Existence and characterization of attractors for a nonlocal reaction-di§usion equation with an energy functional
Abstract
In this paper we study a nonlocal reaction-di§usion equation in which the di§usion depends on the gradient of the solution. Firstly, we prove the existence and uniqueness of regular and strong solutions. Secondly, we obtain the existence of global attractors in both situations under rather weak assumptions by deÖning a multivalued semiáow (which is a semigroup in the particular situation when uniqueness of the Cauchy problem is satisÖed). Thirdly, we characterize the attractor either as the unstable manifold of the set of stationary points or as the stable one when we consider solutions only in the set of bounded complete trajectories.
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Existence and characterization of attractors for a nonlocal reaction-di¤usion equation with an energy functional R. Caballero1, P. Marín-Rubio2and José Valero1 1Centro de Investigación Operativa, Universidad Miguel Hernández de Elche, Avda. Universidad s/n, 03202, Elche (Alicante), Spain 2Dpto. Ecuaciones Diferenciales y Análisis Numérico, Universidad de Sevilla, C/Tar…a, 41012-Sevilla, Spain Dedicated to the memory of Russell Johnson Abstract In this paper we study a nonlocal reaction-di¤usion equation in which the di¤usion depends on the gradient of the solution. Firstly, we prove the existence and uniqueness of regular and strong solutions. Secondly, we obtain the existence of global attractors in both situations under rather weak assumptions by de…ning a multivalued semi‡ow (which is a semigroup in the particular situation when uniqueness of the Cauchy problem is satis…ed). Thirdly, we characterize the attractor either as the unstable manifold of the set of stationary points or as the stable one when we consider solutions only in the set of bounded complete trajectories. Keywords: reaction-di¤usion equations, nonlocal equations, global attractors, multivalued dynamical systems, structure of the attractor AMS Subject Classi…cation (2010): 35B40, 35B41, 35B51, 35K55, 35K57 1 Introduction In real applications there might exist several nonlocal e¤ects that in‡uence the evolution of a system. For instance, usually we do not have enough information about the systems under study and its features at every point. In reality, the measurements are not made pointwise but through some local average. This is just one possible reason of introducing nonlocal terms in models. Actually, during the last decades many mathematicians have been studying nonlocal problems motivated by its various applications in physics, biology or population dynamics [13, 14, 15, 16, 17, 27]. For instance, let consider the problem of …nding a function u(t; x)such that 8 < : uta(Ru(t; x)dx)u=g(t; u);in (0;1); u= 0 on @(0;1); u(0) = u0in : (1) Here is a bounded open subset in Rn,n1;with smooth boundary and ais some function from R to (0;+1). In such equation ucould describe the density of a population subject to spreading. The di¤usion coe¢ cient ais then supposed to depend on the entire population in the domain rather than on the local density. A wide literature with signi…cant results about (1) have been developed during the last few decades (see for example [14, 17, 27]). However, it is possible to distinguish two basic cases of the following more general equation 8 < : uta(u)u=g(t; u); t > 0; x 2; u= 0;on @(0;1); u(0; x) = u0(x)x2: 1
Some authors consider adepending on a linear functional l(u), i.e., a(u) = a(l(u)) with l(u) = Z (x)u(x; t)dx; where (x)is a given function in L2(). For g(t; u) = f(t)the existence and uniqueness of solutions and their asymptotic behavior are studied for example in [15, 16, 18, 32]. For g(t; u) = f(u) + h(t) the existence, uniqueness and asymptotic behaviour of solutions is studied in [1, 6, 8, 9]. Moreover, the authors prove the existence of pullback attractors in L2() and H1 0(). Extensions in this direction for equations governed by the p-laplacian operator instead of the laplacian operator are given in [7, 10], whereas nonclassical di¤usion equations are considered in [29]. On the other hand, it is possible to consider a function asuch that a(u) = a(kuk2 H1 0). The existence and uniqueness of solutions of the following problem 8 < : uta(kuk2 H1 0)u=f; t > 0; x 2; u= 0;on @(0;1); u(0; x) = u0(x)x2: is proved in [32, 19], where f2L2(); u02H1 0() and a=a(s)is a continuous function such that 0< m a(s)M: By this way, in this paper the following problem is considered 8 < : uta(kuk2 H1 0)u=f(u) + h(t);in (0;1); u= 0 on @(0;1); u(0; x) = u0(x)in ; (2) where h2L2(0; T;L2());for all T > 0; a :R+!R+is a continuous function such that a(s)m > 0 and fis a continuous function satisfying standard dissipative and growth conditions (see (7) below). The aim of this paper is three-fold. First, we will prove the existence of solutions for problem (2) under di¤erent assumptions on the nonlinear function f. Second, we will obtain the existence of attractors for the semi‡ows generated by either regular or strong solutions in the autonomous case, that is, when hdoes not depend on time. Third, we establish that the global attractor can be characterized by the unstable manifold of the set of stationary points. It is important to notice that the proof of this last fact requires the existence of a Lyapunov function on the attractor, and for this aim the term a(kuk2 H1 0)is crucial. In the case when a(u) = a(l(u)) it is not known whether such a function exists or not. We prove the existence of strong solutions by assuming that either the function fis continuously di¤erentiable and f0(s)or a more strict growth condition on f: Supposing additionaly that the function ahas sublinear growth we prove the existence of regular solutions as well. Moreover, when f0(s)and the function s7! as2sis non-decreasing, uniqueness is proved. When studying the asymptotic behaviour of solutions, new challenging di¢ culties arise for problem (2). For this problem we consider the autonomous situation, that is, h2L2() does not depend on t. If uniqueness holds, then we de…ne classical semigroups (one for regular solutions and one for strong solutions) and prove the existence of the global attractor. Under some extra assumptions on the functions a; h we are able to obtain that the global attractor is bounded in H2() and L1(). If uniqueness is not known to be true, then we have to de…ne a (possibly) multivalued semi‡ow. Then the existence of the global attractor is proved for regular solutions in the topology of the space L2() and for strong solutions in the topology of the space H1 0() (or H1 0() \Lp()), extending in this way the known results for the local problem [21]. The structure of the global attractor is an important feature as it gives us an insight into the long-term dynamics of the solutions. In the multivalued situation it is a challenging problem that has not been completely understood yet. So far in the local case several results in this direction have been obtained for reaction-di¤usion equations without uniqueness [2, 5, 21, 22]. 2
In our nonlocal problem for both situations (for regular and strong solutions) we are able under some conditions to de…ne a Lyapunov function on the attractor and to prove that it is characterized as the unstable set of the stationary points (denoted by Mu(R)). Also, the attractor is equal to the stable set of the stationary points when we consider solutions only in the set of bounded complete trajectories (denoted by Ms(R)). 2 Existence of solutions Throughout this paper we will denote by kkXthe norm in the Banach space X: We consider the following nonlocal reaction-di¤usion equation 8 < : uta(kuk2 H1 0)u=f(u) + h(t);in (0;1); u= 0 on @(0;1); u(0; x) = u0(x)in ; (3) where is a bounded open set of Rnwith smooth boundary @: Let us consider the following conditions on the functions a; f; h : h2L2(0; T;L2()) 8T > 0;(4) a2C(R+); f 2C(R);(5) a(s)m > 0;(6) 2jsjpf(s)s1jsjp;(7) where m; 1; 2>0and 0; p 2. Observe that then there exists C > 0such that jf(s)j C(1 + jsjp1)8s2R;(8) and that the function F(s) := Rs 0f(r)dr satis…es e2jsjpe F(s)ee1jsjp(9) for certain positive constants ei; i = 1;2;and e0;and jF(s)j e C(1 + jsjp)8s2R:(10) Conditions (4)-(7) will be always assumed throughout the paper. Sometimes, some of the following additional assumptions will also be used: f2C1(R)be such that f0(s); 8s2R;(11) p2n2 n2;if n3;(12) a(s)M1+M2s; 8s0;(13) s7! a(s2)sis non-decreasing, (14) a()2C1R+and a0(s)0;8s0;(15) for some constants M1; M2; 0. Remark 1 a0(s)0implies that (14) holds, so condition (15) is stronger than (14). Assumption (14) is used to prove uniqueness of solutions. Assumption (15) is used to obtain the H2() regularity of the global attractor. 3
De…nition 2 A weak solution to (3) is a function u()such that u2L1(0; T ;L2())\L2(0; T ;H1 0())\ Lp(0; T;Lp()) for any T > 0and satis…es the equality d dt(u; v) + a(ku(t)k2 H1 0)(ru(t);rv)=(f(u(t)); v)+(h(t); v)8v2H1 0() \Lp();(16) in the sense of scalar distributions. Here, we denote by (;)the inner product in L2() (or L2()dfor d2N) and also the duality product between Lp() and Lq() (where qis the conjugate exponent of p, that is, q=p=(p1)). The duality between H1 0() and H1() will be denoted by h,i : We need to guarantee that the initial condition of the problem makes sense for a weak solution. This can be achieved in a standard way assuming that the function ahas an upper bound, that is, there exists M > 0such that a(s)Mfor all s0:(17) Indeed, if uis a weak solution to (3), taking into account (8) and (17) it follows that ut=a(kuk2 H1 0)u+f(u) + h2L2(0; T;H1()) + Lq(0; T;Lq()):(18) Therefore, by [12, p.33] u2C([0; T]; L2()) and the initial condition makes sense when u02L2(). For the operator A=, thanks to the assumptions on the domain , it is well known that D(A) = H2() \H1 0() [30, Proposition 6.19]. De…nition 3 A regular solution to (3) is a weak solution with the extra regularity u2L1("; T;H1 0()) and u2L2("; T;D(A)) for any 0< " < T: Remark 4 Since du dt 2Lq("; T;Lq()) for any regular solution, in this case equality (16) is equivalent to the following one: ZT "Z du (t; x) dt (t; x)dxdt ZT " a(ku(t)k2 H1 0)Z udxdt (19) =ZT "Z f(u(t; x)) (t; x)dxdt +ZT "Z h(t; x)(t; x)dxdt; for all 0< " < T and 2Lp(0; T;Lp()) : Lemma 5 Let u2Lp("; T;X),du dt 2Lq("; T;X0)for all 0< " < T, where Xis a re‡exive and separable Banach space and X0denotes its dual space. Assume that 2C(R+)is such that 2 W1;1("; T; [("); (T)]) and 0< (")< (T)for all 0< " < T. Then w() = u(()) 2Lp("; T ;X), dw dt 2Lq("; T;X0);for all 0< " < T, and dw dt (t) = du dt ((t)) d dt (t)for a.a. t > 0:(20) Proof. We …x arbitrary 0<"<T. There exists a sequence un2C1([("); (T)]; X)such that un!uin Lp(("); (T); X)and dun dt !du dt in Lq(("); (T); X0)[20, Chapter IV]. We de…ne wn(t) = un((t)). Following the same proof of [4, Corollary VIII.10] we obtain that wn()2W1;1("; T;X)and dwn dt (t) = dun dt ((t)) d dt (t)for a.a. t > 0: It is clear that wn!win Lp("; T;X)and dun dt (()) !du dt (()) in Lq("; T;X0). Passing to the limit we obtain that dw dt () = du dt (()) d dt () 4
in the sense of distributions D0(0;+1;X). As du dt (()) d dt ()2Lq("; T;X0),dw dt 2Lq("; T;X0)and (20) holds true. We would like to avoid abeing uniformly bounded by above (i.e. to relax assumption (17)). We can still prove the continuity in L2() of ufor regular solutions by assuming that ahas at most linear growth. Lemma 6 Assume that conditions (4)-(7), (13) hold. Then any regular solution satis…es that u2 C([0; T]; L2()) for all T > 0:Moreover, w(t) = u1(t), where (t) = Rt 0a(ku(s)k2 H1 0)ds, is a regular solution to the problem 8 > > < > > : wtw=f(w) + h(t) a(kwk2 H1 0);in (0;1); w= 0 on @(0;1); w(0; x) = u0(x)in : (21) Proof. Condition (13) guarantees that a(ku()k2 H1 0)2L1(0; T)if u2L20; T;H1 0(). We make the following time rescaling u(t; x) = w((t); x): As a(ku()k2 H1 0)2L1(0; T), the function t7! (t)is continuous and () = 1()satis…es the conditions of Lemma 5. It is clear that the function w(t; x) = u(1(t); x)belongs to the space L1(0; T;L2()) \ L2(0; T;H1 0()) \Lp(0; T;Lp()) and also to the spaces L1("; T;H1 0()) and L2("; T;D(A)) for any 0< " < T. Moreover, du dt 2Lq("; T;Lq()) and Lemma 5 give dw dt 2Lq("; T;Lq()) and dw dt (t) = du dt 1(t)d dt1(t) = du dt 1(t)1 akw(t))k2 H1 0;for a.a. t: (22) Equality (19) implies that du dt 1(t)aku(1(t))k2 H1 0u1(t)=fu1(t)+h(t);for a.a. t > 0; so (22) gives dw dt (t)w(t) = f(w(t)) a(kw(t)k2 H1 0)+h(t) a(kw(t)k2 H1 0)for a.a. t > 0. Hence, wis a regular solution to problem (21). Since 0<1 a(s)1 m, we obtain that dw dt 2L2(0; T;H1()) + Lq(0; T;Lq()): Therefore, w2C([0; T]; L2());so that u2C([0; T]; L2()): Remark 7 Under assumptions (4)-(7) any regular solution u()satis…es that du dt 2Lq("; T;Lq()) for all 0< " < T. Then by [12, p.33] u2C(["; T]; L2()),t7! ku(t)k2is absolutely continuous on ["; T] and d dt ku(t)k2 L2= 2 du dt ; ufor a.a. t > ": If the initial condition belongs to H1 0() \Lp(), we can de…ne strong solutions as well. 5
De…nition 8 A strong solution to (3) is a weak solution with the extra regularity u2L1(0; T;H1 0() \ Lp()),u2L2(0; T;D(A)) and du dt 2L20; T;L2()for any T > 0: We observe that if uis a strong solution, then u2C([0; T]; H1 0()) (see [31, p.102]). Also, u2 L1(0; T;Lp()) and u2C([0; T]; L2()) imply that u2Cw([0; T]; Lp()) (see [33, p.263]). Thus, an initial condition in H1 0()\Lp() makes sense. Also, the equality f(u) = utakuk2 H1 0uhimplies that f(u)2L20; T;L2() Also, if uis a regular solution such that du dt 2L2"; T;L2()for all 0< " < T, then u2 C((0; T]; H1 0()): The phase space for regular solutions will be L2(), whereas for strong solutions we will use the space H1() \Lp() (or just H1 0() when H1 0() Lp()). The following results will be proved in Theorems 9, 10, 11, 12, 14: Conditions (4)-(7), (11), (13) imply the existence of at least one regular solution for any u02L2(). If, in addition, (14) holds, then it is the unique regular solution. Conditions (4)-(7), (11) imply the existence of at least one strong solution for any u02H1 0() \ Lp(). If, in addition, (14) holds, then it is the unique strong solution. Conditions (4)-(7), (12) imply the existence of at least one strong solution for any u02H1 0(). Conditions (4)-(7), (12), (13) imply the existence of at least one regular solution for any u02L2(). To start with we prove the existence of regular solutions for initial conditions in L2() : Theorem 9 Assume that conditions (4)-(7), (11) and (13) hold. Then, for any u02L2() there exists at least one regular solution to (3). Proof. We will prove the result by compactness and using Faedo-Galerkin approximations. Consider a …xed value T > 0:Let fwjgj1be a sequence of eigenfunctions of in H1 0() with homogeneous Dirichlet boundary conditions, which forms a special basis of L2(). If is a bounded regular domain, then it is well known that fwjg H1 0() \Lp() and that for the set Vn=span[w1; : : : ; wn]we have that [n2NVnis dense in L2() and also in H1 0() \Lp() [25]:As usual, Pnwill be the orthogonal projection in L2(), that is zn:= Pnz= n X j=1 (z; wj)wj; and jwill be the eigenvalues associated to the egienfunctions wj. For each integer n1, we consider the Galerkin approximations un(t) = n X j=1 nj(t)wj; which satisfy the following nonlinear ODE system (d dt(un; wi) + a(kunk2 H1 0)(run;rwi)=(f(un); wi)+(h; wi)8i= 1; : : : ; n; un(0) = Pnu0: (23) where Pnu0!u0in L2(). Since (23) can be written in the normal form with a continuous right-hand side, this Cauchy problem possesses a solution on some interval [0; tn). We claim that for any T > 0such a solution can be extended to the whole interval [0; T];which follows from a priori estimates in the space L2() of the sequence fung. Multiplying by ni(t)and summing from i= 1 to n, we obtain 1 2 d dtkun(t)k2 L2+a(kunk2 H1 0)kun(t)k2 H1 0= (f(un(t)); un(t)) + (h; un(t)) for a:e: t 2(0; tn):(24) 6
Using (7) and the Young and Poincaré inequalities we deduce that (f(un(t)); un(t)) jj 1kun(t)kp Lp; (h(t); un(t)) m 2kun(t)k2 H1 0+1 21mkh(t)k2 L2: Hence, from (24) it follows that 1 2 d dtkun(t)k2 L2+m 2kun(t)k2 H1 0+1kun(t)kp Lpjj+1 21mkh(t)k2 L2for a.e. t2(0; tn):(25) Then, integrating (25) from 0to t2(0; tn)we deduce 1 2kun(t)k2 L2+m 2Zt 0 kun(s)k2 H1 0ds +1Zt 0 kun(s)kp Lpds jjt+1 21mZt 0 kh(s)k2 L2ds +1 2kun(0)k2 L2TK2+K3(T) + 1 2kun(0)k2 L2: (26) Therefore, the sequence fungis well de…ned and bounded in L1(0; T;L2())\L2(0; T ;H1 0())\Lp(0; T;Lp()). Also, fungis bounded in L2(0; T;H1()): On the other hand, by (8) it follows that ZT 0Z jf(u(x; t))jqdxdt 2q1Cq(jjT+ZT 0 ku(t)kp Lpdt); with 1 p+1 q= 1:Hence, since fungis bounded in Lp(0; T;Lp()),ff(un)gis bounded in Lq(0; T;Lq()). On the other hand, multiplying (23) by ini(t)and summing from i= 1 to n, we obtain 1 2 d dtkunk2 H1 0+mkunk2 L2 hf(un);uni+ (h(t);un)kunk2 H1 0+1 2mkh(t)k2 L2+m 2kunk2 L2: Integrating the previous expression between sand t, with 0< s tT; and using (11) we have 1 2kun(t)k2 H1 0+m 2Zt s kun(r)k2 L2dr ZT 0 kun(r)k2 H1 0dr +1 2kun(s)k2 H1 0+1 2mZt s kh(r)k2 L2dr: (27) Now, integrating in sbetween 0and t, it follows that tkun(t)k2 H1 0(2T + 1) ZT 0 kun(r)k2 H1 0dr +K3(T)T: Hence, kun(t)k2 H1 02T + 1 "ZT 0 kun(r)k2 H1 0dr +K3(T)T ":(28) for all t2["; T]with "2(0; T):From the last inequality and (26) we deduce that fkun(t)kH1 0gis uniformly bounded in ["; T]and by the continuity of the function awe get that fa(kun(t)k2 H1 0)gis bounded in ["; T ]. Also, it follows that fungis bounded in L1("; T;H1 0()):(29) On the other hand, taking s="and t=Tin (27), by (26) we obtain that fungis bounded in L2("; T;D(A));(30) so fungand fa(kunk2 H1 0)ungare bounded in L2("; T;L2()). Thus, dun dt is bounded in Lq("; T;Lq()):(31) 7
Therefore, there exists u2L1("; T;H1 0()) \L2(0; T;H1 0()) \L1(0; T;L2()) \L2("; T;D(A)) \ Lp(0; T;Lp()) such that du dt 2Lq("; T;Lq()) and a subsequence fung, relabelled the same, such that un * u in L1("; T;H1 0()); un * u in L1(0; T;L2()); un* u in L2(0; T;H1 0()); un* u in Lp(0; T;Lp()); un* u in L2("; T;D(A)); dun dt *du dt in Lq("; T;Lq()); f(un)* in Lq(0; T;Lq()); a(kunk2 H1 0) * b in L1("; T); (32) for any 0< " < T, where *means weak convergence and *weak star convergence. Moreover, by (30)-(31) the Aubin-Lions Compactness Lemma gives that un!uin L2("; T;H1 0()), so un(t)!u(t)in H1 0() a.e. on ("; T)for any " > 0. Consequently, there exists a subsequence fung, relabelled the same, such that un(t; x)!u(t; x)a.e. in (0; T). Also, we know that Pnf(un)* (see [30, p.224]). Since fis continuous, it follows that f(un(t; x)) !f(u(t; x)) a.e. in (0; T ). Therefore, in view of (32), by [26, Lemma 1.3] we have that =f(u). As a consequence, by the continuity of a, we get that a(kun(t)k2 H1 0)!a(ku(t)k2 H1 0)a.e. on ("; T): Since the sequence is bounded, by the Lebesgue theorem this convergence takes place in L2("; T)and b=a(kuk2 H1 0)on ("; T). Thus, a(kunk2 H1 0)un* a(kuk2 H1 0)u; in L2("; T;L2()):(33) Finally, since fwigis dense in H1 0() \Lp(), in view of (32) and (33), we can pass to the limit in (23) and conclude that (16) holds for all v2H1 0() \Lp(). To conclude the proof, we have to check that u(0) = u0. Indeed, let be 2C1([0; T ]); H1 0()\Lp()), with (T) = 0,(0) 6= 0. We consider the functions w(t) = u(1(t)); wn(t) = un1 n(t)(here n(t) = Rt 0a(kun(r)k2 H1 0dr)), which by Lemma 6 are regular solutions to problem (21) with initial conditions w(0) = u0and to the corresponding Galerkin approximations with initial condition wn(0) = un(0) = Pnu0, respectively. Since dw dt 2L2(0; T;H1())+Lq(0; T;Lq()), we can multiply the equation in (21) by and integrate by parts in the tvariable to obtain that ZT 0 ((w(t); 0(t)) hw(t); (t)i)dt =ZT 0 f(w(t)) + h(t) a(kw(t)k2 H1 0); (t)!dt + (w(0) ; (0)) ;(34) ZT 0 ((wn(t); 0(t)) hwn(t); (t)i)dt =ZT 0 Pnf(wn(t)) + Pnh(t) a(kwn(t)k2 H1 0); (t)!dt + (wn(0) ; (0)) : (35) We can easily obtain by the previous convergences and (6) that wn* w in L20; T;H1 0(); wn*win L20; T;H1(); Pnf(wn(t)) + Pnh(t) a(kwn(t)k2 H1 0)*f(w(t)) + h(t) a(kw(t)k2 H1 0)in Lq(0; T;Lq()) : 8
Passing to the limit in (35), taking in to account (34) and bearing in mind wn(0) = Pnu0!u0we get (w(0) ; (0)) = (u0; (0)) : Since (0) 2H1 0() \Lp() is arbitrary, we infer that w(0) = u(0) = u0. Hence, uis a regular solution to (3) satisfying u(0) = u0: Second, we will prove the existence of strong solutions for initial conditions in H1 0() \Lp():In this case, we do not need to impose the upper bound (13) of the function a: Theorem 10 Suppose that conditions (4)-(7) and (11) are ful…lled. Then, for any u02H1 0() \Lp() there exists at least a strong solution to (3). Proof. We consider, as in Theorem 9, the Galerkin approximations fungand an element ufor which (32) holds. Under the aforementioned conditions, we will obtain that unconverges to a strong solution to (3). In this proof it is important to observe that Pnu0!u0in the spaces H1 0() and Lp() [30, p.199 and 220]. Thus, the sequences kPnu0kH1 0and kPnu0kLpare bounded. First, we multiply the equation in (23) by dun dt to obtain kd dtun(t)k2 L2+a(kunk2 H1 0)1 2 d dtkunk2 H1 0=d dt Z F(un)dx + (h(t);dun dt ): Introducing A(s) = Zs 0 a(r)dr; (36) we have 1 2kd dtun(t)k2 L2+d dt 1 2A(kunk2 H1 0)Z F(un(t))dx1 2kh(t)k2 L2:(37) Now, integrating (37) we have 1 2Zt 0 kd dsun(s)k2 L2ds +1 2A(kun(t)k2 H1 0)Z F(un(t))dx 1 2A(kun(0)k2 H1 0)Z F(un(0))dx +1 2Zt 0 kh(s)k2 L2ds: From (6) and (9) we get m 2kun(t)k2 H1 0+e1kun(t)kp Lp+1 2Zt 0 kd dsun(s)k2 L2ds 1 2A(kun(0)k2 H1 0) + e2kun(0)kp Lp+K(T): (38) Now, from (38) we obtain that dun dt is bounded in L2(0; T;L2());(39) so dun dt *du dt in L2(0; T;L2()):(40) On the other hand, the embedding H1 0() L2() and the Aubin-Lion Compactness Lemma imply that un!uin L2(0; T;L2()): Hence, un!ufor a.e. (x; t)2(0; T): 9
Proof. The following calculations are formal but can be justi…ed by the Galerkin approximations. Arguing as in (52)-(55) we obtain the existence of a constant Csuch that kTr(1; u (0))k2 H1 0+kTr(1; u (0))kp LpC(1 + ku(0)k2 L2): Hence, the semigroup property Tr(t+ 1; u0) = Tr(1; Tr(t; u0)) and (62) imply that kTr(t+ 1; u0)k2 H1 0+kTr(t+ 1; u0)kp LpC(1 + K2)8tt0(R); if ku0kL2R, which proves the statement. Theorem 17 Let (5)-(7), (11), (13), (14) and (60). Then the equation (3) has a connected global compact attractor Ar, which is bounded in H1 0() \Lp(). Proof. Since a bounded set in H1 0() is relatively compact in L2() which is a connected space, the result follows from Theorem 10.5 in [30] and Proposition 16. We will also obtain the boundedness of the attractor in the spaces L1() and H2(). First, we recall that a function :R!L2() is a complete trajectory of the semigroup Trif (t) = Tr(ts; (s)) for any ts.is bounded if the set [s2R(s)is bounded. It is well known [24] that the global attractor is characterized by Ar=f(0) : is a bounded complete trajectoryg:(66) Theorem 18 Let (5)-(7), (11), (13), (14) and (60) hold. Then the global attractor Aris bounded in L1(), provided that h2L1(). Proof. We de…ne v+= maxfv; 0g; v=maxfv; 0g:We multiply equation (3) by (uM)+for some appropriate constant Mand integrate over to obtain 1 2 d dt Z j(uM)+j2dx +a(ku(t)k2 H1 0)Z jr(uM)+j2dx =Z (f(u(t)) + h)(uM)+dx; where we have used the equality 1 2 d dt Rj(uM)+j2dx = (ut;(uM)+);which is proved by regularization. Since h2L1(), by (7) we deduce that (f(u) + h)ueejujp: It follows that f(u) + h0when u(e e)1=p =M: Therefore, we have (f(u) + h)(uM)+0: Thus, by (6) and the the Poincaré inequality, we deduce that d dt Z j(uM)+j2dx 2m1Z j(uM)+j2dx: Using the Gronwall inequality, we have Z j(u(t)M)+j2dx e2m(t)Z j(u()M)+j2dx: For any y2 Arthere is by (66) a bounded complete trajectory such that (0) = y. Then taking t= 0 and ! 1 in the last inequality, we obtain y(x) = (0; x)M; for a.a. x2. The same arguments can be applied to (uM), which shows that kykL1M; 8y2 Ar: If we assume (15), then it is possible to show that the global attractor is more regular. 16
Proposition 19 Let (5)-(7), (11), (13) and (60) hold. If, additionally, (15) is satis…ed, then there exists an absorbing set in H2() and the global attractor is bounded in H2(). Proof. We will prove the existence of an absorbing set in H2(). The boundedness of the global attractor in this space follows then immediately. We proceed formally, but the estimates can be justi…ed via Galerkin approximations. Let u(t) = Tr(t; u0)with ku0kL2R: First, we di¤erentiate the equation with respect to t utt a0(kuk2 H1 0)d dtkuk2 H1 0ua(kuk2 H1 0)ut=f0(u)ut: Multiplying by utwe get 1 2 d dtkutk2 L2+1 2a0(kuk2 H1 0)( d dtkuk2 H1 0)2+a(kuk2 H1 0)kutk2 H1 0=Z f0(u)(ut)2dx: (67) By (6), a0(s)0and f0(s)we obtain 1 2 d dtkutk2 L2+mkutk2 H1 0kutk2 L2:(68) Second, multiplying (3) by utand reordering terms, we obtain d dt a(kuk2 H1 0) 2kuk2 H1 0Z F(u)dx Z h(x)udx!+kutk2 L2=a0(kuk2 H1 0) 2kuk2 H1 0 d dtkuk2 H1 0:(69) Proposition 16 implies that a0(kzk2 H1 0):= supjsjMa0(s2) if zbelongs to the absorbing set in H1 0(). On the other hand, multiplying the equation by uand using Proposition 16, we obtain d dtkuk2 H1 0+mku(t)k2 L22ku(t)k2 H1 0+1 mkhk2 L2K18tt1(R): Hence, by (69) and Proposition 16, it follows d dt a(kuk2 H1 0) 2kuk2 H1 0Z F(u)dx Z h(x)udx!+kutk2 L2 2K1M2;8tt1(R):(70) Multiplying both sides of the inequality f0(s)by sand integrating between 0and s, we obtain sf(s) F(s) + s2 2; 8s2R:(71) Moreover, integrating f0(s)twice between 0and s, we infer F(s) 2s2+Cs; 8s2R:(72) Now, we multiply (3) by uand integrate between tand t+ 1 to obtain 1 2ku(t+ 1)k2 L2+Zt+1 ta(kuk2 H1 0)ku(s)k2 H1 0Z f(u)udx Z h(x)udxds =1 2ku(t)k2 L2:(73) From (71), (73) and Proposition 15 it follows Zt+1 t a(kuk2 H1 0) 2kuk2 H1 0Z F(u)dx Z h(x)udx!ds 1 2ku(t)k2 L2+ 2Zt+1 t kuk2 L2ds e L8tt0 17
The last inequality allows us to apply the Uniform Gronwall Lemma to (70) in order to obtain a(kuk2 H1 0) 2kuk2 H1 0Z F(u)dx Z h(x)udx e L+ 2K1M28tt1+ 1:(74) Using (6) and (72) we get a(kuk2 H1 0) 2kuk2 H1 0Z F(u)dx Z h(x)udx 2kuk2 L2e CkukL2:(75) Now, integrating (70) from tto t+ 1, using (74), (75), by Proposition 15 we have Zt+1 t kusk2 L2ds e L+K1M2+ 2K2+e CK =1;8tt1+ 1:(76) Hence, the last equation allow us to apply to (68) the Uniform Gronwall Lemma [34] to obtain kdu dt (t)k2 L22;8tt1+ 2:(77) Finally, we multiply (3) by uand use (6) to obtain m 2kuk2 L2kuk2 H1 0+1 mkhk2 L2+1 mkutk2 L2: Thus, by Proposition 16 and (77), we deduce that ku(t)k2 H238tt1+ 2: 3.1.2 Abstract theory of attractors for multivalued semi‡ows Prior to studying the case of non-uniqueness, we recall some well-known results concerning the structure of attractors for multivalued semi‡ows. Consider a metric space (X; d)and a family of functions RC(R+;X). Denote by P(X)the class of nonempty subsets of X. Then we de…ne the multivalued map G:R+X!P(X)associated with the family Ras follows G(t; u0) = fu(t) : u()2 R; u(0) = u0g:(78) In this abstract setting, the multivalued map Gis expected to satisfy some properties that …t in the framework of multivalued dynamical systems. The …rst concept is given now. De…nition 20 A multivalued map G:R+X!P(X)is a multivalued semi‡ow (or m-semi‡ow) if G(0; x) = xfor all x2Xand G(t+s; x)G(t; G(s; x)) for all t; s 0and x2X. If the above is not only an inclusion, but an equality, it is said that the m-semi‡ow is strict. Once a multivalued semi‡ow is de…ned, we recall the following concepts. De…nition 21 A map :R!Xis called a complete trajectory of R(resp. of G) if (+h)j[0;1)2 R for all h2R(resp. if (t+s)2G(t; (s)) for all s2Rand t0). A point z2Xis a …xed point of Rif '()z2 R. The set of all …xed points will be denoted by RR. A point z2Xis a stationary point of Gif z2G(t; z)for all t0. De…nition 22 Given an m-semi‡ow Ga set BXis said to be negatively (positively) invariant if BG(t; B)(G(t; B)B) for all t0, and strictly invariant (or, simply, invariant) if it is both negatively and positively invariant. The set Bis said to be weakly invariant if for any x2Bthere exists a complete trajectory of R contained in Bsuch that (0) = x. We observe that weak invariance implies negative invariance. 18
De…nition 23 A set A Xis called a global attractor for the m-semi‡ow Gif it is negatively invariant and it attracts all bounded subsets, i.e., distX(G(t; B);A)!0as t!+1. Remark 24 When Ais compact, it is the minimal closed attracting set [28, Remark 5]. In order to obtain a detailed characterization of the internal structure of a global attractor, we introduce an axiomatic set of properties on the set R. (K1) For any x2Xthere exists at least one element '2 R such that '(0) = x. (K2) '() := '(+)2 R for any 0and '2 R (translation property). (K3) Let '1; '22 R be such that '2(0) = '1(s)for some s > 0. Then, the function 'de…ned by '(t) = '1(t) 0 ts; '2(ts)st; belongs to R(concatenation property). (K4) For any sequence f'ngRsuch that 'n(0) !x0in X, there exist a subsequence f'nkgand '2 R such that 'nk(t)!'(t)for all t0. Remark 25 If in assumption (K1), for every x2X, there exists a unique '2 R such that '(0) = x, then the set f'2 R :'(0) = xgconsists of a single trajectory ', and the equality G(t; x) = '(t)de…nes a classical semigroup G:R+X!X: It is immediate to observe [11, Proposition 2] or [23, Lemma 9] that Rful…lling (K1) and (K2) gives rise to an m-semi‡ow Gthrough (78), and if besides (K3) holds, then this m-semi‡ow is strict. In such a case, a global bounded attractor, supposing that it exists, is strictly invariant [28, Remark 8]. Several properties concerning …xed points, complete trajectories and global attractors are summarized in the following results [21]. Lemma 26 Let (K1)-(K2) be satis…ed. Then every …xed point (resp. complete trajectory) of Ris also a …xed point (resp. complete trajectory) of G. If Rful…lls (K1)-(K4), then the …xed points of Rand Gcoincide. Besides, a map :R!Xis a complete trajectory of Rif and only if it is continuous and a complete trajectory of G. The standard well-known result in the single-valued case for describing the attractor as the union of bounded complete trajectories (see [24]) reads in the multivalued case as follows. Theorem 27 Consider Rsatisfying (K1) and (K2) and either (K3) or (K4). Assume that Gpossesses a compact global attractor A. Then A=f(0) : 2Kg=[t2Rf(t) : 2Kg;(79) where Kdenotes the set of all bounded complete trajectories in R. Hence, Ais weakly invariant. We …nish this section by stating a general result about the existence of attractors. We recall that the map t7! G(t; x)is upper semicontinuous if for any x2Xand any neighborhood O(G(t; x)) in Xthere exists > 0such that if d(y; x)< , then G(t; y)O. Theorem 28 [28, Theorem 4 and Remark 8] Let the map t7! G(t; x)be upper semicontinuous with closed values. If there exists a compact attracting set K, that is, distX(G(t; B); K)!0;as t!+1; for any bounded set B, then Gpossesses a global compact attractor A, which is the minimal closed attracting set. If, moreover, Gis strict, then Ais invariant. We observe that, although in the papers [28], [21] the space Xis assumed to be complete, the results are true in a non-complete space. 19
3.1.3 The case of non-uniqueness If we do not assume the additional assumptions on the function a()of Section 3.1.1 ensuring uniqueness of the Cauchy problem, we have to de…ne a multivalued semi‡ow. We have two possibilities: either to consider the conditions of Theorem 9 with an extra growth assumption or to use the conditions of Theorem 12. If we assume conditions (5)-(7), (12), (13) and (60), then by Theorem 12 for any u02L2() there exists at least one regular solution and (45) implies that f(u)2L2("; T;L2()) for any regular solution, so du dt 2L2("; T;L2()) as well. In this case, as H1 0() Lp(), we have that u2C((0;+1); H1 0()) C((0;+1); Lp()) : If we assume conditions (5)-(7), (11), (13) and (60) as well, then we known by Theorem 9 that for any u02L2() there exists at least one regular solution. In order to obtain the necessary estimates leading to the existence of a global attractor, we need to ensure that du dt 2L2("; T;L2());for all 0< " < T; (80) holds, as by [31, p.102] we obtain that d dtkuk2 H1 0= 2(u; ut)for a.a. t: (81) and u2C((0;+1); H1 0()): We note that the set of regular solutions of that kind is non-empty, as using inequalities (52)-(56) in the proof of Theorem 9 we prove that the regular solution satis…es (80). We also observe that we can force all the regular solutions to satisfy du dt 2L2("; T;L2()) with an additional assumption on the constant p, which is weaker than (12). This is achieved by obtaining that f(u)2L2("; T;L2()), which can be done by using an interpolation inequality. Indeed, for u2 L1("; T;H1 0()) \L2("; T;D(A)) we have the interpolation inequality kuk2(+1) L2(+1)(";T ;L2(+1)()) kuk2 L1(";T ;Lp1())kuk2 L2(";T ;Lp2());(82) where =4 n2; p1=2n n2; p2=2n n4, provided that n > 4; < 2; p1= 4; p2=4 2if n= 4; = 3; p1= 6; p2= +1if n= 3; and 0is arbitrary for n= 1;2:We have used the embeddings H1 0() Lp1() ; H2() Lp2() and [35, Lemma II.4.1, p. 72]. Thus, (8) implies that f(u)2L2("; T;L2()) if p+ 2 (83) and also that kf(u)k2 L2(";T ;L2()) =ZT "Z jf(u(x; t))j2dxdt C1+C2ZT "Z ju(x; t)j2(+1)dxdt: (84) Condition (83) also implies H1 0() Lp(), so u2C((0;+1); Lp()). Another necessary property to obtain estimates is the fact that t7! RF(u(t))dx is absolutely continuous on ["; T]for all 0< " < T and d dt Z F(u(t))dx =f(u(t)) ;du dt (t);for a.a. t > 0:(85) This can be proved by regularization in both situations by using the regularity of regular solutions and either (45) or (84). Therefore, under either the conditions of Theorem 9 with the extra assumption (83) or the conditions of Theorem 12 we de…ne the set R=K+ r:= fu() : uis a regular solution of (3)g: 20
We de…ne the (possibly multivalued) map Gr:R+L2() !P(L2()) by Gr(t; u0) = fu(t) : u2K+ rand u(0) = u0g: With respect to the axiomatic properties (K1) (K4) given above, we observe that obviously (K1) is true, and (K2) can be proved easily using equality (19). Therefore, Gris a multivalued semi‡ow by the results of the previous section. In this case we are not able to prove (K3), so Grcould be non-strict. Further we will prove that (K4) holds true. Lemma 29 Let us assume (5)-(7), (13) and (60). Additionally, assume one of the following assumptions: 1. (11) and (83) hold; 2. (12) is true. Given a sequence fung K+ rsuch that un(0) !u0weakly in L2(), there exists a subsequence of fung(relabeled the same) and u2K+ r, satisfying u(0) = u0, such that un(t)!u(t)strongly in H1 0() 8t > 0: Proof. We take an arbitrary T > 0. Arguing as in the proof of Theorem 9 we obtain the existence of a subsequence of unsuch that fungis bounded in L1(0; T;L2()); fungis bounded in Lp(0; T;Lp()); ff(un)gis bounded in Lq(0; T;Lq()): (86) The only di¤erence is that we obtain inequality (26) in an arbitrary interval ["; T ]and then pass to the limit as "!0(see the proof of Proposition 15). Since dun dt 2L2("; T;L2());for any " > 0, we have that u2C((0; T]; H1 0()) and we know that (81), (85) are true. Therefore, arguing as in the proofs of Theorems 9 and 12 and using (84) and (45) there exists u2L1("; T;L2()) \L2(0; T;H1 0()) and a subsequence fung, relabelled the same, such that un * u in L1(0; T;L2()) un * u in L1("; T;H1 0()) un* u in L2(0; T;H1 0()) un* u in Lp(0; T;Lp()) un* u in L2("; T;D(A)); dun dt *du dt in L2("; T;L2()) f(un)* f(u)in Lq(0; T;Lq()); f(un)* f(u)in L2("; T;L2()) ; a(kunk2 H1 0)un* a(kuk2 H1 0)uin L2("; T;L2()): (87) In view of (87), the Aubin-Lions Compactness Lemma gives un!uin L2("; T;H1 0()):(88) Since the sequence fungis equicontinuous in L2() on ["; T ]and bounded in C(["; T ]; H1 0()), by the compact embedding H1 0() L2() and the Ascoli-Arzelà theorem, a subsequence ful…lls un!uin C(["; T]; L2()); un(t)* u(t)in H1 0() 8t2["; T]: 21
By a similar argument as in the proof of Theorem 9 we establish that u2K+ r; u (0) = u0: Finally, we shall prove that un(t)!u(t)in H1 0() for all t2["; T]. Multiplying (3) by un tand using (36), (81), and (85) we obtain 1 2 dun dt 2 L2 +d dt 1 2A(kun(t)k2 H1 0Z F(un(t))dx1 2khk2 L2=D: Thus, 1 2A(kun(t)k2 H1 0)Z F(un(t))dx 1 2A(kun(s)k2 H1 0)Z F(un(s))dx +D(ts); t s" > 0: The same inequality is valid for the limit function u():We observe that the map y7! RF(y(x))dx is continuous in the topology of H1 0(), which follows easily from H1 0() Lp() and (10) using Lebesgue’s theorem. Hence, the functions Jn(t) = 1 2A(kun(t)k2 H1 0)RF(un(t))dx Dt; J(t) = 1 2A(ku(t)k2 H1 0) RF(u(t))dx Dt are continuous and non-increasing in ["; T]:Moreover, from (88) we deduce that Jn(t)!J(t)for a.e. t2("; T ):Take "<tm< T such that tm!Tand Jn(tm)!J(tm)for all m: Then Jn(T)J(T)Jn(tm)J(T) jJn(tm)J(tm)j+jJ(tm)J(T)j: For any > 0there exist m()and N(m()) such that Jn(T)J(T)if nN: Then lim sup Jn(T) J(T);so lim sup kun(T)k2 H1 0 ku(T)k2 H1 0(see the explanation below). As un(T)!u(T)weakly in H1 0() implies lim inf kun(T)k2 H1 0 ku(T)k2 H1 0;we obtain kun(T)k2 H1 0! ku(T)k2 H1 0; so that un(T)!u(T)strongly in H1 0(): In order to …nish the proof rigorously, we have to justify that lim sup Jn(T)J(T)implies the inequality lim sup kun(T)k2 H1 0 ku(T)k2 H1 0:First, we observe that by (10) we have Z F(un(T; x))dxCZ (1 + jun(T; x)jp)dx; so the boundedness of un(T)in Lp() implies that RF(un(T; x))dx < 1. Also, (9) gives F(un(T; x)) e, so by Fatou’s lemma we obtain lim inf Z F(un(T; x))dxZ lim inf (F(un(T; x))) dx =Z F(u(T; x))dx; where we have used that F(un(T; x)! F(u(T; x)) for a.a. x2. By contradiction let us assume that lim sup kun(T)kH1 0>ku(T)k. Then using the continuity of the function A(s)we have lim sup 1 2Akun(T)k2 H1 0Z F(un(T; x))dx lim sup 1 2Zkun(T)k2 H1 0 0 a(s)ds + lim inf Z F(un(T; x))dx 1 2Zlim sup kun(T)k2 H1 0 0 a(s)ds Z F(un(T; x))dx >1 2Zku(T)k2 H1 0 0 a(s)ds Z F(un(T; x))dx; which is a contradiction with lim sup Jn(T)J(T). 22
Corollary 30 Assume the conditions of Lemma 29. Then the set K+ rsatis…es condition (K4). Proposition 31 Assume the conditions of Lemma 29. The multivalued semi‡ow Gris upper semicontinuous for all t0, that is, for any neighborhood O(Gr(t; u0)) in L2() there exists > 0such that if ku0v0k< , then Gr(t; v0)O. Also, it has compact values. Proof. We argue by contradiction. Assume that there exists t0; u02L2(), a neighbourhood O(Gr(t; u0)) and a sequence fyngwhich ful…lls that each yn2Gr(t; un 0), where un 0converges strongly to u0in L2();and yn=2O(Gr(t; un)) for all n2N. Since yn2Gr(t; un 0)for all n, there exists un2K+ r, un(0) = un 0;such that yn=un(t). Now, since fun 0gis a convergent sequence of initial data, making use of Lemma 29 there exists a subsequence of fungwhich converges to a function u2K+ r. Hence, yn!y2Gr(t; u0). This is a contradiction because yn=2O(Gr(t; u0)) for any n2N. Proposition 32 Assume the conditions of Lemma 29. Then there exists an absorbing set B1for Gr, which is compact in H1 0() and Lp(). Proof. Reasoning as in Proposition 15, we obtain an absorbing set B0in L2() : Let K > 0be such that kyk Kfor all y2B0. Since du dt 2L"; T;L2()and (85) holds, we are allowed to multiply (3) by ut, use (81) and argue as in (52)-(55) to obtain the existence of a constant C such that ku(1) k2 H1 0+ku(1)kp LpC(1 + ku(0)k2 L2);(89) for any regular solution u()with initial condition u(0). For any u02L2() with ku0kL2Rand any u2K+ rsuch that u(0) = u0, the semi‡ow property Gr(t+ 1; u0)Gr(1; Gr(t; u0)) and Gr(t; u0)B0, if tt0(R);imply that ku(t+ 1) k2 H1 0+ku(t+ 1)kp LpC(1 + K2)8tt0(R): Then there exists M > 0such that the closed ball BMin H1 0() centered at 0with radius Mis absorbing for Gr. By Lemma 29 the set B1=Gr(1; BM)is an absorbing set which is compact in H1 0(). The embedding H1 0() Lp() implies that it is compact in Lp() as well. Theorem 33 Assume the conditions of Lemma 29. Then the multivalued semi‡ow Grpossesses a global compact attractor Ar. Moreover, for any set Bbounded in L2() we have distH1 0(Gr(t; B);Ar)!0as t! 1:(90) Also Aris compact in H1 0() and Lp(). Proof. From Propositions 31 and 32 we deduce that the multivalued semi‡ow Gris upper semicontinuous with closed values and the existence of an absorbing which is compact in H1 0() and Lp(). Therefore, by Theorem 28 the existence of the global attractor and its compactness in H1 0() and Lp() follow. The proof of (90) is analogous to that in Theorem 29 in [21]. The set of all complete trajectories of K+ r(see De…nition 21) will be denoted by Fr. Moreover, we write Kras the set of all complete trajectories which are bounded in L2(), and K1 ras the ones bounded in H1 0(). Lemma 34 Assume the conditions of Lemma 29. Then the sets de…ned above coincide, that is, Kr=K1 r. Proof. Let ()2Kr. Then there is Csuch that k(t)kL2Cfor any t2R. Let u() = (+) for any , which is a regular solution. Since du dt 2L2("; T;L2());for any " > 0, the equality (81) holds true. Also, (85) is satis…ed. Therefore, we can multiply the equation in (3) by utand apply again similar arguments as in Theorem 12 to deduce that ku(t+r)k2 H1 0K1(T) (1 + ku(0)k2 L2) r+K2(T)for any 0< r < T: (91) 23
Denote B=[t2R(t). Therefore, BGr(1; B) and (91) implies that Bis bounded in H1 0(), so ()2K1 r. The other inclusion is obvious. In view of Corollary 30 and Theorem 27, the global attractor is characterized in terms of bounded complete trajectories: Ar=f(0) : ()2Krg=f(0) : ()2K1 rg =[ t2R f(t) : ()2Krg=[ t2R f(t) : ()2K1 rg:(92) The set RK+ rwas de…ned in the previous section as the set of …xed points of K+ r, which means that z2RK+ rif the function u()de…ned by u(t) = z; for all t0;belongs to K+ r:This set can be characterized as follows. Lemma 35 Assume the conditions of Lemma 29. Let Rbe the set of z2H2() \H1 0() such that a(kzk2 H1 0)z=f(z) + hin L2() :(93) Then RK+ r=R: Proof. If z2RK+ r, then u(t)z2K+ r. Thus, u()satis…es (19) and du dt = 0 in L20; T;L2(), so (93) is satis…ed. Let z2R. Then the map u(t)zsatis…es (93) for any t0and du dt = 0 in L20; T;L2(), so (19) holds true. The following result is proved exactly as Theorem 18. Theorem 36 Assume the conditions of Lemma 29. Then the global attractor Ais bounded in L1(), provided that h2L1(). We are now ready to obtain the characterization of the global attractor. Theorem 37 Assume the conditions of Lemma 29. Then it holds that Ar=Mu r(R) = Ms r(R); where Ms r(R) = fz:9()2Kr; (0) = z; dist L2()((t);R)!0; t !+1g;(94) Mu r(R) = fz:9()2Fr; (0) = z; dist L2()((t);R)!0; t ! 1g:(95) Remark 38 In the de…nition of Mu r(R)we can replace Frby Kr. Also, as the global attractor Ais compact in H1 0(), in the de…nitions of Ms r(R)and Mu r(R);it is equivalent to write H1 0() instead of L2() : Proof. We consider the function E:Ar!R E(y) = 1 2A(kyk2 H1 0)Z F(y(x))dx Z h(x)y(x)dx; (96) where A(r) = Rr 0a(s)ds. We observe that E(y)is continuous in H1 0(). Indeed, the maps y7! 1 2A(kyk2 H1 0); y 7! Rh(x)y(x)dx are obviously continuous in H1 0(). On the other hand, both conditions (12) and (83) imply that H1 0() Lp(), so making use of the Lebesgue theorem the continuity of y7! RF(y(x))dx follows as well. 24
Since du dt 2L2"; T;L2()and (85) holds for any u2K+ rand 0< " < T, we obtain the energy equality Zt s kd dru(r)k2 L2dr +E(u(t)) = E(u(s)) for all ts > 0:(97) Hence, E(u(t)) is non-increasing and, by (6) and (9), bounded from below. Thus, E(u(t)) !l, as t!+1, for some l2R. Let x2 Arand (0) = x, where 2Kr. We reason by contradiction, so let suppose that there exists " > 0and a sequence (tn); tn!+1, such that dist L2()((tn);R)> ": In view of Theorem 33, Aris compact in H1 0(), so we can take a converging subsequence (relabeled the same) such that (tn)!yin H1 0(), where tn!+1. Since the function E:H1 0() !Ris continuous, it follows that E(y) = l. We obtain a contradiction by proving that y2R. In view of Lemma 29, there exists v2K+ rand a subsequence vn() = (+tn)such that v(0) = yand vn(t)!v(t) = zin H1 0() for t > 0. Thus, E(vn(t)) !E(z)implies that E(z) = l. Also, v()satis…es the energy equality for all 0st, so that l+Zt 0 kvrk2 L2dr =E(z) + Zt 0 kvrk2 L2dr =E(v(0)) = E(y) = l: Therefore, dv dt (t)=0for a.a. t, and then by Lemma 35 we have y2RK+ r=R. As a consequence, ArMs r(R). The converse inclusion follows from (92). For the second equality we observe that for any 2Frthe energy equality (97) is satis…ed for all 1 < s t: Let x2 Arand let 2Kr=K1 r(cf. Lemma 34) be such that (0) = x. Since the second term of the energy function is bounded from above by (9), E((t)) !l, as t! 1, for some l2R. We reason as before, so let suppose that there exists " > 0and a sequence (tn),tn! 1, such that dist L2()((tn);R)> "; and we have that (tn)!yin H1 0(),E(y) = l. Moreover, for a …xed t > 0, there exists v2K+ rand a subsequence of vn() = ( tn)(relabeled the same) such that v(0) = yand vn(t)!v(t) = zin H1 0(). Therefore, E(vn(t)) !E(z)implies that E(z) = land reasoning as before we get a contradiction since it follows that y2R. Hence, ArMu r(R)and the converse inclusion follows from (92). We can improve the regularity of the global attractor of the semigroup Trof Section 3.1.1 and obtain its characterization Lemma 39 Let the conditions of Theorem 17 hold. Then the global attractor Arof the semigroup Tris compact in H1 0(), bounded in Lp() and the convergence takes place in the topology of H1 0(), that is, distH1 0()(Tr(t; B);A)!0;as t!+1; for any set Bbounded in L2() : Proof. The estimates of Lemma 29 can be justi…ed for Trvia Galerkin approximations, so in this case we do not need to impose assumption (83) in order to use (85). Thus, the proof follows the same lines as in Proposition 32 and Theorem 33. Proposition 40 Let the conditions of Theorem 17 hold. Also, assume one of the following conditions: 1. h2L1() ; 2. p2n n2if n3: 25
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