16) ϑ(cid:17) 0 1 χ +Bχ+W′(χ)=1− , in V′ (2.6) t ϑ (in order to get the Riesz map B, χ has been added and subtracted from the left hand side, and −χ has been included in W′). Next, we define the associated energy functional as: 1 1 E =E(ϑ,χ):= ϑ−logϑ+ |χ|2+ |∇χ|2+W(χ) . (2.7) ZΩ(cid:16) 2 2 (cid:17) We immediately observe that E is finite and bounded from below on the “energy space” X := (ϑ,χ):ϑ∈L1(Ω), ϑ>0 a.e. in Ω, logϑ∈L1(Ω), χ∈V, W(χ)∈L1(Ω) . (2.8) E (cid:8) (cid:9) Nevertheless, due to the lack of coercivity (and consequently of compactness) in ϑ (the finiteness of energyonlyimpliesthatϑ∈L1(Ω)), noexistenceresultisknown,uptoourknowledge,fordatalying just in X . Namely, noting asProblem(P)the coupling of (2.5)–(2.6) (intended to holdfor a.e.value E of time in (0,∞)) with the initial condition ϑ| =ϑ , χ| =χ , a.e. in Ω, (2.9) t=0 0 t=0 0 we have the following result, proved in [10, Thm. 2.1] (see also [11, Prop. 2.1]): 3 Theorem 2.1. Let (2.3)–(2.4) hold and let (ϑ ,χ ) ∈ X . Let, in addition ϑ ∈ Lp(Ω) for some 0 0 E 0 p>6/5. Then, there exists one and only one couple (ϑ,χ) solving Problem (P) and such that ϑ∈H1(0,T;H−1(Ω))∩L∞(0,T;Lp(Ω)), ϑ>0 a.e. in Ω×(0,T), (2.10) 1−1/ϑ ∈L2(0,T;V ), (2.11) 0 χ(cid:0) ∈H1(0(cid:1),T;H)∩C0([0,T];V)∩L2(0,T;H2(Ω)), (2.12) hold for all T >0. Such a couple will be called a “solution” in the sequel. Since we need to control uniformly in time the “large values” of the temperature, we have to ask a bit more summability on ϑ and a bit more regularity on χ . Correspondingly, we will also get 0 0 some more regularity than (2.10)–(2.12). Namely, we set X := (ϑ,χ)∈X :ϑ∈Lp(Ω), χ∈V , (2.13) E 3+ǫ (cid:8) 2 (cid:9) where we assume that ǫ∈(0,1), p>3. (2.14) Actually, we need ǫ>0 in orderto ensure that χ staysin L∞(Ω), while the higher summability ofϑ 0 seems necessary to get a uniform in time estimate for ϑ(t). We remark that the set X, which of course has no linear structure, can be endowed with a complete metric which makes it a suitable phase space for the associated dynamical process. As in [22] (see also [24, 26]), we can take d (ϑ ,χ ),(ϑ ,χ ) :=kϑ −ϑ k +kχ −χ k X 1 1 2 2 1 2 Lp(Ω) 1 2 3+ǫ (cid:0) (cid:1) 2 +klog−ϑ −log−ϑ k +kβ(χ )−β(χ )k , (2.15) 1 2 L1(Ω) 1 2 L1(Ω) where (·)− denotes negative part (notice, however, that the latter term could be omitted since it is dominated by the second one due to (2.3) and the continuous embedding V ⊂ L∞(Ω)). Corre- 3+ǫ 2 spondingly, we take initial data such that (ϑ ,χ )∈X. (2.16) 0 0 Inthesequel,wewilldenotebyS(t)the semigroupoperatorassociatingto (ϑ ,χ )the corresponding 0 0 solutionevaluatedattimet. TheproofthatS(·)fulfillstheusualpropertiesofacontinuoussemigroup onX ismoreorlessstandardandcanbecarriedoutalongthelines,e.g.,of[22,Sec.4]. Hence,weomit the details. Instead, we focus on regularization properties of S(t). The key step of our investigation is the following Theorem 2.2. Let (2.3)–(2.4), (2.14) hold and let B be a set of initial data bounded in X. More precisely, let D stand for the d -radius of the set, namely 0 X D := sup d (ϑ ,χ ),(1,0) . (2.17) 0 X 0 0 (ϑ0,χ0)∈B (cid:0) (cid:1) Then, letting (ϑ ,χ )∈B and (ϑ(t),χ(t)):=S(t)(ϑ ,χ ), there exist a time T >0 and a constant 0 0 0 0 ∞ Q depending only on D , such that, for all t≥T , there holds ∞ 0 ∞ kϑ(t)kV∩L∞(Ω)+kϑ−1(t)kV∩L∞(Ω) ≤Q∞, (2.18) kχ(t)k ≤Q . (2.19) H2(Ω) ∞ Remark 2.3. Suitably modifying the proofs,onecouldshow thatany strictly positive time couldbe takenasT . Weomittheproofofthisfactsinceitwouldinvolvefurthertechnicalcomplications. We ∞ just notice that the quantity Q in (2.18)–(2.19) would then depend on T and explode as T ց0. ∞ ∞ ∞ Notice that the bounds (2.18)–(2.19) are somehow weaker than a true dissipative estimate. Nevertheless,theywillsufficefortheproofofourmainresult(forthedefinitionoftheglobalattractor we refer to the monograpgh[30]): 4 Theorem 2.4. Let the assumptions of Theorem2.2 hold. Then, the semigroupS(·) associatedwith Problem (P) admits the global attractor A, which is compact in X. More precisely, ∃cA >0: kϑkV∩L∞(Ω)+kϑ−1kV∩L∞(Ω)+kχkH2(Ω) ≤cA ∀(ϑ,χ)∈A. (2.20) Finally, we can prove existence of an exponential attractor: Theorem 2.5. Let the assumptions of Theorem 2.2 hold. Then, the semigroup S(·) associated with Problem(P)admitsanexponentialattractorM. Moreprecisely,M isacompactsetofX, whichhas finite fractal dimension in V′×H, such that for any bounded set B ⊂X there holds 0 dist(S(t)B,M)≤Q(D )e−κt, ∀t≥0, (2.21) 0 where dist represents the unilateral Hausdorff distance of sets with respect to the (product) norm in V′ ×H, κ > 0 is independent of B, Q is a monotone function, and D is the X-radius of B given 0 0 by (2.17). Remark 2.6. As noted in the Introduction, we use the (rather weak) topology of V′ ×H since it 0 seems difficult to prove a contractive estimate in a better norm. Further comments will be given at the end of the proof (cf. Remark 3.6 below). 3 Proofs Inwhatfollows,thesymbolsc,κ,andc ,i≥0,willdenotepositiveconstantsdependingonW,Ω,and i independent of the initial datum and of time. The values of c and κ are allowed to vary even within the same line. Moreover, Q : R+ → R+ denotes a generic monotone function. Capital letters like C or C will be used to indicate constant which have other dependencies (in most cases, on the initial i datum). Finally, the symbol c will denote some embedding constants depending only on the set Ω. Ω Proof of Theorem 2.2. The basic idea to provethe uniformbounds (2.18)–(2.19) is to combine an estimate in a small interval [0,T ], where T depends on D , with a further uniform estimate holding 0 0 0 on [T ,∞). This procedure requires a number of steps, which are carried out below. Notice that 0 some parts of the procedure might have a formalcharacterin the presentregularitysetting (e.g., test functionscouldbenotregularenough). However,alltheprocedurecouldbestandardlymaderigorous by working on some approximation and then passing to the limit (notice that the solution is known to be unique). We omit the details of this straighforwardargument, for brevity. First estimate. We start by deriving the energy estimate. Testing (2.5) by 1−1/ϑ, we have d 1 2 1 ϑ−logϑ + 1− =− χ ,1− . (3.1) dtZΩ(cid:0) (cid:1) (cid:13)(cid:13) ϑ(cid:13)(cid:13)V0 D t ϑE Next, multiplying (2.6) by χ , we obtain (cid:13) (cid:13) t d 1 1 1 |χ|2+ |∇χ|2+W(χ) +kχ k2 = χ ,1− , (3.2) dtZΩ(cid:16)2 2 (cid:17) t D t ϑE whence, summing (3.1) and (3.2) and recalling (2.7), d 1 2 E + 1− +kχ k2 =0. (3.3) t dt (cid:13)(cid:13) ϑ(cid:13)(cid:13)V0 (cid:13) (cid:13) In particular, integrating from 0 to an arbitraryt>0 we have t 1 2 E(t)+ 1− +kχ k2 ≤E(0)≤Q(D ). (3.4) Z0 (cid:16)(cid:13)(cid:13) ϑ(cid:13)(cid:13)V0 t (cid:17) 0 (cid:13) (cid:13) Second estimate. We test (2.6) by 2B1+2ǫχt. Thanks to ǫ < 1 and using Poincar´e’s and Young’s inequalities, we obtain d 1 2 1 2 kχk2 +kχ k2 ≤c 1− −W′(χ) ≤c 1− +ckW′(χ)k2. (3.5) dt 3+2ǫ t 1+2ǫ (cid:13)(cid:13) ϑ (cid:13)(cid:13)1+2ǫ 1(cid:13)(cid:13) ϑ(cid:13)(cid:13)V0 V (cid:13) (cid:13) (cid:13) (cid:13) 5 Then, we note that, by (2.3) and the continuous embedding V ⊂L∞(Ω), 3+ǫ 2 kW′(χ)k2 =kW′(χ)k2+ W′′(χ)∇χ 2 ≤ 1+k∇χk2 Q kχk2 ≤Q kχk2 . (3.6) V Z L∞(Ω) 3+ǫ Ω(cid:12) (cid:12) (cid:0) (cid:1) (cid:0) (cid:1) (cid:0) 2 (cid:1) (cid:12) (cid:12) Next, let us compute (3.5) plus c ×(3.3). Using also (3.6), we arrive at 1 d kχk2 +c E +kχ k2 ≤Q kχk2 . (3.7) dt 3+ǫ 1 t 1+ǫ 3+ǫ (cid:2) 2 (cid:3) 2 (cid:0) 2 (cid:1) Thus, noting as Ψ the quantity in square brackets on the left hand side, and using the comparison principle for ODE’s, it follows that there exists a time T > 0, depending on D in a monotonically 0 0 decreasing way, such that kΨkL∞(0,T0) ≤Q(D0), (3.8) whence, integrating (3.7) in time over (0,T ), and recalling (3.4), 0 kχk +kχ k ≤Q(D ). (3.9) L∞ 0,T0;V3+ǫ t L2 0,T0;V1+ǫ 0 (cid:0) 2 (cid:1) (cid:0) 2 (cid:1) In particular, by the continuous embedding V ⊂L3(Ω), we have 1+ǫ 2 kχ k ≤Q(D ). (3.10) t L2(0,T0;L3(Ω)) 0 Third estimate. Let us note that, since ϑ solves (2.5), it is ϑ > 0 a.e. in Ω×(0,∞) and ϑ = 1 a.e. on Γ×(0,∞). Thus, we can test (2.5) by ϑp−1 −1 (p given by (2.14)), which (at least in an approximation) lies in V for a.e. t∈(0,∞). We then get 0 ddtZΩ(cid:16)1pϑp−ϑ(cid:17)+ 4(p(p−−21)2)(cid:13)∇ϑp−22(cid:13)2 ≤−ZΩ(cid:0)ϑp−1−1)χt (3.11) (cid:13) (cid:13) and we estimate the right hand side as follows: −Z ϑp−1−1)χt ≤kχtkL3(Ω) ϑp−22 L6(Ω) ϑp2 +c 1+kχtk2 Ω(cid:0) (cid:13) (cid:13) (cid:13) (cid:13) (cid:0) (cid:1) (cid:13) (cid:13) (cid:13) (cid:13) ≤ σp ϑp−22 2L6(Ω)+cσpkχtk2L3(Ω) ϑp2 2+c 1+kχtk2 (cid:13) (cid:13) (cid:13) (cid:13) (cid:0) (cid:1) ≤ σp(cid:13)∇ϑp−2(cid:13)2 2+ σp +cσpkχtk2L3((cid:13)Ω)kϑ(cid:13)kpLp(Ω)+c 1+kχtk2L3(Ω) , (3.12) (cid:13) (cid:13) (cid:0) (cid:1) (cid:13) (cid:13) whereσ >0denotesa“small”constant,independentofp,tobechosenattheend,andcorrespondingly c >0 depends on the same quantities as the generic c and, additionally, on the final choice of σ. In σ fact, passing from row to row, we allow σ to “incorporate” embedding constants. We used here the continuous embedding V ⊂ L6(Ω) and the Young and Poincar´e inequalities. Although p is a fixed value (cf. (2.14)), here and below we emphasize the dependence on p of the estimates, since they will be readily repeated with different exponents. Adding2×(3.1)(wherethetermontherighthandsideissplitviaYoung’sinequality)to(3.11), multiplying the result by p, and taking σ small enough, we then obtain ddtZ ϑp+p(ϑ−2logϑ) +κ ∇ϑp−22 2 ≤cp+p2c2kχtk2L3(Ω) 1+kϑkpLp(Ω) (3.13) Ω(cid:2) (cid:3) (cid:13) (cid:13) (cid:0) (cid:1) (cid:13) (cid:13) for some c >0. Now, let us set 2 m:=c kχ k2 , so that, by (3.10), kmk ≤Q(D ). (3.14) 2 t L3(Ω) L1(0,T0) 0 Defining Y as 1 plus the integral on the left hand side of (3.13), we then have dY +κ ∇ϑp−22 2 ≤cp+p2m 1+kϑkp ≤cp+p2mY, (3.15) dt Lp(Ω) (cid:13) (cid:13) (cid:0) (cid:1) (cid:13) (cid:13) 6 whence, recalling (3.10) and (2.13) and using Gronwall’s Lemma, kϑk ≤Q(D ). (3.16) L∞ 0,T0;Lp(Ω) 0 (cid:0) (cid:1) Fourth estimate. We test (2.5) by 2tϑ /ϑ2; next, we differentiate (2.6) in time and test the result t by 2tχ . Taking the sum and noting that two terms cancel, we get t d 1 2 ϑ2 1 2 tkχ k2+t ∇ +2t t +2tkχ k2 ≤(1+2λt)kχ k2+ ∇ . (3.17) dt(cid:16) t (cid:13)(cid:13) ϑ(cid:13)(cid:13) (cid:17) ZΩ ϑ2 t V t (cid:13)(cid:13) ϑ(cid:13)(cid:13) (cid:13) (cid:13) (cid:13) (cid:13) Then, integrating over (0,T ) and using (3.4), we obtain 0 χ (T ) 2+ ∇ 1 2 ≤Q(D ) 1+ 1 ≤Q(D ); (3.18) t 0 0 0 (cid:13) (cid:13) (cid:13) ϑ(T0)(cid:13) (cid:16) T0(cid:17) (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) actually, (T )−1 depends increasingly on D . 0 0 Fifth estimate. Up to now, we got uniform bounds in the (small) time interval [0,T ]. Our aim 0 is now to get uniform estimates on [T ,∞). First, we essentially repeat the previous estimate, but 0 without the weight t. This gives, of course, d 1 2 ϑ2 kχ k2+ ∇ +2 t +2kχ k2 ≤2λkχ k2, (3.19) dt(cid:16) t (cid:13)(cid:13) ϑ(cid:13)(cid:13) (cid:17) ZΩ ϑ2 t V t (cid:13) (cid:13) whence, integrating over (T ,t) for arbitraryt≥T , 0 0 χ (t) 2+ ∇ 1 2+2 t (logϑ) (s) 2+ χ (s) 2 ds (cid:13)(cid:13) t ≤(cid:13)(cid:13) χ ((cid:13)(cid:13)(cid:13)T )ϑ2(t+)(cid:13)(cid:13)(cid:13) ∇ 1ZT0(cid:16)2(cid:13)(cid:13)+2λ tt kχ(cid:13)(cid:13)(s)k(cid:13)(cid:13)2 dts≤(cid:13)(cid:13)VQ(cid:17)(D ), (3.20) (cid:13)(cid:13) t 0 (cid:13)(cid:13) (cid:13)(cid:13)(cid:13) ϑ(T0)(cid:13)(cid:13)(cid:13) ZT0 t 0 where we used (3.18) and (3.4) to control the terms on the right hand side. Sixth estimate. We repeat the Third estimate restarting from T . Let us notice that, by (3.14), 0 (3.20) and the continuous embedding V ⊂L6(Ω), m=c kχ k2 satisfies now M :=kmk ≤Q(D ). (3.21) 2 t L3(Ω) L1(0,∞) 0 Thus, by the continuous embedding V ⊂L6(Ω), (3.15) takes the form d Y +κkϑkp−2 ≤cp+p2mY. (3.22) dt L3p−6(Ω) Let us now notice that, thanks to p>3, we can write p−2 p−2 kϑkpL−p(2Ω) =(cid:16)ZΩϑp(cid:17) p ≤(cid:16)(cid:13)ϑp(cid:13)L3pp−6(Ω)(cid:13)1kL23pp−−66(Ω)(cid:17) p =kϑkpL−3p2−6(Ω)|Ω|2p3−p(cid:13)6 ≤(cid:13)cΩkϑkpL−3p(cid:13)2−6(Ω). (3.23) Moreover,we have p−2 kϑkpL−p(2Ω) =(cid:16)Y−1−pZΩ(ϑ−2logϑ)(cid:17) p ≥cYp−p2 −pQ(D0). (3.24) Thus, (3.22) gives dY+κYp−p2 ≤p2mY +pQ(D0), (3.25) dt so that, for H:=logY, dH+e−2pH κ−pQ(D0)e−(p−p2)H ≤p2m. (3.26) dt (cid:2) (cid:3) 7 Noting as Σ the quantity in square brackets, an easy computation shows that p p Σ≥0⇔H≥ log Q(D ) =:ζ. (3.27) 0 p−2 (cid:16)κ (cid:17) Consequently, it is not difficult to obtain ∞ kHkL∞(T0,∞) ≤max ζ,H(T0) +p2Z m(s) ds, (3.28) (cid:8) (cid:9) 0 so that, being by (3.27) and (3.16), exp(ζ)≤pQ(D ), exp(H(T ))=Y(T )≤pQ(D ), (3.29) 0 0 0 0 and using (3.21), we readily get kϑkpL∞(T0,∞;Lp(Ω)) ≤kYkL∞(T0,∞) ≤exp max{ζ,H(T0)} exp p2M (cid:0) (cid:1) (cid:0) (cid:1) ≤pQ(D )exp p2M , (3.30) 0 (cid:0) (cid:1) whence, clearly, kϑkL∞(T0,∞;Lp(Ω)) ≤Q(D0). (3.31) Suitable time integrations of (3.22) permit us collect what we have proved so far in a Lemma 3.1. Under the assumptions of Theorem 2.2, there exist a time T > 0 and a quantity 0 Q >0, both depending on D , such that, for all t≥T ≥T , 0 0 0 t+1 kϑ(t)kp + kϑ(s)kp−2 ds≤Q , (3.32) Lp(Ω) Z L3p−6(Ω) 0 t t kϑ(s)kp−2 ds≤Q +Q (t−T). (3.33) Z L3p−6(Ω) 0 0 T Our next aim is to extend (3.32) and (3.33) to any finite exponent. Namely, we have Lemma 3.2. Underthe assumptionsof Theorem2.2,for allq ∈[p,∞)there exista time T >0and q a quantity Q >0, both depending on D and q, such that, for all t≥T ≥T , q 0 q t+1 kϑ(t)kq + kϑ(s)kq−2 ds≤Q , (3.34) Lq(Ω) Z L3q−6(Ω) q t t kϑ(s)kq−2 ds≤Q +Q (t−T). (3.35) Z L3q−6(Ω) q q T Proof. It suffices to iterate finitely many times the procedure in the Sixth Estimate. Namely, setting p :=p, we observe that, by (3.32) and interpolation, there follows 0 sup kϑk ≤Q(D ), (3.36) Lpi(t,t+1;Lpi(Ω)) 0 t∈[T0,∞) where we have set (for i=1, at least in the meanwhile) 5p −6 i−1 p := . (3.37) i 3 Note that p >p since p >3. Then, we repeat the argument leading to (3.22), but with p in place 1 0 0 1 of p. This gives (for i=1 and with obvious meaning of Y ) i d Y +ckϑkpi−2 ≤cp +p2mY . (3.38) dt i L3pi−6(Ω) i i i Noting that both m and Y are summable on time intervals of finite length thanks to (3.21) and, i respectively, (3.36), we can use the uniform Gronwall Lemma (cf., e.g., [30, Lemma III.1.1]), that gives kϑ(t)kpi ≤Y (t)≤Q (D ) ∀t≥T :=T +1, (3.39) Lpi(Ω) i i 0 i i−1 8 with obvious meaning of Q (D ). Thus, suitable integrations in time of (3.38) give the analogue of i 0 (3.32)and(3.33)withp inplaceofq. Toget(3.34)and(3.35),itthensufficestoproceedbyiteration 1 oni untilp is largerthanq. Notice thatsince a finite numberofsteps is sufficient, we do nothaveto i take care of the dependence on i of the quantities Q and T (both, in fact, would explode if infinite i i iterations were needed). The proof of the Lemma is concluded. A similar property holds also for the inverse temperature: Lemma 3.3. Setting u:=ϑ−1, under the assumptions of Theorem 2.2, for all q ∈[1,∞) there exist a time T >0 and a quantity Q >0, both depending on D and q (and possibly largerfrom those in q q 0 the previous Lemma) such that, for all t≥T ≥T , q t+1 ku(t)kq + ku(s)kq+2 ds≤Q , (3.40) Lq(Ω) Z L3q+6(Ω) q t t ku(s)kq+2 ds≤Q +Q (t−T). (3.41) Z L3q+6(Ω) q q T Proof. Note thatwe alreadyknowthe bound ofthe firsttermin(3.40)for q =6 thanks to(3.20) and the continuous embedding V ⊂ L6(Ω). Then, we proceed essentially as in the Third estimate, i.e., for a generic q ≥6, we multiply (2.5) by 1−uq+1. In place of (3.11), we get ddtZΩ(cid:16)1quq+ϑ(cid:17)+ 4(q(q++21)2)(cid:13)∇uq+22(cid:13)2 ≤−ZΩ(cid:0)1−uq+1)χt, (3.42) (cid:13) (cid:13) so that, estimating the right hand side as in (3.12), we infer d Z +ckukq+2 ≤cq+q2mZ , where Z := uq+qϑ (3.43) dt q L3q+6(Ω) q q Z Ω(cid:0) (cid:1) and m is as in (3.14) (possibly for a different value of c ). At this point, noticing that the exponents 2 are even better than in (3.22), the proof can be completed by mimicking the arguments in the Sixth estimate and in the proof of Lemma 3.2. Lemma 3.4. Under the assumptions of Theorem 2.2, there exist a time T and a quantity Q , both ∗ ∗ depending on D , such that, for all t≥T , 0 ∗ kχ (t)k2 ≤Q . (3.44) t L24/5(Ω) ∗ Proof. The exponent 24/5 in (3.44) is chosen just for later convenience. In fact, (3.44) can be proved for any exponent strictly smaller than 6. Differentiating in time (2.6), we have ϑ χ +Bχ =Φ:=−W′′(χ)χ + t (3.45) tt t t ϑ2 and we claim that, for any ν ∈(0,1), we can choose T >0 and Q >0, both depending only on D ν ν 0 and ν, such that ∞ kΦ(s)k2 ds≤Q(D ). (3.46) Z H−ν(Ω) 0 Tν Actually, recalling (3.20) and applying standard regularity results to (2.6) (seen here as a time- dependent family of elliptic equations), it follows that kχ(t)k ≤Q(D ) for all t≥T . (3.47) H2(Ω) 0 0 Thus, by (2.3), the continuous embedding H2(Ω)⊂L∞(Ω), and (3.4), ∞ kW′′(χ(s))χ (s)k2 ds≤Q(D ). (3.48) Z t 0 T0 9 Analogously, using the first integral bound in (3.20), the bound of the first term in (3.40) with q sufficiently large (depending on ν), and elementary interpolation, it is not difficult to get, for some T′ >0 and Q >0, ν ν ∞ ϑ (s) 2 t ds≤Q (D ). (3.49) ZTν′ (cid:13)(cid:13)ϑ2(s)(cid:13)(cid:13)L3+62ν(Ω) ν 0 (cid:13) (cid:13) Thus, thanks to (3.48), (3.49)and the continuousembedding L3+62ν(Ω)⊂H−ν(Ω), we see that(3.46) holds for any T ≥T′. Now, let us observe that, by the bound of the second integral term in (3.20), ν ν T ∈[T′,T′ +1] can be chosen such that ν ν ν kχ (T )k2 ≤ckχ (T )k2 ≤Q(D ). (3.50) t ν H1−ν(Ω) t ν V 0 Then,applyingthestandardlinearparabolicHilberttheorytotheequation(3.45)onthetimeinterval [T ,∞) andwith the initialcondition χ (T ), andusing (3.46), we have(possibly for a different value ν t ν of Q ) ν kχtkC0([Tν,∞);H1−ν(Ω))+kχtkL2(Tν,∞;H2−ν(Ω)) ≤Qν(D0), (3.51) whence the assert follows from the continuous embedding H1−ν(Ω) ⊂ L24/5(Ω), which holds for ν small enough. End of proof of Theorem 2.2. We use a modified Alikakos-Moser[2] iteration scheme similar to that in [14], but suitably adapted in order to obtain time regularization effects. Similar procedures have been proved to be effective in other recent papers, cf. [8, 25]. Asafirststep,wecomebackto(3.11), wherethe exponentpis substitutedbyanumber q to i be chosenlater. Since weneedinfinitely manyiterations,nowthe righthandside hasto be estimated more carefully. Namely, we have −Z ϑqi−1−1)χt ≤kχtkL24/5(Ω) ϑqi2−2 L6(Ω) ϑq2i L8/5(Ω)+c 1+kχtk2 Ω(cid:0) (cid:13) (cid:13) (cid:13) (cid:13) (cid:0) (cid:1) (cid:13) (cid:13) (cid:13) (cid:13) ≤ qσ ∇ϑqi2−2 2+ qσ +cσqiQ∗kϑkqLi4qi/5(Ω)+C0, (3.52) i(cid:13) (cid:13) i (cid:13) (cid:13) where Q is exactly the same quantity as in (3.44) and the constant C depends on D and is inde- ∗ 0 0 pendent of q . Thus, possibly modifying C , in place of (3.22) we get i 0 d Y +κkϑkqi−2 ≤cq2Q kϑkqi +C q , (3.53) dt i L3qi−6(Ω) i ∗ L4qi/5(Ω) 0 i where it is worth noting that kϑkqi ≤Y :=1+ ϑqi +q (ϑ−2logϑ) ≤kϑkqi +C q , (3.54) Lqi(Ω) i Z i Lqi(Ω) 1 i Ω(cid:2) (cid:3) where C is another quantity depending only on D . Moreover,(3.53) gives 1 0 dY +κkϑkqi−2 ≤cq2Q kϑkq5i+2 kϑk45qi−2 +C q , (3.55) dt i L3qi−6(Ω) i ∗ L45qi(Ω) L125qi−6(Ω) 0 i provided that 12q /5−6≥4q /5, which is true for all i∈N if q is large enough and we set i i 0 5 q = q , ∀i≥1. (3.56) i i−1 4 The choice (3.56) permits to rewrite (3.55) in the form dY +κkϑkqi−2 ≤c q2Q Y14+25qikϑkqi−1−2 +C q , (3.57) dt i L3qi−6(Ω) 3 i ∗ i−1 L3qi−1−6(Ω) 0 i for some c >0. Let us now define 3 ∞ ∞ τ :=i−2, so that τ := τ = i−2 <∞. (3.58) i ∞ i X X i=1 i=1 10